Body:
$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.

key points: This is supposed to be solved with knowledge of just L'hospital's rule.
*The value of limit i.e, k is finite and non zero

I have tried manipulation using logarithm which didn't turn out to be of good use. I think multiplying by conjugate should work but that just  gives complex derivatives  and doesn't seem to lead to any answer.
Am I missing some basic point or is there a specific procedure for this problem.

Body:
Set $1/x=h$ to find

$$S=\lim_{h\to0^+}\left(\dfrac{(1+5h+3h^4)^c}{h^{4c}}-\dfrac1h\right)$$

If $4c=1+d, d>0$

$$S=\lim_{h\to0^+}\left(\dfrac{(1+5h+3h^4)^{\frac{d+1}4}}{h^{1+d}}-\dfrac1h\right)=\lim_{h\to0^+}\dfrac{(1+5h+3h^4)^{\frac{d+1}4}-h^d}{h^{1+d}}=?$$

What if $4c=1-d, d>0?$

So, we need to have $4c=1$

$$S=\lim_{h\to0^+}\left(\dfrac{(1+h(5+3h^3))^{1/4}-1}h\right)$$

Use Binomial series  to find $$S=\dfrac54$$


问:
Prove there exist $2 \times 2$ matrix $B$ such that $B^3=A^2$, for every $A$.
Prove the following, for every $2 \times 2$ matrix $A$ with complex entries, there exist $2 \times 2$ matrix $B$ with complex entries, such that $$B^3=A^2$$ (working in the field of complex numbers)

My attempt so far, take the Jordan decomposition of $A$ to get $$A=UJU^{-1}$$
then $$A^2=UJ^2 U^{-1}$$

I have  $J=D+S$ where $D$ is diagonal and $A$ is nilpotent, for $2 \times 2$ matrices the only nilpotent matrix is $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$
Then $$J^2=(D+S)^2=D^2+DS+SD$$
Now chose $G$ such that $$G^3=J^2$$ (can I do that?) then finally I have $$B^3=UG^3U^{-1} =UJ^2U^{-1}=A$$.

I am stuck at this point.

Is this the correct approach?
What should I do next?

答:
Working with Jordan forms basically does it.

If $J = \operatorname{diag}(a,d)$ is diagonal, then just set $G = \operatorname{diag}((a^2)^{1/3}, (d^2)^{1/3}).$

If $J$ isn't diagonal, then $J = aI + S$ where $S=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, so $$J^2 = a^2I + 2aS.$$  Now, setting $G = (a^2)^{1/3}I + xS$ and setting equal the coefficients of $S$ in $G^3 = J^2$ gives $3((a^2)^{1/3})^2x = 2a$, which may be solved for $x$ if $a\neq 0$ and any $x$ works if $a=0$.

ContentLicense: CC BY-SA 4.0


问:
Find all the functions $f:\left[0,1\right]\rightarrow\mathbb{R}$ so that ${x}'f\left(x\right)\geq 2xf\left(x^{2}\right)$

Find all the functions $f:\left [ 0, 1 \right ]\rightarrow\mathbb{R}$ on the interval $\left [ 0, 1 \right ]$ so that
$${x}'f\left ( x \right )\geq {\left ( x^{2} \right )}'f\left ( x^{2} \right ),$$
in other words, that $f\left ( x \right )\geq 2xf\left ( x^{2} \right ).$

I simplify the original problem, I think we need to find the increasing function $f\left ( x \right )= {g}'\left ( x \right )$ since $x\geq x^{2}$ but are these all ? And how to find at least one function $g\left ( x \right )$ like that.. is it necessary to add the condition that the function $f$ is continuous ? Thanks for helping.

答:
A discontinuous solution is $f:[0,1]\to[0,\infty)$ given by $$f(x) =\begin{cases} \frac{-1}{\log x} & x\in(0,1), \\0  &x=0\text{ or } 1.\end{cases}$$Then there is equality at $x=0$ and $x=1$, and for $x\in(0,1)$, $f(x^2) = \frac{-1}{2\log x} = \frac12 f(x)$, so it is an exact solution to
$$ f(x) = 2f(x^2)$$
since  for $x\in (0,1)$, $2\ge 2x$ and $f\ge 0$, it solves the inequality
$$ f(x) \ge 2xf(x^2).$$
The same proof also shows that for any
$a,b\ge 0$, the following function is also a solution: $$f(x) =\begin{cases} \frac{-a}{\log x} & x\in(0,1), \\b  &x=0, \\ 0 &x=1.\end{cases}$$

In fact, we can create many more solutions by using the extra space in the inequality $2x\le2$ better. Given a positive solution to $f(x)\ge 2f(x^2)$ if $h$ is a positive solution to $h(x)\ge x h(x^2)$ then $\tilde f:= hf$ is another solution to the inequality in the question. It is easy to show that we can use $h(x)=x^{r}$ for $r\ge -1$ and $h(x)=(1-x)^s$ for $s\le0$.

Furthermore, it is also similarly easy to show that $1/\log(1-x)$ is a negative solution to $2f(x^2)\le f(x)$. Due to the sign change, the relations for $r$ and $s$ in determining if $x^r$ and $(1-x)^s$ can be used are flipped.

Finally the inequality is additive in the function $f$. Thus, if $f_1$ and $f_2$ are solutions then so is $f_1+f_2$.

So for instance, another solution is $$\frac{x^{-3}(1-x)^{100}}{\log(1-x)}-\frac{(1-x)^{-2}}{\log x}$$
This one blows up as you approach both endpoints (after suitable redefinition at the endpoints so that they are indeed functions on $[0,1]$).


问:
Why is $ y $ the argument of $ e^z $?
I'm having a hard time understanding how come $y$ is the argument of $e^z$. I know $e^x$ is the real part and it makes sense why this is the modulus. But all proofs i've found about the $ e^z=e^xe^{iy} $, just doesn't make any sense to me.

Truly hope some of you can clarify this.

I need a thoroughly guidance, sense this is a key element for later uses in the complex numbers.

Thanks in advance.

答:
Let $z = x+iy$, then $e^z = e^{x+iy}=e^xe^{iy} = e^x(\cos y + i\sin y)$ by Euler.
So the modulus is $e^x$ and the argument is $y$, not $e^{iy}$.

Note that this means that lines parallel to the real-axis (so with constant imaginary part $y$) are mapped by $e^z$ to straight half-lines through the origin... (but never achieving $0$ of course).

Title: Why is $ y $ the argument of $ e^z $?


问:
A simple probability problem
I'm trying to solve the following problem:

Given a person, the probability of it being a male is 30%, and the probability of a male being taller than 160cm is 60%. For a female person, the probability of being shorter than 160cm is 70%. Given that a person is taller than 160cm, what's the probability of this person being a male ?

My attempt of a solution is:

Probability of a person being taller than 160cm: $P(\text{tot}) = P(\text{m}) + P(\text{f})$,
where:

$P(\text{m}) = P(\text{male}) \cdot P(\text{male} > 160\text{cm}) = 0.3  \cdot  0.6 = 0.18$

$P(\text{f}) = P(\text{female}) \cdot P(\text{female} > 160\text{cm}) = 0.7  \cdot 0.3 = 0.21$

so we have $P(\text{tot}) = 0.39$

than the probability asked by the problem should be $P\text{(m})/P(\text{tot}) \approx 0.46$.

答:
You can divide it as follows: If there are $100$ people in the population,

 $30$ are male, $70$ are female.

 Out of those thirty males, $60\%$, i.e. $18$ are taller than $160$ cm.

 Out of the $70$ females, $70\%$, i.e. $49$ are shorter than $160$ cm (other $21$ are taller, assuming no one's height is $160$ cm)

 Now there are $18+21=39$ tall people. Out of those, $18$ are male. So the required probability is $${18\over39}={6\over13}\approx0.46$$
It's a lot easier doing it this way instead of involving sets; but use whatever works for you

Title: A simple probability problem


问:
calculating the area of a circle by integrating over straight lines

The Problem

Imagine having a circle of radius $R$. We all know the area of this circle is $A=\pi R^2$ and the circumference is $2\pi R$. I would like to know why this approach for calculating the area is wrong, and why it gives me the circumference instead:

If I have a line of length $R$, I can rotate it around its edge a full rotation ($2\pi$ rad) and it would cover the whole circle, meaning it should give me the area, so we have the area be equal to $A=2\pi R$ or alternatively $A=\int\limits _{0}^{2\pi}Rd\theta =2\pi R$.

Note: By rotating the line I mean taking the original line (purple in the image) and then placing its end (which isn't in the center of the circle) on some other point on the circumference of the circle, as seen in the picture (the red lines), doing so for every possible point on the circle would look like the entire area of the circle is colored red, since its all covered by the rotations of this one purple line.

Note: I tried to look at other similar SE questions such as this one, but they didn't answer my question well.

Context

The reason I thought of this is because a few days ago I tried to calculate the electric field of a ring centered at $\left(0,0,0\right)$ with some inner radius $R_1$  and outer radius $R_2$ and charge density $\sigma = \frac{\lambda}{r^{3}}$, at the point $\left(0,0,z\right)$ for some $z$. since the ring is symmetrical in the $x,y$ directions the field has affect in the $z$ direction at that point. I thought about finding the field one "line" in the ring causes, and multiply that by $2 \pi$, but I found this approach to be incorrect.

答:
Consider in your illustration, an outer ring of some width $h$. The "spokes" covering it have an area of $9hw$, where $w$ is the width of each spoke. But near the center, spokes covering another ring of width $h$ would also have an area of $9hw$, even though the area of that ring is less. So it ends up weighing the area near the center disproportionately more than the area close to the edge of the circle.

You can do what Yves Daoust suggested of adding up triangles, or you could add up rings. For the latter, you can add up the areas of small rings (which is the circumference of the ring times $dr$) as $$\int_0^R 2\pi r dr$$


问:
Can you use an equals sign if the answer is rounded to two decimal places?
For my latest maths PSMT, the teacher has specified that equals signs should only be used if the values are actually equal. Does that mean I should use an approximation sign if I'm rounding the values to two decimal places?

This is what I've been doing so far...

\begin{align}
&x = -5.057588.... \\
&x = -5.06\;\text{(2 d. p.)} \\
\end{align}

Should I be using an approximation sign instead? Should I remove the $\text{(2 d. p.)}$ at the end?

答:
If you intend to be strictly precise, then :

\begin{align}
&x \approx -5.057588.... \\
&x \approx  -5.06\;\text{(2 d. p.)} \\
\end{align}

because at base those numbers are only ever approximations.

On the other hand, saying "$x$ is equal to $-5.06$ correct to $3$ significant figures" could be interpreted as needing an equals sign, so I believe either could be used here. I anticipate disagreement with other contributors as to which is strictly correct, either $x = -5.06 \ \text {(to $3$ s.f.)}$ or $x \approx -5.06 \ \text {(to $3$ s.f.)}$

Also please note that there are purists who say that "$\text {(to $2$ d.p.)}$" presupposes that you do not round the last place, but that it should be "$x \approx -5.05 \ \text {(to $2$ d.p.)}$" because you are quoting the decimal places, not the significant figures of accuracy.

Hence it may be a good habit to get into to prefer the use of significant figures when discussing the precision and/or accuracy of a number, and not decimal places. The former is more general and is unchanged when changing the units of measurement: $5.06 \, \mathrm {cm} \ \text {(to $3$ s.f.)}$ is the same thing as $50.6 \, \mathrm {mm} \ \text {(to $3$ s.f.)}$ is the same thing as $0.00506 \, \mathrm {m} \ \text {(to $3$ s.f.)}$, but you can't say the same about the decimal places.

However, it all depends on whether you are using mathematics as a professional in an engineering field, or whether you are merely using it to teach schoolchildren. I believe the goals are different.


答:
Let $T$ (tall) be the event that a person is taller than 160 cm. $M$ to be male and $F$ to be female. So we know $P(T| M)= 0.6$ and $P(M)=0.3$ and $P(T|F)=0.3$. $P(F)=0.7$ follows from the assumption that the total population consists only of male and female so that $P(M) + P(F)=1$

We are asked for $$P(M|T) = \frac{P(M \cap T)}{P(T)}$$

$$P(T) = P(T|M)P(M) + P(T|F)P(F) = 0.6 \times 0.3 + 0.3 \times 0.7$$

while $$P(M \cap T) = P(T|M)P(M) = 0.6 \times 0.3$$

substituting we get $P(M|T) = \frac{18}{39}$, which is about the same as your answer, obtained in a different way.

Title: A simple probability problem

Body:
I'm trying to solve the following problem:

Given a person, the probability of it being a male is 30%, and the probability of a male being taller than 160cm is 60%. For a female person, the probability of being shorter than 160cm is 70%. Given that a person is taller than 160cm, what's the probability of this person being a male ?

My attempt of a solution is:

Probability of a person being taller than 160cm: $P(\text{tot}) = P(\text{m}) + P(\text{f})$,
where:

$P(\text{m}) = P(\text{male}) \cdot P(\text{male} > 160\text{cm}) = 0.3  \cdot  0.6 = 0.18$

$P(\text{f}) = P(\text{female}) \cdot P(\text{female} > 160\text{cm}) = 0.7  \cdot 0.3 = 0.21$

so we have $P(\text{tot}) = 0.39$

than the probability asked by the problem should be $P\text{(m})/P(\text{tot}) \approx 0.46$.


问:
Find the critical points and say whether they are maxima, minima or saddle points
I have this problem:

Find the critical points and say whether they are maxima, minima or saddle points

$$f(x,y)=x^2y(2−x−y)$$

My answer:

$f_x = xy (4-3 x-2 y) $

$f_y = -x^2 (-2+x+2 y)$     then

$xy (4-3 x-2 y)=0 , -x^2 (-2+x+2 y)=0$ the solutions are:

1)x=0 and $y \in{} \mathbb{R}$

2)x=1 and y=1/2

3)x=2 and y=0

I have three critical points,

$f_{xx} = -2 y (-2+3 x+y) $

$f_{yy} = -2 x^2 $

$f_{xy} = x (4-3 x-4 y) $, using

$D=f_{xx}f_{yy}-(f_{xy})^2 = -16 x^2+24 x^3-9 x^4+24 x^2 y-12 x^3 y-12 x^2 y^2$

x=1 , y=1/2 then D=0 but $f_{xx}=-2y (-2+3 x+y)= -2/3$ concluding (1,1/2) is a maxima point

x=2 , y=0 then D=-16 we say that (2,0) is a saddle points

x=0 , $y \in{} \mathbb{R}$ then D=0 the point  (x,a) with $a \in{} \mathbb{R}$ is a saddle points because for example if x=0 , a=-8 $fxx=-160$ (maxima) but if x=0 , a=1 then $f_{xx}=2$ (minima)

Is my exercise correct?

答:
Look at the Hessian matrix and consider its eigenvalues. If the determinant is negative, then you get a saddle point. If the determinant is zero, the test is inconclusive; if the determinant is positive and $f_{xx} > 0$, then you get a minima; and if the determinant is positive and $f_{xx} < 0$ , then you have a maxima.
Now, can you see whether you are correct?

Hessian matrix is just the matrix

\begin{pmatrix}
f_{xx} & f_{xy} \\
f_{yx} & f_{yy} \\
\end{pmatrix}

ContentLicense: CC BY-SA 3.0


问:
A sequence is defined by $x_1=1$, $x_{n+1}= \sqrt{8+x_n}$. Prove each term of the sequence after the first is greater than the preceding term.
I tried to use mathematical induction, but letting $n=k$ and $n=k+1$ is getting nowhere. I also tried squaring both sides to get rid of the square root but that doesn't seem to help either. Need some hints...

After reading some hints,
For $n=k$, assume $x_k > x_{k-1}$

$\sqrt{8+x_{k-1}} > \sqrt{8+x_{k-2}}$

Prove it is true for $k+1$,

$x_k = \sqrt{8+x_{k-1}}$

Squaring both sides,

$8+x_k > 8+x_{k-1}$

$x_k > x_{k-1}$

Since it is true for $n=k+1$, it is also true for $n=k$.

Seems right?

Body:
$$n=1: \;\; x_n = x_1 =1<3=\sqrt{8+1}=\sqrt{8+x_n}=x_{n+1}$$
S.t. it holds for $n$.
$$\begin{aligned}n \mapsto n+1: \;\;\;& x_{n+1} \;<\; x_{n+2} \\[5mm] \iff\;\;& x_{n+1} < \sqrt{8 + x_{n+1}} \\[5mm] \iff\;\;& \sqrt{8+x_n} \;<\; \sqrt{8+\sqrt{8+x_n}}\\[5mm] \iff\;\;& 8 + x_n \;<\; 8 + \sqrt{8 + x_n} \\[5mm] \iff\;\;& x_n \;<\; \sqrt{8+x_n} \;\; \iff \;\; x_n \;<\;x_{n+1}\end{aligned} $$
S.t. it is also true for $n+1$, therefore completing the proof by induction.


问:
$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.

$\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.

key points: This is supposed to be solved with knowledge of just L'hospital's rule.
*The value of limit i.e, k is finite and non zero

I have tried manipulation using logarithm which didn't turn out to be of good use. I think multiplying by conjugate should work but that just  gives complex derivatives  and doesn't seem to lead to any answer.
Am I missing some basic point or is there a specific procedure for this problem.

答:
$$L=\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$$.
Here $c$ needs to be $1/4$ so that asymptotically the first part in above go as $x$, hence
$$L=\lim_{x\to \infty} (x^4 +5x^3+3)^{1/4}-x=k$$.
$$L=\lim_{x\to \infty} x (1 +5/x+3/x^4)^{1/4}-x=k$$.
Apply binomial expansion
$(1+z)^{p}=1+pz+O(z^2)+...$ id $|z|<<1$
$$L=\lim_{x \to \infty} (x+5/4+3/(4x^3)+...)-x=5/4=k$$


问:
How does Sobolev norm change with respect to domain size?
Suppose $u\in H^s(0,r)$, let's say $s\ge 0$. It then follows that $u|_{(0,r')}\in H^s(0,r')$ for any $r'<r$.  But is it true that $|u|_{H^s(0,r)}\rightarrow 0$ as $r\rightarrow 0$?  How about negative s, too?

答:
I believe it is true using Lebesgue dominated convergence for $s \geq 0$. For exemple for $s=1$ you write, for $r' < r$ :

$$\int_{B_{r'}} |u|^2 \ \mathrm{d}x + \int_{B_{r'}} |\nabla u|^2 \ \mathrm{d}x = \int_{B_{r}} \mathcal{1}_{B_{r'}}|u|^2 \ \mathrm{d}x + \int_{B_{r}} \mathcal{1}_{B_{r'}}|\nabla u|^2 \ \mathrm{d}x $$
and you use Lebesgue dominated convergence when $r' \rightarrow 0$ on each integrals.


问:
How can I work this limit exercise?
Consider points M, N and O with coordinates (1, 0), (0, 1) and (0, 0), respectively, additionally P point with coordinates (x, y) in the graph $y = \sqrt{x}$ . Calculate:

(a) $\lim_{x \to 0^{+}} \frac{perimeter \bigtriangleup NOP}{perimeter \bigtriangleup MOP}$

(b) $\lim_{x \to 0^{+}} \frac{area \bigtriangleup NOP}{area \bigtriangleup MOP}$

One of my first ideas for this exercise, after drawing the points M, N, O and the $y = \sqrt{x}$ graph, was to see the formulas of perimeter and area for triangles.

perimeter $\bigtriangleup $: is the sum of all its three sides.

area $\bigtriangleup $: is always half the product of the height and base

However, I do not find how to use the formulas of perimeter and area of a triangle with a point P = (x, y)  in the graph $y = \sqrt{x}$.

答:
$perimeter \bigtriangleup NOP = 1(NO) + \sqrt{x^2+(\sqrt{x}-1)^2}(NP) + \sqrt{x^2+(\sqrt{x})^2}(OP)$

$perimeter \bigtriangleup MOP = 1(MO) + \sqrt{(x-1)^2+(\sqrt{x})^2}(MP) + \sqrt{x^2+(\sqrt{x})^2}(OP)$

$area \bigtriangleup MOP = \frac{1}{2} * 1 * \sqrt{x}$

$area \bigtriangleup NOP = \frac{1}{2} * 1 * x$

I think you can finish remain.

Title: How can I work this limit exercise?


问:
Is learning (theoretical) physics useful/important for a mathematician?
I'm starting to read The Princeton Companion to Mathematics, at the beginning it says:

A proper appreciation of
  pure mathematics requires some knowledge of applied
  mathematics and theoretical physics.

Some of my professors have told me that modern Mathematics require some knowledge about Quantum Mechanics and theoretical Physics.

I attended the second and third day of the José Adem Memorial Lecture Series by Matilde Marcolli at the beginning of this year, they were about Number theory, Quantum statistical Mechanics and Quantum Field theory. I did not understand a single word but somehow the few things I understood had a big impression on me.

Please give me some examples of pure Mathematics that require/use Physics.

答: No, one does not need to know anything about physics to learn mathematics. But learning theoretical physics has at least two advantages. Theoretical physics could be seen as a set of good examples of mathematics, so this has some didactic value. And, if you are a math researcher, a lot of new math were born out of physical problems! So it's probably a good idea to keep an eye out for what's going on in the world of physics.</p>


问:
Number Of Solutions $X^{2}=X$
The equation $x^{2}=x$ $,x\in \mathbb{R}$ has only the solutions $x=0$ and $x=1$.

For a $2 \times 2$ matrix X, how many solutions does $X^{2}=X$ have?

What if $X$ is symmetric?

答:
Infinitely many solutions. And a lot of structure. Since you said "symmetric", I'll detail the real case. But note that the argument of the first paragraph below shows that the solutions in $M_n(K)$, for any field $K$, split by diagonalization into $n+1$ similarity orbits under $GL(n,K)$ of the obvious diagonal solutions.

1) Elements $p=p^2$ are called idempotents. Equivalently, these are the diagonalizable (just think about the minimal polynomial) matrices with spectrum in $\{0,1\}$. An idempotent is characterized by the decomposition of the vector space into the direct sum of its range and its nullspace. In $M_n$ ($\mathbb{R}$ or $\mathbb{C}$) in general, they split in $n+1$ connected components according to their rank, which is also equal to their trace. Each component corresponds to a similarity orbit. The natural representatives are the diagonal idempotents $0_n$ and $I_n$ (which are alone in their orbits), and $(1,\ldots,1,0,\ldots,0)$ with $k$ $1$'s, $1\leq k\leq n-1$ (whose orbit is a manifold of dimension $2k(n-k)$).

In $M_2(\mathbb{R})$, there are therefore three components: $\{0_2\}$, $\{I_2\}$, and the rank one idempotents. That is, the $2\times 2$ matrices whose characteristic polynomial is $X^2-X$:
$$
\pmatrix{a&b\\c&d}\qquad a+d=1\qquad ad-bc=0
$$
I let you work on these two equations to realize that this manifold is in affine bijection with the one-sheet hyperboloid. If you want a parametrization, here is a rational one for all but a subset of them of topological dimension one:
$$
\pmatrix{\frac{1}{1+st} &\frac{s}{1+st}\\\frac{t}{1+st}&\frac{st}{1+st}}\qquad (s,t)\in\mathbb{R}^2\setminus\{1+st=0\}.
$$

2) Elements $p=p^*=p^2$ are called projections (=self-adjoint idempotents) in operator algebras. They are characterized by their range solely, as their nullspace is the orthogonal of their range. Again, they split into $n+1$ components according to their rank. The rank $k$ component is called the Grassmannian $G(k,n)$ and has dimension $k(n-k)$, half (we dropped the nullspace) of the dimension of the corresponding idempotent component in which it lies as a submanifold.

In $M_2(\mathbb{R})$, we still have three components. I let you check that the nontrivial one, rank one projections, can be parametrized by
$$
\pmatrix{\cos^2\theta&\cos\theta\sin\theta\\ \cos\theta\sin\theta&\sin^2\theta}\qquad \theta\in [0,\pi].
$$
It should not surprize you that we recover the complex unit circle. These are the one-dimensional subspaces of $\mathbb{R}^2$. That is the projective line. Note that unlike rank one idempotents, it is now compact.

Title: Number Of Solutions $X^{2}=X$


答:
First you may find $c$ by considering that

$$0=\lim_{x\to \infty}\frac kx =\lim_{x\to \infty}\frac{(x^4+5x^3+3)^c - x}{x}= \lim_{x\to \infty}\left(x^{4c-1}(1+\frac 5x+\frac 3{x^4})^c - 1\right)$$

Hence, $\lim_{x\to \infty }x^{4c-1}=1\Leftrightarrow c=\frac 14$.

Now, you can use L'Hospital to find

\begin{eqnarray*}\sqrt[4]{x^4+5x^3+3} - x
& = & \frac{\sqrt[4]{1+\frac 5x + \frac 3{x^4}}-1}{\frac 1x} \\
& \stackrel{t=\frac 1x}{=} &  \frac{\sqrt[4]{1+5t + 3t^4}-1}{t} \\
& \stackrel{L'Hosp.,\; t\to 0^+}{\sim} & \frac{5+12t^3}{4\sqrt[4]{(1+5t + 3t^4)^3}}\\
& \stackrel{t\to 0^+}{\longrightarrow} & \frac 54
\end{eqnarray*}

Title: $\lim_{x\to \infty} (x^4 +5x^3+3)^c-x=k$. Find $k$ and $c$.


问:
Bounded random variables $X,Y$ satisfying $\mathbb{E}(X^mY^n) = \mathbb{E}(X^m)\mathbb{E}(Y^n)$ for every $m, n\in\mathbb{N}$ are independent

Suppose $X, Y$ are bounded random variables and we have that for every $m, n$ positive integers, $\mathbb{E}[X^mY^n] = \mathbb{E}[X^m]\mathbb{E}[Y^n]$. Then show that $X, Y$ are independent.

I have some idea of how this is supposed to go:

Using linearity, etc, $\mathbb{E}[f(X)g(Y)]$ = $\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ for all polynomials $f, g$.

After this we use limit theorems to extend this across all measurable functions, and thus characteristic functions, which will let us conclude that for every measurable set $A, B$, we have that $\mathbb{E}[1_{X \in A}\cdot 1_{Y \in B}] = \mathbb{E}[1_{X \in A}\cdot 1_{Y\in B}]$ which reduces to $P(X \in A, Y \in B) = P(X \in A)P(Y \in B)$ which gives independence.

But all the limit identities and how to use them has never been natural to me, so I wanted to verify and ask for help for the details.

First, for continuous functions $f, g$, we use the Stone-Weierstrass theorem to get polynomials sequences $f_n, g_n$ that converge uniformly to $f, g$ on the bounded interval that $X, Y$ belong to. Then $\mathbb{E}[f_n(X)] \rightarrow \mathbb{E}[f(X)]$, and similar for $Y, g$. Which theorem exactly are we using here for this convergence (assuming I'm not doing something completely wrong of course); dominated convergence theorem? Once we get this, we also then demonstrate that $\mathbb{E}[f_n(X)g_n(Y)] \rightarrow \mathbb{E}[f(X)g(Y)]$ which requires showing $f_ng_n \rightarrow fg$, but this just levarages standard analysis techniques and uses the uniform convergence, am I correct?

Next we want to extend this to all measurable functions $f, g$. For this we use the idea that there are sequences $f_n, g_n$ that will converge a.e pointwise to $f, g$. Is this enough to conclude that $\mathbb{E}[f_n(X)] \rightarrow \mathbb{E}[f(X)]$, and similar for $Y, g$? It seems we need stronger assumptions here. I'm not entirely sure how valid this last step is and would appreciate some detail.

答:
Your strategy almost works, except that not every simple function is the pointwise limit of continuous functions (e.g., $1_{\mathbb{Q}\cap[0,1]}$ since it has Baire class 2), so you need to find a way around that.  The usual way (asked before many times, see e.g. here, here and here) is to go via the characteristic function $\varphi_{\vec{X}}(\vec\theta):=\mathbb{E}[\exp(i\vec{\theta}\cdot\vec{X})]$ or its cousin the moment generating function, but let me describe a more measure-theoretic way with the monotone class theorem.

Let's start from the beginning.  We could transfer the problem to $\Omega=[-M,M]^2$ and $X,Y$ being the coordinate function without any loss of generality.

Stone-Weierstrass and uniform convergence: if $X_n\to X$ uniformly, then $\mathbb{E}X_n\to\mathbb{E}X$ since $\mathbb{E}\lvert X_n-X\rvert\leq\sup\lvert X_n-X\rvert\to 0$.  In particular, if $f,g$ are continuous, then there exists $f_n,g_n$ polynomials converging uniformly to $f,g$.  Then $(x,y)\mapsto f_n(x)g_n(y)\to f(x)g(y)$ uniformly and thus $\mathbb{E}[f_n(X)g_n(Y)]\to\mathbb{E}[f(X)g(Y)]$.  Hence $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ for all continuous $f,g$.

Then we appeal to

Monotone Class Theorem (for Functions) Let $\mathcal{K}$ be a collection of bounded $\mathbb{R}$-valued functions on $\Omega$ closed under multiplication, and $\mathcal{B}=\sigma(\mathcal{K})$.  Suppose $\mathcal{H}\supseteq\mathcal{K}$ is a vector space over $\mathbb{R}$ of bounded $\mathbb{R}$-valued functions containing $1_\Omega$ and closed under uniform limit of nonnegative functions nondecreasing to a bounded function.  Then $\mathcal{H}$ contains $\mathcal{B}_b$, the set of all bounded $\mathcal{B}$-measurable $\mathbb{R}$-valued functions.

Here we use $\mathcal{K}_1$ is our polynomials on $[-M,M]$ and $\mathcal{H}_1$ the set of all functions $f$ such that $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ for all $g\in\mathcal{K}_1$.  By MCT, $\mathcal{H}_1$ contains all bounded measurable functions on $[-M,M]$.  Now MCT again, with $\mathcal{K}_2=\mathcal{B}_b$ and $\mathcal{H}_2$ being those $g$ satisfying $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ for all $f\in\mathcal{B}_b$ gives all $g\in\mathcal{B}_b$ too.


问:
Injectivity of a map between manifolds
I'm learning the concepts of immersions at the moment. However I'm a bit confused when they define an immersion as a function $f: X\rightarrow Y$ where $X$ and $Y$ are manifolds with dim$X <$ dim$Y$ such that $df_x: T_x(X)\rightarrow T_y(Y)$ is injective.

I was wondering why don't we let $f$ be injective and say that's the best case we can get for the condition dim$X <$ dim$Y$(since under this condition we can't apply the inverse function theorem)?

Also does injectivity of $df_x$ inply the injectivity of $f$ (it seems that I can't prove it)?

How should we picture immersion as (something like the tangent space of $X$ always "immerses" into the tangent space of $Y$)?

Thanks for everyone's help!

答:
I think a glimpse on the wikipedia's article helps. Immersions usually are not injective because the image can appear "knotted" in the target space.

I do not think you need $\dim X<\dim Y$ in general. You can define it for $\dim X=\dim Y$, it is only because we need $df_{p}$ to have rank equal to $\dim X$ that made you "need" $\dim X\le \dim Y$.

I think you can find in classical differential topology/manifold books (like Boothby's ) that an immersion is locally an injection map $$\mathbb{R}^{n}\times \{0\}\rightarrow \mathbb{R}^{m+n}$$You can attempt to prove this via inverse function theorem or implicit function theorem. The proof is quite standard.

Title: Injectivity of a map between manifolds


问:
Does variation of parameters always work?
I want to solve the differential equation $y'-2y=e^x$.

I solved the homogenous differential equation and I got that $y_0(x)=e^{2x}\cdot c$, where $c$ is a real constant.

Now I want to apply variation of parameters. So, $y_p=c(x)\cdot e^{2x}$ and $y'_p=c'(x)\cdot e^{2x}+2\cdot c(x)e^{2x}$ and if I plug it into my equation I get that $c'(x)\cdot e^{2x}=e^x$, so $c'(x)=e^{-x}$ and $c(x)=-e^{-x}$. But this doesn't verify my equation. What did I do wrong?

答: The particular solution is not $c(x)$ but $y_p = e^{2x}c(x) = -e^{x}$, which verifies the equation.


问:
Proposition IV.2.3 in Hartshorne

(2.2) Here was referring to the exact sequence
$$0\to f^*\Omega_Y\to\Omega_X\to \Omega_{X/Y}\to 0.$$
My concern is: as $\Omega_{X/Y}\cong\mathscr{O}_R$, I cannot see why tensoring with $\Omega_X^{-1}$ preserves $\mathscr{O}_R$ ONLY but not for the others.

Thanks in advance for answering.

答:
I think it is a consequence of the three following points:

If $\mathcal{F}$ has a discrete support then $\mathcal{F}\otimes\mathcal{G}$ too

If $\mathcal{L}$ is an invertible sheaf and $P$ is a point $(\mathcal{F}\otimes\mathcal{L})_P=\mathcal{F}_P$

For coherents sheaves, if $\mathrm{Supp}(\mathcal{F})=\mathrm{Supp}(\mathcal{G})=Z$ and $\mathcal{F}_{|Z}\simeq\mathcal{G}_{|Z}$ then $\mathcal{F}\simeq\mathcal{G}$

Title: Proposition IV.2.3 in Hartshorne


问:
If $a^Tz \ge a^Tx$ for all $z \in dom f$ then $x \not \in dom f$.

$a^Tz \ge a^Tx$ for all $z \in \text{dom} f$ then $x \not \in dom f$. Where $f: R^n \to R$ is a convex function.

The question is why $a^Tz \ge a^Tx$ should imply that $x \not \in \text{intdom} f$?

答:
You need additional assumption that $a \ne 0$.

Guide:

If $x$ is in the interior, then $x - \delta a$ is in the domain for sufficiently small positive $\delta$.

Try to use this to argue for contradiction.


问:
Combinatorics problem. Are these two approaches equivalent?
The problem goes like this: 7 accidents ocurred in a week, what is the probability that 2 days of the week had 2 accidents each and 3 days had 1 accident each.

So I thought about this in two different ways, getting two different results.

1st approach: I assign a number of accidents to each day of the week. The sample space consists of sequences like $100024$ which means $1$ accident on Sunday, $2$ on Friday and $4$ on Saturday. I will omit my calculations for now, but I got $1716$ of these sequences. Favorable cases are the permutations of $2211100$, and I got $\frac{7!}{2!3!2!} = 210$ of them. The probability is therefore $\frac{210}{1716} \simeq 0.122$

2nd approach: I have a sequence of accidents $a_1, ...,  a_7$ and I assign a day of the week to each of them. The sample space consists of sequences like $Su, Mo, Su, Th, W, W, Su$ which means that $3$ accidents ocurred on Sunday, $1$ on Monday, $1$ on Thursday, $2$ on Wednesday. There are $7^7$ of this sequences, and favorable cases are $\frac{7!}{2!3!2!} \frac{7!}{2!2!1!1!1!0!0!}$. The probability is therefore $\simeq 0,321$.

Although it is certainly possible that I got my calculations wrong, I don't believe so because I simulated each approach in Python, and computer results agree with my own calculations. I think that there is a conceptual difference between these two approaches that I am not seeing. Are these approaches equivalent?

答: The first approach is wrong, because the $1716$ sequences are not equiprobable. E.g., the sequence $a:=(7,0,0,0,0,0,0)$ is much less probable than the sequence $b:=(1,1,1,1,1,1,1)$, because in $a$ you want every accident at the day of your liking, namely Sunday, while in $b$ there is no prescription which of the seven accidents happened on which day.


答:
The simple solution of the linear ODE $$y'-2y=e^x \implies y=e^{2x}\int e^{-2x} e^x dx+C e^{2x} \implies y(x)=-e^{x}+Ce^{2x}$$

The solution of homogeneous part is $y=P e^{2x}$. Let us vary $P$ as $P(x)$  and put in $y'-2y=e^{x}$ to get $$P'=e^{-x} \implies P(x)=-e^{-x} +D \implies y(x)=-e^{x}+De^{2x}.$$
The same as above.

Title: Does variation of parameters always work?

Body:
I want to solve the differential equation $y'-2y=e^x$.

I solved the homogenous differential equation and I got that $y_0(x)=e^{2x}\cdot c$, where $c$ is a real constant.

Now I want to apply variation of parameters. So, $y_p=c(x)\cdot e^{2x}$ and $y'_p=c'(x)\cdot e^{2x}+2\cdot c(x)e^{2x}$ and if I plug it into my equation I get that $c'(x)\cdot e^{2x}=e^x$, so $c'(x)=e^{-x}$ and $c(x)=-e^{-x}$. But this doesn't verify my equation. What did I do wrong?


问:
Is the image of closed unit ball under the linear operator closed?
I have the linear operator $A:l_1 \rightarrow L_3(0,+\infty)$ which is defined with the formula $(Ax)(t)=\sum\limits_{k=1}^{+\infty} \frac{x(k)}{\sqrt{k}+\sqrt{t}} $ for all $t\in (0,+\infty)$ and for all $x\in l_1$. The full task sounds like this: find the adjoint operator $A^*$ for the operator $A$, then check if the set $A(B_1(0))$ is totally bounded and closed in $L_3(0,+\infty)$. I have found the adjoint operator, then using the criterion of Kolmogorov and Frechet I proved total boundedness of $A(B_1(0))$. But I do not know how to check this set for closedness. It seems difficult to check it with the definition. Also, I cannot see what theorems I may apply. This task is given in the topic about adjoint operators, so maybe I should use some results about adjoint operators, but I am not sure about it.

答:
I agree with the answer of @thing, but I allow myself to post my own solution, which I reached last night. My solution starts with a similar idea, but more difficult.

Firstly, we take a point $y_0\in Cl(A(B_1^{l_1}(0)))$. So there is a sequence $\{x_n\}\subseteq B^{l_1}_1(0)$ such that $||Ax_n -y_0||_{L_3} \rightarrow 0$ with $n\rightarrow +\infty$. Then we notice that $l_1\subseteq l_{10/9}$. Because for all $n\in \mathbb{N}$ we have that $||x_n||_{10/9}\leq ||x_n||_1\leq 1$, the sequence $\{x_n\}$ is bounded in $l_{10/9}$. Since $l_{10/9}$ is reflexive and separable by the Banach-Tihonov theorem there is a subsequence $\{x_{n_j}\}$, which weakly in the meaning of $l_{10/9}$ converges to some $x_0\in l_{10/9}$.

From weak convergences of $\{x_{n_j}\}$ to $x_0$ we have that for all $k\in \mathbb{N}$ that $x_{n_j}(k)\rightarrow x_0(k)$ with $ j\rightarrow +\infty $. For all $K\in \mathbb{N}$ we have that $\sum\limits_{k=1}^K |x_0(k)|=\lim\limits_{j\rightarrow +\infty}
\sum\limits_{k=1}^K |x_{n_j}(k)|\leq 1$, so $\sum\limits_{k=1}^{+\infty} |x_0(k)|\leq 1$. It proves that $x_0\in B^{l_1}_1(0)$.

Now the most strange step. I want to prove that $A$ can be considered as a continuous linear operator from $l_{10/9}$ to $L_3(0,+\infty)$. Let $x\in B^{l_{10/9}}_1(0)$. With simple estimations and Helder's inequality it is true that \begin{equation} \int\limits_0^{+\infty} \left|\sum\limits_{k=1}^{+\infty} \frac{x(k)}{\sqrt{k}+\sqrt{t}}\right|^3 dt\leq \int\limits_0^{+\infty} \left(\sum\limits_{k=1}^{+\infty} \frac{|x(k)|}{\sqrt{k}+\sqrt{t}}\right)^3 dt=\sum\limits_{m=0}^{+\infty} \int\limits_{m}^{m+1} \left(\sum\limits_{k=1}^{+\infty} \frac{|x(k)|}{\sqrt{k}+\sqrt{t}}\right)^3 dt\leq\end{equation}
\begin{equation} \leq\sum\limits_{m=0}^{+\infty} \left(\sum\limits_{k=1}^{+\infty} \frac{|x(k)|}{\sqrt{k}+\sqrt{m}}\right)^3 \leq \sum\limits_{m=0}^{+\infty} \left( \left(\sum\limits_{k=1}^{+\infty} |x(k)|^{10/9}\right)^{9/10} \left(\sum\limits_{k=1}^{+\infty} \frac{1}{(\sqrt{k}+\sqrt{m})^{10}}\right)^{1/10} \right)^3.\end{equation}

If $m>0$ we have that $\sum\limits_{k=1}^{+\infty} \frac{1}{(\sqrt{k}+\sqrt{m})^{10}}\leq\sum\limits_{k=1}^{+\infty} \int\limits_{k-1}^k  \frac{1}{(\sqrt{t}+\sqrt{m})^{10}} dt=\int\limits_0^{+\infty} \frac{dt}{(\sqrt{t}+\sqrt{m})^{10}}=\frac{1}{36m^4}$. So we have that \begin{equation} \sum\limits_{m=0}^{+\infty} \left( \left(\sum\limits_{k=1}^{+\infty} |x(k)|^{10/9}\right)^{9/10} \left(\sum\limits_{k=1}^{+\infty} \frac{1}{(\sqrt{k}+\sqrt{m})^{10}}\right)^{1/10} \right)^3 \leq \left(\sum\limits_{k=1}^{+\infty} \frac{1}{k^5} \right)^3+ \left(\sum\limits_{m=1}^{+\infty} \frac{1}{(36m^4)^{3/10}}\right)^3 <+\infty.\end{equation} The last number is finite and does not depend on $x$. So $A$ is bounded linear operator from $l_{10/9}$ to $L_3(0,+\infty)$.

Since $x_{n_j}$ converges weakly in the meaning of $l_{10/9}$ we have for all $f\in (L_3(0,+\infty))^*$ that $\lim\limits_{j\rightarrow +\infty} f(A(x_{n_j}))-f(A(x_0))=\lim\limits_{j\rightarrow +\infty}(f\circ A)(x_{n_j}-x_0)=0$, because $f\circ A$ is linear functional on $l_{10/9}$. So, $Ax_{n_j}$ weakly converges to $Ax_0$.

After all, we have strongly converging to $y_0$ sequence $Ax_n$ and it's weakly converging to $Ax_0$ subsequence $Ax_{n_j}$. That implies that $y_0=Ax_0$.

Because $y_0\in B^{l_1}_1(0)$ we have that $y_0\in A(B^{l_1}_1(0))$. So, $A(B^{l_1}_1(0))$ is closed. Q.E.D.


问:
Division Ring and Integral Domain
In abstract algebra, there are division ring, integral domain and field.

Division Ring is a Ring when all elements are unit.

Integral Domain is a Ring with no div '0' and commutative.

Is there any example of

Division Ring, but not I.D.

I.D but not Division Ring

I can't literally imagine any of them.

Only thing I can guess is if I.D. is finite, then it is a field so 2. is an infinite one.

答:
Units are no zero divisors.
Let $u$ be a unit with $uv=0$.
Then $v = 1 v = (u^{-1}u)v = u^{-1}(uv) = u^{-1}0=0$, since $0$ is absorbing.

Thus each (commutative) D.R. is an I.D.

Conversely, not every I.D. is a D.R.

Title: Division Ring and Integral Domain


问:
Connected ball $B(x,r)$ in an arbitrary space $\Omega$
Are balls always connected? Or is there a condition on a ball $B(x,r)$ to be connected? Where $x$ is an arbitrary point in an arbitrary space $\Omega$, and $r>0$

答: All balls in $\Bbb Q$ (usual metric) are disconnected. In a real linear space with a norm-induced metric all balls will be convex and so path-connected. This includes spaces like $\Bbb R^n$ etc.


答:
A division ring is not necessarily an integral domain. An integral domain is a commutative ring by definition. A division ring is not necessarily commutative.

The set of integers is an integral domain which is not a division ring. For example, $2 x = 1$ has no solution for $x$ in the integers.

Thanks to Jyrki Lahtonen for correcting my initial answer.

But please note: apparently not all authorities demand that an integral domain is commutative. If you are using such a definition of integral domain, then yes, it is true that all division rings are also integral domains.

I will do some research to find out what sources use such a definition for an integral domain.

EDIT: Found it. According to Wikipedia:

J.C. McConnell and J.C. Robson "Noncommutative Noetherian Rings" (Graduate Studies in Mathematics Vol. 30, AMS)

does not insist that an integral domain necessarily has to be commutative.

Title: Division Ring and Integral Domain

Body:
In abstract algebra, there are division ring, integral domain and field.

Division Ring is a Ring when all elements are unit.

Integral Domain is a Ring with no div '0' and commutative.

Is there any example of

Division Ring, but not I.D.

I.D but not Division Ring

I can't literally imagine any of them.

Only thing I can guess is if I.D. is finite, then it is a field so 2. is an infinite one.


问:
Simplified expression for $\left| \dfrac{1}{1+z^{9}}\right| \leq \dfrac{C}{|z^9|}$ for $|z| > 1$?
It is clear that the poles lies on the unit circle, wherefore I think such expression is possible since the denominator will never be zero. In the best of cases, I would assert $|z^9| \leq |1 + z^9|$ and so forth,

$$\left|\dfrac{1}{1+z^9} \right| \leq \left|\dfrac{1}{z^9}\right|$$ for $|z|>1$, however I am not sure if this is right. However, I think it is possible to instead have,
$$\left|\dfrac{1}{1+z^9} \right| \leq C\left|\dfrac{1}{z^9}\right|$$ for $|z|>1$ and $C< \infty$.

Thanks in advance for any hints on how I can take it from here.

答: There is no such constant $C$. If you let $z $ tend to a $9-$th root of $-1$ through values outside the unit disk you will immediately get a contradiction to the inequality.  More specifically, put $z=-1-\frac  1n$ in the  inequality and let $n \to \infty$ to get a  contradiction.


问:
On localization and integral domains
I'm attempting to solve exercise 5 from chapter 3 in Atiyah-McDonald's text on commutative algebra; in particular, it asks if 'being an integral domain' is a local property. The counter-example I found is the following, and was interested in asking whether it made any sense.

Consider the ring $A=\mathbb{Q}[x,y]/(xy)$; evidently, $A$ is not an integral domain, as it is a quotient of $\mathbb{Q}[x,y]$ by an ideal which is not prime. Because prime ideals in $A$ are such that their contraction in $\mathbb{Q}[x,y]$ is a prime ideal that contains $(xy)$, the only possible prime ideals in $A$ are $p=(x)/(xy)$ and $q=(y)/(xy)$; one can verify directly that these are both in fact prime.

My intuition behind this exercise is that $A_p$ should be an integral domain, because the expression $\frac{x+(xy)}{1}\cdot\frac{y+(xy)}{1}=0$ fails to be a product of two non zero elements in $A_p$, seeing as $\frac{x+(xy)}{1}=0$ in the localization, becuase $y+(xy)\in (A-p)$ and $(x+(xy))(y+(xy))=0$.

I'm not quite sure if this observation is enough to assert that $A_p$ (and hence, by symmetry, $A_q$ as well) is an integral domain. I would greately appreciate any advice or suggestion which may help, thanks a lot!

答: Your intuition that $A_p$ and $A_q$ are domains is correct - a little more pushing will show that $A_p\cong k[x]_{(x)}$. Unfortunately, your claim that $p$ and $q$ are the only primes is false: $(x,y)$ is a prime ideal in $A$ which is not of the form you claim (there are many more).

Title: On localization and integral domains


问:
Proving that $f(x)= \frac{1}{x}$ is not integrable on $[0, 1]$
Given
$$
f(x)=\begin{cases}
      \frac{1}{x} &, 0<x\leq 1 \\
      0 &, x=0
   \end{cases}
$$
I'm trying to prove that $f$ is not integrable, is my solution is valid ?

So suppose $f$ is integrable, then there should exist a constant function $g=M$ so that $f(x)\leq M,\forall x$ and thus $\int^1_0 f{\rm d}x\leq \int^1_0 M{\rm d}x$ or $\int^1_0f{\rm d}x\leq M.$ However, this is clearly false since we can always find a value of $x$ sufficiently small such that $\frac{1}{x}>M$ for any value of $M.$

Is is correct ?

答:
I would like to offer a solution from a little more general perspective.
Lets check when does $\int_0^1{x^\alpha}dx$ exists. This happens when $\lim_{t\to0}{\int_t^1{x^\alpha}dx}$ exists. We'll get:

$\lim_{t\to0}\left[\frac{x^{\alpha+1}}{\alpha+1}\right]_t^1 = \frac{1}{\alpha+1}-\lim_{t\to0}\frac{t^{\alpha+1}}{\alpha+1}$ and the second argument diverges if $t$ goes to the denominator, hence we must have $\alpha>-1$


问:
Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$

Calculate the sum of all irrational roots of
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$

I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}- xy+ y^{2} \right ),$ but have had no meaningful results.

Body:
Problem. Half-solve
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$

Substitute $y^{3}= 8x- 3$ in the original equation_
$$\left ( 4y- 8x^{3}- 3 \right )+ \left ( y^{3}- 8x+ 3 \right )= \left ( y- 2x \right )\left ( 4x^{2}+ 2xy+ y^{2}+ 4 \right )= 0$$
With $y= 2x,$ we have the equivalent one
$$8x^{3}- x+ 3=\!\left ( 2x- 1 \right )\!\left ( 4x^{2}+ 2x- 3 \right )=\!0\!\Rightarrow\exists_{x_{1}, x_{2}\in\left ( {\mathbb{Q}}' \right )^{2}}x_{1}+ x_{2}=\!-\frac{1}{2}$$
With $4x^{2}+ 2xy+ y^{2}+ 4= 0,$ there are no irrational roots, because
$$4x^{2}+ 2xy+ y^{2}+ 4= y^{6}+ 4y^{4}+ 6y^{3}+ 16y^{2}+ 12y+ 73= 0\;{\rm illegal}$$


答:
Cobsider $(n)_q:=\frac{q^n-1}{q-1} $, the so called q-analogue of $n$. Observe that this is a polynomial of degree n-1 in the variable q.
Define now the q-factorial and the q-binomial as follow:

$$(n)_q!:= \prod_{i=1}^n (i)_q  \qquad \binom{n}{k}_q:=\frac{(n)_q!}{(k)_q! \,(n-k)_q!}$$

It is easy to prove that both are polynomials, for example using the identity

$$
\binom{n}{ k}_q = q^k \binom{n-1}{k}_q + \binom{n-1}{k-1}_q
$$

What you need follows from the evaluation in 2 on the q-binomial.

Title: Prove that the sequence $ \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1) } { (2^1-1)(2^2-1)...(2^k-1) }$ is an integer

Body:
For any given integer $n\ge 1$ and $k \in\{1, 2, \dots, n\},$

$$ F_n(k) = \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1)   } { (2^1-1)(2^2-1)...(2^k-1) } $$

For example, if $n = 1,$ then $k \in \{1\}$ and $$F_1(1) = \frac{2^1-1}{2^1-1} $$

For $n=3,$ then $k\in\{1, 2, 3\}$ and
$$ F_3(1) = \frac{2^3-1}{2^1-1}  = 7 $$
$$ F_3(2) = \frac{(2^3-1)(2^2-1)}{(2^1-1)(2^2-1)} =7 $$
$$ F_3(3) = \frac{(2^3-1)(2^2-1)(2^1-1)}{(2^1-1)(2^2-1)(2^3-1)} =1 $$


答:
The exponent $9$ is immaterial and you are essentially asking for an upper bound on

$$\left|\frac z{z+1}\right|$$ for $|z|>1$. Even in the reals, this does not work.

Title: Simplified expression for $\left| \dfrac{1}{1+z^{9}}\right| \leq \dfrac{C}{|z^9|}$ for $|z| > 1$?

Body:
It is clear that the poles lies on the unit circle, wherefore I think such expression is possible since the denominator will never be zero. In the best of cases, I would assert $|z^9| \leq |1 + z^9|$ and so forth,

$$\left|\dfrac{1}{1+z^9} \right| \leq \left|\dfrac{1}{z^9}\right|$$ for $|z|>1$, however I am not sure if this is right. However, I think it is possible to instead have,
$$\left|\dfrac{1}{1+z^9} \right| \leq C\left|\dfrac{1}{z^9}\right|$$ for $|z|>1$ and $C< \infty$.

Thanks in advance for any hints on how I can take it from here.


答:
Hint:

$$\int_1^a \frac 1t dt+\int_1^b \frac 1t dt = \int_1^{ab} \frac 1t dt.$$

Big hint:

 $$\int_{\frac 12}^1 \frac 1t dt +\int_{\frac 12}^1 \frac 1t dt =\int_{\frac 14}^1 \frac 1t dt$$

Title: Proving that $f(x)= \frac{1}{x}$ is not integrable on $[0, 1]$

Body:
Given
$$
f(x)=\begin{cases}
      \frac{1}{x} &, 0<x\leq 1 \\
      0 &, x=0
   \end{cases}
$$
I'm trying to prove that $f$ is not integrable, is my solution is valid ?

So suppose $f$ is integrable, then there should exist a constant function $g=M$ so that $f(x)\leq M,\forall x$ and thus $\int^1_0 f{\rm d}x\leq \int^1_0 M{\rm d}x$ or $\int^1_0f{\rm d}x\leq M.$ However, this is clearly false since we can always find a value of $x$ sufficiently small such that $\frac{1}{x}>M$ for any value of $M.$

Is is correct ?


问:
Finding the Mean for Normal Distribution
I'm a little stuck trying to find the mean of this practice normal distribution question.

Standard deviation $= 300$

$20$% of workers are paid less than $1500$

So I created the probability like this

$$P(x < 1500) = 0.2$$

But I'm unsure what I'm meant to do next

Any help is appreciated!

答:
Go back to the standard normal distribution, with mean $0$ and $\sigma=1$. If $X$ is normally distributed, $\frac{X-\mu}{\sigma}$ is standard-normally distributed and for $\Phi$ (the cdf for the SND) tables exist.

From this table I get that $\Phi(-0.84) \approx 0.2005$ which is a nice approximation to $0.2$. Many calculators can compute $\Phi^{-1}$ as well, nowadays (in my old high school days we had to do it with tables though).

So we know that when $\frac{1500-\mu}{300}$ should be about $-0.84$ to get the correct probability. Now compute $\mu$.


问:
Finding an orthonormal basis z of $\Re^n$ with respect to which the matrix $A_z$ is diagonal
Let $\textbf{A}:\Re^n \rightarrow \Re^n$ be given by $\textbf{A} \vec{x}=\vec{x}+l(\vec{a},\vec{x})\vec{a}$ where $||\vec{a}|| = 1, \vec{a} \in \Re^n, l \in \Re.$ And let $\textbf{A}$ be orthogonal for $l=0 \vee l=-2$. Find an orthonormal basis z of $\Re^n$ with respect to which the matrix $A_z$ is diagonal.

What I've got so far: for the first part of the basis we can take $\vec{a}$ as it's on $\Re^n$ and has length 1. Next we take $\vec{a}^\bot$ which forms a basis of $\Re^n$ together with $\vec{a}$. It can be normalized by dividing it by its length. Is this right so far? And I don't know how to show if it's diagonal to $A_z$.

答:
The matrix $A_z$ will be diagonal if and only if each basis element of $z$ is an eigenvector of $\bf A$.

To include $a$ into the basis $z$ is a good start, since ${\bf A}(a)=(1+l)a$.

Note, however, that $a^\perp=\{x\in\Bbb R^n:(a,x)=0\}$ is a whole subspace of dimension $n-1$.

But, luckily any vector $x\in a^\perp$ is an eigenvector of $\bf A$, because then ${\bf A}(x)=x$.

So, we can extend the single vector $a$ by an arbitrary orthonormal basis of $a^\perp$, and this will work.


问:
An elegant solution to the classic Monty Hall problem.
Think of the game as drawing a ball, blindfolded, from a box, which contains 1 white ball and 2 black balls, except that every time after you pick a ball, a black ball will be thrown away by the host, so only one ball remains in the box.

If you stick with the ball that you pick every time you play the game, regardless of what the host does afterwards, then in the long run you will end up having a white ball in your hand 1/3 of the time and a black one 2/3 of the time.

However, if you choose to switch every time regardless of the colour you picked, then the times when you should end up having a white ball become black and vice versa. So, in the long run, you will end up having a black ball in your hand 1/3 of the time and a white one 2/3 of the time.

Any constructive criticism is welcomed! Thank you!

答:
Edit

Re the comment of aschepler, you are forced to assume that you were blindfolded before picking the ball, and therefore never knew the color of the ball that you had picked.

I agree that this does analogize to the Monty Hall problem, which I will explain below.  My only gripe with the query, as it is presented, is that the last paragraph, while accurate and valid is somewhat convoluted:

However, if you choose to switch every time regardless of the colour you picked, then the times when you should end up having a white ball become black and vice versa. So, in the long run, you will end up having a black ball in your hand 1/3 of the time and a white one 2/3 of the time.

In the normal Monty Hall scenario, (1/3) of the time, you correctly picked the right door, and (2/3) of the time you did not.  Always switching :

In the (1/3) of the time that you initially picked correctly, you have turned victory into defeat.

In the (2/3) of the time that you initially picked incorrectly, you have turned defeat into victory.

Your statement that

...then the times when you should end up having a white ball become black and vice versa...

is somewhat convoluted.  I would have expressed it as (assuming that you always switch):

the (1/3) of the time that you picked a white ball, you have turned victory into defeat.

the (2/3) of the time that you picked a black ball, you have turned defeat into victory.

Beyond that, in the case where you pick Door #1, Monty Hall's showing that behind Door #2 is a goat does roughly analogize to the host selecting a black ball from the ones remaining and discarding it.

The key points here are that when Door #1 has the grand prize, Monty has a choice of which Door (#2 or #3) to show.  Similarly, when your initial selection is the white ball, the Host has a choice of which black ball to discard.

Alternatively, if Door #1 does not have the grand prize then, (assuming that Door #3 does have the grand prize), Monty has no choice but to reveal Door #2.  Similarly, if your initial selection is a black ball, the host has no choice but to discard the lone remaining black ball.


答:
As already explained in the other answers there is no such $C$ as long as you wish $|z|>\color{blue}{1}$.

But if you can relax the lower bound to $|z|^9 >\color{blue}{1+\epsilon}$ for a fixed $\epsilon >0$, you can find a corresponding $C_{\epsilon}$ using the reverse triangle inequality:

$$\frac 1{|z^9+1|}= \frac 1{|z|^9\left|1+\frac 1{z^9}\right|}$$$$\leq \frac 1{|z|^9\left|1-\frac 1{|z|^9}\right|}\stackrel{|z|^9>1+\epsilon}{<}\underbrace{\frac{\epsilon}{1+\epsilon}}_{=C_{\epsilon}}\cdot \frac 1{|z|^9}$$


问:
Proving that there is a path in set.
Given that $\underline Y$ is a subspace of $\underline ℝ^2$ and defined as follows:

$$I = [0,1]$$

$$X = (\{1\} × I) ∪ \left(I × \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)\right)$$

$$Y = X \setminus \{ (0.5, 0) \}$$

How can one prove that there is no path from $(0,0)$ to $(1,0)$ using sequences?

To show the contradiction I started by assuming that such a map exists:

$$f: \underline{I} \rightarrow \underline Y \ \text{continuous map}$$

$$f(0) = (0,0),\ \ f(1) = (1,0)$$

If $f$ is continuous, then it must hold that $f^{-1}[\ [0,0.5)\times \{0\} ] = A$ is a closed set. So $A$ contains its maximum $m$. From here on I'm not sure how to proceed. I wanted to somehow show that there is a sequence that due to $m$ being in $A$ cannot converge to $f(m) = (x,0)$.

答:
Your space $X$ is a variant of the comb space. Many questions in this forum deal with it. Let us write
$$X =  \{1\} \times I \cup I \times N$$
with $N = \{0\} \cup \left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}$. For later use we define $N_m = \left\{ \frac{1}{n} \,\Bigg\vert\, n \ge m \right\}$ and $N^*_m = \left\{ \frac{1}{n} \,\Bigg\vert\, n < m \right\}$.

My approach is this.

Consider a continuous $f :  I \to X$. We claim that $f(I)$ intersects only finitely many of the sets $B_n = [0,\frac{3}{4}] \times \{ \frac{1}{n}\}$.

Otherwise there would exist a strictly increasing sequence $(n_k)$ of integers such that $D_k  = f(I) \cap B_{n_k} \ne \emptyset$ for all $k$. Pick $t_k \in I$ such that $f(t_k) \in D_k$. The sequence  $(t_k)$ has a convergent subsequence; so let us assume w.l.o.g. that $t_k \to \tau \in I$. By continuity $f(t_k) \to f(\tau)$. The second coordinate $f_2(t_k)$ of $f(t_k)$ is  $\frac{1}{n_k}$, thus $f_2(t_k) \to 0$. Since the first coordinate $f_1(t_k)$ of $f(t_k)$ is always in $[0,\frac{3}{4}]$, we conclude that $f(\tau) = (\theta,0)$ for some $\theta \in [0,\frac{3}{4}]$. Obviuosly $t_k \ne \tau$ for all $k$. But $U = [0,1) \times \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)$ is an open neigborhood of $f(\tau) = (\theta,0)$ in $X$, thus there exists $\epsilon > 0$ such that $f(t) \in U$ for $\lvert t - \tau  \rvert < \epsilon$. Choose $k$ such that $\lvert t_k - \tau  \rvert < \epsilon$. Hence we get a path in $U$ connecting $f(\tau)$ and $f(t_k)$. This a contradiction because these points lie in different path components of $U$.

Now assume there is a path in $Y$ from $(0,0)$ to $(1,0)$. Then there also is a path $f$ from $(0,0)$ to $(1,1)$ because $(1,0)$ and $(1,1)$ lie in the same path component of $Y$. We know that $f(I) \subset Y_m = \{1\} \times I \cup I \times N^*_m \cup [\frac{3}{4},1] \times N_m \cup (I \setminus \{\frac{1}{2} \}) \times \{0\}$ for sufficiently large $m$. But $Y' =  \{1\} \times I \cup [\frac{3}{4},1] \times N \cup (I \setminus \{\frac{1}{2} \}) \times \{0\}$ is a retract of $Y$, hence $f$ induces a path $f'$ in $Y'$ from $(0,0)$ to $(1,1)$. The coordinate function $f_1'$ is a continuos real-valued function such that $f_1'(0) = 0, \ f_1'(1) = 1$. The IVT says that $f_1'(t) = \frac{1}{2}$ for some $t$ between $0$ and $1$. This is a contradiction because $Y'$ does not contain any point whose first coordinate is $\frac{1}{2}$.

Title: Proving that there is a path in set.


问:
Grade Probability based on Binomial Distribution
The student knows answers to 75 out of 100 questions in the exam. He picks 3 random questions of the 100 and answers. Three correct answers yield grade A, two correct answers yield grade B, 1 correct answer yields C and 0 yields D.

What's the probability that 2 out of 5 students will get A, 2 will get B and 1 will get C?

I am using the following approach:

since the student knows 75 out of a hundred questions. probability of getting a known question in the exam would be p = $\frac{3}{4}$

Using binomial distribution I have calculated the probability of getting grade A, B and C as follows:

P(A) = ${5 \choose 2} \times p^2 \times q^3 = \frac{27}{64} \approx 0.42$

P(B) = ${5 \choose 2} \times p^2 \times q^3 = \frac{27}{64} \approx 0.42$

P(C) = ${5 \choose 1} \times p^2 \times q^3 = \frac{9}{64} \approx 0.14$

Now to get the probability of 2 grade A, 2 grade B and 2 grade C students out of 5 should be:

P(X) = ${5 \choose 2} p_A^2 (1 - p_A^3) \times {5 \choose 2} p_B^2 (1 - p_B^3) \times {5 \choose 1} p_C (1 - p_C^3) \approx 0.04$

But this is reported as wrong. Can anyone suggest what could be the problem?

答:
I was mistakenly using the binomial distribution whereas I need to use the Multinomial Distribution:

P(X) = $\frac{n!}{n_1! \times n_2! \times n_3!} p_1^{n_1} \times p_2^{n_2} \times p_3^{n_3}$

Now
$ p_1 = 0.42; n_1 = 2 \\
p_2 = 0.42; n_2 = 2 \\
p_3 = 0.14; n_3 = 1$

Hence

P(X) = $\frac{5!}{2! \times 2! \times 1!} \times 0.42^2 \times 0.42^2 \times 0.14 \approx 0.13$


问:
Let 1 ,… ,  be independent and identically distributed (i.i.d.) random variables with distribution function
Apologies for the formatting. I didn't really know how to format this.

Let $_1,\ldots, _n$ be independent and identically distributed (i.i.d.) random variables with
distribution function :
$$P(Y_i \le y \mid \alpha,\beta) = \begin{cases}0 \quad \text{if } y < 0 \\
(y/\beta)^\alpha \quad \text {if } 0 \le y \le \beta \\
1 \quad \text{ if } y > \beta\end{cases} $$

where the parameters $\alpha$ and $\beta$ are positive.

Find the Maximum Likelihood Estimators (MLEs) of $\alpha$ and $\beta$.

I integrated the original function. to get $y^{\alpha+1}/ [(\beta^\alpha)(\alpha+1)]$
I take the likelihood function and then take its logarithm to get

$-\alpha n\times \ln(\beta) - n\times\ln(\alpha+1) + (\alpha+1) \sum_1^n \ln(y_i)$

I then take the partials in terms of alpha and beta
When I take it in terms of Beta I get
$0=-\alpha n/\beta$
I do not know what to do with this if I can do anything at all

Similarly when I take the partials in terms of alpha I get a similar problem.  I don't know what to do with it. Would appreciate it if someone could show me how to get to the solution.

答:
As hinted in the comments, to compute the density, you have to differentiate the distribution function :
$$f_i(y\mid\alpha,\beta) := \partial_y \mathbb P(Y_i \le y \mid \alpha,\beta) =\alpha  \frac{ y^{\alpha-1}}{\beta^\alpha} \mathbf{1}_{0\le y \le \beta} $$
Now, the likelihood is given by the product of the conditional densities taken at each observation, and as is common, we'll rather consider the log-likelihood to convert the product into a sum :
$$ \begin{align}\mathcal L(\alpha,\beta) &= \ln\left(\prod_i^n f_i(Y_i\mid\alpha,\beta)\right) \\
&= \ln\left(\prod_i^n \alpha  \frac{ Y_i^{\alpha-1}}{\beta^\alpha} \mathbf{1}_{0\le Y_i \le \beta}\right) \\
&= n\ln\left(\frac{\alpha}{\beta^\alpha}\right) + (\alpha - 1)\sum_i^n \ln(Y_i \mathbf{1}_{0\le Y_i \le \beta}) \end{align} $$

The MLEs of $\alpha$ and $\beta$ are the values $\hat \alpha$ and $\hat \beta$ which maximize $\mathcal L(\alpha, \beta)$

For $\beta$ : We see that, for the likelihood not to be equal to $- \infty$, we necessarily need $\hat \beta \ge \max\{Y_1,\ldots,Y_n\}$. Since $\mathcal L(\alpha,\beta)$ is a non increasing function of $\beta$, we see that $$\hat \beta = \max\{Y_1,\ldots,Y_n\}$$ (the minimal acceptable value) is the MLE of $\beta$.

For $\alpha$ : $\partial_\alpha \mathcal L(\alpha, \beta) = \frac{n}{\alpha} - n\ln(\beta) + \sum \ln(Y_i \mathbf{1}_{0\le Y_i\le \beta})$

We find $$\hat \alpha = \frac{n}{\ln\left(\frac{{\hat \beta}^n}{\prod_i^n Y_i}\right)}$$


问:
Finding an angle in a circle with triangles
good day! Excuse me, I am new to finding angles using a circle, and I don't know how to find X using trigonometric functions and everything else about triangle angles. What is the 100° talking about? Or, how can one proof that BE is parallel to AC? (Maybe it isn't!) What I know is the angle of ACD is going to be 78. Any help, guidance, approach would be warmly welcomed.

答:
Hint:

Work with multiple isosceles triangles with one of the vertices being at the center of the circle. Start with $\triangle AOB$ and we know $\angle AOB = 100^0$, can you find $\angle OAB$? Then can you go on to find $\angle OCA$ and $\angle OCE$?

Now using the fact that $FC$ is tangent to the circle, you should be able to find $\angle FCE$.

Then can you find $\angle FEC$ given $ABEC$ is a cyclic quadrilateral?

Now you know two angles of $\triangle FCE$. So you know $\angle X$.


问:
Continuous random vector with joint pdf (find marginal and validity)
Let (X,Y) be a continuous random vector with joint probability density function
$$f_{X,Y}(x,y)=\left\{\begin{matrix}
6xy & for \; 0<x<1, 0<y<\sqrt{x}\\
 0& otherwise
\end{matrix}\right.$$

Show that it is a valid joint pdf and that the marginal pdf's are given by
$$f_{X}(x)=\left\{\begin{matrix}
3x^2 & for \; 0<x<1\\
 0& otherwise
\end{matrix}\right.$$
and
$$f_{Y}(y)=\left\{\begin{matrix}
3y(1-y^4) & for \; 0<y<1\\
 0& otherwise
\end{matrix}\right.$$

My try

I would think that to show that it is indeed a valid joint pdf I would have to show the double integral over both variables from $-\infty$ to $\infty$ is equal to 1. Would this be it?

For the marginal pdf's I am thinking that I should take the integral from $0$ to $1$ with respect to the variable of the pdf that I am not interested in. I.e for pdf for $X$ i want to integrate over $y$. Would this be correct?
I need this for the next part to find $P(Y>X^2).

答:
$f_{XY}(x,y)$ is a valid pdf because $f(x,y)\geq 0$ $\forall (x,y)$ and

$$\int_0^1 6x\left[\int_0^{\sqrt{x}}y dy\right]dx=1$$

To calculate the marginals, simply integrate $f(x,y)$ w.r.t. the other variable

$$f_Y(y)=\int_{y^2}^1 6xy dx$$

$$f_X(x)=\int_0^{\sqrt{x}} 6xy dy$$

Here is a drawing of the suppport; this will help to understand the solution


问:
Intuition of Algebraic Partition of Unity
In Vakil's Rising Sea (Nov. 18, 2017 Draft) on p. 130 - 131, he gives a proof that his definition of the structure sheaf on affine schemes indeed defines a sheaf. He coins this proof an argument by "partition of unity". In fact, he writes that the name is due to Serre. (In Vakil's book it appears again, e.g. on p. 161 and p. 203.) My questions are:

Why is this called partition of unity? Does it - or in which way does it - resemble the topological/analytical partition of unity?

Does anyone have a reference where Serre said this?

For those who do not have Vakil's argument in front of their eyes, let me sketch it.

Definition. Let $A$ be a ring and we define $\mathcal{O}_{\operatorname{Spec}{A}}(D(f)) = A_f$ with the obvious restriction map.

Claim. Suppose $\operatorname{Spec}{A} = \bigcup_{i=1}^n D(f_i)$ for this argument. We want to show that this satisfies the base gluability property.

Proof. Let $\frac{a_i}{f_i^{\ell_i}} \in A_{f_i}$ be elements which agree on overlaps $A_{f_i f_j}$. After some algebraic manipulations (please trust me on this or you may read it in Vakil's book), we can rewrite $\frac{a_i}{f_i^{\ell_i}} = \frac{b_i}{h_i}$ with overlap condition being $h_j b_i = h_i b_j$ and $\operatorname{Spec}{A} = \bigcup_{i=1}^n D(h_i)$.

The latter implies $1 = \sum_{i=1}^n r_i h_i$ for some $r_i \in A$. The key step is to define $$r = \sum_{i=1}^n r_i b_i.$$ This is an element in $A$ and restricts to $\frac{b_i}{h_i}$ since the overlap condition gives $$ rh_j = \sum_{i=1}^n r_i b_i h_j = \sum_{i=1}^n r_i h_i b_j = b_j. $$

This leaves me with one last question: What motivates the definition of $r = \sum_{i=1}^n r_i b_i$?

答:
The short answer is that, yes, this is basically the same kind of partition of unity argument as one sees in analysis, with some adaptations to handle the coarseness of the Zariski topology and the fact that one is not working with actual functions taking values in some field.

Suppose, however, that $A$ is a finitely generated integral domain over an algebraically closed field $k$.
Then, for every maximal ideal $\mathfrak{m} \subseteq A$, we have an evaluation homomorphism $\operatorname{ev}_\mathfrak{m} : A / \mathfrak{m} \to k$, and the Jacobson radical of $A$ is trivial, so we can think of $A$ as a ring of $k$-valued functions: simply replace each $f \in A$ with the $k$-valued function $\mathfrak{m} \mapsto \operatorname{ev}_\mathfrak{m} (f)$ defined on the set of maximal ideals of $A$.
In this scenario the analogy between the analytic and algebraic situations is clear to see.

Recall that the problem is to find $r \in A$ such that $r = b_i / h_i$ on each $U_i = \mathrm{D} (h_i)$.
In the analytic situation, we would seek $c_i \in A$ such that the closure of $\mathrm{D} (c_i)$ is contained in $U_i$ and $\sum_i c_i = 1$, so that extending $c_i b_i / h_i$ by $0$ outside $U_i$ defines a continuous (or smooth or analytic etc.) function on the whole space.
This is not reasonable in the algebraic situation because open sets in the Zariski topology are "often" dense!
If we could find such a partition of unity then the formula $\sum_i c_i b_i / h_i$ would be a sum of functions defined on the whole space and it is easy to see that it satisfies the equations we want.

Since extension by zero is not feasible in the algebraic situation, we must find a workaround.
We can clear denominators in the problem statement: it is sufficient to find $r \in A$ such that $h_i r = b_i$ (on the whole space).
If we could find $r_i \in A$ such that $\sum_i r_i h_i = 1$, then
$$r = \sum_i r_i h_i r = \sum_i r_i b_i$$
but we do indeed have such $r_i \in A$ because the $U_i$ constitute a cover.
One then notes that the RHS of the equation is already a well-defined function on the whole space and checks that it satisfies the equations we require.
If we define $c_i = r_i h_i$ then we have $\sum_i c_i = 1$ as in the analytic situation, and we have $r_i b_i = c_i b_i / h_i$ on $U_i$, so in some sense we do get $r = \sum_i c_i b_i / h_i$ – but this should not be taken too literally, and one should note that no extension by zero has happened here.


问:
Problem $13.7$ in Apostol's Mathematical Analysis.
I am looking at problem $13.7$ in Apostol's Mathematical Analysis.

Let $D=\{|z|<1\}$ and let $f=u+iv$ be such that

$(\rm i)$ $u,v\in{\mathcal C}^1(D)$

$(\rm ii)$ $f$ is continuous on $\overline D$

$(\rm iii)$ $f(z)=z$ on $\partial D$.

$(\rm iv)$ $J_f(z)>0$ if $z\in D$.

Show that

$(1)$ For every open $\Omega\subseteq D$, $f(\Omega)$ is open in $f(D)$.

$(2)$ $f(D)$ is an open ball of radius $1$.

$(3)$ For each $w\in f(D)$, $D\cap f^{-1}(w)$ is finite.

I have only proven $(1)$:

PROOF1 Since $u,v$ are of class $\mathcal C^1$ over $D$, so is $f$, and since the Jacobian doesn't vanish, the inverse function theorem says that for each $z\in D$ there exists an open nbhd $N_z$ such that $f\mid_{N_z}$ is a diffeomorphism of class $C_1$, thus $f\mid_{N_z}$ is a homeomorphism, so it is an open map. Let $\Omega\subseteq D$ be open, and choose $z\in\Omega$. Then there exists $N_z$ as above. Let $\widetilde N_z=N_z\cap \Omega$. This is open and lies inside $N_z$ so its image under $f\mid_{N_z}$ is open by the above. Then $f(\Omega)=f(\Omega)\cap f(D)=\bigcup_{z\in\Omega}f\left(N_z\right)\cap f(D)$ is the intersection of the union of open sets with $f(D)$, so it is open in $f(D)$. $\blacktriangle$.

PROOF2 Suppose that $f(D)\not\subseteq D$. Then certainly $f(\bar D)\not\subseteq \bar D$. Thus, there exists $q\in\bar D$ such that $|f(q)|>1$. But $|f|$ is continuous on the compact $\bar D$; thus it attains its maximum on some $p\in\bar D$. But we cannot have $p\in\partial D$ since we would have $|f(p)|=1$, so it must be the case $p\in D$. Let $K=\overline B$ where $B=B(0,r)\; ;\; r=|f(p)|$. Then $f(\bar D)\subseteq \bar B$, and $f(p)\in\partial B$. Let $N_p$ be an open nbhd of $p$ in $D$. Then we see $f(N_p)$ is not open in $f(D)$, contrary to what was proven in $(1)$. It follows that $f(D)\subseteq D$.

PROOF3 Suppose for the sake of a contradiction that there exists $w\in f(D)=D$such that $D \cap f^{-1}(w)$ is infinite. Since this is a subset of the compact $\bar D$, it must have an accumulation point in $\alpha \in \bar D$. Suppose first that $\alpha\in D$. Then for each $\epsilon >0$ the ball $B(\alpha;\epsilon)$ contains infinitely many points of $\{z\in D:f(z)=w\}$, in contradiction with the fact $f$ is a local diffeomorphism in $\alpha$. Suppose $\alpha\in \partial D$. Then $f(\alpha)=\alpha$, and by continuity, $f(\alpha)=w$, which is absurd since $D\cap\partial D=\varnothing$. $\blacktriangle$.

Chapter $13$ is "Implicit functions and extremum problems". I am not sure what theorems to use to prove $(2)$, and I haven't thought about $(3)$, so I will worry about $(3)$ later. I want to prove $(2)$ now. Any hints?

ADD If $A$ is any set, $\partial A$ denotes its boundary and $\overline A$ its closure.

答:
Here's an approach to prove $f(D) = D$. First show that $f(D) \subset D$ as follows: The fact that $f(D)$ is open means that $|f|$ cannot attain a maximum on $D$ and must therefore attain a maximum on $\partial D$; use condition (iii) to conclude that $|f(z)|< 1$ for $z\in D$. Then show that $D\subset f(D)$, using the fact that $f(\overline{D})$ is a compact set that is at the same time equal to $f(D)\cup \partial D$. (Therefore, $\partial f(D)\subset \partial D$, meaning that $f(D)$ must contain every point of $D$.)

Here is a little more detail about the proof of $D\subset f(D)$, as requested in the comments. Note that $f(\overline{D}) = f(D)\cup \partial D$ is compact, hence closed, so it contains $\overline{f(D)} = f(D)\cup\partial f(D)$. As $f(D)$ is open, it is disjoint from its boundary, so $f(D)\cup \partial f(D)\subset f(D)\cup \partial D$ implies $\partial f(D)\subset \partial D$. Then $f(D)\cap D = \overline{f(D)}\cap D$, meaning that $f(D)\cap D$ is both closed and open relative to $D$; since $D$ is connected and $f(D)\cap D$ is nonempty, it must be the case that $f(D)\cap D = D$, or equivalently that $D\subset f(D)$.


问:
Can the closure of a set be written as the intersection of open neighborhoods in a non-metrizable space?
Let $\Omega$ be a topological space. If $A\subseteq\Omega$, then $$\overline A=\bigcap_{\substack{B\supseteq A\\B\text{ is closed}}}B\tag1.$$ If the topology on $\Omega$ is generated by a metric $d$, then $$B_\varepsilon:=\left\{x\in\Omega:d(x,A)<\varepsilon\right\}$$ is open for all $\varepsilon>0$ and $$\overline A=\bigcap_{\varepsilon>0}B_\varepsilon=\bigcap_{n\in\mathbb N}B_{\frac1n}\tag2.$$

By $(2)$, the closure of an arbitrary subset of a metric space can be written as the (countable) intersection of open neighborhoods. Does the same hold in an arbitrary topological space?

答:
No. Consider a set $X$ with $p \in X$ and $\vert X \vert > 1$ with the excluded point topology w.r.t. $p$. Then $\{p\}$ is closed since its complement is open, but the intersection of all of its neighbourhood is $X$, since it is the only set containing $p$ and being open i.e. the closure is strictly contained in the intersection of open neighbourhoods.

It can fail in the other direction as well, as can be shown with the particular point topology w.r.t. $p$. i.e. the intersection of open neighbourhoods can be strictly contained in the closure.


问:
$f(x-a)+f(x-b)$ vs $f(x-a)f(x-b)$
Consider the condition $f(x)=0$ if $x<c$ and now we wish to determine the values of $x$ for which :

$f(x-a)f(x-b)$ is equal to $0$, for $a,b\in\mathbb{R}$

$f(x-a)+f(x-b)$ is equal to $0$, for $a,b\in\mathbb{R}$

Now I thought of it this way:

For $(1)$: if $x<c$ then we assign a new value of $x$ as $x\leftarrow x-a$ for $f(x-a)$ and $x\leftarrow x-b$ for $f(x-b)$ so we get $x-a<c$ and $x-b<c$ this would mean $x<a+c$ and $x<b+c$. Therefore, we have:
$$
(x<a+c)\cap(x<b+c)
$$
Therefore, $x<\min(a+c,b+c)$.

For $(2)$: Same thing, we get $x-a<c$ or $x-b<c$ this would mean $x<a+c$ or $x<b+c$. Therefore, we have:
$$
(x<a+c)\cup(x<b+c)
$$
Therefore, $x<\max(a+c,b+c)$.

Is my reasoning right?

Is this also true for negative real numbers?

答:
Based on your response to my comment/questions, it seems that you made a mistake in part 2.  If I am not mistaken, the constraint of

$$f(x-a)+f(x-b) = 0 \tag1$$

must be interpreted as

$$f(x-a) = 0 ~\textbf{and}~ f(x-b) = 0.\tag2$$

At least this is the impression that I got when I indicated that no other information is known about $f(x)$ except that

$$x < c \implies f(x) = 0. \tag3$$

This means, that in trying to guarantee that equation (1) above is satisfied, since the only information given is represented by equation (3) above, you are forced to interpret part 2 of the problem as requiring that constraint (2) above is satisfied.

This means that you must have:

$$(x-a) < c \implies x < (a+c) ~~~\textbf{and}~~~
(x-b) < c \implies x < (b+c).$$

Title: $f(x-a)+f(x-b)$ vs $f(x-a)f(x-b)$


问:
Intuition: Why do we divide by $\sqrt{n}$ instead of $n$ in the Central Limit Theorem?
The law of large numbers for example is very straightforward to understand. You sum up the random variables, and divide them by their number. You take the almost sure limit of the arithmetic mean and end up with the expectation. What's not to love?

Now for the Central Limit Theorem (CLT) we consider convergence in distribution so I realize the logic is different. Let me state a simple version for reference:

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of i.i.d. random variables with $E[X_1]=0$ and >$Var(X_1) = \sigma^2$ where $0< \sigma <\infty$. Then we have that

$$ G_n := \frac{1}{\sqrt{n \sigma^2}} \sum_{k = 1}^{n} X_k  \overset{d}{\longrightarrow} G$$

where $G \sim N(0,1)$

I have seen more than one proof for the CLT but they are quite technical. For a long time I brushed this issue aside but it is repeating itself in several other theorems I encountered (e.g. Donsker's theorem for partial sum processes) so by now I really want to understand. Is there an intuition as to where the square root is coming from?

答:
You want $G_n$ to be a normalized random variable. i.e. $E(G_n) = 0$ and $\text{Var}(G_n) = 1$. This is done by translating $\sum_{i = 1}^n X_i$ by $n E(X_i)$ and by dividing by $\sqrt{n \sigma^2}$, since variance is quadratic.

So what you get is

$$ E(G_n) = E(\frac{\sum_{i = 1}^n X_i - n E(X_i)}{\sqrt{n \sigma^2}}) = \frac{n E(X_i) - n E(X_i)}{\sqrt{n \sigma^2}} = 0$$

$$ \text{Var}(G_n) = \text{Var}(\frac{\sum_{i = 1}^n X_i - n E(X_i)}{\sqrt{n \sigma^2}}) = \frac{n \text{Var}(X_i) - 0}{n \sigma^2} = 1$$

If you did not normalize the random variable you really could not expect any kind of convergence. In the case of the law of large numbers, the variances go to $0$ (which is why you expect it to converge to a constant).

$$ \text{Var}(\frac{\sum_{i = 1}^n X_i}{n}) = \frac{ n \text{Var}(X_i)}{n^2} \rightarrow 0$$

More generally, if you want to look for interesting limit distributions the question of "how much" one needs to normalize is very important. If you converge "too little" the sequence will not converge. If you normalize "too much" (as in the law of large numbers) the sequence will converge to something constant (which can still give meaningful information). If you converge "just right" as in the CLT you get a meaningful distribution.

There are more than one meaningful normalization factors though. For example the normalization factor $\frac{1}{\sqrt{2n \ln(\ln(n))}}$ gives the law of iterated logarithm from which one can also extract information.


问:
How to calculate: $\sum_{n=1}^{\infty} n a^n$
I've tried to calculate this sum:

$$\sum_{n=1}^{\infty} n a^n$$

The point of this is to try to work out the "mean" term in an exponentially decaying average.

I've done the following:

$$\text{let }x = \sum_{n=1}^{\infty} n a^n$$
$$x = a + a \sum_{n=1}^{\infty} (n+1) a^n$$
$$x = a + a (\sum_{n=1}^{\infty} n a^n + \sum_{n=1}^{\infty} a^n)$$
$$x = a + a (x + \sum_{n=1}^{\infty} a^n)$$
$$x = a + ax + a\sum_{n=1}^{\infty} a^n$$
$$(1-a)x = a + a\sum_{n=1}^{\infty} a^n$$

Lets try to work out the $\sum_{n=1}^{\infty} a^n$ part:

$$let y = \sum_{n=1}^{\infty} a^n$$
$$y = a + a \sum_{n=1}^{\infty} a^n$$
$$y = a + ay$$
$$y - ay = a$$
$$y(1-a) = a$$
$$y = a/(1-a)$$

Substitute y back in:

$$(1-a)x = a + a*(a/(1-a))$$
$$(1-a)^2 x = a(1-a) + a^2$$
$$(1-a)^2 x = a - a^2 + a^2$$
$$(1-a)^2 x = a$$
$$x = a/(1-a)^2$$

Is this right, and if so is there a shorter way?

Edit:

To actually calculate the "mean" term of a exponential moving average we need to keep in mind that terms are weighted at the level of $(1-a)$. i.e. for $a=1$ there is no decay, for $a=0$ only the most recent term counts.

So the above result we need to multiply by $(1-a)$ to get the result:

Exponential moving average "mean term" = $a/(1-a)$

This gives the results, for $a=0$, the mean term is the "0th term" (none other are used) whereas for $a=0.5$ the mean term is the "1st term" (i.e. after the current term).

答:
We give a mean proof, at least for the case $0\lt a\lt 1$. Suppose that we toss a coin that has probability $a$ of landing heads, and probability $1-a$ of landing heads. Let $X$ be the number of tosses until the first tail. Then $X=1$ with probability $1-a$, $X=2$ with probability $a(1-a)$, $X=3$ with probability $a^2(1-a)$, and so on. Thus
$$E(X)=(1-a)+2a(1-a)+3a^2(1-a)+4a^3(1-a)\cdots.\tag{$1$}$$
Note that by a standard convergence test, $E(X)$ is finite.

Let $b=E(X)$. On the first toss, we get a tail with probability $1-a$. In that case, $X=1$. If on the first toss we get a head, it has been a "wasted" toss, and the expected number of tosses until the first tail is $1+b$. Thus
$$b=(1-a)(1)+a(1+b).$$
Solve for $b$. We get $b=\dfrac{1}{1-a}$.

But the desired sum $a+2a^2+3a^3+\cdots$ is $\dfrac{a}{1-a}$ times the sum in $(1)$. Thus the desired sum is $\dfrac{a}{(1-a)^2}$.


问:
find a statistic with variance equal to the cramer rao bound
I have found the CR-bound for the paremter of the exponential distribution $f(x) = 1/\lambda \exp(-x/\lambda)$

where $Var(T)\geq \frac{\lambda^2}{n}$.

I know how to the find the minimal sufficient statistics by ratio of likelihood functions:  $\frac{L(X,\theta)}{L(Y,\theta)}, X = [X_1,...,X_n]',X_i,Y_i \sim_{i.i.d} Exp(\lambda)$,
but i don't understands what it means and the significance to get a statistic with variance equal to the CR bound.

答:
Let's consider the following statistics:

$$T=\overline{X}_n$$

It is well known and very easy to prove that

$$\mathbb{E}[T]=\lambda$$

Thus $T$ is unibased.

Now you can calculate variance for the sample mean and compare it with the lowest bound you stated.

$$\mathbb{V}[T]=\frac{\lambda^2}{n}$$

I edited you question because it is $\mathbb{V}[T]\geq \frac{\lambda^2}{n}$ where $T$ is an unbiased estimator and not $\mathbb{V}[\lambda]$ as you stated

Edit answering to your comments

$$\mathbb{E}[T]=\frac{1}{n}\mathbb{E}[\Sigma_i X_i]=\frac{1}{n}\cdot n \lambda=\lambda$$

If you do not want to do all the calculation there is also a Necessary and Sufficient condition which grants that the CR inequality is valid with $=$

$$\Sigma_i \frac{\partial}{\partial \lambda}\log f(x_i,\lambda)=b(n,\lambda)[T-g(\lambda)]$$

If you can factorize in this way, you can substitute $\geq$ with $=$ in CR inequality... (and of course you can, in your case)


问:
What language is accepted by the following dfa (english)?

I feel like it accepts either the string {0}, or repeated {10} only. I tried making it have odd/even number of 0s and 1s but it still doesn't work. I know my answer is wrong but I really don't know how to read this one DFA.

(Sorry my english isn't good)

答:
If the accepting states are the double circled ones:

$0$ moves us to an accepting state.
Anything after the initial $0$ will move us to the "central state" which is a black hole: we never leave and can no longer reach an accepting state anymore.

Downstairs it's more interesting: $11$ gets us to the black hole, so no accepted string can start with $11$, $10$ does bring is into an accepting state; after that we should have no $0$ but $10$ again to bring us back. $11$ is going to the black hole again.

So I get that $\{0\} \cup 10(10)\ast$ is the accepted language for this DFA.


问:
Prove equality of linear maps
Let $V,W$ be $K$-vector spaces and $S$ a $K$-basis of $V$. Suppose that $V$ is ﬁnite dimensional.

Let $\varphi_1,\varphi_2: V → W$ be $K$-linear maps. Suppose $\varphi_1(s) = \varphi_2(s)$ for all $s ∈ S$. Prove: $\varphi_1 = \varphi_2$.

For every $s ∈ S$, let $w_s ∈ W$. Prove that there is a unique $K$-linear map $\varphi : V → W$ such that $\varphi(s) = w_s$.

My Thinking:

Suppose $\varphi_1(s) = x$, then $\varphi_2(s) =x \:$ for all $s \in S$. Then we have $\varphi_1^{-1}(x) =s\:$ and $\:\varphi_2^{-1}(x)=s \:$ for all $x\in W $. So $\varphi_1 = \varphi_2.$

By 1) we have that $\varphi$ is unique.

I appreciate the help.

答:
Your reasoning for 1 seems a little bit off. For the fact that the maps need not be invertible so your preimage could contain multiple elements, furthermore you then prove something odd, you say that the preimage of every element in $W$ is equal to s, which clearly does not always have to hold(take for example $V=W$ and the identity map on $V$). You have to show that $\phi_1$ and $\phi_2$ map every element of $V$ to the same element in $W$, i.e. $$\forall v\in V: \phi_1(v)=\phi_2(v)$$I will try to help you understand my proof if you have any questions, so here we go:

Assume that for all $s\in S$ we know $\phi_1(s)=\phi_2(s)$, then take any $v\in V$. As $S$ is a basis we know there exist $\alpha_s\in K$ such that $v=\sum_{s\in S}\alpha_s s$. Now as the $\phi$'s are linear we can use the linearity together with the base representation to deduce the following $$\phi_1(v)\overset{\textrm{base representation}}{=}\phi_1(\sum_{s\in S}\alpha_s s)\overset{\textrm{linearity}}{=}\sum_{s\in S}\alpha_s \phi_1(s)\overset{\textrm{assumption}}{=}\sum_{s\in S}\alpha_s \phi_2(s)\overset{\textrm{linearity}}{=}\phi_2(\sum_{s\in S}\alpha_s s)\overset{\textrm{base representation}}{=}\phi_2(v)$$
Now as $v\in V$ was arbitrary we know that $\phi_1=\phi_2$.

As you already noted correctly the uniqueness of such a map follows from 1) but we still have to show the existence! Showing that there exists something unique always is a two part exercise. As the first part is already done in 1) we can concentrate on the second one. As we know that $S$ is a basis of $V$ the idea how to construct such a map is somewhat natural. If $v=\sum_{s\in S}\alpha_s s$ is a base representation of $v\in V$ then we just define $$\phi(v)=\sum_{s\in S}\alpha_s w_s$$ So we just define $\phi$ on the basis of $V$. Now we still need to prove that this map is linear as we want to construct a linear map, which i am sure you can do on your own so i will leave this as an exercise to you.

Title: Prove equality of linear maps


问:
How to integrate $\int \frac{\sqrt{a^2-x^2}}{x}dx$?
Ok! So I tried putting $x=a\cos(t)$ and my answer came out as $a\sin(t)-a\log{(\sec(t)+\tan(t))}$. But the answer given is $(a^2-x^2)+a\frac{\log{(\sqrt{a-(a^2-x^2)})}}{x}$. I'd like to know where am I going wrong? Thanks in advance!

答:
Hint: rewrite the integral as $\displaystyle\int \frac{\sqrt{a^2(1-\frac{x^2}{a^2})}}{x}dx$ and take $\displaystyle \frac{x}{a}=\sin{t}$.

So this implies that: $$dx=\cos{t}dt\implies \int \frac{\sqrt{a^2(1-\frac{x^2}{a^2})}}{x}dx=\int \frac{\sqrt{a^2(1-(\sin{t})^2)}}{\sin{t}}\cos{t}dt=\int \frac{\sqrt{a^2}\cos{(t)}}{\sin{(t)}}\cos{(t)}dt=\int \sqrt{a^2}\cot{(t)}\cos{(t)}dt$$
Now rewriting $\cos{(t)}\cot{(t)}$ as $\csc{(t)}-\sin{(t)}$ we have:
$$\sqrt{a^2}\Big(-\int \sin{(t)}dt+\int \csc{(t)}dt\Big)=\sqrt{a^2}\Big(-\cos{t}-\log{(\cot(t) + \csc(t))}\Big) $$
Now $t=\arcsin{\frac{x}{a}}\implies a\Big(-\cos{(\arcsin{\frac{x}{a}})}-\log{(\cot(\arcsin{\frac{x}{a}}) + \csc(\arcsin{\frac{x}{a}}))}\Big) $.

Since  $\cos(\arcsin{\frac{x}{a}})=\sqrt{1-\frac{x^2}{a^2}}$ and $\cot(\arcsin{\frac{x}{a}})=\sqrt{-\frac{x^2}{a^2}}=\frac{\sqrt{1-\frac{x^2}{a^2}}}{\frac{x}{a}}$ and $\csc(\arcsin{(\frac{x}{a}}))=\frac{1}{\frac{x}{a}}$, it remains:
$$\sqrt{a^2-x^2}-a\log\Bigg({\frac{a\big(\sqrt{1-\frac{x^2}{a^2}}+1\big)}{x}}\Bigg)+c$$


问:
Help prove that $\lim_{n \to \infty} \sum_{k = 1}^{n/2} (\frac{ne}{k})^k(1-1.8\frac{\ln(n)}{n})^{k(n-k)} = 0$
I am trying to prove that
$$\lim_{n \to \infty} \sum_{k = 1}^{n/2} (\frac{ne}{k})^k(1-1.8\frac{\ln(n)}{n})^{k(n-k)} = 0,$$
but did not manage to get far. I tried using the same techniques as proposed in
How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$? and $\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\ln k}{n} ( 1-\{\frac{n}{k} \} ) ( (1-\frac{k}{n} \{\frac{n}{k}\} )-\frac{1}{2})$, but did not see how these could be expanded to work in my case. Any hints in how to solve the limit are very welcome. I am especially struggling with the fact that the summation only goes to $n/2$, not to $n$.

答:
The summands can be bounded as follows
\begin{align*}
& \le \left( {\frac{{ne}}{k}} \right)^k \exp \left( { - 1.8\frac{{\log n}}{n}k(n - k)} \right) = \left( {\frac{e}{k}} \right)^k \exp \left( {k\left( {1 - 1.8\frac{{n - k}}{n}} \right)\log n} \right).
\end{align*}
For $1 \le k \le \left[ {0.4n} \right]$ the right-hand side is at most
$$
\left( {\frac{e}{k}} \right)^k \exp \left( { - \frac{1}{2}\log n} \right) = \left( {\frac{e}{k}} \right)^k \frac{1}{{\sqrt n }},
$$
whereas for $\left[ {0.4n} \right] < k \le \left[ {0.5n} \right]$, it is at most
$$
\left( {\frac{e}{k}} \right)^k \exp \left( {\frac{n}{{20}}\log n} \right) \le \left( {\frac{e}{{0.4n}}} \right)^{0.4n} \exp \left( {\frac{n}{{20}}\log n} \right) = n^{ - 7n/20} \left( {\frac{e}{{0.4}}} \right)^{0.4n}  < n^{ - 7n/20} 3^n .
$$So your sum (which is positive) can be bounded from above by
$$
\sum\limits_{k = 1}^{\left[ {0.4n} \right]} {\left( {\frac{e}{k}} \right)^k \frac{1}{{\sqrt n }}}  + \sum\limits_{k = \left[ {0.4n} \right] + 1}^{\left[ {0.5n} \right]} {n^{ - 7n/20} 3^n }  = \mathcal{O}(1)\frac{1}{{\sqrt n }} + \mathcal{O}(1)n^{1 - 7n/20} 3^n  = \mathcal{O}(1)\frac{1}{{\sqrt n }}.
$$


问:
What is the relation?
The function $x^2 = y\quad$ limits two areas $A$ and $B$:

$A$ is further limited with the line $x= a$, $a\gt 0$.
$A$ rotates around the $x$-axis, which gives Volume $A = Va$.

$B$ is limited with the line $y=b$, $b\gt 0$.
$B$ rotates around the $y$-axis, which gives Volume $B = Vb$.

What are the relations between $a$ and $b$, when $Vb = Va$?

..................

I have come to the solution that:

$Vb = (\pi b^2 )/ 2$

$Va = (\pi a^5) / 5$

so the relation between them is:

$2.5b^2 = a^5$

Is that the final solution or is it more?

答:
Without a drawing or a more detailed description, I cannot be certain.  But under the reasonable interpretation of what you wrote, your conclusion is absolutely correct.  Maybe, since $a$ and $b$ are positive, it might be slightly better to say that
$$b=a^2\sqrt{\frac{2a}{5}}$$</p>

Title: What is the relation?


答:
No, the property that every closed set is a countable intersection of open sets (which is what your property comes down to) is called being perfectly normal (usually in combination with being $T_4$ as well). I've also seen it called a $G_\delta$ space. All metric spaces have this property, but not all spaces do:

The Michael line, and the lexicographically ordered square are some examples of non-metrisable but nice (herditarily normal etc) spaces that do not satisfy this. And $[0,1]^I$ with uncountable $I$ is compact Hausdorff where even singleton sets are not $G_\delta$'s.

Title: Can the closure of a set be written as the intersection of open neighborhoods in a non-metrizable space?

Body:
Let $\Omega$ be a topological space. If $A\subseteq\Omega$, then $$\overline A=\bigcap_{\substack{B\supseteq A\\B\text{ is closed}}}B\tag1.$$ If the topology on $\Omega$ is generated by a metric $d$, then $$B_\varepsilon:=\left\{x\in\Omega:d(x,A)<\varepsilon\right\}$$ is open for all $\varepsilon>0$ and $$\overline A=\bigcap_{\varepsilon>0}B_\varepsilon=\bigcap_{n\in\mathbb N}B_{\frac1n}\tag2.$$

By $(2)$, the closure of an arbitrary subset of a metric space can be written as the (countable) intersection of open neighborhoods. Does the same hold in an arbitrary topological space?


问:
Difference between $\mathscr B$ and $\mathscr B[b_1, ... , b_k]$ in first order logic.
This is from "Introduction to Mathmatical logic-forth edition" by Elliot Mendelson.

In page 60 of the book, it says:

A wf is true for the interpretation M (written $\vDash _M\mathscr B $) if and only if every sequence in $\Sigma$ satisfies $\mathscr B$.

But in page 62 it says:

...we shall say that every $k$-tuple $\langle b_1, ... , b_k \rangle$ in this
relation satisfies $\mathscr B(x_{i_1}, ... , x_{i_k})$ in the interpretation $M$; this will be written as $\vDash_M \mathscr B[b_1, ... , b_k]$

My question is, what is the difference between $\mathscr B$ and $\mathscr B[b_1, ... , b_k]$?

Lets say $\mathscr B[b_1, ... , b_k] = \mathscr C$ .In the book,  $\vDash _M\mathscr B $ means that every possible denumerable sequence of the elements of $D$ (here $D$ is the "Domain of disclosure" in $M$) satisfies $\mathscr B$. So, because the  $\vDash$ notation is used in that way, I assume that  $\mathscr C$ is a first order logic wf and $\vDash_M  \mathscr C$ means that
"every possible denumerable sequence of the elements of $D$ (here $D$ is the "Domain of disclosure" in $M$) satisfies $\mathscr C$ . But the problem is , I have no idea what this wf $\mathscr C$ actually is.

答:
The basic definition is that of the satisfaction relation (page 60), that means:

"intuitively, a sequence $s = (s_1, s_2, \ldots)$ satisfies a wf $\mathscr B$ if and only if, when, for each $i$, we replace all free occurrences of $x_i$ (if any) in $\mathscr B$ by a symbol representing [the objcet of the domain of the interpretation] $s_i$, the resulting proposition is true under the given interpretation."

The relation is expressed by the following symbol: $s \vDash_M \mathscr B$.

Consider a simple example in the language of arithmetic. Formula $(x_1=0)$ will be satisfied in $\mathbb N$ by the sequence $s$ such that $s(x_1)=0$.

Now, the example of page 62 consider the general case of a formula $\mathscr B(x_{i_1},\ldots,x_{i_k})$ with $k$ free variables (listed in increasing order; see the specifications of the language: page 57).

Again, consider as formula $\mathscr B (x_1,x_2)$ the simple example: $(x_1+x_2 \le 5)$.

This formula "specifies" a binary relation in $\mathbb N \times \mathbb N$, i.e. the set of all pairs of natural numbers $n,m$ such that their sum is less-or-equal to five. Thus, $(1,1)$ and $(3,2)$ will belong to that set of pairs, because in each case the two numbers of the pairs satisfy the formula.

According to the previous definition, we have that for sequence $s_1$ such that $s_1(x_1)=s_1(x_2)=1$ and $s_2$ such that $s_2(x_1)=3$ and $s_2(x_2)=2$ we have:

$s_1 \vDash_{\mathbb N} (x_1+x_2 \le 5) \text { and } s_2 \vDash_{\mathbb N} (x_1+x_2 \le 5)$.

If we call $R$ the binary relation on $\mathbb N$ such that: $R = \{ (n,m) \mid n+m \le 5 \}$, we have that

every $2$-tuple (pair) $\langle b_1, b_2 \rangle$ in the relation $R$ satisfies formula $\mathscr B(x_1,x_2)$ in the interpretation $\mathbb N$; this will be written as

$\vDash_{\mathbb N} \mathscr B[b_1,b_2]$.

Now,

what is the difference between $\mathscr B$ and $\mathscr B[b_1,\ldots,b_k]$?

The first one is a symbol in the meta-language denoting a formula with its free variables: $\mathscr B(x_{i_1},\ldots,x_{i_k})$.

The expression $\vDash_M \mathscr B[b_1,\ldots,b_k]$ instead, is an expression of the meta-language referrin to a property of the "interpreted formula": it is an "extension" of the original notion of satisfaction and means that the sequence $s$ that maps the variables $(x_{i_1},\ldots,x_{i_k})$ to a $k$-uple $(b_1,\ldots,b_k)$, i.e. where $s(x_{i_j})=b_j$, satisfies formula $\mathscr B$ in the interpretation $M$.


问:
Given the lenght of the twine, how one could determine the sizes of letters in a word made of this twine?
I have a twine 5 meters long. I would like to arrange the word "MARYSIA" from that tape, keeping all letters the same size, it is ok to cut the twine. I want to choose well known Arial typeface, so I could easily determine the ratios of the letters. How I could determine the size of the letters? First, it could be height, then all other dimensions from the ratios. It could also be an iterative way.

When I am thinking about this problem, I would start with one letter. I would the twine on the floor and would form the letter M (it would be big, 5 meters is a lot of the twine). Then I could shrink the first letter "M" and take the spare twine to form "A", and so on. Finally, I would get all letters in the same size.

Please, how to proceed with such a task only theoretically? Just using pencil and paper (and finally a computer, I can do some calculations using JavaScript)

答:
Start from the fact that the length of each string needed for a specific letter scales linearly with its height $h$ and also linearly with its width $w$. Keeping the aspect ration this means
$$
l_i(\gamma h,\gamma w)  = \gamma^2 l_i(h,w)
$$
Doubling the size means you need four times the length. You can think of starting with fixed $h$ and $w$ and adjusting $\gamma$ until
$$
\sum_i l_i = 5m
$$
From this you can calculate $\gamma$
$$
5m = \sum_i l_i(\gamma h,\gamma w) = \gamma^2 \sum_i l_i(h,w)
$$
since you know the initial length $\sum_i l_i(h,w)$. This leads to
$$
\tilde{l}_i = l_i(\gamma h,\gamma w) = \frac{5m}{\sum_i l_i} l_i
$$
where the $l_i$'s are the initial lengths of letter $i$ and the $\tilde{l}_i$'s are the letters you want to use.


问:
Why are abelian groups of interest? What is their usefulness?
I am reading about Abelian groups

So apparently it is a set, with an associative binary operation, and identity element, an inverse operation and the binary operation must also be symmetric.

But it is not clear to me how they are useful. Trying to find why they are important it seems they arise as "additive structures" in various systems but this is too abstract for me.

Could someone give some less formal/more practical application or usages showing what is important about abelian groups (layman's terms basically)?

答:
When considering the notion of Abelian groups, you are dealing with an algebraic abstraction: you will not find in mathematics any structures that are just Abelian groups without being something more specific, just like you won't find any animals that are mammals without belonging to some more specific species. The importance of an abstract notion is not that it describes something that cannot be described in a more specific way, but that the notion captures common properties of distinct structures in a useful way. Useful here means that it can explain certain properties common to all these distinct structures that follows the exact same pattern of explanation, even though the structures may be quite different. The same principle holds for all kinds of abstract notions, like (to stay within algebra) fields, vector spaces, rings, groups and whatnot.

The importance of the notion of Abelian groups then is that for certain purposes it captures a sweet spot: it is sufficiently restrictive to prove properties that do not hold for more general groups, while it is not so restrictive that it excludes numerous examples which, for these specific purposes, behave in a very similar way. It is not immediately obvious that Abelian groups do serve such a purpose, but experience shows that on one hand one very often encounters examples of Abelian groups, and on the other hand very many interesting mathematical questions can be treated by arguments that apply uniformly across all these examples, while they are not applicable to the even broader class of groups in general. Basically all structures where one can do the customary additive bookkeeping, adding and subtracting items at will, without constraints on the possibility to perform these operations and with associativity and commutativity guaranteeing that the order of such operations does not matter, are Abelian groups.


问:
Normal operator matrix norm
I have some troubles to show that the operator norm of a normal operator is always equal to its largest eigenvalue, how can I proof this? Does anybody of you have a hint? My problem is, that I do not know where to use that this operator is normal? </p>

答:
I'm assuming we are talking finite dimension here, since you say "matrix norm". </p>

<p>Being normal, your operator is unitary diagonalizable. And the operator norm of a diagonal operator is easily seen to be the largest eigenvalue in absolute value. </p>

Title: Normal operator matrix norm

Body:
I'm assuming we are talking finite dimension here, since you say "matrix norm".

Being normal, your operator is unitary diagonalizable. And the operator norm of a diagonal operator is easily seen to be the largest eigenvalue in absolute value.


问:
What is your best way to memorize results for common arguments of trigonometric functions?
Especially looking at sine and cosine here. I'm trying to learn them by heart and it's quite unintuitive for me. If you guys could tell me about your way, be it by drawing up something or a table, I'd be glad to hear you guys suggestions :))</p>

答: The values for sine of 0, 30, 45, 60, and 90 degrees (of course in "real" maths we don't use degrees, but this is how I learnt it when I first encountered trigonometric functions) are $\frac12\sqrt0$, $\frac12\sqrt1$, $\frac12\sqrt2$, $\frac12\sqrt3$, $\frac12\sqrt4$. For cosine you go in the opposite direction.