问:
Variance for linear combination of coefficients
Why does:

$Var(aX)=a^2Var(X)$

and

$Var(aX\pm bY)=a^2var(X)+b^2var(Y)\pm 2ab Cov(X,Y)$?

I've looked at similar stackexchange links, but haven't been able to find a "simple" answer without extensive math proof.

答:
Simply use the definition of Variance expanding

$$\mathbb{V}[aX+bY]=\mathbb{E}[(aX+bY)^2]-(\mathbb{E}[aX+bY])^2$$

$$\mathbb{E}[a^2X^2+2abXY+b^2Y^2]-a^2\mathbb{E}^2[X]-b^2\mathbb{E}^2[Y]-2ab\mathbb{E}[X]\mathbb{E}[Y]$$

Using linearity of the expectation and re-organizing the addends you get

$$a^2[\mathbb{E}(X^2)-\mathbb{E}^2(X)]+b^2[\mathbb{E}(Y^2)-\mathbb{E}^2(Y)]+2ab[\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)]$$

Now you are almost done

$X$ and $Y$ are rv's thus Capital letters are needed

The second statement is valid also for $V(X-Y)$ thus I amended it.


问:
What do we mean visually by complex eigen values of a matrix? Intuition behind the rotation of space using a 2x2 matrix and eigen values?
The rotation matrix on $\Bbb R^2$ , the Euclidean plane given by $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ has two imaginary eigen values $i$ and $-i$. The definition of eigen vectors are those vectors $x$ that are parallel to $x$  [i.e. $Ax= \lambda x$].

Here the definition says $Ax=ix$ or $Ax=-ix$, since multiplying by $ i $ rotates my space by ninety degrees , does that essentially mean that we see two vectors in complex planes that are perpendicular to each other as being parallel to each other at the same time. What intuition am I missing here?

答:
We have an intuition of what "parallel" means in a real inner product space. Algebraically, it means one vector is a scalar multiple of another vector. This can get tricky, though, if our scalars are complex.

A big reason this gets tricky is because not only can there be a complex inner product $\langle u,v\rangle$ (which gives complex values), but there is an induced real inner product $\mathrm{Re}\langle u,v\rangle$. For instance, the standard complex inner product space $\mathbb{C}^n$ may be reinterpreted as the real inner product space $\mathbb{R}^{2n}$. So for instance, this means that within $\mathbb{C}^1$ itself, any two complex numbers are "parallel" because they are all multiples of each other, but interpreted as a real inner product space this is no longer true - e.g. $1$ and $i$ are perpendicular, not parallel.

When you say the matrix $A$ rotates a vector by $90^{\circ}$ and is thus perpendicular, you are describing that from the point of view of the real inner product on $\mathbb{C}^2$ (or, of course, the real inner product subspace $\mathbb{R}^2$), but it is no longer true if we treat $\mathbb{C}^2$ as a complex inner product space, as an eigenvector $x$ and $Ax=ix$ are now "parallel" in the sense that algebraically they are scalar multiples of each other.

ContentLicense: CC BY-SA 4.0


问:
Critical points of the moment map
Let $(M,\omega)$ be a symplectic compact connexe manifold endowed with a hamiltonian action of a torus $T$. Let $\mu : M \longrightarrow {Lie(T)}^*,$  be a moment map associated to this action.

We say that a point $m \in M $ is critical point of $\mu$, if $T_m \mu : T_m M \longrightarrow {Lie(T)}^*$,  the tangent map of $\mu$ is not surjective.

How can one prove that the above definition of critical points of the moment map $\mu$ is equivalent to the following statements:

(1).  $m \in M $ is critical point of $\mu$ if and only if the stabilizer of m, stab(m), contains a sub-torus of T of dimension 1.

(2). The tangent map $T_m \mu$ is not surjective if there exists an element $X \in Lie(T)-\lbrace 0 \rbrace$, such that $\langle T_m \mu, X \rangle = 0$.

Any feedback would be appreciated.

答:
We have $(ker T_m\mu)^\perp =T_mO$, if $O$ is the orbit of the action of the torus, and $\perp$ denote the symplectic orthogonal.

let us accept this basic fact. then the equivalence is obvious : $T_m\mu$ is not surjective iff the co-dimension of its kernel is smalle than the dimension of the image, namely the dimension of the torus, and the dimension of the orbit is smaller than the dimension of the torus iff the stabilizer contains a sub-torus.

The proof of the "basic fact" is well known : $T_mO$ is the set of fondamental vectors, or the image by the transpose of the derivative of the moment moment : $(T \mu )^t= T_mO$. the result follows as the image of the transpose of a linear map is just the orthogonal of the kernel of this map.

Title: Critical points of the moment map


问:
Solving $e^{0.0002n} + e^{0.0003n} = 0.01$
I need to solve an exponent equation which is:

$$
        e^{0.0002n} + e^{0.0003n} = 0.01
$$

Please help me with this equation.

Thanks.

Body:
Let $\;x=e^{-0.0001n}\;$.

The equation

$e^{0.0002n} + e^{0.0003n} = 0.01$

turns into

$x^3-100x-100=0$.

The only positive solution is

$x=\dfrac{20}{3}\sqrt{3}\cos\left(\dfrac{\theta}{3}\right)\simeq10.46680532$

where $\;\theta=\cos^{-1}\left(\dfrac{3\sqrt{3}}{20}\right)\simeq74.94135263^\text{o}.$

Hence

$n=-10000\ln\left[\dfrac{20}{3}\sqrt{3}\cos\left(\dfrac{\theta}{3}\right)\right]\simeq-23482.08851$

where $\;\theta=\cos^{-1}\left(\dfrac{3\sqrt{3}}{20}\right)\simeq74.94135263^\text{o}.$


问:
Limit by polar coordinates
Hi I need some help with this limit:
$$
\lim_{(x,y) \to (0,0)}  \frac{x^4y}{x^2+(x+y)^2}$$
I used polar coordinates and got that
$$
\frac{x^4y}{x^2+(x+y)^2}= \frac{r^3\cdot(\cos φ)^4  \sin φ } {1+ (\cosφ)^2+\sin(2φ)}
$$
I didn't know how to proceed from this.

答:
As an alternative, without polar coordinates, since $(x+y)^2 \ge 0$, we have that

$$x=0 \,(y\neq 0)\implies  \frac{x^4y}{x^2+(x+y)^2}=0$$

$$x\neq 0 \implies \left|\frac{x^4y}{x^2+(x+y)^2}\right|\le\left|\frac{x^4y}{x^2}\right| =x^2|y| \to 0$$

Note that using polar coordinates, to conclude by squeeze theorem again, we need to show that

$$f(\theta)=\frac{(\cos φ)^4  \sin φ } { (\cosφ)^2+2\sin φ\cos φ +1}$$

is bounded which leads to the same previous method.

Title: Limit by polar coordinates


问:
Prove that $x < y$ implies $x^{1/n} < y^{1/n}$.
Let $x$ and $y$ be positive real numbers and let $m$ and $n$ be positive integers.

Prove that $x < y$ implies $x^{1/n} < y^{1/n}$.

I have so far figured out two different approaches to prove this.

By induction:

Base case: $P(1)\iff x^{1/1} < y^{1/1} \iff x<y$

Inductive step: $P(n+1)\iff x^{1/n+1} < y^{1/n+1} $

But then I'm stuck from here as I 'm not sure on how to solve $1/n+1$.

Another approach is to use direct proof:

By rules of inequalities we know

$x<y\implies x-x < y-x$

$\implies xy < yy$

$\implies xx < xy$

Hence $x^{1/n} < y^{1/n}$
as
$(x^1)^{1/n} < (y^1)^{1/n}$

May I ask how can I solve for $1/n+1$ at the first approach and is the second approach correct at all?

答: One can argue this way: Let if possible, there exists some $n$ such that $x^{\frac{1}{n}}\geqslant y^{\frac{1}{n}}$ then take nth power both sides and we get $x\geqslant y$. Everything is fine with inequality as $x$ and $y$ are positive.


问:
Is $f$ necessarily measurable?
(1) Suppose a function $f$ has a [Lebesgue] measurable domain and is continuous except at a finite number of points. Is $f$ is necessarily [Lebesgue] measurable?

Comments For (1), If $f$ is defined on a [Lebesgue] measurable set $E$ and is continuous except for a finite number of values say $x_1,x_2, ... x_n$ are those points of discontinuity, then can't we describe the pre-images of $f$ just as a finite union of of the pre-images of the collection of continuous functions $\{f_i\}_{i=1}^{n}$, where each $f_i$ is defined up between each point of discontinuity of $f$? Or am I missing something here?

(2) Suppose the function $f$ is defined on a measurable set $E$ and has the property that $\{x \in E | f(x) > c\}$ is measurable for each rational number $c$. Is $f$ necessarily [Lebesgue] measurable?

 Comments I got this one, thanks to everyone who commented.

Any hints would be appreciated.
The text being used is Royden-Fitzpatrick 4th Edition.

答:
(2) It is true, this not contradicts the proposition 1 (Page 54), just is an equivalent form of proposition 1.

Suppose $\{x\in E : f(x)>c\}$ measurable for each rational $c$. Let $r\in\mathbb{R}$, then exist a sequence $\{c_n\}$ of rational numbers such that $c_n\to r$ and $c_n\geq r$ for all $n\in\mathbb{N}$. Note that

$$A=\{x\in E : f(x)>r\}=\{x\in E : f(x)\leq r\}^c=\left(\bigcap\{x\in E : f(x)\leq c_n\}\right)^c=\bigcup \{x\in E : f(x)> c_n\}$$

Hence $A$ is measurable. Try (1) by yourself.

(1) When you say "where each $f_i$ is defined up between each point of discontinuity of $f$" you are wrong, $E$ does not necessarily has a good order, we only know that $E$ is a topological space. If you don't assume that each open set in a topology space must be measurable, (1) is obviously false (indeed, proposition 3 in page 55 Royden 4th ed. will fail). The correct word in this Royden Chapter is Lebesgue Measurable functions instead of measurable function, then we know that $E\subset\mathbb{R}$.

If $E\subset\mathbb{R}$ (1) is true. Instead of considering $f_n$ consider simply $f|\{E\setminus{x_1,\ldots,x_n}\}$, define $f(x_i)=y_i\in\mathbb{R}$. What happen when an open interval $I$ contains some of $y_i$? (in this case is much easier work with $y_n$ rather than $x_n$)

How fails "preimage of open set is open" when a continuous function $f\colon\mathbb{R}\to\mathbb{R}$ has finitely many discontinuities?. Note that the only case of preimage of an open interval that is not an open set correspond to closed intervals, semi-open intervals (for example in $[a,b)$ form) or numerable union of them, so that not only the function $f$ is measurable, it is also Borel measurable (preimage of open sets are Borel Measurable).

In any topological space $(X,\tau)$ where each open set is measurable, and consider as measurable sets the Borel $\sigma$-algebra (the smallest sigma algebra containing the open sets) I don't know if (1) is true.

ContentLicense: CC BY-SA 3.0

Title: Is $f$ necessarily measurable?


问:
Can a projection operator increase the $L_\infty$ norm?
Suppose $P$ is a $n\times n$ matrix of real values satisfying $P^2=P$ so that $P$ is a projection matrix. Is it possible that

$$\|Px\|_\infty > \|x\|_\infty$$

for some $x\in \mathbb{R}^n$? Here $\|y\|_\infty := \max_i |y_i|~.$

If possible, please give an example. If not possible, how can I prove it?

答:
Take the following matrix $A$ written in the canonical basis of $\mathbb{R}^2$:

\begin{pmatrix}
3 & -6  \\
1 & -2
\end{pmatrix}
It is idempotent ($A^2=A$) and if you take the vector $v = \begin{pmatrix}
1 \\
0
\end{pmatrix}$ then $\vert\vert v \vert\vert_{\infty} = 1$ and $\vert\vert Av \vert\vert_{\infty} = 3$.


答:
HINT:

$$x^{\frac{1}{n}}>y^{\frac{1}{n}}$$
What happens if we raise everything to the power of
$$\frac{n}{n+1}\;\;?$$

Body:
Let $x$ and $y$ be positive real numbers and let $m$ and $n$ be positive integers.

Prove that $x < y$ implies $x^{1/n} < y^{1/n}$.

I have so far figured out two different approaches to prove this.

By induction:

Base case: $P(1)\iff x^{1/1} < y^{1/1} \iff x<y$

Inductive step: $P(n+1)\iff x^{1/n+1} < y^{1/n+1} $

But then I'm stuck from here as I 'm not sure on how to solve $1/n+1$.

Another approach is to use direct proof:

By rules of inequalities we know

$x<y\implies x-x < y-x$

$\implies xy < yy$

$\implies xx < xy$

Hence $x^{1/n} < y^{1/n}$
as
$(x^1)^{1/n} < (y^1)^{1/n}$

May I ask how can I solve for $1/n+1$ at the first approach and is the second approach correct at all?


问:
Selecting of students under restraints

A delegation of $4$ students is to be selected from $12$ students. In how many ways can the delegation be selected if $2$ particular students wish to be together and other $2$ wish not to be together in the delegation?

A B wish to be together C D not together.

My attempt:

CASE$1$

A B are included (remaining $10$, $2$ to choose), C is included D excluded: $\binom11\times\binom81$

CASE$2$

A B are included (remaining $10, 2$ to choose), D is included C excluded:$\binom11\times\binom81$

CASE$3$

A B are both excluded(remaining $10, 4$ to choose), C is included D excluded: $\binom11\times\binom94$

CASE$4$

A B are both excluded (remaining $10, 4$ to choose), D is included C excluded: $\binom11\times\binom93$

Edit

CASE$5$

A B are both excluded(remaining $10, 4$ to choose), D is excluded C excluded: $\binom11\times\binom84$

CASE$6$

A B are both included(remaining $10, 2$ to choose), D is excluded C excluded: $\binom11\times\binom82$

Total: $324$

This apparently gave the wrong answer and I feel I've done a mistake in selecting. I don't know which case is extra.

答: In addition to the cases you were missing, your answers for case $3$ and $4$ are wrong. AB are excluded and either one of CD is included. So out of the remaining $8$ students, you want to select $3$. This will give you the correct answer of $226$.

Title: Selecting of students under restraints


问:
Find the last three digits of $17^{102}$.

Find the last three digits of $17^{102}$.

That is we have to find out $17^{102}\equiv ?\pmod{1000}$.

Now if we solve it through general congruence relation , then the process becomes very difficult , computational and laborious. Again from Euler's theorem , $$17^{\phi(1000)}\equiv 1 \pmod{1000}.$$ But $\phi(1000)=400(>102)$. So it can't help me.

Does there any simplest way to solve this congruence relation easily ?

Body:
The units digit of $17^1$ is $7$.

The units digit of $17^2$ is $9$ ($7 \times 7 = 49$).

The units digit of $17^3$ is $3$ ($7 \times 9 = 63$).

The units digit of $17^4$ is $1$ ($7 \times 3 = 21$).

Calculating,

$\; 17^4 = 289 \cdot 289 = (280+9)^2 = 280^2 + 18 \cdot 280 + 81 \equiv$

$\quad\quad\quad 400 + 800 + 600 + 640 + 81 \equiv 521 \pmod{1000}$

Calculating,

$\; 17^{102} \equiv 17^2 \cdot 521^{25} \equiv 289 \cdot (1 + 520)^{25}  \equiv 289 \cdot (1 + 2^3 \cdot 5^1 \cdot 13^1)^{25} \equiv$

$\quad\quad\quad\; \, \; 289 \; \bigr(1 + \binom{25}{1}\,2^3 \cdot 5^1 \cdot 13^1 + \binom{25}{2}\,2^6 \cdot 5^2 \cdot 13^2 + 0\bigr) \equiv$

$\quad\quad\quad\; \, \;  289 \; \bigr(1 + 0 + 0 + 0 \bigr) \equiv 289 \pmod{1000}$


问:
Is there a simple closed curve in the plane whose length is six meters and bounds an area of three square meters?
I have tried it for a rectangle if I could find sides of rectangle whose length is six meters and area is three square meters. Let $x$ and $y$ be the length and breadth of rectangle. The product of $x$ and $y$ is three square meters which is area and perimeter is twice the sum of length and breadth is six meters. On calculating, I got the complex values of $x$ and $y$. whether I have tried according to the statement of question else if I am wrong in solving the problem. Then please guide which will be the correct simple closed curve for this condition.

答:
To bound an area of $3$ square metres the least possible curve length is such that the curve is a circle, by the isoperimetric inequality. This circle will have a radius of $\sqrt{\frac3\pi}$ metres and a length of $2\pi\sqrt{\frac3\pi}=6.1399\dots$ metres. Since we only have $6$ metres to work with, the task is impossible.

For the same reason, we cannot enclose $4$ square metres with a $7$-metre curve, nor $8$ with $10$, but we can enclose $5$ with $8$.


问:
Notation $\ln|x|$ vs. $\ln(x)$
Years ago, when I was taught about the function $\ln$ (logarithm in base $e$), all of my teachers and our book, too, insisted that we should write input of this function inside the absolute value notation and I am doing this since now. But, now when I am reading some university books or some answers on this site, I see in most answers, people write $\ln(x)$  using parentheses. It's been a question for a long time to me why people just use parentheses instead and how it is not wrong conventionally? I am sure if I used $\ln(x)$ in high school, it would've always been possible to get a minus point! I am a university student now. Can I write $\ln(x)$ safely and is it conventionally acceptable in mathematics? (I mean, in general, for $\ln(f(x))$ not only $\ln(x)$)

答:
The domain of $\ln$ is $(0,\infty)$ and if $x\in(0,\infty)$, then $\ln(x)=\ln|x|$. And if $x\notin(0,+\infty)$, $\ln(x)$ is undefined.

I suspect that you were told something a bit different, namely that a primitive of $\frac1x$ is $\ln|x|$. That's another matter, since $\frac1x$ is defined for every $x\ne0$, and therefore, if we want to work with a primitive of $\frac1x$, it must also be defined for every $x\ne0$ too.

Title: Notation $\ln|x|$ vs. $\ln(x)$


答:
Your book is an outlier. $\ln(x)$ is different from $\ln|x|$; the latter is practically only encountered as the indefinite integral of $\frac1x$ from $0$ and is the result of precomposing the absolute value function to $\ln(x)$.

An increasing number of sources I read, and my own answers on this site, are lazy enough to go all the way and omit brackets: $\ln x$.

Title: Notation $\ln|x|$ vs. $\ln(x)$

Body: Years ago, when I was taught about the function $\ln$ (logarithm in base $e$), all of my teachers and our book, too, insisted that we should write input of this function inside the absolute value notation and I am doing this since now. But, now when I am reading some university books or some answers on this site, I see in most answers, people write $\ln(x)$  using parentheses. It's been a question for a long time to me why people just use parentheses instead and how it is not wrong conventionally? I am sure if I used $\ln(x)$ in high school, it would've always been possible to get a minus point! I am a university student now. Can I write $\ln(x)$ safely and is it conventionally acceptable in mathematics? (I mean, in general, for $\ln(f(x))$ not only $\ln(x)$)


问:
Evaluating integral similar to $\sin^{-1}(3x)$
I was just doing revision for an upcoming exam and I came across a question I do not know how to answer. It seems pretty simple, however I am having a blank moment. If anyone knows how to solve that I would really appreciate the help.

The question:

It is given that y = $\sin^{-1}(ax)$, where a is a positive constant and $-\frac{1}{a}\leq x\leq \frac{1}{x}$.

a) Find $\frac{dy}{dx}$ in terms of y and a.

b) Hence find $\frac{dy}{dx}$ in terms of x and a.

c) Hence, or otherwise, find $\int \frac{1}{\sqrt{1-9x^{2}}}dx$.

d) Find the exact value of $\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$

I have finished parts a to c (workings below) but cannot seem to solve part d (I do realize it's probably relatively simple, however I have just spent good 45 minutes trying to solve it or find the answer and I can't seem to solve it)

$$
y=\sin^{-1}(ax)\\\sin(y)=ax\\\frac{dy}{dx}\cos(y)=a\\\Rightarrow\frac{dy}{dx}=\frac{a}{\cos(y)}=\frac{a}{\sqrt{1-\sin^{2}(y)}}=\frac{a}{\sqrt{1-a^{2}x^{2}}}\\\int\frac{1}{\sqrt{1-9x^{2}}}dx = \frac{1}{3}\int{\frac{3}{\sqrt{1-9x^{2}}}}dx=\frac{1}{3}\sin^{-1}(3x)+c$$

Any help would be greatly appreciated.

Body:
use this formula
$$\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin \frac{x}{a}+C$$
$$\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$$
$$\left(\frac{3x}{6}\sqrt{1-9x^{2}}+\frac{1}{6}\arcsin\frac{3x}{1} \right)_{0}^{1/6}$$
$$\left(\frac{x}{2}\sqrt{1-9x^{2}}+\frac{1}{6}\arcsin\frac{3x}{1} \right)_{0}^{1/6}$$
$$(\frac{1}{12}\sqrt{1-\frac{9}{36}}+\frac{1}{6}\arcsin\frac{1}{2})-0$$
$$\frac{\sqrt{27}}{72}+\frac{\pi}{36}$$
$$\dfrac1{72} \left(2 \pi + 3 \sqrt{3} \right)$$


问:
A question regarding properties of Algebraic Integers
I am trying questions of Field theory from Thomas Hungerford and couldn't solve this question exercise 18 of Section 5.1 .

I have done (a), (b) but couldn't do any of parts of (c) , (d) .

I tried using definition of algebraic integer by assuming a polynomial p(x) in $\mathbb{Z}(x) $ and trying to prove that there will exist an polynomial in $\mathbb{Z}(x) $ with required properties but couldn't for any parts of (c) and (d).

Can you please tell how to solve any part of (c) or(d) . Rest I would like to try by myself.

Thanks!!

答:
For the part (c), consider $P(X-n)$ and $n^d P(\frac{X}{n})$ where $P$ is a monic polynomial of $\mathbb{Z}[X]$ vanishing $u$ and $d$ its degree.

The results of the question (d) is well-known but I don't know if there is simple answer. The classical proof consists in proving that $u$ is integral over $\mathbb{Z}$ if and only if the subring $\mathbb{Z}[u]$ is finitely generated over $\mathbb{Z}$. There exists an alternative proof based on the resultant which produce an explicit equation, see the question How to prove that the sum and product of two algebraic numbers is algebraic?.


问:
Let $f(x)$ be a polynomial in $\mathbb{Q}[x]$. Prove that if $a + \sqrt2 b$ is a zero of $f(x)$ then so is $a - \sqrt2b$.

Let $f(x)$ be a polynomial in $\mathbb{Q}[x]$. Prove that if $a + \sqrt2 b$ is a zero of $f(x)$ then so is $a - \sqrt2b$.

Can I treat $a + \sqrt2 b$ like a complex number? Then I can do something like

let $f(x) = \sum a_ix^i$, and  $z = a+\sqrt2b$

Then $f(\bar z) = \sum a_i \bar z^i = ... = \bar{f(z)}$.

If not.. what else can I try?

答:
Hint: employ conjugation  as here or, if that's unknown, give a direct proof, e.g. as below.

$a\pm\sqrt2\,b\,$ is a root of $f(x)\!\iff\! \pm\sqrt 2\,$ is a root of $\,f(a+bx)=: \bar f(x)\in\Bbb Q[x]$

Thus, equivalently, we seek  $\,\bar f(\sqrt 2) = 0\,\Rightarrow\, \bar f(-\sqrt 2) = 0$.

Now split $\bar f(x) = g(x^2) + h(x^2)\,x\,\ $ into even + odd parts $ $
(cf. here and here)

so $\ 0 = \bar f(\sqrt 2) = g(2)\ +\ h(2) \sqrt 2\,\Rightarrow\, \color{#c00}{h(2)\!=\!0}\Rightarrow \color{#c00}{g(2)\!=\!0},\,$ by $\,g(2),h(2)\in\Bbb Q\not\ni \sqrt 2$

thus $\,\ \bar f(-\sqrt 2) =  \color{#c00}{g(2)\ -\ h(2)} \sqrt 2 = 0. \ \ \ \rm\small QED$


问:
Complex derivative - how is this result obtained?
I am quite new to the field of complex algebra. During my studies, I have come across the following term:

$$(z - \bar{z}) H(z)$$

where $z$ is a complex variable, $\bar{z}$ its complex conjugate, and $H(z)$ is a complex function depending on $z$. The authors take the complex derivative with respect to $z$ and obtain:

$$\frac{\partial}{\partial z}\left( (z - \bar{z}) H(z) \right) = H(z) + z\frac{\partial H(z)}{\partial z} - \bar{z}\frac{\partial H(z)}{\partial z}$$

Now my question: How was this result obtained?

My own approach would look like this:

$$\frac{\partial}{\partial z}\left( (z - \bar{z}) H(z) \right)$$
Pull $H(z)$ inside the paranthesis
$$\frac{\partial}{\partial z}\left( zH(z) - \bar{z}H(z)  \right)$$
And split up the terms
$$\frac{\partial}{\partial z}\left( zH(z)\right) - \frac{\partial}{\partial z}\left(\bar{z}H(z)  \right)$$
Now apply the chain rule $\frac{\partial}{\partial x}a(x)b(x) = a'(x)b(x) + a(x)b'(x)$:

$$\frac{\partial}{\partial z} zH(z) +  z\frac{\partial}{\partial z}H(z) - \frac{\partial}{\partial z}\bar{z}H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$

Now, I expect $\frac{\partial}{\partial z} z= \frac{\partial}{\partial z} \bar{z} = 1$:

$$H(z) +  z\frac{\partial}{\partial z}H(z) - H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$

which should simplify to:

$$z\frac{\partial}{\partial z}H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$

That's not the result the authors obtained (the term $H(x)$ is missing). A possible explanation might be that $\frac{\partial}{\partial z} \bar{z}=0$, but I would be surprised if $\bar{z}$ is independent of $z$. Do you know where my mistake is? I have least confidence in the second-to-last step.

答: Indeed $\frac{\partial}{\partial z} \bar{z} = 0$. Remember that $\frac{\partial}{\partial z}$ is defined as $\frac{\partial}{\partial z} = \frac{\partial}{\partial x} - i\frac{\partial}{\partial y}$,  so $\frac{\partial}{\partial z} \bar{z} = 1 + i^2 = 0$.


答:
You may try to think to "i" not as a number, but rather as a funcion.

In complex plane, the multiplication by $i$ is a rotation of 90 degrees. In $\mathbb C^2$, the multiplication by $i$ acts as a rotation on each cohordinate. So we can identify the multiplication by $i$ with a function $R_i:\mathbb C^2\to \mathbb C^2$.

Your matrix $A$ acts on $\mathbb R^2$ which you can think sitting inside $\mathbb C^2$. And in fact $A$ acts on $\mathbb C^2$. So you may ask if the matrix $A$ and the multiplication by $i$ act in the same way on some vector of $\mathbb C^2$. In other words, you may ask if there is $X\in\mathbb C^2$ so that $AX=R_i(X)$. A complex eigenvector of $A$ with eigenvalue $i$, is then a vector $X\in\mathbb C^2$ so that $AX$ gives the same result that the multiplication by $i$.
The equation $AX=iX$ now reads as $AX=R_i(X)$.

Note that if you identify $\mathbb C^2$ with $\mathbb R^4$, then the matrix $A$ becomes $\begin{pmatrix}0&0&-1&0\\0&0&0&-1\\1&0&0&0\\0&1&0&0\end{pmatrix}$ and the function $R_i$ is given by the multiplication by the matrix $R=\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}$

So in $\mathbb C^2=\mathbb R^4$ the problem of searching for eigenvectors with eigenvalue $i$ reduce to solve $AX=RX$ which is equivalent to search eigenvector with eigenvalue $1$ of $R^{-1}A$. That is to say $X$ so that $R^{-1}AX=X$.

Title: What do we mean visually by complex eigen values of a matrix? Intuition behind the rotation of space using a 2x2 matrix and eigen values?

Body:
The rotation matrix on $\Bbb R^2$ , the Euclidean plane given by $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ has two imaginary eigen values $i$ and $-i$. The definition of eigen vectors are those vectors $x$ that are parallel to $x$  [i.e. $Ax= \lambda x$].

Here the definition says $Ax=ix$ or $Ax=-ix$, since multiplying by $ i $ rotates my space by ninety degrees , does that essentially mean that we see two vectors in complex planes that are perpendicular to each other as being parallel to each other at the same time. What intuition am I missing here?


问:
Why is the limit of $\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$?
$$\lim_{n\to+\infty}\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$$

I have used the fact exponentials grow faster than polynomials for this question but was wondering whether there was an algebraic way e.g. with the ratio lemma.

答:
According to the suggestion by Jyrki we have

$$\frac{11^n n^4 + 9^n n^9} {7^{2n} +1} \le \frac{11^n n^9 + 11^n n^9} {7^{2n} }=2\cdot\frac{ 11^n n^9} {49^n }=2\cdot\frac{ n^9} {\left(\frac{49}{11}\right)^n }\le 2 \cdot \frac{n^9}{4^n}$$

Refer also to the related

How to prove that exponential grows faster than polynomial?


问:
Functors that aren't continuous/smooth
Let $\mathbf{FinVec}$ denote the category of finite-dimensional real vector spaces and linear maps, let $F\colon\mathbf{FinVec}\rightarrow\mathbf{FinVec}$ be a functor. For each pair of objects $V,W$, the Hom-set $\operatorname{Hom}(V,W)$ is itself a finite-dimensional vector space and we equip it with the unique structure of a topological space/of a smooth manifold that is compatible with its vector space structure. The composition maps are then continuous/smooth. In this way, we view $\mathbf{FinVec}$ as enriched over the category $\mathbf{Top}$/$\mathbf{Diff}$ respectively. It then makes sense to ask whether the functor $F$ is enriched over $\mathbf{Top}$/$\mathbf{Diff}$. That is just to ask whether the induced map $F\colon\operatorname{Hom}(V,W)\rightarrow\operatorname{Hom}(FV,FW)$ is continuous/smooth for all $V,W$.

(Irrelevant to the rest of the question, but the reason this is a useful notion is that it is the natural hypothesis assuring that the functor induces another functor on the category of topological/smooth vector bundles, applying the original functor fiberwise.)

In the case $V=W$, by functoriality, we can restrict $F$ to a group homomorphism $\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$. If the functor is continuous/smooth, so is this map. Note that $\operatorname{GL}(V)$ is a Lie group. It is a general result that a measurable group homomorphism between Lie groups is automatically smooth. This begs the following questions:

What's an argument for the existence of functors that aren't continuous? Can we construct one? Given the above, it either happens that the restrictions to $\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$ are continuous and continuity fails elsewhere, or the induced maps will be non-measurable, so that the construction will necessarily have to involve choice to some extent.

Are there continuous functors that aren't smooth? Given the above, a continuous functor will have smooth restrictions to $\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$, but smoothness may fail elsewhere.

答:
Reinstate Monica gives a construction in the comments to the question I linked in the comments (I can't close as a duplicate because that question has no upvoted or accepted answers): first, the category $\text{FinVec}(\mathbb{C})$ of finite-dimensional complex vector spaces has discontinuous endofunctors given by pulling back along discontinuous automorphisms $\mathbb{C} \to \mathbb{C}$. Second, these give rise to discontinuous endofunctors of $\text{FinVect}(\mathbb{R})$ given by complexifying, applying one of these automorphisms, then restricting back down to a real vector space of twice the dimension.

I suspect every continuous endofunctor is smooth, and in fact I suspect every continuous endofunctor is a (possibly infinite) sum of Schur functors, and hence even polynomial (by which I mean the induced maps $\text{Hom}(V, W) \to \text{Hom}(F(V), F(W))$ are polynomial; there are a few other possible definitions and I don't think they agree in general). The "possibly infinite" is to handle examples like the exterior algebra $V \mapsto \wedge^{\bullet}(V)$.

Title: Functors that aren't continuous/smooth


问:
Doubt on the proof of inequality involving a Linear Transformation
There's one exercise on Spivak's Calculus on Manifolds that he uses all the time in the text. The exercise is: prove that if $T: \mathbb{R}^n \to \mathbb{R}^m$ is linear, then there's some $M \in \mathbb{R}$ such that for all $h \in \mathbb{R}^n$ we have $|T(h)|\leq M|h|$.

My attempt to prove his was the following: consider the canonical basis of $\mathbb{R}^n$ $\{e_i\}$ and the canonical bases of $\mathbb{R}^m$ $\{f_j$}. Then if $h \in \mathbb{R}^n$ we have $h=\sum_{i=1}^{n} h^i e_i$ and for each $e_i$ we have $T(e_i) = \sum_{j=1}^{m}a_{ij}f_j$. This implies that:

$$T(h)=\sum_{i=1}^{n}h^iT(e_i)=\sum_{j=1}^m\left(\sum_{i=1}^{n}h^ia_{ij}\right)f_j$$

Now, using $|\cdot|:\mathbb{R}^n \to \mathbb{R}$ the euclidean norm we find that:

$$|T(h)|^2=\sum_{j=1}^{m}\left(\sum_{i=1}^{n}h^ia_{ij}\right)^2$$

Since this is a sum of positive numbers and since the square root function is always increasing this satisfies:

$$|T(h)| \leq \left(\sum_{i=1}^{n}h^ia_{ij}\right)$$

Now taking $M=\min\{a_{ij} : 1 \leq i \leq n, \ 1 \leq j \leq m\}$ we have:

$$|T(h)|\leq M\sum_{i=1}^{n}h^i$$

Until there I'm fine, I just don't know how to conclude that this is equivalent as what's being purposed. If $h^i > 1$ for all $i$, it's obvious that $(h^i)^2>h^i$ and so the sum of all $h^i$ is less or equal than the sum of the squares. But if $0 < h^i < 1$ then we have $(h^i)^2<h^i$ so that this argument no longer holds.

Can someone give a help with this?

答:
Observe that if ${e_i}$, $1 \leq i \leq n$ is the canonical basis of $\mathbb{R}^n$ and $x = \lambda_1x_1+\ldots+\lambda_nx_n$ then $||x||\geq|\lambda_i|$ for all $i$. Thus, $|\lambda_1|+\ldots+|\lambda_n| \leq n||x||$.

Now set $M' := \max\{||T(e_1)||,\ldots,||T(e_n)||\}$ and $M := nM'$, then you have \begin{align}
       ||T(x)||
    =& ||T(\lambda_1e_1+\ldots+\lambda_ne_e)||
    =  ||\lambda_1T(e_1)+\ldots+\lambda_nT(e_n)|| \\
 \leq& |\lambda_1|\cdot||T(e_1)|| +\ldots+ |\lambda_1|\cdot||T(e_1)||
 \leq  M'(|\lambda_1|+\ldots+|\lambda_n|)
 \leq  M'n||x|| \\
 \leq&  M||x|| \text{.}
\end{align}


问:
Prove by induction $\ n^{\log_2(n)} \le e^n $
prove for any $\ n \ge 1 $

$\ n^{\log_2(n)} \le e^n $

Since both positive for every $\ n \ge 1 $ I was thinking of taking the $\ln $ of each function and then:

$\ln(n^{\log_2(n)}) \le \ln(e^n) \Rightarrow \log+2(n) \cdot \ln(n) \le n$

and for $\ n = 1 $  can see that $\log_2(1) \cdot 0 \le 1 $

the problem is I can't really do much with the function for $ n+1$

$\log_2(n+1) \cdot \ln(n+1) = ? $

can't really transform it to anything else.. I 've also tried taking the $\log_2 $ instead of $\ln $ but still..

答: Do you want to use induction only? Otherwise one can consider the function $f(x)= x-(lnx)^2$ in $x>1$  and see that the function is non-decreasing by using derivative test. Since $x>1$ and $f$ is non-decreasing, hence, $x-(lnx)^2>1>0$. Now take exponential both sides which will preserve the inequailty as exponential is increasing function and then you get the desired result.


答:
$\overline z$ is treated as a variable independent of $z$. Of course, it's not really independent, since there's a clearly defined one-to-one correspondence between complex numbers and their conjugates. But some formalisms of complex analysis pretend that it is not so. Which is why I personally dislike those formalisms.

But there is a way to make this rigorous. We can note that if a function $f(x+\mathrm iy)$ is complex differentiable, then $\partial_z f=\partial_x f=-\mathrm i\partial_y f$. From this, we can conclude that
$$\partial_z f=\frac{1}{2}(\partial_x -\mathrm i\partial_y )f.$$
We can also note that we can apply the differential operator on the right to functions which are not complex differentiable. So it's a generalization of the complex derivative operator, and we define $\partial_z:=\frac{1}{2}(\partial_x-\mathrm i\partial_y)$. For complex-differentiable functions, this still only returns the complex derivative. But it will also return something for non-differentiable functions, for instance, $\partial_z\overline z=0$. Using the old meaning where $\partial_z$ is strictly the complex derivative, $\partial_z\overline z$ simply didn't exist, since $\overline z$ is not complex-differentiable. So you wouldn't have been able to use any of the rules for differentiation in the first place.

Title: Complex derivative - how is this result obtained?

Body:
I am quite new to the field of complex algebra. During my studies, I have come across the following term:

$$(z - \bar{z}) H(z)$$

where $z$ is a complex variable, $\bar{z}$ its complex conjugate, and $H(z)$ is a complex function depending on $z$. The authors take the complex derivative with respect to $z$ and obtain:

$$\frac{\partial}{\partial z}\left( (z - \bar{z}) H(z) \right) = H(z) + z\frac{\partial H(z)}{\partial z} - \bar{z}\frac{\partial H(z)}{\partial z}$$

Now my question: How was this result obtained?

My own approach would look like this:

$$\frac{\partial}{\partial z}\left( (z - \bar{z}) H(z) \right)$$
Pull $H(z)$ inside the paranthesis
$$\frac{\partial}{\partial z}\left( zH(z) - \bar{z}H(z)  \right)$$
And split up the terms
$$\frac{\partial}{\partial z}\left( zH(z)\right) - \frac{\partial}{\partial z}\left(\bar{z}H(z)  \right)$$
Now apply the chain rule $\frac{\partial}{\partial x}a(x)b(x) = a'(x)b(x) + a(x)b'(x)$:

$$\frac{\partial}{\partial z} zH(z) +  z\frac{\partial}{\partial z}H(z) - \frac{\partial}{\partial z}\bar{z}H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$

Now, I expect $\frac{\partial}{\partial z} z= \frac{\partial}{\partial z} \bar{z} = 1$:

$$H(z) +  z\frac{\partial}{\partial z}H(z) - H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$

which should simplify to:

$$z\frac{\partial}{\partial z}H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$

That's not the result the authors obtained (the term $H(x)$ is missing). A possible explanation might be that $\frac{\partial}{\partial z} \bar{z}=0$, but I would be surprised if $\bar{z}$ is independent of $z$. Do you know where my mistake is? I have least confidence in the second-to-last step.


问:
Cubic diophantine equation in two variables
I am trying to solve the Diophantine equation $xy^2 + 2xy + x - 243y = 0$.

I simplified it to $x(y^2 +2y +1) = 243y$ but I am stuck on what to do now.  Any help would be appreciated.

答:
If $x$ and $y$ are integers satisfying your equation, then also
$$3^5y=243y=x(y^2+2y+1)=x(y+1)^2.$$
Because $y$ and $y+1$ have no common divisor, it follows that $(y+1)^2$ divides $3^5$. Can you finish from here?

 It follows that $(y+1)^2=3^{2k}$ for some $k\in\{0,1,2\}$ and so $y=-1\pm3^k$, and $$x=\frac{3^5y}{(y+1)^2}=3^{5-2k}y.$$ Checking these few values of $k$ yields the following six solutions $(x,y)$: $$(0,0),\quad(-486,-2),\quad(54,2),\quad(-108,-4),\quad(24,8),\quad(-30,-10).$$


问:
Borel operator calculus of compound function
I am studying operator calculus of complex-valued bounded Borel measurable function. In our textbook it is induced by Gelfand representation over $A_N$, which is the smallest C* algebra generated by normal operator $N$. It turns out that
$$\phi (\psi(N))=(\phi \circ \psi) (N), \forall \phi, \psi \text{ continuous.}$$
which can be shown using Gelfand representation. Does it still correct in the case $\phi, \psi$ is bounded Borel measurable? I have seen the spectrum decomposition in this case, but don’t know if it can helps.
Thanks in advance.

答:
I know at least one variation that is true: Consider the following theorem from Murphy's excellent text "$C^*$-algebras and operator theory":

Theorem 2.5.7 (p73): Let $u$ be a normal operator on the Hilbert space $H$ and let $g: \mathbb{C}\to \mathbb{C}$ be a continuous function. Then $(g\circ f)(u) = g(f(u))$ for all $f \in B_\infty(\sigma(u)).$


问:
How many ways to choose a 5 player basketball team
How many ways are there to choose $5$ players between $12$ players?
I know this part:

$C(12,5)$.

But part $b$ asks:
How many of our choices include strongest and weakest players?
And I don't know how to solve part $b$.

Please help me.

答: Part (b) says: there are two people called strongest and weakest, and they must be on the team. The other three slots may be filled by any of the "middle" ten players, so the answer is $\binom{10}3$.


问:
What can be said about $p\in Spec(R)$ when $R_p$ is a field?
What can be said about $p\in Spec(R)$ when $R_p$ is a field? Especially when $R$ is local noetherian</p>

答:
Obviously $p$ must be a minimal prime, in other words have height zero.

But minimality is not sufficient  : if $k$ is a field and $R=k[\epsilon]=k[t]/(T^2)$, the ideal $p=(\epsilon )$ is minimal but $R_p=R$ is not a field.

However if $R$ is reduced and $p$ is a minimal prime, then $R_p$ is reduced (as a localization of a reduced ring) and $pR_p\subset R_p$ is the only prime ideal of $R_p$.

It is thus zero since it equals the nilradical of the reduced ring $R_p$, and so $R_p$ is a domain.

But a domain of dimension zero is a field, since the prime  $(0)$ is maximal by zero-dimensionality.

We have thus proved :

Proposition:

In a reduced ring $R$ the localization $R_p$ of $R$ at a prime is a field if and only if $p$ is a minimal prime.


问:
Parameter Elimination with Products of Sines and Cosines in N-dimensions
Given N coordinate dimensions (the $x$ 's) and M parameters (the $u$ 's) with M = N - 1

$x^1 = a ~cos~u^1$

$x^2 = a ~sin~u^1~cos~u^2$

...

$x^{N-1} = a ~sin~u^1~sin~u^2~sin~u^3 ... sin~u^{N-2} ~cos~u^{N-1}$

$x^N = a ~sin~u^1~sin~u^2~sin~u^3 ... sin~u^{N-2} ~sin~u^{N-1}$

We want to do parameter elimination to come up with the $F$ such that

$F(x^1,x^2,...,x^N) = 0$.

Can someone show me how to do this? This is just for my interest in learning, not for a homework. It comes from the first chapter of Synge and Schild Tensor Calculus (1949 - a Dover reprint). Thank you.

答:
$Σ^N_{i=1} (x^i)^2 ~−~a^2 = 0$

Using the relation $sin^2x+cos^2x = 1$, and a bunch of factoring and simplifying. Easier to see by starting with the higher order terms rather than the lower order ones. Solved with the comment/hint of @player3236.


问:
Variance for linear combination of coefficients
Why does:

$Var(aX)=a^2Var(X)$

and

$Var(aX\pm bY)=a^2var(X)+b^2var(Y)\pm 2ab Cov(X,Y)$?

I've looked at similar stackexchange links, but haven't been able to find a "simple" answer without extensive math proof.

答:
Be aware that: $$\mathsf{Var}(Z)=\mathsf{Cov}(Z,Z)\tag1$$

where by definition: $$\mathsf{Cov}(X,Y)=\mathbb E(X-\mathbb EX)(Y-\mathbb EY)\tag2$$

Based on $(2)$ that we find that:

$\mathsf{Cov}$ is symmetric in the sense that $\mathsf{Cov}(X,Y)=\mathsf{Cov}(Y,X)$

$\mathsf{Cov}(aX,Y)=a\mathsf{Cov}(X,Y)$

$\mathsf{Cov}(X+Z,Y)=\mathsf{Cov}(X,Y)+\mathsf{Cov}(Z,Y)$

This together makes it possible to prove that $\mathsf{Cov}$ is bilinear so that:

$\mathsf{Cov}(aX,aX)=a^2\mathsf{Cov}(X,X)$

$\mathsf{Cov}(aX+bY,aX+bY)=a^2\mathsf{Cov}(X,X)+2ab\mathsf{Cov}(X,Y)+b^2\mathsf{Cov}(Y,Y)$

$\mathsf{Cov}(aX-bY,aX-bY)=a^2\mathsf{Cov}(X,X)-2ab\mathsf{Cov}(X,Y)+b^2\mathsf{Cov}(Y,Y)$

Now apply $(1)$ in those rules.


问:
Solving for a coordinate along an axis given a triangle (3 coordinates)
Not sure what the best, or most descriptive title is but here is my problem:

I have added a picture to help clarify it. I am trying to solve for the diamond position, more specificly I really only need the x (assuming x is red) distance to the line P1->P2. In the picture P2 is just Y translated relative to 0/0 but it may not always be the case.

How can I solve for this distance/position with respect to P1/P2 (they are not always fixed) but relative to 0/0.

答:
Since $\triangle OXP_2\sim\triangle AP_1P_2$,

\begin{align}
\frac{|XO|}{|P_2O|}
&=
\frac{|P_1A|}{|P_2A|}
,\\
X_x&=
\frac{|P_{2y}|\cdot|P_{1x}|}{|P_{1y}-P_{2y}|}
.
\end{align}


问:
Find the last three digits of $383^{101}$
We have to find out $383^{101} \equiv ? \pmod {1000}$.
I know that $383^2 ≡ 689 \pmod {1000}$
$383^5≡143 \pmod {1000}$

I know that $ϕ(1000)=400 >101 $  from Euler.
It definitely can't help me.
I don't know how to continue.
I can't use the Chinese Remainder Theorem.

Body:
The units digit of $383^1$ is $3$.

The units digit of $383^2$ is $9$ ($3 \times 3 = 9$).

The units digit of $383^3$ is $7$ ($3 \times 9 = 27$).

The units digit of $383^4$ is $1$ ($3 \times 7 = 21$).

Calculating,

$\; 383^4 \equiv 721 \pmod{1000}$

Calculating,

$\; 383^{101} \equiv 383 \cdot 721^{25} \equiv 383 \cdot (1 + 720)^{25}  \equiv 383 \cdot (1 + 2^4 \cdot 3^2 \cdot 5^1)^{25} \equiv$

$\quad\quad\quad\quad  \; 383 \; \bigr(1 + \binom{25}{1}\,2^4 \cdot 3^2 \cdot 5^1  + \binom{25}{2}\,2^8 \cdot 3^4 \cdot 5^2  + 0\bigr) \equiv$

$\quad\quad\quad\quad  \; 383 \; \bigr(1 + 0 + 0 + 0 \bigr) \equiv 383 \pmod{1000}$


答:
We can divide $f$ by $\left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right) = x^2-2ax+a^2-2b^2 \in \Bbb{Q}[x]$ to obtain
$$f(x) = \left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right)q(x) + cx+d$$
for some $q \in \Bbb{Q}[x]$ and $c,d \in \Bbb{Q}$. Plugging in $x = a+\sqrt{2}b$ gives
$$0 = f(a+\sqrt{2}b) = c(a+\sqrt{2}b) + d = (ca+d) + \sqrt{2}cb$$
which by irrationality of $\sqrt{2}$ implies $ca+d = cb = 0$. Assuming $b \ne 0$, this implies $c = 0$ and then $d = 0$. Therefore $$f(x) = \left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right)q(x)$$
so in particular $f(a-\sqrt{2}b) = 0$.

Body:
Let $f(x)$ be a polynomial in $\mathbb{Q}[x]$. Prove that if $a + \sqrt2 b$ is a zero of $f(x)$ then so is $a - \sqrt2b$.

Can I treat $a + \sqrt2 b$ like a complex number? Then I can do something like

let $f(x) = \sum a_ix^i$, and  $z = a+\sqrt2b$

Then $f(\bar z) = \sum a_i \bar z^i = ... = \bar{f(z)}$.

If not.. what else can I try?


问:
How to find deg 4 and deg 5 irreducible polynomials over $\mathbb{F}_3$?

Give examples of irreducible polynomials $f(x)$ and $g(x)$ in $\mathbb{F}_3[x]$ such that $\deg f(x)=4$ and $\deg g(x)=5$.

I can find irreducible polynomials of degree $2$ or $3$ by taking such polynomials which has no zeros in $\mathbb{F}_3$. But for $\deg 4$ and $\deg 5$ I don't know how to find a polynomial and then show that it will be irreducible.

Please suggest me any tips.

Thanks.

答:
A polynomial of degree $4$ or $5$ is irreducible if it has no linear or quadratic factor. There aren't many linear and quadratic polynomials over $\Bbb{F}_3$, so list them all and then just pick polynomials of degrees $4$ and $5$ that isn't divisible by any of them.

Of course you only need to consider monic polynomials. There are precisely $3$ linear monic polynomials over $\Bbb{F}_3$, and $9$ quadratic monic polynomials over $\Bbb{F}_3$, and only $3$ of them are irreducible. So there are only $9$ irreducible factors to avoid.

Alternatively you could take a more abstract approach, which may seem like magic if you aren't already very familiar with this topic:

Because $\Bbb{F}_3$ contains no primitive fifth root of unity, the fifth cyclotomic polynomial
$$\Phi_5=x^4+x^3+x^2+x+1$$
is irreducible over $\Bbb{F}_3$.

A similar trick works for degree $5$; there is of course no cyclotomic polynomial of degree $5$, but because $\Bbb{F}_3$ has no primitive eleventh root of unity, the eleventh cyclotomic polynomial $\Phi_{11}$ is irreducible over $\Bbb{F}_3$. Then for any root $\zeta$ of $\Phi_{11}$ the minimal polynomial of $\zeta+\zeta^{-1}$ is irreducible of degree $5$, and the minimial polynomial is
$$x^5+x^4-4x^3-3x^2+3x+1.$$


问:
proving that a certain function is not differentiable at $(0,0)$
Let's consider the function given by:
$$
F\left( {x,y} \right) =
\begin{cases}
\frac{{xy}}{{\sqrt {x^2  + y^2 } }}\text{ when }\,\,\,\,(x,y) \ne (0,0)
 \\
 \\
 \\
 \\
\,\,\,\,\,\,0\hspace{10ex}\text{ if } \,\,\,\,\,\,\,\,\,(x,y)=(0,0)
\end{cases}
$$

I proved a lot of things of this function, the partial derivated exist everywhere, in particular at $(0,0)$ , But I want to prove that $F$ is not differentiable at $(0,0)$, I I tried by definition but it does not work. Please help me !!

答:
Correct hint: If you've computed the partial derivatives at $(0,0)$, you know they're both $0$. So, if $f$ were differentiable, you'd have
$$\lim_{(x,y)\to (0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$
Is this so?

Alternatively, if $f$ were differentiable, the directional derivative in every direction $\vec v$ would have to be given by $\nabla f(0,0)\cdot\vec v$.


问:
If x tends to 0,then what does cx tend to where c is a constant?
I saw in a math solution that $\lim_{x\to0}(1+2x)^{\frac{1}{2x}(10)}=e^{10}$.But here we are using $x$ tends to $0$ but not $2x$ tends to $0$. Does that mean when $x$ tends to $0$ then $cx$ tends to $0$ when $c$ is a constant? If so, why?

Thanks.

答:
Yes, if $x\rightarrow 0$ then $cx\rightarrow 0$ too since $c$ is a constant. We have $\lim_{x\rightarrow 0}cx=c\lim_{x\rightarrow 0}x=c\cdot0=0$ for any $c\in\mathbb R$. See here for properties of limits.

Indeed setting $y=2x$ then we have $$\lim_{x\rightarrow 0}\big(1+2x\big)^{\frac{10}{2x}}=\lim_{y\rightarrow 0}\big(1+y\big)^{\frac{10}{y}}$$
$$\lim_{y\rightarrow 0}\big[\big(1+y\big)^{\frac{1}{y}}\big]^{10}$$
$$=\big[\lim_{y\rightarrow 0}\big(1+y\big)^{\frac{1}{y}}\big]^{10}=e^{10}$$

by the power rule where we have $\lim_{y\rightarrow 0}\big(1+y\big)^{\frac{1}{y}}=e$


答:
From your expression, $7^{2n}= 49^n$ seems to be the dominant term. We can divide both numerator and denominator by $49^n$:

$$a_n = \frac{n^4 \left(\frac{11}{49} \right)^n + n^9\left(\frac{9}{49} \right)^n }{1 + \left(\frac{1}{49}\right)^n}= \frac{\dfrac{n^4}{\left(\frac{49}{11} \right)^n}+ \dfrac{n^9}{\left(\frac{49}{9} \right)^n}}{1 + \left(\dfrac{1}{49} \right)^n}$$

Since $\left(\left(\frac{1}{49} \right)^n\right)$ is a basic null sequence and by the reciprocal rule, $\left(\left(\frac{49}{11} \right)^n\right), \left(\left(\frac{49}{9} \right)^n\right)$ both tend to infinity which makes $\left(\dfrac{n^4}{\left(\frac{49}{11}\right)^n}\right)$ and $\left(\dfrac{n^9}{\left(\frac{49}{9}\right)^n}\right)$ both basic null sequences.
$$\lim_{n \to \infty} a_n =  \frac{0 + 0}{1 + 0} = 0$$

Alternatively, you can also say that $\left(n^4\left(\frac{11}{49} \right)^n\right)$ and  $\left(n^9\left(\frac{9}{49} \right)^n\right)$ are basic null sequences.

Title: Why is the limit of $\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$?

Body:
$$\lim_{n\to+\infty}\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$$

I have used the fact exponentials grow faster than polynomials for this question but was wondering whether there was an algebraic way e.g. with the ratio lemma.


问:
Show level surfaces are perpendicular at point $p$.

Show that the level surfaces $f(x,y,z)=0$ and $g(x,y,z)=0$ are perpendicular at a point p where $∇f(p)≠ 0$ and $∇g(p)≠ 0$ if and only if $(\frac{∂f}{∂x})(\frac{∂g}{∂x})+(\frac{∂f}{∂y})(\frac{∂g}{∂y})+(\frac{∂f}{∂z})(\frac{∂g}{∂z})=0$ at $p$.

Show that the surfaces $z^2=x^2+y^2$ and $x^2+y^2+z^2=a^2$ (with $a$ constant) are perpendicular at every point of intersection.

What I have tried

since $∇f(p)≠ 0$ and $∇g(p)≠ 0$ then

$N_g=(\frac{∂g}{∂x},\frac{∂g}{∂y},\frac{∂g}{∂z})$ and $N_f=(\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z})$ (both at $p$)

So, for the surfaces to be perpendicular, the normal lines need to be perpendicular, hence,

(at $p$) $0=Ng\cdot Nf=(\frac{∂g}{∂x},\frac{∂g}{∂y},\frac{∂g}{∂z}).(\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z})=(\frac{∂f}{∂x})(\frac{∂g}{∂x})+(\frac{∂f}{∂y})(\frac{∂g}{∂y})+(\frac{∂f}{∂z})(\frac{∂g}{∂z})$ as required. This seems like I am on the right lines, however I am a bit uneasy about the point at $p$.

Part 2 For the second part I tried to find the normal lines and do the dot product with them both but it did not work out to be zero. For $z^2=x^2+y^2$, I solved for $z$ and used the formula for a graph $(−∂z/∂x,∂z/∂y,1)$ and got $\left(\frac{−x}{\sqrt{(x^2+y^2)}},\frac{−y}{\sqrt{(x^2+y^2)}},1\right)$ and $\left(\frac{x}{\sqrt{(a^2−x^2−y^2)}},\frac{y}{\sqrt{(a^2−x^2−y^2)}},1\right)$ for $x^2+y^2+z^2=a^2$. Dot product of these two don't yield zero, this is where I get stuck.

答:
Answer to part $1$ is correct.

Since the point $p$ lies on both surfaces $z^2=x^2+y^2$ and $x^2+y^2+z^2=a^2$, the dot product$$\begin{align*}&\left(\frac{−x}{\sqrt{x^2+y^2}},\frac{−y}{\sqrt{x^2+y^2}},1\right)\cdot\left(\frac{x}{\sqrt{a^2−x^2−y^2}},\frac{y}{\sqrt{a^2−x^2−y^2}},1\right)\\&=\left(-\frac x{|z|},-\frac y{|z|},1\right)\cdot\left(\frac x{|z|},\frac y{|z|},1\right)\\&=\frac{z^2-x^2-y^2}{z^2}\\&=0.\end{align*}$$Note that you could have also used the previous result by taking the surfaces to be the level curves $f(x,y,z)=z^2-x^2-y^2=0$ and $g(x,y,z)=x^2+y^2+z^2-a^2=0$ and using the result directly.


答:
Do some basic asymptotic analysis:

It is easy to check that $9^nn^9=o\bigl(11^nn^4\bigr)$, so the numerator is asymptotically equivalent to $11^nn^4$ and
$$\frac{9^nn^9+11^nn^4}{7^{2n}+1}\sim_\infty\frac{11^nn^4}{7^{2n}}=\Bigl(\frac{11}{49}\Bigr)^n n^4\to 0.$$


问:
How to show this inequality between those products?
How to show that :

$$\forall n\in\mathbb N^*,\forall (a_1,\ldots,a_n)\in([1;+\infty[)^n, \prod_{k=1}^{n}(1+a_k)\le2^{n-1}(1+\prod_{k=1}^{n}a_k)$$

I tried by induction but i don't know how to find this.

答: Hint. If $a,\,b \geqslant 1$ then $(1+a)(1+b) \leqslant 2(1+ab).$


问:
Why isn't the perfect closure separable?
Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, a graduate course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

答:
Assume $K=F(x)$ and let $f\in F[X]$ be the minimal polynomial of $x$ over $F$.
One then knows that $f(X)=g(X^{p^m})$ for some $m\in\mathbb{N}$ and a separable
polynomial $g\in F[X]$. Thus $K_s =F(x^{p^m})$.

If there exists an intermediate field $L$ such that $L/F$ is purely inseparable and
$K/L$ is separable, then $[K:L]=[K_s:F]$ since $L$ and $K_s$ are linearly disjoint
over $F$.

Consequently the minimal polynomial $h\in L[X]$ of $x$ over $L$ has the same degree
as $g$.

Now $f=hh_1$ in $L[X]$. Since $x$ is a simple root of $h$ put a $p^m$-fold root of $f$
one concludes $f=h^{p^m}h_2$. Comparing degrees yields $h^{p^m}=g(X^{p^m})$.
This shows that a necessary condition for $L$ to exist is that the coefficients of $g$
are $p^m$-th powers in $K$.

I guess this is sufficient too - have no time to think about it now. And I guess that what
I just wrote down is what mephisto meant ...

By the way: if $K/F$ is normal, then $L$ exists: it is the fixed field of $\mathrm{Aut (K/F)}$.


问:
Find the residue of the function $\frac{1 - \cos z}{z^{3} (z-3)}$
 I need to find the residue of $\dfrac{1 - \cos z}{z^{3} (z-3)}$ at all its singular points. Is this correct? Also, are there removable singularities? Thanks in advance

答:
Note that
$$\dfrac{1-\cos(z)}{z^3(z-3)} = \dfrac{2 \sin^2(z/2)}{z^3(z-3)} = \dfrac12 \dfrac{\left(\dfrac{\sin^2(z/2)}{(z/2)^2}\right)}{z(z-3)}$$
Can you conclude from this?</p>


问:
Mathematical induction Divisibility proof Help [My proof does not match my teacher's solution]
https://imgur.com/a/SkOdYlE

Synopsis: while proving $7$ divides $\,4^n-(-3)^n = : f(n)\,$ by induction on $\,n,\,$ my teacher and I got different expressions in the result of the induction step, namely

$\, f(k+1) = 7(4b+(-3)^k) \ $ by the teacher, vs.

$\, f(k+1) = 7(4^k - 3b) \ $ by me.

What did I do wrong?

I have spent 2 hours on this but don't know what I did wrong. I know my teacher worked it out differently but I want to know what error did I make on my solution because my teachers proof does not match mine.

thanks

答: Hint: both are the same by $\ \overbrace{4^k\!-(-3)^k = \color{#c00}{4b + 3b}}^{\!\!\!\!\!\!\!\!\large {\rm induction\ hypothesis}\ f(k)\, =\, \color{#c00}{7b}\!\!\!\!\!\!\!\!\!}\, \Rightarrow\,\overbrace{4^k-3b}^{\large \rm you} = \overbrace{4b+(-3)^k}^{\large \rm teacher} $


问:
Decide convexity
Decide whether the following function is convex
$$
f^{*}(x)=\sup_{y \in \operatorname{Dom}(f)} \left\{ x^T y-f(y) \right\}, f(y):\mathbb{R}^n \xrightarrow{} \mathbb{R}.
$$
Where
$$
\operatorname{Dom} (f)=\left\{ y: f(y) < \infty, y \in \mathbb{R}^n \right\}.
$$
Can you give me some hint or theorem which is suitable for my problem?

答:
you can use the definition of convexity :

$$\forall (x,y)\in\mathbb{R}^{n}\times \mathbb{R}^{n}, \forall t\in [0,1 ]\\
f^*((1-t)x+ty)\leq (1-t)f^*(x)+tf^*(y)$$

You can just develop the expression of $f^*$ and use the fact that $sup_z\{a(z)+b(z)\}\leq sup_z\{a(z)\}+sup_z\{b(z)\}$

Title: Decide convexity


问:
Prove that $x < y$ implies $x^{1/n} < y^{1/n}$.
Let $x$ and $y$ be positive real numbers and let $m$ and $n$ be positive integers.

Prove that $x < y$ implies $x^{1/n} < y^{1/n}$.

I have so far figured out two different approaches to prove this.

By induction:

Base case: $P(1)\iff x^{1/1} < y^{1/1} \iff x<y$

Inductive step: $P(n+1)\iff x^{1/n+1} < y^{1/n+1} $

But then I'm stuck from here as I 'm not sure on how to solve $1/n+1$.

Another approach is to use direct proof:

By rules of inequalities we know

$x<y\implies x-x < y-x$

$\implies xy < yy$

$\implies xx < xy$

Hence $x^{1/n} < y^{1/n}$
as
$(x^1)^{1/n} < (y^1)^{1/n}$

May I ask how can I solve for $1/n+1$ at the first approach and is the second approach correct at all?

答:
First we prove a simple property. Given $a,b,c,d$ are positive real numbers

(eqn 1)

$a \ge c$ and $b \ge d \implies ab \ge cd$

This is pretty trivial but isn't usually given as one of the standard inequality rules.

$\boldsymbol{Proof}$:

Given $a \ge c$ and $b \ge d$

Since $a \ge c$, we know that $ad \ge cd$

Since $b \ge d$, we know that $ab \ge ad$

So by transitivity we know that $ab \ge cd$

$\boldsymbol{Induction}$:

Now we can use induction to prove that given e and f are positive real numbers

(eqn 2)

$e \ge f \implies e^n \ge f^n$

(eqn 3)

Given $e \ge f$

Base case, n=1 is trivial:

Since $e \ge f$

$e^1 \ge f^1$

So true for n=1.

Assume true for $n=k$

(eqn 4)

$e^k \ge f^k$

Now we prove it's true for $n=k+1$

Using (3) and (4) and rule (1), we know that

$e^k(e) \ge f^k(f)$

$e^{k+1} \ge f^{k+1}$

That completes our proof of eqn 2.

$\boldsymbol{Substitution}$:

Now we can substitute $e = x^{1/n}$ and $f= y^{1/n}$ into eqn 2, where x and y are positive real numbers. Since x and y are positive, e and f are positive (we are taking the positive principal nth root).

$x^{1/n} \ge y^{1/n} \implies (x^{1/n})^n \ge (y^{1/n})^n$

$x^{1/n} \ge y^{1/n} \implies x \ge y$

Now we can take the contrapositive:

$x < y \implies x^{1/n} < y^{1/n}$


问:
What is the definition of $\text{faithfully flat}$ map of rings?
What is the definition of a faithfully flat map of ring ?

I know that a map of rings $A \to B$ is flat if it makes $B$ a flat $A$-module.

A morphism of schemes is faithfully flat if it is both surjective and flat.

So in that sense the map $A \to B$ of rings is faithfully flat if it is flat and surjective.

Is it the definition ?

答:
A map of rings $A\longrightarrow B$ is flat if it makes $B$ a flat $A$-module.

A map of rings $A\longrightarrow B$ is faithfully flat if it makes $B$ a faithfully flat $A$-module.


问:
Laguerre polynomial question
Can someone help me with this

$$\frac{1}{1-t}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{n=\infty}L_{n}(x)\frac{t^{n}}{n!}$$

The author said that we should just expand it but I don't understand how and what $L_{n}$ is equal to.

Since there is $\frac{t^{n}}{n!}$ I guess the exponential function should be expanded but I don't know what to do with $\frac{1}{1-t}$ neither what is the full expression for the $L_{n}$. Thank you for your help, :)

答:
$L_n^{(0)}(x)$ which is also written $L_n(x)$ is  $L_n(x) = \sum_{k=0}^n (-1)^k {n \choose k}\frac{x^k}{k!}$. The expression in the question is an exponential generating function. Try to show that coefficients of like powers of $t$ on the left and right sides of the equation are equal. You can multiply the right side by 1 - t.

The equation in the question is incorrect. the correct generating function is not an exponential one as noted in red. Please see the Digital Library of Mathematical Functions, equation 18.12.13. I suggest you try some other examples first.

Title: Laguerre polynomial question


问:
How to take the derivative of the following expression
I am taking macro course this Fall and my calculus is quite rusty. So in the lecture notes they derive the following:
$$
\begin{split}
MPL&=\frac{dY}{dL}\\&=\frac{d(ALf(k))}{dL}\\&=Af(k)+ALf′(k)(−K)/(L^2A)\\&=A(f(k)−kf′(k))\\&=w
\end{split}
$$
Specifically, I don't quite understand how they got $Af(k) + ALf'(k) (-K)/(L^2A)$ from $d(ALf(k))/dL?$

Let me also clarify that k = K/AL

Honestly, I am stuck at this point and guess will be doing a lot of differentiation like this during the entire course. Would really appreciate if someone could clarify the differentiation part.

答:
I interpret that $A$ is a constant. We want to calculate the following expression:
$$
\frac{d(A\cdot L \cdot f(k))}{d L}
$$
Notice that there are three terms which are multiplied by each other, and therefore we can use the product rule:
$$
\frac{d(A\cdot L \cdot f(k))}{d L} =
\color{blue}{\frac{dA}{dL}}\cdot L \cdot f(k)
 + A \cdot \color{blue}{ \frac{dL}{dL}} \cdot f(k)
+A \cdot L \cdot \color{blue}{ \frac{d f(k)}{dL}}
$$
Notice how each terms in the product is separately differentiated and then the results are added together. A few observations are required here:

$\frac{dA}{dL}=0$, because $A$ is a constant

$\frac{dL}{dL}=1$

The term $\frac{df(k)}{dL}$ is a tricky one, because $k$ is a function of $L$. Therefore, we must "multiply and divide by $dk$", to rewrite it as
$$
\frac{df(k)}{dL} = \frac{df(k)}{dk}\cdot \frac{dk}{dL}
$$
This is a term that we can actually compute.

We need to compute $\frac{dk}{dL}$. We know that $k = \frac{K}{AL}$, so
$$
\frac{dk}{dL} = - \frac{K}{AL^2}
$$
(Does this seem odd? Compare to $f(x)= \frac{1}{x}\Rightarrow f'(x) = - \frac{1}{x^2}$)

Now, we can just plug in the results!
$$
\begin{split}
\frac{d(A\cdot L \cdot f(k))}{d k} &=
 0
 + A \cdot 1 \cdot f(k)
+A \cdot L \cdot \left( - \frac{K}{AL^2} \right)\\
\end{split}
$$
Can you finish it from here?


答:
In this case there is used the so called 'chain rule': if $f(x)$ is depending from the variable $x$, but $x(t)$ depends from $t$, then $f(x(t))$ depends from $t$, and it has sense to talk about the derivative $\frac{df}{dt}$ which is calculated by $$f'(t)=\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}.$$
You also need to have in mind the formula $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).$

On your case, since $k=\frac{K}{AL},$ $k$ depends on L and so does $f(k).$ Then you have:

$$\frac{dALf(k)}{dL}=AL'f(k)+AL\frac{df(k)}{dL}=Af(k)+AL\frac{df(k)}{dk}\frac{dk}{dL}=Af(k)+ALf'(k)\frac{-K}{AL^2}.$$

Title: How to take the derivative of the following expression

Body:
I am taking macro course this Fall and my calculus is quite rusty. So in the lecture notes they derive the following:
$$
\begin{split}
MPL&=\frac{dY}{dL}\\&=\frac{d(ALf(k))}{dL}\\&=Af(k)+ALf′(k)(−K)/(L^2A)\\&=A(f(k)−kf′(k))\\&=w
\end{split}
$$
Specifically, I don't quite understand how they got $Af(k) + ALf'(k) (-K)/(L^2A)$ from $d(ALf(k))/dL?$

Let me also clarify that k = K/AL

Honestly, I am stuck at this point and guess will be doing a lot of differentiation like this during the entire course. Would really appreciate if someone could clarify the differentiation part.


问:
How to compare $\root 3 \of 2 + \root \of 3$ and $3$?

How to compare $\root 3 \of 2 + \root \of 3$ and $3$?

I tried to make them as 6th root, but it doesn't work. I'm completely lost.

Body:
Your problem is $\root 3\of 2 +\sqrt 3\; ?\; 3$, where you have to determine if $?$ is $>$, $<$ or $=$.

$$\root 3 \of 2\;?\;(3-\sqrt 3)$$
$$ 2\;?\;(3-\sqrt 3)^3 = 54-30\sqrt 3$$
$$ 1\;?\; 27-15\sqrt 3$$
$$ 15\sqrt 3\;?\;26$$
$$(15\sqrt 3)^2\;?\;26^2$$
$$225\times 3 = 675\;?\;676$$

So $?$ is $<$, that is $\root 3\of 2 +\sqrt 3< 3$.


答:
Using the following definition of Laguerre polynomials
\begin{eqnarray*}
L_{n}=\sum_{k=0}^{k=n}\frac{(-1)^{k}}{k!}  {n\choose k}  x^{k}
\end{eqnarray*}
Now we should have ... (See https://en.wikipedia.org/wiki/Laguerre_polynomials#Recursive_definition,_closed_form,_and_generating_function)
\begin{eqnarray*}
\sum_{n=0}^{\infty}L_{n}(x)\frac{t^{n}}{\color{red}{1}} &= &  \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^{k}}{k!}  {n\choose k}  x^{k} t^{n} \\
&= &  \sum_{k=0}^{\infty}  \sum_{n=k}^{\infty }\frac{(-1)^{k}}{k!}  {n\choose k}  x^{k} t^{n}\\
&= &  \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} x^{k}\sum_{n=k}^{\infty }  {n\choose k}   t^{n}\\
&= &  \sum_{k=0}^{\infty}  \frac{(-1)^{k}}{k!}    x^{k} \frac{t^{k}}{(1-t)^{k+1}} \\
&= & \frac{1}{1-t}  \sum_{k=0}^{\infty}  \frac{(-1)^{k}}{k!}  \left(   \frac{xt}{1-t}  \right)^{k} \\
&= & \frac{e^{-\frac{xt}{1-t}}}{1-t}.
\end{eqnarray*}

Title: Laguerre polynomial question

Body:
Can someone help me with this

$$\frac{1}{1-t}e^{-\frac{xt}{1-t}}=\sum_{n=0}^{n=\infty}L_{n}(x)\frac{t^{n}}{n!}$$

The author said that we should just expand it but I don't understand how and what $L_{n}$ is equal to.

Since there is $\frac{t^{n}}{n!}$ I guess the exponential function should be expanded but I don't know what to do with $\frac{1}{1-t}$ neither what is the full expression for the $L_{n}$. Thank you for your help, :)


答:
To compare $\sqrt[3]{2} + \sqrt 3$ and $3$ better compare

$\sqrt[3]{2} $ and $3 - \sqrt 3$

cube both numbers

$2$ and $54-30 \sqrt{3}$

now subtract $2$ and add $30 \sqrt{3}$ to both sides

$30 \sqrt{3}$ and $52$

square both sides

$2700<2704$ thus $\sqrt[3]{2} + \sqrt 3<3$

Body:
How to compare $\root 3 \of 2 + \root \of 3$ and $3$?

I tried to make them as 6th root, but it doesn't work. I'm completely lost.


问:
Find rotations about x,y,z such that
I have a vector $(0,0,1)$ and it is rotated about x, then rotated around y and then rotated around the z axis to a new point $(x,y,z)$. In other words.

$\begin{bmatrix} x \\ y \\ z\end{bmatrix} = R_z(\alpha) R_y(\beta) R_x(\gamma) \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$

Where $R_x, R_y, R_z$ are rotation matrices. How can I find the rotation angles $\alpha, \beta, \gamma$ in terms of final point coordinates $x,y,z$?

答:
First find the azimuth $\theta$ and elevation $\phi$ and radius $r$ of the final point $(x,y,z)$. Then rotate the the initial point in the x,y plane by the $\theta$. Then rotate that point around the axis perpendicular to the line at the angle of $\theta$. Let this second axis of rotation be defined by the unit vector $u = (u_x, u_y, u_z)$.

First rotation about z axis

$R_z = \left(\begin{array}{ccc} \cos\left(\mathrm{\theta}\right) & -\sin\left(\mathrm{\theta}\right) & 0\\ \sin\left(\mathrm{\theta}\right) & \cos\left(\mathrm{\theta}\right) & 0\\ 0 & 0 & 1 \end{array}\right)$

Second rotation about u axis. In this case the axis of rotation is defined by: $u_x = -\sin(\theta)$, $u_y = \cos(\theta)$, $u_z = 0$.

$R_u=\left(\begin{array}{ccc} \cos\left(\mathrm{\phi}\right)-{u_{x}}^2\,\left(\cos\left(\mathrm{\phi}\right)-1\right) & -u_{z}\,\sin\left(\mathrm{\phi}\right)-u_{x}\,u_{y}\,\left(\cos\left(\mathrm{\phi}\right)-1\right) & u_{y}\,\sin\left(\mathrm{\phi}\right)-u_{x}\,u_{z}\,\left(\cos\left(\mathrm{\phi}\right)-1\right)\\ u_{z}\,\sin\left(\mathrm{\phi}\right)-u_{x}\,u_{y}\,\left(\cos\left(\mathrm{\phi}\right)-1\right) & \cos\left(\mathrm{\phi}\right)-{u_{y}}^2\,\left(\cos\left(\mathrm{\phi}\right)-1\right) & -u_{x}\,\sin\left(\mathrm{\phi}\right)-u_{y}\,u_{z}\,\left(\cos\left(\mathrm{\phi}\right)-1\right)\\ -u_{y}\,\sin\left(\mathrm{\phi}\right)-u_{x}\,u_{z}\,\left(\cos\left(\mathrm{\phi}\right)-1\right) & u_{x}\,\sin\left(\mathrm{\phi}\right)-u_{y}\,u_{z}\,\left(\cos\left(\mathrm{\phi}\right)-1\right) & \cos\left(\mathrm{\phi}\right)-{u_{z}}^2\,\left(\cos\left(\mathrm{\phi}\right)-1\right) \end{array}\right)$

Then to map the vector $(0,0,1)$ to the point $(x,y,z)$ I need to do the following:

${\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}} = R_u R_z {\begin{pmatrix}
0\\
0\\
1
\end{pmatrix}}$

Because the matrices only do rotations, this formula will only work if the magnitude of the vector $(x,y,z)$ is 1

Title: Find rotations about x,y,z such that


问:
How many ways are there of choosing $k$ numbers from $\{1, . . . , n\}$ if $1$ and $2$ can’t both be chosen?
How many ways are there of choosing $k$ numbers from $\{1, . . . , n\}$ if $1$ and $2$ can’t both be chosen? (Suppose $n, k ≥ 2, n ≥ k$).

My answer is $2\binom{n-1}{k} - \binom{n-2}{k}$. First, I counted sets without $1$, then without $2$, then found their sum and, as the formulas are the same, it's just $2$ multiplied by the formula.  Then I subtracted all sets with both $1$ and $2$, as they was already counted. Is it correct?

答:
Your answer is correct. Here are some alternate, equivalent expressions.

You can also find it like this: select $k$ numbers from $n$ numbers and remove the selections that contain both $1$ and $2$. Such selections contain $k-2$ numbers from the remaining $n-2$ numbers. Thus the answer can be written as$$\binom nk-\binom{n-2}{k-2}$$

Alternatively, consider two cases:

$1$ is selected: Then $2$ is not selected. So we have $\binom{n-2}{k-1}$ selections.

$1$ is not selected: $2$ may or may not be selected. We have $\binom{n-1}k$ selections.

This gives total selections$$\binom{n-2}{k-1}+\binom{n-1}k$$
You can prove that all $3$ expressions are identical using Pascal's identity.

A suggestion for asking future questions to confirm your answers: explain your logic in a few lines so that we know how you obtained your expression. I solved the question in a different manner and obtained a different looking but numerically equal expression. It took me some effort to infer that your and my expressions are identical. Also use the solution-verification tag.


问:
Prove ${x \choose k}1/x^k < {y \choose k}1/y^k$ given that $x < y$ and $x, y \in \mathbb{N}$
I am struggling with this proof because of the $1 / x^k$ and $1 / y^k$ part. Proving the inequality of the combinations just boils down to axioms, but I'm not sure how to prove the entire statement. I tried assuming that $y = x + a$ for some $a \in \mathbb{N}$ and then using induction for $k+1$, but I don't know if that's the correct way to do it.

答:
Rewrite the inequality as
$$
\frac{y^kx(x-1)\ldots(x-k+1)}{k!} < \frac{x^ky(y-1)\ldots(y-k+1)}{k!}.
$$
Cancelling $k!$ on both sides and rearranging gives
$$
\left(1-\frac{1}{x}\right)\ldots\left(1-(\frac{k-1}{x})\right) < \left(1-\frac{1}{y}\right)\ldots\left(1-(\frac{k-1}{y})\right),
$$
which holds because $x < y$.

(I assumed that $k$ is a natural number that is not more than $x$ or $y$.)


问:
If $f$ surjects from $A$ to $B$ then there is a bijection $\phi$ from a subset of $A$ to $B$
Let $A$ be a set and let $ f: A \to B $ be a surjective function. Prove that there exists a subset $ C \subseteq A $ and a function $ \phi: C \to B $ such that $ \phi $ is bijective.

I did it by giving particular examples, giving $ A = \{1,2,3 \} $ and $ B = \{1,2 \} $ and the function $f(x)=\begin{cases}
 1& \text{ if } x=1 \\
 1& \text{ if } x=2 \\
 2& \text{ if } x=3
\end{cases}$ it is clearly a surjective function. Now if $ C = A \setminus \{1 \} $ what is desired remains. Guiding me from particular examples, I thought to consider the set $ C $ of all the images that are repeated because the function is not necessarily injective, as in the previous example removing 1, since it repeated with 2. But I don't know if the idea is well and I don't know how to express that set either. I tried as follows, $ C = A\setminus \{f (a) = f (b) \wedge a \neq b \mid \forall a, b \in A \} $ but I don't know if it's ok. Help.

答:
Let's define in $A$ the equivalence relation $a\sim a' \stackrel{(def.)}{\iff} f(a)=f(a')$.

Claim.
The map $\psi_f\colon (A/\sim)\to B$, defined by $\psi_f([a]_\sim):=f(a)$, is well-defined and bijective.

Proof.

Good definition: $a'\in [a]_\sim \Rightarrow [a']_\sim= [a]_\sim \Rightarrow\psi_f([a']_\sim)=\psi_f([a]_\sim)=f(a)$;

Surjectivity: by the surjectivity of $f$ and the definition of $\psi_f$, $\forall b\in B, \exists a\in A\mid b=f(a)=\psi_f([a]_\sim)$;

Injectivity: $\psi_f([a]_\sim)=\psi_f([a']_\sim)\Rightarrow f(a)=f(a')\Rightarrow a\sim a'\Rightarrow [a]_\sim=[a']_\sim\space$.

$\Box$

Moreover, by the Axiom of Choice, there is a bijection $\varphi\colon (A/\sim)\to R$, where $R\subseteq A$ is a set of representatives (see e.g. here). Therefore, your sought bijection is $\phi\colon R\space(\subseteq A)\to B$ defined by $\phi:=\psi_f\circ \varphi^{-1}$:

$\color{white}{............................}$


答:
We have
$$
\frac{\sqrt[3]{2}}{3} < \frac{42}{100} = \frac{21}{50},
\quad
\frac{\sqrt 3}{3} < \frac{58}{100} = \frac{29}{50}
\implies
\frac{\sqrt[3]{2} + \sqrt 3}{3} = \frac{\sqrt[3]{2}}{3} + \frac{\sqrt 3}{3} < 1
$$
Prove this by removing the radicals and cross multiplying.

These approximations come from the decimal values.

For the sake of curiosity, the simplest approximations that work in the argument above come from their continued fractions:
$$
\frac{\sqrt[3]{2}}{3} < \frac{8}{19},
\quad
\frac{\sqrt 3}{3} < \frac{11}{19}
$$

Title: How to compare $\root 3 \of 2 + \root \of 3$ and $3$?


问:
Infinite set as a countably infinite union of infinite disjoint subsets
I am struggliglng with an exercise as the title.

I have been able to get to a proof, if the set is infinitely countable, but I am having problems on infinite uncountable sets.

Any thoughts on how to think about it?

EDITED: Added the word 'disjoint' and "countably"

答:
Given an infinite set $S$, for any element $x\in S$ the subset $S\setminus\{x\}$ is also infinite. By assumption, $S$ has infinitely many elements, therefore the union
$$\bigcup_{x\in S}(S\setminus\{x\})=S$$
is an infinite union of infinite subsets of $S$.

Edit:

With the added requirement of countably many disjoint sets, the following needs the axiom of choice (via the well-ordering theorem).

According to the well-ordering theorem, the set $S$ is in bijection with a von-Neumann ordinal. Be $\phi:\alpha\to S$ such a bijection.

Since you say you already have the proof for countable $S$, let's further assume that $S$ (and therefore $\alpha$) is uncountable.

Now it is possible to write each ordinal uniquely as sum $\lambda + n$ of a non-successor (i.e. zero or limit) ordinal $\lambda$ and a finite ordinal $n$.

Define
$$\alpha_n = \{\rho\in\alpha | \rho = \lambda + n \text{ for some non-successor ordinal $\lambda$}\}$$

Clearly there are countably many such $\alpha_n$, one for each $n\in\omega$. Moreover, quite obviously the different $\alpha_n$ are disjoint, their union is $\alpha$, and each of them is clearly infinite (because otherwise $\alpha$ would be countable).

Now we get the desired union for $S$ by simply using the images of $\alpha_n$ under the bijection $\phi$:
$$S = \bigcup_{n\in\omega} \{\phi(x)|x\in \alpha_n\}$$


问:
Lower bound on number of disjoint spherical caps that can be packed on hypersphere
Let $S^n = \{x \in \mathbb{R^{n+1}: \ ||x||_2=1}\}$ be the L2 unit sphere in $\mathbb{R^{n+1}}$.

I saw the following result (https://ocw.mit.edu/courses/brain-and-cognitive-sciences/9-520-statistical-learning-theory-and-applications-spring-2006/lecture-notes/class22.pdf ,  page 28):

You can pack $(\frac{1}{\epsilon})^{n+1}$ disjoint spherical caps of radius $\epsilon$ on the surface of $S^n$.

The statement does not include a proof, so I assume it shouldn't be too hard to derive.

Thanks for the help!

答:
I went through the presentation linked in the OP. The claim referenced is on slide 28 of 51:

Interpreted literally, this claim is false (and the exponent on $\epsilon$ should be $d-1$, since she was talking about the surface of the sphere rather than its interior), but the $\Omega(d \log 1/\epsilon)$ suggests that it's meant to be interpreted as an average order of growth, and maybe she just forgot to put the Big O notation1. A more precise statement would be something like, "For small enough $\epsilon$, the number $N(\epsilon)$ of spherical caps of radius $\epsilon$ that I can pack onto the $d$-dimensional unit sphere $S^{d-1} \subset \Bbb{R}^d$ is asymptotically $$N(\epsilon) \sim \frac{C_d}{\epsilon^{d-1}},$$ where $C_d$ is a constant depending only on the dimension of the space."

1  She could also have been confusing it with the fact that one can pack exactly $1/\epsilon^d$ cubes of sidelength $\epsilon$ into the unit cube $[0, 1]^d \subset \Bbb{R}^d$.


问:
What's the Fourier transform of $e^{-x^{2}-y^2-xy}$
I'm asked to calculate the Fourier transform of $f(x)=e^{-x^{2}-y^2-xy}$ and I have no idea how to do this. Can someone help me?

答:
Correcting error.
If you mean 2 dim. Fourier transform as mentioned in comment by reuns and in the format of $F(s,w)=\frac 1{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\exp(\!-\!isx\!-\!iwy)\:dx\:dy\;$ so $\;f(x,y)=\frac1{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(s,w)\exp(isx\!+\!iwy)\:ds\:dw\;$ and you really meant $f(x,y)=e^{(-x^2-y^2\!-xy)}\;$ then the answer should be $F(s,w)\!=\!\frac1{\sqrt{3}}\exp({\frac{sw-s^2-w^2}3})\;$ but if you really meant only a function of the one variable x with y as a cst or parameter in the format $F(s)\!=\!\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)\exp(\!-\!isx)\:dx\;$ so $\;f(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(s)\exp(isx)\:ds\;$ then $F(s)\!=\!\ \frac1{\sqrt{2}}\exp({\frac{2isy-s^2-3y^2}4})\;$ Very briefly to get the answer as mentioned in comments you first complete a square to rid the linear term in the exponent so it's only a function of one quadratic variable but in the completing of the square there will also be another factor considered constant {since it is not the integration variable atleast not in the first integration} function, (...) generally complex, of form $e^{(...)}$ times a gaussian integral of form $\int_{-\infty}^{\infty}e^{-ax^2}\:dx\;$ which would assume you can do. Or in a little more detail in this particular case use $\exp{(ax^2\!+\!bx)}=\exp{(-\frac{b^2}{4a})}\exp{\left (a(x\!+\!\frac b{2a})^2\right )}$, and substitute say $z=x\!+\!\frac b{2a}$, noting $dx\!=\!dz$ , so then take $\exp{(-\frac{b^2}{4a})}\int_{-\infty}^{\infty}\exp{(az^2)}\:dz\;$ but there is a little more to it such as also multiplying by a phase factor and perhaps another factor which result from details to consider when doing contour integration since at first integration $b\!=\!-y\!-\!is$ is complex so will also need to do some complex analysis which I won't go into here.


问:
Domain and Codomain after adding function
I have two functions $H_1,H_2:\mathbb{R} \to \mathbb{R}$ and I want to define a new function $g(x)=H_1(x) + H_2(x)$. Would the domain and codomain off this function still be all the real numbers? Thanks

答:
Domain of the sum of two functions is the intersection of the domains of each function, since $H_1(x)+H_2(x)$ only makes sense when both $H_1(x)$ and $H_2(x)$ are defined. In this case, the intersection of $\Bbb R$ and $\Bbb R$ is $\Bbb R$.

Both $H_1,H_2$ are real valued, so $H_1+H_2$ is also real valued. You can keep the codomain as $\Bbb R$ but remember that the codomain can also be a superset of $\Bbb R$ (or sometimes even a proper subset depending upon $H_1,H_2$) since a function does not need to attain all values in its codomain.


问:
Finding the polygon that encompasses some points, but others must be outside
I have a convex/concave hull problem that must include some vertices, but not all of them. Some points must be outside of the polygon and some are optional.

I took a look at Graham's scan algorithm, but it's not exactly what I'm looking for. Since I have these optional vertices and these points that must be outside the resulting polygon.

For example:

Points that must be included: (0,0),(10,4),(0,5),(2,4).

Points that are optional: (2,3),(4,3).

Points that must be outside: (3,2).

The resulting polygon would be with the points: (0,0),(2,3),(10,4),(0,5).

If I were to just use the first 3 points, (3,2) would be inside. So we use the optional (2,3), but we don't need (4,3).

So this would be the result: (Red line marks an invalid path taken)

Is there an algorithm that can solve this problem already? I can only find algorithms that would include everything.

Thank you! :)

答:
Just an update. I haven't found anything that helped me, but this is how I solved the problem:

I have organization groups that dictates the order of points I'm traveling. Let's call it PointGroup.

The polygon in the original example would be a group with this points in this order: (0,0),(2,3),(10,4),(0,5). The last points connects with the first point as well.

The algorithm:

Create several PointGroups, each containing a different only one mandatory point. So far the perimeter is 0, and it's not a polygon.

Pick the PointGroup with smallest perimeter.

Check if:

It contains at least 3 points (is it a polygon?).

All mandatory points are in the group.

A point that needs to be outside is inside. (https://en.wikipedia.org/wiki/Point_in_polygon)

If the condition fails:

Create several child PointGroups, each by adding a single point. I try every possibility using mandatory and possible points.

Return to 2.

If the condition succeeds, this PointGroup is the answer to the problem.

Not sure if this is optimal, but it works.


问:
vector sum/subtraction magnitude
I got stuck in a multiple choice assignment.

The angle between $u,v$ is $120$ degrees, with $||u||$ = 4, $||v||$ = 3.

I need to calculate $||(u - v)||$ and $||(u + v)||$. I used $(||u||^ 2 + ||v||^ 2-2 ||u|| \ ||v|| \cos(x))^ 2$ but still not sure.

答:
$u \cdot v = \lVert u \rVert \lVert v \rVert \cos \theta$

$u \cdot v = 12 \cos 120^0 = -6$

$\lVert u + v \rVert^2=(u+v)\cdot(u+v)=\lVert u\rVert^2+\lVert v\rVert^2+2(u\cdot v) = 3^2 + 4^2 + 2 \times (-6) = 13$

So, $\lVert u + v \rVert = \sqrt {13}$

$\lVert u - v \rVert^2=(u-v)\cdot(u-v)=\lVert u\rVert^2+\lVert v\rVert^2 - 2(u\cdot v) = 3^2 + 4^2 - 2 \times (-6) = 37$

So, $\lVert u - v \rVert = \sqrt {37}$

Title: vector sum/subtraction magnitude


问:
An old multiplication technique and its reverse for Integer Factoring
An ancient Indian multiplication technique is as follows:

$$\array{
  a=107 & +7 & (\text{base}\space r=100)\\
  b=113 & +13 \\
  \hline
  (a+b-r)=(107 + 13) & (7 \times 13) & \space\text{or}\\
  (b+a-r)=(113 + 7) & (7 \times 13) \\
  \hline
  120 &  91 \\
  \hline
  120 \times r + 91 = 12,091 & = 107 \times 113
}$$

Let $a,b \in Z$ and we want to compute the product $z = ab$.

We write $a, b$ in the first column one below the other. We choose a base $r > 0$, in this case $100$ and write the excess or deficiency of $a,b$ with respect to $r$ in the second column. In this case, the excess over $r=100$ for $107$ is $+7$ and for $113$ is $+13$.
We then compute the diagonal sum, it doesn't matter which diagonal as both will sum to the same number. In this case it is $107+13 = 113+7 = 120$. We then compute the product of the excess (or deficiency), in this case $7 \times 13 = 91$. The required product is

$$z = (a+b-r)r + (a-r)(b-r) = 120*100 + 91 = 12091$$

As we can see, a convenient choice of the base $r$ helps us write the product easily in terms of the diagonal sum and the product of the excess (or deficiency).

If we look at the reverse problem of factoring $12091$, we could again choose the base $r = 100$. It then gives us

$$12091 = 91 \mod 100 + \bigg\lfloor {12091 \over 100} \bigg\rfloor \times 100 = 91 + 120 \times 100$$

Factoring the smaller number $91$ into $7 \times 13$ helps us determine the factors of $12091$ as $r + 7 = 107$ and $r + 13 = 113$.

This works only because the base $r$ is special where $(a-r)(b-r) < r$. It will work for other $r$, but we will have to deal with the carry i.e., $\bigg\lfloor {(a-r)(b-r) \over r} \bigg\rfloor$

Question:

1. The case of $0 \le (a-r)(b-r) \lt r$

Let $\bigg\lfloor {z \over r} \bigg\rfloor = a + b - r$.

Given $z$ could we determine an $r$ such that $(a-r)(b-r) < r$ without any additional knowledge of anything about $a, b$ and with $(a-r)(b-r) \ge 0$?

Is such an $r$ guaranteed to exist?

Note that if $(a-r)(b-r) = 0$ then $r$ is a divisor of $z$

2. The case of $(a-r)(b-r) \ge r$

Let $\bigg\lfloor {z \over r} \bigg\rfloor = a + b - r + k$ for some integer $k = \bigg\lfloor {(a-r)(b-r) \over r} \bigg\rfloor$.

$$(a-r)(b-r) = kr + (z \mod r)$$

If case (1) is not possible for a given $z$ (i.e, $r$ satisfying the criteria does not exist), could we find an $r, k$ that satisfies case (2)?

As an aside, the equality condition $(a-r)(b-r) = r$ gives us $r$ as a trivial divisor of $z$.

If we are able to choose such an $r$ (or $r, k$ for case (2)) depending on $z$ alone, we will have a fast factoring algorithm.

Update (Oct 21, 2020):

There is an interesting empirical observation from the plot of $\lfloor {z \over r} \rfloor + r$ and its relationship to the sum $a+b, z = ab$ with $a,b$ positive. Plot shown below for $z=12091, a+b=220$.

Here is a closeup of the range where the curve hits minimum.

This means, if we pick a reasonably good estimate for $\lfloor {z \over r} \rfloor + r$ with $r = r_e$ being the corresponding value of $r$ and also pick random $r_0$ and $r_1$ such that $r_0 < r_e < r_1$, we can then determine the minimum $\lfloor {z \over r} \rfloor + r$ using successive bisection of the range $[r_0, r_1]$ since the minimum value of $\lfloor {z \over r} \rfloor + r$ must lie in either $r \in [r_0, r_{mid}]$ or $r \in [r_{mid}, r_1]$ where $r_{mid} = {r_0 + r_1 \over 2}$.

The initial estimate for $r_e$ could be:

$$r_e = {{2 + \lfloor {z \over 2} \rfloor} \over 2}$$

With successive iterations, we narrow the range until we are left with a range of width 1 $(r_1 - r_0 = 1)$ and one of the range bounds must result in the minimum for $\lfloor {z \over r} \rfloor + r$. This would require $O(\log(z))$ steps.

Conjecture C1: The conjecture is that the actual value of $a+b$ lies within a bounded distance from the minimum of $\lfloor {z \over r} \rfloor + r$ for composite $z$.

I am not sure if this conjecture is true. It would be very interesting if this conjecture is indeed true and the bounded distance is small relative to complexity of other factoring algorithms.

For $z$ prime, we get a plot as given below (example for $z = 4397$, prime):

As we can observe, for prime $z$, the distance between the sum of divisors (i.e, $a+b = 1+z$ is maximal from the minimum of $\lfloor {z \over r} \rfloor + r$.

For $z = 3 \times \ 443 \times 617 = 819993$, the plot of $\lfloor {z \over r} \rfloor + r$ is given below:

Here's the closeup of the plot of $\lfloor {z \over r} \rfloor + r$ for $z = 3 \times \ 443 \times 617 = 819993$

If $z$ is composite, there should be a minimal sum of divisors $a+b$ among all possible combination of divisors $a,b$ and if the conjecture is true, is the minimal $a+b$ at a bounded distance from the minimum of $\lfloor {z \over r} \rfloor + r$ that gives us an effective search procedure for $a+b$?

Additional conjectures and comments:

Conjecture C2: The minimal sum of divisors $a+b \ge \min(\lfloor {z \over r} \rfloor + r)$

Observation O1: Since $z$ is assumed to be odd, the divisors $a,b$ are both odd. Therefore, the sum of divisors $a+b$ is even. Assuming Goldbach's conjecture to be true, we should be able to partition $a+b = 2u = P+Q$ with $P,Q$ prime.

Conjecture C3: The value of $z \mod r$ for $r$ such that $\lfloor {z \over r} \rfloor + r)$ is minimum yields a non-trivial divisor of $z$ in $GCD(z \mod r, z)$.

Empirical data for Conjecture C3 in update section (dated Oct 23, 2020) below.
[Update: Oct 23, 2020 - $\color{red}{\text {The Conjecture C3 is false.}}$]

Counterexample is $z = 991 \times 443 = 439013$. Minimum value of $\lfloor {z \over r} \rfloor + r)$ is $1325$ and $r \in [640, 686]$ and none of the $z mod r$ have a $GCD(z mod r, z) \ne 1$. Therefore, Conjecture C3 is false.

Update: Oct 22, 2020

Theorem: Conjecture C2 is true

Proof: Conjecture C2 states that The minimal sum of divisors $a+b \ge \min(\lfloor {z \over r} \rfloor + r)$.

Let $⌊z/r_m ⌋+r_m=\min(⌊z/r⌋+r)$ be the minimum value.
$$z=⌊z/r_m ⌋ r_m+(z \mod r_m)$$
Also, let $(z/a).a+0$ be the divisor decomposition of $z$. Here $a=r$ and $b=z/r=⌊z/r⌋$. This also holds if we swap $b=r,a=⌊z/r⌋$. This gives the equality
$$a+b=⌊z/r⌋+r$$
If we choose an $r=r_m$, such that $r_m≠a$ and $r_m≠z/a$, we have
$$⌊z/r_m ⌋=(a+b-r+k)=(a+b-r_m+k)$$
where,

$k$ is the carry $⌊(a-r_m )(b-r_m )/r_m ⌋$ and

$z \mod r_m$ is the remainder $(a-r_m )(b-r_m )  \mod r_m$

Therefore,
$$a+b=⌊z/r_m ⌋+r_m+k$$
$$a+b=⌊z/r_m ⌋+r_m+⌊(a-r_m )(b-r_m )/r_m ⌋$$
$$a+b>⌊z/r_m ⌋+r_m$$

So, we have shown that $a+b≥⌊z/r⌋+r$ for any $r$.

Since, $\min(⌊z/r⌋+r)$ is the minimum value of $⌊z/r⌋+r$, we have
$$a+b≥ \min(⌊z/r⌋+r)$$

Hence the proof.

Update: Oct 23, 2020

Conjecture C3: The value of $z \mod r$ for $r$ such that $\lfloor {z \over r} \rfloor + r$ is minimum yields a non-trivial divisor of $z$ in $GCD(z \mod r, z)$.

[Update: Oct 23, 2020 - $\color{red}{\text {The Conjecture C3 is false.}}$. See counterexample above. $z = 439013 = 991 \times 443$. Leaving past update as is for posterity.]

Here is some empirical data for this conjecture:

This table is for $z=12091=103\times117$. The minimum of $\lfloor {z \over r} \rfloor + r$ is $219$. The values of $r$ that give this minimum value for $\lfloor {z \over r} \rfloor + r)$ are $r \in [108,112]$. For $r = 112$, we have $z \mod r = 107$ and $GCD(107, 12091) = 107$, a non-trivial divisor of $z$.

This table is for $z=13733=31\times443$. The minimum of $\lfloor {z \over r} \rfloor + r$ is $234$. The values of $r$ that give this minimum value for $\lfloor {z \over r} \rfloor + r)$ are $r \in [109,126]$. For $r \in \{110,124\}$, we have $z \mod r = 93$ and $GCD(93, 13733) = 31$, a non-trivial divisor of $z$.

[End of Conjecture C3 (proven false)]

Update: Nov 2, 2020

I found an interesting linkage between $\lfloor {z \over r} \rfloor + r$ and the Digital Root of a number defined in terms of the floor function. The digial root of an integer $n$ in base $b$ is denoted by $dr_b(n)$

$$dr_b(n) = {n - (b-1)\bigg\lfloor {n - 1 \over b - 1} \bigg\rfloor}$$

So,

$$n = dr_b(n) + {(b-1)\bigg\lfloor {n - 1 \over b - 1} \bigg\rfloor}$$

Choosing $n - 1 = z, b - 1 = r$, we get

$$z + 1 = dr_{r+1}(z + 1) + {r\bigg\lfloor {z \over r} \bigg\rfloor}$$

Adding $r^2$ both sides,

$$z + 1 + r^2 = dr_{r+1}(z + 1) + {r\bigg\lfloor {z \over r} \bigg\rfloor} + r^2$$

$$z + 1 - dr_{r+1}(z + 1) + r^2 = {r\bigg(\bigg\lfloor {z \over r} \bigg\rfloor + r\bigg)}$$

$$\implies \bigg\lfloor {z \over r} \bigg\rfloor + r = {z + 1 - dr_{r+1}(z + 1) + r^2 \over r}$$

Also observe that

$$z = (dr_{r+1}(z + 1) - 1) + {r\bigg\lfloor {z \over r} \bigg\rfloor}$$

$$\implies z \equiv (dr_{r+1}(z + 1) - 1) \mod r$$

If $r$ is a factor of $z$ then

$$z \equiv 0 \equiv (dr_{r+1}(z + 1) - 1) \mod r$$

$$\implies dr_{r+1}(z + 1) \equiv 1 \mod r$$

This might be an interesting lead to pursue as there is probably a recurrence relation we can form between digital roots in successive bases $r$ and $r+1$ or $r+1$ and $r-k$ that helps recover $r-k$ as a factor that gets $z \mod (r-k) \equiv 0$.

答:
This answer proves the following claims :

Claim 1 : $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\begin{cases}2\lfloor{\sqrt z}\rfloor&\text{if $\{\sqrt z\}\lt\frac 12$ and $\lfloor \sqrt z\rfloor\gt\dfrac{\{\sqrt z\}^2}{1-2\{\sqrt z\}}$}
\\\\2\lfloor{\sqrt z}\rfloor+1&\text{otherwise}
\end{cases}$$
where $\{x\}$ denotes the fractional part of $x$.

Claim 2 : Conjecture C1 is true.

Claim 1 : $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\begin{cases}2\lfloor{\sqrt z}\rfloor&\text{if $\{\sqrt z\}\lt\frac 12$ and $\lfloor \sqrt z\rfloor\gt\dfrac{\{\sqrt z\}^2}{1-2\{\sqrt z\}}$}
\\\\2\lfloor{\sqrt z}\rfloor+1&\text{otherwise}
\end{cases}$$
where $\{x\}$ denotes the fractional part of $x$.

Proof :

Using that $x-1\lt \lfloor x\rfloor \le x$ and AM-GM inequality, we have
$$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{r_m}\right\rfloor+r_m\gt \frac{z}{r_m}+r_m-1\ge 2\sqrt{z}-1\tag1$$

Also, if $\sqrt{z}=n+a$ where $n\in\mathbb Z$ and $0\le a\lt 1$, we have
$$\begin{align}\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1&=\left\lfloor\dfrac{n^2+2an+a^2}{n+1}\right\rfloor+n+1
\\\\&=\left\lfloor n+2a-1+\frac{(1-a)^2}{n+1}\right\rfloor+n+1
\\\\&=2\lfloor\sqrt z\rfloor+\left\lfloor 2a+\frac{(1-a)^2}{n+1}\right\rfloor\end{align}$$

We can say that $2a+\frac{(1-a)^2}{n+1}\lt 2$ always holds since
$$2a+\frac{(1-a)^2}{n+1}\ge 2\implies \frac{(1-a)^2}{n+1}\ge 2(1-a)\implies \frac{1-a}{n+1}\ge 2\implies -a\ge 2n+1$$which is impossible.

Case 1 : If $a\lt \frac 12$ and $n\gt\frac{a^2}{1-2a}$, then we have
$$(2\sqrt z-1)-(2\lfloor\sqrt z\rfloor-1)=2n+2a-1-2n+1=2a\ge 0\implies 2\sqrt z-1\ge 2\lfloor\sqrt z\rfloor-1$$
and
$$(2\sqrt z-1)-2\lfloor\sqrt z\rfloor=2a-1\lt 0\implies 2\sqrt z-1\lt 2\lfloor\sqrt z\rfloor$$
and
$$2a+\frac{(1-a)^2}{n+1}\lt 1\iff 2an+2a+1-2a+a^2\lt n+1\iff n\gt\frac{a^2}{1-2a}$$

So, in this case, it follows from $(1)$ that $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1=2\lfloor{\sqrt z}\rfloor$$

Case 2 : If $a\lt \frac 12$ and $n\le\frac{a^2}{1-2a}$ ($\iff n\le 2an+a^2$), then we have
$$(2\sqrt z-1)-(2\lfloor\sqrt z\rfloor-1)=2a\ge 0\implies 2\sqrt z-1\ge 2\lfloor\sqrt z\rfloor-1$$
and
$$(2\sqrt z-1)-2\lfloor\sqrt z\rfloor=2a-1\lt 0\implies 2\sqrt z-1\lt 2\lfloor\sqrt z\rfloor$$
and
$$2a+\frac{(1-a)^2}{n+1}\ge 1\iff 2an+2a+1-2a+a^2\ge n+1\iff n\le\frac{a^2}{1-2a}$$

For any integer $c$, we have
$$\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+c}\right\rfloor+\lfloor \sqrt z\rfloor+c=\left\lfloor\dfrac{n^2+2an+a^2}{n+c}\right\rfloor+n+c=2n+\left\lfloor 2a+\frac{(a-c)^2}{n+c}\right\rfloor$$

Here, suppose that
$$2a+\frac{(a-c)^2}{n+c}\lt 1$$
Then, we have $$2an+2ac+a^2-2ac+c^2\lt n+c\implies 2an+a^2\lt n+c-c^2$$
$$\implies n\le 2an+a^2\lt n+c-c^2\implies n\lt n+c-c^2\implies c(c-1)\lt 0$$
which contradicts that $c$ is an integer.

So, we see that if $a\lt \frac 12$ and $n\le\frac{a^2}{1-2a}$, then there is no $r$ such that $\lfloor\frac zr\rfloor+r=2\lfloor\sqrt z\rfloor$.

So, in this case, it follows from $(1)$ that $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1=2\lfloor{\sqrt z}\rfloor+1$$

Case 3 : If $a\ge \frac 12$, then we have
$$(2\sqrt z-1)-2\lfloor\sqrt z\rfloor=2a-1\ge 0\implies 2\sqrt z-1\ge 2\lfloor\sqrt z\rfloor$$
and
$$(2\sqrt z-1)-(2\lfloor\sqrt z\rfloor+1)=2(a-1)\lt 0\implies 2\sqrt z-1\lt 2\lfloor\sqrt z\rfloor+1$$
and
$$2a+\frac{(1-a)^2}{n+1}\ge 1\iff 2an+2a+1-2a+a^2\ge n+1\iff a^2\ge \underbrace{(1-2a)}_{\le 0}n$$which always holds.

So, in this case, it follows from $(1)$ that $$\min\bigg(\left\lfloor\dfrac{z}{r}\right\rfloor+r\bigg)=\left\lfloor\dfrac{z}{\lfloor \sqrt z\rfloor+1}\right\rfloor+\lfloor \sqrt z\rfloor+1=2\lfloor{\sqrt z}\rfloor+1.\quad\blacksquare$$

Claim 2 :  Conjecture C1 is true.

Proof :

We may suppose that $3\le a\le \sqrt z$ from which we have
$$3\le a\le z\implies (3a-z)(a-3)\le 0\implies 3a^2+3z\le az+9a\implies a+\frac za\le \frac z3+3$$
we get
$$a+b=a+\frac za\le \frac z3+3\tag2$$

It follows from Claim 1 that
$$-\min\bigg(\left\lfloor\frac zr\right\rfloor+r\bigg)\le -2\lfloor\sqrt z\rfloor\tag3$$

Finally, from $(2)(3)$, we have
$$(a+b)-\min\bigg(\left\lfloor\frac zr\right\rfloor+r\bigg)\le \frac z3-2\lfloor\sqrt z\rfloor+3.\quad\blacksquare$$


问:
Which functors agree only on objects?
Is there a term to describe when two functors agree on objects but generally differ on morphisms? I would appreciate any elaboration about what this structure implies.

Thanks!

答:
Let $C$ the category with 1 object and 2 non-identity morphism $a$ and $b$ such that $a\circ b=b\circ a =id$ (a non-trivial isomorphism). Then we have two functors $F,G:C\to C$ that necessarily agree on objects, but differ on the morphisms. Namely, take $F$ to be the identity functor (the non-identity morphisms go to themselves), and $G$ takes each non-identity morphisms to the other.

This works for instance for any group with a non trivial isomorphism. In fact, as have been said in the comments, all non-identity group homomorphisms yield a functor with the characteristics that you are looking for.

Title: Which functors agree only on objects?


问:
Prove $\int_{0}^{1}\!f(x)\cos(k\pi x)dx>0$
I have a positive continuous function $f(x)$ that is strictly monotonically decreasing on the interval $(0,1)$. That is, $f(x)>0$ and $f'(x)<0$ for $x\in(0,1)$. I want to show that the integral $$I = \int_{0}^{1}\! f(x)\cos(k\pi x)dx$$
is always positive, for all $k = 0,1,2,...$

Edit:
As pointed out in the comments, it's not possible to show this for general $f(x)$, but I have numerical eveidence to suggest that this is true for $$f(x) = {\frac {\left( 3\,{\epsilon}^{2}-\cos \left( \pi\,x
  \right) +1 \right) }{ \left( 2\,{\epsilon}^{2}
  -\cos \left( \pi\,(x) \right) +1 \right) ^{3/2}}}$$
where $0<\epsilon<1$.

答:
$I$ can be negative. For instance, with $k=2$, take the function $g(x)=1$ if $x\le\frac34$, and $g(x)=0$ if $x>\frac34$. Then
$$\int_0^1 g(x)\cos(2\pi x)dx=\int_0^\frac34\cos(2\pi x)dx=-\frac{1}{2\pi}$$
Of course, $g$ doesn't satisfy your conditions; but you can easily construct a continuous, differentiable, strictly decreasing $f$ that does satisfy them, and is sufficiently close to $g$ that the integral is still negative.

You can use this idea to construct a counter-example for any integer $k\ge 2$ $-$ just replace $\frac34$ with $\frac{3}{2k}$.


问:
How to determian the linear independence of linear functionals?
Say, we have a vector space $V$ (assume dim $>=3$), over the field of rational numbers with the basis $v_1...v_n$. Now, if we define the linear functionals $g_i$ as $g_i(v_j) = j+(i-1)*(-1)^{ij}$, for $i,j\in[1,n]$, what is the dim of span{$g_1,....,g_n$}?

I know the key is to find out the LI element number among such $g_i$, however, are there any techniques on such $g_i$?

And by the way, what is the intersection of those $g_i$'s Kernels, aka is there some basis $v_i$ such that for all $g_i$ we have $g_i=0$?

答:
(Promoting my comment to an answer.)

Since you have finite dimensions, there is some row-vector representation of each $g_i$. (There is a theorem for this, but I forgot the name of it.)

You can initially assume that $V$ has the standard basis to determine the dimension of the span of the functionals, since a change in basis won't change the dimension. In other words, start by stuffing $g_i(v_j)$ into a row vector to represent $g_i$. This will be sufficient to determine linear independence of $\{g_i\}_{i=0}^n$.


问:
Limit multivarible
How do I solve this limit?

$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$

This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$

The first one is a standard limit with value one, but I'n not sure about the other term.

答:
Yes your idea is very good to put away the $\sin$ term, now we have that, for example by spherical coordinates

$$\frac{xyz}{x^2+y^2+z^2} =r \cos \theta\sin \theta\cos^2 \phi\sin \phi $$

or as an alternative by AM-GM we can use that

$$x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2y^2z^2} \implies \left|\frac{xyz}{x^2+y^2+z^2}\right| \le \frac{|xyz|^\frac13}{3}$$

concluding in both cases by squeeze theorem.

Title: Limit multivarible


问:
How to compute conditional expectation of a discrete random variable with respect to a continuous one?
while learning about conditional expectation $\operatorname{E}(X\mid Y)$, with $X$ and $Y$ being random variables, I read the corresponding Wikipedia page, where explicit formulas are given for computation of the cases where $X$ and $Y$ are both continuous, both discrete or $X$ is continuous and $Y$ is discrete (continuous here means absolutely continuous with respect to Lebesgue measure).

How about the remaining case where $X$ is discrete but $Y$ is continuous? Is there some explicit formula in terms of the pmf of $X$ and pdf of $Y$?

答:
You can think of the conditional expectation of a random variable $X$ with finite expectation given another random variable $Y$ as the expectation of $X$ if $Y$ "is known", roughly speaking we treat $Y$ as a constant. With that in mind, \begin{align*}\textbf{E}[X\mid Y=y]=\sum_{x\in\mathbb{Z}}xf_{X\mid Y}(x\mid y)\end{align*} would make sense.

By $f_{X\mid Y}(x\mid y)$ I mean the "conditional PMF of $X$ given $Y=y$", that is $f_{X\mid Y}(x\mid y):=\frac{f_{X,Y}(x,y)}{f_Y(y)}$, where $f_Y$ denotes the PDF of $Y$ and we define $f_{X,Y}(x,y):=\frac{\mathrm d~~}{\mathrm dy}\textbf{P}(X=x,Y\leqslant y)$.

We can check this using the definition of the conditional expectation: Obviously $\textbf{E}[X\mid Y=y]$ is $Y$-measurable and for every measurable $A\in\sigma(Y)$ it holds that $$\begin{align*}\int_{A}\mathbf E[X\mid Y]\mathrm dP&=\int_{y\in Y(A)}f_Y(y)\sum_{x\in\mathbb{Z}}x f_{X\mid Y}(x\mid y)\;\mathrm dy\\&=\int_{y\in Y(A)}\sum_{x\in\mathbb{Z}}xf_{X,Y}(x,y)\;\mathrm dy\\&=\sum_{x\in\mathbb{Z}}x\textbf{P}(X=x,Y\in Y(A))\\&=\int_{A}X\;\mathrm dP.\end{align*}$$ For the first and last equality we have used the Trafo-Theorem. The second follows from the fact that $Y$ is continuous and the third should follow from the Dominated Convergence Theorem. So I think this should work.