问:
Why Integral domains haven't an unified definition?
We can define integral domains as:

rings without zero divisors

commutative rings without zero divisors

commutative rings with identity and without zero divisors

I don't know why integral domains don't have an unified definition, maybe due to historical reasons? theoretically what's the benefit to use the definition 1 instead 3 for instance?

REMARK

I noticed that when I was studying Hungerford's book and Bhattacharya's book.

Thanks in advance

答:
"Why don't we all agree on a single definition for X" has the universal answer:

Different people are interested in exploring theory under different conditions, and definitions are a matter of convenience. It's natural then that one condition (like not having nonzero zero divisors) is used, it'll propagate under the same name to different schools assuming different conditions, and wind up with those extra conditions attached. (So maybe "yes, historical reasons!" is a good short answer.)

This is especially the case for "identity" and "commutativity" in ring theory. Certain schools may adopt one or the other or both, but there isn't any reason to foist either of those conventions upon other people who want to study rings without zero divisors.


问:
Question about the proof for "The rational line Q does not have the completeness property"
Referring to the attached png file, I have questions about the inequality:

$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$

How is this inequality derived?  How did the n get into the inequality?  Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$, and later in the proof, $2q^2/(p^2 - 2q^2)$ come from?   There were no derivations for these 3 expressions.  Could someone please provide the derivations?

答:
Yeah....

So we have $(\frac pq)^2 < 2$ and we want to show $\frac pq$ is not an upper bound so we need to show there is $S > \frac pq$ so that $(\frac pq)^2 < S^2 < 2$.

$S$ is rational, so there is a rational $r$ so that $S =r\frac pq$ and $r > 1$.

We need to show that $r^2\frac {p^2}{q^2} < 2$ or that $r^2 p^2 < 2q^2$ for some rational $r$.

Now we know $\frac {p^2}{q^2} < 2$ so $2q^2 > p^2$ and so we need to show that

$(r^2 -1) p^2 < 2q^2-p^2$ or that $r^2 -1 < \frac {2q^2-p^2}{p^2}$.

Now the RHS of that is a constant fixed positive (albeit it possibly quite small) so we need to show that we can have an $r^2-1$ that is smaller than that.

And it is sufficient to show we can have $r$ so that $r^2$ is arbitrarily close to $1$.

Now we usually can show rationals are arbitrarily close but showing that if $n$ is arbitrarily large then $\frac 1n$ is arbitrarily small.

So if we let $r = 1+ \frac 1n$ the want to show, we can choose and $n$ so that

$(1 + \frac 1n)^2 -1 = \frac 2n + \frac 1{n^2} < \frac {2p^2-p^2}{p^2}$ we will be done.

And that's a matter of making $\frac 2n + \frac 1{n^2} = \frac {2n+1}{n^2}$ arbitrarily small

or

making $\frac {n^2 }{2n+1}$ arbitrarily large.  Or at least, $\frac {n^2}{n+1} > \frac {p^2}{2q^2 -p^2}$.

We can do that as $\frac {n^2}{n+1} \to \infty$

And that's it:  If $\frac {n^2}{n+1} > \frac {p^2}{2q^2 -p^2}$ it's just algebraic manipluation that

For $r = 1+\frac 1n = \frac {n+1}n$ then $r\frac pq = \frac {p(n+1)}{qn}$ when squared will be less than $2$.

$(r\frac pq)^2 = (1+\frac 1n)^2\frac {p^2}{q^2}=$

$(1 + \frac {2n+1}{n^2})\frac {p^2}{q^2}=$

$\frac {p^2}{q^2} + \frac {2n+1}{n^2}\frac {p^2}{q^2} < $

$\frac {p^2}{q^2} + \frac {2q^2-p^2}{p^2}\frac {p^2}{q^2} =$
$\frac {p^2}{q^2} + \frac {2q^2 -p^2}{q^2}=$

$\frac {p^2 + 2q^2 -p^2}{q^2}=\frac {2q^2}{q^2}=2$.

We are done.

.....

But this was a bass-ackward proof in that we started with what we needed and worked our way back to show we can have it.  That's not an invalid way of doing a proof because we never assumed what we wanted to prove, it's okay to say "If we need this, we can do that by this, and, hey, we can do that" but it's not an ideal argument.  We like proofs to be direct and forward.

So we start with:  "We can always find an $n$ so that $\frac {p^2}{2q^2 - q^2} < \frac {n^2}{2n+1}$".

To which the reader reacts with ".... well, sure, but where the #@%! did that come from"

And the proofer can say "It doesn't matter.  We can do it, right?"

And the reader says "well, sure, but..."

And the proofer goes "ninh, ninh, ni..."

And the reader goes "...but..."

And the proofer goes "NINH! .... so if we do that then....."  And lo and behold, it works.

ContentLicense: CC BY-SA 4.0


问:
Derivative by $a$ of the solution to $x’ = x^2 - x\cos t$, $x(0) = a$
Question:
Find the derivative with respect to $a$ of the solution to the following IVP:
\begin{align}
x’ &= x^2 - x\cos t \\
x(0) &= a
\end{align}
at $a = 0$.

What I tried:

I'm not sure how to solve the quation explicitly (I am not sure I need to).

Yet, if I try to solve, I don't successeed.

Separation of variables is a problem because I have
$x\cos t$.

Linear of course not, Bernoulli doesn't work, substitioution I don't see how.

I would like a hint please.

Thank you.

答:
If $y=\frac{\partial x}{\partial a}$, then $$y'=2xy-y\cos t,~~~ y(0)=1$$ in addition to the ODE for $x$.

Now the reference solution $x$ is for the initial value $x(0)=0$, which gives the trivial solution $x(t)=0$. So it remains to solve
$$
y'=-y\cos t,~~~y(0)=1.
$$


问:
Can we find an unbiased estimator for $\theta$ which reaches the Cramer Rao Lower Bound?
Let $f(x)=\theta x^{-(\theta+1)}$ for $x>1$, where $\theta>1$ is an unknown parameter. Can we find an unbiased estimator for $\theta$ in this case which reaches the Cramer Rao Lower Bound?

My attempt:

To find such an estimator, I think I first need to find a sufficient statistic. After that, I think the Rao-Blackwell theorem  should be applied.
Therefore, I tried to use the following theorem: (Screenshot from the Wikipedia website).

In my case this would imply $\prod\limits_{i=1}^n \theta x_i^{-\theta-1}=\prod\limits_{i=1}^n x_i^{-1}\theta x_i^{-\theta}$.

I am in doubt whether this is in the form required by theorem or not (I would say no). Is it in the required form? If not, would there be an other way to factor the function properly?

答:
You can always find an estimator which attains the Cramer Rao (Fischer Information) lower bound by scaling your MLE estimator to attain the right variance. However, this risks making the estimator biased. Sometimes, the unbiased estimator has minimum variance but other times an unbiased one does. Try calculating the CR bound for your function and the variance of your estimator.

To be clear, sufficiency is a wholly different concept. Sufficiency says that you can't squeeze out extra information about your statistic. For example if you have 10 observations but only use 6 of them to estimate the mean that would be not sufficient since there's an extra 4 observations you didn't use.

The concept you want is efficiency which is a measure of the mean square error or the variance. The more efficient an estimator is the closer it's variance is to the CR bound.

ContentLicense: CC BY-SA 3.0


问:
3-variable 3-equations problem

For positive reals $x,y,z$, if $xy+yz+xz=\sqrt{\frac pq}$, given
$$x^2+xy+y^2=2$$
$$y^2+yz+z^2=1$$
$$z^2+zx+x^2=3$$
find $p-q$, where $p$ and $q$ are relatively prime integers.

I tried grouping the required sum by adding all the equations but it led me to nowhere. I tried applying basic AM-GM inequality to conclude that the sum is less than or equal to $2$, but this is of no use since $x,y,z$ are positive reals, not just positive integers.

I have exhausted all my ideas. Any hint/help is appreciated!

答:
Actually there is a geometric solution to this. Consider a point $P$ in $\mathbb{R}^2$ and points $A$, $B$, $C$ with $PA=x$, $PB=y$, $PC=z$ and $\angle APB=\angle APC=\angle BPC=120^{\circ}$.

If you apply the cosine theorem you get that the triangle $\triangle ABC$ has sides $BC=1$, $AC=\sqrt{3}$, $AB=\sqrt{2}$. Notice that it is a right triangle since $AC^2=BC^2+AB^2$.

Computing the area in two ways $xy+yz+zx=\cfrac{4}{\sqrt{3}}\cdot \operatorname{Area}(\triangle ABC)=\cfrac{2\sqrt{2}}{\sqrt{3}}$

As a remark the numbers are quite incidental here; you can pick your favorite numbers for the system and do the same construction and instead use Herron formula for the area to compute $xy+yz+zx$.

Title: 3-variable 3-equations problem


问:
Sum of powers in multiple ways
I came across this little tidbit in a book with no reference or extra explanation.
We have
$$1^n+6^n+8^n=2^n+4^n+9^n\qquad \forall 1\leq n\leq 2,$$
$$1^n+5^n+8^n+12^n=2^n+3^n+10^n+11^n\qquad \forall 1\leq n\leq 3,$$ and
$$1^n+5^n+8^n+12^n+18^n+19^n=2^n+3^n+9^n+13^n+16^n+20^n\qquad \forall 1\leq n\leq 4.$$

I have no idea how to search for this, and whatever I have tried has turned up empty. Does anyone know if this has a name of some kind? If this is a duplicate, I couldn't find it.

More specifically, and what I care about, can we continue this? In other words, for every $N$, do there exist distinct $a_1,\ldots,a_r,b_1,\ldots,b_r$ such that $\sum a_i^n=\sum b_i^n$ for all $1\leq n\leq N$?

答:
Turned into an answer and added more details on request.

This type of equations is called Multigrade Equations, or more specifically, the Prouhet-Tarry-Escott Problem. (Links are to their respective MathWorld articles). Six of a kind . contains some more discussion of this problem.

As to whether a solution exists for each $N$, this answer contains a constructive theorem by Tarry-Escott which proves the affirmative, with $r = 2^N$. Hence the interest in the problem is to find ideal solutions, where $r = M+1$ as they are minimal.

Title: Sum of powers in multiple ways


问:
How does a morphism in a distinguished triangle induce an element of $\operatorname{Ext}^{2}$
On p. 260 of Deligne and Illusie's paper (https://eudml.org/doc/143480) it says:

Let $e(K)\in\operatorname{Ext}^{2}(H^{1}(K),H^{0}(K))$ be the class defined by the degree $1$ morphism in the distinguished triangle
$$H^{0}(K)\longrightarrow K\longrightarrow H^{1}K[-1]\longrightarrow^{+1}$$

What is meant by "the class defined by the degree 1 morphism"? How does a degree-1 morphism in a distinguished triangle define an element of $\operatorname{Ext}^{2}$? Sorry if that question is stupid, I am only a beginner when it comes to triangulated categories. Can someone please explain it to me in as much detail as possible?

答:
A morphism of degree 1 from $A$ to $B$ is by definition the same as a morphism $A \to B[1]$. In your case, this is an element in
$$
\mathrm{Hom}(H^1(K)[-1],H^0(K)[1]) \cong
\mathrm{Hom}(H^1(K),H^0(K)[2]) \cong
\mathrm{Ext}^2(H^1(K),H^0(K)).
$$


问:
Finding the Bound of the Error in Interpolation
The function $f(x)$ has the following values known at the points: $f(x_1) = 2, \quad f(x_2) = 8 \quad \text{and} \quad f(x_3) = -2$.

I've also been told that I can assume that $f$ and its derivatives satisfy the following bounds for all $x$: $$|f(x)| \le 10, \,\, |f'(x)| \le 20, \,\, |f''(x)| \le 30, \,\,
  |f'''(x)| \le 50 \text{ and } |f''''(x)| \le 100.$$

I want to be able to bound the error in approximating the polynomial $f(x)$ at $x=2$, for two different methods I have used to approximate the polynomial:

Lagrange Polynomial Interpolation: $P(x) = -2x^2 + 11x -7$

Piecewise Linear Interpolation: $S(x) = 3x - 1$ upon the interval $[1,3]$

I know that the formula for the error in polynomial interpolation is given as: $$f(x) - P(x) = \frac{f^{(n)}(\xi)}{n!} \prod_{k=1}^n (x-x_k).$$

But how can I use this formula to $\underline{bound}$ the error for these two types of interpolations? Any guidance would be appreciated.

答:
In your case $n=3$, so
$$|f(x) - P(x)| = \frac{\left|f^{(3)}(\xi)\right|}{3!} \prod_{k=1}^3 |x-x_k|\le\frac {50}6 \prod_{k=1}^3 |x-x_k|$$

Since you know that $1 \le x \le 3$, you can also find bounds on the $|x-x_k|$ elements, but since you haven't told us what the $x_i$ are, we cannot take it any farther.


问:
Is the real locus of an elliptic curve the intersection of a torus with a plane?
In Lawrence Washington's book Elliptic Curves: Number Theory and Criptography I read that if $E$ is an elliptic curve defined over the real numbers $\mathbb{R}$ then the set of real points $E(\mathbb{R})$ can be obtained as the intersection of the torus of complex points $E(\mathbb{C})$ and a plane. The following is the relevant page of the book.

Basically it says that if the plane passes through the hole in the torus, then the real locus of the elliptic curve looks like the following

and if not, it looks like this one

So passing through the hole in the middle seems to determine whether the graph of the elliptic curve has two real components or just one. I have never seen such a claim before in other books about elliptic curves, and I'm really curious about it. Thus I have a couple of questions about this.

Can this claim about the real locus being the intersection of a torus with a plane be made precise somehow, and if so can anyone please provide an explanation about it?

If this is possible, can an explicit example be given?

Notes:

I have to say that I'm really confused about this because the correspondence of an elliptic curve with a torus (when considering the complex points $E(\mathbb{C})$) is given by an isomorphism with the complex modulo a lattice, so even thinking about intersecting "the torus" with a plane seems rather odd to me.

The reference to section 9.3 in the book does not seem to clarify this, it basically deals with the identification of the elliptic curve with the torus.

Thank you very much for any help with these questions.

答:
The complex points of the elliptic curve lie in the projective plane over $\mathbb C$.  Supposing that the coefficients of the elliptic curve are in $\mathbb R$,
it then makes perfect sense to intersect the curve with the projective plane over $\mathbb R$.  If we forget the point at infinity, then we can just think that
we are intersecting a certain punctured torus in $\mathbb C^2$ with $\mathbb R^2$.

Of course, the torus is not embedded as neatly as in the picture you posted (or,
more accurately, it is not so much a question of "neatness" of the embedding, but rather the fact that the torus is embedded holomorphically in $\mathbb C^2$,
which as a real manifold is of dimension four),
but the description of the intersection is nevertheless correct.

Added: David Speyer's answer here makes an important point about the topology of the intersection, which I didn't address in my answer.


问:
Integral equation that's cant solve... Need a hand
Help me solve this integral equation, I'm having some troubles... I need to use the Fredholm method for second kind integral equations.

$$\phi(x)= \sin(x)+ \lambda \int_{0}^{\pi}\cos(2x+y)\phi (y)dy$$

Thanks

答:
I don't know the method you mention, but notice that you can differentiate twice to get
$$\phi''(x)=-\sin x-4\lambda\int_0^\pi dy\ \phi(y)\cos(2x+y),$$
so that
$$\phi''(x)+4\phi(x)=3\sin x.$$
This is easily solved as
$$\phi(x)=\sin x+A\cos(2x)+B\sin(2x)$$
with constants of integration $A$ and $B$. Now plug $\phi$ back in to the integral and solve for $A$ and $B$ in terms of $\lambda$. I get
$$A=\frac{6\pi\lambda^2}{8\lambda^2-9}\qquad{\rm and}\qquad B=\frac{9\pi\lambda}{2(8\lambda^2-9)}.$$</p>


问:
MCQ on Linear Algebra. A is a $5\times5$ matrix with real entries and x is a $5\times1$ vector. What is correct about the set $\{Ax,A^2x..A^5x\}$?
The options given are :

a. Linear dependence/ independence cannot be determined without knowing the whole matrix A.

b. The vectors are linearly dependent

c. The vectors are linearly independent iff A is symmetric

d. The vectors are linearly independent iff A is PD (Positive Definite)

e. The vectors are linearly independent

For background, I tried to apply the principle of Krylov subspaces here and ended up with a solution that the vectors are almost always linearly independent, albeit with some exceptions. Some exceptions among diagonalizable matrices occur when A has an eigenvalue of 0,1, or a root of unity and x is an associated eigenvector.

So I am now confused as to which option to select. Any help appreciated.

答:
The statement that is the least false is a.

A matrix $A$ will be such that there exists an $x$ for which $\{Ax,\dots,A^5 x\}$ is linearly independent if and only if $A$ is invertible and non-derogatory. For such a matrix $A$, a vector $x$ will have this property if and only if it is an "$A$-cyclic vector", i.e. if the set $\{x,Ax,A^2x,\dots\}$ spans $\Bbb R^5$.

Statement b is false because the vectors are independent, for instance, when $A$ an invertible diagonal matrix and $x = (1,\dots,1)^T$. Statement c is false: the vectors are not independent if $A$ is the identity matrix. Statements d and e are false for the same reason.


问:
Achilles and the tortoise problem
You know the story - Achilles and the tortoise are having a race. The tortoise is given a head start of 100m. After some time, Achilles will arrive at where the tortoise was at, but the tortoise will have moved further. Question: find how long it will take for Achilles to catch up to the tortoise in seconds, and how far Achilles will travel when he reaches the tortoise in meters. Achilles runs at $10m/s$ and the tortoise $0.2m/s$. Use an infinite series.

Intuition

Obviously I would have to use infinite series to solve. As I see it, I should create two series: one for Achilles and one for the tortoise. For the turtle, it would roughly be $100 + \sum_{n=1}^{\infty}a_n$. The 100 is the 100 meter head start, and $a_n$ would be the series that represents the movement of the tortoise. For Achilles it would be similar.

Goal

The goal is to find $n$, which would depict the time in seconds, where Achilles will catch up to the tortoise. To do this, I would have to equate Achilles' series and the tortoise's series, evaluate each, and solve for n. To find the total distance traveled by Achilles, it would be substituting $n$ into the equation and evaluating the series.

Answers

So, any answers for $n$? I'm mainly confused as to how to formulate the infinite series, and how to solve for each series (which would involve seeing if it converges and such).

答:
You may regard this "answer" as an extended comment, since I have not proposed to answer the question in the way you asked for it to be answered.
But I would prefer that this answer be viewed as a hint (though still not a complete answer).

For the turtle, it would roughly be $100 + \sum_{n=1}^{\infty}a_n$. The 100 is the 100 meter head start, and $a_n$ would be the series that represents the movement of the tortoise.

There's no need to do anything "roughly". The formula
$100 + \sum_{n=1}^{\infty}a_n$ is an exact formula, provided that you have a way to determine what $a_n$ will be for every $n.$

The goal is to find $n$, which would depict the time in seconds, where Achilles will catch up to the tortoise.

This is a conceptual error. A second is just an arbitrary unit of time.
If you were told that Achilles ran at $600$ meters per minute and the tortoise
moves at $12$ meters per minute, does that mean $n$ should depict the time in minutes when Achilles catches up with the tortoise?
(This is the exact same problem as the original one, but with the speeds expressed in different units.)

The only thing $n$ has to represent in $100 + \sum_{n=1}^{\infty}a_n$
is the number of steps of some process. You can make the number of seconds in each step be whatever is suitable to the process you're trying to describe.

If you describe the race in the usual "paradoxical" way,
the first step is for Achilles to reach the tortoises' starting point.
This takes $10$ seconds.
But now the tortoise is $2$ meters away, so the second step is for Achilles to run another $2$ meters. This takes $2/10$ of a second.
In the second step the tortoise moves another $4/100$ meter, so the third step is for Achilles to run yet another $4/100$ meter, which takes another $4/1000$ second.

So we have:

For $n=1,$ $a_n = 2,$ total elapsed time so far is $10$ seconds.

For $n=2,$ $a_n = 0.04,$ total elapsed time so far is $10.2$ seconds.

For $n=3,$ $a_n = 0.0008,$ total elapsed time so far is $10.204$ seconds.

And so forth. There is no particular $n$ at which Achilles catches the tortoise;
each $n$ just adds more decimal places to the repeating decimal that describes the exact amount of time required to catch the tortoise.
The time at which Achilles catches the tortoise is the limit of the elapsed time as $n$ goes to infinity.

The solution without infinite series is much simpler, as this other answer has already shown.

Title: Achilles and the tortoise problem


问:
Procedure or technique or reasoning to get $2+4h$?
I have solved for my students of an high school this simple trigonometric equation:

$$\tan(\pi+6x)=-\tan(2x)\tag 1$$

The $(1)$ is equivalent to (I remember also that $\tan(\alpha)=\tan(\mathbb Z\pi+\alpha)$)

$$\tan(\pi+6x)=\tan(-2x)\tag 2 \iff x=\frac{\pi}{8}k^*, \quad k^*=k-1\in\Bbb Z$$

But the solution of the textbook is $x=k\pi/8$ with $k\neq 4h+2$.

How can I find the value $\color{red}{4h+2}$?

If I calculate the domain I will have

$$\begin{cases}
 x \neq -\dfrac \pi{12}+\Bbb Z\dfrac \pi{6}\\[0.5em] \tag 3
 x\neq \dfrac \pi4+\Bbb Z\dfrac\pi2
\end{cases}$$

I have done the tests and the condition $4h+2$ is equivalent to $(3)$. Just a curiosity looking the 2nd negation of the $(3)$ I have the denominator $4$ and $2$.

答:
The constrains come from the domain of tangent function. The domains of $\tan x$ excludes $k\pi+\frac{\pi}{2}$,  there  are constrains placed on $\tan2x$ and $\tan 6x$, where $k,m,n,h \in \mathbb Z$.

For $\tan 2x$, we have $$2x \neq m\pi +\frac{\pi}{2}$$ with $x\neq  \frac{(2m+1)\pi}{4}$.

For $\tan  6x$, we have $$6x \neq n\pi+\frac{\pi}{2}$$ with  $x \neq  \frac{(2n+1)\pi}{12}$.

Or

$k\pi/8 \neq\frac{(2m+1)\pi}{4}$, $k\neq  2(2m+1) =4m+2, \quad (A)$;

$k\pi/8 \neq\frac{(2n+1)\pi}{12}$, $k\neq  \frac{2(2m+1)}{3} \equiv \frac{4m+2}{3} \quad (B)$.

If (A) is satisfied, so is (B).

Therefore, The solutions are

$$x = \frac{k\pi}{8}$$ with $k\neq 4h+2$


问:
Showing that if $u$ is a real-valued harmonic function then for any real $c$ we have that $u^{-1}(c)$ is unbounded
I have the following homework question:

Let $u$ be a non-constant real-valued harmonic function in
  $\mathbb{C}$. Prove that $u^{-1}(c)$ is unbounded for every real
  number $c$

There is a hint that say we may use the result of the following exercise:

Prove that a positive harmonic function in $\mathbb{C}$ is
  non-constant

This exercise is taken from

Donald Sarason Notes on Complex Function Theory (page 89)

What I tried:

I don't really know how to start, but a reasonable beginning can be:

Assume by negation that there is $c\in\mathbb{R}$ s.t $$u^{-1}(c)\subseteq D(0,M)$$
where $$D(0,M)=\{z\in\mathbb{C}:\,|z|<M\}$$

We know that there is a harmonic conjugate $v$ s.t $$f=u+iv\in H(D(0,M))$$

I don't know how to move forward, or how to use the hint (since in
this question $u$ need not be a positive function).

Can someone please direct me in the right direction ?

答:
I think you mean a positive harmonic function is constant. See Does there exist a harmonic function in the whold plane that is postive everywhere?

Also, you should be able to use the fact that $u^{-1}(c)$ is bounded to conclude that, by continuity, everything falls on one side of $c$. This can be manipulated into giving you a positive harmonic function.


问:
a.s convergence of a series
$(X_n)_n$ is a sequence of independent random variable, such that $E[X_n]=0.$ Let $Y \in L^2$ such that for all $n \in \mathbb{N}^*,Y-\sum_{k=1}^nX_k$ is independent of $(X_1,...,X_n).$

Prove that for every $n \in \mathbb{N}^*,X_n \in L^2$ and that $\sum_{k=1}^nX_k$ converges a.s.

Since $Y=Y-\sum_{k=1}^nX_k+\sum_{k=1}^nX_k \in L^2,$ by independence $X_n \in L^2.$
To prove that $\sum_{k=1}^nX_k$ converges a.s, it's sufficient to prove the convergence of $\sum_nVar(X_n),$ also we can derive from above, that $\sum_{k=1}^nVar(X_k)=\sum_{k=1}^nE[X_kY].$

Any suggestions, hints how to continue?

答:
If $Y=Z+W$ where $Y \in L^2$ and $Z,W$ are independent, then $Z,W \in L^2$ and consequently Var$(Y)=$Var$(Z)+$Var$(W)$. The first assertion follows from Fubini's Theorem: We may assume $Z,W$ have mean $0$.  Writing
$\mu, \nu$ for the laws of $Z,W$ respectively, we have
$$E(Y^2)=\int \int (z+w)^2 \, du(z)\,  d\nu(w)$$
so $Z+w \in L^2$ for $\nu$-almost all $w$, whence $Z \in L^2$.
In the question we get Var$(X_1+...+X_n) \le$Var$(Y)$.

Title: a.s convergence of a series


问:
Drawing the grid without lifting the pen
The below shape consist of $24$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw?
$1)25\quad\quad\quad\quad\quad\quad2)26\quad\quad\quad\quad\quad\quad3)27\quad\quad\quad\quad\quad\quad4)28\quad\quad\quad\quad\quad\quad5)29\quad\quad\quad\quad\quad\quad$

I think we can do it somehow by segments with the length $28$ (by passing twice the four middle segments on each side of outer big square).

I'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?

答:
You will be able to draw all the edges of the graph without lifting the pen and drawing every edge exactly once (starting and ending at the same node), iff there exists an Euler circuit in the graph.

We shall use the following theorem: a connected graph has an Euler circuit iff all the nodes of the graph has even degree.

The given grid graph does not have all nodes with even degree. Let's first find the nodes with odd degrees, as shown in the next figure. Notice that the nodes B,C, E,I, H,L and N,O have odd degrees (namely 3).

Let's add 4 additional (red) edges to the grid graph as shown in the next figure to make all the nodes have even degrees.

Now, in the augmented graph above, by the theorem, it must have an Euler circuit, so we shall be able to draw all the edges of the graph exactly once without lifting the pen.

We shall use Flury's algorithm to find an Euler circuit in the augmented graph first. The key idea is that when you have a choice between a bridge and a non-bridge, always choose the non-bridge (don't burn the bridges).

Since the augmented graph has all nodes with even degrees, we can start at any node, let's start at B, a node with degree 4 in the augmented graph.

Using the algorithm, the following animation shows how to construct an Euler circuit in the augmented graph.

Now, let's find where the additional edges (ones not present in the original graph) were used in the circuit found in the augmented graph.

Let's go back to our original graph by replacing those edges by the ones in the original graph, we had exactly 4 such additional edges needed to make all nodes in the original graph to have even degrees.

It implies that in order to have the Euler circuit in the original graph we need 4 additional edges (requiring 24+4=28 edges). It also means that in the original graph we need to traverse the corresponding 4 edges twice, in order to draw all the edges without lifting the pen, as shown below:

[EDIT]

If we don't need to start and end at the same node, it's sufficient to have an Euler trail instead of a circuit, needing one less edge: $28-1=27$ edges (only 3 edge segments, namely, EI, ON and HL are needed to be traversed twice, leading to minimum length of the line as $24+3=27$).

In this case the trail shown above starts at $B$ but ends at $C$ (instead of $B$ as before), since the edge $BC$ has already been visited once, we don't need to visit it again (as shown in the next figure).


答:
If you like a more algebraic answer, consider the following linear combination of the three equations in the order written above $(x-y)(1)+(y-z)(2)+(z-x)(3)$. The LHS of the resultant equation vanishes and we get a linear relation:

$$ 2(x-y)+(y-z)+3(z-x)=0\iff z=\frac{x+y}{2}$$

Under this light we see that $\Pi=xy+xz+yz=6z^2-2$ because in fact equation $(1)$ can be rewritten in the form

$$x^2+xy+y^2-2=0\iff xy=4z^2-2$$

Thus we only need to compute $z^2$. In the following denote $\rho=\frac{x-y}{2}$. Then consider the combination $(3)-(2)$ we get

$$x^2-y^2+(x-y)z=2\iff z\rho=\frac{1}{3}$$

and finally we note that we can write

$$x^2+xy+y^2=(x+y)^2-xy=3z^2+\rho^2=2$$

The last two equations can be solved by substitution of the 1st into the 2nd, which yields the equation

$$3z^4-2z^2+\frac{1}{9}=0\iff 6z^2=2\pm\sqrt{\frac{8}{3}}$$

which since $\Pi>0$ yields immediately

$$\Pi=\sqrt{\frac{8}{3}}\Rightarrow p=8~,~q=3$$

and thus $p-q=5$.

Title: 3-variable 3-equations problem

Body:
For positive reals $x,y,z$, if $xy+yz+xz=\sqrt{\frac pq}$, given
$$x^2+xy+y^2=2$$
$$y^2+yz+z^2=1$$
$$z^2+zx+x^2=3$$
find $p-q$, where $p$ and $q$ are relatively prime integers.

I tried grouping the required sum by adding all the equations but it led me to nowhere. I tried applying basic AM-GM inequality to conclude that the sum is less than or equal to $2$, but this is of no use since $x,y,z$ are positive reals, not just positive integers.

I have exhausted all my ideas. Any hint/help is appreciated!


答: Item $(e)$ is false by taking $x=0$. This would make $\{Ax,\ldots ,A^5x\}=\{0\}$ which is clearly linearly dependent. To see why items $(b)-(d)$ fail, take $x=e_1$, and consider the $5\times 5$ matrix $A$ defined by $$A=\begin{pmatrix}0&0&0&0&-1\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\end{pmatrix}$$ It is easily verified that $\{Ax,A^2x,\ldots,A^5x\}$ is linearly independent. However, $A$ is not symmetric nor positive definite as $A$ has an eigenvalue of $-1$.

Title: MCQ on Linear Algebra. A is a $5\times5$ matrix with real entries and x is a $5\times1$ vector. What is correct about the set $\{Ax,A^2x..A^5x\}$?

Body:
The options given are :

a. Linear dependence/ independence cannot be determined without knowing the whole matrix A.

b. The vectors are linearly dependent

c. The vectors are linearly independent iff A is symmetric

d. The vectors are linearly independent iff A is PD (Positive Definite)

e. The vectors are linearly independent

For background, I tried to apply the principle of Krylov subspaces here and ended up with a solution that the vectors are almost always linearly independent, albeit with some exceptions. Some exceptions among diagonalizable matrices occur when A has an eigenvalue of 0,1, or a root of unity and x is an associated eigenvector.

So I am now confused as to which option to select. Any help appreciated.


问:
Calculating integral (actually calculating Fourier transform)
I want to calculate the Fourier transform of the function $ f\left(t\right)=e^{-t^{2}} $

The answer should be $$ \mathscr{F}\left(f\right)[\omega]=\frac{1}{2\sqrt{\pi}}e^{-\frac{\omega^{2}}{4}} $$

I solved it in 2 different ways (and got the same result), but I doubt my second way because I cannot explain in 100%.

So the first way is to note that both $$ \intop_{-\infty}^{\infty}|f\left(t\right)|dt<\infty,\thinspace\thinspace\thinspace\intop_{-\infty}^{\infty}t|f\left(t\right)|dt<\infty $$

And thus by a well known theorem $$ \frac{d}{d\omega}\mathscr{F}\left(f\right)[\omega]=\mathscr{F}\left(g\right)[\omega] $$

Where $ g\left(t\right)=-itf\left(t\right) $.

Using integration by parts, we find that $$ \frac{d}{d\omega}\mathscr{F}\left(f\right)[\omega]=-\frac{\omega}{2}\mathscr{F}\left(f\right)[\omega] $$

Solving this differential equation we find: $$ \mathscr{F}\left(f\right)[\omega]=Ce^{-\frac{\omega^{2}}{4}} $$

And for $\omega=0 $ we have $$ \mathscr{F}\left(f\right)[0]=\frac{1}{2\pi}\intop_{-\infty}^{\infty}f\left(t\right)dt=\frac{1}{2\pi}\intop_{-\infty}^{\infty}e^{-t^{2}}dt=\frac{1}{2\sqrt{\pi}} $$

And hence we found $ C $.

This way require a lot of details, and I have much easier one.

The second way:

Note that :

$$ \mathscr{F}\left(f\right)[\omega]=\frac{1}{2\pi}\intop_{-\infty}^{\infty}e^{-t^{2}}e^{-i\omega t}dt=\frac{1}{2\pi}\intop_{-\infty}^{\infty}e^{-\left(t^{2}+i\omega t\right)}dt=\frac{1}{2\sqrt{\pi}}e^{-\frac{\omega^{2}}{4}}\intop_{-\infty}^{\infty}\frac{\sqrt{2}}{\sqrt{2\pi}}e^{-\left(t+\frac{i\omega}{2}\right)^{2}}dt $$

And note that the density of a normal distributed random variable is given by $ N\left(\mu,\sigma\right)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{\left(x-\mu\right)^{2}}{2\sigma^{2}}} $

So we got $ \frac{1}{2\sqrt{\pi}}e^{-\frac{\omega^{2}}{4}} $ multiplied by an integral (over the whole space) over a density of random distributed random variable with $ N\left(-\frac{i\omega}{2},\frac{1}{\sqrt{2}}\right) $.

We got the same result.

Finally, my question is:

When I learned probability theory, we never mentioned complex numbers, So my argument that this is an integral over a density of a normal distribution, may be not acceptable.

Is it legit?

Thanks in advance.

答:
If you know some complex analysis, you can fill in the details of the second argument. For simplicity, I'm going to omit all constant factors. Here's the claim:

For any $\beta\in\Bbb{R}$, we have $\int_{\Bbb{R}}e^{-t^2}\,dt = \int_{\Bbb{R}}e^{-(t+i\beta)^2}\,dt$.

To prove this, consider for each $R>0$, the closed rectangular loop $\gamma_R$ oriented counter-clockwise passing through the four points $-R,R, R+i\beta, -R+i\beta$ (draw a picture, and maybe you would like to assume $\beta>0$ as well for ease of visualization). Then, the function $z\mapsto e^{-z^2}$ is holomorphic on $\Bbb{C}$ so by Cauchy's theorem, we have
\begin{align}
\int_{\gamma_R}e^{-z^2}\,dz &= 0.
\end{align}
Writing this out explicitly, we have
\begin{align}
\int_{-R}^{R}e^{-x^2}\,dx + \int_{0}^{\beta}e^{-(R+iy)^2}\,dy - \int_{-R}^Re^{-(x+i\beta)^2}\,dx -\int_{0}^{\beta}e^{-(-R+iy)^2}\,dy &= 0.
\end{align}
Here, the $+,-$ signs take the orientation the the curve $\gamma_R$ into account (again draw a picture). I leave it to you to justify why as $R\to \infty$, the two integrals $\int_0^{\beta}$ vanish, thereby leaving us with the equality
\begin{align}
\int_{\Bbb{R}}e^{-t^2}\,dt =\int_{\Bbb{R}}e^{-(t+i\beta)^2}\,dt.
\end{align}


问:
Minimal polynomial of $\alpha \in \mathbb{Q}(\sqrt[3]{d})$
Given some cubefree $d$, is there a general way to determine the minimal polynomial over $\mathbb{Q}$ of $\alpha=1+\sqrt[3]{d}+\sqrt[3]{d^2} \in \mathbb{Q}(\sqrt[3]{d})$? Or more generally, $\alpha=a+b\sqrt[3]{d}+c\sqrt[3]{d^2}, a,b,c \in \mathbb{Q}$?

答:
Yes, You have that
\begin{equation}
\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{d})
\end{equation}
$\alpha$ is algebraic, since that
\begin{equation}
\alpha=\sqrt[3]{d}\Rightarrow \alpha^3-d=0
\end{equation}
Then the minimal polynomial are $p(x)=x^3-d$, the $[\mathbb{Q}(\sqrt[3]{d}):\mathbb{Q}]=3$, the  a basis $\mathbb{Q}$ over $\mathbb{Q}(\sqrt[3]{d})$ is:

\begin{equation}
\lbrace 1,\sqrt[3]{d},(\sqrt[3]{d})^2\rbrace
\end{equation}
so
\begin{equation}
\mathbb{Q}(\sqrt[3]{d})=\lbrace a+b\sqrt[3]{d}+c(\sqrt[3]{d})^2\big{|}a,b,c\in\mathbb{Q}\rbrace
\end{equation}


问:
Prove that the given sequence contains all natural numbers
A sequence $\{a_n\}$ is defined in the following manner. Let $a_1$ be any arbitrary natural number. Let $S_n=a_1+a_2+…a_n$. Define $a_{n+1}$ to be the smallest natural number $k$ such that $\gcd(S_n,k)=1$ and $a_i \neq k$ for $i=1,2,…,n$. Prove that every natural number appears in this sequence.

My work so far has just been some casework showing that $2$, $3$ and $4$ will always be in the sequence. Not sure how to extend this to all numbers- I tried induction, but that didn't work. Let the induction hypothesis be: at some point all numbers up to $n-1$ have appeared. Then if $n$ is prime then it's quite easy to show that at some point, $S$ will be divisible by $n$ and we are done. But it's not that simple when $n$ has more than one prime divisor.

答:
This is not a complete solution.

For $a_1 = 1$.

Hint: List out the first 20 terms.

Find a pattern. The pattern doesn't start immediately, which is why I asked for 20 terms.

That should be enough for you to guess what $a_i$ is, then prove it.

Here's my guess of how the general case will proceed:

Show that for any $a_1$, there exists an $N=4k+1$ such that the first $N$ terms in the sequence are the first $N$ integers.

Then, the result follows from $a_1 = 1$.


问:
partial ordering in weight lattice
Is there a command in Sage to find out the set of dominant weights less than or equal to a given dominant weight in the weight lattice ? Where the ordering is the partial ordering in the weight lattice: $\lambda \leq \mu$ if and only if $\mu-\lambda$ is a nonnegative integral linear combination of positive roots.

答:
something like this maybe</p>
```<code>sage: RS=RootSystem(['A',3])
sage: WL=RS.weight_lattice()
sage: RL=RS.root_lattice()
sage: l1,l2,l3=WL.gens()
sage: def neighbors(l):
....:     for r in RL.gens():
....:         delta = l - r
....:         if delta.is_dominant():
....:             yield delta
....:
sage: X = RecursivelyEnumeratedSet([3*(l1+l2)], neighbors)
sage: list(X)
[3*Lambda[1] + 3*Lambda[2],
 Lambda[1] + 4*Lambda[2],
 4*Lambda[1] + Lambda[2] + Lambda[3],
 2*Lambda[1] + 2*Lambda[2] + Lambda[3],
 3*Lambda[2] + Lambda[3],
 3*Lambda[1] + 2*Lambda[3],
 Lambda[1] + Lambda[2] + 2*Lambda[3],
 3*Lambda[1] + Lambda[2],
 Lambda[1] + 2*Lambda[2],
 2*Lambda[1] + Lambda[3],
 Lambda[2] + Lambda[3]]
</code>```

Title: partial ordering in weight lattice

Body:
something like this maybe

sage: RS=RootSystem(['A',3])
sage: WL=RS.weight_lattice()
sage: RL=RS.root_lattice()
sage: l1,l2,l3=WL.gens()
sage: def neighbors(l):
....:     for r in RL.gens():
....:         delta = l - r
....:         if delta.is_dominant():
....:             yield delta
....:
sage: X = RecursivelyEnumeratedSet([3*(l1+l2)], neighbors)
sage: list(X)
[3*Lambda[1] + 3*Lambda[2],
 Lambda[1] + 4*Lambda[2],
 4*Lambda[1] + Lambda[2] + Lambda[3],
 2*Lambda[1] + 2*Lambda[2] + Lambda[3],
 3*Lambda[2] + Lambda[3],
 3*Lambda[1] + 2*Lambda[3],
 Lambda[1] + Lambda[2] + 2*Lambda[3],
 3*Lambda[1] + Lambda[2],
 Lambda[1] + 2*Lambda[2],
 2*Lambda[1] + Lambda[3],
 Lambda[2] + Lambda[3]]


问:
Find all real $x$ and $y$ that satisfy $(x^3+y^3)(x^2+y^2)=2(x+y)=2$
I need help solving.

Find positive real numbers $x$ and $y$ that satisfies the equation:
$$(x^3+y^3)(x^2+y^2)=2(x+y)=2$$

I think that in these type of equations I have try to factor out in one side of the equation and look the prime factorization of the number on the right side; and, after that, divide into cases. But I can't seem to quite do it in this one.

答: By looking at the second equation $2 (x+y) = 2$, we see that $x+y=1$.  Now you might substitute $y = 1-x$ into the other equation and solve for $x$.


问:
Looking for general approach to find number of $(x,y,z)$ so that $x+y=z$
How many $(x,y,z)$ exist so that $x+y=z$ holds for $x<y<z\le10$ ($x,y,z\in \mathbb{Z}^+$ ) ?
$1)15\quad\quad\quad\quad\quad\quad2)20\quad\quad\quad\quad\quad\quad3)40\quad\quad\quad\quad\quad\quad4)45\quad\quad\quad\quad\quad\quad5)50\quad\quad\quad\quad\quad\quad$

I solved this problem with counting all the possibilities: $z$ should be at least $3$ and for $z=3$ the only possibility is $3=1+2$ I continued to count for other values of $z$:
$$z=3\to1\quad z=4\to1\quad z=5\to2\quad z=6\to2\quad z=7\to3\quad z=8\to3\quad z=9\to4\quad z=10\to4$$
So the answer is $1+1+2+2+3+3+4+4=20$.

But I wonder is it possible to solve this problem without counting all the possibilities like I did?  for example I don' know what to do if it was $x<y<z\le100$.

答:
Show that for fixed $z$, there are $ \lfloor (z-1) / 2 \rfloor $ solutions.

Hence, if we want to find $ z \leq 100$, then just add up the terms.


答:
One of the ways would be to first look at the number of positive integer solutions of $x+y \leq 10$. Taking $x = a + 1, y = b  +1$, it is equivalent to,

Number of non-negative solutions of $a + b + c = 8$ and that is $\displaystyle {10 \choose 2} = 45$. We discard solutions where $a = b = 0$ (as $x+y \gt 2)$ and we discard in general all solutions where $a = b $ (i.e $x = y$). That gives us $40$ solutions where $x \ne y$. This includes solutions where $x \gt y$ and $y \gt x$ and due to symmetry, the solutions that we are interested in is $\frac{40}{2} = 20$.

Title: Looking for general approach to find number of $(x,y,z)$ so that $x+y=z$

Body:
How many $(x,y,z)$ exist so that $x+y=z$ holds for $x<y<z\le10$ ($x,y,z\in \mathbb{Z}^+$ ) ?
$1)15\quad\quad\quad\quad\quad\quad2)20\quad\quad\quad\quad\quad\quad3)40\quad\quad\quad\quad\quad\quad4)45\quad\quad\quad\quad\quad\quad5)50\quad\quad\quad\quad\quad\quad$

I solved this problem with counting all the possibilities: $z$ should be at least $3$ and for $z=3$ the only possibility is $3=1+2$ I continued to count for other values of $z$:
$$z=3\to1\quad z=4\to1\quad z=5\to2\quad z=6\to2\quad z=7\to3\quad z=8\to3\quad z=9\to4\quad z=10\to4$$
So the answer is $1+1+2+2+3+3+4+4=20$.

But I wonder is it possible to solve this problem without counting all the possibilities like I did?  for example I don' know what to do if it was $x<y<z\le100$.


问:
Can all points in the plane be represented like this?
Solving a task regaring affine geometry, I've come across a problem:

Is it  true that, for every point $(x,y)\in \mathbb{R}^2$, there  exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that

$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$

I don't have any idea how to prove this or find a counterexample. All hints appreciated!

答:
This can be written as
\begin{align}
&\left(\begin{array}{cr}
\tilde{x} \\ \tilde{y}
\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cr}
1 &-1 \\ 1 & 1
\end{array}\right)\left(\begin{array}{cr}
x \\ y
\end{array}\right)=
\left(\begin{array}{cr}
t-1 & 0 \\ 0 & t+1
\end{array}\right)\left(\begin{array}{c}
\cos\left(\alpha+{\pi}/{4}\right) \\ \sin\left(\alpha+{\pi}/{4}\right)
\end{array}\right).\tag{1}
\end{align}
Since $(\tilde{x},\tilde{y})$ are obtained by rotation of $(x,y)$ by $\pi/4$, it suffices to show that any point in $\mathbb{R}^2$ can be represented by the right side of (1). This is equivalent to showing that for any point $P=(x,y)$ with $x,y\neq 0$ one can always find $t\neq \pm1$ such that the ellipse
$$\left(\frac{x}{t-1}\right)^2+\left(\frac{y}{t+1}\right)^2=1 \tag{2}$$
passes through $P$ (which is true - think on what happens to this ellipse when $t$ continuously decreases from $\infty$ to $1$, and then from $1$ to $-1$).</p>


答: Hint: expand out $\cos(2x+y)$, and you see that the right side will always be a linear combination of $\sin(x)$, $\cos(2x)$ and $\sin(2x)$.   So the left side...</p>

Title: Integral equation that's cant solve... Need a hand

Body:
Help me solve this integral equation, I'm having some troubles... I need to use the Fredholm method for second kind integral equations.

$$\phi(x)= \sin(x)+ \lambda \int_{0}^{\pi}\cos(2x+y)\phi (y)dy$$

Thanks


答:
NOT AN ANSWER

Here for everyone to see is the first 100 elements of this sequence, with $a_1=1$ (generated with Python)

```<code>[  1.   2.   4.   3.   7.   5.   9.   6.   8.  11.  13.  10.  12.  15.
  17.  14.  16.  19.  21.  18.  20.  23.  25.  22.  24.  27.  29.  26.
  28.  31.  33.  30.  32.  35.  37.  34.  36.  39.  41.  38.  40.  43.
  45.  42.  44.  47.  49.  46.  48.  51.  53.  50.  52.  55.  57.  54.
  56.  59.  61.  58.  60.  63.  65.  62.  64.  67.  69.  66.  68.  71.
  73.  70.  72.  75.  77.  74.  76.  79.  81.  78.  80.  83.  85.  82.
  84.  87.  89.  86.  88.  91.  93.  90.  92.  95.  97.  94.  96.  99.
 101.  98.]
</code>```

If you want to generate lists of your own, here is the code to do so:

```<code>import numpy as np
N=100
a=np.zeros(N)
a[0]=3
for n in range(1,N):
    S=np.sum(a[:n])
    k_found=False
    k=2
    while k_found==False:
        in_a=True
        while in_a==True:
            if k in a:
                k+=1
            else:
                in_a=False
        if np.gcd(int(S),int(k))==1:
            a[n]=k
            k_found=True
        else:
            k+=1
print(a)
</code>```

Here a[0] is the starting element (I have used $0$ because of the way Python indexes lists) and N is the number of terms you wish to compute.

Title: Prove that the given sequence contains all natural numbers

Body:
A sequence $\{a_n\}$ is defined in the following manner. Let $a_1$ be any arbitrary natural number. Let $S_n=a_1+a_2+…a_n$. Define $a_{n+1}$ to be the smallest natural number $k$ such that $\gcd(S_n,k)=1$ and $a_i \neq k$ for $i=1,2,…,n$. Prove that every natural number appears in this sequence.

My work so far has just been some casework showing that $2$, $3$ and $4$ will always be in the sequence. Not sure how to extend this to all numbers- I tried induction, but that didn't work. Let the induction hypothesis be: at some point all numbers up to $n-1$ have appeared. Then if $n$ is prime then it's quite easy to show that at some point, $S$ will be divisible by $n$ and we are done. But it's not that simple when $n$ has more than one prime divisor.

Body:
NOT AN ANSWER

Here for everyone to see is the first 100 elements of this sequence, with $a_1=1$ (generated with Python)

[  1.   2.   4.   3.   7.   5.   9.   6.   8.  11.  13.  10.  12.  15.
  17.  14.  16.  19.  21.  18.  20.  23.  25.  22.  24.  27.  29.  26.
  28.  31.  33.  30.  32.  35.  37.  34.  36.  39.  41.  38.  40.  43.
  45.  42.  44.  47.  49.  46.  48.  51.  53.  50.  52.  55.  57.  54.
  56.  59.  61.  58.  60.  63.  65.  62.  64.  67.  69.  66.  68.  71.
  73.  70.  72.  75.  77.  74.  76.  79.  81.  78.  80.  83.  85.  82.
  84.  87.  89.  86.  88.  91.  93.  90.  92.  95.  97.  94.  96.  99.
 101.  98.]

If you want to generate lists of your own, here is the code to do so:

import numpy as np
N=100
a=np.zeros(N)
a[0]=3
for n in range(1,N):
    S=np.sum(a[:n])
    k_found=False
    k=2
    while k_found==False:
        in_a=True
        while in_a==True:
            if k in a:
                k+=1
            else:
                in_a=False
        if np.gcd(int(S),int(k))==1:
            a[n]=k
            k_found=True
        else:
            k+=1
print(a)

Here a[0] is the starting element (I have used $0$ because of the way Python indexes lists) and N is the number of terms you wish to compute.


问:
How to solve this PDE $(x-y)\partial_x\partial_yu-\partial_xu+\partial_yu=0$
How do I solve this partial differential equation
$$(x-y)\partial_x\partial_yu-\partial_xu+\partial_yu=0$$

I know how to solve
$$\partial_x\partial_yu=0$$

The solution to this should be $u=f(x)+g(y)$, however I don't know how to solve it when there are other terms in the equation. Should I go for some $\xi(x,y),\eta(x,y)$ substitution or is there some other way ?

答: Let $z:=x-y$ so $\partial_z=\partial_x-\partial_y$ and $-z\partial_xv-v=0$ with $v:=\partial_zu$, i.e. $v=f(z)\exp(-x/z)$ with $f$ arbitrary. The "integration constant" in the indefinite integral $u=\int f(z)\exp(-x/z)dz$ is an arbitrary function of $x$.


问:
Lower limit of supremum of sequence of continuous functions
Let $X\subset\mathbb{R}$ and let $f_n:X\to\mathbb{R}$, $n\in\mathbb{N}$, by the way $\exists M\geq 0$ so that $|f_n(x)|\leq M$ for all $n\in\mathbb{N}$ and $x\in X$.

Prove that
$\liminf\limits_{n\to\infty}{(\sup\limits_{x\in X}{f_n(x)})}\geq\sup\limits_{x\in X}{(\liminf\limits_{n\to\infty}{f_n(x)})}$.

I've no idea how to prove this.

答:
Let $g(x)=\liminf_{n\rightarrow\infty}f_{n}(x)\in\mathbb{R}$ (it
is real-valued because $f_{n}$ is uniformly bounded). Let $y\in X$
and $\varepsilon>0$ be arbitrary. Since $\lim_{n\rightarrow\infty}\inf_{k\geq n}f_{k}(y)>g(y)-\varepsilon$,
there exists $N$ such that $\inf_{k\geq n}f_{k}(y)>g(y)-\varepsilon$
whenever $n\geq N$. In particular, $f_{k}(y)>g(y)-\varepsilon$ whenever
$k\geq N$. Clearly, for each $k\geq N$, $\sup_{x\in X}f_{k}(x)\geq f_{k}(y)>g(y)-\varepsilon$.
Therefore, for any $n\geq N$, $\inf_{k\geq n}\left(\sup_{x\in X}f_{k}(x)\right)\geq g(y)-\varepsilon.$
It follows that $\lim_{n\rightarrow\infty}\inf_{k\geq n}\left(\sup_{x\in X}f_{k}(x)\right)\geq g(y)-\varepsilon$.
That is, $\liminf_{n\rightarrow\infty}\left(\sup_{x\in X}f_{n}(x)\right)\geq g(y)-\varepsilon$.
Since $\varepsilon$ is arbitrary, we conclude that $\liminf_{n\rightarrow\infty}\left(\sup_{x\in X}f_{n}(x)\right)\geq g(y).$
Finally, since $y\in X$ is arbitrary too, we have that $\liminf_{n\rightarrow\infty}\left(\sup_{x\in X}f_{n}(x)\right)\geq\sup_{y\in X}g(y)=\sup_{x\in X}\left(\liminf_{n\rightarrow\infty}f_{n}(x)\right).$


问:
Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$
I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appearance. What is more amusing is the claim, without providing the proof, that is has the close-form result
\begin{align}
\frac{\sqrt{8\sqrt2+2\sqrt7}}{7^{3/4} }\tanh^{-1}  \frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 +\sqrt7}
+\frac{\sqrt{8\sqrt2-2\sqrt7}}{7^{3/4} }\tan^{-1}\frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 - \sqrt7}
\end{align}
I was able to verify it numerically; yet rather curious in how it could ever be derived. Based on my knowledge and experience, I do not assume it would be easy.

I was able to find an unsolved post here from a long time ago, which only reexpresses the corresponding  indefinite integral via the tangent half-angle substitution as
$$\int\frac{2t(1-t)}
{2 t^4-3 t^3+3 t^2+t+1}dt$$
but gives up due to the complexity in partial fractionalization. I am doubtful that this approach would lead to the explicit expression claimed above. I am interested in any suitable methods producing the close-form.

答:
Let's simply assume the roots are known and are $$\rho_1,\rho_2,\rho_3,\rho_4$$ and that way we evaluate

$$\int \dfrac{2t(1-t)}{(t-\rho_1)(t-\rho_2)(t-\rho_3)(t-\rho_4)}dt.$$

By fraction decomposition,

$$2t(1-t)=A(t-\rho_2)(t-\rho_3)(t-\rho_4)+B(t-\rho_1)(t-\rho_3)(t-\rho_4)+C(t-\rho_1)(t-\rho_2)(t-\rho_4)+D(t-\rho_1)(t-\rho_2)(t-\rho_3)$$ or,

$$A=\dfrac{2\rho_1(1-\rho_1)}{(\rho_1-\rho_2)(\rho_1-\rho_3)(\rho_1-\rho_4)}$$
$$B=\dfrac{2\rho_2(1-\rho_2)}{(\rho_2-\rho_1)(\rho_2-\rho_3)(\rho_2-\rho_4)}$$
$$C=\dfrac{2\rho_3(1-\rho_3)}{(\rho_3-\rho_1)(\rho_3-\rho_2)(\rho_3-\rho_4)}$$
$$D=\dfrac{2\rho_4(1-\rho_4)}{(\rho_4-\rho_1)(\rho_4-\rho_2)(\rho_4-\rho_3)}$$
and so
$$\int\frac{2t(1-t)}{2 t^4-3 t^3+3 t^2+t+1}dt = \dfrac{2\rho_1(1-\rho_1)}{(\rho_1-\rho_2)(\rho_1-\rho_3)(\rho_1-\rho_4)}\log|t-\rho_1|+\dfrac{2\rho_2(1-\rho_2)}{(\rho_2-\rho_1)(\rho_2-\rho_3)(\rho_2-\rho_4)}\log|t-\rho_2|+\dfrac{2\rho_3(1-\rho_3)}{(\rho_3-\rho_1)(\rho_3-\rho_2)(\rho_3-\rho_4)}\log|t-\rho_3|+\dfrac{2\rho_4(1-\rho_4)}{(\rho_4-\rho_1)(\rho_4-\rho_2)(\rho_4-\rho_3)}\log|t-\rho_4|+c$$ Now what remains are the roots given by

$$2 t^4-3 t^3+3 t^2+t+1=0.$$ I tried other substitutions which inevitably led to the same form.

Edit: I did not have the patience to evaluate the roots or to then compute the coefficients. However, wolfram kindly provides the simple expressions for the roots as

$$\dfrac{3+\epsilon \sqrt{7}i}{8}+ \dfrac{\varepsilon}{2}\sqrt{-\frac{7}{8}+\epsilon \frac{11\sqrt{7}i}{8}}$$ where $\epsilon,\varepsilon =\pm 1$ independently.


问:
Probability combined with permutations.
Suppose we select a permutation of the sequence 1,...,n, uniformly at random, where n ≥ 2 is even. What is the probability that none of numbers 1, . . . , n/2 stays in its place? Justify your answer.

I am struggling as to where to start any suggestions would be appreciated.

答:
Applying the inclusion exclusion principle one ends up with the expression:
$$
\frac1{n!}\sum_{i=0}^{n/2}(-1)^i\binom{n/2}i(n-i)!
$$

In particular for $n=10$ the probability is $\frac{18089}{30240}\approx0.598181$ in a good agreement with numerical result of the previous answer.

Title: Probability combined with permutations.


问:
Kalman filtering: Processing all measurements together vs processing them sequentially
If I have $m$ measurements to estimate an $n $ dimensional state vector, and I am using Kalman filter to do the filtering, then:
Should I put all the $m$ measurements together in the measurement matrix (measurement transformation matrix ) and perform the filtering or, should I filter each measurement sequentially? Please provide some supporting explanation for your choice.

For e.g: Let $m = 2$ and $n = 3$. The state vector is 3 dimensional and we need to use the two measurements to get the posterior estimate the state vector. Now I can use one of these two methods:

Use all these measurements together to form a gain matrix of size $3 \times 2$.

Use one measurement at one time and perform the filtering two times. The gain matrix will be $3\times 1$ in this case.

Which of the two methods is a better choice?

答:
Short Answer:

This is an interesting question that I have wondered myself, but I have not worked out this problem. My intuition says that both approaches are equivalent as long as the measurements are uncorrelated. The reason is that if the measurements are correlated, then you must include the off diagonal terms of the covariance matrix. This wouldn't be possible if the updates are performed one after another.

However, we can know once and for all if we work out the equations, which shouldn't be too difficult. Here, we will consider scalar measurements for simplicity, but the concept applied for vector measurements as well.

Long Answer:

I suggest looking at the information form of the Kalman filter (KF) and extended Kalman filter (EKF) for more insight in this respect. The information matrix (i.e., inverse of covariance matrix) of the information form can be written in a single equation as follows where the time indices are omitted for clarity:
$$
\begin{align}
  P^{-1} = (F P^{-1}F^T + Q)^{-1} + H^TR^{-1}H
\end{align}
$$
where $P$ is the covariance matrix, $F$ and $H$ are the Jacobians for the process and observation models, and $Q$ and $R$ are the covariance matrices for the process and observation noises.

Case 1 (Sequential Updates)

Let $A = \frac{\partial h_A}{\partial x_i}$ represent the Jacobian for the first measurement with measurement function $h_A$, and let $B = \frac{\partial h_B}{\partial x_i}$ represent the Jacobian for the second measurement with measurement function $h_B$. If we perform sequential updates, then we have the following:
$$
\begin{align}
  P^{-1} = (F P^{-1}F^T + Q)^{-1} + A^TR_A^{-1}A + B^TR_B^{-1}B
\end{align}
$$
Now, assume the measurement is a scalar, then
$$
  A^T R^{-1}_A A =
  \begin{bmatrix}
  a_{00} & \cdots  & a_{0n} \\
  \vdots   & \ddots  & \vdots \\
  a_{n0} & \cdots  & a_{nn}\\
  \end{bmatrix}, \;\;\;\;
  B^T R^{-1}_B B =
  \begin{bmatrix}
  b_{00} & \cdots  & b_{0n} \\
  \vdots   & \ddots  & \vdots \\
  b_{n0} & \cdots  & b_{nn}\\
  \end{bmatrix}
$$
where
$$a_{ij} = R^{-1}_A \frac{\partial h_{A}}{\partial x_i} \frac{\partial h_{A}}{\partial x_j}, \;\;\; b_{ij} = R^{-1}_A \frac{\partial h_{B}}{\partial x_i} \frac{\partial h_{B}}{\partial x_j}$$
Therefore,
$$
\begin{align}
A^T R^{-1} A + B^T R^{-1} B =
  \begin{bmatrix}
  a_{00} + b_{00} & \cdots  & a_{0n} + b_{0n} \\
  \vdots   & \ddots  & \vdots \\
  a_{n0} + b_{n0} & \cdots  & a_{nn} + b_{nn}\\
  \end{bmatrix}
\end{align}
$$

Case 2 (Simultaneous Updates)
Let $h = [h_A, h_B]^T$ represent the measurement function, but in this case, the measurement function is a vector-valued function; therefore, the Jacobian and is given as follows:
$$
H =
\begin{bmatrix}
  \frac{\partial h_{A}}{\partial x_0} & \frac{\partial h_{A}}{\partial x_1} & \cdots \\
  \frac{\partial h_{B}}{\partial x_0} & \frac{\partial h_{B}}{\partial x_1} & \cdots \\
\end{bmatrix}
$$
Now,
$$
H^T R^{-1} H =
  \begin{bmatrix}
  H_{00} & \cdots  & H_{0n} \\
  \vdots   & \ddots  & \vdots \\
  H_{n0} & \cdots  & H_{nn}\\
  \end{bmatrix}
$$
where
$$
H_{ij} = \frac{\partial h_{A}}{\partial x_j} \bigr(R^{-1}_{A} \frac{\partial h_{A}}{\partial x_i} + R^{-1}_{BA} \frac{\partial h_{B}}{\partial x_i}) + \frac{\partial h_{B}}{\partial x_j} \bigr(R^{-1}_{AB} \frac{\partial h_{A}}{\partial x_i} + R^{-1}_{B} \frac{\partial h_{B}}{\partial x_i})
$$
where (with slight abuse of notation)
$$
R^{-1} = \begin{bmatrix} R^{-1}_A & R^{-1}_{AB} \\ R^{-1}_{BA} & R^{-1}_B \end{bmatrix}
$$

So, if the measurements are uncorrelated (i.e., $R^{-1}_{AB} = R^{-1}_{BA} = 0$), then $H^T R^{-1} H = A^T R^{-1} A + B^T R^{-1} B$, which means that applying the updates sequentially or simultaneously givens the same results. However, in general, you cannot apply the updates sequentially for correlated measurements.


问:
Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series)

Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series).

I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.

I tried some things that didn't work so I won't post them.

Can you please explain me how do I find the sum of this series using Parseval's identity?

Thanks

答:
The series $\sum_{n\ge 1}\frac{1}{n^4}$ also appears in the following funny way. Consider the operator ${\rm d}^2/{\rm d}x^2$ on $[0,\pi]$ s.t. Dirichlet boundary conditions. Find Green's function $G(x,s)$. Find the normalized in $L^2[0.\pi]$ eigenfunctions $u_n(x)$ of the operator; the eigenvalues are $\lambda_n=n^2$.

Evaluate the integral $\int_0^\pi\int_0^\pi G^2(x,s){\rm d}x~{\rm d}s$ in two ways, a) using the explicit formula for $G(x,s)$ and b) using the representation $G(x,s)=\sum_{n\ge 1}\frac{u_n(x)u_n(s)}{\lambda_n}$.

After simplifications, you will find $\sum_{n\ge 1}\frac{1}{n^4}=\frac{\pi^4}{90}$.


问:
How did Le Verrier calculate Neptune's position?
In the Wikipdia article on Neptune the discovery is described as a mathematical achievement:

Subsequent observations revealed substantial deviations from the
  tables, leading Bouvard to hypothesize that an unknown body was
  perturbing the orbit through gravitational interaction. In 1843,
  John Couch Adams began work on the orbit of Uranus using the data he
  had. Via Cambridge Observatory director James Challis, he requested
  extra data from Sir George Airy, the Astronomer Royal, who supplied it
  in February 1844. Adams continued to work in 1845–46 and produced
  several different estimates of a new planet.
  [...]
  In 1845–46, Urbain Le Verrier, independently of Adams, developed his
  own calculations [...]. Neptune was discovered within 1° of where Le
  Verrier had predicted it
  to be, and about 12° from Adams' prediction.

Q1: Which method did Le Verrier employ to calculate Neptune's position with such accuracy?

Q2: How would it be done with today's tools?

Edit: Springer has the chapter in the answer available for download.

Edit 2: since the work of Le Verrier seems beyond the scope of a stackexchange question, is it possible to explain the general approach?

Quote from the article above:

Jean-Baptiste Biot attempted to explain Le Verrier's methods in six
  papers in *Journal des Savants (October 1846, pp. 577–596; November
  1846, pp. 641–664; December 1846, pp. 750–768; January 1847, pp.
  18–35; February 1847, pp. 65–86; March 1847, pp. 182–187). Arrived at
  the third paper, he writes: “As I progress in the task I have
  undertaken, the difficulty of the subject seems to increase.”

答:
There are some very nice illustrations and a detailed explanation on the equations that led to the discovery of Neptune in a thread about solar system dynamics. Here is the link:

http://forum.spaceengine.org/viewtopic.php?t=446&start=60#p22455

Since the thread is long, I cannot transcribe the entirety of it. They discuss the concept of perturbations on the angular momentum of a planet caused by the gravity of another nearby planet. And by some calculations and beautiful charts, pretty much explain how the two mathematicians were able to predict the existence of an unknown planet. I enjoyed reading it very much and highly recommend it to the future visitor of this question.


答:
Obviously $x+y=1$ and using $x^3+y^3=(x+y)(x^2-xy+y^2)$ we get
$$\begin{align}
&&(x^3+y^3)(x^2+y^2) &&=2 \\
&\iff&(x^2-xy+y^2)(x^2+y^2) &&=2 \\
&\iff&\left((x+y)^2-3xy\right)\;\left((x+y)^2-2xy\right) &&=2 \\
&\iff&(1-3xy)(1-2xy) &&=2
\end{align}$$

Put $xy=t$ we get a quadratic equation in t with solution $t_1 = 1$ and $t_2= -\frac{1}{6}$.
The system $xy=1$ and $x+y=1$ has no solution, while $xy=-\frac{1}{6}$ and $x+y=1$ results in a
quadratic equation $6x^2-6x-1=0$ with gives the final solution
$$x=\frac{3+\sqrt{15}}{6}\; \text {and} \; y=\frac{3-\sqrt{15}}{6}$$ or with x and y exchanged.

Title: Find all real $x$ and $y$ that satisfy $(x^3+y^3)(x^2+y^2)=2(x+y)=2$

Body:
I need help solving.

Find positive real numbers $x$ and $y$ that satisfies the equation:
$$(x^3+y^3)(x^2+y^2)=2(x+y)=2$$

I think that in these type of equations I have try to factor out in one side of the equation and look the prime factorization of the number on the right side; and, after that, divide into cases. But I can't seem to quite do it in this one.


问:
Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$
Let $T$ be a linear operator on a finite diemnsional inner product space $V$.

Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$.

Proof

 Let $\beta=\{v_1,\dots,v_n\}$ be an orthonormal basis for $V$ so $T(\beta)$.

$\langle T(v_i),T(v_j)\rangle=\langle v_i,v_j\rangle =\delta_{ij}$ , therefore $T(\beta)$ is an orthonormal basis for $V$.

The first equality is from the equivalent theorem $\langle T(x),T(y)\rangle=\langle x,y\rangle$.

I wonder why the last sentence means that $T(\beta)$ is an orthonormal basis for $V$.

They are linearly independent, of course, but they do span $V$? How can show that?

Sorry to make you be confused. I omitted this information : $TT^*=T^*T=I$.

答: So, now that you've added the hypothesis that $T$ is unitary, it makes sense. You've shown $T$ takes an orthonormal set of vectors to another orthonormal set. If $\{v_j\}$ is a basis, then $\dim V=n$, and so the $n$ linearly independent vectors $T(v_j)$ must likewise be a basis, as they span an $n$-dimensional subspace of $V$.</p>


问:
Can a submanifold of a flat Manifold be curved?
Consider the Minkowski space time

\begin{equation}
ds^2= -dt^2 + dx^2 + dy^2 + dz^2$
\end{equation}

Can we such a coordinate transformation on just the spatial part of the above metric (and not on the time part) such that the metric changes to something like

\begin{equation}
ds^2= -dt^2 + “ curved { } { } spatial  { } { } part”
\end{equation}

such that the original flat (Riemann Tensor vanishing indentically) 4d Manifold have a curved( non vanishing Riemann Tensor) 3d spatial submanifold.

Is such a case possible. Can a flat (Riemann Tensor vanishing indentically) manifold have a curved ( non vanishing Riemann Tensor) submanifold?

Can a flat 4d manifold be foliated with curved ( non vanishing Riemann Tensor) hypersurfaces ?

答:
Curvature is independent of coordinate system. If a metric is flat, it is flat. (For example, in $\Bbb R^2$, we have the standard metric $ds^2=dx^2+dy^2$; when we switch to polar coordinates, we have $ds^2=dr^2+r^2\,d\theta^2$. Christoffel symbols may be nonzero, but the curvature is still $0$.)

Of course flat spaces have submanifolds of all sorts of curvature. Just take any surfaces you wish sitting in $\Bbb R^3$; the same holds in all dimensions. It all then depends on the second fundamental form, which is not intrinsic to the submanifold. And you can certainly foliate flat space by non-flat hypersurfaces, yes.


答: Note that with $x, y$ positive and $x+y=1$ you can get $0\lt x^3\lt x^2 \lt x$ and $0\lt y^3\lt y^2\lt y$


问:
Nontrivial trivial integrals
Consider two propositions in geometry:

Circumscribe a right circular cylinder about a sphere.  The surface area of the cylinder between any two planes orthogonal to the cylinder's axis equals the surface area of the sphere between those two planes.  (Archimedes showed this.)

Let a curve running from the south pole to the north pole on a perfectly spherical earth meet every meridian of longitude at the same angle $\gamma$.  Then the length of the curve is proportional to $\sec\gamma$.  (In particular, it's infinite if $\gamma$ is a right angle, and one could even go beyond a right angle and consider it an oriented length, so that the oriented length is negative if the curve goes from the north pole to the south pole.  In that case, the infinite length would be the $\infty$ that's at both ends of the real line rather than $+\infty$ or $-\infty$, so the length depends continuously on $\gamma$.)

Both propositions admit the same kind of proof: For the first, consider what happens when the distance between the two planes is infinitesimal; then show that if it works then, then it also works for larger-than-infinitesimal distances.  For the second, show that the length of an infinitesimal increment of the curve is $\sec\gamma$ times the latitude-component of the distance, just by drawing an infinitesimal right triangle.

You might think that in both cases one is finding an integral, but it's an integral of a constant: just the constant itself times the length of the interval over which one integrates.  So it's as if you don't need to know about integrals and you certainly don't need to find any antiderivatives, but the method of proof doesn't work for larger-than-infinitesimal quantities.

One would like to say things like "energy is force times distance; it is only when the force is not constant that one must use integrals and multiply the force at an instant by the infinitesimal distance and then integrate".  But if proving the force is constant itself requires an infinitesimal viewpoint, then despite the triviality of computing the integral by multiplying two numbers, it seems necessary to the logic of the argument to view it as an integral.

So:

What other instances besides these two are known?  (I suspect there are many in geometry.)

What else of interest can be said about this phenomenon?

答: How about Mamikon's Sweeping-Tangent Theorem? The simplest example is the area of the annular ring between two concentric circles, where it can be shown that the area only depends on the length of the chord $a$. Thus one can imagine the inner circle shrinking to a point, and simply compute the area of an ordinary circle.

Title: Nontrivial trivial integrals


问:
How do I find the sum of a Fourier series around x=0?
For some reason, I haven't written it down, and the lecturer hasn't done it in his notes either, so it's probably in a lecture, which I could rewatch, but that means using up to 9hrs of my time trying to find that little bit about summation of Fourier series.

I got $-1+\sum_{m=0}^∞\frac{-12(-1)^{2m+1}+12}{(2m+1)^2π^2}cos(\frac{(2m+1)πx}{3})$

This should equal to $2x+2$ between $-3$ and (but not including) $0$, and $-2x+2$ between $0$ and (but not including) $3$.

I am asked to find the sum around $x=0$, except from understanding that x is 0, what do I do, is there any special way to calculate the sum of Fourier series?

答: The function is continous at $x=0$ (or the corresponding left and right limits are equal). Therefore the Fourier series converges to the value of the function at this point.


问:
Jordan canonical form of a square matrix with rational entries
Let $$A\in M_n(\mathbb{Q})$$ and $B$ it's Jordan canonical form. Is it true that B must have only rational entries? What if $A\in M_n(\mathbb{Z})$?

答:
The answer is no. As an example, the matrix
$$
A = \pmatrix{1 & 1\\1 & 0}
$$
has the Jordan form
$$
B = \frac 12\pmatrix{1 + \sqrt{5} & 0\\ 0 & 1 - \sqrt{5}}.
$$
It is true, however, that if $A$ is rational, then the Frobenius normal form of $A$ will have rational entries.


问:
How do you determine where the cdf has a jump?
I am trying to figure out where the cdf has a jump. I believe there is a jump at $x=\frac{1}{2}$, but it is obviously wrong and I am not sure why that is the case. I would appreciate if someone could provide an explanation on how it is done for any random variable and how the sketch looks like.

$Y(\omega)=\omega$ for $\omega\in[0,\frac{1}{2}]$ and $Y(\omega)=\frac{1}{4}$ for $\omega\in(\frac{1}{2},1]$

答:
For $0 \le x \le 1/2$, $\mathbb P(\omega \le 1/2, X \le x) = \mathbb P(\omega \le x) = x$.  Then there is an additional probability of $1/2$ that $X = 1/4$ with $\omega > 1/2$.  So the cdf is
$$ F(x) = \mathbb P(Y \le x) = \cases{0 & if $x < 0$\cr
                            x & if $0 \le x < 1/4$\cr
                             1/2 + x & if $1/4 \le x \le 1/2$\cr
                              1 & if $x > 1/2$\cr}$$


问:
Value of a zero-sum matrix games, if one matrix has smaller entries than another
Suppose that we have two zero-sum games represented by matrices $A = (a_{ij})$ and $B = (b_{ij})$. I am curious to know that if we assume $a_{ij} \leq b_{ij}$ for all $i,j$, does this imply that the value of nash equilibria in game $A$ has a smaller value than equilibria in game $B$?

This seems intuitive to me but I have not been able to find any information, nor prove or disprove the statement.

答:
We can write $B=A+E$ where $E$ is a matrix of non-negative numbers. If Player 1 (rows) plays the optimal strategy of the game $A$ (namely $x_A^*$) in the game $B$, and player 2 plays $y$, the payoff is:

$$x_A^*By=x_A^*(A+E)y=x_A^*Ay+x_A^*Ey\geq v_A + (x_A^*Ey)$$

where $v_A$ is the value of the game $A$. The second term is non-negative, since all entries in $E,x_A^*$ and $y$ are non-negative.
Thus, we found a strategy of Player 1 that garantuees at least $v_A$ in the game $B$. The optimal strategy in $B$ should get at least that, so $v_B\geq v_A$.


问:
Resultant of two polynomials in $K[X,Y]$ is a polynomial in $K[X]$ with degree less than the product of the original ones
I'm trying to prove the following:

Let $f, g$ be two polynomials in $K[X,Y]$ with $K$ a field and consider them as polynomials in one indeterminate with coefficientes in$ K[X]$, ie, as elements of $K[X][Y]$, then the resultant of these two polynomials would be another polynomial $d \in K[X]$ and I want to show that $\operatorname{deg}(d) \leq \operatorname{deg}(f)\operatorname{deg}(g)$.

I found it in a textbook on algebraic geometry as part of some proof but there is no hint or reference on how to prove it and I have no idea on how to get the result.

Could someone give a proof about this result?

Thanks a lot.

答:
Expand $f$ and $g$ as polynomials in $y$ as follows:

$$f= a_0y^d+a_1y^{d-1}+\cdots+a_d$$
$$g= b_0y^e+b_1y^{e-1}+\cdots+b_e.$$

Then the resultant of these two polynomials is given by the determinant of the Sylvester matrix. Here's an example when $f$ has degree three in $y$ and $g$ has degree two in $y$:

$$S=
\begin{pmatrix} a_0 & & b_0 & & \\
 a_1 & a_0 & b_1 & b_0 & \\
 a_2 & a_1 & b_2 & b_1 & b_0 \\
 a_3 & a_2 & & b_2 & b_1 \\
 & a_3 & & & b_2 \end{pmatrix}$$

Now recall that $\renewcommand{\s}{\sigma}\det S = \sum_{\s\in S_{d+e}} |\s|\prod s_{i\s(i)}$. I claim that when $s_{ij}\neq 0$, $\deg_x s_{ij} \leq \deg(f)-d+i-j$ or $\deg_x s_{ij} \leq \deg(g)-e+i-j+e=\deg(g)+i-j$ depending on whether $j\leq e$ or not. Therefore $$\deg_x \prod s_{i\s(i)} \leq e(\deg(f)-d)+d(\deg(g)) + \sum i-\s(i).$$ As $\sum i-\s(i) =0$ and $e(\deg(f)-d) + d(\deg(g)) \leq \deg(f)\deg(g)$, we have the result.


问:
If $A\subset A'$ are Boolean algebras, can the Stone space of $A$ be embedded in the Stone space of $A'$?
I think the following is true but I can't find a reference. If $A$ and $A'$ are Boolean algebras, and $A \subset A'$, then $S_A$ can be embedded in $S_{A'}$, where $S_A$ and $S_{A'}$ are the associated Stone spaces.

If anyone can point me in the right direction I'd be very grateful.

Greg

答: This is false.  For instance, let $A'=P(\mathbb{N})$ and let $A$ be the subalgebra consisting of finite or cofinite sets.  Then $S_A\cong\mathbb{N}\cup\{\infty\}$ and $S_{A'}\cong\beta\mathbb{N}$, but $\mathbb{N}\cup\{\infty\}$ does not embed in $\beta\mathbb{N}$.  (As a sketch of how to prove this, suppose $i:\mathbb{N}\to\beta\mathbb{N}$ is an embedding.  Then we can find disjoint subsets $A_n\subset\mathbb{N}$ such that $A_n$ is in the ultrafilter $i(n)$ for each $n$.  This means the set $A=\bigcup_n A_{2n}$ is in $i(n)$ iff $n$ is even, and so the sequence $(i(n))$ cannot converge in $\beta\mathbb{N}$ and $i$ cannot be extended to $\infty$.)


问:
Describe ideals of a ring
I have difficulties with a task in linear algebra.

R is a ring and R = {a, b, c, d}

These are the tables for + and . in R:

```<code> +| a b c d              .| a b c d
 -------------           -------------
 a| a b c d              a| a a a a
 b| b a d c              b| a b c d
 c| c d a b              c| a c c a
 d| d c b a              d| a d a d
</code>```

I know the definition for ideal and I understand why for example we take c and we look at the . table for the row and column of the element c to find out which elements stand there - because a.x and x.a should stay in the ideal.

But I don't quite understand the role of the + table here. We have that a-x should be in the ideal, too. So what? When we see that multiplying c to whatever element we get a or c, we should look if c+a and a+c gives us a or c, too? I can't quite get this.

Can you please help me? Thanks very much in advance!

答:
The addition table tells us that $a$ is the additive identity, i.e. $a=0$. The multiplication table tells us that $b$ is the multiplicative identity, i.e. $b=1$ so $R$ is a commutative ring with identity and we immediately get that $R$ and $(a)$ are ideals.

What are the non-trivial additive subgroups? We notice that $c+c=a=0$ and $d+d=a=0$ so $\{a,c\}$ and $\{a,d\}$ are additive subgroups. Are they ideals? A quick look at the multiplication table shows they are.

Are there any others? Since $b=1$ we know any ideal containing $b$ is the entire ring and since $c+d=b$ we know any any ideal containing both $c$ and $d$ must be the entire ring.

Therefore, the ideals are $R$, $(a)$, $(c)$ and $(d)$.

Title: Describe ideals of a ring


问:
Deduction in PA that irreducible implies prime
I'm currently showing a few things about $\mathbb{N}$ in the proof assistant Coq, with the goal to take these proofs and turn them into deductions in the first-order theory of PA.

One of the open goals right now is the implication "irreducible $\Rightarrow$ prime". Looking for proofs on this site and elsewhere on the web, I have so far only found proofs using the $\operatorname{gcd}$ plus Bézout's Lemma or ideals, everything taking place in $\mathbb{Z}$. The fact that they are shown for $\mathbb{Z}$ is not per se a problem, since it should be possible to show something like this:
$$
PA \vdash \forall x y ~ \exists ab ~~a x = by \, + \, \operatorname{gcd}(x,y) ~\lor ~ ay = bx \, + \, \operatorname{gcd}(x,y)
$$
which would be a translation of Bézout into PA. One could then continue from there.

However I feel that PA should be strong enough to circumvent the necessity to define and develop results about the $\operatorname{gcd}$ just to prove this implication. Since my attempts so far have not been fruitful, any hint for a different approach would be very welcome.

 Bonus points if the proof is in Heyting Artihmetic instead of PA.

答:
I now found a proof in this document (Proposition $1.4$) which uses strong induction.
The statement on which the induction is used can be simplified slightly, which I will present here for completeness. The proof also needs Euclid's lemma and the following fact about divisibility which can be shown by induction on $a$:

Lemma: $~~~n \vert (an + b) ~\Longleftrightarrow~ n\vert b$.

Proof: Assume that $x$ is irreducible and try to show $x \vert a b ~\Rightarrow~ x \vert a ~\lor ~ x \vert b~$ by strong induction on $a$. We distinguish two cases:

($x \leq a$) By Euclid we have $a = d x +r$, with $r < x$ implying $r < a$. Since we have
$$
x \vert ab ~\Longrightarrow~ x \vert (dxb + r b) ~\stackrel{\text{Lemma}}{\Longrightarrow}~ x \vert rb
$$
the induction hypothesis can be applied on $r$ giving us $~x \vert r ~\lor~ x \vert b~$, which shows  $~x \vert a ~\lor~ x \vert b$.

($x > a$) It is trivial for $a = 0$, so assume $a > 0$. In this case Euclid gives us $x = da +r$ with $r<a$. Since $~x \vert ab~$ there is a $k$ with $kx = ab$. So we have
$$
 x \vert xb ~\Longrightarrow ~ x \vert (dab + rb) ~\Longrightarrow~ x \vert (dkx + rb)
~\stackrel{\text{Lemma}}{\Longrightarrow}~ x \vert rb
$$
again making it possible to apply the induction hypothesis on $r$ to get $~x \vert r ~\lor~ x \vert b~$. If $~x\vert b~$ we are done.
If $~x \vert r~$ we must actually have $r = 0$ since $r < a < x$. Thus $x = da$ and by the irreducibility of $x$ we get $a = x ~\lor~ a = 1$ showing $~ x \vert a ~\lor~ x \vert b$ respectively.

tl;dr of the proof: Depending on whether $x \leq a$ or $x > a$ we do the divisions $a / x$ or $x / a$ with remainder and use strong induction on the remainder.


问:
Is there a name for this problem or any close problem?
I'm trying to figure out if there is a common name for the following optimization problem of computing products of subsequences of a sequence:

Given a set $S$ of $m$ sets $\{S_1,\dots, S_m\}$, where $S_i = \{j_0, \dots, j_{k_i}\}$ s.t. each $j_\ell$ is an index i.e. $0\leq j_\ell \leq n$, and a sequence of elements $X = (x_0,\dots, x_n)$, find the minimal number of multiplications it will take to compute all sets, assuming that for computing each set $S_i$ we need to multiply all elements with the corresponding indices i.e. $X(S_i) = \prod_{j \in S_i} x_j$.

For example, suppose we have $S= \{S_1,S_2,S_3,S_4\}$:

$S_1 = \{0,1,3,4\} \\ S_2 = \{0,1,2,3,4\} \\S_3 = \{0,1,2,5,6\}\\ S_4 = \{0,1,3,4,5,6\}$

The minimal number of multiplications will be $8$ (as far as I can tell):

$y_1 =x_0x_1\\
y_2 = y_1x_2\\
y_3 =x_3x_4\\
y_4 = x_5x_6\\
X(S_1) = y_1y_3\\
X(S_2) = y_2y_3\\
X(S_3) = y_2y_4\\
X(S_4) = X(S_1)y_4$

It's not exactly a set cover problem, though it's very similar. Generalized matrix chain multiplication is close too, seems even a better fit than the set cover, but maybe there is something else. I was wondering if some problem in graphs might be a better match.

Do you have any idea, what this problem might be called? Or to which one is it the closest?

答:
Not sure if this problem has a name, but you can solve it via integer linear programming as follows, where I have changed the meaning of $S$, $x$, and $y$.  For each subset $S \subseteq \{0,\dots,6\}$, let binary decision variable $x_S$ indicate whether subset $S$ is formed.  The problem is to minimize $\sum\limits_{S:|S|>1} x_S$ subject to
\begin{align}
x_S &= 1 &&\text{for the (four) target subsets $S$} \tag1\\
x_S &\le \sum_{\emptyset \not= T \subsetneq S} x_T x_{S \setminus T} &&\text{for all $S$ such that $|S|>1$} \tag2
\end{align}
Constraint $(1)$ forces the formation of the specified target subsets.
Constraint $(2)$ forces each formed subset to arise from the disjoint union of two other proper subsets.  You can linearize this constraint by introducing a new binary variable $y_{T,S \setminus T}$ and replacing $(2)$ with
\begin{align}
x_S &\le \sum_{\emptyset \not= T \subsetneq S} y_{T,S \setminus T} &&\text{for all $S$ such that $|S|>1$} \tag{2a} \\
y_{T,S \setminus T} &\le x_T &&\text{for all $S$ and $T$ such that $\emptyset \not= T \subsetneq S$ and $|S|>1$} \tag{2b} \\
y_{T,S \setminus T} &\le x_{S \setminus T} &&\text{for all $S$ and $T$ such that $\emptyset \not= T \subsetneq S$ and $|S|>1$} \tag{2c}
\end{align}
For your four target sets, the optimal objective value is indeed $8$, attained by taking
$x_S = 1$ for all $S$ with $|S|\le 1$,
$$x_{01}=x_{34}=x_{56}=x_{0156}=x_{0134}=x_{01234}=x_{01256}=x_{013456}=1,$$
and all other $x_S=0$.

In terms of graph theory, you can interpret $x_S$ as a node variable and $y_{T,S \setminus T}$ as an edge variable in an undirected graph.  The problem seems like a generalization of minimum edge cover for which only the target nodes need to be covered.


问:
Understanding bicomplex numbers
I found by chance, the set of Bicomplex numbers. These numbers took particularly my attention because of their similarity to my previous personal research and question. I should say that I can't really understand the fact that $j^2=+1$ (and must of other abstract algebra) without using matricial interpretations. When I look a bit on the Bicomplex numbers, the think that surprised me a lot was the fact that $ij=ji=k$ and $k^2=-1$. Because using matricial representations, we get :
$$ij=\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}$$
$$ji=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
So, $ij$ should be different then $ji$. Then if we define $k=ij$ or $k=ji$, we get both way $$k^2= I$$
Looked at the Wikipedia article and did some internet search but couldn't find a matricial representation for the bicomplex $k$. So I wanna ask some clarifications about where from the relations $ij=ji$ and $k^2=-1$ come from.</p>

答:
In Mathematica, the following code assd support for the split-complex unity J along with complex unity I. This allows to calculate any expressions with tessarines:

```<code> $Pre = If[FreeQ[#, J], #, Module[{tmp},

 tmp = Evaluate[
    MatrixFunction[Function[J, #], {{0, 1}, {1, 0}}]] //
   FullSimplify;
       tmp /. {{a_, b_}, {b_, a_}} -> a + J b]] &;

J^I

Out= 1/2 (1 + (-1)^I) + 1/2 J (1 - Cosh[Pi] + Sinh[Pi])
</code>```

Title: Understanding bicomplex numbers

Body:
In Mathematica, the following code assd support for the split-complex unity J along with complex unity I. This allows to calculate any expressions with tessarines:

 $Pre = If[FreeQ[#, J], #, Module[{tmp},

 tmp = Evaluate[
    MatrixFunction[Function[J, #], {{0, 1}, {1, 0}}]] //
   FullSimplify;
       tmp /. {{a_, b_}, {b_, a_}} -> a + J b]] &;

J^I

Out= 1/2 (1 + (-1)^I) + 1/2 J (1 - Cosh[Pi] + Sinh[Pi])


问:
Question about expectation and conditional expectation in pmf
I've been trying this question:

A discrete random variable has probability mass function:

$$
p(x) := \left\{
        \begin{array}{ll}
            \left(\frac{2}{5}\right)^x\frac{3}{5} & \quad x=0,1,2,\dots \\
             0 & \quad o.w.
        \end{array}
    \right.
$$

Let $A$ be the event that $X < 2$ and $B$ be the event that $X$ is even (including $0$).

These are the questions. I solved the first three, stuck on the last one.

What is the conditional probability of $A$ given $B$?

This is $P(A\mid B) = 0.84$.

What is the conditional probability of $B$ given $A$?

This is $P(B\mid A) = 0.71$.

Find the expected value of $X$

$$E(X) = 0.67.$$

I got this by computing the summation of $x\left(\frac{2}{5}\right)^x\frac{3}{5}$ from $0$ to $\infty$.

Find the conditional expected value of $X$ given $A^c$.

$E(X | A^c)$

This is where I'm stuck. I got the probabaility of the complement of $A$ to be $4/25$ $(U1 / 1-r)$ but I'm not sure how to use it to get the expected value they want here.

答:
First observe that you rv is a geometric  $Geo\left(\frac{3}{5}\right)$ counting the failures before the first success.

1)2) and 3) are correct but observe that

The expectation is given by $\frac{1-p}{p}=\frac{\frac{2}{5}}{\frac{3}{5}}=\frac{2}{3}$

For 4), without doing a lot of calculation, just remembering that this distribution is memoryless:

$$\mathbb{E}[X|X\geq 2]=\mathbb{E}[X]+2=2+\frac{2}{3}=\frac{8}{3}$$


问:
Intersection of principal filters
Prove that intersection of principal filters on $I$ is still a principal filter.

Let S be a nonempty set and C be a nonempty subset. Then {C} is a filter base. The filter it generates (i.e., the collection of all subsets containing C) is called the principal filter generated by C. Sorry for posting in this way, I understand it's not polite, but could you give any tip-off.

答: Take any two non-empty subsets $A,B$ of $I$, and let $\mathcal E,\mathcal F$ the principal filters on $I$ that they generate. Then $\mathcal E\cap\mathcal F$ is the set of subsets $C$ of $I$ that contain both $A$ and $B$ (that is, having both $A$ and $B$ as subsets). If $C\subseteq I$ is such that $A,B\subseteq C,$ what can you say with regard to $A\cup B$ and $C$? If $A\cup B\subseteq C,$ what can you say with regard to $A,B$ and $C$?</p>

Title: Intersection of principal filters


问:
Paradox: Roots of a polynomial require less information to express than coefficients?
A somewhat information theoretical paradox occurred to me, and I was wondering if anyone could resolve it.

Let $p(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0 = (x - r_0) \cdots (x - r_{n-1})$ be a degree $n$ polynomial with leading coefficient $1$. Clearly, the polynomial can be specified exactly by its $n$ coefficients $c=\{c_{n-1}, \ldots, c_0\}$ OR by its $n$ roots $r=\{r_{n-1}, \ldots, r_0\}$.

So the roots and the coefficients contain the same information. However, it takes less information to specify the roots, because their order doesn't matter. (i.e. the roots of the polynomial require $\lg(n!)$ bits less information to specify than the coefficients).

Isn't this a paradox? Or is my logic off somewhere?

Edit: To clarify, all values belong to any algebraically closed field (such as the complex numbers). And note that the leading coefficient is specified to be 1, meaning that there is absolutely a one-to-one correspondence between the $n$ remaining coefficients $c$ and the $n$ roots $r$.

答:
There is no paradox.

You cannot reason with real coefficients, because they convey infinite information and you can add them extra bits at no cost. You have to reason with coefficients having a finite representation, hence a finite number of polynomials. Let $p$ this number.

The information content of a polynomial chosen among $p$ is not the total number of bits of the coefficients, but just $\lg(p)$, assuming equiprobable polynomials.

There is only one way to produce the roots in sorted order, then this yields $\lg(p)$ bits of information again. If you enforce a specific order, you increase the amount of information by $\lg(n!)$. So supplying the roots as a tuple in fact takes more information than as a set.


问:
Probability of winning a patience game
The Game:
There is the following patience game:
One starts with an ordinary deck of cards, i.e. 52 cards and divides them equally into 13 piles of 4 face-down cards, each pile being labelled by a card value, i.e. one pile corresponding to aces, one to 2's etc.. After setting the game up, one takes a card from the top of the kings pile and looks at its value. This card is then placed face up under its corresponding pile, i.e. if it is a queen one would place it under the queens-pile. Next, the top card of this particular pile one has just added to, is picked up and one repeats the procedure, placing it under the pile corresponding to its value.

The game ends when the pile one has added to has no more face-down cards to pick up.

The Objective:
To win the game, one needs to turn over every card before the game ends. This means that the last card to be turned over is the fourth king, and that all other cards are turned over before the fourth king is.

The Problem:
Since this game is deterministic, i.e. its outcome cannot be changed once the cards are separated into piles, I have always wondered what the probability of winning this game is. Furthermore, when exploring this question it is also interesting to think about the chances of winning the analogue game with a different amount of piles, i.e. possible values of cards, and colors, i.e. how large the individual stacks are.

My Progress:
I have no experience working on problems like this one, so my progress is very very limited. Through my efforts I have only been able to find that the probability of winning the a game with $n$ stacks with 1 card is $1/(n!)$, as well as some other combination of very small stacks and colors (simply through working it out explicitly, not by noticing any general pattern).
Since winning the game is in some sense "finding a loop" through the entire deck of cards, I have also thought about trying to analyse when smaller loops exist (for example all 3's being in the 3 pile). In this sense, if there is more than one "loop", the game is destined to fail, so analysing them might solve the problem. However, I have not found a way to properly grapple with this idea in a productive way.

It would be very helpful if anyone could give me some Ideas or perhaps resources to read about problems like this.

答: The game ends when you pick a card and it is the fourth king. So you win if the last card drawn is a king. The set of cards that haven’t been turned round yet is arranged in random order, so the chance of the last card being a king is one in thirteen.</p>

Title: Probability of winning a patience game


问:
Closed Form of the Product of Two Summations with the Same Indices
Does a closed form exist for the following product of two summations with the same indices?

Assuming $a_i$, $b_i \in (-\infty, \infty)$, $i = \{1, 2, ..., n\}$, we have:
$$
\sum_{i=1}^{n} \frac{a_i^2}{(b_i - a_i)^2} \sum_{i=1}^{n} \frac{(b_i - a_i)^2}{2a_i^2}
$$

What I have so far, is that since there are cross-product terms, I will need to rewrite the summation as the following :

$$
\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{a_i^2}{(b_i - a_i)^2}  \frac{(b_j - a_j)^2}{2a_j^2}
$$

It is evident that the portion of the summation, whenever $i = j$, would sum up to $\frac{n}{2}$. What confuses me are the cross-terms, which do not seem to converge to some closed form or a constant.

However, since the indices are not the same in the two fractions, is there nothing I can do to the summation to reduce it further into a closed form? (particularly, in the context of the question I am working on, this summation is supposed to converge to some sort of a constant over 2).

答:
The question is stated in an overcomplicated way. Setting
$$x_i = \frac{a_i^2}{(b_i-a_i)^2},$$
You are asking for

$$\frac12 \sum_{i=1}^n x_i \sum_{i=1}^n \frac1{x_i}=\frac12 (n + \sum_{i\neq j} \frac{x_i}{x_j})$$


问:
Approximating the roots of $\epsilon^{2}x^{3}+x+1$
I saw the following in my lecture notes, and I am having difficulties
verifying the steps taken.

The question is:

Assuming $0<\epsilon\ll1$ find all the roots of the polynomial
  $$\epsilon^{2}x^{3}+x+1$$ which are $O(1)$ up to a precision of
  $O(\epsilon^{2})$

and the solution given was

Assume that $x=O(1)$ and that $$x(\epsilon)=x_{0}+\epsilon
 x_{1}+O(\epsilon^{2})$$ Then by setting it in the equation and letting
  $\epsilon\to0$ we get $$x_{0}=-1,x_{1}=0$$

Hence $x(\epsilon)=-1+O(\epsilon^{2})$

I have two questions:

Where did we use the assumption that $x=O(1)$

How did they get $$x_{0}=-1,x_{1}=0 ?$$

When I did the step setting
it in the equation and letting $\epsilon\to0$ I got $$x_{0}+1+O(\epsilon^{2})=0$$
and so I don't know anything about $x_{1}$.

Should I ignore $O(\epsilon^{2})$
and from that I should get $x_{0}=-1$

答:
This is not exactly as it is written above, but my approach would be as follows.

First, assume that $x(\epsilon) = O(1)$. Then we can say $x(\epsilon) = x_0 + f(\epsilon)$, where $f(\epsilon) = o(1)$ is some function in $\epsilon$, and $x_0$ is constant in $\epsilon$. Filling this in, and using that $x(\epsilon)$ is a root of the equation, we get
$$\epsilon^2 (x_0 + f(\epsilon))^3 + (x_0 + f(\epsilon)) + 1 = 0.$$
Expanding the cubic, we get
$$\left(\epsilon^2 x_0^3 + 3 \epsilon^2 x_0^2 f(\epsilon) + 3 \epsilon^2 x_0 f(\epsilon)^2 + \epsilon^2 f(\epsilon)^3\right) + \left(x_0 + f(\epsilon)\right) + 1 = 0.$$
Now, note that $f(\epsilon) = o(1)$, so also $f(\epsilon)^2 = o(1)$ and $f(\epsilon)^3 = o(1)$. Further note that $x_0 = O(1)$, so also $x_0^2 = O(1)$ and $x_0^3 = O(1)$. So the second, third and fourth terms above are all $o(\epsilon^2)$. So if we combine all these $o(\epsilon^2)$-terms into one $o(\epsilon^2)$-term, we get
$$\epsilon^2 x_0^3 + o(\epsilon^2) + x_0 + f(\epsilon) + 1 = 0.$$
Rewriting this slightly, to group the terms of the same order, we get
$$(x_0 + 1) + \left(f(\epsilon) + x_0^3 \epsilon^2 + o(\epsilon^2)\right) = 0.$$
Equating the proper order terms on both sides, we must therefore have $x_0 = -1$ and $f(\epsilon) = -x_0^3 \epsilon^2 + o(\epsilon^2) = \epsilon^2 + o(\epsilon^2)$, from which we may conclude that $x(\epsilon) = -1 + \epsilon^2 + o(\epsilon^2)$.

If we do not assume $x(\epsilon) = O(1)$, then we get more solutions. In particular, the two other solutions are of the form
$$\begin{align} x(\epsilon) &= \frac{\pm i}{\epsilon} + O(1) \end{align}$$
For these solutions, we cannot say that $x(\epsilon) = x_0 + o(1)$ for some constant $x_0$, so we did not find these solutions with the above method.


问:
Epsilon-delta proofs
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I am getting back to $\delta-\epsilon$ arguments after a break, so I'd like ask if my $\delta$-response to the below $\epsilon$-challenges is technically correct and rigorous.

Problem 4.2.2. For each stated limit, find the largest possible $\delta-$neighbourhood that is a proper response to the given $\epsilon$-challenge.

(a) $\lim_{x \to 3}(5x - 6) = 9$, where $\epsilon = 1$

(b) $\lim_{x \to 4}\sqrt{x} = 2$, where $\epsilon = 1$

(c) $\lim_{x \to \pi} [[x]] = 3$, where $\epsilon = 1$. (The function $[[x]]$ returns the greatest integer less than or equal to $x$)

(d) $\lim_{x \to \pi} [[x]] = 3$, where $\epsilon = 0.01$

Proof.

(a) We would like to make the distance $\absval{(5x - 6) - 9}<1$, so $\absval{5x - 15} < 1$, thus $\absval{x - 3} < 1/5$. Thus, $\delta = 1/5$.

(b) The expression $\absval{\sqrt{x} - 2}$ can be written as,

\begin{align*}
 \sqrt{x} - 2 &= (\sqrt{x} - 2) \times \frac{(\sqrt{x} + 2)}{(\sqrt{x} + 2)}\\
 &= \frac{\absval{(x - 4)}}{\absval{\sqrt{x} + 2}}\\
 &\le \frac{\absval{(x - 4)}}{2}\\
 &< \frac{\delta}{2}
\end{align*}

So, if we pick $\delta = 2$, the distance $\absval{\sqrt{x} - 2}$ would be smaller than $\epsilon = 1$.

(c) Consider the expression $\absval{[[x]] - 3}$.

We are interested to make the distance $\absval{[[x]] - 3}$ smaller than $\epsilon = 1$. So, $2 < [[x]] < 4$. Therefore, $[[x]]=3$. This is true, if and only if, $3 \le x < 4$. Consequently, $3 - \pi \le x - \pi < 4 - \pi$. This inequality is satisfied, if the distance $\absval{x - \pi} < \pi - 3$. So, $\delta = \pi - 3$.

(d) Again, $\delta = \pi - 3$ is a proper response to the given $\epsilon-$ challenge.

答:
all good except part (b), which is wrong because you are requested to find the largest $\delta$, which must be $(\delta = 3)$, since
$|\sqrt{4-3} - 2| = 1.$

Title: Epsilon-delta proofs


问:
Finding complex roots $z^3 = 4\sqrt{2}(-1 + i)$
I'm trying to find the complex roots of $z^3 = 4\sqrt{2}(-1 + i)$.

Here is what I've done:

I've figured that this should be equivalent to

$$z = 2(cos(-\frac{\pi}{12} + \frac{2k\pi}{3}) + isin(-\frac{\pi}{12} + \frac{2k\pi}{3}))$$

$$root_1 = 2e^{-\frac{\pi}{12}i}$$
$$root_2 = 2e^{\frac{7\pi}{12}i}$$
$$root_2 = 2e^{\frac{15\pi}{12}i}$$

Yet I'm not sure if my answer is correct, since the actual answer looks very different:

$root_1 = \sqrt{2}(1+i)$

$root_2 = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i + \frac{\sqrt{6}}{2}i$

$root_3 = -\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i - \frac{\sqrt{2}}{2}i$

Any help would be appreciated, thanks in advance!

答:
There is slight mistake. Note that $\arg (-1+i) = 3\pi /4$, not $-\pi/4$.

Hence for $k=0,1,2$
$$z=2 \exp \left\{ i\left( \frac{\pi}{4} + \frac{2k\pi}{3}\right) \right\} $$

Now you can evaluate.


答:
The argument of your $z$ equals $3\pi/4.$ So, the roots should be $2\exp(i\pi/4), 2 \exp(i(\pi/4 + 2\pi/3)), 2\exp(i(\pi/4 + 4\pi/3)).$ The arguments are

$\pi/4, 11\pi/12, 17\pi/12$

Title: Finding complex roots $z^3 = 4\sqrt{2}(-1 + i)$

Body:
I'm trying to find the complex roots of $z^3 = 4\sqrt{2}(-1 + i)$.

Here is what I've done:

I've figured that this should be equivalent to

$$z = 2(cos(-\frac{\pi}{12} + \frac{2k\pi}{3}) + isin(-\frac{\pi}{12} + \frac{2k\pi}{3}))$$

$$root_1 = 2e^{-\frac{\pi}{12}i}$$
$$root_2 = 2e^{\frac{7\pi}{12}i}$$
$$root_2 = 2e^{\frac{15\pi}{12}i}$$

Yet I'm not sure if my answer is correct, since the actual answer looks very different:

$root_1 = \sqrt{2}(1+i)$

$root_2 = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i + \frac{\sqrt{6}}{2}i$

$root_3 = -\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i - \frac{\sqrt{2}}{2}i$

Any help would be appreciated, thanks in advance!