答:
The approach for (a) should work, but you have forgotten to subtract cases where there are two different repeated letters. There are three patterns for this: AABB, ABAB, ABBA. For each there are $5\times 4$ choices for the letters, so this is an extra $60$ to subtract.

The problems with the first attempt for (b) aren't quite as the other answer suggests. You do actually count AABB type possibilities, since case 1 counts cases where the first two are equal and different to the third, and the fourth can be anything (equal to the first or third or neither). Then case 2 counts cases where the middle two are equal, and the first and fourth are not the same as the middle two (but can be the same as each other). Case 3 seems to be the same as case 1 in reverse, in which case you have actually now counted AABB twice.

What's missing are the cases where there are more than two consecutive equal letters: $(5\times1\times 1\times 4)+(5\times 4\times1\times 1)+(5\times1\times1\times1)=45$.

Together with the fact that you have overcounted AABB-type ($5\times 1\times 4\times 1=20$) this should give the same answer as your other approach.

Title: Four-letter word contains no two consecutive equal letters.

Body:
This is taken from the book on Combinatorics, by Daniel Marcus.

Request vetting:

A16: Find the probability that a four-letter word that uses letters from $A,B,C,D,E$ contains (a) no repeated letters; (b) no two consecutive equal letters.

(a)
Total (Inclusion) ways to form four-letter word from five letters: $5^4$.

Ways to have repetitions of $A$:

Two $A$: Choose two positions for $A$ in $\binom{4}2= 6$ ways. Rest two positions can be filled in : $4\times 3= 12$ ways. So, by product rule: $72$ words. Also, the same for other $4$ letters, leading to $72\times 5= 360$ words.

Three $A$: Choose three positions for $A$ in $\binom{4}3= 4$ ways. Rest one position can be filled in : $4$ ways. So, by product rule: $16$ words. Also, the same for other $4$ letters, leading to $16\times 5= 80$ words.

Four $A$: Choose all four positions for $A$ in $\binom{4}4= 1$ ways. Also, the same for other $4$ letters, leading to $5$ words.

Sum of above three cases:
$360+80+5= 445.$

Left are: $5^4 - 445= 180.$

$p= \frac{180}{625}$

(b) no two consecutive same letters' case.

Approach#1:

To get, need subtract from $5^4$  the chances of having two consecutive same letters. This is divided into cases:

Case 1: Position #1,2 are the same, and position #4 can match too :  $5\times 1\times 4\times 5= 100$ cases.

Case 2: position #2,3 are the same: $5\times 4\times 1 \times 4= 80$ cases.

Case 3: Position #3,4 are the same, and position #1 can match too:
$5\times 4\times 1\times 5=100$

So, get : $625- 280= 345$ cases.

Approach#2:

Alternatively, can 'directly' calculate by having $5$ choices for the first position, then $4$ choices, then again need eliminate one choice (of the second element) to have $4$ choices for the third position, and same for the fourth position. This leads to: $5\times 4^3= 320$ choices possible.

The second approach seems correct, but not clear why first approach is faulty.

Seems there are too many factors to consider in the first approach & hence error-prone (to miss some case(s) as here). So, second approach seems better.


问:
What explains the bizarre behavior of the partial sums of this series?
In an attempt help answer the question of a fellow member, I recently tried to evaluate the integral

$$\int_{0}^{2}\frac{1-\cos(y^5)}{y^3}\text{ }dy$$

I figured I was unlikely to find a closed-form expression for the integral using only basic integration techniques, so I instead opted to find a series expression for it (maybe I can manipulate the series into a more familiar one?). Using the Maclaurin series of $\cos$ and some algebraic manipulation, I concluded that

\begin{align*}
\frac{1-\cos(y^5)}{y^3} &= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{y^{10n-3}}{(2n)!}\\
&= \frac{y^7}{2!}-\frac{y^{17}}{4!}+\frac{y^{27}}{6!}-\frac{y^{37}}{8!}+\cdots
\end{align*}

and thus that

\begin{align*}
\int_{0}^{2}\frac{1-\cos(y^5)}{y^3}\text{ }dy &= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{2^{10n-2}}{(2n)!(10n-2)}\\
&= \frac{1}{8}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1024^n}{(2n)!(5n-1)}
\end{align*}

Not the prettiest series I've seen, but definitely not the ugliest!

Not seeing any tricks I could do to manipulate the series into one I'm familiar with, I figured the closed-form I was hoping for was out of the equation, so I settled with estimating the integral using the series I derived.

This is where things get interesting. Using Desmos for computation, I noticed that the partial sums $\{s_k\}$ quickly reached the billions in absolute value (according to the site, this first occurs at $k=11$), so I assumed I ran into a convergence issue. However, after raising the upper limit for $k$ to $50$ and varying $k$, I noticed that for $k\geq 19$, the partial sums quickly decreased in absolute value until $k=39$, where $s_k$ finally started approximating the integral rather well (the integral is approximately $0.47659$).

As you can probably guess, I'd like to know why this is happening. Most of the series I've worked with only need a few terms to approximate their sum with an error of at most $1$, even $0.1$ at times. This series isn't that much different than them, so I find it odd for it to behave so erratically and take so long to settle. Any assistance is appreciated.

答:
Since it is an alternating series, you want to know $n$ such that$$\frac 18\frac{1024^n}{(2n)!\,(5n-1)}\leq 10^{-k}$$ that is to say
$$(2n)!\,(5n-1)- \frac 18 1024^n\, 10^{k}\geq 0 \tag 1$$
Taking logarithms, using Stirling approximation and continuing with Taylor series
$$2n(\log(n)-a)+\frac32 \log(n)+b+O\left(\frac{1}{n}\right)=0$$

where $a=1+4 \log (2) $ and $b=(1-k) \log (10))+\frac{1}{2}\log (\pi )+3 \log (2)$.

Ignoring the higher order terms and neglecting the $\frac32 \log(n)$, we can approximate the solution and get
$$n_0=-\frac{b}{2 W\left(-\frac{b}{2} e^{-a} \right)}$$ where $W(.)$ is Lambert function. This is an overestimate of the solution.

Trying for $k=10$, this would give $n=51.7838$ while the "exact" solution would be $49.2429$.

Taking into account the $\frac32 \log(n)$ term, we cannot get an explicit solution but the result would be $n=49.2414$

Edit

We can have a better approximation  using the fact that, for all $n$,
$$\left(2 n+\frac{3}{2}\right)! > (2n)!\, (5n-1)$$ and rewrite the equation $(1)$ as
$$\left(2 n+\frac{3}{2}\right)! =(32)^{\left(2 n+\frac{3}{2}\right)}\frac {10^k}{1024 \sqrt{2}}$$ which is very similar to the question. Adapting @robjohn's beautiful approximation, we have
$$n \sim 16\,e^{1+W(t)}-1 $$ where$$t=\frac {2k \log(10)-(27\log(2)+\log(\pi))}{64 e}$$ For the worked example, this would give $n=48.6269$. Summing then $49$ terms, we have $1.76\times 10^{-11}<10^{-10}$ while one less term would give $1.74\times 10^{-10}>10^{-10}$

ContentLicense: CC BY-SA 4.0


问:
Find the ratio of Surface area$^3$ and volume$^2$ of an sphere
IF surface are and volume of a sphere are S and V respectively,then value of $$\dfrac {S^3}{V^2} $$

a) $32$ unit;
b) $9$ unit;
c)$18$ unit ;
d)$27$ unit;

I know the formula $S=4\pi r^2$ and $V=\dfrac43\cdot\pi r^3$.I tried to solve it and my answer is $36\pi$ which seems to nowhere of any option.

答: Your expression for the ratio is wrong, the ratio, $R$ is simply: $R=S/V=3/r$, which you cannot find without $r$. But if you want to find $S^3/V^2=36 \pi $, see the other answer. Like the comments said, there is something wrong in the options you were given as possible answers</p>

ContentLicense: CC BY-SA 3.0


问:
Exercise 6.17 from Isaac's Character Theory of Finite Groups.
I am trying to solve the following problem in Isaac's Character Theory of Finite Groups and I am really sorry if this is a very simple question.

6.17: Let $N$ be a normal subgroup of $G$ with $G/N$ cyclic. Let $\eta\in \operatorname{Irr}(N)$ be invariant in $G$ and assume that $(\eta(1),|G:N|)=1.$ Show that $\eta$ is extendible to $G.$

Q1: The hint is to show that $G/\ker(\det\eta)$ is abelian. But I don't know how to prove the hint or how to solve the problem using it. I think proving that $\det \eta$ would be enough but I don't know how to do that.

Q2: There's a note right below the problem saying that this is true even without the assumption $(\eta(1),|G:N|)=1.$ Could someone please give me a hint for this? My only experience in Character Theory is reading up to chapter 6 in Isaac's book so I'm not sure if I know enough material to solve this without this assumption.

Thank you!!

答:
All credits go to prof Holt (+1 from me), but here is how I solved it a long time ago (hope it is still correct...). We need a lemma (actually more than we need, but this is not be found in Isaacs's book, I follow his notation and also the formulation of Problem (6.17) ...)

Lemma Let $N \unlhd G$ and $\vartheta \in Irr(N)$ be $G$-invariant. Put $\lambda=det(\vartheta)$. Then the following hold.

$(a)$ $\lambda$ is a $G$-invariant linear character of $N$

$(b)$ $ker(\vartheta) \subseteq ker(\lambda)$ and both subgroups are normal in $G$

$(c)$ $[G,N] \subseteq ker(\lambda)$ or equivalently, $N/ker(\lambda) \subseteq Z(G/ker(\lambda))$

$(d)$ $N/ker(\lambda)$ is cyclic and $|N:ker(\lambda)|=o(\lambda)$.

Proof (a) For all $g \in G, n \in N$, we have $\lambda^g(n)=\lambda(gng^{-1})=det(\vartheta)(gng^{-1})=det(\vartheta^g)(n)= \text { ($\vartheta$ is $G$-invariant) }det(\vartheta)(n) =\lambda(n)$.

(b) Let $\frak{T}$ be a representation of $N$ affording $\vartheta$ and let $n \in ker(\vartheta)$. Then $\frak{T}$$(n)=I$, and hence det($\frak{T}$$(n))=1$, so $ker(\vartheta) \subseteq ker(\lambda)$. For all $g \in G, n \in ker(\lambda)$, we have, using (a), $\lambda(gng^{-1})=\lambda^g(n)=\lambda(n)=1$, whence $gng^{-1} \in ker(\lambda)$. A similar argument shows that $ker(\vartheta) \unlhd G$.

(c) For all $g \in G, n \in N$ we have $\lambda([g,n])=\lambda(g^{-1}n^{-1}gn)=\lambda(g^{-1}n^{-1}g)\lambda(n)=\lambda^{g^{-1}}(n^{-1})\lambda(n)= \text { (by (a)) }\lambda(n^{-1})\lambda(n)=\lambda(1)=1.$

(d) Observe that since $\lambda$ is a linear character of $N$, $Z(\lambda)=N$. Now by Lemma(2.27)(d) (M.I. Isaacs, CTFG) $N/ker(\lambda)$ is cyclic. $\lambda$ is a faithful character of $N/ker(\lambda)$, so $o(\lambda)=|N:ker(\lambda)|. \square$

Now we are ready to prove the Problem(6.17): the Lemma(c) tells us that $N/ker(\lambda) \subseteq Z(G/ker(\lambda)) \subseteq G/ker(\lambda)$. But since $G/N$ is cyclic, it follows that $\overline{G}/Z(\overline{G})$ is cyclic (here $\overline{\cdot}$ denotes modding out by $ker(\lambda))$. This implies that $\overline{G}=G/ker(det(\vartheta))$ is abelian. Owing to Theorem (6.25) in Isaacs' book, the conclusion now follows, since $\overline{N}$ is a subgroup of the abelian group $\overline{G}$, and $\lambda$ can be viewed as a character of $\overline{N}$. (See also (5.5) Corollary in Isaacs' book)

The fact that the condition $gcd(\vartheta(1),|G:N|)=1$ can be dropped is explained in Chapter $11$, see Corollary (11.22). The theory here depends on projective representations, group extensions and hence a bit of cohomology theory (the so-called Schur multiplier $M(G)=H^2(G,\mathbb{C}^*)$ plays a role here, "measuring" certain extensions of the group $G$).

The theory amounts to another powerful tool called character triples: an object $(G,N,\vartheta)$ is called a character triple, when $N \unlhd G$ and $\vartheta \in Irr(N)$ is $G$-invariant. Without going into the details (see Isaacs' CTFG page 186, and even better Isaacs' latest book Characters of Solvable Groups (CSG), where he, starting on p. 14, explains the essentials of the theory in 2.5 pages (!) (OK, for proofs he refers to his book CTFG)) one can define "isomorphisms" between character triples. A central theorem is then that every character triple $(G,N,\vartheta)$ admits an isomorphic one $(G^*,N^*,\vartheta^*)$, with $\vartheta^*$ faithful and linear and $N^* \subseteq Z(G^*)$. Here $G/N \cong G^*/N^*$.
So the power of this tool lies in the fact that a certain assertion is made in the normal character world, then this is transposed to the $*$-world, where life is much easier so to speak, prove it there and cycle it back to the original assertion. The same tool allows you to drop the solvability condition in Problem (6.7), for example.

For further (historical) reading I suggest the aforementioned CSG and I.M. Isaacs, Characters of solvable and symplectic groups, Am. J. Math 95 (1973) 594-635, first couple of chapters. Here character triple isomorphisms are introduced. The rudiments of character triple theory was originally developed by Everett C. Dade if I am not mistaken.


Title: Find the ratio of Surface area$^3$ and volume$^2$ of an sphere

Body:
IF surface are and volume of a sphere are S and V respectively,then value of $$\dfrac {S^3}{V^2} $$

a) $32$ unit;
b) $9$ unit;
c)$18$ unit ;
d)$27$ unit;

I know the formula $S=4\pi r^2$ and $V=\dfrac43\cdot\pi r^3$.I tried to solve it and my answer is $36\pi$ which seems to nowhere of any option.

Body:
$S=4\pi r^2$ and $V=\frac{4}{3}\pi r^3$

$$S^3/V^2=\frac{(4\pi r^2)^3}{(\frac{4}{3}\pi r^3)^2}=\frac{64\pi^3r^6}{\frac{16\pi^2r^6}{9}}=36\pi$$


问:
Equivalence classes of $x^{2}\equiv y^{2} (\operatorname{mod} 4)$
Question: Let $S\subseteq \mathbb{Z}\times\mathbb{Z}$ be the equivalnce relation
$$
x^{2}\equiv y^{2} (\operatorname{mod}4)
$$
Find all the equivalence classes of $S$ on $\mathbb{Z}$.

My Answer: Let $z\in\mathbb{Z}$.
The euivalence class of $S$ on $z$ will be the set of all $x$ that satisfy
$$
z^{2} - x^{2} = 4m\Leftrightarrow x^{2} = z^{2} - 4m.
$$
If $z$ is even, in other words $z = 2c$, then $x^{2} = 4c^{2} - 4m = 0 - 4(c^{2} + m)$ which is the equivalence class of 0.
Therefore, the equivalence class of every even integer is the same as the equivalence class of zero.

Respectively, if $z$ is odd, in other words $z = 2c + 1$, then $x^{2} = 4c^{2} + 4c + 1 + 4m = 1 + 4(c^{2} + c + m)$, which is the equivalence class of 1.
Therefore, the equivalence class of every odd integer is the same as the equivalence class of 1.

So all the equivalence classes of $S$ on $\mathbb{Z}$ are given by the following
\begin{align*}
    [2n]_{S} = [0]_{S} &= \{x\in\mathbb{Z}\mid 0 - x^{2} = 4m, m\in\mathbb{Z}\} = \{\ldots, -8, -4, -2, 0, 2, 4, 8, \ldots\}\\
    [2n + 1]_{S} = [1]_{S} &= \{x\in\mathbb{Z}\mid 1 - x^{2} = 4m, m\in\mathbb{Z}\} = \{\ldots, -9, -5, -3, -1, 1, 3, 5, 9, \ldots\}
\end{align*}

Is my thought process correct? Does my answer satisfy the question? Thank you in advance and sorry for any spelling error, I'm translating from Greek:)

答:
You are correct in the two equivalence classes, but you can simplify the presentation a bit:
$$[0]_S=\{2x:x\in\mathbb Z\}$$
$$[1]_S=\{2x+1:x\in\mathbb Z\}$$
One class is the even integers, one is the odd integers, both are disjoint and both unify to the integers.


问:
Projecting a grid on a 2d plane
Unsure if this is the right network to post this, but maybe someone can help...

I'm working on a personal project where, with the help of AI, I want to automatically estimate the perspective of a given image.

I've been able to programatically estimate the position of the horizon, as seen below. This is based on how "quickly" cars farther away shrink in the image.

Not sure the above is useful, but now I want to draw a grid on the plane. Something like:

This is where it's getting too complicated for me. I have no idea how to solve this. Not the drawing per se, but the mathematics behind it.

I first thought I needed to find the plane equation based on the coordinates of the bottom left and bottom right corners of each box, since these are the ones intersecting the plane (the tires touching the road).

I did manage to find the equation ax + by + cy + d, but I am not sure I am on the right track. Looking for some guidance if anyone has any ideas.

EDIT:

Just to clarify the unknowns:

Distance to any object or horizon.

Physical size of objects (in meters or otherwise)

Height or angle of camera

答:
Projecting back to 3D-space

First, we have to set a couple of (somewhat arbitrary) constants in order to be able to translate bounding box sizes and 2D-positions to real world 3D-positions. For example
$$
\newcommand{\cm}{\operatorname{cm}}
\begin{align}
\operatorname{carwidth}&=200\cm\\
\operatorname{screen}&=30\cm\times20\cm\\
\operatorname{screendist}&=60\cm\\
\end{align}
$$
So here we assume that the focal point of the eye watching "through" the window of the screen is placed at a distance of $60\cm$ from the center of the screen. Let the eye be the origin of the coordinate system $(0,0,0)$

Now suppose some car has a bounding box with bottom corners at $C_1=(x_1, y_1)$ and $C_2=(x_1+10\cm,y_2)$ in centered screen coordinates. Then we have the two embedded 3D-points:
$$
\begin{align}
C_1&=(x_1,y_1,60\cm)\\
C_2&=(x_1+10\cm,y_2,60\cm)
\end{align}
$$
And we can simply rescale those positions by a factor:
$$
\frac{200\cm}{10\cm}=20
$$
in order to get the corresponding positions of the car in 3D-space:
$$
\begin{align}
P_1&=(20x_1,20y_1,1200\cm)\\
P_2&=(20x_1+200\cm,20y_2,1200\cm)\\
\end{align}
$$

NOTE: Here we have assumed that each bottom point of a car is at the same $z$-distance to the observer. If we use the top points too, we might be able to derive inclinations of cars too, but that becomes a different model.

Estimating the common plane

Suppose we have done this to all the bottom point of a series of cars. Now we want to determine the approximate common plane for those. First we compute the mean of all the points (since any square minimizing solution will pass through the mean):
$$
M:=\frac1N \sum_{i=1}^N P_i
$$
Then we minimize the square distances:
$$
Squares(a,b,c)=\sum\frac{(ax_i+by_i+cz_i-aM_x-bM_y-cM_z)^2}{a^2+b^2+c^2}
$$
for some limited search space of potential normal vectors $\langle a,b,c\rangle$. For instance we can choose the serach space to be $\langle a, 1, c\rangle$, so that we only consider normal vectors pointing at least partly upwards. This should be possible to either solve or approximate. This should provide you with a plane equation:
$$
ax+by+cz+d=0
$$
where $d=-aM_x-bM_y-cM_z$.

Projecting the resulting grid

Again, we need to make a choice. For example, do you want the grid in 3D-space to be $100\cm\times 100\cm$. Once this has been decided, all you need to know is that lines in 3D are projected to becomes lines in 2D, so just compute some lattice points in the plane in 3D-space, project those to 2D-space by shortening by a factor that renders the z-coordinate equal to $60cm$ and draw a grid line through them on the screen.

Title: Projecting a grid on a 2d plane


问:
Book Suggestions for an Introduction to Measure Theory
Couldn't find this question asked anywhere on the site, so here it is! Do you guys have any recommendations for someone being introduced to measure theory and lebesgue integrals?

A mentor has suggested a book that's in french, but unfortunately I don't know french (heck I barely know english) - so english books only please!

Thanks in advanced!

EDIT: Did not realize this had been asked here but I'm going to leave this question open to see if there are any newer books (that question was originally asked in 2011). If I am doing something wrong here, just give me a gentle shoutout in the comments and I'll be understanding and close this

答:
You might want to take a look at Schillings Measures, Integrals, and Martingales.

It's a great introductory text for Measure Theory, gentle, but rigorous. The author's website has solutions to the book, as well as Errata, etc. You can take a look at the table of contents in the link given above.


问:
Number of paths of length $5$ in a $5$-regular graph with $m$ edges and girth at least $6$.
Let $G$ be a $5$-regular graph with $m$ edges and girth at least $6$. To find the number of paths of length $5$.

A regular graph is a graph where each vertex has the same number of neighbours.

The girth of a graph is the length of a shortest cycle contained in the graph.

答:
By the handshaking lemma there are $\frac{2m}5$ vertices. Since the girth is at least $6$, we won't have to worry about running back onto ourselves; choosing a vertex and picking a neighbour five times will yield all paths exactly twice (we may trace the same path both ways), so the number of paths is
$$\frac12\cdot\frac{2m}5\cdot5\cdot4\cdot4\cdot4\cdot4=256m$$


问:
Show that the genus of the projective curve $x^n_1 = x_2x^{n-1}_0 - x^n_2$ is $(n-1)(n-2)/2$.
Suppose that we have a curve $X \subset \mathbb{P}^2$ given by $x^n_1 = x_2x^{n-1}_0 - x^n_2$ .  How do we show that the curve has genus $(n-1)(n-2)/2$ (whenever the curve is smooth).

I should use the fact that the sequence
$$
0 \to \mathcal{F}(X) \stackrel{\iota}{\to} \mathcal{F}(U_1) \oplus \mathcal{F}(U_2)  \stackrel{\delta}{\to} \mathcal{F}( U_1 \cap U_2)
$$
is exact. Where $\mathcal{F} =\mathcal{O}_X(D)$ for some divisor $D$  on $X$ and $\{ U_1, U_2\}$ an affine cover of $X$ (however, $X$ should be irreducible I think).  Moreover $\iota(f) =( f|_{U_1}, f|_{U_2})$  and $ \delta(f_1,f_2) = f_1|_{U_1 \cap U_2} - f_2|_{U_1 \cap U_2}$ .

There is a hint: compute the cokernel of $\delta$ using bases for the infinite-dimensional vector spaces $\mathcal{O}_X(X_1)$, $\mathcal{O}_X(X_2)$ and $\mathcal{O}_X(X_{12})$ that are as simple as possible. (As a sanity check on your computation: the kernel of $\delta$ should be one-dimensional!)

Any help on how to proceed or how to solve it would really be appreciated!

答:
First, considering the general open cover of $\mathbb{P}^{2}$ and setting $X_{i}=X\cap D(x_{i}) \ \ (i=0,1,2)$ it can be easily seen (by writing down the equation) that $X_{1}\subseteq X_{2}$ and so $\{X_{0},X_{2}\}$ is an open covering of $X$. Therefore, the genus of $X$ can be computed by the sheaf exact sequence

\begin{equation*}
    0\longrightarrow {\cal O}_{X}(0)(X)\longrightarrow {\cal O}_{X}(0)(X_{0})\oplus
{\cal O}_{X}(0)(X_{2})\longrightarrow {\cal O}_{X}(0)(X_{0}\cap X_{2}).
\end{equation*}

But there is a (natural) isomorphism of sheaves of ${\cal O}_{X}$-modules ${\cal O}_{X}(0)\xrightarrow{\sim}{\cal O}_{X}$ and so we may compute the genus of $X$ by computing the
dimension of the cokernel of the map $\delta$ in the following exact sequence:

\begin{equation*}
    0\longrightarrow {\cal O}_{X}(X)\longrightarrow {\cal O}_{X}(X_{0})\oplus
{\cal O}_{X}(X_{2})\overset{\delta}{\longrightarrow} {\cal O}_{X}(X_{02}),
\end{equation*}

where $X_{02}:=X_{0}\cap X_{2}$ and $\delta(f_{0},f_{2})=f_{0}|_{X_{02}}-f_{2}|_{X_{02}}$
for each $f_{0}\in{\cal O}(X_{0})$ and $f_{2}\in{\cal O}(X_{2})$.

We will compute
$\dim_{k}(\mathrm{coker}\delta)$ by precisely finding a basis of $\mathrm{coker}\delta$.

Note that by dehomogenization we  have that the isomorphisms of varieties
\begin{equation*}
    X_{0}\cong Z(f_{0}) \ \ \ \text{and} \ \ \ X_{2}\cong Z(f_{2}),
\end{equation*}
where $f_{0}=x_{01}^{n}-x_{02}+x_{02}^{n}\in k[x_{01},x_{02}]$ and
$f_{2}=x_{21}^{n}-x_{20}^{n-1}+1\in k[x_{20},x_{21}]$. Hence, we have that
\begin{equation*}
    {\cal O}_{X}(X_{0})\cong k[x_{01},x_{02}]/(f_{0}) \ \ \ \ \text{and} \ \ \ \
    {\cal O}_{X}(X_{2})\cong k[x_{20},x_{21}]/(f_{2}).
\end{equation*}

Before we start, observe that a system of generators for ${\cal O}_{X}(X_{0})$ is

\begin{equation*}
    T_{0}=\{x_{01}^{i}x_{02}^{j}:0\leq i<n,j\geq 0\}
\end{equation*}

and for ${\cal O}_{X}(X_{2})$ is

\begin{equation}
    T_{2}=\{x_{20}^{i}x_{21}^{j}:i\geq 0,0\leq j<n\}
\end{equation}

The above follows directly from the fact that in ${\cal O}_{X}(X_{0})$ we have the relation $x_{01}^{n}=x_{02}-x_{02}^{n}$ and in ${\cal O}_{X}(X_{2})$ the relation $x_{21}^{n}=x_{20}^{n-1}-1$. Hence, $\mathrm{im}\delta$ is generated by the image $T$ of
$T_{0}\cup T_{2}$ under the map $\delta$.

We now proceed with the construction of a basis for $\mathrm{coker}\delta$. We have that
\begin{equation*}
    {\cal O}_{X}(X_{02})\cong k[x_{01}.x_{02},x_{20}]/(f_{0},x_{02}x_{20}-1),
\end{equation*}
i.e. ${\cal O}_{X}(X_{02})$ results by inverting $x_{02}(=x_{2}/x_{0})$ in ${\cal O}_{X}(X_{0})$. Therefore, it can be readily seen that taking into consideration the relations in ${\cal O}_{X})(X_{02})$ and the generating set $T$ of $\mathrm{im}\delta$ the set

\begin{equation*}
    S=\{x_{01}^{i}x_{20}^{j}:0\leq i<n,j>0\}.
\end{equation*}

generates ${\cal O}_{X}(X_{02})$ and so also $\mathrm{coker}\delta$,
$S\cap T=\emptyset$ and is linearly independent in ${\cal O}_{X}(X_{02})$.
Now by considering the other relation $x_{21}=x_{01}x_{20}$ in ${\cal O}_{X}(X_{02})$
it can (again) be readily seen that we may further refine $S$ into

\begin{equation*}
    S'=\{x_{01}^{i}x_{20}^{j}:0\leq i<n,0<j<i\},
\end{equation*}

which has cardinality $|S'|=(n-1)(n-2)/2$ and its image forms a basis of $\mathrm{coker}\delta$.


问:
How can I compute rotation angle of an image?
An image is translated with $x_0$ and $y_0$ (vertical and horizontal displacements), rotated with an angle $θ$, and then is scaled on both vertical and horizontal directions with the scaling parameters $α$ and $β$. The resulting composed geometric transform in homogeneous matrix form is
$$\begin{pmatrix}\sqrt 3 & 1 & 20 \\ -1 & \sqrt 3 & 10 \\ 0 & 0 & 1\end{pmatrix}$$

I need to compute the rotation angle $θ$, the scaling parameters $α$ and $β$, and the vertical and horizontal displacements $x_0$ and $y_0$.

答:
You are using three matrices:
$$\text{Traslating matrix: }
T_{(x_0,y_0)} = \begin{pmatrix} 1 & 0 & x_0 \\\ 0 & 1 & y_0 \\\ 0 & 0 & 1 \end{pmatrix},$$
$$\text{Rotating matrix: }
R_\theta = \begin{pmatrix} \cos\theta & \sin\theta & 0 \\\ -\sin\theta & \cos\theta & 0 \\\ 0 & 0 & 1 \end{pmatrix},$$
$$\text{Scaling matrix: }
S_{(\alpha,\beta)} = \begin{pmatrix} \alpha & 0 & 0 \\\ 0 & \beta & 0 \\\ 0 & 0 & 1 \end{pmatrix}.$$

You just need to solve the system given by:
$$S_{(\alpha,\beta)} R_\theta T_{(x_0,y_0)} = \begin{pmatrix} \sqrt{3} & 1 & 20 \\\ -1 & \sqrt{3} & 10 \\\ 0 & 0 & 1 \end{pmatrix}.$$


问:
How to find number of injections from a finite set A to finite set B where |A| < |B|?
I need to find number of injections from a finite set A to finite set B where |A| < |B|. How could i do that? I don't understand what does |A| and |B| mean</p>

答:
$|A|$ means the cardinality of set $A$. We have $|B|$ options for the first element of $A$, $|B|-1$ options after that for the second element and so on (if we want to make this an injection), so the number of injections is
$$|B|(|B|-1)\cdots(|B|-|A|+1)=\frac{|B|!}{(|B|-|A|)!}$$


问:
On a limit concerning the exponent of a negative number
A question requires you to find out the greatest value for a for which a complicated limit(with variables in both the base and its exponent) equals $\frac{1}{4}$. Now, the limit simplifies down to $(\frac{sinh-ah}{h+sinh})^{1+\sqrt{1+h}}$, where h approaches 0. By direct substitution, this is $(\frac{1-a}{2})^{2}$

The answer is 0, and not 2, as the exponential function isn't defined as having a negative base, which it would if a was 2.

Apparently the limit doesn't exist due to the surrounding values of 2, and not when the exponent is exactly 2, as the expression obviously exists at that exact point and equals $\frac{1}{4}$.

So is it due to the fact that as the limit approaches the exact value, there are multiple points where the expression becomes complex?

答:
The answer to the question posed in the last paragraph is, It depends on your definition of limit. It also depends on what you consider the "natural" domain of a function when the formula used to define it doesn't always make sense.

If $1-a\over2$ is negative, then $\sin h-ah\over h+\sin h$ will be negative for all values of $h$ near $0$, in which case the exponential function is rarely real (only when the exponent $1+\sqrt{1+h}$ is rational with odd denominator), so the function is undefined at most points. If your definition of limit requires the function to be defined at all points in a (punctured) neighborhood of the point being tended to, then that's enough to conclude the limit does not exist (in which case the answer to the question in the OP's last paragraph is simply Yes).

But if your definition of limit requires only that the function be defined somewhere in each (punctured) neighborhood of the point being tended to, then you have to look a bit more closely. Now the natural domain of the function $(\sin h-ah)/(h+\sin h)^{1+\sqrt{1+h}}$ (for $h$ near $0$) consists of points for which $1+\sqrt{1+h}=m/n$ with $n$ odd, so we can ask if we have a well-defined limit as $m/n\to2$.

We don't.

We don't because there are two ways to approach $2$ using fractions with odd denominator. If we approach $2$ using fractions with even numerator, e.g., ${m\over n}={2\cdot3^k+2\over3^k}$, then we indeed get $\left(1-a\over2\right)^2$. But if we approach $2$ using fractions with odd denominator, then (when $1-a\over2$ is negative) we get $-\left(1-a\over2\right)^2$. Since these values are different, the limit does not exist.

In sum, the reason the limit does not exist when $(1-a)/2$ is negative is related to the fact that the expression takes complex values at multiple points, but the precise reason is a bit more complicated, depending on how you define things.


问:
Why isometric isomorphic between Banach spaces means we can identify them?
Is the "isometric" part really necessary? For what reason is that?

Eg. we prove that there is an isometric isomorphism between $(L^p)'$ and $L^q$ ($(p,q)$ conjugate) and then we identify them together as the same space. If they were isomorphic but not necessarily isometric, could we not identify them?

答:
Each mathematical theory study their own objects and maps between them. In our case these are Banach spaces and bounded linear maps. We choose linear maps because we want them to preserve linear structure of Banach spaces. We choose them bounded to nicely interact with the norm structure of Banach spaces. Once we choose our objects and transformation between them we want to classify our objects. There mathematicians start to invent different properties which can be used for classification. The more properties two Banach spaces have in common the more they are similar.

Now we want to know what are the maps that preserve a particular property. The best maps in Bаnach space theory are

isomorphisms, i.e. bijective bounded linear operators with bounded inverse

isometric isomorphisms, i.e. isomorphisms that preserve norms

The first type of isomorphisms preserve the following properties

complementability

separablity

completeness

type/cotype

reflexivity

and many other that depend on the topology generated by the norm. Since topology of Banach spaces can be generated by different norms isomorphisms of Banach spaces can not catch all properties of the norm structure of Banach space.

In this case the second type of isomorphisms will be useful. They preserve all properties mentioned above plus

1-complementabilty

extreme points

distances between susbsets

and others that depend on the metric generated by norm.

These types of isomorphisms give two ways of identification of Banach spaces - tolerant and more rigorous. For example $\ell_1^n$ and $\ell_2^n$ are two $n$-dimensional Banach spaces that isomorphic but not isometrically isomorphic.

Note that isomorphisms of Banach spaces catch only those properties of spaces that correspond to their Banach space structure. For example the spaces of convergent sequences $c$ is isomorphic as Banach space to the space $c_0$ of sequences converging to $0$. But as Banach algebras (Banach spaces with continuous multiplication of vectors) they are not isomorphic, because $c$ is unital and $c_0$ is not.


问:
Proof that correlation coefficient $\rho(X,Y) = 1$ given continuous random variable X and $Y := g(X)$
Question: "Let X be a continuous random variable, and $Y := g(X)$ where $g: \mathbb{R} \rightarrow \mathbb{R}$ is a non-decreasing, Borel-regular function.

Prove that the correlation coefficient $\rho(X,Y) := \frac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}} = 1$
"

My approach: So if we show that Cov(X,Y) = $\sqrt{\text{Var(X) Var(Y)}}$, then we're done.

I wrote out:

Var(X) Var(Y) $= (\mathbb{E}(X^2)-\mathbb{E}(X)^2)(\mathbb{E}(Y^2)-\mathbb{E}(Y)^2)$ and expanded it.

Then I did the same for: Cov(X,Y) $= \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)$.

And then, $\text{Cov(X,Y)}^2 = \left( \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y) \right)^2$.

But I'm not sure how to proceed from here. Any help would be appreciated, thank you!

答: This is not true. Let $X \sim U[0,1]$ and $y=X^{2}=g(x)$ where $g(x) =x^{2}$ for $x \geq 0$ and $g(x)=0$ for $x <0$. I will let you compute the correlation coefficient and see that it is not $1$.


问:
Conditional Probability with Poisson's variable
I've been having difficulty with the following problem.

A driver is circling a parking lot searching for a parking space. The parking spaces distribute in a poisson's distribution of λ=0.3.

Knowing that the driver has found a parking space, what's the probability that that would be the only parking space?

I tried approaching the problem via conditional probability saying that $P(s=1|s>=1)$ but I can't figure out what to do next.

Thank you.

Body:
Note that

$$P(S|S>0)=\frac{1}{1-e^{-0.3}}\frac{e^{-0.3}0.3^s}{s!}$$

Thus

$$P(S=1|S>0)=\frac{0.3}{e^{0.3}-1}\approx 85.75\%$$


问:
How does the matrix logarithm change when a matrix is extended with zero rows and columns?
Imagine the matrix $A$ and the extended matrix B:
\begin{equation*}
B =
\begin{pmatrix}
A & 0 \\
0 & 0
\end{pmatrix}
\end{equation*}

Now how does the logarithm of both matrices relate?
I have done some testing using from scipy.linalg import logm.

if you try:

```<code>    A= np.array([[0.572, 0.180, 0.814, 0.085, 0.065],
           [0.462 , 0.462, 0.167, 0.027, 0.997],
           [0.735, 0.919, 0.353, 0.939, 0.895],
           [0.763, 0.931, 0.511, 0.450, 0.186],
           [0.144, 0.428 , 0.314, 0.077, 0.081]])

    B= np.array([[0.572, 0.180, 0.814, 0.085, 0.065, 0.0],
           [0.462 , 0.462, 0.167, 0.027, 0.997, 0.0],
           [0.735, 0.919, 0.353, 0.939, 0.895, 0.0],
           [0.763, 0.931, 0.511, 0.450, 0.186, 0.0],
           [0.144, 0.428 , 0.314, 0.077, 0.081, 0.0],
            [0.0, 0.0 , 0.0, 0.0, 0.0, 0.0]])
</code>```

you will notice that the difference between logm(A) and logm(B) is only a row and a column of zeros except for the corner value which becomes always -4.60517019e+01.

Similar observation can be noticed if you manipulate the identity matrix and set one or more of the diagonals to zero, the logm would produce -4.60517019e+01 in the corresponding diagonals.

答:
Even in the complex setting, a matrix has a logarithm iff it is invertible. Since $B$ has a row and column of zeros, it is not invertible and has no matrix logarithm at all.

$-46.0517109$ should have tipped you off that SciPy is cheating: it is $\log10^{-20}$. It has been deliberately added to produce a result, which is therefore incorrect.


问:
Finding a function $f:(0, \infty) \to(0, \infty)$ satisfying the differential equation ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$

Let $f:(0, \infty) \to(0, \infty)$ be a differentiable function such that ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$ where $a$ is positive constant and $f'(1) = 1$, $f'(2) = 2$, then find the value of $f(5)$.

Since the arguments of $f$ and $f'$ are different, it is not possible for me to solve the differential equation directly. By hit and trial, I got the function as $\frac{x^2}2$, but cannot solve it mathematically. Can someone please provide me some hints for this?

答:
Let $\varphi(x) := f(x) f\left(\frac{a}{x}\right)$. Differentiating we get, for every $x > 0$,
$$
\varphi'(x) = f'(x) f\left(\frac{a}{x}\right)+f(x) \left[f\left(\frac{a}{x}\right)\right]'
= \frac{a}{x f\left(\frac{a}{x}\right)} f\left(\frac{a}{x}\right)- \frac{a}{x^2} f(x) \frac{x}{f(x)} = 0.
$$
Hence there exists $c > 0$ such that $\varphi(x) = c$ for every $x > 0$.
Since
$$
f'(x) = \frac{a}{x f(a/x)} = \frac{a}{c} \cdot \frac{f(x)}{x}
$$
integrating we get
$$
f(x) = f(1) x^{a/c}.
$$
Substituting in $f'(x) = \frac{a}{c} f(1) x^{a/c - 1}$ the values $x=1$ and $x=2$ we finally get $a/c = 2$ and $f(1) = 1/2$, so that $a=2$ and
$f(x) = x^2/2$.


问:
Why do you have to do division in these trigonomic equations?
For example $\sin 2x= 1$,
the unit circle gives all these answers for $\sin x = 1 $:
$$
\frac{\pi}{2} + n 2 \pi,\quad  \text{(so n times 2pi as you can rotate endlessly)}
$$
but since there is 2x you have to divide 1/2pi by 2, thus 1/4pi, but why do you have to divide n*2pi by 2 as well?

The answer of the equation is apparently 1/4pi + n*pi

Can someone explain this (visually with the unit circle) why all the solutions of 1/4pi + n*2pi are wrong?

Also when you have sin x +1/3pi = 0

the solution is x= n*pi - 1/3pi

which is very clear from me as you have to balance the extra 1/3pi out, so you subtract 1/3pi from all the solutions.

(or compensate/cancel out, whatever english term fits best).

But when it comes to sin2x= y

I dont understand the balancing of division. Is there a way to explain why (preferably with the unit circle) ?

Thanks in advance

答:
$\sin (2x) = 1 \\\Rightarrow 2x = \dfrac{\pi}{2} + 2n\pi\\
\Rightarrow x = ?$


问:
Maximum area of circles inscribed in sectors of the unit circle

We divide the unit circle into equal sectors and inscribe circles within each sector.
Assuming we make smaller and smaller sectors, what does the total area of the inscribed circles converges to?

I think I have solved it, but since I'm not a mathematician, I wonder if this is correct.

I began by asking: What is the total area of all the circles in a single sector?

We will start with the unit circle as the outer perimeter, with $\theta$ being half of the subtended angle, and $r_0$ being the radius of the first ring of circles, $r_1$ being the radius of the second ring of circles, and so on.

Since the unit circle has a radius of 1, we can immediately deduce the following:
$$r_{0} = ({\sin{\theta} \over 1+\sin{\theta}})$$
and we also know that:
$$r_{n+1} = r_{n} ({1-\sin{\theta} \over 1+\sin{\theta}})$$
that means that:
$$r_{1} = ({\sin{\theta} \over 1+\sin{\theta}}) ({1-\sin{\theta} \over 1+\sin{\theta}}) = {(\sin{\theta})({1-\sin{\theta})} \over (1+\sin{\theta})^2}$$
and:
$$r_{2} = {(\sin{\theta})({1-\sin{\theta})^2} \over (1+\sin{\theta})^3}$$
so generally we can say that:
$$r_{n} = {(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}}$$
the area of a single circle in a sector is:
$$A_{n} = \pi r_{n}^2 = \pi ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2$$
so the area of a single sector of circles will be:
$$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 $$
and since we're dealing with two or more sectors, we can say that:
$$ 0 < \theta \leq {\pi \over 2} $$
so this means the sum of areas of all the circles inscribed in a sector converges to:
$$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 = {\pi \sin{\theta} \over 4}$$

I then continued with: What is the total area of all the circles in all the sectors?

Assuming we have $k$ sectors, then $\theta = {\pi \over k}$ and our sum would be:
$$A_{k} = {k A_{s}} = {\pi k \sin{\pi \over k} \over 4}$$
and if we take the limit of $k$, we will get:
$$\lim_{k \to \infty} {\pi k \sin{\pi \over k} \over 4} = {\pi^2 \over 4} \tag*{$\blacksquare$}$$

Is this valid?

答:
The first question is "What is the total area of all the circles in a single sector?".

If "the total area of all the circles in a single sector" means $\displaystyle\lim_{n\to\infty}\sum_{i=0}^{n} \pi r_i^2$, then I think that you've correctly got $\displaystyle\lim_{n\to\infty}\sum_{i=0}^{n} \pi r_i^2=\dfrac{\pi \sin{\theta}}4$.

The second question is "What is the total area of all the circles in all the sectors?".

I think that the meaning of "the total area of all the circles in all the sectors" is ambiguous since here we have two variables.

Let $$f(n,k):=k\cdot \frac{\pi\sin\frac{\pi}{k}}{4}\bigg(1-\bigg(\frac{1-\sin\frac{\pi}{k}}{1+\sin\frac{\pi}{k}}\bigg)^{2n}\bigg)$$ be the sum of the areas of the $nk$ circles where we have $k$ sectors each of which has $n$ circles.

If "the total area of all the circles in all the sectors" means $\displaystyle\lim_{k\to\infty}\lim_{n\to\infty}f(n,k)$ (first take $n\to\infty$ and then take $k\to\infty$), then your argument looks valid to me, and I think that you've correctly got $\displaystyle\lim_{k\to\infty}\lim_{n\to\infty}f(n,k)=\dfrac{\pi^2}{4}$.

If "the total area of all the circles in all the sectors" means $\displaystyle\lim_{n\to\infty}\lim_{k\to\infty}f(n,k)$ (first take $k\to\infty$ and then take $n\to\infty$), then your argument is not valid since $\displaystyle\lim_{n\to\infty}\lim_{k\to\infty}f(n,k)=0$.

If "the total area of all the circles in all the sectors" means $\displaystyle\lim_{(n,k)\to (\infty,\infty)}f(n,k)$, then your argument is not valid since $\displaystyle\lim_{(n,k)\to (\infty,\infty)}f(n,k)$ does not exist. To see this, we set $n=ak$ to have
$$\lim_{(n,k)\to (\infty,\infty)}f(n,k)=\lim_{k\to\infty}\frac{\pi^2}{4}\cdot \frac{\sin\frac{\pi}{k}}{\frac{\pi}{k}}\bigg(1-\bigg(\bigg(1+\frac{1}{\frac{1-\sin\frac{\pi}{k}}{2\sin\frac{\pi}{k}}}\bigg)^{\frac{1-\sin\frac{\pi}{k}}{2\sin\frac{\pi}{k}}}\bigg)^{\frac{2}{1-\sin\frac{\pi}{k}}\cdot\frac{\sin\frac{\pi}{k}}{\frac{\pi}{k}}\cdot(-2a\pi)}\bigg)$$This equals $\dfrac{\pi^2}{4}(1-e^{-4a\pi})$ which depends on $a$. So, $\displaystyle\lim_{(n,k)\to (\infty,\infty)}f(n,k)$ does not exist.

Some comments :

If we assume that we have $n$ sectors instead of $k$ sectors, then "the total area of all the circles in all the sectors" should be $\displaystyle\lim_{n\to\infty}f(n,n)=\dfrac{\pi^2}{4}(1-e^{-4\pi})$.

If we assume that we have $2n$ sectors instead of $k$ sectors, then "the total area of all the circles in all the sectors" should be $\displaystyle\lim_{n\to\infty}f(n,2n)=\dfrac{\pi^2}{4}(1-e^{-2\pi})$.

If we assume that we have $mn$ sectors instead of $k$ sectors where $m$ is a fixed positive integer which does not depend on $n$, then "the total area of all the circles in all the sectors" should be $\displaystyle\lim_{n\to\infty}f(n,mn)=\dfrac{\pi^2}{4}(1-e^{-\frac{4\pi}{m}})$.


答: Using $f'\left(\dfrac{a}{x}\right)=\dfrac{x}{f(x)}$, and performing the substitution $x \to \dfrac{a}{x}$, we get $f'(x)=\dfrac{a}{xf\left(\dfrac{a}{x}\right)}$, Now differentiating both sides yields $$f''(x)=\dfrac{-a\left(f\left(\dfrac{a}{x}\right)-\dfrac{a}{f(x)}\right)}{x^2f^2\left(\dfrac{a}{x}\right)}$$ Now substituting the value of $f\left(\dfrac{a}{x}\right)$, we get our final DE as $$\dfrac{f''(x)}{f'(x)}=\dfrac{1}{x}-\dfrac{f'(x)}{f(x)}$$Now integrating this, and using initial conditions, one finds out that $f'(x)=\dfrac{x}{f(x)}$, which implies $f(x)=\dfrac{x^2}{2} \implies f(5)=12.5.$

Title: Finding a function $f:(0, \infty) \to(0, \infty)$ satisfying the differential equation ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$

Body:
Let $f:(0, \infty) \to(0, \infty)$ be a differentiable function such that ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$ where $a$ is positive constant and $f'(1) = 1$, $f'(2) = 2$, then find the value of $f(5)$.

Since the arguments of $f$ and $f'$ are different, it is not possible for me to solve the differential equation directly. By hit and trial, I got the function as $\frac{x^2}2$, but cannot solve it mathematically. Can someone please provide me some hints for this?


答:
The correct solution $\frac \pi4+n\pi$ covers
$$ \ldots,\frac\pi4-2\pi,\frac\pi4-\pi,\frac \pi4,\frac\pi4+\pi,\frac\pi4+2\pi,\frac\pi4+3\pi,\ldots$$
whereas $\frac \pi4+2n\pi$ covers (only)
$$ \ldots,\frac\pi4-2\pi,\frac \pi4,\frac\pi4+2\pi,\frac\pi4+4\pi,\ldots$$
So you found some solution (asyou would have if you had written $\frac\pi4$ only), but not all.

Title: Why do you have to do division in these trigonomic equations?

Body:
For example $\sin 2x= 1$,
the unit circle gives all these answers for $\sin x = 1 $:
$$
\frac{\pi}{2} + n 2 \pi,\quad  \text{(so n times 2pi as you can rotate endlessly)}
$$
but since there is 2x you have to divide 1/2pi by 2, thus 1/4pi, but why do you have to divide n*2pi by 2 as well?

The answer of the equation is apparently 1/4pi + n*pi

Can someone explain this (visually with the unit circle) why all the solutions of 1/4pi + n*2pi are wrong?

Also when you have sin x +1/3pi = 0

the solution is x= n*pi - 1/3pi

which is very clear from me as you have to balance the extra 1/3pi out, so you subtract 1/3pi from all the solutions.

(or compensate/cancel out, whatever english term fits best).

But when it comes to sin2x= y

I dont understand the balancing of division. Is there a way to explain why (preferably with the unit circle) ?

Thanks in advance


答:
The solutions of

$$\sin y=1$$ are $$y=\frac\pi2+2k\pi.$$

And if $y=2x$, mechanically

$$2x=\frac\pi2+2k\pi\iff x=\frac\pi4+k\pi.$$

This is elementary arithmetic.


问:
How can I draw a polygon from a set of angles?
In recreational mathematics, polytans are polygons formed by edge-connecting isosceles right triangles.  Order-n polytans are those constructed from n such triangles.  My question is this:

Given a set of n angles and an N x N grid, how do I draw a graph using a set of angles? I am trying to find if the angles form the perimeter of a polytan (i.e., given a starting point on the plane, does progressively iterating through the angles get me back to the point of origin?).  For example, given these three sets: {{270, 315, 315}, {315, 270, 315}, {315, 315, 270}}, the output would show two isosceles right triangles: {270, 315, 315} and {315, 315, 270}, and one non-polygon: {315, 270, 315}, as seen in the image below.  In the image, the initial reference line is {{0,0}->{0,1} (on the y-axis), and the point of origin is {0,1}.

I need the constructions to be graphs so that I can check whether two polytans of the same order are identical, but just in different orientations (like the two triangles in the image).  Vertex count, vertex degree, edge-connectedness, etc., would also be useful.

答: It appears that you are also constraining the side length to be $1$ if the side is horizontal or vertical and $\sqrt 2$ if it is diagonal.  If so, you can just keep track of the lattice points and see if the polygon closes.  For the first, you have points $(0,0), (0,1), (1,1), (0,0)$.  You can note that the last point is the same as the first to see that the polygon closes.  To check if two are the same, you can put the two $270$ degree turns at the origin, rotate one polygon around (there are only four choices) and see if the other vertices match.</p>


问:
Why “characteristic zero” and not “infinite characteristic”?
The characteristic of a ring (with unity, say) is the smallest positive number $n$ such that $$\underbrace{1 + 1 + \cdots + 1}_{n \text{ times}} = 0,$$ provided such an $n$ exists. Otherwise, we define it to be $0$.

But why characteristic zero? Why do we not define it to be $\infty$ instead? Under this alternative definition, the characteristic of a ring is simply the “order” of the additive cyclic group generated by the unit element $1$.

My feeling is that there is a precise and convincing explanation for the common convention, but none comes to mind.  I couldn't find the answer in the Wikipedia article either.

答: This is all just a convention, and I'm adding an answer $8$ years late because nobody else has pointed out yet that many years ago some people did use the term "characteristic $\infty$". Schilling does this in his book "Theory of Valuations" (AMS, 1950). Either way, you get used to it.


问:
Changing variables of integration (transfer lemma)
This is a lemma that I know as "Transfer Lemma" in Integration Theory or in Probability theory. Although I state it in general form, this lemma seems to be essential when studying conditional expectations formally.

Let $(\mathcal{X}, \mathcal{F}, \mu)$ be a measure space and $(\mathcal{Y},\mathcal{B})$ be a measurable space. $f: (\mathcal{X}, \mathcal{F}, \mu) \to (\mathcal{Y},\mathcal{B})$ is a measurable map. Let $f_* \mu (A):= \mu(\{ f^{-1}(A)\})$ for $A \in \mathcal{B}$ be a pushed forward measure. Let $\phi: (\mathcal{Y},\mathcal{B}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be measurable. Then, the following hold.

(1) If $\phi$ is non-negative, then $\phi \circ f$ is also non-negative, and $\int_\mathcal{X} \phi \circ f d \mu= \int_\mathcal{Y} \phi d (f_* \mu)$

(2) $\phi$ is integrable with respect to $f_* \mu \iff$ $\phi \circ f$ is integrable with respect to $\mu$, and in this case $\int_\mathcal{X} \phi \circ f d \mu= \int_\mathcal{Y} \phi d (f_* \mu)$.

Question: I know that the above holds, but does $\int_\mathcal{X} \phi \circ f d \mu= \int_\mathcal{Y} \phi d (f_* \mu)$ also hold for all measurable $\phi$? It seems that it does, but I am unable to come up with a proof. Would anyone have a nice proof to this? (Please consider writing it out in detail; I get easily confused if the answers are too compact!)

答:
Theorem: A measurable function $g$ on $X_2$ is integrable with respect to the pushforward measure $f_*(μ)$ if and only if the composition $g \circ f$ is integrable with respect to the  measure $μ$. In that case, the integrals coincide, i.e.,

$$\int_{X_2} g \, d(f_* \mu) = \int_{X_1} g \circ f \, d\mu$$

So if one of the function $g$ or $g \circ f$ is integrable (with respect to the corresponding measure), then by the above the other function is integrable as well, and the equality holds.

Edit: The question arises, wether the equality also holds in the case, when

the LHS and RHS are exist

but $g$ and $g \circ f$ are both non-integrable.

The answer is affirmative. We will consider this in the following:

Recall the definition of the integral of a real valued function $f$. Denote the positive and negative parts of $f$ with $f^+$ and $f^-$. If at least one of $\int f^+$ and $\int f^-$ is finite, we define
$$\int f = \int f^+ - \int f^-$$
(If $\int f^+$ and $\int f^-$ are both finite, we say that $f$ is integrable.)

In the following we denote the push forward measure with $\nu\equiv f_* \mu$. By assumption 1), the following integrals exist

$$ \int_{X_2} g \, d\nu = \int_{X_2} g^+ \, d\nu - \int_{X_2} g^- \, d\nu =G^+ -G^-$$

$$ \int_{X_1} g \circ f \, d\mu =  \int_{X_1} g^+ \circ f \, d\mu  - \int_{X_1} g^- \circ f \, d\mu =F^+ - F^-$$

(where we used that $(g \circ f)^+=g^+ \circ f$), which means that at least one of $G^+$ and $G^-$ is finite. Assumtion 2) means that at least one of $G^+$ and $G^-$ is infinite. This means that exactly one of $G^+$ and $G^-$ is finite and the other one is infinite. By similar argument the same is true for $F^+$ and $F^-$. It is easy to see that $G^+$ and $F^+$ are finite or infinite simultaneously. (Indeed, assume that $G^+$ is finite, then by Theorem it follows, that $F^+$ is finite as well.) Thus RHS and LHS are plus infinite or minus infinite in the same time. This completes the proof.


问:
Antiderivative of a rational trigonometric function
I'm looking for some help in solving the following integral:

$$\int{\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)+A}}}d\theta$$

I've seen some similar cases with elliptical integrals, but even in this case, I couldn't solve the antiderivative above. It would be great to solve it analytically, but any help is appreciated.

答: $$\frac{\sin^2\theta}{\sqrt{\cos^2\theta+a}}=\frac{a+1}{\sqrt{\cos^2\theta+a}}-{\sqrt{\cos^2\theta+a}}$$ shows the decomposition in elliptic integrals of the first and second kind.


问:
Why is Brouwer’s Fixed Point Theorem considered a result of algebraic topology?
Brouwer’s Fixed Point Theorem and some generalisations can be proved using properties of Barycentric subdivisions and easy special cases of Sperner’s Lemma relating to them plus basic properties of metric spaces and the fact that every compact convex subset of $\mathbb{R}^n$ is homeomorphic to an $m$-simplex for $m\leq n$ which can be proved by projecting both onto the unit sphere and rotating. For example, see the Wikipedia entry for its generalisation The Kakutani Fixed Point Theorem which gives a sketch. Not so much as a homotopy, let alone any algebra. Why then is Brouwer’s Fixed Point Theorem considered part of algebraic topology when it is not about the objects of algebraic topology and it does not require the methods of algebraic topology?

答:
It is a result of algebraic topology because it can be easily proved by the machinery of algebraic topology (using homology groups or homotopy groups of spheres). However, there are alternative proofs based on simplicial triangulations, Sperner's lemma etc. By the way, the study of simplicial complexes was one of the starting points of algebraic topology (simplicial homology).

A similar example is the Fundamental Theorem of Algebra. There are many proofs using different approaches, for example via complex analysis, via algebraic topology (using e.g. the fundamental group), etc. In my opinion it is "a child of many parents" which doesn't make it wrong to call it, say, a result of complex analysis.


问:
Four-letter word contains no two consecutive equal letters.
This is taken from the book on Combinatorics, by Daniel Marcus.

Request vetting:

A16: Find the probability that a four-letter word that uses letters from $A,B,C,D,E$ contains (a) no repeated letters; (b) no two consecutive equal letters.

(a)
Total (Inclusion) ways to form four-letter word from five letters: $5^4$.

Ways to have repetitions of $A$:

Two $A$: Choose two positions for $A$ in $\binom{4}2= 6$ ways. Rest two positions can be filled in : $4\times 3= 12$ ways. So, by product rule: $72$ words. Also, the same for other $4$ letters, leading to $72\times 5= 360$ words.

Three $A$: Choose three positions for $A$ in $\binom{4}3= 4$ ways. Rest one position can be filled in : $4$ ways. So, by product rule: $16$ words. Also, the same for other $4$ letters, leading to $16\times 5= 80$ words.

Four $A$: Choose all four positions for $A$ in $\binom{4}4= 1$ ways. Also, the same for other $4$ letters, leading to $5$ words.

Sum of above three cases:
$360+80+5= 445.$

Left are: $5^4 - 445= 180.$

$p= \frac{180}{625}$

(b) no two consecutive same letters' case.

Approach#1:

To get, need subtract from $5^4$  the chances of having two consecutive same letters. This is divided into cases:

Case 1: Position #1,2 are the same, and position #4 can match too :  $5\times 1\times 4\times 5= 100$ cases.

Case 2: position #2,3 are the same: $5\times 4\times 1 \times 4= 80$ cases.

Case 3: Position #3,4 are the same, and position #1 can match too:
$5\times 4\times 1\times 5=100$

So, get : $625- 280= 345$ cases.

Approach#2:

Alternatively, can 'directly' calculate by having $5$ choices for the first position, then $4$ choices, then again need eliminate one choice (of the second element) to have $4$ choices for the third position, and same for the fourth position. This leads to: $5\times 4^3= 320$ choices possible.

The second approach seems correct, but not clear why first approach is faulty.

Seems there are too many factors to consider in the first approach & hence error-prone (to miss some case(s) as here). So, second approach seems better.

答:
some of  your answers are incorrect, and some also use an unnecessarily complex approach.

(a) no. of words with no repeats is simply $P^5_4 = 5\cdot4\cdot3\cdot2 =120$

(b) no. of words with no two consecutive equal letters = $5\cdot4\cdot4\cdot4 = 320\quad$

You can then compute probabilities in each case by dividing by $5^4$


问:
A boundary version of Cauchy's theorem
I am looking for a reference for the following theorem (or something like it) that is not Kodaira's book.

Let $D$ be a domain and $\overline{D}$ be it's closure. Suppose that $f:\overline{D} \rightarrow \mathbb{C}$ is holomorphic in $D$ and continuous on $\overline{D}$. If the boundary of $D$, denoted $\partial D$ is composed of piecewise $C^1$ curves then
    $$f(w) = \frac{1}{2\pi i}\oint_{\partial D}\frac{f(z)}{z-w}dz$$
    for all $w\in D$.

This is stronger than the classical Cauchy integral theorem, because here we can allow the path we integrate over to be the boundary. This appears in Kodaira's book, but the book seems sloppily written and I found an egregious error in the first few pages, so I'd like another reference.

This does not appear in Ahlfors or any other standard text I looked at. I am looking for any proof that extends to usual Cauchy's theorem to allow paths that lie on the boundary of a region, instead of requiring them to lie entirely in the open region like the usual version does.

I managed to find a paper that sketches a high-powered proof when the boundary is a rectifiable Jordan arc (or some multiply connected region where each boundary component is) and gives a reference to a more elementary proof of the same result. I think a simpler proof should be possible in this case, where the boundary is assumed to be $C^1$.

This is closely related to an old question that did not get much attention.

答:
I guess here that you mean that the boundary of $D$ is a finite number of pairwise disjoint curves. In this case, this Cauchy formula follows directly from Mergelyan's Theorem about uniform rational approximation of holomorphic functions. Indeed, the formula holds for rational functions with poles outside $\overline{D}$, and Mergelyan's Theorem says that any $f$ continuous on $\overline{D}$, holomorphic in $D$, can be approximated uniformly on $\overline{D}$ by rational functions with poles outside $\overline{D}$.

It also works under the weaker hypothesis that the curves are rectifiable.

Title: A boundary version of Cauchy's theorem


问:
Find the formula for the linear part of a curve
The function $f(x) = -log_2(1- (1-2^{-x})^m)$ seems to behave linearly as $x$ goes to infinity ($m$ is a natural integer)

Here are different values of $m$: red(up) is 30, blue is 80 and green(down) is 160

I try to find the equation $f(x) = ax +b$ for the linear part of the curve

For $a$ I think that I need to solve $a =  \lim_{x\to\infty} ( - log_2(1- (1-2^{-x-\Delta})^m) + log_2(1- (1-2^{-x})^m) )/\Delta $

From the looks of the curve i think $a= 1$, but I don't know how to proove it...

For $b$ (supposing $a = 1$) I start to solve $x + b = \lim_{x\to\infty} log_2(1- (1-2^{-x})^m) $

But I am unable to express $b$ as a function of $m$...

Can someone help me ? I am not good enough with the interactions between limits and logarithms, and I am stuck after a few steps of the resolutions...  Thanks in advance !

答:
Using the binomial theorem,

$$1-(1-2^{-x})^m=m2^{-x}-\frac{m(m-1)}22^{-2x}+\frac{m(m-1)(m-2)}62^{-3x}-\cdots\approx m2^{-x}.$$

Then taking the base-$2$ cologarithm,

$$x-\log_2m.$$


问:
prove by induction that $P\left(\bigcup\limits_{i=1}^{n} E_i\right) = 1-\prod\limits_{i=1}^{n}(1-P(E_i))$, $E_1,E_2,\ldots , E_i$ independent
Suppose $E_1,E_2,\ldots , E_i$ are independent events. prove by induction that $$P\left(\bigcup\limits_{i=1}^{n} E_i\right) = 1-\prod\limits_{i=1}^{n}(1-P(E_i))$$

The first step is easy. For $n=1$ we have
$$P\left(\bigcup\limits_{i=1}^{1} E_i\right) = P(E_1)= 1-(1-P(E_i))=1-\prod\limits_{i=1}^{1}(1-P(E_i))$$
I don't really see how to continue from here though. Any help would be appreciated!

答:
I figured it out so I thought I might as well post it.

We note that:
$$P\left(\bigcup\limits_{i=1}^{1} E_i\right) = P(E_1)= 1-(1-P(E_i))=1-\prod\limits_{i=1}^{1}(1-P(E_i))$$
Now suppose for a certain $k \in \mathbb{N}$ that:
$$P\left(\bigcup\limits_{i=1}^{k} E_i\right) = 1-\prod\limits_{i=1}^{k}(1-P(E_i))$$
So
$$P\left(\left(\bigcup\limits_{i=1}^{k} E_i\right)^c\right)=\prod\limits_{i=1}^{k}(1-P(E_i))$$
Now for $k+1$ we have:
$$P\left(\bigcup\limits_{i=1}^{k+1} E_i\right)=P\left(E_{k+1}\cup\left(\bigcup\limits_{i=1}^{k} E_i\right)\right)$$
Now we use that $P(A\cap B) = P(A)P(B)$ for independent $A$ and $B$, and the fact that if all $E_i$ are independent then so are the events $E_i^c$:
$$P\left(E_{k+1}^c\cap (\left(\bigcup\limits_{i=1}^{k} E_i\right)^c\right)=(1-P(E_{k+1}))\prod\limits_{i=1}^{k}(1-P(E_i))=\prod\limits_{i=1}^{k+1}(1-P(E_i))$$
Finally we use the fact that $P(A ^c\cap B^c)=P(A\cup B)^c$ so $$P\left(E_{k+1}^c\cap (\left(\bigcup\limits_{i=1}^{k} E_i\right)^c\right) = 1 - P\left(E_{k+1}\cup\left(\bigcup\limits_{i=1}^{k} E_i\right)\right)=1 - P\left(\bigcup\limits_{i=1}^{k+1} E_i\right)$$
Combining these results we have
$$1 - P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = \prod\limits_{i=1}^{k+1}(1-P(E_i))$$
or $$ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = 1-\prod\limits_{i=1}^{k+1}(1-P(E_i)).$$
Now by induction we have
$$ P\left(\bigcup\limits_{i=1}^{n} E_i\right) = 1-\prod\limits_{i=1}^{n}(1-P(E_i))$$
for all $n \in \mathbb{N}$.


答:
A bit of softening up first:
$$\frac{\sin^2x}{\sqrt{\cos^2x+A}}=\frac1{\sqrt{1+A}}\frac{\sin^2x}{\sqrt{1-\frac1{1+A}\sin^2x}}$$
Then, by Byrd and Friemdan 281.01,
$$\int_0^\varphi\frac{\sin^2x}{\sqrt{\cos^2x+A}}\,dx=\frac1{\sqrt{1+A}}\int_0^\varphi\frac{\sin^2x}{\sqrt{1-\frac1{1+A}\sin^2x}}\,dx$$
$$=\frac1{\sqrt{1+A}}\int_0^{F(\varphi,m=1/(1+A))}\operatorname{sn}^2u\,du$$
By B&F 310.02 this in turn evaluates as
$$\frac1{\sqrt{1+A}}\cdot\frac1{1/(1+A)}(F(\varphi,m)-E(\varphi,m))=\sqrt{1+A}(F(\varphi,m)-E(\varphi,m))$$

In some references like the DLMF, the difference $F(\varphi,m)-E(\varphi,m)$ is denoted in terms of a new function $D(\varphi,m)$:
$$D(\varphi,m)\equiv\int_0^\varphi\frac{\sin^2x}{\sqrt{1-m\sin^2x}}\,dx=\frac1m(F(\varphi,m)-E(\varphi,m))$$
This is because the difference can be evaluated in one step (avoiding a subtraction) using the Carlson symmetric form as
$$D(\varphi,m)=\frac{\sin^3\varphi}3R_D(\cos^2\varphi,1-m\sin^2\varphi,1)$$
The original integral expressed in this symmetric form is
$$\int_0^\varphi\frac{\sin^2x}{\sqrt{\cos^2x+A}}\,dx=\frac1{\sqrt{1+A}}D\left(\varphi,\frac1{1+A}\right)$$
$$=\frac{\sin^3\varphi}{3\sqrt{1+A}}R_D\left(\cos^2\varphi,1-\frac1{1+A}\sin^2\varphi,1\right)\qquad|\varphi|\le\frac\pi2$$

Title: Antiderivative of a rational trigonometric function

Body:
I'm looking for some help in solving the following integral:

$$\int{\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)+A}}}d\theta$$

I've seen some similar cases with elliptical integrals, but even in this case, I couldn't solve the antiderivative above. It would be great to solve it analytically, but any help is appreciated.


问:
Evaluating $\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k$
Let $k,m,n \in \mathbb{Z}$ and consider $\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k$. I was wondering why
$$
\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k =
\begin{cases}
m , \ \ \ \ \ \ n/m \in \mathbb{Z}\\
0 , \ \ \ \ \ \ \ n/m \notin \mathbb{Z}
\end{cases}
$$
I understand this is actually a geometric series, so
$$\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k = \frac{1- e^{-2in}}{1- e^{-2in/m}}.$$
This fraction can only be zero if and only if $e^{-2\pi in} =1$. But I don't see the reason why this is true. Furthermore, how come that the sum is equal to $m$ when $n/m \in \mathbb{Z}$. Does it mean that $\left(e^{-2\pi in/m}\right)^k = 1$ when $n/m \in \mathbb{Z}$?

答: To prove $\sum_{k=0}^{m-1}(e^{-2\pi in/m})^k=m[m|n]$ (for brevity I've used an Iverson bracket), separately consider the case $m|n$ for which the sum is $\sum_k1=m$, and the case$$m\nmid n\implies e^{-2\pi in/m}\ne1\implies\frac{\color{blue}{1-e^{-2\pi in}}}{\color{red}{1-e^{-2\pi in/m}}}=0,$$since the blue part is $0$ but the red part isn't.


问:
Does strictly upper triangularity of $PAP^{-1}$ imply strictly upper triangularity of $PBP^{-1}$?
$A$ and $B$ are nilpotent matrices in $\mathbb M_n(F)$ that commute. Let $P$ be an invertible matrix such that $PAP^{-1}$ is strictly upper triangular. Does it also imply that $PBP^{-1}$ is also strictly upper triangular?

I was trying to prove that $A+B$ is nilpotent and can't think of a proof or a contradiction

答:
No. Since $A$ and $B$ commute (and they have full spectra over $F$), they can be simultaneously triangualarised by some similarity transform. However, that doesn't mean any similarity transform that triangularises $A$ will also triangularise $B$. For instance, consider
$$
X=\pmatrix{1&-1\\ 1&-1},A=\pmatrix{0&X\\ 0&0},B=\pmatrix{X&X\\ X&X} \text{ and } P=I_4.
$$


问:
2 values expected value problem
A bank account has 1000 dollars and get's a 5% interest daily. Every day the account has $0.405$ probability to lose $20$ dollars.

What's the mean of the amount of money in the acount after 2 days?

I thought about calculating $((1000-20*0.405)*1.05-20*0.405)*1.05$

But I think it might be wrong because the probabiltity changes depending on the previous day.
I mean that the probability to lose in both days is $0.405^2$, one day lose and the other no lose would be $0.405*(1-0.405)$ to my understaing, so I would need to take it into acount somehow but don't know how.

thank you.

答:
There are four possibilities.

Loses 20\$ on the first day but not on the second day

Loses 20\$ on the second day but not on the first day

Loses 20\$ on both days

Loses nothing on both day

Calculate the amount in each case, $A_1, A_2, A_3$ and $A_4$. Now to get the mean value, you just multiple each amount with its probability of happening.
$$A = A_1\cdot 0.405\cdot (1-0.405) + A_2\cdot (1-0.405)\cdot 0.405 + A_3\cdot0.405\cdot0.405 + A_4(1-0.405)(1-0.405)$$

Title: 2 values expected value problem


答:
Two days are a small horizon. The random part is modelled by the binomial tree:

```<code>                        * [L] (1000-20)*r - 20 ... and finally ((1000-20)*r -20)*r
                      /
       * [L] 1000-20  - * [K] (1000-20)*r      ... and finally (1000-20)*r*r
      /
1000 *                 * [L] 1000 * r - 20     ... and finally (1000*r -20)*r
      \              /
       * [K] 1000    - * [K] 1000 * r          ... and finally 1000*r*r
</code>```

with [L] used for LOSE, and [K] for KEEP.
Let $p$ be the probability for losing, i.e. for the [L] ramifications, and $q$ the complementary probability.

Above, the multiplication rate is (EDIT: adjusted after the edit in the OP)
$$
r = 1 + 5\% =1.05\ .
$$
The mean amount of money after two days is thus:
$$
\begin{aligned}
M
&=
p^2\cdot((1000-20)r-20)r
+pq\cdot(1000-20)r^2
\\
&\qquad\qquad
+pq\cdot(1000r-20)r
+q^2\cdot 1000r^2
\\
&=1000r^2-20p^2(r+r^2)-20pq(r+r^2)
\\
&=1000r^2-20p(r+r^2)(p+q)
\\
&=1000r^2-20p(r+r^2)
\ .
\end{aligned}
$$

Later EDIT:

An other way to think about this is as in the OP. The mean value after two days is by linearity $1000r^2$ plus mean value of the possible loss.

The probability to lose $20\$$ in the first day is $p$, and the corresponding mean is $p\cdot  r^2\cdot 20\$$, using the convention that we "first lose, then use the daily rate".

The probability to lose $20\$$ in the second day is $p$ again, and the corresponding mean is $p\cdot  r\cdot 20\$$, using the same convention.

So the mean, computed as in the OP comes to the same value:
$$1000 r^2 -20pr^2-20pr\qquad\text{(dollars)}\ .$$

Title: 2 values expected value problem

Body:
A bank account has 1000 dollars and get's a 5% interest daily. Every day the account has $0.405$ probability to lose $20$ dollars.

What's the mean of the amount of money in the acount after 2 days?

I thought about calculating $((1000-20*0.405)*1.05-20*0.405)*1.05$

But I think it might be wrong because the probabiltity changes depending on the previous day.
I mean that the probability to lose in both days is $0.405^2$, one day lose and the other no lose would be $0.405*(1-0.405)$ to my understaing, so I would need to take it into acount somehow but don't know how.

thank you.

Body:
Two days are a small horizon. The random part is modelled by the binomial tree:

                        * [L] (1000-20)*r - 20 ... and finally ((1000-20)*r -20)*r
                      /
       * [L] 1000-20  - * [K] (1000-20)*r      ... and finally (1000-20)*r*r
      /
1000 *                 * [L] 1000 * r - 20     ... and finally (1000*r -20)*r
      \              /
       * [K] 1000    - * [K] 1000 * r          ... and finally 1000*r*r

with [L] used for LOSE, and [K] for KEEP.
Let $p$ be the probability for losing, i.e. for the [L] ramifications, and $q$ the complementary probability.

Above, the multiplication rate is (EDIT: adjusted after the edit in the OP)
$$
r = 1 + 5\% =1.05\ .
$$
The mean amount of money after two days is thus:
$$
\begin{aligned}
M
&=
p^2\cdot((1000-20)r-20)r
+pq\cdot(1000-20)r^2
\\
&\qquad\qquad
+pq\cdot(1000r-20)r
+q^2\cdot 1000r^2
\\
&=1000r^2-20p^2(r+r^2)-20pq(r+r^2)
\\
&=1000r^2-20p(r+r^2)(p+q)
\\
&=1000r^2-20p(r+r^2)
\ .
\end{aligned}
$$

Later EDIT:

An other way to think about this is as in the OP. The mean value after two days is by linearity $1000r^2$ plus mean value of the possible loss.

The probability to lose $20\$$ in the first day is $p$, and the corresponding mean is $p\cdot  r^2\cdot 20\$$, using the convention that we "first lose, then use the daily rate".

The probability to lose $20\$$ in the second day is $p$ again, and the corresponding mean is $p\cdot  r\cdot 20\$$, using the same convention.

So the mean, computed as in the OP comes to the same value:
$$1000 r^2 -20pr^2-20pr\qquad\text{(dollars)}\ .$$


问:
Limes of Lebesgue integral
So here's my measure theory problem

Prove, that if $\mu(X) < \infty$, $f_{n}$ and $f$ are non-negative and bounded by common constant L, and $f_{n} \rightarrow f$ by measure, then $$\lim\limits_{n\to\infty} \int f_{n} \;d\mu = \int f\; d\mu$$
I don't think I should solve by like I normally would in analysis, but I might be wrong. And help would be appreciated.

答:
Define $g_n = f_n - f$ and note that $g_n$ converges to $0$ in measure if $f_n$ converges to $f$ in measure. Furthermore $|g_n|\leq |f_n|+|f| \leq 2L$. Showing that $\int f_n d\mu \rightarrow \int f d\mu$ is equivalent to showing that $\int g_n d\mu \rightarrow 0$.

Now for any $\epsilon > 0$
\begin{align*} |\int g_n d\mu| &\leq  \int | g_n | d\mu \\ &=  \int_{\{|g_n|> \epsilon  \}} |g_n| d\mu + \int_{\{|g_n| \leq \epsilon\}}|g_n|d\mu \\
&\leq 2L \cdot \mu(\{ |g_n|> \epsilon\}) + \epsilon \mu(X),
\end{align*}
where $\mu(\{|g_n| > \epsilon\}) \rightarrow 0$ for $n\rightarrow \infty$ by definition of convergence in measure. Thus we conclude that the right hand side can be made arbitrarily small and thus $|\int g_n d\mu| \rightarrow 0$ as desired.

Title: Limes of Lebesgue integral


问:
The condition for the equations $\vec r \times \vec a=\vec b$ and $\vec r \times \vec c=\vec d$ to be consistent is?
This is an MCQ type question

A) $\vec b.\vec c=\vec a.\vec d$

B) $\vec a. \vec b =\vec c. \vec d$

C) $\vec b.\vec c+\vec a. \vec d=0$

D) $\vec a. \vec b +\vec c.\vec d =0$

Since $\vec a$ is perpendicular to $\vec b$ and same for c and d, both options B and D can be correct.

But surprisingly, the correct answer is option C

I don’t how that is possible. Am I missing something here?

答: Hint: $a.(r \times c) = -c.(r \times a)$ look up triple scalar product https://en.wikipedia.org/wiki/Triple_product


答:
Reasons to why only $B$ and $D$ being true doesn't necessarily prove that the equations hold.

$\vec{a}\cdot\vec{b} = \vec{c}\cdot\vec{d} $ doesn't necessarily mean they're equal to zero.

Similarly, $\vec{a}\cdot\vec{b} $ can be any non-zero scalar $k$ and $\vec{c}\cdot\vec{d} = -k$. They both add to zero, but individually non-zero.

Now, $\vec{c}\cdot(\vec{r}\times\vec{a}) = [ \vec c \ \vec r \ \vec a] = \vec{b}\cdot\vec{c}$

and $\vec{a}\cdot(\vec{r}\times\vec{c}) = [ \vec a \ \vec r \ \vec c] = - [ \vec c \ \vec r \ \vec a] = \vec{a}\cdot\vec{d} = -\vec{b}\cdot\vec{c}$.

So, $\vec{b}\cdot\vec{c} + \vec{a}\cdot\vec{d}=0$

Title: The condition for the equations $\vec r \times \vec a=\vec b$ and $\vec r \times \vec c=\vec d$ to be consistent is?

Body:
This is an MCQ type question

A) $\vec b.\vec c=\vec a.\vec d$

B) $\vec a. \vec b =\vec c. \vec d$

C) $\vec b.\vec c+\vec a. \vec d=0$

D) $\vec a. \vec b +\vec c.\vec d =0$

Since $\vec a$ is perpendicular to $\vec b$ and same for c and d, both options B and D can be correct.

But surprisingly, the correct answer is option C

I don’t how that is possible. Am I missing something here?


问:
Need help in logic and statements
I am stuck with some questions. Please help me out. Thanks.

If $P(x):x^2 < 12$, then $P(1.5)$ is a statement  (I think yes. As the Universe of $x$ is not given but can be taken as set of Real Numbers.)

If $Q(n): n+3=6$, then  $Q(m)$ is the statement   (I think no. As $m$ is a variable and we can not have variables or undetermined values in a statement.)

If $P(y): 1+2+3+\ldots+y =0$, then $P(5)$ is the statement    (I think yes.)

If $Q(m):m \leq 3^m$, then $Q(k)$ is the statement.   (I think no. As $m$ is a variable and we can not have variables or undetermined values in a statement.)

Where Statement is a sentence that evaluates to either true or false but not both.

答:
A CW post to eliminate this question from the Unanswered queue.

As has been remarked in the comments, your suggested answers and accompanying reasoning, which yield the answers:

Yes, No, Yes, No

are all correct. Cheers!

Title: Need help in logic and statements


问:
Expected number of coins tossed
If probability of head for a coin toss is $p$, then what is the expected number of tosses required to get a head?

My answer: P(H) = $p$ (probability of one toss), P(TH)=$p^2$ (probability of two tosses), P(TTH) = $p^3$ and so on.
So, expected number of tosses = $E = 1*p+2*p^2+3*p^3+...+\infty $

$Ep = p^2 + 2p^3 +....+\infty$

$E(1-p)=p+p^2+p^3+...+\infty = \frac{p}{1-p}$

So, $E = \frac{p}{(1-p)^2}$

I have heard someone telling expectation in this case is $\frac{1}{p}$, but I could derive above. Please let me know if I am wrong anywhere.

答:
Your expression for $E$ is not quite right; getting the first head down the line implies getting tails beforehand. It should be
$$E=1(p)+2(1-p)p+3(1-p)^2p+\cdots$$
and when you work it out it should give $\frac1p$.

Title: Expected number of coins tossed


答: The probability that you need $m$ tosses to get a head (i.e. a head on the last toss and tails on other tosses) is $(1-p)^{m-1}p$ for $m>0$. The required expectation is$$\sum_{m=1}^\infty mp(1-p)^{m-1}=1/p$$The number of tosses needed to get a head follows geometric distribution.

Title: Expected number of coins tossed

Body:
If probability of head for a coin toss is $p$, then what is the expected number of tosses required to get a head?

My answer: P(H) = $p$ (probability of one toss), P(TH)=$p^2$ (probability of two tosses), P(TTH) = $p^3$ and so on.
So, expected number of tosses = $E = 1*p+2*p^2+3*p^3+...+\infty $

$Ep = p^2 + 2p^3 +....+\infty$

$E(1-p)=p+p^2+p^3+...+\infty = \frac{p}{1-p}$

So, $E = \frac{p}{(1-p)^2}$

I have heard someone telling expectation in this case is $\frac{1}{p}$, but I could derive above. Please let me know if I am wrong anywhere.


问:
Find the limit of the following sequence
$$\lim_{n\to\infty} (\frac{1^{10} + 2^{10} + 3^{10} + ... + n^{10}}{n^{10}} - \frac{n}{11})$$

So I am aware that Stolz theorem may be used here but I don't know how. What I tried doing is I combined the two fractions:

$$ (\frac{11\dot(1^{10} + 2^{10} + 3^{10} + ... + n^{10}) - n^{11}}{11n^{10}})$$

But I can't apply Stolz theorem to this because we have two sequences in the numerator. Moreover the bigger problem is the $-$ sign next to $n^{11}$. Any help would be appreciated!

答:
Hint: Faulhaber's formula ... https://en.wikipedia.org/wiki/Faulhaber%27s_formula
\begin{eqnarray*}
\sum_{k=1}^{n} k^p= \frac{n^{p+1}}{p+1} + \color{red}{\frac{1}{2}} n^p + \text{lower order terms.}
\end{eqnarray*}


问:
Can a bounded function be monotonically increasing for all x>0?
Can we have a function such that it is always increasing for all x>0 and is also bounded above by some real number? How do we prove or disprove this statement?(The above statement is not a problem itself, this concept is required in another problem, so I want some help regarding it!)

答: Consider a function such as $f(x)=1-e^{-x}$, or $g(x)=\arctan{(x)}$, or $h(x)=1-\frac{1}{x}$. These all satisfy the requirement you've given, being bounded above by $1, \frac{\pi}{2}, 1$, respectively.


问:
Adjoint question in the Category

If $\langle F,G,φ \rangle : X \to A$ is adjunction with $G$ full and every unit $η_x$ a monic, then every $η_x$ is also epi.

Some similar questions like this, maybe I do not catch the key, who can help me?

答: Not an answer but you can prove similar sentences in a general setting: see  exercise EQUIII at page 55 of 'Toposes, Triples and Theories' by Barr-Wells here. Hint for the hint: the dual hint proves the dual statement.</p>

Title: Adjoint question in the Category


问:
4 equations with set of numbers
I must make 1 addition (x+y=z), 1 subtraction (x-y=z), 1 multiplication (x*y=z), and 1 division (x/y=z) equation with the following numbers. All the numbers must be used to fill x, y, and z of each equation. x, y, and z can consist be 1, 2, or 3-digit numbers.

0-4 available

1-9 available

2-1 available

3-2 available

5-1 available

6-1 avilalble

7-0 available

8-1 available

9-1 available

Thanks!

答: There is no magic here, just guess and try.  I find $11+18=29, 11-6=5, 100*0=0,31/1=31$ as one of what must be many solutions.</p>

Title: 4 equations with set of numbers


问:
A convex function property
Consider $\mathbb{R}^n$ with the canonical scalar product $(\cdot,\cdot)$
and the associated euclidean norm.
Let $f: \mathbb{R}^n \to \mathbb{R}$ of class $\mathcal{C}^1$ be $\omega$-convex, i.e.
$\forall u,v \in \mathbb{R}^n$: $(\nabla f(u)- \nabla f(v), u-v) \geq \omega \|u-v\|^2$.

I have to show that $\forall x \in \mathbb{R}^n$ : $f(x) \geq \frac{\omega}{2} \|x\|^2 - \| \nabla f(0) \| \|x\| + f(0)$.

I started by introducing the function $g(t)=f(tx)$ for $t \in [0,1]$. Applying the Taylor expansion on g gave me:
$g(1) = g(0) + \frac{1-0}{1!} g'(0)$

Then I've calculated the derivative of g, I've found that $g'(t) = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(xt) \cdot x_i$. Thus $g'(0) = (\nabla f(0),x)$.

Finally I've found
$f(x) = f(0) + (\nabla f(0), x)$ but I don't know how to proceed to get the wanted result.

答:
Working with the function $g$ is the right way. We find

$$
f(x)=g(1)=g(0)+\int_0^1 g'(t)\,\mathrm{d}t=f(0)+\int_0^1 (\nabla f(tx),x)\,\mathrm{d}t.
$$

On the other hand, by assumption,

$$
(\nabla f(tx)-\nabla f(0),tx)\geq \omega \|tx\|^2,
$$

which gives

$$
(\nabla f(tx),x)\geq (\nabla f(0),x)+\omega t \|x\|^2\geq -\|\nabla f(0)\|\|x\|+\omega t \|x\|^2.
$$

Now just insert that inequality in the above integral and be done.

Title: A convex function property


问:
How to prove that in an arbitrary graph the number of vertices with odd degree is even
I know that degree of a vertex of an (undirected) graph is the number of edges incident to the vertex. I need to prove that in an arbitrary graph the number of vertices with odd degree is even. How to formally prove it?</p>

答: The sum of degrees of all vertices is even. Thus the number of odd-degree vertices must be even, since otherwise the first sum is odd (even-degree vertices do not affect the parity of the first sum).</p>


答: I proved this many hard ways. Years later I found this. This is the best way!</p>

Title: How to prove that in an arbitrary graph the number of vertices with odd degree is even

Body: I know that degree of a vertex of an (undirected) graph is the number of edges incident to the vertex. I need to prove that in an arbitrary graph the number of vertices with odd degree is even. How to formally prove it?</p>


问:
What will this integral solve to mainly?
What will the following integral give for variable x and y :
$$\int d(xy)$$
$$\int xyd(xy)$$
This is a followup on What do these mean pdv . vdp . d(pv)?.

I posted this by mistake.
Just realized its like asking
$$\int udu$$

答:
$$\int d(xy) = xy + C$$
$$\int xyd(xy) = \frac{x^2y^2}{2} + C$$

You might be confused because you see two variables instead of one, but you may notice that this is nothing new by putting $t=xy$ to get the following integrals:
$$\int dt \text{ and } \int t dt$$


问:
prove that $P_{N}$ is a projection is to say $P_{N}^2=P_{N}$
Let $\{u_1,u_2,...\}$ an orthonormal system in a space with internal product $E$, $\bigcup_{N}$ is the space generated by $\{u_1,u_2,...,u_N\}$ and the projection operator $P_{n}f$ is defined as
$$P_{N}f= \sum_{k=1} ^{N} \langle f, u_{k} \rangle u_k  $$
prove that $P_{N}$ is a projection is to say $P_{N}^2=P_{N}$

we know that $P_{N}^2f= \left(\sum_{k=1} ^{N} \langle f, u_{k} \rangle u_k\right)^2$

but I don't see an easy way to eliminate this square and reach equality, can you give me a suggestion to illuminate the way?

答:
Your $P_n^2f$ is wrong (and have no sens, because the operation $u_ku_j$ is not well defined).
\begin{align*}
P_n^2f&=\sum_{k=1}^n\left<P_nf,u_k\right>u_k\\
&=\sum_{k=1}^n\sum_{j=1}^n\left<f,u_j\right>\underbrace{\left<u_j,u_k\right>}_{=\delta _{kj}}u_k\\
&=\sum_{k=1}^n\left<f,u_k\right>u_k\\
&=P_nf,
\end{align*}

where $\delta _{kj}=1$ if $j=k$, and $0$ otherwise.


问:
Solve an equation; Ax=b regarding matrices
I'm studying to linear algebra in community college, and I'm having a hard time with this exercise because I don't really understand what we are doing.

We have a 3x3 matrix A, and we know that the eigenspaces of that matrix are E(1,A)=Sp(v1,v2)  and E(2,A)=Sp(v3) where v1=(1,0,1) v2=(0,1,0) and v3=(1,0,-1).

Then we are supposed to solve an equation of the form Ax=b where b is a vector b=3,4,5.

Am I supposed to create a matrix out of the v1,v2,v3 vectors and then somehow solve x from that? Also i dont know what matrix A is, We just assume to know its eigenspaces (and values), could I solve what matrix A is from the eigenspaces (I dont know how to do that lol).

Any help with this problem would be greatly appreciated, since I don't really know what steps to take and where.

答:
Seems that, from your problem statement:

$$Av_1 = v_1 \,\,\text{  and  }\,\, A v_2 = v_2 \,\, \text{  and  }\,\,  Av_3 = 2v_3.$$

Also, it should be clear that $v_1,v_2,v_3$ span $\mathbb{R}^3$ so

$$b= c_1 v_1 + c_2 v_2 + c_3 v_3$$ for some $c_1,c_2,c_3$ not all zero.

Find the $c_j$'s .  Then

$$Ab = A(c_1v_1 + c_2v_2 + c_3 v_3)= c_1 Av_1 + c_2 Av_2 + c_3 Av_3=c_1v_1+c_2v_2+ 2c_3 v_3.$$


问:
if $|f''(x)|\le |f'(x)|,\forall x\in R$ show that either $f'(x)>0$ or $f'(x)<0,\forall x\in R$
Edit:let $f:R\to R$ be twice differenriable on $R$, such that
$$|f''(x)|\le |f'(x)|,\forall  x\in R$$
show that either:
$f'(x)>0$
or $f'(x)<0,\forall x\in R$

since
$$0\ge (f''(x))^2-(f'(x))^2=(f''(x)-f'(x))(f''(x)+f'(x))$$
so
$$f"(x)+f'(x)\ge 0,\mbox{and} f''(x)-f'(x)\le 0$$
or
$$f"(x)+f'(x)\le 0,\mbox{and} f''(x)-f'(x)\ge 0$$

答:
Let $g=f'$. Then $g$ is once differentiable, we have $|g'| \leq |g|$
on $\mathbb R$, and we wish to show that $g$ always keeps the same sign.

Suppose that $g$ is nonzero, so that there is a $x_0\in{\mathbb R}$ with
$g(x_0)\neq 0$. Replacing $g$ with $x\mapsto g(x_0+x)$, we may assume without loss that $x_0=0$, so that $g(0)\neq 0$.

Replacing $g$ with $-g$, we may further assume that $g(0) \gt 0$.

Let $$A=\bigg\lbrace x \gt 0 \ \bigg| \ \forall t\in [0,x], \ g(t) \gt 0 \bigg\rbrace \tag{1}$$.

Then $A$ is a nonempty subset of $[0,\infty)$. Let $a=\sup(A)$ (note that $a$ can be $\infty$). By the construction of $A$, we have

$$[0,a) \subseteq A \tag{2}$$

For $t\in [0,a)$, we then have $-1 \leq \frac{g'(t)}{g(t)} \leq 1$, and integrating, we deduce $-t \leq \ln(g(t))-\ln(g(0)) \leq t$, whence

$g(t) \geq g(0)e^{-t} \ (t\in[0,a]) \label{3}\tag{3}$

If $a$ were finite, by \eqref{3} above and the continuity of $g$ we would have
$g(a) \geq g(0)e^{-a}$, so that $g$ would be positive on a neighborhood of $a$,
contradicting the definition of $a$. So $a=\infty$, and hence $g(t) \geq g(0)e^{-t}$ for all $t\geq 0$. A symmetric argument shows that $g(t) \geq g(0)e^{t}$ for all $t\leq 0$. This finishes the proof.


问:
Maximum value of $4|\cos x|-3|\sin x|$
How will I find the maximum of $4|\cos x|-3|\sin x|$

The absolute value confuses me, The maximum value would be 5 if it were not there.

答:
The maximum of your expression cannot exceed $4$, which is obtained when $4|\cos x|$ is maximized and $3|\sin x|$ is minimized independently.

In this case, at $x=n\pi~(n\in\Bbb Z)$, both maximization of the first term and minimization of the second term happen simultaneously. So the maximum value is indeed $4$.

Title: Maximum value of $4|\cos x|-3|\sin x|$


答:
$|\cos (x)| = 1$(max value)  for all $x = n\pi, n\in \Bbb Z$

So, $4|\cos (x)| = 4$ is the maximum value possible of the first term.

$3|\sin x| \ge 0$. So, we need the term $3|\sin x|$ to have the minimum value possible as it is being subtracted from the first term and that value is zero. This again occurs at $x = n\pi,  n\in \Bbb Z$.

So, $4|\cos x| - 3|\sin x|$ attains a max. value of $4-0 = 4$ at $x = n\pi,  n\in \Bbb Z$.

Title: Maximum value of $4|\cos x|-3|\sin x|$

Body:
How will I find the maximum of $4|\cos x|-3|\sin x|$

The absolute value confuses me, The maximum value would be 5 if it were not there.


问:
Self intersection of zero section of decomposable ruled surface
Let $L_1, L_2$ be two line bundles over a curve $C$. Then we can consider the ruled surface $S=\Bbb P (L_1\oplus L_2)$ over $C$. Let $s_0$ and $s_\infty$ denote the zero section and the infinity section, respectively. In this case is there a way to express the self intersections of $s_0$ and $s_\infty$ by the degrees of $L_1$ and $L_2$?

答:
The normal bundle for a section is isomorphic to the restriction of the relative tangent bundle. The relative tangent bundle $\mathcal{T}$ fits into the relative Euler exact sequence
$$
0 \to \mathcal{O} \to \pi^*(L_1 \oplus L_2) \otimes \mathcal{O}(1) \to \mathcal{T} \to 0,
$$
where $\pi$ is the projection and $\mathcal{O}(1)$ is the Grothendieck bundle. If you restrict this to the section, say, $C_1$ and use the fact that the pullback of $\mathcal(1)$ is, say, $L_1^\vee$, you obtain the exact sequence
$$
0 \to \mathcal{O} \to (L_1 \oplus L_2) \otimes L_1^\vee \to \mathcal{T}\vert_{C_1} \to 0,
$$
and it follows that
$$
\mathcal{N}_{C_1/S} \cong \mathcal{T}\vert_{C_1} \cong L_2 \otimes L_1^\vee.
$$


问:
Finding $b$ such that $y=mx+c$ is tangential to $y=b^x$.
I was thinking about the following problem:

Find $b$ such that $y=mx+c$ is tangential to $y=b^x$.

My work so far:

To satisfy the condition we need to find $x$ such that $\frac d{dx}\left[b^x\right]=\frac d{dx}[mx+c]$. Solving the derivatives, we get $\ln b\cdot b^x=m\Rightarrow b^x=\frac m{\ln b}\Rightarrow x=\log_b\left(\frac m{\ln b}\right)=\frac{\ln \left(\frac m{\ln b}\right)}{\ln b}=\frac{\ln m-\ln\ln b}{\ln b}$.

At the same time, we have $b^x=mx+c$. Using the substitution $u=mx+c$ and noting $mx=u-c\Rightarrow x=\frac1mu-\frac cm$, we have:

$$b^{\frac um-\frac cm}=u$$

$$b^{-\frac cm}=u\left(\frac1b\right)^\frac um$$
$$ue^{\ln\left(\frac1b\right)\frac um}=ue^{-\frac{\ln b}mu}=b^{-\frac cm}$$
$$-\frac{\ln b}mue^{\ln\left(\frac1b\right)\frac um}=ue^{-\frac{\ln b}mu}=-b^{-\frac cm}\frac{\ln b}m$$
Let $z=-\frac{\ln b}mu$
$$ze^z=-b^{-\frac cm}\frac{\ln b}m$$
$$z=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$-\frac{\ln b}mu=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$-\frac{\ln b}mu=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$u=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$mx+c=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$mx=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$x=-\frac 1{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac cm$$
Putting the two equations together, we have:
$$\frac{\ln m-\ln\ln b}{\ln b}=-\frac 1{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac cm$$
$$\ln m-\ln\ln b=-W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac{c\ln b}m$$
$$\ln m-\ln\ln b+\frac{c\ln b}m=-W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$\ln\ln b-\ln m-\frac{c\ln b}m=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)e^{\ln\ln b-\ln m-\frac{c\ln b}m}=-b^{-\frac cm}\frac{\ln b}m$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)e^{\ln\ln b-\ln m-\frac cm\ln b}=-b^{-\frac cm}\frac{\ln b}m$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)\frac{\ln b}mb^{-\frac cm}=-b^{-\frac cm}\frac{\ln b}m$$
$$\ln\ln b-\ln m-\frac{c\ln b}m=-1$$
$$\ln\ln b-\frac{c\ln b}m=\ln m-1$$
$$e^{\ln\ln b-\ln b\frac cm}=e^{\ln m-1}$$
$$\ln b\cdot b^{-\frac cm}=\frac me$$

I was not able to get any further than this. The equation is probably not solvable through elementary means but I would be interested in a closed solution using special functions (such as the Lambert W function) as well as a derivation.

答:
Any point  on $y=b^x$ can be $P(t,b^t)$

The equation of the tangent at $P$ will be $$\dfrac{y-b^t}{x-t}=b^t\ln b$$

$$\iff y=x(b^t\ln b)+b^t(1-t\ln b)$$

Comparing with $y=mx+c$

$$m=b^t\ln b, c=b^t(1-t\ln b)$$

$$\ln m= t\ln b\cdot \ln(\ln b)\implies t=?$$

$$\dfrac cm=\dfrac{1-t\ln b}{\ln b}\implies \ln b=?, b=?$$


Title: Maximum value of $4|\cos x|-3|\sin x|$

Body: $a=|sin x|,b=|\cos x|$ where $a,b\in[0,1]$ we have to maximise $$4a-3b=4a-3\sqrt{1-a^2}=f(a)$$ but $$f'(a)=4+\frac{3a}{\sqrt{1-a^2}}>0$$ hence $$f(a)\le f(1)=4$$


问:
Let X be a set and let $ \tau_1 $ and $ \tau_2 $ topologies over X. Show that if (X, $ \tau_1 $) e (X, $ \tau_2 $) are compact Hausdorff spaces...
i)Let X be a set and let $ \tau_1 $ and $ \tau_2 $ topologies over X. Show that if (X, $ \tau_1 $) e (X, $ \tau_2 $) are compact Hausdorff spaces, then $ \tau_1 $ = $ \tau_2 $ or $ \tau_1 $ and $ \tau_2 $ are incomparable.

ii) Let {$ X_i: i \in I $} be a family of compact Hausdorff spaces . Show that $ \sigma $ is a topology over the set $ \prod_ {i \in I} X_i $ that makes all projections and ($\prod_ {i \in I} X_i, \sigma $) is compact, so $ \sigma $ matches the product topology on $ \prod_ {i \in I} X_i $.

My attempt at i) went like this

Assume that $\tau_1 \subset \tau_2$. Then the identity map: (X,$\tau_2$)→(X,$\tau_1$)is continuous. Let C be closed in $\tau_2$, so, X\C $\in \tau_2$. Since $\tau_2$ is compact,C is compact as a subspace of $\tau_2$. Because continuous maps map compact sets to compact sets, C is also compact as a subspace of $\tau_1$. But $\tau_1$ is Hausdorff, so this means that C is closed in $\tau_1$. We showed that every closed set in $\tau_2$ is closed in $\tau_1$, so $\tau_2 \subset \tau_1$ and therefore $\tau_1 =\tau_2$.
Similarly if we instead assume $\tau_2 \subset \tau_1$, the same argument also shows $\tau_1 =\tau_2$.

I'm having bigger troubles with ii)

Thanks in advance for any help.

答:
Well, in order to apply $i)$, you just need to show that $\sigma$ is Hausdorff (the product topology is compact by Tychonoff).

We just want to prove that if all the projections are continuous and each $X_i$ is Hausdorff, then $\sigma$ is Hausdorff.

Indeed, let $x\neq y\in \prod_{i\in I} X_i$ and fix some $j$ such that $x_j\neq y_j$. Since $X_j$ is Hausdorff, there exist open sets $U,V\subseteq X_j$ such that $U\cap V$ is empty and $x_j\in U$ and $y_j\in V$. Now, let $\pi_j:\prod_{i\in I}X_i\to X_j$ denote the corresponding projection. Then, $\pi_j^{-1}(U)\cap \pi_j^{-1}(V)=\emptyset$ and by continuity of $\pi_j$, both of these sets are open. Furthermore, clearly $x\in \pi_j^{-1}(U)$ and $y\in \pi_j^{-1}(V)$. Thus, $\sigma$ is Hausdorff.


问:
What does this Set actually mean/contain?
I'm having trouble understanding the following set:

$$\mathbb{Z}[\sqrt{2}] := \{ a + \sqrt{2}b : a,b \in \mathbb{Z}\}$$ What elements does it actually contain? I thought $\mathbb{Z}$ only contains integers, so what does the root of $2$ behind $\mathbb{Z}$ mean?

Thank you.

答:
You are right: $\mathbb{Z}$ only contains integers. The set displayed here, though, is a subset of $\mathbb{R}$ and can thus contain $\sqrt{2}$.

The notation says it contains all elements of the form $a+\sqrt{2}b$, where $a$ and $b$ are integers. Let me give you a few examples:

$a=b=1$: Then $a+\sqrt{2}b  = 1+\sqrt{2}$.

$a=b=0$: Then $a+\sqrt{2}b  = 0$.

$a=3, b=-2$: Then $a+\sqrt{2}b  = 3-2\sqrt{2}$.

$b = 0$: Then $a+\sqrt{2}b = a$ and because of this any integer is in your set. But as you have seen also nonintegers like $1+\sqrt{2}$ can be part of that set.

I hope this helps clearing things up a bit.


问:
Discrete functor preserves serre fibrations
Consider $Top$ equipped with the Serre's structure and $Set$ with the trivial model structure (All maps fibrations and cofibrations, bijections weak equivalences) and the functor
$D: Set \to Top$ that equips each set with the discrete topology

How can I show that $D$ preserves fibrations? (The fibrations in are the morphisms satisfying the right lifting property with respect to $\{i_0 : D^n \to D^n \times I | \; n > 0\}$ )

答:
Since the discrete space functor $\textbf{Set} \to \textbf{Top}$ is a fully faithful embedding, you can just forget about the functor and prove the following instead:

Every map between discrete spaces is a fibration.

You can use the fact that the spaces $D^n$ and $D^n \times I$ are connected to prove this.

The above argument is morally the same as the following adjointness argument. The discrete space functor $\textbf{Set} \to \textbf{Top}$ has a left adjoint, namely the connected components functor $\pi_0 : \textbf{Top} \to \textbf{Set}$. For formal (= abstract nonsense) reasons, a right adjoint functor preserves fibrations if and only if its left adjoint preserves trivial cofibrations. But $\pi_0 : \textbf{Top} \to \textbf{Set}$ sends weak homotopy equivalences to bijections, so the discrete space functor $\textbf{Set} \to \textbf{Top}$ preserves fibrations.


问:
Expected number of coins tossed
If probability of head for a coin toss is $p$, then what is the expected number of tosses required to get a head?

My answer: P(H) = $p$ (probability of one toss), P(TH)=$p^2$ (probability of two tosses), P(TTH) = $p^3$ and so on.
So, expected number of tosses = $E = 1*p+2*p^2+3*p^3+...+\infty $

$Ep = p^2 + 2p^3 +....+\infty$

$E(1-p)=p+p^2+p^3+...+\infty = \frac{p}{1-p}$

So, $E = \frac{p}{(1-p)^2}$

I have heard someone telling expectation in this case is $\frac{1}{p}$, but I could derive above. Please let me know if I am wrong anywhere.

答:
Your way of proceeding is ok, let us go the same way and get the claimed answer, after a minor bug-fix, which is $P(\underbrace{T\dots T}_{n\text{ times}}H)=q^np$. (Well, please do not write the $\infty$ as a "final term" in the series.)

Start with the formula for $E$ from the OP, using $p$ as the probability for $H$, and the complementary $q=1-p$ for the probability of one $T$ in a simple experiment (one toss):
$$
\begin{aligned}
E &= 1\cdot P(H)+2\cdot P(TH)+3\cdot P(TTH)+4\cdot P(TTTH)+\dots
\\
&=p + 2qp +3q^2 p + 4q^3p+\dots
\\
&=p(1 + 2q+3q^2+4q^3+\dots)\ ,\text{ and after multiplicatioin by $q$}\\
qE &= p(q+2q^2+3q^3+4q^4+\dots)\ ,\text{ so in difference}\\
(1-q)E &= p(1 + q+ q^2+q^3+\dots)=p\cdot\frac 1{1-q}=p\cdot\frac 1p=1\ ,\text{ giving}
\\
E&=\frac 1{1-q}=\frac 1p\ .
\end{aligned}
$$

Title: Expected number of coins tossed


问:
Equivalent of absolute value when referring to reciprocals
I'm currently unaware of terminology or notation that refers to "some function" of a rational number similarly to how "absolute value" relates to a number.  Let me explain.

If one wants to state that a number $x$ must be between $-a$ and $a$ (for some positive value $a$), one can state that $-a < x < a$.  This can also be stated as $|x| < a$, or "the absolute value of $x$ must be less than $a$."

If one wants to state that a number $x$ must be between $\frac{1}{a}$ and $a$, one can state that $\frac{1}{a} < x < a$.  How can this also be stated in a more compact way, similar to the notation/terminology for the absolute value of a number?  Something like $f(x) < a$, or "the magic ratio function of $x$ must be less than $a$."

(As an additional thought, similarly to how the absolute value can be considered the "distance from zero", this unknown function I am trying to find could be considered the "scaling factor from 1.")

答:
If $x$ is between $a$ and $b$, then

$a<x<b \implies \frac{a+b}{2}-\frac{b-a}{2}<x<\frac{a+b}{2}+\frac{b-a}{2}$
$\implies -\frac{b-a}{2}<x-\frac{a+b}{2}<\frac{b-a}{2}$

Therefore
$$\bigg|x-\frac{a+b}{2}\bigg|<\frac{b-a}{2}$$


问:
Minimizing Mean Squared Error for Exponential Function
I have a function that I'm trying to model using an exponential function and I'm trying to determine the constants for the exponential. I know I could optimize it using trial-and-error in R or another language, but I'd like to learn an analytic solution.

I figured that minimizing mean squared error would be the way to go about this, so I have:

$$\underset{r,k}{\operatorname{argmin}}\sum_{t=0}^T (s(t)-\hat{s}(t\mid r,k))^2$$

Where $s(t)$ is the function that I'm trying to model and $\hat{s}(t)$ is $(1+r)^{t+k}$ .

The only constraint I could come up with was $r>0$. I suppose I could also assume $k>0$ for now.

I learned about Kuhn-Tucker constraints as an extension to Lagrange constraints (which I already know) but I wasn't able to solve it.

Am I even going about this the right way? If I am, how can I solve this problem?

Thanks in advance!

答: Your setup is fine.  This sort of problem will not (usually) have an analytic solution.  You have a two-dimensional non-linear minimization problem.  There are  many numeric routines that can solve this in libraries, and they are discussed in any numerical analysis text.  They really consist of informed trial and error, where the informed part comes from keeping track of past trials to build up information about the error function.</p>


问:
$\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}$ equals to $1$ or $e^2$?
$$\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}$$ equals to $1$ or $e^2$?

My Attempt

\begin{align}
\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}
    &= e^{ \displaystyle \lim_{x\to\infty} 2x(x-\sqrt{x^2-2})}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{2x(x^2-(x^2-2))}{x+\sqrt{x^2-2}}}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{4x}{x+\sqrt{x^2-2}}}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{4}{1+\sqrt{1-\frac{2}{x^2}}}}\\
    &=e^2
\end{align}

However, Mathematica Alpha gives a different result which is $1$. Which is the correct one?
If the Mathematica version is correct, how is its answer derived?

Edit

The rule that I am using here is

$$
\lim_{x\to a} (1+f(x))^{g(x)} = e^{\lim_{x\to a} f(x)g(x)}
$$

I am not sure, the constraint for this rule is

both $f(x)$ and $g(x)$ must approach zero or

both $f(x)$ and $g(x)$ must approach infinity

答: Note that $$a^b = e^{b\ln a}$$ So we have $$ (2x+1)^{x-\sqrt{x^2-2}} = e^{({x-\sqrt{x^2-2}})\ln(2x+1)}$$ Also we have $$({x-\sqrt{x^2-2}})\ln(2x+1) = \frac{2\ln(2x+1)}{x+\sqrt{x^2-2}}$$Computing the limit using L'Hôpital's rule gives $$\lim_{x \to \infty}\frac{2\ln(2x+1)}{x+\sqrt{x^2-2}} = \lim_{x \to \infty} \frac{\frac{4}{2x+1}}{1 + \frac{x}{\sqrt{x^2 -2}}}$$ Also we know that $$\lim_{x \to \infty} \frac{4}{2x+1} = 0$$ $$\lim_{x \to \infty} 1 + \frac{x}{\sqrt{x^2 -2}} = \lim_{x \to \infty} 1 + \frac{1}{\sqrt{1 -\frac{2}{x^2}}} = 2$$ So we conclude that $\lim_{x \to \infty} e^{({x-\sqrt{x^2-2}})\ln(2x+1)} =  e^{\lim_{x \to \infty} ({x-\sqrt{x^2-2}})\ln(2x+1)} = e^0 = 1$