问:
If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$
I'm trying to prove the following:

If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n<\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2<\infty$, then $\sum_{n=1}^\infty a_n^2 <\infty$.

The context here is functional analysis homework, in the subject of Hilbert spaces.

Here's what I've thought:

Let $f=(a_n)>0$. Then the problem reads: if $\int f\overline{g}<\infty$ for all $g>0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.

I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle <\infty$ for all $g>0,g\in \ell^2$ then $f\in \ell^2$.

I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P

From there, I don't know what else to do. I thought of Hölder's inequality which in this context states:
$$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$

but it's not useful here.

答:
Following the idea left by david as an answer, I'll post a detailed solution.

Let $T_n:\ell^2 \to \mathbb{C}, T_n(c)=\sum_{j=1}^n a_j c_j$. Then clearly $T_n \in (\ell^2)^*$ for every $n$.

I claim that for every $c\in \ell^2$, the limit $\lim_n T_n(c)=\sum_{j=1}^\infty a_jc_j$ exists.

Indeed, we know it does for $c\in \ell^2$ such that $c(n)\geq 0$ for all $n$. But then, for an arbitrary $c\in \ell^2$, we can decompose $c$ as:

$c=(\mbox{Re }  c)^+ - (\mbox{Re } c)^- + i\left( (\mbox{Im }c)^+ - (\mbox{Im } c)^-\right)$

which proves the claim.

As a consequence we have that $\sup_n \lVert T_n(c) \rVert <\infty$ for all $c\in \ell^2$, but then by the uniform boundedness principle, $\sup_n \lVert T_n \rVert <\infty$.

Now, $T_n(c) = \langle c, a^{(n)}\rangle_{\ell^2}$ where $a^{(n)}(j)=\begin{cases} a_j & \mbox{if }j\leq n \\ 0 & \mbox{if } j>n \end{cases}$.

By Riesz' representation theorem on Hilbert spaces, we know that $\lVert T_n \rVert = \lVert a^{(n)}\rVert_2= \left( \sum_{j=1}^n a_j^2 \right)^{\frac{1}{2}}$.

To conclude, since $a_n\geq 0$ for all $n$, we have that $\left( \sum_{n=1}^\infty a_n^2 \right)^{\frac{1}{2}} = \sup_n \lVert a^{(n)} \rVert_2 = \sup_n \lVert T_n \rVert < \infty$, and then $\sum_{n=1}^\infty a_n^2 <\infty$.


问:
How to Show that (S V R) logically follows from the following knowledge base.
Not (Not Q) AND Z

• Not W

• (Not W And Q) -> Not P

• (W AND Z) -> S

• Q -> (S AND P)

• (P AND Q) -> R

答:
The statement to be proved is $ S \vee R $

$ \neg (\neg Q) \land Z \implies Q \land Z \implies $ Q is tautology, Z is tautology

$ Q \rightarrow S \land P \implies $ S is tautology, P is tautology

Since we already know P and Q are tautologies, R is also a tautology from the last statement.

However, there is a contradiction between the third and fifth premise so this conclusion may not be useful.

ContentLicense: CC BY-SA 4.0


问:
Probability events | Show that $P(A \cap B) = 0.257379$
Considering 2 events $A$ and $B$ in sample space $S$. Suppose

\begin{align}
P(A \cup B) = 0.7949 \\
P(B^c |A) = 0.5835 \\
P(A^c |B) = 0.4074 \\
P(A^c \cap B^c) = 0.2051.
\end{align}

Show that $P(AB) = 0.257379$ (correct solution).

My wrong attempt

$1-P(A^c \cap B^c)- P(B^c |A) = 0.3784$

Please could you show how the answer $\left( P(A \cap B) = 0.257379\right)$ was found using the given probabilities

答:
We have:

$$P(B|A) = 1- P(B^c|A) = 0.4165 = \frac {P(A \cap B)}{P(A)}$$

$$P(A|B) = 1- P(A^c|B) = 0.5926 = \frac {P(A \cap B)}{P(B)}$$

$$P(A \cup B) = P(A)+P(B)-P(A\cap B) = 0.7949$$

Can you use the equations above to solve for $P(A \cap B)$? Try to express $P(A)$ and $P(B)$ in terms of $P(A \cap B)$ and substitute them into the last equation.


问:
Ambiguity in the evaluation of a real integral multiplied by a complex number
Let's consider the following integral, where x is a real value: $\int_{0}^{1}ln(i)x^{2}dx=i(\frac{\pi}{2}+2k\pi)\int_{0}^{1}x^{2}dx$. Since $ln(i)=i(\frac{\pi}{2}+2k\pi)$, k being an integer, what value should I consider?

答:
In complex analysis, the logarithm is what we call a multivalued function, and you need to do what is called a branch cut in order for it to be a well defined function. The "standard" is what we call the principal branch, often denoted $\text{Log}$, where the argument is chosen within the interval $(-\pi, \pi]$. However, other choices may be beneficial or even necessary in certain situations.

You can read more about the complex logarithm here: https://en.wikipedia.org/wiki/Complex_logarithm


答:
small hint: the last equality follows from the first, so you can ignore it.

Calculate
$$P(B\mid A) = 1 - P(B^c\mid A) = 1- 0.5835= 0.4165\\
P(A \mid B) = 1 - P(A^c\mid B) = 1 - 0.4074=0.5926$$

Now, using the basic formula for the conditional probability we get
$$\frac{P(AB)}{P(A)} = 0.4165\\
\frac{P(AB)}{P(B)} =0.5926$$
and so
$$P(A) = \frac{1}{0.4165} P(AB)\\
P(B) = \frac{1}{0.5926}P(AB)$$

Now use also the fact
$$P(A\cup B) = P(A) +P(B) - P(AB)$$
given to be $0.7949$.  From the last three equalities you should be able to get $P(AB)$.

Title: Probability events | Show that $P(A \cap B) = 0.257379$

Body:
Considering 2 events $A$ and $B$ in sample space $S$. Suppose

\begin{align}
P(A \cup B) = 0.7949 \\
P(B^c |A) = 0.5835 \\
P(A^c |B) = 0.4074 \\
P(A^c \cap B^c) = 0.2051.
\end{align}

Show that $P(AB) = 0.257379$ (correct solution).

My wrong attempt

$1-P(A^c \cap B^c)- P(B^c |A) = 0.3784$

Please could you show how the answer $\left( P(A \cap B) = 0.257379\right)$ was found using the given probabilities


问:
sequence of functions, limits and uniform convergence
Let $f_n(x)=\frac{x}{1+nx^2}.$

(a) Find $f(x)=\lim_{n\to\infty}f_n(x)$ and $g(x)=\lim_{n\to\infty}f'_n(x)$. For which x, if any, is it true that $f'(x)=g(x)$?

(b) Prove that $|f_n(x)|\leq\frac{1}{2\sqrt{n}}$ for all $x\in R$. Do the $f_n$ converge uniformly? Why or why not?

For(a), I calculate the limit of $f_n$ and $f'_n$, and I got zero for both. Does that mean for every x, $f'(x)=g(x)$?

For(b), I have a hint to find the local extrema. But I still have no idea how to do that.

答:
$\lim f_n(x)=0$ for all $x$ but $\lim f_n'(x)=0$ only for $x \neq0$. For $x=0$ the limit becomes $1$. Hence it is not true that $f'=g$.

$f_n'(x)=0$ gives $1-nx^{2}=0$ or $x=\pm \frac 1 {\sqrt n}$. The maximum value of $f_n(x)$ is attained at these points and the maximum value of $|f_n(x)|$ is $\frac  1 {2\sqrt n}$. Hence $|f_n(x)| \leq \frac  1 {2\sqrt n}$ which proves that $f_n \to 0$ uniformly.


Title: sequence of functions, limits and uniform convergence

Body:
Let $f_n(x)=\frac{x}{1+nx^2}.$

(a) Find $f(x)=\lim_{n\to\infty}f_n(x)$ and $g(x)=\lim_{n\to\infty}f'_n(x)$. For which x, if any, is it true that $f'(x)=g(x)$?

(b) Prove that $|f_n(x)|\leq\frac{1}{2\sqrt{n}}$ for all $x\in R$. Do the $f_n$ converge uniformly? Why or why not?

For(a), I calculate the limit of $f_n$ and $f'_n$, and I got zero for both. Does that mean for every x, $f'(x)=g(x)$?

For(b), I have a hint to find the local extrema. But I still have no idea how to do that.

Body: (b): $|f_n(x)|\leq\frac{1}{2\sqrt{n}} \iff 1-2\sqrt{n}|x| +n|x|^2 \ge 0 \iff (1-\sqrt{n}|x|)^2 \ge 0.$


问:
Given vectors OA and OB calculate vector AB
So the teacher said it's like this OB - OA = AB and I don't understand it. Why aren't we doing this instead: OA - OB? aren't they the same thing?</p>

答:
Ask Tom and Jerry to go from O to B. Tell them that the start point is the tail and the end point is the head of the vector or arrow.

Tom goes directly to B and Jerry detours via point A before heading to B. Whether OB or OA+AB their effective effective displacement has remained the same.


问:
Find the maximum $r$ such that $(\beta_i,\,\beta_j)<0$, $\;1\leq i<j\leq r$.

Let $\mathcal{L}$ be an $n$-dimension real Euclid space. Find the maximum $r$ such that there are $r$ vectors $\beta_1,\cdots,\,\beta_r$ in $\mathcal{L}$ satisfying
$$
(\beta_i,\,\beta_j)<0,\quad 1\leq i < j  \leq r.
$$

I have tried to consider the case that $\beta_1,\cdots,\;\beta_r$ are dependent and the case that $r=2,\,3$, however, for technical reason, I couldn't get any hint from trying these.

答:
Since the inner products are only considered pairwise let us restrict to unit vectors.

Let $u_1,\ldots,u_m$ be unit vectors such that
$$
(u_i,u_j)\leq \delta<0,\quad 1\leq i<j\leq r.
$$
Then
$$
0\leq \left|\sum_{i=1}^m u_i\right|^2=(\sum_{i=1}^m u_i,\sum_{j=1}^m u_i)=\sum_{j=1}^m |u_i|^2+2\sum_{i<j} (u_i,u_j)
$$
$$
\leq m + m(m-1) \delta
$$
giving $$m\leq 1-\frac{1}{\delta}.$$

Edit: The result I quoted is from Bollobas' little white book Combinatorics, Set Systems and Hypergraphs.

There is a counterpart result which shows if a set $S$ of $n+r$ nonzero (not necessarily unit) vectors in $n$ dimensions satisfy
$$
(x_i,x_j)\leq 0,\quad 1\leq i<j\leq n+r
$$
then $r\geq n,$ and if $r=n,$ then the space $\mathbb{R}^n$ has an orthonormal basis such that for all $1\leq i\leq n,$
$$
S=\{\lambda_1 a_1,\mu_1 a_1, \lambda_2 a_2, \mu_2 a_2,\ldots,\lambda_n a_n, \mu_n a_n\}
$$
where $\mu_i<0<\lambda_i.$ If the inner products are strictly negative, and we let $\delta\rightarrow 0,$ I think it can be argued by continuity that we end up in this case. Thus, really the smallest $\delta$ must be $-1/2n$ or something like that.

Another way of thinking of it, is if the inner product $(u_i,u_j)$ is arbitrarily near zero and negative, that will force some $(u_i,u_j')$ to tend to be positive. So a fourth moment sum might make this idea clearer, but I have no time now.


问:
Annulus Theorem
I'm trying to read Rolfsen's "Knots and Links" and I'm a little discouraged that I can't do one of the first and seemingly more important exercises.  The question is

Use the Schoenflies theorem (every topological imbedding $S^1 \rightarrow \mathbb{R}^2$ is the boundary of a 2-disk) to prove the annulus theorem: Given two disjoint imbeddings of $S^1$ in $\mathbb{R}^2$, where one is "inside" the other (I.e. one is in the bounded component of the complement of the other), then the space "between" the two curves is homemorphic to $S^1 \times [0,1]$.

There is a hint that says "connect the two curves with arcs".  Unfortunately I can't say much about "what I've tried." At this point in the book, we know only that all simple closed curves in $\mathbb{R}^2$ or $S^2$ are "equivalent" in the sense that there exists a homeomorphism of the ambient space taking one to the other.  This allows us to assume one of the curves is the standardly imbedded $S^1$. After that I'm totally unsure of what to do. I would appreciate any hints or advice.  Thank you for reading my question.

答:
I haven't thought this through carefully, but just following the hint:

choose a point $a$ on the outer cirlce and a point $b$ on the inner circle.  Join $a$ and $b$ with an arc that doesn't cross the two circles.  Now "thicken up" this arc very slightly (and thicken up each of $a$ and $b$ to a very short piece of circular arc on their respective circles) so that now we have to the two circles with a very thing rectangle sitting between them.  (I think that making this "thickened" picture rigorous is probably part of what is involved in translating the hint into a full argument.)

Now the consider the boundary of the union of the two circles and this rectangle.  If you are picturing what I want you to picture, you will see that it is a circle.  (Just to be sure: if you take an annulus and snip out a small rectangle joining the inner and outer annulus, you get a kind of "sliced" annulus which is homeomorphic to a disk with circle as boundary.  But you can verify that the boundary of the region I am describing is a circle without knowing that the region inbetween the two circles is an annulus just follow the outer circle around from the top left corner of the rectangle all the way around to the top right corner, then move along the right boundary of the rectangle to its bottom right corner, now go back around the inner circle to the bottom left corner of the rectangle, and now go back up the left edge of the rectangle to get back where you started: you've described a circle.)  So it bounds a disk.
So the region between the two circles, with a thin rectangle removed, is a disk.  Gluing back in the rectangle should make into an annulus, as required.

ContentLicense: CC BY-SA 3.0

Title: Annulus Theorem


问:
Problems in solving Number Theory problems (specifically Congruences)
I am very new to Number Theory, I understood the basic concepts and theorems but what I am struggling at is that I am not able to apply them on questions, and basically I am not able to process the problem in 'number theory' theorems rather I am thinking from an absolute manner.

I have solved around 15-20 divisibility problems of Number Theory and I still have no idea how you deal with congruences in a real problem, for example:

Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.

My approach to this problem was by simplifying the expression as much as I can but it involved congruences, which didn't even strike me. Even when I am reading congruence questions' answers, I am not able to simply understand

Now I have a number of questions:

first, does it take time to develop that 'intuition' while solving problems?

second, can elementary number theory be mastered through sheer practice?

third, is there a good resource to study and practice congruences?

答:
Of course, SunShine's answer is great but you can do this by a simple induction without the need for the binomial theorem. The following might give you some guidance on how to think about these problems (as that was part of your original question):

The case for $n = 1$ is clear.

Then $169 | 3^{3n+3} -26n-27$ for some $n$.

Since the inductive step is to show that $P(n) \implies P(n+1)$ (this is done by replacing $n+1$ with $n$), we must show that $169 | 3^{3n+6} -26n-53$.

Since $169|3^{3n+3} -26n-27$ and we must get the right exponent for the term with base $3$, we want to multiply by $3^3$. This is allowed since it is obvious that if $a|b$, then $a|bc$.

Hence, $169|(3^{3n+3} -26n-27)3^{3}=3^{3n+6}-26\cdot27n - 27^{2}$. To get $169$ to divide our desired expression, we must figure out a way to convert one to the other using congruences. It makes sense then to check $27^2 \pmod {169}$.

You may find this is conveniently congruent to $53 \pmod {169}$ either by multiplying it out, or if you really want to use the binomial theorem the fact that $27^{2} = (2\cdot 13 +1)^{2}$ (But it is easy with exponent $2$).

So we have reduced down to the statement that $169|(3^{3n+3} -26n-27)3^{3}=3^{3n+6}-26\cdot27n - 53$.

By either doing the simple computation $26\cdot 27 \pmod {169}$ or using that $n(n+1)=n^{2}+n$,

then $26\cdot 27 = 26^{2}+27 = (13\cdot2)^{2} + 27 = 13^{2}2^{2} + 27 = 169\cdot2^{2} + 27 \equiv 27 \pmod {169}$.

Altogether, this means that $169 | 3^{3n+6} -26n-53$ and so we have proved that $P(n) \implies P(n+1)$.


问:
How do I notate "the x-coördinate of $G^{-1}(z)$ for $z = G(x, y) : \mathbb{N}\times\mathbb{N} \to \mathbb{N}$
My question is pretty much as given in the title, I have a function that bijectively maps $\mathbb{N}$ to $\mathbb{N}\times\mathbb{N}$, and I now wish to define a function which I, as a part-time programmer would write as
$$K(x) = 2^{G^{-1}(z).x}\cdot 3^{G^{-1}(z).y}$$
However, this clearly cannot be mathematically correct since in mathematics maps do not return named tuples, they return ordered sets, so I could write
$$(x, y) = G^{-1}(z) $$
but also
$$(y, x) = G^{-1}(z) $$
or even
$$(\mu, \aleph) = G^{-1}(z) $$
In what mathematical way could I write that I want two raised to a power equal to the variable that would have been first in ordering when taking $G(x, y) = z$ multiplied by three raised to a power equal to the parameter that would have been second in order when taking $G(x, y) = z$?

答:
In general, when you have a function to a direct product $f : Z \to X \times Y$, it is common to denote $f_1 : Z \to X$ and $f_2 : Z \to Y$ the components, so that $f(z) = (f_1(z), f_2(z))$. So you could write $(G^{-1})_1(z)$ instead of $G^{-1}(z).x$.

Another possibility is to give a name to the projections $X \times Y \to X$ and $X \times Y \to Y$. A common choice is $\pi_1, \pi_2$ or $p_1, p_2$. And then $G^{-1}(z).x$ becomes $(\pi_1 \circ G^{-1})(z)$. Or omitting the symbol for function composition, $\pi_1 G^{-1}(z)$.


问:
Question about the Z
I am confused while self-studying statistics.

It is about the standard score. I have no problem in understanding the formula $Z = \dfrac {X - \mu}{\sigma}$ and how it is used on area finding under a normal distribution curve.

Later on, when I come to Z-score test of a significant level, the formula Z becomes $Z = \dfrac {X - \mu}{\sigma} \times \sqrt n$. This probably is due to the central limit theorem, I guess.

My first question is “why the Z is treated differently?”

I also notice that some use $Z = \dfrac {X - \mu}{\sigma} \times \sqrt n$ but some use $\dfrac {X + 0.5 - \mu}{\sigma} \times \sqrt n$ (for the one-sided test). I understand that the +0.5 is the continuous correction for the discrete case.

My second question is “shouldn’t the answer for the same question be unique?” Or “the former is an acceptable answer while the latter will provide a more accurate estimate?”

答:
Later on, when I come to Z-score test of a significant level, the
formula Z becomes $Z=\frac{X-\mu}{\sigma}\sqrt{n}$

This formula is incorrect. You find

$$Z=\frac{\overline{X}_n-\mu}{\sigma}\sqrt{n}$$

This because the sample mean $\overline{X}_n$ is approx (with CLT) Gaussian with mean=$\mu$ and variance=$\frac{\sigma^2}{n}$

For the second question yes, the result with the correction factor is more accurate when approximating a discrete distribution with a continuous one.

Note that you can have $\pm0.5$ as continuous correction factor; this depends on the situation you are dealing with

Title: Question about the Z


问:
Show that an embedding from a compact, smooth manifold to a connected smooth manifold is a diffeomorphism.
Let $M,N$ be nonempty, smooth manifold of the same dimension.  Let $N$ be compact and $M$ be connected.  Suppose that $f:N\rightarrow M$ is an embedding.  We want to show that $f$ is a diffeomorphism.

Since $f$ is an embedding, then we know that

$f(N)\subset M$ is a smooth submanifold, and

$f:N\rightarrow f(N)$ is a diffeomorphism. That is we identify the domain $N$ with its image $f(N)\subset M$ such that $N\subset M$.

We would like to show that $f$ is a diffeomorphism.  So, we need to show $f$ is an invertible smooth map.

答:
As described in you question, since $f:N \to f(N)$ is a diffeomorphism, it suffices to show that $f$ is surjective.

The surjectivity comes from the topological structure of $M$ and $N$. Since $M$ is connected, the goal is to show that $f(N)$ is both open and closed in $M$.

Since $f:N \to f(N)$ is a diffeomorphism, then $f^{-1}$ is continuous, which implies that $f(N)$ is open.

Since $f$ is continuous and $N$ is compact, we have that $f(N)$ is compact in $M$. Since $M$ is a Hausdorff space as a manifold, we know that $f(N)$ is closed in $M$.


问:
find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$
find all polynomials that satisfy this functional equation
$$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$

I write the polynomial
$$P(x)=\sum_{k=0}^n a_k x^k$$
but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be zero, except the fourth coefficient; $P(x)=a x^4$.
But when I re-substitute it into the equation I found that it's inconsistent but it make sense only if $a=0$.

Body:
If you use the brute force and write
$$P(x)=\sum_{k=0}^n a_k\, x^k$$
$$ (x-16)P(2x)-16(x-1)P(x)=0$$
and solve for one sigle $a_k$ at the time, you will have succesively
$$a_1=-\frac{15 a_0}{16}\qquad a_2=\frac{35 a_0}{128}\qquad a_3=-\frac{15 a_0}{512}\qquad a_4=\frac{a_0}{1024}$$ and, for $k>4$, $a_k=0$.

So
$$P(x)=a_0\left(1-\frac{15 x}{16}+\frac{35 x^2}{128}-\frac{15
   x^3}{512}+\frac{x^4}{1024} \right)$$ that is to say
$$P(x)=\frac{a_0}{1024}(x-16) (x-8) (x-4) (x-2)$$


问:
A question about inverse modulo a number.
If $xy \equiv 1 \pmod a$, can I write $x \equiv \frac{1}{y} \pmod a$.

I was thinking about this while solving IMO 2005 P4, so in particular can I write $$2^{p-2} + 3^{p-2} + 6^{p-2}\equiv\frac{1}{2} + \frac{1}{3} + \frac{1}{6} \pmod p?$$

答:
As long as $\gcd(a,y)=1$, yes, you could. If not, then you can't.

Find some integer solution to $my+na=1$, say by the extended Euclidean algorithm. Modulo $a$ this yields $my\equiv 1$, which means that $m$ corresponds to $\frac 1y$.

As noted in the comments above, though, it is more common in modular arithmetic to write $y^{-1}$.


问:
How do you get $(-1)^n$ when you have $e^{in\pi}$
I am having a problem understanding this.

In which cases can you turn $e^{in\pi}$ in to $(-1)^{n}$?

What happens when it instead is $e^{-in\pi}$?

答: Using euler`s identity $e^{i\pi}=-1$ and $(e^{i\pi})^{n}=(-1)^{n}$


问:
Showing that $ \sum \limits_{m=1}^{n} b_m x_{m-n}~\to~ ab$ as $n~\to~\infty$
If $x_n ~\to ~a$ as $n~ \to~ \infty$

Does:

$ \sum \limits_{m=1}^{n} b_m x_{n-m}~\to~ ab$ as $n~\to~\infty$?

$b_m ~\geq~0$ and $ b~\equiv~ \sum \limits_{m=1}^{\infty} b_m < \infty$

My attempt:

$\lim\limits_{n\to\infty}x_n~=~a$

I would like to have something like this I think

$$\inf x_n \sum_{m=1}^n b_m~\leq~\sum b_m x_{n-m}~\leq ~\sup x_n \sum_{m=1}^n b_m$$

Then taking the limits will squeeze it in to $ab$, I am having trouble getting this form, is there any easier way to do this?

i.e. $$\sum_{m=1}^n b_m\inf x_{n-m} ~\leq~\sum b_m x_{n-m}~\leq ~\sum_{m=1}^n b_m\sup x_{n-m} $$

Then $\sup x_{n-m} \leq \sup x_n$ and $\inf x_{n-m} \geq \inf x_n$

Is $b_m \geq 0$ necessary?

答:
Write
$$
\begin{align*}
\left| \sum_{m=1}^n b_m x_{n-m+1} - ab \right|
&= \left| \sum_{m=1}^n b_m x_{n-m+1} - a\sum_{m=1}^n b_m + a\sum_{m=1}^n b_m - ab \right|\\
&\leq \left|\sum_{m=1}^n b_m x_{n-m+1} - a\sum_{m=1}^n b_m \right| + \left|a\sum_{m=1}^n b_m - ab \right| \\
&\leq \left|\sum_{m=1}^n b_m \left( x_{n-m+1} - a\right) \right| + \left|a\left( \sum_{m=1}^n b_m - b\right)\right| \\
&\leq \left|\sum_{m=1}^n b_{n-m+1} \left( x_m - a\right) \right| + |a|\left| \sum_{m=1}^n b_m - b\right|          \tag{$\dagger$}
\end{align*}
$$
The second term goes to $0$ by convergence of the series $\sum b_m$, so it only remains to deal with the first one. Define the sequence $(a_n)_n$ by $a_n = x_n -a$ (so that it converges to $0$), and fix any $\varepsilon > 0$: there exists $N_\varepsilon\in\mathbb{N}$ such that $\forall n \geq N_\varepsilon$, $|a_n|\leq \varepsilon$.

Now, for any $n \geq N_\varepsilon$,
$$
\begin{align*}
\Delta_n\stackrel{\rm{}def}{=}\left|\sum_{m=1}^n b_{n-m+1} \left( x_m - a\right) \right|
&= \left|\sum_{m=1}^n b_{n-m+1} a_m \right|
= \left|\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1} a_m + \sum_{N_\varepsilon}^{n} b_{n-m+1} a_m \right| \\
&\leq \left|\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1} a_m\right| + \left|\sum_{N_\varepsilon}^{n} b_{n-m+1} a_m \right| \\
&\leq \sum_{m=1}^{N_\varepsilon-1} \left|b_{n-m+1} a_m\right| + \sum_{N_\varepsilon}^{n} \left|b_{n-m+1} a_m \right| \\
&\leq \underbrace{\max_{1\leq k \leq N_\varepsilon-1}|a_k|}_{\alpha_{\varepsilon}}\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}  + \varepsilon\sum_{N_\varepsilon}^{n} b_{n-m+1} \qquad ((b_k)\text{ non-negative})\\
&= \alpha_{\varepsilon}\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}  + \varepsilon\sum_{N_\varepsilon}^{n} b_{n-m+1}
\end{align*}
$$
Finally, $0\leq \sum_{N_\varepsilon}^{n} b_{n-m+1} \leq \sum_{1}^{\infty} b_{n} = b$ (it actually converges when $n\to+\infty$ (monotone convergence) to some $b_\varepsilon\leq b$); while $\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}$ is the sum of constantly many ($N_\varepsilon-1$ to be precise) terms, each of them going to $0$ as $n\to+\infty$ (since $\sum b_k$ converges, $b_k\to 0$), and therefore $\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}\xrightarrow[n\to\infty]{}0$. Therefore, there exists $N^\prime_\varepsilon \geq N_\varepsilon$ such that for all $n\geq N^\prime_\varepsilon$, $\alpha_\varepsilon\sum_{m=1}^{N_\varepsilon-1} b_{n-m+1}\leq b\varepsilon$:
$$
\forall n\geq N^\prime_\varepsilon,\ \Delta_n\leq 2b\varepsilon
$$
i.e. $\Delta_n\xrightarrow[n\to\infty]{}0$. Thus the first term of ($\dagger$) goes to $0$ as well, proving that
$$
\left| \sum_{m=1}^n b_m x_{n-m+1} - ab \right|\xrightarrow[n\to\infty]{}0
$$


答: If $z\in\Bbb C$ and $n\in\Bbb Z$, then $e^{nz}=(e^z)^n$. So$$(\forall n\in\Bbb Z):e^{\pi in}=(e^{\pi i})^n=(-1)^n.$$

Title: How do you get $(-1)^n$ when you have $e^{in\pi}$

Body:
I am having a problem understanding this.

In which cases can you turn $e^{in\pi}$ in to $(-1)^{n}$?

What happens when it instead is $e^{-in\pi}$?


问:
Show that $A$ is diagonalisable when $P(A)$ is diagonalisable and $P'(A)$ is invertible
Let $n \geq 1$ and $A \in M_n(\mathbb{C})$.

Assume there exists $P \in \mathbb{C}[X]$ such that:

$P(A)$ is diagonalisable

$P'(A)$ is invertible

I have to show that $A$ is diagonalisable.

My try:

I solved the problem when $\textrm{deg}(P) \leq 1$. I also know that $A$ is trigonalisable, then I deduced that: $\forall \lambda \in \textrm{Sp}(A), P'(\lambda) \neq 0$. Also, there exists a diagonal matrix $D$ and $S \in GL_n(\mathbb{C})$ such that $P(A) = SDS^{-1}$. But I can't say anything more. Any help is welcome.

答:
Since $S^{-1}p(A)S = p(S^{-1}AS)$, we can assume that $A$ is in Jordan normal form.
Hence, it is sufficient to consider only one Jordan block $J$ to eigenvalue $\lambda$ of $A$.
According to (matrix function of Jordan block), we have
$$
p(J) = \pmatrix{ p(\lambda) & p'(\lambda) & \dots\\0 & p(\lambda) & p'(\lambda) & \dots\\&\ddots&\ddots&\ddots},
\quad
p'(J) = \pmatrix{ p'(\lambda) & p''(\lambda) & \dots\\0 & p'(\lambda) & p''(\lambda) & \dots\\&\ddots&\ddots&\ddots}.
$$
By assumption, $p'(A)$ and hence $p'(J)$ are invertible, so $p'(\lambda)\ne0$. But $p(A)$ and hence $p(J)$ are diagonalizable. This implies that the Jordan block $J$ is of size $1\times1$ (or $p'(\lambda)=0$, which is impossible).

This shows that all Jordan blocks of $A$ are of size $1\times 1$, and $A$ is diagonalizable.


问:
$f : \mathbb{R} \to \mathbb{R}$ is continuous iff $\{(x,y):y\lt f(x)\}$ and $\{(x,y):y\gt f(x)\}$ are both open sets in $\mathbb R^2$
Given $f: \mathbb{R} \to \mathbb{R}$ define two sets $A$ and $B$ as $ A = \{(x, y) : y < f(x)\}$ and $B = \{ (x, y) : y > f(x)\}$ in $\mathbb{R^2}$. Show that $f$ is continuous iff  both $A$ and $B$ are open.

I'm not sure if this follows from the definition of continuity. Can someone help?

答:
$\Rightarrow:$ Suppose that $f$ is continuous. Observe that
\begin{eqnarray*}
A & = & \cup_{r\in\mathbb{Q}}\{(x,y)\mid y<r<f(x)\}\\
 & = & \cup_{r\in\mathbb{Q}}\{(x,y)\mid y<r\}\bigcap\{(x,y)\mid r<f(x)\}\\
 & = & \cup_{r\in\mathbb{Q}}\mathbb{R}\times(-\infty,r)\bigcap f^{-1}(r,\infty)\times\mathbb{R},
\end{eqnarray*}
which is an open subset of $\mathbb{R}^{2}$. Similarly, we can prove
that $B$ is open.

$\Leftarrow:$ Suppose that $A$ and $B$ are open. Let $x_{0}\in\mathbb{R}$
be arbitrary. Let $\varepsilon>0$. Denote $p_{1}=(x_{0},f(x_{0})-\varepsilon)$
and $p_{2}=(x_{0},f(x_{0})+\varepsilon)$. Observe that $p_{1}\in A$.
Since $A$ is open, there exists $r_{1}>0$ such that $B(p_{1},r_{1})\subseteq A$.
Similarly, $p_{2}\in B$, so there exists $r_{2}>0$ such that $B(p_{2},r_{2})\subseteq B$.
Let $r=\min(r_{1},r_{2})$. Let $x\in(x_{0}-r,x_{0}+r)$. Since $d((x,f(x_{0})-\varepsilon),p_{1})<r$,
so $(x,f(x_{0})-\varepsilon)\in B(p_{1},r_{1})\subseteq A$. Hence,
$f(x_{0})-\varepsilon<f(x)$. Similarly, $d((x,f(x_{0})+\varepsilon),p_{2})<r$
implies that $(x,f(x_{0})+\varepsilon)\in B(p_{2},r_{2})\subseteq B$
. Hence $f(x_{0})+\varepsilon>f(x)$. Combining, we have $f(x)\in(f(x_{0})-\varepsilon,f(x_{0})+\varepsilon)$.
This show that $f$ is continuous at $x_{0}$. Since $x_{0}$ is arbitrary,
$f$ is continuous.


问:
Summation evaluation
If n takes values less than $-1$ how do I evaluate the following summation:

$$y[n]= \sum_{k=n+1}^{\infty} a^{n-k}$$

Do I need to switch signs of n because n takes only negative values?
I'm sorry if this question seems stupid....I'm not that well versed with summation

Edit: a is a number greater than 1

The way I tried to solve it is:

$$y[n]= a^n((\sum_{k=n+1}^{-1} a^{-k})+(\sum_{k=0}^{\infty} a^{-k}))$$

where the second summation evaluates to $$\frac{1}{1-a^{-1}}$$

but somehow the first summation seems to give me a problem

答:
As $a^{n-k}=\dfrac{1}{a^{k-n}}$, your sum is
$$y[n]= \sum_{k=n+1}^{\infty} a^{n-k}=\sum_{k-n=1}^{\infty}\left(\frac{1}{a}\right)^{k-n}=\sum_{m=1}^{\infty}\left(\frac{1}{a}\right)^{m}$$ (where in the last equality i've change $k-n$ by $m$). So you have a geometric series (independent of $n$) and you probably  can solve it by your own.

Title: Summation evaluation


答: You evaluate it as it is: as $\lim_{h\to\infty}S_h$, where $S_\bullet$ is the sequence $S_\bullet:\{x\in\Bbb Z\,:\, x\ge n+1\}\to\Bbb R$ such that $S_{n+1}=a^{n-(n+1)}$ and, $\forall m\in\{x\in \Bbb Z\,:\, x\ge n+1\}$, $S(m+1)=a^{n-m-1}+S_m$.

Title: Summation evaluation

Body:
If n takes values less than $-1$ how do I evaluate the following summation:

$$y[n]= \sum_{k=n+1}^{\infty} a^{n-k}$$

Do I need to switch signs of n because n takes only negative values?
I'm sorry if this question seems stupid....I'm not that well versed with summation

Edit: a is a number greater than 1

The way I tried to solve it is:

$$y[n]= a^n((\sum_{k=n+1}^{-1} a^{-k})+(\sum_{k=0}^{\infty} a^{-k}))$$

where the second summation evaluates to $$\frac{1}{1-a^{-1}}$$

but somehow the first summation seems to give me a problem


问:
Compute the spectral measure of an operator $(T\mathbf{a})_{n}=a_{n+1}-a_{n}$ on the space of square summable sequences
Consider the space $\mathcal{H}$ of square summable sequences $\mathbf{a}=\{a_{n}\}_{n=-\infty}^{\infty}$. The operator is defined by $$(T\mathbf{a})_{n}=a_{n+1}-a_{n}.$$

I want to compute the spectral resolution of this operator. From this website, https://encyclopediaofmath.org/wiki/Spectral_resolution, the spectrum measure $\mu$ can define resolution by $\mu((-\infty, t))$. So I guess the key is to get the spectral measure of this operator.

Edit 1: I think I got at least the spectrum

Given $\Psi\in\ell^{2}(\mathbb{Z})$, we can define $\hat{\Psi}\in L^{2}([0,2\pi))$ by $\hat{\Psi}(\xi):=\sum_{n\in\mathbb{Z}}e^{in\xi}\Psi_{n},$ with $$\Psi_{n}=\dfrac{1}{2\pi}\int_{0}^{2\pi}e^{in\xi}d\xi.$$

Then, the map $$U:\ell^{2}(\mathbb{Z})\longrightarrow L^{2}([0,2\pi),\dfrac{d\xi}{2\pi}),\ \ \Psi\longrightarrow_{U}\hat{\Psi},$$ is unitary.

We want to know what does $UTU^{-1}$ give us. Take $f(\xi)\in L^{2}([0,2\pi))$, then $$(TU^{-1}f)(n)=\dfrac{1}{2\pi}\int_{0}^{2\pi}(e^{-i(n+1)\xi}-e^{-in\xi})f(\xi)d\xi=\dfrac{1}{2\pi}\int_{0}^{2\pi}e^{-in\xi}(e^{-i\xi}-1)f(\xi)d\xi,$$ which implies that $$(UTU^{-1})(\xi)=(e^{-i\xi}-1)f(\xi).$$ Now we have the form of our operator in Fourier space, we are equipped to determine its spectrum.

Look at the resolvent:
\begin{align*}
R(T):=\{\lambda:(\lambda-T)\ \text{is invertible}\}&=\{\lambda:\lambda-e^{-i\xi}+1\neq 0\ \ \text{for all}\ \ \xi\in [0,2\pi)\}\\
&=\{\lambda:\lambda\neq e^{-i\xi}-1\ \ \ \text{for all}\ \ \xi\in [0,2\pi)\}\\
&=\mathbb{C}\setminus \{|\lambda+1|=1\}.
\end{align*}

This implies that the spectrum is $$\sigma(T)= \{|\lambda+1|=1\}.$$ We denote this set to be $\mathbb{S}_{1}^{*}(-1,1)$.

Am I correct?

Edit 2: I possibly get the spectral measure. But it confuses me

Now, for $f\in C(\mathbb{S}^{*}(-1,1))$ in Fourier space $f(T)$ is just multiplication by $f(e^{-i\xi}-1)$. Fix $\Psi\in\ell^{2}(\mathbb{Z})$, and let $\hat{\Psi}$ be the corresponding function in $L^{2}([0,2\pi))$. Then
\begin{align*}
\int f(T)d\mu_{\Psi}=\langle \Psi|f(T)\psi\rangle_{\ell^{2}}&=\langle U^{-1}U\Psi|f(T)U^{-1}U\Psi\rangle_{\ell{^2}}\\
&=\Big\langle U\Psi\Bigg|\Big(Uf(T)U^{-1}\Big)U\Psi\rangle_{L^{2}}\\
&=\langle \hat{\Psi}|f(e^{-i\xi}-1)\hat{\Psi}\rangle_{L^{2}}\\
&=\dfrac{1}{2\pi}\int_{0}^{2\pi}\overline{\hat{\Psi}}f(e^{-i\xi}-1)\hat{\Psi}d\xi\\
&=\dfrac{1}{2\pi}\int_{0}^{2\pi}f(e^{-i\xi}-1)|\hat{\Psi}(\xi)|^{2}d\xi.
\end{align*}

Write $\lambda:=e^{-i\xi}-1$, then $d\xi=\frac{d\lambda}{-e^{-i\xi}}$ and $\xi=i\log(\lambda+1)$ where $\log$ is taking over the principal branch (it's okay if just a general $\log$ since $\xi\in [0,2\pi]$). Then,
\begin{align*}
\int fd\mu_{\Psi}&=\dfrac{1}{2\pi}\int_{\mathbb{S}_{1}^{*}(-1,0)}f(\lambda)|\hat{\Psi}(i\log(\lambda+1))|^{2}\dfrac{d\lambda}{-e^{-i\xi}},
\end{align*}
therefore, $$d\mu_{\Psi}=|\hat{\Psi}(i\log(\lambda+1))|^{2}\dfrac{d\lambda}{-e^{-i\xi}}\Bigg|_{\mathbb{S}_{1}^{*}(-1,0)}.$$

According to this website, https://encyclopediaofmath.org/wiki/Spectral_resolution, the spectrum measure $\mu$ can define resolution by $\mu((-\infty, t))$. But you see my spectrum density is restricted to the unit circle. I am not sure how to go back to the real line.

Are the above things correct? How can I proceed?

答:
HINT:

Enough to get the spectrum of $T(e_n)= e_{n-1}$, where $(e_n)_{n\in \mathbb{Z}}$ is the orthnormal basis of $\mathcal{H}$.  Now, $\mathcal{H}$ is isometric with $L^2(S^1)$ with the norm
$$f\mapsto \frac{1}{2\pi} \int_{0}^{2\pi} |f(e^{it}|^2 dt $$
with
$$e_n \mapsto \epsilon_n$$
with $\epsilon_n(t) = e^{i n t}$.
(this is the inverse Fourier transform)

Now, on the $L^2(S^1)$, $T$ becomes the map $\epsilon_{n}\mapsto \epsilon_{n-1}$, that is, the multiplication by the function $\epsilon_{-1}(t)= e^{-i t}$


问:
Prove that inner space is incomplete and operator is bounded
Let's consider the following space:

$$ X = \{f \in C^\infty [0,1]: f(0)=f(1)=0   \}$$

with an inner product given by

$$ \langle f,g \rangle = \int_0^1 f'(t) \overline{g'(t)}dt. $$

Moreover, let's fix $h \in C^\infty [0,1]$ and define operator $M_h:X \rightarrow X$ given by
$$M_h(f) = h(t)f(t)$$
for $f \in X$ and $t \in [0,1]$.

First question is to show that space $X$ with this inner product is not a Hilbert space (motivation being that the completion of this space is the Sobolev space $H^1_0[0,1]$). My idea was the following: I wanted to find a sequence of smooth functions similar to the bump function https://en.wikipedia.org/wiki/Bump_function , for example given by
$$ f(t) = \begin{cases}
\operatorname{exp}\Big(\frac{-(t-\frac{1}{2})^2}{\frac{1}{2}^2 - (t-\frac{1}{2})^2}\Big),  & t \in (0,1) \\
0, & t \in \{0,1\}
\end{cases}$$

such that the mass becomes more and more concentrated in the middle (but the area of the integral of the differences keeps being small, so it would be a Cauchy sequence), and this sequence would converge to a non-continuous (but integrable) function. Unfortunately I'm not sure if this is a reasonable approach or how to explicitly describe this, since I feel like functions from the space $X$ have very strong properties so it's hard to find a sequence "breaking" it. I would appreciate any feedback about it, if possible.

The second question is to show that operator $M_h$ is bounded. I managed to show this (I will write $h$ and $f$ instead of $h(t)$ and $f(t)$ for better clarity): denote $M = \max\{M_1, M_2\}$ where
$$ M_1 = \max_{t \in [0,1]} |h(t)|$$
$$ M_2 = \max_{t \in [0,1]} |h'(t)|.$$

Then
\begin{align}
\Vert M_h \Vert^2 &= \Vert hf \Vert = \int_0^1 (hf)'\overline{(hf)'}dt =\int_0^1 |(hf)'|^2 \\&= \int_0^1 |h'f + hf'|^2 \leq \int_0^1 (|h|^2+|h'|^2)(|f|^2 + |f'|^ 2 )dt \\
&\leq 2M^2 \Big( \int_0^1 |f|^2 dt + \int_0^1 |f'|^2 dt \Big) \\
&=2M^2 \Big( \int_0^1 |f|^2 dt + \Vert f \Vert ^2 \Big)
\end{align}
where the first inequality comes from Cauchy-Schwartz lemma. Now, I would like to do something about the remaining integral, namely $\int_0^1 |f|^2 dt$ but here I'm stuck. Could someone share a hint if this approach makes sense, assuming my previous calculations were correct?

Thanks!

答:
You wrote $ \int_0^1 |f|^2 dt + \| f\|$ for $ \int_0^1 |f|^2 dt + \| f\|^{2}$

$f(x)=\int_0^{x} f'(t)dt$ so $|f(x)| \leq \int_0^{1}|f'(t)| dt\leq \sqrt {\int_0^{1}|f'(t)|^{2} dt}$ This implies that $\int_0^{1}|f(t)|^{2} dt \leq \int_0^{1}|f'(t)|^{2} dt=\|f\|^{2}$.

Here is a different approach to the first part. Define $T(f)=f(\frac 1 2)=\int_0^{\frac 1 2} f'(t)dt$. This is clearly a continuous linear functional. If the space is complete then we can apply Riesz Theorem to conclude that there exists $g \in X$ with $Tf=\int_0^{1}f'(t)g'(t)dt$ for all $f \in X$. Thus $\int_0^{1} f'(t)[g'(t)-\chi_{(0,\frac 1 2)}] dt=0$ for all $f \in X$. This implies that $g'(t)=\chi_{(0,\frac 1 2)}$ a.e. but this contradicts the fact that $g'$ is continuous.


问:
Transference principle
How can I proof this:

Show that if $\sigma$ is a $\forall\exists-$sentence in the language $\mathcal{L} = \{<\}$ which is true for all finite linear orders then $\sigma$ is true in any lineal order.

I have tried to use the union of elemental chains but I am not sure if lineal orders is the crecent union of finite lineal orders. I also tried to use the transference principle from $\forall$ to $\exists$ and viceversa but I do not get to any place

答: Let $L$ be a linear order and let $\forall \bar{x} \exists \bar{y} \phi(\bar{x}, \bar{y})$ be an $\forall \exists$-sentence that is true in every finite linear order (so $\phi(\bar{x}, \bar{y})$ is quantifier-free). Now let $\bar{a}$ be an arbitrary tuple in $L$, matching the length of $\bar{x}$. We write $A$ for the substructure of $L$ with universe $\bar{a}$. So $A$ is a finite linear order. By assumption we have that $A \models \forall \bar{x} \exists \bar{y} \phi(\bar{x}, \bar{y})$, so in particular $A \models \exists \bar{y} \phi(\bar{a}, \bar{y})$. Since $\exists \bar{y} \phi(\bar{a}, \bar{y})$ is an $\exists$-formula (with parameters from $A$) it is preserved upwards by the substructure relation, so we conclude that $L \models \exists \bar{y} \phi(\bar{a}, \bar{y})$. As $\bar{a}$ was arbitrary we thus have $L \models \forall \bar{x} \exists \bar{y} \phi(\bar{x}, \bar{y})$, as required.

Title: Transference principle


答:
A number divisible by $169=13^2$ should be divisible by $13$: and sure enough, $3^3=27\equiv1\pmod{13}$, so you get
$$
3^{3n+3}-26n-27\equiv 27^{n+1}-1\equiv 1-1\equiv0\pmod{13}
$$
Of course, this doesn't help in proving the statement, but it's a step forward.

Our statement can be rewritten as
$$
169 \mid (27^{n+1}-26(n+1)-1)
$$
and we can change $n+1$ into $n$: it doesn't harm, because
$$
169 \mid (27^{n}-26n-1)
$$
certainly holds for $n=0$. Good, now we can go by induction. Conpute
$$
(27^{n+1}-26(n+1)-1)-(27^n-26n-1)=27^{n+1}-27^n-26=26\cdot(27^n-1)
$$
Now we just need to show that this number is divisible by $169$, which is equivalent to $27^n-1$ being divisible by $13$. OK, we already know this.

Title: Problems in solving Number Theory problems (specifically Congruences)

Body:
I am very new to Number Theory, I understood the basic concepts and theorems but what I am struggling at is that I am not able to apply them on questions, and basically I am not able to process the problem in 'number theory' theorems rather I am thinking from an absolute manner.

I have solved around 15-20 divisibility problems of Number Theory and I still have no idea how you deal with congruences in a real problem, for example:

Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.

My approach to this problem was by simplifying the expression as much as I can but it involved congruences, which didn't even strike me. Even when I am reading congruence questions' answers, I am not able to simply understand

Now I have a number of questions:

first, does it take time to develop that 'intuition' while solving problems?

second, can elementary number theory be mastered through sheer practice?

third, is there a good resource to study and practice congruences?


问:
Substitution - what's the technical name of the inference rule?
Suppose the following are written down in some context.
$$3x^2 < y$$
$$x^2=xy-1$$

Then we may deduce (also within that context) that
$$3(xy-1) < y$$

What is the technical name of this inference rule?

答:
$$\text{Substitution}\quad\quad\quad$$

If $f(x,y) = g(x,y)$, then replacing any occurrence of $f(x,y)$ by $g(x,y)$ is replacing one equivalent expression, with another. One might think of this as employing the axioms of equality.

That is, it is not so much an "inference rule" as it is an equality or identity. It is using, within the context in which the identification between $\;x^2 = xy - 1\;$ is valid, Leibniz's law, "the principle of substitutivity".

$$\color{blue}{\bf x^2 = xy - 1} \iff \color{red}{\bf 3}\color{blue}{\bf x^2} = \color{red}{\bf 3}(\color{blue}{\bf xy - 1})$$

$$\color{red}{\bf 3}\color{blue}{\bf x^2} \lt y \iff \color{red}{\bf 3}(\color{blue}{\bf xy - 1}) \lt y$$


问:
How many times is $f^{-1}$ differentbable if $f$ is differentiable $m$ times at $x$?
Let $f$ be continuously differentiable $m$ times at $x \in \mathbb{R}$. Does it then follow that $f^{-1}$ (assuming it exists) is also continuously differentiable $m$ times at $f(x)$? This is true for $m=1$ but does it hold for any $m$? Thank you!

答:
Hint

What about $f(x)=x^3$ and $m=1$ ?

Body:
Hint

What about $f(x)=x^3$ and $m=1$ ?


答:
No, it is not true for $m=1$.
$f(x)=x^3$ is infinitely many times differentiable at 0, and its inverse function $\sqrt[3] x$ is not differentiable at 0. There can be a problem with zero derivatives.

Title: How many times is $f^{-1}$ differentbable if $f$ is differentiable $m$ times at $x$?

Body: Let $f$ be continuously differentiable $m$ times at $x \in \mathbb{R}$. Does it then follow that $f^{-1}$ (assuming it exists) is also continuously differentiable $m$ times at $f(x)$? This is true for $m=1$ but does it hold for any $m$? Thank you!


问:
Very simple question about tensor product of modules
I'm sure that this is a silly question, however I couldn't find any clarify on the internet. Take two copies of the $\mathbb Z$-module $\mathbb Z$ (i.e. of $\mathbb Z$ as abelian group) and consider their tensor product. We have that $M:=\mathbb Z \otimes_\mathbb Z \mathbb Z$ is again a $\mathbb Z$-module, so an abelian group. I have some problems in identifying the equivalence class of $0_M$: in fact, one has that $$(0,1)=(0\cdot 0,1)\sim(0,0\cdot1)=(0,0),$$ and similarly $(1,0)\sim (0,0)$. I'm thinking a couple $(a,b)$ as an element of $\mathbb Z \times \mathbb Z$ and the dot indicates the product in both modules (that actually is the usual product in $\mathbb Z$). However the contradiction is that $(0,1)+(1,0)=(1,1)$, that is not  in the equivalence class of $0_M$. What am I missing? Thank you in advance

答:
You seem to be confusing tensor products with usual products (which are the same as sums) of two modules. An element of $M\otimes M$ is not a pair of elements of $M$, but an equivalence class of the free module with $M \times M$ as a free basis.

In the tensor product, $0_\mathbb Z \otimes 1_\mathbb Z = 1_\mathbb Z\otimes 0_\mathbb Z = 0_{\mathbb Z \otimes \mathbb Z}$ is just the zero element. However, $1_\mathbb Z \otimes 1_\mathbb Z$ is distinct from zero.

Note that in the tensor product addition is not pairwise, i.e. $a \otimes b + c \otimes d \neq (a + c) \otimes (b + d)$. Rather, $a \otimes b + c \otimes d$ is just some element of the tensor product. What is true is that e.g. $a \otimes c + b \otimes c = (a + b) \otimes c$, so $\otimes$ behaves much like multiplication.


答:
Whatever the value of $n$ your sum is

$$a^{-1}+a^{-2}+a^{-3}+\cdots$$ so why worry ?

Title: Summation evaluation


问:
Prove that a polygon with nonnegative area is determined by at least three points.
How do you prove this statement in geometry?

A polygon with nonnegative area can't be formed with fewer than 3 points.

答: Well, if it's fewer than $3$ it must be $0$, $1$ or $2$...  </p>


问:
One to one function
I am stuck on this problem can someone solve it for me?

Show that the function $f(x) = \dfrac{3^x}{3^x + 1}$ is one-to-one

答:
To prove that a function $f: A \to B$ is one-to-one we need to show that
$\require{cancel}$
$$ \forall x,y \in A: f(x)=f(y) \implies x=y $$

So, take $f(x)=\dfrac{3^x}{3^x + 1}$ and plug $x,y$:

$$ f(x) = f(y) \\
\dfrac{3^x}{3^x + 1} = \dfrac{3^y}{3^y + 1} \\
3^x(3^y + 1) = 3^y(3^x +1) \\
\cancel{3^{x+y}} + 3^x = \cancel{3^{x+y}} + 3^y \\
3^x = 3^y \implies x=y $$

So, $f$ is one-to-one.

Title: One to one function


问:
complex numbers exponential form
I wish to show that $\cos^2(\frac{\pi}{5})+\cos^2(\frac{3\pi}{5})=\frac{3}{4}$

I know the solutions to $z^5+1=0$ are $-1$, $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$ and that

$z^5+1=(z+1)(z^4-z^3+z^2-z+1)$

which means that the solutions to $z^4-z^3+z^2-z+1=0$ are $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$

and so $z^4-z^3+z^2-z+1=(z-e^{i\frac{\pi}{5}})$$(z-e^{-i\frac{\pi}{5}})$ $(z-e^{i\frac{3\pi}{5}})$$(z-e^{i\frac{-3\pi}{5}})$

but I am not sure where to go from there.

答:
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$

Divide both sides by $z^2$ and replace $2u=z+\dfrac1z$

The roots of $$(2u)^2-2-2u+1=0$$ are $$\cos\frac\pi5,\cos\frac{3\pi}5$$

Now use https://mathworld.wolfram.com/VietasFormulas.html

Title: complex numbers exponential form


问:
Completely stuck at integrating $\int x\sqrt{\cos x}\,dx$
I am currently learning Calculus and I just started studying Integrals. I want to find the general indefinite integral $$\int x\sqrt{\cos x}\,dx$$

I have tried using some substitution such as letting $u=\cos x$ or $u=\sqrt x$, etc. but it seems like none of them is effective here.

I have also tried using online integral calculators to solve this question, but all I got is something like "steps are currently not supported for this problem."

Since even an integral calculator couldn't handle this, I wondered if this question can be solved. Therefore I came back to MathSE to seek help.

Is it solvable? If so, then how? Thanks in advance.

Body:
Another solution kindly suggested by @Raymond Manzoni
$$I=\int x\sqrt{\cos (x)}\,dx=\int x\sqrt{1-2 \sin ^2\left(\frac{x}{2}\right)}\,dx$$
Let $x=2y$
$$I=4 \int y \sqrt{1-2 \sin ^2(y)}\,dy$$ Integration by parts
$$u=y \qquad v'=\sqrt{1-2 \sin ^2(y)}$$ gives
$$\frac 14 I=y\,E(y|2)-\int E(y|2)\,dy$$
Using the expansion
$$E(y|2)=y-\frac{y^3}{3}-\frac{y^5}{30}-\frac{19 y^7}{630}-\frac{559 y^9}{22680}-\frac{2651
   y^{11}}{113400}+O\left(y^{13}\right)$$ the remaining integral can be approximated integrating termwise.

Without any special function
$$x\sqrt{\cos (x)}=x-\frac{x^3}{4}-\frac{x^5}{96}-\frac{19 x^7}{5760}-\frac{559
   x^9}{645120}-\frac{29161 x^{11}}{116121600}+O\left(x^{13}\right)$$
$$\int x\sqrt{\cos (x)}\,dx=\frac{x^2}{2}-\frac{x^4}{16}-\frac{x^6}{576}-\frac{19 x^8}{46080}-\frac{559
   x^{10}}{6451200}-\frac{29161 x^{12}}{1393459200}+O\left(x^{14}\right)$$


问:
Which is the better approximation to $e$?
Let $a_n = (1+1/n)^n$
and $b_n = (1+1/n)^{n+1}$.
Both $a_n \to e$
and $b_n \to e$,
and
$a_n < e < b_n$.

A better approximation to $e$
is known to be
$c_n = (1+1/n)^{n+1/2}
= \sqrt{a_n b_n}
$,
the geometric mean aof $a_n$ and $b_n$.

My question is:
how about $d_n = (a_n+b_n)/2$,
the arithmetic mean of $a_n$ and $b_n$?

Is $d_n$ a better approximation to $e$
than $c_n$?

More precisely,
let $r_n = \frac{c_n-e}{d_n-e}$.
Does $\lim_{n \to \infty} r_n$
exist? If so, is it $0$,
$\infty$, or a finite value?
If the limit is finite, what is it?

Is there some other mean of $a_n$ and $b_n$
(such as the harmonic mean)
which does better than either?

No, I haven't tried to solve these yet.
I thought they would be interesting questions.

Extra points if the answer
does $not$ use the expansions of
$\ln(1\pm x)$ or $e^x$.

答:
We only need third-order approximations to settle this:

$$\ln a_n = n \ln(1 + 1/n) = n(1/n - 1/2n^2 + 1/3n^3 + O(1/n^4)) = 1 - 1/2n + 1/3n^2 + O(1/n^3).$$
$$\ln b_n = (n+1)(1/n - 1/2n^2 + 1/3n^3 + O(1/n^4)) = 1 + 1/2n - 1/6n^2 + O(1/n^3).$$
$$\ln c_n = \tfrac12 (\ln a_n + \ln b_n) = 1 + 1/12n^2 + O(1/n^3).$$

To get asymptotics for $d_n$, expand $a_n = e \exp(\ln a_n - 1) = e( 1 - 1/2n + 11/24n^2 + O(1/n^3))$ and
$b_n = e \exp(\ln b_n - 1) = e(1 + 1/2n - 1/24n^2 + O(1/n^3))$, to see that

$$d_n = e(1 + 5/24n^2 + O(1/n^3)),\text{ while }c_n = e(1 + 1/12n^2 + O(1/n^3)).$$

This explains why $c_n$ is the better approximation by a factor of $2/5$.

An even better approximation would be $(\tfrac12a_n^p + \tfrac12b_n^p)^{1/p}$ for $p=-2/3$, which cancels the second-order term.  Maple calculations show that this in fact also cancels the third-order term and that the coefficient of $n^{-4}$ is quite small (on the order of $1/10000$), so this approximation is surprisingly good!

For $n=100$ we get $2.718281828463159$, good to 10 decimal places and gunning for the 11th.


问:
Distribution algorithm from rectangle to sine
I am trying to distribute a constant value along a vector according to sine distribution. I would like to obtain something like this</p>
```<code>given S = 16        ## constant value, sum of the vector elements
given length = 7    ## length of the vector

result = [1 2 3 4 3 2 1]
</code>```

答:
Consider $\;s_L(k):= \sin\left(\left(k-\frac 12\right)\frac{\pi}L\right)$

then you probably want something like :

$\displaystyle\left(\frac {S\;\cdot s_L(k)}{\sum_{i=1}^L s_L(i)}\right)_{\; k=1\cdots L}$

Example : for $\,L=7\,$ and $\,S=16\,$ we get :
$ [0.79225, 2.2198, 3.2078, 3.5603, 3.2078, 2.2198, 0.79225]$

Which is your result if rounded to the nearest integer.


问:
Prove $\ln(e^{x_1^2+x_2^2}+e^{\sqrt{x_1^2+1}})$ is convex
prove $\ln(e^{x_1^2+x_2^2}+e^{\sqrt{x_1^2+1}})$ is convex over $\mathbb{R}^2$

I know that theorem

(i) Let $f : \mathbb{R}^n → \mathbb{R}$ be a convex function which is nondecreasing with respect
to each of its variables separately; that is, for any i ∈ {1,2,... ,n} and fixed
$x_1 x_2,..., x_{i−1}, x_{i+1},..., x_n,$ the one-dimensional function
$g_i(y) = f (x_1, x_2,..., x_{i−1}, y, x_{i+1},..., x_n)$
is nondecreasing with respect to y. Let $h_1, h_2,..., h_n : \mathbb{R}^p → \mathbb{R}$ be convex
functions. Prove that the composite function
$r(z_1, z_2,..., z_p ) = f (h_1(z_1, z_2,..., z_p ),..., h_n(z_1, z_2,..., z_p))$
is convex.

I thought about setting $h_1(x_1,x_2)=e^{x_1^2+x_2^2},h_2(x_1,x_2)=e^{\sqrt{x_1^2+1}}$ and $g(x_1,x_2)=\ln(x_1+x_2)$ but $g$ is not convex.

any hints?

答: You can use that the sum of two log-convex functions is log-convex and apply that to $h_1$ and $h_2$.


问:
Constructing an example that supports the Propositional Interpolation Theorem
I'm trying to construct a simple proof of the Propositional Interpolation Theorem. For the following, let $At(\phi)$ be the set of sentence symbols that occur in a sentence $\phi$. Suppose that $\psi$ is a tautological consequence of $\phi$, but neither $\neg\phi$ nor $\psi$ is a tautology, and so $\psi$ is not a tautological consequence of $\phi$ for trivial reasons. I want to show that there is some sentence $\gamma$ such that

i) $\gamma$ is a tautological consequence of $\phi$.

ii) $\psi$ is a tautological consequence of $\gamma$

iii) $At(\gamma)\subseteq At(\phi)\cap At(\psi)$

From iii) $\gamma$ must be constructed from the sentence symbols found both in $\phi$ and $\psi$, so I first showed that $At(\phi)\cap At(\psi)\neq\emptyset$. Since neither $\neg\phi$ nor $\psi$ is a tautology, there must exist truth assignments $V,W$ such that $V(\neg\phi)=F$ and $W(\psi)=F$ and from this $V(\phi)=T$ and $W(\phi)=F$. If $At(\phi)\cap At(\psi)=\emptyset$ we may define a truth assignment for sentence symbols $p_i$,

$U(p_i)=V(p_i)$ if $p_i\in At(\phi)$, $U(p_i)=W(p_i)$ if $p_i\in At(\psi)$.

But then $U(\phi)=V(\phi)=T$, but $U(\psi)=W(\psi)=F$, which contradicts the fact that $\psi$ is a tautological consequence of $\phi$.

My question now is, is there some sort of method to construct $\gamma$ such that conditions i) and ii) are satisfied?

答: You can find some other proofs of the Craig interpolation theorem in this paper by Krajíček.</p>

ContentLicense: CC BY-SA 2.5


问:
Is a square a special case of a diamond (rhombus)?

Is the diamond shape (rhombus) necessarily different from the square (sides not perpendicular, different lengths of the diagonals), or is the square a special case of a diamond?

Sometimes I see a square (with right angles) that is called a diamond just because of being rotated by 45 degrees from the position where two of the sides were horizontal and the other two vertical, but I doubt that just this rotation (which is an isometric transformation, without any deformation) is sufficient for the square to become another shape.

答:
Mathematically speaking, something is called a rhombus, when it has four courners and all edges have the same length.

Something is called a square, when it has four courners, all edges have the same length and the four interior angles are $90°$.

So yes, a square is a special case of a rhombus.


问:
How to show $A$ is compact in $\Bbb{R}$ with standard topology?
Let $A = \{0\} \cup \{ \frac{1}{n}| n \in \Bbb{N}\} \subseteq \Bbb{R}$. I am struggling with showing whether $A$ is compact or not. I do not even know how to start.
Is $u=\{0\} \cup \left\{\left(\frac{1}{n+1}, \frac{1}{n-1}\right)| n \in \Bbb{N}\right\} \cap \{(1/2, 2)\}$ an open cover of A with no finite subcover? If so, what is the next step?

答: Hint : Compacts in $\mathbb{R}$ are precisely the closed and bounded sets. $A$ is clearly bounded. What about $L=\lim_{n}\frac{1}{n}$ ? Does $L\in A$ ?


问:
problem is about Eigen space, eigenvalues, eigenvectors and diagonalization
You are given a square matrix $A$ and that

• $\det (A) =  0$,
• The characteristic polynomial of $A$ has degree $5$,
• If we add $4$ to the diagonal of the matrix, the resulting matrix has $2$ pivots,
• The linearly independent vectors $u$ and $v$ satisfy the equation,
$A(u+v)= 7u+7v$

By using the information above determine answers to the following questions. Please write up your
solutions clearly and carefully, giving justification for each of your steps.

Can you identify any eigenvalues or eigenvectors in this problem?

What is the characteristic polynomial? How do you know?

Is the matrix $A$ diagonalizable? Why or why not? If it is diagonalizable, do you have enough
information to determine $P$ or $D$ where $A= PDP^{-1}$ ?

答:
Given That The Matrix has Characteristics equation of degree 5 so Order of Matrix Is 5 .
Again  Given that when you add 4 to Diagonal element Then matrix has 2 pivots that is  rank of Matrix become 2 and Nulity is 3
I.e. $(A+4I)$ has Nulity 3  that Implies - 4 is an eigenvalue with GM 3 so AM of -4 is  greater than equal to 3 and we have 3 LI eigenvectors for -4

Again  $A(u+v) = 7u +7v$ where $u$ and $v$ are LI so $u$ and $v$ are 2 LI eigenvectors corresponding to eigenvalue $7$

Hence We have Total 5 LI eigenvectors corresponding to $-4$ and $7$ .  So matrix is Diagonizable  hence AM= GM for each eigenvalues ..

So characteristics Polynomial is  $(x+4)^3(x-7)^2 =0$
Again We have nothing more than this So we can't find eigenvectors here. And So we can't find the Change of Basis matrix P .

And Diagonal matrix here is. diag$( -4, -4,-4,7,7)$

Hope this Was useful to You


问:
Probability density function $\max\{2,X\}$
Let $X$ be a random variable with a PDF $f(x)=\frac{1}{9}x^2\mathbb{1}_{(0,3)}(x)$. Find the distribution and PDF of $Y=\max\{2,X\}$ .

$F_X(t)=\int\limits_0^tf(x)dx=\int\limits_0^t\frac{1}{9}x^2dx=\frac{t^3}{27}$

$Y=\begin{cases}
                    2 &\text{for $X \in (0,2)$} \\
                    X &\text{for $X \in [2,3)$}
\end{cases}$

$F_Y(t)=\Bbb P(Y \leq t)=\Bbb P(\max\{2,X\}\leq t)=\Bbb P(2\leq t, X\leq t)= \ ?$

PDF:

$f_Y(t)=\begin{cases}
                    C &\text{for $X \in (0,2)$} \\
                    \frac{1}{9}x^2 &\text{for $X \in [2,3)$}
\end{cases}$

$\int\limits_0^2Cdt+\int\limits_2^3\frac{1}{9}t^2dt=2C+\frac{19}{27}=1$, then:
$C=\frac{4}{27}$

答:
The pdf is correct and you can find the CDF as well by integrating the pdf, but if you wish to find $\Pr\{\max(X,2)\le t\}$ directly, I suggest you the following way:
$$
\Pr\{\max(X,2)\le t\}{=
\Pr\{\max(X,2)\le t|X<2\}\Pr\{X<2\}\\+
\Pr\{\max(X,2)\le t|X\ge 2\}\Pr\{X\ge 2\}
}
$$


答:
First observe that Y-support is the following: $y \in[2;3]$

Then observe that $P(Y=2)=F_X(2)=\int_0^2 f(x)dx=\frac{8}{27}$

Concluding, Y-distribution is the following

$$ F_Y(y) =
\begin{cases}
0,  & \text{if $y<2$} \\
\frac{8}{27},  & \text{if $y=2$} \\
\frac{y^3}{27},  & \text{if $2<y<3$} \\
1,  & \text{if $y\geq 3$}
\end{cases}$$

As you can see the distribution function has "a jump" in $Y=2$ thus the rv is not absolutely continuous (it is a mixed distribution)

So technically the PDF (as it is defined only for continuous rv) does not exist.

Actually there is a generalization of density called "mixed density", or "mixed PDF". You can calculate it by derivating F where the variable is continuous and calculating $p(y_0)=F_Y(y_0^+)-F_Y(y_0^-)$ where Y is discrete (in your case in $Y=2$)

Here is a drawing of the transformation function. The line in red is $Y=max(2;X)$

As you can see, when $X \in (0;2]$, a mass of probability $p=\frac{8}{27}$ is concentrated in a singleton: $Y=2$. This means that the resulting Law of Y is discret in this point...in the rest of Y-support, Y is continuous being $Y=X$

Title: Probability density function $\max\{2,X\}$

Body:
Let $X$ be a random variable with a PDF $f(x)=\frac{1}{9}x^2\mathbb{1}_{(0,3)}(x)$. Find the distribution and PDF of $Y=\max\{2,X\}$ .

$F_X(t)=\int\limits_0^tf(x)dx=\int\limits_0^t\frac{1}{9}x^2dx=\frac{t^3}{27}$

$Y=\begin{cases}
                    2 &\text{for $X \in (0,2)$} \\
                    X &\text{for $X \in [2,3)$}
\end{cases}$

$F_Y(t)=\Bbb P(Y \leq t)=\Bbb P(\max\{2,X\}\leq t)=\Bbb P(2\leq t, X\leq t)= \ ?$

PDF:

$f_Y(t)=\begin{cases}
                    C &\text{for $X \in (0,2)$} \\
                    \frac{1}{9}x^2 &\text{for $X \in [2,3)$}
\end{cases}$

$\int\limits_0^2Cdt+\int\limits_2^3\frac{1}{9}t^2dt=2C+\frac{19}{27}=1$, then:
$C=\frac{4}{27}$


问:
A question about measurable and simple functions- measure theory
Let $(X,A), (Y,B)$ be measurable spaces.
$\mu$ is a measure on $A$, and $h:X\to Y$ a measurable function.
Define $\eta: B\to [0,\infty]$ such as $\forall  E\in B$ , $\eta(E)= \mu(h^{-1}(E))$.

A. Show that $\eta$ is well defined and is a measure on $B$.

B.Prove: $\forall E\in B \int_{Y} \chi_E d\eta= \int_X (\chi_E\circ h) d\mu$.

C.Let $\phi : Y\to [0,\infty)$ be a simple measurable function. Prove
$\int_{Y} \phi d\eta= \int_X (\phi\circ h) d\mu$.

D.Let $f:Y\to\ [0,\infty]$ be a measurable function. Prove
$\int_{Y}f d\eta= int_X (f\circ h) d\mu$.

My solution:

A.$\eta(\emptyset)=\mu(h^{-1}(\emptyset))$ then I did not manage to show that it is exactly $\mu(\emptyset)=0$.
Let $E_1,E_2,\ldots \in B$ nonoverlapping sets so $\eta(\cup_{i=1}^{\infty}(E_i))=\mu(h^-1(\cup_{i=1}^{\infty}(E_i)))=\mu(\cup_{i=1}^{\infty} h^-1(E_i))= \sum_i \mu(h^-1(E_i))=\sum_i \eta(E_i)$.
What does it mean that $\eta$ is well defined?

B.$\forall E\in B,  \mu(E)=\int_{E} 1 d\mu=\int_Y \chi_E d\mu$ I did not succeed to finish it!.

C. We can write $\phi=\sum_{j=1}^{n} a_{j} \chi_{E_j}$ where $E_j$ are measurable.
Then $\int_Y \phi d\eta= \int_Y \sum_{j=1}^{n} a_{j} \chi_{E_j} d\eta$ = ($\phi_n$ is a sequence of measurable positive functions)= $\sum_{j=1}^{n} a_{j} \int_Y \chi_{E_j} d\eta$  =(by part b) =
$\sum_{j=1}^{n} (a_{j} \int_X (\chi_{E_j}\circ h)) d\mu = \int_X (\phi\circ h) d\mu$.

D. f is measurable therefore by a theorem, there is an increasing monotone sequence of simple measurable sets $\phi_n$ such that:
$lim_{n\to \infty}\phi_n=f$ and by using the monotone theorem of Lebesgue we get

$\int_Y f d\eta= lim_{n\to \infty} \int _Y \phi_n d\eta$= (by  part  c)= $lim_{n\to \infty} \int_X (\phi_n\circ h) d\eta$ = ($\phi_n\circ h$ is measurable as a composition of two measurable functions)= $\int_X lim_{n\to \infty} (\phi_n\circ h) d\eta= \int_X (f\circ h ) d\eta$.

I will be glad if you can help in the points which I did not succeed, and tell me if what i did manage was okay.

答:
For part A, that $\eta $ is well defined means that the expression $\eta (E)$ makes sense, and it does because $h^{-1}(E)$ is $\mu$-measurable. By the other side the preimage of the empty set is empty for every function, just note that
$$
h^{-1}(A):=\{x:h(x)\in A\}
$$
So, when $A=\emptyset $ then $h^{-1}(\emptyset )=\emptyset $.

For the part B note that
$$
(\chi _E\circ h)(x)=1\iff h(x)\in E\iff x\in h^{-1}(E)\iff \chi _{h^{-1}(E)}(x)=1
$$

so $\chi _E\circ h=\chi _{h^{-1}(E)}$, and so the relation between the integrals is trivially true in view of the definition of $\eta $.


答:
This is a nice exercice for practicing with definitions.

So $X$ is compact if EVERY open cover has a finite sub cover. In order tho prove that $A$ is compact, you then have to start with a generic open cover of $A$ (you cannot choose the cover because the proof has to work for every cover). Say $U=\{U_i\}_{i\in I}$ is an open cover. Then there is at least one of the $U_i$, say $U_{i_0}$ which contains $0$. Since $U$ is open for the standard topology, and contains $0$, it must also contain an interval $(-\epsilon,\epsilon)$, which contains all but finitely many points of $A$. (Precisely, it contains all points $1/n$ for wich $n>1/\epsilon$). For any of those finitely many points $1/n$ left outside $U_{i_0}$ there is an $U_{i_n}$ containing it. So the family formed by $U_0$ and those finitely many $U_{i_n}$ is a finite cover of $A$.

Title: How to show $A$ is compact in $\Bbb{R}$ with standard topology?

Body:
Let $A = \{0\} \cup \{ \frac{1}{n}| n \in \Bbb{N}\} \subseteq \Bbb{R}$. I am struggling with showing whether $A$ is compact or not. I do not even know how to start.
Is $u=\{0\} \cup \left\{\left(\frac{1}{n+1}, \frac{1}{n-1}\right)| n \in \Bbb{N}\right\} \cap \{(1/2, 2)\}$ an open cover of A with no finite subcover? If so, what is the next step?


问:
Contour Integral and Complex Identity
While studying I came across this problem:

(a) For $z=x+iy$, show that

$$|\cos \pi z|^2=\frac{1}{2}(\cos(2\pi x)+\cosh(2\pi y))$$

(b) For a positive integer let $\gamma N$ be the square connecting the points $\pm N\pm Ni$ oriented in positive direction. Show that

$$|\cos \pi z|\geq 1$$

for all $z$ on $\gamma N$.

(c) Making use of the contour integral

$$\int_{\gamma N} \frac{1}{z^3\cos \pi z}dz$$

evaluate the sum of the series

$$\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)^3}$$

I wish I could show more effort, but I honestly don't know where to begin or what the connection between parts is. ANY help appreciated.

答:
For (a), use the cosine addition formula: $\cos{(a+i b)} = \cos{a} \cos{i b} - \sin{a} \sin{i b}$. Note that $\cos{i b} = (e^{i (i b)} + e^{-i (i b)})/2 = (e^{-b}+e^b)/2 = \cosh{b}$.  Similarly, you can show that $\sin{i b} = i \sinh{b}$.  Then

$$\cos{(a+i b)} = \cos{a} \cosh{b} - i \sin{a} \sinh{b}$$

so that

$$\begin{align}|\cos{\pi z}|^2 &= \cos^2{\pi x} \, \cosh^2{\pi y} + \sin^2{\pi x} \, \sinh^2{\pi y}\\ &= \left (\frac{1+\cos{2 \pi x}}{2} \right )\left (\frac{1+\cosh{2 \pi y}}{2} \right )+\left (\frac{1-\cos{2 \pi x}}{2} \right )\left (\frac{\cosh{2 \pi y}-1}{2} \right )\\ &= \frac12 (\cos{2 \pi x} + \cosh{2 \pi y})\end{align}$$

Note that, on $\gamma_N$,

$$|\cos{\pi z}|^2 = \frac12 (1+\cosh{2 \pi N}) = \cosh^2{\pi N} \ge 1$$

Therefore, because $|\cos{\pi z}|\ge 1$:

$$\left | \int_{\gamma_N} \frac{dz}{z^3 \cos{\pi z}} \right | \le \frac{8 N}{N^3} = \frac{8}{N^2}$$

Therefore, the integral vanishes as $N \to \infty$.  In that case, the sum of the residues of the poles of the integrand also vanish.  There is a pole of order 3 at the origin, and poles at $z=(2 n+1)/2$ for $n \in \mathbb{Z}$.  For the pole at the origin, the residue is

$$\frac12 \left[\frac{d^2}{dz^2} \sec{\pi z}\right ]_{z=0} = \frac{\pi^2}{2}$$

For the pole at $z=(2 n+1)/2$, the residue is

$$\frac{2^3}{(2 n+1)^3 (-\pi \sin{(n+1/2)\pi})} = -\frac{(-1)^{n} 8}{\pi (2 n+1)^3}$$

Summing over all $n$, we express that the sum of the residues is zero as

$$-\frac{8}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(2 n+1)^3} + \frac{\pi^2}{2} = 0 $$

Note that the sum is even in $n$; this the sum from $(-\infty,\infty)$ is twice the sum from $[0,\infty)$.  Therefore

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^3} = \frac{\pi^3}{32}$$

Title: Contour Integral and Complex Identity


问:
Please help with this approximation problem
Question : If you approximate $2.7145$ to $2.715$ then write the percentage relative error.

A) $0.000106$%

B)$0.00106$%

C) $0.0106$%

D) $0.106$%

I tried this by using the following approach :
$$100 \times\frac {\text{assumed value }- \text{ original value }} {\text{assumed value }}$$
but I did not get any results. Please anyone help me understand how to solve.

Body:
Percent Error =

$ \frac {Vobserved - Vtrue} {Vtrue}
=
\frac{2.715 - 2.7145}
{2.7145}
=
\frac {0.00049999999999972} { 2.7145} * 100
=  0.018419598452744 % $


答: The standard terminology in plane geometry (if the English language even has one) is that a square is a rhombus which is also a rectangle; equivalently, in a more exhotic and minimalistic way, a kite which is also a rectangle. Of course plane geometry is invariant by the action of the group of plane isometries, and therefore it isn't concerned with the graphical rendering of a square in an "organized" space where one or more directions are more important than the others.</p>

Title: Is a square a special case of a diamond (rhombus)?

Body:
Is the diamond shape (rhombus) necessarily different from the square (sides not perpendicular, different lengths of the diagonals), or is the square a special case of a diamond?

Sometimes I see a square (with right angles) that is called a diamond just because of being rotated by 45 degrees from the position where two of the sides were horizontal and the other two vertical, but I doubt that just this rotation (which is an isometric transformation, without any deformation) is sufficient for the square to become another shape.


答:
It is easy to see that

$$f'(x)= \frac{3^x \ln(3)}{(3^x+1)^2}.$$

Hence $f'(x) >0$ for all $x$. The mean value theorem shows then, that $f$ is one-to- one.

Title: One to one function

Body:
I am stuck on this problem can someone solve it for me?

Show that the function $f(x) = \dfrac{3^x}{3^x + 1}$ is one-to-one


问:
Calculate $P(0)$ knowing condition that polynomial satisfies.
Polynomial $P(x)$ is of degree $n$ and it satisfies:

$P(k)=\frac1{k}$ for $k=1,2,4,...,2^n$

Calculate $P(0)$.

$P(x)=a_n x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$

$0=k\times P(k)-1=a_nk^n+...+a_0=a_n(x-1)(x-2)...(x-k)$

$-1=0 \times P(0)-1=a_n\times-1\times-2\times...\times-k=a_n\times(-1)^{n+1}\times2^{\frac{n(n+1)}{2}}$

$a_n=\frac{(-1)^{n}}{2^{\frac{n(n+1)}{2}}}$

答:
To summarize the rather lengthy discussion in the comments:

First define $Q(x)=xP(x)$.  We know, of course, that $$Q(k)=1\quad \text{for}\quad k\in\{1, 2, 4,\cdots, 2^n\}$$

Since $Q(x)-1$ has degree $n+1$ and we have $n+1$ zeroes we have $$Q(x)-1=\lambda\prod_{i=0}^n(x-2^i)$$
for some constant $\lambda$.  We can compute $\lambda$ by noting that $Q(0)=0$ so $$-1=\lambda\prod_{i=0}^n(-2^i)\implies \lambda=\frac {(-1)^n}{\prod_{i=0}^n2^i}$$

So we know $Q(x)$ explicitly, hence we know $P(x)$.  To compute $P(0)$ we remark that $P(0)$ is the coefficient of $x$ in $Q(x)=xP(x)$. This quickly returns $$\boxed{P(0)= \left(1+\frac 12 + \frac 14+\cdots +\frac 1{2^n}\right)= \left(2-\frac 1{2^n}\right)}$$ where, at the last stage, we have used the standard method for evaluating Geometric Series.

Note:  I suggest checking the arithmetic here.  In particular, check the signs. This is the sort of problem for which it is annoyingly easy to drop or add a sign.


答:
The given set is $A=\{0\} \cup \big\{ \frac1n : n \in \mathbb{N} \big\}$

To show that $A$ is compact set in $\mathbb{R}$, we need to show that any open cover of $A$ has a finite subcover.

Consider any open cover of $A= \bigcup_{i \in I} U_i$. Then since $0 \in A$, then for some $i_0 \in I, 0 \in U_{i_0}$. since $U_{i_0}$ is an open set. Hence there exists $\epsilon>0$ such that $(-\epsilon,\epsilon ) \subset U_{i_0}$.

Archimedean Property

For any $\epsilon >0$, there exists $n \in \mathbb{N}$ such that $\frac1n < \epsilon$

Using the Archimedean Property, choose an $n_0$ such that $\frac{1}{n_0} < \epsilon$. Thus for $n \ge n_0$ we have $\frac1n \le \frac{1}{n_0} < \epsilon$. Thus all but finite number of elements of $A$ are contained in $(-\epsilon,\epsilon)$. Thus they are all contained in $U_{i_0}$. Now choose a finite number of open sets from $U_i$ to create a finite subcover of $A$


问:
Probability, basic statistics
I am new at this subject, I have a question, I am working on it for couple hours but I can not trust my answers could you please help me,

```<code>The probability that a certain type of electronic component
will fail during the first hour of operation is 0.05. If 200 components are tested
independently,
1. What is the expected number of defective components in a given hour?
2. What is the probability that the number of defective components is at most
two?
</code>```

My answer is for the first one; Fail: 0.005 so %5 and for 200 hundred answer is 10 ? But I don't have any idea for the second

Title: Probability, basic statistics

Body:
probability all good + probability exactly 1 bad + probability exactly 2 bad.

$$\left[(0.95)^{(200)}\right] + \left[\binom{200}{1}(0.05)^{(1)}(0.95)^{(199)}\right] + \left[\binom{200}{2}(0.05)^{(2)}(0.95)^{(198)}\right].$$


问:
Composition of a vector measure and a linear functional
Assume that $X$ is topological vector space over the field $\mathbb{R}$. Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a nonempty set $T$. We say that
$$m: \mathcal{M}\to X$$
is a vector measure if
$$m(E_1\cup E_2)=m(E_1)+m(E_2)$$
whenever $E_1$ and $E_2$ are disjoint members from $\mathcal{M}$.

Question. Why is it that for each $x'\in X'$, the composition map
$$x'\circ m: \mathcal{M}\to \mathbb{R}$$
is a measure on $\mathcal{M}$ whenever the vector measure $m:\mathcal{M}\to X$ is countably additive?

In the above, $X'$ denotes the space of all continuous linear functionals on $X$. So far, I can show that $(x'\circ m)(\varnothing)=0$. What bothers me is to show that $x'\circ m$ is countably additive. Any tips? Thanks in advance...

答: You should double check the meaning of linear functionals to see the answer to your question. $m$ is countable additive, so $x$ applied to a linear combination gives....


问:
Dont see why this linear algebra equation is correct
From the wikipedia page on Bayesian Linear Regression

A prior $\rho(\beta, \sigma^2)$ is conjugate to this likelihood function if it has the same functional form with respect to $\beta$ and $\sigma$. Since the log-likelihood is quadratic in $\beta$, the log-likelihood is re-written such that the likelihood becomes normal in $(\beta - \hat{\beta})$. Write

$$
(\mathbf{y} - \mathbf{X}\beta)^T (\mathbf{y} - \mathbf{X}\beta) = (\mathbf{y} - \mathbf{X} \hat{\beta})^T (\mathbf{y} - \mathbf{X}\hat{\beta}) + (\beta - \hat{\beta}) \mathbf{X}^T\mathbf{X} (\beta - \hat{\beta})
$$

It kind of just throws this out without showing why it is true.

答:
First do a Taylor expansion (the function is quadratic in $\beta$):
$$
(y-X\beta)^T(y-X\beta) = (y-X\hat\beta)^T(y-X\hat\beta)
+ 2 (y-X\hat\beta)^TX(\beta-\hat \beta) + (\beta-\hat\beta)^TX^TX(\beta-\hat\beta).$$
Since $\hat\beta$ satisfies $X^TX\beta = X^Ty$, the middle term vanishes.
This is because $\hat\beta$ is a least-squares solution, i.e., it solves
$$
\min_\beta (y-X\beta)^T(y-X\beta)
$$


问:
how to solve this integral that arises from keplers equation
How do I solve the indefinite integral given below

thanks!

Body:
$$\phi=\int \frac{M dr}{r\sqrt{2Er^2+2kr-M^2}}$$
Let $r=1/x$, then
$$\phi=-\int \frac{M dx}{\sqrt{2E+2kx-M^2x^2}}=-\int \frac{M dx}{\sqrt{A^2-(Mx-k)^2}}, A^2=2E+k^2/M^2$$
$$\implies \phi=-\sin ^{-1} \frac{Mx-k}{A}+B, x=1/r$$


问:
Prove that $\lim\sup a_n\leq \sup a_n$
So here's one question which I came across some time ago.

Given any sequence $a_n$ in $\mathbb{R}$, I need to show that $\lim\sup a_n\leq\sup a_n$.

My approach: As the sequence $a_n\in\mathbb{R}$, thus to talk about the $\lim\sup a_n$, I assumed that $a_n$ is bounded. I defined
$$b_n:=\sup\{a_k:k\geq n\}$$ i.e.
$$\begin{align}b_1&=\sup\{a_1,a_2,...\}\\b_2&=\sup\{a_2,a_3,...\}\\b_3&=\sup\{a_3,a_4,...\}\\.&\\.\end{align}$$
Now, I can say that $<b_n>$ is a decreasing sequence. Thus, we have, $b_1\geq b_2\geq b_3\geq ...$ and so on.

Now, like we had defined $b_n$,  we get that, $\sup a_n=b_1$.

I then did this:
$$\sup b_1=\sup\{b_1,b_2,...\}\geq\inf\{b_1,b_2,...\}\implies\sup a_n\geq\lim\sup a_n$$
The reason why I did this is because $<b_n>$ is a decreasing sequence. But I am a bit doubtful if I can write $\sup b_1=\sup\{b_1,b_2,...\}$. Can anyone please verify this?

Also, is there any alternate way of proving this inequality? Any help would be very much appreciable.

答:
For an easier proof : For all $n\in\mathbb N$,$$\sup_{k\geq n}a_k\leq \sup_{m\in\mathbb N}a_m.$$
taking $n\to \infty $ yields the wished result.

Title: Prove that $\lim\sup a_n\leq \sup a_n$


问:
Solving over-determined systems of quadratics in 3 dimensions
I have a 3x3 symmetric matrix $M$ with known components $M_{ij}$. I'd like to find a vector $x$ which has the following relationship with $M$

$$ M_{ij} = x^T E_{ij} x$$

Where $E_{ij}$ is also a 3x3 matrix with known components

$$ E_{ij} = \begin{bmatrix}
\epsilon_i^0 \epsilon_j^0 & \epsilon_i^0 \epsilon_j^1 & \epsilon_i^0 \epsilon_j^2\\
\epsilon_i^1 \epsilon_j^0 & \epsilon_i^1 \epsilon_j^1 & \epsilon_i^1 \epsilon_j^2\\
\epsilon_i^2 \epsilon_j^0 & \epsilon_i^2 \epsilon_j^1 & \epsilon_i^2 \epsilon_j^2\\
\end{bmatrix} $$

The $\epsilon_m^n$ are almost Kronecker delta symbols, i.e. small if $m \ne n$, nearly 1 otherwise. In addition, I know the length of the 3 vector $x$, so if I break this down, I have 7 equations to solve for 3 unknowns (six from the distinct $M_{ij}$, and one $|x| = r$).

If I expand those 6 equations out, they yield a generic system of quadratics in 3 variables (all possible terms present), which seems to not have a general solution based on this question.

However, there are a few key differences. My problem has more than double the number of conditions than variables and I have the extra 7th condition for the length, also I'm only interested in the case for 3 dimensions.

If I set the off-diagonal $\epsilon$ to 0, I get an approximation that is not good enough unfortunately. I'm currently trying to throw away terms with two small $\epsilon$ (i.e. two off diagonals), but haven't had too much luck with that either so far. Does anyone have any ideas about how one could solve this system?  I'm thinking there might be some way to get an iterative approximation?

Thanks!

答:
My problem has more than double the number of conditions than variables and I have the extra 7th condition for the length

This is very restrictive. Indeed, let $x=(x_0, x_1, x_2)$ and $\epsilon_k=(\epsilon_k^0, \epsilon_k^1, \epsilon_k^2)$ for each $k=1,2,3$. Then for each $i$, $j$ we have

$$M_{ij}=\sum_{m, n=0}^2 x_mx_n\epsilon^m_i\epsilon^n_j=\left(\sum_{m=0}^2 x_m\epsilon^m_i\right)
\left(\sum_{n=0}^2 x_n\epsilon^n_j\right)=(x,\epsilon_i)(x, \epsilon_j) .$$

That is $M=XX^t$, where $X=((x,\epsilon_0), (x,\epsilon_1), (x,\epsilon_2))^t$. In particular, the matrix $M$ should have rank at most $1$ . Moreover, for each $i$ we should have $M_{ii}=(x,\epsilon_i)^2\ge 0$. Then  $(x,\epsilon_i)=\pm\sqrt{ M_{ii}}$. We should chose the signs assuring $\operatorname{sign}(x,\epsilon_i)\cdot   \operatorname{sign}(x,\epsilon_i)= \operatorname{sign} M_{ij}$ for each distinct $i$ and $j$. Satifying the conditions we obtain at most eight possiblities for the matrix $X$.

Let $F$ be a $3\times 3$ matrix whose rows from the top to the bottom are $\epsilon_0$, $\epsilon_1$, and $\epsilon_2$. Then $Fx=X$. When $\epsilon_m^n$ are sufficiently close to
Kronecker $\delta$-symbols then the matrix $F$ is non-singular by Levy–Desplanques theorem, so then $x=F^{-1}X$.


问:
$f(n) = \Omega(n), g(n) = \mathcal O(n), h(n)= \theta(n) \implies f(n).g(n)+h(n)=\Omega(n)$

Let $f(n) = \Omega(n), g(n) = \mathcal O(n), h(n)= \theta(n) \implies f(n).g(n)+h(n)=\Omega(n)$

We get

$g(n) \leq [n ==h(n)] \leq f(n)$

$f(n).g(n)+h(n) \geq n+n \cdot g(n)$

But problem gives upper bound of g but not lower bound. To get greatest lower bound, i can put $g(n)$'s max value as $n$. Then $f(n).g(n)+h(n) \geq n+n\cdot n = n^2 \geq n$ PLease correct me if wrong.

答:
I assume that there is an implicit assumption that $g(n)\geq 0$; without this the statement would not be correct, e.g. by taking $f(n)=h(n)$ and $g(n)=-1$.

With this assumption, we simply have $f(n)g(n)+h(n)\geq h(n)=\Omega(n)$.

(The fact that $f(n)$ and $h(n)$ are positive, at least for $n$ sufficiently large, does follow from the given conditions.)

Incidentally, when you talk about "putting $g(n)$'s greatest value as $n$", don't forget the constant: we only know $g(n)\leq cn$ for some $c>0$ (and $n$ sufficiently large).


问:
Arc length parameter s
Consider the metric $$ds^2 = \frac{dx^2+dy^2}{y^2}.$$
Assume $R>0, a\in\mathbb{R}$. Consider the curve $$\gamma(\theta)=(a+R\sin\theta,R\cos\theta)$$
for $-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}$. Find the arclength parameter s. How would I go about doing this? Thank you in advance! </p>

答:
Since for $y>0$,

$$ds^2=\frac{dx^2+dy^2}{y^2} \iff ds=\frac{\sqrt{(dx/d\phi)^2+(dy/d\phi)^2}}{y}d\phi,$$

the arclength parameter is

$$s(\theta)=\int_0^\theta ds=\int_0^\theta \frac{\sqrt{\left(\frac{d}{d\phi}(a+R\sin\phi)\right)^2+\left(\frac{d}{d\phi}(R\cos\phi)\right)^2}}{R\cos\phi}d\phi.$$

This boils down to finding the antiderivative of $\sec\phi$.

Title: Arc length parameter s


问:
Meaning of $f(n) = \mathcal {O}(g(n)), g(n) \neq \mathcal O(f(n)) $

Meaning of $f(n) = \mathcal {O}(g(n)), g(n) \neq \mathcal O(f(n)) $

First condition says f(n) is upper bounded by g(n) ie.,  $f(n) \leq c g(n),$ for some c , $n_0 >0$

Second condition says that $g(n) \nleq c' f(n)$ for some $c', n_0' >0$

It means $g(n) >c'f(n)$ But 1st condition says $f(n) \leq cg(n)$.It means overall f(n) < g(n)

But why can we just write it as single one condition - 2nd condition. We dont need 1st condition to get f(n)<g(n) right. Please correct me if i am thinking wrongly

答:
No, the second condition is not enough.

For example, consider $f(n) = n^3$ for all $n$; $g(n) = n^4$ for $n$ even, $g(n) = n^2$ for $n$ odd.  Then $f(n) \neq \mathcal {O}(g(n))$ and $g(n) \neq \mathcal O(f(n))$ .

$f(n) = \mathcal {O}(g(n)), g(n) \neq \mathcal O(f(n))$  means:
$$
\limsup_{n \to +\infty} \frac{f(n)}{g(n)} < + \infty, \quad\text{but}\quad
\limsup_{n \to +\infty} \frac{g(n)}{f(n)} = +\infty .
$$
Stated another way (assuming $f(n), g(n)$ are both positive)
$$
0 = \liminf_{n \to +\infty}\frac{f(n)}{g(n)} \le \limsup_{n \to +\infty}\frac{f(n)}{g(n)} < + \infty.
$$


问:
Is this true $f(n)+\mathcal O(f(n)) = \Theta(f(n))$?

Is this true $f(n)+\mathcal O(f(n)) = \Theta(f(n))$?

f(n) + {set of all functions upper bounded by f(n)} = function upper bounded by f(n) and so $\mathcal O$ follows from this.

But how lower bound is also f(n) from which we can infer $\Theta $

$f(n)+\mathcal O(f(n))  \geq ?$

In above statement in Qn, if i replace $\mathcal O $ with $\mathcal o$ then how the statement is true? kindly elaborate

答:
No, it is entirely possible that $g\in O(f)$ and $f+g\in o(f)$. Namely, with $g=-f$.

It is however true that if $g\in o(f)$, then $f+g\in \Theta(f)$. In fact, eventually $\lvert g\rvert\le \frac12\lvert f\rvert$ and therefore, eventually, $$\frac12\lvert f\rvert\le\lvert f\rvert-\lvert g\rvert\le\lvert f+g\rvert\le\lvert f\rvert+\lvert g\rvert\le\frac32\lvert f\rvert$$


问:
Calculating banking rates - Equation
[Note: I have no mathematical background at all and as this is my first time posting here, let me know if I forget anything in my post, especially regarding the tags as I can't just pick up "equation" and I don't understand the popular tags' meaning.]

I try to calculate a rate for banking purpose.
This formula gives me the monthly payment I'll be entitled to reimburse each month :

$M = \frac{C\cdot\frac{t}{12}}{1-\left(1+\frac{t}{12}\right)^{-12n}}$

Given :

C = 200000 (if I loan 200k€)

t = 0,02 (at a rate of 2%)

n = 20 (over 20 years)

M = ? (I'll have to pay M €/months)

$M = \frac{200000*\frac{0,02}{12}}{1-(1+\frac{0,02}{12})^-12*20} = 1011.77 \approx 1012€/\text{month}$

Now my problem is : if I change the $C$ value to $199000$ how can I get the new $t$ from the following equation ?

$1012 = \frac{199000\cdot\frac{t}{12}}{1-(1+\frac{t}{12})^{-12\cdot20}}$

I know the answer is $t = 2,05%$ but I have no idea how to calculate it so I can't integrate it into my code. Thanks for your help.

答:
Since you are very close to the first situation, use it for starting Newton method.

Its iterates will be
$$\left(
\begin{array}{cc}
 0 & 0.02000000000 \\
 1 & 0.02056153090 \\
 2 & 0.02056062632
\end{array}
\right)$$

Edit

Since, in comments, you confirmed that $M$ is exactly the same is both cases, then you want to solve for $i_2$ the more general problem
$$C_1 \frac {i_1} {1-(1+i_1)^{-m}}=C_2 \frac {i_2} {1-(1+i_2)^{-m}}$$ where $i_k=\frac {t_k}{12}$ and $m=n t$.

Using Taylor expansion for the rhs around $i_2=i_1$ we have
$$\frac {i_2} {1-(1+i_2)^{-m}}=\frac{i_1}{1-(i_1+1)^{-m}}+(i_2-i_1) \left(\frac{1}{1-(i_1+1)^{-m}}-\frac{i_1 m
   (i_1+1)^{m-1}}{\left((i_1+1)^m-1\right)^2}\right)+O\left((i_2-i_1)^2\right)$$

Ignoring the higher order terms, we just need to solve for $i_2$ the equation
$$0=-\frac{i_1 (i_1+1)^m (C_1-C_2)}{(i_1+1)^m-1}+C_2\frac{ (i_1+1)^{m-1} \left((i_1+1)^{m+1}-
   (i_1( m+1)+1)\right)}{\left((i_1+1)^m-1\right)^2}(i_2-i_1)$$
which gives, as an approximation,
$$i_2\sim i_1 \frac {C_1-C_2}{C_2}\frac {i_1 (i_1+1) \left((i_1+1)^m-1\right) } {(i_1+1)^{m+1}-i_1(m+1)-1 }$$

Applied to your numbers, this gives $t_2=0.0205368$ which does not look too bad.

Title: Calculating banking rates - Equation


问:
divisibility observation in particular patterns
Please look at the following observation made by trial and error.

Let us take some $2$-digit numbers like $12, 15, 24,\dots$

$12 = 1 *  2 = 2 \implies 2|12 $ ($2$ divides $12$)

$15 = 1 * 5 = 5 \implies 5|15$

for $3$ digit numbers:
$315 = 3 * 15 = 45 \implies 45|315$

for $4$-digit numbers:
$1352 = 13 * 52 = 676 \implies 676|1352$

Like this kind of applicability is applicable only for some rare such numbers.
I would like to know from this observation the following:

1) In what cases the $2$-digit, $3$-digit cases we can write? Is there any formal method to justify this observation?

2) Is there any particular formula or method to generate such pairs?

Thanks in advance!


答: Let us start from the formula which you use, $\Delta(\alpha)=(-1)^{n(n-1)/2}N(f'(\alpha))$, where $N$ is the norm map of $\mathbf Q(\alpha)/\mathbf Q$ (note that you don't even need $\alpha$ to be an algebraic integer); this comes simply from the calculation of a Vandermonde determinant, see e.g. D. Marcus' "Number Fields", chap. 2, thm. 8. In the particular case of a trinomial $f(X)=X^n +aX+b$, put $\beta=f'(\alpha)=n\alpha^{n-1}+a=-(n-1)a-nb\alpha^{-1}$ because $f(\alpha)=0$, so that $\alpha= -nb(\beta+(n-1)a)^{-1}$. The minimal polynomial of $\beta$ is the numerator of $f(-nb(Y+(n-1)a)^{-1}))$, which is (all calculations done), $(Y+(n-1)a)^n-na(Y+(n-1)a)^{n-1}+(-1)^nb^{n-1}$. The norm of $\beta$ is $(-1)^n$ times the constant term of this polynomial, i.e. $n^n b^{n-1}+(-1)^{n-1}(n-1)^{n-1} a^n$, and this gives the desired formula for $\Delta(\alpha)$. You can check that for $n=2$ (resp. $3$), you recover the usual expressions $a^2-4b$ (resp. $-(4a^3+27b^2)$) .

Title: Finding the formula for the discriminate of an algebraic integer

Body:
Let $n \geq 2$, and $\alpha$ be an algebraic integer with minimal polynomial
$$f(x) = x^n + ax + b, \, \, \, a, b \in \mathbb{Z}.$$
Show that
$$\Delta(\alpha) = (-1)^{\frac{n(n-1)}{2}}((-1)^{n-1}(n - 1)^{n -1}a^n + n^n b^{n-1}
).$$

** So since I know that $\Delta(\alpha) = (-1)^{\frac{n(n-1)}{2}} \prod_{j = 1}^n f'(\alpha_j )$, I just need to show $\prod_{j = 1}^n f'(\alpha_j ) = (-1)^{n-1} (n - 1)^{n -1} a^n + n^n b^{n-1}$.

So for practice I first tried for $n  =2$ and got $\alpha$ has minimal polynomial, $$x^2 + ax + b = (x - \alpha_1)(x - \alpha_2) = x^2 -(\alpha_1 + \alpha_2)x + \alpha_1\alpha_2 \, \, \, \, \, \, \, \, \, \, (1).$$
So $f'(\alpha_1)f'(\alpha_2) = (2\alpha_1^{2 - 1} + a)(2\alpha_2^{2 - 1} + a) = (2\alpha_1 + a)(2\alpha_2 + a)$
$= 4\alpha_1\alpha_2 + 2a(\alpha_1 + \alpha_2) + a^2 =  4b + 2a(\alpha_1 + \alpha_2) + a^2 = 4b + 2a(-a) + a^2$
$= 4b -2a^2 + a^2 = -a^2 + 4b = (-1)^{2-1}
(2 - 1)^{2 -1}
a^2 + 2^2
b^{2-1}$, where equalities $4$ and $5$ follow from $(1)$.

Then I tried for $n = 3$ and got $\alpha$ has minimal polynomial, $$x^3 + ax + b = (x - \alpha_1)(x - \alpha_2)(x - \alpha_3)$$
$$= x^3 -(\alpha_1 + \alpha_2 + \alpha_3)x^2 + (\alpha_1\alpha_2+ \alpha_1\alpha_3 + \alpha_2 \alpha_3)x - \alpha_1\alpha_2\alpha_3 \, \, \, \, \, \, \, \, \, \, (2).$$
So $f'(\alpha_1)f'(\alpha_2)f'(\alpha_3) = (3\alpha_1^{3 - 1} + a)(3\alpha_2^{3 - 1} + a)(3\alpha_3^{3 - 1} + a)$
$= (3\alpha_1^2 + a)(3\alpha_2^2 + a)(3\alpha_3^2 + a)$
$= 27\alpha_1^2\alpha_2^2\alpha_3^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2) + a^3$
$= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2) + a^3$
$= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2 + 0) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$
$= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2 + 2(0)\alpha_1\alpha_2\alpha_3) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$
$= 27b^2 + 9a(\alpha_1^2\alpha_2^2 + \alpha_1^2\alpha_3^2 + \alpha_2^2\alpha_3^2 + 2(\alpha_1 + \alpha_2 + \alpha_3)\alpha_1\alpha_2\alpha_3) + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$
$= 27b^2 + 9a(\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 )^2 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$
$= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 ) + a^3$
$= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - 0 ) + a^3$
$= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - 0^2 ) + a^3$
$= 27b^2 + 9a^3 + 3a^2(\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - (\alpha_1 + \alpha_2 + \alpha_3)^2) + a^3$
$= 27b^2 + 9a^3 + 3a^2(- 2(\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3) ) + a^3$
$= 27b^2 + 9a^3 + 3a^2(- 2(a) ) + a^3$
$= 27b^2 + 9a^3 - 6a^3 + a^3 = 27b^2 + 4a^3$
$= (-1)^{3-1}
(3 - 1)^{3 -1}
a^3 + 3^3
b^{3-1}$, where equalities $4$, $7$, $9$, $12$, and $14$ follow from $(2)$.

**At this point I was hoping I was ready for the general case so I started off with
Now suppose for some $n \geq 2$, and $\alpha$ has minimal polynomial,
$$x^n + ax + b = (x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_n)$$
$$= x^n - \sum_{i = 1}^n \alpha_ix^{n -1} + \sum_{i = 1}^{n - 1} \sum_{j > i}^n \alpha_i \alpha_jx^{n -2} - \sum_{i = 1}^{n - 2}\sum_{j > i}^{n - 1}\sum_{k > j}^n \alpha_i\alpha_j \alpha_k x^{n - 3} +$$
$$\cdots + (-1)^{n - 1}\sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n\alpha_{i_1}\alpha_{i_2}\cdots\alpha_{i_{n-1}}x + (-1)^n\alpha_1\alpha_2\cdots\alpha_n \, \, \, \, \, (3).$$

So $\prod_{j = 1}^n f'(\alpha_j ) = (n\alpha_1^{n-1} + a)(n\alpha_2^{n - 1} + a)\cdots (n\alpha_n^{n - 1} + a)$

$= n\alpha_1^{n - 1}n\alpha_2^{n - 1}\cdots n\alpha_n^{n-1} + \sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n n\alpha_{i_1}^{n - 1}n\alpha_{i_2}^{n - 1} \cdots n\alpha_{i_{n - 1}}^{n - 1}a$
$+ \sum_{i_1 = 1}^3\sum_{i_2 > i_1}^4 \cdots \sum_{i_{n - 2} > i_{n - 3}}^n n\alpha_{i_1}^{n - 1}n\alpha_{i_2}^{n - 1} \cdots n\alpha_{i_{n - 2}}^{n - 1}a^2 +$
$ \cdots + \sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n n\alpha_{i_1}^{n - 1}n\alpha_{i_2}^{n-1} \cdots \alpha_{i_{n + 1 - j}}^{n - 1}a^{j - 1} +$
$\cdots + \sum_{i = 1}^n n\alpha_i^{n - 1}a^{n - 1} + a^n$

$= n^n(\alpha_1\alpha_2\cdots\alpha_n)^{n - 1} + n^{n - 1}a \sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 1}})^{n - 1}$

$+ n^{n - 2}a^2\sum_{i_1 = 1}^3\sum_{i_2 > i_1}^4 \cdots \sum_{i_{n - 2} > i_{n - 3}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 2}})^{n - 1} +$
$\cdots + n^{n + 1 - j} a^{j - 1}\sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1} +$
$\cdots + na^{n - 1}\sum_{i = 1}^n \alpha_i^{n - 1} + a^n$

$ = n^nb^{n - 1}  + n^{n - 1}a \sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 1}})^{n - 1}$
$+ n^{n - 2}a^2\sum_{i_1 = 1}^3\sum_{i_2 > i_1}^4 \cdots \sum_{i_{n - 2} > i_{n - 3}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n - 2}})^{n - 1} +$
$\cdots + n^{n + 1 - j} a^{j - 1}\sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1} +$
$\cdots + na^{n - 1}\sum_{i = 1}^n \alpha_i^{n - 1} + a^n$

Now I am pretty sure that the idea is to use the fact that from $(3)$ we have $(-1)^{n - 1}\sum_{i_1 = 1}^2\sum_{i_2 > i_1}^3 \cdots \sum_{i_{n - 1} > i_{n - 2}}^n\alpha_{i_1}\alpha_{i_2}\cdots\alpha_{i_{n-1}} = a$ and the coefficients in front of $x^{n-1}, x^{n - 2}, \cdots, x^2$ are zero, to some how rewrite $ \sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1}$ as a bunch of products and sums of the coefficients on the right hand side of $(3)$ to get in such a way to get $ \sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1} = (-1)^{n -1}(_{n - 1}$C$_{j - 2})(-1)^{j - 2)a^{n  + 1 - j}$, and then use the binomial theorem to finish.

The problem is every time I try to play with  reweighting $\sum_{i_1}^j\sum_{i_2 >i_1}^{j + 1}\cdots \sum_{i_{n + 1 - j} - i_{n - j}}^n (\alpha_{i_1}\alpha_{i_2} \cdots \alpha_{i_{n + 1 - j}})^{n - 1}$ I get an algebraic nightmare


答:
The following are just some ideas, I think a similar argument works but it seems iffy/wrong:

If $\lim  \sup a_{n} > \sup a_{n}$, then $\lim  \sup a_{n} - \sup a_{n} > \epsilon_{1}$ for some $\epsilon_{1} > 0$.

By definition $\lim  \sup a_{n} = L$ such that $|\sup a_{m} - L| <
\epsilon$ for every $\epsilon > 0$ where $m \geq n$.

For this last inequality set $\epsilon := \epsilon_{1}$.

Then $|\sup a_{m} - L| <
\epsilon_{1}$ and $L - \sup a_{n} > \epsilon_{1}$ so that $|\sup a_{m} - L| < L -\sup a_{m}$

By supposition we had that $L = \lim  \sup a_{n} > \sup a_{n}$ so that $|\sup a_{m} - L| = L-\sup a_{m} < L -\sup a_{m}$, a contradiction.

Therefore it must be that $\lim \sup a_{m} < \sup a_{m}$.

Title: Prove that $\lim\sup a_n\leq \sup a_n$

Body:
So here's one question which I came across some time ago.

Given any sequence $a_n$ in $\mathbb{R}$, I need to show that $\lim\sup a_n\leq\sup a_n$.

My approach: As the sequence $a_n\in\mathbb{R}$, thus to talk about the $\lim\sup a_n$, I assumed that $a_n$ is bounded. I defined
$$b_n:=\sup\{a_k:k\geq n\}$$ i.e.
$$\begin{align}b_1&=\sup\{a_1,a_2,...\}\\b_2&=\sup\{a_2,a_3,...\}\\b_3&=\sup\{a_3,a_4,...\}\\.&\\.\end{align}$$
Now, I can say that $<b_n>$ is a decreasing sequence. Thus, we have, $b_1\geq b_2\geq b_3\geq ...$ and so on.

Now, like we had defined $b_n$,  we get that, $\sup a_n=b_1$.

I then did this:
$$\sup b_1=\sup\{b_1,b_2,...\}\geq\inf\{b_1,b_2,...\}\implies\sup a_n\geq\lim\sup a_n$$
The reason why I did this is because $<b_n>$ is a decreasing sequence. But I am a bit doubtful if I can write $\sup b_1=\sup\{b_1,b_2,...\}$. Can anyone please verify this?

Also, is there any alternate way of proving this inequality? Any help would be very much appreciable.


问:
Is the complex given series convergent or divergent? Give a reason. Show details.
Is the given series convergent or divergent? Give a reason. Show details.

$$\sum_{n=2}^{\infty} \frac{(-i)^n}{ln \ n}$$

So maybe I'll try using the ratio test?

So the series converges if $$\left| \frac{z_{n+1}}{z_n} \right| < 1$$

So I have that $$z_n = \frac{(-i)^{n+1}}{ln \ (n + 1)} \cdot \frac{ln \ n}{(-i)^n}$$

So I think the lns cancel right so all we're left with is an -i in the numerator? Is that right? Is that less than 1 so does it converge?

答:
The series converges by virtue of alternating series test.

Your series actually represents two alternating series, one with a real component, and one with an imaginary component.

Both series converge because $(1/\log ~n)$ is strictly decreasing to zero.


答:
we can write
$$\sum _{n=2}^{\infty } \frac{(-i)^n}{\log (n)}=\sum _{n=1}^{\infty } \frac{(-i)^{2 n}}{\log (2 n)}+\sum _{n=1}^{\infty } \frac{(-i)^{2 n+1}}{\log (2 n+1)}$$
The first one can be rewritten as
$$\sum _{n=1}^{\infty } \frac{(-1)^{ n}}{\log (2 n)}$$
which converges for the Leibniz criterion because general term tends to zero

The other series can be written as
$$\sum _{n=1}^{\infty } \frac{(-i)^{2 n+1}}{\log (2 n+1)}=i\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{\log (2 n+1)}$$
which converges for the same test as above.

Title: Is the complex given series convergent or divergent? Give a reason. Show details.

Body:
Is the given series convergent or divergent? Give a reason. Show details.

$$\sum_{n=2}^{\infty} \frac{(-i)^n}{ln \ n}$$

So maybe I'll try using the ratio test?

So the series converges if $$\left| \frac{z_{n+1}}{z_n} \right| < 1$$

So I have that $$z_n = \frac{(-i)^{n+1}}{ln \ (n + 1)} \cdot \frac{ln \ n}{(-i)^n}$$

So I think the lns cancel right so all we're left with is an -i in the numerator? Is that right? Is that less than 1 so does it converge?


问:
Solving system of differential equation with laplace transformation.
I've got a very simple question which I'm failing to solve .
Here it is:

$x'=4x+2y, y'=2x+4y, x(0) =-1, y(0)=2$

I'm just starting to learn Laplace Transformation and still very weak at it.
Can anyone help me by showing steps which will lead me to the final solution?

答: $(x+y)'=6(x+y)\implies x+y=ce^{6t}$. So $x'=4x+2(ce^{6t}-x)=2x+2ce^{6t}$. Write $x=c_0e^{2t}+c_1e^{6t}$, then $4c_1=2c\implies c_1=\dfrac c2$. Hence $x=c_0e^{2t}+\dfrac c2e^{6t}$, which determines the solution for y. The constants may be determined by plugging in the initial values.


问:
SVD: Matrix size mismatch: Part 1
I'm reading this article.

In the "Proof" section, there is an equation:

$A^T = (USV^T)^T = VS^TU^T = VSU^T$

I don't understand the $VS^TU^T = VSU^T$ part.

Is this $VSU^T$ simply a typo?

If we borrow the following matrix A from the "Example" section:
$$
A=
\begin{pmatrix}
3 & 2 & 2\\\ 2 & 3 & -2
\end{pmatrix}
$$
, the shape of S would be 2 x 3:
$$
S=
\begin{pmatrix}
5 & 0 & 0\\\ 0 & 3 & 0
\end{pmatrix}
$$

So, $VS^TU^T$ works because (3 x 3)@(3 x 2)@(2 x 2),

while $VSU^T$ doesn't work because (3 x 3)@(2 x 3)@(2 x 2).

Am I correct, or missing something?

答:
It isn't a typo. It is a genuine mistake. $A^TA$ is equal to $VS^TSV^T$, not $VS^2V^T$. Since $S$ in general is not a square matrix, it cannot be squared in the first place.

In fact, not only is the equality $A^TA=VS^2V^T$ wrong, the author's attempt to prove the existence of SVD is also wrong.

In essence, the author tries to prove that if $A=USV^T$, then each column of $V$ is necessarily an eigenvector of $A^TA$ and each column of $U$ is necessarily an eigenvector of $AA^T$. However, he hasn't justifed why $U$ and $V$ exist. That is, he hasn't justified why we can always find an orthonormal set of eigenvectors of $A^TA$ or $AA^T$.

But there is an even bigger issue. The author tries to solve for $V$ and $U$ separately without considering the consistency between the two solutions. However, given an orthogonal matrix $V$ whose columns are eigenvectors of $A^TA$ and given an orthogonal matrix $U$ whose columns are eigenvectors of $AA^T$, it is not necessarily true that $S=U^TAV$ is a diagonal matrix with a nonnegative diagonal. For example, pick any orthogonal matrix $A\ne I$. Then $A^TA=AA^T=I$. If we take $U=V=S=I$, then $A^TAV=S^TSV$ and $AA^TU=SS^TU$. Yet, $A\ne I=USV^T$.

Title: SVD: Matrix size mismatch: Part 1


问:
Question about $\frac{1}{\pi} \int_{-\pi}^{\pi} te^{int} dt$ to $\frac{-2i(-1)^{n}}{n}$.
How do you get

$\frac{1}{\pi} \int_{-\pi}^{\pi} te^{int} dt$

to $\frac{-2i(-1)^{n}}{n}$.

The furthest i come to is:

$\frac{1}{\pi}((\frac{\pi e^{in\pi}+\pi e^{-in\pi}}{in}) -\frac{1}{in}(\frac{e^{in\pi}-e^{-in\pi})}{in}))$

答:
$$I=\frac{1}{\pi}\int_{-\pi}^{\pi}t e^{int} dt$$
Integration by parts leads to
$$I=\frac{1}{\pi}\left (t \frac{e^{int}}{in}+\frac{e^{int}}{n^2}\right)_{-\pi}^{\pi}$$
$$I=\frac{\pi}{in\pi}(e^{in\pi}+e^{-in\pi})+\frac{1}{n^2 \pi} (e^{in\pi}-e^{-in\pi})$$
$$I=\frac{2\cos(n\pi)}{in}+\frac{2i\sin n\pi}{n^2\pi}=\frac{-2i(-1)^n}{n}.$$
As $\cos n\pi=(-1)^n, \sin n\pi =0$


问:
The interval [0,1] with additional modulo 1
The interval $[0, 1]$ is an abelian group with addition modulo $1$.
Let $H$ be a proper subgroup of $[0, 1]$, which is closed as subset of $[0,1]$.
Show $H$ is finite.

I assumed $H$ is infinite: since $H$ is closed, all limit points of $H$ are in $H$.

Then let $x$ be in $[0, 1]\setminus H=S$ ($S$ is open since $H$ is closed) so there exists an open set $U_x$ which contains $x$ in it and $U_x \subset S$.

Let $\{x_n \}_n$ be a sequence converging to $x$. Thus for all $\epsilon>0$ there exists a $N$ natural number so that for all $n>N$ $|x_n-x|<\epsilon$.

For some $\epsilon_1 >0$ $B(x,\epsilon_1)$ is contained in $U_x$, then if I prove that $x$ is an limit point of $H$ then I can say $H$ is dense (because I express a generic element of $[0, 1]$ as belonging to $H$ or as limit point of $H$) then closure of $H$ will equal to $[0, 1]$ and this would contradict $H$ being proper subgroup.

But I couldn't find a way to show that.

答:
A canonical answer.

It suffices to show that each closed infinite subgroup $H$ of $G=[0,1]$ coincides with $G$.  Suppose to the contrary that $G\setminus H$ is non-empty. Since $H$ is closed, $G\setminus H$ contains an interval $I$ of length $\varepsilon>0$. By the pigeonhole principle there exist elements $x<y$ of $H$ such that $y-x<\varepsilon$. Then $n(y-x)\in I$ for some natural $n$. But $n(y-x)\in H$, a contradiction.