问:
Collecting math websites
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.

答: http://www.algebra.com is a very good site for algebra problems and solutions.</p>

Title: Collecting math websites


问:
Show $x+\frac{\lambda}x \geq 2\sqrt{\lambda}$ all $x,\lambda>0$

For $\lambda>0 $ and $x > 0$,
  $$x+\frac{\lambda}x \geq 2\sqrt{\lambda}$$

I tried to let function $g(x) =$ the difference of them and then find $g'(x) = 0$. With the given $x$, I can get the min point in $g(x)$ and find out $g(x)$ is greater than zero and finally rearrange it.

Seems not work though..

答:
As noted in the comments, the claim is false. Now, if it were

$$\frac{x+\lambda}{\sqrt x}\ge2\sqrt\lambda\iff (x-\lambda)^2\ge 0$$

and since the right inequality is trivial we're done....if ....

Edit: What you edited in your question has a solution very close to the above one:

$$x+\frac\lambda x\ge 2\sqrt\lambda\iff x^2+2\lambda+\frac{\lambda^2}{x^2}\ge4\lambda\iff\left(x-\frac\lambda{x}\right)^2\ge 0$$

ContentLicense: CC BY-SA 3.0


问:
The Question about Euler's Totient Function : How phi(i*p) = p*phi(i)? Here, p is prime and p divides i, phi-> Euler totient function
I have a question about the Euler totient function. I am new to the number theory and i don't know where to start to prove this. If $p$ is prime and $p$ divides $i$,

$$ Φ (i\cdot p) = p \cdot Φ (i),$$

where $Φ (n)$ is Euler Totient Function.

I know as $p$ divides $i$, they($i$ and $p$) will not be co-prime so I can't apply multiplicative property here but I am unable to understand how this result came.

答: An intuitive way of thinking of this is that, because $p$ divides $i$, the numbers counted in $\phi(i)$ are already coprime to $p$ (and thus to every power of $p$). So when you multiply by $p$, you retain all those coprime numbers in every block of $i$ numbers up to the new value $ip$, $\phi(i)$ coprime values in each of $p$ blocks; giving $\phi(ip) = p\cdot\phi(i)$.

ContentLicense: CC BY-SA 4.0


问:
Cyclic totally ramified Galois extension of of non-archimedean fields
Let $K \subset L := K(a)$ be a simple totally ramified
extension of non-archimedean
local fields of degree $n$ generated by a
$n$-th root of $K$; ie $a$ is a root of irreducible polynomial
$X^n- b \in K[X]$.

Additionally, we impose the condition

$$\vert k \vert =1 \operatorname{ mod } n$$

for the cardinality of the residue field $k$ of $K$.

I want to check that $L/K$ is Galois and
has cyclic Galois group. The Galois problem I was
able to solve myself but I don't know how to show that
the Galois group is cyclic.

on Galois: $K$ has characteristic zero, because it's a local field,
therefore the extension is separable. In order to check that
it's Galois, we have to check that it's normal. equivalently,
$L$ contains all roots of $X^n- b \in K[X]$.

My key observation was that $L$ contains all roots of
$X^n-1$, because the condition
$\vert k \vert =1 \operatorname{ mod } n$ is equivalent
to that one that $\vert k \vert-1$ is divisible by $n$ and
therefore $k^{\times}$ contains all $n$-th root. By Hensel's lemma
these roots can be lifted to $n$-th roots in $K$ and
obviously the roots of $X^n- b \in K[X]$ are $\zeta_n^m a,
m=0,1,..., n-1$.

Therefore $L/K$ is Galois. Why is it cyclic?

答:
If a primitive root of unity $\zeta_n\in F$ and $a^n\in F$ then $F(a)/F$ is separable because $a$ is a root of the separable polynomial $x^n-a^n$, which splits completely in $F(a)$ so $F(a)/F$ is Galois,

It is cyclic because its automorphisms are of the form $\sigma : a\to \zeta_n^{\phi(\sigma)} a$ making $Gal(F(a)/F)$ a subgroup of $\Bbb{Z}/n\Bbb{Z}$.

What Nico said is the converse: every degree $d|n$ cyclic extension of $F$ is of the form $F(c)/F$ with $c^n\in F$, one extension per cyclic subgroup of $F^\times/F^{\times n}$.


问:
How to prove that triangle-free graph with $t^2$ vertices contains set of $t$ vertices that form an independent set
Let $G$ be a triangle-free graph with $t^2$ vertices. Prove that $G$ contains a set of $t$ vertices that form an independent set. I don't know how to start this proof. Please help.

Observation: If $G$ has a vertex $v$ of degree at least $t$, then the neighbors of $v$ form an independent set of size $t$, since an edge within the neighborhood will create a triangle. So, then if it is not the case, every vertex has degree at most $t-1$. How do we construct an independent set?

答:
It's enough to show that any $n$-vertex triangle-free graph $G$ has an independent set of size at least $\sqrt{n}$.

Any graph $H$ (triangle-free or not) with $n$ vertices and maximum degree $\Delta=\Delta(H)$ has an independent set of size at least $\left\lceil n/(\Delta+1) \right\rceil$. Just consider the greedy algorithm in order to get a maximal independent set $I(H)$, which in each step search only in the non-neighborhood $\overline{N}(v)$ of a arbitrary first vertex $v$.
$$
\text{Choose } v\in V(H)\\
I(H)\leftarrow \{v\}\cup I(\overline{N}(v))
$$
Note that in each step the algorithm consider at most $\Delta+1$ vertices, therefore it needs to take at least  $\left\lceil n/(\Delta+1) \right\rceil$ steps to stop, i.e.,
$$
|I(H)|\geq \left\lceil n/(\Delta+1) \right\rceil.
$$

On the other hand, if $G$ is a $n$-vertex triangle-free graph, either there is a vertex with at least $\sqrt{n}$ neighbors, in which case those neighbors are an independent set, or $\Delta(G)\leq\sqrt{n}-1$, in which case any maximal independent set $I(G)$  has size at least $\left\lceil n/(\Delta+1) \right\rceil \geq \left\lceil n/\sqrt{n} \right\rceil \geq \sqrt{n}$.


问:
evaluating $\int_0^{\infty}\frac{e^{-t-\frac{x}{t}}}{t} dt$
I got to this integral, while proving some theorem in statistics:

$$\int_0^\infty \frac{e^{-t-\frac{x}{t}}}{t} \mathop{dt}$$

I have trouble evaluating it. I tried partial integration, tried substitution with some polynomial and some trigonometric functions. None of them helped, and Wolfram can't compute it either. Do you have a hint on how to solve this?

答: It worked on WolframAlpha after I hit the "Extended computation time" button:


问:
prove that $S$ is a subgroup of $G$
QUESTION: Let $S$ be a subset of $G$ such that the identity element $1 \in S$. Assume that the subsets $aS := \{as \mid s \in
S\} \subseteq G$ for $a \in G$ form a partition of $G$. Prove that $S$ is a subgroup of $G$.

This is my solution thus far. Can someone check if it is remotely correct and where I can improve?

答: Your equation involving $\{1\}$ is unnecessary and seems wrong. (Did you mean $\{x\}\cup\dots$ instead of $\{1\}\cup\dots$ ?) For that step, I'd just write "we have $x=x1\in xS$". Also, you might want to note that $S=1S$ is of the form $aS$ in order to apply the fact about the partition. Besides that, LGTM.

Title: prove that $S$ is a subgroup of $G$


问:
Is there a homeomorphism between the $n$-dimensional cube to the $n$-dimensional simplex that doesn't create hard corners?
Is there a homeomorphism between the $n$-dimensional cube to the $n$-dimensional simplex that doesn't create hard corners?  This is related to a similar question found here.  However, the mappings in response to that question create corners.  Specifically, I've been playing with schemes where

Set a direction to squeeze the point $x\in[0,1]^n$, $v$

Project $x$ into the nullspace of $v$, $y = x - \frac{v^Tx}{v^Tv}v$

Draw a line between $x$ and $y$, $l(\alpha) =  \alpha(y-x)+x$

$l(0)=x$, so we know it's in the hypercube.  Find
$$
  \alpha_l = \arg\max\{\alpha\geq 0:l(\alpha)\in[0,1]^n\}
  $$
$$
  \alpha_r = \arg\min\{\alpha\leq 0:l(\alpha)\in[0,1]^n\}
  $$
Basically, how far left and right we can travel and stay inside the cube

Define the squeeze factor
$$
  \alpha = \frac{\alpha_l}{\alpha_l-\alpha_r}
  $$
Basically, how far between the leftmost and rightmost points we are along the direction $v$.

Determine a new leftmost and rightmost point on the simplex,
$$
  z_l = l(\alpha_l)
  $$
$$
  z_r = \frac{1-\sum x_i}{\sum (y_i-x_i)}
  $$
It turns out the left most point is the same for the cube and the simplex.

Use the squeeze factor to determine the new mapping for $x$
$$
  w=\alpha(z_r-z_l)+z_l
  $$

Anyway, that was a little circuitous, but it works.  It turns out if $v=(1,\dots,1)$, we get the first answer to the linked question above.  If we set $v = x$, we get the second scheme.

Alright, so what's the problem?  This mapping makes something that kind of looks like a reentrant corner on smooth areas mapped to the simplex.  For example, if we set $v=(1,\dots,1)$ and map a circle centered at $(0.5,0.5)$ of radius $0.25$ to the simplex, we go from

to

Basically, we get a heart shaped object with these hard corners.  I've tried a bunch of different schemes for choosing $v$ and I keep getting corners, which is undesirable.  I really want something that's mapped smoothly to the simplex.

Anyway, is there a smooth map from the hypercube to the simplex where where something like a circle mapped from the hypercube to the simplex won't have these corners?

Edit 1

I think @Del has a good answer and here's what it looks like.  Given a rectangular grid

The scheme transforms this into

which shows how the grid is squished to fit into the simplex.  It turns out this is smooth for the use case I described above.  We can see this with the deformed circle

Anyway, thanks for the help!

答: See my answer to the duplicate question here.


问:
Orthonormal Matrix and Transpose.
If Q is an orthogonal matrix that has orthonormal vectors as columns, I can total understand the following result
$$Q^TQ=I$$
but $QQ^T=I$ only if Q is a square matrix and not if Q is a rectangular matrix.
I cannot understand why the second result is true. When I'm doing $QQ^T=I$ I'm not taking dot product of orthonormal vectors and yet it yields as identity matrix for a square matrix. How is this possible?

Edit: Orthonormal changed to Orthogonal

答:
Whenever $Q$ is not a square matrix , the result may not be true.

For example,
take $$Q=
    \begin{pmatrix}
    1 & 0 \\
    0 & 1 \\
    0 & 0 \\
    \end{pmatrix}
$$

Clearly then , $$Q^{T} Q=
    \begin{pmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{pmatrix}
$$
But, $$Q Q^{T}=
    \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 0 \\
    \end{pmatrix}
$$

If $Q$ is a square matrix such that $Q^{T}Q=I$

Clearly then $Q$ is an invertible matrix i.e. $Q^{-1}$ exists.

So, now $Q=QI $
$\implies Q=Q(Q^{T}Q)$
$\implies Q=(QQ^{T})Q$
$\implies (I-QQ^{T})Q=0$
$\implies QQ^{T}=I (\text{as, $Q$ is invertible })$

Title: Orthonormal Matrix and Transpose.


答:
Bringing $2\sqrt{\lambda}$ to the left we get

$$x-2\sqrt{\lambda}+\frac{\lambda}{x}=(\sqrt{x}-\sqrt{\frac{\lambda}{x}})^2\ge 0$$

So the claim is true.

Title: Show $x+\frac{\lambda}x \geq 2\sqrt{\lambda}$ all $x,\lambda>0$

Body:
For $\lambda>0 $ and $x > 0$,
  $$x+\frac{\lambda}x \geq 2\sqrt{\lambda}$$

I tried to let function $g(x) =$ the difference of them and then find $g'(x) = 0$. With the given $x$, I can get the min point in $g(x)$ and find out $g(x)$ is greater than zero and finally rearrange it.

Seems not work though..


问:
Proving set identities involving conditionals.
How can I prove the following conditional set identity?

For any three sets A, B, C we have $A\triangle(B\triangle C)=\varnothing \rightarrow(A\cup B\cup C)=((A\cap B)\setminus C)\cup((A\cap C)\setminus B)\cup((B\cap C)\setminus A)$

答:
How can I prove the following conditional set identity?

A conditional proof involves deriving the consequent under the assumption of the antecedent.

Our consequent is an equality between sets.   These are derived by showing that each side of the equality is a subset of the other side.   That is done by demonstrating that any element in a subset is an element in the superset.

Our antecedent is $A\triangle (B\triangle C)=\varnothing$.   Using this requires applying the definition of symmetric difference, and identifying what it means for a set to be equal to the emptyset.


问:
I am trying to prove gcd$(8,20) = 4$
I am trying to prove $\gcd(8,20) = 4$.

I am not quite sure how to go about it, but I think I need to prove that $\gcd(8,20)$ is not equal to $1$.
I have also set up $S = \{k \in \mathbb{N} : k = 8x + 10y \text{ for some } x, y \in \mathbb{Z}\}$. I am trying to somehow show that $4 \in S$.

Any suggestions on how I can prove my approaches? This was a personal challenge exercise, so I am just trying to further my learning.
I did take a look at other relevant posts on here, but they are too complex or advanced in comparison to where I am at. I just learnt how to do basic proofs with axioms and proofs for natural numbers. This exercise is given near the end of practice exercises after introduction on the Well Ordering Principle and [GCD].

Thanks for any help or solutions posted!

答: Since working with specific numbers, you may prove by division into cases, showing every number greater than 4 will not divide both, or use an algorithm that has been proven to give correct results, such as the Euclidean algorithm: https://en.wikipedia.org/wiki/Euclidean_algorithm.</p>

Title: I am trying to prove gcd$(8,20) = 4$


答:
Let us first state division Algorithm:  Let a, b are two integers > 0. Then there is unique integer q and r such that ${b=aq+r}$ where ${0 \leq < r <a}$ , so ${20 = 8 \cdot 2 + 4}$

Theorem: Let a ,b are two integers > 0, and  suppose ${b=aq+r}$ for some integers q and r. then GCD(b, a)= GCD(a,r) -> GCD(20,8)=GCD(8,4), r divides a, hence GCD=4

Title: I am trying to prove gcd$(8,20) = 4$

Body:
I am trying to prove $\gcd(8,20) = 4$.

I am not quite sure how to go about it, but I think I need to prove that $\gcd(8,20)$ is not equal to $1$.
I have also set up $S = \{k \in \mathbb{N} : k = 8x + 10y \text{ for some } x, y \in \mathbb{Z}\}$. I am trying to somehow show that $4 \in S$.

Any suggestions on how I can prove my approaches? This was a personal challenge exercise, so I am just trying to further my learning.
I did take a look at other relevant posts on here, but they are too complex or advanced in comparison to where I am at. I just learnt how to do basic proofs with axioms and proofs for natural numbers. This exercise is given near the end of practice exercises after introduction on the Well Ordering Principle and [GCD].

Thanks for any help or solutions posted!


问:
How to determine a $n\times n$ matrix $H$ has non trivial kernel?
We know $\vec{u}$ is a non-zero vector, and the equation of $H$ is:

$$I-\frac{\mathbb{uu}^T}{||\mathbb{u}||^2_2}$$

Is there any slick&quick method besides defining $\vec{u}$ to be $(u_1,u_2,...,u_n)^T$ then extend the whole matrix to solve the question?

答: $(I-uu^T/(u^Tu))v = 0$ if and only if $v = u(u^Tv)/(u^Tu)$, so the kernel is precisely the span of $u$.


问:
Is this equation of the tangent line correct?
This is a 2 part question, I figured out the first part easily but don't understand what to plug in to get the second part.

PART a:
Use the definition of derivative to calculate the f '(x) for f(x)= sqrt[x]

I got 1/2sqrt[x] as the derivative.

PART b
Use the result in (a), find the equation of the tangent line to the curve of
f (x)=sqrt[x] at x=4.

Plugging the numbers into the formula y-y1=m(x-x1) I get y-2=1/4(x-4), is this correct or have I made an error?

答:
The general formula for a straight line, as you mentioned, is
$$y=m(x-x_1)+y_1$$ where $m$ is the slope and $y_1$ is the value for $y$ when $x=x_1$. A more convenient notation, is writing $y$ as a function of "$x$" ($y=T(x)$. I use "$T$" for tangent) so we'd have
$$T(x)=m(x-x_1)+T(x_1)$$
(notice that we have $y_1=T(x_1)$ since, as I just stated, "$y_1$ is the value for $y$ when $x=x_1$", that is, $T(x_1)$).

But we want $T$ to be a function (a straight line) that describes the tangent at a point, say $\bar x$, for a function $f$ so it ought be the case that $T(\bar x)=f(\bar x)$. To ensure this, we set $\bar x=x_1$ (now, we have a line $T$ that passes through $(\bar x, f(\bar x))$, the desired point). All that's left is the slope $m$. As I'm sure you are aware, the slope of our desired tangent is $f'(\bar x)$ (so $m=f'(\bar x)$). Now we have
$$T(x)=f'(\bar x)(x-\bar x)+f(\bar x) \tag{$\star$}$$
This is the general formula for the tangent line of a function $f$ at $\bar x$.

If we let $\bar x= 4$ and $f(x)=\sqrt x$ then $\displaystyle f'(x)=\frac{1}{2\sqrt x}$ and we have a tangent line
$$T(x)=\frac{1}{2\sqrt{4}}(x-4)+\sqrt4 .$$
Make sure you undertsand $(\star)$ and remember it.


问:
Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?
Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?

I came across this when studying linear transformations in these notes which says:

The similarity group is obtained from the affine group by requiring that $A$ be orthogonal: $A^TA=\lambda^2I$

Can't seem to wrap my head around this one.

答:
Let $\mathbf {x, y} \in \mathbb{R}^n$ be unit vectors.

Then, the angle $\theta_1$ between them is given by $ \cos \theta_1 = \mathbf{x^\top y}$.

Also, angle between $\mathbf{Ax}$ and $\mathbf{Ay}$ is given by :

$\cos \theta_2 = \dfrac{(\mathbf{Ax})^\top (\mathbf{Ay})}{\|\mathbf{Ax}\|\|\mathbf{Ay}\|}$.

Now, $\|\mathbf{Ax}\|_2^2 = (\mathbf{Ax})^\top(\mathbf{Ax}) = \mathbf{x^\top A^\top Ax} = \mathbf{\lambda ^2 x^\top x}$.

Thus, $\mathbf{\|Ax\|} = \mathbf{|\lambda|}$.

$\implies \cos \theta_2 = \mathbf{\dfrac{\lambda^2 x^ \top y}{\lambda^2}} = \mathbf{x^ \top y} \implies \theta _1 = \theta _2$.

Thus, $\mathbf{A}$ preserves angles and is a similarity transformation.


问:
Does $\lim_{e \rightarrow 0}$ $ \int^{+e}_{-e} f(x) dx \rightarrow 0$ for all finite $f(x)$?
Does $\lim_{e \rightarrow 0}$ $ \int^{+e}_{-e} f(x) dx \rightarrow 0$ for all finite $f(x)$? And, if so, why? This comes up when showing that Schrodinger's Equation is continuous.

答:
Yes. Note that if $f$ is Riemann integrable on $[-\varepsilon,\varepsilon]$, then in particular $f$ is bounded on this interval, and we have the inequality:
$$\left|\int_{-\varepsilon}^\varepsilon f(x)\,dx\right|\leq\sup_{|x|\leq\varepsilon}|f(x)|\int_{-\varepsilon}^\varepsilon\,dx=\sup_{|x|\leq\varepsilon}|f(x)|\cdot2\varepsilon.$$
As $\varepsilon\to0$, the rightmost quantity vanishes, and so the desired result follows from the squeeze theorem.


答:
Orthogonality is forced by requiring a scalar multiple of the identity; if any two columns $a_i, a_j$ were not orthogonal, there would be a nonzero entry in the corresponding entries $b_{ij}, b_{ji}$ of the product.

Orthogonality results in the absence of any shear/twist in the transformation, restricting it to only reflection, rotation and translation.

The eigenvalue squares as scalars allow for non-normality, as it is possible to scale the original space without affecting the angles between lines in it, provided all axes of the space are scaled by the same amount. If $A$ was instead orthonormal, you force a lack of scaling as well.

See also this question, for which answers explain a converse point, why similarity transformation is a subtype of affine transformation.

Body:
Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?

I came across this when studying linear transformations in these notes which says:

The similarity group is obtained from the affine group by requiring that $A$ be orthogonal: $A^TA=\lambda^2I$

Can't seem to wrap my head around this one.


问:
Coupon Collector Problem with a constant probability of getting a coupon
My question includes a simple modification to well-known Coupon Collector Problem.

In this version there is a constant probability of getting a coupon at each trial. How this changes the original problem?

Denote by $T_m$ the number of trial required to collect all m coupons. Denote by $p$ the probability of getting a coupon at a trial.

I think the expectation of $T_m$ is simply $$\rm E[T_m] = \frac{m}{p} \sum_{I=1}^m \frac{1}{i}.$$

However, I am not sure about the distribution of $T_m$. I think it may include a Binomial distribution. Can you help about that?

Thanks in advance

答:
Look at the accepted answer here to see how to compute the distribution in the standard problem.  For the modified problem it should be straightforward to generalize this.

We can imagine that we have $m$ ordinary coupons, each of which occurs with probability $\frac pn$, and one blank coupon, which occurs with probability $1=p$.  For $M>=m$ we want the probability $\pi_M$ that in $M$ trials we collect all $m$ non-blank coupons.  (If we want the probability that it takes exactly $M$ trials to get all the non-blank-coupons, this is simply $\pi_M-\pi_{M-1}$.)

This is the sum from $k=0$ to $k=M-m$ that we collect exactly $k$ blank coupons, and in the other $M-k$ trails, we collect only non-blank coupons, and get at least one of each.  The second part of this is just an ordinary coupon collector's problem.

This will be a messy expression.  I don't know if it simplifies at all.


问:
The average time for a spider to reach the top of a cliff
The following question was asked in an theoretical computer science entrance exam in India:

A spider is at the bottom of a cliff, and is n inches from the top.
  Every step it takes brings it one inch closer to the top with
  probability $1/3$, and one inch away from the top with probability
  $2/3$, unless it is at the bottom in which case, it always gets one
  inch closer. What is the expected number of steps for the spider to
  reach the top as a function of $n$?

How do we solve this problem? The recurrence for the expectation seems tricky because of the boundary conditions :(

答:
The recurrence relation can be thought as,

Tn = (1/3)(Tn-1 + 1) + (2/3)(Tn-1 + 1 + Tn - Tn-2)

On further simplification

Tn = 3* Tn-1 - 2*Tn-2 + 3

The intuition is: To find expected number of steps to reach nth inch we first need to reach (n-1)th inch now after reaching (n-1)th inch we need 1 more step with probability 1/3 to complete nth inch this "move forward" step incurred additional cost of 1 otherwise with probability 2/3 we have to move 1 inch behind from (n-1)th inch so this "move back" step incurred additional cost 1 so we are now at (n-2)th inch, expected number of steps to complete n inches is Tn and expected number of steps to complete n-2 inches is Tn-2 so after reaching (n-2)th inch we require Tn-Tn-2 additional cost to complete n inches.
Hope this helps


问:
Find bound for $\sum_{n=0}^\infty\frac{(-1)^n sin \pi s}{n!(n+1-s)\pi}$
We have the sum
$$\sum_{n=0}^\infty\frac{(-1)^n \sin \pi s}{n!(n+1-s)\pi}$$ for all $s$ with $|im(s)|\le1$
This is a step from the proof of growth order of Gamma function by E.Stein in his Complex analysis.

Here for each $|im(s)\le1|$ exists integer $k$ such that $k-1/2\le Re(s)<k+1/2$ (here I only need to consider $k>0$, otherwise trivial). Now truncate this sum into $$\sum_{n\neq k-1}^\infty\frac{(-1)^n\sin \pi s}{n!(n+1-s)\pi}+\frac{(-1)^{k-1} \sin \pi s}{（k-1）!(k-s)\pi}.$$
For the first term, it is clear that $1/2 \le n+1-s$ therefore the whole term is bounded by $C|sin \pi s|$ for some constant $C$ and therefore bounded by $Ce^{\pi|s|}$. But for the second term, I know $sin\pi s$ also has a simple zero at $k$， so $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}=\frac{(-1)^{k-1} h(s)}{（k-1)!\pi}$ for some analytic $h$, with $h$ bounded by $C_s$ on $\{(x,y): k-1/2\le x<k+1/2, |y|\le1 \}$ but $C_s$ depends on $k$ and $k$ depends on $s$, so $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}$ is bounded by $C_s$ is saying nothing but $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}$ well defined on $\{(x,y):|y|\le1\}$. Is there a way to find the bound independent of $s$ for $\frac{(-1)^{k-1} sin \pi s}{（k-1）!(k-s)\pi}$?

答: Here is an another answer: Note $\frac{sins}{s}$ is bounded on $\{(x,y):|y\le 1\}$ due to periodicity of $sin s$. Therefore, $|\frac{sin\pi s}{s-k}|$ is also uniformly bounded on $\{(x,y):|y\le 1\}$.


问:
I need visual examples of topological concepts
I'm trying to understand basic concepts of topology, unfortunately I'm a very visual person, and as much documentation there is on how to come up with closed/open/clopen/etc. There are very few visual examples (using actual sets of integers, shapes, etc). So it's very hard for me to understand. Not everyone learns through text or mathematical definitions, so I figured this could help other people that have the same difficulties I have. Thanks for your help!

Can anyone get an example of a topology over a set.

I need one for a topology over a set, an open set, closed set, and clopen set. It needs to be something real. Like, say a set contained numbers { 1,2,3,4 }

It doesn't have to be that set, but it has to be a actual set of something (numbers, letters, geometric shapes), no labels (set X is the union of set Y and set Z), etc.

答:
Knowledge comes in the doing. So, in order to understand topological definitions, it would be wise to study a specific real-world example, and a great real-world example is the celebrated topological proof (by Furstenberg/פורסטנברג) of the infinitude of primes. Here’s the Wikipedia article about it.

Also, I realize that it is somewhat tangential to the thrust of your question, but I feel I should mention the classic book “Mathematical Snapshots” by Hugo Steinhaus. Here’s a review from the Amazon site:

Numerous photographs and diagrams help explain and illustrate mathematical phenomena in this series of thought-provoking expositions. Ranging from simple puzzles and games to more advanced problems, topics include the psychology of lottery players, the arrangement of chromosomes in a human cell, new and larger prime numbers, the fair division of a cake, how to find the shortest possible way to link a dozen locations by rail, and many other absorbing conundrums. A fascinating glimpse into the world of numbers and their uses. 1969 edition. 391 black-and-white illus.

Regards,
Mike Jones


问:
Number of bits in a stirling number
I know that the number of bit required for $N!$ is $\Theta(n\log n)$ by the stirling approximation. But what is the number of bits required for a stirling number of the first kind of the form $\left[\begin{smallmatrix} n \\ k \end{smallmatrix}\right]$?

I know that we can upper bound the stirling numbers since, $\sum_{k=0}^{n}\left[\begin{smallmatrix} n \\ k \end{smallmatrix}\right] = n!$

答: It's a good question. I don't know of a nice satisfying answer for the full two-parameter space. Wikipedia has good introductory discussion. You'd probably have to dive into papers to see what cases are well known if you're not interested in one of the simple ones they list.</p>

Title: Number of bits in a stirling number


问:
Why that element belongs to FC subgroup?
Let $\Delta$ denote Finite Conjugate subgroup of group G. K - a field

Then

The photo above is from Passman book "Infinite group rings".

I dont understand why elements $u_{i}$ belong to $\Delta$.

Could You tell me that?

I understand everything else on that photo.

答:
Since $\theta(\alpha)$ and $\theta(\gamma)$ belong to the image of $\theta$ which is contained in (equal to, actually) $K[\Delta]$, their supports must be in $\Delta$.

More explicitly, write $\theta(\alpha) = \sum_{x\in\Delta} k_x x$ and $\theta(\gamma) = \sum_{x\in\Delta} \ell_x x$. Then $\{u_1, u_2, \ldots, u_r\}$ is just the set of $x\in\Delta$ such that $k_x\neq 0$ or $\ell_x \neq 0$ in those two sums.


问:
Proving a general result about an iterated function
I recently found a question that stated as follows:

Let $f:\mathbb N\to\mathbb N$ be defined as follows: $$f(n)=2n+1-2^{\lfloor\log_2n\rfloor+1}$$ and let $f^\alpha(n)=f(f^{\alpha-1}(n))$. Let $t(n)$ be the smallest positive integer such that there exists a positive integer $N$ such that $f^{t(n)}(n)=f^{t(n)+N}(n)$. Determine $$\sum^{2^{2021}}_{n=2^{2020}}f^{t(n)}(n)\pmod{1009}$$

I was able to solve the problem as follows:

First, $f(n)\leq n$. To show this, we first note that to minimize $2^{\lfloor\log_2n\rfloor+1}-n$, we want $n$ to be as close to a power of $2$ as possible without exceeding said power of $2$. This is because provided $2^a\leq n<2^{a+1}$, $2^{\lfloor\log_2n\rfloor+1}=2^{a+1}$, i.e. the value of $2^{\lfloor\log_2n\rfloor+1}$ does not change regardless of where $n$ is in the interval $[2^a, 2^{a+1}-1]$. Within this interval, we can rewrite $f(n)$ as $2n-c$ where $c=2^{a+1}-1$. Hence, we have:

$$f(n)\leq n$$
$$\Rightarrow 2n-c\leq n$$
$$\Rightarrow n\leq c$$

As $n$ is bounded above by $2^{a+1}-1=c$, this inequality is always true. We note that the equality holds when $n$ is of the form $2^m-1$ for some positive integer $m$, i.e. all integers of the form $2^m-1$ are fixed points of $f$. Finally, $N=1$. If $N=2$, then $f^{t(n)}(n)>f^{t(n)+1}(n)$ by the above inequality ($f^{t(n)}(n)\neq f^{t(n)+1}(n)$ because $N\neq1$). We also have $f^{t(n)+2}(n)=f^{t(n)}(n)>f^{t(n)+1}(n)$ by definition of $N$. However, $f^{t(n)+2}(n)>f^{t(n)+1}(n)$, a contradiction. Therefore, $N\neq2$ and if $N\neq1$, then $f^{t(n)+2}(n)<f^{t(n)}(n)$. By induction, we can show $N\neq3,4\cdots$ and hence $N=1$.

Thus, we are interested in the fixed points of $f(n)$. We first note that $f\left(2^b\right)=1$ for all positive integer $b$. This is because $f\left(2^b\right)=2\left(2^b\right)+1-2^{\left\lfloor\log_22^b\right\rfloor+1}=2^{b+1}+1-2^{b+1}=1$. We also note that $1=2^1-1$ and is hence a fixed point of $f$ Hence $f\left(2^{2021}\right)=1$ and $\sum^{2^{2021}}_{n=2^{2020}}f^{t(n)}(n)\pmod{1009}=\sum^{2^{2021}-1}_{n=2^{2020}}f^{t(n)}(n)+1\pmod{1009}$.

We can try looking at which fixed point each number in the interval $[16, 31]$ "reach" via iterated composition to find a pattern. In the below graphs, I showed the sequence formed by applying $f$ to numbers in $[16, 31]$ multiple times. Numbers in the interval $[16, 31]$ are marked with $\color{blue}{\text{blue}}$, numbers that are fixed points are marked with $\color{red}{\text{red}}$, and numbers that are both fixed points and lie in the interval $[16, 31]$ are marked with $\color{purple}{\text{purple}}$.

$$\begin{align*}
&\color{purple}{31} \\
&\color{blue}{30},\color{blue}{29},\color{blue}{27},\color{blue}{23},\color{red}{15} \\
&\color{blue}{28},\color{blue}{25},\color{blue}{19},\color{red}7 \\
&\color{blue}{26},\color{blue}{21},11,\color{red}7 \\
&\color{blue}{24},\color{blue}{17},\color{red}3 \\
&\color{blue}{22},13,11,\color{red}7 \\
&\color{blue}{20},9,\color{red}3 \\
&\color{blue}{18},5,\color{red}3 \\
&\color{blue}{16},\color{red}1
\end{align*}$$

So in total, $1$ number reaches $2^5-1$, $4$ numbers reach $2^4-1$, $6$ numbers reach $2^3-1$, $4$ numbers reach $2^2-1$, and $1$ number reaches $2^1-1$. This looks suspiciously like the binomial coefficients.

This led to the following hypothesis:

Within the interval $\left[2^a, 2^{a+1}-1\right]$, the number of numbers that reach the fixed point $2^b-1$ with $b$ being a positive integer satisfying $b\leq a+1$ is $\binom a{b-1}$.

The hypothesis applied to to the interval $[16, 31]$ (as shown above) and $[8, 15]$. Not wanting to spend too much time on this problem, I decided that I was going to assume this hypothesis was true.

Hence,

$$\sum^{2^{2021}-1}_{n=2^{2020}}f^{t(n)}(n)=\sum^{2020}_{k=0}\binom{2020}k2^{k+1}-1=2\sum^{2020}_{k=0}\binom{2020}k2^k-\sum^{2020}_{k=0}\binom{2020}k$$

Applying Binomial theorem, we know $\sum^{2020}_{k=0}\binom{2020}k2^k=(1+2)^{2020}=3^{2020}$ and $\sum^{2020}_{k=0}\binom{2020}k=2^{2020}$. We also note that $1009$ is prime, and thus we can use Fermat's little Theorem.

Thus,

$$\sum^{2^{2021}-1}_{n=2^{2020}}f^{t(n)}(n)+1\pmod{1009}=2\cdot 3^{2020}-2^{2020}+1\pmod{1009}=2\cdot \left(3^{1009}\right)^2\cdot 3^2-\left(2^{1009}\right)^2\cdot 2^2+1\pmod{1009}=2\cdot3^2\cdot3^2-2^2\cdot2^2+1\pmod{1009}=2\cdot3^4-2^4+1\pmod{1009}=162-16+1\pmod{1009}=147\pmod{1009}$$

Though $147$ is the correct answer, my solution does not feel very satisfying. I had to assume my hypothesis was true and in the end I still am not sure why this hypothesis applied. Therefore, I would like to ask why my hypothesis is true, i.e.

Let $f:\mathbb N\to\mathbb N$ be defined as follows: $$f(n)=2n+1-2^{\lfloor\log_2n\rfloor+1}$$ and let $f^\alpha(n)=f(f^{\alpha-1}(n))$. Let $t(n)$ be the smallest positive integer such that $f^{t(n)+1}(n)=f^{t(n)}(n)$. Show that, $\forall a\in\mathbb{Z}_+\forall b\in\mathbb{Z}_+$ s.t. $b\leq a+1$, we have $|\{n\in \mathbb{Z}, n\in[2^a, 2^{a+1}-1]: f^{t(n)}(n)=2^b-1\}|=\binom a{b-1}$

答:
(Note: This answer uses binary numbers)

This answer provides an interesting perspective on the definition of $f(n)$ and how we should think of $n$.

Essentially, suppose $n$ is a binary string and the digit $1$ of $n$ appears $y$ times at $a_1, a_2, \cdots, a_y$, i.e.

$$n=\sum^y_{k=1}10^{a_k}$$

We know $f(n)=10n+1-10^{\left\lfloor\log_{10}n\right\rfloor+1}$. We note that $\left\lfloor\log_{10}n\right\rfloor+1$ is the position of the largest digit in $n$, plus $1$.

So we have:

$$f(n)=10\sum^y_{k=1}10^{a_k}+1-10^{a_y+1}=\sum^{y-1}_{k=1}10^{a_k+1}+\sum^0_{k=0}10^k$$
$$f^2(n)=10\sum^{y-1}_{k=1}10^{a_k+1}+10\sum^0_{k=0}10^k+1-10^{a_{y-1}+2}=\sum^{y-2}_{k=1}10^{a_k+2}+\sum^1_{k=0}10^k$$
$$\vdots$$
$$f^m(n)=\sum^{y-m}_{k=1}10^{a_k+m}+\sum^{m-1}_{k=0}10^k, m<y$$
$$\vdots$$
$$f^y(n)=\sum^{y-y}_{k=1}10^{a_k+y}+\sum^{y-1}_{k=0}10^k=\sum^{y-1}_{k=0}10^k=10^y-1$$
We note that $10^y-1$ is a fixed point of $f$. Therefore, $t(n)=y$ and $f^{t(n)}(n)=10^y-1$.

My conjecture earlier asks about numbers in the interval $[10^a, 10^{a+1}-1]$. This is equivalent to asking about asking about all $a+1$ digit numbers.

We know that, for $f^{t(n)}=10^b-1$, $n$ must contain exactly $b$ $1$s (and $a-b+1$ $0$s), and the first digit must be $1$, so there are in total $\frac{(b-1+a-b+1)!}{(b-1)!(a-b+1)!}=\binom a{b-1}$ such $n$. Therefore the conjecture is correct.


问:
product with a constant term and a divergent series
In doing some calculation, I am getting a step as

$k$(a divergent series over reals)$=0$ where $k$ is a real constant. Can I conclude that $k=0$?

Infact I have like this:

$k(\sum r^n)=0$ with $r>1$. Can I conclude $k=0$?

答:
Yes, evidently you can. Because the sum of a strictly divergent series is never zero.
And as your series appears to be a geometric series, the sum will always be positive.</p>


问:
A uniformly continuous semigroup has a unique infinitesimal generator
I'm studying the theory of semigroups from Pazy's book. I'm struggling to understand the proof of a theorem stating that

Theorem 1.2. Let $X$ be a banach space. A linear operator $A:X\rightarrow X$ is the infinitesimal generator of a uniformly continuous semigroup iff $A$ is a bounded linear operator.

The proof of $\Leftarrow$ is pretty straight forward by setting
\begin{equation}
T(t)=e^{tA}:=\sum_{n=0}^{\infty}\frac{(tA)^n}{n!}.
\end{equation}

But I don't understand a part in $\Rightarrow$ proof, where we let $T(t)$ be a uniformly continuous semigroup of bounded linear operators on $X$ and some $\rho>0$ be such that $\|I-\rho^{-1}\int_0^\rho T(s)ds\|<1$ so that $\int_0^\rho T(s)ds$ is invertible. We can show that the infinitesimal generator $A:=\lim_{h\searrow 0}h^{-1}(T(h)-I)$ converges in $\mathcal{L}(X)$:
\begin{align}
h^{-1}(T(h)-I)\int_0^\rho T(s)ds = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)
\end{align}
and therefore
\begin{equation}
h^{-1}(T(h)-I) = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)\bigg(\int_0^\rho T(s)ds\bigg)^{-1}.
\end{equation}
We have $h^{-1}(T(h)-I)\rightarrow (T(\rho)-I)(\int_0^\rho T(s)ds)^{-1}$ in $\mathcal{L}(X)$ as $h\searrow 0$.

By definition of a semigroup, the infinitesimal generator of $T(t)$ is uniquely defined. But it seems to me that there are several ways to construct a different $\int_0^\rho T(s)ds$ by fixing $\rho>0$ at another value.

So my question: how can we show that $(T(\rho)-I)(\int_0^\rho T(s)ds)^{-1}$ is independent of $\rho$ described above?

答:
Realized I was basically writing an answer in comments, so here's the comments as an answer.

Note that $\rho$ appears in two places in your formula (in $T(\rho)$ and in the integral you constructed, to the $−1$ power); so it is not formally inconceivable that the $\rho$- dependence might "cancel out." Running the construction with $T(t)=e^{At}$ for bounded $A$ will show this more explicitly.

For example in the one dimensional case if $T(t) = e^{at}$ is the semigroup, and writing $r$ in place of $\rho$, then $\int_0^r T(s) \, ds = (e^{ar} - 1)/a$, so $(T(r)-1) * (\int_0^r T(s) ds)^{-1} = (e^{ar} - 1) * a/(e^{ar} - 1) = a$.

And of course in general, the very presence of $\rho$ in as a limit of something an operator limit involving only $T(h)$ and $h$ (which do not depend on $\rho$) also shows that the choice of $\rho$ won't matter, subject to the hypotheses. It's just uniqueness of the generator.


问:
If $g:X\to Y$ be a one-to-one correspondence and $X$ is finite, then $Y$ is finite too.

Corollary. Let $g:X\longrightarrow Y$ be a one-to-one correspondence. If the set $X$ is finite, then $Y$ is finite.

Proof. Exercise.

Book :"Set Theory With Applications" by You-Feng Lin and Shwu Yeng T.Lin.

My try:

We know we have a one-to-one correspondence as:
$$g:X→Y$$
And also we can define a function as:
$$f:X→X$$

So :$$g∘f∘g^{-1}:Y→Y$$
We have:
$$g∘f∘g^{-1}(Y)=g∘f(X)$$
$$=g(f(X))$$

And because X is not Dedekind infinite so $f(X)=X$ and:
$$g(f(X))=g(X)=Y$$

And by the way:
$$g∘f∘g^{-1}(Y)=Y$$

So Y is finite.

Is that correct?

答:
Well, you don't need repeat the proof of Theorem 2 of your textbook's chapter titled "Denumerable & Nondenumerable sets" which says that "Let $g:X \to Y$ be an one-to-one correspondence. If $X$ is infinite, then $Y$ is infinite".

Suppose you have $g:X\to Y$ an one-to-one correspondence and $X$ is finite. If $Y$ was infinite, then $g^{-1}:Y\to X$ is an one-to-one corrrespondence and as a consequence of Theorem 2, $X$ must be infinite which would contradict our assumption. Hence, $Y$ must be finite.


问:
Independence of a random variable $X$ from itself
In our lecture on probability, my professor made the comment that "a random variable X is not independent from itself." (Here he was specifically talking about discrete random variables.) I asked him why that was true. (My intuition for two counterexamples are $X \equiv 0$ and $X$ s.t. $$m_X(x) = \begin{cases}1, &\text{ if } x = x_0\\ 0, &\text{ if }x \neq x_0.)\end{cases}$$

In these cases, it seems that $\mathbb{P}(X \leq x_1 , X \leq x_2) = \mathbb{P}(X \leq x_1) \cdot \mathbb{P}(X \leq x_2)$.

My professor's response was, "The independence from or dependence of $X$ on itself depends on the definition of the joint distribution function $m_{X,X}$, which is essentially arbitrary."

Can someone help me to understand this?

答:
Suppose X is an event from a sample space S, and suppose that X is independent of itself. This means that P(X/X)=P(X). But by definition,

```<code>           P(X∩X)=P(X)P(X/X)= P(X)P(X)=P(X)^2
</code>```

but P(X∩X)=P(X), since X∩X=X. Therefore, P(X)=P(X/X)=P(X)P(X)=P(X)^2.

This result only can occur when P(X)=1 or P(X)=0. In conclusion, an event X from a sample space S can be independent of itself only if P(X)=1 or P(X)=0.

Body:
Suppose X is an event from a sample space S, and suppose that X is independent of itself. This means that P(X/X)=P(X). But by definition,

           P(X∩X)=P(X)P(X/X)= P(X)P(X)=P(X)^2

but P(X∩X)=P(X), since X∩X=X. Therefore, P(X)=P(X/X)=P(X)P(X)=P(X)^2.

This result only can occur when P(X)=1 or P(X)=0. In conclusion, an event X from a sample space S can be independent of itself only if P(X)=1 or P(X)=0.


答:
Let $A=RU$ be the polar decomposition of $A$, where $R$ is positive semi-definite and $U$ is unitary.

Then $R^2=A^TA=\lambda^2I$; being diagonalizable  this forces $R=|\lambda|I$ and thus $A=|\lambda|U$, a similarity transformation.

Title: Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?


问:
The Question about Euler's Totient Function : How phi(i*p) = p*phi(i)? Here, p is prime and p divides i, phi-> Euler totient function
I have a question about the Euler totient function. I am new to the number theory and i don't know where to start to prove this. If $p$ is prime and $p$ divides $i$,

$$ Φ (i\cdot p) = p \cdot Φ (i),$$

where $Φ (n)$ is Euler Totient Function.

I know as $p$ divides $i$, they($i$ and $p$) will not be co-prime so I can't apply multiplicative property here but I am unable to understand how this result came.

答:
If $p$ is prime and $n\in \Bbb Z^+$ then $\Phi (p^{n+1})=p^n(p-1)=\Phi (p^n)\cdot p.$

Let $0\ne i=jp^n$ where $j\in \Bbb Z$ and $n\in \Bbb Z^+$ and $p\not | \,j.$ Then $j, p^{n+1}$ are co-prime, and $j,p^n$ are also co-prime. So $$\Phi (ip)=\Phi (jp^{n+1})=\Phi (j)\cdot  \Phi (p^{n+1})=\Phi (j)\cdot \Phi (p^n)\cdot p=\Phi (jp^n)\cdot p=\Phi (i)\cdot p.$$

The 2nd equality in the line above is because $j,p^{n+1}$ are co-prime. The 3rd is by my top line. The 4th is because $j,p^n$ are co-prime.

For a proof of the top line, consider that $\{m\in \Bbb Z^+: m\le p^n\land \gcd (m,p^n)>1\}=$ $\{m\in \Bbb Z^+: m\le p^n\land p|\,m\}=$ $\{pm':p^{n-1}\ge m'\in \Bbb Z^+\}$ is a set with $p^{n-1}$ members, so $\Phi (p^n)=p^n-p^{n-1}=p^{n-1}(p-1).$


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
A proof is a completely convincing argument.  Thus, a proof of the Pythagorian theorem would be a completely convincing argument that the Pythagorian relation is correct as stated.  The notion of "proof" is arguably more fundamental than this or that axiom system or system of formal logic. This point of view is that of Errett Bishop.  It is the underlying theme of his book 1967 "Foundations of Constructive Analysis"
(for a review see http://www.ams.org/journals/bull/1970-76-02/S0002-9904-1970-12455-7/home.html as well as http://www.jstor.org/stable/2314383?origin=crossref).</p>

Title: What is a proof?


问:
Units, determinants and irreducibility.
Here is the question I want to answer:

Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ ( the coordinate ring of the unit circle).

Show that $B^* = \mathbb R ^*$.

I know that Since $y$ is integral over $\mathbb R[x]$ there is a ring inclusion $B \rightarrow M_2(\mathbb R[x]),$ called a representation of $B.$ And I found this representation of $B$ because I know that $B = \mathbb R[x] \oplus \mathbb R[x]y$ and multiplication by $y$ is a map of $B$ as an $\mathbb R[x]$-module. And here is this representation $$\phi:B \rightarrow M_{2}(\mathbb R[x]), \phi (a_{1} + a_{2}y) =  \begin{pmatrix}
a_{1} & a_{2}(1-x^2) \\
a_{2} & a_{1}
\end{pmatrix} $$

Then I was thinking about using the determinant to solve my questions and I guess I should work in a way similar to $\mathbb Z \sqrt {3}$ but still I am confused about the details.

Any help will be appreciated.

答:
If $a_1+a_2y$ is a unit, then $\phi(a_1+a_2y)$ must be invertible, so $\det\phi(a_1+a_2y)=a_1^2+a_2^2(x^2-1)$ needs to be a unit of $\mathbb{R}[x]$, i.e., an element of $\mathbb{R}^\times$.  If $a_1a_2\neq 0$, write
\begin{align*}
a_1&=a_{1,m}x^m+a_{1,m-1}x^{m-1}+\dots+a_{1,0}\\
a_2&=a_{2,n}x^n+a_{2,n-1}x^{n-1}+\dots+a_{2,0}
\end{align*}
with $a_{i,j}\in\mathbb{R}$, $a_{1,n}a_{2,m}\neq 0$.  Then $a_1^2+a_2^2(x^2-1)$ has top degree (in $x$) piece
$$
\begin{cases}
a_{1,m}^2x^{2m} & \text{if }m>n+1\\
a_{2,n}^2x^{2n} & \text{if }m<n+1\\
(a_{1,m}^2+a_{2,n}^2)x^{2m} & \text{if }m=n+1
\end{cases}
$$
Can you finish it from here?

Title: Units, determinants and irreducibility.


问:
Finding population median and IQR(x)

I have done the following for the median

$2(x-1) = .50$

results in $x = 1.25$

For the quartiles, I have $q_{1}$--> $2(x-1)=.25$ which results in $q_{1} = 1.125$

for $q_{3}$--> $2(x-1)=.75$ which results in $q_{3}=1.375$ and then subtract them to get $.25$
Please correct me if I am wrong! thank you in advance

答:
An approach using a beta distribution. Let $Y \sim \mathsf{Beta}(2,1).$
Then $X = Y+1.$  Results are the same as in @heropup's Answer (+1).

The quartiles of $Y$ are as follows (using R) are
$1/2$ for the lower quartile, $\sqrt{2}/2$ for the
median, and $\sqrt{3}/2$ for the upper quartile.
The IQR is $(\sqrt{3} - 1)/2.$
Also, the lower
quartile for $X$ is $1.5$ and the IQR for $X$ is
the same as for $Y.$ [In R, qbeta is the quantile function (inverse CDF) of a beta distribution.]

```<code>qbeta(c(1,2,3)/4, 2, 1)
[1] 0.5000000 0.7071068 0.8660254
sqrt(1:3)/2
[1] 0.5000000 0.7071068 0.8660254
</code>```

Simulation: By simulation of a million realizations of $X$ in R, we
can get good approximations to the quartiles, accurate
to about three significant digits--and a figure for
illustration.

```<code>set.seed(223)
x = 1 + rbeta(10^6, 2, 1)
median(x);  IQR(x)
[1] 1.706684    # aprx 1.7071
[1] 0.3666178   # aprx 0.8660
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.001   1.499   1.707   1.666   1.866   2.000

hdr = "Simulated values of 1+BETA(2,1) with Quartiles"
hist(x, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dbeta(x-1, 2, 1), add=T, lwd=3, col="orange")
 abline(v=quantile(x, c(1:3)/4), col="darkgreen", lwd=2)
</code>```

Title: Finding population median and IQR(x)

Body:
An approach using a beta distribution. Let $Y \sim \mathsf{Beta}(2,1).$
Then $X = Y+1.$  Results are the same as in @heropup's Answer (+1).

The quartiles of $Y$ are as follows (using R) are
$1/2$ for the lower quartile, $\sqrt{2}/2$ for the
median, and $\sqrt{3}/2$ for the upper quartile.
The IQR is $(\sqrt{3} - 1)/2.$
Also, the lower
quartile for $X$ is $1.5$ and the IQR for $X$ is
the same as for $Y.$ [In R, qbeta is the quantile function (inverse CDF) of a beta distribution.]

qbeta(c(1,2,3)/4, 2, 1)
[1] 0.5000000 0.7071068 0.8660254
sqrt(1:3)/2
[1] 0.5000000 0.7071068 0.8660254

Simulation: By simulation of a million realizations of $X$ in R, we
can get good approximations to the quartiles, accurate
to about three significant digits--and a figure for
illustration.

set.seed(223)
x = 1 + rbeta(10^6, 2, 1)
median(x);  IQR(x)
[1] 1.706684    # aprx 1.7071
[1] 0.3666178   # aprx 0.8660
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.001   1.499   1.707   1.666   1.866   2.000

hdr = "Simulated values of 1+BETA(2,1) with Quartiles"
hist(x, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dbeta(x-1, 2, 1), add=T, lwd=3, col="orange")
 abline(v=quantile(x, c(1:3)/4), col="darkgreen", lwd=2)


问:
What's wrong in this $\epsilon-\delta$ argument?
In Spivak's Calculus, one exercise asks whether the following is true:

Let $f$ and $g$ be functions such that $f(x) < g(x)$, for all $x$.
Does it follow that $\lim\limits_{x \to a} f(x) < \lim\limits_{x \to
> a} g(x)$?

The previous result holds if the signs $<$ get replaced by $\leq$, but it turns out this is not true in general for strict inequality. However I "proved" it was true. Obviously my argument is wrong, but it is not clear to me where lies the mistake, so I am requesting your help to figure it out.

First I envisioned using one neat trick I found in Terence Tao's blog (second paragraph in item 2), namely that to prove a quantity $x$ vanishes one can prove $\lvert x \rvert \leq \epsilon$, for every $\epsilon > 0$.

So my argument goes as follows: let $f$ and $g$ be functions as in the statement above. Then $\lim\limits_{x \to a} f(x) \leq \lim\limits_{x \to a} g(x)$. We show that equality leads to a contradiction.

If $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = m$ and $\epsilon > 0$, then there are $\delta_1, \delta_2 > 0$ such that

if $0 < \lvert x-a \rvert < \delta_1$, we have $\lvert f(x) - m \rvert < \cfrac{\epsilon}{2}$;

if $0 < \lvert x-a \rvert < \delta_2$, we have $\lvert g(x) - m \rvert < \cfrac{\epsilon}{2}$.

Now, if $0 < \lvert x-a \rvert < \delta$, where $\delta$ equals the smallest number between $\delta_1$ and $\delta_2$, then
$$\lvert g(x) - f(x) \rvert = \lvert g(x) -m + m - f(x) \rvert \leq \lvert g(x) -m \rvert + \lvert m - f(x) \rvert < \cfrac{\epsilon}{2} + \cfrac{\epsilon}{2} = \epsilon$$.

This implies, by Prof. Tao's trick, that $f(x) = g(x)$; this is impossible since $f(x) < g(x)$ for all $x$, so we conclude $\lim\limits_{x \to a} f(x) < \lim\limits_{x \to a} g(x)$.

Where's the error? Thanks in advance.

答: You can conclude that whenever $x$ satisfies $0<|x-a|<\delta$, you have $|g(x)-f(x)|<\varepsilon$. This is not enough to establish that $f(x)=g(x)$. In order to conclude that $f(x)=g(x)$, you'd need to know that for $x$ fixed, $|f(x)-g(x)|<\varepsilon$ for every $\varepsilon>0$. But this might not be true - there's no guarantee in general that for a smaller choice of $\varepsilon$ (say $\varepsilon/2$), the same $\delta$ is still small enough so that $|x-a|<\delta\implies|f(x)-g(x)|<\varepsilon/2$. You might need a smaller $\delta$, say $\delta'$, and it might not be the case that $|x-a|<\delta'$ anymore. (In fact, unless $f$ and $g$ are constant in a deleted neighborhood of $a$, this will never be the case.)


问:
Understanding Cayley-Hamilton Theorem for Modules
I'm trying to understand the Cayley-Hamilton Theorem for modules, here is what I'm having trouble with.

PROPOSITION (Cayley-Hamilton) Suppose $I \subseteq A$ is an ideal and $\phi:M→M$ is  an A-module  homomorphism,  where $M$ is  an $A$-module  generated  by $\{x_1,...,x_n\}$. And suppose $\phi(M) \subseteq IM$. Then $\phi$ satisfies an identity $\phi^n+a_1 \phi^{n−1}+···+a_n = 0 \in End(M)$,with $a_j \in I^j$.

The proof goes as follows.

PROOF. Write $\phi x_i= \sum_j a_{ij} x_j$, and form the $n \times n$ matrix $(\phi Id_n−(a_{ij}))$ over the ring $A[\phi]$.  This matrix acts on $M^n$, with $(\phi Id_n−(a_{ij}))(x_1,...,x_n)^t= (0)^t$. There is the adjugate matrix $(\phi Id_n−(a_{ij}))^*$ with $(\phi Id_n−(a_{ij}))^*(\phi Id_n−(a_{ij}))=det(\phi Id_n−(a_{ij}))Id_n$. It follows that $det(\phi Id_n−(a_{ij}))$ annihilates all $x_j$, hence is the 0 endomorphism of M.  Expanding the determinant shows the coefficient of $\phi^{n-j}$ belongs to $I^j$.

The part I'm having trouble with is the following:

-From the fact that $(\phi Id_n - (a_{ij}))(\vec{x})^t = (\vec{0})^t$ has a non-zero solution $(x_1 \ldots x_n)^t$, can we not conclude $det(\phi Id_n−(a_{ij}))$ is the zero homomorphism?

-How does it follow from $(\phi Id_n−(a_{ij}))^*(\phi Id_n−(a_{ij}))=det(\phi Id_n−(a_{ij}))Id_n$ that $det(\phi Id_n−(a_{ij}))$ annihlates all $x_i$?

-Why is the assumption $\phi(M) \subseteq IM$ neccessary? (The notes I'm reading from sets $\phi \colon M \to IM$, but I believe there's no difference if we set $\phi \colon M \to M$ and $\phi(M) \subseteq IM$.

Thanks for your help!

答:
1/ For your third question: Yes, $\phi \colon M \to IM$ is the same as $\phi \colon M \to M$ and $\phi(M) \subseteq IM$. The reason why we need this assumption is because we want $a_{ij} \in I$ so that the coefficients of $\phi^{n-j}$ belongs to $I^j$ (in the determinant the coefficient of $\phi^{n-j}$ is a product of $j$ elements $a_{ij}$).

2/ For your second question: since $(\phi Id_n−(a_{ij}))(x_1,...,x_n)^t= (0)^t$ if we apply $(\phi Id_n−(a_{ij}))^*$ to the left we will get
$$[det(\phi Id_n−(a_{ij}))Id_n](x_1,...,x_n)^t = [(\phi Id_n−(a_{ij}))^*(\phi Id_n−(a_{ij}))](x_1,...,x_n)^t = (0)^t$$
Note that
$$[det(\phi Id_n−(a_{ij}))Id_n](x_1,...,x_n)^t = det(\phi Id_n−(a_{ij})) \left( Id_n(x_1,...,x_n)^t \right) = det(\phi Id_n−(a_{ij}))(x_1,...,x_n)^t$$
Therefore each entry of the last vector must be $0$, i.e, $det(\phi Id_n−(a_{ij}))x_j =0$ for all $j$.

3/ I'm not fully understand your first question since it was proved in the proof (by using the above).


问:
More complicated ways of solving a problem
Feynman says the following,

we shall first study the equation,
  $$d^2x/dt^2 = -x$$
  [...] We assume that in the meantime the Mathematics Department has brought forth a function
  which, when differentiated twice, is equal to itself with a minus sign. (There are, of
  course, ways of getting at this function in a direct fashion, but they are more complicated > than already knowing what the answer is.) The function is $x= \cos t$.

What are these more complicated fashions and where can I learn more about them?

答:
What are these more complicated fashions?

Incomplete list, partly based on comments:

Solve $x''+x=0$ by forming the characteristic equation $\lambda^2+1=0$, finding $\lambda=\pm i$, then massaging $x=c_1 e^{i t}+c_2e^{-it}$ into the more digestable form $x=a\cos t+b\sin t$

Laplace-transform $x''+x=0$ into $s^2 X+X = as+b$ where $a,b$ come from initial conditions at $t=0$. Solve: $X=\dfrac{as}{s^2+1}+\dfrac{b}{s^2+1}$. Invert the Laplace transform:  $x=a\cos t+b\sin t$.

Look for solution in the form of a power series $x=\sum_{n=0}^\infty a_n t^n$. When plugged into the equation, this will yield $n(n-1)a_n+a_{n-2}=0$ for $n\ge 2$. This will allow you to write down the entire series in terms of the first two coefficiets $a_0,a_1$. (It helps to change the notation by letting $b_n=n!a_n$.)

Rewrite the equation in the integral form $x(t)=a+bt-\int_0^t (t-s)x(s)\,ds$, where $a,b$ correspond to the initial conditions. Then run the Picard iteration starting from $x_0(t)=0$. This gives $x_1(t)=a+bt$, $x_2(t)=a+bt-\frac{a}{2}t^2-\frac{b}{6}t^3$, and so on... the result will be the same power series as in method 3.

and where can I learn more about them?

In any book on ordinary differential equations, such as the free textbook by  Jiří Lebl.


问:
Solve $\int_{0}^{\pi/2} \arccos\left( \frac{\cos(x)}{1+2\cos(x)} \right) \mathrm dx$

$$\int_{0}^{\pi/2} \cos^{-1}\left( \dfrac{\cos(x)}{1+2\cos(x)} \right) \,dx$$

The final answer is: $\dfrac{5\pi^2}{24}$

答:
Here is the method to evaluate this integral.

Source:-  Some very challenging calculus problems by Joseph Breen


问:
Singularities implies divergence of a Taylor series at that point?
Does a power series solution to an ODE of variable coefficients with no singularities converges everywhere?. Taylors series have a radius of convergence at least to the nearest singularity, at this point the series diverges. Can we regard them as the same?</p>

答:
A Taylor series of infinite radius of convergence corresponds to an entire function, i.e. a function analytic on all $\mathbb C$.  A linear differential equation whose coefficients are analytic in a region of the complex plane, and whose leading coefficient is $1$,
has solutions that are analytic in that region.  In particular if the coefficients
are entire and the leading coefficient is $1$, the solutions are entire, and their Taylor series converge everywhere.

You do need that restriction on the leading coefficient: e.g. $(x-p) y'(x) + y(x) = 0$ has solutions $y = c/(x-p)$ that have a pole at $x=p$.

It is not true that the series always diverges at a singularity.  For example, a homogeneous linear differential equation always has at least one entire solution, $0$.


问:
How do I show that $\sum_{1\leq b<n}\{\frac{ab}{n}\}=\frac{n-1}{2}$
Let $a$ be coprime to $n$.

How do I show that $\sum_{1\leq b<n}\{\frac{ab}{n}\}=\frac{n-1}{2}$

where $\{x\}$ denotes the fractional part of $x$.

I'm not sure how to start this, any help would be great!

答:
Since $a$ is coprime to $n$, the equation $ab=c\pmod{n}$ can be inverted to $b=a^{-1}c\pmod{n}$ for any $c=0, \ldots,n-1$. So there is a 1-1 correspondence between $b$ and $c$; also $b=0$ corresponds to $c=0$.

Thus $$\sum_{1\le b<n}\{\frac{ ab}{n}\}=\sum_{1\le c<n}\{\frac{c}{n}\}=\frac{1}{n}\sum_{1\le c<n}c=\frac{1}{n}\frac{(n-1)n}{2}$$


答:
Your $\epsilon-\delta$ argument is correct, but for the trick to work, your $x$ must be fixed. If a fixed non-negative quantity is $\leq \epsilon$ for any arbitrary $\epsilon>0$, then the quantity must be $0$. However, in this case $|g(x)-f(x)|$ is changing as $x$ takes any value in $(0,\delta)$.

You showed $|g(x)-f(x)|<\epsilon$ when $x$ is in the neighborhood $0<|x-a|<\delta$. This just means you can make $g(x)$ and $f(x)$ arbitrarily close when $x$ is not too far away from $a$, which is consistent with $\displaystyle\lim_{x\to a} f(x)=\lim_{x\to a} g(x)$.

Title: What's wrong in this $\epsilon-\delta$ argument?

Body:
In Spivak's Calculus, one exercise asks whether the following is true:

Let $f$ and $g$ be functions such that $f(x) < g(x)$, for all $x$.
Does it follow that $\lim\limits_{x \to a} f(x) < \lim\limits_{x \to
> a} g(x)$?

The previous result holds if the signs $<$ get replaced by $\leq$, but it turns out this is not true in general for strict inequality. However I "proved" it was true. Obviously my argument is wrong, but it is not clear to me where lies the mistake, so I am requesting your help to figure it out.

First I envisioned using one neat trick I found in Terence Tao's blog (second paragraph in item 2), namely that to prove a quantity $x$ vanishes one can prove $\lvert x \rvert \leq \epsilon$, for every $\epsilon > 0$.

So my argument goes as follows: let $f$ and $g$ be functions as in the statement above. Then $\lim\limits_{x \to a} f(x) \leq \lim\limits_{x \to a} g(x)$. We show that equality leads to a contradiction.

If $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = m$ and $\epsilon > 0$, then there are $\delta_1, \delta_2 > 0$ such that

if $0 < \lvert x-a \rvert < \delta_1$, we have $\lvert f(x) - m \rvert < \cfrac{\epsilon}{2}$;

if $0 < \lvert x-a \rvert < \delta_2$, we have $\lvert g(x) - m \rvert < \cfrac{\epsilon}{2}$.

Now, if $0 < \lvert x-a \rvert < \delta$, where $\delta$ equals the smallest number between $\delta_1$ and $\delta_2$, then
$$\lvert g(x) - f(x) \rvert = \lvert g(x) -m + m - f(x) \rvert \leq \lvert g(x) -m \rvert + \lvert m - f(x) \rvert < \cfrac{\epsilon}{2} + \cfrac{\epsilon}{2} = \epsilon$$.

This implies, by Prof. Tao's trick, that $f(x) = g(x)$; this is impossible since $f(x) < g(x)$ for all $x$, so we conclude $\lim\limits_{x \to a} f(x) < \lim\limits_{x \to a} g(x)$.

Where's the error? Thanks in advance.


问:
Possible point of inflection?
Heyy guys! I have been solving multiple kinds of curve sketching questions when it comes using the first derivative and second derivative. I was curious to know why we call the point of inflection a possible point of inflection. Is there a time when the point of inflection may not really be a point of inflection when solving for my second derivative at 0? Thank you in advance!!</p>

答:
You only have an actual point of inflection where the second derivative is zero and changes sign.

It is possible that $f''(x) = 0 $ and there is no point of inflection.

A simple example is $f(x)=x^4$ whose second derivative is always positive. $f''(0)=0$ but $(0,0)$ is not a point of inflection.

Title: Possible point of inflection?


问:
Strassen's Laser Method Technique AND Tensors in matrix multiplication algorithms

I understand the first algorithm presented by Strassen in 1968, for fast matrix multiplication. This was the first improvement to the naive approach of multiplying matrices. Thereafter, he went on to introduce the 'Laser Method' technique. What is this technique exactly?  How do we really go about using this approach?

I have been referring to a lot of papers for this. But, no paper exactly describes the technique appropriately. Some of the papers are referred to are:

http://www.maths.ed.ac.uk/assets/files/pgrexternalfiles/theses/probability/stothers.pdf

I also don't well understand the use of Border rank in the matrix multiplication algorithms (both Strassen's and Winograd's via Arithmetic progressions) Can anyone help me with this as well?

http://www.cs.umd.edu/~gasarch/TOPICS/ramsey/matrixmult.pdf

Thanks!

答: The laser method (introduced in Strassen, Crelle 1987) and border rank and its uses (introduced in Bini, Capovani, Lotti, & Romani, Inform. Proc. Lett. 1979) are very nicely explained in Bläser's graduate survey.</p>


问:
Pointwise convergence of sine series of $x^{-2}$
I was wondering if the sine series of $x^{-2}$ converges pointwise on the open interval $(0,1)$. </p>

<p>What is the most general criterion to ensure pointwise convergence of a Fourier series? </p>

答:
Some results about pointwise convergence: list in Wikipedia, notes on the Dini test. There is no such thing as the most general criterion.

The function $x^{-2}$ on the interval $(0,1)$ does not have a Fourier series, since it is not integrable. So the question of its convergence is moot.


问:
prove sum of elements of subsets are different
The set S contains 18 distinct (but unknown) four-digit positive integers. Linda calculates the sum of the elements in each non-empty subset of S. Prove that it is impossible for all of these sums to be distinct.</p>

答: The number of subsets of $S$ is $2^{|S|}$ where $|S|$ is the number of elements in $S$. Excluding the empty set, Linda calculated $2^{18}-1=262143$ sums. The smallest possible sum is $1000$, which corresponds to the singleton subset $\{1000\}$. The largest possible sum is $9982+9983+\cdots + 9999=179829$, which corresponds to the subset $\{9982,9983,\cdots,9999\}$. All possible values for these sums are between $1000$ and $179829$. There are $178829$ of them, but usually less because you may not be able to reach the minimum $1000$ and maximum $179829$ all the time. But Linda calculated $262143$ sums. By the pigeonhole principle, at least two of these sums are the same.


问:
Prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist.
I need to prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist, and request your help verifying my argument is correct. I've got somewhat decent  at making $\epsilon-\delta$ arguments, but proving a limit does not exist has been a bit challenging actually.

We need to prove $\lim\limits_{x \to \infty} x \sin^2 x \neq \ell$, for every $\ell \in \mathbb{R}$, so take $\ell \in \mathbb{R}$. The result follows if we can find some $\epsilon > 0$ such that for every $N \in \mathbb{R}$, there is some real $x$ such that $x > N$ but $\lvert x \sin^2 x - \ell \rvert \geq \epsilon$.

If $\epsilon =\lvert \ell \rvert + 1$, then for every $N \in \mathbb{R}$ we can find $x_0 > \max(N, \quad 2\lvert \ell \rvert + 1)$ such that $x_0 = \cfrac{\pi}{2} + 2\pi k$, for some big enough $k \in \mathbb{Z}$. For this $x_0$ we have that
$$x_0 \sin^2 x_0 = x_0 \cdot 1 = x_0 > 2\lvert \ell \rvert + 1 \geq 1 + \lvert \ell \rvert + \ell$$.
This implies $x_0 \sin^2 x_0 - \ell = x_0 - \ell > 1 + \lvert \ell \rvert = \epsilon$ and the proof is complete.

Any comments or suggestions for improvement are very appreciated.

答:
Your argument is fine, though as mentioned in the comments you don't address the possibility that the limit is infinite.

Here is a slightly simpler idea. Given any $k>0$, then $x_k=\frac\pi2+2k\pi>k$ and $x_k\sin^2x_k=x_k>k$. Also $z_k=2k\pi>k$ and $z_k\sin^2z_k=0$. So
$$
\lim_{k\to\infty}x_k\sin^2x_k=\infty,
$$
while
$$
\lim_{k\to\infty}z_k\sin^2z_k=0.
$$
This precludes the existence of the limit (finite or infinite), since if it existed all sequences that go to infinity would give the same limit.


问:
Prove or disprove $\frac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)} < (n-1)!$
Is it correct that for any  positive integers $x,n$, that $\frac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)} < (n-1)!$ where lcm is the least common multiple.

I ask because I find this relationship very interesting but I haven't seen it stated in this form.

I have seen a related inequality that:

$$\frac{\text{lcm}(x,x+1, \dots, x+n)}{x} \ge {{x+n}\choose{n}} = \frac{(x+n)!}{x!n!} $$

Which rearranges to this:

$$\frac{n!}{x} \ge \frac{(x+n)!}{(x!)\text{lcm}(x,x+1,\dots,x+n)}$$

Or even closer:

$$n! \ge \frac{(x+n)!}{((x-1)!)\text{lcm}(x,x+1,\dots,x+n)}$$

So that:

$$n! \ge \frac{(x+n)!}{(x!)\text{lcm}(x+1,\dots,x+n)}$$

This appears to me to be a stronger result than the one I am asking about. I am not clear how to derive my result from this stronger result.

On the other hand, I am able to justify my result independently of this equation.  Here's my argument:

(1)  Let $f_n(x) = \dfrac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)}$

(2)  No prime greater than $n-1$ divides $f_n(x)$ since:

Assume that a prime $p>n$ divides $x+c$ and $x+d$ with $0 < c < d \le n$.  It follows that $p | (x+d - x+c) = d - c < n$ which is impossible.

(3) For each prime $p < n$ that divides $f_n(x)$, we can use Legendre's Formula to get this result (since we are dividing by the least common multiple):

$$v_p\left(\frac{(x+n)!}{(x!)\text{lcm}(x+1,\dots,x+n)}\right) = \sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}-1\right\rfloor < v_p((n-1)!) = \sum_{i=1}^{\infty}\left\lfloor\frac{n-1}{p^i}\right\rfloor$$

where $v_p(x)$ is the largest power of $p$ that is less than or equal to $x$

Note: It is based on $n$ instead of $x+n$ because $p^t$ is necessarily less than $n$

Is my reasoning correct?  Is there a straight forward way to derive this result from the first equation?  Is the argument that I present using Legendre's Formula valid?  If valid, can it be improved or simplified?  If not valid, what was my mistake?

答:
The answer from mathlove gives several equality counter-examples to your conjecture. I prove below that changing your conjecture's $\lt$ to $\le$ is sufficient to make it correct.

As for your proof, the left part of your expression in step $(3)$ is

$$v_p\left(\frac{(x+n)!}{(x!)\operatorname{lcm}(x+1,\dots,x+n)}\right) = \sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}-1\right\rfloor \tag{1}\label{eq1A}$$

With $i$ large enough so $p^i \gt n$, then $-1 \lt \frac{n}{p_i} - 1 \lt 0 \implies \left\lfloor\frac{n}{p^i}-1\right\rfloor = -1$. Thus, $\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}-1\right\rfloor = -\infty$. Also, I believe your note of "It is based on $n$ instead of $x+n$ because $p^t$ is necessarily less than $n$" means you tried using

$$\left\lfloor\frac{x+n}{p^{i}}\right\rfloor - \left\lfloor\frac{x}{p^{i}}\right\rfloor = \left\lfloor\frac{n}{p^{i}}\right\rfloor \tag{2}\label{eq2A}$$

However, even with your restriction, this is not always true. For example, with $t = i = 1$, $p = 3$, $n = 4$ and $x = 5$, the left side of \eqref{eq2A} becomes $\left\lfloor\frac{9}{3}\right\rfloor - \left\lfloor\frac{5}{3}\right\rfloor = 3 - 1 = 2$ while the right side becomes $\left\lfloor\frac{4}{3}\right\rfloor = 1$.

In addition, I assume the $-1$ term on the right side in \eqref{eq1A} comes from the $\operatorname{lcm}$ having its exponent of $p$ be the maximum among the prime factorizations of the integers in $[x+1,x+n]$. However, the $-1$ term should only be used up to that exponent, not to $\infty$, since this causes the sum in \eqref{eq1A} to go to $-\infty$, as I explained earlier.

I don't see any direct way to use what you're trying to do. Instead, your conjecture's $\leq$ version can be proven using an equivalent form of your inequality,

$$\begin{equation}\begin{aligned}
\frac{(x + n)!}{(x!)\operatorname{lcm}(x + 1, \, \ldots, \, x + n)} & \le (n-1)! \iff \\
\frac{(x + n)!}{(x!)(n-1)!} & \le \operatorname{lcm}(x + 1, \, \ldots, \, x + n) \iff \\
n\left(\frac{(x + n)!}{(x!)(n!)}\right) & \le \operatorname{lcm}(x + 1, \, \ldots, \, x + n) \iff \\
n\binom{x + n}{n} & \le \operatorname{lcm}(x + 1, \, \dots, \, x + n)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Note this form is related to that in your question Reasoning about least common multiples and their relationship to binomial coefficients, and my answer there is also fairly similar to this proof. First, for somewhat simpler algebra, let

$$b = \binom{x + n}{n}, \; c = \operatorname{lcm}(x + 1, \, \dots, \, x + n) \tag{4}\label{eq4A}$$

Using the prime factorization exponents based on the definition of $\operatorname{lcm}$, and the $p$-adic order function, then for any prime $p$, let

$$d = \nu_{p}(n) \tag{5}\label{eq5A}$$

$$e = \nu_{p}(b) \tag{6}\label{eq6A}$$

$$f = \nu_{p}(c) = \max_{x + 1 \, \le \, i \, \le x + n}(\nu_{p}(i)) \tag{7}\label{eq7A}$$

By proving

$$d + e \le f \tag{8}\label{eq8A}$$

for all primes $p$, then \eqref{eq3A} will be true. There are $2$ basic cases to consider:

Case #$1$: $e = 0$

Since $[x + 1, x + n]$ is a set of $n$ consecutive integers, they represent a complete residue system of integers modulo $n$, so there's a $1 \le i \le n$ where $x + i \equiv 0 \pmod n \implies x + i = kn, \; k \in \mathbb{N}$. Thus, for this $i$ and $k$,

$$d = \nu_p(n) \le \nu_p(kn) = \nu_p(x + i) \le f \tag{9}\label{eq9A}$$

which shows \eqref{eq8A} holds.

Case #$2$: $e > 0$

Note Kummer's theorem states $e$ (as defined in \eqref{eq6A}) is the number of carries when $x$ is added to $n$ in each prime base $p$. Thus, there must be at least one carry in this case.

Define for any integer $y \ge 0$ it's standard base $p$ representation as

$$y = \sum_{i=0}^{\infty}a_i p^{i}, \; 0 \le a_i \le p - 1 \tag{10}\label{eq10A}$$

and the following function to specify the requested digit

$$g(y, i) = a_i \tag{11}\label{eq11A}$$

Thus, from \eqref{eq5A},

$$g(n, i) = 0 \; \forall \; 0 \le i \le d - 1 \tag{12}\label{eq12A}$$

i.e., the lowest $d$ digits of $n$ in base $p$ are $0$. This means none of these digits can be involved in a carry when $x$ is added to $n$, so the first possible such digit is $g(n, d)$. Since there's at least one carry, there's a maximum index involved. Let $h \ge d$ be the index of the digit where the highest carry comes from. This means the next higher index digit is increased by $1$, i.e.,

$$\begin{equation}\begin{aligned}
& g(n + x, h + 1) = g(n, h + 1) + g(x, h + 1) + 1 \implies \\
& g(n + x, h + 1) \ge g(x, h + 1) + 1
\end{aligned}\end{equation}\tag{13}\label{eq13A}$$

Note $g(n, h + 1) + g(x, h + 1) + 1 \lt p$ because it can't require a carry since index $h$ is where the highest index of any carry comes from. Also, \eqref{eq13A} shows there's an integer in $[x + 1, x + n]$ where the coefficient at index $h + 1$ increases by $1$ to $g(n + x, h + 1)$, so all lower coefficients then are $0$, i.e., there's a $1 \le i \le n$ where

$$g(n + i, j) = 0 \; \forall \; 0 \le j \le h \implies \nu_p(n + i) \ge h + 1 \tag{14}\label{eq14A}$$

Using \eqref{eq7A}, this means

$$\nu_p(n + i) \le f \implies h + 1 \le f \tag{15}\label{eq15A}$$

The carries occurred only in the digits with indices between $d$ and $h$, inclusive, so the number of carries, i.e., $e$ (as Kummer's theorem mentioned earlier shows), is less or equal to the number of indices in that range. This gives

$$e \le h - d + 1 \implies d + e \le h + 1 \tag{16}\label{eq16A}$$

Putting \eqref{eq15A} and \eqref{eq16A} together gives \eqref{eq8A}.


答:
That only shows it for that $\epsilon$.  To show it for any other $\epsilon$ you might need another smaller $\delta$.

So for $0 < |x-a| < \delta \implies |f(x) -g(x)| < \epsilon$ for ALL $\epsilon$s you might need that $0 < |x-a| < \delta$ for ALL $\delta$.  But that would mean $0 < |x-a| = 0$.

And it is true that for all $x$ where $0 < |x-a| = 0$ we will have $f(x)= g(x)$ but... there aren't any $x$ where $0 < |x-a| = 0$ so that is only vacuously true.


问:
What is the approach to find the expression of the partial sum of a series?
This question is about computing the limit of a series as $n$ goes to $\infty$.
The series $(x_n)$ is defined as:
$$\sum_{i=0}^n \frac{1}{n^2+i}$$

So the $n$th partial sum will be given by: $\frac{1}{n^2+1} + \frac{1}{n^2+2} + \frac{1}{n^2+3} + ... + \frac{1}{n^2+n}$

So $x_1 = \frac{1}{1^2 + 1}$,
$x_2 = \frac{1}{2^2 + 1} + \frac{1}{2^2 + 2}$,
$x_3 = \frac{1}{3^2 + 1} + \frac{1}{3^2 + 2} + \frac{1}{3^2 + 3}$, and so on.

I have thought about partial fraction decomposition and trying to form a telescoping series, but have been unable to re-write the expression in a way that gives such a series. I have a feeling that the limit of this series is zero, but to show it, I will need to find an expression for the $n$th partial sum. If the above idea is wrong, can I have some hints as to how to approach this problem?

答:
I have a feeling that the limit of this series is zero, but to show it, I will need to find an expression for the  $n$ th partial sum.

Your feeling is right, but you don't need an explicit expression for the sum in order to prove it. Just note that
$$
 0 < \sum_{i=0}^n \frac{1}{n^2+i} < \sum_{i=0}^n \frac{1}{n^2} = \frac {n+1}{n^2}
$$

What happens for $n \to \infty$?

Remark: $\sum_{i=0}^n \frac{1}{n^2+i}$ is not the partial sum of an infinite series because the terms which are added depend on $n$.


问:
Proving that the cross product of two injective functions is bijective.
I've been asked to prove the following property:

If $f$ and $g$ are injective then $F3$ is bijective, where $F3$ is given by $A×B→f(A)×g(B), (x,y)⟼(f(x),g(y))$

Here's what I have so far...

To prove that this is bijective, we must show that it is both injective and surjective:

Injectivity:

We must prove that if $(f(x), g(x)) = (f(y), g(y))$, then $x = y$

Suppose that $(f(x), g(x)) = (f(y), g(y))$

By the equality of ordered pairs, we know that $f(x) = f(y)$ and that $g(x) = g(y) $

$f$ is injective according to the hypothesis and $f(x) = f(y)$, implying that $x = y$

Surjectivity:
???

I'm not too sure how to prove the surjective nature of the cross product and am therefore stuck. I was thinking of finding the inverse in order to show surjectivity, but am not too show how to proceed.

答:
Suppose $f: A \to X$ is injective.  And $g:B\to Y$ is injective.

Note if we define $H:A\times B \to X\times Y$ via $H(a,b) = (f(a),g(b))$ then $H$ is might NOT be surjective.  But that was not the question that was asked.

We aren't defining $H:A\times B \to X \times Y$; we are defining $F3:A\times B \to f(A)\times f(B)$.  And $f(A)\times f(B) $ is not the same codomain as $X \times Y$ would be.   We do have $f(A)\times f(B)\subset X\times Y$ but it could be that that $f(A)\times f(B) \subsetneq X\times Y$, a proper subset.

If $f$ is not surjective then there will be some $x \in X$ where there is no $a\in A$ so that $f(a) =x$.  But then $x \in X$ but $x$ is NOT in $f(A)$. And $f(A)\subsetneq X$.

So if $f$ or $g$ are not both surjective then $f(A)\times f(B) \subsetneq X\times Y$.

Now... the proof that $F3: A\times B \to f(A)\times f(B)$ is downright trivial!

If $(x,y) \in f(A)\times f(B)$ then $x \in f(A)$ and $y \in f(B)$.

But if $x \in f(A) \subset X$ then there is an $\alpha \in A$ so that $f(\alpha) = x$.  And if $y \in f(B)$ then there is a $\beta \in A$ so that $f(\beta) = y$.

And so ..... $F3(\alpha, \beta) = (f(\alpha), f(\beta)) = (x,y)$.

So $F3$ is surjective.


问:
Taylor expansion of $(1+x)^α$ to binomial series – why does the remainder term converge?
For $α ∈ ℝ$ the function $g_α \colon B_1(0) → ℝ, x ↦ (1+x)^α$ is $C^∞$ and $g_α^{(n)}(x) = n! \tbinom{α}{n}(1+x)^{α-n}$, where $\tbinom{α}{n} = \frac{α(α-1)\cdots(α-n+1)}{n!}$ is the generalized binomial coefficient.
I want to show that $g_α$ has a Taylor expansion $g_α (x) = \sum_{k=0}^∞ \tbinom{α}{k}x^k$. I can easily show that the series really converges for $|x| < 1$, but can I use this fact to show that the remainder term
$$R_n(x) = \int_0^x \frac{(x-t)^n}{n!} (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1} dt$$
or in Lagrange form
$$\quad R_n(x) = \frac{x^{n+1}}{(n+1)!} (n+1)! \binom{α}{n+1}(1+ξ)^{α-n-1} = x^{n+1} \binom{α}{n+1}(1+ξ)^{α-n-1}$$
converges to zero for $|x| < 1$?

How else can I show that the remainder term converges to zero for $|x| < 1$?

Progress: Using the Lagrange form, for $x ∈ (-1,1)$ and $n ∈ ℕ$ there is an $h=h(n,x) ∈ (0,1)$ such that:

$$|R_n(x)| = \Big| x^{n+1} \binom{α}{n+1}(1+hx)^{α-n-1} \Big| = \Big| \binom{α}{n+1}\big(\frac{x}{1+hx}\big)^{n+1}(1+ξ)^{α} \Big| $$
If $x > 0$, then $\big|\frac{x}{1+hx}\big| < {\big|\frac{x}{1+x}\big|}< 1$ and so for $q := \big|\frac{x}{1+x}\big|$, since $\sum_{k=0}^∞ \tbinom{α}{n} q^n$ converges:
$$|R_n(x)| ≤ \Big| \binom{α}{n+1} · q^{n+1} ·(1+ξ)^{α} \Big| \overset{n → ∞}{\longrightarrow} 0$$
The problem is still open for $x < 0$.
Mhenni suggested to use Stirling approximation, but I want to do this more elementary and I think it is possible.

答:
Again, I think that I got it. One needs to proceed differently:

Let $x ∈ ℝ$ such that $|x| < 1$.
For $t ∈ ℝ$ between $x$ and zero, one has $0 ≤ |t| ≤ |x| < 1$ as well as $|x-t| < |1+t|$, because:

for $x > 0$ one has $|x-t| = x - t < 1 < 1 + t = |1 + t|$, and

for $x < 0$ one has $|x-t| = t - x < t + 1 = |1 + t|$, since $-x < 1$.

Therefore $\big|\frac{x-t}{1+t}\big|$ assumes a maximum $q<1$ for $t$ on the compact interval between zero and $x$.
Then:
\begin{align*}
 |R_n(x)| &= \Big| \int_0^x \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) dt \Big|\\
 &≤ \Bigg| \int_0^x \Big| \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) \Big| dt \Bigg| \\
 &\overset{(1)}{=} \Bigg| \int_0^x \Big| \frac{(x-t)^n}{(1+t)^n} ·(α-n) · \binom{α}{n} · (1+t)^{α-1} \Big| dt \Bigg|\\
 &\overset{(2)}{≤} n · \binom{α}{n} ·|q|^n · \Bigg| \int_0^x |α/n -1| ·|1+t|^{α-1} dt\Bigg| \\
 &\overset{(3)}{≤} n · \binom{α}{n} · |q|^n · (|α| +1) · C \overset{n → ∞}{\longrightarrow} 0,
\end{align*}
where the convergence follows from $|(n+1)\tbinom{α}{n+1}\big/n\tbinom{α}{n}|\overset{n → ∞}{\longrightarrow} 1$ and $|q| < 1$, and

(1) $g_α^{(n)}(t) = (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1}$ and $\tbinom{α}{n+1} = \tfrac{α-n}{n+1} \tbinom{α}{n}$,

(2) $\big|\frac{x-t}{1+t}\big|^n ≤ |q|^n$, and

(3) $|α/n - 1| ≤ |α| + 1$, and $C := \Big| \int_0^x |1+t|^{α-1} dt\Big|$


问:
How to get the minimum cost for a set of two oranges?
The problem is as follows:

Jenny goes to the groceries supermarket and she notices the following;
In the fruit section, for one kilogram of oranges, six small or four
large ones are obtained, in addition the price of one kilogram of
oranges varies from $4$ usd to $6$ usd. Find the minimum cost for
Jenny to buy a dozen large oranges and two dozen small oranges.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{27 USD}\\
2.&\textrm{28 USD}\\
3.&\textrm{30 USD}\\
4.&\textrm{29 USD}\\
\end{array}$

How exactly should this problem be solved?. My initial thought goes as this. In order to minimize the cost, I would want to get the least price and maximize the number of oranges. But in this case the number is fixed because it mentions that a dozen of the larger ones is required and two dozen of the small ones are needed.

Then I made the following convertion factors:
I'm using $l$ for large oranges and $s$ for small oranges.

$1\,kg=4l=6s$

Since it is requested that it is needed two dozen of the small ones then:

$2\times 12\times\frac{6s}{1\,kg}=144\frac{s}{kg}$

For the bigger ones it is needed just one dozen:

$12\times \frac{4l}{kg}=48\frac{l}{kg}$

Then this means in order to ensure the least prize is to use the small prize.

Thus:

$144\times 4 + 48 \times 6 = 576+288=864\,\textrm{usd}$

But this answer does not appear in the alternatives. What exactly went wrong here and more importantly why? it would help me a lot a wordy explanation or the logic which should be used in this scenario because I'm confused.

Which part did not went right in my strategy?. Can someone help me here?.

答: As you have stated, we have $ 1kg = 4l = 6s$, where $s$ denotes the small oranges, $l$ denotes the large ones. Now to buy two dozen ($2 \cdot 12$) small oranges, we need to buy $24s$, that is $4$kg. Similarly, to obtain $1$ dozen of large oranges $l$, we need to buy $12l$, i.e.  $ 3$kg. If we assume we get the best price, that is $4$ usd per kg, we will pay $7 \cdot 4 = 28$ usd minimum.


答:
For the limit itself, you aleady received a good answer.

We can approximate the partial sums using harmonic numbers since
$$S_n=\sum_{i=0}^n \frac{1}{n^2+i}=H_{n^2+n}-H_{n^2-1}$$ Now, using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120
   p^4}+O\left(\frac{1}{p^6}\right)$$ apply it twice and continue with Taylor series. This could give
$$S_n=\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right)$$ Compute for $n=10$; the exact value is
$$S_{10}=\frac{61961120721796411}{590910221462160600}\approx 0.1048571$$ while the above truncated series gives
$$S_{10}\sim \frac{12583}{120000}\approx 0.1048583$$

Title: What is the approach to find the expression of the partial sum of a series?

Body:
This question is about computing the limit of a series as $n$ goes to $\infty$.
The series $(x_n)$ is defined as:
$$\sum_{i=0}^n \frac{1}{n^2+i}$$

So the $n$th partial sum will be given by: $\frac{1}{n^2+1} + \frac{1}{n^2+2} + \frac{1}{n^2+3} + ... + \frac{1}{n^2+n}$

So $x_1 = \frac{1}{1^2 + 1}$,
$x_2 = \frac{1}{2^2 + 1} + \frac{1}{2^2 + 2}$,
$x_3 = \frac{1}{3^2 + 1} + \frac{1}{3^2 + 2} + \frac{1}{3^2 + 3}$, and so on.

I have thought about partial fraction decomposition and trying to form a telescoping series, but have been unable to re-write the expression in a way that gives such a series. I have a feeling that the limit of this series is zero, but to show it, I will need to find an expression for the $n$th partial sum. If the above idea is wrong, can I have some hints as to how to approach this problem?


问:
Binomial distribution greater than
If $n = 6$ and $p = 0.50$, what is the probability that $x ≥ 1$?

$P(x ≥ 1 | n = 6 \,\text{and}\, p = 0.50) = ?$

Attempt:

$p=0.5$

$n=6$

$x=1 \,\text{or}\, 2$ etc

$(\frac{n!}{x!(n-x)!})(p^x(1-p)^{n-x})$

So I calculated 0.0938 for 1 and 2,3 etc etc but none of the answers were right, what am i missing and need to do to arive at the answer?

Title: Binomial distribution greater than

Body: $$P(X\geq 1)=1-P(X=0)=1-0.5^6=0.984375$$


问:
Tails of Fourier Transformed family of functions
I am reading a thesis where on page 39, Definition 4, $\epsilon$-oscillatory is defined as a property for a family of functions $\{f_{\epsilon}\}_{0<\epsilon<1}$ in $L^2(\mathbb{R}^d)$ to have if $$\lim_{R\to\infty}\limsup_{\epsilon\to 0}\int_{\{\left|\xi\right|>R\}}\frac{1}{\epsilon^d}\left|\widehat{f_{\epsilon}}\left(\frac{\xi}{\epsilon}\right)\right|^2d\xi=0.$$

It is mentioned soon after that for this property to hold it suffices that the weak derivative $\{\epsilon\nabla f_{\epsilon}\}$ is bounded in $L^2$. I cannot connect the two. The most I could do was the following

$$\sup_{\epsilon}||\epsilon\nabla f_{\epsilon}||_{L^2}<C \implies \sup_{\epsilon}||\epsilon\widehat{\nabla f_{\epsilon}}||_{L^2}<C\implies\sup_{\epsilon}||\epsilon\xi \widehat{f_{\epsilon}}||_{L^2}<C$$

$$\implies \sup_{\epsilon}\int_{\mathbb{R}^d}\frac{1}{\epsilon^d}\left|p\widehat{f_{\epsilon}}\left(\frac{p}{\epsilon}\right)\right|^2dp<C$$ after a change of variables $\epsilon\xi=p$. Could someone please help me in figuring out the rest?

答:
In the first displayed equation, the substitution $t=\frac{\xi}\varepsilon$ reduces us to show that
$$\lim_{R\to\infty}\limsup_{\varepsilon\to 0}\int_{\{|t|>\frac R\varepsilon\}}|\widehat{f_\varepsilon}(t)|^2dt=0.$$
Rewriting
$$\int_{\{|t|>\frac R\varepsilon\}}|\widehat{f_\varepsilon}(t)|^2dt=\int_{\{|t|>\frac R\varepsilon\}}|t|^2|\widehat{f_\varepsilon}(t)|^2\frac 1{|t|^2}dt,$$
and and noticing that over the set of integration, $\frac 1{|t|^2}\leqslant\frac{\varepsilon^2}{R^2}$, we get
$$\int_{\{|t|>\frac R\varepsilon\}}|\widehat{f_\varepsilon}(t)|^2dt\leqslant \frac{\varepsilon^2}{R^2}\int_{\Bbb R^d}|t|^2|\widehat{f_\varepsilon}(t)|^2dt\leqslant\frac 1{R^2}\sup_{\varepsilon}\lVert\varepsilon \xi\widehat f_{\varepsilon}\rVert^2.$$ </p>


问:
Given series $\sum a_n,\sum b_n$ converge absolutely show that $\sum c_n$ where $c_n=a_n\cdot b_n$ converges.

Given series $\sum a_n,\sum b_n$ converge absolutely to $a,b$ show that $\sum c_n$ where $c_n=a_n\cdot b_n$ converges absolutely and converges to $a\cdot b$

I believe I can show that $\sum c_n$ converges by comparison since $\sum \vert c_n\vert=\sum \vert a_n\vert\vert b_n\vert\leq \sum \vert a_n\vert\sum\vert b_n\vert$

But my issue is showing the convergence is $a\cdot b$

I want to show that $\lim_{n\to \infty}\sum_{i=0}^n \vert c_n\vert =\lim_{n\to \infty} \sum_{i=0}^n \vert a_i\vert\vert b_i\vert=ab$

But the same thing I used before no longer seems to work because I get $\lim_{n\to \infty} \sum_{i=0}^n\vert a_i\vert\vert b_i\vert\leq \lim_{n\to \infty} \sum_{i=0}^n \vert a_i\vert \sum_{i=0}^n \vert b_i\vert$

Which I don't believe is what I want, I would need to somehow show that $ab$ is also a lower bound for the limit.

答:
$\sum a_nb_n=ab$ is false. Take $a_n=b_n=0$ for $n \geq 2$ and you will easily get a counter-example.

[Any choice of $a_1,a_2,b_1,b_2$ in $(0,\infty)$ will do].


问:
How many subsets $S \subseteq [10]$ are there such that $|S| = 5$ and $S$ contains either $1$ or $2$ or both?
I'm not quite sure on what expression to use. So far, I have $\binom{8}{4} + \binom{8}{4} + \binom{8}{3}$. My thinking with the 8 choose 4 is to choose 4 other numbers from the set, with either the 1 or 2 already in it. Then, the 8 choose 3 is to choose three numbers in a set with both 1 and 2 already in it. Is this the right thinking or is there a better method?

答:
Yes, this is correct. Another way to think about it, which is common whenever there is a requirement like "containing 1 or 2 or both" is to think about which subsets do not satisfy this and use the complement rule.

There are $\binom{10}{5}$ subsets in total, and if a subset does not satisfy "containing 1 or 2 or both", then that subset contains neither 1 nor 2. There are $\binom{8}{5}$ such subests.

Therefore, we get $\binom{10}{5}-\binom{8}{5}$. You'll see that this and your answer both equal 196.


问:
Find all the possible values of $\int_{\gamma} \frac{1}{z^2+1}dz$
a) Let $D = \{z\in\mathbb{C}: z \neq \pm i\}$ and let $\gamma$ be a closed contour in D. Find all the possible values of $\int_{\gamma} \frac{1}{z^2+1}dz$.

b) If $\sigma$ is a contour from  0 to 1,, find all possible values of $\int_{\sigma} \frac{1}{z^2+1}dz$

I am stuck on a) I am not sure which $\gamma$ to choose. (I cannot use the Residue theorem, I did not learn it yet)

but for b) I chose $\sigma(t) = t$ for $t \in [0,1]$ which then when I solved the integral I got $\int_{\sigma} \frac{1}{z^2+1}dz = \pi/4$

答: Note that$$\frac1{z^2+1}=\frac{i/2}{z+i}-\frac{i/2}{z-i}.$$So,$$\oint_\gamma\frac1{z^2+1}\,\mathrm dz=2\pi i\left(\frac i2\operatorname{ind}_\gamma(-i)-\frac i2\operatorname{ind}_\gamma(i)\right).$$Since each of the numbers $\operatorname{ind}_\gamma(-i)$ and $\operatorname{ind}_\gamma(i)$ can be any integer, $\oint_\gamma\frac1{z^2+1}\,\mathrm dz$ can be any integer multiple of $\pi$.


问:
Does all proofs by contrapositive have a direct proof that is similar in length?
I know this may be a question not suitable to ask here or is a duplicate, but I'm just wondering if all proofs by contrapositives can be rephrased into a direct proof, without using the fact that they are logically equivalent?

Are there instances when a proof by contrapositive is not too hard, but direct proof is way more involved or literally impossible to do?

答: Some existence theorems go into that direction: First one assumes that a solution does not exist, then use a contradiction argument using some deep (and non-constructive) theorem, like for instance Brouwer fixed point theorem or Hahn-Banach theorem.</p>


问:
algebra and geometry question
A cylinder has a circumference of $4$, height of $3$, A is directly above B, and the distance from B to C along the circumference is $1$. The shortest distance through the cylinder from point A to C is same as $\displaystyle \sqrt {\frac{M+Nπ^2}{Pπ^2}}$ where M, N and P are all positive integers.I need to find the smallest value of the sum of M, N, P.

I drew the bottom of the cylinder as a circle and drew a square inside it, I also drew a line through the square representing the diameter of the circle.
Then I calculated the diameter so I can find the distance of the straight line BC.

$4/π = 1.273 = \sqrt{2x^2}, x = 0.798$. So that means $AC = \sqrt{3^2 + 0.798^2} = 3.104$

Now the part I'm stuck on is where I have to find the value of M,N, and P. I have no clue what I should do here. Any help would be much appreciated.

答:
Given the circumference is $4$ and the distance from $B$ to $C$ along circumference is $1$, $\angle BOC = 90^0$.

$\displaystyle OB = OC = \frac{2}{\pi}$. So, $\displaystyle BC =  \frac{2 \sqrt2}{\pi}$.

We also know that $AB = 3$ and $AB \perp BC$ (the question could have been written more clearly that $B$ and $C$ are on the circumference at the base of the cylinder and $A$ is above $B$ on the circumference at the top of the cylinder).

$AC^2 = AB^2 + BC^2 = 9 + \displaystyle \frac{8}{\pi^2}$

$AC = \displaystyle \sqrt{\frac{8 + 9 \pi^2}{\pi^2}} $

Hence $M = 8, N = 9, P = 1$ i.e. $M+N+P = 18$.

Title: algebra and geometry question


答: Have a look at $$a_n=\begin{cases}2^{-n}&n\text{ odd}\\0&n\text{ even}\end{cases}$$ and $$b_n=\begin{cases}2^{-n}&n\text { even}\\0&n\text{ odd}\end{cases},$$ which makes $c_n=0$.

Title: Given series $\sum a_n,\sum b_n$ converge absolutely show that $\sum c_n$ where $c_n=a_n\cdot b_n$ converges.

Body:
Given series $\sum a_n,\sum b_n$ converge absolutely to $a,b$ show that $\sum c_n$ where $c_n=a_n\cdot b_n$ converges absolutely and converges to $a\cdot b$

I believe I can show that $\sum c_n$ converges by comparison since $\sum \vert c_n\vert=\sum \vert a_n\vert\vert b_n\vert\leq \sum \vert a_n\vert\sum\vert b_n\vert$

But my issue is showing the convergence is $a\cdot b$

I want to show that $\lim_{n\to \infty}\sum_{i=0}^n \vert c_n\vert =\lim_{n\to \infty} \sum_{i=0}^n \vert a_i\vert\vert b_i\vert=ab$

But the same thing I used before no longer seems to work because I get $\lim_{n\to \infty} \sum_{i=0}^n\vert a_i\vert\vert b_i\vert\leq \lim_{n\to \infty} \sum_{i=0}^n \vert a_i\vert \sum_{i=0}^n \vert b_i\vert$

Which I don't believe is what I want, I would need to somehow show that $ab$ is also a lower bound for the limit.


问:
Changing the angle / curvature of a mesh created using python and parametric equations
Changing the angle / curvature of a mesh created using python and parametric equations.

I can create a Hyperboloid using Python and parametric equations

Python script formula snippet:

```<code>for i in range (0, num_x):
        for j in range(0,num_y):
            # nomalize range

            u = 8*(i/num_x-1/2)
            v = 2*math.pi*(j/(num_y-1)-1/2)

            x = math.sqrt(1+u**2)*math.cos(v)*uval_a/2
            y = math.sqrt(1+u**2)*math.sin(v)*uval_a/2
            z = uval_b*(u)/(2*math.pi/1)
            # uval_a changes the diameter of center hole
            # uval_b changes the height

            vert = (x,y,z)
            verts.append(vert)
</code>```

My goal is to create a mesh which curves left and curves right using Python and Parametric Equations, similar to the image below.

I want to say that the issue is with the U and V variables but I'm not exactly sure how to change these to cause those types of curvatures in the mesh.

答:
This is not Python, but Mathematica... Not what you wanted but it is another example exercise to see how  the parameters are written differently or how their domain limits are set.

```<code>ParametricPlot3D[{(4 - Sqrt[2 + u^2]) Cos[
    u - v], (4 - Sqrt[2 + u^2]) Sin[u - v], (u + v)^2}, {u, -1,
  1}, {v, -2, 2}, Boxed -> False, Axes -> None]
ParametricPlot3D[{Sqrt[1 + u^2] Cos[v], Sqrt[1 + u^2] Sin[v],
  u}, {u, -4, 4}, {v, -Pi, Pi}, Boxed -> False, Axes -> None]
</code>```

Body:
This is not Python, but Mathematica... Not what you wanted but it is another example exercise to see how  the parameters are written differently or how their domain limits are set.

ParametricPlot3D[{(4 - Sqrt[2 + u^2]) Cos[
    u - v], (4 - Sqrt[2 + u^2]) Sin[u - v], (u + v)^2}, {u, -1,
  1}, {v, -2, 2}, Boxed -> False, Axes -> None]
ParametricPlot3D[{Sqrt[1 + u^2] Cos[v], Sqrt[1 + u^2] Sin[v],
  u}, {u, -4, 4}, {v, -Pi, Pi}, Boxed -> False, Axes -> None]


问:
dimension of kernel of linear transformation on vector space of matrix
Let $M_n(\mathbb{C})$ be the set of $n \times n$ matrices thought of as vector space of $n^2$ dimension over $\mathbb{C}$. Let $F_A: M_n(\mathbb{C}) \rightarrow M_n(\mathbb{C})$ be defined by $B \mapsto AB$.  Show that $\ker(F_A)$ for any $A$ has dimension $kn$ for some $k \in \mathbb{N}$

What I have tried that is that since $\ker(F_A)$ is finite dimensional, it is isomorphic to a subspace spanned by some of the canonical basis of $M_n(\Bbb C)$, which I call $B_{ij}$, then I showed that $AB_{ij}=0$ iff $i^{th}$ column of $B$ is 0, which tells me that the $i^{th}$ row of any matrix in the kernel must be free, but then I cannot finish the argument.

答:
First note that if $A$ is invertible, the only $B$ in the kernel is the zero matrix. This is because we can multiply $A^{-1}$ on both sides of $AB=0$ to get $B=0$. In this case, $\ker(F_A)=0$ has dimension $0$. The other extreme is when $A$ is the zero matrix. In this case any $B$ satisfies $AB=0$, and $\ker(F_A)=n^2$.

Now suppose that $A$ has rank $n-k$, that is, $k$ of the columns of $A$ can be written as linear combinations of the other linearly independent $n-k$ columns. We claim that $\ker(F_A)$ has dimension $kn$. Let us denote the columns of $A$ as $a_1,a_2,\dots,a_n$ and the $(i,j)$-entry of $B$ as $b_{ij}$. To find the dimension of the kernel, we solve for $B$, that is, look for restrictions on its entries $b_{ij}$. For every $j$ between $1$ and $n$, by the definition of matrix multiplication, we have
\begin{equation}
\sum_{i=1}^n b_{ij} a_i = 0,
\end{equation}
where the $0$ means the zero column, or more precisely, the $j$-th column of $0=AB$.

The previous sum is a system of $n$ equations with $n$ variables/unknowns, the unknowns being the $b_{ij}$. We write the $k$ dependent columns of $A$ as linear combinations of the other $n-k$ columns. Then the system becomes an overdetermined system with $k$ free variables. We can let the $k$ free variables to have any value, and the other $(n-k)$ variables $b_{ij}$ are determined uniquely after that because once we have chosen values for the free variables the system is $n-k$ by $n-k$ with invertible coefficient matrix. We can do this for every column $j$ of $B$, so the dimension of the kernel is $kn$.

Furthermore, to find a basis for the kernel, let one of the free variables be $1$ and the others $0$, and then solve for the other $b_{ij}$. The resulting matrix is one basis element of the kernel.

For example, if $n=4$ and the last two columns of $A$ are dependent on the first two, we can choose the last two rows of $B$ freely, and the first two rows of $B$ will be determined uniquely once the last two rows are specified.


问:
Algebraic argument for why any $A_5$ in $S_6$ can be extended to an $S_5$ in $S_6$
It is well known that $S_5$ is a subgroup of $S_6$ in a way that acts transitively on $6$ points, a surprising fact related to the outer automorphism of $S_6$. One can see this using the icosahedron: its rotational symmetry group is isomorphic to $A_5$, and acts transitively on the $6$ axes of the icosahedron that go through opposite vertices. One can show that this exotic $A_5$ in $S_6$ extends to an $S_5$ in $S_6$ (see this excellent page by John Baez), using some geometric observations.

Is there a purely algebraic way to see that any $A_5$ including into $S_6$ extends to an inclusion of $S_5$ in $S_6$, that avoids working directly with the icosahedron? If such an argument were slick enough, I might prefer it to the way Baez gives (not that I don't love the icosahedron!)

答: Suppose we're given a transitive action of $A_5$ on a set $X$ of size 6.  Then the point stabilizers have order 10, which forces them to be the usual copies of $D_{10}$ in $A_5$; equivalently, these are the normalizers of the six Sylow 5-subgroups.  But this implies that $X$ is isomorphic as an $A_5$-set to the set $Syl_5(A_5)$ consisting of Sylow 5-subgroups of $A_5$ under conjugation.  Once you know this, it's clear that the $A_5$-action extends to an $S_5$-action:  $A_5$ and $S_5$ have the same Sylow 5-subgroups, so $S_5$ also acts on them by conjugation.


问:
Is $\sup|f(x)|_{\{x \in \Bbb R: |f(x)|\leq a\}}=a?$
Let $f: \Bbb R \to \Bbb R$ be a function. Let $a \geq 0$ and
$$A=\{x \in \Bbb R : |f(x)|\leq a\}$$

Question: Is $\sup_A |f(x)| = a$?

I was just able to prove that $\sup_A |f(x)|\leq a?$, since $|f(x)|\leq a$ for all $x \in A$, but I can't get the reverse inequality.

答: Hint: take $f$ defined as $\forall x: f(x)=0$ and $a=1$.