问:
Formal Rules for Categorical Duality
EDIT: My question is not about what categorical duality is. I generally understand the idea, but my problem is that for every theorem, finding the dual theorem requires us to think about what it means, so that we can translate it into the opposite category by reversing the arrows. I am looking for a concrete set of "algebraic manipulations" that lets us take a statement to its dual, without understanding what anything means. In the example of negating a logical statement (in the final paragraph of the question), we can swap the symbols around without knowing what $P$ is, and more generally without understanding what things mean.

Duality is very important in category theory, but as far as I have seen, statements that "follow by duality" are often taken to be obvious, even though in reality there are lots of details to be checked. People seem to follow some informal procedure of replacing all instances of "left" with "right", "right" with "left", "limit" with "colimit" and so on.

This is reminiscent of the method for negating logical statements. For example, the statement $(\forall x, \exists y: P(x,y))$ has negation $(\exists x:\forall y, \neg P(x,y))$. We can obtain this negation by swapping $(\forall)$ with $(\exists)$, $(:)$ with $(,)$, and $(P)$ with $(\neg P)$. The set of rules for converting a statement of this form is well-known and easy to implement. It is also completely rigorous because we can actually prove in general that the rules apply.

Is there a similar set of rules for converting categorical statements into their duals?

答:
Inductive definition of duality operator $\_{}˘$

Here is a fragment of the categorial language,

```<code>|  src f | tgt f |
|  Idₐ   | f ∘ g |
|  p + q | p × q |
</code>```

Then, we can define the dual $e ˘$ of an expression $e$ as follows.

```<code>(src f)˘ = tgt (f ˘)
(tgt f)˘ = src (f ˘)
(Idₐ)˘   = Idₐ
(p + q)˘ = p ˘ × q ˘
(p × q)˘ = p ˘ + q ˘
p ˘      = p   provided p is an object OR a variable name
</code>```

By induction (over the possible fragement of terms), one can show that this operator is self-inverse:
$e ˘ ˘ = e$.

Example 1: Duality switches arrow argument positions

Then, we can apply these rules as follows:
For any morphism expression $f$,

```<code>  (f : A ⟶ B)˘
≡     Arrow is defined in terms of source, target
  (src f = A  ∧  tgt f = B)˘
≡     Distribute duality over non-categoiral terms
  (src f)˘ = A ˘  ∧  (tgt f)˘ = B ˘
≡     Dual of an object is just the object unchanged
  (src f)˘ = A  ∧  (tgt f)˘ = B
≡     Duality slips src and tgt
  tgt (f ˘) = A  ∧  src (f ˘) = B
≡    Arrow notation
  f˘ : B ⟶ A
</code>```

Example 2: Duality flips terminals and initials

```<code>   ( is terminal)˘
 ≡   Expand definitions
   (∀ X • ∃₁ f • f : X ⟶ )˘
 ≡   Distribute duality over non-categorial operators
   ∀ X • ∃₁ f • (f : X ⟶ )˘
 ≡   Above calculation: Duality flips arrow arguments
   ∀ X • ∃₁ f • f˘ :  ⟶ X
 ≡   Duality does nothing to variable names
   ∀ X • ∃₁ f • f :  ⟶ X
 ≡   Foldup using definitions
   I is initial
</code>```

Duality is functorial

Moreover, notice that _˘ :  ⟶  is an identity-on-objects functor ;-)

Title: Formal Rules for Categorical Duality

Body:
Inductive definition of duality operator $\_{}˘$

Here is a fragment of the categorial language,

|  src f | tgt f |
|  Idₐ   | f ∘ g |
|  p + q | p × q |

Then, we can define the dual $e ˘$ of an expression $e$ as follows.

(src f)˘ = tgt (f ˘)
(tgt f)˘ = src (f ˘)
(Idₐ)˘   = Idₐ
(p + q)˘ = p ˘ × q ˘
(p × q)˘ = p ˘ + q ˘
p ˘      = p   provided p is an object OR a variable name

By induction (over the possible fragement of terms), one can show that this operator is self-inverse:
$e ˘ ˘ = e$.

Example 1: Duality switches arrow argument positions

Then, we can apply these rules as follows:
For any morphism expression $f$,

  (f : A ⟶ B)˘
≡     Arrow is defined in terms of source, target
  (src f = A  ∧  tgt f = B)˘
≡     Distribute duality over non-categoiral terms
  (src f)˘ = A ˘  ∧  (tgt f)˘ = B ˘
≡     Dual of an object is just the object unchanged
  (src f)˘ = A  ∧  (tgt f)˘ = B
≡     Duality slips src and tgt
  tgt (f ˘) = A  ∧  src (f ˘) = B
≡    Arrow notation
  f˘ : B ⟶ A

Example 2: Duality flips terminals and initials

   ( is terminal)˘
 ≡   Expand definitions
   (∀ X • ∃₁ f • f : X ⟶ )˘
 ≡   Distribute duality over non-categorial operators
   ∀ X • ∃₁ f • (f : X ⟶ )˘
 ≡   Above calculation: Duality flips arrow arguments
   ∀ X • ∃₁ f • f˘ :  ⟶ X
 ≡   Duality does nothing to variable names
   ∀ X • ∃₁ f • f :  ⟶ X
 ≡   Foldup using definitions
   I is initial

Duality is functorial

Moreover, notice that _˘ :  ⟶  is an identity-on-objects functor ;-)


问:
Can there exist an $\omega$-Moti Gitik model?
Moti Gitik had prove that there exists a model of ZF in which every aleph bigger than $\aleph_0$ is of $\aleph_0$ cofinality, call such models as Moti Gitik models.

Supposing that there exists an $\omega$-model of ZF. Would it follow that there can exist an $\omega$-Moti Gitik model.

An $\omega$-model is a model in which all naturals are the standard naturals as externally seen.

答:
Gitik's model is obtained by forcing methods, these do not change the ordinals, so in particular preserve things like $\omega$-model.

However, the requirements for Gitik's model are currently not known to be weaker than "a proper class of strongly compact cardinals", which exceed "there is an $\omega$-model" by a lot. And while it is conjectured that a similar result can be obtained with a lot less than even a single strongly compact, we do know that it implies there are many Woodin cardinals in an inner model, so you cannot bring this result anywhere near "there is an $\omega$-model".

ContentLicense: CC BY-SA 4.0


问:
Taylor formula for two variables mapping
Let $f\in \mathcal{C}^2(\mathbb{R}^d,\mathbb{R}^d)$, I know that in the case of one dimension :

$$f(x,b)-f(a,b)=\int_0^1\partial_x f(a+\alpha(x-a),b)\times(x-a)d\alpha.$$

I wonder if there's any formula for $2$ variables :

$$f(x,y)-f(a,b)=\int_0^1\int_0^1\partial_x f(a+\alpha(x-a),b)\times(x-a)+\partial_y f(a,b+\beta(y-b))\times(y-b) d\beta d\alpha.$$
Is this formula true ?

答:
The second equation in your post is not true. The correct one is
\begin{align}
  & f(x,y)-f(a,b)\\
= & \int_0^1 [\partial_x f(a+\alpha(x-a),b+\alpha(y-b))\times(x-a)
+ \partial_y f(a+\alpha(x-a),b+\alpha(y-b))\times(y-b) ]  d\alpha.
\end{align}

Proof
\begin{align}
& \int_0^1 [\partial_x f(a+\alpha(x-a),b+\alpha(y-b))\times(x-a)
+ \partial_y f(a+\alpha(x-a),b+\alpha(y-b))\times(y-b) ] d\alpha\\
& = \int_0^1 \frac{\partial f}{\partial \alpha}(a+\alpha(x-a),b+\alpha(y-b)) d\alpha\\
& = f((a+\alpha(x-a),b+\alpha(y-b))|_0^1\\
& = f(x,y)-f(a,b).
\end{align}


问:
How to find 3 x 3 matrix inverses
Is there a way of finding the inverse of a $3 \times 3$ matrix without forming an augmented matrix with the identity matrix? Also, is there a quick way of checking that a  $3 \times 3$ matrix's inverse exists, without trying to compute it? Thanks.</p>

答:
In order to check invertibility, you compute a determinant. If it's zero, the matrix is not invertible.

Check here wiki, section "Analytic solution".

ContentLicense: CC BY-SA 3.0

Title: How to find 3 x 3 matrix inverses


问:
Let $T =\{X\in P(\mathbb{Q}) | X\cup \mathbb{N} = \mathbb{Q} \}$ the cardinality of $T$ is $\aleph$?
Let $T =\{X\in P(\mathbb{Q}) | X\cup \mathbb{N} = \mathbb{Q} \}$ the cardinality  of $T$ is $\aleph$ ? prove

attempt

we know that $X\in P(\mathbb{Q})$ so, $X\cup \mathbb{N} = \mathbb{Q}$ create an infinity numbers

答:
It is clear from your definition that $T = \{ A \cup (\mathbb Q \setminus \mathbb N) \, | \, A \in P(\mathbb N) \}$. It can obviously be put in bijection with $P(\mathbb N)$ via the bijection $f: T \to P(\mathbb N)$ that sends $X$ to $X \cap \mathbb N$.

Hope that helps,


问:
Simplifying a set (Venn Diagram)
I am practicing for a math exam about Venn Diagrams. How would you simplify this set? Thanks
$$(A \cap B) \cup (B' \cap A) $$

答:
The maroon area is $A\cap B$, i.e. both $A$ and $B$.

The yellow area is $B’ \cap A$, i.e. $A$ but not $B$.

Their union is $A$, as from the picture.

Title: Simplifying a set (Venn Diagram)


问:
$G$ is cyclic of order $10$, describe all quotients of $G$
I have the following task:

$G$ is cyclic of order $10$, describe all quotients of $G$

My answer is:
I know that $G \cong \mathbb{Z}/10\mathbb{Z}$ so we have 10 quotients

$[0]_{10}$ which is the neutral element

$[1]_{10}$ of order 10, $[2]_{10}$ of order 5, $[3]_{10}$ of order 10, $[4]_{10}$ of order 5, $[5]_{10}$ of order 4, $[6]_{10}$ of order 5, $[7]_{10}$ of order 10, $[8]_{10}$ of order 5, $[9]_{10}$ of order 10

Where [1],[3],[7],[9] are generators and cyclic and therefore abelian.

Is this correct and can more be said about these quotient groups?

答:
No, it is not correct. If $G$ is generated by $a$, then its quotient groups are:

$G/\{e\}\simeq G$;

$G/\{e,a^5\}\simeq\Bbb Z_5$;

$G/\{e,a^2,a^4,a^6,a^8\}\simeq\Bbb Z_2$;

$G/G$, which is the trivial group.


问:
GP problem to calculate sum
I have to form a $f(n)$ to calculate the sum of n-terms of the progression $5, 55, 555, 5555, ...$

My attempt:

$$5+55+555+5555+...=\frac{5}{9}(9+99+999+9999+...)=\frac{5}{9}((10-1)+(10^{2}-1)+...+(10^n-1))$$

But I am not sure how to proceed from here. I know that this is related to GP. A major help would be identifying the common ratio here.

Title: GP problem to calculate sum

Body: I've got it. $$\frac{5}{9}((10-1)+(10^{2}-1)+...+(10^n-1))=\frac{5}{9}((10+10^2+...+10^n)-n)$$ Now it's easy to calculate the common ratio for $10+10^2+...+10^n$ as Peter suggested. $$\frac{50}{81}(10^n-1)-\frac{5n}{9}$$ In general, for the series $m+\overline{mm}+\overline{mmm}+\overline{mmmm}+...$, the sum is $$\frac{10m}{81}(10^n-1)-\frac{mn}{9}$$


答:
Hint: by distributivity (and commutativity):
$$(A\cap B)\cup(A\cap B')=A\cap(B\cup B')$$

Title: Simplifying a set (Venn Diagram)

Body:
I am practicing for a math exam about Venn Diagrams. How would you simplify this set? Thanks
$$(A \cap B) \cup (B' \cap A) $$


问:
Given certain conditions, prove a function G(x) is always equal to 4.
Theres a question I've been having trouble with:

$G(x)$ is the function $|x+2|+|x-2|$. Show that if $-2<x<2$ then $G(x) = 4$.

Any help would be greatly appreciated. No calculus please :D

Body:
If $x>-2$, then $x+2>0$, thus $|x+2|=x+2$.

Likewise, if $x<2$, then $x-2<0$, thus $|x-2| = -(x-2) = 2-x$.

Thus, for $-2<x<2$ we have $|x+2|+|x-2| = x+2+2-x = 4$.


问:
How to determine the cardinality of this set?
I would like to determine the cardinality of this set (given that $a,b \in \mathbb{N}$ and we do not know which one is larger):

$\Big\{\frac{1}{1},\frac{2}{1},\frac{3}{1},...,\frac{a}{1},\frac{1}{2},\frac{2}{2},\frac{3}{2},...,\frac{a}{2},...,\frac{1}{b},\frac{2}{b},\frac{3}{b},...,\frac{a}{b} \Big\}$?

I am aware of the existence of https://oeis.org/A018805

and that we may use this to compute the cardinality of

$\Big\{ \frac{1}{1},\frac{2}{1},...,\frac{k}{1},\frac{1}{2},...,\frac{k}{2},...,\frac{1}{k},...,\frac{k}{k} \Big\}$. I am sort of confused if this may be used to compute the first set I described, or if we would have to come up with an additional rule.

This was my idea: we need to look at two different cases: $a>b$ and $a<b$ since $a=b$ case has been done already directly by the given function in A018805.

Now, the question is, would this also work for the cases $a>b$ and $a<b$? Would we have to use something additional?

答:
The cardinality of your set equals the cardinality of all pairs  $(x,y)\in \{1,\ldots, a\}\times \{1,\ldots, b\}$ with $\gcd(x,y)=1$.

An upper bound is of course $ab$, but we have to subtract the $\lfloor \frac a2\rfloor\cdot \lfloor \frac b2\rfloor$ pairs where both $x$ and $y$ are even, as well as the $\lfloor \frac a3\rfloor\cdot \lfloor \frac b3\rfloor$ pairs where $x$ and $y$ are multiples of $3$. Unfortunately, this subtracts multiples of $6$ twice, so we have to add back $\lfloor \frac a6\rfloor\cdot\lfloor \frac b6\rfloor$. Ultimately, we end up with this expression:
$$ \sum_{d=1}^\infty\mu(d)\left\lfloor \frac ad\right\rfloor\left\lfloor \frac bd\right\rfloor$$


答:
Any quotient of a cyclic group is cyclic.  One way to see it is that it is a homomorphic image (of the cyclic group).

Furthermore, any homomorphic image has order dividing the order of the group (first isomorphism theorem).

Finally it is easy to see that all the cyclic groups of orders dividing, in this case $10$, do occur as quotients.

As far as what you did, a bunch of them are the same.  So there are only four.

Body:
I have the following task:

$G$ is cyclic of order $10$, describe all quotients of $G$

My answer is:
I know that $G \cong \mathbb{Z}/10\mathbb{Z}$ so we have 10 quotients

$[0]_{10}$ which is the neutral element

$[1]_{10}$ of order 10, $[2]_{10}$ of order 5, $[3]_{10}$ of order 10, $[4]_{10}$ of order 5, $[5]_{10}$ of order 4, $[6]_{10}$ of order 5, $[7]_{10}$ of order 10, $[8]_{10}$ of order 5, $[9]_{10}$ of order 10

Where [1],[3],[7],[9] are generators and cyclic and therefore abelian.

Is this correct and can more be said about these quotient groups?


Title: Show by mathematical induction that if $n$ is a positive integer, then $(2n)!\lt 2^{2n}(n!)^{2}.$

Body:
I want a proof verification.

Prove that $$\boxed{ n\in \mathbb{Z}^{+} \Rightarrow (2n)!\lt 2^{2n}(n!)^{2}}$$ by mathematical induction.

Since $n=1$ implies that $(2\cdot 1)!=2\lt 2^{2\cdot 1}\cdot (1!)^2=2^{2}\cdot 1=4$ the base case holds.

Assume that $(2n)!\lt 2^{2n}(n!)^{2}$ is true for $n$. We now wish to show that this inequality is also true for $n+1$, namely that $$(2(n+1))!\lt 2^{2(n+1)}((n+1)!)^{2}.$$

Multiply the inequality $(2n)!\lt 2^{2n}(n!)^{2}$ by $(2(n+1))^{2}$ and we get that $$(2n)!\cdot (2(n+1))^{2}\lt 2^{2n}(n!)^{2}\cdot (2(n+1))^{2} \Leftrightarrow
\\
(2n)!\cdot (2(n+1))^{2}=(2(n+1))!\cdot 2(n+1)\lt 2^{2n}(n!)^{2}\cdot (2(n+1))^{2}=2^{2n}\cdot 2^{2}\cdot (n!\cdot (n+1))^{2}=2^{2(n+1)}((n+1)!)^{2}.$$ That is $$(2(n+1))!\cdot 2(n+1)\lt 2^{2(n+1)}((n+1)!)^{2}.$$ Since $2(n+1)\gt 0$, $\forall n\geq 0$ we have that $$(2(n+1))!\lt (2(n+1))!\cdot 2(n+1)\lt 2^{2(n+1)}((n+1)!)^{2},$$ so $$(2(n+1))!\lt 2^{2(n+1)}((n+1)!)^{2}$$ which we wanted to show.

Body:
Another proof :

$$\frac{(2n)!}{(n!)^2}=\binom{2n}{n}< \sum\limits_{k=0}^{2n} \binom{2n}{k} = 2^{2n}$$


问:
Solving Linear Differential Systems
I have no knowledge about solving these type of problems.
Neither youtube or the other websites have anything about it.

Could you help me ?

$x' - y' - 2x + 4y = t$

$x' + y' -x - y = 1$

答:
If we add the two equations, we get

$$x' = \dfrac{1}{2}\left(3x - 3 y + t + 1\right)$$

If we subtract the first from the second, we get

$$y' = \dfrac{1}{2}\left(-x + 5 y - t + 1\right)$$

We can write this as the non-homogeneous system, $X' = AX + f$ as

$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \dfrac{1}{2}\begin{bmatrix} 3 & -3 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} t+1 \\ -t+1 \end{bmatrix} $$

Title: Solving Linear Differential Systems


问:
predicate logic | truth value of a sentence
what is the truth value of the next sentence:
$$\forall x\exists y(x\nsubseteq y)$$ In the world of: $P(\mathbb{R})$

答:
The negation of this assertion is
$$\exists x \forall y, x\subset y$$

And this is true since for $x=\emptyset$, $\forall y,\emptyset \subset y$.

So the assertion is false.


问:
Prove that the following equality holds in a group
Let $G$ be a group (not necessary abelian) generated by $a, b$ that has a group presentation: $$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$Then how can I prove that $(a^sb^t)(a^ub^v) = a^xb^y$, where $x$ is the remainder of $s + u(r^t)$ when divided by $m$, and $y$ is the remainder of $t + v$ when divided by $n$? It seems that it can be solved by changing $b^{...}a^{...}$ form into $a^{...}b^{...}$ form successively using the condition $ba = a^rb$, but I don't understand precisely how to do this.

答:
First note that
$$ ba^{k+1}=baa^k=a^rba^k$$
so that by induction on $k$,
$$ ba^k=a^{rk}b.$$
Next,
$$ b^{l+1}a^k=b^nba^k=b^la^{rk}b$$
so that by induction on $l$,
$$ b^la^k=a^{r^lk}b^l.$$
Apply this to $a^sb^ta^ub^v$ and note that you may reduce exponents of $a$ modulo $m$ and exponents of $b$ modulo $n$.


答:
It is enough to show $b^t a^u=a^{u(r^t)} b^t$, which is equivalent to $b^t a^u b^{-t}=a^{u(r^t)}$. After we show $b^t a b^{-t}=a^{r^t}$, we take $u$ power of both sides and finish the proof.

Now enough to show $b^t a=a^{r^t} b^t$, and it comes easily from repeated use of $ba=a^r b$.

+

We get the last part basically from the idea you have mentioned. We start from the left hand side and switch $a,b$ so that $b$ move right each step. I think this is the most intuitive one.

However, I think induction on $t$ would be clearer. For $t=1$ it is obvious. To proceed from $t$ to $t+1$, we only have to show $b a^{r^{t}}=a^{r^{t+1}} b$. But it is equivalent to $a^{r^{t}}=b^{-1} a^{r^{t+1}} b$, and we obtain it by taking $r^t$ power of given relation $a=b^{-1} a^{r} b$.

Title: Prove that the following equality holds in a group

Body: Let $G$ be a group (not necessary abelian) generated by $a, b$ that has a group presentation: $$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$Then how can I prove that $(a^sb^t)(a^ub^v) = a^xb^y$, where $x$ is the remainder of $s + u(r^t)$ when divided by $m$, and $y$ is the remainder of $t + v$ when divided by $n$? It seems that it can be solved by changing $b^{...}a^{...}$ form into $a^{...}b^{...}$ form successively using the condition $ba = a^rb$, but I don't understand precisely how to do this.


问:
The Krull Dimension of the Completion of a Module
Let $R$ be a Noetherian local ring, and $M$ a finitely generated $R$-module.

I am trying to show the following:

$\dim_R(M)=\dim_{\widehat{R}}(\widehat{M})$

I have found a reference from this question that the result holds for $M=R$.

Then my proof would be complete if I could show that $\text{Ann}_{\widehat{R}}(\widehat{M})=\widehat{\text{Ann}_R(M)}$, since localisation is exact and so the completion of $R/\text{Ann}_R(M)$ is isomorphic to the quotient of the completions.

It is clear that $\widehat{\text{Ann}_R(M)}\subseteq\text{Ann}_{\widehat{R}}(\widehat{M})$, but I have been unable to show the converse.

Is my claim true, or is there a different way to prove this result?

Any help would be much appreciated.

Update:

I have found a reference, Corollary 2.1.8(a) of Cohen-Macaulay rings by Bruns and Herzog, for the original result I am trying to prove (though not my statement about annihilators).

Their proof is that the extension $R\to\widehat{R}$ is flat and local, and $\widehat{M}=M\otimes\widehat{R}$ since $M$ is finitely generated.

However I'm unsure how this proves the claim, and would appreciate any clarification.

答:
Here's a useful lemma.

Let $R$ be a commutative ring and $M$ is a $R$-module. Assume $R\rightarrow S$ is flat. Consider $m\otimes 1\in M\otimes S$.  Then $\operatorname{Ann}_{S}(m\otimes 1)=\operatorname{Ann}_R(m)S$.

Proof: One considers the exact sequence $$0\rightarrow \operatorname{Ann}_R(m)\rightarrow R\xrightarrow{1\mapsto m} M$$ Using the fact thatt $-\otimes_R S$ is an exact functor, we get the exact sequence $$0\rightarrow \operatorname{Ann}_R(m)\otimes_R S\rightarrow R\otimes_R S\rightarrow M\otimes_R S$$ This gives the exact sequence $$0\rightarrow \operatorname{Ann}_R(m) S\rightarrow S\xrightarrow{1\mapsto m\otimes 1} M\otimes_R S$$ after identifying $R\otimes_R S\cong S$ and the ideal $\operatorname {Ann}_R(m)\otimes_R S$ with $\operatorname{Ann}_R(m) S$ under this isomorphism.

If $M$ is a finite $R$ module with generators say $m_1,m_2,\dots, m_k$, then $Ann_R(M)=\bigcap_{i=1}^k\operatorname{Ann}_R(m_i)$ and $Ann_S(M\otimes_R S)=\bigcap_{i=1}^k\operatorname{Ann}_S(m_i\otimes 1)=\bigcap_{i=1}^k\operatorname{Ann}_R(m_i) S = \left (\bigcap_{i=1}^k\operatorname{Ann}_R(m_i)  \right )S=\operatorname{Ann}_R(M)S$ Note that the last equality follows again from flatness since for 2 ideals $I,J$ of $R$, we have the exact sequence $$0\rightarrow I\cap J \rightarrow R \rightarrow R/I \oplus R/J$$ Tensoring by $S$ we get the exact sequence $$0\rightarrow (I\cap J)\otimes_R S \rightarrow R\otimes_R S\rightarrow \left ( R/I \oplus R/J \right )\otimes_R S $$ which is the exact sequence $0\rightarrow (I\cap J)S\rightarrow S\rightarrow S/IS \oplus S/JS$ forcing $(I\cap J)S=IS\cap JS$.


问:
What is the PDF of the sum of two dependent variables?
Let $X \sim \mathcal{N}(\mu_1, \sigma_1)$ and $Y \sim \mathcal{N}(\mu_2, \sigma_2)$ denote two independent normal random variables, then what is the PDF of $Z=(X+Y)^2-X^2=2XY+Y^2$?

The problem is hard due to that $XY$ and $Y^2$ are not independent, since they both depend on $Y$. So I decided to use the bivariate transformation technique to solve the problem.

Here is what I have tried so far. Since $X \sim \mathcal{N}(\mu_1, \sigma_1)$ and $Y \sim \mathcal{N}(\mu_2, \sigma_2)$ are independent, the joint PDF of (X, Y) is
\begin{align}
f_{X,Y}(x, y) = f_X(x)f_Y(y).
\end{align}
Now define $U=2XY+Y^2$ and $V=Y$, thus $X=\frac{U-V^2}{2V}$ and
\begin{align}
f_{U,V}(u, v) = f_{X,Y}(\frac{u-v^2}{2v}, v) = f_X(\frac{u-v^2}{2v})f_Y(v).
\end{align}
Therefore, we can compute the marginal as
\begin{align}
f_Z(z) =& f_U(u) \\
=& \int_{-\infty}^{\infty}f_{U,V}(u, v)dv \\
=& \int_{-\infty}^{\infty} f_X(\frac{u-v^2}{2v})f_Y(v) \\
=& \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} \exp\left(-\frac{\left(\frac{u-v^2}{2v} - \mu_1\right)^2}{2\sigma_1^2} -\frac{\left(v - \mu_2\right)^2}{2\sigma_2^2} \right) dv\\
=& \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} \exp\left(-\frac{\left(\frac{u}{2v}-\frac{v}{2} - \mu_1\right)^2}{2\sigma_1^2} -\frac{\left(v - \mu_2\right)^2}{2\sigma_2^2} \right) dv\\
=& \text{I can't sovle it}.
\end{align}

As you can see, the integral is too hard and I have totally no idea how to solve it. Did I do something wrong? or it is ok but I just can't solve the integral. em...well

答:
If you write $Z$ as
$
\begin{bmatrix}
X, Y
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\
1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
X \\
Y \\
\end{bmatrix}
$, where $X$ and $Y$ are uncorrelated Normal variables, it fits the definition of Generalized Chi-Square.


问:
System feedback: why does not output feedback change observability of a (linear time-invariant) system?
Consider teh following system written on state-space form:

\begin{align}
\dot{x}(t) &= Ax(t) + Bu(t) \\
y(t) &= Cx(t)
\end{align}

where $x\in\mathbb{R}^{n}$, $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times k}$, $u\in\mathbb{R}^{k}$, $C\in\mathbb{R}^{m\times n}$ and $y\in\mathbb{R}^{m}$. When we want to stabilize such a system we may use some feedback. If we have information about the whole state $x$, then we call it state feedback. If only the output $y$ is measurable, then we may use this information to create some feedback, and we call it output feedback. Mathematically we have that

\begin{align}
\text{state feedback}:\quad u(t) = Kx(t) + v(t), \\
\text{output feedback}:\quad u(t) = Ky(t) + v(t).
\end{align}

Here $K$ is some matrix that we design. As I see it, state feedback is just a special case of output feedback since $Ky = KCx$. Now to my question, it apperently holds that under any such feedback the reachability, or controllability, of the system remains unchanged. Is it also the case that observablity for the system is the same, after a feedback has been applied? This claim is made in my book, at least that output feedback leaves system observability the same, but should not this then be also the case for state feedback? How do we see this, that observability for the new system with output feedback is the same as for the original system without feedback?

答:
In order to show this one could use the Hautus test to check for observability

$$
\text{rank}\!
\begin{bmatrix}
C \\
A - \lambda\,I
\end{bmatrix} = n, \forall\ \lambda \in \mathbb{C}. \tag{1}
$$

When using output feedback one can substitute $A$ with $A - B\,K\,C$, such that the system is observable if

$$
\text{rank}\!
\begin{bmatrix}
C \\
A - B\,K\,C - \lambda\,I
\end{bmatrix} = n, \forall\ \lambda \in \mathbb{C}. \tag{2}
$$

However, the matrix in $(2)$, from which you need to calculate the rank, can also be written as

$$
\begin{bmatrix}
C \\
A - B\,K\,C - \lambda\,I
\end{bmatrix} =
\begin{bmatrix}
I & 0 \\
-B\,K & I
\end{bmatrix}
\begin{bmatrix}
C \\
A - \lambda\,I
\end{bmatrix}, \tag{3}
$$

whose most right matrix is the same as in $(1)$ and the matrix multiplied by it is always full rank, due to its lower block triangular structure and the fact that the matrices on the diagonal are full rank. Therefore, the condition for observability of $(3)$ is identical to that of $(2)$.

This same factorization can not be done in general when using state feedback, since then you would need to use $B\,K$ instead of $B\,K\,C$.


问:
Why do many textbooks on Bayes' Theorem include the frequency of the disease in examples on the reliability of medical tests?
A "standard" example of Bayes Theorem goes something like the following:

In any given year, 1% of the population will get disease X. A particular test will detect the disease in 90% of individuals who have the disease but has a 5% false positive rate. If you have a family history of X, your chances of getting the disease are 10% higher than they would have been otherwise.

Virtually all explanations I've seen of Bayes' Theorem will include all of those facts in their formulation of the probability. It makes perfect sense to me to account for patient-specific factors like family history, and it also makes perfect sense to me to include information on the overall reliability of the test. I'm struggling to understand the relevance of the fact that 1% of the population will get disease X, though. In particular, that fact is presumably true for all patients who receive the test; that being the case, wouldn't Bayes' Theorem imply that the actual probability of a false positive is much higher than 5% (and that one of the numbers is therefore wrong)?

Alternatively, why doesn't the 5% figure already account for that fact? Given that the 5% figure was presumably calculated directly from the data, wouldn't Bayes' Theorem effectively be contradicting the data in this case?

答:
Further to user856's explanation in the comments, here's a complementary answer.

The way to frame/interpret medical tests in general is to understand them as updating one's level of certainty that the patient has the disease:

without a medical-test result, the disease prevalence (a measure of disease frequency) can
be taken as the patient's probability of having the disease;

however, in the context of a medical-test result, the aforementioned probability has changed: its updated value depends not just on the disease prevalence (as before), but now also on the test's sensitivity (true positive rate) and specificity (true negative rate). In other words, our knowledge of said probability has been refined.

p:  disease prevalence and other (prior) risk factors

v:  test sensitivity

f:  test specificity

D:  Diseased

H:  Healthy

+:  Positive test result

-:  Negative test result

The abovementioned probabilities are

the positive predictive value, i.e., the probability that the patient is indeed Diseased given a positive test result $$P(D|+)=\frac{P(D+)}{P(D+)+P(H+)}=\frac{pv}{pv+(1-p)(1-f)},$$

the false omission rate, i.e., the probability that the patient is actually Diseased given a negative test result $$P(D|-)=\frac{P(D-)}{P(D-)+P(H-)}=\frac{p(1-v)}{p(1-v)+(1-p)f}.$$

Thus, a screening test's predictive values $P(D|+)\,$ & $\,P(H|-)$ and overall accuracy $$P(D+)+P(H-)=pv+(1-p)f$$ depend on both its technical characteristics (sensitivity and specificity) and the population that it is being used on (disease prevalence). In particular:

unless the test has 100% sensitivity, its number of false-negative results is proportional to the disease prevalence $p;$

unless the test has 100% specificity, its number of false-positive results is proportional to $(1-p).$

N.B. The OP mentions “test reliability”, but that’s a separate issue, since reliability typically refers to consistency across retakes of a test’s results.

Here is a glossary.
$$\\$$
Finally, here is a concrete extended example (based on actual data, and simplistically assuming that successive tests are independent of one another) to put all this in context:

Due to the low disease prevalence,

the PCR and rapid tests have a positive predictive value of only
$4\%$ and $17\%$ respectively,

whereas their negative predictive value are both almost $100\%;$

the tests' overall accuracy are $95\%$ and $99\%$ respectively.


问:
Uniform convergence of sum in given interval
Let $a>0$ and observe $$\ln(a) - \sum_{n=1}^{\infty} \frac{1}{n}a^{-n}x^n$$

Show that the above sum converges uniformly in $[-a,b]$ with $-a<b<a$

Here's my current approach: For $x \in (-a,b]$ we see that $$\left|\sum_{n=1}^{\infty} \frac{1}{n}a^{-n}x^n\right|\le\left|\sum_{n=1}^{\infty} a^{-n}x^n\right|=\left|\sum_{n=1}^{\infty} \left(\frac{x}{a}\right)^n\right|$$ which a geometric sequance, that do convergence since $\left(\frac{x}{a}\right)^n<1 \ \forall x$ in the given interval. Therefore, by the Weierstrass-M test, the original sum converges uniformly. But here is my question, is this correct, and if so, how can I show uniform convergence in the endpoint $-a$

In a previous task, I've shown that the sum converges pointwise in $[-a,a)$

答:
Put $f_n(x) = \ln (a) - \sum_{i=1}^{n} \frac{1}{n}a^{-n}x^n $. You've showed that $f_n$ converges uniformly in $(-a, b]$. On the other hand, \begin{align}\lim_{n \to \infty} f_n(-a)&=\ln(a) - \sum_{n=1}^{\infty}\frac{1}{n}a^{-n}(-a)^n  \\ &= \ln(a) + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \\&= \ln(a)+\ln(2)=\ln(2a)\end{align}

Now define $f : [-a, b] \to \mathbb{R}$ by $f(x)=\lim_{n\to \infty}f_n(x)$ for $-a < x \leq b$ and $f(-a)=\ln(2a)$. We shall show that $f_n$ converges uniformly to $f$ in $[-a, b]$.

Let $\epsilon>0$ be given. Since $f_n$ converges uniformly to $f$ in $(-a, b]$, there exists $N_1 \in \mathbb N$ such that $n \geq N_1$ implies $|f_n(x)-f(x)| < \epsilon $ for all $x\in (-a, b]$. Moreover, $\lim_{n \to \infty}f_n(-a) = f(-a)$ guarantees the existence of $N_2 \in \mathbb N$ such that $n \geq N_2$ implies $|f_n(-a) - f(-a)| < \epsilon$. Thus $n \geq \max(N_1, N_2)$ implies $|f_n(x) - f(x)| < \epsilon$ for all $x \in [-a, b]$. That is, $f_n$ converges uniformly to $f$ in $[-a, b]$.


问:
Bounds on $S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+\dots+\frac{1}{3001}$

$S =  \frac{1}{1001} +   \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$.

Prove that $\dfrac{29}{27}<S<\dfrac{7}{6}$.

My Attempt:

$S<\dfrac{500}{1000} +   \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$

(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)

Using AM-HM inequality gave me $S > 1$, but the bounds are weak.

Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$

Inequality with sum of inverses of consecutive numbers

The answers to these questions are nice, but the bounds are weak.

Any help without calculus and without calculations involving calculators would be appreciated.

(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))

答:
Upper bound $\dfrac{7}{6}$

Your question describes a method of breaking the range up into sub-ranges, and then estimating the sum over a sub-range by noting that each term is $\le$ the first term. But we can do much better if we estimate this sum using a linear approximation instead. Then we only need three sub-ranges to establish the upper bound of $\frac76$.

Consider a 'graph' of the function $$f(k)=\frac{1}{k}$$ for integers $k\ge 1$. If we look at any interval $m\le k\le n$ with $1\le m\le n-2$, we can compare it with the straight-line function joining points $(m,\frac{1}{m})$ and $(n,\frac{1}{n})$. We can define this function explicitly if you want, as:
$$g(k)=\frac{1}{n-m}\left(\frac{k-m}{n}+\frac{n-k}{m}\right)$$
Then $f(m)=g(m),f(n)=g(n)$; and for $m<k<n,f(k)<g(k)$ (because $f$ is a convex function).

So $$\sum_{k=m}^nf(k)<\sum_{k=m}^ng(k)$$
But the $g(k)$ are in arithmetic progression, so the RHS is half the sum of the first and the last terms multiplied by the number of terms, i.e.
$$\frac{n-m+1}{2}\left(\frac{1}{m}+\frac{1}{n}\right)$$

We divide the range into three equal sub-ranges of size $667$. This gives us:
$$\begin{align}
\sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\
& <\frac{667}{2}\left(\frac{1}{1001}+\frac{1}{1667}\right)+\frac{667}{2}\left(\frac{1}{1668}+\frac{1}{2334}\right)+\frac{667}{2}\left(\frac{1}{2335}+\frac{1}{3001}\right)\\
& <\frac{667}{2}\left(\frac{1}{1000.5}+\frac{1}{1667.5}+\frac{1}{1667.5}+\frac{1}{2334.5}+\frac{1}{2334.5}+\frac{1}{3001.5}\right)\\
& =\frac13+\frac15+\frac15+\frac17+\frac17+\frac19\\
& =\frac{356}{315}\\
& <\frac{7}{6}\end{align}$$

where we have used the fact that $\frac{1}{m}+\frac{1}{n}<\frac{1}{m-\frac12}+\frac{1}{n+\frac12}$ if $m<n$, again by convexity of $f(k)$.

Note that if the last term were $\frac{1}{3000}$, we could do this with just two sub-ranges, and we would get exactly $\frac{7}{6}$ as a bound. So perhaps I have missed an easier way.

Lower bound $\dfrac{29}{27}$

This time we use the convexity of the function to estimate the sum over a sub-range from below. The sum of $\frac{1}{k}$ over the range $m\le k\le n$ is greater than the number of elements multiplied by the middle element if the range contains an odd number of elements. So we get
$$\begin{align}
\sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\
& > 667\left(\frac{1}{1334}+\frac{1}{2001}+\frac{1}{2668}\right)\\
& = \frac12+\frac13+\frac14\\
& = \frac{13}{12}\\
& >\frac{29}{27}
\end{align}$$


问:
How to find inverse of the function $f(x)=\sin(x)\ln(x)$
My friend asked me to solve it, but I can't.

If $f(x)=\sin(x)\ln(x)$, what is $f^{-1}(x)$?

I have no idea how to find the solution. I try to find

$$\frac{dx}{dy}=\frac{1}{\frac{\sin(x)}{x}+\ln(x)\cos(x)}$$

and try to solve it for $x$ by some replacing and other things, but I failed.

Can anyone help? Thanks to all.

答: The function fails the horizontal line test for one, very badly in fact. One to one states that for any $x$ and $y$ in the domain of the function, that $f(x) = f(y) \Rightarrow x = y$. that is to each point in the domain there exists a unique point in the range. Many functions can be made to be one to one $(1-1)$ by restricting the interval over which values are taken, for example the inverse trig functions and the square root function (any even root). We typically only take the positive square root because otherwise the function would have two answers and each x wouldn't have a unique y value (The so called vertical line test, which is why generally if functions fail the horizontal line test they don't typically have an inverse). If you graph this function it looks very much like a growing sinusoidal shape this cannot be restricted uniquely in a manner that makes an inverse definable.</p>


问:
Show that the ring of polynomials $k[x_{1},...,x_{n}]$ is an integral extension of the ring of symmetric polynomials

I want to prove that $A= k[x_{1},...,x_{n}]$, where $k$ is a field, is an integral extension of the subring $B$ consisting of the symmetric polynomials in $n$ variables.

It is sufficient by finding an expression with coefficients in B that nulls when evaluated the variables $x_{i}$ for every $i=1,...,n$. I tried it but I can’t find a suitable polynomial the satisfies this. Some help?

答:
This question may have interest for others visiting this site.

In fact the result is more general: If $H$ is any finite group acting on a commutative ring $A$ and if $A^H$ is the set of elements $a\in A$ with $ha=a$ for all $h\in H$
it follows $A^H \subseteq A$ is an integral extension of rings. Let $H:=\{h_1,..,h_n\}$
and let $a\in A$. Let

$f_a(t):=(t-h_1a)\cdots (t-h_na)\in A[T]$.

It follows

$hf_a(t)=(t-(hh_1)a)\cdots (t-(hh_n)a)=f_a(t)$

for any $h\in G$, since the element $h$ permutes the set $\{h_1,..,h_n\}$. Hence $f_a(t)\in A^H[t]$. The polynomial
$f_a(t)$ is monic and $f_a(a)=0$ since there is an $i$ with $h_ia=ea=a$. Hence $a$ is integral over $A^H$ for any $a\in A$.

Question: "It is sufficient to find an expression with coefficients in B that is zero when evaluated in the variables $x_i$ for every $i=1,...,n$. I tried it, but I can’t find a suitable polynomial that satisfies this. Some help?"

Answer: The polynomial $f_a(t)$ defined above gives a general answer to your question for an arbitrary finite group $H$ and commutative unital ring $A$.

Example: Let $S_n$ be the symmetric group on $n$ elements and let $k:=\mathbb{Z}$ be the ring of integers. Let

$S_n:=\{\sigma_1,..,\sigma_l\}$

and let $a(x_1,..,x_n)\in A:=\mathbb{Z}[x_1,..,x_n].$

Define

$f_a(t):=(t-\sigma_1a)\cdots (t-\sigma_la) \in B:=A^{S_n}[t]$.

It follows by the above argument that $f_a(t)$ is a monic polynomial and that $f_a(a)=0$. Hence the result is true over the ring of integers $\mathbb{Z}$:
The ring extension

$\mathbb{Z}[x_1,..,x_n]^{S_n} \subseteq \mathbb{Z}[x_1,..,x_n]$

is an integral extension.


答:
The first step is to get the system into the form $\vec{x}' = A\vec{x} + \vec{g}(t)$, where $\vec{x} = \begin{bmatrix}x\\y\end{bmatrix}$ and $A$ is a matrix.

If we add the two equations together, we get that $2x' - 3x + 3y = t+1$, which we can rearrange to get $x' = \dfrac{3}{2}x - \dfrac{3}{2}y + t + 1$.

Similarly, if we subtract the first equation from the second, we get $2y' + x - 5y = 1 - t$, which can be rearranged to get $y' = -\dfrac{1}{2}x+\dfrac{5}{2}y-t+1$.

These two equations can be rewritten as $\vec{x}'=\begin{bmatrix}\dfrac{3}{2}&-\dfrac{3}{2}\\-\dfrac{1}{2}&\dfrac{5}{2}\end{bmatrix}\vec{x}+\begin{bmatrix}t+1\\-t+1\end{bmatrix}$.

This is still a linear equation, so the principle of superposition applies. So, first we will look for solutions to the complementary equation, $\vec{x}' = \begin{bmatrix}\dfrac{3}{2}&-\dfrac{3}{2}\\-\dfrac{1}{2}&\dfrac{5}{2}\end{bmatrix}\vec{x}$.

Like with the equations you've studied in the past, it makes sense here to assume that the solution should be exponential because the derivative is some kind of multiple of the original function.

So, let's assume the form $\vec{x} = \vec{\eta}e^{rt}$, where $\vec{\eta}$ is constant. Taking the derivative gives us $\vec{x}' = r\vec{\eta}e^{rt}$, so plugging this into our equation gives us $r\vec{\eta}e^{rt} = A\cdot(\vec{\eta}e^{rt})$.

We can take advantage of the linearity of matrix multiplication by pulling out our scalar $e^{rt}$, yielding $e^{rt} \cdot r\vec{\eta} = e^{rt} \cdot A\vec{\eta}$. The exponential function is never zero, so we can divide both sides by $e^{rt}$ to get $r\vec{\eta} = A\vec{\eta}$. In linear algebra terminology, this means that $r$ must be an eigenvalue of $A$, and $\vec{\eta}$ is a corresponding eigenvector.

Now let's do some linear algebra. If $r\vec{\eta} = A\vec{\eta}$, then $A\vec{\eta}-r\vec{\eta} = \vec{0}$. Because matrix multiplication is distributive, we can rewrite this as $(A - rI)\vec{\eta} = \vec{0}$, where $I$ is the identity matrix.

$\vec{\eta} = \vec{0}$ is a trivial solution to this equation for any $A$ and $r$, but it's not very helpful because this would lead to $\vec{x} = \vec{0}$, a trivial solution to our original equation. So, we only want solutions with $\vec{\eta} \neq \vec{0}$. By the Invertible Matrix Theorem, this means that our matrix $(A - rI)$ must be non-invertible, which implies that $det(A - rI) = 0$, and conversely, if $det(A - rI) = 0$, then there is some non-zero $\vec{\eta}$ such that $(A - rI)\vec{\eta} = \vec{0}$.

So, the eigenvalues of $A$ are all of the values of $r$ such that $det(A - rI) = 0$:

$$\begin{vmatrix}\dfrac{3}{2}-r & -\dfrac{3}{2} \\-\dfrac{1}{2} &\dfrac{5}{2} - r\end{vmatrix} = 0 \to (r - \dfrac{3}{2})(r - \dfrac{5}{2}) - \dfrac{3}{4} = 0 \to r^2 - 4r + 3 = 0 \to r_1 = 1, r_2=3$$.

Now we need to find the corresponding eigenvectors. For $r_1 = 1$:

$$(A - rI)\vec{\eta} = \vec{0} \to \begin{bmatrix}\dfrac{1}{2} & -\dfrac{3}{2} \\-\dfrac{1}{2} &\dfrac{3}{2}\end{bmatrix}\begin{bmatrix}\eta_1\\ \eta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\to\dfrac{1}{2}\eta_1 - \dfrac{3}{2}\eta_2 = 0 \to \eta_1 = 3\eta_2$$

This equation has infinite solutions, but since we're going to multiply by an arbitrary constant later, we just need one. I typically like to let the smaller entry be $1$, so $\vec{\eta}_1 = \begin{bmatrix}3\\1\end{bmatrix}$.

Solving similarly for $r_2 = 3$ gives us $\vec{\eta}_2 = \begin{bmatrix}1\\-1\end{bmatrix}$. We are now ready to write our complementary solution:

$$\vec{x}_c = c_1\begin{bmatrix}3\\1\end{bmatrix}e^t + c_2\begin{bmatrix}1\\-1\end{bmatrix}e^{3t}$$

Now we just need to find our particular solution. Just like with a single equation, we can do this with the method of variation of parameters, although the procedure is a bit more complicated. Let's define the fundamental matrix of the system, $Y$, as the matrix where each column of the matrix is one of our homogeneous solutions. It follows from that definition and the definition of the homogeneous solutions that $Y' = AY$.

Now, let's define $\vec{y}_p = Y\vec{u}$ for some vector-valued function $\vec{u}$. From the product rule, $\vec{y}_p' = Y'\vec{u} + Y\vec{u}'$. Plugging this into our equation:

$$Y'\vec{u} + Y\vec{u}'= A(Y\vec{u}) + \vec{g}(t) \to (AY)\vec{u} + Y\vec{u}' = A(Y\vec{u}) + \vec{g}(t)$$

Matrix multiplication is associative, so $(AY)\vec{u} = A(Y\vec{u}),$ and by subtracting that from both sides it follows that $Y\vec{u}' = \vec{g}(t)$ and $\vec{u}' = Y^{-1}\vec{g}(t)$, provided that $Y$ is invertible, which follows from the linear independence of the fundamental solutions. (and my precious Invertible Matrix Theorem)

Here, we have $Y = \begin{bmatrix}3e^t&e^{3t}\\e^t&-e^{3t}\end{bmatrix}$. Sparing the details, we then have:

$$Y^{-1} = \begin{bmatrix}\dfrac{1}{4}e^{-t}&\dfrac{1}{4}e^{-t}\\\dfrac{1}{4}e^{-3t}&-\dfrac{3}{4}e^{-3t}\end{bmatrix}, \vec{u}' = \begin{bmatrix}\dfrac{1}{2}e^{-t}\\te^{-3t}-\dfrac{1}{2}e^{-3t}\end{bmatrix}, \vec{u} = \begin{bmatrix}-\dfrac{1}{2}e^{-t}\\\dfrac{-6t+1}{18}e^{-3t}\end{bmatrix}, \vec{y}_p = \begin{bmatrix}\dfrac{1}{3}t - \dfrac{13}{9}\\ -\dfrac{1}{3}t - \dfrac{5}{9}\end{bmatrix}$$

Note that we can neglect the constants when integrating $\vec{u}'$ to get $\vec{u}$ because any additional constant would simply lead to an additional term of the same form as our complementary solution, so it would essentially just correspond to different values for our arbitrary constants in the complementary solution.

So, our final solution is $\vec{x}=c_1\begin{bmatrix}3\\1\end{bmatrix}e^t + c_2\begin{bmatrix}1\\-1\end{bmatrix}e^{3t}+\begin{bmatrix}\dfrac{1}{3}t - \dfrac{13}{9}\\ -\dfrac{1}{3}t - \dfrac{5}{9}\end{bmatrix}$

Title: Solving Linear Differential Systems

Body:
I have no knowledge about solving these type of problems.
Neither youtube or the other websites have anything about it.

Could you help me ?

$x' - y' - 2x + 4y = t$

$x' + y' -x - y = 1$


Title: Bounds on $S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+\dots+\frac{1}{3001}$

Body:
$S =  \frac{1}{1001} +   \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$.

Prove that $\dfrac{29}{27}<S<\dfrac{7}{6}$.

My Attempt:

$S<\dfrac{500}{1000} +   \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$

(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)

Using AM-HM inequality gave me $S > 1$, but the bounds are weak.

Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$

Inequality with sum of inverses of consecutive numbers

The answers to these questions are nice, but the bounds are weak.

Any help without calculus and without calculations involving calculators would be appreciated.

(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))

Body:
Well not an answer but an idea to inspire someone :

I remenber the Gauss's solution of the first case concerning the Faulhaber's formula :

We have :

$$\frac{1}{1001}+\frac{1}{3001}>\frac{1}{1002}+\frac{1}{3000}>\cdots>\frac{1}{2002}+\frac{1}{2000}$$

Wich gives :

$$S<\frac{1000}{1001}+\frac{1000}{3001}+\frac{1}{2001}$$

Now if we divide (3001-1001) by four we get :

$$\frac{500}{2500}+\frac{500}{2502}+\frac{1}{2501}+\frac{499}{1499}+\frac{499}{1501}+\frac{1}{1500}<S<\frac{500}{1001}+\frac{500}{2000}+\frac{1}{1500}+\frac{499}{2001}+\frac{499}{3001}+\frac{1}{2500}$$


答:
$$\sum_{n=1}^{2001}\frac{1}{1000\left(1+\frac{n}{1000}\right)} =\int_{0}^{2}\frac{1}{\left(1+x\right)}dx \approx \ln3$$

$$\frac{29}{27} < \ln3 <\frac{7}{6}$$

Used:

Summation to integration when step size is small


Title: Calculate $\sum_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}$

Body:
Find $\sum_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}$

here's first attempt

\begin{eqnarray}
\sum_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}&=&\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n-1}{(n)!}\\
&=&\sum_{n=1}^{\infty}(-1)^{n}\frac{1-n}{(n)!}\\
&=&\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{(n)!}-\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{(n)!}
\end{eqnarray}
By $$e^{-x}=\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}x^n$$we have
$$e^{-1}-1=\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{(n)!}$$
so we get $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}=2e^{-1}-1$$

What are other possible approaches?
thanks

Body:
$$\frac{e^x -1}{x}=\sum_0^{\infty} \frac{x^{n}}{(n+1)!} $$
Differentiate:

$$\frac{xe^x -(e^x-1)}{x^2}=\sum_0^{\infty} \frac{nx^{n-1}}{(n+1)!}$$

Set $x=-1$:

$$1-2e^{-1} = \sum_1^{\infty} \frac{(-1)^{n-1} n}{(n+1)!} $$

Multiply both sides by $-1$:

$$2e^{-1}-1 =\sum_1^{\infty} \frac{(-1)^n n}{(n+1)!} $$


问:
Noah and his ark - inequalities
By spending an emerald, Noah turns a swan into a frog or a duck into a lizard, and by spending a ruby, he turns a swan into a duck or a frog into a lizard. Today he spent 20 emeralds and 23 rubies, and the number of frogs increased by 5. How and how much can the number of ducks change? (Give all possible results and explain why).

I found this in a Facebook group.

Let's name $S_0, F_0, D_0, L_0$ the initial number of each kind and $S_e, F_e, D_e, L_e$ their numbers after the exchange with the use of emeralds.
Clearly we have, for 1 emerald, $S_0-S_n = F_n-F_0$, therefore for $x$ emeralds, $x(S_0-S_n) = x(F_n-F_0)$.

Also for $y$ emeralds,
$y(D_0-D_n) = y(L_n-L_0)$.

We have $x+y=20$.

We get similar equations for the rubies:
$z(S_0-S_m) = z(D_m-D_0)$.

$w(F_0-F_m) = w(L_m-L_0)$.

$z+w=23$.

The difference in number of frogs after all transformations is 5 but I don't know how to take into account the consecutive transformations; it is getting very complicated.

答:
This story is a reported fact. Therefore we may assume that there are enough animals of all sorts around to make it possible.

I denote the number of animals $\>a$...$\ $ turned to $\>b$...$\ $ by $\vec{ab}$. According to the story we have
$$\vec{sf}+\vec{dl}=20,\quad \vec{sd}+\vec{fl}=23,\quad\vec{sf}-\vec{fl}=\Delta f=5$$
and therefore
$$\Delta d=\vec{sd}-\vec{dl}=(23-\vec{fl})-(20-\vec{sf})=3+(\vec{sf}-\vec{fl})=8\ .$$

Title: Noah and his ark - inequalities


问:
$\int_0^\infty x^n e^{-x^{12}\sin^2x}dx$ is divergent for any natural numbers $n$?
$\int_0^\infty x^n e^{-x^{12}\sin^2x}dx$ is divergent for any natural numbers $n$? My attempt: Cauchy criteria near $2k\pi$ for integers $k$.

Body:
$$\begin{align}\int_{k\pi}^{(k+1)\pi}e^{-x^{12}\sin^2x}\,\mathrm dx
&= \int_0^\pi e^{-(k\pi+x)^{12}\sin^2x}\,\mathrm dx \\
&> \int_0^\pi e^{-((k+1)\pi)^{12}x^2}\,\mathrm dx \\
&= \int_0^\pi e^{-\pi^{10}((k+1)^6x)^2}\,\mathrm dx \\
&> \int_0^{\frac1{(k+1)^6}} e^{-\pi^{10}((k+1)^6x)^2}\,\mathrm dx \\
&> \int_0^{\frac1{(k+1)^6}} e^{-\pi^{10}}\,\mathrm dx \\
&=\frac{e^{-\pi^{10}}}{(k+1)^6}
\end{align} $$
so that you will readily find that the integral diverges at least for $n\ge 6$.


问:
Reflexive, $s$, $t$ relations
$A=\{1,2,3,4\}$. Determine with reasons whether $R$ is reflexive, symmetric or transitive.

$R=\left\{(1,1),(1,2),(2,1),(2,2)\right\}$

How is this done?

Reflexive must contain every element to itself. Therefore it is not reflexive as there is no $(3,3)$ and $(4,4)$? Or is it the same as symmetric because $(1,2)$ and $(2,1)$ are elements of $R$.

Not transitive because there are not three elements with a transitive link e.g. $(1,2)$, $(2,3)$ and $(1,3)$.

答: You are right for symmetry and reflexivity, but wrong for transitivity. Indeed transitivity means: for every $(a,b)$ and $(b,c)$ in $A\times A$, $\textbf{IF}$ $(a,b)$ and $(b,c)$ are in $R$, then $(a,c)$ is in R. So not-transitive means that there exist $(a,b)$ and $(b,c)$ in $R$ such that $(a,c)$ is not in $R$. Now, can you find such $(a,b)$ and $(b,c)$ in $R$, with $(a,c)$ not in $R$? No, you can't, so $R$ is transitive.</p>

Title: Reflexive, $s$, $t$ relations


Body:
Almots the same as you did, maybe a bit clearer.

We have $\frac{n}{(n+1)!} = \frac{1}{n!}\cdot\frac{n}{n+1}=\frac{1}{n!}\cdot (1-\frac{1}{n+1}) = \frac{1}{n!}-\frac{1}{(n+1)!}$.

So
$$\sum\limits_{n=1}^{+\infty} \frac{n}{(n+1)!} = \sum\limits_{n=1}^{+\infty} \frac{(-1)^n}{n!} - \sum\limits_{n=1}^{+\infty} \frac{(-1)^n}{(n+1)!} = \sum\limits_{n=0}^{+\infty} \frac{(-1)^n}{n!} - 1 + \sum\limits_{n=0}^{+\infty} \frac{(-1)^n}{n!} -1+1$$
$$=2e^{-1} -1$$


问:
Transitive probabiliity
Let's say I have 3 objects that can belong to some type (color, size, whatever you like).
However, I'm not certain (drunk or blind, lol) whether or not each pair of objects belongs to the same type. The probability $p_{12}$ measures this uncertainty, namely, tells me how frequently the objects 1 and 2 are indeed of the same type.

My question is: what are the probabilities (which should sum up to 1) that all of the objects are of different type, encoded as $P_{123}$, two of the same type but the third is not, encoded as $P_{(12)3}$, $P_{1(23)}$ and $P_{(13)2}$, and that all of them are of the same type, i.e., $P_{(123)}$.

The problem here is that this "sameness" is actually an equivalence relation, which means that if 1~2 and 2~3, then definitely 1~3, which should affect the probabilities.

答:
This can not be determined from the given data.

To see this, consider two scenarios.

$I$.  Suppose that each object is one of three types, independently and uniformly.  In that case, $P_{ij}=\frac 13$ for $i\neq j$.  In that case, the probability that all three match is, of course, $\frac 19$.

$II$.   Suppose there are only two types, $H, T$ and that the objects might be $HHT, HTH, THH$ with equal probability.  In that case we again have $P_{ij}=\frac 13$ for $i\neq j$.  But now it is impossible for the three to match.

If it troubles you that we have three types in the first scenario but only two in the second, rewrite the second to have cases $AAB, BBC, CAA$, for example.  That version has the nice property that, as in $I$, each object is equally likely to be any of the three types (though, of course, the probabilities are far from independent).

Title: Transitive probabiliity


问:
Infinite sum of matrix traces
I am trying to work out the following sum where C is an n by n matrix:
$$S=\sum_{k=1}^\infty{\ln \left|e^{C^k} \right|}$$
The matrix $C$ has the property that $\operatorname{tr}(C^{2m-1})=0 $ when $m\in \mathbb{N}$ therefore I figured the sum should be equal to:
$$S=\sum_{k=1}^\infty{\ln \left|e^{C^{2k}} \right|}$$
Another property is that all the values in C are natural numbers and their sum is less than 2n.

From here I dont know where to go, but I tried the following:
$$S={\ln \left|\prod_{k=1}^{\infty} e^{C^{2k}} \right|}={\ln \left| e^{\sum_{k=1}^\infty{C^{2k}}} \right|}$$ Can I now treat the sum as a geometric series?

答:
You will obtain
$$
S=\sum_{k=1}^\infty \ln|e^{C^k}|=\sum_{k=1}^\infty {\rm Tr}(C^k)
=\sum_{k=1}^\infty {\rm Tr}(C^{2k})=\sum_{k=1}^\infty \sum_{i=1}^N \lambda_i^{2k}
=\sum_{i=1}^N\frac{\lambda^2_i}{1-\lambda^2_i}
$$
and a property
$$\sum_{i=1}^N \lambda_i^{2k+1}=0$$
where $N$ is the dimension of the square matrix $C$ , $\lambda_i$ are its eigenvalues. Without any more conditions, you cannot simplify it further.

Title: Infinite sum of matrix traces


问:
Calculation of the error for the weighted arithmetic mean
I am trying to derive the error for the weighted arithmetic mean
$$
\langle A \rangle=\frac{\sum_{i=1}^N \frac{A_i}{\sigma_{A_i}^2}}{\sum_{i=1}^N \frac{1}{\sigma_{A_i}^2}}.
$$
of a Data-Set $A_i$ with the errors $\sigma_{A_i}$.

With the gaußian propagation of uncertainty for a function $f(x_1,...,x_N)$
$$
\sigma_f=\sqrt{\sum_{j=1}^{N}\left(\frac{\partial f}{\partial x_j}\right)^2\sigma_{x_j}^2}\,\,\,\,\,\,(1)
$$
we get (here: $f\equiv\langle A \rangle$, $x_i\equiv A_i$) ($\sigma_i:=\sigma_{A_i}$, sums and products go from 1 to N):
$$
\sigma_{\langle A \rangle}\stackrel{(1)}{=}\sqrt{\sum_j\left(\frac{1}{\sigma_j\sum_i \frac{1}{\sigma_i^2}}\right)^2}=\sqrt{\sum_j\frac{1}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}}}\sqrt{\frac{1}{\sum_i \frac{1}{\sigma_i^2}}}\stackrel{(2)}{=}\sqrt{\frac{1}{\sum_i \frac{1}{\sigma_i^2}}}.
$$
This is the correct answer. So the first root must be equal to 1 $\forall \{\sigma_i|i=1,...,N\}$ (for all "error lists")(I have computed this in Python and it is correct!). I managed to bring the radicand on one denominator, but I still don't get to the conclusion I want:
$$
\sum_j\frac{1}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}}
    =\frac{1}{\prod_k \left( \sigma_k^2 \sum_i \frac{1}{\sigma_i^2}\right)}
    \sum_j \frac{\prod_k \left( \sigma_k^2 \sum_i \frac{1}{\sigma_i^2}\right)}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}}
    =\frac{\sum_j \prod_{k\neq j} \left(\sigma_k^2 \sum_i \frac{1}{\sigma_i^2} \right)}
    {\prod_k \left( \sigma_k^2 \sum_i \frac{1}{\sigma_i^2}\right)}
    \stackrel{?}{=}1.\,\,\,\,\,\,\, (2)
$$
Question: How do I show that Equation (2) is equal to 1?

答:
I have the answer now. I overcomplicated it by trying to put it on one denominator.
$$
\sum_j\frac{1}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}}=\sum_j\frac{1}{\sigma_j^2}\left(\sum_i \frac{1}{\sigma_i^2}\right)^{-1}=1
$$

OwnerDisplayName: Rodrigo


问:
Proving $\arctan x$ has no extreme points.

How can I prove the function $f\left(x\right) = \arctan\left(x\right)$ fulfils these requirements?
I was thinking about using supremum/infimum as I know the function is bounded by $-\frac{\pi}{2}\ ,\frac{\pi}{\ 2}$, but how do I show that they're not part of the image?

答:
Arctan is strictly increasing, so its sup and inf are its limits in $\pm \infty$, so $\pm \frac{\pi}{2}$.

But we can't have Arctan $x$ $=\pm\frac{\pi}{2}$, since $\tan \pm \frac{\pi}{2}$ is not defined.


问:
Calculate $I= \iiint\limits_{\Omega}2\,{\rm d}x\,{\rm d}y\,{\rm d}z$

Calculate
$$I= \iiint\limits_{\Omega}2\,{\rm d}x\,{\rm d}y\,{\rm d}z$$
with
$$\Omega: \left\{\begin{matrix}
\left ( \frac{x}{3} \right )^{2}+ \left ( \frac{y}{2} \right )^{2}+ z^{2} & \leq & 1\\
z & \geq & 0
\end{matrix}\right.$$

My solution is:

Let $u= x/3, v= y/2, w=z.$ Using Jacobi's formula with:
$$\left | J \right |= \begin{vmatrix}
{x_{u}}' & {x_{v}}' & {x_{w}}'\\
{y_{u}}' & {y_{v}}' & {y_{w}}'\\
{z_{u}}' & {z_{v}}' & {z_{w}}'
\end{vmatrix}= \begin{vmatrix}
3 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1
\end{vmatrix}= 6\Rightarrow I= \iiint\limits_{\Omega}2\cdot 6{\rm d}u{\rm d}v{\rm d}w= 12\cdot \frac{1}{2}V= 6\frac{4\pi r^{3}}{3}= 8\pi$$

But seems like this way's not useful if I replace $x^{2}/9$ by $x.$ So I want to see another way to solve this problem without Jacobi's formula. Maybe I can understand when we should use Jacobi's formula. I used these to prep for my test. I need to the help, thank you.

答:
Your $\Omega$ is half of the ellipsoid $E$ with semiaxes $3$, $2$, $1$. It follows that
$$\int_\Omega 2\>{\rm dvol}={\rm vol}(E)=3\cdot2\cdot1\cdot{4\pi\over3}=8\pi\ .$$


问:
The notation of the second fundamental theorem of Calculus
I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.

the way the theorem is described is:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$

and it was told that the meaning is that the derivative of an integral of a function is the function itself.

I don't get how you can get that from this. the expression that I would think suggests this is:

$\Large\frac{d}{dx}(\int f(x)\,dt) =  f(x)$

so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.

another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function.
this is also something I don't understand how the FToC2 suggests of?

to me, it seems like what this means:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$

is how a very small change in $x$ affects that area under $f(t)$ between $a$ (a constant) and $x$. how do I get from that to the right interpretation?

答:
The two are the same statements just expressed in different ways:

Use your first theorem as $\newcommand{\d}{\text{d}}$
$$\int_a^b f(x)\,\text{d}x=F(b)-F(a)$$
Set $b=x$ and take derivatives:
$$\frac{\text{d}}{\text{d}x}\int_a^xf(x)\,\d x=f(x)-0=f(x)$$
which is the mathematical expression of the quoted sentence of the theorem.

The point is, the antiderivative $F(x)=\displaystyle\int f(x)\,\d x$ attains a constant value for the second limit and is the same function at the first limit; the derivative of constant is zero, which produces the result.

You may wonder why it is expressed so indirectly.

The first and second theorem have many different expressions. For example, an other version of the second theorem states

If $f$ is a function and $c$ is any constant, then there exists an antiderivative $A$ such that $A(c)=0$ and is given by
$$A(x)=\int_c^xf(x)\,\d x$$

So, don't worry about the wording. Just understand the meaning!

Hope this helps. Ask anything if not clear :)

EDIT

You may wish to visit this as well.


问:
Why aren't the integrand $(-y\,dx+x\,dy)$ path independent?
Use Green's theorem to prove: if the vertices of the polygon in counterclockwise order are $(x_i,y_i)$, $i=1(1)n$ then the area of the polygon is
$$A=\frac 12 \sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$ $x_{n+1}=x_1, y_{n+1}=y_1$.

Now the statement of Green's theorem:
${\displaystyle \int_{\partial D}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dx\,dy}.$

It is clear that we have to take $L=-y,M=x$ the confusion occurs while calculating ${\displaystyle \int_{(x_i,y_i)}^{(x_{i+1},y_{i+1})}(L\,dx+M\,dy)}.$ I am telling that I got the answer by taking this path $L_i:(tx_{i+1}+(1-t)x_{i},ty_{i+1}+(1-t)y_{i})$.

I attempted to calculate the integration over the path "$L_i^1(x_{i},y_{i})$ to $(x_{i+1},y_{i})$" and then $L_i^2:(x_{i+1},y_{i})$ to $(x_{i+1},y_{i+1})$. I think that the integration

$${\displaystyle \int_{(x_i,y_i)}^{(x_{i+1},y_{i+1})}(L\,dx+M\,dy):=\int_{L_i}(L\,dx+M\,dy)=\int_{L_i^1}(L\,dx+M\,dy)+ \int_{L_i^2}(L\,dx+M\,dy)}$$ because the the integration is independent of path but

$$\int_{L_i^1}(-y\,dx+x\,dy)+ \int_{L_i^2}(-y\,dx+x\,dy)=x_{i+1}y_{i+1}+x_iy_i-2x_{i+1}y_i$$
and $$\int_{L_i}(-y\,dx+x\,dy)=x_{i}y_{i+1}-x_{i+1}y_i.$$

Can you please help me, why aren't they same?

答:
The issue is that the vector field $\vec{F} = \left<-y , x \right>$ isn't conservative.

We can show this using Clairaut's theorem: if there was some $f$ such that $f_x = -y$ and $f_y = x$ then $f_{xy} = -1$, but $f_{yx} = 1$. Therefore there can be no $f$ such that $\vec{F} = \nabla f$ and $\vec{F}$ is not conservative.

Also, this is to be expected because if the integral were path-independent, then your total area would always have to equal zero.


答:
In fact it has an infinity of solutions for any real value of $y=f(x)$ as may be seen on this picture by Alpha (nothing special happens near $0$ because of the equivalence to $x\ln(x)$) :

You may find solutions numerically by iterations.

For example using Newton-Raphson applied to $\log(x)\sin(x)=y$ we get the first solution illustrated (at the left) in the case $y=1$ :
$$x_{n+1}=x_n-\frac{\sin(x_n)\log(x_n)-y}{\cos(x_n)\log(x_n)+\sin(x_n)/x_n},\quad x_0=7$$
$$s_1\doteq 6.83056530451751861265\cdots$$

Title: How to find inverse of the function $f(x)=\sin(x)\ln(x)$

Body:
My friend asked me to solve it, but I can't.

If $f(x)=\sin(x)\ln(x)$, what is $f^{-1}(x)$?

I have no idea how to find the solution. I try to find

$$\frac{dx}{dy}=\frac{1}{\frac{\sin(x)}{x}+\ln(x)\cos(x)}$$

and try to solve it for $x$ by some replacing and other things, but I failed.

Can anyone help? Thanks to all.


问:
Papa Rudin - Prologue Theorem
My question concerns a step in the proof of the first theorem in Rudin's Real and Complex Analysis.

From Real and Complex Analysis (p. 2-3):

There exists a positive number $\pi$ such that $e^z = 1$ if
and only if $\frac{z}{2\pi i}$ is an integer.

Now the relevant part of the proof:

It follows that $e^{\pi i} = i^2 = -1$, $e^{2\pi i} = (-1)^2 = 1$, and then $e^{2\pi i n} = 1$ for
every integer n. Also: \begin{equation} \tag{10} e^{z+2\pi i} = e^z \end{equation}If $z = x + iy$, $x$ and $y$ real, then $e^z$ = $e^xe^{iy}$; hence $\lvert e^z \rvert = e^x$. If $e^z = 1$, we therefore must have $e^x = 1$, so that $x = 0$.

Now I understand everything up to this point. The next step is where I don't understand:

To prove that $\frac{y}{2\pi}$ must be an integer, it is enough to show that $e^y \neq 1$ if $0 < y < 2\pi$, by (10).

Why is this the case? If $\frac{y}{2\pi}$ is an integer then we have $e^{iy} = e^{2\pi i \frac{y}{2\pi}} = 1$.

答:
The author makes a proof by contraposition.

First, since $e^{y+2\pi i} = e^y$, you can suppose that $y\in [0,2\pi]$.

You want to prove that if $e^iy=1$ then $\frac{y}{2\pi}$ is an integer.

So its the same as proving that if $\frac{y}{2\pi}$ is not an integer (i.e. $0<y<2\pi$ since $y\in [0,2\pi]$) then $e^i y \neq 1$.

Title: Papa Rudin - Prologue Theorem


问:
Let $U=\Bbb{Z} \times (-\frac12,\frac12) = \{(m,y) \mid m\in \Bbb{Z}, |y| < \frac12 \}$. Show that $U$ is an open set in the metric space $(A, d_A).$

Let $A= \Bbb{Z} \times \Bbb{R} \subset \Bbb{R}^2$ and let $d_A$ be the Euclidean metric restricted to $A$. Let $U=\Bbb{Z} \times (-\frac12,\frac12) = \{(m,y) \mid m\in \Bbb{Z}, |y| < \frac12 \}$. Show that $U$ is an open set in the metric space $(A, d_A).$

My idea was to define $U_0 = \{(0,y) \mid |y| < \frac12\}$ and show that this is open and since an arbitary union of open sets are open I would have that $U$ is open since $U = \bigcup_{i\in \Bbb{Z}}U_i$. I cannot seem to be able to show that $U_0$ would be open.

If I pick a point $x=(0, x_0) \in U_0$ and construct a ball $B(x,r)$ with $r = \frac12-x_0 >0$ I should be using this to show that $B(x,r) \subset U_0$? I'm getting very confused about what should be a subset of what etc in order to get this to work...

答:
Let $(k,x)\in U$. Since $V:=(k-\frac{1}{4},k+\frac{1}{4})\times (-\frac{1}{2},\frac{1}{2})$ is open in $\mathbb R^2$, there is $\delta >0$ s.t.  $B_\delta (k,x)\subset V$. In particular, since $V\cap A\subset U$, $$B_{\delta }(k,x)\cap A\subset V\cap A\subset U,$$
and thus, $U$ is open.


问:
What are the number of ways to pack apples and oranges in 6 baskets with restrictions?
There are 6 baskets numbered 1,2,3,4,5,6. Each basket needs to be filled up with fruits of the same kind (either apple or oranges). These are the conditions which needs to be satisfied:

At least 1 basket contains apples

If there are more than one basket that contains apples, these apple baskets must be numbered consecutively

The total number of ways this can be done is?

My attempt:

I used combination with repetitions. Since there are 2 objects and 6 bins, you just plug in these values into the formula which is ${n+r-1 \choose r}$ with n bins and r objects, which is ${6+2-1 \choose 2}$ = 21.

I'm not sure if this is the correct answer, since it's too straightforward and didn't take into account the restrictions. I don't know how to approach it otherwise.

答:
If only one basket contains apples, then you have $6$ ways of choosing the basket.

If two apple baskets, then they are either $1,2$ or $2,3$ or $\dots$ or $5,6$, so $5$ ways.

If three apple baskets, then all of them must be consecutive, so $1,2,3$ or $2,3,4$ or $\dots$ or $4,5,6$ i.e $4$ ways.

Similarly, for $k$ apple baskets, there are $7-k$ ways, so the required number of ways are $$\sum_{k=1}^{6} (7-k) =21$$


答: $\mathbb{Z}$ is an open set of $\mathbb{Z}$ and $(-1/2,1/2)$ is an open set of $\mathbb{R}$. So the product is opened in $\mathbb{Z}\times \mathbb{R}$.

Body:
Let $A= \Bbb{Z} \times \Bbb{R} \subset \Bbb{R}^2$ and let $d_A$ be the Euclidean metric restricted to $A$. Let $U=\Bbb{Z} \times (-\frac12,\frac12) = \{(m,y) \mid m\in \Bbb{Z}, |y| < \frac12 \}$. Show that $U$ is an open set in the metric space $(A, d_A).$

My idea was to define $U_0 = \{(0,y) \mid |y| < \frac12\}$ and show that this is open and since an arbitary union of open sets are open I would have that $U$ is open since $U = \bigcup_{i\in \Bbb{Z}}U_i$. I cannot seem to be able to show that $U_0$ would be open.

If I pick a point $x=(0, x_0) \in U_0$ and construct a ball $B(x,r)$ with $r = \frac12-x_0 >0$ I should be using this to show that $B(x,r) \subset U_0$? I'm getting very confused about what should be a subset of what etc in order to get this to work...


问:
Given certain conditions, prove a function G(x) is always equal to 4.
Theres a question I've been having trouble with:

$G(x)$ is the function $|x+2|+|x-2|$. Show that if $-2<x<2$ then $G(x) = 4$.

Any help would be greatly appreciated. No calculus please :D

答:
For problems with absolute values, break into cases. Remember,
$$|y|=\begin{cases}\hphantom{-}y & \text{if }y\geq 0,\\ -y & \text{if }y<0.\end{cases}$$
Thus,
$$\begin{align*}
|x+2|&=\begin{cases}\hphantom{-(}x+2 & \text{if }x+2\geq 0,\\ -(x+2) & \text{if }x+2<0\end{cases}\\\\\\
&=\begin{cases}\hphantom{-}x+2 & \text{if }x\geq -2,\\ -x-2 & \text{if }x<-2\end{cases}
\end{align*}$$
and
$$\begin{align*}
|x-2|&=\begin{cases}\hphantom{-(}x-2 & \text{if }x-2\geq 0,\\ -(x-2) & \text{if }x-2<0\end{cases}\\\\\\
&=\begin{cases}\hphantom{-}x-2 & \text{if }x\geq 2,\\ -x+2 & \text{if }x<2.\end{cases}
\end{align*}$$
Putting it all together,
$$\begin{align*}
G(x)&=|x+2|+|x-2|\\\\\\
&=\begin{cases}
\hphantom{-}x+2+|x-2| &\text{if }x\geq -2=\begin{cases}\hphantom{-}x+2+x-2 &\text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}x+2-x+2 &\text{if }x\geq -2\textbf{ and }x<2
\end{cases}\\\\\\\\\\\\\\\\
-x-2+|x-2| &\text{if }x< -2=\begin{cases}-x-2+x-2 &\text{if }x<-2\textbf{ and }x\geq 2,\\
-x-2-x+2 &\text{if }x<-2\textbf{ and }x<2
\end{cases}
\end{cases}\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}4 & \text{if }x\geq -2\textbf{ and }x<2,\\
-4 & \text{if }x<-2 \textbf{ and }x\geq 2\; (\color{red}{\text{impossible}}),\\
-2x & \text{if }x<-2\textbf{ and }x<2
\end{cases}\\\\\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq 2,\\
\hphantom{-}4 & \text{if }-2\leq x<2,\\
-2x & \text{if }x<-2.
\end{cases}
\end{align*}$$</p>


问:
A question about what does $|A|^{<\omega}$ mean for a set
I was reading this answer https://math.stackexchange.com/a/346249/629594 and I am not really sure what $|A|^{<\omega}$ means (I am not so familiar with things that concern ordinals and cardinals). I think that it means the cardianlity of the set $A^k$, where $k$ is a finite cardinal. Am I right? Furthermore, could we write $|A|^{<|\mathbb{N}|}$ instead of $|A|^{<\omega}$?

答:
By $A^{< \omega}$ we mean the set of all finite sequences in $A$. By $|A|$ we mean the cardinality of $A$. Assuming the axiom of choice, which people usually do, this cardinality is represented by an ordinal: the least ordinal that is in bijective correspondence with $A$. So technically $|A|^{< \omega}$ would be the set of all finite sequences in that ordinal, as is mentioned in the comments.

Via the bijection between $A$ and $|A|$ you might as well think of $|A|^{<\omega}$ as the set of all finite sequences in $A$. The point of writing $|A|^{<\omega}$ instead of $A^{<\omega}$ is that we want to stipulate that we ultimately only interested in cardinalities.

This is also true in the context of the question you linked: ultimately they just want $|A|^{<\omega} = |A|$, so they are really only interested in the cardinality of $|A|^{<\omega}$. Perhaps it would be more accurate to write $||A|^{<\omega}|$, but that becomes messy so convention is to not do this. We could also write $|A^{<\omega}|$, but for some reason this does not seem to be standard.

Edit: I left some remaining points in your original question untouched. So it is not quite $A^k$ for some finite $k$, but rather $A^{<\omega} = \bigcup_{k < \omega} A^k$. The second point: yes $|A|^{<\omega} = |A|^{<|\mathbb{N}|}$, this would just be different notation for the same thing (since $|\mathbb{N}| = \omega$), but this notation is again a bit more messy and nonstandard.


问:
problems on Lebesgue integral

1) Given a measure space, $f$ a non-negative measurable function and $A$ in the $\sigma$-algebra such that $\mu(A)=0$, prove that $\displaystyle\int_{A} f\;d\mu=0$ .

My try:

$$0\leq \int_{A}f\;d\mu=\int fI_{A}\;d\mu\leq \int \sup{f(A)}I_{A}\;d\mu=\sup{f(A)}\int I_{A}\;d\mu=\sup{f(A)}\mu(A)=0\;,$$
then $\displaystyle\int_{A}f\;d\mu=0\,.$

2) Given a measure space, let $f$ be an integrable function. Show that $\displaystyle \int_{\{|f|>t\}} |f|\;d\mu\rightarrow 0$ as $t\rightarrow\infty$ .

My try: since $f$ is integrable, $|f|\lt \infty$ almost everywhere, $\mu (\{|f|>t\}) = \mu(\Omega)- \mu(\{|f|\leq t\})$, and as $t\rightarrow\infty$,  $\mu (\{|f|>t\})<\epsilon$, then
$$\int_{\{|f|>t\}} |f|\;d\mu=\int |f|I_{\{|f|>t\}}\;d\mu\leq\max\{|f|\}\int I_{\{|f|>t\}}=\max\{|f|\}\;\mu(\{|f|>t\})<\max\{|f|\}\;\epsilon\;.$$
Then it's done; I can take $\epsilon=\frac1t$.

Well, both demonstrations seem right for me. I will be thankful if someone tells me what's wrong on them.

Greetings.

答:
1) I think a more natural argument would be to first prove the result in the case where $f$ is a simple function, then extend via the monotone convergence theorem. That being said, I like your solution, although something about it seems weird.

2) This argument fails if $\max|f|$ doesn't exists; i.e. if $\sup|f| = \infty$. I recommend proving by contradiction: If the result fails, then there is $\epsilon>0$ and a sequence $t_1 < t_2 < t_3 <\cdots$ with $t_i \to \infty$ such that $$\int_{\{|f|>t_{i}\}} |f| \, du > \epsilon$$ for all $i$. Using the monotone convergence theorem, you can then show that $$\sum^{\infty}_{i=k} \int_{\{t_i < |f| \leq t_{i+1}\}} |f|\, du = \int_{\{|f|>t_k\}}|f|\, du >\epsilon$$ for every $k$, so that sum on the left-hand side must be infinite (for each $k$), contrary to $f$'s integrability.

Title: problems on Lebesgue integral


答:
Suppose $ f $ is continuous on $ I \subset\mathbb{R}$, let $ a,x_{0}\in I $.

Let $ \varepsilon >0 $, there exists some $ \eta >0 $, such that $ \left(\forall y\in\left(x_{0}-\eta,x_{0}+\eta\right)\cap I\right),\ \left\vert f\left(y\right)-f\left(x_{0}\right)\right\vert<\varepsilon$

We have for any $ x\in \left(x_{0}-\eta,x_{0}+\eta\right)\cap I $ : \begin{aligned}\left|\int_{a}^{x}{f\left(y\right)\mathrm{d}y}-\int_{a}^{x_{0}}{f\left(y\right)\mathrm{d}y}-\left(x-x_{0}\right)f\left(x_{0}\right)\right|&=\left|\int_{x_{0}}^{x}{\left(f\left(y\right)-f\left(x_{0}\right)\right)\mathrm{d}y}\right|\\&\leq\int_{\min\left(x,x_{0}\right)}^{\max\left(x,x_{0}\right)}{\left|f\left(y\right)-f\left(x_{0}\right)\right|\mathrm{d}y}\\ &\leq\left|x-x_{0}\right|\varepsilon\end{aligned}

Since $ x_{0} $ is arbitrary, for any $ x_{0}\in I $, we have : $$ \int_{a}^{x}{f\left(y\right)\mathrm{d}y}=\int_{a}^{x_{0}}{f\left(y\right)\mathrm{d}y}+\left(x-x_{0}\right)f\left(x_{0}\right)+\underset{\overset{x\to x_{0}}{}}{\scriptsize{\mathcal{O}}}\left(x-x_{0}\right) $$

Which means $ x\mapsto f\left(x\right) $ is the derivative of $ x\mapsto\int_{a}^{x}{f\left(y\right)\mathrm{d}y} $.

Title: The notation of the second fundamental theorem of Calculus

Body:
I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.

the way the theorem is described is:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$

and it was told that the meaning is that the derivative of an integral of a function is the function itself.

I don't get how you can get that from this. the expression that I would think suggests this is:

$\Large\frac{d}{dx}(\int f(x)\,dt) =  f(x)$

so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.

another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function.
this is also something I don't understand how the FToC2 suggests of?

to me, it seems like what this means:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$

is how a very small change in $x$ affects that area under $f(t)$ between $a$ (a constant) and $x$. how do I get from that to the right interpretation?


问:
Why is better to work with the spectrum of prime ideals than with the maximal one, for example in the definition of affine scheme.
When we have an algebraic variety we can identify the points of the variety with maximal ideals of the coordinate ring.

I would like to know why it is more natural to define the main structure of the theory of schemes, the affine scheme, with prime ideals and not with maximal ones.

When Grothendieck was creating theory of schemes, why did he decide to work with the normal spectrum instead of the maximal one?

(As you can see I dont have an strong background of Algebraic Geometry, I just want to have some intuition)

In which sense the schemes generalize the notion of variety and why is better to work with this notion?

答:
In this thread

https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961

you will find a construction of a locally ringed space $(X^m, \mathcal{O}_{X^m})$
from any scheme $X$ of finite type over a field $k$ or the ring if integers $\mathbb{Z}$. The topological space $X^m$ is by definition the space of closed points in $X$ with the induced topology. This construction allows you to speak of nilpotent elements without using prime ideals.

Here I include the post linked to above:

In this thread

https://mathoverflow.net/questions/55244/why-must-nilpotent-elements-be-allowed-in-modern-algebraic-geometry/378811#378811

you find the following construction: Let $k$ be a field or the ring $\mathbb{Z}$ of integers and let $X$ be a scheme of finite type over $k$. By this we mean $X$ has a finite open affine cover $X=\cup_{i=1}^n Spec(A_i)$ where $A_i$ is a finitely generated $k$-algebra for all $i$.

Definition 1. Let $X^m$ be the set of closed points in $X$ with $u:X^m \rightarrow X$ the canonical inclusion map and with the induced topology. Let $\mathcal{O}_{X^m}:=u^{-1}(\mathcal{O}_{X})$ be the topological inverse of $\mathcal{O}_X$.

It follows the pair $(X^m, \mathcal{O}_{X^m})$ is a locally ringed space. In the above thread I explain why $X^m$ is a locally ringed space that is "similar" to a "classical algebraic variety" as defined in Hartshornes book in chapter I as the set of zeros of an ideal in a polynomial ring in a finite set of variables. The difference is that Hartshorne starts with a fixed algebraically closed field $k$ and an ideal $I:=\{f_1,..,f_n\} \subseteq B:=k[x_1,..,x_n]$ in a polynomial ring $B$ in a finite set of variables.

Example: Hartshorne defines $V(I)$ as the "set of $n$-tuples" $\{t:=(t_1,..,t_n)\in k^n\}$ such that $f(t_1,..,t_n)=0$ for all poynomials in $I$. Hence Hartshornes $n$-tuples $t$ have their coefficients in the base field $k$. Since the field $k$ is algebraically closed it follows the set of maximal ideals $\mathfrak{m}$ in $X^m$ all have residue field $k$. Hence for an algebraically closed field $k$ it follows $X^m=V(I)$ if $X:=Spec(A)$ and $A:=B/I$. The above Definition 1 makes sense for any Hilbert-Jacobson ring $k$. You need the property that any prime ideal $\mathfrak{p}$ in $k$ is the intersection of maximal ideals.

Example. Let $A:=\mathbb{R}[x]$ and let $C:=Spec(A)$. It follows $C^m$ is the set of irreducible polynomials in $A$. A polynomial $p(x)$ in $A$ is irreducible iff $p(x)=x-r$ with $r\in \mathbb{R}$ or $p(x)=(x-z)(x-\overline{z})=x^2-4ax+a^2+b^2$ with $a,b\in \mathbb{R}$ and $b \ne 0$. In the above thread I explain how you may use nilpotent elements to Taylor expand sections of $\mathcal{O}_{C^m}(U)$. To explain Taylor expansion for sections of sheaves to students using real curves is "easier to understand" and "more intuitive". Sometimes students have problems understanding the field of complex numbers.

The structure sheaf $\mathcal{O}_{X^m}$ has the property

P1. $\Gamma(X^m, \mathcal{O}_{X^m})=i^{-1}(\mathcal{O}_X)(X^m):=lim_{X^m \subseteq U}\mathcal{O}_X(U) \cong \Gamma(X, \mathcal{O}_X)$,

hence $\mathcal{O}_{X^m}$ and $\mathcal{O}_X$ have the same global sections. Hence if $X:=Spec(A)$ it follows $\Gamma(X^m, \mathcal{O}_{X^m})=A$ and you recover the ring $A$ from the locally ringed space  $(X^m, \mathcal{O}_{X^m})$.

Example: If $S:=A[x_1,..,x_n]/I$ where $I$ is a homogeneous ideal and $A$ is a Hilbert-Jacobson ring we may define $X^m \subseteq X:=Proj(S)$. If $\mathcal{E}$ is any finite rank locally trivial  $\mathcal{O}_X$-module, we may define $\mathbb{P}(\mathcal{E}^*)^m \subseteq \mathbb{P}(\mathcal{E}^*)$.
If $I \subseteq A\otimes A$ is the ideal of the diagonal and if

$\mathcal{P}(l):=Spec(A\otimes A/I^{l+1})$

we may define $\mathcal{P}(l)^m$. It has the property that

$\Gamma(\mathcal{P}(l), \mathcal{O}_{\mathcal{P}(l)})=A\otimes A/I^{l+1}$.

The canonical surjective map $m: A\otimes A/I^{l+1} \rightarrow A$ gives a one-to-one correspondence between $\mathcal{P}(l)^m$ and $Spec(A)^m$. Hence $\mathcal{P}(l)^m$ has the same points as $Spec(A)^m$ but it has non-trivial nilpotent elements in the structure sheaf - it is a "classical algebraic variety" with a non-reduced structure sheaf.

Comment: "Of course, there are some problems with this approach: The class of all fields is not a set. Technically, we can limit ourselves to some very large set of "test fields". So this can be swept under the rug."

Remark: The ring $A$ can be an arbitrary finitely generated $k$-algebra and it can be non-reduced. Hence this gives a way of introducing nilpotent elements for "classical algebraic varietiwith this. Moreover you don't use prime ideals, hence the construction gives an answer to the original question: Yes you may do this if the base ring is  a finitely generated ring over a field or the integers $\mathbb{Z}$.

Example. Let $(A,\mathfrak{m})$ be a local ring and let $X:=Spec(A)$ with $X^m:=\{\mathfrak{m}\}$ the one-point space with inclusion map

$i: X^m \rightarrow X$

and structure sheaf $\mathcal{O}_{X^m}:=i^{-1}(\mathcal{O}_X)$. By definition

$\Gamma(X^m, \mathcal{O}_{X^m}):=lim_{\mathfrak{m}\in U}\mathcal{O}_X(U)\cong \mathcal{O}_{X,\mathfrak{m}} \cong A_{\mathfrak{m}}\cong A$

since $A-\mathfrak{m}$ consists of units. Hence the global sections of the structure sheaf of $X^m$ recover the ring $A$. It follows you recover the ring $A$ from the locally ringed space $(X^m, \mathcal{O}_{X^m})$, and this construction can be done for an arbitrary local ring.

Let $\phi: (A, \mathfrak{m})\rightarrow (B, \mathfrak{n})$ be a map of local rings ($\phi^{-1}(\mathfrak{n})=\mathfrak{m}$). It follows we get a map of ringed spaces

$(\phi^m, (\phi^m)^{\#}): (X^m, \mathcal{O}_{X^m})\rightarrow (Y^m, \mathcal{O}_{Y^m})$.

The map

$(\phi^m)^{\#}: \mathcal{O}_{Y^m}\rightarrow (\phi^m)_*\mathcal{O}_{X^m}$

recovers the map $\phi$. Here $X^m:=Spec(B)^m$ and $Y^m:=Spec(A)^m$.
Hence the ringed space $(X^m, \mathcal{O}_{X^m})$ recovers the local ring $(B, \mathfrak{n})$ and maps of local rings. As a topological space $X^m$ is not very interesting, but the
algebraic structure is captured by the structure sheaf $\mathcal{O}_{X^m}$.

Your question: "In which sense the schemes generalize the notion of variety and why is better to work with this notion?"

Answer: In general the notion "scheme" is better since it gives a definition valid for arbitrary commutative unital rings. The definition given above was a suggestion in response to a question on MO on the problem of defining algebraic varieties/schemes (including non-reduced schemes) without using prime ideals and where the definition was "more intuitive". With the definition given above it follows a point $x\in X^m$ corresponds to a solution of a system of polynomial equations over a field.

There were several suggestions, and one of these suggestions used "classes". With the definition given above you avoid using classes: You work with the "set of closed points" in a fixed scheme $X$, hence there is no non-standard set theory involved. The idea was that this definition would be easier to understand for students in algebraic geometry.


答:
Your answer is correct: to visualize this, let's use bars ($|$) to symbolize the baskets and plusses ($+$) to symbolize our possible choices.

We have six baskets, and we want to find how many ways we can select a set of consecutive baskets to put the apples in, given that there must be at least one basket with apples in it.

Consider the following arrangement of bars and plusses: $||+||+||$. We could take this to mean that the baskets inside the plusses will get apples and the rest will get oranges. The baskets inside the plusses will always be consecutive, so as long as we don't place our two plusses in the same location, we will meet both requirements.

There are $7$ possible spots for the plusses (we include the ends outside the bars) and we need to place $2$ plusses without replacement, and the order doesn't matter, so our final answer is $\binom{7}{2}$.

Title: What are the number of ways to pack apples and oranges in 6 baskets with restrictions?

Body:
There are 6 baskets numbered 1,2,3,4,5,6. Each basket needs to be filled up with fruits of the same kind (either apple or oranges). These are the conditions which needs to be satisfied:

At least 1 basket contains apples

If there are more than one basket that contains apples, these apple baskets must be numbered consecutively

The total number of ways this can be done is?

My attempt:

I used combination with repetitions. Since there are 2 objects and 6 bins, you just plug in these values into the formula which is ${n+r-1 \choose r}$ with n bins and r objects, which is ${6+2-1 \choose 2}$ = 21.

I'm not sure if this is the correct answer, since it's too straightforward and didn't take into account the restrictions. I don't know how to approach it otherwise.


问:
Finding Periodic/Fixed Points in the Julia Sets close to the Period-3 Cardioid
The first image below shows the Julia Set at $-1.749512 + 0i$ (close to the base of the Period-3 Cardioid), and I'd like to find the periodic point located at where the white arrow is pointing at. However, this is my first exposure to Julia Sets beyond the Period-1 Cardioid and the Period-2 Main Disk of the Mandelbrot, and my methods of calculating periodic points don't seem to work here.

Working off my question here, I concluded that I needed to use $\frac{1 - \sqrt{1 - 4c}}{2}$ to find the desired point - and this worked as intended for all Julia Sets close to the Period-1 and 2 components of the Mandelbrot. Applying that to this Julia Set, however, centers at the white arrow in the second image (I understand why this is, but I don't know how to work around it).

I also tried using the following formula for finding pre-images of periodic points (which I learned from this question), but the found point never falls on the center, regardless of the combination of pluses and minuses I use...

$$z_{-1}^*(t)=\pm\sqrt{z_0^*(t)-c(t)}=\pm\sqrt{\frac{1\mp\sqrt{1-4c(t)}}{2}-c(t)}$$

Edit:

Thanks to Mark in the comments, I figured it might be beneficial if I just show what I'm looking for - given that I don't quite know which terms to use (although I'm pretty confident I'm looking for fixed points, based off of this question). Here are some videos showing the intended effect on the main cardioid and disk of the Mandelbrot Set (using the algebraic solutions to $z_{p+1}=z_p^2+c$), and what I'd like to replicate with the period-3 cardioid:

Main Cardioid ($z_0=\frac{1 - \sqrt{1 - 4c}}{2}$ - THIS is what I'd like to replicate with the Period-3 Cardioid): https://youtu.be/C22P6ScEU1I

Main Disk ($z_0=\frac{1 - \sqrt{1 - 4c}}{2}$ - NOT what I'm looking for): https://youtu.be/MUuuf8K4SiQ

Main Disk ($z_1=\frac{-1 - \sqrt{-3 - 4c}}{2}$): https://youtu.be/3RuNjWLHuWA

Period-3 Cardioid (THIS is the video I'd like to apply the previous effect on): https://youtu.be/G1tkBMshd2k

Judging by the trend in the videos, I'd like to think that finding the algebraic solution to $z_{p+1}=z_p^2+c$ when $p=2$ would get me the points I'm looking for, but given that Wolfram Alpha has trouble with this (or I'm just not understanding how I can use the answer), I'm unsure of how to get a solution I can work with:

https://www.wolframalpha.com/input/?i=solve+for+z%3A+%28%28%28%28%28%28z%5E2%29+%2B+c%29%5E2%29+%2B+c%29%5E2%29+-+z+%2B+c%29

答:
For the promenade around the main cardioid, the point in the Julia set you want is a fixed point $z = f_c(z)$.  By tuning (renormalization), the point you want in the "little Julia set" at the center of the Julia set is a periodic point $z = f_c^p(z)$, in this particular case $p = 3$.  You can use Newton's method to find the root numerically, the degree is too high for symbolic solution via radicals.  You need a good initial guess for Newton's method to converge to the "correct" tuned fixed point (periodic point), as there is another one at the "tip" of the little Julia set (just as there two fixed points for $z = f_c(z)$), and moreover the fixed points $z = f_c(z)$ are also (unwanted) solutions for $z = f_c^p(z)$.

Here is a small C99 program using my mandelbrot-numerics library:

```<code>/*
gcc julia-spiral-center.c `PKG_CONFIG_PATH=${HOME}/opt/lib/pkgconfig pkg-config --cflags --libs mandelbrot-numerics` -lm -fopenmp -O3 -std=c99 -Wall -Wextra -pedantic
LD_LIBRARY_PATH=${HOME}/opt/lib/ ./a.out |
ffmpeg -framerate 25 -i - -pix_fmt yuv420p -profile:v high -level:v 4.1 -crf:v 20 -movflags +faststart julia-spiral-center.mp4 -y
*/

#include <complex.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

#ifndef M_PI
#define M_PI 3.141592653589793
#endif

// git clone https://code.mathr.co.uk/mandelbrot-numerics.git
// make -C mandelbrot-numerics/c/lib prefix=${HOME}/opt install
#include <mandelbrot-numerics.h>

double cnorm(double _Complex z)
{
  double x = creal(z);
  double y = cimag(z);
  return x * x + y * y;
}

int main(int argc, char **argv)
{
  // silence -Wunused-parameter
  (void) argc;
  (void) argv;
  // number of animation frames
  const int frames = 25 * 30 * 1;
  // parameters for Julia set graphics rendering
  const int iterations = 1024;
  const double escape_radius_squared = 1 << 24;
  const int width = 640;
  const int height = 360;
  unsigned char *pgm = malloc(width * height);
  // how far from the target component to place the moving c
  // 1.0 is the boundary of the component
  const double radius = 1.25;
  // period of the target component
  const int period = 3;
  // these don't need to be exact, they will be changed by m_d_interior()
  double _Complex z = 0;
  double _Complex c = -1.754;
  // maximum number of steps of Newton's method
  const int steps = 64;
  // needs a good initial guess to converge to the correct attractor
  double _Complex w = -0.1 - 0.01 * I;
  for (int frame = 0; frame < frames; ++frame)
  {
    // path around the exterior of the component
    double t = (frame + 0.5) / frames;
    double _Complex q = radius * cexp(2 * M_PI * I * t);
    // find c,z such that f_c^p(z) = z and d/dz f_c^p(z) = q
    m_d_interior(&z, &c, z, c, q, period, steps);
    // find w such that f_c^p(w) = w
    // use previous frame's w as initial guess for continuity
    m_d_attractor(&w, w, c, period, steps);
    // scaling for view
    double r = cabs(w);
    // render Julia set
    #pragma omp parallel for
    for (int j = 0; j < height; ++j)
    {
      for (int i = 0; i < width; ++i)
      {
        // view centered on w
        double _Complex z0 =
          ( ((i + 0.5) / width - 0.5) * 2 * width / height
            - I * ((j + 0.5) / height - 0.5) * 2 ) * r + w;
        // initialize derivative with pixel spacing
        double _Complex dz = 2 * r / height;
        // unescaped pixels will be grey
        pgm[j * width + i] = 128;
        for (int k = 0; k < iterations; ++k)
        {
          dz = 2 * z0 * dz;
          z0 = z0 * z0 + c;
          if (cnorm(z0) > escape_radius_squared)
          {
            // compute exterior distance estimate
            double de = 2 * cabs(z0) * log(cabs(z0)) / cabs(dz);
            // colour boundary based on distance estimate
            pgm[j * width + i] = 255 * tanh(de);
            break;
          }
        }
      }
    }
    // output PGM stream to stdout
    fprintf(stdout, "P5\n%d %d\n255\n", width, height);
    fwrite(pgm, width * height, 1, stdout);
    fflush(stdout);
  }
  // cleanup
  free(pgm);
  return 0;
}
</code>```

The source code for the two functions I used from the library can be browsed at:

m_d_interior.c

m_d_attractor.c

They are both applications of Newton's root finding method, in 2 complex variables and 1 complex variables respectively.

You can view the video rendered by this program.

Body:
For the promenade around the main cardioid, the point in the Julia set you want is a fixed point $z = f_c(z)$.  By tuning (renormalization), the point you want in the "little Julia set" at the center of the Julia set is a periodic point $z = f_c^p(z)$, in this particular case $p = 3$.  You can use Newton's method to find the root numerically, the degree is too high for symbolic solution via radicals.  You need a good initial guess for Newton's method to converge to the "correct" tuned fixed point (periodic point), as there is another one at the "tip" of the little Julia set (just as there two fixed points for $z = f_c(z)$), and moreover the fixed points $z = f_c(z)$ are also (unwanted) solutions for $z = f_c^p(z)$.

Here is a small C99 program using my mandelbrot-numerics library:

/*
gcc julia-spiral-center.c `PKG_CONFIG_PATH=${HOME}/opt/lib/pkgconfig pkg-config --cflags --libs mandelbrot-numerics` -lm -fopenmp -O3 -std=c99 -Wall -Wextra -pedantic
LD_LIBRARY_PATH=${HOME}/opt/lib/ ./a.out |
ffmpeg -framerate 25 -i - -pix_fmt yuv420p -profile:v high -level:v 4.1 -crf:v 20 -movflags +faststart julia-spiral-center.mp4 -y
*/

#include <complex.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

#ifndef M_PI
#define M_PI 3.141592653589793
#endif

// git clone https://code.mathr.co.uk/mandelbrot-numerics.git
// make -C mandelbrot-numerics/c/lib prefix=${HOME}/opt install
#include <mandelbrot-numerics.h>

double cnorm(double _Complex z)
{
  double x = creal(z);
  double y = cimag(z);
  return x * x + y * y;
}

int main(int argc, char **argv)
{
  // silence -Wunused-parameter
  (void) argc;
  (void) argv;
  // number of animation frames
  const int frames = 25 * 30 * 1;
  // parameters for Julia set graphics rendering
  const int iterations = 1024;
  const double escape_radius_squared = 1 << 24;
  const int width = 640;
  const int height = 360;
  unsigned char *pgm = malloc(width * height);
  // how far from the target component to place the moving c
  // 1.0 is the boundary of the component
  const double radius = 1.25;
  // period of the target component
  const int period = 3;
  // these don't need to be exact, they will be changed by m_d_interior()
  double _Complex z = 0;
  double _Complex c = -1.754;
  // maximum number of steps of Newton's method
  const int steps = 64;
  // needs a good initial guess to converge to the correct attractor
  double _Complex w = -0.1 - 0.01 * I;
  for (int frame = 0; frame < frames; ++frame)
  {
    // path around the exterior of the component
    double t = (frame + 0.5) / frames;
    double _Complex q = radius * cexp(2 * M_PI * I * t);
    // find c,z such that f_c^p(z) = z and d/dz f_c^p(z) = q
    m_d_interior(&z, &c, z, c, q, period, steps);
    // find w such that f_c^p(w) = w
    // use previous frame's w as initial guess for continuity
    m_d_attractor(&w, w, c, period, steps);
    // scaling for view
    double r = cabs(w);
    // render Julia set
    #pragma omp parallel for
    for (int j = 0; j < height; ++j)
    {
      for (int i = 0; i < width; ++i)
      {
        // view centered on w
        double _Complex z0 =
          ( ((i + 0.5) / width - 0.5) * 2 * width / height
            - I * ((j + 0.5) / height - 0.5) * 2 ) * r + w;
        // initialize derivative with pixel spacing
        double _Complex dz = 2 * r / height;
        // unescaped pixels will be grey
        pgm[j * width + i] = 128;
        for (int k = 0; k < iterations; ++k)
        {
          dz = 2 * z0 * dz;
          z0 = z0 * z0 + c;
          if (cnorm(z0) > escape_radius_squared)
          {
            // compute exterior distance estimate
            double de = 2 * cabs(z0) * log(cabs(z0)) / cabs(dz);
            // colour boundary based on distance estimate
            pgm[j * width + i] = 255 * tanh(de);
            break;
          }
        }
      }
    }
    // output PGM stream to stdout
    fprintf(stdout, "P5\n%d %d\n255\n", width, height);
    fwrite(pgm, width * height, 1, stdout);
    fflush(stdout);
  }
  // cleanup
  free(pgm);
  return 0;
}

The source code for the two functions I used from the library can be browsed at:

m_d_interior.c

m_d_attractor.c

They are both applications of Newton's root finding method, in 2 complex variables and 1 complex variables respectively.

You can view the video rendered by this program.


问:
$\mathbb{C\otimes_QC} $ is not Noether ring, but $\operatorname{Spec}(\mathbb{C\otimes_QC})$ is Noether as topological space
I once heard that $\mathbb{C}\otimes_\mathbb{Q}\mathbb{C} $ is not Noether ring, but $\operatorname{Spec}(\mathbb{C}\otimes_\mathbb{Q}\mathbb{C})$ is Noether as topological space.

This is probably a last in my memory, but I want to check whether this is true or not.
Thank you for your help.

答:
Let $$I_n = (\zeta_{3^n}\otimes 1-1\otimes \zeta_{3^n})\subset \mathbb{C}\otimes_\mathbb{Q}\mathbb{C}$$ It is a strictly increasing chain of ideals.

$$\{ P\in Spec(\mathbb{C}\otimes_\mathbb{Q}\mathbb{C}), I_n\subset P\}$$ is a strictly decreasing chain of closed subsets.


问:
Rank $r$ matrix $U$ such that $AU=UB$
Is the following equivalence true?

Let $A, B$ two $n \times n$ matrices with real entries and $1\leq r \leq n$. The following are equivalent:

$A, B$ share $r$ eigenvectors (counting multiplicity).

There exists a matrix $U$ (of size $n \times n$), of rank $r$ such that $AU=UB$.

In the case $r=1$, then for example one can take two eigenvectors ($x$ for $A$, $y$ for $B$), and let $U=xy^T$. It might be possible to prove the above equivalence by induction.

答:
The implication 2) => 1) is not always true.

Let $A=\text{diag}(1,1,0,0)$, $B=\text{diag}(2,3,0,0)$ and $U=\text{diag}(0,0,1,1)$

It is clear that rank(U)=2 and AU=UB=0

But A and B do not share 2 distincts eigenvalues.

Title: Rank $r$ matrix $U$ such that $AU=UB$


问:
Is my solution correct? Generating functions question: How many non-negative solutions does the equation $x_1+x_2+x_3+x_4+x_5+x_6=12$ have?
so we began studying this subject, and I tried solving this question:  How many non-negative and whole ($\in \Bbb Z$) solutions does the equation $x_1+x_2+x_3+x_4+x_5+x_6=12$ have?

I would like to know if my solution is correct.

For non negative solutions it means $0\leq x_1, x_2, x_3, x_4, x_5, x_6$, so I built the next function:

$A(x)=(1+x^1+x^2+x^3+...)^6$, and due to the formula of the sum of an infinite geometric sequence:

$A(x)= {1 \over (1-x)^6}= \sum_{n=0}^∞ {n+6-1 \choose 6-1}x^n$.

Since we need the sum of 12, we need the coefficient of $x^{12}$, so I place n=12, and got the number of solutions to be 6188.

Now this seemed a bit weird to me. Is it so?

Thanks for any input!

答:
I assume you look for integer non-negative solutions. Try to write your problem as follows:  represent $x_i$ as a set  of $1$ (i.e. if $x_1=3$, then write it as $111$.) Then put your signs of $+$. In the end, you have 12 symbols of $1$ ($x_i$ add to $12$) and 5 symbols of $+$ (you have 6 variables). Now, how many ways do you have to choose 5 symbols out of 12+5=17? Of course, $C_{17}^5=6188$.

For example, a solution $5+3+2+2+0+0=12$ looks like $11111+111+11+11++=12$.