问:
Sheldon Axler Linear Algebra Done Right: Confused about "smallest containing subspaces."
I have a few questions regarding Axler's proofs that concern "smallest containing subspaces": in particular, 1.39.

I understand we want to show that $U_1 + \cdots + U_m$ = the smallest subspace of V containing $U_1, \cdots, U_m$ by showing that each set is a subset of the other set. What I don't understand, however, is why Axler says "Conversely, every subspace of V containing $U_1 \cdots U_m$ contains $U_1 + \cdots + U_m$)" because isn't the point to compare $U_1 + \cdots + U_m$ with the SMALLEST subspace, not an arbitrary subspace?

答: If $V$ is a subspace containing $U_1,\ldots,U_m$, then $V\supset U_1+U_2+\cdots+U_m$. In particular, the smallest subspace containing $U_1,\ldots,U_m$ will also contain $U_1+U_2+\cdots+U_m$. But $U_1+U_2+\cdots+U_m$ is a subspace itself, which contains $U_1,\ldots,U_m$. So, $U_1+U_2+\cdots+U_m$ contains the smallest subspace containing $U_1,\ldots,U_m$, and therefore they're equal.


问:
geometric distribution on interval
Is it possible to define a geometric distributed variable with parameter p on an interval $[a,b]$, such that

$$ c \cdot \sum_{j=a}^b (1-p)^j p=1$$
with a scale factor $c$.
Is that reasonable?

答:
It is possible to define it, although the name "geometric distribution" may be confusing, since it commonly refers to a distribution on a different domain. "Geometric distribution on a bounded domain" seems correct – maybe it's used somewhere. You might also call it a "geometrically varying" distribution if you like, since subsequent probabilities are in a fixed ratio, just like for the ordinary geometric distribution.

Such a distribution can in fact be rewritten in the form $\mathrm{p}(i \vert \lambda) = \frac{1}{Z(\lambda)} \exp(\lambda\ i)$ with
$$Z(\lambda):=\frac{\exp[(b+1/2)\ \lambda]-\exp[(a-1/2)\ \lambda]}{\exp(\lambda/2)-\exp(-\lambda/2)}\ ,$$
obtainable using the expression of the partial sum of a geometric series.
Note that the parameter $\lambda \equiv \ln(1-p)$ is determined by the mean of the distribution, and vice versa. I don't think there's an analytic expression giving $\lambda$ as a function of the mean, but I'm not sure to be honest.

Also, the mean (ie, first moment) is a sufficient statistics because of the special exponential form (see ref. below). So this is also a maximum-entropy distribution with the mean (ie, first moment) as constraint.

I think that the parameter $p$ in your expression ceases to have the specific meaning it has for the ordinary geometric distribution.

Whether this distribution is reasonable or not depends completely on the specific problem you're applying it to – for example, can you consider the mean to be a sufficient statistics, in your problem?

See eg Bernardo, Smith: Bayesian Theory (Wiley 2000), § 4.5.

ContentLicense: CC BY-SA 4.0

Title: geometric distribution on interval


问:
For natural $n$, prove $\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$

Prove that, for a natural number $n$,
  $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$

This follows from a continued fraction identity for which, I think, there is a lengthy proof. But, I thought, that there may be a direct geometric or another proof. Constructing a polynomial with the sines and tangents roots may be helpful.

答:
The proof for this should be identical to the one for:
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$</p>

ContentLicense: CC BY-SA 3.0


问:
Find $P(X+Y > \frac{3}{2})$ given that X is uniformly distributed on [0,1] and the conditional distribution of Y given X=x is uniform on [1-x,1].
Suppose $X$ has the uniform distribution on the interval [0,1] and that the distribution of random variable $Y$ given that $X=x$ is uniform on the interval $[1-x,1].$

Find the probability that $P(X+Y > \frac{3}{2})$ given to 4 decimal places.

So far this is what I have:

$P(X+Y > \frac{3}{2}) = 1 - P(X+Y < \frac{3}{2})= 1 - P(Y < \frac{3}{2} - X)$

$P(Y < \frac{3}{2} - X) = \int_0^1 \int_{1-x}^{\frac{3}{2}-x} f_{XY}(x,y) \,dy \,dx = \int_0^1 \int_{1-x}^{\frac{3}{2}-x} \frac{1}{x} \,dy \,dx $

$= \int_0^1 [\frac{y}{x}]_{1-x}^{\frac{3}{2}-x} \,dx = \int_0^1 \frac{1}{2x} \, dx = [\frac{1}{2}ln(2x)]_0^1$

As $ln(2x)$ can't be evaluated at $0$ I can't go any further.

Have I done anything wrong and what is a better method to answer this question so that this problem doesn't occur?

Note: I calculated $f_{XY}(x,y)$ from the distributions of $f_X(x)$ and $f_{Y|X}(y|x)$ as $f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x).$

答: Your entire joint distribution is supported on the unit square $[0,1]\times[0,1]$. You are interested in the probability of a $(X,Y)$ being on a corner of the square, namely $X+Y\ge 3/2$. The integration limits there are $1/2$ to $1$ in $X$, and $3/2-x$ to $1$ in the $Y$. This should solve your problem.


问:
Prove or disprove: for an odd $a\in\mathbb{Z}$, $1010∣a(2+3+5+⋯+p_{π(a)})$
I am looking for a way to prove or disprove that $1010∣a(2+3+5+⋯+p_{π(a)})$ for an odd natural number $a$. I haven't progressed much further.

答:
To summarize the discussion in the comments:

It is not clear what the question is asking. One interpretation follows from the remark that, for $a>1$, there are only finitely many values of $n$ for which $a$ might be the second largest divisor.  Indeed, the good $n$ are $\{2a, 3a, \cdots, p_{\pi(a)}a\}$  where, as usual, $\pi(a)$ denotes the number of primes less than or equal to  $a$ and, for $i\in \mathbb N$, $p_i$ denotes the $i^{th}$ prime.

We then interpret the question as:  Is there an odd number $a$ such that $1010$ divides the sum $a(2+\cdots +p_{\pi(a)})$?

The answer to that is yes.

To construct an example we consider the odd multiples of $505$.  Indeed, for such $a$, we only need the sum of the primes to be even.  That is equivalent to requiring that $\pi(a)$ be odd and we check (numerically) that this is the case for $a=5\times 505$.  Of course, there may be a smaller value of $a$ that works, but any example suffices to prove existence.

Note on the calculation:  I used Wolfram Alpha, but the numbers here are small enough that a table of the primes would suffice.  Note that $\pi(5\times 505)=369$ so your table need only cover the first few hundred primes.

I stress that we are not sure we are reading the problem correctly, but the above interpretation and calculation represents our best guess.

Several users were involved in this work, notably @player3236 and @BarryCipra.  I will, accordingly, post this as a Community solution.


答:
I get
\begin{eqnarray}
P[X+Y > {3 \over 2}] &=& \int_{x={1 \over 2}}^1 \int_{y={3 \over 2}-x}^1 {1 \over x} dy dx \\
&=& \int_{x={1 \over 2}}^1   (1-{1 \over 2x}) dx \\
&=& (x-{ 1\over 2} \ln x) \mid_{1 \over 2}^1 \\
&=& {1 \over 2}(1+\ln {1 \over 2}) \\
&\approx& 0.1534
\end{eqnarray}

Where did the bounds come from?

First, note that the distribution is supported in $[0,1]^2$.
So we are interested in the set
$D=\{(x,y)| x+y > {3 \over 2} \} \cap [0,1]^2$.

Note that if $x \le {1 \over 2}$, then $x+y \le {3 \over 2}$ so
we can write
$D = \{(x,y)| x+y > {3 \over 2} , x \in ({1 \over 2}, 1], y \in [0,1] \} = \{(x,y)| x+y > {3 \over 2} , x \in ({1 \over 2}, 1], y \in ({3 \over 2} -x,1] \} $.

Just as a check, if $x \in ({1 \over 2}, 1]$ and $y \in ({3 \over 2} -x,1]$ then $y \in (1-x,1]$, so for $(x,y) \in D$, the partial pdf for $Y$ is $ {1 \over 1-(1-x)} = { 1\over x}$ and the partial for $X$ is $1$.

Title: Find $P(X+Y > \frac{3}{2})$ given that X is uniformly distributed on [0,1] and the conditional distribution of Y given X=x is uniform on [1-x,1].

Body:
Suppose $X$ has the uniform distribution on the interval [0,1] and that the distribution of random variable $Y$ given that $X=x$ is uniform on the interval $[1-x,1].$

Find the probability that $P(X+Y > \frac{3}{2})$ given to 4 decimal places.

So far this is what I have:

$P(X+Y > \frac{3}{2}) = 1 - P(X+Y < \frac{3}{2})= 1 - P(Y < \frac{3}{2} - X)$

$P(Y < \frac{3}{2} - X) = \int_0^1 \int_{1-x}^{\frac{3}{2}-x} f_{XY}(x,y) \,dy \,dx = \int_0^1 \int_{1-x}^{\frac{3}{2}-x} \frac{1}{x} \,dy \,dx $

$= \int_0^1 [\frac{y}{x}]_{1-x}^{\frac{3}{2}-x} \,dx = \int_0^1 \frac{1}{2x} \, dx = [\frac{1}{2}ln(2x)]_0^1$

As $ln(2x)$ can't be evaluated at $0$ I can't go any further.

Have I done anything wrong and what is a better method to answer this question so that this problem doesn't occur?

Note: I calculated $f_{XY}(x,y)$ from the distributions of $f_X(x)$ and $f_{Y|X}(y|x)$ as $f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x).$


问:
Definite integral of $\frac{1}{a\sin x+b}$ with upper limit $\pi$
I am really confused about the following, I hope it's not a stupid question with an obvious answer.

I want to solve the definite integral

$$\int\limits_c^\pi\frac{1}{a\sin(x)+b}dx$$

with $a>0$, $b>0$, $b>a$, $\pi/2\leq c <\pi$. Computer algebra systems (CAS) such as Mathematica and Matlab (symbolic math toolbox) can only solve the definite integral when the stated assumptions are provided, otherwise they don't yield a solution. (This is a general question about math, not the implementation details of CAS).

This seems weird to me, because the indefinite integral calculates to

$$\int\frac{1}{a\sin(x)+b}dx=\frac{2 \arctan\left(\frac{a+b \tan \left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}},$$
and what is most important about this result, as far as I know, is that it holds regardless of any constraints on the constants. (Apart from the fact that the term inside the square root has to be $>0$ in order to avoid a complex solution). So what I'd do next is to insert my limits $c$ and $\pi$ into the antiderivative, but it's obvious that inserting $x=\pi$ into this antiderivative gives us issues, because $\tan(\pi/2)$ is undefined.

My questions are: why can the definite integral be solved under the stated constraints $a>0$, $b>0$, $b>a$, $\pi/2\leq c <\pi$, and how does one arrive at this solution, given that the antiderivative seems undefined at $x=\pi$?

A few thoughts:

Is there a different way to arrive at an equivalent formulation of the antiderivative which doesn't have this issue, either by performing the antiderivative calculation differently, or maybe by somehow simplifying $\arctan(d+e\tan(x/2))$ after it's calculated? After all, $\arctan$ is just the inverse function of $\tan$.

Maybe we can work with the antiderivative provided above, knowing that because of the fact that the lower integration limit $c$ is smaller than $\pi$ we approach the upper limit $\pi$ from below, so we know that $\tan(\pi/2)$ equals infinity and we're basically calculating $\arctan(\infty)$, which is $\pi/2$? That seems quite hand-waving to me, though. But at least that seems to yield the correct result for the upper limit, by the way.

(Why is the antiderivative undefined at all at some arguments? The original function $1/(a\sin(x)+b)$ with the stated assumptions is always $>0$ and continuous, so it's not obvious to me why it should not be possible to calculate the area under the curve within any interval.)

Thank you for your help, any insight is greatly appreciated!

答:
While, $F(x) = \frac{2 \arctan\left(\frac{a+b \tan \left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}}
$  may not be defined at $x = \pi, \lim_\limits{x\to\pi} F(x)$ is defined and we can use that to evaluate the function at the upper bound.


问:
is this solution of $\int_0^\infty \sin(z^2) dz $ valid?
Is this method valid?

We have $\displaystyle \int_0^\infty e^{-u^2}du = \frac{\sqrt \pi}{2}$.

Let $u = \frac{1+i}{\sqrt 2}x$, we get $\displaystyle \frac{1 + i}{\sqrt 2 }\int_0^{\infty }e^{-i x^2} dx = \frac{\sqrt \pi }{2 } \implies \int_0^\infty e^{-ix^2}dx = \frac{1-i}{2\sqrt 2 } \sqrt{\pi}$.

Comparing real and imaginary parts we get $\displaystyle \int_0^{\infty}\sin(x^2)dx = \int_0^{\infty}\cos(x^2)dx = \frac{\sqrt \pi }{2 \sqrt 2}$.

答:
The method works, but in present form it lacks justification. The key step is to prove that

the integral of $\exp(-z^2)$ along the circular arc from $R$ to $Re^{\pi i/4}$ tends to $0$ as $R\to \infty$.

Once you have this, the Cauchy integral theorem tells you that
$$\int_0^R \exp(-(re^{\pi i /4})^2)\,e^{\pi i/4}dr - \int_0^R \exp(-r^2)\,dr \to 0$$
as $  R\to\infty $. It follows that
$$\int_0^\infty \exp(-r^2 i )\,dr =  \frac{\sqrt{\pi}}{2}e^{-\pi i /4}$$
which is the desired result.

It remains to prove the statement emphasized above. First,
$$\left| \int_0^{\pi/4} \exp(-R^2 e^{2i\theta} )\,R\,d\theta \right| \le
R\int_0^{\pi/4} \exp(-R^2 \cos 2\theta ) \,d\theta
\tag2$$
Next,  split the interval of integration into two subintervals at $\pi/4-R^{-3/2}$. The short interval contributes at most $R\cdot R^{-3/2}=R^{-1/2}$. On the long one the integrand is bounded by $\exp(-R^2 \cos (\pi/2-2R^{-3/2}) )$ where $$-R^2 \cos (\pi/2-2R^{-3/2})= - R^2 \sin (2R^{-3/2})\sim -2R^{1/2}$$
and of course $R\exp(-2R^{1/2})\to 0$ as $R\to \infty$.


问:
Finding the exact area bound between the curves $kx^2$, and $\sqrt {kx}$, where $K$ is not given.

I attempted to solve this question as you can see above (in the second image) by integrating the two functions from 0, where the area begins, and then the point where they equal each other, point Q. However I have absolutely no idea if this is the correct way to go about it since I cant find any similar worked questions in my textbooks or on the internet.

Did I completely go about this the wrong way? Have I written down the question wrong and its actually just unsolvable? I am not sure, and I would appreciate some help.

答:
1st of all the subtraction is in the wrong direction because for small numbers $x^2$ is less than $\sqrt{x}$.  Also you need to integrate to the upper bound Q.  You can solve for Q based on k by solving for the value of x where they cross.

$$kx^2 = \sqrt{kx}$$

$$kx^2 = \sqrt{k}\sqrt{x}$$
$$x^{3/2} = \frac{\sqrt{k}}{k} = \frac{1}{\sqrt{k}}$$
$$ Q = x = (k^{-1/2})^{2/3} = k^{-1/3}$$

Now just finish by doing the integrals.


问:
Show $ρ( A_i \bigcup B_j ) = ρ( A_i - B_j ) + ρ(B_j - A_i ) + ρ( A_i \bigcap B_j ) $ finite
If  $\mu$ and $\nu$  are $\sigma$-finite measures on $(\mathscr \Omega , \mathscr F )$
and   $\nu$ is absolutely continuous with respect to $\mu$ : ( $\nu << \mu $ )

and let $ ρ  = \mu + \nu $

I want to prove that $ρ$ is  $\sigma$-finite and that  $\nu << ρ $ :

I tried the below:

$\text {  There exists   } \{A_{n}\}   \text { , such that  }\mu(A_n)<\infty  \text { and } \bigcup_{n=1}^\infty A_n = \Omega $ .

$\text { Also}, \text {  There exists   } \{B_{n}\}   \text { , such that  }\nu(B_n)<\infty  \text { and } \bigcup_{n=1}^\infty B_n = \Omega $ .

$\text { Then , for } \bigcup_{i=1}^\infty \bigcup_{j=1}^\infty (A_i \bigcup B_j)   = \Omega $ ,

$ρ( A_i \bigcup B_j ) = ρ( A_i - B_j ) +  ρ(B_j -  A_i )   + ρ( A_i \bigcap B_j ) $

$\text {at this point , it is enough to show that each of the above terms are finite ,
             because that would mean } ρ( A_i \bigcup B_j )<\infty   \text {  so ,   ρ is }  \sigma-finite. $

How can we show that$\ ρ( A_i - B_j ) \text { and }  ρ(B_j -  A_i ) \text { are finite} $?

答:
I understand that $\mu$ and $\nu$ are non-negative measures.

Actually, you don't need the condition  $\nu \ll \mu $.

Let $\mu$ and $\nu$  are $\sigma$-finite measures on $(\mathscr \Omega , \mathscr F )$ and let $ \rho = \mu + \nu $. Then $\rho$ is $\sigma$-finite and $\mu \ll \rho$ and $\nu \ll \rho$.

Proof:
Since $\mu$ is a $\sigma$-finite measure, there are $\{A_{n}\}_n $ such that $\mu(A_n)<\infty$ and $\bigcup_{n=0}^\infty A_n = \Omega $.

Since $\nu$ is a $\sigma$-finite measure, there are $\{B_{n}\}_n $ such that $\nu(B_n)<\infty$ and $\bigcup_{n=0}^\infty B_n = \Omega $.

Now consider $\{A_i\cap B_j : i, j \in \mathbb{N}\}$. This is a countable collection of set in $\mathscr F $, so we can enumerate it as $\{E_n\}_n$. Then we have:

For all $n \in \mathbb{N}$, there are $i, j \in \mathbb{N}$ such that $E_n=A_i\cap B_j$ and so
$$\rho(En) = \mu(A_i\cap B_j) + \nu(A_i\cap B_j) \leqslant  \mu(A_i) + \nu(B_j) <+\infty$$

and
$$ \bigcup_{n=0}^\infty E_n  = \bigcup_{i,j=0}^\infty (A_i\cap B_j)= \bigcup_{i=0}^\infty \bigcup_{j=0}^\infty (A_i\cap B_j)= \bigcup_{i=0}^\infty \left [\bigcup_{j=0}^\infty (A_i\cap B_j) \right ]= \bigcup_{i=0}^\infty (A_i \cap \Omega)  = \Omega$$

So $\rho$ is $\sigma$-finite.

Now, for all $E \in \mathscr F$, if $\rho(E)=0$ then $\mu(E)+\nu(E)=0$ and so $\mu(E)=0$ and $\nu(E)=0$. So, $\mu \ll \rho$ and $\nu \ll \rho$.


问:
Factoring $5x^4-10x^2+1$ given its roots
After a long and tedious calculation I found the roots of $5x^4-10x^2+1$ to be $\sqrt{1-\frac{2}{\sqrt{5}}}$, $-\sqrt{1-\frac{2}{\sqrt{5}}}$, $\sqrt{1+\frac{2}{\sqrt{5}}}$ and $-\sqrt{1+\frac{2}{\sqrt{5}}}$. But now how can I factor the polynomial with those roots ?

答:
Let's call the roots $r_1, r_2, r_3, r_4$. Since all roots are distinct and the polynomial is of degree four, we know that $(x - r_i)$ is a factor of the polynomial for every $i$. If we just multiply those factors we get a monic polynomial, so we just need to scale it to match the leading coefficient of the original, and we get

$$p(x) = 5(x-r_1)(x-r_2)(x-r_3)(x-r_4).$$

If you would have had less roots than the degree of the polynomial, you'd have to find the multiplicity of each of them, and there are various ways to do that.


答:
A polynomial can always be factored with respect to its roots as
$$a_n x^n + \cdots + a_0 = A(x-p_1)(x-p_2)\cdots(x-p_n)$$
where $a_i$ are the coefficients and $p_i$ are the roots.

What must $A$ be? Note that $A$ is the coefficient of the term $\overbrace{x\cdot x\cdots x}^{n\text{ times}}=x^n$, so $A=a_n$.

Title: Factoring $5x^4-10x^2+1$ given its roots

Body: After a long and tedious calculation I found the roots of $5x^4-10x^2+1$ to be $\sqrt{1-\frac{2}{\sqrt{5}}}$, $-\sqrt{1-\frac{2}{\sqrt{5}}}$, $\sqrt{1+\frac{2}{\sqrt{5}}}$ and $-\sqrt{1+\frac{2}{\sqrt{5}}}$. But now how can I factor the polynomial with those roots ?


问:
How to find if a sequence is converging or diverging?
The question is as follows: Determine if the sequence:

$$\{1/1,\ 1/3,\ 1/2,\ 1/4,\ 1/3,\ 1/5,\ 1/4,\ 1/6,\ \ldots\}$$

diverges or converges. If it is convergent, find the limit.

I am not sure where to begin, I'd appreciate it if anybody could help me on this one.

Thankyou

答:
As noted above, for all $n \in \mathbb N$:

$$a_{2n}=\frac1{n+1} \qquad \text{and}\qquad a_{2n+1}=\frac1{n+3}.$$

Or equivalently:

$$a_n = \left. \begin{cases}
\frac 1 {1+n/2} = \frac 2 {n+2},    &\text{if } n \text{ even}, \\
\frac 1 {3+\frac{n-1}2} = \frac 2 {n+4},  &\text{if } n \text{ odd}
\\\end{cases} \right\}\leq \frac2n.$$

(We just cut off the extra bit on the denominator.)

I.e., for all $n \in \mathbb N$, $$ 0 \le a_n \le \frac 2 n$$ and the squeeze theorem finishes it.


Body:
hint

With $ t=x^2 $, the polynom becomes

$$5x^4-10x^2+1=$$
$$5t^2-10t+1=$$

$$5\Bigl(t-(1+\frac{1}{\sqrt{5}})\Bigr)\Bigl(t-(1-\frac{1}{\sqrt{5}})\Bigr)$$

Replace $ t $ by $ x^2 $ and use
$$x^2-a^2=(x-a)(x+a)$$


问:
Infinite sum of semisimple Ideals of a Lie algebra
Following the paper by Kubok (Infinite-dimensional Lie algebras with null Jacobson radical), we have the following definitions. Let $L$ be a Lie algebra. If every finite subset of $L$ is contained in a finite-dimensional subalgebra (resp. a solvable subalgebra) of $L$ then $L$ is called locally finite (resp. locally solvable). For a locally finite Lie algebra $L$, $\sigma(L)$ is the maximal locally
solvable ideal of $L$ and $L$ is said to be semisimple if $\sigma(L)=0$.

Suppose that $L$ is an arbitrary Lie algebra over a field of characteristic zero and that $L=\sum_{i\in\mathcal I}L_i$, where each $L_i$ is a finite-dimensional semisimple ideal of $L$.

If $\mathcal I$ is finite then $L$ is semisimple. In order to show this, it is enough to verify the case $L=L_1+L_2$ and apply induction. As $L_1$ is semisimple the Killing form $\kappa(\cdot,\cdot)$ of $L$ restrict $L_1$ is non-degenerated and thus $L=L_1\oplus L_1^\perp$, where $L_1^\perp=\{x\in L\mid \kappa(x,y)=0, \mbox{ for } y\in L_1 \}$. Now, since $L=L_1+L_2$ it follows that
$$L_1^\perp \cong L/L_1=(L_1+L_2)/L_1\cong L_2/(L_1\cap L_2)$$is semisimple. Hence $L=L_1\oplus L_1^\perp$ is semisimple.

But, what happens if $\mathcal I$ is infinite? Is $L$ semisimple? I believe that the answer is no, but I can't find a counterexample. Is there some condition on $L$ or $L_i$ that guarantees the semisimplicity of $L$?

I appreciate any help.

答:
Each $L_i$, being finite-dimensional semisimple, is a finite direct sum of simple ideals, say

$$L_i = \bigoplus_{j \in J(i)} S_j$$

where for each $i \in \mathcal I$, $J(i)$ is some finite set.

Now if $A, B$ are simple ideals of any Lie algebra, then either $A \cap B = \{0\}$ or $A=B$.

Consequently, we can w.l.o.g. assume that $S_j \cap S_k =\{0\}$ if $j \in J(i_1)$ and $k \in J(i_2)$ for $i_1 \neq i_2 \in \mathcal I$, consequently $L_{i_1} \cap L_{i_2} = \{0\}$ for all $i_1 \neq i_2 \in \mathcal I$ and we have

$$\displaystyle L = \sum L_i = \bigoplus L_i= \bigoplus_{i \in I} \bigoplus_{j \in J(i)} S_j = \bigoplus_{j\in \coprod_{i\in \mathcal I} J(i)} S_j$$

with all the $S_j$ being simple (and finite-dimensional) ideals and the direct sum being one of Lie algebras. Frome here it should be easy to describe all ideals of $L$ and in particular see that the only solvable ideal of $L$, hence $\sigma(L)$, is zero.


问:
Can countability coexist with infinity?
This question concerns the countability of the real numbers.
First I will show how I count the numbers between 0 and 1 on the real line.
It is done by reversing digits behind the coma, so
that e.g. 0,761 maps to 167. Obviously this is 1 to 1 mapping,
but there are infinite number of those unique mappings depending on the chosen reasonable radix.
In decimal number system I could count numbers like this:

$
0 , 0.1 , 0.2 , 0.3 ... 0.9 , \\
0.01 , 0.11 , 0.21 , 0.31 , ... , 0.91 , \\
0.02 , 0.12 , 0.22 , 0.32 , ... , 0.92 , \\
0.03 , 0.13 , 0.23 , 0.33 , \dots , 0.93 , \\
\vdots \\
0.09 , 0.19 , 0.29 , 0.39 , \dots , 0.99 , \\
0.001 , 0.101 , 0.201 , 0.301 , \dots , 0.901 , \\
\vdots \\
0.002 , \dots , 0.902, \\
\vdots \\
0.092, \dots , 0.992 , \\
0.003 , \dots , 0.903 , \\
\vdots \\
0.004, \dots , 0,904,\\
\vdots \\
\infty \\
$

Now, given that I can "succeed" to count to infinity, I would also count all irrational numbers.
There is no reason to haste. But then all irrational numbers are "somewhere" in the infinity.
So either counting to infinity allows me to write irrational numbers backwards, or infinity and countability can not coexist.
Which one is true ? How does your solution compare to rational numbers ?

答:
You have only counted the numbers whose decimal expansion is finite. This covers absolutely no irrational number, and in fact not even all the rational numbers as well.

Furthermore there is absolutely no reason to expect that this sort of process is continuous. That is to say, the set of finite strings of integers is countable, but the set of infinite strings is uncountable. Even if you can approximate one infinite string by its finite initial segments with a countable process, you can't approximate all the infinite strings at the same time with finite approximation with a countable process.

Title: Can countability coexist with infinity?


问:
Show that the set $A$ is inductive
Prove that for $m,n\in \mathbb{N}$, $m>n$ it holds that $m-n\in \mathbb{N}$.

Hint: Consider $A=\left \{n\in \mathbb{N}\mid \forall m\in \mathbb{N}, m>n:m-n \in \mathbb{N}\right \}$ and show that $A$ is inductive.

$$$$

I have done the following:

$1\in A$ : Let $m > 1$ and $m-1=r \Rightarrow m = 1 + r$ for some $r \in \mathbb{N}$ . We have that since $\mathbb{N}$ is an inductive set and $r\in \mathbb{N}$ then $r+1\in \mathbb{N}$, i.e. $m\in \mathbb{N}$.

Let $n\in A$. Then let $m > n$ and $m = n + r$ for some $r \in \mathbb{N}$ then $m \in \mathbb{N}$. We want to prove that $n+1\in A$, i.e. that if $m' > n+1$ and $m' = (n+1) + r$ for some $r \in \mathbb{N}$ then $m'\in \mathbb{N}$.

From $m'>n+1$ we have that $m'-1>n$. So we get that $m'-1\in A$. Then $m'-1=n+r$ for some $r\in \mathbb{N}$, so $m'=(n+1)+r$, for the same $r\in \mathbb{N}$ and so $n+1\in A$.

$$$$

Is that correct? Or am I confusing the meaning of the set $A$ ?

答:
You seem to be confusing the point is to prove $k\in A$ you must prove for every natural $m > k$ than $m-k = r$ or $m = r+k$ then you must prove that it is !!$r$!!! that is a natural number; not !!$m$!! that is a natural number.

So to prove $1 \in A$ you must prove that if $m > 1$ and $m\in \mathbb N$ that $m-1\in \mathbb N$.  By definition if $m\in \mathbb N$ then either $m = 1$ or $m$ is the successor of a natural number $r$.  That is $m = r+1$, for some $r\in \mathbb N$.

And as $m$ isn't equal to $1$ then $m$ is a successor of some $r$.  And $m-1 = r\in \mathbb N$.  So that proves that $m-1 \in \mathbb N$ for all $m \in \mathbb N; m > 1$ and therefore than $1\in A$.

And to prove $n\in A\implies n+1 \in A$, you know that $n\in A$ so for all $m> n$ then $m-n \in \mathbb N$.  So for all $m > n+1$ (no need to use a second variable-- we never took a specific $m$; we said all $m > n$ and now we are refering to all $m > n+1$. we have by trasitivity that $m > n$ therefore $m-n \in \mathbb N$.  So either $m-n = 1$ or $m-n$ is a successor to some $r \in \mathbb N$. That is $m-n = r+1$.

If $m-n= 1$ then $m=n+1$ and we do not have that $m > n+1$.  So $m-n \ne 1$.  So $m-n=1 + r$ for some $r\in \mathbb N$.  So $m -n - 1 = m-(n+1) = r\in \mathbb R$. So for any $m > n+1$ we have shown that $m-(n+1) \in \mathbb N$.  Therefore $n+1 \in A$.

So $A$ is inductive.

So as $1 \in A$ and $A$ is inductiove then $\mathbb N\subset A$.

So for any $n \in \mathbb M$ and any $m > n$ then $n \in A$ so $m-n \in \mathbb N$.

And that proves our statement.

Title: Show that the set $A$ is inductive


问:
If angles $a+b+c=\pi$ radians then $\sin (a)\leq \sin (b) + \sin (c)$
If angles $a+b+c=\pi$ radians then $\sin (a)\leq \sin (b) + \sin (c)$

How can I prove that? I know that angle $b+c=d $ if $d $ is the exterior angle of a and $a<d$

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain why you are stuck.)

Hint: $\sin a = \sin (b+c) = \ldots  \leq \sin b + \sin c$.

Use the fact that $ \sin b, \sin c \geq 0$, and $ \cos b, \cos c \leq 1$.


答:
I think it's easiest to write this as a double integral. So,  first you need to find the points where $kx^2 = \sqrt{kx}$. Square both sides and factorise to arrive at $kx(kx^3 - 1) = 0$, which has solutions when $x = 0$ and when $x = k^{-1/3}$. Then write your region in the $x-y$ plane as

$$
D = \{ (x,y)\in \mathbb{R}^2 : 0 \leq x \leq k^{-1/3}, kx^2 \leq y  \leq \sqrt{kx}\}
$$

Now you can write your area as

$$
R = \int\int_D dxdy = \int_0^{k^{-1/3}}\int_{kx^2}^{\sqrt{kx}}dy dx  = \int_0^{k^{-1/3}}\sqrt{kx} -kx^2dx = \frac{1}{3}
$$
I assume you know how to solve the last integral to get $\frac{1}{3}$.

So we see in the end that the answer is independent of $k$ for all $k>0$, perhaps surprisingly.

Title: Finding the exact area bound between the curves $kx^2$, and $\sqrt {kx}$, where $K$ is not given.

Body:
I attempted to solve this question as you can see above (in the second image) by integrating the two functions from 0, where the area begins, and then the point where they equal each other, point Q. However I have absolutely no idea if this is the correct way to go about it since I cant find any similar worked questions in my textbooks or on the internet.

Did I completely go about this the wrong way? Have I written down the question wrong and its actually just unsolvable? I am not sure, and I would appreciate some help.


答:
I would go about the interval bounderies in a different way.
First, notice that $k$ must be non negative, otherwise $g(x)$ is not well defind.

Next, notice that $f(0) = k \cdot 0^2 = 0$ and $g(0) = \sqrt{k \cdot 0} = 0$.
Moreover, both functions are strictly montone increasing on $[0, \infty)$. Thus, we can easily calculate their intersect in terms of $k$. Notice, that finding $Q$ was given as a hint in the exercise itself.
Finding the intersect:
$$f(x) = g(x) \Leftrightarrow kx^2 = \sqrt{kx} \Leftrightarrow k^2x^4 = kx \Leftrightarrow x^3 = \frac{1}{k} \Leftrightarrow x = (\frac{1}{k})^{\frac{1}{3}}$$

Now, notice that on the interval $(0, (\frac{1}{k})^{\frac{1}{3}})$ $g(x) > f(x)$

Thus, to calculate your area, set up your integrals as follows:

$$\int_0^{(\frac{1}{k})^{\frac{1}{3}}} g(x)dx - \int_0^{(\frac{1}{k})^{\frac{1}{3}}} f(x)dx$$

Now, simply integrate and you recieve your area.


答:
Just out of curiosity I wrote a program to find the smallest $a$. It's very little work because I answered another question a few days ago and it also involves Prime Sums and a bit programming.

Anyway here's the program and the output:

```<code>is.prime <- function(n) n == 2L || all(n %% 2L:max(2,floor(sqrt(n))) != 0)

N=10^4;
M=1010;
temp=rep(NA,N)

for(i in 1:N) temp[i]=as.integer(is.prime(i));

temp[1]=0; # exclude 1
pn=which(temp==1); # prime number list

pcf=rep(NA,N); # prime counting function
for(i in 1:N) pcf[i] = sum(pn<=i)

pnr=pn %% M; # avoid overflow

S = cumsum(pnr); # cumulative sum

P = rep(NA, N);
P[1]=1;
for(a in 2:N)
{
    P[a]=a*S[pcf[a]];
}
indx = which((P %% M==0))
indx[which(indx%%2==1)]
</code>```

[1] 2525 4545 4949 6363 7575 8181 8585 9191 9805 9975 9985 9995

You can see that $2525=5\cdot 505$ is indeed the smallest, followed by another multiple of $505$.

Title: Prove or disprove: for an odd $a\in\mathbb{Z}$, $1010∣a(2+3+5+⋯+p_{π(a)})$

Body: I am looking for a way to prove or disprove that $1010∣a(2+3+5+⋯+p_{π(a)})$ for an odd natural number $a$. I haven't progressed much further.

Body:
Just out of curiosity I wrote a program to find the smallest $a$. It's very little work because I answered another question a few days ago and it also involves Prime Sums and a bit programming.

Anyway here's the program and the output:

is.prime <- function(n) n == 2L || all(n %% 2L:max(2,floor(sqrt(n))) != 0)

N=10^4;
M=1010;
temp=rep(NA,N)

for(i in 1:N) temp[i]=as.integer(is.prime(i));

temp[1]=0; # exclude 1
pn=which(temp==1); # prime number list

pcf=rep(NA,N); # prime counting function
for(i in 1:N) pcf[i] = sum(pn<=i)

pnr=pn %% M; # avoid overflow

S = cumsum(pnr); # cumulative sum

P = rep(NA, N);
P[1]=1;
for(a in 2:N)
{
    P[a]=a*S[pcf[a]];
}
indx = which((P %% M==0))
indx[which(indx%%2==1)]

[1] 2525 4545 4949 6363 7575 8181 8585 9191 9805 9975 9985 9995

You can see that $2525=5\cdot 505$ is indeed the smallest, followed by another multiple of $505$.


问:
Is $X = (n-1)^n + n$ always composite for $n \geq 4 \in \Bbb Z$?
$$\textit{Proposition}: \text{Given} \ X = (n-1)^n + n,\ \forall \ n \in \Bbb Z, \ n \geq 4, X \ \text{is composite.}$$

This came up sort of at random on a mailing list and it's jammed its way into being stuck in my brain. It's easy to test that it's true for the first few hundred thousand integers, suggesting it's likely true.

There are a few things that are easy to prove:

$X$ is always odd: if $(n-1)^n$ is odd, $n$ is even and vice versa.

$(n - 1) \equiv -1 \pmod n  $. Since $n \equiv 0 \pmod n$, we can see that $X \equiv (-1)^n \pmod n $. Therefore, $X \equiv -1 \pmod n$ for odd $n$, and $X \equiv 1 \pmod n $ for even $n$.

If $n \equiv 2 \pmod 3$, then $n-1 \equiv 1 \pmod 3$, so $(n-1)^n \equiv 1 \pmod 3$. If we add $n \equiv 2 \pmod 3$, we have $0 \pmod 3$, so $3 \mid X$. So a third of possible values for $n$ are easily shown to be composite, namely $n = 3m + 2$.

If $n \equiv \{4,13\} \pmod {20}$, then $5 \mid X$, as follows:

If $n \equiv 4 \pmod {20}$ , then $4 \mid n$, $n \equiv 4 \pmod{10}$, and $(n - 1) \equiv 3 \pmod{10}$. Then we have $4^3+4 \equiv 5 \pmod{10}$

If $n \equiv 13 \pmod {20}$, then $4 \mid (n-1)$, $(n-1) \equiv 12 \pmod{20}$, and $(n-1) \equiv 2 \pmod {10}$.

Additionally, $n \equiv 1 \pmod 4$ and $n \equiv 3 \pmod{10}$.

Therefore, in decimal, $(10k + 2)^{4m+1} \equiv 2 \pmod{10}$. Adding $n \equiv 3 \pmod{10}$ gives $X \equiv 5 \pmod{10}$.

That takes care of 40% of the integers. But (1) I feel like we ought not to have to go through the integers modulus by modulus, and (2) well, we have a lot of integers left.

Some other interesting tidbits that may or may not help:

For $n$ even and $n \not\equiv 2 \pmod 3$, $X$ is often divisible by $(n+1)$, or by $p$ if $(n+1) = p^2$, $p$ prime. Up to $n = 72$, this fails for $n = 34, 54, 64$.

For  $n \not\equiv 2 \pmod 3$, $X$ is sometimes divisible by $(2n-1)$. Again up to $n = 72$, this is true for even $n = 6, 10, 22, 30, 34, 42, 54, 66, 70$. (Interesting, for all of those, $n \equiv 2 \pmod 4$. Just noticing that.) It's true for odd $n = 9, 21, 37, 45, 49, 57$... which are all $n \equiv 1 \pmod 4$.

Of course, being sometimes divisible by those numbers doesn't tell us why they're divisible by them. I tried using partial binomial expansions (learning about Stirling numbers along the way) but couldn't get anything useful. Has anyone seen or worked with this proposition before? Is it proven and I just can't find it anywhere?

答:
I wanted to follow up on this a bit as I've done some more... research? Trial and error? One or the other. In case anyone is still following this!

First, I want to acknowledge René Gy above, for this lovely use of Fermat's Little Theorem, showing that if $n + 1$ is prime, $n + 1 \mid X$:

$$ (n-1)^n + n = ((n+1) - 2)^{((n+1) - 1)} + (n+1) -1 $$

For $n+1 = p$ prime, this is:

$$ (p-2)^{p-1} + p - 1 \equiv (-2)^{p-1} - 1 \pmod p$$

If this expression is $0$, that is, if $(n+1) = p \mid X$--then we have

$$ (-2)^{p-1} \equiv 1 \pmod p$$

This is Fermat's Little Theorem, indicating that it is true for all primes coprime to $2$, that is, all odd primes. This clears up the abundance of $p+1$ in our factorizations, and also indicates that if $n+1$ is prime, then X is composite.

The abundance of divisibility by $2n -1$ is, I suspect, explainable in the same way. It looks like for all numbers of the form $4k+1$ or $4k+2$, $2n-1 \mid X$ iff $2n-1$ is prime.

I've also done some trial and error following up on the modulo $20$ results I described originally. My thought was "Hey, $20$ is a pronic number (i.e., $n(n+1)$), I wonder what I can do with other pronics!" And it turns out that for each pronic number where $n+1$ is prime, we can find a set of residues modulo $p(p-1)$ that are divisible by $p$. These will obviously include $p-1$, but each set includes quite a few more. I don't think listing all of them here is really necessary, but I'll list a few. Sage/Python code if you wish to mess around with it is here: https://pastebin.com/Y24BmLs6

Divisible by $p$ for $5 \leq p \leq 13$ with modulus $M$:

$$p = 5; M = 20 \implies 5 \mid \{\mathbf{4}, 13\} \pmod{20}\\
p = 7; M = 42 \implies 7 \mid \{\mathbf{6}, 25\} \pmod{42}\\
p = 11; M = 110 \implies 11 \mid \{6, \mathbf{10}, 18, 61, 65, 74, 97\} \pmod{110}\\
p = 13; M = 156 \implies 13 \mid \{\mathbf{12}, 85, 94, 95, 99\} \pmod{156}$$

Residues covered by $n = p-1$ are in bold. The abundance of residues a bit above half the modulus is intriguing, and continues for larger moduli.

Edit: OK, now this is the sort of pattern I hadn't expected to find at all...

All those cases where $n \equiv 2 \pmod 3$ and therefore we can just kinda skip them? I noticed that all of them have squares in their factorizations. But it turns out they're not just any squares, they're the squares of the [centered] hexagonal numbers $3m^2 + 3m + 1$. More specifically:

$$ k \in \mathbb{N}, n = 3k + 2  \iff (3k^2 + 3k + 1)^2 \mid X = (n-1)^n + n $$

So far as I can tell, (1) almost all other $X$ are squarefree (the exceptions being $6, 64, 73$ so far), and (2) I have no clue why or how or what's going on here. It's baffling.


问:
Possible Characterization of localization of $\mathbb{Q}[x, y]$
So I wish to know if it is possible to Characterize the localization of $\mathbb{Q}[x, y]$. The motivation behind this question is that I was asked to determine (True or false) that every subring of $\mathbb{Q}(x,y)$ containing $\mathbb{Q}[x, y]$ is some form of localization. I believe it to be false; and was hoping that some knowing some form of characterization can help me in coming up with a counter-example.

While I'm at it; any insights behind my question is appreciated.

Cheers

答:
Every nontrivial localization of a ring introduces new units. Your goal is to find a ring extension $\Bbb Q[x,y]\subset R \subset\Bbb Q(x,y)$ so that $R$ has no units other than $\Bbb Q$. Hint: blowups. A full solution is under the spoiler.

 Consider $\Bbb Q[x,y,x/y]$ - this is the coordinate algebra of one of the affine patches of the blowup of $\Bbb A_\Bbb Q^2$ at the origin. It fits the role of $R$ from before, but it has no new units: it's isomorphic to the polynomial ring on the variables $y,x/y$, and the only units in a polynomial ring are the invertible constants.


问:
Do uniformly continuous functions map complete sets to complete sets?
Let $f: (M, d) \rightarrow (N, \rho)$ be uniformly continuous.
Prove or disprove that if M is complete, then $f(M)$ is complete.

If I am asking a previously posted question, please accept my apologies and tell me to bugger off.  I saw a similar problem but the solution was dealing with a Bi-Lipschitz function or some such business.

I believe this statement to be true and here is a rough sketch of my reasoning:

Since $f$ is uniformly continuous, then $f$ maps Cauchy to Cauchy.
Let $(x_n)$ be a Cauchy sequence in $M$.  Since $M$ is complete, $x_n \rightarrow x \in M$.  Again, because of $f$'s uniform continuity, we now have $(f(x_n))$ is Cauchy in $N$ and $f(x_n) \rightarrow f(x) \in N$.  Thus $N$ is complete.

By the way, I am studying for an exam.  This is certainly not homework.  I gladly accept your criticisms.  Thank you in advance for your help.

答: Let $M:=\Bbb R$ and $N:=(-\pi/2,\pi/2)$ and $f:=\arctan$. This is uniformly continuous (and, also a homeomorphism), but $N$ is not complete.</p>


问:
rank of a matrix if one component of each row is a linear combination of the others
I am sure that this problem is really simple but now I can not see the solution. Let $A$ be a matrix $m\times m$ with full rank. Then, we change all the components $A(j,j)$ by $-\sum_{k,k\neq j} A(j,k)$, where $j$ is the row number. I have made simulations with random matrix and after the change the rank of the new matrix is $m-1$. Why?. I see that in every row we have a term that is a linear combination of the term from the other rows. But I do not see how to demonstrate the reduction of the rank.

答:
Denote column $i$ by $A_i$. The matrix has full rank, so it's columns are linearly independent. Namely $ \sum_i A_i \neq 0$, where $0$ is the zero vector.

Now for every diagonal entry $A(j,j)$, replace it with negative the sum of all other entries in that row, call this new matrix $B$.  Then clearly $ \sum_i B_i = 0$ (can you see why?).  This means that the columns are no longer linearly independent.  In other words, $A$ has reduced in rank when it becomes $B$:
$$ rank(B) < m . $$

To compute the rank exactly, we apply elementary column operations to kill the last column. Recall that the rank of a matrix is unchanged by elementary operations on rows or columns:
$$ \begin{cases} \mbox{Add column } 1 \mbox{ to column } m \\
 \mbox{Add column } 2 \mbox{ to column } m \\
 \dots \\
\mbox{Add column } m-1 \mbox{ to column } m \end{cases}$$

Check yourself that $B$ now has columns $B_1 , B_2 , B_{m-1}, 0 $.  But the first $m-1$ columns are linearly independent, since the columns of $A$ were linearly independent. To see this, assume there were coefficients $\beta_1, ..., \beta_{m-1}$, not all $0$, such that
$$ \sum_{i=1}^{m-1} \beta_i B_i = 0 .$$
This would mean that the first row of $B$ sums to $0$:
$$ \beta_1 B(1,1)  + \beta_2 B(1,2) + \dots + \beta_{m-1} B(1, m-1) = 0  $$
$$ \implies - \beta_1 \left( \sum_{k=2}^{m} A(1,k) \right)  + \beta_2 A(1,2) + \dots + \beta_{m-1} A(1, m-1) = 0  $$
$$ \implies  - \beta_1 A(1,m) + \sum_{k=1}^{m-1} \beta_k A(1,k) = 0 $$

By full rankness of $A$, this last equality is true if and only if the coefficients were all $0$, a contradiction.

Therefore the first $m-1$ columns of $B$ are linearly independent. Namely,
$$ rank(B)=m-1 .$$


问:
Integral of $\sin(x)/x$ from $-400$ to $400$.
I recently saw this on Glassdoor as an interview question.

Is there any way of performing this integral analytically? I cannot seem to do it by substitution or by parts. Given it asks for the definite integral, I assume it is possible?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{400}{\sin\pars{x} \over x}\,\dd x} =
\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x -
\int_{400}^{\infty}{\sin\pars{x} \over x}\,\dd x
\\[5mm] = &\
{\pi \over 2} - {\cos\pars{400} \over 400} +
\int_{400}^{\infty}{\cos\pars{x} \over x^{2}}\,\dd x
\\[5mm] = &\
\underbrace{{\pi \over 2} - {\cos\pars{400} \over 400} -
{\sin\pars{400} \over 160000}}
_{\ds{1.5721148\color{red}{8\ldots}}} +
2\int_{400}^{\infty}{\sin\pars{x} \over x^{3}}\,\dd x
\end{align}
$\ds{{\tt Mathematica}: 1.5721148\,692738117518013214479640847648
\ldots}$


问:
Showing that not every pair of points in a two sheet hyperboloid can be joined by a curve - solution verification & feedback
I'm working through Multivariable Mathematics by Shifrin currently on chapter 2 and I've encountered the following question:

I've done the first part and got that $X$ is a hyperboloid of one sheet and $Y$ is a hyperboloid of two sheets. The difficulty I've had is in coming up with a solution for part b). This is the thought process I've had up to this point:

Looking at the hyperboloid of two sheets you see there is a disconnect between two sets of points. In a previous question the idea of collinearity was used. So I thought to apply it here. To test out the idea I found two points one in the top sheet and one in the bottom sheet. After some algebra I defined:
$A = (0,\sqrt{3},2)$ to be the top sheet point and $B = (0,\sqrt{3},-2)$ to be the bottom sheet point (i.e $A \in X$ and $B \in Y$).

The idea of collinearity states that two points are collinear if there exists a scalar, call it $\lambda$ such that $\lambda A = B$. In this specific case that would mean:

$$\lambda (0,\sqrt{3},2) = (0,\sqrt{3},-2)$$

Where $|x| \leq \sqrt{3}$

I chose to solve this system in a matrix form:
$$
\begin{bmatrix}
0 & 0 \\
\sqrt{3} & \sqrt{3} \\
2 & -2 \\
\end{bmatrix}
\text{using row reduction for good measure we arrive at:}\
\begin{bmatrix}
0 & 0 \\
1 & 1 \\
0 & -4 \\
\end{bmatrix}
$$

Which implies that there does not exist a solution. Now I attempted to extend this idea to a general case of:

$$
\lambda\begin{bmatrix}
x \\
\pm \sqrt{(z^{2}-1)-x^{2}} \\
\pm \sqrt{x^{2}+y^{2}+1}  \\
\end{bmatrix}
=
\begin{bmatrix}
x \\
\pm \sqrt{(z^{2}-1)-x^{2}} \\
\pm \sqrt{x^{2}+y^{2}+1}  \\
\end{bmatrix}
$$

Whereby if I choose the 2nd components to have the same sign , the 3rd component on each side to have opposite signs and restrict $|x| \leq \sqrt{z^{2}-1}$, I'll have the sort of conditions to establish what I want. So that would give me a set up of:

$$
\lambda\begin{bmatrix}
x \\
 \sqrt{(z^{2}-1)-x^{2}} \\
 \sqrt{x^{2}+y^{2}+1}  \\
\end{bmatrix}
=
\begin{bmatrix}
x \\
 \sqrt{(z^{2}-1)-x^{2}} \\
- \sqrt{x^{2}+y^{2}+1}  \\
\end{bmatrix}
$$

My questions are the following:

If my presentation is right, how do I treat the argument in the general case? Here it is easy to see that the 3rd component is different but that is only "talking" about it. I haven't really argued it rigourously.

I feel there is an easier argument to the one I'm attempting to present what would it be if it is the case?

Just to confirm the understanding of collinear, if this wasn't a straight line would the same idea stand about collinearity? i.e instead of a "line" as a straight line the line could be any type of curve.

答: Collinearity only applies to straight lines, so there's a problem with your argument.  However, a much easier argument is available.  There are no points on the two-sheeted hyperbola with $z=0$.  You have found a point on the surface with $z>0$ and also one with $z<0$.  Argue that by the intermediate value theorem, any curve connecting these points must pass through $z=0$ and therefore cannot lie entirely on the surface.


问:
Break a stick into two pieces, and break the shorter one again, what's the pdf of the shortest pieces.
Break a stick of unit length at a uniformly chosen random point. Then take the shorter of the two pieces and break it into two again at a uniformly chosen random point. Let X denote the shortest of the final three pieces. Find the probability density function of X.

I let the shorter one on the first break be x, which should be between (0, $1/2$), so the pdf should be
$$f(x)=2, 0<y<\frac{1}{2}$$
then the conditional probability of $f(x|y)$ should be (since it is another uniform distribution)
$$f(x|y)=\frac{1}{y/2-0}=\frac{2}{y}$$
by applying conditional probability formular
$$f_X(x)=\int_0^{1/2}f_{X|Y}(x|y)f_Y{y}dy=\int_0^{1/2}\frac{2}{y}\times2dy=4ln(y)|^{1/2}_0$$

I stopped here since it is impossible.
Can someone help?

答: The last integral should be taken from $x$ to $1/2$, since $f(x \mid y)$ is zero when $x > y$.


问:
Why does the limit definition of e fail?
I'm trying to calculate the following limit:
$$\lim_{x\to +\infty} \frac{(1+\frac{1}{x})^{x^2}}{e^x}$$
I tried to use the fact that $e^x = \lim_{x\to +\infty}(1+\frac{1}{x})^{x}$, but this gives $\frac {e^x}{e^x} = 1,$ which is not the answer. WolframAlpha says the answer should be $\frac {1}{\sqrt{e}}$. I was able to get the right answer using Taylor series of $\ln(1+ \frac{1}{x})$, but I can't find a good reason why my first approach fails. Can someone please help? Thanks!

答:
hint

You get a false result because the exponent is $ x^2 $ , not $ x $.

$$(1+\frac 1x)^{x^2}=e^{x^2\ln(1+\frac 1x)}$$

$$=e^{x^2\Bigl(\frac 1x -\frac{1}{2x^2}(1+\epsilon(x))\Bigr)}$$

with $$\lim_{x\to+\infty}\epsilon(x)=0$$


问:
Chinese Remainder Theorem, what to do when no inverse?
I have the three equations:

Using the Chinese Remainder Theorem I have found that 10.15.84 are coprime, however they are not pairwise co-prime.

I have worked out the following:

However, none of these values have multiplicative inverses. Where do I go from here?

答:
I prefer to separate the system into the prime power factors in the first place.

$$x\equiv 3\pmod{10}\iff x\equiv 1\pmod{2}\text{ and } x\equiv 3\pmod{5}$$

$$x\equiv 8\hspace{3pt}(mod\hspace{3pt} 15) \iff x\equiv 2\hspace{3pt} (mod\hspace{3pt} 3) \text{ and }x\equiv 3 \hspace{3pt}(mod\hspace{3pt} 5)$$

$$x\equiv 5\hspace{3pt}(mod\hspace{3pt} 84) \iff x\equiv 2\hspace{3pt} (mod\hspace{3pt} 3) \text{, }x\equiv 1 \hspace{3pt}(mod\hspace{3pt} 4)\text{ and }x\equiv 5 \hspace{3pt}(mod\hspace{3pt} 7)$$

So, your initial system is equivalent to:

$$x\equiv 1\hspace{3pt} (mod\hspace{3pt} 4)$$
$$x\equiv 2\hspace{3pt} (mod\hspace{3pt} 3)$$
$$x\equiv 3 \hspace{3pt}(mod\hspace{3pt} 5)$$
$$x\equiv 5 \hspace{3pt}(mod\hspace{3pt} 7)$$


答:
The Chinese remainder theorem works for pairwise co-prime moduli,

and, as you noted, yours are not pairwise co-prime.  Now

$x\equiv3\bmod10\iff x\equiv1\bmod2$ and $x\equiv3\bmod5$,

$x\equiv8\bmod15\iff x\equiv2\bmod3$ and $x\equiv3\bmod5$, and

$x\equiv5\bmod84\iff x\equiv2\bmod3, x\equiv1\bmod4$ and $x\equiv5\bmod7$.

Putting this all together, your system is satisfied when

$x\equiv3\bmod5, x\equiv2\bmod3, x\equiv5\bmod7, $ and $x\equiv1\bmod4$,

which you solved (per comments) using CRT to get $x\equiv173\bmod420$.

Title: Chinese Remainder Theorem, what to do when no inverse?

Body:
I have the three equations:

Using the Chinese Remainder Theorem I have found that 10.15.84 are coprime, however they are not pairwise co-prime.

I have worked out the following:

However, none of these values have multiplicative inverses. Where do I go from here?


问:
Power Series Without Differentiating
As the title suggests, I want to find a way to show that for $|x|<1: $ \begin{equation}
S=\sum\limits_{n=1}^{\infty}\frac{x^n}{n}=\ln\bigg(\frac{1}{1-x}\bigg) \end{equation}   without taking the derivative of the power series, or any calculus. I have tried for several hours and gotten nowhere. I've mainly tried transforming the equation into something of the form $S(x-1)$ which gives us \begin{equation}
S=\sum\limits_{n=2}^{\infty}\frac {n\cdot x^n}{(n-1)\cdot (x-1)}  - \sum\limits_{n=2}^{\infty}\frac{(n-1) \cdot x^n}{n\cdot (x-1)} - 1 -\frac 1{x-1} \end{equation}
Don't really know how useful this is (if I haven't messed up my algebra). Any ideas?

答:
The geometric series converges for $|x| < 1$, which is given by
\begin{align*}
\sum_{n=0}^{\infty}x^{n} = 1 + x + x^{2} + x^{3} + \ldots = \frac{1}{1-x}
\end{align*}
If we integrate both sides (which is possible due to the properties of real-analytic functions), one gets that
\begin{align*}
\ln\left(\frac{1}{1-x}\right) = -\ln(1-x)= \int_{0}^{x}\frac{\mathrm{d}z}{1-z} = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \ldots = \sum_{n=1}^{\infty}\frac{x^{n}}{n}
\end{align*}

Hopefully this helps!

Title: Power Series Without Differentiating


问:
Can Dirichlet characters mod q only agree on the same congruence class?
In my lecture notes for an analytic number theory module, the lecturer (I think) implicitly uses that, for $(a,q)=1$, if $\chi(a)=\chi(n)$ for all Dirichlet characters $\chi\mod q$, then $n\equiv a \mod q$. Looking at the table of the first few characters on Wikipedia, I can believe this is true. So, is this true? And if so, how would I prove it?

答:
In addition to @reuns' to-the-point answer, the is a slight generalization that perhaps shows what's really at play.

E.g., for a finite abelian group $A$, the collection $\mathbb C(A)$ of complex-valued functions on $A$ has basis consisting of the group homomorphisms $A\to \mathbb C^\times$.

And, part of the point (to my mind) is that there is a (reasonable) proof that does not depend on the structure theorem for finitely-generated abelian groups. Rather, we use the spectral theorem for mutually commuting unitary operators on finite-dimensional (complex) vector spaces. With these hints, it's not hard to contrive such a proof.


答:
The short answer is that $\lim_{x \to \infty} g(x, f(x)) \ne \lim_{x \to \infty}g (x, \lim_{x \to \infty} f(x))$. (Here, $g(x,y) = \frac{y^x}{e^x}$ and $f(x)=(1+1/x)^x$.) See grand_chat's answer for a very stark example of why taking limits of parts of the function before taking the limit as a whole is "illegal."

The long answer is that although $(1+1/x)^x$ does get close to $e$ as $x \to \infty$, you cannot simply replace it with $e$ (before taking the limit again) because you don't know how the discrepancy between $e$ and $(1+1/x)^x$ will interact with the power of $x$ on the outside and the rest of the expression.

$$((1+1/x)^x)^x = [e + \underbrace{((1+1/x)^x - e)}_{\text{error}}]^x$$

In principle, you have to understand the asymptotics of the error term. If the error vanishes extremely quickly, it might be ok to ignore it. However, as you see in this example, this [negative] error term is significant enough that it "slows down" the numerator to $e^{x-1/2}$ in some sense.

Long story short, make sure each step you take is a valid application of a limit property.

Title: Why does the limit definition of e fail?

Body:
I'm trying to calculate the following limit:
$$\lim_{x\to +\infty} \frac{(1+\frac{1}{x})^{x^2}}{e^x}$$
I tried to use the fact that $e^x = \lim_{x\to +\infty}(1+\frac{1}{x})^{x}$, but this gives $\frac {e^x}{e^x} = 1,$ which is not the answer. WolframAlpha says the answer should be $\frac {1}{\sqrt{e}}$. I was able to get the right answer using Taylor series of $\ln(1+ \frac{1}{x})$, but I can't find a good reason why my first approach fails. Can someone please help? Thanks!


答:
Take $f:\mathbb R\longrightarrow \mathbb R $, $f(x)=\arctan x$. Then
$$f(\mathbb R)=(-\frac\pi2,\frac\pi2)$$
and $f'(x)=\dfrac{1}{1+x^2}\leq 1$ which implies that $f$ is uniformly continuous.

However, $\mathbb R$ is complete, while $f(\mathbb R)=(-\frac\pi2,\frac\pi2)$ is not complete

Title: Do uniformly continuous functions map complete sets to complete sets?

Body:
Let $f: (M, d) \rightarrow (N, \rho)$ be uniformly continuous.
Prove or disprove that if M is complete, then $f(M)$ is complete.

If I am asking a previously posted question, please accept my apologies and tell me to bugger off.  I saw a similar problem but the solution was dealing with a Bi-Lipschitz function or some such business.

I believe this statement to be true and here is a rough sketch of my reasoning:

Since $f$ is uniformly continuous, then $f$ maps Cauchy to Cauchy.
Let $(x_n)$ be a Cauchy sequence in $M$.  Since $M$ is complete, $x_n \rightarrow x \in M$.  Again, because of $f$'s uniform continuity, we now have $(f(x_n))$ is Cauchy in $N$ and $f(x_n) \rightarrow f(x) \in N$.  Thus $N$ is complete.

By the way, I am studying for an exam.  This is certainly not homework.  I gladly accept your criticisms.  Thank you in advance for your help.


问:
How to prove gradient with exponent
To be honest I'm stucking as far as proofs are concerned so I hope you were to push me a little.

I got to show:

$\quad f(h x) = h^k  f(x)  \quad \Leftrightarrow \quad \nabla f(x) \cdot x = k  f(x)$

(Given: $f: \mathbb{R}^n \rightarrow \mathbb{R},\quad x \in \mathbb{R}^n,\quad k\in\mathbb{R},\quad h\in\mathbb{R}$)

答:
Suggestion:

For each $x\in\mathbb{R}^n$ consider the function $\phi_x:\mathbb{R}\rightarrow\mathbb{R}^n$
$$
\phi_x(h)=h\,x$$

Define $F(h)=f\circ\phi_x (h)=f(h\,x)$

Use the chain rule:

$$F'(h)=f'(\phi_x(h))\phi'_x(h)$$

On the one hand,
$$F'(h)=\nabla f(hx) \cdot x=\nabla f(hx)\cdot x$$
On the other hand,

$F(h)=h^k f(x)$ and so
$$F'(h)=k h^{k-1} f(x)$$

Consequently

$$ \nabla f(hx)\cdot x = k h^{k-1} f(x)$$

In particular, for $h=1$ we get

$$ \nabla f(x)\cdot x = k f(x)$$

Title: How to prove gradient with exponent


问:
Lagrangian equation compute conserved angular and linear momentum.
Question

Consider the following Lagrangian for a particle of mass $m$ ($c$ is a constant)

$L = −mc \sqrt{c^2 - r' .r'} $

Show that it is invariant under translations, rotations and compute the conserved angular
momentum $L$, linear momentum $p$ and energy.

What I've Done:

To show the translation $r$ -> $r + a$ so $(r + a)' = r'$ since $a$ is a constant.

For the rotation put $O$ which represents an orthogonal rotation matrix so  $r$ -> $Or$ so $(Or)' = Or'$then $Or'.Or'$ = $r'^T O^T . Or' = r'^T.r' = r'.r'$

Im not sure if what I did is right but I think it is so if you could just check that it would be helpful and help me with the computing the momentum and energy thank you.

答:
What you've done to show invariance under translations and rotations is correct.

Then you can calculate the momentum, angular momentum and energy as follows. I've used a dot instead of a prime for the time derivative.
We have $$\mathcal{L} = - mc^2 \sqrt{1-\frac{\dot{r}^2}{c^2}}.$$
Then the momentum is given by $$p = \frac{\partial \mathcal{L}}{\partial \dot{r}} = \frac{m \dot{r}}{\sqrt{1-\frac{\dot{r}^2}{c^2}}}$$

The Euler Lagrange equation is then$$\dot{p} = \frac{\partial \mathcal{L}}{\partial r}.$$ This Lagrangian doesn't depend on $r$, so $\dot{p} = 0$ and $p$ is conserved.

Then the angular momentum is given by
$$L = r \times p = \frac{m }{\sqrt{1-\frac{\dot{r}^2}{c^2}}}r\times \dot{r}.$$To see that $L$ is conserved, note that $$\dot{L}=\dot{r}\times p + r\times \dot{p}.$$ We already know $p$ is conserved so the second term is zero, and the first term is zero because $p$ is proportional to $\dot{r}$.

The energy is given by the Hamiltonian;
$$E = \mathcal{H} =\dot{r}\cdot p - \mathcal{L} = \frac{m c^2}{\sqrt{1-\frac{\dot{r}^2}{c^2}}}.$$

To show that the energy is conserved, we'll rewrite in terms of $p$ instead of $\dot{r}$. To do this, first define the Lorentz factor $\gamma := \frac{1}{\sqrt{1-\frac{\dot{r}^2}{c^2}}}$. Then rearrange to see that
$
\dot{r}^2 = c^2\frac{\gamma^2-1}{\gamma^2}
$.Now square the relation between $p$ and $\dot{r}$, and substitute for $\dot{r}^2$  in terms of $\gamma$ to give that $p^2 = m^2c^2(\gamma^2-1).$ Solve this equation for $\gamma$ to see that$$\frac{p^2}{m^2c^2} + 1
 = \gamma^2.$$ Finally, substitute this into the expression for the energy to arrive at Einstein's famous mass-shell relationship,$$E^2 = m^2c^4 \gamma^2 = m^2c^4 + p^2c^2.$$ In the particle's reference frame $p=0$ and hence $E=mc^2$. We can also see now that the energy is manifestly conserved, because it's a function only of $p$ and $p$ is conserved.


答:
Your first approach uses the reasoning

$(1+\frac1x)^x$ is approximately $e$, so $(1+\frac1x)^{x^2}=[(1+\frac1x)^x]^x$ behaves like $e^x$.

This is illegal because you're letting $x\to\infty$ in your inner expression while preventing $x\to\infty$ in the outer expression. There's no rule that allows you to do that -- you can't hold $x$ fixed in one part of your expression while letting it tend to infinity in another part.

As a simpler example, it is incorrect to compute $\lim_{n\to\infty}1$ by writing $1=\frac1n\cdot n$, then observing that $\frac1n\to0$ as $n\to\infty$, and deducing that
$$
\lim_{n\to\infty}1 = \lim_{n\to\infty}\left[ \left(\lim_{n\to\infty}\frac 1n\right)\cdot n \right]= \lim_{n\to\infty}(0\cdot n) = 0.
$$

Another simpler, but relevant example: $(1+\frac1x)$ tends to $1$ as $x\to\infty$, but it doesn't follow that
$$\lim_{x\to\infty}\left(1+\frac1x\right)^x = \lim_{x\to\infty}\left[\lim_{x\to\infty}\left(1+\frac1x\right)\right]^x=\lim_{x\to\infty} 1^x = 1.
$$


问:
Find the minimal polynomial of $e^{i\frac{\pi}{6}} - \sqrt{3}$ over $\mathbb{Q}$
I have problems with finding the minimal polynomial of $e^{i\frac{\pi}{6}} -\sqrt{3}$ over $\mathbb{Q}$. Let $\omega = e^{i\frac{\pi}{6}}$. I have already shown that $[\mathbb{Q}(\omega):\mathbb{Q}] = 4$ and that $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}] = 2$, so by degree formula it follows that $[\mathbb{Q}(\omega - \sqrt{3}):\mathbb{Q}] \geq 4$ and it can not be $6$.

Hints are greatly appreciated!

答:
But what is $e^{i\pi/6}-\sqrt3$?
$$e^{i\pi/6}=\frac{\sqrt3}2+\frac12i$$
Thefefore
$$e^{i\pi/6}-\sqrt3=-\frac{\sqrt3}2+\frac12i=e^{5i\pi/6}$$
and the minimal polynomial is $\Phi_{12}(x)$ or $x^4-x^2+1$.


问:
Can all linear operators be represented as functions?
Can we correspond any linear operator $L$ to a function $\phi(x)$ such that

$$L f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\phi(\omega) \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

?

For instance,

$$f(x)\to f(x): \phi(\omega)=1$$

$$f(x)\to f'(x): \phi(\omega)=-i\omega$$

$$f(x)\to \int f(x) dx: \phi(\omega)=\frac1{-i\omega}$$

$$f(x)\to \Delta f(x): \phi(\omega)=e^{-i\omega}-1$$

$$f(x)\to f(x+1)-f(x-1): \phi(\omega)=-2i\sin \omega$$

etc. Can we find such $\phi(\omega)$ for any linear operator?

答:
If well defined, your linear operator $$L f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\phi(\omega) \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$ will satisfy

$$L(f(.+a))(x)=L(f)(x+a)$$ this is called a convolution operator. For example $L(f)(x)= f(x+\frac1{1+x^2})$ is not one.

If a convolution operator $T$ sends $L^2(\Bbb{R})\to L^2(\Bbb{R})$ and $\forall f,\|T f\|_2\le C \|f\|_2$ then (the restriction to $L^2(\Bbb{R})$ of) $T$ is given by a $\phi\in L^\infty(\Bbb{R})$. Trying to generalize this is the point of functional analysis and distribution theory.


问:
Can countability coexist with infinity?
This question concerns the countability of the real numbers.
First I will show how I count the numbers between 0 and 1 on the real line.
It is done by reversing digits behind the coma, so
that e.g. 0,761 maps to 167. Obviously this is 1 to 1 mapping,
but there are infinite number of those unique mappings depending on the chosen reasonable radix.
In decimal number system I could count numbers like this:

$
0 , 0.1 , 0.2 , 0.3 ... 0.9 , \\
0.01 , 0.11 , 0.21 , 0.31 , ... , 0.91 , \\
0.02 , 0.12 , 0.22 , 0.32 , ... , 0.92 , \\
0.03 , 0.13 , 0.23 , 0.33 , \dots , 0.93 , \\
\vdots \\
0.09 , 0.19 , 0.29 , 0.39 , \dots , 0.99 , \\
0.001 , 0.101 , 0.201 , 0.301 , \dots , 0.901 , \\
\vdots \\
0.002 , \dots , 0.902, \\
\vdots \\
0.092, \dots , 0.992 , \\
0.003 , \dots , 0.903 , \\
\vdots \\
0.004, \dots , 0,904,\\
\vdots \\
\infty \\
$

Now, given that I can "succeed" to count to infinity, I would also count all irrational numbers.
There is no reason to haste. But then all irrational numbers are "somewhere" in the infinity.
So either counting to infinity allows me to write irrational numbers backwards, or infinity and countability can not coexist.
Which one is true ? How does your solution compare to rational numbers ?

答: To extend a little Peter's comment, you are assuming that you have the decimal expansion of all real numbers. Even if you find a way to extend your method to reals with infinite decimal places (such as $\pi$), there are still a countable number of them. You are missing  actually a number of reals that are uncountable and cannot be written with any algorithm: the uncomputable numbers (that is: most of the real numbers!)</p>

Title: Can countability coexist with infinity?


问:
Strange Summation in calculus
I have a assignment in calculus 1 to do and it has a problem involving a limit and a sum, but I have no idea how to do it. This is the problem:

$$\lim_{n\to\infty}\left(n\left(\frac1{2n+1}+\frac1{2n+3}+\ldots+\frac1{4n-1}\right)\right)\,.$$

答:
HINT: Note that $$4n-1=(2n+1)+(2n-2)=(2n+1)+2(n-1)\,,$$ so there are $n$ fractions in the expression. The largest is $\frac1{2n+1}$, and the smallest is $\frac1{4n-1}$, so

$$\frac14<\frac{n}{4n-1}\le\frac1{2n+1}+\ldots+\frac1{4n-1}\le\frac{n}{2n+1}<\frac12\,.$$

Now multiply by $n$ and think about what happens as $n$ increases.

Title: Strange Summation in calculus


问:
Prove Matching of Size K
The problem is as follows: Given a graph with n vertices, each having degree >= 2k, prove that a matching of size >= k exists. (n > 2k > 0)

My initial approach has been that if we can just find a path within the graph of length 2k, then simply alternating the edges in that path would result in a matching of size k. But now I'm not sure how to prove that such a path of length 2k exists.

答: Use induction. If you pick any edge $vw$, then deleting $v$ and $w$ from $G$ produces a graph with minimum degree $2k-2$.

Title: Prove Matching of Size K


问:
$(f(X_n))_{n \in \mathbb{N}_0}$ martingale implies $f'' = 0$?
I want to understand why the following statement holds:  $f \in \mathcal{C}^2(\mathbb{R})$. If for every martingale $(X_n)_{n \in \mathbb{N}_0}$, $(f(X_n))_{n \in \mathbb{N}_0}$ is also a martingale, then the second derivative of $f$ is zero.

答: Suppose that $x<y<z$ and $\lambda\in(0,1)$ is such that $\lambda x+(1-\lambda z) = y$. Let $X_1$ be a random variable with $P[X_1 = x]=\lambda = 1-P[X_1=z]$. Put $X_0=y$. Then $(X_0,X_1,X_1,\ldots)$ is a martingale. By the hypothesis, $f(\lambda x+(1-\lambda) z) = f(y) = \lambda f(x)+(1-\lambda)f(z)$. Vary $x,z$, and $\lambda$ and you see that $f(x) = mx+b$.


问:
(0-1) knapsack polytope
$$P = \{x \in \{0,1\}^4 : 4x_1 + 3x_2 + 2x_3 + x_4 \leq 4\}$$

I have to disaggregate the main constraint in $P$ by using minimal cover extension procedure.
I have found that minimal cover is: $\{2,3,4\}$, $\{1, 2\}$, $\{1,3\}$, $\{1,4\}$?
I'm not sure if I'm correct so far, please help!

答: You want to find minimal solutions that satisfy $4x_1 + 3x_2 + 2x_3 + x_4 > 4$.  Your last three are correct, but $\{2,3,4\}$ is not minimal.

Title: (0-1) knapsack polytope


问:
Formula for exponentially increasing series
Given $a,$ $n,$ and $b$ are all constants, how would one solve for $x$ in the exponentially increasing series?

Here is the formula:

$$x+x(a)+x(a^2)+x(a^3)+...x(a^n) = b$$

答:
Factor out the $x$.

To sum $1 + a + \cdots + a^n$ in closed form, think about how to factor $a^{n+1} -1$.


问:
Do primes of the form $n^2+1$, proportional to $p_n^2$, converge?
An open and famously difficult problem is to (dis)prove the infinitude of primes of the form $n^2+1$. My question is related to these primes:

Suppose there are infinitely many primes of the form $n^2+1$. Let $p_k$ denote the $k$-th prime and $s_k$ the $k$-th prime of the form $n^2+1$. My question is: does
$$\lim_{k \to \infty} s_k/p_k^2$$
Exist, and if so, what does it converge to?

I conjecture that it does, though I had an unusual amount of trouble finding information on the asymptotics for primes of the form $n^2+1$. I suspect a work of Bateman and Horn to be useful here, but it's a little high-level for me.

I computed the first ~two million or so primes of the form $n^2+1$ and plotted $s_k/p_k^2$, for your leisure.

答:
There is a conjectured asymptotic formula (indeed from the Bateman–Horn conjecture) for the number of positive integers $k\le x$ for which $k^2+1$ is prime, of the form $\sim Cx/(\log x)$ for a particular positive constant $C$. This implies that the $n$th such integer $k$ has size asymptotically equal to $\sim \frac1C n(\log n)$, and thus that the $n$th such prime $k^2+1$ has size asymptotically equal to $\sim \frac1{C^2} n^2(\log n)^2$. Since $p_n \sim n\log n$ by the prime number theorem, we see that $s_n/p_n^2$ conjecturally tends to the constant $\frac1{C^2}$.

I can't recover it now, but the constant $C$ can be estimated explicitly, so that should give a numerical value for the constant $\frac1{C^2}$ that could be checked against your numerical explorations.


问:
Show that there exists a holomorphic function $f:C\backslash[0,1] \to{C}$ such that $\exp(f(z))=\frac{z-1}{z}$ for all $z \in \Bbb{C}\backslash[0,1]$
The hint for the problem says use the Schwartz reflection principle. Can I get some clues on how to proceed?</p>

答:
More detail here. Two ways, one geometry, the other analytic.

Geometry argument

Let $w = \frac{z-1}{z}$ (here $z \in \bar{\mathbb{C}}$; when $z=\infty$, the definition of $w$ should be understood to mean $w\Big|_{z=\infty}=1$). $w$ is a Mobious transformation of $z$. These transformations map lines or circles in $\bar{\mathbb{C}}$ to lines or circles. Being a continuous map, it maps connected sets to connected sets.

Note: In this context, when I said "lines" I mean, of course, a line in  $\mathbb{C}$, viewed as $\mathbb{R}^2$, union the point at infinity. Technically, the context is the projective line over $\mathbb{C}$.

So one can argue this way (draw 2 planes and arrows to follow):

The segment (of the real axis) [0,1] is contained on the real axis. We can pick 3 points on this line (I mean "line"!) and see where they are map. Take $z=0,1,\infty$, which are on the real line. They are mapped respectively to $w=\infty, 0,1$, and there is only one line/circle passing through these points: the real line.
If we remove $z=0,1,\infty$ from the real line, we end up with 3 connected segments, and by continuity they are map to connected pieces of the line removing $w=\infty, 0, 1$.

If $z$ is real and large (positive), on the connected piece of the real line from $1$ to $\infty$, then $w = \frac{z-1}{z}$ is near $1-$, so the connected piece $(1,+\infty)$ is mapped to the connected piece to the left of $1$, i.e. to $(0,1)$.

If $z$ is real and large negative, on the connected piece of the real line from $0$ to $\infty$, then $w = \frac{z-1}{z}$ is near $1+$, so the connected piece $(-\infty, 0)$ is mapped to the connected piece to the right of $1$, i.e. to $(1,+\infty)$.

Note: here $+\infty$ and $-\infty$ are the same point. I am using $\pm$ to indicate the direction of approaching the point from the line, in a similar way to may use of $1-$ and $1+$, they are the same point, and the sign indicates a direction of approaching it from within the line.

Finally, the remaining connected piece, namely the segment $(0,1)$ must be mapped to the remaining piece, namely the half line $(-\infty,0)$.

Etc.

Analytic argument

If $w = \frac{z-1}{z}$, then $z = \frac{1}{1-w}$. For this approach we need to define the segment $[0,1]$ in terms of linear equations (i.e.: equations of the form $Re(\bar{\zeta}z) = c$, etc.) and linear inequalities.

The segment is defined as the simultaneous solution of:

$Im(z) = 0$

$0 \le Re(z)$

$Re(z) \le 1$

So, we write $z$ in terms of $w$, and simplify the expression to get the constraints on $w$.

Here follow the steps (here $z$ is generic, we handle the exceptions (when some denominator is 0, etc) by taking the closure at the end):

$Im(z)=0$

$Im( \frac{1}{1-w} ) = 0$

$Im( \frac{1 - \bar{w}}{|1-w|^2} )= 0$ iff $ \frac{Im(1 - \bar{w})}{|1-w|^2} = 0$, because $|1-w|^2$ is real

$ Im(w) = 0$ because $Im(1-\bar{w}) = 0 - Im(\bar{w}) = Im(w)$

This shows the line $Im(z)=0$ is mapped to the line $Im(w)=0$.

Now the inequalities. I'll do one only:
$Re(z) \le 1$

$Re( \frac{1}{1-w} ) \le 1$

$Re( \frac{1 - \bar{w}}{|1-w|^2} ) \le 1$

$Re( 1 - \bar{w}| ) \le |1-w|^2$ because  $Re( \frac{1 - \bar{w}}{|1-w|^2} ) = \frac{Re( 1 - \bar{w})}{|1-w|^2}$ and $|1-w|^2 > 0$, so passing it to the other side does not change the direction of the inequality

$1 - Re(w) \le 1 + |w|^2 - 2Re(w)$ because $Re(\bar{w})=Re(w)$, and for any $a,b \in \mathbb{C}$, $|a-b|^2 = |a|^2 + |b|^2 - 2Re(\bar{a}b)$

$0 \le |w|^2 - Re(w)$

now we complete the square:

$0 \le |w|^2 - Re(w) = |w|^2 - 2Re(\frac{1}{2}w) + \frac{1}{2}^2 - \frac{1}{4} = |w - \frac{1}{2}|^2 - \frac{1}{4}$

$\frac{1}{2} \le |w - \frac{1}{2}|$

thus we get that the half plane $Re(z) \le 1$ is mapped to the exterior of the circle centered at $\frac{1}{2}$ of radious $\frac{1}{2}$

With similar steps one can get that $0 \le Re(z)$ is mapped to $Re(w) \le 1$.

Thus, the 3 equations defining the segment in terms of $z$, when expressed in terms of $w$ are:

$Im(w) = 0$

$\frac{1}{2} \le |w - \frac{1}{2}|
3) $Re(w) \le 1$

etc.

Note: the value of $w$ for $z=0, \infty$ is derived by continuity, and not by the formula of the Moebious transformation. Those points are analyzed either explicitly or implicitly (by continuity, taking the closure of the set we found).


问:
Break a stick into two pieces, and break the shorter one again, what's the pdf of the shortest pieces.
Break a stick of unit length at a uniformly chosen random point. Then take the shorter of the two pieces and break it into two again at a uniformly chosen random point. Let X denote the shortest of the final three pieces. Find the probability density function of X.

I let the shorter one on the first break be x, which should be between (0, $1/2$), so the pdf should be
$$f(x)=2, 0<y<\frac{1}{2}$$
then the conditional probability of $f(x|y)$ should be (since it is another uniform distribution)
$$f(x|y)=\frac{1}{y/2-0}=\frac{2}{y}$$
by applying conditional probability formular
$$f_X(x)=\int_0^{1/2}f_{X|Y}(x|y)f_Y{y}dy=\int_0^{1/2}\frac{2}{y}\times2dy=4ln(y)|^{1/2}_0$$

I stopped here since it is impossible.
Can someone help?

答:
$$p(y) = 8 - 32 y\quad {\rm for}\ 0\leq y \leq 1/4$$

and $0$ otherwise.

After the first break the short leg is uniformly likely between $0 \leftrightarrow 1/2$, as shown by the range of the abscissa in the graph.

After this short piece is broken, it longer remaining part is uniformly distributed in $1/4 \leftrightarrow 1/2$.  The other (shorter) part obeys a triangle distribution between $0 \leftrightarrow 1/4$.  So just write down a triangle distribution that goes between $0 \leftrightarrow 1/4$.


问:
Is the set of allowed sequences for the golden mean shift uncountable?
Consider infinite sequences of two symbols, $L$ and $R$. The set of infinite sequences of these two symbols that do not contain consecutive $R$'s, i.e. $LRRL...$, is called the golden mean shift space.

Is the set of all such sequences uncountable?

COMMENT: Standard diagonalization arguments don't work, since you can produce forbidden sequences through the diagonalization process. However, there are certain properties that lead me to believe that this set is uncountable—for example, that there are an infinite number of sequences that contain $n$ $R$'s for every integer $n\geq 1$.

答:
Actually a standard diagonalization argument will work here - we just have to be careful how to set it up.

Before presenting a diagonal argument which works, let me give a simpler proof. To show that the set in question - which I'll call "$F$" - is uncountable, it will be enough to find an injection $2^\omega\rightarrow F$ (here $2^\omega$ is the set of infinite binary sequences, which the standard diagonal argument shows is uncountable). The idea is to use a coding process which transforms bits in the input sequence to blocks of bits, rather than individual bits, in the output sequence. Specifically, given finite sequences $\sigma_0,\sigma_1$ of $R$s and $L$s, define a function from infinite binary sequences to infinite $R/L$-sequences as follows: $$f_{\sigma_0,\sigma_1}(\alpha)=\sigma_{\alpha(0)}{}^\smallfrown \sigma_{\alpha(1)}{}^\smallfrown \sigma_{\alpha(2)}{}^\smallfrown ...$$ where "${}^\smallfrown$" denotes concatenation. For example, we have $$f_{\color{red}{RLR}, \color{green}{LRL}}(01010101...)=\color{red}{RLR}\color{green}{LRL}\color{red}{RLR}\color{green}{LRL}\color{red}{RLR}\color{green}{LRL}...$$

Now not every function of this form is nice: some $f_{\sigma_0,\sigma_1}$s are not injective, and some do not always output elements of $F$. Can you find a choice of $\sigma_0,\sigma_1$ such that $f_{\sigma_0,\sigma_1}$ is an injection from $2^\omega$ to $F$?

 Take for example $\sigma_0=LLL$ and $\sigma_1=LRL$ - the $L$s on each end of each string prevent double $R$s, and $f_{LLL,LRL}$ is clearly injective.

Now let me package that as a diagonal argument:

Suppose we have a sequence $(\alpha_i)_{i\in\mathbb{N}}$ of elements of $F$; we want to build an element $\beta$ of $F$ not in this sequence. To do this we'll block diagonalize: rather than have the $i$th bit of $\beta$ "attack" the $i$th bit of $\alpha_i$, we'll have the $i$th block of bits of $\beta$ "attack" the $i$th block of bits of $\alpha_i$. The possible behaviors of $\beta$ will be chosen analogously to the argument above, to make sure that we don't produce a bad sequence at the end.

For example, the choice of $\sigma_0$ and $\sigma_1$ above suggests the following block diagonalization:

 Let $\tau_i=LLL$ if $\langle\alpha_i(3i),\alpha_i(3i+1), \alpha_i(3i+2)\rangle= LRL$ and let $\tau_i=LRL$ otherwise. Then set $$\beta=\tau_0{}^\smallfrown\tau_1{}^\smallfrown\tau_2{}^\smallfrown...$$

As in the usual argument, we can now show that $\beta\not=\alpha_i$ for each $i\in\mathbb{N}$.


问:
Is my justification right to evaluate the limit?
I want to evaluate :

$$\lim_{x\to 0^+}\left(\frac1e(1+x)^{\tfrac1x}\right)^{\tfrac{1}{x^2}}$$

We know $\lim_{x\to 0^+}(1+x)^{\tfrac1x}$  is other form of $\lim_{u\to\infty}(1+\frac1u)^u=e$ which is increasing on $(0,\infty)$ it means at the infinity we get $e^-$ (the value is a litter less than $e$) therefor in the original limit we have $(1^-)^{\infty}$ therefore the limit equals to $0$.
Is my justification right?

答:
Another good way to think is :

Using Logarithm Identity - $ \ \ \displaystyle e^{\ln a} = a. \ \ $Thus,

$$\displaystyle  \left ( \frac {1}{e} (1+x)^{1/x}  \right )^{1/x^2} = \ \ e^{\displaystyle \frac {\ln (1+x)-x }{x^3}} .$$

Using Taylor Expansion, $ \ \ \displaystyle \ln (1+x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4} + ... $

Thus expression gets simplified to $ \ \ \displaystyle e^{ 1/3 - 1/2x }, $ as rest of the terms equals zero according to the limit.

$$\displaystyle \lim _{x \rightarrow 0} e^{ 1/3 - 1/2x } = 0.$$


问:
approximate the integral of sin(t)/t from 0 to 1
The problem asks me to use my 7th degree maclaurin approximation (x-x^3/3!+x^5/5!-x^7/7!) to approximate (sin(t)/t)dt. I have no clue how to go about this I know that the integral of sin(t) from 0 to 1 is equal to the integral of the maclaurin polynomial from 0 to 1. But I don't know what to do with the extra 1/t.</p>

Body:
Hint:

You have that $$\sin(t)\approx t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}$$ so: $$\frac{\sin (t)}{t}\approx 1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}$$


问:
show that the series $ \sum^{\infty}_{n=1} \frac{z^n}{n^{3/2}}$ converges uniformly for $|z|< 1$

show that the series $ \sum^{\infty}_{n=1} \frac{z^n}{n^{3/2}}$ converges uniformly for $|z|< 1$

I don't know how to begin this problem, can someone tell me how can i demonstrate this.

答:
HINT: For all $|z| \le 1$:

$$\left|\sum_{n=N}^{\infty} \frac{z^n}{n^{3/2}} \right| \ \le \ \sum_{n=N}^{\infty} \frac{|z|^n}{n^{3/2}} \ \le \ \sum_{n=N}^{\infty} \frac{1}{n^{3/2}}.$$

What do we know about $\sum_{n=N}^{\infty} \frac{1}{n^a}$ for all constants $a$ at least say 5/4?


问:
How to derive the dual for LP like this?
I know how to derive dual for normal LPs, but what if we are unlikely to have something like this:

```<code>maximize z
s.t.   z < 3y-2
       1 < y < 2
</code>```

, where the constraints are not directly related to the objective but related indeed. In this case, how can we compute dual for it?

Currently, the way I solve it is to ignore the second constraint and convert the original one into:

```<code>minimize -z
s.t.   z < 3y-2
       1 < y < 2
</code>```

Firstly, I have this Lagrangian:

$L(z,\lambda) = -z + \lambda (z-3y+2)$.

Then take derivative w.r.t. $z$, and set it to be $0$:

$\frac{\partial L}{\partial z} = -1 + \lambda = 0$.

Finally, plugging in $\lambda=1$ into Lagrangian, we have:

$g(\lambda) = L(z*, \lambda) = -3y+2$.

Even if $g$ is not a function in terms of  $\lambda$, but let's just write it in this way. Then I don't what to do next.

Thank you very much!

答:
Rewrite as:

```<code>maximize 0y + 1z
s.t.    -3y + 1z <= -2
        -1y + 0z <= -1
         1y + 0z <=  2
</code>```

Then the dual is:

```<code>minimize -2t - 1u + 2v
s.t.     -3t - 1u + 1v = 0
          1t + 0u + 0v = 1
           t,   u,   v >= 0
</code>```

Equivalently:

```<code>minimize -u + 2v - 2
s.t.     -u +  v = 3
          u,   v >= 0
</code>```

Equivalently:

```<code>minimize  u + 4
s.t.      u >= 0
</code>```

Hence $(t,u,v)=(1,0,3)$ is the unique optimal dual solution, and complementary slackness yields optimal primal solution $(y,z)=(2,4)$.

Body:
Rewrite as:

maximize 0y + 1z
s.t.    -3y + 1z <= -2
        -1y + 0z <= -1
         1y + 0z <=  2

Then the dual is:

minimize -2t - 1u + 2v
s.t.     -3t - 1u + 1v = 0
          1t + 0u + 0v = 1
           t,   u,   v >= 0

Equivalently:

minimize -u + 2v - 2
s.t.     -u +  v = 3
          u,   v >= 0

Equivalently:

minimize  u + 4
s.t.      u >= 0

Hence $(t,u,v)=(1,0,3)$ is the unique optimal dual solution, and complementary slackness yields optimal primal solution $(y,z)=(2,4)$.


答:
Hint:

Writing in the form $\sum^{\infty}_{n=0} \frac{z^n}{n^{3/2}}=\sum^{\infty}_{n=0} a_nz^n,$ we see that $a_n=\frac{1}{n^{3/2}}$.

The radius of convergence is given by $R=\frac{1}{\limsup |a_n|^{1/n}}$.

Then, for any $z$ with $|z|\leq r<R$, the series will converge uniformly.

Body:
show that the series $ \sum^{\infty}_{n=1} \frac{z^n}{n^{3/2}}$ converges uniformly for $|z|< 1$

I don't know how to begin this problem, can someone tell me how can i demonstrate this.


问:
How can I show that $x^4+6$ is reducible over $\mathbb{R}$?
How can I show that the polynomial $x^4 + 6$ is reducible over $\mathbb{R}$ without explicitly finding factors?

I was trying to find a non-prime ideal that would generate it but I'm kind of lost as to how to proceed. Is there some sort of criterion that will allow me to show that it's reducible in $\mathbb{R}$?

答:
Hint: Fundamental Theorem of Algebra.

Then use the fact that if a complex number $z$ is a root of a real polynomial, then $\bar z$ is also a root.

It follows that in $\Bbb R$ only the polynomials of degree one, and the quadratic polynomials with negative discriminants are the irreducible.


问:
How many tuples $( A , B , C ) $ of positive integers satisfy $A × B + C = N$
This is a programming question I found on Atcoder and I am trying to solve it just using mathematics. I want a simple formula to find number of tuples without using a computer. So I derived a formula to calculate it: $Sn = \sum_{x=1}^n [\frac{n-1}{x}]$

But this formula is so complicated. The best approximation I found is $Sn \approx  [(n-1) ln(n)]$ is there any way to simplify it even more ? I need the formula to work for all positive integers up to $10^6$.

答:
First count the cases with $A=B$: There are $$\tag1\lfloor \sqrt{N-1}\rfloor$$ of these.

Next count the cases with $A<B$: For each $A$, there are $\lfloor\frac {N-1}A\rfloor-A$. So all together contribute
$$\tag2 \sum_{A=1}^{\lfloor\sqrt{N-1}\rfloor}\left(\left\lfloor\frac{N-1}A\right\rfloor-A\right)$$
Finally, the cases with $A>B$ are the same as $(2)$.

This is now only $O(\sqrt N)$ times as opposed to the $O(N)$ in your formula.


问:
Texts on the theory of partial algebras.
In all the texts on universal algebra I have read, I have not seen any mention of partial operations. Partial operations are sometimes important in mathematics, like the operation of division on real numbers, or the operation of subtraction on naturals. Is there a book or paper somewhere that develops the theory of algebras with partial operations, like universal algebra texts do with algebras with total operations?</p>

答:
I've been interested in similar questions. I haven't come across any such book.

However, there is the notion of an inverse semigroup which model the algebra of partial maps on a space by partial compositions and there are books devoted to that. For example:

Petrich, Inverse Semigroups

Amd

Lawson, Inverse Semigroups: The Theory of Partial Symmetries

Unfortunately, I haven't looked at either (they are on my to-read list) so please don't take this as a recommendation. They can be categorified into Partial categories. There is also the paper:

Ruy Exel, Partial actions of groups and actions of semigroups

Which might be worth taking a look at. It's available on the Arxiv as article 9511003


问:
Finding a constant using joint conditional PMF
Nick shops for statistics books for $H$ hours, where $H$ is a random variable, equally likely to be $1,2,3$ or $4$. The number of books $B$ he buys is random and depends on how long he shops according to the joint conditional PMF
$$Pr(B=b|H=h) = \frac{c}{h}, \text{ for } b= 1, . . . , h$$
for some constant $c$

Problem: Find a value for $c$

From the definition of joint conditional PMF
\begin{align*}
P_{B|H}(B=b|H=h) &= \frac{P_{B,H}(B=b,H=h)}{P_H(H=h)} \\
\frac{c}{h} &= \frac{P_{B,H}(B=b,H=h)}{P_H(H=h)} \\
\frac{c}{h} &= \frac{P_{B,H}(B=b,H=h)}{\frac{1}{4}} && \text{since $H$ is equally like to be $1,2,3,4$}
\end{align*}

I'm not sure how to calculate $P_{B,H}(B=b,H=h)$ or if my approach is correct.

答: Just sum the equation you got over $b$ from $1$ to $h$. You get $h\frac c h=\frac {P(H=h)} {1/4}=1$. So $c=1$.


问:
Automorphism group of direct product of groups, proof
I need a proof of the following fact: if a group $G=M \times N$ is the direct product of subgroups $M$ and $N$ such that $\lvert M \rvert$ and $\lvert N \rvert$ are relatively prime, then $\mathrm{Aut}(G) = \mathrm{Aut}(M) \times \mathrm{Aut}(N)$.

This is the way the result is recalled on some lecture notes I am reading, I think $M$ and $N$ are implicitly supposed finite (maybe $G$ also?) otherwise this does not make sense.

This should be rather standard (it is used but not proven here), but I've taken a quick look on Rotman, Roman and Robinson but none of them seems to prove this. Do you have a reference for the proof/can you share here a proof you know?

Thanks

答:
Let $n=|M|, m=|M|$ there exists integers $a,b$ such that $am+bn=1$. Let $x\in M$, write $f(x,1)=(u,v)$. We have $f((x,1)^{am})=f(x^{am},1)=f(1,1)=(1,1)=f(u^{am},v^{am})$.

We have $u^{am}=(u^m)^a=1$, $v^{am}=v^{1-bn}=vv^{-bn}=v$, we deduce that $(1,1)=(1,v), v=1$.
Similarly, we show that for every $y\in N$, $f(1,y)=(1,l)$. This implies that $f(x,y)=(g(x),h(y))$ and $f^{-1}(x,y)=(g'(x),h'(y))$. $g'$ is the inverse of $g$ and $h'$ the inverse of $h$.


问:
$L^p$-space is a Hilbert space if and only if $p=2$
Inspired by $\ell_p$ is Hilbert if and only if $p=2$, I try to prove that a $L^p$-space (provided with the standard norm) is a Hilbert space if and only if $p=2$. I already know that every $L^p$-space is a Banach space with respect to the standard norm.

This is what I got so far.

To prove

Let $(S, \Sigma, \mu)$ be a measure space and assume that $\mu$ is a positive, $\sigma$-finite measure that is not the trivial measure. Let $p\in [1, +\infty]$. The normed space $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} )$ is a Hilbert space if and only if $p=2$.

Proof

Case 1: $p=2$

The standard inner product $\langle \cdot , \cdot \rangle _{L^2}$ induces the standard norm $\| \cdot \| _{L^2}$, so $(L^2 (S, \Sigma , \mu), \| \cdot \| _{L^2})$ is a Hilbert space.

Case 2: $p\in [1, \infty) \backslash \{ 2 \}$

Assume $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} ) $ is a Hilbert space. Then it is also an inner product space, so it must satisfy the parallelogram rule:
\begin{align}
\forall f,g \in L^p(S,\Sigma , \mu): \ \ \| f + g \| ^2 + \| f -g \| ^2 = 2 ( \| f \| ^2 + \| g \| ^2 ).
\end{align}
Let $A, B$ be disjoint measurable sets such that $0 < \mu (A), \ \mu (B) < \infty$. Define $f_p, g_p \in
\mathcal{L} ^p (S, \Sigma, \mu)$ by
\begin{align*} f_p  := \frac{1}{(\mu (A) ) ^{1/p}} 1 _A \geq 0 \ \ \ \text{ and } \ \ \ g_p  := \frac{1}{(\mu (B) ) ^{1/p}} 1 _B  \geq 0.\end{align*}
Doing some calculations gives us \begin{align*} 2 \left ( \| f_p \| _{L^p} ^2  + \| g_p \| _{L^p} ^2 \right ) = 4 \ \ \text{ and } \ \ \| f_p + g_p \| _{L^p} ^2 + \| f_p - g_p \| _{L^p} ^2 = 2 \cdot 2 ^{2/p}. \end{align*} The functions $f_p$ and $g_p$ do not satisfy the parallelogram rule, since $4\neq 2 \cdot 2 ^{2/p}$ (remember that $p\neq 2$). But this contradicts our earlier conclusion that all $f,g\in L^p (S, \Sigma , \mu)$ must satisfy the parallelogram rule. Therefore, our assumption that $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} ) $ is a Hilbert space is wrong. Hence, $(L^p(S,\Sigma , \mu), \| \cdot \| _{L^p} ) $ is not a Hilbert space.

Case 3: $p=\infty$

Assume $(L^{\infty} (S,\Sigma , \mu), \| \cdot \| _{L^{\infty}} ) $ is a Hilbert space. Then it is also an inner product space, so it must satisfy the parallelogram rule.

Let $A$ and $B$ be disjoint measurable sets which are not null sets and consider $F=1 _A$ and $G=1_B$. Then, we have
\begin{align*}
\| F + G \| _{L^{\infty}} ^2 + \| F - G \| _{L^{\infty}} ^2 = 1 + 1 = 2 \neq 4 = 2(1+1) = 2( \| F \| _{L^{\infty}} ^2 + \| G \| _{L^{\infty}} ^2 ).
\end{align*}
But this contradicts our earlier conclusion that all $f,g\in L^{\infty} (S, \Sigma , \mu)$ must satisfy the parallelogram rule. Therefore, our assumption that $(L^{\infty} (S,\Sigma , \mu), \| \cdot \| _{L^{\infty}} ) $ is a Hilbert space is wrong. Hence, $(L^{\infty} (S,\Sigma , \mu), \| \cdot \| _{L^{\infty}} ) $ is not a Hilbert space.

Question

I am not completely sure about the following statements "Let $A, B$ be disjoint measurable sets such that $0 < \mu (A), \ \mu (B) < \infty$" and "Let $A$ and $B$ be disjoint measurable sets which are not null sets". How do I know for sure that such measurable sets $A,B$ actually exist? Should I make more assumptions about the measure space in order to make these arguments work?

答:
Here is an interesting characterization, from which one can see that for the Lebesgue measure, all $L^p$ spaces with $p\ne 2$ are not Hilbert.
https://www.docdroid.net/Ib35oGd/lp-pdf


问:
prove by induction that $(6\,/\,7^n)-1, \forall n \geq 1$
I'm trying resolve this exercise but I cannot resolve. Can you help me?
Follow the question.

Prove by induction that: $(6\,/\,7^n)-1, \forall n \geq 1$

Thanks!

Edit:

The question was wrong.

About this question, I dont know start with this problem. I'm newbie in discrete.

答:
Hint

For the inductive step, take $7^{n+1}-1 = 7(7^n)-1$

Now add a "clever form of $0$"


Title: prove by induction that $(6\,/\,7^n)-1, \forall n \geq 1$

Body:
I'm trying resolve this exercise but I cannot resolve. Can you help me?
Follow the question.

Prove by induction that: $(6\,/\,7^n)-1, \forall n \geq 1$

Thanks!

Edit:

The question was wrong.

About this question, I dont know start with this problem. I'm newbie in discrete.

Body: Hint : $7^{k+1}-1 = 7\times 7^k-1=6\times 7^k + 7^k - 1$


问:
REVISITED $^2$: Does a solution in $\mathbb{R}^n$ imply a solution in $\mathbb{Q}^n$?
Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $B ∈ \mathbb{Q}^m$. Suppose that the system of linear
equations $AX = B$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution
in $\mathbb{Q}^n$?

Where do I start?

I feel I can use this to help:

$$x_i=\frac{1}{a'_{ii}}\left(b'_i - \sum_{j=i+1}^{k} a'_{ij} x_j \right).$$

Here is a thought that just crossed my mind, but I'm not sure if this is legal:

Suppose $Y\in \{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $Y'\in\mathbb{Q}^n$ are solutions to $AX=B$. Then $AY=AY'=B$, or $A(Y-Y')=A(Y+(-Y'))=0$, but  addition between $\{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $\mathbb{Q}^n$ is . . .

答:
I think the question is: if $A\in M_{m\times n}$ and $B\in\mathbb{Q}^m$ and $AX=B$ for some $X\in\mathbb{R}^n$, does there exist an $X'\in\mathbb{Q}^n$ so that $AX'=B$?

Assume the $m$ rows of $A$ are independent. If not, choose a subset of the rows of $A$ that form a basis of the row space of $A$ and the corresponding subset of the rows of $B$. Show that the removed rows can always be put back and  will satisfy any solution that applies to the basis rows.

Since the rows of $A$ are independent, show that $AA^T\in M_{m\times m}$ is invertible.

What can be said of $A^T(AA^T)^{-1}B$?

Full answer

 I think the question is: if $A\in M_{m\times n}$ and $B\in\mathbb{Q}^m$ and $AX=B$ for some $X\in\mathbb{R}^n$, does there exist an $X'\in\mathbb{Q}^n$ so that $AX'=B$?

 Assume the $m$ rows of $A$ are independent. If not, choose a subset of the rows of $A$ that form a basis of the row space of $A$ and the corresponding subset of the rows of $B$. The removed rows can always be put back and, because there is a solution $X$, the removed rows will satisfy any solution that applies to the basis rows. This is because the same combination of basis rows in $A$ that make up a removed row in $A$ will make up the corresponding removed row of $B$ from the corresponding basis rows of $B$.

 Since the rows of $A$ are independent, we have that $AA^T\in M_{m\times m}$ is invertible: suppose that $xAA^t=0$, then $|xA|^2=xAA^Tx^T=0\implies xA=0\implies x=0$.

 Set $X'=A^T(AA^T)^{-1}B\in\mathbb{Q}^n$, then $AX'=AA^T(AA^T)^{-1}B=B$.


问:
Integral of $\sin(x)/x$ from $-400$ to $400$.
I recently saw this on Glassdoor as an interview question.

Is there any way of performing this integral analytically? I cannot seem to do it by substitution or by parts. Given it asks for the definite integral, I assume it is possible?

答:
As TonyK mentioned, a good estimate is . By abuse of notation, we expect that the error is $\mathcal{O}\left(1/400\right)$, meaning that it should be on the order of $1/400$ (because there are no other relevant length scales) with higher order corrections in powers of $1/400$. This is indeed what Felix Marin has shown in his answer, i.e. you basically have an infinite series of corrections, with the leading order term being roughly $1/400$.

This analysis needs to be slightly modified when trying to estimate $\int_{-N}^N \frac{\sin x}{x}\, dx$, $N\gg 1$ whenever $N\approx (2m+1)\pi/2$ for some $m\in \mathbb{N}$. Indeed, if that is the case, then $\cos N\approx 0$ and the leading order term is not anymore $\mathcal{O}\left(1/N\right)$, as that term will vanish, but now $\mathcal{O}\left(1/N^2\right)$.


问:
About convergence a.s of iid random variable sequence
$$
\mbox{I want to prove that}\quad
\frac{\max\left\{X_1,\ldots,X_n\right\}}{n} \to 0
$$
almost sure
$\iff E\left\lvert X_{1}\right\rvert < \infty$ when $\left\lbrace X_{n}\right\rbrace_{n}$ are random independent and identically distributed variables. I could show the $\Leftarrow$ implication but have no idea for the other.

答:
The positivity condition is missing. If $X_n$ can be negative, then any sequence of real negative random variables satisfies the first claim while the second claim does not always hold.

Given the positivy of $(X_n)$, the other direction can be proven as follows:

$\mathbb{E}( |X_1| ) = +\infty \iff \sum_{n \ge 1} \mathbb{P}(|X_n| \ge n) = +\infty \underbrace{\iff}_{\text{ Borel Cantelli}}  1= \mathbb{P}( \limsup \frac{|X_n|}{n} \ge 1) \iff \limsup \frac{|X_n|}{n} \ge 1 \text{ a.s} $

From which, we imply the other direction.


问:
Proof of: If $x_0\in \mathbb R^n$ is a local minimum of $f$, then $\nabla f(x_0) = 0$.

Let $f \colon \mathbb R^n\to\mathbb R$ be a differentiable function.
If $x_0\in \mathbb R^n$ is a local minimum of $f$, then $\nabla f(x_0) = 0$.

Where can I find a proof for this theorem? This is a theorem for max/min in calculus of several variables.

Here is my attempt:

Let $x_0$ = $[x_1,x_2,\ldots, x_n]$
Let $g_i(x) = f(x_0+(x-x_i)e_i)$ where $e_i$ is the $i$-th standard basis vector of dimension $n$.

Since $f$ has local min at $x_0$, then $g_i$ has local minimum at $x_i$.
So by Fermat's theorem, $g'(x_i)= 0$ which is equal to $f_{x_i}(x_0)$.
Therefore $f_{x_i}(x_0) = 0$ which is what you wanted to show.
Is this right?

答:
Do you know the proof for $n=1$? Can you try to mimic it for more variables, say $n=2$? Since $\nabla f(t)$ is a vector what you want to prove is that $\frac{\partial f}{\partial x_i}(t)=0$ for each $i$. That is why you need to mimic the $n=1$ proof, mostly.

Recall that for the $n=1$, we prove that $$f'(t)\leq 0$$ and $$f'(t)\geq 0$$ by looking at $x\to t^{+}$ and $x\to t^{-}$. You should do the same in each $$\frac{\partial f}{\partial x_i}(t)=\lim_{h\to 0}\frac{f(t_1,\dots,t_i+h,\dots,t_n)-f(t_1,\dots,t_n)}h$$

ADD Suppose $f:\Bbb R\to \Bbb R$ is differentiable and $f$ has a local minimum in $t=0$. Then $f'(t)=0$.

P Since $f$ has a local minimum at $t=0$, for suitably small $h$, $$f(t+h)-f(t)\geq 0$$

If $h>0$ then this gives  $$\frac{f(t+h)-f(t)}{h}\geq 0$$

While if $h<0$ we get $$\frac{f(t+h)-f(t)}{h}\leq 0$$

Since $f'$ exist, the side limits also exist and equal $f'(t)$. From the above we conclude $f'(t)\geq 0$ and $f'(t)\leq 0$, so that $f'(t)=0 \;\;\blacktriangle$.

Now, just apply that coordinatewise, and you're done.


问:
Set of open sets notation
While studing from Nakahara's book of Differential Geometry, I reached the part where topological spaces are introduced, and to define a topological space $\mathcal{T}$, they take it as a collection of subsets of a given set $X$, such that $\mathcal{T} = \left\{ U_i | i \in I \right\}$. This may be a stupid question, but I have not been able to figure what does the set $I$ is, to which $i$ belongs to. I have thought it to be the set of all possible subsets of $X$, or the set that contains the element indexes of the subsets of $X$, but these do not sound correct to me at all.

答:
A topology $\mathcal T$ on a set $X$ is a set of subsets of $X$ satisfying the well-known axioms for "open sets". Thus, letting $\mathfrak P(X)$ denote the powerset of $X$, we have $\mathcal T \subset \mathfrak P(X)$.

In my opininon there is no benefit to regard a topology as an indexed collection $\{U_i \mid  i\in I\}$. Writing it in this form means that we are given a set $I$ of indices and for each $i \in I$ an element $U_i \in \mathfrak P(X)$. The index set  $I$ can be any set. You should be aware that such an indexed collection of sets $U_i$ formally is a function $\phi : I \to \mathfrak P(X)$ where we write $U_i = \phi(i)$. Note that this allows $U_i = U_{i'}$ for $i \ne i'$, i.e. your indexed family of sets $U_i$ may have multiple occurences of the same set.

If we simply say that $\mathcal T \subset \mathfrak P(X)$, then we can regard $\mathcal T $ as indexed collection by taking $I = \mathcal T$ and letting $\phi : \mathcal T \to \mathfrak P(X)$ be the inclusion, i.e. $\phi(U) = U$. This may be called the "self-indexing" of $\mathcal T$.

Anyway, both approaches are correct, thus it is a matter of taste which you prefer.

Title: Set of open sets notation


问:
Proof about $T^2 = T\circ T$
I have the following statement to prove:

Proof if $T$ is a linear map defined as $T: V\to V$ with $Dim(V)=n$
and $T^2 = T \circ T=I_V$ with $I_V$ the identity linear map, then $Ker(T-I_V) \cap Ker(T+I_V) = \{0\}$

My attempt was:

If $\lambda_1, \lambda_2$ are different eigenvalues, then we have that $W_{\lambda_1} \cap W_{\lambda_2} = \{0\}$ and since $W_{\lambda_1}=Ker(T-\lambda_1I_n)$, $W_{\lambda_2}=Ker(T-\lambda_2I_n)$ , this will equivalent to prove that there exists eigenvalues equal to $\lambda_1=1, \lambda_2=-1$.

So i need to prove that there exists at least one $v_1,v_2 \in V$ such that $T(v_1)=v$ and $T(v_2)=-v_2$.

But from here i can't do more.

I tried to find from the information $T(v)\cdot T(v) = T(T(v))$ that $T(v)=v$ but i could not.

答: T is a symetry of V, then $\mu_{T} = X^2-1 = (X-1)(X+1)$. And $(X-1)\wedge (X+1)=1$. By Kernels' lemma, we have : $E= Ker(T-I_{V}) \bigoplus Ker(T+I_{V})$. The conclusion is clear now.

Title: Proof about $T^2 = T\circ T$