问:
Complex Solutions to Polynomials
I'm trying to use topology to prove that:

$z^{n} + a_{n-1}z^{n-1} + ... + a_{1}z + a_{0} = 0$

has a solution in $\mathbb{C}$ if and only if, for each positive real number $c$, the equation

$z^n + \frac{a_{n-1}}{c}z^{n-1} + ... + \frac{a_1}{c^{n-1}}z + \frac{a_0}{c^n} = 0 $

has a solution in $\mathbb{C}$

But I'm having trouble getting going on it.

From experimentation I think there must be some function that takes c, and the roots of the second question, and returns the roots of the first equation (or vice versa) but I'm not seeing why that would be the case.

Hints?

答: HINT: $z_0$ is a soloution of the second equation if and only if of $cz_0$ is a soloution of the first equation.</p>

Title: Complex Solutions to Polynomials


问:
Given Probability
When run, a computer program chooses the word "expensive'' with probability 2/3 and otherwise chooses the word "free''. It then selects a letter uniformly at random from the word it chose and outputs that letter. Given the program outputs "e'', what is the probability that it chose the word "free''?

The answer is 3/7 but I couldn't come up with this. I got a 1/5. Could anyone help?
Thanks

答:
The probability that the program chose expensive and then an E is $\frac23×\frac39=\frac29$. The probability of the same, but having chosen free first, is $\frac13×\frac24=\frac16$. By total probability, the desired result is
$$\frac{1/6}{2/9+1/6}=\frac37$$

ContentLicense: CC BY-SA 4.0

Title: Given Probability


问:
Probability that in a group of 25 people there are at least 10 unique birth months
I am trying to find the probability that in a group of 25 people there are at least 10 unique birth months using the principle of inclusion exclusion.

Principle of inclusion exclusion

$\bigcup_{i=1}^{n} A_{i}\left|=\sum_{i=1}^{n}\right| A_{i}\left|-\sum_{1 \leq i<j \leq n}\right| A_{i} \cap A_{j}\left|+\sum_{1 \leq i<j<k \leq n}\right| A_{i} \cap A_{j} \cap A_{k}\left|-\cdots+(-1)^{n-1}\right| A_{1} \cap \cdots \cap A_{n} \mid$

I approached the problem as follows.

Let $M_1,... M_{12}$ represent the months January, ..., December.

Let $M_k^{c}$ be the event that no one is born in $M_k$. Then $P(M_k)^{c} = (11/12)^{25}$. Similarly, the probability that no one is born in any of x months is $((12-x)/12)^{25}$.

Using the principle of inclusion exclusion,

$$P(M_{1}^{c} U M_{2}^{c} U ... U M_{10}^{c}) = {10 \choose 1}(11/12)^{25} - {10 \choose 2}(10/12)^{25} +{10 \choose 3}(9/12)^{25} - {10 \choose 4}(8/12)^{25} + ... -{10 \choose 10}(2/12)^{25}$$

The probability that at least one person is born in each of the months $M_{1}$ through $M_{10}$ is 1 - $P(M_{1}^{c} U M_{2}^{c} U ... U M_{10}^{c})$.

I wrote the following Matlab code to evaluate this expression. The output is approximately 0.25.

```<code>fprintf("%.60f\n",findp2(25))
function output = findp2(p)
    %the output is the probability that there are at least 10 unique
    %birthmonth among p people
pc = 0;
for k = 1:10
pc = pc + (-1)^(k-1)*nchoosek(10,k)*((12-k)/12)^(p);
end
format long
output = 1-pc;
end
</code>```

Because this is the probability for 10 specific months, I multiplied this by ${12 \choose 2}$ to obtain 16.5 as the probability that there are at least 10 unique birth months among 25 people.

Can someone explain what I am doing wrong?

答:
Using PIE

Number of ways to assign $n$ birth months to $k$ people  $ = \displaystyle {12 \choose n}\sum \limits_{i=0}^{n-1} {(-1)^i} {n \choose i} (n-i)^{k}$

Here $k = 25, n = 10, 11, 12$.

So for $n = 10$

$P(10) = \displaystyle \frac{{12 \choose 10}\sum \limits_{i=0}^{9} {(-1)^i} {10 \choose i} (10-i)^{25}}{12^{25}} \approx 0.302$

Similarly, $P(11) \approx 0.403, P(12) \approx 0.182$

So $P (\geq 10) \approx  0.887$


问:
Integral representation of Hermite polynomial
I'm studying Merzbacher's Quantum Mechanics. In his treatment of harmonic oscillators, he introduces Hermite polynomials and their properties. He mentions that the Hermite polynomials (in the form such that $H_n$ has the coefficient of the highest power term equal to $2^n$) can be represented as
$$
H_n(x) = \frac{2^n}{\sqrt{\pi}} \int_{-\infty}^{\infty} (x+iu)^n e^{-u^2}\; du,
$$
for $n=0, 1, 2, \ldots$.

He justifies this by noting that the above satisfy, first, the relation
$$
H_n'(x) = 2nH_{n-1}(x)
$$
for all $n\ge1$, and, second, that their values agree with the Hermite polynomials at $x=0$.

Question: I'm unsure how the above two conditions are sufficient for any family of functions (indexed by $n=0, 1, 2, \ldots$) to be equal to the Hermite polynomials. I'm looking for a proof.

答:
We proceed by induction. If $p_0(x)$ is a polynomials such that
$p\,_0'(x)=0$ then $p_0(x)$ is a constant and since
$p_0(0)=H_0(0)$ then $p_0(x)=H_0(x).$ Now suppose that $p_n(x)$
is a polynomial such that $p\,_n'(x) = 2nH_{n-1}(x).$
But also $H_n(x)$ satisfies that equation, thus they differ by a constant.
The condition that $p_n(0)=H_n(0)$ implies that the constant is $0$ and thus
$p_n(x)=H_n(x).$ By induction this is true for all $n$.


问:
How is $g(12)(34)g^{-1} = (g(1) g(2))(g(3)g(4))$
Firstly I think (please confirm) the notation $g(x)$ (for $x \in \{1,2,3,4\}$) caters to the position $y \in \{1,2,3,4\}$ such that $x \to y$.

Is it true that $g((12)) = (g(1) g(2))$? I can not wrap around my head if this is even defined. If so, an explanation would be really helpful.

How is $g(12)(34)g^{-1} = (g(1) g(2))(g(3)g(4))$ where $g \in S_4$?

I am fine if hints or heuristics are provided so I can think how it works. A definition for $(1)$ would enable me to think $(2)$.

答:
In general, when you conjugate a $k$-cycle (say $(12\dots k)$) by a permutation $\sigma$, you get $(\sigma(1)\sigma(2)\dots\sigma(k))$.

Secondly, conjugation by a permutation is a homomorphism.


问:
f(a) = inverse of a is an isomorphism iff a group G is Abelian

$G$ is a group and $f:G \rightarrow G$ is a function defined as $f(a)=a^{-1}$ where $a^{-1}$ is the inverse of $a$ under the group operation. Prove that $f$ is an isomorphism if and only if $G$ is abelian.

I understand that I have to prove $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. How might I do that?

Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra

答:
$f\colon G → G$ defined by $f(x)=x^-1$
$f$ isomorphism $⇒$ f is a 1-1 homomorphism
$∀ x,y∈G, xy = ((xy)^{-1})^{-1} = ((y^{-1})(x^{-1}))^{-1}
 = (f(y)(x))^{-1} = f(yx)^{-1}$  (∵$f$ is a homomorphism)
$= ((yx)^{-1})^{-1}$ (by mapping defining)
$= yx (∵ (a^{-1})^{-1} = a)
∴ xy = yx ⇒ G$ is abelian

ContentLicense: CC BY-SA 3.0


问:
A simple problem about partition function and Young diagram
How many pairs (λ,μ) of partitions of integers are there such that λ⊢n,and the Young diagram of μ is obtained from the Young diagram of λ by adding a single square?</p>

答: This is A000070 in the OEIS, following the explanations given there by Jon Perry or Thomas Wieder. (Both are pretty clearly equivalent to your problem but are not stated exactly the same way.) In particular, if we call the answer to your question $f(n)$, then $f(n) = \sum_{k=1}^n p(k)$ where $p(k)$ is the number of partitions of $k$.</p>


答:
Notice that
$$(cz)^{n} + a_{n-1}(cz)^{n-1} + ... + a_{1}(cz) + a_{0} = c^nz^{n} + a_{n-1}c^{n-1}z^{n-1} + ... + a_{1}cz + a_{0}$$

Now divide through by $c^n$ to obtain $z^n + \frac{a_{n-1}}{c}z^{n-1} + ... + \frac{a_1}{c^{n-1}}z + \frac{a_0}{c^n}$.

Writing $p_1, p_2$ for the two polynomials of interest, we have $p_1(cz)/c^n = p_2(z)$.  So the existence of a root $w$ for $p_1$ implies the existence of a root $w/c$ for $p_2$.

(Thinking of each of these polynomials as functions from $\mathbb{C}$ to $\mathbb{C}$, they only differ by some pre- and post-composition with some scaling.  Since the scaling preserves $0$, it doesn't affect the existence of roots.)

Title: Complex Solutions to Polynomials

Body:
I'm trying to use topology to prove that:

$z^{n} + a_{n-1}z^{n-1} + ... + a_{1}z + a_{0} = 0$

has a solution in $\mathbb{C}$ if and only if, for each positive real number $c$, the equation

$z^n + \frac{a_{n-1}}{c}z^{n-1} + ... + \frac{a_1}{c^{n-1}}z + \frac{a_0}{c^n} = 0 $

has a solution in $\mathbb{C}$

But I'm having trouble getting going on it.

From experimentation I think there must be some function that takes c, and the roots of the second question, and returns the roots of the first equation (or vice versa) but I'm not seeing why that would be the case.

Hints?


问:
Two specific irrational numbers summing to rational number that is not zero?
There already exist pages that show how two irrational numbers can sum to a rational number. However, is there an actual example of this? What are some of irrational numbers $x$ and $y$ such that $x + y$ is rational and $x ≠ -y$ (for an example $\sqrt{2}$ and -$\sqrt{2}$ doesn't count)?

If there is an example of an irrational number $y$ for every irrational number $x$ that will sum to a rational number, does this property apply to all irrational numbers?

答:
Here is a simple example:
$$\frac 1{\sqrt n-1}-\frac1{\sqrt n+1}=\frac 2{n-1},$$
so choose $n$ such that $\sqrt n$ is irrational.


问:
Splitting of the tangent bundle and Euler characteristic of surfaces
Let $M$ be a be a closed orientable surface, and suppose that its tangent bundle $TM$ splits into a direct sum of line bundles.

How to prove that $M$'s Euler characteristic is zero?

Unfortunately, I am entirely unfamiliar with Characteristic Classes, but I would accept a solution based on properties of these objects;

if there is  a reference for a book where I could find this claim or details of a proof, that would be great.

答:
Characteristic class argument: The Euler class of a vector bundle vanishes if there is a nonvanishing section. The Euler class of the tangent bundle evaluated on the fundamental class is the Euler characteristic. If we have a trivial subbundle, we automatically have a section. We know that we have a 1-dimensional subbundle. Since we are orientable, this implies this subbundle is either trivial or both it and its complement are nonorientable. In the first case, this gives the Euler characteristic 0 by before.

In the second case, see Jason's answer.

Singularity argument: A vector field on a surface has the sum of the indices of the singularities equal to its Euler characteristic. Since a nonvanishing section of the tangent bundle is a vector field with no singularities, we deduce that if either of the bundles is trivial, the Euler characteristic of the surface vanishes.

In case both are nontrivial, we can pullback to a cover as in Jason's answer.


问:
Looking for ONLY a hint on how to do this question
"Consider the polytope in $\mathbb{R}^4$ generated by taking the convex hull of the points $(\pm 1,0,0,0),(0,\pm 1,0,0),(0,0,\pm 1,0),$ and $(0,0,0,\pm 1)$. Describe all of its faces. How many are there in total?"

My issue is simply that I cannot see how you can do this (count the number of faces in 4-D) since I do not know how the vertices are connected.  For instance, is a vertex connect to $3$ other vertexes etc.

I am only looking for a hint here.

答:
Hint: Technically, the polytope is one of its own "faces"; I'll refer to the rest of the faces as "proper" faces.

Each proper face corresponds to a set $S$ of vertices. Note that if the convex hull of $S$ intersects the interior of the polytope, then $S$ does not lie within a face. Note that if the affine subspace "spanned" by a subset is the same for sets $S_1,S_2$, then these subsets correspond to the same face.

With that said: consider pairs of vertices. Then consider subsets of $3$ non co-linear vertices. Then consider subsets of $4$ non co-planar vertices. Among these sets, which have a convex hull that intersects the interior? Which sets correspond to the same face?


问:
Relatively simple proof that real-valued Dirichlet eta function is increasing, and also a query of the function at / about $x = 0$
This question/ series of questions arose from just curiosity of alternating series.

The Dirichlet eta function is usually defined for all complex input (in the branch of maths known as analytic number theory).

For this question, define the real-valued Dirichlet eta function for $x \in (0, \infty):$ $$\eta\ (x) = \sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^x} = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} + ... $$

$\eta(1)$ is known to be $\ln(2)$, and this thread concludes that $\eta(\frac{1}{2}) \approx 0.6048986434$

In fact, it appears that the concavity of the Dirichlet eta function was only proved as recently as 2015:

https://www.sciencedirect.com/science/article/pii/S0022314X15000220

I'm interested in whether there's a relatively nice and simple proof that $\eta\ (x)$ is increasing on $x \in (0, \infty)$

My other query is that, surely $\eta\ (x)$ is undefined at $ x = 0 $ because $1 - 1 + 1 - 1 + ...$ does not converge. However, several sources state that $ \eta\ (0) = 1/2$. For example, at the bottom of the Wolfram/Mathworld page on it:

Who is wrong, me or them? (Probably me, but why?)

And finally, assuming I'm right on this, is there a relatively nice and simple proof that $ \lim_{ x \to 0^+ } \eta\ (x) = 1/2$ ?

Progress on first question:

$\eta\ (x) = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} + ... = 1 - 2^{-x} + 3^{-x} - 4^{-x} + ....$

This is increasing $ \impliedby \eta'\ (x) > 0$, so if we can show $\eta'\ (x) > 0 $ for $x \in (0, \infty)$, then we are done.

$\eta'\ (x) = \frac{d}{dx}(1 - 2^{-x} + 3^{-x} - 4^{-x} + ....) = 2^{-x}ln(2) - 3^{-x}ln(3) + 4^{-x}ln(4) + .... = \frac{ln(2)}{2^x} - \frac{ln(3)}{3^x} + \frac{ln(4)}{4^x} - \frac{ln(5)}{5^x} ...$ How do I show this converges to a positive number for all $x>0$?

答:
Answer to the first part: "Surely $\eta (x)$ is undefined at $x=0$ because $1−1+1−1+... $ does not converge."

By usual way we do series, the sequence of partial sums, the sequence does not converge. However, there are other ways to carry out limits. Wikipedia states that:

The Cesàro sum is defined as the limit, as n tends to infinity, of the
sequence of arithmetic means of the first n partial sums of the
series.

The Abel summation seems to be perhaps more relevant, but I don't understand it at this time.

See:

en.wikipedia.org/wiki/Dirichlet_eta_function#Particular_values and en.wikipedia.org/wiki/Divergent_series#Abel_summation and
en.wikipedia.org/wiki/Ces%C3%A0ro_summation

$$$$

Answer to the second part: See:

https://mathoverflow.net/questions/180716/prove-that-the-dirichlet-eta-function-is-monotonic

However, if someone finds a simpler proof, continuing with my differentiation attempt for example, then of course by all means feel free to post an answer. Edit: in fact, I've asked the derivative method as a separate question here.


问:
$(\cdot)^r$ is continuous function for $r \in \mathbb{R}_{\ge 0}$
I'm trying to prove this proposition, and getting nowhere: $(\cdot)^a$ is continuous function $\mathbb{R}_{>0} \to \mathbb{R}$ for $a \in \mathbb{R}_{\ge 0}$.

I'm also having difficulty manipulating the function $(\cdot)^a$; the only definition I know of this is through the exponential function, which is not ideal for proving things like this. Would anyone have a nice proof of the above proposition from first principles? I tend to get confused when the answers are compact, so please consider writing things out please!

答:
A shortcut: $x^a=e^{a\ln x}$ so it s enough to show that $x\to e^x$ is continuous.

By definition, $e^x=\underset{n\to \infty}\lim\left(1+\frac xn\right)^n$ and it is easy to show that this sequence is increasing (apply Bernoulli's inequality). Then, $e^x\ge \left(1+\frac xn\right)^n\ge  1+x.$ From this we get also, $e^{-x}\ge 1-x$ so $e^x\le \frac{1}{1-x}$ if $x<1.$ Hence, $1+x\le e^x\le \frac{1}{1-x}$ if $0<x<1.$

Using the above, as soon as $|x-x_0|<1$, we get

$1+(x-x_0)\le e^{x-x_0}\le \frac{1}{1-(x-x_0)}\Rightarrow x-x_0\le e^{x-x_0}-1 \le \frac{1}{1-(x-x_0)}-1$

so $|e^x-e^{x_0}|=e^{x_0}|e^{x-x_0}-1|\to 0$ as $x\to x_0.$


问:
$A^3+A=0$ We need to show $\mathrm{rank}(A)=2$

Let $A\ne 0$ be a $3\times 3$ matrix with real entries such that $A^3+A=0$. We need to show $\mathrm{rank}(A)=2$.

$\det A(A^2+I)=0\Rightarrow\det A=0\Rightarrow \mathrm{rank}(A)<3$,  Suppose $\mathrm{rank}(A)=1$, Then I showed one matrix with rank $1$ which do not satisfies the given relation, is my answer is ok? Thank you for help and discussion

答:
The "standard" way to solve this kind of problems is to examine the minimal polynomial of $A$. Yet, for this particular problem, there are other approaches as well. Here is one of them:

By considering the determinants of both sides of $A^3 = -A$, argue that $\operatorname{rank}(A)\ne3$.

If $\operatorname{rank}(A)=1$, then $A=uv^T$ for some nonzero vectors $u,v\in\mathbb{R}^3$. Show that $A^3+A$ is a nonzero scalar multiple of $A$ and hence it cannot be zero.

As $\operatorname{rank}(A)\ne3,1$ and $A\ne0$ by assumption, the only possibility is that $\operatorname{rank}(A)=2$.


问:
How many triangles are in this figure?
How many triangles are there in the given figure?

I counted 12 but the solution manual says 14. If you have any tips on counting triangles in figures like this please advise. I have to prepare for aptitude test.

答: There are, indeed, 14 triangles:

Title: How many triangles are in this figure?

Body: There are, indeed, 14 triangles:


答:
To complement Connor's answer:

Classifing spaces argument:  Because $M$ is orientable, its tangent bundle is classified by a map $\phi:M\rightarrow BSO(2)=\mathbb{C}P^\infty$.  The Euler class of the bundle (which gives $\pm \chi(M)$ when paired with a fundamental class) is by definition the element $\phi^\ast(z)$, where $H^\ast(\mathbb{C}P^\infty;\mathbb{Z})\cong \mathbb{Z}[z]$.  Identifying $H^2(M;\mathbb{Z})$ with $\mathbb{Z}$, $\phi^\ast(z)$ can directly be interpreted as $\pm\chi(M)$.

So, we need to show that the condition that the tangent bundle splits implies that $\phi^\ast(z) = 0$.

Now a splitting of the tangent bundle implies a reduction of the structure group to $B:=BS(O(1)\times O(1))\subseteq BSO(2)$.  That is, $\phi$ factors through the inclusion $B\rightarrow BSO(2)$.  Since $B\simeq \mathbb{R}P^\infty$, and $H^2(\mathbb{R}P^\infty;\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$, we see that on degree $2$ cohomology, $\phi^\ast$ factors as a composition $\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}$.  Since any homomorphism $\mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}$ is trivial, we deduce $\phi^\ast(z) = 0$ as desired.

By the way, the assumption that $M$ is orientable is not needed.  For, if $M$ is non-orientable, and $\pi:\widetilde{M}\rightarrow M$ is the orientation double covering, then the fact that $T\widetilde{M}\cong \pi^\ast TM$ shows that splitting of $TM$ gives a splitting of $\widetilde{M}$.  By the orientable case, $\chi(\widetilde{M}) = 0$, from which it follows that $\chi(M) = 0$ as well.

Title: Splitting of the tangent bundle and Euler characteristic of surfaces

Body:
Let $M$ be a be a closed orientable surface, and suppose that its tangent bundle $TM$ splits into a direct sum of line bundles.

How to prove that $M$'s Euler characteristic is zero?

Unfortunately, I am entirely unfamiliar with Characteristic Classes, but I would accept a solution based on properties of these objects;

if there is  a reference for a book where I could find this claim or details of a proof, that would be great.


问:
Volume of solid rotated about the x-axis
I am to find the volume of the area $R$ bounded by the curve $x=y^2+2$, $y=x-4$ and $y=0$.
I have already found the points of intersection by first setting the lines equal to each other and used the quadratic formula:
\begin{align*}
    y^2+2=y+4
    \\-y^2+y+2=0
    \\
    \\y_{1,2}=\frac{1\pm3}{2}\\
    \\y_1 = 2 \\y_2=-1
\end{align*}

With regard to y:

\begin{align*}
A=\int_0^2{(y+4)-(y^2+2)}dy\\
A=\left[(2\cdot2)+\frac{2^2}{2}-\frac{2^3}{3}\right]-\left[(2\cdot0)-\frac{0^2}{2}-\frac{0^3}{3}\right]\\
A=4+2-\frac{8}{3}\\
A=\frac{10}{3}
\end{align*}

Then I tried finding the volume of $R$:

We have:

$f(y)=2+y-y^2 dy $
\begin{align*}
V= 2\pi\int_a^byf(y) dy\\
=2\pi\left(\int_2^4 y^3+2ydy\right)+2\pi\left(\int_4^6(y^3+2y)-(y^2+4y)dy\right)\\
=4\pi\left(\left[\frac{y^4}{4}+y^2\right]_2^4 + \left[\left(\frac{y^4}{4}+y^2\right)-\left(\frac{y^3}{3}+2y^2\right)\right]_4^6\right)\\
=4\pi \cdot 290\\
= \frac{\pi}{2}(145)
\end{align*}

But the right answer should be $16\frac{\pi}{3}$. What am I doing wrong?

答:
Assuming $R$ is the region in the first quadrant (i.e. with $x\ge0$ and $y\ge0$), then the volume - using cylindrical shells - is

$$\begin{align}
V&=2\pi\int_0^2y((y+4)-(y^2+2))\,\mathrm dy\\[1ex]
&=2\pi\int_0^2(2y+y^2-y^3)\,\mathrm dy\\[1ex]
&=2\pi\left[y^2+\frac{y^3}3-\frac{y^4}4\right]_0^2\\[1ex]
&=2\pi\left(2^2+\frac{2^3}3-\frac{2^4}4\right)=\boxed{\frac{16\pi}3}
\end{align}$$

Your error is in the limits of integration. You seem to be using the bounds for the variable $x\in[2,6]$, not $y\in[0,2]$.

If you did want to set up an integral using the bounds for $x$, you can use disks and washers:

$$\begin{align}
V&=\pi\left(\int_2^4\left(\sqrt{x-2}\right)^2\,\mathrm dx+\int_4^6\left(\left(\sqrt{x-2}\right)^2-(x-4)^2\right)\,\mathrm dx\right)
\end{align}$$

and you would find this to have the same value.


问:
Proving whether or not a certain recursively defined sequence is a Cauchy sequence or not
Question:

Let $\{x_{n}\}$ be a recursively defined sequence defined as:
$$x_{1} = \frac{1}{2}$$
$$x_{n+1} = \frac{1-x_{n}}{4}$$

I want to show first that $|x_{n+1}-x_{n}| \le \frac{1}{4}|x_{n}-x_{n-1}|$. Using this, I have to prove that the sequence is Cauchy and then find its limit. Here is what I have done:

Part 1:

$$|x_{n+1}-x_{n}| = \left|\frac{1-x_{n}}{4} - \frac{1-x_{n-1}}{4}\right| \\
=\left|\frac{-x_{n}}{4} + \frac{x_{n-1}}{4}\right|
= \frac{1}{4}|-x_{n}+x_{n+1}|$$

But, I do not know how to go from this equality to the inequality that the question has. How may I manipulate the equation so that the $=$ changes to a $\le$?

Part 2:

Now as for the second subpart where I have to show that the sequence is Cauchy, I am not sure where to start, so I started with the definition of Cauchy sequence, which is as follows:

A sequence $\{a_{n}\}$ is called a Cauchy sequence if $\forall \epsilon>0 \exists      N \in \mathbb{N}$ s.t. if $m,n \ge N$, then
$$|a_{m}-a_{n}| < \epsilon$$
But the problem I face is that if I try to use this definition and try to apply it here, I am getting that:
$$x_{m}-x_{n} = \left|\frac{-x_{m-1}}{4}+\frac{x_{n-1}}{4}\right|$$
But after this, how can I use my previous result to prove that this is always bounded by $\epsilon$?

Could someone help me?

答:
You have found $|x_{n+1}-x_n| =\frac14|-x_n+x_{n-1}| = \frac14|x_n-x_{n-1}|$, which satisfies the inequality. If you insist, write $|x_{n+1}-x_n| \le |x_{n+1}-x_n| = \frac14|x_n-x_{n-1}|$ and now you have the inequality.

Using this we see that for any $n$:

$$|x_{n+1}-x_n|\le\frac14|x_n-x_{n-1}| \le \frac1{4^2}|x_{n-1}-x_{n-2}| \le\cdots\le\frac1{4^{n-1}}|x_2-x_1|=\frac1{2^{2n-2}}\cdot\frac38 = \frac3{2^{2n+1}}$$

The "$\cdots$" can be made rigorous via induction on $n$.

Now suppose $m > n$. Then

\begin{align}|x_m-x_n| &\le |x_m-x_{m-1}| + |x_{m-1}-x_{m-2}| + \cdots + |x_{n+1}-x_n|\\
&=\sum_{k=n}^{m-1}|x_{k+1}-x_k|\\
&\le\sum_{k=n}^{m-1}\frac{3}{2^{2k+1}}\\
&<\sum_{k=n}^{\infty}\frac{3}{2^{2k+1}}\\
&=\frac {3}{2^{2n+1}}\div\left(1-\frac14\right)\\
&=\frac1{2^{2n-1}}\end{align}

Using this upper bound, it should be easy to find the $N \in \mathbb N$ for any arbitrary $\epsilon >0$.


问:
Find the integrating factor of the differential equation $(y\log y)dx = (\log y-x)dy$
I am familiar with linear differential equations of the form
$$\frac{dy}{dx} + P(x)y = Q(x)$$

But this equation doesn’t seem to be of that form, or any other ones I know. It could be converted into the linear form, but I don’t know how. How should I simplify?

答:
$$(y\log y)dx = (\log y-x)dy$$
Rewrite the DE as:
$$\log y\,dx = (\log y-x)\dfrac 1 ydy$$
$$\log y \,dx = (\log y-x)d ( \log y)$$
$$\log y \,dx +x d \log y= \log yd ( \log y)$$
$$d(x\log y )= \log yd ( \log y)$$
$$d(x\log y )= \dfrac {d ( \log^2 y)}2$$
Inetegrate.


问:
$P\left( n,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ Disaggregating Tail of Poisson
I have a Poisson tail $P\left( x,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ which is sum of two independent Poisson distribution with rate $\lambda_1$ and $\lambda_2$. I am trying to write the distribution tail  (probability of $x$ events or more) in terms of two Poisson tails. So I am trying to disaggregate it.

where: $P\left( x,\lambda T \right)=\sum\limits_{i=x}^{\infty }{p\left( i,\lambda T \right)}$

Thanks!

答:
If I understand your question, yes you can. It is a well known result that the sum of two independent Poisson random variables is another Poisson random variable whose parameter is the sum of the constituent parameters. One can show this in a straight forward way, but it is easier to do so via mgfs:
If $X_1 \sim $Poisson$(\lambda_1)$ and $X_2 \sim $Poisson$(\lambda_2)$ is independent of $X_1$ then the mgf for $X_i$ is $\mathbb{E}[e^{sX_i}] = e^{\lambda_i(e^s-1)}$ and so
$$
  \mathbb{E}[e^{s(X_1 + X_2)}]
= \mathbb{E}[e^{sX_1}]\mathbb{E}[e^{s X_2}]
= e^{\lambda_1(e^s-1)}e^{\lambda_2(e^s-1)}
= e^{(\lambda_1 + \lambda_2)(e^s-1)}.
$$</p>


问:
How do I know the number of cards which have different colors from a set?
The problem is as follows:

In an porcelain jar there is a set of $2016$ cards. These cards have a
numbered printed on their faces and those are $1$ to $2016$, assume
there is no repetition. These cards have a peculiar characteristic.
The cards such that the sum of their numbers are the same are of the
same color, and those that have these different sums are of a
different color. Using this information find the number of different
colored cards there are inside the jar.

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{26 cards}\\
2.&\textrm{27 cards}\\
3.&\textrm{28 cards}\\
4.&\textrm{29 cards}\\
5.&\textrm{25 cards}\\
\end{array}$

I'm a bit confused with this specific question. How exactly should I assess what is it intended?.

From reading at the problem, my interpretation is that there are two kinds of sets of cards.

One group has cards whose numbers add up to the same amount as their partner. And these happen to be the same color

The other group is the opposite. They do not have a pair that adds up to the same number as the numbers that appear to be written on it. And these happen to have a different color

Is this part correct?. If that's so. Then it comes into play the challenging part. How to distinguish which from which set?

What I attempted to diagnose first was the extreme situation:

This happens with $1999$

Since the maximum is $2016$ there will not be $9991$. In other words there is no reverse order. But that was the only card which I identified which must be a different color.

But how about the card below that number:

$1998$, it just happens that $1+9+9+8=27$

and this also occurs with $999$, as $9+9+9=27$. Hence, what to do here?. Does it exist a way to find which group and how many of these cards must have a different color?.

But again, is this the right way to approach this problem?.

Since I feel lost in this question. I'd like someone could help me with the most detailed explanation as possible and be include a step by step analysis of what should be done to solve this problem. Can someone help me?.

答:
The way I understood the problem statement is the way you sketched it; the cards would be coloured according to the sum of their digits. For example, the cards labeled 1, 10, 100 and 1000 would have the same colour and any cards with different sums would have to be coloured differently. For example, you would need a new colour for the cards numbered $2, 11, 101$ etc.

For the case $a = 1, b \in \{ 0,..., 9 \}, c \in \{0, ...,9 \}, d \in \{0, ..., 9\}$ you can get at most $a+b+c+d = 1+27 = 28$. No number abcd with  $2016 \geq abcd >2000$ or $abcd < 1000$ can get a higher sum so this means that you need at most $28$ colours to group your cards by the sum of their digits. On the other hand it seems that by choosing b, c , d in the above case with $a=1$ suitably you can sum to any number between 1 and 28, so that there would be in fact cards of $28$ different colours in the jar.


问:
Give the context-free grammar that generates the pushdown automaton language
The pushdown automaton :

From Here i figured out that having an idea of the language is a good first step so i came to the conclusion that the language represented by this pushdown automaton is something that looks like this :
$a^n b(b^*aa^*bb^*)^*a^n :n \in\mathbb{N}$ (correct me if i am wrong)

But from there i am kinda lost on how to approach the problem to figure out the CFG, in class were not really told a precise technique to do this so i would like to know how someone would proceed to solve this problem.

Thanks a lot.

EDIT my attempt: could someone validate if it's good or what i am missing thanks

$S \Rightarrow aSa$

$S \Rightarrow B$

$S \Rightarrow bB$

$B \Rightarrow aC$

$C \Rightarrow \epsilon$

$C \Rightarrow bB$

$B \Rightarrow b$

答:
I won't give a full solution, but I will give some pointers towards the answer. Note that if $X$ is a grammar for a language, and we want to wrap it with the same number of $a$s on either side (possibly none), then the rule
$$Y\to X\;|\;aYa$$
will do that for us.

Also if $X$ is a grammar for a language $L$, the grammar for $L^*$
is
$$Y\to \varepsilon \;|\; XY.$$

Dealing with a $+$ in the language is easy. If $B$ and $C$ are grammars for $L$ and $M$, then the grammar for $L+M$ is
$$A\to B\;|\; C.$$

You should assign variables that build small parts of your language, like $a^*$ and $b^*$ and piece them together to make the full grammar. I hope this helps.


问:
Show that the finite product of compact (or connected) spaces are compact (or connected) without homeomorphisms
I know this question has been asked many times before. The problem is, every answer I have found so far invokes homeomorphisms. My class has not discussed these, and my professor said this problem is possible without homeomorphisms.

Is anyone able to shine some light on how I'd do this without homeomorphisms?

And if possible, I'd also like a brief explanation of what a homeomorphism is for my future.

To write it here, the two problems are:

Show that the finite product of compact spaces are compact.

Show that the finite product of connected spaces are connected.

Thank you!

答:
It's enough to show the product of 2 compact spaces is compact, and the general case is given by induction.

Let $X,Y$ be compact. Note that for any $y\in Y$ the map $i_{y} : X \rightarrow X \times Y$ given by $i_{y} (x) = (x,y)$ is continuous, and so its' image - $X \times \{y\}$  - is compact as the continuous image of a compact space, for any $y \in Y$. The same argument shows $\{x\} \times Y$ is compact for any choice of $x \in X$. Let $\{ u_{\alpha} \}$ be an open cover of $X \times Y$. For any $y \in Y$, this is also an open cover of $X \times \{y\}$ so it admits a finite subcover $u_{\alpha_{1}} (y), ... , u_{\alpha_{n(y)}} (y)$. Let $U(y) = \cup_{i=1}^{n(y)} u_{\alpha_{i}} (y)$. Note that the set $\{ U(y) | y \in Y$ is an open cover of $ \{x\} times Y$ for arbitrary choice of $x \in X$, so it admits a finite subcover, $U(y_{1}), ... , U(y_{m})$. It is now easy to verify that $\{ u_{\alpha_{k}}(y_{j}) | 1\leq j \leq m, 1 \leq m \leq j(y) \}$ is a finite subcover of $X \times Y$.


问:
Deriving a general formula for the number of onto functions from the given theorem
I want to use this:

Let $A_1, A_2, \ldots, A_n$ be $n$ sets in a universe $U$ of $N$ elements. Let $S_1 = \sum_jN(A_j), \ S_2 = \sum_{i, j}N(A_iA_j), \ldots, S_k = \sum N(A_{j_1}A_{j_2\ldots A_{j_k}}), \ldots, S_n = N(A_1A_2\ldots A_n)$

Then $$ N(\overline A_1\overline A_2\ldots \overline A_n) = N(U) - S_1 + S_2 - S_3 + \ldots + (-1)^kS_k+ \ldots + (-1)^nS_n$$

to derive this:

The number of onto functions $\{1, 2, 3, \ldots, r\} \to \{1, 2, 3, \ldots, n\}$ is $$\sum_{j=0}^n\binom nj(-1)^j(n - j)^r$$.

So,

Let $S_0 = N(U).$ Then $$N(U) - S_1 + S_2 - S_3 + \ldots + (-1)^kS_k+ \ldots + (-1)^nS_n \\ = \sum_{k=0}^n(-1)^kS_k \\ = \sum_{k=0}^n(-1)^k\sum N(A_{j_1}A_{j_2\ldots A_{j_k}}) \\  = \sum_{k=0}^n(-1)^k \binom nkN(A_{j_1}A_{j_2\ldots A_{j_k}}) \\ = \sum_{k=0}^n(-1)^k \binom nkN(A_{j_k})$$

Now let $A_{j_k}$ be the set of functions that miss $k$ elements.  Then $N(A_{j_k})$ is the number of such functions which is given by $(n - k)^r.$

The derivation above seems to work iff $N(A_{j_1}A_{j_2\ldots A_{j_k}}) = N(A_{j_k})$ which I am not sure is true. If so, how do we prove that? Thanks.

答: Notice that $N(A_{j_k})$ has $(n-1)^r$ because you are missing just $1$ element. Instead, $N(A_{j_1}A_{j_2}\cdots A_{j_k})$ is $(n-k)^r$ because you are missing now $k$ elements instead. They are not equal is just that the last equality is not true.


问:
For all closed convex curves, let central convex curves circumscribe them, what is the minimum of $B/A$? ($A$ and $B$ are corresponding areas)
I guess the answer is $2$, but I cannot prove it.

I mean supremum of the minimum $B/A$ among all $A$.

Notice:

Central convex curve means the convex curve is symmetric about a point in it, that is, when you rotate the curve 180 degrees about the point, it comes back to the original curve.

For the triangle, which is considered the most asymmetric, $2$ is a lower bound. That is why I think $2$ is a lower bound for all other convex curves.

I need to calculate a minimum $r=B/A$ such that for all of the closed convex curves，there always exists a centrally symmetric convex curve (with area $A\cdot r$) circumscribed about it. A lot of thanks.

答:
I understand your question to be asking about the supremum over all planar convex sets $A$ of the ratio $\text{Area}(B)/\text{Area}(A)$, where $B$ is chosen to be a centrally-symmetric curve containing $A$ of minimal area.

This supremum is $2$. To attain this bound, take $A$ to be any triangle; any possible $B$ must contain the inversion of $A$ about its center, so $B$ has area at least that of the best convex hull of $A$ and a translation of $-A$. But for a triangle, such convex hulls can be pretty easily analyzed to see that their area is always at least twice that of the starting triangle. (Hint: fix a side of the triangle, and consider the area of $B$ as a function of the altitude between the side of $A$ and the corresponding side of $-A$.)

To see that we can obtain a ratio of at most $2$ for any convex set $A$, inscribe $A$ in a rectangle with one side parallel to its diameter (i.e., the line between two maximally-distant points on $A$). By considering the points where $A$ touches the sides of this rectangle and the fact that it is convex, we can see that $A$ contains a quadrilateral with area exactly half that of the rectangle, so the rectangle is of area at most twice that of $A$. But rectangles are centrally symmetric, so we are done.


问:
Computing real definite integral using residue theorem
I am trying to solve integral $\int_{-∞}^\infty \frac{cos(x)dx}{(x^2+a^2)(x^2+b^2)}$ using residue theorem. I found example of that with very similair integral, which is $\int_{0}^\infty \frac{cos(x)dx}{x^2+b^2}$. Everything would be clear, except one thing. When we substitute x $\to$ z and make our complex function, in the example cos(x) is replaced by $e^{ix}$. Why only $e^{ix}$, and not $\frac{e^{ix} + e^{-ix}}{2}$, since cos(x) = $\frac{e^{ix} + e^{-ix}}{2}$? Does this mean, that I should use this substitution on my integral, too?

答:
Note that $\cos(x)=\text{Re}(e^{ix})$.  Hence, we can write

$$\int_{-\infty}^\infty \frac{\cos(x)}{(x^2+a^2)(x^2+b^2)}\,dx=\operatorname{Re}\left(\int_{-\infty}^\infty \frac{e^{ix}}{(x^2+a^2)(x^2+b^2)}\,dx\right)\tag1$$

which facilitates the analysis.  To see this, we evaluate the contour integral $I(a,b)$, $ab\ne0$, given by

$$I(a,b)=\oint_C \frac{e^{iz}}{(z^2+a^2)(z^2+b^2)}\,dx\tag2$$

where $C$ is the closed contour comprised of $(i)$ the real line segment from $-R$ to $R$ and $(ii)$ the semi-circular arc centered at $z=0$ with radius $R$ from $R$ to $-R$.

We evaluate $(2)$ by application of the residue theorem.  Then, we let $R\to \infty$ and take the real part to find the value of $(1)$

$$\begin{align}\int_{-\infty}^\infty \frac{\cos(x)}{(x^2+a^2)(x^2+b^2)}\,dx&=\operatorname{Re}\left(2\pi i \operatorname{Res}\left(\frac{e^{iz}}{(z^2+a^2)(z^2+b^2)}, z=i|a|\right)\\+2\pi i \operatorname{Res}\left(\frac{e^{iz}}{(z^2+a^2)(z^2+b^2)}, z=i|b|\right)\right)
\end{align}$$

And the rest is left to the interested reader.


问:
Nearest Kronecker Product of a commutation matrix
Given a commutation matrix $K$ of the size $n²$ by $n²$. How can I find matrix $A$ and $B$ both of the size
$n$ by $n$ such that the equation below is (approximately) satisfied by finding its Nearest Kronecker Product.

$K = A ⊗ B$.

答:
Write $A = \sum_{i,j} a_{ij}e_i e_j^T$, $B = \sum_{i,j} b_{ij}e_ie_j^T$.

First, we note that
$$
K = A \otimes B \iff\\
K = \sum_{i,j,p,q} a_{ij}b_{pq} (e_ie_j^T) \otimes (e_pe_q^T) \iff\\
K = \sum_{i,j,p,q} a_{ij}b_{pq} (e_i \otimes e_p)(e_j \otimes e_q)^T \iff\\
\operatorname{vec}(K) = \sum_{i,j,p,q} a_{ij}b_{pq} (e_j \otimes e_q \otimes e_i \otimes e_p) \iff\\
P\operatorname{vec}(K) = \sum_{i,j,p,q} a_{ij}b_{pq} (e_j \otimes e_i \otimes e_q \otimes e_p) \iff\\
\operatorname{vec}^{-1}[P\operatorname{vec}(K)] = \sum_{i,j,p,q} a_{ij}b_{pq} (e_j \otimes e_i)(e_q \otimes e_p)^T \iff\\
\operatorname{vec}^{-1}[P\operatorname{vec}(K)] = \operatorname{vec}(A)\operatorname{vec}(B)^T
$$
For the permutation matrix $P$ satisfying
$$
P(u \otimes v \otimes x \otimes y) = u \otimes x \otimes v \otimes y.
$$

The commutation matrix $K = K^{(n,n)}$ can be rewritten as
$$
K = \sum_{i,j = 1}^n (e_i \otimes e_j)(e_j \otimes e_i)^T,
$$
so that
$$
\operatorname{vec}(K) = \sum_{i,j} e_j \otimes e_i \otimes e_i \otimes e_j \implies\\
P \operatorname{vec}(K) = \sum_{i,j} e_j \otimes e_i \otimes e_i \otimes e_j
= \operatorname{vec}(K) \implies\\
\operatorname{vec}^{-1}(P\operatorname{vec}(K)) = K.
$$
All together, we are looking for $A,B$ that give us the best (approximate) solution to
$$
K = \operatorname{vec}(A) \operatorname{vec}(B)^T.
$$
Because $K$ is an orthogonal matrix, the EYM theorem tells us that every rank-$1$ matrix with spectral norm $1$ is an equally good approximation for $K$. In other words, for every $A,B$ we have
$$
\|K - A \otimes B\|_F^2 \geq n^2 - 1,
$$
and $\geq$ can be replaced with $=$ if and only if $\sum_{i,j,p,q} a_{ij}^2 b_{pq}^2 = 1.$


答:
With appropriate substitution(s) one can extract a separable ODE.

$$y\log y\,\mathrm dx=(\log y-x)\,\mathrm dy$$

Substitute $v(x)=\log y(x)\iff y(x)=e^{v(x)}$ and $\mathrm dv=\frac{\mathrm dy}y$:

$$\begin{align}
ve^v\,\mathrm dx&=(v-x)e^v\,\mathrm dv\\
v\,\mathrm dx&=(v-x)\,\mathrm dv\\[1ex]
\frac{\mathrm dv}{\mathrm dx}&=\frac v{v-x}
\end{align}$$

Substitute $w(x)=v(x)-x\iff v(x)=w(x)+x$ and $\mathrm dw=\mathrm dv-1$:

$$\begin{align}
\frac{\mathrm dw}{\mathrm dx}+1&=\frac{w-x}w\\[1ex]
\frac{\mathrm dw}{\mathrm dx}&=\frac xw\\[1ex]
w\,\mathrm dw&=x\,\mathrm dx
\end{align}$$

Now integrate both sides and solve:

$$\begin{align}
\frac{w^2}2&=\frac{x^2}2+C\\[1ex]
w^2&=x^2+C\\[1ex]
(v-x)^2&=x^2+C\\[1ex]
(\log y-x)^2&=x^2+C
\end{align}$$

Title: Find the integrating factor of the differential equation $(y\log y)dx = (\log y-x)dy$

Body:
I am familiar with linear differential equations of the form
$$\frac{dy}{dx} + P(x)y = Q(x)$$

But this equation doesn’t seem to be of that form, or any other ones I know. It could be converted into the linear form, but I don’t know how. How should I simplify?


问:
optimal way to play a game with uniform random variables and opportunity to redraw
Suppose you are playing a game where you and an opponent both receive a uniform, random number in interval $[0,1]$. Both of you can choose whether or not to get a redraw after receiving the first number, but neither person knows whether the opponent re-draws, or what the values of the non-final draws are. You can only redraw at most once. In the end, the final numbers are compared, and whoever has the largest number wins. I want to figure out what is the probability I will win?

My intuition is that the strategy should be some number $a$ such that I will redraw if my first number is smaller than $a$. If I assume the opponent has the same strategy but with a number $b$, I can separate into two cases: $a \leq b$ or not. In the former case, there are 3 cases for my final result: my final result is less than a, my final result is in $[a, b]$, or my final result is in $[b, 1]$. The same is true for my opponent giving a total of 9 scenarios between the two of us. If both our answers are in the same interval, I think there is a 50% chance I will win. Otherwise, the intervals are disjoint, and I have either a 0 or 100% chance I will win. Since all 9 of these scenarios are also disjoint, I can add these probabilities together to get an answer for the probability I will win in terms of $a$ and $b$, but when I take a derivative to find a maxima, the answer for $a$ still depends on $b$.

I'm having trouble figuring out what $a$ should be since I don't have a game theory background.

答:
I modeled a discrete approximation of this in Excel as follows, and it looks interesting. The Excel file is here. While this doesn’t answer your question, it suggests the optimal choice for $a$ is over $1/2$, and that there is only one value of $a$ for which your opponent has no winning strategy. (If I had to guess, I’d guess it’s 0.625.)

The game I modeled is the following one:

Aisha and Bob each spin a 26-equal-segment wheel labeled a-z twice. Each player has a secret threshold letter (shown in the example below with Aisha's as m, Bob's as r; these correspond to your $a$ and $b$). If the first spin result is at least the threshold, the player's first spin counts (orange shading); otherwise, the player's second spin counts (pink shading). The winner is the player whose result is highest (closest to z). [I used letters instead of equally-spaced numbers in $[0,1]$ to make it easier to look at and understand.]

The top $26$ by $26$ grid shows Aisha's result for each of the $26^2=676$ equally likely possibilities for the two spins. The bottom grid shows Bob's result.

Columns AO and AP show the distribution of results for each player.

The matrix of numbers at top right tabulates the number of winning games for Aisha among each of the $676^2$ equally likely results of each player spinning twice.

Let me explain the blue square, at the intersection of the m-column and the q-row. Aisha’s result is q $38/676$ of the time, and Bob's is m $17/676$ of the time. So the game outcome is q vs m (Aisha wins) $38x17 = 646$ times (out of all $676^2$ games), so we count $646$ wins for Aisha.

Summing over all $676^2$ games, Aisha wins $0.4734$ of the time with these secret thresholds (m for Aisha, r for Bob). Bob wins $0.4820$ of the time, giving Bob an edge of $0.00853$ wins per game.

What was initially counter-intuitive for me is that given a specific secret threshold for Aisha (say m), the optimal threshold for Bob isn't obvious. (For Aisha's m, Bob's best threshold is p. If Aisha's threshold is e or f, Bob's should be n.)

If Aisha's threshold is q, however, it looks like Bob has no winning strategy, but he can break even by using the same threshold. The lack of a winning strategy isn't surprising, since matching thresholds make it a symmetric game, but the fact that the only threshold that can't be beat is q was unexpected.

Perhaps you or someone else will try to prove that $a=0.625$ is the optimal strategy, guaranteeing at least a break-even game.


问:
Producing a large list of irreducible polynomials of given degree over a specific finite field using a CAS?
I'm curious if there are computer algebra systems that allows one to easily produce a list of irreducible polynomials of a given degree $n$ in $\mathbb{F}_p[X]$, where $p$ is a specific prime. Does Sage or GAP have any built-in capabilities to return this list? I've been trying to see if either of these can return of a list of polynomials of given degree, and then filtering said list by some built in function or method such as is_irreducible, but can't find anything in the documentation.

I can do it in a round about way constructing $X^{p^n}-X$ over $\mathbb{F}_p$ and factoring, but this does not return very convenient results, and produces many factors of lower degree. For example, I can find the $116$ irreducible polynomials of degree $6$ in $\mathbb{F}_3[X]$ with

```<code>sage: R = GF(3)['x']
sage: factor(R(x^(3^6)-x))
</code>```

and reading off the factors of degree $6$, but it's not in a convenient list form, and also produces all irreducible polynomials of degrees $1,2,3$ as well.

答:
GAP has a function AllIrreducibleMonicPolynomials(<degree>,<field>) that does exactly what you want. In your example:

```<code>gap> l:=AllIrreducibleMonicPolynomials(6,GF(3));
[ x^6+x^5+x^4+Z(3)^0, x^6-x^5+x^4+Z(3)^0, x^6-x^4+Z(3)^0, x^6-x^5+x^3+Z(3)^0,
  x^6+x^4+x^3+Z(3)^0, x^6-x^5-x^4+x^3+Z(3)^0, x^6+x^5-x^3+Z(3)^0,
  [ ... and so on]
</code>```

The function is used for example in enumerating conjugacy classes of GL.

Body:
GAP has a function AllIrreducibleMonicPolynomials(<degree>,<field>) that does exactly what you want. In your example:

gap> l:=AllIrreducibleMonicPolynomials(6,GF(3));
[ x^6+x^5+x^4+Z(3)^0, x^6-x^5+x^4+Z(3)^0, x^6-x^4+Z(3)^0, x^6-x^5+x^3+Z(3)^0,
  x^6+x^4+x^3+Z(3)^0, x^6-x^5-x^4+x^3+Z(3)^0, x^6+x^5-x^3+Z(3)^0,
  [ ... and so on]

The function is used for example in enumerating conjugacy classes of GL.


问:
Fitting circle in $f(x)=x^3-3x^2$ what is the radius?

The whole problem is the above.

I tried to solve it analytically but I couldnot find enough equation to solve it.

Assume the center is $(a,-r)$ where $r$ is the radius of the circle.

The points on the sphere $(x-a)^2+(y+r)^2=r^2$

and $(x,f(x))$ must satisfy it so if we denote the intersection points $(x_1,y_1),(x_2,y_2)$:

$$(x_1-a)^2+(x_1^3-3x_1^2+r)^2=r^2$$
$$(x_2-a)^2+(x_2^3-3x_2^2+r)^2=r^2$$

Are two equation with 4 unknown.

Can you give me elegant relations between these two unknowns?

答:
The following is only a hint, not an answer.

The rest of the equations involves the derivative, since the line connecting the center with a tangent point is perpendicular to the tangent line. Hence for $i=1,2$:

$$\frac{-1}{f'(x_i)} = \frac {-r-f(x_i)}{a-x_i}$$

which gives two extra equations

$$x_i^3-3x_i^2+r =\frac {x_i-a}{-3x_i(x_i-1)}$$

Familiar terms surface, but I'm not sure how solving these four equations can be elegant.


问:
Hausdorff topological group is topologically isomorphic to a subgroup closed of a Hausdorff topological group
I am studying group topology and came across this exercise

Show that every Hausdorff topological group is topologically isomorphic to a subgroup
closed of a Hausdorff topological group arcwise connected and locally arcwise connected.

In this question Every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups. the following theorem is mentioned

Theorem: Every Hausdorff topological group G is topologically isomorphic to a subgroup of the group of isometries Is(M) of some metric space M, where Is(M) is taken with the topology of pointwise convergence.

is it possible to use this theorem to prove the exercise? otherwise, how could you prove this?

Any hints would be appreciated!

答:
Here is a solution. Although I can't imagine that it is what the instructor had in mind for the exercise, it does completely destroy the problem.

It's a standard fact that any space $X$ can be embedded as a closed subspace of a contractible space. The usual construction is to use the cone $CX=X\times[0,1]/X\times\{1\}$ and embed $X$ as $X\times\{0\}$. While this is satsifactory for many applications, it has many faults. For one thing it does not preserve subspaces. Another is that it does not preserve separation properties past $T_2$. More relevantly for us is that $CX$ need not be locally contractible, and that $CX$ carries no group structure.

Here is a construction which remedies some of these defects. In particular it will embed any (Hausdorff) topological group into a contractible, locally contractible (Hausdorff) topological group. Note that every contractible space is path-connected. In the Hausdorff case we can replace 'path' everywhere with 'arc' (this is really a consequence of the Hahn–Mazurkiewicz Theorem, although see here for some details).

Let $X$ be a space. A right-continuous step function in $X$ is a map $f:[0,1)\rightarrow X$ for which there is a finite partition $t_0=0<t_1<\dots<t_n<1=t_{n+1}$ of $[0,1)$ such that $f$ is constant on $[t_i,t_{i+1})$ for each $i=0,\dots,n$. Let $EX$ denote the set of all right-continuous step functions $[0,1)\rightarrow X$.

For $t_0<t_1\in [0,1]$, $U\subset X$ open and $\epsilon>0$ let $N_\epsilon(t_0<t_1,U)\subseteq EX$ be the set of all $f\in EX$ with the property that the set $\{t\in [t_0,t_1)\mid f(t)\not\in U\}$ has Lebesgue measure $<\epsilon$. Topologise $EX$ by giving it the subbase $\{N_\epsilon(t_0<t_1,U)\mid t_0<t_1\in[0,1),\;U\subseteq X\;\text{open},\;\epsilon>0\}$. Note that a function $f\in EX$ has a neighbourhood subbase consisting of those sets $N_\epsilon(t_0<t_1,U)$ where $f$ is constant on $[t_0,t_1)$ and $f(t_0)\in U$.

There is a function $i_X:X\rightarrow EX$ which sends a point $x\in X$ to the constant step function function at $x$.

Let $X$ be a nonempty space. Then $EX$ is contractible and locally contractible. The map $i_X:X\rightarrow EX$ is an embedding, which is closed if $X$ is Hausdorff. If $X$ is $T_i$ for some $i\in\{0,1,2,3\frac{1}{2}\}$, then $EX$ if $T_i$. If $X$ is completely regular, then $EX$ is completely regular. If $X$ is first-countable/second-countable/separable/metrisable, then $EX$ is first-countable/separable/metrisable.

It's worth recording that $EX$ does not have all good properties that $X$ may have. The space $EX$ need not be normal, paracompact, locally compact, completely metrisable, or finite-dimensional, even when $X$ is.

The construction is functorial. A map $\alpha:X\rightarrow Y$ induces $E\alpha:EX\rightarrow EY$, $f\mapsto \alpha\circ f$, which is continuous and satisfies $E\alpha \circ i_X=i_Y\circ\alpha$. It can be shown that if $\alpha$ is an embedding, then so is $E\alpha$.

For nonempty spaces $X,Y$, the natural map $E(X\times Y)\rightarrow EX\times EY$  is a homeomorphism.

Now let $G$ be a topological group. The multiplication $m:G\times G\rightarrow G$ induces a map
$$\mu:EG\times EG\cong E(G\times G)\xrightarrow{Em}EG$$
and similarly the inversion $G\rightarrow G$, $g\mapsto g^{-1}$ gives rise to $\iota:EG\rightarrow EG$. It is easy to use the functorality to see that $\mu$ furnishes $EG$ with a continuous multiplication for which $\iota$ is a continuous inverse. Moreover, with these definitions the map $i_G:G\rightarrow EG$ is a homomorphism.

In summary;

Let $G$ be a topological group. Then $G$ embeds into a contractible, locally contractible topological group $EG$. If $G$ is Hausdorff, then so is $EG$, and moreover $G$ is closed in $EG$ in this case. If $G$ is abelian/divisible/torsion/torsion-free, then so is $EG$.

So, as promised, the exercise has been completely demolished. To keep the length somewhat sane I haven't included too many details. If you would like to follow them up, the construction is due to R. Brown and S. Morris in the joint paper
Embeddings in contractible or compact objects*, Coll. Math. 38 (1978), 213-222. Some further details are found in a follow up paper of the second author. (Edit: The topology I give above differs from that given in the reference. While I checked many of the details with my description, I have quoted many unchecked. You should believe the Brown-Morris paper before me.)

The construction has applications in topology, but was actually inspired by the group-theoretic problem. I believe it was in fact S. Hartman and J. Mycielski's paper On the imbedding of topological groups into connected topological groups Coll. Math. 5 (1958) 167-169, which inspired the construction.

In fact, given the name of the paper, I'd recommend you might like to start with this earlier paper :P.


问:
If $A$ is orthogonal, for any vector $x$ such that $Ax = b$, $\Vert x \Vert = \Vert b \Vert$
Is this statment true: For any vector $x$ such that $Ax = b$, $\Vert x \Vert = \Vert b \Vert$, if $A$ is orthogonal.

I was working on a proof for my linear algebra class, when I noticed that the entire proof could be reduced to simple algebraic work conditional on the following statement being true:

$$\text{For any vector $x$ such that $Ax = b$, $\Vert x \Vert_2 = \Vert b \Vert_2$, if $A$ is orthogonal.}$$

I have not found a straight answer to this question. I believe the statement is true but I am not sure about how to prove it.

Thanks in advance.

答:
Recall that $A$ is orthogonal if $A^TA = I$. Now look at $\Vert b \Vert_2^2$. Hence, we get that
$$\Vert b \Vert_2^2 = b^Tb = (Ax)^T (Ax) = x^T\underbrace{A^TA}_Ix = x^Tx = \Vert x \Vert_2^2$$</p>


问:
What is the minimal value of $a$ that will give me the greater result?
$(a+\frac{1}{8})^0*n$ $+$ $(a+\frac{1}{8})^1*n$ $+$$(a+\frac{1}{8})^2*n$ $+$ $(a+\frac{1}{8})^3*n$ + ... $+$ $(a+\frac{1}{8})^{log_8(n)}*n$

What is the minimal value of $a$ that will give me the greater result ?

I noticed that if $a=\frac{7}{8}$ then I can get $nlog(n)$.

I can use a hint.

Thank you!

答:
Simplify by $x=a+\frac{1}{8}$ to get $f_n(x)=n\sum_{k=0}^b x^k +nx^{log_8(n)}=\frac{1-x^{b+1}}{1-x}+nx^{log_8(n)}$ where $b=$ integer $\le log_8(n)$.

Set $f'_n(x)=0$ to find min for $x$.


问:
Evaluate $\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$, for integrable $f$
Let $f : [0, 1] \to \mathbb{R}$ be Riemann integrable. Evaluate
$$\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$$
I have tried applying by parts by assuming
$$\\ \int_{0}^1 f(x) \ dx = F(1)-F(0)$$
but it doesn't seem to get me anywhere. Do I have to use partitions and if so how do I go about it?

答:
Since $f$ is integrable, say $\int_0^1 f(x)\,dx = a$, it is bounded, say $m\le f(x)\le M$ for $x\in [0,1]$. In particular, $g(x) = f(x)-m \ge 0$. Then we have
$$
\int _0^1 x^{n+1} f(x)\,dx = \int _0^1 m x^{n+1}\,dx + \int_0^1 (f(x)-m)x^{n+1}\,dx
$$The first integral clearly vanishes in the limit. For the second integral, we can use the First MVT for integrals to deduce
$$
\int _0^1 (f(x)-m)x^{n+1}\,dx = c^{n+1} \int _0^1 f(x)-m\,dx = (a-m)c^{n+1},
$$for $c\in(0,1)$; this also vanishes in the limit.


问:
Borel $\sigma$-algebra of metric subspace
I am supposed to show that the Borel $\sigma$-algebra (i.e. the algebra generated by open sets under complementation and arbitrary unions) of a metrix subspace $Y$ of metric space $X$ is the same as the intersection of the original algebra with the subspace, i.e. $B_Y=\{A\cap Y:A\in B_X\}$. I am having some trouble with this and would appreciate some pointers, if not outright answers.

The right-to-left direction seemingly works by considering the inclusion map $\iota:Y\to X$, which is continuous (I think) and therefore measureable. Therefore, it takes elements of the original algebra $A\in B_X$ to elements of the subspace algebra $i^{-1}A\in B_Y$, but $i^{-1}A=A\cap Y$, therefore the RHS is a subset of $B_Y$. For the left-to-right direction I believe I would have to show that every open set in $Y$ is of the form $A\cap Y$, where $A$ is an open set in $X$ and I'm not quite sure how to do that.

答: Let $(X, \tau_X)$ a topological space and $Y \subseteq X$, the classical topology given to $Y$ is called the subspace topology $\tau_Y = \{Y \cap U | U \in \tau_X \}$. Now let's show that the induced metric on $Y$ is the same as the one induced by $X$. The topology induced by the metric of $X$ on $Y$ is generated by the basis of balls $B(y, r)$ with $y \in Y$ end $r \in \mathbb{R}$. If the ball lies in $Y$ then obviously $B(y, r) \cap Y = B(y, r)$. If $B(y, r)$ lies not entirely in $Y$ then obviously you have to take the intersection of the ball with $Y$ so again clearly both topologies are equivalent.


问:
Is my proof correct? (minimal distance between compact sets)
I'm working out the following problem form Ahlfors' Complex Analysis text:

"Let $X$ and $Y$ be compact sets in a complete metric space $(S,d)$. Prove that there exist $x \in X,y \in Y$ such that $d(x,y)$ is a minimum."

My attempt:

Define $E:=\{d(x,y): x \in X,y \in Y \}$. We will prove that $E \subset \mathbb R$ is compact.

Firstly we will prove that $E$ is bounded:

$X$ is compact, and therefore it is bounded. That is, there exist $x_0 \in X,r_1>0$ such that $d(x,x_0)<r_1$ for all $x \in X$. $Y$ is also compact, and similarly there exist $y_0 \in Y,r_2>0$ such that $d(y,y_0)<r_2$ for all $y \in Y$.

Now, for any $d(x,y) \in E$, we have $$d(x,y) \leq d(x,x_0)+d(x_0,y_0)+d(y,y_0)<r_1+d(x_0,y_0)+r_2 =:M.$$ This proves that $E$ is bounded.

And now we will prove that $E$ is also closed:

Let $a_n=d(x_n,y_n)$ be sequence in $E$, which is convergent in $\mathbb R$. We will prove that its limit $a:=\lim_{n \to \infty} a_n$ is in $E$.

$(x_n)$ is a sequence in the compact set $X$, therefore it admits a convergent subsequence $(x_{n_k}) \to \bar{x}$. $(y_{n_k})$ is a sequence in the compact set $Y$, and therefore it admits a convergent subsequence $(y_{n_{k_l}}) \to \bar{y}$. The sequence $(x_{n_{k_l}},y_{n_{k_l}})_l$ converges to $(\bar{x},\bar{y})$ in the product space, and the continuity of the metric gives $a=\lim_{l \to \infty} a_{n_{k_l}}=\lim_{l \to \infty} d(x_{n_{k_l}},y_{n_{k_l}})=d(\bar{x},\bar{y})$. This shows that $a \in E$ as required.

In summary I have shown that $E$ is compact, and from a well-known theorem it has a minimum.

Is this proof OK? I think that the completeness of $S$ is unnecessary.

Thank you.

答:
I just wanted to mention that your characterization of compactness is not quite correct. Compact is equivalent to closed and bounded in $\mathbb{R}^n$, but it is not true in general. In the case of metric spaces,a set is compact if and only if it is complete and totally bounded.

However, in this case is enough to see that $E$ is closed as Plop mentioned in his answer so you are fine. Another way to see it is to consider the function $d:S\times S\rightarrow\mathbb{R}$, which is continuous, so it must take a minimum on the compact set $X\times Y\subset S \times S$.


问:
Deriving the solution to Bernoulli differential equation using integrating factors
This is problem 20 chapter 1.9 in Differential Equations and Their Applications - Martin Braun (4:th ed).

The Bernoulli differential equation is given by

$$(dy/dt) + a(t)y = b(t)y^{n} \tag{1}.$$

Multiplying both sides by $\mu(t) = \exp(\int a(t) \ dt)$ we can reduce it to

$$\frac{d}{dt}(\mu (t) y) = b(t)\mu (t)y^{n} \tag{2}.$$

My task is to find an appropriate integrating factor to find the general solution of $(1)$ using $(2)$.

The book suggest to divide by a expression of $y$ which gives me the idea of rewriting $(2)$ to

$$-b(t)\mu(t)y + y^{1-n}\frac{d}{dt}(\mu(t)y) = 0 \tag{3}.$$

I know the equation

$$M(t,y)+N(t,y)\frac{dy}{dt} = 0 \tag{4}$$

has a solution in case the partial derivatives satisfy $\partial M / \partial y = \partial N / \partial t$, or, using integration factors has a solution if the expression

$$\frac{\partial M / \partial y - \partial N / \partial t}{N} \tag{5}$$

is a function $R(t)$ (i.e only dependent on $t$, in which case the integrating factor is $c(t) = \exp(\int R(t) \ dt)$).

I am unsure exactly how to transform $(3)$ into $(4)$. If for example I make the substitution $z = \mu(t)y$ then we get
$$-bz(t,y) + z^{1-n}(t,y)\mu^{n-1}(t,y)\frac{dz(t,y)}{dt} = 0 \tag{6}$$
But this seems incorrect as it is not really on the form of $(4)$.

My substitution in $(6)$ does not seem correct. How can I otherwise proceed?

答:
You already have the integrating factor ($\mu(t) = \exp(\int a(t) \ dt)$) and it depends on $a(t)$ then the differential equation becomes separable.
$$\frac{d}{dt}(\mu (t) y) = b(t)\mu (t)y^{n}$$
Divide by $y^n \mu ^{n}$:
$$\dfrac {1}{(\mu (t) y)^n} \frac{d}{dt} {(\mu (t) y)} = b(t)\mu^{1-n} (t) $$
$$\dfrac {d {(\mu (t) y)} }{(\mu (t) y)^n}  = b(t)\mu^{1-n} (t)dt $$
Integrate both sides.
$$\int \dfrac {d(\mu (t) y)}{(\mu (t) y)^n} = \int b(t)\mu^{1-n} (t) dt$$


问:
Prove that the series $\sum\limits_{n=2}^{\infty}(-1)^n\frac{\ln(n)}{n^x}$ converges to a positive real number for all $x > 0$
Prove that the series $\sum\limits_{n=2}^{\infty} (-1)^n\frac{\ln(n)}{n^x}=\frac{\ln(2)}{2^x} - \frac{\ln(3)}{3^x} + \frac{\ln(4)}{4^x} - \frac{\ln(5)}{5^x} + ...$ converges to a positive real number for all $x > 0$.

This question arose from this one:

Relatively simple proof that real-valued Dirichlet eta function is increasing, and also a query of the function at / about $x = 0$

Update:

There is a number $\alpha \in (1,2)$ such that for each $x > \alpha,$ the absolute sequence $(a_n)_{n\geq2}  = \left(\frac{\ln(n)}{n^x}\right)_{n\geq2} \ $is strictly decreasing. So for $x > \alpha$, the alternating series converges to a positive number.

$x \leq \alpha, \ $ in particular when $x << 1$ is more difficult.

I investigated the series for different value of $x$ below $0$ and it seems that the obvious approach is to show that $\exists$ an odd number $j$ such that $\frac{\ln(2)}{2^x} - \frac{\ln(3)}{3^x} + \frac{\ln(4)}{4^x} - \frac{\ln(5)}{5^x} + ... + \frac{\ln(j-1)}{(j-1)^x} - \frac{\ln(j)}{j^x} > 0,$ and that the absolute sequence $\left(\frac{\ln(n)}{n^x}\right)_{n\geq j}$ is strictly decreasing. I think that, even for the lowest value of $j$ for a given $x$, the absolute sequence will be decreasing. However, this may need to be proven as well.

$x = \frac{12}{20} \implies j = 51$

You can certainly check with a program, terms for "large enough $x$ " i.e. for $x \in (0.5, \alpha)$, but for very small positive $x$ you cannot check with a program. So we probably have to prove it using by rearranging the first $j$ terms and using inequalities to show that the sum of the first $j$ terms is $>0$. And yes, you also need to prove that the absolute sequence is strictly decreasing after the $j-th$ term also, but I don't think this will be as difficult as proving the sum of the first $j$ terms is $>0$.

I have also done a table showing x and values of $j$ where the sum is first positive.

\begin{array}{|c|c|c|c|}
\hline
x& j\\ \hline
 1&5 \\ \hline
 \frac{19}{20}&7 \\ \hline
 \frac{18}{20}&7 \\ \hline
 \frac{17}{20}&9 \\ \hline
 \frac{16}{20}&11 \\ \hline
 \frac{15}{20}&15 \\ \hline
 \frac{14}{20}&21 \\ \hline
 \frac{13}{20}&31 \\ \hline
 \frac{12}{20}&51 \\ \hline
 \frac{11}{20}&89 \\ \hline
 \frac{10}{20}&181 \\ \hline
 \frac{9}{20}&441 \\ \hline
 \frac{8}{20}&1395 \\ \hline
 \frac{7}{20}&6483 \\ \hline
 \frac{6}{20}&54805 \\ \hline
\end{array}

I made this table by manually inputting values into WA, so it may in fact be incorrect. A Python / C++ program or something may ensure more accurate results?

The bottom line so far: we need to prove $ \exists j \in \mathbb{N}$ for every $x$ with $0< x <\alpha$.

答:
CW

positivity is immediate when $x>1$

Relation to zeta

alright, there is a functional equation for the derivative,


问:
$(\log_23) ^ x + (\log_35) ^ x = 2 (\log_34) ^ x$
So, I have encountered this equation and I do not know how to solve it.
$$(\log_23) ^ x + (\log_35) ^ x = 2 (\log_34) ^ x$$
My first idea was dividing the equation by $\log_34$ and getting into something like $\left(\frac{\log_23}{\log_34}\right)^ x + \left(\frac{\log_35}{\log_34}\right)  ^ x = 2$. The obvious solution is $x = 0$ and I tried proving that LHS is smaller than $1$ if $x \lt 1 $ and bigger if $x > 1$, but realized $\frac{\log_23}{\log_34} \lt 1$, so my proof does n

答:
$$\left(\frac{\log _3 5}{\log _3 4}\right)^x+\left(\frac{\log _2 3}{\log _3 4}\right)^x=2$$
has only the root $x=0$.

Define the function $$f(x)=\left(\frac{\log _3 5}{\log _3 4}\right)^x+\left(\frac{\log _2 3}{\log _3 4}\right)^x$$
$f(x)$ is always increasing because, approximating to simplify the writing, we have
$$f(x)\approx 1.16^x+1.26^x\to f'(x)\approx 0.15\cdot 1.16^x+0.23\cdot1.26^x>0, \forall x\in\mathbb{R}$$
Therefore before and after the root $x=0$ there are no other roots.


问:
countable union of closed paracompact subspaces
This is an exercise of Munkres topology section 41.

Let $X$ be a regular space. If $X$ is a countable union of closed paracompact subspaces of $X$ whose interiors cover $X$, show $X$ is paracompact.

I Know this lemma:

Let $X$ be a regular space. Then TFAE (note: a refinement must itself also be a cover).

Every open cover of $X$ has a locally finite open refinement. (i.e. $X$ is paracompact)

Every open cover of $X$ has a $\sigma$-locally finite open refinement (where a family of sets is  $\sigma$-locally finite iff it is a countable union of locally finite families)

Every open cover of $X$ has a locally finite refinement (of any kind).

Every open cover of $X$ has a locally finite closed refinement.

The following is what I tried.

Denote each closed parcompact subspace $U_{n}$. Think about an open covering $\mathscr{A}$ of $X$. To prove paracompactness, I should prove that $\mathscr{A}$ satisfies $1$ of the above lemma. However, because $X$ is a 'countable' union of $U_{n}$, I think that the way to the answer is proving that $\mathscr{A}$ satisfies $2$.

But I can't proceed any longer.

答:
Let $\mathcal{U}$ be an open covering of a regular space $X=\bigcup_\mathbb{N}X_n$ which is a union of countably many closed paracompact subspaces $X_n\subseteq X$ whose interiors $(X_n)^\circ$ cover $X$.

For each $n\in\mathbb{N}$ the family $\mathcal{U}_n=\{U\cap X_n\}_{U\in\mathcal{U}}$ is an open covering of the paracompact $X_n$. Thus there is a family $\mathcal{V}_n'$ of open subsets of $X$ which has the property that $\{V\cap X_n\}_{V\in\mathcal{V}'_n}$ is a locally-finite open refinement (in $X_n)$ of $\mathcal{U}_n$. Write $\mathcal{V}_n=\{V\cap(X_n)^\circ\}_{V\in\mathcal{V}'_n}$ to obtain a locally-finite family of open subsets of $X$ which covers the interior $(X_n)^\circ$ and refines $\mathcal{U}_n$.

The collection
$$\mathcal{V}=\bigcup_\mathbb{N}\mathcal{V}_n$$
is now a $\sigma$-locally-finite open refinement of $\mathcal{U}$.

Appeal to part $2$ of your theorem to complete.


问:
Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational?
$$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$

Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.

I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$

Sorry if this is a stupid question.

Body:
Note that the expression
$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$
is a special case $(m=1)$ of the identity
$$s=\sqrt[3]{\left(3m-1\right) -\sqrt{ (8m-3)m^2}}
 + \sqrt[3]{\left(3m-1\right) +\sqrt{ (8m-3)m^2}} =1
$$
To verify, take cubic power to get
$(s-1)(s^2+s+6m-2)=0$ and in turn
$s=1$. Then, write the identity in the desired form
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
$$x= 3m-1,\>\>\>\>\>y=(8m-3)m^2,\>\>\>\>\> m=1,2,3... $$
So, there are infinite possibilities, with the first few cases listed below
\begin{align}
\sqrt[3]{2-\sqrt{5}} + \sqrt[3]{2+\sqrt{5} } & =1 ,\>\>\>m=1\\
\sqrt[3]{5-\sqrt{52}} + \sqrt[3]{5+\sqrt{52} } & =1 ,\>\>\> m=2\\
\sqrt[3]{8-\sqrt{189}} + \sqrt[3]{8+\sqrt{189} } & = 1 ,\>\>\>m=3\\
 \sqrt[3]{11-\sqrt{464}} + \sqrt[3]{11+\sqrt{464} } & = 1 ,\>\>\>m=4\\
\sqrt[3]{14-\sqrt{925}} + \sqrt[3]{14+\sqrt{925} } & = 1 ,\>\>\>m=5\\
...&= 1,\>\>\>...
\end{align}


问:
The best constants $c$ and $C$ such that $c||x||_p \leq ||x||_q\leq C||x||_p$
Taken from Conway's A course in Functional Analysis Chapter 3 Section 1 Problem 3

Problem statement: For $1\leq p < \infty$ and $x = (x_1,...,x_d)$ in $\mathbb{F}^d$, define $||x||_p = \left[\sum_{j = 1}^d|x_j|^p\right]^{1/p}$; define $||x||_\infty = \sup\{|x|_j: 1\leq j \leq d\}$. Show that all of these norms are equivalent. For $1\leq p, q \leq \infty$, what are the best constants $c$ and $C$ such that $c||x||_p \leq ||x||_q\leq C||x||_p$ for all $x$ in $\mathbb{F}^d$.

I went straight to finding constants $c$ and $C$ such that $c||x||_p \leq ||x||_q\leq C||x||_p$ for all $x$ in $\mathbb{F}^d$ since the equivalence of th enormous would immediately follow from this. However, I'm not even sure where to begin. I thought of trying to let $x_j$ all be the same for a possible candidate for $c$ or $C$,but this doesn't seem too promising.

Any tips would be greatly appreciated!

答:
Even though it is tempting, I do not think that Hölder's inequality will help you in this case. Consider this rather obvious inequality (remember that $\lvert x_j \rvert \leq \lVert x \rVert_\infty$ for all $j$):
$$
\lVert x \rVert_p = \left (\sum_{j = 1}^d \lvert x_j \rvert \right)^{\frac{1}{p}} \leq \left( d \lVert x \rVert_\infty^p \right)^\frac{1}{p} = d^\frac{1}{p} \lVert x \rVert_\infty
$$
Now note that there has to be some $\ell \in \lbrace 1, ..., d \rbrace$ such that $\lVert x \rVert_\infty = \lvert x_\ell \rvert$. So:
$$
\lVert x \rVert_\infty = \lvert x_\ell \rvert = (\lvert x_\ell \rvert^q)^\frac{1}{q} \leq \left( \sum_{j = 1}^d \lvert x_j \rvert^q \right)^\frac{1}{q} = \lVert x \rVert_q
$$
If we combine all of this, we get
$$
c \lVert x \rVert_p \leq \lVert x \rVert_\infty \leq  \lVert x \rVert_q
$$
for $c := \frac{1}{d^\frac{1}{p}}$. If you reapply the above reasoning, you get:
$$
\lVert x \rVert_q \leq C \lVert x \rVert_p
$$
for $C := d^\frac{1}{q}$. This means:
$$
c \lVert x \rVert_p \leq \lVert x \rVert_\infty \leq  \lVert x \rVert_q \leq C \lVert x \rVert_p
$$
for the already defined constants $c, C > 0$. This proves the equivalence of the norms $\lVert \cdot \rVert_r$, $r \in [1, \infty]$.

I have not quite yet figured out whether these are the best constants.


问:
Prove $A=A^\perp$
If $A$ is a n by n matrix such that $rank(A)=1$ and $Null(A)=Col(A)^\perp$
Prove $A=A^\perp$

I already proved that $dim(Col(A)^\perp)=1$ and $dim(Null(A))=n-1$, but then what comes next?

I am also thinking that if A is orthogonally diagonalizable, then $A=A^\perp$, but how to prove this?

答:
Since $A$ has rank 1, it is of the form $A = uv^\top$. Then $\operatorname{Null}(A) = \{v\}^\perp$ and $\operatorname{Col}(A)^\perp = \{u\}^\perp$. Since $\operatorname{Null}(A) = \operatorname{Col}(A)^\perp$, it follows that $u$ is parallel to $v$.

So now $A = c(uu^\top)$ for some scalar $c$ and it follows that $A$ is symmetric.

Title: Prove $A=A^\perp$


问:
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$
Please help me to find a closed form for the infinite product
$$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$
where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$  is the hyperbolic tangent.</p>

Body:
For $x < 1$, we have the Taylor series expansion:
$$f(x):= \frac{-1}{4} \log \left(-  \frac{x - x^{-1}}{x + x^{-1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots $$

Then

$$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots
= \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + \frac{x^{10}}{10} + \ldots  $$
$$= - \frac{1}{2} \log(1 - x^2).$$

Now let $x = e^{-2}$. Then

$$ \log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) =
\frac{1}{2^n} \log \left( \frac{e^{2^n} - e^{-2^n}}{e^{2^n} + e^{-2^n}}\right)
$$
$$= \frac{-4}{2^n} f(e^{-2^n}) =  \frac{-4}{2^{n}} f(x^{2^{n-1}}),$$

Hence summing over all $n \ge 1$, we see that, if the product is $P$, then

$$\log P =  -4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n-1}})
= -2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1 - x^2),$$

and thus

$$P = \exp \log(1 - x^2) = 1 - x^2 = 1 - e^{-4}.$$


问:
Understanding change of variables
So there's this one thing I can't wrap my head around.

The problem is:

Use the change of variables $x = \sinh u$ to compute the indefinite integral
\begin{equation*}
    \int \frac{dx}{\sqrt{1+x^2}}
\end{equation*}

Solution:

\begin{align*}
    \frac{dx}{du}=\sinh'x=\cosh x
\end{align*}
$$\iff$$
\begin{align*}
    \int \frac{\cosh x}{\sqrt{1+\sinh^2x}}\,du = \int \frac{\cosh x}{\sqrt{\cosh^2x}}\,du = \int\frac{\cosh x}{\cosh x}\,du= \int 1 \,du = u + C = \sinh^{-1}x+C
\end{align*}

and I understand every step except the first one, $\frac{dx}{du}=\sinh'x$. I'm just generally confused when it comes to change of variables and how to use the Leibniz notation to understand what's going on.

答:
Note that if $x=\sinh(u)$, so you have $$\frac{dx}{du}=\frac{d}{du}\left(\sinh(u) \right)=\cosh(u)$$
It's the same that you write $$x=\sinh(u) \implies x'=\cosh(u)$$
However, the second notation it's not recommended for when you are integrating because this may be confused you.

Title: Understanding change of variables


问:
Is it possible to draw a letter "T" on every rational number such that they do not intersect?
The upper part of the letter "T"-s are closed and bounded non degenerate intervals, and the perpendicular part (which is also bounded, closed, non degenerate) must be inside the upper part (cannot be at the edges).

I have proved that there are at most countable disjoint "T"-s in the plane. Also, if the perpendicular part of the T could be at the edges, then it is possible to erect such T-s. (The height of the T's follow a strictly monotone decreasing function, and the perp. part is at the left edge, then the width of the upper part can be arbitrary).

Any hints, ideas or solutions are welcome! I have been going crazy over this.

答:
As hinted at by Will's comment, we can easily place the "T"'s inside of Ford circles. The wiki on Ford Circles gives that the Ford circle at $p/q$ denoted $C[p/q]$ is disjoint or tangent to every other Ford circle. So if we just make the height of the "T" at $p/q$, lets say $T[p/q]$, the same as the radius of the Ford Circle, $(1/2q^2)$ and the length of the top part of the "T", half this radius, $(1/4q^2)$, then they will not touch one another. In the end, we can define the set $$T[p/q]= \{(p/q,y): 0\leq y\leq (1/2q^2) \}\cup \{(x,(1/2q^2)): (p/q-1/8q^2)\leq x\leq (p/q+1/8q^2) \}$$
as described. It should be noted that $p/q$ is the reduced fraction of the rational number, and we can take $0=0/1$


问:
Triple integral - possible change of variables? Tried using cylindrical coordinates.
Let $R =\{(x,y,z) \in \mathbb{R^3} \mid 1 \leq x^2 + y^2 \leq 16, 0 \leq z \leq y + 4\}$.

Let $$I = \int_{R} (x-y) \,dV.$$
Calculate $I$ to 3 decimal places.

I have tried converting to cylindrical coordinates, however I get a negative answer. What am I doing wrong?

So far I have:

$1 \leq x^2 + y^2 \leq 16 \implies 1 \leq r \leq 4$ and $ 0 \leq z \leq r\sin(\theta) + 4$ from $y = r\sin(\theta)$ and $x^2 + y^2 =r^2$

$$ (x-y) \,dV = r(r\cos\theta - r\sin\theta) \,dz \,dr \,d\theta $$

Using a diagram of the region $R$ I have concluded that theta should be between $0$ and $\pi$.

I have included the Jacobian and my resulting triple integral is completely correct as checked by an online calculator. However, these limits and change of variables still leads to a negative definite integral.

Are my limits incorrect? Is the change of variables incorrect?  Should I use a different change of variables?

$$I = \int_{\theta =0}^\pi \int_{r=1}^4 \int_0^{r\sin(\theta)+4} r^2(\cos\theta- \sin\theta) \,dz \,dr \,d\theta = -\frac{255\pi}{8} $$

答:
Your are integrating over a cylinder with axial hole ($1 \leq r \leq 4$).

It is bound above by the plane $z = y + 4$ and below by $z = 0$. Please note the plane $z = y + 4$ truncates the cylinder (with a slant) such that $z = 0$ at $y = -4$ and $z = 8$ at $y = 4$ (So clearly $\theta$ should go from $0$ to $2 \pi$). Note that $z$ varies with $y$ and for a given $y$, $z$ remains constant wrt. $x$.

Now if you visualize the truncated cylinder, the volume of the cylinder is less in third and fourth quadrant (below $y$ axis) and is increasing as we get into positive $y$ (first and second quadrant).

Please also consider the function ($x-y$) that you are integrating over this cylinder and how the function is set up in different quadrants.

So I am not surprised with a negative value from integration.

Lastly, limits for $\theta$ should be $0$ to $2 \pi$.

$I = \displaystyle \int_{0}^{2\pi}\int_{r=1}^{4}\int_0^{4+rsin(\theta)} r^2(cos\theta- sin\theta) \, dz \, dr \, d\theta$


问:
Finding the the number of Lattice paths with three steps.
We define a nice path of length as a path on a lattice from $(0,0)$ to $(n,n)$ such as every step in it is: $r=(0,1)$ , $u=(1,0)$ and $v=(1,2)$, and it does not go under the line $y=x$.

A. Write a formula for the generating function that counts the number of the nice paths.

B.Solve the above formula and find an expression for the number of these paths of length n.

C.Find the number of $(1,2)$ steps in all paths of length n.

I started with drawing the possible paths for $n=1,2,3,4$, getting that we have $1,3,9,27$ paths (respectively) if I did really succeed to count all the possible paths for each $n$.

So, then I concluded that each legal path (which is mentioned above) is:

Every path = $\epsilon$ or v(every path) u (every path) or r(every path) u (every path).

Whhere $\epsilon$ denote the empty path.

If $A(x)$ the generating function and x denote the number of steps then I get:

$A(x)=1+(xA(x))(xA(x))+(xA(x))(xA(x))$.
$A(x)=1+x^2A(x)^2+x^2A(x)^2$

In B. I can do this after extracting $A(x)$
With finding the coefficient of $x^n$ in $A(x)$.

C. I don't understand if this is related to statistics on the number of (1,2) steps..

I'm not sure if my way of thinking is correct or not, I discovered that my way is not accurate. I'll be glad if you can correct me, and explain it.

答:
Unfortunately, there are complications that make it not quite so straightforward as you've written. Most notably, what you have is the generating function for the number of paths with $n$ steps — but since you have one step that's longer than the others, that won't get you the number of paths to $(n,n)$. Instead, try writing a recurrence relation for $A_n$ in terms of smaller $A_i$ using the same approach that you sketched out loosely; you should find a very similar answer. (I suspect you're missing a factor of $x$ one way or the other in one of your terms, but writing the relation explicitly should make it clear.)

For (c), you'll need to expand your generating-function approach a bit. Let $A_{n,k}$ be the number of paths from $(0,0)$ to $(n,n)$ that contain $k$ (1,2)-steps. Then with a similar approach to what you've done for determining the number of steps you should be able to generate a recurrence relation for $A_{n,k}$, or equivalently an equation for the generating function $A(x,y)=\sum_n\sum_kA_{n,k}x^ny^k$. Once you've got this, the quantity that (c) asks for is $B_n=\sum_kkA_{n,k}$ (note that $A_n$ is just $\sum_kA_{n,k}$ without the multiplying factor). As a further hint if you aren't familiar with the particular manipulation that's relevant here, consider the effect of the $\frac{\partial}{\partial y}$ operator on $A(x,y)$.


问:
Find the flux of the vector field out of the region $T$ bounded by $z = x^2 + y^2$ and $z = 4$. Please help
Find the flux of the vector field $$\vec{F}(x,y,z)= <x,y,3>$$
out of the region $T$ bounded by $z = x^2 + y^2$ and $z = 4$.

It says the unit vector $n$ is $<0,0,1>$, but how do you find it?

答:
Make use of Gauss divergence theorem
$$\int_{S} \vec{F}.\hat{n} dS = \int_V \vec{\nabla}\cdot \vec{F} dV$$
to simplify your computation.

Note that $\vec{\nabla} \cdot \vec{F} = 2$. Hence, $$\int_{S} \vec{F}.\hat{n} dS = \int_V \vec{\nabla}\cdot \vec{F} dV = 2 \int_V dV = 2\int_0^4 \pi z dz = \left. \pi z^2 \right \vert_0^4 = 16 \pi$$


问:
Strange point lies on common tangent of 9-point circle and incircle

Let $ABC$ be a triangle, with medial triangle $DEF$ and intouch triangle $PQR$. Let $J$ be the midpoint of $\overline{AD}$, and let $BJ$ meet $AP$ at $K$. Let $X$ be the point on ray $\overrightarrow{CB}$ such that $CX=CA$. Let the line through $K$ parallel to $BC$ meet $AX$ at $U$. Let $RU$ meet $BC$ at $T$. Prove that $T$ lies on the common tangent of the nine-point circle and the incircle of $\triangle ABC$.

I have a proof with barycentric coordinates. We can compute $K=(2(s-b):s-c:s-b)$, $X=(0:b:a-b)$, $U=(a-b+c:b:a-b)$ and $T=(0:a-c:a-b)$. The common tangent of the nine-point circle and incircle has equation $$(a-b)(a-c)x+(b-c)(b-a)y+(b-c)(a-c)z=0,$$ and the result follows.

However, as you might guess, I'm interested in a synthetic solution to this problem.

答:
The question has been asked and answered at Art of Problem Solving.  User amar_04 gives a synthetic proof.  Reproduced here, in case the URL breaks.

Just use Menelaus repeatedly

$\textbf{LEMMA 1:-}$   The Tangent to the Incircle at the Feuerbach
Point and the Nagel Line of $\Delta ABC$ are Isotomic Lines.
Proof:- See Theorem 10 here

$\textbf{LEMMA 2:-}$ Let $\tau_1,\tau_2$ be two Isotomic Lines in
$\angle A$ cutting $BC$ at $X,Y$. Then $BX=CY$

Proof:-   Let $\tau_1$ cut $\overline{AB},\overline{AC}$ at $\{P,M\}$ and let $\tau_2$ cut $\overline{AB},\overline{AC}$ at $Q,N$.
Apply Menelaus Theorem on $\Delta ABC$ cut by the transversals
$\tau_1,\tau_2$
$$\begin{align*}\frac{CX}{XB}\cdot\frac{BP}{PA}\cdot\frac{AM}{MC}=-1=\frac{BY}{YC}\cdot\frac{CN}{NA}\cdot\frac{AQ}{QB}&\implies\frac{CX}{XB}=\frac{BY}{YC}\\ &\implies\frac{CB}{XB}=\frac{CB}{YC}\\ &\implies
BX=CY\end{align*}$$

$\textbf{LEMMA 3:-}$  If the Tangent $(\ell)$ to the Incircle at the
Feuerbach point of $\Delta ABC$ cuts $\overline{BC}$ at $X$. Then
$BX=\frac{a(b-a)}{b+c-2a}$

Proof:-   Let $G,N_a$ be the Centroid and Nagel Point of $\Delta ABC$ and let $\overline{AG}\cap\overline{BC}=\{M\}$ and
$\overline{AN_a}\cap\overline{BC}=\{P\}$. Let the Nagel Line of
$\Delta ABC$ cut $\overline{BC}$ at $Y$. Now applying Menelaus Theorem
on $\Delta AMP$ cut by the Nagel Line gives

\begin{align*}\frac{PY}{YM}\cdot\frac{MG}{GA}\cdot\frac{AN_a}{N_aP}=-1
 &\implies\frac{PY}{YM}=\frac{2(s-a)}{a}\\
 &\implies\frac{PM}{MY}=\frac{2s-3a}{a}\\
 &\implies\frac{\frac{a}{2}-(s-b)}{MY}=\frac{2s-3a}{a}\\ &\implies
 MY=\frac{ab-ac}{2(b+c-2a)}\\ &\implies CY=MY+CM\\
 &=\frac{ab-ac}{2(b+c-2a)}+\frac{a}{2}\\
 &=\frac{a(b-a)}{b+c-2a}\end{align*}
Now by $\textbf{LEMMA 1}$ we get
that $\ell$ and the Nagel Line are Isotomic Lines and if
$\ell\cap\overline{BC}=X$. Then from $\textbf{LEMMA 2}$ we get
$CY=BX=\frac{a(b-a)}{b+c-2a}$.

Now come back to the Problem at hand. Applying Menelaus Theorem on
$\Delta ADP$ cut by the transversal $\overline{BI}$ we get
$$\frac{DB}{BP}\cdot\frac{PK}{AK}\cdot\frac{AJ}{JD}=-1\implies\frac{AK}{PK}=\frac{a}{2(s-b)}=\frac{AU}{XU}$$
Again by Menelaus Theorem we get
$$\frac{XT}{TB}\cdot\frac{BR}{RA}\cdot\frac{AU}{XU}=-1\implies
> \frac{BX}{TB}=\frac{2s-3a}{a}\implies TB=\frac{a(b-a)}{b+c-2a}$$ So by
$\textbf{LEMMA 3}$ get that $T\in$ on the Tangent to the Incircle at
the Feuerbach Point. $\blacksquare$


问:
How to prove that frac($\sqrt{4^n-1}$) is greater than 1/2 for any natural n
I would like to prove that the fractional part of $\sqrt{4^n-1}$ is greater than 1/2 for any natural n.

Just to clarify- $frac(x)=$ $x$$-$$\lfloor x \rfloor$.

e.g frac($\pi$) = $\pi - 3$ .

My Attempts:

Induction - for n=1 $frac($$\sqrt{3}$$)>0.5$

Assuming $frac($$\sqrt{4^n-1}$$)>0.5,$ I could not prove that $frac($$\sqrt{4^m-1}$ $)>0.5.$$(m=n+1)$

Declare the follow sequence $a$n= $frac($$\sqrt{4^n-1}$

And prove it's limit is 1 (I believe it converges to 1), I tried to prove it using the limit's definition but again I got stuck.

Would love to get some help with my attempts here or  to hear of a different method. Thanks in advance.

答:
Note that $\sqrt {4^n-1}$ is very close to $\sqrt {4^n} = 2^n$. If we can somehow show that $\sqrt {4^n-1} > 2^n-\frac12$ then $\operatorname {frac} (\sqrt{4^n-1}) > \frac12$. But

$$\left(2^n-\frac12\right)^2 = 4^n-2^n+\frac14 < 4^n-2^n+1 \le 4^n-1$$

The last inequality holds true for $n \ge 1$.


问:
Classify the odd primes $q$ such that a NEGATIVE number is a quadratic residue $\mod{q}$
Suppose we are given $y < -1$. I wish to classify all primes $q$ such that $y$ is a quadratic residue $\pmod{q}$, i.e. such that there exists a number $x$ satisfying $$y \equiv x^2 \pmod{q}.$$

How would I go about doing this? I'm aware that if instead $y>1$ then I can write $y$ as a product of primes and use the law of quadratic reciprocity. However, this doesn't seem to be applicable if $y<-1$, as the Legendre/Jacobi symbol requires a positive number in the denominator, and since we do not know $q$, it seems (although I may be wrong) we cannot find a meaningful positive representative modulo $q$.

答: There are $\frac{p-1}{2}$ quadratic residues between $1$ and $p-1$. Let $x$ be any of these, and let $y=x-17p$. Since $x$ is a quadratic residue of $p$, so is $y$. </p>


答: Note that$$2^n-\sqrt{4^n-1}=\frac1{2^n+\sqrt{4^n-1}}<\frac1{2^n}\leqslant\frac12.$$Can you take it from here?

Title: How to prove that frac($\sqrt{4^n-1}$) is greater than 1/2 for any natural n

Body:
I would like to prove that the fractional part of $\sqrt{4^n-1}$ is greater than 1/2 for any natural n.

Just to clarify- $frac(x)=$ $x$$-$$\lfloor x \rfloor$.

e.g frac($\pi$) = $\pi - 3$ .

My Attempts:

Induction - for n=1 $frac($$\sqrt{3}$$)>0.5$

Assuming $frac($$\sqrt{4^n-1}$$)>0.5,$ I could not prove that $frac($$\sqrt{4^m-1}$ $)>0.5.$$(m=n+1)$

Declare the follow sequence $a$n= $frac($$\sqrt{4^n-1}$

And prove it's limit is 1 (I believe it converges to 1), I tried to prove it using the limit's definition but again I got stuck.

Would love to get some help with my attempts here or  to hear of a different method. Thanks in advance.


问:
Write $h \in V^*$, defined by $h(p(x)) = p'(x)$ as a linear combination of the functionals in $\beta^*$.
Let $\beta^*$={f1, f2} be a basis of $V^*$, where $V= P_1(\mathbb{R})$ and $f_1(p(x))=p(1)$ and $f_2(p(x))=p(2)$.
Write $h \in V^*$, defined by $h(p(x)) = p'(x)$ as a linear combination of the functionals in
$\beta^*$.

I've been stuck on this homework problem for a while no and am just stumped on what to do, any help is appreciated.

答: Observe that if $p(x)=ax+b$ then $p'(x)=a=(2a+b)-(a+b) =p(2)-p(1)$.


问:
How do I change the position of a quarter circle on the x-axis.
I have an equation of a quarter circle that looks like this:
$$y=(3^2-x^2)^{1/2}{{0<x<3}}$$
And I know that I can change the position on the y-axis by adding
$$+3$$
or whatever number instead of "3", but how would I change the position of the Quarter Circle on the x-axis?

答: As you've mentioned, if you want to change the position on the $y$-axis you add to the right hand side, or equivalently replace $y$ with $y - b$.  So, writing $$y - b=(3^2-x^2)^{1/2}\quad \text{for}\quad 0<x<3$$ gives you the upper-right quarter of a circle of radius $3$ centered at the point $(0,b).$  To control the $x$ coordinate of the center we can replace $x$ with $x - a$, which gives us  $$y - b=(3^2-(x-a)^2)^{1/2}\quad \text{for}\quad 0<x-a<3$$ or equivalently, $$y - b=(3^2-(x-a)^2)^{1/2}\quad \text{for}\quad a<x<a+3.$$  This gives you the upper-right quarter of a circle of radius $3$ centered at the point $(a,b)$.  See this Desmos Plot for an interactive version.


问:
If-then statement problems

Your guarantee is good only if you bought your CD player less than 90 days ago.

If your guarantee is good, then you must have bought your CD player less than 90 days ago.

Can we also write the if-then statement of the sentence of this link like the following:

if  you have bought a CD player less than 90 days ago then the guarantee
is good

If I am wrong then please correct me.

Edit: Thanks for the answers

答: You are not correct. "If $P$, then $Q$" is equivalent to "$P$ only if $Q$" ($P \rightarrow Q$ In symbolic notation). In this case, $P \equiv$ "Your guarantee is good" and $Q \equiv$ "You bought your CD player less than 90 days ago". So, the given statements are equivalent to $P \rightarrow Q$, but the one you stated is equivalent to $Q \rightarrow P$.

Title: If-then statement problems


答:
I think it's clearer if you use differentials: remember the differential of a function $f$ of the variable $u$ is defined as the formal expression
$$\mathrm df=f'(u)\,\mathrm du, $$
where $\mathrm du$ (the differential of the variable) represents symbolically a small increment of the variable, and $\mathrm df$ (the differential of the function) is an approximation of the corresponding increment of $f(u)$, as measured along the tangent line.

This being said, the substitution $x=\sinh u$ yields the differential of $x$:
$$\mathrm dx=\cosh u\,\mathrm du, $$
and we have to express the whole differential form $\frac{\mathrm dx}{\sqrt{1+x^2}}$ in function of $u$ and its differential. Therefore, the integral becomes
$$\int\frac{\mathrm dx}{\sqrt{1+x^2}}\to \int\frac{\cosh u\,\mathrm du}{\sqrt{1+\sinh^2u}}=\int\frac{\cosh u\,\mathrm du}{\cosh u}=\int\mathrm du=u=\operatorname{argsinh}x.$$

Title: Understanding change of variables

Body:
So there's this one thing I can't wrap my head around.

The problem is:

Use the change of variables $x = \sinh u$ to compute the indefinite integral
\begin{equation*}
    \int \frac{dx}{\sqrt{1+x^2}}
\end{equation*}

Solution:

\begin{align*}
    \frac{dx}{du}=\sinh'x=\cosh x
\end{align*}
$$\iff$$
\begin{align*}
    \int \frac{\cosh x}{\sqrt{1+\sinh^2x}}\,du = \int \frac{\cosh x}{\sqrt{\cosh^2x}}\,du = \int\frac{\cosh x}{\cosh x}\,du= \int 1 \,du = u + C = \sinh^{-1}x+C
\end{align*}

and I understand every step except the first one, $\frac{dx}{du}=\sinh'x$. I'm just generally confused when it comes to change of variables and how to use the Leibniz notation to understand what's going on.


问:
How to show that the general orthogonal group $O(n,1)$ has four components?
Define a bilinear form on $\mathbb R^{n+1}$ by $(x,y)=\sum\limits_{i=1}^n x_iy_i-x_{n+1}y_{n+1}$, the general orthogonal group $O(n,1)$ is defined to be $\left\{B\in GL(n+1,\mathbb R)\mid(Bx,By)=(x,y),\forall x,y\in\mathbb R^{n+1}\right\}$, where $(x,y)=\langle x,gy\rangle,\, g=\operatorname{diag}\{1,\cdots,1,-1\}$ and $\langle\cdot,\cdot\rangle$ denotes the usual Euclidean inner product. Thus $B\in O(n,1)\Leftrightarrow B^tgB=g$. This is shown as a result in Brian C. Hall's "Lie Groups, Lie Algebras and Representations", but I hope to get a proof that $O(n,1)$ has four components.

答:
Denote the standard basis of $\mathbb R^{n+1}$ by $\{e_1,\ldots,e_{n+1}\}$. $O(n,1)$ has four connected components:

$\{B\in O(n,\mathbb R):\det(B)=1,\ (Be_{n+1},e_{n+1})>0\}$,

$\{B\in O(n,\mathbb R):\det(B)=1,\ (Be_{n+1},e_{n+1})<0\}$,

$\{B\in O(n,\mathbb R):\det(B)=-1,\ (Be_{n+1},e_{n+1})>0\}$,

$\{B\in O(n,\mathbb R):\det(B)=-1,\ (Be_{n+1},e_{n+1})<0\}$.

In fact, each member of $O(n,1)$ can be expressed as
\begin{align}
B&=\begin{bmatrix}Q\\ &1\end{bmatrix}
\begin{bmatrix}
I_n+\left(\sqrt{1+c^2}-1\right)uu^T&cu\\ cu^T&\sqrt{1+c^2}
\end{bmatrix}
\begin{bmatrix}I_n\\ &t\end{bmatrix}\\
&=\begin{bmatrix}
Q\left[I_n+\left(\sqrt{1+c^2}-1\right)uu^T\right]&tcQu\\ cu^T&t\sqrt{1+c^2}
\end{bmatrix}\tag{0}
\end{align}
where $t\in\{1,-1\},\,Q\in O(n),\,c\in\mathbb R$ and $u$ is a unit vector in $\mathbb R^n$. The sign of $(Be_{n+1},e_{n+1})$ is $-t$ while $\det(B)=t\det(Q)$. With this characterisation of $B$, the path-connectedness of each of the four aforementioned sets is evident.

To prove that $B$ can be written in the above form, let us partition $B$ as $\pmatrix{X&v\\ w^t&\lambda}$, where $\lambda$ is a scalar and $X,v,w$ are matrices or vectors of appropriate sizes. The equation $B^tgB=g$ can then be rewritten as
\begin{align}
&X^tX=I+ww^t,\tag{1}\\
&X^tv=\lambda w,\tag{2}\\
&\lambda^2=1+\|v\|^2.\tag{3}
\end{align}
By $(3)$, $\lambda$ is never zero. Moreover, the RHS of $(1)$ is positive definite. Therefore $X$ is nonsingular and $(2),(3)$ implies that $v$ and $w$ are either both zero or both nonzero.

If $v=w=0$, clearly $X$ is a real orthogonal matrix and $\lambda=\pm1$. Therefore $B$ takes the form of $(0)$. Now suppose $v,w\ne0$. The positive definite square root of $I+ww^t$ is $I+\left(\sqrt{\|w\|^2+1}-1\right)\frac{ww^t}{\|w\|^2}$. Therefore $(1)$ is equivalent to
$$
X=Q\left[I+\left(\sqrt{\|w\|^2+1}-1\right)\frac{ww^t}{\|w\|^2}\right]\tag{4}
$$
for some real orthogonal matrix $Q$. Substitute this into $(2)$, we obtain
$$
\left[I+\left(\sqrt{\|w\|^2+1}-1\right)\frac{ww^t}{\|w\|^2}\right]Q^tv=\lambda w\tag{5}
$$
and hence
$$
Q^tv=tw\tag{6}
$$
for some nonzero scalar $t$. Substitute this back into $(5)$, we obtain $(t\sqrt{\|w\|^2+1})w=\lambda u$. Hence
$$
\lambda=t\sqrt{\|w\|^2+1}.\tag{7}
$$
Since $\|v\|=\|Q^tv\|=\|tw\|$, by $(3)$ we have $t^2(\|w\|^2+1)=1+t^2\|w\|^2$, i.e. $t=\pm1$. Now, $w=cu$ for some scalar $c$ and unit vector $u$. Then by $(4),(6)$ and $(7)$, we see that $B=\pmatrix{X&v\\ w^t&\lambda}$ takes the form of $(0)$.


答:
The statement that you presented is not equivalent to the one presented in the picture.

Note the following. Let $p$ and $q$ be any statements. The statement $p \implies q$ means “if $p$ then $q$”, or equivalently, “$p$ only if $q$.” (Note that this is what is written in the picture).

Your statement is $q \implies p.$ And you can easily see that they are not equivalent.

If $v(p) = T$ and $v(q) = F,$ then $v(p \implies q) = F$ but $v(q \implies p) = T.$ So $p \implies q$ and $q \implies p$ are not equivalent statements.

In short, you have the following.

$$\begin{align}
p \implies q & \quad \text{If $p,$ then $q$}\\
& \quad\text{$p$ only if $q$}\\
\quad \\
q \implies p & \quad \text{If $q,$ then $p$}\\
& \quad \text{$q$ only if $p$}
\end{align}$$

Title: If-then statement problems

Body:
Your guarantee is good only if you bought your CD player less than 90 days ago.

If your guarantee is good, then you must have bought your CD player less than 90 days ago.

Can we also write the if-then statement of the sentence of this link like the following:

if  you have bought a CD player less than 90 days ago then the guarantee
is good

If I am wrong then please correct me.

Edit: Thanks for the answers


问:
Combined probability of drawing king after ace
I'm studying for a math test and I found an interesting question:
Assume someone is drawing cards from 32-card deck. How likely is he to get a pair of king and ace after each other. He isn't returning the cards back.

I wanted to solve this by $\frac{4}{32} * \frac{4}{31}$ which would result in $\frac{1}{62}$. But these are questions which already pre-solved and the expected answer should be $\frac{1}{14384}$

How should I approach this?

答: I think your approach was (almost) correct, the first draw must be a king and you have 4 out of 32 possibilities, the second must be again a king, but there are only 3 left so you have 3 out of 31. As Brian Tung noticed $\frac{1}{14384}$ = $\frac{60}{32 \cdot 31 \cdot 30 \cdot 29}$ which means that you probably draw 4 cards. If the third card shall be an ace you would have 4 out of 30 and for another ace you would have 3 out of 29. In total the probability would be $\frac{144}{14384}$ (144 = 4 x 3 x 4 x 3). I don't see a simple question where the answer would be $\frac{60}{14384}$ because 60 = 5 x 12, which means at some point you should have the choice of 5 cards!? Really strange ...


问:
Proof of the product rule in probability theory for causal independence
There is a common attitude in the text books on probability that the so-called product rule is an obvious property, when events are independent, i.e., $P(A\cap B)=P(A)P(B)$ when $A$ and $B$ are independent events. Yet, this is NOT an axiom that a probability must satisfy, nor it is a property that follows from the axioms. That is because there is no definition of the concept of independent events in the axioms.

Of course, the product rule is the actual definition of the (stochastic) independence of two events, if one wants to evade the concept of conditional probability. So you, clearly, see the chicken-and-the-egg conundrum.

Still, we continue to use the rule because we intuitively feel the independence property is true, which seems natural in those cases when our random experiment is obtained as a succession of two (simpler) random experiments which are causal independent.

And now my question: how is this idea (of causal independence) formalized in the context of the axiomatic approach to probabilities such that we can prove from here the product rule?

Body: $(|)=(),(|)=\frac{(∩)}{()}→(∩)=(|)()→(∩)=()()$


问:
$d\mid p\!-\!1\Rightarrow x^d-1\pmod{\!p^2}$ has exactly $d$ roots for odd prime $p$
I'm trying to figure out the number of solutions to the congruence equation $x^d \equiv1 \pmod{p^2}$ where $p$ is prime and $d\mid p-1$.

For the congruence equation ${x^d}\equiv1 \pmod p$ where $p$ is prime and $d\mid p-1$ I've shown that there are exactly $d$ solutions modulo $p$.

I'm trying to use the above result to extend it to the higher power. I'm aware of something called Hensel's Lemma which says if a polynomial has a simple root modulo a prime  $p$, then this root corresponds to a unique root of the same equation modulo any higher power of $p$. We can 'lift' the root iteratively up to higher powers.

I'm unsure exactly how this process works. All I have to start with is that I assume a solution of the form $m+np$ solves the 'base' congruence and I'm trying to somehow extend it to $p^2$.

Any help would be appreciated.
Thanks.

答:
Taking Discrete Logarithm with respect to a primitive root $a\pmod {p^n}$ for integer $n\ge1,$

$$d\cdot ind_ax\equiv0\pmod {\phi(p^n)}\implies d\cdot ind_ax\equiv0\pmod {p^{n-1}(p-1)}$$

As $d|(p-1),(d,p-1)=d$ and $(p,p-1)=1\implies (d,p)=1\implies (d,p^{n-1})=d$

$\implies (d,p^{n-1}(p-1))=d$

Using Linear congruence theorem,

as $d|0,$  there will be  has exactly $d$ solutions.


问:
Show that a set S is a closed subset of C([0,1])
S = $\{\frac{sin(nx)}{1 + n} \ s.t. \ n \in N, x \in [0,1]\}$

Show that S is a closed subset of C([0,1])

My instructor told me there was a "way of thinking of this," and that if you don't see it right away, you won't be able to get the solution. I must be not thinking of it correctly as I am not quite sure how to go about it. Does anyone have a way of getting started that could help guide me?

At the time I do not have a solution attempt. I really am not even convinced this is continuous right now since Sin(nx) could be -1, which is outside of [0,1]. So I'm not even convinced this is continuous at the moment.

Thank you!

答:
On any metric space, if $(x_n)_{n\in\Bbb N}$ is a convergent sequence, if $x$ is its limit, and if $x\ne x_n$ for any $n\in\Bbb N$, then the set $\{x_n\mid n\in\Bbb N\}$ is not a closed set.

So, if you are working with the distance $d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$, then, since$$\left(\frac{\sin(nx)}{1+n}\right)_{n\in\Bbb N}$$converges to the null function, and since none of the functions $x\mapsto\frac{\sin(nx)}{1+n}$ is the null function, the set $S$ is actually not a closed set.