问:
What's the correct null and alternative hypothesis?
For the scenario below, write down the most appropriate null and alternative hypotheses that you would use for statistical inference on the population mean. Write down the critical value that you would use to run your test. Choose a significance level for the test of 5%.

Mercury levels higher than two parts per billion in drinking water are unsafe. You are responsible for determining whether drinking water in a given city is safe, based on 100 water samples that have been taken from different areas of that city.

My idea is that the null is that Mercury level is three or higher than three, which means that drinking water is unsafe. The alternative is that Mercury level is less than three, which means that water is safe for drinking. If we find evidence against null, we can reject null and conclude that water is safe for drinking. However, in group discussions, my classmates get different ideas from me.

答:
In NHST, at the end of the day, you will either have statistical evidence for the alternative hypothesis, or nothing. So the alternative hypothesis is better something you are interested in showing!

If you want to be able to tell people with confidence: "Yes, you can drink this water", choose $H_1\colon \mu < 2$. If you want to be able to tell the city with confidence "You have a mercury problem", choose $H_1\colon \mu > 2$. An appropriate null-hypothesis to be tested against is $H_0: \mu=2$ in both cases.

The text "You are responsible determining whether drinking water in a given city is safe" indicates that you want to be able to tell people with confidence: "Yes, you can drink this water", and you choose $H_1\colon \mu < 2$.


问:
Difference of Poisson r.v's. Is there a simpler way?
Here's a problem I'm having trouble with (section on Poisson r.v.'s):

Suppose that when a baby is born, the probability it's a boy is $0,52$ and the probability that it's a girl is $0,48$. On some day, 100 babies were born in a hospital. What's the probability that more girls were born? Here's my attempt:

Let $B$ be the amount of boys born, and $G$ the amount of girls. We want to find $P(G-B>0)$. The two r.v.s are Poisson distributed: $G \sim Poiss(48), B \sim Poiss(52)$. This would mean having to find the distribution of $G-B$, the difference of two Poisson r.v's with different parameters - according to Wikipedia that's something called a Skellam distribution, and it looks complicated, which makes me think that there's probably a simpler way. But I don't know how to do it. Any tips would be appreciated.

答:
Let $g$ and $b$ be the parameters for the $G$ and $B$ distributed as Poisson. Since you know that there are $100$ babies born, you can use conditioning as follows:

$P(G > B)=\sum_{k=0}^{49} [P(B=k) | P(G+B=100)]$

Now,

\begin{align*}
 P(B=k) | P(G+B=100)&=\frac{P(B=k)*P(G=100-k)}{P(G+B=100}) \\
 &=\frac{\frac{e^{-b}b^{k}e^{-g}g^{100-k}}{k!(100-k)!}}{\frac{e^{-(b+g)}(b+g)^{100}}{100!}}
\end{align*}

The above can be simplified to a Binomial distribution  $ \sim \operatorname{B} \left({100,\frac{b}{b+g}}\right)$

So in essence you need to calculate:

$\sum_{k=0}^{49} \operatorname{B} \left({100,\frac{b}{b+g}}\right)$ which is equivalent to calculating the CDF of the binomial distribution. You can lookup the calculations involving Binomial CDFs here.

ContentLicense: CC BY-SA 3.0


问:
In $O(2)$ a left-invariant form is not right invariant
We know that the space of left-invariant $1$-forms in $O(2)$ has dimension $1$ and we have a smooth group homomorphism $\lambda :O(2)\rightarrow \mathbb{R}-\{0\}$, where if we have that $\mu$ is a left-invariant $1$-form then $R_h^*\mu =\lambda(h)\mu$, where $h\in O(2)$. I also know that the expression for $\lambda$ is $\lambda(h)=det(Ad(h^{-1}))$, and that since $O(2)$ is a lie group $|\lambda(h)|=1,\forall h\in G$. Now I wanted to show that there is a value of $\lambda$ such that $\lambda(h)=-1$. Is there any good way of doing this without having to look at the local representation of the function $Ad(h)$? I wanted to avoid this since I am not interily sure what are the local coordinates for $O(2)$, since the way I proved it was a submanifold was using the regular value theorem. Any help is aprecciated.

答:
I'm not 100 percent sure about it, but here's what I think happens.

$O(2)$ is compact so the image of $\lambda$ is a compact subgroup of $\mathbb{R}^*$, so it's either $\{1\}$ or $\{\pm 1\}$.

$O(2)$ has two connected components, and the identity component is abelian so the restriction of $\lambda$ to it is abelian. So we're looking for an $h$ which is outside the  component of the identity. The simplest such one is h=$\begin{bmatrix} 1 & 0\\0 & -1\end{bmatrix}$.

Now, if we look at tangent vectors instead of $1$-forms (the dual), we find that the actions of $(L_h)_*$ and $(R_h)_*$ as maps $T_{I_2}\,O(2) \rightarrow T_h\,O(2)$ are opposites one of the other. So, for any $1$-form $\tau$, $((L_h)^*\tau)_e=-((R_h)^*\tau)_e$. In particular, if $\tau$ is left-invariant (and doesn't vanish at $e$), the LHS is $\tau_e$, so this forces $\lambda(h)=-1$.

ContentLicense: CC BY-SA 4.0


问:
Finding the linear span of a polynomial with a condition
I'm trying to find the linear span of the following set:

$M = \{p(x)∈ \mathbb R_4 [x] : p(x)=p(x-1),  x∈\mathbb R\} $

I've tried taking a vector from $M$ and applying the condition on it. Like that:

Let:
$p(x) ∈ M$

$p(x) = p(x-1)$

$a + bx + cx^2 + dx^3 = a + b(x-1) + c(x-1)^2 + d(x-1)^3 $

$b(1) + c(2x-1) + d(3x^2 -3x +1) = 0$

$(b-c+d) + (2c-3d)x + (3d)x^2 = 0$

Now, if I open an equation system based on the variables of the polynomial, I get that b, c and d are equal to 0.
What does that mean in terms of the linear span?
Am I trying to find it wrong?

答: Let V = $\{P(x) \in \mathbb{R}_4[x]: P(x) = P(x-1) \}$, what you showed means that $V \subset \text{Span}\{ 1 \}$, (where in this case 1 is the polynomial, not the scalar). On the other hand, any constant polynomial is in V. Hence $V = \text{Span}\{ 1 \}$, the subspace of constant polynomials.


问:
Limit of a bounded sequence
The $\lim_{n\to\infty} (\frac{\sin(n)+7}{15})^{n^2}=0$.

This is due by the fact that $|a_n|=|\frac{\sin(n)+7}{15}|<1$?

答:
we know that

$-1\leq \sin n \leq 1 \Rightarrow \frac{1}{4} \leq\frac{2+\sin n}{4} \leq \frac{3}{4}$

which means

$(\frac{1}{4})^n \leq (\frac{2+\sin n}{4})^n \leq ( \frac{3}{4})^n$ and

$lim_{n \to \infty}(\frac{1}{4})^n=lim_{n \to \infty}( \frac{3}{4})^n=0$ according to sandwich theorem

$lim_{n \to \infty}(\frac{2+\sin n}{4})^n=0 $

Title: Limit of a bounded sequence


问:
Drawing exactly 2 queens by picking a card, not replacing it, and picking another
What I am thinking is for part "a" to do (4/52)*(3/51) since the card is not being replaced. But the word "exactly" is throwing me off. Am I on the right track?

Pick a card; keep it; pick another card.
Find the probability of getting

a. exactly 2 queens.______

b. exactly one queen.______

c. no queens.________

答:
First, one thing to note is that the question indicates no replacement this is usually a good hint for you to think about the hypergeometric distribution as that is the only probability distribution with no replacement.

In general and as indicated by Wikipedia, the distribution takes the following form:

$\frac{{K \choose k} {N-K \choose n-k}}{N \choose n} $

Where:

$N$ is the population size,

$K$ is the number of success states in the population,

$n$ is the number of draws (i.e. quantity drawn in each trial),

$k$ is the number of observed successes

Applying this formula for your problem we get the following general formula:
$\frac{{4 \choose k} {48 \choose 2-k}}{52 \choose 2}$ for $k=0,1,2$

$N$ and $n$ are fixed since we are always talking $2$ cards out of $52$ and $K$ is fixed since in any standard deck of cards there is always $4$ queens.

Now going back to your main question and applying the formula, we get the following:

1.$\frac{{4 \choose 2} {48 \choose 0}}{52 \choose 2}$ since both our cards are queen we are not picking anything from the other $48$.

2.$\frac{{4 \choose 1} {48 \choose 1}}{52 \choose 2}$ since we are picking one queen out of the $4$ we have and $1$ other card out of the $48$ we have.

3.$\frac{{4 \choose 0} {48 \choose 2}}{52 \choose 2}$ since we are picking no queens and both our cards are from the $4$8 we have that aren't queens.


问:
Let $A$ be a matrix sized $p\times p$, where $2\le p$. Using recurrence relations, describe $A^k$.
Let $A$ be a matrix sized $p\times p$, Where $2\le p$.
The matrix values in the main diagonal are $0$ and the rest are $1$'s.

Example for $A$ where $p=5$:
$$\begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ \end{bmatrix}
$$

Using reccurences relations, describe $A^k$.

Body:
Let $B$ be the matrix with all entries equal to $1$. Then $B^n=p^{n-1}B$ for $n\geq1$. We have $A=B-I$. So
$$
A^n=(B-I)^n=(-1)^nI+\sum_{k=1}^n\,{n\choose k}\,B^k\,(-1)^{n-k}=(-1)^nI+\sum_{k=1}^n\,{n\choose k}\,p^{k-1}B\,(-1)^{n-k}\\
=(-1)^nI+B\,\sum_{k=1}^n\,{n\choose k}\,p^{k-1}\,(-1)^{n-k}=(-1)^nI+B\,\frac1p\,\sum_{k=1}^n\,{n\choose k}\,p^{k}\,(-1)^{n-k}\\
=(-1)^nI+B\,\frac1p\,[(p-1)^n-(-1)^n].
$$
Using that $B=A+I$, we get
$$
A^n=(-1)^nI+\frac1p\,[(p-1)^n-(-1)^n]\,(A+I).
$$


问:
Difference of two (possibly dependent) variables is Gaussian iff the variables are Gaussian?
In probability, a reknown result is that the sum (or the difference) between to Gaussian random variables is in turn normally distributed. What about the converse? If the difference of two (possibly dependent) random variables $X_1$ and $X_2$  is normally distributed, must $X_1$ and $X_2$ be Gaussian random variables? The answer is positive in the case where $X_1$ and $X_2$ are independent

https://en.wikipedia.org/wiki/Cram%C3%A9r%27s_decomposition_theorem

But what if they're dependent?

答: If $X$ is normally distributed and $Y$ is any random variable then the difference $(X+Y)-Y$ is normal. So the answer is NO.


问:
Adding two points on a elliptic curve over $\operatorname{GF}(16)$ constructed using $p(x) = x^4+x+1$
E is the elliptic curve over P(GF($16$)) satisfying

$$y^2 + xy = x^3+β^4x^2+1$$.
$$\text{Let} \quad β= x\pmod {p(x)}.$$

I know that $(β^5,β^3)$ is a point on the curve. I want to find $13(β^5,β^3)$.

I am using the minimal addition chain for $13 (1,2,4,8,9,13)$ to find this.

But how do I add $(β^5,β^3)$ to itself to find $2(β^5,β^3)$?

I do not understand how to add two points on elliptic curve.

答:
OK, let us do the computations explicitly, this is often a question when starting to use elliptic curves, without having the whole algebraic geometry needed for it in the blood.

I will use sage to illustrate and confirm the computations.

First of all, let us introduce the elliptic curve in sage.

```<code>R.<X> = PolynomialRing(GF(2))
F.<b> = GF(2^4, modulus=X^4+X+1)
E = EllipticCurve(F, [1, b^4, 0, 0, 1])
print(E)
</code>```

The print explains E:

```<code>Elliptic Curve defined by y^2 + x*y = x^3 + (b+1)*x^2 + 1
    over Finite Field in b of size 2^4
</code>```

Now let us consider the given point $P$, and ask for its multiples. It is not fair in an exam, but let us have them first, then we recover the values with bare hands. The sage interpreter gives:

```<code>sage: P = E.point( (b^5, b^3) )
sage: # this was also checking that P is a point on the curve
sage: for k in range(P.order()):
....:     print(f"{k:>2}P = {k*P}")
....:
 0P = (0 : 1 : 0)
 1P = (b^2 + b : b^3 : 1)
 2P = (1 : b^3 + b^2 + 1 : 1)
 3P = (b^2 + b + 1 : b : 1)
 4P = (0 : 1 : 1)
 5P = (b^2 + b + 1 : b^2 + 1 : 1)
 6P = (1 : b^3 + b^2 : 1)
 7P = (b^2 + b : b^3 + b^2 + b : 1)
sage: P.order()
8
</code>```

So $P$ is a point of order $8$, and thus $13P=5P$ with the above value.

Let us compute $2P$ explicitly. For this, consider the "line through $P$ and $P$", i.e. the tangent to the elliptic curve $E$ in $P$ in the ambient affine space parametrized with unknowns $x,y$. Formally:
$$
\begin{aligned}
&\text{Curve:} & 0 &=y^2 + xy + x^3 + b^4x^2 + 1 \ ,\\
&\text{Formal differential:} & 0 &= 0 + y\; dx+ x\; dy + x^2\; dx + 0 + 0 \ ,\\
&\text{Taken in $P=(x_0,y_0)=(b^5,b^3)$:} & 0 &= y_0(x-x_0)+x_0(y-y_0)+x_0^2(x-x_0) \ ,\\
&\text{Separate $(y-y_0)$ and $(x-x_0)$:}
&&  (y_0+x_0^2)(x-x_0) = x_0(y-y_0) \ ,\\
&\text{So the slope in $(x_0,y_0)$ is:}
& m &=\frac{y-y_0}{x-x_0} =\frac{y_0+x_0^2}{x_0}=\frac{y_0}{x_0}+x_0 \ ,\\
&\text{So using $(b^5,b^3)$ explicitly:} &
m&=\frac{b^3}{b^5}+ b^5=\frac 1{b^2}+b^5=(b^3+b^2+1)+b^2+b\\&&&=b^3+b+1 \ .
\end{aligned}
$$
We have used $2=0$ in $F$. (So $3=1$, $-1=1$, et caetera.)

Now we "draw" the line with equation $(y-y_0)=m(x-x_0)$ in the ambient space, and intersect it with the elliptic curve. The theorem of Bezout predicts $3$ solutions for the system
$$
\begin{aligned}
0 &=y^2 + xy + x^3 + b^4x^2 + 1 \ ,\\
y &= b^3 +m(x-b^5)\ ,
\end{aligned}
$$
and we already know "two" of them (with multiplicity), corresponding to the double point $P$ in the intersection. The elimination of $y$ gives (note that in characteristic two squaring is simpler, $(A\pm B)^2=A^2+B^2$,)
$$
0 = b^6 + m^2(x^2+b^{10}) + xb^3 + mx(x+b^5) + x^3 + b^4x^2 + 1 \ .
$$
From this equation we need only the part in $x^3$ and $x^2$,
$$
x^3 + x^2(m^2+m+b^4) +\dots =0\ ,
$$
since Vieta gives the third point $x'$ of intersection, the point which satisfies
$$
x'+x_0+x_0 =-(m^2+m+b^4)\ .
$$
So
$$x'=m^2+m+b^4=1\ .$$
The corresponding $y'$ is
$$
y'=b^3+m(x'-b^5)=b^3 +(b^3+b+1)(1+b^5)=b^3+b^2\ .
$$
To obtain $2P$, we have to compute the opposite of the obtained point $Q=(x',y')=(1,b^3+b^2)$. This point is $Q'=\ominus Q=\ominus(x', y')=(1,b^3+b^2+1)$, the only other point with the same $x=x'$ component. This confirms the sage computation for $2P$.

Similarly, one computes $4P=2(2P)$, $8P=2(4P)$, and so on. To get $4P$ start with the point $(x_0,y_0)=(1, b^3+b^2+1)=(1,b^{13})$, the slope in this point is
$$
m = \frac{y_0}{x_0}+x_0=(b^3+b^2+1)+1=b^3+b^2\ ,
$$
we build again the same system of two equations, Bezout tells there are three solutions, we eliminate $y$ again, the solution $x$ is a solution for
$$
x^3 + x^2(m^2+m+b^4)+\dots = 0\ ,
$$
so this solution is
$$
x'' = m^2+m+b^2=0\ .
$$
The corresponding $y$-value is $y''=y_0+m(x''-x_0)=(b^3+b^2+1)+(b^3+b^2)(0-1)=1$. So $4P=\ominus(0,1)=(0,1)$. Here, $\ominus(0,1)$ is the only "other point" with the same $x$-component, zero, where "other" means also considering the multiplicity. Indeed, setting $x=0$ in the equation of the curve we get $y^2=1$, i.e. $(y-1)^2=0$.

Since $4P$ already lives over $\Bbb F_2$, it is simple to compute $8P$. We start with $(x_0,y_0)=(0,1)$, corresponding to $4P$, while building the tangent in this point we obtain the line $x=0$. Two points of intersection of this line with the curve are known, $4P$ and $4P$, the third point is the infinity point, since the system with the two equations
$$
\begin{aligned}
0 &=y^2 + xy + x^3 + b^4x^2 + 1 \ ,\\
0 &= x\ ,
\end{aligned}
$$
has no further solution in the affine $(x,y)$-plane.

We obtain $9P=8P+P=O+P=P$, and need to compute $13P=9P+4P=P+4P$. I will stop here....

(Once understood, the computation principle is simple. There is no reason to do this effort with bare hands in the present era, one should rather put the accent as a human on the structure.)

Body:
OK, let us do the computations explicitly, this is often a question when starting to use elliptic curves, without having the whole algebraic geometry needed for it in the blood.

I will use sage to illustrate and confirm the computations.

First of all, let us introduce the elliptic curve in sage.

R.<X> = PolynomialRing(GF(2))
F.<b> = GF(2^4, modulus=X^4+X+1)
E = EllipticCurve(F, [1, b^4, 0, 0, 1])
print(E)

The print explains E:

Elliptic Curve defined by y^2 + x*y = x^3 + (b+1)*x^2 + 1
    over Finite Field in b of size 2^4

Now let us consider the given point $P$, and ask for its multiples. It is not fair in an exam, but let us have them first, then we recover the values with bare hands. The sage interpreter gives:

sage: P = E.point( (b^5, b^3) )
sage: # this was also checking that P is a point on the curve
sage: for k in range(P.order()):
....:     print(f"{k:>2}P = {k*P}")
....:
 0P = (0 : 1 : 0)
 1P = (b^2 + b : b^3 : 1)
 2P = (1 : b^3 + b^2 + 1 : 1)
 3P = (b^2 + b + 1 : b : 1)
 4P = (0 : 1 : 1)
 5P = (b^2 + b + 1 : b^2 + 1 : 1)
 6P = (1 : b^3 + b^2 : 1)
 7P = (b^2 + b : b^3 + b^2 + b : 1)
sage: P.order()
8

So $P$ is a point of order $8$, and thus $13P=5P$ with the above value.

Let us compute $2P$ explicitly. For this, consider the "line through $P$ and $P$", i.e. the tangent to the elliptic curve $E$ in $P$ in the ambient affine space parametrized with unknowns $x,y$. Formally:
$$
\begin{aligned}
&\text{Curve:} & 0 &=y^2 + xy + x^3 + b^4x^2 + 1 \ ,\\
&\text{Formal differential:} & 0 &= 0 + y\; dx+ x\; dy + x^2\; dx + 0 + 0 \ ,\\
&\text{Taken in $P=(x_0,y_0)=(b^5,b^3)$:} & 0 &= y_0(x-x_0)+x_0(y-y_0)+x_0^2(x-x_0) \ ,\\
&\text{Separate $(y-y_0)$ and $(x-x_0)$:}
&&  (y_0+x_0^2)(x-x_0) = x_0(y-y_0) \ ,\\
&\text{So the slope in $(x_0,y_0)$ is:}
& m &=\frac{y-y_0}{x-x_0} =\frac{y_0+x_0^2}{x_0}=\frac{y_0}{x_0}+x_0 \ ,\\
&\text{So using $(b^5,b^3)$ explicitly:} &
m&=\frac{b^3}{b^5}+ b^5=\frac 1{b^2}+b^5=(b^3+b^2+1)+b^2+b\\&&&=b^3+b+1 \ .
\end{aligned}
$$
We have used $2=0$ in $F$. (So $3=1$, $-1=1$, et caetera.)

Now we "draw" the line with equation $(y-y_0)=m(x-x_0)$ in the ambient space, and intersect it with the elliptic curve. The theorem of Bezout predicts $3$ solutions for the system
$$
\begin{aligned}
0 &=y^2 + xy + x^3 + b^4x^2 + 1 \ ,\\
y &= b^3 +m(x-b^5)\ ,
\end{aligned}
$$
and we already know "two" of them (with multiplicity), corresponding to the double point $P$ in the intersection. The elimination of $y$ gives (note that in characteristic two squaring is simpler, $(A\pm B)^2=A^2+B^2$,)
$$
0 = b^6 + m^2(x^2+b^{10}) + xb^3 + mx(x+b^5) + x^3 + b^4x^2 + 1 \ .
$$
From this equation we need only the part in $x^3$ and $x^2$,
$$
x^3 + x^2(m^2+m+b^4) +\dots =0\ ,
$$
since Vieta gives the third point $x'$ of intersection, the point which satisfies
$$
x'+x_0+x_0 =-(m^2+m+b^4)\ .
$$
So
$$x'=m^2+m+b^4=1\ .$$
The corresponding $y'$ is
$$
y'=b^3+m(x'-b^5)=b^3 +(b^3+b+1)(1+b^5)=b^3+b^2\ .
$$
To obtain $2P$, we have to compute the opposite of the obtained point $Q=(x',y')=(1,b^3+b^2)$. This point is $Q'=\ominus Q=\ominus(x', y')=(1,b^3+b^2+1)$, the only other point with the same $x=x'$ component. This confirms the sage computation for $2P$.

Similarly, one computes $4P=2(2P)$, $8P=2(4P)$, and so on. To get $4P$ start with the point $(x_0,y_0)=(1, b^3+b^2+1)=(1,b^{13})$, the slope in this point is
$$
m = \frac{y_0}{x_0}+x_0=(b^3+b^2+1)+1=b^3+b^2\ ,
$$
we build again the same system of two equations, Bezout tells there are three solutions, we eliminate $y$ again, the solution $x$ is a solution for
$$
x^3 + x^2(m^2+m+b^4)+\dots = 0\ ,
$$
so this solution is
$$
x'' = m^2+m+b^2=0\ .
$$
The corresponding $y$-value is $y''=y_0+m(x''-x_0)=(b^3+b^2+1)+(b^3+b^2)(0-1)=1$. So $4P=\ominus(0,1)=(0,1)$. Here, $\ominus(0,1)$ is the only "other point" with the same $x$-component, zero, where "other" means also considering the multiplicity. Indeed, setting $x=0$ in the equation of the curve we get $y^2=1$, i.e. $(y-1)^2=0$.

Since $4P$ already lives over $\Bbb F_2$, it is simple to compute $8P$. We start with $(x_0,y_0)=(0,1)$, corresponding to $4P$, while building the tangent in this point we obtain the line $x=0$. Two points of intersection of this line with the curve are known, $4P$ and $4P$, the third point is the infinity point, since the system with the two equations
$$
\begin{aligned}
0 &=y^2 + xy + x^3 + b^4x^2 + 1 \ ,\\
0 &= x\ ,
\end{aligned}
$$
has no further solution in the affine $(x,y)$-plane.

We obtain $9P=8P+P=O+P=P$, and need to compute $13P=9P+4P=P+4P$. I will stop here....

(Once understood, the computation principle is simple. There is no reason to do this effort with bare hands in the present era, one should rather put the accent as a human on the structure.)


问:
For any prime p>5 proving the existence of consecutive quadratic residues and consecutive quadratic non residues
This question was asked in an assignment which I am solving and I couldn't solve it.

Prove that for any prime $p>5$ there exist integers $1\leq a,b \leq  p-1$ for which $\binom{a}{p}=\binom{a+1}{p}=1$ and $\binom{b}{p}=\binom{b+1}{p} =-1$.

Quadratic residues and non-residues are equal in number , if I prove that any 2 residues are consecutive then non-residues will be automatically proved consecutive.

But I am unable to prove and 2 residues to be consecutive. For theory, I am studying David M Burton's book.

Can you please help?

答:
Here is a rather interesting solution involving a bit of combinatorical tricks:

We wil start with some basic theory:

Lemma $(1)$:

For any odd prime $p$, there are exactly $\frac{p-1}{2}$ quadratic residues (greater than $0$) and exactly $\frac{p-1}{2}$ non-quadratic residues.

Lemma $(2)$:

For any integer $n$, $n^2$ is a quadratic residue$\pmod{p}$. in other words, $\big(\frac{n^2}{p}\big)=1$ (I used Legendre's symbol here)

Lets now start the proof!

Consider the residues$\pmod{p}$: $(1,2,...,p-1)$. Assign $1$ to a residue $k$ if it is a quadratic residue, and $0$ if it is not a quadratic residue. We will now get a new sequence of $1$s and $0$s out of the originl sequence.

As an example, for $p=7$ we would get $(1,2,3,4,5,6)\mapsto(1,1,0,1,0,0)$.

Observe that to prove your question, we must show that for any prime $p\geq 5$, in this sequence, there are $2$ consecutive $1$s and $2$ consecutive $0$s (at least).

Because we have a sequence of $p-1$ $1$s and $0$s and there are exactly $\frac{p-1}{2}$ $1$s and $\frac{p-1}{2}$ $0$s $($this is an implication of lemma $(1)$, because there are $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ non-quadratic residues$)$ we can easily see that if there are 2 consecutive $0$s, then there must be $2$ consecutive $1$s too.

Suppose for this prime $p$ there are no $2$ consecutive $1$s or $0$s. Then, using lemma $(2)$, because $1$ is a quadratic residue, our sequence should look like this: $(1,0,1,0,......,0)$, in other words all odd numbers from $1$ to $p-1$ are quadratic residues and all even numbers from $1$ to $p-1$ are non-quadratid residues. But using lemma $(2)$ again, we know $4$ is a quadratic residue, thus leding in a contradiction for $p\geq 5$ $(p$ must be $\geq 5$ so $4\leq p-1$).

So for $p\geq 5$ we will always find $2$ consecutive $1$s and $2$ consecutive $0$s.


问:
Integration of $(z-a)^n$
Let $C$ be the circle of radius $r$ centered around $a\in\mathbb{C}$, we're interested in finding the integral
$$I_n = \int_C \frac{1}{(z-a)^n}\ \mathrm{d}z.$$
We know that $C$ can be parameterized by $C(t) = a+re^{it}$ where $t\in[0,2\pi]$. In the case that $n=1$, I can solve that $I_n=1$ as
$$I_1 = \int_0^{2\pi}\frac{1}{(a+re^{it})-a}(ire^{it}) = 2\pi i.$$
However, for $n\ne1$, we have
$$I_n = \frac{i}{r^{n-1}}\int_0^{2\pi}\frac{1}{(e^{it})^{n-1}}.$$
I know that is supposed to be zero, and I can prove it by using De Moivre's formula by expanding the exponential function. But I want to know if there's another way to proceed without invoking additional theorems? Thanks!

Title: Integration of $(z-a)^n$

Body:
For $n\ne1$,

$$\int_0^{2\pi}\frac{ie^{it}}{(e^{it})^n}dt=i\int_0^{2\pi}e^{i(1-n)t}dt=\left.\frac{e^{i(1-n)t}}{1-n}\right|_0^{2\pi}=0.$$

The case of $n=1$ is trivial.


问:
Equivalent condition for tightness of probability measures on a Banach space
Let $E$ be a $\mathbb R$-Banach space and $\mathcal M_1(E)$ denote the set of probability measures on $\mathcal B(E)$. Remember that $\mathcal F\subseteq\mathcal M_1(E)$ is called tight if $$\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:\sup_{\mu\in\mathcal F}\mu(K^c)<\varepsilon\tag1.$$

How can we show that $\mathcal F$ is tight if and only if for all $\varphi\in E'$, the family $\left\{\mu\circ\varphi^{-1}:\mu\in\mathcal F\right\}$ is tight and for all $\varepsilon>0$, there is a finite-dimensional subspace $L$ of $E$ with $$\inf_{\mu\in\mathcal F}\mu\left(N_\varepsilon(L)\right)>1-\varepsilon\tag2,$$ where $$N_\varepsilon(L):=\bigcup_{x\in L}B_\varepsilon(x)$$ and $B_\varepsilon(x)$ denotes the open ball with radius $\varepsilon$ around $x$ for $x\in E$?

The claim is made in Example 2.3.2 of Bogachev's Weak Convergence of Measures. He claims that the equivalence follows from the following fact:

Lemma: Let $(X,d)$ be a complete metric space. Then $\mathcal H\subseteq\mathcal M_1(X)$ is tight if and only if $$\forall\varepsilon>0:\forall\delta>0:\exists U\in\mathcal U_\delta(E):\sup_{\mu\in\mathcal F}\mu(U^c)<\varepsilon\tag3$$, where $$\mathcal U_\delta(B):=\left\{\bigcup_{i=1}^kB_\delta(x_i):k\in\mathbb N\text{ and }x_1,\ldots,x_k\in B\right\}$$ for $B\subseteq E$.

The lemma is really easy to prove, since the completeness assumption ensures that every totally bounded set is relatively compact.

However, it's absolutely not clear to me how we can prove the desired claim. It's not even clear to me in which direction the lemma should be used.

Clearly, if $L$ is a finite-dimensional subspace of $E$, then $L=\operatorname{span}\left\{x_1,\ldots,x_k\right\}$ for some $x_1,\ldots,x_k\in E$ and $k:=\dim L$. We somehow need to use this ...

答:
Let's prove the reverse implication (the forward one is easier -- tightness follows from the fact that for any $\ell \in E'$ and any compact $K \in E$, $l(K)$ is compact; the other condition follows from the fact that any compact subset of $E$ is in a finite union of balls of radius $\epsilon$).

Let $\epsilon,\delta > 0$. We know that for any $0 < \eta < min(\epsilon,\delta)$, there is a finite-dimensional subspace $L$ with dimension $d$ (depending on $\eta$) such that for any $\mu \in \mathcal{F}$ (this is always the quantification on $\mu$, ie $\mu$ is always any element in $\mathcal{F}$ and nothing depends on $\mu$), $\mu(N_{\eta}(L)) \geq 1-\eta$.

(We specify $\eta$ later without circular dependence).

Let $\phi_1,\ldots,\phi_d \in E'$ be such that they generate $L'$. We know that there is some $K > 0$ such that $\mu(\{|\phi_i| \geq K\}) \leq \frac{\eta}{d}$. Let now $N^L$ be the following seminorm on $E$: $N^L(x) = \|(\phi_i(x))_{1 \leq i \leq d}\|_{\infty}$. $N^L$ generates the topology of $L$ and we know that $\mu(E \backslash B_{N^L}(0,K)) \leq \eta$.

It follows that $\mu(N_{\eta}(L) \cap B_{N^L}(0,K)) \geq 1-2\eta$. In particular, if $2\eta < \delta$, then $N_{\eta}(L) \cap B_{N^L}(0,K)$ can be covered by finitely many balls with center in $L$ with radius $\delta$ (consider a covering of $L \cap B_{N^L}(0,K)$ with balls of radius $\delta/2$, and consider the balls with the same centers but radius $\delta$ instead).

So we have a finite reunion $U$ of balls of radius $\delta$ such that $\mu(U^c) \leq 2\eta$. So if we choose $\eta < \frac{\epsilon}{2}$ and $\eta < \frac{\delta}{2}$, we are done, and the criterion from the lemma applies.


问:
Intuitive understanding of the derivatives of $\sin x$ and $\cos x$
One of the first things ever taught in a differential calculus class:

The derivative of $\sin x$ is $\cos x$.

The derivative of $\cos x$ is $-\sin x$.

This leads to a rather neat (and convenient?) chain of derivatives:

```
sin(x)
cos(x)
-sin(x)
-cos(x)
sin(x)
...
```

An analysis of the shape of their graphs confirms some points; for example, when $\sin x$ is at a maximum, $\cos x$ is zero and moving downwards; when $\cos x$  is at a maximum, $\sin x$ is zero and moving upwards.  But these "matching points" only work for multiples of $\pi/4$.

Let us move back towards the original definition(s) of sine and cosine:

At the most basic level, $\sin x$ is defined as -- for a right triangle with internal angle $x$ -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.

To generalize this to the domain of all real numbers, $\sin x$ was then defined as the Y-coordinate of a point on the unit circle that is an angle $x$ from the positive X-axis.

The definition of $\cos x$ was then made the same way, but with adj/hyp and the X-coordinate, as we all know.

Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!"

That is, from the unit circle definition alone.  Or, even more amazingly, the right triangle definition alone.  Ignoring graphical analysis of their plot.

In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?"

答:
This isn't exactly what you asked, but look at the Taylor series for the polynomials:

$$ \sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\text{ for all } x\!$$

$$\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\text{ for all } x\! $$

The relationships between the derivatives are clear from this.


问:
Given $f(x) =x^2+bx+c$, $f(1) = 2$, and $f(-1) = 12$, how do I get $b$ and $c$?
Is there a different way to get $b$ and $c$ values and then the value of $f(2)$?

Here's how my book does it:

Given that
$$f(x) =x^2+bx+c, \qquad f(1) = 2,\qquad f(-1) = 12$$
we see
$$
f(1) = 2 \implies 1^2+b+c=2\implies b+c = 1\\
f(-1) = 12 \implies (-1)^2+b(-1)+c = 12 \implies -b+c = 11
$$
and therefore
$$
b+c = 1, \qquad -b+c = 11.
$$
Thus
$$
2c = 12 \implies c = 6 \implies b = -5
$$

and then

$$
f(2) = 2^2-5\cdot2+6 = 0
$$
$$
f(2) = 0.
$$

I'm sorry if this topic is "too easy". I'm just beginning this math.

(sorry for my bad english, not my native language)

答:
There is nothing wrong with this. Just plug in the conditions given... If you're given $f(5)=3$, this means that if you put $5$ in place of $x$, your expression should end up giving you $3$.

You'll need two of these conditions to get $2$ equations having $b$ and $c$ as variables. You solve them simultaneously to get the values of $b$ and $c$.

And then once you have a completely "known" expression for $f(x)$, you can find $f(a)$ for any $x=a$.


问:
Finding the linear span of complex numbers over the real number field.
I'm trying to find the linear span of the following group:

$L=\{(z_1,z_2,z_3 )∈C^3 |  z_2=\bar z_1   \}$

Over the field:
$F=\mathbb R$

I'm a bit confused as to what the linear span of a group of complex numbers could be over the real number field.

If anyone could explain how to approach this problem, I'd appreciate it.

答: The set $L$ is already a real vector space, since the sum of any two element of $L$ is again an element of $L$ and the product of an element of $L$ by a real number is again an element of $L$. Therefore, $\operatorname{span}(L)=L$.


问:
How do I estimate $\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt$
Question: How do I estimate the following integral? $$\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt$$

Attempt: Setting $u=\log t$, we get

$$\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt = \int_{\log 2}^{\infty} \frac{|\log u|}{u^2}\, du$$ and then I am stuck.

答:
Integration by substitution $t=e^u;\;dt=e^u du;\;\log t = u$
$$\int \frac{\log (\log t)}{t \log ^2 t} \, dt=\int\frac{ \log u}{u^2}\,du=-\frac{\log u+1}{u}+C$$

$$\int \frac{\log (\log t)}{t \log ^2 t} \, dt=-\frac{\log (\log t)+1}{\log t}+C$$
$$\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt=\int_{2}^{e} \left(-\frac{\log \log t}{t\, (\log t)^2}\right) \, dt+\int_{2}^{\infty} \frac{\log \log t}{t\, (\log t)^2} \, dt=$$
$$=\left[\frac{\log (\log t)+1}{\log t}\right]_2^e+\left[-\frac{\log (\log t)+1}{\log t}\right]_e^{\infty}=$$
$$=1-\frac{1+\log (\log 2)}{\log 2}-\underset{m\to \infty }{\text{lim}}\frac{\log (\log m)+1}{\log m}+1=$$
$$=2-\frac{1+\log (\log 2)}{\log 2}\approx 1.086$$


问:
How to find all functions $f:\mathbb R\to\mathbb R$ such that $\forall a,b\in\mathbb R$: $f(a)+f\big(a+f(b)\big)=b+f\big(f(a)+f^2(b)\big)$

Find all functions $ f : \mathbb R \to \mathbb R $ such that for all $ a , b \in \mathbb R$:
$$ f ( a ) + f \big( a + f ( b ) \big) = b + f \big( f ( a ) + f ^ 2 ( b ) \big) \text . $$

Here, for any $ n \in \mathbb N $, $ f ^ n $ denotes the $ n $-th iteration of $ f $.

My ideas so far:

I substituted $ ( 0 , x ) $ that yields:
$$ f ( 0 ) + f ^ 2 ( x ) = x + f \big( f ( 0 ) + f ^ 2 ( x ) \big) \text . \tag 1 \label 1 $$

Let's say that $ a , b \in \mathbb R $, $ a \ne b $ and $ f ( a ) = f ( b ) $.  Then the LHS does not change value with $ a , b $ but the RHS does. That is a contradiction and thus $ f ( a ) = f ( b ) \implies a = b $. The function is therefore injective.

If we substitute $ \big( x , f ( x ) \big) $ we can cross the functions like so:
$$ f \big( x + f ^ 2 ( x ) \big) = f \big( f ( x ) + f ^ 3 ( x ) \big) \text ; $$
$$ \therefore \quad x + f ^ 2 ( x ) = f ( x ) + f ^ 3 ( x ) \text . \tag 2 \label 2 $$

If we assume $ f ( 0 ) = 0 $, we have $ f ^ 2 ( x ) = x + f ^ 3 ( x ) $.

Using \eqref{2} we get $ f ( x ) = 2 x $. However, this does not satisfy the functional equation and thus we can conclude that $ f ( 0 ) \ne 0 $.

I also noticed that if you substitute $ f ( x ) $ for $ x $, you get $ f ( x ) + f ^ 3 ( x ) =f ^ 2 ( x ) + f ^ 4 ( x ) $ and expressing $ f ^ 3 ( x ) $ from \eqref{2} we get an intresting result: $ f ^ 4 ( x ) = x $ which means the function is iterative with a cycle of $ 4 $ (or $ 2 $ or $ 1 $).

I am not sure how to continue or what substitution should I try next.

答:
You can show that there is no function $ f : \mathbb A \to \mathbb A $ satisfying
$$ f ( x ) + f \big( x + f ( y ) \big) = y + f \Big( f ( x ) + f \big( f ( y ) \big) \Big) \tag 0 \label 0 $$
for all $ x , y \in \mathbb A $, where $ ( \mathbb A , + ) $ is any abelian group with the neutral element $ 0 $ and the inverse function $ - $, such that there is $ b \in \mathbb A $ with $ 5 b \ne 0 $. As you are interested in $ \mathbb A = \mathbb R $ with the group operation $ + $ taken to be the usual addition of real numbers, this would be the case, since any nonzero real number can be chosen as $ b $.

To see this, substitute $ f ( x ) $ for $ x $, and see that
$$ f \big( f ( x ) \big) - y = f \Big( f \big( f ( x ) \big) + f \big( f ( y ) \big) \Big) - f \big( f ( x ) + f ( y ) \big) \\
= f \Big( f \big( f ( y ) \big) + f \big( f ( x ) \big) \Big) - f \big( f ( y ) + f ( x ) \big) = f \big( f ( y ) \big) - x \text , $$
which in particular shows
$$ f \big( f ( x ) \big) = f \big( f ( 0 ) \big) - x \text . \tag 1 \label{1a} $$
Putting $ x = 0 $ in \eqref{0} and using \eqref{1a} you have
$$ f ( 0 ) + f \big( f ( 0 ) \big) - y = y + f \Big( f ( 0 ) + f \big( f ( 0 ) \big) - y \Big) \text , $$
which by letting $ a = f ( 0 ) + f \big( f ( 0 ) \big) $ and substituting $ - x + a $ for $ y $ shows that
$$ f ( x ) = 2 x - a \text . \tag 2 \label{2a} $$
Using \eqref{1a} and \eqref{2a} you get $ 5 x = 0 $ for all $ x \in \mathbb A $, and in particular for $ x = b $, which is a contradiction.

In case every $ b \in \mathbb A $ is of order $ 5 $, choosing any $ a \in \mathbb A $ and taking $ f $ to be of the form \eqref{2a}, the equation \eqref{0} will be satisfied for all $ x , y \in \mathbb A $. To see this, use \eqref{2a} to rewrite \eqref{0} as
$$ 2 x - a + 2 ( x + 2 y - a ) - a = y + 2 \big( 2 x - a + 2 ( 2 y - a ) - a \big) - a \text , $$
or equivalently
$$ 4 x + 4 y - 4 a = 4 x + 9 y - 9 a \text , $$
which is true since $ 5 y = 0 $ and $ 5 a = 0 $. As the derivations resulting in \eqref{2a} were valid for any abelian group (independent of the order of its elemnts), we've characterized all the solutions it this case. Examples of abelian groups in which the order of every element is $ 5 $ are trivial groups, the cyclic group $ \mathbb Z _ 5 $ and the direct product of $ \mathbb Z _ 5 $ with itself.


问:
residue class rings - value calculations with modulo
computer science studies - residue class rings.

I have to calculate the following expression in $ \mathbb{Z_9}$

and the other one in $ \mathbb{Z_{16}}$

my question is how one should handle $-1, 23, 10$ in the "powers"?

are they really powers or do they have another meaning perhaps?

Thank you for your help

答:
In a ring, $x^n$ means the power. So you have, for instance, $x=[5]_9$ as an element of $\mathbb{Z}/9\mathbb{Z}$ and you raise it to the power $10$.

What can you do here? You know that $\gcd(5,9)=1$ and therefore Euler-Fermat applies, yielding
$$
5^{\varphi(9)}=5^6\equiv1\pmod{9}
$$
Therefore
$$
[5]_9^{10}=[5_9]^6[5]_9^4=[5]_9^4=([5]_9^2)^2)=[7]_9^2=[4]_9
$$
What about $[7]_{16}^{-1}$? Hint: $7\cdot7=49$.


问:
How to show that if $p(A) = 0 \implies p(\lambda_0)=0$?
Let $V$ be a finite dimensional vector space and $V \ne \{ 0 \}, A\in L(V), \lambda_0 \in \sigma(A)$. If $p(\lambda)$ is an arbitrary polynomial for which the following applies: $p(A) = 0  $,  prove that $p(\lambda_0)=0$

$L(V)$ = The set of all linear mappings (linear operators) from V to V

$\lambda_0$ = eigenvalue

$\sigma(A)$ = spectrum of A

If I am correct this would involve the Hamilton-Cayley theorem but I don't know how to apply it. I know that $\lambda_0$ is a zero of the polynomial $det(A-\lambda I)$ but I don't even know where to start.

Edit (hint from mrf):

Since $V$ is finite, we can mark the degree of the polynomial with n $\implies p(\lambda) = \alpha_n\lambda^n + \alpha_{n-1}\lambda^{n-1} + ...  +  \alpha_2\lambda^2  + \alpha_1\lambda + \alpha_0$

So $p(A) = \alpha_nA^n + \alpha_{n-1}A^{n-1} + ... + \alpha_2A^2 + \alpha_1A + \alpha_0I$

Let $v$ be an corresponding eigenvector to the eigenvalue $\lambda_0 \implies Av = \lambda_0v$

$p(A)v = (\alpha_nA^n + \alpha_{n-1}A^{n-1} + ... + \alpha_2A^2 + \alpha_1A + \alpha_0I)v $

$p(A)v = \alpha_nA^nv + \alpha_{n-1}A^{n-1}v + ... + \alpha_2A^2v + \alpha_1Av + \alpha_0Iv$

If $\lambda \in \sigma(A) \implies \lambda^k \in \sigma(A^k)$

Proof:

$\lambda \in \sigma(A) \implies \exists v \in V$ so that $Av = \lambda v$

$Av = \lambda v \hspace{10px}| A$

$A^2v = A(Av) = A(\lambda v) =$ A is linear $ =\lambda(Av) = \lambda^2v$

$A^2v = \lambda^2v \hspace{10px}| A$

$A^3v = A(A^2v) = A(\lambda^2 v) =$ A is linear $ =\lambda^2(Av) = \lambda^3v$

And by induction we get: $A^kv = \lambda^k v$

$\implies$

$p(A)v = \alpha_n\lambda_0^nv + \alpha_{n-1}\lambda_0^{n-1}v + ... + \alpha_2\lambda_0^2v + \alpha_1\lambda_0v + \alpha_0v \hspace{10px}| :v$

$p(A) = \alpha_n\lambda_0^n + \alpha_{n-1}\lambda_0^{n-1} + ... + \alpha_2\lambda_0^2 + \alpha_1\lambda_0 + \alpha_0$

$0 = \alpha_n\lambda_0^n + \alpha_{n-1}\lambda_0^{n-1} + ... + \alpha_2\lambda_0^2 + \alpha_1\lambda_0 + \alpha_0$

答:
This is true in a much more general context.

If you have some (unital) algebra $\mathbf A$ over arbitrary field $\bf K$ (it could be a Banach algebra or finite dimensional, but not necessarily) and some $A\in {\bf A}$ and a polynomial $p\in {\bf K}[X]$ such that $p(A)=0$, then for any $\lambda$ if $p(\lambda)\neq 0$, then $(A-\lambda\cdot{\bf 1})$ is invertible in ${\bf A}$.

To see this, notice that you have $-p(\lambda\cdot {\bf 1})=p(A)-p(\lambda \cdot {\bf 1})=(A-\lambda\cdot {\bf 1})g(A)$ for some polynomial $g\in {\bf K}[X]$. In particular ${\bf 1}=(A-\lambda\cdot {\bf 1})(-p(\lambda)^{-1}g(A))$.

In fact, we could replace ${\bf K}$ with some arbitrary (commutative unital) ring if we require that $p(\lambda)$ be invertible and not just nonzero.


问:
tolerance dimensions arithmetic or how to add or subtract values with probability distributions?
This is a follow up to this question. Imagin two physical objects A and B:

```
   <img src="https://i.stack.imgur.com/YMgvV.png" width="500">
```

with nominal heights of $l_A$ and $l_B$. These parts are being mass-produced so their heights have a probability distribution (e.g., Natural/Gaussian distributions), with standard deviations of $\sigma_A$ and $\sigma_B$. Now I want to know what are the expected probability distributions if the blow measurements:

```
   <img src="https://i.stack.imgur.com/Q7lF8.png" width="500">
```

In other words, what are the standard deviations of the values $l_1$ and $l_2$ from their nominal values $l_A + l_B$ and $l_A - l_B$? Is it correct to assume $\sigma_1 = \sigma_2 = \sigma_A + \sigma_B$?

P.S. Dear moderators, I have no clue if I have used the right terms or if I have chosen the right tags here. Your help in that regard would be highly appreciated.

答:
Just be sure to distinguish between tolerances (absolute max and min values) - which would simply add and subtract - and the measured and calculated standard deviation for a set of data - which would mean 99.7% of data falls with 3x the std deviation (give a normal distribution).

I.e. 5  ± 1 means min 4 and max.6.
5 with σ = 1 implies the mean value is 5 and most of the data would sit between 2 and 8 given a normal distribution.


问:
An argument for getting $L^2$ convergence from $L^1$ convergence and boundedness in $L^2$
This is a theorem proof from Protter's Stochastic Integration and Differential Equations on the completeness of the space $\mathcal{H}^2$ of semimartingales.

Setting: Let $A$ be a (predictable) integrable variation (IV) process with $A_0 = 0$, and with the $L^2$ norm on a probability space we have $$\Vert \int_0^\infty |dA_s| \vert_{L^2} < \infty.$$

We would like to show that this norm space is complete.

Suppose $(A^n)$ is a Cauchy sequence of (predictable) IV processes in $\Vert \cdot \Vert_2$ where $\Vert A \Vert_p = \Vert \int_0^\infty |dA_s| \Vert_{L^p}, p \ge 1$. To show $(A^n)$ converges it suffices to show a subsequence converges. Therefore, without loss of generality we can assume $\sum_n \Vert A^n \Vert_2 < \infty$.

Then $\sum A^n$ converges in $\Vert \cdot \Vert_1$ to a limit $A$. Moreover $$\lim_{m \to \infty} \sum_{n \ge m} \int_0^\infty |dA_s^n| = 0$$ in $L^1$ and is dominated in $L^2$ by $\sum_n \int_0^\infty |dA_s^n|.$ Therefore $\sum A^n$ converges to the limit $A$ in $\Vert \cdot \Vert_2$ as well, and there is a subsequence converging almost surely.

I cannot follow the arguments in bold above. Firstly, how do we get that $\sum A^n$ converges in $\Vert \cdot \Vert_1$ given $\sum_n \Vert A^n \Vert_2 < \infty$? And why do we get that $\sum_{n \ge m} \int_0^\infty |dA_s^n| \to 0$ as $m\to \infty$ in $L^1$?

Finally, how do we get convergence in $\Vert \cdot \Vert_2$ from these results?

答:
There's a lot wrong with this. First of all, that $A^n$ is Cauchy does not imply that we can have $\sum_{n}\|A^n\|_{2} < \infty$ after passing to a subsequence. Consider a constant sequence $A^n = A$ as a counterexample. Secondly, we will not be able to conclude the convergence of $\sum_{n\in\mathbb{N}} A^n$ in any norm unless we have $A^n \to 0$.

What we can do is find a subsequence such that $\sum_{n\in\mathbb{N}} \|A^{n+1}-A^{n}\|_{2} < \infty$. We may then write
$$ A^{n} = A^{1} + \sum_{i=1}^{n-1} (A^{i+1} - A^{i}).$$
Note that for any $t > 0$ we have
$$ |A_t^{i+1}-A_t^{i}|=\left|\int_{0}^{t} dA^{i+1}_s - \int_{0}^{t} dA^{i}_s\right| \leq \int_{0}^{t} |d(A^{i+1}-A^{i})_s|.$$
It follows that
$$ \left\|\sum_{i=1}^{\infty} \sup_{t\geq 0} |A^{i+1}-A^{i}| \right\|_{L^2} \leq \left\|\sum_{i=1}^{\infty} \int_{0}^{\infty} |d(A^{i+1}-A^{i})_s| \right\|_{L^2} \leq \sum_{i=1}^{\infty}\|A^{i+1} - A^{i}\|_2.$$
This shows that $\sum_{i=1}^{n-1} (A^{i+1}-A^{i})$ converges almost surely uniformly as $n\to\infty$ to some predictable $\bar{A}$ with $\|\bar{A}\|_{2} < \infty$. Letting $A := \bar{A}-A^1$, we have that $A^n \to A$ almost surely uniformly on $[0,\infty)$. By Fatou's lemma we obtain
$$ \|A^{n} - A\|_2 \leq \liminf_{m\to\infty} \|A^{n} - A^{m}\|_{2} \leq \liminf_{m\to\infty} \|\sum_{i=n}^{m} (A^{i+1} - A^{i})\|_2 \leq \sum_{i=n}^{\infty} \|A^{i+1}-A^{i}\|_2,$$
which vanishes as $n\to \infty$.


问:
Prove $\lim_{x\to 0}$ of piecewise $f(x)=1$ if $x=1/n$ where $n\in \mathbb{Z}$ or $0$ otherwise on $[-1,1]$ doesn't exist.
$$f(x)=\begin{cases} 1 & \text{for } x=\frac{1}{n}, \; n\in\mathbb{Z}\\0 & \text{otherwise.}\end{cases}$$

I want to prove with the Epsilon-Delta definition  that the limit as $x \to 0$  doesn't exist so I can say there is an essential discontinuity at $x=0$.

My reasoning is:

$\lim_{x\to 0}  f(x) \not= L$

Given any $\epsilon > 0$ , we can take  $\delta >0$ for which:

There exists $0 < |x| < \delta$ $\to$  $|f(x)-L|$ $≥$ $\epsilon$

So I'm trying to say that there are values of $x$ within $(-\delta, \delta)$ for which  $|f(x)-L|$ $≥$ $\epsilon$. I think what I wrote might be saying that, for any $\delta$ i choose, all values within $(-\delta, \delta)$ will result in  $|f(x)-L|$ $≥$ $\epsilon$, which I think isn't true no matter how infinitely small our chosen $\delta$ is.

How do I properly prove this?

答:
To prove that the limit doesn't exist, you shouldn't start with Given any $\epsilon \gt 0$... but by finding an epsilon for which the definition of the limit won't be satisfied.

So suppose that the limit exists and its value is $L$. Now take $\epsilon = 1/3$. We have to prove that whatever $\delta \gt 0$ we pick, it exists $0 \lt \vert x \vert \lt \delta$ such that $\vert f(x) - L \vert \ge \epsilon = 1/3$.

Now it is quite simple and you can separate two cases.

Case 1: $L \lt 1/2$

Then you can find $n \in \mathbb N$ such that $0 \lt 1/n \lt \delta$ and you have
$$\vert f(1/n) -L \vert = \vert 1 -L \vert \gt 1/2 \gt 1/3.$$

Case 2: $L \ge 1/2$

In that case pick $n \in \mathbb N$ such that $0 \lt \sqrt 2/n \lt \delta$. Now
$$\vert f(\sqrt2/n) -L \vert = \vert 0 -L \vert \ge 1/2 \gt 1/3.$$

Note: I picked up $\epsilon = 1/3$ because in all neighborhood of $0$ $f$ takes values at distance equal to $1$. And so, whatever $L$ would be, some function values would be at more than $1/3$ distance from the potential limit.


问:
Evaluation of $\lim_{x\to 0^+}\frac{(1+x)^x - 1}{x^2}$
I'm interested in evaluating mentioned limit without using l'Hopital's rule, if possible.
What I'm thinking of is doing the following:

Let $$f(x) = \frac{(1+x)^x - 1}{x^2} = \frac{\left((1+x)^\frac{1}{x}\right)^{x^2} - 1}{x^2}$$

Now if we define:
$$g(x) = (1+x)^\frac{1}{x} \Rightarrow \lim_{x\to 0^+}g(x) = e > 0$$
and $$h(x) = x^2 \Rightarrow \lim_{x\to 0^+}h(x) = 0$$
then $f$ becomes a compound function $f(x) = \frac{g(x)^{h(x)} - 1}{h(x)}$

Now I'd like to argue that since $g(x) \xrightarrow{x\to 0} e > 0$, then $$\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}\frac{g(x)^{h(x)} - 1}{h(x)} \stackrel{?}{=} \lim_{x\to 0^+}\frac{\Big(\displaystyle\lim_{x\to 0^+}g(x)\Big)^{h(x)} - 1}{h(x)} = \lim_{x\to 0^+}\frac{e^{h(x)} - 1}{h(x)} = 1$$

However, the step with the question mark is troublesome for me and I struggled to prove it. Which got me thinking, is such statement even true or possible in this case? Or were such cases namely the very reason why L'Hopital's rule was discovered and is applicable?

Any clarification is greatly appreciated! Thanks a lot in advance!

答:
In my opinion, whenever it is possible at the price of a low effort, it is always good to go beyonf the limit itself and, for that, composition of Taylor series is easy.

Considering
$$y=\frac{(1+x)^x - 1}{x^2}$$
$$a=(1+x)^x\implies \log(a)=x \log(1+x)=x \left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)$$
$$\log(a)=x^2-\frac{x^3}{2}+\frac{x^4}{3}+O\left(x^5\right)$$
$$a=e^{\log(a)}=1+x^2-\frac{x^3}{2}+\frac{5 x^4}{6}+O\left(x^5\right)$$
$$y=\frac{a-1}{x^2}=1-\frac{x}{2}+\frac{5 x^2}{6}+O\left(x^3\right)$$


问:
When can we inscribed sphere in tetrahedron?
I'm wondering is there any conditions to do that. I believe propably there are but what ?</p>

答:
Given any tetrahedron with vertices $v_1,\ldots,v_4$. For $k = 1,\ldots 4$, let

$h_k$ be the distance of $v_k$ to its opposite face.

$\Delta_k$ be the area of the face opposite to $v_k$.

$V$ and $\Delta = \sum_{k=1}^4 \Delta_k$ be the volume and surface area of the tetrahedron.

Consider following area weighted average of the vertices.
$$v = \frac{1}{\Delta}\sum_{k=1}^4 \Delta_k v_k$$
It is easy to see its distance to the face  opposite to vertex $v_k$ equals to
$$\frac1\Delta \Delta_k h_k = \frac{3V}{\Delta}$$
Since this is independent of $k$, this point $v$ is equal distance to all faces of the tetrahedron.
You can inscribe a sphere of radius $\frac{3V}{\Delta}$, centered at $v$ to the tetrahedron.


问:
Prime decomposition in algebraic integer rings
Let $\mathcal{O}_K \subset \mathcal{O}_L$ be algebraic integer rings and $\theta \in \mathcal{O}_L$ such that $L=K( \theta )$. Let $\mathfrak{F}_\theta$ be the unique maximal ideal of $\mathcal{O}_L$ contained in $\mathcal{O}_K[\theta]$. We know that if a prime $\mathfrak{p}$ of $\mathcal{O}_K$ is prime to $\mathfrak{F}_\theta$, then we can get the prime decomposition of $\mathfrak{p}$ in $\mathcal{O}_L$ using the minimal polynomial of $\theta$ over $K$.

My question is that is there any conclusion saying that we can get the prime decomposition of all primes of $\mathcal{O}_K$ using this method by changing the $\theta$.

答:
First let me clarify what the $\mathfrak{F}_\theta$ is. $\mathfrak{F}_\theta=\{ \alpha \in \mathcal{O}_K[\theta] \mid \alpha \mathcal{O}_L \subset \mathcal{O}_K[\theta] \}$ is the largest ideal of $\mathcal{O}_L$ contained in $\mathcal{O}_K[\theta]$.

Then, if $K=\mathbb{Q}$ and $\mathcal{O}_K=\mathbb{Z}$, the condition $p$ prime to $\mathfrak{F}_\theta$ is equivalent to $p \nmid |\mathcal{O}_L/\mathbb{Z}[\theta]|$. For the necessity, if $p$ is prime to $\mathfrak{F}_\theta$, then $p \nmid |\mathcal{O}_L/\mathfrak{F}_\theta |$. Hence $p \nmid |\mathcal{O}_L/\mathbb{Z}[\theta]|$ since $|\mathcal{O}_L/\mathbb{Z}[\theta]| \mid |\mathcal{O}_L/\mathfrak{F}_\theta |$. For the sufficiency, if  $p \nmid |\mathcal{O}_L/\mathbb{Z}[\theta]|$, then the map from $\mathcal{O}_L/\mathbb{Z}[\theta]$ to itself which sends every element to its multiple by $p$ is an isomorphism of the additive group. Then $p \mathcal{O}_K + \mathbb{Z}[\theta]=\mathcal{O}_K$. Therefore, $p \mathcal{O}_K + \mathfrak{F}_\theta=\mathcal{O}_K$, that is, $p$ prime to $\mathfrak{F}_\theta$.

Next is our counterexample. Let $L=\mathbb{Q}(\theta)$ where $\theta^3+\theta^2-2\theta+8=0$. As in the handout mentioned above, $2 \mid |\mathcal{O}_L/\mathbb{Z}[\alpha] |$ for every integral primitive element $\alpha$. Hence $2$ cannot be decomposed using this method in $\mathcal{O}_L$.
This shows that the general answer is NO. As mentioned in the handout, the prime decomposition lies somewhat deeper than factoring in $\mathbb{F}_q[X]$.


问:
The index of $\xi_4^*$ in $\xi_4$
Just seeing if i'm right:

With the set of solutions for $z^4=1$: $\xi_4=\{1,i,-1,-i\}$, one can construct the group of the $4$th roots of unity: $(\xi_4,\cdot_\mathbb{C})$ and its multiplicative subgroup $(\xi_4^*,\cdot_\mathbb{C})$ with $\xi_4^*=\{1,-1\}$.

What is the Index $[ \xi_4 : \xi_4^* ]$?

I'd say through Lagrange:
$$ [ \xi_4 : \xi_4^* ] =\frac{|\xi_4|}{|\xi_4^*|}=\frac{4}{2}=2.$$
Or counting the cosets of $\xi_4^*$:
$$1\cdot\xi_4^*=\{1,-1\}= -1\cdot\xi_4^*,\quad i\cdot\xi_4^*=\{i,-i\}= -i\cdot\xi_4^*$$
we get two.

Is this correct?

答:
Yes, indeed, the index $$[\xi_4:\xi^*_4] = \frac{|\xi_4|}{|\xi^*_4|} = \dfrac 4 2 = 2$$

Alternatively, as you've shown, the index $=$ the number of cosets of $\;\xi^*_4 \leq \xi_4 = 2$.

REMARK:

When you're dealing with a finite group, as in this case, the first "method" of computing the index of a subgroup with respect to its containing group is probably quickest. (Indeed, the first method works strictly for finite groups!)

Title: The index of $\xi_4^*$ in $\xi_4$


问:
Structure of Column Sums of Real Orthonormal Matrices
Suppose I have a square real orthonormal matrix $A \in O(D)$. I'd like to understand what structure exists in the set of column sums of $A$.

For instance, $O(2)$ can be parameterized by a single scalar. To see why, consider $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$. Since the first column must have unit norm, $c = \sqrt{1 - a^2}$. Since the second column must be orthogonal to the first column and must also have unit norm, $b = -c$ and $d = a$. Consequently, $A = \begin{bmatrix} a & -\sqrt{1 - a^2}\\ \sqrt{1 - a^2} & a \end{bmatrix}$ and the column sums are $a + \sqrt{1 - a^2}$ and $a - \sqrt{1 - a^2}$. When I plot the column sums as a function of $a$, I observe these nice curves:

My question is: how does this structure generalize to $O(D)$? Is some quantity conserved? If I order the column sums in decreasing order, does some relationship between them hold?

Maybe what I'd like is some theorem that states "if the previous columns' sums were $A, B, C,...$ then the next column's sum is is equal to $Z$ / bounded between $[-X, Y]$"

答:
Knowing that the set of all possible column-sum-vectors is a sphere essentially answers all possible questions you could want to ask about such vectors. Specifically, we have:

Let $S(n)$ be the set of column-sum-vectors of orthogonal matrices in $O(n)$. Then $S(n)$ is equal to the sphere of radius $\sqrt n$ centered at the origin.

From the comments:

can I say anything beyond that? Since the vectors are orthonormal, it suggests that fixing one (or several) severely limits what remaining points on the sphere can be chosen.

Bringing in the hypothesis that the vectors are orthonormal cannot possibly get you any stronger results, since that hypothesis is embedded in the theorem that the set of all column-sum-vectors is a sphere. So yes, fixing one or several coordinates restricts the others - but it restricts them only and precisely in that they must be chosen so that the resulting point ends up on a sphere. There is no point trying to get any further restrictions, as the result is that $S(n)$ is equal to a sphere - not a subset of it, and not a superset of it, but equal. Therefore the restriction is as tight as it gets.

For example:

You can parameterize $S(n)$, using any standard parameterization of a sphere.

Yes, if you fix the first $k$ coordinates, this restricts the remaining coordinates since the whole vector must end up on a sphere. Specifically, the remaining coordinates $a_{k+1}, ..., a_n$ must be chosen so that
$$a_{k+1}^2+...+a_n^2=n-(a_1^2+...+a_k^2)$$
In other words, if $r^2=a_1^2+...+a_k^2$, the remaining coordinates must be chosen from a sphere of radius $\sqrt{n-r^2}$ in $(n-k)$-dimensional space.


问:
Can a limit of an integral be moved inside the integral?
After coming across this question: How to verify this limit, I have the following question:

When taking the limit of an integral, is it valid to move the limit inside the integral, providing the limit does not affect the limits of integration?

For instance in the question, the OP is trying to determine that:

$$\lim_{n\to\infty}\int_{0}^{1}{\frac{dx}{(1+\frac{x}{n})^{n}}}=1-\rm{e}^{-1}$$

The answers to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is:

$$\lim_{n\to\infty}{\int_{0}^{1}{\frac{dx}{(1+\frac{x}{n})^{n}}}}=\int_{0}^{1}{\lim_{n\to\infty}\frac{dx}{(1+\frac{x}{n})^{n}}}=\int_{0}^{1}{\frac{dx}{{\rm{e}}^x}}=\rm{e}^{0}-\rm{e}^{-1}=1-\rm{e}^{-1}$$

As required. So is this a valid technique, or is it just coincidental that this works?

Check :

$\displaystyle\begin{align}\lim\limits_{n\to\infty}\int_0^1&\dfrac{dx}{\left(1+\dfrac xn\right)^n}=\lim\limits_{n\to\infty}\left[\dfrac{-n}{n-1}\left(1+\dfrac xn\right)^{1-n}\,\right]_0^1=\\
&=\lim\limits_{n\to\infty}\dfrac{-n}{n-1}\left[\left(1+\dfrac1n\right)^{1-n}-1\right]=\\
&=\lim\limits_{n\to\infty}\dfrac n{n-1}\left[1-\left(1+\dfrac1n\right)^{1-n}\,\right]=\\
&=\lim\limits_{n\to\infty}\dfrac n{n-1}\left[1-\dfrac{\left(1+\dfrac1n\right)}{\left(1+\dfrac1n\right)^n}\,\right]=\\
&=1\cdot\left[1-\dfrac1{\rm{e}}\right]=\\
&=1-\rm{e}^{-1}
\end{align}$

$\displaystyle\begin{align}\int_0^1 \lim\limits_{n\to\infty}\dfrac{dx}{\left(1+\dfrac xn\right)^n}&=\int_0^1\dfrac{dx}{{\rm{e}}^x}=\bigg[\!-{\rm{e}}^{-x}\bigg]_0^1=\\
&=\rm{e}^0-\rm{e}^{-1}=1-\rm{e}^{-1}
\end{align}$

答:
There's a recent paper that exactly answers this question.

Kamihigashi, T. Interchanging a limit and an integral: necessary and sufficient conditions. J Inequal Appl 2020, 243 (2020).

https://doi.org/10.1186/s13660-020-02502-w

The main result gives a necessary and sufficient condition under which the limit can be moved inside the integral. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability.


问:
Do two matrices commute only if one matrix is the inverse of the other?
Is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? The only way I could believe this was the 2 commutative type properties I know of:
$$AA^{-1}=A^{-1}A=I$$
and
$$adj A*A=A*adjA = |A|I$$
where $adj A$ is the adjoint of matrix A.

The first equation when expanded with $A^{-1}=\frac{adjA}{|A|}$ you end up getting the second equation. So both are the same equation. These are the only two properties that explains commutativity as far as I'm aware of(are there any more?) and my conclusion was based on these only. Also multiplication with identity matrix is also commutative but any other counter cases? So is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? Thanks!!

答: Not at all. For instance, the identity matrix commutes with every other square matrix of the same size. As another example, the matrices $\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right]$ commute, but none of them is a multiple of the other one.


问:
Integration of $(z-a)^n$
Let $C$ be the circle of radius $r$ centered around $a\in\mathbb{C}$, we're interested in finding the integral
$$I_n = \int_C \frac{1}{(z-a)^n}\ \mathrm{d}z.$$
We know that $C$ can be parameterized by $C(t) = a+re^{it}$ where $t\in[0,2\pi]$. In the case that $n=1$, I can solve that $I_n=1$ as
$$I_1 = \int_0^{2\pi}\frac{1}{(a+re^{it})-a}(ire^{it}) = 2\pi i.$$
However, for $n\ne1$, we have
$$I_n = \frac{i}{r^{n-1}}\int_0^{2\pi}\frac{1}{(e^{it})^{n-1}}.$$
I know that is supposed to be zero, and I can prove it by using De Moivre's formula by expanding the exponential function. But I want to know if there's another way to proceed without invoking additional theorems? Thanks!

答: Hint: for $n \ne 1$, the function $\frac{1}{(z-a)^n}$ has an anti-derivative on $\mathbb C \setminus \{a\}.$

Title: Integration of $(z-a)^n$


问:
If there is a linear map from V to W where the dimension of W is greater than V, how to you map the basis?
There are problems in linear algebra where the proofs go like

Suppose we have $\dim V \leq \dim W$. Since $V$ and $W$ are finite-dimensional, there exist a basis of $V$ and a basis of $W$. Define a linear map $T:V \to W$ by $$T(a_1v_1+ \dots +a_nv_n) = a_1w_1+ \dots +a_mw_m + \dots. $$

The part I don't understand is how you can map each basis element of $V$ to a basis element of $W$ when $W$ has a higher dimension than $V$. For example, let $\dim V = 3$ and $\dim W = 4$, then $v_1$ maps to $w_1$, $v_2$ maps to $w_2$, and so on, but what about $w_4$?

答: Each basis element $v_i\in V$ gets mapped to an element $w_i\in W$, and each $w_i$ can be expressed as a linear combination of basis elements of $W$, but the images of the basis elements of $V$ cannot comprise a basis of $W$, because the dimension of $W$ is higher.


答: Yes, it looks good.  There are $2$ cosets of the subgroup $\xi_4^*$ in $\xi_4$.</p>

Title: The index of $\xi_4^*$ in $\xi_4$

Body:
Just seeing if i'm right:

With the set of solutions for $z^4=1$: $\xi_4=\{1,i,-1,-i\}$, one can construct the group of the $4$th roots of unity: $(\xi_4,\cdot_\mathbb{C})$ and its multiplicative subgroup $(\xi_4^*,\cdot_\mathbb{C})$ with $\xi_4^*=\{1,-1\}$.

What is the Index $[ \xi_4 : \xi_4^* ]$?

I'd say through Lagrange:
$$ [ \xi_4 : \xi_4^* ] =\frac{|\xi_4|}{|\xi_4^*|}=\frac{4}{2}=2.$$
Or counting the cosets of $\xi_4^*$:
$$1\cdot\xi_4^*=\{1,-1\}= -1\cdot\xi_4^*,\quad i\cdot\xi_4^*=\{i,-i\}= -i\cdot\xi_4^*$$
we get two.

Is this correct?


问:
What an implication means when stating the solutions of an equation?
I am having trouble understanding what the following statement means: $$x^2=4 \implies x=2 \hspace{1em} \text{or} \hspace{1em} x=-2.$$

How can we relate this statement to a material conditional?

I am trying to see it as a material conditional. So on the left hand side we have the equation $A(x)$ which takes different truth values for different $x$. The same goes for the right hand side which we can denote it as $B(x)$. Now does the original statement means that $$\forall x \in \mathbb{R}: A(x) \implies B(x).$$

If I substitute the values of $x$ in these two expressions then the conditionals are all true. So how can such a statement make sense when we are trying to solve an equation or when stating the solutions of the equation? Does it mean that in the case that the left hand is true then the value "we substituted" was $2$ or $-2$ in order to make the right hand side also true?

答:
Related post : What's the difference between material implication and logical implication?

An " implies" sentence is a material conditional sentence that is logically true.

One can use $X \implies Y$ to express the fact that , not only it is not (factually)  the case that  X is true and Y false, but also that it cannot ( logically) be the case.

For example " $x$ lives in Florida " materially implies " $x$ lives in the USA" for all $x$ ( indeed, factually, it is the case for no $x$ that $x$ lives in Florida while $x$ does not live in the USA).  But the former does not logically imply the second, for it could be the case one lives in Florida while not living in the USA (in case Florida had not been joinded to the Union, which is a logically possible situation).

So one cannot write : $  Florida(x) \implies USA(x) $ , you cannot deduce the consequent from the antecedent , using logic alone

First there is a material conditional formula that is not a sentence

$x^2 = 4 \rightarrow (x=2 \lor x= -2)$.

Second, you turn it into a sentence using a quantifier

$\forall(x) [x^2 = 4 \rightarrow (x=2 \lor x= -2)]$.

This sentence can be true or false.

At this stage, you still have a material conditional.

Third, you manage to prove the consequent under the antecedent taken as hypothesis. It means that the consequent logically follows from the antecedent.

You can express this as:

$x^2 = 4 \implies ( x=2 \lor x= -2) $$\space$( $\space$ for all $x$)

meaning that there is no possible case in which $x^2 = 4$ is true and $( x=2 \lor x= -2)$ is false.

In equation solving, we want the next step to follow logically from the previous one, so the arrrow is understood as a material conditional that holds in all possible cases , that is as " implies " or " $\implies$, or " logical implication".


答:
Any two diagonal matrices commute, because:

$$\begin{bmatrix}a_1&0&\cdots&0\\0&a_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&a_n\end{bmatrix}\cdot\begin{bmatrix}b_1&0&\cdots&0\\0&b_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&b_n\end{bmatrix}=\begin{bmatrix}a_1b_1&0&\cdots&0\\0&a_2b_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&a_nb_n\end{bmatrix}$$

and "ordinary" multiplication commutes.

Now, if a matrix $P$ is an arbitrary invertible matrix, and $D_1$ and $D_2$ are arbitrary diagonal matrices, this means that the matrix $P^{-1}D_1P$ will also commute with $P^{-1}D_2P$ because:

$$(P^{-1}D_1P)(P^{-1}D_2P)=P^{-1}D_1D_2P=P^{-1}D_2D_1P=(P^{-1}D_2P)(P^{-1}D_1P)$$

Title: Do two matrices commute only if one matrix is the inverse of the other?

Body:
Is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? The only way I could believe this was the 2 commutative type properties I know of:
$$AA^{-1}=A^{-1}A=I$$
and
$$adj A*A=A*adjA = |A|I$$
where $adj A$ is the adjoint of matrix A.

The first equation when expanded with $A^{-1}=\frac{adjA}{|A|}$ you end up getting the second equation. So both are the same equation. These are the only two properties that explains commutativity as far as I'm aware of(are there any more?) and my conclusion was based on these only. Also multiplication with identity matrix is also commutative but any other counter cases? So is it safe to say matrix multiplication is commutative ONLY if one matrix is the inverse of the other? Thanks!!


问:
Limit of a bounded sequence
The $\lim_{n\to\infty} (\frac{\sin(n)+7}{15})^{n^2}=0$.

This is due by the fact that $|a_n|=|\frac{\sin(n)+7}{15}|<1$?

答:
We can use the triangle inequality
$$0 \le \left|\frac{\sin(n)+7}{15}\right|^{n^2} \le \left(\left|\frac{\sin(n)}{15}\right| + \frac{7}{15}\right)^{n^2} \le \left(\frac 12\right)^{n^2}$$
Therefore $a_n \to 0, \text{ as } n \to \infty$.

BTW you changed your sequence?

Title: Limit of a bounded sequence


问:
Verify if polynomial in $\mathbb R^n$ is a square
Let $f(x_1, \ldots, x_n)$ be a polynomial of degree 2 on $\mathbb R^n$, which can be written as
$$
f(x_1, \ldots, x_n) = \sum_{i=1}^n\alpha_i x_i^2+\sum_{i=1}^n\sum_{j=i+1}^n\beta_{ij}x_ix_j,
$$
where the coefficients $\{\alpha_i\}_{i=1}^n$ and $\{\beta_{ij}\}_{i=1,j>i}^n$ are real numbers.

I'm wondering if there is a way to verify whether $f$ can be written as a square, i.e., if there exist real coefficients $\{\gamma_i\}_{i=1}^n$ such that
$$
f(x_1, \ldots, x_n) = \left(\sum_{i=1}^n \gamma_i x_i\right)^2.
$$

答:
Let us denote $A$ as a matrix that represents $f$ w.r.t the standard basis, meaning we have $f(x)= x^T A x$. Note $A$ isn't unique, and one might alter one element $i,j$ but change $j,i$ accordingly and we get the same polynomial.

Let's denote $c=[\gamma_1\ \cdots\ \gamma_n]^T$, then the second expression is $\left ( c^T x \right )^2$. Let's equate both ways of expressing $f$ and see what needs to happen
$$x^T A x=(c^T x)^2 = (c^T x) (c^T x) = (x^T c)(c^T x) = x^T (cc^T)x$$
This means $A$ and $cc^T$ represent the same quadratic form. Using the fact $cc^T$ is symmetric, we can conclude that it is the symmetrization of $A$, meaning it is the only symmetric matrix which represents $f$
$$cc^T = \frac{A+A^T}{2}:=B$$
The right hand side is a matrix with $\alpha_i$s on the diagonal and $\frac{\beta_{ij}}{2}$ at elements $ij$ and $ji$. So to conclude, $f$ is a square of a linear combination iff its symmetric represntation is of the shape $$B = cc^T=\begin{bmatrix}
\gamma_1 \gamma_1 &  & \gamma_1 \gamma_n\\
 & \ddots & \\
\gamma_n \gamma_1 &  & \gamma_n \gamma_n
\end{bmatrix}$$
Note we can solve this system under the assumption $\gamma_i \neq 0$ since otherwise we get a polynomial of less than $n$ variables. and we would have seen that in $f$, so I guess that you could denote $B$'s columns and get
$$c = \frac{b_1}{\gamma_1} = \cdots = \frac{b_n}{\gamma_n}$$
From this you can get $c$ up to scaling $\alpha$.


问:
How to determine number with same amount of odd and even divisors
With given number N, how to determine first number after N with same amount of odd and even divisors?
For example if we have N=1, then next number we are searching for is :

```<code>2
because divisors:
odd : 1
even : 2
</code>```

I figured out that this special number can't be odd and obviously it can't be prime.
I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number.
Cheers

答:
If $2n$ is a divisor of $2m$1, then

$$
2n \mid 2m \Leftrightarrow n \mid m
$$

In other words, $2n$ is a divisor of $2m$ if and only if $n$ is a divisor of $m$. So for every even divisor of $2m$ there is a divisor of $m$, and for every divisor of $m$ there is an even divisor of $2m$. Now we conclude that the number of even divisors of $2m$ is equal to the number of divisors of $m$ (#).

Now if $2n - 1$ is a divisor of $2m$, then

$$
2n - 1 \mid 2m \Leftrightarrow 2n - 1 \mid m
$$

In other words, $2n - 1$ is a divisor of $2m$ if and only if $2n - 1$ is a divisor of $m$.  So for every odd divisor of $2m$ there is an odd divisor of $m$ and for every odd divisor of $m$ there is an odd divisor of $2m$. Now we conclude that the number of odd divisors of $2m$ is equal to the number of odd divisors of $m$ (##).

If the number of even divisors of $2m$ is equal to the number of odd divisors of it, from ## and # we conclude that the number of divisors of $m$ is equal to the number of odd divisors of it, which means all of the divisors of $m$ are odd, so $m$ is odd.

So if the number of even divisors and odd divisors of $2m$ are equal, then $m$ is odd.

\begin{align}
2m &= 2(2n - 1) \\
&= 4n - 2 \\
&= 2 + 4(n - 1) \\
&= a_1 + d(n - 1), n \in \mathbb{N}
\end{align}

Every term of the sequence of every other even number starting with $2$ has the property.

1.As you said the number has to be even so I wrote $2m$.


答:
For any matrix $A$, the matrix $A^n$ whatever be $n$ satisfies $A A^n = A^n A$.

Also for any two real numbers $a, b$ the matrices $1\quad a\choose 0\quad 1$ and $1\quad b\choose 0\quad 1$ can be multiplied in any order to get the same product.


问:
Concrete example: closed subspace of finite codimension admits complement
This is an example from Section 2.4 of Brezis's "functional analysis":

Suppose $E$ is a Banach space, and let $N \subset E^*$ of dimension $p$. Then $G := \{x \in E: \langle f, x \rangle = 0 \ \forall f \in N \}$ is closed and of codimension $p$. Indeed, let $f_1, \cdots, f_p$ be a basis for $N$. Then there exists $e_1, \cdots, e_p \in E$ such that $\langle f_i, e_j \rangle = \delta_{ij}$. It is easy to check that $e_i$'s are linearly independent and $\text{span}\{e_1, \cdots, e_p\}$ is a complement of $G$.

Question: I could show the existence of $e_{i}$, and they are linearly independent. But I don't see $\text{span}\{e_1, \cdots, e_p\}$ is a complement of $G$. How could I check it? Could anyone give me some hint on this?

答:
Write $$H=\operatorname{span}\{e_1,\ldots,e_p\}.$$

If $x\in G\cap H$, then $x=\sum_1^px_je_j$; using that $x\in G$, you get that $0=f_j(x)=x_j$. So $x=0$. That is, $G\cap H=0$.

Given any $x\in E$, we have $y=x-\sum_jf_j(x)e_j\in G$. So $$x=y+\sum_jf_j(x)e_j\in G+H.$$

The idea is that the map $P:x\longmapsto \sum_jf_j(x)e_j$ is a projection.


问:
Show that the subgroup $s(K) \subset H \rtimes_{\alpha} K$ is normal if and only if $\alpha: K \rightarrow Aut(H)$ is the trivial homomorphism.
Show that the subgroup $s(K) \subset H \rtimes_{\alpha} K$ is normal if and only if $\alpha: K \rightarrow Aut(H)$ is the trivial homomorphism, where $s : K \rightarrow H \rtimes_{\alpha} K$ is given by $s(k) = (e,k)$.

What's confusing about this question is that $\alpha: K \rightarrow Aut(H)$ is given by $\alpha(k)(h) = khk^{-1}$. So two elements (k and h) are taken to one element in Aut(H), right? In other words, in a regular homomorphism $f: X \rightarrow Y$, we would take one element from X and end up with one element in Y, right? So if we wanted to prove that f was the trivial homomorphism, we would just need to show that, for any $x \in X$, $f(x)=e$. But this question is a bit confusing because we take two elements and end up with one. So how is it possible to show that $\alpha(k)(h)$ is the trivial homomorphism? Do I fix h and just change k? Or do I fix k and change h? Or do I not fix anything? In other words, how does one usually deal with these types of homomorphism with more than one element?

Thank you in advance

答:
If both $H$ and $K$ are normal, then you have two normal subgroups such that $HK=KH=G$ and $H\cap K=\{e\}$. This is to say that $G=H\times K$. This forces the multiplication in $G$ to be defined as $(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$, that is, $\alpha=id$.

The converse is trivial.


问:
Show that if $\vert Y \vert \le c$ a.s. then $ \vert E\{Y |G\}\vert \le c$ a.s. also.
I was trying to prove that if $\vert Y \vert \le c$ a.s. then $ \vert E\{Y |G\}\vert  \le c$ a.s. also by contradiction.

If I assume $\vert Y \vert \le c$ a.s and $P(\vert E\{Y |G\}\vert  > c)>0$ then

$P(\vert E\{Y |G\}\vert  > c,c\ge \vert Y \vert)>0$ since  $\vert Y \vert \le c$ a.s. and we can conclude that $P(\vert E\{Y |G\}\vert  >\vert Y \vert)>0$.

Another alternate idea that I had was
Since conditional expectation  E{Y |G} takes the values $E[Y\mathbb{1}_A]$ as $A$ varies in $G$ we can conclude that
$Y\mathbb{1}_A \le Y \le \vert Y \vert \le c $ we have that for all $A$ in $G$ we have $E[Y\mathbb{1}_A] \le c$ and hence  $\vert E\{Y |G\} \vert \le E\{\vert Y \vert |G\} \le c$

答: Your argumentgs don't seem rigorous. For  a correct proof you have to use the definition of conditional expectation. Let $Z=E[Y|\mathcal G]$. By definition $\int_AZdP=\int_A YdP$ for all $A \in \mathcal G$. Let $A=\{Z >c\}$. If $P(A) >0$ then $c P(A) <\int_A Z dP =\int_A YdP \leq cP(A)$ which is a contradiction. This proves that $P(A)=0$ so $Z \leq c $ a.s.. Similarly $Z >-c$ a.s.


问:
Does $1-ab\geq (1-a)(1-b)$ hold, for $a,b\in [0,1]$?
Does this inequality hold, for any $a,b\in[0,1]$

$$1-ab\geq (1-a)(1-b)?$$

I'm don't have idea to conclude $1-ab\geq 1-a-b+ab = (1-a)(1-b)$.

Anyone can prove (or disprove) it?

答:
$1-ab \ge 1-a \ge (1-a)(1-b).\blacksquare$

Update: I found this is equivalent to that of @Albus Dumbledore and @Tony Ip.


问:
Finding a function which is distance $1$ from $x^2$ along its normals
So I recently thought of a problem, which I have tried to solve many ways but couldn't. It's very simple to describe...

Find the function which is distance $1$ from parabola $y=x^2$ along its normals. Visual of what I mean.

This one of the things I've tried...

This function would get the corresponding $x$ value on the function we want to find from an $x$ value on curve $y=x^2$. The inverse would give the corresponding $x $ value on $x^2$ for a given $x$ value which would make it trivial to determine the $y$ value.
$$
f(x)=x+\sin(\arctan(2x))
$$
This can be resolved to this...
$$
f(x)=x+\frac{2x\sqrt{1+4x^2}}{1+4x^2}
$$
However I don't know how to get the inverse of this function.

答:
I have found  parametric equations of the locus

$$
\begin{cases}
x=2 t^3-\frac{8 t^5}{4 t^2+1}-\frac{2 t^3}{4 t^2+1}+t +\frac{2 t}{\sqrt{4 t^2+1}}\\
y= \frac{4 t^4+t^2-\sqrt{4 t^2+1}}{4 t^2+1}\\
\end{cases}
$$


问:
A question involving complex numbers and polynomials
Let $\iota$ be a root of the equation $x^2+1=0$ and let $\omega$ be a root of the equation $x^2+x+1=0$. Construct a polynomial

$$f(x)=a_0+a_1x+\dots+a_nx^n$$

where $a_0,a_1,\dots,a_n$ are all integers such that $f(\iota+\omega)=0$.

My approach was to substitute $(\iota + \omega)$ in place of $x$ and try some expansion but that resulted in nothing.
This question was asked in a 10 marks question of a very popular entrance examination at the 10+2 level.
A rigorous answer is appreciable.
Thank you.

答:
Let $\bar\omega$ be the other root of $x^2+x+1$.  Compute $$(x-i-\omega)(x-i-\bar\omega)(x+i-\omega)(x+i-\bar\omega).\tag{1}$$

The first factor of (1) is $x-i-\omega$, so it should be clear that when $x=i+\omega$, the whole product is zero  as required.  The only remaining issue is whether the coefficients of (1) will be real integers, which we can verify by direct computation.

To do this, it will help to observe that  $$(x-\omega)(x-\bar\omega) = x^2+x+1\tag{2}.$$ Now group expression (1) above as follows:

$$((x-i)-\omega)((x-i)-\bar\omega)\cdot((x+i)-\omega)((x+i)-\bar\omega)$$

Then by (2), the product of the left two factors  is $$(x-i)^2 + (x-i)+1$$ and the product of the right two factors is $$(x+i)^2 + (x+i)+1.$$

Can you take it from there?


问:
Can this function be represented? Set theory.
I want to know how you can represent this dependency using the operations of

intersection

union

complement

cartesian product

symmetric difference

etc

$$X1 \: ??? \: X2 = Y$$

```
+----+----+---+
| X1 | X2 | Y |
+====+====+===+
| Ø  | A  | Ø |
+----+----+---+
| B  | A  | A |
+----+----+---+
```

A, B - nonempty sets

答:
Using only unions, intersections and complements, this cannot be done.

Namely, each formula using only those operations can be converted to its "disjunctive normal form", which will be a disjunction of zero to four of the following terms $X_1\cap X_2$, $X_1\cap X_2^c$, $X_1^c\cap X_2$, $X_1^c\cap X_2^c$. On the universal set $\{0,1,2,3\}$ and the $2$-element sets $X_1=\{1,2\}, X_2=\{1,3\}$ you verify that the formula providing $Y=X_2$ as a result must be $(X_1\cap X_2)\cup(X_1^c\cap X_2)$, which is (by the way) equivalent to "$X_2$" standing alone.

But then this formula does not work for $X_1=\emptyset$ and nonempty $X_2$, as it will still return a nonempty set ($X_2$).

However, if "etc." means that we can use *Cartesian product" ($\times$) and "projection" ($\pi_1$ and $\pi_2$), then you can have the formula doing what you want, and it is going to be $Y=\pi_2(X_1\times X_2)$, i.e. $Y=\{y\in X_2\mid (\exists x\in X_1)((x,y)\in X_1\times X_2)\}$.

Update You removed "etc." and added that you allow for Cartesian product but not for projections. This makes it a different question. I will keep this answer and see if you will additionally allow for projections later on. Next time, please don't change the essence of your question once you've posted it.


问:
polynomials with integer coefficients part 2
Let $p(x)=x^7+x^6+b_5x^5+...+b_1x+b_0$ and $q(x)=x^5+c_4x^4+...+c_1x^+c_0$ be polinomial with integer coefficients.Assume that p(i)=q(i) for integers i=1,2,..,6.Thus there exists a negative integer r such that p(r)=q(r).

答:
$$\implies p(i)-q(i)=(i-1)(i-2)(i-3)(i-4)(i-5)(i-6)(i+s). s\in I^{+}$$
$7th$ degree polynomial equation $r(x)=p(x)-q(x)$ has integer coefficients, since it has 6 positive integer roots (1 to 6). So its 7th rood needs to be a negative integer and further as sum of the roots of $r(x)=0$ is $-1$, so the 7th root is $-22 (s=22).$


问:
How to find the limit for the quotient of the least number $K_n$ such that the partial sum of the harmonic series $\geq n$
Let $$S_n=1+1/2+\cdots+1/n.$$ Denote by $K_n$ the least subscript $k$ such that $S_k\geq n$. Find the limit $$\lim_{n\to\infty}\frac{K_{n+1}}{K_n}\quad ?$$</p>

答:
We know that $H_n=\ln n + \gamma +\epsilon(n)$, where $\epsilon(n)\approx \frac{1}{2n}$ and in any case $\epsilon(n)\rightarrow 0$ as $n\rightarrow \infty$.  If $m=H_n$ we may as a first approximation solve as $n=e^{m-\gamma}$.  Hence the desired limit is $$\lim_{m\rightarrow \infty} \frac{e^{m+1-\gamma}}{e^{m-\gamma}}=e$$

For a second approximation, $m=\gamma + \ln n +\frac{1}{2n}=\gamma+\ln n+\ln e^{\frac{1}{2n}}=\gamma+\ln ne^{\frac{1}{2n}}$.  This may be rearranged as $ne^{\frac{1}{2n}}=e^{m-\gamma}$.  This has solution $$n=-\frac{1}{2W(-e^{\gamma-m}/2)}$$
where $W$ is the Lambert function.  Hence the desired limit is now $$\lim_{m\rightarrow \infty}\frac{W(-e^{\gamma-m}/2)}{W(-e^{\gamma-m-1}/2)}=e$$

Although not a proof, this is compelling enough that I'm not going to think about the next error term.


问:
Finding double integrals using polar coordinates.
So Im new to using polar coordinates to work out double integrals and was wondering if anyone could help me.

So I am looking at the question:

$I_{6}=\int\int_{Q}x/\pi(x^{2}+y^{2}) dxdy$ where $Q=\{(x,y)|(x-2)^{2}+y^{2}\leq 4, y\geq0\}$

Looking at the first equation:

$(x-2)^{2}+y^{2} \leq4$   which gives
$x^{2} + y^{2} = 4x$

and in polar form:

$r^{2}cos^{2}\theta +r^{2}sin^{2}\theta =4rcos\theta$

which simplifies to :

$r^{2}=4rcos\theta$ which equals $r=4cos\theta$

So my first integral limits will be between $0$ and $4cos\theta$

So the first part of the equation looks like:

$\int_{0}^{4cos\theta} (rcos\theta)/\pi(r^2cos^2\theta+r^2sin^2\theta) rdr$

Which simplifies to:

$\int_{0}^{4cos\theta}(cos\theta/\pi) dr$

which gives me: $\large[(cos\theta/\pi)r\large]_{0}^{4cos\theta}$

and gives:

$4cos^2\theta/\pi$

but im unsure on what my next set of limits will be. I know from a previous question that ive asked that I need to draw a diagram with the centre of the circle at (2,0) but I dont undestand how to utilise this to get my second limit.

Any help would be greately appreciated.

答: If you draw a picture of the circular region you will see that the given region is $\{r,\theta): r\leq 4\cos \theta, 0\leq \theta \leq \frac {\pi}2\}$. [ $\theta$ ranges from $0$ to $\frac {\pi}2$].


问:
Proof of Hardy inequality in $\mathbb{R}^n$
I've often seen people use the inequality
$$\int_{\mathbb R^3} \frac{|u(x)|^2}{|x|^2}\,dx  \leq 4\int_{\mathbb R^3}|\nabla u(x)|^2\,dx,\qquad u\in C_0^\infty(\mathbb R^3) $$
without proof, refering to it as "Hardy's inequality".

I struggled to find a direct proof of this in the literature and couldn't prove it myself. Does anyone know a straightforward proof of this or a book in which Hardy's inequality in this form is proved?

I would also be interested in the general form of this inequality, i.e. what happens if one replaces $\mathbb R^3$ with $\mathbb R^n$?

答:
I learned the following integration by parts method to prove Hardy inequalities from Luca Fanelli.

Assume that all functions are real-valued and define operators on $C^{\infty}_c(\mathbb R^d\setminus\{0\})$ by
$$
Su(x)=\frac{xu(x)}{|x|^2},\qquad Au(x)=\nabla u(x).$$
Notice that the sought inequality reads
$$
\int_{\mathbb R^d} \lvert Su(x)\rvert^2\,dx \le C\int_{\mathbb R^d} \lvert Au(x)\rvert^2\, dx.$$
So, for a $k\in \mathbb R$ yet to be determined, consider
$$\begin{split}
0&\le \int_{\mathbb R^d} \lvert Su+kAu\rvert^2\, dx\\&=\int_{\mathbb R^d} \lvert Su(x)\rvert^2\, dx +k^2\int_{\mathbb R^d} \lvert Au(x)\rvert^2\, dx+2k\int_{\mathbb R^d} Su(x)\cdot Au(x)\, dx.
\end{split}$$
The last summand is
$$
\int_{\mathbb R^d} Su\cdot Au\, dx=\int_{\mathbb R^d}  \frac{x}{\lvert x\rvert^2}\cdot u\nabla u\, dx =\int_{\mathbb R^d} \frac{x}{\lvert x \rvert^2}\cdot \nabla \left(\frac{u^2}{2}\right)\, dx, $$
and since
$$\nabla \cdot \frac{x}{\lvert x\rvert^2}=\frac{d-2}{\lvert x \rvert^2}, $$
integration by parts reveals that
$$\tag{*}
\int_{\mathbb R^d} Su(x)\cdot Au(x)\, dx = -\frac{d-2}{2}\int_{\mathbb R^d} \lvert Su(x)\rvert^2\,dx.$$
Going back to the first inequality, we have that
$$
0\le k^2\int_{\mathbb R^d} \lvert Au(x)\rvert^2\, dx+(1-(d-2)k)\int_{\mathbb R^d} \lvert Su(x)\rvert^2\, dx, $$
which is a second-degree polynomial in $k$, and it attains its minimal value for $k=\frac{d-2}{2}\frac{\int \lvert Su\rvert^2}{\int \lvert Au\rvert^2}$. Plugging this into the last inequality finally yields
$$
\int_{\mathbb R^d} \lvert Su(x)\rvert^2\, dx \le \left( \frac{2}{d-2}\right)^2\int_{\mathbb R^d} \lvert Au(x)\rvert^2\, dx, $$
which is the result we wanted. $\Box$

Remark 1. The multiplicative constant $\left( \frac{2}{d-2}\right)^2$ turns out to be sharp, but proving this needs further work.

Remark 2. This method of proof is interesting because it has some robustness. Changing the definition of $S$ and $A$ may yield more inequalities, provided that some suitable replacement of the identity  (*) holds. For example, taking
$$
Su(x)=\frac{u(x)}{\lvert x \rvert}, \qquad Au(x)=\frac{x}{\lvert x\rvert}\cdot \nabla u(x)$$
yields the following refinement of the Hardy inequality;
$$
\int_{\mathbb R^d} \frac{u^2(x)}{\lvert x \rvert^2}, dx \le \left( \frac{2}{d-2}\right)^2\int_{\mathbb R^d} \lvert \partial_r u(x)\rvert^2\, dx, $$
where $\partial_r$ stands for the radial derivative. This inequality is tighter than the previous one, because the right-hand side is smaller; indeed, here there are no angular derivatives.


问:
Using functions within differential equations

Let y be a differentiable function in $ \Bbb R$ which can be factorised as y = f(x)·sin(x) for some differentiable function f(x) in $ \Bbb R$.

Show that the function y satisfies the differential equation
$$ y' = \Biggl[\frac {f'(x)}{f(x)}
+ cot(x)\Biggr]·y $$

I am not sure how to go about approaching this problem.  I started by taking it at face value and getting $$ y' = f'(x)sin(x) + f(x)cos(x) $$ and then plugging it back into the equation.  I ended up with $$ \frac{f'(x)f'(x)sin(x)}{f(x)} + 2f'(x)cos(x) + \frac{f(x)cos(x)}{cot(x)} $$
This does not seem to be the way to go about solving the problem.

Guidance on how to solve it would be much appreciated.

答:
$$y' = \Biggl[\frac {f'(x)}{f(x)}
+ \cot(x)\Biggr]·y$$
$$\dfrac {y'}y -\frac {f'(x)}{f(x)}
= \cot(x)$$
$$(\ln y -\ln f(x))'
= \cot(x)$$
$$(\ln \dfrac y {f(x)})'
= \cot(x)$$
Since $y=f(x)\sin x$:
$$(\ln \sin x)'
= \cot(x)$$
$$ \dfrac {\cos x} { \sin x}
= \cot(x)$$
Which is true.


问:
determine with proof, all the positive integers $n$ for which $n$ is not the square of any integer and $[\sqrt{n}]^3$ divides $n^2$
determine with proof, all the positive integers $n$ for which:

$n$ is not the square of any integer and,

$[\sqrt{n}]^3$ divides $n^2$. where $[x]$ is the greatest integer less than or equal to $x$.

My Approach:

Well I tried a few test cases and I concluded that no prime number greater than 4 holds true to this statement. Except that, I couldn't recognize a pattern or make a solution on my own so please help me out.

答:
Consider $k$ such that $k^2<n<(k+1)^2$. Then $\lfloor n\rfloor=k$ and $n=k^2+i$, $1\leq i\leq 2k$. We get the following:

$$k^3|(k^2+i)^2\Leftrightarrow k^3|k^4+i(2k^2+i)\Leftrightarrow k^3|i(2k^2+i)$$

$\text{If }k>1$

Consider a prime $p$ that divides $k$ and let $v_p(k)$ the exponent of $p$ in the factorisation of $k$. We must have $$v_p(k^3)\leq v_p(i(2k^2+i))\Leftrightarrow3\cdot v_p(k)\leq v_p(i)+v_p(2k^2+i)$$ Suppose $v_p(i)\leq v_p(k)$. Then, obivously, $v_p(2k^2+i)=v_p(i)$ so $v_p(i)+v_p(2k^2+i)\leq 2\cdot v_p(k)$, which would lead to $3\cdot v_p(k)\leq2\cdot v_p(k)$ which is a contradiction ($p$ divides $k$ so $v_p(k)\geq 1$).

Thus, for any $p $that divides $k$, $v_p(i)>v_p(k)$ so $k|i$ and $k\neq i$. However, $i\leq 2k$, so this implies $i=2k$ (so $n=k^2+2k$) which gives us $$k^3|2k(2k^2+2k)\Leftrightarrow k^3|4k^3+4k^2\Leftrightarrow k^3|4k^2 \Leftrightarrow k|4$$

Thus, just check cases $k=2,4$, which give $n=8,24$

$\text{If }k=1$

Then this gives us $n=2,3$ and again, just check those cases.


问:
Variance of random variables with unfair coins
I have the following problem:

We toss two unfair coins each 5 times. First, we 5 times toss a coin that has probability $\frac{1}{4}$ for heads (H), followed by tossing 5 times a coin that has probability $\frac{3}{4}$ for H. All coin tosses can be assumed to be independent. Let $X$ denote the total number of heads. Compute Var($X$).

The following is my computation:

Let $X_{1}$ and $X_{2}$ be the number of heads for the first coin and the second coin respectively. Since these events are independent we have,
\begin{equation}\text{Var}(X)=\text{Var}(X_{1})+\text{Var}(X_{2})\end{equation}
So now I compute the separate variances,
\begin{equation}\text{Var}(X_{1})=\mathbb{E}[X_{1}^{2}]-(\mathbb{E}[X_{1}])^{2}\end{equation}
\begin{equation}\text{Var}(X_{1})=5^{2}\cdot\frac{1}{4}-(5\cdot\frac{1}{4})^{2}\end{equation}
\begin{equation}\text{Var}(X_{1})=\frac{25}{4}-\frac{25}{16}=\frac{75}{16}\end{equation}
and for $X_{2}$,
\begin{equation}\text{Var}(X_{2})=\mathbb{E}[X_{2}^{2}]-(\mathbb{E}[X_{2}])^{2}\end{equation}
\begin{equation}\text{Var}(X_{2})=5^{2}\cdot\frac{3}{4}-(5\cdot\frac{3}{4})^{2}\end{equation}
\begin{equation}\text{Var}(X_{2})=\frac{75}{4}-\frac{225}{16}=\frac{75}{16}\end{equation}
So now to conclude I compute,
\begin{equation}\text{Var}(X)=\frac{75}{16}+\frac{75}{16}=\frac{75}{8}\end{equation}

The teacher did not give any solutions for our exercises so the reason I put this question on here is so that someone can check my work. Could someone do that?

答:
no, it's wrong.

Setting (as you did)

$$X_1\perp\!\!\!\!\!\!\perp X_2$$

You have that $X_1\sim Bin(5;0.25)$ and $X_1\sim Bin(5;0.75)$. Knowing that for a binomial rv its variance is $n p(1-p)$, the total variance is

$$\mathbb{V}[X]=5\times\frac{1}{4}\times\frac{3}{4}+5\times\frac{1}{4}\times\frac{3}{4}=\frac{15}{8}$$

FYK,  this is the distribution of Heads

I calculate it manually...it is not difficult, if you want to do an exercise...


问:
Relation between maximal matching and minimum vertex cover

Let $M$ be any maximal matching of $G(V,E)$ and $S^*$ be the minimum vertex cover of $G(V,E)$, how to show $|S^{*}|\leq 2|M|$ where $G(V,E)$ is any unweighted undirected graph

I know $|M|\leq|M^*|\leq2|M|$ if we take $M$ to be the maximum matching of $G(V,E)$, and I thought if I can show $|S^{*}|\leq |M^*|$ then done, but then I found it actually is $|M^*|\leq|S^{*}|$! Someone can help me? Thank you!

答:
Hint: Let $V_M$ be the set of vertices incident with edges of $M$. Since $M$ is maximal, every edge of the graph must be incident to a vertex of $V_m$.

Full proof (click spoiler):

 By the hint above, we see that $V_m$ must be a vertex cover of the graph, and $|V_m| = 2|M|$. So the smallest vertex cover $S^*$ has at most $2|M|$ vertices.


问:
Does the definition of countable ordinals require the power set axiom?
I am trying to understand the consequences of the different axioms of ZFC. In particular, I was trying to understand what you get on ZFC-power set (ZFC minus the power set axiom). If you have any references that I could read please let me know. In particular, I have a question. From the definition of ordinal number (for instance, Jech, p.19 and above) I believe that you do not need the power set axiom to define infinite ordinals beyond $\omega$, but I am not completely sure. If you don't need it, which is the largest ordinal that you can reach without using power set? can you reach $\omega_1$?</p>

答:
You may be very interested in the following paper:

Victoria Gitman, Joel David Hamkins, Thomas A. Johnstone, What is the theory ZFC without power set? (arXiv, 2011)

Note that you cannot prove the existence of $\omega_1$ either, this is because $H(\omega_1)$, the set of the hereditarily countable sets contain all the countable ordinals, and it is a model of $\sf ZFC-Pwr$, but $\omega_1\notin H(\omega_1)$ because it is not hereditarily countable.


问:
A local domain $(R,\mathfrak m)$ which is not a field, then for its fraction field $K$ we have $\mathfrak m K=K$.

If $(R,\mathfrak m)$ is a local domain which is  not a field, then for
its fraction field $K$ we have $\mathfrak m K=K$.

Amusingly, this proves that $K$ is not a finitely generated
$R$-module.

I know this gives counterexample to NYK, my question is,

$1$.How can I get $\mathfrak m K=K$?

$2$.Why this proves that $K$ is not a finitely generated $R$-module?

Thank you for your kind help.

答: We have $\mathfrak{m}K \subset K$. Conversely, if $R$ is not a field then $\mathfrak{m} \neq 0$, and thus if you pick a nonzero $m \in \mathfrak{m}$, then $m$ acts invertibly on $K$ (because $m, m^{-1} \in K$). The reason that $K$ is not finitely generated then is because it would contradict Nakayama's lemma (maybe there's a more direct way of seeing this though), which in this case says that there should exist some nonzero $r \in R$ such that $rK = 0$ (which clearly can't happen).


问:
Claim of convexity of unitaries with $\|U\|_\infty \leq 1$ for a minimax theorem
The context of this question is the step to go from (24) to (25) here.

One considers an optimization problem of the following form
$$\inf_{\rho \in S}\sup_{U}\ (\cdot)$$

where $S$ is a convex set and $U$ is a unitary matrix such that $\|U\|_\infty\leq 1$. It is claimed that the supremum is taken over a convex set and hence, one can use a minimax theorem to commute the $\inf$ and the $\sup$ i.e.

$$\inf_{\rho \in S}\sup_{U}\ (\cdot) = \sup_{U}\inf_{\rho \in S}\ (\cdot)$$

I don't understand why the set

$$\{U\ \vert \ U^\dagger U  = I, \|U\|_\infty\leq 1\}$$

is convex. A linear combination of unitaries is not a unitary. So why do they claim that the variation is over a convex set and why can one use a minimax theorem here?

答: They claim something different: First, they solve the inner problem and compute the sup. Then they observe that the inner problem can be replaced by taking the supremum over $\{U: \ \|U\|_\infty\le 1\}$, which is convex. So no unitary matrices involved anymore.


问:
Matrix exponential of a skew-symmetric Toeplitz matrix
From today's exam:

Given the following matrix,
$$B = \begin{pmatrix} 0 & -2/3 & 1/3\\ 2/3 & 0 & -2/3\\ -1/3 & 2/3 & 0\end{pmatrix}$$ Prove that $$\exp(aB) = I + \sin(a) B + (1 -\cos(a)) B^2$$

I have tried expanding the exponential
$$\exp(aB)=I+aB+\frac{a^2}{2}B^2+\frac{a^3}{3!}B^3 + \cdots$$

But could not manage to find the $B^k$ expression.
Can anyone help with this?

答:
The characteristic polynomial of the given matrix is $x^3 +x$. Hence by Cayley-Hamilton theorem,
\begin{align*}
B^3+B&=0\\
B^3 &= -B\\
B^4 &= -B^2\\
B^5 &= B\\
\vdots\\
\exp(aB)&= I + aB+\frac{a^2B^2}{2!}+\frac{a^3B^3}{3!}+\cdots\\
\exp(aB) &= I + \left(a-\frac{a^3}{3!}+\cdots \right) B + \left( \frac{a^2}{2!}-\frac{a^4}{4!}+\cdots \right)B^2  &&\square\\
\end{align*}


问:
Is it possible that the union of a Bernstein set and a singleton isn't a Bernstein set?
Since the construction of a partition of two Bernstein sets is almost identical to  that of a partition of three in an uncountable Polish space. It's possible that the union of a Bernstein set and a singleton is  a Bernstein set. But is it possible otherwise, that is, the union of a Bernstein set and a singleton contains an uncountable closed set ? </p>

答:
No.

To see this, notice that if you remove a point from a perfect set, what remains will still contain a perfect set, so the intersection of a Bernstein set with a perfect set is not just nonempty, but also infinite. In fact, it's dense (in the perfect set) by a similar argument.

Therefore, removing (or adding) a single point does not turn a Bernstein set to a non-Bernstein set.

Using measure arguments you can show that the intersection is uncountable: the intersection of a Bernstein set with any perfect set has full outer measure with respect to any continuous probability measure on the perfect set, so it can't be countable.

(I vaguely recall that any Bernstein set is a union of $\mathfrak c$-many Bernstein sets, so that the intersection is actually of cardinality of the continuum, but I don't remember the proof, exactly, and might be mixing something up.)


问:
What are some books on history and development of calculus?
I am well versed with modern calculus but I want to know how people in history developed the subject. I want to learn how Newton and Leibniz came up with it, what Greeks did with the method of exhaustion et cetera</p>

答:
The Calculus Gallery: Masterpieces from Newton to Lebesgue by William Dunham.

It's easy to follow, provides the most important examples or methods by Newton, Weirstrass, Cauchy, Cantor and so on. It also explains some basic concepts, but I like it mostly for its easiness to understand and follow.

Link here: https://www.amazon.com/Calculus-Gallery-Masterpieces-Newton-Lebesgue/dp/0691136262/ref=sr_1_1?ie=UTF8&s=books&qid=1273235330&sr=8-1


答:
Actually, for any nonzero ideal $I$ of an integral domain (whether local or not), we have $IK=K$ since a field has only two ideals: $\{0\}$ and $K$.

If $K$ were a finitely generated $R$-module  ($R$ local), then, by Nakayama's lemma, $\; \mathfrak mK=K\implies K=\{0\}$. However, in a field $1\ne 0$.

Title: A local domain $(R,\mathfrak m)$ which is not a field, then for its fraction field $K$ we have $\mathfrak m K=K$.

Body:
If $(R,\mathfrak m)$ is a local domain which is  not a field, then for
its fraction field $K$ we have $\mathfrak m K=K$.

Amusingly, this proves that $K$ is not a finitely generated
$R$-module.

I know this gives counterexample to NYK, my question is,

$1$.How can I get $\mathfrak m K=K$?

$2$.Why this proves that $K$ is not a finitely generated $R$-module?

Thank you for your kind help.


问:
Prove $x_n=n^2-n$ sequence converges to $\infty$
Prove $x_n=n^2-n$ sequence converges to $\infty$

Here is my proof, I proved this using two attempt. Is there anything wrong please tell me?

$1^{st}$ attempt,

Let $ M \in \mathbb{R}$ and $M>0$,

$$x_n>M\Longleftrightarrow n^2-n>M\Longleftrightarrow n>\sqrt{M+\frac{1}{4}}+\frac{1}{2} $$

By the Archimedian property  $N \in\mathbb{N}$ such that

$$N>\sqrt{M+\frac{1}{4}}+\frac{1}{2}$$

Then $x_n>M$ whenever $n>N$

therefore $x_n=n^2-n$ sequence converges to $\infty$

$2^{nd}$ attempt,

Let $ M \in \mathbb{R}$ and $M>0$,

For lower bonded, $x_n=n^2-n\geqslant n^2-\frac{n^2}{2}=\frac{n^2}{2}$ when $n\geqslant2 ,$

take $N > max\{2, \sqrt{2M}\}$. Then $n>N$ implies $n>2$ and $n>\sqrt{2M}  $ thus for $n>N$ we have $x_n>M$

$x_n>M$ whenever $n>N$

therefore $x_n=n^2-n$ sequence converges to $\infty$

答:
Your two proofs are perfectly correct. With experience, you will be able to omit some details. For exemple, writing
$$ n^2 - n \geq \frac{n^2}{2} \quad \text{ for } n \geq 2$$
hence
$$n^2 - n \longrightarrow \infty$$
will be enough. But to begin, writing all the details is a good way to progress.