问:
The consistency of PA is falsifiable. Can the same be said of its soundness?
The statement that 'PA is consistent' is absolutely falsifiable (a term I just made up), in the sense that if it is false, then we can demonstrate its falseness independently of any metatheory (just deduce a contradiction within PA).

Can something similar be said of the statement that 'PA is sound'? I think not, because to even make sense of the claim that PA is sound, we need a metatheory (although see here).

Clarification. Let me attempt to clarify what I mean by "absolutely falsifiable." If we're given a formal system with a recursively ennumerable set of axioms, then if someone claims that "the string $\phi$ is not deducible in the system," well this is a falsifiable claim. In the sense that, if the claim is false, in other words if $\phi$ is deducible in the system, then we can verify this within the rules of the system, simply by deducing $\phi$.

For instance, if I claim that "ZFC cannot prove that the Ackermann function is defined for all inputs," well this is falsifiable: because, if it is a falsehood, then you can confirm that its a falsehood by using ZFC to prove that the Ackermann function is, in fact, defined for all inputs. In fact, since its already known that ZFC proves that the Ackermann function is total, you already know the claim is a falsehood.

On the other hand, if I simply claim that "the Ackermann function is not defined for all inputs" without reference to a foundational system, well this is a more problematic claim. If its true, we can never confirm this, because we have to check infinitely many inputs to make sure the program halts. If its false, then for some input, we'll have to wait an infinite amount of time before we can verify that, indeed, its not going to halt. So the claim that "the Ackermann function is not defined for all inputs" is not falsifiable (and neither is its negation).

答:
There is indeed a significant difference between proving a theory is consistent, and proving that all of its theorems are true, which is what the question calls "soundness".

The consistency of a theory is a universal statement that can be concretely disproved by giving a counterexample. Soundness cannot be disproved in that way. The argument that it cannot is necessarily heuristic.

The first key point is that any consistent theory of arithmetic which is sufficiently strong will be sound for atomic sentences - so it will not prove anything like "$2 + 2 = 5$" that could be used to concretely see the theory is unsound. This is because any is  strong theory already proves every true atomic sentence, so it can only prove a false one if it is inconsistent.

So if PA is consistent and unsound, the false sentence $\phi$ that it proves would have to involve at least one quantifier. But then, to show PA is unsound, we would have to show that $\phi$ is true. In general, $\phi$ might have a very large number of nested quantifiers; it is known that for each $n$ there is a theory that is unsound but is sound for all formulas that have no more than $n$ quantifiers. So there is no reason to think that the falsity of $\phi$ would be intuitively clear or obvious. In general we would need to give an argument in some (meta)theory that $\phi$ is false.

Wrapping up point 1: if $\phi$ is provable in PA, that can be demonstrated concretely by giving a proof of it in PA.  But, to argue PA is unsound, we also need to argue that $\phi$ is false. That additional step is what distinguishes proving unsoundness from merely proving inconsistency.

As a second point: for any sound, effective theory $T$ of arithmetic, the theory $S = T + \lnot \text{Con}(T)$ is also consistent, and is "sound relative to $T$" in the sense that whenever $S$ proves a sentence, $T$ cannot disprove that sentence. So, if we had some effective way of listing all the "obvious" facts about the natural numbers, would could call that $T$, and there would be some other theory $S$ that is unsound although our "obvious" facts cannot verify the unsoundness.

As I said, these two points provide only a heuristic argument, but together they give a lot of evidence that there is no reason to expect that a concrete (finitistic) demonstration of the unsoundness of an unsound theory of arithmetic can always be provided. As usual, we cannot make that statement precise without formalizing the metatheory.

Here are two definitions, as mentioned in the comments.

A first-order theory $T$ is called sound if all the theorems of $T$ are true. Here "true" means "true in the intended interpretation of $T$", which is usually determined from context. For a theory in first-order logic, because the inference rules all preserve truth, soundness of a theory is equivalent to all the axioms of the theory being true in the intended interpretation.  This sense of soundness is closely related to the traditional notion of a sound argument: an argument is sound if its form is valid and all of its premises are true. On the other hand one can present valid arguments that use false hypotheses; the difference is that sound arguments establish the truth of their conclusions, while valid arguments may not.  Similarly, sound theories establish the truth of their theorems, while unsound theories do not.

Soundness of a logic is a more delicate matter. A general (two-valued) logic has a collection $\mathcal{L}$ of sentences, a collection $\mathcal{D}$ of inference rules, and a collection $\mathcal{I}$ of interpretations, where each interpretation includes a function from $\mathcal{L}$ to $\{T,F\}$ which tells whether each sentence is true or false in that interpretation. The logic $(\mathcal{L}, \mathcal{D}, \mathcal{I})$ is sound if, whenever a sentence $\phi \in \mathcal{L}$ is provable from some set $H$ of hypotheses, then every interpretation in $\mathcal{I}$ that makes every sentence in $H$ true also makes $\phi$ true.

The difference is that a general logic may have no sense of an "intended interpretation", so soundness of a logic merely states that the inference rules preserve truth in all the interpretations in $\mathcal{I}$. But in the case of foundational theories, such as Peano arithmetic, we already have an intended interpretation in mind, and so for soundness of a theory we focus on truth in the intended interpretation rather than truth in all interpretations.


问:
Looking for an intuitive explanation why the row rank is equal to the column rank for a matrix
I am looking for an intuitive explanation as to why/how row rank of a matrix = column rank. I've read the proof on Wikipedia and I understand the proof, but I don't "get it". Can someone help me out with this ?

I find it hard to wrap my head around the idea of how the column space and the row space is related at a fundamental level.

答:
Consider an $ m \times n $ matrix $ A = (a_{ij}) $. By row reduction, there exist elementary matrices $ E_1, \cdots ,E_p$ such that $ E_p \cdots E_1 A $ is a step matrix $ A' $.

[The first non-zero entry of a row, if exists, is called a pivot of that row. For an $ m \times n $ matrix $ A $, we'll denote it's rows by $ A_{1*}, \cdots, A_{m*} $ and columns by $ A_{*1}, \cdots, A_{*n} $. A matrix $ A $ is in step form when it's zero rows (if any) are at the bottom, and pivot indices $ j_1, \cdots, j_r $ of non-zero rows $ A_{1*}, \cdots , A_{r*} $ satisfy $ j_1 < j_2 < \cdots < j_r $ . Step matrices are also called row echelon matrices]

Say $ A' $ has $ r $ non-zero rows with pivot indices $ j_1 < \cdots < j_r $.

For any elementary matrix $ E $, $ A \mapsto EA $ preserves row space (i.e. span of matrix rows). So especially row rank (i.e. dimension of row space) of $ A $ is row rank of $ A' $, which is $ r $.

Now let's think of the column rank of $ A $. Here we'll be considering only columns, so let's take $ A_j $ to just mean column $ A_{*j} $. For any elementary matrix $ E $, $ ( A_{i_1}, \cdots, A_{i_k} ) $ is a basis of column space of $ A $ if and only if $ ( EA_{i_1}, \cdots, EA_{i_k} ) $ is a basis of column space of $ EA $, so we need only focus on the column space of $ A' $. But $ ( A'_{j_1}, \cdots, A'_{j_r} ) $ is a basis of column space of $ A' $, hence $ (A_{j_1}, \cdots, A_{j_r}) $ is a basis of column space of $ A $. Especially column rank of $ A $ is also $ r $.

To summarise, an elementary row operation on $ A $ preserves both row space and dimension of column space, so we need only look at row and column ranks of its step form $ A' $. But both of these are just the number of pivots of $ A' $.

ContentLicense: CC BY-SA 4.0


问:
Let $A$ an integral domain. Show that the zero ring has only one associate: himself.
I did the following demonstration: Would it be possible to give me a feedbak if i'm to wrong.

$A$  must be the zero ring.

Here's a proof.

By definition, if $a$ and $b$ are associates then each is a unit multiple of the other. In particular for this special $a$ we have that $b$ can be anything. Take $b=0$. Then $a=cb=0$ for some unit $c$, hence $a=0$. This uses the fact that $a$ is a multiple of $b$.

To finish the proof that $A$ is the zero ring, note that no matter what $b$ we choose, it must be a multiple of $a$. But $a=0$, hence $b=0$, so $0$ is the only element of $A$.

$\square$

答:
I think you have the right ideas, but your presentation is a little bit disorganized. You're also a little bit confused about what you're trying to prove; you want to show that if $A$ in an integral domain, then $0$ has only itself as an associate. There's lots of ways to word something like this, but here is what I would write.

Suppose $A$ is an integral domain and let $r$ be an associate of $0$. If $A$ is the zero ring then the statement is obviously true. Suppose $A \neq 0$. Then there exists some unit $u$ such that $0 = ur$. Since $A$ is an integral domain this implies that $u =0$ or $r=0$ and since units are nonzero (here we are using that $A \neq 0$ implicitly), we have that $r=0$. Thus the only associate of $0$ is $0$, which was the desired result.

Note that your demonstration shows that any associate of $0$ must be $0$, not that every element in the ring is $0$ because you started by assuming that $a$ and $b$ are associates. This is an important distinction and one you should make sure you understand well.


答: Your proof is correct. However, I would explain in more details why $f(x) \notin Y_2$ is equivalent to $x \notin f^{-1}(Y_2)$. The reason is that the domain of $f$ is $X$, because $f$ is a function. This equivalence would no longer be true if $f$ were only a partial function.

Title: Given a function $f:X \to Y$,show that $f^{-1}(Y_1-Y_2)=f^{-1}(Y_1)-f^{-1}(Y_2)$ Where $Y_1,Y_2 \subseteq Y$.

Body:
Given a function $f:X \to Y$,show that $$f^{-1}(Y_1-Y_2)=f^{-1}(Y_1)-f^{-1}(Y_2)$$

Where $Y_1,Y_2 \subseteq Y$.

$$ x \in f^{-1}(Y_1-Y_2)$$
$$ \iff f(x) \in (Y_1-Y_2)$$
$$ \iff f(x) \in Y_1 \wedge  f(x) \notin Y_2 $$
$$ \iff x\in f^{-1}(Y_1) \wedge  x\notin f^{-1}(Y_2)$$
$$ \iff x\in f^{-1}(Y_1)- f^{-1}(Y_2)$$

Is my proof true?


问:
Amateur equating powers question
I will admit to this being a rather amateur question that is fueled by total ignorance but can someone explain how i=-5/2,   j=1/2, and k=1/2 please? Thank you so so much in advance!

答:
From the exponents of M we have $k=j$ and substituting into the other two equation gives:

$$ 0=i+5j$$

$$1=-i-3j$$

Can you end it now?

Title: Amateur equating powers question


问:
How to tell if a Linear Function is Orthogonal over some Inner Product
Let $V$ be a finite dimensional $K$-Vector Space ($K= \Bbb R$ or $\Bbb C$). I'm trying to characterize all linear transformations $f:V \rightarrow V$ such that $f$ is orthogonal (or unitary) for some inner product over $V$. Said in another way, given a linear function $f:V \rightarrow V$, how can I tell if there exists an inner product over $V$ that makes $f$ and orthogonal (or unitary) transformation.

For example, a necessary condition is that $| \det (f)|=1$ . Another necessary condition is that if $\lambda \in \Bbb C$ is an eigenvalue of $f$ then $|\lambda|=1$.

A necessary and sufficient condition for such an inner product to exist is "There exists a bases $B$ in $V$ such that the columns of $|f|_B$ form an orthonormal basis over $\Bbb R^n$ (or $\Bbb C^n$) with the usual inner product". This is not something hard to prove. So we could change the question and ask the following

Let $K= \Bbb R$ or $\Bbb C$, we say that a matrix $M\in K^{n \times n}$ is orthonormal if the columns of $M$ form an orthonormal basis over $K^n$ with the cannonical inner product. Given a matrix $A\in K^{n \times n}$, how can we tell if there is an invertible matrix $C\in K^{n \times n}$ such that $C \cdot A \cdot C^{-1}$ is an orthonormal matrix?

Im more interested in working over $K=\Bbb R$ but maybe is easier to work over $K=\Bbb C$ so we can use the Jordan Normal Form.

答:
A matrix $A$ is orthogonal over some inner product when it is diagonalizable with eigenvalues of unit length. Diagonalizability can be checked from its minimal polynomial, so this is not simply a restatement of the problem.

$A$ is orthogonal over some inner product $S$ when $A^*SA=S$ (as in the usual proof when $S=I$). For $S$ to be an inner product it must be positive definite, $S=C^*C$, with $C$ invertible. Hence $$A^*C^*CA=C^*C\implies (CAC^{-1})^*=(CAC^{-1})^{-1}\implies A=C^{-1}UC$$ for some unitary $U$. Since $U$ is diagonalizable this shows that $A$ is also diagonalizable, with the same eigenvalues as $U$, namely of unit length.

Conversely, if $A=C^{-1}e^{iD}C$ then $$A^*(C^*C)A=C^*e^{-iD}C^{-*}C^*CC^{-1}e^{iD}C=(C^*C)$$ so $S=C^*C$ is the required inner product.


问:
Does $\int_{-\infty}^\infty \frac{1}{x-a}dx$ converge?
I don't think this integral converges:

$$\int_{-\infty}^\infty \frac{1}{x-a}dx\qquad x,a\in\mathbb R$$

The integrand is unbounded.

I keep getting undefined natural logarithms and non-existent limits (infinites) whenever I try to work this integral around.

Can you confirm or prove otherwise?

Thank you!

答:
It diverges at $x\to+\infty$.  It diverges at $x \to-\infty$ it diverges at $x \to a$.  Summary: it diverges.

Physicists may investigate a "principal value" for this, but in this forum we do not consider that unless the problem specifically mentions it.


问:
Convergence of p-series over subset of the naturals
This was a question from an exercise in a first semester Algebra class. The question is as follows:

Let $S \subseteq \mathbb{N}$ and suppose that $|\{n \leq x: n \in S\}| = O ((\log(x))^d)$ for some $d \geq 1$. Show that $$ \sum_{s \in S} \frac{1}{s^{\epsilon}}$$ converges for each $\epsilon > 0$.

My attempt:

I first tried to solve a simpler version of this problem in which we consider a subset $|\{n \leq x: n \in S\}| = O ((\log(x)))$. In this case, the series must converge because we can find a subsequence of partial sums of a geometric series with r < 1 that bounds the partial sums from above. I believe this is possible since the terms eventually become less dense in the reals than the terms of the aforementioned geometric series.

The problem arrives really when we have to prove that this result holds for all $d \geq 1$. I really don't know how to analyze this polylogarithmic growth of the cardinality of the subset.

We also proved earlier that $$ \prod _{n=1}^{\infty }(1+a_{n})$$
converges if and only if the series
$$ \sum _{n=1}^{\infty }a_{n}$$
converges for a sequence of positive reals $\{ a_n\}$. However, I don't see how converting this sum to a product really helps.

答:
This is a nice application of Abel's summation formula:

Indeed, define the indicator function $1_S \colon \mathbb{N} \to \{0,1\}$ of $S$ by $1_S(x) = 1$ if $x \in S$ and $1_S(x) = 0$ if $x \not\in S$. Now consider the partial sum
$$\sum_{s \in S, s \leq x} \frac{1}{s^{\varepsilon}} = \sum_{1 \leq n \leq x} 1_S(n) \cdot \frac{1}{n^{\varepsilon}} =^{\text{Abel}} A(x) \cdot \frac{1}{x^{\varepsilon}} + \varepsilon \int_{1}^x A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y,$$
where $A(x) := \sum_{1 \leq n \leq x} 1_S(n) = |\{n \leq x \colon n \in S\}|$. (Here we used that $-\varepsilon/y^{\varepsilon+1}$ is the derivative of $1/y^{\varepsilon}$.)

Now it remains to check that
$$\lim_{x \to \infty} \Big( A(x) \cdot \frac{1}{x^{\varepsilon}} + \varepsilon \int_{1}^x A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y \Big) < \infty$$
for every $\varepsilon > 0$. For the proof, we need two ingredients: On the one hand, our assumption says that
$$A(x) \leq C_0 \cdot \log(x)^d$$
for every sufficiently large $x > 0$ and some constant $C_0 > 0$. On the other hand, it is true that $\log(x) \in \mathcal{O}(x^{\delta})$ (as $x \to \infty$) for any $\delta > 0$. (You can prove this by computing $\lim_{x \to \infty} \log(x)/x^{\delta}$ using L'Hospital's rule.) As a consequence, we also have $\log(x)^d \in \mathcal{O}(x^{\delta})$ (as $x \to \infty$) for any $\delta > 0$ and $d \geq 1$.

The above suggests that there exists a constant $C_2 > 0$ such that for all $x > C_2$, we have
$$A(x) \leq C_0 \cdot \log(x)^d$$
and
$$\log(x)^d \leq C_1 \cdot x^{\varepsilon/2} \text{ for some constant } C_1 > 0.$$
Therefore,

$$\lim_{x \to \infty} A(x) \cdot \frac{1}{x^{\varepsilon}} \leq C_0C_1 \cdot \lim_{x \to \infty} \frac{x^{\varepsilon/2}}{x^{\varepsilon}} = C_0C_1 \cdot \lim_{x \to \infty} \frac{1}{x^{\varepsilon/2}} = 0$$
and
$$\lim_{x \to \infty} \varepsilon \int_{1}^x A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y = \int_{1}^{C_2} A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y + \lim_{x \to \infty} \varepsilon \int_{C_2}^x A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y \leq $$
$$\int_{1}^{C_2} A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y + \varepsilon C_0 C_1 \lim_{x \to \infty} \int_{C_2}^x \frac{y^{\varepsilon/2}}{y^{\varepsilon+1}} \text{ } \mathrm{d}y = \int_{1}^{C_2} A(y) \frac{1}{y^{\varepsilon+1}} \text{ } \mathrm{d}y + \varepsilon C_0 C_1 \int_{C_2}^\infty \frac{1}{y^{\varepsilon/2+1}} \text{ } \mathrm{d}y < \infty$$
(because $1 + \varepsilon/2 > 1$, suggesting that the integral $\int_{C_2}^\infty$ converges).


问:
Normal approximation to Poisson Distribution
Using Stirling's formula, show that, if $\lambda \rightarrow \infty$, then for fixed $\alpha < \beta$
\begin{equation}
    \sum_{\lambda + \alpha \sqrt{\lambda}< k < \lambda + \beta \sqrt{\lambda}}p(k; \lambda) \rightarrow \mathcal{R}(\beta) - \mathcal{R}(\alpha)
\end{equation}

where $\mathcal{R}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-y^2/2}dy$

This is what I have thus far:
By definition we have $p(k; \lambda) = e^{-\lambda} \frac{\lambda^k}{k!}\sim e^{-\lambda} \frac{\lambda^k}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^k}$ using $\lambda \rightarrow \infty$ and Stirling's Approximation. We note that when $\lambda \rightarrow \infty$, we must have $e^{-\lambda} \sim 1 + \frac{1}{\lambda}$. Thus,    $p(k; \lambda) =\left(1 + \frac{1}{\lambda}\right) \frac{\lambda^k}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^k}$.

I'm not sure if I'm walking down a dead end, and would appreciate some feedback.

答: To prove$$\lim_{\lambda\to\inf}\left(\sum_{\frac{k-\lambda}{\sqrt{\lambda}}\in[\alpha,\,\beta],\,k\in\Bbb Z^\ast}\frac{\lambda^ke^{-\lambda}}{k!}\right)=\int_\alpha^\beta\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy,$$we use a discrete version of Laplace's method, together with the limit-of-sums definition of a Riemann integral. Since$$-\ln p(k)=\lambda-k\ln \lambda+\ln k!\in\lambda-k\ln\lambda+(k+\tfrac12)\ln k-k+\ln\sqrt{2\pi}+O(\tfrac1k),$$the first two derivatives thereof with respect to $k$, treated as if continuous, are$$-\ln\lambda+\ln k+\tfrac{1}{2k}+O(\tfrac{1}{k^2}),\,\tfrac1k-\tfrac{1}{2k^2}+O\left(\tfrac{1}{k^3}\right).$$To a first-order approximation, $-\ln p$ is minimized at $k\approx\lambda$ with second derivative $\approx\tfrac{1}{\lambda}$, so$$-\ln p\approx-\ln p(\lambda)+\tfrac{(k-\lambda)^2}{2\lambda}.$$This approximates the distribution as Gaussian, so unitarity fixes the proportionality constant. With $y:=\tfrac{k-\lambda}{\sqrt{\lambda}}\approx N(0,\,1)$, the sum operator $\sum_{\alpha\le y\le\beta}$ is applied to approximately $\frac{1}{\sqrt{2\pi}}e^{-y^2/2}$, giving the desired result.


问:
Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC
As title,

I could only figure out that

$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$

But couldn't get how to derive $2(a+b+c)$.

答:
Not sure if there is a more straightforward solution but in meantime,

$3 \sin\frac{A}{2} = 4 \sin\frac{B}{2} = 5 \sin\frac{C}{2} = r$

$\frac{C}{2} = 90^0 - \frac{A+B}{2}$

So, $\cos \frac{A+B}{2} = \frac{r}{5}$

Expanding and writing values in terms of $r$ and squaring both sides, we simplify it to a cubic

$\frac{\sqrt {9-r^2} \times \sqrt {16-r^2}}{12} - \frac{r^2}{12} = \frac{r}{5}$

$25(9-r^2)(16-r^2) = (5r^2 + 12r)^2$

$120r^3 + 769r^2 - 3600 = 0$.

As $0 \lt r \lt 3$, we get $r \approx 1.9$ (with help from WolframAlpha)

From here on, we get sides as $\approx 5.842, 6.947, 8.145$.


问:
Notation for set of equivalence classes
Suppose we define an equivalence relation $\sim$ on a set $S$. The exact details of the relation and checking that it verifies reflexivity, symmetry, and transitivity is less important, so let's assume it satisfies those properties. Then $S/\sim$, read "S mod $\sim$ denotes the set of equivalence classes.

I am trying to make sense of this notation, specifically the use of "mod." In a quotient $R/S$, we say that $r_1, r_2 \in R$ are equivalent if $r_1 - r_2 \in S$. Similarly, in $\mathbb{Z}/2\mathbb{Z}$, $a$ is equivalent to $b$ if $a - b \in 2\mathbb{Z}$. I do not see any parallel in general to $S/\sim$. The only similarity is that they are equivalent "with respect to $\sim$."

Am I missing something? Is there a better way to think of this notation or even what it means to "quotient by $\sim$"?

答: There is in fact a much more general answer to your question that covers quotient structures in general. First, read this post, which explains how we can form quotient structures using a suitable partition of the larger structure into parts that respects the structure (the operations and predicates). Notice that it uniformly explains the isomorphism theorems for all kinds of structures. Next, observe that we write $S/\sim$ when $\sim$ is an equivalence relation that partitions $S$, but we also write $S/T$ when $T$ is an ideal of $S$, meaning that special part referred to in the linked post that we can use to construct a structure-respecting partition $S$. That is precisely why we write $ℤ/nℤ$, and $G/H$ for groups $G,H$ with $H ◁ G$, and $V/W$ for vector spaces $V,W$ with $W ⊆ V$, and $R/I$ for ring $R$ and ideal $I$ of $R$, and so on...

Title: Notation for set of equivalence classes


问:
Is the disc minus a countable set of points connected? Path-connected?
I am trying to solve exercises from Pugh's Real Mathematical Analysis 2nd ed. and there is a question related to the connectedness of a disc. The question statement is as follows:

Is the disc minus a countable set of points connected? Path-connected? What about the sphere or the torus instead of the disc?

Intuitively, I think the disc is connected and also path-connected after removing those points. However, I would like to see proof of this. Can someone help me with that?

答: Given two points in the disk, there exist an uncountable family of paths connecting them, pairwise disjoint (except the endpoints).  Subtracting a countable set will still leave at least one of these paths undisturbed.</p>


问:
Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$
Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?

I've calculated that the recurrence relation for this integral is:

$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$

So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.

Can this be useful here?

I would appreciate all your help.

答:
A related technique. You can use integration by parts technique by letting $u=e^{x}$ which leads to

$$ I_n = \left( 2\,n+1 \right)  \left( {\frac {{{\rm e}}}{n+1}}-{\frac {{
{\rm e}}}{2+3\,n+{n}^{2}}}+\int _{0}^{1}\!{\frac {{x}^{n+2}{
{\rm e}^{x}}}{ \left( n+2 \right)  \left( n+1 \right) }}{dx} \right) .$$

$$\implies \lim_{n\to \infty}I_n = 2 \,\rm{e} + 0 + \lim_{n\to \infty } \int _{0}^{1}\!{\frac {(2n+1){x}^{n+2}{
{\rm e}^{x}}}{ \left( n+2 \right)  \left( n+1 \right) }}{dx} $$

$$ \implies \lim_{ n\to \infty } = 2 \rm e . $$

Note that, the change of the limit with integral is due to the uniform convergence of the sequence $$ \frac {(2n+1){x}^{n+2}}{ \left( n+2 \right)  \left( n+1 \right) }. $$

Here is a technique for proving uniform convergence.

Added: Integration by parts,

$$ \int u\, dv = u\,v -\int v \,du. $$

So, in your case $ u = e^{x} $ and $ dv = x^n dx $.

ContentLicense: CC BY-SA 3.0


问:
Problem 25 (c) Kosniowski. A subset homeomorphic to the open disc is not a open neighbourhood of any of its points.
25.1 (c) Prove that if a subset $W$ of $\mathbb{R}^3$ is homeomorphic to the open disc $\mathring{D}^2$ then $W$ is not an open neighbourhood in $\mathbb{R}^3$ of any of its points .
Hint: If $W$ is an open neighbourhood in $\mathbb{R}^3$ of $w\in W$ then there is a subset $U_1\subseteq W$ with $w\in U_1$ and $U_1\cong \mathring{D}^3$ y definition.

Hi. I cannot understand this problem. It corresponds to the section on fundamental groups.
The idea I have is to calculate the fundamental group of $W$ which is isomorphic to the fundamental group of the open disk $\mathring{D}^2$ and also calculate the fundamental group of $U_1$ which is isomorphic to the fundamental group of $\mathring{D}^3$ and then obtain a contradiction but I don't know if it is a good way.

Assuming the hint. What would be the idea of contradiction? (I ask to try to prove it)

Actualization 1. Assuming the hint. I have the following idea:

exists $\phi:U_1\to\mathring{D}^3$  homeomorphism

exists $\psi:W\to\mathring{D}^2$ homeomorphism then $\left.\psi\right|_{U_1}:U_1\to \psi(U_1)$ is a homeomorphism.

Now, $\left.\psi\right|_{U_1}\circ\phi^{-1}:\mathring{D}^3\to \psi(U_1)\subset \mathring{D}^2$ is a homeomorphism and I don't know how to conclude anymore

答:
Building on @Daniel Apsley's interesting comment, we may consider the following.

Suppose $\phi:W\to B^2$ is the homeomorphism, then $\phi|_{U_1-\{w\}}$ is a homeomoprhism, too. By the assumption (the hint), $\phi(U_1)\approx B^3$, so $\pi_1(\phi(U_1)-\phi(\{w\}))\cong\pi_1(B^3-\{0\})\cong\{1\}\implies$ simply connected.

On the other hand, we have $\phi(U_1)-\phi(\{w\})\subset B^2\subset\Bbb{R}^2$. This space cannot be simply connected (if you've seen a proof, then this is immediate)
If $\phi(U_1)-\phi(\{w\})$ were simply connected, then every map $f:S^1\to \phi(U_1)-\phi(\{w\})$ can be extended to $\tilde{f}:D^2\to\phi(U_1)-\phi(\{w\})$, so it suffices to find one counterexample to this statement. Actually, there is a simple and natural one, just have a try before revealing the spoiler. ;)

 Consider the map \begin{align}f:S^1\to\phi(U_1)-\phi(\{w\}),\\ \begin{pmatrix}\cos(\theta)\\\sin(\theta)\end{pmatrix}\mapsto \phi(w)+\begin{pmatrix}\epsilon\cos(\theta)\\\epsilon\sin(\theta)\end{pmatrix}\end{align} where $\epsilon$ is carefully chosen so that the radius fits in the target space. Then, one may notice that there exists another map \begin{align}f^{-1}:y\mapsto \dfrac{y-\phi(w)}{||y-\phi(w)||}\end{align} that acts as an inverse since $(f^{-1}\circ f)(x)=x$ by direct computation. Then, the continuous map $f^{-1}\circ \tilde{f}$ induces $f^{-1}_*\circ\tilde{f}_*$, which $\twoheadrightarrow$. But this is impossible because $f^{-1}_*\circ\tilde{f}_*:\pi_1(D^2)\to\pi_1(S^1)\not\cong\{1\}$.


答:
Let $A$ be a countable set and $a,b\in\Bbb D\setminus A$. A line that passes between $a$ and $b$ intersects $\Bbb D$ in continuum-many points and each such point $p$ gives rise to a path $apb$ in $\Bbb D$, consiusting  of two line segments. Any distinct two such paths have only $a,b$ in common and only countably many of them pass through a point $\in A$. Hence there are still uncountably many such paths left that are all insider $\Bbb D\setminus A$.

The same works for any connected manifold of dimension $\ge 2$, where you can start with uncountably many disjoint (up to ends) paths between given endpoints

Body:
I am trying to solve exercises from Pugh's Real Mathematical Analysis 2nd ed. and there is a question related to the connectedness of a disc. The question statement is as follows:

Is the disc minus a countable set of points connected? Path-connected? What about the sphere or the torus instead of the disc?

Intuitively, I think the disc is connected and also path-connected after removing those points. However, I would like to see proof of this. Can someone help me with that?


答:
Any path-connected space $X$ such that for every $x \neq y \in X$ there are uncountably many paths $p_i: [0,1] \to X, i \in I$ from $x$ to $y$ such that $p_i[(0,1)] \cap p_{j}[(0,1)] = \emptyset$ for all $i \neq j \in I$ (so the paths only intersect at end points $x$ and $y$) has the property that $X\setminus D$ is still path-connected when $D$ is countable. (Because at least one of the uncountably many  paths must miss $D$).

This holds for spaces like $R^n, (n >1)$ and the disk too.

Title: Is the disc minus a countable set of points connected? Path-connected?


问:
Mapping Cylinder Neighbourhoods: Examples
So, yesterday I was trying to do Exercise 2.2.35 from Hatcher's Algebraic Topology, which goes like this:

35. Use the Mayer-Vietoris sequence to show that a nonorientable closed surface, or more generally a finite simplicial complex $X$ for which $H_1 (X)$ contains torsion, cannot be embedded as a subspace of $\mathbb{R}^3$ in such a way as to have a neighbourhood homeomorphic to the mapping cylinder of some map from a closed orientable surface to $X$. [This assumption on a neighbourhood is in fact not needed if one deduces the result from Alexander duality in §3.3.]

I found myself rather confused by this question. Sure, when you took the stuff in the square brackets into account, it started to make sense, what we were being asked to prove was simply a weaker version of the statement that closed non-orientable surfaces--like for instance, the Klein bottle--cannot be embedded as subspaces in three-dimensional space. Fair enough.

But this whole notion of a "neighbourhood homeomorphic to the mapping cylinder of some map from a closed orientable surface to $X$" just confused the living daylights out of me. Specifically, I could think of no surface $Y$, be it orientable, nonorientable, open, or closed, of any sort that could be embedded in $\mathbb{R}^3$ in such a way as to have a neighbourhood homeomorphic to the mapping cylinder of a closed orientable surface to $Y$.

In fact, the whole thing seemed a bit bizarre, since it appeared to me that a mapping cylinder from one surface to another ought to be homeomorphic to a volume, and thus trivially it ought to be something that you couldn't embed in a surface.

Nevertheless, looking around online, I found that this isn't some newfangled invention of Hatcher's purely for this question, and that there exists a notion of a "mapping cylinder neighbourhood" in topology.

Problem is, searching around, I cannot find any decent examples of a mapping cylinder neighbourhood that I can really visualize.

Thus, my question is, can somebody please give me an example of a surface $Y$, be it orientable, nonorientable, open, or closed, of any sort that can be embedded in $\mathbb{R}^3$ in such a way as to have a neighbourhood homeomorphic to the mapping cylinder of a closed orientable surface to $Y$.

I look forward to your answers.

答:
Well, I think that I have sorted it out, and that, at the end of the day, I merely had misunderstood what Hatcher was asking for. When he says "embedded in $\mathbb{R}^3$ in such a way as to have a neighbourhood homeomorphic to the mapping cylinder of a closed orientable surface to $X$", he doesn't mean that there exists a neighbourhood contained in $X$ that is homeomorphic to the mapping cylinder of a closed orientable surface to $X$, rather, he means that there exists a neighbourhood in $\mathbb{R}^3$ "surrounding" $X$ that is homeomorphic to the mapping cylinder of a closed orientable surface to $X$.

Then all of a sudden the question makes sense, at the very least.


问:
If $X_n \to X$ in probability and $E[X_n] < \infty$, is $E[X] < \infty$?
I am trying to prove that

If $X_n \sim N(\mu_n, \sigma_n^2)$ converges to $X$ in probability, then $X$ is also normal, and the convergence is indeed in $L^2$.

by following the argument in https://stats.stackexchange.com/questions/24362/convergence-in-probability-and-l-2-for-normal-random-variables and without using the fact that $X_n \to X$ in distribution and tightness of $X_n$.

I believe it boils down to say $E[X] := \mu < \infty$. If that's the case, I could consider the characteristic function of $X_n - \mu_n\sim N(0,\sigma_n^2)$, and show the limit $X-\mu$ has normal distribution $N(0,\sigma^2)$. Could anyone give me some hint on this?

答: Using characteristic functions we see that $e^{i\mu_n t} e^{-t^{2}\sigma_n^{2}/2} \to Ee^{itX}$ for all $t$. Take absolute values to see that $\sigma_n $ converges. Then note that convergence of $e^{i\mu_n t}$ for every $t$ implies convergence of $(\mu_n)$. Hence $Ee^{itX} =e^{i\mu t} e^{-t^{2}\sigma^{2}/2}$, so $X$ is normal.


问:
Integrals related to the modified Bessel functions of the second kind
The problem arises from calculating the spatial dependence of the electric fields of a disk charge (charge uniformly distributed on a disk in the x-y plane). I can find the field in the x-y plane is scaled by:
$$ F_1(b,r)=\frac{1}{2\pi} \int_0^b dx \int_0^{2\pi} d\theta x K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right) $$
with $b$ the radius of the disk and $r$ the distance from the center of the disk.The modifield Bessel function $K_0(x)$ comes from the Fourier transform of the point-charge's field in the z-direction. The non-trivial solusion of $F_1$ is:
$$ F_2(b,r)=1-b K_1(b) I_0(r) $$
for $r \leq b$ and
$$ F_2(b,r)=b K_0(r) I_1(b) $$
for $ r > b$. But I cannot figure out how to achieve $F_2$ from $F_1$.

Here $K_0(x)$, $K_1(x)$, $I_0(x)$, and $I_1(x)$ are the modified Bessel function of the second and first kinds, respectively.

Later I found that the problem can be reduced to finding the solution of
$$ F_3(b,r)=\int_0^{2\pi} d\theta K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right) $$
The non-trivial solution of $F_3$ is
$$ F_4(b,r) = 2\pi \left[ \theta(r-b) K_0(r) I_0(b) + \theta(b-r) K_0(b) I_0(r) \right] $$
with $ \theta(x)$ the Heaviside step function.

One can numerically test the identity of $F_1$ and $F_2$, and also of $F_3$ and $F_4$. I tried to find the answer from handbooks of integrals and special functions but was not successful.

Clearly, $F_1$ and $F_3$ are well-defined problems of mathematics, independent of the physical problems they are extracted from. So I raise the question here and hope someone could help me find the techniques necessary to do the integrals.

答:
This result can be derived from the Graff addition theorem for the modified Bessel function (eq. (8) in Watson, XI, 11.3) which reads, for $K_0$
\begin{equation}
 K_0\left(\sqrt{ Z^2+z^2-2Zz\cos\phi} \right)=\sum_{m=-\infty}^\infty K_m(Z)I_m(z)\cos m\phi
\end{equation}
which is valid for $\left|z\right|<\left|Z\right|$.

For $0\le r\le b$, we first decompose the integral $F_1(b,r)$ into
\begin{align}
 F_1(b,r)&=\frac{1}{2\pi} \int_0^b  \int_0^{2\pi}  K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right)\,xdxd\theta \\
        &=\frac{1}{2\pi}  \int_0^{2\pi} d\theta\left[\int_0^r+\int_r^b\right]K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right)\,xdx
\end{align}
For the first $x-$integral, we choose $z=x,Z=r$ in the Graff representation and $z=r,Z=x$ for the second. Only the terms with $m=0$ survive after the angular integration, then
\begin{equation}
 F_1(b,r)=K_0(r)\int_0^rI_0(x)\,xdx+I_0(r)\int_r^bK_0(x)\,xdx
\end{equation}
These integrals are readily done (DLMF):
\begin{equation}
 F_1(b,r)=rK_0(r)I_1(r)-bI_0(r)K_1(b)+rI_0(r)K_1(r)
\end{equation}
From the Wronskian of the modified Bessel functions,
\begin{equation}
 I_{\nu}\left(z\right)K_{\nu+1}\left(z\right)+I_{\nu+1}\left(z\right)K_{\nu}\left(z\right)=1/z
\end{equation}
We deduce
\begin{equation}
 F_1(b,r)=1-bI_0(r)K_1(b)=F_2(b,r)
\end{equation}
The same decomposition gives also the identity between $F_3$ and $F4$.


问:
Meaning of $\cap_{i=1}^{\infty}$
I have seen cases where it refers to infinity, for example, the infinite de morgan's rule that obsoletes mathematical induction.

I have also seen cases where it refers to each member of natural number such as infinite intersection of sets with finite elements.

So my question is, if I see $\cap_{i=1}^{\infty}$, under what circumstances should I interpret it as

$P(\infty)$; or

$\forall n\in N$

I have not studied much set theory and I just want to know the bare minimum to thrive in real analysis. Thanks in advance.

答:
Let $\{A_i\}_{i\in\mathcal I}$ a collection of sets in $X$ where $\mathcal I$ is an indice set (not necessarily countable). Then, $$\bigcap_{i\in \mathcal I}A_i=\{x\in X\mid \forall i\in \mathcal I, x\in A_i\},$$
and
$$\bigcup_{i\in\mathcal I}A_i=\{x\in X\mid \exists i\in \mathcal I:x\in A_i\}.$$

For example, $$\bigcap_{n\in\mathbb N}\left(-\frac{1}{n},\frac{1}{n}\right)=\{0\},$$
and $$\bigcup_{n\in\mathbb N}\left[-1+\frac{1}{n},1-\frac{1}{n}\right]=(-1,1).$$

Title: Meaning of $\cap_{i=1}^{\infty}$


问:
What is the domain of the Hermite operator?
Good evening, the question is the following: what is the "natural" subspace of $L^2(\mathbb{R}^d)$ that provides the domain of the operator $H=-\Delta+|x|^2$? I mean, $H$ is an unbounded operator on $L^2$ and it is surely defined on the dense subset of $L^2$ given by the Schwartz functions, $\mathcal{S}$. However, it may be difficult to prove that $H$ is self-adjoint on $\mathcal{S}$ (as in the definition of self-adjointness for unbounded operators). I guessed that the domain of $H$ should sound like this: $D(H)$ is the set of those functions $f\in L^2(\mathbb{R}^d)$ s.t.

the partial derivatives of $f$ exist a.e. and they are $L^2$ functions;

the same for the second derivatives;

some further conditions on the decay of $f$, $\nabla f$ and $\Delta f$ in order to integrate by parts and "cancel the boundary term".

This would prove easily at least that $H$ is symmetric. The self-adjointness may still be problematic.

Thank you in advance.

答:
The Hermite operator is essentially self-adjoint on $D(H)=C_c^\infty(\mathbb{R}^d)$: If $\psi\in D(H^\ast)$, then
$$
\langle \phi, H^\ast \psi\rangle=\langle H\phi,\psi\rangle=-\int (\Delta \phi)\psi+\int|x|^2\phi\psi
$$
for $\phi\in C_c^\infty(\mathbb{R}^d)$.

Thus $\Delta \psi\in L^2_{\mathrm{loc}}$ (distributional Laplacian) and $H^\ast\psi=-\Delta \psi+|x|^2\psi$. By elliptic regularity, $\psi\in W^{2,2}_{\mathrm{loc}}$. It is not hard to see that in fact $D(H^\ast)=\{\psi\in L^2\cap W^{2,2}_{\mathrm{loc}}\mid -\Delta \psi+|x|^2\psi\in L^2\}$.

For essential self-adjointness it is sufficient to prove that $\ker(H^\ast+1)=\{0\}$. For that purpose let $\psi\in \ker(H^\ast+1)$ and $\phi\in C_c^\infty(\mathbb{R}^d)$ with $0\leq \phi\leq 1$, $\phi=1$ on $B_1(0)$ and $\operatorname{supp}\phi\subset B_2(0)$. Set $\phi_R(x)=\phi(x/R)$.

As $\psi\in W^{2,2}_{\mathrm{loc}}$ we have $\psi\phi_R\in W^{2,2}$. In particular,
$$
\langle H^\ast(\psi\phi_R),\psi\phi_R\rangle=\int|\nabla(\phi_R\psi)|^2+|x|^2(\phi_R\psi)^2\geq 0.
$$
A little computation with the Leibniz rule shows
$$
\langle (H^\ast+1)(\psi\phi_R),\psi\phi_R\rangle=\mathrm{Re}\langle \phi_R(H^\ast+1)\psi,\phi_R\psi\rangle+\int |\nabla \phi_R|^2\psi^2=\int|\nabla \phi_R|^2\psi^2.
$$
Thus
$$
\int_{B_R(0)}\psi^2\leq \langle (H^\ast+1)(\phi_R \psi),\phi_R\psi\rangle=\int|\nabla \phi_R|^2\psi^2\leq \|\nabla \phi_R\|_\infty^2\int \psi^2.
$$
Letting $R\to \infty$, we obtain $\psi=0$.

Note that this argument works not only for $-\Delta+|x|^2$, but more generally for $-\Delta+V$ with lower bounded $V\in L^2_{\mathrm{loc}}$.


问:
Linear independence is hereditary
If we have a set of functions $f_i\quad, i\in\mathbb{N}$, then is it true that if $f_i$ is linearly independent on an interval $J$ then, it is also linearly independent on $I$ where  $I\subseteq J$? Is the reverse statement also true?

I think the answer is yes. But I saw somewhere that the result is not true. Any counterexamples? Thanks beforehand.

答:
The first assertion is not true. Let $I\subset J, I\not =J$. Take any functions $g_i\colon J\setminus I\to \Bbb R$ which are linearly indepedent and define $ f_i\colon J\to\Bbb R$ by $f_i(x):=g_i(x)$ for $x\in J\setminus
I$ and by $f_i(x):=0$ for $x\in I$. Then the $ f_i$ are linearly independent as functions defined on $J$ but
dependent as functions defined on $I$.

The reverse statement however is true (and used above). Let the functions $ h_i\colon J\to\Bbb R$ be given, which are linearly independent when restricted to $I$. Let moreover $\sum_i \lambda_ih_i(x)=0$ for all $x\in J$. Then this is also true for all $x\in I$. But then, by assumption, all $\lambda_i=0$ implying that the $ h_i$ are linearly independent as functions defined on $J$.

Title: Linear independence is hereditary


问:
Null-Space Projection with a Long Sparse Matrix
I want to solve the classical projection problem: given $x_0:n\times 1$ and a sparse matrix $C: m \times n$, solve for:
$$
x = \operatorname{argmin}|x-x_0|^2,\ s.t. \\
Cx = 0
$$
The three usual ways to do this require either:

Sparse QR for $C^T$, or

Sparse LDLT for $C\cdot C^T$, or,

Solving the KKT equation with Lagrange multipliers.

The problem that I am facing is that $m\gg n$, and that both $n$ and $m$ can be pretty large, where $C$ is still unique in that it has a big (right) null space. Therefore, (2) fails since $C\cdot C^T$ is not invertible (and huge besides), and (3) fails because $C$ has non-independent rows, requiring (1), which is alas very expensive for a matrix this size. Is there a gradient descent way to do this? or some direct method with alternative decompositions?

答:
Expanding on my comment: Let $$\widehat{x} = \text{argmin}\|x-x_0\|_2^2 \quad \text{s.t.} \quad Cx = 0, \quad (1)$$ and for $\rho > 0$, let $$\widehat{x}_{\rho} = \text{argmin}\|x-x_0\|_2^2 + \rho\|Cx\|_2^2. \quad (2)$$

To solve $(2)$, we can use gradient descent. Initialize $x^{(0)}_{\rho} = $ some guess, and iterate $$x^{(k+1)}_{\rho} = x^{(k)}_{\rho} - \gamma\left(2(x^{(k)}_{\rho}-x_0)+2\rho C^TCx^{(k)}_{\rho}\right).$$ Note that each iteration requires multiplying a vector by $C$, multiplying the result by $C^T$, and then a few vector operations. If $C$ is sparse, each iteration should be fairly quick. Assuming you choose the stepsize $\gamma > 0$ well, these iterations $x^{(k)}_{\rho}$ should converge to $\widehat{x}_{\rho}$ reasonably fast.

I believe it can be shown that $\displaystyle\lim_{\rho \to \infty}\widehat{x}_{\rho} = \widehat{x}$, i.e. the solution to problem $(2)$ converges to the solution to problem $(1)$ as $\rho \to \infty$. So for large enough $\rho$, solving $(2)$ should give a good approximation to the solution to $(1)$.

One method to solve $(1)$ is to pick a sequence of values $\rho_1 < \rho_2 < \ldots$, and for each $\rho_i$, use gradient descent to find $\widehat{x}_{\rho_i}$ where $x^{(0)}_{\rho_i} = \widehat{x}_{\rho_{i-1}}$ is used as the initial guess (i.e. use the solution for the previous value of $\rho$ to initialize gradient descent for the current value of $\rho$.)


问:
limit of $\ln(\frac{x^2+y^2}{x^2-y^2})$ as $x $and $y$ tends $0$
I tried to find this limit using polar representation $x=r \cos\theta $ and $y=r \sin \theta$
i got something like $\ln (\frac{1}{\cos2 \theta})$ which makes me feel im in wrong way !

is my approach correct ?? and why alpha wolfram said its not exists if so

答:
Before calculating you should get some intuition about how the function behaves in the neighbourhood of 0 and whether the limit will depend on the trajectory you choose to converge to zero.
What will be the limit of $(x_n, y_n)$ where $x_n = 0$ and $y_n = \frac{1}{n}$?
What is the limit if $x_n = \frac{1}{n}$, but $y_n = \frac{1}{2n}$?
What happens if we try to do $x_n = y_n = \frac{1}{n}$?
This should give you a hint of what's going on


问:
Show that $M+(r)$ is an ideal of $R$
This is the question from abstract algebra.

Suppose $R$ is a commutative ring with 1, and $M$ is a maximal ideal of $R$.
Take $\bar{r} \in R/M$, $r$ is not contained in $M$.

Show that $M+(r)=M+rR$ is an ideal of $R$.

Showing the additive subgroup is easy.

And, take any $a \in R$, then $a(M+rR)=aM+arR \subset M+rR$.(*)

In (*), $aM \subset M$ is obvious. But how can I complete that...?

答:
To finish the proof, we want to show $a(M + (r)) \subseteq M + (r)$.

Let $m + xr \in M + (r)$. Then $a(m+xr) = am + axr$. But $am \in M$ since $M$ is an ideal, and $axr \in (r)$. So $am + axr \in M + (r)$, as needed.

As is pointed out in the comments, we never used maximality of $M$, principality of $(r)$, or $r \not \in M$. The same proof shows that for any ideals $I$ and $J$ of $R$, the set $I + J$ is again an ideal. In fact, it is the smallest ideal containing both $I$ and $J$.

I hope this helps ^_^

Title: Show that $M+(r)$ is an ideal of $R$


答:
If the limit exists, then you have that :

$\underset{x,y\rightarrow 0}{\text{lim}} \ln \frac{x^2 + y^2}{x^2 - y^2} = \underset{x\rightarrow 0}{\text{lim}} \underset{y\rightarrow 0}{\text{lim}} \ln \frac{x^2 + y^2}{x^2 - y^2} = \underset{y\rightarrow 0}{\text{lim}} \underset{x\rightarrow 0}{\text{lim}} \ln \frac{x^2 + y^2}{x^2 - y^2}$.

But $\underset{x\rightarrow 0}{\text{lim}} \frac{x^2 + y^2}{x^2 - y^2} = -1$, and $\ln -1$ is not defined, so the limit does not exist.

Title: limit of $\ln(\frac{x^2+y^2}{x^2-y^2})$ as $x $and $y$ tends $0$

Body:
I tried to find this limit using polar representation $x=r \cos\theta $ and $y=r \sin \theta$
i got something like $\ln (\frac{1}{\cos2 \theta})$ which makes me feel im in wrong way !

is my approach correct ?? and why alpha wolfram said its not exists if so


答:
Here is an alternative argument. Note that for every $a\in [0,1)$,
$$\frac{(1-a^{n+1})e^a}{n+1}=e^a\int_a^1x^ndx\le\int_a^1x^ne^xdx\le\int_0^1x^ne^xdx\le e \int_0^1x^ndx=\frac{e}{n+1}.$$
Multiplying the inequality above by $2n+1$ and letting $n\to\infty$, it follows that:
$$2e^a\le\liminf_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le\limsup_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le 2e.$$
Since $a\in [0,1)$ is arbitrary, we can conclude that
$$\lim_{n\to\infty}(2n+1)\int_0^1x^ne^xdx=2e.$$</p>

Title: Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$

Body:
Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?

I've calculated that the recurrence relation for this integral is:

$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$

So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.

Can this be useful here?

I would appreciate all your help.


问:
How to obtain arc length of this curve??
In order to get the arc length we must have $r$ as a function of $\theta$.
But how can I find arc length of curve which is drawn between $r$ and $\theta$ if $r$ is an implicit function of $\theta$ like this:

\begin{align}
\theta = \frac{a\cdot \left( b - \sqrt{r^2 - rf+p^2} \right) }{r}
\end{align}

Where $f$ and $p$ are constants.

答: Since $ds^2=dr^2+r^2d\theta^2$, when $\theta$ is given as a function of $r$ the arc length is$$\int ds=\int\sqrt{1+r^2(d\theta/dr)^2}\;dr.$$In this case,$$\begin{align}r\frac{d\theta}{dr}&=-\frac{a}{r}\left(b+\frac{fr/2-p^2}{\sqrt{r^2-fr+p^2}}\right)\\\implies\int ds&=\int\sqrt{1+\frac{a^2}{r^2}\left(b^2+\frac{(fr/2-p^2)^2}{r^2-fr+p^2}+\frac{b(fr-2p^2)}{\sqrt{r^2-fr+p^2}}\right)}\;dr.\end{align}$$However, I suspect you may need to evaluate such an arc length numerically.


问:
Given a linear operator $T:\mathbb R^3 \to \mathbb R^3$ with $(x_1,x_2,x_3) \mapsto (3x_1,x_1-x_2,2x_1+x_2+x_3)$. Show that $(T^2-I)(T-3I)=0$
Given a linear operator $T:\mathbb R^3 \to \mathbb R^3$ with $(x_1,x_2,x_3) \mapsto (3x_1,x_1-x_2,2x_1+x_2+x_3)$.

Show that $$(T^2-I)(T-3I)=0$$

The linear operator $T$ is a function such that $$
\begin{bmatrix}
x_1  \\
x_2\\
x_3 \\
\end{bmatrix}
\quad
 \mapsto P\begin{bmatrix}
x_1  \\
x_2\\
x_3 \\
\end{bmatrix}
\quad$$
Where $$P=
\begin{bmatrix}
3& 0 & 0 \\
1 & -1 & 0\\
2 & 1 & 1\\
\end{bmatrix}
\quad
$$

Moreover $$T^2(x_1,x_2,x_3)=T(3x_1,x_1-x_2,2x_1+x_2+x_3)=(9x_1,2x_1+x_2,9x_1+x_3)$$
$$I(x_1,x_2,x_3)=(x_1,x_2,x_3)$$
$$T^2(x_1,x_2,x_3)-I(x_1,x_2,x_3)=(8x_1,2x_1,9x_1)$$
$$T(x_1,x_2,x_3)-3I(x_1,x_2,x_3)=(0,x_1-4x_2,2x_1+x_2-2x_3)$$

But I don't understand what does $(T^2-I)(T-3I)$ means and that's why I can't show the desired result,besides is the matrix $P$ helpful here?

答: The given $T$ has a triangular matrix. In this case the diagonal of $T$ shows the eigenvalues. It follows that $T$ can be diagonalized with the numbers $3$, $-1$, $1$ in the diagonal. Since the polynomial $p(z):=(z^2-1)(z-3)$ vanishes at these three places it follows that $p(T)=0$.


问:
Question on binomial theorem: Find the term independent of $x$ in the expansion of $(2+x)\left(2x+\frac{1}{x}\right)^5$

Find the term independent of $x$ in the expansion of $$(2+x)\left(2x+\frac{1}{x}\right)^5$$

I am only able to do this with one binomial, but when there are two I do not totally understand the process to do this.

答:
You can make life a bit easier in first writing

$$(2+x)\left(2x+\frac 1x\right)^5 = \frac 1{x^5}\color{blue}{(2+x)(2x^2+1)^5}$$

Now, you see you need only calculate the coefficient of $x^5$ in $(2+x)(2x^2+1)^5$. This is easy since $(2x^2+1)^5$ contains only even powers of $x$ but you are looking for $x^5$. Hence using binomial expansion you get

\begin{eqnarray*}[x^5](2+x)(2x^2+1)^5
& = & [x^5](2+x)\binom 52(2x^2)^2 \\
& = & \binom 52 \cdot 4 = 40
\end{eqnarray*}


问:
what does "in wide sense" mean?
I came across the statement "the sequence increases(in wide sense)".

So my doubt is what does author mean by wide sense?I came across this in number theory book

答: The sequence increases in wide sense means that it is not strictly increasing ( $a_n>a_{n-1})$ but it is non-decreasing $a_n\geq a_{n-1}$</p>

Title: what does "in wide sense" mean?


问:
How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$
How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$

Starting from the first expression, I couldn't figure out how to reach the conclusion. I replaced 1 by $\cos^2(x) + \sin^2(x) $ hoping to factor it but to no avail.

答:
Write $u = \cos x$, $v=\sin x$. From $u^2 + v^2=1$ we get  $u$, $v$ $\le 1$  so $u^3\le u^2$, $v^3 \le v^2$, and adding up we get $u^3 + v^3\le 1$. We have equality if and only if we have equality in the previous equalities, therefore $(u,v)=(1,0)$ or $(0,1)$.

It is interesting to sketch the curves $u^2+v^2=1$ and $u^3+v^3=1$. From the above, the function $u^3+v^3$ has maximum value $1$ on $u^2+v^2=1$, so the two curves are tangent at $(1,0)$ and $(0,1)$. In other words, the multiplicities of the intersection at $(1,0)$, $(0,1)$ are $2$. Bézout's theorem tells us there should be two more intersection points. They are complex,  $(-1+\frac{i}{\sqrt{2}},-1-\frac{i}{\sqrt{2}})$ and its conjugate $(-1-\frac{i}{\sqrt{2}},-1+\frac{i}{\sqrt{2}})$


问:
How does the manifold topology affect the properties of the metric tensor?
Consider a smooth manifold. Its atlas uniquely specifies the manifold topology. Now suppose we want to define a (pseudo-)Riemannian metric tensor on such manifold. Are there any constraints on the properties of the metric tensor, coming from the underlying manifold topology?

As an example, consider a smooth manifold with manifold topology $S^2\times R^2$. Am I allowed to specify ANY (pseudo-)Riemannian metric on it?

答:
There is a huge interaction between the Riemannian metric of a manifold and its topology. This interaction appears through its curvature.

A first example is the well-known Gauss-Bonnet formula. Given a Riemannian surface $(\Sigma,g)$, denote by $K$ its Gaussian curvature. Then:

$$\int_\Sigma K = 2\pi\chi(\Sigma)$$
Thus, the metric tensor is forced to have a global behaviour controlled by the topology of the surface. As an application, there does not exist any metric with negative curvature on the sphere or on the torus.

Another well-known interaction is the following. If $(M,g)$ is a Riemannian manifold with non-positive sectionnal curvature, then its universal cover is diffeomorphic to $\mathbb{R}^n$. This implies that a $n$ dimensional sphere does not have any metric of non-positive sectionnal curvature.

A last interaction is more subtle: the bahaviour of the curvature affects the aglebraic properties of the fundamental group of a compact manifold. On some positive assumptions on the curvature, the fundamental group has to have polynomial growth, while with some negative assumptions, it has at least exponential growth.

Note that every manifold can be endowed a Riemannian metric. The case of pseudo-Riemannian metrics is not the same because one cannot glue local pseudo-Riemannian metrics with partition of unity and expect the result to be a pseudo-Riemannian metric. For example, this theorem shows that the Lorentzian case is restricted to a few examples: every noncompact connected smooth manifold admits a Lorentz metric, and a compact connected smooth manifold admits a Lorentz metric if and only if its Euler characteristic is zero.

In your example, the manifold $S^2\times\mathbb{R}^2$ is simply connected and is not diffeomorphic to $\mathbb{R}^4$. Hence, it does not admit a metric with negative curvature.


问:
How does the second derivative imply concavity?
consider this equation, $$y=x(400-x)$$the second derivative of this equation is $$y''=-2$$ As far as I know, a negative sign in the second derivative indicates the curve will concave down. As it is a constant I think it says that the curve concaves down all the time. Which means the tangent line will always lie above the function's graph.

QUESTION 1: How can second derivative define concavity? Surely it's not like mathematicians came up with this out of nowhere.

QUESTION 2: What is the significance of knowing where the graph concaves up or down? There may be many but I just to know the absolute basic ones.

EDIT: I messed up badly with concavity last time. But wikipedia made my views clear about concavity. So I edited my questions.

答:
To add to @B.Goddard's answer, we can think write the first derivative as:

$$ y' = \tan \theta(x)$$

Differentiating this,

$$ y'' = \sec^2 \theta \frac{d \theta}{dx}$$

Notice that since $ \sec^2 \theta \geq 0$, the sign of $y'' $ is controlled by $ \frac{ d \theta}{dx}$.

If $ \frac{d \theta}{dx} \geq 0 $ , it means that as you move to the right, the tangent line is rotating counter clockwise.

If $ \frac{ d \theta}{dx} \leq 0 $ , it means that as you move to the right, the tangent line is rotating clockwise.

We can think of the tangent line to a curve, doing a translational + rotational motion as we move along the curve to further understand this.


问:
does a graph with two vertices of the same degree automatically have a non-trivial automorphism
Hi I am trying to proof that all graphs with 4 vertices have a non-trivial automorphism group. I constructed a two case proof (connected and unconnected graphs) and said that if at least two vertices have the same degree the graph has a non-trivial automorphism since those two vertices can swap.

I realise this is wrong as this does not necessarily maintain the adjacency of the graph but don't know how else I would complete this proof?

any advice would be appreciated

答:
Welcome to MSE ^_^

Hint: Consider any of the following asymmetric graphs

Notice each of these has two vertices of the same degree, yet none of them have any nontrivial automorphisms (do you see why?). The idea is that it's not enough to swap two vertices $x$ and $y$. To be an automorphism we must the swap things next to $x$ with things next to $y$. But there might be some obstruction at this next stage. Or, similarly, at some stage many vertices away from $x$ and $y$. There is a more detailed explanation for the top left graph above below the fold, but you should try to work it out yourself too.

 In the top left graph, for instance, we can't swap the two vertices of degree $1$ because we would then be forced to swap their (unique!) adjacent vertices too. Yet the two adjacent vertices have different degrees, and so cannot be swapped. Every such graph has a similar argument.

I hope this helps ^_^


问:
The Riemann hypothesis as a statement about natural numbers
In this answer it is claimed that the Riemann hypothesis can be expressed as a statement about natural numbers.  How would that look like?</p>

答:
He's probably referring to either Robin's theorem (most likely), which establishes that Riemann hypothesis is equivalent to the fact that for all $n\ge5041$, $$\sigma(n)<e^\gamma n\log\log n$$ where $\sigma(n)=\sum_{d\mid n} d$ and $\gamma$ is Euler-Mascheroni constant, or to this result by Lagarias which establishes that Riemann hypothesis is equivalent to the statement that, for all $n\ge2$, $$\sigma(n)<H_n+e^{H_n}\log H_n$$

where $H_n=\sum_{k=1}^n\frac1k$.

Added: For instance, refer to the Theorem 3.2 in Lagarias' paper (an improvement on prop. 1 section 4 of Robin's paper):

The negation of Riemann hypothesis implies that there are some $C>0$ and some $0<\beta<\frac12$ such that $\frac{\sigma(n)}{n\log\log n}\ge e^\gamma+\frac C{(\log n)^\beta}$ holds frequently as $n\to\infty$.

As a consequence, Riemann hypothesis is false if and only if there is some integer $m\ge5041$ (in fact, infinitely many) such that $\sigma(m)>e^\gamma m\log\log m$. A purported such number $m$ may be (theoretically) checked to satisfy this inequality in finitely many passages of arithmetic involving rational numbers.


问:
distributive law of vector subspace
Let $V$,$W$,$F$ be subspaces of vector space $L$.
I want to find examples of subspaces $V$,$W$,$F$ , which do not satisfy

（$V$＋$W$）∩ $F$ ＝（ $V$∩$F$）＋（$W$∩$F$） （distributive law）

Are there good examples?

Thank you for your help.

答: Sure. Let $k$ be the field that you are working with and take $L=k^2$, $V=k\times\{0\}$, $W=\{0\}\times k$, and $F=\{(x,x)\mid x\in k\}$. Then $F\cap V=F\cap W=0_L$.

Title: distributive law of vector subspace


问:
Name for this Product-Rule Generalisation
I've stumbled on a nice identity which seems like it would be well-known, and I would like to know more about it, but I can only express it in non-standard notation so I can't look it up. Let $M$ be an $a$ by $b$ matrix and $N$ be an $a$ by $c$ matrix so that $M^T$ and $N$ are compatible. Then we have a generalisation of the dot product $\cdot$ given by by $M\cdot N=M^TN$. Suppose that $f,g:\mathbb{R}^m\to\mathbb{R}^n$ are differentiable. Then for $x\in\mathbb{R}^m$, $d{(f\cdot g)}_x=df_x\cdot g(x)+dg_x\cdot f(x)$. To verify the identity consider the case $m=1$ and see that this is just a restatement of the partial derivative product rule gives this identity, then generalise the proof of the dot product rule for $n=1$. Note that, unlike in the special cases, $\cdot$ is not commutative, so the only cosmetic change that can be made is swapping the summands.

答:
The particular phrasing you've brought up in terms of transposes does not have a name, as far as I know. This is because it's a special case of the
Leibniz Rule for products of matrices.

That is, for matrices $\mathbf{A}(x)$ and $\mathbf{B}(x)$, we have
$d(\mathbf{A} \mathbf{B}) = \mathbf{A} \frac{d\mathbf{B}}{dx} + \frac{d\mathbf{A}}{dx} \mathbf{B}$.

This is because the standard proof of the Leibniz Rule relies only on the commutativity of $+$, not of multiplication! This means the identity is still true in the case of matrices (as you've noticed).

You might be interested more generally in matrix calculus, which explores other identities of this form. This particular identity can be found in the "matrix by scalar identities" section


问:
Linear Approximation of a product of functions
If

$f(x) = g(x)h(x)$

does the linear approximation of $f(x)$ equals the linear approximation of $g(x)$ times the linear approximation of $h(x)$?

if it is true:

what is the proof?

is it true for quadratic approximations as well?

答:
Let $F:x\mapsto ax+b$ be the linear approximation of $f$ around $0$, and $G:x\mapsto cx+d$ be the one of $g$.
Then we have $FG:x\mapsto (ax+b)(cx+d) = ac x^2 +(bc+ad)x + bd$ which is clearly not linear.

But actually the linear approximation of $fg$ is $x\mapsto (bc+ad)x + bd$ since you just need to get rid of the $x^2$ term.

For quadratic approximations, it's the same : you can multiply the two approximations, but then you need to get rid of all the terms with $x^k$ where $k>4$.

So to answer your question, except for some particular cases, it is false.


问:
Finding the maximum amount of positive numbers in a sequence
Let $(a_1,...,a_{100})$ be real numbers satisfying :
$\forall i\in\mathbb{Z}/100\mathbb{Z}$, $a_i > a_{i+1}+a_{i+2}$. What is the maximum amount of strictly positive real numbers contained in the sequence $(a_i)_{1\le i\le100}$.

The obvious issue is that if you wish for all the reals to be positive, then they must be decreasing and so when you loop around you'll meet a problem. I feel like it should be possible to alternate between positive and negative numbers, but I'm not sure how to show this.

答:
This is essentially taken from Combinatorics on AoPS.

There can be at most $50$ positive numbers because otherwise a there would be two positive numbers $a_{i+1}, a_{i+2}$ immediately following a non-positive number $a_i$.

If there are $50$ positive numbers then the $a_i$ must be alternatingly positive and non-positive. Without loss of generality assume that $a_i > 0$ for odd $i$. Then $a_i > a_{i+2}$ for even $i$ and
$$
 a_{100} > a_2 > a_4 > \ldots > a_{98} > a_{100}
$$
gives a contradiction.

So there can be at most $49$ strictly positive numbers. An example with $49$ positive numbers is
$$
(-101, 1, -103, 1, -105, 1, \ldots, -197, 1, -199, -99)
$$
i.e.
$$
 a_i = \begin{cases}
 -100-i & \text{ if $i$ is odd,} \\
 +1 & \text{ if $i < 100 $ is even,} \\
 -99 & \text{ if $i = 100$.}
\end{cases}
$$

In the same way one can show that if $n$ is even and $(a_1, a_2, \ldots, a_n)$ are real numbers satisfying $a_i > a_{i+1}+a_{i+2}$ for all $i$ (with wrapping around indices) then there are at most $n/2-1$ strictly positive numbers among these.

An example with $n/2-1$ strictly positive numbers is
$$
 (-n-1, 1, -n-3, 1, -n-5, 1, \ldots, -2n+3, 1, -2n+1, -n+1)
$$
i.e.
i.e.
$$
 a_i = \begin{cases}
 -n-i & \text{ if $i$ is odd,} \\
 +1 & \text{ if $i < n $ is even,} \\
 -n+1 & \text{ if $i = n$.}
\end{cases}
$$


问:
Convergence of p-series over subset of the naturals
This was a question from an exercise in a first semester Algebra class. The question is as follows:

Let $S \subseteq \mathbb{N}$ and suppose that $|\{n \leq x: n \in S\}| = O ((\log(x))^d)$ for some $d \geq 1$. Show that $$ \sum_{s \in S} \frac{1}{s^{\epsilon}}$$ converges for each $\epsilon > 0$.

My attempt:

I first tried to solve a simpler version of this problem in which we consider a subset $|\{n \leq x: n \in S\}| = O ((\log(x)))$. In this case, the series must converge because we can find a subsequence of partial sums of a geometric series with r < 1 that bounds the partial sums from above. I believe this is possible since the terms eventually become less dense in the reals than the terms of the aforementioned geometric series.

The problem arrives really when we have to prove that this result holds for all $d \geq 1$. I really don't know how to analyze this polylogarithmic growth of the cardinality of the subset.

We also proved earlier that $$ \prod _{n=1}^{\infty }(1+a_{n})$$
converges if and only if the series
$$ \sum _{n=1}^{\infty }a_{n}$$
converges for a sequence of positive reals $\{ a_n\}$. However, I don't see how converting this sum to a product really helps.

答:
Here’s another proof.

For each $n \geq 0$, let $k_n$ be the number of elements $x \in S \cap [e^n,e^{n+1})$, so that $k_n \leq C(n+1)^d$ for some positive constants $C,d$.

Then, for each $n \geq 1$, $r > 0$,

$$S_n := \sum_{s \in S, s< e^n}{s^{-r}} \leq\sum_{p=0}^{n-1}{\sum_{s \in S, e^p \leq s < e^{p+1}}{s^{-r}}},$$ so

$$S_n \leq\sum_{p=0}^{n-1}{\sum_{s \in S,e^p \leq s < e^{p+1}}{e^{-rp}}},$$ which simplifies to

$$S_n \leq\sum_{p=0}^{n-1}{k_pe^{-rp}},$$ and thus by what we assumed on $S$,

$$S_n \leq C\sum_{p=0}^{n-1}{(p+1)^de^{-rp}},$$

and the last sum is convergent, hence the conclusion.


问:
Adaptedness of random step process
Suppose you are given a random stem process of the form
$$
f(t) = \sum_{j=1}^{n-1}\eta_j \mathbb{1}_{[t_j, t_{j+1})}(t),
$$
where $0 = t_0 < t_1 < \cdots < t_n$ and $\eta_1, \eta_2, \cdots \eta_{j-1}$ are square integrable random variables. The book I am reading claims that if $\eta_j$ are $\mathcal{F}_{t_j}$-measurable, then $f(t)$ is adapted to the filtration $\mathcal{F_t}$. Why is this true? Note that filtrations here are defined with respect to the Wiener process, i.e.,
$$
 \mathcal{F}_s = \sigma\{W(r): 0 \le r \le s\}.
$$

答:
Suppose that $t_{j_0}\leqslant t<t_{j_0+1}$ for some $j_0\in\{0,\dots,n-1\}$. Then
$f(t)=\eta_{j_0}$ if $j_0\geqslant 1$ and $0$ if $j_0=0$; in any case, $f(t)$ is $\mathcal F_{t_{j_0}}$-measurable and the inclusion $\mathcal F_{t_{j_0}}\subset \mathcal F_t$ guarantees that $f(t)$ is $\mathcal F_t$-measurable.

Title: Adaptedness of random step process


问:
Complex components of the first-kind elliptic integral
I have written a monograph called Elliptic Integrals and Functions, which contains a sizeable collection of formulas and results relating to the named mathematical topics. At the time of this question, however, the monograph was missing a result on the following problem.

Assume $0<m<1$. For an arbitrary complex number $\gamma$, decompose $F(\gamma,m)$ into real and imaginary parts.

(The argument convention, as with all my other questions and answers on the MSE, follows Mathematica/mpmath convention; $m$ is the parameter.)

Because of the addition formulas – section 2.5 in the monograph – as soon as I can write
$$F(\gamma,m)=F(\alpha,m)+F(i\beta,m)$$
where $\alpha$ and $\beta$ are real, I solve the above problem as well as the corresponding problems for the other two kinds of elliptic integrals because the extra parts in the addition formulas can themselves be decomposed (though writing out the full formula would be tedious).

Without loss of generality we can take $|\operatorname{Re}(\gamma)|<\frac\pi2$ (if it is exactly $\frac\pi2$, the formula below doesn't work and my answer here applies). Abramowitz and Stegun 17.4.11 gives the following solution: suppose $\gamma=a+bi$, then letting
$$B=(1-m)-\cot^2a-m\sinh^2b\csc^2a\qquad C=(m-1)\cot^2a$$
we have
$$\alpha=\tan^{-1}\sqrt{\frac2{-B+\sqrt{B^2-4C}}}\qquad\beta=\sinh^{-1}\sqrt{\frac{\tan^2a\cot^2\alpha-1}m}$$
and the signs of $\alpha$ and $\beta$ (which sign for the outer square roots) match those of $a$ and $b$ respectively.

Like the rest of A&S, no proof is given inside. The references, though numbered, are actually in alphabetical order and give no further leads. How can the above solution be shown and can it be simplified in any way?

答:
Let $c=\sin^2a$, $d=\sinh^2b$, $A=\sin^2\alpha$ and $B=\sinh^2\beta$. The expression for $\sin\gamma$ in the addition formula yields the following two equations in real numbers, corresponding to real and imaginary parts:
$$c(1+d)(1+mAB)^2=A(1+B)(1+mB)$$
$$(1-c)d(1+mAB)^2=B(1-A)(1-mA)$$
A resultant computation (eliminating $A$ and $B$ individually) shows that $A$ and $B$ are the smaller roots of
$$(d+1)mA^2-(1+(c+d)m)A+c$$
$$(c-1)mB^2-(1-(c+d)m)B+d$$
Still the solution requires solving a quadratic, but I find this more symmetric than the A&S solution. This has been put into the EIF update.


问:
Why injective matrix is bounded below?
Problem: Let $A$ be a matrix with $m$ rows and $n$ columns. The matrix defines a linear operator, also denoted by $A$,
\begin{align}
A: \mathbb R^n & \longrightarrow \mathbb R^m\\
x & \longmapsto Ax
\end{align}
Assume that $n$ columns of matrix $A$ are independent (this is equivalent to say that operator $A$ is injective, see the 3rd property here), then operator $A$ is bounded below, i.e. there exists a positive constant $c>0$ such that
$$||Ax||\geq c||x||, \hspace{0.5cm} \forall x\in \mathbb R^n.$$
where $||\cdot||$ is the Euclidean norm.

Motivation: Above problem is a special case (finite dimensional case) of Theorem 2.5 in this book, which stated that: "A continuous operator between Banach spaces is bounded below iff it is injective and has closed range." Regarding our problem, we know that:

The operator $A$ is continuous,

$\mathbb R^n$ and $\mathbb R^m$ are Banach spaces,

The operator $A$ trivially has closed range since we are working with finite-dimensional spaces.

Thus, the statement in our problem should hold true. In the proof of Theorem 2.5, they used the Open Mapping Theorem, which is a technical theorem! I am wondering are there any simpler proof with intuition and without Functional analysis for our special case. Thank you in advance!

答:
Yes. If there exists no $c$ such that $\|Ax\|\geq c\|x\|$ for all $x$, then there exists a sequence $x_1,x_2,\dotsc$ such that $\|x_i\|=1$ for all $i$, and $\|Ax_n\|\to 0$ as $n\to \infty$. Since $\mathbb{R}^n$ is finite-dimensional, its unit ball is compact, hence some subsequence $(x_{k_i})_i$ tends to a limit $x$ in the unit ball (by compactness). But $A$ is bounded and linear, hence continuous, so we get

$$\|Ax\|=\lim_i \|Ax_{k_i}\| =0$$

i.e. $Ax=0$ for an $x$ with $\|x\|=1$, which contradicts the fact that the columns of $A$ are independent.

Title: Why injective matrix is bounded below?


问:
What are the eigenvalues of this Hamiltonian?
\begin{equation}
H=E_{0} \sum_{m=-\infty}^{\infty}(|m\rangle\langle m+1|+\mathrm{h.c.})
\end{equation}
I tried to solve this equation as follows(h.c. means dagger or hermitian conjugate and m kets form orthonormal basis):

\begin{equation}
H|\psi\rangle=\lambda|\psi\rangle
\end{equation}
\begin{equation}
|\psi\rangle=\sum_{m^{\prime}=-\infty}^{\infty} \psi_{m^{\prime}}|{m^{\prime}}\rangle
\end{equation}
Which gives me:
\begin{equation}
\sum_{m^{\prime}} E_{0} \psi_{m^{\prime}}\left(\left|m^{\prime}-1\right\rangle+\left|m^{\prime}+1\right\rangle\right)=\sum_{m^{\prime}} \lambda \psi_{m^{\prime}}\left|m^{\prime}\right\rangle
\end{equation}
From which (please verify) I get:
\begin{equation}
\psi_{m^{\prime}+2}=\frac{\lambda}{E_{0}} \psi_{m^{\prime}+1}-\psi_{m^{\prime}}
\end{equation}
I could not advance here onwards.

This recursion can grow from both sides which is fine, I assumed 0th and 1st psi to be some constants but it's without pattern, I am not getting how to proceed further. Help!

答:
Regarding your recursion relation: it is linear and thus may be solved by an ansatz of the form $\psi_m = a^{rm}$; $a, r$ unknown. We know that plane waves are going to feature$^\dagger$ in the solution so take $\psi_m = e^{ikm}$, Substituting in

$$e^{ikm}e^{2ik}=e^{ikm}e^{ik} \frac{\lambda}{E_0}-e^{ikm}$$

Dividing by $e^{ikm}e^{ik}$ and rearranging

$$\frac{\lambda}{E_0}= e^{ik}+e^{-ik}=2\cos(k)$$

A very general tip for such Hamiltonians: if they have terms such as $\left|n\right>\left<n+a\right|$, then they are likely diagonal in the Fourier transformed basis  $\left|k \right>=\sum_m e^{ikm} \left| m \right> $. So usually we would start with the plane wave eigenstates, rather than the general ones you used$^\ddagger$:

$$\left|\psi_k \right>=\sum_m e^{ikm} \left| m \right> $$

Applying $H$ to this, we find (after exploiting orthogonality, and $E_0=1$ for simplicity)

$$H \left|\psi_k \right>= \sum_m e^{ikm}\left( \left|m+1\right>+\left|m-1\right> \right)$$

Then relabeling summation indices

$$H \left|\psi_k \right>= \sum_m e^{ikm} \left( e^{ik}+e^{-ik}\right) \left|m\right> $$

$$H \left|\psi_k \right>=2\cos(k) \sum_m e^{ikm}\left|m\right> = 2\cos(k) \left|\psi_k\right>$$

From which we can read off the eigenvalues.

$\dagger$ We should really label these as $\psi_{k,m}$ or $\psi_m(k)$ or something similar, the eigenstates are labelled by $k$. Similarly $\lambda=\lambda(k)$.

$\ddagger$ To justify this, we could rewrite $H$ in the Fourier transformed basis, given by the inverse Fourier transforms
$$\left|m \right>=\sum_k e^{-ikm} \left| k \right> $$
And, taking the adjoint
$$\left<m \right|=\sum_k e^{ikm} \left< k \right| $$
Substitute these into $H$, you'll have three summations, two will be removed using orthogonality, and you should end up with exactly the same thing as with the plane wave ansatz.


问:
Given a vector space $V$, show that the following statements are equivalent.
Given a subset $W$ of $V$ then I want show that,

$\forall v \in V, w \in W$ $\exists \lambda \in \mathbb{R}$ such that $w + \alpha v \in W$ for any $0 < \alpha < \lambda$

iff

$\forall v \in V, w \in W$ $\exists \alpha \in \mathbb{R}$ such that $\lambda w + (1-\lambda) \alpha v \in W$ for any $0 < \lambda < 1 $

答:
first suppose $\forall w \in W ,v\in V\ \exists \lambda $ such that $w+\alpha v \in W\ \forall  0<\alpha<\lambda$ consider $\lambda <1$ so :

$w+\alpha (1-\lambda )(\frac{-w}{\alpha}+v)\in W $ so $w-w+\lambda w+\alpha(1-\lambda )v \in V$ $\to$ $\lambda w +\alpha (1-\lambda)v \in V$

on the other hand :

if $\forall w \in W , v\in V\ \exists \alpha $ such that $\lambda w+(1-\lambda )\alpha v \in V$ $\forall 0<\lambda<1$

consider $\lambda =\epsilon $ so $\epsilon w +(1- \epsilon )\alpha (\frac{w}{\alpha}+\frac{v}{1-\epsilon})\in W$ so $w+\alpha v \in W$


问:
To what ring is $\mathbb{Z}[X,Y,Z]/(X-Y, X^3-Z)$ isomorphic?
The problem:

Let $(\mathbb{Z}[x,y,z],+,\cdot)$ be the ring of polynomials with coefficients in $\mathbb{Z}$ in the variables $x$, $y$ and $z$ and the obvious operations $+$ and $\cdot$. Let $(x-y, x^3-z)$ be the ideal generated by $x-y$ and $x^3-z$. Find the ring to which $\mathbb{Z}[x,y,z]/(x-y,x^3-z)$ is isomorphic.

A similar question has been asked before: Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?. Using the same approach, we set $x=y$ and $x^3=z$ and substitute for $y$ and $z$. Then we find $\mathbb{Z}[x,x,x^3]$ which is just $\mathbb{Z}[x]$. I'm trying to find out why this approach would work, i.e., formalise it in a proper proof. My idea is to use the first isomorphism theorem, but I get stuck near the end. I would very much appreciate if someone could help fill in this gap, check my proof for mistakes or provide additional (intuitive) information as to why the above method works.

My proof:

Consider the map $\varphi : \mathbb{Z}[x,y,z] \to \mathbb{Z}[x] : p(x,y,z) \mapsto p(x,x,x^3)$.

This is a homomorphism, since it obeys the relations $\varphi(p) + \varphi(q) = \varphi(p+q)$ and $\varphi(p) \cdot \varphi(q) = \varphi(p \cdot q)$ for all $p,q \in \mathbb{Z}[x,y,z]$ (verifying these is quite trivial).

Now take any $q \in \mathbb{Z}[x]$ and define a $p \in \mathbb{Z}[x,y,z]$ such that $p(x,y,z) = q(x)$. Since $\varphi(p(x,y,z)) = p(x,x,x^3) = q(x)$, $\varphi$ is surjective and it follows that $\mathbb{Z}[x] \subseteq \varphi(\mathbb{Z}[x,y,z])$. It is obvious that $\varphi(\mathbb{Z}[x,y,z]) \subseteq \mathbb{Z}[x]$ and, therefore, we must have $\varphi(\mathbb{Z}[x,y,z]) = \mathbb{Z}[x]$.

Lastly, we must show that $\ker(\varphi) = (x-y, x^3-z)$. For brevity's sake I will denote $I := (x-y,x^3-z).$

Let $p \in I$. Then there must exist $q_1, q_2 \in \mathbb{Z}[x,y,z]$ such that $p = q_1(x-y) + q_2(x^3-z)$. It follows that:
\begin{align*}\varphi(p(x,y,z)) &= \varphi(q_1(x,y,z)(x-y) + q_2(x,y,z)(x^3-z))\\ &= q_1(x,x,x^3)(x-x) + q_2(x,x,x^3)(x^3-x^3) = 0.\end{align*}
Thus $p \in \ker(\varphi)$ and $I \subseteq \ker(\varphi)$, since $p$ was chosen arbitrarily in $I$.

Now, take a $q \in \ker(\varphi)$. Then $\varphi(q(x,y,z)) = q(x,x,x^3) = 0$. Somehow we must get that $q \in I$ so that we can conclude $\ker(\varphi) \subseteq I$ and thus $I = \ker(\varphi)$, but I seem to get stuck on every path. Proof by contradiction or by contraposition also seems fruitless. What am I missing?

With all of the above, we may apply the first isomorphism theorem to conclude that:
$$\mathbb{Z}[x,y,z]/I = \mathbb{Z}[x,y,z]/\ker(\varphi) \cong \varphi(\mathbb{Z}[x,y,z]) = \mathbb{Z}[x]$$

答:
Write $\mathbb{Z}[X]= R$ ring ( comm, $1$). If $a_1$, $\ldots$, $a_n$ are in $R$ then
the evaluation map at $(a_1, \ldots, a_n)$,
$$f(X_1, \ldots, X_n) \mapsto f(a_1, \ldots, a_n)$$
from  $R[X_1, \ldots, X_n]$  to $R$ is surjective. Clearly the kernel contains the ideal $I= \langle X_1-a_1, \ldots, X_n - a_n\rangle$. In fact it equals $I$. Indeed, for every $f\in R[X_1, \ldots, X_n]$ we have $f(X_1, \ldots, X_n) \equiv f(a_1, \ldots, a_n)  \mod I$, since $X_i \equiv a_i \mod I$ (@Thorgott:'s observation)- so  $f(a_1, \ldots, a_n) = 0$ iff $f(X_1, \ldots, X_n) \in \langle X_1-a_1, \ldots, X_n - a_n\rangle$, a multidimensional Bézout's theorem.  We conclude that the evaluation map induces the isomorphism
$$R[X_1, \ldots, X_n]/\langle X_1-a_1, \ldots, X_n - a_n\rangle \stackrel{\sim}{\rightarrow} R$$


问:
Calculating an integral using the Laplace transform
I'm looking for a "polite" way to calculate this integral using Laplace transform:

$$ \int_0^{+\infty} \frac{e^{-ax} - e^{-bx} }{x} dx. $$

Now the impolite way is to invoke a famous theorem from basic differential equations:
$$ \mathcal{L}( f(x)/x) = \int_s^\infty F(t) dt,$$

Where $ F$ is the LT of $f$. Setting $f(x)/x= 1/x$ in this theorem (and recalling the definition of LT) gives
$$ \int_0^\infty \frac{e^{-ax}}{x} =  \infty - \ln(a).$$
Similarly
$$ \int_0^\infty \frac{e^{-bx}}{x} = \infty - \ln(b).$$

Putting the results together and cancelling infinity gives the final result $\ln \frac{b}{a}$ which I know for a fact is the true value of the integral. But I also know too well that cancelling infinity is not allowed. So what is the trick?

答:
$$g(x) = \frac{e^{-ax} - e^{-bx} }{x} = \frac{(1-e^{-bx}) - (1-e^{-ax}) }{x}= \frac{f_b(x)}{x} - \frac{f_a(x)}{x}$$ where

$f_c(x) = (1- e^{-cx})$ for $c \gt 0$. Then denoting the step function by $u$ we have $\mathcal L(f_c \cdot u)(s) = \frac{c}{s(s+c)}$ and

$$ \begin{aligned}\mathcal{L}( g)(s) &= \int_s^\infty F(t) dt = \int_s^\infty \left(\frac{b}{t(t+b)}-\frac{a}{t(t+a)}\right) \ dt\\
&= \int_s^\infty \left(\frac{1}{t+a}-\frac{1}{t+b}\right) \ dt\\
&= \ln\left(\frac{s+b}{s+a}\right)
\end{aligned}$$

The desired equality is finally obtained by plugging in $s=0$ in previous equation.


问:
How to calculate $x_{n}= \sqrt{n+ \sqrt{n- 1+ \sqrt{n- 2+ \sqrt{\cdots + \sqrt{2+ \sqrt{1}}}}}}$ without using surd signs

How to calculate:
$$x_{n}= \sqrt{n+ \sqrt{n- 1+ \sqrt{n- 2+ \sqrt{\cdots + \sqrt{2+ \sqrt{1}}}}}}$$
without using surd signs.

My attempt:
I saw that
$$x^{2}_{n}= n+ x_{n- 1}$$
Therefore
$$x= 1+ \frac{n}{x}$$
So
$$x= 1+ \frac{n}{1+ \frac{n- 1}{1+ \frac{n- 2}{1+ \frac{n- 3}{\ddots }}}}$$
But it seems like no much useful for small $n.$ I need to your helps to get it. Thanks a real lot !

答:
First you can't apply the limit to

$$x_n^2=n+x_{n-1} \ \ (1)$$

to find the limit of $(x_n)$ because you have the divergent term $n$.

Note that

$$ \forall n,\ x_n >0 $$

So from $(1)$ you get

$$ \forall n , \ x_n > \sqrt{n}$$

So the sequence diverges.

Secondly closed form doesn't match well with the combination of plus sign and square-root so it is very unlikely to find an easier closed form.


问:
Fourier series at discontinuity: can the left and right limits be retrieved from the Fourier series?
I understand that the Fourier Series for a discontinuous function, one with a jump discontinuity such as the sawtooth wave, converges to the average of the left and right limits at the discontinuity.

Is there some way by which the left and right limits themselves can be read out from the Fourier series representation?

That is, a way to write these quantities
$$\lim_{x\rightarrow x_0^+} \sum_{n=-\infty}^{\infty} c_n \exp(i n x)$$
$$\lim_{x\rightarrow x_0^-} \sum_{n=-\infty}^{\infty} c_n \exp(i n x)$$
directly as series over $c_n$, without the limits.

答:
When you have a single jump discontinuity then you can work numerically. Say, you have a periodic function $f$ which is smooth, apart from a jump at the points $2k\pi$. Let
$$J(t):={\pi-t\over2}\quad(0<t<2\pi),\qquad J(t+2\pi)\equiv J(t)\ ,$$
be your favorite jump function. Then
$$J(t)\rightsquigarrow\sum_{k=1}^\infty{\sin(kt)\over k}\ .$$
The auxiliary function
$$g(t):=f(t)-c J(t)$$
is then everywhere smooth for a suitable choice of $c$, hence has Fourier coefficients going to $0$ much more rapidly than ${1\over k}$. It follows that the Fourier sine coefficients of $f$ decay as fast as $b_k\approx{c\over k}$ $(k\gg1)$. This allows to obtain an estimate for $c$, maybe doing some minimization work involving many coefficients. The jump height then is $\pi c$.


问:
Prove divisibility using linear congruences
I need to prove that:

$$10|(53^{53} - 33^{33})$$

I can and should only use linear congruences ($a \equiv b \mod n$) - how can I do this?

答:
Hint:

$53 \equiv 3 (\mod 10) \implies53^2 \equiv -1(\mod 10) \implies 53^{32} \equiv 1(\mod 10)$

$33 \equiv 3 (\mod 10) \implies33^2 \equiv-1(\mod 10) \implies 33^{32} \equiv 1\mod 10)$

$53^{33} \equiv \dots (\mod 10)$ and $33^{33} \equiv \dots(\mod 10)$ (Fill in the blanks)


问:
Prove the only normal unipotent subgroup $G$ is trivial
Let $G$ be an algebraic subgroup of $GL_n(k)$ which acts irreducibly on $V = k^n$, via the natural morphism $G \hookrightarrow GL_n(k)$, where $k= \overline{k}$.

How to prove that  the only normal unipotent subgroup $G$ is trivial one?

Let $N$ be the normal unipotent subgroup of $G$, we can also assume that $N$ is subgroup of $U_n(k)$, but the irreducibility condition confused me, I didn't konw how to use it to solve this question.

Any help will be appreicate.

答:
If $N$ is a unipotent group, there exists $x\in V$ such that for every $x\in V$, $n(x)=x$ (see the reference). Let $g\in G$ and $n\in N$, $n(g(x))=g(n'(x))$, $n'\in N$, we deduce that $n(g(x))=g(x)$ and $n$ is the identity since $G$ acts irreducibly on $V$.

https://mathoverflow.net/questions/54161/reference-request-representations-of-unipotent-groups-have-a-fixed-point


问:
Parameterizing Geodesics on the Sphere in Polar Coordinates
I seem to have this seemingly trivial Problem, but can't figure it out.

Situation:
I have my Unit Sphere, $S^2$ defined as a Riemannian Manifold. Parameterizing Geodesics (Great circles) on this sphere is absolutely no Problem, IF I embed it in $\mathbb{R}^3$. BUT, I don't want to do this.

What I really want to do, is use riemannian polar coordinates. Now genereally, this isn't a problem, as I could just take any point on the geodesic and call it $(0,0)$ (with $(r,\varphi)$ being my riemannian polar coordinates). And all the lines going through the origin would be my great circles.

However, The origin of the exponential map I'm using does not lie on the geodesic of interest. I do however know 2 points (say $P_1, P_2$) and their polar coordinates.

So basically, I have $\text{exp}_p^{-1} : S^2\to \mathbb{R}^2$ and my two points in $\mathbb{R}^2$ defined by $(r_1,\varphi_1)$ and $(r_2,\varphi_2)$.

I'm looking for some parameterisation to connect these two points via a great circle. And I just can't think of anything sensible.

Any help would be highly appreciated.

Edit: To clarify, I have a working method using an $\mathbb{R}^3$ embedding, but it's very bulky and "unpretty".

Further, I'm using this geodesic to "bound" one of my integrals and shifting the Origin is therefore not an option

答:
Why don't you just use spherical trigonometry? You have a "base triangle" with vertices $O$, $A_1$ and $A_2$, sides $O A_i$ of length $r_i$ and an angle $\alpha:=|\phi_2-\phi_1|$ between them. Now consider an arbitrary point $P$ on the third side $A_1 A_2$. The "line" $OP\ $ has length $r$ and encloses angles $\alpha_i$ with the sides  $OA_i$. The formulas of spherical trigonometry will give you an equation connecting $r$ and the $\alpha_i$.

Let $s_i$ be the lengths of the sides $A_i P$. Then
$$\cos s_i=\cos r_i \cos r +\sin r_i \sin r \cos\alpha_i\qquad(i=1,2)$$
and
$$\cos(s_1+s_2)=\cos r_1\cos r_2+\sin r_1\sin r_2 \cos(\alpha_1+\alpha_2)\ .$$
Eliminate $s_1$ and $s_2$ from these three equations to get the desired result. Hint: Square the identity $\sin s_1 \sin s_2=\cos s_1 \cos s_2-\cos(s_1+s_2)$ and replace $\sin^2 s_i$ by $1-\cos^2 s_i$.


答:
$(53)^{4k}\equiv 1 \mod 10 \Rightarrow (53)^{53}=(53)^{52}\times53 \equiv 1\times53\equiv 3 \mod 10$

similarly $33^{4k} \equiv 1 \mod 10 \Rightarrow 33^{33} \equiv 3 \mod 10$

So the result follows.

Reason: $(10k+3)^4 \equiv 3^4\equiv  1 \mod 10\Rightarrow (10k+3)^{4l}\equiv 1 \mod 10$

Here $k,l\in \mathbb{N}$

Title: Prove divisibility using linear congruences

Body:
I need to prove that:

$$10|(53^{53} - 33^{33})$$

I can and should only use linear congruences ($a \equiv b \mod n$) - how can I do this?


问:
Finding the standard deviation from data grouped by intervals

Starting Monthly Salary
Number of Graduates

1,001 - 1,400
1

1,401 - 1,800
11

1,801 - 2,200
14

2,201 - 2,600
38

2,601 - 3,000
36

Total
100

What I have done, is averaged out each value in Monthly Salary column and continuing from there by calculating the deviation at the end

Calculating all 100 of these values is tedious:

(1001 + 1400) / 2 = 1200.50

(1401 + 1800) / 2 = 1600.50

(1401 + 1800) / 2 = 1600.50

(1401 + 1800) / 2 = 1600.50

Continued for all 100 values to the calculate the deviation for each...

Then calculating the standard deviation with σ=√((Σ(x-µ)^2)/2). This formula once everything has been entered in will look really nasty. Is there an easier (or cleaner) solution to finding the (approximate) standard deviation of the monthly starting salaries above?

Body:
$$\sqrt{ \Sigma_i x_i^2 p(x_i)-(\Sigma_i x_ip(x_i))^2}$$

$x_i  $
$p(x_i)$
$xp(x_i)$
$x^2p(x_i) $

1,200.5
0.01
12.01
14,412.00

1,600.5
0.11
176.06
281,776.03

2,000.5
0.14
280.07
560,280.04

2,400.5
0.38
912.19
2,189,712.10

2,800.5
0.36
1,008.18
2,823,408.09

Total
1.00
2,388.50
5,869,588.25

$$\sigma=\sqrt{5,689,588.25-2388.5^2}=405.78$$


问:
How to find the cost of a tv set when it is given as an algebraic expression?
The problem is as follows:

Gabriel wants to purchase a new TV for her mother as a Christmas gift. The cost is $2P$ thousand dollars. In order to fulfill this wish he begins saving his earnings as a pharmacist clerk for six months. Assume that the money he earns is the same each month and $P$ to be the sum of the coefficients of the rational integer algebraic expression and not null of the form

$$M(x,y)=\left(\frac{n+4}{3}\right)x^{5-n}y^{4-m}-2x^{n-3}y^{m-4}+(n-3)y^{n+2}$$

If Gabriel decides to buy the gift after $4$ months instead. By how much additional dollars should Gabriel have to save each month with respect of his original plan so that he can buy the tv?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{300 dollars}\\
2.&\textrm{400 dollars}\\
3.&\textrm{500 dollars}\\
4.&\textrm{200 dollars}\\
\end{array}$

How exactly should be the relationship between the given condition of the integer rational integer algebraic expression should be used?. Can someone help me here?.

So far what I've been able to spot was this:

$P=\frac{n+4}{3}-2+(n-3)$

$P=\frac{4n-11}{3}$

But that's the part where I'm stuck, where exactly should I go from here?. I don't know what else to do with the information related with those $4$ months.

答:
Say he saves $s$ for $6$ months. Then $$6s = 2000P \implies s=\frac{1000}{3} P$$

Say he saves $s’$ for $4$ months. Then $$4s’=2000P \implies s’=500P$$

Then $$s’-s = \frac{500}{3} P = 500 \times \frac{4n-11}{9} $$ Now, this must be equal to exactly one of $200,300,400,500$. Set $s’-s = 100a$ with $a\in\{2,3,4,5\}$ so that $$5(4n-11) = 9a \implies n= \frac{9a+55}{20} $$ $n$ must be an integer, so $20$ must divide $9a+55$. That means $9a+55$ must end in a $0$, and that only happens for $a=5$ and so $$s’-s =100\cdot 5 =500$$


问:
How can I show this field extension equality?
How can I show this field extension equality   $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?</p>

答:
The easier direction is $\mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$ Can you tell us why this is true?

Now we have $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$

Since $\sqrt{2}$ is a root of the degree 2 polynomial $X^2-2$ which is irreducible over $\mathbb{Q}$ (because if it were reducible, it would have a linear factor corresponding to a rational root which the rational root theorem rules outs) we have $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]=2.$

Now $\sqrt{3}$ is a root of the degree 2 polynomial $X^2-3.$ Let's check this is irreducible over $\mathbb{Q}(\sqrt{2}).$ It is reducible if and only if it has a root in $\mathbb{Q}(\sqrt{2}).$ Again by the rational root theorem, it has no roots in $\mathbb{Q}.$ So if there is a root $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ it must have $b\neq 0$ and satisfy $(a+b\sqrt{2})^2-3 = a^2+2ab\sqrt{2}+2b-3=0.$ If $a=0$ then $2b^2=3$ so $b=\pm \sqrt{3/2}$ which is rational if and only if $2b = \pm \sqrt{6}$ is rational. It is not hard to show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square which $6$ is not, so $a+b\sqrt{2}$ is a root of $X^2-3$ then $a$ and $b$ must both be non-zero. But then we can rearrange to get $\sqrt{2} = \dfrac{3-2b+a^2}{2ab}$ and the right hand side is rational while the left is not. So $X^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$ so $[\mathbb{Q}(\sqrt{2})(\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2.$

Thus $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2\cdot 2=4.$$

Now we find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}.$ The natural guess is $$(X-\sqrt{2}-\sqrt{3})(X-\sqrt{2}+\sqrt{3})(X+\sqrt{2}-\sqrt{3})(X+\sqrt{2}+\sqrt{3})=X^4+2X^2+25.$$ Let's check it is irreducible to confirm. Gauss's lemma allows us to simply check it is irreducible over $\mathbb{Z}.$ The rational root theorem gives $\pm 1, \pm 5, \pm 25$ as the only possibilities for integer roots, none of which work. So any factorization of $X^2+2X^2+25$ has no linear factors, and thus is of the form $$X^4+2X^2+25 = (X^2+aX+b)(X^2+cX+d)$$ where $a,b,c,d\in \mathbb{Z}.$ Equating coefficients of powers of $X$ gives $a=-c, ac+b+d=2, ad+bc=0$ and $bd=25.$ If $a=-c=0$ then $b+d=2$ and $bd=25$ which implies $4 = (b+d)^2 = b^2+2bd + d^2 = b^2+d^2+50 \implies b^2+d^2<0.$ Thus $a=-c\neq 0$ and from $ad+bc=0$ we get $b=d$ and from $bd=25$ we conclude $b=d=\pm 5.$ This makes $ac+b+d=2$ becomes $-a^2 = 2\pm 10$ which is impossible.

Thus $$ [\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4.$$

Putting everything together, $$ 4 = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})][\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]\cdot 4
$$

so $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]=1$$ which implies $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3}).$


问:
Exponential map is diffeomorphism on a neighborhood in TM
Let $M$ be a smooth Riemannian manifold and let $TM$ be the tangent bundle. For each $p \in M$, it is known that there is a neighbourhood $\mathcal{U}$ of $(p,0)$ in $TM$ such that the exponential map $exp: \mathcal{U} \rightarrow M$ is well-defined and smooth on $\mathcal{U}$.

My question: is it possible to find a neighbourhood $\mathcal{V}$ of $(p,0)$ in $TM$ such that $exp: \mathcal{V} \rightarrow M$ is diffeomorphism?

It is also known that if we only stick to $T_{p}(M)$, which is a subspace of $TM$, then we can find a neighbourhood $\mathcal{W}$ of $0$ in $T_{p}(M)$ such that $exp: \mathcal{W} \rightarrow M$ is differomorphism. The difference here in my question is that the neighborhood $\mathcal{V}$  contains vectors in different tangent spaces, not only in $T_{p}(M)$.

Thank.

答: It is not always possible to find such a neighborhood. Suppose that $M=S^1$, it is not possible to find an a neighborhood diffeomorphic to $S^1$, since such a neighborhood is a connected interval, which is open or has a boundary.


问:
If $\mu$ is a measure on a metric vector space and the translations of $\mu$ by $x$ and $y$ coincide, then $x=y$
Let $\mu$ be a finite signed measure on the Borel $\sigma$-algebra $\mathcal B(E)$ on a metric vector space $E$ and $x,y\in E$ with $$\mu\ast\delta_x=\mu\ast\delta_y\tag1,$$ where "$\ast$" denotes the convolution operator.

Under which assumptions on $\mu$ can we conclude $x=y$?

I've read that the claim is true when $\mu$ is a probability measure, but I don't know why. We clearly need to assume that $\mu$ is nontrivial, since otherwise the claim is obviously wrong.

It might be useful to note that if $$\tau_z:E\to E\;,\;\;\;w\mapsto w+z$$ denotes the right shift by $z\in E$, then $\mu\ast\delta_z=\tau_z(\mu)$. So, $(1)$ means $$\mu(B-x)=\mu(B-y)\;\;\;\text{for all }B\in\mathcal B(E).$$

Most crucially, if $\mu$ is "translation invariant", which precisely means that $\tau_z(\mu)=\mu$ for all $z\in E$, then the desired implication obviously cannot hold. And the trace of the Lebesgue measure on $\mathcal B(\mathbb R)$ on $[0,1]$ is clearly a translation invariant probability measure ... What am I missing?

The claim can be found in the proof of Corollary 2.3.5 of Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions:

答:
Let $E$ a bounded measurable set with $\mu(E)>0$. You have that $\mu*\delta_{x-y} =\mu$.
Let $z=x-y$. If $n$ is big enough, $E\cap (E+nz)=\emptyset$, and by hypothesis $\mu(E+nz)=\mu(E)$. So the $E+knz$, $k\in \mathbf{N}$, are disjoint sets with the same measure, and that's impossible for a finite measure.


问:
Open and closed functions, is there an equivalent, easier way to check?
I know that an open function is a function that maps every open set of the domain to some open set. So it is easy to prove that a function is not open just by giving a counter example. However, I have some difficulties while checking whether a function is open or not, since there are lots of different types of open sets, especially in a metric space. In particular, even in $\mathbb{R}$ there are lots of open sets, not only intervals I mean. Here is my question, Is it possible to change the condition to have some equivalent one. For instance, would it be sufficient to check only open balls, or closed balls for closed mappings. My question is general, so I ask the same for closed maps. Thanks!

答:
For open sets it is enough to check it for open balls (or any family of sets such that every open set is a union of some members of the family).

For closed sets such  a criterion usually fails. For example the map $(x,y) \to x$ from $\mathbb R^{2}$ to $\mathbb R$  maps closed balls to closed sets, but it is not a closed map since the image of $\{(x,y): xy=1\}$ is not closed.


问:
If and only if condition for UFD
This is a question in abstract algebra.

Question : Let $D$ : integral domain. Show that $D$ is a UFD if and only is every nonzero element is a product of finitely many prime elements.

But I think this question is wrong. If the condition on right has 'uniquely', it would be right.

(But it is ACTUALLY the definition of UFD...)

Is this question wrong? Or something I misunderstood...?

答:
The problem, as stated, is correct.

As has been mentioned in the comments, you seem to be confusing primes with irreducibles. This is an extremely common mistake, but one which you should be cognizant of!

As for the problem itself, here's a hint:

Can you show that if an element of an integral domain admits a prime factorization, then this factorization must be unique (up to order and units, of course)?

Once you have this, since every element admits a prime factorization in our ring of interest, we see that every element admits a unique prime factorization. So we have UFD-ness.

There are lots of examples of non-UFDs where every element admits a (non-unique!) factorization into irreducibles. Every noetherian ring has this property, for instance. Factorization into primes is a stronger condition, though (as this question shows).

I hope this helps ^_^

Title: If and only if condition for UFD


Body:
$53^{53}\equiv3^{53}\equiv{(3^{4})}^{13}\cdot3\equiv3\mod 10$.

$33^{33}\equiv3^{33}\equiv{(3^{4})}^{8}\cdot3\equiv3\mod 10$.

because $a^{\varphi(n)}\equiv1\mod n$ and $\varphi(10)=4$, and thus $3^4\equiv1\mod 10$.

Therefore $53^{53}-33^{33}\equiv3-3\equiv0\mod 10$


问:
why the restriction of a normal operator is normal?
can anyone explain why it's obvious that a normal operator restricted to a invariant subspace is still normal?

答:
Other way to see it as follows:
Suppose $T$ is normal and
$$TT^*(v)=T^*T(v)$$, this is true for all $v \in V$,
Now,
Claim: $T_{|W}:W \rightarrow W$ is normal. So,
$$\langle T^*Tw,w' \rangle=\langle TT^*w,w' \rangle $$
for all $w,w' \in W$.
So,
$$T^*Tw=TT^*w$$ for all $w \in W$.
Hence $T_{|W}$ is normal.