问:
Proofs involving totally disconnected space
Below are some proofs involving totally disconnected sets. It would be great if the community can check if the first four are correct, however the last two are the hardest and ones I am stuck on.

"A topological space(X,τ)is said to be totally disconnected if every non-empty connected subset is a singleton set"

i)(X,τ) is totally disconnected if and only if for each a∈X, the connected component of a={a}.

Connected component of a is defined to be the union of all connected sets that contains a and it is the largest connected set that contains a.
Forward: Assume all non empty connected subsets are singletons. Then since singletons are connected, for each a in X, the largest connected set that contains a is just {a} so the connected component of each a in X is just {a}

Backward: Assume for each a∈X, the connected component of a={a} Then there cannot be a connected set that is not a singleton, for if there is a connected set that contains more than one point, then the connected component of a point in that set would contain more than one point which is a contradiction. So all non empty connected subset of X are singletons.

ii)The set Q of all rational numbers with the usual topology is totally disconnected.

It has been previously proven that Q as a subspace of R is not connected. Furthermore if it were possible for a subset of Q with more than one point to be connected, we can always take an irrational in between any two rationals and find a separation.

iii) Indiscrete spaces with more than one point are not totally disconnected.

If an in-discrete space has two points or more, every subset is connected because it is impossible to find open disjoint open sets U and V that can separate the set since the open open sets are the entire set X and the empty set.

iv) Every subspace of a totally disconnected space is totally disconnected

Take a subset A of X and form the subspace topology. Assume A is not totally disconnected, that is assume there is a subset C in A with the subspace topology with more than one point that is connected. Since X is totally disconnected, there must be a separation of C in X. Lets call the open sets U and V that separate C in X. U and V in the subspace topology A forms a separation of C, so C cannot be connected in A, so the only connected subsets of A are the singletons.

v) Every countable subspace of R2 is totally disconnected
If a subspace or R2 is countable, then every element is open so any subset with more than one point is separable.

vi) The Sorgenfrey line is totally disconnected.

Open sets in the Sorgenfrey line are [a,b) and unions those intervals. Any interval of the form [a,b) can be separated by taking a point between a and b, so those intervals are not connected in the Sorgenfrey line. Intervals of the form (a,b) are also open in the Sorgenfrey line and those can be seperated by (a,c), [c,b).

For intervals of the form [a,b] and (a,b], I am unsure how to find a separation for those.

(iv) If f is a continuous mapping of R into Q, prove that there exists a c∈Q such that f(x)=c, for all x∈R; that is, the only continuous functions from R to Q are the constant functions

Using the topological definition of continuity, pre-image of open sets are open. Constant functions are easily seen to be continuous, but how to show that there are the only continuous functions from R to Q?
I can try a proof by contradiction and suppose the mapping is not constant, but it seems difficult showing how that leads to a contradiction. Also, since Q is totally disconnected, I assume we are suppose to somehow use that in the proof. The image of a connected set is connected and preserved by continuity. However R is not connected...

答:
$(i)$ quite is trivial: if $X$ is totally disconnected the components can only be singletons because they are non-empty connected subsets. And if $C \subseteq X$ is non-empty connected it is a subset of some component and thus a singleton (for the reverse).

$(ii)$ needs only the second part of your argument: if $C \subseteq \Bbb Q$ and $q_1 < q_2$ exist in $C$, we can find an irrational $s$ inbetween and then $\{(\leftarrow,s) \cap \Bbb Q, (s,\rightarrow) \cap \Bbb Q\}$ disconnects ($\Bbb Q$ and) $C$. So any subset of $\Bbb Q$ with two or more points is disconnected.

$(iii)$ Any indiscrete space is connected by definition almost, so if has two or more points $X$ itself is a counterexample to being totally disconnected.

$(iv)$ follows directly from the definition as connectedness is absolute: if $C \subseteq Y (\subseteq X)$ were connected and had more than two points, it directly would contradict total disconnectedness of $X$.

$(v)$ Needs a different argument. A connected subset of $\Bbb R$ is either a singleton or an interval. Suppose $A \subseteq \Bbb R^2$ is connected and countable. Then $\pi_x[A]$ is connected but cannot be an interval as $A$ is only countable, so it is a singleton. Likewise for $\pi_y[A]$, so $A$ is a singleton.
It's not true that every (or any) singleton in $A$ is open, we also have sets like $\Bbb Q^2$.

$(vi)$ uses that a set $[a,b)$ in the Sorgenfrey line $\Bbb S$ is both open and closed (clopen): open as it's basic open, closed as $b$ is not an adherence point of it (and $[a,b]$ is also closed, as it refines the Euclidean topology). Also sets like $[a,\rightarrow)$ are clopen too. So if $C$ has two or more points as a subset of $\Bbb S$, say $x < y$, pick $z$ inbetween and $\{[z,\rightarrow), (\leftarrow,z)\}$ disconnects ($\Bbb S$ and) $C$. Quite similar in idea to $\Bbb Q$ or $\Bbb P$ (the irrationals) that also have a base of clopen sets.

In general, by the same proof essentially, any $T_0$ space with a base of clopen sets (a so-called zero-dimensional space) is totally disconnected, provided it has at least two points. (we cannot omit $T_0$ due to the indiscrete example).

$(vii)$ is simpler: if $f:\Bbb R \to \Bbb Q$ is continuous, $f[\Bbb R]$ is connected (by continuity) and non-empty in $\Bbb Q$, which is totally disconnected so the image is a singleton and $f$ is constant.


问:
Conditioning a martingale increment by earlier increments
I have a $L^1$ - martingale ($E[|X|]<\infty$) defined on $(\Omega,\mathcal F , \mathbb P)$, with constant expectation $EX_t$, and I have to prove that $$E\{(X_v-X_u)|(X_t-X_s)\}=0$$ for $0\le s<t\le u<v$.
$$$$
Can I consider $X_t-X_s$ as a filtration, then apply the linearity property?</p>

答:
Notice that $X_t-X_s$ is $\mathcal F_u$-measurable.  Therefore by the tower property we a have

$$\begin{array}{rl}E(X_v|X_t-X_s) &= E(E(X_v|\mathcal F_u)|(X_t-X_s)) \\ &=E(X_u|(X_t-X_s)).\end{array}$$
Hence
$$\begin{array}{rl}E(X_v-X_u|X_t-X_s) &=E(X_v|X_t-X_s)-E(X_u|(X_t-X_s)) \\&=0.\end{array}$$

ContentLicense: CC BY-SA 3.0


问:
Solving difficult exponent equation $x^a-x^b=c$
I have the following equation (derived from a bunch of steps prior but this is the last step that matters)
$$
x^{\frac{p^2+p+1}{p^2}}-x^{\frac{p+1}{p}} = \log(p)
$$
and need to find $x$ given a value of $p$.

It is also sufficient to simplify the notation down to the equivalent form
$$
x^a-x^b = c
$$

This can obviously be calculated numerically, but I would like to have an analytic solution, or at least $\color{blue}{\textrm{a lower bound approximation}}$ which is analytic.

I attempted to get it into a form that could be solved but got stuck (even when using Lambert W functions and such).

If there exists no analytical solution that is also acceptable (albiet sad) to learn

Edit:

This problem actually arises because I wanted to find the maximum value $x$ that satisfies the inequality
$$
x^{\frac{p^2+p+1}{p^2}}-x^{\frac{p+1}{p}} < \log(p)
$$
where $p\in\mathbb{Z}\geq 3$, which is why a lower bound of the value of $x$ that satisfies the inequality would be acceptable as well (and prefered over an upper bound!)

答:
The initial problem is very interesting for numerical analysis.

Consider that you look for the zero of function
$$f(x)=x^{\frac{p^2+p+1}{p^2}}-x^{\frac{p+1}{p}} - \log(p)$$

As @Yuval commented, the solution has to be something like $x=kp^2$ and a quick look at a few numerical results suggest $k\sim \frac 12$.

Making one single iteration of Newton method gives, as an estimate,
$$\color{blue}{x_1=\frac 12 p^2+\frac{2^{\frac{p^2+p+1}{p^2}} p^{2-\frac{2}{p}} \log
   (p)+\left(2^{\frac{1}{p^2}}-p^{\frac{2}{p^2}}\right) p^4}{2
   \left(p^{\frac{2}{p^2}} \left(p^2+p+1\right)-2^{\frac{1}{p^2}} p (p+1)\right)}}\tag 1$$

Checking for a few values of $p$, some results
$$\left(
\begin{array}{ccc}
p & x_1 & x \\
 5  & 10.4182 &  10.3150 \\
 10 & 41.9375 &  41.7121 \\
 15 & 97.4125 &  97.1441 \\
 20 & 177.728 &  177.450 \\
 25 & 283.307 &  283.035 \\
 30 & 414.389 &  414.132 \\
 35 & 571.127 &  570.890 \\
 40 & 753.622 &  753.406 \\
 45 & 961.947 &  961.754 \\
 50 & 1196.15 &  1195.98 \\
 55 & 1456.29 &  1456.14 \\
 60 & 1742.37 &  1742.24 \\
 65 & 2054.43 &  2054.33 \\
 70 & 2392.49 &  2392.40 \\
 75 & 2756.57 &  2756.49 \\
 80 & 3146.66 &  3146.61 \\
 85 & 3562.80 &  3562.76 \\
 90 & 4004.98 &  4004.95 \\
 95 & 4473.20 &  4473.18 \\
 100 & 4967.49 &  4967.48
\end{array}
\right)$$ which does not look too bad.

We could do better using one iteration of Halley or Householder mathod. The formulae will not be printed here (too long and very massy). To give an idea, for $p=5$, the first iteration of Halley method would give $10.3241$ reducing the $1.000$% relative error to $0.088$% (a factor of $\sim 11.5$). The first iteration of Householder method would give $10.3159$.

Edit

If instead of $f(x)$, we consider the function
$$g(x)=\log\Bigg[x^{\frac{p^2+p+1}{p^2}}\Bigg]-\log\Bigg[x^{\frac{p+1}{p}} + \log(p)\Bigg]$$ Let $x=kp^2$ and expand as series for infinitely large $p$, we have
$$g(x)=\frac{\frac{(2 k-1) \log (p)}{k}+\log (k)}{p^2}+\frac{\log (p) \log \left(k
   p^2\right)}{k p^3}+O\left(\frac{1}{p^4}\right)$$ Cancelling the first term gives
$$k=\frac{\log (p)}{W\left(p^2 \log (p)\right)}\implies \color{blue}{x_0=\frac{p^2\log (p)}{W\left(p^2 \log (p)\right)}}$$ which shows an horizontal asymptote $k_\infty=\frac 12$.

Now, one iteration of Newton method gives $x_1$ which is as good as the one given by $(1)$
$$\color{blue}{x_1=x_0+\frac{p^2 x_0^{-1/p}}{\left(p^2+p+1\right) x_0^{\frac{1}{p^2}}-p (p+1)} \left(\log (p)-x_0^{\frac{1}{p}+1} \left(x_0^{\frac{1}{p^2}}-1\right)\right)}$$

Using it for $p=123.456$, $x_0=8150.37$, $x_1=7636.04$. Now, Newton iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 2 & 7636.0400 \\
 3 & 7634.2367 \\
 4 & 7634.2371
\end{array}
\right)$$

ContentLicense: CC BY-SA 4.0


问:
Difference of extremely large squares
This was an extremely difficult question I've encountered in a practice Olympiad test, and I'm not just asking for the answer, I'm asking for the method of solving this and similar problems, so here's the gist:

Let $n$ be the number:

$$(\underbrace{777777777\ldots 777}_{4020~7\text{'s}})^2 - (\underbrace{888888888\ldots 888}_{4006~8\text{'s}})^2$$

So the aim of this question is to find the remainder when $n$ is divided by $11$

Can someone help me understand how to simplify this down and the method and steps for solving this and other similar problems since this might come up again and there is no use to me when I only know the answer and not the strategies involved in getting the answer, Thanks.

答:
All numbers composed by an even number of repeated digit are divisible by $11$, so are their squares and obviously the difference between two of them.

Example. $7777=11\cdot 707, \;7777^2=11^2\cdot 707^2$

The answer is zero.

Title: Difference of extremely large squares


问:
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$?
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?

I need to solve it using combinatorics.
I've started with calculating the number of possible solutions  without restrictions and I am stuck.

答:
Let M be an event such that at least one of a, b, c, d is equal to  . Let N be the event that either of a,b,c,d are greater than  .

Let K be the total number of solutions to the equation .

We need to find

Now,

K=  where

Now to find M, we take M as a union of events , where

.

Using Inclusion-Exclusion

Now to find N, we take N as a union of events as well such that

where

Using Inclusion- Exclusion

Now

Now we just need to find


问:
Proof and transformations
So I got this problem:

I have made a proof, but I'm not sure if it is correct.

$$T(\vec{x})=x_1\cdot T(\vec{e_1})+x_2\cdot T(\vec{e_2})+x_3\cdot T(\vec{e_3})$$

$$S(\vec{x})=x_1\cdot S(\vec{e_1})+x_2\cdot S(\vec{e_2})+x_3\cdot S(\vec{e_3})$$

Then I take:

$T(\vec{x})-S(\vec{x})=x_1\cdot T(\vec{e_1})+x_2\cdot T(\vec{e_2})+x_3\cdot T(\vec{e_3})-x_1\cdot S(\vec{e_1})-x_2\cdot S(\vec{e_2})-x_3\cdot S(\vec{e_3})$$

That can be written as:

$T(\vec{x})-S(\vec{x})=x_1( T(\vec{e_1})-S(\vec{e_1}) )+x_2( T(\vec{e_2})-S(\vec{e_2}) )+x_3( T(\vec{e_3})-S(\vec{e_3}) )$

$T(\vec{x})-S(\vec{x})=0  \rightarrow S(\vec{x})=T(\vec{x})$

Does this seem correct? I'm not a good at proofs so thats the reason why I ask.

答:
Yes, it is correct, but you should have added that $\vec x$ is an arbitrary vector of $\Bbb R^3$ and that $\vec x=x_1\overrightarrow{e_1}+x_2\overrightarrow{e_2}+x_3\overrightarrow{e_3}$.

However, I think that the proof is too long. After having written those expressions for $T\left(\vec x\right)$ and $S\left(\vec x\right)$, you could simply say that, since $T\left(\overrightarrow{e_1}\right)=S\left(\overrightarrow{e_1}\right)$, $T\left(\overrightarrow{e_2}\right)=S\left(\overrightarrow{e_2}\right)$, and $T\left(\overrightarrow{e_3}\right)=S\left(\overrightarrow{e_3}\right)$, the RHS of both expressions are equal.

Title: Proof and transformations


问:
Why does $A_1(T_p \mathbb{R}^n)={T_p}^{*}(\mathbb{R}^n)$
Hi i am reading about manifolds and there it is mentioned that $A_1(T_p \mathbb{R}^n)={T_p}^{*}(\mathbb{R}^n)$ where $A_k(V)$ is the space of alternating k-linear functions. Why does the above equality holds? Does the equality hold in general that is $A_k(T_p \mathbb{R}^n)={T_p}^{*}(\mathbb{R}^n)$. I know that on the rhs we have the cotangent space at $p$. Is this by definition or is there a proof for this? Is this because the basis for both lhs and rhs is same?

答: $A_1(\mathbb{R}^n)$ is the space of all alternating 1-linear functions. That is functions from $V\to \mathbb{R}$ and that we know equals to the dual space $V^*$. Hence proved


问:
Topology on a 3-elements set: how are the definitions via open set and via neighborhood consitent with each other?
Given the set of 3 points X={A, B, C}, wikipedia says that the collection N={{},{A},{A,B,C}} is a topology on X according to the so-called open-sets definition. I.e., N is a topology because: (i) the empty set and X both belong to N; (ii) any arbitrary union of elements of N belongs to N; (iii) arbitrary intersection of elements of N also belongs to N.

But the definition via neighborhoods says that every superset of a neighborhood of a point in X must also be a neighborhood of this point. So if {A} is a neighborhood of A, then {A,B} and {A,C} should also be a neighborhood of A.

To me it seems that the collection N={{},{A},{A,B,C}} is a topology according to the open-sets definition, but not according to the neighborhoods definition. Could anyone explain this seeming "inconsistency"?

P.s.: I am not a mathematician, so please go easy on terminology.

答:
Indeed for $X=\{a,b,c\}$ the collection $\mathcal{T}=\{\{a\},\{a,b,c\},\emptyset\}$ is a topology that obeys the usual open set axioms.

It's not how a topology via neighbourhoods is defined, for that we have to specify a set of neighbourhoods for each point. For this topology (using that a neighbourhood of $x$ is any set that contains an open set containing $x$) the equivalent description in terms of neighbourhoods per point is

$$\mathcal{N}(a)=\{\{a\},\{a,b\},\{a,c\},\{a,b,c\}\}, \mathcal{N}(b)=\{\{a,b,c\}\}, \mathcal{N}(c)=\{\{a,b,c\}\}$$

but if you define the space that way, you have to check a different set of axioms that are defined in terms of such neighbourhood systems. Now I know these are satisfied already because I start with a standard open-set topology.

Both ways of looking at it give the same notions of open/closed/continuity/convergence etc., so in practice we use whatever description is more convenient.

But do note that the descriptions have different "type": one is a single collection of subsets we call open, the other a function that assigns to each point a collection of subsets (the neighbourhoods of that points).


答:
If you read the whole section, before the definition of Topology via open sets you can read the followin about the definition via neighbourhoods

Given such a structure, a subset $U$ of $X$ is defined to be open if $U$ is a neighbourhood of all points in $U$.

In particular, this implies that not every neighbourhood is an open set. It seems to me that you are assumming that a neighbourhood is the same thing a superset, but you actually need to specify what the neighbourhoods of a points are. Neighbourhoods are always supersets of the points the are associated to, but not every superset must be a neighbourhood.

If you take as neighbourhoods of each point all its supersets, you can check that in your example you also get that the subsets $\{A\}$ and $\{A,B,C\}$ are open with respect to the definition of open sets in terms of neighbourhoods given above.

Title: Topology on a 3-elements set: how are the definitions via open set and via neighborhood consitent with each other?

Body:
Given the set of 3 points X={A, B, C}, wikipedia says that the collection N={{},{A},{A,B,C}} is a topology on X according to the so-called open-sets definition. I.e., N is a topology because: (i) the empty set and X both belong to N; (ii) any arbitrary union of elements of N belongs to N; (iii) arbitrary intersection of elements of N also belongs to N.

But the definition via neighborhoods says that every superset of a neighborhood of a point in X must also be a neighborhood of this point. So if {A} is a neighborhood of A, then {A,B} and {A,C} should also be a neighborhood of A.

To me it seems that the collection N={{},{A},{A,B,C}} is a topology according to the open-sets definition, but not according to the neighborhoods definition. Could anyone explain this seeming "inconsistency"?

P.s.: I am not a mathematician, so please go easy on terminology.


问:
Sheaf of $1$-forms on affine opens
First some definitions:

Let $k$ be algebraically closed. Let $X$ be a topological space. We define an algebraic variety $X$ to be a $k$-space $(X,\mathcal O_X)$, such that for each $x\in X$, we have an open $U\subset X$ with $x\in U$, such that $(U,\mathcal O_X\vert_U)\cong (Z,\mathcal O_Z)$, where $Z\subset\mathbb A^n$ is an (embedded) affine variety, and $\mathcal O_Z$ is its sheaf of regular functions. In short: $X$ is locally affine.

An (embedded) affine variety $Z\subset\mathbb A^n$ is a closed subset of $\mathbb A^n$ (with the Zariski topology), and for open $U\subset Z$, $\mathcal O_Z(U)$ is the $k$-algebra consisting of functions $U\to k$ that locally can be written as $f/g$ with $f,g\in k[x_1,\dots,x_n]$.

Given an algebraic variety $X$ we have the so-called sheaf of 1-forms $\Omega^1_X$, which is defined to be the sheafification of the presheaf $U\mapsto \Omega^1_{\mathcal O_X(U)}$, where $\Omega^1_{\mathcal O_X(U)}$ is the module of Kähler differentials of $\mathcal O_X(U)$ (the regular functions on $U$).

Now, my lecture notes state the following two results without proof:

For an affine open $U\subset X$, we have $\Omega^1_X(U)\cong\Omega^1_{\mathcal O_X(U)}$.

For affine opens $V\subset U\subset X$, we have $\Omega^1_{\mathcal O_X(V)}\cong\Omega^1_{\mathcal O_X(U)}\otimes_{\mathcal O_X(U)}\mathcal O_X(V)$, and furthermore, the restriction map $\Omega^1_X(U)\to\Omega^1_X(V)$ is given by $\omega\mapsto\omega\otimes 1$.

When they say the restriction map is given this way, do they mean that we have the following commutative diagram:

Here, the restriction mapping $\Omega^1_{\mathcal O_X(U)}\to\Omega^1_{\mathcal O_X(V)}$ is the one induced from $\iota^*$ (exactly such that this diagram commutes).

答:
Your question: "For an affine open $U \subseteq X$, we have

$$\text{O1}. \text{   }\Omega^1_X(U) \cong \Omega^1_{\mathcal{O}_X(U)}.$$

For affine open sets  $V\subseteq U \subseteq X$ we have

$$\text{O2}. \text{   }  \Omega^1_{\mathcal{O}_X(V)} \cong \Omega^1_{\mathcal{O}_X(U)}\otimes_{\mathcal{O}_X(U)} \mathcal{O}_X(V),$$

and furthermore, the restriction map $\Omega^1_X(U) \rightarrow \Omega^1_X(V)$ is given by $ω \rightarrow ω\otimes 1$.

It seems to me the following is true:

Question O1. Since your sheaf $\Omega^1_X$ is defined as the sheafification $F^{+}$ of the pre-sheaf $F(U):= \Omega^1_{\mathcal{O}_X(U)}$, it seems to me $F^{+}$ and $F$ should have she same sections on affine open subsets $U:=Spec(A) \subseteq X$. Hence

$$F(U)=F^{+}(U) \cong \Omega^1_{A/k}:=\Omega^1_{\mathcal{O}_X(U)}.$$

Question O2. The construction of the module of differentials in functorial in the sense that for any open subscheme $U \subseteq X$ it follows $i^*(\Omega^1_{X/k}) \cong \Omega^1_{U/k}$ where $i: U \rightarrow X$ is the inclusion  map.

Moreover for any pair of maps of commutative rings $A \rightarrow B$ and $A \rightarrow S$
let $B_S :=S\otimes_A B$. There is (Matsumura's book "Commutative ring theory", Exercise 25.4 page 198) a canonical isomorphism

$$ B_S \otimes_B \Omega^1_{B/A} \cong \Omega^1_{B_S/S}.$$

Hence for any inclusion of basic open sets $D(f) \subseteq D(g) \subseteq U:=Spec(A)$
you should get the following:

$$(\Omega^1_{D(g)/k})_{D(f)}:= \Omega^1_{A_g/k} \otimes_{A_g} A_f \cong \Omega^1_{A_{fg}/k}\cong \Omega^1_{A_f/k}.$$

By $(\Omega^1_{D(g)/k})_{D(f)}$ I mean the restriction of $\Omega^1_{D(g)/k}$
to the open set $D(f)$. Here we have used that $D(f)\cap D(g):=D(fg) =D(f)$ since $D(f) \subseteq D(g)$.
The restriction map

$$\rho: \Omega^1_{D(g)/k}\rightarrow (\Omega^1_{D(g)/k})_{D(f)} := \Omega^1_{D(g)/k}\otimes_{A_g} A_f\cong \Omega^1_{D(f)/k}$$

is the canonical map $\rho(\omega):=\omega \otimes 1$ where $1\in A_f$ is the multiplicative identity. Formula O2 should follow since $U$ and $V$ have open covers on the form $D(f)$.

Title: Sheaf of $1$-forms on affine opens


问:
Linearization of a SDE and comparison to deterministic setting
I am a bit puzzled having no experience in this area on the following problem.

It is well know that if we consider a system as:

\begin{equation}
\dot x = f(x(t)), x \in R
\end{equation}

with $f$ of class $C^{\infty}$ it possible to linearize the dynamic around the equlibria (and to be honest along a trajectory with appropriate modifications) via the computation of the Jacobian:

\begin{equation}
\delta \dot x = \dfrac{\partial f(x)}{\partial x}|_{x=x^*}, \quad \delta x = x-x^*
\end{equation}

But how can a similar reasoning apply in the following stochastic setting? Consider for example:

\begin{equation}
dX = f(X)dt + \sqrt{\beta \cdot X}dW, \quad x \in R,\beta \in R, dW \sim \mathcal{N}(0,\,dt)\,.
\end{equation}

where $dW$ is a Wiener process.

I imagine that Jacobian remains the same for $f(X)$ part but how is the term $\sqrt{\beta \cdot X}$ handled?

答:
I have found a solution to my specific problem however it definitely does not suite a formal treatment.

Specifically, I was investigating how to deal with the afore mentioned nonlinear diffusion terms in order to compute a linear model to be used by a Kalman filter.

It appears that it is a common practice - yet not mathematically formal - that the diffusion term is evaluated at the mean value or equilibria under considerations. See this book- Applied Stochastic Differential Equations Simo Särkkä and Arno Solin -, algorithm 9.4 p. 173.

This approximation results in having once again the Jacobian for the drift term $f(X)$ but the a diffusion term that varies with the equilibria/trajectory of the mean under consideration.


问:
How to prove $\int_{0}^{\pi}\frac{\sin^n{x}}{(1+r^2-2r\cos{x})^{(n+2)/2}}\,\mathrm dx=\frac{1}{1-r^2}\int_{0}^{\pi}\sin^n{x}\,\mathrm dx$

Prove that

$$\displaystyle\int_{0}^{\pi}\dfrac{\sin^n{x}}{(1+r^2-2r\cos{x})^{(n+2)/2}}\,\mathrm dx=\dfrac{1}{1-r^2}\int_{0}^{\pi}\sin^n{x}\,\mathrm dx,$$
where $n\ge 1, n\in \Bbb N,|r|<1.$

I think use
$$\dfrac{1-r^2}{1-2r\cos{x}+r^2}=1+2\displaystyle\sum_{m=1}^{\infty}r^m\cos{(mx)},$$
where $|r|<1.$

But it is very ugly.

答: The integral in the LHS is a partial case of formula 3.665.2 from Gradshteyn & Ryzhik. The integral in the RHS also is well-known (ibid 3.621.1 ). I don't find any motivation to reinvent the wheel.</p>


问:
Find the area of the shaded region.

Grade 10 question, but got confused on how to solve?

答:
Hint: Subtract the area of the triangle from the area of the trapezoid. Area of a trapezoid is:  $$\frac{(upper base+lowerbase)\cdot height}{2}$$
Area of a triangle is:
$$\frac{base\cdot height}{2}$$

Title: Find the area of the shaded region.


问:
How many ways to read the palindromic word ROTATOR in this design?
The problem is as follows:

The figure from below shows the word $\textrm{ROTATOR}$ arranged in a
peculiar way. How many ways can this word be read assuming the equal
least distance from one letter to another?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{490 ways}\\
2.&\textrm{480 ways}\\
3.&\textrm{245 ways}\\
4.&\textrm{400 ways}\\
\end{array}$

I noticed that the word is palindromic, hence it can be read back and forth, thus this means that I should account for these possibilites.

In order to keep the right track for this purpose I used an auxiliary numbers atop the letters to account for these as shown in the diagram from below.

After doing all of that I reached the conclusion that:

$\textrm{ways}=(74+74+96)\times 2 =488$

But this doesn't appear in any of the alternatives. Did I mess up something or what? Can someone help me here? Please I require a step by step explanation as I feel lost if my method did worked out properly?

答:
As @Peter hinted in a comment, we have to count for paths going from left to right, right to left as well as left to central $A$ column and back to left and so on.

Notation : $^{m}T^{n}$ indicates $T$ was first reached in $n$ number of ways and later $m$ total number of ways. Read right number first, left number second always.

Left to right paths (unidirectional):
$$
R \quad R \quad R\\
O^{1} \quad O^{2} \quad O^{2} \quad O^{1} \\
T^{3} \quad T^{4} \quad T^{3} \\
A^{3} \quad A^{7} \quad A^{7} \quad A^{3} \\
T^{3} \quad T^{10} \quad T^{14} \quad T^{10} \quad T^{3} \\
O^{13} \quad O^{24} \quad O^{24} \quad O^{13} \\
R^{37} \quad R^{48} \quad R^{37}
$$

Left + Right $= 2(37+48+37)=2\cdot 122 = 244$

Left to center and back to left : (top to bottom to top)
$$
^{34}R^{0} \quad ^{48}R^{0} \quad ^{34}R^{0}\\
^{10}O^{\color{red}{1}} \quad ^{24}O^{\color{red}{2}} \quad ^{24}O^{\color{red}{2}} \quad ^{10}O^{\color{red}{1}} \\
^{10}T^{3} \quad ^{14}T^{4} \quad ^{10}T^{3} \\
A^{3} \quad\quad A^{7} \quad\quad A^{7} \quad\quad A^{3}
$$

Left to center, back to left : $34+48+34=116$

Right to center and back to right : (bottom to top to bottom)
$$
A^{4} \quad\quad A^{7} \quad\quad A^{7} \quad\quad A^{4} \\
^{4}T^{1} \quad ^{11}T^{3} \quad ^{14}T^{4} \quad ^{11}T^{3} \quad ^{4}T^{1} \\
^{15}O^{\color{red}{1}} \quad ^{25}O^{\color{red}{2}} \quad ^{25}O^{\color{red}{2}} \quad ^{15}O^{\color{red}{1}} \\
^{40}R^{0} \quad ^{50}R^{0} \quad ^{40}R^{0}
$$

Right to center, back to right : $40+50+40=130$

Total ways : $244+116+130=490$


问:
How does linear algebra over the octonions and other division algebras work?
An interesting question, which has been discussed in many forms on this site, is how many results from the study of linear algebra over vector spaces carries over when we allow the scalars to form an algebraic structure more general than a field.

For example, if you stop requiring multiplicative commutativity and allow the scalars to form an arbitrary division ring, then a surprising amount of structure carries over unchanged, as discussed here: you lose the notion of the determinant and eigenvalues become more subtle, but you still have unique basis cardinality, Gaussian elimination, the Rouché-Capelli theorem, and matrix representations of arbitrarily linear maps between finite-dimensional modules. A similar story applies if you allow the scalars to form an arbitrary commutative ring. However, all hell can break loose if you drop commutativity and division: a module over an arbitrary ring does not necessary have an invariant basis number, so much of the structure of linear algebra over fields immediately falls apart.

What happens if you further generalize the scalars' division ring to be a division algebra that is not associative? How much structure do you lose if you give up associativity but preserve division?

To be concrete, consider generalized modules with the octonions as their "scalars". (A precise definition of what I mean is given in this question, together with the requirement $(rs)\cdot x = r \cdot (s \cdot x)$.)

Do these modules have a notion of invariant basis number?

If so, can arbitrary linear maps between finitely generated such generalized modules be represented by matrices? Are the matrix elements given by the usual formula $A(\hat{e}_j) = \sum_i A_{ij} \hat{e}_i$?

答:
You definitely want to change your definition of what a module is, since

$(r s) * x = r * (s * x)$  for all $r, s$ in skew field $K$, $x$ in module $M$

is clearly a version of the associative law, which doesn't work for non-associative alternative algebras (like alternativ-körper / alternative skew fields like the octonions).
I'm assuming you know about the commutator

$[a, b] = a b - b a$

and the associator

$[a, b, c] = (a b) c - a (b c)$

already. Sometimes the associator is written as $\{a, b, c\}$ instead, if $[a, b, c]$ is already used for the "jacobian" $[[a, b], c] + [[b, c], a] + [[c, a], b]$. But let's not complicate things too much.

Let's say that the above mentioned left module law is part of the definition of an associative left module.
For an alternative left module we instead want (works in characteristic not 2):

$[r, s, x] + [s, r, x] = 0$,

where $[r, s, x]$ is the left "associator" $(r s) * x - r * (s * x)$, which vanishes for associative modules.
In characteristic 2 (yes there are split octonions in char 2 and other finite characteristics, not really fields, but almost, they are composition algebras) we instead have to use

$[r, r, x] = 0$

since this gives us

$0 = [r + s, r + s, x] = [r, r, x] + [r, s, x] + [s, r, x] + [s, s, x] = [r, s, x] + [s, r, x]$ as desired.

The other direction doesn't hold though. This because in characteristic 2,
a + b = 0 i.e. a = -b is the same as a = b, not forcing
a = -a => a = 0, thus the desired left alternativity
r^2 * x = r * (r * x) is not guaranteed otherwise.

Similarly for an alternative right module we use:
[x, r, s] + [x, s, r] = 0 using the right "associator" [x, r, s] = (x * r) * s - x * (r s)
or
[x, r, r] = 0.

I have know idea if the notion of an alternative "middle module":
[s, x, r] + [r, x, s] = 0 or [r, x, r] = 0
is useful in and of itself, but the middle "associator" [s, x, r] = (s * x) * r - s * (x * r) is definitely useful.

For an alternative bi-module we need more, namely the law
[x, r, s] = [r, s, x] = [s, x, r]
which is not deducible from the above.
This law together with either one of left alternativity, right alternativity or middle alternativity (plus distributivity), give us the other above laws, and a feasible notion of alternative bi-module we seek.

We could go one step further and define a central field of our alternative skew field, that our module must be a vector space over. Or, if we like, a nuclear skew field that our module must be an associative module over.
Nuclear elements n are such that for all r, s :

(n r) s = n (r s)  {left nuclear},

(r n) s = r (n s)  {middle nuclear} and

(r s) n = r (s n)  {right nuclear}.

All such elements n together form the left nucleus, middle nucleus or right nucleus of a Moufang loop or alternative algebra,
and all such elements n satisfying all these three conditions (the intersection of these nuclei) give us simply the nucleus of the Moufang loop or alternative algebra, this nucleus is always a group or associative algebra, respectively;
by using similar laws for certain nuclear scalars acting on the module, we ensure that our base skew field (if any) of certain such scalars for our alternative algebra is functioning as a base skew field for our module as a bi-vector space (left and right vector space, being an associative bi-module) over this skew field. This idea is probably pretty alien if not combined with the next one.
Central elements c are such that for all r :

c r = r c

; these form the commutative center.
The intersection of the nucleus and the commutative center form simply the center, which is always an Abelian group or commutative associative algebra, depending on context;
by using similar laws for (nuclear) central scalars acting on the module we ensure that our base field of certain such scalars for our alternative algebra is functioning as a base field for our module as a vector space over this field.

In the ordinary field case we get (for a, b in base field F; r, s in algebra K; x in module M):

(a r) * x = a (r * x) = r * (a x)

and

(a x) * r = a (x * r) = x * (a r)

for all base field scalars a, algebra scalars r, and module elements x.
Noting that a r = r a, (a r) s = a (r s), (r a) s = r (a s), (r s) a = r (s a), (a b) x = a (b x) and redundantly a x = x a.

In the skew field case we get (for a, b in base skew field F; r, s in algebra K; x in module M):

(a r) * x = a (r * x)

(r a) * x = r * (a x)

(r * x) a = r * (x a)

(a x) * r = a (x * r)

(x a) * r = x * (a r)

(x * r) a = x * (r a).

Noting that (a r) s = a (r s), (r a) s = r (a s), (r s) a = r (s a), (a b) x = a (b x), (a x) b = a (x b) = a x b and (x a) b = x (a b).

Note that the action of an associative algebra over an alternative (left, right, bi-) module may not be associative, and the action of a commutative associative algebra over an associative (left, right, bi-) module may not be commutative.

Thus e.g. the quaternions, as well as the complement of the complex numbers in the quaternions, are (non-commutative) associative bi-modules for the complex numbers.
And the octonions, as well as the complement of the quaternions in the octonions, are (non-associative) bi-modules for the quaternions.
The octonions, the octonion complement of the quaternions, and the octonion complement of the complex numbers are (non-associative) bi-modules for the complex numbers as well.
Using either the real numbers or the rational numbers (or any field in between) as the base field.

This paper is a wee little bit sloppy but still rigorous enough to give a good notion about what bimodules over octonions are, or ought to be:
https://arxiv.org/abs/2007.05375


答:
Note that $(\underbrace{777777777\ldots 777}_{4020~7\text{'s}})^2$ can be written as $S = \sum_{i=0}^{4020} 7 \cdot 10^i $.

Now, one can express $7 \cdot 10^i \text{ mod } 11$ as $7 \cdot (-1)^i$, since $10 = -1  \text{ mod } 11$. Now, replacing this in $S$ we obtain

$$S = \sum_{i=0}^{4020} 7 \cdot (-1)^i = 7 \cdot \sum_{i=0}^{4020} (-1)^i = 7 \cdot 0 = 0$$

One can argue the same for the second sum $F = (\underbrace{888888888\ldots 888}_{4006~8\text{'s}})$, and conclude that $$(\underbrace{777777777\ldots 777}_{4020~7\text{'s}})^2 - (\underbrace{888888888\ldots 888}_{4006~8\text{'s}})^2 = 0^2 - 0^2 \text{ mod } 11 = 0 \text{ mod } 11$$

Therefore, the remainder is $0$

Title: Difference of extremely large squares

Body:
This was an extremely difficult question I've encountered in a practice Olympiad test, and I'm not just asking for the answer, I'm asking for the method of solving this and similar problems, so here's the gist:

Let $n$ be the number:

$$(\underbrace{777777777\ldots 777}_{4020~7\text{'s}})^2 - (\underbrace{888888888\ldots 888}_{4006~8\text{'s}})^2$$

So the aim of this question is to find the remainder when $n$ is divided by $11$

Can someone help me understand how to simplify this down and the method and steps for solving this and other similar problems since this might come up again and there is no use to me when I only know the answer and not the strategies involved in getting the answer, Thanks.


问:
Expected winnings in a game with buckets of prizes
Imagine a game where there are $3$ buckets ($1$, $2$, and $3$) full of an unlimited amount of prizes. You choose a bucket at random and pull out a prize - the prize you pull out has a chance of having "choose another" written on it as well, in which case you once again choose a bucket at random and pull out a prize. Any given prize has a chance of giving you the opportunity of choosing another. If you pull out a prize that doesn't say "choose another" then the game ends.

I want to know what the expected total winnings from this game is.

Say bucket $1$ gives an average prize of $x_1$, bucket $2$ an average prize of $x_2$ and bucket $3$ an average prize of $x_3$, and the corresponding probabilities of giving you another go are $p_1$, $p_2$ and $p_3$.

I thought about setting up an absorbing Markov chain with transition matrix:
$$\begin{pmatrix}\frac{p_1}{3} & \frac{p_1}{3} & \frac{p_1}{3} & 1-p_1 & 0 & 0 \\
\frac{p_2}{3} & \frac{p_2}{3} & \frac{p_2}{3} & 0 & 1-p_2 & 0\\
\frac{p_3}{3} & \frac{p_3}{3} & \frac{p_3}{3} & 0 & 0 & 1-p_3 \\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 1\end{pmatrix}$$

Where the states in order across the top are you pick from bucket $1$, you pick from bucket $2$, you pick from bucket $3$, you finish the game on bucket $1$, you finish the game on bucket $2$ and you finish the game on bucket $3$.

I then tried to do some stuff from the wikipedia page https://en.wikipedia.org/wiki/Absorbing_Markov_chain, but then got stuck.

The matrix $N$ there can give me the probability that I end at a particular bucket, and also the average number of steps at each bucket, but this didn't seem to help me when I tried this experiment in some Python code with actual numbers.

Any help would be appreciated.

答:
Assuming that buckets are chosen uniformly at random, then we can say:

Let $W$ be the expected total winnings

By the law of total expectation, conditioning on the first step of your process, we have

$W = \frac{1}{3}(x_1 + p_1W) + \frac{1}{3}(x_2 + p_2W)+ \frac{1}{3}(x_3 + p_3 W)$

Rearranging for $W$ it follows that
$W = \frac{x_1 + x_2 + x_3}{3 - p_1 - p_2 - p_3}$


问:
An inequality for an increasing, concave function
Let $f:[0,\infty)\rightarrow[0,\infty)$ be a smooth function such that $f(0)=0$,
$f''<0$,  $f'(x)>0$ for $x\in(0,N)$ and $f'(N)=0$. I would like to show that
$$
\max_{x\in(0,\infty)}\bigg(\sum_{k=1}^N \frac{1}{1+|f(k)-x|}\bigg)=\sum_{k=1}^N \frac{1}{1+f(N)-f(k)},
$$
i.e. the maximum is attained at $x=f(N)$.

The geometric considerations seem to confirm the validity of the above claim, but I don't know how to formalize the proof. Any hints will be appreciated!

答:
I think $N = 3$, $f(x) = x \, (2 N - x)$ yields a counterexample. Denote
$$
g(y) := \sum_{k = 1}^N \frac{1}{1 + |f(k) - y|}.$$

Then,
$$
1.1181 \approx g(f(2)) > g(f(3)) \approx 1.0928.$$

(Actually, one has to modify $f$ slightly to obtain a smooth function $f \colon [0,\infty) \to [0,\infty)$, but this can be done without changing $f(1), f(2)$ and $f(3)$)


问:
Determinant of product of symplectic matrices
In optical ray tracing it's possible to use symplectic matrices. I have a problem with them.
If a matrix $M$ is symplectic, this means that for $M$ the following equation hols:
$$M^T\Omega M=\Omega$$
where
$$\Omega =
 \begin{pmatrix}
  0 & I_n & \\
-I_n & 0
 \end{pmatrix}$$
The determinant of $M$ is one:
$$det(M)=1$$
If I have a product of symplectic matrices:
$$M_t=\prod_{k=1}^NM_k$$
the determinant of the product is the same way $1$.
So, how is it possible to prove:
$$det(M_t)=1$$
Thanks</p>

答: Use $det(AB...Z)=det(A)det(B)...det(Z)$ and the fact that $det(M_k)=1$.</p>


问:
What is my mistake in this determinant of order $n$?
Consider real numbers $a_i,b_i\in\mathbb{R}$ and the following determinant

$$ \alpha_n= \begin{vmatrix} a_1 + b_1 & b_1 & b_1 & b_1&\cdots & b_1 \\
b_2 & a_2 + b_2 & b_2 & b_2 & \cdots & b_2 \\
 b_3 &   b_3 & a_3 + b_3 & b_3 & \cdots & b_3 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
b_n & b_n & b_n &  b_n & \cdots & a_n + b_n\end{vmatrix} $$
My attempt was to substitute column 1 with column 1 - column 2 $(C_1 \to C_1 - C_2)$. Then do the same with others columns: $C_2 \to C_2 - C_3$,..., $C_n\to C_n - C_1$. After all these transformations, arise this determinant
$$ \alpha_n =\begin{vmatrix}
a_1   &0& 0 &0&\cdots & -a_1 \\
-a_2 &a_2 & 0 & 0 & \cdots & 0 \\
0 &  -a_3 & a_3  & 0 & \cdots &0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 &  0 & \cdots & a_n \end{vmatrix}  $$
So it seems that $\alpha_n$ only depends on $a_i$. However, computing the case $n=2$, gives
$$ \alpha_2 = \begin{vmatrix}
a_1 +b_1  &b_1 \\ b_2 & a_2 + b_2 \end{vmatrix} = a_1a_2 + a_1b_2 + a_2b_1. $$
What am I doing wrong? How can I compute a closed form for $\alpha_n$? Thank you in advance.

答:
As your error was already clarified in other answers here only a hint to solution will be given.

I will assume $a_i\ne0$. Observe that your matrix is in form:
$$
\alpha=A+u^Tv,
$$
where
$$A=\operatorname {diag}(a_1,a_2,\dots,a_n),\quad u=(b_1,b_2,\dots,b_n),\quad
v=(1,1,\dots,1).
$$
Then by matrix determinant lemma:
$$
\det (\alpha)=(1+vA^{-1}u^T)\det (A)=\left(1+\sum_i\frac {b_i }{a_i}\right)\prod_i a_i.
$$


答:
The straightforward way to get the result is to compute the coefficient of $X^{12}$ in $(1+X+X^2+X^4+X^5+X^6)^4$. Though computing that coefficient basically amounts to counting solutions explicitly in a systematic manner, it is not so hard to do even by hand: $$(1+X+X^2+X^4+X^5+X^6)^2=\\1+2X+3X^2+2X^3+3X^4+4X^5+6X^6+4X^7+3X^8+2X^9+3X^{10}+2X^{11}+X^{12},$$
and from there one easily sees the required coefficient is $2\times(1^2+2^2+3^2+2^2+3^2+4^2)+6^2=122$.

I can also propose a more ad hoc method that might be slightly easier (because fewer and smaller numbers are involved) if one were not allowed to use pen and paper. Most of the solutions will involve two of $a,b,c,d$ being${}<3$ and the other two being${}>3$, the exceptions being the permutations of $(0,4,4,4)$ and of $(2,2,2,6)$, which are $8$ solutions. For the remaining solutions that do have a $2$-$2$ split, there are $\binom42=6$ ways to choose the larger positions, and after we have attributed the minimum of $4$ units to those two positions, we must still distribute the remaining $12-8=4$ units over the $4$ positions, while never putting more than $2$ of those units together. By an argument similar to the one used above, that amounts to finding the coefficient of $X^4$ in $(1+X+X^2)^4$, and from $(1+X+X^2)^2=1+2X+3X^2+2X^3+X^4$ one finds that the number is $1^2+2^2+3^2+2^2+1^2=19$. All in all $6\times19+8=122$ solutions are possible.

Title: If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$?

Body:
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?

I need to solve it using combinatorics.
I've started with calculating the number of possible solutions  without restrictions and I am stuck.


答:
I am not sure what exactly you are asking. If you ask why $\det \prod_{k=1}^N M_k=1$ when $\det M_1,\ldots,\det M_N=1$, this is because the determinant function is multiplicative (i.e. $\det AB=(\det A)(\det B)$). If you are asking why a sympletic matrix has determinant $1$, this has been explained in the related Wikipedia article. If you are asking why $\prod_{k=1}^N M_k$ is sympletic, this is because
\begin{align*}
(M_1\cdots M_k)^T\Omega(M_1\cdots M_k)
&=(M_k^T\cdots M_1^T)\Omega(M_1\cdots M_k) \\
&=M_k^T\left(M_{k-1}\left(\cdots\left(M_2^T(M_1^T\Omega M_1)M_2\right)\cdots \right)M_{k-1}\right)M_k\\
&=\Omega.
\end{align*}</p>

Title: Determinant of product of symplectic matrices

Body:
In optical ray tracing it's possible to use symplectic matrices. I have a problem with them.
If a matrix $M$ is symplectic, this means that for $M$ the following equation hols:
$$M^T\Omega M=\Omega$$
where
$$\Omega =
 \begin{pmatrix}
  0 & I_n & \\
-I_n & 0
 \end{pmatrix}$$
The determinant of $M$ is one:
$$det(M)=1$$
If I have a product of symplectic matrices:
$$M_t=\prod_{k=1}^NM_k$$
the determinant of the product is the same way $1$.
So, how is it possible to prove:
$$det(M_t)=1$$
Thanks</p>


问:
When dealing with an improper antiderivative, why do you take the limit from the left?
The example I have is $$\left[\tan x+\frac{1}{\cos x}\right]_0^{\frac{\pi }{2}}$$

In my mind, you treat it as a limit as $x \rightarrow \frac{\pi}{2}$

However, apparently it its $$\lim _{x\to \left(\frac{\pi }{2}\right)^-}\left(\tan x+\frac{1}{\cos x}\right)$$ because "you approach $\frac{π}{2}$ from the left".

Why is that so? The source is here: https://www.youtube.com/watch?v=TJzAp_NSsbM&t=0s (timestamp 4:45)

答:
Imagine that you decide to compute the integral as the Reimann sum

$$\lim_{n \to \infty} \sum_{k=1}^n g(k) \times \frac{\pi/2}{n},$$

where $g(k)$ is the value of the function $f(x)$ at $x= (k−1) \frac{π/2}{n}$.

Then in this Reimann sum, the value of $f(x)$ at any value $x \geq (\pi/2)$ is totally irrelevant.

Consequently, when examining the limit as $x$ approaches $\pi/2$ of the anti-derivative $F(x)$, you dismiss the limit as $x$ approaches $\pi/2$ from above as totally irrelevant.


问:
Can any smooth even function $f: \mathbb{R} \to \mathbb{R}$ be written as a smooth function of $x^2$?
If $f: \mathbb{R} \to \mathbb{R}$ is a smooth ($C^\infty$) even function, then the Taylor expansion at any point contains only even powers of $x$. Is it true that $f$ can be written as a smooth function of $x^2$? Similarly, if $f$ is an odd function, the Taylor expansion around any point has a factor of $x$ that can be factored out. In this case, can $f$ always be written as $x\cdot g(x)$ for some smooth $g$?

答:
Yes.

This is a theorem by Whitney, have a look to this page.


问:
Orthogonal complement of continuous functions on $(0,1)$ where $f(\frac{1}{2}) = 0$

What is the orthogonal complement of $Y = \{ f \in C(0,1):f(1/2) = 0 \}$?

What about $X = \{1, x\}$ in $C[0,1]$

The dot product is naturally $\langle f, g \rangle = \int_0^1f(x) \overline{g(x)}dx$.

I had no idea how to tackle this question from my exam. Judging by the points possible it wasn't that hard.

I felt like the $X^{{\perp}}$ would be $Span(\{0, x^2, x^3, \dots \})$ but wasn't able to prove it rigourously.

I appreciate your help.

答: The orthogonal complement is $\{0\}$ because $Y$ is dense. Given any continuous function $f$ let $f_n(x)=f(x)$ for $|x-\frac 1 2| >\frac 1 n$, $f_n(\frac  1 2)=0$ and let $f_n$ be linear in $[\frac 1 2 -\frac 1 n, \frac  1 2]$ and $[\frac 1 2, \frac 1 n+ \frac  1 2]$. Then $f_n \in Y$ and $f_n \to f$ in the norm induced by the inner product.


答:
Here is a slightly easier argument.

Let's just count how many ways there are to go from an R to an A. The R's on one side give this:

$$R^1 \quad R^1 \quad R^1\\
O^1 \quad O^2 \quad O^2 \quad O^1 \\
T^3 \quad T^4 \quad T^3 \\
A^3 \quad A^7 \quad A^7 \quad A^3$$

Starting at the R's on the other end gives this:

$$A^4 \quad A^7 \quad A^7 \quad A^4 \\
T^1 \quad T^3 \quad T^4 \quad T^3 \quad T^1 \\
O^1 \quad O^2 \quad O^2 \quad O^1 \\
R^1 \quad R^1 \quad R^1$$

Adding those together, there are $7$, $14$, $14$, and $7$ ways to go from any R to each of the central A's. Conversely there are the same number of ways to go from each of those A's back to any R. Combining any R-to-A path with any A-to-R path from the same A we get $7\cdot7+14\cdot14+14\cdot14+7\cdot7 = 490$.

Title: How many ways to read the palindromic word ROTATOR in this design?

Body:
The problem is as follows:

The figure from below shows the word $\textrm{ROTATOR}$ arranged in a
peculiar way. How many ways can this word be read assuming the equal
least distance from one letter to another?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{490 ways}\\
2.&\textrm{480 ways}\\
3.&\textrm{245 ways}\\
4.&\textrm{400 ways}\\
\end{array}$

I noticed that the word is palindromic, hence it can be read back and forth, thus this means that I should account for these possibilites.

In order to keep the right track for this purpose I used an auxiliary numbers atop the letters to account for these as shown in the diagram from below.

After doing all of that I reached the conclusion that:

$\textrm{ways}=(74+74+96)\times 2 =488$

But this doesn't appear in any of the alternatives. Did I mess up something or what? Can someone help me here? Please I require a step by step explanation as I feel lost if my method did worked out properly?


问:
Is is true that if $H$ entails $H'$, then $P(H\mid E)\le P(H'\mid E)$ for any $E$?
Is it a theorem in probability calculus that if $H$ entails $H'$, then $P(H\mid E)\le P(H'\mid E)$ for any $E$? If yes, how do I prove this?

Thanks!

答:
$H \subseteq H'$

$\implies H \cap E \subseteq H' \cap E$

$\implies \mathbb{P}(H \cap E) \leq \mathbb{P}(H' \cap E)$

$\implies \frac{\mathbb{P}(H \cap E)}{\mathbb{P}(E)} \leq \frac{\mathbb{P}(H' \cap E)}{\mathbb{P}(E)}$ (Provided that $\mathbb{P}(E) \neq 0$)

$\implies \mathbb{P}(H | E) \leq \mathbb{P}(H' | E)$


问:
Prove that $(x+1)^x \over x^x $ $=e$ as $x$ approaches infinity.
I graphed this function on desmos and noticed that its value got close to $2.7$. However, the graph soon cut off because the numbers were too big for desmos to calculate. So, here is my attempt. We have $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$ and by Lhopital's rule, we have $\lim \limits_{x \to \infty} {(x+1)^x\ln(x+1)+1 \over x^x\ln(x)}$. I soon realized, using Wolfram that this won't work.

I start at the beginning with $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Let this limit be $L$. So, we have $L = \lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Taking natural log of both sides, we get $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$. I know that $x+1 \over x$ will get close to $1$ as $x$ gets very big, so $\ln({x+1 \over x})$ will approach $0$ when $x$ gets very big. Thus this limit $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$ is a $0*\infty$ case, so I rewrite the limit as $\ln(L) = \lim \limits_{x \to \infty} {x \over \frac1{\ln({(x+1) \over x})}}$. I apply Lhopitals rule, but I end up with another complicated expression on the bottom using Wolfram, that is also an undefined limit. I have tried LHopital's rule twice and a "clever" manipulation, but the problem still won't yield, could someone please help? Thanks.

答:
First note that
$$\frac{(x+1)^x}{x^x}=\left(\frac{x+1}{x}\right)^x=\left(1+\frac{1}{x}\right)^x$$Even if you don't want to take
$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x$$
as your definition of $e$, we can prove it is equal to $e$ by using L'Hopital's rule, as follows. I will use the notation $\exp(x)$ instead of $e^x$ to make the working more clear. By the way, I think you made a small error anyway: $\frac{d}{dx}x^x=x^x(\ln x+1)$, not $x^x\ln x$.
$$\begin{align}
\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x&=\lim_{x\to\infty}\exp\left(\ln\left(1+\frac{1}{x}\right)^x\right)\\
&=\lim_{x\to\infty}\exp\left(x\ln\left(1+\frac{1}{x}\right)\right)\\
&=\exp\left(\lim_{x\to\infty}x\ln\left(1+\frac{1}{x}\right)\right)\\
&=\exp\left(\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)\\
\end{align}$$
Now we are in a position to use L'Hopital's rule!

Hence we have
$$\begin{align}
\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x&=\exp\left(\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)\\
&=\exp\left(\lim_{x\to\infty}\frac{-\frac{1}{x^2}/(1+\frac{1}{x})}{-\frac{1}{x^2}}\right)\\
&=\exp\left(\lim_{x\to\infty}( 1+\frac{1}{x})\right)\\
&=\exp(1)=e
\end{align}
$$
as required! I hope that was helpful. If you have any questions, please don't hesitate to ask :)


问:
Prove $\lim \frac{cos(x)-1}{x} = 0$ without l'Hopital
While deriving $\frac{\rm{d}}{\rm{dx}}\rm{sin}(x)$, using the definition of the derivative and expanding $\rm{sin}(x+h)$ leads to

$\frac{\rm{d}}{\rm{dx}}\rm{sin}(x) = \rm{sin}(x)\lim_{h\to 0} \frac{cos(h)-1}{h} + \rm{cos}(x)lim_{h\to 0}\frac{\rm{sin}(h)}{h}$

The second limit can be evaluated by applying the squeeze theorem, and if I wasn't going from first principles I could use l'Hopital's rule for the first limit, but that would be circular logic.

How can I evaluate $\lim_{h\to 0} \frac{cos(h)-1}{h}$ without l'Hopital?

答:
$$\frac{\cos(h)-1}{h}=\frac{\cos(h)-\cos(0)}{h-0} \longrightarrow \cos'(0)=0$$

(if you don't know yet that $\cos'=-\sin$, you can still get that $\cos'(0)=0$ by noticing that $\cos$ has a local maximum at $x=0$, since $\cos(0)=1=\max_{x \in \mathbb{R}} \cos(x)$)


问:
When will the hour hand be on a whole number if it jerks and it's speed is increasing?
I was recently introduced to modular arithmetic and was playing around with clock related questions, when I thought, if we ignore the minute-hand and second-hand, the hour hand will be on a whole number every 60 minutes.

But if the hour hand jumps directly a certain distance, like say 3 minutes every 1 minute (let's call it $n$), we can see, it will be on a whole number every 20 minutes as $60 \div 3 = 20$. And if 60 was not divisible by $n$, then we can multiply it by 60, for example, if $n=7$, then the answer will be $7 \times 60 = 420$ . So, the first time the hour hand will be on a whole number will be in 60 minutes (or 60 steps) and it will be on the number 7 on the clock.

Now what if the hour-hand gains a constant speed every step? For example, 3 + 5 + 7 + 9 + ...

As in, on the first step it covers 3 minutes, on second step it covers another 5 minutes, so a total of 8 minutes in two steps, and it continues on, going around the 12-hour clock until it reaches exactly a whole number.

I tried to solve it as follows:

$3 + 5 + 7 + 9 + 11 + 13 + ...$ upto  m steps

taking 3 from all numbers gives

$3m + (0 + 2 + 4 + 6 + 8 + ...$ upto m steps)

$= 3m + 2 (0 + 1 + 2 + 3 + 4 + ...$ upto m steps) (taking 2 common)

$= 3m + 2 (1 + 2 + 3 + 4 + ...$ upto (m-1) steps)

$ = 3m + \frac{2(m-1)(m)}{2} $

$ = 3m + (m-1)m$

$ = 3m + m^2 - m$

$ = m^2 + 2m$

Which should be equal to $60n$ I think, (I'm a little confused at this point), but then that gives me a single equation with 2 variables. How do I solve this?

Do I need to put every possible value one by one or there can be a generalized formula here, which can help me to come up with a solution in least number of steps with other values of n, and 12 or 60 as well?

(Note: As you can probably see, I'm not very good at maths, so please try to keep it simple, while giving as much information as possible. I'm willing to study further if needed. Thank you).

答:
To answer your mathematical question; you want to find the smallest positive integer $m$ such that
$$m^2+2m=60n,$$
for some integer $n$. This will be the number of steps $m$ until you reach the first whole number $n$. Phrased in terms of modular arithmetic, you are looking to solve the congruence
$$m^2+2m\equiv0\pmod{60}.$$
A very general and powerful observation is that if $m$ is a solution to the congruence, then so is $m\pm60$. So if any solution exist, then can keep adding or subtracting $60$ to get a solution in the range $0\leq m<60$. This already greatly limits the search for a solution.

In this case, because $60$ is not a very large number, and the left hand side is a fairly simple function, by some educated trial and error you can find that
$$10^2+2\cdot10\equiv0\pmod{60},$$
which shows that after $10$ steps you reach the number $2$. Then it is not hard to check that no smaller positive integer $m$ satisfies the congruence.

Here is an approach that is applicable more generally: If $n$ is a positive integer with prime factorization $n=\prod_pp^{n_p}$, then by the Chinese remainder theorem the solutions to a congruence
$$f(m)\equiv0\pmod{n},$$
correspond bijectively to the simultaneous solutions of the system of congruences
$$f(m)\equiv0\pmod{p^{n_p}}\ \text{ for each prime $p$ dividing $n$}.$$
In this case we have $60=2^2\times3\times5$ and so it suffices to solve
\begin{eqnarray*}
m^2+2m&\equiv0\pmod{2^2},\\
m^2+2m&\equiv0\pmod{3},\\
m^2+2m&\equiv0\pmod{5}.
\end{eqnarray*}
The moduli are now so small that it is easy to simply check all possibilities; we find that
\begin{eqnarray*}
m&\equiv0,2\pmod{2^2},\\
m&\equiv0,1\pmod{3},\\
m&\equiv0,3\pmod{5}.
\end{eqnarray*}
These congruences give $2\times2\times2=8$ solutions to the original congruence, and they are
$$m\equiv0,10,18,28,30,40,48,58\pmod{60}.$$
This shows that indeed $10$ is the smallest positive integer that satisfies the congruence.

If you would like to remove 'guessing' or 'trial and error' completely, then there are general methods to solve the system of simple congruences above. These use the (rather amazing) fact that modulo prime numbers you can use a lot of algebraic tricks you know for real numbers. For example, the quadratic formula still works modulo every prime number $p>2$. So the congruence
$$m^2+2m\equiv0\pmod{3},$$
leads to the solutions
$$m\equiv\frac{-2\pm\sqrt{2^2-4\cdot1\cdot0}}{2\cdot1}\equiv\frac{-2\pm\sqrt{4}}{2}\equiv-1\pm1\pmod{3},$$
where of course $-1+1\equiv0\pmod{3}$ and $-1-1\equiv-2\equiv1\pmod{3}$. You can solve the equation modulo $5$ in the exact same way.


答:
You can do
$$
\frac{\cos h-1}h=\frac{\cos^2h-1}{h(\cos h+1)}=-\sin h \,\frac{\sin h}h\,\frac1{\cos h+1}.
$$
The nontrivial one is always $\displaystyle\frac{\sin h}h$.

Title: Prove $\lim \frac{cos(x)-1}{x} = 0$ without l'Hopital

Body:
While deriving $\frac{\rm{d}}{\rm{dx}}\rm{sin}(x)$, using the definition of the derivative and expanding $\rm{sin}(x+h)$ leads to

$\frac{\rm{d}}{\rm{dx}}\rm{sin}(x) = \rm{sin}(x)\lim_{h\to 0} \frac{cos(h)-1}{h} + \rm{cos}(x)lim_{h\to 0}\frac{\rm{sin}(h)}{h}$

The second limit can be evaluated by applying the squeeze theorem, and if I wasn't going from first principles I could use l'Hopital's rule for the first limit, but that would be circular logic.

How can I evaluate $\lim_{h\to 0} \frac{cos(h)-1}{h}$ without l'Hopital?


问:
The solution of this Cauchy–Riemann equations.
・$f(z)=f(x+iy)=e^xu(y)+ie^xv(y)$.

・$f(z)$ is regular.

・$f(x)=e^x.$ (When $y=0$, $f(z)=e^x$.)

Under this condition, determine $u(y), v(y).$

The answer is following.

From Cauchy–Riemann equations, $u(y)=v'(y), u'(y)=-v(y).$

So, $(u(y)^2+v(y)^2)'=2u(y)u'(y)+2v(y)v'(y)=0.$

Thus $u(y)^2+v(y)^2=k$, and from $f(x)=e^x,$ $u(0)=1, v(0)=0.$ Then $k=1.$

Therefore, $u(y)^2+v(y)^2=1$.

The answer is $u(y)=\cos y, v(y)=\sin y.$ ■

But I'm not convinced. This answer says that
$u(y)^2+v(y)^2=1 \Longrightarrow u(y)=\cos y, v(y)=\sin y.$ Aren't there other $u(y),v(y)$ ?  Only $u(y)=\cos y, v(y)=\sin y$  is solution?

答: I'm not convinced either. But the answer is correct. Note that $u''=-v'=-u$ and therefore $u(x)=\alpha\cos(x)+\beta\sin(x)$ for some constants $\alpha$ and $\beta$, and then $v=-u'=\alpha\sin(x)-\beta\cos(x)$. But $\alpha=u(0)=1$ and $\beta=v(0)=0$.


问:
Computing the appproximation numbers of an operator $T:\ell_2\longrightarrow\ell_2$
I'm practicing for an operator theory exam, and doing exercises from previous exams. One of them is this, it is a test and it is assumed that there is only one correct solution.

Let $T:\ell_2\longrightarrow\ell_2$ be a linear bounded operator defined as
$$T\Bigl(\sum_{n=1}^\infty\langle x,e_n\rangle e_n\Bigr)=\sum_{n=1}^\infty\langle x,e_{2n}\rangle e_n,$$
where
$$e_n=(0,0,\ldots,\underbrace{1}_n,\ldots),\ \forall n\in\mathbb{N}.$$
Show which of the following statements is true:

(a) $T'=T$.

(b) $a_n(T)=1,\ \forall n\in\mathbb{N}$.

(c) $T\circ T(x)=\sum_{n=1}^\infty\bigl(\langle x,e_{2n}\rangle\bigr)^2e_n$.

In this case, $T'$ is the transpose operator of $T$ and $a_n$ is the $n$-th approximation number, defined as
$$a_n(T):=\inf\bigl\{\|T-S\|:\ S\in\mathcal{L}(\ell_2),\  \dim S(\ell_2)<n\bigr\}.\tag{1}$$

If I'm not wrong, I can write, for simplicity
$$T(x_n)=(x_2,x_4,\ldots,x_{2n},\ldots)$$
and calculate the transpose operator
$$T'(y_n)=(0,y_1,0,y_2,\ldots,0,y_n,\ldots)$$
so the first option is false. The third is also trivially false, which leaves me as a good option the second one.

But how can I prove it directly? I am unable to apply the given definition $(1)$. I have an additional lemma from Diestel, Jarchow and Tonge, Absolutely Summing Operators, which says:

Let the compact operator $u:H_1\longrightarrow H_2$ be represented as
$$u=\sum_{n=1}^\infty\tau_n\langle\cdot,e_n\rangle f_n,$$
where $(e_n)$ is an orthonormal sequence in $H_1$, $(f_n)$ is an orthonormal sequence in $H_2$, an $(\tau_n)$ is a null sequence of scalars which satisfy
$$0\leq\tau_{n+1}\leq\tau_n,\ \forall n\in\mathbb{N}.\tag{2}$$
Then, $\tau_n=a_n(u)$, for all $n\in\mathbb{N}$.

My class notes relax this conditions a bit and do not require that $(\tau_n)$ be a null sequence, only that the condition $(2)$ be met.

Anyway, I don't think this can be applied in this case, because $T$ is not a compact operator.

Any idea how I can continue?

答:
Take $S=0$ then $\|T-S\|=\|T\|=1$ and $a_n(T)\le1$.

Take $S$ with $\dim \ S(l^2)<n$ and $\|S-T\|= 1-\epsilon$ for $\epsilon\ge0$. Take $y\in R(S)^\perp$, $\|y\|=1$, which is possible as the range of $S$ is finite-dimensional. Then by definition of $T$ there is $x$ with $\|x\|=\|y\|=1$ and $Tx=y$.
Then
$$
(1-\epsilon)^2 \ge \|(S-T)x\|^2 = \|Sx\|^2+\|Tx\|^2 + 2Re(\langle Sx,Tx\rangle)
=\|Sx\|^2 +1.
$$
This implies $0\le\|Sx\|^2 \le \epsilon^2 -2\epsilon = \epsilon(1-2\epsilon)$, and $\epsilon\le\frac12$, so $a_n(T)\ge\frac12$.
In addition,
$$
1=\|Tx\|\le \|Sx\| +\|(S-T)x\| \le \sqrt{\epsilon^2 -2\epsilon} + 1-\epsilon,
$$
which implies $\epsilon \le \sqrt{\epsilon^2 -2\epsilon}$. And $\epsilon$ has to be zero, and $a_n(T)\ge1$.


问:
What is degree of freedom in statistics?
In statistics, degree of  freedom is widely used in regression analysis, ANOVA and so on. But, what is degree of  freedom ?

Wikipedia said that

The number of degrees of freedom is the number of values in the final  calculation of a statistic that are free to vary.

Mathematically, degrees of freedom is the number of dimension of the  domain of a random vector, or essentially the number of 'free'
  components: how many components need to be known before the vector is
  fully determined.

However, it is still hard for me to understand the concept intuitively.

Question:

Could anyone provide a intuitive explain to the concept and anything can help me understand?

Thanks!

Update:

I'm NOT asking how to calculate degree of freedom.

Let me give an example:

For Chi-squared distribution, different degree of freedom produces different probability density function. Could you explain it intuitively?

答:
To understand the concept intuitively, ask the question of why it is needed. And start from a simpler question, why is the variance computed as $V = \frac 1 {N-1} \sum (x_i-\bar x)^2$ with a $N-1$ as denominator.

The fact is that you are interested in the variance of the population, not of the one of sample. Now obviously the sample is less dispersed than the population (because it is very likely that your sample missed a few of the extreme values), so the variance computed on the sample is lower than the variance of the population.

You have to correct the bias.

Intuitively, the observed average $\bar x = \frac 1 N \sum x_i$ is not exact but only an approximation of the population mean. The variance of this approximation should be added to the observed variance on the sample in order to to get the best approximate of the population sample.
Now this variance can be computed: $\sigma^2(\bar X) = \frac 1 N \sigma^2(X)$ (using the iid if the $X_i$). So the sample variance is $1 -\frac 1 N$ the variance of the population.

Formally (in case you need to read twice the previous reasoning), compute the expected value of $\sum (X_i-\bar X)^2$. You will find $(N-1) \sigma^2$ rather than $N \sigma^2$, hence the population variance is $\frac N {N-1}$ the sample variance (as claimed).

When you follow the computations, you start by replacing $\bar X$ by its definition $\bar X = \frac 1 N \sum X_i$, develop the squares, expand the sum, and then one of the term disappears. Namely $N\bar X=\sum X_i$ appears twice with opposite sign, negative in the double product $2 \bar X X$ and positive in the square $\bar X^2$. So $\sum (X_i-\bar X)^2$ is the sum of $N-1$ terms equal in expectation.

This is because the $X_i$ are not independent but linked by one relation $\sum X_i=N \bar X$. In general, if you know the $X_i$ are linked by $p$ independent linear relations $f_j$, then you can cancel out $p$ terms out of the sum $\sum (X_i-f_j(X_i))^2$. Hence the unbiased estimator, $\sum (X_i-f_j(X_i))^2 \approx \frac N {N-p} \sum (x_i-f_j(x_i))^2$.

In regression, ANOVA, etc, the independents relations are not so independent because it is often supposed that the sum of the independents variables (causes) have the same average than the dependent variable (effect). $\sum a_i \bar X_i = \bar Y$. Hence the degree of freedom $N-1$, $p-1$ and $N-p$ and unbiasing factors $\frac N {N-1}$, $\frac p {p-1}$ and $\frac N {N-p}$  for $SS_{Total}$, $SS_{Model}$ and $SS_{Error}$ respectively.

In two words, the degree of freedom is the number of independent relationships linking a set of variables, taking into account the variable introduced for intermediary estimators.


问:
Uniform convergence with exponential function
I have to study the uniform convergence of the following function for $x \geq 0$:

$$f_n(x)=\frac{ne^x+xe^{-x}}{n+x}$$ the pointwise limit is $f(x)=e^x$. For the uniform convergence i had thought to do so:

$$|f_n(x)-f(x)|=\left|\frac{xe^{-x}-xe^x}{n+x}\right|=\frac{x(e^x-e^{-x})}{n+x}$$ now, since $x \geq 0$ i can say:

$$\frac{x(e^x-e^{-x})}{n+x} \leq \frac{x(e^x-e^{-x})}{n}$$ now I don't know how to go on. Can I say that the last limit goes to $0$ as n goes to $+\infty$, so there's uniform convergence for $x \geq 0$? For $x$ too large, even the numerator goes to $+\infty$, so I don't know what to do.

I would also like to know if there's uniform convergence in $[0, b]$ with $b < + \infty$.

答: If the convegence is uniform then  there exists $m$ such that $|f_n(x)-f(x)| <1$ for all $n >m$ and and for all $x \geq 0$. Take $n=m+1$.  You get a contradiction by letting $x \to \infty$. Note that $\frac x {n+x} \to 1$ and $e^{x}-e^{-x} \to \infty -0=\infty$.


答:
It is easy to see that $\lim_{n\to\infty}\bigl|f_n(n)-f(n)\bigr|=\infty$. Therefore, the convergence is not uniform.

On the other hand, the convergence is uniform in intervals of the type $[0,b]$. You can check that$$f(x)-f_n(x)=2\frac{\sinh(x)x}{n+x}$$and that tharefore$$f'(x)-f_n'(x)=2\frac{n\sinh(x)+x(n+x)\cosh(x)}{(n+x)^2}>0.$$So, the maximim of $f-f_n$ on $[0,b]$ is $f(b)-f_n(b)=2\frac{\sinh(b)b}{n+b}$. So, since $\lim_{n\to\infty}2\frac{\sinh(b)b}{n+b}=0$, the convergence is uniform.

Title: Uniform convergence with exponential function

Body:
I have to study the uniform convergence of the following function for $x \geq 0$:

$$f_n(x)=\frac{ne^x+xe^{-x}}{n+x}$$ the pointwise limit is $f(x)=e^x$. For the uniform convergence i had thought to do so:

$$|f_n(x)-f(x)|=\left|\frac{xe^{-x}-xe^x}{n+x}\right|=\frac{x(e^x-e^{-x})}{n+x}$$ now, since $x \geq 0$ i can say:

$$\frac{x(e^x-e^{-x})}{n+x} \leq \frac{x(e^x-e^{-x})}{n}$$ now I don't know how to go on. Can I say that the last limit goes to $0$ as n goes to $+\infty$, so there's uniform convergence for $x \geq 0$? For $x$ too large, even the numerator goes to $+\infty$, so I don't know what to do.

I would also like to know if there's uniform convergence in $[0, b]$ with $b < + \infty$.


问:
Finding number of tangents to a circle using homogenization

The equation of circle $x^2 + 2gx + 2fy + y^2 =c^2$ , find the pair of tangents from origin.

For applying the result on page-108 of this Differential calculus for beginners by Joseph Edwards (here), I homogenize the equation:

$$ x^2 + 2gxz + 2fyz + y^2 - c^2 z^2=0$$

And then I take partial with $z$

$$ 2gx + 2fy -2c^2 z=0$$

Or,

$$ gx +fy -c^2 z=0$$

Setting $z=1$

$$ gx +fy -c^2=0$$

This would suggest that there is only one tangent from the origin when there are possible with two tangents from origin.. but that's not always the case, there maybe a tangent pair. So, where did I go wrong?

答:
As the book says, you need to calculate partial derivatives $f_x, f_y, f_z$ of your $f(x,y,z)$ function (with a homogenizing variable $z$ added) and plug them as 'coefficients' to obtain the equation of a tangent: $$X\,f_x+Y\,f_y+z\,f_z=0$$
Of course, the equation will have parameters $x,y$ in it, which are coordinates of the point of tangency.

What you did is taking $f_z=0$ with $z=1$.


答:
Let's rewrite your espression by adding and subtracting 1:

$$ (m+1) ^2- 1 \equiv 0 \pmod{60}$$

Which is equivalent to

$$ (m+1) ^2 \equiv 1 \pmod{60}$$

So we are looking for a perfect square that is just 1 next a multiple of 60. Let's change perspective: we are looking for a number 1 next to a multiple of 60 that is a perfect square. So let's begin by listing the numbers of the form $60n+1$ and see if we can spot a perfect square: $1, 61,121, \ldots$ Here it is! Thus $(m+1) ^2 = 121$ is the minimal solution, which gives $m=10$.

Title: When will the hour hand be on a whole number if it jerks and it's speed is increasing?

Body:
I was recently introduced to modular arithmetic and was playing around with clock related questions, when I thought, if we ignore the minute-hand and second-hand, the hour hand will be on a whole number every 60 minutes.

But if the hour hand jumps directly a certain distance, like say 3 minutes every 1 minute (let's call it $n$), we can see, it will be on a whole number every 20 minutes as $60 \div 3 = 20$. And if 60 was not divisible by $n$, then we can multiply it by 60, for example, if $n=7$, then the answer will be $7 \times 60 = 420$ . So, the first time the hour hand will be on a whole number will be in 60 minutes (or 60 steps) and it will be on the number 7 on the clock.

Now what if the hour-hand gains a constant speed every step? For example, 3 + 5 + 7 + 9 + ...

As in, on the first step it covers 3 minutes, on second step it covers another 5 minutes, so a total of 8 minutes in two steps, and it continues on, going around the 12-hour clock until it reaches exactly a whole number.

I tried to solve it as follows:

$3 + 5 + 7 + 9 + 11 + 13 + ...$ upto  m steps

taking 3 from all numbers gives

$3m + (0 + 2 + 4 + 6 + 8 + ...$ upto m steps)

$= 3m + 2 (0 + 1 + 2 + 3 + 4 + ...$ upto m steps) (taking 2 common)

$= 3m + 2 (1 + 2 + 3 + 4 + ...$ upto (m-1) steps)

$ = 3m + \frac{2(m-1)(m)}{2} $

$ = 3m + (m-1)m$

$ = 3m + m^2 - m$

$ = m^2 + 2m$

Which should be equal to $60n$ I think, (I'm a little confused at this point), but then that gives me a single equation with 2 variables. How do I solve this?

Do I need to put every possible value one by one or there can be a generalized formula here, which can help me to come up with a solution in least number of steps with other values of n, and 12 or 60 as well?

(Note: As you can probably see, I'm not very good at maths, so please try to keep it simple, while giving as much information as possible. I'm willing to study further if needed. Thank you).


问:
Is there a simple sequence with bi-alternating parity?
Any polynomial $p$ over $\mathbf Z$ has either alternating or constant parity, i.e., either
$p(n)$ is even for all $n$, $p(n)$ is odd for all $n$, or $p(n)$ alternates; odd, even, odd, even etc.
This follows just by noting that any polynomial $a_dn^d + \cdots + a_1n+a_0$ equals
$$n(a_dn^{d-1}+\cdots+a_1)+a_0,$$
so for even $n$, the parity of $p(n)$ is fixed (it is the same as that of $a_0$), and for odd $n$, the parity is also fixed (it is the same as that of $a_0+a_1+\cdots+a_d$).

I would like a simple expression which has bi-alternating parity, i.e., some expression $f(n)$ whose parity for successive $n$ looks like:
$$\text{even, even, odd, odd, even, even, odd, odd},\dots$$

What is the "simplest" kind of expression I can create which has this property? It cannot be polynomial by the reasoning above, but maybe something algebraic (involving radicals)? I would like to avoid floors/ceilings and have something nicely differentiable.

答:
Probably one of the simplest such sequences is the sequence of triangular numbers, $T_n =  \frac{n(n+1)}{2}$. For nonnegative integers it starts as

$$0, 1, 3, 6, 10, 15, 21, \ldots$$

It is odd when $n = 1, 2 \pmod 4$ and even when $n = 3, 0 \pmod 4$. Note that it can be given as a polynomial expression with coefficients over $\mathbb{Q}$ (but not over $\mathbb{Z}$, as your question shows).


答:
$$x^2+y^2+2 f y+2 g x=c^2\tag{1}$$

The equation of the circle $(1)$ can be written as
$$(x+g)^2+(y+f)^2=c^2+f^2+g^2\tag{2}$$
The center of the circle is $C(-g;\;-f)$ and the squared radius $r^2=c^2+f^2+g^2$

The squared distance from the center to the origin is
$CO^2=g^2+f^2$ therefore we have

$CO^2=g^2+f^2 <r^2$ that is, the origin is internal and so zero tangents, unless $c=0$ which means that the circle passes through the origin.

TL;DR

To have a tangent passing through $O$ must be $c=0$.

So $(1)$ becomes
$$x^2+y^2+2 f y+2 g x=0\tag{1'}$$
A line through the origin has equation $y=mx$

it is tangent to the circle $(1')$ if the equation
$$x^2+(mx)^2+2f(mx)+2gx=0$$
$$\left(m^2+1\right) x^2 +2x (f m+ g)=0\tag{3}$$
has a double root. That is if the discriminant of $(3)$ is zero.

which means
$$(fm+g)^2=0\to m=-\frac{g}{f}$$

The tangent equation is
$$y=-\frac{g}{f}\,x\to gx+fy=0$$

Edit

If the point is external, then the line we get intersects the circle in the tangent points, as the book you have linked says "the $(n-1)th degree curve passes through the points of contact of the tangents"

If the circle is $(x-5)^2+(y-3)^2=1$ and we follow the procedure we get the line $5 x + 3 y=33$ as you can see in the picture below.

In the second picture I draw a more intriguing example. I leave to you  the  details.

Title: Finding number of tangents to a circle using homogenization

Body:
The equation of circle $x^2 + 2gx + 2fy + y^2 =c^2$ , find the pair of tangents from origin.

For applying the result on page-108 of this Differential calculus for beginners by Joseph Edwards (here), I homogenize the equation:

$$ x^2 + 2gxz + 2fyz + y^2 - c^2 z^2=0$$

And then I take partial with $z$

$$ 2gx + 2fy -2c^2 z=0$$

Or,

$$ gx +fy -c^2 z=0$$

Setting $z=1$

$$ gx +fy -c^2=0$$

This would suggest that there is only one tangent from the origin when there are possible with two tangents from origin.. but that's not always the case, there maybe a tangent pair. So, where did I go wrong?


问:
Interesting Theorems on Finitely Generated Abelian Groups?
First time teaching algebraic topology, probably gonna be related to most of my questions on here for a while.

I was wondering if anyone knows of particularly interesting theorems or examples from the theory of finitely generated abelian groups, beyond the standard auxiliary stuff and the classification/primary decomposition.  By auxiliary I mean stuff like the basic results on subgroups, quotients, torsion/rank uniqueness and change of base.  Doesn't have to be something that can quickly be PROVEN from first principles, but as long as it's something digestible and believable (or even better, unbelievable).  Perhaps certain actions on manifolds or trees or the like?  Something for topology-minded folks rather than algebra-minded would be ideal, but whatever you think is cool I'd love to hear about!

It could also be surprising internal/structural features, or results about their automorphism groups.

Thanks, hope someone finds this topic interesting!

答:
A Theorem of Gromov states that: If $G$ is quasi-isometric to $\mathbb{Z}^{n}$ then $G$ has a finite index subgroup isomorphic to $\mathbb{Z}^{n}$. (From a geometric information for the group we get algebraic information.) Therefore $\mathbb{Z}^{n}$  is quasi-isometrically rigid. The proof for general $n$ is quite difficult but for $n=1$ is far easier.

Also an interesting topic is the mapping class group $Mod(M)$ (the group of symmetries of a topological surface $M$). If $M$ is compact and orientable then $Mod(M)$ is generated by Dehn twists. Now two examples are if $M$ is the annulus $S^{1}\times[0,1]$ then $Mod(M)\cong \mathbb{Z}$ and if $M$ is a pair of pants (a sphere with three disks removed) then $Mod(M)\cong \mathbb{Z}^{3}$.


问:
What mathematical operation replaces series of multiplication

Technical details

because I might misunderstand my problem...

I have made an algorithm that calculates, what camera view frustum is needed to view a mesh so that it would be fully visible in the camera view, but as little as possible outside of it.

```<code>Left view - Side view preview
Right view - Desired effect

Red box - Mesh bounds (MMB)
Blue lines - View frustum lines
Green lines - Far view frustum
White line - Center of the red Box from camera
</code>```

Mathematics problem

To calculate the effect (right image) I took most outside points of the bounding box and calculated its position in the Viewport (2D) coordinates system ($\Bbb V$). Then for each vertex of the $\Bbb{V}$, I multiplied it by Left ($\mathbf l$), Right ($\mathbf r$), Top ($\mathbf t$), and Bottom ($\mathbf b$) separately, where $\mathbf l$, $\mathbf r$, $\mathbf t$, $\mathbf b$ are used to create frustum matrix which is later passed to rendering.

To achieve the results I'm looking for I need to multiply $\mathbf l$, $\mathbf r$, $\mathbf t$, $\mathbf b$ multiple times

Additional notation notes:

$\Bbb V$ - Viewport values are [0,1] when inside view of the camera

$\Bbb{Vl}$ - Viewport left corner on X axis position

$\Bbb{Vr}$ - Viewport right corner on X axis position $\Bbb{Vr} = \Bbb{Vl} + \Bbb{V} width$

$\Bbb{Vt}$ - Viewport top corner on Y axis position

$\Bbb{Vb}$ - Viewport bottom corner on Y axis position $\Bbb{Vb} = \Bbb{Vt} + \Bbb{V} height$

Example values:

Starting values
Final values

$\Bbb{Vl}$ = 0.85627
$\Bbb{Vl}$ = 1

$\Bbb{Vr}$ = 0.14078
$\Bbb{Vr}$ = 0

$\Bbb{Vt}$ = 0.72571
$\Bbb{Vt}$ = 1

$\Bbb{Vb}$ = 0.27459
$\Bbb{Vb}$ = 0

$\Bbb{l}$ = -0.01
$\Bbb{l}$ = -0.0072

$\Bbb{r}$ = 0.01
$\Bbb{r}$ = 0.0072

$\Bbb{t}$ = 0.01
$\Bbb{t}$ = 0.0045

$\Bbb{b}$ = -0.01
$\Bbb{b}$ = -0.045

Current solution

Note: Each multiplication is stepped - so $\Bbb{V}$ position is recalculated after mutliplication
$$\mathbf{l} = \prod\mathbf{l}*(1-\Bbb{Vr})  $$
$$\mathbf{r} = \prod\mathbf{r}*(\Bbb{Vl})  $$
$$\mathbf{t} = \prod\mathbf{t}*(\Bbb{Vt})  $$
$$\mathbf{b} = \prod\mathbf{b}*(1-\Bbb{Vb})  $$

Main Question - Is there any way to replace (infinity) multiplications to get proper values of $\mathbf l$, $\mathbf r$, $\mathbf t$, and $\mathbf b$?

Secondary Question - If anwser still will be multiplication, is there any way to fix the issue when $\mathbf l$, $\mathbf r$, $\mathbf t$, or $\mathbf b$ become 0?

Sorry for the wrong notation or any problems with understanding my question...

EDIT:
Video to showcase how it "should work"
Video - As you can see it works because there are around 60 multiplications per second and for my use case I need to make it to one to make my solution optimized

EDIT 2:

Single side value progression example

Top value = $\Bbb{Vr}$ => Approaching 0

Bottom value = $\Bbb{l}$ => Approaching ~0,002691

答:
So I figured this out (but have no idea why this works)

Instead of relying on Viewport values alone - I decided to calculate Viewport to Frustrum at distance (White box - rectV value) and then use Far plane as reference.

Array elements of rectV and farV are points in WorldSpace of each corner

```<code>                Left = rectV[2].x / rectV[2].z / (farV[2].x / farV[2].z) * Right;
                Right = rectV[0].x / rectV[0].z / (farV[0].x / farV[0].z) * Left;
                Top = rectV[0].y / rectV[0].z / (farV[0].y / farV[0].z) * Bottom;
                Bottom = rectV[1].y / rectV[1].z / (farV[1].y / farV[1].z) * Top;
</code>```

If someone smarter then have an idea of why this returns the proper value, and why the order of those calculations do not matter please let me know, I would love to know why...

Body:
So I figured this out (but have no idea why this works)

Instead of relying on Viewport values alone - I decided to calculate Viewport to Frustrum at distance (White box - rectV value) and then use Far plane as reference.

Array elements of rectV and farV are points in WorldSpace of each corner

                Left = rectV[2].x / rectV[2].z / (farV[2].x / farV[2].z) * Right;
                Right = rectV[0].x / rectV[0].z / (farV[0].x / farV[0].z) * Left;
                Top = rectV[0].y / rectV[0].z / (farV[0].y / farV[0].z) * Bottom;
                Bottom = rectV[1].y / rectV[1].z / (farV[1].y / farV[1].z) * Top;

If someone smarter then have an idea of why this returns the proper value, and why the order of those calculations do not matter please let me know, I would love to know why...


问:
Find the probability $\mathbb{P}(U\leq z|U+M=y)$ given $U,M\sim\mathcal{N}(0,1)$
I came here to verify whether my solution is correct or not. I'm not sure if what I'm doing is valid:

Let $U,M$ such that $U,M\sim\mathcal{N}(0,1)$. Assume $U,M$ are statistically independent. Define:

$$\Phi(x)\triangleq\int_{-\infty}^{x}f(t)\ \text{d}t\qquad f(t)\triangleq \frac{1}{\sqrt{2\pi}}e^{-t^2/2}$$

Find the probability $p\triangleq\mathbb{P}(U\leq z|U+M=y)$.

My Attempt:

$$p\underset{(\star)}=\int_{-\infty}^{\infty}\mathbb{P}(U\leq z|U+M=y,M=m)f(m)\ \text{d}m\underset{(\star\star)}=\int_{-\infty}^{\infty}\mathbb{P}(U\leq z|U=y-m)f(m)\ \text{d}m = \\= \int_{y-z}^{\infty}f(m)\ \text{d}m=1-\Phi(y-z)$$

What I'm most unsure about is $(\star)$. $(\star\star)$ is justified (I hope) since $U,M$ are statistically independent.

Thanks!

答:
The conditional distribution of $U$ given $U+M$ is given by:
$$
U\mid (U+M)=y\sim N(y/2,1/2).
$$


问:
Additive group of rational numbers

Let $\mathbb{Z}[\frac{1}{p}]$ be the additive group of rational numbers of the form $mp^n$ where $m$, $n$ are elements of $\mathbb{Z}$ and $p$ is a fixed prime. Describe $\text{End}(\mathbb{Z}[\frac{1}{p})]$ and $\text{Aut}(\mathbb{Z}[\frac{1}{p}])$.

Unfortunately I have never done this type of exercise and do not know where to start: I have great difficulties in general because I do not know how to describe these types of groups. Can you help?

Sorry if my English is not exactly correct.

答:
${\mathbb Q}_p = {\mathbb Z} [1/p]$, right? You changed the notation...

Hi, the endomorphisms $E$ have to map the numbers $mp^n$ to some other numbers $mp^n$. Because it's an additive group and we consider homomorphisms, it's enough to figure out what the generators $p^n$ are mapped to for various numbers of $n$. The multiples by $m$ are obtained by the homomorphism conditions.

Moreover, also because it is a homomorphism,
$$E(p\cdot p^n)  = E(p^n+\cdots p^n)$$
must be equal to
$$E(p^n)+E(p^n)+\dots + E(p^n) = p\cdot E(p^n)$$
This proves that the information about $E(p^{n})$ and $E(p^{n+1})$ is not independent at all; the latter must be $p$ times the former. We may go in both directions; the former is $1/p$ times the latter.

It follows that the endomorphism is totally determined by $E(p^0)=E(1)$ which has to be an element of ${\mathbb Q}_p$. Note that ${\mathbb Q}_p$ was an additive group so $E(1)$ doesn't have to be $1$. It follows that the endomorphisms as a set are ${\mathbb Q}_p$ and the the endomorphisms, like always, are a semigroup, which is the multiplicative semigroup of these elements, ${\mathbb Q}_p^\times$. The superscript $\times$ means the multiplication. So ${\rm End} {\mathbb Q}_p \equiv {\mathbb Q}_p$.

The automorphisms are those endomorphisms that are one-to-one and therefore invertible. What can $E(1)$ be so that $E(mp^n)$ covers the whole ${\mathbb Q}_p$? If
$$E(1) = mp^n $$
then
$$ p = (E(1)/m)^{1/n} $$
For which $m,n$ is the expression above in ${\mathbb Q}_p$ for any $E(1)$ in ${\mathbb Q}_p$?
The $n$th root is clearly too bad and sends us away from ${\mathbb Q}_p$. It can only work for $n=\pm 1$ if $m$ is allowed to be something else than a power of $p$. At any rate, I think that $E(1)$ has to be of the $\pm p^n$ form. The exponent $n$ is an integer so I think that
$${\rm Aut} {\mathbb Q}_p \equiv {\mathbb Z} \times {\mathbb Z}_2 $$
It's the direct product of the integers (with addition, the group shifting things by a power of $p$) and the two-element group (responsible for the overall sign). (Thanks for the fix, no semidirect here.)

Title: Additive group of rational numbers


问:
Estimate the Nevanlinna counting function for zeros of a polynomial
Let $P$ be a polynomial of degree $q,$ say $$P(z)= c_{q}z^{q} + c_{q-1}z^{q-1}+ \cdots + c_0,$$ where $c_q\neq 0.$

Then it is known that $$N\left(r, \frac{1}{P}\right)= q\log{r}$$

I wish to verify this.

For a meromorphic function $f$ on $|z|\leq r\leq\infty,$ the Nevanlinna counting function is  defined as follows:

For $0<t<r,$ we have
$$N\left(r,f\right)= \int_{0}^{r}\frac{n(t,f)-n(0,f)}{t} dt + n(0,f)\log{r},$$ where $n(t,f)$ counts the poles of $f$ in $|z|\leq t;$ $n(0,f)$ counts the poles of $f$ at the origin.

This means that the quantity $N(r,f)$ is a counting function of poles of $f$ inside $|z|\leq r$ and so the quantity $N(r,1/f)$ counts the zeros of $f$ inside $|z|\leq r$

Now given that $P$ is a polynomial of degree $q$ so it is going to have at most $q$ zeros inside $|z|\leq r.$ However I fail to understand how exactly we arive at the formula that $$N\left(r, \frac{1}{P}\right)= q\log{r}$$

Any help is largely appreciated.

答:
$N\left(r, \frac{1}{P}\right)= q\log{r}$ can not hold for all values of $r$, e.g. because $N(r, f) = n(0, f) \log r$ for small $r$. Exact equality for all $r > 0$ holds only if $P$ is a monomial, i.e. $P(z) = c_q z^q$.

What you have is asymptotic equality for $r \to \infty$: All zeros of $P$ are in some disk of radius $R$. For $r > R$ is $n(t, 1/P) = q$ and therefore
$$
\begin{align}
N\left(r,\frac 1P\right)-N\left(R,\frac 1P\right)&=  \int_{R}^{r}\frac{q-n(0,1/P)}{t} dt+ n(0,1/P)\log{r} - n(0, 1/P) \log R \\
&=  q (\log r - \log R) \, .
\end{align}
$$
It follows that
$$
 N\left(r,\frac 1P\right) = q \log r + \text{const} \sim q \log r \, .
$$


问:
Locally convex implies convex?
There is a question about convex set.

Let $\Omega\subset\mathbb R^n$ be an open, connected set. If for any $x\in\overline{\Omega}$, there is a neighborhood $U_x$ such that $\Omega\cap U_x$ is convex, then $\Omega$ is convex.

Intuitively, if $\Omega$ is not convex, there must be a point on $\partial\Omega$ such that $\Omega$ is not convex locally. But I have no idea how to write it clearly. Thanks to your help.

答:
Let $B_r(x) \subset \mathbb R^n$ denote the open ball with center $x \in \mathbb R^n$ and radius $r > 0$. For $x,y \in \mathbb R^n$ and $t \in  \mathbb R$ define $L(x,y;t) = (1-t)x + ty$ and $L(x,y) = \{ L(x,y;t) \mid  t \in [0,1] \}$ = line segment from $x$ to $y$.

A polygonal chain in $\mathbb  R^n$ is a set of the form
$$P(x_1,\ldots,x_n) = \bigcup_{i=1}^{n-1} L(x_i,x_{i+1})$$
The $x_i$ are called its vertices. We also say that $P(x_1,\ldots,x_n)$ is polygonal chain from $x_1$ to $x_n$.

Lemma 1. If $\Omega \subset \mathbb R^n$ is open and connected, then for any two $x,y \in \Omega$ there exists a polygonal chain $P \subset \Omega$ from $x$ to $y$.

Proof. Let $x \in \Omega$ and let $U$ be the set of points $y \in \Omega$ which admit a polygonal chain $P \subset \Omega$ from $x$ to $y$. Then $U$ is open: Let $y \in U$ and $P(x_1,\ldots, x_n) \subset \Omega$ be a polygonal chain from $x$ to $y$. We have $B_r(y) \subset \Omega$ for some $r > 0$, thus for each $y' \in B_r(y)$ the set $P(x_1,\ldots, x_n,y') \subset \Omega$ is a polygonal chain from $x$ to $y'$. Also the complement $V = \Omega \setminus U$ is open: For $y \in V$ we have $B_r(y) \subset \Omega$ for some $r > 0$, and if $y' \in U$ for some $y' \in B_r(y)$, then there is a polygonal chain $P(x_1,\ldots, x_n) \subset \Omega$ from $x$ to $y'$, hence $P(x_1,\ldots, x_n,y) \subset \Omega$ is a polygonal chain from $x$ to $y$, a contradiction. The connectivity of $\Omega$ shows that $U = \Omega$.

Lemma 2. Let $\Omega \subset \mathbb R^n$ be regular (which means that it is open, connected and satisfies the local convexity property of the question) and let $P(x,y,z) \subset \Omega$ be a polygonal chain with three vertices. Then also line segment $L(x,z)$ is contained in $\Omega$.

Proof. Assume that $L(x,z)$ intersects the closed set $C = \mathbb R^n \setminus \Omega$. In particular it is impossible that  $y \in L$ and we conclude that $x, y, z$ lie in a (unique) two-dimensional affine plane $E \subset \mathbb R^n$.

For $t \in [0,1]$ let $T(x, t) \subset E$ denote the solid triangle with vertices $x, y, y_t =  L(y,z;t)$. For $t = 0$ it is degenerate to the line segment $L(x,y)$. The line segment $L(x,y_t)$ is an edge of $T(x,t)$ and we have $T(x,t) = \bigcup_{s \in [0,t]} L(x,y_s)$. Define
$$A = \{ t \in [0,1] \mid T(x, t) \subset \Omega \} , \tau = \sup A .$$
We have $0 \in A$, thus $\tau \in [0,1]$. We claim that $A$ is open in $[0,1]$. Let $t \in A$. Then $[0,t] \subset A$ since $T(x,s) \subset T(x,t)$ for $0 \le s \le t$. The line segment $L(x,y_t)$ has positive distance $d > 0$ from $C$. For suffienctly small $\epsilon > 0$ we see that for all $s \in (t - \epsilon, t + \epsilon)$ each point in $L(x,y_s)$ has distance $< d/2$ from $L(x,y_t)$. Hence $L(x,y_s)$ has positive distance from $C$ for $s \in (t - \epsilon, t + \epsilon)$ which implies that $L(x, y_s) \subset \Omega$ for $s \in (t - \epsilon, t + \epsilon)$. We conclude that $T(x,t') =  \bigcup_{s \in [0,t']} L(x,y_s) \subset \Omega$ for $t' \in [0,1] \cap (t - \epsilon,t + \epsilon)$.

By definition $[0,\tau) \subset A \subset [0,\tau]$. We claim that $\tau \notin A$, i.e. $A = [0,\tau)$. If $\tau = 1$ this is clear since $1 \notin A$ and follows from the fact that $A$ is open if $\tau < 1$.

This means that $\tau$ has the property $T'(x,\tau) =  T(x,\tau) \setminus L(x,y_\tau) \subset \Omega$ and $C_\tau = L(x,y_\tau) \cap C \ne \emptyset$. Therefore $C_\tau \subset \overline \Omega$, but $C_\tau \cap \Omega = \emptyset$. Now define
$$\sigma = \inf\{ s \in [0,1] \mid L(x,y_\tau;s) \in C_\tau\} .$$
Clearly $0 < \sigma < 1$ and $\xi = L(x,y_\tau;\sigma)$ is the point on $L(x,y_\tau) \cap C \subset \overline \Omega$ which has the smallest distance to $x$. Let $U$ be an  open neigborhood of $\xi$ such that $U' = U \cap \Omega$ is convex. There exists $\sigma' \in (0,\sigma)$ such that $\xi'  = L(x,y_\tau;\sigma') \in U'$. We find $r > 0$ such that $B_r(\xi') \subset U'$. The set $B_r(\xi')$ contains a point $\xi'' \in E \setminus T(x,\tau)$. We have $\xi'' \in U'$. The ray which starts at $\xi''$ and goes through $\xi$ contains a point $\eta \in T'(x,\tau) \cap U'$. Since $U'$ is convex, we get $\xi \in L(\xi'',\eta) \subset U' \subset \Omega$, a contradiction since $\xi \in C_r \subset C = \mathbb R^n \setminus \Omega$.

Corollary: Each regular $\Omega \subset \mathbb R^n$ is convex.

Proof: Let $x,y \in \Omega$. Let $P(x_1,\ldots, x_n) \subset \Omega$ be a polygonal chain from $x$ to $y$ with $n$ minimal. Assume $n > 2$. Then Lemma 2 gives us a polygonal chain in $\Omega$ from $x$ to $y$ with $n-1$ vertices, a contradiction. Thus $n =2$, i.e. $L(x,y) \subset \Omega$.

Title: Locally convex implies convex?


问:
If $\langle x,y \rangle=\|x\|\|y\|$ then $y=0$ or $x=\alpha y$

Let X be a (complex) inner product space.
Proof that if for some $x,y \in X$ we have  $\langle x,y \rangle=\|x\|\|y\|$ then $y=0$ or there exists complex $\alpha$ such that $x=\alpha y$

I have tried to proof it using different properties of inner product but that led me nowhere. I appreciate any hints how to solve it.

答:
Case 1: $y=0$, then we are done.

Case 2: let $y \ne 0$ and put $ \alpha = \frac{||x||}{||y||}$ and compute $||x-\alpha y||^2$ with the inner product ( $||z||^2=\langle z,z \rangle$). See what happens !


问:
Hyperbolic Riemannian Metric
Ergodic Theory with a view towards Number Theory, Chapter 9, Page 277, 278.

According to the attached image (below) it seems that $v$ and $w$ should be of the form $v=(z,\alpha)$ and $w=(z,\beta)$, where $\alpha$ and $\beta$ are two complex number and $z\in \mathbb{H}$ ($z$ is also a complex number).

The questions are:

How is the inner product $\langle \cdot,\cdot \rangle_z$ defined?

Is not it supposed to be a real inner product? (...dentos the usual inner product in $\mathbb{C}$ under the identification of ....).

A convenient model for the hyperbolic plane is the upper half-plane
$$\mathbb{H}=\{x+\mathrm{i}y\in\mathbb{C}\mid y>0\}$$
with the hyperbolic metric. To define this metric, we need to introduce the tangent bundle* $\mathrm{T}\mathbb{H}=\mathbb{H}\times\mathbb{C}$ comprising the disjoint union of the tangent planes $\mathrm{T}_z\mathbb{H}=\{z\}\times\mathbb{C}$ for all $z\in\mathbb{H}$. One should think of $\mathrm{T}_z\mathbb{H}$ as a plane touching $\mathbb{H}$ tangentially at $z$ and having no other intersection with $\mathbb{H}$. This suggests that $\mathrm{T}_z\mathbb{H}$ is the natural space for derivatives in the following sense. If $\phi:[0,1]\longrightarrow\mathbb{H}$ is differentiable at $t\in[0,1]$ with $\phi(t)=z$, then we define the derivative of $\phi$ at $t$ by
$$\mathrm{D}\phi(t)=(\phi(t),\phi'(t))\in\mathrm{T}_z\mathbb{H}.$$

Here $\phi'(t)$ is the derivative of $\phi$ as a map into $\mathbb{C}$. We give $\mathrm{T}_z\mathbb{H}$ the structure of a vector space inherited from the second component in $\mathrm{T}_z\mathbb{H}=\{z\}\times\mathbb{C}$.

The hyperbolic Riemannian metric is defined as the collection of inner products*
$$\langle v,w\rangle_z=\frac{1}{y^2}(v\cdot w)$$
for $z=x+\mathrm{i}y\in\mathbb{H}$ and $v,w\in\mathrm{T}_z\mathbb{H}$. Here $(v\cdot w)$ denotes the usual inner product in $\mathbb{C}$ under the identification of $\mathbb{C}$ with $\mathbb{R}^2$ as real vector spaces.

The hyperbolic Riemannian metric induces the hyperbolic metric $\mathsf{d}(\cdot,\cdot)$ mentioned above as follows. If $\phi:[0,1]\longrightarrow\mathbb{H}$ is a continuous piecewise differentiable curve (we will refer to these as paths), then its length is defined by
$$\mathrm{L}(\phi)=\int_0^1\|\mathrm{D}\phi(t)\|_{\phi(t)}\mathrm{d}t.$$
where $\|\mathrm{D}\phi(t)\|_{\phi(t)}$ denotes the length of the tangent vector
$$\mathrm{D}(\phi(t),\phi'(t))\in\mathrm{T}_{\phi(t)}\mathbb{H}$$
with respect to the norm derived from $\langle\cdot,\cdot\rangle_{\phi(t)}$. We will refer to as the speed of the path $\phi$ at time $t$. The hyperbolic distance is now defined as
$$\mathsf{d}(z_0,z_1)=\inf_\phi\mathrm{L}(\phi)$$
where the infimum is taken over all continuous piecewise differentiable curves $\phi$ with $\phi(0)=z_0$ and $\phi(1)=z_1$. It may be checked that this does indeed define a metric on $\mathbb{H}$.

(*) The tangent bundle can be defined abstractly on any manifold, but for our purposes we may think of it as the space in which derivatives live and use an ad hoc definition.

I do appreciate any help could be provided.

答:
The inner product $\langle v,w\rangle_z$ for $v=(z,\alpha)$ and $w=(z,\beta)$ two tangent vectors in $T_z\mathbb{H}$ is defined as
$$
\langle v,w\rangle_z=\frac{\operatorname{Re}(\alpha\bar\beta)}{y^2}
$$
However, since we typically identify tangent spaces of a linear space $V$ with itself, we drop the $z$ and identify $v\in T_z\mathbb{H}$ with the complex number also called $v$ and similarly $w$ to get $v\cdot w=\operatorname{Re}(v\bar{w})$ instead of $\operatorname{Re}(\alpha\bar\beta)$.

Title: Hyperbolic Riemannian Metric


问:
Proving that operator is linear if it is unitary and surjective

Let $X$ be unitary space. Show that surjective operator $T: X\to X$ such that $\langle Tx, Ty\rangle = \langle x, y \rangle \forall x, y \in X$ is linear.

The only thing I was able to conclude (trivial stuff) is that
$$
\langle Tx, Ty\rangle = \langle x, y \rangle \implies \langle Tx, Tx\rangle = \langle x, x \rangle \implies \sqrt{\langle Tx, Tx\rangle} = \sqrt{\langle x, x \rangle} \\
\implies \| Tx\| = \|x\|
$$
so $T$ is isometry.

To show that it is linear I want to have:
$$
T(ax) = aT(x) \\
T(x+y) = T(x)+T(y)
$$

答:
For example, use the following: Let $x,y\in X$. By surjectivity, there is a $z\in X$ such that $y=Tz$. Then

$$\langle T(\alpha x),y\rangle = \langle T(\alpha x), Tz\rangle =\langle \alpha x, z\rangle=\alpha\langle x, z\rangle = \alpha\langle Tx, y\rangle$$

Then we have for all $y$

$$\langle (T(\alpha x)-\alpha Tx),y\rangle=0$$

By non-degeneracy, it follows that $T(\alpha x)-\alpha Tx=0$. I think the second property can be shown in a similar way.


问:
Power Series representation around a=0
Given a function $f(x)=\frac{1}{1-x}$ that has a power series representation around the point $a=0$ given by $f(x)=\sum_k x^k$. Does this allow one to conclude that $\frac25=f(-\frac32)=\sum(-1)^k\frac{3^k}{2^k}$?

答:
Each power series has some radius of convergence $R$, (which in this case, it is easy to see that it is $1$). The convergence radius is a positive real number $R$ which devides the real number line in two distinct regions:

For $|x| \lt R$, the series converges (actually it converges absolutely, which is a stronger kind of convergence), and for $|x| \gt R$ it diverges (does not converge). This fact is true for any power series. What may vary from a power series to the next is what happens when $|x| = R$. However, in this case, since $R = 1$, you actually know that the series does not converge in $x = -3/2$ (by what I stated above). Therefore, you definitely don't have the equality to which you referred.

Hope this helps :).

Title: Power Series representation around a=0