问:
Consistency of theories with circles
We define the circle with the center point - $(a,b)$ and his radius - $r>0$ as:
$$B(a,b,r)=\{\langle x,y \rangle\ \in \mathbb{R}^2:(x-a)^2+(y-b)^2\le r^2\}$$
Let $A$ be the set of the plane's circles:
$$A=\{B(a,b,r):a,b\in \mathbb{R},r\in(0,\infty )\}$$
Let $\mathcal{L}$ be the vocabulary which contains a binary relation $S$, and in addition contains for each circle - $B(a,b,r)\in A$, a constant symbol $\bar c_{a,b,r}$. Let $M$ a model to interpret $\mathcal{L}$, defined by:$$M=\langle A,S\rangle$$ When $S$ defined to be the $\subseteq$ binary relation between the circles ($S(x,y)$ meaning that $x\subseteq y$). The model also interpret $\bar c_{a,b,r}$ as a circle $B(a,b,r)$. Let $T=Th(M)$ the theory of $M$. We define $T_1$, $T_2$ in the vocabulary $\mathcal{L}^+=\mathcal{L}\cup\{d\}$, where $d$ is a new constant symbol:
$$T_1=T\cup\{\forall x [S(x,d)]\}$$
$$T_2=T\cup\{S(d,c_{a,b,r}):a,b\in \mathbb{R},r\in(0,\infty )\}$$
Now, for $T_1, \ T_2$ I need to determine if:

The theory is inconsistent

The theory is consistent, and it has a model which is an expansion of $M$ to $\mathcal{L}^+$.

The theory is consistent, but it does not have a model which is an expansion of $M$ to $\mathcal{L}^+$.

$\textbf{My attempt}$: I wanted to think about the meaning before diving into giving $d$ interpretation and so on.

Therefore, for $T_1$: the additional sentence to $T$ means that for every circle $x$ is within the biggest circle. Now in the world discussion of $B(a,b,r)$, there isn't the biggest one, so if I am appealing to the compatible theorem, I can choose a circle that will contain all finite groups of circles within him. Therefore $T_1$ is consistent but she doesn't have an expansion of $M$ to $\mathcal{L}^+$.

For $T_2$: the additional sentence to $T$ tell us that there is the smallest circle, which impossible, also if we applying the compatible theorem because finite groups of circles can scatter around the cartesian coordinate system, such that we won't be able to choose one that contained in all circles. Hence, we have that $T_2$ is inconsistent.

I am not sure if my way is correct. Thus, I will be glad to get some help from you. Thank you!

答:
Both theories are inconsistent.

For $T_1$, here's a hint: If $N\models T_1$, then $N$ is a model of $T$ in which $d$ is an $S$-maximal element. But a model of $T$ cannot have an $S$-maximal element. Why not?

For $T_2$, your answer is correct, but I would expect a more rigorous explanation. You wrote:

"if we applying the compatible theorem because finite groups of circles can scatter around the cartesian coordinate system, such that we won't be able to choose one that contained in all circles."

What is the "compatible theorem"? Do you mean the compactness theorem? What does "finite groups of circles can scatter around the cartesian coordinate system" mean precisely?

One way to give a very clear explanation is to find finitely many sentences in $T$ and finitely many of the new axioms $S(d,c_{a,b,r})$ in $T_2$ which are inconsistent. (Because of the compactness theorem, we know that we can always explain inconsistency of $T_2$ by finitely many axioms.)  In fact, in  this  case a single sentence of $T$ and $2$ of the new axioms in $T_2$ will be enough.

It's surprising to me that your question asks about $$T_2=T\cup\{S(d,c_{a,b,r}):a,b\in \mathbb{R},r\in(0,\infty )\}$$ instead of $$T_2'=T\cup\{S(c_{a,b,r},d):a,b\in \mathbb{R},r\in(0,\infty )\}.$$ In fact, $T_2'$ is consistent. Can you prove this?

Taken together, the fact that $T_1$ is inconsistent but $T_2'$ is consistent makes a nice point: The theory $T$ rules out having a circle which contains all other circles, but it's possible to find a model with a circle $d$ which contains all of the standard circles (the elements named by constants in the standard model $M$).

Title: Consistency of theories with circles


问:
$C^{1}[a,b]$ equipped with the norm given by $\lVert x\rVert _{\infty} = \sup_{t\in [0, 1]} \lvert x(t) \rvert$ is an incomplete normed space.
I have to show that the real linear space $C^{1}[a,b]$ of all continuously differential functions defined on $[0, 1]$ equipped with the norm given by $\lVert x\rVert _{\infty} = \sup_{t\in [0, 1]} \lvert x(t) \rvert$ is an incomplete normed space.

I have taken sequence $\{x_{n}\} = \sqrt{t^2 +\frac{1}{n}}$ in $C^{1}[a,b]$.

Please tell me whether my procedure to show that $\{x_{n}\}$ is Cauchy in $C^{1}[a,b]$ is correct or not.
Here is my attempt.

Let $n \geq m$

$\lVert x_n - x_m\rVert  = \lVert \sqrt{t^2 +\frac{1}{n}} - \sqrt{t^2 +\frac{1}{m}}\rVert
 = \frac { \frac{1}{n} - \frac{1}{m}}{\sqrt{t^2 +\frac{1}{n}} + \sqrt{t^2 +\frac{1}{m}}}\leq \frac{1}{n} +  \frac{1}{m} \leq \frac{1}{2m}\leq \frac{1}{m} \leq \epsilon $

Further I have shown that $\{x_{n}\}$ converges pointwise to $x(t) =  \frac{1}{\sqrt n}$. But $x(t)$ is not differentiable at $t = 0$ and hence the proof.

Also could anybody provide me any other example?

Thanks.

答:
It does look like there are some typos. For instance, if $n \geq m$, then $\frac{1}{m} \geq \frac{1}{n}$ and so your fraction is negative. This is bad, since it's supposed to be equal to the absolute value of something! The next inequality also seems fishy, since if it were true we would have $$\frac{1}{m} - \frac{1}{n} \leq \bigg(\frac{1}{n} + \frac{1}{m}\bigg)\bigg(\sqrt{\frac{1}{n}} + \sqrt{\frac{1}{m}}\bigg).$$ If we let $n \rightarrow \infty$, this will end up being $$\frac{1}{m} \leq \frac{1}{m\sqrt{m}}.$$ So $\sqrt{m} \leq 1$, which doesn't hold up.

But your idea works fine. Try this: you know that $t^2 + \frac{1}{n}$ converges uniformly to $t^2$. You also know that $\sqrt{x}$ is uniformly continuous on intervals $[0,A]$. Perhaps use these to show that $\sqrt{t^2 + \frac{1}{n}}$ converges uniformly to $\sqrt{t^2} = |t|$? Then you do not even need to bother showing it's Cauchy, since all convergent sequences are automatically Cauchy!

Here's my example: Let $g_n$ be the sequence of functions defined by $$g_n(x) = \begin{cases} 0 &\mbox{ if } -1 \leq x \leq \frac{-1}{n} \\
\frac{n}{2}x + \frac{1}{2} &\mbox{ if } \frac{-1}{n} \leq x \leq \frac{1}{n} \\
1 &\mbox{ if } \frac{1}{n} \leq x \leq 1 \end{cases}.$$

If I've done it right, this should be a piecewise function defined by staying at zero until you hit $\frac{-1}{n}$, going linearly up to 1 until you hit $\frac{1}{n}$, and then staying at 1. These functions are continuous, and so $$f_n(x) = \int_{-1}^x g_n(t) dt$$ are all differentiable.

However, the limit of the $g_n$'s is the step function $g = \chi_{[0,1]}$, whose integral $$f(x) = \int_{-1}^x g(t) dt$$ is not differentiable. But $$\|f_n - f\| \rightarrow 0.$$ So $f_n$ converges in the $C^0$ norm and is therefore Cauchy. The idea is to make the bad behavior "one level down" and integrate it.

ContentLicense: CC BY-SA 3.0


问:
Linear system of inequalities.
Given linear system:

$$4+\delta_1-\delta_2\ge 0 \\ 2-\delta_1+\delta_2 \ge 0 \\ 1+\delta_1-\delta_2+\delta_3 \ge 0$$

How from there it follows that $\delta_2=\delta_3 = 0$ and $\delta_1\neq0$?

答: The feasible set is something prismatic ...

ContentLicense: CC BY-SA 4.0

Title: Linear system of inequalities.


问:
Isomorphisms of CW Coverings
This is an exercise I've been working on from Hatcher (1.3.32):

Consider covering spaces $p: \tilde{X} \to X$ with $\tilde{X}$ and $X$ connected CW complexes, the cells of $\tilde{X}$ projecting homeomorphically onto $X$ cells. Restricting $p$ to the 1-skeleton then gives a covering space $\tilde{X}_1^1 \to X^1$. Show the following:

(a) Two such covering spaces $\tilde{X}_1 \to X$ and $\tilde{X}_2 \to X$ are isomorphic iff the restrictions $\tilde{X}_1^1 \to X$ and $\tilde{X}_2^1 \to X$ are isomorphic.

(b) $\tilde{X} \to X$ is normal iff $\tilde{X}^1 \to X^1$ is normal.

(c) The groups of deck transformations of $\tilde{X} \to X$ and $\tilde{X}^1_1 \to X^1$ are isomorphic via the restriction map.

For (a) and (b), one direction of the iff is trivial.

For (b), I think I can show the converse: Assume $\tilde{X}^1 \to X^1$ is normal. Then given $\tilde{x_1}, \tilde{x}_2 \in \tilde{X}$ such that $p(\tilde{x}_1) = p(\tilde{x}_2) := x \in X$, find a path $\ell$ in $X$ from $x$ to a point $x_0$ in the 1-skeleton. Then there exist unique lifts $\tilde{\ell}_1$ and $\tilde{\ell}_2$ beginning at $\tilde{x}_1$ and $\tilde{x}_2$ respectively, and ending at some points $\hat{x}_1$ and $\hat{x}_2$ in the 1-skeleton of $\tilde{X}$. By assumption, there's a deck transformation of the 1-skeleton taking $\hat{x}_1$ to $\hat{x}_2$. By (a), this would correspond to a deck transformation $h$ of $\tilde{X}$. Uniqueness of lifts and the fact that $ph = p$ would force this deck transformation to take $\tilde{x_1}$ to $\tilde{x_2}$.

For (c), clearly the restriction map from deck transformations of $\tilde{X}$ to deck transformations of $\tilde{X}^1$ is injective since a deck transformation (of a connected space) is determined by its action on a point. Preservation of compositions also clear. Surjectivity would follow from (a).

So what's left is proving the nontrivial implication in (a): given a covering isomorphism $\phi: \tilde{X}_1^1 \to \tilde{X}_2^1$, how can I produce an isomorphism $\psi: \tilde{X}_1 \to \tilde{X}_2$?

I want $\psi |_{\tilde{X}_1^1} = \phi$. The space $\tilde{X}_1$ (and $\tilde{X}_2$) is built up by attaching disks; at the first step, disks $D_\alpha$ are added via attaching maps $\theta_\alpha: \partial D_\alpha \to \tilde{X}_1^1$.

This is where I get stuck. I think I want to argue that $\tilde{X}_2$ is built up from its 1-skeleton by attaching disks via maps $\psi \circ \theta_\alpha$, but I'm not sure...

Alternatively, I thought about trying an argument based on the fact the the fundamental group of a CW complex is the group of its one skeleton modded out by relations from its 2-skeleton, and using the fact that coverings are isomorphic iff the projections of their fundamental groups are the same, but I couldn't see my way through that argument either.

Any advice on how to proceed would be appreciated. Thanks.

答:
There is a simple approach to the problem (part (a) non trivial direction).

Suppose there is a graph isomorphism $\phi: {X_1}^1 \to {X_2}^2$ between the 1-skeletons.  We need to extend this isomorphism to the whole CW complex covering. For that we start with a 2-skeleton, say ${X_i}^2; \, i=1,2.$

Let $x$ be a point in the 2-skeleton ${X_1}^2 $ and $y$ be a point in 1-skeleton ${X_1}^1$, then there is path $\gamma$ from $y$ to $x$. So, $p(\gamma)$ is path in X with starting point in $X^1$.

Under the other covering map $q$, lift this path $p(\gamma)$, say $\gamma_q$, to starting point $\phi(y) \in {X_2}^1$. Now this lift is unique by unique lifting criterion. So, define extension $\phi : {X_1}^2 \to {X_2}^2$ by $\phi(x)= \gamma_q(1)$.

A similar approach can be used to extend map $\phi : X_1 \to X_2$.

Title: Isomorphisms of CW Coverings


问:
Does a rectangle's diagonal to its perimeter have a ratio, like Pi?
The ratio of a circle's diameter to circumference merited its very own greek letter, Pi.

I have been wondering if there is such a similar, or important, ratio in a rectangle, i.e. its diagonal to perimeter?

Would I be right in guessing that it's because it's not a single value, but a more complex formula involving trigonometry? Or perhaps not a widely used one?

Many thanks in advance.

答:
You can see with a couple examples that there is no such "rectangle $\pi$" - or more precisely, that different rectangles can have different perimeter/diagonal ratios.

First, consider a square. By the Pythagorean theorem we know that the perimeter/diagonal ratio of a square is just ${4\over \sqrt{2}}=2\sqrt{2}.$

Now consider a "really really thin" rectangle - e.g. a rectangle of length $1$ and height $\epsilon$ for some very small positive $\epsilon$. The length of a diagonal is "basically" $1$ and the perimeter is "basically" $2$, so we get a perimeter/diagonal ratio of "basically" $2$. Of course that's not immediately rigorous, but it should still be pretty convincing - and you can now verify it by e.g. seeing what happens if we set $\epsilon={1\over 2}$.

That said, any two similar rectangles will share the same perimeter/diagonal ratio. For example, "$\pi$ for squares" makes sense and is $2\sqrt{2}$, per the above.

Note that any two circles are automatically similar to each other. The picture becomes nicer if we think of the analogy $$\mbox{circles:ellipses::squares:rectangles.}$$ Just as with rectangles, we can find ellipses with different perimeter/"maximal-diameter" ratios - but any two similar ellipses (e.g. two circles) will have the same ratio.


问:
How can this integral be computed?
How can this integral be computed?
$$\int _1^b\:\left(x^2-1\right)^pdx$$
Where $p>0$ but different than $1$ and $b\:>\:1$.
When I say computed I mean in any relevant way, may it be a formula, a series, with a computer algorithm etc.
When $p =\frac{1}{2}$ I was able to using trig sub and when $p$ was natural I was able to do so using the binomial theorem. What about other cases?

答:
If $p\in\mathbb{C}$ one could use the extended binomial series
$$(x^2-1)^{p}
=\sum_{k=0}^{\infty}\binom{p}{k}x^{2(\alpha - k)}(-1)^{k}\quad (2)$$
to split the integral up. This series converges for all $x\in \mathbb {R}$
with $x>0$ and
$\left|{\tfrac {1}{x}}\right|<1.$

The binomial coefficient is defined for complex values as
$$\binom\alpha z=\frac{1}{(\alpha+1)\,\mathrm B(z+1,\alpha-z+1)}=\frac{\Gamma(\alpha+1)}{\Gamma(z+1)\,\Gamma(\alpha-z+1)},$$
where $\mathrm{B}$ denotes the Euler-Beta function and $\Gamma$ the Gamma-function.

Title: How can this integral be computed?


问:
measure defined in terms of an integral
Here's a problem from a midterm I took 2 decades ago!  Let $A$ be a Borel measurable set on the real line and $\lambda$ be the Lebesgue measure. Define a measure $\mu(A)=\int_A f d\lambda$ where $f(x)=1/x^2$ for $x\ne 0$ and $f(0)=\infty$.

Is $\mu$ finite? $\sigma$-finite?

Now I know it's not finite, since if $A$ is any interval containing $0$ then the integral is infinite.  I'm not sure about $\sigma$-finite though. Any interval that does not contain $0$ will have finite measure, and there's a countable number of these to cover everything but $\{0\}$, so it seems to me that the answer depends on the following....

What is $\mu(\{0\})$?

答:
OK - I was thrown off by the facts that

$f$ takes values on the extended reals and

$f\chi_A$ is not necessarily integrable

but I should have gone back to basics.
Here's my attempt at filling in the details:

Despite 1. and 2. above, $f\chi_A$ is still measurable for all Borel measurable sets $A$.  In particular, $f\chi_{\{0\}}$ is Borel measurable, and can be approximated by simple functions $s_n$ such that $0 \le s_n \le f\chi_{\{0\}}$ (take $s_n = n\chi_{\{0\}}$ for example).  Since the support of $s_n$ is a Borel set of measure $0$, $\int_{\mathbb{R}} s_n=0$ for all $n$, and $\int_{\mathbb{R}} f\chi_{\{0\}} = sup \int_{\mathbb{R}} s_n=0$.  This shows that $\mu$ is absolutely continuous with respect to $\lambda$.

Title: measure defined in terms of an integral


问:
How to prove an identity by induction
I am working on the following problem:
$\text{For $n$ $\ge$ 1, Prove the following:}$

$n(1+x)^{n-1} = \sum_{k=1}^{n} k {n\choose k}x^{k-1} $

$\text{Deduce that: } \sum_{k=1}^{n}k{n \choose k}=n2^{n-1} $

I followed the proof by induction as follows:
$\text{Base Case: let n=1 , both RHS and LHS will evaluate to 1}$

$\text{Assume true for n=l. Now consider n=l+1}$

$(l+1)(1+x)^l=\sum_{k=1}^{l+1}k {l+1 \choose k} x^{k-1}$

$(l+1)(\sum_{k=0}^{l} {l \choose k} x^k)=\sum_{k=1}^{l+1}k {l+1 \choose k} x^{k-1}$

After this, I am kinda lost. any ideas on how to continue the proof?

答:
From the inductive step we have $(1+x)^{l-1}=\frac1l\sum_{k=1}^lk\binom lkx^{k-1}$.

Thus$$\begin{align*}(l+1)(1+x)^l&=(1+x)\left[(l+1)(1+x)^{l-1}\right]\\&=(1+x)\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]\\&=\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]+x\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]\end{align*}$$The coefficient of $x^{k},1\le k\le l-1$,$$a_k=\left(\frac{l+1}l\right)\left[(k+1)\binom l{k+1}+k\binom lk\right]=(k+1)\binom{l+1}{k+1}$$which matches with the statement we need to prove. Similarly you can check the coefficients of $x^{0},x^l$.

Title: How to prove an identity by induction


问:
Exercise 1.2.1 in Shafarevich's Basic Algebraic Geometry I
I'm working on the following question from Shafarevich's Basic Algebraic Geometry I and wanted to check that I understand what's going on.

The set $X\subset{\mathbb A}^2$ is defined by the equation $f:x^2+y^2=1$ and $g:x=1$. Find the ideal $\mathfrak{U}_X$. Is it true that $\mathfrak{U}_X = (f,g)$?

First, is $X$ defined as the common zeros of $f$ and $g$? That is what I would assume but in this case, $X$ would only be a single point $(1,0)$ and that doesn't sound right for this problem.

If $X$ just is a single point, I'm not exactly sure what $\mathfrak{U}_X$ is. I would be surprised if it was $(f,g)$ but I am not sure how to prove that there are more polynomials in $\mathfrak{U}_X$ than there are in $(f,g)$. Could I say for example that $y$ is not in $(f,g)$ because elements of $(f,g)$ have even degrees as polynomials with coefficients in $k[x]$?

答:
You have the right idea. As $X$ consists of a single point, it should not be hard to give explicit generators for $\mathfrak{U}_X$. If you want to get hands-on with the details, you can start by noting that
$$\mathfrak{U}_X\supset(f,g)=(x^2+y^2-1,x-1)=(y^2,x-1).$$
Can you find explicit generators for $\mathfrak{U}_X$ from here? Can you show that the inclusion above is not an equality?


问:
Categorize circle on a plane with "left" and "right" relative to a line.
CONTEXT:

I'm doing a simulation in Object Oriented Programming. I have a bunch of circles with properties:

Location = Vector with x and y coordinate

Angle = Vector détermining "where" the circle is looking at

DetectionRadius = A bigger circle, we want to interact only with others circle located within that circle.

So i have my central circle and secondary circle all around the plane. I have access to all circle properties in an array. My question is:

PROBLEM:

How can i determine wich circle is on the left side of my central circle, and wich are on the right side ?

Here is an image to understand better what i mean: Image1

(Principal circle is in the center, black line is the angle vector (we can extend it to a line if needed), blue circle is the detection radius and other green circle are secondary circles)

I already know wich secondary circle are in the principal circle's detection radius. I just need to know wich are on the left, wich are on the right.

Thank you !

答:
If you take the "viewpoint" vector $\vec v$ and construct a unit-length vector orthogonal to this, you can take the (relative) vector position of the centres of the other circles $\{\vec c_i\}$ and get not only whether they are left or right of the line, but whether they cut the line of sight.

To get an anticlockwise (left-turned) orthogonal vector, you simply need to switch $\vec v=(x,y)$ coordinates of the viewpoint vector to $\vec v^\perp=(-y, x)$. Divide both ordinates by the magnitude $|\vec v|$ to get your unit test vector $\vec t = \frac{\vec v^\perp}{|v|}$.

Then take the dot product of each circle vector to be assessed to get a corresponding set of (scalar) values $\{\vec t\cdot \vec c_i\}$. Dot product is efficient to calculate: $(a,b)\cdot(c,d) = ac+bd$. A positive value means that circle $i$ is to the left; a negative value is to the right, and the absolute value gives you how far away from the line it is - so comparing to the secondary circle radius will tell you if it cuts the line.


问:
How many ways are there of selecting a collection of 70 balls from the box?

A box contains 30 red, 40 blue, and 50 white balls; balls of the same colour are indistinguishable. How many ways are there of selecting a collection of 70 balls from the box?

my work is using the generating function:

let $G(x)=(1+x+......+x^{30} )(1+x+.......+x^{40})(1+x+.......+x^{50})$,then  is to find the coefficient of $x^{70}$

I expand the $G(x)$,i was having a taugh time doing the calculate , and i get the coefficient of $x^{70}$ is 1050

I want to know whether the answer is correct and i want to find a easier way to solve the question.

答:
"Is the answer correct?"  No, it looks like you have an arithmetic error.  I get an answer of $1061$.  wolfram

"Is there an easier way to solve the question?"  Yes.  Approach with stars-and-bars and inclusion-exclusion over the events that you have used too many of a particular type of ball, noting that it is impossible to have too many of multiple colors simultaneously so inclusion-exclusion can end after only one round.

$$\binom{70+3-1}{3-1}-\binom{39+3-1}{3-1}-\binom{29+3-1}{3-1}-\binom{19+3-1}{3-1}=1061$$


问:
Calculate third vertex of a triangle from limited information
This should be a simple problem to solve, but I cannot seem to get my head around it. I would like to find the third vertex of the triangle shown below $(x_i, y_i)$ given the following information:

I know the position of two vertices of a triangle: $(x_a, y_a)$ and $(x_b, y_b)$.

I can calculate the angle $\theta=\arctan\left(\frac{y_b-y_a}{x_b-x_a}\right) $

I can also calculate the length of one side of the triangle: $\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}$.

The other two sides are of length $s_b\Delta t$ and $s_a\Delta t$ where $s_a$ and $s_b$ are both known, but $\Delta t$ is not.

I also know the angle $\phi$, as shown.

What isn't clear to me is if I have enough information for a solution. Can this be solved and, if so, can you point me in the right direction of how to do so? (My old math teacher's voice is in my head muttering "Law of cosines!", which I could apply to get angles within my triangle but I'm drawing a blank on how to proceed from there.) Any advice would be greatly appreciated!

答:
We would like to emphasize that the information given by the OP is enough to solve this problem.

Let us describe what is known as the circle of Apollonius. Given one side of a triangle and the ratio of the lengths of the other two sides, the locus of the third vertex is the Apollonius circle. Sounds familiar? Well, this is exactly what you need to construct to find the sought third vertex mentioned in your problem statement. Probably, you may not need the construction illustrated in $\mathrm{Fig.\space 1}$, if you decide to determine the third vertex analytically. However, we recommend you to study it, because it sheds light on the problem you are trying to solve.

Draw the line segment $AB$ to make an angle $\theta$ with the horizontal as requested in the problem. We assume for the time being $s_A > s_B$. Find the point $E$ on $AB$, such that $AE\space \mathbf{:}\space EB = s_A\space \mathbf{:}\space s_B$. Draw a line segment $BD$, so that $BD > BA$. After joining $AD$, construct a parallel line to it through $E$ to intersect $BD$ at $E$. Draw a circle $\Gamma_1$ with radius $BF$ and center at $B$. Draw the circle $\Gamma_2$ with radius $FD$ and center at $A$. These two circles intersect each other at two points, but we need only one of them, say $G$. Join $EG$ and construct its perpendicular bisector to meet the extended $AB$ beyond $B$ at $O$. This, in fact, is the center of the mentioned Apollonius circle. Complete the construction by drawing a circle having the radius $OE$.

Finally, draw a line through $B$ to make the given angle $\phi$ with the horizontal to meet the Apollonius circle at $C=\left(x_C,\space y_C\right)$, which is the sought third vertex of the triangle.

Please note that, if $s_A < s_B$, the center of the Apollonius circle is located on the extended $BA$ beyond $A$.

Here is the recipe for analytical determination of the third vertex. As shown in $\mathrm{Fig.\space 2}$, we let $\measuredangle CAB = \omega$ and $\measuredangle BCA = \beta$, where $\beta = 180^o + \phi - \theta - \omega$. You have already derived the following couple of equations and, therefore, they need no explanations.

$$AB =\sqrt{\left(x_A – x_B\right)^2 + \left(y_A – y_B\right)^2} \tag{1}$$
$$\theta = \tan^{-1} \left(\frac{y_A – y_B}{x_A – x_B}\right) \tag{2}$$

We apply the law of sines to the triangle $ABC$.
$$\frac{CA}{\sin\left(\theta - \phi\right)} = \frac{AB}{\sin\left(\beta\right)} = \frac{BC}{\sin\left(\omega\right)} \tag{3}$$

$$\therefore\quad \frac{BC}{CA} = \frac{s_B\Delta t}{s_A\Delta t} = \frac{\sin \left(\omega\right)}{\sin\left(\theta - \phi\right)} \quad\rightarrow\quad \omega = \sin^{-1}\left(\frac{s_B}{s_A}\sin\left(\theta - \phi\right)\right). \tag{4}$$

Similarly, from (3) we have,
$$BC=AB\frac{\sin \left(\omega\right)}{\sin \left(\theta + \omega - \phi\right)}. \tag{5}$$

Since $AB$ and $\omega$ are known to us from(1) and (4) respectively, we can determine $BC$ from (5). Once we know $BC$, we can find the $x$- and $y$-coordinates of $C$ from
$$x_C = x_B – BC\cos\left(\phi\right)\quad\space\mathrm{and}\space\quad y_C = y_B – BC\sin\left(\phi\right).$$


问:
Parametric integration negative area?
I know there is a question very similar to mine already here Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?
, but I am still a bit unsure after reading the answers as to when the integral gives you a positive area and when it gives you a negative area. My text book says that the area is positive if t traces out the curve clockwise and negative for anticlockwise, however I think there is more going on with regards to whether the curve is above or below the x axis.

Since $A=\int_a^b y\frac{dx}{dt}dt$, my guess is that you get a positive answer if either the graph is above the x axis (y is positive) and the point is moving to the right (dx/dt positive) or the graph is below the x axis and the point is moving to the left. Then you get a negative answer if either the graph is below the x axis but is being traced out moving to the right, or the graph is above the x axis but is being traced out moving to the left.

Was my book wrong? Is my thinking correct?

Thank you in advance :)

答: The integral works because the assumption is that the integral is around an enclosed 2D area where the outside curve does not cross itself.  Thus it really doesn't matter where the area is with respect to the origin of the space, because the sum will always be the same unless we rotate the curve, then things change.  The equation works because as we march to the right (positive X), we are adding the area and as we march to the left (negative X), we are subtracting the area.  Then clockwise curves will yield A greater than zero and counter clockwise will yield A less than zero.</p>

Title: Parametric integration negative area?


问:
probability: choose 2r shoes from n pairs of shoes
A closet contains n pairs of shoes. If 2r shoes are chosen at random (2r<n), then we know that there are $2^{2r} {n\choose 2r}$ different cases in the event "there are no matching pairs in our selected sample". Now consider a different number ${2n\choose 1}\times {2n-2\choose 1}\times \dots {2n-2(r-1)\choose 1}$, is there an event that contain this number of different cases? Intuitively describe such an event.
Thanks!

答:
Hint 1: Imagine pairs of shoes lined up as shown

$A \;B \;C \;D \;E \;F \;G \;H \;I \;J \;K \;L ...$

$A \;B \;C \;D \;E \;F \;G \;H \;I \;J \;K \;L ...$

What did you do in the first case, and what are you doing that is different in the second ?

Hint 2: Take it that you are drawing $6$ shoes from the $12$ pairs

 You should be able to see that with the first formula, you are just selecting $6$ non-matching shoes, whereas with the second formula, you are arranging them in all possible ways. If you couldn't get it by intuition, verify by computations !


问:
Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.

Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.

The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
    x &= 1\\
    x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\
    x &= (-1)^{2/9} \\
    x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\
    x &= (-1)^{4/9} \\
    x &= -(-1)^{5/9} \\
    x &= (-1)^{2/3} \\
    x &= -(-1)^{7/9} \\
    x &= (-1)^{8/9}
\end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?

答:
Your elements are basically $(e^{ik\pi/9})_{k=0}^8$. They form the cyclic group of $9$ elements under ordinary multiplication. The most obvious generator is $r=e^{i\pi/9}$ (with $k=1$).

However, any such element with $k$ relatively prime to $9$ (e.g. $4,5,8$) would generate the group as well. For example, starting with $r^4$, you get a cycle
$$
\left< r^4,r^8,r^{12} = r^3,r^7,r^{11}=r^2,r^6,r^{10} = r,r^5, r^9 = 1 \right>
$$


问:
$f(x,y)=\langle y- \cos y, x \sin y\rangle$
$f(x,y)=\langle y-\cos y,x\sin y\rangle$

$C$ is the circle $(x-3)^2 + (y+4)^2 = 4$ orientated clockwise.

Relevant theorems:

Green's theorem (this is under the Green's theorem section of our book).

What I Have Tried

Using Green's theorem, I have come up with the double integral $\iint ( \sin y - 1) \,dA$.  Beyond here, I can't figure out what to do for my limits of integration.  I have come to the conclusion that polar might be the way to go, but am at a loss of how to proceed with that, since I have a circle whose center is not the origin.

If it matters, this is out of Stewart Calculus Edition 7.  It is chapter 16.4 question 13.  The answer is $4 \pi$.

答: I must say that your differentiation went wrong near the beginning of the problem, and thus this problem is a little over complicated from that. The y derivative of the first term in the vector field is $ 1 + \sin y$, and the x derivative of the second term is $ \sin y $ the difference of these two is not $\sin y + 1$.</p>


答:
Note that

$x^9=1=\cos 2n\pi +\sin 2n\pi=e^{i2n\pi}$

$\implies x=e^{i2n\pi/9}; n=0,1,2,...,8$ are  roots.

Assume $e^{i2\pi/9}=a$. Then the powers of it coprime to $9$ viz. $a,a^2,a^4,a^5,a^7 $ and $a^8$ are generators.

Title: Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.

Body:
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.

The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
    x &= 1\\
    x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\
    x &= (-1)^{2/9} \\
    x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\
    x &= (-1)^{4/9} \\
    x &= -(-1)^{5/9} \\
    x &= (-1)^{2/3} \\
    x &= -(-1)^{7/9} \\
    x &= (-1)^{8/9}
\end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?


问:
Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$
\begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}

Did I do the right way with those step above?

I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$

Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.

Please help me!!?

答: Just a small note to add that your approach of factoring $2017x^{2017}-x^{2016}-\ldots-x-1$ by breaking up the leading term and then splitting the rest is a clever one that also works here; a little careful examination and some summation notation will show what's going on. What you're left with is $(x^{2017}-x^{2016})+(x^{2017}-x^{2015})+\ldots+(x^{2017}-1)$ $= \sum_{i=0}^{2016}(x^{2017}-x^i)$. Now, we can take $j=2017-i$ and write the sum in terms of $j$: $\sum_{j=1}^{2017} (x^{2017}-x^{2017-j})$ $=\sum_{j=1}^{2017}x^{2017-j}(x^j-1)$ Factoring $x-1$ out of this is pretty straightforward; we get $(x-1)\sum_{j=1}^{2017}x^{2017-j}\left(\sum_{k=0}^{j-1}x^k\right)$. Now the polynomial multiplying $x-1$ can be summed explicitly easily enough through a bit more index juggling, but it's even easier to just plug in $x=1$ and evaluate, getting $\sum_{j=1}^{2017}\left(\sum_{k=0}^{j-1}1\right)$ $=\sum_{j=1}^{2017}j$ $=\frac{2017\cdot 2018}{2}$.


答:
One can simplify the generating function calculation significantly.  We have
$$
\begin{align}G(x)&=\frac{1-x^{31}}{1-x}\frac{1-x^{41}}{1-x}\frac{1-x^{51}}{1-x}\\
&=(1-x^{31})(1-x^{41})(1-x^{51})(1-x)^{-3}\\
&=(1-x^{31})(1-x^{41})(1-x^{51})\sum_{k=0}^\infty\binom{k+2}2x^k
\end{align}$$
by the binomial formula.

We can ignore terms of degree $>70$, so $$\begin{align}
[x^{70}]G(x)&=[x^{70}](1-x^{31}-x^{41}-x^{51})\sum_{k=0}^\infty\binom{k+2}2x^k\\
&=\binom{72}2-\binom{41}2-\binom{31}2-\binom{21}2\\
&=1061
\end{align}$$
where $[x^n]$ is the coefficient extractor operator.

Body:
A box contains 30 red, 40 blue, and 50 white balls; balls of the same colour are indistinguishable. How many ways are there of selecting a collection of 70 balls from the box?

my work is using the generating function:

let $G(x)=(1+x+......+x^{30} )(1+x+.......+x^{40})(1+x+.......+x^{50})$,then  is to find the coefficient of $x^{70}$

I expand the $G(x)$,i was having a taugh time doing the calculate , and i get the coefficient of $x^{70}$ is 1050

I want to know whether the answer is correct and i want to find a easier way to solve the question.


问:
Map $z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it },t\in\mathbb{R}$ maps the real axis $(-\infty,\infty)$ onto the unit circle $\mid z\mid=1$.

Show that the mapping
$$
z^2 = \frac{\frac{1}{2}+it}{\frac{1}{2}-it}, \quad t\in\mathbb{R}
$$
maps the real axis $(-\infty,\infty)$ to the unit circle $|z|=1$.

My try-
$$z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it  }$$
$$|z^2|=\frac{|\frac{1}{2}+it|}{ |\frac{1}{2}-it|  }$$
$$|z^2|=\frac{\sqrt{1/4+t^2}}{  \sqrt{1/4+t^2} }$$
$$|z^2|=1$$
$$|z|^2=1$$
$$|z|=1$$...

答:
You argument is partially correct. You did show that $f$ maps $\mathbb{R}$ into a subset of the unit circle, but the problem requires to show the entire unit circle is covered by the image. In other words, if you look at $f:\mathbb{R} \to D_1(0)$, it would be onto (surjective).

To do this, let $z \in D_1(0)$ and show that there is $t \in \mathbb{R}$ such that $f(t)=z$.


问:
On the finite algebra $\Bbb F_q[X]_{<n}$ over $\Bbb F_q$
Let us take the vector space of all the polynomials of degree less than $n$ over the finite field $\Bbb F_q$,
\begin{eqnarray*}
\Bbb F_q[X]_{<n} &:=& \{ f(X)\in \Bbb F_q[X]:\deg f(X) <n\} \leq \Bbb F_q[X].
\end{eqnarray*}

Clearly, this is not a ring under the usual multiplication of polynomials. On the other hand, let us endow the abelian group $(\Bbb F_q[X]_{<n},+)$ with the following multiplication:
\begin{align*}
\odot: \Bbb F_q[X]_{<n}\times \Bbb F_q[X]_{<n} \longrightarrow \Bbb F_q[X]_{<n},\quad (f(X),g(X))\longmapsto f(X)\odot g(X):=r(X),
\end{align*}
where $r(X)\in \Bbb F_q[X]$ is the  remainder of the polynomial $f(X)g(X)\in \Bbb F_q[X]$, when divided by $X^n-1$.
This multiplication is known as multiplication modulo $\boldsymbol{ X^n-1}$. Then, $(\Bbb F_q[X]_{<n},+,\odot)$ forms a commutative ring with $1_{\Bbb F_q[X]_{<n}}\in \Bbb F_q[X]_{<n}$.

But I stack on the rest of the properties in order to show the algebra structure.

So, please let me expose my thoughts. For the associativity of $\odot$,
\begin{equation}
(f(X)\odot g(X))\odot h(X)=f(X)\odot (g(X)\odot h(X)) \tag{$*$}
\end{equation}
my thought is to apply the Division Algorigthm. That is, by dividing $f(X)g(X)$ by $X^n-1$, there exists unique $q(X),r(X)\in \Bbb F_q[X]$ such that $f(X)g(X)=(X^n-1)q(X)+r(X)$, where $r(X)=0$ or $\deg r(X)<n$. Thus, $f(X)\odot g(X)=r(X)$. Then, $(f(X)\odot g(X))\odot h(X)=r(X)\odot h(X)$. Thus, in the same spirit, there exists $q'(X),r'(X)\in \Bbb F_q[X]$ such that
$r(X)h(X)=q'(X)(X^n-1)+r'(X)$, where $r'(X)=0$ or $\deg r'(X)<n$. Then,
\begin{eqnarray*}
(f(X)\odot g(X))\odot h(X) &=& r'(X) \\
&=& r(X)h(X)-q'(X)(X^n-1) \\
&=& (f(X)g(X)-(X^n-1)q(X))h(X)-q'(X)(X^n-1)\\
&=& f(X)g(X)h(X)-(X^n-1)q(X)h(X)-q'(X)(X^n-1)\\
&=& f(X)g(X)h(X)-(X^n-1)(q(X)h(X)+q'(X)).
\end{eqnarray*}
The latter is of degree $<n$ and as we can see it is the remainder of the division of $f(X)g(X)h(X)$ by $X^n-1$, and it is unique. By working likewise in the RHS of $(*)$, we obtain the same result. Hence the equality holds.

Is that argument complete and correct? Do we work likewise for the rest axioms and, furthermore, is there any faster way?

答:
There is a faster way.  I think it is worth learning about quotients of rings by ideals.  In your case, there is an isomorphism* of $\mathbb{F}_q[X]_{<n}$ and the quotient ring $\mathbb{F}_q[X]/I$ where $I$ is the principal ideal generated by $x^n$.  The general theory then tells you immediately (just from coset business) that $\mathbb{F}_q[X]/I$ is a commutative ring with identity because the "mother" $\mathbb{F}_q[X]$ is.  This way you can avoid checking all the axioms and put all the work on the abstract quotient ring construction.

*The correspondence is that the polynomial $a_0 + a_1x + \cdots + a_{n-1}x^{n-1} \in  \mathbb{F}_q[X]_{<n}$ maps to the coset $a_0 + a_1x + \cdots + a_{n-1}x^{n-1}  + I$.


答:
The map is multivalued as $z^2=1$ at $t=0$ gives $z=\pm 1$ lying on unit circle $\lvert z\vert=1$.

Added 1-The partial proof you have done shows that the map is onto. But it cannot be a bijection as it is not a single valued function.

Added 2- For another proof, note that

$t=\frac{z^2-1}{2i(z^2+1)}$, where $z=x+iy, x,y\in\mathbb R$
After splitting the RHS into real and imaginary parts, you can equate the imaginary part of $RHS=0$ (Why?) to get the result $x^2+y^2=1$.

Title: Map $z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it },t\in\mathbb{R}$ maps the real axis $(-\infty,\infty)$ onto the unit circle $\mid z\mid=1$.

Body:
Show that the mapping
$$
z^2 = \frac{\frac{1}{2}+it}{\frac{1}{2}-it}, \quad t\in\mathbb{R}
$$
maps the real axis $(-\infty,\infty)$ to the unit circle $|z|=1$.

My try-
$$z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it  }$$
$$|z^2|=\frac{|\frac{1}{2}+it|}{ |\frac{1}{2}-it|  }$$
$$|z^2|=\frac{\sqrt{1/4+t^2}}{  \sqrt{1/4+t^2} }$$
$$|z^2|=1$$
$$|z|^2=1$$
$$|z|=1$$...


Title: How to prove an identity by induction

Body:
I am working on the following problem:
$\text{For $n$ $\ge$ 1, Prove the following:}$

$n(1+x)^{n-1} = \sum_{k=1}^{n} k {n\choose k}x^{k-1} $

$\text{Deduce that: } \sum_{k=1}^{n}k{n \choose k}=n2^{n-1} $

I followed the proof by induction as follows:
$\text{Base Case: let n=1 , both RHS and LHS will evaluate to 1}$

$\text{Assume true for n=l. Now consider n=l+1}$

$(l+1)(1+x)^l=\sum_{k=1}^{l+1}k {l+1 \choose k} x^{k-1}$

$(l+1)(\sum_{k=0}^{l} {l \choose k} x^k)=\sum_{k=1}^{l+1}k {l+1 \choose k} x^{k-1}$

After this, I am kinda lost. any ideas on how to continue the proof?

Body:
Well,  $(n+1)(1+x)^n = (1+x)\frac{n+1}n\cdot n(1+x)^{n-1}= (1+x)\frac{n+1}n\cdot \sum\limits_{k=1}^nk{n\choose k}x^{k-1}=$

$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1}+ x\sum\limits_{k=1}^nk{n\choose k}x^{k-1})=$

$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1} + \sum\limits_{k=1}^nk{n\choose k}x^{k})=$

$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1} + \sum\limits_{k=2}^{n+1}(k-1){n\choose k-1}x^{k-1})=$

$\frac {n+1}n(1{n\choose 1}x^0 +\sum\limits_{k=2}^n[k{n\choose k}+(k-1){n\choose k-1}]x^{k-1} + n{n\choose n}x^n)=$

$\frac {n+1}n\cdot n + \frac {n+1}n\sum\limits_{k=2}^n[k\frac {n!}{k!(n-k)!}+(k-1)\frac {n!}{(k-1)!(n-k+1)!}]x^{k-1} +\frac {n+1}n\cdot nx^n)=$ ...

$(n+1) + \frac {n+1}n\sum\limits_{k=2}^n[\frac {n!(n-k+1)}{(k-1)!(n-k+1)!}+\frac {n!(k-1)}{(k-1)!(n-k+1)!}]x^{k-1}   + (n+1)x^n =$

$k{n+1\choose k}x^{k-1}|_{k=1} +\frac {n+1}n\sum\limits_{k=2}^n\frac{n!n}{{(k-1)!(n-k+1)!}}x^{k-1} +k{n+1\choose k}x^{k-1}|_{k=n+1}=$

$k{n+1\choose k}x^{k-1}|_{k=1} +\sum\limits_{k=2}^n\frac{(n+1)!}{{(k-1)!(n-k+1)!}}x^{k-1} +k{n+1\choose k}x^{k-1}|_{k=n+1}=$

$k{n+1\choose k}x^{k-1}|_{k=1} +\sum\limits_{k=2}^nk\frac{(n+1)!}{{k!(n-k+1)!}}x^{k-1} +k{n+1\choose k}x^{k-1}|_{k=n+1}=$

$\sum\limits_{k=1}^{n+1}k{n+1\choose k}x^{k-1}$


问:
Let $F$ be a closed subset of a metric space $X$. Does there exist a continuous function $g : X \to R$ such that $F = g^{-1}({0})$?
Let $F$ be a closed subset of a metric space $X$. Does there exist a continuous function
$g : X \to \mathbb{R}$ such that $F = g^{-1}({0})$? I am trying to produce a counterexample to this. Any ideas would be really helpful

答:
For $F=\emptyset$ we can take $g: x \to \Bbb R$ to be a constant function with value $1$, so that $g^{-1}[\{0\}] = \emptyset = F$. So we can assume that $F \neq \emptyset$.

Let $f(x) = \inf \{d(x,y): y \in F\}$, which is well-defined as the infimum of a non-emptys set that is bounded below (by $0$) in $\Bbb R$.

Then it's not too hard to see that $$\forall x,x' \in X: |f(x)-f(x')| \le d(x,x')$$ which makes $f$ uniformly continuous.

And it's also clear that any $x$ which is a limit point of $F$ obeys $d(x,F)=0$ and the same holds for all $x \in F$, so that $F= f^{-1}[\{0\}]$. So use $g=f$, the distance to the set function.


问:
How to prove that $X$ is not rational $X = Z - \pi \times Y$ where $Y$ is rational , $Z$ is an integer, $\times$ means multiplication
Excuse me for a silly question like this.
I am 60 years old retired engineer and
want to learn some basic math I di did not learn earlier.

I know an example where $(x + y\pi)$ can be an integer,
where $y$ is rational ,  * means multiplication
But how to prove the general case
if $x$ is always rational ,
always non-rational
or both depending upon the case.

Thanks

答:
Suppose for a contradiction $x$ is rational. Then $x=a/b$ where $a,b$ are coprime integers and $b\neq0$.  Also assume $y=c/d$ for coprime integers $c$ and $d$, and furthermore assume $c$ is not zero (so $y$ is not zero). Then,
$$ x = z - \pi y $$
if and only if
$$ \frac{a}{b} = z - \pi \frac{c}{d} $$
Multiply by $bd$:
$$ ad = zbd - \pi cb $$
Then,
$$ \pi cb = zbd - ad $$
Divide by $cb$ (this is possible since both are nonzero):
$$ \pi = \frac{zbd-ad}{cb} $$
This implies that $\pi$ is rational which is a contradiction.


问:
Find the remainder by using Fermat's little theorem
Fermat's little theorem states that if $p$ is a prime number and $\operatorname{gcd}(a,p)=1$, then $a^{p-1} -1$ s a multiple of $p$.

Here's my attempt to find the remainder when $5^{99999}$ is divided by 53.

$5^{99999} = 5^{52}\equiv 1\pmod{53} \implies 5^{99999}\equiv5^{99999\bmod52}=5^{3}.$

Is this the correct approach? How do I proceed from here?

答:
You have just proved that $5^{99999} \equiv 5^3$ $(mod\space53)$. They have the same remainder when divided by $53$. So now you just need to find the remainder of $125$ when divided by $53$.

There is also an extra $5^{99999}$ at the beginning of the third line


问:
At $x_0 \in \partial \Omega$, $\text{Hess}u(x_0) = \Delta u(x_0) (\nu \otimes \nu)$, $\nu$ outer unit normal
Let $x_0 \in \partial \Omega$ and $\nu(x_{0})$ is the exterior unit normal at $x_0$, with $\nu_1(x_0) > 0$, where $u \in C^{2}(\overline{\Omega_{\epsilon}})$ and $\Omega_{\epsilon}=\Omega \cap \{|x-x_0|<\epsilon\}$, $u>0$ in $\Omega$ and $u=0$ on $\partial \Omega \cap \{|x-x_0|<\epsilon\}$. We know that $u_{x_1}(x_0) = 0$. Why does the following hold?

$$
\Delta u = - f(0) > 0 \implies u_{x_{i},x_{j}}=-f(0)\nu_{i}\nu_{j} \quad \text{ at } x_0
$$

Moreover, in their paper, Gidas, Ni and Nirenberg claim, right before what inspires the question, that $\nabla u(x_0) = 0$. It's a mistake, isn't it? Up to my understanding, we only know that $u_1(x_0) = 0$.

This expression appears specifically in the last part of the demonstration of Lemma 2.1 of this article https://projecteuclid.org/download/pdf_1/euclid.cmp/1103905359 of Gidas-Ni-Nirenberg's and I think that it is related to the second directional derivative and Hessian matrix.

答:
The following argument is due to Adam Rudnik and Vitor Borges, making use of the hint by Gio67

Since the Laplace operator is invariant by rotations and translations, and since every tangential derivative is $0$ we can write it with respect to the basis $\{t_1, \ldots, t_{N - 1}, \nu\}$ as
$$
\Delta u(x_0) = u_{\nu \nu}.
$$
But
$$
u_{ij} = [u_{ij}](e_i, e_j) = [u_{ij}] (\nu_i \nu, \nu_j \nu) = \nu_i \nu_j [u_{ij}](\nu, \nu) = \nu_i \nu_j u_{\nu \nu} = \nu_i \nu_j \Delta u(x_0),
$$


问:
Which of the following is an eigenvector of a matrix
So i just finished a linear algebra exam but because of technical issues im gonna do it again! I wanted to ask a question i found there! So if i remember it correctly the question goes like this:

There is a matrix \begin{pmatrix}0 & 0 & 3 & 2\\\ 0 & 4&0&0 \\ 5 &0 &0 &0\\ 0&0&0&0\end{pmatrix}

If i remember correctly this matrix is generated from the elementary basis e1,e2,e3,e4 with each being e1(1,0,0,0,),e2(0,1,0,0),e3(0,0,1,0) and e4(0,0,0,1) !

If i remember correctly the question was which of the elementary basis serves as an eigenvector of the matrix!

Can you help me understand HOW to solve this and if possible an answer of this particular problme! I probably have the exam tomorrow so i would be really helpful if you can help me!

答:
The punchline is that elementary basis vectors get acted upon by matrices in very predictable ways... namely it returns the corresponding column in the matrix.

The question is then, does there exist a column in the matrix which is a scalar multiple of its corresponding elementary basis vector?  The answer is yes... the second column is $4$ times the second elementary basis vector.

That is to say, $\begin{bmatrix}0&0&3&2\\0&4&0&0\\5&0&0&0\\0&0&0&0\end{bmatrix}\begin{bmatrix}0\\1\\0\\0\end{bmatrix} = \begin{bmatrix}0\\4\\0\\0\end{bmatrix}=4\cdot \begin{bmatrix}0\\1\\0\\0\end{bmatrix}$, the matrix acts on the vector the same way as scalar multiplication does on the vector.


问:
What is the answer to this limit
what is the limit value of the power series:

$$ \lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$$

where $m>1$.

答:
My solution is also similar. Let $\left\{ {a \atop b} \right\}$ be the Stirling number of the second kind, as Mhenni Benghorbal pointed out. Then from the identity

$$ x^{n} = \sum_{j=0}^{n} \left\{ {n \atop j} \right\} (x)_{j}, $$

where $(x)_{j} = x (x-1) \cdots (x-j+1)$ is the failing factorial, we obtain

$$ x^{n} = \sum_{j=0}^{n} \left\{ {n+1 \atop j+1} \right\} (x-1)_{j}. $$

This allow us to write

\begin{align*}
\frac{1}{k^{k-m}}
& = \frac{k^{m}}{k^k} = \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \frac{\Gamma(k)}{k^{k}} \\
&= \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \int_{0}^{\infty} t^{k-1}e^{-kt} \, dt.
\end{align*}

Here, we use the convention that $\frac{1}{j!}$ is zero when $j < 0$, which is compatible with the fact that the reciprocal of the Gamma function is an entire function with zeros at nonpositive integers. Then

\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k}
&= \sum_{k=1}^{\infty} (-1)^{k-1} x^{k} \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \int_{0}^{\infty} t^{k-1}e^{-kt} \, dt \\
&= x \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( \sum_{k=1}^{\infty}  \frac{(-1)^{k-1}}{(k-1-j)!} \left( x t e^{-t} \right)^{k-1} \right) \, e^{-t} \, dt \\
&= x \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( \sum_{k=0}^{\infty}  \frac{(-1)^{k+j}}{k!} \left( x t e^{-t} \right)^{k+j} \right) \, e^{-t} \, dt \\
&= x \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( x t e^{-t} \right)^{j} \exp \left( x t e^{-t} \right) \, e^{-t} \, dt \\
&= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} x^{j+1} \int_{0}^{\infty} t^{j} e^{-tj} \exp \left( x t e^{-t} \right) \, e^{-t} \, dt.
\end{align*}

Here, I considered the negation of your sum in order for typographical simplicity. Anyway, with the substitution $u = e^{-t}$, we obtain

\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k}
&= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} x^{j+1} \int_{0}^{1} u^{xu} (-u \log u)^{j} \, du. \tag{1}
\end{align*}

To analyze the behavior of

$$ f_{j}(x) = x^{j+1} \int_{0}^{1} u^{xu} (-u \log u)^{j} \, du, $$

we introduce two functions $g_0 : [0, 1/e] \to [0, 1/e]$ and $g_1 : [0, 1/e] \to [1/e, 1]$ ad follows:

The graph of $y = -x \log x$ on $[0, 1]$ can be divided into two parts:

the increasing part and the decreasing part. Thus on each part, we can define the inverse function $g_0$ and $g_1$ as follows:

It is clear from the slope of the graph $y = - x \log x$ that $g_{0}'(0) = 0$ and $g_{1}' (0) = -1$. Also, obviously both $g_{0}'(y)$ and $g_{1}'(y)$ are integrable on $[0, 1/e]$. Keeping these in mind, we make the change of variable as follows:

\begin{align*}
f_{j}(x)
&= x^{j+1} \left[ \int_{0}^{1/e} e^{x u \log u} (-u \log u)^{j} \, du + \int_{1/e}^{1} e^{x u \log u} (-u \log u)^{j} \, du \right] \\
&= x^{j+1} \left[ \int_{0}^{1/e} y^{j} e^{-xy} g_{0}'(y) \, dy + \int_{0}^{1/e} y^{j} e^{-xy} (- g_{1}'(y) ) \, dy \right] \\
&= \int_{0}^{\infty} x^{j+1} y^{j} e^{-xy} \left( g_{0}'(y) \mathrm{1}_{[0, 1/e]}(y) \right) \, dy + \int_{0}^{\infty} x^{j+1} y^{j} e^{-xy} \left( - g_{1}'(y) \mathrm{1}_{[0, 1/e]}(y) \right) \, dy
\end{align*}

Here, observe that the family of mappings

$$ y \mapsto \frac{1}{j!} x^{j+1} y^{j} e^{-xy} $$

serves as an approximation to the identity as $x \to \infty$. This proves that

$$ f_{j}(x) \xrightarrow{x\to\infty} j! (g_{0}'(0) - g_{1}'(0)) = j!. $$

So taking limit $x \to \infty$ to the identity $(1)$, we obtain

\begin{align*}
\lim_{x\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k}
&= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} j!.
\end{align*}

But it is known that the last sum is equal to

$$ \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} j! = \delta_{0m} = \begin{cases} 1 & m = 0 \\ 0 & m \geq 1 \end{cases}. $$

Therefore we have

$$ \lim_{x\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{k-m}} x^{k} = -\delta_{0m}. $$

The following graphs corresponds to $m = 1, \cdots, 5$ in zigzag order.

Added. Approximation to the identity is a way of justifying the concept of the Dirac delta. Consider a family of functions $\{ K_{t}(x) \}$ (parametrized by $t$ in this example) such that

$$ \int_{\Bbb{R}} K_{t}(x) \,dx = q \quad \text{and} \quad \lim_{t\to\infty} \int_{|x|\geq \eta} K_{t}(x) \, dx = 0 \quad \text{for} \ \eta > 0. $$

That is, we cant interpret $K_{t}(x)$ as a mass density with the unit total mass such that the mass concentrates to the origin as $t \to \infty$. So it is natural to expect that $K_{t}(x)$ converges to the Dirac delta $\delta(x)$ in some sense as $t \to \infty$. The theory of approximation to the identity justifies this naive idea under various assumptions and in various mode of convergence.

In our problem,

$$ K_{x}(y) = x f(xy) \quad \text{where} \quad f(y) = \frac{y^{j}}{j!}e^{-y} \quad \text{on} \ [0, \infty). $$

Thus we expect that as $x \to \infty$,

\begin{align*}
f_{j}(x)
&= j! \int_{0}^{\infty} \left\{ g_{0}'(y) - g_{1}'(y) \right\} \mathbf{1}_{[0, 1/e]}(y) K_{x}(y) \, dy \\
&= j! \int_{0}^{\infty} \left\{ g_{0}'(y) - g_{1}'(y) \right\} \mathbf{1}_{[0, 1/e]}(y) \delta(y) \, dy \\
&= j! \left\{ g_{0}'(0) - g_{1}'(0) \right\} = j!.
\end{align*}

Indeed, some simple estimation proves this.

Title: What is the answer to this limit


问:
Proof $C = A \cdot B \cdot A^{\top}$ nonsingular if $A$ has full rank.
Given are the matrices $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times n}$. $B$ is known to be symmetric positive definite.

I made the following observation: The matrix $C = A \cdot B \cdot A^{\top} \in \mathbb{R}^{m \times m}$ is singular for $\text{rank}\,{(A)} < m$. Example:

$$
C = \begin{pmatrix} 1 & 2 \\ 4 & 8 \end{pmatrix} \cdot \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 4 & 8 \end{pmatrix}^{\top} = \begin{pmatrix} 25 & 100 \\ 100 & 400 \end{pmatrix}
$$

with $\det C = 25 \cdot 400 - 100^2 = 0$.

Is there a proof for the claim: If $A$ has maximal rank, then C is nonsingular?

Any hints or help is highly appreciated.

答:
Because $B$ is symmetric and positive definite, there exists $D$ with $B=DD^T$ with $D$ symmetric and positive definite. Then
$$
C=ADD^TA^T=AD(AD)^T.
$$
From $\operatorname{null}(X)=\operatorname{null}(XX^T)$, and the rank-nullity theorem, the rank of $C$ is equal to the rank of $AD$; as $D$ is invertible,  this is equal to the rank of $A$.

So $C$ will be non-singular precisely when $A$ has  rank $m$. For the rank of $A$ to be equal to $m$ and maximal, you need $m\leq n$.


问:
What happens when C is only partly within the vector field?
So I have the following question where $$\vec{G}=(x,y)/\sqrt{x^2+y^2-1}$$ is a vector field.
I have to calculate the line integral for $$\int_{\mathrm{C}}\vec{G}\, d\vec{r}\, $$ where $C$ is the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.

I know that $\vec{G}$ has the domain $x^2+y^2 \gt 1,$ (meaning that the circle with radius 1 is not part of the domain) and the ellipse has radius 2 with x-component and 3 for y-component.
I have calculated param $x=2cost$ and $y=3sint$ where $0\le t\le 2\pi$.
So does this mean that only a part of the ellipse is within $\vec{G}$?

How do I calculate the line integral in this case?

答:
You do not need to parametrize this, nor do you need Green's theorem.  By usual integration tricks, your function $\vec{G}$ has a potential:
$$
g(x,y) = \sqrt{x^2+y^2-1}.
$$
You can check that $\vec{G} = \nabla g$, and this is valid over a domain that "encloses" your ellipse in a nice, legal way.  Now use the fundamental theorem:
$$
\int_C \vec{G} \cdot d\vec{r} = g(2,0) -g(2,0) = 0
$$
as $C$ is a closed curve.


问:
Quick Average Speed Question
I was studying for a test and found the following prompt (similar to another question that was asked, I just had one question that had a number of different prompts that centered around the main prompt):

"A student programs a robot to move across a flat surface over the time interval $0≤t≤20$  seconds. The position of the robot at time $t$ seconds is given by the parametric equations  $x(t)=3t+\sin(t)$ and $y(t)=3\cos(t−\sin(t))$, where  $x(t)$ and $y(t)$ are measured in meters. The derivatives of these functions are given by $x′(t)=3+\cos(t)$ and $y(t)=−3\sin(t-\sin(t))⋅(1−\cos(t))$. Find the average speed of the robot from time $t=0$ seconds to time $t=20$ seconds."

I guess I'm confused about what the integral would look like in this specific context. I've done average speed problems before and I understood what the formulas and set-ups looked like then, but this specific problem seems to be giving me some trouble. Any help would be appreciated!

答:
The instantaneous speed is $s(t)=\sqrt{x'(t)^2+y'(t)^2}$, and the average speed is $\frac{1}{T}\int_0^Ts(t)\,\mathrm d t$, where $T=20 \,\text{s}$.
Does it make sense?

Title: Quick Average Speed Question


问:
Series expansion of the power 1/4
There exists an inequality that links the expression
$$\sum_{k\ge0}(-1)^{k+j} (2\mu)^{k}\binom{1/4}{k+j-1}$$
where $\mu <1/2$ and $j\ge 2$ with the expression
$$\sum_{k\ge 0 }(-1)^k (2\mu)^k \binom{1/4}{k}=(1-2\mu)^{1/4}$$
?
I would like to have that the first expression is less than the second one.

Title: Series expansion of the power 1/4

Body:
Let $m=k+j-1$: For $j\gt 1$

$
\sum_{k\ge 0}(-1)^{k+j}(2\mu)^k\binom{1/4}{k+j-1}=\sum_{m\ge j-1}(-1)^{m+1}(2\mu)^{m-j+1}\binom{1/4}{m}=$

$-(2\mu)^{j-1}\bigg((1-2\mu)^{1/4}-\sum_{m=0}^{j-2}(-1)^m(2\mu)^m\binom{1/4}{m}\bigg)$


问:
evaluating $\int_0^{\infty}\frac{e^{-t-\frac{x}{t}}}{t} dt$
I got to this integral, while proving some theorem in statistics:

$$\int_0^\infty \frac{e^{-t-\frac{x}{t}}}{t} \mathop{dt}$$

I have trouble evaluating it. I tried partial integration, tried substitution with some polynomial and some trigonometric functions. None of them helped, and Wolfram can't compute it either. Do you have a hint on how to solve this?

答:
With a little help from Maple, the integral is

$$\int_0^\infty \frac{e^{-t-\frac{x}{t}}}{t}\,dt = 2K_0(2\sqrt{x}),$$

where $K_0$ is the modified Bessel function of the second kind of order zero.


答:
The binomial theorem gives another way. Let $x=1+\varepsilon$ with $0<|\varepsilon|\ll1$. Then
$$x^n=1+n\varepsilon+\tfrac12n(n-1)\varepsilon^2+O(\varepsilon^3),\quad x^{n+1}=1+(n+1)\varepsilon+\tfrac12(n+1)n\varepsilon^2+O(\varepsilon^3).$$Hence
$$\frac n{1-x^n}-\frac{n+1}{1-x^{n+1}}=\frac1\varepsilon\left(\frac{-1}{1+\tfrac12(n-1)\varepsilon+O(\varepsilon^2)}-\frac{-1}{1+\tfrac12n\varepsilon+O(\varepsilon^2)}\right)$$
$$=-\frac1\varepsilon\frac{\tfrac12\varepsilon+O(\varepsilon^2)}{1+n\varepsilon+O(\varepsilon^2)}=-\frac{\tfrac12+O(\varepsilon)}{1+O(\varepsilon)}.$$
Therefore the limit as $\varepsilon\to0$ is $-\frac12$.

Title: Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$

Body:
\begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}

Did I do the right way with those step above?

I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$

Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.

Please help me!!?


问:
Does this equation have an explicit solution?
I have ended up needing to solve the following equation:
\begin{equation}
\frac{p \left(n (1-p)^n-2 (1-p)^n+n+2\right)+2
   \left((1-p)^n-1\right) -n p^2 }{(n-2)(p-1)} =  \frac{n p-1}{(n-1)}
\end{equation}
where $n$ is an integer that is strictly greater than $3$, and $p$ is a real number in the  $(0, 1)$ interval. While I can solve this for $p$ after plugging in particular values of $n$, I have not been able to solve it for $p$ in general. Is this possible in principle?

Body:
In principle is not possible. I made Mathematica solve the equation for $p\in(0,1)$ for a few hundred of $n$-value, then looked for a fit and found
$$p_n=\frac{5.7358}{6.25529 n-0.424686}$$
which is quite accurate
$$
\begin{array}{r|r|r|r}
 n & actual\;root & p_n&error \;percentage (\%)\\
\hline
 4 & 0.233196 & 0.233196 & 1.52\times 10^{-6} \\
 204 & 0.00449659 & 0.00449636 & 0.00525814 \\
 404 & 0.00227021 & 0.00227006 & 0.00644436 \\
 604 & 0.00151841 & 0.0015183 & 0.00684953 \\
 804 & 0.00114066 & 0.00114058 & 0.00705397 \\
 1004 & 0.000913426 & 0.00091336 & 0.00717725 \\
\end{array}
$$


问:
Number of submodules of a module with non-isomorphic composition factors
Let $A$ be a finite dimensional algebra over $\mathbb{C}$ and let $M$ be a finite-dimensional $A$-module.

Suppose further that $M$ is semisimple with composition factors $S_1, \dots, S_k$, where $S_i \not\cong S_j$ for all $i \neq j$. I'm attempting to show that $M$ has only finitely many submodules.

By semisimplicity we can write $M = \bigoplus_{i=1}^k M_i$ and from the Jordan-Holder theorem applied to the composition series $0 \subset M_1 \subset M_1 \oplus M_2 \subset \ldots \subset M_1 \oplus M_2 \oplus \ldots \oplus M_k$ we can arrange that $M_i \cong S_i$ after reordering.

My suspicion is that any submodule is of the form $\bigoplus_{i \in I} M_i$ for some $I \subseteq \{1, \ldots, k\}$ so that there are $2^k$ submodules. I've tried a number of different ways to prove this but haven't quite been able to get the result I'm looking for.

Let $N$ be a submodule of $M$. My first thought was to consider a composition series for $M$ containing $N$. I think this gets us our result only up to isomorphism, which isn't quite strong enough.

My next thought was to consider the projection maps $\pi_i \colon N \to M_i$. If $\ker\pi_t = N$ then $N \subset \bigoplus_{i \neq t}M_i$ and the result follows by induction on $k$.

Now suppose that $\ker\pi_i \neq N$ for all $1 \leq i \leq k$. Because each $M_i$ is simple we then have $N / \ker \pi_i \cong M_i$ for each $i$. My hope was to show that if $N$ is too small then we must, by some counting argument, have that two of the kernels are the same, therefore contradicting the non-isomorphism of the $M_i$. I'd then conclude that $N = M$ and be done.

As one avenue, by altering our induction to be on $k + \dim N$ we can write each kernel as a direct sum of the $M_i$. My hope was that each kernel is then "determined" in some sense by the fact that it misses its corresponding simple module. I tried using the maximality of the kernels in $N$ but that didn't get me anywhere.

That's about as far as I've managed to get. Am I along the right lines with any of this or am I exploring a dead end? This problem came up in the context of an introductory representation theory course so maybe the machinery of representations could be helpful, but I can't see how.

答:
A bit of a late reply to this question, however I'm answering this for my benefit, more than anything else.

It's fairly intuitive to say that a direct sum of any combination of these $S_{i}$ is a submodule of M (we know we have $2^k$ of these because the set of combinations is just a power set of k elements, which always has that many elements.) What you need is that any submodule of M sits in a composition series of $M$, so can be represented as some direct sum of composition factors.

So, let $A$ be a $K$-algebra, let $V$ be a finite dimensional $A$-module, and let $L{\subset}V$ be a submodule.

Consider $V/L$ and $L$. If both are simple, then we have that

${0} \subset L \subset V$

is a composition series of $V$. If they're both zero we have a trivial composition series for $V$. If they are not simple we find some module $W$ such that

${0} \subset L \subset W \subset V$.

If $L$, $W/L$ and $V/W$ are simple then we have done it. Otherwise we keep taking refinements in this manner (if L is non-simple we can take the composition series for that and stick it in, in the manner of the proof of the Jordan Holder theorem) until we have our simple composition factors. We know this procedure terminates because V is finite dimensional.

So then applying this we have that all submodules of M must lie in a composition series, and as such must be of this form.


问:
Is there any specific meaning of the formula 2*(max-min)/(max+min)?
I'm a student who studies machine learning.

When I studied this paper(https://dl.acm.org/doi/pdf/10.1145/335191.335388),

I saw the following formula and explanation.

$$(direct_{max} - direct_{min})/direct_{mean} = (indirect_{max} - indirect_{min})/indirect_{mean}$$

The $direct_{mean}$ is the mean value of $direct_{max}$ and $direct_{min}$.

And $indirect_{mean}$ is also the mean value of $indirect_{max}$ and $indirect_{min}$.

In this case, that paper said the fluctuate amount is the same between direct and indirect.

But I don't understand one thing.

Is there any specific meaning of the below formula?

$$2* (value_{max} - value_{min})/(value_{max} + value_{min})$$

I don't know why it has the meaning as the amount of fluctuating.

Thank you for reading my question.

答:
You can rewrite the formula as follows:

$$ \frac{2*(\text{Max}-\text{Min})}{\text{Max}+\text{Min}} $$
$$ = \frac{\text{Max}-\text{Min}}{\frac{\text{Max}+\text{Min}}{2}} $$
$$ = \frac{\text{Range}}{\text{AVG(Min, Max)}} $$

The numerator is the range, and the denominator is the average of the max and min. You can think of that numerator as a measure of spread, and the denominator as a measure of central tendency. So, this ratio is similar to other dimensionless/normalized metrics that measure noise/signal, like $\frac{\sigma}{\mu}$ or $\frac{\text{IQR}}{\text{Median}}$.

EDIT: As noted in comments on the original question, this is a rough normalized measure of spread unless we know a lot more about the distribution. If the distribution of observed values is approximately symmetric and unimodal, then that average is likely close to the mean or median. But, if the distribution is skewed or multimodal, then the average of the min and max can be wildly different from most actual observations.


问:
How to prove this vector identity $\nabla \cdot \Big( ({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u}\Big) $?
While I was studying on analysis of the NSE and MHD problem, I encountered the following identity.

\begin{eqnarray}
\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)=Q({\bf u},{\bf u})-\dfrac{1}{2} {\bf u} \cdot \nabla (\nabla \cdot {\bf u})+\dfrac{1}{2}|\nabla \cdot {\bf u}|^2,  \hspace{2cm}(1)
\end{eqnarray}
where $$Q({\bf u},{\bf u}):=\nabla {\bf u}: (\nabla {\bf u})^T=\sum_{i,j} \dfrac{\partial u_i}{\partial x_j}\dfrac{\partial u_j}{\partial x_i}, \quad d=2,3.$$
I need to generalize the previous identity as

\begin{eqnarray}
\nabla \cdot \Big(({\bf u}\cdot\nabla) {\bf v}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf v})\Big)&=&Q({\bf u},{\bf v})+C_1{\bf u} \cdot \nabla (\nabla \cdot {\bf v})\nonumber\\&&+C_2{\bf v} \cdot \nabla (\nabla \cdot {\bf u})+C_3(\nabla \cdot {\bf u})  (\nabla \cdot {\bf v}),\hspace{2.5cm} (2)
\end{eqnarray}
where $C_1, C_2, C_3$ are unknown coefficients. I need to find them. I believe that, if I can prove the equation (1), I can get the equation (2) in a similar way. Hence, first I tried to prove the equation (1).

My attempt at a solution:
\begin{eqnarray}
\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)=\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}\Big)+ \nabla \cdot \Big(\dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big). \hspace{2.5cm} (3)
\end{eqnarray}
By using
$\nabla \cdot \Big(\psi {\bf u}\Big)=\psi \nabla \cdot {\bf u} +(\nabla \psi)\cdot {\bf u}$, the first term on the right hand side of (3) can be organized as
\begin{eqnarray}
\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}\Big)= ({\bf u}\cdot\nabla ) \nabla \cdot {\bf u} +\Big(\nabla ({\bf u}\cdot\nabla )\Big)\cdot {\bf u}  \qquad \qquad \quad
\end{eqnarray}
Utilizing $\nabla ({\bf u}\cdot\nabla )=\nabla (\nabla \cdot {\bf u} )$ and $\nabla (\nabla \cdot {\bf u} )=\nabla \times (\nabla \times {\bf u} )+\Delta {\bf u}$, we get
\begin{eqnarray}
\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}\Big)= ({\bf u}\cdot\nabla ) \nabla \cdot {\bf u} +\nabla \times (\nabla \times {\bf u} )\cdot{\bf u} +\Delta {\bf u} \cdot {\bf u}\hspace{3.5cm} (4)
\end{eqnarray}

In a similar manner, the second term on the right-hand side of (3) can be written as
\begin{eqnarray}
\nabla \cdot \Big(\dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)&=&\dfrac{1}{2} (\nabla \cdot {\bf u}) \nabla \cdot {\bf u} +\dfrac{1}{2}(\nabla (\nabla \cdot {\bf u}))\cdot {\bf u} \nonumber\\
&=&\dfrac{1}{2} |\nabla \cdot {\bf u})|^2 +\dfrac{1}{2}\nabla \times (\nabla \times {\bf u} )\cdot {\bf u} +\dfrac{1}{2}\Delta {\bf u}\cdot {\bf u} \hspace{3cm}(5)
\end{eqnarray}
Inserting (4) and (5) into (3) produces

\begin{eqnarray}
\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)&=& ({\bf u}\cdot\nabla ) \nabla \cdot {\bf u} +\nabla \times (\nabla \times {\bf u} )\cdot{\bf u} +\Delta {\bf u} \cdot {\bf u}\nonumber\\
&&+\dfrac{1}{2} |\nabla \cdot {\bf u})|^2 +\dfrac{1}{2}\nabla \times (\nabla \times {\bf u} )\cdot {\bf u} +\dfrac{1}{2}\Delta {\bf u}\cdot {\bf u}
\end{eqnarray}

I'm stuck here. How can I continue?

答: @Sal says the identity may need a sign change for the first term on each side, so I'll prove that modified version. (I'll delete this answer if it becomes clear the real issue isn't those signs, but they would fix it.) Summing over repeated indices (all but the leftmost denoting a derivative), the left-hand side is$$(-u_ju_{ij}+\tfrac12u_{jj}u_i)_i=-u_{ji}u_{ij}-u_ju_{iji}+\tfrac12u_{jji}u_i+\tfrac12u_{jj}u_{ii},$$while the right-hand side is$$-u_{ij}u_{ji}-\tfrac12u_iu_{jji}+\tfrac12u_{ii}u_{jj},$$so left minus right is$$-u_ju_{iji}+u_{jji}u_i.$$But $u_{jji}u_i=u_{iij}u_j=u_{iji}u_j$, first by relabelling, then commuting derivatives.


答:
Expand out the product and cancel common factors:

$$\prod_{k=j}^n\frac{p_{k+1}}{p_k}=\frac{p_{j+1}p_{j+2}\cdots p_{n}p_{n+1}}{p_{j}p_{j+1}\cdots p_{n-1}p_{n}}=\frac{\color{red}{p_{j+1}}p_{j+2}\color{red}{\cdots p_{n}}p_{n+1}}{p_{j}\color{red}{p_{j+1}}\cdots p_{n-1}\color{red}{p_{n}}}=\frac{p_{n+1}}{p_j}$$

Note that this has nothing to do with $p_k$ being prime.

Title: Proving $\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_j}\!\!,\;\;1\le j\!<\!n$

Body:
Let $p_n$ denote the $n$th prime number.

How could one prove that:

$$\prod \limits_ {k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_j}\!\!,\;\;1\le j\!<\!n$$

Examples:

$n=3812,\;j=81\qquad\implies\quad\large{\prod \limits _{k=81}^{3812} \,\frac{p_{k+1}}{p_k}= \frac{p_{3813}}{p_{81}} = \frac{35897}{419}}$

$n=20019,\;j=1\qquad\implies\quad\large{\prod \limits _{k=1}^{20019} \frac{p_{k+1}}{p_k} = \frac{p_{20020}}{p_{1}} = \frac{224993}{2}}$

$n=129181,\;j=35\quad\implies\quad\large{\prod \limits _{k=35}^{129181} \!\!\frac{p_{k+1}}{p_k}= \frac{p_{129182}}{p_{35}} = \frac{1715059}{149}}$


问:
How do i step by step evaluate this Complex number expression?
$$-\frac32(-\sqrt{3i})$$

I know that|$-\sqrt 3+i|$ can be evaluated to 2. Is it some way i can use that to evaluate the expression?

答:
I think by evaluating this expression you mean to bring it into cartesian form $a+bi,\,a,b\in\mathbb{R}$. Let's evaluate
$$\sqrt{3i}=\sqrt{3}\cdot\sqrt{i}$$
first. This is not an equation that's why it only has one simplification, if we consider the principal square root. One usually would find the square root of $i$ by the polar form. I have the feeling you are new to complex numbers so I will choose an easier way to follow.

Let's say that the square root of $i$ will equal $a+bi$. Now we have to find out $a$ and $b$:
\begin{align}
\sqrt i &= a+bi\\
i&=(a+bi)^2\\
0+1\cdot i&=a^2-b^2+2abi.
\end{align}
If we compare real and imaginary parts we get $a^2-b^2=0$ and $2abi=1$. This is an easy $2\times2$ linear system of equations. I think you should be able to solve that and get to
$$a=\frac{1}{\sqrt 2},\quad b= \frac{1}{\sqrt 2}.$$ Remember how we defined $a$ and $b$. In conclusion we have
$$\sqrt i = \frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i.$$
Let's get back to our original equation. The negative signs eliminate each other and we are left with
$$\frac{3}{2}\sqrt 3\left(\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i\right)=\frac{3\sqrt 3}{2\sqrt 2}+\frac{3\sqrt 3}{2\sqrt 2}i.$$


问:
Quasi-coherent sheaf which is a vector bundle on curves.
This question is inspired by this question. Given a quasi-coherent sheaf on a smooth variety $X$ such that its restrictions to curves are finite dimensional vector bundles. Does it follow that the sheaf is necessarily a finite dimensional vector bundle?

答:
In the case where $\mathscr E$ is coherent, you can re-use the arguments in the answer to the question you cited. First of all, there is an equivalence of categories between the category of vector bundles over $X$ (i.e. algebraic varieties $E$ with a projection $E \to X$ which locally on $X$ looks like the projection $X \times \mathbb A^n \to X$) and the category of locally free sheaves over $X$, the equivalence being given by sending a vector bundle to its sheaf of sections, and in the other direction, using the relative spec of the symmetric algebra of your locally free sheaf to build an (algebraic) vector bundle out of it. This equivalence respects all sorts of things (tensor products, hom-sheaves, pullbacks, etc.) so you can re-formulate the question as:

Given a quasi-coherent sheaf $\mathscr E$ such that its restriction to curves are locally free of finite constant rank, is $\mathscr E$ locally free of finite constant rank?

If you assume $X$ integral and smooth, then all you have to show is that $\mathscr E$ is such that the function $x \mapsto \dim \mathscr E_x/\mathfrak m_x \mathscr E_x$ is constant on the set of closed points of $X$.

If you pass a smooth connected curve between any two closed points $x,y \in X$, as in the argument in the answer of the question you quoted, $\mathscr E|_C$ is locally free of constant rank (because smoothness implies the curve is irreducible), so since any two points can be joined by such a curve, $x \mapsto \dim \mathscr E_x/\mathfrak m_x \mathscr E_x$ is constant.

I am trying to think of a counter-example in the case where $\mathscr E$ is quasi-coherent but not coherent.

Hope that helps,


问:
Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.

Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.

The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
    x &= 1\\
    x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\
    x &= (-1)^{2/9} \\
    x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\
    x &= (-1)^{4/9} \\
    x &= -(-1)^{5/9} \\
    x &= (-1)^{2/3} \\
    x &= -(-1)^{7/9} \\
    x &= (-1)^{8/9}
\end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?

答:
The multiplicative group of $9$-th roots is isomorphic to the additive group $\mathbf Z/9\mathbf Z$, under the isomorphism:
$$\mathbf Z/9\mathbf Z\longrightarrow U_9,\quad k\bmod9\longmapsto\mathrm e^{\tfrac{2ik\pi}9}.$$
Under this isomorphism, generators of $\mathbf Z/9\mathbf Z$ map onto generators of $U_9$, i.e. primitive $9$-th roots of unity. Therefore it is enough to know the generators of $\mathbf Z/9\mathbf Z$, which is standard.


问:
If $f$ is integrable, how to show that $\sum_{n=1}^\infty 2^n\lambda(\{\lvert f\rvert\ge 2^n\})<\infty$
I want to show the following converse of an (apparently) common integration theory exercise.

Suppose $f:[0,1]\rightarrow\mathbb R$ is integrable, then
$$\sum_{n=1}^\infty 2^n\lambda(\{\lvert f\rvert\ge 2^n\})<\infty.$$

Since series convergence is annoying to prove, I thought the best way to tackle this problem would be by assuming that the series doesn't converge, then showing that $f$ can't be integrable. However, I couldn't get much from the series not converging, as the terms of the series might well go to $0$ even if the series doesn't converge.

Is there something I'm missing to make this approach work? Or is a direct proof easier somehow?

答: Let $f_n=\sum_{n=1}^{n=N}2^n\lambda(\{|f|\geq 2^n\}$$f_n\leq |f|$ and is measurable implies that $\mu(f_n)\leq \mu(|f)$. The sequence $\mu(f_n)$ is a bounded increasing sequence, it converges.


问:
What kind of probability distribution is this?
Not sure what kind of probability distribution deals with summation of results:

"A fair dice is thrown repeatedly until the sum of throws exceeds 6. Let the random variable X be the number of throws needed to obtain this. Find the probability mass function of X."

So far, I'm thinking that this might be a geometric distribution basing it on the unfixed number of trails needed to obtain the first success (sum = 6), yet I might be wrong. It starts getting confusing for me when I look at the possible probabilities of X.

答:
I don't think it has a name.

X is a discrete random variable that can take the values in {2, 3, 4, 5, 6, 7}.

I think you just have to find $P[X=x]$ for each of these $x$.

These might help:

i) for $x>2$, $P[X=x]=P[X\le x]-P[X\le x-1]$

ii) the probability $X\le x$ is the probability that the sum of the rolls on $x$ die are greater than 6.


问:
Generalising $\sum_{r=0}^{n}r^{k}{n\choose r}$ for $k\in \mathbb{Z^{+}}$
I am trying to generalise the closed form expression for $k$ of the following sum using diffferentiation techniques.

$$\sum_{r=0}^{n}r^{k}{n\choose r}$$

We know the following results through trivial application of differentiation followed by multiplication at each step, and at the final step plugging in $x=1$.

$$\begin{aligned}\sum_{r=0}^{n}r{n\choose r}&=n2^{n-1}\\ \sum_{r=0}^{n}r^2{n\choose r}&=n(n+1)2^{n-2}\\\sum_{r=0}^{n}r^{3}{n\choose r}&=n^{2}(n+3)2^{n-3}\end{aligned}$$

Clearly there seems to be some patterns, as for the exponent, the degree of the polynomial function. I am not able to extend a similar pattern for $k=4$. Any ideas are appreciated. Thanks.

答:
For every nonnegative integer $k$, and every integer $m\geq k$, you have $\frac{d^k}{dx^k}(1+x)^m=\frac{m!}{(m-k)!}(1+x)^{m-k}$ and
\begin{eqnarray*}
\frac{d^k}{dx^k}(1+x)^m&=&
\frac{d^k}{dx^k}\sum_{s=0}^m\binom{m}{s}x^s\\
&=&\sum_{s=0}^m\binom{m}{s}\frac{d^k}{dx^k}x^s\\
&=&\sum_{s=k}^m\binom{m}{s}\frac{s!}{(s-k)!}x^{s-k}\\
&=&\sum_{r=0}^n\binom{n}{r}\frac{(r+k)!}{r!}x^r\\
\end{eqnarray*}
Now note that
$$\frac{(r+k)!}{r!}=\prod_{i=0}^{k-1}(r+i)=r^k+\sum_{j=1}^ks_jr^{k-j},$$
where $s_j$ denotes the $j$-th symmetric polynomial in $k$ variables evaluated at $\{1,2,\ldots,k\}$. It follows that
\begin{eqnarray*}
\sum_{r=0}^nr^k\binom{n}{r}x^r&=&\frac{(n+k)!}{n!}(1+x)^n-\sum_{r=0}^n\binom{n}{r}\left(\sum_{j=0}^{k-1}s_jr^{k-j}\right)x^r\\
&=&\frac{(n+k)!}{n!}(1+x)^n-\sum_{j=1}^ks_j\sum_{r=0}^nr^{k-j}\binom{n}{r}x^r,
\end{eqnarray*}
and so plugging in $x=1$ and setting $S(n,k):=\sum_{r=0}^nr^k\binom{n}{r}$ shows that
$$S(n,k)=\sum_{r=0}^nr^k\binom{n}{r}=\frac{(n+k)!}{n!}2^n-\sum_{j=1}^ks_jS(n,k-j).$$


Body: $$\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_n} \times \frac{p_{n}}{p_{n-1}} \dots \frac{p_{j+2}}{p_{j+1}}\times \frac{p_{j+1}}{p_j}=\frac{p_{n+1}}{p_j}$$


问:
Are the events $S>s$ and $T=1$ are independent
$$X,Y\sim \exp(\lambda)$$

$$S=\min(X,Y)$$
$$T=\mathbf1_{X\le Y}=\begin{Bmatrix}
1 & \text{ if } X\le Y \\
0 & \text{ if } X\gt Y
\end{Bmatrix}$$

I have that $P(S>s)=e^{-2\lambda s}$ (so $S\sim \exp(2\lambda)$) and that $P(T=1)=\frac{1}{2}$

I am now struggling to find out how to show their independence. Using $P(S>s,T=1)=P(S>s)P(T=1)$, I attempted to work out the left hand side of the equation to show it was equal to $\frac{1}{2}e^{-2\lambda s}$ as follows:

$$\begin{align}
P(S>s,T=1)&=P(X>s,Y>s,X\le Y) \\
&=P(X>s|Y>s,X\le Y)+P(Y>s|X\le Y)\\
&=P(X>s|X\le Y)+P(Y>s|X\le Y)\\
&=\frac{P(X>s,X\le Y)+P(Y>s,X\le Y)}{P(X\le Y)}
\end{align}$$

However, I could not figure out how to work out $P(X>s,X\le Y)$ and $P(Y>s,X\le Y)$

I was thinking something along the lines of:

$$P(X>s,X\le Y)=\int_{s}^{\infty}\int_{x}^{\infty}2\lambda e^{-2\lambda s}\mathbf1_{X\le Y}dydx$$ but this diverges so it has left me a bit stumped about where to go from here. Any clues to help me get over this hurdle would be much appreciated! Or if I'm completely barking up the wrong tree it would be useful if I could be pointed in the right direction :)

答:
You should have
\begin{align} P(X>s,X\le Y) &= \int_0^\infty dx \int_0^\infty dy\, \rho_X(x)\rho_Y(y) {\bf 1}_{x>s} {\bf 1}_{y\ge x} = \\
&= \int_s^\infty dx \int_x^\infty dy\, \rho_X(x)\rho_Y(y) = \\
&= \int_s^\infty dx\, \rho_X(x) P(Y\ge x) = \\
&= \int_s^\infty dx\,\lambda e^{-\lambda x} \cdot e^{-\lambda x}= \\
&= \lambda \int_s^\infty dx\,e^{-2\lambda x} = \\
&= \lambda \frac{e^{-2\lambda s}}{2\lambda} = \\
&= \frac12 e^{-2\lambda s}
\end{align}


问:
Is a subset contained in a union of its superset?
For context, I'm taking an introductory real analysis course, and our current topic is intervals. One of the questions requires to prove that, for every $x, y$ in some real interval $I$, with $x<y$, the interval $[x,y]$ is also contained in $I$.

I'm taking a shortcut by proving it's a subset of an open interval $I = (a,b)$, then arguing that $[x,y] \subset (I \cup G)$, for some arbitrary set $G \subset \mathbb R$. From there, changing $G$ should span all the cases I'm looking for.

My question: is the provided argument sufficient? Is it a non-trivial fact in set theory? Or am I wrongly using my thesis to prove the hypothesis? Thanks in advance!

答:
I'm going to answer the question in your title, and I hope that it gives you some insight. Yes, a subset is contained in any union of its superset with a different set.

I would say this is a trivial set theoretic fact, however if you don't trust a statement it is always worth going about a proof for it because maths is all about proving to ourselves that we're right. Here's a proof for it:

Lets take $A \subset B$ but $A \not \subset B \cup C$ for some set $C$. This implies there exists $a \in A$ such that $a \in B$ (by the first fact) but $a \not \in B \cup C$. This is a very clear contradiction.

I question you saying you're going to "prove it's a subset of the open interval $I = (a,b)$." Surely doing that is what you're trying to do? You can then just union on the end points to get the half open and closed intervals.


问:
Linearly independent sets and span (polynomials)
We have the set $S = \{1,x-1,(x-1)^2,\dots,(x-1)^n\}$. They ask to proof whether or not it is linearly independent and find a basis of $span(S)$. I have proved it is linearly independent by the following method (which I don't quite know if it is correct):

We consider the linear combination

$a_0 + a_1(x-1) + a_2(x-1)^2 + \dots + a_n(x-1)^n = 0$

which has to be satisfied for all $x \in \mathbb{R}$. Then, differentiating both sides $n$ times, we have:

$a_1x + 2a_2(x-1) + \dots +na_n(x-1)^{n-1}=0 $

$\vdots$

$a_{n-1}x(n-1)! + a_n(x-1)n!=0$

$a_nx\cdot n!=0$

So, from the last equation, we obtain $a_n=0$, which implies that $a_{n-1}=0$ and so on until we get to $a_0 =0$. Therefore, we obtain that $S$ is linearly independent. Am I right? If so, what would be a basis for the $span(S)$, wouldn't it be $S$ itself?

I would really appreciate any kind of help. Besides, if anyone notices any flaw in my proof it will be welcomed!

Thank you!

答:
Perhaps this might help. Consider the polynomial

$$p(x)=a_0 + a_1(x-1) + a_2(x-1)^2 + \dots + a_n(x-1)^n = 0$$

This means that $p(t)=0$, where $t=x-1$ for every $t\in \mathbb{R}$. Note that a non-constant polynomial has at most a finite number of roots. One of these roots is zero; dividing by this root and invoking the same argument, we therefore have that $p(t)$ is a constant polynomial. In particular, it is the zero polynomial.


问:
Boundedness of the spectral radius of matrices $A^h$ as $h\to 0.$
I need to know if the following matrix has a bounded spectral radius $\rho(A)$, as $h\to 0:$
$$A^h=\frac{1}{h^2}\begin{pmatrix}
h^2-2h-2&2&0&0\dots &0\\
1&h^2-2&1&0\dots &0\\
0&1&h^2-2&1\dots &0\\
\dots&\dots&\dots&\dots\\
\dots&\dots& 1&h^2-2&1\\
\dots&\dots&\dots& 2&h^2-2h-2
\end{pmatrix}$$
where this is $(n+2)\times (n+2)$ matrix with $h = \dfrac{1}{n+1}.$ If anyone is curious, this came from an attempted finite difference method on the equation:
$$u''+u = f(x),\quad 0<x<1$$
$$u'(0) = u(0), u'(1) = u(1).$$
As this is neither circulant, nor Toeplitz matrix, I have not found a nice way to compute its eigenvalues; hence the spectral radius.

This matrix is almost tridiagonal, so when I wrote it as $A^h = B+C$ with $B = \frac{1}{h^2}\text{tridiag}(1,-2,1)$ and $C$ some really sparse matrix, $\rho(B)$ is bounded so I have some reason to hope that $\rho(A)$ also stays bounded.

答:
The eigenvalue equation in the middle rows reads as
$$
u_{k-1}+(h^2-2)u_k+u_{k+1}=\lambda h^2u_k,~~~ k=1,...,n.
$$
This has the general solution
$$
u_k=aq^k+bq^{-k}
$$
where $q$ solves the quadratic characteristic equation
$$
q^2+((1-λ)h^2-2)q+1=0\iff q+q^{-1}=2+(λ-1)h^2
$$
Your first and last line result from realizing the boundary conditions via ghost cells $u_{-1}$ and $u_{n+2}$, where the boundary conditions then read as
$$
u_1-u_{-1}=2hu_0\iff a(q-2h-q^{-1})+b(q^{-1}-2h-q)=0\\
u_{n+2}-u_n=2hu_{n+1}\iff aq^{n+1}(q-2h-q^{-1})+bq^{-n-1}(q^{-1}-2h-q)=0
$$
which requires $q^{2(n+1)}=1\iff q=\exp(i\frac{k\pi}{n+1})=\exp(i\pi h·k)$. For the corresponding eigenvalues one gets
$$
λ_k=\frac{2\cos(\pi h·k)-2}{h^2}-1, ~~~k=0,...,n+1
$$
For small $k$ one has $λ_k\approx -\pi^2k^2-1$, the largest is $λ_{n+1}=-4(n+1)^2-1$.

For the stability of the method one is probably more interested in that the smallest eigenvalue has an absolute value larger than $1$, so the matrices $A^h$ stay away from the set of singular matrices. That the condition number diverges is the reason that one goes to more structured representations of the function like multi-grid or wavelet methods.


答:
There's nothing mysterious going on here.  It's just a telescoping product.  You can prove these finite sums by induction.

The general case:

$$\prod_{i=j}^n \frac{a_{i+1}}{a_i} = \frac{a_{n+1}}{a_j}$$

Proof:

The proof will proceed by induction.  The base case $n=j$ is obvious.

Suppose our series does work for $n$.  Then $$\prod_{i=j}^{n+1} \frac{a_{i+1}}{a_i} = \prod_{i=j}^{n} \frac{a_{i+1}}{a_i} \cdot \frac{a_{n+2}}{a_{n+1}} = \frac{a_{n+1}}{a_j} \frac{a_{n+2}}{a_{n+1}} = \frac{a_{n+2}}{a_{j}}$$


问:
All the lines on the Segre quadric

Find all the lines on the quadric surface in $\mathbb P^3$ defined by the equation
   $$xw=yz$$
   (with the homogeneous coordinates $[x:y:z:w]$ of course).

Now it is well known that by the Segre embedding there is an isomorphism $\mathbb P^1 \times \mathbb P^1$ with our quadric. Moreover the images $\{[u_0,v_0]\}\times \mathbb P^1$ and $\mathbb P^1 \times \{[s_0,t_0]\}$ give two rulings of lines of our quadric. I have heard however from my teacher these are all the lines on the quadric. Is there any way to show this that is not too theoretically involved?

答:
Note that all the structure on $\mathbb{P}^1\times \mathbb{P}^1$ comes from the Segre embedding, given by
\begin{align*}
\Sigma_1^1:\mathbb{P}^1\times \mathbb{P}^1 &\rightarrow \mathbb{V}(X_0X_3-X_1X_2)\subset \mathbb{P}^3\\
([x:y],[z:w])&\mapsto [xz:xw:yz:yw].
\end{align*}
This is definitionally an isomorphism. The line through the points $([x:y],[z:w])$ and $([x':y']:[z':w'])$ gets mapped under $\Sigma^1_1$ to $$\ell = \{[xz:xw:yz:yw]\cdot s+[xz:xw:yz:yw]\cdot t\;|\; [s:t]\in \mathbb{P}^1\}.$$ If this lies entirely in the surface, we have
\begin{align*}
\forall [s:t]\in \mathbb{P}^1, \quad (xzs+x'z't)(yws+y'w't)-(xws+x'w't)(yzs+y'w't)&\equiv 0 \\
\Rightarrow st(xy'-x'y)(zw'-w'z)&\equiv 0
\end{align*}
Hence the line lies in the surface if and only if one of the following holds:

$xy'-x'y=0$, i.e. $[x:y]=[x':y']=[\lambda:\mu]$. Then the line is equal to $\{[\lambda:\mu]\}\times \mathbb{P}^1 $, and $\Sigma^1_1$ maps it to $\mathbb{V}(\lambda X_2 - \mu X_0, \lambda X_3-\mu X_1)$.

$zw'-w'z=0$, i.e. $[z:w]=[z':w']=[\lambda:\mu]$. Then the line is equal to $\mathbb{P}^1\times\{[\lambda:\mu]\}$, and $\Sigma^1_1$ maps it to $\mathbb{V}(\lambda X_1-\mu X_0, \lambda X_3-\mu X_2)$.

This shows that the two families above together constitute all the lines on the Segre quadric.

Title: All the lines on the Segre quadric


问:
Complementary book for Conway's book "A Course in Functional Analysis"
I am first year graduate student and I have taken Functional Analysis course this semester. The main book of teaching is Conway's book "A Course in Functional Analysis" but the lecturer is notorious for giving exams from questions that are not exercises of the book he's teaching. So I need an accompanying book for functional analysis with lots of cool problems that covers same materials of Conway's book and is on graduate level. Any suggestions would be much appreciated.</p>

答:
Below are some books that I use sometimes:

"Functional Analysis" written by Theo Bühler and Dietmar A. Salamon: https://people.math.ethz.ch/~salamon/PREPRINTS/funcana-ams.pdf

"Topics in Linear and Nonlinear Functional Analysis" written by Gerald Teschl: https://www.mat.univie.ac.at/~gerald/ftp/book-fa/fa.pdf

"Linear Functional Analysis" written by Hans Wilhelm Alt: https://link.springer.com/book/10.1007/978-1-4471-7280-2

"Functional Analysis - An Introductory Course" written by Sergei Ovchinnikov: https://link.springer.com/book/10.1007/978-3-319-91512-8

"An Introductory Course in Functional Analysis" written by Adam Bowers and Nigel J. Kalton: https://link.springer.com/book/10.1007/978-1-4939-1945-1

The book I most recommend is the following:

"Functional Analysis for the Applied Sciences" written by Gheorghe Moroşanu: https://link.springer.com/book/10.1007/978-3-030-27153-4

Although there is "for the Applied Sciences" in the title of the previous book, the book is rigorous and very good for beginners. The theorems have proofs and almost all exercises are solved at the end of the book.


问:
Changing the time in Brownian motion with a function?
Let $(B_t)$ be a Brownian motion on some probability space. Let $f:\mathbb R \rightarrow \mathbb R^+$ be a continuous function different from the identity. What is "$B_{f(t)}$" ? Clearly it is not a Brownian motion anymore. How can we rigorously define it and show that it exists (or not)?

What is there to know about "$B_{f(t)}$" ? Does it even make sense if $f$ is not monotone ? Can we determine its distribution ? Can we write it as $B_{f(t)}=g(B_t)$ for some function $g$ ?  How can we simulate one starting from $B_t$ ?

答:
I am merely a beginner on that subject, so this is by no means a comprehensive answer, but I'll still attempt to gather some elements that you may find useful :

First off, $B_{f(t)}$ as you said will not be a Brownian Motion, but as long as $f(t)$ is non-negative, the random variable $B_{f(t)}$ is still well-defined and we can still manipulate it as usual. In other words, the family of random variables $\{Y_t := B_{f(t)}, t\in\mathbb R \}$ defines a stochastic process on the image of $\mathbb R$ by $f$. It could be reindexed as $\{Y_T := B_T, T\in f(\mathbb R)\}$, and for all $T \in f(\mathbb R)$ we would have that $Y_T$ is a Gaussian random variable with mean $0$ and variance $T$.

Supposing that $f$ is continuous, we know that the image is some interval $f(\mathbb R) = [t_1,t_2]$ with $t_1,t_2 \ge 0$ and possibly $t_2 = \infty$. So essentially what you'd be looking at is a Brownian motion "truncated" to the interval $[t_1,t_2]$.

As I said, $B_{f(t)}$ only needs $f$ to be non-negative for it to be well-defined. By definition of $B$, we thus have that $B_{f(t)}$ is normal, centered and of variance $f(t)$ for all $t$. It thus follows that we can compute the process' mean $\mathbb E[B_{f(t)}] = 0$ and covariance function $\mathbb E \left[B_{f(t)}B_{f(s)}\right] = f(t)\wedge f(s)$. Since the original Brownian motion is a Gaussian process, it follows that the time-changed Brownian motion also is a Gaussian process, and we can then deduce its distribution from its mean and covariance function.

There are two well-known examples (that I can think of) of time-changing of Brownian motions which yield remarkable results :

If $c > 0$, then $B_{ct} \over \sqrt c$ is also a Brownian motion. That means that $B_{ct}$ is a Brownian motion "scaled" by a factor $\sqrt c$.

$tB_{1/t}$ is a Brownian Motion as well. This can be seen as a "time inversion" of the original Brownian motion.

Note that we cannot write something such as, e.g. , "$B_{ct} = \sqrt c B_t$" because this does not hold, they just have the same law (on the same filtration). I don't have any knowledge of cases where you could write $B_{f(t)} = g(B_t)$.

Finally, one very important results that is quite relevant to what you're asking for is the Dubins-Schwarz theorem. It states that every continuous local martingale is a time-changed Brownian motion, or more precisely :

Dubins-Schwarz theorem : Let $M$ be a continuous local martingale with respect to a filtration $(\mathcal{F}_t)_{t\ge0}$ such that $M_0 = 0$ and $\langle M\rangle_\infty = \infty$ a.s.. Then, let for all $t \ge 0$
$$T_t=\inf\{s\geq0:\langle M\rangle_s>t\}=\langle M\rangle_t^{-1} $$
It holds that :

${W={(M_{\langle M\rangle_t^{-1}})}_{t\geq0}}$ is a Brownian motion with respect to the time-stopped filtration ${{(\mathcal{F}_{T_t})}_{t\geq0}}$

${{(W_{\langle M\rangle_t})}_{t\geq0}={(M_t)}_{t\geq0}}$

So for some quite specific functions $f$ (which, in fact, also depend on $\omega$), we know that $W_{\langle M\rangle_t} := W_{f(t)}$ is, in fact, the original martingale $M$ (careful with the filtrations here) ! I recommend you give a read to both Djalil Chafaï's and George Lowther's notes to learn more about these results, and time changed Brownian motions in general.

Hope that helps.


问:
Find a constant $c$ that best approximates $X$ in $L^1$.
Well the excercise ask me:

Find a constant $c$ that best approximates $X$ in $L^1$ where $X = 2 \cdot \mathbb{1}_{[0,1/3]} - \mathbb{1}_{[2/3,1]}$.

Well I know that the result is 0 but it is impossible for me to reach  obtain $c = 0$.

My idea is the following:

$\text{min}_{c \in \mathbb{R}}$ $\mathbb{E}(|X-c|)$ = $\text{min}_{c \in \mathbb{R}}$ $\int_{0}^{1/3} |2-c|dx + \int_{1/3}^{2/3} |-c|dx + \int_{2/3}^1 |-1-c|dx$ = $\text{min}_{c \in \mathbb{R}}$ $\big{(} \frac{1}{3} (|2-c|+|c|+|1+c|) \big{)}$

And now we can see 4 different cases looking to the value of $c$:

If $c \in (-\infty,-1] \Longrightarrow  \text{min}_{c \in \mathbb{R}} \ \frac{1-3c}{3}$

If $c \in (-1,0] \Longrightarrow \text{min}_{c \in \mathbb{R}} \ \frac{3-c}{3}$

If $c \in (0,2] \Longrightarrow \text{min}_{c \in \mathbb{R}} \ \frac{3+c}{3}$

If $c \in (2,\infty) \Longrightarrow \text{min}_{c \in \mathbb{R}} \ \frac{-1+3c}{3}$

Well from this point I don't know how can I continue, because I have 2 options:

Aplying the notion of minimum by derivating, but as every function is linear the derivate will be something like $a = 0$ with $a \in \mathbb{R}$, so it will be impossible to obtain $c=0$.

Supossing that $\mathbb{E}(|X-c|)=0$ and obtaining the value $c$ for each case. But none give $c=0$ (the mean of the values of each case gives c=0, but I don't know if it is the way).

I don't know how can I continue and obtain $c=0$.

答: You have done the right thing so far. You don't need calculus to minimize $\big{(} \frac{1}{3} (|2-c|+|c|+|1+c|) \big{)}$, just sketch it. The minimum occurs at $c=0$. See, e.g. https://www.wolframalpha.com/input/?i=1%2F3*%28%7C2-c%7C%2B%7Cc%7C%2B%7C1%2Bc%7C%29C for a sketch.


问:
Are all sets in Set, with the same cardinality, isomorphic?
It is said, that all singleton sets are isomorphic(and terminal objects).

I wonder if this is true for any cardinality in the category of sets.

For example, are all two element sets isomorphic in the catergory of sets?

If not, why does this property only hold for singleton sets?

答:
By definition, two sets $A,B$ have the same cardinality if they are bijective to each other, i.e. there exist mutually inverse functions $f:A \to B$ and $g:B \to A$. This is precisely the definition of an isomorphism in $\mathbf{Set}$.

Singletons then of course turn out to be a special case. It seems however that you are reading a passage about terminal objects of which singletons form an example in $\mathbf{Set}$. Since all terminal objects are canonically isomorphic (due to universal property nonsense), the isomorphism immediately follows here.


答:
Let $ n\geq 2 $.

It is possible to prove that $ \mathcal{B}=\left(1,X,\cdots, X\left(X-1\right)\cdots\left(X-n+1\right)\right) $ is a basis of $ \mathbb{R}_{n}\left[X\right] $. To do so, we just need to prove the linear independence of $ 1 $, $ X $, $\cdots $, $X-n+1 $.

Anyway, that means there exists an unique $ \left(n+1\right)-$tuple $ \left(\alpha_{n,k}\right)_{0\leq k\leq n} $ such that : $$ X^{n}=\sum_{k=0}^{n}{\alpha_{n,k}X\cdots \left(X-k+1\right)} $$

Therefore if $ n,k\in\mathbb{N} $, we have for all $ i\in\mathbb{N} $ : $$ i^{k}=\sum_{j=0}^{k}{\alpha_{k,j}i\cdots\left(i-j+1\right)} $$

Thus : \begin{aligned}\sum_{i=0}^{n}{i^{k}\binom{n}{i}}&=\sum_{i=0}^{n}{\sum_{j=0}^{k}{\alpha_{k,j}i\cdots\left(i-j+1\right)\binom{n}{i}}}\\ &=\sum_{j=0}^{k}{\alpha_{k,j}\sum_{i=j}^{n}{i\cdots\left(i-j+1\right)\binom{n}{i}}}\\ &=\sum_{j=0}^{k}{\alpha_{k,j}\sum_{i=j}^{n}{n\cdots\left(n-j+1\right)\binom{n-j}{i-j}}}\\ \sum_{i=0}^{n}{i^{k}\binom{n}{i}}&=\sum_{j=0}^{k}{\alpha_{k,j}n\cdots\left(n-j+1\right)2^{n-j}}\end{aligned}

The $ \left\lbrace\alpha_{k,j}\right\rbrace_{0\leq j\leq k} $ are easy to find in particular cases, calculating the sum should be easy using the previous result, after finding the $ \alpha_{k,j} $.

Their might be a recurrence relation between the $ \alpha_{k,j} $. Finding it will help us to generalise further.

Title: Generalising $\sum_{r=0}^{n}r^{k}{n\choose r}$ for $k\in \mathbb{Z^{+}}$

Body:
I am trying to generalise the closed form expression for $k$ of the following sum using diffferentiation techniques.

$$\sum_{r=0}^{n}r^{k}{n\choose r}$$

We know the following results through trivial application of differentiation followed by multiplication at each step, and at the final step plugging in $x=1$.

$$\begin{aligned}\sum_{r=0}^{n}r{n\choose r}&=n2^{n-1}\\ \sum_{r=0}^{n}r^2{n\choose r}&=n(n+1)2^{n-2}\\\sum_{r=0}^{n}r^{3}{n\choose r}&=n^{2}(n+3)2^{n-3}\end{aligned}$$

Clearly there seems to be some patterns, as for the exponent, the degree of the polynomial function. I am not able to extend a similar pattern for $k=4$. Any ideas are appreciated. Thanks.


问:
Does a rectangle's diagonal to its perimeter have a ratio, like Pi?
The ratio of a circle's diameter to circumference merited its very own greek letter, Pi.

I have been wondering if there is such a similar, or important, ratio in a rectangle, i.e. its diagonal to perimeter?

Would I be right in guessing that it's because it's not a single value, but a more complex formula involving trigonometry? Or perhaps not a widely used one?

Many thanks in advance.

答:
For any regular n-gon define its radius as the radius of its incircle and $\pi_n$ as the ratio of its perimeter and its diameter.  Then, of course, by definition, is $2\pi_nr$ and it's easy to show that its area equals $\pi_nr^2$.

For example, $\pi_3=3\sqrt3$ and $\pi_4=4$; in general $\pi_n=n\cdot\tan(\pi/n)$.


问:
A question about $\det(A^TA)$
Assume that $A\in\mathbb{R}^{m\times n}$, where $m\leq n$. The aim is to find a lower and an upper bound for $\mathrm{det}(A^TA)$. What can be said about this issue?

答:
PS: I think the singular value decomposition of $A$ can help us to find an answer to this question. In fact, let us assume that $A=U\Sigma V^T$ in which $U\in\mathbb{R}^{m\times m}$ and $V\in\mathbb{R}^{n\times n}$ are orthonormal matrices, i.e., $U^TU=I_m$ and $V^TV=I_n$. Moreover, the matrix $\Sigma$ is defined as
$$\Sigma=diag(\sigma_1,\sigma_2,\ldots,\sigma_r,0,\ldots,0)_{m\times n}\quad\text{s.t.}\quad \sigma_1\geq\sigma_2\geq\cdots\geq\sigma_r,$$
in which $\sigma_1$ is the largest singular value of $A$. Following this discussion, we have
$$\mathrm{det}(A^TA)=\mathrm{det}(V\Sigma^TU^TU\Sigma V^T)=\mathrm{det}(V\Sigma^T\Sigma V^T)=\mathrm{det}(\Sigma^T\Sigma)=\prod_{i=1}^r \sigma_i^2\leq\sigma_1^{2r}.$$
On the other hand, since $0\leq\mathrm{det}(A^TA)$, we can say that
$$0\leq\mathrm{det}(A^TA)\leq \sigma_1^{2r}.$$

For this reason, $0$ and $\sigma_1^{2r}$ can be considered respectively as a lower bound and an upper bound for $\mathrm{det}(A^TA)$. Is it possible to find other upper bounds for $\mathrm{det}(A^TA)$?

Title: A question about $\det(A^TA)$