问:
Is there a $3$-category in which natural transformations between functors of small categories are $3$-morphisms?
I'm not sure if this question is well-posed, but what I'm getting at is that in, e.g. the category of groups, the $1$-morphisms are group homomorphisms. Then if we look at $\mathbf{Grp}$ from within $\mathbf{Cat}$, the $1$-morphisms are functors and the $2$-morphisms are natural transformations. But it feels like in something we ought to be three levels up. Does there exist a $3$-category in which this makes sense?

答:
I am not sure what you are asking for. In the the  strict 2-category $\mathbf{Cat}$ of small categories whose objects are small categories, $1$-morphisms are functors and $2$-morphisms are natural transformations, we cannot make sense of non-trivial three morphisms.

We can remedy this by going into the next homotopy level $\mathbf{2Cat}$ whose

objects are $2$-categories;

$1$-morphisms are $2$-functors;

$2$-morphisms are lax natural transformations;

$3$-morphisms are modifications (morphisms between natural transformations).

If this is what you are asking for, I am glad to add more details to my answer.


问:
Conditions given in theorem 3.42 in Rudin's Principal of Mathematical Analysis
The theorem states that if partial sums $A_n$ of series $\sum a_n$ are bounded and $(b_n)$ is a monotonically decreasing sequence such that $b_n\to 0$, then the series $\sum a_nb_n$ converges.

I tried to prove it as follows:

Since $A_n$'s are bounded, there exists $M\gt 0$ such that $|A_n|\lt M$ for all $n\in \mathbb N$. Since $(b_n)$ converges, for every $\epsilon\gt 0, \exists N:\forall q\ge p\ge N,$ we have $b_p-b_q\lt \frac \epsilon {2M}$.

We have, $|\sum_{n=p}^q a_nb_n|=|\sum_{n=p}^{q-1}A_n(b_n-b_{n+1})+A_qb_q-A_{p-1}b_p|\lt M |\sum_{n=p}^{q-1}(b_n-b_{n+1})+b_p-b_q|=2M(b_p-b_q)\lt \epsilon \tag 1$

Hence, by Cauchy Criterion, the series $\sum a_nb_n$ converges.

Please note that in the above proof, I have nowhere used $b_n\to 0$. I have only used the fact that the sequence $(b_n)$ is monotonically decreasing and convergent.

My question is why $b_n$ must converge to $0$? Clearly, from the proof above, convergence of $(b_n)$ is enough, irrespective of to what $\alpha\in \mathbb R$, $(b_n)$ converges.

Edit: Suppose that $(b_n)$ is a sequence of non-negative nos. We have $-M\lt A_q\lt M\implies -Mb_q\lt A_qb_q\lt Mb_q$. Similarly, $-Mb_p\lt A_{p-1}b_p\lt Mb_p$. It follows that $-M(b_p-b_q)\lt A_{p-1}b_p-A_qb_q\lt M(b_p-b_q)\implies |A_{p-1}b_p-A_qb_q|\lt M(b_p-b_q)$ so $(1)$ holds. Is it necessary for $b_n\to 0$?

答:
Your edit clarifies the issue in your proof; indeed, if that proof were valid, why, any bounded set of numbers would have to be equal to one another! (set $b_p = b_q = 1$)

The issue is how you are subtracting the inequalities: If $a_l < a < a_u$ and $b_l < b < b_u$, then $a_l - b_u < a - b < a_u - b_l$ and not $a_l - b_l < a - b < a_l - b_l$. Intuitively, the largest $a - b$ can be is if $a$ is its largest, and $b$ is its smallest value.

ContentLicense: CC BY-SA 4.0


问:
Every closed point is open in a scheme locally of finite type over a field
Let $X$ be a scheme locally of finite type over a field $k$, then we know the closed points in $X$ is very dense.

If we assume that every closed point is also open in $X$, then is $X$ a discrete scheme?

Thanks in advance.

答:
Here's the relevant topological fact: Let $X$ be a topological space such that (1) every closed point of $X$ is open, and (2) every nonempty closed subset of $X$ contains a closed point. Then $X$ is discrete. Indeed, for any $x \in X$, the closure $\overline{\{x\}}$ contains a closed point $y$, which is also an open point, so $\overline{\{x\}} \setminus \{y\}$ is closed, which means $x = y$ by the definition of closure. Thus every point of $X$ is closed and open, so $X$ is discrete.

The only algebraic input needed here is that locally Noetherian schemes (which includes schemes locally of finite type over a field) have the property that every nonempty closed subset contains a closed point. (For reference, this is Lemma 28.5.9 (tag 02IL) in the Stacks Project.)


问:
How to show that condition number of $QA$ is equal to condition number of $A$
If $A$ is an invertible matrix and $Q$ is an orthonormal matrix, show that
$$k(QA) = k(A).$$

Hint: $k(A) = \frac{\sigma_{1}}{\sigma_{n}}$ (the ratio of the largest and smallest eigenvalues).

答:
using polar form, $A$ may be written uniquely as

$A=UP$ where $P$ is (real symmetric) Positive Definite. $P$ contains the singular values of $A$.

$\big(QA\big)=Q\big(A\big)=Q\big(UP\big) = \big(QU\big)P$

and $P$ has the singular values for $\big(QA\big)$ so $\big(QA\big)$ and A have the same singular values.

note: this assumes $Q$ is a real orthogonal matrix.  There really isn't such thing as an 'orthonormal matrix'.


问:
Optimization approaches to solving PDEs
In modern numerical methods, a PDE is often recast into the form of a variational problem, which is sometimes equivalent to a minimization problem.
However in my courses on numerical analysis (say, finite element methods) the focus is not (apparently) on developing optimization techniques to minimize the arosen energy functional, but rather on approximating the variational problem on a smaller subspace.

Are there interesting approaches that focus on the minimization of the energy directly? Is research being done in this field, and could you maybe provide some reference?

答:
A good start to gather some insight into PDE is partial_differential_equation. This shows up that still the knowledge based start is most important. Identify what the PDE under consideration is. Retrieve the historical solution and improvements and then pose your own specifics in details.

There are existence and uniqueness consideration and a wellposed examination needed.

Following the successful and skillfull classification there is for a mathematician in need of reputation the calculation of analytically closed solution. Based on this for inhomogenities a variation problem is to be solved.

If this is not possible some numerical experiments can be conducted. Most of these method solve the pde on finite points and lead to an interpolation function as a result.

The energy method can be used to verify well-posedness of initial-boundary-value-problems. That is a subset of pde and one that does fit into the finite kompartment methodologies too. Energy principles in structural mechanics lists some of the successful applications. This again follows the knowlegde approach. The examples are of eternal importance but the references in the article are a little outdated.

Whether pde or energy principle is easier or less complex in the solution depends on the CAS or FEM system in use. Many of these have the knwolegde incorporated into their codes and the learning curves are steep and their are tutorials available that recompile the knowledge to the extent to provides solutions in these systems.

If there is a trend than in modularity. Most designers of course know that their may be two or more to be successfully sold to the interested ambitioned scientist. A comparison of both article shows that the pde methodologies are described to some depth while the energy principle is not. Why is this the case? The pde methodologies are used more often and considered more comfortable. They are indeed posed more to the community. The energy principle on the other is an overview page and have below each specification another more detailed page for the problem categories covered.

It is necessary to use such great tools a mind maps on the wikipedia graph of linkes articles to gain more overview. The energy principles does no exist independent of the pde page but the path is longer than reproduced in the given question. This path has to be studied and that is the designers of modern pde courses much to much content. They are in need to over the steep learning of the FEM systems or programmin languages too.

The interested and ambitioned student can use the literature to fill in the gaps. Put aside the question of moderness of the literature. The wikipedia links offer a sorrow and convinient introduction and give opportunity to speed plenty of time to start with correspondences up to cohesion.

Title: Optimization approaches to solving PDEs


问:
How can we motivate homotopy in homological algebra?
I'm writing some notes on homological algebra for some readers which are experienced in algebra (had a course in commutative algebra using categorical notions) but who don't necessarily know algebraic topology. With this in mind, I began to wonder how could we motivate the definition of homotopy between two maps of complexes purely algebraically.

While it is true that homotopy is important because homotopic maps induce the same morphism in cohomology or because the homotopic category is useful for constructing the derived category (and is a useful approximation thereof),  it is not clear to me how someone "inventing" (without all the roots in algebraic topology) homological algebra might arrive at such a formula (the one defining homotopy).

What I have in mind is something in the lines of: "this doesn't really work but this formula (the one defining homotopy) is precisely what we need for it to work".

答:
A short exact sequence is a useful presentation when speaking about the relation of an object $B$ with a subobject $A$. Sometimes the short exact sequence
$$0\to A\xrightarrow{i} B\xrightarrow{p} C\to 0$$
splits, and $A$ is now a direct summand of $B$. This is a nice property which is preserved by any additive functor, that is, for any $F$, the sequence
$$0\to F(A)\to F(B)\to F(C)\to 0$$ is exact. On the other hand, if the short exact sequence does not split, then the above sequence is not exact anymore in general.

Split exact sequence is generally defined in term of a section $s:C\to B$. But there are other equivalent conditions. In particular, writing $p:B\to C$, we see that $p(x-sp(x))=0$ so $x-sp(x)\in A$. Thus, the map $id-sp$ factor through the inclusion $i:A\to B$, i.e. can be written $id-sp=ir$ for a so-called retraction $r:B\to A$. Put differently $id=sp+ir$.

Conversely, if $id=sp+ir$ for some maps, then applying $p$ on the left $p=psp+pir=psp+0=psp$. So $(id-ps)p=0$ and since $p$ is onto $id-ps=0$ In other words $s$ is a section of $p$. Dually $r$ is a retraction of $i$.

The concept of split exact sequence can be generalized for any long exact sequence
$$...\to A^{k-1}\xrightarrow{d^{k-1}}A^k\xrightarrow{d^k}A^k\to ...$$in several equivalent ways :

every short exact sequence we get by splitting the long exact sequence is split.

every map $d^k$ induces an isomorphism between a direct summand of $A^k$ onto $\operatorname{im}(d^k)$.

There are maps $f^k:A^k\to A^{k-1}$ such that $id_{A^k}=f^{k+1}d^k+d^{k-1}f^k$.

Let us prove these equivalences :

$1.\Rightarrow 2.$ The short exact sequences are $0\to Z^k\xrightarrow{i^k} A^k\xrightarrow{p^k} Z^{k+1}\to 0$ with $Z^k=\ker(d^k)=\operatorname{im}(d^{k-1})$ and the differential is $d^k=i^{k+1}p^k:A^k\to Z^{k+1}\to A^{k+1}$. If this splits, then this means that $A^k=Z^k\oplus Y^k$ for a direct summand $Y^k$ and the differential $d^k$ induces an isomorphism $Y^k\simeq Z^{k+1}$.

$2.\Rightarrow 3.$ Let us write $Y^k$ for the direct summand. Its complement is necessarily $Z^k$ since by hypothesis $Y^k\hookrightarrow A^k/Z^k\simeq Z^{k+1}$ is an isomorphism. Write $f^k:A^k\to Z^k\simeq Y^{k-1}\to A^{k-1}$. Then by construction $d^{k-1}f^k$ is the projection $A^k\to Z^k\to A^k$ and $f^{k+1}d^k$ is the projection $A^k\to Y^k\to A^k$. It follows that $f^{k+1}d^k+d^{k-1}f^k=id_{A^k}$.

$3.\Rightarrow 1.$ Write $s^k=f^{k+1}i^{k+1}:Z^{k+1}\to A^{k+1}\to A^k$. We compute $$i^{k+1}p^ks^k=d^ks^k=d^kf^{k+1}i^{k+1}=(id_{A^{k+1}}-f^{k+2}d^{k+1})i^{k+1}=i^{k+1}-f^{k+2}d^{k+1}i^{k+1}=i^{k+1}$$ (the last equality follows from $d^{k+1}i^{k+1}=0$). Since $i^{k+1}$ is mono, $p^ks^k=id_{Z^{k+1}}$. Hence every short exact sequence splits.

The last condition actually implies that the long sequence is exact and is open to generalization : replace $id_{A^k}$ by any morphism $\varphi$ of complexes. If $\varphi^k=f^{k+1}d^k+d^{k-1}f^k$, then $\varphi^k$ induces the zero map in homology. (And if $\varphi^k=id_{A^k}$ then the identity is zero in homology which is only possible if the complex has no homology).

Use substraction, that is replace $id_{A^k}$ by $\varphi^k-\psi^k$ and you have a condition, stable by any additive functor, for two morphisms to induce the same map in homology.

Use composition, that is replace $id_{A^k}$ by $id_{A^k}-\varphi^k\psi^k$ and $id_{B^k}-\psi^k\varphi^k$ and you have a condition for two morphisms of complexes $\varphi, \psi$ to induce isomorphisms (inverse to each other) in homology.


问:
Geodesic and Euler - Lagrange equation
If we have a metric $g_{\mu \nu}$, defined in a Riemannian manifold we can write the equation of the geodesic:
$$\frac{d^2x^\mu}{dt^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{dt}\frac{dx^\beta}{dt}$$ in which $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols. The geodesic in that metric can be obtained also using the Euler - Lagrange equations:
$$\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}=0$$
Where the $q_k$ are the generalized coordinates. How can I find the Lagrangian $L$ in order to find the same geodesic obtained using the affine connection symbols?</p>

答:
The Lagrangian is the standard measurement of a distance. That is
$$
   L=-\kappa\int ds
$$
being $\kappa>0$ a constant, that in your case is
$$
   S=-\kappa\int\sqrt{g_{\mu\nu}dx^\mu dx^\nu}.
$$
Then you are supposed to parametrize the coordinates as $x_\mu=x_\mu(t)$ and the computation becomes a standard one. Notice that, for a Minkowski metric, this reduces to the action of a relativistic free particle.</p>

ContentLicense: CC BY-SA 3.0

Title: Geodesic and Euler - Lagrange equation


问:
Open and closed sets in $\mathbb{Q}$
Let $E$ be all $p \in \mathbb{Q}$ such that $2<p^2<3$. Could someone quickly check the following?

$E$ is closed in $\mathbb{Q}$ because $E^c$ is open: consider $x \in E^c$ so $x^2 \geq 3$ or $x^2 \leq 2$, and $x > \sqrt{3}$ or $x<\sqrt{2}$. If $x > \sqrt{3}$ then by the density of the rationals in $\mathbb{R}$ there exists $x>z>\sqrt{3}$ and with $r=|x-z|$, we have $N_r(x) \subset E$ so $x$ is an interior point and $E^c$ is open so $E$ is closed. If  $x<\sqrt{2}$ then by the density of the rationals in $\mathbb{R}$ there exists $x<z<\sqrt{2}$ and with $r=|x-z|$, we have $N_r(x) \subset E$ so $x$ is an interior point and $E^c$ is open so $E$ is closed.

$E$ is also open in $\mathbb{Q}$ because for some $x \in E$, we know $2<x^2 < 3$ so $\sqrt{2}<x < \sqrt{3}$ and then similarly find $y$ and $z$ such that $\sqrt{2} < y < x < z< \sqrt{3}$ and take $r = \min\{y,z\}$ and $N_r(x) \subset E$. So $E$ is open.

We know compact subsets of metric spaces are closed so then since $E$ is open it is not compact.

答:
Here's an alternative proof to show that $E$ is clopen in $\Bbb Q$.

$E$ is closed (in the subspace topology inherited from $\Bbb R$) because $E = \left ( \left [\sqrt 2, \sqrt 3 \right ] \cup \left [ - \sqrt 3, - \sqrt 2\right ] \right ) \cap \Bbb Q$, and $E$ is open (in the same topology) because $E = \left ( \left (\sqrt 2, \sqrt 3 \right ) \cup \left ( - \sqrt 3, - \sqrt 2\right ) \right ) \cap \Bbb Q$.

To show that $E$ is not compact, note that $\left \{\left ( \left( -\sqrt 3, - \sqrt 2 \right) \cup \left (\sqrt 2, \sqrt 3 - \frac 1n \right ) \right) \cap \Bbb Q \mid n \in \Bbb N, n \geq 4 \right \}$ is an open cover of $E$ with no finite subcover.

Title: Open and closed sets in $\mathbb{Q}$


问:
Gram-Schmidt orthonormal basis for x1 + x2+ x3 =0
I need to find an orthonormal basis for the plane x1 + x2 + x3 = 0 by using Gram-Schmidt on the vectors $$ \begin{pmatrix} 1 \\-1 \\0 \end{pmatrix}, \begin{pmatrix} 0 \\1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\0 \\-1 \end{pmatrix}$$

I know how to use Gram-Schmidt on a basis that contains a few vectors, but I don't know what to do with that plane equation. Can anyone help me?

答: Any two of your vectors span the subspace but are not orthogonal.  If you look at the Graham-Schmidt process you only subtract multiples of the earlier vectors at each stage.  If all your vectors start out within a subspace, as they are here, the ones that come out of the process will also be within the subspace.  So just do the process as normal on the any two of the given vectors and you will be fine.</p>


问:
If two random variables are defined on the same probability space do they have the same distribution?
Let's say we have two random variables $X$ and $Y$ both defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. So $X$ and $Y$ are measurable functions from $\Omega$ to $\mathbb{R}$.

Do $X$ and $Y$ necessarily have the same distribution? Can we for example have $X\sim Bern(p)$ and $Y\sim N(0,1)$? Or would we then say that $X$ and $Y$ have for example the measure $\mathbb{P}_X$ and $\mathbb{P}_Y$ respectively?

答: Simple example: Two coins, one biased $P(heads)=.75$ and the other not $P(heads)=.5$.  Same sample spaces for both.


问:
Reference on Fixed Point Theory
Could you please recommend me some references for Fixed Point Theory? I am after references containing the main fixed point theorems, their variants, and sufficient (neccessary?) conditions for uniqueness of the fixed points. Mainly I've been looking at Brouwer's and Kakutani's, but any others like Banach's are welcome. I searched here in stack exchange but I don't think I saw much covering these aspects in a comprehensive manner.</p>

答:
Try these books:

Fixed Point Theory and Applications by Agarwal et al.

Iterative Approximation of Fixed Points by Berinde

Title: Reference on Fixed Point Theory


问:
Can similar convex heptagons tile the plane?
It is a fairly straightforward matter to apply Euler's formula $V-E+F=2$ for planar graphs to see that congruent convex heptagons cannot tile the plane. The graph associated to the heptagons within distance $R$ of the origin, with edges between heptagons that overlap along an edge, has degree $\ge7$ at every "interior" vertex but (for sufficiently large $R$) insufficiently many exterior vertices to balance things out. This can be extended to any family of convex heptagons whose area and perimeter are bounded - see e.g. this thread on the /r/mathriddles subreddit for a more detailed proof.

I am curious whether it is possible to tile the plane with convex heptagons which are similar to each other; I suspect it is impossible, but the above sort of argument seems to fail, because we cannot assume boundedness.

Note that there do exist tilings of the plane by heptagons if we relax these boundedness constraints; for instance, one can take a tiling of the hyperbolic plane by heptagons in the disc model and apply a map $(r,\theta)\mapsto(\frac{r}{1-r},\theta)$ to yield a tiling of the whole plane (if we straighten out the edges after applying the transformation).

Is it possible to tile the plane with similar convex heptagons? If the answer is "no", I am interested in the case of finitely many heptagon shapes, which I suspect is also impossible.

Edit: Edward H.'s answer made me realize that there are some additional tiling conditions that seem relevant when allowing for arbitrarily-scaled pieces.

Say that a tiling is lower-bounded if there is some $\epsilon>0$ such that every tile contains a ball of radius $\epsilon$.

Say that a tiling is locally finite if the boundary of every tile is covered by finitely many other tiles.

Say that a tiling is traversable if you can reach any tile from any other tile by crossing over shared edges.

Say that a tiling is complete if every point in the plane is in the closure of a tile (as opposed to being in the closure of the union of all tiles).

I'm most interested in the case of tilings that are locally finite, traversable, and complete, but I'm also curious to hear about possibility/impossibility results for any of the above restrictions. (I'm also interested in any implications between them - I think that being lower-bounded implies the other three conditions, but I don't have a proof yet.)

答:
Here is a fractal sort of tiling I've found with a single heptagon (up to similarity) which is traversable, but not lower-bounded, locally finite, or complete.

It's not obvious from the picture that this really is traversable, but the idea is to note that after adding the red and blue heptagons (as well as the geometric series of tiles directly above the middle of the blue heptagons), we can subdivide all leftover spaces into $45-45-90$ triangles such that every new tile added touches a larger tile along the hypotenuse of the triangle, and that we can inductively add new tiles to the voids left by these tiles to preserve the connectedness of the adjacency graph.

This doesn't really capture the sort of thing I'm hoping to find in the post, but I thought I would post it since it's still on the Pareto frontier of satisfiable properties for a tiling thus far, and might prompt someone else to discover a better tiling.


问:
Non-zero coefficient in a transcendence proof
I am studying the proof of the simple version of Lindemann's Theorem.

Theorem.  If $\alpha$ is a non-zero algebraic number, then $e^\alpha$ is transcendental.

Apologies for the long read, but to explain my question I need to give a very brief summary of the first part of the proof.

Let $\alpha$ be algebraic with conjugates $\alpha_1,\ldots,\alpha_l$.  Suppose $e^\alpha$ is algebraic with minimal polynomial
$$p(z)=c_0+c_1z+\cdots+c_mz^m$$
over the integers.  We expand
$$p(e^{\alpha_1})\cdots p(e^{\alpha_l})\tag{1}$$
as a sum
$$\sum_\beta a_\beta e^\beta\ ,$$
where each $\beta$ is an integer linear combination of the $\alpha_j$.  By collecting terms we obtain a similar sum in which all the $\beta$ are distinct.  Now it is clear that one of the $\beta$ values is zero, so that we have an integer term $c_0^l$.  However it is possible that more than one $\beta$ is zero, so after collecting terms we have something like
$$a_0+a_1e^{\beta_1}+\cdots\ ,\tag{2}$$
where the $\beta_j$ are non-zero.  If $a_0\ne0$ then the proof proceeds quite smoothly, but if $a_0=0$ then we have to modify the preceding expression... but I don't need to tell you about that.

My question: is this modification ever actually necessary?  That is, can $a_0$ ever be zero?  To be quite clear about it -

Can anyone provide a specific example of an algebraic number $\alpha$ and an irreducible polynomial $p$ such that when we expand $(1)$ we obtain an expression like $(2)$ with $a_0=0$?

答:
Possible example (some details to work out): Let $p(z)=144z^5-27z^3-110z^2-84z-133$ which I expect to be irreducible over the rationals (should be a simple check left to the OP) and let $\alpha_1,..\alpha_5$ the roots.

Then since the sum of the roots is zero (and again I expect no other integral linear relation between them though it is something to be checked), when one does $p(e^{\alpha_1})\cdots p(e^{\alpha_5})$, one gets a free term which is $144^5-27^5-110^5-84^5-133^5=0$ per the famous Lander and Parkin counterexample to Euler's conjecture


问:
Solve $y' = y^2x$
Given the differential equation $y' = y^2x$, what are the solutions to y? Obviously this is not a linear differential equation, which I'm used to solving. I checked if it is separable, which would make it easier to solve, and I ended up with the solution $y= \frac{2}{x^2+c}$, where c is the constant. Are there more solutions to this I'm missing?

答:
By inspection, $y=0$ is a solution. Then

$$\frac{y'}{y^2}=x$$ and by integration

$$-\frac1y=\frac{x^2}2+c.$$

Title: Solve $y' = y^2x$


问:
Show that $(1+\frac{1}{n^2})^n \rightarrow 1$ as $n \rightarrow \infty$

Show that $\left(1+\frac{1}{n^2}\right)^n \rightarrow 1$ as $n \rightarrow \infty$.
hint: $a^2-b^2=(a-b)(a+b)$

I basically have to show that $$\left|\left(1+\frac{1}{n^2}\right)^n-1\right| < \epsilon$$ but how do I choose $\epsilon$? Can anyone help me with that proof?

答:
This is equivalent to showing that the natural log of your sequence goes to 0 (Due to continuity allowing you to exchange the limit and the log)

Looking at the log,  we get
$$n \ln (1 +\frac 1 {n^2})=\frac {\ln (1 +\frac 1 {n^2})}{\frac 1 n} $$
now you have an indeterminate form 0/0,  so you COULD use L'hospitals, but it's much easier to do a variable substitution $u=\frac 1 n$ to transform your limit into
$$\frac {\ln (1 + u^2)} u \text { as }u\to \infty $$
Now you can use L'hospitals a lot easier.  Can you see how to finish from here?


答:
Consider the formula provided by this answer:

$\sum_{0\leq j\leq n/2}\binom{n}{j}x^j = (x+1)^n - \binom{n}{\left\lfloor\frac{n}{2}\right\rfloor + 1} x^{\left\lfloor\frac{n}{2}\right\rfloor + 1} {_2F_1}\left(1,\left\lfloor\frac{n}{2}\right\rfloor - n + 1;\left\lfloor \frac{n}{2}\right\rfloor + 2;-x\right),$

where $_2F_1$ is the hypergeometric function.

Let $f(x)$ be the function above with $n=2k$, that is

$$f(x) = \sum_{j=0}^{k}\binom{2k}j x^j = (x+1)^{2k} - \binom{2k}{k + 1} x^{k + 1} {_2F_1}\left(1,1-k;k + 2;-x\right).$$

Now write $(x+1)^{2k} = \sum_{j=0}^{2k}\binom{2k}j x^{j}$, so that

$$\begin{align}
(x+1)^{2k} - f(x)
&= \sum_{j=k+1}^{2k}\binom{2k}j x^j
\\&= \sum_{i=1}^{k}\binom{2k}{k+i} x^{k+i}
= \sum_{i=1}^{k}\binom{2k}{k-i} x^{k+i}
\end{align}$$

With $h(x) = (x+1)^{2k} - f(x) = \binom{2k}{k + 1} x^{k + 1} {_2F_1}\left(1,1-k;k + 2;-x\right)$, we hence have

$$x^{1-k}h(x) = \sum_{i=1}^{k}\binom{2k}{k-i} x^{i+1}.$$

Differentiating we get

$$(1-k)x^{-k}h(x) + x^{1-k}h'(x) = \sum_{i=1}^{k}(i+1)\binom{2k}{k-i} x^i,$$

so that your sum can be obtained by evaluating the LHS at $x=1$:

$$(1-k)h(1) + h'(1).$$

This yields a complicated expression in terms of hypergeometric functions:

$$\begin{align}
&&(1-k)&&\binom{2k}{k+1}&&{_2F_1}\left(1,1-k;k + 2;-1\right)
\\+\,\,\,&&(k+1)&&\binom{2k}{k+1}&&{_2F_1}\left(1,1-k;k + 2;-1\right)
\\+\,\,\,&&\frac{k-1}{k+2}&&\binom{2k}{k+1}&&{_2F_1}\left(2,2-k;k + 3;-1\right)
\end{align}
\\= \binom{2k}{k+1}\left(2{_2F_1}\left(1,1-k;k + 2;-1\right) + \frac{k-1}{k+2}{_2F_1}\left(2,2-k;k + 3;-1\right)\right).$$

You can check on WolframAlpha that the formula provided above and the expression in the opening question yield the same results.

Title: Calculate $2 C_{n}^{k-1} + 3 C_{n}^{k-2} + 4 C_{n}^{k-3} + ... + (k + 1)C_{n}^{0}$

Body:
Are there any easy ways to calculate this?

$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = 2 C_{2k}^{k-1} + 3 C_{2k}^{k-2} + 4 C_{2k}^{k-3} + ... + (k + 1) C_{2k}^{0} $$

I tried to "turn" the expression:

$$ \sum_{i=1}^{k} (i + 1) \cdot C_{2k}^{k-i} = n \cdot 2^{n-2} - \frac{1}{2} \sum_{i=1}^{k} (2n -i) \cdot C_{2k}^{k-i} $$

But, it doesn't help me.

Thanks for the help!


问:
Showing compactness that involves unilateral shift
Let $V:l^2(\mathbb{N})\to l^2(\mathbb{N})$ be the unilateral shift, the unique bounded operator on $l^2(\mathbb{N})$ that satisfies
$$
Ve_n=e_{n+1}
$$
, where $(e_n)_{n\in \mathbb{N}}$ is an orthonormal basis.

Show that $V^*V-VV^*$ is compact.

Attempt:

I tried to show $V^*e_n = e_{n-1}$ for $n>1$ and $V^*e_1=e_1$

We have that $\langle V^*e_n, e_m \rangle =\langle e_n, Ve_m \rangle  =\langle e_n, e_{m+1} \rangle = \delta_{n,m+1}$.

Now  $\langle V^*e_n,x\rangle = \langle e_n, Vx \rangle  =\langle e_n, V\sum_{k\in \mathbb{N}  }x_ke_k \rangle =\langle e_n, \sum_{k\in \mathbb{N}  }x_kVe_k \rangle=\langle e_n, \sum_{k\in \mathbb{N}  }x_ke_{k+1} \rangle $.

Then I got stuck here... I feel like it's some simple manipulation of inner product, but I just couldn't get what I wanted.

Also, to show compactness, I want to use that finite dimensional range implies compact operator.

So I have to show  $V^*V-VV^*$ has a finite rank.

If we prove that  $V^*e_n = e_{n-1}$ for $n>1$, then we have $V^*V-VV^* = I - VV^*$.

Then I got stuck proving that $I - VV^*$ has a finite rank.

Could someone help me to finish this?

Thank you!

答:
What you got wrong is that $V^*e_1=0$.

You can easily show that
$$
(I-VV^*)e_n=\begin{cases} e_1,& \ n=1\\[0.3cm] 0,&\ n>1\end{cases}
$$
Then its obvious that $I-VV^* $ has finite rank.


问:
Truncating a Time-Series Graph
I'm looking for a way to truncate outliers in a time-series graph.

Some context:
I'm plotting two different metrics on a single graph.  It's important to understand what portion of "Metric A" takes up respective to "Metric B".  "Metric B" can at times spike to a high value thus increasing the Y-Axis scale which then minimizes the visibility of Metric "A".

Are there any good formulas that help normalize this data or visual treatments that can make the graphic honest as well as accessible?

See screenshot for more context.  The blue area is "Metric B".

Thanks!

答: If Metric B spikes up too much (say to more than twice the maximum of Metric A), you can display it as a vertical blue column with a jagged break in the middle to indicated it goes off the top of the graph.  I have also seen it with a jagged break and an arrow on the top.  You can also list the specific data point to show how tall the spike is.</p>

Title: Truncating a Time-Series Graph


问:
Find a lower bound for a function with a cosine.
I have a function $F(z_1,z_2) := z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2$.

I need to show that $\forall \alpha > 0\, \exists \gamma > 0\, \forall z\, \|z\| \geq \alpha \implies F(z_1,z_2) \geq \gamma $.

Aside from using Lagrange multipliers for this function $F$ with restriction $z_1^2 + z_2^2 \geq \alpha^2$ that give a basically unsolvable (?) equation over $z_2$:
$-(\alpha^2 - z_2^2)\sin(2z_2) + \frac{3}{2}z_2 -z_2\cos^2(z_2) = 0$ do you guys have any other ideas I could use here?

I'm asking this because this is an example taken from a book so I feel like it should not be that hard, maybe some classic estimates or inequalities would work here, but I don't know many. I have tried Cauchy–Schwarz but that's basically the extents of my knowledge.

答:
You can do this with two cases.

Case 1. $z_2 <\pi/4$: Use the identity $\cos^2(x) = \frac12 (1 + \cos(2x))$. This allows to write
$$
F(z_1,z_2) = z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2 = \frac12 z_1^2 + \frac12 z_1^2 \cos(2 z_2)+\frac{3}{4}z_2^2\\
\ge \frac12 z_1^2 + \frac{1}{2}z_2^2 = \frac12 \|z\|^2 \ge \frac12 \alpha^2
$$
Case 2. $z_2 \ge \pi/4$:
$$
F(z_1,z_2) = z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2  \ge \frac{3}{4}z_2^2 \ge \frac{3 \pi^2}{64}
$$

Clearly, if $\alpha < \pi/4$, then case 2 never arises.

If no notion of $z_2$ is available, the  required bound is $F(z_1,z_2) \ge \gamma = \min \{ \frac12 \alpha^2, \frac{3 \pi^2}{64} \}$.


问:
Computing $\int \cos^3(x)\,dx$
So I'm trying to figure out how the reduction formula can help me solve integrands like $\int \cos^3(x)\,dx$.
I've figured that if I'd use integration by parts, I would get to the following:
$$\int u\,dv = uv - \int v\,du$$
$$u = \cos^m(x)$$
$$du = -m\sin(x)\cos^{m-1}\,dx$$
$$dv = dx$$
$$v = x$$

$$\int \cos^m(x)\,dx = x\cos^m  + \int x \cdot m\cdot \sin(x)\cos^{m-1}\, dx$$

Which 1. doesn't seem to really help me and 2. doesn't seem to be true.

Would really appreciate some help, thanks in advance!

Title: Computing $\int \cos^3(x)\,dx$

Body:
$$\int \cos^3(x)\,dx= \int (1- \sin^2(x))\cos(x)\,dx$$

and let $u= \sin(x)$.


答:
Given that $z = \cos{θ} + i \sin{θ}$ where $z \neq -1$, show that $\frac{2}{1+z} = 1 - i \tan(\frac{θ}{2})$

First, I would note that $\tan\left(\frac{θ}{2}\right)=\frac{\sin\left(\frac{θ}{2}\right)}{\cos\left(\frac{θ}{2}\right)}$ and $z =e^{i\theta}= \cos{θ} + i \sin{θ}$ , so $$\frac{2\cos\left(\frac{θ}{2}\right)}{1+e^{i\theta}}=\cos\left(\frac{θ}{2}\right)-i\sin\left(\frac{θ}{2}\right)$$

Next, there are a few properties and a formula we should know.

$\sin(-x)=-\sin(x)$

$\cos(-x)=\cos(x)$

De Moivre's Formula: $[\cos(\theta)+\sin(\theta)]^n=\cos(n\theta)+\sin(n\theta)$

In our scenario, we note that $\cos\left(\frac{θ}{2}\right)=\cos\left(-\frac{θ}{2}\right)$ and analogously $-\sin\left(\frac{θ}{2}\right)=\sin\left(-\frac{θ}{2}\right)$. Hence, by applying De Moivre's law, $\cos\left(-\frac{θ}{2}\right)+i\sin\left(-\frac{θ}{2}\right)\rightarrow \left[\cos\theta+i\sin\theta\right]^{-\frac{1}{2}}$

Using the following properties, we can rewrite the equation as such: $$\frac{2\cos\left(\frac{θ}{2}\right)}{1+e^{i\theta}}=\left[\cos(\theta)+i\sin(\theta)\right]^{-\frac{1}{2}}\\
\frac{2\cos(\frac{θ}{2})}{1+e^{i\theta}}=e^{i(-\frac{\theta}{2})}\\2\cos\left(\frac{\theta}{2}\right)=e^{i(-\frac{\theta}{2})}+e^{i\frac{\theta}{2}}$$

Expanding: $$2\cos\left(\frac{\theta}{2}\right)=\left(\cos\frac{-\theta}{2}+i\sin\frac{-\theta}{2}\right)+\left(cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})\right)\\
2\cos\left(\frac{\theta}{2}\right)=\cos\frac{\theta}{2}-i\sin\frac{\theta}{2}+\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right)\\
\therefore 2\cos\left(\frac{\theta}{2}\right)=2\cos\left(\frac{\theta}{2}\right)$$

Body:
Given that $z = \cos{θ} + i \sin{θ}$ where z ≠ -1, show that $\frac{2}{1+z} = 1 - i \tan(\frac{θ}{2})$

$$ \frac{2}{1+z} = \frac{2}{1 +\cos\,(θ) + i\,sin\,(θ)} $$

$$\text{since} \quad z \,=\, \cos,(θ) + i\,\sin\,(θ) $$

$$ \frac{2}{1 + \cos\,(θ) + i\,\sin\,(θ)} \,*\, \frac{1 + \cos\,(θ) − i\,\sin\,(θ)}{1 + \cos\,(θ) − i\,\sin\,(θ)} $$

Which can be re-written as:
$$ \frac{2\,[\,1 + cos\,(θ) − i\,sin\,(θ)\,]}{[\,1 + cos\,(θ)\,]^2 - i^2\,sin^2 (θ)} $$

$$ \frac{2\,[\,1 + \cos\,(θ) − i\,\sin\,(θ)\,]}{1 + 2\,\cos\,(θ) + \cos^2 (θ) + \sin^2 (θ)} $$

Using... $\quad cos^2 (θ) + sin^2 (θ) = 1 $

$$ \frac{2\,[\,1 + \cos\,(θ) − i\,\sin\,(θ)\,]}{2 + 2\,\cos\,(θ)} \,=\, \frac{2\,[\,1 + \cos\,(θ) − i\,sin\,(θ)\,]}{2\,[\,1 + \cos\,(θ)]} \,=\, \frac{1 + \cos\,(θ) − i\,\sin\,(θ)}{1 + \cos\,(θ)} $$

Using... $\quad 1 + \cos(θ) \,≡\, 2\,\cos^2 (\frac{θ}{2})$

And... $\quad sin (θ) \,≡\, 2\,\sin (\frac{θ}{2})\,\cos (\frac{θ}{2}) $

$$ \frac{2\,cos\frac{θ}{2}\,[\,\cos (\frac{θ}{2}) − i\,\sin (\frac{θ}{2})\,]\,}{2\,\cos^2 (\frac{θ}{2})} \,=\, \frac{\cos (\frac{θ}{2}) − i\sin (\frac{θ}{2})}{\cos (\frac{θ}{2})} \,=\, 1 - i\,\tan (\frac{θ}{2})$$

A related question on Mathematics Stack Exchange can be found here.


答:
This is a standard exercise.
$$\int \cos^3{x}\,dx =\int \cos^2{x} \cos{x}\,dx =\int (1-\sin^{2}{x}) \cos{x}\,dx=$$
$$=\int \cos{x}\,dx-\int \sin^{2}{x} \cos{x}\,dx=\sin{x}-\frac{\sin^{3}{x}}{3}+C.
$$

This method is useful in finding $\int \cos^m{x}\,dx$ and $\int \sin^m{x}\,dx$ whenever $m$ is an odd integer.

Title: Computing $\int \cos^3(x)\,dx$

Body:
So I'm trying to figure out how the reduction formula can help me solve integrands like $\int \cos^3(x)\,dx$.
I've figured that if I'd use integration by parts, I would get to the following:
$$\int u\,dv = uv - \int v\,du$$
$$u = \cos^m(x)$$
$$du = -m\sin(x)\cos^{m-1}\,dx$$
$$dv = dx$$
$$v = x$$

$$\int \cos^m(x)\,dx = x\cos^m  + \int x \cdot m\cdot \sin(x)\cos^{m-1}\, dx$$

Which 1. doesn't seem to really help me and 2. doesn't seem to be true.

Would really appreciate some help, thanks in advance!


问:
Slope of tangent in $(x,y)$ on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with initial conditions
I came across an exercise with a sample solution that I unfortunately don't fully understand given that it's shortened.

Let $(x,y)$ be a point on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with $y\neq0$. Show that the slope of the tangent on $K$ in $(x,y)$ is $-\frac{x}{y}$ and find two solutions for the differential equation $y'=-\frac{x}{y}$ that adhere to the initial conditions $y(1)=1$ or $y(-1)=2$.

The sample solution is as follows:
$$x^2+y^2=r^2$$
$$2x+2y\frac{dy}{dx}=0$$
$$\Rightarrow\frac{dy}{dx}=-\frac{x}{y}$$
$$\int y\;dy=\int -x\;dx$$
$$y^2=c-x^2$$
$$c=2\;or\;c=5$$
$$\Rightarrow x^2+y^2=2\;or\;x^2+y^2=5$$

First of all I'm interested in line 2.

From what I could find and understand, we come to this point through implicit differentiation of line 1. I assume the $r^2$ disappeared because it is treated as a constant but why exactly is that the case?

Furthermore, why is the constant of integration omitted on one side of the equation and how exactly do we find the solutions $2$ and $5$ for it?

Thank you very much in advance!

答:
I purpose another solution using the parametric representation of the circle. Consider the mapping $c :[0,2\pi) \to \mathbb{R}^2$ given by:

$$c(t)=(r\cos t, r\sin t)$$

It is obvious that $c([0,2\pi))$ is your circle, notice that the interval is open in $2\pi$ so that we don't pass through the point $(1,0)$ twice. Now, $c$ is the parametric representation of your circle, and so the derivative at a point is the tangent vector at the point. We have:

$$c'(t)=(-r\sin t,r \cos t)$$

Now fix $t_0$ so you fix a point $(x_0,y_0) = c(t_0)$ on the circle. Now the tangent line at $t_0$ is the line with direction vector $c'(t_0)$ and with starting point $c(t_0)$. The slope of that line is just the component of the tangent vector on the $y$ direction divided by the component of that vector in the $x$ direction, so we have:

$$m = \frac{\phantom{-}r\cos t_0}{- r\sin t_0 } = -\cot t_0$$

However, as you probably can see from the initial construction of $c(t)$, the value of $t$ is related to the point $(x,y)$ on the circle by $t = \arctan(y/x)$ so that:

$$m = - \frac{x_0}{y_0}$$

And this is the slope of the tangent to the circle at the point $(x_0,y_0)$.


问:
Is there a $3$-category in which natural transformations between functors of small categories are $3$-morphisms?
I'm not sure if this question is well-posed, but what I'm getting at is that in, e.g. the category of groups, the $1$-morphisms are group homomorphisms. Then if we look at $\mathbf{Grp}$ from within $\mathbf{Cat}$, the $1$-morphisms are functors and the $2$-morphisms are natural transformations. But it feels like in something we ought to be three levels up. Does there exist a $3$-category in which this makes sense?

答:
The $2$-category of (small) categories $\mathbf{Cat}$ is a monoidal $2$-category under the cartesian product ($C \times D$), and so it could also be treated as a single-object (weak) $3$-category whose $1$-morphisms, $2$-morphisms, and $3$-morphisms are categories, functors, and natural transformations respectively.

Alternatively, we could instead use $C+D$ (the disjoint union), $C*D$ (the join, where there is exactly one arrow from each object of $C$ to each object of $D$), or $C \otimes D$ (the tensor product whose corresponding internal hom has arbitrary families of arrows $F(X) \to G(X)$ (without requiring naturality) as arrows $F \to G$) for the monoidal structure.


问:
If $A$ and $B$ are $n$ by $n$ matrices, show that $\operatorname{Null}A =\operatorname{Null}B$ if and only if $A= UB$ for some invertible $U$
I know that this has been proven before on this forum, but I need help formalizing a different approach: $\DeclareMathOperator{\Null}{Null}$

$[\leftarrow]$ If $A = UB$ then $\Null(A)=\Null(B)$. Let $x \in \Null(A)$ then
The previous statement is justified since $U$ is invertible hence $x\in \Null(B)$ and $\Null(A) \subseteq \Null(B)$. Make a similar argument for $y \in \Null(B)$, and we can conclude that $\Null(A) = \Null(B)$

$[\rightarrow]$ This direction I am not so sure If $\Null(A) = \Null(B)$ and $\operatorname{rank}A=\operatorname{rank}B$ (from the fact they are $n×n$ matrices), then the RREF of the two matrices must be the same. There is a $E_1A=R = E_2B \rightarrow A=E_1^{-1}E_2B\;$ QED

Can someone help confirm or edit the issues in this approach?

答: If I were your teacher I would prefer to see more detail.  Instead of just saying "since U is invertible hence x∈Null(B) and Null(A)⊆Null(B)" say "suppose $x\in Null(B)$ then Bx= 0 so Ax= UBx= U0= 0 and $x\in Null(A)$.  If $x\in Null(A)$ then Ax= 0 so $Bx= U^{-1}Ax= U^{-1}0= 0$ and $x\in Null(B)$."


答:
If you really need to use the hint, go to complex numbers.
$$\left( 1+ \frac{1}{n^2}\right)^n = \left( 1+ \frac{i}{n}\right)^n \left( 1- \frac{i}{n}\right)^n$$
Using the limit
$$\lim_{n \to \infty} \left( 1+ \frac{\alpha}{n}\right)^n = e^{\alpha}$$
valid for all $\alpha \in \Bbb C$, as $n \to \infty$ you get
$$e^i \cdot e^{-i} = 1$$

Title: Show that $(1+\frac{1}{n^2})^n \rightarrow 1$ as $n \rightarrow \infty$

Body:
Show that $\left(1+\frac{1}{n^2}\right)^n \rightarrow 1$ as $n \rightarrow \infty$.
hint: $a^2-b^2=(a-b)(a+b)$

I basically have to show that $$\left|\left(1+\frac{1}{n^2}\right)^n-1\right| < \epsilon$$ but how do I choose $\epsilon$? Can anyone help me with that proof?


问:
Bounds of determinant of symmetric positive definite integer matrix
Given an invertible integer matrix $X \in \mathbb{Z}^n_+$, we set $A = X^T X + 1$ with $1$ denoting the identity matrix. Then, with the help of the Hadamard inequality, it is easy to find an upper bound on the determinant. But what is the lower bound of the determinant?

Since the matrix is positive definite (because $X$ is invertible), the matrix $A$ is invertible, hence, $\det A \neq 0$. It is integer and therefore we obtain $\det A \geq 1$. But so far I did not find a matrix $X$ such that $\det A = 1$. Does there exist a better bound on the determinant or does somebody know an example in order to verify that the bound is strict?

答: The matrix $X^tX$ is positive semidefinite so all it's eigenvalues $\lambda_i$, $1\leq i\leq n$  are nonnegative. Moreover since $X$ is invertible with integer entries we have $\det(X^tX)=1$, thus $\lambda_1\lambda_2\ldots \lambda_n=1$. We have that $\det(A)=(1+\lambda_1)(1+\lambda_2)\ldots (1+\lambda_n)$. For any nonnegative $x$ the inequality $1+x\geq 2\sqrt{x}$ holds so $\det(A)\geq 2^n$. There is no equality since this would imply $X=I_n$ and you assumed that $X$ has positive entries.


答:
According to the properties of limits, limit of a function raised to an exponent is a limit of a function raised to an exponent, and also, you can try to think how to get an $e$ in the limit

$$e=\lim_{x\rightarrow \infty} (1 + \frac{1}{x})^x$$

Here are the steps if you want to see...

 $$\lim_{n \rightarrow \infty} (1+\frac{1}{n^2})^n= (\lim_{n \rightarrow \infty} (1 + \frac{1}{n^2})^{n^2})^{\lim_{n \rightarrow \infty}\frac{1}{n}} = \lim_{n \rightarrow \infty} (e^{\frac{1}{n}}) = 1$$


Title: Computing $\int \cos^3(x)\,dx$

Body:
$$\cos(3x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)$$
$$=(2\cos^2(x)-1)\cos(x)-2\cos(x)\sin^2(x)$$

$$=4\cos^3(x)-3\cos(x)$$

So

$$4\cos^3(x)=\cos(3x)+3\cos(x)$$
and

$$4\int \cos^3(x)dx=\frac{\sin(3x)}{3}+3\sin(x)+C$$


答:
It is simpler to use the standard method: linearising  $\cos^3x=\frac14\bigl(\cos3x+3\cos x)$, which yields instantly
$$\int\cos^3x\,\mathrm dx=\frac14\biggl(\int\cos 3x\,\mathrm dx+3\int\cos x\,\mathrm dx\biggr).$$

Title: Computing $\int \cos^3(x)\,dx$


答:
$$S=2 {n \choose {k-1}}+3{n \choose {k-2}}+4{n \choose {k-3}}+....+k{n \choose 1}+(k+1){n \choose 0}$$

Now writing Sum $S$ in reverse order,

$$S=(k+1){n \choose 0}+k{n \choose 1}+....+3{n \choose {k-2}}+2 {n \choose {k-1}}$$

$$\implies S=k \bigg[{n \choose 0}+{n \choose 1}+....+{n \choose {k-2}}+{n \choose {k-1}}\bigg]+{n \choose 0}-\bigg[1{n \choose 2}+2{n \choose 3}+....+(k-3){n \choose {k-2}}+(k-2){n \choose {k-1}}\bigg]$$

Let $$P={n \choose 0}+{n \choose 1}+....+{n \choose {k-2}}+{n \choose {k-1}}$$

$$Q=1{n \choose 2}+2{n \choose 3}+....+(k-3){n \choose {k-2}}+(k-2){n \choose {k-1}}$$

As $$(1+x)^{n}={n \choose 0}+{n \choose 1}x+{n \choose 2}x^2+....+ {n \choose {k-1}}x^{k-1}+{n \choose k}x^{k}+....+{n \choose n}x^{n-1}$$...$(1)$

On putting $x=1$ we get,

$$P=2^{n}-\sum_{r=0}^{n-k}{n \choose k+r}$$

To get Sum $Q$ We will First Divide Equation $1$ by $x$ then differentiate w.r.t $x$ and then Put $x=1$ we have $Q$ as

$$Q=\Big[n(2)^{n-1}-2^{n}\Big]+1-\sum_{r=0}^{n-k}{n \choose k+r}(k+r-1)$$

So our Sum $S$ Becomes

$$S=k \bigg[2^{n}-\sum_{r=0}^{n-k}{n \choose k+r}\bigg]+1-\Bigg[\Big[n(2)^{n-1}-2^{n}\Big]+1-\sum_{r=0}^{n-k}{n \choose k+r}(k+r-1)\Bigg]$$

On simplifying $S$ we have,

$$S=2^{n}\Big[k+1-\frac{n}{2}\Big]+\sum_{r=0}^{n-k}{n \choose k+r}(r-1)$$


问:
Sobolev embeddings *lowering* the order of integrability
All Sobolev embeddings I am aware of increase the order of integrability. What about embeddings into spaces whose order of integrability is lower? Say, is it possible to find $k$ large enough such that $H^k(\Omega)$ embeds into $L^1(\Omega)$? If $\Omega$ is bounded, this is not a problem, since $H^1(\Omega)\hookrightarrow L^2(\Omega)\hookrightarrow L^1(\Omega)$. But what about general, possibly unbounded $\Omega$? Say, $\Omega\subset \mathbb R^3$ with $C^2$-boundary.

答:
This is not true in general, by considering functions of the form
$$ u_{a}(x) = (1+|x|^2)^{-a/2} $$
for $a>0.$ Note each $u_{a}$ is smooth in $\Bbb R^n$ and a computation shows that $|\partial^{\beta}u_a(x)| \leq C\, u_a(x)$ for any multi-index $\beta,$ where $C=C(n,a,\beta)$ is a constant. Also we have
$$ \int_{\Bbb R^n} |u_{\alpha}|^p \,\mathrm{d}x = \omega_n \int_0^{\infty} \frac{r^{n-1}}{(1+r^2)^{ap/2}} \,\mathrm{d}x, $$
which is finite if and only if $a > \frac np.$

Now let $\Omega \subset \Bbb R^n$ be any domain such that for some $c>0$ we have $|\Omega \cap B_R| \geq c R^n$ for all $R>0;$ then we also have $u_a \in L^p(\Omega)$ if and only if $a>\frac np.$ Then given $q>p$ (say $q=2, p=1$) we can pick $a \in (n/q,n/p)$ and consider $u_a.$ By above we can show $u_a \in W^{k,q}(\Omega)$ for any $k \geq 0,$ but $u_a \not\in L^p(\Omega).$


问:
Of two variables, which affects 'y' more?
I have two equations:

(1) $\displaystyle y = \frac {0.0060}{k}$

(2) $\displaystyle y= \frac{0.00016}{m}$

In the first equation, $k$ is held constant and in the second equation, $m$ is held constant.

How do I determine which variable has a greater affect on $y$?

E.g., how do I determine if doubling $k$ as a greater effect on $y$ than doubling $m$ and so forth?

Thanks!

答: Dear user, both relationships are called "inverse proportionality", so they have exactly the same effect: doubling of $k$ in the first case as well as doubling of $m$ in the second case reduces $y$ to one-half of its value.</p>


问:
Pmf for winning $£10$ for each flip up to and including the first head
A sponsor provides two prizes for a raffle. The first prize winner gets to choose a probability $p$ from $\left[\dfrac{1}{e},1-\dfrac{1}{e}\right]$. A sequence of independent coin flips with probability $p$ for a head are then made. The winner receives $£10$ for each flip up to and including the first head. The same coin is tossed in another independent sequence and the second prize winner receives $£5$ for each flip up to and including the first head in this sequence.

(i) Write down the p.m.f.s of the amounts of the two payments $U$ for the first prize and $V$ for the second prize.

I don't know how to write down the pmfs so that we can differentiate between the first and second prize winner.

答:
Given $p$, and for $n\in\mathbb{N}^*$, what is the probability that the first player wins £$10n$? (each possible outcome for his reward being of that form).
In other terms, what is the probability that in the first sequence of $p$-biased coin tosses, the first $n-1$ tosses are tails, and the $n^\text{th}$ is head?</p>


问:
Inverse Function Multi-Variable
I have the following function:

\begin{array}{rccc}
  \xi\colon&\mathbb{R}^2 &\longrightarrow&\mathbb{R}^3\\
   &(u,v) &\mapsto&(u,v^2,v^3).
\end{array}

I need to show this function is a homeomorphism, i.e., $\xi$ has an inverse map which is continuous.

I really do not know where to begin.

Any hints?

答: Take$$U=\xi(\Bbb R^2)=\left\{(x,y,z)\in\Bbb R^3\,\middle|\,y^3=z^2\right\}.$$Then $\xi$ is a homeomorphism from $\Bbb R^2$ onto $U$, whose inverse is$$\begin{array}{rccc}\xi^{-1}\colon&U&\longrightarrow&\Bbb R^2\\&(x,y,z)&\mapsto&\left(x,\sqrt[3]z\right),\end{array}$$which is continuous.

Title: Inverse Function Multi-Variable


问:
An identity related to the second-order Eulerian numbers.
Recently, some of the remarkable properties of second-order
Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$  A340556 have been proved on MSE [  a ,
 b ,  c  ].

But there are also other notable identities to which these numbers lead.
For instance, we also noticed the following identity, which we haven't seen
elsewhere (reference?):

$$ \sum_{j=0}^{k} \binom{n-j}{n-k}
\left\langle\!\! \left\langle n\atop  j\right\rangle\!\! \right\rangle \,=\,
\sum_{j=0}^k (-1)^{j+k} \binom{n+k}{n+j} \left\{ n+j \atop  j\right \} \quad ( n \ge 0) $$

If we use the notation $ \operatorname{W}_{n, k} $ for these numbers,
we can also introduce the corresponding polynomials.

$$ \operatorname{W}_{n}(x) = \sum_{k=0}^n \operatorname{W}_{n, k} x^k \quad ( n \ge 0) $$

So far so routine, but then came the surprise:

$$ 3^n W_n\left(-\frac13\right) \, = \, 2^n \left\langle\!\!\left\langle - \frac{1}{2} \right\rangle\!\!\right\rangle_n  \quad ( n \ge 0) $$

On the right side are the numbers we recently asked about their
combinatorial significance!

Should I trust this strange equation?

答:
In trying to verify the identity

$$\sum_{j=0}^{k} {n-j \choose n-k}
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
=  \sum_{j=0}^k (-1)^{j+k} {n+k \choose n+j}
\left\{ n+j \atop  j\right \}$$

we quote from MSE
the following identity

$$\left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle =
\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j}.$$

We get for the LHS

$$\sum_{j=0}^{k} {n-j \choose n-k}
\sum_{p=0}^j (-1)^{j-p} {2n+1\choose j-p} {n+p\brace p}
\\ = \sum_{p=0}^k {n+p\brace p}
\sum_{j=p}^k (-1)^{j-p} {2n+1\choose j-p} {n-j\choose n-k}.$$

The inner sum is
$$\sum_{j=0}^{k-p} (-1)^j {2n+1\choose j}
{n-j-p\choose n-k}
\\ = \sum_{j=0}^{k-p} (-1)^j {2n+1\choose j}
{n-j-p\choose k-p-j}
\\ =  [z^{k-p}] (1+z)^{n-p}
\sum_{j=0}^{k-p} (-1)^j {2n+1\choose j}
\frac{z^j}{(1+z)^j}.$$

Here the coefficient extractor enforces the upper limit of the sum
and we may extend $j$ to infinity:

$$[z^{k-p}] (1+z)^{n-p}
\sum_{j\ge 0} (-1)^j {2n+1\choose j}
\frac{z^j}{(1+z)^j}
\\ =  [z^{k-p}] (1+z)^{n-p}
\left(1-\frac{z}{1+z}\right)^{2n+1}
= [z^{k-p}]
\frac{1}{(1+z)^{n+p+1}}
\\ = (-1)^{k-p} {k-p+n+p\choose n+p}
= (-1)^{k-p} {n+k\choose n+p}.$$

Introducing the leading term,

$$\sum_{p=0}^k (-1)^{k+p}
{n+k\choose n+p} {n+p\brace p}$$

This is the claim.

 As for the polynomials we find

$$\sum_{k=0}^n x^k
\sum_{j=0}^{k} {n-j \choose n-k}
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
= \sum_{j=0}^n
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
\sum_{k=j}^n {n-j\choose n-k} x^k
\\ = \sum_{j=0}^n
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle x^j
\sum_{k=0}^{n-j} {n-j\choose n-j-k} x^k
= \sum_{j=0}^n
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle x^j
\sum_{k=0}^{n-j} {n-j\choose k} x^k
\\ = \sum_{j=0}^n
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle x^j
(1+x)^{n-j}
= (1+x)^n \sum_{j=0}^n
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
\frac{x^j}{(1+x)^j}.$$

Multiplying by $3^n$ and evaluating at $x=-1/3$ we obtain

$$3^n \frac{2^n}{3^n}
\sum_{j=0}^n
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
\frac{(-1/3)^j}{(2/3)^j}
= 2^n
\sum_{j=0}^n
\left\langle\!\! \left\langle n\atop j
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^j$$

as claimed.

Remark. Maybe we can pause here for a few days. Thanks!


问:
Reordering of summation symbols
How can I manipulate the expression $$\sum_{i=1}^m (\sum_{j=0}^n a_jx_i^j)^2$$ into $$\sum_{j=0}^n\sum_{k=0}^na_ja_k(\sum_{i=1}^mx_i^{j+k}) $$
This is a part of the application of least squares to polynomial regression. $m $ is the number of data points and $n$ is the degree of the polynomial. These expressions were taken from 'Numerical Analysis' 9th edition by  Richard L. Burden and  J. Douglas Faires.

答:
$$\sum_{i=1}^m \left(\sum_{j=0}^n a_jx_i^j\right)^2$$

$$=\sum_{i=1}^m \sum_{j=0}^n \sum_{k=0}^n a_jx_i^j a_k x_i^k$$

$$=\sum_{j=0}^n\sum_{k=0}^na_ja_k\left(\sum_{i=1}^mx_i^{j+k}\right)  $$

Reordering summation symbols in the last step is okay since these are finite sums

Title: Reordering of summation symbols


问:
What is correct answer for this algebraic expression?
If $(x+2)$ and $(x-2)$ are factors of $ax^2+2x-3x^2+bx-4$, then what would be the value of $(a+b)$ from following given options and how to arrive at it?

a) -7

b)  7

c)  14

d)  -8

Body:
Hint: $ $ by the  Factor Theorem, the roots $\,x = 2,\,-2,\,$ yield factors $\,x\!-\!2,\,x\!+\!2,\,$ so

$$ (a\!-\!3)\, x^2 + (b\!+\!2)\, x - 4\, =\, c\,(x\!-\!2)(x\!+\!2)\, =\, c\,x^2 -4\,c $$

Comparing coefficients $\ {-}4 = -4c,\ \ b\!+\!2 = 0,\ a\!-\!3 = c,\ $ so $\ c,b,a\, =\,\ldots$

Alternatively, evaluate at $\,x = \pm 2,\,$ then solve the two equations for $\, a,b.$


问:
Solve for $p$ in $\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$
I did a question $\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$, and evaluating this is divergent integral yes? Then as a general form $\int_{0}^{1} \frac{1}{x^p}\,dx$, $p \in \mathbb{R}$, what values of $p$ can give me $\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$? This is a easy integral to calculate, make it $\int_{0}^{1}x^{-p}dx$ and calculate, etc. Then how do I get this to solve $p$? I am using the fundamental theorem of calculus and confused here.

答: It's divergent for $p\leq 1$.   Otherwise by the fundamental theorem of calculus,  you get $\frac {x^{-p+1}} {-p+1}$.  Plug in your 1 and 0 and then set equal to 4/3 should let you solve for p.


Title: Solve for $p$ in $\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$

Body: I did a question $\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$, and evaluating this is divergent integral yes? Then as a general form $\int_{0}^{1} \frac{1}{x^p}\,dx$, $p \in \mathbb{R}$, what values of $p$ can give me $\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$? This is a easy integral to calculate, make it $\int_{0}^{1}x^{-p}dx$ and calculate, etc. Then how do I get this to solve $p$? I am using the fundamental theorem of calculus and confused here.

Body:
Actually,$$\int_0^1x^{-1/2}\,\mathrm dx=\left[2x^{1/2}\right]_{x=0}^{x=1}=2.$$

On the other hand,$$\int_0^1x^{-p}\,\mathrm dx=\left[\frac{x^{1-p}}{1-p}\right]_{x=0}^{x=1}=\frac1{1-p}.$$So, take $p=\frac14$.


问:
How to calculate $\iiint\limits_V z^2 dxdydz$, where $V$ is determined by a ball and another constraint
How to calculate $\iiint\limits_V z^2 dxdydz$, where $V$ is determined by $x^2+y^2+z^2 \le r^2$ and $x^2+y^2+z^2 \le 2rz$?

The solution seems odd to me because they only consider $0\le z \le r$.

Solution:

Let $S: x^2 + y^2 \le 2rz -z^2 (0 \le z \le r/2); Q: x^2+y^2 \le r^2 - z^2 (r/2 \le z \le r)$. Then,

$$\begin{split}
\iiint\limits_V z^2 dxdydz &= \int_0^{r/2}dz \iint\limits_S z^2 dxdy + \int_{r/2}^r dz \iint\limits_Q z^2 dxdy
\\&= \pi\int_0^{r/2} z^2(2rz-z^2)dz + \pi \int_{r/2}^rz^2(r^2 - z^2)dz
\end{split}$$

答:
The first ball, $x^2+y^2+z^2\le r^2$, is centered at the origin $(0,0,0)$ with radius $r$.

The second ball, $x^2+y^2+z^2\le 2rz$, or upon completing the square with respect to $z$, $x^2+y^2+(z-r)^2\le r^2$, is centered at $(0,0,r)$ and also has radius $r$.

The spheres that bound the balls intersect at

$$x^2+y^2+(z-r)^2=x^2+y^2+z^2\implies r^2-2rz=r(r-2z)=0\implies z=\frac r2$$

which corresponds to a circle in the plane $z=\frac r2$ with equation

$$x^2+y^2+\left(\frac r2\right)^2=r^2\implies x^2+y^2=\frac{3r^2}4$$

i.e. a circle centered at $\left(0,0,\frac r2\right)$ with radius $\frac{\sqrt3\,r}2$ and parallel to the $(x,y)$ plane.

So, the region $V$ is the set of points

$$V=\left\{(x,y,z) \mathrel{\bigg|} -\frac{\sqrt3}2\le x\le \frac{\sqrt3}2\,\land\,-\sqrt{\frac{3r^2}4-x^2}\le y\le\sqrt{\frac{3r^2}4-x^2}\,\land\,r-\sqrt{r^2-x^2-y^2}\le z\le\sqrt{r^2-x^2-y^2}\right\}$$

That is, $z$ is bounded below by the bottom half of the upper sphere, and bounded above by the top half of the lower sphere.

Integrating in rectangular coordinates seems like too much of a hassle to me, so I suggest converting to cylindrical:

$$\begin{cases}x=u\cos v\\y=u\sin v\\z=w\end{cases}$$

Then the integration region is

$$V'=\left\{(u,v,w) \mathrel{\bigg|} 0\le u\le\frac{\sqrt3\,r}2 \,\land\, 0\le v\le2\pi \,\land\, r-\sqrt{r^2-u^2}\le w\le\sqrt{r^2-u^2}\right\}$$

and the integral is

$$\begin{align}
\iiint_Vz^2\,\mathrm dx\,\mathrm dy\,\mathrm dz&=\iiint_{V'}w^2u\,\mathrm du\,\mathrm dv\,\mathrm dw\\[1ex]
&=\int_0^{2\pi}\int_0^{\frac{\sqrt3\,r}2}\int_{r-\sqrt{r^2-u^2}}^{\sqrt{r^2-u^2}}w^2u\,\mathrm dw\,\mathrm du\,\mathrm dv
\end{align}$$


答:
Start with the antiderivative, assuming that $p\ne 1$ $$\int x^{-p}\,dx = \frac{1}{1-p}x^{1-p} + C.$$  Then it must be the case that $$\frac{1}{1-p}x^{1-p}\bigg|_{0}^{1} = \frac{1}{1-p}\left(1-\lim_{x\to 0}x^{1-p}\right).$$  Now, in order for $\lim_{x\to 0}x^{1-p}$ to exist we need $1-p>0,$ in which case the limit is zero.  So, we need two conditions:
\begin{align}
p&<1\\[5pt]
\frac{1}{1-p} &= \frac{4}{3}.
\end{align}

We can calculate $p$ from the second condition to get $p = 1/4$.  This also satisfies the condition that $p<1$.  To confirm that we have the correct result:
$$\int_{0}^{1}x^{-1/4}\,dx = \frac{4}{3}x^{3/4}\bigg|_{0}^{1} = \frac{4}{3} - 0 = \frac{4}{3}.$$


Body:
$\int_0^1{\frac{1}{x^\frac{1}{2}}}dx$ is not divergent.

$\int_0^1{\frac{1}{x^\frac{1}{2}}}dx=\int_0^1 x^{-\frac{1}{2}}dx=2x^\frac{1}{2}|_0^1=2$

Similarly solving: $\int_0^1 \frac{1}{x^p}dx=\frac{4}{3}$

$\int_0^1 \frac{1}{x^p}dx=\frac{x^{p-1}}{1-p}|_0^1=
\frac{1}{1-p}=\frac{4}{3}$

So, $p=\frac{1}{4}$


答:
First of all $$\int_{0}^{1}\frac{1}{x^\alpha}$$ is divergent if $\alpha>1$, so the first case cited by you (for $\alpha=\frac{1}{2}$) corresponds to convergence.

Now let's observe that:
$\displaystyle\int_{0}^{1}\frac{1}{x^\alpha}=\displaystyle\lim_{x\to 0^+} \int_{x}^{1}\frac{1}{x^\alpha}=\lim_{x\to 0^+} \frac{1}{1-\alpha}(1-x^{1-\alpha})=\frac{1}{1-\alpha}$ when $\alpha<1$.

So: $\displaystyle\frac{1}{1-\alpha}=\frac{4}{3}$ when $\displaystyle\alpha=\frac{1}{4}$


问:
Triangular vs Normal distribution
I'm trying to approximate a standard normal distribution with a triangular distribution. What parameters of the triangular distribution (min, max and mode) are more suitable? Thank you</p>

答:
from your Triangle Distribution you have a (min), (max), and (mode).

in a Normal Distribution you have a (mean) and (standard deviation) abbreviated as (sd) .
in the Normal Distribution the (mean) is equal to the (mode) due to the symmetry of the model.

to move from a Triangle model to a Normal model we need to estimate the (mean) and the (standard deviation) from the Triangle models parameters

for the (mean):

it is reasonable to use your (mode) of the Triangle Distribution as your (mean) for your Normal Distribution because they are equal under the Normal model.  NOTE at this point one should be aware they maybe breaking a symmetry assumption in their normal model.

a second reasonable estimate for the (mean) can come from the (midpoint) of the (min) and the (max) where (midpoint)=((max)-(min))/2.

the (mean) of the Normal distribution represents the central tendency. a (midpoint) and a (mode) are also measures of central tendency thus they are candidates that we have readily available from the Triangle Distribution.

a third method will to be use the (mean) of the Triangle Distribution and use it to estimate the mean a Normal Distribution.  the (mean) of a triangle distribution is (mean_tri)=((min)+(max)+(mode))/3

method 1 (central tendency):

(mean)=(mode)

or

method 2 (central tendency):

(midpoint)=((max)-(min))/2

(mean)=(midpoint)

or

method 3 (match moments):

(mean_tri)=((min)+(max)+(mode))/3

(mean)=(mean_tri)

for the (standard deviation):

a Triangle distribution gives us a (min) and a (max).  We can take the difference to find a range value (range)=(max)-(min).  a function of the range can provide an estimate of the standard deviation of a Normal distribution f(range)=(sd). It is up to the user to determine a reasonable function at this point.  example: four standard deviation covers 95% of a Normal Distribution's data and would result in an estimated standard devastation of (sd)=(range)/4, while a six standard deviation range covers 99% and would result in an estimate of (sd)=(range)/6.

Another method will be to take the (standard deviation) of the Triangle Distribution and use that to estimate the (standard distribution) of the Normal Distribution. the (sd) of a triangle distribution is (sd_tri)=sqrt(((min)^2+(max)^2+(mode)^2-(min)(max)-(min)(mode)-(max)(mode))/18)

method 1 (range coverage):

try setting

(range)=(max)-(min)

and using:

(sd)=(range)/2  (65% normal coverage)  (fattest of the three models)

or

(sd)=(range)/4  (95% normal coverage) (middle width of the three models)

or

(sd)=(range)/6  (99% normal coverage) (skinniest of the three models)

method 2 (match central moments):

(sd_tri)=sqrt(((min)^2+(max)^2+(mode)^2-(min)(max)-(min)(mode)-(max)(mode))/18)

(sd)=(sd_tri)

when estimating your (mean) and (standard deviation) you may want to consider how much confidence you had in your Triangle's coverage of the population and if the original distribution was skewed.  Did the (min) and (max) result from a very small sample or a very large sample?  If i am skewed, do i want to focus in on the mode or do i want to go for more coverage?

Title: Triangular vs Normal distribution


答: What do you get from the functional equation, $\Gamma'(1)/\Gamma(1)=-\gamma$ and $\Gamma(s+1)=s\Gamma(s)$

Title: Questions on evaluation of $\frac{\zeta'(s)}{\zeta(s)}$ at integer values of s

Body:
Question (1): Evaluation at non-negative even integers

Is it true that

$$\frac{\zeta '(2 n)}{\zeta (2 n)}=f(n)\tag{1}$$

where

$$f(n)=-H_{2 n-1}-\frac{\zeta'(1-2 n)}{\zeta(1-2 n)}+\gamma+\log(2 \pi)\tag{2}$$

and $n\in \mathbb{Z}\land n\geq 0$?

Question (2): Evaluation at negative odd integers

Is it also true that

$$\frac{\zeta '(-2 n-1)}{\zeta (-2 n-1)}=g(n)\tag{3}$$

where

$$g(n)=-H_{2 n+1}-\frac{\zeta '(2 n+2)}{\zeta (2 n+2)}+\gamma +\log (2 \pi )\tag{4}$$

and $n\in \mathbb{Z}\land n\geq 0$?

The last column in the following table illustrates the error in formula (1) above associated with the evaluation of $\frac{\zeta'(s)}{\zeta(s)}$ at non-negative even integers. At $n=0$ the function $f(n)$ must be evaluated as $\underset{\epsilon\to 0}{\text{lim}}f(\epsilon)$.

$$\begin{array}{ccccc}
 n & 2 n & \frac{\zeta'(2 n)}{\zeta(2 n)} & f(n) & f(n)-\frac{\zeta'(2 n)}{\zeta(2 n)} \\
 0 & 0 & \log (2 \pi ) & \log (2 \pi ) & 0. \\
 1 & 2 & \frac{6 \zeta '(2)}{\pi ^2} & 12 \left(\frac{1}{12}-\log (A)\right)-1+\gamma+\log (2 \pi ) & 0. \\
 2 & 4 & \frac{90 \zeta '(4)}{\pi ^4} & -120 \zeta '(-3)-\frac{11}{6}+\gamma +\log (2 \pi ) & \text{5.551115123125783$\grave{ }$*${}^{\wedge}$-16} \\
 3 & 6 & \frac{945 \zeta '(6)}{\pi ^6} & 252 \zeta '(-5)-\frac{137}{60}+\gamma +\log (2 \pi ) & \text{2.220446049250313$\grave{ }$*${}^{\wedge}$-16} \\
 4 & 8 & \frac{9450 \zeta '(8)}{\pi ^8} & -240 \zeta '(-7)-\frac{363}{140}+\gamma +\log (2 \pi ) & \text{3.0531133177191805$\grave{ }$*${}^{\wedge}$-16} \\
 5 & 10 & \frac{93555 \zeta '(10)}{\pi ^{10}} & 132 \zeta '(-9)-\frac{7129}{2520}+\gamma +\log (2 \pi ) & \text{3.3306690738754696$\grave{ }$*${}^{\wedge}$-16} \\
 6 & 12 & \frac{638512875 \zeta '(12)}{691 \pi ^{12}} & -\frac{32760}{691} \zeta '(-11)-\frac{83711}{27720}+\gamma +\log (2 \pi ) & \text{6.661338147750939$\grave{ }$*${}^{\wedge}$-16} \\
 7 & 14 & \frac{18243225 \zeta '(14)}{2 \pi ^{14}} & 12 \zeta '(-13)-\frac{1145993}{360360}+\gamma +\log (2 \pi ) & \text{1.2212453270876722$\grave{ }$*${}^{\wedge}$-15} \\
 8 & 16 & \frac{325641566250 \zeta '(16)}{3617 \pi ^{16}} & -\frac{8160 \zeta '(-15)}{3617}-\frac{1195757}{360360}+\gamma +\log (2 \pi ) & \text{1.9984014443252818$\grave{ }$*${}^{\wedge}$-15} \\
 9 & 18 & \frac{38979295480125 \zeta '(18)}{43867 \pi ^{18}} & \frac{14364 \zeta '(-17)}{43867}-\frac{42142223}{12252240}+\gamma +\log (2 \pi ) & \text{3.1086244689504383$\grave{ }$*${}^{\wedge}$-15} \\
 10 & 20 & \frac{1531329465290625 \zeta '(20)}{174611 \pi ^{20}} & -\frac{6600 \zeta '(-19)}{174611}-\frac{275295799}{77597520}+\gamma +\log (2 \pi ) & \text{2.6645352591003757$\grave{ }$*${}^{\wedge}$-15} \\
\end{array}$$

The last column in the following table illustrates the error in formula (3) above associated with the evaluation of $\frac{\zeta'(s)}{\zeta(s)}$ at negative odd integers.

$$\begin{array}{ccccc}
 n & -2n-1 & \frac{\zeta'(-2n-1)}{\zeta(-2n-1)} & g(n) & g(n)-\frac{\zeta'(-2n-1)}{\zeta(-2n-1)} \\
 0 & -1 & -12 \left(\frac{1}{12}-\log (A)\right) & -\frac{6 \zeta '(2)}{\pi ^2}-1+\gamma+\log (2 \pi ) & 0. \\
 1 & -3 & 120 \zeta '(-3) & -\frac{90 \zeta '(4)}{\pi ^4}-\frac{11}{6}+\gamma +\log (2 \pi ) & \text{5.551115123125783$\grave{ }$*${}^{\wedge}$-16} \\
 2 & -5 & -252 \zeta '(-5) & -\frac{945 \zeta '(6)}{\pi ^6}-\frac{137}{60}+\gamma +\log (2 \pi ) & \text{2.220446049250313$\grave{ }$*${}^{\wedge}$-16} \\
 3 & -7 & 240 \zeta '(-7) & -\frac{9450 \zeta '(8)}{\pi ^8}-\frac{363}{140}+\gamma +\log (2 \pi ) & \text{3.0531133177191805$\grave{ }$*${}^{\wedge}$-16} \\
 4 & -9 & -132 \zeta '(-9) & -\frac{93555 \zeta '(10)}{\pi ^{10}}-\frac{7129}{2520}+\gamma +\log (2 \pi ) & \text{3.3306690738754696$\grave{ }$*${}^{\wedge}$-16} \\
 5 & -11 & \frac{32760}{691} \zeta '(-11) & -\frac{638512875 \zeta '(12)}{691 \pi ^{12}}-\frac{83711}{27720}+\gamma +\log (2 \pi ) & \text{6.661338147750939$\grave{ }$*${}^{\wedge}$-16} \\
 6 & -13 & -12 \zeta '(-13) & -\frac{18243225 \zeta '(14)}{2 \pi ^{14}}-\frac{1145993}{360360}+\gamma +\log (2 \pi ) & \text{1.2212453270876722$\grave{ }$*${}^{\wedge}$-15} \\
 7 & -15 & \frac{8160 \zeta '(-15)}{3617} & -\frac{325641566250 \zeta '(16)}{3617 \pi ^{16}}-\frac{1195757}{360360}+\gamma +\log (2 \pi ) & \text{1.9984014443252818$\grave{ }$*${}^{\wedge}$-15} \\
 8 & -17 & -\frac{14364 \zeta '(-17)}{43867} & -\frac{38979295480125 \zeta '(18)}{43867 \pi ^{18}}-\frac{42142223}{12252240}+\gamma +\log (2 \pi ) & \text{3.1086244689504383$\grave{ }$*${}^{\wedge}$-15} \\
 9 & -19 & \frac{6600 \zeta '(-19)}{174611} & -\frac{1531329465290625 \zeta '(20)}{174611 \pi ^{20}}-\frac{275295799}{77597520}+\gamma +\log (2 \pi ) & \text{2.6645352591003757$\grave{ }$*${}^{\wedge}$-15} \\
 10 & -21 & -\frac{276 \zeta '(-21)}{77683} & -\frac{13447856940643125 \zeta '(22)}{155366 \pi ^{22}}-\frac{18858053}{5173168}+\gamma +\log (2 \pi ) & \text{8.881784197001252$\grave{ }$*${}^{\wedge}$-15} \\
\end{array}$$


问:
Let {$b_k$} be a sequence of nonnegative real numbers that converges to b and suppose that b>0. Prove that {$\sqrt{b_k}$} converges to $\sqrt{b}$.
So far for my proof I have:

Let {$b_k$} be a sequence of nonnegative real numbers that converges to b and suppose that b>0.
Then, given $\epsilon >0$ there exists a N $\in \mathbb{N}$ such that $k \geq N$ $\Rightarrow$ $|b_k-b|< \epsilon$. Now, consider {$\sqrt{b_k}$}. We need to show that {$\sqrt{b_k}$} $\rightarrow \sqrt{b}$.
Then, $|(\sqrt{b_k})^2-(\sqrt{b})^2| < \epsilon$ for all $k \geq N$.

After this point I am stuck. I not sure how to manipulate the expression so that I can arrive at getting some form of $\epsilon < k$.

答:
$$|\sqrt{b_k}-\sqrt{b}|=|\frac{b_k-b}{\sqrt{b_k}+\sqrt{b}}|\leq\frac{|b_k-b|}{\sqrt{b}}$$

Explanation

For the equality, I used the formula $(x-y)(x+y)=x^2-y^2$,
and that for positive numbers $a>0$ we have $(\sqrt{a})^2=a$.
For the inequality, I used the fact that $\sqrt{b_k}+\sqrt{b}\geq \sqrt{b}$ (can you see why this inequality is true?) and also the fact that if $a,b,c$ are positive numbers such that $b\geq c$ then $\frac{a}{b}\leq\frac{a}{c}$.


问:
If two random variables are defined on the same probability space do they have the same distribution?
Let's say we have two random variables $X$ and $Y$ both defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. So $X$ and $Y$ are measurable functions from $\Omega$ to $\mathbb{R}$.

Do $X$ and $Y$ necessarily have the same distribution? Can we for example have $X\sim Bern(p)$ and $Y\sim N(0,1)$? Or would we then say that $X$ and $Y$ have for example the measure $\mathbb{P}_X$ and $\mathbb{P}_Y$ respectively?

答:
You can easily construct such $X$ and $Y$. Take $(\Omega,\mathcal{F},\mathsf{P})=((0,1),\mathcal{B}_{(0,1)},\lambda)$, where $\lambda$ is the Lebesgue measure, and set
$$
X(\omega):=1\{\omega\le p\},
$$
and
$$
Y(\omega):=\Phi^{-1}(\omega),
$$
where $\Phi$ is the standard normal cdf. Then
$$
\mathsf{P}(X=1)=\lambda(\{\omega\le p\})=p,\quad \mathsf{P}(X=0)=1-p,
$$
and
$$
\mathsf{P}(Y\le y)=\lambda(\{\Phi^{-1}(\omega)\le y\})=\lambda(\{\omega\le \Phi(y)\})=\Phi(y).
$$
That is, $X\sim\text{Bern}(p)$ and $Y\sim N(0,1)$.

An interesting related question is whether one can construct independent random variables on the same probability space (s.t. $P_{X,Y}=P_X\otimes P_Y$).


答:
Since the distribution is discrete (only multiples of 10 or 5 are possible), we have:
$$f_U(n10)=P(\mbox{n-1 times tail followed by one time head})=p^{(n-1)}(1-p)$$
For the same reason is:
$$f_V(n5)=p^n*(1-p)$$
This question is highly related to the geometric distribution. In fact it is almost the same.

Since $E(U)=10\frac{1}{p}$ (use the mean of the geometric distribution to show this), we have that E(U) is maximal when p is minimal (1/e). Since U and V are independent (we toss again for the second winner), we have $E(U/V)=E(U)/E(V)=10\frac{1}{p}\frac{1}{5/p}=2$ and thus constant. This is intuitive because you would guess that the first person gets 2 times as much as the second person.

Title: Pmf for winning $£10$ for each flip up to and including the first head

Body:
A sponsor provides two prizes for a raffle. The first prize winner gets to choose a probability $p$ from $\left[\dfrac{1}{e},1-\dfrac{1}{e}\right]$. A sequence of independent coin flips with probability $p$ for a head are then made. The winner receives $£10$ for each flip up to and including the first head. The same coin is tossed in another independent sequence and the second prize winner receives $£5$ for each flip up to and including the first head in this sequence.

(i) Write down the p.m.f.s of the amounts of the two payments $U$ for the first prize and $V$ for the second prize.

I don't know how to write down the pmfs so that we can differentiate between the first and second prize winner.


问:
Is there a natural intermediate version of PA?
Below, "logic" means "regular logic containing first-order logic" in the sense of Ebbinghaus/Flum/Thomas.

Setup

For a logic $\mathcal{L}$, let $\mathcal{PA}(\mathcal{L})$ be the set of $\mathcal{L}$-sentences in the language of arithmetic consisting of:

the ordered semiring axioms, and

for each $\mathcal{L}$-formula $\varphi(x,y_1,...,y_n)$ the induction instance $$\forall y_1,...,y_n[[\varphi(0,y_1,...,y_n)\wedge\forall x(\varphi(x,y_1,...,y_n)\rightarrow\varphi(x+1,y_1,...,y_n))]\rightarrow\forall x\varphi(x,y_1,...,y_n)].$$

(Note that even if the new logic $\mathcal{L}$ has additional types of variable - e.g. "set" variables - the corresponding induction instances will only allow "number" variables $y_1,...,y_n$ for the parameters.)

For example, $\mathcal{PA}(FOL)$ is just the usual (first-order) PA, and $\mathcal{PA}(SOL)$ characterizes the standard model $\mathbb{N}$ up to isomorphism. Also, we always have $\mathbb{N}\models_\mathcal{L}\mathcal{PA}(\mathcal{L})$.

Note that it's not quite true that $\mathcal{PA}(SOL)$ "is" second-order PA as usually phrased - induction-wise, we still have a scheme as opposed to a single sentence. However, induction applied to "has finitely many predecessors" gets the job done. (If we run the analogous construction with ZFC in place of PA, though, things seem more interesting ....)

Question

Say that a logic $\mathcal{L}$ is PA-intermediate if we have $PA<\mathcal{PA}(\mathcal{L})<Th_{FOL}(\mathbb{N})$ in the following sense:

There is a first-order sentence $\varphi$ such that $PA\not\models\varphi$ but $\mathcal{PA}(\mathcal{L})\models\varphi$.

There is a first-order sentence $\theta$ such that $\mathbb{N}\models\theta$ but $\mathcal{PA}(\mathcal{L})\not\models\theta$.

Is there a "natural" PA-intermediate logic?

(This is obviously fuzzy; for precision, I'll interpret "natural" as "has appeared in at least two different papers whose respective authorsets are incomparable.")

Partial progress

A couple non-examples:

Infinitary logic $\mathcal{L}_{\omega_1,\omega}$ is not PA-intermediate, since in fact $\mathcal{PA}(\mathcal{L}_{\omega_1,\omega})$ pins down $\mathbb{N}$ up to isomorphism (consider $\forall x(\bigvee_{i\in\omega}x=1+...+1\mbox{ ($i$ times)}$). So both second-order and infinitary logic bring in too much extra power.

Adding an equicardinality quantifier ($Ix(\varphi(x);\psi(x))$ = "As many $x$ satisfy $\varphi$ as $\psi$") also results in pinning down $\mathbb{N}$ up to isomorphism (consider "$\neg$ As many $x$ are $<k$ as are $<k+1$"). If we add the weak equicardinality quantifer $Qx\varphi(x) = Ix(\varphi(x);\neg\varphi(x))$, on the other hand, we wind up with a conservative extension of $PA$. That said, the latter is stronger semantically: no countable nonstandard model of $PA$ satisfies $\mathcal{PA}(FOL[Q])$ (to prove conservativity over $PA$ we look at the $\omega_1$-like models).

Fragments of second-order logic don't seem to help either: both $\mathcal{PA}(\Pi^1_1)$ and $\mathcal{PA}(\Sigma^1_1)$ pin down $\mathbb{N}$ (see here).

The proof of Lindstrom's theorem yields a weak negative result: if $\mathcal{L}$ is a logic strictly stronger than first-order logic with the downward Lowenheim-Skolem property, then we can whip up an $\mathcal{L}$-sentence $\varphi$ and an appropriate tuple of formulas $\Theta$ such that $\varphi$ is satisfiable and $\Theta^M\cong\mathbb{N}$ in every $M\models\varphi$. So an example for the above question would have to either allow nasty implicit definability shenanigans or lack the downward Lowenheim-Skolem property. This rules out a whole additional slew of candidates. That said, there is a surprising amount of variety among logics without the downward Lowenheim-Skolem property even in relatively concrete contexts - e.g. there is a compact logic strictly stronger than FOL on countable structures.

答:
This is a continuation of my answer to your easier version of this question on MathOverflow. As I wrote out this argument, I realized that it's not quite as easy as I thought when I originally wrote my comment on the other post, and I can now only claim a weaker result, but it is still sufficient to answer this question.

In the MO answer, we already established that $\mathrm{Con}(\mathsf{PA})$ is a first-order sentence that $\mathsf{PA}(Q_{\mathrm{Brch}})$ entails but which is not entailed by $\mathsf{PA}$.

I will show that there is a c.e. extension $\mathsf{Z}_2^{US}$ of $\mathsf{Z}_2$ (in a bigger language) with the property that if $(M,S)$ is a model of $\mathsf{Z}_2^{US}$, then there is a (first-order) elementary extension $(M',S') \succeq (M,S)$ such that $M' \models \mathsf{PA}(Q_{\mathrm{Brch}})$. By Gödel, this establishes that $\mathsf{PA}(Q_{\mathrm{Brch}}) \not \models \mathrm{Con}(\mathsf{Z}_2^{US})$. (I will denote models of second-order arithmetic like $(M,S)$, with $M$ the numbers sort and $S$ the set sort.)

This argument is partially based on the proof of compactness for $\mathcal{L}(Q_{\mathrm{Brch}})$ in Section 3.1 of this paper of Mekler and Shelah. Although this argument is actually easier, because we can lean on the coding structure of arithmetic.

Let $\mathcal{L}^U$ be the language of second order arithmetic augmented with a single new unary predicate on the set sort. Let $\mathcal{L}^{US}$ be this language enriched with Skolem functions in both sorts. Let $\mathsf{Z}_2^{US}$ be the $\mathcal{L}^{US}$ theory which consists of

$\mathsf{Z}_2$ with comprehension extended to all $\mathcal{L}^{US}$-formulas,

the assertion that $U$ is an (internal) non-principal ultrafilter on the number sort (i.e., any set of the form $[a,\infty)$ satisfies $U$; if $X$ satisfies $U$, then any superset of $X$ satisfies $U$; and if $X$ and $Y$ satisfy $U$, then $X\cap Y$ satisfies $U$), and

the assertion that the Skolem functions in $\mathcal{L}^{US}$ are Skolem functions.

By taking the standard model of $\mathsf{Z}_2$ and appending a non-principal ultrafilter and Skolem functions, we can see that $\mathsf{Z}_2^{US}$ is consistent. It is clearly a c.e. theory, and by Gödel we have that $\mathsf{Z}_2^{US}\not \vdash \mathrm{Con}(\mathsf{Z}_2^{US})$.

Note that for any $(M,S) \models \mathsf{Z}_2^{US}$, the predicate $U$ defines a complete type over $MS$ in the number sort. Call this type $p|_{MS}$. Note furthermore that if $a$ realizes this type, then every element of the Skolem hull of $MSa$ realizes an $MS$-definable type over $MS$.

Fix $(M,S)\models \mathsf{Z}_2^{US}$. Fix an uncountable regular cardinal $\kappa > |MS|$. Let $(M_0,S_0) = (M,S)$. For each positive ordinal $\alpha < \kappa$, given $(M_\beta,S_\beta)$ for $\beta < \alpha$,

if $\alpha = \gamma +1$, then let $(M_\alpha,S_\alpha)$ be the Skolem hull of $M_\gamma S_\gamma a$, where $a$ is a realization of $p|_{M_\gamma S_\gamma}$ and

if $\alpha$ is a limit ordinal, let $(M_\alpha,S_\alpha) = \bigcup_{\beta < \alpha} (M_\beta,S_\beta)$.

Finally, let $(M',S') = \bigcup_{\alpha < \kappa} (M_\alpha,S_\alpha)$. By a not-too-difficult induction argument, one can show that for any $\alpha$, and any $c \in (M',S')$, the type of $c$ over $M_\alpha S_\alpha$ is $M_\alpha S_\alpha$-definable. Note also that for each $\alpha < \kappa$, $|M_\alpha S_\alpha| < \kappa$.

Fact. For any formula $\varphi(x,y,\bar{z})$, there is a formula $\psi(x,y,\bar{z})$ such that for any $\bar{a}$, if $\varphi(x,y,\bar{a})$ defines a directed partial order on the number sort with no largest element, then $\psi(x,y,\bar{a})$ defines a function from the number sort to itself which is strictly order preserving (from the ordinary order to $\varphi(x,y,\bar{a})$) and which is cofinal in the partial order.

Proof. Easy induction argument. $\square$

This implies that in each $(M_{\alpha+1},S_{\alpha+1})$, for any directed $M_{\alpha}S_\alpha$-definable partial order $<$ on $M_\alpha$, there is some $a \in M_{\alpha+1}$ such is $<$-greater than every element of $M_{\alpha}$.

Lemma. If $(<_1,<_2,r)$ are a pair of $M'S'$-definable partial orders and an $M'S'$-definable function on the number sort that form a level tree and if $B \subseteq M'$ is a branch of this tree, then $B \in S'$ (in the sense that there is some element of $S'$ with the same members as $B$.

Proof. First note that since this is a level tree, definable initial segments of $B$ are cofinal in $B$ (consider sets of the form $x <_1 a$). For any $M_\alpha$, we can find $b \in B$ such that the rank $r(b)$ is greater than any of the ranks of elements of $M_\alpha \cap B$. This implies that $M_\alpha \cap B = M_\alpha \cap (-\infty, b]$ (since branches are linearly ordered). By one of the above comments, the type of $b$ over $M_\alpha S_\alpha$ is $M_\alpha S_\alpha$-definable. This implies that the set $M_\alpha \cap (-\infty,b]$ (and therefore $M_\alpha \cap B$) is actually definable in $(M_\alpha,S_\alpha)$.  Since formulas only use finitely many parameters, this implies that for each limit ordinal $\alpha<\kappa$, there is a strictly smaller ordinal $\beta_\alpha$ such that $M_\alpha \cap B$ is definable in $M_\alpha$ with parameters from $M_{\beta_\alpha} S_{\beta_\alpha}$. By Fodor's lemma, there is a single $\delta <\kappa$ such that $M_\alpha \cap B$ is definable with parameters from $M_\delta S_\delta$ for a stationary (and therefore unbounded) set of $\alpha$'s. By the pigeonhole principle, there must be a single formula $\varphi(x,\bar{b})$ with $\bar{b} \in M_\delta S_\delta$ such that for an unbounded set of $\alpha$'s, $\varphi(M_\alpha, \bar{b}) = M_\alpha\cap B$. Therefore, by elementarity, $\varphi(x,\bar{b})$ actually defines $B$ in $(M',S')$. $\square$

For any formula $\varphi(x,\bar{y})$, we have that for any $\bar{b}$, there is a set $X \in S'$ such that $X = \varphi(M',\bar{b})$. (This is just comprehension.) This implies that given any $\mathcal{L}(Q_{\mathrm{Brch}})$ formula $\psi(\bar{x})$ and any $\bar{a} \in M'$, $M' \models \psi(\bar{a})$ if and only if $(M',S')$ satisfies the internal interpretation of it. Therefore, since no $M'S'$-definable subset of $M'$ is a cut, we have that $M' \models \mathsf{PA}(Q_{\mathrm{Brch}})$. If we start with a model $(M,S)$ of $\mathsf{Z}_2^{SU} + \neg \mathrm{Con}(\mathsf{Z}_2^{SU})$, then we get an $M' \models \mathsf{PA}(Q_{\mathrm{Brch}})$ which does not satisfy true arithmetic, establishing that $\mathcal{L}(Q_{\mathrm{Brch}})$ is $\mathsf{PA}$-intermediate.

I vaguely seem to recall that there is actually a forcing-style reduction of of 'second-order arithmetic + an ultrafilter predicate' to $\mathsf{Z}_2$ which gives a relative consistency argument, but that doesn't seem to be enough to get consistency of $\mathsf{Z}_2^{SU}$ from consistency of $\mathsf{Z}_2$.


问:
gen solution of second order differential equation
can anyone please start me off on how to go about solving this differential equation.
I'm unsure of which method to use.

$$ (1-x^2)\frac{\mathrm d^2y}{\mathrm d x^2} + 4x \frac{\mathrm dy}{\mathrm dx} = 0. $$
thanks in advance

答:
Hint: Separation of variables should work: Let $z = \frac{\mathrm dy}{\mathrm dx}$, then your equation reads $\def\d{\mathrm d}$
$$ (1-x^2)\frac{\d z}{\d x} = -4x\, z \iff \frac{\d z}z = -\frac{4x}{1-x^2} $$
Now integrate.


问:
We remove four corners of a $n\times n$ square. For which $n$ can we fill it with $L$ tetrominoes?
We take away four corners of a $n\times n$ square. For which $n$ can we fill it with $L$ tetrominoes?

My attempt: Because we only have tetrominoes then $4 \mid n^2-4$ then we have $2 \mid n$, which means $n$ is even. If we color it line by line it shows that the black and white colors are the same number. Any $L$-tetromino can fill $1$ white and $3$ blacks or $3$ whites and $1$ black. From that, we can know that the two kinds of $L$_tetrominos are equal. Then we can know that the number of tetrominos are even. From that we know that $8 \mid n^2-4$.Which gives us:

$n^2\equiv 4\pmod 8 \Rightarrow n \equiv 2,6 \pmod 8 \Rightarrow n \equiv 2 \pmod 4$

Now we have to show that for any $n=4k+2 (k \in \mathbb{N})$ it is possible to fill or give another condition for $n$.

答:
As you have shown in the original post, such tilings must be of the form $4k+2$ for some positive integer $k$ (if $k=0$ the resulting shape is empty). So we just have to show that all such squares are possible.

Here is a diagram illustrating one possible tiling of a $(4k+2)\times (4k+2)$ rectangle for all $k\ge 1$ (in the image I have used $k=3$):

We tile the left and right three columns with a sort of overlapping arrangement, as shown; each of these arrangements uses $4k$ tiles. Then, since two L-tetrominoes form a $4\times 2$ rectangle, we can tile any rectangle with even dimensions and one side a multiple of $4$; this gets us the green and blue regions, with $2k$ and $4k^2-6k$ tiles respectively. (The total is $4k^2+4k$ tiles, as desired.)


问:
Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$

Suppose $p$ is an odd prime. Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$, where $1$ is the identity.

I know that the conjugate of power is the power of conjugate, so $yx^{k}y^{-1} = (x^{p+1})^{k}$, and used this to obtain $yx^{-1} = x^{-p-1} y$, which will allow me to switch $y$ and $x^{-1}$ in each $yx^{-1}$. However, I still cannot figure out where to go from here. Any help is appreciated.

Body:
$yx^k = x^{k(p+1)}y$, so $y^a x^k = x^{k(p+1)^a} y^a$, so:

$$\begin{array}{rcl}
(yx^{-1})^p
&=& y x^{-1} y x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)}y^2 x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)-(p+1)^2} y^3 x^{-1} \cdots yx^{-1} \\
&=& \cdots \\
&=& x^{-(p+1)-(p+1)^2-\cdots-(p+1)^{p-1}} y^p \\
&=& x^{-(p+1)-(2p+1)-\cdots-((p-1)p+1)} x^p \\
&=& x^{-\frac{(p-1)p}2 p - p} x^p \\
&=& x^{-\frac{(p-1)}2 p^2} \\
&=& 1 \\
\end{array}$$

where $x^{(p+1)^n} = x^{1+np+O(p^2)} = x^{1+np}$ because $x^{p^2}=1$.


问:
How to find basis of the image of a matrix
I need to find the basis of the image of the linear transformation given by the following matrix:

$ \begin{pmatrix} 1 & 2 & 3\\ 0 & 1 & 3 \end{pmatrix} $

I've tried looking online, but have become more confused.

I thought that you would just find a maximal linearly independent subset of the column vectors and then that would be the basis of the image (in which case it would be $\begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $\begin{pmatrix} 2 \\ 1 \end{pmatrix} $ or $\begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $\begin{pmatrix} 3 \\ 3 \end{pmatrix} $).

But I've heard that you have to first row reduce the matrix - why?

答:
First of all, it's not unique, so it should be a basis, not the basis.

You are correct: a maximal linearly independent set of columns form a basis.  In this case any two columns are linearly independent, while of course all three are not,
so any two columns form a basis.  You don't need to row reduce the matrix, that's just a way of ensuring
the column you choose are linearly independent.


答:
From $yxy^{-1} = x^{p+1}$ it follows that $yx^{-1}y^{-1} = (x^{p+1})^{-1}$, so $yx^{-1} = x^{-p-1}y$. We can now prove by an easy induction on $k$ that
$$
(yx^{-1})^k = x^{-p(1+2+\ldots+k)-k}y^k.
$$
For $k=p$, we obtain
$$
(yx^{-1})^p = x^{-p(1+2+\ldots+p)-p}y^p = x^{-p}y^p = 1
$$
since $p(1+2+\ldots+p) = p \cdot \frac{p(p+1)}{2}$ is a multiple of $p^2$.

Body:
Suppose $p$ is an odd prime. Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$, where $1$ is the identity.

I know that the conjugate of power is the power of conjugate, so $yx^{k}y^{-1} = (x^{p+1})^{k}$, and used this to obtain $yx^{-1} = x^{-p-1} y$, which will allow me to switch $y$ and $x^{-1}$ in each $yx^{-1}$. However, I still cannot figure out where to go from here. Any help is appreciated.


问:
Show that $\mathbb{P}\left(\bigcap\limits_{n=1}^{\infty}A_n\right)=0$
My question is:  Be $(\Omega,F,P)$  is a probability space and $A_1,A_2..∈F$

Show that if:
$$  \ P\left( A_n\right) = \frac{2^ne^{-2}}{n!}$$

So $\mathbb{P}\left(\bigcap\limits_{n=1}^{\infty}A_n\right)=0$

答: Hint: $\Bbb P\left(\bigcap_{n=1}^\infty A_n\right) \le \Bbb P(A_n)$ for any $n$, so $\Bbb P\left(\bigcap_{n=1}^\infty A_n\right) \le \inf \{ \Bbb P(A_n) \mid n \}$, so it suffices to show that $\dfrac{2^ne^{-2}}{n!} \to 0$ as $n \to \infty$.


问:
If $f,g\in L^2(\mathbb{R})$ then $\lim_{n\to \infty}\int_{\mathbb{R}}f(x)g(x+n)=0$
I don't know if this is true or false. I tried counterexamples but couldn't find any so then I tried to prove this with some theorems regarding convolutions because it looks a little bit like a convolution but couldn't find any that I could use. Can someone give me a hint?</p>

答:
The basics of the Fourier transform, $F,$ on $L^2(\mathbb R)$ lead to an nice solution: We have

$$ \int_{\mathbb R} f(x)g(x+n)\,dx = \int_{\mathbb R} F(f)(t)F(g(x+n))(t)\,dt$$ $$ = \int_{\mathbb R} F(f)(t)\cdot F(g)(t)e^{int}\,dt.$$

Now $F(f),F(g)\in L^2,$ so $F(f)\cdot F(g)\in L^1.$ Hence by the Riemann Lebesgue lemma, the last integral $\to 0$ as $n\to \infty,$ and we're done.


问:
What is the price of the apples and how many were they brought?
The problem is as follows:

The cook said "I paid $12$ usd for the apples I bought" and they gave
me two apples because I claimed that the apples were too small, that
made me pay one dollar less per dozen. How many apples did the cook
bring?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{16 apples}\\
2.&\textrm{18 apples}\\
3.&\textrm{12 apples}\\
4.&\textrm{15 apples}\\
\end{array}$

This problem is the typical word problem regarding a linear equation but I'm confused because I don't know the right intepretation.

The way how I put the equations is as follows.

The labels are $x$ the number of apples and $y$ the price of each apple.

Thus:

$12+2y=xy$

Because it mentions that a dozen price was one dollar less.

$y=12(y-1)$

The replacing this in the earlier equation yields.

$x=13$

This means that she ended up with $13$ plus the two additional apples that she had thus is $15$.

To me the most confusing part was to assume:

$y=12(y-1)$

I don't know if this is correct, thus I'd like this to be discussed in the answer please. The reason for which I am confused is that,

I think that the new prize of the apples (which ended up being used is one dollar less).

I don't know if this is the right answer, or if my math was correctly used. Hence I would like someone could answer in a more logical way because I'm confused.

答:
Let $p$ be the price per dozen. After the reduction we have that

the number of apples is $$\left(\frac{12}{p-1}\right)\cdot 12=\frac{144}{p-1}$$
which is the number of apples they had got without the reduction plus $2$ apples.
$$\left(\frac{12}{p}\right)\cdot 12+2=\frac{144}{p}+2$$
We have the equation
$$\frac{144}{p-1}=\frac{144}{p}+2\to (p=-8);\;p=9$$
So they payed the apples $\$8$ per dozen, as they spent $\$12$ they got $144/8=18$ apples.


答:
Here is how I would solve it. Say the cook came back home with $x$ apples. But they bought only $(x-2)$ apples and paid $12$ dollars for it. The other two apples came free.

So original price per apple $\displaystyle   = \frac{12}{x-2}$

But since they finally got $x$ apples, now the effective price per apple is $ = \displaystyle \frac{12}{x}$

For each apple, they paid $\displaystyle \frac{1}{12}$ dollar less ($1$ dollar less per dozen).

So, $\displaystyle \frac{12}{x-2} - \frac{12}{x} = \frac{1}{12}$

$\implies x^2-2x = 288$

Or, $(x-1)^2 = 289 = 17^2 \implies x = 18$.

So the cook brought $18$ apples.

Title: What is the price of the apples and how many were they brought?

Body:
The problem is as follows:

The cook said "I paid $12$ usd for the apples I bought" and they gave
me two apples because I claimed that the apples were too small, that
made me pay one dollar less per dozen. How many apples did the cook
bring?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{16 apples}\\
2.&\textrm{18 apples}\\
3.&\textrm{12 apples}\\
4.&\textrm{15 apples}\\
\end{array}$

This problem is the typical word problem regarding a linear equation but I'm confused because I don't know the right intepretation.

The way how I put the equations is as follows.

The labels are $x$ the number of apples and $y$ the price of each apple.

Thus:

$12+2y=xy$

Because it mentions that a dozen price was one dollar less.

$y=12(y-1)$

The replacing this in the earlier equation yields.

$x=13$

This means that she ended up with $13$ plus the two additional apples that she had thus is $15$.

To me the most confusing part was to assume:

$y=12(y-1)$

I don't know if this is correct, thus I'd like this to be discussed in the answer please. The reason for which I am confused is that,

I think that the new prize of the apples (which ended up being used is one dollar less).

I don't know if this is the right answer, or if my math was correctly used. Hence I would like someone could answer in a more logical way because I'm confused.


问:
Regularity conditions for curves in Riemann surfaces
So, I am studying complex analyisis in one variable from various books (Rudin, Forster and Gamelin) and each one of them uses a particular regularity condition to prove their theorems, and I am not sure if they are equivalent.

A point $p$ in the boundary of a Riemann Surface Y is regular if there is some continuous function $u : \bar{Y} \cap V \to \mathbb{R}$ where $V$ is a neighbourhood of $p$ such that $u$ is subharmonic in $Y \cap V$, $u(p)=0$ and $u(x)<0$ otherwise (Forster)

A point $p$ in the boundary of an open subset $U$ of $\mathbb{C}$ is simple if for every sequence $\{z_n\}$ in $U$ converging to $p$ there is some path $\gamma : [0,1] \to U$ and points $t_1<t_2< \ldots \to 1$ such that $\gamma (t_n)=z_n$ (Rudin)

An curve $\gamma$ contained in the boundary of a subset $U$ of $\mathbb{C}$ is a free analytic arc if for every point $p \in \gamma$ there is some conformal map from the unit disk to some neighbourhood $V$ of $p$ that maps the real line to $\gamma \cap V$ and the upper half plane to $U \cap V$ (Gamelin)

And they use these definitions to prove:

The Dirichlet boundary problem can be solved on open subsets whose boundaries consist of regular points.

Conformal maps from a simply connected, bounded subset $U$ of $\mathbb{C}$ to the unit disk whose boundary points are simple can be extended conformally to the $\bar{U}$.

One can use Schwarz reflection principle and construct Green functions for domains with free analytic boundary (ie, each connected component of the boundary is a free analytic closed curve).

Is there any equivalence between the definitions? I already know that free analytic boundary implies regular boundary, and Caratheodory's theorem, ie, all the boundry points of a simply connected domain are simple if and only if the boundary is a Jordan curve, but cannot find more connections.

Of course, there is some ambiguity here because the definition of free analytic arcs depends on a neigbourhood of the boundary, so the question is not about the boundary of a Riemann surface alone, but about the boundary of a subset of a Riemann surface, because I would like to be able to construct an exhaustion of every Riemann surface using Runge compacts with the nicest properties on their boundary.

答:
First of all, note that the subharmonic function in the definition of regular point in Forster is required to be negative, not positive, away from the point.

Every boundary point of a simply connected region $\Omega\subsetneq\mathbb{C}^*$ (the Riemann sphere) is regular (provided $\partial\Omega$ contains more than one point), even if its boundary points are not simple and $\partial \Omega$ does not contain any analytic arc. This is not hard, and is proven e.g. in Donald Marshall's Complex analysis, chapter 13 section 2.
Similarly, there are simply connected regions on which every point is simple and yet the boundary does not contain any analytic arc: it suffices to take a jordan region bounded by an Osgood curve.

Thus, $(2)\not\rightarrow (1), (2)\not\rightarrow (3)$. Similarly, there are regions such that every boundary point is simple but not every point is regular: just take $\mathbb{C}-\{0\}$.  The maximum principle implies that $0$ cannot be a regular point (in general, isolated points cannot be regular). So there is in general no relation between $(1)$ and $(2)$. On the other hand it is quite easy to see how $(1)$ and $(2)$ are implied by $(3)$, as you already noticed.


Title: If $0 < p < 1$, then does $\int_{0}^{p}\frac{1}{x^p}dx =-\frac{1}{(p-1)(p^{p-1})}$.

Body: If $0 < p < 1$, then does $\int_{0}^{p}\frac{1}{x^p}dx =-\frac{1}{(p-1)(p^{p-1})}$ For example, if $p=0.5$ then $\int_{0}^{0.5}\frac{1}{x^{0.5}}dx =-\frac{1}{(0.5-1)({0.5}^{0.5-1})}$. I know $\int_{0}^{p}\frac{1}{x^p}dx = \frac{1}{1-p}$, but it's impossible to make this into the $-\frac{1}{(p-1)(p^{p-1})}$?

Body: $$\int_{0}^p \frac{1}{x^p} \ dx = \frac{x^{-p+1}}{-p+1}\Big]^{x=p}_{x=0}=\frac{p^{1-p}}{1-p}=\frac{1}{(1-p)p^{p-1}}$$


问:
Integrating Normal distribution exponentially decaying in 2nd dimension
Firs, I am struggling with a problem in physics, but the issue lies with the math, so I think it fits here.

Now, let me explain, I have some  normal/Gaussian distributions (of density but that's not relevant to the math) with known $\sigma$ and known height $a$ like: sketch1
they can be described with : $f(x) = a\exp(-\frac{x}{2\sigma^2})$

Integrating $\int f(x) dx = a\sqrt{2\pi\sigma^2}$

What this means for me is that for known area $A$ under the Gaussian (my measurement is proportional to) and known $\sigma$ I can calculate the height $a$. This is still fine, now my issue begins, the distribution is evolving in 2nd dimension (let's call it $z$) this evolution is an exponential decay, like so sketch2 The decay can be described with $\exp(-z*c)$ with $c$ some positive constant.

So, I think the function can be described as $f(x,z) = a\exp(-\frac{x}{2\sigma^2})\exp(-z*c)$,
now when I "measure" the area $A$ under such curve with constant $z$ I can easily get the height $a$ for given z with $a = \sqrt{2\pi\sigma^2}/A$. The problem is that the "measurement" does not integrate along constant $z$ but rather at an angle $\alpha$ like: sketch3 red lines are the integration paths.

Now, I don't know anymore how to relate the $A$ under such a curve to the central height $a(x = 0, z = z)$. The angle $\alpha$ between the $z$ axis and the integration path is known, I would like to express the result in terms of the $\alpha$

Any help would be appricieted.

答:
If I understand correctly, you want to integrate

$$
f(x,z)=a \exp \left(-\frac{x^2}{2 \sigma}-cz \right)
$$

Over some given path $z(x)=\alpha + \beta x$. Along the path, substitute $z(x)$ into $f$, and note the line element is $dl=dx \sqrt{1+\beta^2}$. All together, we have a Gaussian integral

$$
\int\limits_{-\infty}^\infty dx \ \sqrt{1+\beta^2} a  \exp\left(-\frac{x^2}{2 \sigma}-c \alpha-c \beta x \right)= a \sqrt{2 \pi \sigma (1+\beta^2)} \exp \left(-c \alpha +\frac{c^2 \beta^2 \sigma}{2} \right)
$$

Note my $\alpha$ is not the path angle, but you should be able to relate $\alpha$ and $\beta$ to it.


问:
Is there a series expansion formula for $\prod_{j=1}^{n} (a_j - b_j)$?
I would like to expand the product $\prod_{j=1}^{n} (a_j - b_j)$ into a summation. Since there is some $n$ number of factors, I cannot simply 'foil' out the expansion. Checking for any specific value of $n$ by hand, there always seems to such a polynomial so I am simply stuck on navigating Pi-notation.

Is there a series expansion formula for $\prod_{j=1}^{n} (a_j - b_j)$?

答:
$$ \prod_{j=1}^n (a_j - b_j) = \sum_{S \subseteq \{1,\ldots,n\}} \prod_{j \in S} a_j \prod_{j \in \{1,\ldots,n\} \backslash S} (-b_j)$$

That is, there is a term in the sum for every subset $S$ of the indices $1, \ldots, n$.  That term consists of the product of $a_j$ for $j$ in that subset, times the product of $-b_j$ for $j$ not in that subset.