问:
How is the Dirac delta $\delta (x^3)$ different from $\delta(x)$?
How is Dirac delta $\delta (x^3 )$ different from $\delta (x)$? It is my understanding that
$$
\delta [\psi]:=\langle\delta,\psi\rangle=\int_{\Omega}\delta(x)\psi(x)\,dx=
\begin{cases}
\psi(0) & \text{if $0\in \Omega$}\\
0 & \text{otherwise}
\end{cases}
$$
But $x^3=0$ iff $x=0$ so does this not imply,
$$\delta(x^3) = \delta(x) = 0$$
Apologies if I am completely misunderstanding the topic, I am very new to distributions.

答:
Let $\psi$ be a test function, i.e. $\psi \in \mathcal{C}^{\infty} _{c}(\mathbb{R})$, then:
$$
\int \delta(x^3)\psi (x) dx = \int \delta(y) \psi (y^{\frac{1}{3}}) \frac{1}{3y^{\frac{2}{3}}} dy
$$
Note that our new test function $\psi (y^{\frac{1}{3}}) \frac{1}{3y^{2/3}}$ is not smooth, hence we cannot define $\delta(x^3)$ as a distribution. In general, if $u$ is a distribution, then $u \circ g$ is a distribution as well if both $g$ and its inverse are smooth. In such a case: $\langle u \circ g, \psi \rangle := \langle u, \psi \circ g^{-1} |\text{det } dg^{-1}| \rangle$. Please note that the integration above is an abuse of notation.


问:
How to find the average compound interest for all years, knowing the starting and ending values, as well as the number of years
For example:

Company X shares in 2016 - 38 USD

5 years later, in 2021 - 66.95 USD

38 + 12% (4.56 USD) + 12% (5.10 USD) + 12% (5.72 USD) + 12% (6.40 USD) + 12% (7.17 USD) = 66.95 USD

What formula should I use to know that the average compound interest is 12% for each year?

答:
Compound interest formula with unknown interest rate is:

CI = ((FV / PV)^(1/n) − 1) * 100

Where:

CI - AVG Compound Interest (what we need to find)

FV - Future Value (66.95)

PV = Present Value (38)

n = Number of Periods (5 years)

What we got:

((66.95 / 38)^1/5 − 1) * 100 = 11,9969 (AVG %)

ContentLicense: CC BY-SA 4.0


问:
Expected Value of a quite complex coin flipping game.
I have a rather complex game, whose expected value I need to find.
Given is a binary random value with probability for success p (lets say an unfair coin with probability of landing heads p and tails 1-p, p<<0,5)
The rules of the games are the following: you start by betting 1 euro, a multiplier = 1 and 15 tosses.
Each time you land heads (less likely option) you receive 15 more tosses (added to the remaining one you have) and your multiplier increases by 1. Landing tails doesn't give you anything, you've just lost one toss. Your multiplier can grow up to 5, and each time you land a head after achieving a multiplier of 5 adds new 15 tosses, but does not increase the multiplier further.
I am seeking for the expected multiplier of the game.

I started by building a Markov Chain table A[i,j] for the probability of spin j to have multiplier i:
A[1,1] = 1,
A[1,2] = 1-p,
A[2,2] = p,
A[1,3] = (1-p)*(1-p),
etc... We have 15 initial spins then each
A[i, 15+i] = 0.
But I am still stuck of deriving the Avg MP from the table.
Any help will be much appreciated!

Thanks!

EDIT:
I forgot to say, at each toss you win your bet x Multiplier, so actually, I am looking for the EV of the game, but I think it can be presented as finding the average MP of the game and the average number of tosses, so in the end the EV should be their product. I know how to find the Avg number of tosses, and my main issue is the AVG MP.

答:
Solution:
Turned out the Markov Chain approach was pretty successful after all:

Calculating the probabilities for the first 60 tosses (taking into account that A[i, 15+i] = 0) and summing them per columns (per multiplier) one can find the average number of tosses with the respective MP. Instead of calculating ton of probabilities to gain an approximation for the last MP 5, one can simply derive its average number of tosses as subtracting the sum of the first 4 from the expected number of tosses (That is 15/(1- 15*p), coming from the infinite geometric series formula). Lastly simply dividing all average number of tosses per MP by the expected number of tosses, one gets the probabilities for each Mp and the average MP is simply their expectation.

P.P.It's a naive approach 'that gives you the answer, but the reasoning behind it is much more beautiful, so thanks to user for shedding light on it!

Blockquote


问:
How to calculate limit $\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$ without using L'Hospital rule
I have to calculate next limit without using L'Hospital rule:
$$\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$$
I came this far with that $$1+\frac{1}{e}\lim_{t\rightarrow1}\frac{\ln(t)}{t-1}$$
I know that result for this limit using the L'Hospital rule is $1+\frac{1}{e}$
Any suggestions how to finish this?

答:
So you just need to calculate $ \lim\limits_{x\to 0}{\frac{\ln{\left(1+x\right)}}{x}} $ whithout using L'hopital. We'll not be using series expansion either.

Let $ x\in\left(-1,1\right)\setminus\left\lbrace 0\right\rbrace $. We have : \begin{aligned} \frac{\ln{\left(1+x\right)}}{x}&=\int_{0}^{1}{\frac{\mathrm{d}y}{1+xy}}\\&=1-x\int_{0}^{1}{\frac{y}{1+xy}\,\mathrm{d}y} \end{aligned}

Since : \begin{aligned} \left|x\int_{0}^{1}{\frac{y}{1+xy}\,\mathrm{d}y}\right|\leq\left|x\right|\int_{0}^{1}{\frac{y}{\left|1+xy\right|}\,\mathrm{d}y}&\leq\left|x\right|\int_{0}^{1}{\frac{y}{\left|1-\left|x\right|y\right|}\,\mathrm{d}y}\\&\leq\frac{\left|x\right|}{1-\left|x\right|}\int_{0}^{1}{y\,\mathrm{d}y}=\frac{\left|x\right|}{2\left(1-\left|x\right|\right)}\underset{x\to 0}{\longrightarrow}0 \end{aligned}

Then $ \lim\limits_{x\to 0}{x\int_{0}^{1}{\frac{y}{1+xy}\,\mathrm{d}y}}=0 $, which means : $$ \lim_{x\to 0}{\frac{\ln{\left(1+x\right)}}{x}}=1 $$


问:
Solving an equation with the same variable twice
This may be kind of obvious, but our finance lecturer just assumes we know how to do stuff like this and the slides give no explanation. Jumping from question to answer

How do I get $r$ from this?

$27.397= \displaystyle \frac{1- \displaystyle \frac{1}{(1+r)^{36}} }{r}$

Answer:

$r=1.558\%$

答:
You want to find the root $r>0$ of
$$
f(r) = \bigg[1 - \frac{1}{{(1 + r)^{36} }}\bigg]\bigg/r - 27.397,
$$
that is the $r > 0$ for which $f(r)=0$.
Using Wims Function Calculator, I get $r \approx 0.0155764414039$, which is what you want.</p>

ContentLicense: CC BY-SA 3.0


问:
Minimum of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$

If $a,b,c\in\mathbb{R}$ and $a+b+c=1$, then what is the minimum value
of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$.

I used Wolfram Alpha and it says the minimum value is $\dfrac{44}{27}$ for $\left(a,b,c\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. Obviously $uvw$ method doesn't help. and since we have absolute value, I think The Buffalo doesn't help.

I wrote $ab+bc+ca$ this way: $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

We know that
$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le\frac{a+b+c}{3},$$
so
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\Rightarrow ab+bc+ca\ge9abc$$

I don't know if it helps (or is even true).

I think finding the minimum of $\left(1-\left(ab+bc+ca\right)\right)^2+\left(1-abc\right)^2$ has the same procedure. If anyone knows how to find it, it helps a lot (I think).

答:
I don't have a full solution either, but I thought I'd sketch out the theory regarding cubics this problem touches on.

The monic polynomial with roots $a,\,b,\,c$ is $t^3-t^2+yt-z$ with $y:=ab+bc+ca,\,z:=abc$. We want to minimise $\Sigma:=|1-y|+|1-z|$ subject to$$\Delta_3:=y^2-4y^3-4z-27z^2+18yz\ge0.$$This quadratic in $z$ can only be $\ge0$ if it has real roots $z_\pm:=\frac{9y-2\pm\sqrt{\Delta}}{27}$ with$$\Delta:=4(1-9y+36y^2-36y^3),$$whence $\Delta_3\ge0\iff z\in[z_-,\,z_+]$. You can show $\Delta$ is a cubic in $y$ which is $\ge0$ iff$$y\le y_0:=\frac{2+\sum_\pm\sqrt[3]{2\pm\sqrt{3}}}{6}\approx0.7.$$ The minimum of $\Sigma$ will occur at $\Delta_3=0$, i.e. $z=z_\pm(y)$.

Case I $z<1$ so $\Sigma=2-y-z$. Since $z_-\le\tfrac{9y-2}{27}<0.2$, for case I we minimise $2-y-z_+=\frac{56-36y-\sqrt{\Delta}}{27}$. This is an exercise in calculus.

Case II $z>1$ so $\Sigma=z_+-y=\tfrac{-18y-2+\sqrt{\Delta}}{27}$; again, there's a calculus exercise.

Both of the above calculus exercises require us to solve cubics.


问:
Is there a Prime number test that has no false positives
The primality tests that I have found, i.e. Fermat and Miller-Rabin don't return any false negatives, but do sometimes return false positives. That is, some composites are incorrectly decided as primes by these algorithms, but primes are always decided as primes.

Is there an efficient (polynomial time in length of the number) primality test that works the other way? That is, it will always correctly decide whether composite numbers are composite, but can sometimes mistake a prime number for a composite number?

答:
Each of these are primality proof methods.  They have no false positives.  They are not probabilistic tests.

For small inputs, very fast methods such as BPSW and deterministic Miller-Rabin (verified correct for all 64-bit inputs).

BLS75 methods, from the 1975 paper.  Some of these were known and used hundreds of years ago.  The downside is that they rely on partial factoring of either n-1, n+1, or both.  Sometimes this is very fast, but for general inputs this eventually becomes performance limited by factoring.  Can easily produce a certificate so others can verify the result.  Almost certainly not fast enough for general purpose 1024-bit primality testing.

APR-CL, early 1980s.  Quite good in practice and works for numbers of a few thousand digits.  It is not polynomial time but for practically computable numbers will have a smaller exponent than AKS.  Like AKS, doesn't produce a certificate, so the recipient just has to trust that it's right.

ECPP, late 1990s.  The fastest algorithm for very large inputs in practice.  Regularly proves results with 10-30k digits.  Randomized polynomial time.  Produces a certificate so third parties can quickly validate the primality (e.g. there are 3+ verifiers for Primo style certificates, each written completely independently, so really you're only relying on the actual mathematics in the original papers to be correct).

AKS, 2002.  Deterministic polynomial time for all inputs.  Much easier to code than the above, though remarkably full of places to trip up.  In practice, much, much slower than APR-CL and ECPP, as expected from their complexity classes  No certificate is produced, so you just have to trust the implementation.

In summary, if you're writing an academic paper, just reference AKS as being polynomial time and move on.  If you actually want to compute something, use one of the other methods.  Particularly APR-CL and ECPP which will be polynomial time in practice with a lower exponent than AKS.  Both APR-CL and ECPP, with decent implementations, are usable for current cryptographic sizes, proving 1024-bit primes in under a second, 2048-bit primes in 10-60 seconds.

Pari/GP supports all the methods other than AKS (which as they explain in their FAQ, currently has no practical reason to implement).  There are AKS programs for Pari/GP if you insist for some reason, though they are millions of times slower than the others.

I'm harping on AKS because two other answers reference it as the right solution.  I think they got really caught up on your mention of polynomial time.

There is another discussion to be had regarding the value of primality proving, especially in the absence of certificates.  For instance, I personally estimate the probability that an AKS software implementation has a subtle bug is substantially higher than a simple, good, well-implemented probable prime test (e.g. BPSW plus some extra random Miller-Rabin bases) has let a composite pass through.


问:
Proof of Lie's theorem using theorem 4.1 in Humphreys
I'm studying Humphreys' book 'Introduction to Lie Algebras and Representation Theory'
First here is theorem 4.1.

Theorem. Let $L$ be a solvable subalgebra of $\mathfrak gl(V)$, $V$ finite dimensional. if $V \neq 0$, then $V$ contains a common eigenvector for all the endomorphisms in $L$.

And here is Corollary A.

Coroallary A. (Lie's Theorem). Let $L$ be a solvable subalgebra of $\mathfrak gl(V)$, dim$V=n \lt \infty$. Then $L$ stabilizes some flag in $V$.

And the book states that we can prove Cor.A. by using the above theorem along with induction on dim$V$. Here is my attempt.

If dim$V$=1, then the only flag is $0 \subset V$ and it is stabilized clearly by $L$. Now assume dim$V=n$. By the main theorem, there exists $v \in V$ such that $x.v=\lambda(x)v$ for all $x\in L$.  Then $<v>$ is a subspace which is invariant under $L$. Decompose $V$ as $V=<v> \oplus V'$. Let $L'=L\cap \mathfrak gl(V')$. Clearly, $L'$ is solvable in $\mathfrak gl(V')$. Since dim$V'\lt$ dim$V$, by induction hypothesis, there is a flag of $V'$ stabilized by $L'$.

Here is where I stucked. How can I extend that flag of $V'$ stabilized by $L'$ to a flag of $V$ stabilized by L? I found other questions about this problem, but I still don't get it.

Thanks for any help in advance.

答:
Let $\rho\colon L\rightarrow \mathfrak{gl}(V)$ be a Lie algebra representation of a solvable Lie algebra $L$. We'll use induction over $\dim V$ to show that there is a $L$-invariant flag $$0=V_0\subset V_1 \subset \cdots
\subset V_n=V$$
in $V$ such that $\dim V_j=j$. Choosing the basis elements for $V$ as $v_i\in V_i$, the claim follows
because of $\rho(x) (V_i)\subset V_i$. For $V=0$ there is nothing to show. So let $\dim V\ge 1$. By the Theorem
there is a $v\in V$, $v\neq 0$ with $L.v\subset kv$. It follows that $W=kv$ is a $1$-dimensional $L$-submodule.
By applying the induction hypothesis to the quotient module $V/W$ we find there an $L$-invariant flag
$0=W_1 \subset \cdots \subset W_n$ with $\dim W_j=j-1$. Let $\pi\colon V\rightarrow V/W$ be the quotient map. Then $V_0=0$
and $V_j=\pi^{-1}(W_j)$ defines a $L$-invariant flag in $V$ with $\dim V_j=j$.


问:
Inequality for standard normal distribution
Suppose $X\sim N(0,\,1)$ and $c, \, y\in\Bbb R$ with $c>0$. Prove that:

$$|P(X<y)-P(X<cy)| \leq \frac{|c-1|\max(1,1/c)}{\sqrt{2\pi e}} $$

答:
WLOG, suppose $c>1$ (for $c<1$, we work with $\frac{1}{c}$ instead) and $y>0$ (for $y<0$, we work with $-X$).

The left hand side becomes
$$
\begin{align}
P(X<cy)-P(X<y) &= P(y<X<cy) \\
&= \int_y^{cy}\frac{e^{-\frac{-t^2}{2}}}{\sqrt{2\pi} }dt \\
&\le \int_y^{cy}\frac{e^{-\frac{-y^2}{2}}}{\sqrt{2\pi} }dt =(c-1)y\frac{e^{-\frac{-y^2}{2}}}{\sqrt{2\pi} }\\
\end{align}
$$
But we have, according to Wolfram Alpha here
$$ye^{-\frac{-y^2}{2}} < \frac{1}{\sqrt{e}}$$

Hence,The left hand side is less or equal than $(c-1)\frac{1}{\sqrt{2\pi e}}$  (QED).


问:
Field of fractions of $R = \{ \frac{a}{b} | \ a \in \mathbb{Z}, b \in \mathbb{Z} \setminus p\mathbb{Z} \}$.
Let $p$ be a prime number. Define the subring $R = \{ \frac{a}{b} | \  a \in \mathbb{Z}, b \in \mathbb{Z} \setminus p\mathbb{Z} \}$ of $\mathbb{Q}$. Can someone explain to me why the set of rationals $\mathbb{Q}$ is the field of fractions of $R$ ? And also, why does $p$ have to be a prime number ? I haven't studied alot of ring theory yet so an explanation based on only the definition of field of fractions would be useful.

Thanks in advance !

答:
The field of fractions of $R$ is the field $\{\frac{a}{b}\mid a,b\in R, b\neq 0\}$ under the relation of equivalence of fractions, i.e. $\frac{a}{b}=\frac{c}{d}$ iff $ad=bc$ (this products in $R$), with the usual operations of addition and multiplication of fractions.

Let $a=\frac{x}{y}$ and $b=\frac{z}{t}\in R$. We can see that the fraction $\frac{a}{b}$ is equivalent to the fraction $\frac{\frac{xt}{y}}{\frac{z}{1}}\in Frac(R)$, which is identified with the fraction $\frac{xt}{yz}\in \mathbb{Q}$. Indeed, we have that these two fractions are equivalent iff $a\frac{z}{1}=b\frac{xt}{y}$ in $R$, but with the usual operations we have $b\frac{xt}{y}=\frac{z}{t}\frac{xt}{y}$. As you might expect, $t$ cancels, but you can show this using the definition of equivalence of fractions, i.e. $\frac{zxt}{ty}$ is equivalent to $\frac{zx}{y}=a\frac{z}{1}$ in $R$.

This means that any fraction of elements of $R$ can be seen a a rational number after the identification. Therefore, $Frac(R)$ is a non-trival (find two fractions that are not equivalent to show this) subfield of the rationals, and you can use that the only such subfield is $\mathbb{Q}$ itself. Or artenatively, given a rational number $\frac{r}{s}$ you can find an element of $Frac(R)$ that can be identified with it, for instance $\frac{\frac{r}{1}}{\frac{s}{1}}$.


问:
Banach subsequence converges
Show that every sequence in a Banach space such that $\{x_n\} \rightarrow 0$ has a subsequence $\{x_{n_p}\}$ such that $\sum_{p=1}^{\infty} x_{n_p} $ converges by showing $S_N = \sum_{p=1}^{N} x_{n_p}$ is a Cauchy sequence.

The convergent to the zero vector is what is confusing -- typically it is just to some other element? How do I incorporate this?

答: Hint: Since $\lim\limits_{n\to\infty}x_n=0$, then there exist subsequence $\{n_p:p\in\mathbb{N}\}\subset\mathbb{N}$ such that $\Vert x_{n_p}\Vert\leq 2^{-p}$.

Title: Banach subsequence converges


问:
Using generalized logistic curve to create a mathematical model from data.

The first row is time and the second row is height of a plant. We need to use generalized logistic curve to model the behavior of the plant. The equation of the logistic curve is :
$$N = \frac{N_*}{1+(N_*/N_0-1)e^{-a_0t}}$$ where $N_*$ is the maximum height of the plant or in other words, the supremum of the logistic function.

$a_0$ is how fast the function increases. My question is the following:

I know how to determine $N_*$, we just make up a number that is above $251$ and less than $251 + (251 - 247) $. But how to exactly determine $a_0$. I am quite lost here. How I would do it is to calculate every $a_0$ for every time and height, with equation $a_0 = \frac{1}{t}\ln(\frac{N}{N_0})$. In this equation $N$ would be the difference between heights, so let's say between $18$ and $33$ the $N$ for time equals $1$ would be $33 - 18$. Is the process here of creating a table of $\sum_{k = 1}^{10}(10 - i)$ column elements of differences, so that we take multiple $N_0$ until we run out of data from the primary table in the picture (so until $N_0 = 9$). Then we can use a formula for $a_0 = \frac{\sum a_{0l}*t_l}{\sum t_l}$ where $l$ is the number of elements.

Is my procedure for creating a logistic out of the data provided correct?

答:
The answer by @DanielJ.Greenhoe is very complete but it plays along with the initial premisse that you just pick $N_*$ in a compatible way. However, there is no reason why the minimisation process is not performed for the parameters $(N_*, a_0)$, instead of just $a_0$.

This leeds to the set of parameters $N_* = 255.844, a_0 = 0.65392$.


问:
Finding the centre of mass of two right cones joined combined
The following question is given in my textbook

Two uniform cones with base radius $r$ are joined together by their plane faces.Their lines of symmetry are aligned, the height of one cone is $6r$ and the other is $2r$. Given that the smaller cone is $50\%$ denser than the larger, find the distance of the centre of mass from their joint plane face.

The way I worked on it is by first drawing a simple diagram

Then, finding the total volume and mass of the individual cones, for the larger cone:

$$v_1=\frac{1}{3} \pi r^2 (6r)=2\pi r^3$$
$$m_1=2\pi r^3 \rho$$

For the smaller cone:

$$v_2= \frac{1}{3} \pi r^2 (2r)=\frac{2}{3} \pi r^3$$
$$m_2=\pi r^3 \rho$$

Then proceeding to find the $\bar{x}$ of each using the result that:

For a right cone the centre of mass lies $(\frac{1}{4}h,0)$ measured from the vertex and $(\frac{3}{4}h,0)$ from the base

For the larger cone:

$$\bar{x}=\frac{1}{4}(6r)=\frac{3}{2}r$$

For the smaller cone:

$$\bar{x}=\frac{1}{4}(2r)=\frac{1}{2}r$$

Next calculating the center of mass for the whole shape:

$$\bar{x}=\frac{(2\pi r^3 \rho)(\frac{3}{2}r)+ (\pi r^3 \rho)(\frac{1}{2}r)}{\pi r^3 \rho + 2\pi r^3 \rho}=\frac{7}{6}r$$

However the answer given in the book is that the distance is $\frac{5}{6}r$ I do not know how they obtained this answer, is there a problem in my working and how do I proceed with solving this correctly?

答:
I think you have to use a better notation since you do not distinguish the 3 centers of mass.

I'll resume what you have:

Supposing you place your coordinate system on the heigths of the two cones with the origin on the base of the two cones. Then

$$\begin{cases}
v_1 = 2\pi r^3 \\
m_1=2\pi r^3 \rho_1 \\
\bar{x_1} = -\frac{3}{2}r
\end{cases}
\qquad
\begin{cases}
v_2= \frac{2}{3} \pi r^3 \\
m_2= \pi r^3 \rho_1 \\
\bar{x}_2 = \frac{1}{2}r
\end{cases}$$

Now as you wrote:

$$\bar{x}=\frac{m_1 \bar{x_1} + m_2 \bar{x_2}}{m_1 + m_2} = \frac{(2\pi r^3 \rho_1)(-\frac{3}{2}r)+ (\pi r^3 \rho_1)(\frac{1}{2}r)}{\pi r^3 \rho_1 + 2\pi r^3 \rho_1}=  \frac{(-3 + \frac{1}{2}) r}{1 + 2} = -\frac{5}{6}r$$

Please also note that

For a right cone the centre of mass lies on the height at $z = \frac{1}{4}h$ measured from the base


答:
Let $p = a + b + c = 1$, $q = ab + bc + ca$ and $r = abc$.

Fact 1: $q^2 \ge 3pr$.

(The proof is given at the end.)

Fact 2: $p^2 \ge 3q$.

(Proof: $p^2 - 3q = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \ge 0$.)

We split into four cases:

$q \ge 0$:  By Fact 2, we have $q \le \frac{1}{3}$. By Facts 1-2, we have $r\le \frac{1}{27}$.
We have
$|1- q| + |1 - r| = 1 - q + 1 - r = 2 - q - r \ge 2 - \frac{1}{3} - \frac{1}{27} = \frac{44}{27}$ with equality if $a = b = c = \frac{1}{3}$.

$q < 0$ and $r > 1$: By Fact 1, we have $q^2 \ge 3$. Thus, $q < -\sqrt{3}$. We have
$|1-q| + |1 - r| = 1 - q + r - 1 = -q + r > \sqrt{3} + 1 > \frac{44}{27}$.

$q < 0$ and $0\le r \le 1$: By Fact 1, we have $q^2 \ge 3r$. Thus, $q \le -\sqrt{3r}$. We have
$|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 + \sqrt{3r} - r \ge 2 > \frac{44}{27}$.

$q < 0$ and $r < 0$: We have
$|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 > \frac{44}{27}$.

Thus, the minimum is $\frac{44}{27}$.

$\phantom{2}$

Proof of Fact 1: We have
\begin{align}
q^2 &= (ab)^2 + (bc)^2 + (ca)^2 + 2abc(a+b+c)\\
&\ge ab\cdot bc + bc \cdot ca + ca \cdot ab + 2abc(a+b+c) \\
&= 3abc(a+b+c)\\
&= 3pr
\end{align}
where we have used
\begin{align}
&(ab)^2 + (bc)^2 + (ca)^2 - ab\cdot bc - bc \cdot ca - ca \cdot ab\\
=\ & \frac{1}{2}[(ab - bc)^2 + (bc - ca)^2 + (ca - ab)^2]\\
\ge \ & 0.
\end{align}

Title: Minimum of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$

Body:
If $a,b,c\in\mathbb{R}$ and $a+b+c=1$, then what is the minimum value
of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$.

I used Wolfram Alpha and it says the minimum value is $\dfrac{44}{27}$ for $\left(a,b,c\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. Obviously $uvw$ method doesn't help. and since we have absolute value, I think The Buffalo doesn't help.

I wrote $ab+bc+ca$ this way: $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

We know that
$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le\frac{a+b+c}{3},$$
so
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\Rightarrow ab+bc+ca\ge9abc$$

I don't know if it helps (or is even true).

I think finding the minimum of $\left(1-\left(ab+bc+ca\right)\right)^2+\left(1-abc\right)^2$ has the same procedure. If anyone knows how to find it, it helps a lot (I think).


问:
linear inequalities in semi-definite program
A standard form of SDP is

$\min_x c^Tx$ s.t. $F(x)\succcurlyeq 0$

where $F(x)=F_0+\sum_{i=1}^m F_m$ and $F_0,F_1,...,F_m$ are symmetric matrices and $\succcurlyeq 0$ indicates that the matrix is positive semidefinite.

It is stated in many notes (such as https://inst.eecs.berkeley.edu/~ee227a/fa10/login/l_conic_sdp.html under the section special cases) that a linear inequality (where inequality corresponding to a vector denotes component-wise inequality) can be re-formulated as an SDP inequality by defining a diagonal matrix with diagonal elements as the components of the vector.

My doubt is suppose in a problem, we have a large number of component-wise inequalities and we reformulate it as an SDP problem, is it even computationally feasible (assuming the constraint also has an SDP inequality, thus, there is a dire need for SDP formulation). I am asking this because in such a scenario we are faced with a large number of sparse matrices and the algorithm may take a lot of memory and time to solve. Or is there any better way to deal with such situations?

答: In an actual solver you don't formulate linear inequalities using a diagonal matrix which you work with in the SDP cone, but you work with the elementwise (and socp) cone explicitly. The fact that you could do it using a diagonal matrix is just a way to convince the reader that elementwise constraints are easily dealt with both in theory and practice.</p>


问:
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$?
calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you

答:
Your approach can be fixed easily. You have to evaluate the limit of $n(S_n-(1/3))$ and you have also shown that $$\frac{n} {n+1}R_n<S_n<R_n$$ where $R_n=n^{-1}\sum_{k=1}^{n}f(k/n)$ is the Riemann sum for $f(x) =x^2$ over $[0,1]$.

It is well known that $$n\left(R_n-\int_{0}^{1}f(x)\,dx\right)\to\frac{f(1)-f(0)}{2}=\frac{1}{2}\tag{1}$$ Our job is complete if we can evaluate the limit $L$ of $n(S_n-R_n)$ and our desired limit will be $L+1/2$.

But $$n(S_n-R_n) =-\sum_{k=1}^{n}\frac{k/n^2}{1+(k/n^2)}\cdot\frac{k^2}{n^2}$$ The sum on right can again be squeezed (as in question) to get the limit $L=-1/4$. The desired limit is thus $1/4$.

Moral of the Story: Do not deviate even an iota from the hypotheses and conclusion of a theorem. If you do that try to do further analysis and prove the desired changes. The result (equation $(1)$ above) you use deals with a specific Riemann sum and that can't be replaced by any similar looking sum (even having same limit).


问:
Lagrange equations of motion of very unstable robot
I'm trying to control a robot and I have somewhere bug and I can't found it. I would be grateful if is there anybody who would check my equations. The robot diagram can be found here 1.
I would like to obtain differential equations of the robot by using the Lagrange equation of motion. Here is what have I done.

$$x_1 = X + L_2\cos\theta_2 + P_1\cos (\theta_1 + \theta_2) $$
$$  z_1 = Z + L_2\sin\theta_2 + P_1\sin (\theta_1 + \theta_2)$$

$$\dot{x}_1 = - L_2\sin{\theta}_2\dot{\theta}_2 - P_1\sin(\theta_1 + \theta_2)(\dot{\theta}_1 + \dot{\theta}_2) $$
$$  z_1 = L_2\cos{\theta_2}\dot{\theta_2} + P_1\cos(\theta_1 + \theta_2)(\dot{\theta}_1 + \dot{\theta}_2)$$

$$x_2 = X + P_2\cos\theta_2 $$
$$z_2 = Z + P_2\sin\theta_2 $$

$$\dot{x}_2 = -P_2\sin\theta_2\dot{\theta}_2 $$
$$\dot{z}_2 = P_2\cos\theta_2\dot{\theta}_2 $$

$$x_m = X + L_2\cos\theta_2 + P_m\cos (\theta_1 + \theta_2 + \theta_m) $$
$$  z_m = Z + L_2\sin\theta_2 + P_m\sin (\theta_1 + \theta_2 + \theta_m)$$

$$\dot{x}_m = - L_2\sin\theta_2\dot{\theta}_2 - P_m\sin(\theta_1 + \theta_2 + \theta_m)(\dot{\theta}_1 + \dot{\theta}_2 + \dot{\theta}_m)$$
$$  \dot{z}_m = L_2\cos\theta_2\dot{\theta}_2 + P_m\cos(\theta_1 + \theta_2 + \theta_m)(\dot{\theta}_1 + \dot{\theta}_2 + \dot{\theta}_m)$$

Potential energy:

$$T_1 = \frac{1}{2}M_1(\dot{x}_1^2 + \dot{y}_1^2) + \frac{1}{2}I_1(\dot{\theta_1} + \dot{\theta_2})^2$$

$$T_2 = \frac{1}{2}M_2(\dot{x}_2^2 + \dot{y}_2^2) + \frac{1}{2}I_2\dot{\theta}_2^2$$

$$T_m = \frac{1}{2}M_m(\dot{x}_m^2 + \dot{y}_m^2) + \frac{1}{2}I_m(\dot{\theta}_1 + \dot{\theta}_2 + \dot{\theta}_m)^2$$

Kinetic energy:
Gravity which is pointing downward
$$ U_1 = M_1gy_1 $$
$$ U_2 = M_2gy_2 $$
$$ U_m = M_mgy_m $$

Spring with stiffness coefficient K:
$$ \Delta L_p = \sqrt{L_m^2 + L_k^2 - 2L_kL_m\cos(\theta_m + \theta_1-\pi))} - L_{p0} $$
$$ Up  = 1/2K({\Delta L_p})^2 $$

Where coefficents are:

$M_1$: Mass of link 1 in center of gravity

$M_2$: Mass of link 2 in center of gravity

$M_m$: Mass of link m in center of gravity

$L_1$: Length of link 1

$L_2$: Length of link 2

$L_m$: Length of link m

$P_1$: Position of center of mass of link 1

$P_2$: Position of center of mass of link 2

$P_m$: Position of center of mass of link m

$L_{p0}$: Initial lenght of the spring

Generalized coordiantes are angles $\vec{q} = [\theta_1, \theta_2, \theta_m]$, $\vec{\dot{q}} = [\dot{\theta}_1, \dot{\theta}_2, \dot{\theta}_m]$.

Generalize forces $Q = [0 0 \tau_m]$ is torque on motor which is connected in $\theta_m$ angle.

Lagrange equation of motion:

$$L = T - U$$
$$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = Q $$

from this i will obtain 3 very long equations, so I will not include them. I done this in Matlab. After that I used substitution:

$$\dot{\Theta}_1 = \ddot{\theta}_1$$
$$\Theta_1 = \dot{\theta}_1$$

$$\dot{\Theta}_2 = \ddot{\theta}_2$$
$$\Theta_2 = \dot{\theta}_2$$

$$\dot{\Theta}_m = \ddot{\theta}_m$$
$$\Theta_m = \dot{\theta}_m$$

and solved them to receive:
$$\dot{\theta}_1 = \Theta_1$$
$$\dot{\theta}_2 = \Theta_2$$
$$\dot{\theta}_m = \Theta_m$$
$$\dot{\Theta}_1 = f_1(\theta_1, \theta_2, \theta_m, \Theta_1, \Theta_2, \Theta_m, \tau_m)$$
$$\dot{\Theta}_2 = f_1(\theta_1, \theta_2, \theta_m, \Theta_1, \Theta_2, \Theta_m, \tau_m)$$
$$\dot{\Theta}_m = f_1(\theta_1, \theta_2, \theta_m, \Theta_1, \Theta_2, \Theta_m, \tau_m)$$

When I try to solve this in ode45 in Matlab it acts unrealistically. When I try linearization is an arbitrary working point then matrix A and B have numbers in 1e4. I tried many types of control but all failed. I check everything but I'm unable to say whenever are the equations correct.

Thank you for your help.

答:
Follows a MATHEMATICA script which simulates the movement equations for a set of parameters specified in parms

```<code>p0 = {X, Z};
p1 = p0 + L2 {Cos[theta2[t]], Sin[theta2[t]]} + P1 {Cos[theta1[t] + theta2[t]], Sin[theta1[t] + theta2[t]]};
p2 = p0 + P2 {Cos[theta2[t]], Sin[theta2[t]]};
pm = p0 + L2 {Cos[theta2[t]], Sin[theta2[t]]} + Pm {Cos[theta1[t] + theta2[t] + thetam[t]], Sin[theta1[t] + theta2[t] + thetam[t]]};
v1 = D[p1, t];
v2 = D[p2, t];
vm = D[pm, t];
T1 = 1/2 M1 p1.p1 + 1/2 I1 D[theta1[t] + theta2[t], t]^2;
T2 = 1/2 M2 p2.p2 + 1/2 I2 D[theta2[t], t]^2;
Tm = 1/2 Mm pm.pm + 1/2 Im D[theta1[t] + theta2[t] + thetam[t], t]^2;
T = T1 + T2 + Tm;
delLp = Sqrt[Lm^2 + Lk^2 - 2 Lm Lk Cos[thetam[t] + theta1[t] - Pi]] - Lp0;
V = g (M1 p1 + M2 p2 + Mm pm).{0, 1} + 1/2 K0 delLp^2;
L = T - V;
vars = {theta1[t], theta2[t], thetam[t]};
torque = {0, 0, tau};
equsmov = D[Grad[L, D[vars, t]], t] - Grad[L, vars] - torque;

parms = {L2 -> 1, P2 -> 1/2, L1 -> 1/2, P1 -> 0.5, Lm -> 1/4, Pm -> 1/8, Lk -> 3/4, I1 -> 1, I2 -> 2, Im -> 1/2, K0 -> 1000, Lp0 -> 1/2, g -> 10, M1 -> 1, M2 -> 2, Mm -> 1/2, X -> 0, Z -> 0};

ode = Thread[(equsmov /. parms /. tau -> 0) == 0];

tmax = 10;
cinits = {theta1[0] == 0.01, theta2[0] == -Pi/2, thetam[0] == 0, theta1'[0] == 0, theta2'[0] == 0, thetam'[0] == 0};
solthetas = NDSolve[Join[ode, cinits], vars, {t, 0, tmax}][[1]];

ParametricPlot[Evaluate[{p1, p2, pm} /. parms /. solthetas], {t, 0, tmax}]
</code>```

Follows a simulation plot showing in abscense of input, and hanging, the pendular behavior.

Body:
Follows a MATHEMATICA script which simulates the movement equations for a set of parameters specified in parms

p0 = {X, Z};
p1 = p0 + L2 {Cos[theta2[t]], Sin[theta2[t]]} + P1 {Cos[theta1[t] + theta2[t]], Sin[theta1[t] + theta2[t]]};
p2 = p0 + P2 {Cos[theta2[t]], Sin[theta2[t]]};
pm = p0 + L2 {Cos[theta2[t]], Sin[theta2[t]]} + Pm {Cos[theta1[t] + theta2[t] + thetam[t]], Sin[theta1[t] + theta2[t] + thetam[t]]};
v1 = D[p1, t];
v2 = D[p2, t];
vm = D[pm, t];
T1 = 1/2 M1 p1.p1 + 1/2 I1 D[theta1[t] + theta2[t], t]^2;
T2 = 1/2 M2 p2.p2 + 1/2 I2 D[theta2[t], t]^2;
Tm = 1/2 Mm pm.pm + 1/2 Im D[theta1[t] + theta2[t] + thetam[t], t]^2;
T = T1 + T2 + Tm;
delLp = Sqrt[Lm^2 + Lk^2 - 2 Lm Lk Cos[thetam[t] + theta1[t] - Pi]] - Lp0;
V = g (M1 p1 + M2 p2 + Mm pm).{0, 1} + 1/2 K0 delLp^2;
L = T - V;
vars = {theta1[t], theta2[t], thetam[t]};
torque = {0, 0, tau};
equsmov = D[Grad[L, D[vars, t]], t] - Grad[L, vars] - torque;

parms = {L2 -> 1, P2 -> 1/2, L1 -> 1/2, P1 -> 0.5, Lm -> 1/4, Pm -> 1/8, Lk -> 3/4, I1 -> 1, I2 -> 2, Im -> 1/2, K0 -> 1000, Lp0 -> 1/2, g -> 10, M1 -> 1, M2 -> 2, Mm -> 1/2, X -> 0, Z -> 0};

ode = Thread[(equsmov /. parms /. tau -> 0) == 0];

tmax = 10;
cinits = {theta1[0] == 0.01, theta2[0] == -Pi/2, thetam[0] == 0, theta1'[0] == 0, theta2'[0] == 0, thetam'[0] == 0};
solthetas = NDSolve[Join[ode, cinits], vars, {t, 0, tmax}][[1]];

ParametricPlot[Evaluate[{p1, p2, pm} /. parms /. solthetas], {t, 0, tmax}]

Follows a simulation plot showing in abscense of input, and hanging, the pendular behavior.


问:
Is $A=\{x\in[-2,2]:x\cos x\ge\frac{1}{2}\sin x \}$ compact?

Is $A=\{x\in[-2,2]:x\cos x\ge\frac{1}{2}\sin x \}$ compact?

Clearly, $A$ is bounded, so if I want to show that $A$ is compact, then I need to show that $A$ is closed. For any sequence $(a_n)$ in $A$ and $a_n\to a$, I need to show that $a\in A$. But I think I can not check all such sequence, so I guess $A$ is not compact and it suffices to find a sequence $(a_n)$ in $A$ and $a_n\to a$ but $a\notin A$. Hope someone could give me some hints.

答: $A$ is the intersection of the inverse image of the closed set $[0,+\infty)$ by the  continuous function $f(x)=x\cos x-\frac 12\sin x\,$ with the closed interval $[-2,2]$, so it is closed.


问:
For what $a$ the sequence $2x_{n+1}=x^2_n+a$ has finite limit?
Let $(x_n)$ be the sequence defined recursively by the following formular $$x_0=0, \ 2x_{n+1}=x^2_n+a$$
Find $a$ such that the above sequence has finite limit?

With all of my effort, there is still one case left, say, $a<-3$, in which I don't even know how to prove or disprove the existence of the limit.

答:
You could say $2x=x^2+a$ and then you get
the discriminante $D=4-4a$.

Now for $a < 1$ you get 2 real solutions.
For $a = 1$ you get 1 real solution and for $a > 1$ you get none.

Moreover you have to show that the sequence is monotonously falling and limited for certain intervalls.


问:
Functions with property $1<2f(x)-f(x-1)-f(x+1)<2$
Let $x>0$ and let $f(x)$ be an increasing function on the domain $x>0$. Are there any $f(x)$ with the property  $$1<2f(x)-f(x-1)-f(x+1)<2?$$
Since $f(x-1)+f(x+1)<2f(x)$, $f(x)$ is a concave function. But I don't know how to construct $f(x)$ satisfying above inequality. Do you know how to find such $f(x)$ or are there some references on this kind of problems? Thanks!!

答:
Here is a positive result complementing the negative result shown in the other answers : while it is true that there is no solution on $[0,\infty)$, on the other hand there is a solution on $[0,T]$ for every $T\gt 0$.

Indeed, let $f(x)=\frac{3}{4}\big(-x^2+2Tx\big)$. Then $f'(x)=\frac{3}{2}(T-x)$ so $f$ is increasing on $[0,T]$. On the other hand, $2f(x)-f(x-1)-f(x+1)=\frac{3}{2}$ for every $x$.


问:
Factorize certain polynomials as product of irreducibles
I'm trying to solve this question from my abstract algebra's course. I know it's a very trivial question that can be solved using Ruffini like we are taught before college, but my question is how can I solve it using "ring of polynomial" theory. The question goes like this:

Factorize the following polynomials as product of irreducibles:

$X^5 − 2X^4 − 12X + 24$ in $\mathbb{Z}[X]$

$X^4+X^2+1$ in $(\mathbb{Z}/2\mathbb{Z})[X]$

I know about ring theory and some ring of polynomials theory, but I don't see what can I do here to approach this kind of problem using abstract algebra methods (instead of Ruffini, because I guess I'm not supposed to solve it that way after all the theory I've studied, where this method is never introduced). How can I solve this? Any help will be appreciated, thanks in advance.

答:
In view of $GF(2)[x]$ with $GF(2) = \Bbb Z/2\Bbb Z$, the irreducible polynomials of degree at most 4 are

$x,x+1,x^2+x+1,x^3+x+1,x^3+x^2+1, x^4+x+1, x^4+x^3+1, x^4+x^3+x^2+x+1$.

This should suffice to factorize $x^4+x^2+1$ which is reducible.


问:
Solution verification: Showing that $\|A\|_\infty = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$ for $A \in \Bbb R^{n\times n}$
$
\newcommand{\norm}[1]{\| #1 \|}
\newcommand{\inorm}[1]{\norm{#1}_\infty}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\para}[1]{\left( #1 \right)}
\newcommand{\R}{\mathbb{R}}
$Context: This ultimately ties back to a homework assignment. (Specifically it's from Fundamentals of Matrix Calculations by Watkins, Exercise $2.1.30$ from $\S2.1$ but that's not really a huge deal, seems like a fairly common exercise.)

The goal is to show that, knowing matrix $p$-norms are induced by the corresponding vector norms, $\forall A \in \R^{n\times n}$,
$$\|A\|_\infty = \max \limits_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$$
I have an approach and was just curious as to how valid it is; I just don't feel very sure of myself. I feel like I made some sort of small-yet-critical mistake somewhere, but I can't figure out where...

My Attempt:

Ultimately, we will show that

$$\max \limits_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \le \|A\|_\infty \le \max \limits_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$$

which will let me conclude with the desired equality ($a \le b \le a \implies a=b$).

From the definition of the $\infty$-norm, with $A := (a_{i,j})_{1 \le i,j \le n} \in \R^{n \times n}$ and $x := (x_i)_{1 \le i \le n} \in \R^n$,
$$
\inorm{Ax} = \max_{1 \le i \le n} \abs{ \sum_{k=1}^n a_{i,k} x_k }
$$
Applying the triangle inequality yields
$$
\inorm{Ax} \le \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k} x_k|  = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \cdot | x_k|
$$
Note that
$$
|x_k| \le \max_{1 \le j \le n} |x_j| =: \norm{x}_\infty
$$
and thus
$$
\inorm{Ax} \le  \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \cdot \norm{x}_\infty
$$
$\norm{x}_\infty$ is independent of $i$, so we factor it out and conclude
$$
\inorm{Ax} \le \norm{x}_\infty \para{ \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| }
$$
Thus,
$$
\inorm{A} = \max_{x \ne \vec 0} \frac{\inorm{Ax}}{\inorm{x}} \le \frac{\displaystyle \norm{x}_\infty \para{ \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| }}{\norm{x}_\infty} = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|
$$
Thus, with $\inorm{A}$ less than or equal to the desired expression, we need to find an $\hat x$ such that equality is achieved. Suppose in row $r$ the maximum is achieved; for every $k$, let
$$
\hat x_k = \begin{cases}
+1 & \text{if } a_{r,k} \ge 0 \\
-1 & \text{if } a_{r,k} < 0
\end{cases} = \mathrm{sign}(a_{r,k})
$$
Define $\hat x := (\hat x_i)_{1 \le i \le n}$ as defined above; clearly $\inorm{\hat x}=1$. This ensures $a_{r,k}\hat x_k = |a_{r,k}|$. Then
\begin{align*}
\norm{A}_\infty
&=  \max_{x \ne \vec 0} \frac{\inorm{Ax}}{\inorm{x}}  \tag{def. of matrix norm} \\
&\ge \frac{\inorm{A \hat x}}{\inorm{\hat x}} \tag{def. of maximum}\\
&= \inorm{A \hat x} \tag{$\inorm{\hat x} = 1$} \\
&= \max_{1 \le i \le n} \abs{ \sum_{k=1}^n a_{i,k} \hat x_k } \tag{definition}\\
&=  \abs{ \sum_{k=1}^n a_{r,k} \hat x_k } \tag{definition of $r$}\\
&=  \abs{ \sum_{k=1}^n |a_{r,k}| } \tag{choice of $\hat x_k$, $r$}\\
&=   \sum_{k=1}^n |a_{r,k}|  \tag{$|z| \ge 0 \implies \sum_i |z_i| \ge 0$}\\
&= \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \tag{def. of $r$}
\end{align*}
Thus what we have seen is that
$$
\max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \le \norm{A}_\infty \le \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|
$$
thus letting us conclude equality:
$$
\norm{A}_\infty = \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|
$$

答:
Here is an issue. In the part of showing $\big\lVert A \big\rVert_\infty \geq \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}|$ you've write the equality$$
\max_{1 \le i \le n}  \Bigg\lvert \sum_{k=1}^n a_{i,k} \hat x_k \Bigg\rvert
=  \Bigg\lvert \sum_{k=1}^n a_{r,k} \hat x_k   \Bigg\rvert
$$

is true because of the definition of $r$. Recall that $r$ is taken as an  integer between $1$ and $n$ such that $$\max_{1\leq i\leq n} \sum_{k=1}^n |a_{i,k}|= \sum_{k=1}^n |a_{r,k}|$$holds, so the equality does not follow immediately. I think it would be better to write as follows:

\begin{align*}
& \max_{1 \le i \le n} \Bigg\lvert \sum_{k=1}^n a_{i,k} \hat x_k \Bigg\rvert \\
& \geq  \Bigg\lvert \sum_{k=1}^n a_{r,k} \hat x_k \Bigg\rvert \tag{definition of maximum}\\
&=  \Bigg\lvert \sum_{k=1}^n |a_{r,k}| \Bigg\rvert \tag{choice of $\hat x_k$, $r$}\\
&=   \sum_{k=1}^n |a_{r,k}|  \\
&= \max_{1 \le i \le n} \sum_{k=1}^n |a_{i,k}| \tag{definition of $r$}
\end{align*}
Other parts seem good to me.


问:
Representing multivariable functions as vectors
I’m going through some Linear Algebra material and started to wonder that if we can represent functions as vectors for example the elements of $C([0,1])$. These are all usually presented with the case where $f: \mathbb{R} \to \mathbb{R}$ so that all the vectors $\textbf{f} = f(x_1)e_1 + \cdots + f(x_n)e_n$. So what if I have functions of the form $f: \mathbb{R}^n \to \mathbb{R}^m$? Initially I started to think that this would force $\textbf{f}$ to be a matrix instead of a vector?

答:
Of course if $f$ is a function from $\mathbb R^n$ to $\mathbb R^m$, it is a vector $$\underline {f}(x_1,\dots, x_n)=\begin{pmatrix}f_1(x_1,\dots,x_n)\\\vdots\\f_m(x_1,\dots, x_n) \end{pmatrix},$$ where the functions $f_1\,\dots,f_m$ are function from $\mathbb R^n$ to $\mathbb R$.

An example could be the function $\underline f:\mathbb R^3\to \mathbb R^2$ such that
$$\underline f(x,y,z)=\begin{pmatrix}e^{x+y}-y+\sin z\\x^2-e^y+3\cos z \end{pmatrix}$$
but I didn't understand what you wrote about the representation of such a function by a matrix.

If a function is $\underline f:\mathbb R^n\to \mathbb R^m$, the image of $\underline f$ is a vector in $\mathbb R^m$, whose components are the functions $f_1(x_1,\dots,x_n),f_2(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n)$ and its "derivative" corresponds to
$$\operatorname{M}(m\times n,\mathbb R)\ni J_{\underline f}(x_1,\dots,x_n)=\begin{pmatrix}\nabla f_1(x_1,\dots,x_n)\\\nabla f_2(x_1,\dots,x_n)\\\vdots\\\nabla f_m(x_1,\dots,x_n)\end{pmatrix}.$$


问:
Finding primitive element for $\mathbb{Q}(\omega_3, \omega_7)$ and calculating a minimal polynomial.
Consider the field $\mathbb{Q}(\omega_3, \omega_7)$ with $\omega_n = e^{2 \pi i/n} = cos(\frac{2 \pi}{n}) + i sin(\frac{2 \pi}{n})$ the cyclotomic root of unity.

According to the theorem of the primitive element, if $\mathbb{F}$ is a field with characteristic 0 and there are 2 elements $a,b$ that are algebraic over the field $\mathbb{F}$, then there exists an $\alpha$ such that: $\mathbb{F}(a,b) = \mathbb{F}(\alpha)$.

a) Find a primitive element $\alpha$ for $\mathbb{Q}(\omega_3, \omega_7)$.

b) Calculate $[\mathbb{Q}(\omega_3, \omega_7): \mathbb{Q}]$.

c) What is the minimal polynomial of $\omega_7$ over $\mathbb{Q}(\omega_3)$.

For a) i tried to use a similar method as Primitive element of the extension $\mathbb Q(\sqrt{2},\sqrt{3})$ over $\mathbb Q$ . In other words I am trying to find an element $\alpha$ so that with simple calculations with $\alpha$, I can get $\omega_3$ and $\omega_7$ seperately.

$\alpha = \omega_3 + \omega_7$ hasn't worked for me.

$\alpha = \omega_3 \cdot \omega_7 = e^{\frac{10}{21}(2 \pi i)}$ looks like a good idea since then $\alpha^{21/70} = \omega_7$ and $\alpha^{21/30} = \omega_3$ but I'm not a 100% sure if this is correct. Since  using this alpha, for any $n \in \mathbb{N}_0$, it holds that $\alpha^{21/n*10} = \omega_n$ so my instinct says that the field $\mathbb{Q}(\omega_3, \omega_7)$ is contained in $\mathbb{Q}(\alpha)$ but not that they are equal.

So now I think maybe the 'simple calculation' has to be a linear combination for them to be equal, but I don't know what would work.

Can someone help me further finding $\alpha$.

For b) I want to use the fact that 3 and 7 are prime, I can use the product rule and the fact that $X^n-1$ is the minimal polynomial of $\mathbb{Q}$, and get that both 3 and 7 are divisors. Which gives me that it has to be 21.

For c) I think that $X^7-1$ will hold since that it is the minimal polynomial of $\omega_7$ over $\mathbb{Q}$ and since gcd(7,3) = 1, that means that the basis $\{ 1, \omega_3, \omega_3^2\}$ doesn't contain $\omega_7$.

Is my reasoning in b and c correct?

答:
For a): $\alpha=\omega_3\omega_7$ works as a primitive element but not for the reason you mentioned as we cannot take fractional powers in fields in general. We can argue as follows: Let $\beta=\alpha^{-2}=\exp(2\pi i/21)$. Then have $\omega_3=\beta^7,\omega_7=\beta^3$ and we get $\Bbb Q(\omega_3,\omega_7)\subseteq\Bbb Q(\alpha)$ and hence equality. Basically the reason here is that $10$ and $21$ are coprime, so $\alpha$ is a primitive $21$-th root of unity.

For b): This is not correct. The minimal polynomial of $\omega_3$ over $\Bbb Q$ is $x^2+x+1$, so $[\Bbb Q(\omega_3):\Bbb Q]=2$ and similarly for $\omega_7$. Since $\alpha$ is a primitive $21$-th root of unity we get $[\Bbb Q(\alpha):\Bbb Q]=\varphi(21)=12$ (this is not that trivial, see for example Lang's Algebra, chapter VI, 3 on 'Roots of unity').

For c): It is correct that the minimal polynomial of $\omega_7$ over $\Bbb Q$ (which is $x^6+x^5+\dots+x+1$) is the same as the minimal polynomial over $\Bbb Q(\omega_3)$. This follows from the fact that $\Bbb Q(\omega_3)$ and $\Bbb Q(\omega_7)$ are linear disjoint over $\Bbb Q$ (since $3$ and $7$ are coprime). We can see this for example by looking at the degree $$[\Bbb Q(\omega_3):\Bbb Q][\Bbb Q(\omega_7):\Bbb Q]=2\cdot 6=12=[\Bbb Q(\beta):\Bbb Q]=[\Bbb Q(\omega_3,\omega_7):\Bbb Q]$$


问:
let $\sum_{k=0}^{n}|f_{k}|^{2}=g$ then if g is constant then $\forall k\in[n].f_{k}$ is constant
let $\sum_{k=0}^{n}|f_{k}|^{2}=g$ and then if g is constant then if $\left\{ f_{k}\right\} _{k=1}^{n}$ is holomorphic then : $\forall k\in[n].f_{k}$  is constant.
My try:
Each of the $f_k$ is has composition to 2 harmonic function $u+iv$ therefore its 2nd derivative is zero.
therefore and because they have the same derivative they must be linear. but linear function accept constant in the entire plane iff they are constant function therefore $f_k must be constant.

答:
Assume that $g$ is constant. Let $\langle \cdot, \cdot \rangle$ be the usual Hermitian form on $\mathbb C^n$. Take any $z_0$ in the domain.  Let $f(z) = (f_1(z), \ldots, f_n(z))$.  Then $z \mapsto \langle f(z), f(z_0) \rangle$ is holomorphic and by the Schwarz inequality:

$$ \lvert \langle f(z), f(z_0) \rangle \rvert \leq g$$

with equality at $z=z_0$. By the maximum principle $\langle f(z), f(z_0) \rangle = g$ on the connected component of the domain that contains $z_0$.  And again by Schwarz (equality if and only if vectors are dependent) $f(z) = g(z) f(z_0)$ for some holomorphic $g$ with $\lvert g(z)\rvert = 1$. Then $g$ is constant and $f(z) = f(z_0)$.  So $f$ is itself constant.


问:
Is this piecewise-defined function on $\mathbb{R}^2$ continuous at $(0,0)$? What about differentiable?
Is this function is a differentiable function, a continuous function at the point $(0,0)$?
How to show that ?

$$
f(x,y)=\begin{cases}
\frac{x^{3}y+xy^{3}}{x^{2}+y^{4}} &\text{if }(x,y)\neq (0,0),\\
0 & \text{if }(x,y)=(0,0)
\end{cases}
$$

Thanks a lot!

答:
This is an unusual question, but a great one. @guru's method still applies. You should be able to argue continuity.

For differentiability, note that both partials at the origin are $0$. So you need to consider
$$\lim_{(x,y)\to (0,0} \frac {f(x,y)}{\sqrt{x^2+y^2}} \overset{?}= 0\,.$$
The polar coordinates approach should address this as well.


问:
Solving difference equation by using z-transform and z-transform confusion
Use the $z$-transform to solve the following difference equation:
$$y[n+2]+0.5y[n+1]+0.06y[n]=x[n]$$
where $x[n]=(-3)^{-n}u[n]-(-3)^{-n}u[n-2]$.

The first thing to notice is that $x[0]=1$, $x[1]=-\dfrac{1}{3}$, and the rest is all $0$. So, I can just apply the definition of the $z$-transform, namely:
$$Z\{x[n]\}=\sum_{n=0}^\infty x[n]z^{-n}=1-\dfrac{z^{-1}}{3}$$
which is what first confuses me, because when I calculate:
$$Z\{(-3)^{-n}u[n]-(-3)^{-n}u[n-2]\}=Z\{(-3)^{-n}u[n]\}-Z\{(-3)^{-n}u[n-2]\}=\dfrac{3z}{3z+1}-\dfrac{1}{9z^2+3z}\neq 1-\dfrac{z^{-1}}{3}$$
*My calculation of $Z\{(-3)^{-n}u[n-2]\}$ can be verified here.

So, both ways of getting the $z$-transform should yield the same transform (I believe); I don't know what mistake I am making, because I get different transforms. Does anyone know what I am doing wrong?

Thanks

答:
Regarding the fractions, first factor $3z$ from the second term

$$ \frac{3z}{3z+1}-\frac{1}{9z^2+3z}=\frac{3z}{3z+1}-\frac{1}{3z(3z+1)}$$

Now multiply the first by $1=3z/3z$
$$ =\frac{(3z)3z}{3z(3z+1)}-\frac{1}{3z(3z+1)}$$

Combine fractions, write numerator as difference of squares

$$ =\frac{9z^2-1}{3z(3z+1)}=\frac{(3z-1)(3z+1)}{3z(3z+1)}=\frac{3z-1}{3z}$$

Finally we have

$$ \frac{3z}{3z+1}-\frac{1}{9z^2+3z}=1-\frac{1}{3z} \tag*{$\blacksquare$}$$


答:
$2$  is an obvious root of $P(X)=X^5 − 2X^4 − 12X + 24$, hence $P(X)=(X-2)Q(X)$. Try to apply Eisenstein Criterion to $Q(X)$.

$\mathbf Z/2\mathbf Z$ has characteristic $2$, hence $\mathbf Z[X]$ too, so that $x\mapsto x^2$ is a ring homomorphism (the Frobenius homomorphism).

Title: Factorize certain polynomials as product of irreducibles

Body:
I'm trying to solve this question from my abstract algebra's course. I know it's a very trivial question that can be solved using Ruffini like we are taught before college, but my question is how can I solve it using "ring of polynomial" theory. The question goes like this:

Factorize the following polynomials as product of irreducibles:

$X^5 − 2X^4 − 12X + 24$ in $\mathbb{Z}[X]$

$X^4+X^2+1$ in $(\mathbb{Z}/2\mathbb{Z})[X]$

I know about ring theory and some ring of polynomials theory, but I don't see what can I do here to approach this kind of problem using abstract algebra methods (instead of Ruffini, because I guess I'm not supposed to solve it that way after all the theory I've studied, where this method is never introduced). How can I solve this? Any help will be appreciated, thanks in advance.


问:
Cartesian Products of Union is equal to union of cartesian products
I am looking at the proof of "Cavalieris Principle" in my textbook.
We have two $\sigma$-finite measure spaces $(X,\mathcal{E},\mu),(Y,\mathcal{F},\mathcal{v})$. Since they are $\sigma$-finite we can choose sequences $(A_n)$ and $(B_n)$ from $\mathcal{E}$ and $\mathcal{F}$ respectively such that
$$
\cup_{n \in \mathbb{N}} A_n =X, \quad \cup_{n \in \mathbb{N}} B_n =Y, \quad \mu(A_n),\mathcal{v}(B_n)<\infty $$
If we define $H_n = A_n \times B_n$ for all n, will we then have that $\cup_{n \in \mathbb{N}} H_n =X \times Y$ no matter what, or do we have to assume anything about the sequences? E.g. do $(A_n),(B_n)$ have to be increasing sequences, which I know we can choose since the measure spaces are $\sigma$-finite?

答:
You need to make assumptions about the sequences of the sets.

Let's just look at these: $A_1 = B_2 = [0, 1]$ and all other elements of the sequences are empty sets.

Then all $H_n = \emptyset$ (and so is their union) but we have $X \times Y = [0, 1]^2$.


问:
Converting between different ways of representing integer partitions
I'm aware of two ways of representing integer partitions, and I'm trying to understand how to write the mapping between then in terms of mathematical notation.

Firstly, we can say that an integer parition of $n$ is a sequence $\lambda = (\lambda_1,\ldots,\lambda_k)$ such that $\lambda_i \in \mathbb{N}_{+}$, $\lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_k$ and $\sum_{i = 1}^k \lambda_i = n$.

Or equivalently, we can say that an integer partition of $n$ is an $n$-tuple $\eta \in \{0,1,...,n\}^n$ such that $\sum_{i = 1}^n i \,\eta_i = n.$

I understand both definitions fine, and I've found it straightforward to write an algorithm in Mathematica to convert from one to the other. Starting with $\lambda$, I count the number of each element of size $i$ from $1$ up to $n$ and assign the value to $\eta_i$. Starting from $\eta$, I count $i$ down from $n$ to $1$ and put $\eta_i$ of the number $i$ into the sequence $\lambda.$

For the application I'm currently working with, it's more natural to use the second representation. I've recently been directed to findstat.org for a list of possible properties of integer paritions, however they use the first representation. I don't understand how to write the operations that I use programatically to map between the two representations in terms of mathematical notation, can someone please help to show me how to write down this mapping?

答:
We can provide a mapping by introducing an intermediate step taking the ring of formal power series $\mathbb{Z}[[X]]$ with integer coefficients.

We consider the general case and as small example a partition of $11$:
\begin{align*}
4+3+3+1=11
\end{align*}

We have a sequence $\lambda\in\mathbb{N}^k$:
\begin{align*}
&\lambda=\left(\lambda_i\right)_{1\leq i\leq k}&\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_k&\qquad\qquad\\
&&\sum_{i=1}^k\lambda_i=n\qquad\qquad&\qquad\qquad\\
\\
&\lambda=(4,3,3,1)&4\geq 3\geq 3\geq 1\qquad&\qquad\qquad\\
&&4+3+3+1=11&\qquad\qquad
\end{align*}
and a sequence $\eta\in\{0,1,\ldots,n\}^n$:
\begin{align*}
&\eta=\left(\eta_j\right)_{1\leq j\leq n}&\sum_{j=1}^nj\eta_j=n\quad\qquad\qquad&\qquad\qquad\\
\\
&\eta=(1,0,2,1,0,0,0,0,0,0,0)&\qquad1\cdot1+2\cdot 3+1\cdot 4 = 11&\qquad\qquad\\
\end{align*}

We define a mapping $\phi$:
\begin{align*}
&\phi: \mathbb{N}^k\rightarrow \mathbb{Z}[[X]]\\
&\phi(\lambda)=\phi\left(\left(\lambda_i\right)_{1\leq i\leq k}\right)=\sum_{i=1}^kx^{\lambda_i}\\
\\
&\phi: \mathbb{N}^4\rightarrow \mathbb{Z}[[X]]\\
&\phi(\lambda)=\phi\left(\color{blue}{\left(4,3,3,1\right)}\right)=x^4+x^3+x^3+x^1\tag{1}\\
&\quad\qquad\qquad\qquad\qquad=x+2x^3+x^4\tag{2}\\
\end{align*}

We observe the power series representation shows the structure of $\lambda$ in the representation (1). But we also have the structure of $\eta$ when looking at the coefficients of representation (2). In the following it is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We define a mapping $\psi$:
\begin{align*}
&\psi:\mathbb{Z}[[X]]\rightarrow \mathbb{N}^n\\
&\psi\left(L(x)\right)=\left([x^i]L(x)\right)_{1\leq i\leq n}\\
\\
&\psi: \mathbb{Z}[[X]]\rightarrow \mathbb{N}^{11}\\
&\psi\left(x+2x^3+x^4\right)=\left([x^i]\left(x+2x^3+x^4\right)\right)_{1\leq i\leq 11}\\
&\qquad\qquad\qquad\qquad=\color{blue}{\left(1,0,2,1,0,0,0,0,0,0,0\right)}\tag{3}
\end{align*}

We conclude from (1) - (3) the mapping $\phi\circ\psi$:
\begin{align*}
&\color{blue}{\phi\circ\psi: \mathbb{N}^k\rightarrow \mathbb{N}^n}\\
&\color{blue}{\left(\phi\circ\psi\right)\left(\left(\lambda_i\right)_{1\leq i\leq k}\right)
\rightarrow \left([x^j]\sum_{i=1}^kx^{\lambda_i}\right)_{1\leq j\leq n}}
\end{align*}
does the job.


问:
How to use "results from partial fractions"?
Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies

$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$

The denominator can be factored into $(1-2x)(1+x)^2$. Using results from partial fractions, it can be shown that there exists constants $C_1,C_2,C_3$ such that

$$A(x)=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$$

Determine these constants and find $a_5$ using this new expression.

What exactly are "results from partial fractions" and how should I use them?

答:
The main conclusion that drives the method of partial fractions is that any rational function of the form $\rm p(x)/q(x)$ where $\rm q$ factors into irreducibles as $\rm q(x)=w_1(x)^{e_1}\cdots w_n(x)^{e_n}$ can be represented in either of the equivalent forms

$$\rm \frac{p(x)}{q(x)}=a_0(x)+\frac{a_1(x)}{w_1(x)^{e_1}}+\cdots+\frac{a_n(x)}{w_n(x)^{e_n}}$$

$$\rm =a_0(x)+\left[\frac{b_{11}}{w_1(x)}+\frac{b_{12}}{w_1(x)^2}+\cdots+\frac{b_{1e_1}}{w_1(x)^{e_1}}\right]+\cdots+\left[\frac{b_{n1}}{w_n(x)}+\frac{b_{n2}}{w_n(x)^2}+\cdots+\frac{b_{ne_n}}{w_n(x)^{e_n}}\right]$$

for some polynomials $\rm a_i(x)$ with $\rm \deg a_i(x)<e_i\deg w_i(x)$ for each $\rm i=1,\cdots,n$ and constants $\rm b_{ij}$.

To use this theorem in practice, you set the desired equality and then solve for the unknown coefficients using the appropriate techniques from linear algebra.

Given some abstract algebra and field theory as prerequisites, there is a very nice algebraic way of phrasing this theorem. The connection is similar to how a low-tech version of Sun-Ze (better known as the Chinese Remainder Theorem) describes existence and uniqueness of solutions to systems of congruences and the algebraic version describes a ring decomposition.

Let $F$ be a field and $T$ an indeterminate. Let $\mathsf{Irr}$ be the set of monic irreducibles in $F[T]$. We denote by $F(T)_\pi$ and $F[T]_\pi$ the ring completions with respect to the $(\pi(T))$-adic topology. Then

$$\frac{F(T)}{F[T]}\cong\bigoplus_{\pi\in\mathsf{Irr}}\frac{F(T)_\pi}{F[T]_\pi}$$

as additive quotient groups. There is also a number field analogue (specializing to $\bf Q$):

$$\frac{\bf Q}{\bf Z}\cong\bigoplus_{p~\text{prime}}\frac{{\bf Q}_p}{{\bf Z}_p},$$

which amounts to the $p$-primary decomposition of it as an abelian group; the direct summands are called the Prufer $p$-groups ${\bf Z}_{p^\infty}$ and can be alternately characterized as ${\bf Z}[p^{-1}]/{\bf Z}=\varinjlim{\bf Z}/p^n{\bf Z}$.


问:
How to find the parametric vector $r(u,v)$?
find the  area of that portion of the paraboloid $x^2 + z^2 = 2ay$ which is cut off by the
plane $y = a$

My Question : How to find the parametric vector  $r(u,v)$  of the given paraboloid cut off by the plane $y = a$?

My attempt :I was thinking about $x = r \cos θ$, $y = r \sin θ$ but it doesn't work

答:
It is the circle $x^2+z^2=2a^2$ in the plane $y=a$

$\left(a\sqrt{2} \cos t,a,a\sqrt{2} \sin t\right);\;t\in [0,2\pi]$

It's area is $2a^2\pi$


问:
Show $|A| = |B| = n$ are normal
Suppose $G$ is a finite group, $A,B$ are subgroups of $G$ such that they are the only subgroups of order $n \in \mathbb{Z}^+$.

Suppose $X$ is the set of subgroups of $G$ such that $G$ acts on $X$ by conjugation of subgroups, i.e.
$g\cdot K = \{gkg^{−1} : k \in K\}$, if $g \in G$ and $K \in X$.

Suppose that $|G|$ is not even. Prove that $A$ and $B$ are both normal in $G$.

I have worked out that if $K \in X$ then, if $A \in {\rm Orb}_G(K)$ then $|A| = |K|$.

I want to use the fact that $|{\rm Orb}_G(K)| = {|G|\over|{\rm Stab}_G(K)|}$ from the Orbit-Stabiliser Theorem but not sure where to go with it.

答: As you say, $A \in Orb_G(K)$ implies $|A| = |K|$. Therefore $|Orb_G(A)|$ is either equal to $1$ or $2$, since $B$ is the only other subgroup which could be in the orbit of $A$. If $|Orb_G(A)| = 1$ then $gAg^{-1} = A$ for all $g \in G$ so $A$ is normal, and $B$ must be too since if $B$ is not in the orbit of $A$ then $A$ is not in the orbit of $B$. However, if $|G|$ is odd then $|Orb_G(A)|$ cannot be equal to $2$ because that would imply $|Stab_G(A)| = \frac{|G|}{2}$.

Title: Show $|A| = |B| = n$ are normal


问:
Prove one of these linear transformations doesn't exist
I recently came across this problem:

Let $V$ and $W$ be the subspaces
$$V=\mathrm{Lin}\left\{\left(\begin{array}{r} 1\\ 0\\ -1\end{array}\right),\left(\begin{array}{r} 1\\ -1\\ 0\end{array}\right)\right\}\subset\mathbb{R}^3,$$
$$W=\mathrm{Lin}\left\{\left(\begin{array}{r} 1\\ 3\\ 1\\0 \end{array}\right),\left(\begin{array}{r} -2\\ 5\\ 0\\ 1\end{array}\right)\right\}\subset\mathbb{R}^4.$$
Consider the possibility of a linear transformation $T$ and a linear transformation $S$ such that
$$T:\mathbb{R}^a\longrightarrow\mathbb{R}^b\ \text{with}\ N(T)=V\ \text{and}\ R(T)=W;$$
$$S:\mathbb{R}^c\longrightarrow\mathbb{R}^d\ \text{with}\ N(S)=W\ \text{and}\ R(S)=V.$$
Show that one of these, $S$ or $T$, cannot exists. Then find a matrix $A_S$ or $A_T$ representing the other linear transformation (with respect to the standard basis in each Euclidean space).

Check you answer by row reducing the matrix and finding the null space and range of the linear transformation.

And can't find an actual solution.

$A_T$ must be a $b\times a$ matrix since for some $x_{a\times 1}$ we have $A_T{_{b\times a}}x_{a\times1} = [x]_T{_{b\times1}}$. Then I applied the same reasoning on $S$. $A_S$ must be a $d\times c$ matrix since for some $x$ that is $c\times 1$ we have $A_S{_{d\times c}}x_{c\times1} = [x]_S{_{d\times1}}$. I thought of using the rank-nullity theorem to find $a$, $b$, $c$ and $d$.

$\dim(N(T)) + \dim(R(T)) = \dim(R^{a}) \implies 2 + 2 = a \implies a = 4$, so $A_T{_{b\times4}}$ and
$\dim(N(S)) + \dim(R(S)) = \dim(R^{c}) \implies 2 + 2 = c \implies c = 4$, so $A_S{_{d\times4}}$

but I can't think of more operations to prove that $S$ and $T$ can't coexist.

答:
Theorem. If $T\in\mathcal{L}(V_1,V_2)$ and $V_1$ is finite-dimensional then $\dim V_1=\dim(N(T))+\dim(R(T))$.

Clearly $\dim V=\dim W=2$. If $N(T)=V$ then $V\subset\mathbb{R}^a$ so $a=3$. If that were the case, we would have
$$3=\dim\mathbb{R}^3=\dim N(T)+\dim R(T)=2+2=4,$$
which is not possible. Then, no such $T$ can exists.

So $S$ exists, and by the argument above must be $S:\mathbb{R}^4\longrightarrow\mathbb{R}^3$. For constructing such $S$, you can choose two more (independent) vectors to complete a basis for $\mathbb{R}^4$. For example:
$$\mathcal{B}=\left\{\left(\begin{array}{r} 1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{r} 0\\ 1\\ 0\\ 0\end{array}\right),\left(\begin{array}{r} 1\\ 3\\ 1\\0 \end{array}\right),\left(\begin{array}{r} -2\\ 5\\ 0\\ 1\end{array}\right)\right\}.$$
And you can set
$$S\left(\begin{array}{r} 1\\ 0\\ 0\\ 0\end{array}\right)=\left(\begin{array}{r} 1\\ 0\\ -1\end{array}\right),\ S\left(\begin{array}{r} 0\\ 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{r} 1\\ -1\\ 0\end{array}\right),\ S\left(\begin{array}{r} 1\\ 3\\ 1\\0\end{array}\right)=S\left(\begin{array}{r} -2\\ 5\\ 0\\ 1\end{array}\right)=\left(\begin{array}{r} 0\\ 0\\ 0\end{array}\right).$$
Now you have the desired $S$, but you need to apply a change of basis which I left for you.


问:
Every ring homomorphism to the integers is surjective?
Let $R \neq 0$ be a commutative, unital ring and $f: R \to \mathbb{Z}$ be a ring homomorphism.

Does this force $f$ to be surjective?

Attempt of Proof:
By the universal property of $\mathbb{Z}$, there is a unique ring homomorphism $g: \mathbb{Z} \to R$. But then $f \circ g: \mathbb{Z} \to \mathbb{Z}$ is the unique ring homomorphism which is already given by $id: \mathbb{Z} \to \mathbb{Z}$. Thus $f \circ g = id$ and therefore $f$ must be surjective.

Does this proof hold? It seems weird that I can nowhere find this result...

答:
If $f$ is a ring homomorphism, then for the unit elements $f(1_R) = 1_Z$.

For completeness define of multiples of ring elements (unit element is only require):

For each natural number $n$, write $n\cdot 1_R = 1_R+\ldots+1_R$ $n$-times and for each negative integer $n = -m$, where $m>0$, write $n\cdot 1_R = -(m\cdot 1_R)$.

Then for each integer $n$, $f(n\cdot 1_R) = n\cdot f(1_R) = n\cdot 1_Z$.
It follows immediately that the mapping is onto.


问:
Math behind Maximum-Product in Rod-Cutting Problem
I am self-learning Data Structures and Algorithms and I came across a problem recently.

Given a rod of length n, find the optimal way to cut the rod into smaller rods in order to maximize product of profit of each of the smaller rod. Assume each rod of length i has profit i.

For example, consider a rod of length $4$. The maximum product will be $4$ ( $=2\times2$ or not cutting it ).

For another example, a rod of length $6$ has maximum product $9$ ( $=3\times3$ ).

I have designed my program and successfully solved this problem. When I looked into the problem more, I found a pattern:

The maximum product can be obtained be repeatedly cutting parts of length $3$ while length is greater than $4$, keeping the last part as length of $2$, $3$ or $4$.

For example,

Length $14$: maximum product $= 3\times3\times3\times3\times\color{blue}{2}=162$.

Length $15$: maximum product $= 3\times3\times3\times3\times\color{blue}{3}=243$.

Length $16$: maximum product $= 3\times3\times3\times3\times\color{blue}{4}=324$.

This fact is interesting, but I'm not able to prove it mathematically. Any suggestions, hints or answers will be appreciated. Thanks for reading this post.

Source: https://www.techiedelight.com/maximum-product-rod-cutting/

答:
This problem is essentially IMO $1976/P4$.

The idea is that using maximum number of $3$'s is most efficient.
Since larger numbers can be broken down into smaller ones, keeping the sum same but increasing the product.

For example, if any part is $5$ it can be broken down into a $3$ and $2$ increasing the product contribution from $5$ to $6$. A $6$ into two $3$'s - product contribution : $6$ to $9$. A $7$ into $3,4$ or $3,2,2$ - product contribution : $7$ to $12$.

The reason is simple inequality :
$$xy \ge (x+y)\cdot 1 \iff (x-1)(y-1)\ge 1$$

That is, it is beneficial to convert a single part $(x+y)$ into two parts $x$ and $y$ as long as $x,y \ge 2$. Now for size $6=2+2+2=3+3$. But $$3\times 3 > 2 \times 2 \times 2$$

So we observe use of maximum $3$'s is optimal.

For $n \equiv 0 \pmod 3$, all parts should be $3$'s. For $n \equiv 1 \pmod 3$,  $n \equiv 2 \pmod 3$ all $3$'s except for one $4$ and one $2$ can be used respectively.


答: Then it is continuous at 0 through similar reasoning. Don't know about differentiability it is a bit tedious so I didnt do it you have to work it out in a similar way.</p>

Title: Is this piecewise-defined function on $\mathbb{R}^2$ continuous at $(0,0)$? What about differentiable?

Body:
Is this function is a differentiable function, a continuous function at the point $(0,0)$?
How to show that ?

$$
f(x,y)=\begin{cases}
\frac{x^{3}y+xy^{3}}{x^{2}+y^{4}} &\text{if }(x,y)\neq (0,0),\\
0 & \text{if }(x,y)=(0,0)
\end{cases}
$$

Thanks a lot!


问:
For what $a$ the sequence $2x_{n+1}=x^2_n+a$ has finite limit?
Let $(x_n)$ be the sequence defined recursively by the following formular $$x_0=0, \ 2x_{n+1}=x^2_n+a$$
Find $a$ such that the above sequence has finite limit?

With all of my effort, there is still one case left, say, $a<-3$, in which I don't even know how to prove or disprove the existence of the limit.

答:
The limit of the sequence, if it exists, is a fixed point of $g(x)= \frac 12 x^2 +\frac{a}{2}$. Assuming that we are speaking of sequences of real numbers, $g$ only has fixed points if $a\leq 1$, and those fixed points are $1 \pm \sqrt{1-a}$.

Numerically, it seems that the convergence is attained only if $a \leq 1$ and $a > -3$ (aprox.). I used as a convergence criteria that the difference between consecutive iterations becomes less than $10^{-6}$ in less that 1000 iterations.

This is consistent with the fact that the derivative of $g$ on the stable fixed point (one is always unstable) is less then one (in absolute value) if and only if $a \in ]-3,1]$.


问:
Vector bundles of rank $q+k$ in $K(Y)$, where $Y$ is a projective scheme of dimension $q$ over some field.

Given a vector bundle $G$ of rank $q+k$ on a projective scheme $Y$ of dim $q$ over some field,show that there exists an injection $0\rightarrow O_Y^k\rightarrow G(n)$ for some sufficiently large $n$.

I met this question when considering $[G(n)]$ in $K(Y)$.Thanks for your help.

答:
The comment for @random123 's answer is too long so I write it as another answer here.

$B := \lbrace (x, v_1 \wedge \dots \wedge v_p) \in Y \times \mathbb{P}(V_n) \big| \text{ such that } \varphi_x(v_1 \wedge \dots \wedge v_p) = 0 \rbrace$ might be
$B := \lbrace (x, v) \in Y \times \mathbb{P}(\wedge^p V_n) \big| \text{ such that } \varphi_x(v) = 0 \rbrace$,where $v=\sum_ia_iv_{(1,i)}\wedge \dots \wedge v_{(p,i)}$ and the $a_i$'s are the homogenous coordinates of $v\in \mathbb{P}(\wedge^p V_n)$

In the last paragraph of the answer "We get that there exists $v_1, v_2, \dots, v_p \in V_n$ such that $\varphi_x(v_1 \wedge \dots \wedge v_p) \neq 0$ for any $x \in X(k)$",I think the following statement is implicitly used:

Any non-empty open dense $U\subseteq \mathbb{P}(\wedge^p V_n)$ contains some $w_1\wedge\dots \wedge w_p$.


问:
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.

答:
The Quotient ring ${Z_2[x]/(x^4+x^2+x)}$ contains below elements.$${\{1,x,x^2,x^3,x^2+x,x^3+x^2,x^3+x^2+x,x+x^3\}}$$ if we square each of them, then we can see that for ${(x^2+x)^2=x}$, so  ${\sqrt{x}=(x^2+x)=p }$(as defined above by Teresa)

now ${(x^3+x+1)^2= x^6+x^2+1= x^3+x^2+x+x^2+1=x^3+x+1}$


答:
It's evident that if the sequence has a limit, this limit must satisfy the following equation:
$$2t=t^2 +a$$
and then if the sequence converge, it should converge to one of the 2 fixed point $p_1=1 +\sqrt{1-a}$,$p_2=1 -\sqrt{1-a}$.

Let's denote function $f(x)=\frac{1}{2}(x^2+a)$ with $a<-3$  (and so $x_{n-1}=f(x_n)$), we have
$$f'(p_1)=1 +\sqrt{1-a} >1$$
$$f'(p_1)=1 -\sqrt{1-a}<-1$$
then $|f'(p_{1,2})|>1$, according to this post, the both fixed points $p_1$ and $p_2$ are unstable. Because $x_0 \notin \{\pm p_{1,2}\}$ (4 values $p_1,p2,-p_1,-p_2$), then the sequence will never be able to hit the fixed points and so doesn't converge.

Conclusion: the sequence $x_n$ doesn't converge for $a<-3$.

PS: why these conditions $x_0 \notin \{\pm p_{1,2}\}$  are enough to conclude the divergence of the sequence? Let's prove by contractiction. Suppose $k$ is the smallest value that $x_k \in \{\pm p_{1,2}\}$. Then we have $f(x_{k-1})=f(p)$ with $p \in \{\pm p_{1,2}\}$. But this implies that $\frac{1}{2}(x_{k-1}^2 +a)= \frac{1}{2}(p^2+a)$ or $x_{k-1} \in \{\pm p_{1,2}\}$, hence $k-1$ which is smaller than $k$ is also satisfy the assumption(=> contradiction).

Title: For what $a$ the sequence $2x_{n+1}=x^2_n+a$ has finite limit?

Body:
Let $(x_n)$ be the sequence defined recursively by the following formular $$x_0=0, \ 2x_{n+1}=x^2_n+a$$
Find $a$ such that the above sequence has finite limit?

With all of my effort, there is still one case left, say, $a<-3$, in which I don't even know how to prove or disprove the existence of the limit.


问:
Counting subsets in a set
Twelve philosophers sit around a table. In front of everyone of them there is a plate, between every two plates there is a fork and in the middle of the table there's a big bowl of spaghetti. Their time passes by in the following manner: they think, become hungry and reach for the spaghetti. But everyone of them needs two forks to eat. It's obvious that two philosophers, who are side by side, cannot eat at the same time. There is also a maximum number of people, who can eat at the same time - 6. What is the count of all 5-philosopher subsets, whose members can eat at the same time?

Any help would be greatly appreciated. I struggle to find any solution.

答: With 6 philosophers eating, there are six gaps, each of size 1.  With 5 philosophers eating, there are 5 gaps, of total size 7.  Hence there can be four gaps of size 1 and one gap of size 3, or three gaps of size 1 and two gaps of size 2.  This should get you started.</p>

Title: Counting subsets in a set


问:
How to intuitively understand the concept of Hand, Deck , Card and exponential family
I am reading Generating Functionology by Herbert S. Wilf in around page -75, he defines the following objects:

Definition. A card C(S, p) is a pair consisting of a finite set S (the ‘label
set’) of positive integers, and a picture $p \in P$. The weight of C is $n = |S|$.
A card of weight n is called standard if its label set is $[n]$.*

Definition. The weight of a hand is the sum of the weights of the cards
in the hand.

Definition. A relabeling of a card C(S, p) with a set S0 is defined if $|S| =|S
'|$, and it is the card $C(S', p)$. If $S0 = [|S|] $then we have the standard
relabeling of the card.

Definition. A deck $D $is a finite set of standard cards whose weights are
all the same and whose pictures are all different. The weight of the deck is
the common weight of all of the cards in the deck.

Definition. An exponential family F is a collection of decks $D_1, D_2, . . .$
where for each $n = 1, 2, . . ., $the deck $D_n$ is of weight $n.$

These were a lot of things defined at once, and I feel it quite hard to digest them. I seek an answer which gives a mental model for thinking about the objects defined, like a simple intuitive view of all of them.

I think the particular point troubling me after reading this stack is about the concept of label sets

答:
Introduction

The most fundamental connection between the fields of analysis and combinatorics is the notion of a generating function. The fact that counts of certain combinatorial sets can become coefficients of a series is important , because on one side lies the discreteness of set counting techniques, and on the other side lies the continuous notions of differentiation/integration, and we can swiftly move between these notions by considering the generating function.

The concepts presented, of a card, hand, deck, and an exponential family are meant to capture an entire counting problem.

A counting problem is the problem of counting the number of elements of a set, given a description of that set in some way. For example, consider the set of subsets of $[n]$ of size $k\leq n$. A counting problem would be : find this number for all $n$ and $k \leq n$.  The answer is $\frac{n!}{(n-k)!k!}$ for a given $n$ and $k$, but to give the answer for every $n,k$ all at once I can just give you a generating function, and you just find the coefficient of some term depending on $n$ and $k$, you will land up with the answer for that $n$ and $k$. Of course, estimates, asymptotics and other properties of the count sequence will be easy to analyze once you have the generating function.

However, every counting problem boils down to counting the sets consisting of building blocks. This is because, if you can count the building block sets, then every other set can be built out of them, and then using unions, intersections, cartesian products etc. you can easily find the size of another set.

Now, roughly this is the break-up :

Card : A fundamental building block.

Hand : A set built out of cards.

Deck : A set of building blocks, in which each element shares something in common with the others.

Exponential family : The entire set of building blocks, given the entire set of decks.

Card

A card is essentially a set and a picture. Now what is this picture, and what is this label? Essentially, the label is a subset of positive numbers, so you can think of the label as being some set of numbers to associate to the card.

What about the picture? The picture is just a way to associate the labels together to create the ultimate object we want to count. For example, if we want to count graphs, then a label set would label the vertices, but how they are connected can only be specified by a picture. Similarly, if we were counting words, then a label set would tells us what letters are used, but what order they come in is down to a picture.

So, the object that is represented by a card, is a set of numbers and then how the object is created by patching that set of numbers together. The word picture is deliberately used in an abstract sense, because the object we want to count could be very abstract itself! (The number of ways I wish to dress my doll up for a kitty party with the dormouse)

Now, it turns out that a card should be fundamental, much like an element is to chemistry : it should not be "breakable" into simpler mathematical objects that aren't cards already.

The card is then meant to be the building block for the objects that we ultimately wish to count : essentially, our objects of interest are built out of cards. Much like a hand is made of playing cards, which brings us to

Hand

A hand is the mathematical object we want to count. It is created by taking different appropriately cards together, so that the pictures combine and give us a picture of the hand.

The reason for calling it a hand is from the fact that in card games, a set of cards is kept in ones hand during the game, so we call it a hand.

Now, the starting point of solving a counting problem is expressing every object you want to count, as a hand. For example, for counting permutations, you know that a permutation can be written as a product of disjoint cycles (which commute with each other, so there's no order involved in the product), therefore a permutation is hand where each card is a cyclic permutation. Similarly, every finite graph is a disjoint graph-union of connected graphs, so a finite graph is a hand with each card being a connected graph.

So we are counting hands : a hand is made by a mix-and-match of various cards of different types. So it is only fair that we combine the cards of various types together. Which brings us to a

Deck

A deck is a collection of cards that share the same type.

What is the fundamental reason for putting these together? It is because sizes of sub-collections share an obvious relation to the collection which they create by free-choice : by multiplication.

To give an example, say I'm dressing my doll for a party! I have four hair styles to choose from (though I prefer letting hers loose, but that's out because it's not allowed in this party), three skirts (red,red and red in color, but different shades of red. Also every skirt comes with a top as well!) and two shoes (both heeled, she wants to impress her doll friend). Knowing that her ultimate look is a mix-and-match of a hair style, a skirt and a shoe, it is important that I sort the shoes, skirts and hair styles (abstractly) into different piles and count them differently so I can find the number of ways in which I can dress my doll.

So counting the fundamentals , with a mix-and-match means I'm counting what I want eventually. But sorting the fundamentals is important, which is what a deck is doing. A set of decks together creates the entire setting of the problem ,the

Exponential family

An exponential family is a family of decks, so basically every single deck available to you. Note that this completely describes all the fundamental building blocks of the objects you want to count.

For example, knowing how to dress my doll means I need to know everything about what I have to dress her up with : that's the exponential family for that problem.

Thus, at the outset of any counting problem is the exponential family, the notions are near inseparable.

How it all fits together

So to solve a counting problem (from scratch):

Think about what a card is for the problem : the ultimate fundamental object which combines and creates every kind of object you want to count. This really depends on the situation : for example, for permutations we have cycles, but for bipartite graphs, a card is actually a $2$-colored connected labeled bipartite graph (explained on page $88$).

Put the cards into decks : this is fairly natural, in the sense that cards from the same deck can't be mixed.

Creating the exponential family by putting the decks you need together. Sometimes you won't need certain decks : for example, if you choose to count permutations with some property which don't have length bigger than $30$, you can discard decks whose weight is bigger than $30$.

Now, you can count what you want, the setup is made.

Reasons for introducing these terms

There are a variety of reasons why we choose to begin with a high amount of generality. Some of these are :

The fact that we can choose to move to a specific problem, by making it a sub-instance of a larger problem, is a great advantage because we keep the framework while changing the
problem. For example, counting permutations and counting permutations of size less than $30$ both come under this framework, where for the second case you can throw out all decks of size more than $30$.

The fact that within the framework, it is not only the hands we can count now. We can count the deck sizes if we know the size of the hands as well! (Using an inversion formula)

Last but not the least, the framework is flexible in the sense that a hand is picked to be a decomposition of $[n]$, but there might be an order to picking the cards as well , so next you could define a hand by not just a set of cards but rather an ordered list of cards. Then the notion of counting changes, but the framework changes ever so slightly to accommodate it. Beginning in generality, means you don't have to stretch to accommodate even far-off looking counting problems.

Short note on this last point : a hand in the given case is formed by putting together a set of cards. If the order of the cards mattered, we would not study $\mathcal{H}(x,y)$ by that expression, but rather by a different expression. The point is, the framework still accommodates all of this.


问:
$X$ and $Y$ independent random variables such that $X$ and $X+Y$ be independent ($X$ not constant)
Question :

Let's $X$ and $Y$ be two random variables such that $X$ and $Y$ are independent, and $X$ not constant.

Can $X$ and $X+Y$ be independent? If yes, when?

($X$ and $Y$ don't necessarily have an expected value)

My attempt :

$X$ and $X+Y$ independent would mean for all $x\in X(\Omega)$ and $z\in (X+Y)(\Omega)$, $P(X=x,X+Y=z)=P(X=x)P(X+Y=z)$ and we know (by independence of $X$ and $Y$) that $P(X=x,X+Y=z)=P(X=x)P(Y=z-x)$.

So we must have, for all $x$ such that $P(X=x)\neq 0$,
$$P(X+Y=z)=P(Y=z-x)$$
And $P(X+Y=z) = \sum\limits_{x\in X(\Omega), y\in Y(\Omega), \ P(X=x)\neq 0 \text{ and } x+y=z} P(X=x)P(Y=z-x)$.
So if $P(X+Y=z)\neq 0$, $\sum\limits_{x\in X(\Omega), y\in Y(\Omega), \ P(X=x)\neq 0 \text{ and } x+y=z} P(X=x) = 1$.

So for all $z$ such that $P(X+Y=z)\neq 0$ :
$$\forall x\in X(\Omega), P(X=x)\neq 0, \exists y\in Y(\Omega), x+y=z$$

But then I don't really know what it implies, and can't find an example.

Can someone help me?

Edit :

Mrsami has given a beautiful answer in the case where $X$ has finite second moments. But the case where this is not true is still not solved.
Every one that has an idea to tackle it (with basic tools if possible) is welcome.

答:
Suppose $X$ and $X+Y$ are independent. Then
$$
E(X(X+Y))=E(X)E(X+Y)=(EX)^2+E(X)E(Y)
$$

On the other hand
$$
E(X(X+Y))=E(X^2+XY)=E(X^2)+E(XY)=E(X^2)+E(X)E(Y)
$$

Necessarily, one must have
$$
(EX)^2=E(X^2)
$$
i.e.,
$$
\textrm{Var}(X)=E((X-EX)^2)=E(X^2)-(EX)^2=0
$$
which implies that $X$ is a constant (almost surely). A contradiction.

Notes.

This answer assumes all the random variables have finite second moments (and hence are absolutely integrable). In general, pathological (such as not absolutely integrable) random variables are handled in an ad hoc way. I believe you will need to search among the usual pathological examples such as Cauchy distributions. But I don't have such an example off-hand.


问:
Gradient direction / descent
it was a while ago I read multivariable calculus so I need to refresh certain results.

Given $ f:R^n\to R $, at a local stationary point $ x $ the gradient is $ \nabla f(x) = 0 $. However, given the fact that the gradient points at the direction which $f$ increases the most, how come the gradient is zero at a local minima?

Also, about Gradient descent we use that fact to find a local minima, as per saying that if $ \nabla f$ points in the direction with maximum increase, $-\nabla f $ points in the direction of maximum decrease.

How is that equivalent?

答:
You are right in saying that gradient points in direction where $\nabla f$ increases the most and when $f(x)$ is decreasing we have that $\nabla f(x)$ is negative.

A $\nabla f(x) = 0$ at a local minima or a local $\bf{maxima}$ (or an inflection, but we can ignore it for now)!

Why does gradient descent take us to local minima ?

Well because gradient descent is pushing in the direction of $-\nabla f(x)$ !!

$$a_{n+1} = a_n + \lambda (-\nabla f(x))$$

Your each subsequent step $a_{n+1}$ is $\lambda$  sized stride in direction opposite of the steepest increase or in other words in the direction of steepest decrease.

Title: Gradient direction / descent


问:
$\mathcal{B}(\mathbb{R})$-measurable vs. $\mathcal{B}(\overline{\mathbb{R}})$-measurable
Let $(E,\mathcal{E})$ a measurable space and $f \colon E \to \mathbb{R}$ a function. Is it right, that $f$ is $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable iff $i \circ f$ is $\mathcal{E}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable, where $i$ is the inclusion $\mathbb{R} \to \overline{\mathbb{R}}$? ($\mathcal{B}(\mathbb{R}) $ is the Borel $\sigma$-Algebra)

答:
Yes, it is true.

If $f$ is $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable, then, for all $E \in  \mathcal{B}(\overline{\mathbb{R}})$,
$$ (i \circ f)^{-1}(E) =f^{-1}(i^{-1}(E))= f^{-1}(E \cap \Bbb R) \in \mathcal{E}$$
(because $E \cap \Bbb R \in \mathcal{B}(\mathbb{R})$). So, $i \circ f$ is $\mathcal{E}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable

If $i \circ f$ is $\mathcal{E}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable, then, for all $A \in  \mathcal{B}(\mathbb{R})$, we have that $A \in  \mathcal{B}(\overline{\mathbb{R}})$ and $i^{-1}(A) =A$. So
$$ f^{-1}(A) = f^{-1}(i^{-1}(A))= (i \circ f)^{-1}(A)\in \mathcal{E}$$
So, $f$ is $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable


问:
If $\phi(a)$ is invertible then $a$ is invertible
Let $R$ and $S$ be rings and $\phi: R \to S$ is a surjective ring homomorphism.

If $\phi(a)$ is invertible then $a$ is invertible

Is this correct? We know $\phi(1_R)=1_S$ but I am confused thats what I tried:

Let $\phi(a)$ be invertible and its inverse be the $\phi(b)$. Then:

$\phi(a)\phi(b)=1_S$

$\phi(ab)=1_S$

I am stuck here. Can we say $ab=1_R$? I think we cant but I couldnt find any counterexample.

答: Hint: Let $p$ be a prime number and $\mathbb{Z}\rightarrow \mathbb{Z}/p$ the quotient, for example take $p=3$, $2$ is not invertible, but $p(2)$ is invertible.


答:
The image of a ring homomorphism (preserving $1$) is a subring of the codomain.

Since the only subring of $\mathbb{Z}$ is $\mathbb{Z}$ itself, the statement is proved.

The same is true for ring homomorphisms to $\mathbb{Z}/n\mathbb{Z}$.

Your proof is good as well, but overkill.

Title: Every ring homomorphism to the integers is surjective?

Body:
Let $R \neq 0$ be a commutative, unital ring and $f: R \to \mathbb{Z}$ be a ring homomorphism.

Does this force $f$ to be surjective?

Attempt of Proof:
By the universal property of $\mathbb{Z}$, there is a unique ring homomorphism $g: \mathbb{Z} \to R$. But then $f \circ g: \mathbb{Z} \to \mathbb{Z}$ is the unique ring homomorphism which is already given by $id: \mathbb{Z} \to \mathbb{Z}$. Thus $f \circ g = id$ and therefore $f$ must be surjective.

Does this proof hold? It seems weird that I can nowhere find this result...


问:
If a function f is differentiable then its derivate is Borel measurable
I need help solving this interesting result:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be differentiable. Prove that $f'$ Borel measurable.

I tried to start with, if $f$ is differentiable then $f'(x)$ exists for all $x∈ \mathbb{R}$.

Thank you in advance.

答:
You could also say that:
f=differentiable => f=continuous.
Since f=differentiable => $\exists$f '(x) = $\lim_{h\to 0}$$\frac{f(x+h)-f(x)}{h}$ =>  f '(x) = lim$_n$$\frac{f(x+1/n)-f(x)}{1/n}$.

Then from Heine's deffinition (?)* $\forall$(x$_n$)$_n$$\in$$\mathbb{R}$ and $\epsilon$>0: x$_n$ $\rightarrow$ x$_0$ $\Rightarrow$ $\sigma$(f(x$_n$),f(x)) < $\epsilon$.

Suppose f$_n$(x) = $\frac{f(x+1/n)-f(x)}{1/n}$, f=continuous (so f(x+1/n)=continuous $\forall$n). Therefore so far we have: $\forall$x>0 and $\forall$$\epsilon$>0 $\exists$n$_0$$\in$$\mathbb{N}$ so that as n$\geq$n$_0$ then $\sigma$(f(x$_n$),f(x)) < $\epsilon$ $\Rightarrow$  (f$_n$)$_n$ $\rightarrow$ f p.w.

Then f=Baire-1 => f=Borel-meas. (this because f=lim$_n$f$_n$ $\Rightarrow$ f=limsup$_n$f$_n$, for (f$_n$)$_n$ = seqeuence of meassurable functions)

*I dont know how it is called in English. In Measure Theory or Real Analysis we call it in Greek "Arxi/Archi tis Metaforas" (= principle of transport).

PS:I'm an undergraduate student of Mathematics Dpt in University of Patras, Greece. Therefore there may be many mistakes (in english and in mathematical structure)
I hope i helped.


问:
Showing that, if $7\sin^2(x) + \sin(x)\cos(x) = 6$, then $\tan^2(x) + \tan(x) - 6 = 0$
The question is:

Given that:
$$7\sin^2(x) + \sin(x)\cos(x) = 6$$
Show that:
$$\tan^2(x) + \tan(x) - 6 = 0$$

I don't know if there's a proof I'm missing when trying to complete this. I've tried both the $\sin(x)/\cos(x) = \tan(x)$ and the $\sin^2(x) + \cos^2(x) = 1$, and neither of them have got me closer to the answer.

Any help would be appreciated because it's driving me insane not knowing how to do this. I'm not sure if I'm missing something in the proofs I've been using or if I need a different one entirely.

答:
Write as

$$7\sin^2 x + \sin x \cos x = 6(\sin^2 x+\cos^2 x)$$

Simplify to

$$\sin^2 x + \sin x \cos x - 6\cos^2 x=0$$

Show that $\cos x \neq 0$. Hint : $\cos^2 x = 0 \Rightarrow \sin^2 x = 1$.

Divide throughout by $\cos^2 x$,

$$\tan^2 x + \tan x - 6 = 0$$


问:
Finding $\int_0^{\infty } \frac{\tan ^{-1}(a x)-\tan ^{-1}(b x)}{x} \, dx$

Prove that
$$\int_0^{\infty } \frac{\tan ^{-1}(a x)-\tan ^{-1}(b x)}{x} \, dx=\frac{1}{2} \pi  \log \left(\frac{a}{b}\right);\;0<a<b$$

I do not have a clue on how to solve this. In class I have only been taught how to find converging diverging and I do not know how I could apply it here. I can understand how the numerator could give $\frac {\pi}{2}$ but then there are two arc tans so I do not how how I would solve it. And I do not understand why it is $log(a/b)$. And as usual the solution provided by the professor is inadequate as all it tells is to find the limit of arc tan tending to 0 once and tending infinity once and the next step he did says hence proved. Any help is appreciated.

答:
Note

\begin{align}
&\int_0^\infty \frac{\tan^{-1}ax- \tan^{-1}
bx}x dx \\= &\int_0^\infty \int_b^a \frac1{1+x^2t^2}dtdx
=\frac\pi2 \int_b^a\frac 1tdt = \frac\pi2 \ln \frac ab
\end{align}


问:
Rewrite $3^{4x-5}=38$ as common log and then use a calculator to solve
I am given the expression $3^{4x-5}=38$ and asked to rewrite in common log to isolate the variable $x$, then to solve using a calculator.

I am struggling to to get the $x$ on it's own. My attempt:

$$3^{4x-5}=38$$

Rewrite lhs as a log

$$log_3(4x-5)=38$$

Rewrite lhs into common log base 10:

$$\frac{log(4x-5)}{log(3)}=38$$

This is as far as I can go. How can I isolate x here?

[Edit]

I rewrote to this, does it look right?

$$3^{4x-5}=38$$
$$log_3(38)=4x-4$$

My textbook says I am specifically to rewrite using the common log:
$$\frac{log(38)}{log(3)}+5=4x$$
$$x=\frac{(\frac{log(38)}{log(3)}+5))}{4}$$

答: Hint: $\log a^b = b \log a$ and not $\log_ba$


问:
Find a gcd of two polynomials from $\mathbb{Z}[i]$
I'm trying to solve this problem from my abstract algebra course:

Being $z_1=39-8i$ and $z_2=7+i$ elements from the ring of Gauss
integers, $\mathbb{Z}[i]=\{a+bi \mid a,b\in\mathbb{Z}\}$. Find a
greater common divisor of both terms, $d=\text{gdc}(z_1,z_2)$, and find
elements $x_1,x_2\in\mathbb{Z}[i]$ such that $d=x_1z_1 + x_2z_2$.

I have no idea how to proceed with this kind of exercises, I've seen other ones similar to this one in my book and want to find a method to approach these kind of exercises. I decided to post this one in particular because the ring $\mathbb{Z}[i]$ seemed especially weird but interenting, so I assumed this must be one of the most difficult ones of this kind of problems.

How can I solve this? Any help will be appreciated, thanks in advance.

答:
I'd work in the ring $R = \Bbb Z[x]/\langle x^2+1\rangle = \{ax+b\mid a,b\in\Bbb Z\}$, where $x^2+1 = 0$ in $R$.

Then division with rest gives

$-8x+39 = (-8)(x+7) + 95$, i.e., $95 = 1\cdot (-8x+39) + 8\cdot (x+7)$.

Then over the rationals: $1 = \frac{1}{95}(-8x+39) + \frac{8}{95}(x+7)$.


问:
Field of fractions of $R = \{ \frac{a}{b} | \ a \in \mathbb{Z}, b \in \mathbb{Z} \setminus p\mathbb{Z} \}$.
Let $p$ be a prime number. Define the subring $R = \{ \frac{a}{b} | \  a \in \mathbb{Z}, b \in \mathbb{Z} \setminus p\mathbb{Z} \}$ of $\mathbb{Q}$. Can someone explain to me why the set of rationals $\mathbb{Q}$ is the field of fractions of $R$ ? And also, why does $p$ have to be a prime number ? I haven't studied alot of ring theory yet so an explanation based on only the definition of field of fractions would be useful.

Thanks in advance !

答:
Note that the most you can say is that “the field of fractions of $R$ is isomorphic to $\mathbb{Q}$”.

Trying to look at the problem from a more general point of view may help.

You're given a subring $R$ of a field $F$. I want to show that $\operatorname{Frac}(R)$ can be naturally embedded in $F$ as a subfield containing $R$.

Let's take $a,b\in R$, with $b\ne0$; denoting as usual by $a/b$ the fraction in $\operatorname{Frac}(R)$, we can think to send it to $ab^{-1}\in F$.

Now, is $a/b\mapsto ab^{-1}$ a ring homomorphism? First we need to show it's well defined.

Saying that $a/b=c/d$ amounts to $ad=bc$. Then
$$
ab^{-1}=add^{-1}b^{-1}=bcd^{-1}b^{-1}=cd^{-1}
$$
and we're done. The other verifications are easy.

So we have $f\colon\operatorname{Frac}(R)\to F$ such that $f(a/1)=a$, for $a\in R$. Hence the image of $f$ is (a realization of) the field of fractions.

In your case, (this realization of) the field of fractions of $R$ is a subfield of $\mathbb{Q}$ and there are not many of them: $\mathbb{Q}$ has a single subfield, namely itself.


问:
Every ring homomorphism to the integers is surjective?
Let $R \neq 0$ be a commutative, unital ring and $f: R \to \mathbb{Z}$ be a ring homomorphism.

Does this force $f$ to be surjective?

Attempt of Proof:
By the universal property of $\mathbb{Z}$, there is a unique ring homomorphism $g: \mathbb{Z} \to R$. But then $f \circ g: \mathbb{Z} \to \mathbb{Z}$ is the unique ring homomorphism which is already given by $id: \mathbb{Z} \to \mathbb{Z}$. Thus $f \circ g = id$ and therefore $f$ must be surjective.

Does this proof hold? It seems weird that I can nowhere find this result...

答:
You certainly assume that a ring homomorphism between unital rings preserves the multiplicative identity.

However, all what we need to show that $f(R) =\mathbb Z$ is the fact that $1 \in f(R)$ - which is an immediate consequence of $f$ being a ring homomorphism. Beyond this fact the multiplicative structure of $R$ is completely irrelevant, your result is true also for non-commutative unital rings $R$. In fact the essence is this:

A group homomorphisms $\phi : G \to \mathbb Z$ is surjective if and only $1 \in \phi(G)$.

One half is trivial. For the other part, let $g \in G$ such that $\phi(g) = 1$. Then clearly $\phi(g^n) = n $.


问:
Question about a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$
This is from Ross's elementary analysis book. The statement is if $a,b \in \mathbb{R}$ such that $a<b$ then there exists a rational $r \in \mathbb{Q}$ such that $a<r0$ then by the archimidean property there exists a natural number call it $n$ such that $n(b-a)>1$. Now we must prove there exists an integer $m$ such that $an<m<bn$. By the archimidean property again there exists an integer $k$ such that $k>\max(|an|, |bn|)$ so that $-k<an<bn<k$. Then the set $$\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$$ is finite and nonempty and we can set $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$. Then $an < m$ but $m - 1 \leq an$. Also, we have $m = (m - 1) + 1 \leq an + 1 < an + (bn - an) = bn$.

Comment: I get up to the point how the author sets$ k$ to be the integer that is larger than both $an$ and $bn$. And since $a<b$ we get the bounded inequality $-k<an<bn<k$. From here on this is where I get confused.  He creates a set  $\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which he calls finite which I see and nonempty (I'm guessing since this set has a least upper bound $k$ thus it is nonempty). But then he lets  $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which follows $an<m$ but $m - 1 \leq an$. I don't get how the author gets to that point?

答:
The set is finite because there are at most $2k$ elements in it. It is non-empty because by assumption $k$ is in it. By the least upper bound property for $\mathbb Z$, since this set is bounded below, it admits a minimum. The minimum $m$ is in that set, but by minimality, $m-1$ is not in that set, because otherwise $m-1$ would be an element of the set smaller than its minimum. This is why $an < m$ (i.e. $m$ is in the set) but $m-1 \le an$ (i.e. $m-1$ is not in that set).

Hope that helps,