问:
Is there a general procedure to generate theorems from axioms by hand?
I am checking a book of Jan Łukasiewicz "Elements of Mathmatical Logic".

In one section of the book (from page 42 I think), he worked around an axiom system like this one:

Axiom system of $L$:

(B1) $(\lnot B \rightarrow B) \rightarrow B$

(B2) $B \rightarrow (\lnot B \rightarrow C)$

(B3) $(B \rightarrow C) \rightarrow ((C \rightarrow D) \rightarrow (B \rightarrow D))$

the only rule of inference is modus ponens.

Then , like magic , he derived over 50 theorems from them(I can't even finish reading them)! I saw he used a technique like this:

Choose a thesis $\to$ substitute letters $\to$ left one is a thesis $\to$ right one is a new thesis (by MP)

The tricky part is substitution (for me).How did he(Jan Łukasiewicz) substitute just the right wf for the letters so that the left one becomes a thesis.Or did he checked all the substitution by hand? How long did it take him to do something like this?

答:
I remember a similar sort of amazement with Lukasiewcz when first reading that text.  I can't answer all of your questions.  I have other ones like "how did he have the confidence that his preferred three axiom set would work in the first place?"

No, there is not a general procedure to generate theorems from axioms by hand.

Suppose that there existed some general procedure.  Then, for all axioms we could generate at least one theorem.  But, consider the axiom CCNppp and suppose that is the only axiom we have.  We'll need to find two theorems of the type CCNppp in order to detach a theorem, since we can only use the rule of detachment, and the only theorem we know of is CCNppp.  One will have form C$\alpha$$\beta$, which can get called the major premise, and the other $\alpha$, which can get called the minor premise.  But, since C$\alpha$$\beta$ has the type CCNppp, this implies that $\alpha$ has the type CNpp also.  Any formula of the type CCNppp does not have a negation symbol in the second position, and cannot since substitution never produces a shorter meaningful expression than we started with.

We could also consider a system with the axioms CCpqCCqrCpr, CpCNpq, and CCNppp where CCNppp is always the major premise.  But then CCpqCCqrCpr and CpCNpq would have to have the type CNpp.  They don't since the second symbol of CCpqCCqrCpr is not an 'N', nor a variable, and because the consequent of meaningful expressions of the type CpCNpq is longer than the antecedent, while the antecedent of meaningful expressions of the type CNpp is longer than the consequent, and all substitutions for the meaningful expressions CpCNpq and CNpp preserve the relationship in length between the antecedent and consequent of the meaningful expressions.  That relationship gets preserved for CNpp since substitution is uniformly applied across a meaningful expression to all similar symbols of the same type, and for CpCNpq by a similar argument while also noting that any substitution in the consequent will preserve the property of the antecedent being shorter than the consequent.

Given that we've proved soundness, we could also argue that since CNpp is not a tautology, it is not possible to ever use CCNppp as a major premise.

How did Lukasiewicz find his substitutions?  I don't know.  My recollection of trying to redo the substitutions is that I found two mistakes.  Lukasiewicz does not also always find the most general possible theorem derivable from the axiom(s)/theorem(s) he uses when making a detachment.  What I do know is that in J. J. Zeman's book Modal logic: The Lewis-modal systems, in the introduction he uses diagrammatic techniques which the writing suggests to me that he got from A. N. Prior, and Prior collaborated with Meredith who studied with Lukasiewicz.

As an example, let's say we define a new system with the following formation rules.

'0' is a meaningful expression which is a constant.

The numerals '3' through '9' are meaningful expressions.

If 'x' is a meaningful expression, then '1x' is a meaningful expression.

If 'x' and 'y' are meaningful expressions, then '2xy' is a meaningful expression.

Nothing else is a meaningful expression in this language.

Suppose we have 2223403 as thesis, with the rule of inference '2xy' and 'x' as theses, allows us to detach 'y' as a thesis.  Can we derive anything from 2223403?  We can at least substitute numerals for '3' and '4' such that we'll have two meaningful expressions of the type 2228908 such that both of the expressions do not have any numerals in common which succeed '2' in the numeral sequence '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'.  2223403 and 2225605 are two such expressions.

Here's a diagram which helps to find a substitution, I think.

```<code>2 2 2 3     4 0 3
  | | |     | -
  | | ----- - |
  2 2 2 5 6 0 5
</code>```

Here we have a type to guide finding the major premise above the minor premise.  Thus, we need to find some form which matches or unifies 22340 with 2225605.

A collection of symbols like

```<code>|
-
</code>```

suggests a relationship where some sub-expression in the upper expression gets substituted with something in the lower expression.

A collection of symbols like

```<code>-
|
</code>```

suggests a relations where some sub-expression the lower expression gets substituted with something in the upper expression.

Substitutions getting applied to the shortest sub-expressions first, I think, makes for a good general rule.  In the above, due to left-to-right reading, I suppose, it might feel useful to substitution '3' in the upper with '256' in the lower.  But, '5' is in '256' and appears later in 2225605.  So, it makes more sense to apply substitutions for the shortest meaningful expressions first.

Thus, we substitute '4' with '0', and '5' with '0'.  And '3' with '206', since the '5' in '256' needed substituted with '0'.

So, the major premise we find from 2223403 is 22220600206, and the minor premise from 2223403 is not 2225606, but instead 2220600.  Thus, we can detach 206.  I'll note that 2223403 can get interpreted as meaning "if an arbitrary conditional implies a constant false proposition, then it's antecedent is true" and 206 can get interpreted as meaning "if a constant false proposition, then any proposition follows."

I suggest that to find a thesis from two theses, you first make sure that the theses you will use to find a derivation have completely distinct variables.  Then, try to make diagrams as above as aids and use as little substitution as possible.

Body:
I remember a similar sort of amazement with Lukasiewcz when first reading that text.  I can't answer all of your questions.  I have other ones like "how did he have the confidence that his preferred three axiom set would work in the first place?"

No, there is not a general procedure to generate theorems from axioms by hand.

Suppose that there existed some general procedure.  Then, for all axioms we could generate at least one theorem.  But, consider the axiom CCNppp and suppose that is the only axiom we have.  We'll need to find two theorems of the type CCNppp in order to detach a theorem, since we can only use the rule of detachment, and the only theorem we know of is CCNppp.  One will have form C$\alpha$$\beta$, which can get called the major premise, and the other $\alpha$, which can get called the minor premise.  But, since C$\alpha$$\beta$ has the type CCNppp, this implies that $\alpha$ has the type CNpp also.  Any formula of the type CCNppp does not have a negation symbol in the second position, and cannot since substitution never produces a shorter meaningful expression than we started with.

We could also consider a system with the axioms CCpqCCqrCpr, CpCNpq, and CCNppp where CCNppp is always the major premise.  But then CCpqCCqrCpr and CpCNpq would have to have the type CNpp.  They don't since the second symbol of CCpqCCqrCpr is not an 'N', nor a variable, and because the consequent of meaningful expressions of the type CpCNpq is longer than the antecedent, while the antecedent of meaningful expressions of the type CNpp is longer than the consequent, and all substitutions for the meaningful expressions CpCNpq and CNpp preserve the relationship in length between the antecedent and consequent of the meaningful expressions.  That relationship gets preserved for CNpp since substitution is uniformly applied across a meaningful expression to all similar symbols of the same type, and for CpCNpq by a similar argument while also noting that any substitution in the consequent will preserve the property of the antecedent being shorter than the consequent.

Given that we've proved soundness, we could also argue that since CNpp is not a tautology, it is not possible to ever use CCNppp as a major premise.

How did Lukasiewicz find his substitutions?  I don't know.  My recollection of trying to redo the substitutions is that I found two mistakes.  Lukasiewicz does not also always find the most general possible theorem derivable from the axiom(s)/theorem(s) he uses when making a detachment.  What I do know is that in J. J. Zeman's book Modal logic: The Lewis-modal systems, in the introduction he uses diagrammatic techniques which the writing suggests to me that he got from A. N. Prior, and Prior collaborated with Meredith who studied with Lukasiewicz.

As an example, let's say we define a new system with the following formation rules.

'0' is a meaningful expression which is a constant.

The numerals '3' through '9' are meaningful expressions.

If 'x' is a meaningful expression, then '1x' is a meaningful expression.

If 'x' and 'y' are meaningful expressions, then '2xy' is a meaningful expression.

Nothing else is a meaningful expression in this language.

Suppose we have 2223403 as thesis, with the rule of inference '2xy' and 'x' as theses, allows us to detach 'y' as a thesis.  Can we derive anything from 2223403?  We can at least substitute numerals for '3' and '4' such that we'll have two meaningful expressions of the type 2228908 such that both of the expressions do not have any numerals in common which succeed '2' in the numeral sequence '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'.  2223403 and 2225605 are two such expressions.

Here's a diagram which helps to find a substitution, I think.

2 2 2 3     4 0 3
  | | |     | -
  | | ----- - |
  2 2 2 5 6 0 5

Here we have a type to guide finding the major premise above the minor premise.  Thus, we need to find some form which matches or unifies 22340 with 2225605.

A collection of symbols like

|
-

suggests a relationship where some sub-expression in the upper expression gets substituted with something in the lower expression.

A collection of symbols like

-
|

suggests a relations where some sub-expression the lower expression gets substituted with something in the upper expression.

Substitutions getting applied to the shortest sub-expressions first, I think, makes for a good general rule.  In the above, due to left-to-right reading, I suppose, it might feel useful to substitution '3' in the upper with '256' in the lower.  But, '5' is in '256' and appears later in 2225605.  So, it makes more sense to apply substitutions for the shortest meaningful expressions first.

Thus, we substitute '4' with '0', and '5' with '0'.  And '3' with '206', since the '5' in '256' needed substituted with '0'.

So, the major premise we find from 2223403 is 22220600206, and the minor premise from 2223403 is not 2225606, but instead 2220600.  Thus, we can detach 206.  I'll note that 2223403 can get interpreted as meaning "if an arbitrary conditional implies a constant false proposition, then it's antecedent is true" and 206 can get interpreted as meaning "if a constant false proposition, then any proposition follows."

I suggest that to find a thesis from two theses, you first make sure that the theses you will use to find a derivation have completely distinct variables.  Then, try to make diagrams as above as aids and use as little substitution as possible.


问:
Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
Let's say I have one point that will be taken randomly from a normal distribution with mean $\mu_1$ and standard deviation $\sigma_1$.  Let's say I have another point that is taken much in the same way from another normal distribution with mean $\mu_2$ and standard deviation $\sigma_2$;.

How can I compute the probability, given $\mu_1$, $\mu_2$, $\sigma_1$, and $\sigma_2$, that my first point will be larger than the second?

I am sort of interested in the reasoning behind an "analytic" answer (or as analytic as you can possibly get with the normal distribution, which isn't that much), but I am more importantly looking for an algorithm of computing this probability, as it will be used in a simulation/model.

Does anyone know where I could get started on reasoning through this?

Note: For actual computation, having a table of values of the % of the curve within a given multiple of the standard deviation is feasible in my situation.

答:
Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then,
$$
{\rm P}(X_1  > X_2 ) = {\rm P}(X_1  - X_2  > 0) = 1 - {\rm P}(X_1  - X_2  \le 0).
$$
Now, by independence, $X_1 - X_2$ is normally distributed with mean
$$
\mu := {\rm E}(X_1 - X_2) = \mu_1 - \mu_2
$$
and variance
$$
\sigma^2 := {\rm Var}(X_1 - X_2) = \sigma_1^2 + \sigma_2^2.
$$
Hence,
$$
\frac{{X_1  - X_2  - \mu}}{{\sigma}} \sim {\rm N}(0,1),
$$
and so
$$
{\rm P}(X_1  - X_2  \le 0) = {\rm P}\bigg(\frac{{X_1  - X_2  - \mu }}{\sigma } \le \frac{{0 - \mu }}{\sigma }\bigg) = \Phi \Big(  \frac{-\mu }{\sigma }\Big),
$$
where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Thus,
$$
{\rm P}(X_1  > X_2 )  = 1 - {\rm P}(X_1  - X_2  \le 0) = 1 - \Phi \Big(  \frac{-\mu }{\sigma }\Big).
$$</p>

ContentLicense: CC BY-SA 3.0


问:
Linear Algebra, Parseval's Identity
How does one go about proving Parseval's identity?

Let ${v_1, v_2, ..., v_n}$ be an orthonormal basis for a a finite-dimensional inner product space $V$ over some field $F$. For any $x, y$ in $V$, prove:
  $\langle x, y \rangle$ $=$ $\sum\limits_{i = 1}^{n} \langle x, v_i \rangle \overline {\langle y, v_i \rangle}$.

Attempt: using $ x = \sum\limits_{i = 1}^{n} \langle x, v_i \rangle v_i$ and $ y = \sum\limits_{i = 1}^{n} \langle y, v_i \rangle v_i$

But I don't really know how to proceed from here.

答:
You were actually pretty close. With
$$x=\sum_i\langle x,v_i\rangle v_i$$
and
$$y=\sum_j\langle y,v_j\rangle v_j$$
you get

$$\langle x,y\rangle=\sum_i\langle x,v_i\rangle\sum_j
\overline{\langle y,v_j\rangle} \langle v_i,v_j\rangle =
\sum_i\langle x,v_i\rangle \overline{\langle y,v_i\rangle}$$

since $\langle v_i,v_j\rangle=\delta_{ij}$.

Title: Linear Algebra, Parseval's Identity


问:
3rd degree polynomial & solving for $f$
I am trying to find a closed form solution for $f$ in the following equation:

$0 = \mu - f (\mu^2 + \sigma^2) + f^2\lambda_3 - f^3\lambda_4$

This is an excerpt from the following SSRN paper which derives the Kelly Criterion to include the 3rd and 4th moments (which are typically omitted from the common $\mu/\sigma^2$ derivation. Derivation for the solution is not provided.

https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2956161

Please help!

答:
The polynomial is in the form where you can apply Cardano's formula. You can write:
$$t^{3}+pt+q=0$$
Where we applied a change of variable of the form:
$$f=t+\frac{\lambda_{3}}{3\lambda_{4}}$$
Once you do so, the term of degree $2$ will disappear, so that you can proceed.
Now you want to compute:
$$\Delta=\Big(\frac{q}{2}\Big)^{2}+\Big(\frac{p}{3}\Big)^{3}$$
The first solution is:
$$u=\Big(-\frac{q}{2}+\sqrt{\Delta} \Big)^{\frac{1}{3}}+\Big(-\frac{q}{2}-\sqrt{\Delta} \Big)^{\frac{1}{3}}$$
By First solution I mean the one you derive directly from this formula. If $\Delta>0$ then it will give you a real value, thus by dividing by $(f-u)$ you will find the other $2$ solutions (which can be complex or not).

If $\Delta<0$ then the same formula actually holds, but it will give you $2$ conjugated complex numbers, and it requires you to know how to handle complex roots.

Notice this is nothing original by me, it is how historically cubic equations were solved. To see documentation on it, you can search for Cardano's Formula.

ContentLicense: CC BY-SA 4.0

Title: 3rd degree polynomial & solving for $f$


问:
What exactly is a "successor" in the Extended Gale-Shapley algorithm?

Background

In the main Gale-Shapley algorithm, a Stable Matching (if one exists) within the conventional Stable Marriage Problem is achieved as follows:

```<code>INITIALIZE M to be an empty matching
WHILE (some man m is free)
    w <- first woman on m's list to whom m has not yet proposed
IF (w is unmatched)
    Add m-w to matching M
ELSE IF (w prefers m to current partner m')
    Replace m' - w with m - w in matching M
ELSE //w is already matched
</code>```

In the Extended Gale-Shapley algorithm, a weakly stable matching  from the Stable Marriage Problem with Indifference is achieved as follows:

```<code>INITIALIZE M to be an empty matching
WHILE (some man m is free)
    w <- first woman on m’s list
    Add m-w to matching M
    IF(some man m' is engaged to w)
        Remove m'- w from M
    FOR EACH successor m’' of m on w’s list
        Remove m'' - w from M
end
</code>```

The first difference I notice is that adding a pair to Matching M is automatic, and removal happens after that fact, rather then an active check to see if w is matched in the original algorithm.  However, I have trouble understanding what a "successor" is in this context.  What exactly is a successor in the case of ties, and how does the algorithm know how many w's or m's are part of a tie?  This confusion is what lead me to ask:

Question

What exactly is a "successor" in the context of the Extended Gale Shapley algorithm?

答:
As we learn from reading the paper, Wikipedia doesn't really know what's going on here.

The actual way to find weakly stable matchings is this:

Break all ties arbitrarily.

Use the Gale-Shapley algorithm.

The result is a matching that's stable with respect to the tiebroken preferences; this means that it must be weakly stable even without the tiebreaks. (A counterexample to weak stability is a man $m$ and woman $w$ that strictly prefer each other to their partners, and this strict preference is unaffected by breaking ties.)

The extended Gale-Shapley algorithm in the Wikipedia article is quoted directly from Irving's paper. It is still meant for solving the stable matching problem in the case with no ties, so $m'$ is a successor of $m$ on $w$'s list if $w$ prefers $m$ to $m'$, with no discussion of what happens if preferences are tied.

In fact, in all cases, Gale-Shapley and extended Gale-Shapley output exactly the same result; extended Gale-Shapley just does different book-keeping to get there. It deletes edges where Gale-Shapley just keeps track of which women a man has already proposed to, but ordinary Gale-Shapley would never want to use those edges. This may save a few checks down the line, but both algorithms are $O(n^2)$ in any case, so it doesn't save too many.

The extended Gale-Shapley algorithm exists to simplify the exposition in Irving's paper. Irving's other algorithms (the ones quoted in the rest of the Wikipedia article) are refinements of it: they also proceed by deleting edges. Those algorithms are meant to deal with ties, and are careful to only talk about "strict successors". If there are in fact no ties, those algorithms will end up taking the same actions as extended Gale-Shapley (and deleting the same edges).


问:
These functions are very similar. Is there an explanation?
Blue: $x \ln x-x$

Brown: $\ln \frac{\Gamma(x+1/2)}{\sqrt{\pi}}$

答: Well, I post my own answer. The real reason is that (up to some constant terms and factors), the first function is $D^{-1}D^{-1}[\frac1x]$ and the second function is $D^{-1}\Delta^{-1}[\frac1x]$. So, in one case we replace one normal integration operator with discrete integration. There is no surprise, the functions should be similar.


问:
Exponent equation with common power
Solve for $x$ in
$$\log_{2}(2^{x-1}+3^{x+1}) = 2x-\log_{2}(3^x)$$
I simplified by doing:
$$\log_{2}(3^x \cdot 2^{x-1} + 3^{2x+1}) = 2x$$
$$\frac{6^x}{2} + 3^{2x+1} = 2^{2x}$$
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$

Where do I go from here? I tried moving more numbers around, but I haven't been able to get any closer to solving for x. Any help is appreciated.

Title: Exponent equation with common power

Body:
Put $3^x=a$ and $2^x=b$. Then

$$6a^2+ab-2b^2=0$$
$$\Rightarrow 6a^2+4ab-3ab-2b^2=0$$
$$\Rightarrow (3a+2b)(2a-b)=0$$

Now $3a+2b$ is always positive. So $2a-b=0$.

Therefore, $$2\cdot 3^x = 2^x$$
$$\Rightarrow x=\log_{2/3} 2=\frac{1}{\log_2 (2/3)}$$


问:
Why is this proof of Goldbach's Conjecture flawed?
I know just as much as the next guy that this will be a far way off the mark, which is why I have phrased the question as "why is this wrong", not "is this wrong" - so let's do what mathematicians do best and poke holes in this (admittedly short) "proof" of the old-Gold conjecture :)

The conjecture goes like this, we all know:

$\forall n \in \mathbb{N}$, there exist two primes $p_a$ and $p_b$ such that $p_a+p_b=2n$. (statement 1)

For a prime $n$ (let's call it $p_N$ for simplicity), the solution to (1) is trivial, namely $p_N+p_N=2n$.

For a composite $n>3$, without loss of generality, let us say that $2\lt p_a \lt n \lt p_b \lt 2n-2$, and start by assuming the negation, namely:

$\exists n$ such that no $p_a,p_b$ satisfy (1). (statement 2)

Taking (1) mod $p_a$ and mod $p_b$,

$p_a\equiv2n\not\equiv0$ (mod $p_b$), and $p_b\equiv2n\not\equiv0$ (mod $p_a$). The non-equivalence to zero is guranteed by the bounds on a composite $n$, because neither $p_a$ nor $p_b$ can be equal to an even prime nor to $n$ itself. If, say, $p_a \mid n$, then $n=p_a +m\cdot p_a$, which would make $p_b=p_a+2m\cdot p_a$, which is clearly false by the definition of a prime. An equivalent, symmetrical argument can be made for if $p_b\mid n$.

In accordance with (2), we can see that gcd$(n,p_a) = 1$, and similarly that gcd$(n,p_b) = 1$, for all $p_a,p_b\in(2,2n-2)$. However, if gcd$(x,p_k)=1 ,\forall p_k\le \sqrt{x}$, that implies $x$ is prime. It is apparent that $\sqrt{n}\in(2,2n-2)$, and thus implies that n is prime. This contradicts with our assumption that n is composite, namely "If a composite number does not satisfy Goldbach's conjecture, it is a prime number, which satisfies Goldbach's conjecture".

QED?

P.S. Thank you to the MSE community at large for contuing to entertain "proofs" on long-standing questions. It is thanks to people like you that maniacs like me who believe they can solve these questions can avoid embarassing themselves to their PhD committee!

答:
The mistake is "for all $p_a, p_b \in (2,2n-2)$", and is a case of obfuscated context.

While it is true that if $\gcd (x,p_k) = 1, \forall p_k \le \sqrt x$ then $x$ is prime, not all primes in between $2$ and $2n-2$ is coprime to $n$. As shown in the first long paragragh, only primes satisfying statement 1 is coprime to $n$. If a prime $p < \sqrt n$ does not satisfy statement 1, i.e. $2n- p$ is composite, we cannot draw any conclusion on $\gcd(p,n)$. Any odd prime divisor of $n$ provides a counterexample.


问:
What functions are continuous in the Michael line?
The Michael line is given by the basis $B=\tau \cup \{\{x\}| x \in (\mathbb{R}-\mathbb{Q})\}$ where $\tau$ is the usual topology on $\mathbb{R}$. The Michael line has more open sets than the usual topology on $\mathbb{R}$, so I know that any functions that are continuous in the usual topology are also continuous in the Michael line. Which other functions are continuous in the Michael line?

答:
Let $\tau$ be the usual topology on $\Bbb R$, and let $\tau_M$ be the Michael line topology. Let $\Bbb P=\Bbb R\setminus\Bbb Q$. Let $X$ be any space, and let $f:\Bbb R\to X$. Each $x\in\Bbb P$ is isolated in $\tau_M$, so $f$ is automatically $\tau_M$ continuous at each $x\in\Bbb P$, and it’s easy to check that $f$ is $\tau_M$-continuous at a point $x\in\Bbb Q$ if and only if it is $\tau$-continuous at $x$, so we can conclude that $f$ is $\tau_M$-continuous if and only if it is $\tau$-continuous at each rational.

In particular, let $f:\Bbb R\to X$ be $\tau$-continuous, and let $F$ be a $\tau$-closed subset of $\Bbb P$. Then every function $g:\Bbb R\to X$ such that $$g\upharpoonright(\Bbb R\setminus F)=f\upharpoonright(\Bbb R\setminus F)$$ is $\tau_M$-continuous.


答:
I also think there is a problem here:

If, say, $p_a∣n$, then $n=p_a+m\cdot p_a$, which would make $p_b=p_a+2m\cdot p_a$

Since we are assuming that there are no $p_a, p_b$ satisfying $p_a+p_b=2n$:

if $p_b$ is another prime, then $p_a+p_b\neq 2n$, so in the quoted line it actually follows that $p_b\neq p_a+2m\cdot p_a$ (but as you noted, this is already impossible as $p_b$ should be prime and we haven't gained anything)

Title: Why is this proof of Goldbach's Conjecture flawed?

Body:
I know just as much as the next guy that this will be a far way off the mark, which is why I have phrased the question as "why is this wrong", not "is this wrong" - so let's do what mathematicians do best and poke holes in this (admittedly short) "proof" of the old-Gold conjecture :)

The conjecture goes like this, we all know:

$\forall n \in \mathbb{N}$, there exist two primes $p_a$ and $p_b$ such that $p_a+p_b=2n$. (statement 1)

For a prime $n$ (let's call it $p_N$ for simplicity), the solution to (1) is trivial, namely $p_N+p_N=2n$.

For a composite $n>3$, without loss of generality, let us say that $2\lt p_a \lt n \lt p_b \lt 2n-2$, and start by assuming the negation, namely:

$\exists n$ such that no $p_a,p_b$ satisfy (1). (statement 2)

Taking (1) mod $p_a$ and mod $p_b$,

$p_a\equiv2n\not\equiv0$ (mod $p_b$), and $p_b\equiv2n\not\equiv0$ (mod $p_a$). The non-equivalence to zero is guranteed by the bounds on a composite $n$, because neither $p_a$ nor $p_b$ can be equal to an even prime nor to $n$ itself. If, say, $p_a \mid n$, then $n=p_a +m\cdot p_a$, which would make $p_b=p_a+2m\cdot p_a$, which is clearly false by the definition of a prime. An equivalent, symmetrical argument can be made for if $p_b\mid n$.

In accordance with (2), we can see that gcd$(n,p_a) = 1$, and similarly that gcd$(n,p_b) = 1$, for all $p_a,p_b\in(2,2n-2)$. However, if gcd$(x,p_k)=1 ,\forall p_k\le \sqrt{x}$, that implies $x$ is prime. It is apparent that $\sqrt{n}\in(2,2n-2)$, and thus implies that n is prime. This contradicts with our assumption that n is composite, namely "If a composite number does not satisfy Goldbach's conjecture, it is a prime number, which satisfies Goldbach's conjecture".

QED?

P.S. Thank you to the MSE community at large for contuing to entertain "proofs" on long-standing questions. It is thanks to people like you that maniacs like me who believe they can solve these questions can avoid embarassing themselves to their PhD committee!


问:
What is the Fourier transformation in the k-space?
On wikipedia, we see the Fourier transformation examples in the case of the ordinary frequency and angular frequency scenario

https://en.wikipedia.org/wiki/Fourier_transform

But what is the transformation in the "k space" ? (with $k=2\pi/\lambda$)
Is it given somewhere ?

答:
It's the angular frequency case. Whether you want it to be unitary or not is your decision.

A physicist would write $e^{-i\omega t}$ or $e^{-kx}$ for the factors in the time and angular frequency case, and the space and space and angular wave number cases, respectively.


问:
Question about proof of MGF of sample mean (trivial double-check)
There is a part of the proof that I am unsure about for the following theorem:

Say you take a random sample of a random variable $X$ with probability distribution $\mathbb{P}$, i.e. $X\sim\mathbb{P}$. Now, suppose that this random variable $X$ has a Moment Generating Function $M_x$ on an interval $I$.

What this theorem says is that $M_{\bar{X}}(t)$, the Moment Generating Function of the sample mean at $t$, is equal to $(M_X(t/n))^n$, the Moment Generating Function of the actual random variable at $t$ divided by the number of samples, to the power of the number of samples, for any $t/n$ in the interval $I$.

Here is the proof: fix an arbitrary $t$ so that $t/n\in I$. Then $M_{\bar{X}}(t)=\mathrm{E}(e^{t\bar{X}})=\Pi_{i=1}^n\mathrm{E}(e^{t/n X_i})=(M_X(t/n))^n$.

Straightforward enough, but could somebody double-check my logic behind the last equal sign? Is it just the definition of a random sample being a collection of i.i.d. random variables?

Also, is defining $t/n\in I$ related to how the MGF must exist in a neighborhood of 0?

答:
\begin{align}
M_{\bar{X}}(t)
&= E[e^{t\bar{X}}] & \text{defn. of $M_{\bar{X}}$}
\\
&= E\left[\prod_{i=1}^n e^{\frac{t}{n} X_i}\right] & \text{$\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i$}
\\
&= \prod_{i=1}^n E[e^{\frac{t}{n} X_i}] & \text{$X_i$ are independent}
\\
&= \prod_{i=1}^n M_{X_i}(t/n) & \text{defn. of $M_{X_i}$}
\\
&= (M_{X}(t/n))^n & \text{$X_i$ are identically distributed, so $M_{X_i} = M_X$}
\end{align}


问:
Proving $\cos(a+b)=\cos a\cos b-\sin a\sin b$ starting from $\sin(a+b)$
The question is to prove the compound angle identity $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ starting from the $\sin$ compound angle identity. Before this, the task wants me to show that $\sin(\frac \pi 2 - x) = \cos(x)$ and I did not have any problems there. I guess I have to use this fact somehow so thats what I've tried:

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$
$$\sin\left(\frac \pi 2 - (a+b)\right)=\sin\left(\frac \pi 2\right)\cos(a+b)-\cos\left(\frac \pi 2\right)\sin(a+b)$$
$$\cos(a+b)=1\cos(a+b)-0\sin(a+b)$$
$$\cos(a+b)=\cos(a+b)$$
And that brings me nowhere. How can I prove this identity in the required way?

答: Direct calculation:$$\begin{align}\cos(a+b)&=\sin(\tfrac{\pi}{2}-a-b)\\&=\sin(\tfrac{\pi}{2}-a)\cos(-b)+\cos(\tfrac{\pi}{2}-a)\sin(-b)\\&=\cos a\cos b-\sin a\sin b.\end{align}$$


问:
I want to prove completeness of $C[a,b]$ with metric induced by max norm, but need help
I added a picture, please check it.

Can someone please explain the last three line of the proof?

I thought by finding a cauchy sequence that converges to a continuous function we are done with proving completeness of $C[a,b]$ with metric induced by max norm. Why did they add the last three lines "highlighted" and how do I prove them? Please, help, thanks.

答: The proof begins with a particular case: it assumes that the norm $$\|f_{n+k}-f_{k}\|_{\infty}\leq a_{k}$$ is controlled by $a_{k}$ real number which is the addend of a converging sum $\sum_{k}a_{k}$. The last $3$ lines are telling you how to adapt the particular case to the general one.


Title: Exponent equation with common power

Body:
Solve for $x$ in
$$\log_{2}(2^{x-1}+3^{x+1}) = 2x-\log_{2}(3^x)$$
I simplified by doing:
$$\log_{2}(3^x \cdot 2^{x-1} + 3^{2x+1}) = 2x$$
$$\frac{6^x}{2} + 3^{2x+1} = 2^{2x}$$
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$

Where do I go from here? I tried moving more numbers around, but I haven't been able to get any closer to solving for x. Any help is appreciated.

Body:
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$
$$2^x\cdot 3^x+2 \cdot 3^{2x+1} - 2^{2x+1}=0$$
divide by $2^x\cdot 3^x$
$$1+\frac{2 \cdot 3^{2x+1}}{2^x\cdot 3^x}-\frac{2^{2x+1}}{2^x\cdot 3^x}=0$$
$$1+6 \cdot\frac{ 3^{x}}{2^x}-2\cdot\frac{2^{x}}{ 3^x}=0$$
Set $\left(\frac23\right)^x=z$
$$1+\frac{6}{z}-2z=0$$
$$2z^2-z-6=0\to z_1=2;\;z_2=-\frac{3}{2}$$
$$\left(\frac23\right)^x=2\to x\ln \frac23=\ln 2\to x=\frac{\ln  2}{\ln 2 -\ln 3}$$
$\left(\frac23\right)^x=-\frac{3}{2}$ has no real solutions


问:
Trigonometric system with different coefficients
I have a trigonometric system as below and I want to solve it for $x$.

\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x)  = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}

Can anyone suggest a solution?
Regards.

答:
Rewrite the two equations as
$$\sqrt{B^2+C^2}\cos(2x-a) = D-A\cos x $$
$$\sqrt{B^2+C^2}\sin(2x-a) = F+A\sin x $$
with $a = \tan^{-1} \frac CB$. Square and add the two equations
$$B^2+C^2= D^2+F^2 +A^2 +2A(F \sin x - D\cos x)\\
= D^2+F^2 +A^2 +2A\sqrt{D^2+F^2} \sin (x-b)$$
with $b= \tan^{-1} \frac DF$. Now, it should be straightforward.


答: Let $(f_n)_{n\in\Bbb N}$ be a Cauchy sequence of elements of $C[a,b]$. There is a $N_1\in\Bbb N$ such that$$m,n\geqslant N_1\implies\|f_m-f_n\|_\max<1.$$And there is a $N_2\in\Bbb N$ such that$$m,n\geqslant N_2\implies\|f_m-f_n\|_\max<\frac12.$$You can assume, without loss of generality, that $N_2>N_1$. Now, there is a $N_3\in\Bbb N$ such that$$m,n\geqslant N_3\implies\|f_m-f_n\|_\max<\frac1{2^2}$$and you can assume, without loss of generality, that $N_3>N_2$. But then $(f_{N_k})_{k\in\Bbb N}$ is a subsequence of $(f_n)_{n\in\Bbb N}$ and$$(\forall k\in\Bbb N):\left\|f_{N_{k+1}}-f_{N_k}\right\|<\frac1{2^{k-1}}.$$So, by the first part of the proof, $\left(f_{N_k}\right)_{k\in\Bbb N}$ converges uniformly. In particular, the sequence $(f_n)_{n\in\Bbb N}$ has a convergent subsequence. But then $(f_n)_{n\in\Bbb N}$  itself is convergent, since, as the author remarks, in order to prove that a Cauchy sequence converges it is enough to prove that it has a convergent subsequence.

Title: I want to prove completeness of $C[a,b]$ with metric induced by max norm, but need help

Body:
I added a picture, please check it.

Can someone please explain the last three line of the proof?

I thought by finding a cauchy sequence that converges to a continuous function we are done with proving completeness of $C[a,b]$ with metric induced by max norm. Why did they add the last three lines "highlighted" and how do I prove them? Please, help, thanks.


答: This is an example of physics. $k$ is the wave vector. Most problems are easier to solve in the "k space". Maybe take a look at https://en.wikipedia.org/wiki/Wave_vector

Title: What is the Fourier transformation in the k-space?

Body:
On wikipedia, we see the Fourier transformation examples in the case of the ordinary frequency and angular frequency scenario

https://en.wikipedia.org/wiki/Fourier_transform

But what is the transformation in the "k space" ? (with $k=2\pi/\lambda$)
Is it given somewhere ?


问:
What is the relation between the linear transformation B and C?
If  $\left\{ \left. {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right\} \right.$and $\left\{ \left. {{y}_{1}},{{y}_{2}},...,{{y}_{n}} \right\} \right.$ are basis of a vector space V ,and for every $j=1,2,...n$

$B{{x}_{j}}=\sum\limits_{i=i}^{n}{{{a}_{ij}}{{x}_{i}}}$    and $C{{y}_{j}}=\sum\limits_{i=i}^{n}{{{a}_{ij}}{{y}_{i}}}$.
What is the relation between the linear transformation B and C ?

my intuition is that they are similar .I tried to subtrabct B and C .Thats so far

答:
If $\varphi: V\to V$ is the linear transformation defined by $\varphi(x_i)=y_i$, then
\begin{equation}
\varphi^{-1}\circ C\circ\varphi(x_j) = \varphi^{-1}\circ C y_j = \varphi^{-1}\left(\sum_j a_{i j} y_i\right) = \sum_j a_{i j} x_i = B x_j
\end{equation}
Hence
\begin{equation}
B = \varphi^{-1}\circ C\circ\varphi
\end{equation}


问:
Property of trace in optimization problem
If I consider the following optimization problem

\begin{align}
&\min_{\phi \in \mathbb{R}^d} \|X- \phi \phi^T X\|^2_F\\
&\text{s.t. } \, \, \, \phi^T\phi=1
\end{align}

where $X$ is a $d \times n$ matrix and $\|\cdot\|_F$ denotes the frobenius norm. I want to show that this is just an eigenvalue problem.

With some few steps I can write $\|X- \phi \phi^T X\|^2_F = \operatorname{tr}(X^TX - XX^T\phi\phi^T)$ that I now consider as the objective function. So writing the lagrangian (and using linearity of trace) I get

\begin{equation}
\mathcal{L} = \operatorname{tr}(X^TX) - \operatorname{tr}(XX^T\phi\phi^T) -\lambda(\phi^T\phi-1).
\end{equation}

Then,

\begin{equation}
 \mathbb{R}^d \ni\frac{\partial\mathcal{L}}{\partial\phi} = -\operatorname{Tr}(2XX^T\phi) - 2\lambda\phi.
\end{equation}

where I've used again linearity of trace and then commutation with derivative.

\begin{equation}
\frac{d(\operatorname{tr}(f(X))}{dX} = \operatorname{Tr}\left(\frac{df(X)}{dX}\right)
\end{equation}

The little thing that makes me wonder is that now the object I have inside the trace has dimensions $d \times 1$ (it's a vector), so the Trace isn't defined? Or is it just equal to its argument? And if yes why? Thanks!

EDIT:

Thank you, a way that I have found to justify in a more rigorous way that works for me is the following:

\begin{equation}
\frac{\partial tr(XX^T\phi\phi^T)}{\partial \phi} = \frac{\partial}{\partial\phi}\sum_{i,j}x_{ij}^2\phi_j^2 = \sum_{i,j}2x_{i,j}^2\phi_j = 2XX^T\phi
\end{equation}

The key thing here is that the trace of a product of matrices can be seen as the the sum of entry-wise products of their elements i.e.
\begin{equation}
tr(A^TB) = \sum_{i,j}(A\circ B)_{i,j}
\end{equation}

where $\circ$ denotes the Hadamard product.

答:
Since $\phi$ is just a column-vector, you can circumvent the headache of assigning a data-type to a matrix-by-matrix derivative if you rewrite the original expression as
$$
\begin{align}
\|X - \phi\phi^T X\|_F^2 &= \operatorname{tr}[(X - \phi\phi^TX)^T(X - \phi\phi^TX)]
\\ & = \operatorname{tr}(X^TX) - 2\operatorname{tr}(X^T\phi\phi^TX) + \operatorname{tr}(X^T\phi\phi^T\phi\phi^TX)
\\ & = \operatorname{tr}(X^TX) - \operatorname{tr}(XX^T\phi\phi^T)
\\ & = \operatorname{tr}(X^TX) - \operatorname{tr}(\phi^TXX^T\phi)
= \operatorname{tr}(X^TX) - \phi^TXX^T\phi.
\end{align}
$$
In other words, it is equivalent to consider the optimization problem
$$
\max_{\phi \in \Bbb R^d} \phi^T XX^T\phi \quad \text{s.t.} \quad \phi^T\phi = 1.
$$
The fact that this maximum is the maximal eigenvalue of $XX^T$ (attained with $\phi$ equal to the corresponding eigenvector) is known as the "Rayleigh-Ritz theorem", but if you wanted you could derive this result using Lagrange multipliers.


问:
How does $\lim_{x \to \infty} (f(x) - 2x) = 0$ imply that $\lim_{y \to \infty} (g(y) - \frac{y}{2}) = 0$ without doing a calculation?
The original function is

$f(x) = \sqrt{4x^2 + 1}$ for $x \in [0,\infty)$

and the inverse is

$g(y)=\frac{\sqrt{x^2 - 1}}{2}$

答:
hint

we use the limit composition.

$$\lim_{x\to\color{red}{+\infty}}(f(x)-2x)=0$$

and
$$\lim_{y\to +\infty}g(y)=\color{red}{+\infty}$$

$$\implies$$

$$\lim_{y\to+\infty}(f(g(y))-2g(y))=0$$

but

$$f(g(y))-2g(y)=2(\frac y2-g(y))$$


问:
Unsure how to proceed in showing that sin(x/n)n converges uniformly
Currently I am trying to solve essentially the same problem as presented in uniform convergence of n sin(x/n) with the exception that the interval is restricted to [0, 1]. I have not made any progress beyond $\left|n\sin(x/n) - x\right| \leq \max_{x \in [0, 1]}\left|n\sin(x/n) - x\right|$ as the derivative $\frac{n\sin(x/n) - x}{\left|n\sin(x/n) - x\right|}$ does not provide anyting useful and I do not know any upperbound for the expression inside the absolute value. Also, I did not get much out of the two answers in the aforementioned link as (to me) it seems that one answers shows that the function does not converge uniformly and one shows that it does.

How should I proceed?

答:
Hint: For $x\ne 0,$

$$n\sin(x/n) - x = x\left(\frac{\sin(x/n)}{x/n}-1\right).$$


问:
Jacobian matrix of $F:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$
Compute a matrix representative of $F_*$, where $F$ is

$$F:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$$

$$(u,v)\to u\times v$$

and $u\times v$ is the vector cross product of $u$ and $v$.

We can take the charts $(U,\phi)=(\mathbb{R}^3 \times \mathbb{R}^3, Id_{\mathbb{R}^3 } \times Id_{\mathbb{R}^3})$ and $(V,\psi)=(\mathbb{R}^3, Id_{\mathbb{R}^3})$. And then we have

$[F_*]=$ Jacobian matrix.

But I don't know how to compute the Jacobian matrix of this map.

Attempt:

Since $(u^1,u^2,u^3)\times (v^1,v^2,v^3)=(u^2 v^3-u^3v^2,u^1 v^3-u^3v^2, u^1 v^2-u^2 v^1)$

Our matrix would be

\begin{bmatrix}
0 & v^3 & -v^2 & 0 & -u^3 & u^2\\
-v^3 & 0 & v^1 & u^3 & 0 & -u^1 \\
v^2 & -v^1 & 0 & -u^2 & u^1 & 0
\end{bmatrix}

Any help would be appreciated. Thanks

答:
Since you're dealing with the identity chart, this becomes a simple multivariable calculus problem, so you can forget all about manifolds. $F$ is a bilinear map so the derivative is very simple. Bilinear maps should be thought of as products, so we're just going to apply the product rule here: given a point $(u,v)\in\Bbb{R}^3\times \Bbb{R}^3$ and $(\xi,\eta)\in\Bbb{R}^3\times \Bbb{R}^3$, we have
\begin{align}
DF_{(u,v)}[\xi,\eta]&= F(u,\eta)+F(\xi,v)\\
&=u \times \eta + \xi\times v
\end{align}
(The really short way of summarizing all of this is to say $d(u\times v)=u\times dv+ du\times v$). This is now a linear function of $(\xi,\eta)$, so finding the matrix representation means we successively plug in basis vectors. For example, calculate the following six quantities:

$DF_{(u,v)}[(1,0,0,0,0,0)] = u\times (1,0,0)+(0,0,0)\times v = u\times (1,0,0)=(0,u_3,-u_2)$

$DF_{(u,v)}[(0,1,0,0,0,0)]$

$DF_{(u,v)}[(0,0,1,0,0,0)]$

$DF_{(u,v)}[(0,0,0,1,0,0)]$

$DF_{(u,v)}[(0,0,0,0,1,0)]$

$DF_{(u,v)}[(0,0,0,0,0,1)]$

I leave the other computations to you. For each of these you should get certain elements of $\Bbb{R}^3$ as I've demonstrated for the first one. Now, take these $6$ elements of $\Bbb{R}^3$ and arrange them in columns to form a $3\times 6$ matrix. That is the matrix representation of $DF_{(u,v)}$ in terms of standard bases, and equivalently it is also the matrix representation of $F_{*,(u,v)}$ in terms of the corresponding bases.


问:
The inequality in applying convergence/divergence tests
When applying a convergence or divergence test, does it matter if $<$ is used instead of $\leq$?

For example, consider the series $\sum_{n=1}^{\infty} \sqrt{n+1} - \sqrt{n}$.

The direct comparison test: If $0 \leq a_n \leq b_n$ for all $n > N \in \mathbb{Z}^+$, then,

$\sum a_n$ converges if $\sum b_n$ converges and $\sum b_n$ diverges if $\sum a_n$ diverges.

Applying it:

Notice that for $n >$ some $N$, $n > \sqrt{n+1} > \sqrt{n+1} - \sqrt{n}$

$\implies 0 < \frac{1}{n} < \frac{1}{\sqrt{n+1} - \sqrt{n}} <  \sqrt{n+1} - \sqrt{n}$

We know that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, so $\sum_{n=1}^{\infty} \sqrt{n+1} - \sqrt{n}$ also diverges.

Is the above correct? Is it correct to apply the test using $<$ instead of $\leq$ even though the direct comparison test uses $\leq$? Or must the inequality not be strict? Does this apply to all the convergence/divergence test that utilize inequalities? Any insight is much appreciated.

答:
You are not using $<$ instead of $\leqslant$. You proved that you always have $\frac1n<\sqrt{n+1}-\sqrt n$; therefore, you also have $\frac1n\leqslant\sqrt{n+1}-\sqrt n$ and so the direct comparison test can be used here.

On the other hand, in the case of the root test, you cannot replace $\lim_{n\to\infty}\sqrt[n]{|a_n|}<1$ by $\lim_{n\to\infty}\sqrt[n]{|a_n|}\leqslant1$; a similar remark applies to the quotient test.


问:
Limit to infinity of recurrence relation
Looking at some recurrence relations such as $a_{n+1}=\frac{(1-a_n)}{2}$ the limit as $n$ tends to infinity can be solved by solving $x=\frac{(1-x)}{2}$. My question is whether a similar method can be used with:$$a_{n+1}=\frac{a_n}{\operatorname{gcd}(2^{a_n},a_n)}+1$$
When $a_{0}$ is even.

答:
Let $(a_n) $ be defined by
$$a_0\in I \;\text{where I is an interval }$$
$$(\forall n\in\Bbb N)\;\;a_{n+1}=f(a_n)$$
The sequence will be well defined if $ f(I)\subset I$.

If $ I $ is closed, and $ \lim_{n\to+\infty}a_n=L $ and $ f $ is continuous at $ L $ THEN
$$f(L)=L$$

In your case, the function does not seem to be continuous.


问:
What's the meaning of $S:\{(x,y,z)|x+y+z=16\ \cap x^2+y^2\leq 16\}$
$S:\{(x,y,z)|x+y+z=16\ \cap x^2+y^2\leq 16\}$

What's the meaning of this S obejct graph wise?

答:
$x^2+y^2\le 16$ is the set of the points on the surface of the cylinder and its internal points.

$x+y+z=16$ is the set of points which lie on a plane.

The intersection is an ellipse and its internal points.

$$...$$


问:
Formulating an if-else statement with linear programming
How can I formulate the following statement with linear programming?

If $i \neq j$ then $A[i] \neq A[j]$.

$A$ is a one-dimensional array of size $N$ and $i$ and $j$ are integer indexes of it, both in range $0..N$.

答:
Let binary decision variable $x_{i,k}$ indicate whether $A[i] = k$, and impose linear constraints
\begin{align}
\sum_k x_{i,k} &= 1 &&\text{for all $i$} \tag1\\
\sum_i x_{i,k} &= 1 &&\text{for all $k$} \tag2\\
\end{align}
Constraint $(1)$ assigns exactly one value to $A[i]$, and
constraint $(2)$ assigns each value exactly once.


问:
L'Hospital's rule simplification with log
$\lim_{n\to \infty}\frac{n^2 + 4n}{\log_2(n)}$ applying L'Hospital's rule equals $\lim_{n\to \infty}\frac{2n + 4}{\frac{1}{n}}$.

Now my question is, if I have $\frac{2\infty+4}{\frac{1}{\infty}} = \frac{2\infty+4}{\frac{1}{0}} = \frac{2\infty+4}{0} = 0$ wouldn't that be wrong? Should I simplify first so I would have, $\lim_{n\to \infty}\frac{2n + 4}{\frac{1}{n}} = \lim_{n\to \infty}\frac{n(2n + 4)}{1} = \infty$ ?

Body:
Hint

First,

$$(\log_2(n))'=\frac{1}{n(\ln 2)}.$$

Second,

$$\lim_{n\to \infty} \frac{2n+4}{1/n}=\lim_{n\to \infty} (2n^2+4n).$$


问:
Solving the system with $y=x$ and $ y=(1-40x)^{-1/51.33}-1$

Solve for $(x,y)$:
\begin{align}
y &= x\\
y &= (1-40x)^{-1/51.33}-1\
\end{align}

I don't know if finding a closed-form solution is possible, have been trying for almost half an hour but still get nothing.

Please tell me some idea about how to solving it, or if it is possible to do.

Edit: I did manage to get the answer via Geogebra, but I would love to see how people solve it analytically.

Body:
$$y = x \land   y = (1-40x)^{-1/51.33}-1\\
\implies (1-40x)^{-1/51.33}-x=1\\
\implies  \frac{1}{(1 - 40 x)^{1/51.33}} = x+ 1 \qquad 40x\ne1\\
\implies 1=(x+ 1)(1 - 40 x)^{1/51.33}    \\
$$
At this point I must defer to WolframAlpha
here
for $\quad x=y=0\land x=y = 0.00999525$


问:
Recalculating a point onto a custom plane

This is my coordinate system:

______

Essentially, I want to know if it's possible to recalculate a point 'A' from 1 plane (ZY plane) onto another plane (B plane) more efficiently than the way I'm current doing so.

The way I'm current recalculating point A to lie on the B plane:

get the vector projection of point B onto the XZ plane

B_XZ_PROJ = [B.x, 0, B.z]

then I get the angle (theta) between The -Z axis and B_XZ_PROJ

then using Rodrigues' rotation formula, I rotate point A around the +Y axis by theta to get point A'

How do I recalculate A onto plane B directly without using this rotation method?

from reading online, my guess to this approach would be:

create the B plane equation using 3 points: +Y unit vector, origin, point B

substitute components of A into this plane equation to get A' on plane B

is this the correct thinking?

答:
There does exist a more efficient method.  If $\vec{p}$ is a 3D point on a source plane, we can obtain the planar coordinates $(u, v)$ via
$$\vec{p} = \vec{o} + u \, \vec{u} + v \, \vec{v} \quad \iff \quad \left\lbrace \begin{aligned}
u &= \vec{p} \cdot \vec{U} + U_0 = ( \vec{p} - \vec{o} ) \cdot \vec{U} \\
v &= \vec{p} \cdot \vec{V} + V_0 = ( \vec{p} - \vec{o} ) \cdot \vec{V} \\
\end{aligned} \right.$$
and use those coordinates to get the corresponding point on the second plane,
$$\begin{aligned}
\vec{p}^\prime &= \vec{o}^\prime + u \, \vec{u}^\prime + v \, \vec{v}^\prime \\
~ &= \vec{o}^\prime + \bigl((\vec{p} - \vec{o})\cdot\vec{U}\bigr)\vec{u}^\prime  + \bigl((\vec{p} - \vec{o})\cdot\vec{V}\bigr)\vec{v}^\prime \\
\end{aligned}$$
where$$\begin{array}{c|l|l}
\text{Variable} & \text{Description} \\
\hline
\vec{p} & \text{ point on original plane } \\
\vec{p}^\prime & \text{ point on other plane } \\
\vec{o} & \text{ origin on original plane } \\
\vec{o}^\prime & \text{ origin on second plane } \\
\vec{u} & u \text{ unit vector on original plane } \\
\vec{v} & v \text{ unit vector on original plane } \\
\vec{u}^\prime & u \text{ unit vector on second plane } \\
\vec{v}^\prime & v \text{ unit vector on second plane } \\
U_0     & u \text{ origin constant}, ~ - \vec{o} \cdot \vec{U} \\
V_0     & v \text{ origin constant}, ~ - \vec{o} \cdot \vec{V} \\
\vec{U} & \text{ inverse } u \text{ axis vector } \\
\vec{V} & \text{ inverse } v \text{ axis vector } \\
\end{array}$$
The $(u, v)$ coordinates correspond to 3D vectors
$$\begin{array}{c|l}
(u, v) & \text{ 3D Vector } \\
\hline
(0, 0) & \vec{o} \\
(1, 0) & \vec{o} + \vec{u} \\
(0, 1) & \vec{o} + \vec{v} \\
(1, 1) & \vec{o} + \vec{u} + \vec{v} \\
\end{array}$$
The inverse axis vectors $\vec{U}$ and $\vec{V}$ are a bit complicated to compute, however; we have three equations and only two variables, and numerical stability should also be considered.  Here is one way, derived from dropping the least useful Cartesian coordinate axis:

Let
$$\vec{u} = \left[ \begin{matrix} u_x \\ u_y \\ u_z \end{matrix} \right] , \quad \vec{v} = \left[ \begin{matrix} v_x \\ v_y \\ v_z \end{matrix} \right] , \quad
\vec{U} = \left[ \begin{matrix} U_x \\ U_y \\ U_z \end{matrix} \right] , \quad \vec{V} = \left[ \begin{matrix} V_x \\ V_y \\ V_z \end{matrix} \right]$$

Calculate $$\begin{aligned}
d_{x y} &= u_x v_y - u_y v_x \\
d_{x z} &= u_z v_x - u_x v_z \\
d_{y z} &= u_y v_z - u_z v_y \\
\end{aligned}$$ from the $\vec{u}$ and $\vec{v}$ we wish to invert.

If $d_{x y}$ has the largest magnitude (absolute value, $\lvert d_{x y} \rvert \ge \lvert d_{x z} \rvert$ and $\lvert d_{x y} \rvert \ge \lvert d_{y z} \lvert$), then $$\vec{U} = \left[ \begin{matrix} v_y / d_{x y} \\ - v_x / d_{x y} \\ 0 \end{matrix} \right], \quad \vec{V} = \left[ \begin{matrix} -u_y / d_{x y} \\ u_x / d_{x y} \\ 0 \end{matrix} \right]$$

Otherwise, if $d_{x z}$ has the largest magnitude ($\lvert d_{x z} \rvert \gt \lvert d_{x y} \rvert$, $\lvert d_{x z} \rvert \ge \lvert d_{y z} \rvert$), then $$\vec{U} = \left[ \begin{matrix} -v_z / d_{x z} \\ 0 \\ v_x / d_{x z} \end{matrix} \right], \quad \vec{V} = \left[ \begin{matrix} u_z / d_{x z} \\ 0 \\ -u_x / d_{x z} \end{matrix} \right]$$

Otherwise, $d_{y z}$ has the largest magnitude ($\lvert d_{y z} \rvert \gt \lvert d_{x y} \rvert$, $\lvert d_{y z} \rvert \gt \lvert d_{x z} \rvert$), and $$\vec{U} = \left[ \begin{matrix} 0 \\ v_z / d_{y z} \\ -v_y / d_{y z} \end{matrix} \right], \quad \vec{V} = \left[ \begin{matrix} 0 \\ -u_z / d_{y z} \\ u_y / d_{y z} \end{matrix} \right]$$

Obviously, $\vec{U}$ and $\vec{V}$ only need to be recomputed whenever the corresponding $\vec{u}$ and/or $\vec{v}$ change.

Mapping any cuboid to $(u, v, w)$ is actually simpler, because the forward transformation is $$\left[\begin{matrix} x \\ y \\ z \end{matrix}\right] = \left[\begin{matrix} u_x & v_x & w_x \\ u_y & v_y & w_y \\ u_z & v_z & w_z \end{matrix}\right] \left[\begin{matrix} u \\ v \\ w \end{matrix}\right] + \left[\begin{matrix} o_x \\ o_y \\ o_z \end{matrix}\right]$$ and we can just invert the matrix for the inverse,
$$\left[\begin{matrix} u \\ v \\ w \end{matrix}\right] = \frac{1}{d} \left[\begin{matrix}
v_y w_z - v_z w_y & v_z w_x - v_x w_z & v_x w_y - v_y w_x \\
u_z w_y - u_y w_z & u_x w_z - u_z w_x & u_y w_x - u_x w_y \\
u_y v_z - u_z v_y & u_z v_x - u_x v_z & u_x v_y - u_y v_x \\
\end{matrix}\right] \left[\begin{matrix} x - o_x \\ y - o_y \\ z - o_z \end{matrix}\right]$$
where $d$ is the triple product of the three basis vectors, $$\begin{aligned} d = \vec{u} \times \vec{v} \cdot \vec{w} & = u_x v_y w_z - u_y v_x w_z \\
~ & - u_x v_z w_y + u_z v_x w_y \\
~ & + u_y v_z w_x - u_z v_y w_x \\
\end{aligned}$$
If you define $\vec{w}$ perpendicular to the plane, you can use this to calculate the $(u, v)$ coordinates of the point on the plane where the plane normal passes through the specified point, with $w/\lVert\vec{w}\rVert$ being the signed distance along the normal; positive if in the same direction as $\vec{w}$, negative if in the opposite direction.

(Again, $\vec{o} = (o_x, o_y, o_z)$ is the origin in $(u, v, w)$ coordinates, and $\vec{u} = (u_x, u_y, u_z)$, $\vec{v} = (v_x, v_y, v_z)$, and $\vec{w} = (w_x, w_y, w_z)$ the 3D basis vectors for the $(u, v, w)$ coordinate system. $(u=0, v=0, w=0)$ is at $\vec{o}$, $(1, 0, 0)$ is at $\vec{o}+\vec{u}$, $(1,1,1)$ is at $\vec{o}+\vec{u}+\vec{v}+\vec{w}$, and so on.)


问:
Why S4 logic should have kripke models with reflexive and transitive relations?
From here and here you can see that in S4 modal logic we should have Kripke models (S,R,V) which have a relationship R on their set of states S so that R is reflexive and transitive on S!
I know that the set of axioms we obliged on our logic forces us to have our models this way but actually, I don't know how and why!

答:
Welcome to MSE!

It helps to think about what the bonus axioms are saying.

Axiom $T$ says $\square \varphi \to \varphi$

Axiom $4$ says $\square \varphi \to \square \square \varphi$

What do these conditions say on frames?

Let's start with $T$. We know that for any valuation we like, we need to guarantee that $\square p \to p$ holds at every world. That is, no matter what the valuation, if $w$ only sees $p$-worlds, then $w$ must be a $p$-world too.

First, it should be clear that reflexive frames have this property. Since $w$ sees itself, if every world $w$ can see is a $p$-world, then $w$ itself must be a $p$ world.

But now say you look at a frame that isn't reflexive. Then fix a world $w$ that doesn't see itself. Now look at the valuation where $w$ thinks $\lnot p$ and every other world thinks $p$. Then since $w$ doesn't see itself, $w$ must think that $\square p$ is true! So $w$ thinks $\square p$ and $\lnot p$, and axiom $T$ fails!

So the frames validating $T$ are exactly the reflexive ones.

Let's move on to $4$. This says that if everywhere I can get in 1 step is a $p$-world, then everywhere I can get in $2$ steps is a $p$-world too.

Again, it should be clear that transitive frames have this property. After all, the definition of transitivity is that anywhere I can get in multiple steps is somewhere I can already get in 1 step.

As a (fun?) exercise, can you show that any nontransitive frame doesnt have this property? You'll want to argue like we did before with reflexivity. Fix a nontransitive frame. Then you can find worlds $x, y, z$ with $xRy$, $yRz$, but $x \not R z$. Can you cook up a valuation which makes axiom $4$ fail on this frame?

I hope this helps ^_^


问:
Stoke's Thm. $\vec F = (x+y^2, y+z^2, z+x^2)$ and $S$ is the triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$
I am trying to evaluate

$$\int _C \vec F \cdot d\vec r$$

using Stoke's Thm. where $\vec F = (x+y^2, y+z^2, z+x^2)$ and $S$ is the triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$

So, I can see why Stoke's Thm. is easier than working on the line integral, and I believe that the expression is equal to

$$\int \int_S \operatorname{curl} \vec F \cdot d\vec S = \int \int_S (\operatorname{curl} \vec F)\cdot \vec n \space dS$$

and because $\operatorname{curl} \vec F = -2(x,y,z)$ and the normal vector to the triangle is $(1,1,1)$,

the inside portion simplifies to $-2$ so all I have to do is to evaluate

$$\int \int_S dS$$

This is where I am a bit confused.

I though that this expression is equivalent to the surface area which is an equilateral triangle with side $\sqrt{2}$

whose area must be $\frac{\sqrt{3}}{2}$, so I want to say that the answer is $-\sqrt{3}$,

but supposedly it should be $-1$.

One possible thing that I thought of was that the projection of $S$ onto the $xy$-plane is a right triangle with area $\frac{1}{2}$ which would make the answer $-1$.

Can anyone help me out with this problem?

答:
Remember that in the formula

$$ \iint_S \vec V \cdot d \vec S = \iint_S \vec V \cdot \vec n \, dS $$

the normal vector $n$ must be a unit normal vector. Your normal vector $(1,1,1)$ has magnitude $\sqrt{3}$, so the unit normal is $\vec n = \frac{1}{\sqrt{3}}(1,1,1)$. This seems to be your only mistake.


问:
A problem involving centralizer and cycles
I have just begun studying the group $S_n$ and I am having trouble with the cycle notation so this problem seems a bit hard. Any help will be deeply appreciated:

If $a \in S_n$ is an $n$-cycle or an $(n-1)$-cycle then the centralizer of $a$ is equal to the cyclic group $\langle a\rangle$.

答:
Let $a$ be an $m$-cycle. The number of conjugates of $a$ is the number of $m$-cycles, which is:
$$
\frac{n(n-1)\cdots(n-m+1)}{m}
$$

By the orbit-stabilizer theorem, the number of conjugates of $a$ is:
$$
\frac{\left|S_n\right|}{\left|C_{S_n}(a)\right|} = \frac{n!}{\left|C_{S_n}(a)\right|}
$$

Where $C_{S_n}(a)$ is the centralizer of $a$ in $S_n$. It follows that:
$$
\left|C_{S_n}(a)\right| = m (n-m)! \tag{1}
$$

Since $a$ commutes with all elements in the cyclic group $\langle a \rangle$, we have:
$$
\langle a \rangle \le C_{S_n}(a) \tag{2}
$$

For $m \in \{n-1, n\}$, $(1)$ gives $\left|C_{S_n}(a)\right| = m$. Since $\left|\langle a \rangle\right| = m$, we have equality in $(2)$.


问:
How to calculate lim n × sin(1/(n^2)) when n goes to +infinity
I found this problem:

$ \lim\limits_{n \to +\infty} n \sin \left( \frac{1}{n^2} \right)$

Can anyone help me?

答:
hint

$$n\sin(\frac{1}{n^2})=\frac 1n\frac{\sin(\frac{1}{n^2})}{\frac{1}{n^2}}$$

Or use the inequality

$$(\forall X\in\Bbb R)\; \; |\sin(X)|\le |X|$$
to get

$$(\forall n>0)\; |n\sin(\frac{1}{n^2})|\le \frac 1n$$

So, your limit is zero.


问:
Proving that there is a unique real number x such that for every real number $y$, $xy+x-4=4y$
The question is from problem 2 of chapter 3.6 of Velleman's How To Prove It (2nd edition).
The question has been asked by another user before, but I couldn't find any thread that addresses the specific problem I have with my proof/problem.

My proof:

Existence
Let $x = 4$, then $xy+x-4 = 4y+4-4 = 4y$

Uniqueness
Let a and y be arbitrary and suppose $∀(+−4=4)$ and $∀(b+b−4=4)$.

From $∀(+−4=4)$, Suppose $y = b$, then
$ab + a -4 = 4b$.

Then from $∀(b+b−4=4)$, suppose y = a, thus $ba+b-4 = 4a$.

Subtracting $ba+b-4 = 4a$ from $ab + a -4 = 4b$, we get $4b-4a = 0$, thus, $a = b$.

Since $a$ and $b$ are arbitrary, we can conclude that $a$ = $b$ = $x$ = $4$. Thus, $x$ is unique.

Q.E.D.

My issue with the proof:

I think this is correct, but something about the existence part of the proof is bothering me. Specifically the fact that
we infer $x = 4$ by solving $xy+x-4=4y$ for $x$ like so:

$xy+x-4=4y$ iff

$x(y+1)=4y+4$ iff

$x = 4(y+1)/(y+1)$.

The $(y+1)$ factors cancel out and we are left with $4$. However, isn't $x$ undefined for $y=-1$?

So how is the statement "There exists a real number $x$ for all $y$ such that $xy+x-4=4y$" actually true? (As $x$ is undefined for $y = -1$?)
Plugging in $-1$ for $y$ and $4$ for $x$ still makes the equation work, so I think the existence part of my proof is possibly correct, but how?

答:
All of the other answers, while not incorrect, have failed to address the fundamental misunderstanding in your question.

Under the heading "My issue with the proof," you write that in the existence part of the proof, "… we infer $x=4$ by solving …."  But that's not true.  Go back and reread your proof.  In the existence part of the proof, you do not infer that $x=4$.  Rather, you define $x$ to be 4, and then you do the algebra to show that this value of $x$ works.  That algebra is correct for all values of $y$, including $y = -1$.  So there is no problem with the proof.

What is worrying you is not the proof, but the scratch work you did in figuring out the proof.  In that scratch work you did a step that would not be correct for $y = -1$.  But that is irrelevant for judging the correctness of the proof.  You could have come up with this proof by simply having a flash of insight that $x=4$ would work, or by making a lucky guess.  In judging the correctness of the proof, it doesn't matter how you thought of the choice $x=4$; all that matters is whether your proof that this choice works is correct, and it is.

In How To Prove It, go back and reread the paragraph before Example 3.3.3.


答:
You have a sequence of integers.  So it can converge only if it is eventually constant.  That is, it can converge to $L$ only if
$$
L=\frac{L}{\operatorname{gcd}(2^{L},L)}+1
\tag1$$
So in this case, your answer is YES.

Note $(1)$ means $L=2$.  So if your sequence converges, then it converges to $2$.  However, this argument does not show that it does converge.  For that, argue that $a_{n+1} < a_n$ if $a_n > 2$.

Title: Limit to infinity of recurrence relation

Body:
Looking at some recurrence relations such as $a_{n+1}=\frac{(1-a_n)}{2}$ the limit as $n$ tends to infinity can be solved by solving $x=\frac{(1-x)}{2}$. My question is whether a similar method can be used with:$$a_{n+1}=\frac{a_n}{\operatorname{gcd}(2^{a_n},a_n)}+1$$
When $a_{0}$ is even.


答:
HallaSurvivor's answer is absolutely right. Let me add a bit more, however, about the general context surrounding it.

The relevant connection between modal axiom systems and Kripke frame properties is very particular:

Definition:

A Kripke frame $(W,R)$ validates a modal sentence $\sigma$ iff for every valuation $\nu$ the sentence $\sigma$ is true at every world according to $(W,R,\nu)$.

A class $\mathfrak{F}$ of Kripke frames validates a modal sentence $\sigma$ iff every frame in $\mathfrak{F}$ validates $\sigma$.

The precise theorem connecting $S4$ and the class of transitive and reflexive frames, then, is the following multi-part result:

Theorem:

A frame $(W,R)$ validates every theorem of $S4$ iff that frame is reflexive and transitive.

If $\sigma$ is not a theorem of $S4$, then there is a reflexive and transitive frame not validating $\sigma$

It is in this sense that "$S4\approx$ reflexive + transitive."

In their answer above, HallaSurvivor proved the first bulletpoint (actually they proved more than that - they analyzed the role of reflexivity on its own and transitivity on its own). And this was the point relevant to your question.

The second bulletpoint is a bit more subtle. Basically, it says that $S4$ completely captures the nature of reflexive and transitive frames: there's nothing "missing" from $S4$ in this regard. So we really have a very tight connection between a modal theory ($S4$) and a frame property (reflexive-and-transitive)!

(In more technical language, the first bulletpoint says that $S4$ is sound with respect to the reflexive-and-transitive frame semantics (right-to-left direction) and moreover this is the maximal frame semantics with respect to which $S4$ is sound (left-to-right direction), while the second bulletpoint says that $S4$ is complete with respect to the reflexive-and-transitive frame semantics.)

It's separately worth noting that we can rephrase things in terms of operators:

Frames-to-theories: Given a class of frames $\mathfrak{F}$, let $Th(\mathfrak{F})$ be the set of modal sentences validated by $\mathfrak{F}$.

Theories-to-frames: Given a set of modal sentences $T$, let $Fr(T)$ be the set of frames validating every sentence in $T$.

We always have $Th(Fr(T))\supseteq T$ and $Fr(Th(\mathfrak{F}))\supseteq \mathfrak{F}$. However, in general these may be strict inclusions. The two-part theorem above can be reformulated as:

Let $\overline{S4}$ be the set of theorems of $S4$ and let $\mathfrak{X}$ be the class of reflexive and transitive Kripke frames. Then $$Th(Fr(\overline{S4}))=\overline{S4}\quad\mbox{and}\quad Fr(Th(\mathfrak{X}))=\mathfrak{X}.$$

Think of this as saying that the connection between (the theorems of) $S4$ and the class of reflexive and transitive Kripke frames is as good as we could hope - there's no "loss of information" in replacing one with the other, in a precise sense.

(The term "Galois connection" is relevant here, and it's a good exercise to try and figure out in what sense the pair of operators $(Th,\mathfrak{F})$ forms a Galois connection.)

Title: Why S4 logic should have kripke models with reflexive and transitive relations?

Body:
From here and here you can see that in S4 modal logic we should have Kripke models (S,R,V) which have a relationship R on their set of states S so that R is reflexive and transitive on S!
I know that the set of axioms we obliged on our logic forces us to have our models this way but actually, I don't know how and why!


答: Remember that $\mathbf n$ must be a unit normal vector. Your normal vector has magnitude $\sqrt 3$.

Title: Stoke's Thm. $\vec F = (x+y^2, y+z^2, z+x^2)$ and $S$ is the triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$

Body:
I am trying to evaluate

$$\int _C \vec F \cdot d\vec r$$

using Stoke's Thm. where $\vec F = (x+y^2, y+z^2, z+x^2)$ and $S$ is the triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$

So, I can see why Stoke's Thm. is easier than working on the line integral, and I believe that the expression is equal to

$$\int \int_S \operatorname{curl} \vec F \cdot d\vec S = \int \int_S (\operatorname{curl} \vec F)\cdot \vec n \space dS$$

and because $\operatorname{curl} \vec F = -2(x,y,z)$ and the normal vector to the triangle is $(1,1,1)$,

the inside portion simplifies to $-2$ so all I have to do is to evaluate

$$\int \int_S dS$$

This is where I am a bit confused.

I though that this expression is equivalent to the surface area which is an equilateral triangle with side $\sqrt{2}$

whose area must be $\frac{\sqrt{3}}{2}$, so I want to say that the answer is $-\sqrt{3}$,

but supposedly it should be $-1$.

One possible thing that I thought of was that the projection of $S$ onto the $xy$-plane is a right triangle with area $\frac{1}{2}$ which would make the answer $-1$.

Can anyone help me out with this problem?


问:
Determining if $f(x,y)$ can be expanded to be continuous on whole $\mathbb{R}^{2}$
I'm trying to determine if the function

$$f(x,y)=\frac{\sin(xy)}{xy+x^3y^3}$$

can be expanded so that it becomes continuous on the whole $\mathbb{R}^{2}$-plane. I'm aware of that the function isn't defined for when $xy= 0$, so in order to become continuous everywhere the function needs to be expanded to be defined, and continuous when we're located on either axis.

Up to this point I've only worked with problems where the function isn't defined on a single point $(a,b)$. When that was the case it was quite simple to solve since I could switch to polar coordinates and expand the function so that $f(a,b) = \lim_{(x,y)\to (a,b)} f(x,y)$. However, now when I have multiple points, I can't see how I can solve it the same way.

答:
The easiest way for this is studying the limit of $f$ on the points where $xy=0$, and as for the single-point case, just define the extension on those points as the value of the limit, whenever exists.
For studying that limit it might be usefull noting that: $$f(x,y)=\dfrac{\sin(xy)}{xy}\cdot\dfrac{1}{1+x^2y^2}$$
And considering the change of variable $t=xy$.


问:
How to sample points on a triangle surface in 3D?
To take random uniform points inside a triangle Triangle Point Picking method is used. But this is for 2D points, how can I take random points from a triangle that is defined by 3 arbitrary 3D points?

In other words, let's say I have 3 points in 3D space, and these points define a plane, how can I take random points on the plane such that my points are uniformly sampled inside the triangle that is defined by these 3 points?

Thanks in advance..

答:
As mentioned in the comments, this formulation is not guaranteed to give you points on the triangle. The correct formulation is as follow:

$P = (1 - \sqrt{a})v1 + (\sqrt{a} (1 - b))v2 + (b \sqrt{a})v3$

where $a, b \sim U[0, 1]$.


问:
Proof of Convergence of Series?
I am trying to show that the infinite sum (for a fixed complex number $z$)
$$\sum_{n=1}^\infty\frac{z}{(z-n)n}$$
converges absolutely. My idea is to compare it to the series $$1+1/4+1/9+\cdots,$$ but for some reason I cannot seem to figure out how to handle the asymptotics. It seems like "in the long run" the terms of the series get closer together, so I thought of proving a general result like that, but then I realized that the harmonic series and $0+0+0+\cdots$ fits the same property.

Could someone supply the technical details?

答: Since$$\lim_{n\to\infty}\frac{\left|\frac z{(z-n)n}\right|}{\frac1{n^2}}=\lim_{n\to\infty}\left|\frac{n^2z}{nz-n^2}\right|=|z|,$$then, since the series $\sum_{n=1}^\infty\frac1{n^2}$ converges, your series converges absolutely.

Title: Proof of Convergence of Series?


问:
Bilinear form matrix coordinates
I don't know how to solve this problem - I have to transform the coordinates of a bilinear form that has matrix
$$
        \begin{pmatrix}
        1 & 2 & -2 \\
        2 & -2 & 3 \\
        -2 & 3 & 2 \\
        \end{pmatrix}
$$
in base
$$
        \begin{pmatrix}
        -3 & 1 & 0 \\
        5 & -2 & 3 \\
        2 & 0 & -1 \\
        \end{pmatrix}
$$
To matrix of the same bilinear form in the base
$$
        \begin{pmatrix}
        0 & 1 & 0 \\
        1 & 0 & 1 \\
        2 & 1 & 1 \\
        \end{pmatrix}
$$
How do I do it? Should there be scalar products of the base vectors in the matrix(but then I do not understand what is there the first base for) or can I just find a transition matrix from one base to the other and just multipy it?

Thank you.

答: Yes, you need the change of basis matrix $P$ whose columns are the coefficients of the new basis vectors when you express them as linear combinations of the old. I'll let you figure out how to get that from the two matrices directly. Then the expression for the bilinear form in the new basis should be $P^TAP$, where its matrix representation in the old basis is $A$. (To convince yourself of this, remember that the coordinate vector of a vector $x$ with respect to the new basis will be $x'=Px$. So $x^TAy = (Px')^TA'(Py') = \dots$.)

Title: Bilinear form matrix coordinates


问:
Surface integral of a scalar function.
I was doing this question,

Evaluate the surface integral $\int \int_S 3z(d\sigma)$, where the surface $S$ is the portion of the plane $2x+y+z=2$ lying in the first octant.

The way I approached this question was that I took small elemental triangles being projected from the XY-plane, like this,

Here I multiplied the length of hypotenuse $\sqrt {x^2+y^2}$ with $dz$ and then substituted $x=\frac{2-z}{2}$ and $y=2-z$, and the resulting expression is,

$S=3\int_0^2z\sqrt{\frac {5(2-z)^2}{4}}(dz)$

However, after evaluating the integral, I am getting an incorrect answer of $2\sqrt 5$. However, the correct answer is $2\sqrt 6$.
So please help me spot my mistake. Any help would be appreciated!

答:
The integral is indeed $2\sqrt6$, derived here using this well-known formula. With $z=f(x,y)=2-2x-y$, we have

$$\begin{align}
\iint_S 3z\,\mathrm d\sigma&=\int_0^1\int_0^{2-2x} 3(2-2x-y)\sqrt{1+{f_x}^2+{f_y}^2}\,\mathrm dy\,\mathrm dx\\[1ex]
&=\int_0^1\int_0^{2-2x}3(2-2x-y)\sqrt{1+(-2)^2+(-1)^2}\,\mathrm dy\,\mathrm dx\\[1ex]
&=3\sqrt6\int_0^1\int_0^{2-2x}(2-2x-y)\,\mathrm dy\,\mathrm dx\\[1ex]
&=3\sqrt6\times\frac23=\boxed{2\sqrt6}
\end{align}$$

Title: Surface integral of a scalar function.


答: I believe you shouldn’t use L’hôpital’s here. Logarithms grow really slow towards infinity. Whilst squares grow reasonably fast. This can be written as $\log_n(x) = o(x^2)$. Therefore, we get fast growth over slow growth (towards inifnity) which yields infinity.

Title: L'Hospital's rule simplification with log

Body:
$\lim_{n\to \infty}\frac{n^2 + 4n}{\log_2(n)}$ applying L'Hospital's rule equals $\lim_{n\to \infty}\frac{2n + 4}{\frac{1}{n}}$.

Now my question is, if I have $\frac{2\infty+4}{\frac{1}{\infty}} = \frac{2\infty+4}{\frac{1}{0}} = \frac{2\infty+4}{0} = 0$ wouldn't that be wrong? Should I simplify first so I would have, $\lim_{n\to \infty}\frac{2n + 4}{\frac{1}{n}} = \lim_{n\to \infty}\frac{n(2n + 4)}{1} = \infty$ ?


问:
The operator $U$ is unitary
If $A$ is a selft-adjoint operator, then the operator $U$ defined by
\begin{eqnarray*}
U=(A-iI)(A+iI)^{-1}
\end{eqnarray*}
Is unitary.

I know if $U$ is unitary then $U^{*}U=UU^{*}=I$ But when i tried to compute:
\begin{eqnarray*}
U^{*}U&=&{((A+iI)^{-1})}^{*}(A-iI)^{*}(A-iI)(A+iI)^{-1}\\\
&=&{((A+iI)^{-1})}^{*}(A^{*}-iI^{*})(A-iI)(A+iI)^{-1}
\end{eqnarray*}
or I tried to compute $UU^{*}$:
\begin{eqnarray*}
UU^{*}&=&{(A-iI)(A+iI)^{-1}((A+iI)^{-1})}^{*}(A-iI)^{*}\\\
&=&{(A-iI)(A+iI)^{-1}((A+iI)^{-1})}^{*}(A^{*}-iI^{*})
\end{eqnarray*}
I don't know how can I continue with the hypothesis that $A$ is selft-adjoin operator and the way that I have to treat the adjoint inverse. Can you give some hint to continue? Thank you

Title: The operator $U$ is unitary

Body: HINT: Use the fact that $(S^*)^{-1} = (S^{-1})^*$ and $(iI)^* = -i I$.


问:
Joint filtration of two right-continuous filtrations is right-continuous?
Assume we are given two right-continuous filtrations $\mathbb{F}^X=(\mathcal{F}_t^X)$ and $\mathbb{F}^Y=(\mathcal{F}_t^Y)$ both generated by some stochastic processes $X, Y$. Do we have that the joint filtration $\mathbb{F}^{X,Y}$ is right-continuous?

答:
(I assume $X$ and $Y$ are real-valued and right continuous.) Not necessarily. For example, the event
$$
\eqalign{
C
&=\cap_n[\{\omega: X_t(\omega)-Y_t(\omega)>0\hbox{ for some }t\in(0,1/n)\}\cr
&\phantom{bbbbbbb}\cap\{\omega: X_t(\omega)-Y_t(\omega)<0\hbox{ for some }t\in(0,1/n)\}],\cr
}
$$
that $X-Y$ changes sign infinitely often just after time 0, is $\mathcal F^{X,Y}_{0+}$-measurable, but not $\mathcal F^X_0\vee \mathcal F^Y_0$-measurable.

Nonetheless, the  filtration $\mathcal F_t^X\vee\mathcal F_t^Y$, $t\ge 0$, is right continuous (modulo null sets) when $X$ and $Y$ are independent.


问:
Does a predictable process have countably many discontinuities
Suppose $\mathbb{F}$ is a filtration on the complete probability space $(\Omega, \mathcal{F},\mathbb{P})$ and the function $V$ is $\mathbb{F}$-predictable. That means that $V$ is measurable w.r.t the sigma algebra generated by the left continuous functions.

I wonder whether this gives the opportunity to state something on the regularity of predictable functions, in particular:

Is there a representation/modification of $V$ which consists of left-continuous functions?

Can I state, that $V$ has only countable many jumps (either based on the first point or the fact that the sigma algebra is generated by left-continuous functions)?

答: The process $V_t=1_{\{0\}}(B_t)$ is optional (hence predictable) with respect to the filtration of the Brownian motion $B$, but $t\mapsto V_t$ is a.s. discontinuous at every point of the (a.s. perfect) set $\mathcal Z:=\{t\ge 0: B_t=0\}$.


问:
Group equals union of three subgroups
Suppose $G$ is finite and $G=H\cup K\cup L$ for proper subgroups $H,K,L$. Show that $|G:H|=|G:K|=|G:L|=2$.

What I did: so if some of $H,K,L$ is contained in another, then we have $G$ being a union of two proper subgroups, which is impossible due to another result. So none of $H,K,L$ is contained is each other. If some element $h$ belongs to only $H$ and $k$ belongs to only $K$, then its product $hk$ cannot be in $H$ or $K$, so must belong to only $L$.

答:
At least one of $H, K$ and $L$ must have index $2$ in $G$ by counting. Let's say $|G:H| = 2$. Then, $H$ is normal in $G$. Then, $|K:K\cap H| = 2 = |L:L\cap H|$.

Now, for $G = H \cup (K - K\cap H) \cup (L - L\cap H)$, this union must be disjoint by counting and we must have $|G:K| = 2 = |G:L|$.

Title: Group equals union of three subgroups


问:
Properties of Laplacian matrix
Prove that the Laplacian matrix $L$ of a graph $G$ satisfies the following:

For every vector $v \in \mathbf{R}^n$ we have $$v^TLv=\frac{1}{2}\sum_{i,j=1}^nw_{i,j}(v_i-v_j)^2$$

$L$ is symmetric and positive semi-definite.

The smallest eigenvalue of $L$ is $0$, the corresponding eigenvector is the constant vector $1$.

$L$ has $n$ no-negative eigenvalues $0=\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n$.

I don't know the proof for 3 and 4

答:
If it can be useful for anybody, for 1) assuming you have an undirected graph with N nodes, that $w_{i,j}$ are the coefficients of the weight matrix $W$, $d_{i,j}$ the ones of the degree matrix $D$ and that the Laplacian matrix equals to $D-W$ :

$$v^TLv=\sum_{i,j \leq N} l_{i,j}f_if_j=\sum_{i,j \leq N} (d_{i,j}-w_{i,j})f_if_j=\frac{1}{2} (\sum_id_if_i^2 - 2\sum_{i,j}w_{i,j}f_if_j+\sum_jd_jf_j^2)$$
because $D$ is diagonal. Now we can use the fact that the sum of each row of $W$ equals to the degree of the matrix ie $d_i = \sum_j w_{i,j}$ :
$$v^TLv=\frac{1}{2} (\sum_i \sum_j w_{i,j}f_i^2 - 2\sum_{i,j}w_{i,j}f_if_j+\sum_j \sum_i w_{i,j}f_j^2)=\frac{1}{2}\sum_{i,j}w_{i,j}(fi-fj)^2$$

For 4) L, as said before L is semi-definite positive, it's spectrum is in $R^+$

Title: Properties of Laplacian matrix


问:
Sheaf cohomology of the structure sheaf of the grasmannian G(2,7)
How can I calculate $H^i(Gr(2,7),\mathcal{O}_{Gr(2,7)})$ over the base field $\mathbb{C}$?

Or in general $H^i(Gr(k,n),\mathcal{O}_{Gr(k,n)})$?

The first idea came to my mind is using Plücker embedding $i:Gr(2,7)\hookrightarrow\mathbb{P}^{20}$ and the long exact sequence associated to the ideal sheaf exact sequence $$0\rightarrow\mathcal{I}_{Gr(2,7)}\rightarrow\mathcal{O}_{\mathbb{P}^{20}}\rightarrow{i_*}\mathcal{O}_{Gr(2,7)}\rightarrow{0}\,.$$

But I don't know how to calculate $H^i(Gr(2,7),\mathcal{I}_{Gr(2,7)})$ either.

答:
(For some reason, my perfectly valid answer was deleted without notifications. More than 5 people said that this does not provide an answer to the question. This is quite surprising as the theorem literally does provide the answer to the question.)

The original answer : this follows from the Borel-Weil theorem. Indeed, $L = \mathcal O_{Gr(2,7)}$ is the line bundle associated to the trivial representation. Hence $H^0(X,L) = \mathbb C$ and $H^i(X,L) = 0$ for $i>0$. This works for any flag variety $X = G/P$ where $G$ is a semisimple algebraic group, and $P$ is a parabolic subgroup.


问:
sequence limits proof by definition
If I have $\displaystyle\lim_{x \to +\infty} a(x)=5 $. and $\displaystyle\lim_{x \to +\infty} b(x)=+\infty $.
How can I prove that $\displaystyle\lim_{x \to +\infty} (a*b)= +\infty $ by definition.

Can I just put n1 for the first sequence and n2 for the second and say n0 = max (n1, n2) and that applies to all members?

答:
Since $\lim_{x \to \infty} a(x) = 5$ and $\lim_{x \to \infty} b(x) = +\infty$, there is $K_1 >0$ such that $x > K_1$ implies $a(x) > 1$ and $x > K_2$ implies $b(x) > 0$. Therefore, for $x > \max\{K_1, K_2\}$,
$$b(x) < a(x)b(x).$$
Try to conclude from that.

Title: sequence limits proof by definition


答:
You can decompose the terms  $\frac{z}{(z-n)n}$ as $\frac{1}{z-n} + \frac{1}{n}$. So it becomes $$\sum_{n = 1}^{\infty} \frac{1}{z-n} + \frac{1}{n}$$

For $z=0$, it can be seen that series converges to 0 and for any $z \in \mathbb{Z}^+$ the series diverges. For other values of fixed $z$, it can be observed that these terms cancel each other out in the tail (when $n \gg z$).

Title: Proof of Convergence of Series?

Body:
I am trying to show that the infinite sum (for a fixed complex number $z$)
$$\sum_{n=1}^\infty\frac{z}{(z-n)n}$$
converges absolutely. My idea is to compare it to the series $$1+1/4+1/9+\cdots,$$ but for some reason I cannot seem to figure out how to handle the asymptotics. It seems like "in the long run" the terms of the series get closer together, so I thought of proving a general result like that, but then I realized that the harmonic series and $0+0+0+\cdots$ fits the same property.

Could someone supply the technical details?


问:
Four Real roots of sextic polynomial
Given a general sextic polynomial with real coefficients

$$f(x)=x^6+a_5 x^5+a_4 x^4+a_3 x^3+a_2 x^2+a_1 x+a_0$$

is there a way to determine whether there are 1) six real roots; 2) four real roots; 3) no real roots just based on the $\{a_i\}$ coefficients of the polynomial? I am particularly interested in case 2). That is to say, what condition should we have on $\{a_i\}$ to ensure that four real roots exist?

答: For $|x|\gt A=1+|a_0|+|a_1|+..+|a_5|$, the polynomial is always positive.  Sturm sequence then can be used to determine number of real roots $\gt -(A+1)$.

Title: Four Real roots of sextic polynomial


答:
By definition: $$\lim_{n\to\infty}b_n=\infty \Longleftrightarrow \forall M\in\mathbb{R} \ \exists n(M)\in\mathbb{N} \text{ such that } b_n>M \  \ \ \forall n\geq n(M) \\ \lim_{n\to\infty}a_n=5 \Longleftrightarrow \forall \epsilon\in\mathbb{R}^+ \ \exists n(\epsilon)\in\mathbb{N} \text{ such that } 5-\epsilon<a_n<5+\epsilon \ \ \ \forall n\geq n(\epsilon)$$
Then set $\epsilon:=4$. We can choose $n(4)\in\mathbb{N}$ such that $a_n>1=5-\epsilon$ for all $n\geq n(4)$. Also, for every $M\in\mathbb{R}$ we can choose $n(M)\in\mathbb{N}$ such that $b_n>M$ for all $n\geq n(M)$. Then, choosing $n\geq \max\{n(4),n(M)\}$, we have that: $$b_n\cdot a_n>M\cdot 1=M$$
Which is the definition of $\lim_{n\to\infty} b_n\cdot a_n=\infty$.

Title: sequence limits proof by definition

Body:
If I have $\displaystyle\lim_{x \to +\infty} a(x)=5 $. and $\displaystyle\lim_{x \to +\infty} b(x)=+\infty $.
How can I prove that $\displaystyle\lim_{x \to +\infty} (a*b)= +\infty $ by definition.

Can I just put n1 for the first sequence and n2 for the second and say n0 = max (n1, n2) and that applies to all members?


问:
Set of functions that form a basis for a Topology
I'm trying to prove the following statement: Define $X=\lbrace f| f: [0,1]\rightarrow[0,1]\text{ is a function}\rbrace$. For each subset $A$ of $[0,1]$, define
\begin{equation}
B_A=\lbrace f\in X|f(x)=0, \forall x \in A \rbrace.
\end{equation}
Prove that the set $B=\lbrace B_A|A \subseteq [0,1]\rbrace$ is a basis for a topology on $X$.

I tried doing the following: let $f \in X$ with $f \in B_{A_1} \cap B_{A_2}$. Then I need to prove that there is a basis element $B_A$ in $B$ such that $f \in B_A \subseteq B_{A_1} \cap B_{A_2}$. I got stucked in here: if $f \in B_{A_1} \cap B_{A_2}$ then
\begin{equation}
f(x)=0, \forall x\in B_{A_1} \cap B_{A_2}.
\end{equation}
How can I proceed from here? Only hints, not answers, please.

答:
By the definitions, for all $A_1,A_2$ we have: $$B_{A_1 \cup A_2} = B_{A_1} \cap B_{A_2}$$

Proof If $f \in B_{A_1 \cup A_2}$ then for all $x \in A_1 \cup A_2$ we have $f(x)=0$ and so in particular this holds for all $x \in A_1$ and $x \in A_2$ (which are subsets of the union), so $x$ is in the right hand side; and conversely, if $f \in B_{A_1} \cap B_{A_2}$ we have both that $f \in B_{A_1}$ so $f(x)= 0$ for all $x \in A_1$ and also $f \in B_{A_2}$ so that $f(x)=0$ for all $x \in A_2$ as well, and it follows that $f(x)=0$ on $A_1 \cup A_2$.QED

Finally note that $B_\emptyset= X$, voidly.


问:
Field extension
there is for example field $GF(2^4)=GF(16)$. Is $GF(16)$ a subfield of itself? Following this definition http://mathworld.wolfram.com/Subfield.html there is nothing written that subfield must contain less elements than original field.</p>

答: Yes, a field is (trivially) a subfield of itself.  If we want to clearly state that something is a subfield of F and not F itself, we might say "proper subfield".</p>

Title: Field extension


问:
Absolute convergence of a real series
I need to show that the following series:

$\sum_{i=1}^n (-1)^n\dfrac{x^2+n}{n^2} $

Is uniformly convergent on any bounded interval, but not absolutely convergent for any real $x$. My first thought was to use the Weierstrass M-Test, however this is pointless as if the above series "passed" the test, it would have to be absolutely convergent.

In assessing uniform convergence, I considered partial sums. Is it possible to show that the partial sums can get arbitrarily close together, (in a Cauchy sense) which would then imply the series converges uniformly?

答:
Note that in absolute value, we have, since all but $(-1)^n$ is positive:

$$\sum_{n=1}^N\frac{x^2+n}{n^2}$$

But this is

$$\sum_{n=1}^N\frac{x^2}{n^2}+\sum_{n=1}^N\frac{1}{n}$$

The first term converges, but the second term doesn't. Note this tells the $M$ test is not applicable.

On the other hand, your sum is $${f_N}\left( x \right) = \sum\limits_{n = 1}^N {{{( - 1)}^n}} \frac{{{x^2} + n}}{{{n^2}}} = \sum\limits_{n = 1}^N {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}} + \sum\limits_{n= 1}^N {{{( - 1)}^n}} \frac{1}{n}$$

This converges pointwisely to

$$\mathop {\lim }\limits_{N \to \infty } {f_N}\left( x \right) = \sum\limits_{n = 1}^\infty  {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}} + \sum\limits_{n = 1}^\infty  {{{( - 1)}^n}} \frac{1}{n} =  - \frac{{{x^2}{\pi ^2}}}{{12}} + \log 2$$
Now, look at the difference; we have $$\displaylines{
  \left| {f\left( x \right) - {f_N}\left( x \right)} \right| = \left| {\sum\limits_{n = N + 1}^\infty  {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}} + \sum\limits_{n = N + 1}^\infty  {{{( - 1)}^n}} \frac{1}{n}} \right| \cr
   \leqslant \left| {\sum\limits_{n = N + 1}^\infty  {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}}} \right| + \left| {\sum\limits_{n = N + 1}^\infty  {{{( - 1)}^n}} \frac{1}{n}} \right| \cr
   \leqslant {x^2}\left| {\sum\limits_{n = N + 1}^\infty  {\frac{1}{{{n^2}}}} } \right| + \left| {\sum\limits_{n = N + 1}^\infty  {{{( - 1)}^n}} \frac{1}{n}} \right| \cr
   \leqslant M\varepsilon  + \varepsilon  = \left( {1 + M} \right)\varepsilon  \cr} $$

where $M$ is a bound for $x^2$ on any bounded interval, for sufficiently large $N$.

Title: Absolute convergence of a real series


问:
How to calculate $\lim_{n\to\infty} \frac{2^n}{n!}$?
How to calculate $\lim_{n\to\infty} \frac{2^n}{n!}$?
I tried using $\lim \frac{x^n}{a^x} = 0$ but it didn't work

答:
We have
$$2^n = (1 + 1)^n = \sum_{k = 0}^n\frac{n!}{k!(n-k)!}.$$
Try to conclude from that.


答:
The limit is $0$, since factorial growth is faster than exponential growth.

$$ 2^n/n! = (2/1)*(2/2)*\ldots*(2/n)$$
$$ \leq 2 * 2/n$$

This quantity can clearly be made arbitrarily small for sufficiently large n, hence the limit is 0 (since $2^n/n!$ is positive)

Body:
How to calculate $\lim_{n\to\infty} \frac{2^n}{n!}$?
I tried using $\lim \frac{x^n}{a^x} = 0$ but it didn't work


问:
It isn't clear what objects of $\mathbb{C}[x_{1},\dots,x_{n}]$ and $\dfrac{\mathbb{C}[x_{1},\dots,x_{n}]}{\mathbb{I}(V)}$ are isomorphic to me.
Page 22 of Smith. An Invitation to Algebraic Geometry says that "Obviously, restriction defines a surjective ring homomorphism $\mathbb{C}[x_{1},\dots,x_{n}] \rightarrow \mathbb{C}[x_{1},\dots,x_{n}]|_{V}$ with kernel precisely the ideal of functions $\mathbb{I}(V)$ vanishing on $V$. So the coordinate ring $\mathbb{C}[x_{1},\dots,x_{n}]|_{V}$ is isomorphic to the ring $\dfrac{\mathbb{C}[x_{1},\dots,x_{n}]}{\mathbb{I}(V)}$ in a natural way. Equivalence classes in $\mathbb{C}(V)$ correspond to functions on $V$."

The ring homomorphism has a kernel that is mapped to $0$ yet the isomorphism appears to be between equivalence classes of functions that have the same value restricted to $V$. What objects are isomorphic between the two rings and can I have another example of two such isomorphic objects?

答:
Let $R = \mathbb C[x_1, ... , x_n]$.  They are thinking of $R$ in two ways, first as an abstract ring, and second as a ring of polynomial functions $\mathbb C^n \rightarrow \mathbb C$.  It is in the second way that they are making sense of $R|_V$.  By definition, $R|_V$ is the ring of functions $V \rightarrow \mathbb C$ obtained by restricting polynomial functions $\mathbb C^n \rightarrow \mathbb C$ to $V$.

For $R|_V$, you don't have to worry about equivalence classes of functions.  The elements of $R|_V$ are literally functions from $V$ to $\mathbb C$.


问:
MATLAB Vector Assignment of Arbitrary Length
Greetings: let's suppose I have a loop that will execute 20 times, and within that loop I'll obtain a vector of information whose length will increase, each iteration, from 2 to 20. Now suppose I have a matrix of dimension 20, and during each iteration, I'd like to store that vector to one row of the matrix. For n = 2, the vector is length 2, and when I assign it to the first row of the matrix, call it A, I might try something like A(1,:) = p. But, since the length of the first row is 2o, and length of p is 2, there's a mismatch. Any clever suggestions on how to get around this issue.

Thanks so much...

John Sevic

答:
This can be accomplished as follows:

```<code>P = zeros(19,20); %Preallocate memory

for i = 2:20
    p = someFunction();
    P(1:i,i-1)=p;
end
</code>```

This will store each vector p in a row of P.

Body:
This can be accomplished as follows:

P = zeros(19,20); %Preallocate memory

for i = 2:20
    p = someFunction();
    P(1:i,i-1)=p;
end

This will store each vector p in a row of P.


问:
What is the value of a delta function?
So my understanding is that a delta function $\delta(t)$ has the value of $\infty$ at $t = 0$ and $\delta(t) = 0$ everywhere else. Essentially, it's just one big spike on a graph right?

In my signal processing lecture, I came across the following:

Here, we arrive at $X_1[k] = \frac{1}{2}\delta [k - 1] + \frac{1}{2}\delta [k + 1]$, which means  $X_1[k]$ only has non-zero value at $k - 1$ and $k + 1$. However, using my definition of a delta function, shouldn't $X_1[k = \pm 1] = \infty$?

This isn't the case when you look at the graph. It shows $X_1[k = \pm 1] = \frac{1}{2}$, so what's wrong? Is there some other delta function I just don't know?

答:
Two answers here.

First, the answer from the classical point of view.

Delta function is defined as something which has the following property:

$$\int_{-\infty}^\infty \delta(x) f(x)dx=f(0)$$

It is not a function but some special symbol.

Now, $a \delta(x)$ is defined as heaving the following property:

$$\int_{-\infty}^\infty a\delta(x) f(x)dx=a f(0)$$

That's it.

Actually one can consider delta function as a function that takes values from the set of divergent integrals rather than real numbers.

By definition via Fourier transform,

$$\delta (x-\alpha )={\frac {1}{2\pi }}\int _{-\infty }^{\infty } \cos(px-p\alpha ) dp$$

This gives formally $\delta(0)=\frac1\pi \int_0^\infty dx$
You can consider this divergent integral the value of Dirac Delta at zero. It is not a real number though.

This way, $a\delta(0)$ would be $\frac1\pi \int_0^\infty a dx$, also a divergent integral.

Title: What is the value of a delta function?


问:
Are the natural numbers closed under exponentiation?
Is it true that $\forall n \in \mathbb Z^+ \land \forall i \in \mathbb Z \land i \geqslant 0: n^i \in \mathbb Z^+$?

The reason I ask is because this would allow you to create a countably infinite set $$\forall k \in \mathbb Z \land k \geqslant 0, \{a_k \in \mathbb Z^+ : a_k = 10^k\}$$

This set would contain elements of the form $\{1, 10, 100, 1000, \ldots\}$, i.e. each element $a_k$ of the set would be a $1$ followed by $k$ zeros, up to an infinite amount of zeros. But a natural number cannot have an infinite number of digits, so that would seem to say that the natural numbers are not closed under exponentiation. Am I getting something wrong here?

答:
They are closed under exponentiation, the contradiction you gave isn't a real contradiction.

When you say that there can be "up to an infinite amount" of zeroes, that just means that there is no upper bound on the number of trailing zeroes. However any specific number you choose from that set will have some finite number of trailing zeroes.

It's similar to how the natural numbers "go to infinity", even though every natural number you choose will have to be finite.


问:
Convergence of recurrence relation
How can I prove that the following recurrence relation converges to 2:
$$a_{n+1}=\frac{3a_n}{\operatorname{gcd}(2^{a_n},a_n)}+1$$
When $a_0$ is even.
According to a similar post it can be easily proven if $a_n$ is to converge it will converge to 2 however I don't know how I can prove its convergence. In case it is useful, a continuous form of the recurrence is:
$$a(x+1)=e^{x}\frac{\sin(2\pi x)}{2\pi}\sum_{k=1}^{\infty}\frac{e^{-k}f(k)}{x-k}$$
Where $f(x)$ is:
$$f\left(x\right)=\frac{3x}{\gcd\left(2^{x},x\right)}+1$$

答:
The exponent of $2$ appearing in the factorization of $a_n$ will always be smaller than $a_n$ (Cantor's theorem).

This means that, if we write $a_n = 2^v m$ with $m$ odd, then $\gcd(2^{a_n}, a_n)$ is simply $2^v$, and the right hand side of the formula is $m + 1$, which is again even.

Given that $a_0$ is even, we immediately know that all $a_n$ are even.

Now it only remains to remark that $a_{n + 1}$ is strictly smaller than $a_n$ whenever $a_n > 2$ is even.

Title: Convergence of recurrence relation


问:
How many possible solutions are to the inequation $x_1+ ... +x_{n} \leq k$?
Given $\forall i, i=1....n$ | $s.t$  $x_i\in\mathbb{N}$ .

1.How many possible solutions are to the equation: $  x_1+x_2+...x_{n}\leq k$ ?

2.If $x_i\in\mathbb{Z}$?

3.If $x_i\leq i$ $ \forall i$ ?

*Note: $0\notin\mathbb{N} $

答:
The question 3 may be interesting and here is the solution for rather arbitrary restriction $x_i\le r_i$. To satisfy the inequality
$$  x_1+x_2+...x_{n}\leq k$$
we introduce an additional dummy variable $x_0$ and count the number of solutions to the equation:
$$
x_0+x_1+\cdots+x_n=k,
$$
where $x_0$ is non-negative and the other $x_i$ are positive integer numbers.

In terms of generating functions we are looking for the coefficient at $x^k$ in the product:
$$\begin{align}
\frac1{1-x}x^n\prod_{i=1}^n\sum_{j=0}^{r_i-1}x^j
&=\frac1{1-x}x^n\prod_{i=1}^n\frac{1-x^{r_i}}{1-x}\\
&=\frac{x^n\prod_{i=1}^n(1-x^{r_i})}{(1-x)^{n+1}}\\
&=x^n\prod_{i=1}^n(1-x^{r_i})\sum_{j=0}^\infty\binom{-n-1}{j}(-x)^j\\
&=x^n\prod_{i=1}^n(1-x^{r_i})\sum_{j=0}^\infty\binom{n+j}{j}x^{j}\\
&=\sum_{\boldsymbol{\mu}}(-1)^{|\boldsymbol{\mu}|}x^{\boldsymbol{\mu}\cdot\boldsymbol{r}}
\sum_{j=0}^\infty\binom{n+j}{n}x^{n+j},
\end{align}$$
which is
$$
\sum_{\boldsymbol{\mu}}(-1)^{|\boldsymbol{\mu}|}\binom{k-\boldsymbol{\mu}\cdot\boldsymbol{r}}{n},\tag1
$$
where $\boldsymbol{\mu}$ are all $2^n$ binary vectors of the length $n$, $|\boldsymbol{\mu}|=\sum_{i=1}^n\mu_i$ and $\boldsymbol{r}=(r_1,r_2,\dots,r_n)$.

In (1) the equality $\binom mn=0$ is assumed for $m<n$.


问:
Linear and Affine functions
Wanted to be clear on my understanding on affine functions, and if we have a matrix $A \in \mathbb{R}^{n \times m}$ and a vector $c \in \mathbb{R}^{n}$, I know an affine map is given by:
$$f: \mathbb{R}^m \rightarrow \mathbb{R}^n, f(x) = Ax + c  $$
and if $c$ happens to be the $0$ vector, then the map/function would be linear. So going based of this definition and my understanding of it, I would imagine the following three properties to be true?

For every $\gamma \in \mathbb{R}$ and the vectors $v_1, v_2 \in
   R^{m}$, a linear function should satisfy the property: $f(\gamma v_1
 + v_2) = \gamma f(v_1) + f(v_2)$.

Assuming $m=n$, and $A$ nonsingular, then a linear function should
preserve the length of the argument $||f(x)|| = ||x||$.

Assuming $m=n$, and $A$ nonsingular, then an affine function should
be nonsingular/invertible and its inverse is of the form/given by the function,
$f^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^n, f^{-1}(x) = A^{-1}(x
   - c)$

答:
First point

This is correct.

As you correctly pointed out, if $f$ is linear, then $f(x)=Ax$ for some $n\times m$ matrix $A$.

In this case we have
\begin{align}
f(\gamma v_1+v_2) & = A(\gamma v_1+v_2) \\
                  & = A(\gamma v_1)+Av_2 \\
                  & = \gamma Av_1+Av_2 \\
                  & = \gamma f(v_1)+f(v_2).
\end{align}

Hint for the second point

Consider $f\colon\mathbb{R}^3\to \mathbb{R}^3$ defined by $f(x)=Ax$ where
$$
A=
\begin{pmatrix}
k & 0 & 0 \\
0 & k & 0\\
0 & 0 & k
\end{pmatrix}
$$
and $k\neq 0,1$.

Will this preserve length? Where will the vector $$\begin{pmatrix}1 \\ 0 \\0\end{pmatrix}$$ get mapped? And what will the length of its image be?

Third point

Provided $A$ is invertible, yes, the affine function will be invertible with the inverse you described.

Title: Linear and Affine functions


答:
By the limit comparison test, the series does not converge absolutely at any point because the harmonic series diverges and for all $x$, $\lim\limits_{n\to\infty}\dfrac{\dfrac{x^2+n}{n^2}}{\dfrac{1}{n}}=1$.

The series converges uniformly on bounded intervals because $\sum\limits_{n=1}^\infty\dfrac{(-1)^n}{n}$ converges everywhere uniformly (there's no $x$ in it), whereas on the interval $[-M,M]$, the Weierstrass $M$ test can be applied with $\dfrac{x^2}{n^2}\leq \dfrac{M^2}{n^2}$, showing that $\sum\limits_{n=1}^\infty\dfrac{(-1)^nx^2}{n^2}$ converges absolutely and uniformly on $[-M,M]$.

Title: Absolute convergence of a real series

Body:
I need to show that the following series:

$\sum_{i=1}^n (-1)^n\dfrac{x^2+n}{n^2} $

Is uniformly convergent on any bounded interval, but not absolutely convergent for any real $x$. My first thought was to use the Weierstrass M-Test, however this is pointless as if the above series "passed" the test, it would have to be absolutely convergent.

In assessing uniform convergence, I considered partial sums. Is it possible to show that the partial sums can get arbitrarily close together, (in a Cauchy sense) which would then imply the series converges uniformly?


问:
Coin toss game for choosing two numbers a,b and other player predicting the greater among the two. Find winning strategy
$\textit{Question:}$

Given a coin with probability p of landing heads after being flipped, pick two numbers
a, b such that a $\neq$ b, and associate “heads” to one of them and “tails” to the other. Flip
the coin and give the number associated to the flip to your friend. For what values of p is
there a strategy for your friend so that the probability of predicting whether the number
that is given is bigger than the number that is not given is strictly greater than 1/2 despite
being allowed to know how you select the numbers a and b?

Stuck with this problem recently. A strategy definitely exists for certain values of $p$, that's what my prof told. Can't get my head around it though.

答:
The case when $p=1/2$ is a classic puzzle. WLOG, $a<b$.

For any real number $x$, define $S_x$ to be the following strategy:

If the number you are told is less than $x$, guess that it is lower than the other number.

Otherwise, guess that the number you are told is higher than the other number.

Note that $S_x$ always gives a win whenever $a< x \le b$. If $x\le a$, then you always guess "higher" and win with probability $1/2$. Similarly, if $x>b$, you always guess "lower" and  win with probability $1/2$.

Therefore, in order for your friend to ensure they have an edge, all they need to do is randomly choose $x$ so that $x\in (a,b]$ with some nonzero probability. If $a,b$ are integers, this can be done by using a geometric random variable to choose a random integer $n$, then let $x=n+\frac12$. If $a$ and $b$ are real numbers, you can instead choose a random rational number by writing all the rational numbers in a list and using a geometric random variable to choose an element of this list.

I am not sure what to do when $p\neq 1/2$. Certainly when $p=0$ or $p=1$, there is no way to get an edge. I suspect that the same is true for all other $p\neq 1/2$, but I am not sure.