答:
This is good, creative thinking about number systems. But here's the hitch: you have to follow through with the consequences of your definitions. If $z = 1/0$ as you suggest, then it must be that $z \cdot 0 = 1$.

Why is this a problem? Well, you are probably already familiar with the idea that $a \cdot 0 = 0$ for any number $a$. Incidentally, this is a consequence of the fact that $0$ is the additive identity and that your operations of multiplication and addition satisfying a distributive law. To wit,

$$
a \cdot 0 = a \cdot 0 + a \cdot 0 - a \cdot 0 = a \cdot (0 + 0) - a \cdot 0
= a \cdot 0 - a \cdot 0 = 0
$$
With $a = z$, your special reciprocal of $0$, you now have that $z \cdot 0 = 1$ and $z \cdot 0 = 0$. Thus, $1 = 0$.

Once you establish that $1 = 0$ in your number system, then any number is equal to any other number! Consider two numbers $b$ and $c$, and calculate:
$$
b = b + 0 = b + 0 \cdot (c-b) = b + 1 \cdot (c-b) = c.
$$

All your numbers collapse into a singularity. Oh noes!

Title: Why can't we have a third number line for dividing by zero?

Body: Originally, taking the square root of a negative number wasn't possible, but then Rafael Bombelli added an imaginary number axis to allow that. Why, then, is there no 3rd dimensional number line to allow division by 0? If there was, you could create a unit z, where z is 1/0. Therefore, 5/0 could be 5z, etc., and multiplying by its reciprocal is a 90° rotation to the real number axis, just like squaring i causes a 90° rotation to the real axis.


问:
Solving functional equation $f(x)=3f(x+1)-3f(x+2)$
It is given that a function f(x) satisfy:
$$f(x)=3f(x+1)-3f(x+2)\quad \text{ and } \quad f(3)=3^{1000}$$ then find value of $f(2019)$.

I further wanted to ask that is there some general method to solve such equation. The method that I know to solve such questions is to substitute $x$ with $x+1$ in equation and there by making new equation which is
$$ f(x+1)=3f(x+2)-3f(x+3)$$ then again substitute $x$ with $x+2$ in original equation and make new equation $$f(x+2)=3f(x+3)-3f(x+4)$$ Do this for a couple of times and then on combining the equations, in most of the question we get some relation like f(x) = f(x+a) but that does not work here. Please share your ideas on how to solve such questions.

答: Note that, we can rewrite the given recurrence such that $$af(n+1)-f(n)=b(af(n)-f(n-1))$$ with complex numbers $a=\sqrt{3}e^{i\pi/6}, b=\dfrac{\sqrt{3}}{3}e^{i\pi/6}.$ Inductively we have that $$af(n+1)-f(n)=b^{k+1}(af(n-k)-f(n-k-1))$$ for all $k\in\mathbb{N}.$ In particular, this quantity equals to $b^{n}(af(1)-f(0)).$ Now note that $$a^{n+1}f(n+1)-a^nf(n)=(ab)^{n}A,$$ where $A=af(1)-f(0)$ is a constant. By telescoping we can obtain $$a^{n+1}f(n+1)-f(0)=A\sum_{k=0}^n e^{i\pi k/3}=A\left(\dfrac{1-\omega e^{i\pi n/3}}{1- \omega} \right)$$ where $\omega=e^{i\pi /3}$ is the third root of $-1.$ Here RHS is a periodic function of $n$ with period $6,$ and hence $$f(n)=-\dfrac{1}{27}f(n-6)=\dfrac{(-1)^k}{3^{3k}}f(n-6k)$$ for all $n, k\in \mathbb{N}.$ In particular, for $n=2019, k=336$ we get the desire answer.

ContentLicense: CC BY-SA 4.0


问:
If $I\pi ^{-k}\subseteq D$ for all $k\leq m$ but $I\pi ^{-m-1}\nsubseteq D$, then $I\pi ^{-m}\nsubseteq (\pi )$
I'm working through a proof and I've got stuck on one detail, it seems like it is supposed to be totally obvious, but need help to figure out why.

Let $D$ be a Noetherian integrally closed domain with unique prime ideal $(\pi)$. For an ideal $I$ of $D$ we have that $I\pi ^{-k} \subseteq D$ for all $k\leq m$ but $I\pi ^{-m-1}\nsubseteq D$. I wonder how to make the conclusion that $I\pi ^{-m}\nsubseteq (\pi )$.

This is used in proving that every ideal in the ring $D$ above is a power of $(\pi )$. I hope I didn't leave out any crucial information...

答:
Since $D$ is a domain, one should say that $(\pi)$ is the unique non-zero prime ideal. If so, $D$ has dimension $1$ and thus is a Dedekind domain (by definition).

One knows that in a Dedekind domain an ideal is always a product of prime ideals. Thus, a fortiori, the products $(\pi)^n=(\pi^n)$ are all the non-zero ideals that there might be.

This is the approach taken, for instance, by P. Samuel in his classic Théorie Algébrique des Nombres

ContentLicense: CC BY-SA 3.0


问:
Decidability of bi-cartesian closed categories
This is actually in reference to the question posed here
https://stackoverflow.com/posts/66285948/edit
but is more appropriate as a question to be posed on a non-coding site.

I provide a partial answer, but not the full answer, relaying the full question here as an unresolved question. Perhaps others can take the baton.

The question is: what decision procedures are there, for bi-cartesian closed categories? (A secondary question is: what normal order reduction strategies are there for them?)

The following observations will help illustrate why the initial object is being excluded in the reference cited in the original link. Things start to collapse when initial objects are combined with exponentials; or terminal objects with co-exponentials. This collapse becomes complete when both exponentials and co-exponentials are present; in which case, the decideability problem becomes trivial, reducing to provability in a logic, while the "normal form" reduction problem becomes more difficult - ironically.

In a category, an initial object $0$, terminal object $1$, product $A,B ↦ A×B$, co-product $A,B ↦ A+B$, exponential $A,B ↦ B^A$ and co-exponential $A,B ↦ A_B$ (if we may be excused in denoting it like so) correspond respectively to the constants & connectives in logic $⊥$ (false), $⊤$ (true), $A,B ↦ A∧B$ (and), $A,B ↦ A∨B$ (or), $A,B ↦ A⊃B$ (only if) and $A,B ↦ A⊂B$ (unless) (again: if we may be excused in denoting it this way).

This correspondence is made more precise in the following way: if the pre-order $A ≤ B ⇔ ∃f: A → B$ is defined on objects, and if objects are identified up to equivalence ($A ≡ B$ if and only if $A ≤ B$ and $B ≤ A$), then $⊤$, $⊥$, $∧$ and $∨$ are the lattice top, bottom, meet and join respectively, while $⊃$ and $⊂$ are respectively the implication and co-implication operators respectively of a Heyting and co-Heyting lattice.

If all of the following $\{ ∧, ⊃, ⊥ \}$ are present together then so is $⊤$, and then the category practically reduces to a logic. More precisely: it has a negation connective, defined up to equivalence by $¬A ≡ A ⊃ ⊥$ and the full subcategory of the regular objects (i.e. those objects $A$ for which $A$ and $¬¬A$ are equivalent) is a partial order and, in fact, reduces to a Boolean algebra (thus making all such equivalences as also isomorphisms).

That is, if $A$, $B$ are regular objects and $f: A → B$, and $f': A → B$, then $f = f'$; and such $f$ exists if and only if $A ≤ B$. Since $¬¬¬A$ and $¬A$ are isomorphic, then the regular objects comprise all the negations and the negative objects $¬A$ form a subcategory that is actually a Boolean logic.

A categorical algebra that includes either of the full sets of connectives $\{ ∧, ⊃, ⊥ \}$ or $\{ ∨, ⊂, ⊤ \}$ therefore contains a categorical algebra for Boolean algebra, itself, as a subalgebra. One such algebra was, in fact, given by Spencer-Brown in the "Laws of Form" though he did not recognize that it was a categorical algebra; since category theory was virtually unknown at the time and categorical algebras probably weren't even known of at all, at that time. Spencer-Brown actually defined a groupoid for Boolean algebra, not just a category, since his arrows were all reversible and were actually isomorphisms.

It can be described equivalently as a braided monoidal category, whose objects are strings, with the product operator $x,y ↦ xy$ is denoted by juxtaposition, the terminal object by the empty string $λ$. The category is endowed with an extra operator (denoted by circling the operand) which could be denoted here $x ↦ ❨x❩$. The isomorphisms he defined were Position $❨❨x❩x❩ ↔ λ$, Transposition $❨❨xz❩❨yz❩❩ ↔ ❨❨x❩❨y❩❩z$ and an implied isomorphism $xy ↔ yx$ for commutativity - which is already part of the definition for braided monoidal categories. It is coherent (only one morphism per arrow), and a decision procedure was given on Spencer-Brown's Laws of Form. The functor $x ↦ ❨x❩$ is more properly denoted an "anti-functor" since it maps the category to its reverse.

If Transposition is replaced by morphisms for Number $xx ↔ x$, Order $❨❨x❩❩ ↔ x$ and Ortho $❨xy❩❨y❩ ↔ ❨y❩$, then this suffices to characterize orthocomplemented lattices. I don't know how the procedure specified by Spencer-Brown can be adapted to this generalization. Nor am I even sure if it collapses to a coherent category.

The partial "collapse" can be seen directly as follows, where we'll consider the combination $\{ ∧, ⊃, ⊥ \}$ that combines exponentials and initial objects: Cartesian closed categories with initial objects.

Each connective can be defined equivalently as a functor on the category that comes equipped with a natural bijection that provides the categorical algebra for the connective and proof rules for the corresponding logic.

For the following, adopt the following conventions. For a category $$, let $||$ denote its objects and for $A, B ∈ ||$, let $(A,B)$ denote the set of morphisms for the arrow $A → B$. Let $$ denote the unit category with $|| = \{0\}$ and $(0,0) = \{1_0\}$.

For a category $$, let $^+$ denote the reversal of category $$; where $|^+| = ||$ and for $A, B ∈ |^+|$,
$$^+(A,B) = \{ f ¦ A ← B: f \} = \{ f ¦ f: B → A \} = (B,A);$$
composition defined in $^+$ as the reversal of composition in $$. Finally, for categories $$ and $$, let $×$ denote the direct their product; where $|×| = ||×||$, and for $(A,A'), (B,B') ∈ |×|$,
$$×((A,A'),(B,B')) = (A,B) × (A',B'),$$
with composition defined componentwise.

If $$ is the category in question, then the functors include some or all of $⊤, ⊥:  → $; $∧, ∨: × → $; $⊃:^+× → $; $⊂:×^+ → $; and the natural bijections corresponding to each one are given, respectively, by
$$C ∈ |^+| ↦ (C,⊤) ⇔ (0,0),$$
$$(A,B,C) ∈ |××^+| ↦ (C,A∧B) ⇔ (C,A)×(C,B),$$
$$(A,B,C) ∈ |^+××^+| ↦ (C,A⊃B) ⇔ (C∧A,B),$$
$$D ∈ || ↦ (⊤,D) ⇔ (0,0),$$
$$(A,B,D) ∈ |××| ↦ (A∨B,D) ⇔ (A,D)×(B,D),$$
$$(A,B,D) ∈ |^+××| ↦ (A⊂B,D) ⇔ (A,B∨D).$$

In detail, for $∧$, $⊃$, $⊤$ and $⊥$, we can write them as
$$(f: C → A, g: C → B ⇔ h: C → A∧B) ⇒ (<f,g> = h ⇔ ⊤h = f, ⊥h = g),$$
$$(f: C∧A → B ⇔ h: C → A⊃B) ⇒ (⋀f = h ⇔ ⋁h = f),$$
$$f: C → ⊤ ⇔ f = <>_C,$$
$$g: ⊥ → D ⇔ g = []_D,$$
(where I'm operator-overloading by also using $⊤$ and $⊥$ as prefix operators to denote the components of the reverse direction of the natural bijection for $∧$).

The last two conditions are where the "collapse" starts to take place.

In the cases of $⊤$ and $⊥$, "naturality" of the bijections is given by the conditions just stated above, while for $∧$, it is given equivalently by the conditions that
$$⊤(h∘m) = ⊤h∘m, ⊥(h∘m) = ⊥h∘m,$$
where $m: C' → C$. So, we could just as well write $⊤h = ⊤1_{A∧B}∘h$ and $⊥h = ⊥1_{A∧B}∘h$ (taking $1_{A∧B}$ and $h$, respectively, in place of $h$ and $m$).

First, $⊥∧A ≤ B$ is always true and there is a unique $f: ⊥∧A → B$, since
$$f: ⊥∧A → B ⇔ ⋀f: ⊥ → A⊃B ⇔ ⋀f = []_{A⊃B} ⇔ f = ⋁[]_{A⊃B}.$$

Second, by this uniqueness, since
$$⊤1_{⊥∧A}: ⊥∧A → ⊥, []_{⊥∧A}: ⊥ → ⊥∧A,$$
and thus
$$[]_{⊥∧A}∘⊤1_{⊥∧A}: ⊥∧A → ⊥∧A;$$
and since
$$1_{⊥∧A}: ⊥∧A → ⊥∧A,$$
then it follows that
$$[]_{⊥∧A}∘⊤1_{⊥∧A} = ⋁[]_{A⊃(⊥∧A)} = 1_{⊥∧A}.$$

Third, since
$$⊥1_{⊥∧A}: ⊥∧A → A; []_{⊥∧A}: ⊥ → ⊥∧A,$$
then
$$⊥1_{⊥∧A}∘[]_{⊥∧A}: ⊥ → A$$
thus
$$⊥1_{⊥∧A}∘[]_{⊥∧A} = []_A.$$

Fourth, suppose $A ≤ ⊥$ and that $f: A → ⊥$. Then
$$⊤1_{⊥∧A}∘<f,1_A> = ⊤(1_{⊥∧A}∘<f,1_A>) = ⊤<f,1_A> = f.$$
$$⊥1_{⊥∧A}∘<f,1_A> = ⊥(1_{⊥∧A}∘<f,1_A>) = ⊥<f,1_A> = 1_A.$$
Thus
$$[]_A∘f = ⊥1_{⊥∧A}∘[]_{⊥∧A}∘⊤1_{⊥∧A}∘<f,1_A> = ⊥1_{⊥∧A}∘1_{⊥∧A}∘<f,1_A> = ⊥1_{⊥∧A}∘<f,1_A> = 1_A,$$
while $f∘[]_A: ⊥ → ⊥$ and $1_⊥: ⊥ → ⊥$ therefore $f∘[]_A = []_⊥ = 1_⊥$.

Therefore, $f = {([]_A)}^{-1}$, thus making the morphism $f: A → ⊥$ is unique.

When $A ≤ ¬B ≡ B ⊃ ⊥$, then there is a unique $f: A → ¬B$, since
$$f: A → ¬B = B ⊃ ⊥ ⇔ ⋁f: A∧B → ⊥ ⇔ ⋁f = {([]_{A∧B})}^{-1} ⇔ f = ⋀{([]_{A∧B})}^{-1}.$$

It is therefore also possible to define $⊤ ≡ ⊥ ⊃ ⊥$, thereby making the morphisms $f: A → ⊤$ unique, since
$$f: A → ⊤ = ⊥ ⊃ ⊥ ⇔ ⋁f: A∧⊥ → ⊥ ⇔ ⋁f = {[]_{A∧⊥}}^{-1} = ⊥1_{A∧⊥} ⇔ f = ⋀⊥1_{A∧⊥},$$
and we could therefore define $<>_A ≡  ⋀⊥1_{A∧⊥} = {[]_{A∧⊥}}^{-1}$. So, it is indeed a Cartesian closed category, as I alluded to above.

答: (This comment is also posted at stack overflow).  Thanks for your reply! In the time since I asked it, I believe that Gabriel Scherer gave an affirmative answer to this question in his paper arxiv.org/abs/1610.01213</p>


问:
Large Intersection Multiplicity
A cubic curve, say, $x^3+y^3=1$ and some quadratic curve $f(x,y)=0$ generally have six intersection points in $\mathbb{CP}^2$.

Question: If all the intersection points coincide, what will be the exact form of $f(x,y)$?

答:
I haven't ever seen a proof of this, but it seems that curves that intersect with multiplicity 2 or higher have the same tangent line, and multiplicity 3 or higher have same second derivative and so on.

From that, the only points where a quadratic curve could have a intersection of higher multiplicity than 4 with a cubic curve are the points of the cubic curve with a vanishing third derivative, because a quadratic curve always has a vanishing third derivative.

Finding the third derivative of x^3+y^3-1=0 is difficult, but with some approximation algorithms I find that the quadratic curve  0.1893004115226*x^2+0.3127572016461*x+2.8683127572018=y  approximately intersects the curve x^3+y^3-18=0 with a multiplicity of at least 4 at a point near (-2,3).

Title: Large Intersection Multiplicity


问:
How to determine the logical form of a statement?
I'm having trouble understanding how to determine the logical form of a statement.

Here's my current understanding (I'm not sure if all of them are correct though):

p is a propositional variable means that p can be any proposition, so the Truth Value of p has 2 possibilities: T or F(but we don't know which one yet until we assign p to be a certain statement). I think it's kind of analogous to numerical variables, like let x be a natural number, then the "evenness" of x has 2 possibilities: even or odd (but we don't know which one yet until we assign a value to x, like 4 or 123...)

A logical form is an expression consist of the component statement variables and the logical connectives, and the truth table of logical form shows all the possible truth values of the form for all combinations of Truth Values of the statement variables. I like to think of the truth table for the logical form to be what all statements of that logical form mean logically (replacing the propositional variables with the respective component statements, of course).

If my understanding of propositional variable and logical form is correct, then here's what I don't understand: when determining the logical form of a statement, the general procedure is to first identify the component statements and give them names as variables, then translate the whole statement into an expression consist of the propositional variables and appropriate logical connectives. For example, say "the car is black and the bike is white". When identifying the component statement, the solution always says that denote p:= the car is black and q:= the bike is white. How can that work?

I thought what I should do is to completely remove the component statements and replace them with 2 propositional variables p and q, which results in "p and q", and then set up the truth table for that logical form and get to the logical form which is p∧q. If we denote p and q to be the component statements, then p and q are not propositional variables anymore, right? therefore, there will be no truth table for the initial statement (because "p and q" now is not a logical form, it's a statement so it can only have a single truth value). Then how can we determine what is the logical connective for the initial statement's logical form?

I guess I also want to ask is that can there be a truth table for a statement? I thought truth tables only exist for logical forms.*

答:
truth tables are for logical forms.

statements can only be true or false at one time

if the structure of a statement can be expressed in a logical form containing one or more logical connectives, then there will be more than 2 rows to the truth table of that logical form. However the statement will only correspond to one row of the truth table for which the component statements have the correct truth value


问:
Dirichlet series for $\zeta(s) \zeta(2s)\zeta(3s) \zeta(6s)^{-1}$
Let $\kappa(n)$ be defined, for $n=p_1^{a_1}\dots p_k^{a_k}$, by $\kappa(n)=a_1\cdot a_2\dots \cdot a_k$. Prove that
$$
\sum_{n=1}^{\infty}\frac{\kappa(n)}{n^s}=\frac{\zeta(s)\cdot \zeta(2s) \cdot \zeta(3s)}{\zeta(6s)}.
$$

I would like to prove this statement, however I'm not sure what to do with the $\zeta(n)$ function. Are there any useful properties I can use?

答:
Another slightly different approach is to use the fact that the function $\kappa$ is multiplicative and so we have the Euler Product for the Dirichlet series:

$$\sum_{n=1}^\infty \frac{\kappa(n)}{n^s} = \prod_{p} \sum_{n=0}^\infty \frac{\kappa(p^n)}{p^{ns}} = \prod_p \left(1 + \sum_{n=1}^\infty np^{-ns}\right) = \prod_p \left(1 + \frac{p^{-s}}{(1-p^{-s})^2}\right) = \prod_p \frac{1 - p^{-s} + p^{-2s}}{(1-p^{-s})^2}$$

Now using that $\zeta(s) = \prod_p \frac1{1-p^{-s}}$ you get:

$$\frac{\zeta(s)\zeta(2s)\zeta(3s)}{\zeta(6s)} = \prod_{p} \frac{1-p^{-6s}}{(1-p^{-s})(1-p^{-2s})(1-p^{-3s})} = \prod_p \frac{1 - p^{-s} + p^{-2s}}{(1-p^{-s})^2}$$

As the two Euler products agree you have the wanted identity.

REMARK: This is Exercise $9$ of Chapter $11$ in the well-known Apostol's book "Introduction to Analytic Number Theory" $(1976)$. You can find the proof for the Euler product formula for a Dirichelt series involving a multiplicative function in that chapter.


问:
Show $x_{n+1} = 1 + \sqrt{x_n}$ is upper-bounded and increasing
Given $x_1 = 1$, it is clear that $x_n \ge 1$ for any $n \in \Bbb{N}$. I know that {$x_n$} is obviously increasing and limit as $n \to \infty$ is $\frac{3 + \sqrt{5}}{2}$, so {$x_n$} has an upper bound also. I seem to have trouble proving $x_{n+1} / x_n \ge 1$ and don't even have any idea on how to prove the sequence is upper-bounded. Can you guys give me some hints?

答:
Let $f(x)=1+\sqrt{x}$, so $f(\phi^2)=\phi^2$ where $\phi=(1+\sqrt{5})/2$.
Since $f$ is strictly increasing and $x_1<\phi^2$,  it follows by induction that $f(x_n) <\phi^2$ for all $n \ge 1$. Also, since $x_1<x_2$, we infer by induction that $x_n<x_{n+1}$ for all $n \ge 1$. The induction step is simply an application of $f$ to both sides.


问:
Find a continuous map $f(x_i)=t_i$
Let $X$ be a Tychonoff space and $\mathbb{T}$ the circle group. If $x_1,...,x_n\in X$ and $t_1,...,t_n\in \mathbb{T}$ then there exists a continuous map $f:X\to \mathbb{T}$ such that $f(x_i)=t_i$ for each $i$.

How can I prove this? By the way, maybe this result is some more general, I think we can put $Y$ any topological space instead of $\mathbb{T} $ or something like that.

Any hint? Thanks.

答:
Since a Tychonoff space is completely regular and finite sets are closed, for each $j$ there is $g_j: X \to \mathbb R$ such that
$g_j(x_j) =  1$ and $g_j(x_k) = 0$ for $k \ne j$.  Consider $\exp(i \sum_j c_j g_j)$ for suitable $c_j$.

You can't generalize this too far: in particular it doesn't work if $X$ is connected and $Y$ is not.

Title: Find a continuous map $f(x_i)=t_i$


答:
This is a difference equation. In the case of a linear, constant coefficient difference equation we make the following guess:
$$ f(x) = r^x $$
Then
$$ f(x+1) r\times r^x$$
and
$$ f(x+2) = r^2\times r^x $$
Replacing in the difference equation
$$ r^x = 3\times r\times r^x - 3\times r^2\times r^x $$
Factoring out the $r^x$ and canceling
$$ 1 = 3\times r - 3\times r^2 $$
$$ 3\times r^2 - 3\times r + 1 = 0 $$
The solutions of this quadratic equation are
$$ r_{1,2} = 0.5 \pm \frac{\sqrt{3}}{6}i $$
Replacing in our guess:
$$ f(x)= \left(\sqrt{\frac13}\right)\left[C_1\times \cos(1.51312532 \times t) + C_2\times\sin(1.51312532 \times t) \right]  $$
This is the general solution to the functional equation. However, there are two indeterminate constants and only one condition $f(3)=3^{1000}$ so I think that this method does not work here. Because imposing that condition would not yield the values of $C_1$ and $C_2$. Therefore, I think that the way of solving this is the one that Gareth Ma sketched.

Title: Solving functional equation $f(x)=3f(x+1)-3f(x+2)$

Body:
It is given that a function f(x) satisfy:
$$f(x)=3f(x+1)-3f(x+2)\quad \text{ and } \quad f(3)=3^{1000}$$ then find value of $f(2019)$.

I further wanted to ask that is there some general method to solve such equation. The method that I know to solve such questions is to substitute $x$ with $x+1$ in equation and there by making new equation which is
$$ f(x+1)=3f(x+2)-3f(x+3)$$ then again substitute $x$ with $x+2$ in original equation and make new equation $$f(x+2)=3f(x+3)-3f(x+4)$$ Do this for a couple of times and then on combining the equations, in most of the question we get some relation like f(x) = f(x+a) but that does not work here. Please share your ideas on how to solve such questions.


问:
Dealing with SDE involving $\max\{X_t,0\}\mathrm{d}W_t$
I am considering the SDE $\mathrm{d}X_t=\ln(1+X_t^2)\mathrm{d}t+\max\{X_t,0\}\mathrm{d}W_t$, with the initial condition $X_0=a$. I am posed with the following daunting tasks:

prove $\exists$ a unique strong solution $\forall$ $a$;

show that $X_t>0$ (respectively $X_t<0$) almost surely $\forall$ $t\geq0$ if $a>0$ (respectively $a<0$), and that the solutions are deterministic in these cases as well;

solve the SDE explicitly for $a=0$.

I am bamboozled by this SDE, particularly because I am unsure how to deal with the $\max\{X_t,0\}$ term since it is not differentiable (all I know is that it is convex). For the first part I know I should be looking at linear growth and the Lipschitz condition, but I haven’t been able to show it. Not to mention, $\ln(1+x^2)$ is not bounded. This leaves me unable to approach any of the above tasks. Any guidance and help is greatly appreciated!

答:
We can use Theorem 5.2.1 in Oksendal (http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf) which gives us the conditions for existence and uniqueness of a solution. In your case $b(t,x)= \ln(1+x^2)$ and $\sigma(t,x) = \max\{x,0\}$. Now $b,\sigma$ are clearly measurable functions since they are continuous. Further $$\Big|\frac{\partial}{\partial x}b(t,x)\Big| = \Big|\frac{2x}{1+x^2}\Big|\leq 1$$
Thus $|b(t,x)-b(t,y)|\leq |x-y|$ using the Mean Value Theorem. You can also directly calculated that $|\sigma(t,x)-\sigma(t,y)|\leq |x-y|$ by considering the possible cases. This shows that the Lipschitz constraint is satisfied. At the same time
$$|\sigma(t,x)|\leq |x| \leq 1+|x|$$
and
$$|b(t,x)|=|\ln(1+x^2)|=|\ln(1+x^2) - \ln(1+0)|=|b(t,x)-b(t,0)|\leq |x-0|=|x|\leq 1+|x| $$
Since the initial condition is a constant function, then the initial condition is independent of $\mathcal{F}_\infty= \sigma(\{W_t:t\geq0\})$. Thus we can use Theorem 5.2.1 to know that the SDE you have given has a unique solution $X_t$ for any constant $a$. Further $t\to X_t$ is a.s. a continuous function by the same theorem.

I need to think about this one, I will come and put some more work into it tomorrow but I will continue under the assumption that it is true....

To prove this, we will use Lemma 8.1.4 from Oksendal to solve this. What the lemma states is that if $g$ is bounded and continuous, then $u(a)=\mathbb{E}^a[g(X_t)]$ is continuous in $a$ for any fixed $x$. Now in $1)$ we showed that $t\to X_t$ is a.s. continuous and in $2)$ we showed that $X_t$ was deterministic. Thus on any compact interval $[0,T]$, the function $t\to X_t$ is bounded. Suppose that $X_t\in [0,M]$ for all $t\in[0,T]$, then we may find a continuous function $g_M$ such that $g_M(y)=y$ for $y\in[0,M]$ and is bounded outside of this interval. Thus $u(a)=\mathbb{E}^a[g_M(X_t)] =\mathbb{E}^a[X_t]=X_t^a $ is a continuous function in $a$ for all $a$. Here $X_t^a$ just indicates that $X_0=a$ and we again used that the process was deterministic. Now in $2)$ we showed that if $a < 0$, then $X_t < 0$ for all $t$. Thus $X_t^a$ must satisfy the differential equation
$$dX_t = \ln(1+X^2)dt $$
This is equivalent to the ODE
$$\frac{dx}{dt} = \ln(1+x^2) $$
with initial condition $x(0)=a$. If we can solve this ODE, then take the limit at $a\to 0^-$, then we should get the solution at $a=0$.


问:
Why don't introductory calculus textbooks usually introduce the notion of differentiability on closed intervals?
The title says it all. I find the apparent neglect of this idea rather unfortunate because (1) the notion of a function being differentiable on a closed interval is intuitively reasonable (and ought to be discussed), and (2) several theorems from calculus can be strengthened if they're stated in terms of closed interval differentiability.

Before proceeding further, it's worth taking time to make sure we agree on what we mean we say that a function is "differentiable over a closed interval". Here's the definition I had in mind:

A function $f:[a,b]\to\mathbb{R}$ is differentiable over $[a,b]$ if it is differentiable (in the ordinary sense) over $(a,b)$, right-differentiable at $a$, and left-differentiable at $b$.

As you can probably guess, a function $f$ is said to be left-differentiable at a number $x_0$ if the limit

$$\lim_{h\to 0^-}\frac{f(x_0+h)-f(x_0)}{h}$$

exists, with a similar definition applying to right-differentiability. The limits are then defined to be the left-hand and right-hand derivatives of $f$, respectively.

With that said, here's an example of a theorem (FTC1) that, under its usual hypotheses, can be strengthened (albeit slightly) if the notion of closed-interval differentiability is used:

If $f:[a,b]\to\mathbb{R}$ is continuous, then the function $F:[a,b]\to\mathbb{R}$ defined by $F(x)=\int_{a}^{x}f(t)\text{ }dt$ is differentiable over $[a,b]$. Moreover, for all $x\in[a,b]$, $F'(x)=f(x)$ (at the endpoints, $F'(x)$ is understood to be a left or right-hand derivative)

The return you get by applying this notion to this example is obviously minimal. That said, I truly believe that making these extra definitions is worthwhile anyway, namely because the resulting concepts strongly align with our intuitions.

To bring everything back together, I'll restate my question one last time: why don't introductory calculus textbooks usually introduce the notion of differentiability on closed intervals? Any and all responses are greatly appreciated.

Note: the "theorems" I mentioned earlier do not include existence theorems like Rolle's theorem or the MVT. I understand why these only require differentiability in the interior and why we shouldn't strengthen that requirement.

答:
Of course one could do this, it's convention. One issue I can see with your definition is as follows:

Consider $f: \mathbb R \rightarrow \mathbb R, f(x):=\lvert x \rvert $. With your definition, we would now say to the students "$f$ is differentiable on $(− \infty,0]$ and on $[0,\infty)$", but "$f$ is not differentiable on $(-\infty, \infty)$".

Whereas, as long as we insist the endpoints are special and say "$f$ is differentiable on both $(−\infty,0)$ and $(0,\infty)$, and also has a left and a right derivative at $0$", then anyone would see a red flag before concluding it were differentiable on all of $\mathbb R$.

Actually, this "gluing-compatibility" in my eyes is one of the strongest points in favour of definining differentiability on open sets. If $f$ is differentiable on $U$ and differentiable on $V$, then it should be differentiable on $U \cup V$. That is a beautiful property that breaks down too often when one extends the notion of differentiability to all kinds of non-open sets.


问:
Solve : $x^6 - 12x^5 + ax^4 + bx^3 + cx^2 + dx + 64 =0$ has all positive roots then the values of $a , b , c$ and $d$ are
My try : Tried to solve by relation between coefficient and roots of n- degree equations but unable to proceed  because of the variables. But just found the interesting thing that the signs of $a , b , c$ and $d$ are as same as in relation.

Answer : $a= 60, b= -160 ,c= 240 , d= -192$

答:
@farruhota 's approach is correct, but instead of expanding $\left(x-2\right)^{6}$ ,what you can do is use a little bit of permutation and combination: here's how:
let the roots be A,B,C,D,E,F.
now AB+BC+CD......(roots taken 2 at a time)= $\frac{a}{1}$
thus, $\frac{\left(AB+BC+CD......\right)}{15}\ge\left[\left(ABCDEF\right)^{5}\right]^{\frac{1}{15}}$

How 15, well six roots taken two at a time, then total possibilities are 6C2 = 15 and each root appears 5 times, so $\left(ABCDEF\right)^{5}$

Now, plug in the value of ABCDEF=64 (product of roots) and when you solve it you'll get AB+BC+CD......=a $\ge$ 60, but since roots are equal, equality will follow so you get a=60


答:
From the recurrence , we can deduce that
$$x_2 = 2$$
$$x_3 = 1 + \sqrt{2} $$
$$ x_4 = 1 + \sqrt{1+\sqrt{2}} $$
$$ x_5 = 1 + \sqrt{1+\sqrt{1+\sqrt{2}}} $$
You can continue . Let $z = x_n$ as $ n \to \infty $ you can write
$$ z = 1 + \sqrt{z} $$
Solving you get that $z = \frac{3 + \sqrt{5}}{2}$ ($z$ is positive real) .
Also to prove $$ \frac{x_n}{x_{n-1}} \geq 1 $$
You can write it as
$$ 1 + \sqrt{x_n} \geq x_n $$
$$ 0 \geq x_n^2 -2x_n + 1 $$
Which is true from the limit as let $$f(x) = x^2 -3x +1$$ You can write it as $$f(x) = \left(x - \frac{3 + \sqrt{5}}{2}\right)\left(x - \frac{3 - \sqrt{5}}{2}\right) $$
and for $f(x) \leq 0$ , $x \in [\frac{3 - \sqrt{5}}{2} , \frac{3 + \sqrt{5}}{2}] $ .

Title: Show $x_{n+1} = 1 + \sqrt{x_n}$ is upper-bounded and increasing

Body: Given $x_1 = 1$, it is clear that $x_n \ge 1$ for any $n \in \Bbb{N}$. I know that {$x_n$} is obviously increasing and limit as $n \to \infty$ is $\frac{3 + \sqrt{5}}{2}$, so {$x_n$} has an upper bound also. I seem to have trouble proving $x_{n+1} / x_n \ge 1$ and don't even have any idea on how to prove the sequence is upper-bounded. Can you guys give me some hints?


问:
Polar Coordinate function of a Straight Line
I was having some problem when trying to come out a polar coordinate function with straight line equation.

I know it is not good to post images here, but please bear with me as the question requires us to solve the equation from the straight line in the image.

What I have done is I tried to come out with an implicit function for the straight line L.

```<code>2x + 3y - 6 = 0
</code>```

But then I not sure how can I continue from there to come out with a polar coordinate function. Any ideas?

Thanks!

答:
Given the linear equation in two intercept form, $\frac{x}{a}+\frac{y}{b}=1$.

The equivalent polar form is $r=\frac{a*b*\sec(θ)}{b+\tan(θ)}$. Note, however, that $r=\frac{a*b}{b*\cos(θ)+a*\sin(θ)}$ requires one more variable lookup but is well-defined when $θ=π*n, n\in\mathbb{Z}$, instead of involving a division by $0$.


问:
Why does $|dz|=-ir\frac{dz}{z}$ when $|z|=r$?
Sometimes I want to compute a line integral over some circle $|z|=r$, where I have $|dz|$ instead of $dz$ given to me.

Reparametrizing with $z=re^{it}$, it follows that $dz=rie^{it}dt=izdt$. But I always read that
$$
|dz|=|iz|dt=|z|dt=|z|\frac{dz}{zi}
$$
so $|dz|=-ir\frac{dz}{z}$. In the first equality, why does $|dz|=|iz|dt$ instead of $|iz||dt|$? Why doesn't the absolute value extend to the $dt$ as well?

答: Note that $t$ is already a real number and it's conventional to integrate with an increasing $t$, i.e.$\int_{a}^{b} f(t)dt$ with $a\leq b$.


问:
Prove $0 \leq \frac{b-a}{1-ab} \leq 1$ if $ 0 \leq a \leq b \leq 1$

Prove  that $0 \leq \frac{b-a}{1-ab} \leq 1$  if $ 0 \leq a \leq b \leq 1$

It is a pretty easy exersice. I´m starting learning inequalities and  I cant find how prove it.

I want prove that $0 \leq \frac{b-a}{1-ab}$

For it I say that $b-a  \geq 0$ since $a \leq b$, in addition $1-ab$ cannot be zero. it is $a \neq b^{-1}$
then $1-ab<1$ since $1<ab+1$ $a =b \neq 0$ (otherwise $a=b=0$ implies $ 0 \leq\frac{b-a}{1-ab}$)
therefore $\frac{b-a}{1-ab} \geq 0$

Now I want prove that $\frac{b-a}{1-ab} \leq 1$
for this I tryed use the absolute value to the expression, but I fail.
Too try use a triangle inequality, but too fail.
Any hint or advice was very helpful, if someone can tell or explain  how to solve this kind of problems in math was very helpful, because I´m not good for think in tricks for the inequality problems.

答:
If the problem says $a<b$ then we are only left to prove $\dfrac{b-a}{1-ab}<1$
or
$$ 1-\dfrac{b-a}{1-ab} = \dfrac{1-ab+a-b}{1-ab} = \dfrac{(1+a)(1-b)}{1-ab} $$
And we are done


答:
just rearrange the right part of the inequality (excluding $ab=1$):

$$\frac{b-a}{1-ab} \leq 1 \Leftrightarrow b-a\leq 1-ab \Leftrightarrow b(1+a)\leq 1+a$$
which is obviously true under the given conditions.

Title: Prove $0 \leq \frac{b-a}{1-ab} \leq 1$ if $ 0 \leq a \leq b \leq 1$

Body:
Prove  that $0 \leq \frac{b-a}{1-ab} \leq 1$  if $ 0 \leq a \leq b \leq 1$

It is a pretty easy exersice. I´m starting learning inequalities and  I cant find how prove it.

I want prove that $0 \leq \frac{b-a}{1-ab}$

For it I say that $b-a  \geq 0$ since $a \leq b$, in addition $1-ab$ cannot be zero. it is $a \neq b^{-1}$
then $1-ab<1$ since $1<ab+1$ $a =b \neq 0$ (otherwise $a=b=0$ implies $ 0 \leq\frac{b-a}{1-ab}$)
therefore $\frac{b-a}{1-ab} \geq 0$

Now I want prove that $\frac{b-a}{1-ab} \leq 1$
for this I tryed use the absolute value to the expression, but I fail.
Too try use a triangle inequality, but too fail.
Any hint or advice was very helpful, if someone can tell or explain  how to solve this kind of problems in math was very helpful, because I´m not good for think in tricks for the inequality problems.


答:
$$ 1+a = 1+a $$
$$ \Rightarrow b(1+a) \le 1+a $$
$$ \Rightarrow b+ab \le 1+a $$
$$ \Rightarrow b-a \le 1-ab $$
$$ \Rightarrow \frac{b-a}{1-ab} \le 1 $$


问:
Most general $A \subseteq \mathbb R$ to define derivative of $f: A \to \mathbb R$?
I was looking up the definition of the derivative in several books, and what was making me uneasy was the first sentence, generally along the lines of "let $f$ be defined on...". They don't seem to be able to agree on what $f$ should be defined on. Most books say "interval"; one says "open set"; another just says "let $f$ be a real-valued function"; etc. So I decided to do a little investigation.

At the end of the day, we always define the derivative as the limit of a difference quotient (two versions), so we need to look into limits of functions. It turns out that the limit of $f$ at $a$ is only defined for $a \in (\operatorname{dom} f)'$ (we'll denote the set of cluster points of $A$ by $A'$ for convenience), because otherwise, the $0 < |x - a| < \delta$ part of the definition can be made false by choosing $\delta$ small enough to isolate $a$, thereby making the implication vacuously true, which gives non-unique limits, but we don't want that. So at this point, we know that $a$ must be a cluster point of the domain of the difference quotient.

A slight digression: It's easy to show that if $B$ is a finite set, then $A' = (A \cup B)' = (A \setminus B)'$. (That is, "adding" or "subtracting" a finite number of points doesn't affect the "stickiness" of a set.)

Let $A \subseteq \mathbb R$ and let $f: A \to \mathbb R$. For each $a \in A$, we can define a difference quotient function $q_a: A \setminus \{a\} \to \mathbb R$ by $$q_a(x) = \frac{f(x) - f(a)}{x - a}.$$

Now, (in a rather perverted pseudo-self-referential manner), we can define $$f': \{a \in A \cap A': \lim_{x \to a} q_a(x) \in \mathbb R \} \to \mathbb R \qquad \text{by} \qquad f'(a) = \lim_{x \to a} q_a(x).$$

We need $a \in A$ because $f(a)$ is needed to evaluate $q_a$, and we need $a \in A' = (A \setminus \{a\})' = (\operatorname{dom} q_a)'$ for the limit to be defined. So really, derivatives can be defined on sets that are much more general than intervals: we really only need $\operatorname{dom} f \cap (\operatorname{dom} f)' \neq \varnothing$ in order to talk about the derivative of $f$ at some point (of course, the derivative itself need not exist, but at least we can talk about it not existing).

And then I started getting worried... The function $f: \mathbb R \to \mathbb R$, defined by $$f(x) = \begin{cases} x, & x \in \mathbb Q, \\ 0, & x \notin \mathbb Q, \end{cases}$$ has no derivative anywhere. But if we restrict the domain to $\mathbb Q$, then suddenly every point is differentiable with derivative $1$? Now I'm starting to wonder if I made some stupid mistake in the development above, but I can't seem to find it.

So the question is, is the bolded statement above true?

答:
Consider the statement:

Let $A, B$ be subsets of the domain of $f$. If $f$ is differentiable on $A$ and on $B$, then $f$ is differentiable on $A \cup B$.

Sounds reasonable. It is true as long as we define differentiability only on open sets, but becomes wrong as soon as we allow even just endpoints (which are of course accumulation points) of intervals: Consider $f(x) =\lvert x \rvert$ and $A=(-\infty, 0]$, $B=[0, \infty)$.

Of course one can have reason to disregard this issue. But one should be aware of it.


问:
How do you find the second constant in a parabolic pde solution?
Okay, so I am given the following Parabolic PDE:

$y^2u_{xx}-2xyu_{xy}+x^2u_{yy}=x^{-1}y^2u_x+y^{-1}x^2u_y$

I find the characteristics to be:

$\frac{dy}{dx} = \frac{-x}{y}$ and therefore $\psi(x,y) = x^2+y^2$

Now, I have been told that I need to choose the other constant, $\eta$. In the answer for the question, the lecturer decides to choose $\eta(x,y) = y^2$ but gives no reason as to why he does. I decided to choose $\eta(x,y) = x^2$ but this does not give the same answer.

Is there a method on how to know what value to choose for the other constant?

答:
The change of variables should be non-singular, so you should be able to reconstruct the original independent variables $x$ and $y$ from the new ones. So your choice is also valid! It might lead to a different answer, that's ok, they're both valid, just with different lower order terms. Just make sure you did indeed end up with the (same) canonical form.

The reason you have this degree of freedom is because the characteristic equation has a zero determinant. So, just like when a quadratic equation has a zero determinant there's one ("double") root, so too here, there is only one root, which leaves a degree of freedom in choosing the second variable.


问:
The Lefschetz Number: A Geometrical Interpretation
Well, my question is rather brief and straightforward: does anyone know of any good geometrical interpretation of the Lefschetz number, specifically, if $f$ is a continuous map $X \rightarrow X$, $X$ being a compact triangulable space, then the Lefschetz number is defined by
$$\Lambda_f : = \sum_{k \geq 0} (-1)^k \text{tr} (f_* : H_k(X) \rightarrow H_k(X)).$$
I just kind of like having these constructs be slightly more tangible in my mind than mere abstract formulae.

Look forward to your answers!

答:
If you allow me to assume that permutations are geometric enough, we can  give a geometric relationship between trace and fixed points. Let $\sigma \in S_n$ be a permutation, and let $P_\sigma$ be its corresponding permutation matrix. A 1 along the diagonal of $P_\sigma$ corresponds to an index $k$ such that $\sigma(k) = k$, i.e a fixed point of $\sigma$. We see that $\text{tr}(P_\sigma)$ is counting the fixed points of $\sigma$.

With this in mind, we can follow this sketch to show that
$$f\ \text{has no fixed points}\ \implies \Lambda_f = 0.$$

First, after possibly subdividing the triangulation of $X$, we can use simplicial approximation to argue that $f$ is homotopic to a simplicial map $g$ that fixes no simplices. By the above comments, the induced simplicial chain maps $g_* : C_k(X) \to C_k(X)$ all have trace $0$. It follows that
$$\sum_{j \ge 0} (-1)^j \text{tr}(g_*: C_k(X) \to C_k(X) ) = 0.$$
This looks like the Lefschetz number! In fact, we can use some homological algebra (which goes by the Hopf Trace formula here) to argue
$$\sum_{j \ge 0} (-1)^j \text{tr}(g_*: C_k(X) \to C_k(X) ) = \sum_{j \ge 0} (-1)^j \text{tr}(g_*: H_k(X) \to H_k(X) ) = \Lambda_g = \Lambda_f.$$
The last equality follows from the homotopy invariance of homology.


答:
For boundedness from above just note that

$$x_n \leq 4\Rightarrow x_{n+1} = 1+\sqrt{x_n} \leq 1+\sqrt 4 =3 \leq 4$$

Now, the claim follows immediately by induction.


问:
Solving the functional equation $f(x)=f(1/x)$ for $x>0$
I am trying to solve the functional equation
$$f(x)=f\left(\frac{1}{x}\right),\quad x>0$$
for a real-valued function $f$, with $f(1)=2$. We may assume $f$ is continuous or infinitely differentiable. There is, obviously, the solution $f(x)=2$ for all $x>0$, but I have been unable to show that this is the only solution. By differentiating the original equation, one easily that $f^{(2n+1)}(1)=0$, where $f^{(k)}$ denotes the $k$th derivative of $f$, and $n$ is a non-negative integer. I cannot see this helping, though. By evaluating $f$ on $f(x)$ and applying the equation which we want it to satisfy, one gets that $f$ must satisfy
$$f\left(\frac{1}{f(x)}\right)=f\left(f\left(\frac{1}{x}\right)\right).$$
This doesn't seem to help either. Substituting values on the original equation yields nothing. Another observation is that
$$\lim_{x\to0+}f(x)=\lim_{x\to0+}f\left(\frac{1}{x}\right)=\lim_{x\to+\infty}f(x),$$
if the limit even exists. I know these are very superficial observations, but I am inexperienced with functional equations. Any hints?

答: Since $x\gt0,$ there is $t\in\mathbb{R}$ such that $x=e^t$ which transform your equation to $f(e^t)=f(e^{-t}).$ Hence the functional equation only tells us that $f\circ\exp$ is even. So, take any even function $g$ and let $$f(x)=g(\ln(x)).$$


问:
Finding Modulus and Argument for Complex Number $z$
Assume the Complex Number $z$ has argument $$\pi $$ and modulus $$\sqrt{3} $$ how would i find argument and modulus for $$-2iz$$ i already know the correct answer for this(it's from an old test) but i don't know how its calculated. I know how to find modulus and argument but im not sure about this exact math problem. Any help is greatly appreciated, please tell me if i should provide more info.

答:
We can compute the desired modulus and argument without evaluating $z$ (even though as indicated by @Chris, it is straightforward to do so):

Since $\arg(-2i)=-\frac\pi2$ and $\arg(z)=\pi,$

then working modulo $(-\pi,\pi],$ the principal argument of
$(-2i\;z)$ is $$(-\frac\pi2)+(\pi)=\frac\pi2.$$
Similarly, since $\arg(2i)=\frac\pi2,$

then working modulo $(-\pi,\pi],$ the principal argument of $(2i\;z)$
is $$(\frac\pi2)+(\pi)=-\frac\pi2.$$

Since $|-2i|=2,\;\;|2i|=2\;$ and $|z|=\sqrt{3},$ then
$$\left|-2i\;z\right|=2\sqrt{3}$$ and
$$\left|2i\;z\right|=2\sqrt{3}.$$


问:
$λ > 0$ such that $2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$ is true for all positive integers n.
Determine the value of the constant $λ > 0$ such that $2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$ is true for all positive integers n.

I tried equating it to n =  1 and found that $\lfloor \lambda \rfloor = 0 , 2$

Then I equated it to n =  2 and found out that $\lfloor \lambda\rfloor = 2$

I use the original equation to get (let x =  fractional part of lambda)

$2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$

=  $2\lfloor{(2+x) n}\rfloor = 1 - n + \lfloor (2+x) \lfloor(2+x) n \rfloor\rfloor$

->$4n+2\lfloor{nx}\rfloor = 1 - n + \lfloor (2+x)(2n+ \lfloor nx \rfloor\rfloor$

->$4n+2\lfloor{nx}\rfloor = 1 - n + \lfloor 4n +2\lfloor nx \rfloor + 2nx +x\lfloor nx \rfloor\rfloor$

->  $n-1 = \lfloor 2nx + x\lfloor nx \rfloor \rfloor$.

I do not know how to proceed, I tried doing $n-1<y< n$ but it didn't give me anything

答:
$$ \lfloor \lambda \lfloor \lambda n \rfloor \rfloor
   =2\lfloor{\lambda n}\rfloor + n - 1 $$

$$ 2\lfloor{\lambda n}\rfloor + n - 1 \le
   \lambda \lfloor \lambda n \rfloor \lt
   2\lfloor{\lambda n} \rfloor + n$$

$$ n - 1 \le (\lambda - 2) \lfloor \lambda n \rfloor \lt n$$

Since this has to be true for all $n$, we must have $\lambda > 2$.

So $$    (\lambda - 2)(\lambda n - 1)
     <   (\lambda - 2)\lfloor \lambda n \rfloor
     \le (\lambda - 2)\lambda n$$

Then

$$ n - 1 \le (\lambda - 2) \lfloor \lambda n \rfloor \le (\lambda - 2)\lambda n$$
$$ n - 1 \le (\lambda - 2)\lambda n$$
$$ n - 1 \le \lambda^2 n - 2\lambda n$$
$$ \lambda^2 n - 2\lambda n + n \ge 2n-1 $$
$$ (\lambda - 1)^2 n \ge 2n-1$$
$$\lambda \ge 1 + \sqrt{2 - \frac 1n} $$

Letting $n \to \infty$, we find $$\lambda \ge 1 + \sqrt 2 $$

Also, since

$$    (\lambda - 2)(\lambda n - 1)
     <   (\lambda - 2)\lfloor \lambda n \rfloor < n$$

Then

$$ (\lambda - 2)(\lambda n - 1) < n$$
$$\lambda^2n - (2n+1)\lambda < n - 2$$
$$\lambda^2 - \left(2+\frac 1n \right)\lambda < 1 - \frac 2n$$

Letting $n \to \infty$,

$$\lambda^2 - 2\lambda \le 1$$
$$\lambda^2 - 2\lambda + 1 \le 2$$
$$(\lambda - 1)^2 \le 2$$
$$\lambda \le 1+\sqrt 2$$

So $\lambda = 1 + \sqrt 2$


问:
$|\phi(G):\phi(H)|$ divides $|G:H|$

Let $\phi$ be a homomorphism defined on a finite group $G$, and let $H\subseteq G$. Show that $|\phi(G):\phi(H)|$ divides $|G:H|$.

Not quite sure where to start on this one. We have a theorem saying that $\phi(H)$ is a subgroup of $\phi(G)$, but what is $|\phi(H)|$? If $H$ contains the kernel $N$ of $\phi$, then I think everything behaves nicely and $|\phi(G):\phi(H)|=|G:H|$. But if $N\not\subseteq H$, I don't know what I can conclude.

答:
Hint: $\phi(H) = \phi(NH)$ and $|\phi(G):\phi(NH)| = |G: NH|$.

Note that finiteness of $G$ is not necessary, we only need $|G: H|$ to be finite.

Title: $|\phi(G):\phi(H)|$ divides $|G:H|$


问:
Show that the quadratic function $g(x)= 2x^2 - 4kx + k^2$ intersects the $x$-axis for all values of $k$.
Should I use $b^2 - 4ac$?
Is the answer $8k^2 > 0$?

答:
You'll have two real solutions when $b^2-4ac>0$, like you found. But, you can still have one real solution when $b^2-4ac=0$.

So if $b^2-4ac \ge 0$ for all values of $k$, then, you'll have at least one real solution for any value of $k$.

In this case, you have $b^2-4ac = 8k^2$, and since $8>0$ and $k^2 \ge 0$ (no matter what value of $k$ you plug in), you've shown that $b^2-4ac \ge 0$ for all values of $k$, and you're done!


问:
Differentiate vector transpose using rules
I am referring to Tom Minka's Old and New Matrix Algebra Useful for Statistics. I don't have the book by Magnus & Neudecker so I can't refer to the details of the theory.

Regarding rules (6): $d(XY) = (dX)Y + X(dY)$ and (12): $dX^*=(dX)^*$, I am not clear how to apply them. My notation used is numerator layout, i.e. $\dfrac{dx}{dx} = I$

Question 1.

$f(x)=x^Tx$ , $\dfrac{df}{dx}=2x^T$

However, if I use $\dfrac{df}{dx}= x^T\dfrac{dx}{dx} + \dfrac{dx^T}{dx}x$, firstly, $\dfrac{dx^T}{dx}$ is $1^T$? Second, according to rule (12), $\dfrac{dx^T}{dx} = (\dfrac{dx}{dx})^T = I^T = I $?

Question 2.

$f(x) = x^TAx$

$\dfrac{df}{dx}=x^T\dfrac{dAx}{dx}+ \dfrac{dx^T}{dx}(Ax) = x^TA + ???$

$???$ is supposed to be $x^TA^T$, however, it seems to me no matter $\dfrac{dx^T}{dx}$ equals $1^T$ or $I$ it does not give the expected result.

答:
Let's use the numerator-layout notation. First note that $\frac{dx}{dx}=I$ but $\frac{dx^T}{dx}=\begin{bmatrix}\begin{pmatrix}1&0&...&0\end{pmatrix},\begin{pmatrix}0&1&0&...&0\end{pmatrix},...,\begin{pmatrix}0&0&...&0&1\end{pmatrix}\end{bmatrix}$, a tensor, technically 1 x n x n. In denominator layout fashion, $\frac{dx^T}{dx}=\left[\begin{pmatrix}1\\0\\...\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\...\\0\end{pmatrix},...,\begin{pmatrix}0\\...\\0\\1\end{pmatrix}\right]$, a n x 1 x n tensor. It is possible to imagine it as a 3D matrix with the entries behind one another rather than listed liked this.

The inner product is symmetric, e.g. $x^Ty=y^Tx=\langle x, y\rangle$. We have the following four scenarios directly applying the derivative to $x^Tx$ and $x^TAx$:

$$\begin{matrix}&\text{denominator layout}&\text{numerator layout}\\
\frac{d}{dx}x^Tx&\frac{dx}{dx}x+\frac{dx}{dx}x=2x&x^T\frac{dx}{dx}+x^T\frac{dx}{dx}=2x^T\\
\frac{d}{dx}x^TAx&\frac{dAx}{dx}x+\frac{dx}{dx}Ax=(A^T+A)x&x^T\frac{dAx}{dx}+x^TA^T\frac{dx}{dx}x^T(A+A^T)\end{matrix}$$

Therefore the rule is $\frac{d}{dx}\langle x, y\rangle=\frac{dx}{dx}y+\frac{dy}{dx}x$ in denominator layout and $\frac{d}{dx}\langle x, y\rangle=x^T\frac{dy}{dx}+y^T\frac{dx}{dx}$ in numerator layout. Multiplying through by $dx$ suggests that $d\langle x, y\rangle=(dx)y+(dy)x$ in denominator layout but $d\langle x, y\rangle=x^T(dy)+y^T(dx)$ in numerator layout. Therefore it is doubtful that $d\langle x, y\rangle$ is a scalar and you could freely take the transpose.

Conclusion

You should directly use the product rule.


Title: Show that the quadratic function $g(x)= 2x^2 - 4kx + k^2$ intersects the $x$-axis for all values of $k$.

Body:
Should I use $b^2 - 4ac$?
Is the answer $8k^2 > 0$?

Body:
HINT

What about completing squares?

\begin{align*}
2x^{2} - 4kx + k^{2} & = 2(x^{2} - 2kx) + k^{2}\\\\
& = 2(x^{2} - 2kx + k^{2} - k^{2}) + k^{2}\\\\
& = 2(x-k)^{2} - k^{2}
\end{align*}

Can you take it from here?


问:
condition for postive symmetric operator to be essentially self-adjoint

Suppose $A$ is a densely defined symmetric operator on Hilbert space which is positive. (a) prove $||(A+I)\phi||^2\geq ||\phi||^2+||A\phi||^2$ (b) Show $Ran(A+I)$ is closed if $A$ is a closed operator  (c)Prove that A is essentially self-adjoint iff $A^*\psi=-\psi$ has no non-zero solution.

I got stuck on part (c). We know  $(A\pm iI)$ has dense image iff $A$ is essentially image. Also, I am thinking about applying part (b) result to $A^*+I$. Anyone could possibly give me a hint?

答:
Theorem: Let $A : \mathcal{D}(A)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ be a densely-defined symmetric linear operator on a Hilbert space $\mathcal{H}$, and suppose that $\langle Ax,x\rangle \ge 0$ for all $x\in\mathcal{H}$. Then $A$ is essentially self-adjoint iff the range of $A+I$ is dense in $\mathcal{H}$.

The range of $A+I$ is not dense iff $\langle (A+I)x,y\rangle=0$ for some $y\ne 0$, which is equivalent to the existence of $y\in\mathcal{D}(A^*)\setminus\{0\}$ and $(A^*+I)y=0$.


答:
The important thing is finiteness, then we have that

$$|G:H|=|G|/|H|\text{,}$$

$$|\phi(G):\phi(H)|=|\phi(G)|/|\phi(H)|$$

and

$$|H:H\cap\ker\phi|=|H|/|H\cap\ker\phi|\text{.}$$

Here, remember that by the first isomorphism theorem and finiteness

$$|\phi(G)|=|G:\ker\phi|=|G|/|\ker \phi|$$

and

$$|\phi(H)|=|H:H\cap\ker\phi|=|H|/|H\cap \ker \phi|\text{.}$$

Now, do you see that everything is reduced to divide and rearrange terms in order to obtain the wanted divisibility?

Title: $|\phi(G):\phi(H)|$ divides $|G:H|$

Body:
Let $\phi$ be a homomorphism defined on a finite group $G$, and let $H\subseteq G$. Show that $|\phi(G):\phi(H)|$ divides $|G:H|$.

Not quite sure where to start on this one. We have a theorem saying that $\phi(H)$ is a subgroup of $\phi(G)$, but what is $|\phi(H)|$? If $H$ contains the kernel $N$ of $\phi$, then I think everything behaves nicely and $|\phi(G):\phi(H)|=|G:H|$. But if $N\not\subseteq H$, I don't know what I can conclude.


问:
Expected maximum deviation of lattice path
Suppose there is an election with two candidates, Alice and Bob. Each candidate receives $n$ votes. As the votes are counted in a random order, what is the expected value of the maximum difference between Alice and Bob's partial totals? Note that this is process is equivalent to choosing a random lattice path from $(0,0)$ to $(n,n)$, equally likely from all $\binom{2n}n$ possibilities, and looking at the maximum distance of this path from the diagonal $y=x$ (up to a factor of $\sqrt{2}$).

I heard the problem from this blog post, where it is phrased in terms of sock matching. I ask this question because I expect it has a nice approximate answer; namely, the author of that post crunched the numbers and found that the expected value is well approximated by $\sqrt{\frac{3}2n}$, as shown in the data below. I wonder if anyone knows if this problem appears in the literature, and if they can prove or refute this claim?

Here is what I know:

Let $M$ be the maximum difference in votes. In this other answer, I showed that for any $m\ge 1$, we have
$$
P(M<m) = \frac1{\binom{2n}n}\sum_{k=-\lfloor n/m\rfloor}^{\lfloor n/m\rfloor}(-1)^{k}\binom{2n}{n+km}\tag1
$$
You can then use this to compute $E[M]$, via $E[M]=\sum_{m=1}^n \big(1-P(M<m)\big)$. Furthermore, the expression in $(1)$ direcctly implies
$$
P(M<m)=\frac1{\binom{2n}n}\cdot\frac2m\cdot 2^{2n}\sum_{k=1}^{\lfloor m/2\rfloor} \bigg[\cos\left(\frac{(2k-1)\pi}{2m}\right)\bigg]^{2n}
$$
by using a roots-of-unity filter on the generating function $(x^{1/2}+x^{-1/2})^{2n}$. As $n\to \infty$, the main contribution is from the $k=1$ term, so $P(M<m)$ grows like $\frac1{\binom{2n}n}\cdot\frac2m\bigg[2\cos\left(\frac{\pi}{2m}\right)\bigg]^{2n}$.

A somewhat related problem is finding the expected height of a random Dyck path of order $n$, which is just a random lattice path conditioned on staying above the diagonal $y=x$. In $[1]$, de Bruijn, Knuth, and Rice show that the expected height of a Dyck path is asymptotically $\sqrt{\pi n}-\frac12$. Perhaps something similar to their methods works here.

Bruijn, de, N. G., Knuth, D. E., & Rice, S. O. (1972). The average height of planted plane trees. In R. C. Read
(Ed.), Graph Theory and Computing (pp. 15-22). Academic Press Inc. [PDF]

答:
It actually turns out that
$$
\boxed{E[M_n]={\log 2}\;(\pi n)^{1/2}-\tfrac12+O(n^{-1/2}).}
$$
Here is a very rough outline of the proof. Using the reflection principle, for any $k\ge 1$,
$$
P(M\ge k)=\sum_{j\ge 1}(-1)^{j+1}\binom{2n}{n+kj}\Big/\binom{2n}n
$$
Then, when computing $E[M_n]=\sum_{k\ge 1}P(M_n\ge k)$, you can collect like terms, and find there is a positive contribution of $\binom{2n}{n+k}$ for each odd divisor of $k$, and a negative one for each even divisor. Letting $d(k)$ be the number of divisors of $k$, the net number of times $\binom{2n}{n+k}$ added in is therefore $d(k)$ when $k$ is odd, and $d(k)-2\cdot d(k/2)$ when $k$ is even.
\begin{align}
E[M_n]
&=2\sum_{k\ge 1} d(k){\binom{2n}{n+k}}\Big/{\binom{2n}{n}}-4\sum_{k\ge 1} d(k){\binom{2n}{n+2k}}\Big/{\binom{2n}{n}}
\\&\approx 2\sum_{k\ge 1}d(k)\exp(-k^2/n)\quad\quad-4\sum_{k\ge 1}d(k)\exp(-4k^2/n)
\end{align}
The first sum is handled in the Knuth/de Bruijn/Rice paper, and they use complex analysis to show that, for all $m>0$,
$$
\sum_{k\ge 1}d(k)\exp(-k^2/n)
= (\pi n)^{1/2}\bigg(\tfrac14\log n+\tfrac34\gamma -\tfrac12\log 2\bigg)
+\tfrac14+O(n^{-m}).
$$
In their notation, this sum is $g_0(n)$, defined in equation $(26)$, with the approximation appearing at $(32)$. You can then tweak their calculations to show
$$
\sum_{k\ge 1}d(k)\exp(-4k^2/n)
= (\pi n/4)^{1/2}\bigg(\tfrac14\log (n/4)+\tfrac34\gamma -\tfrac12\log 2\bigg)
+\tfrac14+O(n^{-m}).
$$


问:
Name of vector which includes value and its derivatives
What would I call a vector that describes a quantity (Temperature in this case) and its first and second derivatives at a specific location? It seems that a vector like this would have a formal name.

For context, I am solving a BVP for temperature distribution through a cylinder and want to describe the value of the Temperature, its derivative, and second derivative in one vector.

答:
If you have $f:\mathbb R^3\rightarrow \mathbb R$, then the first derivative is called gradient $\nabla f$ and the second derivative is called the Hessian $\nabla^2 f$. The gradient is the direction of quickest increase in temperature (and the negative of the gradient is the direction of quickest decrease). The Hessian is symmetric. The gradient is 3x1 and the hessian is 3x3. I do not believe there is a term that includes all three in one vector, and the dimensions are different so it would be hard to put in one vector. For exapmle, the temperature is just a number. So you probably want something like a list. The first element of the list will be a number, the second element of the list will be a vector, and the third element will be a matrix. There is no name for this as far as I know.

If $f(x,y,z):R^3\rightarrow R$, then $\nabla f=\begin{pmatrix}\frac{\partial f}{\partial x}\\\frac{\partial f}{\partial y}\\\frac{\partial f}{\partial z}\end{pmatrix}$ and
$$h(f(x,y,z))=\begin{pmatrix}\frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial x\partial y}&\frac{\partial^2 f}{\partial x\partial z}\\\frac{\partial^2 f}{\partial y\partial x}&\frac{\partial^2 f}{\partial y^2}&\frac{\partial^2 f}{\partial y\partial z}\\\frac{\partial^2 f}{\partial z\partial x}&\frac{\partial^2 f}{\partial z\partial y}&\frac{\partial^2 f}{\partial z^2}\end{pmatrix}$$


问:
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$?
Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.

However, I just cannot get a proper proof.

Body:
Note, for $x\ge 1$
$$x>\sqrt{x^2-1}$$
$$4x>2x+2\sqrt{x^2-1}=(\sqrt{x+1}+\sqrt{x-1})^2$$
$$2\sqrt x>\sqrt{x+1}+\sqrt{x-1} $$
$$\sqrt x -\sqrt{x-1}> \sqrt{x+1}-\sqrt x$$


问:
Independent and uniformly distributed on $(\frac{1}{2},1]$
I have two random variables $X,Y$ which are independent and uniformly distributed on $(\frac{1}{2},1]$. Then I consider two more random variables, $D=|X-Y|$ and $Z=\log\frac{X}{Y}$. I would like to calculate both, the disitrbution functions $F_D(t), F_Z(t)$ and the the density functions $f_D(t),f_Z(t)$

To do that I think the first thing we need to do is to evaluate the density of the common distribution of $X$ and $Y$, but I do not know how to do that.

The only thing which is clear to me is the density and distribution function of $X$ and $Y$ because we know that they are uniform.

EDIT: Please read my own answer to this question. I need someone who can show me my claculation mistakes.

答:
I thought it makes more sense to write an answer myself, but I still need someones help to complete the calculations. Here is my summary:

For $F_D(t)=\int_{\frac{1}{2}}^1\int_{-t+y}^{t+y}f_X(x) dx f_Y(y) dy=\int_{\frac{1}{2}}^1\int_{-t+y}^{t+y} 2*2 dx dy=4t$ for $x\in(1/2,1]$

For $f_D(t)=P(D=t)=\int_{\frac{1}{2}}^1 \int_{\frac{1}{2}}^1 f_X f_Ydx dy=4*1/2*1/2=1$

For $F_Z(t)=\int_\mathbb R\int_0^{y e^t}f_X dx f_Y dy=\int_{\frac{1}{2}}^1\int_0^{y e^t} 2 *2 dx dy=e^t $

For $f_Z(t)=P(Z=t)=\int\int_{\{(x,y): x/y=e^t\}} f_X*f_Y dx dy=?$

Question: May you help me not doing some calculation mistakes, because I think the bounds from the integrals are false. I understood the idea of the example, but still have difficulties getting the correct numbers, especially in the second case (where does $\log(2)$ appears in wolfies answer?)


答:
Alternative approach that avoids Calculus:

square both sides.

LHS squared is $(2x-1) - 2\sqrt{x(x-1)}$

RHS squared is $(2x+1) - 2\sqrt{x(x+1)}$

So, taking LHS - RHS, the question reduces to whether

$-2 + 2\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 0.$

This is equivalent to asking whether

$\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 1.$

Edit

Stealing a caveat from Quanto's answer, note that when $x\geq 1$, the LHS above is clearly positive.

Again squaring both sides, the problem reduces to asking whether

$x[2x - 2\sqrt{x^2 -1}] > 1.$

This is answered by applying the scalar of $[2x + 2\sqrt{x^2 -1}],$ which is a positive scalar,
to both sides, so that the problem is transformed to asking whether

$$x[2x - 2\sqrt{x^2 -1}][2x + 2\sqrt{x^2 -1}] > [2x + 2\sqrt{x^2 -1}].\tag1$$

Since the LHS of equation (1) above simplifies to $4x$,

and since it is clear that $2\sqrt{x^2 - 1} < 2x$

the problem is resolved.

Title: How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$?

Body:
Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.

However, I just cannot get a proper proof.


问:
Does the complex function $f(z)$ have to be bounded on $|z|<1$?
Take the complex function $f(z)$ =$ \sum_{n=0} ^{\infty} z^n$ = $\frac{1}{1-z}$

I'm trying to make an argument to say that this function is bounded in the disc $|z|<1$. Originally I thought it should be true but I'm not sure if I'm actually able to prove this.

So firstly I looked at the fact that Cauchy-Hadamard guarantees absolute convergence for all z in the interior of the disc $|z|=1$. Therefore we know that $|f(z)|$= $ |\sum_{n=0} ^{\infty} z^n|$ $\leq$ $ \sum_{n=0} ^{\infty} |z|^n$ =$\frac{1}{1-|z|}$

The last expression can be made arbitrarily large since $|z|<1$ so that means that the fraction will tend to infinity, making it unbounded. Is this wrong? Did I miss something?

Are all complex functions represented by a Taylor Series with a finite radius of convergence unbounded on the interior of the disc of convergence?

答:
A power series $f(z) = \sum_{n=0}^\infty a_n z^n$ with finite convergence radius $R> 0$ can be bounded or unbounded on $B_R(0)$. For example:

$\sum_{n=1}^\infty \frac{1}{n^2}z^n$ is has convergence radius $1$ and is bounded on $B_1(0)$.

$\sum_{n=1}^\infty z^n$ is has convergence radius $1$ and is unbounded on $B_1(0)$.

Some (necessary, but not sufficient) criteria are:

If $\sum_{n=0}^\infty |a_n| R^n < \infty $ then $f$ is bounded on $B_R(0)$.

If all $a_n$ are non-negative real numbers and $\sum_{n=0}^\infty a_n R^n = \infty $ then $f$ is unbounded on $B_R(0)$, that is a consequence of Abel's theorem.


答:
These rules pertain to differentials not to gradients.

Let's use them properly, starting with your second example function.
$$\eqalign{
 f_2 &= x^TAx \\
df_2 &= dx^TAx+x^TA\,dx \\
   &= (Ax)^Tdx+(A^Tx)^Tdx \\
   &= (Ax+A^Tx)^Tdx \\
\frac{\partial f_2}{\partial x} &= (Ax+A^Tx) \\
}$$
Setting $A=I$ turns this into your first function. Therefore
$$\eqalign{
\frac{\partial f_1}{\partial x} &= (Ix+I^Tx) \;=\; 2x \\\\
}$$
There are no corresponding rules for gradients, because a gradient operation changes a vector into a matrix, and matrix multiplication is not commutative. Trying to apply the rules to gradients produces nonsense, as you have discovered.

Title: Differentiate vector transpose using rules

Body:
I am referring to Tom Minka's Old and New Matrix Algebra Useful for Statistics. I don't have the book by Magnus & Neudecker so I can't refer to the details of the theory.

Regarding rules (6): $d(XY) = (dX)Y + X(dY)$ and (12): $dX^*=(dX)^*$, I am not clear how to apply them. My notation used is numerator layout, i.e. $\dfrac{dx}{dx} = I$

Question 1.

$f(x)=x^Tx$ , $\dfrac{df}{dx}=2x^T$

However, if I use $\dfrac{df}{dx}= x^T\dfrac{dx}{dx} + \dfrac{dx^T}{dx}x$, firstly, $\dfrac{dx^T}{dx}$ is $1^T$? Second, according to rule (12), $\dfrac{dx^T}{dx} = (\dfrac{dx}{dx})^T = I^T = I $?

Question 2.

$f(x) = x^TAx$

$\dfrac{df}{dx}=x^T\dfrac{dAx}{dx}+ \dfrac{dx^T}{dx}(Ax) = x^TA + ???$

$???$ is supposed to be $x^TA^T$, however, it seems to me no matter $\dfrac{dx^T}{dx}$ equals $1^T$ or $I$ it does not give the expected result.


答:
\begin{align}
   \sqrt{x} - \sqrt{x-1}
   &= \dfrac{(\sqrt{x} - \sqrt{x-1})(\sqrt{x} + \sqrt{x-1})}{\sqrt{x} + \sqrt{x-1})} \\
   &= \dfrac{x-(x-1)}{\sqrt{x} + \sqrt{x-1}} \\
   &= \dfrac{1}{\sqrt{x} + \sqrt{x-1}}
\end{align}

\begin{align}
   \sqrt{x+1} - \sqrt{x}
   &= \dfrac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x})} \\
   &= \dfrac{(x+1)-x}{\sqrt{x+1} + \sqrt{x}} \\
   &= \dfrac{1}{\sqrt{x+1} + \sqrt{x}}
\end{align}

So
\begin{align}
   \sqrt{x+1} &> \sqrt{x-1} \\
   \sqrt{x+1} + \sqrt x &> \sqrt x + \sqrt{x-1} \\
   \left(\dfrac{1}{\sqrt{x+1} + \sqrt x}\right)
      &< \left(\dfrac{1}{\sqrt x + \sqrt{x-1}}\right) \\
   \sqrt{x+1} - \sqrt x &< \sqrt x - \sqrt{x-1} \\
\end{align}


问:
If a two variable smooth function has two global minima, will it necessarily have a third critical point?
Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a  $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$, and then imagine an endless rain pouring with water level rising steadily on the entire plane.

Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima.

Those two points are connected by a compact line segment $K$. As a continuous function, $f$ attains a maximum value $M$ on the set $K$. This means that when the water level has reached $M$, the two lakes will have been merged.

The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$.

It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$. But, can we prove that one always exists?

Follow-ups:

Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ($S^1\times S^1$) or the surface of a sphere ($S^2$). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima?

Similarly, what if we have local minima instead of global?

If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded - not sure that would be at all relevant, but who knows).

答:
In 2019, I posted an answer to a relevant question. See: Can a multivariate function only have local minimum?, and Can a smooth function with compact sublevel sets only admit local minimizers?

In [1], some examples are given.

The function $f(x, y) = (x^2-1)^2 + (x^2y-x-1)^2$ has exactly two stationary points $(-1, 0), \ (1, 2)$ which are both strictly local minima (also are both global minima). There is no another stationary point.

The function $f(x,y) = -\mathrm{e}^{-x} (x\mathrm{e}^{-x} + \cos y)$ has infinitely many strictly local minima. There is no another stationary point.

Reference

[1] Alan Durfee, Nathan Kronefeld, Heidi Munson, Jeff Roy and Ina Westby,
“Counting Critical Points of Real Polynomials in Two Variables,”, The American Mathematical Monthly, Vol. 100, No. 3 (Mar., 1993), pp. 255-271.


答:
$$\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}\quad| +\sqrt{x}$$
$$2\sqrt{x} > \sqrt{x-1}+\sqrt{x+1} \quad|^2$$
$$4x>2x+ 2\sqrt{(x-1)(x+1)} \quad| -2x $$
$$2x>2\sqrt{(x-1)(x+1)} \quad| /2$$
$$x=\sqrt{x^2-1} \quad|^2$$
$$x^2>x^2-1 \quad|-x^2$$
$$0>-1$$
The transfoemations are valid in both directions, top down and bottom up


问:
Does it matter whether we consider field extensions as true supersets of a field?
I am trying to wrap my head around field extensions, and am using Pinter's abstract algebra. He presents a basic theorem on field extensions, namely that given any field $F$, and any polynomial $a(x)$ over $F$, we can always find a field $E$ containing $F$, and some element $c \in E$ s.t. $c$ is a root of $a(x)$ w.r.t. the operations in $E$.

As presented, the theorem seems to imply that $F$ is a 'genuine' subset of $E$. Also prior to stating and proving this theorem, we developed some theory where we assume a priori that we already have some extension $E$, with $F \subset E$.

The reason I am wondering, is that the proof we are presented with doesn't quite show that given any $F$ we have an extension $E$ with $F \subset E$. Rather, we show that actually, $F \cong H$, where $H$ is actually the subfield of $F[x]/(p(x))$ containing all the constant polynomials, and $p(x)$ is an irreducible factor of $a(x)$ in $F[x]$. So we actually prove that $F$ is isomorphic to another field $H$, and that this field $H$ has an extension, $F[x]/(p(x))$.

I assumed that since $F$ and $H$ are isomorphic, then defining a field $E_2$ such that $F \subset E_2$ and $E_2 \cong F[x]/(p(x))$ would be very easy. However I ran into problems actually coming up with this set and isomorphism. I took $E_2 = (E_1 \setminus H) \cup F$, and defined addition and multiplication on $E_2$ in a 'natural' way, but there were so many cases to consider that I struggled to actually prove it was even a field, specifically, with proving additive associativity in the most general case.

This leaves me with two questions: (i) can we simply consider $F$ and $H$ to be the same? That is, that $F[x]/(p(x))$ really is just an extension of $F$?
(ii) Is there an easy way to get a 'true' extension of $F$ easily from the extension of $H$?

答:
(i) NO.  (ii) NO.  (iii) Yes.

$\;\;\;\;\;$ By definition, a field $\Bbb E$ is an $\it{extension}$ $\it{field}$ of a field $\Bbb F$, i.e. $(\Bbb E/\Bbb F)$, when $\Bbb F$ is a $\Bbb E$ subring. Isomorphic fields are NOT necessarily equal. In the OP, $\Bbb F$ and $\Bbb H$ are not equal and should not be formally treated as such. It is also untrue that $\frac{\Bbb F[x]}{\langle p(x)\rangle}$ is an extension field of $\Bbb F$ or, equivalently, that $\Bbb F$ is a subfield of $\frac{\Bbb F[x]}{\langle p(x)\rangle}$  although $$\Bbb F\approx\widehat{\Bbb F}:=\{\alpha+\langle p(x)\rangle:\alpha\in\Bbb F\}$$
$\;\;\;\;\;$$\;\;\;\;\;$ As for the fundamental theorem from the OP, called Kronecker's theorem, given a $\Bbb F[x]$ irreducible $p(x)$ we can easily construct an extension field $\Bbb E$ of $\Bbb F$ in which $p(x)$ has a root by "identifying" $\Bbb F$ with $\widehat{\Bbb F}$. Let $$\Bbb E=(\frac{\Bbb F[x]}{\langle p(x)\rangle}\setminus \widehat{\Bbb F})\cup\Bbb F$$ Define a bijection $\psi:\Bbb E\to\frac{\Bbb F[x]}{\langle p(x\rangle)}$ by $\psi|_{\Bbb E\setminus\Bbb F}=\text{id}$ and $\psi|_{\Bbb F}:\Bbb F\rightarrow_{\small\approx}\widehat{\Bbb F}$. Finally, $$u\mathbf +v:=\psi^{-1}(\psi(u)+\psi(v))\;\;\;\;\;\;u\mathbf\cdot v:=\psi^{-1}(\psi(u)\cdot\psi(v))\;\;\;\;\;$$ defines on $\Bbb E$ a ring structure under which $\psi$ is a ring isomorphism and thus $\Bbb E/\Bbb F$.


问:
Is it possible to find a basis of $\mathbb{R}^4$ with only two vectors?
One of the exercises I did recently is the following:

Find a basis of the subspace of $\mathbb{R}^4$ consisiting of all vectors of the form:
$$ \left[
\begin{array}{c}
  x_1\\
  -5x_1 + x_2\\
  5x_1 - 8x_2\\
  -9x_1 + 7x_2
\end{array}
\right] $$

The solution to this problem is $\{(1,-5,5,-9), (0,1,-8,7)\}$.

How is it possible to have a basis of $\mathbb{R}^4$ with only 2 vectors? Don't I need 4 of them? With 2 vectors, the most I can get is a basis for $\mathbb{R}^2$, because they only span a plane, or am I completely wrong?

答:
Just  look carefully the statement  " find a basis  of the subspace of $ R^4$ "
You are to find the basis for the subspace $W$ which has dimension 2 ( look carefully)


问:
Find value of n in C(n,r), given r (combinations)
Hi
Working with combinatorial notation nCr = (n!)/(n!(n-r)!)
Note nCr is written as C(n,r) and C(n,r) = C(n,n-r)

The question is:

Find the value(s) of n if: C(n,3) = C(n,45)

Plugging that into combinatorial notation I get $\frac{n!}{3!(n-3)!} = \frac{n!}{45!(n-45)!}$
Cross multiplying and then dividing both sides by n! I'm left with $(45!)(n-45)! = (3!)(n-3)!$
This is where I'm stuck.

Anyone can show a way to proceed? Thx

答:
$45*44*...*4=(n-3)*(n-4)*...*(n-46)$

$$n-3=45\\
n-4=44\\
...\\
n-46=4$$

n=48 is certainly one possibility.
Now assume that you switch one of the equations such as the first two
$$n-3=44\\
n-4=45\\
...\\
n-46=4$$

You will always find there is no solution. So 48 is the only answer.

$$\frac{48!}{3!45!}=\frac{48!}{45!3!}$$


问:
Proof verification: $su(2)$ is not isomorphic to $sl(2,R)$
I’m trying to prove Lie algebra $sl(2,R)$ is not isomorphic to $su(2)$ by showing that all Lie algebra isomorphism will have determinant $0$, hence a contradiction. However, my computation gives a ridiculous result that said such isomorphism always exist. Please help me checking what’s wrong here.

I take the usual basis of $sl(2,R)$ such that
$$[X_1,X_2]=X_2$$ $$[X_1,X_3]=-X_3$$ $$[X_2,X_3]=2X_1$$
And the usual basis of $su(2)$ such that
$$[Y_i,Y_j]=\epsilon_{ijk}Y_k$$
Now suppose $sl(2,R)$ is isomorphic to $su(2)$, then there must exist a set of basis in $sl(2,R)$, which can be written as
$$X’_1=a_1X_1+b_1X_2+c_1X_3$$
$$X’_2=a_2X_1+b_2X_2+c_2X_3$$
$$X’_3=a_3X_1+b_3X_2+c_3X_3$$
satisfies
$$[X’_i,X’_j]=\epsilon_{ijk}X’_k$$

We then have
$$[a_1X_1+b_1X_2+c_1X_3,a_2X_1+b_2X_2+c_2X_3]\\=a_1b_2[X_1,X_2]+a_1c_2[X_1,X_3]+b_1a_2[X_2,X_1]+b_1c_2[X_2,X_3]+c_1a_2[X_3,X_1]+c_1b_2[X_3,X_2]\\=a_1b_2X_2+a_1c_2(-X_3)+b_1a_2(-X_2)+b_1c_2(2X_1)+c_1a_2(X_3)+c_1b_2(-2X_1)\\=(2b_1c_2-2c_1b_2)X_1+(a_1b_2-b_1a_2)X_2+(c_1a_2-a_1c_2)X_3$$
Which implies
$$a_3=2b_1c_2-2b_2c_1$$
$$b_3=a_1b_2-a_2b_1$$
$$c_3=a_2c_1-a_1c_2$$

The other $6$ conditions are obtained just by permutation $(1,2,3)$.

By above relationships, I finally get
$$det(a_i,b_j,c_k)=\frac{1}{2}a_1^2+2b_1c_1= \frac{1}{2}a_2^2+2b_2c_2= \frac{1}{2}a_3^2+2b_3c_3\\=\frac{1}{2}a_1^2+\frac{1}{2}a_2^2+\frac{1}{2}a_3^2=b_1c_1+b_2c_2+b_3c_3 $$
by all possible cofactors expansions. This result seems not going to give $det(a_i,b_j,c_k)=0$. I’m not sure what is going wrong here. I have checked my calculations more than $10$ times to make sure nothing goes wrong. Please help me, really thanks!

答:
Your calculations are entirely correct. They give no contradiction at all. They are just not quite finished, because over many fields, including $\mathbb R$, one can show with further calculations that your set of nine equations in nine variables has no solutions except all $a_i=b_i=c_i=0$ which of course we should exclude via the  tenth equation $$\det(\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3}) \neq 0.$$

Indeed, assume our variables $a_i,b_i,c_i$ take values in a field $K$ ($\mathrm{char}(K)\neq 2$) where there exist $x \in K, y \in K^*$ such that $x^2+y^2=-1$. Then

$$\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3} =\pmatrix{-1/y&x/2y&x/2y\\0&1/2&-1/2\\-x/y&-1/2y&-1/2y} $$

is a solution to the system. In particular if $K$ contains an element $i$ with $i^2=-1$ (like $K= \mathbb C$), we can choose $x=0, y=i$ and have the solution

$$\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3} =\pmatrix{i&0&0\\0&1/2&-1/2\\0&i/2&i/2} .$$

So the existence or non-existence of solutions depends on the arithmetic of the field $K$ (which otherwise can be any field of characteristic $\neq 2$). The precise statement is:

Theorem: For a non-trivial solution to the equations to exist, it is necessary and sufficient that $-1$ is a sum of two squares in $K$.

(This condition is sometimes phrased as "the level (French: niveau, German: Stufe) of $K$ is $\le 2$". It trivially includes the case that $-1$ is itself a square in $K$, but is strictly weaker than that: e.g. in all $p$-adic fields $\mathbb Q_p$ with $p \equiv 3$ mod $4$, one can write $-1$ as sum of two squares, but it is no square itself.)

Well, the examples above showed sufficiency, and what we want to see now is the "necessary" part (which obviously excludes any solutions for $K=\mathbb R$ or $\mathbb Q$). The existing solutions above, plus considerations of $ad$-eigenvalues of elements in the Lie algebra, guided me first to the following lemmata which exclude some special cases. They can quickly be derived from what you have. One just has to know what to look for.

Lemma 1: Assume there exists a non-trivial solution $\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3}$ with some $b_i=0$ or $c_i=0$. Then $a_i^2=-1$.

To prove this e.g. for $b_1$, insert the equations for $b_2, b_3$ into the one for $a_1$ and compare with $$b_2c_2+b_3c_3 = \frac{1}{2}a_1^2 + \underbrace{b_1c_1}_0$$ via your equations for the determinant.

(Conceptual reason: If $b_i$ or $c_i$ is zero, the $ad$-eigenvalues of $a_iX_1+b_iX_2+c_iX_3$ are just the square roots of $a$. But the $ad$-eigenvalues of all our $Y_i$ are $\pm\sqrt{-1}$.) $\qquad \square$

Likewise, one can derive (but we will not use it later)

Lemma 1': Assume there exists a non-trivial solution $\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3}$ with some $a_i=0$. Then for $\{j,k\} = \{1,2,3\}\setminus\{i\}$, we have $a_j^2+a_k^2=-1$.

To prove this e.g. for $a_1$, insert the equations for $b_2, b_3$ into the one for $b_1$ which w.l.o.g. (lemma 1) is $\neq 0$.

(Conceptual reason: Now we compute $\pm\sqrt{-1}$, the $ad$-eigenvalues of $a_iX_1+b_iX_2+c_iX_3$, as $\pm 2\sqrt{b_ic_i}$. Via the determinant equations, the square of this is $=a_j^2+a_k^2$.)  $\qquad \square$

One would hope that with enough effort, one can derive a similar contradiction in general; however, I have not been successful with that. So instead I tried to reduce to these special cases and got

Lemma 2: Let $m, n \in K$ with $m^2+n^2=1$. Then $$Y_1' := mY_1+nY_2, \qquad Y'_2:= nY_1-mY_2, \qquad Y'_3 := -Y_3$$ also satisfy $[Y'_i, Y'_j] = \epsilon_{ijk}Y'_k$, or in other words
$$\pmatrix{a'_1&b'_1&c'_1\\a'_2&b'_2&c'_2\\a'_3&b'_3&c'_3} := \pmatrix{ma_1+na_2&mb_1+nb_2&mc_1+nc_2\\na_1-ma_2&nb_1-mb_2&nc_1-mc_2\\-a_3&-b_3&-c_3}$$
is another solution to your equations.

Proof: Direct check. I derived this by just starting with $Y'_1 := mY_1+nY_2$ and computing what restrictions we need to keep the relations. In hindsight, over $K=\mathbb R$, this is just choosing an angle $\theta$, setting $m=\cos(\theta), n=\sin(\theta)$, and applying one of the simplest non-trivial isometry $\pmatrix{\cos(\theta)&\sin(\theta)&0\\ \sin(\theta)&-\cos(\theta)&0\\0&0&-1} \in SO(3)$ to the basis $Y_1,Y_2, Y_3$. However, to deal with far more general fields than $\mathbb R$, the above purely algebraic formulation will be handy.

But first let's quickly prove

Proposition: For $K= \mathbb R$, there are no non-trivial solutions.

Proof: Assume there is one. Because $-1$ is not a square in $\mathbb R$, by Lemma 1 we can assume w.l.o.g. that all $b_i \neq 0$. Then apply Lemma 2 with $m := \dfrac{-b_2}{\sqrt{b_1^2+b_2^2}}, n := \dfrac{b_1}{\sqrt{b_1^2+b_2^2}}$ which gives us a new solution, this time with $b'_1 =0$ but $b'_3 \neq 0$, contradicting Lemma 1. QED.

The same idea, but with a little more work, finishes the

Proof of the Theorem: So let $-1$ be not a sum of two squares in $K$, but assume there is a non-trivial solution. Again via Lemma 1 we can w.l.o.g. assume all $b_i \neq 0$. We further have $b_1^2+b_2^2 \neq 0$ (otherwise $-1 = \left(\dfrac{b_1}{b_2}\right)^2$). Consider $L := K(\sqrt{b_1^2+b_2^2})$ which is either $=K$ itself (like it was for $K=\mathbb R$, then the proof finishes like above) or a quadratic extension of $K$.

I claim that in this case, $-1$ is still not a square in $L$. For if it were, then there would exist $d,e \in K$ such that $-1= (d+e\sqrt{b_1^2+b_2^2})^2 = d^2 + (eb_1)^2 +(eb_2)^2 +2de\sqrt{b_1^2+b_2^2}$. Since the last summand must be $=0$, either $d$ or $e$ must be $=0$, and in both cases we would have written $-1$ as a sum of two squares in $K$, which we had assumed not to be the case. (Actually, one can even show that if $-1$ is a sum of three squares, then it can already be written as sum of two squares, but we don't need that here.)

But now our assumed non-trivial solution is one over $L$ as well, and we can apply Lemma 2 over that field with its elements $m := \dfrac{-b_2}{\sqrt{b_1^2+b_2^2}}, n := \dfrac{b_1}{\sqrt{b_1^2+b_2^2}}$, giving a new solution in $L$ with $b'_1=0$, which via Lemma 1 contradicts what we just established, that $-1$ is not a square in $L$. QED.

For a conceptual reason why solutions of quadratic forms over $K$ pop up here, compare e.g. https://math.stackexchange.com/a/3863613/96384: The relations describing the $X_i$, i.e. what we call $\mathfrak{sl}_2(K)$, correspond to the split quaternion algebra $\left(\dfrac{1,1}{K}\right) \simeq M_2(K)$; the relations describing the $Y_i$, i.e. what we call $\mathfrak{su}_2$, correspond to the quaternion algebra $\left(\dfrac{-1,-1}{K}\right)$; then the question is equivalent to whether the quaternion algebra $\left(\dfrac{-1,-1}{K}\right)$ is split, which is equivalent to whether $-1$ is a norm of the field extension $K(\sqrt{-1})\vert K$, which is equivalent to whether the quadratic form $X^2+Y^2+Z^2$ has a non-trivial zero in $K^3$, which is equivalent to the level of $K$ being $\le 2$.

Finally, to link this to the easy argument via the Killing form suggested by some people in the comments, the Killing form of $\mathfrak{sl}_2(K)$ is isotropic (and hence the unique (up to scalar multiple) isotropic form in three variables over $K$), whereas the Killing form of $\mathfrak{su}_2$ as described is $-X^2-Y^2-Z^2$; and again we see where the condition on $K$ comes from.


问:
Connection of weak and pointwise convergence
Suppose $Y_n:=v_n^{-1}\sum_{i=1}^n X_i\xrightarrow{}Z\sim N(0,1)$ in distribution where $(X_n)_{n\in\mathbb{N}}$ is a stationary sequence of real random variables with finite variance and $v_l$ is a appropriate standardisation.

Is it possible to assume that there exists a random variable $\tilde{Z}$ with $Y_n\xrightarrow{}\tilde{Z}\sim N(0,1)$ pointwise? Or do we have to change the probability spaces where $Y_n$ and $Z$ originate from as in Skorohod's representation theorem?

Thanks!

答:
In general, convergence in distribution tells us nothing about the actual behavior of the random variable itself. Example:
let $X_n\sim N(0,1)$, and $Y_n:= (-1)^n X_n$ . Then $Y_n\overset{D}{\to} X$ , but $P(\omega: \text{$Y_n(w)$  converges } ) = 0$.


答:
We know by identity that

$$\binom{n}{k} = \binom{n}{n-k} $$

So we can see that to find values of $n$ such that

$$ \binom{n}{3} = \binom{n}{45} $$

We could say that $k = 3$ and $n-k = 45$ which gives us $n=48$.

Title: Find value of n in C(n,r), given r (combinations)

Body:
Hi
Working with combinatorial notation nCr = (n!)/(n!(n-r)!)
Note nCr is written as C(n,r) and C(n,r) = C(n,n-r)

The question is:

Find the value(s) of n if: C(n,3) = C(n,45)

Plugging that into combinatorial notation I get $\frac{n!}{3!(n-3)!} = \frac{n!}{45!(n-45)!}$
Cross multiplying and then dividing both sides by n! I'm left with $(45!)(n-45)! = (3!)(n-3)!$
This is where I'm stuck.

Anyone can show a way to proceed? Thx


问:
Is there another proof for Euler–Mascheroni Constant?

Problem

Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent.

One Proof

This proof is based on the following inequality

$$\frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}$$

where $n=1,2,\cdots$, which will be used repeatedly.

On one hand, we obtain that $$\ln 2-\ln 1<1,~~\ln 3-\ln 2<\frac{1}{2},~~\ln 4-\ln 3<\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n<\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)<1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)>\frac{1}{n+1}>0.$$ This shows that $x_n$ is bounded below.

On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}>0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists.

Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.

答:
It follows from Riemann-Stieltjes integral that

$$
\sum_{k=1}^n\frac1k=1+\int_1^n{\mathrm d\lfloor x\rfloor\over x}=\log n+1-\int_1^n{\{x\}\over x^2}\mathrm dx
$$

Consequently, all we need is to show that the integral

$$
\int_1^n{\{x\}\over x^2}\mathrm dx
$$

converges as $n\to\infty$, and this trivially follows from the fact that $\{x\}\le1$.


问:
Connected sum of handlebodies has the homotopy type of a $1$-dimensional CW-complex
Let $M^3$ and $N^3$ be two compact connected $3$-dimensional handlebodies. It is easy to see that they have the same homotopy type of $1$-dimensional CW complexes. Is it true that their connected sum $M \# N$ has this same property (not boundary connected sum)?

答: It does not. That sphere in $M \# N$, which results from gluing $M - \text{(open ball)}$ and $N - \text{(open ball)}$, has nontrivial homology class in $H_2(M \# N)$.


问:
Definition of localization of rings
I'm trying to understand this definition of Hungerford's book:

The definition is simple, I think I understood what the author means, but...

What is $P_P$? because we will have $P_P=S^{-1}P$, with $S=P-P=\varnothing$.

What is $S^{^-1}P$, with $S=\varnothing$?

I'm sure it should be a silly think

I need help.

Thanks in advance.

答: No: $P_P$ is $S^{-1}P$ where still $S=R\setminus P$. In other words, it is the ideal generated by (the elements in) $P$ in the localized ring $R_P$.</p>

Title: Definition of localization of rings


答:
C(n,3)=C(n,n-3)=C(n,45)

From above n-3=45, n=48 is an obvious solution.

Now let 48+k, k>0 be a solution too. Then you get:

45!(3+k)!=3!(45+k)!

1=[46/4]*[47/5]...*[(45+k)/(3+k)]

But it is obvious RHS>LHS. So k>0 cant be true. Now assume solution is of form 48-k,k>0. Then:

45!(3-k)!=3!(45-k)!

1=[(3-k+1)/(45-k+1)]*[(3-k+2)/(45-k+2)]...

Now RHS<LHS. So this is not possible either. Therefore 48 is the only solution.


问:
What is the formula for the nth number in this series.
$1,3,7,9,11,13,17,19,21\dots$

Basically all the numbers that end in the digits $1,3,7,9$

I am working on a formula for approximating how many factors I have to test to find if a large number is prime. So for example to test the number $229,597$. How many possible factors will I have to check?

So the convention is to take the square root of $229,597$ which is approx $479$. Then I take $(\frac{479}{10}) \times (4) - 1$. I do this because out of every ten numbers there are 4 numbers that end in $1,3,7,9$. I subtract the 1 because prime numbers also have 1 as a factor.

So   $(\frac{479}{10}) \times (4) - 1  = 190$ factors to check to see if $229,597$ is prime.

Then I look at the series $1,3,7,9,11,13,17\dots$ to start checking for possible factors. But how do I find the nth number in this series?

答:
One possible formula is $$f(n) = \frac{1}{4}\left(5(2n-1) + (-1)^{n+1} + 2 \cos \frac{n \pi}{2} - 2 \sin \frac{n \pi}{2}\right), \quad n = 1, 2, \ldots.$$  This corresponds to the recurrence relation
$$\begin{align}
f(n) &= f(n-1) + f(n-4) - f(n-5), \\
f(1) &= 1, \\
f(2) &= 3, \\
f(3) &= 7, \\
f(4) &= 9, \\
f(5) &= 11.
\end{align}$$
I don't know why you would need it, though.


答:
This is not a proof.

When $x$ is large
$$\Big[\sqrt{x} - \sqrt{x-1}\Big]-\Big[\sqrt{x+1} - \sqrt{x}\Big]=\frac 1{4 x^{\frac 32}}\sum_{n=0}^\infty \frac {a_n}{x^{2n}}$$  all coefficients being positive make the decreasing sequence
$$\left\{1,\frac{5}{16},\frac{21}{128},\frac{429}{4096},\frac{2431}{32768},\frac{29393
   }{524288},\frac{185725}{4194304},\cdots\right\}$$


问:
$\operatorname{GL}_n(k)/k^\times \cong\operatorname{SL}_n(k)$
In a lot of references, people mention that $\operatorname{SL}_n(k)$ is a normal subgroup of $\operatorname{GL}_n(k)$ and $\operatorname{GL}_n(k)/\operatorname{SL}_n(k) \cong k^\times$ via the determinant map.

But I wonder if there is anything wrong with the following similar map

$$\operatorname{GL}_n(k)\to\operatorname{SL}_n(k), A\mapsto \frac{1}{\det A}A$$

This map is well-defined since $\det A\neq 0$ when $A\in\operatorname{GL}_n(k)$. The multiplication is preserved since $\det AB=\det A \det B$. This map is clearly surjective. So it induces $\operatorname{GL}_n(k)/k^\times \cong\operatorname{SL}_n(k)$.

Is there anything wrong here? I'm just surprised that I cannot find any source mentionging this one.

答: As pointed out in the comments, the given map does not map into $SL_n(k)$, and in general these groups are not isomorphic. For instance, if we consider the case of Lie groups ($k=\mathbb{C}$ or $k=\mathbb{R}$) then there is at most one map with prescribed differential at the identity. Any isomorphism would have the identity as differential at the identity, like the natural map $SL_n(\mathbb{C}) \to GL_n(\mathbb{C})/k^*$. However, this map has a finite kernel, namely the $n$th roots of unity times the identity (if $k=\mathbb{R}$, take $n$ even and $\pm I$). By uniqueness of maps corresponding to identity differential, there is no isomorphism between the groups.


问:
Primes Equidistribute in Ray Class Groups
Fix $K$ a number field and a modulus $\mathfrak m.$ Theory around the class number formula tells us that primes $\mathfrak p$ of $K$ equidistribute over all ideal classes in the normal class group, $J_K^1/P_K^{1,1}.$ (Here we use Dirichlet density.) Is it in general true that primes equidistribute over ideal classes in $J_K^{\mathfrak m}/P_K^{\mathfrak m,1}$?

Additionally, is there a corresponding "ray class number formula"? I am aware of the typical ray class number formula in terms of the class number (say, Milne V.1.7) but am interested if there's a residue of some $\zeta$-function around here.

答: After doing some thinking, I realized the answer is "yes" for somewhat unsatisfying reasons. Simply apply Chebotarev to the needed ray class field to get the result.</p>


答:
You can look at OEIS A000040 or any other list of primes. This does not include $1,9,15,25,27\ldots$ from your list because those are not primes.

You can generate the list of primes yourself using the sieve of Erastothenes.  For a long list of primes this will be faster than downloading it from the internet.  This is much faster than using all the numbers equivalent to $1,3,7,9 \pmod {10}$ because there are many fewer primes.  Roughly speaking, the number of primes below $n$ is $\frac n{\log n}$, which becomes less than $0.4n$ for $n \gt 13$

Added:  to answer the question as asked, if you start counting from $0$ the $n^{th}$ term in the series is $$10\cdot \left\lfloor \frac n4 \right\rfloor+\begin{cases} 1&n\equiv 0 \pmod 4\\3&n\equiv 1 \pmod 4\\7&n \equiv 2 \pmod 4\\9&n \equiv 3 \pmod 4 \end {cases}$$

Title: What is the formula for the nth number in this series.

Body:
$1,3,7,9,11,13,17,19,21\dots$

Basically all the numbers that end in the digits $1,3,7,9$

I am working on a formula for approximating how many factors I have to test to find if a large number is prime. So for example to test the number $229,597$. How many possible factors will I have to check?

So the convention is to take the square root of $229,597$ which is approx $479$. Then I take $(\frac{479}{10}) \times (4) - 1$. I do this because out of every ten numbers there are 4 numbers that end in $1,3,7,9$. I subtract the 1 because prime numbers also have 1 as a factor.

So   $(\frac{479}{10}) \times (4) - 1  = 190$ factors to check to see if $229,597$ is prime.

Then I look at the series $1,3,7,9,11,13,17\dots$ to start checking for possible factors. But how do I find the nth number in this series?


问:
What does it mean to equip a subset with the subspace topology?
This goes for equipping any set with any topology. Suppose that I want to equip a subset $A \subseteq X$ where $X$ is a topological space, with the subspace topology. I know that by definition of the subspace topology that $\tau_A = \{U \cap A | U \in \tau\}$. But what does this do to $A$?

答:
To talk about a topological space, you need a set and a topology on that set.

If you are given a topological space $X$, you're actually given a pair $(X, \mathcal T_X).$

Now, if you're given a subset $A \subseteq X$, you want to get a topological space out of it. To say that "$A$ is equipped with the subspace topology", we mean that we're looking at the topological space $(A, \mathcal T_A)$, where $\mathcal T_A$ is as you described.

Note that this mainly stems from the abuse of notation we do when we say something like "Let $X$ be a topological space", instead of using $(X, \mathcal T_X)$.


答:
A topology $\tau$ on a set $X$ is merely a collection of subsets of $X$ that conforms to certain rules (rules which have been chosen for their utility with respect to proving results elsewhere in mathematics$^\dagger$).

Suppose we have a set $X$ and a topology $\tau \subset \mathcal{P}(X)$. Then, for any subset $Y \subset X$, there always exists, whether explicitly acknowledged or not, what is called the subspace topology, $\tau_{_Y}$, on that subset, which is always defined with respect to an original topology, and whose open sets are, namely, where $Y$ intersects with each of the original open sets. This is to say: $U \cap Y$ is an element of $\tau_{_Y}$ whenever $U \subset X$ is an element of $\tau$.

$\underline{\textbf{Exercise}}$: Verify that $\tau_{_Y}$ as we have defined it is indeed a legitimate topology on $Y$, i.e. that it conforms to the rules a collection of subsets must follow to be a topology.

But the existence of $\tau_{_Y}$ doesn't "verb" anything to $Y$, which still exists independently as itself.

$^\dagger$If we know a given object is a particular kind of topological space, then everything we have proven about that kind of topological space can be automatically applied to that object. This is relevant because it turns out that a number of objects one studies in mathematics are topological spaces in disguise (with some overlying structure) such as your metric spaces from analysis. Cf. my post here along these lines; the motivation for why the topological axioms have been chosen as such is often overlooked at least at the start of the intro course, and I feel that if the motivation were properly addressed from the start, everything would become clear faster.

The idea of a subspace topology provides the rigorous framework upon which statements like this one from calculus can be made:

If a function $f:\mathbb{R} \to \mathbb{R}$ is continuous, then so too is the restriction of that function to a particular interval, say to $[0,1]$.

Here, it is implied that $[0,1]$ is inheriting the subspace topology from $\mathbb{R}$, which is endowed with the so-called Euclidean topology.

Title: What does it mean to equip a subset with the subspace topology?

Body: This goes for equipping any set with any topology. Suppose that I want to equip a subset $A \subseteq X$ where $X$ is a topological space, with the subspace topology. I know that by definition of the subspace topology that $\tau_A = \{U \cap A | U \in \tau\}$. But what does this do to $A$?


答:
$S$ is not $P \backslash P$ - it's still $R \backslash P$.

Look at an example: consider the localization $\mathbb{Z}_{(p)} = \{\frac{a}{b} : p \nmid b\, \; a,b \in \mathbb{Z}\} \subseteq \mathbb{Q}$ of $\mathbb{Z}$. The unique maximal ideal of this ring is $$p \mathbb{Z}_{(p)} = \{\frac{p \cdot a}{b} : p \nmid b, \; a,b \in \mathbb{Z}\}.$$

Title: Definition of localization of rings

Body:
I'm trying to understand this definition of Hungerford's book:

The definition is simple, I think I understood what the author means, but...

What is $P_P$? because we will have $P_P=S^{-1}P$, with $S=P-P=\varnothing$.

What is $S^{^-1}P$, with $S=\varnothing$?

I'm sure it should be a silly think

I need help.

Thanks in advance.


问:
absolute maximum and minimum over closed triangular region

Absolute maximum and absolute minimum value of $f(x,y)=20-8x+12y$ over the closed triangular region formed by the points $(0,0),(12 ,0),(12,16).$

What i try: firstci calculate partial derivative

$f_{x}(x,y)=-8$ and $f_{y}(x,y)=12$

Here partial derivativecare constant.

Means there is no critical point for function.

Now we will calculate the value of $f(x,y)$ in closed region

$\bullet$ Along $x$ axis (where $0\leq x\leq 12$)

$f(x,0)=20-8x$ for $x\in[0,12]$

So for maximum , put $x=0$ we have $f(0,0)=20$

For minimum , put $x=12$ we have $f(12,0)=-76$

$\bullet$ along parallel to $y$ i.e alnog $x=12$ line axis

$f(12,y)=12y-76$ where $0\leq y\leq 16$

For maximum , put $y=16$ we have $f(12,16)=12\cdot 16-76=$

For minimum , put $y=0$ we have $f(12,0)=-76$

Now we will calculate along line $y=(4/3)x$

Put int $f(x,y)=20-8x+12y=20+8x$

How do i solve it after that Help me please , thanks

答:
The approach used by Michael Rosenberg looks at the line intersection of the "level surfaces" $ \ f(x,y) \ = \ 20 \ - \ 8x \ + \ 12y \ \ = \ \ k \ $ with the specified triangle.  Another way to examine this is to consider "slices" through that triangle parallel to one of the coordinate axes; here, we will use slices $ \ y \ = \ Y \ $ and investigate the behavior of the function along them.  (Using "vertical slices" $ \ x \ = \ X \ $ will lead to a similar discussion.)

For any level surface of $ \ f \ $ , the plane described inclines "downward" with increasing $ \ x \ $  and upward with increasing $ \ y \ $ , so we might reasonably expect that the values of $ \ f \ $ are smaller near the vertex $ \ (12,0) \ $ and larger near the other vertices.  Along any slice $ \ y \ = \ Y \ $ , our function becomes $ \ f(x,Y) \ = \ (20 \ + \ 12Y) \ - \ 8x \ $ ; as this is a linear function, there is no relative extremum along the slice, so we only need to examine the endpoints.

The side connecting the origin to $ \ (12,16) \ $ lies on the line $ \ y \ = \ \frac{4}{3} \ x \ $ . so each of our slices covers the interval $ \left[ \ \frac{3}{4} Y \ , \ 12 \ \right]    $ for $ \ x \ . $  The values of $ \ f \ $ at the endpoints are then

$$  f \left(\frac{3}{4} Y \ , \ Y \right) \ = \ \ (20 \ + \ 12Y) \ - \ 8·\frac{3}{4} Y \ \ = \ \ 20 \ + \ 6Y \ \ , $$

$$  f \left(12 \ , \ Y \right) \ = \ \ (20 \ + \ 12Y) \ - \ 8·12 \ \ = \ \  12Y \ - \ 76 \ \ . $$

We see at once that $ f (0 \ , \ 0) \ =  \ 20  \ $ and $ \ f (12 \ , \ 0) \ =  \ -76  \ \ . $

We could then either use the coordinates of the third vertex, or simply note that the ends of the slices meet where the function values are equal; this suffices because both functions are increasing with increasing $ \ Y \ . $  Thus,

$$  f \left(\frac{3}{4} Y \ , \ Y \right) \ = \ \ f \left(12 \ , \ Y \right) \ \ \ \Rightarrow \ \ \ 20 \ + \ 6Y \ \ =  \ \  12Y \ - \ 76 $$

$$   \Rightarrow \ \ \ 96  \ \ =  \ \  6Y \ \ \ \Rightarrow \ \ \  Y \  =  \ 16 \ \ \ . $$

For the third vertex, we have

$$  f (12 \ , \ 16 ) \ \ =  \ \  12·16 \ - \ 76 \ \ = \ \ 116 \ \ . $$

So the absolute minimum for the triangular region occurs at  $ \ f (12 \ , \ 0) \ =  \ -76  \   $ and the absolute maximum at $ \ f (12 \ , \ 16) \ =  \ 116  \ \ . $


问:
The length of projection of the line segment joining the points $(1, –1, 0)$ and $(–1, 0, 1)$ to the plane $2x + y + 6z = 1$ is equal to?

The length of projection of the line segment joining the points $(1, –1, 0)$ and $(–1, 0, 1)$ to the plane $2x + y + 6z = 1$ is equal to?

I am confused if i have to find the length of perpendicular from line segment to plane or the length of projection

答:
Hint: For answering the problem, find the direction vector of the line joining the points, then find the angle between the line and the normal to the plane and if that angle is $\theta$, what would be the angle between the line and the plane?

Hence what would be the projection length?