答:
To try and furnish you with some hints, as you request:

first of all, the only proper and rigorous way to define your set $A$ -- for which I will use another symbol, due to reasons of personal syntactic preference -- is as follows:

$$H\colon=\left\{x \in G \mid\ (\exists n, t, m)\left(n \in \mathbb{N} \wedge t \in S^n \wedge m \in \mathbb{Z}^n \wedge x=\prod_{k=1}^nt_k^{m_k}\right)\right\},$$
where it must be specified that by default any product of the form $\displaystyle\prod_{k=p}^qy_k$ -- of a family $y \in G^{[p, q]}$ indexed by the natural interval $[p, q]$ -- is considered with respect to the standard order on the specified interval $[p, q]$ (this needs to be mentioned because in general within a not necessarily commutative monoid $M$ one needs to specify a total order $T$ on the index set $I$ of any family $x \in M^{(I)}$ of finite support, family the product of which one would like to consider; in a general monoid, different total orders on the same index set easily lead to different values of the product of one and the same family).

second, we introduce $H'\colon=\langle S \rangle=\displaystyle\bigcap_{\substack{F \leqslant G \\ F \supseteq S}} F$, by definition the subgroup generated by $S$. As you remarked, the inclusion $H \subseteq H'$ is easy to obtain, since $H'$ being a subgroup is multiplicatively stable and also closed with respect to inverses.

in order to infer the reverse inclusion, you must follow the indications in the above comment of Arturo Magidin and first establish that $H \leqslant G$ is itself a subgroup such that $S \subseteq H$. The fact that $S$ is included in $H$ is a trivial observation, since for any $t \in S$ you have the family $s\colon=\{(1, t)\} \in S^{1}$ together with the family of exponents $m\colon=\{(1, 1)\} \in \mathbb{Z}^1$ such that $t=\displaystyle\prod_{k=1}^1s_k^{m_k}$.

as to why $H$ is a subgroup, notice first it is clearly nonempty because $1_G \in H$ (the unity can be expressed as the product of the empty family of elements of $S$). It is multiplicatively stable because $\displaystyle\prod_{k=1}^ms_k^{p_k}\displaystyle\prod_{k=1}^nt_k^{q_k}$ -- with both families $s \in S^m, t \in S^n$ consisting of elements of $S$ -- can be equivalently expressed as the product of a family in $S^{m+n}$ (I will let you figure out how) and finally closure with respect to inverses is due to a similar token: in general you have $\left(\displaystyle\prod_{k=1}^nt_k^{m_k}\right)^{-1}=\displaystyle\prod_{k=1}^nt_{n+1-k}^{-m_{n+1-k}}$, and as long as $t \in S^n$ is a family of elements of $S$ the "reversed" family $t'\colon=\left(t_{n+1-k}\right)_{1 \leqslant k \leqslant n} \in S^n$ clearly also is a family of elements of $S$.

Title: Subgroup generated by a subset of the group G

Body:
I'm trying to prove the set $A = \{a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n}\mid m_i \in \mathbb{Z} \textrm{ and } 1 \leq i \leq n \}$ is equal to the set $\displaystyle \bigcap_{H \in \mathbb{\mathscr{L}}} H$
where $\mathscr{L}$ is the collection of all subgroups $H$ of $G$ that contains the set $S = \{a_1, a_2, \cdots, a_n\}$

For the first inclusion it was okay, it's easy to show that

$A \subset \displaystyle \bigcap_{H \in \mathbb{\mathscr{L}}} H$ , since $S \subset H$ and $H$ is a subgroup, so we can operate the elements $a_1, \cdots, a_n$ and find another element in $H$.

But I'm having some issues in the second inclusion.

We consider some $k \in \displaystyle \bigcap_{H \in \mathbb{\mathscr{L}}} H$, if $k \in A$, it means that $A$ is a subgroup of $G$ that contains $S.$

My attempt was the One-step test for subgroups,

$x, y \in A \implies xy^{-1} \in A$, so $A \leq G$

We consider $x,y \in A$, it follows that

$x = a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n}$

$y = a_1^{l_1}a_2^{l_2}\cdots a_n^{l_n}$, with $m_i, l_i \in \mathbb{Z}$

Because of the sock-shoes property we have that

$y^{-1} = a_n^{-l_n}a_{n-1}^{-l_{n-1}}\cdots a_1^{-l_1}$

Then we have to show that

$xy^{-1} \in A$, I got stuck here, because for me it seems that I need some commutative property here.

That's the right way to solve this problem? if it's not, how can I show it? Don't give the answer please, just some tips.


问:
Simple Math Problem
A bat and ball cost \$1.10.

The bat costs a dollar more then the ball.

How much does the ball cost?

If this is not the correct place to ask a question like this please tell me and I will remove it :)

The ball costs \$0.05, not \$0.10

答: I'd like to suggest that the naive guess described at the beginning of java's answer will quickly lead to the correct answer if people would just remember the maxim drilled into us by our elementary-school teachers: Check your answer!  If one guesses that the ball costs \$0.10, then checking would lead one to see that the bat costs \$1.10 and therefore the total is \$1.20.  Not only does one then see that the answer is wrong, but I think one also sees why it's wrong; the guessed price of the ball has contributed to the price of the bat too, and as a result, the excess over \$1.00 is twice what it should be.  So correct for that doubling by halving the guess.  I think this approach could be used by people who know no algebra but have good common sense.</p>

ContentLicense: CC BY-SA 3.0

Title: Simple Math Problem


问:
Name of product of dimensions of matrix
We have a matrix of dimensions $m×n$.

Does the product $p=m*n$ have a special name?

It is of course number of items of matrix.

If it has no special name in math, how would you call it if you have to refer to it be words.

Like capacity of matrix, for example, but it is quite ridiculous.

Same thing for a grid... if we have a grid of m rows and n columns how would you call p=m*n?

答: There seems to be no special name for the number of entries in a matrix. But you could still define it as $p=m*n$.

ContentLicense: CC BY-SA 4.0

Title: Name of product of dimensions of matrix


问:
Analytic Topology for Algebraic Topology
I have done basic topology till compactness and connectedness and I want to take the course on Algebraic Topology. I was wondering if Analytic topology is a must do or not and if I should be well versed with Cohomology before starting Algebraic Topology course. Some details:

My course on Analytic topology has:
Separation axioms. Subbases. Lindelöf and countably compact spaces. Seperable spaces. Filters and ultrafilters. Convergence in terms of filters. Tychonoff's theorem. Compactifications, in particular the Alexandroff One-Point Compactification and the Stone-Čech Compactification. Proper maps. Completeness, connectedness and local connectedness. Components and quasi-components. Urysohn's metrization theorem. Paracompactness. Stone's Theorem; that metric spaces are paracompact. Totally disconnected compact spaces and Stone duality.
Algebraic Topology has:
Brief introduction to categories and functors. Applications of homology theory: Invariance of dimension, Brower fixed point theorem.

Chain complexes of free Abelian groups and their homology. Short exact sequences. of chain complexes, the induced long exact sequence in homology, and naturality. The snake lemma, the five lemma, splitting properties for short exact sequences.

Simplicial homology via Delta complexes.

Singular homology of topological spaces, and functoriality. Relative homology. Chain homotopies, homotopy equivalences. Homotopy invariance and excision (details of proofs not examinable). Retractions, deformation retractions, quotients.

Mayer-Vietoris Sequence. Wedge sums, cones, suspensions, connected sums.

Degree of a self-map of a sphere. Application: the hairy ball theorem.

Cell complexes and cellular homology. Equivalence of simplicial, cellular and singular homology.

Cochains and cohomology of spaces. Cup products.

Künneth Theorem (without proof). Euler characteristic. Ext and Tor groups via free resolutions. (Co)homology with different coefficients. The Universal Coefficient Theorem (proof not examinable).

Topological manifolds and orientability. The fundamental class of an orientable, closed manifold and the degree of a map between manifolds of the same dimension. Poincaré duality (proof not examinable). Manifolds with boundary and Poincaré-Lefschetz duality (proof not examinable). Brief discussion of locally finite homology, and cohomology with compact supports. Cap product.

Alexander duality. Applications: knot complements, Jordan curve theorem.

答:
You do not need to be familiar with the contents of the analytic topology syllabus or cohomology before taking this algebraic topology course. This analytic topology course looks to be the sort of thing that is useful for analysts, logicians, and some topologists.

The one thing I would suggest when you sign up for algebraic topology is to ask the instructor if you are expected to be familiar with the fundamental group; it doesn't look necessary, but it's worth asking.


问:
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$
I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$

Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.

答:
Given by a CAS.

Isuppose that is missing the condition $a_3=0$.

Starting from $n=5$, the sequence
$$\left\{\frac{137}{25},\frac{1009}{150},\frac{17953}{2450},\frac{151717}{19600},\frac{
   170875}{21168},\frac{1474379}{176400},\frac{3751927}{435600},\frac{20228477}{22869
   00}\right\}$$ is not recognized by $OEIS$.

Being totally stuck, I used a CAS without conditions and got, after a lot of simplifications,
$$a_n=\frac{(-3 n^4+22 n^3-69 n^2+194 n+288+96 C) } {4 (n-3) (n-2) (n-1) n }+$$ $$\frac{12 (n-4) (n+1) \left(n^2-3 n+6\right) H_{n+1} } {4 (n-3) (n-2) (n-1) n }$$

For $n=0,1,2,3$, this leads to indeterminate forms. For $n=4$
$$a_4=C+\frac{25}{4}=0 \implies C=-\frac{25}{4}$$ leading to
$$a_n=(n-4)\frac{12 (n+1) \left(n^2-3 n+6\right) H_{n+1}-(n-3) \left(3 n^2-n+26\right) } {4 (n-3) (n-2) (n-1) n }$$ which is identical to what @Raffaele wrote in a comment.

Asymptotically,
$$a_n=3 \left(\gamma -\frac{1}{4}\right)+3 \log (n)+\frac{11}{2 n}-\frac{25}{4
   n^2}+O\left(\frac{1}{n^3}\right)$$

I would really like to know how, starting from scratch, we could arrive to the result.


问:
Is there a case where $\int_{0}^{1} p^4(x) dx>0$ but $\int_{0}^{1} x p^2(x)dx=0$?
Is there a case where $$\int_{0}^{1} p^4(x) dx>0$$ but $$\int_{0}^{1} x  p^2(x)dx=0$$, where $p(x)$ is a polynomial? Can anyone prove it? Thanks.

Additionally, here are some of my thoughts,   $\int_{0}^{1} p^4(x) dx>0$ would imply that  $ p^2(x) dx>0$, then $xp^2(x)>0 ,\forall x \in (0,1]$. Can I do something like:$$\int_{0+}^{1} xp^2(x) dx>0$$

Hence, there are no such $p(x)$?

答:
If $\int_{0}^{1} p^4(x) dx>0$, then $p(x)$ is not identically zero on $[0,1]$.

For $x \in [0,1]$, $x\geq 0$ and $p^2(x) \geq 0$ so $xp^2(x) \geq 0$. As a result, $\int_{0}^{1} xp^2(x) dx>0$ holds. If not, then $\int_{0}^{1} xp^2(x) dx=0$ (it can't be negative, so this is the only option), which forces $p(x)=0$ on $[0,1]$ which is a contradiction to $\int_{0}^{1} p^4(x) dx>0$ as explained earlier.


问:
Formal Proof - Premises & Conclusion (Rules of Inference & Equivalence)
So I have gathered/learned a total of 8 different rules of inference & 10 rules of equivalence for proofs: making a total of 18 proofs (Modus Ponens, Modus Tollens, Disjunctive Syllogism, Hypothetical Syllogism, Conjunction, Addition, Simplification, Constructive Dilemma, De Morgan's Law, Association, Distribution, Commutativity, Double Negation, Contraposition, Material Implication, Material Equivalence, Expotation, and Tautology). I want to turn the following premises GIVEN into a conclusion using the rules I know and mentioned.

Premises:

$(G \wedge I) \implies H$

$(I \implies H) \implies F$

Conclusion [What I want] : $G\implies F$

My Progress :

$(G \wedge I) \implies H$

$(I \implies H) \implies F\qquad\qquad\qquad\qquad [ G \implies F]$

$G \implies (I \implies H)\qquad\qquad\qquad\qquad [1, $exp]

$\sim(I \implies H) \vee F\qquad\qquad\qquad\qquad\quad\;    [2, $Impl]

Not sure what else to do.

答:
Using the first premise, we can find an equivalence :
$$
(G \wedge I) \implies H \equiv (G\implies H)\vee (I\implies H)
$$
We have that from second premise $(I \implies H) \implies F$ so notice that :
$$
[(I \implies H)\wedge[(I \implies H) \implies F]]\implies F\qquad\text{[Modus Ponens]}
$$
Therefore :
$$
(G \wedge I) \implies H \equiv (G\implies F)\vee F$$
Which means $G\implies F$.


问:
Does $f(z)=\frac{2z-1}{z^2(z-1)^2}$ have an antiderivative in given domain.
I have the following problem I'm having trouble with :

Question :

Let $f(z)=\frac{2z-1}{z^2(z-1)^2}$ find the domain of $f$ : $D_f$ , does $f$ have an antiderivative in $D_f$

Attempts :

$D_f= \mathbb{C} $ \ {$0,1$}

First thing I did was rewriting  $f(z)$ using partial fractions I got :

$f(z)=\frac{2z-1}{z^2(z-1)^2} = \frac{1}{(z-1)^2}- \frac{1}{z^2}$ , but I got nothing( I thought using the linearity rules of integration proving that both $\frac{1}{z^2}$
and $\frac{1}{(z-1)^2}$ have one )

Or maybe should I use the residue theorem :  $\int_\gamma f(z)~\mathrm dz=2\pi i \sum_{k=1}^n\operatorname{Res}( f, z_k )\,\mathrm{Ind}_\gamma(z_k).$ ?

Any hint or help would be a lot appreciated , thanks in advance

答: It is immediately obvious that $\frac{1}{(z-1)^2} - \frac{1}{z^2}$ has antiderivative $\frac{1}{z} - \frac{1}{z-1}$, which is well defined on $D_f$. Lets say it isnt - the residue theorem tells us (since the residue of f around both of its’ poles is 0) that the integral of f on any closed curve in $D_f$ is 0 - which means f has an antiderivative in $D_f$


问:
Simple COVID Probability Question
This is a relatively simple question that I have been struggling to answer for some reason. When conducting a thought experiment, I was wondering what the approximate probability of catching COVID would be if one were to host a 15 person house gathering. Just as a reference, if approximately 3% of the U.S has either come into contact with the virus or has it, and we were to account for 15 people in a confined space (approximately 4 independent families), how would one calculate the probability of one person catching the virus in a 15 person get-together?

Also, more broadly, what is the general structure of a statistical problem that outlines this kind of an issue? Is there a method or formulaic approach used to calculate questions like these? (I'm new to statistics and learning more each day!)

答:
There are a few things to consider:

The epidemic reproduction rates: Basic reproduction rate, $R_0$, Effective reproduction rate, $R = R_0 x$, where $x$ is the susceptible population

The infection probability distribution as a function of age. These distributions are available from WHO, US CDC and other national health bodies.

The infection probability distribution due to co-morbidity factors. These distributions are available from various clinical research studies. Of course, there are many co-morbidity factors and research is evolving. for eg: co-morbidity involving diabetes, hypertension, respiratory diseases etc.,

Grouping of family units and distribution of member age within family groups. If you set a maximum number of families (unspecified in your case) and a maximum attendee count (15 in your case), you can calculate the combinations of different family unit sizes. This is essentially a partitioning with constraints subproblem. This can be done using text book methods of counting.

Proportion of attendees who have recovered from the disease

The probability distribution of encountering individuals who have already recovered from the disease. There are two theories you could operate under: (a) recovered individuals do not get infected again (b) recovered individuals lose immunity after a period of $t$. This research is also evolving and no evidence yet. But, current belief is scenario (a)

The probability distribution of disease recurrence in recovered individuals. This is if you want to factor scenario (b) above

The probability distribution of asymptomatic individuals who don't show symptoms but are carriers of the disease. Some data is available.

The disease thresholds - Herd immunity and Herd immunity thresholds

The probability distribution of disease immunity in a population. i.e., people who may never contract the disease. This distribution is not known currently and will be known only when large number of studies are done.

Overall disease prevalence vs. population in the geographic area under consideration. This data varies from country to country and state to state and available from the local/national health bodies. This matters because the infection curves are believed to be in different stages in different countries (eg: Singapore vs. India vs. China vs. New Zealand vs. US etc.,)

Detectability of the disease: How accurate are the tests that show a negative result and the probability of having an individual who tested false negative attending the event

There are probably other factors that you want to consider as well in your model. Since the disease is still evolving, we may have to operate with some uncertainty or incompleteness in the model.

The general technique involves determining the probability distribution for the various events and then combining them using tools such as:

Inclusion-Exclusion principle and counting

Bayes' theorem for conditional probability

Bernoulli trials and Poisson sampling

References:

Rothman KJ, Lash T, Greenland S. Modern Epidemiology (3rd ed.), Lippincott Williams & Wilkins, 2013

Coggon D, Rose G, Barker DJP. Epidemiology for the uninitiated, fourth edition, The BMJ. URL: https://www.bmj.com/about-bmj/resources-readers/publications/epidemiology-uninitiated

Title: Simple COVID Probability Question


问:
connections on an embedded sumanifolds of Euclidean space
I would like to see how to derive the connection 1 form on the bundle of oriented orthonormal frames for a manifold embedded in Euclidean space.

It seems that the geometry resides in a collection of 1 forms on the frame bundle the arise naturally from the embedding. Given a frame (e1,...,en) over a point in the manifold, map it to e1. This is a smooth map of the frame bundle into Euclidean space. The 1 forms are then defined as

wi = de1.ei 1=2,...n

The action of the rotation group on the frame bundle leaves these forms invariant because it preserves the Euclidean inner product. That is

dL(e1).L(ei) = Ld(e1).L(ei) = de1.ei

It is tempting to think that these 1 forms are the components of a connection 1 form but I do not not see how to formalize it.

答:
No, you're confusing the frame bundle and the sphere bundle. For the $\mathfrak{so}(n)$-valued $1$-form on the frame bundle, you need
$$\omega_{ij} = de_i\cdot e_j\,, \quad 1\le i,j \le n\,.$$</p>


问:
Matrices commute if and only if they share a common basis of eigenvectors?
I've come across a paper that mentions the fact that matrices commute if and only if they share a common basis of eigenvectors. Where can I find a proof of this statement?</p>

答:
An elementary argument.

Summary: show that each eigenspace of $A$ has a basis such that each basis vector is contained in one of the eigenspace of $B$. This basis is then the simultaneous common basis we are looking for.

Suppose $A,B$ are both diagonalizable and they commute.

Now let $E_{\lambda_i}$ be eigenspaces of $A$ for each distinct eigenvalue $\lambda_i$ of $A$.

Now let $F_{s_i}$ be eigenspaces of $B$ for each distinct eigenvalue $s_i$ of $B$.

Now I claim that $E_{\lambda_i}$ (of say dimension $m$) has a basis $v_1^i,...,v_m^i\in E_{\lambda_i}$ such that each $v_r^i$ is in one of $B$'s engenspace $F_{s_j}$--this would imply these $v_r^i$ are eigenvectors of $B$ and $A$ simultaneously. Apply this to all eigenspaces $E_{\lambda_i}, i=1,...,n$. The collection of all $v_r^i$ then becomes a common basis for $A$ and $B$ as required.

To show this claim, first pick arbitrary basis $w_1,...,w_m$ of $E_{\lambda_i}$. Each $w_i$ can be written as sum of vectors where each vector is in one of $B$'s engenspace $F_{s_j}$. This is a subtle point so let me repeat: for each $i=1,...,m,$ $w_i=z_1^i+...+z_{l_i}^i, l_i\le m$ and $z_k^i\in F_{s_j}$ for some $j$. This is trivially true because direct sum of $B$'s engenspaces is the entire space.

Now we make a second claim that all $z_k^i\in E_{\lambda_i}$. Then the collection of all $z_k^i$ span $E_{\lambda_i}$ and thus the collection can be reduced to a basis $v_1,...,v_m$ where each $v_j$ is contained in $E_{\lambda_i}$ as required by the first claim.

Note that $B$ is invariant to $E_{\lambda_i}$ since $A,B$ commute. The second claim follows from: $\sum_{i=1}^N z_i \in S$ where $z_i$ are eigenvectors of distinct eigenvalues of $B$ and $S$ is a subspace to which $B$ is invariant, then $z_i\in S,\forall i$. We check this by induction on $N$. It is trivially true for $N=1$. Then suppose $Bz_1=\lambda z_1$. Since $\lambda(z_1+...+z_N)\in S$ and $B(z_1+...+z_N)\in S$, we have $B(z_1+...+z_N)-\lambda(z_1+...+z_N)=a_2z_2+...+a_Nz_N\in S$ for some constant $a_i\neq 0$--the constants are non-zero because we assumed $z_i$ all have distinct eigenvalues. Then apply inductive hypothesis $z_2,...,z_N\in S$. This would imply $z_1\in S$ as well. This finishes the proof.


问:
$V = \text{null}T \oplus \text{range}T$ for all $T\in \mathcal{L}(V)$ is not true - what is the intuition?
$V = \text{null}T \oplus \text{range}T$ for all $T\in \mathcal{L}(V)$ is not true. What is the intuition behind this? I know that if $\dim V = n$, then $V = \text{null}T^n \oplus \text{range}T^n$ holds.

A good way to check the claim is to find a counterexample. Let's say $V = \mathbb{R}^2$ and $T:V\to V$ such that $T(x,y) = (y,0)$. Here, $\text{null}T = \{(x,y)\in \mathbb{R}^2: y=0\} = \text{range}T$, so $\text{null}T + \text{range}T$ is not a direct sum.

答:
The rank-nullity theorem implies the dimensions of the kernel and range sum to $n$, so the two subspaces sum to $V$ if and only if they sum directly, which in turn is equivalent to
$$\operatorname{null} T \cap \operatorname{range} T = \{0\}.$$
However, $T$ can map $V$ into any subspace it wants; there's nothing to stop it from mapping non-zero vectors into its own kernel. In the example you give, the range and the kernel are the same, so $T$ maps vectors exclusively into its own kernel.

"Most" operators will satisfy $\operatorname{null} T \oplus \operatorname{range} T = V$ (indeed, most are invertible, and thus satisfy $\operatorname{null} T = \{0\}$ and $\operatorname{range} T = V$), but some will inevitably have their nullspace and range intersect. Note that, for such vectors, we have a non-zero vector $v \in \operatorname{null} T \cap \operatorname{range} T$. As $v \in \operatorname{range} T$, there exists some $w \in V$ such that $Tw = v$ (necessarily, $w \neq 0$). As $Tw = v \in \operatorname{null} T$, we have $T^2 w = 0$. So, $Tw = v \neq 0$, but $T^2w = Tv = 0$, so
$$w \in \operatorname{null} T^2 \setminus \operatorname{null} T.$$
This implies that the generalised eigenspace corresponding to eigenvalue $0$ is strictly larger than the eigenspace corresponding to eigenvalue $0$ (a.k.a. the nullspace of $T$). It's also not difficult to prove the converse of this too.


问:
solving linear recurrence - general solution confusion
I've been trying to get my head around this for days. I understand what is going on with the calculation of a linear recurrence and I also understand how the characteristic is obtained.

What is confusion me is the general solution.

The general solution to the recurrence $at(n) + bt(n-1) ct(n-2) = 0$

For a unique root $r$, the general solution is: $t(n) = (A + Bn)  r^n$. For two distinct roots $r_1$ and $r_2$, the general solution is: $t(n) = Ar_1^n + Br_2^n$

What are $A$ and $B$? What do they represent in the linear recurrence and in the general solution

I just can't see it and I need to get it so I can start my discrete math assignment ...

Thanks heaps in advance :)

答:
A nice way of looking at this is using the techniques in Wilf's "generatingfunctionology" (2nd edition if free on the 'net).

Start with:
$$
u_{n + 2} + a u_{n + 1} + b u_n = c_n
$$
and starting conditoons $a_0$, $a_1$.

Define generating functions $U(z) = \sum_{n \ge 0} u_n z^n$ and $C(z) = \sum_{n \ge 0} c_n z^n$. You'll have to get $C(z)$ somehow.

Multiply the recurrence by $z^n$ and add for $n \ge 0$, identifying the resulting sums gives:
$$
\frac{U(z) - a_0 - a_1 z}{z^2} + a \frac{U(z) - a_0}{z} + b U(z) = C(z)
$$
Solve for $U(z)$, and split in partial fractions. This will very often give terms of the forms:
$$
\frac{1}{(1 - \alpha z)^m}
  = \sum_{n \ge 0} \binom{-m}{n} (-\alpha)^n z^n
  = \sum_{n \ge 0} \binom{n + m - 1}{m - 1} \alpha^n z^n
$$
Here $\binom{n + m - 1}{m - 1}$ is just a $m - 1$-degree polynomial in $n$:
$$
\binom{n + m - 1}{m - 1}
  = \frac{(n + m - 1) (n + m - 2) \ldots (n + 1)}{(m - 1)!}
$$
You get $u_n$ by picking the coefficient of $z^n$ en each of the sums.

There definitely is a relation between the coefficients in $u_n$, but there isn't any simple formula.


问:
Problem with right angle triangle, circumscribed circle, tangent and the half of its height
(An interesting problem inspired by this one but still different. And, no, I'm not looking for your help to solve a detail here in order to provide a full solution elsewhere. I'll stop here).

A right-angled triangle ABM is given ($\angle M = 90^\circ)$. A tangent $t$ is drawn at point $M$ to the circumscribed circle with center $O$. Perpendicular bisector $OE$ of side $AM$ meets tangent $t$ at point $D$. Prove that line $DB$ cuts triangle height $MH$ in half ($CM=CH$)

Ok, one way to solve this is to use analytical geometry. It's doable but tedious. More ellegant way is to solve the problem by using complex geometry and I will demonstrate that solution here:

We'll take circumscribed circle to be unit circle. Let's assign the following complex numbers to various points: $z_M=z,z_0=0,z_A=-1,z_B=1$. It's a well know formula that:

$$z_D=\frac{2z_Mz_A}{z_M+z_A}=\frac{-2z}{z-1}$$

Also notice that:

$$Re(C)=Re(M)=\frac{z+\bar{z}}{2}$$

$$Im(C)=\frac12 Im(M)=\frac12\frac1{2i}(z-\bar{z})$$

$$z_C=Re(C)+i\cdot Im(C)=\frac{z+\bar{z}}{2}+\frac14(z-\bar{z})=\frac 14(3z+\bar z)$$

The last thing we have to prove is that points $D,C,B$ are colinear which is true if:

$$z_D-z_B=k(z_C-z_B)$$

...for some real $k$. In other words:

$$\frac{-2z}{z-1}-1=k(\frac 14(3z+\bar z)-1)\iff$$

$$\frac{-2z-(z-1)}{z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{3z+\bar z-4}{l}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$k=\frac4l, \quad l=(z-1)(\bar z - 1)\in R$$

Question: I think we can prove this with Euclid too, without complex or any other numbers. Can someone come up with one such solution?

答:
extend $BM$ until it meets the  line $AD$ at $X$ since $\angle AMX=90$

$AD$ is  $\frac{AX} {2}$  by homothety  $CH$ is $\frac{MH} {2}$


问:
Combinatorics: Systems of distinct representatives
I have two related(but separate) questions that are both heading toward something I am trying to prove(I think anyway).

Let's assume that $A_1,A_2,...,A_n$ is a collection of subsets of $S$ such that $|A_i|=m$ for all $i$ and every element in the set $S$ belongs to exactly $m$ of the sets $A_1,A_2,...,A_n$. Prove that the sets $A_1,A_2,...,A_n$ have a system of distinct representatives.

Obviously I am supposed to prove that the criterion in Hall's theorem holds. And I see that $k\leq k\cdot m = \sum_{i=1}^{k}|A_i|$, but I cant really make the needed connection between the union $|\bigcup_{i=1}^{k}A_i|$ and $k$.

Let's assume now that $|S|=m\cdot n$ and that $S=A_1\cup A_2\cup...\cup A_m=B_1\cup B_2\cup...\cup B_m$, where $|A_i|=|B_i|=n$ for all $i$. Prove that it is possible to re-index the sets $A_i$ such that $A_i\cap B_i=\emptyset$ for all $i$.

I don't really know where to start with this one and I am not entirely sure if it is even true but I think it is. Any help would be appreciated!

答:
$$\sum_{i=1}^{k}|A_{i}|=mk$$
Note that every element in $\bigcup_{i=1}^{k}A_{i}$ appears at most $m$ times in the left hand side of the above equation, i.e. it is contained in at most $m$ of the sets $A_{1},\cdots,A_{k}$, which guarantees that
$$\sum_{i=1}^{k}|A_{i}|\leq  m\left|\bigcup_{i=1}^{k}A_{i}\right|.$$
Combine this inequality with the equation, and we get the desired
$$\left|\bigcup_{i=1}^{k}A_{i}\right|\geq k.$$
A good view of the question is think of the elements and the sets as vertices of a bipartite graph, in which an element-representing vertex is connected by an edge to a set-representing vertex iff the element is contained in the set. The result we've just got translates to the fact that a $m$-regular bipartite graph always has a complete matching.


问:
Polynomial approximation of $\sqrt{\log(1/x)}$ or bound on coefficients.
I want to obtain a polynomial approximation (in a closed form) of the function $\sqrt{\log(1/x)}$ centered in any point $c \in (0, 1]$  i.e. the series $S(x-c)= \sum_n^\infty a_k (x-c)^k$. For convenience, we take $c=1/e$.

If this is too difficult to obtain, I am fine with a bound on the coefficients, i.e. I want to prove (or have a counterexample) that each $|a_k| \leq |b_k|$, where $b_k$ is the coefficient of the polynomial approximation of $\log(1/x)$.

Recall that the Taylor approximation of $\log(1/x)$ centered in a generic $x_0=c$ is:
$$S^2(x-c)=\sum_{k\geq1} \frac{(-1)^k c^{-k}(x-c)^k}{k} + \log(1/c)$$
For the case of $x_0=1/e$, we have the more known:
$$S^2(x-c)=\sum_{k\geq1} \frac{(-e)^{k}}{k} (x-\frac{1}{e})^k + 1.$$
From this, we formulate that:
$$S^2(x-c) = \sum_{k=0}^\infty \left( \sum_{l=0}^k a_l a_{k-l} \right) (x-c)^k .$$

We obtain the following relation of the coefficients $a_k$ of $S(x-c)$:
$$
     \begin{cases}
       a_0^2 = 1, \\
       \sum_{l=0}^k a_l a_{k-l} \frac{(-e)^{k}}{k}  = 2a_0a_k + \sum_{l=1}^{k-1}a_la{k-l}
     \end{cases}
$$
From the first equation we obtain that $a_0 = 1$.
From the second equation we obtain a recursive definition of $a_k$, as
$$a_k= \frac{1}{2a_0}\left[\frac{-c^{-k}}{k} -  \sum_{l=1}^{k-1} a_l a_{k-l} \right].$$
which is equivalent to
$$a_k= \frac{1}{2a_0}\left[b_k -  \sum_{l=1}^{k-1} a_l a_{k-l} \right].$$

From here, it is quite difficult to proceed. Experimentally, it is possible to see that the signs of the coefficients are alternating. This makes the bound more difficult to prove.

Similar question:

Series approximations of $\sqrt{-\log(x)}$

答:
Too long for a comment.

Built around $x=\frac 1e$ (which is convenient)$$\sqrt{\log \left(\frac{1}{x}\right)}=\sum_{n=0}^\infty (-1)^n e^n a_n\left(x-\frac{1}{e}\right)^n$$ and the $a_n$ form the sequence
$$\left\{1,\frac{1}{2},\frac{1}{8},\frac{5}{48},\frac{25}{384},\frac{209}{3840},\frac{1
   961}{46080},\frac{2621}{71680},\frac{21407}{688128},\frac{1700267}{61931520},\frac
   {30128123}{1238630400}\right\}$$ (numerators and denominators are unknown in $OEIS$

On the other side
$$\log \left(\frac{1}{x}\right)=\sum_{n=0}^\infty (-1)^n \frac{e^n} n\left(x-\frac{1}{e}\right)^n$$


问:
Algorithm for solving a linear system of equations with external constraints?
I am trying to device an algorithm for rapidly solving systems of linear equations/inequalities with constraints, without necessarily relying on existing LP algorithms, such as Simplex. The reason I do not necessarily want to use existing algorithms is because I am only looking to quickly find a general solution to the system, NOT an optimized one.

Some background notes on the algorithm itself:

The algorithm is developed with Java integers as target datatype, and all involved values thus need to respect the constraints imposed by Java on such values. In particular, this means that all such values need to be in the range [2^32-1, 2^32].

While the target values have to be integers, the values may be treated as real numbers during the computational part of the algorithm for the sake of simplicity, and just be rounded of in the result set (i.e. LP relaxation may be used).

Finally, an important basic assumption of the algorithm (due to the context it will be used in) is that a solution ALWAYS exists - there is hence no need to check for feasibility.

Consider the trivial example (a, b and c are Java integers):

a <= b

b >= c

Let MIN and MAX be the minimum and maximum possible values of Java integers, respectively. We thus add the following constraints to the system:

a <= MAX

a >= MIN

b <= MAX

b >= MIN

c <= MAX

c >= MIN

Now, the system above can be transformed to an equivalent system of equations by introducing dummy variables:

a - b + s1 = 0

b - c - s2 = c

a + s3     = MAX

a - s4     = MIN

b + s5     = MAX

b - s6     = MIN

c + s7     = MAX

c - s8     = MIN

In the above, the additional restriction is imposed that s1...s8 are all >= 0.

Now, this is where I am stuck, and my question is this: what method should I use to solve the above system, seeing as I have to account for the implied constrains on the dummy variables? From what I gather, it would be possible for me to apply the initial phase of the  Simplex algorithm (i.e. using the algorithm itself in order to find an initial tableaux), but I do not know if this is the best way to go with performance in mind, since it still involves optimization, and hence potentially several steps of computation.

I am unfortunately not very math savvy, so are there any other theories for systems of equations which I can apply to reach a solution quickly? Thanks in advance!

答: As mentioned in the comments, this document under point (2) described a solution algorithm for the problem posed.</p>


问:
How do you define “independence” in combinatorics?
I feel like most definitions of “independence” are circular. Consider how we count the number of cards in a standard deck of cards: $|S \times R| = |S||R|$, where $S$ is the set of suits, and $R$ is the set of ranks. That is, $$S = \{\text{Hearts, Diamonds, Spades, Clubs}\}$$ and $$R = \{ 2,3,\dots,\text{King},\text{Ace} \}.$$ We know that $|S| = 4$ and $|R| = 13$, so $|S \times R| = |S||R| = 4 \cdot 13 = 52.$ In such an example, we define that they are independent, because they are disjoint subsets. But do we know that? In this example, it is “obvious.” What about examples where it is not obvious?

答:
Let's say you have a structure $S$. This structure is a combination of a few Attributes, each from a certain set of possible values.

If we let $A_1,...,A_n$ be the sets of the possible values for the corresponding attributes, then we can, pretty general, define the structure $S$ as follows:

$$
S:= \{\pmatrix{x_1\\\vdots\\x_n}\in ⨉_{i=1}^n A_i\mid P\pmatrix{x_1\\\vdots\\x_n}   \}
$$

Where $P$ is a predicate, i.e. it models our constraints, on which combinations of attributes are allowed.

Our goal, as usual in combinatorics, is determining $|S|$.

We say that the structure $S$ is independent of an attribute $A_i$ is (in the combinatoric sense), if:
$$
\forall x_1\in A_1,...,x_n\in A_n,y_i\in A_i:\qquad P(x_1,...,x_i,...,x_n) = P(x_1,...,y_i,...,x_n)
$$

Simply put this means that we don't need to look at the value of attribute $A_i$ to find out whether a specific instance of the structure is valid.

We therefore can fix a specific $x_i\in A_i$ (which exactly we choose doesn't matter), and define
$$P':  ⨉_{k=1\\i\neq i}^n A_k\to \{\text{True},\text{False}\}$$
via
$$P'(x_1,...,x_{i-1},x_{i+1},...,x_{n}) = P(x_1,...,x_{i-1},x_i,x_{i+1},...,x_{n})$$

In terms of the cardinality, this then means the following:
$$
|S| =|A_i|\cdot  |\{\pmatrix{x_1\\...\\x_{i-1}\\x_{i+1}\\...\\x_{n}}\in ⨉_{k=1\\k\neq i}^n A_k\mid P'\pmatrix{x_1\\...\\x_{i-1}\\x_{i+1}\\...\\x_{n}}   \}|
$$

As for every tuple $\pmatrix{x_1\\...\\x_{i-1}\\x_{i+1}\\...\\x_{n}}$, the structure instance $\pmatrix{x_1\\...\\x_{i-1}\\x_i\\x_{i+1}\\...\\x_{n}}$ is either valid for all choises of $x_i$, or for none of it.

To finish this, let's look at your example.

Our structure is the set of valid cards, where each card has the two attributes suit and rank.

Therefore we have $n=2$, $A_1:= \{\text{Hearts, Diamonds, Spades, Clubs}\}, A_2:=\{ 2,3,\dots,\text{King},\text{Ace} \}$.

Since we have no restrictions on our set members of $S$ besides that we have to pick from every attribute, we have further $P(x_1,x_2)=\text{True}$.

So, our structure $S$ is in this case independent of both $A_1$ and $A_2$, and therefore we have $|S| = |A_1|\cdot |A_2|$


问:
Non-continuity of partial derivatives of $\frac{xy}{(x^2+y^2)}$
Continuity of a function's partial derivatives implies differentiability of a function. However, I'm a bit confused applying this to what seems to be a common example brought up in textbooks:

Consider f where:

f(x, y) = $\frac{xy}{x^2+y^2}$ for (x, y) $\ne$ (0, 0) and 0 for (x, y) = (0, 0). Using the definition of the partial derivative, I can obtain that the partial derivatives of f are both 0 at (0, 0).

Furthermore, for (x, y) $\ne$ (0, 0), the partial derivative of f with respect to x is $\frac{y(y^2-x^2)}{(x^2+y^2)^2}$, which is clearly equal to 0 along the x axis where y = 0. Therefor the limit of this partial derivative should approach 0 as (x, y) approaches (0, 0) - I figure that since (0, 0) is the point in question, then I don't need to consider the partial derivative w.r.t x at other y values. This together with the partial derivative equalling 0 at (0, 0) should allow me to conclude that the partial derivative is continuous, no? The same should apply to the partial derivative w.r.t y by symmetry.

But I already know that f isn't differentiable at (0, 0), so I must have made a mistake somewhere. Where did I go wrong?

答: Consider the line $x=y$. Then $$f(x,x)=\frac{x^2}{x^2+x^2}=\frac 12$$The function is not even continuous at $(0,0)$


问:
Combinations from picking r from 2n objects (1...2n), repetitions allowed, if odd object appears, even object must appear?
I am very stuck on this question with combinations. This is where I am at in the problem:

Two cases:

picking an even object:

pick r from n numbered (2, 4, 6 ... 2n)

picking an odd object in which an even object must appear:

I do not know how to approach this point.

Since the two cases are disjoint, we can add them.

I am not sure if I am going in the right direction, any help would be appreciated!

答:
Welcome !

The problem can be easily solved by first counting all possible combinations with repetitions, and subtracting those that consist entirely of odd numbers, thus

${2n+r-1\choose r} - {n+r-1\choose r}$


问:
What is the remainder when $2^{1990}$ is divided by $1990$?
I actually do not have the basic idea on how to approach these type of questions....so please tell me a generalized method about all this too.

It came in RMO, and the question is:

What is the remainder when $2^{1990}$ is divided by $1990$ ?

答:
I was lazy to do the calculations required when applying the Chinese Remainder Theorem, so in the end I managed to come up with the following solution.

First note that $199$ is prime. By Fermat's Little Theorem we have, $$2^{198} \equiv 1\pmod{199} \implies 2^{199} \equiv2\pmod{199} \implies 2^{1990} \equiv 2^{10} \pmod{199}$$

$$ \implies\boxed{199\mid (2^{1990}-2^{10})}$$

We can also observe that $2^{1990}$ and $2^{10}$ have the same last digit, $4$, since $1990 \equiv 10 \equiv 2\pmod{4}$.

$$\implies\boxed{10 \mid (2^{1990}-2^{10})}$$

From the above we conclude that $$1990\mid (2^{1990}-2^{10}) \implies 2^{1990} \equiv 2^{10} \equiv 1024 \pmod{1990}$$

Therefore, the remainder when $2^{1990}$ is divided by $1990$ is $\boxed{1024}$


问:
Please help me in finding the solutions of the Bernoulli D.E.:- y'+(y/3)=(1/3)*(1-2x)*(y^4)
Please Help,
When I am solving the bernoulli differential equation, I am getting only one answer:-  y(x)=[1/((ce^x)-2x-1)]^(1/3), where c is a constant.

But, in the wolfram alpha there are 3 answers for this differential equation which are as follows:-

y(x)=[1/((ce^x)-2x-1)]^(1/3)

y(x)=-((-1)/((ce^x)-2x-1)]^(1/3)

y(x)=[(-1)^2/((ce^x)-2x-1)]^(1/3)

Although all 3 have equal values, but How 3 different solutions are coming?

Body:
$$y'+\frac{y}{3}=\frac{(1-2x)y^4}{3}\implies y^{-4}y'+\frac{y^{-3}}{3}=\frac{(1-2x)}{3}. $$ Let $y^{-3}=v \implies -3y^{-4}y'=v'$, to get a linear ODE
$$v'-v=(2x-1)$$
Its integrating factor is $I=e^{-x}$, then
$$v=e^{x} \int e^{-x} (2x-1) dx+C e^x
 \implies v=-(1+2x)+C e^{x}.$$
Finally, $$y^{-3}=-(1+2x)+Ce^x$$


问:
Is the collection $\mathcal{M}$ of $\mu$-measurable sets maximal so that $\mu|_{\mathcal{M}}$ is a measure?
Let $\mu:2^{X} \to [0, \infty]$ be an outer measure. The collection $\mathcal{M}$ of $\mu$-measurable sets are then defined as those sets $A$ satisfying $\mu(S)=\mu(S \cap A) + \mu(S \setminus A)$ for each $S \subset X$. It's proven in any measure theory course that $\mathcal{M}$ is a $\sigma$-algebra, with $\mu|_{\mathcal{M}}$ a complete measure.

I've always found this to be an elusive definition. This has been discussed elsewhere.

A more natural approach to defining the collection $\mathcal{M}$ of $\mu$-measurable sets, in my mind, is that we want it to satisfy the following property: Let $\mathcal{U}_{\mu} \subset 2^{2^{X}}$ denote the collection of $\sigma$-algebras $\mathcal{F}$ of $X$ with the property that $\mu|_{\mathcal{F}}$ is a complete measure on $\mathcal{F}$. Then $\mathcal{M}$ is inclusion-wise maximal in $\mathcal{U}_{\mu}$. In simple terms, we want $\mathcal{M}$ to be the largest set possible on which $\mu$ is a measure.

This isn't a good definition for $\mathcal{M}$, unfortunately, since we have no guarantee a priori that $\mathcal{U}_{\mu}$ has a unique maximal element.

So I have some questions:

Is $\mathcal{M}$, as defined in the first paragraph, in fact inclusion-wise maximal in $\mathcal{U}_{\mu}$?

Does $\mathcal{U}_{\mu}$ has a unique inclusion-wise maximal element?

If the question to the latter question is "no", then what can we say about inclusion-wise maximal members of $\mathcal{U}_{\mu}$ distinct from $\mathcal{M}$ as defined in paragraph one?

答:
Is $\mathcal{M}$, as defined in the first paragraph, in fact inclusion-wise maximal in $\mathcal{U}_{\mu}$?

Yes, provided subsets of $X$ can be approximated by members of $\mathcal M$, that is for each $S\subset X$ and each $\varepsilon>0$ there exists $S’\in\mathcal M$, $S’\supset S$ such that $\mu(S’)<\mu(S)+\varepsilon$. Indeed, if $A\subset X$ but $A\not\in\mathcal M$ then there exists $S\subset X$ such that $\mu(S)<\mu(S \cap A) + \mu(S \setminus A)$. Pick $S’\in\mathcal M$, $S’\supset S$ such that $\mu(S’)< \mu(S \cap A) + \mu(S \setminus A)$. Then $\mu(S’)< \mu(S’\cap A) + \mu(S’\setminus A)$, so the restriction of $\mu$ on the algebra generated by $\mathcal M$ and $S$ is not additive.

On the other hand, if $\mu$ has no above approximation property then $\mathcal{U}_{\mu}$ can fail to have a unique inclusion-wise maximal element. Indeed, let $X=\{1,2,3\}$. For each $A\subset X$ put
$$\mu(A)=\cases{
0, \mbox{ if }A=\varnothing,\\
1,\mbox{ if }1\le |A|\le 2,\\
2, \mbox{ if }A=X.}
$$
Then $\mathcal M=\{\varnothing, X\}$ but $\mathcal{U}_{\mu}$ has three maximal elements:
$\{\varnothing,\{1\},\{2,3\},X\}$,   $\{\varnothing,\{2\},\{1,3\},X\}$, and $\{\varnothing,\{3\},\{1,2\},X\}$.


问:
Question I had whilst finding the root decomposion of $sl(n,\mathbb{C})$ by a nonmaximal (and $1$-dimensional) semisimple abelian subalgebra.
Consider $L=sl(n,\mathbb{C})$. Let $h = e_{1,1}-e_{2,2}$ and let $H = \operatorname{span}\{h\}$.

$sl(n,\mathbb{C})$ has the basis: $\{e_{i,j} i \neq j\} \cup \{e_{i,i} - e_{i+1,i+1}\}$.

So, I've come to the conclusion that the weight space decomposion $sl(n,\mathbb{C})$ will compose of $H$ and then 4 weight spaces:

$$\alpha_+(c(e_{1,1}-e_{2,2})=c$$

$$\alpha_{+2}(c(e_{1,1}-e_{2,2})=2c$$

$$\alpha_-(c(e_{1,1}-e_{2,2})=-c$$

$$\alpha_{-2}(c(e_{1,1}-e_{2,2})=-2c$$

where in general $\alpha: H \rightarrow \mathbb{C}$ is a weight.

Lets look at the root space $L_{\alpha_+} = \{x \in sl(2,\mathbb{C}): [h,x]=\alpha_+(h)x$ for all $h \in H \}$.

Then we have that, for example, both of the basis vectors $e_{1,6}$ and $e_{1,5}$ are in $L_\alpha$:

$$[h,e_{1,6}] = [c(e_{1,1}-e_{2,2}),e_{1,6}]$$

$$= ce_{1,1}e_{1,6}-ce_{1,6}e_{1,1} - ce_{2,2}e_{1,6} + ce_{1,6}e_{2,2}$$

$$=ce_{1,6}$$

And similarly for $e_{1,5}$. This would mean that the root space of $L_\alpha$ is at least $2$-dimensional. Is this okay in this situation or did I do something wrong? I'm pretty sure if $H$ was a Cartan subalgebra that would mean that $L_{\alpha}$ would be 1 dimensional, right?

答:
Well one would and should not call these weight spaces of an arbitrary abelian (semisimple) subalgebra "root" spaces because the weights do not, in general, form a root system. For more striking examples see https://math.stackexchange.com/a/2112543/96384.

And here, yes, the weight space to the weight you call $\alpha_{+1}$ is spanned by the $e_{1,j}$ with $j \ge 3$ and $e_{i,2}$ with $i \ge 3$, hence has dimension $2n-4$; likewise the weight space to $\alpha_{-1}$ is $(2n-4)$-dimensional. The weight spaces to the weights $\alpha_{\pm2}$ have only dimension $1$ each, and I leave it to you to find a basis for the weight space to the $0$ weight, a.k.a. centraliser of $H$, which must fill up the missing dimensions to $\mathrm{dim}(\mathfrak{sl}_n) =n^2-1$. (A big part of it is made up of $e_{i,j}$ with both $i \neq j \ge 3$, and then see what you can still do on the diagonal.)

By the way, you are sort of describing the decomposition of $\mathfrak{sl}_n$ as an $\mathfrak{sl}_2$-module here for the embedding of $\mathfrak{sl}_2$ into the top left corner of $\mathfrak{sl}_n$. The case $n=3$ of that was recently asked about here, with matching results (note that what's called $V_2$ there contains the weights $-2,0,2$ each with multiplicity $1$; then each of the $V_1$'s contains weights $\pm 1$ each with multiplicity $1$; and one more $V_0$ gives another $1$-dimensional space of weight $0$. Adding up, we have the weights $\pm2$ with dimensions $1$ each, the weights $\pm1$ with dimension $2$ each, and the weight $0$ with dimension $2$).


问:
To use the two-point formula to find the linear equation relating $C$ and $F$:
I've tried to solve a problem which I'm going to give below. What I don't understand is that which variable is dependent and which is independent among $C$ and $F$. I think we can relate $C$ and $F$ using two equations. In one of the equations $C$ can be the dependent variable and $F$ the independent and in the other equation $C$ can be independent and $F$ can be dependent. However, the problem is as follows:

Let $C$ denote Celsius temperature and $F$ Fahrenheit temperature. Thus, $C=0$ and $F=32$  at the freezing point of water, while $C=100$ and $F=212$ at the boiling point of water. Use the two point formula to find the linear equation relating $C$ and $F$.

My solution:

Let $C_1 = 0$ and $C_2 = 100$. Likewise, $F_1=32$ and $F_2=212$. If $P(C_1,F_1)$ and $Q(C_2,F_2)$ are the two points on the line $L$ then we have the following equation for $L$:

$F-F_1= \bigg( \frac{F_2-F_1}{C_2-C_1}\bigg)(C-C_1)$

OR

$F-32= \bigg( \frac{212-32}{100-0}\bigg)(C-0)$

OR

$F = \frac{180}{100}C + 32$

OR

$F = \frac{9}{5}C+32$

Do I need to find the other relation between $C$ and $F$?

答: Your computation and the resulting formula look fine, but there seems to be a problem with the graph.  I think you plotted the intercept 32 on the vertical axis, assuming that the vertical direction represents $F$, but then I think you plotted a slope of $5/9$ rather than $9/5$, assuming that the vertical axis represents $C$ and the horizontal axis represents $F$.  One thing I can say for sure, regardless of which axis represents which variable, is that the graph should go through the point $(-40,-40)$, which your graph doesn't.  The reason is that $-40$ is the (one and only) temperature that comes out the same on the Fahrenheit and Celsius scales.</p>


问:
Can a triangle be divided into four similar triangles such that not all four of the triangles are congruent to each other?
I understand that you can divide a triangle into four congruent triangles by connecting the midpoints of each side. Can ANY NON-EQUILATERAL triangle be divided into four similar triangles with the restriction that not all four triangles can be congruent to each other?
As I explore this question, I keep running into dead ends, and I ask if any of you can help.

EDIT: You guys revealed that there are multiple ways to do this with right triangles. I've been experimenting with a general case and right triangles, but the closest I've gotten is splitting the triangle three times (on triangle ABC, drawing a line from Angle BAC that is perpendicular to Side BC, calling the point of intersection on Line BC Point D, then drawing lines from Angles ADB and ADC to be perpendicular with Lines AB and AC, respectively) yet I cannot prove that the triangles within ACD are similar to the triangles within ABD unless they are all within a right triangle. How to proceed?

答:
To obtain such a figure for a given triangle $\triangle ABC$ with angles $\alpha,\beta,\gamma$ where $|BC|\ne|CA|$ (and therefore $\alpha\ne\beta$),

let $F$ be the intersection of the parallel to $AB$ through $C$ and the parallel to $BC$ through $A$ (so $AFBC$ is a parallelogram),

construct line $\ell$ as tangent to the circumcircle of $ABC$ at $C$,

let $D$ be the intersection of $\ell $ and $AC$,

let $E$ be the intersection of $\ell$ and $BF$.

We have

$\angle ABF = \angle BAC=\alpha$ (alternate angles as $FB\|AC$)

$\angle FAB = \angle CBA=\beta$ (alternate angles as $FA\|BC$)

$\angle BCE =\angle BAC=\alpha$ (inscribed angle theorem / chord-tangent theorem)

$\angle CAD =\pi-\angle FAC=\pi-(\alpha+\beta)=\gamma$ (supplementary angle and angle sum in triangle)

$\angle EBC=\pi-\angle CBF=\pi-(\alpha+\beta)=\gamma$ (supplementary angle and angle sum in triangle)

$\angle DCA = \pi-\angle ACE=\pi-(\alpha+\gamma)=\beta$ (supplementary angle and angle sum in triangle)

$\angle ADC=\alpha$ (angle sum in triangle)

$\angle CEB=\beta$ (angle sum in triangle)

$\angle BFA=\gamma$ (angle sum in triangle)

Thus triangles $ABC$, $DCA$, $CEB$, $BAF$, $DEF$ are all similar.
But they are not all congruent: By comparing the sides opposing angle $\alpha$, we find
$$ {\triangle DCA}:{\triangle ABC}=|CA|:|BC|\ne 1:1$$

Finally, in order to partition $\triangle ABC$ instead of extending it, we need only perform a similarity transformation that maps $\triangle DEF$ to $\triangle ABC$.


问:
Generating Function from using sequence
I am new to generating function and having a hard time figuring out generating function using sequence. The question that I am having trouble on is using the sequence ar = $\sum_{i=0}^{n} \binom{i} {r} $, find the generating function. There is a hint using $\sum_{r>=0}{}$ ar $x^r $. I the r in ar is lower  Can you please help. Thank you.

答:
The generating function for the sequence $\langle a_r:r\ge 0\rangle$ is by definition

$$g(x)=\sum_{r\ge 0}a_rx^r\,;$$

since $a_r=\sum_{i=0}^n\binom{i}r$, we can write this as

$$g(x)=\sum_{r\ge 0}\sum_{i=0}^n\binom{i}rx^r\,.$$

$\binom{i}r=0$ when $i<r$, so every term of $a_r=\sum_{i=0}^n\binom{i}r$ is $0$ when $r>n$. This means that $a_r=0$ when $r>n$, so

$$g(x)=\sum_{r=0}^n\sum_{i=0}^n\binom{i}rx^r\,:$$

it’s actually a polynomial of degree $n$.

$$\begin{align*}
g(x)&=\sum_{r=0}^n\sum_{i=0}^n\binom{i}rx^r\\
&=\sum_{i=0}^n\sum_{r=0}^n\binom{i}rx^r\\
&=\sum_{i=0}^n\sum_{r=0}^i\binom{i}rx^r\cdot 1^{i-r}\\
&=\sum_{i=0}^n(x+1)^i\,.
\end{align*}$$

This is a finite geometric series, so you should be able to express it in closed form; it’s not hard to simplify that closed form to a polynomial.

That approach works even if you know only the most basic facts about binomial coefficients. If you know the hockey stick identity, you can rewrite $g(x)$ as

$$\sum_{i=0}^nx^r\sum_{r=0}^n\binom{i}r$$

and apply the hockey stick identity to get the polynomial directly.

Title: Generating Function from using sequence


问:
$|f(x)| \leq g(x) \forall x$, and $\lim _{x \rightarrow a} g(x)=0$, what is $\lim _{x \rightarrow a} f(x)$? What if $\lim _{x \rightarrow a} g(x)=5$?
$$
\begin{aligned}
&\text { Suppose }|f(x)| \leq g(x) \text { for all } x . \text { What can you conclude about } \lim _{x \rightarrow a} f(x) \text { if } \lim _{x \rightarrow a} g(x)=0 ?\\
&\text { What if } \lim _{x \rightarrow a} g(x)=5 ?
\end{aligned}
$$

答:
HINT:

Can the sketch be useful in finding limit of $f(x)$?


问:
A problem about uniform distribution and Rayleigh distribution
How can I prove this staments?

Let $X$ and $Y$ independent random variables with common distribution $N(0,1)$. Prove that  $R=\sqrt{X^{2}+Y^{2}}$ and $\Theta=\tan^{-1}(Y/X)$ also independent random variables, $\Theta \sim U(0,2\pi)$ and $R$ has Rayleigh's distribution, this's has density $$f_{R}(r)=re^{-r^{2}/2}, r>0$$

My approach: By hypothesis, we have that $$f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)$$and since that $$f_{X}(x)=\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{x^{2}}{2} \right)$$
and $$f_{Y}(y)=\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{y^{2}}{2} \right)$$
So, $$f_{X,Y}(x,y)=\frac{1}{\sqrt{2\pi}}\exp \left(-\frac{x^{2}}{2} \right)\cdot \frac{1}{\sqrt{2\pi}}\exp \left(-\frac{y^{2}}{2} \right)$$
we need to find the PDF of $R=\sqrt{X^{2}+Y^{2}}$.

and we need to prove that  $$f_{R,\Theta}(r,\theta)=f_{r}(r)f_{\Theta}(\theta)$$ where $f$ is de PDF.

But, I don't know how can I write here $f_{R,\Theta}(r,\theta)$ and the marginal functions $f_{R}(r)$ and $f_{\Theta}(\theta)$.

In the previuous problem I proved that $X^{2}+Y^{2}\sim \chi^{2}(2)$. I think that I can use that result, since that $R^{2}=\chi^{2}(2)$. But, I don't sure about my approach.

答:
You have
$$
f_{X,Y}(x,y) = \frac{1}{2\pi}\exp\left ( -\frac{1}{2} (x^2+y^2)\right )
$$
Note further that $X = R \sin \Theta, Y=R \cos \Theta$

Now, we can use the density transform formula to write
$$
f_{R,\Theta}(r,\theta) = f_{X,Y}(r \sin \theta, r\cos\theta)|J|^{-1}
$$
where $|J|^{-1}$ is the inverse Jacobian of the transformation, you should be able to show $|J|^{-1} = r$. Putting this together we have

$$
f_{R,\Theta}(r,\theta) = \frac{1}{2 \pi} r e^{-r^2/2} \mathbb{1}\{r >0\} \mathbb{1}\{0\le\theta \le 2\pi\}
= \left ( \frac{1}{2 \pi} \mathbb{1}\{0\le\theta \le 2\pi\} \right ) \left (r e^{-r^2/2} \mathbb{1}\{r >0\} \right)
$$
and the result follows immediately (independence holds due to the fact that the density factorizes)


问:
What does a joint pmf being uniform over a dotted area mean?
In class, the tabular form used in our examples all had numerical values to them like the one shown:

but the assignment given to us doesn't have any numbers at all. It just says that the joint pmf is uniform over the dotted area and I have no idea what that means. Can someone explain it in an easy way to understand for a beginner stat student like me? Thanks.
I posted the image of the problem below.

答:
A uniform discrete distribution is one where the probability mass of each point in the support is equal to a constant.  Since there are 21 points in that support, the joint probability mass function is:

$$p_{\small X,Y}(x,y)=\tfrac 1{21}\mathbf 1_{\langle x,y\rangle\in{\Bbb N_0}^2,~x+y\leqslant 5}$$


问:
Smoothness in Banach space
I need a reference about a definition. Let $n$ be an integer and $G$ be a group of $H^n$(Sobolev) automorphisms of a vector bundle $E$ on some manifold $M$ and $C$ be the space of connections of class $H^{n-1}$, what is the definition that $G$ acts "smoothly" on $A$?</p>

答:
To talk about smoothness of an action, we need to be sure that the group is a Lie group, and in the case of Sobolev gauge transformations (bundle automorphisms) the smoothness relies on the use of some standard Sobolev theory.

Here's an outline of a special case. Let $G \subset \mathrm{GL}(n, \Bbb R)$ be a Lie group. If $\mathscr{G}_{p,k}$ is the group of $W^{p,k}$ gauge transformations of a principal $G$-bundle $P \longrightarrow M^n$, then it is known that $\mathscr{G}_{p,k}$ is a Banach Lie group for $k - 1 > n/p$ (in your question you have $p = 2$, and in that case $\mathscr{G}_{2,k}$ is a Hilbert Lie group for $k - 1 > n/2$). Hence when $k - 1 > n/p$, $\mathscr{G}_{p,k}$ is a smooth (infinite-dimensional) manifold. Of course, the space $\mathscr{A}_{p,k-1}$ of $W^{p,k-1}$ connections is a smooth Banach manifold (it is an affine space).

Hence when $G \subset \mathrm{GL}(n, \Bbb R)$ and $k - 1 > n/p$, the phrase "$\mathscr{G}_{p,k}$ acts smoothly on $\mathscr{A}_{p,k-1}$" means exactly what you think it should mean: the map
$$\mathscr{A}_{p,k-1} \times \mathscr{G}_{p,k} \longrightarrow \mathscr{A}_{p,k-1},$$
$$(A, g) \mapsto g^\ast A$$
is a smooth map of Banach manifolds.

Here are some references that might be worth checking out for more details:

An Introduction to Gauge Theory by John Morgan, in the book Gauge Theory and the Topology of $4$-Manifolds.

The first few lectures of Tom Mrowka's Master Class at Aarhus (videos included!).

Appendix A of Instantons and Four-Manifolds by Freed and Uhlenbeck. The ideas are similar to the general case, but since this is a book on instantons they assume $M$ is a closed $4$-manifold and $G = \mathrm{SU}(2)$.

Title: Smoothness in Banach space


问:
Does $D_k \mid N_k$ hold at every step of this iterative process involving divisors of odd perfect numbers?
The topic of odd perfect numbers likely needs no introduction.

In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.  Let
$$D(x) = 2x - \sigma(x)$$
denote the deficiency of $x$, and let
$$s(x) = \sigma(x) - x$$
denote the sum of aliquot/proper divisors of $x$.

Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $N = q^k n^2$ is (odd) perfect, it follows that
$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2,$$
from which we have
$$\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}$$
since $\gcd(q^k,\sigma(q^k))=1$.

Letting $N_i$ denote the numerators and $D_i$ denote the denominators of the fractions in the equation above, we get
$$N_0 = \sigma(n^2), D_0 = q^k$$
$$N_1 = 2n^2, D_1 = \sigma(q^k)$$
$$N_2 = N_1 - N_0 = D(n^2), D_2 = D_1 - D_0 = s(q^k)$$
$$N_3 = N_0 - N_2 = 2s(n^2), D_3 = D_0 - D_2 = D(q^k).$$

Note that, we obtain
$$N_4 = N_1 - N_3 = 2D(n^2), D_4 = D_1 - D_3 = 2s(q^k).$$

Here is my question:

In general, if $i \neq j$ are any positive integers (which are both bigger than $1$), then is it guaranteed that
$$\frac{N_k}{D_k} = \frac{N_i - N_j}{D_i - D_j}$$
is always an integer, if $N_0$ and $N_1$ are defined as above?

答:
We can say that $\dfrac{N_i-N_j}{D_i-D_j}\in\mathbb Z$ under the condition that $D_i-D_j\not=0$ instead of $i\not=j$ (see the end of this answer) since we have that if$$\frac BA=\frac DC\in\mathbb Z\quad\text{and}\quad A\not=C$$
then
$$\frac{B-D}{A-C}=\frac{B-\frac BAC}{A-C}=\frac{A(B-\frac BAC)}{A(A-C)}=\frac{B(A-C)}{A(A-C)}=\frac BA\in\mathbb Z$$

Note that $D_i=D_j$ can happen even when $i\not=j$.

For example, if we define $D_5,N_5,D_6$ and $N_6$ as
$$\begin{align}D_5&:=D_0-D_4,\qquad N_5:=N_0-N_4,
\\\\D_6&:=D_3-D_2,\qquad N_6:=N_3-N_2\end{align}$$
then we get $D_5=D_6$, so $\dfrac{N_5-N_6}{D_5-D_6}$ is not defined.


问:
Different methods of computing standard deviation not agreeing
I have been assigned the following problem:
Given numbers  -25, 2, 3, 8, 10, 14, 18, 21, 32

You are told the mean of the numbers is 9.222 and the mean of their
square is 309.666. What is the sample standard deviation?

The answer is given as
$\sqrt{9/8 * (309.666-9.222^2)} =15.896$

This is a new way of calculating standard deviation to me.
So I tried the old way I know with a spreadsheet in yellow.
I get the answer as 44.92

What am I missing?

答: You need to divide $2021.556$ by $9-1=8$ before taking the square root.


问:
Nontrivial h-cobordism
I'm learning the h-cobordism theorem as I want to use it in a talk. I'd like to be able to give an example of an h-cobordism that isn't a cylinder, if possible by drawing a picture. What is the simplest such example?</p>

答:
I learned about such an example recently through a conversation with Kirby:

Given an h-cobordism $W:M_1\to M_0$ between simply-connected 4-manifolds, it is actually a product outside of an "Akbulut cork".

Visual elaboration: There is a contractible 4-manifold $C_1$ with boundary (called the cork) sitting inside of $M_1$, and there is its involution $C_0$ $(\approx C_1)$ sitting inside of $M_0$. Now it turns out that there exists an involution of $\partial C_1$ which doesn't extend to a diffeomorphism of $C_1$. Furthermore, we have a cobordism $A:C_1\to C_0$ which is diffeomorphic to a 5-ball (but not relative-boundary of course).

So the picture is that any such $W$ can be viewed as $(M_1-C_1)\times[0,1]$ outside of $C_1$, and looks like $A$ inside of $C_1$. Amazing!

Title: Nontrivial h-cobordism


问:
Understanding Divergence Test through difference of consecutive partial sums
It's known that the difference between two consecutive partial sums $s_{n+1}$ and $s_n$ should give us the sequence term $a_{n+1}$, i.e. $s_{n+1} - s_n = a_{n+1}$.

If we take a limit of both sides as $n$ approaches infinity, we get:

\begin{equation}
\lim_{n \to \infty} s_{n+1} - s_n = \lim_{n \to \infty} a_{n+1}.
\end{equation}

Let's say it's also known that $\lim_{n \to \infty} a_n = 0$. That would imply that $\lim_{n \to \infty} a_{n+1} = 0$, and $\lim_{n \to \infty} s_{n+1} - s_n = 0$.

This seems to say that $s_{n+1} \approx s_n$ for some large value of $n$ and the sequence of partial sums (infinite sum of $a_n$) converges to some value. Two questions:

Why would the Divergence Test be inconclusive if it seems like the limit of partial sums converges to some value? (I understand that the harmonic series diverges although the limit of $a_n$ goes to zero, so looking for an argument along the lines of partial sums.)

Why is the same limit of $a_n$ going to zero enough to say that an alternating series converges? (i.e. then we have $\lim_{n \to \infty} s_{n+1} - s_n = \lim_{n \to \infty} (-1)^n a_{n+1}$ and the same argument applies).

答:
As the harmonic series illustrates, a sequence's general term can go to zero without its being Cauchy.  It doesn't seem like the partial sums converge, only that the difference of successive partial sums goes to zero.

In the case of an alternating decreasing series, it turns out the series will be Cauchy.  See the (clever) proof of the Leibniz test.  I would reproduce it, but it should be easy to find.  I learned it in Best and Penner's Calculus, a fine book.


答:
$|f(x)| \leq g(x)$ means that $$-g(x) \leq f(x) \leq g(x)$$

If $g(x)\to 0$, then $-g(x)\to 0$ too! Can we use the Sandwich theorem?

However, if $g(x)\to 5$, then $-g(x) \to -5$ and not much can be said about $\lim_{x\to a}f(x)$ in this case.

Title: $|f(x)| \leq g(x) \forall x$, and $\lim _{x \rightarrow a} g(x)=0$, what is $\lim _{x \rightarrow a} f(x)$? What if $\lim _{x \rightarrow a} g(x)=5$?

Body:
$$
\begin{aligned}
&\text { Suppose }|f(x)| \leq g(x) \text { for all } x . \text { What can you conclude about } \lim _{x \rightarrow a} f(x) \text { if } \lim _{x \rightarrow a} g(x)=0 ?\\
&\text { What if } \lim _{x \rightarrow a} g(x)=5 ?
\end{aligned}
$$


问:
Cylindrical Coordinates Question on Derivation(Informal)

In cylindrical coordinates, why do we look at the distance r, which is the length from the origin to the center point of the "cylindrical cuboid's" projection onto the xy plane. In this case, it is to help define the base area $r\Delta\theta$, but I am trying to understand the specific reason for this decision? Why not use "$r - \Delta r /2$" or "$r + \Delta r/2$"?

I am aware that

$\Delta V$

which is defined to be

$ \Delta V = r\Delta \theta \Delta r \Delta z \ $

答:
In the limit $\Delta r\to 0$ the two approaches are identical. When one integrates $r$, you write $$\lim_{\Delta r\to 0}\sum_{r_i=0}^{R-\Delta r}\Delta r$$ in one case, and $$\lim_{\Delta r\to 0}\sum_{r_i0+\Delta r/2}^{R-\Delta r/2}\Delta r$$
Both are equivalent to $$\int_0^R dr$$


问:
Periodic sequence problem
Given sequence $a_n$ defined such that $a_1=3$, $a_{n+1}=\begin{cases}\frac{a_n}{2},\quad 2\mid a_n\\ \frac{a_n+1983}{2},\quad 2\nmid a_n\end{cases}$. Then prove that the sequence $a_n$ is periodic and find the period.

It's easy to prove that $0<a_n<1983$ by induction. By pigeonhole principle, there exist $i,j$ such that $a_i=a_j\implies a_{i+1}=a_{j+1}$. By induction, we can prove $a_{i+k}=a_{j+k},\forall k\in\mathbb{N}$. Otherwise, $a_n\begin{cases}2a_{n+1}, \quad a_{n+1}\le 991\\ 2a_{n+1}-1983, \quad a_{n+1}\ge 992\end{cases}$. So we can prove also $a_{i-k}=a_{j-k} $ for $min(i,j)>k, \forall k\in\mathbb{N}$. So it's periodic. But I can't find the period. In my opinion, the period is $660$. Because $3\mid a_n$ and $0<a_n<1983$. But I can't prove $\forall k, \exists i$ such that $a_i=3k$, Can anyone help me?

答:
As you've noticed, since $3\mid a_1$ and $3\mid 1983$, it follows that $3\mid a_n$ for all $n$. This allows us to simplify the problem by considering the associated sequence defined by $b_n = a_n/3$. We can easily prove by induction that we have $1 \le b_n \le 660$ for all $n$. Since the admissible range of values for $b_n$ is finite, the sequence must be eventually periodic.

Your conjecture that the period is $660$ is in fact true. Showing that the period is $660$ will show that the sequence is not just eventually periodic, but fully periodic (alternatively, as you've noted, this follows from the fact that $b_n$ uniquely determines $b_{n-1}$). The conjecture that the period is $660$, together with the fact that $1 \le b_n \le 660$, motivates looking at the values of the sequence modulo $661$.

Let $[k]$ denote the remainder of $k\in \mathbb{Z}$ modulo $661$, i.e., the unique integer $0 \le [k] < 661$ such that $[k] \equiv k \pmod{661}$. Since $1 \le b_n < 661$, it follows that $b_n = [b_n]$ for all $n\in \mathbb{N}$.

Lemma 1: Let $m \in \mathbb{Z}$ be an even integer. Then $[m/2] = [331m]$.

Proof: Note that $2$ is a unit in $\mathbb{Z}/661\mathbb{Z}$. Indeed, we have $2^{-1} \equiv 331 \pmod{661}$. Therefore we have
$$331m \equiv 331 \cdot \left[2\cdot \left(\frac{m}{2}\right)\right] \equiv [331 \cdot 2]\left(\frac{m}{2}\right)\equiv \frac{m}{2} \pmod{661}.$$
It follows that $[m/2] = [331m]$. $\square$

Lemma 2: For all $n\ge 1$, we have $b_n = [331^{(n-1)}]$.

Proof: Consider the defining recursion
$$b_{n+1} = \begin{cases}b_n/2 & 2 \mid b_n,\\ (b_n + 661)/2 & 2\not\mid b_n.\end{cases}$$
In the first case, we have
$$b_{n+1} = [b_{n+1}] = [b_n/2] = [331b_n].$$
In the second case, we have
$$b_{n+1} = [b_{n+1}] = [(b_n + 661)/2] = [331(b_n + 661)] = [331b_n].$$
In either case, we have $b_{n+1} = [331b_n]$. Starting with $b_1 = 1$, it follows that $b_n = [331^{(n-1)}]$. $\square$

The period of the sequence is therefore the order of $331$ mod $661$. The result then follows by noting $661$ is prime, so that $(\mathbb{Z}/661\mathbb{Z})^{\times} \cong \mathbb{Z}_{660}$ is cyclic, and moreover that $331$ (or equivalently, $2$) is a primitive root modulo $661$. This last fact can be verified with a quick (albeit tedious) calculation.

Title: Periodic sequence problem


问:
Does $\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})}$ converge?
I would like to evaluate the following limit: $$\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})}\ .$$

My working

\begin{align}
\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})} & =
\exp\left(\lim\limits_{x\to3}\ \ln \left[(4 - x)^{\tan (\frac {\pi x} {2})}\right]\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to3}\ \frac {\ln (4 - x)} {\cot \frac {\pi x} {2}}\right)
\\[5 mm] & =
\exp\left[\lim\limits_{x\to3}\ \frac {\frac {-1} {4 - x}} {-\csc^2 (\frac {\pi x} {2})}\right]
\\[5 mm] & =
\exp\left[\lim\limits_{x\to3}\ \frac {\sin^2 (\frac {\pi x} {2})} {4 - x}\right]
\\[5 mm] & =
e
\end{align}

However, I get conflicting results when I try to check my answer.

Firstly, a plot of $y = (4 - x)^{\tan (\frac {\pi x} {2})}$ using a graphing software like Desmos shows that as $x \rightarrow 3$, $y$ converges to some value slightly less than $2$ (not sure what the value is, but it definitely does not look like $2.71828...$).

Secondly, I tried to evaluate the limit directly using software such as Symbolab and it tells me that the limit diverges.

Is my working correct? If not, where did I go wrong? Also, if my answer is indeed correct, why is it that software such as Symbolab cannot evaluate this limit? And why then, does the graph that I plotted not converge to $e$?

Any help/intuition/explanation will be greatly appreciated :)

Edit

So following my accepted answer below, it seems that Desmos was correct and I was careless. However, it seems there still is no explanation as to why Symbolab cannot evaluate this limit. If anyone is familiar with the technicalities of the software and perhaps knows why this is the case, do drop a comment/answer too :)

答:
$$\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})} = e^{\frac{2}{\pi}} \neq e$$

There's a minor mistake where you've applied the chain rule for differentiation - nothing serious. Note that $\frac{d\cot(ax)}{dx} = -a\csc^2(ax)$, you've missed the $a$.


问:
Calculating acceleration based on a starting and ending distance
I have been trying to figure this problem out and just can't.

The answer is supposed to be (A): 1.0 m/s/s.

Can anyone tell me how they got that answer?  I thought this problem was based on the formula: Acceleration $\displaystyle = \frac {(v2 - v1)}{t}$, but I can't get that answer.  What am I doing wrong? Or is the correct answer not listed?  Thanks for any help.

Problem:
  A Pinewood Derby car rolls down a track.   The total distance travel is measured each second.   Use the chart to calculate the acceleration of the car.

Time(s) |  Dist(m)

0       |     0

1       |     1

2       |     3

3       |     6

4       |    10

A) 1.0  meters/sec/sec

B) 1.5  meters/sec/sec

C) 2.0 meters/sec/sec

D) 2.5 meters/sec/sec

答:
OK, I actually think that Ross's answer makes more sense. His assumption is that the last column of your table is the distance travelled between the last two measurements. Also, he assumed (as I did) constant velocity $a$, in which case the distance travelled at time $t$ is

$x = \dfrac{at^2}2$

Now, the last column actually corresponds to

$\Delta x(t) = x(t)-x(t-\Delta t) = a\Delta t(t-\dfrac{\Delta t}2)$,

with $\Delta t$ (time between two successive measurements) is constantly $1\,\text s$, and $\Delta x$ is the distance travelled in this interval of time (in other words: the second column of the table). Then I think you do find that $a = 2\,\text m.\text s^{-2}$.


问:
Evaluate $\int_0^1 {\frac{{{x^{300}}}}{{{x^3} + {x^2} + 1}}} \:dx$
$$\int_0^1 {\frac{{{x^{300}}}}{{{x^3} + {x^2} + 1}}} \:dx$$

My try:
$$\begin{array}{l}{x^{300}} = ({x^3} + {x^2} + 1)q(x) + A{x^2} + Bx + C\\
\Rightarrow \frac{{{x^{300}}}}{{{x^3} + {x^2} + 1}} = q(x) + \frac{{A{x^2} + Bx + C}}{{{x^3} + {x^2} + 1}}\end{array}
$$
Now got stuck here. Also i tried to use $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$, but the denominator is more complicated than the existing one.

EDIT: As per the comments obtained that the integral is not staright forward, i would like to mention that this problem appeared in an Indian competitive exam for research admission. Here is the question asked.

答:
Note: The actual question asks for only for an estimate.

$$0< \int_0^1 \frac{x^{300}}{1+x^2+x^3}dx\leqslant \int_0^1x^{300}dx = \frac1{301} < 0.003333....$$


问:
$(a,b)\in\mathbb{C}(+,.),c\in\mathbb{R}(+,*)$ then $ c\cdot(a,b)=(c\cdot a , c\cdot b)$ or $(c*a,c*b)$?
If I have a vector space of Complex numbers $\mathbb{C}(+,\cdot)$ and $(a,b)\in\mathbb{C}(+,.)$ where $a,b,c \in \mathbb{R}(+, *) $ and  $\mathbb{C}$ be a vector space over $\mathbb{R}$ can we write $$c\cdot(a,b) = (c\cdot a , c\cdot b)\mathrm{~~or~~} c\cdot(a,b) = (c*a,c*b)?$$

答:
If $(a,b)$ belongs to the vector space of complex numbers, $\mathbb{C}(+,\cdot)$ over $\mathbb{R}(+,*)$ then, $a,b\in\mathbb{R}$. For instance, we can say $(a,b)$ represents the complex number $a+b\iota$, where $a,b$ are reals. $c$ belongs to the vector space of real numbers, $\mathbb{R}(+,*)$.

First consider the LHS,$$c\cdot(a,b)\qquad\mathrm{where,~}c\in\mathbb{R}(+,\cdot)\mathrm{~and~}(a,b)\in\mathbb{C}(+,\cdot).$$
We use the notation $``\cdot"$ because we are performing scalar multiplication under $\mathbb{C}(+,\cdot)$. This is equal to
$$(c~\underline{?}~a,c~\underline{?}~b)\qquad\mathrm{where,~}c,a,b\in\mathbb{R}(+,*).$$
Since we are now performing multiplication under $\mathbb{R}(+,*)$, we write
$$c\cdot(a,b)=(c*a,c*b).$$


Title: Does $\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})}$ converge?

Body:
I would like to evaluate the following limit: $$\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})}\ .$$

My working

\begin{align}
\lim\limits_{x\to3}\ (4 - x)^{\tan (\frac {\pi x} {2})} & =
\exp\left(\lim\limits_{x\to3}\ \ln \left[(4 - x)^{\tan (\frac {\pi x} {2})}\right]\right)
\\[5 mm] & =
\exp\left(\lim\limits_{x\to3}\ \frac {\ln (4 - x)} {\cot \frac {\pi x} {2}}\right)
\\[5 mm] & =
\exp\left[\lim\limits_{x\to3}\ \frac {\frac {-1} {4 - x}} {-\csc^2 (\frac {\pi x} {2})}\right]
\\[5 mm] & =
\exp\left[\lim\limits_{x\to3}\ \frac {\sin^2 (\frac {\pi x} {2})} {4 - x}\right]
\\[5 mm] & =
e
\end{align}

However, I get conflicting results when I try to check my answer.

Firstly, a plot of $y = (4 - x)^{\tan (\frac {\pi x} {2})}$ using a graphing software like Desmos shows that as $x \rightarrow 3$, $y$ converges to some value slightly less than $2$ (not sure what the value is, but it definitely does not look like $2.71828...$).

Secondly, I tried to evaluate the limit directly using software such as Symbolab and it tells me that the limit diverges.

Is my working correct? If not, where did I go wrong? Also, if my answer is indeed correct, why is it that software such as Symbolab cannot evaluate this limit? And why then, does the graph that I plotted not converge to $e$?

Any help/intuition/explanation will be greatly appreciated :)

Edit

So following my accepted answer below, it seems that Desmos was correct and I was careless. However, it seems there still is no explanation as to why Symbolab cannot evaluate this limit. If anyone is familiar with the technicalities of the software and perhaps knows why this is the case, do drop a comment/answer too :)

Body:
If $\lim_{x \to a} f(x)^{g)x)} \to 1^{\infty}$, then
$$\lim_{x \to 0} f(x)^g(x)= \exp[\lim_{x \to 0}  g(x)(f(x)-1)].$$
So here $$L=\lim_{x \to 3} (4-x)^{\tan (\pi x/2)}= \exp[\lim_{x \to 3} \tan(\pi x/2)=\exp[\lim_{x \to 3}\frac{(3-x)}{\cot(\pi x/2)}]=e^{2/\pi},$$
By L-Hospital Rule.


问:
Combinatorics: Systems of distinct representatives
I have two related(but separate) questions that are both heading toward something I am trying to prove(I think anyway).

Let's assume that $A_1,A_2,...,A_n$ is a collection of subsets of $S$ such that $|A_i|=m$ for all $i$ and every element in the set $S$ belongs to exactly $m$ of the sets $A_1,A_2,...,A_n$. Prove that the sets $A_1,A_2,...,A_n$ have a system of distinct representatives.

Obviously I am supposed to prove that the criterion in Hall's theorem holds. And I see that $k\leq k\cdot m = \sum_{i=1}^{k}|A_i|$, but I cant really make the needed connection between the union $|\bigcup_{i=1}^{k}A_i|$ and $k$.

Let's assume now that $|S|=m\cdot n$ and that $S=A_1\cup A_2\cup...\cup A_m=B_1\cup B_2\cup...\cup B_m$, where $|A_i|=|B_i|=n$ for all $i$. Prove that it is possible to re-index the sets $A_i$ such that $A_i\cap B_i=\emptyset$ for all $i$.

I don't really know where to start with this one and I am not entirely sure if it is even true but I think it is. Any help would be appreciated!

答:
For your second question, it isn’t always possible to re-index the sets in the desired manner. To see this, let $m=n$ and $S=[n]\times[n]$. If we let $A_k=\{k\}\times[n]$ and $B_k=[n]\times\{k\}$ for each $k\in[n]$, we find that $\langle k,\ell\rangle\in A_k\cap B_\ell$ for each $\langle k,\ell\rangle\in[n]\times[n]$: each $A_k$ has non-empty intersection with each $B_\ell$.

In fact there is a counterexample whenever $n\ge m$. Let $S=[m]\times[n]$, and for $k\in[m]$ let $A_k=\{k\}\times[n]$. For $\ell\in[m]$ let

$$B_\ell=\big([m]\times\{\ell\}\big)\cup\left(\{\ell\}\times\big([n]\setminus[m]\big)\right)\,;$$

clearly $\langle k,\ell\rangle\in A_k\cap B_\ell$ for each $\langle k,\ell\rangle\in[m]\times[m]$.


问:
What *can* Euclid prove?
It is well-known that Euclid's axioms for geometry are not up to modern standards of rigor: in particular, there are a lot of times when he used "obvious" facts about the geometric objects which were neither stated in, nor derivable from, the axioms.

For example, in his very first proposition, Book I Proposition I, he sets out to construct an equilateral triangle. Seems easy enough, right? He gives a construction involving two circles drawn at the ends of a line segment which is to be one side of the triangle and uses this to produce the remaining two sides. Trouble is, he has no axioms which guarantee that the circles will intersect, as they must to produce the third vertex of the triangle! Another, even more subtle, problem is that when you draw the segments to the third vertex, how do you know what you form is in fact a triangle? You need another axiom for that -- specifically, the problem is you don't know the sides don't meet somewhere else than the new point before they get there. This trouble was actually known even back in old Greece: Zeno of Sidon pointed this one out.

So that makes me wonder: just what can you prove from Euclid's axioms, definitions, etc. without resorting to any hidden, "obvious" assumptions? If you can't so much as even get something as basic as whether the points you are attempting to construct by crossing lines and circles really exist, it seems the answer is "not a whole heck of a lot!" Or am I wrong?

答:
Nothing because  the  geometry is axiomatic

and that

```<code>Definition 1 : A point is what the game is drawn for null .
</code>```

Therefore, according to 1 a point has : neither length nor width.

```<code>Definition 2 : A line is a length without a width.
</code>```

But a null part added to itself will always remain identical to itself, namely null ..

in fact,

there is no nul draw

the length of a line is the set of length parts of that line ...

Title: What *can* Euclid prove?

Body:
Nothing because  the  geometry is axiomatic

and that

Definition 1 : A point is what the game is drawn for null .

Therefore, according to 1 a point has : neither length nor width.

Definition 2 : A line is a length without a width.

But a null part added to itself will always remain identical to itself, namely null ..

in fact,

there is no nul draw

the length of a line is the set of length parts of that line ...


问:
$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$, $a_n\in \mathbb Q$ and not $a_n$ not monotonic
How can I construct a sequence $\{a_n\}$ of positive rational numbers, which is not monotonic such that $$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$$
I thought a lot on this question but I always come up with a monotonic sequence. Any help will be appreciated.</p>

答:
As a conditionally convergent series can be rearranged to give any value at will, use your favorite of those series (mine happens to be $\sum_{n \ge 1} \frac{(-1)^n}{n}$). The trick is to pick positive terms until the sum is larger than your target $\pi$ (there is $a_0$), then pick enough negative terms to fall bellow (your $a_1$), and keep going.

No, not exactly a nice set of terms, and it can be considered cheating as you need to know your target beforehand.


问:
Stone-Weierstrass Theorem
In the proof of Stone-Weierstrass theorem provided in Rudin's Principles of Mathematical Analysis, why do we only need to show that there exists a sequence of polynomials $P_n$ that converges uniformly to the continuous complex function on $[0,1]$?</p>

答:
Define $\phi:[a,b]\rightarrow [0,1]$ by
\begin{align*}
   t = \phi(x) = \frac{x-a}{b-a},\quad\text{and}\quad
    x = \phi^{-1}(t) = a + (b-a)t
   \end{align*}
It is continuously increasing function so that $\phi$ is bijective and $\phi^{-1}$ is also continuous. Then let $f(x) = f(\phi^{-1}(t))=g(t)$ so that $g:[0,1]\rightarrow R$ is also continuous by composition theorem. Suppose the theorem holds for $g$. Let $\varepsilon > 0$, there exists $P_n$ such that $\|P_n - g\|_{\infty} < \varepsilon$. Let $t\in[0,1]$, then
\begin{align*}
    P_n(t) =& a_0 + a_1t + a_2 t^2  +\cdots + a_n t^n\\
    =& a_0 + a_1 \phi(x) + a_2 \phi(x)^2  +\cdots + a_n \phi(x)^n\\
    =& a_0 + a_1 \left(\frac{x-a}{b-a} \right) + a_2 \left(\frac{x-a}{b-a} \right)^2  +\cdots + a_n \left(\frac{x-a}{b-a} \right)^n\\
    =&P_n\circ \phi(x)
   \end{align*}
is also a polynomial. Thus, $\|P_n\circ \phi - f\| = \|P_n - f\circ\phi^{-1}\|=\|P_n - g\| < \varepsilon$.

OwnerDisplayName: Paul

ContentLicense: CC BY-SA 2.5

Title: Stone-Weierstrass Theorem


问:
Set Aut F of all field automorphisms is a group
This question is from my abstract algebra course.

Instructor told us to prove this by ourselves.

The set Aut F of all field automorphisms $F \to F$ forms a group under the operation of composition of functions.

I have problem in proving how inverse of a function $\sigma$ exists. Which result I should use?

Can you please tell how to prove that inverse exists.

答:
This is not always the case. For example the mapping
$$
\sigma:\frac{a(t)}{b(t)}\mapsto \frac{a(t^2)}{b(t^2)}
$$
is a non-surjective homomorphism from the field $F=\Bbb{R}(t)$ of rational functions to itself. Here $a(t),b(t)$ are arbitrary polynomials.

However, the following general facts do allow us to conclude this in many special cases:

A homomorphism of fields necessarily has a trivial kernel, and is thus always injective.

A homomorphism of fields is always linear over the prime field $k$. Therefore by rank-nullity an injective homomorphism is always surjective, if $[F:k]<\infty$.

So if $F$ is a finite extension of $\Bbb{Q}$ or $\Bbb{F}_p$ for some prime number $p$, the claim holds automatically.

It is true in some other cases as well. For example in the cases $F=\Bbb{R}$ and $F=\Bbb{Q}_p$ (= the field of $p$-adic numbers) simply because in those cases the identity mapping is the only automorphism.


问:
Formal Proof - Premises & Conclusion (Rules of Inference & Equivalence)
So I have gathered/learned a total of 8 different rules of inference & 10 rules of equivalence for proofs: making a total of 18 proofs (Modus Ponens, Modus Tollens, Disjunctive Syllogism, Hypothetical Syllogism, Conjunction, Addition, Simplification, Constructive Dilemma, De Morgan's Law, Association, Distribution, Commutativity, Double Negation, Contraposition, Material Implication, Material Equivalence, Expotation, and Tautology). I want to turn the following premises GIVEN into a conclusion using the rules I know and mentioned.

Premises:

$(G \wedge I) \implies H$

$(I \implies H) \implies F$

Conclusion [What I want] : $G\implies F$

My Progress :

$(G \wedge I) \implies H$

$(I \implies H) \implies F\qquad\qquad\qquad\qquad [ G \implies F]$

$G \implies (I \implies H)\qquad\qquad\qquad\qquad [1, $exp]

$\sim(I \implies H) \vee F\qquad\qquad\qquad\qquad\quad\;    [2, $Impl]

Not sure what else to do.

答:
So I have gathered/learned a total of 8 different rules of inference & 10 rules of equivalence for proofs: making a total of 18 proofs (Modus Ponens, Modus Tollens, Disjunctive Syllogism, Hypothetical Syllogism, Conjunction, Addition, Simplification, Constructive Dilemma, De Morgan's Law, Association, Distribution, Commutativity, Double Negation, Contraposition, Material Implication, Material Equivalence, Expotation, and Tautology).

Hypothetical Syllogism is the rule of inference you seek:$$\dfrac{\phi\to\psi\qquad \psi\to\rho}{\phi\to\rho}$$


问:
Minimum paths required to cover all vertices of three-way graph.
I want to design fictional fruits that have three properties: color, taste and smell. There are $c$ possible colors, $t$ possible tastes and $s$ possible smells. Further, there is a feasibility matrix describing which colors go with which tastes and another feasibility matrix describing which tastes go with which smells. Together, these attributes form a three-way graph like the one shown in the picture below (here, there are 4 possible colors, 3 tastes and 5 smells). What is the minimum number of fruits I need to create so that all colors, all tastes and all smells are represented at least once? I need to devise an algorithm for this given the two connectivity matrices and prove its optimal.

EDIT: I asked a very similar question on CS stackexchange with some great answers. Check it out as well: https://cs.stackexchange.com/questions/131552/min-path-cover-for-a-three-layer-graph-with-all-paths-traversing-all-layers

My attempt:

I asked a similar question about colors and tastes. In that instances, a min edge cover was sufficient, with each surviving edge becoming one fruit. Now with three attributes, it becomes harder. One solution is to run one min-edge cover for colors and tastes and another for tastes and smells. Then, loop through the tastes and see if it has more colors connected to it or more smells connected to it. Assign fruits numbering the max of the two for that taste and assign each color and smell, repeating the one with smaller connections to that taste as required. This approach is almost certainly not optimal since there are multiple possible solutions for a min-edge cover and the two min-edge covers we ran had no knowledge of each other.

EDIT: here is a toy example demonstrating what we need. We have three colors, two tastes and three smells. The feasibility matrix is shown on the left while the optimal solution is shown on the right. We needed three fruits to cover all colors, tastes and smells. This also demonstrates that the "minimum path cover" algorithm referenced in the answer by Daniel below doesn't apply since it requires the paths to be "vertex-disjoint" i.e. not share any vertices. In the solution on the right, we see that the solution does indeed have two paths that share a vertex, $t_1$.

答: The answer by @Daniel G. nails this. Just adding a picture showing how it works on the toy example I posted in the question. The orange lines are the extra edges we needed to add. The pink lines show a vertex disjoin path cover that maps to a non-vertex disjoint path cover on the original graph.


问:
The Integral that Stumped Feynman?
In "Surely You're Joking, Mr. Feynman!," Nobel-prize winning Physicist Richard Feynman said that he challenged his colleagues to give him an integral that they could evaluate with only complex methods that he could not do with real methods:

One time I boasted, "I can do by other methods any integral anybody else needs contour integration to do."

So Paul [Olum] puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration! He was always deflating me like that. He was a very smart fellow.

Does anyone happen to know what this integral was?

答:
Coincidentally, I just happened to be reading "Genius: The Life and Science of Richard Feynman" a few weeks ago. On page 178, James Gleick writes:

At lunch one day, feeling even more ebullient than usual, he
  challenged the table to a competition. He bet that he could solve any
  problem within sixty seconds, to within ten percent accuracy, that
  could be stated in ten seconds. Ten percent was a broad margin, and
  choosing a suitable problem was hard. Under pressure, his friends
  found themselves unable to stump him. The most challenging problem
  anyone could produce was: Find the tenth binomial coefficient in the
  expansion of $(1 + x)^{20}$. Feynman solved that just before the clock
  ran out. Then Paul Olum spoke up. He had jousted with Feynman before,
  and this time he was ready. The demanded the tangent of ten to the
  hundredth. The competition was over. Feynman would essentially have
  had to divide one by $\pi$ and throw out the first one hundred digits of
  the results - which would mean knowing the one-hundredth decimal digit
  of $\pi$. Even Feynman could not produce that on short notice.

Title: The Integral that Stumped Feynman?


问:
How to manipulate the following integral into a power series?
I have to solve this integral:

$$ \int_0^\infty \frac{x^2}{2(\cosh(x) -1)} \, dx $$

How do I convert this into a power series.

I already know that the result of that integral is $\frac{\pi^2}{3}$. That, at least to me, hints that this is a power series trick of sorts.

Thank you.

答:
$$I=\int \frac{x^2}{\cosh(x) -1} \, dx$$

Trying integration by parts
$$u=x^2 \qquad \text{and} \qquad dv=\frac{dx}{\cosh(x) -1}$$
$$du=2x\,dx\qquad \text{and} \qquad v=-\coth \left(\frac{x}{2}\right)$$
$$I=-x^2 \coth \left(\frac{x}{2}\right)+\int 2 x \coth \left(\frac{x}{2}\right)\,dx$$
Integration by parts again
$$u=2x \qquad \text{and} \qquad dv=\coth \left(\frac{x}{2}\right)\,dx$$
$$du=2 dx \qquad \text{and} \qquad  v=2 \log \left(\sinh \left(\frac{x}{2}\right)\right)$$
$$\int 2 x \coth \left(\frac{x}{2}\right)\,dx=4 x \log \left(\sinh \left(\frac{x}{2}\right)\right)-4\int  \log \left(\sinh \left(\frac{x}{2}\right)\right)\,dx$$ For the last integral, I am stuck but Wolfram Alpha knows !

Comining all of the above plus Wolfram Alpha result, assuming $a>0$
$$J=\frac 12\int_a^\infty \frac{x^2}{\cosh(x) -1} \, dx=$$ $$\frac{1}{2} \left(4 \text{Li}_2\left(e^{-a}\right)-a \left(a+4 \log
   \left(1-e^{-a}\right)-a \coth \left(\frac{a}{2}\right)\right)\right)$$ and a series expansion gives
$$J=\frac{\pi ^2}{3}-a+\frac{a^3}{36}+O\left(a^5\right)$$


问:
Factorizing the determinant
Here, in this question, we have to factorize this determinant. I understood the (a-b), (b-c), (c-a) part but I am not able to understand

how to find if determinant is symmetrical in a, b, c.

I know that rest factor will be of degree 2 as we have got (a-b), (b-c), (c-a), but as here (a^2 + b^2 + c^2) and (ab + bc + ca) are possible factors of degree 2, can't be (a + b + c)^2 will be the 2nd degree factor?

Book HK Das Engineering Mathematics page 388 example no. 36

Book HK Das Engineering Mathematics page 388 example no. 36

答:
(1) If you put $a=b,b=c$ or $c=a$ then, $\Delta=0$.
Being symmetric about $a,b,c$ means that if $(a_0,b_0,c_0)$ is a solution to $\Delta=0$, then $(b_0,a_0,c_0)$ is also a solution and so are all other permutations of $(a_0,b_0,c_0)$. This can be seen easily. For now, we verify it for the $3$ permutations $(a_0,b_0,c_0),(b_0,a_0,c_0),(c_0,b_0,a_0)$:
$$\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=-\begin{vmatrix}1 & 1 & 1\\b_0^2&a_0^2&c_0^2\\b_0^3&a_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\c_0^2&b_0^2&a_0^2\\c_0^3&b_0^3&a_0^3\end{vmatrix}\mathrm{~~and~~}\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=0$$
$$\Rightarrow\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\b_0^2&a_0^2&c_0^2\\b_0^3&a_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\c_0^2&b_0^2&a_0^2\\c_0^3&b_0^3&a_0^3\end{vmatrix}=0$$
$\therefore a=a_0,b=b_0,c=c_0$ and $a=b_0,b=a_0,c=c_0$ and $a=c_0,b=b_0,c=a_0$ are all solutions to $\Delta=0$.

(2) $(a+b+c)^2$ may or may not be a possible factor. Notice that if it is then
$$(a+b+c)^2=1\cdot(a^2+b^2+c^2)+2\cdot(ab+bc+ca).$$
The point of taking $(a^2+b^2+c^2)$ and $(ab+bc+ca)$ is that any $2$ degree factor in terms of $a,b,c$ can be written in the form
$$k\cdot(a^2+b^2+c^2)+l\cdot(ab+bc+ca)$$
whereas, there are possible $2$ degree factors of $\Delta=0$ that may not be multiples of $(a+b+c)^2$.

Title: Factorizing the determinant


答:
From Madhava formula, we have
$$\pi = 4 - \dfrac43 + \dfrac45 - \dfrac47 \pm = \sum_{n=0}^{\infty}\dfrac{(-1)^n 4}{(2n+1)}$$
We can write $\dfrac1{2n+1} = \dfrac{k\sqrt{n}+1}{2n+1} - \dfrac{k\sqrt{n}}{2n+1}$. Hence, we have
$$\pi = \sum_{n=0}^{\infty} \left( (-1)^n\dfrac{\sqrt{n}+4}{2n+1} + (-1)^{n+1}\dfrac{\sqrt{n}}{2n+1}\right) = \sum_{n=0}^{\infty}(-1)^n a_n$$
Writing it out, we see
$$\pi = \dfrac{\sqrt{0} + 4}{2 \cdot 0 + 1} - \dfrac{\sqrt{0}}{2 \cdot 0 + 1} + \dfrac{\sqrt{1}}{2 \cdot 1 + 1} - \dfrac{\sqrt{1} + 4}{2 \cdot 1 + 1} + \dfrac{\sqrt{2} + 4}{2 \cdot 2 + 1} - \dfrac{\sqrt{2}}{2 \cdot 2 + 1} + \dfrac{\sqrt{3}}{2 \cdot 3 + 1} - \dfrac{\sqrt{3} + 4}{2 \cdot 3 + 1} + \cdots$$
where $a_n$'s are as desired, i.e.,
$$a_n = \begin{cases} \dfrac{\sqrt{2k}+4}{2 \cdot (2k)+1} & \text{ if $n = 4k$}\\ \dfrac{\sqrt{2k}}{2 \cdot (2k)+1} &\text{ if $n=4k+1$}\\ \dfrac{\sqrt{2k+1}}{2 \cdot(2k+1)+1} & \text{ if $n=4k+2$}\\ \dfrac{\sqrt{2k+1}+4}{2 \cdot(2k+1)+1} & \text{ if $n=4k+3$}\end{cases}$$
Below is the pictorial representation of the sequence $a_n$. The first $44$ values of this sequence are plotted. As you can see from the figure below, the sequence oscillates up and down and is not monotonic.

Below is the convergence of the partial sum $\displaystyle \sum_{n=0}^N a_n$ to $\pi$. The $\color{red}{\text{red line}}$ is the value of $\pi$ for reference. You can see how the partial sums converge to $\pi$.

I didn't see $a_n \in \mathbb{Q}$. Here is the answer if you want $a_n \in \mathbb{Q}$. $$\dfrac1{2n+1} = \dfrac2{2n+1} - \dfrac1{2n+1}$$
Hence,
$$\pi = \sum_{n=0}^{\infty} \left({(-1)}^n\dfrac{8}{2n+1} + (-1)^{n+1} \dfrac4{2n+1}\right)$$
$$\pi = \dfrac81 - \dfrac41 + \dfrac43 - \dfrac83 + \dfrac85 - \dfrac45 \pm \cdots$$
Now choose $a_n$'s as follows.
$$a_n = \begin{cases}\dfrac{8}{4k+1} & \text{ if $n=4k$}\\ \dfrac4{4k+1} & \text{ if $n=4k+1$}\\ \dfrac4{4k+3} & \text{ if $n=4k+2$}\\ \dfrac8{4k+3} & \text{ if $n=4k+3$}\end{cases}$$

Title: $\sum_{n=0}^{\infty}(-1)^n a_n = \pi$, $a_n\in \mathbb Q$ and not $a_n$ not monotonic

Body:
How can I construct a sequence $\{a_n\}$ of positive rational numbers, which is not monotonic such that $$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$$
I thought a lot on this question but I always come up with a monotonic sequence. Any help will be appreciated.</p>


答:
If $ \sigma \in Aut F$ then $\exists \sigma^{-1} : F \rightarrow F$ which is a bijection.
We have
$$\forall (y_1, y_2) \in F^2 \quad y_1+y_2 = \sigma(\sigma^{-1}(y_1)) + \sigma(\sigma^{-1}(y_2)) = \sigma(\sigma^{-1}(y_1) + \sigma^{-1}(y_2)).$$
Hence
$$\sigma^{-1}(y_1+y_2) = \sigma^{-1}(y_1) + \sigma^{-1}(y_2).$$
And
$\forall \lambda \in \mathbb R \quad \forall y \in F$ we have
$$\lambda y= \lambda\sigma\sigma^{-1}(y) =\sigma(\lambda\sigma^{-1}(y)).$$
Then
$$\sigma^{-1}(\lambda y)=\lambda \sigma^{-1}(y).$$

Title: Set Aut F of all field automorphisms is a group

Body:
This question is from my abstract algebra course.

Instructor told us to prove this by ourselves.

The set Aut F of all field automorphisms $F \to F$ forms a group under the operation of composition of functions.

I have problem in proving how inverse of a function $\sigma$ exists. Which result I should use?

Can you please tell how to prove that inverse exists.


问:
Proving that $Aut_{\mathbb{Q} } { \mathbb{R}} $
The following question is from my Field Theory assignments.

Prove that  $Aut_{\mathbb{Q} } { \mathbb{R}} $  is identity group.

I took $f:\mathbb{R} \to \mathbb{R}$ be an automorphism.
Now  which result I should use to prove that all the automorphism which are $\mathbb{Q}$ -module homomorphisms are identity.

Let $f'$ be such an automorphism and q belongs to $\mathbb{Q}$. So, $f'(q) = qf'(1) =q$.but how to prove that $f'(q) =1$ .

Kindly guide.

答: For every $x$ $f(x)=f(x\cdot1)=f(x)\cdot(1)$. So if $f$ is not constant $=0$, $f(1)=1$.


问:
Probability proof using Bonferroni inequality & union bound
let $\mathcal{A}=\left\{ A_{1},A_{2},\ldots,A_{n}\right\}$ a finite set of events in $\left(\Omega,\mathcal{F},\mathbb{P}\right)$ s.t. $\mathbb{P}\left(A_{1}\cap A_{2}\right)=0$ for all $A_{1},A_{2}\in\mathcal{A}$ and $ \mathbb{P}\left(\underset{A\in\mathcal{A}}{\bigcup}A\right)=1$. Show that every event $B$ apply: $\underset{A\in\mathcal{A}}{\sum}\mathbb{P}\left(B\cap A\right)=\mathbb{P}\left(B\right)$.

I'm trying to prove that $\underset{A\in\mathcal{A}}{\sum}\mathbb{P}\left(B\cap A\right)\geqslant\mathbb{P}\left(B\right)$ using bonferroni's second inequality and $\underset{A\in\mathcal{A}}{\sum}\mathbb{P}\left(B\cap A\right)\leqslant\mathbb{P}\left(B\right)$ using the union bound, but I find myself lost in set-theorem (which I haven't learnt) nuances of unions and intersections

答:
Here are the main steps and I will you do the verifications:

Let $C=(\Omega \setminus \bigcup_i A_i) \bigcup ( \bigcup_{i \neq j} A_i \cap A_j)$.  Then

$P(C)=0$

$C_i \equiv A_i\setminus C$, $1\leq i \leq n$, are disjoint.

$P(B)=P(B\setminus C) \leq P(\bigcup (B \cap  A_i))\leq \sum P(B\cap A_i)$

and

$\sum P(B\cap A_i) =\sum P(B \cap C_i)= P(B \bigcup C_i) \leq P(B)$


答:
We can avoid l'Hospital by

$$(4 - x)^{\tan (\frac {\pi x} {2})}=\left[\left(1 - (x-3)\right])^\frac1{x-3}\right]^{(x-3)\tan (\frac {\pi x} {2})}$$

with

$$\left(1 - (x-3)\right])^\frac1{x-3} \to \frac1e$$

and by $x-3=t \to 0$

$$(x-3)\tan \left(\frac {\pi x} {2}\right)=t\tan \left(\frac {3\pi } {2}+\frac {\pi } {2}t\right)=-\frac{t}{\tan \left(\frac {\pi } {2}t\right)}=-\frac 2 \pi\frac{\frac \pi 2t}{\tan \left(\frac {\pi } {2}t\right)}\to -\frac 2 \pi$$