问:
Average of a pair with respect to a function
I have $200$ units of good A and $200$ units of good B. For each unit, I assign them an integer price $[1, 2, 3, ... 200]$. Each price can only be assigned to one unit of good A and B. For example, 100 could be the price for the 55th unit of good A but it could not simultaneously be the price of the 99th element of good A. However, 100 could be the price for the 55th unit of good A and simultaneously be the price for the 2nd element of good B.

Out of these $200$ from each good, I randomly pick $20$ of good A and $20$ of good B.

From these $20$ of each, I randomly pair together one unit of good A and one unit of good B, $(a_1, b_1)$ and repeat this to $(a_{20}, b_{20})$. How can I prove that there are at least $2$ pairs where the averages of their prices are equal $(\frac{a_1 + b_1}{2} = \frac{a_2 + b_2}{2})$

My thinking:

We can have two functions: $f$ and $g$ where $f$ represents the price for good A and $g$ for good B. I'm thinking that $f$ and $g$ are bijective. How do I go from here?

答:
This need not be true, unless I'm missing something...

Having the same averages is equivalent to having the same sum; I'll speak to that to avoid having to discuss fractions.

So, the question is whether or not you can choose 20 distinct $a_i\in[1,200]$ and 20 distinct $b_i\in[1,200]$ such that $a_1+b_1,\ldots,a_{20}+b_{20}$ are all distinct.

And that, you most certainly can.  Let $a_1=b_1=1$, $a_2=b_2=2$, etc. Then for each $i$ we have $a_i+b_i=2i$, which clearly is distinct for different choices of $i$.


问:
Is there a simpler approach to these system of equations?
I recently came across the following system of equations:

$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2  \\
x^3 + y ^3 + z^3 = 3$$

And I have two questions:

One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :

$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$

Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.

Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.

答:
We will make this into a cubic equation in terms of $z$.  We have $x+y = 1-z$ and $x^2 + y^2 = 2-z^2$.

Factor: $$x^3 + y^3 = ( x + y ) ( x^2 - xy + y^2 ) = ( 1-z ) ( 2-z^2 - xy )$$

$$2xy = ( x + y )^2 - ( x^2 + y^2 ) = ( 1-z )^2 - ( 2 - z^2 ) = 2z^2 - 2z - 1$$

So our third equation becomes (multiplying by $2$):

$$( 1 - z ) ( -4z^2 + 2z + 5 ) + 2z^3 = 6$$

Multiplying both sides by $2$ and expanding the polynomial,

$$6z^3 - 6z^2 - 3z -1 = 0$$

This will have $3$ roots (in the complex numbers), which will be solutions for $x$, $y$, and $z$ (by symmetry).

ContentLicense: CC BY-SA 3.0


问:
How can i prove this identity using the combinatorial method?
$$m\,C(n,m)=n\,C(n-1,m-1)$$
I tried something like how can we choose a group out of n people and then how many ways we can choose the leader , and it worked for the first side but not for the 2nd.
Thank you in advance

答:
You have $n$ people and you are choosing from them team of $m$ players with one captain.

Left-hand side: You pick $m$ players and $C(n,m)$ and then you pick one captain among them.

Right-hand side: You pick a captain and then out of the remaining $n-1$ guys you choose rest of the team $C(n-1,m-1)$.

ContentLicense: CC BY-SA 4.0


问:
probability about the bulb with replacement
I encounter this question

In a lot of 50 light bulbs, there are 2 bad bulbs. If we selected five bulbs and replace it after recording the results.
(a)  Find the probability of at least 1 defective bulb among the 5. (With
replacement)

(b)  How many bulbs at least should he examine so that the probability of
finding at least 1 bad bulb exceeds 80%?

My attempt:
part (a), i find the probability by 1 - ((48C1)^(5) / (50C1)^(5)), is it correct? I'm not sure and think that there may be mistakes in it.

part (b), base on part (a) and let n be the no. of bulbs he should examine at least,
I try to calculate like this: 1 - ((48Cn)^(5) / (50Cn)^(5)) > 80%
((48Cn)^(5) / (50Cn)^(5)) < 0.2
LHS= ((48! / n!(48-n)!)^5) / ((50! / n!(50-n)!)^5)
then I am not sure is it right or not and if it is right how can i continue to find the value of n as i stuck with it.

答:
@JMoravitz has given good advice in Comments. I will give some answers from R. I will leave it to you to see if your work matches my results:

(a) The number $X$ of defective bulbs among five tested with replacement has $X \sim\mathsf{Binom}(n=5, p=2/50),$ You seek $P(X \ge 1) = 1 - P(X=0) = 1 - (24/25)^5 = 0.1846273.$ In R, a binomial PDF is denoted by dbinom. I'll show two computations:

```<code>1 - (48/50)^5
[1] 0.1846273
1 - dbinom(0, 5, 2/50)
[1] 0.1846273
</code>```

(b) Now you want $r$ such that $1 - (24/25)^r > .8$ By using logs you can find that $r=40$ is just enough, as verified in R below.

```<code>1 - dbinom(0, 39, 2/50)
[1] 0.7964936
1 - dbinom(0, 40, 2/50)
[1] 0.8046338
</code>```

In R, one might search among numbers $n = 5$ through $100$
to find the right answer as shown below, where we seek
to make the probability of discovering no bad bulbs to be less than $0.2.$

```<code>n = 5:100
p.0 = dbinom(0, n, 2/50)
r = min(n[p.0 < .2]);  r
[1] 40
</code>```

Note: In practice, it would make no sense to sample with replacement. Sampling without replacement, we would
use a hypergeometric distribution.

Then the probability of at least one bad bulb among $5$ chosen without replacement is

```<code>1 - phyper(0,  2,48,  5)
[1] 0.1918367
</code>```

Then sampling $r = 28$ will be just enough have probability above $0.8$
of finding a bad bulb. For this kind of 'inspection',
sampling without replacement is noticeably more efficient.

```<code>n = 5:50  # only 50 bulbs available
p.0 = dhyper(0, 2, 48, n)
r = min(n[p.0 < .2]);  r
[1] 28    # fewer than for sampling with replacement
1 - dhyper(0, 2, 48, 28)
[1] 0.8114286
1 - dhyper(0, 2, 48, 27)
[1] 0.7934694
</code>```

The relevant hypergeometric probability is

$$P(X = 0) = \frac{{2 \choose 0}{48\choose 28}}{{50\choose 28}} = \frac{22(21)}{50(49)} =    0.1885714 < 0.2.$$

[In computation of hypergeometric probabilities, it is often important to
simplify expressions (as above) in order to avoid factorials
too large for ordinary computer (or calculator) arithmetic
to handle. Or to use logarithms.]

Body:
@JMoravitz has given good advice in Comments. I will give some answers from R. I will leave it to you to see if your work matches my results:

(a) The number $X$ of defective bulbs among five tested with replacement has $X \sim\mathsf{Binom}(n=5, p=2/50),$ You seek $P(X \ge 1) = 1 - P(X=0) = 1 - (24/25)^5 = 0.1846273.$ In R, a binomial PDF is denoted by dbinom. I'll show two computations:

1 - (48/50)^5
[1] 0.1846273
1 - dbinom(0, 5, 2/50)
[1] 0.1846273

(b) Now you want $r$ such that $1 - (24/25)^r > .8$ By using logs you can find that $r=40$ is just enough, as verified in R below.

1 - dbinom(0, 39, 2/50)
[1] 0.7964936
1 - dbinom(0, 40, 2/50)
[1] 0.8046338

In R, one might search among numbers $n = 5$ through $100$
to find the right answer as shown below, where we seek
to make the probability of discovering no bad bulbs to be less than $0.2.$

n = 5:100
p.0 = dbinom(0, n, 2/50)
r = min(n[p.0 < .2]);  r
[1] 40

Note: In practice, it would make no sense to sample with replacement. Sampling without replacement, we would
use a hypergeometric distribution.

Then the probability of at least one bad bulb among $5$ chosen without replacement is

1 - phyper(0,  2,48,  5)
[1] 0.1918367

Then sampling $r = 28$ will be just enough have probability above $0.8$
of finding a bad bulb. For this kind of 'inspection',
sampling without replacement is noticeably more efficient.

n = 5:50  # only 50 bulbs available
p.0 = dhyper(0, 2, 48, n)
r = min(n[p.0 < .2]);  r
[1] 28    # fewer than for sampling with replacement
1 - dhyper(0, 2, 48, 28)
[1] 0.8114286
1 - dhyper(0, 2, 48, 27)
[1] 0.7934694

The relevant hypergeometric probability is

$$P(X = 0) = \frac{{2 \choose 0}{48\choose 28}}{{50\choose 28}} = \frac{22(21)}{50(49)} =    0.1885714 < 0.2.$$

[In computation of hypergeometric probabilities, it is often important to
simplify expressions (as above) in order to avoid factorials
too large for ordinary computer (or calculator) arithmetic
to handle. Or to use logarithms.]


问:
finitely generated subfield of algebraic closure of the finite field with $p$ elements
Let $\mathbb{F}^{\operatorname{alg}}_p$ be the algebraic closure of the finite field with $p$ elements.

I know that any finitely generated subfield of $\mathbb{F}^{\operatorname{alg}}_p $ is contained in some finite sub-union $\displaystyle{\bigcup_{n=1,2,\dots,N} \mathbb{F}_{p^n}}$.

However, I am lacking intuition for why this is so, and I'm having trouble conceiving what a finitely generated subfield of $\mathbb{F}^{\operatorname{alg}}_p $ would look look like.

Though I will continue reading on the matter, any examples of such a finitely generated, and in turn finite, subfield would be greatly appreciated.

答:
What do you mean by $(\mathbf{F}_p^{\mathrm{alg}})^n$? A direct product of $n$ copies of $\mathbf{F}_p^{\mathrm{alg}}$? If you actually just mean a finitely generated subfield of $\mathbf{F}_p^{\mathrm{alg}}$, then this is a general fact about adjoining algebraic elements to a field.

If $F$ is a field and $\alpha_1,\ldots,\alpha_n$ are elements of some extension $K$ of $F$, each of which is algebraic over $F$, then $F(\alpha_1,\ldots,\alpha_n)$ is of finite degree over $F$. This is because you have a tower $F\subseteq F(\alpha_1)\subseteq F(\alpha_1,\alpha_2)\subseteq\cdots\subseteq F(\alpha_1,\ldots,\alpha_n)$, and in each step, you're adjoining a single algebraic element, so you get a finite extension.

In the case of $\mathbf{F}_p^{\mathrm{alg}}$, each of the $\alpha_i$ must lie in some finite extension $F_i$ of $\mathbf{F}_p$ in $\mathbf{F}_p^{\mathrm{alg}}$, and $F_i$ must be of the form $\mathbf{F}_{p^{n_i}}$ for some $n_i\geq 1$ (meaning the unique subfield of $\mathbf{F}_p^{\mathrm{alg}}$ of degree $n_i$ over $\mathbf{F}_p$). But if $n=\max_i n_i$, then each of these $F_i$ is contained in $\mathbf{F}_{p^n}$, so all the $\alpha_i$ lie in $\mathbf{F}_{p^n}$.


问:
Finding a basis of $\mathbb{C}^4$ in which both $\sigma_1\otimes\sigma_1$ and $\sigma_2\otimes\sigma_2$ are diagonal
Let $(\sigma_i)_{i=0,...,3}$ be the Pauli matrices. I'm looking for a way to find a basis of $\mathbb{C}^4$ in which both $\sigma_1\otimes\sigma_1$ and $\sigma_2\otimes\sigma_2$ are diagonal. I thought about finding eigenvectors for them, but this wouldn't be enough it seems. Any hint would be deeply appreciated.

答:
Let $T_1=\sigma_1\otimes\sigma_1$ and $T_2=\sigma_2\otimes\sigma_2$. As $\sigma_1\sigma_2=-\sigma_2\sigma_1$ we see that $T_1$ and $T_2$ commute. We can call upon the fact that commuting diagonalizable matrices are simultaneously diagonalizable, or we can just calculate.

Let us first consider $T_1$. The eigenvectors of $\sigma_1$ are $\binom 1{\pm1}$ belonging to eigenvalues $\lambda_{\pm}=\pm1$. Therefore the eigenspace $V_+$ of $T_1$ belonging to eigenvalue $+1$ is spanned by $v_1:=\binom 11\otimes \binom 11$ and
$v_2:=\binom 1{-1}\otimes \binom 1{-1}$.

Because $T_1$ and $T_2$ commute, the space $V^+$ must be stable under $T_2$. We see that
$\sigma_2\binom 11=\binom{-i}i=-i\binom1{-1}$. Therefore
$$
T_2v_1=\binom{-i}i\otimes \binom{-i}i=(-i)^2\binom 1{-1}\otimes \binom1{-1}=-v_2.
$$
Similarly $\sigma_2\binom1{-1}=\binom i i=i\binom 11$, and thus
$$
T_2v_2=\binom i i\otimes \binom i i = i^2\binom11\otimes\binom11=-v_1.
$$
It follows that

$v_1+v_2$ is an eigenvector of $T_1$ belonging to $+1$ and an eigenvector of $T_2$ belonging to $-1$, and

$v_1-v_2$ is an eigenvector of $T_1$ belonging to $+1$ and an eigenvector of $T_2$ belonging to $+1$.

Similarly, the span $V_-$ of $v_3:=\binom 11\otimes \binom 1{-1}$ and
$v_4:=\binom 1{-1}\otimes \binom 11$ is the 2-dimensional eigenspace of $T_1$
belonging to the eigenvalue $-1$. It, too, must be stable under $T_2$. A similar calculation (verify this, I did not doublecheck) shows that
$$
T_2v_3=v_4\quad\text{and}\quad T_2v_4=v_3.
$$
This implies that

$v_3+v_4$ is an eigenvector of $T_1$ belonging to $-1$ and an eigenvector of $T_2$ belonging to $+1$, and

$v_3-v_4$ is an eigenvector of $T_1$ belonging to $-1$ and an eigenvector of $T_2$ belonging to $-1$.

There you have it. The joint four eigenvectors are $v_1\pm v_2$ and $v_3\pm v_4$. They form a basis of $\Bbb{C}^2\otimes \Bbb{C}^2$. You can identify this space with $\Bbb{C}^4$ by your chosen way.


问:
Computing the Cohomology of a genus-2 surface
I know that it has already been discussed which is the Cohomology of a genus-2 surface but I have a very concrete question about the procedure.

I want to compute the Cohomology of the following genus-2 surface using Mayer-Vietoris:

If I write the Mayer-Vietoris sequence (where $H^j(X;k_{X})=:H^j(X)$), assuming $k$ is a field, I get:
\begin{align}
0\rightarrow H^0(T)\rightarrow H^0(X)\oplus H^0(Y)\rightarrow H^0(S)\rightarrow H^1(T)\rightarrow H^1(X)\oplus H^1(Y)\rightarrow H^1(S)\rightarrow H^2(T)\rightarrow ...
\end{align}
Since $T$ is connected and knowing the cohomology of $X,Y,S$ I get:
\begin{align}
0\rightarrow k\rightarrow k \oplus k\rightarrow k \rightarrow H^1(T)\rightarrow k^4\oplus 0\rightarrow k\rightarrow H^2(T)\rightarrow 0
\end{align}

But now I don't know how to conclude that $H^1(T)=k^4$ and $H^2(T)=k$.

I believe that it follows from knowing how the map $H^1(X)\oplus H^1(Y)\rightarrow H^1(S)$ works, but I am not able to understand this map.

I hope someone can help me.

答: Hint: It might be easier to see the map $S \hookrightarrow X \xrightarrow{\sim} \bigvee_4 \mathbb{S}^1$ on $H_1(- ; k)$, i.e. what happens to the boundary circle of $X$ in the retraction to the wedge of four circles. Then the map on $H^1$ is the dual or transpose.


问:
How to show the sequence $\left\{6(-\frac{5}{6})^n\right\} _{n=1}^\infty$ converges to $0$?
Some quick notes about what I have studied so far about sequences (Note: this is an edit after I accepted the answer and realized my lack of understanding for this simple sequence):

Here I am working with sequences of Real numbers, therefore, for that sequence, I defined a function $a$, such that
$$a: \mathbb{N}\to \mathbb{R}$$
$$a:n  \mapsto a(n) = a_n$$

The sequence is denoted as $\{a_n\}_{n\in\mathbb{N}}$, such that $\mathbb{N} = \{1, 2, 3, ...\}$.

Given sequence $\left\{6(-\frac{5}{6})^n\right\} _{n=1}^\infty$

Once a sequence $\{a_n\}_{n\in\mathbb{N}}$ is convergent for $L\in\mathbb{R}$ when $n\to\infty$, therefore

$$\lim _{n\to\infty} a_n = L$$
this is for only if for a given $\epsilon > 0$, so we can find a $N_0 \in \mathbb{N}$ such that $\forall n > N_0$ we have $|a_n - L| < \epsilon $.
Basically, a sequence is convergent if it tends to a Real number. On the other hand, it is divergent if it does not converge to a Real number. This is the reason why I tried to find the limit but I ended up not being able to solve it.

I know that for the sequence $a_n$, such that $a_n = 6\left(-\dfrac{5}{6}\right)^n$ we have.
$$a_n = 6\left(-\dfrac{5}{6}\right)^n = 6 \cdot\dfrac{(-5)^n}{6^n} = 6^{1-n}\cdot(-5)^n$$

According to Wolfram Alpha, $\lim _{n\to \infty }(6^{1-n}(-5)^n) = 0$ and thus the sequence converges to $0$. But I couldn't solve the limit. Wolfram solves it using exponential but it wasn't a straightforward solution for me.

How can I solve $\lim _{n\to \infty }(6^{1-n}(-5)^n)$ in a simpler and good way?

This might be a  consequence of the first question, but how can I show that the sequence is convergent? What would be another approach?

答: We recognize a geometric sequence. We take $r=-5/6$ and see that $|r| < 1$ so $6 (-5/6)^n \to 0$ (see here for a proof).


问:
Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.

Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.

This problem seems extremely hard until you find out that $f(x)+f(1-x)=1$ and you can then evalute $f(\frac{1}{1996})+f(\frac{1995}{1996}) =1, \dots, f(\frac{998}{1996})+f(1-\frac{998}{1996})=f(\frac12)+f(\frac12)=1 \implies f(\frac{998}{1996})=\frac12$. Thus the sum is just $997+\frac12$. My question is that how can one see something like this from a question like this without having gone through some contest math training where these kinds of "tricks" are taught. Is there some kind of tell here from which one can find this "property"?

答:
There are two aspects which can lead us from
\begin{align*}
f(x)=\frac{9^x}{9^x+3}\tag{1}
\end{align*}
to the functional equation $f(x)+f(1-x)=1$.

Simplification: We can reduce the number of occurrences of $x$ in (1) by factoring out $9^x=3^{2x}$. We obtain
\begin{align*}
f(x)=\frac{9^x}{9^x+3}=\frac{1}{1+3^{1-2x}}\tag{2}
\end{align*}
Known pattern: The right hand side of (2) is a well known pattern which admits the following relationship
\begin{align*}
\frac{1}{1+a}+\frac{1}{1+\frac{1}{a}}=\frac{1}{1+a}+\frac{a}{1+a}=1\tag{3}
\end{align*}

The convenient relationship (3) applies to (2) with a nice twist of
\begin{align*}
\frac{1}{a}=\frac{1}{3^{1-2x}}=3^{-(1-2x)}=3^{1-2(1-x)}
\end{align*}
from which
\begin{align*}
\color{blue}{f(x)+f(1-x)=1}
\end{align*}
follows.


问:
Upward Löwenheim-Skolem Theorem for logics without equality
The usual proof of the upward Löwenheim-Skolem theorem rests on the use of the equality symbol as a primitive symbol, interpreted in every relevant structure as true equality. My question is on how the upward Löwenheim-Skolem theorem has to be modified if equality is not a primitive symbol anymore and, say, just introduced as a binary relation with the relevant axioms expressing that it is a congruence with respect to the other non-logical symbols.

EDIT: Based on the answer of t09l, the above paragraph is best augmented by the following: Is there a meaningful reformulation in contexts without equality or does the literature just not consider upward-type theorems without something like a strong equality?

My question in particular comes from the point of view of many-valued predicate logics where the inclusion of a (crisp) equality primitive creates further expressive strength.

Also I wonder how an exclusion of equality affects the Hanf numbers of extensions of classical first-order logic, i.e. of (e.g.) infinitary first-order logic (without equality).

答:
EDIT: There's a major error here which I will fix later today. (I can't delete this post since it has been accepted; if it is unaccepted, I will delete it until I have time to repair it.)

Per the comments, your main interest is in the behavior of Hanf numbers when we drop equality from a logic. (Below we suppose $\mathcal{L}$ is a "reasonable" logic since we need some meaningful syntax to even talk about equality-free fragments after all.)

For reasonably simple logics - including first-order logic and its infinitary veresions - the situation is quite nice: the Hanf number of the equality-free fragment, the downwardl Lowenheim-Skolem number of the equality-free fragment, and the downward Lowenheim-Skolem number of the original logic all coincide. The argument is as follows:

First, by appropriate "blowing up" we have that for every satisfiable equality-free $\mathcal{L}$-sentence $\sigma$, the class of cardinalities of models of $\sigma$ is closed upwards. So the Hanf number of the equality-free fragment of $\mathcal{L}$ is just the downward Lowenheim-Skolem number of $\mathcal{L}$'s equality-free fragment (= the smallest cardinal $\kappa$ such that every satisfiable sentence has a model of size $\le\kappa$).

Next, for every $\mathcal{L}$-sentence $\varphi$, let $\varphi'$ be the sentence gotten from $\varphi$ by replacing $=$ with $E$ everywhere (for $E$ a fresh binary relation symbol) and adding a clause saying that $E$ is a congruence on the structure (which we can do without equality since reflexivity is expressible as "$\forall x(xEx)$"). For every $\mathcal{M}\models\varphi'$, the quotient $\mathcal{M}/E$ is a model of $\varphi$; this shows that the downward Lowenheim-Skolem number of the equality-free fragment of $\mathcal{L}$ is exactly that of $\mathcal{L}$ itself.

Putting this all together we have, as long as $\mathcal{L}$ is "reasonably simple" (of course I haven't said what that means yet):

The Hanf number of the equality-free fragment of a logic $\mathcal{L}$ is just the downward Lowenheim-Skolem number of $\mathcal{L}$.

In particular, passing from $\mathsf{FOL}$ to the equality-free fragment of $\mathsf{FOL}$ doesn't change the Hanf number.

However, not all logics are so "smooth" when we remove equality. For example, note that the equality-free fragment of second-order logic is as expressive as the whole, since we have $x=y$ iff every equivalence relation $E$ satisfies $xEy$. So here the Hanf number doesn't change at all, even though (contra first-order logic) the "Hanfy" behavior of second-order logic is quite intricate.

So the general picture appears complicated. Certainly as long as our logic "allows blowing-up" in the appropriate sense, as first-order logic does but second-order logic does not, we get the collapse described in the previous section. But in the absence of this the situation is unclear to me.

I suspect, however, that naturally-occurring logics fall into one of the two cases "blow-uppable" or "makes equality redundant," which would explain the absence of much study in the literature.


问:
If $\varphi\in L^p(\mu)'$ and $\varphi(1_B)=\int_Bf\:{\rm d}\mu$, can we deduce $f\in L^p(\mu)$?
Let $(E,\mathcal E,\mu)$ be a finite measure space, $p\ge1$ and $\varphi\in L^p(\mu)'$. Note that $$\nu(B):=\varphi(1_B)\;\;\;\text{for }B\in\mathcal E$$ is a well-defined real-valued measure on $(E,\mathcal E)$ with $\nu\ll\mu$. Hence, $$\nu=f\mu$$ for some $f\in L^1(\mu)$ by the Radon-Nikodým theorem.

Does it somehow follow from $\varphi\in L^p(\mu)'$ that $f\in L^p(\mu)$?

答:
It's standard that $L^{p}(\mu)'$ is isomorphic to $L^{q}(\mu)$, where $q \geq 1$ is the dual exponent satisfying $q^{-1} + p^{-1} = 1$.  Specifically, the map $\Phi : L^{q}(\mu) \to L^{p}(\mu)'$ given by
\begin{equation*}
[\Phi(f)](g) = \int_{E} g(x) f(x) \, \mu(dx)
\end{equation*}
is linear, continuous, and bijective.  This is proved in textbooks on real analysis, often under the name "Riesz Representation Theorem."

Hence, in your case, we can find $f \in L^{q}(\mu)$ so that $\varphi = \Phi(f)$.  Notice $f$ need not be in $L^{p}(\mu)$ (if $p > 2$).


问:
Number of attempts to fully color a disk coloring half of it at a time
Mark precisely one half of the disk by a line that goes through its center.

The angle of the line has a uniform distribution, meaning there is no preferable direction.

You color one half of the disk and keep on with the same coloring scheme.

What is the average number of attempts that you need to color the entire disk?

答:
You need to apply color on average five times (so four or five rotations depending on how you count).

Since we randomly rotate every time WLOG we can assume we always color the right half. I also assume we will always rotate before coloring.

We can represent the uncolored slice simply by a proportion $\alpha$ of the full circle. Let $f(\alpha)$ be the expected number of rotations needed for remaining slice $0 < \alpha \leq \frac{1}{2}$ (the first time around we always color exactly half).

We find that the chance this slice is completely on the right half is $\frac{1}{2} - \alpha$ (consider how much we can rotate $\alpha$ starting from being aligned vertically on one edge to being aligned vertically again on the other edge).  In that case we are done.

We get a similar chance for 'completely on the left half', in which case we made no progress.

In the remaining case we get a partial hit (which has chance $1 - 2\cdot(\frac{1}{2} - \alpha) = 2\alpha$), we can take the mean over all possible proportions that are left (which have uniform equal chance of appearing by rotational symmetry):

\begin{align}
f(\alpha) &= 1 +\left(\frac{1}{2} - \alpha\right) \cdot f(\alpha) + 2\alpha\cdot\lim_{\epsilon \to 0} \frac{1}{\alpha - \epsilon}\left[\int_\epsilon^\alpha f(p)\mathrm{d}p\right]
\end{align}
$$\left(1 + 2\alpha\right)f(\alpha) - 2 = 4\alpha\cdot\lim_{\epsilon \to 0}\frac{1}{\alpha - \epsilon} \left[\int_\epsilon^\alpha f(p)\mathrm{d}p\right]$$

Now as an ansatz*, let $f(\alpha) = 2 + 4\alpha$. Then:

\begin{align}
\left(1 + 2\alpha\right)(2 + 4\alpha) - 2 &= 4\alpha\cdot \lim_{\epsilon \to 0}\frac{1}{\alpha - \epsilon} \left[\int_\epsilon^\alpha (2 + 4p)\mathrm{d}p\right]\\
8\alpha +8\alpha^2 &= 4\alpha\cdot \lim_{\epsilon \to 0}\frac{2\alpha + 2\alpha^2 - 2\epsilon - 2\epsilon^2}{\alpha - \epsilon}\\
8\alpha +8\alpha^2 &=  4\alpha \cdot (2 + 2 \alpha)\quad \checkmark
\end{align}

Thus our final answer is $1 + f\left(\frac{1}{2}\right) = 5$.

My ansatz was generated programmatically:

```<code>import numpy as np
import matplotlib.pyplot as plt

def f(a):
    r = np.random.rand()
    if r < 2*a:
        # Partial hit.
        return 1 + f(np.random.rand() * a)
    if r - 2*a < 0.5 - a:
        # Complete miss.
        return 1 + f(a)

    # Complete hit.
    return 1

def favg(a, it=10000):
    return sum(f(a) for _ in range(it))/it

A = np.linspace(0.0001, 0.5, 100)
fa = np.array([favg(a) for a in A])
plt.plot(A, fa, label="f(a)")
plt.plot(A, 2+4*A, label="2+4a")
plt.legend()
plt.tight_layout()
plt.show()
</code>```

Body:
You need to apply color on average five times (so four or five rotations depending on how you count).

Since we randomly rotate every time WLOG we can assume we always color the right half. I also assume we will always rotate before coloring.

We can represent the uncolored slice simply by a proportion $\alpha$ of the full circle. Let $f(\alpha)$ be the expected number of rotations needed for remaining slice $0 < \alpha \leq \frac{1}{2}$ (the first time around we always color exactly half).

We find that the chance this slice is completely on the right half is $\frac{1}{2} - \alpha$ (consider how much we can rotate $\alpha$ starting from being aligned vertically on one edge to being aligned vertically again on the other edge).  In that case we are done.

We get a similar chance for 'completely on the left half', in which case we made no progress.

In the remaining case we get a partial hit (which has chance $1 - 2\cdot(\frac{1}{2} - \alpha) = 2\alpha$), we can take the mean over all possible proportions that are left (which have uniform equal chance of appearing by rotational symmetry):

\begin{align}
f(\alpha) &= 1 +\left(\frac{1}{2} - \alpha\right) \cdot f(\alpha) + 2\alpha\cdot\lim_{\epsilon \to 0} \frac{1}{\alpha - \epsilon}\left[\int_\epsilon^\alpha f(p)\mathrm{d}p\right]
\end{align}
$$\left(1 + 2\alpha\right)f(\alpha) - 2 = 4\alpha\cdot\lim_{\epsilon \to 0}\frac{1}{\alpha - \epsilon} \left[\int_\epsilon^\alpha f(p)\mathrm{d}p\right]$$

Now as an ansatz*, let $f(\alpha) = 2 + 4\alpha$. Then:

\begin{align}
\left(1 + 2\alpha\right)(2 + 4\alpha) - 2 &= 4\alpha\cdot \lim_{\epsilon \to 0}\frac{1}{\alpha - \epsilon} \left[\int_\epsilon^\alpha (2 + 4p)\mathrm{d}p\right]\\
8\alpha +8\alpha^2 &= 4\alpha\cdot \lim_{\epsilon \to 0}\frac{2\alpha + 2\alpha^2 - 2\epsilon - 2\epsilon^2}{\alpha - \epsilon}\\
8\alpha +8\alpha^2 &=  4\alpha \cdot (2 + 2 \alpha)\quad \checkmark
\end{align}

Thus our final answer is $1 + f\left(\frac{1}{2}\right) = 5$.

My ansatz was generated programmatically:

import numpy as np
import matplotlib.pyplot as plt

def f(a):
    r = np.random.rand()
    if r < 2*a:
        # Partial hit.
        return 1 + f(np.random.rand() * a)
    if r - 2*a < 0.5 - a:
        # Complete miss.
        return 1 + f(a)

    # Complete hit.
    return 1

def favg(a, it=10000):
    return sum(f(a) for _ in range(it))/it

A = np.linspace(0.0001, 0.5, 100)
fa = np.array([favg(a) for a in A])
plt.plot(A, fa, label="f(a)")
plt.plot(A, 2+4*A, label="2+4a")
plt.legend()
plt.tight_layout()
plt.show()


问:
Question: For what value of $c$ will $P(x)= -2x^3+cx^2-5x+2$ have the same remainder when devided by $x+1$ or $x-2$?

For what value of $c$ will $P(x)= -2x^3+cx^2-5x+2$ have the same remainder when devided by $x+1$ or $x-2$?

No idea where to start with this one. Can't use the remainder theorem to find the remainder because of the leading coefficient and can't find the value of c when I have no known remainder. Be nice if the examples in my schools lessons actually included the same kind of questions that are on the tests but I guess that would make to much sense for school.

答:
Remainder theorem:
When $P(x)$ is divided for $(x-k)$, the remainder is $P(k)$

Dividing by $(x+1)$ the remainder is $P(-1)$ and dividing by $(x-2)$ is $P(2)$

So we need to set
$$P(-1)=P(2)\to 9+c=-24 + 4 c\to c=11$$


问:
How many probability measure are there?
I'm new to Probability Theory and I have learned that for a function $\mu$ to be a probability measure, it must satisfy a set of requirements.

I wonder whether for the same measurable space ($i.e.$ when we looking at the same thing/event), can we have 2 different probability measures and why ?

EDIT:
If we can have 2 different probability measures when "measure the same thing", is it counter-intuitive ? (like we have 2 different probability for the same thing to happen)

答:
Certainly we can have many measures on a measurable space. For example, the counting measure assigns measure $n$ to any event of size $n$, and infinite measure otherwise (though that's not a probability measure). The "point mass at $x$" measure, for any $x \in \Omega$, is also a measure - it's even a probability measure, for each $x$. In that case, you're saying that an event only has any chance to happen if it contains one specific outcome, though you're varying which exact outcome it is.

The measure essentially determines the probability distribution. If you change the measure, that means you're redistributing the probability mass between events - for example, maybe you're biasing the dice differently.

There's a useful theorem which tells you when two measures are the same. If they assign finite measure to $\Omega$ (which probability measures always do), and they agree on a $\pi$-system (that is, a collection of sets closed under intersection) generating $\mathcal{F}$, then they agree everywhere. But otherwise you can have a huge variety of measures on the measurable space.

Title: How many probability measure are there?


问:
From Faraday's law in integral form to the differential form(=Maxwell equation) by using Stokes
I want to understand how Stoke's theorem shows that the integral form of Faraday's law:
$$\int_{c(A)} E dr = -\frac{1}{c} \frac{d}{dt} \int_A B ds$$
($A$ is a surface and $c(A)$ its boundary curve) is equivalent to its differential form, i.e. the Maxwell equation
$$\operatorname{rot}E + \frac{1}{c} \frac{d}{dt} B = 0.$$
Now, by applying Stokes theorem we can exchange the line integral on the left by the surface integral
$$\int_A \operatorname{rot} E dA.$$
Then
$$\int_A \operatorname{rot} E dA + \frac{1}{c} \frac{d}{dt} \int_A B dA = \int_A \left(\operatorname{rot} E + \frac{1}{c} \frac{d}{dt} B\right) \cdot dA$$
(what is the rigorous explanation that we are allowed to exchange differentiation by time with the integration here?). If the integrand is zero (i.e. the Maxwell equation holds) then this integral is zero (i.e. Faraday's law in integral form holds). But how do we argue the other way around? Why does it follow here from integral = zero that the integrand = zero?

Thanks.

答:
the integral form of Faraday's law is
$$
\int_{\partial A} \vec E \cdot \mathrm{d}\vec s = -\int\frac{\partial \vec B}{\partial t} \cdot \mathrm{d}\vec A
$$

The form
$$
\int_{\partial A} \vec E \cdot \mathrm{d}\vec s = -\frac{\mathrm{d}}{\mathrm{d}t}\int\vec B \cdot \mathrm{d}\vec A
$$
is usually wrong, if there is a temporal change of A, because
$$
    \frac{\mathrm{d}}{\mathrm{d}t} \int_A \vec{B}\cdot \mathrm{d}\vec{A}=\int_A \frac{\partial \vec{B}}{\partial t} \cdot \mathrm{d}\vec{A} + \oint_{\partial A} (\vec{B} \times \vec{u}) \cdot \mathrm{d}\vec{s} + \int_A (\nabla \cdot \vec{B}) \cdot \vec{u} \cdot \vec{d}\mathbf{A}.
$$
with $\vec u$ being the (local) speed of the area $A$ and its contour line $\partial A$, respectively. The term with $\nabla \cdot \vec B$ does not need to be considered here, because of the Maxwell equation $\nabla \cdot \vec B = 0$.

Note that the field $\vec E$ needs to be measured from the same frame of reference as all other variables.
This means in particular, that an (ideal) conductor moving in space with speed $\vec v$ has an $\vec E$ field of $\vec E=-\vec v \times \vec B$ in the laboratory frame, whereas an observer sitting on the conductor measures an  $\vec E'$ field of $\vec E'=\vec 0$ (the prime ' marks a change of the frame of reference). Many physicists think that the wrong equation is right, because they confuse the different frames of reference.

You will find more information in:
http://de.wikipedia.org/wiki/Elektromagnetische_Induktion#.C3.9Cbergang_von_der_differentiellen_Form_zur_Integralform

and

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

You might want to use google translate to translate the German text.

Best regards


问:
A rational number is given in the form:
A rational number is given in the form: $\frac{p}{q}$; $\frac{p}{q} \in (0,1)$ and $p$ and $q$ are coprime to each other. If $pq= 10!$ , what would be the number of ordered pairs $(p,q)$?

A beginning step would be noticing that $10!$ consist of $4$ primes $(2, 3, 5, 7)$. But how do I proceed? Taking combinations of two numbers from these $4$ primes doesn't seem right and making cases of this also doesn't yield the right result.

答: I believe the answer is 8, but let me go through my reasoning. Since the numbers are coprime, you can't give one prime to both $p$ and $q$, as it seems you have realized. So, we are really counting the number of ways of assigning the primes $\{2,3,5,7\}$ to $p$ and $q$. This is equivalent to picking a subset of these numbers and assigning them to $p$ (the rest need to go to $q$). Thus, this would give us $2^4 = 16$ options. However, given such a pair $(p, q)$, exactly one of $q/p$ and $p/q$ is going to be in the desired range $(0, 1)$. Thus, we cut our options in half and we get 8 such ordered pairs $(p, q)$.

Title: A rational number is given in the form:


问:
How can trigonometric functions such as sine and cosine be expressed as infinite series?
The trigonometric functions being expressed as an infinite series is something I never really understood. I understand that they can be expressed as infinite series but I never actually understood the proof. Can someone explain how we arrive to the following infinite series? I've never seen the derivation.

$${\sin}(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

$${\cos}(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n}}{(2n)!}$$

答: In a calculus course, you will learn about Taylor series.  These are two examples of Taylor series.  So wait until you study calculus; there you will find the derivations for these.</p>


问:
finding the number of triplests (a,b,c) for which (G,*) it is a group
Let the set $ G=\lbrace x\in \mathbb{R}|x^3+ax^2+bx+c=0\rbrace $. Find $p$ which is the number of triplets $(a,b,c)$ from $\mathbb{R}^3$ for which $(G,\cdot)$ it is a group. ($\cdot$ is the multiplication of the real numbers).
I do not know how can be $(G, \cdot)$ a group because I learned only groups defined by a binary operation and I do not see this as $x^3+ax^2+bx+c=0$ a binary operation.

答:
Following TheSilverDoe's hint, one only needs to consider products of three factors equal to $x+1$ or $x-1$ (assuming that there are no non-real complex roots), as well as the  product of $x-1$ with a monic quadratic having a negative discriminant (for the case where there are non-real complex roots).

If $1$ is a real root of multiplicity $3$, then the cubic polynomial must be equal to $(x-1)^3=x^3-3x^2+3x-1$, giving the triplet $(-3,3,-1)$.

If $1$ is a root of multiplicity $2$ and $-1$ is a root of multiplicity $1$, then the cubic polynomial must be equal to $(x-1)^2(x+1)=(x-1)(x-1)(x+1)=(x-1)(x^2-1)=x^3-x^2-x+1$, giving the triplet $(-1,-1,1)$.

If $1$ is a root of multiplicity $1$ and $-1$ is a root of multiplicity $2$, then the cubic polynomial must be equal to $(x-1)(x+1)^2=(x-1)(x+1)(x+1)=(x^2-1)(x+1)=x^3+x^2-x-1$, giving the triplet $(1,-1,-1)$.

If $1$ is a root of multiplicity $1$ and there are also two non-real complex conjugate roots, then the cubic polynomial must be equal to $(x-1)(x^2+dx+e)=x^3+dx^2+ex-x^2-dx-e=x^3+(d-1)x^2+(e-d)x-e$ for some real numbers $d$ and $e$ with $d^2-4e<0$, giving the triplet $(d-1,e-d,-e)$.

So, there are infinitely many triplets $(a,b,c)$ for which $G$ is a group under ordinary multiplication. Of those triplets, only $(-3,3,-1), (-1,-1,1),$ and $(1,-1,-1)$ give a cubic polynomial with no non-real complex roots, and all the other ones give a cubic polynomial with the real root $1$ and two non-real complex conjugate roots.


问:
Why $E|X-Y|^2 = 0$ implies $P(X=Y) = 1$
Why does $$\mathbb{E}|X(t)-Y(t)|^2 = 0$$ imply $$\mathbb{P}(X(t)=Y(t))=1$$
for all $t$.

Idea: By Markov we get $\mathbb{P}(|X-Y|^2 \geq a)=0$, respectively $\mathbb{P}(|X-Y|^2 < a)=1$ with $a$ any real number.
I believe this is a first step, but what is the following step and how is it written mathematically correct?

答:
For every $n \geq 1$, Markov's inequality gives
$$\mathbb{P}\left(|X-Y|^2 \geq \frac{1}{n}\right) = 0$$

So by countable union, you get $$\mathbb{P}\left(\bigcup_{n \geq 1} \left\lbrace |X-Y|^2 \geq \frac{1}{n} \right\rbrace\right)=0$$

which is equivalent to
$$\mathbb{P}(|X-Y|^2 > 0)= 0$$

i.e.
$$\mathbb{P}(|X-Y|^2 =0)= 1$$
i.e.
$$\mathbb{P}(X=Y)= 1$$

Title: Why $E|X-Y|^2 = 0$ implies $P(X=Y) = 1$


答:
Figure out not just the prime factors of $10!$ but the powers as well...i.e. the complete prime factorization of $10!$.

$10!$ is a big (relatively speaking) number, but it's not a monster.  We can express it as $1*2*3*4*5*6*7*8*9*10 = 1*2*3*2^2*5*(2*3)*7*2^3*3^2*(2*5) = 2^8\cdot 3^4\cdot 5^2\cdot 7$

So if $\gcd(p,q) = 1$ and $pq = 2^8\cdot 3^4\cdot 5^2\cdot 7$ then

.... can you take it from here?...

Warning:  The following is free thinking so it will include indirect inefficient reasoning.  The point is to replicate how one would think about solving it; not haw to solve it in hindsight efficiently

$p = 2^a3^b5^c7^d$ and $q= 2^{\alpha}3^{\beta}5^\gamma 7^{\delta}$

Where $a + \alpha = 8; b+\beta = 4; ...etc.....$

But as $\gcd(p,q) = 0$ we either have $a=0$ or $\alpha = 0$ and $b =0$ or $\beta = 0$ etc.

In other word we have $p = 2^{0,8}3^{0,4}5^{0,2}7^{0,1}$ while $q=2^{8,0}3^{4,0}5^{2,0}7^{1,0}$.

So there is two choices for $a:  0$ or $8$; all or nothing.  Same for the other variables.  And $\alpha, \beta, $ etc are completely dependent $a,b,c,d$.

So there are $2^4 = 16$ options.

In hindsight, it be better to answer

If $pq = 10!$ then $q = \frac {10!}p$ so we only have to calculate howmany choices of $p$ there are.

$p = 2^a3^b5^c7^d$ and $q = 2^{8-a}3^{4-b}5^{2-c}7^{1-d}$.  But as $\gcd(p,q)=1$ for each variable either $a =0$ or $8-a=0$ etc so there are $2$ options only.

So $p = 2^{0|8}3^{0|4}5^{0|2}7^{0|1}$ so there are four choices each with $2$ options.

So there $2^4=16$ choices.

.....

Oh.... I overlooked $\frac pq < 1$.

Oh well.  There are $16$ ordered pairs total.  As $\gcd(p,q) = 1$ (or because $10!$ is not a perfect square) we can't have $p < q$ and so there are $\frac {16}2=8$ unordered pairs or in other words pairs where $p < q$.

====

Actually in hind-hindsight it's better to answer as Eyob Tsegaye did and realize it's enought to know $10! =2^m3^n5^s7^t$ and $p$ and $q$ have those powers as "all or nothing" and not have to figure the actual prime factorization.

Title: A rational number is given in the form:

Body:
A rational number is given in the form: $\frac{p}{q}$; $\frac{p}{q} \in (0,1)$ and $p$ and $q$ are coprime to each other. If $pq= 10!$ , what would be the number of ordered pairs $(p,q)$?

A beginning step would be noticing that $10!$ consist of $4$ primes $(2, 3, 5, 7)$. But how do I proceed? Taking combinations of two numbers from these $4$ primes doesn't seem right and making cases of this also doesn't yield the right result.


问:
what is the relationship between vector spaces and rings?
Can you show me an example to show how vector and scalar multiplication works with rings would be really helpful.</p>

答:
Already there are good answers to your (old) question. But none of them provides examples as you asked. So, let me write another answer to partially fulfill that requirement with focus on "algebras". First, let's unwrap the relevant definitions.

A vector space is a a quadruple $(V,\mathbb{F}, +, .)$ consisting of an abelian group $(V, +)$ and field $\mathbb{F}$ acting on it by scalar multiplication $.: \mathbb{F}\times V \to V.$ This scalar multiplication operation is compatible with vector addition and field addition.

A ring is a triple $(R, +,\times),$ where $(R, +)$ is an abelian group. And $(R,\times)$ is a monoid (or a semigroup if the ring is non-unital) with both distributive law, i.e., multiplication is bilinear.

As you can see, there are multiple ways to connect these two concepts. One way is replace (weaker) the field in the definition of a vector space by a ring, i.e. $\mathbb{F}\rightsquigarrow R.$ This is exactly the structure of a modules mentioned in other answers. These are everywhere in mathematics. For example: the set of all smooth vector fields defined on a manifold $M$ form a module over the ring of smooth real valued functions $C^{\infty}(M)$ on $M.$

Another way to connect them is $V\rightsquigarrow R.$ Here we need to be bit careful as we need to add some conditions so that ring multiplication is compatible with field multiplication. Otherwise there can be some exotic algebraic behave miserably. So, let's add two more properties to existing structure:

$a(r_1r_2)=(ar_1)r_2=r_1(a r_2)$ for all $a\in\mathbb{F}$ and $r_1, r_2\in R.$

$(ab)r=a(br)$ for all $a, b\in\mathbb{F}$ and $r\in R.$

Now, we have a vector space equipped with a bilinear product (or ring with a scalar multiplication), known as an (associative and unital) algebra over a field. Form here I will focus on them. There are so many examples for theses in all over the mathematics including:

Square matrices of order $n$ over $\mathbb{F}.$

Polynomial ring $\mathbb{F}[x].$

Set of real-valued continuous functions on a topological space $X.$

Clifford algebras which includes real numbers, complex numbers, bi-complex numbers, quaternions and several other hypercomplex number systems.

Algebras of linear operators are impotant in functional analysis. There the algebra multiplication is given by the composition of operators.

One can further generalize this to  consider more general concept of an associative algebra over a commutative ring, instead of a field. Or replacing the ordinary ring multiplication by a Lie bracket gives a Lie algebra, which is a non-associative algebra, i.e., ring multiplication is not assumed to be associative. This gadget is specially important is differential geometry to describe algebras of vector fields on a differentiable manifold. In general, non-associative algebras include things like octonion, hyperbolic quaternion and Jordan algebras.

On the other hand, by adding more structure into associative algebras over fields we can obtain rich structures like Banach algebras, C*-algebras, Poisson algebras, Hopf algebras and many more.


问:
Is there a simpler approach to these system of equations?
I recently came across the following system of equations:

$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2  \\
x^3 + y ^3 + z^3 = 3$$

And I have two questions:

One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :

$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$

Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.

Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.

答:
For this particular set of three equations, there is no solution. (And therefore there is no solution if you add more equations either.)

From the first equation $x+y+z = 1$ you can eliminate $z$ as $z = 1 - x - y$. Then the second equation $x^2 + y^2 + z^2 = 2$ becomes $x^2 + y^2 + (1 - x - y)^2 = 2$, which being a quadratic expression in $x$ and $y$, defines a conic section in the $x$-$y$ plane, in this case an ellipse. After this, the third equation $x^3 + y^3 + z^3$ which is equivalent to $x^3 + y^3 + (1 - x - y)^3$ defines another curve in the $x$-$y$ plane, which happens not to intersect with the ellipse.

The ellipse in the middle is the curve $x^2 + y^2 + (1-x-y)^2 = 2$; the curves outside but not touching it are $x^3 + y^3 + (1-x-y)^3 = 3$.

Now that we know what to prove (that there is no solution), we can look for an alebraic proof. :-)


问:
A different way to calculate $\int_0^\infty \tfrac{1}{1+x^n} dx$.
Most of you might have stumbled upon this integral, when studying complex analysis, namely via contour integration, but today I asked myself if there is a different way to calculate it. I'm almost sure that I'm not the first one to come up with this technique, but I wanted to share it with you and would like to know if anyone of you knows another approach.

EDIT
I made a mistake regarding the format of this post, so I posted my own approach as an answer now.

答:
Use $x=\tan^{2/n}t$ to write the integral as $$\int_0^{\pi/2}\tfrac2n\tan^{2/n-1}tdt=\tfrac1n\operatorname{B}(\tfrac1n,\,1-\tfrac1n)=\tfrac1n\Gamma(\tfrac1n)\Gamma(1-\tfrac1n)=\tfrac{1}{\operatorname{sinc}\tfrac{\pi}{n}}.$$
You can show $\int_0^1(1-y^k)^{-1/k}dy$ has the same value with $y=\sin^{2/k}t$.


问:
What real values of $c$ makes these two vectors linearly independent?
I had a question from my homework Friday night which stumped me.

It reads, verbatim

Find all real values of $c$ such that the vectors
$$
\begin{bmatrix}2\\c \\0
\end{bmatrix}, \begin{bmatrix}c\\c+3 \\0
\end{bmatrix}
$$

are linearly independent.

I got stumped by this question; after trying to row reduce, I got

\begin{bmatrix}
2c &c^2  & |0 \\
 0&  -\frac{1}{2}c^2+c+3& |0 \\
 0&  0& |0
\end{bmatrix}

I wasn't sure what to do at this point. I see the two quadratics; I'm just unsure of what to do and why.

答:
Suppose the vectors are linearly dependent ie.
$$
\begin{bmatrix}2\\c \\0
\end{bmatrix}=\lambda \begin{bmatrix}c\\c+3 \\0
\end{bmatrix}
$$
with $\lambda \neq 0$. So $c^2 = 2(c+3)$ ie. $c=1 \pm \sqrt{7}$. And if $c=1 \pm \sqrt{7}$ the vectors are linearly dependent. So the vectors are linealy independent iff $c \in \Bbb R - \{1\pm \sqrt{7}\}$.

You can also use row-reduction: let
$$
A=\begin{bmatrix}1 & c/2 \\
0 & c+3-c^2/2
\end{bmatrix},
$$
and use the fact that the vectors are linearly independent iff $\ker A = \{0\}$.


答:
Here is my own answer. Of course, we asumme $n\ge 2$ and I also assume, that you have some knowledge on the Gamma function and its friends. I started by
$$\int_0^\infty \frac{1}{1+x^n}dx =\int_0^\infty \frac{1}{nx^{n-1}}\frac{nx^{n-1}}{1+x^n}dx=\left[\frac{1}{n}x^{-(n-1)}\ln(1+x^n) \right]_0^\infty +\frac{n-1}{n}\int_0^\infty \frac{\ln(1+x^n)}{x^n}dx$$
by integration by parts. One readily checks with L'Hospitals rule that the first term is equal to zero. Now since
$$\left.\frac{d}{dt}(1+x^n)^t\right|_{t=0}=\left.\left((1+x^n)^t \ln(1+x^n)\right)\right|_{t=0}=\ln(1+x^n) $$
by the Leibniz integration rule, we obtain
$$ \int_0^\infty \frac{1}{1+x^n}=\frac{n-1}{n}\frac{d}{dt}\left.\left(\int_0^\infty \frac{(1+x^n)^t}{x^n}dx \right)\right|_{t=0}.
$$
Using the substitution $u=1+x^n$, $(u-1)^{\tfrac{n-1}{n}}=x^{n-1}$ and $dx=\tfrac{1}{n}(u-1)^{-\tfrac{n-1}{n}}du$, we have
$$\int_0^\infty \frac{(1+x^n)^t}{x^n}dx=\frac{1}{n}\int_1^\infty u^t(u-1)^{\tfrac{1-2n}{n}}du.
$$
Substituting $u=\tfrac{1}{v}$, $du=-\tfrac{1}{v^2}dv$ it follows that
$$\int_0^\infty \frac{(1+x^n)^t}{x^n}dx=\frac{1}{n}\int_0^1 v^{-t-2-\tfrac{1-2n}{n}}(1-v)^{\tfrac{1-2n}{n}}dv=\frac{1}{n}B(-t+\tfrac{n-1}{n},-\tfrac{n-1}{n})=\frac{1}{n}\frac{\Gamma(-t+\tfrac{n-1}{n})\Gamma(-\tfrac{n-1}{n})}{\Gamma(-t)},
$$
where $B$ is the Beta function and the well-known connection to the Gamma function was used. Now we "only" have to evaluate the derivate at $t=0$. We have
$$\frac{d}{dt}\left(\frac{\Gamma(-t+\tfrac{n-1}{n})}{\Gamma(-t)}\right)=\frac{\Gamma(-t+\tfrac{n-1}{n})}{\Gamma(-t)}(\psi(-t)-\psi(-t+\tfrac{n-1}{n})),
$$
where $\psi$ is the logarithmic derivative of $\Gamma$ - the digamma function. Since the Gamma function has simple poles at the negative integers including zero, we can't just plug in $t=0$. Now WolframAlpha suggests that the evaluation is in fact $-\Gamma(\tfrac{n-1}{n})$. Does anyone know a simple way to show this? Anyway continuing, we have
$$\int_0^\infty \frac{1}{1+x^n}=\frac{n-1}{n^2}\Gamma(-\tfrac{n-1}{n})(-\Gamma(\tfrac{n-1}{n}))=\frac{1}{n}\Gamma(\tfrac{1}{n})\Gamma(1-\tfrac{1}{n})
$$
and finally by Euler's reflection formula we have
$$\int_0^\infty \frac{1}{1+x^n}dx=\frac{\pi/n}{\sin(\pi/n)}.
$$

Title: A different way to calculate $\int_0^\infty \tfrac{1}{1+x^n} dx$.

Body:
Most of you might have stumbled upon this integral, when studying complex analysis, namely via contour integration, but today I asked myself if there is a different way to calculate it. I'm almost sure that I'm not the first one to come up with this technique, but I wanted to share it with you and would like to know if anyone of you knows another approach.

EDIT
I made a mistake regarding the format of this post, so I posted my own approach as an answer now.


答:
Generally there are uncountably many possible ways to assign probabilities within the axioms of probability. For example in tossing a coin, you can assign to the outcome heads any real number $p\in[0,1]$. Then the outcome tails must be assigned the probability $1-p$ (unless you want to allow for some other outcomes such as the coin lands on its edge; or the coin rolls away and disappears and can't be found.)

However, in practical situations, the preferred probability distribution (or measure) is one that reflects (our view of) reality. We think the coin is fair, so heads and tails are both assigned a probability of $0.5$.  We certainly can assign a fair coin a different probability distribution; but then our probability distribution does not reflect reality, and will not be as useful.

Title: How many probability measure are there?

Body:
I'm new to Probability Theory and I have learned that for a function $\mu$ to be a probability measure, it must satisfy a set of requirements.

I wonder whether for the same measurable space ($i.e.$ when we looking at the same thing/event), can we have 2 different probability measures and why ?

EDIT:
If we can have 2 different probability measures when "measure the same thing", is it counter-intuitive ? (like we have 2 different probability for the same thing to happen)


问:
Wrapped gifts distribution to correct child Probability question
Three different gifts were bought to be distributed to three children. Each gift box was labeled with the recipient’s name. Unfortunately, the gifts were later wrapped, covering the labels. If each child is given one of the three gifts at random to open, what is the probability that no child opens a gift labeled with his or her name? Express your answer as a common fraction

I want to solve this probability question with complementary counting; the probability that each person gets their gifts is 1/3 * 1/2 * 1/1, which is 1/6. Then, can't you do 1 - 1/6, which gives you 5/6. However, the real answer is 1/3. Why?

答:
An arrangement in which no element is in its proper place is called a derangement

For a derangement of $3$ elements, it is obvious by trial and error that out of the $6$ permutations of the "proper" order $ABC$, there are only $2$ derangements, $BCA$ and $CAB$, yielding $Pr = \frac 1 3$

For larger problems, a useful formula you can carry for the number of derangements is

D(n) = $[\frac {n!} e]$ where $[\; ]$ represents the nearest integer function.

Thus number of derangements of $ABCDEF$ will be $[\frac {6!} e] = 265$


问:
Fubini's theorem for Probability measure
Let $(X,B,\mu)$ be a probability measure space. Suppose that $f$ is measurable on $X$.

Let $g(x,y)=f(x)-f(y)$.

Suppose $g $ is $\mu \times \mu$ integrable.

Show $f$ is $\mu$- integrable.

Things I know:

$\mu(X)=1, \mu(\emptyset)=0$.

Since $g$ is $\mu \times \mu$ integrable, by Fubini's Theorem $g_x$ and $g^y$ are integrable for a.e $x$ and $y$.

From here, I'm not sure how to continue.

Any help is appreciated!

Thank you!

答:
From Fubini's theorem existos a $y_0$ such that $f(x)-f(y_0)$ is $\mu-$ integrable

That is $\int |f(x)-f(y_0)| d\mu(x)< \infty$

So $$\int|f(x)|d\mu(x) \leq \int|f(x)-f(y_0)| d\mu(x)+\int|f(y_0)| d\mu(x)$$ $$\int|f(x)-f(y_0)| d\mu(x)+|f(y_0)|<\infty$$


问:
Find the general solution of the differential equation $xy'' + y' = 4x$
I began by rewriting this differential equation as:
$$ \frac { x \frac {d}{dx} w(x) + w(x) } {x} = 4$$
This has the form $y' + P(x)y = Q(x)$ and so $$ \frac {dw} {w} = -P(x)dx$$
Integrating this yields $w= \frac {C}{x}$ and thus $$ \frac {d}{dx}C(x) = 4x$$ so $C(x) = 2x^2 + C$.
But I'm not sure how to proceed with the rest of the solution. Any guidance is greatly appreciated!

答:
I can't follow your steps.

Instead, note that $xy'' + y' = (xy')'$, and the equation $$(xy')' = 4x$$ can be solved for $y$ by integration and basic algebra.


问:
If $(1,1)$ is an eigenvector of $A=\begin{pmatrix}2 &5\\3&k\end{pmatrix}$,then one of the eigenvalues of $A$ is :-
If $(1,1)$ is an eigenvector of $A=\begin{pmatrix}2 &5\\3&k\end{pmatrix}$,then one of the eigenvalues of $A$ is :-

$0,-1,1,2$

Can I get some hints please.

答:
ans: -1
on solving above expression you'll get λ=7 & k=4
put k=4 in given matrics and find the roots of characteristic equation......-1 will be one the roots.</p>


答:
$$x y'' + y' = 4 x$$

The rewrite in the question statement is not correct, but rather it replaces $y'$ with $y$, giving a different differential equation.

One can set $u := y'$, on the other hand and so write the equation as a first order one,
$$x u' + u = 4 x ,$$
and apply the standard technique of looking for an integrating factor.

Hint Alternatively, notice that the left-hand side of the equation can already be written as $$(xy')' = 4 x ,$$
(i.e., we need not introduce an integrating factor), and integrating gives
$$x y ' = 2 x^2 + C,$$
which reduces the problem to rearranging and computing an antiderivative.

Title: Find the general solution of the differential equation $xy'' + y' = 4x$

Body:
I began by rewriting this differential equation as:
$$ \frac { x \frac {d}{dx} w(x) + w(x) } {x} = 4$$
This has the form $y' + P(x)y = Q(x)$ and so $$ \frac {dw} {w} = -P(x)dx$$
Integrating this yields $w= \frac {C}{x}$ and thus $$ \frac {d}{dx}C(x) = 4x$$ so $C(x) = 2x^2 + C$.
But I'm not sure how to proceed with the rest of the solution. Any guidance is greatly appreciated!


答:
$$
\begin{align}
\cos0 & = 1 \\
\cos'0 & = 0 \\
\cos''0 & = -1 \\
\cos'''0 & = 0 \\
\text{Then repeat} & \text{ this sequence of four.}
\end{align}
$$
The problem then is: If $\sum_{n=0}^\infty c_n x^n$ satisfies this same repeating sequence of values of its derivative at $0,$ then what is $c_n \text{?}$ You will find that $c_n$ is $n!$ times the value of the $n$th derivative at $0.$

Title: How can trigonometric functions such as sine and cosine be expressed as infinite series?

Body:
The trigonometric functions being expressed as an infinite series is something I never really understood. I understand that they can be expressed as infinite series but I never actually understood the proof. Can someone explain how we arrive to the following infinite series? I've never seen the derivation.

$${\sin}(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

$${\cos}(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n}}{(2n)!}$$


问:
The natural sequence {$a_n$} is given. Need to prove that from some point all values of $a_n$ will be the same.
Natural number $a_0$ is given. The natural sequence $\{a_n\}$ is given so that $a_n$ is the smallest possible number that satisfies this:   $$\sqrt[n]{a_0\times a_1\times a_2 \times\cdots\times a_n} = \text{natural number}.$$

Prove that after some point all the values of $a_n$ will be the same.

I've done this:

Variable $a_1$ will be $1$ because $a_0\times a_1 \in \mathbb{N}$.
If $a_0$ is a prime  then $a_0\times a_1\times a_2 = x^2$, that means $a_2 = a_0$, and in the same way $a_3 = a_0,\;\cdots,\; a_n=a_0$. That means for prime $a_0$'s it's true.

Then I tried a non-prime example for $a_0$. It was $a_0 = 2^3\times 3$.
And it really gave me that from some point $a_n$-s are the same.
I don't know how to prove this for $a_0 = p_1^{b_1}\times p_2^{b_2}\times\cdots\times p_k^{b_k}$.

答:
This may be a hint and may be a solution.

Mark the $n$-th root by $r_n$:
$r_n=\sqrt[\leftroot{0}\uproot{1}n]{a_0\cdot a_1\cdot\dots\cdot a_n}$.

Note that $r_n$ is a solution for $a_{n+1}$ in sense that if you set $a_{n+1}=r_n$, then the expression $\sqrt[\leftroot{0}\uproot{1}n+1]{a_0\cdot a_1\cdot\dots\cdot a_n\cdot a_{n+1}}$ is equal to a whole (natural) number, which turns out to be exactly $r_n$ again. Now, since you know that you are looking for the smallest possible $a_{n+1}$, you know an upper bound for $a_{n+1}$, and consequently for $r_{n+1}$. What can you now tell about the sequence $\{r_n\}$? Can you compare $r_n$ and $r_{n+1}$? Later, use the fact that every monotone and bounded sequence must converge to see that your sequence also converges, or in other words, becomes constant because that's what convergence is in terms of natural numbers.


答:
I think you've seen the derivation, you just failed to understand it. Naturally, it depends on the definition of the trigonometric functions. Usually, those are introduced geometrically, and with geometry, the addition theorems for $\sin$ and $\cos$ are proved, together with inequalities $$\sin x\le x\le\tan x$$ for $x\in[0,\pi/2)$. From there, you can obtain the derivatives of $\cos$ and $\sin$, and that gives the Taylor series. Known theorems show that the latter are globally convergent. So the logical structure is not all that complicated.

Of course, you can as well start from the series, or from the differential equations, and prove that those functions describe the unit circle (and I've seen textbooks doing that), but it's a bit artificial, since most students learned about trigonometric functions before power series or differential equations.


问:
''Homogeneous'' property on non-singular algebraic curves.
Pick two points $p$ and $q$ on $C$, a non-singular projective curve.  Is an isomorphism $\psi:C\rightarrow C$ such that $\phi(p)=q$?

I know this for automorphisms (in the analytic category) of Riemann surfaces.  Also, every compact riemann surface is algebraic.  So can we turn those holomorphic automorphisms into algebraic ones?

答:
This is very far from true in general. By Hurwitz's automorphisms theorem, any compact Riemann surface (equivalently: any smooth projective algebraic curve over $\mathbb{C}$) of genus $g \geq 2$ has at most $84 (g - 1)$ automorphisms.

On the other hand, for genus 0 and 1, this is true. A genus zero projective curve over an algebraically closed field is isomorphic to $\mathbb{P}^1$, and it's clear that this is homogeneous in this sense (its group of automorphisms is $\operatorname{PGL}_2$, which in fact acts triply transitively on $\mathbb{P}^1$). And in genus 1, choosing a base point makes the curve into an elliptic curve, which itself has a (commutative) group structure, so it's homogeneous just like any other topological group.


问:
non-trivial finite fundamental group
I am a physics student and I want to know about finite fundamental group of subspace of $\mathbb{R}^3 $. Is there a subspace of $\mathbb{R}^3$ with nontrivial finite fundamental group?

I am just familiar with $\pi_1(RP^n)$ for n>1 and $\pi_1 (SO(n))  $ for n>2,  which have nontrivial finite fundamental group and $RP^2$ can not be embedded into $\mathbb{R}^3 $.

Thanks.

答:
The answer to the stated question is "we do not know if there are subsets of $R^3$ with nontrivial finite fundamental groups." It is also unknown if there are subsets of $R^3$ whose fundamental groups contain nontrivial elements of finite order. The latter was discussed in this Mathoverflow question. Below, a group $G$ is called torsion-free if it contains no nontrivial elements of finite order.

My personal take is that the standard invariants of algebraic topology such as fundamental groups, homotopy groups, singular homology groups are designed to work with "nice" spaces and should not be used for arbitrary topological spaces (instead one should use "Chech versions" of the standard invariants). What constitutes "nice" is a bit vague but includes manifolds, simplicial complexes, more generally, cell-complexes.  All these spaces share the property that they are locally contractible. Examples of topological spaces (even among compact subsets of $R^n$) which are not nice are the Cantor set,  Warsaw circle,
Hawaiian earring (and many more). The list of "nice spaces" might include something more general, spaces which are locally path-connected and semi-locally simply connected. The latter combination of conditions appears frequently in the theory of covering spaces (if you read any introductory algebraic topology book you will almost surely encounter these notions). Informally speaking, these two conditions for a topological space $X$ mean:

(a) Any two points in $X$ which are sufficiently close can be connected by a path in $X$ close to these points.

(b) Every sufficiently small loop in $X$ is null-homotopic in $X$.

The examples of "not-nice" spaces I gave above violate one of these two properties.

Here is what I can prove:

Theorem. Suppose that $X\subset R^3$ is a closed subset which is both locally path-connected and semilocally simply connected. Then $G=\pi_1(X,x)$ is torsion-free.

Sketch of the proof. Let $c: S^1\to X$ be a loop in $X$ (based at $x\in X$) representing an element of finite order in $G$. Consider a system of open neighborhoods $U_i$ of $X$ in $R^3$, i.e. open subsets of $R^3$ containing $X$, whose intersection equals $X$:
$$
\bigcap_{i\ge 1} U_i=X.
$$
As I explained in my answer here, fundamental groups of open subsets of $R^3$ are torsion-free. Hence, for each $i$, the group $\pi_1(U_i, x)$ is torsion-free. Since $c$ represents a finite order element of $\pi_1(X,x)$, it also represents a finite order element of $\pi_1(U_i, x)$. Hence, this element of $\pi_1(U_i, x)$ is trivial.

Therefore, for each $i$ there exists a (continuous) extension $f_i: D^2\to U_i$ of $c: S^1\to X$, where $D^2$ is the unit disk with the boundary $S^1$. With a bit more thought, one sees that the images $f_i(D^2)$ can be chosen to lie in a fixed compact subset $K\subset R^3$ (independent of $i$), say, any closed round ball containing $c(S^1)$ would work. Now, using the assumption that $X$ is both locally path-connected and semilocally simply connected, for all sufficiently large $i$, one can "push" $f_i$ into $X$, i.e. find a continuous map $g_i: D^2\to X$ "close" to $f_i$ such that the restriction of $g_i$ to $S^1$ equals $c$. I omit the proof, it imitates the "standard" argument due to Borsuk that locally contractible finite-dimensional compact spaces are absolute neighborhood retracts. (Take a sufficiently fine triangulation of $D^2$. For vertices $v$ of the triangulation define $g_i(v)$ to be a point in $X$ closest to $f_i(v)$. Then use local path-connectedness to $X$ to extend $g_i$ to the edges, then use semilocal simple connectivity of $X$ to extend $g_i$ to the faces of the triangulation.) I do not need infinitely many $g_i$'s, just one, call it $g$. Thus, we found a continuous map $g: D^2\to X$ extending $c: S^1\to X$, hence, $c$ represents the trivial element of $G=\pi_1(X)$. In other words, $G$ is torsion-free. qed

Title: non-trivial finite fundamental group


问:
How to show that this $2n \times n^2$ matrix has rank $2n-1$?
The matrix is fairly messy to present, but quite easy to understand. When $n=3$, the matrix is

\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1
\end{bmatrix}

So, basically, it splits into two parts:

the upper part (from row $1$ to row $n$) has $n$ identity matrices $I_n$.

the lower part is $O_n^1, \ldots, O_n^n$, where each $O_n^i$ is an $n \times n$ matrix whose $i$-th row is all one whereas the other entries are zero.

By some examples and intuition, I am aware that the rank of such a matrix is $2n-1$, but how should I rigorously prove it?

答:
The $2n$th row is the sum of the first $n$ rows minus the sum of rows $n+1$ through $2n-1.$ Thus the rank is at most $2n-1.$

The last $n$ columns in the first $n$ rows and the $0$s in rows $n+1$ through $2n-1$ in the last $n$ columns, show that the only way to make a row of $n^2$ zeros a linear combination of the first $2n-1$ rows is to make the first $n$ coefficients in the linear combination equal to $0.$ But then you can't get $0$s in the first $n^2-n$ columns except by using $0$ as the $(n+1)$th through $(2n-1)$the coefficients. Hence the rows other than the last one are linearly independent.


问:
Derivative of trace involving Hadamard product
Let us assume that $A, S\in\mathbb{R}^{n\times n}$, $U\in\mathbb{R}^{n\times k}$, and $V\in\mathbb{R}^{n\times k}$. I am trying to differentiate the following expression: $$\Phi(U,V)=\mathrm{trace}\left((S\circ A)(S^T\circ(VU^T))\right),$$
with respect to $U$ and $V$, respectively, i.e.:
$$\frac{\partial \Phi}{\partial U}\quad and\quad \frac{\partial \Phi}{\partial V}.$$

in which $\mathrm{trace}(\cdot)$ is the trace of a matrix, and $\circ$ is the Hadamard product.

I appreciate any help.

答:
$\def\T{\operatorname{Tr}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$For
typing convenience, let's define the matrix variable
$$B=S\circ S\circ A$$
and use a colon to denote the trace/Frobenius product
$$\eqalign{
A:B &= \T(A^TB) \\
A:A &= \big\|A\big\|_F^2 \\
}$$
The properties of the trace allow the terms in such a product to be rearranged
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB \\
}$$
Further, the Frobenius and Hadamard products commute
$$A:(B\circ C) = (A\circ B):C$$
Write the cost function using the above notation, then calculate its differential and gradients.
$$\eqalign{
\phi &= (S\circ A):\big(S\circ UV^T\big) \\&= B:UV^T \\
d\phi &= B:\big(dU\,V^T+U\,dV^T\big) \\&= BV:dU \;+\; B^TU:dV \\
\p{\phi}{U} &= BV \qquad\quad \p{\phi}{V} = B^TU \\
}$$


答:
Your solution is not that wrong :
$$xy'' + y' = 4x$$
$$xw' + w = 4x$$
Solve the homogeneous equation:
$$xw'=-w$$
$$\ln w =-\ln x +C$$
$$\implies w=\dfrac {C(x)}{x}$$
Substitute this in the original DE:
$$x(\dfrac {C(x)}{x})'+\dfrac {C(x)}{x}=4x$$
$$x(\dfrac {C'(x)x-C(x)}{x^2})+\dfrac {C(x)}{x}=4x$$
$$C'(x)=4x$$
Integrate:
$$C(x)=2x^2+C$$
Therefore:
$$W=\dfrac {2x^2+C}{x}$$
$$W=2x +\dfrac Cx$$
Since $w=y'$ we have
$$y'=2x +\dfrac Cx$$
Integrate:
$$\boxed {y(x)=x^2 + C_1 \ln x+C_2}$$


答:
Let
$$
A=\begin{bmatrix}
I_n \mid I_n \mid \cdots \mid I_n\\
O^1_n \mid O^2_n \mid \cdots \mid O^n_n\\
\end{bmatrix}.
$$
We're looking for the rank of $A$ so we can do row and column reduction.

Remove the first block-column from the other block-columns

We get
$$
A \sim \begin{bmatrix}
I_n \mid 0 \mid \cdots \mid 0\\
O^1_n \mid O^2_n - O^1_n \mid \cdots \mid O^n_n - O^1_n\\
\end{bmatrix}.
$$

Add the last line to the $n+1$-th line,  the before last to the $n+1$-th line, ..., until you get

$$
A \sim \begin{bmatrix}
I_n \mid 0 \mid \cdots \mid 0\\
O^1_n \mid O^2_n \mid \cdots \mid O^n_n\\
\end{bmatrix}.
$$

Remove the first $n$ lines of the $n+1-$th line

We get
$$
A \sim \begin{bmatrix}
I_n \mid 0 \mid \cdots \mid 0\\
0 \mid O^2_n \mid \cdots \mid O^n_n\\
\end{bmatrix}.
$$

For each $O^k_n$ block (with $k \ge 2$) remove the columns following the first one

So we get
$$
A \sim \begin{bmatrix}
I_n \mid 0 \mid \cdots \mid 0\\
0 \mid U^2_n \mid \cdots \mid U^n_n\\
\end{bmatrix}.
$$
where $U^k_n$ is $n \times n$ matrix with $1$ in the position $(k,1)$ and $0$ elsewhere.

Move the first column of $U_n^n$ to the last column of $A$,
the first column of $U_n^{n-1}$ to the before last column of $A$, ...
(here we're just pushing all the columns to the right).

We can move $I_n$ to the bottom-right (above the previous $1$'s) to finally get
$$
A \sim \begin{bmatrix}
0 \mid 0\\
0 \mid I_{2n-1}
\end{bmatrix}.
$$

It follows that the rank of $A$ equals $2n-1$.

Title: How to show that this $2n \times n^2$ matrix has rank $2n-1$?

Body:
The matrix is fairly messy to present, but quite easy to understand. When $n=3$, the matrix is

\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1
\end{bmatrix}

So, basically, it splits into two parts:

the upper part (from row $1$ to row $n$) has $n$ identity matrices $I_n$.

the lower part is $O_n^1, \ldots, O_n^n$, where each $O_n^i$ is an $n \times n$ matrix whose $i$-th row is all one whereas the other entries are zero.

By some examples and intuition, I am aware that the rank of such a matrix is $2n-1$, but how should I rigorously prove it?


问:
Determine the domain and range of the function? any help would be much appreciated.
Determine the domain and range of the function? any help would be much appreciated.
$y\,\, = \,\,4\cos 3\left( {x\,\, + \,\,1} \right)\,\, - \,\,3$</p>

答:
The function is defined for all $x$. So the domain is the set of all real numbers.

The cosine function takes on all values between $-1$ and $1$, inclusive. So $4\cos(3(x+1))$ takes on all values between $-4$ and $4$. It follows that our function takes on all values between $-7$ and $1$. The range is the interval $[-7,1]$, or equivalently the set of all $x$ such that $-7\le x\le 1$.


问:
Proving the Mean Value Theorem Without Rolle's Theorem
Can someone point me to a proof of the mean value theorem that doesn't directly utilize the results of Rolle's theorem?</p>

答:
Curiosity is awesome and definitely welcomed in maths :)

Let $f \in C^1([a,b], \mathbb{R})$. The Mean Value Theorem asserts the equality $f(b)-f(a) = f'(c)(b-a)$ for some $c \in [a, b]$. Equivalently,

$$
f(b)-f(a) - f'(c)(b-a) = 0.
$$

Let

$$
h(x) := f(b)-f(a) - f'(x)(b-a).
$$

The "standard" argument is to say that, unless $h(a)=0$ or $h(b)=0$, this is equivalent to looking for a (local) extremum $c \in (a, b)$ of the function $x \mapsto (f(b)-f(a))x - f(x)(b-a)$. Indeed, the latter function's derivative is precisely $h$ so then we obtain $h(c) = 0$ as desired. As you pointed out, this is achieved via Rolle's Theorem.

However, one does not need to formulate the problem as an instance of Rolle's Theorem. A more direct proof would only involve $h$. It is easy to show that:

Lemma: If $f \in C([a, b], \mathbb{R})$ and $f$ is differentiable in $(a, b)$ such that $m \le f'(x) \le M$ for all $x \in (a, b)$. Then, $m(b-a) \le f(b) -f(a) \le M(b-a)$.

A fortiori, a function $f \in C^1([a,b], \mathbb{R})$ satisfies the above lemma's hypothesis. Then, for all $x \in [a, b]$

$$
(m - M)(b-a) \le h(x) \le (M-m)(b-a)
$$

Since $(m-M)(b-a) \le 0$ and $(M-m)(b-a) \ge 0$, the Intermediate Value Theorem applied to $h(x)$ shows that there exists $c \in [a, b]$ such that $h(c) = 0$ as desired.

EDIT: the assumption $f \in C^1([a, b], \mathbb{R})$ can be relaxed. Indeed, Darboux's Theorem tells us that if $f \in C([a, b], \mathbb{R})$ is differentiable then for all $y \in [f'(a), f'(b)]$, there exists $c \in (a, b)$ such that $f'(c) = y$.

Therefore, since $0 \in [(m-M)(b-a), (M-m)(b-a)]$, there exists $c \in (a, b)$ such that $h(c) = 0$.


问:
Does convexity of $f(x)$ imply convexity of $f(e^x)$?
Let $0<\epsilon<1$, and let $f:[1-\epsilon,1] \to \mathbb{R}$ be a strictly decreasing, continuous, and strictly convex function. Suppose that $f_-'(1)=0$.

Define $g:[\log(1-\epsilon),0] \to \mathbb{R}$ by setting $g(x)=f(e^x)$. Is $g$ convex on some half-neighbourhood of $0$?

I specifically don't want to assume stronger differentiability assumptions .

The differentiability assumptions do matter here:

If we assume $f,g \in C^2$ and $f''(1)>0$, then $g''(0)=f''(1)>0$ and $g$ is convex.

If we remove the assumption $f_-'(1)=0$, $g$ might be non-convex. Take $f(x)=1-x$. Then $g(x)=1-e^x$, so $g''(x)=-e^x<0$.

答:
No, the given conditions do not imply that $f(e^x)$ is convex in some interval $(-\delta, 0]$. We will give a concrete counterexample below.

Preliminaries/motivation

If $f$ is differentiable then $g(u) = f(e^u)$ is convex iff $h(x) = xf'(x)$ is increasing. We have that $h$ is negative and $h(1) = 0$. Of course that does not imply that $h$ is increasing near $x=1$. But on the other hand, $h$ can not be decreasing near $x=1$ either. Therefore a counterexample must be such that $xf'(x)$ “oscillates” arbitrarily close to $x=1$.

The counterexample

We define $f:(0, 1] \to \Bbb R$ as
$$
 f(x) = -\int_x^1 \frac{h(t)}{t} \, dt
$$
where the function $h:[0, 1] \to \Bbb R$ has the following properties:

$h$ is continuous on $[0, 1]$.

$h$ is negative on $[0, 1)$ and $h(1) = 0$.

$h(x)/x$ is strictly increasing on $(0, 1)$.

There are intervals arbitrary close to $1$ on which $h$ is strictly decreasing.

Then $f'(x) = h(x)/x$ on $(0, 1)$, so that $f$ is strictly decreasing and  strictly convex on $(0, 1]$, with $f_-'(1) = h(1) = 0$. For $g(u) = f(e^u)$ we have
$$
 g'(u) = f'(e^u) e^u = h(e^u)
$$
which is strictly decreasing on intervals arbitrary close to $u=0$, so that $g$ is not convex on any interval $(-\delta, 0]$.

Constructing the function $h$

The idea is to construct $h$ as a piecewise linear function which is alternatingly increasing and decreasing on intervals arbitrarily close to $1$. The slope of the decreasing parts is chosen such that $h(x)/x$ is still increasing.

First we define two sequences recursively:
$$
 x_0 = 0 \, ,\,
 y_n = \frac{x_n+1}{2} \,, \,
 x_{n+1} = \frac{y_n(y_n+3)}{3y_n + 1} \, .
$$
It can be verified that both sequences converge to $1$ and
$$
 0 = x_0 < y_0 < x_1 < y_1 < \ldots < x_n < y_n < x_{n+1} < \ldots
$$
Then $h$ is defined as a linear function on each interval $[x_n, y_n]$ and $[y_n, x_{n+1}]$ with
$$
 h(x_n) = x_n - 1 \, , \, h(y_n) = \frac{y_n-1}{2} \, .
$$
It can be verified that $h(y_n) > h(x_{n+1})$ so that $h$ is strictly decreasing on intervals arbitrary close to $1$.

Also $x-1 \le h(x) \le (x-1)/2$ for all $x \in [0, 1)$, so that $h$ is negative on $[0, 1)$ and can be continuously extended with $h(1) = 0$.

On each interval $[x_n, y_n]$ is $h$ strictly increasing, so that $h(x)/x$ is strictly increasing as well. On each interval $[y_n, x_{n+1}]$ is
$$
 h(x) = \frac{y_n-1}{2} + (x-y_n)\frac{y_n-1}{4y_n}
= \frac{y_n-1}{4} + \frac{y_n-1}{4y_n} x
$$
so that $h(x)/x$ is strictly increasing as well.

Remark

By “smoothing” the function $h$ at all $x_n$ and $y_n$ we can make $f$ infinitely differentiable.


问:
Finding the elements in $\mathbb{Z}[i]/(2+2i)$
I am having some confusion with finding the distinct cosets in $\mathbb{Z}[i]/(2+2i)$.

Attempt: Since $2+2i +(2+2i)=0+(2+2i)$ we can treat the coset representative $2+2i=0 \implies 2i=-2 \implies i=-1$ since cancellation is possible in an integral domain. Thus each element in $\mathbb{Z}[i]/(2+2i)$ can be written in the form $a+(2+2i)$ for some $a \in\{0,1\}$.

Comment: I realize this is incorrect, but do not understand where my approach goes wrong. I believe it is in the cancellation part. However I though cancellation was possible so we can cancel the $2$'s since $\mathbb{Z}$ is an integral domain. What is going on here? What would be the correct way to approach this problem? I know there should be $8$ elements in this ring, however my approach suggests incorrectly that there are only $2$.

Edit: Now I tried this, but still know it is wrong, what is the correct way to do this?

I was thinking $2i=-2 \implies 4=-4 \implies 8=0$ so that the ring is given by $\{I,1+I,2+I,3+I,4+I,...,7+I\}$, however I do not think this is correct. What is wrong with this approach?

Edit: I found an answer somewhere where they said the ring is given by $\{a+bi+(2+2i)|a \in \{0,1,2,3\},b \in\{0,1\}\}$ can someone please help show how to arrive at this answer?

答: You're not using cancellation in $\mathbb{Z}$ or $\mathbb{Z}[i]$, you're trying to use cancellation in $\mathbb{Z}[i]/(2 + 2i)$. (Indeed, note that $2i + 2 = 0$ is false in $\mathbb{Z}[i]$; it's a true equation only in the quotient ring.) How do you know that this quotient ring is an integral domain? You can't use cancellation if you don't know you're working with an integral domain (or at least that the element you're cancelling is a non-zero-divisor).


问:
Follow up question on Folland Prop. 2.34

Suppose $(X,M, \mu)$ and $(Y, N, \nu) $ are complete measures. Let $(X \times Y, L, \lambda)$ be the complete product measure.

If $E \in L$, do we still have $E_x \in N$ for all $x\in X$?

If $A \times B \in L$, do we have $A \in M$ and $B \in N$?

My thinking:

Yes, but I'm not sure about the reason. In the proof of Prop. 2.34 we showed that the set that a) holds is a $\sigma-$ algebra. Not sure if we can do the same in this case.

I'm guessing no. But I'm not sure about this.

Any help is appreciated!

Thanks!

答:
Take  the space $L$ to be the completion of Borel sigma algebra $B_{[0,1]^2}$ on $[0,1]^2$.Note that $B_{[0,1]^2}=B_{[0,1]}\times B_{[0,1]}$ and we use  $m$ the  Lebesgue measure on $[0,1]$

Take a $A \subseteq [0,1]$ a non borel set.Such set $A$ exists and has Lebesgue outer measure zero.

The completion of the above sigma algebra is the sigma algebra of Lebesgue measurable sets with the Lebesgue measure $m_{[0,1]^2}$ which is the complete extension of the measure $m \times m$ on $B_{[0,1]^2}$

Then take $C=\{0\} \times A$.This set has Lebesgue outer measure zero so it belongs  to $L$,but $C_0=A$ is not in $B_{[0,1]}$

I believe,this answers both your questions


问:
How to find $\|A\|\ $?
What will be the operator norm of the matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$ where $a,b,c,d \in \Bbb C\ $?

According to the definition of the operator norm it turns out that $$\|A\|^2 = \sup \left \{ \left (|a|^2 + |c|^2 \right ) |z|^2 + \left (|b|^2 + |d|^2 \right ) |w|^2 + 2\ \mathfrak R\ (a \overline b z \overline w) + 2\ \mathfrak R\ (c \overline d z \overline w) \ \bigg |\ |z|^2 + |w|^2 = 1,\  z,w \in \Bbb C \right \}.$$

Is there any way to simplify the above expression?
Any help will be highly appreciated.

Thanks for your time.

答:
Let $A$ be $n\times n$ matrix, $\|{A}\| = \sup\left\{ \|Ax\| : \|x\| = 1\right\}$. For any $x$ with $\| x\| = 1$ we have
$\|Ax\|^2 = (Ax, Ax) = (A^* A x, x)$, where $A^*$ is conjugate transpose matrix for A.
$A^* A$ is self-adjoint, so $A^* A = U D U^*$ for some unitary matrix $U$ and $D = \mathrm{diag}(\lambda_1, ..., \lambda_n)$, where $\lambda_k$ are non-negative eigenvalues. Hence, $(A^* A x, x) = (U D U^* x, x) = (D U^* x, U^*x)$. Let $y = U^* x$. As $U$ is unitary, $\| U^* x\| = \|x\|$, so $\|Ax\|^2 = (Dy, y)$ and $\|{A}\| = \sup\left\{ \sqrt{(Dy, y)} : \|y\| = 1\right\}$.
$$ (Dy, y) = \lambda_1|y_1|^2 + \lambda_2|y_2|^2 + ... + \lambda_n|y_n|^2 \leq
\underset{1\leq k \leq n}{\max} \lambda_k \cdot \sum_{j=1}^n |y_j|^2 =
\underset{1\leq k \leq n}{\max} \lambda_k \cdot ||y||^2 = \underset{1\leq k \leq n}{\max} \lambda_k$$
Moreover, for $y = (0, ..., 0, 1, 0, ..., 0)$ with $1$ at $k$-th place ($k$ is index of largest eigenvalue) there is an equality.

So, $\|A\| = \sqrt{\underset{1\leq k \leq n}{\max} \lambda_k}$, where $\lambda_k$ are eigenvalues of $A^* A$.

Title: How to find $\|A\|\ $?


问:
Surface area of a torus
One can generate a torus as follows: $\vec{g}=((b+a\cos u)\cos v, (b+a\cos u)\sin v, a \sin u)$. To find its area, we can use a surface integral of the form $S=\iint_{D_{uv}} {\lVert \frac{∂g}{∂u} \times \frac{∂g}{∂v} \rVert \, du \, dv}$. However, in the case of torus the integral becomes seemingly unnecessarily tedious to evaluate. Are there nicer approaches to evaluating the surface area using a surface integral?

答: In fact this problem necessitates only a simple integral, you just have to integrate (b+cos(s))2pia ds from 0 to 2pi. It is a matter of adding an infinity of circles of infinitesimal width ads. The radius of those circles is in average b, but it can be as much as b+a (outer ridge of the torus) and as little as b-a (inner ridge of the torus). For any s, the radius of the circle will be b+cos s, if we choose s so that s=0 corresponds to the position where the radius is maximal. I was very troubled at first to realize that the cosine term becomes zero when we integrate, so that the formula boils down to the absurdly simple 2pia times 2pi*b, big perimeter times small perimeter. We understand why it is so simple if we consider any horizontal slice of the torus. It will be made of two circles, of radius b+cos(s) and b-cos(s), where s is the angle corresponding to one of the two circles (any will do, just keep the same s for both circles). The cosine terms will always cancel! Magic!

Title: Surface area of a torus


答:
Yes, there is. Please, keep in mind that I do not know the definition of operator norm, but if you did that part of job correctly, then there is a simplification.

Firstly, focus on the last half of your expression, the one with real parts. That sum is equal to $\Re(C_3\cdot z\overline{w})$, where $C_3=2(a\overline{b}+c\overline{d})$. Rotating $z$ or $w$ around the zero would affect only this summand, so you are free to multiply $z$ and/or $w$ with appropriate constant $e^{i\varphi}$ to achieve $\Re(C_3\cdot z\overline{w})=|C_3|\cdot|zw|$.

Now, your expression becomes $$\sup\{C_1\cdot|z|^2+C_2\cdot|w|^2+|C_3|\cdot|zw|\mid|z|^2+|w|^2=1,z,w\in\mathbb{C}\}.$$
By agreeing that $C_3$ replaces the $|C_3|$ from now on, your problem reduces to problem of finding the maximum of $$C_1\cdot a^2+C_2\cdot b^2+C_3\cdot ab,\text{  where  }a^2+b^2=1,$$where all numbers involved are real and positive. Now you can apply methods like Lagrangian multipliers to find the maximum and then substitute the expressions behind the constants to get what you want. But it will still be an ugly expression.

Title: How to find $\|A\|\ $?

Body:
What will be the operator norm of the matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$ where $a,b,c,d \in \Bbb C\ $?

According to the definition of the operator norm it turns out that $$\|A\|^2 = \sup \left \{ \left (|a|^2 + |c|^2 \right ) |z|^2 + \left (|b|^2 + |d|^2 \right ) |w|^2 + 2\ \mathfrak R\ (a \overline b z \overline w) + 2\ \mathfrak R\ (c \overline d z \overline w) \ \bigg |\ |z|^2 + |w|^2 = 1,\  z,w \in \Bbb C \right \}.$$

Is there any way to simplify the above expression?
Any help will be highly appreciated.

Thanks for your time.


问:
Proofs involving nested families of sets?
Suppose $\mathscr{A}=\{A_i:i\in\mathbb{N}\}$ is a family of sets such that for all $i,j\in\mathbb{N}$, if $i\le j$ then $A_j\subseteq A_i$. (Such a family is called a nested family of sets).

a) Prove for every $k\in\mathbb{N}$, $\bigcap_{i=1}^{k}A_i=A_k$

b) Prove that $\bigcup_{i=1}^{\infty}A_i=A_1$

Proof of a):

Suppose $x\in\bigcap_{i=1}^{k}A_i$

iff $x\in A_i$ for all $i\in\mathbb{N}$.

iff $x\in A_j$ since $A_i\cap A_j= A_j$, $A_j \subseteq A_i$ and
$A_i=A_j$ since $j\in\mathbb{N}$

iff $x\in A_k$ for all $k\in\mathbb{N}$, since $A_k=A_j\cap A_k$,
$A_k\subseteq A_j$ and $k\in\mathbb{N}$.

Proof of b)

Suppose $x\in\bigcup_{i=1}^{\infty}A_i$. Then $x\in A_1$ since it is
contained in $\bigcup_{i=1}^{\infty}A_i$. Hence
$\bigcup_{i=1}^{\infty}A_i\subseteq A_1$.

Now suppose $x\in A_1$ since $1\le i \le j$ for all
$i,j\in\mathbb{N}$, it follows $1\le i \le j$ for some
$i,j\in\mathbb{N}$. Since $A_j\subseteq A_i\subseteq A_1$, it follows
$x\in A_i$ for some $i\in\mathbb{N}$. Hence $x\in
> \bigcup\limits_{i\in\mathbb{N}}A_i$. Hence
$x\in\bigcup_{i=1}^{\infty}A_i$.

Hence $\bigcup_{i=1}^{\infty}A_i=A_1$

答:
None of the proofs is correct.

a) You mention a $j$ without saying what $j$ is.

b) You claim that every element of $\bigcup_{i\in\Bbb N}A_i$ belongs to $A_1$. That's not obvious.

These statements can be proved as follows:

a) Clearly, $\bigcap_{i=1}^kA_i\subset A_k$. On the other hand, if $x\in A_k$ and if $i\in\{1,2,\ldots,k\}$, then, since $i\leqslant k$, $x\in A_i$. Since this occurs for each $i\in\{1,2,\ldots,k\}$, $x\in\bigcap_{i=1}^kA_i$.

b) Clearly, $A_1\subset\bigcup_{i\in\Bbb N}A_i$. On the other hand, if $x\in\bigcup_{i\in\Bbb N}A_i$, then $x\in A_i$, for some $i\in\Bbb N$. Since $i\geqslant1$, $x\in A_1$.


答:
I’m afraid that neither argument makes sense. In the first argument it is not true that $x\in\bigcap_{i=1}^kA_i$ iff $x\in A_i$ for all $i\in\Bbb N$. For example, let $A_1=\{0\}$, and let $A_i=\varnothing$ for all $i\ge 2$; then $\{A_i:i\in\Bbb N\}$ is a nest. Now take $k=1$: then

$$\bigcap_{i=1}^kA_i=\bigcap_{i=1}^1A_i=A_1=\{0\}\,,$$

so $0\in\bigcap_{i=1}^kA_i$, but clearly $0\notin A_2$ (and in fact $0\notin A_i$ whenever $i\ge 2$).

In the next line you talk about some $A_j$, but you never define it, so nothing that you say about it is meaningful. At the end you conclude that $x\in A_k$ for all $k\in\Bbb N$; this is just a repetition of what you said in the second line, but with a different name for the subscript, and it is every bit as false in general.

In the argument for (b) you start with an $x\in\bigcup_{i=1}^\infty A_i$ and conclude that it must be in $A_1$, since $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Your reasoning here is exactly backwards: the fact that $A_1\subseteq\bigcup_{i=1}^\infty A_I$ tells you that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, not the reverse.

Then when you try to prove the trivial implication that if $x\in A_i$, then $x\in\bigcup_{i=1}^\infty A_i$, you pick some unspecified positive integers $i$ and $j$ with $i\le j$ and claim that the fact that $A_j\subseteq A_i\subseteq A_1$ somehow proves that $x\in A_i$ for some $i\in\Bbb N$. This is a complete non sequitur, and in any case we already know that $x\in A_i$ for some $i\in\Bbb N$, since we assumed at the beginning that $x\in A_1$, and certainly $1\in\Bbb N$.

To prove (a), first observe that if $x\in\bigcap_{i=1}^kA_i$, then $x\in A_k$, so $\bigcap_{i=1}^kA_i\subseteq A_k$. Now suppose that $x\in A_k$, and let $i\in\{1,\ldots,k\}$; then $i\le k$, so $A_k\subseteq A_i$, and therefore $x\in A_i$. Thus, $x\in A_i$ for all $i\in\{1,\ldots,k\}$, and hence $x\in\bigcap_{i=1}^kA_i$. In other words, $A_k\subseteq\bigcap_{i=1}^kA_i$, and it follows that $\bigcap_{i=1}^kA_i=A_k$.

For (b) it’s immediate that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, so $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Now suppose that $x\in\bigcup_{i=1}^\infty A_i$; then there is some $i\in\Bbb N$ such that $x\in A_i$. And $i\ge 1$, so $A_i\subseteq A_1$, so $x\in A_1$. Thus, $\bigcup_{i=1}^\infty A_i\subseteq A_1$, and we conclude that $\bigcup_{i=1}^\infty A_i=A_1$.

Title: Proofs involving nested families of sets?

Body:
Suppose $\mathscr{A}=\{A_i:i\in\mathbb{N}\}$ is a family of sets such that for all $i,j\in\mathbb{N}$, if $i\le j$ then $A_j\subseteq A_i$. (Such a family is called a nested family of sets).

a) Prove for every $k\in\mathbb{N}$, $\bigcap_{i=1}^{k}A_i=A_k$

b) Prove that $\bigcup_{i=1}^{\infty}A_i=A_1$

Proof of a):

Suppose $x\in\bigcap_{i=1}^{k}A_i$

iff $x\in A_i$ for all $i\in\mathbb{N}$.

iff $x\in A_j$ since $A_i\cap A_j= A_j$, $A_j \subseteq A_i$ and
$A_i=A_j$ since $j\in\mathbb{N}$

iff $x\in A_k$ for all $k\in\mathbb{N}$, since $A_k=A_j\cap A_k$,
$A_k\subseteq A_j$ and $k\in\mathbb{N}$.

Proof of b)

Suppose $x\in\bigcup_{i=1}^{\infty}A_i$. Then $x\in A_1$ since it is
contained in $\bigcup_{i=1}^{\infty}A_i$. Hence
$\bigcup_{i=1}^{\infty}A_i\subseteq A_1$.

Now suppose $x\in A_1$ since $1\le i \le j$ for all
$i,j\in\mathbb{N}$, it follows $1\le i \le j$ for some
$i,j\in\mathbb{N}$. Since $A_j\subseteq A_i\subseteq A_1$, it follows
$x\in A_i$ for some $i\in\mathbb{N}$. Hence $x\in
> \bigcup\limits_{i\in\mathbb{N}}A_i$. Hence
$x\in\bigcup_{i=1}^{\infty}A_i$.

Hence $\bigcup_{i=1}^{\infty}A_i=A_1$


问:
What are the units of Singular Value Decomposition components?
I have a symmetric variance/covariance matrix $A$ which is of size (27 x 27). I know that it's rank deficient (rank = 21).

I also know that the units of $A$ are $m^2$. I am trying to use Singular Value Decomposition (SVD) to reduce the dimensionality of the variance/covariance matrix so that it is non-singular and I can use it as a proper variance/covariance matrix to draw random samples from (given a mean $X_0$ of size (27 x 1)).

My plan is to reduce the dimensionality of $A$ to size (21 x 21) ($A^*$) and use this matrix. I know have created a matrix $u$ of size (27 x 1) and $s$ of size (21 x 21) such that $A = u*s*u^T$.

Now, I think that the rectangular matrix $u$ is a mapping of 27 variables in the domain to the 21 variables comprising the subspace. If this is true, I should be able to take $u^T*A*u$ to get $A^*$. This turns into a diagonal (21 x 21) matrix.

My question is:

1.- is this correct until now?

2.- Assuming this is correct, what are the units of $u$ and $s$? I am inclined to believe that $u$ is unit-less (because of my assumption that it is a mapping from one domain to another) and that the units of $s$ are the same as $A$, but I am not sure.

答: $A$ is a map from a vector space $V$ to $W$. Supposing that $V$ has units $v$ and you say $A$ has units $m^2$, it follows that $Av=\sigma u$ for any singular value, so that $\sigma u$ must have units $m^2v$, or if $v$ is unitless, $\sigma$ has units of $m^2$. </p>


答:
I think it makes more sense to call it a 'decreasing family of sets' than a 'nested family of sets' because you can consider all those sets as subsets of $A = \bigcup_{i \in \mathbb N} A_i$, and the power set $\mathcal P(A)$ can be equipped with a partial order where $B \le C$ if and only if $B \subseteq C$. This makes the sequence $\mathscr A$ decreasing in that sense, since $i \le j$ implies $A_i \ge A_j$.

Your proof of (a) shows that you are confused about what the notation actually means. You are right that since $A_i \supseteq A_k$, $x \in \bigcap_{i=1}^k A_i$ if and only if $x \in A_k$, but the formulation is not quite enlightening. Usually you use the axiom of extensionality and try to prove that both sets admit the same elements. Something like this.

($\supseteq$) Since $1 \le i \le k$, we have $A_i \supseteq A_k$ for $i \in \mathbb N$ satisfying $1 \le i \le k$, and therefore $\bigcap_{i=1}^k A_i \supseteq A_k$.

($\subseteq$) We have $\bigcap_{i=1}^k A_i \subseteq A_k$ because $A_k$ is a member of that intersection.

Since both sets contain the same elements, they are equal. Usually the above two sentences suffice, i.e. proving that ($\subseteq$) and ($\supseteq$) hold.

You can do the proof for (b) in the same way. Again, not quite sure what you were trying to say, you seem confused there as well.

($\subseteq$) We have $A_1 \supseteq A_i$ for $i \in \mathbb N$ since $i \ge 1$, and therefore $\bigcup_{i=1}^{\infty} A_i \subseteq A_1$.

($\supseteq$) We have $\bigcup_{i=1}^{\infty} A_i \supseteq A_1$ because $A_1$ is a member of that union.

Hope that helps,


问:
A tree is isomorphic to another tree in which every level α subset of [ωα , ωα+ω)
For a tree of height ω1 with every level countable and no uncountable branch (Aronszajn tree), how can I prove it is order-isomorphic (in the sense there existes a bijection which respects the orders) to another such tree in which every level α is a subset of [ωα , ωα+ω) ?

We can suppose every ordinal β has a height α if β ∈ [ωα , ωα+ω) (so ht(β) is well defined and unique)

What is the isomorphism ?

答:
SKETCH: Let $T$ be the original Aronszajn tree. Each level of $T$ is countable, so if $T_\alpha$ is level $\alpha$ of $T$, there is an injection $f_\alpha:T_\alpha\to[\omega\alpha,\omega\alpha+\omega)$. Let

$$T'=\bigcup_{\alpha<\omega_1}f_\alpha[T_\alpha]\,,$$

and use $T$ and the maps $f_\alpha$ to define the tree order on $T'$ so that $\bigcup_{\alpha<\omega_1}f_\alpha$ is an order-isomorphism. In other words, in essence you start with (what is going to be) the isomorphism and use it to construct an isomorphic tree $T'$ whose levels have the desired property.


答: Write the $x^n$ coefficient in $f(x)$ as $[x^n]f(x)$. You want$$[x^7](1-x^5)^4(1-x)^{-8}=[x^7](1-x)^{-8}-4[x^2](1-x)^{-8}.$$You can do the rest with the generalized binomial theorem.

Title: What is the coefficient of $x^{11}$ in $(x+x^2+x^3+x^4+x^5)^4(1+x+x^2+...)^4$?

Body:
I got
$$(x+x^2+x^3+x^4+x^5)^4=x^4((1-x^5)/1-x)^4$$
and
$$(1+x+x^2+...)^4=1/(1-x)^4,$$
what I should do next? Please help me with it, thank you.


问:
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

答:
What I would try first if I wanted it to be true - no tricks:

$n \ge \sqrt{n + 1} + \sqrt{n} \iff \frac{n}{\sqrt{n}} \ge \frac{\sqrt{n+1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \iff \sqrt{n} \ge \sqrt{1 + \frac{1}{n}} + 1 \Leftarrow n \ge 5$.

Title: Show that $n \ge \sqrt{n+1}+\sqrt{n}$


问:
Is it true that the domain of this function is finite?
(This is a subproblem that came up as I was proving that $\{e: \phi_e(x) \text{ has infinite domain}\}$ is $\Pi_2$-complete.)

Let $A$ be a set of natural numbers and suppose $$n\in A\iff \forall y_1\exists y_2 T(n,y_1,y_2)$$

(where $T$ is a (computable) relation).

Consider this  partial function on pairs of natural numbers:

$$(n,x) \mapsto 1\text{ if }\forall y_1\leq x\exists y_2 T(n,y_1,y_2)\\(n,x)\text{ is undefined otherwise} $$

If $n\in A$, then the above function as a function of $x$ is total. Is it true that if $n\notin A$, then this function (again as a function of $x$) has finite domain? I think so, and I think this is a trivial result but I can't wrap my head around proving it in a formal way.

Here's what I think informally: if $n\notin A$, then the first condition in the definition may be true for small $x$, but starting from some $x_0$, that wouldn't be true. Is this indeed the case? I can't connect this with the negation of $x\in A$ i.e. with $\exists y_1\forall y_2 \neg  T(n,y_2,y_3) $.

Edit: I modified the original function. Does the new function has a finite domain as a function of $x$ when $x\notin A$? If not, I was wondering if there's a way to modify this function further so that (1) it remains computable, (2) if $n\in A$, the domain is still infinite, (3) if $n\notin A$, the domain is finite?

答:
I don't know why you insist that $(n,x)$ not be defined when it doesn't take the value $1$. I prefer to think of it as a total function that takes the value $0$ when $\forall y_1 \le x, \exists y_2 T(n,y_1,y_2)$ is not true, i.e. when $\exists y_1 \le x, \forall y_2, \neg T(n,y_1,y_2)$. Then the question is whether the set of $x$ for which $(n,x)$ maps to $1$ is finite or not.

Since the above function is now defined everywhere, let me call it $F(n,x)$. This function satisfies the following property: if $F(n,x) = 1$ and $x' \le x$, then $F(n,x') = 1$ as well. So there are two possibilities:

Either $F(n,x) = 1$ for infinitely many integers, in which case the existence of the increasing sequence of positive integers $\{x_m\}_{m \in \mathbb N}$ such that $F(n,x_m) = 1$ implies that $F(n,x) = 1$ for all $x \in \mathbb N$. When this is the case, $n \in A$.

Or, $F(n,x) = 1$ for finitely many integers. By the same argument as above, this means that if we denote the largest such integer by $x_0$, then $F(n,1) = F(n,2) = \cdots = F(n,x_0) = 1$, and $F(n,x) = 0$ for $x > x_0$ because $x=x_0 + 1$ would be the first integer for which $\forall y_2, \neg T(n,x_0+1, y_2)$. When this is the case, $n \notin A$.

Hope that helps,


问:
Average of a pair with respect to a function
I have $200$ units of good A and $200$ units of good B. For each unit, I assign them an integer price $[1, 2, 3, ... 200]$. Each price can only be assigned to one unit of good A and B. For example, 100 could be the price for the 55th unit of good A but it could not simultaneously be the price of the 99th element of good A. However, 100 could be the price for the 55th unit of good A and simultaneously be the price for the 2nd element of good B.

Out of these $200$ from each good, I randomly pick $20$ of good A and $20$ of good B.

From these $20$ of each, I randomly pair together one unit of good A and one unit of good B, $(a_1, b_1)$ and repeat this to $(a_{20}, b_{20})$. How can I prove that there are at least $2$ pairs where the averages of their prices are equal $(\frac{a_1 + b_1}{2} = \frac{a_2 + b_2}{2})$

My thinking:

We can have two functions: $f$ and $g$ where $f$ represents the price for good A and $g$ for good B. I'm thinking that $f$ and $g$ are bijective. How do I go from here?

答: Assuming you can pair any elements of your subsets together, there are 20*20=400 distinct pairs you can make from any two selections. However, the sum of any two integers between 1 and 200 is between 2 and 400, so there are only 399 possible sums. By the pigeonhole principle, for any two selections of 20 elements in A and B there must be at least two possible pairs that have the same sum (and thus average).</p>


问:
discrete sum greater than integration
Why for any $j \in N$ (the natural numbers) we have
$\sum_{n = j+1}^{j^2} \frac{1}{n} \geq \int_{j+1}^{j^2 + 1} \frac{1}{x} dx $

答:
More generally, you have that $$\sum_{n=k}^mf(n) \ge \int_k^{m+1}f(x)dx$$ if $f(x)$ is decreasing, and $m \ge k$. This can be seen by splitting the integral into parts as $$\int_k^{m+1}f(x)dx = \sum_{n=k}^m \int_n^{n+1}f(x)dx$$

Then since $f(x)$ is decreasing, you have that $f(n)$ is the maximum value of $f(x)$ for $x \in [n, n+1]$, so the inner integral has an upper bound of $f(n)$. Therefore, the integral has an upper bound of $$\sum_{n=k}^mf(n)$$ which is the original sum.

Title: discrete sum greater than integration