问:
Lines parallel/perpendicular in space & Collinear points
I want to confirm two things :

In space $(\mathbb{R}^3)$ if line $\ell_1$ is perpendicular to line $\ell_2$ and line $\ell_2$ is perpendicular to line $\ell_3$, then $\ell_1$ is parallel to $\ell_3$.

The points $A,B,C$ (pairwise different) are collinear iff the vectors $\vec{AB}, \vec{CB}$ are parallel.

I think the first statement is correct due to the direction vectors and the cross product.

I think the second statement is wrong, since we have that these 3 points are collinear iff $AB + BC = AC$, right?

答:
The first statement is false

$x$ axis is perpendicular to $y$ axis which is perpendicular to $z$ axis. $x$ and $z$ are not parallel.

The second looks correct.

If $A,B,C$ are collinear, then they lie on the same line and $\vec{AB},\vec{CB}$ are parallel.

If $\vec{AB}$ and $\vec{CB}$ are parallel, then there is a scalar $\lambda$ such that

$\vec{AB}=\lambda\vec{CB}$ that is $A=B+\lambda\vec{CB}$ therefore $A,B,C$ are collinear.


问:
Simple ODE question...
I've run up on something that is intuitively obvious but I am having a little trouble seeing formally. Could someone show me exactly why $y^\prime=-2+t-y$ is asymptotic to $y=x-3$. Thanks, in advance for any assistance that you might be able to lend.  </p>

答:
It seems to me that $y'=-2+t-y$ should be $y'=-2+x-y$. Letting $u=y-x+3$, then the DE becomes
$$ u'=-u. $$
Hence $u=Ce^{-x}$. So $\lim_{x\to\infty}u=0$ and hence the solution of $y'=-2+x-y$ is asymptotical to $y=x-3$.</p>

ContentLicense: CC BY-SA 3.0

Title: Simple ODE question...


问:
Dual of Lp is Lq: Detail for Surjectivity?
I am trying to prove $(L^p(X,\mu))^* = L^q(X,\mu)$ (where $p,q$ are conjugate pairs and $1\le p< \infty$) from first principles, and I am stuck on a detail. Let $\Phi: L^q(X,\mu) \to (L^p(X,\mu))^*$ be the map given by

$\Phi(g)(f) := \int_X fg\ d \mu$

I want to prove that $\Phi$ is surjective. Here is my attempt when $X$ is a compact Hausdorff space. Let $F\in (L^p)^*$. Then by Riesz representation theorem for bounded linear functionals, there exists a measure $\nu$ such that

$\Phi(g)(f) = \int_X f \ d \nu$

I want to now apply the Radon-Nikodym theorem to $\nu$ to obtain $g = \frac{d \nu}{d \mu}$. We know that this $g$ is in $L^q$ since Holder's inequality says $||F||_{(L^p)^*} = ||g||_q$. I have two questions with my attempted proof:

My Question.

How (if at all) can we get that $\nu << \mu$ to satisfy the condition for Radon-Nikodym?

How can one generalize this when $X$ is an arbitrary measure space? (According to Bass's book, Riesz representation theorem only applies when the space is compact Hausdorff.)

Note: there are many online sources on proof of the above duality theorem. However, I did not find a proof which used the above approach specifically. (If there is such a resource, then that would be good enough of an answer for me.)

答:
The Riesz representation theorem extends to a measure space $(X, \mathcal{M}, \mu)$, indeed the identification for the dual still holds:

$1<p< \infty$ for a general measure $\mu$ (see Gariepy and Ziemer Theorem 6.43 and Folland Theorem 6.15)

$p=1$ provided $\mu$ is $\sigma-$finite (see Rudin Theorem 6.16)

Let us recall (and to define notation):

Riesz Representation Theorem: Let $1 \leqslant p<\infty$ and denote
$$
p^{\prime}:=\left\{\begin{array}{l}
\frac{p}{p-1} \quad \text { if } 1<p<\infty \\
\infty \text { if } p=1
\end{array} \quad(\text { conjugate exponent of } p)\right.
$$
Then the mapping $T: L^{p^{\prime}}(\Omega) \rightarrow\left(L^{p}(\Omega)\right)^{\prime}$, defined by
$$
\langle T(u), f\rangle_{\left(L^{p}(\Omega)\right)^{\prime} \times L^{p}(\Omega)}:=\int_{\Omega} u f d x \quad \forall f \in L^{p}(\Omega)
$$
is an isometric isomorphism and we will mean this representation by means of the identification
$$
L^{p^{\prime}}(\Omega) \equiv\left(L^{p}(\Omega)\right)^{\prime} \quad \text { if } 1 \leqslant p<\infty
$$

First we prove that $T$ is an isometry.

Then, (what you need, i.e. surjectivity): Suppose first that $|\Omega|<\infty$ (then you can relax it) and let us prove that $T$ is onto, that is, $\forall \phi \in\left(L^{p}(\Omega)\right)^{\prime} \exists u \in L^{p^{\prime}}(\Omega)$ such that
$$
T(u)=\phi \quad \Longleftrightarrow \quad\langle T(u), f\rangle_{\left(L^{p}(\Omega)\right)^{\prime} \times L^{p}(\Omega)}=\phi(f) \quad \forall f \in L^{p}(\Omega)
$$
Because of $|\Omega|<\infty, \chi_{E} \in L^{p}(\Omega)$ whenever $E \in \mathcal{M}:=\mathcal{M}_{n} \cap \Omega, p \in[1, \infty) .$ Define
the set function $\nu: \mathcal{M} \rightarrow \mathbb{R}$
$$
\nu(E):=\phi\left(\chi_{E}\right) \quad E \in \mathcal{M}
$$
Show that
$\begin{array}{ll}\text { (A) } & \nu \text { is a } \sigma \text { -finite, signed measure; }\end{array}$
(B)
$
\nu<<\mathcal{L}^{n} \quad \text { on } \mathcal{M}.
$
Indeed $|\nu(E)|<\infty$ for each $E \in \mathcal{M},$ thus $\nu$ is $\sigma$ -finite. Suppose $\left(E_{h}\right)_{h} \subset \mathcal{M}$ is a disjoint sequence, we have to prove that $\nu\left(\cup_{h=1}^{\infty} E_{h}\right)=\sum_{h=1}^{\infty} \nu\left(E_{h}\right) .$ Set $E:=\cup_{h=1}^{\infty} E_{h}$. Then for any positive integer $m$
$$
\begin{aligned}
\left|\nu(E)-\sum_{h=1}^{m} \nu\left(E_{h}\right)\right| &=\left|\phi\left(\chi_{E}\right)-\sum_{h=1}^{m} \phi\left(\chi_{E_{h}}\right)\right|=\left|\phi\left(\chi_{E}\right)-\phi\left(\sum_{h=1}^{m} \chi_{E_{h}}\right)\right| \\
&=\left|\phi\left(\chi_{E}-\sum_{h=1}^{m} \chi_{E_{h}}\right)\right|=\left|\phi\left(\sum_{h=m+1}^{\infty} \chi_{E_{h}}\right)\right| \\
& \leqslant\|\phi\|_{\left(L^{p}(\Omega)\right)^{\prime}}\left\|\sum_{h=m+1}^{\infty} \chi_{E_{h}}\right\|_{L^{p}(\Omega)}=\|\phi\|_{\left(L^{p}(\Omega)\right)^{\prime}}\left|\cup_{h=m+1}^{\infty} E_{h}\right|^{\frac{1}{p}}
\end{aligned}
$$
and $\left|\cup_{h=m+1}^{\infty} E_{h}\right|=\sum_{h=m+1}^{\infty}\left|E_{h}\right| \rightarrow 0$ as $m \rightarrow \infty$ since $|E|<\infty .$ Thus
$$
\nu(E)=\sum_{h=1}^{\infty} \nu\left(E_{h}\right)
$$
and since the same result holds for any rearrangement of the sequence $\left(E_{h}\right)_{h},$ the series converges asbolutely and (A) follows. Moreover, since
$$
|\nu(E)| \leqslant\|\phi\|_{\left(L^{p}(\Omega)\right)^{\prime}}|E|^{\frac{1}{p}} \quad \forall E \in \mathcal{M}
$$
(B) follows, too (and then you keep going, proving surjectivity).
Warning: Note that if $p=\infty$ in the previous estimate, then (B) need not hold!

I honestly would not use the Riesz representation theorem you mentioned, but the one I stated above.

ContentLicense: CC BY-SA 4.0


问:
Solving $\cos(2\theta)=\cos(\theta)$
The question :

Solve Equation $\cos(2\theta)=\cos(\theta)$ for $0 \le \theta \le 2\pi$ in terms of $\pi$.

My solution: -

Using trigonometry identity , I quickly get
$$2\cos^2(\theta)-1=\cos(\theta)$$
replcae $$\cos(\theta)=a.$$
I'll get quadratic equation of
$$2a^2-a-1=0,$$
where $a$ is $$a = -\frac12 , 1.$$
The result I got is$$\theta=0,{2\over3}\pi,2\pi.$$
The scheme said I'm missing is $\frac43\pi$.

Where am I missing? How to achieve it?

Title: Solving $\cos(2\theta)=\cos(\theta)$

Body:
$$\cos\theta  =c,  \cos 2 \theta =2 c^2-1, \; 2c^2-c-1= (c-1) (2c+1)=0,\quad c= (1, -\dfrac12);\; \theta= (0,2\pi/3,4\pi/3)..$$

and the cyclic co-terminal angles.


问:
Excircle and parallelogram
Can you provide a proof for the following claim:

Claim. Given any parallelogram $ABCD$ and excircle of triangle $\triangle ABC$ oposite to vertex $A$. An arbitrary tangent $t$ is constructed to the excircle. Let $n_1$ , $n_2$ , $n_3$ , $n_4$ be a signed distances from vertices $A$,$B$,$C$,$D$ to tangent line respectively, such that distances to a tangent from points on opposite sides are opposite in sign, while those from points on the same side have the same sign . Then $n_1+n_3=n_2+n_4$ .

I don't have idea how to start the proof. My first guess is to use barycentric coordinates but I am not sure which triangle to choose as the reference triangle. Any hints are welcomed.

EDIT

As DR SK MOBINUL HAQUE suggested in his comment the more general form of this claim holds as well. You can find a GeoGebra applet that demonstrates his generalization here.

答:
Let the feet of perpendiculars from point $A$, $D$, $B$ and $C$ this line be point $E$, $F$, $G$ and $H$. Extend $BG$ through point $G$ to point $L$ such that $GL=DF$.Also, extend $AE$ through point $E$ to point $M$ such that $EM=CH$. Draw $LM$, $MD$ and $MC$.

Observe that,  $EF=GH$. Hence the midpoint of $EH$ and $FG$ are the same.Quadrilaterals $FDGL$ and $MECH$ are parallelograms. $DL$ and $MC$ will pass through this midpoint of $EH$ and $FG$.Also, $MC$ and $DL$ will bisect each other. Thus, quadrilateral $MDCL$ is a parallelogram.

$\Rightarrow ML\parallel CD$

$\Rightarrow ML\parallel AB$

Also, since $AM\parallel LB$, quadrilateral $AMLB$ is a parallelogram.

$\Rightarrow AM=BL$

$\Rightarrow n_{1}+n_{3}=n_{2}+n_{4}$

Title: Excircle and parallelogram


问:
Characterization of the denominator of a polynomial $P \in \mathbb Q[X]$ by looking at roots?
Let $P(X) \in \mathbb Q[X]$ be a monic polynomial of degree $n$ and $a \in \mathbb Z$. We have the implications
$$a P(X) \in \mathbb Z[X] \implies a^n P(X/a) \in \mathbb Z[X] \iff \text{the roots of $P$ lie in } \overline{\mathbb Z}/a \,.$$

I would like to add extra conditions to the statements on the right to make this an equivalence.

This question is a bit vague, naturally, because I don't know what conditions to add. Of course, I don't want anything trivial that amounts to saying that $aP \in \mathbb Z[X]$. My motivation below can help to get a feeling for what type of condition I'd like to see. Something like "$aP$ sends integers to integers" would be acceptable.

Motivation. I started thinking about this when answering (or trying to) the following questions:

If $f(x)\in\mathbb Q[x]$ and $f(f(x)),f(f(f(x)))\in\mathbb Z[x]$ prove that $f(x)\in\mathbb Z[x]$

Can composition of integer polynomial and rational polynomial with a non-integer coefficient result in integer polynomial?

Ideas. Requiring that $aP(\mathbb Z) \subset \mathbb Z$ is not sufficient: take $a = 2$, $P = X^3 + \frac{X^2+X}4 = \frac{(2X)^3 + (2X)^2 + 2 (2X)}{8}$, which has roots in $\overline{\mathbb Z}/2$.

Even imposing the same on all derivatives of $aP$ and imposing that $c-d \mid aP(c) - aP(d)$ is not enough: $a = 2$, $P = X^5 + \frac{X^4+X^2}4$.

答:
Edit: the OP clarified after that the goal was to get $aP\in \Bbb{Z}[x]\iff ...$ not "the roots are in $a^{-1}\Bbb{Z}\iff...$"

For $P=\sum_{n=0}^d c_n X^n =\prod_j (X-\beta_j)\in \Bbb{Q}[X]$ monic then all the $a\beta_j\in \overline{\Bbb{Z}}$ iff $a^d P(X/a)\in \Bbb{Z}[X]$ iff the denominator of $c_n$ divide $a^{d-n}$.

I'd say there is no simpler criterion, the usual statement

$a \beta\in \overline{\Bbb{Z}}$ if $P(\beta)=0$, $P\in \Bbb{Z}[X]$ with leading coefficient $a$

is weaker and cannot be adapted to a converse.

This leads to the $p$-adic Newton polygon: the minimum power of $p$ such that all the $p^k\beta_j$ are in $\overline{\Bbb{Z}_p}$ is the least $k$ such that $v_p(c_n p^{k(d-n)})\ge 0$, allowing $k$ rational we find $\inf v_p(\beta_j)$.


问:
Prove double negation of LEM in intuitionistic logic
I understand that in intuitionistic logic, the law of excluded middle $P \lor \lnot P$ and double negation elimination $\lnot \lnot P \to P$ are not true in general (for every proposition $P$). However, I understand that $\lnot \lnot (P \lor \lnot P)$ and $P \to \lnot \lnot P$ are intuitionistically true in general. Is there an "intuitive" intuitionistic proof for these?

答:
For $P \to \lnot \lnot P$.

Note that negation introduction is "safe" also for Intuitionistic Logic.

According to the BHK interpretation we have:

There is no proof of $\bot$

A proof of $P\to Q$ is a function that converts a proof of $P$ into a proof of $Q$.

Assume we have a proof of $P$ and assume that we have a proof of $P \to \bot$. By Modus Ponens, we get a proof of $\bot$, that is impossible.

Thus, starting from a proof of $P$, we can build a function that converts a proof of $P \to \bot$ into a proof of $\bot$, i.e. into a proof of $(P \to \bot) \to \bot$.

For $¬¬(P∨¬P)$, we have to observe that Intuitionistic rejection of LEM means that LEM is not valid, not that it is false.

In IL we prove that $¬(P∨¬P)$ leads to a contradiction, and this amounts to a prove of $¬¬(P∨¬P)$.

The proof uses negation introduction and disjunction introduction, and both principles are intuitionistically safe.

From an Intuitionistic point of view (Brouwer, 1908), LEM amounts to the (unreliable) principle that every mathematical problem is decidable, while $¬¬(P∨¬P)$ amounts to asserting the principle that no absolutely unsolvable problem exists (Heyting, 1934).


问:
Compute the difference of three series (motivated by an integral formula)
Let
\begin{align}
L&=2^{3 / 2} \pi^{4} \frac{1}{4} \sum_{n, m, k=1}^{\infty} n^{2} m a_{n} a_{m} b_{k}\left(\delta_{m, n+k}+\delta_{n, m+k}-\delta_{k, n+m}\right) \\[5pt]
R_{1}&=2^{3 / 2} \pi^{4} \frac{1}{4} \sum_{n, m, k=1}^{\infty} n^{3 / 2} m^{3 / 2} a_{n} a_{m} b_{k}\left(\delta_{m, n+k}+\delta_{n, m+k}+\delta_{k, n+m}\right) \\[5pt]
R_{2}&=\frac{1}{2} 2^{3 / 2} \pi^{4} \frac{1}{4} \sum_{n, m, k=1}^{\infty} n^{1 / 2} m^{1 / 2} k^{2} a_{n} a_{m} b_{k}\left(\delta_{m, n+k}+\delta_{n, m+k}-\delta_{k, n+m}\right),
\end{align}
where the $\delta$ denote the Kronecker delta.

How can we compute $$L-R_1-R_2 \ ?$$

The motivation for this computation comes from an integration by parts formula analyzed on MathOverflow.

答:
With the help of the Kronecker delta's we can carry out the summation over $k$,
$$L-R_1-R_2=2^{3/2}\pi^4 \frac{1}{4}\sum_{n=1}^\infty\sum_{m=1}^\infty a_na_m\biggl[\bigl(m n^2-(m n)^{3/2}-(m-n)^2 \sqrt{m n}\bigr) b_{|n-m|}$$
$$-\bigl(m n^2+(m n)^{3/2}-(m+n)^2 \sqrt{m n}\bigr) b_{n+m}\biggr].$$
Without further knowledge of the coefficients this cannot be evaluated further.


问:
Regarding $\lim_{n \to \infty} n^{\frac{1}{n}}$
Suppose $\lim_{n \to \infty} n^{\frac{1}{n}} = l \in \mathbb{R}$. The function $f(x) = x^n$ is continuous, then

$$l^n=\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right ) =\lim_{n \to \infty} n = \infty.$$

Then it follows that $l = \infty $. This is false, but I couldn't find which step is wrong. Could you please help me? Thank you for your time.

答: Just note that $\left(1+\frac{\sqrt2}{\sqrt {n-1}}\right)^n\ge1+n\frac{\sqrt 2}{\sqrt n} +\frac{n(n-1)}2\frac 2{n-1}>n$ for $n>1$, so $$1<n^{\frac1n}<1+\frac{\sqrt2}{\sqrt{n-1}}\to 1$$</p>


问:
Difference in calculating continuity and uniform continuity
I've done some exercises on continuity using the $\varepsilon$-$\delta$ definition. Now going through my script I saw that there is a stricter version of continuity called uniform continuity. After watching some videos I think I understand the difference and the importance of the quantifiers and their order.
I've seen many questions here that ask about the intuition behind continuity, but I could not find the difference in proving it.

My question now is: Is the following statement true? Apologies for the informal way of writing this.

When working through an exercise and showing that a function is continuous, we take $|f(x)-f(x_{0})|<\varepsilon$ and then we rearrange things until we find $|x-x_{0}|$. From the definition we know that this is $<\delta$ and we start to "estimate" and calculate some things until we get something of the form: $\delta = \varepsilon\ldots$

In the proof, we then plug it in to "hopefully" get rid of everything and only $\varepsilon$ remains.

Is it true that if only $\varepsilon$ remains, the function is uniformly continuous since it does not depend on the $x_{0}$? And that if the $x_{0}$ were to remain, we have that it is continuous, but not uniformly?

答:
If you are able to find a $\delta>0$ which depends on $\varepsilon$ but not on $x_0$ then, yes, you shall have proved that $f$ is uniformly continuous.

But what if every $\delta$ that you are able to find depends upon $\varepsilon$ and upon $x_0$? Then you shall have proved that $f$ is continuous, but don't jump to the conclusion that it is not uniformly continuous. Take, for instance$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\sqrt x.\end{array}$$If $x_0\in[0,1]$, then$$x-x_0=\left(\sqrt x-\sqrt{x_0}\right)\left(\sqrt x+\sqrt{x_0}\right)$$and therefore\begin{align}\left|f(x)-f(x_0)\right|&=\left|\sqrt x-\sqrt{x_0}\right|\\&=\frac{\left|x-x_0\right|}{\sqrt x+\sqrt{x_0}}\\&\leqslant\frac{\left|x-x_0\right|}{\sqrt{x_0}}\end{align}and therefore you can take $\delta=\sqrt{x_0}\,\varepsilon$. But this only works if $x_0\ne0$ and you shall have to find another $\delta$ for the case in which $x_0=0$ ($\delta=\varepsilon^2$ will do then). So, the $\delta$ does depend upon $\varepsilon$ and upon $x_0$. However, $f$ is uniformly continuous (as is every continuous function form an interval $[a,b]$ into $\Bbb R$). The fact that this approach leads to a $\delta$ which depends upon both $\varepsilon$ and $x_0$ doesn't prove that there is no $\delta$ which depends only upon $\varepsilon$.


问:
Does $x$ irreducible in ring $R$ imply $(x)$ maximal ideal of $R$?
I was studying for my final exam of abstract algebra and, after seeing that $p$ prime element of a ring $R$ is equivalent as saying the ideal $(p)\unlhd R$ is prime, I came up with the assumption that something similar might happend with the irreducible elements and the maximal ideals, but this is not said anywhere in my course book so I may be wrong. So my question is, is this statement true?

Being $R$ a commutative unital ring, and $0\neq x\in\mathbb{R\setminus R^\times}$, then: $$x \text{ irreducible in $R$} \Longleftrightarrow (x) \text{ is maximal ideal of $R$}$$

Maybe this is only verified under certain extra conditions of $R$ (for example, being PID or UFD or some kind of special ring). Is my assumption true in general? If not, could you give me a counter example? It would be nice to have a proof if it's true. Any help will be appreciated, thanks in advance.

答:
$x \in R$ is irreducible if and only if the ideal $(x)$ is maximal among proper principal ideals.  The proof is straightforward.

Thus if there exist proper ideals in $R$ which are not principal, then there will exist irreducible elements whose principal ideals are not maximal ideals.


问:
Why can't a neighbourhood contain the entire plane?
Context:

Let $S$ be the spherical surface of radius 1 with center at the origin of $R^{3}$, let $p=(0,0,1)$ be the north pole of $S,$ and let $f: S-p \rightarrow R^{2}$ be the stereographic projection from $p$ of $S-p$ onto the equatorial plane. Construct a diagram which shows that $f$ is continuous. Over what part of $S-p$ is $f$ a contracting function? Show that $f$ is $1-1,$ and that $f^{-1}: R^{2} \rightarrow S-p$ is also continuous. What is the image under $f$ of the deleted neighborhood $N(p, r, S)-p$ for $r$ -values less than $1 ?$ Why is it impossible to define $f p$ so that the extended function is continuous?

Solution given:

Contracts distances for any pair of points in the lower hemisphere. $f^{-1}$ sends $y \in R^{2}$ into the intersection of the ray py with S. See Fig. S3.
The same picture shows that the inverse is continuous. The image of $N(p, r, S)-p$ is the exterior of a circle with center at the origin. If $f p$ were defined and $f$ were continuous at $p$, then any $N(fp,\epsilon)$ would have to contain the entire exterior of some circle; but this is impossible.

My question:

Why is it impossible for a neighbourhood to contain the entire exterior or the entire plane, excluding the circle itself?

Is it because :

A neighbourhood cannot have infinite radius?

A neighbourhood is an open set, thus in the plane where it exists, there will always be points on the plane that is not in the neighbourhood, in this case any point $q$ such that $d(O,q)=\epsilon=\infty$, assuming there exists the center of this neighbourhood, $O$? (Though the authors have not introduced the concept of open sets)

A neighbourhood is an open disk, while the plane has no definite shape, thus it cannot extend over the whole plane?

Thanks in advance.

Source

Section 1.3, Exercise 3, Chinn, Steenrod: First Concepts of Topology

答:
The keyword here is "any". It is definitely possible for some neighborhood of $f(p)$ to contain the entire plane. But not all of them do.

A neighborhood of a point $p$ is any open set that contains $p$. Since the entire plane is an open set in itself, it is a neighborhood of every point.


问:
What is the discrepancy I am facing in $\lim_{x \to 0} \frac{x \sin x}{1- \cos x}$?
Take a look at the following limit: I am missing something simple but I'm not being able to get it.
$$\lim_{x \to 0} \dfrac{x \sin x}{1- \cos x}$$

We can write the denominator as $2\sin^2x/2$ and the $\sin x$ above it as $2 \sin x/2\cdot \cos x/2$ which gives:
$$\lim_{x \to 0} \dfrac{x \cos x/2}{\sin x/2}$$
which being in $0/0$ form can be operated using L'Hopital's rule. We differentiate and then get $2$.

But, if we take another approach and directly use that rule in the first step, we get-
$$\lim_{x \to 0} \dfrac{x \cos x + \sin x}{\sin x}$$
Here I tried cancelling the sines which gave-
$$\lim_{x \to 0}x \cot x+1=1$$
I know that the first one is right, but I think I am making a very common mistake people make in limits. Is cancelling the sines a wrong step here? In general, what should I do to calculate these limits so that I don't make these mistakes?

答:
Be careful! In fact, $$\lim_{x \rightarrow 0} x \cot x$$ is also an indeterminant form, so a second application of L'Hopital's rule is necessary: $$\lim_{x \rightarrow 0} \frac{x \cos x + \sin x}{\sin x} = \lim_{x \rightarrow 0} \frac{2 \cos x - x \sin x}{\cos x} = 2$$

I always suggest to try direct substitution. In this case, cot 0 is undefined, so you obtain the indeterminant form $$0 \cdot \cot 0 = 0 \cdot \infty$$ Any time you get an infinity in your answer, you need to be careful.


答:
Using your first idea

$$1-\cos(x)=2\sin^2\left(\frac{x}{2}\right).$$
Then $$\frac{x\sin(x)}{2\sin^2(x/2)}=2\cdot \frac{(x/2)^2}{\sin ^2(x/2)}\cdot \frac{\sin(x)}{x}.$$

Using $\sin(x)\sim x$ at $0$ gives the wished answer.

Using your second idea with l'Hospital

You have that \begin{align*}
\lim_{x\to 0}\frac{x\sin(x)}{1-\cos(x)}&=\lim_{x\to 0}\frac{x\cos(x)+\sin(x)}{\sin(x)}\\
&=1+\lim_{x\to 0}\frac{x}{\sin(x)}\cdot \cos(x).
\end{align*}
Using $\sin(x)\sim x$ at $0$, allows you to conclude.

Using Taylor

You have that $\sin(x)=x+o(x)$ and $\cos(x)=1-\frac{x^2}{2}+o(x^2)$

Therefore, $$\frac{x\sin(x)}{1-\cos(x)}=\frac{x^2+o(x^2)}{x^2/2+o(x^2)}=2\cdot \frac{1+o(1)}{1+o(1)}=2+o(1).$$

Title: What is the discrepancy I am facing in $\lim_{x \to 0} \frac{x \sin x}{1- \cos x}$?

Body:
Take a look at the following limit: I am missing something simple but I'm not being able to get it.
$$\lim_{x \to 0} \dfrac{x \sin x}{1- \cos x}$$

We can write the denominator as $2\sin^2x/2$ and the $\sin x$ above it as $2 \sin x/2\cdot \cos x/2$ which gives:
$$\lim_{x \to 0} \dfrac{x \cos x/2}{\sin x/2}$$
which being in $0/0$ form can be operated using L'Hopital's rule. We differentiate and then get $2$.

But, if we take another approach and directly use that rule in the first step, we get-
$$\lim_{x \to 0} \dfrac{x \cos x + \sin x}{\sin x}$$
Here I tried cancelling the sines which gave-
$$\lim_{x \to 0}x \cot x+1=1$$
I know that the first one is right, but I think I am making a very common mistake people make in limits. Is cancelling the sines a wrong step here? In general, what should I do to calculate these limits so that I don't make these mistakes?


答: The simplest counterexample is $R=\mathbb Z[x]$. Then $(x) \subset (2,x)$ is not maximal.

Title: Does $x$ irreducible in ring $R$ imply $(x)$ maximal ideal of $R$?

Body:
I was studying for my final exam of abstract algebra and, after seeing that $p$ prime element of a ring $R$ is equivalent as saying the ideal $(p)\unlhd R$ is prime, I came up with the assumption that something similar might happend with the irreducible elements and the maximal ideals, but this is not said anywhere in my course book so I may be wrong. So my question is, is this statement true?

Being $R$ a commutative unital ring, and $0\neq x\in\mathbb{R\setminus R^\times}$, then: $$x \text{ irreducible in $R$} \Longleftrightarrow (x) \text{ is maximal ideal of $R$}$$

Maybe this is only verified under certain extra conditions of $R$ (for example, being PID or UFD or some kind of special ring). Is my assumption true in general? If not, could you give me a counter example? It would be nice to have a proof if it's true. Any help will be appreciated, thanks in advance.


问:
Mouse trapped in a box with 3 exits - DTMC problem
A mouse is trapped in a box, with 3 exits. Exit 1 leads back to the box in 3 steps,
Exit 2 leads back to the box in 1 step and Exit 3 never leads back to the box. The mouse chooses the exit at random. Define a Markov chain that describes the problem and derive its transition matrix.

Attempt. We need to define $\{X_n\}$ such that $X_{n+1}$ is determined by only
$X_n$ and not its predessecors. $X_n$ being equal to $1$ if the mouse is in the box and $0$ otherwise is not suitable for our problem.

Thank you in advance.

答:
Note that you need to keep track of the location of the mouse, because every chosen exit has different consequences. If we would have that Exit 1 would lead back to the box in one step, the following diagram could be used:

In that case you have four states: $0$ (box), $1$ (exit 1), $2$ (exit 2) and $3$ (exit 3). From state $0$ you go to the three other states with equal probability. If you enter state $3$, you do not return. If you enter state $1$ or $2$, you return back to the box (state $0$).

The problem states that you only return from exit 1 to the box in 3 steps. Therefore, if the mouse chooses exit 1, you have to keep track of the number of steps already taken, to make the chain Markovian. This is not possible with the current states, forcing us to extend the state space with states that keep track of the number of steps if the mouse has chosen exit 1. One way to add a step is via the following scheme, which can be extended to include more steps:


问:
Rationality of negative continued fractions
I was wondering which is the effect of changing additions by substractions on a simple continued fraction, and its mathematical interpretation (if any). For instance,

$$\sqrt{2}=[1;2,2,2,2,\dots]=1+
\cfrac{1}{
  2+\cfrac{1}{
    2+\cfrac{1}{
      2+\cfrac{1}{
        2+\ddots
      }
    }
  }
}$$

And if we change additions for substractions,

$$x=1-
\cfrac{1}{
  2-\cfrac{1}{
    2-\cfrac{1}{
      2-\cfrac{1}{
        2-\ddots
      }
    }
  }
}$$

If we take the partial continued fractions up to $a_k$, we get we get exactly $x=\frac{1}{k}$, and thus the value of this continued fraction would be $0$.

Every simple infinite continued fraction with positive signs is an irrational number, but it seems that the simple infinite continued fractions with negative signs are always rational. Is this true? I would like to know a proof of it in this case, or some counterexample.

Any light on this topic (mathematical interpretation of the change described, and rationality of infinite continued fractions with negative signs) would be welcomed. Thanks in advance!

答:
Regarding the question of rationality of simple negative continued fractions, I have found a counterexample: I was able to represent $\phi$ as a simple negative continued fraction as follows:

$$\phi=2-
\cfrac{1}{
  3-\cfrac{1}{
    3-\cfrac{1}{
      3-\cfrac{1}{
        3-\ddots
      }
    }
  }
}$$

As $\phi$ is irrational, the question is solved.

Additionally, I found this interesting research that I share with you in case you are interested in the relationship between the negative and positive continued fractions:

https://scholarship.claremont.edu/cgi/viewcontent.cgi?filename=2&article=1183&context=hmc_theses&type=additional


问:
If the base change of varieties are isomorphic, are they birationally equivalent?
Let $X,Y$ be smooth projective varieties over a field $k$. Assume that there is a finite Galois extension $L/k$ such that there is an isomorphism $f : X \times_k L \cong Y \times_k L$.

Is it true that $X$ and $Y$ are birationally equivalent over $k$?

Notice that $f$ does not need to be the base change of a morphism $X \to Y$. Also, it does not need to be Galois-equivariant. I don't know if one could see $X$ as a Zariski-dense subscheme of $X \times_k L$, because I only know that fiber products provide projection maps $X \times_k L \to X$ (so it does not go in the right direction...).

答: No. Consider $k=\Bbb R$, $L=\Bbb C$, with $X=\Bbb P^1_{\Bbb R}$ and $Y=\operatorname{Proj} \Bbb R[x,y,z]/(x^2+y^2+z^2)$. After base change to $\Bbb C$, both $X$ and $Y$ are isomorphic to $\Bbb P^1_{\Bbb C}$. On the other hand, the $\Bbb R$-rational points of $X$ are dense while $Y$ has no $\Bbb R$-rational points. Any $\Bbb R$-morphism preserves $\Bbb R$-rational points, so no open subset of $X$ can be isomorphic to any open subset of $Y$.


问:
Extension of the field of scalars.
I want to make sure that I understand this correctly:
$V- \mathbb{Q}$ vector space.
Then $V_\mathbb{R} := V \otimes \mathbb{R} $
is naturally an R-vector space (next to
being a $\mathbb{Q}$-vector space).

If I am not mistaken this can be explained in the following way:
$\mathbb{R}$ is  a $\mathbb{Q}$ vector space, but the basis of $\mathbb{R}$ cannot be handwritten,so we construct $V \otimes_{\mathbb{Q}} \mathbb{R}$ s.t it is generated by $\{v_i\otimes r | v_i \in basis( V) ,r \in \mathbb{R}\}$ indeed if we take any $v\otimes r \in V \otimes_\mathbb{Q} \mathbb{R}$ then:
$$v\otimes r= \sum \gamma _i v_i \otimes r = \sum (v_i \otimes \gamma_i r)=\sum v_i\otimes r'$$
where $v_i$ is basis vectors of $V$ and $\gamma_i \in \mathbb{Q}$. Now,consider $\mathbb{Q}$-linear mapping :
$$V \otimes_\mathbb{Q} \mathbb{R} \rightarrow V_\mathbb{R}$$ defined as
$$v_i \otimes r \rightarrow rv_i$$ then it induces isomorphism between $V_{\mathbb{R}}$ and $V \otimes_\mathbb{Q} \mathbb{R}$. Is there any flaw in my understanding, is there something here which is not obvious and I am skipping? Please help,any help would be very much appreciated!

答:
$V_\mathbb{R} = V \otimes_\mathbb{Q} \mathbb{R}$ is a $\mathbb{R}$-vector space in "the natural way". You do not need to talk about basis elements of neither $V$ nor $\mathbb{R}$ to define the structure.

Any element of of $V_\mathbb{R}$ looks like a sum of terms $v \otimes r$, where $v \in V$ and $r \in \mathbb{R}$. To define a $\mathbb{R}$-vector space structure, we need only define what it means to multiply a vector $v \otimes r$ with another $s \in \mathbb{R}$. But we simply define $$ s \cdot ( v \otimes r) := v \otimes (sr).$$

This makes $V_\mathbb{R}$ a $\mathbb{R}$-vector space. If $v_i \in V$ is a $\mathbb{Q}$-basis for $V$, then (a subset of) $\{v_i \otimes 1\}$  is a $\mathbb{R}$-basis of $V_\mathbb{R}$.

Title: Extension of the field of scalars.


问:
Can $n!$ start with the digits 2020...
Can anyone give me any hints on how to tackle this? I assume the answer is no otherwise why ask the question. I am aware that $k \mid n!$
for all $k\leq n$, and I am aware that divisibility by $3$, $9$ and $11$ in particular have nice and easy tests involving sums of digits etc. I know that $2020 = 2^2 \cdot 5 \cdot 101$ written as a product of primes. However I am not seeing how these things help..

Thanks

Edit: Plot twist... $384!$ begins with the digits $2020$ so now the question is how do we prove this can happen (let's pretend we don't have computers)?

Title: Can $n!$ start with the digits 2020...

Body:
384,8297,8795

Brute force

Select[Table[{FromDigits[Take[IntegerDigits[n!], 4]], n}, {n, 10, 10000}], #[[1]] == 2020 &]


问:
How to apply second geometric form of hahn-banach to prove that exists a functional with norm one and has a specific value at a specific point.
I'm struggling to manage how to solve this problem:
\begin{aligned}
&\text { Let } M \text { be a proper closed subspace of a normed vector space } E \text { . Let } u_{0} \in E \text { such that }\\
&\operatorname{dist}\left(u_{0}, M\right)=\inf _{v \in M}\left\|u_{0}-v\right\|=d>0 . \text { Prove that there exists } \varphi \in E^{\prime} \text { such that } \varphi\left(u_{0}\right)=d\\
&\left.\varphi\right|_{M}=0 \text { and }\|\varphi\|_{E^{\prime}}=1.
\end{aligned}

The main idea that i had to prove this problem is, consider $[\varphi = d]$, and try to prove that is the hyperplan which cuts E in two. Since i have the distance between $u_0$ and $M$, it clearly separates $u_0$ and $M$.  This part $\varphi|_{M}=0$ is already proved. But why $\varphi(u_0)=d$ and its norm is one i'm not able to see it.  I don't know if my idea is correct, i wish some tips to finish this proof.

答:
Consider the quotient space $F=E/M.$ Since $M$ is closed therefore $F$ is an Banach space. Let $\xi : E\to F$, $\xi (u) =[u]_M =u+M$ be a quotient map which is obviously continuous.
Since $||[u_0 ]||_F = d$ therefore by Hahn Banach theorem there exist a $\kappa \in F'$ such that $||\kappa ||=1$ and $\kappa ([u_0 ])= d.$
Now is enough to observe that $\varphi =\kappa \circ \xi $ satisfies the conditions of your exercise.


问:
If $f=g$ almost everywhere, then $\int_E f = \int_E g$.
One task in the book Analysis 2 by Tao is to prove the following Proposition 8.2.6:

Let $(E,\mathcal{E},\mu)$ a measure space and $f,g \colon E \to [0,\infty]$ $\mathcal{E}$-measurable functions on $E$.

If $f=g$ almost everywhere, then $\int_E f = \int_E g$.

My problem is not to prove the claim. For example let $A = \{f \neq g\} \in \mathcal{E}$, then $f = \chi_A f + \chi_{A^c}f$, $g = \chi_A g + \chi_{A^c}g$ and $\chi_{A^c}f = \chi_{A^c}g$. So we have $\int_E f = \int_A f + \int_{A^c} f$, when we know that interchanging sum and integral is possible. Then it's easy to show the rest. ($\chi$ is indicator function)

But the only properties I have available are:

$\int_E f = \sup \left\{ \int_E g : g \text{ is simple, } 0 \leq g \leq f \right\}$

$f_E f = 0 \Leftrightarrow f=0$ almost everywhere

$\int_E cf = c \int_E f$, $c \in \mathbb{R}$, $c \geq 0$

$f \leq g \Rightarrow \int_E f \leq \int_E g$

答:
Hint

Your statement is equivalent to prove that if $N$ is a null set, then $\int_Nf=0$.

First, see what happen with simple function.

Then, if $f\geq 0$, from one of your property, there is a sequence of simple function $(g_n)$ s.t. $$\lim_{n\to \infty }\int_N g_n=\int_N f.$$

Finally, set $f^+(x)=0\vee f(x)$ and $f^-(x)=-(f(x)\wedge 0)$, where $a\wedge b$ denote the minimum of $a$ and $b$ and $a\vee b$ denote the maximum. Remark that $f^\pm \geq 0$. Then, if $f$ is unspecified, write it as $f=f^+-f^-$, and apply the previous step.


问:
If A and B are elementary sets, then A - B is elementary.
Hello I am reviewing my notes, but I do not understand the last step well, could you explain why that happens?

If $A=\bigcup_{k}^{}P_{k}$  and  $B=\bigcup_{j}^{}Q_{j}$

where $P_k,Q_j$ are disjoint rectangles two by two

Demonstration
$$A-B=\bigcup_{k}^{}(P_{k}-B)$$Now I don't understand why$$P_{k}-B=\bigcap_{j}^{n}(P_k-Q_j)$$

I thought
$$P_{k}-B=P_k \bigcap B^{c}=P_k \bigcap (\bigcup_{j}^{}Q_{j})^{c}$$The DeMorgan's$$P_k \bigcap (\bigcap_{j}^{}Q_{j}^{c})=\bigcap_{j}^{}(P_k \bigcap Q_{j}^{c})=\bigcap_{j}^{}(P_k - Q_{j})$$

But,$ P_k \bigcap (\bigcap_{j}^{}Q_{j}^{c})=\bigcap_{j}^{}(P_k \bigcap Q_{j}^{c})$It is right?, why?

答:
Recall that "set difference" (\setminus) is defined as follows: $A \setminus B = A \cap B^c$.

Thus: $$P_k \setminus B=P_k \setminus \bigcup_j Q_j= P_k \cap (\bigcup_j Q_j)^c= [\text { by De Morgan} ] P_k \cap (\bigcap_j Q_j^c)=\bigcap_j(P_k \cap Q_j^c)=\bigcap_j(P_k \setminus Q_j).$$


问:
How to prove this inequality for the given function?
Given function $f(x)=x-\log x$. When $a>1$, there are two roots $r_1(a), r_2(a)$ for equation $f(x)=a$.
We also have $0<r_1(a)<1<r_2(a)$. Now given two real numbers $a>1,b>1$, how to prove $f(r_2(a)r_2(b))>f(r_1(a)r_1(b))$?

Body:
We have,
\begin{align*}
f(r_2(a) r_2(b)) - f(r_1(a) r_1(b)) & = r_2(a) r_2(b) - \log(r_2(a) r_2(b)) - r_1(a) r_1(b) + \log(r_1(a)r_1(b)) \\
& = r_2(a) r_2(b) - \log(r_2(a)) - \log(r_2(b)) - r_1(a) r_1(b) + \log(r_1(a)) + \log(r_1(b)) \\
& = r_2(a) r_2(b) + a - r_2(a) + b - r_2(b) - r_1(a) r_1(b) + r_1(a) - a + r_1(b)-b \\
& = r_2(a) r_2(b)  - r_2(a) - r_2(b) + 1 - 1 - r_1(a) r_1(b) + r_1(a)  + r_1(b) \\
& = (r_2(a) - 1)(r_2(b) - 1) + (1 - r_1(a))(1 - r_1(b))   \\
& > 0,
\end{align*}
as required.


问:
Can a group be thought of as some kind of "rule book" telling how symmetries interact with each other?
I'm new to Abstract Algebra and I'm trying to explain myself what a group really is. So far I've looked into a couple of books, read a little about groups here and there and this is what I've understood so far.

Symmetry: A symmetry of an object $O$ is a bijective function $s:O \to O$ i.e. $s(O)=O$

Intuitively, they are transformations that leave the shape of the object unchanged, they just shuffle around the pieces of the object.

Here, I've taken the object $O$ to be the set $O$ since it is my understanding that everything is a set.

Group: A group $G$ is collection of symmetries of the object $O$ satisfying:

It has a symmetry which does nothing to the object $O$

Every symmetry applied to $O$ can be reversed.

Any sequence of symmetries applied to $O$ is also a symmetry of $O$ present in $G$

Given a sequence of symmetries i.e. $s_1,s_2, s_3$, the order in which you apply this sequence of symmetries to $O$ doesn't matter i.e.
$(s_1 \cdot s_2) \cdot s_3 = s_1 \cdot (s_2 \cdot s_3)$

Hence a Group is merely a "map" or some kind of "guide book" that tells you "how" the symmetries interact with each other and you can just throw away the object, you don't care about it.

Is my intuition so far correct? If there are any flaws in it, please point out. If you can improve on this intuition, please by all means.

Now my questions are,

Why do you not care about the object?

Given a group, how can I find the object that this group describes the symmetries of?

Does every group describe symmetries of some object?

答:
Is my intuition so far correct?

Pretty much, yeah!

Why do you not care about the object?
Two reasons:

Because groups are, by themselves, interesting objects.

Because by not caring about the particular object, everything you discover about its group of symmetries holds for every object with the same symmetry group!

Given a group, how can I find the object that this group describes the symmetries of?
Does every group describe symmetries of some object?

It's... Complicated.


问:
Is a topology a set or a family?
So I've just started learning topology and a lot of the definitions confuse me.
My main problem is that some of the definitions are quite inconsistent to me for example in topology without tears the author says that a topology $\tau$ is a set of subsets of $X$ and then he proceeds with the axioms.
However I've also seen a lot from other sources that a topology is a family of subsets of $X$ and then they proceeds to describe the same axioms. However they do not refer to the topology $\tau$ as being a set at all which confuses me.

I think my confusion is the definition on what a family actually is ive seen lots of confusing definitions on what a family is like it is a surjective function however I cannot seem to grasp the idea.

So why do they define them differently and what is a family?

Thanks in advance.

答:
"A set of sets" is a somewhat abstract notion. It is, formally, what a topology is, but people also need to know how to think about them. Having an informal "hierarchy" of container names gives newcomers some mental intuitive help (or at least, that's the idea).

Thus we get "a family of sets".

Title: Is a topology a set or a family?


问:
Behaviour of $f$ in the neighbourhood of $c$ if $f'(c)= \cdots = f^{(n)}(c)=0$, and $f^{(n+1)}(c) \gt 0$
What can I say about the behaviour of $f$ in the neighbourhood of $c$ if
$f'(c)= \cdots = f^{(n)}(c)=0$, and $f^{(n+1)}(c) \gt 0$? I know the behaviour of $f$ if $n \le 2$, but I do not know how to generalize this. I appreciate your help.</p>

答: $f$ approximately looks like (in the Taylor series sense) $(x-c)^{n+1}$. Conclude from that what you want.</p>


答:
Your intuition is good, but there are some problems. I must admit that I have no idea about the meaning of “Any sequence of symmetries applied to $O$ is also a symmetry of $O$ present in $G$”. Perhaps that you meant that the composition of symmetries present in $G$ also belongs to $G$.

Note that every group $G$ is a set of symmetries of $G$ itself: for every $h\in G$, consider the map$$\begin{array}{ccc}G&\longrightarrow&G\\g&\mapsto&hg.\end{array}$$The set of all these maps is a set of symmetries of $G$. So, yes, in this sense every group is the group of symmetries of some object.

However, thinking this way will perhaps make harder to see that quite often the same group is a group of symmetries of two very distinct objects. For instance, the group of all isometries of a regular triangle and the group of all permutations of the set $\{1,2,3\}$ are the same group (technically speaking, they are isomorphic), although, of course, we have very distinct objects here.

Title: Can a group be thought of as some kind of "rule book" telling how symmetries interact with each other?

Body:
I'm new to Abstract Algebra and I'm trying to explain myself what a group really is. So far I've looked into a couple of books, read a little about groups here and there and this is what I've understood so far.

Symmetry: A symmetry of an object $O$ is a bijective function $s:O \to O$ i.e. $s(O)=O$

Intuitively, they are transformations that leave the shape of the object unchanged, they just shuffle around the pieces of the object.

Here, I've taken the object $O$ to be the set $O$ since it is my understanding that everything is a set.

Group: A group $G$ is collection of symmetries of the object $O$ satisfying:

It has a symmetry which does nothing to the object $O$

Every symmetry applied to $O$ can be reversed.

Any sequence of symmetries applied to $O$ is also a symmetry of $O$ present in $G$

Given a sequence of symmetries i.e. $s_1,s_2, s_3$, the order in which you apply this sequence of symmetries to $O$ doesn't matter i.e.
$(s_1 \cdot s_2) \cdot s_3 = s_1 \cdot (s_2 \cdot s_3)$

Hence a Group is merely a "map" or some kind of "guide book" that tells you "how" the symmetries interact with each other and you can just throw away the object, you don't care about it.

Is my intuition so far correct? If there are any flaws in it, please point out. If you can improve on this intuition, please by all means.

Now my questions are,

Why do you not care about the object?

Given a group, how can I find the object that this group describes the symmetries of?

Does every group describe symmetries of some object?


问:
Can the expression $\log \left[-\frac{b}{k}\right]+2\log \left[\frac{b}{k}\right]-\log \left[-\frac{k}{b}\right])$ be simplified?
Can the following expression be simplified?

\begin{equation}
\log \left[-\frac{b}{k}\right]+2\log \left[\frac{b}{k}\right]-\log \left[-\frac{k}{b}\right]
\end{equation}

given $b,k>0$.

答:
$$ log (-b/k) + 2 log (b/k) - log (-k/b)$$
$$ log (-b/k) + log(b/k)^2 - log(-k/b) $$
$$ log [(-b/k)(b/k)^2]/(-k/b) $$
$$ log(b/k)^4 $$
$$ 4 log b - 4 log k $$

Hope you would have understood

Edit: I am not sure however is the question correct or not because as per log definition $$log _a b$$ in this b should not be negative however for third term you have made b negative


答: What do you think you can say about the behavior of $f(x)=(x-c)^{n+1}$?</p>

Title: Behaviour of $f$ in the neighbourhood of $c$ if $f'(c)= \cdots = f^{(n)}(c)=0$, and $f^{(n+1)}(c) \gt 0$

Body:
What can I say about the behaviour of $f$ in the neighbourhood of $c$ if
$f'(c)= \cdots = f^{(n)}(c)=0$, and $f^{(n+1)}(c) \gt 0$? I know the behaviour of $f$ if $n \le 2$, but I do not know how to generalize this. I appreciate your help.</p>


问:
Prove this formula $\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$
I am trying to use prove, by just simple algebraic manipulation, to prove the equality of this formula.
$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$$

I have been given hints and instructions from this thread

Write using Euler’s identity $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Factor the denominator

Find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again).

This is how I have done:

RHS: $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{1-2r(\dfrac{e^{ix}+e^{-ix}}{2})+r^{2}}=\dfrac{1-r(\dfrac{e^{ix}-e^{-ix}}{2})}{1-re^{ix}-re^{-ix}+r^2}$$

Then from this thread, I have learnt how to factorize the denominator (See Jack D'Aurizio answer, first answer of the thread, first line). The trick is to write $1=e^0$.

$$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(re^{ix})(r-e^{-ix})}$$

The third step is to decompose the fraction:

$$\dfrac{A}{r-e^{ix}}+\dfrac{B}{r-e^{-ix}}=\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(r-e^{ix})(r-e^{-ix})}$$

$$A(r-e^{-ix})+B(r-e^{ix})=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I notice that let $x=0$, we will have

$$A(r-1)+B(r-1)=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I don't know to find $A$ and $B$

Also, could you provide in details how to finish the part "expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again)".

My symbolic manipulation skill is not very good, so a detailed answer is great!

Thanks!

答:
Similar to Buraian's answer, a method I enjoy using is using the equivalent series for $\sin$, and then adding $C+iS$, as follows:

Let
$$\begin{align}
C&=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\
S&=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\
\end{align}$$
Then
$$\begin{align}
C+iS&=1+r(\cos x+i\sin x)+r^2(\cos 2x+i\sin2x)+r^3(\cos3x+i\sin3x)+\cdots\\
&=1+re^{ix}+(re^{ix})^2+(re^{ix})^3+\cdots\\
&=\frac{1}{1-re^{ix}}=\frac{(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{1-r\cos x+ri\sin x}{1-2r\cos x+r^2}
\end{align}$$
Hence, equating real and imaginary parts we obtain
$$\begin{align}
C&=\frac{1-r\cos x}{1-2r\cos x+r^2}\\
S&=\frac{r\sin x}{1-2r\cos x+r^2}\\
\end{align}$$
I hope that was helpful and gives you a new and interesting method for attacking these sorts of problems.


问:
Is it true that any uncountably generated sigma algebra can be generated by a countable subset?
Let $X$ be a set and let $\mathcal C$ be a non-empty collection of subsets of $X$

From my classes I was asked if the following statement is always true:

For any $A ∈ σ(\mathcal C)$, there exists a countable
sub-collection $\mathcal D ⊂ \mathcal C$ such that $A ∈ σ(\mathcal D)$.

This was not difficult to prove true, but I also considered a slight modification to the statement:

There exists a countable sub-collection $\mathcal D ⊂ \mathcal C$ such that, for any
$A ∈ σ(\mathcal C)$, $A ∈ σ(\mathcal D)$

The second statement is equivalent to saying: There exists a countable $\mathcal D ⊂ \mathcal C$ such that $σ(\mathcal C) = σ(\mathcal D)$. My intuition is that this is false but I can't think of any examples at least in the case for $X=\mathbb R$. So my question is this, is the modified statement true for $X=\mathbb R$ and if yes then is it also true for arbitrary $X$.

答:
No, the statement is not even true for $X=[0,1)$. Let $\mathcal{C}= \{\{x\}|\;x\in [0,1]\setminus \mathbb{Q}\}$. We'll prove that $\sigma(\mathcal{C})$ is not countably generated. One may note that every set in this $\sigma$-algebra must have one of the forms
\begin{align}
&\emptyset\\
&\ [0,1)\\
&\bigcup_{n=1}^{\infty}\{x_n\}\\
&\bigcap_{j=1}^{\infty} [0,1)\setminus \{y_j\}\\
&(\bigcup_{n=1}^{\infty}\{x_n\})\cup(\bigcap_{j=1}^{\infty} [0,1)\setminus \{y_j\})
\end{align}
for any choice of sequences of irrationals $x_n$ and $y_j$, since taking unions and complements of the last type of set yields a set of one of the previous forms.

For $x\in [0,1)$, define the atom of x, $A(x):=\cap_{x\in E\in \sigma(\mathcal{C})} E$. If $\sigma(\mathcal{C})$ were countably generated, we would have $A(x)\in \sigma(\mathcal{C})$ for every $x$. To see this, note that if $\sigma(\mathcal{C})=\sigma(\mathcal{D})$, where $\mathcal{D}$ is countable, we would have
$$
A(x)=(\bigcap_{x\in D\in \mathcal{D}} D)\cap (\bigcap_{x\not\in D\in \mathcal{D}} [0,1)\setminus D):=A'(x)\in \sigma(\mathcal{D})
$$
To see this, note that $\mathcal{E}=\{E\subseteq [0,1)|\; x\in E\Longrightarrow A'(x)\subseteq E\}$ is a $\sigma$-algebra containing $\mathcal{D}$.

Now, clearly, for $x\in [0,1)\setminus \mathbb{Q}$, we have $A(x)=\{x\}\in \mathcal{C}$, so assume $x\in \mathbb{Q} \cap  [0,1)$. Using the above characterisation of $\sigma(\mathcal{C})$, we easily get that $A(x)=\mathbb{Q}\cap [0,1)$ and that $\mathbb{Q}\cap [0,1)$ is not an element of $\sigma(\mathcal{C})$. Thus, $\sigma(\mathcal{C})$ cannot be countably generated even though it's a sub-$\sigma$-algebra of the Borel algebra.


问:
Help with the following first-order partial differential equation
I am trying to teach myself how to solve first-order PDEs using the method of characteristics, and I would like to solve the following first-order PDE for $u = u \left(x, y, t \right)$:

\begin{align*}
a \frac{\partial u}{\partial t} + y \frac{\partial u}{\partial x} + b u &= f \left(x, t \right) \tag{1}
\end{align*}

where $a$ and $b$ are constants. From the method of characteristics we get:

\begin{alignat*}{2}
& \frac{dt}{ds} &&= a &&\Rightarrow &&t \left(s \right) &&= as + c_{1} \tag{2}\\
& \frac{d y}{ds} &&= 0 &&\Rightarrow &&y \left(s \right) &&= c_{2} \tag{3}\\
& \frac{d x}{ds} &&= y &&\Rightarrow &&x \left(s \right) &&= y s + c_{3} \tag{4}\\
& \frac{d z}{ds} &&= f \left(x,t \right) - bz &&\Rightarrow &&z \left(s \right) &&= e^{-bs} \left[ \int{e^{bs} f \left(x \left(s \right), t \left(s \right) \right) ds} + c_{4} \right] \tag{5}
\end{alignat*}

However, I am unsure of how to proceed from here to obtain $u \left(x, y, t \right)$. In particular, I am confused regarding how to express the integral in the last equality in terms of $x$ and $t$. Any help on this would be greatly appreciated.

As an aside, I inputted $(1)$ into WolframAlpha, which gave the following:

\begin{align*}
u \left(x,y,t \right) &= e^{\frac{-bx}{y}} \left[\int_{1}^{x}{\frac{e^{\frac{b \xi}{y}} f \left(\xi, t + \frac{a \left(\xi - x \right)}{y} \right)}{y} d \xi} + g \left(y \right) \left(t - \frac{a x}{y} \right) \right] \tag{6}
\end{align*}

Is this correct? I am slightly confused since from $(1)$, $f$ should depend on $x$ and $t$ only; however, from $(6)$ it appears that $f$ depends on $x$, $t$ and $y$. Apologies in advance if these questions sound silly.

Kind regards,
Nathan

答:
The equations $(5)$ and $(6)$ are both correct, however contradictory they may seem.

This is because Eq.$(5)$ is valid only on the characteristic curve but not valid everywhere. Eq.$(6)$ is valid everywhere (any $x,y,t$).

An equivalent but different presentation may be helpful to understand.

$$a \frac{\partial u}{\partial t} + y \frac{\partial u}{\partial x}  =-bu+ f \left(x, t \right) $$
Your system of ODEs is correct. Written on an equivalent form (Charpit-Lagrange) :
$$ds=\frac{dt}{a}=\frac{dx}{y}=\frac{dy}{0}=\frac{du}{-bu+ f \left(x, t \right)}$$
You corectly found the characteristic equation $\quad y=c_2$.

A second characteristic equation comes from solving $\quad \frac{dt}{a}=\frac{dx}{y}=\frac{dx}{c_2}$:
$$t-\frac{ax}{c_2}=c_3.$$
A third characteristic equation comes from solving $\quad \frac{dx}{c_2}=\frac{du}{-bu+ f \left(x, t \right)}$
$$c_2\frac{du}{dx}=-bu+f(x,t)$$
This is a linear first order ODE wrt the unknown function $u(x)$. The solution is:
$$u=e^{-b\frac{x}{c_2}}\int e^{b\frac{x}{c_2}}f(x,t)dx$$
Especially this must be valid on the characteristic curves $y=c_2$ and $t=c_3+\frac{ax}{c_2}$ :
$$u=e^{-b\frac{x}{y}}\int e^{b\frac{x}{y}}f\left(x\:, \left(c_3+\frac{ax}{y} \right)\right)dx$$
With definite integral :
$$u=e^{-b\frac{x}{y}}\left[\int^{\xi=x} e^{b\frac{\xi}{y}}f\left(\xi\:,\left(c_3+\frac{a\xi}{y} \right)\right)d\xi+C \left(c_2,c_3 \right)\right]$$
$$u=e^{-b\frac{x}{y}}\left[\int^x e^{b\frac{\xi}{y}}f\left(\xi\:, \left(t-\frac{ax}{y}+\frac{a\xi}{y} \right)\right)d\xi+C \left(y\:,t-\frac{ax}{y} \right)\right]$$

$$u(x,y,t)=e^{-b\frac{x}{y}}\left[\int^x e^{b\frac{\xi}{y}}f\left(\xi\:, t + \frac{a(\xi-x)}{y}\right)d\xi+C \left(y\:,t-\frac{ax}{y} \right)\right]$$
$C$ is an arbitray function of two variables.

I don't know why WolframAlpha chose $1$  as lower bound for the integral instead $0$ or any other value. So, I let the choice open.


问:
Integration along fibers
The following statements are  from the book heat kernel and dirac operator chapter 1.

" Let $\pi : M \rightarrow B $ be a fiber bundle with n-dimensional fiber, such that both M and B are oriented. If $\alpha \in {A}_c^k(M)$ is a compactly-supported differential form on M, its integral over the fibers of $ M \rightarrow B $ is the differential form $\int\limits_{M/B} \alpha \in A^{k-n}(B)$ such that

\begin{equation}
\int\limits_B (\int\limits_{M/B} \alpha) \wedge \beta = \int\limits_M \alpha \wedge \pi^* \beta \quad  ... (1.15)
\end{equation}
for all differential forms $\beta$ on the base B. We sometimes write $ \pi_* \alpha $ instead of $ \int\limits_{M/B} $. It follows easily from the (1.15) that

\begin{equation}
\pi_*(\alpha \wedge \pi^* \beta ) = \pi_* \alpha \wedge \beta \quad ...(1.16)
\end{equation}
for all $\alpha \in A_c(M) $ and $\beta \in A(\beta).$ "

My questions are the following:

what is the intuition behind defining the integral of $\alpha$ over the fibers by the equation (1.15), and why the notion of integration along fiber  is important ?

how to prove that equation 1.15 implies equation 1.16 ?

答:
For your first question about why integration along the fiber is important, note that it gives a cochain map of degree $-n$,
$$\pi_*: A^k(M) \to A^{k-n}(B).$$
If $M$ was an oriented rank $n$ vector bundle over $B$ then
it can be shown that this map induces isomorphism at the level of cohomology, $H^k(M) \to H^{k-n}(B)$. This is called the Thom isomorphism. You can find a proof in Bott & Tu's classic.

Now for your second part, notice that (1.15) is the defining property for $\pi_*\alpha$. That is if there were any other form $\omega$, which satisfied (1.15): $\int_B \omega\wedge\beta = \int_M \alpha\wedge\pi^*\beta$ for all $\beta$ (of appropriate degree) on the base, then we would have $\omega = \pi_*\alpha$. So, (1.15) uniquely characterizes $\pi_*\alpha$.

T show that $\pi_*(\alpha\wedge\pi^*\beta) = \pi_*\alpha\wedge\beta$, it suffices to show that $\pi_*\alpha \wedge \beta$ indeed satisfies the defining property (1.15) for $\alpha\wedge\pi^*\beta$. That is we need to show that, for any $\phi \in A^*(B)$.

$$
\int_B (\pi_*\alpha\wedge\beta)\wedge\phi= \int_M (\alpha\wedge\pi^*\beta)\wedge\pi^*\phi
$$

but this is obvious because of the defining property of $\pi_*\alpha$ we have,
$$
\int_B (\pi_*\alpha\wedge\beta)\wedge\phi = \int_B \pi_*\alpha\wedge(\beta\wedge\phi) = \int_M \alpha \wedge \pi^*(\beta\wedge\phi) = \int_M (\alpha \wedge \pi^*\beta)\wedge\pi^*\phi.
$$

Lastly, regarding your comment to @Ted about references, I would suggest Bott & Tu's book. They define integration along the fiber more explicitly using local charts (similar to what Ted was suggesting you to do). Then both (1.15) and (1.16) follow easily from that definition. FWIW, these formulas together are called the Projection Formulas (see Proposition 6.15 in Bott & Tu).

Title: Integration along fibers


问:
Proof is needed for a lower bound of the maximal eigen-value of a non-negative, irreducible, integer matrix
$A$ is a non-negative, integer, irreducible, $m$ by $m$ matrix. It is well known (Perron-Frobenius) that $A$ has a positive eigen value (denote it by $\lambda$) with a positive eigen vector ($x$). It is needed to prove that:

$x_{max}/x_{min} \le \lambda^{(m-1)}$

$x_{max}$ denotes the largest element of $x$, $x_{min}$ denotes the smallest
element of $x$.

I proved that when considering A as an adjacency matrix, where there
is an edge from state $u$ to state $v$, the following holds:

$x_v \le \lambda x_u$

(it was the hint in this question). I don't see how to proceed from
here.

答:
Well, I found the answer. It is easy to prove that $x_v \le \lambda x_u$ simply by writing down $Ax=\lambda x$. Now, since $A$ is irreducible, there must exist a power $k \le m$ such that state $u$ is connected to state $v$ by a path of length $k$. $A^k$ represents a graph with paths of length $k$ between the states of the $A$-graph. This matrix has eigenvalue $\lambda^k$ with eigenvector $x$. Thus, we must be able to find indices $u$ and $v$ such that $u$ corresponds to $x_{min}$ and $v$ corresponds to $x_{max}$ for some $k$, so:

$x_{max}/x_{min} \le \lambda^k \le \lambda ^{m-1}$


问:
Is $\{(z_1,z_2)\in \mathbb{C}^2: z_1^2+z_2^2 +1 \ne 0\}$ a domain of holomorphy?
I'm wondering whether $\{(z_1,z_2)\in \mathbb{C}^2: z_1^2+z_2^2 +1 \ne 0\}$ is a domain of holomorphy. It seems that I don't get what this definition really means, so any hints about ways of thinking/intuition would be highly appreciated.

答:
Call $\Omega$ your domain and define $E:=\Bbb C^2\setminus\Omega$.

Then you can prove that $E$ is thin in $\Bbb C^2$ thus every holomorphic mapping on $\Bbb C^2\setminus E=\Omega$ locally bounded (that is $\forall U\subset\subset\Bbb C^2$ such a function is bounded on $U\setminus E$) extends holomorphically across $E$ (hence to an entire function).
As pointed out by AG learner, the mapping $f(z_1,z_2)=1/(z_1^2+z_2^2+1)$ is not locally bounded (not even locally $L^2$, which is a weaker condition, enough to get the conclusion).
So we can't argue that $\Omega$ is a domain of holomorphy so far.

Eventually, the conclusion is given by definition of domain of holomorphy (see book by M. Range "Holomorphic Functions and Integral Representations in Several Complex Variables",II 2.1). The above defined function is holomorphic on the whole $\Omega$ and it is completely singular at every boundary point $p\in E$, hence $\Omega$ IS a domain of holomorphy.


问:
Canonical bundle and skyscraper sheaf
Let $Y$ be a smooth projective variety, $\omega_Y$ its canonical bundle and $k(y)$ the skyscraper sheaf. In a proof in Huybrechts' book, he uses the "restriction" map $r_{y_1,y_2} \colon \omega_Y^k \to \omega_Y^k(y_1) \oplus \omega_Y^k(y_2) \simeq k(y_1) \oplus k(y_2)$, where $y_1$ and $y_2$ are distinct closed points.

I don't understand how the map is defined and why the isomorphism $\omega_Y^k(y_1) \oplus \omega_Y^k(y_2) \simeq k(y_1) \oplus k(y_2)$ holds.

答:
For any affine scheme $X:=Spec(A)$ and any $A$-module $E$ there is a canonical map

$$ \rho_E(x): E \rightarrow E\otimes_A \kappa(x)$$

defined by $\rho_E(x)(e):= e\otimes 1$, where $1\in \kappa(x)$ is the multiplicative identity. Here $\kappa(x)$ is the residue field of the point $x$: If you let $\mathfrak{p}\subseteq A$ be the prime ideal corresponding to $x$ you define $\kappa(x):=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$. It follows $\rho_E(x)$ is a map of $A$-modules, hence you get an induced map of $\mathcal{O}_X$-modules

$$\rho_{\mathcal{E},x}: \mathcal{E} \rightarrow \mathcal{E}(x).$$

If $E:=A$ you get a map

$$\rho_{\mathcal{O}_X,x}: \mathcal{O}_X \rightarrow \underline{\kappa(x)}.$$

The sheaf $\underline{\kappa(x)}$ is the sky-scraper sheaf of the point $x$ (this is defined in Hartshornes book).

If $x,y \in X$ are two points there is a canonical map

$$\rho_{\mathcal{O}_X,(x,y)}: \mathcal{O}_X \rightarrow \underline{\kappa(x)\oplus \kappa(y)}$$

defined using the canonical map

$$ \rho_A(x,y): A \rightarrow \kappa(x)\oplus \kappa(y),$$

where $\rho_A(x,y)(a):=(\rho(x)(a), \rho(y)(a))\in \kappa(x)\oplus \kappa(y).$

Example. If $Y$ is a smooth complex affine algebraic variety of dimension $n$ and $x\in Y$ a (closed) point, it follows the map

$$\rho_{\mathcal{O}_Y,x}: \mathcal{O}_Y \rightarrow \underline{\kappa(x)}$$

evlauates a global section at the point $x$. If $Y:=Spec(A)$ where $A$ is a finitely generated algebra over the complex numbers, and $f\in A$ is a global section of the structure sheaf, it follows the canonical map

$$\rho_{A,x}: A \rightarrow \kappa(x)$$

is the evaluation: If $x\in Y \subseteq \mathbb{C}^n$ where $x:=(x_1,..,x_n)$
it follows $\rho_{A,x}(f):=f(x_1,..,x_n)$ is the value of the function $f$ at $x$. In algebraic geometry you sometimes meet non-closed points, hence the point $x$ may correspond to a non-maximal prime ideal $\mathfrak{p}$ in $A$, and then the "value" of $f$ at $x$ becomes more un-intuitive. It is an element in the residue field  $\kappa(x)$ of $x$, and this field is a transcendental extension of the base field.

When the base field $k$ is non-algebraically closed and $Y$ is a scheme of finite type over $k$ it follows from the HNSS the residue field $\kappa(x)$ at a closed point is a finite extension of the base field $k$. The value $s(x)$ of a global section lives in the residue field $\kappa(x)$ and this field changes with the point. If $k$ is the ring of integers and $x,y$ are two closed points in $Y$
it follows the residue fields $\kappa(x)$ and $\kappa(y)$ may have different characteristic.

If $Y$ is smooth and we define the canonical bundle as follows: $\omega_Y:=\wedge^n \Omega^1_Y$ (see Hartshorne, page 180), and since the rank of the cotangent bundle is $n$, it follows $\omega_Y$ is an invertible sheaf on $Y$. Hence it is locally trivial of rank one, hence you get something similar: There is a map

$$I0,\text{   } \rho_{\omega, x}: \omega_Y \rightarrow \underline{\kappa(x)}$$

Question: "I don't understand how the map is defined and why the isomorphism $ω_Y(y_1)\oplus ω_Y(y_2)\cong k(y_1)\oplus k(y_2)$ holds".

Answer: Since the rank of $\omega_Y$ is one, it follows the fiber $\omega_Y(x)$ is isomorphic to $\kappa(x)$ - the fiber is one-dimensional, hence there is in the case when $Y$ is smooth an isomorphism

$$ \omega_Y(x) \oplus \omega_Y(y) \cong \kappa(x) \oplus \kappa(y).$$

If $\omega^k_Y:=\omega^{\otimes k}_Y$ is the $k$'th tensor product of $\omega_Y$ with itself it follows $\omega^k_Y$ is still an invertible sheaf, and you get a similarly defined isomorphism

$$I1. \text{   } \omega^k_Y(x) \oplus \omega^k_Y(y) \cong \kappa(x)\oplus \kappa(y).$$

Note: The isomorphisms in $I0$ and $I1$ depend on the choice of a local trivialization of $\omega^k_Y$ around the points $x,y$.

Example. If $k:=\mathbb{R}$ is the real numbers and $A:=\mathbb{k}[x]$ is the ring of polynomials with real coefficients you may classify all maximal ideals. A maximal ideal $\mathfrak{m}$ in $A$ on the following form: If the residue field is the field of real numbers $k$ it follows $\mathfrak{m}_1=(x-r)$ with $r\in k$ a real number. If the residue field is the field $K:=\mathbb{C}$ of complex numbers, it follows $\mathfrak{m}_2$ is on the form
$\mathfrak{m}_2=(p_z(x))$ where $z:=a+ib \in K-k$ is a non real complex number
and

$$p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2.$$

In the case $1$ you get an evaluation map

$$\rho_1: A \rightarrow \kappa(\mathfrak{m_1})$$

defined by $\rho_1(f(x)):=f(r)$. In the case $2$ you get an evaluation map

$$\rho_2: A \rightarrow \kappa(\mathfrak{m}_2)$$

and if you want an explicit complex number you must choose an isomorphism $i:\kappa(\mathfrak{m}_2)\cong \mathbb{C}$. Here you have two choices: $i(x):=z$ or $i(x):=\overline{z}$. Once you have made a choice you get $\rho_2(f(x)):=f(i(x))\in \mathbb{C}$. Hence for a point that is not rational over $k$ your "value" and "evaluation map" depends on the choice of an isomorphism $i$. I believe this is written out in detail in Hartshornes book "Algebraic geometry".

When doing algebraic geometry you are seldom interested in the "value" of a function $f(x)$ or a section $s(x)$ in the fiber. What you are studying is where the function (or section) is zero, and this does not depend on the choice of an isomorphism $i$. Hence the fact that you have to choose an isomorphism $i$ above does not create problems.

Some people argue that to introduce non-maximal prime ideals is "less intuitive", but as you can observe above: If you work over a non-algebraically closed field, the elements in the structure sheaf do not give rise to functions in the classical sense.

Title: Canonical bundle and skyscraper sheaf


答:
$f$ has a strict ascneding saddle point if $n$ is even and a strict local minimum if $n$ is odd. You should know this for the case $n\le2$. To go from $n$ to $n+1$, you know by induction that $f'$ has a strict minimum or strict ascending saddle point at $c$, that is $f'$ is strictly positive for $x>c$, $x\approx c$ and either strictly positive or negative for $x<c$, $x\approx c$, from which you conclude that $f$ has a saddle point or minimum, respectively.

Another proof: Show that $g(x)=\frac{f(x)-f(c)}{(x-c)^2}$ is continuous at $c$, and look at $g'(c),\ldots , g^{(n-1)}(c)$. What can you conclude about $f$?


问:
How to solve equation with multiple variables with each of them having different "weight"?
I need to solve equation where the goal is to buy specific number of snacks for the lowest price.

There are 7 types of snack packages, each of them contains different number of snacks and has different price, here is the list:

A: 0.0051 $ / 4 snacks

B: 0.0102 $ / 4 snacks

C: 0.0204 $ / 8 snacks

D: 0.0408 $ / 17 snacks

E: 0.0816 $ / 35 snacks

F: 0.1632 $ / 58 snacks

G: 0.3264 $ / 58 snacks

Now I need to know, how to get the best "combo of packages" for lowest price when buying 10 snacks, 20 snacks..... up to 100 snacks. (When 10 is needed, buying some extra is not problem - such as buying 12 instead of 10.)

I tried bruteforcing (such as starting from maximum number of cheapest package and then combining with other packages), but I think there mast be some smarter way to solve this. :)

答:
You want to minimize the total cost, i.e.,

$$0.0051A+0.0102B+0.0204C+0.0408D+0.0816E+0.01632F+0.3264G$$

under the condition that

$$ 4A+4B+8C+17D+25E+58F+58G\geq T,$$

where $T$ is your target number of snacks. Given prices and quantities of snaks per package, the solution will be such that $G=0$ (because $F$ includes the same number of snacks and it's half the price) and $C>0$ implies that the solution is not unique (as $B$ results in the same price per snack).


问:
Nice description of the Heyting implication for clopen upsets
By the Priestley duality, we know that a lattice can be represented as the clopen upsets of its prime filters space $X$, namely via the map
$$ \eta: a \in L \longmapsto \lbrace x \in X \mid x \ni a \rbrace. $$

Now, Pultr and Sichler gave a characterisation of the Priestley spaces which were dual to a frame (the adherence of an open upset must be an open upset). In particular, every frame is a Heyting algebra with the implication defined as
$$ a \rightarrow b = \bigvee \lbrace c \mid a \wedge c \leq b \rbrace. $$

Now my question is: is there a nice description of $\eta(a\rightarrow b)$ in terms of $\eta(a)$ and $\eta(b)$ ?

答:
Yes there is:
$$
\eta(a \to b) = X - \downarrow(\eta(a) - \eta(b)),
$$
where $\downarrow A$ for $A \subseteq X$ means the downward closure of $A$.

Let us prove this. First note that a prime filter $F$ is in $X - \downarrow(\eta(a) - \eta(b))$ if and only if there is no prime filter $F' \supseteq F$ such that $a \in F'$ while $b \not \in F'$. Or put differently: for every prime filter $F' \supseteq F$ we have that $a \in F'$ implies $b \in F'$. We will show that this happens if and only if $a \to b \in F$.

If $a \to b \in F$ and $F' \supseteq F$ with $a \in F'$ then $a \wedge (a \to b) \in F'$. As $a \wedge (a \to b) = a \wedge b \leq b$, we conclude that $b \in F'$, as required.

For the other direction, we consider
$$
F' = \{ d : c \wedge a \leq d \text{ for some } c \in F \}.
$$
So either $F'$ is the entire lattice or $F'$ is the prime filter generated by $F \cup \{a\}$. Either way we have $b \in F'$. So $a \wedge c \leq b$ for some $c \in F$ and hence $c \leq a \to b$. Since $F$ is a filter we conclude that indeed $a \to b \in F$.


问:
How can I prove the following conditional Expectation?
A sequence of rv $X_{n}$ i.i.d for $n \in \mathbb{N}$ with $S_{0}=0$ and $$S_{n} = \sum_{k=1}^{n}X_{k}$$.

If $k \leq n$ then $$\mathbb{E}[S_{k}|S_{n}] = \frac{k}{n}S_{n}$$

I am thinking that:

$$\mathbb{E}[S_{k}|S_{n}] = \sum_{k=1}^{n} X_{k} \frac{P(X=k)|(P(n-k=n-k)) }{P(n=n)} = \sum_{k=1}^{n} X_{k} \frac{k \cdot 1}{n}= \frac{k}{n}S_{n}$$

答: By linearity, it suffices to show that $\mathsf{E}[X_i\mid S_n]=S_n/n$ for each $1\le i\le n$. However, the latter equality holds because $X_1,X_2,\ldots,X_n$ are exchangeable r.v.s. (See, e.g., this question for details.)


问:
maximize the values of given variables.
i have function A in two variables A(x,y) whose value is known.
i have been provided with a table having some values of A at a particular value of x and y that is  A(x1,y1) = some integer (for example)

A
x
y

First
value1
value 1

Second
value2
value2

(table is just to explain data format)

I need to find the value of x and y where the A has the maximum value and also the maximum value of A by using a minimum of data provided that is provided in the table.

So my question is  :

Do I need to formulate the equation for A.

What is the method to solve such a problem.
please mention all method if know them, so that I will find the best one as per the use case.

P.S: I can use computer for calculation if any method requires it.

Thank you!!!

答: I'm not sure I have understood correctly, but can't you just look at all values of the A-column in your data, find the one that is largest, and then read off the x,y values next to it?</p>

Title: maximize the values of given variables.


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

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

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

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

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

答:
Method 2: Two degrees of freedom

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$
$\newcommand{\opair}[2]{\left( #1,#2\right)}$
$\newcommand{\R}{\Bbb{R}}$

Method 1 gave an expression for $\pderiv{}{a_0}N(t)$; here is a similar lemma for $\pderiv{}{N_*}N(t)$:
$$\begin{align*}
  \boxed{\pderiv{}{N_*}N(t)}
    &\eqd \pderiv{}{N_* }\brs{\frac{N_*}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}}}
    && \text{by definition of $N(t)$}
  \\&= \frac{\brs{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}} - N_*\brs{\frac{1}{N_0}}e^{-a_0 t}}
            {\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}^2}
    && \text{by Quotient Rule}
  \\&= \frac{1 - e^{-a_0 t}}
            {\brp{1 + \brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}^2}
  \\&= \brs{\frac{1 - e^{-a_0 t}} {N_*^2}}
       \brs{\frac{N_*}                 {1 + \brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}}^2
  \\&\eqd \boxed{\brs{\frac{1 - e^{-a_0 t}}{N_*^2}} N^2(t)}
    && \text{by definition of $N(t)$}
\end{align*}
$$
Then$\ldots$
$$\begin{align*}
  \boxed{0}
    &= \frac{N_*^2}{2}\cdot0
  \\&=\frac{N_*^2}{2}
       \pderiv{}{N_*}\norm{e}^2
  \\&\eqd \frac{N_*^2}{2}
       \pderiv{}{N_*}\sum_{n=0}^{n=10}e^2(t_n)
    && \text{by definition of $\norm{\cdot}$}
  \\&\eqd \frac{N_*^2}{2}
       \pderiv{}{N_*}\sum_{n=0}^{n=10}\brs{N(t_n)-y_n}^2
    && \text{by definition of $e$}
  \\&= \frac{N_*^2}{2}
       \sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}\pderiv{}{N_h}N(t_n)
    && \text{by Chain Rule}
  \\&= N_*^2
       \sum_{n=0}^{n=10}  \brs{N(t_n)-y_n}
                          \brs{\frac{1 - e^{-a_0 t_n}}{N_h^2}} N^2(t_n)
    && \text{by lemma}
  \\&= \boxed{\sum_{n=0}^{n=10} N^2(t_n) \brs{N(t_n)-y_n} \brs{1 - e^{-a_0 t_n} }}
\end{align*}$$

So now we have two equations in two unknowns ($N_*$ and $a_0$): $\pderiv{}{N_h}\norm{e}^2=0$ and $\pderiv{}{a_0}\norm{e}^2=0$ (from Method 1).
You can use a multi-root solver to find the solution $\opair{N_*}{a_0}$ to the equation pair
$\opair{\pderiv{}{N_h}\norm{e}^2=0}{\pderiv{}{a_0}\norm{e}^2=0}$.
The result using multiroot
from the R package rootSolve
is $\opair{N_*}{a_0}=\opair{255.8436595023}{0.6539203544}$.
This solution is "better" than Method 1 solution in the sense that it results in a lower cost:
$$\begin{array}{|c|rcl|}
  \text{Method} & cost(N_0,N_*,a_0)
\\\hline
\\1             & cost(18,252,0.6631183) &=& 31.32307
\\2             & cost(18,255.8436595023,0.6539203544) &=& 0.7140973
\end{array}$$

Some R code supporting Reproducible Research:

```<code> require(stats);
 require(R.utils);
 require(rootSolve);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 tdata = c(0:10)
 ydata = c(18,33,56,90,130,170,203,225,239,247,251)
 t     = seq( from=min(tdata), to=max(tdata), length=1000 )

#---------------------------------------
# Estimate Function N(t)
#---------------------------------------
 N0 = ydata[1]
 N = function(t,N0,Nh,a0)
 {
   result = Nh / ( 1 + (Nh/N0-1)*exp(-a0*t) )
 }

#---------------------------------------
# Cost Function
#---------------------------------------
 cost = function(N0,Nh,a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) - ydata[i] )^2
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to a0 of Cost Function
#---------------------------------------
 Pcosta0 = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) )^2 *
                   ( N(tdata[i],N0,Nh,a0) - ydata[i] ) *
                   ( tdata[i] * exp(-a0*tdata[i]) )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to Nh of Cost Function
#---------------------------------------
 PcostNh = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( 1 - exp(-a0*tdata[i]) ) *
                   ( N(tdata[i],N0, Nh, a0) )^2 *
                   ( N(tdata[i],N0, Nh, a0)-ydata[i] )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative vector of cost
#---------------------------------------
Pcost = function(x)
{
   N0 = 18
   Nh = x[1]
   a0 = x[2]
   F1 = Pcosta0( N0, Nh, a0 );
   F2 = PcostNh( N0, Nh, a0 );
   result = c(F1, F2);
}

#---------------------------------------
# Calculate roots
#---------------------------------------
 Roots = multiroot( f=Pcost, start=c(ydata[11-1],0.4) );
 Nh = Roots<span class="math-container">$root[1]
 a0 = Roots$</span>root[2]

#---------------------------------------
# Display
#---------------------------------------
 printf("(N0, Nh, a0) = (%.2f, %.10f, %.10f) with estim.precis=%.2e\n", N0, Nh, a0, Roots$estim.precis )
 colors = c( "red" , "blue" );
 traces = c( "N(t)", "data" );
 plot ( t , N(t, N0, Nh, a0), col=colors[1], lwd=2, type='l', xlab="t", ylab="y", ylim=c(0,max(ydata)+10) )
 lines( tdata, ydata        , col=colors[2], lwd=5, type='p' )
 legend("topleft", legend=traces, col=colors, lwd=3, lty=1:1)
 grid()
</code>```

Body:
Method 2: Two degrees of freedom

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$
$\newcommand{\opair}[2]{\left( #1,#2\right)}$
$\newcommand{\R}{\Bbb{R}}$

Method 1 gave an expression for $\pderiv{}{a_0}N(t)$; here is a similar lemma for $\pderiv{}{N_*}N(t)$:
$$\begin{align*}
  \boxed{\pderiv{}{N_*}N(t)}
    &\eqd \pderiv{}{N_* }\brs{\frac{N_*}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}}}
    && \text{by definition of $N(t)$}
  \\&= \frac{\brs{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}} - N_*\brs{\frac{1}{N_0}}e^{-a_0 t}}
            {\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}^2}
    && \text{by Quotient Rule}
  \\&= \frac{1 - e^{-a_0 t}}
            {\brp{1 + \brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}^2}
  \\&= \brs{\frac{1 - e^{-a_0 t}} {N_*^2}}
       \brs{\frac{N_*}                 {1 + \brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}}^2
  \\&\eqd \boxed{\brs{\frac{1 - e^{-a_0 t}}{N_*^2}} N^2(t)}
    && \text{by definition of $N(t)$}
\end{align*}
$$
Then$\ldots$
$$\begin{align*}
  \boxed{0}
    &= \frac{N_*^2}{2}\cdot0
  \\&=\frac{N_*^2}{2}
       \pderiv{}{N_*}\norm{e}^2
  \\&\eqd \frac{N_*^2}{2}
       \pderiv{}{N_*}\sum_{n=0}^{n=10}e^2(t_n)
    && \text{by definition of $\norm{\cdot}$}
  \\&\eqd \frac{N_*^2}{2}
       \pderiv{}{N_*}\sum_{n=0}^{n=10}\brs{N(t_n)-y_n}^2
    && \text{by definition of $e$}
  \\&= \frac{N_*^2}{2}
       \sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}\pderiv{}{N_h}N(t_n)
    && \text{by Chain Rule}
  \\&= N_*^2
       \sum_{n=0}^{n=10}  \brs{N(t_n)-y_n}
                          \brs{\frac{1 - e^{-a_0 t_n}}{N_h^2}} N^2(t_n)
    && \text{by lemma}
  \\&= \boxed{\sum_{n=0}^{n=10} N^2(t_n) \brs{N(t_n)-y_n} \brs{1 - e^{-a_0 t_n} }}
\end{align*}$$

So now we have two equations in two unknowns ($N_*$ and $a_0$): $\pderiv{}{N_h}\norm{e}^2=0$ and $\pderiv{}{a_0}\norm{e}^2=0$ (from Method 1).
You can use a multi-root solver to find the solution $\opair{N_*}{a_0}$ to the equation pair
$\opair{\pderiv{}{N_h}\norm{e}^2=0}{\pderiv{}{a_0}\norm{e}^2=0}$.
The result using multiroot
from the R package rootSolve
is $\opair{N_*}{a_0}=\opair{255.8436595023}{0.6539203544}$.
This solution is "better" than Method 1 solution in the sense that it results in a lower cost:
$$\begin{array}{|c|rcl|}
  \text{Method} & cost(N_0,N_*,a_0)
\\\hline
\\1             & cost(18,252,0.6631183) &=& 31.32307
\\2             & cost(18,255.8436595023,0.6539203544) &=& 0.7140973
\end{array}$$

Some R code supporting Reproducible Research:

 require(stats);
 require(R.utils);
 require(rootSolve);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 tdata = c(0:10)
 ydata = c(18,33,56,90,130,170,203,225,239,247,251)
 t     = seq( from=min(tdata), to=max(tdata), length=1000 )

#---------------------------------------
# Estimate Function N(t)
#---------------------------------------
 N0 = ydata[1]
 N = function(t,N0,Nh,a0)
 {
   result = Nh / ( 1 + (Nh/N0-1)*exp(-a0*t) )
 }

#---------------------------------------
# Cost Function
#---------------------------------------
 cost = function(N0,Nh,a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) - ydata[i] )^2
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to a0 of Cost Function
#---------------------------------------
 Pcosta0 = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) )^2 *
                   ( N(tdata[i],N0,Nh,a0) - ydata[i] ) *
                   ( tdata[i] * exp(-a0*tdata[i]) )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to Nh of Cost Function
#---------------------------------------
 PcostNh = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( 1 - exp(-a0*tdata[i]) ) *
                   ( N(tdata[i],N0, Nh, a0) )^2 *
                   ( N(tdata[i],N0, Nh, a0)-ydata[i] )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative vector of cost
#---------------------------------------
Pcost = function(x)
{
   N0 = 18
   Nh = x[1]
   a0 = x[2]
   F1 = Pcosta0( N0, Nh, a0 );
   F2 = PcostNh( N0, Nh, a0 );
   result = c(F1, F2);
}

#---------------------------------------
# Calculate roots
#---------------------------------------
 Roots = multiroot( f=Pcost, start=c(ydata[11-1],0.4) );
 Nh = Roots$root[1]
 a0 = Roots$root[2]

#---------------------------------------
# Display
#---------------------------------------
 printf("(N0, Nh, a0) = (%.2f, %.10f, %.10f) with estim.precis=%.2e\n", N0, Nh, a0, Roots$estim.precis )
 colors = c( "red" , "blue" );
 traces = c( "N(t)", "data" );
 plot ( t , N(t, N0, Nh, a0), col=colors[1], lwd=2, type='l', xlab="t", ylab="y", ylim=c(0,max(ydata)+10) )
 lines( tdata, ydata        , col=colors[2], lwd=5, type='p' )
 legend("topleft", legend=traces, col=colors, lwd=3, lty=1:1)
 grid()


答:
Method 3: Three degrees of freedom

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$
$\newcommand{\opair}[2]{\left( #1,#2\right)}$
$\newcommand{\otriple}[3]{\left( #1,#2,#3\right)}$
$\newcommand{\R}{\Bbb{R}}$
$\newcommand{\esth}{N}$

Method 1 optimized with respect to $a_0$ and
Method 2 with respect to $\opair{N_*}{a_0}$.
Both methods assume $N_0\eqd y_n[0] \eqd 18$.
However, often in data modeling, it is assumed that
measured data is the actual true value being measured plus some
measurement noise.
As such, it is highly unlikely that the true height of the plant at time $t=0$
is exactly $y_n[0]=18.000\ldots$, but rather $18+n(0)$, where $n(0)$ is
a random noise variable at time $t=0$.

Method 1 gave a lemma for $\pderiv{}{a_0}N(t)$ and
Method 2 for $\pderiv{}{N_*}N(t)$;
here is a similar lemma for $\pderiv{}{N_0}N(t)$:
$$\begin{align*}
      \boxed{\pderiv{}{N_0}\esth(t)}
        &\eqd \pderiv{}{N_0 }\brs{\frac{N_*}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}}}
        && \text{by definition of $\esth(t)$}
      \\&= \frac{ - N_*\brp{\frac{-1}{N_0^2}}e^{-a_0 t}}
                {\brs{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}}^2}
        && \text{by Quotient Rule}
      \\&= \brs{ \frac{e^{-a_0 t}}{N_*N_0^2}}
           \brs{\frac{N_*}{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}}^2
      \\&= \brs{ \frac{e^{-a_0 t}}{N_*N_0^2}}
           \esth^2(t)
        && \text{by definition of $\esth(t)$}
    \end{align*}$$

$$\begin{align*}
    \boxed{0}
      &= \frac{N_*N_0^2}{2}\cdot0
    \\&=\frac{N_*N_0^2}{2}
         \pderiv{}{N_0}\norm{e}^2
    \\&\eqd \frac{N_*N_0^2}{2}
         \pderiv{}{N_0}\sum_{n=0}^{n=10}e^2(t_n)
      && \text{by definition of $\norm{\cdot}$}
    \\&\eqd \frac{N_*N_0^2}{2}
         \pderiv{}{N_0}\sum_{n=0}^{n=10}\brs{\esth(t_n)-y_n}^2
      && \text{by definition of $e$}
    \\&= \frac{N_*N_0^2}{2}
         \sum_{n=0}^{n=10}2\brs{\esth(t_n)-y_n} \pderiv{}{N_0}\esth(t_n)
      && \text{by Chain Rule}
    \\&= \frac{N_*N_0^2}{2}
         \sum_{n=0}^{n=10}2\brs{\esth(t_n)-y_n} \brs{ \frac{e^{-a_0 t_n}}{N_*N_0^2}}\esth^2(t_n)
      && \text{by lemma}
    \\&= \boxed{\sum_{n=0}^{n=10} \esth^2(t_n)\brs{\esth(t_n)-y_n} e^{-a_0 t_n} }
    \end{align*}$$

So now we have a three equation triple
$\otriple{\pderiv{}{N_h}\norm{e}^2=0}{\pderiv{}{N_0}\norm{e}^2=0}{\pderiv{}{a_0}\norm{e}^2=0}$
in a three variable triple $\otriple{N_*}{N_0}{a_0}$.
The result using multiroot
from the R package rootSolve is
$\otriple{N_0}{N_*}{a_0}=\otriple{18.1994673377}{256.0554740324}{0.6508899900}$.
This solution is "better" than both Methods 1 and 2 in the sense that it results in a lower cost:
$$\begin{array}{|c|rcl|}
  \text{Method} & cost(N_0,N_*,a_0)
\\\hline
\\1             & cost(18,252,0.6631183) &=& 31.32307
\\2             & cost(18,255.8436595023,0.6539203544) &=& 0.7140973
\\3             & cost(18.1994673377, 256.0554740324, 0.6508899900) &=& 0.4533383
\end{array}$$

Some R code supporting Reproducible Research:

```<code>#---------------------------------------
# packages
#---------------------------------------
#install.packages("stats");
#install.packages("R.utils");
#install.packages("rootSolve");
 require(stats);
 require(R.utils);
 require(rootSolve);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 tdata = c(0:10)
 ydata = c(18,33,56,90,130,170,203,225,239,247,251)
 t     = seq( from=min(tdata), to=max(tdata), length=1000 )

#---------------------------------------
# Estimate Function N(t)
#---------------------------------------
 N0 = ydata[1]
 N = function(t,N0,Nh,a0)
 {
   result = Nh / ( 1 + (Nh/N0-1)*exp(-a0*t) )
 }

#---------------------------------------
# Cost Function
#---------------------------------------
 cost = function(N0,Nh,a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) - ydata[i] )^2
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to a0 of Cost Function
#---------------------------------------
 Pcosta0 = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) )^2 *
                   ( N(tdata[i],N0,Nh,a0) - ydata[i] ) *
                   ( tdata[i] * exp(-a0*tdata[i]) )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to Nh of Cost Function
#---------------------------------------
 PcostNh = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( 1 - exp(-a0*tdata[i]) ) *
                   ( N(tdata[i],N0, Nh, a0) )^2 *
                   ( N(tdata[i],N0, Nh, a0) - ydata[i] )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to N0 of Cost Function
#---------------------------------------
 PcostN0 = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( exp(-a0*tdata[i]) ) *
                   ( N(tdata[i],N0, Nh, a0) )^2 *
                   ( N(tdata[i],N0, Nh, a0) - ydata[i] )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative vector of cost
#---------------------------------------
Pcost = function(x)
{
   N0 = x[1]
   Nh = x[2]
   a0 = x[3]
   F1 = Pcosta0( N0, Nh, a0 );
   F2 = PcostNh( N0, Nh, a0 );
   F3 = PcostN0( N0, Nh, a0 );
   result = c(F1, F2, F3);
}

#---------------------------------------
# Calculate roots
#---------------------------------------
 Roots = multiroot( f=Pcost, start=c(ydata[1], ydata[11], 0.6) );
 N0 = Roots<span class="math-container">$root[1]
 Nh = Roots$</span>root[2]
 a0 = Roots$root[3]

#---------------------------------------
# Display
#---------------------------------------
 printf("(N0, Nh, a0) = (%.10f, %.10f, %.10f) with estim.precis=%.2e\n", N0, Nh, a0, Roots$estim.precis )
 colors = c( "red" , "blue" );
 traces = c( "N(t)", "data" );
 plot ( t , N(t, N0, Nh, a0), col=colors[1], lwd=2, type='l', xlab="t", ylab="y", ylim=c(0,max(ydata)+10) )
 lines( tdata, ydata        , col=colors[2], lwd=5, type='p' )
 legend("topleft", legend=traces, col=colors, lwd=3, lty=1:1)
 grid()
</code>```

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

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

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

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

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

Body:
Method 3: Three degrees of freedom

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$
$\newcommand{\opair}[2]{\left( #1,#2\right)}$
$\newcommand{\otriple}[3]{\left( #1,#2,#3\right)}$
$\newcommand{\R}{\Bbb{R}}$
$\newcommand{\esth}{N}$

Method 1 optimized with respect to $a_0$ and
Method 2 with respect to $\opair{N_*}{a_0}$.
Both methods assume $N_0\eqd y_n[0] \eqd 18$.
However, often in data modeling, it is assumed that
measured data is the actual true value being measured plus some
measurement noise.
As such, it is highly unlikely that the true height of the plant at time $t=0$
is exactly $y_n[0]=18.000\ldots$, but rather $18+n(0)$, where $n(0)$ is
a random noise variable at time $t=0$.

Method 1 gave a lemma for $\pderiv{}{a_0}N(t)$ and
Method 2 for $\pderiv{}{N_*}N(t)$;
here is a similar lemma for $\pderiv{}{N_0}N(t)$:
$$\begin{align*}
      \boxed{\pderiv{}{N_0}\esth(t)}
        &\eqd \pderiv{}{N_0 }\brs{\frac{N_*}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}}}
        && \text{by definition of $\esth(t)$}
      \\&= \frac{ - N_*\brp{\frac{-1}{N_0^2}}e^{-a_0 t}}
                {\brs{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0 t}}^2}
        && \text{by Quotient Rule}
      \\&= \brs{ \frac{e^{-a_0 t}}{N_*N_0^2}}
           \brs{\frac{N_*}{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0 t}}}^2
      \\&= \brs{ \frac{e^{-a_0 t}}{N_*N_0^2}}
           \esth^2(t)
        && \text{by definition of $\esth(t)$}
    \end{align*}$$

$$\begin{align*}
    \boxed{0}
      &= \frac{N_*N_0^2}{2}\cdot0
    \\&=\frac{N_*N_0^2}{2}
         \pderiv{}{N_0}\norm{e}^2
    \\&\eqd \frac{N_*N_0^2}{2}
         \pderiv{}{N_0}\sum_{n=0}^{n=10}e^2(t_n)
      && \text{by definition of $\norm{\cdot}$}
    \\&\eqd \frac{N_*N_0^2}{2}
         \pderiv{}{N_0}\sum_{n=0}^{n=10}\brs{\esth(t_n)-y_n}^2
      && \text{by definition of $e$}
    \\&= \frac{N_*N_0^2}{2}
         \sum_{n=0}^{n=10}2\brs{\esth(t_n)-y_n} \pderiv{}{N_0}\esth(t_n)
      && \text{by Chain Rule}
    \\&= \frac{N_*N_0^2}{2}
         \sum_{n=0}^{n=10}2\brs{\esth(t_n)-y_n} \brs{ \frac{e^{-a_0 t_n}}{N_*N_0^2}}\esth^2(t_n)
      && \text{by lemma}
    \\&= \boxed{\sum_{n=0}^{n=10} \esth^2(t_n)\brs{\esth(t_n)-y_n} e^{-a_0 t_n} }
    \end{align*}$$

So now we have a three equation triple
$\otriple{\pderiv{}{N_h}\norm{e}^2=0}{\pderiv{}{N_0}\norm{e}^2=0}{\pderiv{}{a_0}\norm{e}^2=0}$
in a three variable triple $\otriple{N_*}{N_0}{a_0}$.
The result using multiroot
from the R package rootSolve is
$\otriple{N_0}{N_*}{a_0}=\otriple{18.1994673377}{256.0554740324}{0.6508899900}$.
This solution is "better" than both Methods 1 and 2 in the sense that it results in a lower cost:
$$\begin{array}{|c|rcl|}
  \text{Method} & cost(N_0,N_*,a_0)
\\\hline
\\1             & cost(18,252,0.6631183) &=& 31.32307
\\2             & cost(18,255.8436595023,0.6539203544) &=& 0.7140973
\\3             & cost(18.1994673377, 256.0554740324, 0.6508899900) &=& 0.4533383
\end{array}$$

Some R code supporting Reproducible Research:

#---------------------------------------
# packages
#---------------------------------------
#install.packages("stats");
#install.packages("R.utils");
#install.packages("rootSolve");
 require(stats);
 require(R.utils);
 require(rootSolve);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 tdata = c(0:10)
 ydata = c(18,33,56,90,130,170,203,225,239,247,251)
 t     = seq( from=min(tdata), to=max(tdata), length=1000 )

#---------------------------------------
# Estimate Function N(t)
#---------------------------------------
 N0 = ydata[1]
 N = function(t,N0,Nh,a0)
 {
   result = Nh / ( 1 + (Nh/N0-1)*exp(-a0*t) )
 }

#---------------------------------------
# Cost Function
#---------------------------------------
 cost = function(N0,Nh,a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) - ydata[i] )^2
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to a0 of Cost Function
#---------------------------------------
 Pcosta0 = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( N(tdata[i],N0,Nh,a0) )^2 *
                   ( N(tdata[i],N0,Nh,a0) - ydata[i] ) *
                   ( tdata[i] * exp(-a0*tdata[i]) )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to Nh of Cost Function
#---------------------------------------
 PcostNh = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( 1 - exp(-a0*tdata[i]) ) *
                   ( N(tdata[i],N0, Nh, a0) )^2 *
                   ( N(tdata[i],N0, Nh, a0) - ydata[i] )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative with respect to N0 of Cost Function
#---------------------------------------
 PcostN0 = function(N0, Nh, a0)
 {
   summ = 0;
   for (i in c(1:length(tdata)))
   {
     summ = summ + ( exp(-a0*tdata[i]) ) *
                   ( N(tdata[i],N0, Nh, a0) )^2 *
                   ( N(tdata[i],N0, Nh, a0) - ydata[i] )
   }
   result = summ
 }

#---------------------------------------
# Partial derivative vector of cost
#---------------------------------------
Pcost = function(x)
{
   N0 = x[1]
   Nh = x[2]
   a0 = x[3]
   F1 = Pcosta0( N0, Nh, a0 );
   F2 = PcostNh( N0, Nh, a0 );
   F3 = PcostN0( N0, Nh, a0 );
   result = c(F1, F2, F3);
}

#---------------------------------------
# Calculate roots
#---------------------------------------
 Roots = multiroot( f=Pcost, start=c(ydata[1], ydata[11], 0.6) );
 N0 = Roots$root[1]
 Nh = Roots$root[2]
 a0 = Roots$root[3]

#---------------------------------------
# Display
#---------------------------------------
 printf("(N0, Nh, a0) = (%.10f, %.10f, %.10f) with estim.precis=%.2e\n", N0, Nh, a0, Roots$estim.precis )
 colors = c( "red" , "blue" );
 traces = c( "N(t)", "data" );
 plot ( t , N(t, N0, Nh, a0), col=colors[1], lwd=2, type='l', xlab="t", ylab="y", ylim=c(0,max(ydata)+10) )
 lines( tdata, ydata        , col=colors[2], lwd=5, type='p' )
 legend("topleft", legend=traces, col=colors, lwd=3, lty=1:1)
 grid()


问:
Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field
I'm reading Galois Theory by Steven H. Weintraub (second edition), and finding that I'm at least somewhat short on the prerequisites. However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof?

Lemma 2.2.3. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field.

Proof. We need to show that any nonzero $r \in R$ has an inverse. Consider $\{1, r, r^2, \cdots\}$. This is an infinite set of elements of $R$, and by hypothesis $R$ is finite dimensional as an $F$-vector space, so this set is linearly dependent. Hence $\sum_{i=0}^n{c_i r^i} = 0$ for some $n$ and some $c_i \in F$ not all zero.

It then goes on to show, given the above, that we can derive an inverse for $r$.

However, if I consider examples like $r = 2 \in Q[\sqrt{2}]$, $r = \sqrt{2} \in Q[\sqrt{2}]$ or $r = 2 \in Q[X]/{<X^2>}$, the set $\{1, r, r^2, ...\}$ doesn't look linearly dependent to me.

I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something?

[Edit] Well yes, I am. Somehow I had managed to discount the possibility of any $c_i$ being negative, despite repeatedly looking at each fragment of the quoted text in an attempt to find what I might be misunderstanding.

答:
This proof is "similar" to some of the above proofs, but not identical:

Lemma 2.2.3. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field.

Proof: Let $0\neq x\in R$ and let $\phi: F[t]\rightarrow k\{1,x,x^2,..,\}$
be the map defined by $\phi(t):=x$. Since $R$ is a ring it follows $\phi$ is a ring homomorphism, and since $dim_F(R) < \infty$ it follows there is an integer $n$ with $x^{n+1}\in F\{1,x,x^2,..,x^n\}$. Hence the map $\phi$ is a surjective map of $F$-algebras with $ker(\phi)=(p(t))$, where $p(t)$ is a non-zero polynomial. Since $R$ is an integral domain and $Im(\phi) \subseteq R$ is a sub ring it follows $F[t]/(p(t))$ is an integral domain, hence $p(t)$ is an irreducible polynomial. Since $p(t) \neq 0$ it follows $(p(t)) \subseteq F[t]$ is a maximal ideal, hence $Im(\phi)\subseteq R$ is a field containing $x$. It follows there is an element $0\neq y\in Im(\phi)$ with $xy=yx=1$ and it follows $R$ is a field. QED

Question: "However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof?
........
It then goes on to show, given the above, that we can derive an inverse for r.
.....
I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something?"

Answer: The claim of the Lemma is correct but the proof you give above is incomplete. Above I give a(nother) compete proof of the Lemma.

PS: I do not have a copy of the book you mention, hence I cannot comment on the proof given in the book.


问:
Let $f(x)=(x-a)g(x)$. Calculate $f '(a)= g(a)$.
Assuming that $g$ is continuos. What is $f(a)$ in this case? Can you give an explanation for this proof? How can it be possible to ask about a function of another variable of an existing function? Can you give other examples? This made me question my understanding of functions.</p>

答:
Another solution might be that if $f(a)=0$, then forming the difference quotient makes $$\frac{f(a+\Delta a)-f(a)}{\Delta a}=\frac{(((a+\Delta a)-a)g(x))-0}{\Delta a}=\frac{\Delta a \ g(x)}{\Delta a}=g(x).$$
(Hello to anybody running across this from OCW.)