问:
Is there a (deep) relationship between these various applications of the exponential function?
Here is a list of some applications of the exponential function.

1) The exponential mapping in Lie theory.

I put this first because my intuition tells me that this must be the most fundamental, or deep, way of thinking about the exponential function.  I have often been misled by my intuition however, and the main reason I feel strongly about this is because of how fundamental I consider Lie theory to be.

2) Fourier Series

3) Roots of Unity

4) Gaussian Distribution

5) Boltzmann Distribution

There are certainly other applications, but it always kind of bothered me that I couldn't use symmetry methods to see how (all of) these applications are related.  Is it possible that there is no way to do this, i.e. that it's just a happy accident that the exponential function has these applications and it is unrelated to any continuous symmetries?

答:
The most important traits of the exponential function that make $e^x$ occur in that much applications are

a) Euler's identity $e^{i\pi} + 1 = 0$

and its implications for the definition of complex trigonimetric functions, i.e.

$$e^{i \cdot x} = \cos x + i \sin x$$

(like Fourier transformation, unit roots)

b) the fact that $\frac{d}{dx}e^x = e^x$

which lets $e^x$ naturally occur in many solutions of differential equations, which are often the most basic formulation of natural laws (growth/decay, gaussian distribution).

ContentLicense: CC BY-SA 2.5


问:
best way to show tr(AB) = tr(BA) for non square A and b matrices
I have matrices $A \in \mathbb{K}^{n \times m}$ and $B \in \mathbb{K}^{m \times n}$

What is the best way to prove that tr(AB) = tr(BA). I found a prove in Matrix Analysis by Horn and Johnson, but they only prove it if $n \leq m$.

答: Either you have $n\leq m$ or $m\leq n$. If it is the latter, exchange $n$ with $m$ and $A$ with $B$, and the same proof works.

ContentLicense: CC BY-SA 4.0


问:
Inscribing Platonic solids in each other: why can't you put a dodecahedron in an octahedron?
Given convex polyhedra $P, Q$, say that one can inscribe $P$ in $Q$ if we can find points on the surface of $Q$ whose convex hull is similar to $P$.

If we restrict $P, Q$ to be Platonic solids, we can achieve every case except inscribing the dodecahedron into the tetrahedron, cube, or octahedron; the other $17$ distinct pairs work.

For the $17$ working cases, there are "nice" constructions, where the solids are positioned in symmetric ways with respect to each other. Consequently, such constructions are pretty easy to verify by a combination of symmetry arguments, direct computation of one or two distances, and arguments from degrees of freedom / continuity. To see examples of these pairs, Moritz Firsching's 2018 paper "Computing maximal copies of polytopes contained in a polytope" (PDF link via arXiv.org) shows instances of maximal containment, most of which have "nice" symmetry (or are obviously given by e.g. taking duals). Perhaps the hardest case to see a "nice" construction for is the cube inside the tetrahedron, pictured below (bold edges tangent to the faces):

For two of the three impossible cases, I have simple proofs of impossibility:

If a dodecahedron were inscribed in a tetrahedron, every face would need to have a pentagon on it, but the dodecahedron doesn't have four mutually disjoint faces.

If a dodecahedron were inscribed in a cube, then any face with at least three of the vertices of the dodecahedron would have to have an entire pentagon on it (and no more of the vertices), since three points determine a plane and the dodecahedron is convex. Then (by simple counting) at least three of the cube's faces must have such a pentagon on them, so two such are orthogonal to each other. But the dodecahedron has no orthogonal faces.

However, I know of no such proof that the dodecahedron cannot be inscribed in the octahedron; it seems fairly plausible, at first glance, that one could mount the dodecahedron with two opposite faces tangent to opposite faces of the octahedron and arrange the other $10$ points across the other $6$ faces of the octahedron somehow. The simplest proof I know involves doing some gross trigonometry involving the necessary side length of such a dodecahedron and the inradius of the regular $10$-gon formed by taking an equatorial cross-section of the dodecahedron in between these two opposite faces.

Is there a conceptually simple proof that this construction is impossible? I realize this is somewhat subjective, but I hope the intended style of argument is clear: reasoning from symmetry and combinatorial incidences, rather than bashing the coordinates of different vertices.

答:
The regular dodecahedron has $20$ vertices; in order to place every vertex on a face of a regular octahedron, since the octahedron has only $8$ faces, by the pigeonhole principle we must put at least three vertices on some face of the octahedron.
This implies one face $A$ of the dodecahedron lies on a face $B$ of the octahedron.

Consider the sphere $S$ inscribed in the dodecahedron.
Sphere $S$ is tangent to face $B$ of the octahedron.
Moreover, whenever three vertices of the dodecahedron lie on one face of the octahedron, a face of the dodecahedron lies on that face of the octahedron as well,
and sphere $S$ also is tangent to that face of the octahedron.
A face of the octahedron to which $S$ is not tangent can have at most two vertices of the dodecahedron.
Moreover, if $S$ is tangent to face $F$ of the octahedron but not to the opposite face, then no vertices of the dodecahedron can be on the opposite face of the octahedron.

Now let's see how many faces of the octahedron $S$ is tangent to.
By symmetries of the octahedron, $S$ can be tangent to exactly one face; tangent to two adjacent faces and no others; tangent to four faces around a vertex and no others with the center of $S$ on the diagonal between two vertices of the octahedron, or tangent to all eight faces and coincident with the inscribed sphere of the octahedron.

Exactly one tangent face is impossible, since $7$ faces with at most two vertices each is only $14$ vertices, whereas the dodecahedron has $15$ vertices that are not on face $B$ of the octahedron.

Exactly two tangent faces implies the two opposite faces have no vertices of the dodecahedron; since the two tangent faces can contain at most $10$ vertices of the dodecahedron, the remaining four faces must contain the other $10$ vertices, which cannot occur with only two vertices per face.

In the case of four tangent faces, each of those faces contains one face of the dodecahedron, and there is also a diametrically opposite parallel face of the dodecahedron. The planes of those diametrically opposite faces, together with the four tangent faces of the original octahedron, form a smaller octahedron tangent to $S$ on all eight faces, therefore with a face of the dodecahedron on each face of the octahedron.
By inspection, it is not possible to place all five vertices of a regular pentagon on the sides of an equilateral triangle. We can place at most four vertices, none of them at a vertex of the triangle, leaving one inside the triangle.
Since each vertex on an edge of a triangle lies on two faces of the octahedron,
we have an average of at least three vertices per face,
which comes to at least $24$ vertices. Since the dodecahedron has only $20$ vertices, this is impossible.

The case of eight tangent faces is the same as the case of four tangent faces, except that we start with an octahedron tangent to $S$ on all faces rather than having to construct a new one.

Hence it is not possible to inscribe a regular dodecahedron in a regular octahedron.


问:
Is a bounded sublinear operator continuous?
We know that, if $T$ is a linear operator between normed spaces $X$ and $Y$, then $T$ is continuous $\iff$ $T$ is bounded.

I was wondering if the result holds when $T$ is a sublinear operator. It seems that the proof used for linear operators still works when you use a sublinear operator, yet in every book I read boundedness doesn't seem to be used as synonymous for continuity, and this sounds strange. So I thought maybe I'm not seeing something.

Any help will be appreciated!

答:
Let $T:X\to Y$ be a sublinear operator between normed spaces, i.e. $\|T(x+y)\|\leq\|Tx\|+\|Ty\|$ and $\|T(\lambda x)\|=|\lambda\|Tx\|$. Note that any sublinear operator maps $0$ to $0$: we have that $0=\lambda\cdot0$ for all scalars so $\|T0\|=\|T(\lambda 0)\|=|\lambda|\|T0\|$ for all scalars, so $\|T0\|=0$.

Assume that $T$ is continuous. Then $T$ is continuous at $0$, so if $\varepsilon>0$ there exists $\delta>0$ so that $\|x\|<\delta\implies\|Tx\|<\varepsilon$.

Apply this for $\varepsilon=1$ and take a $\delta>0$ satisfying the above implication. Now if $x\in X$ is any non-zero vector then $\frac{\delta}{2\|x\|}\cdot x$ has norm smaller than $\delta$, so $\|T(\frac{\delta}{2\|x\|}x)\|<1$. By the second property in the definition of a sublinear operator we have that
$$\frac{\delta}{2\|x\|}\|Tx\|<1$$
so $\|Tx\|\leq\frac{\delta}{2}\|x\|$ and therefore $T$ is bounded. Note that we have only used the second property of the definition to prove this direction.

For the converse however, I cannot see how the proof for linear operators can be mimicked for sublinear ones: to show continuity one must estimate $\|Tx-Tx_0\|$ in terms of $\|x-x_0\|$. In the linear setting, $\|Tx-Tx_0\|=\|T(x-x_0)\|$ and boundedness can be used to immediately derive continuity at $0$, so when $\|x-x_0\|$ becomes small we have that $x-x_0\to0$, so $\|T(x-x_0)\|\to0$. However the crucial equality $\|Tx-Tx_0\|=\|T(x-x_0)\|$ does not hold in the sublinear setting and I cannot see an "over the counter" inequality to be used here for our estimation. Unless I am missing something obvious; I have not been involved with sublinear operators so I am not the best person to speak on this.


问:
Applications of Ramanujan's Master Theorem
Ramanujan's Master Theorem is really neat. Unfortunately, however I have only used it once before, and I want to use it more. I would like a list of integrals to which I may apply this beautiful theorem.

The Theorem: (Taken from Wikipedia)

If $f(x)$ is a complex valued function with a series representation in the form
$$f(x)=\sum_{n\geq0}\frac{\phi(n)}{n!}(-x)^n$$
Then $$\int_0^\infty x^{s-1}f(x)\mathrm dx=\Gamma(s)\phi(-s)$$
Where $\Gamma(s)$ is the Gamma function.

Cheers!

答:
The modified Mellin transform (MMT) pair allows for interpolation of the coefficients of generating functions, often directly connected to sinc and/or Newton interpolation.

First consider the MMT and its inverse

$$\tilde{f}(s) = MMT[f(x)] = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx$$

$$f(x) = MMT^{-1}[\tilde{f}(s)] = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{sin(\pi s)} \tilde{f}(s) \frac{x^{-s}}{(-s)!} \; ds .$$

Then the RMF holds for a class of functions such that

$$f(x) = e^{-a.x} = \sum_{n \geq 0} \frac{(-a.x)^n}{n!} = \sum_{n=0} a_n \frac{(-x)^n}{n!} =
 \sum_{n=0} \tilde{f}(-n) \frac{(-x)^n}{n!} \;  ,$$

that is, such that we may close the complex contour to the left (e.g., in the sense of the limit of a semicircle with its radius expanding to infinity) for $0 < \sigma < 1$ and $0 < x < 1$ when $F(s)$ has no singularities/poles within the contour. This rep allows an extension of the RMT (and the Mellin transform) to cases in which poles are present in $F(s)$ and other ranges of $x$.

Also note (see, e.g., Gelfand and Shilov's "Generalized Functions") the relation

$$D_x^{m+n+1} \; H(x) \frac{x^m}{m!} = H(x) \frac{x^{-n-1}}{(-n-1)!} = \delta^{(n)}(x)$$

reflected in the two (of several) reps of the fractional differintegro op equivalent under analytic continuation

$$\frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = \frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz  = \frac{1}{2\pi i} \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz ,$$

with $H(x)$ the Heaviside step function.

So, under the conditions above,

$$\tilde{f}(-n) = \int_{0}^{\infty} f(x) \; \frac{x^{-n-1}}{(-n-1)!} \; dx = \int_{0}^{\infty} e^{-a. x} \; \delta^{(n)}(x) \; dx = a_n,$$

and this suggests the analytic continuation and relation to umbral calculus

$$\tilde{f}(s) = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx = \int_{0}^{\infty} e^{-a.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (a.)^{-s} = a_{-s}.$$

The iconic guiding example is the  Euler gamma function integral rep with $(a.)^n = a_n = c^n$

$$ (a.)^{-s} = a_{-s}  = c^{-s} = F(s) = MT[f(x)= e^{-c\; x}] = \int_{0}^{\infty} e^{-c \; x} \; \frac{x^{s-1}}{(s-1)!} \; dx = \frac{1}{c^{s}}.$$

Another useful example, which vividly illustrates the relation to the Appell Sheffer sequences of umbral calculus (of which the $x^n$ with e.g.f. $e^{x}$ is the basic example), is the integral rep for (what I call) the Bernoulli function, simply related to the Hurwitz zeta function and generalizing the Bernoulli polynomials,

$$ B_{-s}(z) = (B.(z))^{-s} = \int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$

$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s,z)$$

where the e.g.f. for the Bernoulli polynomials with $(b.)^n = b_n$ the Bernoulli numbers is

$$e^{B.(x)t} = e^{(b.+x)t} = e^{b.t} e^{xt} = \frac{t}{e^t-1} \; e^{xt}.$$

Note that

$$B_n(z) = -n \; \zeta(1-n,z),$$

$$B_n(1) = -n \; \zeta(1-n,1) =-n \; \zeta(1-n) (Riemann) =  (-1)^n B_n(0) = (-1)^n b_n.$$

Through this characterization, it is not too difficult to show that the Bernoulli function inherits all the elegant properties of a regular Appell sequence, such as $D_z \; B_{s}(z) =  s \; B_{s-1}(z)$.

Riemann knew all this stuff. Ramanujan intuited it. Hardy formalized it. I stumbled across it on a journey starting from the ladder ops of QM and a brief comment by my old math prof Stallybrass about the sequence $D^{m+n} H(x) \frac{x^m}{m!}$ in his integral transforms class an eon ago.

For application to defining fractional powers of operators, see my answer and comments therein to the MO-Q "What does the inverse Mellin transform really mean?" and several of my blog posts, such as "The Creation / Raising Operators for Appell Sequences."

Other examples of interpolation of $a_n$ for the exponential generating funcrtion $g(t) = e^{a.t}$ from the MMT of $f(t) = g(-t) =  e^{-a.t}$, or, conversely, surmising the MMT of $f(t)$ from the Taylor series coefficients of $g(t)$ via $a_n \; |_{n \rightarrow -s} =a_{-s} =\tilde{f}(s)$:

1) $\;g(t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \;  \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = g(-t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \;  \frac{t^n}{n!} ),$

$\; \; \; \; \;\tilde{f}(s) =\cos(\pi \frac{s}{2})$ for $0 < Re(s) < 1,$

2) $\;g(t) = \sin(t)= \sum_{n \geq 0} \sin(\pi \frac{n}{2}) \; \frac{t^n}{n!},$

$\; \; \; \; \;f(t) = g(-t) = \sin(-t) = \sum_{n \geq 0} \sin(-\pi \frac{n}{2}) \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) =-\sin(\pi \frac{s}{2})$ for $-1 < Re(s) < 1,$

3)  $\;g(t) = \frac{1}{1-t} = \sum_{n \geq 0}  \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) =(-s)! $ for $0 < Re(s) < 1,$

4)  $\;g(t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!} ,$

$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1-t} = \sum_{n \geq 0}  \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s)=\cos(\pi s) (-s)!$ for $0 < Re(s) < 1,$

5) $\;g(t) = \ln(1-t) = \sum_{n \geq 0}  \; -(n-1)! \; \frac{t^n}{n!} ,$

$\; \; \; \; \;f(t) = \ln(1+t) = -\sum_{n \geq 0} \cos(\pi n) \; (n-1)! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) = -(-s-1)! $ for $-1 < Re(s) < 0,$

6) $\;g(t) =\sum_{n \ge 0} \frac{x^n}{n!} \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = J_0(2 \sqrt{xt}) =\sum_{n \ge 0} (-1)^n \frac{x^n}{n!} \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) = \frac{x^{-s}}{(-s)!}$ for $0 < Re(s) < \frac{3}{4}.$

7) $\;g(t) = e^{-t^2} =\sum_{n \ge 0} \cos(\frac{\pi n}{2}) \; \frac{n!}{(\frac{n}{2})!} \;  \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = g(-t) = e^{-t^2},$

$\; \; \; \; \;\tilde{f}(s) =  \cos(\pi\frac{ s}{2}) \; \frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} \;$ for $ Re(s) > 0.$

8) $\;g(t) =\sum_{n \ge 0} \frac{1}{(1-x)^{n}} \;\frac{t^n}{n!} = e^{\frac{1}{1-x}t}  , $

$\; \; \; \; \;f(t) = \sum_{n \ge 0}  (-1)^n \frac{1}{(1-x)^{n}}\; \frac{t^n}{n!}= e^{-\frac{1}{1-x}t}  ,$

$\; \; \; \; \;\tilde{f}(s) = (1-x)^{s}$ for $ Re(s) > 0.$

I include this last example because it is important in characterizing the distributional expansion of the Heaviside step function and its derivatives--the derivatives of the Dirac delta function--in terms of the superposition of the Laguerre polynomials as demonstrated in the MO-Q "What's the matrix of the logarithm of the derivative operator, $\ln(D)$? What is the role of this operator in various math fields?"

Some other related MO-Qs:

"An analytic continuation of power series coefficients"

"Deriving the functional equation for ζ(s) from summing the powers of the zeros required to count the integers"

"Newton series and Fourier transform - is there an analogy?"


问:
Manifolds with Euler characteristic equal to $\pm 1$
A compact connected smooth surface has Euler characteristic equal to $\pm 1$ if and only it is homeomorphic to the real projective plane or the connected sum of $3$ real projective planes. What are other examples of connected manifolds (of even dimension) that have Euler characteristic equal to $\pm 1$? For instance, one could take products of the above surfaces.

答:
The Euler characteristic of a closed orientable manifold of dimension $4k + 2$ is even, so to find orientable examples, you need to look in dimensions which are a multiple of four.

One infinite collection of examples can be found amongst manifolds of the form $M_{a,b} := a\mathbb{CP}^2\#b(S^1\times S^3)$, i.e. the connected sum of $a$ copies of $\mathbb{CP}^2$ and $b$ copies of $S^1\times S^3$. Such a manifold has Euler characteristic $2 + a - 2b$, so $\chi(M_{2b-1,b}) = 1$ for any $b \geq 1$, and $\chi(M_{2b-3, b}) = -1$ for any $b \geq 2$. You can vary this construction in many ways:

replace some of the $\mathbb{CP}^2$'s with $\overline{\mathbb{CP}^2}$'s,

replace $S^1\times S^3$ with any manifold with Euler characteristic zero, e.g. $T^4$,

go to dimension $n = 8$ or $16$ where you can replace $\mathbb{CP}^2$ with $\mathbb{HP}^2$ or $\mathbb{OP}^2$ respectively, and replace $S^1\times S^3$ with $S^1\times S^{n-1}$.

Because the Euler characteristic is multiplicative, given any two manifolds with Euler characteristic $\pm 1$, their product also has Euler characteristic $\pm 1$. In particular, $M_{1,1}^k = (\mathbb{CP}^2\#(S^1\times S^3))^k$ gives an example of a closed orientable $4k$-manifold with Euler characteric $1$. More generally, if $F \to E \to B$ is a fiber bundle and $\chi(F), \chi(B) \in \{-1, 1\}$, then $\chi(E) = \chi(B)\chi(F) \in \{-1,1\}$.


问:
Show that the set $S = \{(x_1, x_2, x_3) : x_1^2 + x_2^2 = x_3^2, (x_1, x_2, x_3) + (0,0,0)\}$ is not connected
This problem is in my text:

In $R^3$ with the Euclidean metric, show that the set $S = \{(x_1, x_2, x_3) : x_1^2 + x_2^2 = x_3^2, (x_1, x_2, x_3) + (0,0,0)\}$ is not connected.

I'm stuck on the $(x_1, x_2, x_3) + (0,0,0)$ part of the definition of $S$.
This notation doesn't make sense because its not a condition.

Ignoring that part, $S$ would be a conic surface, which would be connected. So what does  $(x_1, x_2, x_3) + (0,0,0)$ mean?

答:
It must be a typo, the only thing I can think of that makes any sort of sense would be the condition
$$ (x_1,x_2,x_3) \neq (0,0,0). $$


问:
A Lie algebra that is neither solvable nor reductive
What is an example of a Lie algebra that is neither solvable nor reductive? More specifically, what is the Lie algebra of smallest dimension which is neither solvable nor reductive?</p>

答:
This answer assumes throughout that the field $\Bbb F$ underlying the Lie algebra has characteristic $0$; probably many of the results hold over (at least most) finite characteristics, but I don't know that case well enough to say much more.

Any finite-dimensional Lie algebra $\mathfrak{g}$ admits a Levi decomposition, that is, a decomposition as a semidirect product $$\mathfrak{r} \ltimes \mathfrak{s} ,$$ where $\mathfrak{r}$ is the radical (maximal solvable ideal) of $\mathfrak{g}$ and $\mathfrak{s}$ is a semisimple algebra (and is determined up to conjugacies of a certain form). In particular, any nonsolvable Lie algebra $\mathfrak{g}$ admits a semisimple subalgebra $\mathfrak{s}$.

The smallest semisimple algebras have dimension $3$, and there is always at least one up to isomorphism, namely $\mathfrak{sl}_2(\Bbb F)$. (If $\Bbb F = \Bbb C$ this is the only one, but if $\Bbb F = \Bbb R$, there is another, namely, $\mathfrak{so}_3(\Bbb R)$.)

There are no semisimple Lie algebras of dimension $4$, so any $4$-dimensional nonsolvable, nonreductive Lie algebra must have the form $\Bbb F \ltimes_\phi \mathfrak{s}$ for some $3$-dimensional semisimple Lie algebra $\mathfrak{s}$ and homomorphism $\phi: \mathfrak{s} \to \mathfrak{der}(\Bbb F) \cong \Bbb F$. But any $1$-dimensional representation of a semisimple Lie algebra is trivial, so the algebra is $\Bbb F \oplus \mathfrak{s}$, which is reductive.

Thus, any Lie algebras neither solvable nor reductive must have dimension at least $5$, and any of dimension $5$ must have the form $\mathfrak{r} \ltimes_\phi \mathfrak{s}$ for some reductive Lie algebra $\mathfrak{r}$, semisimple Lie algebra $\mathfrak{s}$, and homomorphism $\phi : \mathfrak{s} \to \mathfrak{der}(\mathfrak{r})$, where $\dim \mathfrak{r} = 2$.

If $\mathfrak{r}$ is the abelian $2$-dimensional Lie algebra $\Bbb F^2$, we have $\mathfrak{der}(\Bbb F^2) \cong \mathfrak{gl}_2(\Bbb F)$. In fact $\phi(s) \subset \mathfrak{sl}_2(\Bbb F)$; otherwise $\ker \operatorname{tr} \phi$ is a $2$-dimensional semisimple subalgebra of $\mathfrak{s}$, which does not exist. Then by Schur's Lemma, either $\phi(\mathfrak{s}) = 0$ or $\phi(\mathfrak{s}) \cong \mathfrak{sl}_2(\Bbb F)$. In the former case, $\mathfrak{g} \cong \Bbb F^2 \oplus \mathfrak{sl}_2(\Bbb F)$, which is reductive. On the other hand, up to isomorphism there is exactly one irreducible 2-dimensional representation of $\mathfrak{sl}_2(\Bbb F)$, namely, the standard representation. Taking $\phi$ to be that representation gives us a first example of a $5$-dimensional, nonreductive, nonsemisimple Lie algebra,
$$\Bbb F^2 \ltimes \mathfrak{sl}_2(\Bbb F) .$$ We can realize this example as the (indecomposable) matrix Lie algebra
$$\left\{\pmatrix{a&b&x\\c&-a&y\\\cdot&\cdot&\cdot} \right\} \subset \mathfrak{gl}_3(\Bbb F) .$$ In the Winternitz classification of complex Lie algebras this Lie algebra is labeled $[5, 40]$.

If $\mathfrak{r}$ is the nonabelian $2$-dimensional Lie algebra, $\mathfrak{t}$, we are looking for a map $\phi: \mathfrak{s} \to \mathfrak{der}(\mathfrak{t})$. But every derivation of $\mathfrak{t}$ is inner, so $\mathfrak{der}(\mathfrak{t}) \cong \mathfrak{t} / \mathfrak{z}(\mathfrak{t}) \cong \Bbb F$, and again in this case $\phi$ is trivial, giving the example
$$\mathfrak{t} \oplus \mathfrak{s} .$$
Given a faithful matrix representation $\psi : \mathfrak{s} \to \mathfrak{gl}_n(\Bbb F)$, we can realize this example as the matrix Lie algebra
$$\left\{\pmatrix{\phi(X) \\ & a & b \\ & \cdot & \cdot} : X \in \mathfrak{s}; a, b \in \Bbb F\right\} \subset \mathfrak{gl}_n \oplus \mathfrak{gl}_2 \subset \mathfrak{gl}_{n + 2} .$$


问:
Find an example of a regular triangle-free $4$-chromatic graph
Find an example of a regular triangle-free $4$-chromatic graph

I know that for every $k \geq 3$ there exists a triangle-free $k$-chromatic graph.

So if I can find a triangle-free graph $H$ such that $\chi(H)=3$, then I can use the Mycielski construction to obtain a triangle-free graph $G$ such that $\chi(G)=4$. However, the regular part keep getting me stuck. I try some odd cycle, I also tried the Petersen graph but still can't get a regular triangle-free $4$-chromatic graph.

I wonder if anyone can give me a hint, please.

答:
The Gewirtz graph, a strongly regular graph with parameters $(56,10,0,2)$, is another triangle-free graph with chromatic number $\chi=4$.

It is worth noting that Gewirtz graph is not isomorphic to Kneser graph $K(8,3)$. The Gewirtz graph has independence number $\alpha(G) = 16$ (i.e. the cardinality of a largest set of vertices pairwise nonadjacent, e.g. see Brouwer and Van Maldeghem' survey) whereas $\alpha(K(8,3)) = 21$ (the Erdős-Ko-Rado theorem  states that $\alpha(K(n,k)) = \binom{n-1}{k-1}, n\geq 2k $).

ContentLicense: CC BY-SA 3.0


问:
understanding a proof of a function lying in the closure of a polynomial space

Show that $f(x) = |x|$ lies in the closure of $\mathcal{P}[-1, 1]$ in $\mathcal{C}[-1,1]$ (using the supremum metric), where for $a,b\in\mathbb{R}, \mathcal{P}[a,b]$ is the set of polynomials on $[a,b]$ and $\mathcal{C}[a,b]$ is the set of continuous functions on $[a,b].$

Below is a proof of this proposition.

I understand most of it, except the claim that $|\frac{g^{(n+1)}(t)}{(n+1)!}(x-1)^{n+1}| \leq \dfrac{1\cdot 3 \cdots (2n-1)\cdot a^{2n-1}}{2^{n+1}{(n+1)!}}.$ Can someone explain why this inequality holds?

答:
The Lagrange form of the remainder of the Taylor series is
$$R_n(x) =  \frac{g^{n+1}(t(n,x))}{(n+1)!}(x-1)^{n+1}  = \frac{r_n \cdot (x -1)^{n+1}}{(t(n,x)+a^2)^{(2n+1)/2}}$$
where $r_n = \dfrac{1\cdot3 \cdot \ldots \cdot (2n-1)}{2^{n+1}(n+1)!}$ and $t(n,x) \in (x,1)$. We want to show that
$$d_n := \sup_{x \in [0,1]} \lvert R_n(x) \rvert \to 0$$
as $n \to \infty$. But certainly
$$\sup_{x \in [0,1]} \lvert R_n(x) \rvert \ge  \lvert R_n(0) \rvert = \frac{r_n}{(t(n,0)+a^2)^{(2n+1)/2}} .$$
As long as we do not know anything about $t(n,0) \in (0,1)$, we cannot exclude that there exists $b < 1$ such that $t(n,0) + a^2 \le b^2$ for all $n$. In that case
$$d_n \ge \frac{r_n}{b^{2n+1}} .$$
But we have
$$r_n  \ge \dfrac{2 \cdot 4 \cdot \ldots \cdot (2n-2)}{2^{n+1}(n+1)!} = \dfrac{2^{n-1} (n-1)!}{2^{n+1}(n+1)!} = \dfrac{1}{4n(n+1)}$$
and therefore
$$d_n \ge \frac{1}{4n(n+1)b^{2n+1}} .$$
The RHS diverges to $+\infty$.

This does not necessarily mean that the claim $d_n \le \dfrac{1}{2(n+1)}$ is false, but a proof would require to know something about $t(n,0)$, for example that $t(n,0) + a^2 \ge 1$ for all $n$. I doubt that.


问:
Definition of Borsuk graph
Can someone please clarify what exactly the infinite Borsuk graph is?

I am finding 2 different definitions on the internet, some sources say we join to elements of $\mathbb{S}^n$ iff the distance between them is at least $\alpha$ , whereas some other sources suggest the distance should be strictly greater than $\alpha$.

答:
Lovász's 1977 paper Kneser's conjecture, chromatic number, and homotopy attributes the definition of Borsuk's graph to Erdős and Hajnal in the 1967 paper Kromatikus gráfokról ("On chromatic graphs") where they write

Gráfunk szögpontjai a $k$-dimenziós egységgömb pontjai lesznek. Legyen $\varepsilon = \varepsilon(c)$ elegendő kis pozitív szám. Két pontot akkor kötünk össze éllel, ha távolságuk nagyobb, mint $2-\epsilon$.

In Hungarian, "nagyobb, mint $2-\epsilon$" means "greater than $2-\epsilon$" and so the original definition did not join two points at distance exactly $2-\epsilon$.

Lovász himself, on the other hand, writes

Let the vertices of graph $B_k$ be the points of the $k$-sphere $S^k$, two of them being adjacent iff their distance is at least $2-\epsilon$ for some $\epsilon>0$ (i.e., iff they are almost antipodal).

So you can see that the confusion goes back over four decades and that very prestigious graph theorists disagreed on this point.

The reason that they disagreed is that their results about Borsuk's graph don't really depend on whether you say "at least $2-\epsilon$" or "greater than $2-\epsilon$". The graph defined with parameter $\epsilon$ by one definition is sandwiched between the graphs defined with parameters $2\epsilon$ and $\frac12\epsilon$ by the other definition. This means that theorems about one definition carry over to the other definition provided $\epsilon$ is small enough. Specifically, when $\epsilon$ is sufficiently small,

$B_k$ has chromatic number $k+2$ no matter which definition we use;

$B_k$ contains no short odd cycles no matter which definition we use ($\epsilon$ being chosen for the definition of "short" you want);

$m$-vertex subgraphs of $B_k$ have independence number $\ge cm$ for $c<\frac12$ no matter which definition we use ($\epsilon$ being chosen for the $c$ you want);

I don't know all the other interesting properties of Borsuk's graph, but I bet they also keep working no matter which definition we use.

So it never really mattered and that is why the definition is not standardized. I suppose that if you want an authoritative source, there is none more authoritative than Erdős and Hajnal.

Title: Definition of Borsuk graph


问:
Examples of continuous functions that are not homotopic to each other.
Let $f,g:X \to Y$ be continuous functions. Then, consider $F:X \times I \to Y$ defined as follows:$$F(x,t) = f(1-t)+gt$$

How does one check whether $F$ is continuous(w.r.t the product topology)?
Is there any other way to define a homotopy between two functions?

As an example, take $f,g :  \mathbb{R} \to \mathbb{R}$ to be $f(x) = x, g(x) = sin(x)$. Then how does one PROVE that the homotopy map as defined above is continuous?

Moreover, what are some examples of maps that are not homotopic?

答:
The identity map of the circle $\rm{id}:S^1\to S^1$ and a constant map on $S^1$ are continuous but not homotopic.  This is due to a result by Heinz Hopf, from one point of view.

Intuitively the unit circle is not contractible.

By the alluded to result, we can wrap the circle around itself as many times as we like, continuously, and get non-homotopic maps.

The homotopy you refer to is the straight line homotopy.  There are plenty of others.


问:
Probabilistic Inequalities with Lower Bounds
Many inequalities (Markov, Chevyshev, Chernoff) give some sort of upper bound to $P(X \geq a)$ for some random variable $X$. I am trying to find some results that give a lower bound to this probability.

I have found this result, but unfortunately this only holds for $X \geq 0$. Any there any known results that hold more generally?

If it makes any difference, the specific case I'm trying to apply this to has $X$ symmetric and $E(X^n) < \infty$ for all $n \in \mathbb{N}$, but I would be happy to explore any results on this topic.

答:
You may want to look into "anticoncentration inequalities." One standard example is the Paley—Zygmund inequality.
$$
\forall \theta\in[0,1],\; \mathbb{P}\{ X \geq \theta \mathbb{E}[X]\} \geq (1-\theta)^2 \frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]}
$$
where $X$ is any non-negative r.v. For instance, you could apply that to $X := Z^2$ for any r.v. $Z\in L^4$ with mean $0$, to get a lower bound on
$\mathbb{P}\{ |Z| \geq \theta \sqrt{\operatorname{Var}[Z]}\}$.

Another one is the Carbery—Wright theorem, which applies to polynomials functions of a multivariate Gaussian r.v..


问:
Generalized Rationalization in Finite Radical Field Extensions
In the square root case of a radical extension of, say, $\mathbb{Q}$, we have that $\mathbb{Q}(\sqrt{2}) = \{a + b \sqrt{2} | a, b \in \mathbb{Q} \}$.

The only semi-hard axiom to prove is that inverses exist.  We reason that this is a field because the inverse of $a + b \sqrt{2}$ can be found by rationalizing the denominator.  Specifically:

$$ \frac{1}{a + b \sqrt{2}} = \frac{a - b \sqrt{2}}{a^2 - 2b^2} = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2},$$ which is clearly of the form $x + y \sqrt{2}$, for $x, y \in \mathbb{Q}$, by the closure of the rationals on arithmetic operations.

Let's say we want to consider $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$.  My intuition says that this should be something similar -- at least we know that clearly $\{a + b \sqrt{2} + c \sqrt[3]{2} | a, b, c \in \mathbb{Q} \} \subseteq \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ (though I'm not sure exactly).

I'm just learning field extensions, so I'm not sure if this is right, but I believe that you can say $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}]= 2 \cdot 3$ because the degree of the square root extension is 2, and then you can show that $x^3 - 2$, a polynomial of degree 3, is irreducible over $\mathbb{Q}(\sqrt{2})$.

Can we generalize some form of "rationalization" to help us prove that inverses exist "directly" in the way we do from the square root case?  At the very least, does the existence of a finite degree field extension prove that algebraic rationalizations of this form exist?

答:
Rationalization is an instance of Euclid's algorithm, as used to obtain Bezout's identity.

Let $E$ be a field, and let $\alpha \in E$ be algebraic over a field $F \subseteq E$. Then the minimal polynomial $f$ of $\alpha$ over $F$ is irreducible in $F[x]$.

Let $0 \ne \beta \in F(\alpha)$. Then $\beta = g(\alpha)$ for some $g \in F[x]$. Clearly $g$ is not divisible by $f$, as $\beta = g(\alpha) \ne 0$. Since $f$ is irreducible, $\gcd(g, f) = 1$. Euclid yields $u, v \in F[x]$ such that
$$
1 = g u + f v.
$$
Evaluate for $x = \alpha$ to get
$$
1 = g(\alpha) u(\alpha) + f(\alpha) v(\alpha) = g(\alpha) u(\alpha).
$$
So
$$
\frac{1}{\beta} = \frac{1}{g(\alpha)} = {u(\alpha)}
$$
is the required rationalization.


问:
How to find $B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$ from $\sin\omega+\cos\omega=2\sin\phi$ when $\phi$ is not quadrantal?
The problem is as follows:

First let:

$\sin\omega+\cos\omega=2\sin\phi$ and $\phi$ is a not quadrantal
angle.

Then using this relationship find:

$B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

The alternatives on my book are as follows:

$\begin{array}{ll}
1.&\textrm{0.25}\\
2.&\textrm{0.5}\\
3.&\textrm{1}\\
4.&\textrm{2}\\
\end{array}$

Well what I did to attempt to solve this problem from my precalculus workbook was the following:

I did noticed that on the first expression I could do the following,

$\sin\omega+\cos\omega=2\sin\phi$

$\frac{1}{\sqrt{2}}\sin\omega+\frac{1}{\sqrt{2}}\cos\omega=\frac{2}{\sqrt{2}}\sin\phi$

Then using the sum of angles formula:

$\sin(45^\circ+\omega)=\frac{2}{\sqrt{2}}\sin\phi$

But that's where In got stuck. Now what?

I can't spot a way to make the inner part of the function to be equated with both sides of the equation. In other words, the earlier expression's argument cannot be equated as it is because both sides of the equation are different.

Then there is the other part of the equation which I don't understand. What does it mean not a quadrantal angle?. I mean I do know what a quadrantal is one that is in the standard position and has a measure that is a multiple of $90\circ$. So in short it cannot be $90^\circ$, $180^\circ$, $270^\circ$, $360^\circ$ ... and so on.

But I don't know how to include this information in the solution.

The expression from below doesn't seem to be reduced:

$B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

I could only found that I could:

$B=\frac{1}{\sin^2\phi}-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

$B=\frac{1-\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

Can someone help me here with the right approach?. Was my initial intuition right?. I would like someone could help me with that. Please try to include the most step by step as possible so I can catch properly the idea of the necessary manipulation to solve this problem without much fuss.

Can someone help me here please?. I'd really commend someone could guide me in the right path.

答:
First, we have:

$$\sqrt{2}\sin(\phi) = \frac{\sqrt{2}}{2}\sin(\omega) + \frac{\sqrt{2}}{2}\cos(\omega) = \cos\bigg(\frac{\pi}{4}\bigg)\sin(\omega) + \sin\bigg(\frac{\pi}{4}\bigg)\cos(\omega) = \sin\bigg(\frac{\pi}{4}+\omega\bigg)$$

Then:

$$B = \csc^{2}(\phi) -\frac{\cos^{2}\big(\frac{\pi}{4} + \omega\big)}{\sin^{2}(\phi)} =\frac{1}{\sin^{2}(\phi)}-\frac{1 - \sin^{2}\big(\frac{\pi}{4} + \omega\big)}{\sin^{2}(\phi)}=\frac{(\sqrt{2}\sin(\phi))^{2}}{\sin^{2}(\phi)}=\boxed{2}$$


问:
Central limit theorem for dependent Bernoulli random variables on the edges of a sequence of growing hypercubes?
Imagine a growing sequence of hypercube graphs. For each integer n, you have the $n$-hypercube with $2^n$ vertices.
Now, assign $\text{Bernoulli}(1/2)$ i.i.d. random variables to all the vertices.
On each edge, you define a r.v. taking values in $\{0,1\}$ such that it is $1$ if the Bernoulli r.v.s of the two vertices it touches are equal, and it is $0$ otherwise.
So for each given n, you have Bernoulli r.v.s that are correlated on the edges and the correlation decreases with the distance (of the shortest path on the $n$-hypercube).
Note that going from one vertex to another on the hypercube requires that EXACTLY one binary component of the position changes by 1.
For example, with n=3, you can go from (1,1,0) to (1,1,1) because exactly one component (the third one) changes by 1.
This generalizes to all dimensions.
Now, the question is, if you take the sum of the (dependent) Bernoulli r.v.s on the EDGES (not vertices) and normalize it correctly (i.e. you translate by the expectation and divide by the standard deviation), does this normalized sequence converges to a $\text{Normal}(0,1)$ distribution ?

I know there are versions of the CLT for strong mixing sequences, but I didn't find a version that would suit the problem here.

答:
This is not an answer, but the outline of an idea that might work, but which is too long for a comment.

The edge variables corresponding to the $\ 2^{n-1}\ $ edges parallel to a given axis are Bernoulli$\left(\frac{1}{2}\right)$ i.i.d., because none of them have any vertices in common.  Their sum is therefore Binomial$\left(2^{n-1},\frac{1}{2}\right)$—and hence approximately normal for large $\ n\ $—with mean $\ 2^{n-2}\ $ and variance $\ 2^{n-3}\ $.  If $\ S_i\ $ is the sum of the variables corresponding to the edges parallel to the $\ i^\text{th}\ $ axis, then the total sum is the sum $\ \sum_\limits{i=1}^nS_i\ $ of $\ n\ $ approximately normal random variables.  Although $\ S_1,S_2,\dots,S_n\ $ are not independent, they are uncorrelated, because every term in any one of them is independent of any term in any of the others. Therefore $\ \sum_\limits{i=1}^nS_i\ $ has mean $\ n2^{n-2}\ $ and variance $\ n2^{n-3}\ $, and $\ \sum_\limits{i=1}^n\frac{\big(S_i-2^{n-2}\big)}{2^\frac{n-3}{2}}=\frac{\sum_\limits{i=1}^nS_i-n2^{n-2}}{2^\frac{n-3}{2}}\ $ is the sum of $\ n\ $ uncorrelated, approximately standard normal variates. So $\ \frac{\sum_\limits{i=1}^nS_i-n2^{n-2}}{n2^\frac{n-3}{2}}\ $ should also be approximately standard normal.


答:
Continuing from where you left off: you have $\sin (45^\circ + \omega) = \frac 2{\sqrt 2}\sin \phi$ and hence

$$B = \frac {1-\cos^2(45^\circ + \omega)}{\sin^2\phi} = \frac {\sin^2(45^\circ + \omega)}{\sin^2\phi} = \frac {(\frac 2{\sqrt 2}\sin \phi)^2}{\sin^2\phi} = \frac42\frac {\sin^2\phi}{\sin^2\phi} = 2$$

The requirement that $\phi$ is not quadrantal is such that $\sin \phi \ne 0$, which would make $\csc^2 \phi$ and $B$ undefined.

Title: How to find $B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$ from $\sin\omega+\cos\omega=2\sin\phi$ when $\phi$ is not quadrantal?

Body:
The problem is as follows:

First let:

$\sin\omega+\cos\omega=2\sin\phi$ and $\phi$ is a not quadrantal
angle.

Then using this relationship find:

$B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

The alternatives on my book are as follows:

$\begin{array}{ll}
1.&\textrm{0.25}\\
2.&\textrm{0.5}\\
3.&\textrm{1}\\
4.&\textrm{2}\\
\end{array}$

Well what I did to attempt to solve this problem from my precalculus workbook was the following:

I did noticed that on the first expression I could do the following,

$\sin\omega+\cos\omega=2\sin\phi$

$\frac{1}{\sqrt{2}}\sin\omega+\frac{1}{\sqrt{2}}\cos\omega=\frac{2}{\sqrt{2}}\sin\phi$

Then using the sum of angles formula:

$\sin(45^\circ+\omega)=\frac{2}{\sqrt{2}}\sin\phi$

But that's where In got stuck. Now what?

I can't spot a way to make the inner part of the function to be equated with both sides of the equation. In other words, the earlier expression's argument cannot be equated as it is because both sides of the equation are different.

Then there is the other part of the equation which I don't understand. What does it mean not a quadrantal angle?. I mean I do know what a quadrantal is one that is in the standard position and has a measure that is a multiple of $90\circ$. So in short it cannot be $90^\circ$, $180^\circ$, $270^\circ$, $360^\circ$ ... and so on.

But I don't know how to include this information in the solution.

The expression from below doesn't seem to be reduced:

$B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

I could only found that I could:

$B=\frac{1}{\sin^2\phi}-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

$B=\frac{1-\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

Can someone help me here with the right approach?. Was my initial intuition right?. I would like someone could help me with that. Please try to include the most step by step as possible so I can catch properly the idea of the necessary manipulation to solve this problem without much fuss.

Can someone help me here please?. I'd really commend someone could guide me in the right path.


问:
Inverse triangle inequality on $\mathbb{R}^n$
I do know the basic triangle inequality on $\mathbb{R}^n$, however I am not really sure how I can solve the problem below.

$\|x_n - x\| \leq 1  \Rightarrow   \|x_n\| \leq \|x\| + 1$ (By the Triangle Inequality)

Edit: all the answers are very good I flagged Valerin because of the simplicity.

答:
Actually if you are familiar with inverse triangle inequality, you are done!
$$|\|x_n\|-\|x\||\leq \|x_n-x\| \leq 1$$ then $$\|x_n\|-\|x\|\leq 1\Rightarrow \|x_n\|\leq \|x\|+1$$


答:
Here is a method that works in your specific case and also generalizes.

We know that $\mathbb Q$ is a field, so $\mathbb Q[x]$ is a PID. We know that $x^2-2$ is irreducible over $\mathbb Q$ and has $\sqrt 2$ as a root. It is easy to see it divides every polynomial that has $\sqrt 2$ as a root.

Consider the homomorphism $\phi: \mathbb Q[x] \rightarrow \mathbb Q [\sqrt 2]$ sending $x$ to $\sqrt 2$. By a standard theorem on homomorphisms, we have

$$\frac{ Q [x]}{\ker \phi} \cong \mathbb Q[\sqrt 2].$$

We know that the kernel is an ideal. Because $\mathbb Q[x]$ is a PID, it is generated by a single element, and by the considerations above we see this element is $x^2-2$. So

$$\frac{ Q [x]}{(x^2-2)} \cong \mathbb Q[\sqrt 2].$$

Now note that since $\mathbb Q[x]$ is a PID and $x^2-2$ is irreducible, $(x^2-2)$ generates a maximal ideal. So $Q[\sqrt 2]$ is a field and $\mathbb Q[\sqrt 2]=\mathbb Q(\sqrt 2)$. This shows we don't need to add anything more than polynomials in $\sqrt 2$ to get a field.

This argument can be repeated for $\sqrt[3] 2$ over $Q(\sqrt 2)$ almost word for word. You don't actually need to find the polynomial that generates the kernel. All you need to show is that the one that does is irreducible, and I leave that to you. Let me know if you want more detail.

Title: Generalized Rationalization in Finite Radical Field Extensions

Body:
In the square root case of a radical extension of, say, $\mathbb{Q}$, we have that $\mathbb{Q}(\sqrt{2}) = \{a + b \sqrt{2} | a, b \in \mathbb{Q} \}$.

The only semi-hard axiom to prove is that inverses exist.  We reason that this is a field because the inverse of $a + b \sqrt{2}$ can be found by rationalizing the denominator.  Specifically:

$$ \frac{1}{a + b \sqrt{2}} = \frac{a - b \sqrt{2}}{a^2 - 2b^2} = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2},$$ which is clearly of the form $x + y \sqrt{2}$, for $x, y \in \mathbb{Q}$, by the closure of the rationals on arithmetic operations.

Let's say we want to consider $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$.  My intuition says that this should be something similar -- at least we know that clearly $\{a + b \sqrt{2} + c \sqrt[3]{2} | a, b, c \in \mathbb{Q} \} \subseteq \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ (though I'm not sure exactly).

I'm just learning field extensions, so I'm not sure if this is right, but I believe that you can say $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}]= 2 \cdot 3$ because the degree of the square root extension is 2, and then you can show that $x^3 - 2$, a polynomial of degree 3, is irreducible over $\mathbb{Q}(\sqrt{2})$.

Can we generalize some form of "rationalization" to help us prove that inverses exist "directly" in the way we do from the square root case?  At the very least, does the existence of a finite degree field extension prove that algebraic rationalizations of this form exist?


答:
Here it is a proof related to the result in question.

Based on the triangle inequality, notice that
\begin{align*}
\|x_{n}\| = \|x_{n} - x + x\| \leq \|x_{n} - x\| + \|x\| \Rightarrow \|x_{n}\| - \|x\| \leq \|x_{n} - x\|
\end{align*}

At your case, we have
\begin{align*}
\|x_{n}\| - \|x\| \leq \|x_{n} - x\| \leq 1 \Rightarrow \|x_{n}\| \leq \|x\| + 1
\end{align*}
and we are done.

Hopefully this helps!

Title: Inverse triangle inequality on $\mathbb{R}^n$

Body:
I do know the basic triangle inequality on $\mathbb{R}^n$, however I am not really sure how I can solve the problem below.

$\|x_n - x\| \leq 1  \Rightarrow   \|x_n\| \leq \|x\| + 1$ (By the Triangle Inequality)

Edit: all the answers are very good I flagged Valerin because of the simplicity.


问:
Find the cube roots of -8i. Express them in algebraic form.
Would you do this by translating to trig form and then simplifying?</p>

答:
You can also return to the equivalent polynomial problem, since it is clear what one of the roots is:

$$ z^3 \ + \ 8i \ \ = \ \ (z \ - \ 2i ) \ · \ (z^2 \ + \ 2i z \ + \ [2i]^2)  \ \ = \ \ (z \ - \ 2i ) \ · \ (z^2 \ + \ 2i z \ - \ 4) \ \ = \ \ 0 \ \ . $$

The other two roots are given by

$$ z \ \ = \ \ \frac{-2i \ \pm \ \sqrt{[2i]^2 \ -  \ 4 · 1 · (-4)}}{2} \ \ = \ \  \frac{-2i \ \pm \ \sqrt{(-4) \ +  \ 16}}{2} \ \ = \ \ -i \ \pm \ \sqrt{3} \ \ . $$

(This one works out without needing to know how square-roots of complex numbers are found.)


问:
Binomial identity arising from Catalan recurrence.
The following identity features right at the end of the answer to another question I asked: https://math.stackexchange.com/a/4019598/155881.

$$F(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{1- \sqrt{1-4x}}{2x}\right)^n = \sum\limits_{k=0}^{\infty}{n+2k \choose k}x^k$$

Haven't been able to prove this. We know the expansions of both terms on the left side, making the identity:

$$\left(\sum\limits_{j=0}^{\infty} \left(\binom{n+2j-1}{j} - \binom{n+2j-1}{j-1}\right)x^j\right)
\left(\sum\limits_{l=0}^{\infty}{2l \choose l} x^l\right) = \sum\limits_{k=0}^{\infty}{2n+k \choose k}x^k$$

My attempt:
An obvious thing to do is to try a convolution of the left side and equate coefficients of $x^k$ (also using an alternate expression for the first summation in the product of the left side):

$$\sum\limits_{l=0}^k {n+2l-1\choose l}\frac{l}{n+l} {2k-2l\choose k-l} = {2n+k\choose k}$$

Not sure how to make progress on this. Expanding the binomial terms doesn't lead anywhere.

答:
We seek to show that with

$$Q(z) = \frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n$$

we have

$$[z^k] Q(z) = {n+2k\choose k}.$$

Now with the branch cut on $[1/4, \infty)$ for $\sqrt{1-4z}$ we have
analyticity of $Q(z)$ in a neighborhood of the origin (note that the
exponentiated term does not in fact have a pole at $z=0$) and the Cauchy
Coefficient Formula applies. We obtain

$$[z^k] Q(z) = \frac{1}{2\pi i}
\int_{|z|=\varepsilon} \frac{1}{z^{k+1}}
\frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n \; dz.$$

We put $\sqrt{1-4z} = w$ so that  $\frac{1}{\sqrt{1-4z}} \; dz =
-\frac{1}{2} \; dw$  and $z=(1-w^2)/4.$ With $w = 1 - 2z - \cdots$ we get
as the image of $|z|=\varepsilon$ a contour that winds around $w=1$
counterclockwise once and may be  deformed to a circle, so that we obtain

$$[z^k] Q(z) =
-\frac{1}{2}
\frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{k+1}}{(1-w^2)^{k+1}}
(1-w)^n
\frac{1}{2^n} \frac{4^n}{(1-w^2)^n}
\; dw
\\ = \frac{(-1)^k \times 2^{n+2k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{(w-1)^n}{(w^2-1)^{n+k+1}} \; dw
\\ = \frac{(-1)^k \times 2^{n+2k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{k+1}} \frac{1}{(w+1)^{n+k+1}} \; dw
\\ = \frac{(-1)^k \times 2^{k}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{k+1}} \frac{1}{(1+(w-1)/2)^{n+k+1}} \; dw$$

Apply the Cauchy Residue Theorem to get

$$(-1)^k \times 2^k \times (-1)^k \frac{1}{2^k}
{n+k+k\choose n+k} = \bbox[5px,border:2px solid #00A000]{
{n+2k\choose k}}$$

as claimed.


答:
The basic form of the triangle equality is
\begin{equation}
||a+b|| \leqq ||a||+||b||.
\end{equation}

If you replace $a$ for $a-b$, you get
\begin{equation}
||a||-||b||\leqq ||a-b|| \cdots \ast
\end{equation}

If you use $\ast$ for $a=x_n, b=x$, you can see $||x_n||-||x||\leqq ||x_n-x||$ .  And now, $||x_n-x||\leqq 1$ holds. Therefore, $||x_n||-||x||\leqq 1$. So $||x_n||\leqq ||x|| +1$.


问:
Ordered "sums" of natural number k using natural number addends no smaller than 2
Using natural numbers no smaller than 2, we can express the number 3 as one ordered "sum": 3, and we can express the number 6 as 5 ordered "sums": 6 = 4 + 2 = 3 + 3 = 2 + 4 = 2 + 2 + 2.

In how many ways can the number k be expressed as an ordered "sum" with natural number addends no smaller than 2?

答:
For $k$ a positive integer, let $a_k$ denote the number of ways to express $k$ as an ordered sum of integers no smaller than $2$.

If $k\ge3$, then the number of such expressions with first term equal to $2$ is $a_{k-2}$ (deleting the first term is a bijection), and the number of such expressions with first term greater than $2$ is $a_{k-1}$ (subtracting $1$ from the first term is a bijection). Hence the numbers $a_k$ satisfy the Fibonacci recurrence $a_k=a_{k-1}+a_{k-2}$ for $k\ge3$. Since $a_1=0=F_0$ and $a_2=1=F_1$, it follows that $a_k$ is equal to the Fibonacci number $F_{k-1}$ for all $k\ge3$.

P.S. I have been asked to esplain those bijections in greater detail.

For $k\in\mathbb N$ let $A_k$ be the set of all integer sequences $(x_1,\dots,x_m)$ (of any length $m\le n/2$) such that $x_i\ge2$ for each $i$ and $x_1+\cdots+x_m=k$. Thus $a_k=|A_k|$. Let $B_k=\{(x_1,\dots,x_m)\in A_k:x_1=2\}$ and let $C_k=\{(x_1,\dots,x_m)\in A_k:x_1\gt2\}$. Let $b_k=|B_k|$ and $c_k=|C_k|$; then $a_k=b_k+c_k$, since $B_k\cap C_k=\emptyset$ and $B_k\cup C_k=A_k$.

Claim 1. $b_k=a_{k-2}$ for $k\ge3$.

Proof. Define a bijection $f:B_k\to A_{k-2}$ by setting $f(x_1,x_2,\dots,x_m)=(x_2,\dots,x_m)$.

Claim 2. $c_k=a_{k-1}$ for $k\ge3$.

Proof. Define a bijection $g:C_k\to A_{k-1}$ by setting $g(x_1,x_2,\dots,x_m)=(x_1-1,x_2,\dots,x_m)$.

It follows that, for $k\ge3$, we have
$$a_k=b_k+c_k=a_{k-2}+a_{k-1}.$$

Example. $k=7$, $a_7=8=5+3=a_6+a_5$:

$(7)\mapsto(6)$

$(5,2)\mapsto(4,2)$

$(4,3)\mapsto(3,3)$

$(3,4)\mapsto(2,4)$

$(3,2,2)\mapsto(2,2,2)$

$(2,5)\mapsto(5)$

$(2,3,2)\mapsto(3,2)$

$(2,2,3)\mapsto(2,3)$


答:
Define $$ F_n(x):=\sum_{k=0}^\infty c_{n,k}x^k \tag{1} $$ where
$$  c_{n,k}:={n+2k \choose k}. \tag{2}  $$
Verify the binomial identity
$$ c_{n+1,k-1}=c_{n,k}-c_{n-1,k}. \tag{3} $$
Define $$ b:=\frac{1-\sqrt{1-4x}}{2x}, \quad a:=\frac1x-b. \tag{4} $$
Verify that $$ a+b = a\,b = \frac1x, \quad x\,b^2 = b-1. \tag{5} $$
Note that $$ a-b = \frac{\sqrt{1-4x}}x. \tag{6} $$
Equation $(5)$ implies $$ x\,b^{n+1} = b^n-b^{n-1}. \tag{7} $$
Define $$ G_n(x) := \frac{b^n}{x(a-b)}. \tag{8} $$
Equation $(7)$ implies $$ x\,G_{n+1}(x) = G_n(x)-G_{n-1}(x). \tag{9} $$
Verify that the coefficients of the power series $\,G_n(x)\,$
satisfy the same recurrence as equation $(3)$.

Also verify that $$ G_{-1}(x) = F_{-1}(x),\quad G_0(x) = F_0(x). \tag{10} $$
Thus, $\,F_n(x) = G_n(x)\,$ for all integer $\,n\,$ using forward
and backward recursion.

Title: Binomial identity arising from Catalan recurrence.

Body:
The following identity features right at the end of the answer to another question I asked: https://math.stackexchange.com/a/4019598/155881.

$$F(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{1- \sqrt{1-4x}}{2x}\right)^n = \sum\limits_{k=0}^{\infty}{n+2k \choose k}x^k$$

Haven't been able to prove this. We know the expansions of both terms on the left side, making the identity:

$$\left(\sum\limits_{j=0}^{\infty} \left(\binom{n+2j-1}{j} - \binom{n+2j-1}{j-1}\right)x^j\right)
\left(\sum\limits_{l=0}^{\infty}{2l \choose l} x^l\right) = \sum\limits_{k=0}^{\infty}{2n+k \choose k}x^k$$

My attempt:
An obvious thing to do is to try a convolution of the left side and equate coefficients of $x^k$ (also using an alternate expression for the first summation in the product of the left side):

$$\sum\limits_{l=0}^k {n+2l-1\choose l}\frac{l}{n+l} {2k-2l\choose k-l} = {2n+k\choose k}$$

Not sure how to make progress on this. Expanding the binomial terms doesn't lead anywhere.


问:
Is there a geometric construction that finds the intersection of two conics, given only five points on each?
The question: Five points define a conic, so let points $A,B,C,D,E$ and $F,G,H,I,J$ define two conics $c$ and $d$.  Is there a geometric construction that identifies the common points of $c$ and $d$?

(It's easy enough to do it in Geogebra, or algebraically.  What I'm asking for here is a ruler and compass construction.)

Background and context  Given, say, an ellipse $c$ it is not too difficult to construct tangents from a point $P$ to $c$ using just a straightedge.  But this assumes that the ellipse has been drawn as a given.  If only the points $A,B,C,D,E$ on $c$ have been given, you can easily construct the polar of $P$ but then have to construct the intersection of the polar and the ellipse.  This is less straightforward but the construction is described in several 19th century projective geometry texts (e.g. Cremona or Russell),and requires a reference conic (or compass) in addition to a straightedge.

What I haven't been able to find is a construction for the intersection of two ellipses/conics that are not pre-drawn but given only as two sets of five points each.  Hence my question.  It came up in following a certain construction for "imaginary" chords that involved finding the common points of two conical loci (Russell, XXVII 6.).

Update: Given two or three points of intersection it is possible to construct the remaining points of intersection.

Update 2: This problem is discussed in Veblen and Young, Projective Geometry, Vol I, $\S102$.

答: As a simpler counterexample, consider the parabola $y=x^2$ and the circle $y^2=x(1-x)$.  Both have infinitely many constructible points and they intersect at the origin.  But the second point of intersection requires solving $x^4=x(1-x)$ with $x$ nonzero, leading to the irreducible equation $x^3+x-1=0$ whose real root cannot be (Euclidean) constructed.


答:
$||x_n -x|| \le 1$

$||x_n -x|| + ||x|| \le 1+||x|| $

$||(x_n -x)+x|| \le||x_n -x|| + ||x|| \le ||x||+1$

$||x_n||\le ||x||+1$

....

Negative signs always throw me.

But if you $a = m-n$, and $m = a+n$ you can always show the "flipside" to the triangle inequality

Corollary: $||m - n|| \ge ||m|| - ||n||$

Pf:  $||a +n|| \le ||a|| + ||n||$

$||m|| \le ||m-n|| + ||n||$.

$||m||-||n|| \le ||m-n||$.

That always throws me!

... but if if it didn't ....

$||x_n -x|| \le 1\implies$

$||x_n|| - ||x|| \le ||x_n - x|| \le 1\implies$

$||x_n|| \le 1 + ||x||$ would follow easily.


问:
Proving that either $2^n-1 $ or $ 2^n+1$ is not prime
Not sure what approach to take with this:

Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$

答:
Consider $2^n -1 , 2^n$ and $2^n+1$ we know one of them is multiple of 3 as far as $2^n$ cant be divided by 3 so one of $2^n-1$ or $2^n+1$ must divided by 3.

EDIT : for $n =1$ we have this sequence : $1,2,3$ and for $n=2$ we have this sequence : $3,4,5$. as you see we have a number in both series that divided by 3 but because this number is 3 and 3 is prime we must consider $n>2$.


问:
Stuck on Algebra for Summation Problem
I want to find what the exact value of this sum is. What I was given is: $$\sum_{i=1}^n \frac{1}{\sqrt{2i-1} + \sqrt{2i+1}}$$

The only thing I can think of is turning the denominator into the form:

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

I was wondering if I could get a hint on what to do next.

Title: Stuck on Algebra for Summation Problem

Body:
\begin{align*}
\sum_{i=1}^n \frac{1}{\sqrt{2i-1} +\sqrt{2i+1}}&=\frac12\sum_{i=1}^n\left(\sqrt{2i+1}-\sqrt{2i-1} \right)\\
&=\frac12(\sqrt{2n+1}-1).\\
\end{align*}


问:
Probability that a character kills the boss in a 3 versus 1 fight
In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.

Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

答:
I will give a method for obtaining a closed form answer, and will explain why it is impractical to give an "ultimate closed form" in terms of algebraic numbers.

Short answer: if the boss starts with $N$ HP, then the probability that A kills the boss is given by the following formula:

$$p_N = 0.2830 + 2Re((0.1896 + 0.1642i)\cdot (0.9937 - 0.0458i)^N) + 2Re((-0.0017 + 0.1517i)\cdot (0.9803 - 0.0807i)^N + \dots.$$

The above is a finite sum, but has more than one hundred terms. All the numbers appearing are algebraic numbers, approximated to the fourth decimal place.

The main term is $0.2830$ and all other terms converges exponentially to zero.

The calculation:

Let $x_A, y_A$ be the min/max damage of $A$. Also let $d_A$ denote $y_A - x_A + 1$.

Thus in the original question, we have $x_A = 25, y_A = 50, d_A = 26$.

The same notations apply to B and C.

For convenience, we also write $y = y_A + y_B + y_C$, i.e. the max damage that can be done in one round. It is $200$ in the original question.

Let $p_n$ be the probability that A kills the boss, provided that the boss starts with $n$ HP.

Viewing $(p_n)_n$ as a sequence, we may establish a recurrence relation.

For any $n > y$, the boss can definitely survive the first round. For any $a \in [x_A, y_A]$, $b\in [x_B, y_B]$, $c\in [x_C, y_C]$, there is a probability of $\frac 1 {d_A d_B d_C}$ that A, B, C deal damages $a, b, c$ in the first round, respectively. When this happens, the boss will remain $n - (a + b + c)$ HP, and in the rest of the game, A will kill the boss with probability $p_{n - (a + b + c)}$.

Therefore, for $n > y$ we get the following equation:
$$p_n = \sum_{a = x_A}^{y_A}\sum_{b = x_B}^{y_B}\sum_{c = x_C}^{y_C}\frac 1 {d_A d_B d_C}p_{n - (a + b + c)}.$$

This is a linear recurrence relation, and we may apply the standard solving method to it.

Thus we first look at its characteristic polynomial, which is given by:
\begin{eqnarray}
f(T) &=& T^y - \sum_{a, b, c} \frac 1{d_A d_B d_C} T^{y - (a + b + c)}\\
&=& T^y - \frac 1 {d_A d_B d_C}\sum_{a = x_A}^{y_A} T^{y_A - a}\sum_{b = x_B}^{y_B} T^{y_B - b}\sum_{c = x_C}^{y_C} T^{y_C - c}\\
&=& T^y - \frac 1 {d_A d_B d_C} (T^{d_A} - 1)(T^{d_B} - 1)(T^{d_C} - 1) / (T - 1)^3
\end{eqnarray}

If this polynomial only has simple roots $r_1, \dots, r_y$, then there exists $c_1, \dots, c_y$ such that $p_n = \sum_{i = 1}^y c_i r_i^n$ for all $n$.

It is easy to see that $1$ is always a root of $f$, as $(T^{d_A} - 1)/(T - 1)$ evaluates to $d_A$ at $T = 1$, and idem for B, C.

Although this polynomial looks simple, usually $T = 1$ is the only "nice" root, namely $f/(T - 1)$ is quite probably an irreducible polynomial over $\Bbb Q$, and hence the roots $r_i$ are just some algebraic number of very high degree).

From here, it seems inevitable to use some computer algebra system.

The following code can be pasted into this site to calculate the polynomial $f$.

```<code>xA, yA = 25, 50
xB, yB = 30, 70
xC, yC = 10, 80
dA, dB, dC = yA - xA + 1, yB - xB + 1, yC - xC + 1
y = yA + yB + yC

R.<T> = QQ[]
f = T^y - (T^dA - 1)*(T^dB - 1)*(T^dC - 1)/(T - 1)^3 / (dA * dB * dC)
print(f)
</code>```

The polynomial $f$ looks like $T^{200} - \frac 1 {75686} T^{135} - \frac 3{75686} T^{134} - \dots$.

To check that $f$ doesn't have double roots, we calculate the $\gcd$ of $f$ and its derivative:

```<code>print(gcd(f, f.derivative()))
</code>```

The result is $1$, showing that all the $200$ roots of $f$ are different.

We may also try to factorize $f$ over $\Bbb Q$:

```<code>print(f.factor())
</code>```

The result is $(T - 1)(T^{199} + T^{198} + \dots + \frac 2{37843}T + \frac 1{75686})$. The second factor is irreducible, which means that the ultimate "closed formula" answer to this question will have coefficients in some huge number field, of degree at least $199$ (and probably much, much larger, up to $199!$).

This complexity is the nature of the problem, and cannot be avoided.

Now let us try to find all roots of $f$ in $\Bbb C$. We also sort the roots in decreasing order of the norm.

```<code>CC = ComplexField(200)
S.<T> = CC[]
f = S(f)
roots = [r for [r, e] in f.roots()]
roots.sort(key = norm, reverse = True)
print(roots)
</code>```

The list starts with $1.0000, 0.9937 \pm 0.0458i, 0.9803 \pm 0.0807i, \dots$. Thus we see that $1$ is indeed the largest root. I now rename the roots as $r_0 = 1, r_1, r_2, \dots, r_{199}$ in the descending order.

It remains to obtain the coefficients $c_i$ from the initial values by solving a linear equation.

Here my interpretation is that the boss dies as soon as its HP is reduced to $\leq 0$.

```<code>p, q, r = [0], [1], [0]
for n in range(1, y):
    p.append(sum([q[n - j] if j < n else CC(1) for j in range(xA, yA + 1)]) / dA)
    q.append(sum([r[n - j] if j < n else CC(0) for j in range(xB, yB + 1)]) / dB)
    r.append(sum([p[n - j] if j < n else CC(0) for j in range(xC, yC + 1)]) / dC)
m = matrix([[rt^j for rt in roots] for j in range(y)])
c = m.solve_right(vector(p))
</code>```

The list $c_i$ starts with $0.2830, 0.1896 \pm 0.1642i, -0.0017 \pm 0.1517i, \dots$.

This gives our final "closed form" formula $p_n = \sum_i c_i r_i^n$, in complex numbers.

The final step is to put in $n = 1000$:

```<code>N = 1000
print(c.dot_product(vector([rt^N for rt in roots])))
</code>```

Result: $p_{1000} \approx 0.2834657535$.

Body:
I will give a method for obtaining a closed form answer, and will explain why it is impractical to give an "ultimate closed form" in terms of algebraic numbers.

Short answer: if the boss starts with $N$ HP, then the probability that A kills the boss is given by the following formula:

$$p_N = 0.2830 + 2Re((0.1896 + 0.1642i)\cdot (0.9937 - 0.0458i)^N) + 2Re((-0.0017 + 0.1517i)\cdot (0.9803 - 0.0807i)^N + \dots.$$

The above is a finite sum, but has more than one hundred terms. All the numbers appearing are algebraic numbers, approximated to the fourth decimal place.

The main term is $0.2830$ and all other terms converges exponentially to zero.

The calculation:

Let $x_A, y_A$ be the min/max damage of $A$. Also let $d_A$ denote $y_A - x_A + 1$.

Thus in the original question, we have $x_A = 25, y_A = 50, d_A = 26$.

The same notations apply to B and C.

For convenience, we also write $y = y_A + y_B + y_C$, i.e. the max damage that can be done in one round. It is $200$ in the original question.

Let $p_n$ be the probability that A kills the boss, provided that the boss starts with $n$ HP.

Viewing $(p_n)_n$ as a sequence, we may establish a recurrence relation.

For any $n > y$, the boss can definitely survive the first round. For any $a \in [x_A, y_A]$, $b\in [x_B, y_B]$, $c\in [x_C, y_C]$, there is a probability of $\frac 1 {d_A d_B d_C}$ that A, B, C deal damages $a, b, c$ in the first round, respectively. When this happens, the boss will remain $n - (a + b + c)$ HP, and in the rest of the game, A will kill the boss with probability $p_{n - (a + b + c)}$.

Therefore, for $n > y$ we get the following equation:
$$p_n = \sum_{a = x_A}^{y_A}\sum_{b = x_B}^{y_B}\sum_{c = x_C}^{y_C}\frac 1 {d_A d_B d_C}p_{n - (a + b + c)}.$$

This is a linear recurrence relation, and we may apply the standard solving method to it.

Thus we first look at its characteristic polynomial, which is given by:
\begin{eqnarray}
f(T) &=& T^y - \sum_{a, b, c} \frac 1{d_A d_B d_C} T^{y - (a + b + c)}\\
&=& T^y - \frac 1 {d_A d_B d_C}\sum_{a = x_A}^{y_A} T^{y_A - a}\sum_{b = x_B}^{y_B} T^{y_B - b}\sum_{c = x_C}^{y_C} T^{y_C - c}\\
&=& T^y - \frac 1 {d_A d_B d_C} (T^{d_A} - 1)(T^{d_B} - 1)(T^{d_C} - 1) / (T - 1)^3
\end{eqnarray}

If this polynomial only has simple roots $r_1, \dots, r_y$, then there exists $c_1, \dots, c_y$ such that $p_n = \sum_{i = 1}^y c_i r_i^n$ for all $n$.

It is easy to see that $1$ is always a root of $f$, as $(T^{d_A} - 1)/(T - 1)$ evaluates to $d_A$ at $T = 1$, and idem for B, C.

Although this polynomial looks simple, usually $T = 1$ is the only "nice" root, namely $f/(T - 1)$ is quite probably an irreducible polynomial over $\Bbb Q$, and hence the roots $r_i$ are just some algebraic number of very high degree).

From here, it seems inevitable to use some computer algebra system.

The following code can be pasted into this site to calculate the polynomial $f$.

xA, yA = 25, 50
xB, yB = 30, 70
xC, yC = 10, 80
dA, dB, dC = yA - xA + 1, yB - xB + 1, yC - xC + 1
y = yA + yB + yC

R.<T> = QQ[]
f = T^y - (T^dA - 1)*(T^dB - 1)*(T^dC - 1)/(T - 1)^3 / (dA * dB * dC)
print(f)

The polynomial $f$ looks like $T^{200} - \frac 1 {75686} T^{135} - \frac 3{75686} T^{134} - \dots$.

To check that $f$ doesn't have double roots, we calculate the $\gcd$ of $f$ and its derivative:

print(gcd(f, f.derivative()))

The result is $1$, showing that all the $200$ roots of $f$ are different.

We may also try to factorize $f$ over $\Bbb Q$:

print(f.factor())

The result is $(T - 1)(T^{199} + T^{198} + \dots + \frac 2{37843}T + \frac 1{75686})$. The second factor is irreducible, which means that the ultimate "closed formula" answer to this question will have coefficients in some huge number field, of degree at least $199$ (and probably much, much larger, up to $199!$).

This complexity is the nature of the problem, and cannot be avoided.

Now let us try to find all roots of $f$ in $\Bbb C$. We also sort the roots in decreasing order of the norm.

CC = ComplexField(200)
S.<T> = CC[]
f = S(f)
roots = [r for [r, e] in f.roots()]
roots.sort(key = norm, reverse = True)
print(roots)

The list starts with $1.0000, 0.9937 \pm 0.0458i, 0.9803 \pm 0.0807i, \dots$. Thus we see that $1$ is indeed the largest root. I now rename the roots as $r_0 = 1, r_1, r_2, \dots, r_{199}$ in the descending order.

It remains to obtain the coefficients $c_i$ from the initial values by solving a linear equation.

Here my interpretation is that the boss dies as soon as its HP is reduced to $\leq 0$.

p, q, r = [0], [1], [0]
for n in range(1, y):
    p.append(sum([q[n - j] if j < n else CC(1) for j in range(xA, yA + 1)]) / dA)
    q.append(sum([r[n - j] if j < n else CC(0) for j in range(xB, yB + 1)]) / dB)
    r.append(sum([p[n - j] if j < n else CC(0) for j in range(xC, yC + 1)]) / dC)
m = matrix([[rt^j for rt in roots] for j in range(y)])
c = m.solve_right(vector(p))

The list $c_i$ starts with $0.2830, 0.1896 \pm 0.1642i, -0.0017 \pm 0.1517i, \dots$.

This gives our final "closed form" formula $p_n = \sum_i c_i r_i^n$, in complex numbers.

The final step is to put in $n = 1000$:

N = 1000
print(c.dot_product(vector([rt^N for rt in roots])))

Result: $p_{1000} \approx 0.2834657535$.


问:
Integrand is sharply peaked
I have a gamma function,
$$n!=\int_0^\infty e^{-x}x^n \mathrm{d}x$$ and if I take the derivative of the logarithm of the integrand, I get $$\dfrac{\mathrm{d}}{\mathrm{dx}}(n\ln x -x)=\dfrac{n}{x}-1.$$  But why I can argument that the integrand is sharply peaked with the contribution important only near $x=n$?

答:
The gamma function is
$$ \Gamma(n+1) = \int_0^\infty dz e^{-z}z^n.$$
Rewrite the integrand as
$$ \Gamma(n+1) = \int_0^\infty dz \exp\big[-z + n \ln z\big].$$
The integrand goes to $0$ as $z\rightarrow 0$ or $z \rightarrow \infty. $ Between these extremes, the function somewhere finds a maximum. When $n$ is large, the decay toward $0$ on either side of this maximum becomes very rapid (as you could see by analyzing the first derivative). Therefore, for large $n$, the area very near the integrand's maximum becomes the primary contribution to the integral.

This maximum occurs where
$$ 0 = \frac{d}{dz}\exp\big[-z + n \ln z\big] = \big(-1 + \frac{n}{z}\big)
\exp \big[-z + n \ln z\big],$$
which means $z=n$. Typically one would now Taylor expand the argument of the exponential around $z=n$ to second order and then integrate in what's called a Saddle-Point Expansion. This yields the so-called Stirling's Formula for $n!$.

Title: Integrand is sharply peaked


问:
Why are hyperbolic trigonometric functions avoided in (my) high school and early post-secondary school?
I remember seeing hyperbolic trigonometric functions (sinh, cosh, tanh, etc.) in my precalculus textbook back in high school and see them today in my calculus textbook. However, I have not had a formal introduction to them in my education. This may just be with me, but I would like to know if there is a reason as to why they are avoided.</p>

答: One reason may be that the input of hyperbolic functions is often interpreted as the area partially bounded by a hyperbolic curve, while regular trigonometric functions take an angle as an input. Even though hyperbolic functions have lots of nice properties, teachers may not see them as as important as regular trigonometric functions, since the area is useful only in specific scenarios, for example bridge construction.</p>


问:
Using induction on a word-concatenating function

Let $\rho = \{ \cdot \}$ be a signature with a binary functional symbol. We define the $\rho$-structure  $\mathcal{S} := ( S , \cdot^{\mathcal{S}} )$, where $S$ is the set of finite words over a finite alphabet $\Sigma$, such that $S = \Sigma^*$, and $\cdot^{\mathcal{S}}$ the normal concatenatination of words, $x \cdot^{\mathcal{S}} y = xy \in S$, for all $x,y \in S$. Also, let $h: \Sigma \rightarrow \Sigma$ be any bijection.

Show using induction, that the function $f: S \rightarrow S$ is an automorphism:

$$f(w) :=\begin{cases}
       h(a_1) \cdot^{\mathcal{S}} f(a_2 \ldots a_n) & \text{if } w = a_1a_2 \ldots a_n , \text{ with } a_i \in \Sigma , \text{ for all } i \in \{ 1, \ldots , n \} , \\
       \lambda & \text{if } w = \lambda .
    \end{cases}$$

Where $\lambda$ is the empty word.

I'm practically lost and don't know where to start. What I understand:

Since $f$  has to be a automorphism, it also must be an endomorphism and an isomorphism. Being an endomorphism is given, since it maps $S$ to itself.

To show that $f$ is an isomorphism, we need to show that it is injective and surjective.

I'm not sure how to do this with induction and wasn't even able to come up with the induction hypothesis. Since we're trying to prove a property of a function, how could we use $P(n)$ and $P(n+1)$ (in this case $P(w)$ and $P(w \cdot v)$)? I'm pretty sure I'm looking at this the wrong way and would really appreciate any hints.

答:
Welcome to MSE!

In order to work with induction, we need something to induct on. You've correctly identified that one way to do this might be to look at the length of your words. After all, you can think of a word as being either empty (of length $0$) or a shorter word with a new character at the end (so it has length $n+1$). If you're familiar with structural induction we can actually use this to induct directly on the word itself, but for now let's work with the length.

Ok. So we want to show that $f : \Sigma^* \to \Sigma^*$ as defined in your problem is an isomorphism. You're correct that this comes down to showing it is an endomorphism and a bijection, but you've missed an important detail about endomorphism-ness. It's not enough to check that it maps $\Sigma^*$ to $\Sigma^*$! We also need to check that it's a homomorphism. That is, we need to check that $f(u \cdot v) = f(u) \cdot f(v)$.

So, how do we verify these properties? Well, let's see if there are any particularly easy cases to verify... We've been told to think about induction, so why not try it with words of length $0$? That seems like a reasonable base case to me.

Since $\lambda$ is the unique word of length $0$, we want to show that

(homomorphism, base case) $f(\lambda \cdot w) = f(\lambda) \cdot f(w)$ for any word $w$

(bijection, base case) $f \upharpoonright_{\Sigma^0} : \Sigma^0 \to \Sigma^0$ is a bijection

Here we've written $\Sigma^0$ for the words of length $0$. More generally, to check that $f$ is a bijection on all of $\Sigma^*$, it will be enough to show that $f$ is a bijection on our words of length $n$ for each $n$ (do you see why?).

So, let's verify these claims. These are each a simple computation:

$$
f(\lambda \cdot w) \overset{(1)}{=}
f(w) \overset{(2)}{=}
\lambda \cdot f(w) \overset{(3)}{=}
f(\lambda) \cdot f(w)
$$

In step $(1)$ we used the definition of $\cdot$. In step $(2)$ we used the definition of $\cdot$ again (but on the outside this time). In step $(3)$ we use the definition of $f$.

Next we want to check bijectivity on words of length $0$. But since $\lambda$ is the only word of length $0$, and we know that $f(\lambda) = \lambda$ we're done! It's a fairly simple bijection, but it's a bijection nonetheless.

Now it's time for the inductive step. We can now assume (as our Inductive Hypothesis) that

(homomorphism, IH) for words $u$ of length $n$, $f(u \cdot w) = f(u) \cdot f(w)$ for all words $w$.

(bijection, IH) $f \upharpoonright_{\Sigma^n} : \Sigma^n \to \Sigma^n$ is a bijection on words of length $n$.

We want to show that these results continue to hold for words of length $n+1$.

Let's prove bijection together. I'll leave homomorphism to you.

To show $f \upharpoonright_{\Sigma^{n+1}}$ is a bijection, we want to show it is injective and surjective. We might as well start with injective.

Let $au$ and $bv$ be two words of length $n+1$. Here $a, b \in \Sigma$ are single letters, and $u, v \in \Sigma^n$ are the rest of the word. To show injectivity, we want to show that if $f(au) = f(bv)$ then actually $au = bv$.

Now if $f(au) = f(bv)$, what must we know? Well:

$f(au) = h(a) \cdot f(u)$ by the definition of $f$

$f(bv) = h(b) \cdot f(v)$ for the same reason

So $h(a) \cdot f(u) = h(b) \cdot f(v)$. What does this buy us? Well, if two words are the same, then they had better have the same first letter! So $h(a) = h(b)$.

But since $h$ is a bijection, this says $a=b$!

Moreover, if the two words are the same, then the rest of the characters had better be the same too! So $f(u) = f(v)$, and now by applying our inductive hypothesis, we see that $u=v$.

So $a=b$ and $u=v$... Now we're done! That means $au=bv$, and so $f$ is injective on words of length $n+1$.

Surjectivity is similar, so I'll go a bit more quickly through it. Let's let $bv$ be my favorite word of length $n+1$. We want to show it's in the image of $f$.

Well, we know $h$ is a bijection, so we can find a letter $a$ with $h(a) = b$. Similarly, by induction, we know that $f$ is a bijection on words of length $n$, so we can find a word $u$ with $f(u) = v$.

Putting these pieces together gives $f(au) = bv$ (do you see why?) which shows $f$ is also surjective on words of length $n+1$.

Can you use these same ideas to show that $f$ is a homomorphism?

I hope this helps ^_^


问:
Confused about partial derivatives
I am having some issues understanding what should I keep constant and what not in certain cases when I take partial derivatives. Specifically in this kind of situation: say we have a function

$$f(x,y) = x^3+7y^2$$

and we also know that $y=2x+1$ and we need to find the partial derivative with respect to $x$ at a given point, say $\frac{\partial f(2,3)}{\partial x}$. From what I understood, when taking partial derivatives with respect to a variable, you need to keep the other constant, in which case, if I do that above (keeping $y$ constant) I would get, $\frac{\partial f(x,y)}{\partial x}=3x^2$, so I get $12$. However, if I plug $x$ in $y$ explicitly I would get $$f(x,y) = x^3+7(2x+1)^2$$ so,

$$\frac{\partial f(x,y)}{\partial x}=3x^2+28(2x+1)$$

So, I get $152$. What should I do? Thank you!

答:
It's definitely important to make some distinctions with notation. Given a function ${f(x,y)}$, then ${\frac{\partial f}{\partial x}}$ means "treating $y$ as a constant, what is the derivative of $f$ with respect to ${x}$?" <- this does not, on it's own, necessary tell you about how $f$ changes overall with respect to $x$. In order to know this, we must use the multi-variable chain rule, which is:
$$
\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}
$$
notice if ${\frac{dy}{dx}=0}$, i.e. ${x}$ and ${y}$ are independent, then ${\frac{df}{dx}=\frac{\partial f}{\partial x}}$ (i.e. the partial derivative tells us all we need to know about how $f$ changes with respect to $x$).

Title: Confused about partial derivatives


问:
Let f(n) be the number of n step paths from w to itself on the directed graph below. What is f(12)?
I'm stumped on a combinatorics problem and I'd just like some help starting out! The question I have is:

Let f(n) be the number of n step paths from w to itself on the directed graph below. What is f(12)?

I have no clue where to start, so any guidance would be greatly helpful.

Thank you!

答:
HINT: There are two loops, one of length $2$ and one of length $3$. You can take the longer one $0,2$, or $4$ times, but not once or $3$ times. (Why?) The only case that requires a bit more work is the second one: you take the longer loop twice. How many times must you then take the shorter loop? In how many different orders can you go round the loops?

If you will eventually have to come up with more results about the function $f$, it would be a good idea to work out $f(n)$ for $n\le 12$ using similar ideas. By then you should have a much better idea of how it behaves in general.


问:
$n$ people $n$ hats: probability that at most $m$ $(m\leqslant n)$ people get their own hats!
This is a variant of the popular $Hat$-problem. Say there are $n$ people and $n$ hats. Then what is the probability that at most $m$ $(m\leqslant n)$ people get their own hats!

My solution is $\frac{\sum\limits_{k=0}^{m} {n \choose k}D_{n-k}}{n!}$, where $D_{n-k}$ denotes number of derangements of a set with $n-k$ elements.

The argument is that $k$ people, who get their own hats, can be chosen from $n$ people in $n\choose k$ ways and there is only $one$ way that they get their own hats. None of the rest $n-k$ people get their own hats. And hence the derangement factor $D_{n-k}$.

Is my argument valid?

答:
Yes, that is correct.

The ${n \choose k}D_{n-k}$ terms  (the number of ways $k$ people out of $n$ get their own hats) are called rencontres numbers

For given $m$, the probabilities quickly approach a limit, as $n$ increases, of $\frac1e\sum\limits_{k=0}^{m} \frac1{k!}$, a consequence of $\frac{D_n}{n!} \to e^{-1}$.


问:
Can I use Weinstein–Aronszajn identity to show that the eigenvalues of AB and BA are the same?
I have the matrices $A \in \mathbb{K}^{n \times m}$ and $B\in \mathbb{K}^{m \times n}$ .

Can I use the Weinstein–Aronszajn identity to show that $AB$ and $BA$ have the same non-zero eigenvalues? If so, how would I accomplish this? If not, what would be a good way to show my problem?

答:
Yes, we can use Weinstein-Aronszajn identity $\det (I_n - AB) = \det (I_m - BA)$ to show that non-zero eigenvalues of $AB$ and $BA$ coincide.

Proof Let $\lambda$ be a non-zero eigenvalue of $AB$. This means that $\det (\lambda I_n - AB) = 0$, however

$$
\begin{align}
\det (\lambda I_n - AB) &= \lambda^n\det\left(I_n - \left(\frac{1}{\lambda}A\right)B\right) \\
&= \lambda^n\det\left(I_m - B\left(\frac{1}{\lambda}A\right)\right) \\
&= \lambda^{n-m}\det\left(\lambda I_m - BA\right)
\end{align}
$$

and we see that $\det (\lambda I_m - BA) = 0$ which means that $\lambda$ is a non-zero eigenvalue of $BA$. The converse follows by the analogous argument.


问:
$\sum_{i=1}^n x_i \ln(\bar{x})$ vs. $\sum_{i=1}^n x_i \ln(x_i)$
For a set $x = \{x_1, ..., x_n\}$, I need to show that

$$\sum_{i=1}^n x_i \ln(\bar{x}) \leq \sum_{i=1}^n x_i \ln(x_i),$$

where $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the arithmetic mean.

Intuitively, this makes sense. But how do you prove/show it formally?

答:
I am assuming that $x_i\geq 0$ for all $i$, with the convention that $0\log 0=0$, so that the question is well-defined.

Consider the function $f\colon x\in[0,\infty) \mapsto x\ln x$. You can check it is convex.
Then, apply Jensen's inequality to the sum
$$
\frac{1}{n}\sum_{i=1}^n x_i \ln x_i = \frac{1}{n}\sum_{i=1}^n f(x_i)
$$
to get
$$
f\!\left(\frac{1}{n}\sum_{i=1}^nx_i \right) \leq \frac{1}{n}\sum_{i=1}^n f(x_i)
$$


问:
How to find $F=\left(\frac{B}{2}\right)^{2H}-1$ when $H$ is maximum of $W_{1}$ and $B$ is minimum of $W_{2}$?
The problem is as follows:

A pair of wavelenghts are $W_{1}$ and $W_{2}$:

$W_{1}=5\sin\alpha-5+12\cos\alpha$

$W_{2}=3\sin\beta+\sqrt{3}\cos\beta$

Assuming $H$ and $B$ represent the maximum and the minimum value for
$W_{1}$ and $W_{2}$ respectively.

Find:

$F=\left(\frac{B}{2}\right)^{2H}-1$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{6560}\\
2.&\textrm{5860}\\
3.&\textrm{6562}\\
4.&\textrm{5680}\\
\end{array}$

I'm confused exactly how to tackle this problem from my precalculus workbook. What I attempted to do was to use the boundaries of the sine and the cosine function individually to see if I could obtain the maximum and the minimum values:

$-1<\sin\alpha<1$

$-10<5\sin\alpha-5<0$

Then for $\cos\alpha$:

$-1<\cos\alpha<1$

$-12<12\cos\alpha<12$

Then adding both:

$-22<5\sin\alpha-5+12\cos\alpha<12$

Thus the maximum value for the earlier function would be $12$.

Whereas for the other would be:

$-1<\sin\beta<1$

$-3<3\sin\beta<3$

and for the other term:

$-\sqrt{3}<\sqrt{3}\cos\beta<\sqrt{3}$

Then when adding both:

$-3-\sqrt{3}<3\sin\beta+\sqrt{3}\cos\beta<3+\sqrt{3}$

Then the minimum value would be for the function would be $-3-\sqrt{3}$.

Hence:

$B=12$ and $H=-3-\sqrt{3}$

But the problemas has not ended yet, they are asking the result of $F$

Thus:

$F=\left(\frac{-3-\sqrt{3}}{2}\right)^{2(2)}-1$

Then this is where I got stuck. What could possibly be wrong here?. Which part went wrong?. Can someone help me here?. Could it be that the problem lies in the method to get the minimum and the maximum value for those functions?. The sort of approach which I'm looking to get is derivative-free in other words without using derivatives as this problem was intended to be solved relying only using precalculus.

Can someone help me please?. I'd appreciate an answer with an explanation step-by-step. Thanks in advance.

答:
Let's try to make $a\sin x+ b\cos x$ of the form $\boxed{r\sin(x+\theta)}$  and $r>0$

$\begin{align}&5\sin\alpha+12 \cos\alpha = r(\cos\theta\sin\alpha+\sin\theta\cos\alpha) = r\sin(\alpha+\theta) \\&\Rightarrow r^2 = 5^2+12^2 = 13^2 \\&\Rightarrow r = 13 \text{ and  }  \theta =\arctan\frac{12}{5}  \end{align} $

Now,

$W_1= 5\sin\alpha+12 \cos\alpha - 5 = 13\sin(\alpha+\theta) - 5$

Max. value $13\sin(\alpha+\theta)$ can attain is $13$.

So max. value of $W_1$ is $\boxed{H = W_{1_{\max}} = 13-5 = 8}$

Similarly

$\begin{align}&3\sin\beta+ \sqrt3 \cos\beta = r\sin(\beta+\theta') = r(\sin\beta\cos\theta'+\cos\beta\sin\theta') \\\Rightarrow &r = \sqrt{9+3} = 2\sqrt3\text{ and }\theta' = \arctan\frac1{\sqrt3}\end{align} $

So,

$W_2 = 2\sqrt3\sin(\beta+\theta')$

Min. value of $\sin(\beta+\theta') = -1$   and min. value of $W_2$ is $\boxed{B = W_{2_{\min}} = 2\sqrt3 (-1) =-2\sqrt3}$

So, $$F = \left(\frac{-2\sqrt3}{2}\right)^{2(8)}-1 = 3^8-1 = 6560$$


问:
Eigen values of A if principal minors are all positive .
In my book, there is a statement which says "If all the principal minors of A are positive , all the eigen values of A need not be positive ". Here A is a square matrix.

I think if principal minors are all positive then sum of eigen values which is the trace is positive, product of eigen values is positive and also A(i,i) which is element at ith row and ith column is also positive. So, how can any eigen value be negative ?

答:
Having all principal minors positive is the definition of a P-matrix. Hence, while in general the eigenvalues are not positive, the do have localization away from the negative real axis. If $\lambda \in \mathbb{C}$ is an eigenvalue of $A$, then for $n > 1$ we have

$$|\arg{\lambda}| < \pi (1 - 1 / n).$$


Title: Proving that either $2^n-1 $ or $ 2^n+1$ is not prime

Body:
Not sure what approach to take with this:

Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$

Body: $$(4-1)|4^n-1=(2^n-1)(2^n+1)$$ so $3|(2^n-1)$ or $3|(2^n+1)$ since $3$ is prime. Also $(2^n+1)>(2^n-1)>3$.


问:
Let $q(x)=1+x+x^2+...+x^n$. Prove that if $p_1(x),..,p_n(x)$ is any basis for $W$, then $p_1(x),...,p_n(x),q(x)$ is a basis for $P_n(\mathbb{R})$
Let $W$ be a subspace of $P_n(\mathbb{R})$ be defined by $W = \{p(x) \in P_n (\mathbb{R}) |p(1)=0 \}$. Let $q(x)=1+x+x^2+...+x^n$. Prove that if $p_1(x),..,p_n(x)$ is any basis for $W$, then $p_1(x),...,p_n(x),q(x)$ is a basis for $P_n(\mathbb{R})$

My Attempt

First, construct a basis for $W$: $\{(x-1),(x^2-1),...,(x^n-1)\}$. Since both are basis, for $a_1...,a_n$ there is a  $b_1,...,b_n$ in $\mathbb{R}$:

$$a_1(x-1) + ... + a_n(x^n-1)=b_1p_1(x)+...+b_np_n(x)$$

Now I need to show that $p_1(x),...,p_n(x),q(x)$ is both spanning and linearly independent. Let $k \in \mathbb{R}$:

$$b_1p_1(x) + ... + b_np_n(x) + kq(x) = a_1(x-1)+ ... + a_n(x^n-1) + k(1+x+...+x^n) \\
= -a_1 -....-a_n + k + x(a_1+k) + ... + x^n(a_n + k)$$

Hence it must be spanning. Also, the only solution to $= \vec{0}$ is $a_1= ... =a_n=k=0$. Hence is a basis.

Is this a valid proof?

$$$$

答:
Expanding on what I wrote in the comments, sometimes you don't need to show both span and linear independence. In this case, we dont need to show span.

You already noted that $W$ is $n$-dimensional since you produced the basis $(x-1),\ldots,(x^n-1)$. Now given a basis $p_1(x),\ldots,p_n(x)$ for $W$, if we add one more linearly independent vector, we will get a basis for an $n+1$-dimensional subspace of $P_n$. But $P_n$ is $n+1$-dimensional, so we will obtain all of $P_n$. Thus if we can show $q(x)$ is linearly independent from $p_1(x),\ldots,p_n(x)$, we will be done. How do we do this? Well suppose $q(x)$ is a linear combination of the $p_1(x),\ldots,p_n(x)$. This will imply that $q(1)=0$. But obviously $q(1)>0$, a contradiction. Hence the collection $p_1(x),\ldots,p_n(x),q(x)$ is linearly independent, as desired.


问:
linear algebra 1, linear applications
Given the endomorphism f defined by its associated matrix A, A=((a,-3),(-2,-2)) relative to the canonical base R^2. Find the variable a so that f admets a line of fixed points.

Solution is a=7, line is centered at the origin and is directed by the vector (1,2).

Normally, if a was given I could solve this problem- A*x=x - but as there are two variables I don't know how to go about finding a.

答:
Try Eigenvalues and look for a such that eigenvalue ==1

```<code>ev = Eigenvalues[{{a, -3}, {-2, 2}} ]
(*{1/2 (2 + a - Sqrt[28 - 4 a + a^2]),1/2 (2 + a + Sqrt[28 - 4 a + a^2])}*)

Solve[ev[[1]] == 1, a] (*{a->7}*)
Solve[ev[[2]] == 1, a] (*{}*)
</code>```

Only the first eigenvector corresponding to a=7

```<code>Eigenvectors[{{a, -3}, {-2, 2}} ][[1]]/.a->7
(* {1/2,1}*)
</code>```

defines the line of fixpoints

```<code>ParametricPlot[ {1/2,1} t,{t,-1,1}]
</code>```

OwnerDisplayName: reema alhamdan

Title: linear algebra 1, linear applications

Body:
Try Eigenvalues and look for a such that eigenvalue ==1

ev = Eigenvalues[{{a, -3}, {-2, 2}} ]
(*{1/2 (2 + a - Sqrt[28 - 4 a + a^2]),1/2 (2 + a + Sqrt[28 - 4 a + a^2])}*)

Solve[ev[[1]] == 1, a] (*{a->7}*)
Solve[ev[[2]] == 1, a] (*{}*)

Only the first eigenvector corresponding to a=7

Eigenvectors[{{a, -3}, {-2, 2}} ][[1]]/.a->7
(* {1/2,1}*)

defines the line of fixpoints

ParametricPlot[ {1/2,1} t,{t,-1,1}]

OwnerDisplayName: Ulrich Neumann


答:
We are looking for a line, consisting of points (vectors):v that are invariant under multiplication with m: v== m.v. Vectors that are mapped into a mutliple (called Eigenvalue): mul of themself by m are called eigenvectors: m.vec== mul vec. In our case, mul must be: mul ==1. Therefore we are looking for an eigenvector with eigenvalue 1.

Eigenvalues can be calulated by: Det[m- eigenvalue IdentitytMatrix]==0, or with the built in function Eigenvalues. We then determine a so that the eigenvalues is: 1

```<code>m = {{a, -3}, {-2, 2}};
ev1 = Solve[Eigenvalues[m][[1]] == 1, a][[1]]
(*  {a -> 7} *)
</code>```

Setting this value for a in m, we calculate the belonging eigenvectors by:

```<code>m = m /. ev1
Eigenvectors[m]
(* {{-3, 1}, {1, 2}}  *)
</code>```

We see that there are 2 invariant lines, namely: {x,y}== lam {-3,1} and {x,y}== lam {1,2}

Title: linear algebra 1, linear applications

Body:
Given the endomorphism f defined by its associated matrix A, A=((a,-3),(-2,-2)) relative to the canonical base R^2. Find the variable a so that f admets a line of fixed points.

Solution is a=7, line is centered at the origin and is directed by the vector (1,2).

Normally, if a was given I could solve this problem- A*x=x - but as there are two variables I don't know how to go about finding a.

Body:
We are looking for a line, consisting of points (vectors):v that are invariant under multiplication with m: v== m.v. Vectors that are mapped into a mutliple (called Eigenvalue): mul of themself by m are called eigenvectors: m.vec== mul vec. In our case, mul must be: mul ==1. Therefore we are looking for an eigenvector with eigenvalue 1.

Eigenvalues can be calulated by: Det[m- eigenvalue IdentitytMatrix]==0, or with the built in function Eigenvalues. We then determine a so that the eigenvalues is: 1

m = {{a, -3}, {-2, 2}};
ev1 = Solve[Eigenvalues[m][[1]] == 1, a][[1]]
(*  {a -> 7} *)

Setting this value for a in m, we calculate the belonging eigenvectors by:

m = m /. ev1
Eigenvectors[m]
(* {{-3, 1}, {1, 2}}  *)

We see that there are 2 invariant lines, namely: {x,y}== lam {-3,1} and {x,y}== lam {1,2}

OwnerDisplayName: Daniel Huber


答:
The problem as described quite nicely in the OP, ignoring linear algebra, seems a job for SolveAlways.  But we can't completely ignore algebra and need to exclude the singularity at $x=0$ before $x$ can be eliminated by the software without eliminating everything.

```<code>amat = {{a, -3}, {-2, 2}};
x = {x1, x2};

Solve[{amat . x == x, x != 0}, a, x]
(*  {{a -> 7}}  *)
</code>```

Besides, if this is a homework problem, the intention of Mathematica is to take all the drudgery out of having deal with such boring and complicated topics as eigenvectors.  In due fairness to the problem, it's a simple Algebra I/II problem in which the elimination of, say, x2 from the system {a x1 - 3 x2 == x1, -2 x1 + 2 x2 == x2} can practically be done in one's head.  Or for those who shy away from such drudgery (adjust the commented code to suit your personal level of abhorrence):

```<code>Eliminate[{a x1 - 3 x2 == x1, -2 x1 + 2 x2 == x2}, {x2}] (* // Solve *)
(*  a x1 == 7 x1  *)
</code>```

Title: linear algebra 1, linear applications

Body:
The problem as described quite nicely in the OP, ignoring linear algebra, seems a job for SolveAlways.  But we can't completely ignore algebra and need to exclude the singularity at $x=0$ before $x$ can be eliminated by the software without eliminating everything.

amat = {{a, -3}, {-2, 2}};
x = {x1, x2};

Solve[{amat . x == x, x != 0}, a, x]
(*  {{a -> 7}}  *)

Besides, if this is a homework problem, the intention of Mathematica is to take all the drudgery out of having deal with such boring and complicated topics as eigenvectors.  In due fairness to the problem, it's a simple Algebra I/II problem in which the elimination of, say, x2 from the system {a x1 - 3 x2 == x1, -2 x1 + 2 x2 == x2} can practically be done in one's head.  Or for those who shy away from such drudgery (adjust the commented code to suit your personal level of abhorrence):

Eliminate[{a x1 - 3 x2 == x1, -2 x1 + 2 x2 == x2}, {x2}] (* // Solve *)
(*  a x1 == 7 x1  *)

OwnerDisplayName: Michael E2


答:
As $n>2,2^n-1>3$

If $n$ is even, $=2m$(say),
 $2^n-1=2^{2m}-1=4^m-1$ which is divisible by $4-1=3$ as $(a-b)$ divides $a^n-b^n$ for integer $n\ge 0$

If $n$ is odd, $=2m+1$(say),
$2^{2m+1}+1=2^{2m+1}+1^{2m+1}$ which is divisible by $2+1=3$  as $(a+b)$ divides $a^{2m+1}+b^{2m+1}$ for integer $m\ge 0$

Alternatively,

$(2^n-1)(2^n+1)=4^n-1$ is divisible by $4-1=3,$

but $2^n+1>2^n-1>3$

References :  Fermat Prime 1,2,  Mersenne Prime1,2


问:
$S_1, S_2,...,S_k$ are subsets of $V$ such that, prove that $V = \operatorname{span}\{S_1\} + \dotsb + \operatorname{span}\{S_k\}$.

If $x_1,...,x_k$ is a basis for a vector space $V$, and $S_1, S_2,...,S_k$ are subsets of $V$ such that $x_i \in S_i$ for all $i$, prove that $V = \operatorname{span}\{S_1\} + \operatorname{span}\{S_2\} +\dotsb + \operatorname{span}\{S_k\}$.

My Attempt

Since $x_i\in \operatorname{span}\{S_i\}$, for any $c \in \mathbb{R}$ then $cx_i \in \operatorname{span}\{S_i\}$. This makes AT MINIMAL $\operatorname{span}\{x_i,...,x_k\} = V \subseteq \operatorname{span}\{S_1\} + \operatorname{span}\{S_2\} +\dotsb + \operatorname{span}\{S_k\}$. Furthermore, it is clear that $\operatorname{span}\{S_1\} + \operatorname{span}\{S_2\} +\dotsb + \operatorname{span}\{S_k\} \subseteq V$

QED

答: In my calculation $A+B=\{a+b | a\in A, b\in B\}$. Let $v \in V$, then $v=\sum_{i=1}^k c_ix_i$, for some scalers $c_i$'s. By definition $c_ix_i \in span\{S_i\}$. So $v=\sum_{i=1}^k c_ix_i \in \sum_{i=1}^k span\{S_i\}\Rightarrow V \subseteq \sum_{i=1}^k span\{S_i\}$, the other side is obvious. Combining we get the desired result.


问:
Confusion on notation in the definition of reflexivity.
This is a very silly question i assume to most of you but how would I read this

$\forall x \in X : (x,x) \in R$

I don't quite know what the colon indicates and this definition is used to define reflexivity .

Thanks in advance.

答: It is a hold over of the dot notation used in Principia Mathematica, which has mainly been replaced by bracketing.   It is basically just a punctuation mark to visually separate the quantifier's restriction from its scope. $$\forall x\in S:P(x)\\\forall x\in S~\big(P(x)\big)\\\forall x\in S~P(x)$$


问:
Show that X and Y given Z are not conditionally independent.
Given a fair coin and independently flip it three times.

\begin{equation}
  X =
    \begin{cases}
      1\;\text{if the first flip is H}\\
      0\;\text{if the first flip is T}\\
    \end{cases},\;\;\;
Y =
    \begin{cases}
      1\;\text{if the second flip is H}\\
      0\;\text{if the second flip is T}\\
    \end{cases}
\end{equation}

\begin{equation}
 Z=
    \begin{cases}
      0\;\text{if there is no heads in the first two flips}\\
      1\;\text{if there is exactly one head in the first two flips}\\
2\;\text{if there is exactly two heads in the first two flips}
    \end{cases}
\end{equation}

Show that X and Y are not conditionally independent given
Z.

My attempt:

\begin{align*}
            \mathbb{P}(X=x\;|\;Z=z)&=\mathbb{P}(X=0\;|\;Z=0)+\mathbb{P}(X=0\;|\;Z=1)+\mathbb{P}(X=0\;|\;Z=2)\\
            &+\mathbb{P}(X=1\;|\;Z=0)+\mathbb{P}(X=1\;|\;Z=1)+\mathbb{P}(X=1\;|\;Z=2)\\
            &=1+\frac{1}{2}+0+0+\frac{1}{2}+1=3
        \end{align*}

Same process for Y and Z. My concern is that the probability for $\mathbb{P}(X=x\;|\;Z=z)$ is greater than $1$, which shouldn't be the case. Can someone point out what I did wrong in this case?

After I figure out these probabilities, I can show that $$\mathbb{P}(X=x,\;Y=y\;|\;Z=z)\neq\mathbb{P}(X=x\;|\;Z=z)\mathbb{P}(Y=y\;|\;Z=z)$$
to show that they are not independent.

答:
First to point out something you did. Notice your sum was 3. This suggests that there are 3 complete, well, probability spaces in there. Indeed, you can observe that $P(X=0|Z=0)+P(X=1|Z=0)=1+0=1$, $P(X=0|Z=1)+P(X=1|Z=1)=\frac 1 2+\frac 1 2=1$, and $P(X=0|Z=2)+P(X=1|Z=2)=0+1=1$. Look at it for a bit and you will see that conditioned on an event, say Z=0, you range over all the possible values of X, and they sum to 1. This happens for each possible value of Z.

To show that X and Y are not conditionally independent given Z, it suffices to find one instance such that $P(X=x,Y=y|Z=z)\ne P(X=x|Z=z)P(Y=y|Z=z)$. Let's choose $$P(X=1,Y=1|Z=1)=0$$

compared with

$$P(X=1|Z=1)P(Y=1|Z=1)=\frac 1 2\cdot \frac 1 2=\frac 1 4$$

So X and Y are not independent given Z.

Notice that if you had tried $P(X=0,Y=1|Z=0)=0,P(X=0|Z=0)P(Y=1|Z=0)=1*0=0$ you would've found equivalent evaluations of the expressions. But you can't conclude that X and Y are conditionally independent given Z because you need this to hold over all possible combinations of X=x, Y=y, Z=z.


问:
Xn binomial limit to infinity
Let $X_n$ be random variables with

$X_n \sim \mathcal{B}(n, 2/n)$

Find $\lim_{n \to\infty }(P(X_n=2))$

I don't know where to start, any help is appreciated.

答:
HINT

Please write down an explicit form for $f(n) = \mathbb{P}[X_n=2]$.

Then you can compute explicitly what is $\lim_{n \to \infty} f(n)$.

Title: Xn binomial limit to infinity


问:
Regression error function minimization problem

Could somebody please help with this problem. There is a little explanation in this link:
meaning of some lines in a book

But it is still a bit ambiguous to me.

答:
HINT

For easier notation, let $J(u,v) = \frac1n \sum_{i=1}^n (y_i-u-vx_i)^2$ and let $(u^*,v^*)$ denote the values at which $J$ achieves its minimum. Then, since $J$ is smooth, from calculus we know that
$$
\left. \frac{\partial J}{\partial u} \right|_{u = u^*,v = v^*}
= 0 =
\left. \frac{\partial J}{\partial v} \right|_{u = u^*,v = v^*}
$$
Can you find both partial derivatives?