问:
A ring isomorphic to its finite polynomial rings but not to its infinite one.
I was messing with the ring $k[x_1,\dots,x_n,\dots]$ of polynomials in numerable many variables in order to solve an exercise of Atiyah, and the following question came to me and made me curious:

Is there a commutative unitary ring $A$ isomorphic to $A[X]$ but not isomorphic to $A[x_1,\dots,x_n,\dots]$ the ring of polynomials with coefficients in $A$ and numerable many variables?

答:
Let $\Gamma$ be the set of sequences $a_1 a_2 a_3 \cdots $ of nonnegative integers which are eventually constant. A typical element of $\Gamma$ looks like $3921726666666\cdots$. Note that $\Gamma$ is a commutative semigroup under addition. Let $A$ be the semigroup algebra $\mathbb{Z}[\Gamma]$. I claim that $A \cong A[t]$ but $A \not \cong A[t_1, t_2, t_3, \ldots]$.

The semigroup $\Gamma$ is isomorphic to $\mathbb{Z}_{\geq 0} \times \Gamma$, by the map $a_1 a_2 a_3 \cdots  \mapsto (a_1, a_2 a_3 \cdots)$. So $\mathbb{Z}[\Gamma] \cong \mathbb{Z}[\Gamma \times \mathbb{Z}_{\geq 0}]$ or, in other words, $A \cong A[t]$.

Let $z$ denote the sequence $11111\cdots$ and let $x_i$ denote the sequence $000\cdots01000\cdots$ with the lone $1$ in the $i$th position. Note that $\Gamma$ embeds in the free abelian group generated by $z$ and the $x_i$, so $A$ embeds in the Laurent polynomial ring $\mathbb{Z}[z,x_1^{\pm},x_2^{\pm},x_3^{\pm},\ldots]$. In particular, $A$ is an integral domain. In this notation, the isomorphism $A \to A[t_1, t_2, \ldots, t_k]$ sends $z \mapsto z t_1 t_2 \cdots t_k$; it sends $x_i \mapsto t_i$ for $i \leq k$ and $x_i \mapsto x_{i-k}$ for $i >k$.

Now, suppose that $\phi$ is a map $A \to A[t_1, t_2, \ldots ]$ with $\phi(z)$ nonzero. We show that $\phi(A) \subseteq A[t_1, t_2, \ldots, t_N]$ for some $N$. In particular, $\phi$ is not an isomorphism.

Choose $N$ large enough that $\phi(z) \in A[t_1, t_2, \ldots, t_N]$. Since $\phi(x_i)$ divides $\phi(z)$, and $A$ is a domain, we must also have $\phi(x_i) \in A[t_1, \ldots, t_N]$. For any $\gamma$ in $\Gamma$, write $\exp(\gamma)$ for the corresponding element of $A$. For any $\gamma \in \Gamma$, we have $\prod_{i=1}^M x_i^{b_i} \exp(\gamma) = \prod_{i=1}^M x_i^{c_i} \cdot z^d$ for some sufficiently large $M$ and some $d$. Therefore, $\phi(\exp(\gamma))$ can be written as a ratio of $\prod_{i=1}^M \phi(x_i)^{c_i} \cdot \phi(z)^d$ and $\prod_{i=1}^M \phi(x_i)^{b_i}$, both of which are in $A[t_1, \ldots, t_N]$ (using again that $A$ is a domain). So $\phi(\exp(\gamma)) \in A[t_1, \ldots, t_N]$. Since the $\exp(\gamma)$ are a $\mathbb{Z}$ basis for $A$, this shows that $\phi(A) \subseteq A[t_1, \ldots, t_N]$.


问:
What is the intersection of a non-existiant set?
Given $A$, Is the following true?
$$\not \exists B \implies A \cap B = \emptyset$$

答:
Given some set $A$, you can write

$$ \not\exists B, A\cap B = \emptyset$$

The above is a formula expressing that "there is no set $B$ such that its intersection with $A$ is empty". However, the formula is not valid (choose $B=\emptyset$).

For the formula that you write in your question, the missing parentheses could be placed as in $\not\exists B (\implies A\cap B=\emptyset)$ or as in $(\not\exists B)\implies A\cap B=\emptyset$. Either way, none of the two parenthesized formulas make sense.
In the first case, the formula is syntactically not correct. In the second case, the $B$ in $A\cap B$ is not necessarily the same as the $B$ in $\not\exists B$.
In both cases, the formulas make no sense.

ContentLicense: CC BY-SA 4.0


问:
Let $(R, +, \cdot)$ be a ring. Show that if $(R, \cdot)$ is a group, then the ring only contains one element namely zero.

Let $(R, +, \cdot)$ be a ring. Show that if $(R, \cdot)$ is a group, then the ring only contains one element namely zero.

This seems to be due to the fact that the real numbers don't define a group under multiplication since the zero has no inverse? So $(R, \cdot) = \{0\} \implies (R, +, \cdot) = \{0\}$? Is my reasoning correct?

答:
Your reasoning is correct, although you should look at a general ring and not the ring of reals.

A ring is an abelian group with respect to addition. This means that the additive identity has the following property
$$r \cdot 0 = r \cdot (0 + 0) = r \cdot 0 + r \cdot 0 \implies r \cdot 0 = 0$$
Which is the property you mentioned of the reals. After that you correctly concluded that the ring needs to be zero for the group axioms to be fulfilled.


问:
How we can extract a vector space structure from a category with one object?
How can we associate a vector space structure to a category with one object ?
Is there a canonical way of doing this ?</p>

答:
You must be able to add vectors, but generally there is no way to "add" morphisms in a general category with one object. Also, what should the base field be?

Let $C$ be your category. What you could do, is to fix a field $k$, and consider the freely generated vector space $k[Hom(C)]$ (funny notation...). That is, the vector space with elements formal $k$-linear combinations of the morphisms in $C$.

ContentLicense: CC BY-SA 3.0


问:
Expected number of unpecked chicks - NYT article
In this article, the winner of the math competition answered this question correctly:

In a barn, 100 chicks sit peacefully in a circle. Suddenly, each chick randomly pecks the chick immediately to its left or right. What is the expected number of unpecked chicks?

The answer was given as 25. I am interested to know the correct method for solving this problem as well as a general way to find this out for N chicks. I was thinking of trying to solve by figuring out the number of occurrences of (something?) inside a binary string of length 100 (or N), but I don't know if that is the right way to approach it.

答:
I suspect the original answer to the competition question is not correct. I am not a mathematician and I apologize if I am wrong.
Putting aside "how does the chick feel about pecking left and right" because it's completely random as stated, there are in total 6 possible sets of outcome, 4 that were mentioned and 2 additional ones. The last 2 sets represent another "chance" or another possible set of outcomes. The problem here is that we have 100 chicks and not 4.

all chicks peck to the left. 0% of chicks are unpecked.

all chicks peck to the right. 0% of chicks are unpecked.

all chicks, divided in pairs, peck each other, 0% of chicks are unpecked.

chicks are divided in groups of 4, where the pair in the middle pecks each other, while chicks on the edge peck this pair in the middle. 50% of chicks are unpecked.

There are however other possible probability distributions. Two new patterns emerge from this.

chicks are divided into groups of 3, where a pair of chicks pecks each other and one chick from this pair is double pecked by a chick on the LEFT. This chick on the left is unpecked which makes a total of 33%. The last, 100th chick can peck randomly left or right but remains unpecked itself as in example 4). This gives 34 unpecked chicks.

chicks are divided into groups of 3, where a pair of chicks pecks each other and one chick from this pair is double pecked by a chick on the RIGHT. The last, 100th chick can peck randomly left or right but remains unpecked itself as in example 4). This gives 34 unpecked chicks.

A mixture of all those sets of outcomes are, of course, also possible. Now we can calculate the median of these 6 possible outcomes:
(0+0+0+50+34+34)/6=19.666 somewhat lower than the original solution.


问:
$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$
How can I go about computing
$$
\sum_{k\ =\ 1}^{n - 1}
\left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\
{\Large ?}
$$

I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it is the sum of the reciprocals. I've tried using
$$e^{ix}=\cos x+i\sin x,$$
but I didn't get anywhere with that.

Body:
Call $\omega = \exp(2\pi\text i/n)$. You have
$$
\sum_{k=1}^{n-1} \frac  1{1-\omega^k} =
\sum_{k=1}^{n-1} \frac  1{1-\omega^{n-k}}\\
\implies
\sum_{k=1}^{n-1} \frac  1{1-\omega^k}
=
\frac 12 \sum_{k=1}^{n-1} \frac  1{1-\omega^k} + \frac 1{1-\omega^{n-k}}\\
= \frac 12 \sum_{k=1}^{n-1} \frac  {2-\omega^k-\omega^{n-k}}{2-\omega^k-\omega^{n-k}} =\frac{n-1}2
$$


问:
Give an equation of the plane parallel to $3x-12y+4z=0$ and tangent to the surface $x^2+y^2+z^2=676$
Give an equation of the plane parallel to $3x-12y+4z=0$ and tangent to the surface $x^2+y^2+z^2=676$

What I tried:
the normal vector is:
$$
[3,-12, 4]
$$
From the equation:
$$
[2x_0, 2y_0, 2z_0]
$$
It seems, that the tangent point is:
$$
P\left(\frac{3}{2}, -6, 2\right)
$$
but it's not and I don't know why.

Regards.

答:
$P\left(\frac{3}{2}, -6, 2\right)$ is not on the sourface.

Note that the gradient is not necessarily equal to the normal, is just proportional. Solve

$$[2x_0, 2y_0, 2z_0]= \lambda [3,-12, 4]$$

and check for which values of $\lambda$ the solution is on your surface.


问:
Directional derivative of a non differentiable composition of functions

Hi, I'm having a lot of trouble understanding this problem. I don't get part iii) at all. Because h (if I'm not wrong) is h: R^2---> R and g : R^3---> R^2

I don't get how am I going to find the directional derivative in the direction of a vector that lives in R^3. Wouldn't I need the vector to be in R^2 ?? And the point, why is it given with three coordinates?

Sorry, but I'm stuck in this problem and I would really appreciate If you could help me solve it.

I would also like to know if : when I have a composition of functions that is not differentiable everywhere, then to find the directional derivative(at a point where the composition is indeed differentiable) , the only way I can do it is with the limit definition, or could I use the dot product between the gradient and the directional vector?

And to do it through the first way, do I need to previously substitute the inner function output as the input of the outer function and then take the limit, or can I do some kind of chain rule or something? For example,

g(t) = (t,1/t)

f(x,y) = x^2 + y

I want h(x,y) = f(g(t))
H is not differentiable, so how can I calculate the directional derivative of h at the point (1,2) in the direction (1,1)?

Thanks a lot  !

答:
Hi, I'm having a lot of trouble understanding this problem. I don't get part iii) at all. Because h (if I'm not wrong) is h: R^2---> R and g : R^3---> R^2

The function $h = f\circ g $ is a map from $\mathbb R \times \mathbb R \times \mathbb R_0$ to $\mathbb R$ (since $w \neq 0$). First apply $g$ to go from $\mathbb R \times \mathbb R \times \mathbb R_0$ to $\mathbb R^2$ and then apply $f$ to go from $\mathbb R^2$ to $\mathbb R$.  You may compute it's expression as follows

$$ f\circ g(u,v,w) = f(\sin(uv),\frac 1 w \cos (uv)) = \frac{(\sin (uv) - \frac 1 w \cos uv)^2}{\sin^2 uv + \frac{1}{w^2}\cos^2(uv)}$$

Notice that for any $(u,v,w)$ we have $\left(\sin(uv),\frac{1}{w} \cos(uv) \right)\neq (0,0).$

I would also like to know if : when I have a composition of functions that is not differentiable everywhere, then to find the directional derivative(at a point where the composition is indeed differentiable) , the only way I can do it is with the limit definition, or could I use the dot product between the gradient and the directional vector?

If a function $h$ is differentiable at a point $a$ then it is true that the directional derivative in the direction of $v$ is given by the dot product of $v$ and the gradient of $h$ at $a$. If $h$ is not differentiable at $a$ then you should use the limit definition instead.

For your last example with $g(t) = (t,1/t)$ and $f(x,y) = x^2+y$ you wrote

$$ h(x,y) = f \circ g(t)$$

The above makes no sense at all !!!!! What you wrote implies that $f \circ g (t)$ is function of $x,y$ and not of $t$. You should have

$$ f \circ g (t) = f(t,1/t) = t^2 + 1/t$$

Therefore $h = f \circ g$ is a function from $\mathbb R_0 \to \mathbb R$ so speaking of directional derivatives isn't really appropriate here.

I suggest you first try to understand how to compose functions before trying harder problems like these.


问:
Does taking restriction and left derived functors commute?
Let $R$ be a commutative Noetherian ring. Let Mod$(R)$ denote the category of $R$-modules, and mod$(R)$ denote the category of finitely generated $R$-modules; notice that both of these categories are abelian and has enough projective objects.

For a functor $F: \text{Mod}(R) \to \text{Mod}(R)$, if $F(M)\in$mod$(R)$ for every $M\in \text{mod}(R)$, then let $F|_\text{mod}$ denote the restricted functor $F|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$ where for every $M,N \in \text{mod}(R)$,
$F|_\text{mod}(M):=F(M)$ and $F|_\text{mod}(f):= F(f) $ for every $f\in \text{Hom}(M,N)$.

Now let $F: \text{Mod}(R) \to \text{Mod}(R)$ be an $R$-linear right exact functor satisfying $F(\text{mod}(R))\subseteq \text{mod}(R) $. So $F|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$ is also right exact and $R$-linear.

We can consider the left derived functors (https://en.m.wikipedia.org/wiki/Derived_functor) $L_i F: \text{Mod}(R) \to \text{Mod}(R)$ and $L_i (F|_\text{mod}): \text{mod}(R) \to \text{mod}(R)$ . Since

$F(\text{mod}(R))\subseteq \text{mod}(R) $ holds, so

$L_iF(\text{mod}(R))\subseteq \text{mod}(R) $ holds for every
$i\ge 0$, so for every $i\ge 0$, I can consider the restriction
$(L_iF)|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$.

My question is: Do we have an isomorphism of functors $L_i (F|_\text{mod}) \cong (L_iF)|_\text{mod}$ for all $i\ge 0$ ?

Since different choices of projective resolutions ultimately give Isomorphic derived functors, so I can clearly see that $L_i (F|_\text{mod}) (M) \cong (L_iF)|_\text{mod}(M)$ for all $M \in \text{mod}(R)$ and for all $i\ge 0$.  However, I am not sure how to show if the corresponding functors are Isomorphic.

Please help.

答:
I think the easiest way to understand what is going on is to think about the following more general situation.

Proposition.
Let $\mathcal{C}, \mathcal{C}', \mathcal{D}, \mathcal{D}'$ be abelian categories, let $F : \mathcal{C} \to \mathcal{D}$ and $F' : \mathcal{C}' \to \mathcal{D}'$ be right exact functors, and let $I : \mathcal{C}' \to \mathcal{C}$ and $J : \mathcal{D}' \to \mathcal{D}$ be exact functors.
Assuming $\mathcal{C}'$ and $\mathcal{C}$ have enough projectives, if $F I \cong J F'$ (as functors) and $I : \mathcal{C}' \to \mathcal{C}$ preserves projectives, then $(\mathrm{L}_* F) I \cong J (\mathrm{L}_* F')$.

The proof is easy.
The composite functors $F I$ and $J F'$ are right exact and isomorphic by hypothesis, so their left derived functors must also be isomorphic, i.e. $\mathrm{L}_* (F I) \cong \mathrm{L}_* (J F')$.
Since $J$ is exact, we have $\mathrm{L}_* (J F') \cong J (\mathrm{L}_* F')$. (Consider how each one is computed!)
On the other hand, since $I$ is exact and preserves projectives, it preserves projective resolutions in particular, so $\mathrm{L}_* (F I) \cong (\mathrm{L}_* F) I$.
Composing the three isomorphisms, we obtain $(\mathrm{L}_* F) I \cong J (\mathrm{L}_* F')$, as claimed.

The hypotheses of the proposition are not optimal, but the situation you are considering fits.


问:
In logic, what's the difference between not being a Tautology and being a Contradiction?
Given this statement: The negation of a contradiction is a tautology.

Let T(x) mean that x is a tautology and C(x) mean that x is a contradiction.

My response is $\forall$x(C(x)$\rightarrow$ $\neg$ T(x)), however, in the textbook says the correct answer is  $\forall$x(C(x)$\rightarrow$ T( $\neg$x)).

I was wondering why is my answer not the correct case? And what is the difference between $\neg$ T(x) and T( $\neg$x)?

Thanks.

答:
A formula $X$ is a Contradiction when it is always FALSE [example: $p \land \lnot p$].

A formula $X$ is a Tautology when it is always TRUE [example: $p \lor \lnot p$].

If $X$ is a Contradiction, then $\lnot X$ is a Tautology, and vice-versa [$\lnot (p \land \lnot p) \equiv (\lnot p \lor p)$, by De Morgan].

But a formula that is not a Tautology is a formula that is sometime FALSE, like e.g. $p \lor q$, and this does not mean that it is always FALSE.

Conclusion: $¬ \text {Taut}(X)$ and $\text {Taut}( ¬X)$ are not the same.


问:
Distributional solution of PDE with Kronecker symbol
Let $f: \mathbb{R} \to \mathbb{R}^d$ be $C^1$ and $L^2$ (I'm not sure at all that the $C^1$ regularity is relevant).
I'm trying to find a solution $p : \mathbb{R}^d \to \mathbb{R}^d$ to the following differential system, in the distributional sense
$$ \partial_{x_i} p_k = \delta_{ik} f(x), $$
where $\delta_{ik}$ is the Kronecker symbol, $\delta_{ik} = 1$ if $i = k$, $0$ otherwise.

So basically what I want is a function whose divergence is $f$, but such that the other derivatives are all equal to $0$. I've tried numerous approaches, but the fact that $f$ itself depends on $x$ makes the problem highly non-linear. I only need an existence result, I am unsure whether a solution even exists. If a solution exists, I don't need the explicit form of it, although that would be super nice. Any reference or help would be much appreciated.

答:
Let's say $d=2$ for simplicity (higher dimensions work the same way). Distributional derivatives commute, so
$$
\partial_1 f=\partial_1 \partial_2 p_2=\partial_2 \partial_1 p_2=0
$$
and similarly $\partial_2 f=0$. Thus $f$ is constant. If you really want $f\in L^2$, this only leaves us with $f=0$, in which case any constant functions $p_1$, $p_2$ solve your system.

If we forget about the condition $f\in L^2$, then we have $f(x)=a$ for some constant $a$, and it is not hard to see that the only solutions are of the form $p_1(x)=a x_1+d_1$, $p_2(x)=a x_2+d_2$ for constants $d_1,d_2$.


问:
Does $\sum_{n=2}^{\infty}\frac{1}{(\ln n)^2}$ converge or diverge by comparison test?
$$\sum_{n=2}^{\infty}\frac{1}{(\ln n)^2}$$
I was trying to use $\frac{1}{n}\leq\frac{1}{\ln(n)}$ which concludes that $\frac{1}{\ln(n)}$ diverges.

But after this I know I can't power both sides of the inequality.

Any help?

答:
The simplest way to apply the comparison test is to compare:
$$
\frac1 {(\ln n)^2}\ge\frac1 {n\ln n},\text { for } n\ge 2.
$$
The latter series diverges by integral test.

Alternatively one can show repeatedly applying l'Hopital rule that
$$
\forall p>0:\quad \lim_{n\to\infty}\frac{(\ln n)^p}{n}=0,
$$
which means that there exists such $N_p$ that for all $n>N_p $: $(\ln n)^p<n $.


问:
How Do You Expand and Simplify?
How do you expand and simplify $(x + 2y)^2$ or $(7x + 5y)^2$. I don't understand, how would you multiply that by itself?

I mean I can simplify and expand two brackets; e.g. $(x + 3) (x + 5)$ but I'm not sure how to do one bracket with a squared.

答:
Recall that squaring something just means multiplying it by itself.
So $(x+2y)^2 = (x+2y)(x+2y)$</p>

Title: How Do You Expand and Simplify?


问:
Is the zero set of an analytic function closed nowhere dense?
Given a non-constant analytic function $f(x)$ on a domain $D \subseteq \mathbb{R}^n$. I want to prove that

$\mathcal{Z} = \{x \in \mathcal{D} |f(x) = 0 \}$

is closed nowhere dense.

I originally wanted to prove this by contradiction as follows using the identity theorem.

"Assume that $\mathcal{Z}$ is not a closed nowhere dense set, then $\mathcal{Z}$ must contain a limit point", which shows that $f(x)$ is constant zero by the identity theorem and thus contradicts that $f(x)$ is non-constant.

However, I am not sure if the following statement is true:

"Assume that $\mathcal{Z}$ is not a closed nowhere dense set, then $\mathcal{Z}$ must contain a limit point"

If it is true, could you provide one reference?

If not, is $\mathcal{Z}$ closed nowhere dense, and how can I prove it?

答:
Indeed, more is true. It is enough to assume that $\mathcal Z$ is not nowehere dense to show that it has a limit point.

To see it, recall that a set is nowhere dense, by definition, if and only if its closure has empty interior. Then if $\mathcal Z$ is not nowhere dense, its closure has nonempty interior. Any of these points is a limit point of $\mathcal Z$.

EDIT: As Conrad points out in reply to Shi James' comment, to complete the proof for the given $\mathcal Z$, which is a zero set of an analytic (hence continuous) function, we need to notice first that $\mathcal Z$ is necessarily closed.


问:
Showing $\int_{-1}^1\frac{m(2m-1)x^{2m-2}(1-x^{2m})+m^2x^{4m-2}}{(m^2x^{4m-2}+1-x^{2m})\sqrt{1-x^{2m}}}dx=\pi$, algebraically
Is there a nice algebraic way to solve the following geometrically motivated integral?

$$\int_{-1}^1\frac{m(2m-1)x^{2m-2}(1-x^{2m})+m^2x^{4m-2}}{(m^2x^{4m-2}+1-x^{2m})\sqrt{1-x^{2m}}}dx,$$ where $m$ is a positive integer.

In fact, this integral can be shown to be the integral of the curvature of the plane curve $x^{2m}+y^2=1, y\geq0$, which is the angle rotated by the tangent vector of the curve as it traverses along the curve. So this integral is $\pi$, but I would like to see some alternative, algebraic solutions.

Body:
@user10354138' solution:

Denote the integrand by $f(x)$.
We have
\begin{align}
&\frac{\mathrm{d}}{\mathrm{d} x}\arctan\frac{m x^{2m - 1}}{\sqrt{1 - x^{2m}}} \\
=\ & \frac{1}{1 + (\frac{m x^{2m - 1}}{\sqrt{1 - x^{2m}}})^2}
\left(\frac{m(2m - 1) x^{2m - 2}}{\sqrt{1 - x^{2m}}}
+ \frac{m^2x^{4m - 2}}{(1 - x^{2m})\sqrt{1 - x^{2m}}}
\right)\\
=\ & f(x).
\end{align}
Thus,
$$\int_{-1}^1 f(x) \mathrm{d} x =
\arctan\frac{m x^{2m - 1}}{\sqrt{1 - x^{2m}}}\Big\vert_{-1}^1 = \pi.$$


问:
Error numerical differentiation - how to interpret constant/variable
Im basically looking for help explaning the symbols in this problems. Mores specifically: what does $C^3$  and $d$ mean exactly. How do I interpret them?

Let $f ∈ C^3$ [a, b]. Let $d ∈ N$ and let $ξ_1, . . . , ξ_d ∈ [a, b]$. Show that there exists $ξ ∈ [a, b]$ with

$ f''' (ξ) = \frac{1}{d} \sum_{j=1}^{d} f'''(ξ_j) $

答:
$C^3[a,b]$ is the set of triply continuously differentiable functions defined on $[a,b]$ (meaning that $f'''$ is continuous). $d$ is the number of points chosen in $[a,b]$. The theorem is a consequence of the intermediate value theorem, applied to $f'''$.

Let $m$ and $M$ be the smallest and largest of the $f'''(\xi_k)$.

the average of the $f'''(\xi_k)$ is certainly in $[m,M]$.

by the IVT, there is always a $\xi$ such that $f'''(\xi)$ equals a given value in $[m,M]$.


问:
Calculate enclosing pitch angles of a box in a coordinate system given center position c, orientation q(quaternion) and dimensions l, w, h of the box
Illustration of the problem

Given a coordinate system  (which could sit at the center of a laser sensor) and a box with dimensions $l$, $w$, $h$ and position of center $\vec{c}$ ($c_x$, $c_y$, $c_z$) and its orientation as a quaternion $q$ ($q_x$, $q_y$, $q_z$, $q_w$) in that coordinate system I want to find the minimal enclosing pitch angles (from the perspective of the coordinate system) of a box $β_{max}$ and $β_{min}$ (that means the pitch angles of pitch planes that touch the highest projected point and the lowest projected point - from the perspective of the coordinate system in the laser sensor).

This question seems to be very tricky, because how do I know which is the "highest and lowest point" from the perspective of the coordinate system? It obviously depends on the orientation and and the dimensions of the box: it could be the edge of the box or just the corner, but I have no mathematical way of expressing it.

答:
I have found the solution myself. it's surprisingly simple:

We can simplify the problem tremendously due to the following observation of the property of the box:
We observe that regardless of the orientation of the box, the highest or lowest point of the box from the perspective of the coordinate system is either

a single vertex

an edge comprising of multiple vertices (Multiple vertices share the same max/min value)

That means, that in every case at least two of the eight vertices contain implicitly the $β_{max}$ and the $β_{min}$ values of the box.

We compute all eight vertices of the box in absolute coordinates and their pitch angles in the coordinates and set the minimum, maximum value of all the angles to $β_{max}$ and $β_{min}$.


问:
Under what conditions is a ratio between polynomial functions constant?
I am studying ruled surfaces in $\mathbb{R}^{3}$ and I am trying to compare the Gaussian curvature of two such surfaces constructed along the same curve. In order to do so, I end up considering an expression of the form
$$
H(t,u)=\frac{f(t,u)}{g(t,u)},
$$
where both $f(t,u)$ and $g(t,u)$ are nonzero polynomials of equal degree in $u$.

I would like to know whether there is a simple way to examine the dependence of $H$ on $u$. In other words, under what conditions on numerator and denominator is the ratio independent of $u$?

Taking the derivative is not an option because the expression becomes really messy.

答:
It's not clear to me what sort of answer is useful to you. If you have explicit real polynomials $f(t, u), g(t, u)$ whose ratio is independent of $u$, then you could factor them over $\mathbb{R}[t, u]$ (or a computable subfield...), and all the irreducible factors that actually involve $u$ must cancel.

Why? In this case we have $f(t,u)/g(t,u) = f(t,0)/g(t,0)$, so $f(t,u) g(t,0) = g(t,u) f(t,0)$. As stated this is an equality on the level of functions, but since $\mathbb{R}$ is infinite, it's also an equality on the level of formal polynomials. Now take an irreducible factor $p(t, u)$ of $f(t, u)$ that actually depends on $u$. By unique factorization, $p(t, u)$ divides either $g(t, u)$ or $f(t, 0)$. Since it can't divide $f(t, 0)$, it must divide $g(t, u)$.

Another way to say it is that there are polynomials $\alpha(t), \beta(t), h(t, u)$ such that $f(t, u) = \alpha(t) h(t, u)$ and $g(t, u) = \beta(t) h(t, u)$. Note that the obvious constraint that the $u$-degree of $f$ and $g$ is equal is clear here.


Title: How Do You Expand and Simplify?

Body:
How do you expand and simplify $(x + 2y)^2$ or $(7x + 5y)^2$. I don't understand, how would you multiply that by itself?

I mean I can simplify and expand two brackets; e.g. $(x + 3) (x + 5)$ but I'm not sure how to do one bracket with a squared.

Body: Try this: $(x+2y)^2=(x+2y)(x+2y)$


问:
Classifying the stationary point of $h(x,y,z) = 2(x−1)^2 + 3(y−1)^3 + 4(z−1)^4$
Given the function $h(x,y,z) = 2(x−1)^2 + 3(y−1)^3 + 4(z−1)^4$, I have found the only stationary point to be $(1,1,1)$. I then attempted to use the Hessian matrix to find out whether $(1,1,1)$ is a local minimum/maximum/saddle point, however this was inconclusive.

Is there another way I can go about classifying this point?

Any help would be very appreciated! Thanks.

Body:
We have $ h(1,1,1)=0$  and $h(1,y,1) =3(y-1)^3$

Hence

$$h(1,y,1) >0 =h(1,1,1)$$

for $y>1$ and

$$h(1,y,1) <0 =h(1,1,1)$$

for $y<1$ .

Conclusion ?


问:
Sigma algebra generated by the sets $[n,n+1] ,n \in \mathbb{Z}$ on the real line.
The sigma algebra would need to contain complements $[n,n+1]^c = (\infty, n) \cup (n+1, \infty) $, which can be made from countable union of open sets $(n,n+1), n\in \mathbb{Z}$. It also contains singletons $[n-1,n] \cap [n, n+1] = \{n\} $.

My question is: how do I make sure that's all there is to it (or is there more). I'm having trouble since the only way I know how to find out generated sigma algebra's is to 'guess' it. Is there a better way to do this?

答:
Consider sets of the form $\bigcup_{n \in I} (n,n+1) \cup E$ where $I$ and $E$ are subsets of $\mathbb  Z$. This class is clearly closed under countable unions. The complement of $\bigcup_{n \in I} (n,n+1) \cup E$ is $\bigcup_{n \notin I} (n,n+1) \cup (\mathbb Z \setminus E)$. It follows that this class is  a sigma algebra. We can now easily see that it is the smallest sigma algebra containing the intervals $[n, n+1], n
\in \mathbb Z$.

PS: The motivation for this comes from the fact that the sigma algebra generated by a  countable partition $(E_n)$ is the class of all possible unions of the sets $E_n$.


问:
knowing: $ \tan x=2-\sqrt{3}$ , obtain: $ \cos 2x$

Knowing: $$ \tan x=2-\sqrt{3} $$

Obtain: $$\cos2x$$

I tried converting $\tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:

$$\cos^2x-\sin^2x$$

But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?

Taken out of one of the entry tests to Maths in TAU.

Solution:

$$\cos2x=\cos^2x-\sin^2x=\frac{\cos^2x-\sin^2x}{\sin^2x+\cos^2x}:\frac{\cos^2x}{\cos^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$

$$ \frac{1-\tan^2x}{1+\tan^2x}=\frac{-6+4\sqrt{3}}{8-4\sqrt{3}}=\frac{-3+2\sqrt{3}}{4-2\sqrt{3}} $$

答:
Draw a right triangle with the opposite side being $2 - \sqrt3$, and the adjacent side being $1$ so that $\tan x = 2 - \sqrt{3}$. Then the hypotenuse is $\sqrt{1^2 + (2 - \sqrt{3})^2} = \sqrt{1 + 4 - 4\sqrt3 + 3} = \sqrt{8 - 4 \sqrt{3}}$, and so $\cos^2 x = \frac{1}{8 - 4 \sqrt{3}}$ as $\cos x$ is adjacent/hypotenuse.

Now $\cos 2x = 2 \cos^2 x - 1$, which is $\frac{1}{4 - 2 \sqrt3} - 1$ or $\frac{4 + 2 \sqrt3}{4} - \frac{4}{4} = \frac{\sqrt 3}{2}$.


问:
Are the following two inner products on differential forms equal?
There are two inner product on differential forms:

$\langle \alpha,\beta\rangle$ induced from Riemannian metric $g$ by defining on 1-forms as dual of vector fields then extending to all differential forms i.e. $\langle e_{i_1}\wedge \dots\wedge e_{i_k}, e_{j_1}\wedge \dots\wedge e_{j_k}\rangle =\det[\langle e_{i_s}, e_{j_s}\rangle]$.

On compact oriented Riemannian manifold $(M,g)$ $$(\alpha,\beta)=\int_M\alpha\wedge\star\beta.$$

Q1: Are these two inner products on differential forms equal?

Q2: If the answer to Q1 is "NO" then is it important to notice that two operators are adjoint of each other (or an operator is symmetric or self-adjoint)  w.r.t. which metric? e.g. $d$ and $\delta$ that are adjoint w.r.t. second inner product but I don't know it is w.r.t. other one.

答:
Disclaimer: the comment section is overgrowing so here is an answer that I hope will erase all your doubts.

First part:
As said in the comment section, $(1)$ is a inner product on the vector space $\Lambda^p_xM$ while $(2)$ is an inner product on the vector space $\Omega^p(M)$. The link between them is that $(2)$ is obtained by integrating $(1)$ over the whole manifold $M$.

An analog is this: consider the set $M=[0,1]$. Then for each $x\in [0,1], T_x[0,1] = \mathbb{R}$ and one can define an inner product on $T_x[0,1] = \mathbb{R}$ by $\langle a,b \rangle_x = a\times b$. This is $(1)$.

A vector field on $[0,1]$ is just a smooth function $f:[0,1] \to \mathbb{R}$, and $(2)$ is here
$$
\langle f,g\rangle = \int_0^1 f(x) g(x) \mathrm{d}x = \int_0^1 \langle f, g\rangle_x \mathrm{d}x.
$$

Second part:
If $V$ is a vector space and $\langle\cdot,\cdot\rangle$ is an inner product, one can create a Riemannian metric on $V$, thought as a manifold the following way. As a vector space, the tangent bundle of $V$ is trivial:
$$TV = V\times V$$
and one can define the Riemannian metric $g_v = \langle\cdot,\cdot\rangle$ for $v\in V$. It is a constant Riemannian metric because the canonical trivialization makes $g_v$ to be a function independant of $v \in V$.
Take $V = \Omega^p(M)$ and $\langle \alpha,\beta\rangle = \int_M \alpha \wedge \star \beta$. Now, forget that $V$ and $\langle\cdot,\cdot\rangle$ is defined thanks to a Riemannian manifold $(M,g)$ and just look at its structure: it is a vector space with an inner product. Hence, for this inner product $\|\alpha\|$ is a number.

If you really want to think of this construction as a Riemannian manifold, like in the first paragraph, then $\|\alpha\|$ will be a function:
$$
\|\alpha\| : \beta \in \Omega(M)^p \mapsto \|\alpha\|(\beta) = \|\alpha\|\in \mathbb{R}
$$
which is constant and does not take points of $M$ as entries.

Comment: if you really do not understand what I said, here is just a question for you: for $x \in M$, how would you define $\left(\int_M \alpha\wedge \star \beta\right)(x)$?

This is the exact same thing as this question: how would you define $\left(\int_0^1 t^2 \mathrm{d}t\right)\left(\frac{1}{2}\right)$?


问:
Complex eigenvalues and rotation
I'm a little unsure where to start this question off and proceed...

Suppose that $C\in M_2(\mathbb{R})$ has a non-real eigenvalue $\lambda=a+bi$, and suppose that the rotation matrix $$R=\frac{1}{\sqrt{a^2+b^2}} \begin{pmatrix} a & b\\ -b & a \end{pmatrix}$$ has an anticlockwise angle of rotation of $\pi/3$. If the top-most entry of $C^3$ is $-64$ find $a$ and $b$.

I'm given the following hint.

Express $C$ in terms of $\lambda$, $R$, and an invertible matrix $Q \in M_2(\mathbb{R})$, and then take cubes. You will not be able to actually find Q however.

I'm also given the following proposition to use.

Let $A$ be a real $2 \times 2$ matrix that has a non-real complex eigenvalue $\lambda=a+bi$. If w is an eigenvector for $\lambda$, then the real matrix $Q=(\text{Re}(\mathbf{w})\hspace 0.5cm \text{Im} (\mathbf{w}))$ is invertible and $$Q^{-1}AQ=\begin{pmatrix} a & b\\ -b & a \end{pmatrix}=s\begin{pmatrix} \frac{a}{s} & \frac{b}{s}\\ \frac{-b}{s} & \frac{a}{s} \end{pmatrix}$$ where $s=|\lambda|=\sqrt{a^2+b^2}.$

Can someone help me figure how how to solve this? I'm not sure what the eigenvalues and its relation to the rotation matrices have to really do with any of this or how to apply the hint or proposition...

答: Let $\rho=|\lambda|=\sqrt{a^2+b^2}$. Then $\lambda=\rho\left(\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)i\right)$. So, the eigenvalues of $C$ are $\rho\left(\cos\left(-\frac\pi3\right)\pm\sin\left(-\frac\pi3\right)i\right)$. Since they are distinct, $C$ is similar to$$\begin{bmatrix}\rho\left(\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)i\right)&0\\0&\rho\left(\cos\left(-\frac\pi3\right)-\sin\left(-\frac\pi3\right)i\right)\end{bmatrix},$$and therefore $C^3$ is similar to $\left[\begin{smallmatrix}-\rho^3&0\\0&-\rho^3\end{smallmatrix}\right]$. But this is a scalar matrix and therefore it is similar to itself and only to itself. So,$$C^3=\begin{bmatrix}-\rho^3&0\\0&-\rho^3\end{bmatrix},$$which implis that $\rho=4$. So $\{a,b\}=\left\{2+2\sqrt3\,i,2-2\sqrt3\,i\right\}$.

Title: Complex eigenvalues and rotation


答:
Do you know that $a^2=a\times a$?

Furthermore you should know this equality
$$(a+b)^2=a^2+2ab+b^2$$
hence in your case we have
$$(x+2y)^2=x^2+(2\times x\times(2y))+(2y)^2=x^2+4xy+4y^2$$

Title: How Do You Expand and Simplify?


问:
Showing that $\sum_{n=1}^{\infty}2^{-n}(\max\{|x_n|,1\})$ defines a norm
I'm showing that $$\|x\| = \sum_{n=1}^{\infty}2^{-n}(\max\{|x_n|,1\})$$ defines a norm when $x \in A$ and $A=F(\mathbb{N}, \mathbb{R})$ the vector space formed by $x=(x_1,\dots)$, but I have a slight problem with one of the properties.

For positivity I have that clearly $2^{-n} > 0$ and $\max\{|x_n|,1\} \geqslant 1$. So $\|x\| >0$.

Also for scalar multiplication if $p(x)=\|x\|$ $$p(kx) = \sum_{n=1}^{\infty}k2^{-n}(\max\{|x_n|,1\}) = k\sum_{n=1}^{\infty}2^{-n}(\max\{|x_n|,1\}) =|k|\|x\|$$

The problem comes with the third property that $\|x\|=0 \iff $$x=0$. I just showed that $\|x\| >0$ so this cannot be true?

Body:
This not a norm. $\|(\frac 1  2,0,0,..)\|=1$ and $\|4(\frac 1  2,0,0,0...)\|=2+\sum_{n \geq 2} \frac 1 {2^{n}}=\frac  5 2 \neq 4 \|(\frac 1  2,0,0,..)\|$.

You are also right about the fact that $\|x\|\geq \sum 2^{-n}=1$ for all $x$ including $x=(0,0,0,...)$.


问:
Functional equation $f(f(k+1)+3)=k$
Find all functionas $f:\mathbb{Z}\to\mathbb{Z}$ such that $\forall k\in\mathbb{Z}$, $$f(f(k+1)+3)=k$$.
Obvious solution is $f(k)=k-2$. I have found following:
$k+3=f(f(k+1)+3)+3\implies f(k+3)=f(f(f(k+1)+3)+3)=f(k+1)+2\implies f(k)=\begin{cases} k+a,\quad 2\nmid k \\ k+b,\quad 2\mid k\end{cases}$ for some $a, b\in\mathbb{Z}$. And I can't continue after this. Can anyone help me, please?

答:
What you have done is good. We can continue to search for constraints on $a$ and $b$.

First, $a$ must be even: Because, if $a$ was odd, then setting $k=0$ gives
$$0=f(f(\underbrace{1}_\text{odd})+3) = f(1+a+3) = f(\underbrace{4+a}_\text{odd}) = 4+2a \implies a=-2,$$
contradiction.

Now that we know that $a$ is even,
$$0=f(f(\underbrace{1}_\text{odd})+3) = f(1+a+3) = f(\underbrace{4+a}_\text{even}) = 4+a+b \implies b=-a-4.$$

In summary, $a$ is even and
$$
f(k)=\begin{cases} k+a,  & k \text{ odd} \\ k-a-4, & k \text{ even}\end{cases}
$$

(and $a=-2$ gives the "obvious" solution $f(k)=k-2$).

Title: Functional equation $f(f(k+1)+3)=k$


问:
Chebyshev Polynomials and the Hypergeometric Function
A problem related to Chebyshev Polynomials and the Hypergeometric Function involves transformation from one function to another. The task is to transform the Chebyshev polynomial into its correct scaled Hypergeometric Function. $$
T_{2n} (x) = ( - 1)^n {}_2F_1 \left( { - n,n;\tfrac{1}{2};x^2 } \right).
$$
Is there any way for me to make use of the relationship below?
$$
T_n (x) = {}_2F_1 \left( { - n,n;\tfrac{1}{2};\tfrac{{1 - x}}{2}} \right).
$$

答:
The Chebyshev polynomials are defined by
$$
T_n(\cos\theta)=\cos(n\theta).
$$
Hence
$$
T_{2n}(\cos(\theta))=\cos(2n\theta)=T_n(\cos(2\theta)).
$$
We now use the relation $\cos(2\theta)=2\cos(\theta)^2-1$ and write $\cos\theta=x$ to get
$$
T_{2n}(x)=T_n(2x^2-1).
$$
This should prove the claimed relation.

OwnerDisplayName: Lawrence Calixto


问:
Given $\int _0 ^{+ \infty} \frac{1}{(1+x)^a} =1, \ \ a =?$
Given $\int_0^{+ \infty} \frac{1}{(1+x)^a} =1$ what is the value of  $a$?.

I know that $\int_0^{+ \infty} \frac{1}{(1+x)^2} =1$.

Are there any other solutions?

Could you help me?

Body:
Possible solution: $$\int_0^{+ \infty} \frac{1}{(1+x)^a} =1$$
$$\therefore \int_0^{+ \infty} (1+x)^{-a} =1$$
$$ \therefore \frac {(1+x)^{1-a}}{1-a}|_0^{+ \infty} =1$$
$$ \therefore \frac {1-a}{(1+x)^{1-a}}|_0^{+ \infty} =1$$
$$\therefore \frac{1-a}{\infty}-\frac{1-a}{1}=1$$
$$\therefore -(1-a)=1$$
Your answer: $$\therefore a=2$$


问:
ODE solving mistake in homogeneous and particular solution method
When I was solving the ODE $x^{'}+2x=e^t$, $x(0)=5$, if using convolution method I got below derivation:

$x=x(0)e^{-2t}+\int_0^t {e^{-2(t-\tau)} e^{\tau} d\tau}=5e^{-2t} +\frac{1}{3}e^t-\frac{1}{3} \\$

Above seems to be correct since I.C. is satistifed. But when I practice the method of homogeneous with particular solution, I came up with wrong solution:

$x_h(t)=x(0)e^{-2t}=5e^{-2t}\\$

$x_p(t)=\frac{1}{3}e^t\\$

$x(t)=x_h(t)+x_p(t))=5e^{-2t}+\frac{1}{3}e^t\\$

The particular solution $x_p(t)$ was by assuming $x_p(t)=Le^{t}$, and plug it into DE so that:

$x_p^{'}(t)+2x_p(t)=e^t\\$

$Le^t+2Le^t=e^t\\$

$L=\frac{1}{3}\\$

$x_p(t)=\frac{1}{3}e^t$

Hopefully my mistake can be pointed out.

答:
There was right answer in the early reply. Here I pointed my mistake in two processes in the first place:

(1) need to factor $e^{-2t}$ in the integration

$x=x(0)e^{-2t}+\int_0^t {e^{-2(t-\tau)} e^{\tau} d\tau}=5e^{-2t}+e^{-2t}\int_0^te^{3\tau}d\tau=\frac{14}{3}e^{-2t}+\frac{1}{3}e^t $

(2) initial condition applies to final solution, instead of homogeneous solution

$x_h(t)=\alpha e^{-2t}$ where $\alpha$ is unknow yet

$x_p(t)=\frac{1}{3}e^t$

$x(t)=x_h(t)+x_p(t))=\alpha e^{-2t}+\frac{1}{3}e^t$, apply $x(0)=5$ to get
$\alpha=\frac{14}{3}$,

$\Rightarrow x(t)=\frac{14}{3}e^{-2t}+\frac{1}{3}e^t$


问:
Matrix Squareroot Using Single Value Decomposition (SVD)
I'm trying to find the square root of a square matrix(real or complex) that can have complex eigenvalues and eigenvalues. The approach that I'm using requires me to first calculate the Eigenvalues and the Eigenvectors. For this purpose, I'm using LAPACK's zgeev. I get the correct eigens using this routine.

$$A = \begin{bmatrix}1 & -2 \\ 3 & 4\end{bmatrix}$$

by applying LAPACK's zgeev i get eigens as

$$eigenValues = \begin{bmatrix}2.5+1.9365i \\ 2.5-1.9365i\end{bmatrix} eigenVectors=\begin{bmatrix}-0.3873+0.5i & -0.3873-0.5i\\ 0.7746+0i&0.7746+0i\end{bmatrix}$$

Now we know that the square root of the matrix can be calculated as
$$PD^{1/2}P^{-1}$$

Where P is the eigenVectors and D is eigenValues.

The $P^{-1}$ is not the inverse but the Pseudo Inverse that I'm trying to calculate. Now to calculate the Pseudo Inverse of the matrix P, I'm using the Single Value Decomposition. For this purpose, I'm using LAPACK's zgesvd routine that returns back 3 matrices commonly known as USV

$$U = \begin{bmatrix}-0.5532+0.1892i & 0.7676-0.2625i \\ 0.2625+0.7676i & 0.1892+0.5532i\end{bmatrix} S=\begin{bmatrix}1.2777 \\ 0.6063\end{bmatrix}
 \\ V = \begin{bmatrix}0.7071+0i & 0.7071+0i \\ 0.5590+0.433i & -0.5590-0.4330i\end{bmatrix}$$

Problem: The V matrix that comes from LAPACK contains exact values as of Matlab's SVD but the signs are different.
$$LAPACK's\_V = \begin{bmatrix}0.7071+0i & 0.7071+0i \\ 0.5590+0.433i & -0.5590-0.4330i\end{bmatrix} \\ Matlab's\_V = \begin{bmatrix}0.7071+0i & 0.7071+0i \\ 0.5590-0.433i & -0.5590+0.4330i\end{bmatrix}$$

I have tested this on python as well. I'm using Scipy's Linear Algebra module and the V that I get from Scipy's SVD also matches the V of LAPACK. Matlab's SVD is different however but the correct one as it gives me accurate results.

Now I have created the whole pipeline that calculates the Pseudo Inverse using SVD and then I can calculate the Matrix Squareroot using Python, Matlab, and Swift but due to the sign ambiguity in this only matrix V, my results become inconsistent in Python and Swift. But for Matlab, if I follow the exact steps, I get the perfect square root. I have done multiple tests and have come to the conclusion that the root cause of incorrect results is the sign problem in matrix V of Scipy or Swift(LAPACK).

Questions

Please guide me that why is V different? As far as my knowledge goes, Matlab also uses LAPACK at the backend.

What is the most straightforward way to fix this issue?

Is there any other approach to calculate matrix square root or at least the Pseudo Inverse?

I'm very close to solve this square root problem. I just need to fix V.

答: Taking the complex conjugate for the matrices U and V resolved the problem for me. I'm able to calculate the Squareroot of the matrix perfectly.</p>


问:
Projection matrix formula intuition
I completely understand how projection matrix formula: $$P = A(A^TA)^{-1}A^T$$is derived from: $$ A^T(b - A\hat{x} ) = 0$$ but what I don't understand is the "story proof" or the "intuition" of the first formula as a linear transformation to the column space of $A$, as it is supposed to be.

In fact I have three specific questions:

why would someone transform $b$ (as a vector to be projected onto $A$) into the "row space" of $A$ before it can be transformed into the column space of $A$?

what specific transformation does the matrix $A^TA$ encode?

why one should transform the vector to the "inverse of the tranformation $A^TA$" before it can be transformed to the column space of $A$?

答:
here is a beautiful one:

The strategy for finding the projection vector $A\hat{x}$ in the column space of $A$ is to find a vector $p$ that has the same dot products on the columns of $A$, as $b$.

So, first one should find the dot products of $b$ on each column of $A$ through the production of $A^Tb$

Then you want to find the linear combination of columns of $A$ that gives you the same dot products.

First one must find the coefficients of this linear combination. The columns of the matrix $A^TA$ are composed of dot product of each column on the other columns and also on itself. So, the matrix $A^TA$ translates the coefficients of the columns of $A$ to the dot products on each column of $A$. Thus, $(A^TA)^{-1}$ do the reverse. it takes the dot products on each vector and spits out the necessary coefficient of each column in the linear combination. thats exactly what we want.

Remember that by the production of $A^Tb$ we found the dot product of $b$ on each column of $A$. now we want to know which linear combination gives the same dot products. So, we multiply $A^Tb$ by $(A^TA)^{-1}$.

Now we have the coefficients of each column for the linear combination which has the same dot products on the columns of $A$ as $b$. we should simply multiply $(A^TA)^{-1}A^Tb$, which was derived in the previous paragraph, by $A$, because we have the coefficients, so multiply each coefficient to each corresponding column and add them up, thats exactly what $A(A^TA)^{-1}A^Tb$ does.

Title: Projection matrix formula intuition


问:
Biased graph generating processes
Given a random process (algorithm) that is supposed to generate with equal probability graphs from a given class $\Gamma$. Assume the process is not obviously biased, i.e. generating graphs unevenly (giving for example some graph properties a stronger weight).

How can it be checked or proved that the algorithm doesn't generate a finite amount of graphs with some property which in fact almost never occurs in $\Gamma$?

A related question is this one.

答:
The method(s) will very much depend on the problem at hand, let me share few articles to make the point clear:

Finding Hidden Cliques in Linear Time with High Probability (DEKEL
, GUREL-GUREVICH and PERES).

Here they proposed an algorithm to find a planted clique and showed that it works with a probability bounded below by a constant.

Limits of local algorithms over sparse random graphs by GAMARNIK AND SUDAN.

Here they proved that certain algorithms will not succeed to find the largest independent set in sparse random graphs with high probability.

Title: Biased graph generating processes


问:
Let $a,b$ be positive integers. Prove that if for all integers $s,t$ we have $sa+tb \neq \gcd(a,b)$ or $s+t=1$ then we have $a=b$
Question: Let $a,b$ be positive integers. Prove that if for all integers $s,t$ we have $sa+tb \neq \gcd(a,b)$ or $s+t=1$ then we have $a=b$.

Attempt:
Well I think for any integers $a,b$ by Bezout's theorem there must exist integers $s,t$ with $sa+tb=\gcd(a,b)$. So then $s+t=1$.

Then $s = 1-t$ and so

$(1-t)a + tb = \gcd(a,b) \implies a - t(a-b) = \gcd(a,b) $

then $t(a-b) <a$ since the gcd is positive.

From here I'm not really sure how to proceed.

Edit: Now that I actually understand this problem, the problem is saying is all of the Bezout coefficients $s,t$ have difference $1$ then $a=b$. To show this you can use the general form of the Bezout coefficients to choose two different representations and reason about $a-b$.

答:
Since you asked about the "thought process" in Igor's answer, I will show how we can view it simply as a special case of the well-known fact that lines are uniquely determined by two distinct points on the line. In particular this means that any two lines through two distinct points are identical, so have equal slopes.

By hypothesis all $(\ge 2)$ integer points $\,(x,y)\,$ on $\, a y = -bx + d\,$ also lie on the line $\,y  = -x + 1\,$ so these lines are identical, so they have equal  slope, i.e. $\,-b/a = -1,\,$ so $\,b = a,\,$ QED.

The proof in Igor's answer amounts to equating the slopes $\,m,m'\,$ of each line  computed from their two common points $(x,y)$ and $(x',y') = (x+a,y-b),\,$ yielding

$$ m = \dfrac{x'-x}{y'-y} = m'$$

Remark $ $ We don't need to known the values of the two common  integer points $\,(x,y),\,(x',y')\,$ on the lines - only that at least two such points exist. But general linear theory tells us infinitely many integer solutions exist for $\,a y +b x = d,\,$ viz. the general solution is the sum of any particular solution plus the general sum of the associated homogeneous equation $\,ay + bx=0.\,$ A particular solution exists by Bezout's identity for the gcd, and the homogeneous equation has an obvious solution $(x,y) = (a,-b),\,$ so adding it to any particular solution yields a second solution.

If you know linear algebra you may find it instructive to rephrase it in that language, e.g. using the fact that the determinant $\,m'-m\,$ of the system $\, y = m x + b,\ y = m' x + b'\,$ must be zero if the solution is not unique.


问:
the intersection of a real form and a parabolic subalgebra
Let $\mathfrak g$ be a complex  semisimple Lie algebra, $\mathfrak g_0$
a  real form and $\tau$ the conjugation of  $\mathfrak g$ with respect to $\mathfrak g_0$.
Real Lie algebra $\mathfrak g_{0}$ is called a real form of a complex Lie algebra $\mathfrak g$ if $\mathfrak g$ is the complexification of $\mathfrak g_{0}$.
Now we fix a  parabolic subalgebra $\mathfrak p$ of
$\mathfrak g$. I want to show that $\mathfrak g_{0} \cap \mathfrak p$ is a  real form of  $\mathfrak p \cap \tau\mathfrak p$. I'm completely stuck on how to start this question. Any help and comments will be appreciated.

答:
For any $A \in \mathfrak{p} \cap \tau(\mathfrak{p})$ one has $$2 A=(A+\tau(A))+(A-\tau(A)),$$ where the first summand is contained in $\mathfrak{p} \cap \mathfrak{g}_{0}$ and the second one gets into that subspace after multiplication by $i$. In fact, this argument is valid for  any
complex Lie subalgebra $\mathfrak{p} \subset \mathfrak{g}$.


问:
Solving the equation $-e^{-i2k\ell}=\frac{k-1}{k+1}$ for $k$
I am trying to solve the following equation for $k$:
\begin{equation*}
-e^{-i2k\ell}=\frac{k-1}{k+1}\, ,
\end{equation*}
where $\ell$ is a positive number (constant).

I would like to know how the solutions of $k$ look like. I expect it to be an infinite number of solutions. Does anyone have any ideas how to prove or show this?

答:
This would be an interesting extension of the generalized Lambert function in the complex domain.

Let $c=2i \ell$ and write the equation as
$$e^{-c k}=\frac{k-1}{k+1}$$

Now, have a look at equation $(4)$ in the linked paper.


问:
Implicit Function Theorem
a) Show that the equations
\begin{align*}
x^3-y^2+2u-v&=1,\\
x^2+y^3+u-v&=2
\end{align*}
can be solved simultaneously for functions $u=g(x,y),\ v=h(x,y)$ in a neighbourhood of $(1,1)$ such that $g(1,1) = h(1,1) = 1$.

b) Calculate $\frac{dg}{dx}$ and $\frac{dh}{dy}$ at the point $(1,1)$.

For part (a), I got the answer $-1$. For part (b), my $\frac{dg}{dx}$ is $2x - 3x^2$ and $\frac{dh}{dy}$ is $6y^2 + 2y$.

Please let me know if I am wrong and why; and if i'm correct, please let me know too! Much appreciated!

答:
For part (a) you should be finding the functions $g(x,y)$ and $h(x,y)$ by solving for $u$ and $v$ (do this either by the substitution or elimination methods you would have learnt). The answers should involve $x$ and $y$.

For part (b) you then differentiate the above functions with respect to the appropriate variables. Your current answer is close, but not quite right.

Title: Implicit Function Theorem


问:
Is it possible to plug hypercomplex numbers into the Riemann Zeta function?
I'm aware of the more detailed question:

How to raise a number to a quaternion power

However, from a more high-level perspective (read: probably less mathematical, hence my choice for raising this here.) I'm trying to understand this in terms of the Riemann Zeta function and [multivalued functions][2].

From what I understand of the answer given in the question mentioned above it does not make sense to raise a real number to a hypercomplex (quaternion or octonion) power because it produces multiple values.

Doesn't this also apply to complex number exponents?

Isn't there a way to plug hypercomplex numbers into the Riemann Zeta function?

答:
In tessarines the full answer is

$$\zeta ((a+b i)+j (c+d i))=\frac{1}{2} (\zeta (a+c+b i+d i)+\zeta (a-c-d i+b i))+\frac{1}{2} j (\zeta (a+c+b i+d i)-\zeta (a-c-d i+b i))$$

The bicomplex numbers are isomorphic to tessarins, so tessarine $j$ is equal to $-k$ in bicomplex numbers. This is a general formula that can be used for any function defined on complex numbers.


答:
Let $w_n = \frac{1}{(\frac12-z_n)^2}$. You need that
$$ \left|\sum_{n=1}^\infty w_n \right| = \sum_{n-1}^\infty |w_n|$$
First of all, for the right side of this condition to make sense, we need the series $\sum_{n-1}^\infty |w_n|$ to converge. It happens to be a sufficient condition for the series $\sum_{n=1}^\infty w_n $ to converge, so we don't need an additional condition for the convergence of the left sice of the condition.

Generaly we have
$$ \left|\sum_{n=1}^\infty w_n \right| \le \sum_{n-1}^\infty |w_n|$$
(the generalized triangle inequality), with the equality happening only if all $w_n$ are proportional to each other with positive proportionaility coefficientsts, i.e.
$$ \exists \omega\in\mathbb C\, \exists\lambda_n \in \mathbb R_+\, \forall n\in\mathbb N: w_n = \lambda_n\omega$$
Going back to $z_n$ this condition can be rewritten as
$$ \exists \omega\in\mathbb C\, \exists\lambda_n \in \mathbb R_+\, \forall n\in\mathbb N: \frac{1}{(\frac12- z_n)^2} = \lambda_n\omega$$
$$ \exists \omega\in\mathbb C\, \exists\lambda_n \in \mathbb R_+\, \forall n\in\mathbb N : z_n -\frac12 = \pm\left(\frac{1}{\lambda_n\omega}\right)^\frac12$$
$$ \exists \zeta \in\mathbb C\, \exists\mu_n \in \mathbb R\setminus\{0\} \forall n\in\mathbb N : z_n = \frac12 + \mu_n\zeta$$

Let us assume that ${\rm Re}\, \zeta \neq 0$. From the condition $0 < {\rm Re}\,z_n < 1$ we get then
$$ 0 <  \frac12 + \mu_n {\rm Re}\,\zeta < 1 $$
$$ -\frac{1}{|{\rm Re}\,\zeta|} < \mu_n < \frac{1}{|{\rm Re}\,\zeta|}$$
We have then
$$ |z_n-\frac12| = |\mu_n\zeta| < \frac{|\zeta|}{|{\rm Re}\,\zeta|}$$
$$ |w_n| = \frac{1}{|z_n-\frac12|^2} = \frac{1}{|\mu_n\zeta|^2}> \frac{|{\rm Re}\,\zeta|^2}{|\zeta|^2}$$
For such $|w_n|$ the series $\sum_{n=1}^\infty |w_n|$ would not be convergent. Therefore we need ${\rm Re}\, \zeta = 0$. Without a loss of generality we can choose $\zeta = i$, and we have the conditions
$$ z_n = \frac12 + \mu_n i \qquad \text{for some } \mu_n\in\mathbb R\setminus\{0\}$$
$$ \text{series } \sum_{n-1}^\infty |w_n| = \sum_{n-1}^\infty \frac{1}{\mu_n^2} \text{ is convergent}$$

Body:
If $z_n\in \mathbb{C} $ and $0<\Re(z_n)<1, \ \forall n\geq 1  $

Given,
$\left|\sum_{n=1}^{\infty} \frac{1}{\left(\frac{1}{2}-z_n\right)^2}\right|= \sum_{n=1}^{\infty} \frac{1}{\left|\frac{1}{2}-z_n\right|^2}$
then find the condition on $z_n$ so that the above equality holds.

Attempt

Write, $z_n=a_n+ib_n , 0<a_n<1$
$\Rightarrow|\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-z_n)^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{|\frac{1}{2}-z_n|^2}]^2$

$\Rightarrow |\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n-ib_n)^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$
$\Rightarrow  |\sum_{n=1}^{\infty} \frac{(\frac{1}{2}-a_n+ib_n)^2}{[(\frac{1}{2}-a_n)^2+b_n^2]^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$
$\Rightarrow   |\sum_{n=1}^{\infty} \frac{((\frac{1}{2}-a_n)^2-b_n^2)+i2b_n(\frac{1}{2}-a_n)}{[(\frac{1}{2}-a_n)^2+b_n^2]^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$

$\Rightarrow  [\sum_{n=1}^\infty\frac{((\frac{1}{2}-a_n)^2-b_n^2)}{ [ (\frac{1}{2}-a_n)^2+b_n^2]^2  }]^2 + [\sum_{n=1}^\infty\frac{2b_n(\frac{1}{2}-a_n)}{ [ (\frac{1}{2}-a_n)^2+b_n^2]^2  }]^2=  [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$


问:
How to make an injective function $ f $ such that $ f : \left[0,1\right]\rightarrow2^\mathbb{N} $ using binary notation?
$ 2^\mathbb{N} $ is a power set of $ \mathbb{N} $.

By using binary notation, we can take $ x\in\left[0,1\right]$ and $ x=0.a_1a_2a_3\cdots\left(a_n\in\left\{0,1\right\}\right) $

Then we can make a set $ A=f(x) $ as an element of $ 2^\mathbb{N} $ following next rules:

$n\in A $ when $ a_n=1 $ and $ n\notin A $ when $ a_n=0 $

For example,

$$\begin{align*}
0.00000\cdots&\longmapsto\emptyset\\[5pt]
0.10000\cdots&\longmapsto\left\{1\right\}\\[5pt]
0.01000\cdots&\longmapsto\left\{2\right\} \\[5pt]
0.01111\cdots&\longmapsto\left\{2,3,4,\cdots\right\} \\[5pt]
0.10111\cdots&\longmapsto\left\{1,3,4,5,\ \cdots\right\} \\[5pt]
\end{align*}$$

This seems nice. However, since $ 0.10000\cdots=0.01111\cdots $, there exist two $ f(x) $ for same $ x$. I can't fix this problem and I've been thinking of this all night.. How can I make the injective function $ f $ with solving this?

答:
First define an equivalence relation between the binary representations such that $x:=0.a_1a_2 \cdots \sim y:=0.b_1b_2 \cdots \iff x=y$. Then using the function that you defined above we have an injection from $[0,1]/ \sim \, \mapsto 2^{\mathbb{N}}$ where we choose the representative of each equivalence class as the one which has infinitely many 0s (each class will have one expansion ending in all 0s, and one expansion ending in all 1s; choose the first one).

It's important to note that there are only a countable number of binary expansions which end in infinite 1s, and each equivalence class contains at most two elements and so the cardinality of our quotient set is $\aleph$


问:
Continuity of $f(t)$ from $\frac{df}{dt}+kf(t)=\delta (t)$
Consider the differential equation
$$\frac{df}{dt}+kf=\delta (t)$$
where $k$ is constant. I'm asked to give statement about the continuity of $f(t)$ and  $f'(t)$ at $t=0$.

I tried solving the equation using Laplace transform seeing the $\delta $ function, it seems much convenient.
$$\mathcal{L}\left(\frac{df}{dt}+kf\right)=\mathcal{L}(\delta(t))$$
$$s\mathcal{L}(f)-f(0^+)+k\mathcal{L}(f)=1$$
$$\mathcal{L}(f)=\frac{1+f(0^+)}{s+k}$$
$$f(t)=(1+f(0^+))\mathcal{L^{-1}}\left(\frac{1}{s+k}\right)=(1+f(0^+))e^{-kt}$$
But looking on the function, I don't How can I say about it anything. For example
$$\lim_{t\rightarrow 0^+}f(t)=f(0^+)=(1+f(0^+))e^0\Rightarrow 1=0$$
which is complete non-sense. I don't understand what I'm missing. Can any help me figure it out?

答:
On $\Bbb R\setminus\{0\}$, we have to deal with $$ f'(t)+kf(t)=0,$$
i.e., $$ f(t)=Ce^{-kt}$$
where $C$ is locally constant, i.e., ptoentially different constants $C_+$ and $C_-$ for $(-\infty)$ and $(0,\infty)$. Now $f$ is continuous at $0$ only when $C_+=C_-$ (and $=f(0)$). You want to check what happens when $C_+\ne C_-$ and $f'(0)$ only exists in the distributional sense.


问:
Let (X, Y ) have bivariate normal density $f(x, y) = \frac{\sqrt{ab−c^2}}{2π}e^{− \frac{1}{2}(ax^2+by^2+2cxy)}$, find marginal distribution of Y.
Let (X, Y ) have bivariate normal density $f(x, y) = \frac{\sqrt{ab−c^2}}{2π}e^{− \frac{1}{2}(ax^2+by^2+2cxy)}$, how can i calculate marginal distribution of Y from the joint pdf. I know we have to integrate joint pdf $wrt$ x, but i am not able to do it.

答:
Hint:
\begin{align}
&\exp(-ax^2 + by^2+2cyx) = \exp(\frac{c^2y^2}{a})\times \exp(-(\sqrt ax^2 + \frac{2cy\sqrt{a}x}{\sqrt{a}} + \frac{c^2y^2}{a}))\\
&=\exp(\frac{c^2y^2}{a})\times \exp\bigg(-\bigg(\sqrt{a}x+\frac{cy}{\sqrt{a}}\bigg)^2\bigg)
\end{align}


问:
Chain rule of an abstract function
I am reading my notes of the course on calculus in several variables, and appear the following statement.

I we consider the function $U(f(t),t)$, then
$$
\frac{\partial U(f(t),t)}{\partial t} = \frac{\partial U(f(t),t)}{\partial t} + U(f(t),t) \frac{\partial f(t)}{\partial t}.
$$

Is it correct ?

Every time that I tried to manipulate a function in an abstract form, finally I can no obtain the partial derivates. Somebody can help me, please.

答:
You're confusing partial derivatives and derivatives.

Here's how you can avoid these mistakes.

In this exercise you are asked to calculate the derivative of  a composition of functions namely the function $U$ and the function $\gamma : t \rightarrow (f(t),t).$ To compute the derivative you will therefore use the chain rule

$$ \frac{d(U\circ\gamma)}{dt} = \frac{d U}{d\gamma} \frac{d\gamma }{dt}.$$

While doing this the partial derivatives of $U$ will appear. To avoid ambiguity we need to name the variables which $U$ uses as an input.

Let $U : \mathbb R^2 \rightarrow \mathbb R: (x,y) \rightarrow U(x,y).$ This enables us to write down the partial derivatives of U as $\frac{\partial U}{\partial x}$ and $\frac{\partial U}{\partial y}$.

The derivative of $U \circ \gamma$ is therefore

$$ \frac{d}{dt} \big( U(f(t),t) \big) = \frac{\partial U}{\partial x}(f(t),t) \cdot \frac{df}{dt} + \frac{\partial U}{\partial y} (f(t),t) \cdot \frac{dt}{dt}$$

Here the expression $\frac{\partial U}{\partial y} (f(t),t)$ means "the partial derivative of $U$ with respect to $y$ evaluated at the point $(f(t),t)$". You could rewrite this as

$$ \frac{\partial U(x,y)}{\partial y}_{\Big\vert_{\Large{(f(t),t)}}}$$

Title: Chain rule of an abstract function


Title: if $x +2^{1/3}+2^{2/3}=0$ then find the value of $x^3+6$?

Body: $$x=-2^{1/3}-2^{2/3} \implies x^3=-2-4+3.2(x)=-6+6x \implies x^3+6=6x$$


问:
How we can extract a vector space structure from a category with one object?
How can we associate a vector space structure to a category with one object ?
Is there a canonical way of doing this ?</p>

答:
Let's denote $*$ the (unique) object of your category. The (unique) set of maps of this category is the set of endomorphisms $\mathrm{Hom}(*,*)$. By definition of category, this is set is, well, a set. But there are circumstances where it can have more structure.

For instance, if your category is an abelian one, then $\mathrm{Hom}(*,*)$ is an abelian group. This is the case, e.g., for the category of abelian groups, where the set of morphisms between two abelian groups is also an abelian group.

When the sets of morphisms have the structure of $\mathbf{k}$-vector spaces, the category is called $\mathbf{k}$-linear ($\mathbf{k}$ a field). For instance, in the category of $\mathbf{k}$-vector spaces, hom-sets are also $\mathbf{k}$-vector spaces.

Hence, a $\mathbf{k}$-linear category with just one object is essentially the same as a $\mathbf{k}$-vector space. Namely, $\mathrm{Hom}(*,*)$.


问:
Given n points in the plane, no 3 collinear, show that there is a circle through 3 of the points such that none of the points lies inside the circle.
This is a question from the IMO shortlist which I found here.

Given $n > 3$ points in the plane, no 3 collinear, show that there is a circle through 3 of the points such that none of the points lies inside the circle.

So I found this very similar question on stackexchange but it's the exact opposite. That question asks for a circle which contains all the points.

So what I have tried till now is to start with 3 points and draw a circle through them. After that I add one more point inside the circle. I make a new circle with this new point and the two closest points.

If a point lies inside the new circle again I repeat the process until we reach n points and the desired circle.

But the problem is: Sometimes if one of the previous points comes inside the new circle then if we use that point to make a new circle and another point comes inside that circle again then points may keep coming inside the circle infinitely.

So I think I can omit this strategy. So I'm gonna start from three points, make a circle and keep adding points like the previous strategy. And after adding the new point inside the circle I need to make a new circle passing through it. But I need to choose two other points to make the circle with.

So I'm looking for a strategy to decide which other 2 points to choose.

If I can choose the right two points to make the new circle with in each step of this process of adding points, then eventually I will get the desired circle. But this strategy is what I'm looking for.

答:
Consider the closest two points in the set, $M$ and $N$. Obviously if we draw the circle that has those two points on a diameter, there will be no other points inside it (and no other points on its circumference wither, so this is not the circle we are looking for).

Now imagine shifting the centre of the circle away from the midpoint of $MN$, out along the bisector of $MN$, increasing in radius to keep $M$ and $N$ on the circumference. Assuming there are some points on this side of $MN$ (otherwise we move the circle the other way), we will eventually make the circle big enough to touch another point. This enlarged circle then fulfills the condition.


问:
Can you prove Unions and intersections are treated unsymmetrically in the definition of rings?
In the book of measure theory by Paul R.Halmos he said " unions and intersections are treated unsymmetrically in the definition of rings, while for instance, it is true that a ring is closed under the formation of intersections it is not true that a class of sets closed under the formation of intersections and differences is necessarily closed under the formation of unions" .
I was wondering how would you prove that?</p>

答:
You prove that by giving an example of a set of subsets that is closed under intersections and differences but not closed under unions.

E.g. on any set $X$ with at least two points, the family $\{\{x\}: x \in X\} \cup \{\emptyset\}$ is such.

so being closed under union and differences (a ring) implies being closed under intersections, but being closed under intersections and differences does not imply been closed under unions. So we cannot interchange the rôles of union and intersection, which is the point Halmos is making.
He tries to give a "minimal" set of conditions, because that makes it easier (less work) to check that a family is indeed a ring. So here he goes on to note that he also could have asked for closedness under intersections, proper differences and disjoint unions instead. So there are conditions that we can add to closed under intersections to get a ring, but just closed under differences won't do.


问:
How to calculate a variable x that produces non-linear results?
Sorry I don't have a mathematical background so It is hard for me to put the right words on it.

Let me explain what I am trying to do with an example.

Essentially I want to solve x so that it satisfies both situation:

$100 \times x = 0$

$50 \times x = 3$

I would like to know how to solve this type of equation as well as to know the name of the mathematical concepts involved here.

答:
To find value of a single unknown variable only one equation is adequate.

An extra equation is termed redundant.

You have given two.

The first  equation gives as a solution $x = 0 $

The second equation gives as a solution $x = 0.6$

Both cannot be true at the same time. In other words the
both cannot be simultaneously true.


问:
how to calculate the curvature
The Question is: Let $\gamma(t)$ be a regular curve lies on a sphere $S^2$ with center $(0, 0, 0)$ and radius $r$. Show that the curvature of $\gamma$ is non-zero, i.e., $κ \ne 0$.

My question:

We define $k$ as $k=\frac{(||\gamma'' \times \gamma'||)}{|| \gamma' || ^3} $

The equation of the sphere is: $x^2 + y^2 + z^2 = r$

So, we have $\gamma = (\sqrt{r} \cos u \sin v,\sqrt{r} \sin u \sin v, \sqrt{r} \cos v$)

How can I calculate $\gamma'$ and $\gamma''$ ? because $\gamma$ depends on two variables $u,v$

Thank you.

答:
If you don't know much about Riemannian stuff, there is an easy and elementary proof of this fact.

Since $\gamma$ lies on the sphere of radius $r$, you get $\|\gamma(t)\|^2=r$. Taking the derivative of that expression, it follows that $$(1): \qquad \forall t, \langle \gamma(t),\gamma'(t)\rangle =0.$$

Finally, computing the derivative of the last expression will give you : $$(2):\qquad \langle \gamma(t),\gamma''(t)\rangle + \|\gamma'(t)\|^2=0.$$

Now, you want to show that the curvature does not vanish that is : for every $t$, $\|\gamma'(t)\times \gamma''(t)\|\neq 0$ or equivalently that the family  $(\gamma'(t),\gamma''(t))$ is linearly independent at any time $t$.

Assume that there exists $t_0$ such that $\gamma''(t_0)=\lambda \gamma'(t_0)$ with $\lambda\in \mathbb R$. Then looking at $(1)$ and $(2)$ together you get a contradiction because $\gamma'$ is nonvanishing. So $\kappa(t)\neq 0$ for any $t$.

Title: how to calculate the curvature


问:
Solve $(xp-y)^{2}(x^2+y^2)^{-3/2}=a(1+p^2)$
Question is to solve this differential equation:-
$$(xp-y)^{2}(x^2+y^2)^{-3/2}=a(1+p^2)$$where $$p = \frac{dy}{dx}$$.

Question is given in the clairaut's form in my notebook.
But seems some proper substitution would do the trick .
Hints/leads are appreciated.

答:
Let's try to at least get a graphical picture. Transform into polar coordinates $x+iy=e^{iφ}$, and use differentials in a very loose manner, $dx+i\,dy=(dr+ir\,dφ)e^{iφ}$, so that
\begin{align}
(x\,dy-y\,dx)^2r^{-3}&=a(dx^2+dy^2)\tag1
\\
r\,dφ^2&=a(dr^2+r^2dφ^2)\tag2
\\
\frac{r(φ)}{a}-r(φ)^2&=r'(φ)^2 \iff r'(φ)^2+\left(r(φ)-\frac1{2a}\right)^2=\frac1{4a^2} \tag3
\end{align}
Now one can either go the square root way, or go one derivative deeper
$$
\frac{r'(φ)}{a}-2r(φ)r'(φ)^2=2r'(φ)r''(φ)\tag4
$$
so either $r'=0$ via (3) giving the constant solution $r=0$ or the circle $r=\frac1a$; or in the other factor
$$
\frac1{2a}-r=r''.\tag5
$$
This is a harmonic oscillator with solutions
$$
r(φ)=\frac1{2a}+c_1\cosφ+c_2\sinφ\tag6
$$
In view of the second form of (3), this requires $c_1^2+c_2^2=\frac1{4a^2}$ or
$$
r(φ)=\frac1{2a}\left(1+\cos(φ+\psi)\right)\tag7
$$
for some constant phase shift angle $\psi$.


答:
Define
$$ f(x)=\{\,n\in\Bbb N_0\mid \lfloor2^nx\rfloor\text{ is odd}\,\} $$
which has no ambiguity. Note that I used $\Bbb N_0=\{0,1,2,3,\ldots\}$ because I want $f(1)\ne f(0)$. Alternatively, you can work with $\lfloor 2^{n+1}x\rfloor$ or define $f(1)$ instead specifically as a set not reachable otherwise (e.g., $f(1):=\Bbb N$).
At any rate, $f$ is readily shown to be injective - but is not surjective because it leaves out all co-finite subsets (except one picked for $f(1)$).

Title: How to make an injective function $ f $ such that $ f : \left[0,1\right]\rightarrow2^\mathbb{N} $ using binary notation?

Body:
$ 2^\mathbb{N} $ is a power set of $ \mathbb{N} $.

By using binary notation, we can take $ x\in\left[0,1\right]$ and $ x=0.a_1a_2a_3\cdots\left(a_n\in\left\{0,1\right\}\right) $

Then we can make a set $ A=f(x) $ as an element of $ 2^\mathbb{N} $ following next rules:

$n\in A $ when $ a_n=1 $ and $ n\notin A $ when $ a_n=0 $

For example,

$$\begin{align*}
0.00000\cdots&\longmapsto\emptyset\\[5pt]
0.10000\cdots&\longmapsto\left\{1\right\}\\[5pt]
0.01000\cdots&\longmapsto\left\{2\right\} \\[5pt]
0.01111\cdots&\longmapsto\left\{2,3,4,\cdots\right\} \\[5pt]
0.10111\cdots&\longmapsto\left\{1,3,4,5,\ \cdots\right\} \\[5pt]
\end{align*}$$

This seems nice. However, since $ 0.10000\cdots=0.01111\cdots $, there exist two $ f(x) $ for same $ x$. I can't fix this problem and I've been thinking of this all night.. How can I make the injective function $ f $ with solving this?


问:
Showing Stirling's approximation up to a constant
I am trying to derive a 'rough' version of Stirling's approximation for $N!$:
$$
N!\approx c\sqrt{N}\Big(\frac{N}{e}\Big)^N.\tag{1}
$$
I'm aware that Stirling's approximation has $c=\sqrt{2\pi}$, but I'm just trying to obtain (1) at the moment.

What I've done: I have shown already that
$$
\int_1^N\frac{\{x\}}{x}dx=\sum_{n=1}^N\ln n-N\ln N+N-1\tag2
$$
in my question here. By $\{x\}$ I refer to the Fractional Part of $x$, where $\{x\}=x-\lfloor x\rfloor$.

I noticed that by applying $\exp$ to the summation in (2), I get $N!$, so I wonder how I can simplify all of this to show (1).

A hint that I am given is to start by showing that
$$
\int_1^N\frac{\{x\}}{x}dx=\frac{\ln N}2+\mathcal O(1/N)+C,\tag{3}
$$
for some constant $C$. The equality in (3) above looks like it can be simplified somehow to (1), I think, because for example we have
$$
\exp((\ln N)/2)=\exp(\ln(N^{1/2}))=\sqrt{N},
$$
which gets me the $\sqrt N$ factor in (1). But beyond that I am a little stuck on how to piece everything together. Thank you.

答:
Substitute the RHS of (3) into the LHS of (2):
$$
\frac{{\log N}}{2} + C + \mathcal{O}\!\left( {\frac{1}{N}} \right) = \underbrace {\sum\limits_{n = 1}^N {\log n} }_{\log N!} - N\log N + N - 1.
$$
Taking the exponential of both sides gives
$$
\sqrt N e^C e^{\mathcal{O}\left( {\frac{1}{N}} \right)}  = N!N^{ - N} e^N e^{ - 1} .
$$
Re-ordering and calling $c=e^{C+1}$ gives
$$
N! = c\sqrt N \left( {\frac{N}{e}} \right)^N e^{\mathcal{O}\left( {\frac{1}{N}} \right)} .
$$
By the Taylor series of the exponential function
$$
e^{\mathcal{O}\left( {\frac{1}{N}} \right)}  = 1 + \mathcal{O}\!\left( {\frac{1}{N}} \right).
$$
Thus, finally,
$$
N! = c\sqrt N \left( {\frac{N}{e}} \right)^N \left( {1 +\mathcal{O}\!\left( {\frac{1}{N}} \right)} \right).
$$


问:
About a counterexample
If $(X,\mathcal S ) $ is a measurable space and $f:X \to [-\infty ,\infty] $ is a function such that $f^{-1}((a ,\infty)) \in \mathcal S $ for every $a \in R$ . Then $f \  $  is $\ $ $ \mathcal S $ - measurable . I am trying to disprove this statement by searching counterexample .I am trying to construct a characteristic function for counterexample .Can any one give me some counter example if this statement is false and if true then please give me proof .

答: Consider $\mathcal{B}=\{A \subseteq \overline{\Bbb R}=[-\infty,+\infty]\mid f^{-1}[A] \in \mathcal{S}\}$. It's easy to check that $\mathcal{B}$ is a $\sigma$-algebra on $\overline{\Bbb R}$ and it contains all of $\mathcal{B}':=\{(a+\infty)\mid a \in \Bbb R\}$. If $\mathcal{B}'$ generates $\text{Bor}(\overline{\Bbb R})$ then $f$ will be measurable by definition, as then $\text{Bor}(\overline{\Bbb R}) \subseteq \mathcal{B}$ from the given  $\mathcal{B'}\subseteq \mathcal{B}$ inclusion.

Title: About a counterexample


问:
Calculate the measure in degrees of the acute angle formed by $ NMA $?
Consider the $PAT$ triangle such that the angle $ \angle P = 36 $ degrees, $ \angle A = 56 $ degrees and $ PA = 10 $. Knowing that the points $ U $ and $ G $ belong, respectively, to the sides $ TP $ and $ TA $, so that $ PU = AG = 1 $. If $ M $ and $ N $ are the midpoints of the $ PA $ and $ UG $ segments, respectively. What is the measure in degree of the acute angle formed by $ NMA $?

Can anyone give me any tips?

答:
Hint :

Use complex numbers. Let affixes of $A,P$ be $z_A=-5$, $z_P=5$. So $z_M=0$.

$z_G$ would be $-5+e^{i\,56^{\circ}}$. Why? Similarly find affixes of $U,N$.

Argument of $z_N$ will help you compute $\angle NMA$. I got it as a nice integer.


问:
Finding minima and maxima of the map $f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}, x \in [0,1]$
Suppose that $c>0$ and define $f : [0,1] \to \mathbb R,$
$$
f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}
$$
I want to show that $x=1/2$ is a minimizer of this map and the maxima are attained at $x=0$ and $x=1$. The above map seems to be a Jacobi theta function of third kind. I'm not very familiar with special functions so any help or reference would be appreciated!

答:
A partial answer: Let $\phi$ and its Fourier transform $\hat{\phi}$ on ${\Bbb R}$ be sufficiently smooth and decay sufficiently fast at infinity for the following operations to be legal (this is the case e.g. for the Gaussian function). Here is a fast track to obtain the Poisson summation formula:

Define
$f(x) = \sum_{k\in {\Bbb Z}} \phi(x+k)$.
Then $f$ is $1$-periodic and $f(x) = \sum_{m\in {\Bbb Z}} \hat{f}_m  e^{2\pi i m x}$ with

$$ \hat{f}_m = \int_0^1 f(y)  \exp(-2\pi i m y) dy = \sum_k \int_0^1 \phi(y+k) \exp(-2\pi i m y) dy,  $$

and after a change of variable and using periodicity of the exponential:

$$ \hat{f}_m  = \int_{\Bbb R} \phi(t) e^{-2\pi i m t} dt = \hat{\phi}(m) $$

with $\hat{\phi}$ being the Fourier transform on the whole real line and the normalization factor $2\pi$ in the exponential. We have thus deduced the Poisson summation formula:
$$ \sum_{k\in {\Bbb Z}} \phi(x+k) = \sum_{m\in {\Bbb Z}} \hat{\phi}(m)  e^{2\pi i m x}$$
Suppose now that $\hat{\phi}(-m)=\hat{\phi}(m)>0$ for all $m\in {\Bbb Z}$. Then the RHS is real-valued and clearly maximal when $\exp(2\pi i m x)=+1$ for all $m$, which happens precisely when $x\in {\Bbb Z}$.

When $\phi(x)=\exp(-c x^2)$, $c>0$ you have $\hat{\phi}(m)= \sqrt{\pi/c} \exp(- \pi^2m^2/c)$ with the given normalization so the wanted result follows for the maximizer from the above. For the minimizer it is not so clear but plausible that it should happen when $x-\frac12 \in {\Bbb Z}$.


问:
Proof that stochastic process on infinite graph ends in finite step.
Infinite Graph

Let $G$ be an infinite graph that is constructed this way: start with two unconnected nodes $v_1$ and $u_1$. We call this "level 1".

Create two more unconnected nodes $v_2$ and $u_2$. Connect $v_1$ to both of them with directed edges pointing from $v_1$ to $v_2$, and from $v_1$ to $u_2$ respectively. Then, connect $u_1$ to both of them using directed edges in the same way. This is "level 2".

Repeat this process. At level $n$, nodes $v_n$ and $u_n$ are both connected to the each of the next two nodes $v_{n+1}$ and $u_{n+1}$. This results in an infinite connected graph.

Stochastic Process

We define a discrete time process heuristically on this graph. $v_1$ and $u_1$ are initially "infected". This infection only last for one time period. We start at $t = 0$.

At each time step, infected nodes have an independent probability $p$ of passing the infection to their neighbors (infected node must have a directed edge to the target for the target to be a neighbor). If this happens, the neighbor is infected for one time period.

E.g. $v_1$ is connected to $v_2$ and $u_2$. $v_1$ is infected initially. At time $t=0$, $v_1$ has a probability $p$ of infecting its neighbors $v_2$ and $u_2$. Suppose it successfully infects $v_2$, and $u_2$ is never infected by any of its predecessor. At $t=1$, $v_1$ and $u_1$ stops being infectious. And $v_2$ is infectious.

This process is repeated for each level. If a node is infected twice, it is still just "infected", there is no special meaning to a double infection.

Conjecture

This process ends with probability $1$ after a finite number of steps.

How do I prove this? My attempt: The probability that the infection on one level is passed onto the next level is $q = 1 - (1-p)^4 = $ probability that all $4$ attempts to infect fails. For the infection to reach the $k$th level, the probability is $q^k$ and this tends to zero as $k$ tends to infinity?

答:
If you direct the edges then the problem becomes trivial.

Assume inductively that at time $i$ the only nodes that might be infected are $u_i$ and $v_i$.

Then at time $i+1$ nodes $u_{i+1}$ and $v_{i+1}$ may become infected by $u_i$ and $v_i$, but $u_i$ and $v_i$ become cured, and there is no way of reinfecting them from below.

Thus your argument is correct, and the probability that the process is alive at time $i$ is at most $q^i$.

Now, for any natural number $n$, if the process does not terminate after a finite number of levels, then the process much reach level $n$.

Therefore the probablity that the process does not terminate is less than or equal to the probability that the process reaches level $n$.

So if $A$ is the event that the process survives indefinitely we have
$$0\leq\mathbb P(A) \leq q^n$$for every $n\in\mathbb N$. Therefore we must have $\mathbb P(A) = 0$.


答:
I'm going to interpret your question a little differently, since you yourself say that your mathematical understanding is a bit limited. What I think you want is not a variable $x$ but rather some kind of function $f$ which for the value 100 gives you 0, and for the value 50 gives you 3. I.e., we have

$$f(100) = 0$$
$$f(50) = 3$$

Now, there are infinitely many different functions that would satisfy the two requirements you have. But if we for example stick to the simplest solution, a linear function, the solution would be

$$f(x) = \frac{-3x}{50} + 6$$

From this you can find a new value, for example which one corresponds to 75, by:

$$f(75) = \frac{-3 \cdot 75}{50} + 6 = 1.5$$

Title: How to calculate a variable x that produces non-linear results?

Body:
Sorry I don't have a mathematical background so It is hard for me to put the right words on it.

Let me explain what I am trying to do with an example.

Essentially I want to solve x so that it satisfies both situation:

$100 \times x = 0$

$50 \times x = 3$

I would like to know how to solve this type of equation as well as to know the name of the mathematical concepts involved here.


问:
Is it true that for any $n\geq 1$, $(\ln(x))^n < x$?
I try to graph it on Desmos but it does not seem to be the case?</p>

答:
It is false.  Take the natural log of both sides:
$$ \ln{(\ln(x))^n)} < \ln x$$
Now you can take the n out of the exponent:
$$ n\ln{\ln(x)} < \ln x$$
Now let $y = \ln x$:
$$n \ln y < y$$
For any $y > 1$, you can choose an $n$ big enough that the statement is false.


问:
Fit a logistic growth curve with a custom function in R
(Tag suggestions welcome, I was looking for 'logistic-growth' or 'sigmoid-growth' but those tags don't exist)

I'm working through an exercise on this online textbook, 'Try it #3', almost halfway down the page.

Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years $1997$ to $2012$.
\begin{array}{|c|c|c|c|}
\hline
\text{Year}& \text{Seal Population (Thousands)} &\text{Year} & \text{Seal Population (Thousands)} \\ \hline
 1997 & 3,493&2005&19,590 \\ \hline
 1998 & 5,282&2006&21,955\\ \hline
 1999 & 6,357&2007&22,862\\ \hline
 2000 & 9,201&2008&23,869\\ \hline
 2001 & 11,224&2009&24,243\\ \hline
 2002 & 12,964&2010&24,344\\ \hline
 2003 & 16,226&2011&24,919\\ \hline
 2004  & 18,137&2012&25,108\\ \hline
\end{array}

(a) Let $x$ represent time in years starting with $x=0$ for the year $1997$. Let $y$ represent the number of seals in thousands. Use logical regression to fit a model to this data.

(b) Use the model to predict the seal population for the year 2020.

(c) To the nearest whole number, what is the limiting value of this model?

I would like to attempt this with a custom function in R. Here's the same data as part of a data frame:

```<code>ti3_df <- data.frame(
  x = 0:15,
  y = c(3.493, 5.282, 6.357, 9.201, 11.224, 12.964, 16.226, 18.137,
        19.590, 21.955, 22.862, 23.869, 24.243, 24.344, 24.919, 25.108)
)
</code>```

The textbook says that the logistic function is:

$$y=\frac{c}{1+ae^{-bx}}$$

I know that C is the carrying capacity or upper limit. But this is not provided in the question. I don't know if it's actually possible to solve this manually using the formula approach since I'm fitting a curve aka 'modeling' the data rather than calculating an exact point.

The books says 'The initial value of the model is $\frac{c}{1+a}$'

I wanted to build a function of the form:

```<code>my_log_growth_func <- function(c, a, dataframe) {... r code here}
</code>```

Where I would calculate b and x. But I think I'm missing something and am unsure if it's possible to fit a model using the information I have without a calculator or package.

How can I fit a model to these data?

答:
I suppose that you want to adjust $(a,b,c)$ in order to have the best fit.

Since
$$y=\frac{c}{1+ae^{-bx}}$$

as you wrote $c$ is a bit larger than the largest $y$; say $c\sim 25.5$ for an intial guess

for $x=0$ you have  (this is the hint) $y=\frac c{1+a}$; so, for $c=25.5$, this gives a guess $a\sim 6.3$

Now, to get an estimate of $b$ write (for the time being)
$$\frac {c-y} {ay}=e^{-b x}\implies \log \left(\frac{a y}{c-y}\right)= b x$$ which a linear regression without intercept (easy). This would give $b\sim 0.4$.

At this point, we have all required estimates for starting a nonlinear regression. It will works like a charm and, with $R^2=0.999737$ (very high), you should get
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval}
   \\
 a & 6.1137 & 0.2840 & \{5.4948,6.7325\} \\
 b & 0.3852 & 0.0114 & \{0.3603,0.4101\} \\
 c & 25.657 & 0.2345 & \{25.146,26.168\} \\
\end{array}$$

As you can notice, the guesses were quite good.


问:
Showing A-stability for Runge-Kutta-method
For $0\leq \theta \leq 1$ we have given the method
$$
x_n-x_{n-1}=h(\theta f(x_,t_n) + (1-\theta)f(x_{n-1},t_{n-1}))
$$
Now I should formulate this as a Rune-Kutta-Method and show that it is A-stable if $\theta\leq \frac{1}{2}$.

(1.) For the first part I would start by looking at the definition of Runge-Kutta-Methods. Runge-Kutta-methods with $s$ stepts and with constant lenght $h$ for
$$
x'(t)=f(x(t),t), \quad x(t_0)=x_0
$$
have the form
$$
X_i=f(x_{n-1}+h \sum_{j=1}^s a_{ij}X_j,t_{n-1}+hc_i), \quad i=1,...,s \\ x_n=x_{n-1}+h\sum_{i=1}^s b_iX_i
$$
We have also given the assumption that $c_i=\sum_{j=1}^s a_{ij}$. Now from this I would follow that our given method $x_n = x_{n-1} + h((1-\theta)f(x_{n-1},t_{n-1})+ \theta f(x_,t_n))$ has
$$
s=2, \quad b_1 = (1-\theta), \quad b_2=\theta, \quad X_1=f(x_{n-1},t_{n-1}), \quad X_2=f(x_n,t_n)
$$
Now for finding $c_1,c_2$ I would write
$$
(i) \quad f(x_{n-1},t_{n-1})=X_1 = f(x_{n-1}+h(a_{11}X_1 + a_{12}X_2),t_{n-1}+hc_1)\\
(ii) \quad f(x_n,t_n)= X_2 = f(x_{n-1}+h(a_{21}X_1 + a_{22}X_2),t_{n-1}+hc_2)
$$
Now from (i) it follows that $c_1=0$ and from (ii) that $c_2=1$ since $t_n=t_{n-1}+h$. Furthermore from (i) and the assumption $c_i=\sum_{j=1}^s a_{ij}$ which here ist $c_1=a_{11}+a_{12}$ I would have say
$$
x_{n-1} = x_{n-1}+h(a_{11}X_1 + a_{12}X_2) \Leftrightarrow a_{11}X_1 + a_{12}X_2 = 0 \Leftrightarrow a_{11} = a_{12} = 0
$$
Is that correct?
Then also from (ii)
$$
x_n=x_{n-1}+h(a_{21}X_1 + a_{22}X_2)
$$
and using the definition of the given method
$$
x_n = x_{n-1} + h((1-\theta)f(x_{n-1},t_{n-1})+ \theta f(x_,t_n))
$$
I would get $a_{21}=(1-\theta),a_{22}=\theta$
So this would an implicit Runge-Kutta-Method with
$$
b_1=a_{21}=1-\theta, \quad b_2=a_{22}=\theta, \quad a_{11} = a_{12} = 0, \quad c_1=0, \quad c_2=1
$$
Is that correct?

(2.) For this part I would then use $A=(a_{ij})_{i,j=1}^s$ and look at the stability function
$$
R(z)=1+zb^T(I-zA)^{-1} \mathbb{1}
$$
and check when $|R(z)|\leq 1$

Is that ok?

\Edit: I get
$$
R(z)=\frac{1+z(1-\theta)}{1-z\theta}=1+ \frac{z}{1-z\theta}
$$
Is that correct?
But I am struggeling to show that $|R(z)| \leq 1$ for $\theta \geq \frac{1}{2}$

答:
$$
R(z)=\frac{1+z(1-\theta)}{1-z\theta}=1+ \frac{z}{1-z\theta}
$$
From the second term one observes that for $z=x\to-\infty$ or generally for $z\to\infty$ on the Riemann sphere,
$$R(\infty)=1-\frac1θ.$$
So to remain inside the unit circle, $θ>\frac12$. Use the form $\frac{1+θz}{1-θz}$ to transform a point with $Re(z)\le 0$ to a point inside the unit circle.
$$
R(z)=1-\frac1{2θ}+\frac1{2θ}\frac{1+θz}{1-θz}.
$$
So on the negative half plane we get
$$
|R(z)|\le\left|1-\frac1{2θ}\right|+\frac1{2θ}=1.
$$


问:
Linearity of Power of Point

I have 4 questions about this:

1)What is meant by $\mathbb{R}^2 \rightarrow \mathbb{R}$ in this context?

2)What is meant by $C=kA+(1−k)B$?Aren't $A,B,C$ just points in euclidean plane?How can they be related by an equation?

3)Why is it sufficient to show the condition for points $A$ and $B$?

4)What is meant by $F$ is linear?How to prove it and what is it's use?

Sorry if that's too obvious.I am not aware of the terminology used here i guess.

答:
1)What is meant by R2→R in this context?

$F$ is a function whose domain is $\mathbb{R}^2$ and codomain is $\mathbb{R}$. The domain is $\mathbb{R}^2$ because its inputs are points, and the codomain is $\mathbb{R}$ because its outputs are real numbers. Since what $F$ does is ignore the $\omega$ terms in $\mathbb{P}$, it is what is called a `partial function' in programming. Put another way, the domain of $\mathbb{P}$ is $\mathbb{R}^2 \times \mbox{the space of circles}^2$ (sorry, I don't know what space $\omega$ lives in), and its codomain is $\mathbb{R}$. What $F$ does is ignore some of the inputs.

2)What is meant by C=kA+(1−k)B?Aren't A,B,C just points in euclidean plane?How can they be related by an equation?

It's a statement that $C$ is on the line segment between $A$ and $B$. This kind of formula: $kx + (1-k)y$ is called a convex combination of $x$ and $y$. In the context of two points in Euclidean space, it is a way to associate the points on the line thru $x$ and $y$ with values of $k$. If $0 \leq k\leq 1$, then the point is between $x$ and $y$.

3)Why is it sufficient to show the condition for points A and B?

4)What is meant by F is linear?How to prove it and what is it's use?

The answers to 3 and 4 are connected. A function $f$ is linear if $f(a + cb) = f(a) + c f(b)$ no matter what $a$ and $b$ (and $c$) are. So the statement about sufficiency is just a bit of semantic fluff to get the proof started. Any time you want to prove a function is linear, you start with a statement similar to: "Let $a$ and $b$ be in the domain of the function".

Title: Linearity of Power of Point


问:
Can a notation for ordered pairs in ZFC set theory be introduced as a conservative extension of ZFC without junk theorems?
Currently, a standard way to introduce ordered pairs in ZFC set theory is to define
an ordered pair $(a,b)$ as $\{\{a\},\{a,b\}\}$. From this definition one can deduce the main property of ordered pairs:
$(a,b) = (x,y) \rightarrow a=x \,\&\, b=y$.

But some junk theorems can also be proved, for example, $(a,a) = \{\{a\}\}, \{a\} \in (a,b)$, etc.

see also https://mathoverflow.net/questions/90820/set-theories-without-junk-theorems

https://mathoverflow.net/questions/386756/a-better-way-to-introduce-ordered-pairs-in-zf

For a different approach see the paper "Adding an Abstraction Barrier to ZF Set Theory"

https://link.springer.com/content/pdf/10.1007%2F978-3-030-53518-6_6.pdf

An excerpt from it:

" Such ‘accidental theorems’ do not match how people think about the mathematics and also cause practical difficulties when using set theory in machine-assisted theorem proving."

So a natural question arises: is it possible to introduce ordered pairs in a more safe way, without possibility to get
such strange "theorems".

A rather natural approach is to extend ZFC just by adding a new function symbol (say $\pi$, it is usually omitted)
and a defining axiom such as  $\pi(a,b) = \pi(x,y) \rightarrow a=x \,\&\, b=y$
and prove that such theory extension is conservative (without such proof we can get an undesirable result, in a worse case - an inconsistent theory).

So my question:

Is this extension of ZFC (by adding the function symbol $\pi$ and its defining axiom)  conservative?

答: Whatever we want of ordered pairs is absolutely not only the famous implication. I believe that (a,b) being Goedel-constructible in a,b is a sine qua non in set theoretic practice above 5 or so introductory lectures, and so is (a,b) being easily coded by a real provided a,b are reals, and zillion more such properties. So an abstract non-definable Bourbaki-style pair in the language will not work.</p>


问:
Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$

Let $x,y,z\in \mathbb R^+$ prove that:

$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz \le RHS$. How to prove above inequality ? It is too hard for me.

答:
The inequality is homogeneous, so we may assume that $xyz=1$. Then if we expand left and right hand side, we see that we have to show that
$$\sqrt{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 3} \ge 1 + \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
Let
$$u = \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
If we square both sides of the expanded inequality, we get the equivalent inequality
$$u^3 \ge 2u + u^2,$$
or,
$$u^2 - u - 2 \ge 0.$$
Therefore it is enough to show that $u \ge 2$. But this is certainly true since $x^3 + \frac{1}{x^3}, y^3 + \frac{1}{y^3}, z^3 + \frac{1}{z^3} \ge 2$.</p>


答:
Yes, if $y= \frac{c}{1+ ae^{bx}}$ then when x= 0, $y= \frac{c}{1+ a(e^0)}= \frac{c}{1+ a}$.

You want to find values for a, b, and c so that you get approximately the values given.  You realize, I hope, that there is no one "correct" answer.  You can get different answers using different approximation techniques.

Since the problem is to determine three values, a, b, and, c, you need three equations.  I suggest using three points as far apart as possible, the first point, (1997, 3493), so $3493= \frac{c}{1+ ae^{1997b}}$ the middle point, 2005, and the last point, 2012.  Unfortunately, that data seems to have been cut off,

Title: Fit a logistic growth curve with a custom function in R

Body:
(Tag suggestions welcome, I was looking for 'logistic-growth' or 'sigmoid-growth' but those tags don't exist)

I'm working through an exercise on this online textbook, 'Try it #3', almost halfway down the page.

Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years $1997$ to $2012$.
\begin{array}{|c|c|c|c|}
\hline
\text{Year}& \text{Seal Population (Thousands)} &\text{Year} & \text{Seal Population (Thousands)} \\ \hline
 1997 & 3,493&2005&19,590 \\ \hline
 1998 & 5,282&2006&21,955\\ \hline
 1999 & 6,357&2007&22,862\\ \hline
 2000 & 9,201&2008&23,869\\ \hline
 2001 & 11,224&2009&24,243\\ \hline
 2002 & 12,964&2010&24,344\\ \hline
 2003 & 16,226&2011&24,919\\ \hline
 2004  & 18,137&2012&25,108\\ \hline
\end{array}

(a) Let $x$ represent time in years starting with $x=0$ for the year $1997$. Let $y$ represent the number of seals in thousands. Use logical regression to fit a model to this data.

(b) Use the model to predict the seal population for the year 2020.

(c) To the nearest whole number, what is the limiting value of this model?

I would like to attempt this with a custom function in R. Here's the same data as part of a data frame:

```<code>ti3_df <- data.frame(
  x = 0:15,
  y = c(3.493, 5.282, 6.357, 9.201, 11.224, 12.964, 16.226, 18.137,
        19.590, 21.955, 22.862, 23.869, 24.243, 24.344, 24.919, 25.108)
)
</code>```

The textbook says that the logistic function is:

$$y=\frac{c}{1+ae^{-bx}}$$

I know that C is the carrying capacity or upper limit. But this is not provided in the question. I don't know if it's actually possible to solve this manually using the formula approach since I'm fitting a curve aka 'modeling' the data rather than calculating an exact point.

The books says 'The initial value of the model is $\frac{c}{1+a}$'

I wanted to build a function of the form:

```<code>my_log_growth_func <- function(c, a, dataframe) {... r code here}
</code>```

Where I would calculate b and x. But I think I'm missing something and am unsure if it's possible to fit a model using the information I have without a calculator or package.

How can I fit a model to these data?


问:
Probability problem regarding distributions

Here which distribution should I use? I am confused between hypergeometric and multinomial distributions

答:
Obvioulsly it is an Hypergeometric

$$\frac{\binom{100}{15}\binom{80}{35}}{\binom{180}{50}}$$

If you have time to lose you can calculate it manually....

$$\frac{\underbrace{100\times99\times\dots\times86}_{\text{15 times}}\cdot\underbrace{80\times79\times\dots\times46}_{\text{35 times}}}{\underbrace{180\times179\times\dots\times131}_{\text{50 times}}}\cdot\frac{\underbrace{50\times49\times\dots\times36}_{\text{15 times}}}{15!}$$


问:
Guaranteeing that three straight lines form a triangle in the third quadrant
The points of intersection of the following three straight lines form a triangle with all of the vertices in the third quadrant.

$x-2y=1$

$y-4x=3$

$3y+2x=-19$

How can we guarantee that any three lines will intersect to form a triangle with all of its vertices in the third quadrant?

I have explored looking at the gradients, $x$-intercepts and $y$-intercepts, as well as looking at the forming the equations of the lines given the points, but I'm wondering if there is an easier way.

答:
The following analysis is an attempt to show what happens if you look
for an alternative approach to computing the vertices directly.

Constraint-1: $~x-2y=1$

Constraint-2: $~y-4x=3$

Constraint-3: $~3y+2x=-19$

There are two issues:

Show that if $x \geq 0$, it is impossible to find any value
for $y$ that satisfies at least two of the pertinent constraints.

Show that if $y \geq 0$, it is impossible to find any value
for $x$ that satisfies at least two of the pertinent constraints.

Setting $x \geq 0$:

$1 + 2y \geq 0 \implies y \geq (-1/2).$

$y - 3 \geq 0 \implies y \geq 3.$

$-19 - 3y \geq 0 \implies 3y + 19 \leq 0 \implies y \leq (-19/3).$

It is immediate that only the 1st and 2nd constraints need be considered.

At this point, I see no alternative to multiplying the 2nd constraint by 2,
and then comparing it with the 1st constraint.  I consider
this approach out-of-bounds because you are (in effect) descending to
solving for the specific values of $x$, $y$.

Contrast this with the alternative query: show that at least two of
the vertices must lie in the 3rd quadrant.  With the alternative query,
considering the specific assumption of $(x \geq 0)$, it would have been sufficient to show, as I did, that when
$x \geq 0$, Constraint-3 is incompatible with both Constraint-1 and Constraint-2.

Exploring this alternative query, with respect to setting $y \geq 0$:

$x - 1 \geq 0 \implies x \geq 1.$

$3 + 4x \geq 0 \implies x \geq (-3/4).$

$-19 - 2x \geq 0 \implies 2x + 19 \leq 0 \implies x \leq (-19/2).$

Again, Constraint-3 is incompatible with both Constraint-1 and Constraint-2.

Assume that the 3 equations can be solved to generate 3 vertices.  Consider the incompatibility of Constraint-3 with the other two constraints, when either $x \geq 0$ or $y \geq 0$.  Because of these incompatibilities, you immediately know,
without solving for any of the vertices, that at least two of the vertices must lie in the 3rd quadrant.

As I see it, identifying the location of the vertex that corresponds to Constraint-1 and Constraint-2 will require descent into the normal methods.


问:
probability distribution of a random variable that is uniformly distributed in [-1, 1]
If $X$ is a random variable that is uniformly distributed between $-1$ and $1$, find the PDF of $\sqrt{\vert X\vert}$ and the PDF of $-\ln\vert X\vert$.
Solution or an explantion of approach wooulf work.

thank you in advance for your help

答:
For $\sqrt{|X|}$, take a numbber $t\in [0,1]$,
$$
P(\sqrt{|X|} \leq t)=P({|X|} \leq t^2)= P(-t^2\leq{X} \leq t^2) =t^2.
$$
Differentiate and the density is $2t$ in $[0,1]$ and $0$ otherwise.

For $-\ln{|X|}$ take a number $t \in [0,\infty)$,
$$
P(-\ln{|X|}\leq t) = P(|X|\geq e^{-t}) = P(X\geq e^{-t} \text{  or  }X\leq -e^{-t}) =P(X\geq e^{-t} )+(X\leq -e^{-t}) \\=1-e^{-t}.
$$
Differentiate and the density is $e^{-t}$ on $[0,\infty)$ and $0$ otherwise.


问:
Solving the equation $-e^{-i2k\ell}=\frac{k-1}{k+1}$ for $k$
I am trying to solve the following equation for $k$:
\begin{equation*}
-e^{-i2k\ell}=\frac{k-1}{k+1}\, ,
\end{equation*}
where $\ell$ is a positive number (constant).

I would like to know how the solutions of $k$ look like. I expect it to be an infinite number of solutions. Does anyone have any ideas how to prove or show this?

答:
Let's define :

$e^{-c k}=\frac{k-1}{k+1}$

we can write :

\begin{align}
-ce^{-c k}=-\frac{k-1}{k+1}c
& \iff -c(k+1)e^{-c(k+1)}=-(\sqrt k - 1)(\sqrt k + 1)ce^{-c} \\
\\
& \iff -c(k+1)e^{-c(k+1)}= -(\sqrt k - 1)(\sqrt k + 1)ce^{-c} \\
\\
& \iff \ln(-c(k+1)) - c(k+1)=- (\sqrt k - 1)(\sqrt k + 1)ce^c \\
\\
& \iff W(-(\sqrt k - 1)(\sqrt k + 1)ce^c) - c(k+1)= \ln(- (\sqrt k - 1)(\sqrt k + 1)ce^c) \\
\\
& \iff k = -\frac{c + \ln(- (\sqrt k - 1)(\sqrt k + 1)c) - W(-(\sqrt k - 1)(\sqrt k + 1)ce^c) + c }{c} \\
\\
& \iff k = -(2 + \ln(- (\sqrt k - 1)(\sqrt k + 1)c) - W(-(\sqrt k - 1)(\sqrt k + 1)ce^c)) \\
\\
& \iff k = -2 - \ln(- (\sqrt k - 1)(\sqrt k + 1)c) + W(-(\sqrt k - 1)(\sqrt k + 1)ce^c))
\\
\end{align}

This express the multiple solutions using the $W(-(\sqrt k - 1)(\sqrt k + 1)ce^c)$ Lambert function with $c=2i \ell$.