问:
Why this Artin ring is simple?
Suppose we have a non-commutative left Artin ring $R$, such that the subring generated by any nonzero two-sided ideal together with the unity is $R$ itself. I read that $\mathrm{Z}(R)$, which is the center of $R$, forms an integral domain, but why is this true? And how can one proceed to deduce that $R$ is simple? I tried to use DCC on the sets of all two-sided ideals to go further, but don’t see why this implies there can only be extremal two-sided ideals in this case. Any hints on this?

答:
Let $R$ be left artinian such that, for all non-zero two-sided ideals $I$, the subring generated by 1 and $I$ is all of $R$. We will describe all such $R$, including the commutative ones.

Write $J$ for the Jacobson radical of $R$, a two-sided ideal. Suppose first that $J^2$ is non-zero. Then $R$ is generated as a ring by 1 and $J^2$, so $R/J^2$ is generated by 1, and hence is a quotient of $\mathbb Z$. In particular, $J/J^2$ is generated by the image of some integer $a$, so by Nakayama's Lemma $J=Ra$, and hence, using the nilpotency of $J$, we see that $\mathbb Z$ surjects onto $R$. Thus $R=\mathbb Z/n$ for some integer $n$.

Otherwise $J^2=0$. Assume $J\neq0$. Then $R$ is generated by 1 and $J$, so $R/J$ is generated by 1, so is a quotient of $\mathbb Z$. If the image of some $a\in\mathbb Z$ is non-zero and in $J$, then $R$ is generated by $1$ and $Ra$, so by 1, and hence is a quotient of $\mathbb Z$.

Otherwise, the image $A$ of $\mathbb Z$ intersects $J$ trivially, and $R=A\oplus J$. If $e\in A$ is a non-trivial idempotent, then $R$ is generated by 1 and $Re$, so $J=Je$. Similarly $J=J(1-e)$, so $J=0$, a contradiction. Thus $A=\mathbb Z/p$ for some prime $p$. Taking any non-zero $t\in J$ we see that $R=(\mathbb Z/p)[t]/(t^2)$.

Finally, if $J=0$, then $R$ is semisimple artinian, so a product of matrices over division rings. If $R=A\times B$ is a proper product, then $R$ is generated as a ring by 1 and either $A$ or $B$. Thus both $A$ and $B$ are quotients of $\mathbb Z$ by square-free integers. This holds for any such decomposition, so either $R=\mathbb Z/m$ with $m$ square-free, or else $R=(\mathbb Z/p)^2$ for some prime $p$. Note that up to now, all such $R$ are commutative.

The only other case is when $R$ is a matrix ring over a division ring, and so simple. In particular, every such non-commutative $R$ is simple artinian.

In summary, then only such rings are
$$ \mathbb Z/n, \quad (\mathbb Z/p)[t]/(t^2), \quad (\mathbb Z/p)^2, \quad \mathbb M_r(D) $$
with $p,n,r\in\mathbb N$ and $p$ prime, and $D$ a division ring.

Title: Why this Artin ring is simple?


问:
Is the space of subprobability measures weakly sequentially closed?
Let $(E,d)$ be a metric space, $\mathcal M(E)$ denote the space of finite signed measures on $\mathcal B(E)$ and $\mathcal M_{\le1}$ denote the subset of those $\mu\in\mathcal M(E)$ which satisfy $0\le\mu\le1$.

If $\tau$ denotes the topology generated by the weak convergence on $\mathcal M(E)$, are we able to show that $\mathcal M_{\le1}(E)$ is sequentially $\tau$-closed?

Given $(\mu_n)_{n\in\mathbb N}\subseteq\mathcal M_{\le1}(E)$ weakly converging to $\mu\in\mathcal M(E)$, we need to show $\mu\in\mathcal M_{\le1}(E)$.

Thoughts:

(a) Let $\nu=\nu^+-\nu^-$ denote the Jordan decomposition of $\nu\in\mathcal M(E)$ and $|\nu|:=\nu^++\nu^-$. If $(\nu_n)_{n\in\mathbb N}\subseteq\mathcal M(E)$ and $(\nu_n^\pm)_{n\in\mathbb N}$ weakly converges to $\nu_{\pm}\in\mathcal M(E)$ with $\nu_{\pm}\ge0$, it obviously follows that $(\nu_n)_{n\in\mathbb N}$ weakly converges to $\nu_+-\nu_-$ and $(|\nu_n|)_{n\in\mathbb N}$ weakly converges to $\nu_++\nu_-$.

(BTW: If we define $\nu:=\nu_+-\nu_-$, does it necessarily follow that $\nu^\pm=\nu_\pm$? I guess not ... We can simply take $\nu_+=\nu_-$ to see that this is wrong)

(b) However, unless I'm missing something, if $(\eta_n)_{n\in\mathbb N}\subseteq\mathcal M(E)$ weakly converges to $\eta\in\mathcal M(E)$, it seems like it doesn't necessarily follow that $(\eta_n^\pm)_{n\in\mathbb N}$ weakly converges to $\eta^\pm$. Or am I missing something?

答: If $\mu_n \to \mu$ weakly then $\int fd\mu \geq 0$ for all $f \in C_b^{+}(E)$ and this implies that $\mu$ is a positive measure. Clearly $\mu (E)=\lim \mu_n(E) \leq 1$. So the answer is YES.

ContentLicense: CC BY-SA 4.0


问:
Countexample: Sum and Product of two non-zero nilpotent operators is nilpotent?
I am asked to come up with a proof or a counterexample of the following two statements:

The sum of two non-zero nilpotent matrices is nilpotent.

The product of two non-zero nilpotent matrices is nilpotent.

I think both of these claims are false, and a counterexample exists, but I can't seem to come up with one

答:
Both claims are indeed false. you can look at:
$$A_1 =\begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix}\space\space\space
A_2=\begin{pmatrix}
0 & 0\\
1 & 0\\
\end{pmatrix}$$
$$(A_1)^2 = (A_2)^2 = \begin{pmatrix}
0 & 0\\
0 & 0\\
\end{pmatrix}$$
and thus, both matrices are nilpotent, but
$$A_1+A_2 = \begin{pmatrix}
0 & 1\\
1 & 0\\
\end{pmatrix}$$
$$(A_1+A_2)^2 = \begin{pmatrix}
1 & 0\\
0 & 1\\
\end{pmatrix}$$
and thus, for any n $\ge2$ $(A_1+A_2)^n=I$, Which is not nilpotent.

$$A_1A_2 = \begin{pmatrix}
1 & 0\\
0 & 0\\
\end{pmatrix}$$
and for any n $\ge2$ $(A_1A_2)^n=\begin{pmatrix}
1 & 0\\
0 & 0\\
\end{pmatrix}$ , Which again, is not nilpotent.


答:
In fact, the set of sub-probability measures is even $\tau$-closed (and thus, it is inparticular, sequentially $\tau$-closed). This follows from the following two observations (which are essentially the same arguments as used in Kavi Rama Murthy' answer):

A measure $\mu \in \mathcal{M}(E)$ satisfies $\mu \ge 0$ if and only if $\langle \mu, f\rangle \ge 0$ for all $0 \le f \in C_b(E)$.

A measure $0 \le \mu \in \mathcal{M}(E)$ satisfies $\mu(E) = \langle \mu, 1 \rangle$, where $1 \in C_b(E)$ denotes the constant function with values $1$.

Title: Is the space of subprobability measures weakly sequentially closed?

Body:
Let $(E,d)$ be a metric space, $\mathcal M(E)$ denote the space of finite signed measures on $\mathcal B(E)$ and $\mathcal M_{\le1}$ denote the subset of those $\mu\in\mathcal M(E)$ which satisfy $0\le\mu\le1$.

If $\tau$ denotes the topology generated by the weak convergence on $\mathcal M(E)$, are we able to show that $\mathcal M_{\le1}(E)$ is sequentially $\tau$-closed?

Given $(\mu_n)_{n\in\mathbb N}\subseteq\mathcal M_{\le1}(E)$ weakly converging to $\mu\in\mathcal M(E)$, we need to show $\mu\in\mathcal M_{\le1}(E)$.

Thoughts:

(a) Let $\nu=\nu^+-\nu^-$ denote the Jordan decomposition of $\nu\in\mathcal M(E)$ and $|\nu|:=\nu^++\nu^-$. If $(\nu_n)_{n\in\mathbb N}\subseteq\mathcal M(E)$ and $(\nu_n^\pm)_{n\in\mathbb N}$ weakly converges to $\nu_{\pm}\in\mathcal M(E)$ with $\nu_{\pm}\ge0$, it obviously follows that $(\nu_n)_{n\in\mathbb N}$ weakly converges to $\nu_+-\nu_-$ and $(|\nu_n|)_{n\in\mathbb N}$ weakly converges to $\nu_++\nu_-$.

(BTW: If we define $\nu:=\nu_+-\nu_-$, does it necessarily follow that $\nu^\pm=\nu_\pm$? I guess not ... We can simply take $\nu_+=\nu_-$ to see that this is wrong)

(b) However, unless I'm missing something, if $(\eta_n)_{n\in\mathbb N}\subseteq\mathcal M(E)$ weakly converges to $\eta\in\mathcal M(E)$, it seems like it doesn't necessarily follow that $(\eta_n^\pm)_{n\in\mathbb N}$ weakly converges to $\eta^\pm$. Or am I missing something?


问:
Prove that the following lines so described are perpendicular.
If O be any point on the circumcircle of the triangle ABC, and OL be drawn parallel to BC to meet the circumcircle in L. OD and OF are drawn perpendicular to BC and AB respectively, then show that LA will be perpendicular to DF.

The above question is from my self-preparatory material for an exam(JEE). I tried a lot  thinking on this but I don't know from which theorem or property of triangles and circles to start with.
Any suggestion on how to tackle such problems will be appreciated.

答:
I already saw one solution above but I decided to add my solution because why not. :)

Drop a perpendicular from $O$ to segment $AC$. Then points $D, E$ and $F$ are collinear as $DEF$ is Simson's line.

Now I claim that $∠OFP = ∠FAP$.

Note that,

$∠OFP = ∠OFE = ∠OAE = ∠OAC = ∠OLC$ as $FAEO$ is a cyclic quadrilateral and $∠OAC$ and $∠OLC$ subtends minor arc $OC$.

Also note,

$∠FAP = ∠LAB = ∠LCB = ∠OLC$ as $ALBC$ is cyclic and $OL ∥ BC$.

Thus $∠OFP = ∠FAP$. As claimed.

Now notice, $∠OFA = ∠OFP + ∠PFA = 90$. But by the claim, $∠PFA + ∠FAP = 90$

$\implies$  $∠PFA = 90 - ∠FAP$. Thus $\triangle FAP$ is a right triangle. Hence, the result.


问:
Proof/Counterexample: Any two 6x6 matrices are similar of they have the same rank and same minimal polynomial
I am asked to prove or provide a counterexample for this claim:

Any two $6\times6$ matrices are similar of they have the same rank and same minimal polynomial.

I know two matrices are similar if they have the same Jordan form. I have a hunch that they are similar always in this case and no counterexample would exist but I don't know how to prove it.

Thanks in advance.

Edit: I found this answer which is along the similar lines but how to prove that the Jordan forms will be same?

Body:
Are
$$A=\begin{pmatrix}1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1
\end{pmatrix}$$
and
$$B=\begin{pmatrix}1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&1&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1
\end{pmatrix}$$
similar?
Both have rank $6$ and minimal polynomial $X^2-2X+1$. (This works already with $4\times 4$ matrices)


问:
How to develop patience in mathematics?
How  to develop the patience in mathematics?

When  you read the books like Rudin, Conway, etc., it takes a lot of time for one problem; sometimes $10$ hours each.

I usually  devote one  problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution.

Actually, I have no patience and can't wait for long hours for thinking about the problem.

How  can I improve myself and develop the patience?

答:
Briefly skim through the provided answer. Don't memorise it. Come back at a later date and see if you can now solve the problem. If you can, you have internalised the method. If you can't, you still don't understand it fully.

Rinse and repeat.

Skimming through the work of others is a useful research strategy to develop. Without it you will get bogged down reading even one paper that includes original work.

Title: How to develop patience in mathematics?


问:
Let $z\in\mathbb C$, show using induction that the product of the roots of $z^n =1$ is $1$ when $n$ is odd
Let $z\in\mathbb C$, show using induction that the product of the roots of $z^n =1$ is $1$ when $n$ is odd.

I am taking a class, in which we were asked to prove this using an induction. The other material we have learned is De Moivre and trigo representation of complex numbers. I saw other answers, but they are outside of the scope of the class.

答:
The roots of $z^n=1$ are $\exp(k 2 \pi /n)$ for $k=0,\dots,n-1$.

Therefore, their product is $$\exp(\sum_{k=0}^{n-1} k \frac{2 \pi}{n})=\exp(\frac{n(n-1)}{2}\frac{2 \pi}{n})=\exp((n-1)\pi)=1$$ because $n-1$ is even.

The same argument proves that the product is $-1$ when $n$ is even.


问:
Does there exist a $4\times4$ matrix $A$ such that $A$ and $A^2$ are not normal but $A^3$ is normal?
I'm asked to provide a proof or a counterexample for this claim:

Does there exist a $4\times4$ matrix $A$ such that $A$, $A^2$ is not normal, but $A^3$ is normal.

Any ideas are welcome Thanks in advance.

答:
Note: An upper triangular matrix is normal iff it is diagonal.

This suggests
$$\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix}. $$


问:
Simplify with fractional exponents and negative exponents
I am trying to simplify

$$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$

It seems pretty simple at first. I know that a negative exponent means you flip a fraction. So I flip it.

$$  \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$

Now I need to square it, which is tricky because there are a lot of weird rules with squaring. This is probably where I went wrong.

$$  \left(\frac{x^2 y^{-1/2}}{9x^{3/2}y^3}\right)$$

Now I need to try and simplify things. I know that I can get rid of the $x$ on top since there is a larger one on the bottom.

$4/2 - 3/2 = 1/2$

$$  \left(\frac{x^{1/2} y^{-1/2}}{9y^3}\right)$$

Now I need to get rid of the $y$ exponent.

I am not sure how that is possible.

答:
So we have $$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$

Yes you are right you can "flip" the fraction to remove the negative exponent.

$$  \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$

Here you multiply every exponent by 2 to square the expression inside the brackets. This is where you went wrong here.

$$  \left(\frac{x^4 y^{-1}}{9x^{3}y^6}\right)$$

But $y^{-1}=1/y$

so the above is equal to

$$  \left(\frac{x^4}{9x^{3}y^7}\right)$$

Now cancel the $x^3$

$$  \left(\frac{x}{9y^7}\right)$$

And we're done.

ContentLicense: CC BY-SA 3.0


问:
Real-valued analytic function (complex)
I am studying elementary complex variable topic. An exercise in Brown-Churchill book asks us to show that if $f$ is analytic in a domain $D$ and real-valued at all points then $f$ must be constant throughout $D$.

This should be doable by following the proof of showing zero derivative implies constant. However, can I use the fact (which is already proved in the book): If a function and its conjugate are both analytic in $D$, then $f$ is constant there. ???

I am thinking that since in the exercise, $f$ and its conjugate are equal because $f$ is real-valued, then they are both analytic as $f$ is, and the conclusion follows. Is this all right?

Any help is appreciated! Thanks for the guidance!

答:
Here is an application of Cauchy Riemann equations.

Suppose $f = u + iv$ be real valued , analytic function then $v = 0$. So by Cauchy Riemann equation we must have $u_x = u_y = 0$ implying that $u$ is a constant and consequently $f$ is constant.

Title: Real-valued analytic function (complex)


问:
Distance of two points in a rotated rectangle
I have a rectangle which I need to rotate (blue rectangle in the image). The rotation origin is a random point somewhere inside of this rectangle (point Z in the image). The rotation is ALWAYS clockwise. Actually, the angle is set as well but that's not really important.

After the rotation I need to find out the distance between the points A and Z on the x axis and D and Z - on the y axis. So basically I need to find the coordinates of point Z inside the "wrapping" (imaginary) rectangle that would exactly contain the blue rectangle and would not be rotated.

I was trying to translate the rotation center to 0, 0 but I am honestly confused with the swapped axis and the clockwise rotation. How should I go about it?

Any help appreciated! Thanks a lot.

答:
Since we are interested only in the distances related to the rotation center, nothing changes if we move the rotation center to the origin and consider all the coordinates with respect to that point. So consider a "straight" (meaning, parallel to the $y$-axis) line $AD$, so that $A=(x_0, a_0)$ and $D=(x_0, d_0)$. We rotate this line about the origin and call the new line $A'D'$ with coordinates $A'=(a_x, a_y)$ and $D'=(d_x,d_y)$.

Notice that both points $A$ and $D$ travel the angle $\theta$ about the origin. Therefore, it is easy to compute the new coordinates of each point. The standard rotation matrix is
$$
R_{\text{counter-cw}} =
\left[
\begin{array}{cc}
\cos \theta & -\sin{\theta} \\
\sin \theta & \cos \theta
\end{array}
\right]
$$
but this is defined for counterclockwise rotations. Therefore, we have to just flip the sign of the angle (or take the transpose - these are the same in this case), giving
$$
R_{\text{cw}} =
\left[
\begin{array}{cc}
\cos \theta & \sin{\theta} \\
-\sin \theta & \cos \theta
\end{array}
\right]
$$
Now, the new coordinates are obtained by matrix multiplication:
$$
\left(
\begin{array}{c}
a_x \\ a_y
\end{array}
\right)
= R_{\text{cw}}
\left(
\begin{array}{c}
x_0 \\ a_0
\end{array}
\right)
=
\left(
\begin{array}{c}
 x_0 \cos \theta +  a_0\sin \theta \\
- x_0\sin \theta +  a_0\cos \theta
\end{array}
\right)
$$
$$
\left(
\begin{array}{c}
d_x \\ d_y
\end{array}
\right)
= R_{\text{cw}}
\left(
\begin{array}{c}
x_0 \\ d_0
\end{array}
\right)
=
\left(
\begin{array}{c}
 x_0 \cos \theta +  d_0\sin \theta \\
- x_0\sin \theta +  d_0\cos \theta
\end{array}
\right)
$$
You wanted the distance in the $x$-direction, so it's easy to just take the absolute values of the $x$-coordinates of these results.


答:
If $\omega$ is a root of $z^n=1$, then so is $\bar\omega$. Moreover, $\omega\bar\omega=1$. When $n$ is odd, there is only one real root, $z=1$.
Therefore, in the product of all roots, pairs of conjugate roots reduce to $1$, leaving just the real root $1$.

The same argument proves that the product is $-1$ when $n$ is even.

Body:
Let $z\in\mathbb C$, show using induction that the product of the roots of $z^n =1$ is $1$ when $n$ is odd.

I am taking a class, in which we were asked to prove this using an induction. The other material we have learned is De Moivre and trigo representation of complex numbers. I saw other answers, but they are outside of the scope of the class.


问:
Moving exterior differential/derivative inside a wedge product
Assumptions: Let $M$ be smooth $m$-manifold. (If needed: Let $M$ be orientable and then oriented. Let $M$ be compact. Let $(M,g)$ be a Riemannian manifold.)

Let $\Omega^jM$ be the set of smooth $k$-forms on $M$, for $j=0, 1, ..., m$. Let $d_j: \Omega^jM \to \Omega^{j+1}M$ be exterior differential / derivative on $\Omega^jM$ (based on $d: \Omega(M) \to \Omega(M)$, with $\Omega(M)$ $:= \bigoplus_{j=0}^{m} \Omega^jM$).

Let $k \in \{0, 1, ..., m\}$. Let $(\alpha, \gamma) \in \Omega^kM \times \Omega^{m-(k+1)}M$.

Observations:

$d_k \alpha \wedge \gamma$ is a smooth top form (aka smooth $m$-form)

$(-1)^{1+k^2} \alpha \wedge d_{m-(k+1)}\gamma$ is a smooth top form (aka smooth $m$-form)

Question 1: Assuming the above observations are correct, are they equal?

Question 2: In general, can we just move exterior differential/derivative through wedge products and just multiply $(-1)^{\text{something}}$?

Question 3: In anything above, are we assuming any additional things on $M$ like orientable/oriented/compact/Riemannian?

Question 4: If no to question 1, then do each of the 2 forms at least have equal integrals, i.e. the values we get when we plug each into $\int_M$ are equal? Here, we now suppose $M$ is orientable and then oriented and I guess compact (otherwise I guess we have to assume the forms have compact support or something).

Context: This comes from some definitions and propositions leading to Hodge decomposition theorem, including the definition of Hodge star operator, but I'm trying to see if I understand the non-Hodge parts correctly. ($\gamma$ is actually the image of some $\beta \in \Omega^{k+1}M$ under the Hodge-star operator.)

答:
Here is an attempt of an answer.

Question 1 There is no need for an equality like that. What is true is that
$$
d\left(\alpha\wedge \gamma \right) = d\alpha \wedge \gamma + (-1)^{\deg\alpha}\alpha \wedge d\gamma
$$

And assuming your equality to be true will lead to an assumption on $d(\alpha\wedge\gamma)$

Here is a concrete counter-example:
\begin{align}
\alpha &= dx^1 & \gamma = x^2dx^3\wedge\cdots\wedge dx^n \\
d\alpha \wedge \gamma &= 0 & \alpha \wedge d\gamma = dx^1\wedge\cdots\wedge dx^n
\end{align}

Question 2 the answer is no. See above.

Question 3 above, the computations are local, so it does not depend on compactness or orientability: extend the counterexample by zero outside a chart.

Question 4 the answer is still no: in the counterexample above, $d\alpha\wedge \gamma = 0$, thus has zero integral, but $\alpha\wedge d\gamma$ is a volume form on an orientable manifold, it has non-zero integral.

Regarding @JanBohr's answer, (which leads to two self-refereing answers), I have to add that in case $M$ is oriented, then Stokes theorem states that
$$
\int_M d(\alpha\wedge \gamma) = \int_{\partial M} \alpha\wedge \beta
$$
and thus,
$$
\int_M d\alpha \wedge \gamma = (-1)^{\deg \alpha+1}\int_{M}\alpha\wedge d\gamma + \int_{\partial M}\alpha\wedge \gamma
$$
and thus there is (up to sign) an equality as soon as $M$ has no boundary or $\alpha\wedge \gamma$ is zero on $\partial M$.


答:
It may have been easier to simplify the inside first.  In dividing, you subtract exponents.

$$(\frac{3x^\frac32y^3}{x^2y^{-\frac12}})^{-2}=(3x^{\frac32-2}y^{3+\frac12})^{-2}=(3x^{-\frac12}y^\frac72)^{-2}$$

Now to clear the parentheses, multiply each exponent by $-2$ to get

$$3^{-2}x^1y^{-7}=\frac x{9y^7}$$

When you attempted to square everything, you squared only the $3$.

Title: Simplify with fractional exponents and negative exponents

Body:
I am trying to simplify

$$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$

It seems pretty simple at first. I know that a negative exponent means you flip a fraction. So I flip it.

$$  \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$

Now I need to square it, which is tricky because there are a lot of weird rules with squaring. This is probably where I went wrong.

$$  \left(\frac{x^2 y^{-1/2}}{9x^{3/2}y^3}\right)$$

Now I need to try and simplify things. I know that I can get rid of the $x$ on top since there is a larger one on the bottom.

$4/2 - 3/2 = 1/2$

$$  \left(\frac{x^{1/2} y^{-1/2}}{9y^3}\right)$$

Now I need to get rid of the $y$ exponent.

I am not sure how that is possible.


问:
An Inequality with increasing $3$'s and a Fibonacci sequence.

Prove that :- $\frac{\sqrt{n+1}}{n} + \frac{\sqrt{n+4}}{n+3} + \frac{\sqrt{n+7}}{n+6} + \frac{\sqrt{n+10}}{n+9} + \frac{\sqrt{n+13}}{n+12} < \frac{1}{\sqrt{n-1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+5}} + \frac{1}{\sqrt{n+8}} + \frac{1}{\sqrt{n+11}}$ .

What I Tried: I first think that the Right-Hand-Side should contain, $\Big(\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+5}} + \frac{1}{\sqrt{n+8}} + \frac{1}{\sqrt{n+11}}\Big)$ instead of that $(n-1)$ , so I think that is a typo given in the question and that can be ignored.

Other than that, as the title suggests, I can see that the left-hand-side has the numerator and the denominator to be increasing by $3$ , and the terms of the denominator in the right-hand-side, form the fibonacci sequence.

So an usual question comes, can this inequality be generalised to more terms?

As a try for solving it, as I am a novice in Inequality I tried AM-GM and C-S , but they definitely won't work here I guess, because the Inequality seems a bit unusual. Another idea was that I thought of proving each separate term is greater than it's corresponding term in the right hand side, forming a bijection. But that did not work either, as I saw $\frac{\sqrt{n+1}}{n} > \frac{1}{\sqrt{n-1}}$ .

Can anyone help me here? Thank You.

答:
The inequality  is OK, there is no typo.

BlockquoteI thought of proving each separate term is greater than it's corresponding term in the right hand side, forming a bijection. But that did not work either, as $\frac{\sqrt{n+1}}{n} > \frac{1}{\sqrt{n-1}}$.

This luckily is  not true. We have
$$\frac{\sqrt{n+1}}{n} < \frac{1}{\sqrt{n-1}},$$
as
$$\sqrt{n+1}\sqrt{n-1} < n,$$
$$\sqrt{n^2-1}<n.$$

Summing up, we get the inequality.


答:
One of the defining properties of the exterior differential is the Leibniz rule $$d(\alpha\wedge \gamma)=d\alpha\wedge \gamma+(-1)^{k} \alpha\wedge d\gamma,$$ where $k$ is the degree of $\alpha$, see e.g. on wikipedia. This holds true for arbitrary smooth manifolds, no need for a Riemannian metric or orientation. As $k$ and $k^2$ have the same parity, the right hand side in the previous display is exactly the difference between your two $m$-forms. In particular they are equal iff $\alpha \wedge \gamma$ is closed. The integral over both $m$-forms, say if $M$ is oriented and compact, is the same just because the integral of an exact form is zero by Stokes' theorem.

Regarding @DIdier_'s counterexample for question 4: This is a situation where the boundary integral in Stokes' theorem does not vanish (for any smooth domain in $\mathbb{R}^n$). Above I avoid this problem by assuming $M$ to be boundaryless. Another way out is to assume that $\alpha $ and $\gamma$ have compact support in the interior.

Title: Moving exterior differential/derivative inside a wedge product

Body:
Assumptions: Let $M$ be smooth $m$-manifold. (If needed: Let $M$ be orientable and then oriented. Let $M$ be compact. Let $(M,g)$ be a Riemannian manifold.)

Let $\Omega^jM$ be the set of smooth $k$-forms on $M$, for $j=0, 1, ..., m$. Let $d_j: \Omega^jM \to \Omega^{j+1}M$ be exterior differential / derivative on $\Omega^jM$ (based on $d: \Omega(M) \to \Omega(M)$, with $\Omega(M)$ $:= \bigoplus_{j=0}^{m} \Omega^jM$).

Let $k \in \{0, 1, ..., m\}$. Let $(\alpha, \gamma) \in \Omega^kM \times \Omega^{m-(k+1)}M$.

Observations:

$d_k \alpha \wedge \gamma$ is a smooth top form (aka smooth $m$-form)

$(-1)^{1+k^2} \alpha \wedge d_{m-(k+1)}\gamma$ is a smooth top form (aka smooth $m$-form)

Question 1: Assuming the above observations are correct, are they equal?

Question 2: In general, can we just move exterior differential/derivative through wedge products and just multiply $(-1)^{\text{something}}$?

Question 3: In anything above, are we assuming any additional things on $M$ like orientable/oriented/compact/Riemannian?

Question 4: If no to question 1, then do each of the 2 forms at least have equal integrals, i.e. the values we get when we plug each into $\int_M$ are equal? Here, we now suppose $M$ is orientable and then oriented and I guess compact (otherwise I guess we have to assume the forms have compact support or something).

Context: This comes from some definitions and propositions leading to Hodge decomposition theorem, including the definition of Hodge star operator, but I'm trying to see if I understand the non-Hodge parts correctly. ($\gamma$ is actually the image of some $\beta \in \Omega^{k+1}M$ under the Hodge-star operator.)


问:
find $\varepsilon$ such that $P(X\geq a+\varepsilon \land X\leq b - \varepsilon)=0.95$, with $X$ is uniformly distributed.
Sketching the situation using the density function of the uniform distribution on $[a,b]$
its clear to me that is just a $0.025$ of $V:=∫_Ω \mathbb 1_Ω(x)\ dx$ it can't be simply $\varepsilon:= a+b\cdot 0.025$ or can it? Since I'm unsure if this holds when $\Omega \subset \mathbb R^n$. Is the skech even correct?

答: The given probability is just $\frac {(b-\epsilon) -(a+\epsilon)} {b-a}$. Setting this equal to $0.95$ and solving for $\epsilon$ gives $\epsilon=(0.025) (b-a)$.


问:
Convergence of $\sum^{\infty}_{n=1}\frac{2^{-n}}{n^2}z^n$ when $\mid z\mid =2$
I know that it converges everywhere on the circle $\mid z\mid=2$, but how do I show it? Something like comparison test or alternating series test doesn't work. Is there any general theorem to see the convergence on the boundary? Or is it something that is hard to see?

答:
By Cauchy Hadamard, $r=\limsup_{n\to\infty}\dfrac1{\sqrt[n]{\dfrac{2^{-n}}{n^2}}}=\dfrac1{\dfrac12}=2$.

When $z=2$, we get $\zeta(2)=\dfrac{\pi^2}6$.

When $|z|=2$, the sum is absolutely convergent, by comparison with $\zeta(2)$.


问:
Least square method to fit data according to $g(x) = a - x + bx^2$, where $a, b \in \mathbb{R}$
If I have a set of points $(x_i, f(x_i)) \in \mathbb{R^2} $ and I want to use the method of the the least square fitting to calculate the free parameters $a$ and $b$ of the function
$$
g(x) = a - x + bx^2
$$

By looking for the function $g$, one possible basis of function would be:
$$
\phi_1(x) = 1,  \phi_2(x)  = x^2
$$
And the missing function of $g$ would cause a translation of the linear space of the solutions. Is this correct?

I don't know to proceed with this exercise.
Could anyone give me a hint?
Thanks in advance.

答:
You have $n$ data points $(x_i,y_i)$ and you want to minimize
$$\text{SSQ}=\sum_{i=1}^n \big(a-x_i+b x_i^2-y_i\big)^2$$ Start defining two new variables$$z_i=y_i+x_i\qquad \text{and} \qquad  t_i=x_i^2$$ which make
$$\text{SSQ}=\sum_{i=1}^n \big(a+b\, t_i-z_i\big)^2$$  Setting the partial derivatives $\frac{\partial \text{ SSQ}}{\partial a}$ and $\frac{\partial \text{ SSQ}}{\partial b}$ equal to $0$, we have to solve the normal equations
$$n \,a +b \sum_{i=1}^n t_i=\sum_{i=1}^n z_i$$
$$a\sum_{i=1}^n t_i+b\sum_{i=1}^n t_i^2=\sum_{i=1}^n t_iz_i$$ that I shall rewrite as
$$n a +b S_t=S_z$$
$$a S_t+b S_{tt}=S_{tz}$$ and the exact solutions are
$$a=\frac{S_{tt}\, S_z-S_t \,S_{tz}}{n \,S_{tt}-S_t^2}\qquad \text{and} \qquad b=\frac{n\, S_{tz}-S_t\, S_z}{n \,S_{tt}-S_t^2}$$


问:
$P_n$ is a.s. to P, $Q_n$ is L1 to Q. $P(P_n \neq Q_n) \leq 1/n$. Is $P=Q$ a.s.?
$\left\{P_{n}\right\}_{n=1,2, \ldots}$ and $\left\{Q_{n}\right\}_{n=1,2, \ldots}$ are two sequences of r.v.s. For every $n$ , $\mathbb{P}\left(P_{n} \neq Q_{n}\right) \leq 1 / n . P_{n} \stackrel{\text {a.s.}}{\rightarrow} P$ and $Q_{n} \stackrel{L^{1}}{\rightarrow} Q$. Can we conclude that $P=Q$ a.s.?

Is it right?

答: Consider a subsequence of $P_n$, $P_{mn} = P_{mn} - Q_{mn} + Q_{mn}$


Body:
$$ \left(\frac{3x ^\frac{3}{2}y^3}{x^2 y^{\frac{-1}{2}}}\right)^{-2}=(3x^{3/2-2}y^{3-(-1/2)})^{-2}=$$
$$=(3x^{-1/2}y^{7/2)})^{-2}=3^{-2}x^{(-1/2)(-2)}y^{(7/2)(-2))}=$$
$$=3^{-2}x^1y^{(-7)}=\frac{x}{9y^7}$$


答:
That is not a typo. If you think that the inequality holds with $n+1$, it should hold with $n-1$ too (why?). Also, there is nothing related to Fibonacci series here. The Fibonacci series is $1,1,2,3,5,8,13,21,\dots$.

Recheck your claim that $\frac{\sqrt{n+1}}{n}>\frac1{\sqrt{n-1}}$. Its incorrect. Try substituting $n=2$ or $n=3$ and see it for yourself.

Another idea was that I thought of proving each separate term is greater than it's corresponding term in the right hand side.

This idea does work indeed. Continue your approach. Try proving that
\begin{align*}
\frac{\sqrt{n+i+2}}{n+i+1}<\frac1{\sqrt{n+i}}
\end{align*}
for any $i$ for which the denominators are positive and square roots are well defined.

Title: An Inequality with increasing $3$'s and a Fibonacci sequence.

Body:
Prove that :- $\frac{\sqrt{n+1}}{n} + \frac{\sqrt{n+4}}{n+3} + \frac{\sqrt{n+7}}{n+6} + \frac{\sqrt{n+10}}{n+9} + \frac{\sqrt{n+13}}{n+12} < \frac{1}{\sqrt{n-1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+5}} + \frac{1}{\sqrt{n+8}} + \frac{1}{\sqrt{n+11}}$ .

What I Tried: I first think that the Right-Hand-Side should contain, $\Big(\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+5}} + \frac{1}{\sqrt{n+8}} + \frac{1}{\sqrt{n+11}}\Big)$ instead of that $(n-1)$ , so I think that is a typo given in the question and that can be ignored.

Other than that, as the title suggests, I can see that the left-hand-side has the numerator and the denominator to be increasing by $3$ , and the terms of the denominator in the right-hand-side, form the fibonacci sequence.

So an usual question comes, can this inequality be generalised to more terms?

As a try for solving it, as I am a novice in Inequality I tried AM-GM and C-S , but they definitely won't work here I guess, because the Inequality seems a bit unusual. Another idea was that I thought of proving each separate term is greater than it's corresponding term in the right hand side, forming a bijection. But that did not work either, as I saw $\frac{\sqrt{n+1}}{n} > \frac{1}{\sqrt{n-1}}$ .

Can anyone help me here? Thank You.


问:
CLT Application: converge in distribution to N(0,1)
$Y_{1}, Y_{2}, \ldots$ is a seq. of independent r.v.s, which meets the condition $\left|Y_{i}\right| \leq H$ for all $i$ for some positive $H$. If $\sum_{n=1}^{\infty} \operatorname{Var}\left(Y_{n}\right)=\infty .$ Define $S_{n}=Y_{1}+\cdots+Y_{n}$ for $n=1,2, \ldots$

How to prove that
$$
\frac{S_{n}-\mathbb{E}\left[S_{n}\right]}{\sqrt{\operatorname{Var}\left(S_{n}\right)}} \stackrel{d}{\rightarrow} \mathcal{N}(0,1)
$$
where $\mathcal{N}(0,1)$ denotes the standard normal distribution.

My idea:
I want to use Lindeberg-Feller CLT to show this, but I am not sure how to check the condition.

答: You can try $X_n = \frac{Y_n - \mathbb{E}\left[Y_{n}\right]}{Var(S_n)}$, then apply Lindeberg-Feller CLT


问:
Is there any way to calculate $\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$ faster?
During solving a problem I get this expression:

$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$$

To calculate it normally, I do this way:

$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}=2\binom{10}{1}-2\binom{10}{3}+\binom{10}{5}$$
And here we have to evaluate $\binom{10}{3}$ and $\binom{10}{5}$. after all I got $32$ as the result correctly.

My question: Is there any other way to calculate this expression easier (faster)? I suspect there is because the answer is $32=2^5$ and I guess maybe there is other way to calculate it for example by combinations or other ways.

Body:
$$(1+i)^{10} = {10\choose 0} + {10\choose 1}i -{10\choose 2}-{10\choose 3}i +{10\choose 4}+ ... -{10\choose 10} $$

$$(1-i)^{10} = {10\choose 0} - {10\choose 1}i -{10\choose 2}+{10\choose 3}i +{10\choose 4}+...-{10\choose 10}$$

So $$...={(1+i)^{10}-(1-i)^{10}\over 2i} = {1\over 2i}((2i)^5-(-2i)^5) = 32$$


问:
SO(2,1) not connected
I am trying to show that $SO(2,1)$ is not connected but I have no idea where to start really, I know that it is connected if there is a path between any two points. My definition of $SO(2,1)$ is:

$SO(2,1)=\{X\in Mat_3(\mathbb{R}) \mid X^t\eta X=\eta, \ \det(X)=1\}$ where $\eta$ is the matrix defined as: $$\left ( \begin{array}{ccc} 1 &0&0\\0&1&0\\0&0&-1\end{array}\right )$$

Thanks for any help

答:
Consider the transitive action of $O(2,1)$ on $N_{-1}= \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2-z^2=-1 \}$ by left multiplication. For some $x \in N_{-1}$, you get a continuous surjection $$\left\{ \begin{array}{ccc} O(2,1) & \to & N_{-1} \\ M & \mapsto & M \cdot x \end{array} \right. .$$

So you have $SO(2,1) \hookrightarrow O(2,1) \twoheadrightarrow N_{-1}$. But $N_{-1}$ is a two-sheeted hyperboloid, and the image of $SO(2,1)$ in $N_{-1}$ is not contained in one sheet of $N_{-1}$ since $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right) = \left( \begin{matrix} 0 \\ 0 \\ -1 \end{matrix}\right)$.

Because a continuous image of a connected set is connected, $SO(2,1)$ has at least two connected components.

In fact, it can be shown that the connected component of $SO(p,q)$ containing $\operatorname{Id}$ is $$SO_0(p,q)= \left\{ \left( \begin{matrix} A & B \\ C & D \end{matrix} \right) \in SO(p,q) \mid A \in GL_p(\mathbb{R}), \ \det(A) >0 \right\}.$$

Therefore, $SO(p,q)$ is connected iff $p=0$ or $q=0$.

Title: SO(2,1) not connected


问:
Show that This infinite product is entire
This question is from Ponnusamy and silvermann complex analysis and I am making some mistake in this question and unable to find it.

Show that $\prod_{n=1}^{\infty} (1-z/n) e^{z/n +5z^2/n^2}$ is entire.

$1+ f_n (z)= (1-z/n) e^{z/n +5z^2/n^2}$ and finding a upper bound on $f_n (z)$ . $1+ f_n (z) = exp( Log(1-z/n) + z/n+ 5 z^2 /n^2)$ For $|z|\leq R$ then choose N large enough such that |z/n|<1 for all $n \geq N$ and using the identity$ Log(1-z/n) = -\sum_{k=1}^{\infty} 1/k (z/n)^k$ and the fact that|R/n|<1.

So, $|Log (-\sum_{k=1}^{\infty} 1/k (z/n)^k +z/n +5 z^2/n^2) | \leq  -5 -1/2 R^2/n^2 $, where I used Formula of infinite geometric series.

Now $|f_n(z)| \leq exp|e^{-5 -1/2 R^2 /n^2 -1}-1| exp( -5 -1/2 R^2 /n^2)$ -1 and using $e^x -1 \leq x e^x  for all x \geq 0 $ we get $|f_n(z)|\leq 5+ 1/2 R^2/ n^2 $ and if I sum $|f_n (z)| $ from N to $\infty$ i get Infinity as 5 is also there .

So, I think while approximating I am doing some mistake and I am unable to find it. Can you please point that out.

If there is some other method of proving it entire that is also welcome.

Thanks!

答:
Hint.

Theorem: Suppose $C$ is an open subset of $\Bbb C$ and $(f_n)_n$ is a sequence of complex functions that are analytic on $C,$ such that $f_n$ converges uniformly on $D$ to $f$ whenever $D$ is a compact subset of $C$. Then $f$ is analytic on $C.$


问:
Finding matrices for which $x^2+1$ is the minimal polynomial

Show that there is no $A\in M_3(\mathbb{R})$ whose minimal polynomial is $x^2+1$, but there is $B\in M_2(\mathbb{R})$ and $C\in M_3(\mathbb{C})$ whose minimal polynomials are both $x^2+1$. $M_n(\mathbb{F})$ denotes the vector space of $n\times n$ matrices over the field $\mathbb{F}$.

My work:

So if there exists $A\in M_3(\mathbb{R})$ whose minimal polynomial is $x^2+1$, its characteristic polynomial must be of the form $(x-a)(x^2+1)$ for some $a$, since the minimal polynomial divides the characteristic polynomial and the characteristic polynomial has degree $3$. I'm not able to find a contradiction from here.

For $B\in M_2(\mathbb{R})$, by a similar argument, $x^2+1$ must be $B$'s characteristic polynomial. Another thing I noticed is that $B$ cannot be diagonalizable, since if it were, $x^2+1$ should have split into distinct linear factors over $\mathbb{R}$, but that is clearly not the case. Doing a brute force calculation, considering $$B = \left[\begin{matrix} a & b \\ c & d  \end{matrix}\right]$$ we get $x^2 - (a+d)x + ad - bc = x^2 + 1$, so $a+d = 0$ and $ad-bc=1$. This gives $a^2+1 = -bc$. Choosing $(a,b,c,d) = (1,-2,1,-1)$ yields a desirable $B$.

I am planning on doing something similar for $C\in M_3(\mathbb{C})$, but seems like it would be overkill. Is there an easier way around this? It'd be great if someone could help me out with disproving the existence of $A$, proving the existence of $C$, and finding a somewhat better solution for the existence of $B$.

Thanks!

答:
For the 3 by 3 matrix consider
$
M=
\begin{bmatrix}
i&0&0\\
0&i&0\\
0&0&-i
\end{bmatrix}
$
This has minimal polynomial $x^2 +1$


问:
A question regarding the metric space $(\mathbb N , d)$
Give an example of a metric $d$ on $\mathbb N$  so that $n \to 1$  in $(\mathbb N , d)$.

Can anyone please tell me what is $n \to 1$ here ? I have really no clue.

答:
Try the following: Let $\Phi: \mathbb{R} \rightarrow \mathbb{R}$ be
$$
\Phi(x) :=
\begin{cases}
0 & \text{ if } x=1 \\
\exp(-x) & \text{ else}
\end{cases}.
$$
Note that $\Phi$ is injective. We define $d: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{R}$ as $d(n, m) := \lvert \Phi(n)- \Phi(m) \rvert$.

Since $\Phi$ is injective, we have $d(n, m) = 0 \iff \Phi(n) = \Phi(m) \iff n = m$. Symmetry is obvious and the triangle inequality we get from the $\lvert \cdot \rvert$-triangle inequality.
Now let $n$ be very large:
$$
\lim_{n \rightarrow \infty} d(n, 1) = \lim_{n \rightarrow \infty} \lvert \Phi(n) - \Phi(1) \rvert = \lim_{n \rightarrow \infty} \lvert \exp(-n) - 0 \rvert = 0
$$

This is, I admit, not a very elegant solution but it works.

Remark: If $\Phi$ is injective and arbitrary, then $d(x, y) = \lvert \Phi(x) - \Phi(y) \rvert$ always defines a metric on (a subset of) $\mathbb{R}$.


问:
The Schur complement is matrix concave
I am confused regarding the Schur complement. From Boyd and Vandenberghe's Convex Optimization,

Suppose $X \in S^n_{++}$ partitioned as $$X = \begin{bmatrix} A & B\\ B^T & C\\
\end{bmatrix}$$ where $A \in S^k$. The Schur complement of $X$ (with respect to $A$) is $S=C-B^TA^{-1}B$. Show that the Schur complement, viewed as function from $S^n$ into $S^{n-k}$, is matrix concave on $S^n_{++}$.

I believe the goal is to show that $y^TSy\leq0$, where $y\in\mathbb{R}^{n-k}$. So, I follow a similar argument as given in the book (i.e., Example 3.4). I define $h(y,S)=y^TSy$. Then, I think we should consider the hypograph? That is, if the hypograph of $h$ is convex set, then $h$ is concave. Following this idea, I have the following:

$$\begin{align}
\mathbf{hypo}\, h &= \left\{ (y,S,t) \mid y^TSy\geq t \right\} \\ &= \left\{ (y,S,t) \mid \begin{bmatrix} S & y \\ y^T & t \end{bmatrix} \leq 0 \right\}
\end{align}$$

Since the last line can be expressed as a linear matrix inequality (LMI) of $(y,S,t)$, the hypo of $h$ is a convex set. This implies that $S$ is matrix concave? But I can also consider the epigraph isn't it? Then, I'll get a similar LMI with a different inequality. So, does that means it is both convex and concave? Moreover, if $X \in S^n_{++}$, it also implies that $A$ and $S$ is positive definite? Do they play role here? I'm very confused.

答:
You want to show matrix concavity:
$$
S(\theta X_1 + (1-\theta) X_2) \succeq \theta S(X_1) + (1-\theta) S(X_2), \ \  \forall \theta \in [0,1], X_1, X_2 \in S^n_{++}
$$
This is equivalent to showing that the matrix hypograph is a convex set:
$$
\mathbf{hypo}\ S:  =\{ (X,T) \ | \ S(X) \succeq T, \ X \in S^n_{++}, \ T \in S^{n-k} \}
$$
But then we can use properties of the Schur Complement to show for positive definite $X$ we have:
$$
\ S(X) = C - B^T A^{-1} B \succeq T \Leftrightarrow  \begin{bmatrix} A & B\\ B^T & C- T \end{bmatrix}  \succeq 0\Leftrightarrow \ X -  \begin{bmatrix} 0 & 0\\ 0 & T \end{bmatrix} \succeq 0
$$
So let $L$ be the linear map $S^n \times S^{n-k} \rightarrow S^n$:
$$
L(X,T) := X -  \begin{bmatrix} 0 & 0\\ 0 & T \end{bmatrix}
$$
We conclude:
$$
\mathbf{hypo}\ S  =\{ (X,T) \ | \ L(X,T) \in S^n_{+}, \ X \in S^n_{++}, \ T \in S^{n-k} \}
$$
So then we can conclude the hypograph is a convex set because the semidefinite cone is.

Title: The Schur complement is matrix concave


问:
A linear operator is non-negative iff all its eigenvalues are non-negative
I am asked to present a proof/ counterexample:

A linear operator $T$ on a finite-dimensional complex inner-product space is non-negative iff each of its eigenvalues are non-negative.

I think this claim is true. My thoughts:

Let $\lambda$ be an eigenvalue of $T$. Then for some nonzero $x$,
$$
\lambda \|x\|^2 = \langle \lambda x,x \rangle = \langle T x,x\rangle  \ge 0.$$

Am I correct or did I miss something?

Thanks in advance

答: If $T$ is non-negative definite then all the eigen values of $T$ are non-negative, as you have shown. Converse is false. Consider the $2 \times 2$ matrix  $a_{12}=1$ and $a_{ij}=0$ for all other $i,j$. Then the only eigen value is $0$ but this is not non-negative definite.


问:
Find the smallest $4$ digit number which is a factor of $2005^6 - 1$.
Find the smallest $4$ digit number which is a factor of $2005^6 - 1$.

My attempt:

We see that $(2005-1)$ is a factor of $2005^6-1$ by the factorization of $2005^6-1$. Therefore, $1002$ is a factor of $2005^6-1$. I then have to show that $1001$ is not a factor but how do I do that? Should I use the factorization of $x^6-1$?

Body:
Let's try basic algebra to figure this out:

$$ (2,005^6 = 64,966,020,037,537,515,625) - 1 = 64,966,020,037,537,515,624 $$

We already know that this number isn't divisible by $1,000$, so what about the others?

$$ \frac{2,005^6 - 1}{1,001} = 64,901,118,918,618,896.72 $$
$$ \frac{2,005^6 - 1}{1,002} = 64,836,347,342,851,812 $$

The real answer to this is $1,002$ then.


问:
No-Adjacent objects using inclusion exclusion
Teacher X, Y, Z and 6 students sitting in  a row. How many ways can the teachers choose their seats so that there are no 2 adjacent teachers?

I know that you can solve it by: (s) = student. We put 1 student in between XY and YZ to satisfy the condition

(s) X (s) Y (s) Z (s) then using stars and bars to solve $a+b+c+d=4$ then multiplying by 3! because X,Y,Z can be arranged we have: $7C3 * 3!$ or $35*3!$

Now I tried with Inclusion Exclusion,

Choosing any 3 in the 9 spots $9C3$ , This includes the set with at least 1 adjacent pair of teachers.

To get the number of ways for at least 1 adjacent pair of teachers, we pair 2 of them at the start, so we have to choose 2 slots in 8 objects. $8C2$. But $9C3 -8C2$ = 56. So I don't know what I didn't count

答: Say $A$ is a set where $X,Y$ are together nad similary for $B$ and $C$. Then you need $$|U| -|A\cup B\cup C|$$ where $U$ is a set of all confiugration, so $|U| = 9!$. Then $$|A| = |B|=|C| = 2\cdot 8!$$ and $$|A\cap B| = |B\cap C| = |C\cap A |= 2\cdot 7!$$ and $|A\cap B\cap C| = 0$. Now use the PIE.


问:
Counterexample: There exist non-zero nilpotent matrices $A$ and $B$ that $AB$ is non-negative
I am asked to prove or state a counterexample for the following statement:

There exist non-zero nilpotent matrices $A$ and $B$ such that $AB$ is non-negative.

Any ideas would help a lot as I'm not able to come up with a counteraxample. Thanks in advance

答:
Look at

$$A =\begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix},\space\space\space
B=\begin{pmatrix}
0 & 0\\
1 & 0\\
\end{pmatrix}.$$


问:
Find $x$ if $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$

The relation $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$ holds true for all $1.$ $ x\in \mathbb R$, $2.$ $ x\in (-1,\infty) $, $3.$ $ x\in (-\infty,1) $, $4.$ $ x\in (-\infty,2)$

I took RHS=$\frac{\pi}{4}+\tan^{-1}x=\tan^{-1}1+\tan^{-1}x=\tan^{-1}(\frac{1+x}{1-x})$,if $x\gt0, x\lt1$

(using the property $\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac{x+y}{1-xy})$, if $x\gt0,y\gt0,xy\lt1$)

Therefore, my answer is $x\in(0,1)$, which is not an option.

The answer in the key is $2. $ $x\in(-\infty,1)$.

Looks like, they just used $x\lt1$ condition and not $x\gt0$. Why?

答:
A drawing might help.

Construct a right-angle triangle $\triangle ABC$ with sides $\overline{AB} = 1$ and $\overline{CB} = |x|$. Then
$$\measuredangle BAC =\arctan |x|.$$
Produce $AB$ to $D$ so that $\overline{BD} = \frac{|x|}{|y|}$. As a consequence you have
$$\measuredangle ADC = \arctan |y|.$$

By Euclid's Theorem, if $x^2 < \frac{|x|}{|y|}$, i.e. $|x|<\frac1{|y|}$, the angle $\angle ACD$ is obtuse. Draw from $D$ the line perpendicular to $AC$ and let $H$ be the foot of the altitude.

Similarity $\triangle ABC \sim \triangle ADH$ yield $$\overline{DH} = \frac{|x|(|x|+|y|)}{|y|\sqrt{1+x^2}},$$ and $$\overline{CH} = \frac{|x|(1-|xy|)}{|y|\sqrt{1+x^2}}$$.

External angle theorem gives
$$\angle HCD= \arctan |x| + \arctan |y|,$$
i.e.
$$\arctan\frac{|x|+|y|}{1-|xy|}=\arctan |x| + \arctan |y|,$$
valid for $|xy|<1$.
Thus we have, using $\arctan$'s odd simmetry,

If $x>0$ and $0<y<\frac1x$, or if $x<0$ and $\frac1x < y<0$, then $$\arctan\frac{x+y}{1-xy}=\arctan x + \arctan y$$

If $x>0$ and $-\frac1x<y<0$, or if $x<0$, then and $0<y<-\frac1x$ $$\arctan\frac{x-y}{1+xy}=\arctan x - \arctan y$$

You can analyze the situation when $|xy|>1$ in the same manner, considering that now the angle $\angle ACD$ is acute.


问:
Prove that $\iint_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{\frac{f\left( x \right)}{f\left( y \right)}\text{d}x\text{d}y\ge 1}$
Assume $f(x)>0, x\in[0,1]$. Prove that
$$\iint_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{\frac{f\left( x \right)}{f\left( y \right)}\text{d}x\text{d}y\ge 1}
$$

It gives me the sense that the integration of $\dfrac{f(x)}{f(y)}$ will be small in one part of $[0,1]\times [0,1]$, but large in another part of $[0,1]\times [0,1]$ and finally their sum will be more than $1$. But how to formulate this sense formally? I guess we need to make some inequality scalings. Can anyone help?

答:
Suppose $\int_{0}^{1} f$ and $\int_{0}^{1} \frac{1}{f}$ exists and let $I$ be the given integral,

Observe that the given integral can be written as
$$ I =
\left(\int_{0}^{1} f(x) \, dx \right)\left(\int_0^1\frac{1}{f(x)} \, dx\right)
$$

Then by Cauchy Schwartz inequality for integrals we have
$$
I \geq \left(\int_0^1 1\, dx\right)^2 = 1
$$


问:
Prove that if $|A \cup B|=|A \cap B|$, then $|A|=|B|$
Let A and B be sets. Prove that if $|A \cup B|=|A \cap B|$, then $|A|=|B|$

答:
Hint
$$|A \cup B|=|A|+|B|-|A \cap B|$$

if $|A \cup B|=|A \cap B|$ then
$$|A \cap B|=\frac{|A|+|B|}{2}$$

Suppose $|A|>|B|$, then
$$|A \cap B|=\frac{|A|+|B|}{2}>|B|.$$

However, by definition, $A \cap B\subset B$.

Did you get it?


问:
Exponential integral question
How would I solve the following problem?
$$f(x)=\int\!\frac{4}{\sqrt{e^x}}\,dx$$

Using $u$ substitution I have set $u=e^x$ andd $du=e^x dx$

so would I have $$4\int\!\frac{du}{\sqrt{u}}$$

What would I do from here?

答:
Notice: You may use your method, but in doing so, recall that $\color{blue}{u=e^x}$ andd $du=e^x dx \implies  dx = \dfrac{du}{\color{blue}{\bf e^x}} = \dfrac{du}{\color{blue}{\bf u}}$

This gives us

$$4\int \frac{1}{\sqrt{u}}\frac{du}{u}=4\int u^{-3/2}\,du$$

Title: Exponential integral question


答:
WLOG let $|A| \le |B|.$ Then

$$|B| \le |A \cup B| = |A|+|B| - |A \cap B| \le 2|B|-|A \cap B|.$$

With $|A \cup B|=|A \cap B|$ this gives

$$|B|=|A \cup B|.$$

Can you proceed ?

Body: Let A and B be sets. Prove that if $|A \cup B|=|A \cap B|$, then $|A|=|B|$


问:
What is the natural way to make the set of all holomorphic functions into a Banach space?
Let $E$ and $F$ be locally convex spaces, and $U \subseteq E$ an open subset. Let $\mathcal H(U; F )$
denote the vector space of all holomorphic mappings $f : U \to F$ . When $F = \mathbb C$,
we write $\mathcal H(U)$ instead of $\mathcal H(U; F )$,

What is the natural way to make the set of all holomorphic functions into a Banach space?

If $f$ is assumed to be bounded, We endow $\mathcal H(U)$ with the norm $\lVert f\rVert:=\sup_{z\in U}|f(z)|$, which is well-defined since $f$ is bounded.

Now in  general case,I want to know what norm would be suitable in order to make it complete .

Thanks

答:
Not a very useful one: if $E=\mathbb{C}$, if we equip $\mathcal{H}(U)$ with its natural topology (i.e. the compact open topology), we have Montel's theorem,
which implies that a set is compat in $\mathcal{H}(U)$ iff it is closed and bounded. Since clearly $\mathcal{H}(U)$ is an infinite dimensional vector space (it contains the polynomials), it cannot be a Banach space (it is actually an example of a Montel space)

On the other hand, there is a norm which makes it a complete banach space, since
$\mathcal{H}(U)$ has dimension $2^{\aleph_0}$ and it is thus isomorphic (as a vector space) to a Banach space $B$ of the same dimension. Defining $||v||_{\mathcal{H}(U)}:=||f(v)||_B$ gives $\mathcal{H}(U)$ a Banach space structure


问:
Curvature invariant under isometry but not parallel transport?
I have read that the curvature is invariant under isometry in the general, though I'm urrently reading up a little on symmetric spaces and I've seen that we have

$\nabla R=0$ iff the space is locally symmetric.

But isn't this condition just the statement that curvature is invarient under parallel transport, which is an isometry?

I'm guessing that not all spaces are locally symmetric, but can't seem to figure out where I'm making the mistake.

答:
What is true is that if $f : M \to M$ is an isometry, , $f^*g = g$, and then $f^*\nabla = \nabla$ and consequently this leads to $f^*R = R$. So an isometry of $M$ leaves unvariant the curvature.

But this has nothing to do with parallel transport: say $\gamma : I \to M$ is a smooth path and let $P_t$ be the parallel transport from $\gamma(0)$ to $\gamma(t)$. Then:
$$
P_t : \left(T_{\gamma(0)}M,g_{\gamma(0)}\right) \to \left(T_{\gamma(t)}M,g_{\gamma(t)}\right) $$
is a linear isometry. It is not an isometry of $M$.

In fact, on a connected riemannian manifold, any two points can be joined by a smooth path and thus, there are always parallel transport between them. But they may not even exist an isometry of $M$ sending one point to another given one in the general case. This would imply that $M$ is two-point homogeneous and this has huge consequences on the curvature.


问:
Prove $\operatorname{rank}(A^*A)=\operatorname{rank}(A)$
Let $A$ be a linear operator on a finite-dimensional complex inner product space. Then prove that $$\operatorname{rank}(A^*A)=\operatorname{rank}(A)$$ (the * denotes complex conjugate transpose)

My thoughts:

$$0 = ||Ax||^2 = (Ax)^*Ax = \bar x^T(A^* A) x. $$
Hence

$$Ax = 0 \, \text{iff}  \,  A^* Ax = 0. \tag 1$$

I am not able to continue further. Are my steps correct till now? Any help would be appreciated. Thanks in advance.

答: You've just shown that the kernels of $A$ and $A^*A$ coincide, now use Rank-Nullity.


问:
Is the extension $ℚ(\alpha):ℚ(\alpha^2)$ a Galois extension?
Is the extension $ℚ(\alpha):ℚ(\alpha^2)$ Galois? (for $\alpha$ a complex number)

I understand to show an extension is Galois one must show it is normal and separable.

I've never seen an example like this before so I don't know where to start with the proof. Does anyone have any links to similar questions or sites which may help me understand and solve this problem?

Thank you.

答: A finite field extension $L/K$ is Galois if and only if $L$ is a splitting field of some separable polynomial $f\in K[x]$. In your example, this is the case because $\mathbb{Q}(\alpha)$ is the splitting field of $x^2-\alpha^2\in\mathbb{Q}(\alpha^2)[x]$. Well, at least this polynomial is separable if $\alpha\ne 0$. If $\alpha=0$ then it is a trivial extension, so it is obviously Galois.


问:
Why does $(a, f(a))$ in this graph below not have a tangent line?
Why does $(a, f(a))$ in this graph below not have a tangent line?

Thanks a lot!

答:
Let's try an every-day analogy. When you are rolling a (prismatic) object on a plane (your table), the line along which you roll is the tangent at the contact point. See blue object below.

Now if at some point your object "makes an angle" (i.e, has a kink as shown in your snapshot, red object), then when that point touches the table, several inclinations are possible. This means that you do not have a unique line that can be called the tangent to the object at that point. In fact, you'll find that if you stop right when reaching that point as contact point, the inclination will depend on whether you approach from right or left. This illustrates the fact that there is a discontinuity in the tangent vector at that point.

Note: here I am asking you to visualize a 2D problem (tangent to a curve) with an everyday analogy, so a 3D one. For this to work, you have to be able to discard the direction perpendicular to the rolling one, so to have a prismatic object and consider just a vertical slice through it.

Note 2: my everyday analogue works only with convex shapes, for purely physical reasons (physics applies in addition to maths in everyday life!)


问:
Let $f(x)$ be differentiable on for all $x\in\mathbb{R}$, and $|f'(x)|\le2$, for every $x$, and $f(9)=33$, $f(1)=17$. what is $f(5)=?$
I can't get how I can know stuff about $f(5)$ from the information that I got.
I tried to apply Lagrange: $\exists  c\space such\space  that\space 1<c<9 \space \space :f'(c)=\frac{f(9)-f(1)}{9-1} =2 $
 and I have no idea how to reach $f(5)$ from here.

any hints or tips or solutions will be appreciated!

答:
Answer: $$f(5) = \frac{f(1)+f(9)}{2} = \frac{17 + 33}{2} = 25.$$
Justification: Suppose not.  $f(5) = 25 + \epsilon$ for some $\epsilon > 0$.  Invoke MVT: $\exists c_1 \in (1,5): f'(c_1) = \frac{f(5)-f(1)}{5-1} = \frac{25+\epsilon-17}{4} > 2$, contradicting the condition that $|f'(x)| \le 2$ for all $x$ in the domain.
Exercise: Repeat this type of argument for $f$ on $(5,9)$.


答: Well, this is certainly false for the infinite case, take for example $\mathbb{Z}$ and $2\mathbb{Z}$ (the even numbers). But in the finite case it is true, as $A \cap B \subset A \cup B$ so if their cardinal is equal, then $|A\cup B|=|(A\cap B)\cup ((A\cup B)-(A\cap B))|$ and as this sets are finite and that union is disjoint, we have that $|A \cap B| = |A\cup B|=|(A\cap B)|+| (A\cup B)-(A\cap B)|$ which implies that $| (A\cup B)-(A\cap B)|=0$ and so that set is the empty set and every element of B is in the union an so in the intersection (by the emptyness) and therefore in A and similary every element of A is in the union and then in the intersection and therefore it is also in B. Deducing that both sets are equal

Title: Prove that if $|A \cup B|=|A \cap B|$, then $|A|=|B|$


问:
Milnor's exercise: for any manifold $M$, $\mathrm{Hom}(C^\infty(M,\mathbb{R}),\mathbb{R})\cong M$
I am studying some chapters of Kolar-Michor-Slovak, Natural Operations in Differential Geometry, in particular the one on Weil bundles, where they present the famous "Milnor's exercise". They give two similar proofs, one much shorter than the other, supposedly because of omitted details. I would like to understand it better.

So we have to prove the:

Theorem For any manifold $M$, $\mathrm{Hom}(C^\infty(M,\mathbb{R}),\mathbb{R})\cong M$

Where $\mathrm{Hom}$ refers to morphisms of commutative unital algebras and $\cong$ to bijection of sets (I suppose).

Now the short proof goes like this:

Consider an homomorphism of algebras $C^\infty(M,\mathbb{R}\xrightarrow{\phi}\mathbb{R})$, by the usual short exact sequence argument, $\ker(\phi)\leq C^\infty(M,\mathbb{R})$ is an ideal of codimension $1$ (notice that such an homomorphism cannot be the zero map, since $\phi(1_\mathbb{R})=1\in\mathbb{R}$).

Now, consider the following system of subsets of $M$, where $V(f)=f^{-1}(0)$:
$$\{V(f)\mid (M\xrightarrow{f}\mathbb{R})\in \ker(\phi)\}$$
this is clearly a filter of closed sets of $M$, since $V(f^2+g^2)=V(f)\cap V(g)$.

Find a function $M\xrightarrow{f^*}\mathbb{R}\in \ker(\phi)$ which is unbounded on each non-compact closed subset of $M$ (for instance a positive proper function, such as the square of the geodesic distance w.r.t. any Riemannian metric).  Then $V(f^*)$ is a compact set contained in  $\bigcap_{f\in \ker(\phi)} V(f)$, which is then nonempty.

Let $x_0$ be one of its elements.
Then for any $f\in C^\infty(M,\mathbb{R})$ $f-\phi(f)1$ is in $\ker(\phi)$ so $(f-\phi(f)1)(x_0)=0$, and $\phi(f)=f(x_0)$.
So $x_0$ completely determines $\phi$ and the map $\phi\mapsto x_0$ is the desired bijection.

I have fetched the italicised details. What I am having trouble understanding is the bold passage. Moreover I don't see where we use the fact that the given system of sets is a filter and that $\ker(\phi)$ has codimension 1: I suspect these are relevant for the bold passage also.

Any help would be much appreciated.

答:
This is a proof of the "hard part" mentioned in  @Elad's answer.

Lemma.  A smooth manifold $M$ admits a smooth real valued function $f$ such that
$$
  \lim_{x\to \infty }f(x) = \infty ,
  $$
(in the sense that for every $a>0$, there exists a compact subset $K\subseteq M$, such that $f(x)>a$, for every $x\in M\setminus
K$)
if and only if $M$ is $\sigma $-compact.

Proof. For the  "only if" part notice that,  given $f$ as above, one has that $f^{-1}((-\infty,a])$ is compact for every real
number $a$, and
$$
  M= \bigcup_{n\in \mathbb N}f^{-1}((-\infty,n]).
  $$

Conversely, let $\{K_n\}_{n\in \mathbb N}$ be a family of compact subsets of $M$  such that   $M= \bigcup_{n\in \mathbb N}K_n$.  By a
well known trick, based on the local compactness of $M$, we may assume WLOG that each $K_n$ is contained in the interior
of $K_{n+1}$.

By Lemma (1.3.2) in https://www.math.ucla.edu/~petersen/manifolds.pdf (Smooth Urysohn Lemma), for each $n$ in $\mathbb
N$ we may choose a nonegative smooth function $f_n$ vanishing on $K_n$ and taking on the value 1 on $M\setminus \text{int}(K_{n+1})$ (and
hence also on $M\setminus K_{n+1}$).

Defining
$$
  f(x) = \sum_{n=1}^\infty  f_n(x), \quad \forall x\in  M,
  $$
we claim that, for every $x$ in $M$,  there exists   a neighborhood $V$ of $x$,  and a natural number $m$, such that
$f_n=0$ on $V$, for every $n\geq m$.

To see this, find some $i$ such that $x\in  K_i$.   It then suffices to take
$V=\text{int}(K_{i+1})$ and $m=i+1$.

This said, it is clear that the  above function $f$ is well defined and smooth.

To see that  $\lim_{x\to \infty }f(x) = \infty $, observe that if $x\not\in  K_{m+1}$, then  surely $x\not\in K_{n+1}$, for  every $n\leq m$,  whence
$$
  f(x) \geq   \sum_{n=1}^m f_n(x) = m.
  $$
This completes the proof of the Lemma.  QED

Back to the question, let us assume that  $M$ is $\sigma $-compact, so we may pick  $f$  as in the
Lemma.

Setting
$$
  f^*=f-\phi(f)1,
  $$
one has  that $f^*\in \text{Ker}(\phi)$, and clearly also   $\lim_{x\to \infty }f^*(x) = \infty $, so we have that
$$
  V(f^*) =   {f^*}^{-1}(0)
  $$
is compact, as desired.


答:
We have (mean value theorem)

$$| \frac{f(5)-f(1)}{4}|  \le 2.$$

Hence

$$ -8 \le f(5)-17 \le 8.$$

Therefore

$$f(5) \le 25.$$

In a similar way we get from $| \frac{f(9)-f(5)}{4}|  \le 2, $ that

$$f(5) \ge 25.$$

Title: Let $f(x)$ be differentiable on for all $x\in\mathbb{R}$, and $|f'(x)|\le2$, for every $x$, and $f(9)=33$, $f(1)=17$. what is $f(5)=?$

Body:
I can't get how I can know stuff about $f(5)$ from the information that I got.
I tried to apply Lagrange: $\exists  c\space such\space  that\space 1<c<9 \space \space :f'(c)=\frac{f(9)-f(1)}{9-1} =2 $
 and I have no idea how to reach $f(5)$ from here.

any hints or tips or solutions will be appreciated!


问:
why measure theory for probability?
I studied elementary probability theory. For that, density functions were enough. What is a practical necessity to develop measure theory? What is a problem that cannot be solved using elementary density functions?</p>

答:
The standard answer is that measure theory is a more natural framework to work in. After all, in probability theory you are concerned with assigning probabilities to events (sets)... so you are dealing with functions whose inputs are sets and whose outputs are real numbers. This leads to sigma-algebras and measure theory if you want to do rigorous analysis.

But for the more practically-minded, here are two examples where I find measure theory to be more natural than elementary probability theory:

1) Suppose X~Uniform(0,1) and Y=cos(X). What does the joint-density of (X,Y) look like? What is the probability that (X,Y) lies in some set A? This can be handled with delta-functions but personally I find measure theory to be more natural.

2) Suppose you want to talk about choosing a random continuous function (element of C(0,1) say). To define how you make this random choice you would like to give a p.d.f. but what would that look like? (The technical issue here is that this space of continuous functions is infinite dimensional and so Lebesgue measure cannot be defined). This problem is very natural in the field of Stochastic Processes including Financial Mathematics -- a stock price can be thought of as a random function. Under the measure theory framework you talk in terms of probability measures instead of p.d.f.'s and so infinite dimensions do not pose an obstacle.

Title: why measure theory for probability?


问:
Solving the double integral with $|y-2x|\leq 0.1$ as boundary
Question:

Solve the integral of $x$ with boundaries $|y-2x|\leq 0.1,
 0 \leq x \leq 1,
0 \leq y \leq 1 $

What gave me trouble is the absolute value, I tried graphing it online and got a noncontinuous line, how can I proceed?

答:
Consider $y \geq 2x$ and $2x \geq y$ separately.

Please see the sketch. You have to integrate over the area shaded in black.

Integrate over $dx$ first and then over $dy$ so you have to split it into two integrals otherwise you will have to split it into three.

$D1: 0 \leq x \leq \frac{y + 0.1}{2}, 0 \leq y \leq 0.1$

$D2: \frac{y - 0.1}{2} \leq x \leq \frac{y + 0.1}{2}, 0.1 \leq y \leq 1$


问:
"Interpretation" versus "judgement" versus "assertion"?
I believe "assertion" and "judgment" mean the same thing (at least according to Per Martin Lof). I understand that they are meta statements (above that of say formal propositions). How, if at all, do these meanings differ from that of "interpretation", which is also on top of a proposition?</p>

答:
In "mainstream" treatment of mathematical logic, The Turnstile symbol is used in:

$\Gamma \vdash \alpha$

to mean that formula $\alpha$ is derivable from the set $\Gamma$ of premises according to the rules of the logical calculus (aka: proof system), i.e. using logical axioms and Modus Ponens or e.g. Natural Deduction rules.

It derives from Frege (1879) : "...Frege's symbol came to be called the assertion sign.  Frege's notation for a judgement $\vdash A$ can then be read I know $A$ is true."

The Double turnstile symbol is used in:

$\Gamma \vDash \alpha$

to mean that $\alpha$ is a logical consequence of the set $\Gamma$ of premises.

This second notion is defined using the concept of interpretation: the set of the interpretations that make all members of $\Gamma$ true is a subset of the set of the interpretations that make $\alpha$ true.

The two relations are very different but strictly linked: a calculus is said to be sound and complete when:

$\Gamma \vDash \alpha \text {  iff  } \Gamma \vdash \alpha$.

In other words, the caluclus is able to derive from a set of premises all and only the logical consequences of the said premises.

If so, we are licensed to read the judgment $\Gamma \vdash \alpha$ both as "formula $\alpha$ is derivable from premises $\Gamma$" and as "formula $\alpha$ is true in every model of $\Gamma$".

Things are a little bit different with a different approach; see e.g. Per Martin-Lof's Intuitionistic type theory (1984):

[page 2] When we hold a proposition to be true, we make a judgement: $A$ is a proposition, "proposition $A$ is true" is a judgement.

In particular, the premisses and conclusion of a logical inference are judgements.

And

[page 6] a proposition is true if it has a proof, that is, if a proof of it can be given.


问:
How to find $g(x)$ without having boundaries?
Question:

Let $X$ and $Y$ be rvs with
$$
f(i,j)
 = \mathbb{P}[X=i,Y=j]
 = \frac{1}{e} \left(\frac{1}{2^{i+1}} \frac{1}{j!}\right)
$$
and $i,j$ are natural numbers.

Prove that $\mathbb{P}[X=i]=1/2^{i+1}$

Prove $Y$ has a Poisson distribution

For a I was thinking of finding $g(x)$ but I don't have the boundaries of $X$ and $Y$.

答:
HINT You are given that $i,j$ are natural numbers, so $X,Y$ are non-negative integers. To validate this, you need to prove that $\sum_{i,j=0}^\infty f(i,j) = 1$.

Also note that $\mathbb{P}[X=i] = \sum_{j=0}^\infty f(i,j)$...


答:
First there is a typo in your question. The correct joint pmf is the following

$$\mathbb{P}[X=i;Y=j]=e^{-1}\frac{1}{2^{i+1}j!}$$

To find the marginal X sum in Y and viceversa

Thus

$$\mathbb{P}[X=i]=e^{-1}\frac{1}{2^{i+1}}\sum_{j=0}^{\infty}\frac{1}{j!}=e^{-1}\frac{1}{2^{i+1}}e=\frac{1}{2^{i+1}}$$

$$\mathbb{P}[Y=j]=\frac{e^{-1}}{j!}$$

that is

$$Y\sim Pois(1)$$

$$X\sim Geom\Big(\frac{1}{2}\Big)$$

An easy way to solve the problem was to observe that

$$\mathbb{P}[X=x,Y=y]=\Bigg(\frac{1}{2}\Bigg)^{x}\frac{1}{2}\times\frac{e^{-1}1^y}{y!}=\mathbb{P}[X=x]\times \mathbb{P}[Y=y]$$

$x=\{0;1;2;...\}$

$y=\{0;1;2;...\}$

Title: How to find $g(x)$ without having boundaries?

Body:
Question:

Let $X$ and $Y$ be rvs with
$$
f(i,j)
 = \mathbb{P}[X=i,Y=j]
 = \frac{1}{e} \left(\frac{1}{2^{i+1}} \frac{1}{j!}\right)
$$
and $i,j$ are natural numbers.

Prove that $\mathbb{P}[X=i]=1/2^{i+1}$

Prove $Y$ has a Poisson distribution

For a I was thinking of finding $g(x)$ but I don't have the boundaries of $X$ and $Y$.


问:
Show that $c$ is closed in $l^{\infty}$
Let $$c=\{ (a_i)_{i \in \mathbb{N}} \ ; \ \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N} \ , \ \mbox{exist} \ \displaystyle \lim_{i \to \infty}(a_i)\}$$
 $$l^{\infty}=\{ (a_i)_{i \in \mathbb{N}} \ ; \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N}; \  \displaystyle \sup_{i \in \mathbb{N}}|a_i|< \infty \}.$$ Show that $c$ is closed in $l^{\infty}$.

i am trying... taking a sequence of sequence in $c$ but i can't...i show that $c_0$ is closed in $l^{\infty}$.

答:
Let $f : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \limsup\limits_{n \to + \infty} x_n \end{array} \right.$ and $g : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \liminf\limits_{n \to + \infty} x_n \end{array} \right.$.

Show that $f$ and $g$ are continuous, and deduce that $c= \{ x \in \ell^{\infty} \mid f(x)=g(x) \}$ is closed in $\ell^{\infty}$.

Title: Show that $c$ is closed in $l^{\infty}$


问:
How can I simplify $\frac{\left (\sum x_{i} \right )}{\left ( \sum X_{i} \right ) \left (\sum x_{i}^2 \right )}$
How can I simplify
$$\frac{\left (\sum x_{i}  \right )}{\left ( \sum X_{i} \right ) \left (\sum x_{i}^2  \right )}$$

where $x_{i}=X_{i} - \bar{X}$ and $\bar{X}$ is the mean of all $X_{i}$'s

I thought of the following which, I must admit, looks wrong to me:
multiply the numerator and the denominator by $x_{i}$
$$\frac{\left (\sum x_{i} x_{i} \right )}{\left ( \sum X_{i} x_{i} \right ) \left (\sum x_{i}^2  \right )} = \frac{1}{\left (\sum X_{i}x_{i}  \right )}$$

Can anyone help me figure it out?

答:
You can't just multiply by $x_i$ because in the summation is a "moving index", which means that $x_i$ represents actually many values (one value for each $i$), so in order to multiply by $x_i$ you'll have to fix a value for $i$, e.g. $i = i_0$ for which it's easy to see why $\sum_i {X_i x_{i_0}} \neq \sum_i x_i x_{i_0}$. That's why your approach seems wrong.

As @Paresseux Nguyen pointed out, the correct way to approach this is by noting that:

$$
\sum_i x_i = \sum_i X_i - \frac{n}{n} \sum_j X_j = 0
$$

I hope it helps :)


问:
show that $\frac{\cos(3x)-\sin(3x)}{1-2\sin(2x)}=\sin(x)+\cos(x)$
show that $$\frac{\cos 3x -\sin 3x }{1-2\sin 2x}=\sin x + \cos x $$

I tried using compound angle formula first to convert $\sin(3x)$ and $\cos(3x)$ to double and single angles. Then I expanded the double angle but I was not able to simply it any further.

答: Note $$\cos 3x -\sin 3x = (4\cos^3x -3\cos x) -(3\sin x-4\sin^3x)\\  = 4(\sin^3x +\cos^3x) -3(\sin x +\cos x) \\ =(\sin x+\cos x) \left( 4(\sin^2 x +\cos^2 -\sin x \cos x) -3 \right) \\ =(\sin x+\cos x)(1-4\sin x \cos x)$$ while $$1-2\sin2x =1-4\sin x\cos x$$


问:
Actions on left and right cosets of a subgroup of prime index.
This raised up while trying to answer this post.

Let $G$ be a group and $H\le G$. Let be $X:=\{aH, a\in G\}$ and $Y:=\{Ha, a\in G \}$. The maps $\circ: G\times X\to X$ and $\star:G\times Y\to Y$, defined respectively by $g\circ aH:=gaH$ and $g\star Hb:=Hbg^{-1}$, are both left transitive actions, with pointwise stabilizers $\mathrm{Stab}_\circ(aH)=aHa^{-1}$ and $\mathrm{Stab}_\star(Hb)=b^{-1}Hb$, respectively. Is there any group action-theoretic argument to conclude that, if the two orbits have prime size (i.e. $[G:H]=p$) and intersect in more than one point (i.e. $\tilde aH=H\tilde a$ for some $\tilde a\in G\setminus H$), then they coincide (i.e. $H\lhd G$)?

答: Yes, with the same argument: If $aH=Ha$, then $a$ normalizes $H$, so the normalizer of $H$ is strictly larger than $H$ itself. But as the index is prime, the only larger subgroup is $G$.


问:
Struggling to both understand and remember maths!
I hope it is OK to post this question here...

I'm currently in my 2nd year of maths at uni (doing cal 2, discrete, prob/stat) and hadn't studied maths formally for about 15 years beforehand. In my first year I did somewhat OK and scraped a pass in all of my modules (alg 1, calc 1, proofs) with thanks to distance exams. Otherwise, I would have absolutely failed.

I enjoy maths... the whole problem solving/challenging aspect is what drives me. However, I cry almost everyday because I'm so frustrated at how little I actually "get" it and it's actually starting to depress me (I dedicate at least 40 hrs p/week, 7 days a week to it). On the other hand, my boyfriend can look at a question, investigate it a little online and it makes sense to him with little effort.

In terms of learning methods I've found that notes do zilch. Perhaps my notetaking is just bad but I've found it wasted my time. My comp is full of ebooks as recommended by the staff too. So I have many sources of info at my hands but yet my intuition or core understanding is roughly the same as when I started.

Exams are in 5-6 months, I'm already getting myself stressed about it and I have to improve. I know maths is challenging, that's whats great about it! But when you find yourself staring at a question that involves bayesian theorem, wondering why the heck you have to even use division, then there's something missing.

Would it be most effective to simply do as many exercises as possible? How do you conceptualise theorems, etc and apply them to problems?

Any tips would be hugely appreciated :) Thank you in advance.

答:
I am also someone new in formal mathematics, so I hope to be of use.

First, don't compare yourself to others; the best way you can measure your performance is by looking at how much you've grown in math.
My advice is to try to exchange ideas with friends or colleagues; you can also start the homework problems trying to see them from an "informal" point of view and then have a general idea and be able to formally write the solution / proof.

If you have no idea how solving the exercise tries to "discover" what is the meaning of the exercise and thus be able to invest in a specific way on that topic by reading similar proofs or seeing some solved exercises.
Also, take a break from time to time, I understand that you want to level yourself but mental health and well-being is important, so take breaks from time to time and enjoy math.


答:
Also not quite sure whether this is the appropriate place to discuss your problem. However, it is very relatable! During my first year of mathematics I already was coming up with a backup plan (working as a train driver) just in case I would not make it. I kept that backup plan all the way up to receiving all my points in the first year. After that I was still not sure whether I was cut out for it. A few years ago I received my BSC and I will be going for my MSC next years.

Even though you have doubts, if your passion really lies with mathematics, don't give up! You proved that you can do this by scraping a pass in all of your modules in the first year. You could say that is thanks to distance exams. The main reason for you passing your exams is you though. YOU did it, not someone or something else.

There will be some that understand certain concepts of mathematics just by reading about them once and there will be a whole lot more that have to do exercises in order to get a better understanding. Unfortunately that is the truth. What might help is to try to find a study group in which you can comfortably do homework together and talk about these issues. Believe me; you will find that most of your fellow students feel the same way. I know it's rather difficult today due to COVID, but try to find some way online or something. It might help.

Title: Struggling to both understand and remember maths!

Body:
I hope it is OK to post this question here...

I'm currently in my 2nd year of maths at uni (doing cal 2, discrete, prob/stat) and hadn't studied maths formally for about 15 years beforehand. In my first year I did somewhat OK and scraped a pass in all of my modules (alg 1, calc 1, proofs) with thanks to distance exams. Otherwise, I would have absolutely failed.

I enjoy maths... the whole problem solving/challenging aspect is what drives me. However, I cry almost everyday because I'm so frustrated at how little I actually "get" it and it's actually starting to depress me (I dedicate at least 40 hrs p/week, 7 days a week to it). On the other hand, my boyfriend can look at a question, investigate it a little online and it makes sense to him with little effort.

In terms of learning methods I've found that notes do zilch. Perhaps my notetaking is just bad but I've found it wasted my time. My comp is full of ebooks as recommended by the staff too. So I have many sources of info at my hands but yet my intuition or core understanding is roughly the same as when I started.

Exams are in 5-6 months, I'm already getting myself stressed about it and I have to improve. I know maths is challenging, that's whats great about it! But when you find yourself staring at a question that involves bayesian theorem, wondering why the heck you have to even use division, then there's something missing.

Would it be most effective to simply do as many exercises as possible? How do you conceptualise theorems, etc and apply them to problems?

Any tips would be hugely appreciated :) Thank you in advance.


问:
Express $4+\sqrt{-2}$ as a product of irreducibles
This is part of an old Oxford Part A exam paper. (1992 A1)

Suppose we equip $R=\mathbb{Z}[\sqrt{-2}]$ with the Euclidean function $d$ defined by $$d(m+n\sqrt{-2})=|m+n\sqrt{-2}|^2$$
I want to determine the units of $R$ and express $4+\sqrt{-2}$ as a product of irreducibles and to use this to determine how many ideals of $R$ contain $4+\sqrt{-2}$.

Progress

I think I have shown that the only units are $1,-1$, but cannot see how one can show this element is a product of irreducibles. In general, we have that in a Euclidean Domain every element is a finite product of irreducibles, but I cannot see how to calculate them.

Thanks

答:
Hint: The key is using the Euclidean function. Like many subrings of the complex plane, we have Euclidean function you give, $\phi$, such that $\phi(a+bi) = a^2+b^2$. We know that $\phi$ is multiplicative on $\mathbb{C}$ and we can show that in the case of  $\mathbb{Z}[\sqrt{-2}]$, the division algorithm works.

To determine the units of $R$, if $z$ is a unit then $zz^{-1}$=1 so $\phi(z)\phi(z^{-1})=1.$ Thus either $\phi(z)= 1$ or, without loss of generality, $\phi(z)<1$. Could we achieve this in the given ring?

To express $4+\sqrt{-2}$ as a product of irreducibles, if you can't see how to factorise directly, suppose $ab=4+\sqrt{-2}$ (where neither are units). It follows that $\phi(a)\phi(b)=18$. Using that $\mathbb{Z}$ is a UFD, and using our knowledge gained previously about what values of $\phi$ non-unit elements can take, we can determine some possible values of $\phi$ for the factors. Playing around with possible elements of a given "size" should allow you to factorise $4+\sqrt{-2}$ into smaller parts. To show that these each factor is irreducible, suppose that it is expressible as a product of non-unit elements, determine possible the values of $\phi$ the divisors can take and then show that this is not possible.


问:
Finite permutation group whose every non-identity element has a unique fixed point
If I'm not wrong, the following statement is true :

Theorem. Let $E$ be a (finite or infinite) set, let $G$ be a finite subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Then all the non-identity elements of $G$ have the same fixed point. (And thus, if $G$ is not trivial, there is one and only one element $x$ of $E$ that is fixed by every element of $G$. Then $G$ acts freely on $E \setminus \{x\}$.)

I looked for a proof in textbooks and on Internet, but I saw nothing. (Maybe I looked badly.) I found a proof myself (I sketch it below) but it is not very beautiful and I am afraid it is too complicated. Thus my question is : do you know a more direct proof ?

Here is my proof.

$\mathbf{Step 1.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that $G$ is abelian. Then all the non-identity elements of $G$ have the same fixed point.

$\mathbf{Proof.}$ Let $\alpha$ and $\beta$ non-identity elements of $G$. Since $G$ is abelian,
$\alpha ^{-1} \beta \alpha = \beta$.

Applying both members to the unique fixed point $b$ of $\beta$ gives

$\alpha ^{-1} \beta \alpha (b)= b$.

Applying $\alpha$ to both members gives

$\beta \alpha (b) = \alpha (b)$, thus $\alpha (b)$ is a fixed point of  $\beta$. Since $b$ is the only fixed point of $\beta$, we have thus $\alpha (b) = b$, thus $b$ is the fixed point of $\alpha)$, thus $\alpha$ and $\beta$ have the same fixed point. This proves step 1.

$\mathbf{Step. 2.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that $G$ has a nontrivial normal subgroup whose all non-identity elements have the same fixed point. Then all the non-identity elements of $G$ have the same fixed point.

$\mathbf{Proof.}$ By hypothesis, we can choose a nontrivial normal subgroup $H$ of $G$ such that all the non-identity elements of $H$ have the same fixed point.

Choose a non-identity element $\alpha$ of $H$. From the hypotheses,

(1) $\alpha$ has a unique fixed point, say $a$, and

(2) every non-identy element of $H$ has $a$ as unique fixed point.

Let $\gamma$ be a non-identity element of $G$. Since $H$ is normal in $G$, $\gamma ^{-1} \alpha \gamma$ is a non-identity element of $H$, thus, in view of (2),

$\gamma ^{-1} \alpha \gamma (a) = a$. Applying $\gamma$ to both members gives $\alpha \gamma (a) = \gamma (a)$, thus $\gamma (a)$ is a fixed point of $\alpha$. Thus, by (1), $\gamma (a) = a$. By hypothesis, $\gamma$ has only one fixed point, thus our result means that the only fixed point of $\gamma$ is $a$. This is proven for every non-identity element $\gamma$ of $G$, thus step 2 is proven.

$\mathbf{Step. 3.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that there is a generating subset $X$ of $G$ such that all the non-identity elements of $X$ have the same fixed point. Then all the non-identity elements of $G$ have the same fixed point.

$\mathbf{Proof.}$ It is an easy consequence of the fact that every element of $G$ is a product of non-identity elements of $X \cup X^{-1}$.

$\mathbf{Step. 4.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that there are two differet maximal subgroups $M_{1}$ and $M_{2}$ such that

(i)  $M_{1} \cap M_{2} \not= 1$,

(ii) all the non-identity elements of $M_{1}$ have the same fixed point and

(iii) all the non-identity elements of $M_{2}$ have the same fixed point.

Then all the non-identity elements of $G$ have the same fixed point.

$\mathbf{Proof.}$ From hypotheses (i), (ii) and (iii), it results that

(1) all the non-identity elements of $M_{1} \cup M_{2}$ have the same fixed point.

On the other hand, since $M_{1}$ and $M_{2}$ are two different maximal subgroups of $G$, they generate $G$, in other words,

(2) $M_{1} \cup M_{2}$ is a generating subset of $G$.

By (1), (2) and step 3,  all the non-identity elements of $G$ have the same fixed point, thus step 4 is proven.

$\mathbf{Step. 5.}$ Let $E$ be a (finite or infinite) set, let $G$ be a $\mathbf{finite}$ subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Then  all the non-identity elements of $G$ have the same fixed point.

$\mathbf{Proof.}$ Assume, by contradiction, that

(hyp. 1) the statement is false.

Thus, there exist a set $E$ and a finite subgroup $G$ of $S_{E}$ such that  every non-identity element of $G$ has exactly one fixed point and the non-identity elements of $G$ do not all have the same fixed point. Among these subgroups $G$ of $E$, choose $G_{0}$ with the least possible order. Then

(2) $G_{0}$ is a finite subgroup of $S_{E}$,

(3) every non-identity element of $G_{0}$ has a unique fixed point,

(4) the non-identity elements of $G_{0}$ do not all have the same fixed point,

and, in view of the minimality of $\vert G_{0} \vert$,

(5) for every proper subgroup $K$ of $G_{0}$, all non-identity elements of $K$ have the same fixed point.

In view of (3), (4), (5) and step 4,

(6) the maximal subgroups of $G_{0}$ intersect pairwise trivially.

Assume that

(hyp. 7) $G_{0}$ has a normal subgroup $H$ such that $1 < H < G_{0}$.

By (5) (and the assumption $H < G_{0}$), all non-identity elements of $H$ have the same fixed point. Thus, by step 2 (and the assumption $1 < H$), all non-identity elements of $G_{0}$ have the same fixed point. This contradicts (4), thus (hyp. 7) is absurd, thus $G_{0}$ is a simple group. Thus, by (2) and step 1,

(8) $G_{0}$ is a finite non-abelian simple group.

Now, (6) and (8) are incompatible, as proven here :

https://groupprops.subwiki.org/wiki/Finite_and_any_two_maximal_subgroups_intersect_trivially_implies_not_simple_non-abelian

Thus our hypothesis (1) is absurd, so step 5 is proven.

答:
If you are familiar with the theory of finite Frobenius groups, then it might be easier to use that. The action of the group on any non-regular orbit of length greater than $1$ must be as a Frobenius group, and it must be a faithful action. But then the fixed point free elements in the action cannot fix any points, because they cannot also lie in a Frobenius complement.

Note that the result is false for infinite groups, because there are infinite Frobenius groups in which all elements fix a unique point. (See, for example,
Timm von Puttkamer's answer here.)


问:
Simplify with fractional exponents and negative exponents
I am trying to simplify

$$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$

It seems pretty simple at first. I know that a negative exponent means you flip a fraction. So I flip it.

$$  \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$

Now I need to square it, which is tricky because there are a lot of weird rules with squaring. This is probably where I went wrong.

$$  \left(\frac{x^2 y^{-1/2}}{9x^{3/2}y^3}\right)$$

Now I need to try and simplify things. I know that I can get rid of the $x$ on top since there is a larger one on the bottom.

$4/2 - 3/2 = 1/2$

$$  \left(\frac{x^{1/2} y^{-1/2}}{9y^3}\right)$$

Now I need to get rid of the $y$ exponent.

I am not sure how that is possible.

答:
We can flip the equation:

$ \left(\dfrac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $=
$\left(\dfrac{x^2y^{-1/2}}{3x^{3/2}y^3}\right)^2$

Multiplying the exponents:

$\left(\dfrac{x^2y^{-1/2}}{3x^{3/2}y^3}\right)^2$= $\dfrac{x^4y^{-1}}{9x^3y^6}$

We can move the $y^{-1}$ to the bottom and write everything factored:

=  $\dfrac{x·x·x·x}{3·3·x·x·x·y·y·y·y·y·y·y}$

Looking at this, we can cancel x's.

=  $\dfrac{x}{3·3·y·y·y·y·y·y·y}$

Which simplifies to:

= $\dfrac{x}{9y^7}$


问:
Compare $\ln(\pi)$ and $\pi-2$ without calculator
Unlike the famous question of comparing $e^\pi$ and $\pi^e$, which I solved almost instantly, I am stuck with this problem.  My thought was the following.

Since exponential function is order-preserving, we exponentiate both terms and get $\pi$ and $e^{\pi-2}$.  Then we study the function $f(x) = e^{x-2} - x$ or the function $g(x) = \frac{e^{x-2}}{x}$, and compare them with zero and one respectively.  I tried both.  But both involve solving the equation
$$e^{x-2} = x.$$

I tried Lagrange error terms and have
$$f(x) = -1 + \frac{(x-2)^2}{2!} + R_2(x-2),$$
where
$$\frac{(x-2)^3}{3!} \le R_2(x-2) \le \frac{e^{x-2}}{3!} (x-2)^3.$$

It is easy to see that the equation have a root between $3$ and $2 + \sqrt2$.  But I don't know how close it is to $\pi$.  It is to provide some lower bounds since we can plug in some values and calculate to show that $f(x) > 0$ for such values.  But for the upper bound, it is hard to calculate by hands since it has the $e^{x-2}$ factor.  At my best attempt by hand, I showed that $f(3.15) > 0$.  All it entails is that for all $x \ge 3.15$, $e^{x-2}$ is greater than $x$.  But it tells nothing about the other side.

Then I looked at the calculator and find that $e^{\pi-2} < \pi$.

I also tried Newton-Raphson iteration, but it involves a lot of exponentiation which is hard to calculate by hand and also involves approximation by themselves.  And I don't know how fast and close the iteration converges to the true root of the equation.

Any other hint for comparing these two number purely by hand?

答: By hand I took $e^{\pi -2}<e\cdot e^{0.1416}<(2.7183)e^{0.1416}.$ Using $B_1=0.1416=0.1+0.04(1+0.04)$ for manual calculation, I computed, to $5$ decimal places, an upper bound $B_2$   for $(B_1)^2/2$ and an upper bound $B_3$ for $B_1B_2/3 $ and an upper bound $B_4$ for $B_1B_3/4,$ etc., until I was sure that the sum of the remaining terms was less than $0.00005,$ to obtain an upper bound $B$ to $4$ decimal places for $e^{0.1416}.$ Then I multiplied $B\times 2.7183$ and got less than $\pi.$


问:
Possibly incomplete problem - to determine minimal polynomials of projection and reflection maps
Is this problem incomplete? I don't know how to proceed! If it's incomplete, please let me know what details need to be added.

Let $V=U\oplus W$. Let $P_W:V\to W$ be the canonical projection and $R_W:V\to V$ be the reflection w.r.t $W$, i.e. it does $w+u\mapsto w-u$. Compute the minimal polynomials of $P_W$ and $R_W$.

Thank you!

答: Let $\{u_1,\ldots,u_k\}$ be a basis of $U$ and let $\{w_1,\ldots,w_l\}$ be a basis of $W$. Then $B=\{u_1,\ldots,u_k,w_1,\ldots,w_l\}$ is a basis of $V$. Besides, $P_W(u_j)=0$ for each $j\in\{1,2,\ldots,k\}$ and $P_W(w_i)=w_i$ for each $i\in\{1,2,\ldots,l\}$. So, the matrix of $P_W$ with respect to $B$ is a diagonal matrix such that the first $k$ entries of the main diagonal are all equal to $0$ and the remaining ones are all equal to $1$. So, the characteristic polynomial fo $P_W$ is$$(-\lambda)^n(1-\lambda)^l=(-1)^{\dim V}\lambda^n(\lambda-1)^l.$$Can you deal with $R_W$ now?


问:
$A^{k+1} = A$ for some $k\in \mathbb{N}$, then $A$ is diagonalisable

Let $A^{k+1} = A$ for some $k\in \mathbb{N}$. Prove that $A$ is diagonalisable.

 Hint: $$(J_n(\lambda))^{k+1} = (\text{diag}_n(\lambda) + J_n(0))^{k+1}$$

So, I am not sure how to use the hint, and I'm just trying my own thing.

Since $A^{k+1} = A$, $A$ is a root of $p(x) = x(x^k-1)$. The minimal polynomial of $A$, say $m_A(x)$, divides $x(x^k-1)$. What's next? If I can show that $m_A(x)$ splits into distinct linear factors, I'll be done.

Let me know how I can proceed with this argument, and/or use the hint provided.

One thing that I found worth noticing is that $$\text{diag}_n(\lambda)^k  = \text{diag}_n(\lambda^k)$$Does this help?

Thanks!

答:
The statement is not true in general. E.g. $A=\pmatrix{0&1\\ 1&0}=A^3$ is not diagonalisable over $GF(2)$ and $A=\pmatrix{0&1&0\\ 0&0&1\\ 1&0&0}=A^4$ is not diagonalisable over $\mathbb R$. It is true, however, over $\mathbb C$ because the minimal polynomial of $A$, which divides $x^{k+1}-x$, is a product of distinct linear factors.

Alternatively, you may use the given hint. Suppose the contrary that $A$ is not diagonalisable over $\mathbb C$. Then via a similarity transform, we may assume that $A$ is in Jordan form. Therefore it contains a nontrivial Jordan block $J_n(\lambda)$ of some size $n$ such that $J_n(\lambda)^{k+1}=J_n(\lambda)$. But this implies that
$$
\sum_{r=0}^n\binom{n}{r}\lambda^{n-r}J_n(0)^r=\lambda I_n+J_n(0).
$$
By comparing the coefficients of $I_n$ and $J_n(0)$ on both sides, we get $\lambda^n=\lambda$ and $n\lambda^{n-1}=1$. Therefore $|\lambda|=1$ and  $n=1$, but this contradicts the assumption that $J_n(\lambda)$ is a nontrivial Jordan block. Hence $A$ must be diagonalisable.


问:
Basis transformation matrix and eigenvectors
we generally know that if we have some matrix $A$ and we diagonalize it and get a matrix $D$, there is some matrix $P$ such that:

$$ D = P^{-1}AP$$

My question is then, does the matrix P generally contain the eigenvectors corresponding to the eigenvalues on the diagonal of D? If yes, why? I fail to see how that comes out explicitly, but it seems it is that way (is it true only for orthogonal matrices?).

Thank you!

答:
Yes, the matrix $\;P\;$ contains the eigenvectors of the eigenvalues of $\;A\;$ as its columns. In fact, $\;P\;$ is a matrix of change of basis: from the usual one, or the one over which is $\;A\;$ defined, to the basis formed by the linearly independent
eigenvectors of $\;A\;$ eigenvectors.

Remember that a square matrix over any field is diagonalizable iff there exists a basis for the linear space over which we work of eigenvectors of the matrix.


答:
First of all, let me notice that your post sounds as you have the right attitude and motivation for doing maths. Of course, there is no one-fits-all recipe that I can give here, as learning maths is an individual task for everyone. Therefore, I try to focus on things I observed and that were helpful for me, when I started doing maths.

For understanding the topics covered in school-maths it often was sufficient to be talented, which is in high contrast to how it is with university-maths, which is hard-work (that, in the best case, is supported by talent). Hardly anyone understands everything just by hearing/reading about it. Especially in the first semesters of study there are many students pretending that they understand things easily. According to my experiences; be sure that this is not true for 90% of them ;-)

For me, taking notes from lecture is very important and the quality of my notes usually gives immediate feedback of how good I indeed understood the lecture. As many lectures currently are recorded due to the pandemic situation take this as an advantage. Try to watch the lectures first catching the main points and then (if you feel that handwritten notes might be useful for you) try to reproduce the statements and definitions. Writing something down helps you to find out whether you undestood something well or just think that you got the whole idea.

When you learn about a theorem/lemma, try to group the statement in prerequisites and statement. Make yourself clear what is demanded and what is the conclusion of the theorem. Try to find out why you need every single prerequisites, i.e. why does the theorem/the proof not work if you remove this part.

It is hard to learn all the proofs you see during the semester by heart, so try to grasp the idea of the proof, i.e. that what remains if you forget about the technical details. Often, you can find similar ideas in other proofs again and again. So, once you build up a certain "toolbox" of ideas you understand fundamentally, it turns out that other things become easier for you, too. This is also a great strategy to solve assignment problems. Very often assignment problems can be solved if you understood the proof-idea of a theorem in the lecture (of course, this does not work every time, but helps you to get started with math problems).

Finally, math is not just learning definitions and theorems. Most of the time, you are going to spend with solving problems (or, at least, trying to). Again, you have to build up your personal "toolbox" where you store all ideas that worked for previous problems. Adding ideas to your toolbox is nothing but hard work. Try to solve as many problems as possible on your own and if you see solutions from others, try to extract the main idea behind the proof and try to proof it on your own afterwards, without copying the whole solution, as, again, this shows whether you understood the proof or just think that you understood the proof.

Summing up, don't be daunted by the speed others solve (or pretend to solve) problems and grasp ideas. Many people do not admit how much effort it was to get there, as it is more impressive to pretend to be the 'genious'. Don't compare yourself to others, compare with your younger self and think about your progress. For me it sounds like you have a great attitude and a honest fascination about maths, which, in my eyes, is the most important thing.

I am going to finish with a quote from one of my professors, that was helpful for me in my first year of maths: "Maths is about failing 95% of your time and the rest is being incredibly glad and proud that you haven't given up on your way. Mathematicians live from the hope that this 5% of pure happiness is going to outperform the other 95% of suffering." ;-)

I hope that helps at least a bit :)


问:
How many different strings of lights can be created by placing 40 coloured lights on a horizontal string?
How many different strings of lights can be created by placing 40 coloured lights on a horizontal string if 12 of them are red, 6 are blue, 14 are green and 8 are yellow? Assume that lights of the same colour are indistinguishable.</p>

答: This is simply permutations with repeated elements, so directly use the appropriate formula to get $\dfrac{40!}{12!6!14!8!}$


问:
How to reduce system of non linear equations to a polynomial root finding problem?
I have a set of 5 non-linear equations:

$w_0 + w_1 + w_2 = 1$,

$w_0v_0 + w_1v_1 + w_2v_2 = a$,

$w_0v_0^2 + w_1v_1^2 + w_2v_2^2 = a^2 + b^2$,

$w_0v_0^3 + w_1v_1^3 + w_2v_2^3 = a^3 + 3ab^2$,

$w_0v_0^4 + w_1v_1^4 + w_2v_2^4 = a^4 + 6a^2b^2 + 3b^4$,

Where $v_0 = a$.
I want to express the values of $w_0,w_1,w_2,v_1,v_2 $in terms of $a,b,$ and $c = w_1/w_2,$ where c is a root of a polynomial of c parametrized by a and b. I know I can do this for the case of $v_0 = 0$, but I can't manage to do it it for $v_0 = a$. Could anyone help me figure this out, or more generally to reduce the system to a root finding problem of a polynomial (that is not necessarily a polynomial of $c = w_1/w_2)$. Any tips would also be much appreciated.

答:
For $a \ne 0$, Mathematica gives the solutions
$$
(w_1,w_2,w_2,v_0,v_1,v_2) = (0, \frac{1}{2},\frac{1}{2}, a, a \pm b, a \mp b) \, .
$$


问:
Approximation Algorithm question
Define an independent set of a graph $G = (V, E)$ to be a subset $S$ of vertices such that $V-S$ is a vertex cover of $G$. Is every $2$-approximation algorithm for finding a minimum vertex cover also a $2$-approximation algorithm for finding a maximum independent set?

I have read about approximation algorithms, but could relate vertex cover problem to independent set. Do you have any suggestions on why such algorithm can be found or not?

答: Can anyone provide some suggestions to solve this question, please? I provided my progress but it is very insufficient to finish the problem.</p>

Title: Approximation Algorithm question


问:
Integration of $e^{-\langle Ax , x \rangle}$ over $\mathbb{R}^n$

Problem :

If $A_{n \times n}$ is a symmetric, positive-definite matrix, show that : $$\int_{\mathbb{R}^n} e^{-\langle Ax , x \rangle}~ dx = \sqrt{\dfrac{\pi^n}{\det(A)}}$$ where $\langle a , b\rangle$ denotes the inner product of $a$ and $b$.

Approach :

I was approaching the problem using the Change of Variable Formula, using the function $\varphi(x) = A^{-1}x$ . Since $A$ is p.d., I can show that it is invertible. But I can't proceed anymore.

I found a similar-looking problem here, but couldn't understand anything.

答:
Let $v_1,\ldots,v_n$ be an orthonormal basis for the inner product induced by $A$, with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n>0$. We have $\det(A)=\prod_{j=1}^{n}\lambda_j$ and by an isometry
$$ \int_{\mathbb{R}^n}\exp(-x^t A x)\,dx = \int_{\mathbb{R}^n}\exp(-\lambda_1 x_1^2-\ldots-\lambda_n x_n^2)\,dx\stackrel{\text{Fubini}}{=}\prod_{j=1}^{n}\frac{1}{\sqrt{\lambda_j}}\int_{\mathbb{R}}e^{-z^2}\,dz. $$