答:
It may help to consider the similar-sounding definition of a ring as an abelian group $(A,+)$ equipped with an additional operation $*$ satisfying [properties]. As with the "digraph-to-category" situation, not every abelian group can be expanded to a ring; see e.g. here. On the other hand, there may be more than one  way to expand an abelian group into a ring (this is a good exercise).

And this happens already at the level of sets: the fact that

"I can take any set and define group operations on it"

is not something to be taken for granted. For example, it's false for fields, since any finite field has cardinality a prime power; there is no field whose underlying set has $6$ elements, for example. Boolean algebras provide another case, since every finite Boolean algebra has size a power of $2$. The fact that for every $n$ there is a group of cardinality $n$ is something which requires proof. Now that proof is quite easy, so this may be overlooked, but it is definitely a nontrivial feature.

EDIT: I don't understand your update - what exactly is problematic? Not every graph is the underlying graph of a category, but so what?

Again, it might help to consider a perhaps-more-familiar example: think of the functor sending a ring to its underlying abelian group (the "addition-only" part of the ring). Like the $U$ of the question, this functor is neither injective on objects (reflecting the fact that there may be more than one way to expand a given abelian group into a ring) nor surjective on objects (reflecting the fact that there may be no way to expand a given abelian group into a ring). But none of this is an issue.

Title: Confusion about definition of category using directed graph

Body:
A directed graph (diagram scheme, quiver) is a quadruple $(O, A, s, t)$, where $O$ is a set of objects, $A$ is a set of arrows and $s$ and $t$ are two mappings $s, t: A \to O$ ("source" and "target" of arrows respectively).

Then a category is defined as a directed graph with two additional mappings $id: O \to A$ (called identity) and $\circ : A \times_{O} A \to A$ (called composition), following usual rules of associativity, composition with identity etc.

The confusion with that definition for me is that it looks like any other definition in style of "a set with some extra stuff". I mean, for example, I can take any set and define group operations on it. But I can't take any directed graph and define a category structure on it. E.g. let $D = (\{1,2,3\}, \{a,b\}, s, t)$, where $s(a) = 1$, $s(b) = 2$ and $t(a) = 2$ and $t(b) = 3$ (i.e. there's no possible choices for identities and compositions). Identities and compositions restrict the choices of directed graphs.

What's wrong with such reasoning? And what is correct way of thinking about such definition of category?

NB. That's not a question about how to construct a free category on a given directed graph, it's about construction of a category according given definition.

UPDATE. Thanks to the answers, now I see where's the error in my reasoning. But confusion is still there.
Citing Categories for the Working Mathematician:

Every category $C$ determines a graph $UC$ with the same objects and
arrows, forgetting which arrows are composites and which are identities.
Every functor $F: C \to C'$ is also a morphism $UF: UC \to UC'$ between
the corresponding graphs. These observations define the forgetful functor
$U : \mathbf{Cat} \to \mathbf{Grph}$ from small categories to small graphs.

How this underlying graph of category relates with the one mentioned in definition of a category? They can't be the same just because underlying graph doesn't contain identities and compositions.


问:
How to Solve System of Parametric Equations
I am trying to represent two cars approaching an intersection where one car is traveling from a negative point along the x axis and another car is traveling from a negative point up the y-axis.

I have one car whose position is determined by the following:

$$x_1(t) = x_0+v_1*t+1/2*a_1*t^2$$
$$y_1(t) = 0$$

And another car whose position is determined by:

$$x_2(t) = 0$$
$$y_2(t) = y_0+v_2*t+1/2*a_2*t^2$$

How do I tell if there exists a time t for which these two cars collide at the origin with respect to the variables given?

答:
You need to solve the system $$x_1(t)=y_2(t)=0.$$

For convenience, you can form the equation

$$a_2x_1(t)-a_1x_2(t)=0$$ which is linear in $t$.

After finding $t$, plug in the system and check if that $t$ matches.

ContentLicense: CC BY-SA 4.0


问:
Determine how diagonal 2D Vector is
I'm trying to calculate the amount (defined as a value $0.0 \rightarrow 1.0$) that a 2D Vector is diagonal.
for example

A vector of $(1,1)$ is perfectly diagonal so would have a diagonal
value of $1.0$

A vector of $(1,0)$ is perfectly NOT diagonal so would have
a diagonal value of$ 0.0$

I also need to account for all $4$ directions, so
$(-1,1)$ is also perfectly diagonal as well as $(-1,-1)$ and $(1,-1).$

The solution I am looking for is for a piece of software so ideally would be efficent in terms of processing (no sqrts etc).

I have tried a few things (like using the DOT product against a fixed vector $(1,1),$ and getting the value of the DOT product - as a DoT product of $2$ parallel Vectors is $1$)

答:
Probably your easiest method is, for a vector $(x,y)$
$$\dfrac{\min\{|x|,|y|\}}{\max\{|x|,|y|\}}$$

Or algorithmically,

```<code>Diag(x, y)
    a = abs(x)
    b = abs(y)
    if (a > b) then exchange a, b
    if (b == 0) then return 1
    return a / b

</code>```

where I'm considering $(0,0)$ to be "perfectly diagonal", but you can make your own call on that.

Title: Determine how diagonal 2D Vector is

Body:
Probably your easiest method is, for a vector $(x,y)$
$$\dfrac{\min\{|x|,|y|\}}{\max\{|x|,|y|\}}$$

Or algorithmically,

Diag(x, y)
    a = abs(x)
    b = abs(y)
    if (a > b) then exchange a, b
    if (b == 0) then return 1
    return a / b

where I'm considering $(0,0)$ to be "perfectly diagonal", but you can make your own call on that.


问:
$A_i$ are closed. Any compact set intersects with finitely many $A_i$. Is $\bigcup_{i\in\mathbb N} A_i$ closed?
I do not know if this proposition is true.

My guess: yes?

my attempts:

Take a convergent net $a_\lambda\to a_0$ from $\bigcup_{i\in\mathbb N} A_i$.

Suppose there is some $i$ such that for every $\lambda$ there is $\eta\geq\lambda$ such that $a_\eta\in A_i$. Then $a_\eta$ such that $a_\eta\in A_i$ forms a subnet of $a$ and therefore converges to $a_0$. This shows $a_0\in A_i$.

Suppose there is not, so that there are increasing $\lambda_i$ such that element of $A_i$ does not appear after $\lambda_i$. Samely, $a_{\lambda_i}\to a_0$, and therefore $\{a_0,a_{\lambda_1},...\}$ is compact. By the assumption it only intersects with finitely many $A_i$, but this is a contradiction.

I did it! Is this proof correct then?

By the way, is there some weaker condition which makes $\bigcup_{i\in\mathbb N} A_i$ a closed set?

答:
The result is false, I’m afraid.

Let $p$ be a free (i.e., non-principal) ultrafilter on $\Bbb N$, and let $X=\Bbb N\cup\{p\}$. Let

$$\tau=\wp(\Bbb N)\cup\big\{\{p\}\cup A:A\in p\big\}\,;$$

$\tau$ is a topology on $X$. For $n\in\Bbb N$ let $A_n=\{n\}$. The sets $A_n$ are closed, and the only compact sets in $X$ are the finite sets, so every compact subset of $X$ intersects only finitely many of the sets $A_n$. However, $\bigcup_{n\in\Bbb N}A_n=\Bbb N$, which is not closed: $p$ is a limit point of $\Bbb N$.

The problem with your argument is that the sequence $\langle a_{\lambda_i}:i\in\Bbb N\rangle$ need not converge to $a_0$.

Added: To see that every compact subset of $X$ is finite, let $A\subseteq X$ be infinite. Clearly $A$ is not compact if $A\subseteq\Bbb N$, so suppose that $p\in A$. If $A\notin p$, then $\Bbb N\setminus A\in p$, so $X\setminus A$ is an open nbhd of $p$, and

$$\{X\setminus A\}\cup\big\{\{n\}:n\in A\big\}$$

is an open cover of $A$ with no finite subcover. If $A\in p$, let $\{B,C\}$ be a partition of $A$ into two infinite sets. Since $p$ is an ultrafilter, exactly one of $B$ and $C$ belongs to $p$; without loss of generality suppose that $B\in p$. Then

$$\big\{\{p\}\cup B\big\}\cup\big\{\{n\}:n\in A\setminus B\big\}$$

is an open cover of $A$ with no finite subcover.


问:
How to find the value of $m$ given $f_{X,Y}(x,y)$
given $(X,Y)$ random variable 2D continuously with the density function $$f_{X,Y}(x,y)=\begin{cases}\frac{1}{{30}}\cdot x^{m}\cdot ye^{-y} & x \in [0,y] \ , \ y>0 \\ 0 & \textrm{otherwise} \end{cases}$$

I need to find the value of $m$

I know I should use the properties of the density function $$1=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)dxdy$$ $$1=\int_{0}^{y}\int_{-\infty}^{\infty}\frac{1}{{30}}\cdot x^{m}\cdot ye^{-y}dxdy$$

I dont understand how to determine the limits of $x$ from the given about it.

答:
In the computation of $\displaystyle\iint_{\Bbb R^2}f(x,y)\; dx\; dy$ you can use Fubini and

either let $y$ run in the interval $[0,\infty)$, and then $x\in[0,y]$,

or let $x$ run in the interval $[0,\infty)$, and then $y\in[x,\infty)$.

The two ways are:
$$
\begin{aligned}
1 &=
\int_{y\in[0,\infty)}dy\int_{x\in[0,y]}
\frac 1{30}x^m\; y\; e^{-y}\; dx
\\
&=
\int_{y\in[0,\infty)}dy\cdot
\frac 1{30}\cdot\frac {y^{m+1}}{m+1}y\; e^{-y}\\
&=
\frac 1{30(m+1)}\cdot \Gamma(m+3)
\qquad\text{ and if $m$ is integer}
\\
&=
\frac 1{30(m+1)}\cdot (m+2)!
\\
&=\frac 1{30}(m+2)\cdot m!
\ ,
\\[4mm]
1
&=
\int_{x\in[0,\infty)}dx\int_{y\in[x,\infty)}
\frac 1{30}x^m\; y\; e^{-y}\; dy
\\
&=
\int_{x\in[0,\infty)}dx\cdot
\frac 1{30}x^m\; (x+1)\; e^{-x}
\\
&=\frac 1{30}\cdot(\Gamma(m+2)+\Gamma(m+1))
\qquad\text{ and if $m$ is integer}
\\
&=
\frac 1{30}\cdot ((m+1)!+m!)
\\
&=\frac 1{30}\cdot m!\cdot((m+1)+1)
\\
&=\frac 1{30}(m+2)\cdot m!
\ .
\end{aligned}
$$
For $m=3$ we have indeed equality.

The equation to be solved $(m+2)\Gamma(m+1)=30$ has a unique solution $m>0$ since the L.H.S. is an increasing function of $m$.

But note that the $\Gamma$ function comes with "its own caprice", we find for instance...

```<code>? plot(m=-.99, -0.9, (m+2)*gamma(m+1) - 30)

70.426911 "''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''|
          :                                                              |
          |:                                                             |
          |"                                                             |
          |                                                              |
          | x                                                            |
          |                                                              |
          |  x                                                           |
          |   _                                                          |
          |                                                              |
          |    "                                                         |
          |     "                                                        |
          |      "                                                       |
          |       "_                                                     |
          |         "_                                                   |
          |           x_                                                 |
          |             "x_                                              |
          -----------------"x_--------------------------------------------
          |                   ""xx__                                     |
          |                         ""xx___                              |
          |                                """"xxxx____                  |
-19.53514 |............................................""""""xxxxxxxx_____
          -0.99                                                       -0.9
? solve(m=-.99, -0.9, (m+2)*gamma(m+1) - 30)
%21 = -0.96617414943112079757353172624510260510
</code>```

Body:
In the computation of $\displaystyle\iint_{\Bbb R^2}f(x,y)\; dx\; dy$ you can use Fubini and

either let $y$ run in the interval $[0,\infty)$, and then $x\in[0,y]$,

or let $x$ run in the interval $[0,\infty)$, and then $y\in[x,\infty)$.

The two ways are:
$$
\begin{aligned}
1 &=
\int_{y\in[0,\infty)}dy\int_{x\in[0,y]}
\frac 1{30}x^m\; y\; e^{-y}\; dx
\\
&=
\int_{y\in[0,\infty)}dy\cdot
\frac 1{30}\cdot\frac {y^{m+1}}{m+1}y\; e^{-y}\\
&=
\frac 1{30(m+1)}\cdot \Gamma(m+3)
\qquad\text{ and if $m$ is integer}
\\
&=
\frac 1{30(m+1)}\cdot (m+2)!
\\
&=\frac 1{30}(m+2)\cdot m!
\ ,
\\[4mm]
1
&=
\int_{x\in[0,\infty)}dx\int_{y\in[x,\infty)}
\frac 1{30}x^m\; y\; e^{-y}\; dy
\\
&=
\int_{x\in[0,\infty)}dx\cdot
\frac 1{30}x^m\; (x+1)\; e^{-x}
\\
&=\frac 1{30}\cdot(\Gamma(m+2)+\Gamma(m+1))
\qquad\text{ and if $m$ is integer}
\\
&=
\frac 1{30}\cdot ((m+1)!+m!)
\\
&=\frac 1{30}\cdot m!\cdot((m+1)+1)
\\
&=\frac 1{30}(m+2)\cdot m!
\ .
\end{aligned}
$$
For $m=3$ we have indeed equality.

The equation to be solved $(m+2)\Gamma(m+1)=30$ has a unique solution $m>0$ since the L.H.S. is an increasing function of $m$.

But note that the $\Gamma$ function comes with "its own caprice", we find for instance...

? plot(m=-.99, -0.9, (m+2)*gamma(m+1) - 30)

70.426911 "''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''|
          :                                                              |
          |:                                                             |
          |"                                                             |
          |                                                              |
          | x                                                            |
          |                                                              |
          |  x                                                           |
          |   _                                                          |
          |                                                              |
          |    "                                                         |
          |     "                                                        |
          |      "                                                       |
          |       "_                                                     |
          |         "_                                                   |
          |           x_                                                 |
          |             "x_                                              |
          -----------------"x_--------------------------------------------
          |                   ""xx__                                     |
          |                         ""xx___                              |
          |                                """"xxxx____                  |
-19.53514 |............................................""""""xxxxxxxx_____
          -0.99                                                       -0.9
? solve(m=-.99, -0.9, (m+2)*gamma(m+1) - 30)
%21 = -0.96617414943112079757353172624510260510


问:
exponential additive functional equation
Let $S$ be a semigroup with no identity element and $m:S\to \Bbb C$ be given function($m\not\equiv 0$) satisfying the exponential functional equation
$$
m(x+y)=m(x)m(y)
$$
for all $x, y\in S$. Find all solutions $f:S\to \Bbb C$ satisfying
the equation
\begin{equation}
f(x+y)=f(x)m(y)+f(y)m(x) \tag 1
\end{equation}
for all $x, y\in S$.

Remark.  If $S$ is a group, then using the fact that $m(x)\ne 0$ for all $x\in S$ and dividing $(1)$ by $m(x+y)$, we have
$$
f(x)=m(x)A(x)
$$
for all $x\in S$, where $A$ is an additive function.

答:
The subset $I=\{x\in S|m(x)=0\}$ is a prime ideal in $S$ (i.e. $xy\in I \Rightarrow x\in I \vee y\in I$), so $S\setminus I$ is a subsemigroup to which you can apply your reasonings concerning groups.

Addendum: Here is a counter-example showing that generally speaking $f$ has not the form $f(x)=m(x)A(x)$, where $A$ is additive on whole $S$.

Take an arbitrary set $J$ with a fixed element $z$. Let ${\bar J}$ be a set such that $|J|=|{\bar J}|, J\cap {\bar J}=z$, and $x\to {\bar x}$ a bijection from $J$ to ${\bar J}$ (${\bar z}=z$). Let $I=J\cup {\bar J}$ and $S=I\cup \{e\}$ where $e$ is an extra element.

Define the addition:

$$
I+I=z, \ \ e+e=e, \ \ e+x={\bar x}
$$
so that $S$ will be a semigroup.

Set $m(a)=0$ for $a\in I$, $m(e)=1$. Then an arbitrary $f$ with $f(x)=f({\bar x})$ yields the equation (I).

ContentLicense: CC BY-SA 3.0


答:
We can considerably simplify $f(k)$ which makes the proof easy.

We obtain
\begin{align*}
\color{blue}{f(k)}&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{k-b}\right)\prod_{i=1}^b a_{k-b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{b}\right)\prod_{i=1}^{k-b} a_{b+i}\tag{1}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\prod_{i=1}^{k-b} a_{b+i}-\sum_{b=1}^{k-1}a_b\prod_{i=1}^{k-b} a_{b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=2}^{k}\prod_{i=1}^{k-b+1} a_{b-1+i}-\sum_{b=1}^{k-1}\prod_{i=0}^{k-b} a_{b+i}\tag{2}\\
&=\prod_{i=1}^k a_i+\sum_{b=2}^{k}\prod_{i=0}^{k-b} a_{b+i}-\sum_{b=1}^{k-1}\prod_{i=0}^{k-b} a_{b+i}\tag{3}\\
&=\prod_{i=1}^k a_i+a_k-\prod_{i=0}^{k-1} a_{i+1}\tag{4}\\
&=\prod_{i=1}^k a_i+a_k-\prod_{i=1}^{k} a_{i}\tag{5}\\
&\,\,\color{blue}{=a_k}\tag{6}
\end{align*}

Comment:

In (1) we change the order of summation of the sum $b\to k-b$.

In (2) we shift the index of $b$ by one in the left sum and merge the factor $a_b$ into the product of the right sum.

In (3) we shift the index of the product in the left sum by one and observe the sums are telescoping.

In (4) we do the telescoping.

In (5) we shift the index of the right-hand product and simplify in the next line.

With $f(k)=a_k$ the proof is simple, since we have
\begin{align*}
\color{blue}{f(k) a_{k+1} + (1 - a_k)a_{k+1}}&= a_ka_{k+1}+(1-a_k)a_{k+1}=a_{k+1}\\
&\color{blue}{=f(k+1)}
\end{align*}
and the claim follows.

Title: Proving that, if $f(k)=\prod_{i=1}^ka_i+\sum_{b=1}^{k-1}(1-a_{k-b})\prod_{i=1}^ba_{k-b+i}$, then $f(k+1)=f(k)\cdot a_{k+1}+(1 - a_k)a_{k+1}$

Body:
Given that $$f(k) = \prod_{i=1}^k a_i + \sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i}$$ for all $k$ where $(a_1, a_2, a_3, \ldots )$ are random constants, prove that: $$f(k+1) = f(k) \cdot a_{k+1} + (1 - a_k)a_{k+1}$$

So far, my current work is:

$$f(k+1) = \prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1}.$$

So, we have to show that: $$a_{k+1}\prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\prod_{i=1}^{k+1} a_i + a_{k+1}\sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i} + (1-a_k)(a_{k+1}).$$

The $a_{k+1}\prod_{i=1}^{k+1} a_i$ cancel out on both sides, leaving us to prove: $$\sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i} + (1-a_k)(a_{k+1}).$$

Turning the max value of $i$ on the first summation to $k-1$ and changing the second series of products a bit to match the first one gives:

\begin{align}
\sum_{b=1}^{k-1} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} + (1-a_1)\prod_{i=1}^k a_{k-b+i+1} = &\\
a_{k+1} \sum_{b=1}^{k-1} \left( (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i+1} \cdot \frac{a_{k-b+1}}{a_{k+1}}\right)  + (1-a_k)(a_{k+1}).
\end{align}

We can cancel out the $a_{k+1}$ terms on the RHS, leaving us to prove: $$\sum_{b=1}^{k-1} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} + (1-a_1)\prod_{i=1}^k a_{k-b+i+1} = \sum_{b=1}^{k-1} \left( (1-a_{k-b}) a_{k-b+1} \prod_{i=1}^b a_{k-b+i+1}\right)  + (1-a_k)(a_{k+1}).$$

However, I don't know how to continue on from here. Any help would be greatly appreciated.


问:
Let $f:\Bbb R^2 \to \Bbb R$ be ${\cal C}^1$, such that $|f(x) − f(y)| ≤ ||x − y||^2$ for all $x,y\in \Bbb R^2$. Show that $f$ is constant.
Let $f:\Bbb R^2 \to \Bbb R$ be ${\cal C}^1$, such that  $|f(x) − f(y)| ≤ ||x − y||^2$ for all $x,y\in  \Bbb R^2$. Show that $f$ is constant.

I'm not sure about :

$=> \lim_{x\to y} \frac{|f(x) − f(y)|}
{\|x − y\|}  \le  \lim_{x\to y}\|x − y\|$

$=> |f '(x)| \le 0$

Is it true or not ?

答:
This is true more generally for $f : \mathbb{R}^n \to \mathbb{R}^m$ where $m \geqslant 1$.

Since $f \in C^1$ , for all $\epsilon > 0$ there exist $\delta> 0$ such that if $0 < \|h \| < \delta$, then

$$\| f(x+h) - f(x) - Df(x) \cdot h\| < \epsilon\|h\|$$

By the reverse triangle inequality,

$$\|Df(x) \cdot h\| \leqslant \|f(x+h) - f(x)\| + \epsilon \|h\| \leqslant \|h\|^2 + \epsilon \|h\|$$

For any partial derivative $D_jf_i(x)$ where $1 \leqslant j \leqslant n$ and $1 \leqslant i \leqslant m$, we have

$$ |D_jf_i(x)| |h_j|\leqslant \|Df(x) \cdot h\|\leqslant \|h\|^2 + \epsilon\|h\|  $$

and, in particular for $h = (0, \ldots, h_j, \ldots 0)$ where $0 < h_j < \delta$,

$$ |D_jf_i(x)||h_j| \leqslant |h_j|^2 + \epsilon|h_j|  $$

Thus,

$$|D_jf_i(x)| \leqslant \lim_{\|h\| \to 0}(|h_j|+ \epsilon) = \epsilon$$

Since this holds for all $\epsilon >0$, the partial derivatives are zero and $f$ is constant.


问:
Infinite sum $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$ -- does it evaluate to an exact value or approaches a limit
I bumped into the question what does the sum $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$$ evaluate to. It asks whether it evaluates to a value? I'm not sure whether $\frac{1+\sqrt{5}}{2}$ is the exact value or the limit this sum approaches to in infinity? I think the question implied that it's the exact value, but for me it looks more like a limit, and that the denominator under 1 would never be equal to 1+fraction no matter how many steps the fraction takes.

答:
Let the number be $x$.  Then $x = 1+\frac{1}{x}$ or

$$x^2=x+1.$$

This has solutions

$$x = \frac{1 \pm \sqrt{5}}{2}.$$

So if the limit exists, it must be one of those numbers.


问:
Minimizing and eliminating a variable
We want to minimize
$$f(a,b,c)=-\frac{b(3b-4a+1)+a(a+5))}{(c+b+a)(a-1)}$$
subject to the constraints $0\le a\le b\le c\le 1$, and $a + b + c = p$ for $p\in\left[\frac{1}{2},\frac{3}{2}\right]$.

How can we prove that, for all $p\in\left[\frac{1}{2},\frac{3}{2}\right]$, the minimum is attained when $b=c$ ?

答: One approach is to suppose that you have a feasible solution with $b < c$ and perturb the solution to $b+\epsilon$ and $c-\epsilon$ for some $\epsilon \in (0,(c-b)/2]$.  This perturbation preserves feasibility, so if you can show that $f(a,b,c)>f(a,b+\epsilon,c-\epsilon)$, you are done.

Title: Minimizing and eliminating a variable


问:
Implicit Differentiation Help
I have to use implicit differentiation to find $\frac{dy}{dx}$ given:

$$x^2 \cos(y) + \sin(2y) = xy$$

I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck.

Thank you everyone.

Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that

答:
Expanding my comment above. Assume you have a differentiable implicit
function $F(x,y)=0$. Let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0$. This function $f(x)$ does not need to be explicitly known, as pointed out in Christian Blatter's answer.

If you differentiate both sides of $F(x,y)=0$ and apply the chain rule,
 you get the following total derivative with respect to $x$:

$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial x}+\frac{%
\partial F}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}\equiv 0.\qquad (1)$$

Solving $(1)$ for $\frac{\mathrm{d}y}{\mathrm{d}x}$, gives you the following
formula

$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\partial F}{\partial x}/\frac{%
\partial F}{\partial y}.\qquad (2)$$

For $F(x,y)=x^{2}\cos (y)+\sin (2y)-xy$, since the partial derivatives are

$$\frac{\partial F }{\partial x}=2x\cos y-y\quad\text{ and }\qquad \frac{\partial F }{\partial y}
=-x^{2}\sin y+2\cos 2y-x,$$

the derivative of $y=f(x)$ is

$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x\cos y-y}{-x^{2}\sin y+2\cos 2y-x}.$$

As a final note I say that the way I presented the computation is the same as it is explained in  Advanced Calculus book by Angus Taylor. I have seen this computation done here many times without considering partial derivatives explicitly, but it is essential the same as when evaluating those partial derivatives.

Title: Implicit Differentiation Help


问:
Why is the volume of a parallelepiped given three vectors zero?
I'm trying to find the volume of a parallelepiped given three vectors:

$A=(-2,3,1)$

$B=(1,1,2)$

$C=(1,2,3)$

So,

\begin{align*}
V &= \| C \cdot (A \times B)\| \\
 &= \| (1,2,3) \cdot \big((-2,3,1)\times (1,2,3)\big)\| \\
&= \| (1,2,3) \cdot (5,5,-5)\| \\
&= 0 \\
\end{align*}

Why is $0$? I checked the cross and dot products on Symbolab and they're fine. But I don't know what I'm missing. Any help?

答: Not a surprise since $A+7B-5C=0$ which means $A$,$B$ and $C$ are in the same plane.


问:
Is a polynomial expression with unit coefficients guaranteed to be invertible in a ring?
I'm working through a proof of Amitsur's Theorem, that if $\mathbb{K}$ is a field and $A$ a $\mathbb{K}$-algebra with $\dim_{\mathbb{K}}A<|\mathbb{K}|$, then $J(A)$ (the Jacobson radical) is nil (all it's elements are nilpotent).

We let $a\in J(A)$. In the final step of the proof, we arrive at
$$
0 = a^k\underbrace{(b_0 + b_1a + ... + b_m a^m)}_{=: p(a)},\quad b_i\in\mathbb{K}, b_0\neq 0
$$

My prof says $p(a)$ is invertible "since each $b_i$ is a unit", but I fail to see how this is true.

Is it true, in general, that if each $u_i$ is a unit of a ring $R$, and each $u_i$ commutes with $a$ along with every other $u_j$, then $u_0 + u_1a + \cdots + u_n a^n$ is also a unit? I know that if $u\in R^\times$, $a\in R$ is nilpotent, and $ua = au$, then $u + a\in R$, but this cannot be applied in our case as we cannot assume a priori that $a$ is nilpotent (this is what we are trying to show). If the general statement is false, what could be "special" about our $p(a)$ above which guarantees it's a unit?

答:
I see what the issue is. The statement in the question title is not true in general (although I suppose it might be true if $a$ is nilpotent). Instead, the reason $p(a)$ is invertible is because $a\in J(A)$.

Recall that if $a\in J(A)$ then $1 - ra$ is invertible for all $r\in A$. From this we can deduce that $b_0 + ra$ is invertible for all $b_0\in\mathbb{K}\backslash\{0\}$ and $r\in A$. This includes $r = b_1 + b_2a + \cdots + b_m a^{m-1}$, hence $b_0 + (b_1 + \cdots + b_m a^{m-1})a = p(a)$ is invertible.


问:
Finding the maximum of $G(x) = \int_{x}^{x+a} f(t) \,dt$ using FTOC.
Let's discuss with a simple example: $f(t) = -t^2.$

Given $a>0,$ define $G(x) = \int_{x}^{x+a} -t^2 \,dt$.

I'm interested in how to find the value of $x$ that maximises $G(x)$.

The direct method is:

$G(x) = \left[ -\frac13 t^3 \right]_x^{x+a} = -\frac13(x+a)^3 - (-\frac13x^3) = \ ...\ = -a\left[\left(x+\frac{a}{2}\right)^2 + \frac{a^2}{12} \right],$

which is maximised when $x = -\frac{a}{2},$ as expected.

But this method was only easy because I chose $f(t) = -t^2$. In general, it would appear we would have to use numerical methods to optimise $G(x),$ in particular if $G(x)$ is not easily integrable.

However, is there some trick we can use with the FTOC, that tells us the value of $x$ that maximises $G(x)$?

For example, $G(x) = \int_{x}^{x+a} -t^2 \,dt$ achieves a maximum when

$\frac{d}{dx}\left(\int_{x}^{x+a} -t^2 \,dt\right) = 0 \implies \frac{d}{dx}\left(\int_{q}^{x+a} -t^2 \,dt - \int_{q}^{x} -t^2 \,dt\right) = 0 \implies \frac{d}{dx}\left(\int_{q}^{x+a} -t^2 \,dt\right) - \frac{d}{dx}\left(\int_{q}^{x} -t^2 \,dt\right) = 0,$

where q is a sufficiently large negative constant. Now, the second term simplifies to $-x^2$ by FTOC. But I'm not certain if/how the first term reduces, or if there is some other trick I am missing here, which can also be applied for more difficult functions $f(t)$?

答:
As David Mitra points out in the comments, the chain rule one would think "should work" here does in fact work.

I made a silly mistake in my working. I let $u=x+a$ so that $\frac{d}{dx}$ can be interchanged with $\frac{d}{du}$, so:

$\frac{d}{dx}\left(\int_{q}^{x+a} -t^2 \,dt\right) = \frac{d}{du}\left(\int_{q}^{u} -t^2 \,dt\right) = -u^2 = -(x+a)^2$

but somehow I then made the following basic mistake:

$x^2 - (x+a)^2 = x^2 - (x^2 + 2a + a^2) = ... = 0$,

so I missed an $x$ in the term "$2a$" and got a wrong answer. So I thought my method was wrong.

However, that last term should be $x^2 - (x^2 + 2ax + a^2)$ and then setting this to $0$ gives the correct answer of $x = \frac{-a}{2}.$


问:
Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$?
I suspect that

$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
 {{\pi}\over{2}}$$

for $C>0$.

I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?

答:
Let $f : \Bbb{R} \to \Bbb{C}$ be an integrable even function. Then we claim that

$$ \int_{0}^{\infty} f\left( y - \frac{1}{y} \right) \, dy = \int_{0}^{\infty} f(x) \, dx. \tag{1} $$

Indeed, let $I$ denote the LHS of $(1)$. Then by the substitution $y \mapsto y^{-1}$ we have

$$ I = \int_{0}^{\infty} f\left( \frac{1}{y} - y \right) \, \frac{dy}{y^2} = \int_{0}^{\infty} \frac{1}{y^2} f \left( y - \frac{1}{y} \right) \, dy. $$

Thus with the substitution $x = y - y^{-1}$, we have

$$ 2I = \int_{0}^{\infty} \left( 1 + \frac{1}{y^2} \right) f\left( y - \frac{1}{y} \right) \, dy = \int_{-\infty}^{\infty} f(x) \, dx = 2 \int_{0}^{\infty} f(x) \, dx $$

and the conclusion follows.

Now you can plug

$$ f(x) = \frac{1}{C + x^2} $$

to $(1)$ to obtain the conclusion:

$$ \int_{0}^{\infty} \frac{y^2}{Cy^2 + (y^2 - 1)^2} \, dy = \int_{0}^{\infty} \frac{dy}{C + (y - y^{-1})^2} \, dy = \int_{0}^{\infty} \frac{dx}{C + x^2} = \frac{\pi}{2\sqrt{C}}. $$


问:
Evaluate $\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}dx=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}} $

(1) Prove that$$\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}\mathrm{d}x=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}}$$
(2) Prove that$$\int_0^\infty \frac{\mathrm{e}^{-ax^2}-\mathrm{e}^{-bx^2}}{x^2}\mathrm{d}x=\frac{1}{2}\ln\frac{b}{a} $$

I just learned improper integral with parameters and worked out a basic example below
$$
\int_0^\infty \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x}\mathrm{d}x=\ln\frac{b}{a}
$$
But I do not know how to compute the two integrals above. I think the basic example may be of some help and tried integration by parts, inspired by $\int_0^\infty\frac{\sin x}{x}\mathrm{d}x$ and $\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm{d}x$.

Any hint is appreciated.

答:
Integrate (1) by parts with $\frac{dx}{x^2}=-d(\frac1x)$,
\begin{align}
&\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}dx\\
= &2a \int_0^\infty \frac{{e}^{-(a+b)x}-{e}^{-2ax}}{x}dx
 + 2b \int_0^\infty \frac{{e}^{-(a+b)x}-{e}^{-2bx}}{x}dx\\
= &2a \ln \frac{2a}{a+b} + 2b \ln \frac{2b}{a+b}
=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}}
\end{align}
where $
\int_0^\infty \frac{\mathrm{e}^{-px }-\mathrm{e}^{-qx }}{x}\mathrm{d}x=\ln\frac{q}{p}
$ is used.

Integrate (2) by parts
$$\int_0^\infty \frac{{e}^{-ax^2}-{e}^{-bx^2}}{x^2}{d}x=
2\int_0^\infty (b{e}^{-bx^2}-a{e}^{-ax^2})dx=
\sqrt{\pi}(\sqrt b-\sqrt a) $$
where $\int_0^\infty e^{-px^2} =\frac{\sqrt{\pi}}{2\sqrt p}$ is used


问:
Is the limit of a sequence of continuous linear operators in the weak operator topology again a continuous linear operator?
From the Banach-Steinhaus theorem we know that if $(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(X,Y)$, where $X$ is a Banach and $Y$ a normed space, converges in the strong operator topology, then its limit in the strong operator topology is again a bounded linear operator from $X$ to $Y$.

Now I've read that in a Hilbert space $H$ the following stronger result holds: If $(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(H)$ converges in the weak operator topology, then its limit in the weak operator topology is again a bounded linear operator on $H$.

Why is it important that $H$ is a Hilbert space? Doesn't the claim remain true in the previous considered case $(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(X,Y)$, where $X$ is a Banach and $Y$ a normed space?

If $E$ is a normed space, we know that $B\subseteq E$ is bounded if and only if it is weakly bounded. Thus, a weakly convergent sequence is norm bounded.

Shouldn't it immediately follow that if $(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(X,Y)$ is weakly convergent, it is bounded in the strong operator topology and hence bounded in the uniform operator topology by the Banach-Steinhaus theorem?

答:
I think what you are saying is true. Never thought about it since i've always pre-assumed that the weakly-operator limit $A$ of the $A_n's$ was always in $A\in \mathfrak L(X,Y)$. Am writing the argument just to convience ourselfs. Indeed, we only need to assume that $Y$ has a norm, not necessarily a complete one.

So, lets suppose that $A_n\overset{\text{wo}}{\to}A$ in the weak operator topology where $A:X\to Y$ is a linear operator, not necessarily bounded. Convergence in the weak operator topology is described by $h(A_n x)\to h(A x)$ for every $x\in X$ and $h\in Y^*$. This implies that the set $\{A_n x: n\in \mathbb{N}\}$ is weakly bounded in $Y$, hence it is also bounded in $Y$. By the Banach-Steinhaus it follows that $\sup_{n}||A_n||=M<\infty$. Now, for $x\in X$ with $||x||=1$ we have
$$||Ax||=\max_{h\in Y^*,\, ||h||=1}|h(Ax)|$$
So, there is some $||h||=1$ in $Y^*$ such that $||Ax||=|h(Ax)|$. Using the weak convergence for $A_nx$ we end up with
\begin{align}
||Ax||&=|h(Ax)|\\
&=\lim_{n\to \infty}|h(A_nx)|\\
&\leq \underbrace{||h||}_{=1}\liminf_{n\to \infty}||A_n||\cdot \underbrace{||x||}_{=1}
\end{align}
Hence, $||Ax||\leq M$ for every $||x||=1$ and therefore, $||A||\leq M<\infty$.

Edit: (Responding to the comment)

The existence of such $A$ is trickier. To ensure such existence we need another assumption for $Y$, since there is a counter example in here where $X=Y=c_0$. The only natural that i could think while i was trying to prove it is that $Y$ has to be reflexive (from not being a Banach space we went straight out to reflexivity :P). In the case where $X=Y=H$ is a Hilbert space things were slightly more easier since we can identify $H^*$ with $H$ and dont need to mess with the second duals.

The argument in the case where $Y$ is reflexive is the following:

Suppose that $\lim_{n}\langle A_n x, h \rangle$ exists for every $x\in X$ and $h\in Y^*$. For fixed $x\in X$ let $f_x:Y^*\to \mathbb{R}$ defined by
$$\langle h, f_x\rangle =\lim_{n\to \infty}\langle A_n x, h\rangle$$
Its easy to check that $f_x$ is a linear functional and by the previous discussion it is also bounded. Meaning, $f_x \in Y^{**}$. By reflexivity, there is some $y_x\in Y$ such that $\langle h, f_x\rangle =\langle y_x, h\rangle$ for all $h\in Y^*$. Now, let $x\overset{A}{\longmapsto} y_x$. Now, its easy to check that $A:X\to Y$ is a linear operator. By the previous discussion it is also bounded.


答:
If $(A_i)_{i\in I}$ is a family of closed sets and is also locally finite ( every point $x\in X$ is contained in some $U_x$ open such that $U_x$ intersects finitely many of the $A_i$'s ) then $\cup_{i \in I} A_i$ is closed.

For proof: a set is closed if and only it is locally closed ( since it is true for "open").

Now, if your space is also locally compact then OK.

Title: $A_i$ are closed. Any compact set intersects with finitely many $A_i$. Is $\bigcup_{i\in\mathbb N} A_i$ closed?

Body:
I do not know if this proposition is true.

My guess: yes?

my attempts:

Take a convergent net $a_\lambda\to a_0$ from $\bigcup_{i\in\mathbb N} A_i$.

Suppose there is some $i$ such that for every $\lambda$ there is $\eta\geq\lambda$ such that $a_\eta\in A_i$. Then $a_\eta$ such that $a_\eta\in A_i$ forms a subnet of $a$ and therefore converges to $a_0$. This shows $a_0\in A_i$.

Suppose there is not, so that there are increasing $\lambda_i$ such that element of $A_i$ does not appear after $\lambda_i$. Samely, $a_{\lambda_i}\to a_0$, and therefore $\{a_0,a_{\lambda_1},...\}$ is compact. By the assumption it only intersects with finitely many $A_i$, but this is a contradiction.

I did it! Is this proof correct then?

By the way, is there some weaker condition which makes $\bigcup_{i\in\mathbb N} A_i$ a closed set?


问:
Does there exist a function $g(z)$ that agrees with $f(z)$
So I was going over old solutions to my homework for my complex analysis course, and we were given the series $$\sum_{n=0}^{\infty} (n+1)z^n$$ which is convergent in $|z|<1$ as the radius of convergence $= 1$ .The question following asked if there exists a function $g(z)$ that agrees with $f(z)$ when $|z|<1$ and is analytic everywhere except at $z=1$. The answer was yes, but the work my prof did doesn't make sense to me. He said yes, and did this, when $|z|<1$ $$\sum_{n=0}^{\infty} (n+1)z^n = (\sum_{n=0}^{\infty}z^n)^{'} = (\frac{1}{1-z})^{'} = \frac{1}{(z-1)^2}$$ but shouldn't we have this instead: $$\sum_{n=0}^{\infty} (n+1)z^n = (\sum_{n=0}^{\infty}z^{n+1})^{'}$$ If so how do I go about fixing this issue, or am I understanding the professors work wrong?

答:
Normally, when you differentiate a power series, you alter the lower bound to omit the previously constant term that is now zero.  Showing this step, and your professor's change of index:
\begin{align*}
&\sum_{n=0}^\infty (n+1)z^n  &  &\left[{} = 1 + 2z + 3z^2+\cdots\right] \\
&= \sum_{n = 0}^\infty \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1}  &  &\left[{} = 1 + 2z + 3z^2 + \cdots  \right]  \\
&=_{|z| < 1} \frac{\mathrm{d}}{\mathrm{d}z} \sum_{n = -1}^\infty  z^{n+1}  &  &\left[{} = \frac{\mathrm{d}}{\mathrm{d}z} \left(  1 + z + z^2 + \cdots  \right)\right]  \\
&  &  &[n \mapsto m-1]  \\
&= \frac{\mathrm{d}}{\mathrm{d}z} \sum_{m=0}^\infty z^m  &  &\left[{} = \frac{\mathrm{d}}{\mathrm{d}z} \left(  z + z^2 + \cdots  \right)\right]  \\
\end{align*}
Of course, since $\frac{\mathrm{d}}{\mathrm{d}z}1 = 0$, as was referenced in comments to the Question, these are the same.

The change of index (and why including or excluding the zero doesn't matter) is also discussed here.


问:
Mean and Median Calculations
A group of 20 has a mean = 85 and median = 80. A different group of 30 has a mean = 75 and median = 72. Now, what is the mean of and median for the entire group of 50?  </p>

答: Hint: Use the idea of a weighted mean. </p>

Title: Mean and Median Calculations

Body: Hint: Use the idea of a weighted mean.


问:
Striking applications of integration by parts
What are your favorite applications of integration by parts?

(The answers can be as lowbrow or highbrow as you wish. I'd just like to get a bunch of these in one place!)

Thanks for your contributions, in advance!

答:
Integration by parts shows that (modulo a constant) the Fourier transform interchanges differentiation and multiplication by the variable:

$\begin{align*}
f'(x) \rightarrow \widehat{f'}(\xi) & = \int_{\mathbb{R}} f'(x)e^{-2 \pi i x \xi}dx\\
 & = f(x)e^{-2 \pi i x \xi}|_{-\infty}^{\infty} - \int_{\mathbb{R}} f(x) e^{-2 \pi i x \xi} (-2 \pi i \xi) dx \\
& = (2\pi i \xi) \widehat{f}(\xi)
\end{align*}$

where $f(x)e^{-2 \pi i x \xi}|_{-\infty}^{\infty}$ vanishes if $f$ decays fast.

ContentLicense: CC BY-SA 2.5


问:
Compactly embedded in $L^p(0,1)$ but is not a subspace of $C^0[0,1]$
By the Rellich-Kondrachov theorem, one knows that the embedding $H^1(0,1) \subset L^2(0,1)$ is compact.

On another hand, by Sobolev inequalities, one also has $H^1(0,1) \subset C^0[0,1]$ (in fact, even $C^{0,\frac{1}{2}}$ in this one-dimensional case, by using the fundamental theorem of calculus and some Cauchy-Schwartz arguments).

My question is whether there exists some "intermediate subspace" in the following sense.

Namely, does there exists a Hilbert space $H$ which is compactly embedded in $L^p(0,1)$ for some $p\geq 1$, and which is not a subspace of $C^0[0,1]$?

答:
Yes, such Hilbert spaces exist and they are a special case of fractional Sobolev spaces.
For $\alpha\in(0,1/2)$ we have
$H^\alpha(0,1)\subset L^2(0,1)$ by definition,
and one can show that the step function which is $1$ on $(1/2,1)$ and $0$ else
is in $H^\alpha(0,1)$.
Since this function is not continuous,
$H^\alpha(0,1)$ does not embed in $C^0[0,1]$.

See also Proof that the characteristic function of a bounded open set is in $H^{\alpha}$ iff $\alpha < \frac{1}{2}$
and To what fractional Sobolev spaces does the step function belong? (Sobolev-Slobodeckij norm of step function)
for more details.

It is also known that $H^\alpha(0,1)$ embeds compactly into
$L^2(0,1)$ for $\alpha\in (0,1/2)$.
This follows from Theorem 7.1 in this pdf.


问:
Product of two simply connected compact sets in $\mathbb{C}$ is polynomially convex in $\mathbb{C}^2$?
Is it true that the product of two simply connected compact sets in $\mathbb{C}$ is polynomially convex in $\mathbb{C}^2$ ?

Thanks !

答:
Sure, assuming the two compact sets are nice enough (not just simply connected).  Any points that is not in the product has either the first or the second coordinate not in one of these simple connected sets.  So get a polynomial in the first or the second variable that is bigger at that point then on the compact set, and voila.

Of course, you're in trouble if the compact set in $\mathbb{C}$ is simply connected but its complement is not connected (that could happen if you for example loop one of the sides of the graph of $\sin(1/x)$ back on itself, it's simply connected because there is no continuous loop that goes around).

But if you instead do it with compact sets with connected complements, then you're fine.


问:
Why is the answer of laplace of $1 * t^2$ not equal to that you would get from convolution theorem?
So I know that
$$L(t^2) = \frac{2}{s^3}$$
but why is is that, if we use the convolution theorem for $1 \ast t^2$, we get
$$L(1*t^2) = \frac{1}{s}*\frac{2}{s^3}$$

Isn't $1 \ast t^2$ equal to $t^2$? Or am I missing something in the definition?
Sorry if it's a dumb question but I just really can't seem to understand.

(edit: corrected the fraction)

(edit: sorry about that, I just realized I definitely didn't have a good grasp on the definition!)

答:
$$\mathcal {L}(1*t^2) = \mathcal {L} (1) \mathcal {L}(t^2)=\frac{1}{s}\times \dfrac {2!}{s^3}=\dfrac 2 {s^4}$$
Note that the convolution theorem gives us :
$$f(t)=1 *t^2 =\int^t_0 1 \times \tau ^2 d\tau$$
$$f(t)=\dfrac {\tau ^3}3 \bigg |_0^t=\dfrac {t^3}3$$
$$\mathcal {L}(f(t))=\dfrac 13 \dfrac {3!}{s^4}=\dfrac 2 {s^4}$$


问:
Let $f: X\to X$ be continuous map. Show that if $X=[0,1]$, there is a point $x$ such that $f(x)=x$. What happens if $X$ equals $[0,1)$ or $(0,1)$?

Let $f: X\to X$ be continuous map. Show that if $X=[0,1]$, there is a point $x$ such that $f(x)=x$. What happens if $X$ equals $[0,1)$ or $(0,1)$?

Here is my proof: Let $f:[0,1]\to[0,1]$ be continuous. Suppose neither

$0$ or $1$ are fixed points of $f$ because if they are fixed points, there is

nothing to prove. $f(0)\ne0$ and $f(1)\ne1$. Let $g:[0,1]\to[-1,1]$ be defined by

$g(x)=f(x)-x$. Then, $g(0)=f(0)-0$ or $f(0)>0$, $g(1)=f(1)-1<0$ or $f(1)<1$. By

the immediate value theorem, $\exists x\in(0,1)$ such that $g(x)=0$, which follows

that $f(x)-x=0$ or $f(x)=x$. Hence, there is a fixed point $f$ such that

$f(x)=x$ if $X=[0,1]$.

If $x$ equals $[0,1)$ or $(0,1)$, then $f$ can have a discontinuity at the

endpoints, so we cannot apply the immediate value theorem since we can't

find the values at $0$ and $1$.

答:
Your argument for $[0,1]$ is fine. For the other two cases, however, you need to come up with explicit counterexamples. For each of them you could, for instance, let

$$f(x)=\frac{x+1}2\,:$$

clearly $x<\frac{x+1}2<1$ for $0\le x<1$.

Your statement about discontinuities at the endpoints doesn’t make sense: if one of the endpoints of the interval is included in $X$, then $f$ must be continuous there, and if one is not included, then $f$ isn’t even defined there and so cannot be discontinuous there.


问:
What is the definite integral of...
$$\int^L_{-L} x \sin(\frac{\pi nx}{L})$$

I've seen something like this in Fourier theory, but I'm still not sure how to approach this integral. Wolfram Alpha gives me the answer, but no method. Integrate by parts? Substitution?

答:
Integrate by parts letting $u = x, dv = \sin(ax) dx$, and so $du = dx, v = -\cos (ax)/a$ and you have

$$\int x \sin(ax) dx = \int u \cdot dv = uv - \int v \cdot du
  = \frac{-x\cos(ax)}{a} + \frac{1}{a} \int \cos(ax) dx,$$

which should be much easier.

Title: What is the definite integral of...


问:
Estimating values using the binomial theorem trivial
I want to estimate a value such as $101^{1/3}$ to 4 decimal places.

I have derived the taylor expansion for the binomial $(1+x)^n$ with the remainder.

$f(x) = (1+x_0)^s + s(1+x_0)^{s-1} + ... + \frac{s(s-1)(s-2)...(s-n+1)(1+x)^{s-n}}{n!}(x-x_0)^n + R_n$

With $R_n$ = $\frac{f^{(n+1)}(c)}{n!}(x-x_0)^{n+1}$ for $x_0 < c < x$

I want to sub in $100$ into my formula but I can't since |x| < 1. I think I should do some form of a substitution for x or change my binomial.

I know I should use my remainder term and solve $R_n = 10^{-4}$ to gain the value of $n$ I should take when evaluating my sum however I do not know how to actually evaluate my sum. Also another question - what purpose does my $x_0$ actually serve in this instance, can I just set $x_0 = 0$ and use the Maclaurin expansion instead?

答:
$ 101^{\frac13} = (0.808 \times 125)^{\frac13} = 5 \times (1 + (-0.192))^{\frac13}$

and now use the Binomial expansion, which is valid because $| -0.192| < 1.$


问:
Subfield of cyclotomic extension of degree 2 is either $\Bbb Q(\sqrt{p})$ or $\Bbb Q(\sqrt{-p})$.
From my previous post, I conlcude that 'If $\zeta$ is a primitive $p$th root of 1 ($p\geq 3$), then there is a unique subfield $K\subset\Bbb Q(\zeta)$ such that $[K:\Bbb Q] = 2$'. I want to find the field $K$ explicitly. I hear that such $K = \Bbb Q(\sqrt{p})$ or $\Bbb Q(\sqrt{-p})$. Why is that?

I also heard that this is the special case of 'Kroneker-Weber' theorem.

答:
It's not really Kronecker-Weber, which asserts (as a special case) that every quadratic extension of $\mathbb Q$ is a subfield of a cyclotomic field. This is a much easier question.

A non-intuitive, but profoundly illuminating, argument is via Gauss sums. Namely, letting $\chi(a)$ be the quadratic character mod $p$, and $\psi(a)=e^{2\pi ia/p}$, the Gauss sum is $g(\chi)=\sum_{a\mod p} \chi(a)\psi(a)$.

A standard computation gives $|g(\chi)|=\sqrt{p}$, and $g(\chi)^2=\varepsilon\cdot p$, where $\varepsilon=\pm 1$ depending whether $-1$ is a square mod $p$ or not.

(This computation is very standard, but/and is also eminently do-able, once one knows that it is possible.)


问:
Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$
Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$</p>

答:
Ron:

I am not trying to be pretentious, and I realize this is an old post, but if I may offer a little input on the evaluation of your zeta sum.  If I can assume you're still interested at this point.

I believe it was Choi who has done research on these series.

Using the Barnes G function, many series involving zeta have been evaluated.

Here is a little.  The G represents the Barnes G function. The derivation of this can probably be found by searching for Choi's papers on the matter.

$\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)-1}{k+1}=\frac{-1}{2}log(2\pi)+1/2-\log(2)-\int_{0}^{1}log[G(t+2)]dt+\int_{0}^{1}log[G(t+1)]dt$

Note that $\displaystyle\int_{0}^{1}log[G(t+2)]dt=log(2\pi)-1+C$

$\displaystyle\int_{0}^{1}log[G(t+1)]dt=\frac{1}{2}log(2\pi)+C$

Putting this together we find the C cancels and we arrive at the result of $\displaystyle\frac{3}{2}-log(\pi)$

So, taking your log(2) and subtracting this from it:

$\displaystyle log(2)-(3/2-log(\pi))=log(2\pi)-3/2$ as required.

A working knowledge of the Barnes G function will allow one to evaluate some tough series, especially involving log Gamma or zeta.


问:
On the self-adjoint form for the general elliptic PDE
Under what conditions can a general elliptic PDE
$$a_{ij}(x) \frac{\partial^2 u}{\partial x_i \partial x_j} + b_i(x) \frac{\partial u}{\partial x_i} + c(x)u = f(x)\label{a}\tag{A}
$$
be reduced to a self-adjoint form
$$\frac{\partial}{\partial x_i} \left(\tilde{a}_{ij}(x) \frac{\partial u}{\partial x_j} \right) + \tilde{c}(x) u = \tilde{f}(x)\;?\label{b}\tag{B}
$$

答:
In fact, there exists a necessary and sufficient condition in order for an elliptic PDE of the form \eqref{a} to be put in the self-adjoint form \eqref{b}. In order to see this, let's first drop Einstein summation convention and show the summations explicitly. Then, formally (i.e. without considering, for the moment, the differentiability requirements of the functions involved) we have:
$$
\DeclareMathOperator{\divg}{\nabla\cdot}\begin{split}
\divg \left(\sum_{j=1}^n{a}_{ij}(x) \frac{\partial u(x)}{\partial x_j} \right) & = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_i}\left({a}_{ij}(x) \frac{\partial u(x)}{\partial x_j} \right) \\
& = \sum_{i=1}^n\sum_{j=1}^n\left[ \frac{\partial a_{ij}(x)}{\partial x_i } \frac{\partial u(x)}{\partial x_j} + a_{ij}(x) \frac{\partial^2 u(x)}{\partial x_i \partial x_j} \right]
\end{split}
$$
This implies that, for the two forms \eqref{a} and \eqref{b} of a PDE to be equivalent it should be
$$
\sum_{i=1}^n \left[\sum_{j=1}^n \frac{\partial a_{ij}(x)}{\partial x_i } \frac{\partial u(x)}{\partial x_j} - b_i(x)\frac{\partial u(x)}{\partial x_i } \right]=0\label{1}\tag{1}
$$
However, working a bit by using Kronecker's delta $\delta_{ij}$, we have
$$
\begin{split}
\sum_{i=1}^n \left[\sum_{j=1}^n \frac{\partial a_{ij}(x)}{\partial x_i } \frac{\partial u(x)}{\partial x_j} - b_i(x)\frac{\partial u(x)}{\partial x_i } \right] & = \sum_{i=1}^n \left[\sum_{j=1}^n \frac{\partial a_{ij}(x)}{\partial x_i } \frac{\partial u(x)}{\partial x_j} - \delta_{ij} b_i(x)\frac{\partial u(x)}{\partial x_j } \right] \\
& = \sum_{j=1}^n \left[\sum_{i=1}^n \frac{\partial a_{ij}(x)}{\partial x_i } \frac{\partial u(x)}{\partial x_j} - \delta_{ij} b_i(x)\frac{\partial u(x)}{\partial x_j } \right] \\
& = \sum_{j=1}^n \left[\sum_{i=1}^n \frac{\partial a_{ij}(x)}{\partial x_i }  - \delta_{ij} b_j(x) \right]\frac{\partial u(x)}{\partial x_j }
\end{split}
$$
and, due to the arbitrariness of $\frac{\partial u(x)}{\partial x_i }$ for $i=1,\ldots, n$, we finally get the condition
$$
\left[\sum_{i=1}^n \frac{\partial a_{ij}(x)}{\partial x_i }  - \delta_{ij} b_j(x) \right]=0\qquad\forall j=1,\ldots, n \label{2}\tag{2}
$$
which can be given an elegant form by putting $\big(a_{ij}(x)\big)_{i,j=1,\ldots,n}\triangleq \mathbf A(x)$ and $\mathbf b(x)\triangleq {(b_1,\ldots, b_n)}$:
$$
\divg\mathbf{A}(x)= \mathbf b(x)\label{2'}\tag{2'}
$$

Addendum: is it possible to find an equivalent form \eqref{b} for a given non self-adjoint PDE \eqref{a}? (Follow-up to the comments of Fizikus)

Several answers to the problem of finding an equivalent symmetric form for a given PDE have been given by researchers who investigated the "inverse problem of the calculus of variations": this problem asks, for a differential equation (ordinary or partial, linear or nonlinear), to find a Lagrangian functional such that the give DE is its Euler-Lagrange equation. Since for a self-adjoint DE the solution of this problem is straightforward, the researchers sought to find ways of transforming non self-adjoint DE in self-adjoint ones.

In particular Copson [A1], for equations of type \eqref{a} whose coefficient matrix $A(x)$ is a symmetric, constructed a function $\Phi(x)$ and a linear partial differential operator $\mathscr{L}_{\Phi}(x,\partial_i)$ such that the equation
$$
e^{\Phi(x)} \big( \mathscr{L}_{\Phi}(x,\partial_i)u(x) - f(x)\big)=0
$$
has the self-adjoint form \eqref{b}. The Copson construction is explicit: apart from the original paper, it is also described by Filippov ([A2], §11.2 pp. 94-97), who gives also some "symmetrization methods" which work for particular classes or single equations, both linear and non-linear.

Notes

Formula \eqref{2}-\eqref{2'} is seen in classical monographs on the theory of elliptic PDEs ([2], chapter 1, §6 p. 9 and [3] chapter I, §6 p. 12) where it is clearly stated that it is a necessary and sufficient condition for selfadjointness, without proof possibly due to the basic character of the result.

From the calculation above, it can be seen that condition \eqref{2}-\eqref{2'} is meaningful for weakly differentiable matrix functions $\mathbf{A}(x)$ and $\mathbf b(x)$. This implies also that all the above calculations are rigorously valid for an elliptic PDE with weakly differentiable coefficients as, for example, bounded differentiable functions.

Equation \eqref{2}-\eqref{2'} in the homogeneous case ($\mathbf{b}(x)\equiv\mathbf{0}$) was explicitly solved by Bruno Finzi in 1932 (see [1]) for second, third and fourth order symmetric tensors and after by Maria Pastori in 1942 (see [4]) for every general, not necessarily symmetric, $n$-th order tensors. Perhaps their work could be of some interest in calculating "symmetrizing" factors for transforming the form \eqref{a} in the \eqref{b}.

References

[1] Bruno Finzi, "Integrazione delle equazioni indefinite della meccanica dei sistemi continui", (Italian), Atti della Accademia Nazionale dei Lincei, Rendiconti, VI Serie 19, pp. 620-623 (1934), JFM 60.0708.02.

[2] Carlo Miranda, Equazioni alle derivate parziali di tipo ellittico,  (Italian), Ergebnisse der Mathematik und ihrer Grenzgebiete, 2. Heft, Berlin-Göttingen-Heidelberg: Springer-Verlag pp. VIII+222 (1955), MR0087853, Zbl 0065.08503.

[3]  Carlo Miranda (1970) [1955], Partial Differential Equations of Elliptic Type, Ergebnisse der Mathematik und ihrer Grenzgebiete – 2 Folge, Band 2, translated by Motteler, Zane C. (2nd Revised ed.), Berlin – Heidelberg – New York: Springer Verlag, pp. XII+370, doi:10.1007/978-3-642-87773-5, ISBN 978-3-540-04804-6, MR 0284700, Zbl 0198.14101.

[4] Maria Pastori, "Integrale generale dell’equazione $\operatorname{div}\mathsf T=0$ negli spazi euclidei", (Italian)
Rendiconti di Matematica e delle Sue Applicazioni, V Serie, 3, pp. 106-112 (1942), MR0018968 Zbl 0027.13302.

Addendum references

[A1] Edward Thomas Copson, "Partial differential equations and the calculus of variations" Proceedings of the Royal Society of Edinburgh 46, 126-135 (1926), JFM 52.0509.01.

[A2] Vladimir Mikhailovich Filippov,, Variational principles for nonpotential operators. With an appendix by the author and V. M. Savchin, Transl. from the Russian by J. R. Schulenberger. Transl. ed. by Ben Silver, Translations of Mathematical Monographs, 77. Providence, RI: American Mathematical Society (AMS). pp. xiii+239 (1989), ISBN: 0-8218-4529-2. MR1013998, ZBL0682.35006.


问:
Differentiating both sides of a differential equation introduces an extra constant?
$$y'=y^2$$

This differential equation can be solved by dividing both sides with $y^2$ and integrating both sides to obtain $$y=\frac{1}{c_1-x}$$

But if I differentiate both sides instead, I get $$y''=2yy'=2y^3=2y'^\frac{3}{2}$$

This is a Bernoulli DE that can be solved with the substitution $u=y'^\frac{-1}{2}$ and results in

$$y'=\frac{1}{(-x+c_1)^2}$$

$$y=\frac{1}{c_1-x}+c_2$$

Is the method of differentiating both sides of the DE correct?

It resulted in an extra constant and I'm not sure what went wrong.

答: What surprises you? Differentiating the differential equation $y'= 0$ introduces another constant: you have linear solutions in addition to the constants.


问:
Quotient, Tensor Quasicoherent Sheaves on Affine Open Subsets
Quotient and tensor sheaves are defined by sheafification which is somehow implicit. Is there any easy way to handle these constructions?

Could we ignore the sheafification when we are working in affine open subsets?

For example, let $X$ be a scheme and $\mathcal F$ and $\mathcal G$ be quasicoherent sheaves. Let $U\subseteq X$ be an affine open. Is it true that  $(\mathcal F/\mathcal G)(U) = \mathcal F(U)/\mathcal G(U)$ and $\mathcal F\otimes_{\mathcal O_X}\mathcal G(U) = \mathcal F(U)\otimes_{\mathcal O_X(U)}\mathcal G(U)$?

How about other constructions involving sheafification? (for example, image, cokernel, ann, nil and so on)

Thank you in advance.

答:
The key statement to verify is whether the thing you're looking at commutes with localization. If it does, then it commutes with sections on affine opens: start by defining the presheaf which takes the naive value on every open set, and then note that on principal affine opens $D(f)$ you get exactly the value you would expect from the quasi-coherent sheaf associated to the operation on the module of global sections. As the principal affine opens form a basis for the topology on an affine scheme and a sheaf can be specified by its values on a basis, the result follows.

In particular, this means that kernels, cokernels, images, and any other "homological" constructions commute with the associate sheaf functor on an affine scheme because localization is exact. Tensor products and nilradicals commute with localization, so they commute with the associated sheaf functor, too. Annihilators don't always commute with localization, so you need to be more careful with them.

Let me just mention one alternate neat proof that tensor products commute with sheafification, which I originally saw at the Stacks Project: consider the morphism of ringed spaces $\pi:(\operatorname{Spec} R,\mathcal{O}_{\operatorname{Spec} R})\to (\{pt\},R)$. Then the associated sheaf functor is $\pi^*$, which means it commutes with tensor products by the standard proof.


问:
It is true that $ \lim_{x \to a} f(x) = L \Leftrightarrow \lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = L $?
$$ \lim_{x \to a} f(x) = L \Leftrightarrow \lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = L    $$

Do you think it is true? I need to know that to determine these two limits:

$ \lim_{x \to 0} x^{|\tan x|} \overset{??}{=} 1 $

$ \lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^x-1-\ln(1+x))^3} \overset{??}{=} \frac{1}{36} $

It is clear that it is fulfilled when $ x \to 0^+ $, but what about when $ x \to 0^- $?

Thank you all!

答:
hint

For the first, you can only compute $\lim_{x\to 0^+}$ using

$$(\forall x>0)\; x^{|\tan(x)|}=e^{\frac{\sin(x)}{x\cos(x)}x\ln(x)}$$
and the well-known limits.

For the second, you will need Taylor expansion
$$e^x=1+x+\frac{x^2}{2}(1+\epsilon_1(x))$$
$$\ln(1+x)=x-\frac{x^2}{2}(1+\epsilon_2(x))$$
$$\sin(x)=x-\frac{x^3}{6}(1+\epsilon_3(x))$$
and

$$\tanh(x)=x-\frac{x^3}{3}(1+\epsilon_4(x))$$


问:
The boundary behavior of the series $\sum_{n=1}^{\infty}\sin(z^{n})$ on $|z|=1$.
A similar question was asked here: Analytic function in the open unit disc. It was actually partially solved.

The problem is as follows:

Show that the series $\sum_{n=1}^{\infty}\sin(z^{n})$ defines an analytic function in $|z|<1$. What happens on the boundary?

The first part can be easily proved if one knows that $|\sin(z)|\leq C|z|$ for all $|z|\leq 1$. This can be proved if one considers the power series of $\sin(z)/z$, and the module of the power series is always bounded by the power series of $\sin(z)/z$ at $z=i$, so $\sin(i)/i$ is the maximum, which is $C$.

Okay, now, if $z\in\overline{\mathbb{D}}$, then clearly $z^{n}\in\overline{\mathbb{D}}$. Hence, the inequality can be applied to $z^{n}$ so that $$|\sin(z^{n})|\leq C|z^{n}|\ \ \text{for all}\ \ |z|\leq 1.$$

In particular, on $|z|<1$, we have $$\Bigg|\sum_{n=1}^{\infty}\sin(z^{n})\Bigg|\leq\sum_{n=1}^{\infty}|\sin(z^{n})|\leq C\sum_{n=1}^{\infty}|z|^{n}=C\sum_{n=1}^{\infty}r^{n}<\infty,$$ since $r<1$. Thus, the series converges absolutely on $|z|<1$ and thus defines an analytic function on $|z|<1$.

Now, I am having trouble analyzing the boundary behavior. Indeed, the series clearly diverges when $z=1$, since then you have $$\sum_{n=1}^{\infty}\sin(1)=\sin(1)\sum_{n=1}^{\infty}1=\infty.$$ But can we say something more? Like, perhaps $1$ is the only point that this series diverges?

I tried to use the common trick that $z=e^{2\pi \frac{p}{q}i}$ to show the divergence and use the density, but it seems useless..

Any idea? Thanks for any help!

答:
Thanks to Conrad's inspiring comment, I derived the following proof:

We firstly consider $|\sin(z)|$. Note that $$\sin(z)=\sin(x+iy)=\sin(x)\cos(iy)+\cos(x)\sin(iy),$$ and $$e^{iy}=\cos(y)+i\sin(y)\implies e^{-y}=e^{i(iy)}=\cos(iy)+i\sin(iy)$$ $$e^{-iy}=\cos(y)-i\sin(y)\implies e^{y}=e^{-i(iy)}=\cos(iy)-i\sin(iy).$$ From these two, one could derive the formula of $\cos(iy)$ and $\sin(iy)$. (I always do this since I can never remember the formula of $\sinh(\cdot)$ and $\cosh(\cdot)$.

After rearranging the real part and imaginary part and taking $\sin(z)\overline{\sin(z)}$, and using $\cos^{2}(x)=1-\sin^{2}(x)$, one could get $$|\sin(z)|^{2}=\dfrac{e^{2y}-2+e^{-2y}}{4}+\sin^{2}(x)\geq \dfrac{e^{2y}-2+e^{-2y}}{4}+0.$$

Now, for $|z|=1$, we have $x,y\in [-1,1]$. Take $f(y):=\frac{e^{2y}-2+e^{-2y}}{4},$ and compute the critical point of it which is $z_{0}=0$, and one could compare as $$f(0)=\dfrac{1}{4}\ \ \text{but}\ \ f(1)=f(-1)=\dfrac{1}{4}\Bigg(\dfrac{1}{e^{2}}+e^{2}\Bigg)>\dfrac{1}{4}\Bigg(\dfrac{1}{9}+4\Bigg)=\dfrac{37}{36}>\dfrac{1}{4},$$ so the absolute minimum of $f(y)$ in $[-1,1]$ is $\dfrac{1}{4}$.

This implies that for $|z|=1$, $$|sin(z)|^{2}\geq \dfrac{1}{4}\implies |\sin(z)|\geq \dfrac{1}{2}.$$

Note that $|z^{n}|=1\iff |z|=1$, so this inequality can be applied to $z^{n}$ so that $$|\sin(z^{n})|\geq C>0\ \text{on}\ |z|=1.$$

This implies that on $|z|=1$, $$\lim_{n\rightarrow \infty}|\sin(z^{n})|\neq 0\iff\lim_{n\rightarrow\infty}\sin(z^{n})\neq 0,$$ which means that the original series must diverge on the whole $|z|=1$.


问:
Hahn-Banach theorem in separable space
Let $E$ be a normed separable space with infinite dimension.

a) Show that there exist a sequence $(M_k)$ of finite dimension subspaces such that $M_k\subset M_{k+1}$, for all $k$, and $\bigcup_{k=1}^{\infty}M_k$ is a dense subset of $E$.

b) Using Hahn-Banach Theorem, show that there exist a sequence of functions $(\varphi_k)\subset E'$ such that $\Vert \varphi_k \Vert=1$, for all $k$, and $\displaystyle\lim_{k\rightarrow\infty}\varphi_k(x)=0$ for all $x\in E$.

For item a), let $D=\{x_k;k\in\mathbb{N}\}$ a dense subset of $E$ which exists beacuse $E$ is separable. Then $M_k=span\{x_1,...,x_k\}$ solves.

My problem is with item b), can anyone help me with ideas?

答:
Fix $k$. Choose $y\in E\setminus M_k$. On $\operatorname{span}M_k\cup\{y\}$, define a linear functional $\varphi_k$ with $\varphi_k(y)=1$, $\varphi_k(x_j)=0$, $j=1,\ldots,k$. This works because $y$ is linearly independent with $M_k$.

We need to show that $\varphi_k$ is bounded. We have
\begin{align}
\|\varphi_k\|&=\sup\{\frac{|\varphi_k(x+\lambda y)|}{\|x+\lambda y\|}:\ \lambda\in\mathbb C\setminus\{0\},\ x\in M_k\}\\[0.3cm]
&=\sup\{\frac{|\lambda|}{\|x+\lambda y\|}:\ \lambda\in\mathbb C\setminus\{0\},\ x\in M_k\}\\[0.3cm]
&=\sup\{\frac{1}{\|x/\lambda+ y\|}:\ \lambda\in\mathbb C\setminus\{0\},\ x\in M_k\}\\[0.3cm]
&=\sup\{\frac{1}{\|y-x\|}:\ x\in M_k\}\\[0.3cm]
&=\frac1{\inf\{\|y-x\|:\ x\in M_k\}}\\[0.3cm]
&=\frac1{\operatorname{dist}(y,M_k)}<\infty
\end{align}
Knowing that $\varphi_k$ is bounded, we can extend it by Hahn-Banach to $\varphi_k\in E'$, that satisfies $\varphi_k|_{M_k}=0$.

Now, given $x\in E$ and $\varepsilon>0$, there exists $k_0$ such that $\|x-x_{k_0}\|<\varepsilon$. Then, for any $k>k_0$,
$$
|\varphi_k(x)|\leq|\varphi_k(x-x_{k_0})|+|\varphi_k(x_{k_0})|
=|\varphi_k(x-x_{k_0})|\leq\|x-x_{k_0}\|<\varepsilon.
$$
Thus $\lim_k\varphi_k(x)=0$.

Title: Hahn-Banach theorem in separable space


问:
Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$?
I suspect that

$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
 {{\pi}\over{2}}$$

for $C>0$.

I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?

答:
So, you can start out by writing the denominator like this: $ y^4+y^2(C-2)+1=(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})$. This is simply the solution of a quadratic equation (ugly, but it will do). So, now, we want to simplify this expression somehow, like by writing:

$ \frac{y^2}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{A}{y^2-\frac{2-C-\sqrt{C(C-4)}}{2}}+\frac{B}{y^2-\frac{2-C+\sqrt{C(C-4)}}{2}}=\frac{A(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})+B(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{y^2(A+B)-1(A(\frac{2-C-\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}$.

Therefore,we want $A+B=1 \wedge A(\frac{2-C+\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4}}{2})=0$. This is a linear system, that you can solve to find A and B, and then calculate the integral.


问:
radius of nth circle inside a circle of given radius
A circle of radius 100 cm is given. The goal is to place 7 circles of unequal radius inside the initial circle, so that each 3 of circles don't overlap each other and all of them stay inside the initial circle.
The first smaller circle is of radius 50 cm and so on.

I thought it has to do something with Ford circles, but answer r = 14.36 cm for the 7th circle is incorrect.

答:
Hint:

One way is to consider
this construction as a part of the (infinite)
Steiner’s chain of circles defined by two circles,
$\bigcirc O((0,0), R)$ and $\bigcirc o((-r,0),r)$,
with the distance between their centers $d=r$,
where $R=100,\ r=50$.
The other $6$  circles $\bigcirc O_0,\dots,\bigcirc O_5$
are sandwiched between the two,
starting with $\bigcirc O_0((r,0),r)$.

All tangent points of the kissing circles in the chain
are located on the circle (shown green in the picture) centered at $X$ with radius
$r_x=|XO_i|$, $i=0,\dots,5$,

\begin{align}
r_x&=
\sqrt{rR-\frac{rRd^2}{(r+R)^2}}
=\frac{200}3\approx 66.7
,
\end{align}

while all the centers $O_i$, $i=0,\dots,5$,
belong to the ellipse (shown yellow in the picture)
focused at $o,\,O$.

Edit:

Using Descartes' theorem,
for the curvatures
$k_1,k_2,k_3,k_k$
of four mutually tangent circles,
\begin{align}
k_4&=k_1+k_2+k_3\pm 2\sqrt{k_1 k_2+k_2 k_3+k_3 k_1}
\tag{1}\label{1}
.
\end{align}

In this particular arrangement of circles,
\begin{align}
k_1&=-\tfrac1R
,\quad k_2=\tfrac2R
,\quad k_3=k_2=\tfrac2R
\end{align}

since each next circle in the chain is smaller,
the sign in \eqref{1} would be always positive
and we have

\begin{align}
k_4&=\tfrac3R
,\\
k_5&=\tfrac{6}R
,\\
k_6&=\tfrac{11}R
,\\
k_7&=\tfrac{18}R
,\\
k_8&=\tfrac{27}R
,
\end{align}

so the radii of
$\bigcirc O_i$, $i=1,\dots,5$
can be found as

\begin{align}
r_i&=\frac R{i^2+2}
.
\end{align}

For example, the seventh inner circle, the smallest one,
must have the radius

\begin{align}
r_5&=\frac R{27}
\approx 3.7037
.
\end{align}


问:
$F_k|F_n$ given $k | n$
Problem: Given $n, k \in \mathbb{N}$ such that $k | n$, show that $F_k|F_n$.

My attempt: We have that $F_n = |\tau_{n-1}|$ where $|\tau_{n-1}|$ represents the number of ways to tile a row of $n-1$ squares using dominos(which can cover two squares) and monominos(which can cover one square). Similarly, $F_k = |\tau_{k-1}|$. Then since $k|n$, we have that $n = ak$ for some $a \in \mathbb{N}$. So $$|\tau_{n-1}| = |\tau_{k-1}|\cdot|\tau_{n-k}| = |\tau_{k-1}|\cdot |\tau_{(a-1)k}| = |\tau_{k-1}|\cdot|\tau_k|^{a-1},$$
where the last equality is obtained from the product rule. Then this shows that $|\tau_{k-1}|$ divides $|\tau_{n-1}|$, and so $F_k$ divides $F_n$.

Question: Is this approach correct or did I make a mistake somewhere? Thank you.

答:
Hint: You are getting close! Your formula for $\tau_{n-1}$ is incorrect, since it does not account for tilings where a domino covers squares numbered $k-1$ and $k$. So, how many of those tilings did you miss? Adding in the number of missing tilings gives the correct formula:
$$
\tau_{n-1}=\tau_{k-1}\tau_{n-k}+\underline{\hspace{2cm}}
$$
and you can use the corrected formula to prove that $F_{n}$ is a multiple of $F_{k}$.

Further hint, if you need it:

 Start with the special case $n=2k$, then $n=3k$, until the pattern of the proof is clear.

Title: $F_k|F_n$ given $k | n$


问:
An inverse of the Borel–Cantelli lemma (without requiring the independence)
There is a theorem, a converse to the Borel-Cantelli lemma (without requiring the independence), that I learnt from a talk of which I don't know the source:

Let $(f_n)$ be a sequence of nonnegative functions on the probability space $(X,\mathcal{B},\mu)$. Assume that there exists a constant $C>0$ such that for every $N\ge 1$,

$$\sum_{m,n=1}^{N}|\int_X f_m f_n d\mu-\int_X f_m d\mu \int_X f_n d\mu|\le C\sum_{n=1}^N \int_X f_n,$$

then

$$\sum_{n=1}^N \int_X f_n=\infty ~\text{implies}~ \sum_{n=1}^{\infty}f_n(x)=\infty$$

for almost every $x\in X$.

Can anyone show me any reference for this theorem (or a direct proof if not too long)?

答:
Let's

Rewrite in terms of RVs and Expectations.

Weaken the condition.

Suppose $X_1,X_2,\dots$ are nonnegative RVs. Let $S_N = \sum_{n=1}^N X_n$. Suppose that for some constant  $C>0$,

$$(*)\quad \mbox{Var} (S_N)\le C E[S_N]$$
for all $N$.

Note that
\begin{align*} \mbox{Var}(S_N) &= E[ (S_N- E[S_N])^2] \\
& = E[ \sum_{n=1}^N \sum_{m=1}^N (X_n-E[X_n])(X_m-E[X_m]) ] \\
& = \sum_{n=1}^N \sum_{m=1}^N E[X_n X_m] -E[X_n]E[X_m].
\end{align*}

Then $\lim_{N\to\infty} E[S_N]=\infty$ implies $\lim_{N\to\infty} S_N = \infty$, a.s.

The idea is this: use $(*)$ and Chebychev to show that $S_N$ is typically "close" to $E[S_N]$. Let's do it.

$$ P(|S_N  - E[S_N]| > \frac 12  E[S_N]) \overset{\mbox{Chebychev}}{\le} \frac{4\mbox{Var} (S_N)}{E[S_N]^2 }\overset{(*)}{\le} \frac{4C}{E[S_N]}.$$

Now take a subsequence $N_1< N_2 < \dots$ such that $\sum_{k=1}^\infty \frac{1}{E[S_{N_k}]}< \infty$. It follows form Borel-Cantelli that $|S_{N_k} -E[S_{N_k}]|> \frac 12 E[S_{N_k}]$ finitely often a.s. In particular, $S_{N_k} \ge \frac 12 E[S_{N_k}]$, eventually a.s. This implies $\lim_{N\to\infty} S_N =\infty$ a.s.

A comment: the argument shows that we can replace the RHS in $(*)$ by $C (E[S_N])^{\alpha}$ with $\alpha \in [0,2)$.


问:
Covering a rectangular board with Tetrominoes
I am reading about a puzzle question that is about Tetrominoes and proving that if a rectangular board can be covered with T-Tetrominoes the board's number of squares has to be a multiple of 8.

The solution first proves that if a rectangular board can be covered by Tetrominoes one of the sides must be even. This part is clear.

Now the solution is using a coloring "trick" i.e. it colors both the board and the Tetrominoes. The coloring of the tetrominoes is either e.g.

```<code>-----------------------
|-BLACK-|-WHITE-|BLACK|
 -------|-BLACK-|-----
         -------
</code>```

or

```<code>-------------------------
|-WHITE-|-BLACK-|-WHITE-|
 -------|-WHITE-|-------
         -------
</code>```

Also the rectangular board is considered as colored like a chessboard.

The part that I don't understand is the following.

Based on the fact that was proven earlier i.e. that one of the rectangular boards side must be even if the rectangular board can be covered with Tetrominoes at all, the solution consider this implies that the number of black equals the number of white.

This part (black = white) is not obvious to me. Could someone please help?

答: Suppose that one side has $2n$ cells. Then every row parallel to that side has $n$ black and $n$ white cells. Thus, if there are $m$ rows, there are $mn$ black and $mn$ white cells.


问:
Showing continuity between two metric spaces.
Let $K \subset C([0,1])$ such that for each $f \in K$, we have $|f(0)| \leq 10$ and $f$ has Lipschitz constant at most 50. Note that $K$ is in fact a compact subset. This can be shown by the Arzela-Ascoli Theorem but I will omit the details.

We want to show there exists a function $f \in K$ such that $\int_0^1 \cos{f(x)}dx \geq \int_0^1 \cos{g(x)}dx$ for all $g \in K$.

Show: The function $T: K \rightarrow \mathbb{R}$ defined by $T(f) = \int_0^1 \cos{f(x)}$ is continuous.

Since $K$ is compact, by EVT we will get our maximum $f$.

Let $\epsilon > 0$. We need some $\delta > 0$ which we will try to determine after our rough work. But we will assume $\|f-g\|_\infty < \delta$.

Now for our rough work.
\begin{align*}
\Big|\int_0^1\cos{f(x)}dx - \int_0^1\cos{g(x)dx}\Big| &= \Big|\int_0^1(\cos{f(x)} -\cos{g(x)})dx\Big| \\
&\leq \int_0^1|\cos{f(x)} -\cos{g(x)}|dx \\
&\leq \int_0^1\|\cos{f(x)} -\cos{g(x)}\|_\infty dx \\
&= \|\cos{f(x)} -\cos{g(x)}\|_\infty
\end{align*}
But now I'm having trouble using the fact that $\|f-g\|_\infty < \delta$ to complete this proof. Perhaps I'm approaching this wrong. Or I'm overlooking some sort of algebraic trick to get rid of the cosines above. Any help would be appreciated.

答:
You have already done a lot of the work, good job!

Note that $\cos$ is Lipschitz continuous with Lipschitz constant $1$
(This is because the Lipschitz constant can be bounded
by the $\infty$-norm of the derivative).
Thus we have
$$
|\cos f(x)-\cos g(x)| \leq |f(x)-g(x)|
$$
for all $x\in [0,1]$.
Taking the supremum over these $x$ leads to the estimate
with $\|f-g\|_\infty$ which you need.


答: If $n=0,$ this is simple, so suppose not. Use the substitution $$u=\frac{n\pi}lx,$$ so that $$x=\frac{l}{n\pi}u,$$ and so $$\int_{-l}^lx\sin\left(\frac{n\pi}lx\right)\,dx=\frac{l^2}{n^2\pi^2}\int_{-n\pi}^{n\pi}u\sin u\,du,$$ then integrate by parts.</p>

Title: What is the definite integral of...

Body:
$$\int^L_{-L} x \sin(\frac{\pi nx}{L})$$

I've seen something like this in Fourier theory, but I'm still not sure how to approach this integral. Wolfram Alpha gives me the answer, but no method. Integrate by parts? Substitution?


问:
Adjoining an inverse to a central element of an algebra
Given a (not necessarily commutative) algebra $A$, and a central element $c \in A$, is it always possible to enlarge $A$ to a an algebra $A'$ in which $c$ is invertible? I guess one can take a set of generators and relations for $A$, and add a formal inverse $c^{-1}$ along with extra relations making $c^{-1}$ central and satisfying $c^{-1}c = c^{-1}c = 1$. But I guess this is a specific example of a general formal process. What could this process be?

答:
Let $S=\{a^n \mid n \in \Bbb N\}$.
Then $S$ satisfies the Ore condition, so one can form the Ore localization $AS^{-1}$. This is constructed in the same way that localizations of commutative rings work. This is treated in many non-commutative algebra books, e.g. in Lam's "Lectures on Modules and Rings".

In this special case, there's another contruction yielding the same result: we can form the polynomial ring $A[x]$ and divide by the two-sided ideal $(ax-1)A[x]=A[x](ax-1)=A[x](ax-1)A[x]$ to get $A[x]/(ax-1)A[x]$. The class of $x$ will be a formal inverse of $a$.


问:
Let $a$,$b$,$c$ be real numbers.
Let $a$,$b$,$c$ be real numbers such that $a^{4}+b^{4}+c^{4}\leq27$. Prove the inequality:
$$\sum_{cyc}\frac{1}{\sqrt{a^{2}+1}}\geq \frac{3}{2}$$

My solution:
$\frac{1}{\sqrt{a^{2}+1}}\geq\frac{1}{2}\implies a^{2}\leq 3\implies a^{4}\leq 9$.
Similary : $b^{4}\leq9$ and $c^{4}\leq9$.
Then $a^{4}+b^{4}+c^{4}\leq27$ which is obviously true.

My solution is good ?

Title: Let $a$,$b$,$c$ be real numbers.

Body:
Let $x_a = \sqrt{a^2 + 1}$, and similarly for $x_b$ and $x_c$. Then by RMS-HM inequality on $x_a, x_b, x_c$,

$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}}
&\le \sqrt{\frac{x_a^2+x_b^2+x_c^2}3} \\
&= \sqrt{\frac{(a^2+1)+(b^2+1)+(c^2+1)}3}\\
&= \sqrt{\frac{a^2+b^2+c^2}3 + 1}\tag1\\
\end{align*}$$

By RMS-AM inequality on $a^2,b^2,c^2$,

$$\begin{align*}
\frac{a^2+b^2+c^2}3
&\le \sqrt{\frac{a^4+b^4+c^4}3}\\
&\le \sqrt{\frac{27}3}\\
&= 3\tag2\\
\end{align*}$$

Combining results $(1)$ and $(2)$,

$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}} &\le \sqrt{3+1}\\
\sum \frac{1}{\sqrt{a^2+1}} &\ge \frac{3}{2}\\
\end{align*}$$


问:
Generator for $\mathbb{Z}[x]$ ring?
Is the polynomial ring $\mathbb{Z}[x]$ generated exclusively by $\langle 1,x\rangle$? Or has to be generated by more or less elements?
My intuition is that you can generate any of the "$x$'s" polynomials with just multiplying and adding $x$ to the equation but you need the "$1$" to generate the coefficient with degree $0$.

答: Your intuition would be correct. The only way to generate $\mathbb{Z}\subset \mathbb{Z}[x]$ is by having 1 in your ideal. Similarly, you must have x in your ideal to generate all the polynomials.

Title: Generator for $\mathbb{Z}[x]$ ring?


问:
Show that S is non-compact and deduce further that the closed unit ball in X is non-compact.
I have an exercise I am not sure how to solve
The exercise is:
Let X be an infinite dimensional normed vector space over $\mathbb{K}$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$. Define the unit sphere $S=\{x \in X : \|x\|=1\}$

Show that S is non-compact and deduce further that the closed unit ball in X is non-compact.

I am not sure how to show that S is non-compact, but for the second part where I have to deduce that the closed unit ball in X is non-compact (I denote closed unit ball in X by B)  I am thinking that I will show B is compact $\Rightarrow$ X is finite.

I assume that B is compact and then since S is a closed subset of B, then S will be compact and then I can use the first part to say that X may be finite. So now I have shown that B is compact $\Rightarrow$ X is finite, hence if X is infinite then B is non-compact. But I have troubles showing that S is non-compact, so I was hoping that someone could help me with that. We have already shown following:

答:
$(1).$ Suppose $Y$ is a closed vector-subspace of $X$ and $x\in X$ \ $Y.$ Let $d(x,Y)=\inf_{y\in Y}\|x-y\|$. Take $z\in Y$ with $$(*)...\quad \|x-z\|<2d(x,Y).$$ Let $w=(x-z)/\|x-z\|.$ Then we have $$(**)...\quad d(w, Y)=\inf_{y\in Y}\|w-y\|=\|x-z\|^{-1}\inf_{y\in Y}\|\,(x-(z+y\|x-z\|)\,\|.$$ Now $\{z+y\|x-z\|:y\in Y\}=Y.$ So $\inf_{y\in Y} \|(x-(z+y\|x-z\|)\|=\inf_{y\in Y}\|x-y\|=d(x,Y).$ So from $(*)$ & $(**)$ we have $$d(w,Y)=\|x-z\|^{-1}d(x,Y)>\|x-z\|^{-1}\|x-z\|/2=1/2.$$

So if $Y$ is a closed vector-subspace of $X$ and $Y\ne X$ then there exists $w\in X$ \ $Y$ with $\|w\|=1$ and $d(w,Y)>1/2.$

$(2).$ Prove that if $Y$ is a finite-dimensional vector-subspace of $X$ then $Y$ is closed.

$(3).$ If $X$ is infinite-dimensional: Take $y_1\in X$  with $\|y_1\|=1.$ Recursively for $n\in \Bbb N$ let $Y_n=Span(y_1,...,y_n)$ and by $(1)$ & $(2)$ take $y_{n+1}\in X$ \ $Y_n$ with $\|y_{n+1}\|=1$ and $d(y_{n+1}, Y_n)>1/2.$

Let $T=\{y_n:n\in \Bbb N\}.$ Then $T$ is an infinite set and $\|y-y'\|>1/2$ for any  distinct $y,y'\in T.$ So $T$ is an infinite closed discrete topological-subspace of $X.$

Consider $X, S$, and the closed unit ball $B$ as metric spaces with the metric $e(u,v)=\|u-v\|.$ Then $T$ is an infinite closed discrete topological-subspace of $S$ and of $B.$ Any metric space with such a topological-subspace cannot be compact.

[... Any subset of a closed discrete topological-subspace is also closed. So if $U_n=X\setminus \{y_j:j\ge n\}$ then $\{U_n:n\in\Bbb N\}$ is an open cover of $S$ (and of $B$) with no finite subcover...].

Remarks: In $(1)$ we could take $z\in Y$ with $\|x-z\|$ arbitrarily close to $d(x,Y)$ and hence $d(w,Y)$ can be arbitrarily close to $1.$ And if $Y$ is finite-dimensional we can show there exists $z\in Y$ with $d(x,Y)=\|x-z\|$ and hence $d(w, Y)=1$.


问:
Using probability to determine dependent events
In my HS math class, I'm teaching Prob & Stats.  We are studying how the intersection of events can show dependence.  For example, we took a survey of students in the HS:  57% of those surveyed are in Extra-curricular Activities, 65% of those surveyed are on the Honor Roll, and 46% are both.  If these two events were independent, the the "both" (intersection) would be equal to their product: P(EC) * P(HR) = 37%.  Since the observed intersection is higher, these have a positive correlation.

Is this valid statistics?  I.e., is this kind of data collection and statistical analysis something that is done in research, in order to show events are independent or dependent?  I'd like to be able to tell my students that "This is how it is done in the 'Real World.'"  Can anyone give me examples of real-world situations where this is done?

答: The chi-square test uses the idea you mentioned to test for independence of variables. The null hypothesis in the chi-square is that the data are independent and it constructs the numbers one should expect to see along the same lines of what you mentioned in your question. The expected frequencies are then compared with observed to assess if the variables are in fact independent of each other. </p>


问:
Show that the norm of Gaussian prime of form $a + bi$ is always prime.
I want to prove the partial converse of the following theorem:

Let $z \in \mathbb{Z}[i]$. If $N(z)$ - prime, then $z$ is also prime in $\mathbb{Z}[i]$.

I state the converse like this:

Let $z \in \mathbb{Z}[i]$, $z = a + bi$, $a,b \in \mathbb{Z}$ and both non-zero. If $z$ - prime in $\mathbb{Z}[i]$, then $N(z)$ is also prime.

I have tried to assume that $N(z) = a^2 + b^2 = n$ - composite, and then reach a contradiction, but unfourtunately had no success with it. Since $z$ - prime, writing $z = xy$ involves that either $x$ or $y$ is a unit. By multiplicativity of the norm in $\mathbb{Z}[i]$ I have $N(z) = N(x)N(y)$, and though $N(z) = N(x)$ or $N(z) = N(y)$. But from this point I don't know how to proceed and need some hint.

Would appreciate any help, thank you in advance.

答: $z=a+ib$, so $\overline{z}=a-ib$, thus $z\overline{z}=a^2+b^2=N(z)$, showing that $z \mid N(z)$. $z$ is not a unit, so $N(z)$ is not $1$ and we have a prime factorization $N(z)=\prod_{i=1}^n p_i$. We have $z \mid N(z) =\prod_{i=1}^n p_i$. Thus for some $j$, we will have $z \mid p_j$. Set $p=p_j$. We have $zy=p$ for some $y \in \Bbb Z[i]$. Applying conjugates we get $\overline{z} \overline{y} =p$. This shows that both $z$ and $\overline{z}$ divide $p$. As $z$ does not divide $\overline{z}$ (nota bene: this step uses $a,b \neq 0$) and $z$ is prime, we have $\gcd(z,\overline{z})=1$, so the product $z\overline{z}$ must divide $p$ as well. But $z\overline{z}=N(z)$ is an integer, so from $N(z) \mid p$, we get $N(z)=p$ a required.


问:
Finding Maclaurin series for antiderivative, such that series satisfies some initial condition
I'm given the Maclaurin series representation for some $f(x)$. Take, for example:

$$f(x) = \arctan{x^2}$$

I want now to determine the Maclaurin series for $F = \int f(0)$, such that $F(0) = c$. If you like, take $c = 0$, for example.

I'm not sure how to approach this. Need this be different for each $c$?

Body:
If near zero, we have
$$g'(x)=f(x)$$
$$g(0)=C$$
and
$$f(x)=a_0+a_1x+...a_nx^n(1+\epsilon(x))$$
$$=P_n(x)+x^n\epsilon(x)$$

then

$$g(x)=C+\int_0^xP_n(t)dt+x^{n+1}\epsilon(x)$$

for example
$$\cos(x)=1+\int_0^x(-\sin(t))dt$$
$$=1-\int_0^x(t-\frac{t^3}{6}+...)..$$


问:
$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$ for natural numbers $n_1 < n_2 < ... < n_k$
Prove that for natural numbers $n_1 < n_2 < ... < n_k$:

$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$

I have no idea how to do that. I would go for induction, but there is an infromation that $n_1 < n_2 < ... < n_k$ and induction would not recognise that fact.

What's funny is that I think I saw once that exercise on stack exchange but wasn't able to find it this time.

答:
With (double) induction on $k$. Notice that $n_i^3\geq n_i^2\cdot i$ for all $i$.

If $k=1$ the we have $n_1^3\geq n_1^2$ which is true.

Suppose it is true for $k$:
$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$
and we prove it for $k+1$:

\begin{align}n_1^3 + n_2^3 + ... + n_k^3 +n_{k+1}^3 &\geq (n_1 +n_2 + ...+ n_k)^2 + n_{k+1}^3\\\end{align}

We have to check if $$(n_1 +n_2 + ...+ n_k)^2 + n_{k+1}^3 \geq (n_1 +n_2 + ...+ n_{k+1})^2 $$ i.e. $$n_{k+1}^3 \geq 2(n_1 +n_2 + ...+ n_k)n_{k+1}  +n_{k+1}^2 $$ i.e.  $$n_{k+1}(n_{k+1}-1) \geq 2(n_1 +n_2 + ...+ n_k) \;\;\;(*)$$ is true.

Now this $(*)$ can be separately proved again by induction: Since $n_{k+1}\geq n_k+1$ we have $$n_{k+1}(n_{k+1}-1) \geq n_k(n_k+1) = n_k^2+n_k$$ so it is enough to prove $$n_k^2+n_k\geq 2(n_1 +n_2 + ...+ n_k) $$  i.e. $$n_k^2-n_k\geq 2(n_1 +n_2 + ...+ n_{k-1}) $$

which is again true by induction asumption.


问:
limit of a series equal to the value on boundary
Let $f(x)=\sum_{n=0}^{\infty}{a_nx^n}$ be a real series which has nonnegative coefficients and converges for any $|x|<1$. Assume $f(x)$ is also bounded on $(-1,1)$. Can I conclude

$$\lim_{x\to1^-}f(x)=\sum_{n=0}^{\infty}{a_n}?$$

答:
Let $M=\sup_{x\in (-1,1)}|f(x)|<+\infty$, $S_n=\sum_{k=0}^n a_k$, then for $x\in (0,1)$ and $n\geqslant 0$, we have $\displaystyle\sum_{k=0}^n a_kx^k\leqslant f(x)\leqslant M$, taking the limit as $x\rightarrow 1^-$ gives $S_n\leqslant M$. Thus $(S_n)$ is a bounded non decreasing sequence and therefore converges. Moreover,
$$ \sum_{k=0}^n a_kx^k=\sum_{k=0}^n (S_k-S_{k-1})x^k=\sum_{k=0}^nS_k x^k-\sum_{k=0}^{n-1}S_k x^{k+1}=S_nx^n+(1-x)\sum_{k=0}^nS_k x^k $$
Taking the limit as $n\rightarrow +\infty$ gives $\displaystyle f(x)=(1-x)\sum_{k=0}^{+\infty}S_k x^k$. Therefore, if $S=\lim\limits_{n\rightarrow +\infty}S_n$, we have
$$ S-f(x)=(1-x)\sum_{k=0}^{+\infty}(S-S_k)x^k $$
Let $\varepsilon>0$ and $p\geqslant 1$ such that $S-S_k\leqslant\varepsilon$ for $k\geqslant p$ and $\displaystyle\delta\leqslant\frac{\varepsilon}{\sum_{k=0}^{p-1}(S-S_k)}$ then for $x\in (1-\delta,1)$, we have
$$ 0\leqslant S-f(x)\leqslant (1-x)\sum_{k=0}^{p-1}(S-S_k)+\sum_{k=p}^{+\infty}(S-S_k)\leqslant 2\varepsilon $$
Which means that $\displaystyle\lim\limits_{x\rightarrow 1^-}f(x)=S=\sum_{n=0}^{+\infty}a_n$.


问:
Formula is satisfiable iff its existential closure is satisfiable
How can I prove this statement:

A formula is satisfiable iff its existential closure is satisfiable(the formula has $n$ free variables which are $x_1,\dots,x_n$).

A formula $\varphi$ is satisfiable if and only if  there is a structure  $\mathcal{M}$ such that $\mathcal{M} \vDash \varphi$ .
I applied this definition to $\varphi\left(x_{1}, \ldots, x_{n}\right),$

and say there is a tuple $\vec{a}=(a_{1}, \ldots, a_{n})$ in M satisfying $\varphi$ in $\mathcal{M}$ s.t.

$ \mathcal{M} \vDash \varphi\left(a_{1}, \ldots, a_{n}\right)$

And applying this definition to $\exists x_{1}, \ldots x_{n}  \varphi\left(x_{1}, \ldots, x_{n}\right),$

$ \mathcal{M} \vDash \exists x_{1}, \ldots, x_{n}  \varphi\left(x_{1}, \ldots, x_{n}\right)$

and after I say these, which steps should I follow?

答:
Welcome to MSE!

Hint:

Recall a formula $\psi$ is satisfiable if and only if in some model $\mathfrak{M} \models \psi$. This is the definition of satisfiability.

So applying this definition to $\varphi(x_1, \ldots, x_n)$, we see this is asking for

a model $\mathfrak{M}$

elements $a_1, \ldots, a_n \in \mathfrak{M}$

$\mathfrak{M} \models \varphi(a_1, \ldots, a_n)$.

And applying this definition to $\exists x_1, \ldots x_n . \varphi(x_1, \ldots, x_n)$, we see we want

a model $\mathfrak{M}$

$\mathfrak{M} \models \exists x_1, \ldots, x_n . \varphi(x_1, \ldots, x_n)$

Do you see why these are equivalent conditions? It might be worth reviewing the definition of $\mathfrak{M} \models \exists z . \psi(z)$.

I hope this helps ^_^


问:
What's more general than category theory?
First there was arithmetic with numerical calculations (i.e., one unknown on one side of an equation). Then algebra with manipulations of variables (many unknowns anywhere in an equation). Then systems are studied that differ from ordinary arithmetic but share some of the same properties (equations where the unknowns represent all sorts of things - even functional equations) and then these properties are abstracted in abstract algebra and whole classes are studied such as groups and rings. Then category theory studies maps between structures (functorial equations), then n-category theory, then ...

Where do we go now? Is category theory the end of the road for the foreseeable future?
  Is the only way forward to go backwards and generalize in a different direction (like "generalized equations" of optimization or something)?

答: Generalization goes where problems lead it to. </p>


问:
Basic book about mathematical proofs
I'm looking for an understandable book about mathematical proofs. </p>

<p>As an engineer/computer scientist major I'm used to do higher math on a daily base, but whenever I'm asked in an exercise to "show" something I get the "where should I start? and what exactly do they want me to do?" syndrome. </p>

答: I just came across this physics forum with a thread on how to write mathematical proofs.  At first I was going to copy the list of texts there here, sorting out possible duplicates. But there are so many resources listed: books, tutorials, university-sponsored guides, etc. that I'll simply steer you to the site. </p>

Title: Basic book about mathematical proofs


答:
Let $R$ be a ring and let $c$ be a central element of $R$.
We can form the polynomial ring $R[x]$ and its quotient ring
$$
  R_c
  :=
  R[x] / (cx - 1) \,.
$$
We have a homomorphism of rings $i$ from $R$ to $R_c$ given by the composite of the inclusion from $R$ to $R[X]$ and the quotient map from $R[x]$ to $R_c$.
The element $i(c) = [c]$ is a unit in $R_c$ with inverse given by $[x]$ because
$$
  [c] \cdot [x] = [cx] = [1] = 1_{R_c}
$$
and similarly
$$
  [x] \cdot [c] = [xc] = [cx] = [1] = 1_{R_c} .
$$

The pair $(R_c, i)$ is universal with this property, in the following sense:
Let $(S, j)$ be another pair consisting of a ring $S$ and a homomorphism of rings $j$ from $R$ to $S$ such that $j(c)$ is a unit in $S$.
Then there exists a unique homomorphism of rings $f$ from $R_c$ to $S$ with $f \circ i = j$.
In other words, the homomorphism $j$ factors uniquely through the homomorphism $i$.

The ring $R_c$ is in this sense the universal way of making the element $c$ invertible.

So under what conditions can we embed $R$ into a larger ring $S$ such that $c$ becomes a unit in $S$?
Well, we need a pair $(S, j)$ as above such that $j$ is injective. (We refer to this condition as $(*)$.)

If on the one hand such a pair $(S, j)$ exists, then follows from the factorization $j = f \circ i$ that $i$ also needs to be injective.
If on the other hand $i$ is injective, then one can choose the pair $(S, j)$ as $(R_c, i)$.

We hence see that $(*)$ is possible if and only if the homomorphism $i$ is injective.

We now claim that $i$ is injective if and only if $c$ is no zero divisor in $R$.

Suppose first that $i$ is injective.
If $c$ were a zero divisor in $R$, then it would also be a zero divisor in $R_c$.
Indeed, there would exist some nonzero element $r$ of $R$ with $cr = 0$.
But then $i(r)$ would be a nonzero element of $R_c$ (by the injectivity of $i$) such that $i(c) \cdot i(r) = i(cr) = i(0) = 0$, which would mean that $i(c)$ would be a zero divisor.
But this would contradict $i(c)$ being a unit.
This shows that $c$ cannot be a zero divisor if $i$ is injective.

Suppose on the other hand that $c$ is not a zero divisor.
The element $cx - 1$ of $R[x]$ is central, whence the two-sided ideal $(cx - 1)$ is given by
$$
  (cx - 1)
  =
  \{ p \cdot (cx - 1) \mid p \in R[x] \} \,.
$$
Let $r$ be an element of the kernel of $i$.
Then $0 = i(r) = [r]$, which means that $r$ is contained in the two-sided ideal $(cx - 1)$, and is thus of the form
$$
  r = p (cx - 1)
$$
for some polynomial $p$ in $R[x]$.
The leading coefficient of the polynomial $cx - 1$ is $c$, which is not a zero divisor in $R[x]$.
It follows that
$$
  \deg( p (cx - 1) )
  =
  \deg(p) + \deg( cx - 1 )
  =
  \deg(p) + 1 \,.
$$
But $r$ is an element of $R$, and therefore cannot be of degree $\geq 1$ in $R[x]$.
We hence find that $r = 0$ and $p = 0$.
This shows that $\ker(i) = 0$, so that $i$ is injective.

We have overall shown that $R$ can be embedded into a larger ring in which $c$ is invertible if and only if $c$ is not a zero divisor in $R$.
Such a larger ring can then be constructed as the quotient $R[x] / (cx - 1)$, and this is the universal construction for making $c$ invertible.

Title: Adjoining an inverse to a central element of an algebra

Body: Given a (not necessarily commutative) algebra $A$, and a central element $c \in A$, is it always possible to enlarge $A$ to a an algebra $A'$ in which $c$ is invertible? I guess one can take a set of generators and relations for $A$, and add a formal inverse $c^{-1}$ along with extra relations making $c^{-1}$ central and satisfying $c^{-1}c = c^{-1}c = 1$. But I guess this is a specific example of a general formal process. What could this process be?


答: Just integrate term by term, and then set the constant term to $c$. That's all there is to it.

Title: Finding Maclaurin series for antiderivative, such that series satisfies some initial condition

Body:
I'm given the Maclaurin series representation for some $f(x)$. Take, for example:

$$f(x) = \arctan{x^2}$$

I want now to determine the Maclaurin series for $F = \int f(0)$, such that $F(0) = c$. If you like, take $c = 0$, for example.

I'm not sure how to approach this. Need this be different for each $c$?


问:
For fixed $t_0$ and $s<t$, what's $E(1_{t>t_0}\mid \mathcal{F}_s)$
Let $\{\mathcal{F}_s\}$ be a filtration.

For fixed $t_0$ and $s<t$, what's $E(1_{t>t_0}\mid \mathcal{F}_s)$?

And what about $E(1_{t<t_0}\mid \mathcal{F}_s)$?

Edit: I'll edit in my motivation in case it's important.

I was trying to prove that $B_{\min(t,t_0)}$ (where $B_t$ is a brownian motion) is a Martingale, for a fixed $t_0$.

I tried to separate using indicator functions, and in one of the terms I have: $E(B_{t_0}1_{t>t_0}\mid \mathcal{F}_s)$. So, I was thinking that the indicator functions are simply independent. If so, then I would have $E(B_{t_0}\mid \mathcal{F}_s)1_{t>t_0}$. Now, however, I have problem. In the case of $s<t_0$, then $E(B_{t_0}\mid \mathcal{F}_s)=B_s$, which is something I don't want...

答:
Just let $t \geq s$ and calculate

\begin{align*}
\mathbb{E} \left [B_{t \wedge t_{0}}\, | \, \mathcal{F}_{s} \right ]
&= \mathbb{E} \left [B_{t} \mathcal{1}_{t \leq t_{0}} \, | \, \mathcal{F}_{s} \right ] + \mathbb{E} \left [B_{t_{0}} \mathcal{1}_{t > t_{0}} \, | \, \mathcal{F}_{s} \right ] \\
&\overset{(i)}{=} \mathcal{1}_{t \leq t_{0}} \mathbb{E}[B_{t} \, | \, \mathcal{F}_{s}] + \mathcal{1}_{t > t_{0}} \mathbb{E}[B_{t_{0}} \, | \, \mathcal{F}_{s} ] \\
&= \mathcal{1}_{t \leq t_{0}} B_{s} + \mathcal{1}_{t > t_{0}} B_{s \wedge t_{0}} \\
&\overset{(ii)}{=} \mathcal{1}_{t \leq t_{0}} B_{s \wedge t_{0}} + \mathcal{1}_{t > t_{0}} B_{s \wedge t_{0}} \\
&= B_{s \wedge t_{0}}.
\end{align*}

Note that step $(ii)$ follows since if $t_{0}<s$ then $t \geq s > t_{0}$ by assumption and so $1_{t \leq t_{0}} = 0$.

Step $(i)$ follows since $\mathcal{1}_{t \leq t_{0}}(\omega) \equiv \mathcal{1}_{t \leq t_{0}}$ is deterministic and so is $\mathcal{F}_{s} = \sigma(B_{u} \, | u \leq s ) \, $-measurable. In fact, $\mathcal{1}_{t \leq t_{0}}$  is $\mathcal{F}_{0} = \sigma(B_{0}) = \{\emptyset,\Omega \}$-measurable. For example, if say $t$ (which is fixed) is $\leq t_{0}$, then either $\mathcal{1}_{t \leq t_{0}}^{-1}(B) = \Omega$ (if $1 \in B$) and $\mathcal{1}_{t \leq t_{0}}^{-1}(B) = \emptyset$ (if $1 \notin B$).

Feel free to ask if there are any questions.


问:
Is $\text{Hom}_k(A, B)$ a group? What is its group operator?
I want ask a question related to Hom operator.

I'm studying affine group schemes, and in the definition, it says that an affine group scheme is just a functor F from k-algebras to groups. If F is representable by a k-algebra A, then there is a natural correspondence between $F(R)$ and $\text{Hom}_k(A, R)$. Here A and R are k-algebras.

From this definition, I think it implies that $\text{Hom}_k(A, R)$ is a group, because the functor is from k-algebras to groups.

I'm quite confused here, because k-algebra homomorphisms need to map 0 and 1 in $A$ to 0 and 1 in $R$. So $\text{Hom}_k(A, R)$ is not a group in addition (because the zero map is not an algebra homomorphism). It is also not a group in multiplication, because there is no inverse element. So, how is this a group? And if so, what is the group operator? And can you please explain this situation in this definition? (I'm reading chapter 1 in the book "Introduction to Affine Group Schemes" of Waterhouse.)

答:
an affine group scheme is just a functor F from k-algebras to groups.

This is incorrect. An affine group scheme (over $k$) is a functor from (commutative) $k$-algebras to groups such that the underlying functor to sets is representable. (There are non-affine group schemes, such as elliptic curves, and there are also functors to groups whose underlying functor to sets is not representable by a scheme.)

So, how is this a group? And if so, what is the group operator?

$\text{Hom}_k(A, R)$ has no natural group structure, as you say. The group structure is part of the structure of being an affine group scheme, and is extra data not determined by $A$. The definition of representability only refers to a (natural) identification of sets $F(R) \cong \text{Hom}_k(A, R)$. Here are two simple examples:

If $A = k[x]$, then $\text{Hom}_k(k[x], R) \cong R$ can be given a group structure via addition. This produces an affine group scheme structure on the affine line $\mathbb{A}^1 = \text{Spec } k[x]$ called the additive group scheme $\mathbb{G}_a$.

If $A = k[x, x^{-1}]$, then $\text{Hom}_k(k[x, x^{-1}], R) \cong R^{\times}$ can be given a group structure via multiplication. This produces an affine group scheme structure on the punctured affine line $\mathbb{A}^1 \setminus \{ 0 \} = \text{Spec } k[x, x^{-1}]$ called the multiplicative group scheme $\mathbb{G}_m$.

More complicated examples are possible, of course. It's a nice exercise to verify that the functor $R \mapsto GL_n(R)$ is an affine group scheme, where the group operation on $GL_n(R)$ is matrix multiplication.

It's also a nice exercise in using the Yoneda lemma to verify that a (commutative) $k$-algebra $A$ represents an affine group scheme iff $A$ is equipped with the additional structure of a Hopf algebra, where the comultiplication $\Delta : A \to A \otimes_k A$ encodes the multiplication of the group scheme.


问:
Converse to a theorem regarding monoids and their set of invertibles
In this paper, there is a theorem regarding monoids and their set of invertible elements. Let $(M,*,1)$ be a monoid, and let $U$ be the set of invertible elements of $M$. The theorem states that if $aU=Ua$, for every $a$ in $M$, then the collection of sets $\{aU | a \in M\}$ forms a partition of $M$. What about the converse? Is it true that if the collection of sets $\{aU | a \in M\}$ forms a partition of $M$, then every element $a$ of $M$ commutes with $U$? I would be very surprised if there was a counterexample.

答:
Let $M=\{f:X\to X\ $functions$\}$ where, say, let $X=\{1,2\}$ for simplicity.

Then $|M|=4$ and $\ U=\{{\rm id}_X,\ (1\mapsto 2,\,2\mapsto 1)\}$.

The other two elements of $M$ are the constant maps ${\bf 1}:x\mapsto 1$ and ${\bf 2}:x\mapsto 2$.

If $a={\bf 1}$ or $\bf 2$, then $aU=\{a\}$, otherwise $aU=U$.

So these do form a partition of $M$.

However, $Ua=\{{\bf 1},\,{\bf 2}\}\ \ne aU$ for $a={\bf 1},\,{\bf 2}$.