问:
Meaning of $dx$ in integration and differentiation
In differentiation, we denote $dx$ as in infinitesimal change in $x$ and $dy$ the corresponding change.But in integration, we divide $x$ into infinitely many equal intervals and name them $dx$. Then $f(x)dx$ represents the area of an infinitely thin rectangle. From differentiation we know, $dy = f(x)dx$.Why do we substitute then $dy$ in place of $f(x)dx$? I mean $dy$ is change in differentiation. But here in integration we take $dy$ to be the area of each infinitesimal rectangle. Why do we mix them up? Please someone explain with clarity.

答:
The answer is "the fundamental theorem of calculus".  The fundamental theorem of calculus shows the deep connection between slopes and areas.  Because of the fundamental theorem of calculus, $dy$ is both the $y$-component of the slope of $f(x)$ and the area of the individual boxes $f'(x)\,dx$.  That's the fundamental theorem of calculus.

However, I think viewing calculus as fundamentally about slopes and areas is a wrong way of going about it.  I think a simpler way to think about it is this:

Differentiation is about infinitesimal changes

Integration is about adding together infinitesimal changes

Here, you can easily see why integration works: you are adding up changes to get the total change (and the total change will be off from the `true value' by a constant, $+C$).

That is, if I differentiate $x^2$, I will get $2x\,dx$.  This means that the value of the change in $x^2$ depends on both (a) the current value of $x$, and (b) the amount of (infinitesimal) change in $x$.  If I add up all of those changes (I integrate it), I will get the total change.  In other words,

$$\text{total change in }x^2 = \int 2x\,dx$$

If I make a graph of this, it will be the same as the original graph, but perhaps off by a constant - which is precisely what integration does!  So integration gives us a mechanism to add together an infinite number of infinitely small values.

So, how does this relate to areas?

The area under a curve is simply adding together all of the infinitely small rectangles under it.  Each rectangle is $y$ high and $dx$ wide, or $y\,dx$.  Therefore, the sum of all the individual rectangles is $\int y\,dx$.

Now, we know enough to find the relationship between slopes and areas.  To find the slope of an equation, we take the differential and divide by $dx$.  To find the area under the curve of an equation for $y$, we multiply by $y$ and integrate.  Since integration and differentiation are opposites of each other, the process of finding areas is the reverse of the process of finding slopes.


问:
$\max _{x \in \mathbb{R}_{+*}^{3 \times 1}} x_{2}+x_{3}$ subject to $4 x_{1}=\left(x_{2}^{2}+x_{1} x_{3}\right)\left(x_{1}+x_{3}\right)$
I have a problem of maximization that goes as follows:
$$
\max _{x \in \mathbb{R}^{3 \times 1}} x_{2}+x_{3} \text { subject to } 4 x_{1}=\left(x_{2}^{2}+x_{1} x_{3}\right)\left(x_{1}+x_{3}\right), x_{1}>0, x_{2}>0 \text { and } x_{3}>0
$$
I want to go with a classical approach to solve it, and no specific algorithms. First restriction is that I have to solve this problem without thinking about KKT conditions ( I don't even know if it is solvable in that manner ). I wanted use inequalities and try to restrain the domain of the constraints, with no results.

Any suggestions to an approach ??

答:
You can solve it without the KKT conditions if you add slack variables to convert the inequalities into equalities. Then you use Lagrange multipliers.

\begin{align*}
\max\, & x_2 + x_3\\
\textrm{s.t.} &\\
& 4x_1 -(x_2^2+x_1 x_3)(x_1+x_3)=0\\
& x_1 - s_1^2 = 0\\
& x_2 - s_2^2=0\\
& x_3 - s_3^2=0
\end{align*}

ContentLicense: CC BY-SA 4.0


问:
System of equations word problem
Andrea's Sequoia gets 12 miles per gallon while driving in the city, and 20 miles per gallon while on the highway. The last time she filled up the car it took 16 gallons of gas, and she had driven 280 miles since the previous fill-up. How many miles of that 280 were city driving?</p>

答:
HINT: Start by assigning variables to the unknown quantities. Let $x$ be the number of miles that she drove in the city since the previous fill-up, and let $y$ be the number of miles that she drove on the highway in that period. Since she gets $12$ miles per gallon while driving in the city, she must have used $\frac{x}{12}$ gallons while driving in the city. Similarly, she must have used $\frac{y}{20}$ gallons on the highway. She used a total of $16$ gallons, so we have one equation:

$$\frac{x}{12}+\frac{y}{20}=16\;.$$

The other equation is easy to find: she drove a total of $280$ miles in that period, so ... ?

Now you have two linear equations in the unknowns $x$ and $y$, and you can solve them by any of the standard methods.

(It’s possible to set up the problem directly with only a single unknown, say the $x$ of the solution outlined above, but if you’re having difficulty with the setup it’s probably easier to start with one variable for each unknown quantity and then eliminate one.)

ContentLicense: CC BY-SA 3.0

Title: System of equations word problem


问:
An example of one-to-one function to two variables
I was searching for an one-to-one function $f:[0,1]\times[0,1] \rightarrow \mathbb{R}$. I failed to construct one. Neither can I prove that such a function does not exist. Do you have any answer for this?

答:
I wouldn't call it a trivial example, but here is a construction.  Often we make a chain of bijections between different sets.  This can make each step clear but frustrate people who want a clean formula.  I presume you have seen a bijection $[0,1] \leftrightarrow (0,1)$ so I will work with $(0,1)\times (0,1)$.  There are three bijections in the chain, one of which is used twice.

First we go from $(0,1)$ to binary bit strings.  The bulk of this is to match each string with the real number it represents as a binary fraction.  The problem is the pairs of binary strings that match to the same real-the versions terminating in all $0$s and the ones terminating in all $1$s.  You can handwave that away because it is only a countable number of points.  To be explicit, we will use the version that ends in all $0$s to match the corresponding real.  We have now accounted for all the bit strings except all $0$s, all $1$s, and strings of the form $[n]1111111\ldots$ where $[n]$ is the binary expansion of an even natural (including $0$).  That is a set clearly in bijection with the naturals.  Now pick some countable series of points, say those of the form $0.[m]01010101\ldots$ and use the interleaving trick you used to show the even numbers were in bijection with the naturals to account for the missing strings.

Now we go from pairs of binary strings to single binary strings.  We just alternate the bits of the two in the pair to get a single string.

Now we go from the single string to $(0,1)$ using the same bijection we used two paragraphs ago.

Finally we go from $(0,1)$ to $\Bbb R$, say by $x \leftrightarrow \arctan\left(\pi\left(x-\frac 12\right)\right)$


问:
Line Integral calculating the ds
Doesn't change the result, but why in the last steps constant in front of the integral is 9 and not 27? Where did the 9 that came from squaring the dx/dt and dy/dt go?

答:
They forgot it. The first line of step 4 should have been:

$$\text{Area} = \int_{t = 0}^{t = \pi} (3 \sin(t))(3 \cos(t)) \sqrt{(-3\sin(t))^2 + (3\cos(t))^2} $$

Title: Line Integral calculating the ds


答: $dx$ and $dy$ are not infinitesimal changes, this is an old, obsolete legend. Here is the truth: https://en.wikipedia.org/wiki/Differential_of_a_function.

Title: Meaning of $dx$ in integration and differentiation

Body: In differentiation, we denote $dx$ as in infinitesimal change in $x$ and $dy$ the corresponding change.But in integration, we divide $x$ into infinitely many equal intervals and name them $dx$. Then $f(x)dx$ represents the area of an infinitely thin rectangle. From differentiation we know, $dy = f(x)dx$.Why do we substitute then $dy$ in place of $f(x)dx$? I mean $dy$ is change in differentiation. But here in integration we take $dy$ to be the area of each infinitesimal rectangle. Why do we mix them up? Please someone explain with clarity.


问:
A fair six sided dice is rolled 12 times. What is the probability of getting exactly two 6's?
Is there anyone that can show how this problem can be modeled by a binomial distribution?</p>

答:
Let $X$ be the random variable, "number of sixes you get from $12$ rolls of the die".

Each outcome is either a "success" or a "failure": either you get a $6$ or you don't.

The probability of "success", $p = \frac16$, is constant: it does not vary from one roll to another;

The "trials", namely each roll of the dice, are independent from one another;

The number $n\ $of trials, $12$, is fixed/constant (i.e. it is $12$ and cannot be anything else/doesn't change);

Because these assumptions are good, we may model $X$ with the Binomial distribution:

$$X \sim B\left(12, \frac16\right).$$

That's how you model the random variable $X$ with the Binomial distribution.

The probability of getting two $6$'s is $p(X=2),\ $ which can simply be done on your calculator (assuming your calculator has this functionality).

Alternatively to using your calculator, see the formula at the top of here, and put $k=2.$


问:
What is my operator norm (cannot get good enough bounds).
Given a space of square integrable functions $x(t)$ over the interval $[0;1]$ one can introduce a norm

$$\|x(t)\|= \sqrt{\int_0^1 (x(t))^2 \, dt};$$

Then what is a norm of the transformation below (result is also square integrable function, same space, same norm) assuming the definition.

$$\operatorname{A}x(t)= \int_0^ts \cdot x(s) \, ds;$$

Well according to one definition $\|\operatorname{A}\| = \sup_{\|x\|=1}\|\operatorname{A}x(t)\|\ge \|\operatorname{A}1\| =\sqrt{\frac{1}{20}}$, I tried quite a few different functions and ain't getting any better lower bound than this. To be specific I tried $x(t) = t^\alpha/\| t^\alpha \|$ (for $\alpha \in (-0.5;+ \infty)$) - result is $\sqrt{\frac{1}{20}}$ or lower.

According to other equivalent definition I look for infimum of bounding constants
$$\|\operatorname{A}x(t)\| = \sqrt{\int_0^1 \left(\int_0^t s \cdot x(s) \, ds \right)^2 \, dt} \le \|x(t) \| \cdot \sqrt{\int_0^1 \int_0^t s^2 \, ds \, dt} = \sqrt{\frac{1}{12}} \cdot \|x(t) \|;$$

As a result $\sqrt{\frac{1}{20}} \le \|x(t) \| \le \sqrt{\frac{1}{12}}$. How can I improve?

答:
Let $\mu_n$ be the $n$-th positive root of the Bessel function of the first kind $J_{-1/4}$.
Functions $x^{3/2}J_{-1/4}(\mu_n x^2)\,$ are orthogonal for different $n$:
$$
\int_0^1 x^{3} J_{-\frac{1}{4}}(x^2 \mu_m)J_{-\frac{1}{4}}(x^2 \mu_n)\,dx=
\frac12\int_0^1 y J_{-\frac{1}{4}}( \mu_my)J_{-\frac{1}{4}}(\mu_ny)\,dy=
\frac14J_{3/4}^2(\mu_n)\delta_{mn}.
$$
The last equality is formula $(53)$ for $a=1$ from the cited page.
It follows in particular, what $J_{3/4}(\mu_n)\ne0$, $n\in \mathbb N$. So the functions
$$
j_n(x)=\frac{2x^{3/2}J_{-1/4}(\mu_n x^2)}{J_{3/4}(\mu_n)}
$$
form an orthonormal sequence on $[0,1]$: $$(j_m,j_n)=\delta_{mn}.$$
Using Mma for calculations gives
$$
Aj_n(x)=\int_0^xtj_n(t)\,dt=\frac{x^{3/2}J_{3/4}(\mu_n x^2)}{J_{3/4}(\mu_n)},
$$
$$
(Aj_m,Aj_n)=\frac{\delta_{mn}}{4\mu_n^2}.
$$
Now lets expand a function $f\in L_2([0,1])$ into the series of $j_n$:
$$
f(x)=\sum_{n=0}^{\infty}a_nj_n(x).
$$
Up to the transform $x\rightarrow x^2$ it is the Fourier–Bessel series.
Using the orthogonality properties of $j_n$ and $Aj_n$ stated above, we have
$$
\frac{(Af,Af)}{(f,f)}=
\frac{\displaystyle\frac14\sum_{n=1}^{\infty}\frac{a_n^2}{\mu_n^2}}{\displaystyle\sum_{n=1}^{\infty}a_n^2}.
$$
Since the sequence $\mu_n$ is increasing, the maximum of the lhs is attained for $f=j_1$ and
$$
\|A\|=\frac1{2\mu_1}=0.249215\ldots
$$</p>


问:
How do get combination sequence formula?
What would be a closed-form formula that would determine the ith value of the sequence

1, 3, 11, 43, 171...

where each value is one minus the product of the previous value and 4?

Thanks!

答:
The following is a simple way of solving the problem. The recurrence has the shape
$$y_{n+1}=4y_n-1$$

It would be nice if the recurrence looked like
$$u_{n+1}=4u_n$$

Then life would be easy. Unfortunately, we have that pesky $-1$ that spoils things!

So let's try for the next best thing.  Could we transform our recurrence to something of the shape
$$y_{n+1}-a=4(y_n-a)$$
where $a$ is a constant?  Do a bit of arithmetic.  We get $y_{n+1}=4y_n-3a$.  If we choose $a=1/3$, we can get our recurrence into the desired shape.

Thus we can rewrite our recurrence as

$$y_{n+1} -\frac{1}{3}=4\left(y_n -\frac{1}{3}\right)$$

Temporarily, let
$$u_n=y_n-\frac{1}{3}$$
for all $n$. Then
$$u_{n+1}=4u_n$$

Now we have to decide whether we start our indices at $0$ or at $1$.  Like many mathematicians, I prefer to start at $0$.  (Starting at $1$ would not change things very much.)

We have $y_0=1$, so $u_0=1-1/3=2/3$.
And because to get the "next" $u$ we multiply the previous $u$ by $4$, we have
$$u_n=\left(\frac{2}{3}\right)4^n$$
But $y_n=u_n+1/3$. It follows that
$$y_n=\left(\frac{2}{3}\right)4^n +\frac{1}{3}$$

Comment: Essentially the same method works for the recurrence $y_{n+1}=cy_n+d$ where $c$ and $d$ are constants, and the idea can be adapted to deal with many other situations.


问:
estimate population percentage within an interval, given a small sample
Given a small sample from a normally-distributed population, how  do I calculate the confidence that a specified percentage of the population is within some bounds [A,B]?

To make it concrete, if I get a sample of [50,51,52] from a normally-distributed set, I should be able to calculate a fairly high confidence that 50% of the population lies within the range of 0-100, even with such a small sample.

This is certainly related to the "tolerance interval", but differs in an important way. In all of the examples I can find for tolerance intervals, the required percentile and confidence is given, and the interval is found. In my problem, the interval and percentile are given, and I need to find the confidence.

The relevant equation is this one: (Guttman 1970)

$$1 - \gamma = P\left[P(X \geqq t_0) \geqq 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \leqq \sqrt n K\right]$$

With definitions:

$1 - \gamma$ is the confidence

$n$ is the number of samples

$100p$ is the percentage of the population required to be within the interval, as estimated from the sample mean and sample variance.

$t_0 = x - K_{S, z_p}$ is the $(1 - p) 100$th percentile of the standard normal distribution

$T_v^*(\delta)$ is the noncentral Student’s t distribution with $v$ degrees of freedom and noncentrality parameter $\delta$.

This solves the one-sided problem, but I'm having trouble extending this to the two-sided problem. In confidence-interval land, I'd use the fact that $P(t_1 \leqq X \leqq t_2) = 1 - P(t_1 \gt X) - P(X \gt t_2)$, to break this into two one-sided problems, but in tolerance-interval land I need to relate these back to the confidence ($1-\gamma$), and I don't see how.

$$1 - \gamma = P\left[P(t_1 \geqq X \geqq t_2) \geqq 1 - p\right] = ??? $$

If I attempt to turn this into two one-sided problems:

$$1 - \gamma = P\left[1 - P(t_1 \lt X) - P(X \lt t_2) \geqq 1 - p\right] = ??? $$

And I'm utterly stuck there. I don't see how to relate this back to the one-sided tolerance interval solution.

I'm not certain this is useful for people to understand the question, but it might, so I'm putting it in this addenda.

In scipy, I'm able to pretty easily calculate $K$ given $p$ $\gamma$ and $n$ as:

```<code>def K(p, gamma, n):
    from scipy import stats
    return stats.nct.ppf(1-gamma, n-1, sqrt(n) * stats.norm.ppf(1-p)) / sqrt(n)
</code>```

I'm also able to find $\gamma$ given $K$ $p$ and $n$ as:

```<code>def gamma(p, n, K):
    from scipy import stats
    z_p = stats.norm.ppf(1-p)
    return 1 - stats.nct.cdf(sqrt(n) * K, n-1, sqrt(n) * z_p)
</code>```

Much less important, but is this a valid simplification of the Guttman's formula?

$$1 - \gamma = P\left[P(X \geqq t_0) \geqq 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \leqq \sqrt n K\right]$$
$$\gamma = P\left[P(X \geqq t_0) \lt 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \gt \sqrt n K\right]$$
$$\gamma = P\left[P(X \lt t_0) \lt p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \gt \sqrt n K\right]$$

If so, this form seems way easier to understand, to me.

答:
One possibility: you have a certain number of successes and failures, and you want to guess the percentage of the underlying population. You calculate the Agresti-Coull interval (see Brown, Cai, & DasGupta 2001) with the required parameters.

On the bright side this is easy and makes no assumptions about the type of distribution. On the downside it doesn't take advantage of the extra information about where the samples in the interval fall -- if you know something about how they are distributed this might get you a more precise estimate based on the data you have.

Edit: for a brief overview see http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Agresti-Coull_Interval


问:
Indicated quantities for the given group homomorphism
I was exercising for group theory trying to find the indicated quantities for the given homomorphisms. What I am struggling to understand is to find the quantities when it comes to homomorphism from a group product to a group or another group product. Here's the example

Find $Ker(\alpha)$ and $\alpha(-3,2) $ for the homomorphism $\mathbb{Z}$ x $\mathbb{Z} \to \mathbb{Z}$ where $\alpha(1,0) = 3$ and $\alpha(0,1) = -5$

To find $\alpha(-3,2)$ I tried doing $$ \alpha(-3,2) = -3\alpha(1,0)+2\alpha(0,1) = -3*3 + 2*(-5) = -19$$
and for $Ker(\alpha)$ I tried $$ \alpha(a,b) = 0 \implies a\alpha(1,0)+b\alpha(0,1) = 3*a - 5*b = 0$$

So what is Ker now? Maybe some help or suggestions to tackle these type of problems differently would be huge for me.

Thank you and be safe.

答:
You can write what you currently have as
$$\ker(\alpha) = \{(a, b) \in \mathbb Z \times \mathbb Z: 3a - 5b = 0\}.$$

As it turns out, all solutions to this equation are of the form $(5n, 3n)$ for $n \in \mathbb Z$. You can look up the phrase homogeneous linear Diophantine equation like here to get more information on why (or just convince yourself that all of these tuples are solutions). However, these number theory details are disjoint from the group-theoretic questions about homomorphisms that you're asking.

As for this kind of problem in general: the key is that you knew the value of $\alpha$ on a $\mathbb Z$-basis $\{(1, 0), (0, 1)\}$ for $\mathbb Z \times \mathbb Z$. You can then use the homomorphism property to determine $\alpha(a, b)$ for all other $(a, b)$. It's analogous to how in linear algebra we encode homomorphisms (there called linear transformations) by matrices, and matrices are keeping track of where the basis goes. In this case, we'd write
$$\alpha = \begin{pmatrix} 3 & -5 \end{pmatrix}. $$


问:
Proof that the following composite matrix is invertible
Let $B\in \mathbb{R}^{n\times n}$ and $C\in\mathbb{R}^{m\times n}$ (where $m\leq n$) and let $v^TBv > 0$  $\forall v \neq 0$ with $Cv = 0$, where $C$ has full rank.

Proof that $A = \begin{pmatrix}
B & C^T \\
C & 0
\end{pmatrix}$ is an invertible Matrix.

So far, I've concluded that $B$ is positive definite and therefore also $det(B)>0$, which could help prove that $det(A) \neq 0$.
But other than that I had no idea for a solution.

Any help is appreciated, thanks in advance!

答:
This amounts to simply using the Schur complement.

In detail, note that $A$ is invertible if and only if the matrix
$$
\pmatrix{I & 0\\-CB^{-1} & I}
\pmatrix{B & C^T\\ C & 0}
\pmatrix{I & 0\\-CB^{-1} & I}^T = \pmatrix{B&0\\0 & - CB^{-1}C^T}
$$
is invertible. Thus, $A$ is invertible if and only if both $B$ is invertible and $CB^{-1}C^T$ is invertible. Using the fact that $C$ has full row-rank and $B^{-1}$ is positive definite, conclude that $CB^{-1}C^T$ is positive definite and therefore invertible.


问:
Hyperelliptic function addition formula
$$x= \int_{0}^{f(x)} \frac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}\sqrt{1-l^2u^2}}$$

$$f(0)=0$$

If we apply derivative operation for both sides, we get:

$$f'(x)=\sqrt{(1-f^2(x))(1-k^2f^2(x))(1-l^2f^2(x))}$$
My Questions:

What is closed form addition formula of f(x)?

Is $f(x)$  a periodic function or doubly periodic function or not periodic function?

What is the period if it is a periodic function  ( or what are the periods if it is a doubly periodic function)?

My attempt to solve the questions:

Let's define a new function $\phi(x)=\frac{\sin(ax)}{a}$: and it is single periodic function.

$$x= \int_{0}^{\phi(x)} \frac{du}{\sqrt{1-a^2u^2}}$$
$$\phi(0)=0$$
I realized that we can write addition formula of $\phi(x)$ as:

$$\phi(x+y)=\phi(x)\sqrt{(1-a^2\phi^2(y))}+\phi(y)\sqrt{(1-a^2\phi^2(x))}$$

$$\phi(x+y)=\phi(x)\phi'(y)+\phi'(x)\phi(y)$$
Doubly periodic Jacobi elliptic function sn(x) is defined

$$x= \int_{0}^{sn(x)} \frac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}}$$

$$sn(0)=0$$

Addition formula of sn(x):

$$sn(x+y)=\frac{sn(x)\sqrt{(1-sn^2(y))(1-k^2sn^2(y))}+sn(y)\sqrt{(1-sn^2(x))(1-k^2sn^2(x))}}{1-k^2sn^2(x)sn^2(y)}$$

$$sn(x+y)=\frac{sn(x)sn'(y)+sn'(x)sn(y)}{1-k^2sn^2(x)sn^2(y)}$$

Let's define a new function g(x):

$$x= \int_{0}^{g(x)} \frac{du}{\sqrt{1-a^2u^2}\sqrt{1-b^2u^2}}$$
$$g(0)=0$$
I realized that we can write addition formula of g(x) as:

$$g(x+y)=\frac{g(x)\sqrt{(1-a^2g^2(y))(1-b^2g^2(y))}+g(y)\sqrt{(1-a^2g^2(x))(1-b^2g^2(x))}}{1-a^2b^2g^2(x)g^2(y)}$$

$$g(x+y)=\frac{g(x)g'(y)+g'(x)g(y)}{1-a^2b^2g^2(x)g^2(y)}$$

Let's define a new function h(x):
$$x= \int_{0}^{h(x)} \frac{du}{\sqrt{1-a^2u^2}\sqrt{1-b^2u^2}\sqrt{1-c^2u^2}}$$
If we follow the same rule for h(x) as we applied above, so my estimation for addition formula of h(x) is:

$$h(0)=0$$

$h(x+y)=\frac{h(x)\sqrt{(1-a^2h^2(y))(1-b^2h^2(y))(1-c^2h^2(y))}+h(y)\sqrt{(1-a^2h^2(x))(1-b^2h^2(x))(1-c^2h^2(x))}}{1-(a^2b^2+a^2c^2+b^2c^2)h^2(x)h^2(y)+ a^2b^2c^2P(h(x),h(y))}$

$$h(x+y)=\frac{h(x)h'(y)+h'(x)h(y)}{1-(a^2b^2+a^2c^2+b^2c^2)h^2(x)h^2(y)+ a^2b^2c^2P(h(x),h(y))}$$

Where $P(h(x),h(y))$ is symmetric function

To find $P(x,y)$:

If $a=b=c$ then

$$x= \int_{0}^{h(x)} \frac{du}{(1-a^2u^2)\sqrt{1-a^2u^2}}$$

$$x= \frac{h(x)}{\sqrt{1-a^2h^2(x)}}$$

$$h(x)= \frac{x}{\sqrt{1+a^2x^2}}$$

$$h'(x)= \frac{1}{(1+a^2x^2)\sqrt{1+a^2x^2}}$$

$$h(x+y)=\frac{h(x)h'(y)+h'(x)h(y)}{1-3a^4h^2(x)h^2(y)+ a^6P(h(x),h(y))}$$

I have not found a result from this direction yet.

Do you see a solution way for addition formula for h(x) from other methods?

Thanks a lot for helps

EDIT: We may use algebraic transform on some hyper elliptic function and can transfer into Weierstrass ℘ elliptic function.

$$h'(x)=\sqrt{(1-a^2h^2(x))(1-b^2h^2(x))(1-c^2h^2(x))}$$

If we apply the algebraic transform to the hyper elliptic function above, $h(x)=-\dfrac{1}{\sqrt{g(x)}}$

We will get :
 $$\dfrac{g'(x)}{2g(x)\sqrt{g(x)}}=\dfrac{ \sqrt{(g(x)-a^2)(g(x)-b^2)(g(x)-c^2)}}{g(x)\sqrt{g(x)}}$$

$$g'(x)^2=4(g(x)-a^2)(g(x)-b^2)(g(x)-c^2)$$

$$g'(x)^2=4g^3(x)-4(a^2+b^2+c^2)g^2(x)+4(a^2b^2+a^2c^2+b^2c^2)g(x)-4a^2b^2c^2  \tag{1}$$

Then We can consider on the Weierstrass $\wp$ elliptic function $\wp(z, g_2, g_3)$ with the invariants $g_2\in\mathbb{R}$ and $g_3\in\mathbb{R}$:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3$$

$$\wp(x)=g(x)+\beta$$  , where $\beta$ is a constant

$$g'(x)^2 = 4(g(x)+\beta)^3 - g_2 (g(x)+\beta) - g_3$$
$$g'(x)^2 = 4g^3(x)+12\beta g^2(x) - (g_2-12\beta^2)g(x) - (g_3+g_2\beta-4 \beta^3)$$

If we equal in Equation 1 :

We can find $\beta,g_2,g_3$ as $a,b,c$

$$\beta=-\dfrac{(a^2+b^2+c^2)}{3}$$

$$\wp(x+y)+\wp(x)+\wp(y) =\frac{1}{4} \left(\frac{\wp'(x)-\wp'(y)}{\wp(x)-\wp(y)}\right)^2 \tag{3}$$

We know addition formula of  Weierstrass $\wp$ elliptic function.Thus we can find h(x) addition formula as:

$$h(x)=-\dfrac{1}{\sqrt{\wp(x)-\beta}}$$
$$\wp(x)=\dfrac{1}{h(x)^2}+\beta$$

If we use Equation 3 , we can write addition formula of $h(x)$ as:

$$\dfrac{1}{h(x+y)^2}+\dfrac{1}{h(x)^2}+\dfrac{1}{h(y)^2}+3\beta = \left(\frac{\frac{h'(y)}{h(y)^3}-\frac{h'(x)}{h(x)^3}}{\frac{1}{h(x)^2}-\frac{1}{h(y)^2}}\right)^2 \tag{4}$$

$$\dfrac{1}{h(x+y)^2}+\dfrac{1}{h(x)^2}+\dfrac{1}{h(y)^2}+3\beta = \left(\frac{\frac{\sqrt{(1-a^2h^2(y))(1-b^2h^2(y))(1-c^2h^2(y))}}{h(y)^3}-\frac{\sqrt{(1-a^2h^2(x))(1-b^2h^2(x))(1-c^2h^2(x))}}{h(x)^3}}{\frac{1}{h(x)^2}-\frac{1}{h(y)^2}}\right)^2 \tag{5}$$

答:
\begin{align*}
w&=\sqrt{\frac{a^2 b^2 c^2 u^2 v^2 \left(u^2-v^2\right)^2+P^2}{4 a^2b^2c^2u^2 v^2 \left(u^2+v^2-a^2u^2v^2-b^2u^2v^2-c^2u^2v^2\right)+Q^2}}\\
\\
P&=u \sqrt{\left(1-a^2 v^2\right) \left(1-b^2 v^2\right) \left(1-c^2 v^2\right)}+v\sqrt{\left(1-a^2 u^2\right) \left(1-b^2 u^2\right) \left(1-c^2 u^2\right)}\\
Q&=1-a^2b^2 u^2v^2-a^2c^2u^2v^2-b^2c^2u^2v^2
\end{align*}

We assume that $0<a<b<c<1$.

We can transform hyperelliptic integral into compound function of elliptic integral by quadratic transformation.

\begin{align*}
\int_{0}^{x}\frac{1}{\sqrt{\left(1-a^2t^2\right)\left(1-b^2t^2\right)\left(1-c^2t^2\right)}}\mathrm{d}t\quad\qquad\\
\\
=
\frac{1}{\sqrt{c^2-a^2}}\int_{0}^{x\cdot\sqrt{\frac{c^2-a^2}{1-a^2x^2}}}\frac{1}{\sqrt{\left(1-t^2\right)\left(1-\frac{b^2-a^2}{c^2-a^2}t^2\right)}}\mathrm{d}t
\end{align*}

Using the addition theorem of elliptic integral, we have

\begin{align*}
U&=u\sqrt{\left(1-a^2u^2\right)\left(1-b^2v^2\right)\left(1-c^2v^2\right)}+v\sqrt{\left(1-a^2v^2\right)\left(1-b^2u^2\right)\left(1-c^2u^2\right)}\\
V&=1-a^2\left(u^2+v^2\right)+\left(a^2b^2+a^2c^2-b^2c^2\right)u^2v^2
\end{align*}

Hence

$$ h(u+v)=W=\dfrac{U}{\sqrt{a^2U^2+V^2\,}} $$

In addition, by changing the order of the letters ($a,b,c$), the equation still holds

Title: Hyperelliptic function addition formula


问:
Congruence Linear System Solutions
The question states the following:

Prove that the system $x \equiv a \bmod m, x \equiv b \bmod n$ has solution iff $\gcd(m,n) \vert a-b.$

The first implication is immediate since $x-a = km,x-b=qn$, hence subtracting both expressions we get that $a-b = qn - km$, $a-b \in (m,n)$, $\gcd(m,n) \mid (a-b)$. However, I don't know how to prove the reciprocal. Any suggestions?

答:
For the "if" $\Leftarrow$ direction:

Taking $g:=\gcd(m,n)$ we can find $c,d\in \mathbb Z$ such that $cm+dn = g$ by Bézout's identity.

Now $g \mid (a-b)$, so find the remainder $r$ of $a$ divided by $g$ to get $a=kg+r$ and thus also $b = \ell g+r$.

Now take $x=\ell cm+ kdn+r$. Then

$x \equiv \ell cm + k (g-cm) +r \equiv k g+r \equiv a\bmod m$  and

$x \equiv \ell(g-dn)+kdn +r \equiv \ell g+r \equiv b\bmod n$

Basically using the Chinese remainder theorem.

Title: Congruence Linear System Solutions


问:
Does any curve have a constant speed parameterization?
Let $(M,g)$ be a smooth, connected Riemannian manifold. Let $\gamma: [a,b]\rightarrow M$ be a piecewise smooth curve in $M$.

My question: is it always possible to find a parameterization $\phi: [c,d]\rightarrow [a,b]$ such that the curve $\gamma_1:=\gamma \circ \phi$ has a constant speed in the sense that $|\gamma_1'|_g$ is constant on $[c,d]$?

Here, a parameterization is a diffeomorphism as in Lee's book.

I knew that if $\gamma$ is regular, namely $\gamma' \neq 0$ on $[a,b]$ then the answer is positive and a parametrized curve can be made with speed of 1.

答:
Initially Lee defines the concept of a reparametrization for smooth curves, but later he generalizes it to piecewise smooth curves.

A reparametrization of a smooth curve $\gamma : [a, b] \to M$ is a curve  of the form $\tilde \gamma = \gamma \circ \varphi$, where $\varphi : [c, d] \to [a, b]$ is a diffeomorphism. More generally, if $\gamma$ is piecewise smooth, $\varphi$ is allowed to be a homeomorphism whose restriction to each subinterval $[c_{i-1}, c_i]$ is a diffeomorphism onto its image, where $c = c_0 < c_1 < \ldots < c_k = d$ is some finite subdivision of $[c, d]$.

In Riemannian Manifolds: An Introduction to Curvature he states in Exercise 6.2 :

Let $γ : [a, b] → M$ be an admissible curve. Show that there exists a unique forward reparametrization $\tilde γ : [0, l] → M$ of $γ$ such that $\tilde γ$ is a unit speed curve.

An admissible curve is a piecewise regular curve - and that is the essential point. An arbitrary piecewise smooth curve $\gamma$ can have points $t \in [c_{i-1}, c_i]$ for some subinterval such that $\gamma'(t) = 0$, and no reparameterization can achieve $\tilde γ'(t) \ne 0$ (use the chain rule).

Note that in the subdivision points $c_1, \ldots, c_{n-1}$ the curve $\gamma$ may have corners (i.e. the left-hand and right-hand derivatives of $\gamma$ do not coincide at $c_i$). Nevertheless we can reparametrize it to $\tilde γ : [0, l] → M$ such that $\tilde γ$ is a unit speed curve. This means that the unit speed curve can change its direction in $c_i$. As an example take the unit speed curve $\delta : [-1,1] \to \mathbb R^2, \delta(t) = (t/\sqrt 2, \lvert t \rvert/\sqrt 2)$.


问:
Proof of infinitude of primes using ratio of n to its totient
A few preliminaries:

A primorial is the product of the first primes. There are two notations for this ($n\#$ is the product of all primes under $n$, and $p_n\#$ is the product of the first $n$ primes; I use the second notation).

A $k$-rough integer is an integer that only has factors that are larger than or equal to $k$. It is important to note that the minimum difference between consecutive $k$-rough multiples of $k$ is $2k$ when $k$ is prime (see another of my questions here: Minimum difference between consecutive multiples of $k$ that are $k\text{-rough}$).

I came across this gem: On a ratio between a number and it's totient . Consider: the ratio $\frac{n}{\varphi(n)}$
 gives the average gap between the elements of the reduced residue system modulo $n$.

Now for the meat of the question:

Consider the primorial $p_n \#$; using this as the input for the above ratio gives us the average gap between $k$-rough integers in the range $[1, p_n \#]$. Recall that we need only prove a number is not divisible by any prime smaller than or equal to its square root to show that that number is prime. If $k = p_{n+1}$, any $k$-rough integer in the range $(p_n, p_{n+1}^2)$ must be prime.

EDIT (everything below this has been added or edited from the original question for clarity and completeness).

For primorial inputs, we can simplify Euler's totient function to: $\varphi(p_n \#) = \prod_{i=1}^n (p_i - 1)$. This becomes important for the limit below.

Let's take the limit:

$$\lim_{n \to \infty}\left(2p_{n+1}-\frac{p_n\#}{\varphi(p_n\#)}\right)$$

This limit is the difference between the smallest-possible gap between consecutive $k$-rough multiples of $k$ if $k = p_{n+1}$, and the average gap between the elements of the reduced residue system modulo $p_n \#$. We could replace the first part of this limit with $p_n^2 - p_n$, but I find the form written here to be much more strict in range and it accomplishes more or less the same goal.

We can replace the totient function with the simplified form:

$$\lim_{n \to \infty}\left(2p_{n+1}-\frac{p_n\#}{\prod_{i=1}^n (p_i - 1)}\right)$$

I note that the primes are very difficult to work with in calculus, so I re-write the above limit to remove dependence on primes and come up with something very similar which is much easier to work with:

$$\lim_{n \to \infty}\left(2(n+1)-\frac{n!}{\prod_{i=2}^n (i - 1)}\right)$$
$$= \lim_{n \to \infty}\left(2(n+1) - \frac{n!}{(n-1)!}\right)$$
$$= \lim_{n \to \infty}\left(2(n+1) - n \right)$$
$$= \lim_{n \to \infty} n + 2 = \infty$$

I believe this limit stands regardless of the subset of the positive integers we choose to use in the products so long as the same set is used in both functions in the difference (correct me if I'm wrong on that).

So: does this approach stand as a valid proof of the infinitude of the primes? (I guess I should also include in this: is my calculus valid?)

(addendum)

Here is why I'm interested in this: the simplified form of Euler's totient function can also be modified in other ways. For example:

$$(\prod_{i=3}^n (p_i - 2)) - 1$$

...counts the number of translations of the $k$-tuple $(0, 2)$ (twins) among the residues modulo $p_n\#$. In fact, if we pick the right starting point for this product, we can substitute in any $k$ as the subtrahend and start the product out at the first prime that is at least half as large as the diameter (difference between the largest and smallest elements) of our target $k$-tuple. A few other minor details need to be accounted for, such as the initial number of $k$-tuples in the residues, but this product can be used to count the number of any admissible $k$-tuple among the residues modulo primorials that are large enough relative to the target $k$-tuple. The interesting thing is that all of these minor details pretty much account for nothing when $n$ goes to infinity; the limit remains the same.

We know:

there are infinitely many primes, so this means that 1-tuples fall into the provably-prime interval for $k$-rough integers in their corresponding primorial residues, so this happens infinitely often.

the distribution of any $k$-tuple, when $n$ goes to infinity, among the residues modulo primorials, behaves in exactly the same way as the distribution of $1$-tuples.

So: 1-tuples fall into the provably-prime interval infinitely often, and the distribution of these behaves exactly like the distribution of any $k$-tuple among the residues modulo primorials. So: If there are infinitely many 1-tuples, there must also be infinitely many $k$-tuples for any $k$ because any other result would be contradictory to our observation that this limit applies equally well to all $k$-tuples.

All of this seems far too simple for it to have been missed, so I conclude I must be missing something or my logic is incomplete or impossible in some way, and I'm trying to figure out what that is so this can stop driving me nuts.

答: The simplified Euler's totient function you write above can be generalized to all k-tuples. See my recent comment to your older post regarding Modified Euler's Totient function. Once modified, you'll see that it ceases to be, primarily, a k-tuple counting function, because k-tuples (conventionally described in terms of the number of their elements, k, and the differences between elements) lose their unique identity in the formula (which cares nothing for absolute differences); all k-tuples with the same k and same prime factors of the pairwise differences between elements (disregarding repeats of the prime factor for each smallest element in multiple pairs) behave the same (which specifically accounts for the ratios of Hardy Littlewood constants for prime constellations). When generalized, it converts to, primarily, a sifting function (though it can continue to be used to count k-tuples per primorial).</p>


问:
Show that a function is a homeomorphism
Let $f: \mathbb{R} ^2 \longrightarrow (0, \infty) \times \mathbb{R}$ be given by a formula:

$f(x,y)=(e^x, x-y)$

Show that it is a homeomorphism (in standard euclidean metric).

As I understand we have to check continuity ($f$ is continuous, since $f=(f_{1},f_{2}-f_{3})$, where $f_{1}=e^x$,  $f_{2}=x, f_{3}=y$ and every such a function is continuous. Is this reasoning valid?

Then we should show that it is a bijection (seems to be OK).

And finally propose a continuous inverse function. What kind of formula could describe it?

答:
Welcome to MSE. One way to find the inverse function for some $y = f^{-1}(x)$ is first to switch the variables to $x = f(y)$, and then to do some algebra to get $y$ by itself. In a similar vein, we can start with
$$f^{-1}(x_1, x_2) = (y_1, y_2),$$
switch it around to
$$f(y_1, y_2) = (x_1, x_2),$$
and then do some algebra to get $y_1$ and $y_2$ by themselves again.

Explicitly, we're starting with
$$(e^{y_1}, y_1 - y_2) = (x_1, x_2), $$
and this leads to two equations in which you can get $y_1$ and $y_2$ in terms of $x_1$ and $x_2$.

Title: Show that a function is a homeomorphism


问:
if $A=B \cup C$ and $A,B$ are open and $C$ is closed then $C \subset B$?

if $A=B \cup C$ and $A,B$ are open and $C$ is closed then $C \subset B$?

I thought about this statement for another question that I had.

Suppose $A,B,C$ are subsets of $\mathbb{R}^n $ such that $A=B\cup C$ and $A,B$ are open and aren't closed, $C$ is closed and isn't open, does that necessaraly mean that $C\subset B$?

thanks/

答:
Let $D$ denote all of the boundary points of $C$.  At first glance, I guessed that the conjecture is correct, because $C$, which is closed must contain all of $D$.  Therefore $B\cup C$ must contain all of $D$.  However, since $B\cup C$ is specified as open, each point in $D$ can not be a boundary point in $B\cup C$.  This follows, because $B\cup C$, which is open, can not contain any of its boundary points.

This implies that each boundary point in $D$ must be contained in some open neighborhood of $B\cup C$.  This superficially (and wrongly) suggested that all of $C$ must be contained in some open neighborhood of $B$.

There are two things wrong with this suggested conclusion:

Even if it were true that every point $D$ is contained in some open neighborhood around $B$, that does not imply that each
point in $C$ is contained in some open neighborhood around $B$.

A case in point is $B = (-2,-1) \cup (1,2)$ and $C = [(-3/2),(3/2)].$

As has already been demonstrated by the counter example given in Numbra's response, it is feasible that every point in $D$ will also be a boundary point of $B$.  In such a case, since $B$ is open, it won't contain any point in $D$, and therefore can not be a superset to $C$.

In fact, a less extreme counter-example would be if exactly one point $d \in D$ is also a boundary point of $B$.  This (by itself) implies that $d \in C$ and $d \not\in B$, which (again) implies that $B$ is not a superset of $C$.


问:
What books must every math undergraduate read?
I'm still a student, but the same books keep getting named by my tutors (Rudin, Royden).

I've read Baby Rudin and begun Royden though I'm unsure if there are other books that I "should" be working on if I want to study beyond Masters. I'm not there yet as I'm on a four year course and had a gap year between Years 3 and 4.

Please recommend for Algebra, Linear Algebra and Categories - Analysis, Set Theory, Measure theory (an area I have seen too little books dedicated for).

E.g. Spivak is very good for self learning basic real analysis, but Rudin really cuts to the heart.

答:
Alex Youcis's and other users suggestions are pretty good but I think some of the the books that they mentioned are a little difficult for average undergraduate students. But I totally agree with most of their suggestions.

Here are some books which I found really nice and helpful for undergraduate level:

Algebra :

Algebra in Action: A Course in Groups, Rings, and Fields by Shahriar Shahriari

Abstract Algebra: An Introduction with Applications by Derek Robinson

Linear Algebra :

Linear Algebra Done Right by Sheldon Axler

Analysis :

Real Mathematical Analysis by Charles C. Pugh

Measure, Integration and Real analysis by Sheldon Axler (More Advanced)

Complex Analysis :

Complex Analysis by Theodore W. Gamelin

A Course in Complex Analysis and Riemann Surfaces by Wilhelm Schlag (More Advanced)

Algebraic Topology :

Basic Topology by M. A. Armstrong

Introduction to Topological Manifolds by John M. Lee

Algebraic Geometry :

An Invitation to Algebraic Geometry by Karen E. Smith & others

Smooth Manifolds :

An Introduction to Manifolds by Loring W. Tu

Elementary Number Theory :

Elementary Number Theory and Its Applications by Kenneth H. Rosen

Elliptic Curves :

Rational Points on Elliptic Curves by Joseph H. Silverman & John T. Tate

Dynamical Systems :

Differential Equations, Dynamical Systems, and an Introduction to Chaos by Morris W. Hirsch, Stephen Smale & Robert L. Devaney

Complex Dynamics by Lennart Carleson & Theodor W. Gamelin

Dynamics in One Complex Variable by John W. Milnor


问:
Is the support of $\rho_{AB}$ always contained in the support of $\rho_A\otimes\rho_B$?
Given a positive unit trace Hermitian matrix, i.e. a density operator, $\rho_{AB}$ on Hilbert spaces $\mathcal{H}_A\otimes\mathcal{H}_B$. Consider its marginals $\rho_A$, $\rho_B$. Do we have the support of $\rho_{AB}$ always contained in the support of $\rho_A\otimes\rho_B$? If so, what is the universal constant $C$, such as the dimension, such that $C\rho_A\otimes\rho_B\geq\rho_{AB}$?

答:
To answer your first question: yes.

To see why, let $\Pi_A$ and $\Pi_B$ denote the projection operators onto the supports of $\rho_A$ and $\rho_B$ respectively, and let $1_A$ and $1_B$ denote the corresponding identity operators.
Now,
\begin{align*}
\operatorname{Tr}((1_A-\Pi_A)\otimes I_B)\rho) &= \operatorname{Tr}((1_A\otimes1_B)\rho) - \operatorname{Tr}((\Pi_A\otimes I_B)\rho)\\& = \operatorname{Tr}(\rho) - \operatorname{Tr}(\Pi_A\rho_A)\\& = \operatorname{Tr}(\rho) - \operatorname{Tr}(\rho_A)\\& = 0.
\end{align*}
It follows that $((1_A-\Pi_A)\otimes I_B)\rho = \rho((1_A-\Pi_A)\otimes I_B)= 0$ and thus $$(\Pi_A\otimes I_B)\rho = \rho(\Pi_A\otimes I_B)= \rho.$$
Similar analysis shows that
$$(1_A\otimes \Pi_B)\rho = \rho(1_A\otimes \Pi_B)= \rho.$$
It follows that
$$
(\Pi_A\otimes\Pi_B)\rho = \rho (\Pi_A\otimes\Pi_B) = \rho,
$$
which proves the claim that the support of $\rho$ is contained in the support of $\rho_A\otimes\rho_B$ (since $\Pi_A\otimes\Pi_B$ is the projection onto the support of $\rho_A\otimes\rho_B$).

I need to think more about whether or not there exists such a constant $C$.


问:
Invariance of Levi-Civita tensor
I need some help trying to prove the invariance under rotations of the Levi-Civita tensor, I know that the tensor transform using $U\in SO(3)$ from unitary group with $\det(U)=1$, so with this definition of the determinant (I need to use specifically this definition),
$$
\det(U)=\frac{1}{6}\epsilon^{abc}\epsilon_{ijk}U^i_aU^j_bU^k_c\qquad(\text{for U$\in$ SO(3)})=1
$$

So I need to transform the Lev-Civita tensor,

$$
\epsilon^{abc}\rightarrow U^{a'}_iU^{b'}_jU^{c'}_k\epsilon^{abc}\overset{!}{=}\epsilon^{lmn}
$$

I dont know how to relate the two definitions that I have given, but it could be done, I need help for the following steps. Thanks!

答:
As far as I understand you have to prove that Levi-Civita tensor is constant tensor over transformation by U in $SO(3) $ ($det U=1$) and use as a tool for proving a determinant definition as $\det(U)=\frac{1}{6}\epsilon^{abc}\epsilon_{ijk}U^i_aU^j_bU^k_c$.

We identify $\epsilon^{abc}$ as an antisymmetric tensor ($\epsilon^{abc}=0$ if any two indices are the same; for example, $\epsilon^{112}=0$),  $\epsilon^{123}=1$, the rest of the tensor values are equal to $+1$ or $-1$ - depending on the parity of the permutation of the indices (for example, $\epsilon^{132}=-1$).
Now you want to prove that $\epsilon^{abc}\rightarrow U^{i}_aU^{j}_bU^{k}_c\epsilon^{abc}=B(U)^{ijk}$ (some tensor depending on $U$) $=\epsilon^{ijk}$ over transformation by means of any $U$ (in this formula the summation over repeated indices $a, b, c$ is implied).

Let’s check it directly. First of all, $B(U)^{ijk}=-B(U)^{jik}$ over permutation of neighboring indices. Indeed, $U^{j}_aU^{i}_bU^{k}_c\epsilon^{abc}= U^{i}_bU^{j}_aU^{k}_c\epsilon^{abc}$ $=- U^{i}_bU^{j}_aU^{k}_c\epsilon^{bac}$ But $a$ and $b$ are just the (dumb) indices of summation; we can write $a$ instead of $b$ and $b$ instead of $a$ in our expression: $U^{j}_aU^{i}_bU^{k}_c\epsilon^{abc}=- U^{i}_bU^{j}_aU^{k}_c\epsilon^{bac} =- U^{i}_aU^{j}_bU^{k}_c\epsilon^{abc}= -B(U)^{ijk}$. In the same way we can prove that $ B(U)^{jik}$ is the antisymmetric tensor for all its indices.

Now we have to prove that $ B(U)^{123}=1$. Taking into consideration that $a, b, a$ have to be different (otherwise, we get zero due to anti-symmetry of the tensor $\epsilon^{abc}$) we get:
$ B(U)^{123}= U^{1}_aU^{2}_bU^{3}_c\epsilon^{abc}= U^{1}_1(U^{2}_2U^{3}_3- U^{2}_3U^{3}_2)- U^{1}_2(U^{2}_1U^{3}_3- U^{2}_3U^{3}_1)+ U^{1}_3(U^{2}_1U^{3}_2- U^{2}_2U^{3}_1)=det(U)=1$

Tensor $B(U)^{ijk}$ is antisymmetric with values $+1$ or $-1$, therefore $B(U)^{ijk}=\epsilon^{ijk}$

Title: Invariance of Levi-Civita tensor


问:
the relationship between generating function and recurrence
Does the existence of a generating function of a sequence imply the existence of recurrence relation of the same sequence?</p>

答:
No. As Zev Chonoles noted,
you can create a generating function (GF) for
$any$ sequence.

What make a recurrence come from a GF
is that the GF have certain properties.
Some properties that do this are:

The GF is a rational function. This directly results in a recurrence.

The GF satisfies a differential equation. This often, but not always, results in a recurrence.

The GF satisfies a quadratic equation. The Catalan numbers do this.

See Wilf's "generatingfunctionology" for lots more. It is available for free download at http://www.math.upenn.edu/~wilf/DownldGF.html.


问:
Meaning of $dx$ in integration and differentiation
In differentiation, we denote $dx$ as in infinitesimal change in $x$ and $dy$ the corresponding change.But in integration, we divide $x$ into infinitely many equal intervals and name them $dx$. Then $f(x)dx$ represents the area of an infinitely thin rectangle. From differentiation we know, $dy = f(x)dx$.Why do we substitute then $dy$ in place of $f(x)dx$? I mean $dy$ is change in differentiation. But here in integration we take $dy$ to be the area of each infinitesimal rectangle. Why do we mix them up? Please someone explain with clarity.

答:
Without being formal, $dy = f'(x)dx$ not $dy = f(x)dx$

So $dy$ does actually stand for a small change in $y$. Since you're integrating $f'(x)dx$,
it would give you the change in the antiderivative of $f'(x)$, which is $f$ which is equal to $y$.


问:
Change from differentiation wrt to matrix to wrt to inverse of matrix for symmetric matrices
For the rule below:

$$
\frac{\partial J}{\partial \mathbf{A}}= -\mathbf{A}^{-T} \frac{\partial J}{\partial \mathbf{W}} \mathbf{A}^{-T}
$$

where $\mathbf{A}$ is an invertible square matrix, $\mathbf{W}$ is the inverse of $\mathbf{A}$, and J is a function (see end of section 2.2 in matrix cookbook https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf)

Does this rule hold if $\mathbf{A}$ is a symmetric matrix?

答:
If $A$ is symmetric but not invertible, the rule won't hold as the inverse of $A$ is not even defined.

And if $A$ is symmetric invertible... then it is invertible and the formula holds as for any invertible matrix.


问:
Convergence of $\sum_{n=1}^{\infty} \left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n} \right)^{3n - \ln^2n}$
I need to examine convergence of that sum:
$$\sum_{n=1}^{\infty} \left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n} \right)^{3n - \ln^2n}$$

I can rewrite it as:
$$\sum_{n=1}^{\infty} \frac{\left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n}\right)^{3n}}{\left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n}\right)^{\ln^2n}}$$

But then, I can't even get:

$$\displaystyle \lim_{n \to \infty} \left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n} \right)^{3n - \ln^2n} = 0$$

I know that is true but I don't know how to transform that expression to do something with trigonometric functions (how to merege them).

Additionally, I know that ratio test won't help me once I am done with limit. Same for root test. I think that it will be hard to use Abel's and Dirichlet's since trigonometric functions will equal to something periodic ( $\neq $ monotonic).

答:
Apply the root test

We have
$$\underset{n\to \infty }{\text{lim}}\frac{3 n-\log ^2 n}{n}=3$$

and
$$\left| \frac{3 \cos \left(n^2\right)+1}{2 \sin ^2(n)+7}\right|\le \frac47 $$
$$L=\underset{n\to \infty }{\text{limsup}}\left(\left| \frac{3 \cos \left(n^2\right)+1}{2 \sin ^2(n)+7}\right| ^{3 n-\log ^2(n)}\right)^{1/n}=\frac{64}{343}$$

as $L<1$ the series converges absolutely.


问:
lower bound of $n(n-1)(n-2) \dots (n-k)$
Is there any tight lower bound on following expression for $n\geq k$

$$n(n-1)(n-2) \dots (n-k+1)$$

one lower bound I could think is $(n-k+1)^k$. But  I need to find $f$ s.t

$$f(k) n^{k} \leq n(n-1)(n-2) \dots (n-k+1)$$

my attempt using stirling's approximation

$$n(n-1)(n-2) \dots (n-k+1) = k! \binom{n}{k} \geq \sqrt{2\pi}k^{k} \sqrt{k} e^{-k} . (n/k)^{k} = \Big[\sqrt{2\pi k} e^{-k}\Big] n^k $$. I need to know if I can remove $e^{-k}$ in above and get something more like $k^{-c}$, atleast when $k$ is small.  Any pointers and suggestions would be helpful.

答:
$$n(n-1)(n-2)\cdots(n-k)=\frac{n!}{(n-(k+1))!}=\binom{n}{k+1}(k+1)!$$

Now we know that $$\left(\frac{n}{k+1}\right)^{k+1} \leq {n \choose k+1} \leq \left( \frac{en}{k+1}\right)^{k+1}$$
Therefore we get that
$$\left(\frac{n}{k+1}\right)^{k+1}(k+1)!\leq n(n-1)(n-2)\cdots(n-k)\leq\left( \frac{en}{k+1}\right)^{k+1}(k+1)!$$
where $e=\sum_{m=0}^{\infty}\frac{1}{m!}$


问:
Is there a general method to determine the [non-]solutions to $ax^4+bx^2+c=y^2$?
I am looking for general methods to solve, in integers $x$ and $y$, the equation
$$
ax^4+bx^2+c=y^2
$$
where $a,b,c$ are given integers, and by "solve" I mean:

(i) show there are no [non-trivial] solutions; or

(ii) give a complete finite solution; or

(iii) provide a parameterization, recursion, or algorithm by which all solutions may be obtained.

As an example, I believe I have recently solved the case $(a,b,c)=(48,12,1)$, i.e., I think I have shown, using ad hoc elementary methods, that the equation $$48x^4+12x^2+1=y^2$$ has only the trivial solution $(x,y)=(0,\pm 1)$. [n.b. I could post my proof here, but it is ultimately irrelevant to my present question.]

Now I'm wondering if this particular wheel has already been invented.

答:
A general and elementary method for these equations is given in the article Fermat and the difference of two squares, The Mathematical Gazette, Vol96, Number 537,Nov 2012, pp.480-491.

These equations can have a small set of easily determined 'trivial' solutions together with one or more infinite sequences of solutions derived from base solutions with neither $a$ nor $c$ squares.

Although this is proved by elementary means, involving no more than the difference of two squares and Fermat's proof by infinite descent, it is interesting to see the parallels with the Torsion group and Rank of an elliptic curve solution.

The connection with elliptic curves is explored further in the same journal, New points from old, The Mathematical Gazette, Vol96, Number 537, Nov 2012, pp. 492-498, where a group structure is set up directly on these curves without the usual process of conversion into a standard elliptic curve form. (The method described there extends easily to curves with $y^2=ax^4+bx^2+c$ and, providing one knows a rational point, extends to general quartics).


问:
If the set of specific sequences of 0's and 1's is countable then it can be used to count the set of all infinite sequences of 0's and 1's. Right?
I am seeking some help to understand why an infinite sequence of $\{0, 1\}$ is uncountable. While there are similar questions here with detailed answers, I wasn't able to resolve the contradiction below:

Given an alphabet $A$ composed of only $0$ and $1$, the set of all possible sequences of alphabet characters of the form $$a_1, a_2, a_3, ..., a_k,$$ where $a_i \in A$, is countably infinite. How is it possible that at the same time the set of infinite sequences of $0$'s and $1$'s is uncountably infinite?

First, let's prove the following:

Given an alphabet $A$ composed of only $0$ and $1$, the set of all possible sequences of alphabet characters of the form $$a_1, a_2, a_3, ..., a_k,$$ where $a_i \in A$, is countably infinite.

Proof 1

The set that is composed of described sequences, including zero-length sequence (an empty sequence), can be denoted as $A^*$. Let's correspond a zero-length sequence with the number $1$. Then all single-character sequences will take numbers from $2$ to $n + 1$, where $n$ is the number of characters in $A$ (which are unique by alphabet's definition). All two-character sequences will take numbers from $n+2$ to $n^2 + n + 1$. All three character sequences are getting numbers from $n^2 + n + 2$ to $n^3 + n^2 + n + 1$. This pattern can be continued infinitely. Therefore, it is possible to enumerate every element of $A^*$. Thus, I conclude that the set $A^*$ is countably infinite since there is a correspondence between $A^*$ and the set of all natural numbers $\mathbb{N}$.

Contradiction

At the same time, the set of infinite sequences of $0$'s and $1$'s (let's call it $T$) is proved to be uncountably infinite (here is a link to Wikipedia with the proof using Cantor's diagonal argument). If proof 1 is correct, then I can make an infinite sequence of zeros and substitute the first $k$ elements of it with any sequence from $A^*$ of the length $k$ which is enumerated. Since $|k| = |\mathbb{N}|$, with this arrangement, it should be correct (but I guess it is not) to state that now every element of $T$ is numbered, hence $T$ is countably infinite.

Reflection

OBVIOUSLY, there is a problem somewhere in my line of thought, but it is hidden from me. My goal is to properly understand why $A^*$ cannot be used to count all the elements of $T$. The answer to this question will also help me to deepen my understanding of Cantor's diagonal argument and explain why it cannot be applied to $\mathbb{N}$.

I sort of understand that Cantor's diagonal method will not work with $A^*$ because it is impossible to construct an argument that is not counted. But on the other hand, if a set of all infinite sequences of $0$'s and $1$'s is said to be infinite, doesn't it mean that any argument one can make must already be in the set?

Any comments are very much appreciated. Thanks!

答: Notice that no finite sequence of $A^*$ corresponds to the infinite sequence $1,0,1,0,1,0,\dots$. So, for this sequence, there is no element of $A^*$ which corresponds to it. That is the problem with your reasoning. Every sequence that has infinitely many ones can´t be obtained from an element of $A^*$.


答:
The problem is when you say

If proof 1 is correct, then I can make an infinite sequence of zeros and substitute the first $k$ elements of it with any sequence from $A^*$ of the length $k$ which is enumerated. Since $|k| = |\mathbb{N}|$, with this arrangement, it should be correct (but I guess it is not) to state that now every element of $T$ is numbered, hence $T$ is countably infinite.

For one sequence you can indeed make a countable number of approximations by finite sequences. By doing that you won't obtain all sequences of $0,1$.

Title: If the set of specific sequences of 0's and 1's is countable then it can be used to count the set of all infinite sequences of 0's and 1's. Right?

Body:
I am seeking some help to understand why an infinite sequence of $\{0, 1\}$ is uncountable. While there are similar questions here with detailed answers, I wasn't able to resolve the contradiction below:

Given an alphabet $A$ composed of only $0$ and $1$, the set of all possible sequences of alphabet characters of the form $$a_1, a_2, a_3, ..., a_k,$$ where $a_i \in A$, is countably infinite. How is it possible that at the same time the set of infinite sequences of $0$'s and $1$'s is uncountably infinite?

First, let's prove the following:

Given an alphabet $A$ composed of only $0$ and $1$, the set of all possible sequences of alphabet characters of the form $$a_1, a_2, a_3, ..., a_k,$$ where $a_i \in A$, is countably infinite.

Proof 1

The set that is composed of described sequences, including zero-length sequence (an empty sequence), can be denoted as $A^*$. Let's correspond a zero-length sequence with the number $1$. Then all single-character sequences will take numbers from $2$ to $n + 1$, where $n$ is the number of characters in $A$ (which are unique by alphabet's definition). All two-character sequences will take numbers from $n+2$ to $n^2 + n + 1$. All three character sequences are getting numbers from $n^2 + n + 2$ to $n^3 + n^2 + n + 1$. This pattern can be continued infinitely. Therefore, it is possible to enumerate every element of $A^*$. Thus, I conclude that the set $A^*$ is countably infinite since there is a correspondence between $A^*$ and the set of all natural numbers $\mathbb{N}$.

Contradiction

At the same time, the set of infinite sequences of $0$'s and $1$'s (let's call it $T$) is proved to be uncountably infinite (here is a link to Wikipedia with the proof using Cantor's diagonal argument). If proof 1 is correct, then I can make an infinite sequence of zeros and substitute the first $k$ elements of it with any sequence from $A^*$ of the length $k$ which is enumerated. Since $|k| = |\mathbb{N}|$, with this arrangement, it should be correct (but I guess it is not) to state that now every element of $T$ is numbered, hence $T$ is countably infinite.

Reflection

OBVIOUSLY, there is a problem somewhere in my line of thought, but it is hidden from me. My goal is to properly understand why $A^*$ cannot be used to count all the elements of $T$. The answer to this question will also help me to deepen my understanding of Cantor's diagonal argument and explain why it cannot be applied to $\mathbb{N}$.

I sort of understand that Cantor's diagonal method will not work with $A^*$ because it is impossible to construct an argument that is not counted. But on the other hand, if a set of all infinite sequences of $0$'s and $1$'s is said to be infinite, doesn't it mean that any argument one can make must already be in the set?

Any comments are very much appreciated. Thanks!


问:
Equivalence classes of the relation that the largest digit of integer a = largest digit of integer b.
Define the relation $\mathcal{R}$ on the set of all positive integers by: for all positive integers $a$ and $b$, $a\,\mathcal{R}\,b$ if and only if the largest digit of $a$ is equal to the largest digit of $b$. for example, $271\,\mathcal{R}\,770$ because the largest digit of $271$ is $7$ which is also the largest digit of $770$.

a) Find the number of equivalence classes of $\mathcal{R}$.

b) Find and simplify the number of positive integers between $100$ and $1000$ which are in the equivalence class $[271]$.

I have an idea of where to begin but I think I am wrong. For (a) would there be $9$ equivalence classes?
And for (b) I got my answer to be $56$.

答:
Part (a) sounds about right to me. In part (b), your count is a bit too low:

There are 49 numbers of the form $7XY$ with $0\leq X,Y<7$

Another 7 are of the form $77X$ (with $0\leq X<7$)

There is certainly number $777$

This already adds up to $49+7+1=57$ and we didn't include all the possible forms yet.


问:
How to approach this question?
I have been trying this, but I do not understand how to get the gradient of the triple integration.

If a region $V$ bounded by a surface $S$ has a continuous charge (or mass) distribution of density $\rho$, the potential $\phi(P)$ at a point $P$ is defined by $\phi = \iiint_V \frac{\rho dV}{r}$. Deduce the following under suitable assumptions:

$\iint_S \mathbf E \cdot d\mathbf S = 4 \pi \iiint_V \rho d V$, where $\mathbf E = - \nabla \phi$.

$\nabla^2\phi = - 4 \pi \rho$ (Poisson's equation) at all points $P$ where charges exist, and $\nabla^2 \phi = 0$ (Laplace's equation) where no charges exist.

Now, I tried to use Divergence theorem. But I do not understand how do I find gradient of triple integration. Thanking you all.

Here is how I tried.Assuming $\rho$ to be a function of r.

$\phi = \iiint_V \frac{\rho dV}{r}$

$\nabla\phi = \iiint_V \nabla(\frac{\rho }{r})dV$

$\nabla\phi = \iiint_V (\frac{\rho'\mathbf r}{r}+\rho\frac{-\mathbf r}{r^3})dV$

$\nabla^2\phi = \iiint_V \nabla(\frac{\rho'\mathbf r}{r}+\rho\frac{-\mathbf r}{r^3})dV$

$\nabla^2\phi = \iiint_V (\rho''\frac{\mathbf r . \mathbf r}{r^2}+(\rho'\frac{\mathbf r}{r}+\rho(\frac{3}{r^3}+\mathbf r\frac{-3\mathbf r}{r^5})))dV$

$\nabla^2\phi = \iiint_V (\rho''+(\rho'\frac{\mathbf r}{r}))dV$

$\nabla^2\phi = \iiint_V (\rho''+\rho'\hat{r}))dV$

Here, if I assume $\rho$ to be constant then $\nabla^2\phi$ is equal to zero.
But my result are not upto the expectation of the question.
I followed as I was advised but results are unsatisfactory.
If there is any mistake in my calculation ,please correct me.

Thanking you

答:
Hint: For part 1, the idea is to use the divergence theorem to write
$$
\iint_S \mathbf E \cdot d \mathbf S = \iiint_V \nabla \cdot \mathbf E\,dV.
$$
You should be able to simplify the expression $\nabla \cdot \mathbf E = \nabla \cdot (-\nabla \phi) = -\nabla^2 \phi$ to end up with the integrand $4 \pi \rho$.

For part 2, you will have already found that $-\nabla^2 \phi = 4 \pi \rho$ in part 1. Multiplying both sides gives you the desired result $\nabla^2 \phi = -4 \pi \rho$, which simply becomes $\nabla^2\phi = 0$ at points $P$ for which $\rho(P) = 0$.

Title: How to approach this question?


问:
Find the length of an edge in triangle
So my girlfriend is helping a highschool student with geometry, and we have come across what seems like an unsolvable question. I hope I translate the question to English right:

$AD, BE$ and $CF$ are medians in triangle $ABC$ and they intersect in $G$.

$AD$ perpendicular $BE$.

$|AC|=10$ cm.

$\angle AFC=86°$.

$|CG|=|AB|$.

Find the length of $AF$.

It feels like we tried everything and we just can't get it.

答:
Remember that the centroid of the triangle (intersection of the medians) cuts the medians in 2:1 proportions. That is, $CG=2FG$. Using that $CG=AB$, you have that $CF=\frac{3GC}{2}=\frac{3AB}{2}=3AF$.

Now you just have to use the cosine law in the triangle $AFC$:

$100=AC^2=AF^2+CF^2-2AF\cdot CF\cos(86º)=10AF^2-6AF^2\cos(86º)$.

Now just solve for $AF$ in the second degree equation.

Title: Find the length of an edge in triangle


问:
development of $\frac{x-sin x}{x^2}$ in 0
I have to develop the function $\frac{x-sin x}{x^2}$ until the fifth term in the point 0.
But the function is not defined in 0, and also the derivatives are not defined in 0.
In this case is it still possible to find a function that can approximate the given function?

N.B. For development I intend this:
$$  \sum_{k=0}^5 \frac{f^{(k)}}{k!}*(x-x_0)^k  $$

答:
Hint

$f(x) = \frac{x - \sin x}{x}$ is indeed not defined at $0$. $f$ can however be extended by continuity at $0$. You'll see that when you'll perform the expansion. And see that $f(0)=0$ allows to extend $f$ to a continuous map defined on $\mathbb R$.


答:
If you by "develop" mean to take its Taylor expansion, you can certainly do that.

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \mathcal{O}(x^9)$$

$$x-\sin(x) = \frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} + \mathcal{O}(x^9)$$

$$\frac{x-\sin(x)}{x^2} = \frac{x}{3!} - \frac{x^3}{5!} + \frac{x^5}{7!} + \mathcal{O}(x^7)$$

This series will converge to $f$ for all $x \neq 0$. If you redefined $f$ so that $f(0)=0$, the function would be continuous and the series will converge to $f$ for all $x \in \mathbb{R}$.

Title: development of $\frac{x-sin x}{x^2}$ in 0

Body:
I have to develop the function $\frac{x-sin x}{x^2}$ until the fifth term in the point 0.
But the function is not defined in 0, and also the derivatives are not defined in 0.
In this case is it still possible to find a function that can approximate the given function?

N.B. For development I intend this:
$$  \sum_{k=0}^5 \frac{f^{(k)}}{k!}*(x-x_0)^k  $$


问:
Is every injective ring homomorphism an automorphism?
It seems a very interesting, innocuous and dubious result to me. Is the following true?

Consider the injective ring homomorphism $\phi :  R \rightarrow R$. Is it true that $\phi$ is an isomorphism? I am taking ring homomorphism to mean $f(ab)=f(a)f(b)$ and $f(a+b) = f(a)+f(b)$ with $f(1)=1$.

My attempt: $R \cong \phi (R) \subset R$. So, $\phi(R)=R$.

答:
No, consider $R=\Bbb Z[X]$. Let $\phi$ be defined by $X\mapsto X^2$, i.e. $$\phi\left(\sum_{i=0}^n a_iX^i \right)=\sum_{i=0}^n a_iX^{2i}$$ Then $\phi$ is injective but its image does not contain $X$.

It is also not true if we replace 'injective' with 'surjective'. To see this, let $R=\Bbb Z[X_1,X_2,X_3,\dots]$ where $\phi$ is given by $X_1\mapsto 0$ and $X_i\mapsto X_{i-1}$ for $i>1$.

So rings can be isomorphic to both proper subrings and proper quotients.


答: Hint:  You are correct for a.  For b, it would be easier if you explain how you got $56$.  There are more than that.  How many are in the range $700-799?$  Intuitively, class $9$ must be the largest and there are a total of $901$ so $56$ seems small.</p>

Title: Equivalence classes of the relation that the largest digit of integer a = largest digit of integer b.

Body:
Define the relation $\mathcal{R}$ on the set of all positive integers by: for all positive integers $a$ and $b$, $a\,\mathcal{R}\,b$ if and only if the largest digit of $a$ is equal to the largest digit of $b$. for example, $271\,\mathcal{R}\,770$ because the largest digit of $271$ is $7$ which is also the largest digit of $770$.

a) Find the number of equivalence classes of $\mathcal{R}$.

b) Find and simplify the number of positive integers between $100$ and $1000$ which are in the equivalence class $[271]$.

I have an idea of where to begin but I think I am wrong. For (a) would there be $9$ equivalence classes?
And for (b) I got my answer to be $56$.


问:
A simple question about union of finite groups
My simple question is whether a union of finite groups is a finite group, or it isn't necessarily true. For example, if I take a group $A$ such that: $A = A_1\cup A_2\cup A_3 \cup...\cup A_k$, when $A_i, 1\le i\le k$, are finite sets in $P(\mathbb{N})$. Is $A$ a finite set in $P(\mathbb{N})$, or not?

答:
It is straightforward to see that a finite union of finite sets is again a finite set (actually, you can compute the cardinality of the union, see this: https://en.wikipedia.org/wiki/Inclusion–exclusion_principle).

However, it is false in general that the union of two groups is again a group, unless one is contained in the other. Here is a hint for the proof: take two groups $G $and $H$, take two elements $g\in G-H$, $h\in H-G$ and see what the product $gh$ does.


问:
Proof that the following composite matrix is invertible
Let $B\in \mathbb{R}^{n\times n}$ and $C\in\mathbb{R}^{m\times n}$ (where $m\leq n$) and let $v^TBv > 0$  $\forall v \neq 0$ with $Cv = 0$, where $C$ has full rank.

Proof that $A = \begin{pmatrix}
B & C^T \\
C & 0
\end{pmatrix}$ is an invertible Matrix.

So far, I've concluded that $B$ is positive definite and therefore also $det(B)>0$, which could help prove that $det(A) \neq 0$.
But other than that I had no idea for a solution.

Any help is appreciated, thanks in advance!

答:
Take $B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0  \end{bmatrix}$ to see that $B$ does not have to be positive definite (replace the $-1$ with $0$ in $B$ to see non-invertibility of $B$).

To show that $A$ is invertible we will show $Av = 0 \implies v = 0$. We have

$$\begin{bmatrix} B & C^T \\ C & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 $$

gives $Cv_1 = 0$ and $Bv_1 + C^Tv_2 = 0$. Left multiplying the second equation by $v_1^T$ yields

$$v_1^TBv_1 + v_1^TC^Tv_2 = 0 \iff v_1^TBv_1 + (Cv_1)^Tv_2 = 0$$

which gives $v_1^TBv_1 = 0$ since $Cv_1 = 0$. Thus $v_1 = 0$
(otherwise $v_1^TBv_1 > 0$). Plugging this back in gives $C^Tv_2 = 0$ which implies $v_2 = 0$ since $C^T$ is injective (full column rank).

$\square$


答:
Important: the following answer is assuming $|R|<\infty$, otherwise this isn't true in general.

Consider the ring homomorphism $\varphi: R\to R$. Then, if you want to prove it's an isomorphism, you just need to prove it's injective or it's surjective (you don't need to prove both).

Why is this true? Because, since $\varphi$ goes from $R$ to $R$ ($R$ goes to himself), the cardinality of the origin set and the destiny set are the same (because they are the same set). So:

If $\varphi$ is injective, every different $x\in R$ gives a different $\varphi (x)$, in fact $|R|$ different images, but the cardinality of the destiny set is $R$ so it's also surjective, hence a bijection.

If $\varphi$ is surjective, then every $y\in R$ has some preimage $x$, so in order for that to be true you need $|R|$ different preimages, the cardinality of the origin set, so then $\varphi$ is also injective, hence a bijection.

So, when facing an homomorphism that goes from one finite set to himself, you just need to prove one of the two conditions to prove it's a bijection (since the other one will be a direct consequence of the first).

Title: Is every injective ring homomorphism an automorphism?

Body:
It seems a very interesting, innocuous and dubious result to me. Is the following true?

Consider the injective ring homomorphism $\phi :  R \rightarrow R$. Is it true that $\phi$ is an isomorphism? I am taking ring homomorphism to mean $f(ab)=f(a)f(b)$ and $f(a+b) = f(a)+f(b)$ with $f(1)=1$.

My attempt: $R \cong \phi (R) \subset R$. So, $\phi(R)=R$.


问:
Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.

Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$

My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx = \int_0^{\pi/2} \frac{\sin^3x}{\cos^2x + \sin^3x}dx.$$
Therefore
$$2I = \int_0^{\pi/2} \left(\frac{\cos^3x}{\sin^2x + \cos^3x} + \frac{\sin^3x}{\cos^2x + \sin^3x}\right) dx.$$

I stucked here.

答:
Rewrite the integral with $t=\tan\frac x2$
$$I=\frac\pi2-\int_0^{\pi/2} \frac{\sin^2x}{\sin^2x + \cos^3x}dx
=\frac\pi2 +8 \int_0^1 \frac{t^2}{t^6-7t^4-t^2-1} dt$$
Note that $t^6-7t^4-t^2-1$ is cubic in $t^2$, with one real root $r=7.159$ (analytically solvable with the Cadano’s formula). Then, factorize
$$ t^6-7t^4-t^2-1 =(t^2-r)[t^4+(r-7)t^2+1/r]$$
and partially-fractionalize the integrand to proceed

\begin{align}
I &= \frac\pi2+ \frac8{2r^3-7r^2+1}\int_0^1 \left( \frac{r^2}{t^2-r}
+\frac{1-r^2t^2}{ t^4+(r-7)t^2+1/r}\right) dt\\
&= \frac\pi2 +\frac8{2r^3-7r^2+1} \left(- r^{3/2}\coth^{-1} \sqrt{r}
-\frac{r^2-\sqrt r}{2\sqrt{ \frac2{\sqrt r}+r-7 }}  \cot^{-1} \frac{\frac1{\sqrt r} -1}{\sqrt{ \frac2{\sqrt r}+r-7 }}\\
 + \frac{r^2+\sqrt r}{ 2\sqrt{ \frac2{\sqrt r}-r+7 }}\coth^{-1} \frac{\frac1{\sqrt r} +1}{\sqrt{ \frac2{\sqrt r}-r+7 }}\right)\\
\end{align}
where, as mentioned above
$$r= \frac13\left( 7 + \sqrt[3]{388 +12\sqrt{69}}+ \sqrt[3]{388 -12\sqrt{69}}\right)
$$


问:
Eigenvalues and Eigenvectors or Pseudoinverse Matrix
In Gilbert Strang's "Linear Algebra and Learning from Data" he asks the question

Why do $A$ and $A^+$ have the same rank? If $A$ is square, do $A$ and $A^+$ have the same eigenvectors? What are the eigenvalues of $A^+$?

I've managed to answer the first two questions:

Using the SVD one can see that the $r$ positive singular values of $A$ lead to
$r$ positive singular values for $A^+$ via inversion.

$A=\begin{bmatrix}1 & 2  \\ 1 & 2 \end{bmatrix}$ has eigenvectors $[1,1]^T$ and $[2,-1]^T$. $A^+=\begin{bmatrix} 1/10 & 1/10 \\ 1/5 & 1/5 \end{bmatrix}$ doesn't have any of these as eigenvectors.

Regarding the eigenvalues of $A^+$: I tried playing around with some particular matrices, but I couldn't see a simple relationship with the eigenvalues of $A$. I suspect there is no such relationship between the eigenvalues. Is that the case?

Thanks.

答:
I might have a partial answer. Please don't downvote, just mark the mistake if there is one.

Let v be an eigenvector of A and suppose its corresponding eigenvalue $\lambda$ is not zero. Because $\lambda$ is different from zero, we have that v is in the row space of A ($Row(A) = Null(A)^{\perp}$). Page 125 of states "$A^{+}Ax=x$ exactly when x is in the row space". Hence $A^{+}Av=v$. Thus $A^{+}Av=A^{+}\lambda v$ implies $\frac{1}{\lambda}v = A^{+}v$. Which means $\frac{1}{\lambda}$ is an eigenvalue of $A^+$. Note that $A^+v$ makes sense because we are refering to a square matrix A (in order to talk about eigenvalues).

On the other hand, if we let $\lambda = 0$ (recall v cannot equal the zero vector) we then have $A^{+}Av=A^{+}\lambda v= \lambda A^{+}v = 0$. Which implies $\lambda$ is also an eigenvalue of $A^{+}$.

Thus the eigenvalues of $A^{+}$ are the reciprocal of those of A, or zero when it is the case. However i don't know how to show that these are the only eigenvalues. I suppose it is thanks to the definition of pseudoinverse.


答:
Your answer to (a) is correct. There is one equivalence class for each possible largest digit, and since you’re looking only at positive integers, the largest digit must be non-zero.

For (b) you need to count the three-digit numbers whose largest digit is $7$. One of these is $777$. How many are there with exactly two sevens? There are $3$ places to put the digit that isn’t $7$, and it must be one of the seven digits $0,1,2,3,4,5,6$, so there are $3\cdot7=21$ such numbers. OOPS: one of those is $077$, which isn’t in the required range. Thus, there are really only $20$ of them. Now how many are there with exactly one $7$? If the $7$ is the first digit, there are $7^2=49$ ways to fill out the other two places with digits less than $7$. Otherwise the first digit must be one of the six digits $1,2,3,4,5,6$. There are then $2$ places to put the $7$ and $7$ choices for the other digit, for a total of $6\cdot2\cdot7$ possibilities. Thus, there are $49+84=133$ numbers with exactly one $7$. The grand total for (b) is therefore $1+20+133=154$.


问:
Modified weight 2 Eisenstein series is a modular form for $\Gamma_0(N)$
I'm doing exercise 1.2.8(e) in Diamond & Shurman's A First Course in Modular Forms. The problem is to show that $G_{2,N}(\tau) := G_2(\tau)-NG_2(N\tau)$ is in $M_2(\Gamma_0(N))$. To show this, I need to argue that $G_{2,N}$ satisfies $G_{2,N}[\gamma]_2=G_{2,N}$ where $f[\gamma]_2(\tau):=j(\gamma,\tau)^{-2}f(\gamma(\tau))$ is the weight-2 operator for every $\gamma\in\Gamma_0(N)$, and that $G_{2,N}$ is holomorphic on $\mathcal{H}$ and holomorphic at the cusps.

So far, I proven that $G_2(\tau)$ satisfies the transformation formula $$G_2[\gamma]_2(\tau) = G_2(\tau)-\frac{2\pi ic}{c\tau+d},\;\;\;\;\gamma = \begin{bmatrix}a&b\\c&d\end{bmatrix}$$
and that
$$\frac{\pi}{j(\gamma,\tau)^2\Im(\gamma(\tau))}=\frac{\pi}{\Im(\tau)}-\frac{2\pi ic}{c\tau+d},$$
and I've concluded that $G_2(\tau) - \frac{\pi}{\Im(\tau)}$ is weight-2 invariant under $SL_2(\mathbb Z)$.

I have no idea how to conclude that $G_{2,N}(\tau)$ is a modular form for $\Gamma_0(N)$. Specifically, how does the weight-2 operator act on $G_{2,N}$? Is it
$$G_{2,N}[\gamma]_2(\tau) = (c\tau+d)^{-2}(G_2(\gamma(\tau))-NG_2(\gamma(N\tau)))$$
or
$$G_{2,N}[\gamma]_2(\tau) = (c\tau+d)^{-2}(G_2(\gamma(\tau)) - NG_2(N\gamma(\tau))?$$
In either case, $N\gamma(\tau)$ and $\gamma(N\tau)$ are not of the form $\alpha(\tau)$ for some $\alpha\in SL_2(\mathbb Z)$, so how should $G_2(\tau)$ transform?

Moreover, I don't know how to verify that $G_{2,N}$ is holomorphic on $\mathcal{H}$ or at the cusps. I would greatly appreciate detailed answers to both of these.

P.S. This is not homework, I'm simply trying to better understand computational aspects of modular forms.

答:
This is more of an expansion of my comment to the question.

Let $q\in(0,1)$ be the nome corresponding to elliptic modulus $k\in(0,1)$ (in the symbolism of modular forms $q$ takes backstage and is replaced by $\tau$ where $q=\exp(2\pi i\tau)$ and $\tau$ has positive imaginary part).

Let us define elliptic integrals $K, E$ as $$K=K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{1}$$ and $$E=E(k) =\int_0^{\pi /2}\sqrt{1-k^2\sin^2x}\,dx\tag{2}$$ The nome $q$ is related to $k$ via  $$q=\exp\left(-\pi\frac{K(k')} {K(k)} \right), k'=\sqrt{1-k^2}\tag{3}$$ Ramanujan used the function $P(q) $ instead of your $G_2$ defined by $$P(q) =1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{4}$$ The following link of $P(q) $ and elliptic integrals defined above was well known before Ramanujan: $$P(q^2)=\left(\frac{2K}{\pi}\right)^2\left(\frac {3E}{K}+k^2-2\right) \tag{5}$$ For Ramanujan modular forms were roughly those functions $f(q) $ which could be expressed like $(2K/\pi)^w A(k) $ where $w$ is some positive integer or half of a positive integer and $A(k) $ is an algebraic function of $k$ ($w$ being weight of $f(q) $). The exact definition of modular forms which Ramanujan provided could not however be decoded by Berndt and his collaborators.

The term $3E/K$ on right of $(5)$ is what prevents $P(q^2)$ from being a modular form. Fortunately for us Ramanujan found out a way to fix this. He instead considered the function $NP(q^{2N})-P(q^2)$ ($N$ being a positive integer) and proved that it can be expressed in the form desired.

Let us start with another well known function called Dedekind eta function defined by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)\tag{6}$$ Eta function is related to $P(q) $ via $$P(q) =24q\frac{d}{dq}{\log \eta(q)} \tag{7}$$ The following identity $$ \eta(q^2)=2^{-1/3}\sqrt {\frac{2K}{\pi}}(kk')^{1/6}\tag{8}$$ is well known and shows that $\eta(q^2)$ is a modular form of weight $1/2$.

Using $(1),(2),(3)$ it can be proved that $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{9}$$ and then $(5)$ can be established using $(7),(8),(9)$.

Let $l$ be the elliptic modulus corresponding to nome $q^N$ (which corresponds to $N\tau$). The elliptic integrals $K(l), K(l') =K(\sqrt {1-l^2})$  are usually denoted by $L, L'$ so that $q^N=\exp(-\pi L'/L) $. From $(8)$ we get $$\eta(q^{2N})=2^{-1/3}\sqrt{\frac{2L}{\pi}}(ll')^{1/6}\tag{10}$$ From $(7)$ we get $$P(q^2)=12q\frac{d}{dq}\{\log\eta(q^2)\},NP(q^{2N})=12q\frac{d}{dq}\{\log\eta(q^{2N})\}$$ and thus $$F(q) =NP(q^{2N})-P(q^2)=12q\frac{d}{dq}\left(\log\frac{\eta(q^{2N})}{\eta(q^2)}\right)$$ Using $(8),(9),(10)$ we thus get $$F(q) = \left(\frac{2K}{\pi}\right)^2kk'^2\frac{d}{dk}\left(3\log\frac{L}{K}+\log\frac{ll'}{kk'}\right)\tag{11}$$ From the standard theory of modular equations it is known that $l,l'$ as well as $K/L$ are algebraic functions of $k$ and hence from the equation $(11)$ it follows that $F(q) $ is of the form $(2K/\pi)^2A(k)$ where $A(k) $ is an algebraic function of $k$. Thus $F(q)$ is a modular form of weight $2$.

Ramanujan calculated the function $A(k) $ in explicit form for many values of $N$. For small values of $N$ the values given by Ramanujan can be verified with some effort. But no one knows how Ramanujan evaluated them for large $N$.


问:
Function of modulus value
If $f(x)=|x+2|$ and $g(x)=2|x|$, then it is true that $\frac{f(x)+g(x)-|f(x)-g(x)|}{2}=Min(f(x),g(x))$, How do we prove it.

答:
This expression is always true, you can check that :

if $f\leq g$, then :
$$ \frac{f+g-|f-g|}{2}=\frac{f+g-(g-f)}{2}=f$$
Then, if $g\leq$ f :
$$\frac{f+g-|f-g|}{2}=\frac{f+g-(f-g)}{2}=g $$

So you get that :
$$\frac{f+g-|f-g|}{2}=\min(f,g) $$

Title: Function of modulus value


问:
Converges of composition of two complex functions
if $g$ is a continuous function defined on a bounded domain, suppose $f_{n}$ be a sequence of functions defined on the same domain and uniformly convergent to f then is it possible to say that $g(f_{n})$ will be uniformly convergent.

答:
If $g$ is uniformly continuous, it will certainly be true. I doubt you can generally conclude uniform convergence unless you have that assumption on $g$, but I am not completely sure. Here is a proof for the case where $g$ is uniformly continuous:

Let $\epsilon > 0$ and choose $\eta > 0$ such that for all $x, y$

$$|x-y| < \eta \implies |g(x)-g(y)| < \epsilon.$$

Now, since $f_n$ converges uniformly, we may choose $N$ such that for all $n \geq N$

$$|f_n(x)-f(x)|< \eta $$

for all $x$. But then for all $n \geq N$ we have

$$|g(f_n(x))-g(f(x))| < \epsilon,$$

for all $x$, which shows that $g \circ f_n$ converges uniformly to $g \circ f$.

I have omitted the domains of these functions since the question is unclear as to this, but I trust the reader will be able to adjust the argument to fit the domains in question.


问:
Complex integration and Residue theorem. Where did I mistake the calculation?
I have to calculate $\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} (a>0)$ using complex integral.

Let $f(z)=\dfrac{z^{\frac{1}{2}}}{z^2+a^2}$.

Integral Contour ($r<R$);

$C_1 : [r,R], z=t,t:r\to R$

$C_2$ : Circle that has radius $R, z=Re^{i\theta}, \theta : 0\to 2\pi$

$C_3$ : $[R,r], z=te^{2\pi i}, t: R\to r$

$C_4$ : Circle that has raius $r, z= re^{i\theta}, \theta : 2\pi \to 0$

\begin{equation}
\displaystyle\int_{C_2} f(z) dz, \displaystyle\int_{C_4} f(z) dz \to 0\,\,({r\to 0, R\to \infty} )
\end{equation}

\begin{align}
&\displaystyle\int_{C_1} f(z) dz \to \displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx.\\
&\displaystyle\int_{C_3} f(z) dz \to -e^{i\pi}\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx
=\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx.
\end{align}

From Residue Theorem,
\begin{align}
2\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx
=2\pi i[\text{Res}(f, ai)+\text{Res}(f, -ai)]
\end{align}

\begin{align}
&\text{Res}(f, ai)
=\dfrac{(ai)^{\frac{1}{2}}}{2a i}
=\dfrac{e^{\frac{\pi i}{4}}}{2\sqrt{a}i}\\
&\text{Res}(f, -ai)
=\dfrac{(-ai)^{\frac{1}{2}}}{-2a i}
=-\dfrac{e^{-\frac{\pi i}{4}}}{2\sqrt{a}i}\\
\end{align}

Therefore,
\begin{align}
\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx
=\pi i \left[\dfrac{e^{\frac{\pi i}{4}}}{2\sqrt{a}i}-\dfrac{e^{-\frac{\pi i}{4}}}{2\sqrt{a}i}\right]
=\dfrac{\pi }{\sqrt{2a}}i.
\end{align}
So $\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx=0.$

But answer is
$\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx=\dfrac{\pi}{\sqrt{2a}}$. Maybe, my calculation is wrong, but I cannot find where I mistook.

答:
Because you have cut along the positive real axis, the second residue is $-e^{3 \pi i /4}/(2 \sqrt ai)$ leading to,
\begin{align}
2 \int_0^\infty \frac{\sqrt{x}}{x^2 + a^2} dx
&= \frac{2\pi i}{2\sqrt a i}
\Big( e^{\pi i/4} - e^{3 \pi i / 4} \Big) \\
&=\frac{\pi}{\sqrt{a}} \cdot \frac{2}{\sqrt 2}
\end{align}


问:
Lemma for a question based on Greatest Integer function
I was trying to solve the following question.
For a positive real number x, prove that
$$[x]+\frac{[2x]}{2}+\frac{[3x]}{3}+\cdots+\frac{[nx]}{n}\leq [nx]$$where [] represent the Greatest Integer Function.

Here is my approach. I used Hermite's Identity to reduce $[nx]$ to $[x]+[x+\frac{1}{n}]+[x+\frac{2}{n}]+\cdots+[x+\frac{n-1}{n}]$ and came across the following lemma.

Lemma: For positive real x and $k=1,2,\cdots,n$, prove that
$$\left[x+\frac{k-1}{n}\right]\geq \frac{[kx]}{k}$$which solves the question directly

Could anyone please verify this equation. If true, please provide a proof for it as well. I tried using induction but fell short

Thanks a lot for the help!!:)

答:
It is not true, consider for example $x=\frac{1}{2}$, $n=3$ and $k=2$. Then $$\frac{[kx]}{k}=\frac{1}{2}$$
but$$\left[x+\frac{k-1}{n}\right]=\left[\frac{5}{6}\right]=0$$


问:
How to show a subspace is closed in $\ell^2$
Let $F$ be the real sequence which satisfies $ \sup _{n} {|nx_n|}<\infty$ in $\ell^2$. Show that

$F$ are linear subspace of $\ell^2$.

$F$ is closed in $\ell^2$.

My idea:

is trivial.

As is known to all, $\ell^2$ is a Banach space. The subspace of a Banach space is closed  if and only if it is complete (Is this right?).

So we only to prove $F$ is complete subspace. But how can I show that every Cauchy sequence converge in $F$? Or we need to show that the closure of $F$ is itself? Could someone give me some details. Thank you!

答:
The set $F$ is not closed. To see it consider a sequence.

$$x^k =\left(\frac{1}{1} , \frac{1}{2^{\frac{3}{4} }}, \frac{1}{3^{\frac{3}{4} }} , ...,\frac{1}{k^{\frac{3}{4} }} , 0,0,0,....\right)$$

Clearly $$\sup_n |nx^k_n |<\infty$$ for all $k\in\mathbb{N} .$
Hence $x^k\in F.$ Moreover $x^k \to x$ in $\ell^2 $ where $$x= (n^{-\frac{3}{4}})_{n\in\mathbb{N}}\in\ell^2.$$
But $x\notin F.$


问:
Expressing logical conjunction with logical imlication and equivalence.
I've come across an exercise which requires you to express every one of the logical operators $\{\wedge, \Rightarrow, \Leftrightarrow\}$ with the other two - for example,
$$p \Leftrightarrow q \sim (p \Rightarrow q) \wedge (q \Rightarrow p).$$
So far I've only been able to do this with $\Leftrightarrow$, which is straightforward enough, but now I'm stuck.
I don't think I'm allowed to use constants such as '0' or '1'. Is this even possible? Any help (tips or a solution) would be greatly appreciated.

Edit: ok I found the solution for $\wedge$:

$$p \wedge q \sim (p \Leftrightarrow q)\Leftrightarrow ((p \Leftrightarrow q)\Rightarrow p)$$

答:
I think that works:

$$p\Rightarrow q \sim p\wedge q \Leftrightarrow p\\
p\wedge q\sim (p\Rightarrow q)\Leftrightarrow (q\Rightarrow p)$$

(You can check truth tables)


问:
Is separability + connectedness in metric spaces sufficient to guarantee path connectedness? Are these necessary conditions?
Saw someone make this claim, but I'm not certain this is true. Can't think of counterexamples in either direction. Any help would be appreciated.</p>

答: The Topologist's sine curve is a classical counterexample.</p>


问:
Using trigonometric power formulas to derive an identity for $\cos^3(x)$
I am practicing with manipulating sigma notation and binomial coefficients right now. I am using the formula given here to derive the identity for $\cos^3(x)$

The identity for $\cos^3(x)$ is

$$\cos^3(x)=\dfrac{1}{4}[3\cos(x)+\cos(3x)]$$

Now, the formula for deriving the identity of $\cos^{2n+1}(x)$ is

$$\cos^{2n+1}(x)=\dfrac{1}{4^{n}}\sum_{k=0}^{n}\binom{2n+1}{k}\cos[(2n+1-2k)x]$$

Let $n=1$

I have

$$\cos^{3}(x)=\dfrac{1}{4}\sum_{k=0}^{1}\binom{3}{0}\cos(3x)$$

$$=\cos^{3}(x)=\dfrac{1}{4}\sum_{k=0}^{1}2\cos(3x)$$

I know that a summation notation denotes:

$$\sum_{k=0}^{n}x_k=k_0+k_1+k_2+...k_n$$

But $k$ varnishes because I set it as $k=0$, so I think I am wrong there.

I think my calculation is clearly incorrect. Could you show me full works of how to arriving at the final identity using the power reduction formula?

答:
$$\cos^{2n+1}(x)=\frac{1}{4^{n}}\sum_{k=0}^{n}\binom{2n+1}{k}\cos[(2n+1-2k)x]$$
if I plug $n=1$ becomes
$$\cos^3 x=\frac{1}{4}\sum_{k=0}^1\binom{3}{k}\cos(3-2k)x$$
which is
$$\cos^3 x=\frac14\left(\cos 3x+3\cos x\right)$$

edit

I think I know what confuses you
$$\sum_{k=0}^1\binom{3}{k}\cos(3-2k)x$$
means that you sum the quantity when $k=0$ plus when $k=1$ like the following
$$\binom{3}{0}\cos(3-2\cdot 0)x+\binom{3}{1}\cos(3-2\cdot 1)x$$


答:
Complementing mathcounterexamples.net's answer, it's not a necessary condition either. Consider an "uncountable starfish" - take uncountably many copies of the unit interval, and glue them together at their $0$-endpoints. This is path-connected but not separable.

A bit more rigorously: take an uncountable set $A$, and consider the set $X=(A\times(0,1])\sqcup\{*\}$. We put a metric $d$ on $X$ as follows:

For each $x\in X$ we set $d(x,x)=0$.

For each $a\in A, i\in (0,1]$ we set $d(*,\langle a,i\rangle)=d(\langle a,i\rangle, *)=i.$

For each $a\in A$ and each distinct $i,j\in (0,1]$ we set $d(\langle a,i\rangle , \langle a,j\rangle)=\vert i=j\vert$.

Finally, for each distinct $a,b\in A$ and each $i,j\in (0,1]$ we set $d(\langle a,i\rangle ,\langle b,j\rangle)=i+j.$

It's a good exercise to check that this $d$ is in fact a metric on $X$, and that the resulting topology is path-connected. But it's clearly not separable: given a countable set of points $C\subseteq X$, since $A$ is uncountable we can find some $a\in A$ which is not the left coordinate of any point in $C$, and then $\langle a,1\rangle$ is at distance at least one from any point in $C$.

Title: Is separability + connectedness in metric spaces sufficient to guarantee path connectedness? Are these necessary conditions?

Body: Saw someone make this claim, but I'm not certain this is true. Can't think of counterexamples in either direction. Any help would be appreciated.</p>


问:
Volumes of submanifolds with respect to two different riemannian metrics
Suppose we have a compact manifold $M$ and two riemannian metrics $g_1,g_2$. For each riemannian metric we can define the riemannian measure and for each compact submanifold $Y\subset X$ we have the induced metrics on $Y$ and we can compute the volume of $Y$ with respect to this metric. Now I know that for each $x\in M$ and $v\in T_x M$ we have that there exist constants $a,b>0$ such that $ag_x^1(v,v)\leq g_x^2(v,v)\leq bg_x^1(v,v)$ , using the fact that $SM$ is compact since $M$ is compact. Now I would like to see that for the volume of a submanifold with respect to each metric we get a similar inequality, i.e, that $aVol_1(Y)\leq Vol_2(Y)\leq bVol_1(Y) $, and for that I belive it suffices to get such an equality for $\sqrt{G^{\alpha}}$ , where this denotes the determinant of the matrix $g_{ij}^{\alpha}=g(\frac{\partial}{\partial x_i^{\alpha}},\frac{\partial}{\partial x_j^{\alpha}})$, for each one of the metrics. Does anyone know if this is possible and if so if how can one prove it ?

(My interest in this comes from the Yomdin's Theorem for the topological entropy of the geodesic flow)

答:
I am afraid that, while things can be reduced to smooth functions, which allows you to use a standard compactness argument, a substantial amount of technology is needed to set things up propertly. If you don't know the tools, it will be rather hard to get through this, I hope the description is still helpful.

Let me recall that the square root of the determinant of the Gram matrix of a Riemannian metric constitutes the local coordinate expression for a well defined geometric object called a density. (These are the objects that can be integrated without choosing an orientation.). Technicall speaking, such densities are (smooth) sections of a line bundle, and a crucial feature of the volume density of a Riemannian metric is that it is nowhere zero. (This is a condition that makes sense for sections of a line bunndle.) Now it is a general fact that given a nowhere vanishing section of $\sigma$ of a line bundle, any other section of that bundle can uniquely be written as $f\sigma$ for a smooth function $f$. If the second section is nowhere vanishing, too, then $f$ is nowhere vanishing. On a compact manifold, you can then conclude that $f$ has a maximum $b$ and a minimum $a$, which have to have the same sign and for volume densities you conclude that they are both positive since both densities lead to positive integrals. But then of course $a\sigma\leq f\sigma\leq b\sigma$ and monotonicity of the integral implies that volume of your manifold with respect to $f\sigma$ lies between $aV$ and $bV$, where $V$ is the volume with respect to $\sigma$. This explains how to apply compactness arguments to $\sqrt{\det(g_{ij})}$.

To obtain what you are looking for, there is the additional problem that you restrict the metric to submanifolds of smaller dimensions, and it is not easy to see what this does to the volume density. (This was the point-wise problem I mentioned in my comment.) I think the way to proceed is as follows: There is a fiber bundle $Gr(k,TM)\to M$ for which the fiber over $x$ is the Grassmannian of $k$-dimensional subspaces of $T_xM$. (This can be easily constructed as an associated bundle to the frame bundle of $M$.) Since the Grassmannian is compact, also the total space of $Gr(k,TM)$ is compact. You an also construct a line bundle $L\to Gr(k,TM)$ (again as an associated bundle) such that smooth sections of $L$ correspond to a (smooth) family of volume densities on a each $k$ dimensional subspace in each tangent space. Now given a Riemannian metric on $M$, restricting $g_x$ to each $k$-dimensional subspace of $T_xM$ and taking the square root of the determinant of a Gram matrix defines a nowhere vanishing section of $L\to Gr(k,TM)$ which is smooth by construction. Doing this for two metrics, you get a smooth function $f$ relating the two nowhere vanishing sections and by compactness you get a maximum and a minimum.

Now given a $k$-dimensional submanifold $Y\subset M$, you get a canonical embedding $Y\to Gr(k,TM)$ by sending each point $y\in Y$ to the subspace $T_yY\subset T_yM$. The volume density of the restriction of a Riemannian metric to $Y$ can then be obtained as a pullback along this embedding of the section of $L\to Gr(k,TM)$ along this embedding. Doing this for two metrics the sections are thus related by the restriction of $f$ to the image of the emedding, so the global minimum and maximum of $f$ provide bounds for the volume density of $Y$. Integrating, you then get the required bounds on the volumes.


问:
Is the least-squares solution unique?
I am looking for a line closest to $(-5, -2)$, $(-2, 0)$, $(-1, 0)$, $(2, 3)$, $(5, 4)$ using the least square solution. So I set the line as $$ax+by+c=0$$ let $a=1$ (where $a$ is not $0$ obviously) and got

$$\begin{pmatrix}
-2 & 1 \\
0 & 1 \\
0 & 1 \\
3 & 1 \\
4 & 1\\
\end{pmatrix}
\begin{pmatrix}
b \\
c \\
\end{pmatrix}=
\begin{pmatrix}
5 \\
2 \\
1 \\
-2 \\
-5\\
\end{pmatrix}
$$

Then I solved it by multiplying $A^T$on the both sides.

But the $(b, c)$ I got here was different from that of the usual solution using $$y=ax+b$$ Although the two lines are almost identical (which implies I am not that wrong), they are still different. What's the matter?

答:
Imagine you have two data points $(2,4)$ and $(2,6)$. Now, if we put the slope formula into matrix form, we will have
\begin{pmatrix}2 & 1 \\2 & 1 \\\end{pmatrix}
as our $X$ matrix. First column is for the input $X$ and second for the slope "$b$". We set "$b$" as one because we want to give it freedom to move (otherwise, $0$ would kill the "$b$" out of the formula, $mx = y$). Then, we have the vector $(M ,B)$ for our unknown coefficients. These two matrices multiplied will give the output vector $(4 ,6)$.

Now, try to think of how you would graph through these two data points. The problem is that there are infinite solutions depending on what you set the bias "$b$" to be equal to. Why do we have infinite least square solutions? Well, The column space is not full rank and we have a dependency!

This is the most intuitive way to think in my opinion. Once you understand in 2d, you just "rely" it will hold on to $n$ dimensions (and there is no reason to not hold).

Title: Is the least-squares solution unique?


问:
Prove! that $a+(1/a) ≥ 2$ and $a+(1/a) ≤2$
Let $a \in R$

If $a>0$, then $a+\frac1a\geq2$

If $a<0$, then $a+\frac1a\leq2$

This is how someone explained the first one to me but still not really sure about it.

Proof:

$\Longleftrightarrow$$a+\frac1a\geq2$
$\Longleftrightarrow$ the square of any real number is non-negative
  so we have $(a-1)^2\geq0$ (don't understand this part)
$\Longleftrightarrow$ $a^2-2a+1\geq0$
$\Longleftrightarrow$ $a^2+1\geq2a$
$\Longleftrightarrow$ since $a>0$ then so is $
  a+\frac1a≥2$ if $a>0$

答:
Since the above answer already gave you the proof for first one, I'll discuss the second inequality
$$a+\frac1a\leq2$$
This is a wrong inequality, you'll understand that in a moment.
Let's start with the first inequality user61527 used: $$(a - 1)^2 \ge 0$$ This false for $a \lt 0$ because $(a - 1)^2$ can never be $0$ for any negative value. It will be zero only at $a=1$ which is out of domain for $a$ as $a\lt 0$.

So we'll use the following inequality $$(a + 1)^2\ge 0$$ The equality will hold at $a=-1$, now this fits in fomain of $a$
$$a^2 + 2a + 1 \ge 0$$
Dividing by $a$ will alter the inequality since it is a negative number
$$a + 2 + \frac {1}{a}\le 0$$
So the actual inequation for $a\lt 0$ should be $$a + \frac {1}{a}\le -2$$


问:
Prove that a sequence $\{a_n\}_n$ defined by $a_1=-\frac14$ and $-a_{n+1}=\frac{a_na_{n+1}+4}4$ is convergent and find its limit.
I would like to verify my attempt and deduction. The task is as follows:

Prove that a sequence $\{a_n\}_n$ defined by $a_1=-\frac14$ and $$-a_{n+1}=\frac{a_na_{n+1}+4}4$$ is convergent and find its limit.

This is what I have so far:

$$-a_{n+1}=\frac{a_na_{n+1}+4}4\iff a_{n+1}(a_n+4)+4=0\tag 1$$

I computed a few terms:

$a_2(a_1+4)=-4\implies a_2=-\frac{16}{15}\\a_3(a_2+4)=-4\implies a_3=-\frac{15}{11}\\a_4(a_3+4)=-4\implies a_4=-\frac{44}{29}$

I assumed $a_n<0\quad\forall n\in\Bbb N$.

Then, from $(1)$ and $a_{n+1}<0$, it follows

$\begin{aligned}a_{n+1}(a_n+4)&=-4\\\implies a_n+4&>0\\\implies a_n&>-4\end{aligned}$

Then, inductively, if $\,0>a_1>\ldots>a_{m-1}>a_m$ for some $m\in\Bbb N,$ we have $\begin{aligned}a_{m-1}+4&>a_m+4>0\\\implies \frac1{a_{m-1}+4}&<\frac1{a_m+4}\\\implies \color{red}{a_m}=-\frac4{a_{m-1}+4}&>-\frac4{a_m+4}=\color{red}{a_{m+1}}\end{aligned}$

So, the sequence $\{a_n\}_n$ is monotonic and bounded and, therefore, convergent.

Also, we can prove a stronger statement:

$a_n>-2\quad\forall n\in\Bbb N$.

$$\begin{aligned}a_n+4&>-2+4=2>0\\\implies -\frac1{a_n+4}&>-\frac12\\\implies a_{n+1}=-\frac4{a_n+4}&>-\frac42=-2\end{aligned}$$

Plugging the limit into $(1)$, we get
$$L^2+4L+4=(L+2)^2=0\iff L=-2$$

Hence, $\lim\limits_{n\to\infty}a_n=-2$.

Is there any mistake in my assumptions and conclusions and should I do any steps in a different order?

I know I couldn't prove $a_n<0\quad\forall n$ by induction since the function $f:\Bbb R\setminus\{-4\}\to\Bbb R\setminus\{0\}$ defined by $$f(x)=-\frac4{x+4}$$
isn't monotonic on the whole domain, just on $(-\infty,-4)$ and $(-4,+\infty)$ separately.

Also, when I considered writing $a_n=\frac{x_n}{y_n}$ and then
$$\begin{aligned}a_{n+1}&=\frac{x_{n+1}}{y_{n+1}}\\&=-\frac4{\frac{x_n}{y_n}+4}\\&=\frac{-4y_n}{x_n+4y_n}\end{aligned}$$
and assuming $x_{n+1}=-4y_n$ and $y_{n+1}=x_n+4y_n$, I obtained the homogenous recurrence
$$\begin{aligned}y_{n+1}&=-4y_{n-1}+4y_n\\\iff y_{n+1}-4y_n+4y_{n-1}&=0\end{aligned}$$ with a characteristic polynomial
$$\lambda^2-4\lambda+4=(\lambda-2)^2$$ with a multiple root, so I thought I would overcomplicate.

Thank you very much!

Body:
$$-4A_{n+1}=A_n A_{n+1}+4$$

Let $A_n=\frac{B_{n-1}}{B_n}$
$$-4\frac{B_n}{B_{n+1}}=\frac{B_{n-1}}{B_n} \frac{B_n}{B_{n+1}}+4$$
$$\implies 4B_{n+1}+4 B_n+B_{n-1}=0$$
Let $B_n=t^n$, then
$$\implies 4t+4+t^{-1}\implies t=-1/2.$$
Then $B_n=(Pn+Q)(-2)^{-n}$, $$ A_n=-2\frac{P(n-1)+Q}{Pn+Q}=-2\frac{n-1+R}{n+R}$$
$$A_1=-1/4 \implies R=1/7.$$
Finally, we have the solution to $(1)$ as
$$A_n=\frac{12-14n}{7n+1} \implies \lim_{n \to \infty}A_n=-2.$$