问:
The value of Euler function of $\phi(p^k)$ when $p$ is prime and $k$ is a positive integer?
Prove that the value of Euler function of $\phi(p^k)$ when $p$ is prime and $k$ is a positive integer is,

$$\phi(p^k)=p^k-p^{k-1}$$

I saw similar posts for the same questions, but I am still confused about the following.

Question: the divisors of $p^k$ are,

$$1, p, p^2, p^3, \cdots, p^k$$

Those divisors should not be relatively prime to $p^k$ as the $gcd(p^k, divisor_i) \ne 1$. We can see that we have $p^k$ such divisors that are not relatively prime to $p^k$ as I understand it.

Also, we have multiples of $p$ up to $p^k$ cannot be relatively prime to $p^k$ for the same reason as $gcd(p^k, multipleP_i) \ne 1$,

$$p, 2p, 3p, \cdots, p^{k-1}p$$

Clearly we have $p^{k-1}$ of such multiples, so we should have the total number of numbers that are not relatively prime to $p^k$ as (based on how I understand it),

$$\phi(p^k)=p^k-p^{k-1}$$

Question: How do we know that those only multiples of $p$ and divisors of $p^k$ are the only possibilities of the numbers not to be relatively prime to $p^k$? How do we know that there are not other numbers in between that could also be not relatively prime to $p^k$?

答:
Suppose $p$ is prime and $x,k\in \Bbb N.$

(1). If $\gcd(x,p^n)>1$ then for some prime $q$ we have $q\,|\,\gcd(x,p^n)\,|\,p^k,$ so $q|p^k.$ But the only prime divisor of $p^k$ is $p,$ by the uniqueness of prime decomposition, so $q=p,$ so $p=q\,|\,\gcd(x,p^n)\,|\,x,$ so $p|x,$ so $$\exists y\in \Bbb N\,(x=py).$$

(2).  If $\gcd(x,p^n)=1$ then $p\not |\,x\;$ ( otherwise $p>1$ is a common divisor of $x$ and of $p^k$) so $$\neg \exists y\in \Bbb N\,(x=py). $$

Therefore $$\{x\in \Bbb N: x\le p^k\land \gcd (x,p)>1\}=\{x\in \Bbb N: x\le p^k\land p|x\}=$$ $$=\{py: y\in \Bbb N\land py\le p^k\}.$$ This last set (above) has the same number of members as $\{y\in\Bbb N: y\le p^{k-1}\}$ does, which is $p^{k-1}$ members. Therefore $\phi(p^k)=p^k-p^{k-1}.$


问:
Covering map of the Cylinder

Question: Is there a covering map $\mathbb{R}^2 \to S^1 \times \mathbb{R}$ ?

I picture the cylinder $S^1 \times \mathbb{R}$ as a roll of gift wrap that one "rolls the plane" onto (just a thought). We proved that $\mathbb{R} \to S^1, x \mapsto \mathrm{e}^{2\pi\mathrm{i}x}$ is a covering map of the unit circle. But I can't figure out whether the title question is true or not...

答:
What about

$$\varphi : (x,y) \mapsto (e^{2\pi i x},y)?$$

ContentLicense: CC BY-SA 4.0

Title: Covering map of the Cylinder


问:
Need Help to Understand Simple Vector Subspace Proof of Teacher
Given

$X \in M_{nn}(\mathbb{F})$ is a static matrix

$U = \{A \in M_{nn}(\mathbb{F})\ | AX = XA\}$

Proof that $U$ is a subspace of $M_{nn}(\mathbb{F})$.

Teachers proof

... uses three steps to proof that.

a) Show that zero element of $M_{nn}(\mathbb{F})$ is in $U$:

(irrelevant)

b) Show that $\forall u1, u2 \in U: u1+u2 \in U $

(irrelevant)

c) Show that for $a \in \mathbb{F}, A \in U: aA \in U$

Let $A \in U$ and $a \in \mathbb{F}$. Then $(aA)X = a(AX) = a(XA) = X(aA)$. So, $aA \in U$.

What?

Looking at this particular step: $a(AX) = a(XA)$

How can she just state that AX = XA? That is not generally true except for n=1. To me (and maybe I'm wrong -that is why I'm asking here for help), nothing is proven. Maybe the fact that the $A$ that is scaled by $a$ is already in U makes that implicit, but shouldn't that be incorporated into the proof? Sorry in case this is a dumb question!

答: Your teacher stated that $A\in U$, so by definition $AX=XA$. That also $aA\in U$ is what you want to show!


答:
Show that for $a \in \mathbb{F}, A \in U: aA \in U$

The string of symbols above does not quite read as a valid mathematical satatement.

What you professor was saying, be it as it may in the amount of predicate logic you'll want to use, is that for all $a\in\Bbb F$ and for all $B\in U$, it holds that $aB\in U$.

With this premise, $B\in U$ and therefore $B$ satisfies the definitory property of $U$, which is that $BX=XB$.

Title: Need Help to Understand Simple Vector Subspace Proof of Teacher

Body:
Given

$X \in M_{nn}(\mathbb{F})$ is a static matrix

$U = \{A \in M_{nn}(\mathbb{F})\ | AX = XA\}$

Proof that $U$ is a subspace of $M_{nn}(\mathbb{F})$.

Teachers proof

... uses three steps to proof that.

a) Show that zero element of $M_{nn}(\mathbb{F})$ is in $U$:

(irrelevant)

b) Show that $\forall u1, u2 \in U: u1+u2 \in U $

(irrelevant)

c) Show that for $a \in \mathbb{F}, A \in U: aA \in U$

Let $A \in U$ and $a \in \mathbb{F}$. Then $(aA)X = a(AX) = a(XA) = X(aA)$. So, $aA \in U$.

What?

Looking at this particular step: $a(AX) = a(XA)$

How can she just state that AX = XA? That is not generally true except for n=1. To me (and maybe I'm wrong -that is why I'm asking here for help), nothing is proven. Maybe the fact that the $A$ that is scaled by $a$ is already in U makes that implicit, but shouldn't that be incorporated into the proof? Sorry in case this is a dumb question!


问:
Fourier transform of $\text{sinc}^3 {\pi t}$
$$f(t)=\frac{\sin^3(\pi t)}{(\pi t)^3}$$
I want to calculate the Fourier transform.
I can't calculate this integral:
$$\int_0^\infty\frac{\sin^3(\pi t)}{(\pi t)^3}\cos(ut)\,\mathrm{d}t$$</p>

答:
It is tempting to do this integral using the convolution theorem.  Nevertheless, I propose to perform a direct computation using Cauchy's theorem, i.e., extension into the complex plane.  I will then verify the solution using the convolution theorem.

DIRECT EVALUATION

Rewrite the integral as

$$\frac12 \int_{-\infty}^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} e^{i u t}$$

Now use $\sin{\pi t}  = (e^{i \pi t}-e^{-i \pi t})/(2 i)$ to obtain, as the value of the original integral:

$$\frac{i}{16 \pi^3} \int_{-\infty}^{\infty} dt \, \frac{e^{i (u+3 \pi)t}}{t^3}-  i \frac{3}{16 \pi^3}\int_{-\infty}^{\infty} dt \,  \frac{e^{i (u+ \pi)t}}{t^3} +\\ i \frac{3}{16 \pi^3}\int_{-\infty}^{\infty} dt \, \frac{e^{i (u- \pi)t}}{t^3} - \frac{i}{16 \pi^3} \int_{-\infty}^{\infty} dt \, \frac{e^{i (u-3 \pi)t}}{t^3} $$

Now evaluate these integrals by considering contour integrals over the complex variable $z$ in the complex plane:

$$\frac{i}{16 \pi^3} \oint_C dz \frac{e^{i (u+3 \pi)z}}{z^3} -  i \frac{3}{16 \pi^3}\oint_C dz \frac{e^{i (u+ \pi)z}}{z^3} +\\ i \frac{3}{16 \pi^3}\oint_C dz \frac{e^{i (u- \pi)z}}{z^3} - \frac{i}{16 \pi^3} \oint_C dz \frac{e^{i (u-3 \pi)z}}{z^3} $$

where $C$ is a semicircular contour of radius $R$ with a small semicircular indentation of radius $\epsilon$ about the origin.  Whether $C$ extends into the upper or lower half plane depends on the sign of $u+3 \pi$ or $u+\pi$.  For example, consider the first integral, which may be rewritten as

$$\oint_C dz \frac{e^{i (u+3 \pi)z}}{z^3} = \int_{-R}^{-\epsilon} dx \frac{e^{i (u+3 \pi) x}}{x^3} + i \epsilon \int_{\pi}^0 d\phi\, e^{i \phi} \frac{e^{i (u+3 \pi) \epsilon e^{i \phi}}}{\epsilon^3 e^{i 3 \phi}} + \\ \int_{\epsilon}^R dx \frac{e^{i (u+3 \pi) x}}{x^3} + i R \int_0^{\pi} d\phi\, e^{i \phi} \frac{e^{i (u+3 \pi) R e^{i \phi}}}{R^3 e^{i 3 \phi}}$$

when $C$ is in the upper half plane.  In this case, we consider the conditions under which the fourth integral vanishes as $R \to \infty$.  In fact, the integral is bounded by

$$\frac{1}{R^2} \int_0^{\pi} d\phi \,e^{-(u+3 \pi) R \sin{\phi}} \le\frac{2}{R^2} \int_0^{\pi/2} d\phi \, e^{-2 (u+3 \pi) R \phi/\pi} \le \frac{\pi}{R^3 (u+3 \pi)}$$

Note that the integral vanishes as $R \to \infty$ only when $u+3 \pi \gt 0$.

To evaluate the second integral, we must expand the exponential in the numerator to $O\left(\epsilon^2\right)$ (this is analogous to computing a residue).  It is only for that order of expansion at which we get a nonzero contribution to the integral as $\epsilon \to 0$:

$$ i \epsilon \int_{\pi}^0 d\phi\, e^{i \phi} \frac{e^{i (u+3 \pi) \epsilon e^{i \phi}}}{\epsilon^3 e^{i 3 \phi}} = i \pi \frac12 (u+3 \pi)^2 $$

Putting this all together and invoking Cauchy's theorem (i.e., the integral over the closed contour is zero), we get

$$\frac{i}{16 \pi^3} \oint_C dz \frac{e^{i (u+3 \pi)z}}{z^3} = \frac{i}{16 \pi^3}PV\int_{-\infty}^{\infty} dx \frac{e^{i (u+3 \pi)x}}{x^3} - \frac{1}{32\pi^2} (u+3 \pi)^2 = 0$$

where $PV$ denotes the Cauchy principal value of the integral.  When $u+3 \pi \lt 0$, the analysis is the same, except the sign is reversed because we must consider $C$ in the lower half-plane.  Thus we have

$$\frac{i}{16 \pi^3}PV\int_{-\infty}^{\infty} dx \frac{e^{i (u+3 \pi)x}}{x^3} = \frac{1}{32\pi^2} (u+3 \pi)^2 \text{sgn}(u+3 \pi) $$

where sgn is the signum function (i.e., $\pm 1$ according to the sign of its argument).

It should be clear that the exact same analysis extends to the other three integrals so that we may now simply write down the result we seek:

$$\int_0^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} \cos{u t} = \frac{1}{32 \pi^2} \left [(u+3 \pi)^2 \text{sgn}(u+3 \pi) - (u-3 \pi)^2 \text{sgn}(u-3 \pi)  - 3 (u- \pi)^2 \text{sgn}(u- \pi) + 3 (u+ \pi)^2 \text{sgn}(u+ \pi) \right]$$

Note that we no longer need the $PV$ because the integral is defined at $t=0$.  Through a little manipulation, we may rewrite this into a more transparent form:

$$\int_0^{\infty} dt \, \frac{\sin^3{\pi t}}{(\pi t)^3} \cos{u t} = \begin{cases} \\ 0 & u \lt -3 \pi \\ \frac{1}{16 \pi^2} (u+3 \pi)^2 & -3 \pi \lt u \lt -\pi \\ \frac{1}{8 \pi^2} (3 \pi^2-u^2) & -\pi \lt u \lt \pi \\ \frac{1}{16 \pi^2} (u-3 \pi)^2 &\pi \lt u \lt 3 \pi \\ 0 & u \gt 3 \pi\end{cases}$$

Here is a plot of the integral over $u$:

(The gaps are an artifice of Mathematica and are not real.)  It is interesting that the $1/t^3$ behavior of the integrand leads to a discontinuity in the second derivative - just as $1/t$ behavior leads to a discontinuity in the function value and $1/t^2$ behavior leads to discontinuity in the derivative.

CONVOLUTION THEOREM

The convolution theorem states that, for functions $f$ and $g$ defined on $L^2(\mathbb{R})$, we have

$$I(u) = \int_{-\infty}^{\infty} dt \, f(t) g(t) \, e^{i u t} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} du' \, \hat{f}(u') \hat{g}(u-u')$$

where $\hat{f}$ and $\hat{g}$ are the Fourier transforms of $f$ and $g$, respectively.  The FT of a product of two functions is the convolution of the FTs of the functions.

Here, we have

$$f(t) = \frac{\sin{\pi t}}{\pi t}$$

$$g(t) = \frac{\sin^2{\pi t}}{(\pi t)^2}$$

I will assume that the FFTs of these functions are known; that is

$$\hat{f}(u) = \begin{cases}  1 & |u| \lt \pi \\ 0 & |u| > \pi \end{cases}$$

$$\hat{g}(u) = \begin{cases}  1-\frac{|u|}{2 \pi} & |u| \lt 2 \pi \\ 0 & |u| > 2 \pi \end{cases}$$

If the reader so desires, the reader may derive these in a manner similar to the contour integration above.  Note that the ultimate answer will be $1/2$ of this convolution, as the original integral is over $[0,\infty)$.

The problem of evaluating the integral above reduces to evaluating the convolution of two very simple functions having finite support.  The following picture illustrates what needs to be done:

Here, the rectangle (i.e. $\hat{f}(u-u')$) moves from left to right along the $u'$ axis, while $\hat{g}(u')$ stays fixed.  The convolution at a particular value of $u$ is the overlap area.  The form of this overlap area will vary depending on the value of $u$.  For example, it should be clear that the overlap area is zero when $|u| \gt 3 \pi$.  (The near edges of the rectangle lie outside the triangle.)

Now, when $u \in [-3 \pi,-\pi]$, the overlap area is a right triangle extending from the left vertex of the triangle to the right side of the rectangle (see the figure above).  That is,

$$\begin{align}I(u) &= \frac{1}{4 \pi} \int_{-2 \pi}^{u+\pi} du' \left ( 1+ \frac{u'}{2 \pi}\right )\\ &= \frac{u+3 \pi}{4 \pi} + \frac{(u+\pi)^2-4 \pi^2}{16\pi^2}\\&= \frac{1}{16 \pi^2} (u+3 \pi)^2 \end{align} $$

When $u \in [\pi, 3 \pi]$, the situation is the mirror image:

$$\begin{align}I(u) &= \frac{1}{4 \pi} \int_{u- \pi}^{2\pi} du' \left ( 1- \frac{u'}{2 \pi}\right )\\ &= \frac{u-3 \pi}{4 \pi} - \frac{4 \pi^2-(u-\pi)^2}{16\pi^2}\\&= \frac{1}{16 \pi^2} (u-3 \pi)^2 \end{align} $$

Finally, when $u \in [-\pi,\pi]$, the overlap area is bounded on both sides by the rectangle:

$$\begin{align} I(u) &=\frac{1}{4 \pi}\int_{u-\pi}^0 du' \left ( 1+\frac{u'}{2 \pi}\right) +\frac{1}{4 \pi} \int_0^{u+\pi} du' \left ( 1-\frac{u'}{2 \pi}\right) \\ &= \frac{\pi-u}{4 \pi} - \frac{(u-\pi)^2}{8 \pi^2} + \frac{u+\pi}{4 \pi} - \frac{(u+\pi)^2}{8 \pi^2} \\ &= \frac{3 \pi^2-u^2}{8 \pi^2}\end{align}$$

The reader should see that the final result here matches that from the direct evaluation above.

ContentLicense: CC BY-SA 3.0


问:
Does $\sum |a_n|$ Converge?
If $\sum a_n$ converges, then does $\sum |a_n|$ converge as well? This is the same as "absolute convergence", where if $\sum a_n$ converges, then $\sum|a_n|$ might actually diverge, for example, $\sum\frac{(-1)^n}{n}$. So this would serve as a counterexample to the question, and prove it to be wrong, correct?

答: Yes, you are correct. If $\sum a_n$ converges, $\sum |a_n|$ does not have to converge as well.

Title: Does $\sum |a_n|$ Converge?


问:
Confused on proof of closed intervals being compact in order topology
I am looking at the proof of theorem 27.1 in Munkres which states: Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.

In the proof it says first prove the following: If $x$ is a point of $[a,b]$ different from $b$, there is a point $y>x$ of $[a,b]$ such that the interval $[x,y]$ can be covered by at most two elements of $\mathcal{A}$.

I am confused by the last line, when it says the interval $[x,y]$ can be covered by at most two elements of $\mathcal{A}$. Wouldn't the collection $\mathcal{A}$ itself cover the interval $[x,y]$, and $\mathcal{A}$ possibly has more than two elements? What does it mean to say that $[x,y]$ can be covered by at most two elements from $\mathcal{A}$? Does it mean it can be covered by at most two elements from $\mathcal{A}$ such that the union of the two elements is exactly equal to $[x,y]$ and not just a set containing $[x,y]$?

答: It means that either there is an $A\in\mathscr{A}$ such that $[x,y]\subseteq A$, or there are $A_1,A_2\in\mathscr{A}$ such that $[x,y]\subseteq A_1\cup A_2$. In other words, you need at most $2$ elements of $\mathscr{A}$ in order to cover the interval $[x,y]$. The cover definitely does not have to be exact and generally will not be: in general it will cover more than the interval $[x,y]$.


问:
Linear Algebra Gilbert Strang: Predicting Determinant Patterns

Fing $G_2$ and $G_3$ and then by row operations $G_4$. Can you predict $G_n$?

The matrices $G_2, G_3, G_4$ are below.

$$\begin{vmatrix}0 & 1\\
1 & 0
\end{vmatrix}$$

$$\begin{vmatrix}0 & 1 &1 \\
1 & 0 &1
\\
1 &1 &0
\end{vmatrix}$$

$$\begin{vmatrix}0 & 1 &1 &1\\
1 & 0 &1 &1
\\
1 &1 &0 &1
\\ 1 &1 &1 &0
\end{vmatrix}$$

I understand how to calculate the determinants with pivots and cofactors.

The first $3$ make sense:

$$ G_2 = −1, G_3 = 2, G_4 = −3$$

However, I do not yet see how he got

$$G_n = (−1)^{n−1}(n − 1).$$

答:
The formula for $G_n$ has two parts that accomplish two important things. First, disregarding the signs, you can see a pattern in which the determinant of the $n$th matrix has a magnitude of $n-1$. This is where $(n-1)$ comes from.

The second part involves the signs on the determinants. The determinants appear to be positive for odd $n$ and negative for even $n$. When $n$ is odd, $n-1$ is even, and thus $(-1)^{n-1}$ is positive. Likewise, when $n$ is even, $n-1$ is odd, and $(-1)^{n-1}$ is negative.


问:
Learning Complex Geometry - Textbook Recommendation Request
I wish to learn Complex Geometry and am aware of the following books : Huybretchs, Voisin, Griffths-Harris, R O Wells, Demailly. But I am not sure which one or two to choose. I am interested in learning complex analytic & complex algberaic geometry both.

Could somebody please advise me which of these books deal with the same or similar aspects of the subject ? If I am not mistaken, Huybretchs and Voisin deal with similar aspects ? and Huybretchs would be relatively elementary or may be a preparation for Voisin ?

I would like to select two books (preferably from above-listed ones, though other suggestions are welcome) such that the intersection between their contents is minimal and the union maximal.

Any comments about the above-mentioned books will be very helpful.

答:
Having read considerable chunks from Huybrechts, Voisin, Griffiths-Harris and Demailly, I want to suggest a slightly different viewpoint on the relationship between these books that hopefully complements the brilliant answer by Gunnar Þór Magnússon.

Books by Huybrechts and Demailly focus on the theoretical foundation of complex geometry, with the book by Huybrechts being considerably more elementary than the course by Demailly (the latter touches on singular complex spaces, currents and spectral sequences, to name a few). I think it would be a true statement to say that Demailly fully covers the contents of Huybrecths' book, but then provides much more on top of that. In particular, the end of Demailly's course also taps into the research interests of Demailly himself and discusses notions of positivity for vector bundles and $L^2$ estimates. The very last section of the book proves Grauert's Direct Image Theorem, which in and of itself shows how much technique is developed in the book.

Voisin's books go much deeper into Hodge theory, just as the title suggests. The beginning of Volume I covers some of the same foundations as Huybrechts and Demailly up to the discussion of  the Hodge decomposition in cohomology. However, starting with chapter 7 the book goes deeper into Hodge theory proper, and culminates in the discussion of variations of Hodge structures in chapters 9 and 10. The end of Volume I lays ground for the discussion of analytic and algebraic cycles in Volume II, again something that is mentioned in Demailly on only a couple of pages. Volume II then discusses advanced topics close to the research level in variations of Hodge structures (relevant for moduli theory), algebraic cycles and fine topology of algebraic varieties. In conclusion, only about a half of Volume I overlaps with Huybrechts/Demailly.

The book by Griffiths and Harris shines (in my opinion) when it comes to examples and applications of the foundational theory. The material of chapters 0 and 1 is fully covered by Huybrechts/Demailly, and in a more detailed fashion. But chapters 2 and 4 on curves and surfaces contain a wealth of beautiful geometric material that is scattered in the literature, but here you can find it all in one place. This "examples and applications" part overlaps neither with Huybrecths/Demailly nor with Voisin.

PS. I am not really sure whether there exists a better exposition for parts of chapters 3 and 5 of Griffiths-Harris. In case there does not, it should still be considered a go-to resource to learn about applications of currents and residues in complex geometry.


问:
Commutant of a set of operators in $B(H)$
Let $\mathcal{T} \subset B(T)$ be a set of operators.

Then we define the commutatnt $\mathcal{T}' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}\}$.

I'm trying to show $\mathcal{T}' = \mathcal{T}'''$.

Attempt:

By definition, we have that $\mathcal{T}'' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}' \}$, and $\mathcal{T}''' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}''\}$.

"$\subset$":

Let $S \in \mathcal{T}$. Then we have $ST=TS$ for all $T\in \mathcal{T}$.

Then I'm a bit confused on what to do next, because the $\mathcal{T}'''$ I wrote above doesn't really give me a clear idea of what an element in $\mathcal{T}'''$ looks like.

Any help will be appreciated!

Thank you.

答:
Lemma.  Let $X$ be any set  and let $\lozenge$ be a symmetric relation on $X$
(e.g.
$x\mathrel{\lozenge} y \Leftrightarrow x y= yx$, on a ring,  or
$x\mathrel{\lozenge} y \Leftrightarrow x\perp y$, on an inner-product space).  For each subset $S\subseteq X$ define
$$
  S^\lozenge = \{x\in X: x\mathrel{\lozenge} s, \text{ for all $s$ in S} \}.
  $$
Then
$$
  S^\lozenge = ((S^\lozenge)^\lozenge)^\lozenge,
  $$
for any $S$.

Proof.  First notice that
$$
  S\subseteq (S^\lozenge)^\lozenge
  \tag 1
  $$
for a pretty elementary reason (which
nevertheless sounds a bit like a tongue-twister):  every  element in  $S$ is  $\lozenge$ to anything that is $\lozenge$ to every element in $S$.

Plugging in  $S^\lozenge$ in
place of $S$ in (1), we get $S^\lozenge \subseteq  ((S^\lozenge)^\lozenge)^\lozenge$.

Next observe that
$$
  S_1\subseteq S_2 \Rightarrow   S_2^\lozenge\subseteq S_1^\lozenge,
  $$
and if this is applied to (1), we get
$
  ((S^\lozenge)^\lozenge)^\lozenge\subseteq   S^\lozenge.
  $
QED

This is based on this answer.


问:
Are binomial coefficients defined for $k \in \mathbb{R}$
So I've been trying to learn some basic precalc and I got stumped on definition of combinations namely the following confuses me .

I know that The symbol ${n\choose k}$ is read as "$n$ choose $k$." It represents the number of ways to choose $k$ objects from a set of $n$ objects. However in most definitions no restriction on the value of k is made for example what would ${5\choose 2.5}$ be.

furthermore what do we define ${n\choose k}$ if $k<0$

Some help would be greatly appreciated.

Thanks in advance.

答:
The definition of the factorial is commonly extended to reals by means of the Gamma function. The application to the Binomial coefficients is obvious, and directly relates to the Beta integral.

https://en.wikipedia.org/wiki/Beta_function#Properties

The values at integer $k<0$ and $k>n$ are usually defined to be zero, and if I am right, this is consistent with the definition in terms of Gamma.


问:
What is the probability that the coin was tossed $n$ times
Q: A coin is tossed untill k heads has appeared. If a mathematician knows how many heads appeared, can he figure out what is the probability that the coin was tossed $n$ times?

What I tried: The number of heads debends of the number of tosses. So I tried Bayes' theoream $P(X=k|N=n)P(N=n)/P(X=k)$ where $X$ is random variable for heads and $N$ for tosses. $P(X=k|N=n)$ is binomial distributions mass probability function, but i don't know what $P(N=n)$ and $P(X=k)$ are.

Because the last toss is known to be heads, we only focus on the the tosses before the last one. There is $\binom {n-1}{k-1}$ combinations for $k-1$ heads to appear before the last one. So the probability is

$$P(N=n|X=k) = \binom{n-1}{k-1}p^k(1-p)^{n-k}$$

I believe this can be done also with Bayes Theoream. Can someone wiser show how it is done?

答:
Bayes' Theorem would be useful if the experiment consisted of choosing $N$ first, before starting to count the heads, and then flipping the coin exactly $N$ times. Then we'd have

$$
P(N=n|X=k)
=\frac{P(X=k|N=n)P(N=n)}{P(X=k)}
=\frac{\binom nkp^k(1-p)^{n-k}P(N=n)}{\sum_{n=0}^\infty \binom nkp^k(1-p)^{n-k}P(N=n)}.
$$

In this model, the probabilities $P(N=n)$ need to be specified, as they describe how we make our advance choice of how many times to toss the coin.

The experiment you describe is different: $k$ is a fixed parameter and every outcome ends with a total of exactly $k$ heads, so treating the number of heads as a random variable ("$X$") isn't useful. The random variable of interest is $N$: the number of tosses needed to get exactly $k$ heads.

In order to get $N=n$, we need to get exactly $k-1$ heads in the first $n-1$ tosses, in any order, followed by a head on the next toss. There are $\binom{n-1}{k-1}$ different $n$-toss result sequences that meet these conditions, and every such sequence has probability $p^k(1-p)^{n-k}$, so $P(N=n)=\binom{n-1}{k-1}p^k(1-p)^{n-k}$, as you computed.

In other words, $P(N=n)$ in the given experiment is equal to $P(\sum_{i=1}^{n-1}X_i=k,X_n=1)$ in a different experiment that consists of tossing a coin $n$ times, and in this experiment, $X=\sum_{i=1}^{n-1}X_i$ is binomially distributed. But the random variables $X$ and $N$ don't exist in the same experiment, so you can't relate them with conditional probabilities.


问:
Two-parameter Lie transformation from two one-parameter Lie transformations
Define one-parameter Lie transformation which takes values $\Psi(\varepsilon, x)$ to be a transformation which satisfies the following properties, where $\varepsilon \in I \subset \mathbb{R}$ and $x$ is in some suitable space to make it well-defined.

$$ \Psi(\varepsilon_1, \Psi(\varepsilon_2,x)) = \Psi(\varepsilon_1+\varepsilon_2, x) $$
$$ \Psi(0,x) = x $$

My question is whenever you have two such transformations, is it possible to create a two-parameter Lie transformation defined by the following properties, and so that two initial one-parameter Lie transformations are somehow included in the two-parameter transformation (for example, by taking $\Theta(\varepsilon,0,x)$ or $\Theta(0,\delta,x)$)?

$$ \Theta(\varepsilon_1, \delta_1, \Theta(\varepsilon_2, \delta_2, x)) = \Theta(\varepsilon_1+\varepsilon_2, \delta_1+\delta_2, x) $$
$$ \Theta(0,0,x) = x $$

Intuitively, I think it should be true as, for example, for groups, it is easy to define group on Cartesian product by $(g,h) \cdot_{G \times H} (g',h') = (g\cdot_{G}g', h\cdot_H h')$, but I do not see how this translates here. I also think that intuition for wanting this two-parameter group is to define transformation such that if we start from some $x$ then all $y$ that can be reached by finite applications of two one-parameter Lie transformations is the same set as all $y$ that can be reached by newly defined two-parameter transformation.

If possible, can you please provide example construction for $x = (x_1, x_2, x_3) \in \mathbb{R}^3$ where two one-parameter Lie groups are translation and rotation, as given next?

$$ \Psi_1(\varepsilon, x) = (x_1 + \varepsilon, x_2, x_3)$$
$$ \Psi_2(\varepsilon, x) = (x_1 \cos \varepsilon + x_2 \sin \varepsilon, -x_1 \sin \varepsilon + x_2 \cos \varepsilon, x_3) $$

EDIT: It seems that this example is not sufficiently interesting, as I believe it was worked out by user @Cosmas Zachos, and so I provide another one in hope to make it more interesting. These are just two rotations about different axes.

$$ \Psi_1(\varepsilon, x) = (x_1 \cos \varepsilon + x_3 \sin \varepsilon, x_2, -x_1 \sin \varepsilon + x_3 \cos \varepsilon)$$
$$ \Psi_2(\varepsilon, x) = (x_1 \cos \varepsilon + x_2 \sin \varepsilon, -x_1 \sin \varepsilon + x_2 \cos \varepsilon, x_3) $$

答:
I really don't appreciate what you are asking. If you are asking for a simple representation or realization of your three maps, it is evident by inspection that
$$
\Psi(\epsilon, x)= x+\epsilon, \\
\Theta (\epsilon, \delta, (x_1,x_2))= (x_1+\epsilon,x_2+\delta),
$$
1d and 2d translations; and, lastly,
$$
\Psi_1(\epsilon,(x_1,x_2)) =(x_1+\epsilon, x_2) ,\\
\Psi_2( \epsilon, (x_1,x_2))=    \begin{pmatrix}
      \cos \epsilon &\sin\epsilon \\
      -\sin \epsilon &\cos\epsilon
    \end{pmatrix}    \begin{pmatrix}
       x_1\\
     x_2
    \end{pmatrix}  =  \begin{pmatrix}
      x_1 \cos \epsilon +x_2 \sin\epsilon \\
     -x_1  \sin\epsilon +x_2   \cos \epsilon
    \end{pmatrix},
$$
1d translation and 2d rotation respectively. I skipped $x_3$, as it is a superfluous inert parameter here, and need not be written down explicitly.

Response to EDIT in question.  This it a composition of two rotations, a latitude and a longitude,
$$
\Psi_1( \epsilon, (x_1,x_2,x_3))=    \begin{pmatrix}
      \cos \epsilon &0&\sin\epsilon \\ 0&1&0 \\
      -\sin \epsilon &0&\cos\epsilon \\
 \end{pmatrix}    \begin{pmatrix}
       x_1\\
     x_2\\ x_3
    \end{pmatrix}  =  \begin{pmatrix}
      x_1 \cos \epsilon +x_3 \sin\epsilon \\ x_2 \\
     -x_1  \sin\epsilon +x_3   \cos \epsilon\\
    \end{pmatrix},  \\
\Psi_2( \delta, (x_1,x_2,x_3))=    \begin{pmatrix}
      \cos \delta   &\sin \delta &0\\
      -\sin  \delta &\cos \delta &0\\
0&0& 1
    \end{pmatrix}    \begin{pmatrix}
       x_1\\
     x_2\\ x_3
    \end{pmatrix}  =  \begin{pmatrix}
      x_1 \cos \delta  +x_2 \sin\delta   \\
     -x_1  \sin\delta   +x_2   \cos\delta  \\x_3
    \end{pmatrix},
$$
non-commuting rotations about the y and z axes, respectively, so ε is an altitude angle an δ an azimuth.

You may always consider $\Psi_1(\epsilon, \Psi_2(\delta,\vec x  ))\equiv \Theta(\epsilon,\delta, \vec x)$, via matrix multiplication of the above two matrices, to get a general rotation matrix, as you learned in college mechanics.


问:
If $M$ is an $R$-module, when is $A\mapsto Aut(M\otimes_{R}A)$ a representable functor?
Let $R$ be a ring (unitary, commutative, associative) and $M$ an $R$-module. Is the functor
\begin{eqnarray*}
    F & : & {R}\mathbf{-Alg} & \longrightarrow & \mathbf{Grp}\\
       & & A                & \longmapsto     & \operatorname{Aut}(M\otimes_{R}A)
\end{eqnarray*}
representable? I know this is true if $M=R^{n}$ for some $n\in\mathbb{N}$, for then it is represented by the ring
\begin{equation*}
    A[x_{11},...,x_{nn},y]/(\det(x_{ij})y-1)\text{.}
\end{equation*}
I guess (although this is no more than a gut feeling) this should be true at least if $M$ is finitely generated and flat, but I might be wrong.

As usual, a literature reference might also be helpful.

答:
If $M$ is finitely presented and flat, then $F$ is representable; indeed this follows pretty easily from the finitely generated free case.  If $M$ is finitely presented and flat, then it is projective, so it is a direct summand of $R^n$ for some $n$.  This direct sum decomposition can be represented by an idempotent matrix $e\in M_n(R)$.  Then, the group of automorphisms of $M$ can be described as the set of $a\in GL_n(R)$ such that $ea(1-e)=(1-e)ae=0$ and $(1-e)a(1-e)=1-e$ (this just means that the matrix representation of $a$ with respect to the direct sum decomposition $R^n\cong M\oplus N$ has the form $\begin{pmatrix}b & 0 \\ 0 & 1_N\end{pmatrix}$ for some invertible $b$).  This description is still valid after base-changing, so the functor $\operatorname{Aut}(M\otimes-)$ is represented by the quotient of $R[a_{11},\dots,a_{nn},\det^{-1}]$ that imposes the relations $ea(1-e)=(1-e)ae=0$ and $(1-e)a(1-e)=1-e$.

A simple necessary condition for $F$ to be representable is that it must preserve monics.  For instance, this means that if $R$ is a domain and $M$ is a nonzero torsion module then $F$ cannot be representable, by considering what $F$ does to the inclusion from $R$ into its fraction field.

Here is another necessary condition that is a sort of finiteness condition (for instance, it implies $F$ cannot be representable if $R$ is nonzero and $M$ is free of infinite rank).  Suppose $F$ is representable.  Then for each $R$-algebra $A$ and each $m\in M\otimes A$, there must be a finitely generated submodule of $M\otimes A$ that contains the images of $m$ under all the automorphisms of $M\otimes A$.  To prove this, note that by base-changing to $A$, we may assume $A=R$. Let $S=\operatorname{Aut}(M)$ and consider $F(R^S)$.  Since $F$ is representable, it preserves products, so there is an automorphism $\alpha$ of $M\otimes R^S$ which becomes $s$ after applying the $s$th projection $R^S\to R$, for each $s\in S$.  This means that if we write $\alpha(m\otimes 1)$ as a sum $\sum x_i\otimes y_i$ of elementary tensors, we have $s(m)=\sum y_i(s)x_i$ for each $s\in S$.  Thus, all of these images $s(m)$ are contained in the submodule of $M$ generated by the finitely many elements $x_i$.


问:
Check if subgroup is normal based on generators
Let's say we have a group $G=\langle g_1,g_2,g_3\rangle$ for some generators $g_i$. Further, suppose I have some subgroup $\langle h_1,h_2\rangle =H\subset G$ for some generators $h_i$. For $H$ to be normal in $G$ we must have $gHg^{-1}=H$ for all $g\in G$. Is there an easy way to check this computationally? What I mean is, can I do something like check $g_jh_ig_j^{-1}\in H$ for each $i,j$? Or is this not sufficient? If it is sufficient, is there something even easier we can do, like check if each $g_jh_ig_j^{-1}$ is one of the generators $h_k$?

答:
Theorem. Let $G$ be a group, and $N\leq G$ a subgroup. Let $X\subseteq N$ and $Y\subseteq G$ be subsets such that $N=\langle X\rangle$ and $G=\langle Y\rangle$. The following are equivalent:

$N\triangleleft G$.

For all $g\in G$, $gNg^{-1}=N$.

For all $g\in G$, $gNg^{-1}\subseteq N$.

For all $g\in G$ and $n\in N$, $gng^{-1}\in N$.

For all $g\in G$ and all $x\in X$, $gxg^{-1}\in N$.

For all $y\in Y$, $yNy^{-1}=N$.

For all $y\in Y$, $yNy^{-1}\subseteq N$ and $y^{-1}Ny\subseteq N$.

For all $y\in Y$ and $x\in X$, $yxy^{-1}\in N$ and $y^{-1}xy\in N$.

Proof. (1)$\iff$(2) is the usual definition.

Clearly (2)$\implies$(3); to get (3)$\implies$(2), note that for $g\in G$, we have $gNg^{-1}\subseteq N$ from (3), and applying $3$ to $g^{-1}$ we have $g^{-1}Ng\subseteq N$. Left multiplying by $g$ and right multiplying by $g^{-1}$ we get $N\subseteq gNg^{-1}$, giving $gNg^{-1}=N$, thus proving (2).

It is easy to see that (3)$\iff$(4).

(4)$\implies$(5) is immediate. To get (5)$\implies$(4), first note that for all $x\in X$ and $g\in G$, since $gxg^{-1}\in N$ and $N$ is a subgroup, then $gx^{-1}g^{-1}=(gxg^{-1})^{-1}\in N$. Now let $n\in N$. Since $N=\langle X\rangle$, we can write $n = x_1^{\epsilon_1}\cdots x_n^{\epsilon_n}$, where $x_i\in X$ are not necessarily distinct, and $\epsilon_i\in\{1,-1\}$ for all $i$. Then
$$\begin{align*}
gng^{-1} &= g(x_1^{\epsilon_1}\cdots x_n^{\epsilon_n})g^{-1}\\
&= (gx_1^{\epsilon_1}g^{-1})(gx_2^{\epsilon_2}g^{-1})\cdots(gx_n^{\epsilon_n}g^{-1}).
\end{align*}$$
Each $gx_i^{\epsilon_i}g^{-1}\in N$, so this is a product of elements of $N$, and hence lies in $N$. Thus, $gng^{-1}\in N$, proving (4).

(2)$\implies$(6) is immediate. To prove that (6)$\implies$(2), note that from $yNy^{-1}=N$ we get, left multiplying by $y^{-1}$ and right multiplying by $y$, that $N=y^{-1}Ny$. Let $g\in G$; using $G=\langle Y\rangle$, we know that we can write $g=y_1^{\epsilon_1}\cdots y_n^{\epsilon_n}$, with $y_i\in Y$ and $\epsilon_i\in \{1,-1\}$ for all $i$. We also know that $y_j^{\epsilon_j}Ny_j^{-\epsilon_j}=N$. From this, we get
$$\begin{align*}
gNg^{-1} &= (y_1^{\epsilon_1}\cdots y_n^{\epsilon_n})N(y_1^{\epsilon_1}\cdots y_n^{\epsilon_n})^{-1}\\
&= y_1^{\epsilon_1}\cdots y_{n-1}^{\epsilon_{n-1}}(y_n^{\epsilon_n }Ny_n^{-\epsilon_1})y_{n-1}^{-\epsilon_{n-1}}\cdots y_1^{-\epsilon_1}\\
&= y_1^{\epsilon_1}\cdots y_{n-1}^{\epsilon_{n-1}}Ny_{n-1}^{-\epsilon_{n-1}}\cdots y_1^{-\epsilon_1}
\end{align*}$$
and inductively we get $gNg^{-1}=N$, proving 2.

(6)$\iff$(7), since $y^{-1}Ny\subseteq N$ is equivalent, multiplying by $y$ on the left and $y^{-1}$ on the right, to $N\subseteq yNy^{-1}$.

(7)$\implies$(8) is clear. For (8)$\implies$(7), using the same argument to show that (5)$\implies$(4) we see that (8) implies that $yny^{-1}\in N$ and $y^{-1}ny\in N$ for all $n\in N$, giving (7). $\Box$

However, in (7) and (8) it is not, in general, enough to require $yNy^{-1}\subseteq N$ (respectively $yxy^{-1}\in N$). A counterexample can be found in this post.


问:
nonlinear partial-differential-equations
I have a problem with  partial differential equation

$$u_xu_y=xy$$
where $$u(0,y)=-y.$$

Therefore $x_0(s)=0$,$y_0(s)=s$,$u_0(s)=-s.$
I wrote my equation as $F(x,y,u,p,q)=pq-xy=0$ and defined the characteristic equations:
$$\frac{dx}{dt}=q$$
$$\frac{dy}{dt}=p$$
$$\frac{du}{dt}=2pq$$
$$\frac{dp}{dt}=y$$
$$\frac{dq}{dt}=x.$$

I know that
$F(x_0(s),y_0(s),u_0(s),p_0(s),q_0(s))=0$ and $p_0(s)x_0'(s)+q_0(s)y_0'(s)=u_0'(s)$,
then
$p_0(s)q_0(s)=0$ and $q_0(s)=-1$, therefore $p_0(s)=0.$ Then $p(t,s)=0$ and $q(t,s)=-1$ and then I solved characteristic equations: $x(t,s)=-t, y(t,s)=s, u(t,s)=-s.$ But u(x,y)=-y is not a solution, and I don't know what I did wrong.

答:
That's a good start, but the point where you go wrong is when you claim that $p(t,s)$ and $q(t,s)$ are constant functions; that's not correct, since their derivatives are $dp/dt=y$ and $dq/dt=x$, not $dp/dt=0$ and $dq/dt=0$.

To solve the characteristic equations correctly, note that the system decouples into two subsystems that you can solve separately,
$$
\frac{dx}{dt} = q
,\qquad
\frac{dq}{dt} = x
$$
and
$$
\frac{dy}{dt} = p
,\qquad
\frac{dp}{dt} = y
,
$$
together with the last equation for $u(t,s)$, which you can integrate rather easily once you have computed $q(t,s)$ and $p(t,s)$:
$$
\frac{du}{dt} = 2pq
.
$$
And of course you also need to use the right initial conditions at $t=0$ for all these equations, but you have already determined those conditions correctly ($x(0,s)=x_0(s)=0$, etc.), so I'm not writing them here.

There are many ways to solve the subsystems. For example, note that $d^2 x/dt^2=x$. Or use $x+q$ and $x-q$ as new variables.

Do you think you can take it from here?


问:
Volume of parabolic cylinder bound by plane
Make a sketch of the solid in the first octant bounded by the plane

$x + y = 1$

and the parabolic cylinder

$x^{2} + z = 1$

Calculate the volume of the solid..

I have no idea where to even start :(

答:
You want to take a triple integral

$$\iiint 1 \, dx \, dy \, dz$$

where $X$ is your solid.  This makes sense to compute a volume, since you're adding up one unit to your integral per unit value.  So all you need to do is work out what the bounds on those integrals should be so that they exactly cover your solid.  (You may want to change the order of the $dx$, $dy$, and $dz$ around if it helps).


问:
About exercise 1.2 item 1-b in Brezis’ book on functional Analysis
I’m following Brezis’ book on functional analysis and I'm stuck with the following exercise:

Let $E$ be a vector space of dimension $n$ with the $1$-norm, that is, if $(e_i)_{1\leq i \leq n}$ is a basis, then for every $x\in E$ we define $||x||= \sum_{i=1}^{n}|x_i|$ where $x=\sum_{i=1}^n x_ie_i$. Show that for every $x\in E$,  the duality map $F(x)$ is exactly the set of linear functionals $f\in E^*$ such that
$$
f_i=
\begin{cases}
 (\mathrm{sign}(x_i))||x||_1 &\text{if}\, x_i\neq 0\\
 \text{anything in the interval} [-||x||_1, ||x||_1] &\text{if}\, x_i=0
\end{cases}
$$
where $f_i= f(e_i)$.

We know from the previous exercise in the book that the norm in the dual space $E^*$ is given by $||f||= \max|f_i| $.

I can’t really show any progress since I’ve tried all I can think of with no luck. I know the definition of the duality map and have seen the solution to the exercise but I can’t see how to actually obtain it myself. Any help would be much appreciated.

答:
Recall that $f \in F(x)$ means that $\|f\| = \|x\|$ and $\langle f,x \rangle = \|x\|^2$. The first condition states
$$\max_{1 \le i \le n} |f_i|= \|x\|$$
which gives you $-\|x\| \le f_i \le \|x\|$ for all $i$.

The second condition states $$\sum_{i=1}^n f_i x_i = \|x\|^2$$ and in particular
$$ \sum_{i=1}^n \frac{f_i}{\|x\|} x_i = \|x\| = \sum_{i=1}^n |x_i|.$$ Since $$ \frac{f_i}{\|x\|} x_i \le |x_i|$$ for all $i$, to obtain equality in the sum over all indices it must be the case that $\displaystyle \frac{f_i}{\|x\|} x_i = |x_i|$ for all $i$. If $x_i = 0$ this adds no restriction to to $f_i$. If $x_i \not= 0$ it forces $f_i = \mathrm{sgn}(x_i) \|x\|$.


答:
Assume a series of positive terms $a_n$. The corresponding alternating series $a_1-a_2+a_3-\cdots$ can be seen as the series with terms $b_1=a_1-a_2,b_2=a_3-a_4,b_3=a_5-a_6,\cdots$ As those terms are differences, they decrease faster than the $a$'s. E.g.

$$a_n=\dfrac1n$$ and $$b_n=\frac1{(2n-1)(2n)}=\Theta\left(\frac1{n^2}\right).$$

For this reason, we can have a series $a_n$ that diverges, while the $b_n$ converge.

Title: Does $\sum |a_n|$ Converge?

Body: If $\sum a_n$ converges, then does $\sum |a_n|$ converge as well? This is the same as "absolute convergence", where if $\sum a_n$ converges, then $\sum|a_n|$ might actually diverge, for example, $\sum\frac{(-1)^n}{n}$. So this would serve as a counterexample to the question, and prove it to be wrong, correct?


问:
Question about the divergence of a series
I have below 2 very similar questions about divergence of series and sequences:

If $\sum a_n$ diverges, does $\sum |a_n|$ diverge?

If $\sum |a_n|$ diverges, does $\sum a_n$ diverge as well?

This would be sort of opposite to absolute convergence, and would be concerning "absolute divergence". For the first one, I think $\sum a_n$ will diverge, since I have not found a counterexample for it.

For the second one, let $a_n = \frac{(-1)^n}{n}$, then the absolute value of $\sum a_n$ diverges, while the actual sum $\sum a_n$ converges.

答:
If a series converges absolutely, then it converges (you seem to be aware of that). That answers the first question (which is just the contrapositive).

As for the second, certain series converge conditionally, for example
$$
1-1/2+1/3-1/4+ etc.
$$
which means precisely that they, although convergent, are not absolutely convergent. That answers your second question


答:
Both your answers are correct. For the first one, the triangle inequality implies
that
$$|\sum_na_n|\leq\sum_n|a_n|$$
so if a series converges absolutely, it also converges.

Title: Question about the divergence of a series

Body:
I have below 2 very similar questions about divergence of series and sequences:

If $\sum a_n$ diverges, does $\sum |a_n|$ diverge?

If $\sum |a_n|$ diverges, does $\sum a_n$ diverge as well?

This would be sort of opposite to absolute convergence, and would be concerning "absolute divergence". For the first one, I think $\sum a_n$ will diverge, since I have not found a counterexample for it.

For the second one, let $a_n = \frac{(-1)^n}{n}$, then the absolute value of $\sum a_n$ diverges, while the actual sum $\sum a_n$ converges.


答: If $R$ is noetherian and $M$ is finitely generated then the automorphism sheaf is representable if and only if $M$ is projective, see here.

Title: If $M$ is an $R$-module, when is $A\mapsto Aut(M\otimes_{R}A)$ a representable functor?

Body:
Let $R$ be a ring (unitary, commutative, associative) and $M$ an $R$-module. Is the functor
\begin{eqnarray*}
    F & : & {R}\mathbf{-Alg} & \longrightarrow & \mathbf{Grp}\\
       & & A                & \longmapsto     & \operatorname{Aut}(M\otimes_{R}A)
\end{eqnarray*}
representable? I know this is true if $M=R^{n}$ for some $n\in\mathbb{N}$, for then it is represented by the ring
\begin{equation*}
    A[x_{11},...,x_{nn},y]/(\det(x_{ij})y-1)\text{.}
\end{equation*}
I guess (although this is no more than a gut feeling) this should be true at least if $M$ is finitely generated and flat, but I might be wrong.

As usual, a literature reference might also be helpful.


问:
Twice differentiable function from $[0,\infty) \longrightarrow [0,\infty)$ and with $f''(x) < 0$ must be nondecreasing, uniformly differentiable
I've seen variations of this problem around, but none that quite match my situation. I'm asking because the given solution to this problem is much more complicated. Given a function $f:[0, \infty) \longrightarrow [0,\infty)$ with $f''(x) < 0 \; \forall \; x \in [0,\infty)$, I want to show $f$ is non decreasing $(f'(x) \geq 0)$ and $f$ is uniformly continuous.

For a contradiction, suppose $f'(x_0) < 0$. Since $f'(x)$ is always decreasing, for any $y > x_0$ we have $f'(y) < f'(x_0)$. In an integral, this means $$\int_{x_0}^yf'(x)dx \leq f'(x_0)(y-x_0) < 0$$

I have a feeling that's where my mistake is. Using familiar properties of integrals, we apply this to get

$$\int_0^y f'(x)dx = \int_0^{x_0}f'(x)dx + \int_{x_0}^yf'(x)dx \leq \int_0^{x_0}f'(x)dx + f'(x_0)(y-x_0)$$

The FTOC says
$$f(y) - f(0) \leq f(x_0) - f(0) + f'(x_0)(y-x_0)$$
$$f(y) - f(x_0) \leq f'(x_0)(y-x_0)$$

This object on the right goes to $-\infty$ as $y \longrightarrow \infty$, which says there is some $y$ for which $f(y) - f(x_0) < -f(x_0)$. But then $f(y) < 0$. This contradicts the fact that $f$ is always positive. Is this right? Is there a way to do this with MVT and not integrals?

For uniform continuity, choose $\varepsilon > 0 $. Since $f'(x)$ is always decreasing, we have $f'(0) > f'(x) > 0 $ for all $x$. Let $\delta = \varepsilon/f'(0)$. For any pair $x<y$, the MVT gives us a $c \in (x,y)$ so that
$$f(y) - f(x) = f'(c)(y-x)$$

But then $|y-x| < \delta$ implies
$$|f(y)-f(x)| = f'(c)(y-x) < f'(0)\delta = \varepsilon$$

and we're done.

答:
If $g'<0$ then by the MVT, $g$ is strictly deceasing. Applying this with $g=f',$ we see that $f'$ is strictly decreasing. Now if $f'(x_0)<0$ then for any $x>x_0,$ the MVT says there is $y\in (x_0,x)$ with $$f'(x_0)>f'(y)=\frac {f(x)-f(x_0)}{x-x_0},$$ implying $$f(x)<(x-x_0)f'(x_0)+f(x_0).$$ But the RHS above goes to $-\infty$ as $x\to\infty,$ contrary to the requirement that every $f(x)\ge 0$.

For uniform continuity of $f$: We have $0\le f'(y)\le f'(0)$ for all $y\ge 0.$ Given $e>0,$ take $d>0$ such that $d\cdot f'(0)<e.$ Now if $0\le x<x',$ there is $y\in (x,x')$ with $$\left|\frac {f(x)-f(x')}{x-x'}\right|=|f'(y)|=f'(y)\le f'(0). $$ Therefore $$|x-x'|<d\implies |f(x)-f(x')|\le |x-x'|\cdot f'(0)\le d\cdot f'(0)<e.$$


问:
What is the difference between product of two functions and dot product of functions?
Given two functions e.g.:
$$f(x)=\sin(x)
\text{, }
g(x)=2\cos(x)$$

Why can't we write:
$$f\cdot g=\sin(2x)$$

But we use this:
$$(f,g)=\int_{\alpha}^{\beta}f(x)g(x)dx$$
I don't quite understand if those two things are completely different and I'm mixing two totally different concepts.

答:
First operation maps two functions to a function, namely their product.

Second one maps two functions to a number.

So indeed very different operations.


问:
Riemann problem for general Scalar Conservation Law
The Riemann problem for Burgers' equation $u_t +(f(u))_x = 0$, where $f(u)=\frac{1}{2}u^2$, has a shock solution:
$$
u(x,t) =
\left\lbrace
\begin{aligned}
&u_L &&\text{if}\quad x<st \\
&u_R &&\text{if}\quad x>st \, .
\end{aligned}
\right.
$$
with Rankine-Hugoniot jump condition for $u_L > u_R$
$$s = \frac{f (u_R) − f (u_L)}{u_R - u_L} $$

My question is that does this solution satisfy Riemann problem for any general Scalar Conservation Law $u_t +(f(u))_x = 0$?

答: Yes, it does, since it is obviously a solution on either side of the shock and satisfies the Rankine-Hugoniot condition. The nonlinearity $f$ only affects the shock. Of course this is not the case for non-piecewise constant solutions.


问:
(n-1)-dimensional manifold is orientable iff it has smooth outward unit normal vector field
I am reading Spivak Calculus on Manifold Chapter 5 and it has the following argument:

If $M$ is an oriented $(n-1)$-dimensional manifold in $R^n$, a substitute for outward unit normal vectors can be defined, even though M is not necessarily the boundary of an $n$-dimensional manifold. If $[v_1,...,v_{n-1}] = \mu_x$ (which is the orientation of $M$ at $x$), we choose $n(x)$ in $R^n_x$ so that $n(x)$ is a unit vector perpendicular to $M_x$ and $[n(x),v_1,...,v_{n-1}]$ is the usual orientation of $R^n_x$. The vectors $n(x)$ vary continuously on $M$, in an obvious sense.

I understand this approach of defining outward unit normal vectors. But why $n(x)$ vary continuously? Or why $n(x)$ is a smooth vector field?

答: $n(x)$ is uniquely determined up to sign. Now think about the determinant of the matrix with columns $n(x)$, $v_1(x), v_2(x), \dots, v_{n-1}(x)$. The sign on $n(x)$ is chosen to make the determinant positive. Remember that determinant is a continuous (smooth) function of the $n$ column vectors. If $v_i(x)$ vary continuously (smoothly) with $x$ (work in a coordinate chart to make this condition hold), then $n(x)$ must likewise vary continuously to keep the sign of the determinant positive.


问:
Need help understanding Erdős' proof about divergence of $\sum\frac1p$
I'm looking at proofs from Proofs from the Book (Martin Aigner, Günter M. Ziegler). The proof I'm having trouble is the sixth proof of the infinitude of the primes they give (on page 5; although I'll be reconstructing it here). I'll be highlighting the areas I'm having trouble with. They set out to prove two things; namely, and letting $P$ be the set of all primes:

$P$ is infinite.

$\sum_{p \in P} {1 \over p}$ diverges.

They begin by assuming the sum in 2. is convergent and letting $P = \{p_1, p_2, ...\}$ The first problem comes up here. By assuming the given form for $P$, haven't they already assumed that it is infinite? Then, by assumption,

$$\sum_{i > k} {N \over p_i} < {N \over 2}, \ \  (N \in \mathbb{Z}^+). \tag{1}$$

They then define $p_1, p_2, ..., p_k$ to be the small primes, and $p_{k + 1}, p_{k + 2}, ...$ to be the large primes, and define $N_b$ to be the number of positive integers $n \leq N$ which are divisible by at least one of the big primes; and, similarly, $N_s$ the number of positive integers $n \leq N$ which only have small prime divisors. By definition, $N_b + N_s = N$, and the intended contradiction will hinge on showing $N_b + N_s < N$.

Note, since $\left \lfloor {N \over p_i} \right \rfloor$ gives the number of positive integers $n \leq N$ which are multiples of $p_i$,

$$N_b \leq \sum_{i > k} \left \lfloor {N \over p_i} \right \rfloor < {N \over 2}. \tag{2}$$

Then, they let $n = a_n b_n^2$, where $n \leq N$ such that it has only small prime divisors. Here $a_n$ is the square free part. Then, every $a_n$ is a product of different small primes. Assuming that by this it is simply meant that every factor of $a_n$ is distinct, they assert there are exactly $2^k$ different square free parts. Notice, also, $b_n \leq \sqrt{n} \leq \sqrt{N} \implies b_n \leq \sqrt{N}$, and as such,

$$N_s \leq 2^k \sqrt{N}. \tag{3}$$

Here, is the second problem for me. I can see intuitively how $(3)$ holds; but, I'm not really sure, and I'd like to see it through a more rigorous approach, and also, why it's relevant. I'm not really sure, and I'd like a more rigorous approach.

They conclude the proof by showing that, since $(2)$ holds for any $N$, $2^k \sqrt{N} \leq {N \over 2}$ holds for $N = 2^{2k + 2}$. This is the third area I'm having trouble. I'm not really sure why this is relevant and how the contradiction $N_s + N_b < N$ follows from it. Maybe I missed it, but I also didn't see a proof of the infinitude of primes here.

Finally, I have a question as well. Looking back, how could one have guessed that this approach will lead us to a proof of the two claims above? In other words, what's the insight behind taking this particular approach?

Thanks to all in advance!

Note: I wasn't really sure what tags relating to "Proofs" would be appropriate here, so I've only used tags concerning number theory and primes. Please feel free to tag away.

答:
The first issue is just notational.  Just agree to stop the sums and so forth if you run out of primes.  Assuming what you've given here is their notation I'll agree that they probably didn't write the proof in the most reasonable way.

For point (2), there are exactly $2^k$ squarefree numbers whose prime factors are $p_1, \ldots, p_k$, so there are at most $2^k$ squarefree numbers satisfying that condition AND which are less than some number $N$.  And there are at most $\sqrt{N}$ numbers $b$ such that $b^2 < N$, so there are certainly at most $\sqrt{N}$ numbers $b$ satisfying that condition AND such that $b$ is only divisible by $p_1, \ldots, p_k$.

For the rest, just plug all your equations together to get the contradiction you described at the beginning.

As for the motivation, since numbers are made up of prime factors, it makes sense to see what we can learn about them by starting from prime factors and multiplying them out rather than vice versa.  Of course the distributional information we get will usually just be inequalities, so we want to combine them with something that lets us get meaningful results out of inequalities -- in this case, something like the pigeonhole principle.


问:
Finding arbitrary value in joint pmf of discrete random variables.
I am learning about probability and have a relatively good understanding of how random variables work. I am a little confused with this question as I have never seen an arbitrary (k value in this case) value in the pmf of two discrete random variables.

The joint pmf of two discrete random variables X and Y is given by Px,y (x,y) = k(x + 3y).

The support of X is X = {0,1,2}, whereas the support of Y is Y = {0,1,2,3).

The question then continues to ask to find the value of k, obtain the pmf's of both X and Y and determine whether they are independent or not.

I am finding it difficult to understand how to find the value of k in the beginning?

答:
As mentioned in the comments, the sum of a PMF over its support is $1$. Therefore,
$$1=\sum_{x=0}^2\sum_{y=0}^3 P_{X,Y}(x,y) =k \sum_{x=0}^2\sum_{y=0}^3 (x+3y)=66k,$$ from which we deduce $k=\displaystyle\frac{1}{66}$.

Moreover, notice that \begin{align}
P_X(x) &= P(X=x)\\
&=\sum_{y =0}^3 P(X=x, Y=y) &\textrm{(law of total probablity)}\\
&=\sum_{y =0}^3 P_{XY}(x,y).
\end{align}
Similarly, $P_Y(y)=\sum_{x=0}^2 P_{XY}(x,y)$. Hence,
$$P_Y(y)=\sum_{x=0}^2 P_{XY}(x,y)=\sum_{x=0}^2 \frac{1}{66}(x+3y)=\frac{3y}{22}+\frac{1}{22}$$ and
$$P_X(x)=\sum_{y=0}^3 P_{XY}(x,y)=\sum_{y=0}^3\frac{1}{66}(x+3y)=\frac{2x}{33}+\frac{3}{11}.$$

For independence, recall that $X$ and $Y$ are independent iff $P_{X,Y}(x,y)=P_X(x)P_Y(y)$, for all $x$ and $y$.


问:
How to find the maximum likelihood estimator (MLE)?
I apologize in advance for the mess you're about to read.

I have recently been introduced to estimators, and right after the method of moment estimators we have the maximum likelihood estimators. But I'm failing to really understand it. It just seems harder to compute because the likelihood function is a product of mass/density functions.

Moreover, I have a problem which poses the question "Why would it be difficult to find an MLE for $(k,p)$?" given a sample $X_{1},...,X_{n}$ taken from $\operatorname{Bin}(k,p)$ and I have no idea why. Is it because the likelihood function$$L(k,p) = \prod_{i=1}^{n}f_{X}(x_{i};k,p) = \prod_{i=1}^{n}\binom{k}{x_{i}}p^{x_i}(1-p)^{k-x_{i}}$$is hard to compute or maximise? or maybe because of the product of binomial coefficients?

Any help is well appreciated!

答:
Given $k$, the maximum likelihood estimator for $p$ will be $\hat p_k=\frac{\frac1n\sum x_i}{k}$, and using that for $p$, the likelihood for $k$ becomes $$\prod_{i=1}^{n}\binom{k}{x_{i}}\left(\frac{\frac1n\sum x_i}{k}\right)^{x_i}\left(1-\frac{\frac1n\sum x_i}{k}\right)^{k-x_{i}} \\ = \frac{(k!)^n n^{nk}}{k^{nk} \prod x_i!(k-x_i)!}\left(\sum x_i\right)^{\sum x_i}\left(nk-{\sum x_i}\right)^{nk-\sum x_i}$$ which looks difficult to handle even dropping multiplicative terms not depending on $k$. You could perhaps do an integer search, starting at $\max(x_i)$ and working upwards.

You would not expect the result to necessarily be unbiased, though that is a relatively minor issue.  More worrying might be sensitivity to the precise values of the data despite natural random variation.

But I think that is not the real difficulty. There seem to be some observation values for which the the likelihood seem to be an increasing function of $k$ and so the maximum likelihood solution would be $\hat k= \infty$.

As a small contrived example, I suspect that observing $x_1=2, x_2=0$ may lead to this effect. Less artificially, a case of $x_1=42, x_2=56, x_3=40, x_4=41, x_5=54$ seemed to lead to this problem $($generated as one of several binomial samples with $k=100, p=\frac12)$.  I would guess you may have the same issue with a method of moments if the variance is not smaller than the mean of the observations.


问:
Taylor series of $\log x$ at $x=2$
Good night to everyone.

I have a doubt.

Can I obtain the Taylor serie of $g(x) = \log (x)$ centered at $x=2$ starting in the kwon Maclaurin series of $e^x$ or $1/(1-x)?$

Best regards

答:
Hint:

From your post: The Taylor series of $1/(1-x)$ centered at $2$ is known.

Now set $x\to 1-t,$ so that we have $1/t$.  Then integrate on both sides, with the lower bound $1$ and the upper bound $x$.

Title: Taylor series of $\log x$ at $x=2$


问:
Unions of two elementary classes that have special properties
This is a follow up to my previous question on the union of two elementary classes. Define an $\forall$-elementary class as a class $K$ of structures in some fixed language $L$ that can be axiomatized by $\forall$-sentences. An $\exists$-elementary class is defined analogously. It is not too hard to show that the intersection of two $\forall$-elementary classes is itself $\forall$-elementary, and similarly for $\exists$-elementary. What about the union? Is the union of two $\forall$-elementary classes (respectively $\exists$-elementary classes) itself $\forall$-elementary (respectively $\exists$-elementary)?

答:
The disjunction of $\exists$-sentences is equivalent to an $\exists$-sentence and the disjunction of $\forall$-sentences is equivalent to a $\forall$-sentence. So applying the argument of the answer to your previous question, both of the questions here have affirmative answers.

The closure of $\exists$-sentences under disjunction is immediate since $\exists$ distributes over $\vee$. For the $\forall$-case, we use a cute trick: $$[\forall x(P(x))]\vee[\forall x(Q(x))]$$ is equivalent to $$\forall x,y[P(x)\vee Q(y)],$$ which is a $\forall$-sentence.


答:
Let $f(x)=\ln(x+2)$ then $(x+2)f'(x)=1$ and develop for $f(x)=\sum\limits_{n=0}^\infty a_nx^n$

$(x+2)\sum\limits_{n=1}^\infty na_nx^{n-1}=\sum\limits_{n=1}^\infty na_nx^n+\sum\limits_{n=0}^\infty 2(n+1)a_{n+1}x^n=1$

Identify with the constant series $1$ gives $\begin{cases}2a_1=1\\a_{n+1}=-\frac n{2(n+1)}a_n\end{cases}$

Therefore $$a_n=\frac{(-1)^{n-1}}{n2^n}\quad\text{and}\quad a_0=f(0)=\ln(2)$$

Title: Taylor series of $\log x$ at $x=2$

Body:
Good night to everyone.

I have a doubt.

Can I obtain the Taylor serie of $g(x) = \log (x)$ centered at $x=2$ starting in the kwon Maclaurin series of $e^x$ or $1/(1-x)?$

Best regards


问:
If $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$, then $R-R^*$ is an ideal
Suppose that $R$ is a nonzero commutative ring with $1$. Suppose $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$. I'd like to understand why this implies that $R-R^*$ is an ideal.
In particular, I'd like to know why the sum of two non-units is a non-unit.

So suppose $a,b$ are non-units. We'd like to show that $a+b\in R-R^*$ as well. From the assumption, we have that $1+a,1+b\in R^*$, and so $(1+a)(1+b)=1+a+b+ab\in R^*$.

But now I'm stuck. Any suggestions?

答:
Consider a (nonzero) commutative (unital) ring $R$ with multiplicative identity $1_R$ and multiplicative group of units $R^\times$ with the property that $1_R + a$ is a unit whenever $a$ is a non-unit (i.e., whenever $a$ is in $R \setminus R^\times$). We claim that $R \setminus R^\times$ is an ideal.

Observe that the product $ra$ of any element $r$ of $R$ and any non-unit $a$ must be a non-unit. (Why?) Consequently, it remains to show that $R \setminus R^\times$ is closed under addition.

Given non-units $a$ and $b,$ as you have written, we have that $1_R + a$ and $1_R + b$ are units (by hypothesis). Our aim is to show that $a + b$ is a non-unit. On the contrary, assume that $a + b$ is a unit so that there exists a $u$ such that $1_R = (a + b)u = au + bu.$ By rearranging, we have that $bu = 1_R - au.$ But this implies that $bu$ is a unit so that $b$ is a unit --- a contradiction. (Why?)

Closely related to this post is the Jacobson radical of the ring $R,$ i.e., the ideal $$\operatorname{Jac}(R) = \bigcap \{\mathfrak m \,|\, \mathfrak m \text{ is a maximal ideal of } R\}.$$ One can show that $a$ is in $\operatorname{Jac}(R)$ if and only if $1_R + ra$ is a unit for all $r \in R.$ For your case, the non-units of $R$ belong to a single maximal ideal $\mathfrak m,$ hence we have that $\operatorname{Jac}(R) = \mathfrak m = R \setminus R^\times.$


问:
application on Lebesgue Dominated Convergence Theorem
If $\{f_n\}$ is a sequence of measurable functions on the measurable set $E$ such that $f_n \rightarrow f$ converges pointwise almost everywhere on $E$ where $|f| \in L(E)$ "i.e. $|f|$ is Lebesgue integrable on $E$, say, $\int_{E} |f| = a$ and $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)|= b$.

I need to show that $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)-f(x)|$ exists and to find its value.
I can see that $|f_n(x)-f(x)| \leq  |f_n(x)|+|f(x)|$ so passing the integral over $E$ then taking the limit as $n \rightarrow \infty$ proves that the limit of the integral exists, but I am still not sure how to find the limit value, my guess is to apply the Lebesgue Dominated  Convergence Theorem or the general Lebesgue Dominated  Convergence Theorem.

答:
Here is a hint: apply Fatou's lemma to $|f_n - f| + |f| - |f_n| \ge 0$ to find that
$$b-a \le \liminf \int |f_n - f|.$$
Then apply Fatou's lemma to $|f_n| + |f| - |f_n - f|$ to find in turn that
$$\limsup \int |f_n - f| \le b-a.$$


问:
How to understand the integral of $f(x)$ on the interval $[a,b]$.
If I visualize in integral, I imagine it the way I was always taught: as the area under the curve described by $f(x)$, on some interval $[a,b]$. For this area it should not matter whether we integrate from $a$ to $b$ or from $b$ to $a$, the value should be the same.

However, using the fundamental theorem of calculus,

$\int_{b}^{a}f(x)dx=F(a)-F(b)=-[F(b)-F(a)]=-\int_{a}^{b}f(x)dx$,

with $f(x)=\frac{dF}{dx}$.

Is there some better way to visually or conceptually understand an integral instead of "the area under a function", that makes more sense in this aspect?

答:
In order to appreciate the distinction that you are making, you have to get a little algebraic. The integral is unfortunately not totally understandable in an intuitive and visual way. Loosely speaking, the integral is the limit of rectangles under the graph, so we can look at the approximation using rectangles to see how it affects the resulting integral.

We define integrating from a to b as partitioning the interval $[a,b]$ from left to right, so that $x_{i-1}<x_i$, therefore, that factor of $(x_i-x_{i-1})$ is positive.

If we then look at $b\to a$

using the same reasoning, we see that $(x_i-x_{i-1})$ is negative. Therefore, $f(x)(x_i-x_{i-1})$ has the opposite sign depending on the direction of integration.

So really, the issue is about consistency of definition more than anything. Hope that helps!


答:
These are equivalent definitions of local rings.

For a nonzero commutative ring $R$, we show the equivalence of the following statements:

For every $a \in R \backslash R^\times$, we have $1 + a \in R^\times$.

$R$ has a unique maximal ideal.

If $a, b \in R \backslash R^\times$, then $a + b \in R\backslash R^\times$.

We may prove in the following order: $1 \implies 2 \implies 3 \implies 1$.

$1 \implies 2$: let $I$ be a maximal ideal of $R$. We want to show that any proper ideal $J \subseteq R$ is contained in $I$.

Suppose there is a proper ideal $J$ not contained in $I$. Then the ideal $I + J$ must be $R$, as $I$ is maximal. Thus there exist $i \in I$ and $j \in J$ such that $i + j = 1$. But both $-i$ and $1 + (-i) = j$ are non-units, contradiction.

$2 \implies 3$: both $aR$ and $bR$ are proper ideals of $R$, hence are contained in the unique maximal ideal $M$. Therefore their sum is also in $M$ and is not a unit.

$3 \implies 1$: suppose there exists $a \in R$ such that both $a$ and $1 + a$ are non-units. Then $1 = (1 + a) + (-a)$ is a non-unit, contradiction.

Title: If $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$, then $R-R^*$ is an ideal

Body:
Suppose that $R$ is a nonzero commutative ring with $1$. Suppose $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$. I'd like to understand why this implies that $R-R^*$ is an ideal.
In particular, I'd like to know why the sum of two non-units is a non-unit.

So suppose $a,b$ are non-units. We'd like to show that $a+b\in R-R^*$ as well. From the assumption, we have that $1+a,1+b\in R^*$, and so $(1+a)(1+b)=1+a+b+ab\in R^*$.

But now I'm stuck. Any suggestions?


问:
Spot the error in this induction
I'm learning induction. Spot the error in this induction:

Let $P(n): 1 = 2 = 3 = \dots = n$

It's true for $n = 1$, because $1 = 1$, then let's assume $P(n)$ is true *, so we must show that $P(n+1)$ is true. Summing $1$ on every element we will have: $2 = 3 = 4 \dots = n = n + 1$ **

Then, with * and **, we have that $P(n+1)$ is true.

I'm not enterly sure where's the error, but I think the error is in **, because we don't have $1 = n+1$, or maybe, because at the end we have $n = n+1$ that is false.

Well, if there's any missconcept or anything wrong in my thoughts please let me know.

答: This is a disguised version of the famous "all horses are the same colour" proof. The induction step does not actually go through in every case; in particular, it fails if $n = 2$. As written, you have $1 = 1$, adding $1$ to this equation gives you... $2 = 2$. Not $2 = 1$ like you need to complete the argument.

Title: Spot the error in this induction


问:
What is the probability that the coin was tossed $n$ times
Q: A coin is tossed untill k heads has appeared. If a mathematician knows how many heads appeared, can he figure out what is the probability that the coin was tossed $n$ times?

What I tried: The number of heads debends of the number of tosses. So I tried Bayes' theoream $P(X=k|N=n)P(N=n)/P(X=k)$ where $X$ is random variable for heads and $N$ for tosses. $P(X=k|N=n)$ is binomial distributions mass probability function, but i don't know what $P(N=n)$ and $P(X=k)$ are.

Because the last toss is known to be heads, we only focus on the the tosses before the last one. There is $\binom {n-1}{k-1}$ combinations for $k-1$ heads to appear before the last one. So the probability is

$$P(N=n|X=k) = \binom{n-1}{k-1}p^k(1-p)^{n-k}$$

I believe this can be done also with Bayes Theoream. Can someone wiser show how it is done?

答:
The coin toss process can be described in the context of independent and identically distributed (i.i.d.) Bernoulli trials.

Suppose we define a random variable corresponding to the number of trials required to have $k$ successes (i.e. $k$ heads in this setting). We say $N \sim NegBin(k, p)$ where $p$ is the probability of getting a head.

Sample space: $\{k, k+1, k+2, \dots\}$

Probability mass function (pmf): For $n = k, k+1, k+2, \dots,$
\begin{align*}
f(n; k, p) = P(N = n) &= {n - 1 \choose k - 1} p^{k-1} (1-p)^{n-k} \cdot p\\
&= {n - 1 \choose k - 1} p^k (1-p)^{n-k}.
\end{align*}
Reasoning: If $n$ is the number of coin tosses (when $k$ heads were observed), then the first $(n-1)$ tosses resulted in $(k-1)$ heads and the last toss was a head.

Remark: More information on Negative binomial distribution can be found here. Be careful about the different formulations (see 1.3 Alternative formulations).


问:
Show that $G$ satisfies Hall's condition

Show that $G$ satisfies Hall's condition

I knew the definition:

Let $G$ be a bipartite graph with partite sets $U$ and $W$, where $r=|U|\leq|W|$. Then $G$ satisfies Hall's condition if, for every $S\subseteq U$, $|S|\leq|N(S)|$.

But did I need to check it for all subsets manually to answer that question? Because I easily found a perfect match. And Hall condition implies the existence of a perfect match, but I don't know the reverse is true or not? Then how to deal with this question without huge amount of checking?

答: Well, one direction of Hall's Theorem is easy to see: If a bipartite graph $G$ has a perfect matching, then $G$ satisfies Hall's Condition. [The other direction: If $G$ satisfies Hall's Condition, then $G$ has a perfect matching, is the harder direction to see.] However here $G$ does have a perfect matching $M$; in fact $M$ is as follows: $M=\{1a, 2c,3b,4d,5f,6e\}$.


问:
Given a function $u$, how to find an orthogonal matrix $A$ such that each component of $A \nabla u$ has same $L^2$ norm?
Let $u$ be a differentiable function defined in a bounded domain $\Omega\subset \mathbb{R}^n$, with $\nabla u \in L^2(\Omega)$. Show that there is an orthogonal matrix $A$ such that for any $1 \le i \le n$, we have $$\Vert(row_iA) \cdot \nabla u\Vert_{L^2(\Omega)}=\frac{1}{\sqrt{n}}\Vert \nabla u \Vert_{L^2(\Omega)}.$$

In two dimensions, this can be easily proved by setting $A$ to be a rotation matrix with parameter $\theta$, which is solvable due to the above equation. In three dimensions, I cannot find an elegant argument to prove the above statement. Any ideas or comments are really appreciated.

答:
Up to notational differences, your question is summed up in the following:

Proposition. Let $E$ be an $n$-dimensional linear space with scalar product, $(\Omega,\mu)$ a measure space, and $V \colon \Omega \to E$ a function with $\int |V|^2 d\mu = 1$. Then there is an orthonormal basis $e_1,\ldots,e_n$ of $E$ for which
$$
\| V \cdot e_1 \|^2_{L^2} = \| V \cdot e_2 \|^2_{L^2} = \ldots = \| V \cdot e_n \|^2_{L^2} = \frac 1n.
$$

To see this, just take $V$ to be the gradient $\nabla u$ normalized by its $L^2$-norm, and $A$ to be the related change-of-basis matrix.

$\newcommand{\R}{\mathbb{R}}$

Proof. To rephrase the statement, let me introduce the operator
$$
T(e_1,\ldots,e_n) = \left( \| V \cdot e_1 \|^2_{L^2}, \| V \cdot e_2 \|^2_{L^2}, \ldots, \| V \cdot e_n \|^2_{L^2} \right) \in \R^{n}.
$$
More precisely, $T$ maps any orthonormal basis into the $(n-1)$-dimensional simplex
$$
\Delta^{n-1} = \{ x \in \R^n : x_1,\ldots,x_n \ge 0, \ x_1+\ldots+x_n = 1 \}.
$$
Our claim is that $T$ admits the point $p_0 := (\frac 1n,\ldots,\frac 1n)$ as its value.

For $n=2$, this is easily seen by fixing an arbitrary orthonormal basis $e_1,e_2$ and rotating both vectors until we obtain $e_2,-e_1$. Since $\Delta^1$ is just a segment and points $T(e_1,e_2)$ and $T(e_2,-e_1)$ lie on the opposite sides of $p_0$, by continuity the value $p_0$ is also achieved.

For higher dimensions, let us assume that $T$ doesn't take the value $p_0$. Now the domain of $T$ is the set of all orthonormal bases, which can be identified with $O(n)$ (space of all orthogonal transformations). Thus, the domain is a compact set and of course $T$ is continuous, which means that the function
$$
(e_1,\ldots,e_n) \mapsto |T(e_1,\ldots,e_n)-p_0|^2 \in \R
$$
achieves a minimum. Let us consider such minimal basis $e_1,\ldots,e_n$. Of course, $T(e_1,\ldots,e_n) \neq p_0$, and so $\| V \cdot e_i \|^2_{L^2} \neq \| V \cdot e_j \|^2_{L^2}$ for some $i \neq j$. Applying the $2$-dimensional case, we can replace $e_i,e_j$ by an orthonormal basis $\hat{e}_i,\hat{e}_j$ of the same subspace $\operatorname{span}(e_i,e_j)$, satisfying
$$
\| V \cdot \hat{e}_i \|^2_{L^2} + \| V \cdot \hat{e}_j \|^2_{L^2}
= \| V \cdot e_i \|^2_{L^2} + \| V \cdot e_j \|^2_{L^2},
\qquad
\| V \cdot \hat{e}_i \|^2_{L^2} = \| V \cdot \hat{e}_j \|^2_{L^2}.
$$
By strict convexity of the function $x \mapsto (x-\frac 1n)^2$, the pair $(\| V \cdot \hat{e}_i \|^2_{L^2},\| V \cdot \hat{e}_j \|^2_{L^2} )$ is strictly closer to $(\frac 1n,\frac 1n)$ than the pair $(\| V \cdot e_i \|^2_{L^2},\| V \cdot e_j \|^2_{L^2} )$. Considering the whole basis of $E$ with $e_i,e_j$ replaced by $\hat{e}_i,\hat{e}_j$, we obtain a contradiction with minimality of $e_1,\ldots,e_n$.


问:
How do I simplify $8t^7 \left(\sqrt{t }-9\right) + t^8 \left(\frac{1}{2\sqrt{t}}\right)$?
I'm not sure how to go about doing this, expecting a detailed explanation: I already know what the answer should look like after seeing the solution for a similar problem on MyMathLab but I don't know how to reach that form.
$$8t^7 \left(\sqrt{t}-9\right) + t^8 \left(\frac{1}{2\sqrt{t}}\right)$$

The only thing I've done is just distribute: however, that's not the solution the website will accept. After pressing similiar problem on MyMathLab it showed this as the solution's format:

Ignore the derivative part, this step is mostly just all algebra.

答:
It's a bit unclear to me exactly what is causing the problem. But I am assuming it's the square roots? You can rewrite them as normal exponents.

$$
\sqrt{t} = t^{\frac{1}{2}}
$$
And
$$
\frac{1}{2\sqrt{t}} \;=\; \frac{1}{2}\cdot\frac{1}{\sqrt{t}} \;=\; \frac{1}{2}\cdot\frac{1}{t^{\frac{1}{2}}} \;=\; \frac{1}{2}\cdot t^{-\frac{1}{2}}
$$


问:
Are binomial coefficients defined for $k \in \mathbb{R}$
So I've been trying to learn some basic precalc and I got stumped on definition of combinations namely the following confuses me .

I know that The symbol ${n\choose k}$ is read as "$n$ choose $k$." It represents the number of ways to choose $k$ objects from a set of $n$ objects. However in most definitions no restriction on the value of k is made for example what would ${5\choose 2.5}$ be.

furthermore what do we define ${n\choose k}$ if $k<0$

Some help would be greatly appreciated.

Thanks in advance.

答:
The function $\binom{n}{k}$ can be extended to all real (and even complex) $n,k$ by using the Gamma function in place of factorials:
$$
\binom{n}{k}:=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}
$$
This suffices to define $\binom{n}k$ except when $k$ is a negative integer or $n-k$ is a negative integer, since $\Gamma$ has a pole at all non-positive integers. We can fix some of these exceptional cases by using a limit:
$$
\binom{n}{k}:=\lim_{(x,y)\to (n,k)}\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}
$$
It turns out that this limit exists, and is zero, if at least one of the following is true:

$k$ is a negative integer, and $n$ is not a negative integer.

$n-k$ is a negative integer, and $n$ is not a negative integer.

$k$ and $n-k$ are both negative integers.

This only leaves the following two cases, where $\binom{n}k$ has a non-removable discontinuity:

$k$ and $n$ are negative integers, but $n-k$ is a nonnegative integer.

$n-k$ and $n$ are negative integers, but $k$ is a nonnegative integer.

In these cases, we can instead define $\binom{n}{k}$ to be the limit as $(x,y)\to (n,k)$ along a path of a certain slope. That is, we pick a certain real number $\theta$, and define
$$
\def\ep{\varepsilon}
\binom{n}{k}
:=\lim_{\ep\to 0}\frac{\Gamma(n+1+\ep)}{\Gamma(k+1+\theta \ep)\Gamma(n-k+1+(1-\theta)\ep)}
$$
Any choice of $\theta$ leads to a valid extension of the binomial coefficients to the entire real plane with nice properties. Namely, Pascal's rule $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ is now true for all real $n$ and $k$.

The usual choice is $\theta=0$. Then $\binom{n}k$ is zero in case $1$, and is $\frac{n(n-1)\cdots(n-k+1)}{k!}$ in case $2$. With this convention, we have the nice property that $(1+x)^n=\sum_{k= 0}^\infty\binom{n}k x^k$ holds for all complex $n$.

Another nice choice is $\theta=1$, where $\binom{n}k$ is $\frac{n(n-1)\cdots(k+1)}{(n-k)!}$ in case $1$, and is zero in case $2$.

The only choice of $\theta$ which preserves the symmetry property $\binom{n}k=\binom{n}{n-k}$ for all real $n,k$ is $\theta=\frac12$, in which case $\binom{n}k$ is the average of the two previous definitions. In general, different values of $\theta$ lead to weighted averages of the last two cases.


问:
Cardinal of the Special Orthogonal Group $SO_n\left(\mathbb{Z}/2^n\mathbb{Z}\right)$
Let $SO_n(R)$ be the group of matrices with coefficients in a ring $R$ whose determinant is equal to $1$.

I remembered an old question asked at an oral exam :

$$\text{What is the order of the group }SO_n(\mathbb{Z}/2^n\mathbb{Z}) \ ?$$

Is there a way to count them ? I don't see a way to solve this question

答:
Let us count the cardinality $c_n(m) $ of $C_n(m):= SO_m(\mathbb{Z}/2^n\mathbb{Z}) $ by induction on $m$. The base case is easy, since evidently $c_n(1) = 1$. Now consider the action of $C_n(m)$ on  $(\mathbb{Z}/2^n\mathbb{Z}) ^m$ for $m\ge 2$.

The key observation is that the stabilizer of $e_1$ is precisely $C_n(m-1) $. Indeed, you have an orthogonal matrix with first column $(1 \ 0 \ldots \ 0) $, so all the other columns must start with zero and the determinant of the $(m-1) \times (m-1) $ matrix left in the bottom-right must be one.

The orbit of $e_1$ under $SO_m$ are all the vectors
$$(x_1, \ldots, x_m) \in (\mathbb{Z}/2^n\mathbb{Z}) ^m \ : x_1^2+ \ldots + x_m^2 = 1 $$
but up to a square root of 1 in $\mathbb{Z}/2^n\mathbb{Z}$. Let us denote this set by $X_2(n)$. Indeed, $O_m(\mathbb{Z}/2^n \mathbb{Z}) = SO_m(\mathbb{Z}/2^n \mathbb{Z}) \cdot X_2(n) $, so that the orbit under $O_m$ will be the orbit under $SO_m$ times the orbit under scalar multiplication of $X_2(n) $. I don't have any effective ideas at the moment to compute $O_n(\mathbb{Z}/2^n\mathbb{Z}) e_1$, so let's call it $K_{n, m}$. The other involved quantity - $X_2(n) $, is easy to compute: recall that the multiplicative group of $ \mathbb{Z}/2^n \mathbb{Z}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z}) ^{n-2} $ for $n\ge 2$, so that the elements of order 2 are 4 if $n\ge 3$, and $n$ if $n\le 2$.

Since the orbit and the stabilizer have product equal to the cardinality of the group, we have that

$$ c_n(m) = c_n(m-1) \cdot K_{n, m} /x_2(n) $$

This implies that

$$ c_n(m) =\left (  \prod_{r=2}^m K_{n, r} \right ) / (x_2(n) ) ^{m-1}$$

A special case. In the case of $n=1$ the sphere equation is particularly simple, as it would be in any field of characteristics 2. Indeed the square commutes with the sum and we have

$$ (x_1+\ldots+x_m) ^2 = 1$$

which implies that the sum must be an element of $X_2(1) =\{1\}$. If you choose the first $m-1$ elements as you wish in 0,1, you can choose in the right way $x_m$ in a unique way so that the equation holds true. This shows that $K_{n, m}= 2^{m-1} x_2(1) $, so that

$$ c_1(m) =  \prod_{r=2}^m K_{1, r} / (x_2(1) ) ^{m-1} = \prod_{r=2}^m 2^{r-1} = \boxed{ 2^{\binom{m}{2} }}$$

Edit. The last observation makes me suspect that the original question was in the field of characteristics 2 $F_{2^n}$, for which the last result hold but with exponent $n\binom{m}{2} $ (the argument is pretty the same!).


答:
(1) The proof does not assume $P$ is infinite, an explicit upper bound is simply omitted.

(1*) To prove not one but two claims, we choose to prove something equivalent, and we rephrase the problem of proving them both as follows: "either the sum diverges hence primes are infinite, or else we may assume the sum converges and we arrive at a contradiction." It is necessary to set the stage this way so that we are proving something logically equivalent to "(1) and (2)," and the first case being trivial is an artifact of this reformulation.

(2) If $X\to Y$ is surjective then $|X|\ge|Y|$. We have a surjection

$$\overbrace{\{n\le N~{\rm sqrfree~only~divisible~by~small~primes}\}}^{\Large{\rm size~}S\,\le\, 2^k}\times\{1,2,\cdots,\lfloor \sqrt{N}\rfloor\}$$

$$\longrightarrow\underbrace{\{n\le N~{\rm only~divisible~by~small~primes}\}}_{\Large {\rm size}\,=\,N_s}:(s,x)\mapsto sx^2 $$

and hence $2^k\sqrt{N}\ge S \lfloor\sqrt{N}\rfloor\ge N_s$.

(3) Suppose $2^k\sqrt{N}\le \frac{N}{2}$ for some $N$ (choosing $N=2^{2k+2}$ works). Then

$$N=\underbrace{N_s}_{\le N/2}+\underbrace{N_b}_{<N/2}<\frac{N}{2}+\frac{N}{2}=N,$$

i.e. $N<N$, a contradiction.

(4) Instead of there being one overarching insight (other than "count a quantity in two ways that disagree"), the process behind this proof seems to be developed in a number of distinct stages, and each stage progresses to the next one by asking an isolated but natural question to which there are many possible answers but the right configuration of answers yields a complete proof. (Like choosing doors or turns in a maze.) This is very much dependent on guesswork, hindsight, and discarding routes that didn't seem to lead anywhere.

Title: Need help understanding Erdős' proof about divergence of $\sum\frac1p$

Body:
I'm looking at proofs from Proofs from the Book (Martin Aigner, Günter M. Ziegler). The proof I'm having trouble is the sixth proof of the infinitude of the primes they give (on page 5; although I'll be reconstructing it here). I'll be highlighting the areas I'm having trouble with. They set out to prove two things; namely, and letting $P$ be the set of all primes:

$P$ is infinite.

$\sum_{p \in P} {1 \over p}$ diverges.

They begin by assuming the sum in 2. is convergent and letting $P = \{p_1, p_2, ...\}$ The first problem comes up here. By assuming the given form for $P$, haven't they already assumed that it is infinite? Then, by assumption,

$$\sum_{i > k} {N \over p_i} < {N \over 2}, \ \  (N \in \mathbb{Z}^+). \tag{1}$$

They then define $p_1, p_2, ..., p_k$ to be the small primes, and $p_{k + 1}, p_{k + 2}, ...$ to be the large primes, and define $N_b$ to be the number of positive integers $n \leq N$ which are divisible by at least one of the big primes; and, similarly, $N_s$ the number of positive integers $n \leq N$ which only have small prime divisors. By definition, $N_b + N_s = N$, and the intended contradiction will hinge on showing $N_b + N_s < N$.

Note, since $\left \lfloor {N \over p_i} \right \rfloor$ gives the number of positive integers $n \leq N$ which are multiples of $p_i$,

$$N_b \leq \sum_{i > k} \left \lfloor {N \over p_i} \right \rfloor < {N \over 2}. \tag{2}$$

Then, they let $n = a_n b_n^2$, where $n \leq N$ such that it has only small prime divisors. Here $a_n$ is the square free part. Then, every $a_n$ is a product of different small primes. Assuming that by this it is simply meant that every factor of $a_n$ is distinct, they assert there are exactly $2^k$ different square free parts. Notice, also, $b_n \leq \sqrt{n} \leq \sqrt{N} \implies b_n \leq \sqrt{N}$, and as such,

$$N_s \leq 2^k \sqrt{N}. \tag{3}$$

Here, is the second problem for me. I can see intuitively how $(3)$ holds; but, I'm not really sure, and I'd like to see it through a more rigorous approach, and also, why it's relevant. I'm not really sure, and I'd like a more rigorous approach.

They conclude the proof by showing that, since $(2)$ holds for any $N$, $2^k \sqrt{N} \leq {N \over 2}$ holds for $N = 2^{2k + 2}$. This is the third area I'm having trouble. I'm not really sure why this is relevant and how the contradiction $N_s + N_b < N$ follows from it. Maybe I missed it, but I also didn't see a proof of the infinitude of primes here.

Finally, I have a question as well. Looking back, how could one have guessed that this approach will lead us to a proof of the two claims above? In other words, what's the insight behind taking this particular approach?

Thanks to all in advance!

Note: I wasn't really sure what tags relating to "Proofs" would be appropriate here, so I've only used tags concerning number theory and primes. Please feel free to tag away.


问:
Proof that the dimension of a matrix row space is equal to the dimension of its column space
I have the following theorem:

Theorem 3.12. Let A be an m  n matrix. Then the dimension of its row
space is equal to the dimension of its column space.

And the following proof is given:

Proof. Suppose that $\lbrace v_1,v_2,\dots,v_k\rbrace$ is a basis for the column space of $A$.  Then each column of $A$ can be expressed as a linear combination of these vectors; suppose that the $i$-th column $c_i$ is given by $$c_i = \gamma_{1i}v_1+\gamma_{2i}v_2+\dots+\gamma_{ki}v_k$$ Now form two matrices as follows: $B$ is an $m\times k$ matrix whose columns are the basis vectors $v_i$, while $C=(\gamma_{ij})$ is a $k\times n$ matrix whose $i$-th column contains the coefficients $\gamma_{1i},\gamma_{2i,}\dots,\gamma_{ki}$.  It then follows$^7$ that $A=BC$.

However, we can also view the product $A= BC$ as expressing the rows of $A$ as a linear combination of the rows of $C$ with the $i$-th row of $B$ giving the coefficients for the linear combination that determines the $i$-th row of $A$.  Therefore, the rows of $C$ are a spanning set for the row space of $A$, and so the dimension of the row space of $A$ is at most $k$.  We conclude that: $$\dim(\operatorname{rowsp}(A))\leq\dim(\operatorname{colsp}(A))$$
  Applying the same argument to $A^t$, we conclude that:$$\dim(\operatorname{colsp}(A))\leq\dim(\operatorname{rowsp}(A))$$and hence these values are equal

However, I am finding this proof impossible to follow and understand. Can someone please offer an alternative proof or explain what this proof is saying?

Thank you.

答: The long proof by Anakhand, even though perfectly correct, is harder to understand than the proof given in the original question. This is because it gets to the nitty-gritty details of matrix multiplication. Matrix multiplication can be understood in several different ways. One of them is the fact that if a matrix is right-multiplied by a column vector, then the output is a linear combination of columns of the matrix where the coefficients are the vector's coordinates (this is crucial to understanding the proof in the question and is a simple observation). So, if one multiplies 2 matrices, the output is a matrix where each column of the output matrix is a linear combination of the columns of the first matrix and the coefficients are the coordinates of the corresponding vector in the second matrix. On the other hand, if you left-multiply a matrix by a row vector, then the output is a row vector which is a linear combination of rows in the matrix. Similarly when you multiply 2 matrices. If one understands this, the proof in the question becomes much clearer than the proof that goes into the nitty-gritty details of matrix multiplication.</p>


问:
Simply connected implies isomorphism of Lie algebras.
Let $SL(2,C)$  be a special linear group and $SO(1,3)$ the special orthogonal group. We can show that $SL(2,C)$ is a double cover of $SO(1,3)$. Now the book I am reading says that  since $SL(2,C)$ is simple connected their Lie algebras are isomorphic. Why is that simple connectedness implies Lie algebra isomorphism?

答:
The covering map $\pi\colon SL(2,\Bbb C)\longrightarrow SO(1,3)$ induces a Lie algebra homomorphism $\pi^*\colon\mathfrak{sl}(2,\Bbb C)\longrightarrow\mathfrak{so}(1,3)$. Since $\pi$ is surjective, $\pi^*$ is surjective. And since $\ker\pi$ is discrete, $\ker\pi^*=\{0\}$. So, $\pi^*$ is an isomorphism.

You can also say that, since $\pi$ is a covering map, there are neighborhoods $U$ and $V$ of the identity elements of $SL(2,\Bbb C)$ and of $SO(1,3)$ respectively such that the restriction of $\pi$ to $U$ is a diffeomorphism from $U$ onto $V$. But then the derivative of $\pi$ at the identity element of $SL(2,\Bbb C)$ is a vector space isomorphism. But this derivative is precisely $\pi^*$.


问:
What is an epsilon function?
I have ran into it a few time in my books and it seems to be a function that tends  to $0$ as its argument tends to $0$. They are all written in the same notation  ($\varepsilon(x)$) and we do not use indices as all the epsilon functions behave the same way. Is it related to O-notation?

I could not find any useful sources.

A example can be seen in the comments

答:
Most likely it is little-$o$ notation. For example, in John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 69 you can find definition for differentiation for function of one real variable:
$$\Delta y=\frac{dy}{dx}\Delta x + \varepsilon \Delta x$$
where $\varepsilon$ is infinitesimal, as wrote author, i.e. $\varepsilon(\Delta x)\to 0$, when $\Delta x \to 0$.

Farther, in same book, in page 267 for definition for differentiation for multiple variables is used formula:
$$\Delta u = \frac{\partial u}{\partial x}\Delta x+\frac{\partial u}{\partial y}\Delta y +\varepsilon_1 \Delta x + \varepsilon_2 \Delta y$$
where again $\varepsilon_1, \varepsilon_2$ are infinitesimals i.e. tending toward zero as $\Delta x$ and $\Delta y$ tend toward zero.

At end let me bring one version of definition little-$o$:
$$o(f),x\to x_0=\{g: \exists \varepsilon(x), \lim\limits_{x\to x_0} \varepsilon(x) = 0, \exists U_\delta(x_0), \forall x \in U_\delta(x_0), g(x)=f(x) \varepsilon(x) \}$$
where $U_\delta(x_0)$ is neighbourhood of $x_0$. This definition works as for single, so for multiple, real variables.

Title: What is an epsilon function?


问:
Motivation for the definition of quotient space
I'm having some trouble understanding the concept of quotient space of a vector space. The definition my teacher gave us is the following:

Let $V$ be a finite-dimensional vector space over a field $\mathbb K$ and $V' \leq V$.

We can define the following equivalence relation: for any $u,v \in V, u \simeq v$ iff $v -u\in V'$. Let, for any $u\in V$, $[u]=\{v \in V: u \simeq v\}$.

Then we can define: $$V/V':=\{[u],u\in V\}$$

In the first question, my teacher asked us to determine the vector space $\mathbb R^2 / span(1,2)$ and I have no clue how to proceed because I'm still confused about the definition. Is there any intuition on why we define it this way or the meaning of the quotient space? And, how can I determine the space $\mathbb R^2 / span(1,2)$?

答:
The integers modulo a number are defined the same way.  The integers $x$ and $y$ are congruent modulo $4$ if $x - y \in 4 \Bbb{Z}$, that is, if $x-y$ is an integral multiple of $4$.

This definition is really saying this:

let there be a large space, $V$ (the integers in modular arithmetic);

in $V$, let there be a small space $V'$ (the multiples of $4$ in the modular arithmetic example), all of whose elements we view as equivalent to zero in the quotient, "$V/V'$" (or $\Bbb{Z}/4\Bbb{Z}$);

we treat two elements of $V$ as equivalent in the quotient if they differ by an element of $V'$ (a multiple of $4$).

So, in the problem you are given, $V = \Bbb{R}^2$ and $V' = \mathrm{span}\left( (1,2) \right)$.  We check that $\mathrm{span}\left( (1,2) \right)$ is a subspace of $V$, which it is.  We want to treat everything in $V'$ is equivalent to $0$.  (All real multiples of $(1,2)$ are equivalent to $0(1,2) = \vec{0}$.)  One way to draw a sketch of what is going on is the line spanned by $(1,2)$ is being crushed onto the origin of $\Bbb{R}^2$ and every parallel line is being crushed the same way, leaving a line perpendicular to $(1,2)$.

Let's check that.  What is $(2,3)$ equivalent to in the quotient?  $(2,3)$ is equivalent to $y \in \Bbb{R}^2$ if $(2,3) - y \in \mathrm{span}\left( (1,2) \right)$.  There are infinitely many solutions, but a very, very easy one to find is $(2,3) - (1,1) = (1,2) \in \mathrm{span}\left( (1,2) \right)$.

So $(2,3)$ and $(1,1)$ are equivalent in this quotient.  How can we find everything that is equivalent to $(2,3)$ in this quotient?  Notice that every element of $\mathrm{span}\left( (1,2) \right)$ is a $\Bbb{R}$-multiple of $(1,2)$, so
$$  \mathrm{span}\left( (1,2) \right) = \{r(1,2) \mid r \in \Bbb{R} \}  \text{.}  $$
So, we ask for $y = (y_1,y_2)$ such that
$$  (2,3) - (y_1, y_2) = r(1,2)  $$
for some real $r$.  But this is an easy vector equation.  Taken in components, it is the system
\begin{align*}
2 - y_1 &= r  \\
3 - y_2 &= 2r \text{.}
\end{align*}
Eliminating $r$ between these and simplifying a little, we obtain
$$  2y_1 - 1 = y_2  \text{.}  $$
This means every vector $(y_1, 2y_1 - 1)$ is equivalent to $(2,3)$ in this quotient.

(As a quick check, $(2,3)$ should be a vector in the set of equivalences generated as $y_1$ ranges through the reals.  This does happen -- when $y_1 = 2$, $2y_1 - 1 = 3$.)

Of course, the set $\{(y_1, 2y_1 - 1) \mid y_1 \in \Bbb{R}\}$ is a line, so an entire line is equivalent to $(2,3)$.

If you do more examples, you will find that each set of equivalent vectors lie along a line and all of the lines are parallel to the vector $(1,2)$.  Now think of the subspace spanned by $(2,-1)$.  This subspace is perpendicular to $(1,2)$ and no two vectors in $\mathrm{span}\left( (2,-1) \right)$ are equivalent in the quotient.  How would we show that?  We use the definition: two points in that span are equivalent if their difference is in $V'$, i.e., there are reals $r$ and $s$ such that there is a real $t$ where
$$  r(2,-1) - s(2,-1) = t(1,2)  \text{.}  $$
As a system of equations among components,
\begin{align*}
2(r-s) &= t  \\
-1(r-s) &= 2t  \text{.}
\end{align*}
Using the first to replace $t$ in the second,
$$  s - r = 2(2(r-s))  \text{,}  $$
which simplifies to
$$  5s = 5r  \text{.}  $$
So two points in the span of $(2,-1)$ are equivalent only when $r = s$, so when they are both the same point.  Further, when $r = s$,
$$  r(2,-1) - s(2,-1) = t(1,2)  $$
reduces to $0 = t$, so the one point that we could be talking about is the origin.  And, yeah, the origin is equivalent to itself in every quotient and the origin is in every subspace.


Title: Taylor series of $\log x$ at $x=2$

Body:
hint

Let $ f(x)=\ln(x)$.

$$g(x)=\ln(x+2)=\ln(2(1+\frac x2))$$
$$=\ln(2)+\ln(1+\frac x2)$$

$$=\ln(2)+\frac x2-\frac{(\frac x2)^2}{2}+\frac{(\frac x2)^3}{3}-...$$

thus

$$f(x)=g(x-2)$$
$$=\ln(2)+\frac{x-2}{2}-...$$


问:
Showing the partial derivative of Lagrangian
I got the Lagrangian equation
$$L(q,\dot{q})=T-V=\frac{1}2m|(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})|^2-V(\textbf{Rq})$$

where S and R are 3x3 matrices. I have to show that

$$\frac{\partial L(\textbf{q},\dot{\textbf{q}})}{\partial \dot{\textbf{q}}}=m(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})$$
$$\frac{\partial L(\textbf{q},\dot{\textbf{q}})}{\partial \textbf{q}}=-m\textbf{S}(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})-\frac{\partial}{\partial \textbf{q}}V(\textbf{Rq})$$

I am told that it might be convenient to go into coordinates, i.e., work out the k'th Euler-Lagrange equation.

In the first partial derivative it seems that $-V(\textbf{Rq})$ disapears when we take the partial derivative with respect to $\dot{\textbf{q}}$ and i would have guessed that we had to use the chain rule for the squared part but it seems that isnt necessary?

For the second part I cant wrap my head around how you get $-m\textbf{S}(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})$

I also have to show that the Euler-Lagrange equations consequently becomes

$$m(\ddot{\textbf{q}}+2\textbf{S}(\omega)\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{S}(\omega)\textbf{q}+\textbf{S}(\dot{\omega})\textbf{q})+\frac{\partial}{\partial \textbf{q}}V(\textbf{Rq})=0$$

Can anyone help me out here?

Thanks

EDIT
Guy3141 came with a super answer so the question is already partly solved, but..
When I doint the partial derivative for q I get

$$\frac{\partial L(\textbf{q},\dot{\textbf{q}})}{\partial \textbf{q}}=m\textbf{S}(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})-\frac{\partial}{\partial \textbf{q}}V(\textbf{Rq})$$

and I dont get the minus sign in front of the right hand side. Can anyone explain where the minus gets into the equation?

EDIT EDIT
I've been told to write everything in coordinates so that $q=(q_1,q_2,q_3)$ should do the trick and show why the minus sign is there, but I still don't see where the minus sign comes from :( Anyone able to help here?

Thanks

答:
How about we write $$|\dot{\mathbf{q}}+\mathbf{S}q|^2=\langle\dot{\mathbf{q}}+\mathbf{S}q,\dot{\mathbf{q}}+\mathbf{S}q \rangle$$
and use rule for differentiating an inner product.


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
A few steps should allow to determine the surface of a given egg

Take a measuring cup as accurate as possible.

Pour some water in the cup, say half the cup: note the current volume $v_1$

Put the egg inside the cup and it should sink

Note the new volume $v_2$, where $v_2 - v_1$ gives the volume of the egg

consider that to be the equivalent volume of a sphere, where $$v_2 - v_1 = \frac{4}{3}\pi r^3$$ and get $r$ from that.

the surface is $4\pi r^2$


问:
Problem from Zorich's book
Choose real numbers $a$ and $b$ so that the function $f(x)=\cos x - \dfrac{1+ax^2}{1+bx^2}$ is an infinitesimal of highest possible order as $x\to 0$.

My solution: Actually this problem is not so difficult and I guess that I've almost solved it. We'll split the solution into some cases:

If $b=0$ and using Taylor's series one can write $f(x)=x^2(-1/2-a)+o(x^3)$ as $x\to 0$.

So we see that if $a=-1/2$ then $f(x)=o(x^3)$. If $a\neq -1/2$ then $f(x)=o(x)$ as $x\to 0$.

Suppose that $b\neq 0$ then again using Taylor's series one can derive that $$f(x)=\cos x-(1+ax^2)(1+bx^2)^{-1}=x^2\left(-\frac{1}{2}-a+b\right)+x^4\left(\frac{1}{24}-b^2+ab\right)+x^6\left(-\frac{1}{720}-ab^2+b^3\right)+o(x^7).$$

One can show that if $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ then $f(x)=o(x^5)$ as $x\to 0$.

Definition: If $f=o(g)$ and g is itself infinitesimal as $x\to 0$, we say that $f$ is an infinitesimal of higher order than $g$ as $x\to 0$.

Question: Can anyone explain in details which one has the highest possible order? I guess it should be in the case when $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ but cannot explain it in a rigorous way. If someone can explain it in a great detail that would be awesome!

答: You've chosen the unique $a$ and $b$ to make the coefficients of $x^2$ and $x^4$ vanish (I presume — I haven't checked) in the Taylor expansion. If you make any other choices, you'll have $f(x) = O(x^2)$ or $f(x)=O(x^4)$ instead. Getting $f(x)=O(x^6)$ is the optimal result.

Title: Problem from Zorich's book