问:
Positive logical definition of finiteness
In his debate with Cassirer and in some other writing, Heidegger tries to define positively the concept of finiteness for the beings. It is a novel and difficult approach as traditionally, in theology and metaphysic, finiteness of beings is negatively defined as an imperfection or a negation, the absence of infiniteness, as opposed to some perfect and infinite being like god.

I feel that this difficulty is strangely reflected in set theory. I reckon we can naturally associate positive definition with $\Sigma_1$ sets (it abides to our definition if there is one set such that... ) and negative definitions with $\Pi_1$ sets (it abides to our definition if there is not set such that...).

Now for example Dedekind definition of an infinite set is a positive definition, a set is infinite if there is a bijection between itself and a proper subset. Then the natural way of defining finite sets would be non-infinite set which is $\Pi_1$ or a negative definition.

We could also use $\omega$, the smallest set satisfying the axiom of infinity and say that another set is finite if there is no injection from $\omega$ in this set ; but once again its a negative definition.

So formally, what this boils down to is the following :

Is the definition
of finiteness in set theory $\Pi_1$ or $\Delta_1$ ? I've just seen that
it is $\Delta_1$ according to wikipedia, then what is a $\Sigma_1$
definition of it ?Also, have the philosophical implications of this already been discussed ?

And for a subsidiary questions :

Are there models of set theory that "think" one of its element is infinite when it is actually finite ?
Akin to what happens between countable and non-countable sets with the Löwenheim-Skolem theorem.

答:
Yes, finiteness is $\Delta_1$ in set theory.

A set $x$ is finite iff there are $f,\alpha$ such that:

$f$ is a bijection from $x$ to $\alpha$;

$\alpha$ is a nonzero ordinal; and

every ordinal $<\alpha$ has an immediate predecessor.

This is a $\Sigma_1$ definition, the only subtle point being that ordinalhood is $\Delta_1$: to see this, use the "hereditarily transitive set" characterization of ordinals.

Note meanwhile that unlike the $\Sigma_1$ definition above, the $\Pi_1$ definition of finiteness coming from the $\Sigma_1$ definition of infiniteness you give assumes "Dedekind-finite = finite" (the definition of "finite" in $\mathsf{ZF}$ is "in bijection with an ordinal $<\omega$"). While easily provable in $\mathsf{ZFC}$ (or indeed vastly less), this is not a theorem of $\mathsf{ZF}$ alone.

Here's a $\Pi_1$ characterization of finiteness which works in $\mathsf{ZF}$ alone:

A set $x$ is finite iff there is no nonempty successor-closed set of ordinals $I$ together with a set of maps $\{f_a: a\in X\}$ such that each $f_a$ is an injection from $a$ to $x$.

The point is that in $\mathsf{ZF}$, a set is infinite iff every finite ordinal injects into it. In $\mathsf{ZF}$ this falls short of the existence of an injection from $\omega$ (= Dedekind-infiniteness), but is still fundamentally of the same "shape." EDIT: As Asaf shows, Dedekind-finiteness is not $\Delta_1$ in $\mathsf{ZF}$ alone, so we really have divergent complexity behaviors here.

As a coda, note that your question is in terms of the Levy hierarchy. Other notions of complexity can yield more complicated situations - see e.g. here.

As to your subsidiary question, models of $\mathsf{ZFC}$ can only be confused about finiteness in one way: $M\models\mathsf{ZFC}$ may have elements it thinks are finite which are actually infinite, but it cannot have elements it thinks are infinite which are actually finite.

The former point is a consequence of the compactness theorem, and the argument is identical to the existence of nonstandard models of arithmetic. To see the latter point, just note that $M$ is correct about the finiteness of all sets of size $0$ (there's only one of those) and if $M$ is correct about the finiteness of all sets of size $n$ (for $n$ finite) then $m$ is correct about the finiteness of all sets of size $n+1$ (think about what $M$ knows about the impact on cardinality of adding a single element to a set).


问:
How to prove that this function is continuous
I was working on proving that $S^1$\{(1,0)} with the induced topology by $R^2$ ( product topology) is homeomorphic to the interval ]0,2π[..the homeomorphism is the function f(x)=(Cos(x),Sin(x)).. I already proved the bijection but I'm stuck proving the continuity .. Knowing that a function is continuous if the inverse image of every open set is open.
Can anyone help me.

答:
To show that $f$ is a homeomorphism, we can use the following. If $f:X\rightarrow Y$ is a continuous bijection where $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

Then as @blamethelag suggested, you can prove continuity of $f$ by using results of product functions. (Here, I'll write $f(x)=(f_1(x),f_2(x))$.)

Let $U\times V$ be an open subset of $\mathbb{R}^2$. Then for every $x\in f^{-1}(U\times V)$, we have that $x\in f_1^{-1}(U)\cap f_2^{-1}(V)\subseteq f^{-1}(U\times V)$. Since $f_1$ and $f_2$ are continuous, $f_1^{-1}(U)\cap f_2^{-1}(V)$ is open in the domain of $f$, and this shows $f$ is continuous.

ContentLicense: CC BY-SA 4.0


问:
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$?
I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.

Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.

Can you suggest any ideas how to calculate $S$?

答:
let $$y=\sum_{n=1}^{\infty}H^2_{n}x^n$$

then we have
$$y=x+xy+\ln^2{(1-x)}+\int_{0}^{x}\dfrac{\ln{(1-t)}}{t}dt$$

so
$$y=\dfrac{\ln^2{(1-x)}}{1-x}+\sum_{n=1}^{\infty}\left(1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}\right)x^n$$

then you can use:Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

ContentLicense: CC BY-SA 3.0


问:
is every edge of a zonotope a generator?
Given a Zonotope $Z \subset \mathbb{R}^d$, with generators $G \in \mathbb{R}^{n\times d}$.

Every facet is a $d-1$ polytope. And I'm wondering is it true that every edge of the facets is one generator with some positional offset?

Since every vertex is in the form of $\sum_{i \in G} = \lambda_i i, \lambda_i \in \{0, 1\}$, and the edge is difference between two vertices, so every edge has to be some sum of a subset of generators. But does the difference has to be exactly one generator?

答:
The answer is yes. And here is the proof:

If the edge is composed of more than one generator, one can always add an extra vertex correspondingly.


问:
Algebra: manipulation of variables
I apologise maths is not my strong point, there are 2 ways to isolate n

How do I know which is the correct way:

Correct equation:

${n={cv}}$

way 1 divide by n

${c=\frac{n}{v}}$

${\frac{c}{n}=\frac{n}{vn}}$

${\frac{c}{n}={v}}$

${\frac{c}{nc}=\frac{v}{c}}$

${{n}=\frac{v}{c}}$

Way 2 multiply by v:

${c=\frac{n}{v}}$

${cv=\frac{nv}{v}}$

${n={cv}}$

答:
$\dfrac c{cv}=\dfrac 1v$, not $v$ (which you have in the first way)

Addendum in response to question edit:

$\dfrac nv=c\implies\dfrac nc=v\implies \dfrac {nc}c=vc\implies n=vc$

($\dfrac nc=\dfrac c{cv}$ is not correct, nor is $\dfrac{nc}c=\dfrac vc$, nor $n=\dfrac vc$.)

Title: Algebra: manipulation of variables


问:
Proof of the Existence of Integer Solutions for $(f_1)^2 + (f_2)^2 +( f_3)^2... = I^2$

Using fractions($f_n$) where integer $n$ is the number of fractions, prove that the sum of the squares $f_1$ to $f_n$ has no integer($I$) solutions when $n >=  1$ given each fraction has a distinct positive integer numerator and denominator and every fraction is a fully simplified non integer:

$$\left(f_1\right)^2 + \left(f_2\right)^2 + \left(f_3\right)^2 ...  =  I^2$$

I could not find a proof, but using distinct positive integers $a,b,c,d,e,f$ I managed to prove this for the case of
$$\left(\frac{a}{b}\right)^2 + \left(\frac{c}{d}\right) ^2 = \left(\frac{e}{f}\right)^2 $$

by simplifying:

$$\frac{(ad)^2 + (bc)^2}{(bd)^2}  = (\frac{e}{f})^2 $$

Since $(ad)^2 + (bc)^2$  must satisfy a Pythagorean triple, we know that each side length(excluding the hypotenuse) must be the product of two distinct integers, which can be proven through Euclid's formula :

$$ac = (2m)(n)$$
$$bd = (m -n)(m + n)$$

Q.E.D.

Although this still did not prove useful to answering my question, I later found a proof that the sum of the sqaures $f_1$ to $f_2$ has no integer solutions using distinct positive integers $a,b,c,d$ where $(\frac{a}{b})^2 +  (\frac{c}{d} ) ^2 = I^2 $   using the following:

$$\frac {a^2}{b^2} = \frac {I^2d^2 - c^2}{d^2}$$

Both sides are in their simplest form since
$$\gcd(a^2,b^2) = 1 = \gcd(c^2,d^2) = \gcd(I^2d^2 - c^2, d^2)$$

so $a^2 = e^2d^2-c^2, b^2 = d^2$.

Given $b,d > 0$, $b=d$. Therefore no solutions exists.

But this lead me to question what if there were more than $2$ integer fractions? This would mean that
the denominator of some integer fraction could be the product of $2$ or more distinct integers, which suggests that this proof which uses $b = d$ does not nesscacarily need to be true for there to be an integer solution where $n > 2$.

So I made a conjecture that there are no integer solutions for the sum of any number of squared fractions where the numerator and denominator are distinct positive integers and the fractions are fully simplified non-integers.

Edit:

This turned out to be false,

New questions:

Provide a proof

A method to finding integer solutions

(The one that finds the most affective method will be more likely to receive the bounty)

答:
This proof is INVALID, but pieces of it might give insights to the forms of solutions...

In fact as long as just the denominators are distinct, the sum of squares is never an integer, so it can't be the square of an integer.

Let's call the fractions $\frac{a_i}{b_i}$, where $a_i, b_i \in \mathbb{N}, b_i \geq 2$, and we require that for each index $i$, $\gcd(a_i,b_i)=1$ and for all index pairs $i \neq j$ that $b_i \neq b_j$. We look for solutions where:

$$S = \sum_{i=1}^n \frac{a_i^2}{b_i^2} \in \mathbb{Z}$$

Suppose by way of contradiction that the set of such solutions is not empty. Let $N$ be the smallest count with at least one solution using exactly $n=N$ fractions in this set. Then let $B$ be the smallest positive integer which appears as a denominator $b_i$ in any solution of size $N$.

Consider a "minimal" solution which has $n=N$ and has $b_i=B$ for some $i$. Let $p$ be a prime factor of $B$ (which is at least $2$). Let $\beta$ be the set of denominators $b_i$ which are a multiple of $p$. This is not the empty set since $p \mid B \Rightarrow B \in \beta$. If $\beta$ is the entire set $\{b_1, \ldots b_N\}$, then we have

$$ \sum_{i=1}^N \frac{a_i^2}{(b_i/p)^2} = p^2 \sum_{i=1}^N \frac{a_i^2}{b_i^2} = p^2 S \in \mathbb{Z} $$

is a solution with the same $N$, and one of the denominators is $B/p$, but this contradicts the choice of $B$ as the smallest denominator in any solution of size $N$.

If $\beta$ is not the entire set $\{b_1, \ldots b_N\}$, then the set difference $\beta^c = \{b_1, \ldots b_N\} \setminus \beta$ is not empty. The least common multiples of these two subsets of denominators satisfy $p \mid \mathop{\rm lcm}(\beta)$ and $p \nmid \mathop{\rm lcm}(\beta^c)$. Splitting the sum $S$ over these two subsets:

$$ S = \sum_{b_i \in \beta} \frac{a_i^2}{b_i^2}
+ \sum_{b_i \in \beta^c} \frac{a_i^2}{b_i^2} \\
S (\mathop{\rm lcm} \beta^c)^2 = \sum_{b_i \in \beta} \frac{a_i^2 (\mathop{\rm lcm} \beta^c)^2}{b_i^2} + \sum_{b_i \in \beta^c} a_i^2 \left(\frac{\mathop{\rm lcm} \beta^c}{b_i}\right)^2 \\
\sum_{b_i \in \beta} \left(\frac{a_i \mathop{\rm lcm}(\beta^c)}{b_i}\right)^2 = S (\mathop{\rm lcm} \beta^c)^2 - \sum_{b_i \in \beta^c} a_i^2 \left(\frac{\mathop{\rm lcm} \beta^c}{b_i}\right)^2 \in \mathbb{Z} $$

For each addend term in the left side, $p \mid b_i$ by definition of $\beta$. Since $\gcd(a_i, b_i) = 1$, $p \nmid a_i$, and we already noted $p \nmid \mathop{\rm lcm}(\beta^c)$. So the fraction is not an integer. Every $b_i$ in these fractions is a distinct integer, since they're a subsequence of the entire sequence of $N$ different $b_i$. Therefore the left side is a solution to the problem with $n = |\beta| < N$, contradicting the choice of $N$ as minimal.

EDIT: Here's the error. The $b_i$ values are distinct, but they are not necessarily the denominators of the new fractions when reduced to lowest form. The actual reduced denominators could duplicate some values, meaning the formula is not another solution to the problem as I modified it.

All possibilities lead to contradiction, so there are no solutions at all.


问:
Where is the mistake in this use of the derivative rule?
I know that $$\left(\frac{1}{g}\right)’=\frac{-g’}{g^2}$$
But where is the mistake here?
$$\left(\frac{1}{g}\right)’ = (g^{-1})’=-(g^{-2})=\frac{-1}{g^2}$$ is using the power rule here wrong?

答:
You forgot to apply the chain rule. Since
$$
(f \circ g)'(x)=f'(g(x))g'(x) \, ,
$$
we have
\begin{align}
\frac{d}{dx}\left(\frac{1}{g(x)}\right) &= \frac{d}{dx}\left(g(x)^{-1}\right) \\[4pt]
&= -g(x)^{-2} \cdot \color{red}{g'(x)} \\[4pt]
&= -\frac{g'(x)}{g(x)^2} \, .
\end{align}

Alternatively, we could let $y=1/g(x)$ and multiply both sides by $g(x)$:
\begin{align}
yg(x) &= 1 \\[4pt]
\frac{d}{dx}(yg(x)) &= \frac{d}{dx}(1) \\[4pt]
y'g(x)+yg'(x) &= 0 \\[4pt]
y' &= -\frac{yg'(x)}{g(x)}=-\frac{\frac{1}{g(x)}g'(x)}{g(x)}=-\frac{g'(x)}{g(x)^2} \, .
\end{align}


答:
What you've written is true ... if you're differentiating with respect to $g$.

Remember that the prime symbol is generally shorthand for $\frac{d}{dx}$ (or whatever the variable of interest happens to be), and in this case $g$ is not just a variable but is a function of $x$ and so we must apply the chain rule:

$\begin{eqnarray}\frac{d}{dx}\left( \frac{1}{g} \right) & = & \frac{d}{dg} \left( \frac{1}{g} \right) \times \frac{dg}{dx} \mbox{   per the chain rule} \\
& = & \frac{d}{dg} g^{-1} \times \frac{dg}{dx} \\
& = & -g^{-2} \times g' \\
& = & \frac{-g'}{g^2} \end{eqnarray}$

Note that the first term in the second-last line is what you got, but we can't stop there because we've differentiated not with respect to $g$, but to $x$.

Title: Where is the mistake in this use of the derivative rule?

Body:
I know that $$\left(\frac{1}{g}\right)’=\frac{-g’}{g^2}$$
But where is the mistake here?
$$\left(\frac{1}{g}\right)’ = (g^{-1})’=-(g^{-2})=\frac{-1}{g^2}$$ is using the power rule here wrong?


问:
Write the dual LP of the primal LP problem
I have to find the dual of the lp problem given below
Minimise $$z=-x_1+\frac43 x_2$$
subject to∶
$$\begin{array}[t]{l}
2x_1+4x_2\le16\\
-\frac{1}2 x_1-x_2\le4\\
-3x_1+4x_2\ge-24\\
x_1≥0,x_2≤0
\end{array}
$$

Also find the solution to the dual problem without solving it.

答:
The prime has infinite many optimal solution as long as x1 belongs to [8/5,8]. However, the dual solution is unique.

Note that the third constraint of the prime problem has to be tight to achieve the optimal value of -8 (dividing it by 3 and you obtain the objective function). Thus, y3 must not equal to 0.
The first and the second constraints of the prime problem are equivalent, and cannot be tight. Thus, y1=y2=0.


问:
Compute $P(B_{t}>0,B_{s}>0)$ where $B$ is a standard Brownian motion
Compute $P(B_{t}>0,B_{s}>0)$ for $t >s$ where $B$ is a standard BM

My attempt:

$P(B_{t}>0,B_{s}>0)=P(B_{t}-B_{s}+B_{s}>0,B_{s}>0)$ and now considering $X:=B_{t}-B_{s}\sim \mathcal{N}(0,t-s)$ and $Y:=B_{s}\sim\mathcal{N}(0,s)$ such that for $\xi$, $\eta$ standard normal iid random variables we have $X=\sqrt{t-s}\xi$ and $\sqrt{s}\eta$ such that we obtain:

$$ P(B_{t}-B_{s}+B_{s}>0,B_{s}>0)=P( \sqrt{t-s}\xi+\sqrt{s}\eta>0,\sqrt{s}\xi>0)=P( \sqrt{t-s}\xi+\sqrt{s}\eta>0,\xi>0)$$

Now if I want to switch the above into polar coordinates I have:

$$\{(r,\theta):\; \sqrt{t-s}r\cos(\theta)+\sqrt{s}r\sin(\theta)>0,r\cos(\theta)>0\}=\{(r,\theta):\; \sqrt{t-s}r\cos(\theta)+\sqrt{s}r\sin(\theta)>0,\theta \in (-\pi/2,\pi/2)\}\\ = \{(r,\theta):\; -\frac{\sqrt{t-s}}{\sqrt{s}}<\tan(\theta),\theta \in (-\pi/2,\pi/2)\} = \{(r,\theta):\; \arctan(-\frac{\sqrt{t-s}}{\sqrt{s}})<\theta,\theta \in (-\pi/2,\pi/2)\}$$

This looks rather ugly, am I on the right path?

答:
I provide a solution without using changing of variables to polar coordinate (which doesn't lead to a solution IMHO)
\begin{align}
P(B_t >0, B_s>0) &= E(\mathbb{I}_{B_s>0}\mathbb{I}_{B_t>0}) \\
&= E(\mathbb{I}_{B_s>0}\mathbb{I}_{(B_t - B_s)+B_s>0}) \\
&= E\left(E(\mathbb{I}_{B_s>0}\mathbb{I}_{(B_t - B_s)+B_s>0}|\mathcal{F}_s)\right)\\
&= E\left(\mathbb{I}_{B_s>0}E(\mathbb{I}_{(B_t - B_s)>-B_s}|\mathcal{F}_s)\right)\\
&= E\left(\mathbb{I}_{B_s>0}P(\sqrt{t-s}\mathcal{N}(0,1)>-B_s|\mathcal{F}_s)\right)\\
&= E\left(\mathbb{I}_{B_s>0}\Phi(\frac{B_s}{\sqrt{t-s}})\right)\\
&= E\left(\mathbb{I}_{X>0}\Phi(\sqrt{s}\frac{X}{\sqrt{t-s}})\right)\\
&= \color{red}{\int_0^{+\infty}\Phi\left( \sqrt{\frac{s}{t-s}}x \right)\varphi(x)dx}\\
\end{align}

where  $X \sim \mathcal{N}(0,1)$, $\Phi(x)$ and $\varphi(x)$ are CDF and density function of $\mathcal{N}(0,1)$.


问:
studying the function $f(x)= \sqrt{x^2+x}-x$
I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.

In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.

$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0 \Rightarrow   2x+1-2 \sqrt{x^2+x} >0 \Rightarrow   2x+1>2 \sqrt{x^2+x}  \Rightarrow   1>0 $ . So it seems that the derivative is always positive in the domain.

But comparing with the result in wolfram alpha the function is decreasing in $(-\infty, -1]$

答: $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0\iff 2x+1-2\sqrt{x^2+x}>0\iff 2x+1>2\sqrt{x^2+x}\iff \begin{cases}2x+1>0\\  x^2+x> 0\\ (2x+1)^2>4(x^2+x)\end{cases}\iff \begin{cases}x>\frac{-1}{2}\\x> 0 \vee x< -1\\ \forall x \end{cases}\iff x> 0 $$ Be careful to exclude $0$ in order to have the denominator not equal to $0$


问:
A space which is 0-dimensional but not totally disconnected
Notation: A space is zero-dimensional if it is $T_1$ and has a basis consisting of clopen sets.

Is there any topological space which is zero-dimensional but it is not totally disconnected?

答:
Just to be rigorous on the definition we use for dimensionality. Here I use the following definition:

A space is zero-dimensional if it is $T_1$ and has a basis consisting of clopen sets.

Every zero-dimensional space is totally disconnected, just by inspection of the relative definitions. In fact, we can say even more: every space $X$ that admits a one-to-one continuous function into a zero-dimensional space is totally disconnected.

The converse is, however, not true. There is indeed a space that is totally disconnected but not-zero dimensional: $E:=\{x \in \ell^{2} : x_i \in \mathbb{Q} \text{ for every } i \}$. It is known in the literature as the Erdős' space.


问:
If the homotopic category is abelian, then it is semi-simple.
Let $\sf{A}$ be an abelian category and consider its homotopic category $\sf{K}(\sf{A})$. I am writing some notes about homological algebra and, since the reader usually wonders why this isn't generally an abelian category, I want to prove right after its definition that, if it is abelian, then it is semi-simple. (This means that, at this stage, I haven't yet proved that it is triangulated.)

Consider a short exact sequence in $\sf{K}(\sf{A})$:

The proof (which uses the triangulated structure of $\sf{K}(\sf{A})$) in Gelfand-Manin shows that $M^\bullet$ is homotopy equivalent to $L^\bullet\oplus \operatorname{cone}(\varphi^\bullet)$. This means that a splitting $M^\bullet\to L^\bullet$ of the exact sequence above would be given by the composition
$$M^\bullet\to L^\bullet\oplus \operatorname{cone}(\varphi^\bullet) \to L^\bullet,$$
where the first morphism is the (homotopy) inverse of
\begin{align*}
L^i\oplus M^i\oplus L^{i+1} &\to M^i \\
(l,m,l') &\mapsto \varphi^i(l)+m
\end{align*}
and the second is just the usual projection. Now, a morphism $M^\bullet\to L^\bullet\oplus \operatorname{cone}(\varphi^\bullet)$ in $\sf{K}(\sf{A})$ would be an equivalence class of morphisms $M^i\to L^i$, $M^i\to M^i$ and $M^i\to L^{i+1}$. The only reasonable choice seems to take the morphism on the middle to be the identity and the other two to be zero. This would imply that our splitting $M^\bullet\to L^\bullet$ is the zero morphism. But the zero morphism is not usually homotopic to $\operatorname{id}_{L^\bullet}$.

What am I doing wrong and/or how can we prove this result without using the triangulated structure of $\sf{K}(\sf{A})$?

答:
As JHF said in the comments, you shouldn't take a "reasonable" choice. Indeed, if you expect to construct a retraction $M^\bullet\to L^\bullet$ with only a "reasonable" map, the map would not use the fact that the category is semi-simple and would work in any category.

I would like to give a stronger statement : any triangulated category (not just the homotopy category) which is also an abelian category is semi-simple. (Here I use Gelfand-Manin definition of semi-simple : that is that any short exact sequence splits).

To see this, we have the following lemmas

Lemma 1 : Let $A\to B\to C\to A[1]$ be a distinguished triangle, then the following are equivalent :

$C\to A[1]$ is the zero map

$A\to B$ has a retraction

$B\to C$ has a section

There is an isomorphism $B\simeq A\oplus C$ such that $A\to B$ corresponds to the canonical inclusion $A\to A\oplus C$ and $B\to C$ correspond to the canonical projection $A\oplus C\to C$.

Proof :

$2.\Leftrightarrow 3. \Leftrightarrow 4.$ hold in any additive category.

$1.\Rightarrow 2.$ : Apply the functor $Hom(.,A)$, you get a long exact sequence
$$ ...\to Hom(B,A)\to Hom(A,A)\to Hom(C[-1],A)\to ...$$
The map $Hom(A,A)\to Hom(C[-1],A)$ is zero since this is the precomposition with $C[-1]\to A$ which is zero by hypothesis. It follows from exactness that $Hom(B,A)\to Hom(A,A)$ is onto. So $id_A$ has a preimage $r$ which is thus a retraction of $A\to B$.

$2.\Rightarrow 1.$ Since there is a map $B\to A$ such that the composition $A\to B\to A$ is the identity, the map $C[-1]\to A$ can be written as the composition $C[-1]\to A\to B\to A$, but $C[-1]\to A\to B$ is zero, so $C[-1]\to A$ is zero.

Lemma 2 : In a triangulated category, every mono and every epi are split.

Proof : The proof for epi is dual of the proof for mono, so let us prove only the mono case. Consider a monomorphism $i:A\to B$. From the axioms of triangulated categories you can form a distinguished triangle
$$A\rightarrow B\to C\rightarrow A[1]$$
We have $C[-1]\to A\to B$ is the zero map, but since $A\to B$ is mono, $C[-1]\to A$ is zero, and by lemma 1, $A\to B$ has a retraction.

The claim is now straightforward : consider a short exact sequence in a category which is both triangulated and abelian. Consider a short exact sequence :
$$0\to A\to B\to C\to 0$$
Since the category is abelian, $A\to B$ is mono. Since the category is triangulated, by lemma 2, $A\to B$ splits.


问:
show that the power series $\sum_{n=1}^\infty \frac{2^n}{n^n}x^n$ is convergent for all $ x \in \mathbb{R} $
Show that $$\sum_{n=1}^\infty \frac{2^n}{n^n}x^n$$ is convergent for  all $ x \in \mathbb{R} $. So guys so far I have attempted till this step :

$\frac{U_n+1}{U_n} = \frac{(2^{n+1})({x^{n+1}})}{{(n+1)^{n+1}}} \frac{n^n}{2^nx^n} = |2x|\lim\limits_{x \to \infty} (\frac{n}{n+1})^{n}\frac{1}{n+1}$.

I could further express the answer as :

$ \lim\limits_{x \to \infty} \frac{1}{1+\frac{1}{n}^n} \frac{1}{n+1}$, how is this appraching $0$?

答:
Ratio test $\frac{2x}{n}\to 0$ for all $x$ as $x\to \infty$.

Detail for ratio test:   Let $a_n=(\frac{2x}{n})^n$ we have $\frac{a_{n+1}}{a_n}\lt\frac{2x}{n}$, because $a_{n+1}\lt (\frac{2x}{n})^{n+1}$.


问:
Must an exterior angle of a polygon be less than 180 degrees?
Is it true that all exterior angles of any polygon, be it convex, concave, or complex, must be less than 180 degrees? I believe this to be true and cannot think of any counterexample. Am I correct? If so, is there a simple proof or even an intuitive explanation for this?</p>

答: The Exterior Angle is the angle between any side of a shape, and a line extended from the next side. So the interior and exterior angle will add to 180 degrees. For the exterior angle to be 180 degrees, the interior angle must be 0 degrees. So yes, the exterior angle will always be less than 180 degrees.</p>


答: Do the ratio test. $\left|\dfrac{a_{n+1}}{a_n}\right|= \dfrac{2^{n+1}}{(n+1)^{n+1}}\cdot \dfrac{n^n}{2^n}= \dfrac{2}{n+1}\cdot \dfrac{1}{\left(1+\frac{1}{n}\right)^n}\to 0 < 1$. Thus the series converges for all $x$.

Title: show that the power series $\sum_{n=1}^\infty \frac{2^n}{n^n}x^n$ is convergent for all $ x \in \mathbb{R} $

Body:
Show that $$\sum_{n=1}^\infty \frac{2^n}{n^n}x^n$$ is convergent for  all $ x \in \mathbb{R} $. So guys so far I have attempted till this step :

$\frac{U_n+1}{U_n} = \frac{(2^{n+1})({x^{n+1}})}{{(n+1)^{n+1}}} \frac{n^n}{2^nx^n} = |2x|\lim\limits_{x \to \infty} (\frac{n}{n+1})^{n}\frac{1}{n+1}$.

I could further express the answer as :

$ \lim\limits_{x \to \infty} \frac{1}{1+\frac{1}{n}^n} \frac{1}{n+1}$, how is this appraching $0$?


问:
How to prove that $a_n =(1+\frac1n)^n$ for all $n\ge 1$ applies to $a_n=\sum_{k=0}^n \frac1{k!}(1-\frac1n)(1-\frac2n)\cdots(1-\frac{k-1}n)$?
In an assignment for the course Real-Analysis I have to show that for all $n \ge 1$ applies:
$$a_n = \displaystyle \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right)$$
where $a_n$ is given as:
$$a_n =\left(1+\frac1n\right)^n$$

I have to use the binomial formula for this: $(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{k} y^{n-k}$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

Body:
$$LHS = \sum_{k=0}^n \frac{1}{k!} \prod_{j=0}^{k-1}\left(1-\frac{j}{n} \right)$$

$$= \sum_{k=0}^n \frac{1}{k!} \frac{1}{n^{k-1}} \prod_{j=0}^{k-1}\left(n- j \right)$$

$$= \sum_{k=0}^n \frac{1}{n^{k-1}} \left(\frac{1}{k!}\prod_{j=0}^{k-1}\left(n- j \right) \right)$$

Note that $$\frac{1}{k!}\prod_{j=0}^{k-1}\left(n- j \right) = \binom {n}{k}$$

Hence
$$LHS = \sum_{k=0}^n \binom {n}{k} \frac{1}{n^{k-1}}$$

$$= \left(1+\frac{1}{n} \right)^n$$


问:
How many squarings are needed to exceed 2?
Given the base 2 representation of $x\in\mathbb Q$ ($1<x<2$ and has a finite number of digits in base 2), find a number $k\in\mathbb N$ such that $2<x^{2^k}<4$.

The final answer/algorithm is not allowed to use powers, roots and logarithms.

I managed to find a range of possible values for $k$.

Looking at the digits after he decimal point, if the $n$th digit is the first non-zero digit, then squaring this number will necessarily turn at least one of the two preceding digits into a 1 but none of the digits before it. I figured this out by looking at the edge cases: if all the digits after the $n$th digit are zero then the $n-1$ digit becomes 1 when the number is squared, and if all these digits are 1 it is equivalent to a number where only the $n-1$ digit of the fractional part is 1, and squaring this number makes the $n-2$ digit be 1.

So the algorithm for finding the upper and lower limits of $k$ is:

Looking at the digits after the decimal point, count the leading zeros (let's call the result of the count $z$). Then $\lfloor\frac{z+2}{2}\rfloor ≤ k ≤ z+1$.

答:
There is no known solution for all numbers, but I found a quick solution for a significant amount of cases, which is useful for optimizing programs (try this, if it fails do something else, like squaring repeatedly).

Part 1: define edge cases

Let's construct a number whose fractional part has $z$ leading zeros followed by $n$ consecutive ones, and see to what power $2^k$ do we need to raise it to get it between 2 and 4. (This leads to a more complete solution, you'll see later.)

To produce a chain of $n$ ones in base 2, all we need is to subtract 1 from the $n$th power of 2. For example $2^4-1=15$ and the base 2 representation of 15 is $1111$.

The next thing to do is to divide it by $2^n$ to move the 1-digits to the right side of the decimal point. To add $z$ leading zeros, just divide it by $2^z$. And finally, +1 to make it a number between 1 and 2.

$$2 < \left(\frac{2^n-1}{2^{n+z}}+1\right)^{2^k} < 4$$

Part 2: isolate $k$

For brevity, let's call the part in the brackets $m$:

$$2 < m^{2^k} < 4$$

Take the logarithm (base 2) of the inequality:

$$1 < 2^k\log_2(m) < 2$$

Divide by the logarithm:

$$\frac{1}{\log_2(m)} < 2^k < \frac{2}{\log_2(m)}$$

Take the logarithm again:

$$- \log_2(\log_2(m)) < k < 1 - \log_2(\log_2(m))$$

Part 3: find the integer values of $k$

Since $k$ is a whole number, and there is only one valid value of $k$ that solves this inequality at any given point $(z,n)$, we can take the lower limit and round it up:

$$k = \lceil - \log_2(\log_2(m))\rceil = - \lfloor\log_2(\log_2(m))\rfloor = -\operatorname{msb}(\log_2(m))$$

Where $\operatorname{msb}(x)$ is the position of the most significant bit (digit) of $x$, which for $0<\log_2(m)<1$ is negative. (Note that $\lfloor\log_2(x)\rfloor = \operatorname{msb}(x)$ is correct only for this range.)

$1<m<2$, and in this range $\log_2(x) \approx x-1$, but remember that $\log_2(x) > x-1$ in this range, so it will throw the $\operatorname{msb}$ function result off in some cases (it will be too big by +1).

$$k \approx -\operatorname{msb}(m-1) = -\operatorname{msb}\left(\frac{2^n-1}{2^{n+z}}\right) = z+1$$

For very small numbers ($z>2$) the approximation is always wrong, this makes the result reliable in this range. Manually testing some cases of small $z$ and $n$ leads to a complete answer:

$$k =
\begin{cases}
1 & & z = 0\\
z+1 & n = 1 & z > 0\\
z+1 & n = 2 & 1 ≤ z ≤ 2\\
z & n = 2 & z > 2\\
z & n > 2 & z > 0
\end{cases}$$

Part 4: fill in the gaps

Until now we focused on very specific numbers, now let's take care of all the numbers in between.

Let $C(z,n)$ be the set of all numbers whose base 2 representation starts with:

$$1.\underbrace{0000000000}_{z\text{ zeros}}\underbrace{111111111}_{n\text{ ones}}0$$

For a fixed value of $z$, if $k$ is the same for $n=x$ and $n=x+1$, then it is also the same for every number in between (in $C(z,x)$), because if two numbers reach the range between 2 and 4 when raised to the same power, then all the numbers between them will also reach that range with the same power. Otherwise there are 2 possible answers with a difference of 1.

To sum it up, every number $1<x<2$ is in a set of the form $C(z,n)$. By counting the consecutive zeros ($z$) after the decimal point and then the consecutive ones ($n$) we can either know the answer with certainty or at least know that it is one of two numbers, as detailed in this table:


问:
What is $\frac{1}{2+\frac{1}{x}}$ equal to when $x=0$?
$$\frac{1}{2+\frac{1}{x}}$$
What's the output of this function when $x=0$?

I tried graphing it on Desmos and it shows a curve passing through the origin so the output must be equal to zero, however I don't get how it reached this result.

The only way it could have reached it is maybe by changing the equation a little bit:

$$\frac{1\cdot x}{(2+\frac{1}{x})\cdot x}$$ (where $x\neq0$)
$$\frac{x}{2x+{1}}$$

However since we stated in the first step that $x$ can't be equal to zero we can't make it equal to zero again and solve with the simplified version we've just arrived at:

at $x=0$,

$$\frac{0}{0+{1}}=0$$ $\rightarrow$ so this is wrong

How did Desmos then arrive at the answer?

答:
When $x$ is close to zero, $1/x$ is very large. Hence,
$$
\frac{2}{2+\frac{1}{x}}=\frac{2}{1+\text{very large}}
$$
and
$$
\frac{2}{1+\text{very large}}=\text{very close to $0$} \, .
$$
In the language of limits, we can say that
$$
\lim_{x \to 0}\frac{2}{2+\frac{1}{x}}=0 \, .
$$
But this doesn't mean that when $x$ equals $0$,
$$
\frac{2}{2+\frac{1}{x}}=\frac{2}{2+\frac{1}{0}}=0 \, .
$$
The above expression doesn't make any sense, because we are not allowed to divide by zero. What we can say is that as $x$ gets very close to $0$, $2/(2+1/x)$ is also very close to $0$. The graph is misleading: because our eyes cannot discern what happens at a single point, it looks as if the curve crosses the origin, even though it does not.


答: Take the limit as x tends to 0 from above, this will be 0 as 1/x tends to infinity. Then take the limit as x tends to 0 from below, this will also be zero, as 1/x tends to minus infinity. Therefore, despite the fact the function has no value at x = 0, arbitrarily close to x=0 from any direction (0-delta, 0+delta), the function will approach 0.</p>

Title: What is $\frac{1}{2+\frac{1}{x}}$ equal to when $x=0$?

Body:
$$\frac{1}{2+\frac{1}{x}}$$
What's the output of this function when $x=0$?

I tried graphing it on Desmos and it shows a curve passing through the origin so the output must be equal to zero, however I don't get how it reached this result.

The only way it could have reached it is maybe by changing the equation a little bit:

$$\frac{1\cdot x}{(2+\frac{1}{x})\cdot x}$$ (where $x\neq0$)
$$\frac{x}{2x+{1}}$$

However since we stated in the first step that $x$ can't be equal to zero we can't make it equal to zero again and solve with the simplified version we've just arrived at:

at $x=0$,

$$\frac{0}{0+{1}}=0$$ $\rightarrow$ so this is wrong

How did Desmos then arrive at the answer?


答: The function is equal to (2x) / (2x+1) except for x=0 where it is not defined. The function isn’t equal to anything when x=0.</p>


问:
What does this double integral even represent?
I'm preparing for a calculus resit and I'm really confused by this double integral. I somewhat understand where the $\int_{3y}^{12-3y}$ comes from (just expressing $y=\frac{1}{3}x$ and $y=-\frac{1}{3}(x-6)+2$ in terms of x). I also get how to solve the integral (my answer was also 8), but now what this answer actually represents.

$$\int_0^2 \int_{3y}^{12-3y} (y)\ dx\ dy = 8$$

If you plot these two functions, the area enclosed by them and the x-axis is obviously 12 (just the area of two right-angled triangles, and I even checked it with integrals to make sure I wasn't going mad). I believe the double integral represents some sort of volume, but I don't understand what volume, because nowhere in this question a z-axis is mentioned.

I feel like I'm either missing something very fundamental, or this is just a really badly worded/confusing question. Either way, any help would be very much appreciated! :)

答:
Before we consider a double integral lets think about what a single integral represents.
The usual interpretation of a single integral is the area under a curve.
So
$$ \int_a^b f(x) \,\mathrm{d}x $$
represents the area between the $x$ axis (which is just $y=0$), the lines $x=a$, $x=b$ and $y=f(x)$.
This often a valid and useful interpretation but it isn't the only interpretation.
Another interpretation may arise with a more specific function. Say we have some string with length $L$ and we know that its mass per unit length at some point $x$ along the string is given by $\rho(x)$. Then the total mass of the string can be expressed as [1]
$$m = \int_0^L \rho(x)\,\mathrm{d}x $$
To see why this is the case we think about what an integral is, really its just a sum of lots of infinitesimal bits (excuse the hand waving, I'm a physicist).
So a discrete version of the integral above may be
$$ m \approx \sum_i \rho(x_i) \Delta x $$
where we approximate the integral as a sum and we split the string into finite lengths $\Delta x$ and we consider the density at the point $x_i$ within each length to be a good approximation of the density of the whole length.

Personally I think that this interpretation of the integral is better suited for generalisation to higher dimensions.
We can similarly consider a function like $\sigma(x, y)$ which represents the density at a point $(x, y)$ on a plane. Then the integral
$$ \iint_S \sigma(x, y)\,\mathrm{d}x\mathrm{d}y $$
is just the mass of the surface, $S$, over which we are integrating.

There is still a valid interpretation of a double integral as a volume however.
Imagine that the function $f$ represents the height above the $x,y$-plane.
Then the integral [2, 3]
$$ \iint_S f(x, y)\,\mathrm{d}x\mathrm{d}y $$
represents the area between the $x,y$-plane and the function $f$ only above the volume in the $x,y$-plane that we call $S$.
This interpretation can still hold even if we haven't got an explicitly constructed $z$-axis to think of the height as being along since this interpretation is just to aid in our thought process.

Notice how similar this is to the one-dimensional case with all dimensions shifted up by one. Single integrals become double integrals, intervals such as $[a, b]$ become regions of the plane, $S$, and areas, given by single integrals, become volumes, given by double integrals.
The problem comes when we want to evaluate a triple integral and then with the "area under a curve" interpretation we have to consider the 4-dimensional hyper-volume under a volume and I don't know about you but that is not something I can picture.

So to answer your specific question, what does the given integral represent, the answer is: it depends.
This is sort of a rubbish answer but it is true.
If the function we are integrating, in this case simply $y$ is naturally thought of as a height then a volume is a valid interpretation (and probably the best interpretation in this case). If it can be thought of as a density then mass is probably a better interpretation. However both of these are just ways of thinking about the integral and really they are the same, we can view mass as simply a coordinate in some "mass-space" in a similar way to how we might consider momentum space or any other more abstract space.


问:
Proof identity of differential equation
I would appreciate if somebody could help me with the following problem:

Q: $f''(x)$ continuous in $\mathbb{R}$ show that
$$ \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$

答:
Use  L'Hospital's rule

$$ \lim_{h\to 0}\frac{F(h)}{G(h)}=\lim_{h\to 0}\frac{F'(h)}{G'(h)}$$

You can use the rule if you have $\frac{0}{0}$ result.

You need to apply it 2 times

$$ \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}= \lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h}=\lim_{h\to 0}\frac{f''(x+h)+f''(x-h)}{2}=f''(x)$$

Title: Proof identity of differential equation


答:
Additional info so OP understands the language.

When you write:

$\require{cancel}$

$${\frac{c}{n}=\frac{n}{vn}}$$

and you do this:

$${\frac{c}{n}=\frac{\cancel{n}}{v\cancel{n}}}\Rightarrow{\frac{c}{n}=\frac{1}{v}}$$

Then you didn't divide by $n$.
You cancelled out $n$ because it was in the numerator and the denominator as well.
The equation does not change when you do this.

Bonus tip:
Say you have equation $a=bc$

When you divide the only thing you mustn't forget is to state that the variable you divide by cannot be zero. So if you divide by $b$:

$$b\neq0$$

And if you divide by $c$

$$c\neq0$$

etc.

Title: Algebra: manipulation of variables

Body:
I apologise maths is not my strong point, there are 2 ways to isolate n

How do I know which is the correct way:

Correct equation:

${n={cv}}$

way 1 divide by n

${c=\frac{n}{v}}$

${\frac{c}{n}=\frac{n}{vn}}$

${\frac{c}{n}={v}}$

${\frac{c}{nc}=\frac{v}{c}}$

${{n}=\frac{v}{c}}$

Way 2 multiply by v:

${c=\frac{n}{v}}$

${cv=\frac{nv}{v}}$

${n={cv}}$


问:
Are this set and this sigma algebra independent?
Let $(\Omega,\mathcal{F},P)$ be a probability space, $\{\mathcal{A}_i\}_i$ a family of sets and $\mathcal{B}$ another set.
Suppose $\forall i$ $\mathcal{A}_i $ and $\mathcal{B}$ are independent.

Set $\sigma(\{\mathcal{A}_i \})$ the sigma algebra generated by $\{\mathcal{A}_i\}_i$.
Are $\mathcal{B}$ and any set from $\sigma(\{\mathcal{A}_i \})$ independent?

答:
This is true if $(A_i)_i$ is closed under intersection, that is if for all $i,j$, there exists $k$ such that $A_i\cap A_j=A_k$. Otherwise not necessarily.

For example, $\Omega=\{0,1,2,3\}$, $\mathcal F=\mathcal P(\Omega)$,
$$
\mathbb P(\{0\})=\mathbb P(\{1\})=\mathbb P(\{2\})=\mathbb P(\{3\})=\frac14,
$$
$\mathcal A_1=\{0,1\}$, $\mathcal A_2=\{1,3\}$ and $\mathcal B=\{1,2\}$.

Then
$$
\mathbb P(\mathcal A_1\cap\mathcal B)=\mathbb P(\{1\})=\frac14=\frac12\times\frac12=\mathbb P(\{0,1\})\mathbb P(\{1,2\})=\mathbb P(\mathcal A_1)\mathbb P(\mathcal B).
$$

Similarly, $\mathbb P(\mathcal A_2\cap\mathcal B)=\mathbb P(\mathcal A_2)\mathbb P(\mathcal B)$. We have $\{1\}\in\sigma(\mathcal A_1,\mathcal A_2)$, yet
$$
\mathbb P(\{1\}\cap\mathcal B)=\mathbb P(\{1\})=\frac14\neq\frac14\times\frac12=\mathbb P(\{1\})\mathbb P(\mathcal B).
$$


问:
Find all positive solutions for equation:
$nx^{(n+1)}-(n+1)x^n+1=0$

There's nothing told about $n$, I guess $ n \in N $. I would like any kind explanations, thanks! I appreciate your time.

答:
Hint: $ x =1$ solves the equation.
Factoring gives
$$
(x-1)(nx^n - x^{n-1}-x^{n-2}- \dots -x -1) = 0
$$
and you can see $x=1$ is again a zero of the second factor. Factoring further
$$
(x-1)^2(nx^{n-1} + (n-1)x^{n-2}+\dots+2x+1) = 0
$$
and you can conclude that for $n=1,2,\dots$ the equation has a double zero at $1$.</p>


问:
If $f$ is a monotone, then the difference between $f$ and an absolutely continuous function is also monotone.
Let $f: [a,b] \rightarrow \mathbb{R}$ be a bounded variation function.

Let $f_a(x) = f(a)+ \int_a^x f' dm$ be the indefinite Lebesgue integral of $f$.

Let $g= f - f_a$.

Show if $f$ is monotone, then $g$ is also monotone.

Attempt:

Consider $f_a(x) = f(a)+ \int_a^x f' dm$.

We know since $f \in BV[a,b]$, $f'$ is integrable and $\int_a^x f' dm$ is monotone increasing.

So $f_a$ is also monotone.

But we know that the sum of a monotone increasing and monotone decreasing function is not necessarily monotone.

And that's where I got stuck.

I don't have any clues how to use the monotonicity other than this approach.

Any help will be appreciated!
Thank you.

答:
Let's assume $f$ is increasing. Let $y < x$. Then we have
$$
  g(x) - g(y)
  = f(x) - \bigg[f(a) + \int_{a}^{x} f'(s) \mathrm{d} s\bigg]
  - \bigg[f(y) - \bigg[f(a) + \int_{a}^{y} f'(s) \mathrm{d} s\bigg]\bigg].
$$
Distributing all of those minus signs and cancelling, we obtain
$$
  g(x) - g(y) = f(x) - f(y) - \int_{y}^{x} f'(s) \mathrm{d} s.
$$
Now we state a lemma that will finish the proof.

Lemma: Let $I \subseteq \mathbb{R}$ be an interval and let $u \colon I \rightarrow \mathbb{R}$ be a monotone function. Then $u'$ is a Lebesgue measurable function and for every $[a, b] \subseteq I$,
$$
  \int_{a}^{b} |u'(x)| \mathrm{d} x \leq |u(b) - u(a)|.
$$
By this lemma and the fact that $f$ is increasing,
$$
  g(x) - g(y) \geq f(x) - f(y) - \int_{y}^{x} |f'(s)| \mathrm{d} s \geq 0.
$$
For reference, this lemma is Corollary 1.25 in Giovanni Leoni's A First Course in Sobolev Spaces (2nd edition), or Corollary 1.37 in the 1st edition.


问:
Show that $\left| A_n \right| = \sum_{k=1}^{n} (-1)^k \binom{n}{k} B(n - k)$ where $B(n)$ is the nth Bell number.
I am having some trouble solving the following problem:

Let $A_n$ be the set of set partitions of $\{1, . . . , n\}$ without any singleton blocks. Show that $$\left| A_n  \right| = \sum_{k=0}^{n} (-1)^k  \binom{n}{k} B(n - k)$$ where $B(n)$ is the nth Bell number.

Here we cannot use generating functions and our professor told us to apply the principle of inclusion-exclusion. So, I let $A_S$ to be the collection of set partitions of $\{1, . . . , n\}$ where each element of $S$ is in a singleton block and start working from there but I end up with
$$\left| A_{=S}  \right| = \sum_{k=0}^{n-|s|} (-1)^k  \binom{n-|s|}{k} B(n -|s|- k)$$
Can someone help me with this?

答:
Your displayed formula is clearly wrong, since for $n=4$ the righthand side is $-11$.

The inclusion-exclusion principle says that

$$\begin{align*}
\left|\bigcup_{k=1}^nA_{\{k\}}\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}A_{\{k\}}\right|\\
&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}|A_I|\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{n}kB(n-k)\,,
\end{align*}$$

since there are $\binom{n}k$ subsets $S$ of $[n]$ of cardinality $k$ and $B(n-k)$ partitions of $[n]\setminus S$. This is the number of ‘bad’ partitions of $[n]$, i.e., of partitions that have at least one singleton block. There are altogether $B(n)$ partitions of $[n]$, so

$$\begin{align*}
|A_n|&=B(n)-\sum_{k=1}^n(-1)^{k+1}\binom{n}kB(n-k)\\
&=\sum_{k=0}^n(-1)^k\binom{n}kB(n-k)\,.
\end{align*}$$


问:
$\lim_{(x,y)\to(0,0)} \frac{xy}{x-y}$
I was trying to calculate the following limit:
$$\lim_{(x,y)\to(0,0)} \frac{xy}{x-y}$$

I used polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$.
But this gives me: $$\lim_{r\to 0} \frac{r\cos\theta\cdot r\sin\theta}{r\cos\theta-r\sin\theta}=\lim_{r \to 0} r\frac{\cos\theta\sin\theta}{\cos\theta-\sin\theta}=0$$

But Wolfram Alpha says that this limit does not exist. What is the problem in what I did? Is it because of the fact that if $\theta = \frac{\pi}{4}$ I am dividing by $0$? (My guess is that it is in fact the problem and by applying L'Hôpital's rule I will find out the limit does not exist.)

答:
The first problem is that $\frac{xy}{x - y}$ doesn't even make sense if $x = y$, because the denominator is then zero. But even if we exclude the line $x = y$ from the domain, the limit still doesn't exist. Essentially, the problem is that we can approach $(0, 0)$ along a path that approaches the line $x = y$ rapidly enough that $x - y$ approaches zero faster than $xy$, forcing the limit along this path to be nonzero. (For example, $y = x^3 + x$ gives such a path.) On the other hand, the function is constant zero along the line $x = 0$, so if the limit exists, then it has to be zero. Thus the limit can't exist.

Rewriting it in polar coordinates doesn't change this, and the same phenomenon still occurs with that term involving cosines and sines of the angle.

Title: $\lim_{(x,y)\to(0,0)} \frac{xy}{x-y}$


问:
Positive logical definition of finiteness
In his debate with Cassirer and in some other writing, Heidegger tries to define positively the concept of finiteness for the beings. It is a novel and difficult approach as traditionally, in theology and metaphysic, finiteness of beings is negatively defined as an imperfection or a negation, the absence of infiniteness, as opposed to some perfect and infinite being like god.

I feel that this difficulty is strangely reflected in set theory. I reckon we can naturally associate positive definition with $\Sigma_1$ sets (it abides to our definition if there is one set such that... ) and negative definitions with $\Pi_1$ sets (it abides to our definition if there is not set such that...).

Now for example Dedekind definition of an infinite set is a positive definition, a set is infinite if there is a bijection between itself and a proper subset. Then the natural way of defining finite sets would be non-infinite set which is $\Pi_1$ or a negative definition.

We could also use $\omega$, the smallest set satisfying the axiom of infinity and say that another set is finite if there is no injection from $\omega$ in this set ; but once again its a negative definition.

So formally, what this boils down to is the following :

Is the definition
of finiteness in set theory $\Pi_1$ or $\Delta_1$ ? I've just seen that
it is $\Delta_1$ according to wikipedia, then what is a $\Sigma_1$
definition of it ?Also, have the philosophical implications of this already been discussed ?

And for a subsidiary questions :

Are there models of set theory that "think" one of its element is infinite when it is actually finite ?
Akin to what happens between countable and non-countable sets with the Löwenheim-Skolem theorem.

答:
Let me add a remark to Noah's excellent answer.

While $\sf ZFC$ prove that "finite" is equivalent to "Dedekind-finite", and therefore both are $\Delta_1^{\sf ZFC}$ notions; this is not true for $\sf ZF$. Namely, the formula $\varphi(x)$ stating that $x$ is Dedekind-finite is not $\Delta_1^{\sf ZF}$.

If $M$ is a countable transitive model of $\sf ZFC$, then there is a generic extension of $M$, $M[G]$, and an intermediate model of $\sf ZF$, $N$, such that:

$M\subseteq N\subseteq M[G]$ are all transitive.

$x\in N$ and therefore $x\in M[G]$.

$N\models x$ is Dedekind-finite, but $M[G]\models x$ is countable (and therefore Dedekind-infinite).

So, if "Dedekind-finite" was $\Delta_1^{\sf ZF}$, it would be absolute between the transitive models $N$ and $M[G]$. But that's not true, of course. So it's worth keeping track of our assumptions when going between equivalent definitions.

Let me also throw in another $\Pi_1$ definition of finiteness, due to Tarski: $x$ is finite if and only if $(\mathcal P(x),\subseteq)$ is well-founded. Which is to say, for every non-empty $y\subseteq\mathcal P(x)$, there is some $z\in y$ which is $\subseteq$-minimal. Writing this definition out carefully will be $\Pi_1$.


问:
Linearized ordinary differential equation for Hydraulic Mill
The hydraulic mill is consists of two main part; servo valve and hydraulic cylinder.

We will consider flow-control servo valve. The dynamics of the servo valve is
\begin{equation}
\label{1}\tag{1} Q_s=C_Q\pi dx\sqrt{\frac{2}{\rho}(P_s-P_1)}
\end{equation}
Here,

$Q_s$=supply flow,

$C_Q$=flow coefficient,

$\rho$=oil density,

$x$=servo-valve displacement,

$P_s$=supply pressure,

$P_1$=output pressure of the valve.

The equation for the flow of oil to the hydraulic cylinder,
\begin{equation}
\label{2}\tag{2}Q_s=a\dot{y}+\frac{V_1}{\beta}\dot{P_1}
\end{equation}
Here,

$a$=area of the cylinder,

$y$=hydraulic piston displacement,

$V_1$=volume of the primary side of the cylinder,

$\beta$=bulk modulus of the oil,

$P_1$=cylinder pressure on primary side.

Schematic: Hydraulic servo system

Equation \eqref{1} needs to be linearized for both input-flow & output-flow to the cylinder according to valve displacement $x$.

Help me find the ODE for rate of change in cylinder Pressure $\dot{P_1}$ and $\dot{P_2}$.

Body:
Assuming an operating point at $x_0,P_{s_0},P_{1_0}$ we have

regarding $K_0x\sqrt{P_s-P_1}$

$$
K_0 x \sqrt{P_s-P_1} = K_0 x_0\sqrt{P_{s_0}-P_{1_0}}+K_0\sqrt{P_{s_0}-P_{1_0}} (x-x_0)-\frac{K_0 x_0 (P_1-P_{1_0})}{2 \sqrt{P_{s_0}-P_{1_0}}}+\frac{K_0 x_0 (P_s-P_{s_0})}{2
   \sqrt{P_{s_0}-P_{1_0}}}+O_1(|x-x_0|^2)+O_2(|P_s-P_{s_0}|^2)+O_3(|P_1-P_{1_0}|^2)
$$

so near the operating point we have

$$
a\dot{y}+\frac{V_1}{\beta}\dot{P_1} = K_0 x_0\sqrt{P_{s_0}-P_{1_0}}+K_0\sqrt{P_{s_0}-P_{1_0}} (x-x_0)-\frac{K_0 x_0 (P_1-P_{1_0})}{2 \sqrt{P_{s_0}-P_{1_0}}}+\frac{K_0 x_0 (P_s-P_{s_0})}{2
   \sqrt{P_{s_0}-P_{1_0}}}
$$


答: Hint: if you distribute the equation, you get the expression $2 - x^{n} = 0$. </p>

Title: Find all positive solutions for equation:

Body:
$nx^{(n+1)}-(n+1)x^n+1=0$

There's nothing told about $n$, I guess $ n \in N $. I would like any kind explanations, thanks! I appreciate your time.


问:
If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.
If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.

This is what I have so far. We know that $d|a$ and $d|b$ which implies that $d|c$, but that is not enough to show that $gcd(a,c)=gcd(b,c)=d$, only that they are at least $d$. Also $a^2=c^2-b^2$ and $b^2=c^2-a^2$. Any hints or solutions are greatly appreciated.

答:
Hint:

You have found $d|c$ so $d|\gcd(a,c)$ and $\gcd(a,c) = kd$ for some positive integer $k$.

Then $k^2d^2 | a^2$ and $k^2d^2 | c^2$ so $k^2d^2 | (c^2-a^2)$.

This implies ... and ...


问:
studying the function $f(x)= \sqrt{x^2+x}-x$
I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.

In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.

$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0 \Rightarrow   2x+1-2 \sqrt{x^2+x} >0 \Rightarrow   2x+1>2 \sqrt{x^2+x}  \Rightarrow   1>0 $ . So it seems that the derivative is always positive in the domain.

But comparing with the result in wolfram alpha the function is decreasing in $(-\infty, -1]$

答:
Note that the domain of the derivative
$$f^\prime(x)\:=\: \frac{2x+1}{2 \sqrt{x^2+x}}-1$$
is not equal to the domain of $f$ since the denominator zeros $\{-1,0\}\,$ have to be excluded.

The sign analysis is quickly achieved using the AGM inequality:

If $0<x$ then
$$\sqrt{x(x+1)}\:<\:\frac{2x+1}{2}\quad\implies\; 0\;<\;\frac{2x+1}{2 \sqrt{x^2+x}}-1$$
For $x<-1$ one obtains
$$\sqrt{(-x)(-(x+1))}\:<\:\frac{-x-(x+1)}2 \;=\; -\frac{2x+1}{2}\quad\implies\;
f^\prime(x)<-2$$


问:
General solution of $a^x = b^x\cdot(1-c) - c$
I am doing some simple models to estimate the cost of having unrealized taxed investments, but I keep running into equations that have the form of:

$$a^x = b^x\cdot(1-c) - c$$

Or even,

$$a^x = b^x\cdot d - c$$

All of the 'numbers' in the equations are in $(0,\infty)$.

From my understanding, these equations fall into the category of transcendental equations.
I have no idea of to proceed to solve the above equations.
I am not looking just for the solution to the equations, but also want to understand how to solve them, in case I get similar trouble in the future.

答:
A transcendental equation contains a transcendental function that cannot be expressed in finite number of simple algebraic operations, and also such equation has no closed form expression, has no finite number of standard operations.

You can express the equation through a function $f$ and then use some simple calculus methods to at least approximate $x$

$$f(x)=b^x\cdot d - c - a^x \quad f:[0, \infty) \rightarrow \mathbb{R} \hspace{30px} \text{$a,b,c,d$ are also in the domain}$$

$$f(0)=d-c-1\hspace{30px} d\geq1+c \quad \text{or} \quad d< 1+c \\ \text{When $d=1+c$ you have $f(0)=0$}$$

$$f'(x)=d\ln{b}\cdot b^x-\ln{a}\cdot a^x \\[6px] f'(x)=0\implies a^x=d\frac{\ln{b}}{\ln{a}}b^x\\[6px]x=\log_a{d}+\log_a{\ln b}+x\cdot\log_a{b}-\log_a{\ln a}\\[6px] x=m=\frac{\log_a{d}+\log_a{\ln b}-\log_a{\ln a}}{1-\log_a{b}} \\[20px] \text{There exists only one $m$ for which the function is min or max} \\ \text{This is used for the following...}$$

You can then set $f$ to be defined by, $f(x) = d\cdot b^x -a^x$
And $g(x)=d\cdot b^x -c -a^x$

then, $f(n) = 0 \implies a^n=d\cdot b^n \implies n=\log_a{d} + n\log_a{b} \implies n=\frac{\log_a{d}}{1-\log_a{b}}$

If you think about the graph of the functions, the area between $f$ and $\frac{c-0}{t_0-n}(x-n)$ and interval $[t_0, n]$ is useful in the following

Where the second function is the line that goes through point $x:f(x)=0$, (that is point $(n, 0)$) and point $(t_0, f(n) + c)=(t_0, c)$, where $t_0: g(t_0)=0$

So $$\int_{t_0}^n f(x)-c\frac{(x-n)}{t_0-n}dx \approx 0$$

But the bigger $c$ gets, the less approximate it gets, though as $c \rightarrow 0$ both sides are equal.


答: If $f$ divides at least two of $a,\,b,\,c$, $f^2$ divides at least two of $a^2,\,b^2,\,c^2$ and hence all of them, so $f$ divides all of $a,\,b,\,c$. Hence $\gcd(a,\,b),\,\gcd(b,\,c),\,\gcd(c,\,a)$ all divide each other, and are equal.

Body:
If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.

This is what I have so far. We know that $d|a$ and $d|b$ which implies that $d|c$, but that is not enough to show that $gcd(a,c)=gcd(b,c)=d$, only that they are at least $d$. Also $a^2=c^2-b^2$ and $b^2=c^2-a^2$. Any hints or solutions are greatly appreciated.


问:
Some simple questions regarding Modular Spaces, Modular Forms, Hyperbolic Geometry, Projective Special Linear and Modular Groups
Asymptotic Winding of the Geodesic Flow on Modular Surfaces and Continuous Fractions. Y. Guivarc'h and Y. Le Jan.

I have been struggling with the first section (1. Framework and notations) for a long time.

I have some questions, that are:

How is the hyperbolic distance defined to get that the hyperbolic distance between $gK$ and $hK$ is $2 \log(||h^{-1}g||)$?

what do they mean by saying $||\cdot||$ is the euclidean operator norm? How is it defined?

the last line we have $i(c+d)^{-2}$, is not it supposed to be $i(ci+d)^{-2}$?

what is the idea of the last line?

Is $C_K(G)$ the space of all complex-valued continuous maps with compact support defined on $G$?

after how Haar measure is defined, is not it supposed to be for any $g$ such that $g(0)= \alpha$ not $g(0)= \infty$?

$\Sigma$ is finite, Does it mean it is finite with respect to a measure or finite in the usual sense that it is cardinality is finite?

Are not cusps subsets of $\mathbb{Q} \cup\infty$?

How does $\Sigma$ act transitively on $C$?

How is the stabiliser of the cusp defined?

What is the order of the stabiliser?

The space of entire modular forms of dimension $-2$, How is it possible (negative dimension)?

In the end, I would like to enquire regarding the references that might help me understand the first section properly, regardless of being the references of this article or not. (In the worst case, I would be very grateful if one could tell me the references that I need from the image attached below).

答:
(First, this is a somewhat awkward way to present your question... giving us no idea of your personal context, for example, ...)

All that your copied text discusses is absolutely standard "elliptic modular forms, Riemann surfaces" stuff. R. Gunning's classic orange Princeton series book on "modular forms" is terrific.

Or nearly any other of the many introductions in either texts or on-line in the last 60+ years. E.g., my course notes at my web page.

It's not the sort of thing that one absorbs in a few minutes or a few days... And, no, many people have never heard of any of it, etc. But, in fact, this sort of thing is a huge part of contemporary mathematics. Played a role in A. Wiles' proof of Fermat's Last Theorem, and such.

No wonder that a typical undergrad math preparation does not prepare a person for this sort of thing.


问:
Inner product polynomials

Let $V$ be the vector space of real polynomial $\mathbb{R}[x]$ endowed
  with the inner product

$\langle f,g \rangle = \displaystyle\int_{-\infty}^{\infty}
 e^{-|x|}f(x)g(x) \ dx$

By considering the sequence of subspaces $\{V_n\}$ where

$V_n = \{f(x) \in \mathbb{R}[x] : \deg f \leq n \}$

or otherwise, show that there exist unique monic polynomials
  $\phi_n(x)$ for $n \geq 0$ such that

$\displaystyle\int_{-\infty}^{\infty} e^{-|x|}\phi_n(x)g(x) \ dx = 0$

whenever $\deg g < n$, and find $\phi_n(x)$ for $n = 0,1,2.$

What is the coefficient of $x^{2000}$ in $\phi_{2007}(x)$?

I'm having trouble even knowing where to begin with this question, any help appreciated!

答:
$n = 0$: $\phi_n$ being monic implies $\phi_0 = 1$, if $\mathrm{deg}(g)<0$, then $g = 0$, and
$$
\int_{-\infty}^{\infty} e^{-|x|}\phi_0(x)g(x) \, dx = 0
$$
is trivial.

$n= 1$: $\phi_1$ has to be $x$ for being monic, and $g = c$ is a constant, by symmetry of the integral:
$$
\int_{-\infty}^{\infty} e^{-|x|}\phi_1(x) c\, dx = 0
$$
for $e^{-|x|}\phi_1(x)$ is odd.

$n= 2$: let $\phi_2(x) = x^2 + a_1 x +a_2$, $g\in \mathrm{span}\{1,x\}$, by symmetry
$$
\int_{-\infty}^{\infty} e^{-|x|}(x^2 + a_1 x +a_2) x\, dx = 0
$$
implies $a_1= 0$, and
$$
\int_{-\infty}^{\infty} e^{-|x|}(x^2  + a_2)\, dx = 0
$$
yields $a_2 = -2$. Hence $\phi_2(x) = x^2 - 2$ which does not have a degree 1 term.

$n= 2007$: $g\in \mathrm{span}\{1,x,\ldots,x^{2006}\}$, choosing certain $g$ will let you find out certain degree terms in $\phi_{2007}$ is zero by a symmetry argument like above (the integration of an odd function is zero from $-\infty$ to $\infty$, given that it is integrable), the rest is left for you to try.

General result for $\phi_n$, discuss the cases for $n$ being odd and even like above.

Title: Inner product polynomials


答:
Let $d = (a,b)$, $e = (a,c)$, $f=(b,c)$.

We have $d\mid a$ and $d\mid b$, then $d^2 \mid a^2+b^2 = c^2$, and then $d \mid c$.

Since $d \mid a$ and $d \mid c$ then $d \mid (a,c) = e$. So $d \mid e$

In practically the same way you can prove $d \mid f$, $e \mid d$, $e \mid f$, $f \mid d$ and $f \mid e$.

So $d=e=f$.

Title: If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.


问:
Proving $F(x)=F(a)+\sum_{\mu=1}^{n}(x^\mu -a^\mu )H_\mu (x)$
Where $F:R^n \rightarrow R$, $a=(a^1,...,a^\mu)$ and $x=(x^1,...,x^\mu)$.

Also, $H_\mu (a)=\frac{\partial F}{\partial x^\mu}|_{x=a}$

Hi there, this is a problem on General Relativity from Robert Wald. I'm trying to solve it for a few hours but still, it doesn't look that difficult.. I'm pretty sure the fundamental theorem of calculus is a good start:

$F(x)-F(a)=∫_{a}^{x}F′(s)ds$

Then , with $ s=t(x-a)+a$ we have

$F(x)-F(a)=(x-a)∫_{0}^{1}F′[t(x-a)+a]dt$

Which bares a nice similarity! But I don't know how to generalize from there. I was checking up on Stokes theorem to see if there's any connection.

Every time I get stuck on a problem for more than one hour I know it must be something obvious. Weird.

答: I'm sorry for my ignorance, maybe I'm not getting something, but.. isn't it just the Taylor expansion of a function, up to the fist order?</p>


问:
Two rings with the same multiplicative structure but non-isomorphic underlying Abelian groups
I am giving a series of lectures where I introduce some undergraduates to basic ideas from category theory. One of the things I would like to show them is how category theory could be used to make precise some of the intuitive ideas they have been exposed to (e.g., what "natural" means when we speak about the natural isomorphism of a finite dimensional vector space with its double dual, what "universal" really means when we speak of a universal construction, etc.).

One of the subtleties I would like to illustrate with explicit examples is the difference between "property" and "structure". I would really like to show them that "being a preadditive category" (by preadditive here I just mean $Ab$-enriched category, I do not assume existence of a $0$-object) is not a property, so it does not make sense to give me an abstract category and ask me if it is preadditive or not (I can ask whether it can be made into a preadditive category, but even in that case the preadditive structure may not be unique).

To illustrate this, I would like to find some minimal example of two different $Ab$-enrichments of the same abstract category. A good way to do that seems that of finding two (unitary and associative) rings $(R,+,.)$ and $(S,+',*)$, such that $(R,.)\cong (S,*)$ as monoids (i.e., as one-object categories) but $(R,+)\not\cong (S,+')$ as Abelian groups.

Do you have some easy example of two rings with the properties described above?

答:
I think this works. Take $R=\mathbb{Z}[x]$, and $S=\mathbb{F}_3[x]$. They certainly have different additive structure, since $R$ has characteristic $0$ but $S$ has characteristic $3$.

Since they are both UFDs, the multiplicative structure on the nonzero elements will be of the form $U\times M$, where $U$ is the unit group, and $M$ is the free commutative monoid of rank $\kappa$, where $\kappa$ is the number of irreducibles up to associates. The unit group of both $R$ and $S$ is cyclic of order $2$; $\mathbb{Z}[x]$ and $\mathbb{F}_3[x]$ both have $\aleph_0$ non-associate irreducibles. Now adjoin a zero element to both of these to get the full multiplicative structures of $R$ and $S$. So their multiplicative structures are isomorphic.


问:
Prove that there exists an injective function K : S → ℚ given the set S:={s ∈ R: sup {f(x): x < s}≠ inf {f(y): y > s}}
$f : \Bbb R \to \Bbb R$ is an increasing function, which means that $f(x) \leq f(y)$ whenever $x \leq y$.

Define the set
$S:= \{ s \in \Bbb R \mid \sup \{f(x): x \lt s \} \neq \inf \{f(y): y \gt s \} \}$.
Show that there exists an injective function $K : S \to  \Bbb Q.$

The hint provided is to find a rational number $q$ between the sup and the inf, given $s \in S$. My question is: How would I prove the existence of an injective function by looking for this number $q$? I'm not sure what the process would look like..

答: Let $s \in S$. Since $f$ is increasing we have $\sup \{f(x) : x<s\} <\inf \{f(y) >s\}$. Let $q_s$ be  a rational number with $\sup \{f(x) : x<s\} <q_s<\inf \{f(y) >s\}$. Define $K(s)$ to be $q_s$. We only have to check that $K$ is injective. Suppose $s<t$. Then $q_s<\inf \{f(y) : y>s\} \leq \sup \{f(x) : x<t\}<q_t$ where the middle inequality is proves as follows: $\inf \{f(y) : y>s\} \leq f(\frac  {s+t} 2) \leq \sup \{f(x) : x<t\}$ since $f$ is increasing and $s<\frac  {s+t} 2<t$. We have proved that $s<t$ implies $K(s) <K(t)$. This implies that $k$ is injective.


问:
$\tau = \{U \subseteq X: X\setminus U \mbox{ is finite or } p\not\in U \}$ is not metrizable

Let X be a uncountable set and fix $p\in X.$ Define a topology on $X$ by : $\tau = \{U \subseteq X: X\setminus U \mbox{ is finite or } p\not\in U \}$. Show that $(X,\tau)$ is not metrizable.

Every metrizable space is first countable, so I want to prove that $(X,\tau)$ is not first countable. In my intuition, the most attractive point to show that this condition fails is the $p\in X$ itself. So let $\{B_n\}$ be a countable basis at $p$. Then, by definition of $\tau$, the set $X\setminus B_n$ is finite for every $n.$ For every $x\in X, x\neq p$, we have $p\in X\setminus \{x\}$, where $X\setminus \{x\}$ is open in $X$, as its complement is $\{x\}$, which is finite. Therefore, there exists $n_x\in \mathbb N$ such that $$p\in B_{n_x}\subseteq X\setminus \{x\}.$$

Taking complements on both sides, we have $\{x\}\subseteq  X\setminus B_{n_x}$.

I can't proceed from here. I need to find a way to argue, by the uncountability of $X$ and the fact that $\{x\}\subseteq X \setminus B_{n_x} $, that at least one of the $B_{n_x} $ will not be finite. How can I proceed? Thank you.

答:
You started fine, but then you bogged down a bit. For each $n\in\Bbb N$ let $F_n=X\setminus B_n$, and note that $B_n=X\setminus F_n$. Let $C=\bigcup_{n\in\Bbb N}F_n$; each $F_n$ is finite, and there are only countably many of them, so $C$ is countable. Then

$$\bigcap_{n\in\Bbb N}B_n=\bigcap_{n\in\Bbb N}(X\setminus F_n)=X\setminus\bigcup_{n\in\Bbb N}F_n=X\setminus C\,.$$

$C\cup\{p\}$ is still countable, and $X$ is uncountable, so $X\setminus(C\cup\{p\})$ is uncountable and therefore non-empty. Let $x\in X\setminus(C\cup\{p\})$, and let $U=X\setminus\{x\}$.

Show that for each $n\in\Bbb N$, $B_n\nsubseteq U$, and conclude that $\{B_n:n\in\Bbb N\}$ is not a base for $\tau$.


问:
Is the set $\{(z, \bar{z}), z\in\mathbb{C}\} \subseteq \mathbb{C}^2$ Zariski closed?
Question is in the title. It feels like the answer should be no, but I don't see a simple way to prove it.

I mean, if it was Zariski closed, then there would exist a polynomial expression in z and $\bar{z}$ that would be 0 for all $z\in\mathbb{C}$. And it's somewhat clear from a complex analysis perspective that this cannot exist (by considering partial derivatives after z and $\bar{z}$). But it feels like it should be more elementary than this. Am I missing something obvious?

答:
Suppose for contradiction that this set is Zariski closed, i.e. there exist polynomials $p_1, \dots, p_k \in \mathbb{C}[x,y]$ such that $p_i(a,b) = 0$ for all $i$ if and only if $a = \overline{b}$.

Now for each $i$, we have that $p_i(z,z) \in \mathbb{C}[z]$ has infinitely many zeros (e.g. any real number), so $p_i(z,z) = 0$.

Finally, for any $z \in \mathbb{C}$, we have $p_i(z,z) = 0$ for all $i$, so $z = \overline{z}$. This is a contradiction.

Edit: made a silly mistake at first (assumed $k = 1$), should be fixed now.


问:
Equivalent definition of continuity in metric spaces
I am confused at the equivalency of the two statements about continuity in metric spaces.

$f$ is continuous at $x$ if and only if for every $\epsilon>0$ there is a $\delta>0$ such that $x' \in B_\delta(x) \implies f(x')\in B_\epsilon(f(x))$, I know this is equivalent to $d(x,x')<\delta \implies d(f(x),f(x'))<\epsilon$ however I am confused as to why the statement is equivalent to $f(B_\delta(x)) \subset B_\epsilon(f(x))$.

Can someone explain this? Thanks

答:
By definition $x'\in B_\delta(x)$ if and only if $d(x,x')<\delta$, and $f(x')\in B_\epsilon(f(x))$ if and only if $d(f(x),f(x'))<\epsilon$. Thus, the statement

$$x'\in B_\delta(x)\implies f(x')\in B_\epsilon(f(x))$$

is exactly equivalent to the statement

$$d(x,x')<\delta\implies d(f(x),f(x'))<\epsilon\,:$$

the lefthand sides are equivalent, and the righthand sides are equivalent.


问:
Reducing mean deltas in numbers adding up to a certain value
This is likely quite common, I have 12 numbers that add up to 1, in fact they are seasonality numbers, % of occurences per month. I want to reduce the deltas (closer but not to the mean), while still adding up to 1, this would achieve the effect of smoothing the graph and reducing the spikes / uneveness without making it a flat line. Thanks</p>

答:
Instead of $x_1,x_2,\ldots, x_{12}$

you could use  $y_1,y_2,\ldots, y_{12}$

where $y_i=kx_i +\frac{1-k}{12}$ for some constant $k$ of your choice with $0 < k < 1$.

$k$ closer to $0$ will make the pattern almost flat, while $k$ closer to $1$ will be close to the original pattern


问:
How to calculate: $\sum_{n=1}^{\infty} n a^n$
I've tried to calculate this sum:

$$\sum_{n=1}^{\infty} n a^n$$

The point of this is to try to work out the "mean" term in an exponentially decaying average.

I've done the following:

$$\text{let }x = \sum_{n=1}^{\infty} n a^n$$
$$x = a + a \sum_{n=1}^{\infty} (n+1) a^n$$
$$x = a + a (\sum_{n=1}^{\infty} n a^n + \sum_{n=1}^{\infty} a^n)$$
$$x = a + a (x + \sum_{n=1}^{\infty} a^n)$$
$$x = a + ax + a\sum_{n=1}^{\infty} a^n$$
$$(1-a)x = a + a\sum_{n=1}^{\infty} a^n$$

Lets try to work out the $\sum_{n=1}^{\infty} a^n$ part:

$$let y = \sum_{n=1}^{\infty} a^n$$
$$y = a + a \sum_{n=1}^{\infty} a^n$$
$$y = a + ay$$
$$y - ay = a$$
$$y(1-a) = a$$
$$y = a/(1-a)$$

Substitute y back in:

$$(1-a)x = a + a*(a/(1-a))$$
$$(1-a)^2 x = a(1-a) + a^2$$
$$(1-a)^2 x = a - a^2 + a^2$$
$$(1-a)^2 x = a$$
$$x = a/(1-a)^2$$

Is this right, and if so is there a shorter way?

Edit:

To actually calculate the "mean" term of a exponential moving average we need to keep in mind that terms are weighted at the level of $(1-a)$. i.e. for $a=1$ there is no decay, for $a=0$ only the most recent term counts.

So the above result we need to multiply by $(1-a)$ to get the result:

Exponential moving average "mean term" = $a/(1-a)$

This gives the results, for $a=0$, the mean term is the "0th term" (none other are used) whereas for $a=0.5$ the mean term is the "1st term" (i.e. after the current term).

答:
Your approach is really quite good,
and a nice way of discovering the result,
as opposed to deriving an already known result.

The only change I would make is to
make the upper limit of the sum a variable,
and let that go to $\infty$ if you want.
Besides, the finite sum is often useful.

So, let
$S(a, m)=\sum_{n=0}^{m} n a^n$.
I prefer to have the sum start at $0$,
and it doesn't change the result.

Then

$\begin{align}
S(a, m)
&=0+\sum_{n=1}^{m} n a^n\\
&= a\sum_{n=1}^{m} n a^{n-1}\\
&= a\sum_{n=0}^{m-1} (n+1) a^{n}\\
&= a\big(\sum_{n=0}^{m-1} n a^{n} + \sum_{n=0}^{m-1} a^{n}\big)\\
&=a(S(a, m-1)+T(a, m-1))\\
\end{align}
$

where
$T(a, m)= \sum_{n=0}^{m} a^{n}$.

We now do the same thing for $T$,
with the difference that the term for $n=0$ is not $0$.

$\begin{align}
T(a, m)
&=1+\sum_{n=1}^{m}  a^n\\
&= 1+a\sum_{n=1}^{m} a^{n-1}\\
&= 1+a\sum_{n=0}^{m-1} a^{n}\\
&=1+aT(a, m-1)\\
\end{align}
$

Since $T(a, m-1)
= T(a, m)-a^m$,
$T(a, m) = 1+a(T(a, m)-a^m)
=1+aT(a, m)-a^{m+1}$
so $T(a, m) = (a^{m+1}-1)/(a-1)$.

Putting this in the equation for $S$,
$S(a, m) = a(S(a, m-1)+\frac{a^m-1}{a-1})$.

Again, since $S(a, m-1) = S(a, m)-m a^m$,

$\begin{align}
S(a, m) &= a(S(a, m-1)+\frac{a^m-1}{a-1})\\
&= a(S(a, m)-m a^m+\frac{a^m-1}{a-1})\\
&= aS(a, m)+\frac{a(a^m-1)-m a^{m+1}(a-1)}{a-1}\\
&= aS(a, m)+\frac{a^{m+1}-a-m a^{m+2}+m a^{m+1}}{a-1}\\
&= aS(a, m)-\frac{m a^{m+2}+(m+1) a^{m+1}+a}{a-1}\\
\end{align}
$

so
$(a-1)S(a, m) = \frac{m a^{m+2}+(m+1) a^{m+1}+a}{a-1}$
and
$S(a, m) = \frac{m a^{m+2}+(m+1) a^{m+1}+a}{(a-1)^2}$.

If $|a| < 1$, if we let $m \to \infty$ in the expressions
for $T$ and $S$,
we get
$T(a, \infty) = 1/(1-a)$
and
$S(a, \infty) = a/(1-a)^2$.

Note that by writing
$S(a, m)$ (and, similarly, $T(a, m)$)
in two ways in terms of $S(a, m-1)$,
we avoided having to discover
the expression for $S(a, m)$.

Again, this is certainly not the best way to
$verify$ the formulas
for $S(a, m)$ and $T(a, m)$,
but it is a way to $discover$ them.


问:
Simplify a sum with a product and multinomial coefficient
This is a follow-up question to my previous post, where I've got a great help! (I created a new one to avoid editing the original one).
Can the following sum be simplified?

$$
 \sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} \left(\prod_{i=1}^{m}k_i\right) \cdot {B \choose k_1, k_2, \ldots k_m} \cdot {W \choose g_1, g_2, \ldots, g_m}
$$
where $B + W = n$, and $\forall_i: g_i = x - k_i$, where $x = \frac{n}{m}$, and $m\mid n$ and $k_i \geq 1$

I tried to follow the steps suggested in my previous post, but I got stuck. For $m=2$ I've got the following
$$
\sum_{\substack{k_1 + k_2 &=& B \\ k_1, k_2 &\geq& 0}} k_1 \cdot k_2 \cdot {B \choose k_1, k_2} \cdot {2N - B \choose N-k_1, N-k_2} \\ = \sum_{\substack{k_1 + k_2 &=& B \\ k_1, k_2 &\geq& 0}} k_2 \cdot {B \choose k_1-1} \cdot {2N - B \choose N-k_1, N-k_2} \\
= \sum_{k_1 = 0}^B \left(B-k_1 \right) \cdot {B \choose k_1-1} \cdot {2N - B \choose N-k_1} \\
$$
From the previous post I know how to handle the binomial coefficients, but I'm not sure what can I do now with the $B-k_1$ factor inside the sum?

答:
We have
$$
S = \sum\limits_{\left\{ {\matrix{
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr
   {g_{\,1}  + g_{\,2}  + \, \ldots  + \,g_{\,m}  = W}  \cr
 } } \right.} {\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{ B \cr  k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m}  \cr}  \right)
\left( \matrix{W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m}  \cr}  \right)}
$$

plus the added conditions
$$
\left\{ \matrix{
  B + W = n \hfill \cr
  g_{\,i}  = x - k_{\,i}  \hfill \cr
  1 \le k_{\,i}  \hfill \cr}  \right.
$$
Leave apart for the moment the fact that $x=n/m, \; m | n$.

Because of the product, we can consider $0 \le ki$, and for the multinomial not to be null it shall be
$0 \le g_i$.

Thus  the set of conditions becomes (if I understood them properly)
$$
\eqalign{
  & \left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr
   {g_{\,i}  = x - k_{\,i} }  \cr
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr
   {g_{\,1}  + g_{\,2}  + \, \ldots  + \,g_{\,m}  = W}  \cr
 } } \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr
   {g_{\,i}  = x - k_{\,i} }  \cr
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr
   {\left( {x - k_{\,1} } \right) + \left( {x - k_{\,2} } \right)
 + \, \ldots  + \,\left( {x - k_{\,m} } \right) = W}  \cr
 } } \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr
   {g_{\,i}  = x - k_{\,i} }  \cr
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr
   {W + B = mx = n}  \cr
 } } \right. \cr}
$$

Under these conditions the summand can be rewritten as
$$
\eqalign{
  & \left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{ B \cr  k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m}  \cr}  \right)
\left( \matrix{ W \cr  g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m}  \cr}  \right) =   \cr
  &  = \left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right){{B!} \over {k_{\,1} !\,k_{\,2} !\, \cdots \,k_{\,m} !}}
{{\left( {mx - B} \right)!} \over {\left( {x - k_{\,1} } \right)!\,\left( {x - k_{\,2} } \right)!\, \cdots \,\left( {x - k_{\,m} } \right)!}} =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {x!} \right)^{\,m} }}
\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right){{x!} \over {k_{\,1} !\,\left( {x - k_{\,1} } \right)!}}
{{x!} \over {k_{\,2} !\,\left( {x - k_{\,2} } \right)!}} \cdots {{x!} \over {k_{\,m} !\,\left( {x - k_{\,m} } \right)!}} =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {x!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {{{x\left( {x - 1} \right)^{\,\underline {\,k_{\,i}  - 1\,} } } \over {\left( {k_{\,i}  - 1} \right)!\,}}}  =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {{{\left( {x - 1} \right)^{\,\underline {\,k_{\,i}  - 1\,} } } \over {\left( {k_{\,i}  - 1} \right)!\,}}}  =
 {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr  k_{\,i}  - 1 \cr}  \right)}  \cr}
$$
so that the sum becomes
$$
\eqalign{
  & S(B,m,x)\quad \left| {\;1\, \le mx \le B} \right.\quad  =   \cr
 &  = \sum\limits_{\left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr
   {g_{\,i}  = x - k_{\,i} }  \cr
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr
   {g_{\,1}  + g_{\,2}  + \, \ldots  + \,g_{\,m}  = W}  \cr
 } } \right.}
 {\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{B \cr k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m}  \cr}  \right)
\left( \matrix{ W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m}  \cr}  \right)}  =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\sum\limits_{\left\{ {\matrix{
   {\left( {1\, \le } \right)\,k_{\,i} \,\left( { \le \,x} \right)}  \cr
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr
 } } \right.}
 {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr  k_{\,i}  - 1 \cr}  \right)} }  =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\sum\limits_{\left\{ {\matrix{ {\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}  \cr
   {j_{\,1}  + j_{\,2}  + \, \ldots  + \,j_{\,m}  = B - m}  \cr
 } } \right.}
 {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr  j_{\,i}  \cr}  \right)} }  \cr}
$$

Now, since
$$
\eqalign{
  & \left( {\left( {1 + z} \right)^{x - 1} } \right)^{\,m}
  = \prod\limits_{i = 1}^m {\left( {\sum\limits_{\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}
 {\left( \matrix{ x - 1 \cr  j_{\,i}  \cr}  \right)z^{\,j_{\,i} } } } \right)}  =   \cr
  &  = \sum\limits_{0\, \le \,s\, \le \,m\,\left( {x - 1} \right)}
 {\sum\limits_{\left\{ {\matrix{ {\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}  \cr
   {j_{\,1}  + j_{\,2}  + \, \ldots  + \,j_{\,m}  = \,s}  \cr
 } } \right.}
 {\left( {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr j_{\,i}  \cr}  \right)} } \right)z^{\,s} } }  \cr}
$$
we conclude that
$$ \bbox[lightyellow] {
\eqalign{
  & S(B,m,x)\quad \left| {\;1\, \le mx \le B} \right.\quad  =   \cr
 & = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;\;\left[ {z^{\,B - m} } \right]\left( {1 + z} \right)^{m\left( {x - 1} \right)}  =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;\;\left( \matrix{ m\left( {x - 1} \right) \cr  B - m \cr}  \right) =   \cr
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;{{\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {B - m} \right)!\left( {mx - B} \right)!}}\; =   \cr
  &  = {{B!} \over {\left( {B - m} \right)!}}\;
{{\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}\;
 = \left( \matrix{ B \cr  m \cr}  \right)\;
{{m!\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }} \cr}
}$$

--- in reply to your comment  ---

Consider the following simple case
$$
\eqalign{
  & \left( {1 + z} \right)^a \left( {1 + z} \right)^a
  = \left( {\sum\limits_{k_1 } {\left( \matrix{ a \cr  k_1  \cr}  \right)z^{k_1 } } } \right)
\left( {\sum\limits_{k_2 } {\left( \matrix{ a \cr   k_2  \cr}  \right)z^{k_2 } } } \right) =   \cr
  &  = \prod\limits_{i = 1}^2 {\left( {\sum\limits_{k_i }
 {\left( \matrix{a \cr k_i  \cr}  \right)z^{k_i } } } \right)}  =   \cr
  &  = \sum\limits_{k_1 } {\sum\limits_{k_2 }
 {\left( \matrix{ a \cr  k_1  \cr}  \right)\left( \matrix{ a \cr k_2  \cr}  \right)z^{k_1  + k_2 } } }
  = \sum\limits_s {\sum\limits_{k_1  + k_2  = s}
 {\left( \matrix{a \cr k_1  \cr}  \right)\left( \matrix{a \cr   k_2  \cr}  \right)z^s } }  =   \cr
  &  = \sum\limits_s {\left( {\sum\limits_{k_1  + k_2  = s} {\left( {\prod\limits_{i = 1}^2
 {\left( \matrix{ a \cr  k_i  \cr}  \right)} } \right)} } \right)z^s }
  = \sum\limits_s {\left( \matrix{ 2a \cr   s \cr}  \right)z^s }  \cr}
$$
which implies
$$
\sum\limits_{k_1  + k_2  = s} {\left( {\prod\limits_{i = 1}^2
 {\left( \matrix{a \cr  k_i  \cr}  \right)} } \right)}
  = \left( \matrix{ 2a \cr   s \cr}  \right)
$$
and which is just the Vandermonde convolution.


问:
The Zeta Function and the Prime Number Theorem
I an trying to learn more about the $\zeta$ function, and I stumbled upon this very interesting paper https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Riffer-Reinert.pdf and having a few questions on some of the proofs.

$(1)$ In the proof of theorem $2.4$, I am confused about the part that states "Using theorem $3.3$ we see that
$\zeta(z) =$" Since there is no Theorem $3.3$ in the paper.

$(2)$ In the proof of Lemma $3,2$, I don't understand the second equality after "It now follows that". In order for an expression to equal the real part of itself, wouldn't we have to know that the imaginary part is $0$? I don't see why that would be the case here.

$(3)$ In Theorem $4.3$, why is $\zeta$ s homomorphic at $z_0$?

$(4)$ In the proof of Theorem $5.2$, I don't understand how applying Theorem $6.1$ gives us the inequality's it states.

答:
Well, (1) was already addressed by reuns in his first comment: we just have the functional equation withou developing the gamma function. There is certainly no lemma 3.3 in that paper. Their bad...

(2) Here observe we're taking the logarithm of the absolute value (or modulus) of that expression. This means that is the real logarithm function...

(3) is something that I suppose was proved earlier: the zeta function is holomorphic in the shole complex plane except for a simple pole at $\;z=1\;$. Thus, for any element $\;\sigma+it_0\;,\;\;t_0\neq0\;$, the zeta function is analytic there.

Number (4) I don't get it. That paper is a weird one: using future results to deduce stuff. That doesn't seem wise nor even good mathematical writing.


问:
Finding parametrization of curve-cylinder intersection
I want to show that the curve
$$ \gamma(t)=(1+\cos(t),\sin(t),2\sin(t/2)) $$
is the interesection between
\begin{align*}
     x^2+y^2+z^2=4, \newline (x-1)^2+y^2=1.
\end{align*}

The way I wanted to show this was by parametrizing the intersection and then finding a bijective re-parametrization.

So for the intersection curve I substituted
$$ 1=(x-1)^2+y^2=x^2-2x+1+y^2 \iff x^2+y^2=2x $$
into the sphere equation, giving
$$ 2x+z^2=4 \iff x=2-\frac{z^2}{2}. $$
From the ylinder equation I get
$$y=\sqrt{2x-x^2}=\sqrt{4-z^2-4+2z^2-\frac{z^4}{4}}=\pm z\sqrt{1-\frac{z^2}{4}} $$
Now I have the intersection curve
$$ \alpha(t)=(2-\frac{z^2}{2},\pm z\sqrt{1-\frac{z^2}{4}},z). $$

I plotted them in WolframAlpha and they seem to have the same image, but I can't find a proper map $\varphi$ with $\gamma=\alpha\circ\varphi$.

From the last parameter it looks like it should be $\varphi(z)=2\sin(z/2)$, which is correct for the first (and obviously third) entry because
$$ 2-\frac{(2\sin(z/2))^2}{2}=1+\cos(z), $$

but in the second entry I have

$$\pm2\sin(z/2)\sqrt{1-\frac{(2\sin(z/2))^2}{4}}\neq\sin(z).$$

Is there something wrong with my parametrizations?

答:
I would recommend starting with the parametrization of the cylinder given by the second equation:

\begin{align*}
(x-1)^{2} + y^{2} = 1 \Longleftrightarrow
\begin{cases}
x = 1 + \cos(\theta)\\\\
y = \sin(\theta)
\end{cases}
\end{align*}
where $\theta\in[0,2\pi]$. If we insert such relations into the first equation, we get that
\begin{align*}
(1 + \cos(\theta))^{2} + \sin^{2}(\theta) + z^{2} = 4 & \Longleftrightarrow 1 + 2\cos(\theta) + 1 + z^{2} = 4\\\\
& \Longleftrightarrow z^{2} = 2 - 2\cos(\theta)\\\\
& \Longleftrightarrow z = \pm\sqrt{2-2\cos(\theta)}
\end{align*}

But we can go further: according to the double-angle identity, we get that
\begin{align*}
\cos(\theta) & = \cos^{2}(\theta/2) - \sin^{2}(\theta/2) = 1 - 2\sin^{2}(\theta/2) \Rightarrow 2 - 2\cos(\theta) = 4\sin^{2}(\theta/2)
\end{align*}

Finally, we have that
\begin{align*}
z = \pm\sqrt{4\sin^{2}(\theta/2)} = \pm 2\sin(\theta/2)
\end{align*}

and we are done.

Hopefully this helps!


问:
Show that $\left \lfloor\log_{2}\left( \lfloor x \right \rfloor)\right \rfloor=\left \lfloor{\log_{2}(x)}\right \rfloor $
I am having some trouble constructing a proof for this equality. Here is what I have tried so far:

Show that $\left \lfloor\log_{2}\left( \lfloor x \right \rfloor)\right \rfloor=\left \lfloor{\log_{2}(x)}\right \rfloor $

Proof:

$\left\lfloor x\right\rfloor=m\iff x-1<m\leq x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$

Let $x-1<m\leq x \rightarrow\left \lfloor\log_{2}(  m \right) \rfloor=\left \lfloor{\log_{2}(x)}\right \rfloor \;\;\;\;\;\;(2)$

From $(1)$ we can see $x−1<m≤x\rightarrow m=\left\lfloor x\right\rfloor \;\;\;(3)$

Let $m=x\rightarrow \left \lfloor{\log_{2}(x)}\right \rfloor=\left \lfloor{\log_{2}(x)}\right \rfloor \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4)$

These statements are identical, thus $\left \lfloor\log_{2}\left( \lfloor x \right \rfloor)\right \rfloor=\left \lfloor{\log_{2}(x)}\right \rfloor $

$\square$

答:
I think you need to state why if $m \le x$ then $\log_2 m \le \log_2 x$ but why that means there can't be an integer $k$ so that $\log_2 m < k \le \log_2 x$.

The reason for that would be that if $\log_2 m < k \le \log 2 x$ then $m < 2^k \le x$.

As $2^k$ is an integer we must have, in this case, $2^k= x$ and $x$ is an integer and $m = 2^k = x$ and $[\log_2 (x)] = [k] = k = \log_2 x = \log_2 [x]$.

....

TO put it together I'd do the following.

If $x \le 0$ then neither $\log_2 x$ nor $\log_2 [x]$ are defined.

However if $0 < x < 1$ the statement is not true as $\log_2 x$ exist (and is negative) but $\log_2 [x]=\log_2 0 $ is undefined.

Now if $x\ge 1$ then there is a distinct integer $m$ so that $m \le x < m+1$ and $m = [x]$.  Also, as $m \ge 1$ there is unique non-negative integer $k$ so that $2^k \le m \le x < m+1 \le 2^{k+1}$.

ANd now we've nailed it!

$2^k \le [c]\le x < 2^{k+1}$.

So $k \le \log_2 [x] \le \log_2 (x) < k+1$.

So that means $[\log_2(x) ]$ and $[\log_2[x]] $ are both equal to the same integer; $k$.

(In essence.  If $k \ge 0$ then differences between $2^k$ and $2^{k+1}$ is always greater or equal to $1$ (equal only if $k=0$), so the difference between the log (base $2$) of integers must always less than or equal to $1$ (equal only if the integers are $1$ and $2$).


问:
Finding Matrix of Linear Transformation without formula
We are given $[T]_\beta ^\gamma = \begin{pmatrix}1&2\\-1&0\end{pmatrix}$. We are also given that $\beta= \{sin^2x, cos^2x\}$ and $\gamma= \{1,sin^2x-1\}$. We want to find $[T]_\gamma ^\beta$. However, we aren't given a formula for the linear transformation, so I'm not sure how to start this.

答:
Hint:

In general, when $\;\alpha,\beta\;$ basis of a vector space, we have that

$$[T]^\beta_\alpha=\left([T]_\beta^\alpha\right)^{-1}$$

Observe that a linear map (or matrix) mapping a basis to a basis is always an isomorphism (in finite dimensions)


问:
Is it true that $\text {ran} (T^*)$ is closed whenever $\text {ran} (T)$ is closed?
Let $\mathcal H$ be a Hilbert space and $T \in \mathcal L (\mathcal H).$ Suppose that $\text {ran} (T)$ is closed. Does it always mean that $\text {ran} (T^*)$ is also closed?

I am trying to prove that by sequential convergence but I couldn't quite do that. Can anybody please help me in this regard? Thanks for your time.

答:
Yes,  it is.  Let $H_1=\text{Ker}(T)^\perp$ and $H_2=\text{Ran}(T)$.  Also let $\iota _1$ and $\iota _2$ be the inclusions of
$H_1$ and $H_2$ into $H$,  respectively, and observe that $\iota _1^*$ and $\iota _2^*$ are the corresponding projections.

If $T$ has closed range then $\iota _2^*T\iota _1$ is an invertible operator from $H_1$ to $H_2$, whence  its adjoint, namely
$\iota _1^*T^*\iota _2$ is an invertible operator from $H_2$ to $H_1$.

Observing that the range of $T^*$ is contained in $H_1$, we deduce that it is precisely  $H_1$, hence closed.


问:
How to solve a root (eg, $x^{1/3}$) using arithmetic but without using a power?
How can I use the exponential constant and the natural logarithm operation, to express a "root" operation?

Example: $x^{1/3}$.

Is there a combination of operations that are only arithmetic, using natural logarithm and/or the $e$ constant, which can express this without an exponential power operation?

The original expression (e.g. $x^{1/3}$) has only one variable, so the re-formulation should also only have the variable $x$.

Example (incorrect): $x \cdot ln{3}$

I read the answers to How can you express radicals as multiplication/addition? but didn't understand how it applies.

答:
you probably dont mean this,  but I'll write it anyway:

$x^{\frac{1}{3}} = e^{\frac{1}{3} ln(x)} $


问:
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$
Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.

答: Apply Cauchy-Schwarz or Buniakovsky inequality:  $(xy+x+y)^2 = (x\cdot y + 1\cdot x + y\cdot 1)^2 \le (x^2+1^2+y^2)(y^2+x^2+1^2) = (x^2+y^2+1)^2\implies xy+x+y \le x^2+y^2+1$.


问:
finding the perfect cube
I had to find the least positive integer $k$ such that $2^4 \cdot 3^5 \cdot 7 \cdot 11^2 \cdot k$ is a perfect cube which is $2^2 \cdot 3 \cdot 7^2 \cdot 11$.

Now I need to write $2^4 \cdot 3^5 \cdot 7 \cdot 11^2 · k$ as a perfect cube using the value of $k$.

I'm confused as to how to use the value of $k$ to find the perfect square.

答:
In order to be a "perfect cube" the prime factorization of a number must be primes all to a power a multiple of 3.

For example, $2^3= 8$ is a perfect cube but so is $2^6= 64= 4^3$.

Here you are given the number $2^4\cdot 3^5\cdot 7^1\cdot 11^2$.  The smallest multiple of that that is a perfect cube is $2^6\cdot 3^6\cdot 7^3\cdot 11^3$ which means it must be multiplied by $2^2\cdot 3^1\cdot 7^2\cdot 11^1$.

The cube root of $2^6\cdot 3^6\cdot 7^3\cdot 11^3$ is $2^2\cdot 3^2\cdot 7^1\cdot 11^1$.

Title: finding the perfect cube


问:
Relationship between discriminants and smoothness of curves
My understanding of the use of the discriminant in elliptic curve theory is to test whether an elliptic curve in Weierstrass normal form over a field not of characteristic either 2 or 3, $y^{2} = x^{3}+Ax+B$, is smooth, i.e., non-singular. What I am confused about is the relationship between the discriminant and smooth curves. For instance, $y=x^{2}$ over $\mathbb{R}$ has a zero discriminant, yet is a smooth curve.

Why does a zero discriminant of an elliptic curve tell us that the curve is not smooth, and why does this  not seem to apply likewise to other types of curves, such as, quadratics?

Is it perhaps because smoothness implies non-singularity for elliptic curves, but in general this is not the case?

答:
Given an elliptic curve, we mean a smooth projective plan curve of the form $Y^2Z = X^3+ AXZ^2 +BZ^3 \subset\mathbb{P}^2.$

You can check that the point $(0: 1: 0)$ is always smooth, by choosing the chart $y\neq 0$.

Then you work on the chart $z\neq 0$, then the points satisfy $ 0= F(x,y) = -y^2 + x^3 +ax+ b$. Smoothness means the differential doesn't vanish in this case. If $\frac{\partial F}{\partial  y} =0, $ then $y=0. $ Then $0=x^3 + ax + b$, whose partial w.r.t. x zero iff have double roots iff discriminant is zero. So singular iff discriminant = 0.

In conclusion, discriminant, in this case, is related to the part of degree three equations, which can be linked to smoothness. Degree 2 is another case, and it has a different definition of discriminant.


问:
Solving $2 + \frac 1{\cos^2 x} + \frac 5{\cos x} = 0$
Simple but confusing trig question:

$$2 + \frac 1{\cos^2 x} + \frac 5{\cos x} = 0$$

I could not find anything in my textbook similar to this....

答:
Hint: by multiplying both sides by $\cos^2 x$, this is the same equation as
$$
2\cos^2x + 1 + 5\cos x = 0,
$$
which is a quadratic equation in $\cos x$.


问:
Pseudovectors in Geometric / Clifford Algebra
I am studying geometric algebra and I am confused about why pseudovectors are written as single vectors with an $i$ in front.  In other words, for basis vectors $\gamma_{\mu}$ where $\mu = 1,2,3,4$, there are 4 pseudovectors, written as $i \gamma_{\mu}$.

I know that in general, for a space of dimension $N$, pseudovectors are elements of the set of $(N-1)$ fold wedge products, namely $\wedge^{N-1}\mathcal{C}_N$ where $\mathcal{C}$ denotes a Clifford algebra from dimension $N$.

So from this, for a 4 dimensional space, the pseudovectors are trivectors, for example $(\gamma_1\wedge \gamma_2 \wedge \gamma_3)$.  Where does the $i$ come in?  I know that the pseudoscalar is written as an $N$ fold wedge product, so in this case would be a quad-vector, but I don't see how that can be worked in.

I can't seem to find any literature that explains this directly.  I did find something that says there is a one-to-one mapping between $(N-1)$ fold wedge products, and single vectors, but it didn't go much into exactly how this mapping works.

答:
I believe I have gathered enough information from these replies here, other sources, as well as from the replies here:

Hodge star operator

I believe this aggregation of information warrants me answering my own question here:

For a particular 'c-number' from a Clifford Algebra, it can be decomposed into a scalar part, a 1 form part, 2 form part, and so on until you get to an N-form part, which is the pseudoscalar $i$.

The Hodge Star operation is critical in this context it seems.  In Clifford Algebra, the Hodge Star is captured by multiplying by $i$.  This multiplication by $i$ maps a $k$ form into an $N-k$ form by the same kind of steps done here by foghorn (and the steps done in the link above). The important aspect here is that this mapping is one-to-one, meaning there is no ambiguity in transformations between these two regimes.

In 4 dimensions, a c-number has a scalar, vector, bivector, trivector, and pseudoscalar (quad-vector) parts.  A trivector is an 3-form ($k=3$).  Taking the Hodge dual of this (see foghorn above), yields a 1-form (vector) ($N-k = 4-3 = 1$).  Taking the hodge dual again would again yield back the same 3-form you started with, for example:

$$i e^0 =  e^0 e^0 e^1 e^2 e^3 = e^1 e^2 e^3$$

(assuming all basis vectors norm to +1).  Because this mapping is one-to-one, one could instead represent the 3-forms as a vector multiplied by $i$ as shown above.  This allows one to treat the object as if it were a vector, but with a sign change upon conjugation (which is why its called a pseudovector).


问:
Yoneda's lemma: group morphisms give Hopf-algebra morphisms
Let $k$ be a commutative ring. Let $\text{Alg}$ be the category of commutative $k$-algebras and $\text{CHopf}$ the category of commutative Hopf-algebras. Let us also write $[\text{Alg}, \text{Grp}]$ for the category of representable functors $\text{Alg} \to \text{Grp}$ (the morphisms are natural transformations between such functors).

We have an obvious functor
$$\text{CHopf}^{op} \to [\text{Alg}, \text{Grp}]: H \mapsto \text{Hom}_{\text{Alg}}(H,-)$$
since the sets $\text{Hom}_{\text{Alg}}(H,A)$ obtain a group structure via convolution for every algebra $A$. Concretely, the multiplication is defined by
$$f \star g := \mu(f \otimes g)\Delta$$ where $\mu$ is the multiplication on $A$ and $\Delta$ is the comultiplication on $H$. On the level of morphisms, the above functor is defined using the Yoneda lemma.

I'm trying to show that this functor is fully faithful. For this, I need to show that if $\lambda: \text{Hom}_{\text{Alg}}(H,-) \to \text{Hom}_{\text{Alg}}(H',-)$ is a natural transformation that consists of group morphisms and where $H,H'$ are commutative Hopf-algebras, then the morphism
$$\lambda_H(\text{id}_H) \in \text{Hom}_{\text{Alg}}(H',H)$$
is a Hopf-algebra morphism. In particular, I try to check that
$$(\lambda_H(\text{id}_H) \otimes \lambda_H(\text{id}_H))\Delta_{H'}= \Delta_H \lambda_H(\text{id}_H).$$

First of all, is this true? If it is true, how can I show it? I have tried to exploit the fact that $\lambda_H$ is a group morphism but could not conclude.

答:
The condition that $\lambda$ preserves the group structure is equivalent to saying that, for any $A \in \text{Alg}$, the diagram
$\require{AMScd}$
\begin{CD}
\text{Hom}_{\text{Alg}}(H \otimes_{k} H, A) @>{- \circ \Delta_{H}}>> \text{Hom}_{\text{Alg}}(H,A) \\
@V{- \circ (\lambda_{H}(\text{id}_{H}) \otimes \lambda_{H}(\text{id}_{H}))}VV @VV{- \circ \lambda_{H}(\text{id}_{H})}V\\
\text{Hom}_{\text{Alg}}(H' \otimes_{k} H' ,A) @>>{- \circ \Delta_{H'}}> \text{Hom}_{\text{Alg}}(H',A)
\end{CD} commutes ("$- \circ f$" means "precomposition by $f$"). Take $A := H \otimes_{k} H$ above, and consider the image of $\text{id}_{H \otimes_{k} H} \in \text{Hom}_{\text{Alg}}(H \otimes_{k} H,H \otimes_{k} H)$ in $\text{Hom}_{\text{Alg}}(H',H \otimes_{k} H)$; via the bottom-left corner, the image is $(\lambda_{H}(\text{id}_{H}) \otimes \lambda_{H}(\text{id}_{H})) \circ \Delta_{H'}$, and via the top-right corner, the image is $\Delta_{H} \circ \lambda_{H}(\text{id}_{H})$.