问:
Should Two Unrelated Remarks Be Presented in One "Remark" Section, or Two?
Suppose at a certain point in my article I would like to make what are essentially two unrelated observations, where each pertains to the topic of discussion but neither is of sufficient importance to warrant inclusion in the main text.  I could opt to have two remark sections, one immediately following the other with no intervening main text; alternatively, I could have only one remark section, with both remarks being placed there.

Which is considered better form?

答: I think it's clearly better to have two remarks if they're unrelated; you want to be able to refer to them separately (e.g. "oh, see remark 2.4.2" vs. "oh, see remark 2.4.3").</p>


问:
Is there a way to rate how close data is to being sinusoidal?
I am looking for a mathematical method of determining how relatively close a data set is, to fitting a generic sine wave (in other words, $\sin(x)$). In my mind, this would be something like an R squared value for a linear regression, except for a sine wave. I would like to be clear that I am not looking for a way to fit data to a sinusoidal function.

So if my data looked something like the following, how close is that to being ($\sin(x)$). Again, I am not looking to do a curve fit for this data. Also, the data was just plucked from Google images as an example of what I am talking about.

Read the comments below for some additional clarification.

答: One simple approach is to fit it to a sine wave with unknown frequency, amplitude and phase.  That is a three dimensional nonlinear minimization problem.  Depending on your definition of sinusoidal, you might have a constant term as well, for four parameters.  You can then look at what fraction of the variation is explained by the sine wave you found.  How does the amplitude of the error compare to the amplitude of the function you found?  In your figure, the peak to peak amplitude appears to be about $6$.  There is clearly noise in the range of $0.5-1$ peak to peak.  A bad fit to a sine wave might make this worse.  This approach cannot tell the difference between noise and a repetitive signal that is not shaped like a sine wave.  Is that important to you?  You can certainly do a full FFT and see if the energy is in high frequency bins like noise or in low frequency bins because the wave shape is not sinusoidal.

ContentLicense: CC BY-SA 4.0


问:
Shtarkov sum approximation
Consider an alphabet $B = \{1, 2, ..., K \}$ and let $B^n$ be the $n$th fold Cartesian product.

$y^n \in B^n$ denotes a sequence of length $n$ with $y_i \in B$ for $1 \leq i \leq n$.

$\mathcal{P}(B)$ denotes the set of all probability distributions on $B$.

Shtarkov sum is defined as

$$S_n = \sum_{y^n \in B^n } \sup_{p \in \mathcal{P}(B)} p^n(y^n)$$
where $p^n(y^n) = \prod \limits_{i=1}^n p(y_i)$

Can anyone provide some references containing approximations of $S_n$? I am actually trying to see if, for large $n$, something like $$S_n \approx \frac{K - 1}{2} \log (n)$$
has been proven somewhere?

Edit: I found the answer. See below

答:
Paper Reference: A. Orlitsky, N. P. Santhanam and Junan Zhang, "Universal compression of memoryless sources over unknown alphabets"

In this paper, there is equality between $(2)$ and $(4)$ which directly implies the following result about $S_n$: $$\log S_n \approx \frac{K - 1}{2} \log (n)$$

Or more exactly, $$\log S_n = \frac{K - 1}{2} \log \left (\frac{n}{2\pi} \right ) + \log \left( \frac{\Gamma(1/2)^K}{\Gamma(K/2)} \right) + o_K(1) $$

where $o_K(1)$ goes to zero as $n \to \infty$ with rate determined by $K$.

Brief Interpretation:

$\log S_n$ is shown to be equal to universal lossless source coding redundancy of i.i.d. sources taking values on alphabet $B$.

i.i.d. means independent and identically distributed source also called discrete memoryless source in this paper.

universal means the source distribution is unknown but is known to be i.i.d. over the alphabet $B$.

Title: Shtarkov sum approximation


问:
Solving $x + 3^x < 4$ analytically

I am solving the problems in Michael Spivak's Calculus book. In the Prologue chapter, there is the following problem:
$$x + 3^x < 4$$
I can solve this using graphical means, but not analytically. Is it possible to do the latter?

I can arrive at the solution as follows:

$$3^x < 4 - x$$

$$x < \log_3 (4-x)$$

I then draw the graphs of the functions $\log_3 (4-x)$ and $x$, and I can determine that $x$ needs to be smaller than 1 to be smaller than the $\log_3 (4-x)$.

How do I solve this purely analytically?

答: $x+3^{x}$ is a continuous strictly increasing function. Hence $\{x: x+3^{x}<4\}=(-\infty, t)$ where $t$ is the unique real number $t$ such that $t+3^{t}=4$. Note that $t=1$!

Title: Solving $x + 3^x < 4$ analytically


问:
How best we can allocate?
Thank you for this opportunity to share my post here

Advanced thanks to the people who answered my post

Problem: There are $5$ items and cost of these items in dollars are $3$,  $3.2$,  $4.5$,  $2.5$,  $1.7$ respectively. How  can we allocate $100$ dollars to buy the above items in a such way in order to get maximum number of items in our hand.

All types of items should be covered

Repetition of items may be allowed

Maximum number of items should be purchased

Not necessary to spend entire $100$. but, maximum amount should spend near to $100$ or with entire $100$

We need to buy each kind of item as much as possible

How to solve this type of problem?
Kindly answer
Thanks again

答:
You can solve the problem via integer linear programming as follows.  For item $i$, let $c_i$ be the cost, and let integer decision variable $x_i$ be the number bought.  The problem is to maximize $\sum_i x_i$ subject to
\begin{align}
\sum_i c_i x_i &\le 100 \\
x_i &\ge 1 &&\text{for all $i$}
\end{align}
An optimal solution is $x=(1,1,1,1,51)$, with optimal objective value $55$.

If you want the numbers of items bought to vary by no more than $1$, you can impose additional constraint
$$-1 \le x_i - x_j \le 1 \quad \text{for all $i<j$}.$$
An optimal solution is $x=(7,7,6,7,7)$, with optimal objective value $34$.

Title: How best we can allocate?


问:
Polar coordinations - problem with r and $\theta$
let's take a look on Archimedean spiral.
the polar equation is $r = \theta$. click here to look.

but $\tan (\theta) = y/x$ and $r = \sqrt{x^2+y^2}$,

so $r = \theta \rightarrow \tan(\sqrt{x^2+y^2}) = y/x$. click here to look.

but wolfram alpha shows that the equations are different.

where is my mistake? or this is error on wolfram alpha?

thank you.

答:
If you look carefully, it actually does plot the same spiral. The second plot will of course also plot the spiral's 180 degree rotation (since $(-x,-y)$ solves the same equation).

It also looks to be plotting "numerically unstable" concentric circles of radii $\approx\pi(n+\frac 1 2)$. Presumably this is because of tan's singularity there and a numerical loss of precision that causes unexpected equality. That looks wrong, maybe worth a bug report, but shouldn't phase you.

ContentLicense: CC BY-SA 3.0


问:
Show that for each $g \in L^q([0,1])$, $\lim_{n \to \infty} \int_0^1 f_ng = \int_0^1 fg$.
Let $1 < p < \infty$. Let $\{f_n\}$ be a sequence of functions in $L^p([0,1])$ that converges almost everywhere to a function $f \in L^p([0,1])$. Also, suppose there exists a constant $M$ such that $\|f_n\|_p \leq M$ for all $n$.

Show that for each $g \in L^q([0,1])$, where $\frac{1}{p} + \frac{1}{q} = 1$, we have

$$ \lim_{n \to \infty} \int_0^1 f_ng = \int fg.$$

This involves either BCT or DCT and Holder's inequality somewhere, but we cannot figure out how to start. We can bound $\|f_ng\|_1 \leq M\|g\|_q$ by Holder's Inequality, but after that we are stuck. That fact may not even be useful.

答:
First consider a bounded measurable function $g$. Then the conclusion follows from the fact that $\int |f_n-f| \to 0$. This well known fact is proved using uniform integrability of the sequence $(f_n)$. [See below for a proof].

The general case is proved easily using Holder's inequlaity: There exist a bounded measurable function $h$ such that $\int |g-h|^{q} <\epsilon$. Note that $|\int f_ng -\int f_nh | \leq \|f_n\|_p \|g-h\|_2 \leq M \|g-h\|_2$ and $|\int fg -\int fh | \leq \|f\|_p \|g-h\|_2 \leq M \|g-h\|_2$.

Let $f_n \to f$ a.e. on $[0,1]$ and assume that $\|f_n\|_p $ is bounded for some $p \in (1,\infty)$. Then $\int |f_n-f| \to 0$.

Proof: It is clearly enough to prove this when  $f=0$ (since $\|f\|_p$ is necessarily finite by Fatou's Lemma).

Now $$m\{x: |f_n(x) |>C\}) \leq \frac 1 {C^{p}} \int |f_n|^{p}.$$

From Holder's  inequality we get $$\int_{\{x: |f_n(x) |>C\}} |f_n| \leq \|f_n\|_p (m\{x: |f_n(x) |>C\})^{1/q}.$$ It follows that $\int_{\{x: |f_n(x) |>C\}} |f_n|$ can be made less than $\epsilon$ for all $n$ by choosing $C$ large enough. On the other hand $\int_{\{x: |f_n(x) |\leq C\}} |f_n| \to 0$ by DCT.


问:
Why can't differentiability be generalized as nicely as continuity?
The question: Can we define differentiable functions between (some class of) sets, "without $\Bbb R$"* so that it

Reduces to the traditional definition when desired?

Has the same use in at least some of the higher contexts where we would use the present differentiable manifolds?

Motivation/Context:

I was a little bit disappointed when I learned to differentiate on manifolds. Here's how it went.

A younger me was studying metric spaces as the first unit in a topology course when a shiny new generalization of continuity was presented to me. I was thrilled because I could now consider continuity in a whole new sense, on function spaces, finite sets, even the familiar reals with a different metric, etc. Still, I felt like I hadn't escaped the reals (why I wanted to I can't say, but I digress) since I was still measuring distance (and therefore continuity, in my mind) with a real number: $$d: X\to\huge\Bbb R$$

If the reader for some reason shares or at least empathizes with (I honestly can't explain this fixation of mine) the desire to have definitions not appeal to other sets/structures*, then they will understand my excitement in discovering an even more general definition, the standard one on arbitrary topological spaces. I was home free; the new definition was totally untied to the ever-convenient real numbers, and of course I could recover my first (calculus) definition, provided I toplogized $\Bbb R^n$ adequately (and it seemed very reasonable to toplogize $\Bbb R^n$ with Pythagoras' theorem, after all).

Time passed and I kept studying, and through other courses (some simultaneous to topology, some posterior) a new sort of itch began to develop, this time with differentiable functions. On the one hand, I had definitions (types of convergence, compact sets, orientable surfaces, etc.) and theorems (Stone-Weierstrass, Arzelá-Ascoli, Brouwer fixed point, etc.) completely understandable through my new-found topology. On the other hand, the definition of a derivative was still the same as ever, I could not see it nor the subsequent theorems "from high above" as with topological arguments.

But then a new hope (happy may 4th) came with a then distant but closely approaching subject, differential geometry. The prospect of "escaping" once again from the terrestrial concepts seemed very promising, so I decided to look ahead and open up a few books to see if I could finally look down on my old derivative from up top in the conceptual clouds. My expectation was that, just like topology had first to define a generalized "closeness structure" i.e. lay the grounds on which general continuous functions could be defined via open sets, I would now encounter the analogous "differentiable structure" (I had no idea what this should entail but I didn't either for topology so why not imagine it). And so it went: "oh, so you just... and then you take it to $\Bbb R^n$... and you use the same definition of differentiable".

Why is this so? How come we're able to abstract continuity into definitions within the same set, but for differentiability, we have to "pass through" the reals? I realize that this really has to do with why we have to generalize in the first place, so what happens is that the respective generalizations have usefulness in the new contexts, hence the second point in my question statement.

Why I imagine this is plausible, a priori, is because there's a historical standard: start with the low-level definitions $\rightarrow$ uncover some properties $\rightarrow$ realize these are all you wanted anyhow, and redefine as that which possesses the properties. Certainly, derivatives have properties that can be just as well stated for slightly more general sets! (e.g. linearity, but of course this is far from enough). But then, we'll all agree that there's even been a lust for conducting the above process, everywhere possible, so maybe there are very strong obstructions indeed, which inhibit it's being carried out in this case. In this case, I should ask what these obstructions are, or how I should begin identifying them.

Thank you for reading this far if you have, I hope someone can give some insight (or just a reference would be great!).

* If I'm being honest, before asking this I should really answer the question of what on earth I mean, precisely, by "a structure that doesn't appeal to another". First of all, I might come across a new definition that apparently doesn't use $\Bbb R$, but is "isomorphic" to having done so (easy example: calling $\Bbb R$ a different name). Furthermore, I'm always inevitably appealing to (even naïve) set theory, the natural numbers, etc. without any fuss. So, if my qualms are to have a logical meaning, there should be a quantifiable difference in appealing to $\Bbb R$ vs. appealing to set theory and other preexisting constructs. If the respondent can remark on this, super-kudos (and if they can but the answer would be long and on the whole unrelated, say this and I'll post another question).

答:
Here is another definition based on the Carathéodory derivative that seems quite general:

Let $A$ and $B$ be pretopological magmas. Let $\operatorname{Hom}(A, B)$ be the set of continuous (with respect to the given pretopologies) magma homomorphisms from $A$ to $B$. Then $f : A \to B$ is differentiable at $a : A$ iff there exists $\phi : A \rightarrow \operatorname{Hom}(A, B)$ continuous at $a$ such that

$$ f(x) \bullet \phi(x)(a) = f(a) \bullet \phi(x)(x) $$

If such a $\phi$ exists, then $\phi(a)$ is a derivative of $f$ at $a$.

References:

Stephen Kuhn. The Derivative á la Carathéodory. The American Mathematical Monthly. Volume 98, No. 1, pp. 40-44, January 1991.

Ernesto Acosta G. Differentiability in Topological Groups. Soochow Journal of Mathematics. Volume 22, No. 1, pp. 39-48, January 1996.


问:
Restricting split short exact sequence of quasi-coherent sheaves to their coherent sub-sheaves.
Let $X$ be a projective variety, $Z$ a hypersurface section and $U \overset{def}= X \setminus Z$ its complement, an open affine subscheme of $X$. Let $i:U \hookrightarrow X$ be the corresponding open embedding. Given a coherent sheaf $M$ on $U$ we can consider the quasi-coherent sheaf $i_*M$. This seems to have a filtration of coherent sheaves in the following manner: Let's assume we are working on an affine chart and $W=\text{Spec}(R)$ is an affine in this chart and the equation of $Z$ corresponds to the element $f$ in $R$. The quasi-coherent sheaf $i_*M|_{W\cap U}$ on $W$ has this increasing filtration that $M_0=M|_W$, $M_1=\frac 1fM|_W$, $M_2=\frac 1{f^2}M|_W, \ldots$. This filtration seems to glue and give a filtration by coherent sheaves $(i_*M)_0\subset (i_*M)_1 \subset \ldots$ of $i_*M$. (Please correct me if I am wrong). Note that for each $j$, $i^*(i_*M)_j\cong i^*i_*M\cong M$.

Now let's consider a short exact sequence of vector bundles on $X$ like $0\rightarrow E_1 \rightarrow E_2 \rightarrow E_3 \rightarrow 0$. This leads to a short exact sequence of quasi-coherent sheaves $0\rightarrow i_*i^*E_1\rightarrow i_*i^*E_2\rightarrow i_*i^*E_3\rightarrow 0$ which restricts back to the original short exact sequence on $U$. Note that this exact sequence of quasi-coherent sheaves is split. This is because short exact sequence of vector bundles split on $U$. If the contents of the first paragraph is correct, my question is, does this split short exact sequence of quasi-coherent sheaves restrict to a split short exact sequence of coherent sheaves of the form $0\rightarrow (i_*i^*E_1)_j\rightarrow (i_*i^*E_2)_j\rightarrow (i_*i^*E_3)_j\rightarrow 0$ for each $j\geq 0$? (Here $j$ is the same indexing of the filtration defined in the first paragraph.)

答:
What you're doing when you compute $(i_*M)_j$ in the affine setting is that you have a module $M$ and you tensor up with $\frac 1f R$ over $R$, which is an $R$-submodule of $R_f \otimes_R M$. So in other words, $(i_*M)_j = \frac 1{f^j}R \otimes_R M$. I think your question can be converted into a module-theoretic question, i.e. is the $R$-module $\frac 1{f^j}R$ flat. I think your intuition for this statement comes from the fact that $\frac 1{f^j}R$ is an invertible $R$-module, since
$$
f^j R \otimes_R \frac 1{f^j} R \simeq R
$$
via the multiplication map, so that $\frac 1{f^j} R$ is an invertible $R$-module; this would be the module-theoretic equivalent of "twisting" when dealing with sheaves, in some sense I can't put my finger on since it's been a while. I'm not sure, but I think you may need the assumption that $f$ is a non-zero divisor so that the map $R \to R_f$ is injective and the computations in the above statement (i.e. the 'multiplication map') can happen within $R_f$.

Also because it's been a while, I don't feel super confident in my intuition that tells me that the above implies that $f^j R$ is faithfully flat for $j \in \mathbb Z$, but it seems to make sense when $X$ is integral, because then $R$ is integral, and saying that $f^jR$ is flat is basically saying that $f^jR_{\mathfrak p}$ is a free $R_{\mathfrak p}$-module of rank $1$ for any $\mathfrak p \in \mathrm{Spec}(R)$, and the isomorphism $R_{\mathfrak p} \to f^j R_{\mathfrak p}$ can clearly be given by multiplication by $f^j$ when both modules are seen as submodules of the quotient field $Q(R)$ (which is possible when $R$ is integral).

Hope that helps,


问:
When is the Lie bracket the unique intertwiner $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$?
Let $\mathfrak{g}$ be the complexified Lie algebra of a simple compact real Lie group. Then the adjoint and coadjoint representation are isomorphic (see Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices?).

In the representation category of $\mathfrak{g}$, we naturally have an intertwiner $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ (viewing $\mathfrak{g}$ as the adjoint representation) given by the Lie bracket. My question is wether simplicity and compactness are sufficient conditions for the Lie bracket to be the only such intertwiner (up to a constant factor). Phrased differently, when does $F: \mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ with
\begin{align}
F([x,y] \otimes z) + F(y  \otimes [x,z]) = [x,F(y \otimes z)]
\end{align}
for all $x,y,z \in \mathfrak{g}$ imply $F(y \otimes z) = c \cdot [y,z]$, with $c \in \mathbb{C}$?

I already tried to show that $[[y,z],F(y \otimes z)]=0$, but I can't even get that done.

I also tried working in the basis mentioned in the page I reference. Then the formula becomes
\begin{align}
\forall r \in \{1 \ldots \dim \mathfrak{g}\}:    \sum_{n=1}^{\dim \mathfrak{g}} \langle [e_k,e_i]f^{(nj)}_r + [e_k,e_j]f^{(in)}_r + [e_k,e_r]f^{(ij)}_n , e_n \rangle = 0
\end{align}
where
$F(e_i \otimes e_j) = \sum_{k=1}^{\dim \mathfrak{g}} f^{(ij)}_k e_k$, and the inner product is defined as $\langle x,y \rangle = \text{Tr}(\text{Ad}_x \text{Ad}_y^*)$, where the star is just taking a matrix adjoint in the basis $\{e_i\}$. However, I don't know how to proceed from here.

Any help is welcome.

答:
Turning comments into an answer:

For any semisimple complex Lie algebra $\mathfrak g$, if $W$ is any finite-dimensional representation of $\mathfrak g$, then $W$ decomposes into a finite direct sum $W \simeq \bigoplus_i V_i^{m_i}$ where the $V_i$ are mutually isomorphic irreducible (=simple) representations with multiplicities $m_i \ge 1$. Further, if $V$ is any irreducible representation of $\mathfrak g$, then either $$Hom_{\mathfrak g} (V,W) = Hom_{\mathfrak g} (W,V) = 0,$$
in which case $V \not \simeq V_i$ for all $i$; or, if $V \simeq V_i$, then $$dim(Hom_{\mathfrak g} (V,W)) = dim(Hom_{\mathfrak g} (W,V)) = m_i.$$
(This is essentially the definition of semisimple representation plus Schur's Lemma.)

Now in our case we let $\mathfrak g$ be simple, and $V$ the adjoint representation (well-known to be simple). According to https://mathoverflow.net/q/129857/27465 and https://mathoverflow.net/q/130185/27465 (or case-by-case check in LiE), the multiplicity of $V$ in $V \otimes V$ is $1$, except in the case of $\mathfrak g \simeq \mathfrak sl_{n \ge 3}$, in which it is $2$.

(Funnily, the MO posts leave a bit of a gap for showing the multiplicity is $\ge 1$: which is exactly what your question shows, namely, the Lie bracket is an obvious non-trivial element of $Hom_{\mathfrak g}(V \otimes V, V)$.)

So in all cases except $\mathfrak g \simeq sl_{n \ge 3}$, the result is true, again due to Schur's Lemma.

In the case $\mathfrak g \simeq sl_{n \ge 3}$, the missing dimension is given by the intertwiner

$$F_{extra}: x\otimes y \mapsto xy + yx - \frac1n Tr(xy+yx)$$

(the "traceless part of $xy +yx$", as Allen Knutson writes in the MO post; note that this is the composition of the map $x\otimes y \mapsto xy+yx$, whose codomain is $\mathfrak{gl}_n$, with the projection $\mathfrak{gl}_n \twoheadrightarrow \mathfrak{sl}_n, z \mapsto z - \frac1n Tr(z)$).

A little computation indeed shows that $F_{extra}$ is an element of $Hom_{\mathfrak g}(V \otimes V, V)$, and is linearly independent from the intertwiner given by the Lie bracket for all $n \ge 3$ (it does vanish for $n=2$, matching earlier results). What happens there seems to be a decomposition of $V \otimes V$ into symmetric versus alternating tensors and what's special in this case is that there is an extra copy of $V$ within the symmetric ones. If, on the other hand, one tries to imitate the map $F_{extra}$ e.g. in the case of $\mathfrak{so}_n$, it is not even well defined i.e. the image does not lie in $\mathfrak{so}_n$ etc.


问:
Choosing with replacement and subsequent sequencing.
The question I'm facing is a relatively basic combinatorics question, namely:

How many words of exactly four letters can you construct using the letters a,b,c and d?

The straightforward solution is of course 4^4 = 256, as you fill up the four available "empty spaces" with the respective letters, and for each space you have four possibilities.

What I was wondering is how this problem can be solved with the binomial coefficient for choosing with replacement:

$${n+k-1}\choose{k}$$

My reasoning is the following. You have the four letters in a "container", take one out, and put it back (since you can select one letter twice). You repeat this four more times.
The letters you took out must then be arranged in all possible ways to construct a word.

In this sense we would first have $${4+4-1}\choose{4}$$ These are the 35 ways in which one can choose letters with replacement from the four available ones. What I am unable to do is figure out how I can calculate the possible arrangements now, because I don't know how many repeated letters there are in the sample. Is there a way to account for this, to get to the result of 256?

答:
If you must do it the painful way, use [choose]$\times$[permute] and add up, thus

4 of a kind:$\;\binom41 \frac{4!}{4!} = 4$

3-1$\;$of$\;$a$\;$kind:$\;\binom41\binom31 \frac{4!}{3!1!}=48$

2-2$\;$of$\;$a$\;$kind:$\;\binom42 \frac{4!}{2!2!}=36$

2-1-1$\;$of$\;$a$\;$kind:$\;\binom41\binom32 \frac{4!}{2!1!1!}=144$

1-1-1-1$\;$of$\;$a$\;$kind:$\;\binom44 \frac{4!}{1!1!1!1!}=24$, adding up to the expected $256$

Btw, this also shows why stars and bars doesn't give equiprobable outcomes


问:
Solve the integral in the expression $x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt$
I'm following the course MIT 18.03. In [this problem][1], we evaluate the following expression containing an integral:
$$
x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt\tag{1}
$$
which finally gives:
$$
x(t) = \frac{2k \sin(t/2)+ 4k^2\cos(t/2)}{1+4k^2}+ce^{-kt}\tag{2}
$$
One possibility would be to make a substitution $u = \sin(x)$, but more than that I'm stuck

Edit (with the help of the comments)

So we can use the following substitutions:

\begin{align}
u &= \cos(t/2)\tag{3}\\
\implies u' &= 1/2\sin(t/2)\tag{4}\\
v'&= e^{kt}\tag{5}\\
\implies v &= \frac{1}{k} e^{kt}\tag{6}\\
\end{align}

Now using integration by part we can write

\begin{align}
\int v'u &= vu - \int vu'\tag{7}\\
&=\cos(t/2) \cdot \frac{1}{k}e^{kt} - \int \frac{1}{k}e^{kt}\frac{1}{2}\sin (t/2)\tag{8}\\
\end{align}
[1]: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-i-first-order-differential-equations/solution-by-integrating-factors/MIT18_03SCF11_rec_04s5_sol.pdf

答:
To do it faster, consider
$$I =  \int e^{kt}  e^{\frac{i t}{2}}\,dt=\int e^{\left(k+\frac{i}{2}\right) t } \,dt=\frac{2 }{2 k+i}e^{\left(k+\frac{i}{2}\right) t}$$ Now, rewrite the rhs as
$$\frac{2 (2k-i)}{(2 k+i)(2k-i)}e^{k t}e^{\frac{i}{2} t}=\frac{4k-2i}{4k^2+1}\,e^{k t}\, \left(\cos
   \left(\frac{t}{2}\right)+i \sin \left(\frac{t}{2}\right)\right)$$

Expand to isolate the real and imaginary parts to get
$$\Re(I)=\int e^{k t} \cos \left(\frac{t}{2}\right)\,dt=\frac{2 e^{k t} \left(2 k \cos \left(\frac{t}{2}\right)+\sin
   \left(\frac{t}{2}\right)\right)}{4 k^2+1}+C$$
$$\Im(I)=\int e^{k t} \sin \left(\frac{t}{2}\right)\,dt=\frac{2 e^{k t} \left(2 k \sin \left(\frac{t}{2}\right)-\cos
   \left(\frac{t}{2}\right)\right)}{4 k^2+1}+C$$

Now, just multiply $\Re(I)$ by $k \,e^{-kt}$ to obtain $x(t)$.

If you think about it, no need of integration by parts. The only think I used is that
$$\int e^{ax} \,dx=\frac 1 a  e^{ax} $$


问:
Is a finite von Neumann algebra with separable predual always a subalgebra of a type $II_1$ factor with separable predual?
As in the title, given $N$, a finite von Neumann algebra with a separable predual, does there always exist a type $II_1$ factor $M$ with a separable predual such that $N \subseteq M$? I've heard that letting $M = N \bar{\ast} R$, the tracial free product of $N$ and the hyperfinite $II_1$ factor $R$, gives a solution, but I am unable to find a proof of this.

答:
It can be done as you say.

As far as I can tell, the first source (which uses a different technique) is Elliott-Handelman, Math. Scand. 49 (1981), no. 1, 95–98. The requirement of properly infinite commutant is no big deal since it can be achieved by tensoring.

The free-product argument appears for instance in Proposition 8.1 in Bercovici-Collins-Dykema-Li-Timotin, JFA 258 (2010) 1579–1627.


问:
Maintaining Polynomial Irreducibility
Is there a polynomial $f(x) \in \mathbb{Q}[x]$ of degree $d$, such that on changing any $O(\sqrt{d})$ of its coefficients, it still remains irreducible?

By changing, I mean I have no control over what those changed $O(\sqrt{d})$ coefficients would be. Now of course, if the constant term becomes zero after change, then we will have $x$ as a factor, and hence reducible. Say, I am okay with such changes. In other words, is there an explicit polynomial $f(x)$ such that after change, it becomes $x^i.f^{\prime}(x)$, where $f^{\prime}(x)$ is irreducible and $i \geq 0$.

I was thinking to start with a polynomial with different prime numbers as coefficients that satisfies Eisenstein's or some irreducibility criterion, so that it becomes easier to argue irreducibility after change also, by looking at remaining $d-O(\sqrt{d})$ coefficients. But, I was unable to find any irreducibility criterion that is compatible with change of coefficients.

Any positive or negative answer to the above question is appreciated, in fact, for $f \in \mathbb{F}[x]$, where $\mathbb{F}$ is any field. I also don't mind if there is a result which is not precisely $O(\sqrt{d})$ coefficients but is anything of this flavor that after perturbation of some of its coefficients, the polynomial remains irreducible.

答:
Let $d$ be an integer with $d\ge 2$, and let $f\in \mathbb{Z}[x]$ be irreducible with $\deg(f)=d$.

Write $f={\displaystyle{\sum_{k=0}^d a_kx^k}}$.

Let $b_1=a_1-f(1)$, and let $g$ be the polynomial obtained from $f$ by replacing the term $a_1x$ by the term $b_1x$.

Then $g=f+(b_1x-a_1x)$, hence $g(1)=f(1)+(b_1-a_1)=0$, so $g$ is reducible.

Note that in the transition from $f$ to $g$, only one coefficient was changed.

Title: Maintaining Polynomial Irreducibility


问:
If $u\in L^p(R^N)\cap C(R^N)$, then $u\in L^\infty(R^N)$, is the conclusion is true?, thanks!
If $u\in L^p(R^N)\cap C(R^N)$, then $u\in L^\infty(R^N)$, is the conclusion is true?thanks!

答: This is false. Consider a function made up of "spikes": a triangular spike with base the interval $[-1/2,1/2]$ and height $1$, a triangle with base $[1-1/2^3,1+1/2^3]$ and height $2$, a triangle with base $[2-1/2^5,2+1/2^5]$ and height $2^2$ etc. (and zero elsewhere, so continuous). The series of the areas $1/2+1/2^2+\dots$ is convergent so the function is in $L^1$. However, it is clearly not bounded since the heights of the triangles increases to infinity. A similar construction works for any $p\ge 1$.


问:
Average of histogram estimator
I am studying for a statistics course. We would like to estimate an unknown probability density function $f$. Let us consider a random sample $X_1, X_2, \dots, X_n$ drawn from $f$. In the lecture notes, the histogram estimator is defined as
$$\hat{f}_{nha} ^{hist} = \frac{1}{nh} \sum _{k\in \mathbb{Z}} N_{kha} \mathbf{1} (\cdot \in (a+kh, a + (k+1)h] ),$$
where $h>0$ is the bin width, $a\in [0,h)$ a parameter that defines where to include the jumps, $n$ the sample size, $k$ any integer, $N_{kha} := \# \{ i : X_i \in (a+kh, a+(k+1)h] \}$ the number of observations in the interval $(a+kh, a+(k+1)h]$, and where $\mathbf{1} (\cdot \in (a+kh, a + (k+1)h] )$ is an indicator function for the interval $(a+kh, a + (k+1)h]$.

The choice of $a$ seems to be quite arbitrary. However, we can consider the average $\hat f _{nh} = h^{-1} \int _0 ^{h} \hat f ^{hist} _{nha} da$ over all choices of $a$, so that we do not have to make a choice for $a$.

According to the lecture notes it can be shown that
$$\hat f _{nh} (x) = \frac{1}{nh} \sum _{i=1} ^n \left ( 1 - \frac{|X_i - x|}{h} \right ) _+,$$
which is the kernel density estimator with triangular kernel $K(u)=(1-|u|)_+$.

I try to figure out how this can be shown. My idea is to write $N_{kha} = \sum _i \mathbf{1} (X_i \in (a+kh, a + (k+1)h] )$ and to use that only finitely many $N_{kha}$'s are non-zero so that I can pull in the integral into the summations. However, I do not know how to proceed further. Does someone have any hints?

答:
It is possible to view a histogram as an estimator of the density function of the population from which a random sample is taken. There are various rules suggesting the number of histogram bins, based on sample size, it order to make histograms useful for that purpose. [One example is the Freedman-Diaconis rule.]

I'm not sure I follow your notation or the method you are using. If you can edit your question to define $a, n, k, h,$ etc. maybe someone will give a direct answer.

However, I want to make sure you know about kernel density estimators (KDEs), which use data to estimate the population density function--without regard to whatever histogram you choose to make. I will illustrate the use of a KDE so you can see the result, and you can read details online, if interested.

Suppose we have a random sample of size $n = 1000$ from
a population distributed as $\mathsf{Gamma}(\mathrm{shape}=5, \mathrm{rate}=0.1),$ The figure
below shows (a) a density histogram from R of the sample,
(b) the density function (dotted curve) of $\mathsf{Gamma}5, .01),$ which would not be known is a real application, and (c) the default KDE from R for this sample.

```<code>set.seed(2020)
x = rgamma(1000, 5, 0.1)
hist(x, prob=T, br=20, col="skyblue2",
     main="n = 1000: GAMMA(5, 0.1)")
 curve(dgamma(x, 5, 0.1), add=T, lwd=2, lty="dotted")
 lines(density(x), lwd=2, col="orange")
</code>```

As you can see, the KDE is very close to the true population density.
From the Wikipedia page on gamma distributions, you can see that the mode of this distribution is $\delta=40.$ If you were to try to
estimate the mode from the histogram you would guess
somewhere between 30 and 40.

The KDE averages together kernels of a particular shape, which can be specified, to estimate the density function. In R, it consists of 512 x and y coordinates, from which
we can find its mode: about 37.5 (which is close to 40), and other features of the estimate.

```<code>kde = density(x);  kde

Call:
        density.default(x = x)

Data: x (1000 obs.);    Bandwidth 'bw' = 4.725

        x                 y
 Min.   : -7.537   Min.   :1.264e-06
 1st Qu.: 38.811   1st Qu.:1.976e-04
 Median : 85.159   Median :1.714e-03
 Mean   : 85.159   Mean   :5.389e-03
 3rd Qu.:131.506   3rd Qu.:9.915e-03
 Max.   :177.854   Max.   :1.961e-02

mean(kde<span class="math-container">$x[kde$</span>y==max(kde$y)])
[1] 37.45053
</code>```

The figure below shows KDEs for samples of size $n = 200$ and $n = 5000$ from the same distribution.
Not surprisingly results are better for larger sample sizes. At left, tick marks along the horizontal axis show
actual positions of the 100 data values [made with rug(x)]; at right, the population density curve is not shown because it is
difficult to distinguish from the KDE. The R code for the figure is similar to the code shown above.

References: This Wikipedia article gives an introductory discussion of KDEs, and many references; publications of Silverman and Wahba may be of particular interest.

Title: Average of histogram estimator

Body:
It is possible to view a histogram as an estimator of the density function of the population from which a random sample is taken. There are various rules suggesting the number of histogram bins, based on sample size, it order to make histograms useful for that purpose. [One example is the Freedman-Diaconis rule.]

I'm not sure I follow your notation or the method you are using. If you can edit your question to define $a, n, k, h,$ etc. maybe someone will give a direct answer.

However, I want to make sure you know about kernel density estimators (KDEs), which use data to estimate the population density function--without regard to whatever histogram you choose to make. I will illustrate the use of a KDE so you can see the result, and you can read details online, if interested.

Suppose we have a random sample of size $n = 1000$ from
a population distributed as $\mathsf{Gamma}(\mathrm{shape}=5, \mathrm{rate}=0.1),$ The figure
below shows (a) a density histogram from R of the sample,
(b) the density function (dotted curve) of $\mathsf{Gamma}5, .01),$ which would not be known is a real application, and (c) the default KDE from R for this sample.

set.seed(2020)
x = rgamma(1000, 5, 0.1)
hist(x, prob=T, br=20, col="skyblue2",
     main="n = 1000: GAMMA(5, 0.1)")
 curve(dgamma(x, 5, 0.1), add=T, lwd=2, lty="dotted")
 lines(density(x), lwd=2, col="orange")

As you can see, the KDE is very close to the true population density.
From the Wikipedia page on gamma distributions, you can see that the mode of this distribution is $\delta=40.$ If you were to try to
estimate the mode from the histogram you would guess
somewhere between 30 and 40.

The KDE averages together kernels of a particular shape, which can be specified, to estimate the density function. In R, it consists of 512 x and y coordinates, from which
we can find its mode: about 37.5 (which is close to 40), and other features of the estimate.

kde = density(x);  kde

Call:
        density.default(x = x)

Data: x (1000 obs.);    Bandwidth 'bw' = 4.725

        x                 y
 Min.   : -7.537   Min.   :1.264e-06
 1st Qu.: 38.811   1st Qu.:1.976e-04
 Median : 85.159   Median :1.714e-03
 Mean   : 85.159   Mean   :5.389e-03
 3rd Qu.:131.506   3rd Qu.:9.915e-03
 Max.   :177.854   Max.   :1.961e-02

mean(kde$x[kde$y==max(kde$y)])
[1] 37.45053

The figure below shows KDEs for samples of size $n = 200$ and $n = 5000$ from the same distribution.
Not surprisingly results are better for larger sample sizes. At left, tick marks along the horizontal axis show
actual positions of the 100 data values [made with rug(x)]; at right, the population density curve is not shown because it is
difficult to distinguish from the KDE. The R code for the figure is similar to the code shown above.

References: This Wikipedia article gives an introductory discussion of KDEs, and many references; publications of Silverman and Wahba may be of particular interest.


问:
Find $\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$
Find $$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$$

One easy step is replacing $\sin t = t$, since $\lim_{t \rightarrow 0} \sin t / t = 1$. How do I continue with the rest?

答:
Use the fundamental theorem of calculus and L'Hospital rule
$$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx=\lim_{t \rightarrow 0} \frac{\frac d{dt}\,\int_{0}^{t}\sqrt{1+\sin(x^2)}\,dx}{\frac d{dt}\,\sin(t) }$$ that is to say
$$\lim_{t \rightarrow 0} \frac{\sqrt{1+\sin(t^2)}}{\cos(t)}$$


答:
When data is this irregular,
I first do some
fairly simple-minded smoothing
such as a moving average.
Estimate the location of the peaks.
If irregular data
causes two peaks to be too close,
take their average as the peak location.

I would then look at the
distances between successive peaks.
If this are approximately equal,
this gives an estimate
of the period.
Call this $p$.

I would then look at
the maximum and minimum values -
call them $a$ (min) and $b$ (max).

Then, with
$h = (b-a)/2$ being
an estimate for the amplitude
and
$c = (b+a)/2$ being
an estimate for the center,
an initial estimate for the
fitting curve is
$c+h\cos(2\pi\frac{t-t_0}{p})
$
where $t$ is the time
(x-axis)
and
$t_0$
is the location of
the first peak.

Finally,
put these parameters
($c, h, t_0, p$)
into a nonlinear fitting routine
(least squares is probably
reasonable and available)
as the initial values
for fitting that model function
to the actual (unsmoothed) data.

Then look at the fit.

All this is moderately ad hoc,
but I have done similar things
in the past successfully.

Title: Is there a way to rate how close data is to being sinusoidal?

Body:
I am looking for a mathematical method of determining how relatively close a data set is, to fitting a generic sine wave (in other words, $\sin(x)$). In my mind, this would be something like an R squared value for a linear regression, except for a sine wave. I would like to be clear that I am not looking for a way to fit data to a sinusoidal function.

So if my data looked something like the following, how close is that to being ($\sin(x)$). Again, I am not looking to do a curve fit for this data. Also, the data was just plucked from Google images as an example of what I am talking about.

Read the comments below for some additional clarification.


问:
Solution of Vector Cross Product of Different Vectors
Although the solution of a cross-product identity is crystal clear here, my question differs from the asked ones. Let's assume that I have three force vectors and these vectors are applied at different locations. They are given as:

\begin{align*}
\vec{F_1} = \pmatrix{5 \\ 22 \\ 9}, \vec{F_2} = \pmatrix{8 \\ 5 \\ 11}, \vec{F_3} = \pmatrix{9 \\ 21 \\ 7}
\end{align*}

\begin{align*}
\vec{r_1} = \pmatrix{1 \\ 0.75 \\ -6.23}, \vec{r_2} = \pmatrix{7.12 \\ 5.63 \\ 8.45}, \vec{r_3} = \pmatrix{-9.45 \\ -1.12 \\ 6.11}
\end{align*}

I can find the equivalent force and moment at the origin $(0,0,0)$ as follows:

\begin{align*}
\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} = \pmatrix{ 22 \\ 48 \\ 27}
\end{align*}

\begin{align*}
\vec{M} = \vec{r_1} \times \vec{F_1} + \vec{r_2} \times \vec{F_2} + \vec{r_3} \times \vec{F_3} = \pmatrix{ 27.34 \\ 70.27 \\ -179.56}
\end{align*}

From now on, I would like to find a point (which are infinitely many) to yield zero moment. Moment is a free vector but force is not. Denoting such point at $(x_0,y_0,z_0)$, I need to solve the following equation:

\begin{align*}
-\vec{R} \times \vec{F} + \vec{M} = \vec{0}
\end{align*}

\begin{equation}
\vec{F} \times \vec{R} = - \vec{M}
\end{equation}

$\vec{R}$ being my unknown, this is a vector equation whose solution is actually given as:

\begin{align*}
\vec{R} = \frac{-\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}} + k \vec{F}
\end{align*}

where $k$ is an arbitrary scalar and hence, $\vec{R}$ is infinitely many. However, if you check, $\vec{F}$ is not normal to $\vec{M}$ and for these values I cannot solve the equation because the cross product identity is not satisfied. How come the cross product equation indicates that this is vice versa? What I am missing here?

答:
The cross-product is orthogonal to the two vectors in the product. So for a vector $\vec R$ to exist for which $\vec M = \vec R \times \vec F$, $\vec M$ has to be orthogonal to $\vec F$.

When $\vec F$ and $\vec M$ are orthogonal, then $\vec F$ is on the plane normal to $\vec M$. For any other vector on that plane not parallel to $\vec F$, the cross-product with $\vec F$ will be parallel to $\vec M$, and some scalar multiple will give $\vec M$ itself. Thus there are infinitely many solutions for $\vec R$.

But when $\vec F$ and $\vec M$ are not orthogonal, no such $\vec R$ can exist. There is just no vector that will solve the equation.

Your solution works fine when $\vec F$ is orthogonal to $\vec M$. Because then $\vec F \times (\vec M\times \vec F) = \|\vec F\|^2\vec M$. But if $\vec F$ is not orthogonal to $\vec M$, it is not a multiple of $\vec M$ at all.


问:
Special linear transformations
Special linear transformations are matrices with determinant equal to 1.

What additional properties do such transformations have compared to "regular" linear transformations?

答:
For example, for $A\in SL(2,\mathbb{C})$ one has the property
$$ A+A^{-1}=\mathrm{Tr}\,A\cdot \mathbf{1}_2$$
which is often useful, and also $\mathrm{Tr}\,A=\mathrm{Tr}\,A^{-1}$.</p>

Title: Special linear transformations


问:
Second Derivative of polar function
i'm trying to find $ \frac{d^2\rho}{(d\phi)^2}$ of $\rho=\tan(\phi+\rho)$
in this exercise is not specified if $\rho$ (argument of tangent) is a constant o another variable.

I do $$\rho'=\sec^2(\phi+\rho) $$
then i re-derivate it  $$ \rho''=2 \sec^2(\phi+\rho)\tan(\phi+\rho) $$

In my answer book this must be: $$ −\frac{\cot^3(\phi+\rho)}{\sin(\phi+\rho)}$$

Obviously i make a mistake or a misinterpreted some concept. I need some orientation or a example in order to continue.
Thanks in advance.

答:
Well considering $ 0=\tan(\phi+\rho)-\rho $ as a implicit function
I proceed to derivate respect to $\phi$
$$ 0=\tan(\phi+\rho)-\rho$$
$$ 0=\sec^2(\phi+\rho)((1+\rho')-\rho'$$
$$ 0=\sec^2(\phi+\rho)+sec^2(\phi+\rho)\rho'-\rho'$$
$$ -\sec^2(\phi+\rho)=\rho'(sec^2(\phi+\rho)-1)$$
$$ \rho'=\frac{-\sec^2(\phi+\rho)}{(sec^2(\phi+\rho)-1)}$$
then I go to the second derivative of $\phi$

$$\rho''=\frac{-(2\sec^2(\phi+\rho)\tan((\phi+\rho))(1+\rho')(\sec^2(\phi+\rho)-1)+(2\sec^2(\phi+\rho)\sec^2(\phi+\rho)\tan(\phi+\rho)(1+\rho')}{(\sec^2(\phi+\rho)-1)^2}$$
$$\rho''=\frac{(1+\rho')[-2\sec^2(\phi+\rho)\tan((\phi+\rho))(\sec^2(\phi+\rho)-1)+(2\sec^4(\phi+\rho)\tan(\phi+\rho]}{(\sec^2(\phi+\rho)-1)^2}   $$

$$ \rho''=\frac{(1+\rho')\biggl(-2\sec^2(\phi+\rho)\tan((\phi+\rho))\biggl[(\sec^2(\phi+\rho)-1)+(\sec^2(\phi+\rho)\biggr]\biggr)}{(\sec^2(\phi+\rho)-1)^2} $$

$$ \rho''=\frac{(1+\rho')\biggl(-4\sec^2(\phi+\rho)\tan((\phi+\rho))\biggr)}{(\sec^2(\phi+\rho)-1)} $$
now replacing $\rho'$ in $\rho''$
$$ \rho''=\frac{(1+\frac{-\sec^2(\phi+\rho)}{(sec^2(\phi+\rho)-1)})\biggl(-4\sec^2(\phi+\rho)\tan((\phi+\rho))\biggr)}{(\sec^2(\phi+\rho)-1)} $$

$$ \rho''=\frac{(\frac{-1}{(sec^2(\phi+\rho)-1)})\biggl(-4\sec^2(\phi+\rho)\tan((\phi+\rho))\biggr)}{(\sec^2(\phi+\rho)-1)} $$
Finally
$$ \rho''=\frac{\biggl(4\sec^2(\phi+\rho)\tan((\phi+\rho))\biggr)}{(\sec^2(\phi+\rho)-1)^2} $$

this is more tricky than considering the $\rho$ inside of the argument of tangent like a constant, but still not looking like my suppousted answer. Sorry for the long equation but i want to show every step, so if someone find some error or if i miss something somewere please comment.

Title: Second Derivative of polar function


问:
Show Diagonal Matrix is not possible
Consider the real matrix $\alpha = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$. Show/explain that there does not exist a invertible real matrix $\beta$, so that $\beta ^{-1} \alpha \beta$ is a diagonal matrix

My effort so far

I know that to diagonalize a matirx $A$

Find the eigenvalues of $A$ using the characteristic polynomial.

For each eigenvalue $λ$ of $A$ , compute a basis $B λ$ for the $λ$-eigenspace.

If there are fewer than $n$ total vectors in all of the eigenspace bases $B λ$ , then the matrix is not diagonalizable. I am losing it when it comes to the specific case "Show/explain that there does not exist a invertible real matrix $\beta$, so that $\beta ^{-1} \alpha \beta$ is a diagonal matrix". How would I go about this one in question?

答:
To address your specific question at the end of your post:

You may go about it by first assuming that such a $\beta$ exists and lead this assumption to a contradiction.

But first of all, it seems your question is about whether the given matrix can be diagonalized over $\mathbb{R}$. Because, over $\mathbb{C}$ you will find $3$ linearly independent (complex) eigenvectors.

Now, back to your question:

Assume there is an invertible $\beta = (b_1\: b_2\: b_3)$ with columns $b_1,b_2,b_3 \in \mathbb{R}^3$.  s.t.
$$\beta^{-1}\alpha\beta=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)\Leftrightarrow
\alpha\beta=\beta\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$$

But since $$\beta\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3) = (\lambda_1b_1\: \lambda_2b_2\: \lambda_3b_3)$$ this means
$$\alpha(b_1\: b_2\: b_3)=(\lambda_1b_1\: \lambda_2b_2\: \lambda_3b_3)$$

Hence, the columns of $\beta$ form a basis of real eigenvectors of $\alpha$. But since you find only one linearly independent real eigenvector of $\alpha$, such a $\beta$ does not exist.

Title: Show Diagonal Matrix is not possible


问:
Prove that if $f(x)$ is differentiable, $f(0)=0$ and $f(1)=1$, then $\int_0^1|f(x)-f'(x)|\mathrm dx\geq\frac1{\mathrm e}$.
Prove that if $f(x)$ is continuous and differentiable, $f(0)=0$ and $f(1)=1$, then $\int_0^1|f(x)-f'(x)|\mathrm dx\geq\frac1{\mathrm e}$.

Body:
Let $$F(x)=e^{-x}f(x)$$
We have
$$\frac{1}{e}=F(1)-F(0)=\displaystyle\int_{0}^{1}F^{\prime}(x)dx\leqslant\displaystyle\int_{0}^{1}\left|F^{\prime}(x)\right|dx\leqslant\displaystyle\int_{0}^{1}\left|f(x)-f^{\prime}(x)\right|dx$$


问:
Function such that $f(x) = -1$ for $x < 0,$ and $f(x)=1$ for $x > 0$?
What is a function to returns $-1$ if number is negative, $1$ if positive, and zero if number is equal to 0?

for example:

$$
f(-8) = -1
$$
$$
f(8) = 1
$$
$$
f(0) = 0
$$

for $$x < 0$$ maybe?

$$ f(x) = (-x-(-x-1))\cdot-1 $$

答: This function is the $\text{sgn}$ (sign) function. You can write it as $$\text{sgn}(x)=\begin{cases}\frac x{|x|} &x\neq 0\\ 0 & x=0\end{cases}$$</p>


问:
Solving $\frac1x + \frac1{1-x} > 0$ two ways gives different results, one of which is simply $0<1$. What does this mean?
Given

$$\frac1x + \frac1{1-x} > 0 \tag{1}$$

we can do

$$\frac{(1-x) + x}{x(1-x)} = \frac1{x(1-x)} > 0 \tag{2}$$

So, $x(1-x) > 0$ and we arrive at $x>0$ and $x<1$.

However, if we take another path from $(1)$, we arrive at $0<1$. Here are the steps:

$$\begin{align}
\frac1x &> -\frac{1}{1-x} \tag{3} \\[4pt]
-x &< 1-x \tag{4} \\[4pt]
0 &< 1 \tag{5}
\end{align}$$

This result, $0<1$ seems to imply that $-x$ is always smaller than $1-x$, which is true; but is that what it always means, and given where we started from, is this derivation even correct?

Is there a step that is incorrect here? Why does the first approach differ from the second approach?

答:
The error stems from the fact that you took the reciprocal of both sides of the inequality (and changed the direction of the inequality) when the two sides do not necessarily have the same sign. if $x > x'$, it is not in general true that $1/x < 1/x'$. The latter will only hold if the two sides of the inequality have the same sign.

For example, supposed in your example we set $x=0.5$. As you have found, the inequality should hold for this value of $x$. However, subbing in this value into $1/x > -1/(1-x)$ we find $2 > -2$, which is true, but of course we do not have that $\frac{1}{2} < -\frac{1}{2}$, which is what your second to last step implies.


问:
Let $P$ be a polytope of dimension $d$. Then for some $v \notin$ aff($P$), the pyramid $v \ast P$ has dimension $d+1$.

Let $P$ be a polytope of dimension $d$. Then for some $v \notin$ aff($P$), the pyramid $v \ast P$ has dimension $d+1$.

I’m looking for a straightforward argument to prove this statement. All the arguments I’ve come up with so far involve embedding $P$ in $\mathbb{R}^{d+1}$, and analyzing the number of linearly independent vectors. I’m hoping to find a simpler explanation, if possible.

Here are the definitions of the relevant terms in the course I’m doing:

$\cdot$ A polytope is defined as the convex hull of a finite set of points in some $\mathbb{R}^d$. Alternatively, it is a bounded polyhedron - defined as the intersection of finitely many closed halfspaces in $\mathbb{R}^d$.

$\cdot$ Let $P \subset \mathbb{R}^d$ be a (possibly empty) polytope of dimension $< d$. For a point $v \in \mathbb{R}^d \setminus \text{aff}(P)$, we define the pyramid $v \ast P$ with apex $v$ and base $P$ as the convex hull of $P \cup \{v\}$.

答:
I think if we review a few (standard) definitions that are missing from the Question statement, the sought "straightforward argument to prove" the quoted claim will quickly materialize.

Note the definition of a polytope added to the Question:

A polytope is defined as the convex hull of a finite set of points in some $\mathbb R^d$.

This is satisfactory, provided one knows how to define the convex hull of a finite set $S$ of points in a Euclidean space $\mathbb R^d$, or more generally in a vector space $V$ (or in an affine space).  If one knows what it means for a set (in a vector space) to be convex, then one can say the convex hull of (finite) $S$ is the smallest convex set that contains $S$.

In any case our polytope $P$ will be a convex set in whatever vector space $V$ where it exists.  As we are about to see, that space $V$ plays a critical role in proving the desired claim.

The notation $\operatorname{aff}(P)$ presumably means the affine hull of $P$, namely the smallest affine subset of $V$ that contains $P$.  The affine subsets of $V$ are precisely the translates of its linear (vector) subspaces.  That is:

$$ \operatorname{aff}(P) = \{w\} + W $$

for some vector $w \in V$ and some linear subspace $W \subset V$.

The dimension of $P$ is the same as the dimension of $\operatorname{aff}(P)$, which is defined to be the dimension of the linear subspace $W$ of which it is a translate.

If the dimension of $P$ is $d$, then in order to pick $v \in V \backslash \operatorname{aff}(P)$, $V$ will have to be of dimension greater than $d$.  In other words, if $V$ were the same dimension as $\operatorname{aff}(P) \subset V$, then they would have to be equal as affine subsets, and we would be unable to choose $v\in V$ that is not in $\operatorname{aff}(P)$.

Thus we will assume $\operatorname{dim} V \gt d$ and choose $v \in V \backslash \operatorname{aff}(P)$.  Then, since $\operatorname{aff}(v*P)$ is larger than $\operatorname{aff}(P)$, the dimension of pyramid $v*P$ must be greater than $d$, so its dimension is at least $d+1$.

Why is the dimension of $v*P$ not more than $d+1$?  It suffices to show that $\operatorname{aff}(v*P)$ is contained the translate of a linear subspace of $V$ of dimension not more than $d+1$.  We have all the facts necessary to show this. Since:

$$ P \subset \operatorname{aff}(P) = \{w\} + W $$

we have the pyramid:

$$ v*P \subset \operatorname{aff}(v*P) \subset \{w\} + W' $$

where $W'=\operatorname{span}(\{v-w\} \cup W)$ results from $W$ by extending its basis with vector $v-w$ (so that $v\in \{w\} + W'$).  Hence $\operatorname{dim} W' \le d+1$ and the claim is proved.


问:
The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is
I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$

$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$

$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$

From here I am not able to process further to think of the way that will give me the solution.

答:
$2x + 5y = 99$ is very rare and a strong restriction.

If $2a + 5b=99$ is a solution to find another solution $2(a\pm d) +5(b\mp e) =99=2a+5b$ we must have $2d - 5e = 0$ and that requires that $d$ be divisible by $5$ and $e$ be divisble by $2$.

so if $2a + 5b =99 $ is one answer than any $2(a + 5k) + 5(b-2k) =99$ is also a solution and these are the only solutions.

Now lets find our first solution.

$2x +5y = 99$ so $5y = 99-2x$ and we need the RHS to be divisible by $5$.  We can do that by letting $x = 2$ and $5y = 95$ so $y = 19$.

So if $2x + 5y = 99$ then $x = 2+5k$ and $y = 19 - 2k$ for some integer $k$ and for any integer $k$ $x = 2+5k$ and $y =19-2k$ will be a solution.

So we must have $2+5k \ge 19-2k\ge -20$.  How many integers key can that be true for?

We must have $2+5k \ge 19-2k$ so $7k \ge 17$ so $k \ge 3$.  And we have $19-2k \ge -20$ so $2k \le 39$ so $k \le 19$.  So that's it.

$3\le k \le 19$ and $x = 2+5k$ and $y = 19 - 2k$ are all the solution pairs.  There are $17$ of them:

$(x,y) = (17, 13),(22,11),(27,9).....,(97, -19)$.

All other solutions will either have $y > x$ (such as $(12,15)$) of have $y < -20$ (such as $(102, -21)$.


问:
How to find the solutions to this modular equation: $-2131\cdot 3^x\equiv y\pmod{1763}$?
Consider:

$-2131\times 3^x\equiv y\pmod {(43\cdot 41)}$, with $x$ and $y$ are positive integers.

I found this solution:

$x=2+168\times s$, with $s$ positive integer, for example.

For $x=2$, $y=214$.

Is there any other $y$ for which a solution $x$ exists?

How can I find all the solutions?

Body:
Solution in pari/gp:

erd()=
{
 m= Mod(3, 1763);
 h= znorder(m);
 print("\nMultiplicative order of "m" is "h"\n");
 S= [];
 for(y=0, 1762,
  x= znlog(y/(-2131), m, h);
  if(x, S= concat(S, [[x,y]]))
 );
 S= vecsort(S);
 print("[x, y] = "S)
};

Output:

Multiplicative order of Mod(3, 1763) is 168

[x, y] = [[1, 659], [2, 214], [3, 642], [4, 163], ... [165, 1227], [166, 155], [167, 465]]

Verify for $(x,y)$=[2, 214]:

Mod(-2131,1763)*Mod(3,1763)^(2+168*0)=Mod(214, 1763)

Mod(-2131,1763)*Mod(3,1763)^(2+168*777)=Mod(214, 1763)

i.e. $-2131\cdot 3^{2+168\cdot k}\equiv 214\pmod {1763}$, where $k$=0,1,2,3,...

Verify for $(x,y)$=[167, 465]:

Mod(-2131,1763)*Mod(3,1763)^(167+168*0)=Mod(465, 1763)

Mod(-2131,1763)*Mod(3,1763)^(167+168*77777)=Mod(465, 1763)

i.e. $-2131\cdot 3^{167+168\cdot k}\equiv 465\pmod {1763}$, where $k$=0,1,2,3,...


问:
Are there infinitely many $p$ such that $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ has no integer solutions?
$\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ with $p_k$ as the $k$-th prime has no integer solutions for $9$ values of $k \leq 25$, $17$ values of $k \leq 50$, and $33$ values of $k \leq 100$. This means that roughly $33\%$ of each sample consists of such primes. Though this is a small sample, the ratio of primes with this property to those which do not seems to be consistent. Are there infinitely many primes with this property, and if so, what proportion of primes have it?

答: Write $m^2-n^2=(m+n)(m-n)$  The two factors on the right have the same parity, so any number that is equivalent to $2 \bmod 4$ cannot be expressed this way.  Because you divide the sum by $2$, this is any pair of primes that sum to a number that is equivalent to $4 \bmod 8$.  For example, the sequential primes $5,7$ give $\frac 12(5+7)=6$, which cannot be expressed as $m^2-n^2$ because it only factors into $1,6$ or $2,3$, and neither pair has the same parity.  $\frac 12(7+11)=9$, which we can factor as $1,9$ or $3,3$ and $m=5,n=4$ works for the first and $m=3,n=0$ for the second.  $\frac 12(11+13)=12$, which we can factor as $2,6$ getting $m=4,n=2$.  If the left side is odd, both factors will be odd, while if the left side is a multiple of $4$ there will be at least one factorization with both factors even.


答: Your function is exactly the signum function

Title: Function such that $f(x) = -1$ for $x < 0,$ and $f(x)=1$ for $x > 0$?

Body:
What is a function to returns $-1$ if number is negative, $1$ if positive, and zero if number is equal to 0?

for example:

$$
f(-8) = -1
$$
$$
f(8) = 1
$$
$$
f(0) = 0
$$

for $$x < 0$$ maybe?

$$ f(x) = (-x-(-x-1))\cdot-1 $$


问:
How many 3 digit even numbers are there(No Repetition)?
First find numbers ending with 0

So, 1's place-1 10's place-9 100's place-7 (2 digits are already consumed and 0 can't be used)

So 7*9*1.Im i doing the right thing?

答:
There are some good answers here. But I just want to mention one more approach:

First calculate total 3 digit numbers without repetition: 9 x 9 x 8 = 648

Then calculate 3 digit odd numbers without repetition: 8 x 8 x 5 = 320

Now subtract the number of odd numbers from total: 648 - 320 = 328


答:
$2x + 5y = 99$

$x \geq y \geq - 20$

If $x = y, 7 x = 7 y = 99$ which means for $x \leq 14$, $y \gt x$ so we will have solutions only for $x \gt 14$.

Also, $5y \geq -100 \,$ so $\,2x \leq 199 \implies x \lt 100$

Also, $2x$ is an even number so the only way for $y$ to be integer is when $2x$ ends in $4$ which means $x$ ends in either $2$ or $7$.

So, $x = 17 + 10m, 22 + 10n \lt 100$ ($m, n \geq 0) \,$are the solutions. You can easily count the solutions.

Title: The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is

Body:
I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$

$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$

$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$

From here I am not able to process further to think of the way that will give me the solution.


答:
y must be of the form $y=-(2k+1)5\geq-20$ which gives:

$(2k+1)5= (-5, -15) \geq -20$ $\rightarrow$ $2k+1=1, 3$

So for any $k\geq 0$ there will be an integer solution for x. in fact number of solutions is 2 when y is negative. Examples:

$2k+1=1$ $\rightarrow$ $y=-5>-20$, gives : $2x=25+99$ $\rightarrow$ $x=62$

$2k+1=3$ $\rightarrow$  $y=-15>-20$ ,gives: $2x=99+75$ $\rightarrow$ $x=87$

Now suppose x is negative then we must have:

$2x=-4, -14, -24, -34, -44, -54, -64, -74, -84, -94$

And corresponding y is:

$y=19, 17, 15, 13, 11, 9, 7, 5, 3, 1$

But this is not possible because dute to statement $x>y\geq-20$.

Now suppose x and y are both positive we have:

$x=47, 42, 37, 32, 27, 22,17, 12, 7, 2$

$y=1, 3, 5, 7, 9, 11, 13, 15, 17, 19 $

In this case only following solutions safices the condition:

$(x, y)=(47,1),(42, 3), (37, 5), (32, 7), (27, 9), (22, 11), (17, 13)$

Hence number of solutions is 9.


问:
Scale parameter definition
I have seen two definitions of scale parameter of a distribution: Let $X$ be random variable and $\{{P_{\theta, X }: \theta \in \Theta \subset \mathbb{R}\}}$ be a parametric family of distribution of $X$ indexed by $\theta$

$$1)~~ \theta \text{   is a scale parameter iff} \rightarrow \text{ the distribution of } cX \text{ is in } \{{P_{\theta, X }: \theta \in \Theta \subset \mathbb{R}\}} \text{ for every positive constant c}$$

For example Let $X$ ~ $LogNormal(\theta, 1)~, \theta \in (-\infty, \infty)$ then $cX$ ~ $LogNormal(\theta^*,1)$ where $\theta^* = \theta + ln(c) \in (-\infty, \infty)$. With this definition $\theta$ is a scale parameter.

For the second definition it is required that $\theta$ is positive:

$$2) ~ \theta > 0 \text{ is a scale parameter } iff \rightarrow \text{ the distribution of } X/\theta \text{ is independent of } \theta$$

If $X$ ~ $LogNormal(\theta, 1)~, \theta \in (0, \infty)$ (We restrict the parameter space) then $X/\theta $~ $LogNormal(\theta + ln(1/\theta), 1)$. With this definition $\theta$ is not a scale parameter.

I was wondering which of these two is the correct definition of a scale parameter? Or it depends on the book and the author?

答:
I think both can be seen as expressions of a more general idea of a location-scale family of distributions:

A family of distributions $f(x;\mu,\theta),\; \mu \in \mathbb{R}, \theta > 0$ is a location-scale family iff $g(x):=\frac{1}{\theta}f((x-\mu)/ \theta)$ is free of $\mu, \theta$.

This implies both other definitions

Location-scale families are closed under multiplication

The location and scale parameters can be sent to 0, 1 (respectively) via subtraction and division (repectively)

So, $\theta$ is a scale parameter if it transforms like a the scale parameter in a location-scale family (after "centering" by subtracting mean).

Essentially, $\theta  \propto \sigma$ (the standard deviation of $f(x;\eta, \theta)$

Title: Scale parameter definition


问:
Spectral radius of a block matrix with a block that has spectral radius larger than 1
Suppose we have a $n\times n$ block matrix

$$
X=
\begin{bmatrix}
A_{11} &  \ldots & A_{1n} \\
\vdots & \ddots & \vdots \\
A_{n1} & \ldots & A_{nn}
\end{bmatrix},
$$
where each block $A_{ij}$ is $d_i\times d_j$ and the diagonal blocks $A_{ii}$ are symmetric.

If there exists a block $A_{ii}$ such that $\rho(A_{ii})>1$, can we say anything about $\rho(X)$, for example, $\rho(X)>1$?

For instance, we could think of $X$ as a Jacobian matrix of a vector field that describes a $n$-player optimization dynamics. My conjecture is that the overall dynamics would be unstable, $\rho(X)>1$, if there is a player who is unstable under its own optimization dynamics, $\rho(A_{ii})>1$.

答:
A quick search with math software gave me several counterexamples:

$$\begin{pmatrix}0. & -0.3 \\ 0.8 & 1.1\end{pmatrix} \mapsto \lambda=0.8, 0.3$$
$$\begin{pmatrix}1.2 & -1 & 0.3 \\
 0 & 0 & -0.3 \\
 -1 & 0.5 & -0.5\end{pmatrix} \mapsto \lambda=0.643, 0.461, -0.404$$

One can imagine replacing each number by a block matrix of the same order of magnitude (e.g. $a\mapsto aI$) and the results should still hold.

In the context of dynamics, one variable -- the prey -- may increase exponentially if isolated, but the other variables -- the predators -- may restrict it from growing.


问:
Lebesgue outer measure and complete metric space
Let $A$ be an elementary set. For $E,F\subseteq A$, let $E\sim F$ iff $E\triangle F$ is null. Define $d([E]_{\sim},[E']_{\sim})=m^*(E\triangle E')$. I've shown that $((\mathcal PA)/\sim,d)$ is a metric space, but how do I show that it's complete?

An attempt I had was to take a $d$-Cauchy sequence $([E_n]_{\sim})_{n=1}^{\infty}$ and show that it $d$-converges to $\left[\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}E_k\right]_{\sim}$ or $\left[\bigcap_{j=1}^{\infty}\bigcup_{k=j}^{\infty}E_k\right]_{\sim}$ using the monotone convergence theorem but it didn't work.

This is from Tao's "An Introduction to Measure Theory" p. 35.

答:
Let $(E_n)_{n=1}^{\infty}$ be a $d-$Cauchy sequence. Using recursively the definition of $(E_n)_{n=1}^{\infty}$ being Cauchy we may find a strictly increasing sequence of integers
$$n_1<n_2<...<n_k<n_{k+1}<...$$
such that
$$\tag{1}d(E_n,E_m)=m^*(E_n\triangle E_m)<\frac{1}{2^k}$$
for every $n,m\geq n_k$. Let $E=\liminf_{k}E_{n_k}$. We claim that $E_n \overset{d}{\to}E$. To see this, first observe that
$$\tag{2} E\triangle E_{n_k}\subseteq \bigcup_{m=k}^{\infty}E_{n_m}\triangle E_{n_{m+1}}$$
Using $(1),(2)$ we obtain
\begin{align}
m^*(E\triangle E_{n_k})&\leq \sum_{m=k}^{\infty}m^*(E_{n_m}\triangle E_{n_{m+1}})\\
&\leq \sum_{m=k}^{\infty}\frac{1}{2^m}=\frac{1}{2^{k-1}}
\end{align}
Hence, for every $n\geq n_k$
\begin{align}
d(E_n,E)&=m^*(E_n\triangle E)=m^*\bigl((E_n\triangle E_{n_k})\triangle (E_{n_k}\triangle E)\bigr)\\
&\leq m^*(E_n\triangle E_{n_k})+m^*(E_{n_k}\triangle E)\\
&\leq \frac{1}{2^k}+\frac{1}{2^{k-1}}
\end{align}
The latter shows that $d(E_n,E)=m^*(E_n\triangle E)\to 0$ as $n\to \infty$.


问:
If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.
Let $(G,*)$ be a group. If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.

$a^2*b=b*a \implies ab=a^{-1}*b*a$.
I can't proceed further.

答:
Proof:
As a⁴=e. Then
b=a⁴b
b=(a²)²b
since a²b=ba
b=ba²
Since G is a group
b=a²b
Since
a²b=ba
b=ba=>a=e.


问:
Axiomatic construction of natural numbers and its properties from Zorich's book
I am reading the book of V.A.Zorich Mathematical analysis and trying to follow the formal construction of natural numbers and its properties:

Definition 1: The set $X\subset \mathbb{R}$ is called inductive, if for any $x\in R$ the element $x+1
\in \mathbb{R}$.

Definition 2: The set of natural numbers is defined as the minimal inductive set containing $1$, i.e. $$\mathbb{N}:=\bigcap_{A\in \mathcal A}A,$$ where $\mathcal A$ is the family of all inductive sets containing $1$ and we see that $\mathcal A\neq \varnothing$ because $\mathbb{R}\in \mathcal A$.

Principle of Mathematical Induction: If $E\subset \mathbb{N}$ with $1\in E$ and $\forall x\in E (x+1\in E)$ then $E=\mathbb{N}$.

I was able to show that:

If $m,n \in \mathbb{N}$ then $m+n\in \mathbb{N}$ and $mn\in \mathbb{N}$.

But I have some issues to show the following:

$(n\in \mathbb{N}) \land (n\neq 1) \Rightarrow ((n-1)\in \mathbb{N}).$

Proof: Let $E:=\{n-1\in\mathbb{N}\mid (n\in \mathbb{N}) \land (n\neq 1)\}$ then he show that $E$ satisfies the principle of Induction and hence $E=\mathbb{N}$ and we are done.

But in my opinion I don't think that this formally correct proof because if $E=\mathbb{N}$ then it does not imply our statement.

Can anyone explain me please? Maybe I am misunderstanding smth.

答:
I suspect there is a much cleaner way of showing this:

Zorich actually defines $ E=\{n-1\in\mathbb{R} | n\in \mathbb{N} \text{ and }n\neq 1 \} $. It is straightforward to show that $E$
is inductive and contains $1$. Then we have $\mathbb{N} \subset E$, of course. It remains to show that $E = \mathbb{N}$ (Zorich appears to have omitted this part).

First show that the $n \neq 1$ is equivalent to $ n \ge 2$:

Let $F=\{n \in\mathbb{N} | n = 1 \text{ or } n \ge 2 \}$. It is straightforward to show that $F$ is inductive, $1 \in F$ and since $F \subset \mathbb{N}$ we
see that $F = \mathbb{N}$.
Hence, if $n \in \mathbb{N}$ and $n \neq 1$ we have $n \ge 2$ and so
$N_2 = \{ n | n\in \mathbb{N} \text{ and }n\neq 1 \} = \{ n | n\in \mathbb{N} \text{ and }n \ge 2 \}$.
So, we have $E =\{n-1\in\mathbb{R} | n \in N_2 \} $.

Now show that $n \in N_2$ means that $n-1 \in \mathbb{N}$ (this is essentially 2.):

Define $\sigma:\mathbb{R} \to \mathbb{R}$ by $\sigma(n) = n+1$. From the properties of $+$ on the reals we know that $\sigma$ is a bijection on the reals. It is straightforward to show that $\{ n \in \mathbb{N} | \sigma(n) \in N_2 \}$ is inductive and contains $1$ hence it equals $\mathbb{N}$. In particular, $\sigma: \mathbb{N} \to N_2$ is well defined. To show that $\sigma$ is a bijection it is sufficient to show that $\sigma$ is surjective. Let $G = \{1\} \cup \{\sigma(n) | n \in \mathbb{N}\}$, it is straightforward to show that $1 \in G$ and that $G$ is inductive hence $G = \mathbb{N}$ and so $\{\sigma(n) | n \in \mathbb{N}\} = N_2$.

Finally, $E = \{ \sigma(n)-1 | n \in \mathbb{N} \}  = \{ n+1-1 | n \in \mathbb{N} \}  = \mathbb{N}$.


问:
How to stay organised and productive as a mathematician?
Like first time I try to self study before the lecture then I make notes to have better understanding and comprehension.

Then,I have notes during lecture , else I will feel sleepy and not able to remember things .

Then, again after coming back from lecture I make notes because I usually come with better insights and understanding.

Now, again while revising text I get new things and understanding while browsing and reading online sites and various university lecture notes.

But, now I am not able to understand how should I revise and remember things from various multiple sources.

Prepare for exams and scholarship exams.

答:
I will include what I find works best for me (this is probably not universal for everyone so take it with a pinch of salt).

Maths is no different to any other skill and you learn by doing. If you have a reading list in your course, this is often useful to find a rich source of exercises. Do many problems, even if you don't properly write them all down. You will get much better at a subject by solving problems than by memorising the proofs of all the important theorems. If your course does not have a reading list, you can usually find good recommendations for every subject online.

However, you might be expected to just memorise some proofs. What I usually do for this is typing them in LaTex. Wanting them to look nice usually makes me think a lot more about them, so that I eventually come up with the nicest method while keeping it concise. You will also have a really good set of notes because you will have made them and it will be easier to study everything.

Finally, if you have the time and motivation, I recommend reading a bit further on every topic you study, moving to slightly more advanced content. This will make all of the previous content look easier in comparison as well as force you to understand well everything before it.


问:
Sum of two independent binomial variables
How can I formally prove that the sum of two independent binomial variables X and Y with same parameter p is also a binomial ?</p>

答:
We can prove this using Moment generating function as follows if someone is not comfortable with characterstic functions as answered above:

Let $X \sim B(n_1,p_1)$ and $Y \sim B(n_2,p_2)$ be independent random variables.

We know the MGF of the Binomial distribution is as follows:

$M_X(t)=(q_1+p_1e^t)^{n_{1}},M_Y(t)=(q_2+p_2e^t)^{n_{2}} $

Since X and Y are independent

$M_{X+Y}(t)=M_X(t) \cdot M_y(t) =(q_1+p_1e^t)^{n_{1}} \cdot (q_2+p_2e^t)^{n_{2}}$

We see that we cannot express it in the form $(q+pe^t)^{n}$ and thus by uniqueness property of MGF $X+Y$ is not a binomial variate.
However if we take $p_1=p_2=p$ then we have:

$M_{X+Y}(t)=M_X(t) \cdot M_y(t) =(q+pe^t)^{n_{1}} \cdot (q+pe^t)^{n_{2}}$
$=(q+pe^t)^{n_{1}+n_{2}} $

which is MGF of binomial variate with parameters $(n_1+n_2,p)$


问:
Complex:Schwarz lemma
Let $f:\mathbb{D} \rightarrow \mathbb{D}$ be analytic such that $f(0)=0$ and $f'(0)=0$. Then which of the following are possible values of $f(1/2)=$

i. $\dfrac{1}{4}$

ii.$\dfrac{-1}{4}$

iii.$\dfrac{1}{3}$

iv.$\dfrac{-1}{3}$

By Schwarz lemma, all the options are correct. But does $f'(0)=0$ shorten the possible value list?

答: Let $g(z)=\frac {f(z)} {z^{2}}$. Since $f(0)=f'(0)=0$ it follows that $g$ has  removable singularity at $0$. Now $|g(z)| \leq \frac1 {(1-r)^{2}}$ on $|z|=1-r$. By MMP the inequality holds for all $z$ with $|z| \leq 1-r$. Letting $r \to 0$ we see that $|g(z)| \leq 1$ for all $z$. Thus, $|f(\frac 1 2)| \leq \frac  1 4$ ruling out iii) and 1v). The first two are possible: just look at $f(z)=z^{2}$ and $f(z)=-z^{2}$.

Title: Complex:Schwarz lemma


问:
Evaluating $\lim_{x \to 0} \dfrac{\arctan x}{x}$

Evaluate $$\lim_{x \to 0} \dfrac{\arctan x}{x}$$

I know that it's $1$ using L'Hopital's rule or numerically, but what about algebraically? Thanks in advance and I really appreciate it.

答:
With $y=\tan x$,
$$\lim_{x\to 0}\frac{\arctan x}{x}=\lim_{y\to 0}\frac{y}{\tan y}=\lim_{y\to 0}\frac{y\cos y}{\sin y}=\lim_{y\to 0}\cos y\cdot \lim_{y\to 0}\frac y{\sin y}=\lim_{y\to 0}\frac y{\sin y}=1$$


答: Use the power series expansion of $\arctan$, $\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\ldots$, to get $\arctan(x)/x=1-\frac{x^2}{3}+\frac{x^4}{5}-\ldots$.

Title: Evaluating $\lim_{x \to 0} \dfrac{\arctan x}{x}$

Body:
Evaluate $$\lim_{x \to 0} \dfrac{\arctan x}{x}$$

I know that it's $1$ using L'Hopital's rule or numerically, but what about algebraically? Thanks in advance and I really appreciate it.


答:
You can use the series expansion at $x = 0$

$$\dfrac{\arctan x}{x} = 1 - \dfrac{x^2}{3} + \dfrac{x^4}{5} + O(x^6)$$

Thus
$$\lim_{x \to 0} \dfrac{\arctan x}{x} = \lim_{x \to 0} {1 - \dfrac{x^2}{3} + \dfrac{x^4}{5} + O(x^6)} = 1 $$


问:
What is the integral of three orthonormal basis functions?
The hyperspherical harmonics, given by:

$Z_{l,m}^n(\omega,\theta,\phi)=(-i)^l\frac{2^{l+1/2}l!}{2\pi}\sqrt{(2l+1)\frac{(l-m)!}{(l+m)!}\frac{(n+1)(n-l)!}{(n+l+1)!}}\sin^l(\omega/2)C_{n-l}^{l+1}(\cos(\omega/2))P_l^m(\cos\theta)\exp(im\phi)$

(where $C_{n-l}^{l+1}$ is a Gegenbauer polynomial, and $P_l^m$ is an associated Legendre function), form an orthonormal basis for an expansion of functions on the hypersphere.

As such, the following identity is true:

$\int_\Omega Z_{l,m}^n(\omega,\theta,\phi)Z_{l',m'}^{n'}(\omega,\theta,\phi)d\Omega=\delta_{n,n'}\delta_{l,l'}\delta_{m,m'}$

My question is whether or not the following identity is also true:

$\int_\Omega Z_{l,m}^n(\omega,\theta,\phi)Z_{l',m'}^{n'}(\omega,\theta,\phi)Z_{l'',m''}^{n''}(\omega,\theta,\phi)d\Omega=\delta_{n,n',n''}\delta_{l,l',l''}\delta_{m,m',m''}$

Note: $\Omega$ is the entire domain, i.e., $\omega\in[0,\pi], \theta\in[0,\pi], \phi\in[0,2\pi]$, and $d\Omega=\frac{1}{2}\sin^2(\omega/2)\sin(\theta)d\omega d\theta d\phi$

答: It seems unlikely.  The product of two orthogonal polynomials can have a component of a third. For example, in the Legendre polynomials, $$\int_{-1}^1P_1(x)P_2(x)P_3(x)\;dx=\int_{-1}^1x\frac{3x^2-1}{2}\frac{5x^3-3x}{2}\;dx=\frac{6}{35}$$</p>


问:
Calculating sum of fractions without calculator
Calculate $$\dfrac{1}{25}+\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+\dfrac{1}{29}=?$$ without using calculator

Update: Is there any shortcut to determine the sum of this type of fractional series?

Body:
$$S=\dfrac{1}{25}+\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+\dfrac{1}{29}=\frac 1 {25}\left(1+\sum_{n=1}^4\frac{1}{1+\frac n{25} }\right)$$ Now, by Taylor
$$\frac 1{1+\frac n{25}}=1-\frac{n}{25}+\frac{n^2}{625}-\frac{n^3}{15625}+O\left(n^4\right)$$
$$\sum_{n=1}^4\frac{1}{1+\frac n{25} }=\frac{2276}{625}$$
$$S=\frac 1 {25}\left(1+\frac{2276}{625}\right)=\frac{2901}{15625}=0.1856640$$ while the exact value is
$$S=\frac{1323137}{7125300}\sim 0.1856956$$ Then a relative error of $0.017$%.

Done by hand

Edit

Another solution
$$S=\dfrac{1}{25}+\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+\dfrac{1}{29}$$ $$S=\dfrac{1}{27-2}+\dfrac{1}{27-1}+\dfrac{1}{27}+\dfrac{1}{27+1}+\dfrac{1}{27+2}$$ Factor
$$27S-1=\frac 1 {1-\frac 2 {27}}+\frac 1 {1+\frac 2 {27}}+\frac 1 {1-\frac 1 {27}}+\frac 1 {1+\frac 1 {27}}$$ Now
$$\frac 1{1-\epsilon}+\frac 1{1+\epsilon}=2+2\epsilon^2+O\left(\epsilon^4\right)$$
$$27S-1=2+\frac{8}{729}+2+\frac{2}{729}=\frac{2926}{729}\implies S=\frac{3655}{19683}\sim 0.1856932$$ which is much better (but I prefer to divide by $25$ rather by $27$ even if $27=3^3$).

Still done by hand


问:
series of bounded measurable functions converge implies coefficients convergence
$\{f_n\}$ sequence of non-negative measurable bounded functions on [0,1], $f_n(x) \leq M$ for all $n$  and almost all $x$. $\int_0^1 f_n = 1$ for all $n$. If $\{a_n\}$ be any sequence of non-negative real numbers satisfying $\sum a_n f_n(x) < \infty$ for almost all $x$, show that $\sum a_n < \infty$.

I could not figure out how to do this. Any hint would be appreciated.

答:
Consider $g_k(x)=\sum_{n\geq k}a_nf_n(x)$. This is a series of non-negative measurable functions, hence is non-negative and measurable, and $g_1\geq g_2\geq \dots$. Also, the assumption $\sum_{n\geq 1}a_nf_n(x)<\infty$ a.e implies that $g_k\to 0$ pointwise a.e (if a series is finite, then the tail of the series has to converge to zero... almost by definition).

Now, since $[0,1]$ has finite measure, we can use Egoroff's theorem to deduce that for each $\epsilon>0$, there is a set $E\subset [0,1]$ with $\mu(E)<\epsilon$ and such that $g_k\to 0$ uniformly on $E^c$. In particular, there exists an $N\in\Bbb{N}$ such that $\sup\limits_{x\in E^c}g_N(x)\leq 1$. So, (using monotone convergence to interchange sum and integral)
\begin{align}
\mu(E^c)\geq \int_{E^c}g_N = \sum_{n\geq N}a_n \int_{E^c}f_n \geq \sum_{n\geq N}a_n(1-M\epsilon) \geq \frac{1}{2}\sum_{n\geq N}a_n,
\end{align}
where the last inequality can be achieved by choosing $\epsilon>0$ so small that $1-M\epsilon \geq \frac{1}{2}$. Thus, $\sum_{n\geq N}a_n \leq 2\mu(E^c)<\infty$. Since the tail of the series is finite, the whole series $\sum_{n\geq 1}a_n$ is also finite.

By the way, there's nothing special about $\frac{1}{2}$ in the proof above; all we need is to choose $\epsilon>0$ small enough so that $1-M\epsilon>0$. Also, as the proof shows, there's nothing special about $[0,1]$; the same is true on any finite measure space.

Edit:

Here are some extra comments regarding the thought process of the solution.

Egoroff's theorem is an exceptionally useful technical result that now whenever I see a finite-measure assumption, I immediately think about whether it is helpful for the problem at hand. Once we have that in mind, note that Egoroff requires us to look at sequences of complex-valued functions which converge pointwise a.e. In this problem, we're given information about the sum $f:=\sum\limits_{n\geq 1}a_nf_n$. From here, there are two natural sequences to consider:

First is the partial sums $s_k=\sum\limits_{i=1}^ka_if_i$.

Second is the tail $g_k=\sum\limits_{n\geq k}a_nf_n$, as I considered above.

By definition, we can always say that $s_k\to f$. So, this isn't any extra information for us. Our hypothesis tells us that $f$ is finite a.e. so this gives us the extra information that $g_k\to 0$ a.e. That's why I decided to focus on the sequence $g_k$.

Next, by looking over the proof, we see that the key idea is the lower bound $\int_{E^c}f_n\geq \frac{1}{2}>0$. In other words, the crucial part of this proof is not the assumption that $f_n$'s are uniformly bounded with integral $1$. Rather what we want is that the $f_n$'s are "reasonably-distributed" functions, in the sense that their mass isn't concentrated on sets of arbitrarily small measure. Let us make the following definition:

Let $(X,\mathfrak{M},\mu)$ be a measure space and $\{f_n\}_{n=1}^{\infty}$ be a sequence of non-negative. We shall say $\{f_n\}_{n=1}^{\infty}$ is "reasonably-distributed" with respect to $\mu$ if there is a $\delta>0$ such that for any measurable set $E\subset X$ with $\mu(E)<\delta$, we have $\liminf\limits_{n\to\infty}\int_{E^c}f_n\,d\mu>0$.

The assumptions in the problem, that $\int f_n=1$ and $f_n$'s are uniformly bounded certainly imply the above condition, but not conversely. So, we have a slight strengthening of the result:

Suppose $(X,\mathfrak{M},\mu)$ is a finite measure space, and $\{f_n\}_{n=1}^{\infty}$ is a sequence of non-negative integrable functions which are "reasonably-distributed" with respect to $\mu$. If $\{a_n\}_{n=1}^{\infty}$ is any positive sequence of numbers such that $\sum\limits_{n=1}^{\infty}a_nf_n$ is finite a.e. then $\sum\limits_{n=1}^{\infty}a_n<\infty$.


答:
I did all of the following without writing any "multiplication algorithms" down, and it's based on the assumption that one can "add two numbers" and "multiply any number with a one-digit number" quickly, and we wrote down all previous results for later use.

Notice that
$$27\times29=87\times9=783$$

First we calculate
$$25\times26\times27\times28\times29=25\times8\times7\times13\times783$$
$$=200\times(90\times783+783)=200\times(78300+783-7830)=200\times(79253-8000)$$
$$=200\times71253=14250600$$

Next we calculate
$$26\times27\times28\times29=8\times71253=570024$$

Next we calculate
$$25\times27\times 28\times 29 = 25\times4\times7\times783=100\times5481=548100$$

Next we calculate
$$25\times26\times28\times29=25\times28\times(27-1)\times29$$
$$25\times4\times7\times(783-29)=100\times7\times754=527800$$

Next we calculate
$$25\times26\times27\times29=50\times13\times783=700\times783-50\times783=548100-39150=508950$$

Finally we calculate
$$25\times26\times27\times28=25\times28\times27\times(29-3)$$
$$=25\times4\times7\times(783-81)=700\times702=491400$$

Now addition time,

$$570024+548100+527800+508950+491400=2646274$$

Now the final result,

$$2646274\over 14250600$$

Title: Calculating sum of fractions without calculator

Body:
Calculate $$\dfrac{1}{25}+\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+\dfrac{1}{29}=?$$ without using calculator

Update: Is there any shortcut to determine the sum of this type of fractional series?


问:
Unclear choice of polynomial in Gaussian Quadrature method?
Recently, I have been reading a paper regarding the evaluation of an integral using the Gaussian quadrature method but I cannot understand the rationality behind the weight and orthogonal polynomials the author has chosen.

$\begin{gathered}
  \int\limits_{{s_1}}^{{s_2}} {f\left( x \right)dx}  \approx \frac{{\left( {{s_2} - {s_1}} \right)}}{2}\sum\limits_{s = 1}^{\operatorname{Max} \,\,sample} {\frac{{\pi \sqrt {1 - {v^2}} }}{{\operatorname{Max} \,\,sample}}f\left( u \right)}  \hfill \\
  v = \cos \left( {\frac{{\left( {2s - 1} \right)\pi }}{{2 \times \operatorname{Max} \,\,sample}}} \right) \hfill \\
  u = \frac{{\left( {{s_2} - {s_1}} \right)}}{2}v + \frac{{\left( {{s_2} + {s_1}} \right)}}{2} \hfill \\
\end{gathered} $

I understand that $u = \frac{{\left( {{s_2} - {s_1}} \right)}}{2}v + \frac{{\left( {{s_2} + {s_1}} \right)}}{2}$ is some smart change of variable to move the integration region to $\left[ { - 1,1} \right]$ but I do not know why a trigonometric function such as cos is chosen ?

From my domain knowledge, in this application the function $f\left( x \right)$ is something that look like a Gaussian bell-shaped. Does this knowledge justify the author's choice ?

Please help me understand this !

Thank you

答:
That's Gauss-Chebyshev quadrature and will be the most accurate if $f(x)$ has branch points like $(x-s_1)^{n-1/2}$ and $(s_2-x)^{m-1/2}$ for nonnegative integers $n$ and $m$ at the endpoints. If  $f(x)$ is regular at one or both endpoints, not as great but still can be pretty accurate. For regular $f(x)$ Gauss-Legendre quadrature would be more accurate although it requires more effort to set up. Clenshaw-Curtis quadrature can be easier to set up than Gauss-Legendre in that it uses the straightforward Gauss-Chebyshev quadrature to expand the function $f(x)$ in Chebyshev polynomials and then uses their known integrals. It would take more information to decide whether the author is using an optimal quadrature strategy or something quick and dirty.

EDIT: The issue is that the error in Gaussian quadrature is equal to $$\left(\frac{s_2-s_1}L\right)^{2n+1}C_nf^{(2n)}(\xi)$$ where $n$ is the number of data points, $L$ is the length of the model interval, $2$ for Gauss-Chebyshev quadrature, $s_1<\xi<s_2$, and $C_n$ is a constant depending on the quadrature algorithm and the number of data points. If $f(x)$ looks like $\sqrt{x-s_1}$ or $\sqrt{s_2-x}$ at one of its endpoints then the derivatives of $f(x)$ are unbounded so the error could be anything. Gauss-Chebyshev uses a weight function $w(x)=(1-x^2)^{-1/2}$ so this creates an effective $g(x)=\sqrt{(x-s_1)(s_2-x)}f(x)$ and this is the function whose derivatives should all remain finite at the endpoints if we want to impose a bound on the error of the quadrature as given.

As I mentioned, Gauss-Chebyshev quadrature may even give good results if $g(x)$ has unbounded derivatives but other methods may yield better results with fewer data points.


问:
Whats wrong with my counterexample for an MST CUT?
bevor I start, I want to mention, that there is an quivalent problem here How to I have to understand the Cut in an graph in this case?

But the answer doesn't help me so.

Following Problem:
Given is an Graph $G=(V,E)$, a minimal spanning Tree $G'=(V,E')$ with $k : E \rightarrow \mathbb{R}$.

Let $f = \{x, y\} \in E'$ and let $A$ be the set of all nodes reachable from $x$ in $(V, E' \setminus {f})$. Then for every edge $e \in \delta(A)$ ($\delta(A) := \{ e = \{v, w\} \in E \mid v \in A \text{ and } w \in V \setminus A \}$) it holds that $c(e) \geq c(f)$.

Now i  tired to understand this definition, but if I draw it, I get this
drawed counterexample

What I am I doing wrong

答: In your example you write $A = \{6,7\}$, but this is incorrect: the set of all nodes reachable from $x$ includes (trivially) $x$. So in your case $A = \{4,6,7\}$, and then $\delta(A)$ indeed only contains edges with weight at least that of $f$.


问:
If $\lim\limits_{x \to \pm\infty}f(x)=0$, does it imply that $\lim\limits_{x \to \pm\infty}f'(x)=0$?

Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is everywhere differentiable and

$\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)=0$,

there exists $c \in \mathbb{R}$ such that $f(c) \gt 0$.

Can we say anything about  $\lim_{x \to \infty}f'(x)$ and $\lim_{x \to -\infty}f'(x)$?

I am tempted to say that $\lim_{x \to \infty}f'(x)$ = $\lim_{x \to -\infty}f'(x)=0$.

I started with the following, but I'm not sure this is the correct approach, $$\lim_{x \to \infty}f'(x)= \lim_{x \to \infty}\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}.$$

答: No. Try $f(x)=\sin(x^a)/x$ for various values of $a$.</p>


问:
Degree d integral curves are contained in a "linear $\mathbb P^d$"
This is Vakil 18.6 M, self-study.

We are to show that an integral degree $d$ curve $C$ in $\mathbb P^N_k$ (with $N \geq d$ and $k$ WLOG algebraically closed) is contained in a "linear $\mathbb P^d_k \subset \mathbb P^N_k$."

So, far I see that if you take a hyperplane $H$ in $\mathbb P^N_k$ intersecting $C$ but not containing it, then by Bezout's Theorem (18.6 K in Vakil),

$$\deg C \cap H = \deg C \deg H = d$$

where the degree indicates the leading coefficient of the Hilbert polynomial times $n!$, where $n$ is the dimension of the space. The problem is that I do not see what this tells us geometrically. I can see the leading coefficient of the Hilbert polynomial of $C \cap H$ must have been $d$ to begin with, but this seems purely formal.

I have studied both answers here, but we have not covered general position (the content of the first answer), and I do not follow the conclusion of the second answer, in addition to the fact that it does not seem to use Bezout's Theorem, at least as we have covered it. I suspect neither answer is what Vakil is looking for, and that I am simply missing some geometric realization about degree that would quickly convert what I have done thus far into an answer.

By "a linear $\mathbb P^d_k$" I assume Vakil means a linear space of dimension $d$ in $\mathbb P^N_k$, meaning a closed subscheme cut out by $N - d$ homogeneous linearly independent linear polynomials.

答:
Continued from the comments - we established there that if $C\not\subset H$, then $|C\cap H|\leq d$. If you can only ever get at most $d$ points with a proper intersection, what happens when you pick a hyperplane through $d+1$ points on $C$? Full details under the spoiler.

 Pick $d+1$ distinct points on $C$. If $d<n$, then there is a hyperplane $H$ through these $d+1$ points. If $C\not\subset H$, then $H\cap C$ is a zero-dimensional subscheme of degree $\geq d+1$, contradicting Bezout's theorem. So $C\subset H$. Repeat until $d=n$.


问:
Find the domain for a function with 2 or more variables.
I have the following function:

```<code>f(x,y)=4x+y/(2x-5y).
</code>```

For this function, I have to find the domain of the function, and to represent it graphically.

答:
If $f(x,y):=\dfrac{4x+y}{2x-5y}$, we only have to be sure that the term $2x-5y$ is not equal to zero.

So $5y\ne 2x \iff y\ne\dfrac{2}{5}x\implies \text {dom}(f)=\{(x,y)\in \mathbb R^2:y\ne\dfrac{2}{5}x\}$.

The domain is represented by the set $\mathbb R^2$ \ $r$, where $r\equiv y=\dfrac{2}{5}x$.


问:
Prove that $\triangle AMC$ is an isosceles right triangle

In $\triangle ABC$ construct squares $ABDE$ and $BCFG$. Let $M$ be the midpoint of $EF$ then, prove that $\triangle AMC$ is an isosceles right triangle.

I was able to prove that $\triangle AMC$ is an isosceles only and here is my solution:

The area of $\triangle BAF$ is equal to the area of $\triangle EBC$ so we get
\begin{equation}
AE×AF\sin \angle BAF=CF×CE\sin \angle ECB
\end{equation}
And using law of cosines on $EF$ we get
\begin{equation}
AE^2+AF^2=CF^2+CE^2
\end{equation}
By the length formula of median $AM$ and $CM$ we get that $AM=CM$. I know this solution is kinda rush and not give much detail but I'm not really good at Latex, please don't mind me. Please help prove that it is the right triangle. Thanks!

BTW, $N$ in the photo is the point where $AE$ meet $CF$

答:
Let $P$ and $Q$ be the midpoints of $BF$ and $BE$ respectively. Draw $MQ$, $MP$, $AQ$ and $CP$.

Notice that, $\angle AQM=\angle EQM-\angle EQA=\angle EBP-90=\angle ABC$ [Since $\angle EBA+\angle CBP=90^{\circ}$] and similarly, $\angle MPC=\angle ABC$.

Observe that, in $\triangle AQM$ and $\triangle MPC$,

$\left(i\right)$  $AQ=QB=MP$

$\left(ii\right)$  $QM=BP=PC$

$\left(iii\right)$  $\angle AQM=\angle ABC=\angle MPC$

Hence, $\triangle AQM\cong MPC$ by $S-S-A$ criterion of congruence.

Thus, $AM=MC$ and $\angle AMQ=\angle MCP$.

$\Rightarrow   \angle AMC=360-\angle AMQ-\angle QMP-\angle CMP$

$=\left(180-\angle QMP\right)+\left(180-\angle MCP-\angle CMP\right)$

$=\angle MPB+\angle MPC=\angle BPC=90^{\circ}$

In $\triangle AMC$, $AM=MC$ and $\angle AMC=90^{\circ}$; Therefore, it is an isosceles right triangle.


问:
"Set of all formal products" - what does this mean?
List the set of all formal products of $(1+x^2+x^4)^2(1+x+x^2)^2$ with exponents summing to $4$.

What is this question asking exactly? What is a "formal product"? Does it have anything to do with formal power series? How can be different products of the same expression? What counts as a valid "formal" product of an expression? Obviously if I just simplify the expression I get one thing, not a set. The book seems to not define "formal product", only "product of formal series" which yields a simplified form that I don't think is a set of all possible products.

答:
Multiplying out the given expression you get a certain polynomial in $x$ of degree $12$. The coefficient of $x^4$ in this polynomial collects all the terms you are interested in, but it is a single number, namely $10$. If you want the individual products leading to terms of exponent $4$ exhibited you can do the following: Define
$$a:=(u_01+ u_2 x^2 + u_4x^4)^2(v_01+ v_1 x+ v_2 x^2)^2$$
and determine the coefficient of $x^4$ in the expansion of $a$. Mathematica gives
$$u_2^2v_0^2+ 2u_0u_4v_0^2+2u_0 u_2v_1^2+4u_0u_2 v_0v_2+u_0^2v_2^2\ .$$
From this expression you can immediately deduce in which way the $10$ terms of exponent $4$ come about.</p>


问:
How many variables in the multinomial expansion have an exponent that's different from 2?
Multinominal: $(a + b + c + d)^{10}$

Question: How many variables in the multinomial expansion have an exponent that's different from 2?

I want to solve it using generating functions.

So we can write: $t_1 + t_2 + t_3 + t_4 = 10$, When $t_i$ cannot use $x^2$.

That equals to the generating function:
$$
(1+x+x^3+x^4...)^4 = (1+x+\frac{x^3}{1-x})^4
$$
But now what? How can I keep expending this till I find the coefficient of $x^{10}$?

答:
You can rewrite $$(1+x+x^3+x^4...)^4=(1+x)^4(1+x^3+x^5+x^7+...)^4$$

$$=(x^4+4x^3+6x^2+4x+1)(1+x^3+x^5+x^7+x^9+...)^4$$

Now first look at the $x^4$ term: the only way to add exponent up to $10$ is $4+3+3=10$, therefore we have ${4\choose 2}=6$ ways.

Next look at $4x^3$ term, the only way is $3+7=10$, therefore we have $4{4\choose 1}=16$ ways.

Next look at $6x^2$ term, the only way is $2+3+5=10$, therefore $6{4\choose 1}{3\choose 1}=72$ ways.

Next look at $4x$ term, we have $1+9=1+3+3+3=10$, therefore $4({4\choose 1}+{4\choose 3})=32$ ways.

Next look at $1$ term, we have $0+3+7=0+5+5=10$, therefore ${4\choose1}{3\choose1}+{4\choose 2}=18$ ways.

Finally add them all up, $6+16+72+32+18=144$ ways.


Body:
Multinominal: $(a + b + c + d)^{10}$

Question: How many variables in the multinomial expansion have an exponent that's different from 2?

I want to solve it using generating functions.

So we can write: $t_1 + t_2 + t_3 + t_4 = 10$, When $t_i$ cannot use $x^2$.

That equals to the generating function:
$$
(1+x+x^3+x^4...)^4 = (1+x+\frac{x^3}{1-x})^4
$$
But now what? How can I keep expending this till I find the coefficient of $x^{10}$?

Body:
$(1 + x + x^3 + x^4 + ...)$ can be simplified as $[(1 - x)^{-1} - x^2]$.

Now, we had to find coefficient of $x^{10}$ in ${[(1 - x)^{-1} - x^2]}^4$

${[(1 - x)^{-1} - x^2]}^4$ = $\binom{4}{0}$ $[({1 - x}) ^ {-4}]$ - $\binom{4}{1}$ $[({1 - x}) ^ {-3}]$ $x^2$ + $\binom{4}{2}$ $[({1 - x}) ^ {-2}]$ $x^4$ - $\binom{4}{3}$ $[({1 - x}) ^ {-1}]$ $x^6$ + $\binom{4}{4}$ $x^8$

Hence the answer will be $\binom{13}{3}$ - 4 $\binom{10}{2}$ + 6 $\binom{7}{1}$ - 4 $\binom{4}{0}$ i.e. 144


问:
Find the sum of all the digits of the least positive integer $n$ such that $(P_2 * P_3 * $ ... $ * P_n)$ exceeds $2010$ .

Let $P_k = 1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3}$ , where $k$ is a positive integer. Find the sum of all the digits of the least positive integer $n$ such that $(P_2 * P_3 * $ ... $ * P_n)$ exceeds $2010$ .

What I Tried: I have that :-
$$\rightarrow P_k = 1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3} = \frac{(k - 1)(k + 1)^2}{k^3}$$
But I could not find way to use this. I thought this would give me some way to work on the problem but no.

Basically I have to find the least $n$ such that :-
$$\displaystyle\prod_{k = 2}^{n} \bigg(1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3}\bigg) > 2010$$

I have no idea on how to find $n$ . I used Wolfram Alpha mainly for Trial and Error and to get the solution, and I found $n = 8038$ .

Check this here:- https://www.wolframalpha.com/input/?i=Product+k+%3D+2+to+8038%2C++%281+%2B+1%2Fk+-+1%2Fk%5E2+-+1%2Fk%5E3%29

So the final answer would be $19$ , but mathematically, how can you find $n$?

Can anyone help me?

答:
Hint You already observed the key property

$$ P_k =  \frac{(k - 1)(k + 1)^2}{k^3}$$

Therefore
$$P_2*P_3 *P_4*...*P_n=  \frac{1(3)^2}{2^3} \frac{2(4)^2}{3^3} \frac{3(5)^2}{4^3} ... \frac{(n - 1)(n + 1)^2}{n^3}$$

Now, just cancel as much as you can, what do you have leftover?


答:
You can rewrite the product as telescoping product:

$$\prod_{n=2}^n P_k= \prod_{n=2}^n \left(\frac{k+1}{k}\right)^2\cdot \prod_{n=2}^n \frac{k-1}k = \frac{(n+1)^2}4\cdot \frac 1n >2010$$

$$\Leftrightarrow n+2+\frac 1n >8040$$

I think, now you can take it from there.

Body:
Let $P_k = 1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3}$ , where $k$ is a positive integer. Find the sum of all the digits of the least positive integer $n$ such that $(P_2 * P_3 * $ ... $ * P_n)$ exceeds $2010$ .

What I Tried: I have that :-
$$\rightarrow P_k = 1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3} = \frac{(k - 1)(k + 1)^2}{k^3}$$
But I could not find way to use this. I thought this would give me some way to work on the problem but no.

Basically I have to find the least $n$ such that :-
$$\displaystyle\prod_{k = 2}^{n} \bigg(1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3}\bigg) > 2010$$

I have no idea on how to find $n$ . I used Wolfram Alpha mainly for Trial and Error and to get the solution, and I found $n = 8038$ .

Check this here:- https://www.wolframalpha.com/input/?i=Product+k+%3D+2+to+8038%2C++%281+%2B+1%2Fk+-+1%2Fk%5E2+-+1%2Fk%5E3%29

So the final answer would be $19$ , but mathematically, how can you find $n$?

Can anyone help me?


答: $\lim_{x\to 0}\dfrac{\arctan x}{x}=\lim_{x\to 0}\dfrac{\arctan x-\arctan 0}{x-0}=\dfrac{d\arctan(x)}{dx}(0)=\cos^2(\arctan 0)=1.$


问:
Bounded net of complex numbers
I'm still unfamiliar with nets and I'm trying to get some intuitions about them right. Here is one question I have.

Question Let $\{x_\alpha\}\subseteq\mathbb C$ be a convergent net converging to $x\in \mathbb C$. If $|x_\alpha|\leq 1$ for all $\alpha$, then $|x|\leq 1$.

Attempt Suppose not, i.e., $|x|>1$. Pick an $r>0$ small enough so that the ball $B(x,r)$ centered at $x$ with radius $r$ does not intersect the unit disk $\mathbb D$. Since $B(x,r)$ is open, there exists $\alpha_0$ such that for all $\beta\geq \alpha_0$ $x_\beta\in B(x,r)\subseteq \mathbb C\setminus \mathbb D$. In other words, $|x_\beta|>1$, which is a contradiction.

Is this correct or there is a subtle flaw somewhere?

答: Your argument is indeed correct. It can be used to something more general: if $C$ is closed and $x_\alpha$ is a net from $C$ converging to $x$ then $x \in C$. This can be applied to the closed set $C=\{z \in \Bbb C: |z| \le 1\}$.

Title: Bounded net of complex numbers


问:
Function such that $f(x) = -1$ for $x < 0,$ and $f(x)=1$ for $x > 0$?
What is a function to returns $-1$ if number is negative, $1$ if positive, and zero if number is equal to 0?

for example:

$$
f(-8) = -1
$$
$$
f(8) = 1
$$
$$
f(0) = 0
$$

for $$x < 0$$ maybe?

$$ f(x) = (-x-(-x-1))\cdot-1 $$

答:
As pointed out in the other answers, this function goes by various names.  However, if your question is just "what is such a function?" then the answer is that it is the function you have defined.  Note that

$$f(x) = \begin{cases} -1 & \text{if } x<0\\
1 & \text{if } x>0\\
0 & \text{if } x=0
\end{cases}$$

is a perfectly valid definition of a function on $\mathbb{R}$, because it uniquely specifies what $f(x)$ is for every real number $x$.  Definitions by cases are just as good as any other kind of definition, and I think it is confusing to hide the distinction between the three cases (negative, positive, and zero) by some algebraic trick.


问:
A good practise book suggestion
I am a beginner in Special Functions. So I am looking for a Good Practise/Application  Book  about the  properties of Hypergoemetric Function  and its relations with orthogonal polynomials. I want to learn finding
derivative, contiguous relations, integral representation, differential equalities and transformations practisely. You can suggest a thesis or article
, pdf, too. Thanks for your comments.

答: I don't own it, but Special Functions: A Graduate Text might be worth looking into.</p>

Title: A good practise book suggestion