问:
Analytical description of rotation operator
Let $\delta$ be the rotation with center $(1,2)$, that maps the point $(-1,3)$ to the point $(3,3)$.

(a) Describe $\delta$ analytically and determine the rotation angle.

(b) Let $g$ be the line with equation $x-3y-10=0$. Calculate an equation of the line $\delta (g)$.

(c) Let $\tau$ be the translation that maps $(-1,3)$ to $(3,3)$. What kind of transformation is $\tau\circ\delta$ ?

$$$$

For (a) do we have to write it in the form $\delta (x)=Ax$ ? To find the matrix do we have to shift the center to the origin first?

$$$$

EDIT:

In (c) do we have to calculate $\tau$ ? Or is the composition of a translation and a rotation always a specific transformation?

In this case the rotation $\delta$ and the translation $\tau$ map both the point $(-1,3)$ to $(3,3)$.

Is this important?

答:
Since you've essentially answered this in comments, here it is for the record:
$$
\left[ \begin{array}{rr} -3/5 & 4/5 \\ -4/5 & -3/5 \end{array} \right] \left[ \begin{array}{r} -2 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 2 \\ 1 \end{array} \right]
$$

So $\cos\theta = \dfrac{-3}5$  and $\sin\theta = \dfrac{-4}5.$


问:
solving for matrices
Suppose that

$$A = \begin{pmatrix}3 & 2\\\ 1 & 1\end{pmatrix}$$

Find all real numbers $\lambda$ such that (A − λI2) multiplied by

$$(A - \lambda I)\begin{pmatrix}x_1 \\ x_2\end{pmatrix} = O$$

has a solution which is not $x_1 = x_2 = 0$.
Such real numbers, $\lambda$, are called eigenvalues of $A$ and the non-zero solutions are called eigenvectors of $A$. Please don’t use determinants.

this is one of my worksheet question in which are not marked for the week however im stuck I have used $A − \lambda I$ to obtain $$\begin{pmatrix}3-λ & 2\\\ 1 & 1-λ\end{pmatrix}$$. would I now multiply to get a linear simultaneous equation but this cant happen as the answer is a $2x2$ matrix which would not fit this any help will be appreciated.

答:
It is probably a typo and means $0_{2\times 1}$ not $0_{2\times 2}$. So solve $\begin{pmatrix}3-\lambda&2\\1&1-\lambda\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$.

We get the system of equations $(3-\lambda)x+2y=0,x+(1-\lambda)y=0$. Solve the second one for x and return it in the first equation, to get $(3-\lambda)y(\lambda-1)+2y=0$. Do some manipulations to get $y(-\lambda^2+4\lambda-1)=0$. Apply the quadratic formula to solve for $\lambda$:

$$\begin{split}\lambda&=\frac{-4\pm\sqrt{16-4}}{-2}\\
&=2\mp\sqrt 3\end{split}$$

ContentLicense: CC BY-SA 4.0

Title: solving for matrices


问:
Why do my Stack Exchange reps follow a power law?
I noticed a pattern while looking at my network profile the other day, and I'm wondering if it's a fluke, or if there is something deep to it.

My reps for my top five Stack Exchange communities follow a inverse-square power law pretty neatly.  (Red dots are data; blue curve is a perfect inverse-square law.)

The ratio of my top SE rep to subsequent ones is 1, 4.02, 8.86, 16.58, 21.15.  All but the last are within a few percent of the expected 1, 4, 9, 16, 25 for an inverse-square law.  I'm not a statistician, so I have the following questions:

How can I go about testing whether this is a fluke or not?

Is there a universality argument for why this power law might arise?

Some remarks:

If you go further down my list of SE reps, the power law disappears, because it mostly consists of sites I've only visited or maybe posted once on.  All power laws have a cutoff, though, so this is unsurprising.

I've looked at some other user profiles, and there seem to be different classes of users.  Many are only active on one site, and there is no power law there.  Others are active on multiple sites and do seem to have superficially similar statistics to mine, but I don't know enough to test whether any of this is statistically significant.

答:
Statistical hypothesis testing

To test your hypothesis, you might (unsurprisingly) find statistical hypothesis testing to be a useful paradigm. In your case, your null hypothesis $H_0$ is that your empirical data follows a power law distribution. Your alternative hypothesis $H_a$ would be that it follows some other distribution. Can you reject your null hypothesis?

Your empirical data is a sample, and as such you can compute statistics (just a numeric value, basically) on that sample. You can choose a statistic such that the less likely the sample is, assuming the null hypothesis is true, the "more extreme" (usually, the larger) the statistic.

A given statistic follows a given distribution (by definition). As such, you can compute the probability of a value as extreme or even more extreme for the statistic than that obtained from your sample, given the null hypothesis is true: this is just the complementary CDF (i.e. the tail distribution) of the statistic.

This p-value can also be interpreted as the probability of a type I error, also known as a false positive. In statistical hypothesis testing, a false positive happens when you reject your null hypothesis when it was in fact true. As such, the p-value is a convenient tool for finding out how likely a null hypothesis is to be true (or not!).

It is good scientific practive to fix a statistical significance level $\alpha$ before hand: if the p-value is smaller than it, a type I error is very unlikely, so you can safely reject the null hypothesis (still, you might be wrong, so never think that the p-value is conclusive, definitive evidence!). Likewise, the larger the p-value is, the more likely the null hypothesis is to be true. Probably the most common value for $\alpha$ is $0.05$, but it's really up to you.

Applied statistics

In your case, you want to find out whether your empirical data (your sample) comes from a specific distribution. A fairly standard test for this is the Kolmogorov-Smirnov test (KS test). In this case, the relevant KS test is the one-sample test, since you're comparing a sample to a reference distribution (instead of two samples).

I don't have access to your data, but if you want to test this on your own, many statistical software packages offer functionality to use the KS test to test a hypothesis. SciPy is such a package for Python, which allows you to test the goodness of fit of your power law to the empirical data you have.

As a useful extra tidbit to your problem, the vocabulary revolving around the significance level might be a bit opaque: replacing the significance level $\alpha$ with the confidence level $1 - \alpha$ might make a bit more sense.

A small warning though: your very small number of data points might give you a large p-value despite having a very good fit! This is due to the way the KS statistic is defined.

Universality

It's quite difficult to say much about the universality of your particular case, without looking at data for a wider sample for StackExchange users. StackExchange offers a data explorer which allows you to query the reputation of users on different sites.


问:
Finding the Taylor series for $\ln(\sin(x))$
I'm trying to calculate the Taylor series for $\ln(\sin(x))$, though I'm not sure how to correctly expand the series.

when $a = \frac{\pi}{4}$

Here's what I have so far:

$$\ln(\sin(x))= \ln\left(x-\frac{x^3}{3!}+ \cdots+ (-1)^{n-1}\frac{x^{2n-2}}{(2n-2)!}\right)$$

What are some tips for expanding logarithmic series, for I have many exercises very similar to this and need an approach, though I'm stuck on converting them into series, your help is much appreciated!

As far as I'm aware, I've noticed two approaches thus far.

$\frac{dy}{dx}=\frac{1}{\sin(x)}\cdot \cos(x) = \cot(x) \implies \int\cot(x)$

Given the taylor series is:

If we let $(x-a) = h$

$f(a+h) = f(a) + f'(a)h+\frac{f''(a)}{2!}h^2 ...$

Then the first few series should be:

$\ln(\sin(x)) = \ln(\sin(\frac{\pi}{4}))+\cot(\frac{\pi}{4})(x-\frac{\pi}{4})...$?

答: The problem there is you are still talking the logarithm of an infinite series, so its not actually a Taylor series as such, instead you would need to derive the Taylor series from the start using the general formula for Taylor series. Another problem you will encounter is that $\sin(x)=0$ has infinite repeating solutions due to its periodicity and so your function will also be undefined at all of these points.


问:
Is there a divergent sequence such that $(x_{k+1}-x_k)\rightarrow 0$?
Is there a divergent sequence such that $\lim_{k\rightarrow\infty}(x_{k+1}-x_k)=0$?</p>

答:
We know that $1/k \to 0$ so pick $x_k$ such that $x_{k+1} - x_{k} = 1/k$.

So start with $x_1 = 0$ and then define $x_{k+1} = 1/k + x_{k}$ now you can see this diverges since $\sum_{k = 1}^\infty 1/k$ does.

ContentLicense: CC BY-SA 3.0


问:
Is $g(x)=\int_a^x (x-s)^{-\alpha}f(s) \, ds\in AC[a,b]$ if $f\in L_1[a,b]$ and $\alpha\in (0,1)$?
Let $f\in L_1[a,b]$ and $\alpha\in(0,1).$ Is it true that the function
$$
g(x)=\int_a^x \frac{f(s)}{(x-s)^\alpha}\,ds
$$
is absolutely continuous on $[a,b]$? First, I thought it is true and I tried to show that there exists $\phi\in L_{1}[a,b]$ such that  $$g(x)-g(y)=\int_y^x \phi(s) \, ds$$ but it was very difficult (for me). Then I tried to find a function $f\in L_1 [a,b]$ such that $g\not\in AC[a,b]$ but I couldn't. Any idea please?

答: Consider $\alpha = 1/2$, $f(s)=(x-s)^{-1/2}$.  Then the integral does not even exist.


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
Every time $x$ is incremented by $1,$ $3^x$ gets $3$ times as big.

When $x$ is incremented from $1\,000\,000$ to $1\,000\,001,$ $x^2$ increases by $2\,000\,001$ which may seem big, but is tiny by comparison to $1\,000\,000^2 = 1\,000\,000\,000\,000.$


问:
What is the advantage of LU decomposition over storing the elimination matrix directly?
The answers to this question indicate that one of the advantages of LU decomposition is that the algorithm do not redo the elimination steps.

Why do we insist on storing the factorized form of $L$ and $U$?  Why not just save $E$ and $U$, where

\begin{align}
Ax = b
&\implies EAx = Eb && \text{(multiply both sides by $E$)} \\
&\implies Ux = Eb && \text{(since $EA = U$)}
\end{align}

As you can see, we just need to store $U$ and the elimination matrix $E$, so whenever we have a new matrix $b$ on the right-hand side, we can just multiply it out with $E$, and use back-subsitution to solve for $x$. What's advantage of storing $L$ and $U$ over storing $E$ and $U$?

答:
Note that in your case, the matrix $E$ is $L^{-1}$. So one might just go even further and store $U^{-1}$ as well or even $A^{-1}=U^{-1}L^{-1}$.

So why not just store $A^{-1}$? In a word, accuracy. A matrix $A$ has a so-called condition number $\kappa(A)$. The error $\| x_{\rm computed} - x\|$ in solving $Ax = b$ is roughly $\kappa(A)\times 10^{-16}$ in double precision arithmetic (more generally, $10^{-16}$ is roughly the unit roundoff for your floating point number system). However, even when $\kappa(A)$ is large, conventional $LU$ factorization with back- and forward-substitution tends to be backwards stable in the sense that the residual norm is small: $\|Ax_{\rm computed} - b\| \approx 10^{-16}$. In some applications, a small residual is more important than a small error, so this backwards stability is a desirable property.

Unfortunately, if you compute and store $A^{-1} = U^{-1}L^{-1}$, then the computed solution is (usually) no longer backwards stable and the residual is much larger $\|Ax_{\rm computed} - b\| \approx \kappa(A)\times 10^{-16}$. The error $\| x_{\rm computed} - x\|$ is usually of the same magnitude, but is usually a bit larger than if we just did back substitution.

So if we store $E=L^{-1}$, our residual norm will usually be higher. But not higher like $\|Ax_{\rm computed} - b\| \approx \kappa(A)\times 10^{-16}$, but higher like $\|Ax_{\rm computed} - b\| \approx \kappa(L)\times 10^{-16}$. We pay for the condition number of $L$ not $A$. In practice, $L$ is usually fairly well-conditioned even if $A$ is not. For instance, in some tests I did with a randsvd random matrix $A$ with condition number $10^{14}$, the condition number of $L$ computed with partial pivoting is only like $10^2$. So the residual norm is only like $100\times $ larger than with back- and forward- substitution.

There are other reasons to prefer storing $L$ over $E$. For example, if $L$ is a sparse matrix, $E$ may be less sparse and thus more computational costly to multiply by.

You're usually not losing a lot of accuracy or speed by the approach you advocate, but it's not clear exactly what you're gaining by your approach since you are losing something in accuracy by storing $E$ instead of $L$. Unless you're in a setting where you have a good reason to use $E$ instead of $L$ (which do exist), I would suggest sticking with $L$.


答:
There can be two approaches here, in both, you shall have to make direct use of Taylor's formula:

\begin{align*}f\left( x \right) & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}} \\ & = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + \frac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \frac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \cdots \end{align*}

Method-1:

Take $ f(x) = ln(y) $ with $ y= \sin (x)$ and $ a = \sin(\pi / 4)$ and expand directly

Method-2:

Make use of the fact that:

$$\frac{d}{dx} \ln (\sin x ) =\frac{1}{\sin x} \cdot \cos x = \cot (x)$$

So, you can directly evaluate with $f(x) = \cot x $ and $ a = \pi / 4 $ and integrate from $0$ to $x$

I trust that you will be able to perform the calculation now.

Title: Finding the Taylor series for $\ln(\sin(x))$

Body:
I'm trying to calculate the Taylor series for $\ln(\sin(x))$, though I'm not sure how to correctly expand the series.

when $a = \frac{\pi}{4}$

Here's what I have so far:

$$\ln(\sin(x))= \ln\left(x-\frac{x^3}{3!}+ \cdots+ (-1)^{n-1}\frac{x^{2n-2}}{(2n-2)!}\right)$$

What are some tips for expanding logarithmic series, for I have many exercises very similar to this and need an approach, though I'm stuck on converting them into series, your help is much appreciated!

As far as I'm aware, I've noticed two approaches thus far.

$\frac{dy}{dx}=\frac{1}{\sin(x)}\cdot \cos(x) = \cot(x) \implies \int\cot(x)$

Given the taylor series is:

If we let $(x-a) = h$

$f(a+h) = f(a) + f'(a)h+\frac{f''(a)}{2!}h^2 ...$

Then the first few series should be:

$\ln(\sin(x)) = \ln(\sin(\frac{\pi}{4}))+\cot(\frac{\pi}{4})(x-\frac{\pi}{4})...$?


问:
Classical Tautologies in a Deviant Logic of Paradox
I am  stuck with the following problem. The Logic of Paradox (LP) has the truth values {f,p,t}, designated values {p,t} and the following truth tables:

```<code>   A  ~A   A -> B   A & B   A v B
   t  f    t t t    t p f   t t t
   p  p    p p t    p p f   t p p
   f  t    f p t    f f f   t p f
</code>```

A known theorem between LP and classical logic (M2) is that the Logic of Paradox can reproduce the classical tautologies as follows:

$$\vDash_{LP}\varphi\iff\vDash_{M2}\varphi$$

Now I am experimenting with a deviant Logic of Paradox (LP') where negation and implication is replace by the following truth tables, but conjunction and disjunction are the same as above:

```<code>   A  ~A   A -> B
   t  f    t t t
   p  f    p p t
   f  t    f f t
</code>```

Does $\vDash_{LP'}\varphi\iff\vDash_{M2}\varphi$ still hold?

Remark:

To prove $\vDash_{LP}\varphi\iff\vDash_{M2}\varphi$, first observe that the direction $\vDash_{LP}\varphi\Rightarrow\,\vDash_{M2}\varphi$ is tivial. For the direction $\vDash_{LP}\varphi\Leftarrow\,\vDash_{M2}\varphi$ one can go observe that we have the following valuation identities:

```<code> A -> B = ~A v B
 ~(~ A) = A
 ~(A v B) = ~A & ~B
 ~(A & B) = ~A v ~B
</code>```

So from a formula $\varphi(x_1,\dots,x_n)$ one can obtain a negation normal form formula $\varphi'(x_1,\dots,x_n)$. And then a monotonic formula $\varphi'(x_1,\dots,x_n,y_1,\dots,y_n)$ and the side conditions $x_1\neq y_1,\dots,x_n\neq y_n$. Since the formula is monotonic it will be also satisfied when side condition is violated via $x_j\&y_j$, the satisfaction corresponds to the designation {p,t} and the pair to the truth value p.

For the deviant Logic of Paradox the same negation normal form proof is not anymore so easily obtained, since already the identity ~(~ A) = A fails:

```<code> A ~(~ A)
 t  t
 p  t
 f  f
</code>```

So what would be proof idea for $\vDash_{LP'}\varphi\iff\vDash_{M_2}\varphi$? Or is there a counter example?

答:
$LP'$ has still the same theorems as classical logic.

Intuitively, the reason is that $LP'$ cannot distinguish between the designated truth values: Given a three-valued interpretation $I$, if we change some values $p$ to $t$ or $t$ to $p$ in $I$, then the value of $I$ on a given formula never switches from designated to undesignated (it might switch from $p$ to $t$ or vice versa, while value $f$ remains fixed).

This property can be readily verified for the truth tables. (Do it graphically: if you switch between a $t$-column and a $p$-column in the same row, you either have two designated values or $f$ twice. Same goes for rows.)

By induction this property follows for all formulas. Therefore any $LP'$-countermodel can be reduced to a classical countermodel (just identify the truth values $t$ and $p$). Conversely, every classical countermodel is also a $LP'$-countermodel as the truth tables of $LP'$ extend that of classical logic.


问:
Goldschmidt division example with estimates?
I have found in Gonzalez/Diaz-Herrera/Tucker: Computing Handbook an example for the Goldschmidt division in the case of real numbers. The example is the following:
\begin{align}
N_0 &= N = 0.625  \\
D_0 &= D = 0.75  \\
F_1 &= 1.3& & &\text{Table Look Up} \\
N_1 &= N_0 \cdot F_1 = 0.625 \cdot 1.3 &= 0.8125&  &\text{Multiply} \\
D_1 &= D_0 \cdot F_1 = 0.75 \cdot 1.3 &= 0.975&  &\text{Multiply} \\
F_2 &= 2 - D_1 = 2 - 0.975 &= 1.025&  &\text{Subtract} \\
N_2 &= N_1 \cdot F_2 = 0.8125 \cdot 1.025 &= 0.8328125&  &\text{Multiply} \\
D_2 &= D_1 \cdot F_2 = 0.975 \cdot 1.025 &= 0.999375&  &\text{Multiply} \\
F_3 &= 2 - D_2 = 2 - 0.999375 &= 1.000625&  &\text{Subtract} \\
N_3 &= N_2 \cdot F_3 = 0.8328125 \cdot 1.000625 &= 0.8333330078125&  &\text{Multiply} \\
(D_3 &= D_2 \cdot F_3 = 0.999375 \cdot 1.000625 &= 0.999999609375)&  &\text{(Multiply)} \\
\end{align}

Even though the example is quite simple, I have problems in two parts:

How to get the estimate of 1.3 as the value of F1, it mentions one table, but how to use it?

In the case of division with integers, the procedure is the same? I have been trying to check it up one example for this case, but I only found explanations that use binaries and real numbers. If somebody could post an example of how to do this procedure with integer numbers would be great.

Thanks

答:
Let's try and divide 6250 by 75.

First, almost arbitrarily re-interpret as fixed point numbers 62.50 and 00.75.

Look up the reciprocal of leading digit .7: 1.4 $= F_1$

Products: $N_1 = 62.5 \cdot 1.4 = 87.5$, $D_1 = .75 \cdot 1.4 = 1.05$

$F_2 = 2-1.05 = .95$, $N_2 = 87.5\cdot.95 = 83.12(5)$

($D_2 = 1.05\cdot.95 = .99(75)$, but talking integers, 83 is the quotient)

With integers, one may pre-check the absolute value of the dividend for being smaller than the divisor. To profit most from the table of reciprocals, the operands conventionally are scaled such that the divisor is in a smallish range, its most significant digit in a known position.

Choosing the size of the reciprocal table is a trade-off between small table and small number of iterations.

The most significant part/first few digits of the divisor are used to index the table, yielding a reciprocal with a similar number of digits of resolution.

Thinking of the bit patterns as fixed point numbers with more significant half integers, less significant part fractions is just a convenience - otherwise, I'd have to hand-wave about scaling all the time.


问:
How to approach $\int\limits_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x$
How to approach this integral?

$$\int\limits_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x$$

Body:
Note $ \sin^{-1}e^x+\cos^{-1}e^{x} =\frac\pi2$ and substitute  $y= \tan^{-1}e^a+ \tan^{-1}e^x$, $dy = \frac{dx}{e^x+e^{-x}}$ to integrate

$$\int_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x
=\frac\pi2 \int_{ \tan^{-1}e^a}^{2\tan^{-1}e^a}\frac1ydy=\frac\pi2\ln2
$$


问:
prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

Prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

We can prove this easily by showing $f(1)=f'(1)=0$.

But is there a way to solve this problem without using calculus?

答: You can do long polynomial division and get that$$\frac{nx^{n+1}-(n+1)x^n+1}{x-1}=nx^n-x^{n-1}-x^{n-2}-\cdots-x-1$$(or, of course, you can just compute $(x-1)\times(nx^n-x^{n-1}-x^{n-2}-\cdots-x-1)$). And it is clear that $1$ is a root of $nx^n-x^{n-1}-x^{n-2}-\cdots-x-1$.


答: Let $f:(0,\infty)\to\mathbb R$ be any function such that $\lim\limits_{x\to\infty}f(x)=\infty$ and $\lim\limits_{x\to\infty}f'(x)=0$.  Then $(f(1),f(2),f(3),\ldots)$ is an example.  In particular, you could take $f(x)=\log(x)$ or $f(x)=\sqrt x$.</p>

Title: Is there a divergent sequence such that $(x_{k+1}-x_k)\rightarrow 0$?

Body: Is there a divergent sequence such that $\lim_{k\rightarrow\infty}(x_{k+1}-x_k)=0$?</p>


答:
Here is one way to prove it:

The statement is equivalent to $x^2|f(x+1)$. It is enough to show that $f(x+1)$'s constant term and coefficient of $x$ is $0$. You can calculate this from the binomial theorem.

Title: prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

Body:
Prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

We can prove this easily by showing $f(1)=f'(1)=0$.

But is there a way to solve this problem without using calculus?


问:
Travelling Salesman Problem (TSP): can FN beat NN given these conditions?
Take the traveling salesman problem, but with three slight twists:

You can choose a different start vertex for each of the two algorithms.

Each path from one vertex to another is of unique, arbitrary length (irrespective of the distances between the cities and they don't intersect).

Each algorithm goes to each vertex once and only once (it doesn't return to its starting vertex).

With these conditions, is it possible for a furthest neighbor algorithm to beat a nearest neighbor algorithm.

I've tried using induction, but condition 2 really throws it off. I see this previous post answers a similar question, but mine is slightly different for the above three reasons.

答:
Yes furthest neighbors first can beat nearest neighbors first search. This is simply done by example. Your condition 2 does not restrict anything. Important is that the lengths are unique and can not be changed. And intersections can be eliminated easily under keeping uniqueness if length are assigned and not real.

So in what cases does this preference make a difference?
Only if a mixed choice strategy is best and the graph is best solved with annealing and an optimization function than there will be a difference. Therefore the graph has to provide some arbitrariness. For example two nor more knots are connected by more than on path of different length. The there are removable edges that make fn or nn better.

Look at this Travelling salesman problem for some more indepth cases in the given literature.

Neither fn nor nn are optimal strategies for the general case of the tsp. This indicates that there are either cases where one of the selection of the next path of the other is better. It is always an experiment to calculate a given total path length depending on the choice of fn or nn. This is not so a great influence that avoiding longer pathes taking shorter once. It has no influence if the compare pathes have to be taken to reach each knot.

If the graph has a cover of the knot with only longer and only shorter pathes than this is another trivial case. Than always the selection of the nearest, shortes path is the optimal choice.

There are solvable graphs with only some choice are present. If there is a mixture of such choices then not all edges might be meet and the graph can be unsolvable for tsp. So in graphs with directed edges with different length in each directions are example for the deep need of sorrow calculations and comparisons. Graphs without that are not a problem and often again both perform the very same. This a priori undecidability is a first indicator that different performances come with the given tsp graph. Whether this is just the case for bidirectional differently weighted graphs and the total visit constraints should be clear.

The easiest example is a tsp graph with two knots and one bidirectional path. One direction is longer than the other. Starting from the knot in the shorter direction favours nn, starting in the opposite direction favors fn. This is because the other preference has no solution to compare to. There are cases were fn is better and were nn is better and were both are the same. The start shows that tsp with two successive knots connected by bidirectional pathes gives a longer minimal necessary  description already. So if long-short and short-long appears neither strategy can be applied. Only long-long and short-short can be discussed.


问:
Manifolds with Boundary - I think I misunderstood the definition of boundary points, because I found a contradiction
The following definitions are from the 2nd edition of John Lee's "Introduction to
Smooth Manifolds", page 25.

Definition 1

We will call a chart $(U,\phi)$ an interior
chart if $\phi(U)$ is an open subset of $\mathbf{R}^n$ and a boundary chart if $\phi(U)$ is an open subset
of $\mathbf{H}^n$ such that $\phi(U)\cap\partial\mathbf{H}^n$ is not empty.

Definition 2

A point $p\in M$ is called an interior point of $M$ if it is in the domain of some
interior chart. It is a boundary point of $M$ if it is in the domain of a boundary chart
that sends $p$ to $\partial\mathbf{H}^n$.

Consider the manifold $S^1$. The inverses of
\begin{align}
[0,\pi)&\to S^1\\
x&\mapsto\mathrm{e}^{ix}
\end{align}
and
\begin{align}
[0,\pi)&\to S^1\\
x&\mapsto-\mathrm{e}^{ix}
\end{align}
are boundary charts and according to the definition above, $(-1,0)$ and $(1,0)$ are boundary points of $S^1$.
But the inverse of
\begin{align}
(-\pi,\pi)&\to S^1\\
x&\mapsto\mathrm{e}^{ix}
\end{align}
is an interior chart (smoothly compatible with the other two charts) and $(1,0)$ turns out to be an interior point.

答:
Let $p : [0, \pi) \to S^1,~~ x \mapsto e^{ix}$ be one of the parametrizations you chose, and let $U := p([0, \pi)) \subset S^1$.

Then $p^{-1} : U \to [0, \pi)$ is not a chart because $U$ is not an open subset of $S^1$.


问:
What's wrong with my method for this integral?
So I have to solve
$$\int_0^1\frac{1}{x^{2/3}(1-x)^{1/3}}dx.$$
To do this I made a branch cut from $z=0$ to $z=1$ and took the bone-shaped contour that straddles the real axis, going clockwise. Now on the line above the real axis, I took the angle with respect to $z=0$ to be $0$ and that with $z=1$ to be $\pi$.  Then the corresponding line integral comes out to $e^{-\frac{i\pi}{3}}I$ and so the one below comes out to $-e^{\frac{i\pi}{3}}I$ with the second angle $-\pi$. But this gives me the wrong answer. Most places I found online seem to take the first angle to be $0$ and the second one to go from $0$ to $2\pi$. But why should this make a difference? Why is my way of taking the angles wrong?

答:
Note that your integrand here is discontinuous at $x = 0$ and $x=1$.

Thus, if you take the argument for your contour from $ - \pi $ to $ \pi$, then the arc will pass through a discontinuity and your answer may be incorrect by a constant residue.

This problem is not present when taking the angles $0$ to $ 2 \pi$ as there will not be any integration along a discontinuity.


Body: Define polynomials $p_n$ of degree $n-1$ by$$(x-1)^2p_n(x)=nx^{n+1}-(n+1)x^n+1$$for $n\ge1$, viz.$$p_1=1,\,p_2=2x+1,\,p_{n+2}=1+2xp_{n+1}-x^2p_n.$$


问:
Determining whether a piecewise function is odd or even or neither
I am trying to determine whether my piecewise function is even or odd or neither. If it wasn't a piecewise I would use the trick of subbing in a negative x but when there are two parts to it I don't believe that would work. Is the best way just to observe a sketch of the function?

Cheers

答:
You can still plug in $-x$ and compare. As a simple example, consider the function
$$
f(x)=\begin{cases} 2x, \quad &x<5 \\ 15-x, \quad &x\ge 5 \end{cases}
\tag1$$
Replace $x$ with $-x$ throughout:
$$
f(-x)=\begin{cases} -2x, \quad &-x<5 \\ 15+x, \quad &-x\ge 5 \end{cases}
\tag2$$
Rewrite the conditons on the right in a more digestible form:
$$
f(-x)=\begin{cases} -2x, \quad &x>-5 \\ 15+x, \quad &x\le -5 \end{cases}
\tag3$$
So, $f(-x)$ has a transition at $-5$ instead of $5$. This  already suggests that $f$ is neither even nor odd. But to be absolutely sure, calculate
$$
f(x)+f(-x) =
\begin{cases} 15+3x, \quad &x\le -5 \\ 0 & -5<x<5\\ 15-3x, \quad &x\ge 5 \end{cases}
\tag4$$
Since $f(x)+f(-x)$  is not identically zero, $f$ is not odd. </p>

<p>Similarly,
$$
f(x)-f(-x) =
\begin{cases} x-15, \quad &x\le -5 \\ 4x & -5<x<5\\ 15+x, \quad &x\ge 5 \end{cases}
\tag5$$
Since $f(x)-f(-x)$  is not identically zero, $f$ is not even. </p>

Body:
You can still plug in $-x$ and compare. As a simple example, consider the function
$$
f(x)=\begin{cases} 2x, \quad &x<5 \\ 15-x, \quad &x\ge 5 \end{cases}
\tag1$$
Replace $x$ with $-x$ throughout:
$$
f(-x)=\begin{cases} -2x, \quad &-x<5 \\ 15+x, \quad &-x\ge 5 \end{cases}
\tag2$$
Rewrite the conditons on the right in a more digestible form:
$$
f(-x)=\begin{cases} -2x, \quad &x>-5 \\ 15+x, \quad &x\le -5 \end{cases}
\tag3$$
So, $f(-x)$ has a transition at $-5$ instead of $5$. This  already suggests that $f$ is neither even nor odd. But to be absolutely sure, calculate
$$
f(x)+f(-x) =
\begin{cases} 15+3x, \quad &x\le -5 \\ 0 & -5<x<5\\ 15-3x, \quad &x\ge 5 \end{cases}
\tag4$$
Since $f(x)+f(-x)$  is not identically zero, $f$ is not odd.

Similarly,
$$
f(x)-f(-x) =
\begin{cases} x-15, \quad &x\le -5 \\ 4x & -5<x<5\\ 15+x, \quad &x\ge 5 \end{cases}
\tag5$$
Since $f(x)-f(-x)$  is not identically zero, $f$ is not even.


问:
Which operator $\oplus$ turns a function $f$ into a monoid homomorphism from $(L,*,[])$ to $(\mathbb{R},\oplus,\epsilon)$?
This question is inspired by this question. I do not have a strong background in abstract algebra, so this might be trivial or pointless.

Let $L$ be the set of all lists of finite length and denote $*$ as the list concatenation. Denote $[]\in L$ as the empty list (length $0$). Then $(L,*,[])$ is a monoid under $*$ and with identity $[]$ (because $x*y=y*x$ and $x*[]=[]*x$, $\forall x,y \in L$).

Let us define the operator $[l,i]$ with $l \in L$ and $i\in \{1,\cdots, |l|\}$ which returns the $i$th element from $l$. We simply write $[i]$ if no confusion arises and $l$ is clear from the given context.

Let $(\mathbb{R}, \oplus, \epsilon)$ be a monoid with set $\mathbb{R}$ under some binary operation $\oplus$ and identity element $\epsilon$. By definition:

$\forall a,b,c \in \mathbb{R}: (a\oplus b) \oplus c= a \oplus (b\oplus c)$

$\forall a \in \mathbb{R}: \epsilon \oplus a = a \oplus \epsilon=a$

We define

$$f:L \to \mathbb{R}:[x_1,\cdots,x_n] \mapsto
\left\{\begin{array}{ll} [1]-\sum_{i=2}^n [i], & n \geq 2 \\
         [1], & \text{if } n=1 \\
0 , & \text{otherwise} \end{array} \right. $$

Question 1: Which operator $\oplus$ turns $f$ into a monoid homomorphism from $(L,*,[])$ to $(\mathbb{R},\oplus,\epsilon)$? I have a strong feeling that this is wrong, but how would we disprove it? Moreover, are there any general ways to define monoids in order for a given function to be a homomorphism between them?

The following must hold

$\forall x,y \in L: f(x * y)=f(x) \oplus f(y)$

$f([]) \mapsto \epsilon$

The map $f$ already yields to $f([])=0$, hence $\epsilon=0$. Now, we are limited for the choice of $\oplus$, e.g. $(\mathbb{R}, \cdot)$ is not possible anymore. $(\mathbb{R},+,0)$ forms a monoid, however $f(x * y)=f(x)+f(y)$ is obviously wrong.

Question 2: If we omit the possibility of lists of length zero, and thus removing the identity element from $L$, then $(L,*)$ would form a semigroup. Could we define $\oplus$ such that $(\mathbb{R},\oplus)$ form a semigroup and $f$ (omitting the $n=0$ case of course) is a homomorphism between two semigroups?

答:
This may be a misunderstanding of the notation, but it seems there is no structure on $\mathbb{R}$ making $f$ a monoid homomorphism.

Since $f([a, b]) = a-b$, the only viable $\oplus$ would be the usual subtraction on $\mathbb{R}$, but this is not associative.


问:
Stuck with integral involving Polylogarithms $\int_{0}^{\infty}\frac{2t}{e^{t\pi}+1} \,dt$
For a research work I ended up needing to give a proof of Zeta's trivial zeros, in order to do so I tried using the Abel-Plana formula.

$\zeta(s)=\frac{2^{s-1}}{s-1}-2^{s}\int_{0}^{\infty} \frac{\sin(s\cdot \arctan(t))}{(e^{t\pi}+1)(1+t^2)^{\frac{s}{2}}} \,dt \implies \frac{2^{-2k-1}}{-2k-1}=-2^{-2k}\int_{0}^{\infty} \frac{\sin[2k\cdot  \arctan(t)]}{(e^{t\pi}+1)(1+t^2)^{-k}} \,dt$

In the above equation -2k represents the even negative integer, after some development using the complex form of Sin and Arctan I ended up at the following stalemate.

$\frac{1}{2i}\int_{0}^{\infty}  \frac{(1+it)^{2k} - (1-it)^{2k}}{(e^{t\pi}+1)} \,dt = \frac{1}{4k+2}$

Willing to verify my result for $k=1 \implies \zeta(-2)$ I managed to find the following:

$\int_{0}^{\infty}\frac{2t}{(e^{t\pi}+1)} \,dt=\frac{1}{6}$

However my competences in calculus give me no clue on how to directly justify this last equality.

Body:
More generally, for $a > 1$,

$$ \eqalign{\int_0^\infty \frac{2t}{e^{at}+1}\; dt &= \sum_{k=1}^\infty (-1)^{k-1}
\int_0^\infty \frac{2t}{e^{kat}}\; dt \cr
 &= \sum_{k=1}^\infty \frac{2 (-1)^{k-1}}{a^2 k^2}\cr
&= \frac{\pi^2}{6a^2}}$$


答:
Assuming you can interchange integrals and infinite sums, we have
\begin{align*}
2\int_0^\infty\frac t{e^{\pi t}+1}dt&=2\int_0^\infty t\sum_{n\ge1}(-1)^{n-1}e^{-\pi n t} dt\\
&=2\sum_{n\ge1}(-1)^{n-1}\int_0^\infty te^{\pi n t} dt.\\
\end{align*}
Here, by integration by parts, we have
\begin{align*}
\int_0^\infty te^{\pi n t} dt&=\frac1{\pi n}\int_0^\infty t(e^{\pi n t})' dt\\
&=\frac1{\pi n}\left(\left[te^{\pi nt}\right]_0^\infty-\int_0^\infty e^{\pi n t} dt\right)\\
&=\frac1{\pi^2 n^2}.
\end{align*}
Thus, we have that
\begin{align*}
2\int_0^\infty\frac t{e^{\pi t}+1}dt&=\frac2{\pi^2}\sum_{n\ge1}(-1)^{n-1}\frac1{n^2}.\\
&=\frac2{\pi^2}\frac{\pi^2}{12}=\frac16.
\end{align*}

Title: Stuck with integral involving Polylogarithms $\int_{0}^{\infty}\frac{2t}{e^{t\pi}+1} \,dt$

Body:
For a research work I ended up needing to give a proof of Zeta's trivial zeros, in order to do so I tried using the Abel-Plana formula.

$\zeta(s)=\frac{2^{s-1}}{s-1}-2^{s}\int_{0}^{\infty} \frac{\sin(s\cdot \arctan(t))}{(e^{t\pi}+1)(1+t^2)^{\frac{s}{2}}} \,dt \implies \frac{2^{-2k-1}}{-2k-1}=-2^{-2k}\int_{0}^{\infty} \frac{\sin[2k\cdot  \arctan(t)]}{(e^{t\pi}+1)(1+t^2)^{-k}} \,dt$

In the above equation -2k represents the even negative integer, after some development using the complex form of Sin and Arctan I ended up at the following stalemate.

$\frac{1}{2i}\int_{0}^{\infty}  \frac{(1+it)^{2k} - (1-it)^{2k}}{(e^{t\pi}+1)} \,dt = \frac{1}{4k+2}$

Willing to verify my result for $k=1 \implies \zeta(-2)$ I managed to find the following:

$\int_{0}^{\infty}\frac{2t}{(e^{t\pi}+1)} \,dt=\frac{1}{6}$

However my competences in calculus give me no clue on how to directly justify this last equality.


问:
Is an ideal finitely generated if its radical is finitely generated?
Let $R$ be a commutative ring. If $R$ is not Noetherian, we can ask if some some ideals is finitely generated. For examples:

Is intersection of finitely generated ideals finitely generated? No, see for instance here.

Is radical $\sqrt{I}=\{x \in R \mid x^n \in I \text{ for some } n\ge 1\}$ of a finitely generated ideal $I$ finitely generated? No, see for instance here.

In the light of the previous two (sets of) counter-examples, I believe that claim

If $\sqrt{I}$ is finitely generated, then $I$ is also finitely generated,

is also false, but I wasn't able to construct a counter-example. It would be interesting for me to see counter-examples of various nature.

答:
The ideal $I=(x^2,xy_1,xy_2,xy_3,\ldots)$ of $\Bbb{Q}[x,y_1,y_2,y_3,\ldots]$ is not finitely generated, its radical is $(x)$.

(it is immediate that $I\subset (x), (x)\subset \sqrt{I}$ and $\sqrt{(x)}=(x)$ so $\sqrt{I}=(x)$)


Body:
Let $x=e^{-\pi t}$. Then

$$\int_{0}^{\infty}\frac{2t}{e^{t\pi}+1} \,dt
=-\frac2{\pi^2}\int_0^1 \frac{\ln x}{1+x}dx
=-\frac2{\pi^2}\cdot (-\frac{\pi^2}{12} )= \frac16
$$
where $\int_0^1 \frac{\ln x}{1+x}dx = -\int_0^1 \frac{\ln (1+x)}{x}dx = -\frac{\pi^2}{12}$


问:
Cartesian Product over Symmetric Difference
Let's say I have that $(x,y)\in (A\times B)\Delta (A\times C).$ How would I expand this statement assuming $A\Delta B = (A-B)\cup (B-A)$?

Here is as far as I have gotten. Let $(x,y)\in (A\times B)\Delta (A\times C).$ Since we define $A\Delta B = (A-B)\cup (B-A)$, then $(A\times B)\Delta (A\times C) = [(A\times B)-(A\times C)]\cup [(A\times C)-(A\times B)].$

Is the next step ($x\in A\times B$ and $x\notin A\times C$) or ($x\in A\times C$ and $x\notin A\times B$)? Wouldn't this produce a contradiction where $x\in A$ and $x\notin A$?

答:
We can simplify further to get

\begin{align}
(A\times B)\Delta (A\times C) & = [(A\times B)-(A\times C)]\cup [(A\times C)-(A\times B)] \\
& = [A \times (B-C)] \cup [A\times (C-B)]\\
& = A \times [(B-C) \cup (C-B)] \\
& = A\times (B\Delta C)
\end{align}

So $(x,y)\in (A\times B)\Delta (A\times C)$ implies $x\in A$ and $y\in B\Delta C$.


问:
Can we visualize the closed balls for the space $l^{\infty}$ equipped with the $\sup$ norm
The closed unit balls for the $l^{p}$ in $\mathbb{R}^2$ look like

I want to know could we also visualize the closed balls for the space $l^{\infty}$ equipped with the $\sup$ norm .

Thanks

答:
For any $n$, the set $\{(x_1,\dots,x_n):\sup|x_i|\le 1\}$ is the $n$-dimensional hypercube of sidelength $2$, namely $[-1,1]^n$.

Normally, one does not directly visualize an infinite-dimensional object, but rather infer its geometric properties from a finite-dimensional model. For example: the unit cube of $\ell^\infty$ is not strictly convex (its boundary contains some line segments), has a non-smooth boundary, and infinitely many "vertices" (extreme points) which are the sequences of $\pm 1$.


问:
Evaluating $\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1}$

I am seeking to evaluate the sum $$S=\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1} \\ =(n-1)^2 \binom{n}{1}+(n-3)^2 \binom{n}{3}+(n-5)^2\binom{n}{5}+\cdots$$

I re-wrote the sum as $$S=(n+1)^2\sum_{r \in \mathbb{N}} \binom{n}{2r-1}-4(n+1)\sum_{r\in\mathbb{N}}r\binom{n}{2r-1}+4\sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$$

It is clear from the binomial expansion of $(1+x)^n$, (by plugging $x=\pm1$ ) we get the value of the sum $\displaystyle \sum_{r\in\mathbb{N}}\binom{n}{2r-1}$ as $2^{n-1}$.

Now, after this I can find the value of the sums $\displaystyle \sum_{r\in\mathbb{N}}r\binom{n}{2r-1}$ and  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$ by considering the binomial expansion of $\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$ and substituting $x$ for $\sqrt{x}$ and then differentiating, and by manipulations, I will be able to find the desired sums.

However, this method gets very lengthy, especially when doing differentiation for finding  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$.

I am hoping to see some other different approaches which don't involve so much calculation and are easy to understand.

The final closed form is $n(n+1)2^{n-3}$.

Body:
We can rewrite the sum as:

$$\begin{aligned}S&=\sum_{r \in \mathbb{N}} (n-2r+1)^2 {n \choose 2r-1} \\&= \sum_{r \in \mathbb{N}} (n-2r+1)^2 {n \choose n-2r+1} \\&= \sum_{r \in \mathbb{N}} (n-2r+1)^2 \frac{n!}{(n-2r+1)!(2r-1)!} \\&= \sum_{r \in \mathbb{N}} n(n-2r+1) \frac{(n-1)!}{(n-2r)!(2r-1)!} \\&= \sum_{r \in \mathbb{N}} n(n-2r+1) {n-1 \choose 2r-1} \\&= n^2 \sum_{r \in \mathbb{N}} {n-1 \choose 2r-1} - n \sum_{r \in \mathbb{N}} (2r-1){n-1 \choose 2r-1} \\&= n^2 \frac{2^{n-1}}{2} - n(n-1) \frac{2^{n-2}}{2} \\&= n2^{n-3}(2n-n+1) \\&= n(n+1)2^{n-3}\end{aligned}$$


问:
What is $(7^{2005}-1)/6 \pmod {1000}$?
What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$

My approach:

Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.

So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.

Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.

Any help is appreciated!

答:
We must multiply the modulus by $\,6\,$ to balance the division by $6$, i.e.

$\qquad 6\mid a \,\Rightarrow\,  a/6 \bmod 1000 = (a \bmod 6000)/6\ $ by the mod Distributive Law

$6000\!=\! 2^4\cdot 3\cdot 5^3$  whose totients $2^3,2,100\mid 2000\,$ so $\,\color{#c00}{7^{2000}\!\equiv 1}\bmod 2^4,3,5^3$ so also mod $6000,\,$ so $\bmod 6000\!:\ a\equiv 7^5 \color{#c00}{7^{2000}}-1 \equiv 7^5-1\equiv4806,\,$ so $\bmod 1000\!:\ a/6\equiv 4806/6 \equiv 801$.

Title: What is $(7^{2005}-1)/6 \pmod {1000}$?


问:
Calculate and find the flux integral
Find the flux integral $$\iint_S \operatorname{rot} \vec{F} {N}\,dS$$ where $S$ is the half sphere $x^2+y^2+z^2=4$ and $z \ge 0$ with an aligned unit standard $\vec{N}$ (normal) and $\vec{F} =(3x-y, yz, xy)$.

I know that I should use Stokes theorem and know that rot is the same as curl and $curl \vec{F} = \nabla \times \vec{F}$ = $x-y-y+1$ but I don't know how to continue on from here.
Does working with a half sphere mean that I have 2 normals? How do I calculate the normal?

Edit: I've calculated curl: $\nabla \times F = \det \begin{bmatrix}\mathbf i& \mathbf j& \mathbf k\\\frac {\partial F}{\partial x}&\frac {\partial F}{\partial y})&\frac {\partial F}{\partial z}\\3x-y&yz&xy \end{bmatrix} = (x-y)\mathbf i  -(y\mathbf j + (1)\mathbf k$

答:
As per Stokes' theorem, $\displaystyle \iint_S (\nabla \times  \vec{F}) \cdot \hat{n} \, dS = \int_C \vec {F} \cdot dr \ $, where $C$ is the boundary curve of the surface $S$.

$S$ is $x^2+y^2+z^2 = 4, z \geq 0$. Its boundary curve is circle $x^2+y^2 = 4$ at $z = 0$. Parametrize the boundary curve in polar coordinates as $r(t) = (2 \cos t, 2 \sin t, 0), 0 \leq t \leq 2\pi$.

$\vec{F} = (3x-y, yz, xy)$ so $\vec F(r(t)) = (6 \cos t - 2 \sin t, 0, 4 \sin t \cos t)$

Now find $r'(t)$, do the dot product and complete the line integral. That is same as the flux of $curl \vec F$ over the given hemisphere.

Title: Calculate and find the flux integral


问:
Teacher claims this proof for $\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$ is wrong. Why?
My son's high school teacher says his solution to this proof is wrong because it is not "the right way" and that you have to "start with one side of the equation and prove it is equal to the other". After reviewing it, I disagree. I believe his solution is correct, even if not "the right way", whatever that means. I asked my son how he did it: he cross-multiplied the given identity, simplified it to a known/obvious equality, and then reversed the steps for the proof. This was a graded exam, and the teacher gave him a zero for this problem.

What do you think about my son's solution? Thanks!

Problem: prove the following trigonometric identity
\begin{align*}
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}\ .\\
\end{align*}
Solution:  for all real $\theta$ not equal to an integer multiple of $\pi/2$, we have
\begin{align*}
\cot^2(\theta)&=\cot^2(\theta)\\[8pt]
\frac{\cos^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\frac{1-\sin^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\csc^2(\theta)-1&=\cot^2(\theta)\\[8pt]
\frac{\csc^2(\theta)-1}{\cot(\theta)}&=\cot(\theta)\\[8pt]
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}
\end{align*}

答:
I would assume that the teacher considered it an invalid answer, because your son assumed that the equality is true before showing that it is, if that even makes any sense. This is obviously a guess, but in the mind of the teacher, what he did was not "showing" that the equality is true, but merely "verifying" that it is. In other words, what the teacher likely wanted is for the students to get the answer on the right-hand side by exclusively using the left-hand side. More explicitly, even though the students knew what the end result should look like, they should have pretended they don't, manipulated the left-hand side and exclaimed "Eureka! It's true!" when they got the RHS.

With that logic, all he needed to do was to multiply numerator and denominator by $\csc(\theta)+1$, which immediately yields the desired result after simplifications.


答:
Use the chain rule.

$$ f(x)=\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$$

We want $\frac{d^2 f(\sqrt{x})}{dx^2}$ which can be simplified as:

$$ \frac{d^2 f(\sqrt{x} )}{dx^2} = \frac12  \frac{d}{dx}( \frac{df}{dx}|_{\sqrt{x} } \frac{1}{ \sqrt{x}}) =\frac12 \left[ \frac{d^2f}{dx^2}|_{\sqrt{x}} \frac{1}{2x} -\frac{1}{2} \frac{df}{dx}|_{\sqrt{x}} \frac{1}{x^{\frac32}}\right] $$

Now, here's the interesting part put $x=1$

$$ \frac{d^2 f \circ (\sqrt{x})}{dx^2}|_{x=1}= \frac{1}{4} \left[\frac{d^2 f}{dx^2}|_{1} - \frac{df}{dx}|_{1} \right]$$

And, the right side is now easy to evaluate on the given expression

Title: Evaluating $\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1}$

Body:
I am seeking to evaluate the sum $$S=\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1} \\ =(n-1)^2 \binom{n}{1}+(n-3)^2 \binom{n}{3}+(n-5)^2\binom{n}{5}+\cdots$$

I re-wrote the sum as $$S=(n+1)^2\sum_{r \in \mathbb{N}} \binom{n}{2r-1}-4(n+1)\sum_{r\in\mathbb{N}}r\binom{n}{2r-1}+4\sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$$

It is clear from the binomial expansion of $(1+x)^n$, (by plugging $x=\pm1$ ) we get the value of the sum $\displaystyle \sum_{r\in\mathbb{N}}\binom{n}{2r-1}$ as $2^{n-1}$.

Now, after this I can find the value of the sums $\displaystyle \sum_{r\in\mathbb{N}}r\binom{n}{2r-1}$ and  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$ by considering the binomial expansion of $\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$ and substituting $x$ for $\sqrt{x}$ and then differentiating, and by manipulations, I will be able to find the desired sums.

However, this method gets very lengthy, especially when doing differentiation for finding  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$.

I am hoping to see some other different approaches which don't involve so much calculation and are easy to understand.

The final closed form is $n(n+1)2^{n-3}$.


问:
independence of sets of a sigma algebra
I have been trying to solve this question all day, i am not good at probability theory so i am not able to solve this.

Let (Ω, A, P) be a probability space and A, B, C ∈ A and A, B, C are independent of
each other. \ Show that A and B ∪ C are independent as well.

My approaches are:

using properties of a probability space: the intersection of the disjoint sequence of events is an empty set.

P(A).P(BuC) = P(A).(P(B)+ P(C) - P(BnC))

but P(BnC) = P(empty set) = 0

therefore it remains = P(A).(P(B) + P(C)) \

I don't think this is right. \

looking at properties of a sigma algebra: if A,B,C ∈ A, this means that complements of A, B, C ∈ A, and so are their unions.

P(A) = P(A\B) + P(B\C) + P(AnBnC)

P(BuC)= P(B\C) + P(C\B) + P(BnC)

P(BuC) = P((B\C) + P(BnC)) + (P(C\B) + P(BnC)) - P(BnC)

after combining this, i don't know how to solve it further.

You don't have to give  the answer. An idea of the solution or what properties one should follow is all i am asking.

答:
Let $I \perp \!\!\! \perp J$, $I \neq J, I,J \in \{ A , B, C\}$

We want to show $A \perp \!\!\!\perp B \cup C$.

Hints:

$P(I \cap J) = P(I)P(J)$. This comes from independence

$P(I \cup J) = P(I) + P(J) - P(I \cap J) = P(I) + P(J) - P(I)P(J)$

The spoiler below is the solution

 $$P(A \cap (B \cup C))\\ = P((A\cap B) \cup (A \cap C)) \\ =  P(A\cap B)+P(A\cap C) - P((A\cap B) \cap (A\cap C)))\\ = P(A)P(B) + P(A)P(C) - P(A\cap B \cap C)\\= P(A)P(B) + P(A)P(C) -P(A)P(B)P(C)\\  = P(A)[P(B) + P(C) -P(B)P(C)]\\ = P(A)P(B \cup C)$$Hence, $A \perp \!\!\! \perp B \cup C$.

Title: independence of sets of a sigma algebra


答:
All of these methods of proof are valid, and should be graded as such (unless the question specifically said to use a certain method).

Method $1$ - Start with one side, simplify to the other.

$$\frac{\csc(\theta)-1}{\cot(\theta)}=\frac{(\csc(\theta)-1)(\csc(\theta)+1)}{\cot(\theta)(\csc(\theta)+1)}=\frac{\csc^{2}(\theta)-1}{\cot(\theta)(
\csc(\theta)+1)}=\frac{\cot^{2}(\theta)}{\cot(\theta)(\csc(\theta)+1)}$$$$=\boxed{\frac{\cot(\theta)}{\csc(\theta)+1}}$$

Method $2$ - Start with the asserted identity, simplify to a known identity.

$$\frac{\csc(\theta)-1}{\cot(\theta)}=\frac{\cot(\theta)}{\csc(\theta)+1}\Longleftrightarrow(\csc(\theta)-1)(\csc(\theta)+1)=\cot^{2}(\theta)\Longleftrightarrow\csc^{2}(\theta)-1$$$$=\cot^{2}(\theta)\ \blacksquare$$

Method $3$ - Start with a known identity, turn it into the asserted identity (which is what your son did):

$$\csc^{2}(\theta)-1=(\csc(\theta)+1)(\csc(\theta)-1)=\cot^{2}(\theta)\Longrightarrow\boxed{\frac{\csc(\theta)-1}{\cot(\theta)}=\frac{\cot(\theta)}{\csc(\theta)+1}}$$

That said, however, perhaps the teacher took points off due to the length of your son's proof, which was more roundabout than it could have been. For instance, he could have started with a Pythagorean identity instead of starting with a '$x=x$' identity and using a Pythagorean identity, and took a few less steps in simplifying to the desired identity.

Title: Teacher claims this proof for $\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$ is wrong. Why?

Body:
My son's high school teacher says his solution to this proof is wrong because it is not "the right way" and that you have to "start with one side of the equation and prove it is equal to the other". After reviewing it, I disagree. I believe his solution is correct, even if not "the right way", whatever that means. I asked my son how he did it: he cross-multiplied the given identity, simplified it to a known/obvious equality, and then reversed the steps for the proof. This was a graded exam, and the teacher gave him a zero for this problem.

What do you think about my son's solution? Thanks!

Problem: prove the following trigonometric identity
\begin{align*}
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}\ .\\
\end{align*}
Solution:  for all real $\theta$ not equal to an integer multiple of $\pi/2$, we have
\begin{align*}
\cot^2(\theta)&=\cot^2(\theta)\\[8pt]
\frac{\cos^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\frac{1-\sin^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\csc^2(\theta)-1&=\cot^2(\theta)\\[8pt]
\frac{\csc^2(\theta)-1}{\cot(\theta)}&=\cot(\theta)\\[8pt]
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}
\end{align*}


答:
It's disappointing to see that the teacher believes there is only one way to prove trigonometric identities. Actually, there are a number of approaches, all equally valid and useful in different circumstances. Sometimes it is more elegant to 'start with one side and prove that it is equal to the other', but a proof is a proof.

Your son is actually using quite an advanced technique, where he shows that a statement $P$ implies another statement $Q$, and then shows that all of the steps are invertible, meaning that $P \iff Q$. As long as you are able to show that all of the steps are indeed invertible, this is a very useful method. Let's look at his method in more detail. He started by assuming that
$$
\frac{\csc \theta - 1}{\cot\theta} = \frac{\cot\theta}{\csc\theta+1}
$$
and working from there. If we cross mutliply both sides, look at what happens:
\begin{align}
&\frac{\csc \theta - 1}{\cot\theta} = \frac{\cot\theta}{\csc\theta+1} \\[6pt]
\implies & (\csc\theta-1)(\csc\theta+1) = \cot^2\theta \\[6pt]
\implies & \csc^2\theta-1=\cot^2\theta \\[6pt]
\implies & \frac{1-\sin^2\theta}{\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} \\[6pt]
\implies & \frac{\cos^2\theta}{\sin^2\theta}=\frac{\cos^2\theta}{\sin^2\theta} \\[6pt]
\implies & \cot^2\theta = \cot^2\theta \, .
\end{align}
Now, try reading the proof 'backwards'. Does $\cot^2\theta = \cot^2\theta$ imply $$
\frac{\cos^2\theta}{\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} \, ?
$$
Does $\frac{\cos^2\theta}{\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta}$ imply
$$
\frac{1-\sin^2\theta}{\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} \, ?
$$
Keep going. If the answer to all of these question is 'yes', then we have successfully proven that
$$
\frac{\csc \theta - 1}{\cot\theta} = \frac{\cot\theta}{\csc\theta+1} \iff \cot^2\theta = \cot^2\theta \, .
$$
And since the RHS is an identity, the LHS must also be an identity. There's just one tiny snag to this proof. When we are going in a backwards direction, and conclude that
$$
(\csc\theta-1)(\csc\theta+1) = \cot^2\theta
$$
implies
$$
\frac{\csc \theta - 1}{\cot\theta} = \frac{\cot\theta}{\csc\theta+1} \, ,
$$
we are assuming that both $\cot\theta$ and $\csc\theta + 1$ are non-zero. But, your son directly addresses this by stating at the start of the proof, 'for all real $\theta$ not equal to an integer multiple of $\pi/2$...', meaning that this is never a problem. This kind of attention to detail is very impressive for a high-school student.


问:
Freeness of stalk Implies locally free
Let $ A $ be a Noetherian ring, and $ M $ a finitely generated $ A $ module. Suppose that $  \mathfrak  { p } \in M $ such that $ M_{\mathfrak{p}} $ is free. Show that there is a $ f \in A  \setminus \mathfrak{p} $ such that $ M_{f} $ is free over $ A_{f} $. </p>

<p>P.S. Some related questions are
1) Flatness and Local Freeness
2)  Locally free sheaves, though, both of these don't answer the specific question that I have above, in that I am just looking around one prime (so my module is not projective etc). I have seen this in Vakil but I can't find it at the moment. I will post my proof of the fact above below but I would like to see what are some other ways to do it.  </p>

答:
$\newcommand{\gp}{{\mathfrak p}}$Let $\left\{ \frac{m_1}{s_1}, \cdots, \frac{m_n}{s_n} \right\}$ be an $A_\gp$-basis of $M_\gp$, where $m_i \in M$ and $s_i \in A-\gp$. Observe that $\{m_1, \cdots, m_n\}$ is still a $A_\gp$-basis of $M_\gp$.

This gives an map $A^n \to M$ that localizes to an isomorphism $A_\gp^n \to M_\gp$ at $\gp$. Let $K$ and $Q$ be the kernel and cokernel respectively of the map $A^n \to M$. This gives an exact sequence:
$$0 \to K \to A^n \to M \to Q \to 0$$

Localizing is exact, so we have an exact sequence:
$$0 \to K_\gp \to A_\gp^n \to M_\gp \to Q_\gp \to 0$$

But we know that the map $A_\gp^n \to M_\gp$ is an isomorphism, so we must have $K_\gp = 0$ and $Q_\gp = 0$.

$A$ is noetherian, and $K$ is a submodule of $A^n$, so $K$ is finitely generated, so the fact that $K_\gp = 0$ implies that there is $s \in A-\gp$ such that $K_s = 0$; $M$ is finitely generated, and $Q$ is a quotient of $M$, so $Q$ is finitely generated, so the fact that $Q_\gp = 0$ implies that there is $t \in A-\gp$ such that $Q_t = 0$.

Now take $f = st \in A-\gp$. Localize the exact sequence at $f$ to get:
$$0 \to K_f \to A_f^n \to M_f \to Q_f \to 0$$

But $K_f = (K_s)_t = 0_t = 0$ and $Q_f = (Q_t)_s = 0_s = 0$, so the exact sequence becomes:
$$0 \to 0 \to A_f^n \to M_f \to 0 \to 0$$

So $M_f$ is free over $A_f$ as required.

Title: Freeness of stalk Implies locally free


问:
How do we know if a problem is hardest in NP
I read that the definition of
NP-complete is :
    These are the hardest problems in NP. Such a problem is NP-hard and in NP

How do we know if a problem is hardest in NP, and no harder problem exists.
I understand that let's assume that somehow magically we know that a problem L is hardest in NP and then we can find out more hardest problems H if we can reduce H to L and vice versa.But my question is how does it all begin? How do we know 1 hardest problem to begin with?

Also, to be able to say that something is hardest (or any extreme), we need to know all possible problems in NP and then argue about the hardest.. How do we know all possible NP problems? Is this where turing machine comes useful and by using representation of output string in form of 1 and 0 in output tape, we can theoretically talk about all possible NP problems.

I understand that I may not have been able to articulate my question well - due to confusion.

Thanks,

答:
It all began with the Cook-Levin theorem. The proof of that uses the Turing Machine definition of NP and reduces every problem in NP to SAT.

Any textbook on Computational Complexity will have at least one section devoted to this. I would recommend Papadimitrou's book on Computational Complexity for that. I heard Sipser's book is good too, but I haven't read that.

So to answer your question, yes, one could use (and the above theorem actually does) the Turing Machine to talk about all problems in NP and prove the existence of NP-Hard problems.


问:
Question about circles.
Find the value of x. If necessary, round your answer to the nearest tenth. The figure is not drawn to scale.

AB = 19, BC = 10, and CD = 5

A)23

B)53

C)38

D)58

What theorem should I use? And how do I use that theorem?

Title: Question about circles.

Body:
AC = AB + BC = 19 + 10 = 29

if AC*BC=CD*(CD+X), then 29*10 = 5*(5+x)

290 + 25 + 5x

5x = 265

x = 53


问:
Find all primes such that $\prod_{i=1}^{n} p_i=10\sum_{i=1}^{n}p_i$
Find all primes $p$ (it's not necessary to this primes to be different) such that $$\prod_{i=1}^{n} p_i=10\sum_{i=1}^{n}p_i$$

I’ve realized that $$10\mid \prod_{i=1}^{n} p_i\implies \prod_{i=1}^{n} p_i =2\cdot5...$$

But i don’t know what is next.

答:
Using your observation, the equation can be reduced to (here I replaced $n-2$ with $n$):

$$\prod_{i=1}^n p_i = 7+\sum_{i=1}^np_i$$

It helps to first establish upper and lower bounds for the expressions. The product grows much faster than the sum, so we will find a lower bound for the product and an upper bound for the sum.

Let $p$ be the largest prime among $p_i$. Then we have

$$\prod_{i=1}^n p_i \ge 2^{n-1}p, \quad \sum_{i=1}^np_i \le np$$

therefore there are no solutions if $2^{n-1}p > 7+np$.

Obviously $p\ne 2$ and $n\ne 1$. Since $p\ge 3$, we can check that $n < 4$. This helps reduce the number of cases to check.

For $n=2$, we have the equation $pq = 7 + p + q$, or $(p-1)(q-1)=8$. By consider the factors of $8$, we must have $p=5, q=3$ (and not $p=9, q=2$).

For $n=3$, by a parity argument we see that there is exactly one $2$ in $p_i$.  Hence we have the equation

$$2pq = 7+2 + p + q \leadsto4pq-2p-2q+1 = 18+1 \leadsto(2p-1)(2q-1) = 19$$

but by considering the factors of $19$, we see that a prime solution to $(p,q)$ is impossible.

Therefore the only set of primes satisfying our original equation is $(2,3,5,5)$, with

$$2\times 3 \times 5 \times 5 = 10 (2+3+5+5)$$


问:
Finding the minimum number of terms in an alternating series to be accurate to be accurate to given value
I have the following question here.

How many terms do we need to sum for $\displaystyle \sum_{n=1}^{\infty} \sin\left( \frac{(-1)^{n+1}}{n^2} \right)$ so the error in the sum is less than $0.001$?

My attempt:

I know the alternating series estimation theorem says the following:

If $S=\displaystyle \sum_{n=1}^{\infty} (-1)^nb_n$ is the sum of an alternating series that satisfies:

i) $0\leq b_{n+1} \leq b_n$

ii) $\displaystyle \lim_{n \to \infty} b_n =0$

then $|R_n|=|S-S_n|\leq b_{n+1}$

Cool. So the official solutions does the following:

$\sin\left( \frac{1}{(n+1)^2} \right)<0.001$

$\frac{1}{(n+1)^2} < \arcsin(0.001)$

$n>30.622...$

$n\geq 31$

Why is this allowed though?

$\displaystyle \sum_{n=1}^{\infty} \sin\left( \frac{(-1)^{n+1}}{n^2} \right)$ isn't in the form $S=\displaystyle \sum_{n=1}^{\infty} (-1)^nb_n$ since the Sin term is in the way. Can someone explain how this actually works? Thanks!

答:
As Robert Israel said, $\sin(\frac{(-1)^{n+1}}{n^2}) = (-1)^{n+1} \sin(1/n^2)$.

So the theorem applies: let $b_n = \sin(\frac{1}{n^2})$, and for $n>0$, we know that

$0 \leq \sin(\frac{1}{(n+1)^2}) \leq \sin(\frac{1}{n^2})$

$\sin(\frac{1}{n^2}) \rightarrow 0$ as $n\rightarrow \infty$.

To apply the theorem, you use $b_{n+1} = \sin(1/(n+1)^2)$.


问:
Is $\det(AB) = \det(A)\det(B)$ for entries in commutative ring?
It is well known (and the proof is simple) that $\det(AB)=\det(A)\det(B)$ when $A,B\in M_n(F)$ where $F$ field. But what about for $A,B\in M_n(R)$ where R is any arb. commutative ring?

A field is distinguished from a com. ring by the presence of multiplicative inverses. I feel like this isn't a necessary condition for proving the determinant function is a homomorphism but I'm having a hell of a time trying to do so.

How would I go about proving the above?

答:
Here's a proof based on the abstract definition of the determinant:

Suppose $R$ is a commutative ring and $M$ is an $R$-module. Define $\bigwedge^n M$ to be the quotient of $M^{\otimes_R n}$ by the sub-$R$-module generated by the pure tensors $x_1\otimes ... \otimes x_n$ such that there exist $i\neq j$ with $x_i = x_j$. The image of the pure tensor $x_1\otimes ... \otimes x_n$ in this quotient is usually written as $x_1\wedge ... \wedge x_n$.

Note that by expanding out $(x+y)\otimes (y+x)$, it follows that in this quotient we have relations analogous to $x\wedge y = - y\wedge x$, and more generally, $x_{\sigma(1)}\wedge ... \wedge x_{\sigma(n)} = \epsilon(\sigma) x_1\wedge ... \wedge x_n$, where $\epsilon(\sigma)$ is the signature of the permutation $\sigma$.

Then

Lemma 1: If $M=R^n$ is free of rank $n$, $\bigwedge^k R^n$ is free of rank $\binom{n}{k}$, a basis being given by the $e_{i_1}\wedge ... \wedge e_{i_k}, i_1 <...< i_k$ (the $e_i$'s being the canonical basis of $R^n$)

It's fairly straightforward to prove that they generate $\bigwedge^k R^n$, slightly more complicated to show that they are linearly independent, but they are.

In particular, with $k=n$, $\bigwedge^n R^n$ is free of rank $1$. In this case, one possible way to show that $e_1\wedge ... \wedge e_n$ is not zero is to show that $\det : (R^n)^n \to R$ actually factors through $(R^n)^n \to\bigwedge^n R^n$ (that's because $\det$ is $n$-linear and satisfies the condition that if two of the rows of the matrix are equal, then $\det = 0$) and has value $1$ on $(e_1,...,e_n)$. This is the only case we will need, so this is a possible proof if you already know what $\det$ is.

Note that this proof actually shows more : it shows that $\det : \bigwedge^n R^n\to R$ is an isomorphism.

Lemma 2: Any $R$-linear map $f:M\to N$ induces an $R$-linear map $\bigwedge^n f: \bigwedge^n M\to \bigwedge^n N$ given by $x_1\wedge ... \wedge x_n \mapsto f(x_1)\wedge ... \wedge f(x_n)$. In particular, $\bigwedge^n (f\circ g) = \bigwedge^nf \circ \bigwedge^n g$

This is the key point, and is not hard to show.

Proposition 1: Let $f: R^n\to R^n$ be an $R$-linear map. Then $\bigwedge^n f$ is an $R$-linear endomorphism of a free $R$-module of rank $1$, so it's a homotethy. The homotethy in question is given by $\det(f)$.

Proof: we showed that $\det : \bigwedge^n R^n \to R$ was an isomorphism of $R$-modules, with inverse given by $1\mapsto e_1\wedge ... \wedge e_n$. Therefore it suffices to show that $\det\circ \bigwedge^n f(e_1\wedge...\wedge e_n) = \det(f)$. But that's clear, because $\bigwedge^nf(e_1\wedge...\wedge e_n) = f(e_1)\wedge ...\wedge f(e_n)$ by definition, and $\det(f)$ is by definition $\det(f(e_1),...,f(e_n))$.

Note : this might seem a bit circular to people who define the determinant by this proposition. Here I'm assuming that the OP knows what the determinant is, and proving, based on that, that it's equivalent to this.

Corollary (Conclusion) : $\det(f\circ g) =\det(f)\det(g)$.

Proof: By Proposition 1, $\bigwedge^n (f\circ g)$ is given by multiplication by $\det(f\circ g)$. But by lemma 2, $\bigwedge^n (f\circ g) =\bigwedge^n f\circ \bigwedge^n g$ is given by the composition of multiplication by $\det(f)$, and the multiplication by $\det(g)$, so multiplication by $\det(f)\det(g)$. Therefore, because it is over a free module, $\det(f\circ g) = \det(f)\det(g)$.

Let me make another note: it seems like we did no hard work at all. This is only what it looks like, the hard work is in the definition of the determinant, essentially in showing that $\bigwedge^n R^n$ is in fact free of rank $1$. The proof I gave here was that we had an explicit isomorphism, given by the determinant, this presupposes we know that the determinant exists, is $n$-linear, and satisfies the condition on matrices with two equal rows.


问:
Computing a differential on the loop space of a Symplectic manifold
Suppose we have a symplectic manifold $(M,\omega)$ such that $\omega$ is exact ,i.e, $\omega=d\alpha$.  Now let $L_0$ and $L_1$ be two lagrangian submanifolds and consider $\Omega(L_0,L_1)$ to be the set of smooth paths $\gamma:[0,1]\rightarrow M$ such that $\gamma(0)\in L_0$ and $\gamma(1)\in L_1$.  Now the "tangent space" at a point of this space is just the space of vector fields $\xi(t)$ along that curve, such that $\xi(i)\in T_{\gamma(i)}L_i$ for $i=0,1$.

Define $A:\Omega(L_0,L_1)\rightarrow \mathbb{R}$ as $A(\gamma)=\int_{0}^{1}\gamma^*\alpha$, in local coordinate we have $A(\gamma)=\int_{0}^{1}\alpha(\frac{\partial \gamma}{\partial t})dt$.

Now given a family of smooth almost-complex structures $J_t$  we have a canocanil family of riemannian metrics $g_t$ and can define the inner product in $\Omega(L_0,L_1)$ as $\langle v_1, v_2\rangle=\int_{0}^{1}g_t(v_1(t),v_2(t))dt$.

Now with this I would like to compute the gradient of $A$. My attempt is the following :

$dA_{\gamma}(\xi)=\frac{\partial }{\partial s}|_{s=0}A(c(s))$, where $c(s):(-\epsilon,\epsilon)\rightarrow \Omega(L_0,L_1)$ such that $c(0)=\gamma$ and $\frac{\partial }{\partial s}|_{s=0}c(s,t)=\xi(t)$, and this will be equal to $\int_{0}^{1} \frac{\partial }{\partial s}|_{s=0} \alpha(\frac{c(s,t)}{\partial t})dt$ , and now from here I don't know what to do. I belive the answer is supposed to be that $grad A(\gamma)=J_t\frac{\partial \gamma}{\partial t}$, but I don't know how to get there.

Any help is appreciated. Thanks in advance.

答:
First, recall the following formula:

Proposition. Let $V$ be a smooth manifold, $(\lambda_t)_t$ be a smooth path of $p$-forms of $V$ and $X_t$ be a time-dependent vector field on $V$ whose local flow is $(\phi_t)_t$, then the following holds true:
$$\frac{\operatorname{d}}{\operatorname{d}t}{{\phi_t}^*\lambda_t}_{\vert t=t_0}={\phi_{t_0}}^*\left(\frac{\operatorname{d}}{\operatorname{d}t}{\lambda_t}_{\vert t=t_0}+L_{X_{t_0}}\lambda_t\right).$$

Proof. This follows easily writing the infinetisimal rate of change of $t\mapsto {\phi_t}^*\lambda_t$ at $t_0$. $\Box$

Then, writing $c_s(t)=c(s,t)$ and using Cartan's magic formula for the Lie derivative of time-dependent vector fields (It suffices to show that the formula holds for functions and is well-behaved with respect to exterior derivative and wedge product. A full proof can be found in Lemma B.2. of An introduction to contact topology by H. Geiges) gives:
$$\frac{\partial}{\partial s}{c_s}^*\alpha_{\vert s=0}=\gamma^*L_\xi\alpha=\gamma^*\left(\iota_\xi\operatorname{d}\alpha+\operatorname{d}\iota_{\xi}\alpha\right).$$
Using that $\omega=\operatorname{d}\alpha$, $\xi$ is a section of $\gamma^*TM$ and that the exterior derivative and pullback commute, yields:
$$\frac{\partial}{\partial s}{{c_s}^*\alpha}_{\vert s=0}=\omega(\xi,\dot{\gamma})+\operatorname{d}\iota_\xi\alpha.$$
At last, recalling that $g(\cdot,\cdot)=\omega(\cdot,J\cdot)$ and $J^2=-1$, $\xi(0)\in T_{\gamma(0)}L_0$ and $\xi(1)\in T_{\gamma(1)}L_1$ gives:
$$T_\gamma A(\xi)=\int_0^1g(\xi,-J\dot{\gamma})+\alpha(\xi(1))-\alpha(\xi(0)).$$
I am not able to conclude as $\alpha(\xi(1))-\alpha(\xi(0))$ may not vanish...


问:
In a finite affine space $(\mathbb{Z}/p)^3$, has at most $p^2 +p +1$ lines passing through a point.
It is a simple question not much backgrond information is required. However, I do need ton understand the proof of why.

Prove the finite affine space $(\mathbb{Z}/p)^3$, has only $p^2 +p +1$ lines passing through a point.

Also does this generalise for $(\mathbb{Z}/p)^n$ have a at most  $p^{n-1}+p^{n-2}+\cdots+p+1$ lines passing through a point.

Look forward to your solutions.

答:
Assuming you mean $\mathbb{Z}/p$.

Fix an element $x$. Then there are $p^3-1$ other points $y$ such that $xy$ is a line through $x$. But we have double-counted, because each line goes through other points too.

But of course, the points on each line are just $x$, $x+(x-y)=2x-y$, $3x-2y$, etc. until we get to $px-(p-1)y=y$ and we start to repeat. So each line contains at most $p-1$ distinct points (other than $x$). Are they all distinct? Well, if two of them are equal, then there must be $m,n \in \mathbb{Z}/p$ with $(m-n)(x-y) \equiv 0$, which can only happen if $m=n$ or $x=y$.

So there are $$\frac{p^3-1}{p-1}$$
distinct lines that pass through your given point. I'll leave it to you to see that that's equal to your formula, and how it generalises.


答:
I would at least have skipped the first line and started with $\dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta.$

One reason why an instructor might have qualms about this argument is the frequency with which students do this in the opposite direction, i.e. they write
$$
\require{cancel}
\xcancel{
\begin{align}
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1} \\[8pt]
\frac{\csc^2(\theta)-1}{\cot(\theta)}&=\cot(\theta)\\[8pt]
\csc^2(\theta)-1&=\cot^2(\theta)\\[8pt]
\frac{1-\sin^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\frac{\cos^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\cot^2(\theta)&=\cot^2(\theta)
\end{align}
}
$$
This would be logically correct if one wrote "This equality is true if the one below it is true." on each line. But one must be clear about the correct direction of "If ... then ...". Without those explicit words, the sequence of equalities can be taken to mean "If ... then ..." in just the opposite order.

Often what instructors want is something like this:
$$
\frac{\csc\theta-1}{\cot\theta} = \cdots\cdots\cdots\cdots = \frac{\cot\theta}{\csc\theta+1}.
$$
in which each "equals" sign asserts the equality of things already known to be equal.

I would give full credit to the answer that your son wrote if each line had some brief explanation of how it is deduced from the line before it. For example, he could write this:

$$ \cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta} = \frac{1-\sin^2\theta}{\sin^2\theta} = \csc^2\theta-1. $$ Then, dividing both sides of the equality $\cot^2\theta = \csc^2\theta-1$ by $\cot\theta,$ we get $$ \cot\theta = \frac{\csc^2\theta-1}{\cot\theta}. $$ Finally, dividing both sides of that by $\csc\theta+1$ we get $$ \frac{\csc\theta-1}{\cot\theta} =  \frac{\cot\theta}{\csc\theta+1}. $$

Note that I wrote words above, not just mathematical notation. Well written answers do that, except perhaps in fairly simple cases.


问:
Is every abstract simplicial complex the independence complex of a simple graph?
Given a simple (no loops, no multi-edges) undirected graph $G$ on $n$-vertices, one can assign an abstract simplicial complex known as the independence complex (https://en.m.wikipedia.org/wiki/Independence_complex) on $n$-vertices whose faces are exactly the independent sets of the graph $G$.

My question is: Is every abstract simplicial complex the independence complex of a simple graph?

答: If $X$ is a subset of the vertex set of a graph and if every $2$-element subset of $X$ is independent, then $X$ itself is independent. So the simplicial complex consisting of all the $\leq2$-element subsets of $\{1,2,3\}$ is not the independence complex of any graph.


问:
Change of variables in double integrals intuitively
So I have been trying to understand the intuitive proof of change of variables in double integrals.

Equations $$ x=x(u, v) \ \mathrm{and} \ y=y(u,v) $$
define a mapping from uv-Cartesian plane to points in xy-plane. The mapping is one-to-one.

The differential area element $$ dA = dx dy$$ is bounded in xy-plane by $$ u, \ u+du,\ v, \ \mathrm{and} \  v+dv.$$
Now the question is, what is dA in uv-plane?

In most intuitive proofs it is said that since du and dv are small and the area element is bounded by smooth curves, the area element is approximately a parallelogram and therefore its are is $$ dA = \left|\vec{AC} \times \vec{AB} \right|=\dots= \left|\frac{\partial(x, y)}{\partial(u, v)} \right| du dv,$$
where
$$ \vec{AB}=\frac{\partial x}{\partial u}du\hat{i}+\frac{\partial y}{\partial u}du\hat{j} \ \mathrm{and} \ \vec{AC}=\frac{\partial x}{\partial v}dv\hat{i}+\frac{\partial y}{\partial v}dv\hat{j}.$$

But if we approximate the area element as a parallelogram, shouldn't it be $$ \left \Vert  \vec{AB} \right \Vert = du \ \mathrm{and} \ \left \Vert  \vec{AC} \right \Vert = dv$$
and therefore we would have a different result for dA?

Edit: So my question really is, are $$ du=\left \Vert \frac{\partial x}{\partial u}du\hat{i}+\frac{\partial y}{\partial u}du\hat{j} \right \Vert  \ \mathrm{and} \ dv=\left \Vert \frac{\partial x}{\partial v}dv\hat{i}+\frac{\partial y}{\partial v}dv\hat{j} \right \Vert ?$$

答:
How so? The reason we use the cross product is because the area element spanned by two vectors is given by the cross product of the two. If you do the way you are doing, then you need to take one side as base and find the 'height' of the parellogram, which will end up in the same formula. Here is a more explicit calculation of the proof you mentioned, consider a coordinate system $(x,y)$ and another old coordinate system $(u,v)$ , let's say we have a map from $(u,v)$ system to $(x,y)$ system then we can write:

$$ dx= < \frac{\partial x}{\partial u} du ,   \frac{\partial x}{\partial v} dv>$$

The idea here is the amount of increase in $u$ direction as we increase little bit in x in original system, is just slope of $x$ with $u$ times $u$ and similarly for in v direction

$$ dy = < \frac{\partial y}{\partial u} du , \frac{\partial y}{\partial v} dv>$$

So, it's like we have an area spanned by $dx$ and $dy$ in our original system, then the expression  equivalent area produced in the other system can be found by cross product the two vectors, that is:

$$ |dA| = dx \times dy = < \frac{\partial x}{\partial u} du ,   \frac{\partial x}{\partial v} dv> \times < \frac{\partial y}{\partial u} du , \frac{\partial y}{\partial v} dv>$$


问:
How can I calculate the Eisenstein series of a complex lattice?
Suppose I have a lattice $\Lambda = \mathbb{Z}+\frac{3}{2}i\mathbb{Z}$. How would I go about calculating $G_{2n}(\Lambda)$ for a given $n \in \mathbb{N}$?

答:
Up to the $\zeta(4),\zeta(6)$ constants this is $E_4,E_6$, some modular forms with rational coefficients.

$E_6(i)=0$ and $E_4(i)$ is given in term of the beta function $\int_0^\infty \frac{dx}{\sqrt{x^3+x}}$, then
$E_{2k}(z)\in \Bbb{Z}[E_4(z),E_6(z),\Delta(z)]$ so $E_{2k}(i\frac32)\in \Bbb{Z}[\frac16,E_4(i\frac32),E_6(i\frac32)]$

and $E_4(i\frac32),E_6(i\frac32)$ are algebraic over $\Bbb{Z}[E_6(i)]$, this is because the coefficients of $$\prod_{\gamma \in \Gamma_0(6)\backslash SL_2(\Bbb{Z})} (X-E_6(\frac32 \gamma(z)))$$ are modular forms $\in M_{(d-n)6}(SL_2(\Bbb{Z}))$ with rational coefficients, so they are in $\Bbb{Q}[E_4(z),E_6(z)]$.


问:
What does "uniquely determined" mean in a particular statement?

The objects of a category are in bijective correspondence with the
identity morphisms, which are uniquely determined by the property
that they serve as two-sided identities for composition. Thus, one can
define a category to be a collection of morphisms with a
partially-defined composition operation that has certain special
morphisms, which are used to recognize composable pairs and which serve as two-sided identities.

I know that such expression usually means that the determined object is a function of the other, but I feel that here, there is a logical meaning that I'm ignoring. Because of it, I can't see how this fact implies that such an alternative definition of a category is possible.

答:
I would take it to mean that for each object, the identity morphism is the only morphism that has that property.

I.e. if for every morphism $f$ from an object to itself, one has $\iota \circ f = f = f\circ\iota$ and $\varepsilon\circ f=f = f\circ\varepsilon,$ then $\iota=\varepsilon.$


问:
Continuous Function and Open Subsets in $\mathbb R$
Let $E$ be a subset in $\mathbb R$, $f$ a real-value function on $E$.

Prove that $f$ is continuous on $E\iff$ for every open subset $V$ of $\mathbb R$, $f^{-1}(V)$ in open relative to $E$.

My question is about the ($\Rightarrow$) direction only.

Let $f$ be a continuous function on $E$ and $V$ a open subset on $\mathbb R$.

If $f^{-1}(V)=\{\}$, then it is open.  Suppose that $f^{-1}(V)\neq\{\}$. Let $p\in f^{-1}(V)$.

Then $f(p)\in V$.  Select $\epsilon$ such that $N_\epsilon(f(p))\subset V$.

My question is this.  At this point, we do not know if $p$ is an element of $E$.

If $p\in E$, since $f$ is continuous on $E$, $\exists\delta$ such that $f(x)\in N_\epsilon(f(p))$ for all $x\in N_\delta(p)\cap E$.

Thus $N_\delta(p)\cap E\subset f^{-1}(V)$.

But, suppose that $p\notin E$.  How do I know that the above statement is still true?

I tried the following:

Let $q\in E$ be a point such that $f(q)\in N_\epsilon(f(p))$

Select $\alpha$ such that $N_\alpha(f(q))\in N_\epsilon(f(p))$.

Then $\exists\delta$ such that $f(x)\in N_\alpha(f(q))$ for all $x\in N_\delta(q)\cap E$.

But this only shows that $N_\delta(q)\cap E\subset f^{-1}(V)$, not $N_\delta(p) ....$

I also thought about showing that if $p\notin E$, then $N_\delta(p)\cap E=\{\}$,

but I have no idea about how to do it.

答:
You must change only one step in your proof:

When you say:

"If $f^{-1}(V)=\{\}$ then it is open"

reeplace by

"If $f^{-1}(V)\cap E=\{\}$" then $f^{-1}(V)$ is open relative to $E$".

Then the following line must be

"suppose now $f^{-1}(V)\cap E\neq\{\}$ then exists $p\in f^{-1}(V)\cap E$"

and your problem was solved since $p\in E$.


问:
Normalizing an array so the summation of the array is equal to the size of the array
I am trying to normalize the array, so that the sum of their elements is equal to the size of the array. If $\alpha_i$ are the elements of the array of size $C$, I want to normalize the array, so the property below holds.

Is it possible to normalize the array as described above? If yes, how can I approach the problem. Thank you!

答:
Do it in two steps. First get the original sum of the array, call that $A$ and divide each array entry by $A.$ At this point array sum is $1.$ Second multiply each of the new array's  elements by your desired sum $C.$ After both steps the array sum will be $C.$ You can check this yourself.

Note this method won't work if the original array sum is zero. [Can't divide by zero.]


问:
Basic questions on Turing degrees and the jump operator
I'm trying to understand the notation $0^{(k)}$ (which is related to Turing degrees).

Here's what I understand: we have the equivalence relation on $\mathscr P(\mathbb N)$: $A$ is equivalent to $B$ iff $A\leq_T B$ and $B\leq_T A$.  For a set of natural numbers $A$, let $A'=K^A=\{n\in\mathbb N: \phi_n^A(n)\downarrow\}$. (The notation $\phi_n^A(x)\downarrow$ means if we run program number $n$ on input $x$ and use $A$ as an oracle, then it halts.)

$0$ is the equivalence class of any computable set

$0'$ is the equivalence class of the set $K=\{n\in \mathbb N: \phi_n(n)\downarrow\}$

$0''$ is the equivalence class of the set $K'=K^K=\{n\in\mathbb N: \phi_n^K(n)\downarrow\}$

Is this correct so far?

Then $0'''$ should be the equivalence class of $K''=(K^K)^K=\{n\in\mathbb N:(\phi_n^K)^K(x)\downarrow\}$. Is this right? This is harder to process for me. How should I think of $(\phi_n^K)^K(x)\downarrow$? What does it mean intuitively? (Above I described  $\phi_n^A(n)\downarrow$ as "if we run program number $n$ on input $n$ and use $A$ as an oracle, then it halts", how to describe $(\phi_n^K)^K(x)\downarrow$ in a similar manner?)

And in general, I guess $0^{(n)}$ is the equivalence class of $K^{(n-1)}$.

答: $0'''$ is the equivalence class of the halting problem for computations that use oracle $K^K$. This would be written $K^{(K^K)}$, not $(K^K)^K$. (The latter notation hasn't really been defined. Your attempted definition of it uses the notation $(\phi_n^K)^K$, which hasn't been defined.)


问:
Proving homomorphism from functions
I am really struggling to grasp the concept of homomorphism as it has been poorly explained to me in courses, and I have not been able to find any clear examples of how to actually classify a function correctly. For instance, I have the function:

substract, that takes a list of integers $[x_1,...,x_n]$ as input and computes the result as a function of: $$x_1 - \sum_{i=2}^n x_i,\quad n > 0.$$

So with an example this would be:

$\operatorname{substract}([19,5,8]) = 19 - (5 + 8) = 6$.

But how should I approach showing whether this is a $homomorphism$ or not? From my understanding a $homomorphism$ is suc that I can find an operator $\oplus$ so that:

$f(x_1 * x_2) \mapsto f(x_1) \oplus f(x_2)$

For which $*$ resembles the concatenation operator on list structures.

But what would be the best case to prove an example as the one provided above? I have to classify a series of such functions with various signatures, but I want to understand the HOW I would go about doing this, as I clearly (and hornestly) cannot grasp it.

答:
I have asked similar question here, since I wasn't 100% sure.

In order for $f$ to be a homomorphism, we must define the algebraic structures first, which is straight forward.

Let $L$ be the set of all finite lists and denote the list concatenation operator by $*$ and denote $[]$ as the empty list. Then the triplet $(L,*,[])$ forms a monoid, since

$\forall a,b,c \in L: (a *b)*c=a*(b*c)$ (associativity)

$\forall a \in L: a*[]=[]*a$ (existence of the identity element)

If we omit the existence of the identity element, then $(L \setminus \{[]\},*)$ would form a semigroup.

Since $f:L \to \mathbb{R}$ only holds for $n > 0$, our domain is $L \setminus \{[]\}$, and $(L \setminus \{[]\},*)$ is a semigroup. Now, we need to find a semigroup $(\mathbb{R},\oplus)$ such that $f$ defines a (semigroup) homomorphism.

Thus,

$$f(x*y)=f([x_1,\cdots,x_n]*[y_1,\cdots,y_m]) = f([x_1,\cdots,x_n,y_1,\cdots,y_m]) \\ \iff x_1-(\sum_{i=2}^n x_i + \sum_{j=1}^m y_j)=x_1-(\sum_{i=2}^n x_i + y_1 + \sum_{j=2}^m y_j)=x_1-\sum_{i=2}^n x_i-y_1-\sum_{j=2}^m y_j \\
\iff f(x) - f(y)$$

And therefore, $\oplus$ must be substraction "$-$". However, $(\mathbb{R}, -)$ is not a semigroup, since "$-$" is not associative (here is a proof).

We can conlcude that given $f$ as it is defined, we can not endow $\mathbb{R}$ with a structure which makes $f$ a homomorphism.

Title: Proving homomorphism from functions


答:
Is this correct so far?

Yes.

After that, it looks like you're getting thrown off by the superscript notation. Note that if we write $K^A$ for $A'$, then $A''=(A')'=K^{K^A}$, whereas the notation $(K^K)^A$ is not defined (although it might be reasonable to interpret it as $A''$).

Similarly, instead of $(\phi^K_n)^K$, you should write $\phi^{K^K}_n$, or more clearly, $\phi^{0''}_n$. Here we're considering programs that can query $0''$ as an oracle; these programs not only know which regular programs halt but also which $0'$-oracle programs halt.

We think of  $A'$ as "the halting problem relative to $A$", i.e. (the equivalence class of) the set of programs that halt when run on an oracle machine with oracle $A$. In this sense, $0'''$ is the halting problem relative to $0''$. Every time we apply the jump operator we get a class of sets that is more difficult to compute.

Title: Basic questions on Turing degrees and the jump operator

Body:
I'm trying to understand the notation $0^{(k)}$ (which is related to Turing degrees).

Here's what I understand: we have the equivalence relation on $\mathscr P(\mathbb N)$: $A$ is equivalent to $B$ iff $A\leq_T B$ and $B\leq_T A$.  For a set of natural numbers $A$, let $A'=K^A=\{n\in\mathbb N: \phi_n^A(n)\downarrow\}$. (The notation $\phi_n^A(x)\downarrow$ means if we run program number $n$ on input $x$ and use $A$ as an oracle, then it halts.)

$0$ is the equivalence class of any computable set

$0'$ is the equivalence class of the set $K=\{n\in \mathbb N: \phi_n(n)\downarrow\}$

$0''$ is the equivalence class of the set $K'=K^K=\{n\in\mathbb N: \phi_n^K(n)\downarrow\}$

Is this correct so far?

Then $0'''$ should be the equivalence class of $K''=(K^K)^K=\{n\in\mathbb N:(\phi_n^K)^K(x)\downarrow\}$. Is this right? This is harder to process for me. How should I think of $(\phi_n^K)^K(x)\downarrow$? What does it mean intuitively? (Above I described  $\phi_n^A(n)\downarrow$ as "if we run program number $n$ on input $n$ and use $A$ as an oracle, then it halts", how to describe $(\phi_n^K)^K(x)\downarrow$ in a similar manner?)

And in general, I guess $0^{(n)}$ is the equivalence class of $K^{(n-1)}$.


问:
prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

Prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

We can prove this easily by showing $f(1)=f'(1)=0$.

But is there a way to solve this problem without using calculus?

答:
$f(x) =n((x-1)+1)^{n+1}-(n+1)((x-1)+1)^n+1;$

Binomial expansion:

1)The constant term:

$n-(n+1)+1=0$, i.e. all terms have (at least) a factor $(x-1)$.

2)Coefficient of $(x-1)$:

$n(n+1)-(n+1)n=0$, i.e. all remaining terms have (at least) a factor $(x-1)^2$, and we are done.


问:
Is there a divergent sequence such that $(x_{k+1}-x_k)\rightarrow 0$?
Is there a divergent sequence such that $\lim_{k\rightarrow\infty}(x_{k+1}-x_k)=0$?</p>

答: Define $x_k:=\log k\ (k \ge 1)$, and then $x_{k+1}-x_k=\log(1+\frac{1}{k})\to 0$ as $k\to \infty$</p>


问:
Density of random variable max{ξ1, ξ2}
Let $\xi_1, \xi_2$ be independent and equally distributed absolutely continuous random variables with density $p$.

How can I find the density of the random variable $\max(\xi_1, \xi_2)$.

答:
I am going to assume the support is $\mathbb{R}$ just for my own psychological reasons; modify as necesasry.

$\{\xi \le x \} = \{\xi_1 \le x, \xi_2 \le x \}$ where $\xi = \max(\xi_1,\xi_2)$

Now $\int_{-\infty}^x dF(y)= \int_{-\infty}^x p(y) dy = P(\xi \le x) = P(\xi_1 \le x, \xi_2 \le x ) = P(\xi_1 \le x)P (\xi_2 \le x ) = [F(x)]^2$ where $F$ is the CDF of the distribution.

Can you finish from here?

Again, note that I made an assumption about the support of the random variables which as pointed out in the comments is vital information.

Title: Density of random variable max{ξ1, ξ2}


问:
There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color?
A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?

Here's what I tried:

Case 1: 2 black balls and 1 white ball

8/15 * 7/14 * 7/13 = 392/2730 = 28/195

Case 2: 2 white balls, 1 black ball

7/15 * 6/14 * 8/13 = 336/2730 = 24/195

28/195 + 24/195 = 52/195 = 4/15

My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!

答: What you have written corresponds to the case when first two balls of one color and then the ball of the other color are drawn. Therefore your calculations miss the factor 3 accounting for possible permutations of the balls.</p>


问:
A recurrence of the second-order Eulerian polynomials
Recently, some of the remarkable properties of second-order
Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$  A340556
have been proved on MSE [  a ,
 b ,  c  ]
and in d the second-order Eulerian polynomials
were introduced.
$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n =
\sum_{k=0}^n \left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle \, x^k $$

These polynomials can be computed recursively,  starting $ \left\langle\!\left\langle x \right\rangle\!\right\rangle_{0} \, = \, 1 $, and then proceeding

$$ \left\langle \! \left\langle x \right\rangle \! \right\rangle _n =
x\,(x-1)^{2n}\
\left(\frac{ \left\langle \! \left\langle x \right\rangle \! \right\rangle _{n-1}}{(1-x)^{2n-1}} \right)'
\quad (n \ge 1). $$

With this formula the polynomials can not only be calculated efficiently but also
may simplify the derivation of combinatorial identities.
For instance from Marko Riedel's thorough proof
we know the following identity for Euler's tree function
$\operatorname{T}(z) = - \operatorname{W}(-z)$, where $\operatorname{W}$
is the principal branch of Lambert's function A000169:

$$  m^{m+n} \,  = \,
    m!\, [x]^{m} \frac{\left\langle \! \left\langle \, \operatorname{T}(x) \, \right\rangle \! \right\rangle _n}  {(1-\operatorname{T}(x))^{2n+1}} \quad ( n \ge 0) $$

The reader may enjoy to investigate what happens if we evaluate the function
at $ x = (2 \sqrt{e})^{-1}$, (A225170, A006351).
Is someone willing to show us a proof of the recurrence of the second-order Eulerian polynomials
given above?

答:
This is the same as in the Tree function Eulerian identity.
Observe that

$$x (1-x)^{2n}
\left(\frac{1}{(1-x)^{2n-1}}
\sum_{k=0}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle x^k\right)'.$$

gives with $n\ge 2$ for $\left\langle\! \left\langle x
\right\rangle\! \right\rangle_n$

$$x(1-x)^{2n}
\frac{2n-1}{(1-x)^{2n}}
\sum_{k=0}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle x^k
+x(1-x)^{2n}
\frac{1}{(1-x)^{2n-1}}
\sum_{k=1}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k x^{k-1}
\\ = \sum_{k=0}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle (2n-1) x^{k+1}
+ \sum_{k=1}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k x^k
- \sum_{k=1}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k x^{k+1}
\\ = \sum_{k=1}^{n}
\left\langle\!\! \left\langle n-1 \atop k-1
\right\rangle\!\! \right\rangle (2n-1) x^{k}
+ \sum_{k=1}^{n-1}
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k x^k
- \sum_{k=2}^{n}
\left\langle\!\! \left\langle n-1 \atop k-1
\right\rangle\!\! \right\rangle (k-1) x^{k}.$$

We may lower $k$ to zero in all three terms (zero contribution)
and raise it to $n$ in the second term (also a zero contribution).
We find

$$\sum_{k=0}^n
\left(\left\langle\!\! \left\langle n-1 \atop k-1
\right\rangle\!\! \right\rangle (2n-k) +
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k \right) x^k.$$

Now apply the recurrence for the Eulerian numbers to get

$$\sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle x^k$$

as desired. Note that the case $n=1$ for the initial recursion goes
through by  inspection with $\left\langle\! \left\langle x
\right\rangle\! \right\rangle_0 = 1.$


问:
Show that the equation $\log(1+e^x)=\cos(x)$ has infinitely many negative solutions.
Show that the equation $\log(1+e^x)=\cos(x)$ has infinitely many negative solutions. Find out if there is a positive solution and if it is unique.

From the graph I can see that it has an infinitely many negative solutions and that it has one positive solution. I don't know how to prove it rigorously. Could you help me? Thank you for your time.

答:
If anon's answer seems a little abstract to you, you can argue through inequalities that

for $ \ x < 0 \ , $

$$ \ 0 \ < \ e^x \ < \ 1 \ \Rightarrow \ 1 \ < \ 1 + e^x \ < \ 2 \ $$

$$\Rightarrow \ \ln \ 1 \ = \ 0 \ < \ \ln (1 + e^x) \ < \ \ln 2 \ < 1 \ . $$

So $ \ \ln (1 + e^x) \ $ lies entirely within the bounds of the cosine function (and in fact is always positive), and thus for negative values of  $ \ x \ $ , there will be two solutions of $ \ \ln (1 + e^x) \ = \ \cos x \ $ for every "cycle" of the cosine function on $ \ ( -\infty \ , \ 0 ) \ . $

The existence of a positive solution is found through the Intermediate Value Theorem, as anon has indicated.  To see that it is unique, consider the derivative of the function $ \ \ln (1 + e^x) - \cos x \ $ for  $ \ x > 0 \ $ to show that the function is monotonically increasing.


问:
Expectation of an Expectation
I need to solve two exercises:

Calculate V[ui|xi] using E[yi|xi] and ui = yi - E[yi].

Calculate E[$y^2$|xi].

Information given for the exercise:

E[$u^2$|xi] = V[yi|xi]

E[$u^2$|xi] = V[yi|xi] because E[ui|xi]

There are also a values given for E[yi] = 2λi, V[yi] = 2$λ^2$ and λi = (e$^{xβ}$)/2

What I know:

V[yi|xi] = V[$y^2$|xi] = E[$y^2$|xi] - $(E[y|x])^2$

It would be easy to set the value for E[yi] into u and square it. Afterward set the solution into E[$u^2$|xi] and V[yi|xi] would be solved.

Obviously I wouldn't use E[yi|xi] in this case, so I tried it by rearranging terms, but often end at a point where they cancel each other out and I am back at the beginning.

So my question is: How do I properly derive V[ui|xi] and E[$y^2$|x]. Especially important for me is how to link it to E[yi|xi].

答:
In the part where you write "what I know" it does not make sense to say $V(y|x)=V(y^2|x)$.

$V(u|x)=V(y-E(y)|x)=V(y|x)$ (because E(y) is a constant). This in turn is $E(y^2|x)-\left[E(y|x)\right]^2$.

Also, $E(y^2|x)=V(y|x)+\left[E(y|x)\right]^2$

But, I am not clear how all these conditional expectations and variances tie into the particular values. One way to do it is for (2), $\frac{e^{2x\beta}}{2}+e^{2x\beta}=\frac{3}{2}e^{2x\beta}$. And for (1), we can go directly from $V(y|x)=\frac{e^{2x\beta}}{2}$.

Title: Expectation of an Expectation


答:
As stated in the comments to your question, you have ascertained the probability of drawing two black balls and a white ball in that order and the possibility of drawing two white balls and a black ball in that order.  You have to multiply by $3$ to account for all possible orders of drawing the balls you want.

Here's another approach.  The only other possibility is to draw three balls of the same color.  The probability of drawing three black balls is $\frac{8}{15} \cdot \frac {7}{14} \cdot \frac{6}{13}=\frac{336}{15 \cdot 14 \cdot 13}$ and the probability of drawing three white balls is $\frac{7}{15} \cdot \frac{6}{14} \cdot \frac{5}{13}=\frac{210}{15 \cdot 14 \cdot 13}$, so the probability of one of these two occurrences is $\frac{446}{15 \cdot 14 \cdot 13}=\frac 15$, and the probability that doesn't happen (i.e., you end up with two balls of one color and one ball of another color) must therefore be $1-\frac 15=\frac 45$.

Title: There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color?

Body:
A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?

Here's what I tried:

Case 1: 2 black balls and 1 white ball

8/15 * 7/14 * 7/13 = 392/2730 = 28/195

Case 2: 2 white balls, 1 black ball

7/15 * 6/14 * 8/13 = 336/2730 = 24/195

28/195 + 24/195 = 52/195 = 4/15

My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!


问:
Integration of $\sin(2x)$ via two different methods
When I integrate $\sin(2x)$ first using the trig substitution of $\sin(2x) = 2 \sin x\cos x$ and then using u-substitution I get $\sin^2(x) +C$. Integrating the same expression using u-substitution directly, I get $-\cos(2x)/2+C$.

These answers do not seem to be equivilent.  When I let $x=0$ in the first example's answer I get $=0$ but when I let $x=0$ in the seccond answer I get $-1/2$.

Clearly I must be doing something wrong.  If I had made these integrations definite integrals over $0$  to $\pi/2$ seems like I would have gotten different answsers.

Can you assist?

答:
See this related post. This is a commmon mistake while doing integration.

For example, if $f(x)=g(x)$, then would you say $\displaystyle \int f(x) \mathrm dx=\int g(x) \mathrm dx$?


问:
Vector space and $\mathbb Z_n.$
My professor gave us this as an example of a module (I guess my professor were comparing the linear independence in a vector space to that in a module):

In a vector space $V$ over a field $k,$ we have if $v \in V, v \neq 0.$ Then $$cv = 0 \textbf{  for  } c \in k \Leftrightarrow c = 0.$$ So, $$\{c \in k| cv = 0 \} = \{0\}.$$

$G \cong \mathbb Z_n$ is a $\mathbb Z$-module. $g \in G, g \neq 0, \{m \in \mathbb Z| mg = 0\} \supset n\mathbb Z$ and if $\langle g \rangle = \mathbb Z_n$ then $\{m \in \mathbb Z| mg = 0\} \supset n\mathbb Z.$

And I am for my life do not understand the idea my professor wanted to convey. Could anyone help me in trying to understand my professor mind please?

答:
This example shows that while a single nonzero vector $v$ is always linearly independent ($cv=0\implies c=0$), this doesn't hold for modules over rings.

Indeed, no single element of the $\Bbb Z$-module $\Bbb Z_n$ can be linearly independent, since we have $nx=0$ for every $x\in\Bbb Z_n$.

The last part should say that if $x\in\Bbb Z_n$ generates the whole module $\Bbb Z_n$ (that is, (any representative of) $x$ is coprime to $n$), then
$$\{m\in\Bbb Z:mx=0\}\ =\ n\,\Bbb Z\,.$$

Title: Vector space and $\mathbb Z_n.$