问:
Last digits of a power of 2

Prove that there exists a power of 2 such that the last 1000 digits in its decimal representation are all 1 and 2.

One fact that I think can be used in this problem: if $2^{n}=\cdots dn$ where $d$ is the digit to the left of $n$, then $2^{dn}=\cdots dn$ (A concept that was used in MMO 1978).

Furthermore I have a feeling that reducing $n$ to its binary or ternary base may be of some help.

If one feels that the question is wrong then please mention why this is never possible.

Thanks!

答:
Let $n = 1000$. Since $\Bbb Z/10^n \Bbb Z = (\Bbb Z/2^n \Bbb Z)\times(\Bbb Z/5^n \Bbb Z)$, a first step is to inquire about the possible values of powers of $2$ modulo $2^n$ and $5^n$.

The first one is easy : If $k\ge n$, then $2^k \equiv 0 \pmod {2^n}$. And with enough luck, we may find a solution with $k \ge 1000$ so let's not think about $k<1000$ for now.

As for the second one, one can show that the order of $2$ is $4.5^{n-1}$ in the multiplicative group $(\Bbb Z/5^n \Bbb Z)^*$, and so $2$ generates that group :

First, we check that $2$ is of order $4$ modulo $5$, and that $2^4 = 3.5 + 1 \neq 1 \pmod {5^2}$.
Since $(2^4)^5 \equiv 3.5^2 + 1 \pmod {5^3} \equiv 1 \pmod {5^2}$, $2$ is of order $20$ modulo $5^2$. And so on, $(2^4)^{5^k} \equiv 3.5^{k+1} + 1 \pmod {5^{k+2}} \equiv 1 \pmod 5^{k+1}$, hence $2$ is of order $4.5^k$ modulo $5^{k+1}$.

This shows that the powers of $2$ modulo $5^n$ are exactly the invertible classes modulo $5^n$, i.e. those not divisible by $5$.

Finally we have to show that there is a class modulo $10^n$ containing only the digits $1$ and $2$ such that it is not divisible by $5$ (easy !), and it is divisible by $2^n$

In fact, we can show that forall $x \in \Bbb Z/2^n \Bbb Z$, there is a unique $y \in \Bbb Z/10^n \Bbb Z$ such that $y$ only uses the digits $1$ and $2$ and $y \equiv x \pmod {2^n}$:

This is true for $n=1$, if $x=1 \pmod {2}$ we pick $y=1 \pmod {10}$ and if $x=0 \pmod {2}$ we pick $y=2 \pmod {10}$.
Suppose $n>1$ and let $x' \in \Bbb Z/2^{n-1}\Bbb Z$. Using the induction hypothesis we have a unique $y' \in \Bbb Z/10^{n-1}\Bbb Z$ such that $y' \equiv x' \pmod {2^{n-1}}$. Above $y'$ there are two allowed classes mod $10^n$, $1y$ and $2y$. Since their difference is $10^{n-1} \equiv 2^{n-1} \pmod 2^n$, they are distinct and they give modulo $2^n$ the two classes above $x'$

Applying this proof we get a number $...1212122112$ with $1000$ digits which is divisible by $2^{1000}$ modulo $10^{1000}$, not divisible by $5$, and hence is a power of $2$ modulo $10^{1000}$.

Title: Last digits of a power of 2


问:
Understanding generalizations of the wave equation.
I firstly apologise for my English which is not my mothertongue.I hope this question is not too physiscs related so that my post fits well in here.

I am trying to understand a generalization of the wave equation. Consider all the unknown functions to be at least of class $C^2$ for every possible input.

I firstly introduce the wave equation for the unknown function $u : \mathbb{R}^{n} \times \mathbb{R}_{\geq 0} \to \mathbb{R}$ and $c \in \mathbb{R}_{>0}$, which in cartesian coordinates reads:
\begin{align}
\partial_{tt} u= c^2 \nabla^2 u = c^2 \sum_{1\leq i \leq n} \partial_{x_{i}x_{i}} u.
\end{align}

Now for ease of notation we consider $n=2$. I am aware that the solution depends heavily on the Initial and Boundary Condition but let us consider a possible simple solution described [in Wikipedia][1] which has many properties like being radially symmetric.

Now we focus on the main question. I consider two generalizations.

For the first, we consider $c_{1} \neq c_{2}$ both in $\mathbb{R}_{>0}$ and consider the following PDE for the function $u: \mathbb{R}^2 \to \mathbb{R}$
\begin{align}
\partial_{tt} u= c_{1}^2 \partial_{x_{1}x_{1}}u +c_{2}^2\partial_{x_{2}x_{2}}u.
\end{align}
My first question is... does this PDE still belongs to the class of "Wave Equations"? My first answer would be yes. We could think of $c_{1}$ as the speed of the propagation in $x_{1}$ and $c_{2}$ in $x_{2}$. I would expect, instead of a radially symmetric solution, an elliptic one. To be more precise the image from Wikipedia would propagate ellipses instead of circles. Is this interpretation correct or at least does it make sense?

The second generalization stems out of the first. Define for two matrices $A,B \in \text{Mat}_{3\times 3}(\mathbb{R})$ their anticommutator $$]A,B[ = AB+BA.$$ Now consider a vector-valued function $u: \mathbb{R}^2 \to \mathbb{R}^3$ and the following PDE \begin{align}
\partial_{tt}u = (A \partial_{x_{1}}+B\partial_{x_{2}})^2u = A^2\partial_{x_{1}x_{1}}u+]A,B[\partial_{x_{1}x_{2}}u +B^2 \partial_{x_{2}x_{2}}u
\end{align}
where $]A,B[ \neq 0$.

Does this vector valued PDE still belongs to the broad class of wave equation (my doubts are related to the presence of a mixed derivative term)? If yes, how can I make sense of $A,B$? If not does anyone have any interpretation or any source that I can read?

Thanks in advance for anyone answering!

Edit: Specified regularity of $u$
[1]: https://en.wikipedia.org/wiki/Wave_equation#/media/File:Spherical_wave2.gif

答:
Take $A$ to be 90 deg rotation around the$z$ axis, and $B=A^{-1}$. Then the first component of the pde is
$$u_{1tt}=-u_{1xx}+2u_{1xy}-u_{1yy}.$$
I don't think you would call this a wave equation. For example any function of $x+y$ is a solution.

There might not be a formal definition of wave equation, but this pde is parabolic. See any text or Wikipedia on classification of pdes. Generally, hyperbolic equations are thought of as wave equations.

ContentLicense: CC BY-SA 4.0


问:
Does this property of Radon-Nykodym derivatives exist
Suppose I have $\pi \ll \lambda$ so that
$$
f = \frac{d \pi}{d\lambda}
$$
and $\lambda \ll \eta$ so that
$$
g = \frac{d \lambda}{d \eta}
$$
Can we say that
$$
\frac{d \pi}{d\lambda} \frac{d \lambda}{d\eta} = \frac{d \pi}{d \eta} \qquad ?
$$
and if so, what can we say about this density? And does it have a name?

答:
I do not think this has a specific name, but it is a commonly known result.

Suppose that $\nu$ is a $\sigma$ -finite signed measure and $\mu, \lambda$ are $\sigma$ -finite measures on $(X, \mathcal{M})$ such that $\nu \ll \mu$ and $\mu \ll \lambda$

If $g \in L^{1}(\nu),$ then $g(d \nu / d \mu) \in L^{1}(\mu)$ and
$$
\int g d \nu=\int g \frac{d \nu}{d \mu} d \mu .
$$

We have $\nu \ll \lambda,$ and
$$
\frac{d \nu}{d \lambda}=\frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} \quad \lambda-a . e
$$

I give you here a proof you can find in Folland's book:

By considering $\nu^{+}$ and $\nu^{-}$ separately, we may assume that $\nu \geq 0$. The equation $\int g d \nu=\int g(d \nu / d \mu) d \mu$ is true when $g=\chi_{E}$ by definition of $d \nu / d \mu .$ It is therefore true for simple functions by linearity, then for nonnegative measurable functions by the monotone convergence theorem, and finally for functions in $L^{1}(\nu)$ by linearity again. Replacing $\nu, \mu$ by $\mu, \lambda$ and setting $g=\chi_{E}(d \nu / d \mu),$ we obtain
$$
\nu(E)=\int_{E} \frac{d \nu}{d \mu} d \mu=\int_{E} \frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} d \lambda
$$
for all $E \in \mathcal{M},$ whence $(d \nu / d \lambda)=(d \nu / d \mu)(d \mu / d \lambda) \lambda$ -a.e.


问:
Expected value of number of ice-cream cones
The father promised his daughter to buy from 2 to 5 ice-cream cones. To determine how many cones he will buy, he tosses a coin 3 times, counts the number of heads and adds 2 to that number. What is the expected value of the number of cones?

My solution:

Let $X = {0, 1, 2, 3}$ - the number of heads. Then $C = {2, 3, 4, 5}$ is the number of cones that the father may buy. $P(X = 0, 1, 2, 3) = {1/8, 1/2, 1/4, 1/8}, P(C = 2, 3, 4, 5) = {1/8, 1/2, 1/4, 1/8}$.

Then $EC = 2 * 1/8 + 3 * 1/2 + 4 * 1/4 + 5 * 1/8 = 1/4 + 3/2 + 1 + 5/8 = 2/8 + 12/8 + 8/8 + 5/8 = 27/8$.

However, I was told that this answer is wrong.

Where is the mistake and how to solve that problem?

答:
The probabilities of getting $0,1,2,3$ are slightly off. The change of getting $1$ or $2$ times head is equal. You can see this by writing out the possibilities:

$$HHH, \ HHT, \ HTH,  \ HTT,  \ THH,  \ THT,  \ TTH,  \ TTT$$

Or by noting that the change of a number of times head should equal the change of number of times tails. If you have 1 times head, you have 2 times tails. So

$$P(X=0,1,2,3)=\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}$$


问:
Approximation of an integral
)
I can't get over a step in my teacher's exercise.

$$I(x) = 2\int_{0}^{1} \frac{y^3 + 4y^2}{y^2+4y+5} dy = 2\int_{0}^{1} y dy  -5\int_{0}^{1} \frac{2y}{y^2+4y+5} dy$$

Probably it's very easy..

Thank you :)

ContentLicense: CC BY-SA 3.0

Title: Approximation of an integral

Body:
Observe that:
$$\begin{align*}
2\int_{0}^{1} \frac{y^3 + 4y^2}{y^2+4y+5} dy
&= 2\int_{0}^{1} \frac{y(y^2+4y+5)−5y}{y^2+4y+5} dy \\
&= 2\int_{0}^{1} \left( \frac{y(y^2+4y+5)}{y^2+4y+5} - \frac{5y}{y^2+4y+5}\right) dy \\
&= 2\int_{0}^{1} \left( y - \frac{5y}{y^2+4y+5}\right) dy \\
&= 2\left(\int_{0}^{1}  y~dy - \int_{0}^{1}\frac{5y}{y^2+4y+5}dy \right) \\
&= 2\int_{0}^{1}  y~dy - 2\int_{0}^{1}\frac{5y}{y^2+4y+5}dy \\
&= 2\int_{0}^{1} y~dy  -5\int_{0}^{1} \frac{2y}{y^2+4y+5} dy
\end{align*}$$


问:
Do the Coefficients in a Power Series need to be constants?
Apologies, this is a really basic question but I can't find a clear answer anywhere.

Suppose there is a function $f:\mathbb{R}\to\mathbb{R}$ that is analytic and you express it as a power series around $x_0$,

$$f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$$

are the coefficients $a_n$ required to be strictly constants? In other words, can we instead express them as functions of x?

I've only seen it written that $a_n$ are real numbers, but $f$ maps to the reals, so I'm unsure about whether or not they can be written as functions of $x$.

答:
Yes, $a_n$ are required to be constant in the definition of analytic functions.

In principle, you could consider functions of the form
$$
f(x)=\sum_{n=0}^\infty a_n(x) \cdot (x-x_0)^n,
$$
if you wished. But there's so much room that you could just choose
$$
a_0(x) := f(x), \quad a_1(x),a_2(x),\ldots := 0,
$$
so this form doesn't tell us much.


问:
Multipliciousness within an inner product space.

Question:

Let $V$ be an inner product space and $v,w\in V$. Prove that $\lvert\langle v,w\rangle\rvert=\lVert v\rVert \lVert w\rVert$ if and only if one of the vectors $v$ or $w$ is a multiple of the other.

Attempt:

Assume the identity holds and $y\neq 0$. Let
$$ a=\frac{\langle x,y\rangle }{\lVert y \rVert^2}, $$
and let
$$ z=x-ay. $$
Now, note that $y$ and $z$ are orthogonal because if $z=x-ay$, then by the definition of $a$ we have
$$a=\frac{\langle x,y\rangle }{\lVert y \rVert^2}=\frac{\langle z+ay, y\rangle }{\langle y,y\rangle }=\frac{\langle z,y\rangle }{\langle y,y\rangle }+a\frac{\langle y,y\rangle }{\langle y,y\rangle }=\frac{\langle z,y\rangle }{\langle y,y\rangle }+a,$$
which means that
$$ \frac{\langle z,y\rangle }{\langle y,y\rangle }=0~\text{or}~ \langle z,y\rangle =0,$$
namely $y$ and $z$ are orthogonal. Furthermore, since $\lvert \langle x, y \rangle \rvert= \lVert x \rVert \lVert y \rVert$, then
$$ \frac{\lvert \langle x, y \rangle \rvert}{\lVert y \rVert} =\lVert x \rVert ~\overset{1/\lVert y \rVert}{\implies}~\frac{\lvert \langle x, y \rangle \rvert}{\lVert y \rVert^2} =\frac{\lVert x \rVert}{\lVert y \rVert} =\lvert a \rvert.$$

Lemma: Let $V$ be an inner product space, and suppose that $x$ and $y$ are orthogonal vectors in $V$. Then $\lVert x+y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2$.

Now, we know that
$$ \lVert x \rVert^2 = \lVert ay+z \rVert^2,$$
but, by the lemma, this means that
$$ \lVert ay+z \rVert^2 = \lVert ay \rVert^2 + \lVert z \rVert^2~~\overset{\text{not sure where to go}}{\dots}$$

Note:

I'm trying to follow along with Friedberg's description on pages 337 and 338 in his Linear Algebra:

EDIT$^1$:

OK, so I think I've got an idea. To complete the proof I need to notice that $\frac{\lVert x \rVert}{\lVert y \rVert} =\lvert a \rvert$, which means that $\lVert x \rVert = \lvert a \rvert\lVert y \rVert$ and more importantly that $\lVert x \rVert^2 = \lvert a \rvert^2\lVert y \rVert^2,$ but this is exactly the $\lVert ay\rVert^2$ from the application of the lemma as $\lVert ay\rVert^2=(\lVert ay\rVert)^2=(\lvert a\rvert\lVert y\rVert)^2=\lvert a\rvert^2\lVert y\rVert^2$. This will give the result.

CONCLUSION:

Here is my end proof:

EDIT$^2$:

I found here in E. B. Vinberg's A Course in Linear Algebra a different flavor of this proof (if anybody happens to care):

答:
What you did and showed in your lengthy post looks fine but rather messy and long. The slickest, simplest proof of this I know is:

Let $\,t\in\Bbb R\,$ be some parameter or, if you will, unknown:

$$0\le||\,x+t\langle x,y\rangle y\,||^2=\langle\;x+t\langle x,y\rangle y\,,\,x+t\langle x,y\rangle y\;\rangle=|\langle x,y\rangle|^2||y||^2\,t^2+2|\,\langle x,y\rangle\,|^2 t+||x||^2$$

Now the beauty: the rightmost expression is a quadratic in $\,t\,$ and since it is always non-negative its discriminant $\;\Delta\;$ is non-positive:

$$\Delta:=4\,|\,\langle x,y\rangle\,|^4-4||x||^2||y||^2|\langle x,y\rangle|^2\le 0\implies |\langle x,y\rangle|\le||x||\,||y||$$

and the above already proves the Cauchy-Schwarz-Buniakovski inequality, but we also get:

$$|\langle x,y\rangle |=||x||\,||y||\implies \color{red}{\left\|\,x-\frac{\langle x,y\rangle}{||y||^2}y\right\|}=||x||^2+2\frac{||x||^2\,||y||^2}{||y||^4}||y||^2-2\frac{||x||^2||y||^2}{||y||^2}=0\implies$$

$$\implies x=\frac{\langle x,y\rangle}{||y^2||}y\implies\;x\,,\,y\;\;\text{are linearly dependent}$$

Note: Where does the above red term comes from? From finding out what the unique root of the upper quadratic in $\,t\,$ is: it is $\;\displaystyle{t=-\frac1{||y||^2}}\;$ and substituting $\;\;\ldots\ldots$


问:
Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$.
Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$.

I am thinking about using $Horner's$ $method$ but don't know how.

答:
Let's $P(x) = x^{2n}-nx^{n+1} +nx^{n-1}-1$, we have

$$P'(x) = 2nx^{2n-1}-n(n+1)x^{n}+(n+1)nx^{n-2}$$

$$P''(x) = 2nx^{2n-1}x^{2n-2}-n^2(n+1)x^{n-1}+(n-2)(n-1)nx^{n-3}$$

As $$P(1) = 0$$
$$P'(1) = 0$$
$$P''(1) = 0$$

Then $P(x)$ has root 1 with multiplicity 3

PS: If the OP wants to prove that $P(x)$ has root 1 with multiplicity exactly 3, it suffices to calculate $P'''(x)$ and prove that $P'''(1)  \ne 0$


Body:
Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$.

I am thinking about using $Horner's$ $method$ but don't know how.

Body:
$$x^{2n}-nx^{n+1} +nx^{n-1}-1$$
$$ x^{2n}-1 +nx^{n-1}-n^{n+1}$$
$$(  x^{2n}-1) + (nx^{n-1}-n^{n+1})$$
$$(x-1)(x^{2n-1}+x^{2n-2}+\dots +x+1) + nx^{n-1}( 1-x^2)$$
$$(x-1)(x^{2n-1}+x^{2n-2}+\dots +x+1) - nx^{n-1}( x+1)(x-1)$$
$$(x-1)(x^{2n-1}+x^{2n-2}+\dots +x+1 - nx^{n-1}( x+1) )$$


问:
How to prove that for any real number $a$, $A+aI_n$ is invertible
Let an $n$-order real square matrix $A$ satisfy $A^{2m}+I_n=0$, where $m$ is a positive integer. Prove that for every real number $a$, the matrix $A+aI_n$ is invertible. I don't know how to start.

答:
Hint: Try multiplying your matrix $A + aI$ by
$$
   \sum_{i=0}^{2m-1} (-1)^{i + 1} a^i A^{2m -1 - i}
$$
and simplify. Can you see why the product is invertible?

(Whenever you see something of the form $I + A$ (or even $aI + A$ as in this case), your mind should immediately jump to the Neumann series.)


问:
About Ultrafilters on Infinite Sets

Blackburn's Modal Logic, Ex $2.5.4$: Let $W$ be an infinite set. Recall that $X\subseteq W$ is co-finite iff $W\setminus X$ is finite.

(a) Prove that the collection of co-finite subsets of $W$ has the finite intersection property.

(b) Show that there are ultrafilters over $W$ that do not contain any finite set.

(c) Prove that an ultrafilter is non-principal if and only if it contains only infinite sets
if and only if it contains all co-finite sets.

(d) Prove that any ultrafilter over $W$ has uncountably many elements.

(a) is already answered here. To prove (b), we can prove (c) and argue that non-principal ultrafilters exist. So I shall try $\color{red}{(c)}$ first. I want to show $$\text{non-principal} \stackrel{1}{\Longleftrightarrow} \text{contains only infinite sets} \stackrel{2}{\Longleftrightarrow} \text{contains all co-finite sets}$$
What is a principal ultrafilter? The principal ultrafilter generated by $w$, is $\{X
\in P(W): w\in X\}$.
The second double-implication is easier to prove, so let's do that first. Keep an ultrafilter $u \subset P(W)$ in mind for the entire proof. Suppose $u$ contains only infinite sets, and there is some co-finite set $A$ not contained in $u$. Since $u$ is an ultrafilter, $W\setminus A \in u$, but $W\setminus A$ is finite, which is a contradiction. Suppose $u$ contains all co-finite sets. If possible, let $u$ contain some finite set $B$. Then, $W\setminus B \notin u$. However, $B$ is finite $\implies W\setminus B$ is co-finite. This is a contradiction. So, the second double implication, i.e. $\stackrel{2}{\Longleftrightarrow}$ is proved.

Now, let us turn to $\stackrel{1}{\Longleftrightarrow}$. Suppose $u$ contains only infinite sets. If possible, let $u$ be principal. Then there exists some $w\in W$ such that $u = \{X
\in P(W): w\in X\}$. $w\in \{w\}$, obviously. So, $\{w\} \in u$, but $\{w\}$ is a finite set. This is a contradiction. Thus, $u$ is non-principal. For the other direction, suppose $u$ is non-principal. Suppose, for a contradiction, that $u$ contains some finite set $C$. So, $W\setminus C$ is infinite. Also, let $C = \{c_1,c_2,\ldots,c_n\}$. Stuck here! I want to show that $C\in u$ makes $u$ a principal filter, somehow, in order to find a contradiction. For that, I must find a $w\in W$ so that $\{w\} \in u$. Correct? Please help me out here.

Update for part $\color{red}{(c)}$: If $u$ is an ultrafilter, we know that $X\cup Y \in u$ if and only if $X\in u$ or $Y\in u$. $C\in u$ means that $\{c_1,c_2,\ldots,c_{n-1}\} \cup \{c_{n}\} \in u$. So, either $\{c_1,c_2,\ldots,c_{n-1}\} \in u$ or $\{c_n\}\in u$. If $\{c_n\} \in u$, we are done. If not, we can recursively apply the same argument to $\{c_1,\ldots,c_{n-1}\}$. At every step, we drop an element of this finite set, and so the process terminates. We will be able to find a singleton $\{c_i\}$ so that $\{c_i\}\in u$. This makes $u$ a principal filter (right?), which is the required contradiction.

Once (c) is proved, I must explicitly show that non-principal ultrafilters exist, in order to prove $\color{red}{(b)}$. Any ideas for this? Perhaps the author wanted something else, since they've put (b) before (c) - but I hope this way works too. Please drop some hints or ideas on how to proceed and conclude (b) from here!

Now, for part $\color{red}{(d)}$: Suppose $u\subset P(W)$ has at most countable number of elements. Then, $u$ is either finite or $u = \{X_1,X_2,...\}$. How do I find a contradiction from here? Looks like I'm stuck again!

Thanks a lot for your help. Let me know if any of your solution ideas are different from mine, I'd be happy to see different approaches. Also, please help me complete my solutions, if possible!

答:
For (b): Do you know that any family of sets with FIP extends to an ultrafilter? If so, (b) follows immediately  from (a). If not, prove this using Zorn's Lemma.

For (c): Show that if $X$ and $Y$ are not in the ultrafilter $U$, then $X\cup Y$ is not in $U$. Thus if $U$ is non-principal, it cannot contain any finite union of singletons (any finite set).

For (d): Hint. A union of two countable sets is countable. The powerset of an infinite set is uncountable.

Title: About Ultrafilters on Infinite Sets


问:
Convergence in a metric spaces and infinite sequences
Let $(M,d)$ be a metric space and $Y \subset M$. Prove that, if $y \in \overline {Y}$, exists a sequence $(y_n)$ with $y_n \in Y$ for all $n$ and $y_n \rightarrow y$. Prove that, if $y_n \in Y'$, the sequence is infinite.

How can we use the definition:

If $y \in \overline {Y}$, for all $\epsilon>0$ we have $B(y;\epsilon)\cap Y \ne \emptyset$.

To solve this? I don't know how i can get a sequence from this. I do think that the infinite sequence will be "easy" to solve, cause i'm the list of exercises where this question is, we have a similar question.

答:
For all $n\in\mathbb{N}$, $B(y, \frac{1}{n})\cap Y$ is nonempty so we can choose some $y_n\in B(y, \frac{1}{n})\cap Y$.

Now show that $y_n\to y$


问:
How to proceed with this complex equation?
I don't have experience with solving these types of equations.

After looking around; I learned that I should try substituting a + bi in for a complex number in these types of equations.

$$w^2 + 3w^* + 2 = 0$$

I found that the real part is $a^2 - b^2 + 3a + 2$

I also found the imaginary part to be $2ab - 3b$

I'm not sure how to proceed.

Note, w* is the complex conjugate.

答:
Just for fun, here's another approach:

$$w^2+3w^*+2=0\implies
\begin{cases}
w^*=-(w^2+2)/3\quad\text{(by solving for }w^*)\\
(w^*)^2+3w+2=0\quad\text{(by conjugating)}
\end{cases}$$

and thus

$${(w^2+2)^2\over9}+3w+2=0$$

or, after expanding, simplifying, and factoring,

$$w^4+4w^2+27w+22=(w+1)(w+2)(w^2-3w+11)=0$$

for which the roots are

$$w=-1,-2,\text{ and }{3\pm\sqrt{-35}\over2}$$


问:
Finding general term of sequence
I need to find the general terms of the sequence, $$\{1.25, 2, 3.25, 5, 7.25, \cdots\}$$ and $$\{0, 6, 24, 60, 120, 210, \cdots\}$$

I can't think of any such pattern that would produce these.

答:
For first one, see the differences, the differences are $0.75,1.25,1.75,2.25,\cdots$ and further their differences are of $0.5$. So I hope, you can write the general term from here.

As @Ethan points out in the comments, mathematicians prefer fractions over decimals, one should convert the sequence in the form $$\dfrac{5}{4},\dfrac{8}{4},\dfrac{13}{4},\dfrac{20}{4},\cdots$$ So it is easy to spot that numerators increase by $3,5,7,9,\cdots$ and denominator is constant so it is easy to write the general term from this way also (in fact more easier to see after converting in fractions).

For second one, it is easy to observe that the terms are $1^3-1,2^3-2,3^3-3,4^3-4,\cdots$.

Title: Finding general term of sequence


问:
Questions about the concept of Structure, Model and Formal Language
When we start to define mathematical logic (specifically, propositional, first order, and second order logic) we start defining the concept of a language. At the begining this is done in a purely syntactical way. So far, so good.

Now, in this formalization we have been using what is called "metalogic" and "metamathematics" (we use sets, funcions, recursion, the process of reasoning, etc.). Then we want to attach the meaning of "truth".

(*)In the language of propositional logic we do this by means of the truth tables and the rules of deduction. In the case of first and second order logic we use the concept of "structure" (An actual set where the funcions and relations interact) and also the rules of deduction with quantifiers. Then it comes the concept of "model" and my problem.

As I understand, a model is a structure in wich a system of axioms, expressed in a given language, are satisfied.

Question: Am I right?

But in this case we have defined the term "structure" in terms of the metamathematical sense. So, when we say that we are making a model, we speak also in terms of metamathematics.

Now let's talk about the language of set theory. We built up this language to be sufficiently powerful to express all mathematics.

The axioms of $\mathsf{ZF}$ are then given. But then, to attach meaning to the expressions in the language, we actually need to suppose that there is a model of $\mathsf{ZF}$. It means we need to consider that there exist a structure that satisfies the axioms of $\mathsf{ZF}$ (of course in order to be consistent).

Assuming that the axioms of $\mathsf{ZF}$ are consistent, then from the language of set theory and the axioms of $\mathsf{ZF}$ we can build up all of the mathematics.

Question: given that practically everything can be constructed inside the Language of Set Theory (with $\mathsf{ZF}$) and we use it as such, why people say that it is a metatheory?.

Question: if everything can be expressed by means  of the Language of Set Theory (with $\mathsf{ZF}$), Meaning most of mathematics we practically use, why to consider a different language? Like for example the language of group theory, the language of arithmetic, etc.

Also, when considering a new language, and a system of axioms, we need to talk about a model that satisfies the axioms(in order to be consistent). But such a model is a structure and people use the axioms of set theory ($\mathsf{ZF}$) to build it up. This makes me think of the next question.

Question: when we say that we are modeling, then are we working outside in the methateory or are we working inside the language of the set theory?

I really appreciate any comment about this topic and also any observation that helps me understand this concepts.

答:
You're mistaken about second-order logic. It is just as a syntactical concept as first-order logic, but the logic itself is not as "nice" as first-order logic, and the idea of a second-order variable already refers, to some extend, to the idea of a set.

Question 3: You're half right. The rules of deduction, are in the language, those are syntactical rules because proofs are syntactical. But a model of a theory is an interpretation to the language, where the axioms are true.

Question 2: We refer to $\sf ZF$ as a metatheory because that is the framework in which we can develop logic (syntax and semantics together, even for "large" languages which include uncountably many symbols) and then we work on other theories inside the universe of set theory.

This is the theory of the universe, it's not the theory of calculus, of arithmetics, or groups. It's the underlying theory. And when we argue about groups, or about arithmetics, or about whatever, we argue in those relevant theories. And sometimes we make arguments about those theories, and an argument about a theory is an argument in the meta-theory (hence the prefix meta).

For example the statement: "Group theory does not prove the statement $\forall x\forall y(x*y=y*x)$" is a statement about the theory of groups, and therefore it is a statement in the meta-theory. Whether or not the meta-theory is $\sf ZF$, or something else, is irrelevant at the moment. But it's a statement about the theory of groups.

Question 3: Because the language of set theory is the meta-language. When we want to talk about the theory of fields we need two operations and two constant symbols. That's the language of fields. The language of set theory is the language of the underlying universe.

This question is the same as asking "If the CPU works with opcodes, why do we need C++, Java, Common Lisp, or Haskell?"

Yes, we can certainly express things with $\in$. But that would completely obscure any possible meaning of anything we would like to write, and you'd have to write incredibly long expressions for pretty much anything.

Question 4: If we agree that $\sf ZF$ is the meta-theory, then the arguments are essentially are in the language of set theory. Of course we don't write the in formal expressions using $\in$, but we can. It's just excruciatingly long, and if you're not using a proof assistant it's also useless.

This is why we have English.


问:
Zeta function ratio
Let's the zeta function is defined as
$$
\zeta(s)= \prod_p (1-p^{-s})^{-1}=\prod_p \frac{1}{(1-p^{-s})}
$$
so for a negative therm should be
$$
\zeta(-s)= \prod_p (1-p^{s})^{-1}=\prod_p \frac{1}{(1-p^{s})}
$$
for the Euler product where $p$ is prime and the product is over all the primes.
I can say that the productorial therm could be rewritten as:
$$
(1-p^{-s})=(1-\frac{1}{p^s})=(\frac{p^s-1}{p^s})=(\frac{1-p^s}{-p^s})
$$
So computing the ration between $\zeta(s)$ and $\zeta(-s)$ will result in

$$
\frac{\zeta(s)}{\zeta(-s)}=\frac{\prod_p (1-p^{s})}{\prod_p (1-p^{-s})}=\prod_p\frac{ (1-p^{s})}{ (1-p^{-s})} =\prod_p\frac{ (1-p^{s}) (-p^s)}{ (1-p^{s})}= \prod_p -p^s
$$

That is, the ratio of the zeta function with positive and negative argument $s$ should be equal to the negated complex power $s$ of all the primes.

This is a weird result, but I cannot understand what's wrong in here and especially the meaning of this result.

Sorry for the newbie (mathematical) question.

答:
The analytic continuation of $\zeta(s)$ Dirichlet series and Euler product is exactly the same idea as for the geometric series $$\frac1{1-z}=\sum_{n\ge 0} z^n$$ which is valid only for $|z|<1$.

Your same reasonning would lead to

$$0=\frac1{1-5}+\frac18\frac{1}{1-1/2}= \sum_{n\ge 0} (5^n+\frac18 (1/2)^n)$$
which is obviously wrong.

Title: Zeta function ratio


问:
General approach to finding number of significant figures in mixed operations
There are ways to doing operations with significant figures, but not really for mixed operations. I have looked all over the Internet, Stack Exchange, and the textbook I use. I'm surprised, but in any case, I would like a general approach for dealing with the precision, uncertainty, and accuracy of a calculation.

Each operation has its own set of rules. See below:

Addition/subtraction - Use the significant figures of the number with the fewest decimal places (digits) for the calculation.
$$1.145 + 1.5 = 2.6$$

Multiplication/division - Use the significant figures of the number with the fewest significant figures for the calculation. $$1.145 × 1.50 = 1.72$$

However, problems arise in these cases:

Trigonometry
$$sin(\fracπ{3.1}) = 0.018$$
Here, I used the significant figures of the number with the fewest significant figures for the calculation. In other words, I took 3.1 (2 sig figs) to calculate 0.018.

Logarithms
$$\log_{10}(0.00002734) = \log_{10}(2.734) + \log_{10}(10^{-5}) = 0.46379 + 0.00001 = 0.4630$$
Here, the significant figures of addition turned out to be how you would do significant figures of multiplication/division, but in general, you calculate by calculating the operation as if it were multiplication/division.

But I’m just generally confused. What if you have a situation as I saw above with the logarithms? Or something like this:

$$\frac{[(0.23 × 5) + (8.521)^5]}3$$

There is not much consensus.

Any clarification would be golden.

答: In such a case, I would try the methods of extreme incertitudes. I would compare the result of the calculations made first with the exact calculation, and also of the calculation made with highest possible or maximum value. Here, it would be by replacing $0.23$ by $0.24$ and $8.521$ by $8.522$. As the first (exact) result is $14974$ and the second is $14982$, I would give the result $(1.497±0.001) 10^4$

OwnerDisplayName: Alex


问:
Vector of measurable functions is a measurable function?
Let $f_1,\dots,f_n:X\rightarrow \mathbb{R}$ be measurable functions, where $X$ is endowed with $\sigma$-algebra $\mathcal{F}$ and $\mathbb{R}$ with Borel $\sigma$-algebra $\mathcal{B}^1$. Is it true that $(f_1,\dots,f_n)$ is measurable if $\mathbb{R}^n$ is also endowed with Borel $\sigma$-algebra $\mathcal{B}^2$?

I can see that it holds for sets in $\mathcal{B}^2$ which are product of sets in $\mathcal{B}^1$, but I couldn't prove it for the remaining ones.

答:
Let $\left(X_{1}, \Sigma_{1}\right)$ and $\left(X_{2}, \Sigma_{2}\right)$ be two measurable spaces. The $\sigma$ -algebra for the corresponding product space $X_{1} \times X_{2}$ is called the product $\sigma$ -algebra and is defined by
$\Sigma_{1} \times \Sigma_{2}=\sigma\left(\left\{B_{1} \times B_{2}: B_{1} \in \Sigma_{1}, B_{2} \in \Sigma_{2}\right\}\right)$

The Borel $\sigma$-algebra for $\mathbb{R}^{n}$ is generated by half-infinite rectangles and by finite rectangles. For example,
$$
\mathcal{B}\left(\mathbb{R}^{n}\right)=\sigma\left(\left\{\left(-\infty, b_{1}\right] \times \cdots \times\left(-\infty, b_{n}\right]: b_{i} \in \mathbb{R}\right\}\right)=\sigma\left(\left\{\left(a_{1}, b_{1}\right] \times \cdots \times\left(a_{n}, b_{n}\right]: a_{i}, b_{i} \in \mathbb{R}\right\}\right)
$$

As a result, you see that if $f_1,\dots,f_n:X\rightarrow \mathbb{R}$ are all measurable functions, then the preimage of a Borel set in $\mathbb{R}^{n}$ is sent in a measurebale set in the sigma algebra of $X$.


问:
List of Local to Global principles
What are some of the local to global principles in different areas of mathematics?</p>

答:
Game theory and dynamic programming

The one-shot deviation principle states that a strategy profile of a finite extensive-form game is a subgame perfect equilibrium if and only if there exist no profitable one-shot deviations for each subgame and every player.

A more general statement is the principle of optimality from dynamic programming, which is that an optimal policy is such that whatever the initial state and initial decision are, the remaining decisions must constitute an optimal policy with regard to the state resulting from the first decision.

Title: List of Local to Global principles


问:
Spatial isomorphism between type I factors
I'm reading a paper from Christopher J. Fewster about split property, and I'm trying to catch one of his assertions.

He asserts that every type I factor $M$ (a concrete von Neumann algebra over $\mathcal{H}$) is spatially isomorphic with $B(\mathcal{K})\otimes \mathbb{1}_{\mathcal{K'}}$. Implemented by a isomorphism $U:\mathcal{H}\to \mathcal{K}\otimes \mathcal{K}'$ such that $UMU^{-1}=B(\mathcal{K})\otimes \mathbb{1}_{\mathcal{K'}}$.

Basically I don't underestand who are $\mathcal{K}$, $\mathcal{K}'$ and what is exactly the isomorphism $U$.

In spite of He give us a reference, I failed trying to understand it because it wasn't the explicit construction there.

Many thanks in advance!

答:
We are assuming that $M$ is not degenerate (i.e., it has the same unit as $B(H)$). Because $M$ is type I, it the sot-closed-span of a family $\{E_{kj}\}_{kj\in S}$ of matrix units. In the decomposition you want, $\dim K=|S|$, and $\dim K'=\operatorname{Tr}(E_{11})$. Let $Z$ be any index set with $|Z|=\operatorname{Tr}(E_{11})$.

Explicitly, form an orthonormal basis $\{e_{k\ell}\}_{k,\in S,\ \ell\in Z}$ for $H$, where $E_{kk}H=\operatorname{span}\{e_{k\ell}\}_{\ell\in Z}$, and $E_{kj}e_{j\ell}=e_{k\ell}$. Such a thing exists because $\operatorname{Tr}(E_{kk})=\operatorname{Tr}(E_{11})$ for all $k$ and $\sum_kE_{kk}=1$. Let $K,K'$ be  Hilbert spaces with $\dim K=|S|$, $\dim K'=|Z|$ and orthonormal bases $\{e_{k}\}_{k\in S}$, $\{f_\ell\}_{\ell\in Z}$ respectively. Define $U:H\to K\otimes K'$ by
$$
Ue_{k\ell}=e_k\otimes f_\ell
$$
and extend by linearity. Because $U$ maps an orthonormal basis to an orthonormal basis, it is a unitary.

Let $\{E_{kj}'\}$ be matrix units for $K$, associated with the basis $\{e_k\}$. Then
\begin{align}
UE_{kj}e_{h\ell}&=\delta_{jh}Ue_{k\ell}=\delta_{jh}\,(e_k\otimes f_\ell),
\end{align}
and
$$
(E_{kj}'\otimes 1_{K'})Ue_{h\ell}
=(E_{kj}'\otimes 1_{K'})(e_h\otimes f_\ell)
=\delta_{jk}\,(e_k\otimes f_\ell).
$$
As we get the equality for all elements in the basis,  $UE_{kj}=(E'_{kj}\otimes 1_{K'})U$.  Since $U$ is a unitary, $UE_{kj}U^*=E'_{kj}\otimes 1_{K'}$. As both on $B(H)$ and $B(K)$ the span of the matrix units is sot-dense, this conjugation by $U$ extends to a $*$-isomorphism, and so
$$
UMU^*=B(K)\otimes 1_{K'}.
$$
If anything of the above is hard to follow, consider the case where
$$
M=\left\{\begin{bmatrix}
a&0&0&b&0&0\\
0&a&0&0&b&0\\
0&0&a&0&0&b\\
c&0&0&d&0&0\\
0&c&0&0&d&0\\
0&0&c&0&0&d\\
\end{bmatrix}:\ a,b,c,d\in\mathbb C\right\}.
$$


问:
Show that $f:= id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R}, \tau_d)$ is not continuous
I've got to show that the function $f:= id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R}, \tau_d)$ is not continuous.

$(\mathbb{R}, \tau), (\mathbb{R}, \tau_d)$ are both topologies. $\tau = \{\emptyset\} \cup \{U \subseteq \mathbb{R}: \mathbb{R}\setminus U \text{ is countable}\}$ and $\tau_d$ is the discrete topology.

$id$ is the identity-map.

Maybe I wrongfully understand the problem but it seems really easy to me.

For $f$ to be continuous, I need to show that $f^{-1}(V)$ is open in $\tau$ for any $V\subseteq \mathbb{R}$.

Let $x\in \mathbb{R}$. Since $\tau_d$ is the discrete topology, $\{x\} \subseteq \mathbb{R}$ is open in $\tau_d$. But $f^{-1}(\{x\})=\{x\} \notin \tau$ since $\mathbb{R}\setminus \{x\}$ is not countable.

Therefore $f$ cannot be continuous.

Am I correct?

答:
You are indeed correct. This is a typical example to show how common functions that in highschool we thought to be obviously continuous, may in fact be not continuous if we change the topologies of the domain and the codomain. It is to show how how continuity is a property that crucially depends on the open sets, besides the function as well.

Note as an immediate corollary, that any function is continuous if we endow the domain $\mathbb{R}$ of the function $f : \mathbb{R} \rightarrow \mathbb{R}$ with the discrete topology.


问:
State which of the following Ito integrals exist, calculate its mean and variance.
State if the Ito integral exist for $\int_0^T sin(B_t)dB_t $ calculate its mean and variance.

After applying Ito's lemma this is what I got:

$B_0 = 0$

$sin(B_t) = \int_0^T cos(B_t)dB_t  - \frac 12 \int_0^T sin(B_t)dt $

I am a bit stuck on how to show its mean and variance.

答:
The function $\sin(\cdot)$ is bounded and hence $\int_0^T E(|\sin(B_t)|^2)dt<\infty$. This implies that the Itô integral is a martingale and hence it has $0$ mean.

For the variance apply the Itô isometry

$$E\left(\big|\int_0^T \sin(B_t)dB_t\big|^2\right)=E\left(\int_0^T \sin(B_t)^2dt\right)$$

Then you can apply Tonelli lemma to change the order of integration, and then calculate

$$E(\sin^2(B_t)).$$

To do so use the Euler formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and using the fact that $B_t$ is centered Gaussian with variance $t$ you can solve this expectation using the Characteristic function (I leave you this computations).
Then you can simply compute the time integral to obtain the variance.


问:
Uniform convergence of Taylor series for arcsin(x)
After proving $$  \ \arcsin{x}=\sum_{n=0}^{\infty} \frac {c_n}{2n+1}{x^{2n+1}} $$ $\forall x\in(-1,1)$ where $c_n=\frac {(2n)!}{2^{2n}(n!)^{2}}$

Since $\arcsin{x}$ continuous on $[-1,1]$, explain why the Taylor series for $\arcsin{x}$ converges uniformly to $\arcsin{x}$ on $[-1,1]$.

The context of this question is; after deriving $E_n(x)=1/N!\int_{0}^{x}f^{n+1}(t)(x-t)^ndt $.

I let $f_n(x)= \frac {c_n}{2n+1}{x^{2n+1}}\leq\frac {c_n}{2n+1}{k^{2n+1}}=M_n$ where $k\geq x,  \lvert{k}\rvert<1$

Then I used ratio test on $M_n$ for $(0,1)$ and Raabe's test for the endpoints. For uniform convergence I used M-test. This process yields the proof of uniform convergence.

But I have found this solution to be rather unsatisfactory, Is there an alternate proof of this fact without using a convergence test, for example, using $E_n(x)$?

答:
It is known that
$$
c_n  \le \frac{1}{{\sqrt {\pi n} }}
$$
for all $n\geq 1$. Hence, for $-1\leq x \leq 1$,
\begin{align*}
\left| {\sum\limits_{n = 0}^\infty  {\frac{{c_n }}{{2n + 1}}x^{2n + 1} } } \right| & \le \sum\limits_{n = 0}^\infty  {\frac{{c_n }}{{2n + 1}}\left| x \right|^{2n + 1} }  \le \sum\limits_{n = 0}^\infty  {\frac{{c_n }}{{2n + 1}}} \\ & \le 1 + \sum\limits_{n = 1}^\infty  {\frac{1}{{\sqrt {\pi n} }}\frac{1}{{2n + 1}}}  \le 1 + \frac{1}{{2\sqrt \pi  }}\sum\limits_{n = 1}^\infty  {\frac{1}{{n^{3/2} }}}  <  + \infty .
\end{align*}
Thus, the series converges absolutely and uniformly for all $-1\leq x \leq 1$.


问:
Second Order ODE without $x$
$$x(t)''= {a\over x(t)}(x(t)')^2$$
with $x(1) = 1$ and $x'(1) = 1$

Second order Ode without a independent variable + IVP - how would I approach this considering the complexity of the general solution and the fact that it is in terms of $x$?

I assumed t=t(x) and then substituted x''(t) with $$ -t''/t'^3 $$ and x'(t). at the end I got $$ t = {A\over(1-a)}x^(1-a)+C $$
But From there I am unsure how to apply the first order IVP!
Can't seem to do Mathjax but its x^(1-a)

答:
$$x''= {a\over x}(x')^2$$
$$\dfrac {x''}{x'}= a \dfrac {x'} {x}$$
$$(\ln x')'=a (\ln x)'$$
Integrate.
$$\ln x'=a \ln x+C$$
Apply the given initial condition:
$$\implies C=0$$
$$\ln x'=a \ln x$$
$$x'=x^a$$
This should be easy to integrate.

Title: Second Order ODE without $x$


答:
Since $x(t)$ is on denominator, we search solutions that do not annulate.

Also since $x(1)=1>0$ then $x(t)>0$ for all $t$ and we can set $x(t)=e^{u(t)}$.

Reporting in the ODE we get to $x''=u''e^u+u'^2e^u=\dfrac a{e^u}u'^2e^{2u}\iff u''=(a-1)u'^2$

For $a<1$ we have:

We can rewrite this $-\dfrac{u''}{u'^2}=1-a$ which integrates into $\dfrac 1{u'}=(1-a)t+c_1$

Then $u=\dfrac{\ln((1-a)t+c_1)}{1-a}+c_2$ and finally adapting the constants for a nicer presentation:

$$x(t)=C\ \sqrt[1-a]{(1-a)(t-t_0)}$$

I omit the details for initial conditions: $t_0=a$ and $C=(1-a)^{2a-2}$ and $x(t)=\left(\dfrac{t-a}{1-a}\right)^{1-a}$

I let you treat the cases $a=1$ and $a>1$ on your own.

Title: Second Order ODE without $x$

Body:
$$x(t)''= {a\over x(t)}(x(t)')^2$$
with $x(1) = 1$ and $x'(1) = 1$

Second order Ode without a independent variable + IVP - how would I approach this considering the complexity of the general solution and the fact that it is in terms of $x$?

I assumed t=t(x) and then substituted x''(t) with $$ -t''/t'^3 $$ and x'(t). at the end I got $$ t = {A\over(1-a)}x^(1-a)+C $$
But From there I am unsure how to apply the first order IVP!
Can't seem to do Mathjax but its x^(1-a)


问:
For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$
From Stephen Abbott's Understanding Analysis 1.2.11:

For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$

My try:

$$\forall a\in \Bbb R, \forall b\in \Bbb R, \exists n \in \Bbb N \space \text{that satisfies}: \space a<b \Rightarrow a+\frac 1n<b$$

$\because a<b $

$\therefore\exists m\in\Bbb R, m>0$ such that $b=a+m$.

So the statement we are proving becomes (substitute $b$ by $a+m$):

$$\forall a\in \Bbb R, m \in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}: a<a+m \Rightarrow a+\frac 1n < a+m$$

Which is (take $a$ from both sides):
$$\forall m\in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}:0<m \Rightarrow \frac 1 n<m$$
We write this as: $$\forall m\in \Bbb R, 0<m, \exists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$

We define set $M = \{m\mid M=\Bbb R \cap m>0\}$

And set $N = \{\frac 1n \mid n \in \Bbb N\}$

$\because\forall q \in \Bbb Z, p\in\Bbb Z, \frac p q\in\Bbb Q$

$\space\space\space n\in\Bbb N\subset\Bbb Z$

$\therefore\frac 1n\in\Bbb Q$

Mention that $m\in\Bbb R$

$\because\Bbb Q\subsetneq\Bbb R$

$\therefore\forall n\in\Bbb N, \exists m \in \Bbb R \Rightarrow \frac 1n \geqslant m$

This statement is equivalent to:
$$\forall m\in \Bbb R, 0<m, \nexists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$

Which disproves the statement, which is wrong - What is wrong with my disproof?

Also, I am learning how to use mathematical notations properly (because I am only in year 11 but anyway). Please tell me if there is any error in my expressions.

Thanks a lot!

答:
I think your proof is OK up to the point I write below. Here's a typical trick in doing these things. To prove that:

$\forall m \in \mathbb{R}$ where $m > 0$, then $\exists n \in \mathbb{N}$ so that $\frac{1}{n}<m$.

You want to fix an arbitrary $m > 0$ and show there exhibit some $n \in \mathbb{N}$  so that $\frac{1}{n} < m$, and think about why that is sufficient! Proving this actually requires the Archimedian property, which says for the reals:

$$
\forall x \in \mathbb{R}, \text{ there is a } n \in \mathbb{N} \text{ so that } n > x
$$
and proving this requires the least upper bound property of reals.

Also, with the archimedian property, you can rather shortly conclude the proof.


问:
Solve equation for variable that's both in and out of exponent.
I'm in the process of working on a spreadsheet for a track coach to calculate ASR. link

I managed to get to the point where, given time, I can solve for distance.

However, I'd like to be able to calculate the reverse as well, without having to find the nearest value on the spreadsheet and then referencing the time.

Here is the (manipulated) formula I came up with for time to distance.

$$d=t(r+a*e^(-0.013*t))$$

Can someone help me solve for t?

I know there's an $ln$ involved somewhere, but I don't know how to deal with a variable that's both inside and outside the exponent.

Wolfram Alpha suggests under 'roots' something that looks like t=, but I don't know if that's what I'm looking for.

(not sure on the linear algebra tag, but I guessed 'algebra' was what I'm looking for when it comes to 'solve this equation for this variable)

答:
There is no closed-form solution to this equation.  Numerical solutions (for particular values of the constants) can be found, e.g. using Newton's method, and there can be series solutions. Thus if the $0.013$ is replaced by $s$, there is a series solution in powers of $s$:

$$ t = \frac{d}{a +r}+\frac{d^{2} a}{\left(a +r \right)^{3}} s + \frac{a d^{3} \left(3 a -r \right)}{2 \left(a +r \right)^{5}} s^{2}+ \frac{d^{4} a \left(16 a^{2}-13 a r +r^{2}\right)}{6 \left(a +r \right)^{7}} s^{3}+ \frac{a d^{5} \left(125 a^{3}-171 a^{2} r +39 a r^{2}-r^{3}\right)}{24 \left(a +r \right)^{9}} s^{4}+ \frac{d^{6} a \left(1296 a^{4}-2551 a^{3} r +1091 a^{2} r^{2}-101 a r^{3}+r^{4}\right)}{120\left(a +r \right)^{11}} s^{5}+\ldots
$$
which should converge to a solution for sufficiently small $s$.


问:
Transformation of a normal distribution
I got these 2 models for hourly wage for 2 periods:

The hourly wage for period 1 is normally distributed with mean $µ$ and variance $σ^2$ so $Y_1 \sim N(\mu,\sigma^2)$.

And the hourly wage for period 2 is given by:
$$Y_2=\alpha+\beta Y_1+U$$ where $Y_1$ and $U$ are independent and $U \sim N(0,v^2)$

I have to find the distribution for $Y_2$.

So I think if $Y_1$ is normal distributed and U is normal distributed then we get that $Y_2 \sim N(E(Y_2),var(Y_2))$. Have I understood that correct?

Now I have to check if $Y_1$ and $Y_2$ are independent. When $Y_1$ is included in the model for $Y_1$ I think that they are not independent, but what will the formally argument be?

答:
Yes correct.

$$Y_2\sim N(\alpha+\beta\mu;\beta^2\sigma^2+v^2)$$

To check independence you can check covariance because in a Gaussian model, Incorrelation and independence are equivalent

$$\mathbb{Cov}[Y_1,Y_2]=\mathbb{E}[Y_1Y_2]-\mathbb{E}[Y_1]\cdot \mathbb{E}[Y_2]=\beta\sigma^2$$

thus they are not independent

Title: Transformation of a normal distribution


问:
How do I proceed with this trigonometric integral $∫\sec^3θ\cot θ\; dθ$
I tried to perform trig sub on the equation below and what I got was
$$∫\sec^3θ\cot θ\; dθ$$
I'm not sure if this is correct and I'm also really not sure how to proceed from here.

$$\int{{\sqrt{\ln^2 x+1}\over{x\ln{x}}}\; dx}$$

My solution

答:
Hint:

$$\int \sec^3t\cdot\cot  t\ dt=\int\dfrac{dt}{\sin t\cos^2t}=\int\dfrac{\sin t\ dt}{(1-\cos^2t)\cos^2t}$$

Set $\cos t=u$  then $$\int\dfrac{du}{(1-u^2)u^2}=\int\dfrac{1-u^2+u^2}{(1-u^2)u^2}du=\int\dfrac{du}{u^2}+\dfrac12\int\dfrac{1-u+1+u}{(1-u^2)}\ du$$


问:
Rotation matrix - rotate a ball around a rotating box
I've a 3D box: center point = (a,b,c), width = w, height = h, depth = d.

the center point isn't the origin.

I have a ball on the box(touch each other), its center and radius.

I can rotate the box(around the X axis but its center STAYS the same..) and I want the ball to stay stuck to the box. so the ball needs to be rotated WITH the box.

the angle of the rotation is 45 degrees.

I tried to do this:

I defined the Rotation Matrix around the X axis:

```<code>mat[3][3]
1,    0   ,    0
0, cos(45), -sin(45)
0, sin(45), cos(45)
</code>```

and multiply it by the ball center vector:

```<code>(ball.Center().m_x , ball.Center().m_y, ball.Center().m_z) * mat
</code>```

so I got:

```<code>Point3D new_center(ball.Center().m_x,
                   ball.Center().m_y*cos(45) + ball.Center().m_z*sin(45),
                   -(ball.Center().m_y)*sin(45) + ball.Center().m_z*cos(45));
ball.Center() = new_center;
</code>```

the ball is really rotated when the box is rotated but too far. How can I fix it?

答:
thanks for trying!
I tried something else and it works:

// correcting the ball center to the origin according the box.Center

```<code>    Point3D ball_center_corrected = ball.Center() - box.Center();
</code>```

// rotation_matrix(of X axis) * ball_center_corrected
   // so the rotation is around the X axis

```<code>    Point3D new_center(ball_center_corrected.m_x,
                       ball_center_corrected.m_y*cos(angle) -
                       ball_center_corrected.m_z*sin(angle),
                       ball_center_corrected.m_y*sin(angle) +
                       ball_center_corrected.m_z*cos(angle));
</code>```

// translate the ball center back around the box

```<code>    ball.Center() = new_center + box.Center();
</code>```

Body:
thanks for trying!
I tried something else and it works:

// correcting the ball center to the origin according the box.Center

    Point3D ball_center_corrected = ball.Center() - box.Center();

// rotation_matrix(of X axis) * ball_center_corrected
   // so the rotation is around the X axis

    Point3D new_center(ball_center_corrected.m_x,
                       ball_center_corrected.m_y*cos(angle) -
                       ball_center_corrected.m_z*sin(angle),
                       ball_center_corrected.m_y*sin(angle) +
                       ball_center_corrected.m_z*cos(angle));

// translate the ball center back around the box

    ball.Center() = new_center + box.Center();


问:
Definite integral of Planck's law
I want to be able to solve the following,
\begin{equation}
   \frac{\int_{0}^{\lambda_0} B(\lambda, T)\,d\lambda} {\int_{0}^{\infty} B(\lambda,T)\,d\lambda}
\end{equation}

where $B(\lambda,T)$ represents Planck's law:
\begin{equation}\label{planck}
   B(\lambda,T) = \frac{2 h c^2}{\lambda^5} \frac{1}{\exp{\frac{hc}{\lambda k_B T}} - 1}
\end{equation}

The integral of the denominator results in the Stefan Boltzmann law; however, how would I be able to compute the definite integral in the numerator?

答:
With CAS help I have:

$$\int_0^{\text{$\lambda $0}} \frac{2 h c^2}{\lambda ^5 \left(\exp \left(\frac{h c}{\lambda  \text{kB}
   T}\right)-1\right)} \, d\lambda =\\\int_0^{\text{$\lambda $0}} \frac{\left(2 h c^2\right) \sum _{j=1}^{\infty }
   \exp \left(-\frac{j (h c)}{\lambda  \text{kB} T}\right)}{\lambda ^5} \, d\lambda =\\\sum _{j=1}^{\infty }
   \int_0^{\text{$\lambda $0}} \frac{\left(2 h c^2\right) \exp \left(-\frac{j (h c)}{\lambda  \text{kB}
   T}\right)}{\lambda ^5} \, d\lambda =\sum _{j=1}^{\infty } \frac{2 e^{-\frac{c h j}{\text{kB} T \text{$\lambda
   $0}}} \text{kB} T \left(c^3 h^3 j^3+3 c^2 h^2 j^2 \text{kB} T \text{$\lambda $0}+6 c h j \text{kB}^2 T^2
   \text{$\lambda $0}^2+6 \text{kB}^3 T^3 \text{$\lambda $0}^3\right)}{c^2 h^3 j^4 \text{$\lambda $0}^3}=-\frac{2
   c \text{kB} T \ln \left(1-e^{-\frac{c h}{\text{kB} T \text{$\lambda $0}}}\right)}{\text{$\lambda
   $0}^3}+\frac{6 \text{kB}^2 T^2 \text{Li}_2\left(e^{-\frac{c h}{\text{kB} T \text{$\lambda $0}}}\right)}{h
   \text{$\lambda $0}^2}+\frac{12 \text{kB}^3 T^3 \text{Li}_3\left(e^{-\frac{c h}{\text{kB} T \text{$\lambda
   $0}}}\right)}{c h^2 \text{$\lambda $0}}+\frac{12 \text{kB}^4 T^4 \text{Li}_4\left(e^{-\frac{c h}{\text{kB} T
   \text{$\lambda $0}}}\right)}{c^2 h^3}$$

Mathematica code:

Integrate[(2 h c^2)/\[Lambda]^5*1/( Exp[(h c)/(\[Lambda] kB T)] - 1), {\[Lambda],  0, \[Lambda]0}] == -(( 2 c kB T Log[1 - E^(-((c h)/(kB T \[Lambda]0)))])/\[Lambda]0^3) + ( 6 kB^2 T^2 PolyLog[2, E^(-((c h)/(kB T \[Lambda]0)))])/( h \[Lambda]0^2) + ( 12 kB^3 T^3 PolyLog[3, E^(-((c h)/(kB T \[Lambda]0)))])/( c h^2 \[Lambda]0) + ( 12 kB^4 T^4 PolyLog[4, E^(-((c h)/(kB T \[Lambda]0)))])/(c^2 h^3)

OwnerDisplayName: GitGoodCodes

Title: Definite integral of Planck's law


答:
Part of an answer, too long for a comment.

Your very last statement is true: for every positive real number $m$  there is some integer such that $1/n < m$.

How you know that statement is true depends on what axioms for the real numbers you start with. One common axiom is that for every real number $r$ there is some integer $M > r$.

I think you are creating problems for yourself by thinking that proper mathematics uses lots of symbols. Words are almost always easier to understand and to write. The idea behind this assertion is that no matter how small $b-a$ is there will be a unit fraction smaller still, since the fractions $1/n$ are smaller and smaller the larger $n$. Just turn that idea into words:

Suppose $a < b$. Then $b-a > 0$ so there is an integer $n$ such that
$1/n < b-a$. Then the result follows from simple algebra.

If you need to go all the way back to the Archimedean axiom

Suppose $a < b$. Then $\frac{1}{b-a} > 0$ so there is an integer $n > \frac{1}{b-a} $  Then the result follows from simple algebra.

Body:
From Stephen Abbott's Understanding Analysis 1.2.11:

For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$

My try:

$$\forall a\in \Bbb R, \forall b\in \Bbb R, \exists n \in \Bbb N \space \text{that satisfies}: \space a<b \Rightarrow a+\frac 1n<b$$

$\because a<b $

$\therefore\exists m\in\Bbb R, m>0$ such that $b=a+m$.

So the statement we are proving becomes (substitute $b$ by $a+m$):

$$\forall a\in \Bbb R, m \in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}: a<a+m \Rightarrow a+\frac 1n < a+m$$

Which is (take $a$ from both sides):
$$\forall m\in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}:0<m \Rightarrow \frac 1 n<m$$
We write this as: $$\forall m\in \Bbb R, 0<m, \exists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$

We define set $M = \{m\mid M=\Bbb R \cap m>0\}$

And set $N = \{\frac 1n \mid n \in \Bbb N\}$

$\because\forall q \in \Bbb Z, p\in\Bbb Z, \frac p q\in\Bbb Q$

$\space\space\space n\in\Bbb N\subset\Bbb Z$

$\therefore\frac 1n\in\Bbb Q$

Mention that $m\in\Bbb R$

$\because\Bbb Q\subsetneq\Bbb R$

$\therefore\forall n\in\Bbb N, \exists m \in \Bbb R \Rightarrow \frac 1n \geqslant m$

This statement is equivalent to:
$$\forall m\in \Bbb R, 0<m, \nexists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$

Which disproves the statement, which is wrong - What is wrong with my disproof?

Also, I am learning how to use mathematical notations properly (because I am only in year 11 but anyway). Please tell me if there is any error in my expressions.

Thanks a lot!


问:
Quick question about a proof a series rearrangement tends to $3$

Find a rearrangement of $\displaystyle{\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}}$ that converges to $3$

Solution: Choose $N_1$ s.t. $$1 + \frac 13 + \frac 15 + \ldots + \frac{1}{2N_1 - 1} > 3$$ Next, take enough of the negative terms so that $$\left(1 + \frac 13 + \frac 15 + \ldots + \frac{1}{2N_1 - 1}\right)- \left(\frac 12 + \frac 14 + \frac 16 + \ldots + \frac{1}{2N_2} \right)< 3$$ Now, choose $N_3$ so that it is the first available integer such that  $$\left(1 + \frac 13 + \frac 15 + \ldots + \frac{1}{2N_1 - 1}\right)- \left(\frac 12 + \frac 14 + \frac 16 + \ldots + \frac{1}{2N_2} \right) + \left(\frac{1}{2N_1 + 1} + \frac{1}{2N_1 + 1} + \ldots + \frac{1}{2N_3 - 1}\right) > 3$$ We then add in just enough negative terms to make the next sum of $N_3 + N_4$ terms less than $3$, and so on indefinitely. If $X$ is the rearrangement, we want the swinging radius (from one side of $3$ to the other) of the partial sums of $X$ tend to $0$. From $\frac{1}{2N_2}$ onwards, $\color{red}{\text{the difference between a partial sum and $3$ will always be less than the absolute}}\\ \color{red}{\text{ value of the last term}}$ $\frac{1}{2N_k - 1}$ or $\frac{1}{2N_k}$ and we know that this last term tends to $0$ as we go further out along the original series

How do they know the $\color{red}{\text{assertion in red above}}$ holds? I am not sure it's obvious. Having hard time seeing this.

edit: sorry, some text sticking out the margins. Don't know how to fix it.

答:
The idea should be that the fractions are decreasing in size, and by design:
$$
\left(1 + \frac 13 + \frac 15 + \ldots + \frac{1}{2N_1 - 1}\right)- \left(\frac 12 + \frac 14 + \frac 16 + \ldots + \frac{1}{2N_2} \right)< 3
$$
is less than $\frac 1{2N_2}$ away from $3$, since $\frac1{2N_2}$ is the exact term that shifted the partial sum from being greater than $3$ to less than $3$.

In the same way, $\frac1{2N_3-1}<\frac1{2N_2}$ will be the exact term that flips the inequality again. And so on.

To put it more succinctly:

Any number $x-\frac 1n<3$ that has $x>3$ will be at most $\frac1n$ away from $3$. The same goes in the opposite direction.


问:
Does a global attractor contain every (forward) invariant set?
In Remark 2.1.1 of this work, it is claimed that "the global attractor of a dynamical system contains every invariant set".

It is not completely transparent what their definition of a "(global) attractor" is, but I would like to understand under which assumptions this claim can be verified.

In order to do so, let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. Say that $A\subseteq\Omega$ is$^1$

forward invariant if $A=\tau(A)$;

backward invariant if $A=\tau^{-1}(A)$;

invariant if $A$ is forward and backward invariant.

Moreover, let $E(A)$ denote the basin of attraction for $A$, i.e. the set of all $x\in\Omega$ for which the orbit $$\operatorname{orb}x:\mathbb N_0\to\Omega\;,\;\;\;n\mapsto\tau^n(x)$$ is eventually in every neighborhood of $A$.

If $Q\subseteq\Omega$, let $A_Q:=\bigcap_{n\in\mathbb N_0}\tau^n(Q)$. It's then trivial to see that if $B\subseteq Q$ is forward invariant, then $$B=\bigcap_{n\in\mathbb N_0}\tau^n(\underbrace B_{\subseteq Q})\subseteq A_Q.\tag1$$ So, every forward invariant subset of $Q$ is contained in $A_Q$.

If $A\subseteq\Omega$ is forward invariant and $U$ is a fundamental neighborhood of $A$$^2$, then it is easy to show that $$\overline A=A_U\tag2.$$ Hence, from the former result, we can immediately conclude that every forward invariant subset of $U$ is contained in $\overline A$.

Now, assuming that this particular $A$ is "global" in the sense that $E(A)=\Omega$, are we somehow able to conclude that even every forward invariant subset of $\Omega$ is contained in $\overline A$?

If this is not possible in general, feel free to add further assumptions (e.g. $\tau$ being a homeomorphism as in the reference or $A$ being closed) or to rephrase the claim with "forward invariant" replaced by "backward invariant" (which might also be what the author meant).

$^1$ Clearly, if $\tau$ is injective, then forward invariant implies backward invariant. And if $\tau$ is surjective, then backward invariant implies forward invariant. In particular, if $\tau$ is bijective, ten forward and backward invariance are equivalent.

$^2$ i.e. for every neighborhood $V$ of $A$, there is a $n_0\in\mathbb N_0$ with $\tau^n(U)\subseteq V$ for all $n\ge n_0$.

答: It's not true at all.  Consider a very simple example: $\Omega = \mathbb R$ with $\tau(x) = x/2$.  $\{0\}$ is a global attractor, but $\{2^n: n \in \mathbb Z\}$ is invariant and not contained in the global attractor.


问:
Show that any linear subspace $U \subset V$ of dimension 2 corresponds to a unique line through the origin in $\Lambda^{2}(V)$

Show that any linear subspace $U \subset V$ of dimension 2 corresponds to a unique line through the origin in $\Lambda^{2}(V)$

I am currently starting to learn about affine geometry and I am actually not so familiar with the wedge product. Any guidance would be much appreciated.

答:
Say $\{e_1,e_2\}$ is a basis of $U$. The subspace with basis $e_1\wedge e_2$ is then a one-dimensional subspace of $\Lambda^2(V)$. This subspace is indeed only depending on $U$ since if you take any other basis of $U$, say the basis $\{u,v\}$, where $u=ae_1+be_2$ and $v=ce_1+de_2$, then
$$
u\wedge v = (ae_1+be_2)\wedge (ce_1+de_2) = (ad-bc)e_1\wedge e_2,
$$
where $(ad-bc)\ne 0$, since $\{u,v\}$ forms a basis of $U$.


问:
Complex line bundles over a Riemann surface can be given a holomorphic line bundle structure
In p.20 of this lecture note (link: http://www.math.ubc.ca/~cautis/math428/notes-bundles.pdf), it is written that every complex line bundle over a Riemann surface can be given a holomorphic line bundle structure. How can we prove this? Is there a reference for a proof of this statement?

P.S. I am also curious that for a Riemann surface $C$ and an integer $n$, does there exist a complex line bundle $L\to C$ such that $c_1[L]=n$?

答:
You have to know some (basic) facts about complex/holomorphic line bundles over complex manifolds. I'll try to be much clear as possible.

(1) The first Chern class is a topological invariant for complex line bundles in the sense that it gives a group homomorphism $c_1:H^1(X,\mathcal{O}^*_X)\to H^2(X,\mathbb{Z})$. So we can think of $H^2(X,\mathbb{Z})$ as the group parametrizing complex line bundles.

(2) Holomorphic line bundles are parametrized by $H^1(X,\mathcal{O}^*_X)$

(3) There exists a long exact sequence of cohomology group, which the relevant part (for our purpose) is the following \begin{equation*}\cdots\to H^1(X,\mathcal{O}_X)\to H^1(X,\mathcal{O}_X^*)\overset{c_1}{\to}H^2(X,\mathbb{Z})\to H^2(X,\mathcal{O}_X)\to\cdots
\end{equation*}

But $H^2(X,\mathcal{O}_X)\cong H^{0,2}_{\bar{\partial}}(X)$ and since we are on a Riemann surface there no exist $(0,2)$-forms and therefore $H^{0,2}_{\bar{\partial}}(X)\cong 0$. In particular from this follows that $c_1$ is surjective and by (1),(2) above that any complex line bundle over a Riemann surface admits a holomorphic structure.

For your second question just note that $H^2(X,\mathbb{Z})\cong\mathbb{Z}$, so you can take your favorite integer $n$ and find a complex line bundle $L$ such that $\int_Xc_1(L)=n$. For example if $n>0$ take $\tilde{L}$ to be a degree $1$ line bundle and define $L:=\tilde{L}\underbrace{\otimes\dots\otimes}_{n \ \text{times}}\tilde{L}$ (note that $\mathbb{Z}$ is generated by $1$ and $-1$ as additive group)


问:
Galois group $\operatorname{Gal}(f/\mathbb{Q})$ of the polynomial $f(x)=(x^2+3)(x^2-2)$

Find the Galois group $\operatorname{Gal}(f/\mathbb{Q})$ of the polynomial $f(x)=(x^2+3)(x^2-2)$.

Any explanations during the demonstration, will be appreciated. Thanks!

答:
Hints for you to understand and/or prove:

(1) If $\,\sigma\in\,\text{Gal}\,(f/\Bbb Q)\,$ ,  it must be that

$$\sigma(\sqrt 2)=\pm\sqrt 2\;,\;\;\sigma(i\sqrt 3)=\pm i\sqrt 3$$

(2) We have that $\,x^2+3\in\Bbb Q(\sqrt 2)[x]\,$ is irreducible (there , not only in $\,\Bbb Q[x]\,$ !)

(3)  We have that $\;\{1\,,\,\sqrt 2\,,\,i\sqrt 3\,,\,i\sqrt 6\}\;$ is a basis for $\;\Bbb Q(\sqrt 2,i\sqrt 3)/\Bbb Q\;$


问:
Step functions in $\Bbb R^n$
Let $\phi(x,y)$ be a real and continuous function in $\mathbb{R^2}$ such that $\phi(0,0)=0$ and let $\alpha$ and $\beta$ be two step functions in $\mathbb{R^n}$. How can I prove that $\phi(\alpha,\beta)$ is a step function in $\mathbb{R^n}$?

答:
Since $\alpha$ and $\beta$ are step functions, there exists $a_i,b_i \in \mathbb{R^n}$, $\alpha_{i} \in \mathbb{R}$ such that :

$$\alpha(x) = \sum_{i \in I} \mathcal{1}_{[a_i,b_i]}(x) \alpha_i$$
for all $x \in \mathbb{R}^n$ where $[a_i, b_{i}] \cap [a_j,b_j] = \emptyset$ for $j \neq i$ and $\cup_{i \in I} [a_i,b_i]=\mathbb{R}^n$ (the family of $[a_i,b_i]$ forms a partition of $\mathbb{R}^n$). The sum is taken over a countable set $I$. Using the same kind of notatations for $\beta$ yields :

$$\beta(x) = \sum_{i} \mathcal{1}_{[c_i,d_i]}(x) \beta_i$$

Now we want to evaluate $\phi((\alpha(x),\beta(x))$ for a fixed $x \in \mathbb{R}^n$. Using our notations we have :

$$\phi((\alpha(x),\beta(x))= \phi(\sum_{i} \mathcal{1}_{[a_i, b_i]}(x) \alpha_i ,\sum_{i} \mathcal{1}_{[c_i,d_{i}]}(x) \beta_i)$$
this expression simplifies a lot since there exists $i_0$ and $j_0$ such that $x \in [a_{i_0},b_{i_0}]$ and $x \in [c_{j_0},d_{j_0}]$, therefore we can write :

\begin{align*}
\phi((\alpha(x),\beta(x)) & = \phi(\mathcal{1}_{[a_{i_0}, b_{i_0}]}(x) \alpha_{i_0},\mathcal{1}_{[c_{j_0},d_{j_0}]}(x) \beta_{j_0})\\
& = \phi(\alpha_{i_0},\beta_{j_0})
\end{align*}
and this expression is the same for all $x \in [a_{i_0},b_{i_0}] \cap [c_{j_0},d_{j_0}]$, which can be written as a new interval $[e_i,f_i]$.

Finally, you have the decomposition :

$$\phi((\alpha(x),\beta(x))= \sum_{i} \phi(\alpha_{i_0},\beta_{j_0}) \mathcal{1}_{[e_i,f_i]}(x) $$
and you still have to prove that the new family $[e_i,f_i]$ is indeed a decomposition of $\mathbb{\mathbb{R}^n}$ which should be easy to conclude.

Title: Step functions in $\Bbb R^n$


问:
Expand the function $f (x)=\sin(\frac{3}{x})$ in Fourier series on the interval $[0,10]$
Could you help me understand this a little better?

It asks me if the function $f (x)$ can be expanded in fourier series on the interval $[0,10]$

It confuses me a little more than anything to understand this part. I would like to understand more than anything what kind of conditions a function must fulfill so that it can be expanded in Fourier series. I know that there are expansions for a period $ [- \pi, \pi] $, $ [- L, L] $ and for an arbitrary period $ [a, b] $.

In this case it can be seen that it is an arbitrary period but it is not completely clear to me what must happen for me to say that a function can be expanded in Fouirer series in said interval. I would like to be able to solve this problem but more than anything I would like you to help me with a generalization about how I can know when a function can be expanded in Fourier series according to the interval that they give me.

I hope you can help me.

答:
There are two problems here:

$f(x)$ is not periodic.

$f(x)$ is undefined at $0$. In fact, $f(x)$, though bounded, does not converge at $0$.

The first problem can be bypassed by choosing some period of length greater than $10$. Then the interval in question would fit within one period, and we don't care about its behavior outside that interval. We could choose to extend $f$ to a full period in any way that was convenient for calculation.

The second problem is more significant. Exactly as the question is stated now, it can be easily answered with "no". Because $f$ is not defined at $0$, we cannot express it as a fourier series on $[0,10]$, which includes $0$.

Even if one chooses not to be so pendantic and allow a representation that fails to match $f(x)$ at isolated points, this still doesn't work.

$f(x)$ is not integrable on $[0,10]$. The integral $\int_\epsilon^{10}$ exists for all $\epsilon > 0$, but the limit as $\epsilon \to 0$ doesn't converge. But Fourier series are always integrable, with the integral over a period being the constant term times the period.

Even interpreting the question to refer to formal Fourier series - that is, not requiring that the series converge at all, much less to $f(x)$ - still leaves the answer as "no". The connection of a formal Fourier series to its function is by the integral formula for its coefficients. But the coefficients of the cosine terms for this function are all undefined, since $f(x)\cos Ax$ is not integrable around $0$ for any constant $A$.


问:
Prove the power series $ \sum_{k=1}^{\infty} \frac{x^{k}}{k(k+1)} $ converges uniformly on $-1 \leq x \leq 1$.
I am trying to prove that the power series
$$
\sum_{k=1}^{\infty} \frac{x^{k}}{k(k+1)}
$$
converges uniformly on $-1 \leq x \leq 1$.

I am planning to use the Weierstrass-M test. This means that if
$$
|u_{k}(x)| \leq m_{k}
$$
for all $x$ in some interval $I$, with $m_{k} \geq 0$, and if then
$$
\sum\limits_{k=1}^{\infty} m_{k} <\infty
$$
converges, then the series
$$
\sum\limits_{k=1}^{\infty} u_{k} (x) = f(x)
$$
converges uniformly.

Now I argue as follows:

We look at the hardest part to show convergence, hence that is at the points where the terms are largest, i.e. at $x=-1$ and $x=1$. Now we take $x = 1$. We then find
$$
\left| \frac{x^{k}}{k(k+1)} \right| \leq \left| \frac{1}{k(k+1)} \right| = \left| \frac{1}{k^{2} + k} \right| \leq \left| \frac{1}{k^{2}} \right| = m_{k} ,
$$
which implies
$$
\sum\limits_{k=1}^{\infty} m_{k} =  \sum_{k=1}^{\infty} \frac{1}{k^{2}}  = \frac{\pi^2}{6}< \infty
$$
and we conclude that indeed the power series converges uniformly.

Now my question is: is my reasoning correct? It feels very obvious and I am not sure if what I wrote is valid.

答:
Your reasoning is correct. The exact value of $\sum_{k=1}^\infty m_k$ is not relevant for the M-test, only that the series is convergent.

You could also choose
$$
\left| \frac{x^{k}}{k(k+1)} \right| \leq  \frac{1}{k(k+1)} = m_k
$$
for the comparison, that is a telescoping series:
$$
\sum_{k=1}^\infty m_k = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+1}\right) = 1
$$


问:
From general parametrization to parametrisation as function
Let $\gamma$  be a clesed curve in $\mathbb{R^2}$ given by
$$\begin{array}{ccl}
                              \gamma : [a,b] & \longrightarrow &\mathbb{R^2} \\
                              t    & \longmapsto & \gamma(t)=(x(t),y(t)).
                            \end{array}$$
locally we can parametrise the curve $\gamma$ as a graph of function $(t,f(t))$

My question is how we can express $f(t)$ in terms of $x(t)$ and $y(t)$ ?

答:
If $x(t)$ is invertible on some interval $I\subset[a,b]$, then for all $t\in x(I)$ the point $(t,y(x^{-1}(t)))$ is on $\gamma$ and all points of $\gamma$ where $x(t)\in x(I)$ can be written in this form. Thus for $t\in x(I)$ one should define:
$$f(t)=y(x^{-1}(t)).$$


答:
The splitting field of $f$ is $\mathbb{Q}(\sqrt{2}, i\sqrt{3})$ and since
$$[\mathbb{Q}(\sqrt{2}, i\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}, i\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2\cdot 2=4$$
the Galois group has four element.
Since every $\sigma\in$ Gal$(\mathbb{Q}(\sqrt{2}, i\sqrt{3})/\mathbb{Q})$ must take $\sqrt{2}$ and $i\sqrt{3}$ to another root of Irr$(\sqrt{2}, \mathbb{Q})=x^2-2$ and Irr($i\sqrt{3}, \mathbb{Q})=x^2+3$ respectively, we see that $\sigma(\sqrt{2})=\pm \sqrt{2}$ and $\sigma(i\sqrt{3})=\pm i\sqrt{3}$. Thus every element in Gal$(\mathbb{Q}(\sqrt{2}, i\sqrt{3})/\mathbb{Q})$ has order $1$ or $2$. As the only groups of order $4$ are $\mathbb{Z}_4$ and $V_4$ and our group is not cyclic, it must be that the Galois group of $f$ over $\mathbb{Q}$ is $V_4$.</p>

Body:
Find the Galois group $\operatorname{Gal}(f/\mathbb{Q})$ of the polynomial $f(x)=(x^2+3)(x^2-2)$.

Any explanations during the demonstration, will be appreciated. Thanks!


问:
A minimal maximizing set for a Banach algebra
Let $A$ be a commutative unital Banach Algebra. Let $F$ be a weak*-closed subset of $\Omega(A)$ with the property that for $A$ if for all $x \in A$, $\sup\{
|\hat{x}(\phi)| : \phi \in F\} = \rho(x)$, where $\Omega(A)$ is the set of characters in $A$, $\hat{x} $ is the Gelfand Transform, and $\rho$ is the spectral radius.

Show that $A$ admits a minimal set with respect to inclusion amongst all sets with properties of $F$ for $A$. If $A$ is the disk algebra, what is the minimal set as above?

Things I know:

Since $A$ is a commutative unital Banach Algebra, $\rho(x) = \| \hat{x}\|$.

If $A$ is the disk algebra then the Gelfand transform is the identity map.

Otherwise I have no clue on how to construct the minimal set.

Any help will be appreciated!

答:
To see that a minimal $F$ exists, we can use Zorn's Lemma. We consider the set of all weak$^*$-closed $F\subset \Omega(A)$ (the unit ball of $A^*$) with reverse inclusion as order. If $\{F_j\}$ is a decreasing chain of sets satisfying
$$
\rho(x)=\sup\{|\hat x(\varphi)|:\ \varphi\in F_j\},\qquad x\in A,
$$
then it has the finite-intersection property. As these sets are weak$^*$-compact,  they have the finite intersection property, so $\displaystyle\bigcap_jF_j\ne\emptyset$. So the chain has an upper bound. By Zorn's Lemma, a minimal $F$ exists.

Knowing that the characters are the point evaluations, and since every function in the disk algebra achieves its extremes on the boundary, the obvious candidate for the minimal set is the boundary.

For every $f\in A(\mathbb D)$, we have that $\sigma(f)=f(\overline{\mathbb D})$. So
$$
\rho(f)=\max\{|f(z)|:\ z\in\overline{\mathbb D}\}=\max\{|f(z)|:\ |z|=1\}.
$$
The second equality shows that the unit circle $\mathbb T$ satisfies the "F-condition". It remains to show that any proper subset of $\mathbb T$ does not satisfy the condition.

Let $S\subsetneq\mathbb T$ be closed. So there exists $z_0$ with $|z_0|=1$ and $z_0\not\in S$. Let $g:\mathbb T\to\mathbb C$ be given by $g(z)=z_0-z$. Then $\rho(g)=2$. Write $z_0=e^{i\gamma}$. For any $s=e^{i\theta}\in S$,
$$
|g(s)|=|e^{i\gamma}-e^{i\theta}|=|1-e^{i(\theta-\gamma)}|=2(1-\cos(\theta-\gamma)).
$$
Since $z_0\not\in S$, we get that $|g(s)|<2$ for all $s\in S$. As $S$ is compact, $$\sup\{|g(s)|:\ s\in S\}<2,$$ and so $S$ fails the $F$ property. So indeed $\mathbb T$ is the minimal set.


问:
Nearest semi-orthogonal matrix using the entry-wise $ {\ell}_{1} $ norm
Given an $m \times n$ matrix $M$ ($m \geq n$), the nearest semi-orthogonal matrix problem in $m \times n$ matrix $R$ is

$$\begin{array}{ll} \text{minimize} & \| M - R \|_F\\ \text{subject to} & R^T R = I_n\end{array}$$

A solution can be found by using Lagrangian or polar decomposition, and is known to be

$$\hat{R} := M(M^TM)^{-1/2}$$

If $\|\cdot\|_F$ is replaced by the entry-wise $1$-norm

$$\|A\|_1 := \|\operatorname{vec}(A)\|_1 = \sum_{i,j} |A_{i,j}|$$

the problem becomes

$$\boxed{\begin{array}{ll} \text{minimize} & \| M - R \|_1\\ \text{subject to} & R^T R = I_n\end{array}}$$

What do we know about the solutions in this case? Is $\hat{R}$ still a solution? If the solution is something else, do analytic forms or approximations exist? Any insight or direction to literature is appreciated.

答:
Reformulate the problem by constructing the $R$ matrix in terms of an unconstrained matrix $U$ such that the constraint is always satisfied:
$$R = U(U^TU)^{-1/2}\quad\implies\quad R^TR=I$$
Then optimize with respect to the new unconstrained variable $U.\;$ Since $R$ is independent of $\,\|U\|\,$ controlling the growth of $\,\|U\|\,$ (via normalization) will be important for numerical algorithms.

The (sub)gradient of the Manhattan norm of $X$ is given by
$$\eqalign{
\mu &= \|X\|_1 \\
d\mu &= {\rm sign}(X):dX \;\;\doteq\;\; S:dX \\
}$$
where the colon denotes the trace/Frobenius product, i.e.
$$A:B = {\rm Tr}(A^TB)$$
and the sign function
$${\rm sign}(z) =
\begin{cases}
+{\tt1}\quad{\rm if}\;z\ge 0 \\
-{\tt1}\quad{\rm otherwise} \\
\end{cases}$$
is applied elementwise.

Now calculate the differential of $R$
$$\eqalign{
B &= (U^TU)^{1/2} = B^T
 \qquad&\big\{{\rm square\,root}\big\} \\
B^2 &= U^TU \\
B\,dB+dB\,B &= U^TdU+dU^TU
 \qquad&\big\{{\rm differentials}\big\} \\
(I_n\otimes B+B\otimes I_n)\,db
 &= \Big(I_n\otimes U^T+(U^T\otimes I_n)K\Big)\,du
 \qquad&\big\{{\rm vectorized}\big\} \\
db &= P\,du \\
\\
R &= UB^{-1} \\
dR &= dU\,B^{-1} - UB^{-1}dB\,B^{-1} \\
   &= dU\,B^{-1} - R\,dB\,B^{-1} \\
dr &= (B^{-1}\otimes I_m)\,du - (B^{-1}\otimes R)\,db \\
   &= \Big((B^{-1}\otimes I_m) - (B^{-1}\otimes R)P\Big)\,du \\
   &= Q\,du \\
}$$
where $K$ is the Commutation Matrix associated with the vectorization of a matrix transpose via the Kronecker product.

Finally, let $\,X=(R-M)\,$ and calculate the gradient of the objective function
$$\eqalign{
d\mu &= S:dX \\
  &= S:dR \\
  &= s:dr
 &\qquad\big\{{\rm vectorized}\big\} \\
  &= s:Q\,du \\
  &= Q^Ts:du \\
\frac{\partial\mu}{\partial u}  &= Q^Ts
 &\qquad\big\{{\rm gradient}\big\} \\
\\
}$$
Now use a sub-gradient method initialized with the Frobenius solution
$$\eqalign{
U_0 &= M \\
}$$
and iterated as

$$\eqalign{
B_k &= (U_k^TU_k)^{1/2} \\
R_k &= U_kB_k^{-1}
 \qquad\qquad\qquad\big\{{\rm current/best\,solution}\big\} \\
P_k
 &= (I_n\otimes B_k+B_k\otimes I_n)^{-1}\Big(I_n\otimes U_k^T+(U_k^T\otimes I_n)K\Big) \\
Q_k &= \Big((B_k^{-1}\otimes I_m) - (B_k^{-1}\otimes R_k)P_k\Big) \\
g_k &= Q_k^T\,{\rm vec}\Big({\rm sign}(R_k-M)\Big) \\
w_k &= \operatorname{vec}(U_k) - \lambda_kg_k
 \qquad\;\big\{{\rm subgradient\,step}\big\} \\
u_{k+1} &= \frac{w_k}{\|w_k\|_2}
 \qquad\qquad\qquad\big\{{\rm normalize}\big\} \\
U_{k+1} &= {\rm reshape}\big(\,u_{k+1},\, {\rm shape}(U_k)\,\big) \\
}$$


问:
Why do we use G, H, and 'K' for groups?
In many cases of mathematics, we tend to set several variables in ascending alphabetical order.

$x,y,z \dots$ for normal variables

$A,B,C\dots$ for sets, coefficients, matrices

$V,W\dots$ for vector spaces

${I,J}\dots$ for indices, ideals

$\mathcal{C,D}\dots$ for categories

But when we need 3 groups, we tend to use $G, H$ and suddenly, $K$, Which is unexpected.

Here are some examples that groups are denoted by $K$.

Prove that quotient group K/H is normal subgroup of quotient group G/H

https://en.wikipedia.org/wiki/Group_homomorphism#The_category_of_groups

https://proofwiki.org/wiki/Pullback_of_Quotient_Group_Isomorphism_is_Subgroup

Why do we denote groups by $K$ regardless of the alphabetical order?

答:
I think that $K$ is also used because of the German word kompakt for a compact group. Consider for example the Iwasawa decomposition of a group $G$, which is
$$G=KAN,$$
where $G$ is a connected semisimple real Lie group, $K$ a compact subgroup, $A$ an abelian subgroup and $N$ a nilpotent group. So $A$ for an abelian group is also customary, as is $N$ for a nilpotent group.

On the other hand, $H$ as a group notation doesn't rely on a property, but rather that we need two groups in general for a group homomorphism
$$
\phi\colon G\rightarrow H,
$$
and mathematicians then just proceeded alphabetically.

Title: Why do we use G, H, and 'K' for groups?


问:
Reference request: Variations of inverse function theorem
I have two real and separable Hilbert spaces $X$ and $Y$ as well as a continuously Fréchet differentiable function $f : X \to Y$ that is bijective.
Furthermore, at every point $x \in X$, $D_x f \in \mathcal{L} \left( X, Y \right)$ is continuous, injective and has dense range.
Now, I would love to throw some version of the inverse function theorem on this object to conclude that $f^{-1}$ is (at least) Gâteaux differentiable at every $y \in Y$ in all directions in the range of $D_{f^{-1} \left(y \right)} f$ in order to conclude that
\begin{equation}
D_y f^{-1} =  \left( D_{f^{-1} \left(y \right)} f \right)^{-1}
\end{equation}
but I seem to be unable to find any such theorem or proving it myself.

I am looking for a compilation of inverse function theorems with different assumptions in order to see if I can prove that $f$ perhaps has another nice property allowing me to draw the above conclusion.

Note: I do not expect that $f^{-1}$ is continuous, at least so far I have been unable to prove it.

答:
Under those assumptions, the Gateaux derivative of $f^{-1}$ might not exist. We cannot replace invertibility of the derivative by the weaker assumption of dense range. All the other assumptions do not make up for this.
Here is a somewhat lengthy counterexample.

Take $X$ infinite-dimensional, $X=Y$.
Let $K\in \mathcal L(X,X)$ be compact, injective, with dense range, self-adjoint, and positive semidefinite. Such operators exist. The idea is to build $f$ such that $f'(0)=K$.

For $y\in X$ define
$$
F(x):=\frac12 \langle Kx,x\rangle + \frac13 \|x\|^3 - \langle x,y\rangle.
$$
Due to the assumptions, $F$ is strictly convex, coercive, continuous. Hence $F$ has a uniquely defined global minimum, which is the only solution to $F'(x)=0$.
Hence the equation
$$
Kx + \|x\|^2 x = y
$$
is uniquely solvable for all $y$.  Set
$$
f(x):=Kx + \|x\|^2 x.
$$
Its derivative is given by
$$
f'(x) = K + 2\langle x, \cdot\rangle x + \|x\|^2 Id,
$$
which is positive definite and bijective for all $x\ne0$, injective with dense range for $x=0$.

Now let $y \in X\setminus R(K)$, which exists due to the assumptions on $K$.
Let $x_t$ be defined as $f(x_t)=ty$. Then the difference quotient
$\frac{x_t-0}t=\frac{f^{-1}(ty)-f^{-1}(0)}t$ does not converge for $t\searrow0$, and $f^{-1}$ is not even directionally differentiable at $y=0$. To see this, lets have a look on the equation defining $x_t$:
$$
Kx_t + \|x_t\|^2 x_t =ty.
$$
Divide this by $t$ and write it as an equation in $z_t:=t^{-1}x_t$:
$$
Kz_t + t^2\|z_t\|^2 z_t  =y.
$$
If $z_t \to z$ then passing to the limit $t\searrow0$ in this equation gives $Kz=y$, which is impossible.


问:
Solve equation for variable that's both in and out of exponent.
I'm in the process of working on a spreadsheet for a track coach to calculate ASR. link

I managed to get to the point where, given time, I can solve for distance.

However, I'd like to be able to calculate the reverse as well, without having to find the nearest value on the spreadsheet and then referencing the time.

Here is the (manipulated) formula I came up with for time to distance.

$$d=t(r+a*e^(-0.013*t))$$

Can someone help me solve for t?

I know there's an $ln$ involved somewhere, but I don't know how to deal with a variable that's both inside and outside the exponent.

Wolfram Alpha suggests under 'roots' something that looks like t=, but I don't know if that's what I'm looking for.

(not sure on the linear algebra tag, but I guessed 'algebra' was what I'm looking for when it comes to 'solve this equation for this variable)

答:
From a formal point of view, there is an analytical solution. Using the same notations as the ones used by @Robert Israel , we can write
$$e^{-st}=\frac{d-r t}{a t}$$ and the solution would be given in terms of the generalized Lambert function (have a look at equation $(4)$).

This being said, from a practical point of view, it is not very useful.

Under the same assumptions, what you could do for, at least, the generation of a "reasonable" estimate is to perform one iteration of Newton-like methods (Halley, Householder, ...). Depending on the order $n$ of the method, you would have things like
$$t_{(2)}=\frac{d}{a+r}$$
$$t_{(3)}=\frac{d (a+r)}{a^2-a d s+2 a r+r^2}$$
$$t_{(4)}=\frac{2 \left(a^2 d-a d^2 s+2 a d r+d r^2\right)}{2 a^3-4 a^2 d s+6 a^2 r+a d^2
   s^2-4 a d r s+6 a r^2+2 r^3}$$

Let me try with $s=0.013$, $a=\pi$, $r=e$ and $d=20$. The "exact" solution is $t=4.46864$. The above values would be
$$t_{(2)}=3.41304\qquad t_{(3)}=4.47832\qquad t_{(4)}=4.50790$$

Using $t_0=t_{(2)}$, Newton iterates would be
$$\left(
\begin{array}{cc}
n & t_n \\
 0 & 3.41304 \\
 1 & 4.41371 \\
 2 & 4.46850 \\
 3 & 4.46864
\end{array}
\right)$$


问:
Absolute convergence of $\sum\limits_{k=0}^{\infty}2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$

Let be $\sum\limits_{n=0}^{\infty}a_n$ an absolutely convergent series and $b_k:=2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$. Show that $\sum\limits_{k=0}^{\infty}b_k$ converges absolutely.

答:
If $a_j$ are nonnegative real numbers, then we're free to change the order $\sum\limits_{k=0}^\infty\sum\limits_{j=0}^k=\sum\limits_{j=0}^\infty\sum\limits_{k=j}^\infty$: $$\sum_{k=0}^\infty 2^{-k-1}\sum_{j=0}^k 2^j a_j=\sum_{j=0}^\infty 2^j a_j\underbrace{\sum_{k=j}^\infty 2^{-k-1}}_{=2^{-j}}=\sum_{j=0}^\infty a_j.$$

For the general case, we use $|b_k|\leqslant 2^{-k-1}\sum\limits_{j=0}^k 2^j|a_j|$ and the above to get $\sum\limits_{k=0}^\infty|b_k|\leqslant\sum\limits_{j=0}^\infty|a_j|$.


问:
How do I calculate the volume in square yards for the attached image?
enter image description here

By looking at the link above to see the picture, how do I calculate the volume in square yards?

答:
First, you DON'T "calculate volume in square yards!  You calculate volume in cubic yards.

Here the sides are trapezoids (two parallel edges, two not parallel) with parallel sides of length 12' and 36' and the other two sides of length 60' and 62'.   If you draw a line from the upper left corner perpendicular to the right side it divides the side into a rectangle with base of length 60' and height 12' so area 60(12)= 720 square feet.  The triangle has base of length 60' and height 36- 12= 24' so area (1/2)(60)(24)= 30(24)= 720 square feet.  So the total area of side is 720+ 720= 1440 square feet.

Finally, to get the volume multiply by the 72 feet length.

The volume is 1440(72) cubic feet.  You want cubic yards.  One cubic yard is 27 cubic feet so divide that by 27.  (72 is 89 and 27 is 39 so 1440(72)/27= 1440*8/3.)


问:
what is the compound or unconditional distribution of a hypergeometric with a binomial number of coloured balls
Suppose you have a urn with $N$ black balls, and some number of them $K$ are coloured white with probability $p$. So the number of white balls follows a binomial distribution:

$$
{N \choose K}p^K(1-p)^{N-K}
$$

Then suppose you sample $n$ balls from the urn without replacement. Among them there are $k$ white balls. Conditional on any value of $K$, the distribution would be hypergeometric:

$$
\frac{{ K \choose k } { N- K \choose n - k }}{N \choose n}
$$

But what is the unconditional (aka compound?) distribution if $k$, for all values of $K$?

edit: partial progress

It seems we should start by multiplying the two, as we normally do when when we have two independent events:

$$
\frac{{ K \choose k } { N- K \choose n - k } {N \choose K}p^K(1-p)^{N-K}}{N \choose n}
$$

I'm very rusty with simplifying factorials, but I think this can be simplified to

$$
{ n \choose k }{N-n \choose K-k}p^K(1-p)^{N-K}
$$

But is that correct? does it simplify farther?

edit 2

It does seem to be correct, since according to the Chu-Vandermonde Identity, if we marginalize over all values of $k$, we end up with the original binomial distribution:

$$
\sum_k { n \choose k }{N-n \choose K-k}p^K(1-p)^{N-K} = {N \choose K}p^K(1-p)^{N-K}
$$

But suppose instead we want to marginalize over $K$?

$$
{ n \choose k }\sum_{K=k}^{N-n}{N-n \choose K-k}p^K(1-p)^{N-K}
$$

答:
It seems like the answer is another binomial distribution!

Here's why I think so:

$$
{ n \choose k }\sum_{K=k}^{N-n}{N-n \choose K-k}p^K(1-p)^{N-K}
$$

$$
{ n \choose k }\sum_{K=k}^{N-n}{N-n \choose K-k}p^{K-k}p^k(1-p)^{N-n-(K-k)}(1-p)^{n-k}
$$

$$
{ n \choose k }p^k(1-p)^{n-k}\sum_{K=k}^{N-n}{N-n \choose K-k}p^{K-k}(1-p)^{N-n-(K-k)}
$$

And since we sum over all possible values of $K$, the term in the summation equals 1, which leaves us with:

$$
{ n \choose k }p^k(1-p)^{n-k}
$$

Which of course makes perfect sense. If you take any random subset out of a binomial distribution, it should also be a binomial with a smaller sample size.


问:
How to prove that $0 \leq\lfloor nx \rfloor - n\lfloor x \rfloor \leq n-1$?
$x \in \Bbb R, n \in \mathbb N_{\ne 0}$ , prove the following inequality :

$0 \leq\lfloor nx \rfloor - n\lfloor x \rfloor \leq n-1$

Here's my current try

$\lfloor x \rfloor\leq x < \lfloor x \rfloor +1$

$n\lfloor x \rfloor\leq nx < n\lfloor x \rfloor +n$

$\lfloor n\lfloor x \rfloor \rfloor\leq \lfloor nx \rfloor < \lfloor n\lfloor x \rfloor +n\rfloor$

$n\lfloor x \rfloor\leq \lfloor nx \rfloor < n\lfloor x \rfloor +n$

$0\leq \lfloor nx \rfloor - n\lfloor x \rfloor < n$

And this is where I'm stuck, I don't know how to progress further to get to $n - 1$

答:
Easier to assign variables, to make everything easier to decipher.

I am supposed to disect your work, but it simply gets confusing.

Let $F = \lfloor x\rfloor$ and let $r = x - F \implies 0 \leq r < 1.$

Then $nx = nF + nr,$ and $0 \leq nr < n \implies
0 \leq \lfloor nr\rfloor \leq (n-1)$.

Then $\lfloor nx\rfloor = nF + \lfloor nr\rfloor \implies$

$0 \leq \lfloor nx\rfloor - nF \leq (n-1)$.

Edit

The statement that

$\lfloor nx\rfloor = nF + \lfloor nr\rfloor$

Is based on the idea that if

$A = B + C,$ where $B$ is an integer,
then $\lfloor A\rfloor = B + \lfloor C\rfloor.$


答: Since $⌊nx⌋−n⌊x⌋$ is an integer and is strictly less than $n$ (as you showed), it can be at most $n−1$.

Title: How to prove that $0 \leq\lfloor nx \rfloor - n\lfloor x \rfloor \leq n-1$?

Body:
$x \in \Bbb R, n \in \mathbb N_{\ne 0}$ , prove the following inequality :

$0 \leq\lfloor nx \rfloor - n\lfloor x \rfloor \leq n-1$

Here's my current try

$\lfloor x \rfloor\leq x < \lfloor x \rfloor +1$

$n\lfloor x \rfloor\leq nx < n\lfloor x \rfloor +n$

$\lfloor n\lfloor x \rfloor \rfloor\leq \lfloor nx \rfloor < \lfloor n\lfloor x \rfloor +n\rfloor$

$n\lfloor x \rfloor\leq \lfloor nx \rfloor < n\lfloor x \rfloor +n$

$0\leq \lfloor nx \rfloor - n\lfloor x \rfloor < n$

And this is where I'm stuck, I don't know how to progress further to get to $n - 1$


问:
A matrix calculus problem with vectorization
I'm trying to find out the derivative of $vec(AA^T)$ w.r.t to $vec(A)$, where $A$ is a $m$ by $n$ matrix

I use a simple $3$ by $2$ case and find it entry-wisely.

I guess the answer would be $A^T\otimes I_m+[(I_m\otimes a_1),\ (I_m\otimes a_2),\ \ldots,\  (I_m\otimes a_n)]^T$ where $a_i$'s is columns of $A$ (not sure it is correct)

Is there any systematical way to find out such an expression?

My intuitive thought is to express $vec(AA^T)$ as some differentiable function of $vec(A)$ and utilize some simple matrix calculus rule.

Any tips would be appreciated!

答:
I think that a common way to do this is to use index notation $A_{ij}$ where $i$ goes over rows and $j$ goes over columns of the matrix. Then

$$F(A)_{ij} = \sum_k A_{ik} A_{jk}$$

Notice that the transpose is just swapping the indices in the second $A_{jk}$. Now we can do the derivative. The trick is to use completely different dummy coefficients for the derivative matrix

$$\frac{\partial F_{ij}}{\partial A_{mn}} = \frac{\partial}{\partial A_{mn}} \sum_k A_{ik} A_{jk}$$

Now we will make use of the fact that all coefficients of the matrix are independent. If this is the case, then

$$\frac{\partial A_{ik}}{\partial A_{mn}} = \delta_{im} \delta_{kn}$$

Where $\delta_{ij}$ is the Kronecker delta function, namely, it is equal to 1 if the two coefficients are equal, and 0 otherwise. Thus, using distribution rule for differentiation, we get

$$\frac{\partial F_{ij}}{\partial A_{mn}} = \sum_k(\delta_{im} \delta_{kn} A_{jk} + A_{ik} \delta_{jm} \delta_{kn}) =
A_{jn} \delta_{im} + A_{in} \delta_{jm}
$$

There are tons of matrix calculus rules, but instead of remembering all of them sometimes it is easier to just derive them from scratch using index notation


问:
A property of completely separable mad families
A family of sets $\mathcal{A}\subset[\omega]^\omega$ is called almost disjoint (a.d.) iff $\forall a,b\in\mathcal{A}(a\neq b\rightarrow |a\cap b|<\omega)$ and $\mathcal{A}$ is infinite (as such families turn out to be not that interesting if they are finite).

For an a.d. family $\mathcal{A}$ let $\mathcal{I}(\mathcal{A})$ be the ideal on $\omega$ generated by $\mathcal{A}\cup\{\{n\}:\,n\in\omega\}$ and let $\mathcal{I}^+(\mathcal{A})$ be the corresponding coideal $\mathcal{P}(\omega)\setminus\mathcal{I}(\mathcal{A})$.

An a.d. family is called completely separable (or saturated) if for any $b\in\mathcal{I}^+(\mathcal{A})$  there is $a\in\mathcal{A}$ such that $a\subset b$. It is now easy to see that an infinite completely separable a.d. family must be maximal a.d. (i.e., it is not properly included in another a.d. family) as any $A\subset\omega$ witnessing nonmaximality of $\mathcal{A}$ witnesses that $\mathcal{A}$ is not completely separable.

In [1] and [2] one can read (without proof) that any completely separable a.d. family has the property that for any $b\in\mathcal{I}^+(\mathcal{A})$ there already have to be $2^{\aleph_0}$ many $a\in\mathcal{A}$ s.t. $a\subset b$.

A first try was to take an a.d. family $\mathcal{B}$ on $b$ of size $2^{\aleph_0}$ but since it is not guaranteed that its elements are elements of $\mathcal{I}^+(\mathcal{A})$ one does not get that the elements of $\mathcal{B}$ contain elements of $\mathcal{A}$.

So one way is to find an a.d. family of size $2^{\aleph_0}$ inside $[b]^\omega\cap\mathcal{I}^+(\mathcal{A})$ (which should be possible according to [2]), but I didn't know how this could be done.

Does anybody of you know a proof of this or a reference where one can find a proof?

[1] Dilip Raghavan: A model with no strongly separable almost disjoint families, Israel J. Math., vol. 189 (2012), 39–53. (obtainable from http://www.math.toronto.edu/raghavan/ where it appears as 5 in the papers section)

[2] Saharon Shelah: MAD families and SANE players (preprint obtainable from http://arxiv.org/abs/0904.0816)

答:
Choose distinct sets $A_n\in\mathcal A,A_n\subseteq b$ $(n\in\omega)$. Let $B_n$ be an infinite proper subset of $A_n\setminus(A_0\cup\dots\cup A_{n-1})$. Thus the $B_n$'s are pairwise disjoint, and for $i\in\omega$ we have $A_i\not\subseteq\bigcup\{B_n:n\in\omega\}$. Let $\mathcal N\subset[\omega]^{\omega}$ be an a.d. family with $|\mathcal N|=2^{\aleph_0}$. For each $N\in\mathcal N$, let $B_N=\bigcup\{B_n:n\in N\}$. Then $B_N\in\mathcal I^+(\mathcal A)$, since $B_N$ has infinite intersection with infinitely many elements of $\mathcal A$. For each $N\in\mathcal N$ choose $C_N\in\mathcal A$ so that $C_N\subseteq B_N$. Thus $C_N\ne A_i$ for $i\in \omega$.

I claim that the $C_N$'s are distinct. Assume for a contradiction that $C_N=C_{N'}$ for some $N,N'\in\mathcal N,$ $N\ne N'$. Then $C_N\subseteq B_N\cap B_{N'}=\bigcup\{B_i:i\in N\cap N'\}\subseteq\bigcup\{A_i:i\in N\cap N'\}$. Since $N\cap N'$ is finite, it follows that $C_N\cap A_i$ is infinite for some $i$. Since $C_N,A_i\in\mathcal A$ and $C_N\ne A_i$, this contradicts the fact that $\mathcal A$ is almost disjoint.

Edited to correct an error pointed out in a comment by the original poster.


问:
$\lim_{x \to \infty} f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$ and the function is not uniformly continuous
" Does there exist a continuous function $f: \mathbb{R} \to \mathbb{R}$  so that  $\lim_{x \to \infty} f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$ and the function is not uniformly continuous on $\mathbb{R}$?"

I wanted to prove that this is impossible because it implies that the function has a horizontal asymptote.
I tried to prove:

Take $\epsilon$>0 randomly.

Because $\lim_{x \to \infty} f(x)=1$ there exist an $N\in \mathbb{R}$ so that $|f(x)-1|<\epsilon/2$ for $x>N$.

Because $\lim_{x \to -\infty} f(x)=-1$ there exist an $M\in \mathbb{R}$ so that $|f(x)+1|<\epsilon/2$ for $x<M$.

Because $f$ is continous in $\mathbb{R}$ we take $a \in \mathbb{R}$. There exist a $\delta>0$ so that for $x \in \mathbb{R}$ , $|f(x)-f(a)|<\epsilon$ for $|x-a|<\delta$.

And this is were it gets hard. I thougt you can define a $\delta_{2}=\max\{N,M\}$.
Take $x,y$ random in $\mathbb{R}$ so that $|x-y|<\delta_{2}$. Then $|f(x)-f(y)|=|f(x)-1+1-f(y)|=|f(x)-1|+|1-f(y)|<\epsilon$

I don't know if I defined my $\delta_{2}$ correctly.
Can someone help me how to prove this correctly?

答:
The function $\displaystyle\frac{x^{1/3}}{1+|x^{1/3}|}$ which has a vertical tangent line at the origin.

Ok, this one does not deserve credit, it is wrong, but will edit this to make it right... if possible

So let me make it right. There is no such function.

Proof. Take any $\varepsilon>0.$ Take $M>0$ such that
$|f(x)+1|<\frac\varepsilon3$ if $x\le-M$, and such that
$|f(x)-1|<\frac\varepsilon3$ if $x\ge M$.

Since $[-M,M]$ is compact (i.e., closed and bounded interval in this context), $f$ is uniformly continuous on this interval, and there is $\delta>0$ such that for
$x,x'\in[-M,M]$ if $|x-x'|<\delta$ then we have $|f(x)-f(x')|<\frac\varepsilon3$.

Then for any $x,x'\in(-\infty,\infty)$ if $|x-x'|<\delta$ we have
$|f(x)-f(x')|<\varepsilon$, that is, $f$ must be uniformly continuous on $(-\infty,\infty)$.

The inequality $|f(x)-f(x')|<\varepsilon$ is easy to verify considering cases... will not list them all, but one of the interesting ones (though they all are easy) is say $x\le M \le x'$ (and we may also assume that $\delta<M$ for convenience and without loss of generality).

So assume $x\le M \le x'$ and $|x-x'|<\delta$ (and hence $0<x$).

Then $|f(x)-f(x')|\le$

$\le|f(x)-f(M)|+|f(M)-f(x')|\le$

$\le|f(x)-f(M)|+|f(M)-1|+|f(x')-1|<$

$<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon$.

Thus $|f(x)-f(x')|<\varepsilon$.

Now that I posted my answer and read more of the comments, I see that what @questmath put in a comment, starting with

"$f$ is uniformly continuous on $[M,N]$..."

already provides (a clear enough hint for) the answer. I realize I only provided the details of a similar answer (with my notation being a bit different than what OP was asking, but I will keep it this way).

For that matter I also missed the very first comment, by @Surb (just read it, and understood what it says), already voted up three times because it is right and has what is needed...