问:
How do I show there exist $a,b \in \mathbb{R}$ and $A \in \textit{M}$ such that $f(x)=a \mathbb{1}_{A}(x)+b$?
Let $(X, \textit{M})$ be a measure space. Let $f:X \rightarrow \mathbb{R}$ be a measurable function that actually contains only two values. Prove that there exist $a,b \in \mathbb{R}$ and $A \in \textit{M}$ such that $f(x)=a \mathbb{1}_{A}(x)+b$?

My idea: I know that $f$ is a simple function iff $f$ is measurable and the range of $f$ is a finite subset of $\mathbb{C}$.

By this result, the function $f$ is a simple function.

I also know that $f$ is a simple function if $f$ is a linear combination of the characteristic function of a measurable set.

Can anyone suggest me how do I prove the existence here?

Body:
Let $u,v$ be the values of $f$, write $A=f^{-1}(u)$, $f(x)=(u-v)1_A(x)+v$.

If $x\in A, 1_A(x)=1$, $(u-v)1_A(x)+v=u-v+v=u=f(x)$ if $x$ is not in $A$, $1_A(x)=0$ and $(u-v)1_A(x)+v=v=f(x)$.

Write $u-v=a, v=b, f(x)=a1_A(x)+b$.


问:
Is there a way to manipulate a hyperbola like $\frac{x^2 - 4}{x^2 + 4}$ into a readily sketch-able form?
$\frac{2x}{2x-1}$ cannot be sketched but can be manipulated into  $\frac{2x-1+1}{2x-1}= 1+\frac{1}{2x-1}$ which can be easily sketched.However, is there a way to to this when situations like $\frac{x^2 - 4}{x^2 + 4}$ occur? The only method I know is the lengthy process of y and x int, asymptotes via limits, domain and range, odd or even etc.

答:
Why don't you do the same procedure:
$$\frac{x^2-4}{x^2+4}=1-\frac8{x^2+4}$$
The last term is a Lorentzian function. So this will be like a horizontal line at $y=1$, but has a dip around $0$ to $y=-1$.

ContentLicense: CC BY-SA 4.0


问:
What's the connection between Banach/Hilbert spaces and tools like power series, Fourier transforms etc.
I've learned, abstractly, about Banach and Hilbert spaces, and more concretely about $l^p$ and $L^p$ spaces. I also understand these ideas have something to do with a variety of tools that are useful in physics and engineering, like Power series, Fourier transforms, etc. But no one has really ever spelled out the connection to me.

This is probably a dumb question, but what's the connection?

答: For example, $L^2$ is a very natural domain for the fourier transform, because it's an isomorphism on that space. In other words, the fourier transform is linear, the fourier transform of every function in $L^2$ is also in $L^2$, every function in $L^2$ is the fourier-transform of some other function in $L^2$, and the $L^2$-norm of a function and its fourier transform is always the same. Oh, and it's one-to-one also.

ContentLicense: CC BY-SA 3.0


问:
Show dyadic transpose property $(a \otimes b)^T = b \otimes a$
I want to prove that

$$(a \otimes b)^T = b \otimes a$$ where $\otimes$ is the dyadic product

I think I should just apply the definition. So let $a,b,u,v$ vectors:

$$u \cdot (a \otimes b)^T v=(a \otimes b) u \cdot v = (b \cdot u) a \cdot v$$

Now I don't know how to arrange the terms in order to get the thesis. Any help would be really appreciated

答:
I guess you can go directly by calculations

Let $(a,b) \in \mathbb R^n \times \mathbb R^n$ Then the dyadic product will be as following :

$a \otimes b = \begin{bmatrix} a_1 b_1 && a_1 b_2 && ... && a_1 b_n \\ a_2 b_1 && a_2 b_2 && ... && a_2 b_n\\ . && . &&&&.\\. &&  && .&&. \\a_nb_1 && a_n b_2 &&...&& a_n b_n   \end{bmatrix}$ $........(1)$

The transpose will be

$(a \otimes b)^T =\begin{bmatrix} a_1 b_1 && a_2 b_1 && ... && a_n b_1 \\ a_1 b_2 && a_2 b_2 && ... && a_n b_2\\ . && . &&&&.\\. &&  && .&&. \\a_1 b_n && a_2 b_n &&...&& a_n b_n   \end{bmatrix}$

Since multiplication of scalars is commutative, then we can have $a_i b_j = b_j a_i$. Let's use this property inside the transposed matrix :

$(a \otimes b)^T =\begin{bmatrix} a_1 b_1 && a_2 b_1 && ... && a_n b_1 \\ a_1 b_2 && a_2 b_2 && ... && a_n b_2\\ . && . &&&&.\\. &&  && .&&. \\a_1 b_n && a_2 b_n &&...&& a_n b_n   \end{bmatrix} = \begin{bmatrix} b_1 a_1 && b_1 a_2 && ... && b_1 a_n \\ b_2 a_1  && b_2 a_2 && ... && b_2 a_n\\ . && . &&&&.\\. &&  && .&&. \\b_n a_1  && b_n a_2 &&...&& a_n b_n   \end{bmatrix}$ $........(2)$

If you compare $(2)$ with $(1)$ (which is the definition of a dyadic product), you'll notice that $(2)$ is nothing more than $ b \otimes a$.

Hence, the proof.

$(a \otimes b)^T = b \otimes a$ .


问:
If $f$ is entire and $f(3/n) = 6- \frac{10}{n} - \frac{20}{n^2}$, then find $f$ in explicit form.
I’m not really sure where to start with this problem. I think I might have to use something about the fact that all the zeroes of $f$ are isolated since it’s entire, but i’m not sure. Any help would be appreciated, thanks!

答:
First note that there's an "obvious" function that takes those values at $\frac{3}{n} \colon n \in \mathbb{N}$. You can work out what what that function should be with a bit of clever rearrangement. Let's call this function $g(z)$ and note that it is entire.

Then $f(z)-g(z)$ is an entire function that is zero at each $\frac{3}{n} \colon n \in \mathbb{N}$ (and also at $0$, by continuity). Using the result that you mentioned, $f(z)-g(z)$ is an entire function with non-isolated zeros so is zero everywhere.

Thus $g(z)$ is the only possible function satisfying this property.


答:
You may write $$f(\frac 3 n) = 6 - \frac{10} 3 \cdot \frac 3 n - \frac{20} 9 \left(\frac 3 n\right)^2.$$ This means that, if you define the polynomial function $g(z) = 6 - \frac{10}3 z - \frac{20}9 z^2$, then you have $f(z) - g(z) = 0$ for $z = \frac 3 n$.

Do you know a theorem that allows you to conclude?

Title: If $f$ is entire and $f(3/n) = 6- \frac{10}{n} - \frac{20}{n^2}$, then find $f$ in explicit form.

Body: I’m not really sure where to start with this problem. I think I might have to use something about the fact that all the zeroes of $f$ are isolated since it’s entire, but i’m not sure. Any help would be appreciated, thanks!


问:
what is wrong with Ptolemy's proof of the Euclid's fifth Postulate?
This snapshot is from Heath on Euclid Vol 1 page 205.

Here he is discussing how Ptolemy attempted to prove Euclid's fifth postulate:

Here is the text:

"Let AB, CD be parallel, and let FG meet them. I say that FG does not make the
interior angles on the same side greater than two right angles.

For, if the angles AFG, CGF are greater than two right angles, the
remaining angles BFG, DGF are less than two right angles.

But the same two angles are also greater than two right angles; for AF,
CG are no more parallel than FB, GD, so that, if the straight line falling on
AF, CG makes the interior angles greater than two right angles, the straight line
falling on FB, GD will also make the interior angles greater than two right angles.

But the same angles are also less than two right angles; for the four
angles AFG, CGF, BFG, DGF are equal to four right angles:
which is impossible."

First Ptolemy proves the above and then Euclid's postulate follows.

Then Heath says that Proclus refuted this by pointing out that when Ptolemy says: "for AF and CG are no more parallel than FB and GD, so that, if the straight line falling on AF, CG makes the interior angles greater than two right angles..." he is using the fact that we can only draw a unique parallel line to a given line through a given point and this fact is equivalent to Euclid's fifth (in Euclidean context) so Ptolemy is using what needs to be proved.

My question is this last bit. How does what Ptolemy say imply uniqueness of the parallel line?

答:
Ptolemy's argument is.

Suppose you have a parallel line where the angles on the left are less than two right angles and the angles on the right are more.  Now you can construct a line that has the same angles but mirror reversed, with the angles on the right the same less than two right angles and the angles on the left the same more.  By symmetry both lines are "equally" parallel.

Ptolemy assumes they can't both be parallel so neither are and so unless the angles are equal to two right angles the line won't be parallel.

But what he doesn't consider is ....  maybe they both CAN be parallel.


问:
Find the height drawn from the foot of interior bisector in an obtuse triangle
The problem is as follows:

A $\triangle ABC$ is obtuse on $\angle B$. The interior bisector $BM$ is
traced from point $B$ and as well the altitudes $AN$ and $CQ$
respectively. Assuming that $AN=8\,cm$ and $CQ=12\,cm$. Find the
length of the altitude traced from $M$ in triangle $\triangle{BMC}$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&8\,cm\\
2.&6\,cm\\
3.&5.4\,cm\\
4.&4.8\,cm\\
\end{array}$

The figure from below is the interpretation which I could conclude from reading the word problem, however that's it. I don't know what else can be established from there?.

I think that it requires, similarity or congruence but I cannot say for sure exactly where should it be applied. Please include a drawing or a diagram in the answer.
So far can this be solved relying only in Euclidean postulates?.

答:
First of all, we should all agree that $\overline{MH}=\overline{MG}$ since $\overline{MB}$ is an angle bisector of $\angle ABC$.

Now since $\overline{MH}\parallel\overline{AN}$, we know $\triangle MHC$ and $\triangle ANC$ are similar to each other by A.A.A. Hence we have $\frac{\overline{MH}}{\overline{AN}}=\frac{\overline{MC}}{\overline{AC}}$.

Similarly, since $\overline{MG}\parallel\overline{CQ}$, we know $\triangle AMG$ and $\triangle AQC$ are similar to each other by A.A.A. Hence we have $\frac{\overline{MG}}{\overline{CQ}}=\frac{\overline{MA}}{\overline{AC}}$.

Therefore

$\frac{\overline{MH}}{\overline{AN}}+\frac{\overline{MG}}{\overline{CQ}}=\frac{\overline{MC}}{\overline{AC}}+\frac{\overline{MA}}{\overline{AC}}=1\\
\Longrightarrow \frac{\overline{MH}}{8}+\frac{\overline{MG}}{12}=1\Longrightarrow \frac{1}8+\frac{1}{12}=\frac{1}{\overline{MH}}\Longrightarrow \overline{MH}=\color{red}{4.8\text{ (cm)}}$


问:
Show that every element in the commutator subgroup is a commutator when G/A is cyclic.

Let $A \unlhd G$ be abelian.

  1) If $x \in G$ is fixed, show that the map $\Phi: A \to A$ given by $\Phi_x(a)=a^{-1}x^{-1}ax = [a,x]$ is a group homomorphism.

  2) Suppose that $G/A$ is cyclic and generated by $xA$. Show that every element in the commutator subgroup $G'=[G,G]=< [x,y] \ | \ x,y \in G>$ is a commutator. (Hint: First show that the image of the map in (1) is a normal subgroup of G.)

For (1):

Let $x \in G$ be fixed. Consider $\Phi_x(ab) = (ab)^{-1}x^{-1}(ab)x$.

Since $A \unlhd G$ we have $x^{-1}(ab)x=ab \in A.$ Thus, $(ab)^{-1}x^{-1}(ab)x = (ab)^{-1}(ab)=1$.

Likewise, $\Phi_x(a) \cdot \Phi_x(b) = (a^{-1}x^{-1}ax)(b^{-1}x^{-1}bx)$.

Since $A \unlhd G$ and $a,b \in A$ we have $x^{-1}ax = a$ and $x^{-1}bx = b$. Hence, $(a^{-1}x^{-1}ax)(b^{-1}x^{-1}bx)=a^{-1}(a)b^{-1}(b)=1$. Therefore $\Phi_x(ab) = \Phi_x(a) \cdot \Phi_x(b)$ as required.

For (2):

Now assume $G/A$ is cyclic and generated by $xA$.

First, we will show the image of the map above is a normal subgroup of G.

Notice $\Phi_x (A) = A^{-1}x^{-1}Ax = A^{-1}A = 1$. Therefore, $A^{-1}x^{-1}Ax = 1 \Rightarrow x^{-1}Ax = A$, as desired.

Now I'm not sure how to reach my conclusion. Also, let me know if what I have here so far is accurate. Thanks!

答:
This is an old thread, but nonetheless:
Continuing from Ben Grossmann's response, I proved the result in the following two steps:

1. Show $[x^k,a]\in\mathrm{im}(\Phi)$ for all $a\in A$ and $k\in\mathbb N$.

This is an easy induction (maybe this result is also trivial ... it's late while I'm writing this, so this is how it's going to be).

2. For $g,h\in G$, show $[g,h]\in\mathrm{im}(\Phi)$.

Write $g=x^ka$ and $h=x^\ell b$ with $a,b\in A$ (such a representation comes from $G/A=\langle xA\rangle$).

Disclaimer: I used the definition $[g,h]=ghg^{-1}h^{-1}$ for commutators, which is not the convention the question uses, but again, it's very late and partially I'm writing this here because I know I'm going to need it in a few weeks and would probably forget otherwise. It should be easy to convert my computations to the other convention.

We have $$[g,h]=ghg^{-1}h^{-1}=x^kax^\ell ba^{-1}x^{-k}b^{-1}x^{-\ell}=x^kx^{\ell}x^{-\ell}ax^\ell ba^{-1}x^{-k}b^{-1}x^kx^{-k}x^{-\ell}$$
Since $A$ is abelian, we have $ba^{-1}=a^{-1}b$, so this is equal to $$x^{k+\ell}(x^{-\ell}ax^\ell a^{-1})(bx^{-k}b^{-1}x^k)x^{-(k+\ell)}=x^{k+\ell}[x^{-\ell},a][b,x^{-k}]x^{-(k+\ell)}=x^{k+\ell}[x^{-\ell},a][b,x^{-k}]x^{-(k+\ell)}\\=x^{k+\ell}[x^{-\ell},a][x^{-k},b]^{-1}x^{-(k+\ell)}$$
Note that the two commutators are contained in $\mathrm{im}(\Phi)$ and that $\mathrm{im}(\Phi)$ is a normal subgroup of $G$, so conjugating this with $x^{k+\ell}$ still gives an element in $\mathrm{im}(\Phi)$.

This is the desired result.


答:
First, let $x$ be the length of the altitude traced from $M$ in triangle $\triangle BMC$, and name $R$ the point where this altitude intersects $\overline{BC}$. If we look at $\triangle BMA$, let $y$ be the length of the altitude traced from $M$ to $\overline{AB}$, which intersects this segment at the point $S$. We can see that $x=y$, because $\triangle MRB \cong \triangle MSB$ (using AAS rule).

Looking at $\triangle ANC$ and applying Thales' theorem, we can state that $$\frac{\overline{MR}}{\overline{AN}} = \frac{\overline{MC}}{\overline{AC}} \iff \frac{x}{8} = \frac{\overline{MC}}{\overline{AC}}.$$

Analogously for $\triangle AQC$, we get $$\frac{y}{12} = \frac{\overline{AM}}{\overline{AC}}.$$

Summing up both equations, we get $$\frac{x}{8} + \frac{y}{12} = \frac{\overline{AM} + \overline{MC}}{\overline{AC}}$$ $$\iff \frac{x}{8} + \frac{x}{12} = \frac{\overline{AC}}{\overline{AC}}$$ $$\iff \frac{5}{24}x = 1$$ $$\iff x = \frac{24}{5} = 4.8 \, \textrm{cm}.$$

Title: Find the height drawn from the foot of interior bisector in an obtuse triangle

Body:
The problem is as follows:

A $\triangle ABC$ is obtuse on $\angle B$. The interior bisector $BM$ is
traced from point $B$ and as well the altitudes $AN$ and $CQ$
respectively. Assuming that $AN=8\,cm$ and $CQ=12\,cm$. Find the
length of the altitude traced from $M$ in triangle $\triangle{BMC}$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&8\,cm\\
2.&6\,cm\\
3.&5.4\,cm\\
4.&4.8\,cm\\
\end{array}$

The figure from below is the interpretation which I could conclude from reading the word problem, however that's it. I don't know what else can be established from there?.

I think that it requires, similarity or congruence but I cannot say for sure exactly where should it be applied. Please include a drawing or a diagram in the answer.
So far can this be solved relying only in Euclidean postulates?.


问:
Duality between Morita contexts.
Definition: Let $A,B$ be rings. A Morita context connecting $A$ and $B$ is a sixtuple $(A,B,P,Q,f,g)$ where $P$ is an $(A,B)$-bimodule, $Q$ is a $(B,A)$-bimodule, $f: P \otimes_B Q \to A$ is an $(A,A)$-bimodule morphism and $g: Q \otimes_A P \to B$ is a $(B,B)$-module morphism such that $$q f(p\otimes _B q') = g(q \otimes_A p) q'; \quad pg(q \otimes_A p') = f(p \otimes_B q)p'$$
for all $p,p'\in P'$, $q,q' \in Q$.

Theorem: If $(A,B,P,Q,f,g)$ is a Morita context and $f$ is surjective, then we have the following properties:

(1) $f$ is bijective

(2) $P$ (seen as left $A$-module) and $Q$ (seen as right $A$)-module are finitely generated and projective.

(3) $P \cong _{B}\operatorname{Hom}(Q,B)$ as $(A,B)$-bimodules.

(4) $A\cong \operatorname{End}_B(P)$ as rings

and even more properties hold but I won't write them all down.

Now, if $g$ is surjective, then I expect the following properties to hold:

(1) $g$ is bijective

(2) $Q$ (seen as left $B$-module) and $P$ (seen as right $B$)-module are finitely generated and projective.

(3) $Q \cong _{A}\operatorname{Hom}(P,A)$ as $(B,A)$-bimodules.

(4) $B\cong \operatorname{End}_A(Q)$ as rings

As  mathematicians, we are encouraged to work smart and not too hard. Is there a way to prove these results from the corresponding result about $f$? Is it as simple as observing that if $(A,B,P,Q,f,g)$ is a Morita context, then also $(B,A,Q,P,g,f)$ is a Morita context and we can apply the statement there, or are there subtleties involved that I am not seeing?

答:
Yes, it is as simple as that: Morita context is a symmetric notion.

Just some further notes:

The ingredients of a Morita context give rise to a single ring operation, namely we can consider the generalized matrix ring
$$\pmatrix{A&P\\Q&B}$$
with multiplication given by multiplications in $A$ and $B$, the module actions, and $f$ and $g$.

Moreover, the category of Morita contexts of rings with $1$ is equivalent to the category of 'idempointed' rings $(R,e)$ where $R$ is a ring and $e\in R$ is an idempotent.

The correspondence is given by
$$A=eRe,\ P=eRe',\ Q=e'Re,\ B=e'Re'$$
where $e'=1-e$.

The reverse of $(R,e)$ in this setting is $(R,\,1-e)$.

Title: Duality between Morita contexts.


答:
$\triangle ANB \sim \triangle CQB$ by $AA$ similarity with ratio $$r=\frac{8}{12}=\frac{2}{3}$$

It follows

$$  \dfrac{AB}{CB} = \dfrac{2}{3} = \dfrac{AM}{CM}$$

as angle bisector divides the base into ratio of adjacent sides.

And as $MH || AN$, $$  \dfrac{MH}{AN} = \dfrac{CM}{AC} = \dfrac{CM}{AM + CM} =\dfrac{3}{2 + 3}=\dfrac{3}{5}$$

$$\therefore \boxed{MH =\dfrac{3}{5}AN =4.8}$$


问:
Prove that $A \mathop \triangle C \subseteq (A \mathop \triangle B)\cup (B \mathop \triangle C)$
Suppose A, B, and C are sets, prove that $A \mathop \triangle C \subseteq (A \mathop \triangle B)\cup (B \mathop \triangle C)$

I'm just wondering if this proof is ok, or if I'm overlooking something, thanks!

Proof. Suppose $x \in A \mathop \triangle C$. If $x \in A \mathop \triangle B$, then clearly $x \in (A \mathop \triangle B) \cup (B \mathop \triangle C)$, so we will consider the case when $x \notin A \mathop \triangle B$. Suppose $x \notin A \mathop \triangle B$. This means that $x \in A \leftrightarrow x \in B$. Since $x \in A \mathop \triangle C$, either $x \in A \mathop \backslash C$ or $x \in C \mathop \backslash A$. We will consider both cases.

Case 1: $x \in A \mathop \backslash C$. So $x \in A$ and $x \notin C$. Since $x \in A$ and $x \in A \leftrightarrow x \in B$, it follows that $x \in B$. Then since $x \notin C$, $x \in B \mathop \backslash C$, so $x \in B \mathop \triangle C$. Hence, $x \in (A \mathop \triangle B)\cup (B \mathop \triangle C)$

Case 2: $x \in C \mathop \backslash A$. So $x \in C$ and $x \notin A$. Then since $x \notin A$ and $x \in A \leftrightarrow x \in B$, it follows that $x \notin B$. Thus, $x \in C \mathop \backslash B$, so $x \in B \mathop \triangle C$, and hence, $x \in (A \mathop \triangle B)\cup (B \mathop \triangle C)$.

Edit: Also, I would be interested to see any alternative methods of proving this.

答:
Here is another alternative proof, in my favorite proof format: the calculational Feijen(-Dijkstra-Scholten-Gries-etc.) format.  I will use the following definition of $\Delta$: $$x \in P \Delta Q \equiv x \in P \not\equiv x \in Q$$

Starting with the most complex side, and working on the element level, we try to transform $x \in (A \Delta B) \cup (B \Delta C)$ and work towards $A \Delta C$ using logic:
$$
\begin{align}
& x \in (A \Delta B) \cup (B \Delta C) \\
\equiv & \;\;\;\;\;\text{"definition of $\cup$; above definition of $\Delta$, twice"} \\
& (x \in A \not\equiv x \in B) \lor (x \in B \not\equiv x \in C) \\
\equiv & \;\;\;\;\;\text{"logic: use negation of right side of $\lor$ in left side -- to bring $A$ and $C$ together"} \\
& (x \in A \not\equiv x \in C) \lor (x \in B \not\equiv x \in C) \\
\Leftarrow & \;\;\;\;\;\text{"logic: weakening"} \\
& x \in A \not\equiv x \in C \\
\equiv & \;\;\;\;\;\text{"above definition of $\Delta$"} \\
& x \in A \Delta C \\
\end{align}
$$
By the definition of $\subseteq$ this completes the proof.


问:
Unification: constant, variable, function
I have a small quesion:
Can we unify the following pairs?

p(X, X) and p(a, Z)

Here X is a constant, a is a variable and Z is again a constant.

I think this unification is not possible because we are substituting same constant X for two different identifiers one is 'a' and the other is Z.

What is the general rule for unification of variable and constant?

I think variable can't be substituted for constant but constant can be substituted for a variable.

There is another option i.e. to apply unification to a function

What is the constraint for unify a function with a constant and with a
variable?

One example is:

ancestor(X, Y) and ancestor(bill, father(bill))

X can be replaced for the variable bill and Y can be replaced for father(bill).

Please guide me if my concepts are correct.

Zulfi.

答:
In first-order unification, which you seem to be discussing, that is possible assuming X and Z are variables and a is a constant. I'd suggest you check carefully assumptions about constants, functions, predicates and variables, and do some examples, e.g. on paper with Martelli and Montanari's algorithm or using a logic programming system.

There are various unification algorithms. Unification of p(X, X) and p(a, Z) gives the substitution {a/X, a/Z}. One algorithm is by Martelli and Montanari. It is non-deterministic and involves transformations on multisets of equations, e.g. {p(X, X) = p(a, Z)} => {X = a, X = Z} by term decomposition then {X = a, a = Z} by variable elimination then {X = a, Z = a} by swapping so the variable is on the left side of the equation. It always terminates and produces a most general unifier when one exists that is unique up to renaming of variables.

A variable can be instantiated with a constant but not the converse. In the second example, bill and father(bill) are not variables.

Two functions can be unified if they have have the same function symbol and their subterms can be unified. With most unification algorithms, a function can be unified with a variable if the variable does not occur in the function. This condition is called the "occurs check" and prevents infinite terms, e.g, a substitution {f(X, a),/ X} would give the infinite term f(f(f(..... , a), a), ..., a).


问:
If $A$ has eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, what are the eigenvalues of $(I - A)^{3}$?
Suppose $A$ is a $n \times n$ symmetric real matrix with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, what are the eigenvalues of $(I - A)^{3}$?

Are they $(1 - \lambda_1)^3, (1 - \lambda_2)^3, \ldots, (1 - \lambda_n)^3$? If so, how can I arrive at this conclusion?

答:
If $Ax= \lambda x$ then
\begin{eqnarray*}
(I-A)x &=& (1-\lambda)x \\
(I-A)^3 x &=& (1-\lambda)^3 x. \\
\end{eqnarray*}


问:
Most efficient solution to find polynomial congruence for 0 mod p
I was given the polynomial $$f(x) = x^4 + 2x^3 + 3x^2 + x + 1$$ and told to find $$f(x) \mod 17 = 0 $$ I found the solution to be $$x = 8 + 17n$$ However, I arrived at this solution by computing all of the residues of f(x) mod 17 and then finding where the zero occurred. I was told by the person who gave me the problem that there's a more efficient solution that doesn't involve making a list. I'm quite new to number theory, so I don't know where to look to ask the question in a more advanced way, need guidance to be able to. Thank you kindly if you can.

TL;DR: Looking for a more number theoretic way to solve for x than computing f(x) from 1 to 17

答:
The fastest solution is probably by noticing $f(x) = (x^2 + x + 1)^2 - x$, after which you can apply Euler's Criterion to bound the order of $x \pmod {17}$ and then solve in four cases (either the order is 8, 4, 2, or 1). This solution is pretty similar to the solution given by Bill Dubuque. Here's another interesting solution:

Since $f$ is a polynomial, we can apply finite differences to get a recurrence. To make things nicer, let $a_n = f(n)$. The recurrence is:
$$a_n = 5a_{n-1} - 10a_{n-2} + 10a_{n-3} - 5a_{n-4} + a_{n-5}$$
This is solely based on the fact that $f$ is a fourth degree polynomial. Now we can quickly calculate $f(0)$, $f(1)$, $f(2)$, $f(3)$, $f(4)$:
$$f(0) = 1 \mod 17$$
$$f(1) = 1 + 2 + 3 + 1 + 1 = 8 \mod 17$$
$$f(2) = 16 + 16 + 12 + 2 + 1 = 47 = -4 \mod 17$$
$$f(3) = 81 + 54 + 27 + 3 + 1 = -4 + 3 - 7 + 3 + 1 = -4 \mod 17$$
$$f(4) = 256 + 128 + 48 + 4 + 1 = 1 -8 -3 + 4 + 1= -5 \mod 17$$

Now we can simply calculate the values up to 17 in a table using the recurrence. This isn't actually tedious if you stay organized (make a table with columns of $n$, $a_n$, $5a_n$, and $10a_n$) after which you can skip many multiplications because you've done them before and the only other operation needed is addition.

After this, you will see that 8 is the only value that yields zero mod 17.


问:
Metrizability of product spaces, Horst Herrlich, Topology I
In Horst Herrlich's Topology I, page 117, Statement 4.4.12 the author goes on to prove that product space is metrizable if each component ($\underline X_i$) of the product space is metrizable and the corresp. index set ($I$) is at most countable.

Assuming first the case when $I = N$ is infinitely countable, he then goes on to define such a metric (here each $d_n$ belongs to corresp. $X_n$)

$d((x_n),(y_n)) = max \ \{ \frac 1 n \  d_n(x_n, y_n) | n \in I \}$

If $x = (x_n)$ is element of the product space and $r > 0$, then $\exists m \in N, \forall n \ge m : \frac 1 n < r$

He then goes on to state that it follows:

$S(x,r) = \{ y \in {\displaystyle \prod_{I} X_n} \ | \ d(x,y) < r \ \} = \bigcap \{ p_n^{-1} [S(x_n, n \cdot r)] \ | \ n < m \} \ \ \ [1]$

Here $S(x_n, s) = \{ y \in X_n \ | \ d_n(x_n, y) < s \}$

I really tried to understand but I can't see how [1] just 'follows'. Can someone help in understanding why the equality really holds?

EDIT.
Forgot to mention that  $ diam \ \underline X_i \le 1$

答:
Suppose that $y\in\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}$; then $d_n(x_n,y_n)<nr$ for each $n<m$, so $\frac1n d_n(x_n,y_n)<r$ for each $n<m$. If $n\ge m$, then $\frac1n<r$, so $\frac1n d_n(x_n,y_n)\le\frac1n\cdot1<r$. (You didn’t mention it, but at this point he is clearly assuming that the metrics $d_n$ are bounded by $1$.) Thus, $\frac1n d_n(x_n,y_n)<r$ for all $n\in\Bbb N$, and therefore $$d(x,y)=\max\left\{\frac1n d_n(x_n,y_n):n\in\Bbb N\right\}<r\,,$$ so $y\in S(x,r)$. (The fact that this set has a maximum also depends on the assumption that the metrics $d_n$ have a common upper bound.) This shows that

$$\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}\subseteq S(x,r)\,.$$

Now suppose that $y\in S(x,r)$. Then

$$d(x,y)=\max\left\{\frac1n d_n(x_n,y_n):n\in\Bbb N\right\}<r\,,$$

so $\frac1n d_n(x_n,y_n)<r$ for each $n\in\Bbb N$, and therefore $d_n(x_n,y_n)<nr$ for each $n\in\Bbb N$. In particular, $d_n(x_n,y_n)<nr$ for each $n<m$, so $y_n\in S(x_n,nr)$ for each $n<m$, and therefore $y\in\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}$. Thus,

$$S(x,r)\subseteq\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}\,,$$

so

$$\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n<m\right\}=S(x,r)\,.$$


答:
$\det(I-A-\mu I)=\det((1-\mu)I-A)\Rightarrow$ The eigenvalues $\mu_i$ of $I-A$ are the ones of the form $\mu_i=1-\lambda_i$ with $\lambda_i=1-\mu_i$ being an eigenvalue of $A$.

Now rename $B=I-A$ and consider $B^3$. Take $\mu_i$ an eigenvalue of $B$ and $v_i$ some eigenvector for $\mu_i$. If you consider $\mu^3$ then $B^3v_i=BBBv_i=BB\mu v_i=\mu BBv_i=\mu B\mu v_i=\mu^2B v_i=\mu^3v_i$, so $v_i$ is an eigenvector for $B^3$ with $\mu^3$ as eigenvalue.

We conclude the eigenvalues of $(I-A)^3$ are $(1-\lambda_1)^3,\ldots,(1-\lambda_n)^3$.

Title: If $A$ has eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, what are the eigenvalues of $(I - A)^{3}$?

Body:
Suppose $A$ is a $n \times n$ symmetric real matrix with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, what are the eigenvalues of $(I - A)^{3}$?

Are they $(1 - \lambda_1)^3, (1 - \lambda_2)^3, \ldots, (1 - \lambda_n)^3$? If so, how can I arrive at this conclusion?


问:
Riemann sum proof that if an integral of a non-negative function is $0$ than the function is $0$.
Basically what the title says. I'm aware of other proof techniques that can be used to prove the question.

Question: Suppose that $f$ is a continuous function on $[a,b]$ and that $f(x)≥0$ for all $x∈[a,b]$. Show that if $∫^b_af(x)=0$, then $f(x)=0$ for all $x∈[a,b]$.

Solution: Define an equidistant partition $\tau = \{x_k|k=0,\ldots,n\}$, $x_k=a+\frac{k(b-a)}{n}$, $\Delta x_k=\frac{b-a}{n}$.

My sum is equal to $$\sum^{n}_{k=1}f(x_k)\Delta x_k=\sum^{n}_{k=1}[f(a+\frac{k(b-a)}{n})\frac{b-a}{n}]=\frac{b-a}{n}\sum^{n}_{k=1}f(a+\frac{k(b-a)}{n})\stackrel{!}{=}0$$

Finally, my plan was to evaluate the limit of the sum and to show that if $n \rightarrow \infty$ then $f=0$ at every point of the interval $[a,b]$. Do I have to additionally prove that the function is constant and use this fact to factor it out of the sum? It seems very unelegant. Maybe I'm missing something.

答: I am not sure how much is going to go through without using a proof by contrapositive since any partition you make needs to have a finite number of elements and if you index a sum by the set $\{0,...,n\}$ we don't get information about every $f(x)$, only those $f(x_{k})$ where $k$ is in the index set $\{0,...,n\}$. If you suppose for some $x$ we have $f(x) > 0$ and then create a simple tagged partition where one of your partition elements will be $> 0$ and all the others will be $\geq 0$, then you will eventually get that $∫^b_af(x) > 0$. Your specific approach will not work because all you can prove is that those specific $f(x_{k}) = 0$ and also there is no limiting business for $n$ since $n$ is fixed from the start.


问:
When do projective modules give generators?
I am reading arXiv:math/0111066 [math.RA] and am stuck understanding the proof of Lemma 1.4.  The only part I don't follow is the claim:

If $B$ is a projective (right) module over a ring $R$ which is simple, then $B$ is a generator for $\operatorname{Mod}-R$.

I don't see how to leverage the hypothesis that $R$ is simple.  $B$ is the direct summand of a free module (and free modules generate) but why does that show that $B$ generates (and not the complement of $B$?)

Update: One possible reference is Lang's Algebra section XVII, 7.  Note that $B$ becomes a left $R'(B)=\operatorname{End}_R(B)$ module in a natural way.  Morita's Theorem says that $B$ is a generator if and only if the natural map $R\rightarrow \operatorname{End}_{R'}(B)$ is onto ($B$ is "balanced") and $B$ is finitely generated projective over $R'(B)$.  However, I have no idea how to verify this...

答:
I only know how to prove this if $B$ is finitely generated; fortunately in Lemma 1.4 this follows because $B$ is a direct summand (hence a quotient) of $P$ which is assumed to be finitely generated.

Given that, $B$ is a direct summand of some finite free module $R^n$, hence can be presented as the image of an idempotent $e \in M_n(R)$. In this blog post (skip down to "More explicit Morita equivalences") you can see a proof of the following (this should be standard but I don't know a reference and it was faster to just prove it):

Lemma: $\text{im}(e)$ is a generator iff $e \not \equiv 0 \bmod I$ for any proper two-sided ideal $I$ of $R$.

If $R$ is simple the only such ideal is $I = (0)$ so this condition is automatically satisfied for any nonzero idempotent $e$ and hence for any nonzero finitely generated projective module $B$.

(The blog post is mostly a discussion about different definitions of "generator," which are inequivalent in general but which basically all agree in $\text{Mod}(R)$ and more generally in any cocomplete abelian category. These equivalences allow me to use the weakest definition of generator, which is an object $B$ such that if $C$ is any nonzero object then there's a nonzero map $B \to C$.)


问:
Does the axes of $\mathbb R^n$ have the fixed-point property?
I was studying Croom's Principles of Topology and was asked to decide whether the set
$$A=\{(x_1,x_2)\in\mathbb R^2:x_1=0\text{ or }x_2=0\}$$
has the fixed-point property. I first thought about scaling and rotating, but they all had the origin as a fixed point. So I'm confused if I'm supposed to show that $A$ has the fixed-point property, or I should try harder to come up with a counterexample. Any help would be appreciated.

Fixed-point property. A space $A$ has the fixed-point property if any continuous function $f:A\to A$ has a point $x\in A$ such that $f(x)=x$.

答: Let $f : A \to A, f(x_1,x_2) = (x_1+1,0)$. This map does not have a fixed pioint.


问:
If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$
Matrix Question

If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!

Thanks for help in advance!

Body: $$BA=BAI=BA(ABAB)=B(AA)BAB=BIBAB=(BB)AB=IAB=AB$$


问:
Find all the subspaces of $l_2$ that reduces the right shift operator en $l_2$
Consider the Hilbert Space $l_2$.

Let the right and left shift operator $R$ and $L$ of $l_2$ to $l_2$ $$R(a_0, a_1, a_2,...)=(0, a_0, a_1, a_2,...).$$ $$L(a_0, a_1, a_2,...)=(a_1, a_2, a_3,...).$$

I need to find all the subspaces $W$ of $l_2$ that reduces to $R$ and $L$.

A subspace $W$ reduces to operator $T$ if $T(W)\subseteq W$ and $T(W^\perp)\subseteq W^\perp$.

Since that $R^\ast =L$ and $L^\ast=R$ and

$W$ reduces $T$ $\Leftrightarrow$ $T(W)\subseteq W$ and $T^\ast(W)\subseteq W$.

Then $W$ reduces $R$ if and only if $W$ reduces $L$.

Then I need to find the subspaces $W$ of $l_2$ so that $R(W)\subseteq W$ and $L(W)\subseteq W$ but i don´t have idea how to find this subspaces.

Thank you for your help.

答:
Lemma: The algebra of operators on $\ell^2$ generated by $\{I, L, R\}$ contains all operators whose matrix has
finitely many nonzero entries.

Proof: Denoting the canonical basis of $\ell^2$ by $\{e_n\}_{n=0}^\infty $,  notice that
$$
  p:= I-RL
  $$
is the projection onto the space spanned by $e_0$.  Moreover $R^mpL^n$ is the operator which sends all basis vectors
to zero, except for $e_n$, which is sent to $e_m$.  So the matrix of
$R^mpL^n$  coincides with $e_{m, n}$,  namely the matrix having all entries zero except for the $(m,n)$.th entry
which is 1.

This said, any operator whose matrix $A=(a_{i, j})_{i, j}$  has finitely many nonzero entries can be written as
$$
  \sum_{i,j=0}^N a_{i,j} e_{i,j} =  \sum_{i,j=0}^N a_{i,j} R^ipL^j.
  $$
QED.

Now, if $W$ is a subspace of $\ell^2$ reducing $L$, it is clear that it is invariant under any operator in the above
algebra, and hence also under any finite matrix by the Lemma.

In case $W$ is closed,  then clearly $W$ can only be $\{0\}$ or the whole $\ell^2$.

On the other hand, I suppose there must be a huge number of non-closed spaces invariant under both $R$ and $L$,   and the characterization of these might be
a difficult problem.


问:
Question on continuity and intermediate value theorem
Let $f : [0,1]\to ℝ$ be continuous  and such that $f(0)=f(1)$. Prove that there exists a point $c \in [0, \frac{1}{2} ]$ such that $f(c)=f(c+\frac{1}{2})$.

答:
Hints: suppose $g(x)=f(x+\frac{1}{2})-f(x)$. Now use intermediate value theorem.
Also you can prove it by contradiction suppose equality doesn't hold.


问:
A quick question about Einstein's summation
I am not familiar with Einstein's summation convention. On p.5 of the book Mathematics of Classical and Quantum Physics written by Bryon and Fuller, the authors wrote:
$$\mathbf x\cdot\mathbf y=x_iy_j\delta_{ij}\color{red}{=x_iy_j}=x_jy_j.$$
Why is the sum equal to $x_iy_j$? Since the expression does not contain any repeated indices, is it even a sum?

答:
Where an index is repeated twice, implicitly this means to sum on that index. $x_iy_j$ does not have any implied sum ... it means take the component $x_i$ and multiply it by the component $y_j$.

This could be used to define a matrix
\begin{eqnarray*}
A_{i,j}=x_i y_j.
\end{eqnarray*}


问:
Isomorphisms for a set of functions
I have the following question here.

Define functions $f_1,f_2,f_3 \in F$ by
$$f_1(x)=1,f_2(x)=\cos(x),f_3(x)=\cos\left(x+\frac{\pi}{4}\right)$$
and let $V=\text{Span}(f_1,f_2,f_3)$. Are the spaces $\mathbb{R}^3$ and $V$ isomorphic to each other? If so, provide an isomorphism $\varphi:\mathbb{R}^3 \rightarrow V$. Otherwise, explain why not.

I know the vectors are isomorphic since the dimension of $V$ and $\mathbb{R}^3$ are the same (Since $f_1,f_2,f_3$ are linearly independent, they can't be written in terms of each other so the dimension of $V$ is $3$ as well).

How do I find the actual isomorphism though? We didn't really do many examples of this in class. I know we need to find an invertible linear transformation but I am genuinely stuck on how to do that.

Any help would be greatly appreciated!

答:
Check that whether $f_1,f_2,f_3$ are linearly independent if yes then dimension of $V=3$.
Note that If $V$ and $W$ be finite dimensional vector space then $V\cong W.$
You can find this theorem in each book probably.
Good luck!

Title: Isomorphisms for a set of functions


问:
Finding axes of steepest and shallowest curvature at turning point of a quadratic form
I have a quadratic form, $f(\mathbf{x}) ={1\over 2}\mathbf{x}^{T}\mathbf{A}\mathbf{x} - \mathbf{bx} + c = 0$. The turning point, where $f'(\mathbf{x}) = \mathbf{A}\mathbf{x} - \mathbf{b} = 0,$ can be found using $\mathbf{x}_{\mathrm{T}} = \mathbf{A}^{-1}b$.

I need to find the axes of greatest and least curvature of this paraboloid at the turning point. Are these just the eigenvectors of $\mathbf{A}$ corresponding with the largest and smallest eigenvalue respectively?

The matrix A looks like a Hessian to my untrained eye, based on how it is calculated (as the weighted sum, for each vector in a set of vectors, of the outer product of the vector with itself), which maybe helps with measuring curvature?

答:
Eigenvectors give only the direction of principal axes (for central conics and quadrics) not the principal curvature directions.  There's a paper for finding Gaussian and mean curvatures for implicit surfaces.  The standard results for general quadrics are summarized as follows:

Using matrix representation,

\begin{align}
  0 &= F(x,y,z) \\
  & \equiv \frac{1}{2}
  \begin{pmatrix}
    x & y & z & 1 \\
  \end{pmatrix}
  \begin{pmatrix}
    a & h & g & i \\
    h & b & f & j \\
    g & f & c & k \\
    i & j & k & d
  \end{pmatrix}
  \begin{pmatrix}
    x \\ y \\ z \\ 1
  \end{pmatrix} \\
  \\
  \nabla F &=
  \begin{pmatrix}
    a & h & g & i \\
    h & b & f & j \\
    g & f & c & k
  \end{pmatrix}
  \begin{pmatrix}
    x \\ y \\ z \\ 1
  \end{pmatrix} \\
  \\
  K &= -\frac{1}{|\nabla F|^4} \det
  \begin{pmatrix}
    a & h & g & i \\
    h & b & f & j \\
    g & f & c & k \\
    i & j & k & d
  \end{pmatrix} \\
  \\
  H &=
  \frac{(\nabla F)^T
    \begin{pmatrix}
      a & h & g \\
      h & b & f \\
      g & f & c
    \end{pmatrix}
  (\nabla F)}{2|\nabla F|^3}-
  \frac{a+b+c}{2|\nabla F|} \\
  \\
  \kappa_{\pm} &= H \pm \sqrt{H^2-K}
\end{align}

The principal directions or lines of curvature are not easy to solve, please refer to this paper for your further interest.

For standard paraboloids, the lines of curvature are as follows:

Elliptic paraboloid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{2z}{c}$

$$\boldsymbol{x}=
\begin{pmatrix}
  \frac{a}{c} \sqrt{a^2-b^2} \sinh u \cos v \\
  \frac{b}{c} \sqrt{a^2-b^2} \cosh u \sin v \\
  \frac{a^2-b^2}{4c} (\cosh 2u-\cosh 2v)
\end{pmatrix}$$

Hyperbolic paraboloid $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=\dfrac{2z}{c}$

$$\boldsymbol{x}=
\begin{pmatrix}
  \frac{a}{c} \sqrt{a^2+b^2} \sinh u \cosh v \\
  \frac{b}{c} \sqrt{a^2+b^2} \cosh u \sinh v \\
  \frac{a^2+b^2}{4c} (\cosh 2u-\cosh 2v)
\end{pmatrix}$$

Analogous to quadratic polynomials:

\begin{align}
  \mathbb{A}^{T} &= \mathbb{A} \\
  f(\boldsymbol{x}) &=
  \boldsymbol{x}^{T} \mathbb{A} \, \boldsymbol{x}+
  \boldsymbol{b}^{T} \boldsymbol{x}+c \\
  &=
  (\boldsymbol{x}-\boldsymbol{h})^{T} \mathbb{A}
  (\boldsymbol{x}-\boldsymbol{h})+k \\
   \boldsymbol{h}
  &=
  -\frac{1}{2} \, \mathbb{A}^{-1} \boldsymbol{b} \\
 k &=
  c-\frac{1}{4} \,  \boldsymbol{b}^{T} \mathbb{A}^{-1} \, \boldsymbol{b} \\
  \nabla f(\boldsymbol{x}) &= 2\mathbb{A} \, \boldsymbol{x}+\boldsymbol{b} \\
  \nabla^{2} f(\boldsymbol{x}) &= 2\operatorname{tr}(\mathbb{A})
\end{align}

Note that $\boldsymbol{h}$ is the centre of central quadric (or conic) providing $\det \mathbb{A} \ne 0$.

For paraboloids, $$\det \mathbb{A}=0$$

See also another answer of mine for using Hessian matrix to find the curvature of superellipse.


问:
Does exist a real numbered sequence with a "countable number of limits"?
Does exist a sequence $\{a_n\}$ in the real numbers such that for every element $l$ in a infinite countable set $L$ there is a subsequence of $\{a_n\}$ that converges to $l$?
$ $

(context):
I really do not have much context about this. I just wanted to know if there existed such sequence and couldn't come up with an answer by my own.

答:
If for each of your countably many limits you have a sequence converging to that limit, you can construct $\{a_n\}$ in the form of a triangular array
$$\begin{array}{cccc} a_1 \\ a_2 & a_3 \\ a_4 & a_5 & a_6 \\ \vdots & \vdots & \vdots &  \ddots \end{array}$$
by populating each column with each of your sequences.


答: This is a perfect problem for proving the contrapositive.   Suppose $ f $ is non-negative but not identically zero, and show that the integral is positive.  To accomplish this, let $ x_0 $ be any point where $ f(x_0) = y_0 > 0 $.  By continuity, there exists $ \delta $ such that $ f(x) > y_0 / 2 > 0 $ for all $ x $ with $ |x - x_0| < \delta $.  Now consider any Riemann sum with $ \|\Delta\| < \delta $ and you can see that $ R_*(f, \Delta) > \delta y_0 / 2 $.  Hence the integral itself is greater than or equal to $ \delta y_0 / 2 $.  Does this argument work for you?

Title: Riemann sum proof that if an integral of a non-negative function is $0$ than the function is $0$.

Body:
Basically what the title says. I'm aware of other proof techniques that can be used to prove the question.

Question: Suppose that $f$ is a continuous function on $[a,b]$ and that $f(x)≥0$ for all $x∈[a,b]$. Show that if $∫^b_af(x)=0$, then $f(x)=0$ for all $x∈[a,b]$.

Solution: Define an equidistant partition $\tau = \{x_k|k=0,\ldots,n\}$, $x_k=a+\frac{k(b-a)}{n}$, $\Delta x_k=\frac{b-a}{n}$.

My sum is equal to $$\sum^{n}_{k=1}f(x_k)\Delta x_k=\sum^{n}_{k=1}[f(a+\frac{k(b-a)}{n})\frac{b-a}{n}]=\frac{b-a}{n}\sum^{n}_{k=1}f(a+\frac{k(b-a)}{n})\stackrel{!}{=}0$$

Finally, my plan was to evaluate the limit of the sum and to show that if $n \rightarrow \infty$ then $f=0$ at every point of the interval $[a,b]$. Do I have to additionally prove that the function is constant and use this fact to factor it out of the sum? It seems very unelegant. Maybe I'm missing something.


问:
Is it logical to say that $ 2\over 0$ $\ne$ $ 2\over 0$?
As I was doing a math exercice, I came across a question which I decided to prove by contrapositive. That required me to show that $ f(4-x)$ $\ne$ $f(x)$ - but in both cases the result was $ 2\over 0$.
So my question is; is it correct to say that $ f(4-x)$ $\ne$ $f(x)$ since $ 2\over 0$ is undefined?

答: I am assuming the mathematics you are studying keeps division by zero undefined. So then a statement like $2\over0$ $\neq$ $2\over0$ has no meaning as some of symbols involved have no meaning (depending on the mathematics you are dealing with). More context about the problem might be helpful.


问:
How to prove from ZFC that the intersection of any non-empty set exists?
It is an axiom of ZFC that $\cup x$ exists for any set $x$, where $\cup$ denotes union. But how does one prove that for any non-empty set $x$, the intersection $\cap x$ exists? We need $x$ to be non-empty, because $\cap \emptyset$ would be the set of everything, which does not exist in ZFC.

答:
Assuming $U = \bigcup x$ is defined and using the separation axiom we get
$$
I := \bigcap x, \qquad \forall z:z \in I \leftrightarrow (z \in U)\land (\forall y(y \in x \to z\in y)
$$
where the formula here is $\phi(x,y,z) =\forall y(y \in x \to z\in y)$


答:
It is a subset of $\cup x$. Recall that
$$\cup x=\{ z\mid \exists y(z\in y\wedge y\in x)\}.$$
The intersection is
$$\cap x=\{z\in \cup x\mid \forall y( y\in x\rightarrow z\in y)\}.$$
This is a set by the Axioms of Union and Separation.

In fact, this definition allows you define the intersection of an empty family; the issue is that it is equal to $\varnothing$ (since $\cup\varnothing = \varnothing$), so then you run into issues with associativity.

Title: How to prove from ZFC that the intersection of any non-empty set exists?

Body: It is an axiom of ZFC that $\cup x$ exists for any set $x$, where $\cup$ denotes union. But how does one prove that for any non-empty set $x$, the intersection $\cap x$ exists? We need $x$ to be non-empty, because $\cap \emptyset$ would be the set of everything, which does not exist in ZFC.


问:
Min cost max flow optimization problem
Let $I$ be the set of customers and let $J_i \subseteq J$ be the set of items from which customer $i \in I$ wants to buy only one, where $J$ are the items. Denote $w(i,j) \geq 0$ as the price customer $i$ is willing to pay for item $j$. Assume $|I| = |J| = n$.

I want to find an algorithm which computes the maximum number of customers that get an item at the minimum total cost. I'd like to do this using a min-cost max flow theorem. However, I cannot seem to succeed in constructing a graph for this problem where we can both minimize and maximize functions. Can anyone give a suggestion how I should make this graph?

答:
Besides the customer and items nodes, introduce a source node $s$, sink node $t$, and $|I|+|J|=2n$ edges, each with capacity $1$ and cost $0$:

$(s,i)$ for $i\in I$

$(j,t)$ for $j\in J$

Title: Min cost max flow optimization problem


问:
Asymptotic behavior of tail series $A_n=\sum_{m\ge n+1} \frac{n!}{m!}$ invoked by $e$
Recalling a series, denote
$$
A_n=\sum_{m=n+1}^{\infty} {\frac{n!}{m!}}
$$
which is relative to the tail of $e$, and sometime emerges in some well-known limits like
$$
\lim_{n\to\infty} n\sin(2\pi e n!) = 2\pi
$$
as $n\to\infty$, the very first order of $A_n$ is trivial, since
$$
\frac1{n+1}<A_n<\frac1{n-1}
$$
so
$$
A_n \sim \frac1{n} + o(n^{-1})
$$
when it comes to the higher order, a possible method I used is using the Euler–Maclaurin formula with reciprocal gamma function, which may be an arguably inconvenient approach for this 'simple' series. Here, may I ask for other quick methods to its asymptotic form, some first items of which I found is (may not correct)
$$
A_n \sim \frac1{n} - \frac1{n^3} + \frac1{n^4} + o(n^{-4})
$$
where the squared item is happened to be missing.

答:
We can rewrite $A_n$ in multiple ways
$$
\eqalign{
  & A_{\,n}  = \sum\limits_{n + 1\, \le \,m} {{{n!} \over {m!}}}  = \sum\limits_{0\, \le \,k}
 {{{n!} \over {\left( {n + 1 + k} \right)!}}}  =   \cr
  &  = \sum\limits_{0\, \le \,k} {{1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}}
  = \sum\limits_{0\, \le \,k} {n^{\,\underline {\, - \left( {k + 1} \right)\,} } }  =   \cr
  &  = {1 \over {n + 1}}\sum\limits_{0\, \le \,k} {{1 \over {\left( {n + 2} \right)^{\,\overline {\,k\,} } }}}
  = {1 \over {n + 1}}{}_1F_{\,1} \left( {\left. {\matrix{  1  \cr  {n + 2}  \cr  } \;} \right|\;1} \right) =   \cr
  &  = \Gamma \left( {n + 1} \right)\sum\limits_{0\, \le \,k} {{1 \over {\Gamma \left( {n + 1 + k + 1} \right)}}}
  = \;e\;\gamma \left( {n + 1,1} \right) \cr}
$$

where:

$n^{\,\underline {\,k\,} } ,\quad n^{\,\overline {\,k\,} } $ represent respectively the
Falling and Rising Factorial;

${}_1F_{\,1}$ is the Confluent Hypergeometric Function;

$\gamma(s,z)$ is the Lower Incomplete Gamma Function.

From the expression in the Rising Factorial, inverting $n$ into $1/z$ we get the asymptotics
$$
\begin{array}{l}
 \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}\quad \left| {\;z = } \right.\frac{1}{n}\quad  = \frac{1}{{\left( {\frac{1}{z} + 1} \right)^{\,\overline {\,k + 1\,} } }} =  \\
  = \frac{1}{{\left( {\frac{1}{z} + 1} \right)\left( {\frac{1}{z} + 2} \right) \cdots \left( {\frac{1}{z} + k + 1} \right)}} =  \\
  = \frac{{z^{\,\left( {k + 1} \right)} }}{{\left( {z + 1} \right)\left( {2\,z + 1} \right) \cdots \left( {\left( {k + 1} \right)\,z + 1} \right)}}\quad \left| {\,\left| {\,z\,} \right| < \frac{1}{{k + 1}}} \right.\quad  =  \\
  = z^{\,k + 1} \left( {\sum\limits_{0\, \le \,l_{\,1} \,} {\left( { - z} \right)^{\,\,l_{\,1} } } } \right)\left( {\sum\limits_{0\, \le \,\,\,l_{\,2} \,} {\left( { - 2\,z} \right)^{\,\,l_{\,2} } } } \right) \cdots \left( {\sum\limits_{0\, \le \,\,\,l_{\,k + 1} \,} {\left( { - \left( {k + 1} \right)\,z} \right)^{\,\,l_{\,k + 1} } } } \right) =  \\
  = z^{\,k + 1} \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s} \left( {\sum\limits_{\scriptstyle \left\{ {\begin{array}{*{20}c}
   {0\, \le \,l_{\,j} }  \\
   {l_{\,1}  + l_{\,2}  +  \cdots l_{\,k + 1}  = s}  \\
\end{array}} \right.  \atop
  \scriptstyle \, } {\prod\limits_{1\, \le \,j\, \le \,k + 1} {j^{\,\,l_{\,j} } } } } \right)\;z^{\,\,s} }  =  \\
  = z^{\,\,k + 1} \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s} \left\{ \begin{array}{c}
 s + \,k + 1 \\
 \,k + 1 \\
 \end{array} \right\}\;z^{\,\,s} }  = \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s - \,k - 1} \left\{ \begin{array}{c}
 s \\
 \,k + 1 \\
 \end{array} \right\}\;z^{\,\,s} }  \\
 \end{array}
$$

or more simply
$$
\eqalign{
  & {1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }} = n^{\,\underline {\, - \left( {k + 1} \right)\,} }  =   \cr
  &  = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\, - \left( {k + 1} \right) - s}
 \left[ \matrix{ - \left( {k + 1} \right) \cr   s \cr}  \right]\;n^{\,\,s} }  =   \cr
  &  = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\,k + 1 - s}
 \left\{ \matrix{ - s \cr   k + 1 \cr}  \right\}\;n^{\,\,s} }  =   \cr
  &  = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\,k + 1 - s}
 \left\{ \matrix{ s \cr   k + 1 \cr}  \right\}\;n^{\,\, - \,s} }  \cr}
$$

Thus
$$
\eqalign{
  & A_{\,n}  = \sum\limits_{0\, \le s\,} {\left( {\sum\limits_{0\, \le \,k\,\left( { \le \,s - 1} \right)}
 {\left( { - 1} \right)^{\,\,\,k + 1 - s} \left\{ \matrix{  s \cr  k + 1 \cr}  \right\}\;} } \right)n^{\,\, - \,s} }  =   \cr
  &  = \sum\limits_{0\, \le s\,} {{{c_{\,s} } \over {n^{\,s} }}}   \cr
  & c_{\,s}  = 0,1,0, - 1, - 1,2, - 9,9,50, - 267,413,2180, \cdots  \cr}
$$

But the Stirling Numbers are known to grow very rapidly.

Here in fact is a lin-log graph of the coefficients.

The expression in Incomplete Gamma instead gives
$$
\eqalign{
  & A_{\,n}  = \;e\;\gamma \left( {n + 1,1} \right) \sim   \cr
  &  \sim {{\Gamma \left( {n + 1} \right)e^{\,n + 1} } \over {\sqrt {2\pi } \left( {n + 1} \right)^{\,n + 3/2} }}
\left( {1 + {{11} \over {12\left( {n + 1} \right)}} - {{23} \over {288\left( {n + 1} \right)^{\,2} }}
 + O\left( {{1 \over {n^{\,3} }}} \right)} \right) \cr}
$$


问:
Does unbounded derivative imply No Lipschitz?
We are taught that bounded derivative implies lipschitz. Also, at times, derivative may not exist, yet lipschitz could hold.

I wonder if derivative exists but is unbounded, can we directly say that it is not lipschitz?

答: $$f(x) = x \sin\left(\frac1x\right) \text{ on } (0,1)$$ is a Lipschitz function with unbounded derivative


问:
Recovery of Parameters from Sum of Bernoulli Random Variables with Different Success Probabilities
If you have two Bernoulli random variables, $X$ and $Y$ with success probabilities $a$ and $b$, both independent of each other, and we define a third random variable $Z = X+Y$, is it possible to recover $min\{a,b\}$ and  $max\{a,b\}$ from samples from Z (i.e $Z_{1}, Z_{2}, ..., Z_{n}$)?

I know that it isn't possible to recover either a or b, since Z is defined by the sum of a and b, so I'm lead to believe that you couldn't recover the min or max as doing so would effectively identify/recover a and b.

答:
$Z$ takes values $0,1,2$ with respective probabilities $(1-a)(1-b), a(1-b)+(1-a)b, ab$.

Given i.i.d. samples $Z_1, \ldots, Z_n$, the log likelihood is
$$N_0 \log((1-a)(1-b)) + N_1 \log(a(1-b) + (1-a)b) + N_2 \log(ab)$$
where $N_i = \# \{k : Z_k = i\}$.

In principle you can try to choose $a$ and $b$ to maximize this likelihood, but I am not sure if this is a concave function, and taking derivatives is a little messy. Maybe you or someone else can try digging further.


问:
How do we find the critical points of the function $f(x) = (x-3)(x-19)^{3}$?
I have trouble finding critical points in $(x-3)(x-19)^{3}$ which has critical points of $19$ and $7$ found in the derivative $4(x−19)^{4}(x−7)$. I have only been able to find $19$ and $3$ from the derivative $(x-3)*3(x-19)^{2} + (x-19)^{3}$ but only $19$ was correct. How can I simplify the equation to find the other critical point?

Body:
To begin with, notice that

\begin{align*}
f(x) & = (x-3)(x-19)^{3}\\\\
& = (x - 3 - 16 + 16)(x-19)^{3}\\\\
& = (x-19)(x-19)^{3} + 16(x-19)^{3}\\\\
& = (x-19)^{4} + 16(x-19)^{3}
\end{align*}

Consequently,
\begin{align*}
f'(x) = 4(x-19)^{3} + 48(x-19)^{2} = 0 & \Longleftrightarrow 4(x-19)^{2}(x - 19 + 12) = 0\\\\
& \Longleftrightarrow (x = 19)\vee(x = 7)
\end{align*}

Hopefully this helps.


问:
Implicit partial derivatives
Say I have a function $F(P,V,T) = 0 $.
Using the chain rule: $$dF = \frac{\partial F}{\partial P}dP + \frac{\partial F}{\partial V}dV + \frac{\partial F}{\partial T}dT$$ Now, if P is constant, we have $dP = 0$ and $dF = 0$ of course, because $F$ is a constant (0). We get:
$$dF = \frac{\partial F}{\partial V}dV + \frac{\partial F}{\partial T}dT = 0 \implies (\frac{dV}{dT})_P = - \frac{\frac{\partial F}{\partial T}} {\frac{\partial F}{\partial V}}$$ And that is true generally? Correct me if i'm mistaken please. Now, if we have an explicit expression for $P(T,V)$ we can also say for constant P: $$ (\frac{dV}{dT})_P = - \frac{\frac{\partial P}{\partial T}} {\frac{\partial P}{\partial V}}$$
BUT the results aren't always the same, why is that? Do P,V and T have to be independent of each other for the first derivation to be correct? I'm really confused and will be glad for explanation without getting too rigorous.

Edit: Example (from physics):

Say:$$F(P,V,T)=PV-PNb+\frac{aN^{2}}{V}-\frac{aN^{3}b}{V^{2}}-Nk_{B}T=0$$

We can solve for P:

$$P = N\left(\frac{K_{B}T}{V-Nb}-\frac{aN}{V^{2}}\right)$$

We now get:
$$- \frac{\frac{\partial F}{\partial T}} {\frac{\partial F}{\partial V}} \neq - \frac{\frac{\partial P}{\partial T}} {\frac{\partial P}{\partial V}} $$

答:
Short Answer

They are the same, including in your example. You either did the algebra wrong or forgot one key step at the end.

Long Answer

Yes, your thoughts are correct. If you can write $P$ as a function, then for constant $P$ satisfying $F(P,V,T)=0$, we do have $\dfrac{\mathrm dV}{\mathrm dT}=-\dfrac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}=-\dfrac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$.

Example Case

In your particular example, $F(P,V,T)=PV-PNb+\frac{aN^{2}}{V}-\frac{aN^{3}b}{V^{2}}-Nk_{B}T=0$ we can indeed solve for $P$ to find $P=N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)$.

Then we can calculate the two fractions. For $-\dfrac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}$, we get something like $\dfrac{k_B(V-Nb)V^3}{k_BTV^3-2aN(V-Nb)^2}$. But for $-\dfrac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$ we get something like $\dfrac{k_BNV^3}{aN^2(2Nb-V)+PV^3}$, which looks different. But we have to remember to substitute in the constant value of $P$ from $P=N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)$. Then we get $\dfrac{k_BNV^3}{aN^2(2Nb-V)+N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)V^3}$, which simplifies down to the same $\dfrac{k_B(V-Nb)V^3}{k_BTV^3-2aN(V-Nb)^2}$.

You can verify the calculation with Wolfram/Mathematica code:

Wolfram Language (Mathematica), 458 bytes

```<code>Print[Simplify[Solve[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T == 0, P]]]
Print[Simplify[-D[N (-((kB T)/(b N - V)) - (a N)/V^2), T]/D[N (-((kB T)/(b N - V)) - (a N)/V^2), V]]]
Print[Simplify[-D[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T, T]/D[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T, V]]]
Print["Why Not the same? Substitute in for P."]
Print[Simplify[(kB N V^3)/(a N^2 (2 b N - V) + P V^3) /. P -> N (-((kB T)/(b N - V)) - (a N)/V^2)]]
</code>```

Try it online!

Title: Implicit partial derivatives

Body:
Short Answer

They are the same, including in your example. You either did the algebra wrong or forgot one key step at the end.

Long Answer

Yes, your thoughts are correct. If you can write $P$ as a function, then for constant $P$ satisfying $F(P,V,T)=0$, we do have $\dfrac{\mathrm dV}{\mathrm dT}=-\dfrac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}=-\dfrac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$.

Example Case

In your particular example, $F(P,V,T)=PV-PNb+\frac{aN^{2}}{V}-\frac{aN^{3}b}{V^{2}}-Nk_{B}T=0$ we can indeed solve for $P$ to find $P=N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)$.

Then we can calculate the two fractions. For $-\dfrac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}$, we get something like $\dfrac{k_B(V-Nb)V^3}{k_BTV^3-2aN(V-Nb)^2}$. But for $-\dfrac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$ we get something like $\dfrac{k_BNV^3}{aN^2(2Nb-V)+PV^3}$, which looks different. But we have to remember to substitute in the constant value of $P$ from $P=N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)$. Then we get $\dfrac{k_BNV^3}{aN^2(2Nb-V)+N\left(\dfrac{k_{B}T}{V-Nb}-\dfrac{aN}{V^{2}}\right)V^3}$, which simplifies down to the same $\dfrac{k_B(V-Nb)V^3}{k_BTV^3-2aN(V-Nb)^2}$.

You can verify the calculation with Wolfram/Mathematica code:

Wolfram Language (Mathematica), 458 bytes

Print[Simplify[Solve[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T == 0, P]]]
Print[Simplify[-D[N (-((kB T)/(b N - V)) - (a N)/V^2), T]/D[N (-((kB T)/(b N - V)) - (a N)/V^2), V]]]
Print[Simplify[-D[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T, T]/D[P*V - P*N*b + a*N*N/V - a*N*N*N*b/(V*V) - N*kB*T, V]]]
Print["Why Not the same? Substitute in for P."]
Print[Simplify[(kB N V^3)/(a N^2 (2 b N - V) + P V^3) /. P -> N (-((kB T)/(b N - V)) - (a N)/V^2)]]

Try it online!


Body:
Matrix Question

If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!

Thanks for help in advance!

Body:
$$
\begin{align}
\color{#C00000}{BA}
&=AA(\color{#C00000}{BA})BB\\
&=\color{#00A000}{A}(ABAB)\color{#00A000}{B}\\
&=\color{#00A000}{AB}
\end{align}
$$


问:
Finding conditional density
$X$ is an exponential random variable with $E[X] = \frac{1}{\lambda}$. $Y$ is an exponential random variable such that $E[Y | X = x] = \frac{1}{x}$. Find $f_{X|Y}(x|y)$.

I first find the joint density, which is I think is $f(x,y) = f_{Y|X}(y|x)\cdot f_X(x)$, meaning $f(x,y) = xe^{-xy} \cdot \lambda e^{-\lambda x}$. Then I would have to find the marginal density of $Y$, which is $\int_{0}^{\infty}f\left(x,y\right)dx$. Then, I would have to divide the joint density by the marginal density of $Y$ to get $f_{X|Y}(x|y)$.

Am I on the right track?

答:
Yes.

We have that $X\sim\mathcal{Exp}(\lambda)~$, $~Y\mid X\sim\mathcal{Exp}(X)~$.

Then

$$\begin{align}f_{\small X}(x)&=\lambda\,\mathrm e^{-\lambda x}\mathbf 1_{0\leqslant x}\\[3ex]f_{\small Y\mid X}(y\mid x)&=x\,\mathrm e^{-xy}\mathbf 1_{0\leqslant y~,~0\leqslant x}\\[3ex]f_{\small Y}(y)&=\int_\Bbb R f_{\small Y\mid X}(y\mid s)\,f_{\small Y}(y)\,\mathrm d s\\[1ex]&=\int_0^\infty s\lambda\,\mathrm e^{-(y+\lambda)s}\,\mathbf 1_{0\leqslant y}\,\mathrm d s\\[1ex]&= \lambda\,(y+\lambda)^{-2}\,\mathbf 1_{0\leqslant y}\\[4ex]f_{\small X\mid Y}(x\mid y) &= \dfrac{f_{\small Y\mid X}(y\mid x)\,f_{\small X}(x)}{f_{\small Y}(y)}\\[1ex]&~~\vdots\end{align}$$

Title: Finding conditional density


问:
In which condition，the triangle have the maximum triangle area？
These trangles have the same perimeter 2q，in which condition，can we have the trangle with the maximum area?
I have tried to use the Heron's formula. $S=\sqrt{q(q-a)(q-b)(q-c)}$
I have sought  the partial derivative and tried to find the stationary point but the calculation  is complicated. Is there any strategy can we use to simply the calculation?

答:
By AM-GM $$S\leq\sqrt{q\left(\frac{q-a+q-b+q-c}{3}\right)^3}=\frac{q^2}{3\sqrt3}.$$
The equality occurs for $q-a=q-b=q-c,$ id est, for the equilateral triangle, which says that we got a maximal value.


问:
One parameter solutions to the differential equation $y'(x) = -\frac{x}{1+y}\,e^{-x^2}$.
I've already worked out the solutions to it, 2 explicit functions of x:

y1 = $-1 + \sqrt{c1+1-e^{-x^2}}$

y2 = $-1 + -\sqrt{c1+1-e^{-x^2}}$

What I want to know is,
Can you express the general solution as a genuine one-parameter family of solutions (i.e., with the use of $\pm$) by using the sign-function sgn and the absolute value function?

Body:
$$2y^\prime(1+y)=-2xe^{-x^2}$$

$$\frac{d}{dx}(y^2 +2y) = \frac{d}{dx} e^{-x^2}$$

$$y^2 + 2y = e^{-x^2} +  \text{const.}$$

$$(y+1)^2 = e^{-x^2}+ \text{const.}$$

$$y=\mp\sqrt{e^{-x^2}+c}-1 $$


问:
If $a\equiv b \pmod{n}$ and the integers $a, b$, and $n$ are divisible by $d>0$ then $\frac{a}{d}\equiv\frac{b}{d} \pmod{\frac{n}{d}}$
this is related to my homework right now, i just don't have someone to ask

If $a\equiv b \pmod n$ and the integers $a, b$, and $n$ are divisible by $d>0$ then
$$
\frac{a}{d}\equiv\frac{b}{d} \pmod {n/d}
$$

original screen shot

答: Hint: $a \equiv b \pmod n$ means that there exists an integer $k$ for which $a - b = kn$.


问:
The ring of integers of a cubic number field generated by a root of $x^3-2x-2$

Let $\alpha$ be a root of the polynomial $f(x) = x^3-2x-2$, and let $K = \mathbb{Q}(\alpha)$. Show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha]$.

I know the following
$$\text{disc}(1,\alpha,\alpha^2)=-76$$
$$\text{disc}(1,\alpha,\alpha^2)=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\text{disc}(K)$$
It follows that $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ equals $1$ or $2$. I don't know how to complete from there, any one can help!

Thanks in advance.

答:
There is a tedious way and a slick way.

Tedious way:  Since the index is $1$ or $2$, the quotient $\mathcal O_K/\mathbf Z[\alpha]$ has size $1$ or $2$.  Therefore the group is killed by doubling, so $2\mathcal O_K \subset \mathbf Z[\alpha]$.  Thus $\mathbf Z[\alpha] \subset \mathcal O_K \subset (1/2)\mathbf Z[\alpha]$, which means you can just run through coset representatives $\beta_i$ for $((1/2)\mathbf Z[\alpha])/\mathbf Z[\alpha]$ to see which ones are algebraic integers: if $\beta_i$ is an algebraic integer, then the whole coset $\beta_i + \mathbf Z[\alpha]$ consists of algebraic integers, and if $\beta_i$ isn't an algebraic integer, then nothing in $\beta_i + \mathbf Z[\alpha]$ is an algebraic integer.

As representatives $\beta_i$ use
$(a/2) + (b/2)\alpha + (c/2)\alpha^2$ where $a, b, c \in \{0,1\}$. That is $8$ numbers to check. Well, really it's $7$ numbers to check since there the representative $0$, corresponding to the coset $\mathbf Z[\alpha]$, obviously is inside $\mathcal O_K$.  Not a single one of those $7$ nonzero numbers turns out to be an algebraic integer. For example, $\alpha/2 + \alpha^2/2$ is a root of $x^3 - 2x^2 - x - 1/4$. (This must be the minimal polynomial of $\alpha/2 + \alpha^2/2$ over $\mathbf Q$, granting that it really has that number as a root, since every number outside of $\mathbf Q$ in a cubic field has degree $3$ over $\mathbf Q$.) The only coset in $((1/2)\mathbf Z[\alpha])/\mathbf Z[\alpha]$ consisting of algebraic integers is the zero coset, and that means $\mathcal O_K = \mathbf Z[\alpha]$.

Slick way: the number $\alpha$ is a root of an Eisenstein polynomial at $2$, so $[\mathcal O_K:\mathbf Z[\alpha]]$ is not divisible by $2$ by Theorem 2.3 here. Therefore $[\mathcal O_K:\mathbf Z[\alpha]]$ must be $1$, which means $\mathcal O_K = \mathbf Z[\alpha]$.


问:
Continuous Functions With Integral Existing Only As Limit
Let $\lambda$ denote Lebesgue measure on $\mathbb{R}$. Give an example of a continuous function with $\lim_{t\rightarrow\infty}\int_{[0,t]}f \,d\lambda$ exists but $\int_{[0,\infty)}f \,d\lambda$ is not defined.

I've been thinking about this for a bit now. I think the function $f(x)=\sin(x)/x$ (with $f(0):=0$ works out, but I'm wondering if there are any other simple examples that I'm overlooking.

答:
Any conditionally convergent series will give you examples.

Recally that a series $\sum_{n=1}^\infty a_n$ is conditionally convergent if
\begin{align}
& \lim_{N\to\infty} \sum_{n=1}^N a_n \text{ exists in $\mathbb R$, thus it is not } {+\infty} \text{ or } {-\infty} \\
\text{and } & \lim_{N\to\infty} \sum_{n=1}^N |a_n|=+\infty.
\end{align}
An example is the alternating harmonic series $\sum_{n=1}^\infty(-1)^n/n.$

Now let $f$ be a function from $(0,\infty)$ into $\mathbb R$ for which $\displaystyle \int_n^{n+1} f(x)\, dx = (-1)^n/n.$


问:
Polar coordinate of a circle.
If I have a circle of radius $a$ center at $(a,0)$.
If I want to change coordinate from Cartesian coordinate to polar coordinate. I have convinced myself that $r=2a\cos\theta$ and $\frac{-\pi}{2}\le\theta<\frac{\pi}{2}$.

But what if a circle of radius $a$ center at $(0,a)$?
I know that $r=2a\sin\theta$ but why $0\le\theta<\pi?$ Looking at the graph it seems like $\theta$ can vary from $-\pi$ to $\pi.$

答:
The second circle is in the upper-half plane.  Also, the angle parametrizing the first circle ranged over $\pi$ radians.  Why would the second be any different?

Actually, any interval of length $\pi$ seems to work!

The first circle can be parametrized as:

$$r=2a\cos\theta \text{ and }\phi_0\le\theta<\phi_0+\pi,$$  and the second as

$$r=2a\sin\theta \text{ and }\phi_0\le\theta<\phi_0+\pi,$$

for any $\phi_0.$

Title: Polar coordinate of a circle.


问:
compactness / sequentially compact
I'm looking for two examples:

A space which is compact but not sequentially compact

A space which is sequentially compact but not compact

Explanations why the spaces are compact / not compact and sequentially compact / not sequentially compact would be appreciated. A reference would also be appreciated. So the conclusion would be, that there's no equivalence in general. Of course they are equivalent in a metric space.

math

答:
Example 1 with proof: Stone-Čech Compactification of the Integers $\beta \omega$

Proof: It is compact obviously. We will prove that $\beta\omega$ is not sequentially compact. Note that every infinite set in $\beta\omega$ has $2^\mathfrak c$ cluster points, hence the only convergent sequences in $\beta\omega$ are those which are eventually constant; therefore if $X$ is a subspace of $\beta\omega$  and $X$ is sequentially compact, then $X$ is finite. So $\beta\omega$ cannot be sequentially compact.

Title: compactness / sequentially compact


问:
Does connexity imply reflexivity?
The Wikipedia article on Total Order states, "The connex property also implies reflexivity, i.e., a ≤ a" without giving an explanation (and confusingly uses the "less than or equal to" symbol instead of R). However, there seems to be valid counterexamples; for instance, "less than" or "greater than" would appear to be connex but not reflexive. Is Wikipedia then incorrect, or am I misunderstanding?</p>

答:
You are misunderstanding: the usual $<$ relation on $\Bbb Z$ is not connex. In order for it to be connex, it would have to be true that for all $a,b\in\Bbb Z$, $a<b$ or $b<a$, but if we take $a=b=0$, for instance, neither of these is true.

More generally, suppose that $\preceq$ is a linear order on $A$, and let $a\in A$. $\preceq$ is connex, so either $a\preceq a$, or $a\preceq a$. In other words, $a\preceq a$, and $\preceq$ is therefore reflexive.

Title: Does connexity imply reflexivity?


问:
Limit Proofs Question
Prove the statement using the ε and δ definition of a limit:

$$\lim_{x\to3}  \quad \frac{x}{(3x-10)} = -3  $$

Let $\epsilon>0$ be given.
Find $\delta>0$ which depends on $\epsilon$ so that if $0<|x-3|<\delta$ and solve for $|x-3|$.

Then:

$$|f(x)-3|<\epsilon \text{ iff } |\frac{x}{3x-10} - 3|<\epsilon$$

$$\text{iff } |\frac{x}{3x-10} - \frac{3(3x-10)}{3x-10}| <\epsilon$$

$$\text{iff } |\frac{x-(9x-30)}{3x-10}| <\epsilon$$

$$\text{iff } |\frac{30-8x}{3x-10}|<\epsilon$$

$$\text{iff } |\frac{(-2)(15-4x)}{3x-10}| <\epsilon$$

$$\text{iff } 2 \frac{|15-4x|}{|3x-10|} <\epsilon$$

Arbitarily assume that $δ < 1$.

Then $|x-3|<\delta< 1$ implies that $-1<x-3<1$ and $2<x<4$.

However, range of $x$ values is not appropriate since $f(x)= \frac{x}{3x-10}$ is not defined at $10/3$.
We must simply pick $\delta$ small enough to avoid $x=10/3$. For example, assume that $\delta< 1/4$. Then $|x-3|<δ< 1/4$ implies that $-1/4<x-3<1/4$ and $11/4<x<13/4$ so that $2<|3x-10|<4$ and $1/4<\frac{1}{|3x-10|}<1/2$

It follows that

$$\frac{2|15-4x|}{|3x-10|} < 2|x-3|\frac{1}{2} < \epsilon\text{ (this is where I think I went wrong)}$$

$$\text{iff } 2|x-3| \frac{1}{2} <\epsilon$$

$$\text{iff } |x-3| <\epsilon$$

Now choose $\delta = \min (1/4, \epsilon)$

Then I am not sure where to go from there...

答:
Your very first mistake is that as we want to prove $\lim_{x\to 3}\frac x{3x -10}= \color{red}- 3$ we must find $\delta$ based on

$|f(x) - (\color{red}-3)| < \epsilon$ or

$|f(x) + 3| < \epsilon$

so $|\frac {x}{3x-10} + 3| < \epsilon$

$|\frac {x + 3(3x-10)}{3x-10}| = |\frac {10x -30}{3x-10}|<\epsilon$.

At this point I do something a little different

$|\frac {10x -30}{3x-10}|=|\frac {10x -\frac {10}3*10 + \frac {10}3*10 -30}{3x-10}|=|\frac {10}3 +\frac{3\frac 13}{3x-10}|=|\frac {10}3 +\frac {10}{9x-30}|<\epsilon$

$-\epsilon < \frac {10}3 +\frac {10}{9x-30}< \epsilon$

$-3\frac 13 -\epsilon < \frac {10}{9x-30}< -3\frac 13 +\epsilon$.  If we assume $0 < \epsilon < 3\frac 13$ then all those terms are negative so

$0 < 3\frac 13 - \epsilon < \frac {10}{30-9x} < 3\frac 13 + \epsilon$ and so

$10 -3\epsilon < \frac {10}{10-3x} < 10 + 3\epsilon$

$\frac 1{10-3\epsilon} > \frac {10-3x}{10} > \frac 1{10+3\epsilon}$

$\frac {10}{10-\epsilon} > 10 -3x > \frac {10}{10+3\epsilon}$

$-10 + \frac {10}{10-\epsilon} > -3x> -10 + \frac {10}{10+3\epsilon}$

$\frac {10 - \frac {10}{10-\epsilon}}3 < x < \frac {10 - \frac {10}{10+3\epsilon}}3$

Okay... we can't put it off any longer:

we want $\frac {10 - \frac {10}{10-\epsilon}}3 < 3-\delta < 3-x < 3+ \delta < \frac {10 - \frac {10}{10+3\epsilon}}3$

So we need $\frac {10 - \frac {10}{10-\epsilon}}3 < 3-\delta$ so we solve for $\delta$.  $\delta < 3-\frac {10 - \frac {10}{10-\epsilon}}3=  \frac {10}{30-3\epsilon}-\frac 13$.

And we need $3+ \delta < \frac {10+\frac {10}{10+3\epsilon}  }3$ or $\delta < \frac 13 -\frac{10}{30 +3\epsilon}$.

So if we set $\delta = \min (\frac {10}{30-3\epsilon}-\frac 13, \frac 13 -\frac{10}{30 +3\epsilon})$ and we are done.

Title: Limit Proofs Question


问:
convergence of sequence $a_{n+1}=\sqrt{a_n+2}$
is the sequence $a_{n+1}=\sqrt{(2+a_n)}$ convergent with $a_n=1$?

using induction i was able to prove $a_n<5$ (I choose $5$ randomly) for all n
and also a increasing sequence Which also I proved using induction.
So can I say since a monotonic bounded sequence are convergent can we say above sequence is convergent?

My induction for boundedness
$a_1=1<5$ $a_2=\sqrt(3)<5$

let the statement be true for $1,...n$,
$a_n<5$
$ \implies a_n+2<5+2 \implies \sqrt{a_n+2}<\sqrt{7}\implies a_{n+1}<5$
is this correct?

答:
Your proof that $a$ is bounded is true, but we can make it a bit of simpler.

Also you need to prove that $a$ increases.

By using induction we'll prove that $a_n<2$.

Indeed, $a_1=1<2$ and if $a_n<2$ so
$$a_{n+1}=\sqrt{a_n+2}<\sqrt{2+2}=2.$$
Also, $$a_{n+1}-a_n=\sqrt{2+a_n}-a_n=\frac{(2-a_n)(a_n+1)}{\sqrt{2+a_n}+a_n}>0,$$ which says that $a$ increases.

Thus, there is $\lim\limits_{n\rightarrow+\infty}a_n$ and let this limit be equal to $x$.

Thus, since $f(x)=\sqrt{x}$ is a continuous function, we obtain:
$$x=\sqrt{x+2}$$ or $$x=2.$$
There is also the following way:
$$|a_n-2|=|\sqrt{2+a_{n-1}}-2|=\frac{|a_{n-1}-2|}{\sqrt{2+a_{n-1}}+2}\leq$$
$$\leq\frac{|a_{n-1}-2|}{2}\leq\frac{|a_{n-2}-2|}{2^2}\leq...\leq\frac{|a_1-2|}{2^{n-1}}\rightarrow0,$$
which says that $$\lim_{n\rightarrow+\infty}a_n=2.$$

Here, a it's a name of the sequence. Sequence its a function of a natural argument. $a_n=a(n).$ Like $\sin$ it's a name of the function $f(x)=sin{x}$ and more...


问:
Loss of "significance" in $x-y$, $x=0.3721478693, y =0.3720230572$
If we perform the calculations in a computer with a five digit mantissa then I get:

$$\begin{aligned}
\mathrm{fl}(x) &=0.37215 \\
\mathrm{fl}(y) &=0.37202 \\
\mathrm{fl}(x)-\mathrm{fl}(y) &=0.00013
\end{aligned}$$

The exact calculation are: $$\begin{aligned}
x &=0.3721478693 \\
y &=0.3720230572 \\
x-y &=0.0001248121
\end{aligned}$$

So the relative error will be:
$$\left|\frac{x-y-[\mathrm{fl}(x)-\mathrm{fl}(y)]}{x-y}\right|=\left|\frac{0.0001248121-0.00013}{0.0001248121}\right| \approx 4 \%$$

One say that a loss of significance has occurred. But what have been lost? Is it numbers of significance digits in the result $x-y$? If that is correct, can one say that a loss of 5-6 significances digits has been lost?
Also, how many have been lost: 5 or 6 significant digits?

答:
If the values $x = 0.3721478693$ and $y = 0.3720230572$ are truly exact values--that is, if $x$ is actually $0.372147869300000\ldots$ (continuing with as many zeros as you could ever want to write) and similarly with $y,$
then $x-y = 0.0001248121$ is an exact result too.

The thing is that in practice we usually have to account for the possibility that our given values of $x$ and $y$ are not exact. In many applications this is the situation we find ourselves in most of the time. If $x$ and $y$ are numbers in a digital computer, we might say they are accurate to within the roundoff error of their last digit,
so what we know is that $0.37214786925 \leq x < 0.37214786935.$
Or to put it in a way that doesn't require looking so carefully at so many digits,
$x = 0.3721478693 \pm 5\times10^{11}$ and
$y = 0.3720230572 \pm 5\times10^{11}$.
Hence the most we can say about $x-y$ is that
$$ x - y = 0.0001248121 \pm 1\times10^{10}. $$

Two things just happened there. One is that the absolute maximum size of the error has doubled, but the other thing is that the size of the error relative to the answer is much larger than the sizes of the errors relative to the inputs.
Before subtraction, you had two numbers on the order of magnitude of
$5\times 10^{-1}$ with maximum errors of $5\times 10^{-11}$, a maximum relative error of $10^{-10},$ and afterward you have a number on the order of magnitude of
$1\times 10^{-4}$ with a maximum error of $1\times 10^{-10},$
a maximum relative error of $10^{-6}.$

That is, the subtraction cost you four digits of relative precision, most of which can be attributed to the fact that the first three digits of $x$ and $y$ canceled each other out.

But let's suppose (for sake of the original exercise, which may have made this assumption) that your ten-decimal-digit values of $x$ and $y$ are exact and that the only source of error is rounding them off to five decimal digits in the computer.
You then have $x$ with a relative error

$$ \frac{0.3721478693 - 0.37215}{0.3721478693} \approx -6 \times 10^{-6} $$

and $y$ with a relative error

$$ \frac{0.3720230572- 0.37202}{0.3720230572} \approx 8 \times 10^{-6}, $$

but $x - y$ with a relative error $4 \times 10^{-4}.$
So you have lost almost (but not quite) four decimal digits of precision in the subtraction.

Comparing the number of "significant digits" in $0.0001248121$
(infinitely many if $x$ and $y$ really are exact!)
versus the number of "significant digits" in $0.00013$
tells you nothing about the loss of significance due to the subtraction.
It tells you about the loss of significance due to rounding $x$ and $y$
to just five decimal digits before the subtraction.
What happens in the subtraction $0.37215 - 0.37202$
is that you go from numbers with five significant digits to a number with one or two significant digits, depending on whose rules you follow for counting them.
(Some would not count the leading $1$ as a significant digit.)


问:
proper differentiable maps of manifolds
Is the following statement true?If yes, why?

Let $f: M\to N$ be a proper morphism between smooth manifolds.  Let $x$ be a point of $N$, and $U$ a nbhd of $f^{-1}(x)$ in $M$. Then there exists a nbhd $V$ of $x$ in $N$ such that $f^{-1}(V)\subset U$.

Here, proper means that the preimage of any compact set is compact.

It seems to me this was used in a expository article that I am reading.  If it is true, I expect it to be true for proper morphism of locally compact topological spaces. But for some reason I wasn't able to find a proof.

Thank you.

答:
Suppose not. Then there is a sequence $(y_n)_{n\geq1}$ in $M\setminus U$ such that $f(y_n)\to x$. The set $S=\{f(y_n):n\geq1\}\cup\{x\}$ is compact, so its preimage $f^{-1}(S)$ is also compact. Since the sequence $(y_n)_{n\geq1}$ is contained in $f^{-1}(S)$, we can —by replacing it with one of its subsequences, if needed— assume that in fact $(y_n)_{n\geq1}$ converges to a point $y\in M$.

Can you finish?

(I am using that $x$ has a countable basis of neighborhoods here and that sequences in a compact subset of $M$ have convergent subsequences —to reduce to dealing with sequences— but more technology will remove that in order to generalize this to spaces other that manifolds)

Title: proper differentiable maps of manifolds


问:
Content $0$ implies boundary has content $0$

Problem: If a set $C$ has content $0$, then the boundary must also have content $0$.

The solution say

Solution: Suppose a finite set of rectangle $  U_i =  (a_{i1},b_{i1}) \times \dots   \times (a_{in}, b_{in})$ where $i \in \{1, \dots, m \}$ cover $C$ and have total volume less than $\epsilon/ 2$ where $\epsilon >0.$ Let $V_i = (a_{i1} - \delta ,b_{i1} + \delta)  \times \dots   \times (a_{in} - \delta , b_{in} + \delta)$ where

$$\prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{ij} + 2\delta) - \prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{aj} ) < \epsilon/2m .$$

Or equivalently,

Then the union of $V_i$ cover the boundary of $C$ and have total volume less than $\epsilon$. Hence $\partial C$ is also of content $0$.

This is a really simple question, but does $V_i - U_i$ cover $\partial C$? $V_i - U_i$ means the part where we remove $U_i$ from $V_i$.

Also is the last step because of $$ \sum_{i = 1}^m vol(\cup( V_i - U_i)) \leq \sum_{i = 1}^m\prod (b_{ij} - a_{aj} + 2\delta) - \prod (b_{ij} - a_{aj} ) < \sum_{i = 1}^m \epsilon/2m  = \epsilon /2.$$

or equivalently,

$$ \sum_{i = 1}^m vol(\cup V_i ) \leq \sum_{i = 1}^m\prod (b_{ij} - a_{aj} + 2\delta) < \sum_{i = 1}^m \epsilon/2m  + \prod (b_{ij} - a_{aj} ) < \epsilon /2 + \epsilon/2 = \epsilon. $$

Am I also correct to understand the outline of this proof is

Use the finite  cover  that covers $C$, and take a $\delta$ cover over that finite cover just enough to cover the boundary of $C$.

Using the finite cover of $C$, demand this cover so that it has volume less than $\epsilon/2$. Pick $\delta$ small enough so that we get

$$\prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{ij} + 2\delta) - \prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{aj} ) < \epsilon/2m .$$

Use finite subadditivitly to finish off the estimation.

答:
As this proof seems to raise a lot of questions, here is one that is in my opinion simpler.

If $C$ has content $0$, then for any $\epsilon > 0$ there is a finite collection $\{\bar{R_1},\ldots, \bar{R}_m\}$ of closed rectangles covering $C$ such that

$$\sum_{j=1}^m vol(\bar{R}_j) < \epsilon$$

This is frequently taken as the definition of zero content.  Proof of the equivalence of this definition with one using open rectangles is easy.

To finish, since $F = \bigcup_{j=1}^m \bar{R}_j$ is closed and $C \subset F$, then we must have $\partial C \subset F$. Otherwise, there is a point $x \in \partial C \cap F^c$, and since $F^c$ is open there is an open neighborhood $V_x$  with $x \in V_x \subset F^c$. As $x$ is a boundary point, the neighborhood $V_x$ must contain a point $y \in C\cap F^c$, a contradiction.

Thus, $\partial C \subset \bigcup_{j=1}^m \bar{R}_j$ where $\sum_{j=1}^m vol(\bar{R}_j) < \epsilon$ and $\partial C$ has content zero.


问:
Automorphism Group of Simple Graph
Suppose $G$ is a group. Does there always exist a simple graph $\Gamma$ such that $Aut(\Gamma)=G$,  i. e. such that $G$ is the automorphism group of $\Gamma$?

Given this I'd say no, but I would really appreciate if someone could help figure out an example.

答: I did some Googling and found Frucht's Theorem. The argument seems to involve modifying the Cayley graph of $G$ to remove colors and orientations while not introducing extra automorphisms.

Title: Automorphism Group of Simple Graph


问:
Computing adjoint of differentiation operator
This is a problem I have encountered in my studies and work in unbounded differential operators

Let us define the operator $T = -\frac{d^2}{dx^2}$ as an operator on $L^2(\mathbb{R})$ with domain $C_0 ^{\infty} (\mathbb{R})$ where the domain is the set of infinitely differentiable complex-valued functions on $\mathbb{R}$ with compact support. We are asked to compute the adjoint of this operator along with its domain (which extend the original $T$). Also, we are asked if this operator is essentially self-adjoint.

To be honest, I am new to this area of functional analysis and operator theory and so I find myself struggling, I do not know how to compute the adjoint and its domain. I know a basic criterion of essential self-adjointness is to check if the closure $\bar{T}$ is self-adjoint. I also know $T$ is symmetric (please see below) via integration by parts and using the boundary conditions arising from compact support. I have no idea how to do these things here or in practice in general. I thank all persons who can help with both parts of the problem.

***** Note: please let me clarify a few definitions. If $T$ is a densely defined linear operator on a Hilbert space $H$, the domain $D(T^*)$ is the set of $\phi \in H$ for which there is a $\eta \in H$ with
$$
\langle T\psi,\phi \rangle = \langle \psi,\eta \rangle
$$
for all $\psi \in D(T)$. For each such $\phi \in D(T^*)$ we define $T^* \phi = \eta$, and $T^*$ is called the adjoint of $T$. A densely-defined operator is said to be symmetric if $\langle T\phi,\psi \rangle = \langle \phi,T\psi \rangle$ for all $\phi,\psi \in D(T)$, and in this case $D(T) \subseteq D(T^*)$ and $T=T^*$ on $D(T)$ and $T^*$ is said to extend $T$. A symmetric operator is self-adjoint iff $D(T)=D(T^*)$ and thus $T=T^*$. An operator $T$ is said to be essentially self-adjoint if its closure $\bar{T}$ is self-adjoint, which is equivalent to $\ker(T^* \pm i) = \{0\}$ or $\text{Ran}(T \pm i)$ are dense in $H$.

答:
The adjoint $T^*$ is defined as the set of $g\in L^2(\mathbb{R})$ for which there exists a constant $C_{g}$ such that
$$
               |\langle Tf,g\rangle_{L^2}| \le C_g\|f\|_{L^2},\;\;\; \forall f\in \mathcal{D}(T).
$$
This inequality holds iff there is a unique $T^*g\in L^2$ such that
$$
                      \langle Tf,g\rangle = \langle f,T^*g\rangle,\;\;\; \forall f\in\mathcal{D}(T).
$$
($T^*g$ is unique if it exists because $\mathcal{D}(T)$ is dense in $L^2(\mathbb{R})$.) The Fourier transform $\mathcal{F}$ on $L^2$ can be brought to bear on $|\langle Tf,g\rangle| \le C_g\|f\|_{L^2}$:
$$
             \langle \widehat{Tf},\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\
              \langle -\xi^2\widehat{f},\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\
     \langle\widehat{f},-\xi^2\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\
       \implies \widehat{T^*g}=-\xi^2\widehat{g} \in L^2 \\
        T^*g = -\mathcal{F}^{-1}\xi^2\mathcal{F}g
$$
So the adjoint $T^*$ is fully characterized in terms of the Fourier transform: it is unitarily equivalent to multiplication by $-\xi^2$ in the Fourier domain. Multiplication operators on $L^2(\mathbb{R})$ are self-adjoint.
$$
                   T^*=-\mathcal{F}^{-1}\xi^2\mathcal{F} \\
          \implies T^c = (T^*)^*=-\mathcal{F}^{-1}\xi^2\mathcal{F},
$$
where $T^c$ is the closure of $T$. $T^c$ is self-adjoint because it is unitarily equivalent to a multiplication operator.


问:
When 4 coins are tossed simultaneously, two of them are 1 cent (indistinguishable) and the other two are 5 cents (indistinguishable).
question:
When 4 coins are tossed simultaneously, two of them are 1 cent (indistinguishable) and the other two are 5 cents (indistinguishable).
a) What are the possible results that can be obtained?
b) How many cases are there in which 2 heads and 2 tails come up?

i try do this: do 2 subsets one with two of them are 1 cent and the other are  5 cents maybe could be but here i get lost xd i need help please

答:
a) For each type of coin (pennies and nickels), there are three distinguishable results ($2H$, $1H$ $1T$, $2T$). Since there are two distinguishable types of coins, there are $3\cdot3=\boxed{9}$ total distinguishable results that can be obtained.

b) Assuming that for this part, a "case" still counts as a distinguishable result, we know that in order to get $2H$ and $2T$, either one type of coin is $2H$ and the other type is $2T$ (2 distinguishable cases), or each type of coin results in $1H$ and $1T$ (1 distinguishable case). So the answer to this problem is $\boxed{3}$.


问:
What is the average daily cost for defective components?
A typical day’s production of a certain electronic component is twelve. The probability that one of these components needs rework is $0.11$. Each component needing rework costs $\$100$. What is the average daily cost for defective components?

I understand that we can make Y how many components needed reworking, therefore: Y = [0, 1, 2, ... 12]

Also that the average daily cost for the defective components would equal 100 * Expected Value(Y) But how would I go about finding the probabilities of more than 1 component needing a rework.

答: We're only looking for expected value. The expected number of components that need rework, given that $12$ are produced in a day on average, is $12×0.11=1.32$. Then multiply this by $100$ to get $132$ dollars as the expected cost per day.


问:
Find the formula of the sequence using generating functions
I have the following sequence given:

$$\sum_{k=1}^{n} (-1)^{k}k^{2}$$

How can I do this? The sequence goes like this:

$$-1 + 4 - 9 + 16 - 25 + 36 - ...$$

So it doesn't have any variables inside. It looks like a geometric sequence for me, so the simple known formula would do the whole thing, but I actually doubt it's that simple.

So there are some generating functions that are quite similar, like:

$$\sum (-1)^{n}x^{n} = x - x^{2} + x^{3} - x^{4} + ...$$

But, uhm, well...what's next? The lack of x variable seems a bit strange to me.

答:
Note that the generating function for the sequence $(-1)^k$ is $\frac{1}{1+x}$. Also observe that if $(a_k)$ is any sequence with generating function $f$, then the generating function for the sequence $(ka_k)$ is $x\frac{d}{dx} f$ and the generating function for the sequence $\sum a_k$ is $\frac{f}{1-x}$.
With this in mind, we find the generating function for the sequence $(-1)^k k$ to be
$$x\frac{d}{dx}\frac{1}{1+x}=-\frac{x}{(1+x)^2}$$
and so the generating function for $(-1)^k k^2$ is
$$-x\frac{d}{dx}\frac{x}{(1+x)^2}=-\frac{(1-x)x}{(1+x)^3}$$

Thus, your generating function is
$$-\frac{x}{(1+x)^3}$$

If we use partial fractions, we can write this as
$$\frac{1}{(1+x)^3}-\frac{1}{(1+x)^2}$$
If we use the formla
$$\frac1{(1+x)^k}=\sum_{n\ge 0}\binom{n+k-1}{k-1}(-x)^n$$
we finally get that the sum is
$$\sum_{k=1}^n(-1)^k k^2=(-1)^n\left(\binom{n+2}{2}-\binom{n+1}{1}\right)$$

Edit:
\begin{align}
\frac{f}{1-x}&=\frac{1}{1-x}\sum_{k=0}^{\infty}a_kx^k=(1+x+...)\sum_{k=0}^{\infty}a_kx^k=\\
&=a_0+(a_0+a_1)x+(a_0+a_1+a_2)x^2+...= \\
&=\sum_{k=0}^\infty \left( \sum_{n=0}^k a_n\right)x^k
\end{align}


问:
Prove that the following statements are logically equivalent:

$s_1: A \backslash B = \emptyset$ and $B \backslash A= \emptyset$

$s_2: A \cup B = A \cap B$.

I'm not sure if what I have is a correct way to show this proof, but I have the following so far:

Suppose $A \backslash B = \emptyset$ and $B \backslash A = \emptyset$. We want to show that $A \cup B = A \cap B$.

Take any $x \in A$. If $A \cap B$, then $x \in A$ and $x \in B$.

Since $x \in A$ and $x \in B$, then $A \cap B \subseteq A \cup B$.

I'm not really sure if this is correct or how to progress.

答:
Take any $x \in A$. If $A \cap B$, then $x \in A$ and $x \in B$.

Since $x \in A$ and $x \in B$, then $A \cap B \subseteq A \cup B$.

This is correct, and in fact, it doesn't depend on the assumptions in $s_1$. It's always true that $A \cap B$ is a subset of $A \cup B$.

So now you want to show that $A \cap B \supseteq A \cup B$. Well, consider a Venn diagram with sets A and B. If you draw it, you'll see that there are three parts (not counting the outside): an element is either in $A$ but not $B$, or it's in their intersection, or it's in $B$ but not $A$. So if $x$ is in $A \cup B$, and $A \backslash B$ and $B \backslash A$ are empty, then where specifically does $x$ have to be?

To complete the proof, you have to assume $s_2$ and show that it implies $s_1$. You can do this in a similar way: take an element $x \in A \cup B$. Based on where $x$ has to be according to $s_2$, you can identify parts of the Venn diagram that must be empty.


问:
Evaluate $\int_{-1}^{1} \frac{x^{2016} e^x}{e^x + 1} \ dx$ by splitting the definite integral.
The full question is as follows:

Using
$\int_{-a}^{a}f(x) \ dx = \int_{0}^{a} \left(f(x) + f(-x)\right)dx$, find exact value of the integral $$\int_{-1}^{1} \frac{x^{2016}e^x}{e^x + 1} \ dx$$

Applying the identity given, we get

\begin{align*}
\int_{-1}^{1}\frac{x^{2016}e^x}{e^x + 1} \ dx &= \int_{0}^{1} \left(\frac{x^{2016}e^x}{e^x + 1} + \frac{(-x)^{2016}e^{-x}}{e^{-x}+1}\right)dx \\
&= \int_{0}^{1} \left(\frac{x^{2016}e^x}{e^x + 1} + \frac{x^{2016}e^{x}}{e^{x}+1}\right) dx \\
&= 2 \int_{0}^{1} \frac{x^{2016}e^x}{e^x + 1} \ dx
\end{align*}

I am stuck after this step. Trying various substitutions for $u = e^x$, $u = e^x + 1$ did not seem to work as well.

Is there a mistake in the preliminary steps shown? Any hints would be greatly appreciated!

Body:
Note

\begin{align*}
\int_{-1}^{1}\frac{x^{2016}e^x}{e^x + 1} \ dx &= \int_{0}^{1} \left(\frac{x^{2016}e^x}{e^x + 1} + \frac{(-x)^{2016}e^{-x}}{e^{-x}+1}\right)dx \\
&= \int_{0}^{1} x^{2016} \left(\frac{e^x}{e^x + 1} + \frac{1}{e^{x}+1}\right) dx \\
 &= \int_{0}^{1} x^{2016} \ dx=\frac1{2017}
\end{align*}


问:
Infimum of random variables is a random variable. Clarification of proof.
Let $(\Omega, {\cal A}, {\Bbb P})$ be the probability space. If $X_n$ are random variables then for all $ \{X_n < a\}\in {\cal A}$ and for all $n \in \Bbb N$ and $a \in \Bbb R$,

$$\left\{\inf_n X_n < a\right\} = \bigcup_n \left\{ X_n < a \right\} \in {\cal A}$$

This statement was written in a proof for infimum $X_n$ being a random variable. Can someone please explain why $\left\{\inf\limits_n X_n < a\right\}$ is equal to the union over the set $\{X_n < a\}$?

答:
It's helpful to read the union as "there's some $n$ such that". So the equality you're interested in translates in English to "$\inf X_n < a$ if and only if there's some $n$ such that $X_n < a$."

But that follows easily from the definition of $\inf$. Indeed, suppose $\inf X_n < a$. Since for all $\epsilon > 0$ there's some $n$ such that $X_n - \inf_n X_n < \epsilon$, it follows (with $\epsilon = (a - \inf_n X_n)/2$, for example) that there's some $n$ such that $X_n < a$. Conversely, if there's some $n$ such that $X_n < a$, then clearly $\inf_n X_n < a$ since the $\inf$ is a lower bound.


问:
Continuous Bernoulli Distribution
When studying Bernoulli random variables the following formula is commonly used: $$\Pr(X=x)={n\choose x}p^x(1-p)^{n-x}\,,$$ where $p$ is the probability of a success. Further questions ask to graph the distributions such as the question:

My question is can you construct a continuous bernoulli distribution with the gamma function to derive a continuous ${n\choose x}$ and what is the applications of this continuous choose function in combinatorics?
$${n\choose r}=\frac{n!}{r!(n-r)!}=\frac{\Gamma(n+1)}{\Gamma(r+1)\Gamma(n-r)}$$

答:
Well, there is the Beta Distribution whose probability density function, for strictly positive shape parameters, $\alpha,\beta$, is: $$Y\sim\mathcal{Beta}(\alpha,\beta)\iff f_{\small Y}(y)=\dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\,\Gamma(\beta)}y^{\alpha-1}(1-y)^{\beta-1}\mathbf 1_{0\leqslant y\leqslant 1}$$

Which provides a family of prior probability distributions for binomial distributions in Bayesian inference.

Title: Continuous Bernoulli Distribution


答: $A-B=\phi$ means $A\subseteq B$ while $B-A=\phi$ means $B\subseteq A$ Therefore $B=A=B\cap A=B\cup A$.

Title: Prove that the following statements are logically equivalent:

Body:
$s_1: A \backslash B = \emptyset$ and $B \backslash A= \emptyset$

$s_2: A \cup B = A \cap B$.

I'm not sure if what I have is a correct way to show this proof, but I have the following so far:

Suppose $A \backslash B = \emptyset$ and $B \backslash A = \emptyset$. We want to show that $A \cup B = A \cap B$.

Take any $x \in A$. If $A \cap B$, then $x \in A$ and $x \in B$.

Since $x \in A$ and $x \in B$, then $A \cap B \subseteq A \cup B$.

I'm not really sure if this is correct or how to progress.


问:
Proof that degree of map is sum of its local degrees (prop 2.30 Hatcher)
Hatcher, in his proof of proposition 2.30 refers to the diagram

In his first line he simply says: By excision, the central term $H_{n}(S^{n},S^{n}-f^{-1}(y))$ is the direct sum of the groups $H_{n}(U_{i},U_{i}-x_{i})$ with $k_{i}$ the inclusion of the ith summand and $p_{i}$ the projection onto the $i$th summand.

How does this follow from excision? This is unclear to me.

答: the excision is applied to the triplet $S^n-\cup U_i \subset S^n-f^{-1}(y) \subset S^n$


问:
Show that $dim\ rm{span} \{u_1,...,u_n\} = n-1 $ if and only if $dim\{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0\} = 1$
Struggling with the following question:

Let $n \in \mathbb{N} $ with $n\geq 2$. Let $V$ be a vector space. Let $u_1,...,u_n \in V$. $\\$Show that $\dim\rm{span}  \{u_1,...,u_n\} = n-1 $ if and only if $\dim\{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0\} = 1$$\\$

My "attempt" for the forward direction:

Let $\left \{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0 \right \} = W$

I believe we need to show that $\exists v \in W s.t. v \ne 0$ and $\rm{span}\{v \} = W $

I formed a basis of dimension $n-1$ from $\dim \rm{span} \left \{u_1,...,u_n \right \}$, call it $\beta$.

Now, am I supposed to show $\beta \in W$ and that $\beta$ spans $W$? This is where I'm lost. Any pointers or corrections are greatly appreciated.

Edit: Not allowed to use rank-nullity theorem.

答:
Prove of $\Rightarrow$: From you hypothesis $\dim \rm{span}\{u_1,\ldots,u_n\} = n-1$. There is a linear independent subset that spans this very hyperplane, thus WLOG assume it is $\{u_1, \ldots, u_{n-1}\}$. Hence we have the unique decomposition of $u_n$ under this basis:$u_n = k_1u_1 + \ldots k_{n-1}u_{n-1}$.

Let $v=(k_1, \ldots, k_{n-1}, -1)$, then $v\ne 0$ and $v\in W$. Suppose $w=(l_1, \ldots, l_n)\in W$. If $l_n=0$, the by the definition of $W$ and the linear independence of $u_1, \ldots, u_{n-1}$, $l_1 = \ldots = l_{n-1} = 0$. Otherwise, $l_n \ne 0$, but $l_1 = \ldots = l_{n-1}$  is still uniquely determined by that unique decomposition in the previous paragraph: a multiple of $v$. This is enough to conclude $v$ spans $W$.


问:
Finding density of the distance from an arbitrary starting point to the nearest tree

Trees in a forest are distributed according to a Poisson process. Let $X$ be the distance from an arbitrary starting point to the nearest tree. Given that the average number of trees per square metre is $\lambda,$ find the density $f(x)$ of $X.$

The probability that there are $0$ trees within a circle of radius $x$ from the starting point is $e^{-\lambda \pi x^2},$ since $\lambda \pi x^2$ is the average number of circles that would be expected in that region. But I'm not sure how to relate this to $X,$ the distance from an arbitrary starting point to the nearest tree. Fix a starting point $S.$ If the distance from $S$ to the nearest tree is $x,$ then doesn't that mean that there's at least one tree within that distance of $x$? If so, then the probability of the nearest tree being a distance of $x$ from the starting point should be the probability of at least one tree occurring in the circle of radius $x,$ which is $1-P(no \,trees\, occur).$ This is the cumulative distribution function as it's the sum of the probabilities that the nearest tree is a distance of $0\leq y\leq x$ from the start point, and so to find the density, we just take the derivative.

Is this correct?

答:
What you have said looks correct to me but does not seem answer the original question:

The expected number of trees within $x$ of the point is $\lambda \pi x^2$

The probability there is exactly one tree within $x$ of the point is, as you say,  $\lambda \pi x^2 e^{-\lambda \pi x^2}$ as this is a Poisson process, but this is not particularly helpful

The probability there are no trees within $x$ of the point is $e^{-\lambda \pi x^2}$ as this is a Poisson process

The probability there is at least one tree within $x$ of the point is $1-e^{-\lambda \pi x^2}$ and this is a cumulative probability

The density for the distance of the nearest tree is then $2 \lambda \pi x e^{-\lambda \pi x^2}$ by taking the derivative


问:
Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$
Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$</p>

答:
A related problem. Here is another approach. Recalling the generating function of the harmonic numbers

$$ \sum_{n=1}^{\infty} H_n x^n = \frac{\ln(1-x)}{x-1} \implies (xD)^4\sum_{n=1}^{\infty} H_n x^n=(xD)^4 \frac{\ln(1-x)}{x-1}, $$

$$ \implies \sum_{n=1}^{\infty} H_n n^4 x^n = {\frac {x \left( 1+11\,x+11\,{x}^{2}+{x}^{3} \right) \ln  \left( 1-x
 \right) }{ \left( -1+x \right) ^{5}}}+{\frac {x \left( -1-27\,{x}^{2}
-18\,x-4\,{x}^{3} \right) }{ \left( -1+x \right) ^{5}}}.$$

Substituting $x=-\frac{1}{2}$ in the above identity gives the desired result

$$ \sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}={\frac {28}{243}}-{\frac {10}{81}}\,\ln  \left( \frac{3}{2} \right).  $$