Title: Estimate a value with square roots squared without calculating

Body:
If $n$ is a natural number and $n<(6+\sqrt{29})^2<n+1$, find the value of $n$.

This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $(6+\sqrt{29})^2$. Is there a way to do that?

Body:
Without a calculator,

$$(6+\sqrt{29})^2=36+12\sqrt{29}+29=65+12\sqrt{29}=125+12(\sqrt{29}-5)=125+{48\over\sqrt{29}+5}$$

Now $5\lt\sqrt{29}\lt6$ implies

$$4\lt{48\over11}\lt{48\over\sqrt{29}+5}\lt{48\over10}\lt5$$

Thus

$$129\lt(6+\sqrt{29})^2\lt130$$


问:
Prove / disprove question (just before Calculus exam!)
So I am going to take the final exam for Calculus I.

First, I'd love if some of you have some good questions for me to solve, or some good tips and tricks that I should notice.

Second, I have a prove/disprove question that I am trying to solve, and will appreciate your help!

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function and let $L \in \mathbb{R}$.

If $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=L$, than $\lim_{x \to \infty} \frac{f(x)}{x}=\frac{L}{2}$.

So I have a hint to define a function $g(x)=x f(x)$.

So $g'(x)=f(x)+xf'(x)$.

Also, $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{f(x)}{x}+f'(x)=L$

I wanted to use L'Hopital's rule for $\lim_{x \to \infty} \frac{f(x)}{x}$, but then I have noticed that I do not know if $\lim_{x \to \infty} \frac{f(x)}{x}$ even exists.

Any hints will be appreiciated!

Thanks a lot!

答: Given that $$\lim \frac{g'(x)}{x}=L$$ , now assuming $f(x)\neq0$ then as $x\to\infty$ then $g(x)=xf(x)$ goes to infinity(or negative infinity) thus we can apply L'Hopital's rule , $$\lim \frac{g(x)}{\frac{x^2}2}=L\\\implies\lim \frac{2xf(x)}{x^2}=\lim \frac{2f(x)}{x}=L$$. Now if $f=0$ then the result is obvious.

ContentLicense: CC BY-SA 4.0


答:
You can apply L'Hopital's rule backwards and integrate, given that the limit $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{g'(x)}{x}=L$ exists:

$$L=\lim_{x \to \infty}\frac{\int g'(x)dx}{\int xdx}=\lim_{x \to \infty}\frac{g(x)}{x^2/2}=\lim_{x \to \infty}\frac{2xf(x)}{x^2}=\lim_{x \to \infty}\frac{2f(x)}{x}$$

Then, $$\lim_{x \to \infty}\frac{f(x)}{x}=\frac{L}{2}$$

Title: Prove / disprove question (just before Calculus exam!)

Body:
So I am going to take the final exam for Calculus I.

First, I'd love if some of you have some good questions for me to solve, or some good tips and tricks that I should notice.

Second, I have a prove/disprove question that I am trying to solve, and will appreciate your help!

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function and let $L \in \mathbb{R}$.

If $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=L$, than $\lim_{x \to \infty} \frac{f(x)}{x}=\frac{L}{2}$.

So I have a hint to define a function $g(x)=x f(x)$.

So $g'(x)=f(x)+xf'(x)$.

Also, $\lim_{x \to \infty} \frac{f(x)+xf'(x)}{x}=\lim_{x \to \infty} \frac{f(x)}{x}+f'(x)=L$

I wanted to use L'Hopital's rule for $\lim_{x \to \infty} \frac{f(x)}{x}$, but then I have noticed that I do not know if $\lim_{x \to \infty} \frac{f(x)}{x}$ even exists.

Any hints will be appreiciated!

Thanks a lot!


问:
Taylor series of $e^z$ at point $z_0 = 1+i$
I solved this over a year ago but fail to do so now. I remember that for problems like this one can use $e^z = e^{z_0}e^{z-(z_0)}$.
I can see the answer right away if I had $z_0 = \pi i$. But in my case, with $z_0 = 1+i$, I get stuck or at least I am feeling unsure if this is the best way to write the answer.

$$ \begin{align} e^z & = e^{1+i}e^{z-(1+i)} \\
 & = \sum_{k=0}^{\infty} \frac{e^{1+i}}{k!} (z-(1+i))^k
\end {align}$$

This should be what I was looking for (solved it now that I wrote it down carefully here).
Since I already wrote this, any remarks?

答:
No trick needed. From the derivative of the exponential, the Taylor formula says

$$e^z=\sum_{n=0}^\infty e^{1+i}\frac{(z-(1+i))^n}{n!}.$$


问:
Limit as $x$ goes to the right and at the left of a point of not definition
The function $f(x)=\log(x^4-2x^2+1)+ \frac{-1}{2(x-1)}$ at the right of -1 goes to $-\infty$ and also at the left. But if I plot $f$ in geogebra it does not seem that that $f$ goes to $-\infty$ at the right and the left of -1. Why?

答:
The argument to the logarithm is positive for $x\ne1$ and, since it goes to $0$ for $x\to-1$, you have
$$
\lim_{x\to-1}\log(x^4-2x^2+1)=-\infty
$$
(two-sided limit).

On the other hand
$$
\lim_{x\to-1}\frac{-1}{2(x-1)}=\frac{1}{4}
$$
Thus you can safely state that
$$
\lim_{x\to-1}f(x)=-\infty
$$

Note that, instead,
$$
\lim_{x\to1^-}f(x)=\infty,\qquad \lim_{x\to1^+}f(x)=-\infty
$$


问:
Estimate a value with square roots squared without calculating
If $n$ is a natural number and $n<(6+\sqrt{29})^2<n+1$, find the value of $n$.

This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $(6+\sqrt{29})^2$. Is there a way to do that?

Body:
$5<\sqrt{29}<6$, so $0<6-\sqrt{29}<1$, so $0<(6-\sqrt{29})^2<1$.

$(6+\sqrt{29})^2<(6+\sqrt{29})^2+(6-\sqrt{29})^2<(6+\sqrt{29})^2+1$.

But $(6+\sqrt{29})^2+(6-\sqrt{29})^2=36+12\sqrt{29}+29+36-12\sqrt{29}+29=130$.

So $(6+\sqrt{29})^2<130<(6+\sqrt{29})^2+1$.

So $129<(6+\sqrt{29})^2<130$.


问:
Minimal polynomials of linear transformations classed as scalar polynomials?
I'm currently trying to cover minimal polynomials of linear transformations and their properties, but have become quite confused. I'm all over the basics of matrices, vector spaces, linear transformations and eigenvalues/vectors. The definition given in the notes I'm following is:

"Definition: Let $T:V\rightarrow V$ be a linear transformation from the $\mathbb{F}$-vector space V to itself. Let $m_T$ be the monic polynomial of least degree in the set of polynomials
$\{0 \neq g(x)\in\mathbb{F}[x]|g(T)=0\}$. Call $m_T$ the minimal polynomial of T."

What confuses me is the fact that $g(x)\in\mathbb{F}[x]$, however $T\notin\mathbb{F}$; so why are we talking about g(T)? Shouldn't g be acting over scalars, not matrices?

I was ready to just accept this and then move on, but then later I am encountering situations where multiplication of functions taking T is considered commutative. For instance, an excerpt from the notes:

"The polynomial $f(x)= f_1(x)f_2(x)$ satisfies

$f(T)(v_1+v_2) = f(T)(v_1) + f(T)(v_2)$

$= f_2(T)f_1(T)(v_1)  + f_1(T)f_2(T)(v_2)$"

T is in essence a matrix here, so why is multiplication of functions of T considered commutative? This seems especially blasphemous to me. It seems it all comes back to the classification of $f(x)\in\mathbb{F}[x]$, which appears to be the root of the issue for me.

Any guidance would be greatly appreciated. Thank you!

答: It's all about notation. Given a polynomial $g(x)=\sum_{i=0}^{n}a_ix^i\in\mathbb{F}[x]$, when you write $g(\lambda)$ where $\lambda\in\mathbb{F}$ you're referring to $\sum_{i=0}^{n}a_i\lambda^i\in\mathbb{F}$, which is a scalar because it's the sum and product of scalars. However, given a $\mathbb{F}$-linear map $T:V\rightarrow V$, $g(T)$ denotes a different object: the $\mathbb{F}$-linear map $\sum_{i=0}^{n}a_iT^i$, where $T^i=T\circ\overset{i)}{\dots}\circ T$ is the composition of functions.


问:
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$
How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?

It is part of a question about finding limits at mooculus.

ContentLicense: CC BY-SA 3.0

Body: Hints: $1296=(-6)^4$ and $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})$.


问:
Summation including Stirling numbers of the second kind
I would like to show the following: (I am pretty sure it is correct but was not able to prove it.)

$\begin{eqnarray*}
&&\sum_{j=1}^{n}j\binom{n}{j}S\left( n,j\right) j! =n(n^n-(n-1)^n) \\
&&
\end{eqnarray*}$

where $S(n,j)$ stands for the Stirling numbers of the second kind.

答:
First of all let's change it like the following:

Divide both sides by $n$ and simplify the $\frac{j}{n}$ in the sum in the LHS with $\binom{n}{j}$: $$\frac{j}{n}\cdot\binom{n}{j}=\frac{j\cdot n!}{n\cdot j!\cdot(n-j)!}=\frac{(n-1)!}{(j-1)!\cdot(n-j)!}=\binom{j-1}{n-1}$$

So actually we want to prove the following: $$\displaystyle\sum_{j=1}^{n}{\binom{n-1}{j-1}\cdot j!\cdot S_{(n,j)}}=n^n-(n-1)^n$$

Now assume that $A={\{1,2,\dotsb,n}\}$. Then the RHS is equal to the number of all $n$-tuples of elements of $A$ that contain $1$, since $n^n$ is the number of all $n$-tuples of $A$ and $(n-1)^n$ is the number of all $n$-tuples of $A$ without $1$.

Now we'll prove that the LHS has the same value.

We'll prove that the number of all $n$-tuples of $A$ that have exactly $j$ different elements and have $1$ in them is equal to $\binom{n-1}{j-1}\cdot j!\cdot S_{(n,j)}$ :

Step1: First of all we'll chose the elements we want to use in our $n$-tuple. Including $1$ is one of the conditions so it's one of the elements we choose. Then we want $j-1$ other elements and our choices are $\{2,3,\dotsb,n\}$ so there are $\binom{n-1}{j-1}$ different ways for doing that.

Step2: Now we choose where to position the $j$ elements we picked in step1. For this we have to assign a set of positions of our $n$-tuple to our elements (and these sets shall be non-empty). For this there are exactly $S_{(n,j)}$ different ways. For each partition of $\{1,2,\dotsb,n\}$ (which are our positions in the $n$-tuple) to $j$ non-empty sets, if a position falls into the partition number $i$, then label this position with $i$. At last we have to fill our partitions with our pre-chosen elements! Since there are $j$ different labels of the positions of our $n$-tuple and we have $j$ elements, there are exactly $j!$ different ways to choose which element shall fill all of the positions labeled with each number.

So at last, there are $\binom{n-1}{j-1}$ methods for choosing the elements of our $n$-tuple and $j!\cdot S_{(n,j)}$ methods to place them in our n-tuple; so there are exactly $\binom{n-1}{j-1}\cdot j!\cdot S_{(n,j)}$ pairwise different $n$-tuples that have exactly $j$ different elements and also have the element $\{1\}$. So summing this for different $j$'s, we'll have the number of all different $n$-tuples that have the element $\{1\}$ and hence the proof is complete!


答:
Use the difference of two squares to get:

$$6+\sqrt{29} = \frac{(6+\sqrt{29})(6-\sqrt{29})}{6-\sqrt{29}} = \frac{7}{6 - \sqrt{29}}$$

and so:

$$n<\frac{49}{(6-\sqrt{29})^2}<n+1 \Rightarrow \frac{1}{n+1} < \frac{(6 - \sqrt{29})^2}{49} < \frac{1}{n}.$$

Adding $49$ times this inequality with the original gives:

$$n + \frac{49}{n+1} < (6 + \sqrt{29})^2 + (6 - \sqrt{29})^2 < n+1 + \frac{49}{n}$$
$$\Rightarrow n + \frac{49}{n+1} < 130 < n+1 + \frac{49}{n}$$

From the right-hand side, the minimum value of $n$ which satisfies the inequality is $129$. But from the left-hand side, $n$ cannot be $130$ or greater. Therefore $n = 129$ which is the unique solution.


问:
Upper bound of set
Is it necessary only least upper bound may or may not belongs to the set and not any other upper bound????
Does always upper bound belongs to outside of the set ??</p>

答: If $S$ is a set and $a$ is an upper bound for $S$, but not the least upper bound, then there exists a smaller upper bound $a'$. So for every $x\in S$, we have $x\le a'<a$ and cannot have $a\in S$.

Title: Upper bound of set


问:
Does this IVP have a unique solution for all $x \in \mathbb R$
Is $\displaystyle {dy\over dx}=\sin(y)$ with initial conditions $y(X)=Y$ guaranteed to have a unique solution for all $x\in\mathbb R$? </p>

答:
The equation is equivalent to

$$\frac{dy}{\sin{y}} = dx \implies \int dy \, \csc{y} = x + C$$

where $C$ is a constant of integration.  Performing the integration of the left, we get

$$\log{\tan{\frac{y}{2}}} = x + C$$

$C$ may be determined from the initial condition:

$$C = \log{\tan{\frac{Y}{2}}}-X$$

Now we solve for $y$ in terms of $x$.  Exponentiating, we get

$$\tan{\frac{y}{2}} = e^{x-X} \tan{\frac{Y}{2}} $$

The problem now is that the inverse tangent does not provide a one-to-one mapping, so we would need to restrict the range of $y$ to get a unique solution here.


问:
Infection spread on a torus chessboard
In one of his books, Peter Winkler includes the following problem:

A disease is spreading on a $n\times n$ chessboard as follows: if a healthy cell is neighboring at least 2 infected cells, it becomes infected. Using the property that the perimeter of the infected area never increases, it’s easy to prove that it’s impossible to infect the entire chessboard with fewer than $n$ infected cells.

If the chessboard is a torus, the result no longer holds (verified on some instances of $n$). It seems that $n-1$ is the smallest number of infected cells required to infect the chessboard. The perimeter argument can’t be used here. Does anyone know any other ‘invariant’ that can be used?

Note: two cells are neighbors if they share one side.

答:
Consider separately two ways the infection can grow. When infection spreads to an empty cell, it can be that two of the adjacent infected cells share a corner (which is always the case when there are three or four adjacent infected cells) or not, in which case the two infected cells are both in the same row or column.

Within a row or column, we will say a component is a group of consecutive infected cells in that row or column. (We don't care whether the infected cells are connected via a path that lies outside the row or column.) Our proposed invariant is based on the total number of components among both rows and columns (with an exception noted below).

We will consider growth not in generations, but in individual steps where the infection grows to only a single cell at each step. (The order will not matter unless noted.)

Under the first kind of growth, new components cannot be created. The step can only decrease the number of components in a row or column or both.

Under the second kind of growth, usually the number of components of a single row or column will go down by one (as the components are joined by the infection of the cell between them) and the number of components in the corresponding column or row will either increase by one or stay the same. In this case, then, the total number of components will not increase.

The other case is when the last cell in the row or column is being infected. The first time this happens for rows and the first time for columns, it can indeed increase the number of components by one. However, we not need to concern ourselves with subsequent times. Without loss of generality, consider rows. If one row is already completely infected and a second row is one cell from being completely infected, then either there is a column component that connects these row components or not. If there is a connecting column component, then (by infecting in a different order) the first kind of growth can be used to infect the entire stripe bounded by the rows and the connecting component. (This includes the last cell in this new row which will have been infected without increasing the number of components.) If there is no such connection, then in $n - 1$ columns, there are at least $2$ components each plus at least $1$ component in the last column plus $2$ rows with at least one component. This is at least $2n + 1$ total components. We will not be concerned with cases that have so many components.

If we start with $n - 2$ infected cells, then there are at most $n - 2$ column components and $n - 2$ row components for a total of $2n - 4$ components. At most two new components can be introduced; one by the first full row infection and one by the first full column infection. This gives at most $2n - 2$ components. As this is less than $2n + 1$, any subsequent full row or column infection can be completed by growth of the first kind without increasing the number of components. Thus, the final number of components will be at most $2n - 2$. This is less than the $2n$ of the completely infected board.

Title: Infection spread on a torus chessboard


问:
A projectile has been shot into the atmosphere. Find the polynomial that interpolates the speed and acceleration in function of time.
This is the table (tempo = time, velocidade = speed, aceleracao = acceleration)

I used Newton's interpolation to get the polynomial (5th degree).

I got
$$102 + 138(x-15) + (-124/15)(x-15)^2 + (289/225)(x-15)^2(x-30) + (-0.0689)(x-15)^2(x-30)^2 + (0.0048)(x-15)^2(x-30)^2(x-45)$$

I tested this in my calculator and the values for p(15) and p(30) are correct but p(45) is completely wrong.

Then I got the derivative to test the derivatives
$$14341.5 - 2226.13 x + 122.896 x^2 - 2.8676 x^3 + 0.024 x^4$$

p'(15) and p'(30) are correct, albeit with a slight error, but p'(45) is very off (I got 132 point something).

Is this normal? Is it possible to get a better estimate for this?

My calculations:

Body:
If you do the table correctly, with $p(x_2)=420$, you get
$$\small\begin{array}{c|c|c|}
p[x_0]=102&p[x_0,x_0]=138&p[x_0,x_0,x_1]=-8.2667&p[x_0,x_0,x_1,x_1]=1.28444\\
&p[x_0,x_1]=14&p[x_0,x_1,x_1]=11&p[x_0,x_1,x_1,x_2]=-0.74844\\
p[x_1]=312&p[x_1,x_1]=179&p[x_1,x_1,x_2]=-11.4533&p[x_1,x_1,x_2,x_2]=1.36267\\
&p[x_1,x_2]=7.2&p[x_1,x_2,x_2]=8.9867\\
p[x_2]=420&p[x_2,x_2]=142
\end{array}$$
cont.
$$\small\begin{array}{c|c|c|}
&p[x_0,x_0,x_1,x_1,x_2]= -0.067763&p[x_0,x_0,x_1,x_1,x_2,x_2]=0.004604\\
&p[x_0,x_1,x_1,x_2,x_2]=0.070370&
\end{array}$$


问:
How to solve $3x+\sin x = e^x$
How doese one solve this equation?

$$
3x+\sin x = e^x
$$

I tried graphing it and could only find approximate solutions, not the exact solutions.
My friends said to use Newton-Raphson, Lagrange interpolation, etc., but I don't know what these are as they are much beyond the high school syllabus.

Title: How to solve $3x+\sin x = e^x$

Body:
sorry

f(b)=3(1)+sin(1)−e1= is Not Equal to ---> 1.12...

f(b)=3(1)+sin(1)−e1= is Equal to  ----> 0.299170578

And

f(0.5)=3(0.5)+sin(0.5)−e0.5= is Not Equal to ---> 0.33...>0

f(0.5)=3(0.5)+sin(0.5)−e0.5= is Equal to  ---->   -0.1399947352


问:
What is the motivation behind the definition of the "Neighbourhood Space"?
Bert Mendelson's "Introduction to Topology" (3ed, 1975, Dover pubs) spends a section in both his chapter on metric spaces and that on general topological spaces (chapters 2 and 3 respectively) hammering out the axioms for a neighborhood space.

As follows:

A neighborhood space is a set $S$ such that, for each $x \in S$, there exists a set of subsets $\mathcal N_x$ of $S$ satisfying the following conditions:

$\text N 1$: There exists at least one element in $\mathcal N_x$

$\text N 2$: Each element of $\mathcal N_x$ contains $x$

$\text N 3$: Each superset of $N \in \mathcal N_x$ is also in $\mathcal N_x$

$\text N 4$: The intersection of $2$ elements of $\mathcal N_x$ is also in $\mathcal N_x$

$\text N 5$: There exists $N' \subseteq N \in \mathcal N_x$ which is in $\mathcal N_y$ for each $y \in N'$

He goes on to establish that a neighborhood space is exactly the same as a topological space: the neighborhood space induced by the topological space induced by a neighborhood space is that neighborhood space, and so on.

So the neighborhood space axioms define exactly the same space as the well-known open set axioms of topology.

Proving that a collection of subsets of a set adhere to the neighborhood space axioms is fiddly and tedious, and conceptually non-intuitive. Particularly axiom $N5$, which is horrible.

The question is: is there a good reason for a "neighborhood space" to be defined? Does it have any particular uses that make it essential or worthwhile to study in detail? Or is it just an interesting backwater that Mendelson (and also Willard, it seems, by taking a glance at the Wikipedia page on "Neighborhood system") defines for the fun of it?

Hence, how much effort would one be expected to exert in an attempt to master fluency in handling such constructs?

(No doubt there may be some difficult-to-analyse spaces which are analysed more easily using neighborhood axioms rather than open set axioms -- but are they sufficient bang for the buck?)

答:
You seem to have already made up your mind that this isn't worth your time, so it's hard to know what can be provided as an answer.  But I would say that this is worth your time if for no other reason than you are clearly struggling to translate between two axiomatic systems that describe the same thing.  That is a fundamental mathematical skill and anyone with a Ph.D. will tell you that they needed it, in some form or another, to get where they are.

I personally needed this definition, and at least a dozen other similar ones, for my thesis work.  Let me try to explain why without getting into too much detail.  The key point is this: Different sets of axioms generalize differently, even if the axioms themselves are equivalent.  You are already starting to see this when you call axiom N5 "horrible"; indeed, in this way of viewing topology, filters are natural, but open sets are not.  Obviously, in the most popular definition of a topological space, open sets are taken as fundamental.

Before 1990, rigid analytic geometry (which is very far from being a backwater) was largely studied by means of Grothendieck topologies, which are more or less a big categorical generalization of filters or neighborhood spaces.  This is a bit strange, as there were actual topological spaces.  But these spaces didn't behave in intuitive ways, for example they failed to be locally compact, an important technical property that went through just fine in the more abstract setting.

But Berkovich demonstrated that we can add some points (well, a lot of points) to get a really nice topological space, one that is locally compact and Hausdorff.  Why was this possible, and why did nobody notice this before?

The key lies in something like axiom N5.  You see, traditional non-archimedean spaces have, in a sense, too many open sets.  The whole point of using Grothendieck topologies was to ensure that not every open set containing a point was a neighborhood of that point, by forcing it to contain some "much smaller" neibghborhood (in the literature this is sometimes called overconvergence).

Somehow, mathematicians knew that these open sets were not "open enough", so they had to be somehow excluded.  Now we sort of know why: a (nice) classical non-archimedean space $X$ is a dense subspace of a locally compact Hausdorff space $\hat{X}$ (whose points may be difficult to get a handle on concretely).  In particular, some open sets/coverings in $X$ are special, namely the preimages of open sets/coverings in $\hat{X}$.  In fact, this means that the space $\hat{X}$ was always somehow implicit in a neighborhood system on $X$! (i.e. $\hat{X}$ consists of the points of some topos, blah blah blah)

For me, the moral of this story is that doing mathematics is not just about learning the most popular axioms and drawing conclusions from them.  It's about understanding the interplay between different definitions.  New research is often about forging new connections, and it's held back often precisely because some area is considered "backwater" and so on.  Mathematics can go in or out of style, but it rarely becomes truly obsolete.  Don't just chase style, take the time to understand the various perspectives (even the silly ones) and I can promise you it will pay off.


问:
Trigonometric ratios of angles greater than 90
So, essentially, according to the right triangle definition of trigonometry, trigonometric ratios of acute angles are the ratios of the necessary sides. Then, historically speaking, we derived the sum and difference angle formulae, i.e.,

```<code>                            Sin(A±B) = SinACosB ± SinBCosA
</code>```

Then, perhaps while playing with the equation, we obtain Sinɸ such that ɸ > 90°.

Now, as in the case of acute angles, the ratio of the particular sides seemed to be a constant for a given angle. How do we know that the value of Sin of obtuse angles is a constant, i.e., how did we realise that it is fixed and unchanging like that of acute angles?

Please do not give a unit circle explanation as I know that in it, the trig operators for all angles are expressed as ratios. Is there any proof of my question in the right triangle method?

答:
Consider the obtuse $\triangle$ABC where the line segment $\overline{AB}$ is horizontal and the $\angle$ BAC is obtuse.

Extend the line segment $\overline{AB}$ to the left to point $D$ so that $\angle$ BDC is $90^\circ$.

By convention, $\sin(\angle\text{BAC}) = \sin(\angle\text{DAC})$ and $\cos(\angle\text{BAC}) = - \cos(\angle\text{DAC})$.

Therefore, since $\angle\text{DAC}$ is acute, the previously assumed result that $\sin(\angle\text{DAC})$ is constant kicks in.


问:
Solve $y' = -x, x'' = y$

Find the general solution $(x(t),y(t))$ of the system
$$\begin{cases}
y'(t) + x(t) = 0 \\
x''(t) - y(t) = 0
\end{cases}
$$

It can be easily observed that $$(Ae^{-t}, Ae^{-t})$$
is a solution for any constant $A$. But as $x''$ exists there must be another solution out there. As I am new to this type of problem, how should I proceed?

答:
From the second equation, we have $y' = x'''$.

We substitute the first equation

$$x''' + x = 0$$

Solving this leads to

$$x(t) =  c_1 e^{-t}+c_3 e^{t/2} \cos \left(\frac{\sqrt{3} t}{2}\right)+c_2 e^{t/2} \sin \left(\frac{\sqrt{3} t}{2}\right)$$

From the second equation, we know $y = x''$, so just take the second derivative of $x(t)$ to find $y(t)$.

Title: Solve $y' = -x, x'' = y$


问:
Area of shaded triangular region from absolute value functions
So, I got this question from my teacher. I've tried solving it but to no avail. I cannot work out a method either, which can help me solve this.

So, it gives us two absolute value functions
$g(x) = 4|x-3|+3$
$f(x) = -6|x-3|+9$

There are two points, A & B as shown in the figure given below.
I don't understand, when I find the points of intersection by solving
$g(x) = f(x)$
then what are those $x$ values. Are they the coordinates of point A and B?
And what would be the next steps leading to the complete solution of this question.

I'd appreciate if anyone points out the steps or just solves it for my ease.

Question FIGURE

答:
From your diagram, we can see that we must use A,B, and the vertex of f(x) to calculate the area of our triangle. A,B are the intersection points of f(x) and g(x).

Thus, we set g(x)=f(x) to solve for our intersection points, and we get 4|x−3|+3=−6|x−3|+9. Simplifying, we get 10|x−3|=6, so |x−3|=0.6. The two values that have an absolute value of 0.6 are 0.6 and −0.6. Thus, we have x−3=0.6 or x−3=−0.6. Solving for x, in each case, we get x=2.4,3.6.

We plug these values into either f(x) or g(x), it doesn't matter because both functions are equal at our intersection points. So, we have g(2.4)=4|2.4−3|+3=5.4 and g(3.6)=4|3.6−3|+3=5.4. Thus, our intersection points are (2.4,5.4) and (3.6,5.4).

Now, we have to find the vertex of f(x)=−6|x−3|+9. Looking at our diagram, we see that the vertex of f(x) is when it is the biggest. Since we are subtracting −6|x−3| in f(x), we try to make it as small as possible to make f(x) as big as possible. Since absolute value is never negative, the smallest we can make it is 0. We achieve −6|x−3|=0 when x=3. So, the vertex of f(x) has x-coordinate 3, and the y-coordinate is just f(3)=9 because we made −6|x−3|=0. Thus, the vertex of f(x) is (3,9).

So, we have all the points of our triangle (2.4,5.4),(3.6,5.4),(3,9). The base of our triangle is between the points (2.4,5.4) and (3.6,5.4) and the distance between them is 3.6−2.4=1.2. So b=1.2. The height of our triangle is the distance from (3,9) to our base. The y-coordinate of (3,9) is 9, and the y-coordinate of our base is 5.4. The difference between the y is 3.6, so height h=3.6. Use area of triangle formula A=bh/2 to get A=1.2∗3.6/2=2.16. Thus, the area of the shaded traingle is 2.16.


问:
Existence of polynomials so that inequality holds
Is there a polynomial $P$ such that $|P^2(x)-e^{10x}|<10^{10} \text{ for each }  x\in \mathbb{R} ?$

Is there a polynomial $Q$ such that $|Q^2(x)-e^{10x}|<10^{-10} \text{ for each } x \in [0,1] ?$

I have thought the following: We look at the convergence of $e^{10x}$.

The exponential function is continuous.

I have thought to calculate the supremum of $e^{10x}$ in the given intervals. And then we would look if there is a polynomial function so that the difference of the exponential function and the polynomial tends to $0$ in these intervals.

Is my approach right?

答:
The answer to the first one is negative.
Since $P^2(x)$ is a polynomial and $e^{10x}$ is an exponential function, $\displaystyle\lim_{x\to \infty}{(e^{10x}-P^2(x))}=\infty$.

On the other hand the answer to the second problem is clearly positive since the existance of such $Q$ is equivalent to :

$$-e^{-10}-10^{-10}<Q^2(x)<1+10^{-10}$$

Now take any arbitrary polynomial $P$ and there exists $c\in\mathbb{R}$ (sufficiently big) such that $Q(x)=\frac{P(x)}{c}$ satisfies the condition.


问:
Math.Expectation And Variance of Stationary Series
we have formula of  stationary time series
$$y_t = 0.6\cdot y_{t-1} + 2 + e_t$$
as mathematical expectation is constant, we can do the following:
$$E = 0.6\cdot E + 2 + 0$$
$$E = 5$$
Variance is $\sigma^2 = E(y_t^2)-(E(y_t))^2 = E(y_t^2)-25$

Is it correct?

How to calculate variance until the end?

答:
The mean is correct. You can use the same methodology to compute the variance:$$\begin{align*}\sigma^2 &= V(y_t) \\&= V(0.6\cdot y_{t-1} + 2 + e_t) = 0.6^2\sigma^2 + V(e_t)\end{align*}.$$
Solving for $\sigma^2$ yields $\sigma^2=V(e_t)/0.64$


问:
What is the minimal polynomial of $\alpha = i \sqrt{ 1 + i \sqrt{3}}$ over $\mathbb{Q}$?
What is the minimal polynomial of $\alpha = i \sqrt{ 1 + i \sqrt{3}}$ over $\mathbb{Q}$?

I calculated $\alpha, \alpha^2, ....$ until I saw a linear combination (with coefficients in $\mathbb{Q}$) such that it is equal to 0. I ended up finding the following polynomial:

$f(X) = X^4 + 2X^2 + 4$ which gives:

$f(\alpha) = -2 + 2 i \sqrt{3} + 2(-1 -i \sqrt{3}) + 4 = -4 + 4 - 0$.

I wanted to know if there was a shorter way of finding the minimal polynomial for $\alpha$ because calculating all the powers and then looking for a linear combination requires quite a lot of calculations.

答:
We can just square the equation, rearrange and square again. This gives a polynomial of degree $4$, and we only have to show that it is irreducible.

So we obtain $\alpha^2=-(1+\sqrt{-3})$ and then $(\alpha^2+1)^2=-3$, i.e., $\alpha^4+2\alpha^2+4=0$. Hence the polynomial is $x^4+2x^2+4$.

The polynomial is irreducible over $\Bbb Q$ and $\Bbb Z$. One way to see this is to assume that
$x^4+2x^2+4=(x^2+ax^2+b)(x^2+cx^2+d)$ and compare coefficients. This has no integral solution.


问:
Higher derivatives of multilinear map
Let, $E,F$ are two banach spaces and $V$ be an open subset of $E.$ $f_1,f_2:V\to F$ be twice continuously differentiable function.Let, $H$ be a multilinear map from $F\times F$ to a banach space $G$ and $\phi:U\to G$ defined as $\phi(x)= H(f_1(x),f_2(x)).$Now ,compute $D^2 \phi.$

I can compute $D^1(\phi)$ and that is,
$D^1\phi(x)(z)=
H(Df_1(x)(z),f_2(x))+H(f_1(x),Df_2(x)(z))\text{
 where}~z\in E.$

Now, to compute the second derivative I am facing problem. Any help?

Title: Higher derivatives of multilinear map

Body:
Why don't you continue applying the same rule to each term,

$D^1\phi(x)(z)= H(Df_1(x)(z),f_2(x))+H(f_1(x),Df_2(x)(z))$

$D^2\phi(x)(z,w)=\\= H(D^2f_1(x)(z,w),f_2(x)) + H(Df_1(x)(z),Df_2(x)(w))\\ + H(Df_1(x)(w),Df_2(x)(z)) + H(f_1(x),D^2f_2(x)(z,w))$


问:
Area Between Four Rectangles and a Semi Circle.

Given a halved circle with radius $R$ and four rectangles inside it such that one of the side touches the middle line of the circle. If the area of the four rectangles are maximum, what is the area between the rectangles and the halved circle?

I'm sorry for my bad English so I try to illustrate it such follows.

We need to find the area of grey colored area. I really don't have any idea what to do. If there is only one big rectangle in that semi-circle, I think I know what to do. But what if the rectangles are more than one and the length of each rectangles sides can be vary?

答:
The answer ist going to be symmetric so we will focus on how to maximize the area of two rectangles inside an eighth of a circle. Consider a point $P(x,y)$ inside this sector at the bottom corner of the green rectangle from the figure above. We construct the edges of the two bottom rectangles like this:

Call the width of the bottom rectangle $a$ and the sum of their heights $b$. Then the area of the blue hexalateral is:

$$A=bx+(a-x)y-\frac{b^2}{2}=x\sqrt{r^2-x^2}+y\left(\sqrt{r^2-y^2}-x\right)-\frac{r^2-x^2}{2}$$

In case you aren't familiar with multivariable calculus, partial differentiation works the same as ordinary while treating any other variables as constants. To find the maximum of A, we need the zeros of its partial derivatives:

$$\frac{\partial A}{\partial x}=\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}+x-y=0 \qquad y=\frac{r^2-2y^2}{\sqrt{r^2-x^2}}+x$$
$$\frac{\partial A}{\partial y}=\sqrt{r^2-y^2}-\frac{y^2}{\sqrt{r^2-y^2}}-x=0 \qquad y=\sqrt{\frac{r^2}{2}-\frac{x}{8}\left(x+\sqrt{8r^2+x^2}\right)}$$

Using a numerical method, we obtain $x \approx 0.795r$ and $y \approx 0.359r$, so $A \approx 0.3479r^2$.

So for a semicircle the minimal area not covered by four rectangles is $4r^2\left(\frac{\pi}{8}-A\right) \approx 0.179r^2$.


问:
Needing help with vector spaces
The positive numbers with x + y and ex redefined to equal the usual xy and
x
c do satisfy the eight rules. Test rule 7(e( x + y) = ex + ey) when e = 3, x = 2, y = 1. (Then
x + y = 2 and ex= 8.) Which number acts as the "zero vector"? The answer is the number one, but why? What's going on?

This is a practice problem from Gilbert Strang's linear algebra. Is this question referring to a vector space of 1, where x and y are vectors in a single 1d direction? I'm just very lost with this question(and vector spaces in general) as I'm not used to the conventions in LA and don't really know what I don't know about vector spaces...the book I'm referring to seems to jump straight from the intuition for vector spaces to the solving of problems without teaching the in-betweens. I'm a high school student trying to teach myself LA so I'm quite new to mathematics at the college level. Sorry for the amateurish question.

答:
I think $ex$ is defined by $x^e$

Let us denote by $o$ the zero vector. For each positive number $x$ we get

$$o=x+(-x)=x+x^{-1}=xx^{-1}=1.$$

Title: Needing help with vector spaces


问:
(1, 1) weak inequality for Hardy's integral operator
Consider Hardy's integral operator $T$ defined on $L^1(0, +\infty)$ by
$$ Tf(x) = \frac{1}{x}\int_0^x f(t) \, dt, \quad x \in (0, +\infty), f \in L^1(0, +\infty).$$
I need to find a constant $C > 0$ such that
$$ \lambda m(|Tf| > \lambda) \leq C{\Vert f \Vert}_1 \quad \text{for all } \, \lambda > 0, f \in L^1(0, +\infty)$$
where $m$ denotes the Lebesgue measure.

I don't know how to tackle this problem. Any help will be appreciated.

答:
let $f_0:\mathbb R\to\mathbb R $ be the extension of $f$ by zero. $$|Tf(x)| \le \frac1x\int_0^x |f(x)| dx\le Mf_0(x) := \sup_{I\ni x}\frac1{|I|} \int_I |f_0(x)|dx$$
Since the Hardy-Littlewood maximal operator $M$ is weak $(1,1)$,
$$ m( |Tf|\ge \lambda ) \le m(Mf_0\ge \lambda) \le \frac C \lambda  \|f_0\|_{L^1(\mathbb R)} = \frac C \lambda  \|f\|_{L^1((0,\infty))}. $$


问:
Spectrum of an integral operator
I am trying to find the spectrum of the operator $Tf(x)=\int_{0}^{1}(1+x^2y^2)f(y)dy$ in $L^2(0,1)$.

Thanks a lot for any help.

答:
Hint: $Tf=\lambda f $ is equivalent to $$\lambda f(x)=\int_0^1f(y)dy+x^2\int_0^1y^2f(y)dy$$ Hence either

$\lambda\ne0$ and $f(x)=Ax^2+B$; then the equation above becomes $$\lambda(Ax^2+B)=\tfrac{1}{3}A+B+(\tfrac{1}{5}A+\tfrac{1}{3}B)x^2$$ Comparing coefficients gives $\begin{pmatrix}\frac{1}{5}&\frac{1}{3}\\\frac{1}{3}&1\end{pmatrix}\begin{pmatrix}A\\B\end{pmatrix}=\lambda\begin{pmatrix}A\\B\end{pmatrix}$, so $\lambda$ equals the eigenvalues $\lambda,\mu$ of this matrix and $(A,B)$ their eigenvectors.

$\lambda =0$ then $\int_0^1f=0=\int_0^1y^2f(y)dy$. In this case, there are an infinite number of linearly independent solutions for $f$, in fact any function of the type $f(x):=g(x)-(A+Bx^2)$ for $g\in L^2[0,1]$ can satisfy these two equations if the right $A,B$ are chosen.

Hence the spectrum is $\{0,\lambda,\mu\}$.

Title: Spectrum of an integral operator


答:
With $$e=3, x=2, y=1 $$

we get  $$ x+y = (xy) =2, e(x+y)=6$$

$$ ex+ey = 6(3) = 18$$
Thus the distributive rule fails.
Also $$x+1 = (1x)=x$$
That is $1$ acts as the $0$ vector.

Title: Needing help with vector spaces

Body:
The positive numbers with x + y and ex redefined to equal the usual xy and
x
c do satisfy the eight rules. Test rule 7(e( x + y) = ex + ey) when e = 3, x = 2, y = 1. (Then
x + y = 2 and ex= 8.) Which number acts as the "zero vector"? The answer is the number one, but why? What's going on?

This is a practice problem from Gilbert Strang's linear algebra. Is this question referring to a vector space of 1, where x and y are vectors in a single 1d direction? I'm just very lost with this question(and vector spaces in general) as I'm not used to the conventions in LA and don't really know what I don't know about vector spaces...the book I'm referring to seems to jump straight from the intuition for vector spaces to the solving of problems without teaching the in-betweens. I'm a high school student trying to teach myself LA so I'm quite new to mathematics at the college level. Sorry for the amateurish question.


问:
$F$ is proper if and only if $\operatorname{deg}(f)=\operatorname{deg}(g)$.
Let $f, g \in C[z]$ be two non constant monic polynomials.
Consider the homotopy function $F:[0,1] \times C \rightarrow C$
$$
F(t, z)=(1-t) f(z)+\operatorname{tg}(z)
$$
Show that $F$ is proper if and only if $\operatorname{deg}(f)=\operatorname{deg}(g)$.

$\Leftarrow$ is easy because then $F(t, z)=z^{n}$ for some $n \geq 1$ and it is proper, first it is continuous and for every bounded set, its preimage under $z^{n}$ is bounded because $|z|^{n} \geq|z|$.

About $\Rightarrow$ I need to find a compact set in codomain that its compactness cannot preserved under the homotopy map. First of all since it is continuous closedness will be preserved but I need to find the compact set such that its boundedness should not be preserved.
WLOG assume $n>m$ and let $f=z^{n}, g=z^{m}$ then $F(t, z)=z^{m}\left((1-t) z^{n-m}+t\right)$
I think that finding some patologies with $\epsilon>0$ after analysing the following
Take $B[0, \epsilon] \subset \mathbb{C}$
Then
$$
F^{-1}(B[0, \epsilon])=\left\{(t, z) \in[0,1] \times \mathbb{C}:\left|(1-t) z^{n}+t z^{m}\right| \leq \epsilon\right\}
$$
Stuck here!

答:
Let $\deg f=n > \deg g$ and assume that $F$ is proper so there is $Q>0, F^{-1}(|z| \le 1) \subset [0,1] \times \overline {B(0, Q)}$ or equivalently $|F(t,z)| > 1, |z| >Q, 0 \le t \le 1$

Then let $a_0, b_0$ the free terms of $f,g$ and notice that $ta_0+(1-t)b_0, 0 \le t \le 1$ is the segment joining them in the complex plane (could be degenerate if $a_0=b_0$) so in particular for any $|w| \le 1$ that is not on the segment, we have $|w-ta_0-(1-t)b_0| \ge c(w)>0, 0 \le t \le 1$

Pick a fixed such $w$ and then for fixed $0 <t \le 1$ the polynomial $tf+(1-t)g-w$ has leading coefficient $t$ and free term at least $c=c(w)$ in absolute value so by the usual formulas the product of its $n$ roots is at least $c/t$ in absolute value, so for (at least) one root $|z_t| \ge (c/t)^{1/n} \to \infty, t \to 0^{+}$

In particular, if we pick $t$ small enough, we get $|z_t| >2Q$ and $F(t,z_t)=w, |w| \le 1$ contradicting the above, so we are done!


问:
Applying the CLT on a modified Random variable
X1, X2, X3, . . . independent, U(0, 1) distributed random variables. Prove that if n → ∞, then:

$ Y_n=  \frac{\sum_{k=1}^{n} kX_k - \frac{n^{2}}{4}}{\frac{1}{6}n^{3/2}} \ $ goes to normal distribution N(0,1) in distribution

so what I tried to do is that I defined a new Random Variable $Y=kX$ such that I would apply CLT to it and therefore the new expectation would be E(Y)= $\frac{n}{2}$ and the new variance V(Y)= $\frac{n^{2}}{12}$ and I tried applying the theory but it seems that I am missing something because I couldn't prove the above, can someone please help ?

答:
The new random variable is the following

$$W=\Sigma_{k=1}^n kX_k$$

Now observe that

$$\mathbb{E}[W]=\Sigma_{k=1}^n k\cdot \mathbb{E}[X_1]=\frac{1}{2}\frac{n(n+1)}{2}\approx\xrightarrow{n\rightarrow +\infty}\approx \frac{n^2}{4}$$

$$\mathbb{V}[W]=\Sigma_{k=1}^n k^2\cdot \mathbb{V}[X_1]=\frac{1}{12}\frac{2n^3+3n^2+n}{6}\approx\xrightarrow{n\rightarrow +\infty}\approx \frac{n^3}{6^2}$$

use the CLT to get your result

$$\frac{W-\mathbb{E}[W]}{\sqrt{\mathbb{V}[W]}}\xrightarrow{\mathcal{L}} \Phi_{\{0;1\}}$$


问:
Combinatorics and statistic question (pick all 1s in a hand of 9 cards)
I was going through statistics classes on Khan academy and there was this exercise I was hoping someone could help me with:

we have $36$ unique cards (normal deck, $4$ types - hearts, diamonds etc.). And $9$ different groups of cards (from $1$ to $9$).

a hand is a selection of $9$ cards which can be sorted any way you want

question: what is the probability of getting all $4$ $1$s in a hand? (so $1$ of diamond, $1$ of hearts etc, $4$ total)
link:https://youtu.be/ccrYD6iX_SY

I'm here because I disagree with the Sal's solution and I was hoping someone could tell me where I could be wrong:

we are looking for a probability, and $$
\mathrm{(prob\  of\  needed \ result)\  = \frac{\# (needed \ result\  can\  happen) }{ \# (total
\ results)}}
$$

total results are ok and it is the same in the video

(in python)

n_total_outcomes =
scipy.math.factorial(36)/(scipy.math.factorial(9)*(scipy.math.factorial(36-9)))

but the desired result is where it gets hairy (for me).

Next I decided not to use any special formulas and apply logic: we can get Lucky ~ if we have already picked $\frac 59$ random cards, we have $31$ cards left, and the chance we pick $1$s higher ($\frac 1{31}, \frac 1{30}, \frac 1{29}, \frac 1{28}$) and unlucky or less probable if we start picking the $1$s at the beginning $\frac {1}{36}, \frac{1}{35}, \frac{1}{34}, \frac{1}{33}$.

So I just imagined that these are $2$ side cases ($7\times 10^{-7}$ and $1 \times 10^{-6}$), which is still nowhere close to Sal's solution, where he has very high chance. If you watch his video my beef is that he is picking all four 1s at the beginning and then picking up the rest of the hand. It is nowhere guaranteed you will get all four 1s right away.

If anyone could show me the problem with my logic, it would be much appreciated.

答:
There are two ways of looking at problems of this type. You can consider the final collection of cards OR you can look at the process of selection.

When you say

If you watch his video my beef is that he is picking all four 1s at the beginning and then picking up the rest of the hand

it appears that you are thinking he is studying the process of selection. However, this is not the case; he is just considering the final hand consisting of $4$ 1s and $5$ other cards.

On the other hand you are attempting to consider the process. The way you are doing this would involve you in considering all the different ways the $4$ 1s are selected which you have not done.


问:
Atiyah-Macdonald, Exercise 8.3: Artinian iff finite k-algebra.

Atiyah Macdonald, Exercise 8.3. Let $k$ be a field and $A$ a finitely generated $k$-algebra. Prove that the following are equivalent:

  (1) $A$ is Artinian.

  (2) $A$ is a finite $k$-algebra.

I have a question in the proof of (1$\Rightarrow$2): By using the structure theorem, we may assume that $(A,m)$ is an Artin local ring. Then $A/m$ is a finite algebraic extension of $k$ by Zariski lemma. Since $A$ is Artinian, $m$ is the nilradical of $A$ and thus $m^n=0$ for some $n$. Thus we have a chain $A \supseteq m \supseteq m^2 \supseteq \cdots \supseteq m^n=0$. Since $A$ is Noetherian, $m$ is finitely generated and hence each $m^i/m^{i+1}$ is a finite dimensional $A/m$-vector space, hence a finite dimensional $k$-vector space.

But now how can I deduce that $A$ is a finite dimensional $k$-vector space?

答:
The claim also seems to follow from the Noether normalization lemma:

Let $B := k[x_1, \dotsc, x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal.
Since $A$ is a finitely generated $k$-algebra you may let $A := B/I$. By the Noether normalization lemma it follows that there is a finite set of elements $y_1, \dotsc, y_d \in A$ with $d = \dim(X)$ and the property that the subring $k[y_1, \dotsc, y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1, \dotsc, y_d] \subseteq A$ is an integral extension of rings. If $d  = 0$, it follows from the same lemma the ring extension $k \subseteq A$ is integral and since $A$ is finitely generated as $k$-algebra by the elements $\overline{x_i}$ and since each element $\overline{x_i}$ is integral over $k$, it follows that $\dim_k(A) < \infty$.

Question: “But now how can I deduce that $A$ is a finite dimensional $k$-vector space?”

Answer: It seems from the argument above you can use the Noether normalization lemma to give another proof of your implication, different from the proofs given above. Hence now you have two proofs of your result.


答:
$$E\left\{\sum\limits_{k=1}^{n}kX_k\right\} = \sum\limits_{k=1}^{n}kE\{X_k\} = \frac{1}{2}\sum\limits_{k=1}^{n}k=\frac{n(n+1)}{4}$$

so,

$$ E\{Y_n\} = \dfrac{\frac{n^2+n-n^2}{4}}{\frac{1}{6}n\sqrt{n}} = \frac{3}{2\sqrt{n}} \rightarrow0, \quad \text{as } n\rightarrow\infty$$

and,

$$ E\left\{ \left( \sum\limits_{k=1}^{n}kX_k \right)^2 \right\} =
\sum\limits_{k=1}^{n}\sum\limits_{l=1}^{n}klE\{X_kX_l\} = \sum\limits_{k=1}^{n} k^2 E\{X_k^2\} = \frac{1}{3}\sum\limits_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{18}$$

so,

$$ E\{Y_n^2\}\rightarrow \dfrac{\frac{n^3}{9}}{\frac{n^3}{36}} = 4 $$

Either I have made a mistake, or $Y_n$ tends to $\mathcal{N}(0, 4)$.

Title: Applying the CLT on a modified Random variable

Body:
X1, X2, X3, . . . independent, U(0, 1) distributed random variables. Prove that if n → ∞, then:

$ Y_n=  \frac{\sum_{k=1}^{n} kX_k - \frac{n^{2}}{4}}{\frac{1}{6}n^{3/2}} \ $ goes to normal distribution N(0,1) in distribution

so what I tried to do is that I defined a new Random Variable $Y=kX$ such that I would apply CLT to it and therefore the new expectation would be E(Y)= $\frac{n}{2}$ and the new variance V(Y)= $\frac{n^{2}}{12}$ and I tried applying the theory but it seems that I am missing something because I couldn't prove the above, can someone please help ?


问:
How to Find the Percentage Between Two Ratios?
If 0% = 900 and 100% = 1800, how do you find the percentage of 1200?

Question: 1200 = ?%

ms = milliseconds

In other words, on a timeline from 0ms to 1800ms, if an object weren't to move until 900ms passed, and it would have moved exactly 0% during that time. What percent would it have moved until it stopped at 1200ms?

This formula would be used on a website :)

答:
The percentage can be expressed as a ratio of lengths.

The longer length which represents $100$ percent is $1800 - 900 = 900$ units long, and the shorter length is $1200 - 900 = 300$ units long. Therefore, $1200$ is $\frac{300}{900} = \frac{1}{3}$ of the way to $1800$, which is around $33.3$ percent.

This is equivalent to linear interpolation as you have mentioned: the gradient of the line between the two endpoints, and the gradient of the point $(x,1200)$ and one endpoint must be equal.


问:
Can Ker(T) = Im(T) of a linear transformation of R raised to a power?
So I know this can be true for when T:R -> R, but does this apply for when R is raised to any power? Would this also work if, for example, we had R^3?
Thanks in advance</p>

答: It is true if T is a projection i.e. T2=T.But in general you may take the counterexample T(ϵ1)=ϵ2 and T(ϵ2)=0,on R2 over the field R.Here RangeT=KerT.Although for any finite dimensional vector space V there is a k such that V=R(Tk)⊕N(Tk).</p>


问:
Are preorder DFS and postorder DFS of a graph reverse to each other?
Are preorder DFS and postorder DFS of a graph reverse to each other?

https://en.wikipedia.org/wiki/Depth-first_search#Vertex_orderings says no:

A reverse preordering is the reverse of a preordering, i.e. a list of the vertices in the opposite order of their first visit. Reverse
preordering is not the same as postordering.

A reverse postordering is the reverse of a postordering, i.e. a list of the vertices in the opposite order of their last visit. Reverse
postordering is not the same as preordering.

I believe yes, with an addition to reverse the traversal order between children of a vertex. My reasoning is based on CLRS' Introduction of Algorithms 3ed

22.3 Depth-ﬁrst search

Another important property of depth-ﬁrst search is that discovery and
ﬁnishing times have parenthesis structure. If we represent the
discovery of vertex u with a left parenthesis “(u” and represent its
ﬁnishing by a right parenthesis “u)”, then the history of discoveries
and ﬁnishings makes a well-formed expression in the sense that the
parentheses are properly nested. For example, the depth-ﬁrst search of
Figure 22.5(a) corresponds to the parenthesization shown in Figure
22.5(b). The following theorem provides another way to characterize the parenthesis structure.

Theorem 22.7 (Parenthesis theorem) In any depth-ﬁrst search of a (directed or undirected) graph G = (V,E), for any two vertices u and v, exactly one of the following three conditions holds:

the intervals [u.d, u.f] and [v.d, v.f] are entirely disjoint, and neither u nor v is a descendant of the other in the depth-ﬁrst forest,

the interval [u.d, u.f] is contained entirely within the interval [v.d, v.f], and u is a descendant of v in a depth-ﬁrst tree, or

the interval [v.d, v.f] is contained entirely within the interval [u.d, u.f], and v is a descendant of u in a depth-ﬁrst tree.

Thanks.

答:
The Wikipedia example explains it well : the postorderings always put D at first (so, in reverse order, at the end) while the preorderings always put it in the middle.

Your idea works with trees and with non-oriented graphs, but in oriented graphs you may have multiple paths leading to one node, and in a DFS you'll have to finish exploring that node before going back up ; and later you might have another branch coming to that point that, if you had taken it before the other, would have been prolongated. The preordering and postordering then give different versions.

With non-oriented graphs, you rule out the possibility to come back again to a node, so the set of preorderings and the set of reversed postorderings are the same.

But anyway, in general the question isn't that mush "could I have had the same string with a different algorithm and a different order of visit ?" but more "for a given DFS, that is a given order of visit, how do the preorder and the postorder interact with each other and give me information ?". This leads to a lot of nice algorithms, typically invented by Tarjan in the 70's.


问:
Show that $\bigcup_{n=1}^{\infty} (0, 2+\frac{1}{n}] = (0, 3]$
Show that: $$ \bigcup_{n=1}^{\infty}  (0, 2+ \frac{1}{n}] = (0, 3]$$

The above statement seems quite obvious but i'm unsure how one would prove it.

So far i've done this:

$A_n := (0, 2+ \frac{1}{n}]$  for $n \in N$

$A_{n+1} \subset A_n $    $ ,\forall$ $ n \in N$

So $\bigcup_{n=1}^{\infty} A_n = A_1 = (0, 3] $

I'm not sure if this is the way i'm meant to answer the question.

答:
You asked how we might solve a more general case without relying on the increasingness of the left-hand side.

So: $x \in \bigcup_{n = 1}^{\infty} \left(0, 2+\frac{1}{n}\right]$ if and only if there is $n$ such that $0 < x \le 2 + \frac{1}{n}$.

On the other hand, $x \in (0, 3]$ if and only if $0 < x \le 3$.

$0 < x \le 2 + \frac{1}{n} \implies 0 < x \le 3$: because all $n$ have $0 < 2 + \frac{1}{n} \le 3$, so the left-hand interval is contained in the right-hand interval.

$0 < x \le 3 \implies 0 < x \le 2+\frac{1}{n}$ for some $n$: this is trivial by setting $n = 1$.

The general technique is to identify precisely when something is in the left-hand side, identify precisely when something is in the right-hand side, and then show that the two conditions are equivalent. Your answer takes a handy shortcut by noting that the left-hand side has a sufficiently nice structure that it can be immediately simplified to $(0, 2+1]$.


问:
Projective resolution of tensor product
Let $M,N$ are $R$-modules and $P^\bullet, Q^\bullet$ are their projective resolutions. Can we obtain a projective resolution of $M\otimes N$ using $P^\bullet, Q^\bullet$? If I understand correctly the homology groups of $P^\bullet\otimes Q^\bullet$ are $\mathrm{Tor}(M,N)$ so it's not a projective resolution.

答:
There is a lot going on in your question, so I'll try and address it piece by piece. Lets work over a commutative ring so I dont need to be careful.

$Tor(M,N)$ is a bifunctor, and can be calculated as you probably know, by projective resolutions of $M$ or $N$, tensoring appropriately, and then taking the homology. However what you have described (taking a tensor product of complexes) is the proof of such a statement.

The tensor compled $P^*\otimes Q^*$ is a massive double complex. Depending on your familiarity with spectral sequences, it takes some or a lot of work, to show that since the rows of this double complex are exact, taking homology yields precisely $Tor(M,N)$. However, notice that this does not say that the tensor product complex is not projective or acyclic. Just as when you take a projective resolution, tensor and take homology, "philosophically" you are always left with at least one nonzero homology group (only the higher ones vanish in some cases), namely $Tor_0$ of what you want to calculate, in the case of a double complex, the $P^i$ and $Q^j$ dont contribute their own homology, so you dont need to worry that you're calculating stuff you dont want to.

Now to your actual question: Tensor product of flat objects remain flat. The tensor product of free things remains free (work this out!). The tensor product of projective objects, hence also remains projective!

This all bodes well for us. So now, we have a monstrous double complex, and we want to resolve the object $M\otimes N$. It doesnt quite make sense to say that "the double complex resolves $M\otimes N$", but its not far from the truth: take the total complex of this double complex. That is, direct sum up the objects of a given total degree and form a complex. To make it into a complex you have to be a bit careful: use the sign trick to correct the differentials (see here for instance) and get a complex. This complex is exact (work this out by hand, or use Kunneth). And each of the objects are projective.

Title: Projective resolution of tensor product


问:
The existence of minimum two-variable function
How can we say a function has a global minimum without calculating any critical points or major calculations? I want to know about the right theorem and how do we reduce the theorem to a problem like this.
\begin{align*}f(x,y) = x^{2n} + y^{2n} -nx^2 + 2nxy - ny^2\end{align*}
on $\mathbb{R}^2$ and $n$ be an integer greater than or equal to 2

Theorem: If $f$ is continuous on the region $R$ (the function should also be continuous at the boundary), then the global maximum and global minimum exist, either in the interior or on the boundary.

I know that there is a global minimum and the function is coercive then there is no global maximum. Could you please give me a motivation to answer this question "Is there a maximum/minimum value of $f$? Give the reason"

答:
Denote $f_z = \partial f /\partial z$.

To find convexity of $f$, note that
$$
\vec{\nabla} f^T
 = \begin{bmatrix} f_x \\ f_y \end{bmatrix}
 = \begin{bmatrix} 2nx^{2n-1}-2nx + 2ny \\ 2ny^{2n-1}-2ny + 2nx \end{bmatrix}
 = 2n \begin{bmatrix} x^{2n-1}-x + y \\ y^{2n-1}-y + x \end{bmatrix}
$$
and the Hessian is given by
$$
H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}
  = 2n \begin{bmatrix}
           (2n-1) x^{2n-2}-1 & 1 \\ 1 & (2n-1) y^{2n-2}-1
       \end{bmatrix}
$$
To prove $f$ is convex, it suffices to show $H$ is positive semidefinite, i.e. $u^THu \ge 0$ for any vector $u \in \mathbb{R}^2$. If you let $\vec{u} = (v,w)$, we get
$$
\begin{split}
\vec{u}^TH\vec{u}
 &= 2n \left(v^2 (2n-1) x^{2n-2} + w^2 (2n-1) y^{2n-2}\right) \\
 &= 2n(2n-1) \left(v^2 x^{2n-2} + w^2 y^{2n-2}\right) \\
 &\ge \vec{0},
\end{split}
$$
since it only has square terms, with equality iff $v=w=0$.

Thus, $f$ is convex everywhere.


问:
graph points distance problem.
there are $n \geq 3$ points in plane where each two points have distance at least $1$.
I know there are at most $ 3n - 6 $ pairs with the distance exactly $1$.

I want to prove this fact.

答:
To prove that the number of pairs at distance $\leq1$ is at most $3n-6$ we have to prove that the number of such ordered pairs is at most $6n-12$. That is, calling our set $S$ and $\phi(P)$ the number of points of $S$ at distance 1 from $P$, we have to show that $\sum_{P\in S}\phi(P)\leq 6n-12$.

Now consider $A$, the convex closure of the set $S$. It is a convex poligon, and its vertices are points of $S$, so it has a finite number $n$ of vertices. Call them $P_1,\dots,P_n$, and call $\alpha_n$ the interior angle of the poligon at $P_n$ (for example in the equilateral triangle, $\alpha_n=\frac{\pi}{3}$).

Given some $P_n$, the points at distance $1$ of $P_n$ will have to be in a circumference arc at distance $1$ of $P_n$ and of angle $\alpha_n$. As $\alpha_n<\pi$, there can´t be more than 3 points at distance 1 of $P_n$. That means the $P_n$ have at most 3 adjacent vertices. Thus, if there are 4 or more $P_n$, we have ended because, using that $\phi(P)\leq6$ for all $P$, we have that $\sum_{P\in S}\phi(P)$ will be $\leq6n-12$.

If there are only 3 vertices of $A$, $P_1$, $P_2$ and $P_3$, then $\alpha_1+\alpha_2+\alpha_3=\pi$. If the three angles are $\frac{\pi}{3}$, then $P_1,P_2$ and $P_3$ have at most 2 adjacent points each, and $\sum_{P\in S}\phi(P)\leq 6n-12$. If not, there must be one point with $\alpha_i<\frac{\pi}{3}$ (which can only have one adjacent point), and the other two points can´t have both $\alpha_i\geq\frac{2\pi}{3}$, so they can´t have both more than 2 adjacent points each. So they can´t have more than 5 adjacent points between the two of them. Using this, we reach in this case too the conclusion that $\sum_{P\in S}\phi(P)\leq 6n-12$.

Title: graph points distance problem.


问:
Notation for repeated composition of functions
I have a repeated composition of functions ${T_n}(z) = {\tau _0} \circ {\tau _1} \circ {\tau _2} \circ  \cdots  \circ {\tau _n}(z)$

By analogy with $\sum\limits_{i = 1}^n {} ,\prod\limits_{i = 1}^n {} ,\bigcup\limits_{i = 1}^n {} ,\bigcap\limits_{i = 1}^n {} ,$
I want to write ${T_n}(z) = \left( {\mathop  \circ \limits_{i = 0}^n {\tau _i}} \right)(z)$ or even ${T_n}(z) = {\mathop  \circ \limits_{i = 0}^n {\tau _i}} (z)$. Can I do this?

答:
Iterated summation and product are represented by the Greek uppercase letters sigma and pi. As we can see, there is a clear phoneticity here.

The only phonetic Greek equivalent of 'c' in 'composition' is kappa.

But as uppercase kappa resembles 'K', and is already used for continued fractions in some texts, we might use the lowercase kappa '$\kappa$' for our purpose:

$$\underset{i=1}{\overset{n}{\Huge{\kappa}}} f_i \left( x \right) \equiv f_n \left( f_{n-1} \left( \dots f_2 \left( f_1 \left( x \right) \right) \dots \right) \right)$$

Let's see an example where we can use this notation. Suppose we were to find the derivative of the composition of $n$ functions w.r.t. the variable they act on, say $x$ :

$\frac{d}{dx} f_n \left( f_{n-1} \left( \dots f_2 \left( f_1 \left( x \right) \right) \dots \right) \right)$

The derivative as computed using the chain rule recursively turns out to be,

$f_n^\prime \left( f_{n-1} \left( f_{n-2} \left( \dots f_2 \left( f_1 \left( x \right) \right) \dots \right) \right) \right) \cdot f_{n-1}^\prime \left( f_{n-2} \left( f_{n-3} \left( \dots f_2 \left( f_1 \left( x \right) \right) \dots \right) \right) \right) \dotsm f_2^\prime \left( f_1 \left( x \right) \right) \cdot f_1^\prime \left( x \right)$

But using the notations for iterative product and composition using $\prod$ and $\huge{\kappa}$ respectively, the expressions become much more compact:

$$\displaystyle{ \frac{d}{dx} \: \underset{i=1}{\overset{n}{\Huge{\kappa}}} f_i \left( x \right) = \prod_{i=1}^n f_i^\prime \left( \underset{j=1}{\overset{i-1}{\Huge{\kappa}}} f_j \left( x \right) \right) }$$


问:
Spacing between largest eigenvalues in the GUE
Consider GUE$(N)$, the set of $N \times N$ random Hermitian matrices where the elements on the diagonal are i.i.d. according to $\mathcal{N}(0,1)$ and both the real and imaginary parts of the upper-triangular elements are i.i.d. according to $\mathcal{N}(0,1/2)$. I am interested in the (suitably normalized) probability distribution on the spacing between the two largest (or smallest) eigenvalues as $N \rightarrow \infty$. Specifically I'd like to know $$F_\Delta(s) = \lim_{N \rightarrow \infty} \text{Prob} \left( N^{1/6} \left( \lambda_N - \lambda_{N-1} \right) \leq s \right) \,,$$
where $\lambda_N$ is the largest eigenvalue of a GUE$(N)$ matrix and $\lambda_{N-1}$ is the next-to-largest.

What I've gotten from the literature so far: The distribution of $\lambda_N$ is well-known to follow the Tracy-Widom law: $$\lim_{N \rightarrow \infty} \text{Prob} \left( N^{2/3} \left( \frac{\lambda_N}{\sqrt{N}} - 2 \right) \leq s \right) = F_2(s)$$ where $F_2$ is the Tracy-Widom distribution. From this formula we can expect that the typical spacing between the two largest eigenvalues in the GUE$(N)$ will be of order $N^{-1/6}$, hence the inverse factor in the first formula above. This post by T. Tao discussing this paper focusses on the spacing between two consecutive eigenvalues in the "bulk" of the spectrum, which is not what we're interested in here. In that post he mentions "In the edge case [...] the distribution is given by the famous Tracy-Widom law." However I have not been able to deduce the result. Also in this paper, at the bottom of p.6, it is stated that the joint density for $(\lambda_N,\lambda_{N-1})$ follows a "multivariate Tracy-Widom" distribution (from which we could deduce the distribution on the difference $\lambda_N-\lambda_{N-1}$), referring to this book. However the only remark about this in that book I found was in Remark 3.7.3 on p.134 where one rather vaguely states that some joint distribution for the suitably normalized pair $(\lambda_N,\lambda_{N-1})$ exists.

答: A detailed answer to this question is given in this paper by Perret and Schehr.</p>


问:
a+b=a*b, a-b=? if a and b are positive integers
So this is a question off Facebook (I know). $a, b$ are positive integers.
The answer to the problem turns out to be $a-b=0$ as the only two obvious solutions are $a=b=0, 2$ (and we select two as it can't be zero).

My question is how do we prove this? I was trying

$a+b=ab$

$\dfrac{a}{a-1}=b$

But I can't proceed by substituting this in $a-b$ and solving the polynomial as I would have to first assume that $a-b=0$.

How do we prove this? How can we prove that the only value $a$ and $b$ can take is $2$ (and hence $a-b=0$)? Alternatively, how can we prove that if $a-b$ is non-zero, then $a$ and $b$ aren't integers? How can we prove that there are infinitely many solutions if the constraint is shifted to real numbers?

答:
To solve this question we need another condition.
$$a=b$$
then solve value of a and b
$$a+b=ab$$
replace a to b cause we know $$a=b$$
$$b+b=b*b$$
$$2b=b^2$$
$$b^2=2b$$
$$b^2-2b=0$$
$$b(b-2)=0$$
$$b=(0,2)$$
as we know $$a=b$$
meaning $$a=(0,2)$$
so, $$a-b=0$$


问:
Evaluating a triple integral $\int_0^1 \int_0^x \int_0^{1+x+y} f(x,y,z) dz \, dy\, dx$
\begin{align*}
  \int_0^1 \int_0^x \int_0^{1+x+y} f(x,y,z) dz \, dy\, dx
\end{align*}

where $f(x,y,z)=1$

My work so far

\begin{align*}
  &\qquad= \int_0^1 \int_0^x \left(z\Big|_{z=0}^{z=1+x+y}\right)dy\, dx\\
  &\qquad= \int_0^1 \int_0^x (1+x+y) dy\, dx\\
  &\qquad= \int_0^1 \biggl[y + yx + \frac{y^2}{2}\biggr]_{y=0}^{y=x}  dx\\
  &\qquad= \int_0^1 \biggl(x + x^2 + \frac{x^2}{2}\biggr) dx\\
  &\qquad= \int_0^1 \biggl(x + \frac{3x^2}{2} \biggr) dx\\
  &\qquad= \biggl[\frac{x^2}{2} + \frac{x^3}{2} \biggr]_0^1
  = \frac{1}{2} + \frac{1}{2} = 1
\end{align*}

Would I be on the right path here?

答:
Yes, you did it correctly.

For this particular triple integral, you simply calculate it from the inside iteratively.

Since $f=1$, this integral calculates the volume of the subset in $\mathbb{R}^3$:
$$
S=\{(x,y,z):0<x<1, 0<y<x,0<z<1+x+y\}
$$


问:
Most efficient method for quintic real root isolation?
I have a quintic polynomial whose real roots I would like to isolate. Having seen so many methods on the internet tailored towards different applications, I'm unsure exactly which direction to take.

Options I've seen have been:

Sturm sequences

Sturm sequences combined with the continued-fraction method

Vincent-Collins-Akritas method

Isolator polynomials (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.133.2233&rep=rep1&type=pdf)

Newton interval method

etc. etc.

Which algorithm should I employ to solve my equation with the most efficiency and robustness?

Edit:

Of course, it is often difficult to measure "efficiency" without benchmarks etc. but I am mostly talking about generally-accepted computational superiority i.e. I have often seen that it is rather obvious when certain algorithms are prefered to others, but due to my lack of experience I cannot determine this for myself.

答:
HINT

Making a rough sketch, three range intervals to find number of real roots labelled :

5 roots /inner

3 roots/ peripheral

1 root / outermost

to decide how many real roots have to be iterated.

Several schemes like Newton-Raphson can be employed.


问:
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$
How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?

It is part of a question about finding limits at mooculus.

答:
Here's a harder and similar way, but it still makes sense if you don't have what Julien said memorized.

$$x^4 - 1296 = x^4 - 6^4 \Rightarrow x^4 = 6^4$$Now two roots that seem obvious are $6$ and $-6$. But $x^4 =  x^4 \cdot 1 = x^4 \cdot i^4 = (xi)^4$. So two more roots are $6i$ and $-6i$.

Hence, the factorization is $$\begin{aligned}&(x - 6)(x + 6)(x+6i)(x - 6i) \\  =& (x-6)(x+6)(x^2+36) \\ =& (x+6)(x^3 - 6x^2 +36x-216)\end{aligned}$$


问:
Derivative of the Lagrangian dual: Can I treat the optimal parameter $x^*$ in the dual $g(\lambda) = L(x^*, \lambda)$ as a constant w.r.t $\lambda$?
Suppose I have a standard constrained optimization problem
$$
\min_x f(x) \quad \text{s.t.} h_i(x) \ge 0,
$$
where $f$ and $h_1,...,h_m$ are convex functions $\mathbb{R}^n \to \mathbb{R}$. The dual is given as
$$
g(\lambda) = \min_xL(x, \lambda) = L(x^*, \lambda)
$$
with $L$ the Lagrangian. To obtain the optimal Lagrangian multipliers, I need to solve
$$
\lambda^* = \text{argmax}_\lambda \;g(\lambda).
$$
I would go on and take the derivative w.r.t $\lambda$. My question is now:

Can I treat the optimal parameter $x^*$ in the dual $g(\lambda) = L(x^*, \lambda)$ as a constant w.r.t $\lambda$?

On my first thought: No, since it depends on $\lambda$. (If it didn't we would already have solved the primal problem). But I have found several examples, where we can take the derivative without the substitution of $x^*$ and treat it as a constant. So I was wondering if that is a general rule. If that is somehow true, this would simplify the derivative a lot.

I'll give a concrete example:
Let's solve
$$
\min x^2 \quad \text{s.t.} -x -2 \le 0
$$
The resulting Lagrangian is $ L(x, \lambda) = x^2 + \lambda(-x -2)$.
Taking the derivative w.r.t $x$, and setting it to 0 yields
$ 2x - \lambda = 0 \implies x^* = \lambda / 2$.

The dual is therefore
$$
g(\lambda) = x^{*2} + \lambda(-x^* -2) = \frac 1 4 \lambda^2 + \lambda(-\frac \lambda 2 -2).
$$
Now, when i take the "correct" derivative by taking the derivative of the right side w.r.t. to $\lambda$, I'll get
$$
g'(\lambda) = - \frac \lambda 2 - 2.
$$
This is the same as if I take $x^{*}$ as a constant on the left side and take the derivative, resulting in $$-x^* - 2 = - \frac \lambda 2 - 2.$$

Why does this work?
Any help is much appreciated!

答:
In general, the Lagrangian is
$$L(x,\lambda) = f(x) - \sum_i \lambda_i h_i(x).$$
and $x^*$ satisfies $f'(x^*) - \sum_i \lambda_i h_i'(x^*) = 0$. Assume $x(\lambda)$ is a function that satisfies that equation for all $\lambda$, then
$$g(\lambda) = f(x(\lambda)) - \sum_i \lambda_i h_i(x(\lambda)).$$

By application of the chain rule and the product rule:
\begin{align}
g'(\lambda) &= f'(x(\lambda))x'(\lambda) -\sum_i h_i(x(\lambda))- \sum_i \lambda_i h_i'(x(\lambda))x'(\lambda) \\
&= \left( f'(x(\lambda)) -\sum_i \lambda_i h_i'(x(\lambda)) \right)x'(\lambda) - \sum_i h_i(x(\lambda)) \\
&=  - \sum_i h_i(x(\lambda)),
\end{align}
which is what you get if take the derivative of $g$ without applying the chain rule and plug in $x(\lambda)$.


问:
Equivalent condition to an operator $T$ on a complex Hilbert space being compact
Let $H$ be a complex Hilbert space and let $T:H\rightarrow H$ be a bounded linear map. The problem asks to show

$T$ is compact if and only if for any sequence $x_n$ weakly converging to $0$, $\langle Tx_n,x_n \rangle\rightarrow 0$.

While the forward implication seem to be relatively straightforward, I'm having trouble proving the reverse implication. By reflexivity of $H$ it suffices to show that $\langle Tx_n,x_n\rangle \rightarrow 0$ implies $\|Tx_n\|\rightarrow 0$. Also because the problem sort of suggests that something might be different for real Hilbert spaces, I tried using the polarization identity (but couldn't make it work). Any help would be appreciated. Thanks in advance!

答:
In fact, use spectra decomposition, the following stronger result is also true (for complex case): $T$ is compact if $\langle Te_n, e_n \rangle$ tend to zero for every orthonormal basis $\{e_n\}$, then $T$ is compact.

For the proof, use the fact $$T=(T+T^*)/2+i(T-T^*)/2i,$$ we can assume $T$ is self-adjoint. Then suppose $$T=\int_{\mathbb{R}}x dE$$ is the spectra decomposition of $T$. Argue by contradiction, assume $T$ is not compact.
We claim: there exists $\epsilon > 0$, such that either projection operator $E(-\infty,-\epsilon)$ or $E(\epsilon,+\infty)$ is infinite dimensional. The proof of the claim is again argue by contradiction and approach T by finite-rank operator, which can be found in Conway’s functional analysis Chapter IX, proposition 4.1.
Now set $\epsilon > 0$ as above. We may assume the range of $E(\epsilon,+\infty)$ is infinite dimensional. Choose a orthonormal basis of ${\rm Ran} E$, say $\{ e_n\}$. Use functional calculation, one can check $\langle Te_n,e_n\rangle \geq \epsilon$. Extend $\{e_n\}$ to be a basis of whole $\mathcal{H}$, we get a contradiction.


问:
$K/F$ is a Galois extension, show that any intermediate field $L$ is generated by the traces of elements from $K$ over $L$
This is an exercise in Morandi's Field & Galois Theory. The question has already been answered: if $K/F$ is a Galois extension, show that any intermediate field $L$ is generated by the traces of elements from $K$ over $L$.

Although the answer using Newton's identity is illuminating, it cannot solve the problem when $\operatorname{char}F$ is $p$. By the way, to deal with the case when $\operatorname{char}F$ is 0, considering $\operatorname{Tr}_{K/L}(\alpha)$ for $\alpha \in L$ is quite easy. Futhermore, can the similar statement hold for norm, that is to say, any intermediate field $L$ is generated by the norms of elements from $K$ over $L$, which is the following exercise in the book. The former technique of using Newton's identity cannot work this time and it's hard to disprove this question.

答:
For $K/F$ separable the primitive element theorem gives that

$$K=F(a)=L(a)=L[x]/(f(x)), \qquad f(x)=\prod_{j=1}^d (x-a_j)=\sum_{m=0}^d c_m x^m\in L[x]$$
$$[K:F(c_0,\ldots,c_d)]\le d \quad \implies \quad L=F(c_0,\ldots,c_d)$$

From the Newton identities, if all the traces are $0$ then each $k c_k=0$ so that $f(x)=g(x^p)$ and $K/L$ is not separable.

Whence $K/F$ separable implies that $Tr_{K/L}$ is not identically zero and hence it is surjective $K\to L$.

As you see the Galois hypothesis is superfluous.


问:
What is the motivation behind the definition of the "Neighbourhood Space"?
Bert Mendelson's "Introduction to Topology" (3ed, 1975, Dover pubs) spends a section in both his chapter on metric spaces and that on general topological spaces (chapters 2 and 3 respectively) hammering out the axioms for a neighborhood space.

As follows:

A neighborhood space is a set $S$ such that, for each $x \in S$, there exists a set of subsets $\mathcal N_x$ of $S$ satisfying the following conditions:

$\text N 1$: There exists at least one element in $\mathcal N_x$

$\text N 2$: Each element of $\mathcal N_x$ contains $x$

$\text N 3$: Each superset of $N \in \mathcal N_x$ is also in $\mathcal N_x$

$\text N 4$: The intersection of $2$ elements of $\mathcal N_x$ is also in $\mathcal N_x$

$\text N 5$: There exists $N' \subseteq N \in \mathcal N_x$ which is in $\mathcal N_y$ for each $y \in N'$

He goes on to establish that a neighborhood space is exactly the same as a topological space: the neighborhood space induced by the topological space induced by a neighborhood space is that neighborhood space, and so on.

So the neighborhood space axioms define exactly the same space as the well-known open set axioms of topology.

Proving that a collection of subsets of a set adhere to the neighborhood space axioms is fiddly and tedious, and conceptually non-intuitive. Particularly axiom $N5$, which is horrible.

The question is: is there a good reason for a "neighborhood space" to be defined? Does it have any particular uses that make it essential or worthwhile to study in detail? Or is it just an interesting backwater that Mendelson (and also Willard, it seems, by taking a glance at the Wikipedia page on "Neighborhood system") defines for the fun of it?

Hence, how much effort would one be expected to exert in an attempt to master fluency in handling such constructs?

(No doubt there may be some difficult-to-analyse spaces which are analysed more easily using neighborhood axioms rather than open set axioms -- but are they sufficient bang for the buck?)

答:
This is a just supplement to William Elliot's answer; I furnish some details.

A topological space is nowadays usually defined as a set $X$ with a topology $\mathfrak T$ which is a set of subsets of $X$, called open sets, satisfying the well-known axioms. However, this "modern concept" did not emerge out of nothing, but needed a long time to get its final (if you want: mature) form. Have a look at this short survey. Quotation:

Riesz, in a paper to the International Congress of Mathematics in Rome (1909), disposed of the metric completely and proposed a new axiomatic approach to topology. The definition was based on an set definition of limit points, with no concept of distance. A few years later in 1914 Hausdorff defined neighbourhoods by four axioms so again there were no metric considerations. This work of Riesz and Hausdorff really allows the definition of abstract topological spaces.

After that beginning in the early 20-th century, many classic textbooks picked up these approaches. For example, Alexandroff and Hopf's "Topologie" from 1933 introduced topological space via (in modern terms) a closure operator. If you can read German, have a look at this, page 37. Seifert and Threlfall's "Lehrbuch der Topologie" from 1934 uses neigborhood systems to introduce the concept of a topological space. Wikipedia says It was one of the earliest textbooks on algebraic topology, and was the standard reference on this topic for many years. See here, page 22.

Anyway, there are many equivalent ways to introduce topological spaces. It is a matter of taste which you prefer. But admittedly the open set approach seems to be more elegant than the neigborhood approach. However, it is also a question of tradition and custom.


问:
Does the incenter divide the triangle into a ratio of 2:1?
I'm trying to get the radius of the circle.

First I use the Pythagorean theorem $13^2=5^2+x^2$ and $x=12$. Then I know the height of the triangle. To get the radius i use the Pythagorean theorem $r^2=5^2+(\frac{1}{3}\cdot12)^2$.

But here comes the problem, isn't the red part $\frac{1}{3}$ of the height, as the ratio of the incircle is $2:1$? My classmates don't agree, so who's right?

The triangle:

Edit: Is the black dot in the triangle the incenter or centroid?

答:
The circumradius R can be found (if we let a,b,c be the sides and A=area) with the formula
$$R=\frac{abc}{4A}=\frac{10\cdot13\cdot13}{4(5\cdot12)}
=\frac{1690}{240}\approx 7.041666667$$
The "red" part of your diagram is then
$\quad 12-7.041666667=4.958333333\quad$
so the ratio is more like $\space7:5\space$ than $\space2:1.\space$ The eye views proportions incorrectly, often making lower "heights" appear greater than they are.

Here
and here
and here
are links discussing the circumcircle radius you seek and below is a picture.


问:
Maths geometry past year SAT question having a semicircle inside a triangle

I tried very very hard to find the answer of this question but in vain. Can anyone pls show me how to solve this question?

My working:

If we draw perpendicular from $D$ to $BC$ and mark it $K$ then $BK = R$, if we draw a perpendicular from $E$ to $BC$ and mark the touching of $E$ perpendicular as $L$ then $KL = 2R$.

答: Good. You're on the right track. Now consider triangles $\triangle DBK$ and $\triangle ECL$. They're both right triangles, but what are their other angles? Are they special triangles with memorable trigonometric ratios? Can you form an equation in $R$ for the length of $BC$?


答:
That seems promising. $LC= LE \cot 60° = \frac{R}{\sqrt 3}$ and so $$BC= R+2R +\frac{R}{\sqrt 3} $$

But on the other hand, $BC=26$. So you just need to solve $$R\left(3+\frac{1}{\sqrt 3} \right)= 26$$

Title: Maths geometry past year SAT question having a semicircle inside a triangle

Body:
I tried very very hard to find the answer of this question but in vain. Can anyone pls show me how to solve this question?

My working:

If we draw perpendicular from $D$ to $BC$ and mark it $K$ then $BK = R$, if we draw a perpendicular from $E$ to $BC$ and mark the touching of $E$ perpendicular as $L$ then $KL = 2R$.


问:
Example of a finitely generated flat module which is not free
I couldn't come up with an example of a finitely generated flat module which is not free. I know that over local rings, freeness and flatness are equivalent. So the ring cannot be a local ring. </p>

答: Rings over which every module is flat are called absolutely flat, or Von Neumann regular. If $I$ is any nontrivial ideal of an absolutely flat ring, then $R/I$ is finitely generated flat over $R$, but not free (since $R/I \neq 0$ and $\mathrm{Ann}(R/I)=I \neq 0$). Products of fields are absolutely flat.</p>


问:
A dice containing the numbers 1 to 10 is rolled $n$ times such that the probability that 1 to 5 are all rolled at least once is $50 \%$. What is $n$?
With $n=5$ and $n=6$ the probability is $0.0012$ and $0.0054$ respectively, the difficulty arises when I try to solve for higher numbers of $n$. I solved above by $\frac{5!}{10^5}$ and $\frac{5 \cdot \frac{6!}{2}+5 \cdot 6!}{10^6}$ which isn't or at least not a really obvious closed form. A closed form and its derivation in terms of $n$ (if it exists) would be better though than choosing $n$ that reaches the $50 \%$.

答:
For $i=1,2,3,4,5$ let $E_i$ denote the event that face $i$ is not rolled in $n$ times.

Applying inclusion/exclusion and symmetry we find:
$$P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}^{\complement}\right)=1-P\left(E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\cup E_{5}\right)=$$$$\sum_{k=0}^{5}\binom{5}{k}\left(-1\right)^{k}P\left(\bigcap_{i=1}^{k}E_{i}\right)=\sum_{k=0}^{5}\binom{5}{k}\left(-1\right)^{k}\left(1-\frac{k}{10}\right)^{n}$$

(Substituting $n=5$ and $n=6$ you will find $0.0012$ and $0.0054$ respectively)

Edit: explicit answer to the question in the title is $n=21$ (see comments on this answer).


问:
Intuitive geometric interpretation of Sin(90°)
I know that there can't be two right angles in a right-angled triangle. But, I have seen many proofs that sin 90° = 1.

The proofs make sense analytically, but how do we know that sin(90°) (or, for that matter, sin 0° = 0) is constant as is sin ɸ such that 0° < ɸ < 90°, i.e., in the latter, we know the ratio of opposite side to the hypotenuse is fixed for a given angle but how can this be known in the above case

I have seen the unit circle animations, but historically, before extending trigonometry to the unit circle in the Cartesian plane, we must have figured this out using the standard right triangle definition of trig functions. Please do give an explanation in the right triangle definition.

答:
I assume by right triangle definitions of trig you mean, the rule that with length $1$ hypotenuse:

$$\sin(\theta) = A$$

Where A is the side opposite of theta.

If you use the law of sines on the right triangle you get that

$$\frac{\sin(90)}{H} = \frac{\sin(\theta)}{A}.$$

Plugging in the values we know:

$$\frac{\sin(90)}{1} = \frac{A}{A} = 1.$$

Hopefully I followed all of the necessary restrictions.


问:
Help understanding how to find missing number
$$12 + 12+(y*y*y) = 51$$
I tried working this out and I got that $y = 9$
using that answer I had to figure out what z is equal to $z + z +(z * y) = 25$.
Im struggling in working this out, was my answer my the previous question wrong (y) ?

答:
$$12+12+(y\cdot y\cdot y)=51$$

Subtract $12$ from each side twice

$$y\cdot y\cdot y = 27$$

Simplify as an exponent for my own sanity...

$$y^3=27$$

Note that this is $y\cdot y\cdot y = y^3$, not $y+y+y=3y$.  Perhaps you got addition and multiplication mixed up?

From here, we can take the cubic root of each side:

$$\sqrt[3]{y^3}=\sqrt[3]{27}$$

This simplifies, the cubic root of a (real) number cubed is the original number, $\sqrt[3]{y^3}=y$.  As for the cubic root of $27$, well, you can use a calculator for that or you can recognize the cubic roots of several small examples.  You should hopefully be able to recognize $1^3=1, 2^3=8, 3^3=27, 4^3=64, 5^3=125$ at a minimum since these numbers occur so frequently.  The cubic root of $27$ then is the number who when cubed is equal to $27$ and is then $3$.

$$y = 3$$


问:
Negation of the limit of a function involving infinity
Let $f:[0,\infty)\to\mathbb{R}$ be such that $\lim_{x\to\infty}\lvert f(x)\rvert=\infty$.

I am trying to understand the following:

(1) Why exactly is it, when negating the above, that there exists a sequence $x_{n}\to\infty$ such that $\sup_{n\in\mathbb{N}}\lvert f(x_{n})\rvert<\infty$?

(2) Given that $\sup_{n\in\mathbb{N}}\lvert f(x_{n})\rvert<\infty$, what does it mean that $f$ admits a ''sequential cluster point''?

Help is very much appreciated!

答:
$\newcommand{\R}{\mathbb R}$
$\newcommand{\N}{\mathbb N}$

For simplicity in the notations and the proof, I will assume that $\lim_{x \to +\infty} f(x) = +\infty$ (i.e. I remove the absolute value). Conceptually, this does not make a big difference, and I think that you will easily be able to adapt the ideas to the case where we suppose $|f| \to \infty$

(1) Recall the definitions:

$\lim_{x \to +\infty} f(x) = +\infty \Leftrightarrow \forall A \in \R, \exists N \in \R, \forall x \geq N, f(x) \geq A$

If we know that $f$ does not tend to infinity, then we negate the previous formula to obtain:

$\exists A \in \R, \forall N \in \R, \exists x \geq N, f(x) < A $

Informally, it means that exists some threshold $A$ such that no matter how far you go on the real line, you will always be able to find some point $x$ after that, whose image stays below $A$.

Consider the previous formula. You see that for any $N \in \R$, we get the existence of some $x$. In particular, for any $N \in \N \subset \R$, we get the existence of some point $x_N$ such that $x_N \geq N$ and $f(x_N) < A$.

This defines a sequence $(x_N)_{N \in \N}$ which verifies the following properties:

$x_N \to +\infty$ Indeed, we just said that $x_N \geq N$

$\sup(f(x_N)) < +\infty$ Indeed, we know that $f(x_N) < A$ so obviously $\sup_{N \in \N} f(x_N) \leq A$.

(2) I assume that a "sequential cluster point" is a point $y \in \R$ such that there exist some sequence $(x_n)_{n \in \N}$ with $f(x_n) \to y$

Here, $f$ does have a sequential cluster point. Indeed, consider the sequence $(f(x_N))_{N \in \N}$ that we previously constructed. This is a positive sequence (at least in your case, because you consider the absolute values), and it has a non-infinite upper bound $A$, hence it is bounded. By the Bolzano-Weierstrass theorem, you can extract a convergent subsequence $(f(x_{\phi(N)}))_{N \in \N}$. This sequence is exactly what we wanted, so $f$ has a sequential cluster point.


答:
Laufen,

yes, it's right.

I also checked WolframAlpha:
https://www.wolframalpha.com/widgets/view.jsp?id=3a8a6afa299813f2a9b0d6269f5b9341

Title: Evaluating a triple integral $\int_0^1 \int_0^x \int_0^{1+x+y} f(x,y,z) dz \, dy\, dx$

Body:
\begin{align*}
  \int_0^1 \int_0^x \int_0^{1+x+y} f(x,y,z) dz \, dy\, dx
\end{align*}

where $f(x,y,z)=1$

My work so far

\begin{align*}
  &\qquad= \int_0^1 \int_0^x \left(z\Big|_{z=0}^{z=1+x+y}\right)dy\, dx\\
  &\qquad= \int_0^1 \int_0^x (1+x+y) dy\, dx\\
  &\qquad= \int_0^1 \biggl[y + yx + \frac{y^2}{2}\biggr]_{y=0}^{y=x}  dx\\
  &\qquad= \int_0^1 \biggl(x + x^2 + \frac{x^2}{2}\biggr) dx\\
  &\qquad= \int_0^1 \biggl(x + \frac{3x^2}{2} \biggr) dx\\
  &\qquad= \biggl[\frac{x^2}{2} + \frac{x^3}{2} \biggr]_0^1
  = \frac{1}{2} + \frac{1}{2} = 1
\end{align*}

Would I be on the right path here?


问:
What is the notion for this arrow in category theory?
When reading a CS paper, I encountered an arrow like $\langle f_1, \ldots, f_n \rangle : \langle A_1, \ldots, A_n \rangle  \rightarrow \langle B_1, \ldots, B_n \rangle$.  The author took it for granted to mean a collection of arrows $f_1 : A_1 \rightarrow B_1$, ..., $f_n : A_n \rightarrow B_n$.  I wonder if there is such a notion in category theory.  If it is, I would appreciate a reference.  Thanks.</p>

答: Based on the information provided, the arrow $\langle f_1,\dots,f_n\rangle$ can be regarded as an arrow in the product category $\mathscr D=\mathscr C\times\mathscr C\times\dots\times\mathscr C$ if each arrow $f_i$ originates in $\mathscr C$.</p>


问:
Infinite sum over binomial coefficients up to $\frac{1}{2}n+\frac{1}{2}\sqrt{n}$

I want to evaluate the limit $$L=\lim_{n \rightarrow \infty} \frac{1}{2^n} \times \sum_{k=0}^{\frac{1}{2}n+\frac{1}{2}\sqrt{n}}\binom{n}{k}.$$

Maybe some elementary identies involving the binomial coefficients can help, but the only one I know, which looks similar to mine, is $$\sum_{k=0}^{\frac{1}{2}n}\binom{n}{k}=2^{n-1} \Rightarrow \lim_{n \rightarrow \infty} \frac{1}{2^n} \times \sum_{k=0}^{\frac{1}{2}n}\binom{n}{k}=\frac{1}{2}$$
and that gives me $L \geq \frac{1}{2}$. Also I found a generalized version in an answer of this question on MO.

In the answer it is showen that $L=g(\frac{1}{2})$, but I have no idea what the function $g$ is.

Thanks for your ideas, comments and answers!

答:
The function $g$ is $g(\alpha) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\alpha} e^{-x^{2}/2}dx$. The main idea is using the central limit theorem:

Let $X_i$ be a Bernoulli distribution with $p = 1/2$. Then $\mathbb{E}[X_i] = 1/2$ and $\sigma(X_i) = 1/2$, so we have
$$
\frac{\bar{X} - 1/2}{1/(2\sqrt{n})}\to Z\,(\text{standard normal distribution})
$$
in the sense that
$$
\mathbb{P}\left[\frac{\bar{X} - 1/2}{1/(2\sqrt{n})} \leq \alpha\right] \to \mathbb{P}[Z\leq \alpha] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\alpha}e^{-x^{2}/2}dx
$$
as $n\to\infty$. By the way, $X_1 + \cdots +X_n$ follows a binomial distribution with $n = n$ (sorry for the abusing notation) and $p = 1/2$, so $$
(LHS) = \mathbb{P}\left[X_{1} + \cdots + X_{n} \leq n\left(\frac{1}{2} + \frac{\alpha}{2\sqrt{n}}\right)\right] = \sum_{k=0}^{\frac{n}{2} + \frac{\alpha\sqrt{n}}{2}} \frac{1}{2^{n}}\binom{n}{k}
$$
and we get the result. Note that there's no closed form of $g(1)$ in terms of elementry functions. It is approximately 0.8413...


问:
How to evaluate limiting value of sums of a specific type
We know that if $f$ is integrable in (0,1) then

$$
\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k/n) = \int_{0}^{1}f(x)dx.
$$

Recently I found the following sum

$$
\lim_{n \to \infty} \sum_{k=1}^{n}\frac{k}{n^2 + k^2} = \frac{\ln 2}{2}.
$$

This sum cannot be expressed in the form $\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k/n)$. Rather it is of the form

$$
\lim_{n \to \infty} \frac{1}{n^2}\sum_{k=1}^{n} k f(k/n).
$$

Part 1

Is there any technique to evaluate limits of this form?

Part 2

Generalizing the question one step further, how can we asymptotics of summations of the form

$$
\sum_{k=1}^{n}  f(k)g\Big(\frac{k}{n}\Big).
$$

答:
Let's rewrite your sum. Note, that we have
$$
\frac 1{n^2} \sum_{k=1}^n k\cdot f(k/n) = \frac 1n \sum_{k=1}^n \frac kn \cdot f(k/n)
$$
So, if we define $g \colon [0,1] \to \mathbb R$ by $g(x) = xf(x)$, we get
$$
\frac 1{n^2} \sum_{k=1}^n k\cdot f(k/n) = \frac 1n \sum_{k=1}^n g(k/n), $$
which for $n \to \infty$ converges, as you noted to
$$ \int_0^1 g(x) \, dx = \int_0^1 x\cdot f(x)\, dx. $$</p>


问:
Notation of: a · b = 0 if a, b are orthogonal.
I would like to express the concept a · b = 0 if a, b are orthogonal  in set theory notation (predicate calculus ?).

The nearest I have managed so far is:

If $a\cdot b=0 \space \space\forall\ a,b\subseteq S, \space$then $a,b$ are orthogonal (where $S$ is some previously defined set). This looks horrible and is longer than the original sentence in words.

Can someone help me to make it look better and to correctly represent the original phrase?

答:
$$ \forall a,b \in S \colon a \cdot b = 0 \iff a \perp b $$

Although you might want to use the notation $\langle a,b \rangle$ instead of $a \cdot b$


问:
Questions about probability on defined choices.

A system administrator is going to install three laptops for three new
recruited  employees of his company. He selects 3 laptops from a table
that consists of 15  laptops at which these laptops are nicely
arranged in a row on the table.

i. How many choices does he have?

ii. What is the probability for this to happen if he must select the
first or the fifth or  the last laptop but not any two or all of these
laptops?

For i)
I knew it is nCr which is 15C3 = 455
for ii) I;m clueless except i know the first,fifth and last laptop cannot both be chosen both at the same time and cannot three of them be chosen

Which is cannot choose such combinations:

1st 5th 15th | 5th 1st 15th | 5th 15th 1st| 15th 5th 1st| 15th 1st 5th | 1st 5th xth | 5th 1st xth | 1st 15th xth | 15th 1st xth | 15th 5th xth | 5th 15th xth

I assume that it is 1 minus above combinations? But i don't know how to express it.

Or anything about my concepts are wrong?

答:
You're nearly there.

The number of ways to pick $3$ objects from a set of size $15$ is indeed $C(15,3) = 15!/12!/3! = 455$.

If you must pick exactly $1$ from a subset of size $3$ then you must pick the other $2$ from the rest of the set which is of size $12$. The number of ways is hence:

$C(3,1)\times C(12,2) = (3!/2!/1!)\times (12!/10!/2!) = 3\times 66 = 198.$

The probability that the second event happens randomly is therefore $198/455$, about $0.435$.


问:
Is the following result true about complex numbers?
Is the following result true?

If $i^4 = 1$, $(i^4)^{1/4} = 1^{1/4}$, and$i = 1$, then won't it disprove $i = (-1)^{1/2}$?

If no, give a proper reason?  Can this be a contradiction to complex numbers?

答:
No, because you cannot use the usual rules on exponents when you are working with complex numbers !

The definition $a^b = \underbrace{a \times \dots \times a}_{b \text{ times}}$ is always valid if $b$ is a positive integer, but in order to define what $a^{\frac 1 4}$ means, you need to find something else !

When working over the positive, real numbers, a way to define $a^{\frac 1 b}$ for some positive integer means is to say that it is the only positive real $x$ such that $x^b = a$. However, when working over the complex numbers, this $x$ is no longer unique, because there are non-trivial roots of the unity (numbers that are not $1$, but one of their powers is equal to $1$).

(If you know what the exponential function is)
An even more general way to define exponentiation is to say that $x^\alpha = \exp(\alpha \ln(x))$ for any positive, real number $x$. However, this definition also fails terribly when $x$ is a complex number, as there is no "nice" way of extending $\ln$ to the complex numbers.

In short: the rules have changed, you have to be very careful when manipulating complex numbers along with exponents.


问:
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$
How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?

It is part of a question about finding limits at mooculus.

Body:
$$\begin{array}{r|rrrr|r}
& 1 & 0 & 0 & 0 & -1296 \\
-6 & & -6 & 36 & -216 & 1296 \\
\hline
& 1 & -6 & 36 & -216 & 0
\end{array}
$$

This shows that
$$
x^4-1296=(x+6)(x^3-6x^2+36x-216)
$$

Of course you can also use

$$
a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3)
$$
with $a=x$ and $b=-6$.


问:
Integrate:$ \int_{0}^{1} x^{1/2}(1-x^{2})^{3/2}dx$
Integrate:
$\int_{0}^{1} x^{1/2}(1-x^{2})^{3/2}dx$

I tried substituting $x=\sin \theta$ and ended up with $\int_{0}^{\pi/2} \sqrt{\sin\theta}\cos ^{4}\theta d\theta$. Further substitution made it more complex.
After this what should I do?

答:
See the identities here: https://en.wikipedia.org/wiki/Beta_function

The key identities related to your problem are

\begin{align}
B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} &= \int_0^1 t^{x-1}(1-t)^{y-1} \;dt \\
&= n \int_0^1 t^{nx-1}(1-t^n)^{y-1} \;dt \\
&= 2 \int_0^{\pi/2} (\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\;d\theta.
\end{align}

So your integral is
$$ \frac12 \cdot 2 \int_0^1 x^{2\left(\frac34\right)-1} (1 - x^2)^{\left(\frac52\right)-1} \;dx = \frac12 B\left(\frac34,\frac52\right) = \frac{\Gamma\left(\frac34\right) \Gamma\left(\frac52\right)}{2\Gamma\left(\frac{13}4\right)} = \frac{3\sqrt\pi}8 \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac{13}4\right)}.$$


问:
ODE - getting reciprocal of correct answer
Consider the first order differential equation
$$(1+x^2)\frac{dy}{dx}=x(1-y^2), \quad y(0)=2.$$

I've tried to solve this by separating variables, and I get
$$y=\frac{3(1+x^2)-1}{3(1+x^2)+1}.$$

However, Wolfram Alpha gives the reciprocal of this.

During my solution, I ended up with
\begin{equation}\log|1+y|-\log|1-y|=\log|3(1+x^2)|,\end{equation}
which I simplified to
$$\log\left|\frac{1+y}{1-y}\right|=\log |3(1+x^2)|,$$
and subsequently got my answer.

However, if I had simplified this to
$$\log\left|\frac{1+y}{y-1}\right|=\log |3(1+x^2)|,$$
I get the same answer as WA.

What's going on here? What's wrong with my initial approach?

EDIT

After arriving at
$$\log\left|\frac{1+y}{1-y}\right|=\log |3(1+x^2)|,$$
I wrote
$$e^{\log\left|\frac{1+y}{1-y}\right|}=e^{\log |3(1+x^2)|},$$
and then
$$\frac{1+y}{1-y}=3(1+x^2).$$

Perhaps this is where the issue arises?

答:
You can check that your answer does not satisfy the initial condition $y(0)=2$, but the reciprocal of your answer does.

Your intermediate steps look correct to me. It is a matter of how you "subsequently got your answer".

Notes.

$$
\log\left|\frac{1+y}{1-y}\right|=\log |3(1+x^2)|=\log(3(1+x^2))$$
implies that
$$
\left|\frac{1+y}{1-y}\right|=3(1+x^2)
$$
which implies
$$
\frac{1+y}{1-y}=3(1+x^2),\quad \color{red}{(1+y)(1-y)>0}\tag{1}
$$
or
$$
\frac{1+y}{y-1}=3(1+x^2),\quad \color{red}{(1+y)(1-y)<0}\tag{2}
$$

When you get rid of the absolute value on the left, you need to consider the sign.


问:
Derivate of composition of vector functions
I have troubles to find derivate of composition of two vector functions:$f(x,y)=(x^2,y^2), g(a,b)=(a\cos b,a\sin b)$ where $a>0, b\in(0,2\pi)$. So i need to find $d(f o g)$.
I am not even sure what formula to use, so any link on already done examples or explanation would be helpful.

答:
We have

$$(f \circ g)(a,b)= (a^2 \cos^2 b, a^2 \sin^2 b).$$

Now compute the Jacobian matrix of $f \circ g.$