问:
Trick to change the bounds for the summations
$$\sum_{i=1}^{N-1}\sum_{j=i+1}^{N} |A_i - A_j| $$
$$= \frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N} 2|A_i - A_j| $$
$$= \frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}|A_i - A_j| + |A_j - A_i|$$
making it symmetric and can write that $$\frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}|A_i - A_j| + |A_j - A_i| $$ $$=  \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}|A_i - A_j|$$ My question is, is there a formula based expression that enables you to make the sums of i and j with the same bounds? I'm looking for expressive details into what happened here to get that $$\frac{1}{2}\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}|A_i - A_j| + |A_j - A_i|$$ $$ =  \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}|A_i - A_j|$$

答:
It really is very easy to see if you follow Benjamin Wang’s suggestion in the comments to draw an $N\times N$ square with $|A_i-A_j|$ in position $\langle i,j\rangle$. However, here is an algebraic demonstration.

First note that

$$\begin{align*}
&\sum_{i=1}^{N-1}\sum_{j=i+1}^N\big(|A_i - A_j|+|A_j - A_i|\big)\\
&\qquad=\sum_{1\le i<j\le N}\big(|A_i - A_j|+|A_j - A_i|\big)\\
&\qquad=\sum_{1\le i<j\le N}|A_i-A_j|+\sum_{1\le\color{red}i<\color{blue}j\le n}|\color{blue}{A_j}-\color{red}{A_i}|\\
&\qquad=\sum_{1\le i<j\le N}|A_i-A_j|+\sum_{1\le\color{red}j<\color{blue}i\le n}|\color{blue}{A_i}-\color{red}{A_j}|\,.\tag{1}
\end{align*}$$

In the last step I simply interchanged the names $i$ and $j$ in the second summation. The in each line blue symbols go with the larger index, whatever it’s called, and the red symbols go with the smaller index.

The first summation in $(1)$ covers all pairs $\langle i,j\rangle$ with $i<j$; the second, all pairs $\langle i,j\rangle$ with $i>j$. The pairs $|A_i-A_j|$ with $i=j$ are of course all $0$, so $(1)$ is actually the sum of the quantities $|A_i-A_j|$ over all pairs $\langle i,j\rangle$ with $1\le i,j\le n$, which is

$$\sum_{i=1}^N\sum_{j=1}^N|A_i-A_j|\,.$$

Added: I’m adding on example of the visual approach suggested by Benjamin Wang.

I’ll take $N=6$ for my example. For $1\le i,j\le 6$ let $a_{ij}=|A_i-A_j|$. Now

$$\begin{align*}
\sum_{i=1}^5\sum_{j=i+1}^6\left(|A_i-A_j|+|A_j-A_i|\right)&=\sum_{i=1}^5\sum_{j=i+1}^6(a_{ij}+a_{ji})\\
&=\color{red}{\sum_{i=1}^5\sum_{j=i+1}^6a_{ij}}+\color{blue}{\sum_{i=1}^5\sum_{j=i+1}^6a_{ji}}\,.
\end{align*}$$

In the array below I’ve used the red and blue colors to pick out the terms in those last two summations.

$$\begin{array}{ccc}
a_{11}&\color{red}{a_{12}}&\color{red}{a_{13}}&\color{red}{a_{14}}&\color{red}{a_{15}}&\color{red}{a_{16}}\\
\color{blue}{a_{21}}&a_{22}&\color{red}{a_{23}}&\color{red}{a_{24}}&\color{red}{a_{25}}&\color{red}{a_{26}}\\
\color{blue}{a_{31}}&\color{blue}{a_{32}}&a_{33}&\color{red}{a_{34}}&\color{red}{a_{35}}&\color{red}{a_{36}}\\
\color{blue}{a_{41}}&\color{blue}{a_{42}}&\color{blue}{a_{43}}&a_{44}&\color{red}{a_{45}}&\color{red}{a_{46}}\\
\color{blue}{a_{51}}&\color{blue}{a_{52}}&\color{blue}{a_{53}}&\color{blue}{a_{54}}&a_{55}&\color{red}{a_{56}}\\
\color{blue}{a_{61}}&\color{blue}{a_{62}}&\color{blue}{a_{63}}&\color{blue}{a_{64}}&\color{blue}{a_{65}}&a_{66}\\
\end{array}$$

As you can see, we’re adding up all of the red and blue entries in the array, omitting only the black entries on the diagonal. But $a_{ii}=|A_i-A_i|=0$ for all $i$, so the black entries are all $0$:

$$\begin{array}{ccc}
0&\color{red}{a_{12}}&\color{red}{a_{13}}&\color{red}{a_{14}}&\color{red}{a_{15}}&\color{red}{a_{16}}\\
\color{blue}{a_{21}}&0&\color{red}{a_{23}}&\color{red}{a_{24}}&\color{red}{a_{25}}&\color{red}{a_{26}}\\
\color{blue}{a_{31}}&\color{blue}{a_{32}}&0&\color{red}{a_{34}}&\color{red}{a_{35}}&\color{red}{a_{36}}\\
\color{blue}{a_{41}}&\color{blue}{a_{42}}&\color{blue}{a_{43}}&0&\color{red}{a_{45}}&\color{red}{a_{46}}\\
\color{blue}{a_{51}}&\color{blue}{a_{52}}&\color{blue}{a_{53}}&\color{blue}{a_{54}}&0&\color{red}{a_{56}}\\
\color{blue}{a_{61}}&\color{blue}{a_{62}}&\color{blue}{a_{63}}&\color{blue}{a_{64}}&\color{blue}{a_{65}}&0\\
\end{array}$$

Thus, we can get the same sum by adding up all of the entries in the entire square array, and that sum is

$$\sum_{i=1}^6\sum_{j=1}^6a_{ij}\,.$$

In other words,

$$\sum_{i=1}^5\sum_{j=i+1}^6(a_{ij}+a_{ji})=\sum_{i=1}^6\sum_{j=1}^6a_{ij}\,,$$

i.e.,

$$\sum_{i=1}^5\sum_{j=i+1}^6\left(|A_i-A_j|+|A_j-A_i|\right)=\sum_{i=1}^6\sum_{j=1}^6\left(|A_i-A_j|+|A_j-A_i|\right)\,.$$

And the same argument works for any $N$, not just $6$.


问:
Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions

Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.

My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\sqrt{2}}(x)=\frac{\ln(x)}{\ln(\sqrt{2})}$$
then, we can re-write the equation as $$\left(\frac{\ln(x)}{\ln(2)}\right)^{2}-\left( \frac{\ln(x)}{\ln(\sqrt{2})}\right)^{2}=a-\sqrt{a+\frac{\ln(x)}{\ln(2)}}$$
Let $t:=\ln(x)$ so the equation above we can re-write in terms  of $t$ as $$\left(\frac{1}{\ln(2)}t\right)^{2}-\left( \frac{1}{\ln(\sqrt{2})}t\right)^{2}=a-\sqrt{a+\frac{1}{\ln(2)}t}$$

Then, the idea that I was thinking is squaring both sides of the equation, but there appears an equation of degree 4 that is difficult to solve. How could I continue from there? or maybe a simpler approach?

New approach: Using the point out of @Ross Millikan and since that $$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt{2})}=2\log_{2}(x)$$
and let $t:=\log_{2}(x)$ so the equation can be we-write in terms of $t$ as $$t^{2}-2t=a-\sqrt{a+t} \implies t^{2}-2t\color{blue}{+1}=a-\sqrt{a+t}\color{blue}{+1} \implies (t-1)^{2}=a+1-\sqrt{a+t}$$
or maybe $$ t^{2}-2t=a-\sqrt{a+t} \overset{y^{2}=a+t}{\implies} (y^{2}-a)^{2}-2(y^{2}-a)=a-y^{2}$$
solving a little the expression above we can see that $$y^{4}-2y^{2}a+a^{2}-2y^{2}+2a-a+y^{2}=0$$

how can I solve this?

答:
Somewhat oddly, this becomes easier by first eliminating $a$.  Using your substitution for $t$, let $b=\sqrt{a+t}$.  Then $a=b^2-t$ and the equation becomes
$$t^2 - 2t = b^2 - t - b$$

Or $$
0 = t^2 - b^2 - (t - b)
  = (t-b)(t+b-1)
$$
At this point it might be tempting to get back to $t$ by replacing $b$ with $\sqrt{a+t}$.  Don't do it.  When solving equations with radicals, it's almost always better to solve for the value of the radical instead of solving for the original variable.  The reason is that constraint on $b$ is just $b\ge0$, which is very convenient.  So keep solving for $b$ by getting rid of $t$ with $t = b^2-a$.

$$
\begin{cases}
(b^2 - b - a)(b^2 + b - a - 1)=0\\[2ex]
b\ge0
\end{cases}
$$

The 4 solutions for $2b$ and the requirement for $b\ge0$ are:

$1-\sqrt{1+4a}$,  where $-1\le4a\le0$

$1+\sqrt{1+4a}$,  where $-1\le4a$

$-1+\sqrt{5+4a}$,  where $-4\le4a$

$-1-\sqrt{5+4a}$, never non-negative

Whenever root 1 is valid, so are roots 2 and 3.  So, aside from degenerate roots, the only way to just have two roots is with roots 2 and 3 but not 1.  So, $a>0$

Degenerate Roots

It is important to remember that roots can be degenerate.  While we established that for $-1\le4a\le0$, there are 3 roots, we cannot eliminate that range just yet.  What if there's some value of $a$ in that range where 2 of the 3 roots are equal.

There are two possibilities

$b_1 = b_2$
Then, $4a = -1$ and the third root $b_3$ equals the other two so we will have a triple degenerate root and only 1 distinct root.

$b_3 = b_1$ or $b_3=b_2$

Then $-1+\sqrt{5+4a} = 1 \pm\sqrt{1+4a}$

But, since $$(\sqrt{5+4a}\pm\sqrt{1+4a})(\sqrt{5+4a}\mp\sqrt{1+4a})=4$$

If $\sqrt{5+4a}\pm\sqrt{1+4a} = 2$, then $\sqrt{5+4a}\mp\sqrt{1+4a} = 2$ from which it follows that $\sqrt{1+4a}=0$ and again we would end up with a triple degenerate root.

This leaves us with $a>0$ as the solution

ContentLicense: CC BY-SA 4.0


问:
Solve the system of equations for $\sqrt{xy}$
$$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$

I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.

$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$

$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$

This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.

I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.

答:
$$x + y\sqrt{x} = \frac{95}{8} \tag 1$$
$$y + x\sqrt{y} = \frac{93}{8} \tag 2$$

From $(1)$, you have $y=\frac{95-8 x}{8 \sqrt{x}}$. Plug it in $(2)$ and you need to find the zero of function
$$f(x)=\frac{\sqrt{\frac{95-8 x}{\sqrt{x}}} x}{2 \sqrt{2}}-\sqrt{x}+\frac{95}{8
   \sqrt{x}}-\frac{93}{8}\tag 3$$ which, I agree, is quite ugly.

If you can see is that if $x \to 0^+$, $f(x)\to +\infty$.

Tring for a few values of $x$ you would have the following results
$$\left(
\begin{array}{cc}
 1 & 2.54773 \\
 2 & 0.642631 \\
 3 & 0.289849 \\
 4 & 0.249754 \\
 5 & 0.216853 \\
 6 & 0.0656336 \\
 7 & -0.280515
\end{array}
\right)$$
So, the solution is "close" to $x=6$.

To polish the root, using Newton method with $x_0=6$, we should have the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 6.00000 \\
 1 & 6.27644 \\
 2 & 6.25025 \\
 3 & 6.25000
\end{array}
\right)$$
that is to say $x=\frac {25}4$.

Back to $y=\frac{95-8 x}{8 \sqrt{x}}$, this gives $y=\frac 94$.

I apologize for the typos : I used $938$ instead of $\frac{93}8$


问:
Show: If $C\subseteq\{X,Y\}^*$ is prefix-free and $|C|=2^n$, $\sum\limits_{w\in C}\operatorname{length}(w)\ge n2^n$.
For two letters $X$,$Y$ one can form $2^n$ words (arbitrary sequences composed of the letters $X,Y$) such that none of these words is an initial segment of any other of these words.

An initial segment of a word can be of any length, from a single first letter, to the whole word. Show that the sum of lengths of these words is at least $2^nn$.

PS:

I have tried using Huffman Tree https://en.wikipedia.org/wiki/Huffman_coding as my understanding is that the question is asking is that if a huffman tree has n nodes, the sum of the length of the encoding is >= n*2^n.

However I have not had any luck using this route. I am wondering if I should look to proof by contradiction and/or Induction Method to make another attempt at it.

Any thoughts or leads would be really helpful!

答:
This is actually a special case of a more general result: if $C$ is a set of $m$ words over the alphabet $\{X,Y\}$ such that no word in $C$ is an initial segment of any other word in $C$, then the total length of the words in $C$ is at least $m\lg m$, where $\lg m=\log_2m$ is the binary (base two) logarithm of $m$. This evidently reduces to the desired result when $m=2^n$. I may be missing something simple, but I don’t see a proof of the special case in the question that is any simpler than proofs of the general result. I came up with two proofs of the general result, one using the Kraft-McMillan inequality and one using more elementary ideas.

For convenience let $\ell(w)$ be the length of $w$ for each word $w$ over $\{X,Y\}$, so that the result is that $\sum\limits_{w\in C}\ell(w)\ge m\lg m$.

The Kraft-McMillan inequality says that

$$\sum_{w\in C}2^{-\ell(w)}\le 1\,,$$

and multiplying by $\frac{m}{\sum_{w\in C}2^{-\ell(w)}}$ yields the inequality

$$m\le\frac{m}{\sum\limits_{w\in C}\frac1{2^{\ell(w)}}}\,.\tag{1}$$

The righthand side of $(1)$ is the harmonic mean of the numbers $2^{\ell(w)}$ for $w\in C$, and the harmonic mean of a set of positive real numbers is less than or equal to the geometric mean of those numbers, so

$$m\le\left(\prod_{w\in C}2^{\ell(w)}\right)^{1/m}=2^{\frac1m\sum_{w\in C}\ell(w)}\,.$$

Now take binary logs on both sides:

$$\lg m\le\frac1m\sum_{w\in C}\ell(w)\,,\tag{2}$$

so

$$\sum_{w\in C}\ell(w)\ge m\lg m\,.$$

If one hasn’t already noticed the fact, $(2)$ makes it clear that the result can also be thought of as saying that the mean length of words in $C$ is at least $\lg m$.

If the Kraft-McMillan inequality isn’t available, one can prove the result by induction on $m$. You can easily verify that it holds for $m=1$ and $m=2$. Now assume that the result holds for all such sets of fewer than $m$ words. Arrange the words over the alphabet $\{X,Y\}$ in an infinite binary tree $T$:

```<code>                             *         Level 0
                            / \
                           /   \
                          /     \
                         X       Y     Level 1
                        / \     / \
                       /   \   /   \
                      XX   XY YX   YY  Level 2
</code>```

The left child of a word $w$ is $wX$, and the right child is $wY$. Observe that a word $u$ is a prefix of a word $w$ if and only if the vertex $u$ is on the path from the root $*$ to the word $w$. Thus, the $m$ words must lie on $m$ distinct branches of $T$. If $m=2^n$, for instance, they could (but need not) be the $2^n$ words on Level $n$, in which case their total length would be exactly $n2^n$.

Let $C_X=\{w:Xw\in C\}$ and $C_Y=\{w:Yw\in C\}$; clearly no word in $C_X$ is an initial segment of any other word in $C_X$, and similarly for $C_Y$. Let $m_X=|C_X|$ and $m_Y=|C_Y|$, and assume for now that $m_X\ne 0\ne m_Y$. The induction hypothesis then ensures that

$$\sum_{w\in C_X}\ell(w)\ge m_X\lg m_X$$

and

$$\sum_{w\in C_Y}\ell(w)\ge m_Y\lg m_Y\,.$$

Now $m_X+m_Y=m$, so

$$\begin{align*}
\sum_{w\in C}\ell(w)&=\sum_{w\in C_X}\big(1+\ell(w)\big)+\sum_{w\in C_Y}\big(1+\ell(w)\big)\\
&=m_X+\sum_{w\in C_X}\ell(w)+m_Y+\sum_{w\in C_Y}\ell(w)\\
&=m+\sum_{w\in C_X}\ell(w)+m_Y+\sum_{w\in C_Y}\ell(w)\\
&\ge m+m_X\lg m_X+m_Y\lg m_Y\\
&=m+m_X\lg m_X+(m-m_X)\lg(m-m_X)\,.
\end{align*}$$

It’s an easy exercise in basic calculus to determine that the function

$$f(x)=x\lg x+(a-x)\lg(a-x)$$

has a minimum at $x=\frac{a}2$, so

$$m_X\lg m_X+(m-m_X)\lg(m-m_X)\ge m\lg\frac{m}2=m(\lg m-1)\,,$$

and

$$\sum_{w\in C}\ell(w)\ge m+m(\lg m-1)=m\lg m\,,$$

as desired.

What if one of $m_X$ and $m_Y$ is $0$? In that case the words in $C$ all share a non-empty maximal common initial segment $u$, and since $u$ is maximal, both $uX$ and $uY$ are initial segments of at least one word in $C$. Let $C_0=\{w:uw\in C\}$, the set of words obtained by removing the common initial segment $u$ from the words in $C$. Then

$$\begin{align*}
\sum_{w\in C}\ell(w)&=m\ell(u)+\sum_{w\in C_0}\ell(w)\\
&\ge m\ell(u)+m\lg m\\
&>m\lg m\,.
\end{align*}$$

The result now follows by induction.

Body:
This is actually a special case of a more general result: if $C$ is a set of $m$ words over the alphabet $\{X,Y\}$ such that no word in $C$ is an initial segment of any other word in $C$, then the total length of the words in $C$ is at least $m\lg m$, where $\lg m=\log_2m$ is the binary (base two) logarithm of $m$. This evidently reduces to the desired result when $m=2^n$. I may be missing something simple, but I don’t see a proof of the special case in the question that is any simpler than proofs of the general result. I came up with two proofs of the general result, one using the Kraft-McMillan inequality and one using more elementary ideas.

For convenience let $\ell(w)$ be the length of $w$ for each word $w$ over $\{X,Y\}$, so that the result is that $\sum\limits_{w\in C}\ell(w)\ge m\lg m$.

The Kraft-McMillan inequality says that

$$\sum_{w\in C}2^{-\ell(w)}\le 1\,,$$

and multiplying by $\frac{m}{\sum_{w\in C}2^{-\ell(w)}}$ yields the inequality

$$m\le\frac{m}{\sum\limits_{w\in C}\frac1{2^{\ell(w)}}}\,.\tag{1}$$

The righthand side of $(1)$ is the harmonic mean of the numbers $2^{\ell(w)}$ for $w\in C$, and the harmonic mean of a set of positive real numbers is less than or equal to the geometric mean of those numbers, so

$$m\le\left(\prod_{w\in C}2^{\ell(w)}\right)^{1/m}=2^{\frac1m\sum_{w\in C}\ell(w)}\,.$$

Now take binary logs on both sides:

$$\lg m\le\frac1m\sum_{w\in C}\ell(w)\,,\tag{2}$$

so

$$\sum_{w\in C}\ell(w)\ge m\lg m\,.$$

If one hasn’t already noticed the fact, $(2)$ makes it clear that the result can also be thought of as saying that the mean length of words in $C$ is at least $\lg m$.

If the Kraft-McMillan inequality isn’t available, one can prove the result by induction on $m$. You can easily verify that it holds for $m=1$ and $m=2$. Now assume that the result holds for all such sets of fewer than $m$ words. Arrange the words over the alphabet $\{X,Y\}$ in an infinite binary tree $T$:

                             *         Level 0
                            / \
                           /   \
                          /     \
                         X       Y     Level 1
                        / \     / \
                       /   \   /   \
                      XX   XY YX   YY  Level 2

The left child of a word $w$ is $wX$, and the right child is $wY$. Observe that a word $u$ is a prefix of a word $w$ if and only if the vertex $u$ is on the path from the root $*$ to the word $w$. Thus, the $m$ words must lie on $m$ distinct branches of $T$. If $m=2^n$, for instance, they could (but need not) be the $2^n$ words on Level $n$, in which case their total length would be exactly $n2^n$.

Let $C_X=\{w:Xw\in C\}$ and $C_Y=\{w:Yw\in C\}$; clearly no word in $C_X$ is an initial segment of any other word in $C_X$, and similarly for $C_Y$. Let $m_X=|C_X|$ and $m_Y=|C_Y|$, and assume for now that $m_X\ne 0\ne m_Y$. The induction hypothesis then ensures that

$$\sum_{w\in C_X}\ell(w)\ge m_X\lg m_X$$

and

$$\sum_{w\in C_Y}\ell(w)\ge m_Y\lg m_Y\,.$$

Now $m_X+m_Y=m$, so

$$\begin{align*}
\sum_{w\in C}\ell(w)&=\sum_{w\in C_X}\big(1+\ell(w)\big)+\sum_{w\in C_Y}\big(1+\ell(w)\big)\\
&=m_X+\sum_{w\in C_X}\ell(w)+m_Y+\sum_{w\in C_Y}\ell(w)\\
&=m+\sum_{w\in C_X}\ell(w)+m_Y+\sum_{w\in C_Y}\ell(w)\\
&\ge m+m_X\lg m_X+m_Y\lg m_Y\\
&=m+m_X\lg m_X+(m-m_X)\lg(m-m_X)\,.
\end{align*}$$

It’s an easy exercise in basic calculus to determine that the function

$$f(x)=x\lg x+(a-x)\lg(a-x)$$

has a minimum at $x=\frac{a}2$, so

$$m_X\lg m_X+(m-m_X)\lg(m-m_X)\ge m\lg\frac{m}2=m(\lg m-1)\,,$$

and

$$\sum_{w\in C}\ell(w)\ge m+m(\lg m-1)=m\lg m\,,$$

as desired.

What if one of $m_X$ and $m_Y$ is $0$? In that case the words in $C$ all share a non-empty maximal common initial segment $u$, and since $u$ is maximal, both $uX$ and $uY$ are initial segments of at least one word in $C$. Let $C_0=\{w:uw\in C\}$, the set of words obtained by removing the common initial segment $u$ from the words in $C$. Then

$$\begin{align*}
\sum_{w\in C}\ell(w)&=m\ell(u)+\sum_{w\in C_0}\ell(w)\\
&\ge m\ell(u)+m\lg m\\
&>m\lg m\,.
\end{align*}$$

The result now follows by induction.


问:
For $X\to Y\to S$, how is $X\times_Y X\to X\times_S X$ defined? (Proposition 5.2.5 in Olsson's Algebraic Spaces and Stacks)
Let $R$ denote an etale equivalence relation on a scheme $X \in \operatorname{Sch}/S$.

The statement of Proposition 5.2.5 in Olsson's Algebraic spaces and stacks is as follows:

(i) $X/R$ is an algebraic space, where $X$ is the associated sheaf
with respect to the etale topology of the functor
$(\operatorname{Sch}/S)^{op} \to \operatorname{Set}, T \mapsto
X(T)/R(T)$

(ii) If $Y$ is an algebraic space over $S$, and $X \to Y$ is an etale
surjective morphism, then $R:= X \times_Y X$ is a scheme and the
inclusion $R \hookrightarrow X \times_S X$ is an etale equivalence
relation.

How is the map $R \to X \times_S X$ defined?

答:
Let $f\colon X \to Y$, $g\colon Z \to Y$, and $h\colon Y \to S$ be morphisms in any category with pullbacks. Then we have a commutative diagram
$$\require{AMScd}\begin{CD}
X \times_Y Z @>>> Z \\
@VVV @VV{h\circ g}V \\
X @>{h\circ f}>> S
\end{CD}$$
where the morphisms from $X \times_Y Z$ are the projection maps (which are part of the data of the pullback). Applying the universal property of the pullback $X \times_S Z$ to the above diagram, the maps $X \times_Y Z \to X$ and $X \times_Y Z \to Z$ factor through a unique map
$$X \times_Y Z \to X \times_S Z.$$
This is the map you're asking about (with $X = Z$ in your case).

To get a better sense of what this map looks like in practice, let's consider the case of a concrete category (where we can think of the objects as "sets with some extra structure and properties", and the morphisms as "structure-preserving functions"). The fiber product $X \times_Y Z$ is the subset of $X \times Z$ consisting of pairs $(x, z)$ such that $f(x) = g(z)$. If $f(x) = g(z)$, then certainly $h(f(x)) = h(g(z))$, so the underlying set of $X \times_Y Z$ is a subset of the underlying set of $X \times_S Z$.

The picture is slightly more complicated for schemes since we also have to consider the ringed space structure, but this should at least give some intuition for why this is an inclusion even if we'd need to say a bit more to turn this into a formal proof.


问:
Arbitrarily close uniformly continuous functions
While preparing for my Real Analysis exam I tried to solve the following problem I found:

Problem:

Let $A \subseteq \mathbb{R}$ and suppose that $f: A \rightarrow \mathbb{R}$ has the following property: for each $\varepsilon>0$ there exists a function $g_{\varepsilon}: A \rightarrow \mathbb{R}$ such that $g_{\varepsilon}$ is uniformly continuous on $A$ and $\left|f(x)-g_{\varepsilon}(x)\right|<\varepsilon$ for all $x \in A .$ Prove that $f$ is uniformly continuous on $A$.

Attempt:

Given $A \subset \mathbb{R}$, $f: A \rightarrow \mathbb{R}$ and $\epsilon > 0$ one can obtain a function $g_{\varepsilon}$ which is uniformly continuous on $A$ such that: $$
|f(x)-g_{\varepsilon}(x)|<\frac{\epsilon}{3}, \forall x \in A
$$

Given $y \in A$, one can also obtain that: $$
|f(y)-g_{\varepsilon}(y)|<\frac{\epsilon}{3}, \forall y \in A
$$

Additionally, since $g_{\varepsilon}$ is uniformly continuous, $\forall \epsilon^{'} > 0$ $\exists \delta > 0$ such that $\forall x, y \in A$
$$
\left|g_{\varepsilon}(x)-g_{\varepsilon}(y)\right|<\varepsilon / 3
$$ whenever $|x - y| < \delta$

Now, considering that $$|f(x) - f(y)| = $$  $$= |f(x) - g_{\varepsilon}(x) + g_{\varepsilon}(x) - g_{\varepsilon}(y) + g_{\varepsilon}(y) - f(y)| \leq $$

$$\leq |f(x) - g_{\varepsilon}(x)| + |g_{\varepsilon}(x) - g_{\varepsilon}(y)| + |g_{\varepsilon}(y) - f(y)| < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}$$ $$= \epsilon$$

whenever $|x - y| < \delta$

Therefore, we conclude that f is uniformly continuous.

Questions:

1. Is the solution correct?

2. If the solution is indeed correct, is it well written?

3. If the solution is incorrect, can someone gently me explain why and provide a solution?

Thanks in advance, Lucas!

答:
As Kavi Rama Murthy said, your proof is good. The write up is also generally okay, but does have some issues. These don't affect the quality of the proof, but make it harder to read. (One trivial thing you can do is decide which epsilon you prefer and stick with it. Mixing $\epsilon$ and $\varepsilon$ as you do is a little distracting.)

Given $A \subset \mathbb{R}$, $f: A \rightarrow \mathbb{R}$ and $\varepsilon > 0$ one can obtain a function $g_{\varepsilon}$ which is uniformly continuous on $A$ such that:
$$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$

Good.

Given $y \in A$, one can also obtain that:
$$|f(y)-g_{\varepsilon}(y)|<\frac{\varepsilon}{3}, \forall y \in A$$

There are two problems here. First you introduce an actual variable with the phrase "Given $y \in A$". But then you make no further use of it. No, really. It is not used anywhere in the proof. All of those other "$y$" variables are quantified (that "$\forall y \in A$" is called a relative quantifier - drop the "$\in A$" and it would be a universal quantifier). By quantifying those other uses of $y$, you've made them dummy variables. They only have meaning inside the range of each of their quantifications. Outside those ranges, there is no $y$. Other than this one unnecessary introduction.

The second problem is why Kavi Rama Murthy told you the second sentence should be deleted: This
$$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$
and this
$$|f(y)-g_{\varepsilon}(y)|<\frac{\varepsilon}{3}, \forall y \in A$$
are exactly the same statement. The $x$ and $y$ in them are dummy variables, which are variables used only in support of a notation. The meaning of the notation did not change in the slightest because you used a different letter. So all you are doing in this line is repeating what you said immediately above it.

Additionally, since $g_{\varepsilon}$ is uniformly continuous, $\forall \varepsilon' > 0\ \exists \delta > 0$ such that $\forall x, y \in A$
$$\left|g_{\varepsilon}(x)-g_{\varepsilon}(y)\right|<\varepsilon / 3$$
whenever $|x - y| < \delta$

Other than the unused "$\forall \varepsilon'$", which can be deleted, this is good.

Now, considering that $$|f(x) - f(y)| = $$  $$= |f(x) - g_{\varepsilon}(x) + g_{\varepsilon}(x) - g_{\varepsilon}(y) + g_{\varepsilon}(y) - f(y)| \leq $$ $$\leq |f(x) - g_{\varepsilon}(x)| + |g_{\varepsilon}(x) - g_{\varepsilon}(y)| + |g_{\varepsilon}(y) - f(y)| < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}$$ $$= \varepsilon$$

There are no logic issues here. That both
$$|f(x) - g_{\varepsilon}(x)| < \frac{\varepsilon}{3}$$
and $$|f(y) - g_{\varepsilon}(y)| <\frac{\varepsilon}{3}$$ follow from the single earlier statement of $|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$ since both the $x$ value used here (which is a different variable than the dummy variable in the statement) and the $y$ value used here are in $A$, they satisfy the $\forall$ condition.


问:
An elephant and a lion are currently 1 mile apart. The elephant runs directly away from the lio
An elephant and a lion are currently $1$ mile apart. The elephant runs directly away from the lion at $19$ miles per hour, while the lion runs directly towards the elephant at $24$ miles per hour. How many minutes will it take for the lion to catch the elephant?</p>

答: Every hour, the lion runs 24 miles while the elephant runs 19. Thus, the distance between the two animals closes at a rate of 5 miles every hour. The lion catches the elephant after this distance has closed 1 mile, which takes $\frac{1}{5}$ hours to do, or $\frac{1}{5}\cdot 60 = \boxed{12}$ minutes.

ContentLicense: CC BY-SA 3.0


问:
Showing $H=\{\log a : a\in \Bbb Q, a>0\}$ is a subgroup of $\left \langle \Bbb R,+ \right \rangle$
The Problem: Let $G=\left \langle \Bbb R,+ \right \rangle$ and $H=\{\log a : a\in \Bbb Q, a>0\}$

Show that $H$ is a subgroup of $G$.

My Proof:

Let $\log a$ and $\log b$ be elements in $H$ where $a,b \in \Bbb R$ and $a,b>0$. Then $\log a + \log b=\log ab \in H$ since $ab\in \mathbb Q$ and $ab>0$. Thus $H$ is closed.

Let $-\log a =\log a^{-1} =\log({1\over a}) \in H$, then $\log a + (-\log a)=\log a - \log a=0$. Thus $-\log a \in H$ and $H$ has additive inverses and $H$ is a subgroup of $G$.

Note: I am specifically trying to prove $H$ is a subgroup by showing $H$ is closed and has inverses. What would happen to this result if $a$ was only allowed to be a natural number?

答:
More or less, this proof is correct. My main nitpick is with your wording of showing inverses. You should say

Let $\log a \in H$. Then [...] and thus $-\log a$ (the inverse of $\log a$ under $+$) is in $H$.

As worded, it seems to presuppose that this inverse is in $H$. Some other nitpicks could arise, such as (unless you know $\left \langle \Bbb R,+ \right \rangle$ is an abelian group specifically) needing to show the inverse works on each side.

Also, a minor thing: you should have that $\log a, \log b \in H$ because $a,b$ are specifically in $\Bbb Q^+$ (i.e. are positive rationals, not just real numbers). Perhaps it is also worth noting that you are assuming $\Bbb R, \Bbb Q$ are fields in this proof, with the usual properties - for instance, assumping that $a,b \in \Bbb Q \implies a \cdot b \in \Bbb Q$.

Also, note that you can use the one-step subgroup test instead for a slightly slicker proof, in the sense that

$$\log a + (- \log b) = \log \left( \frac a b \right)$$

which is in $H$ since $a,b > 0 \implies b \ne 0$ and $a/b > 0$.

What would happen to this result if $a$ was only allowed to be a natural number?

Then it would not work since you do not have inverses for every element. For instance, $2 \in \Bbb N \implies \log 2 \in H$ but it has inverse

$$-\log 2 = \log \left( \frac 1 2 \right)$$

which is not in $H$, since $1/2 \not \in \Bbb N$.


问:
Find point at constant distance from parabola
Kindly consider the diagram below, with:

f(x) = x²

x0 given.

d constant, e.g. 1 in the diagram.

y0 unknown.

I would like to calculate point (x0, y0), sitting perpendicular to the curve f(x), at a constant distance of d, given x0. How do I best approach this problem?

To be clear, I do not know x1 in advance. I've merely labeled it for convenience.

答:
I'll work with the parabola $y=x^2$ and the distance $d=1$ for convenience' sake, but the method should be tweakable easily enough.

The tangent vector to the parabola at $x=x_1$ has slope $2x_1$; that is to say, it points in the (unnormalized) direction $\langle 1, 2x_1\rangle$. The orthogonal vector to this (pointing downwards, as in the diagram) then points in the direction $\langle 2x_1, -1\rangle$. The length of this vector is $\sqrt{1+4x_1^2}$, so the normalized vector is $\langle \frac{2x_1}{\sqrt{1+4x_1^2}}, \frac{-1}{\sqrt{1+4x_1^2}}\rangle$, of unit length. The point at distance one from the parabola along this orthogonal line at $\langle x_1, y_1\rangle = \langle x_1, x_1^2\rangle$ is then $\langle x_1+\frac{2x_1}{\sqrt{1+4x_1^2}}, x_1^2-\frac{1}{\sqrt{1+4x_1^2}}\rangle$. We want the $x$ component of this equal to $x_0$, so we set $x_0=x_1+\frac{2x_1}{\sqrt{1+4x_1^2}}$ and solve for $x_1$: $x_0-x_1 = \frac{2x_1}{\sqrt{1+4x_1^2}}$; $(x_0-x_1)^2=\frac{4x_1^2}{1+4x_1^2}$; $(1+4x_1^2)(x_0-x_1)^2-4x_1^2=0$. This is a quartic equation in $x_1$; it can be solved explicitly, but for practical purposes iterative root-finding is likely to yield better results.


问:
Do geometry theorems of n dimension hold true for all topological spaces of n dimension?
Do geometry theorems of n dimension hold true for all topological spaces of n dimension? Let's say that there's a geometric theorem proven for a euclidean space of dimension 2, does that same theorem hold true for all topological spaces or not?</p>

答:
Not. Definitely not. For instance, it's true in dimension 2 euclidean space that any loop can be continuously contracted to a point; but on the surface of the torus, this is definitely not true.

You might say "But that's not really geometric", and you're right. So let's try another: if you take a straight line in euclidean space, there's another straight line parallel to it (i.e., that does not intersect it). Now look at the unit sphere in 3-space, a 2-dimensional manifold. The analog to "straight line" is "great circle". But any two great circles on the sphere intersect.


问:
Examples of logs with other bases than 10
From a teaching perspective, sometimes it can be difficult to explain how logarithms work in Mathematics. I came to the point where I tried to explain binary and hexadecimal to someone who did not have a strong background in Mathematics. Are there some common examples that can be used to explain this?

For example (perhaps this is not the best), but we use tally marks starting from childhood. A complete set marks five tallies and then a new set is made. This could be an example of a log with a base of 5.

答:
The most common scales of non-decimal logrithms are the music scale.

For example, the octave is a doubling of frequency over 12 semitones.  The harmonics are based on integer ratios, where the logrithms of 2, 3, and 5 approximate to 12, 19 and 28 semitones.  One can do things like look at the ratios represented by the black keys or the white keys on a paino keyboard.  The black keys are a more basic set than the white keys (they are all repeated in the white keys, with two additions).

The brightness of stars are in steps of 0.4 dex (ie 5 orders of magnitude = 100), while there is the decibel scale (where the same numbers represent intensity in $10\log_{10}$ vs power in $20\log_{10}$.

The ISO R40 series is a series of decimal preferred numbers, the steps are in terms of $40\log_{10}$, it's very close to the semi-tone scales.

One can, for example, with just rude approximations like $5<6$, and considering a graph of areas of $x=log_2(3)$ vs $y=log_2(5)$, draw the inequality above as a line saying that the point represented by the true value of $log_2(3), log_2(5)$, must be restricted to particular areas above or below a line.  One finds that the thing converges quite rapidly, with inequalities less than 100.


问:
Vector normal to the gradient
I know that the gradient points in the direction of the maximum rate of increase, with the maximum rate given by its magnitude.  Similarly, the maximum rate of decrease is given by the negative of its magnitude.  Does a vector normal to the gradient point in the direction of zero increase?  I.e., if I were standing on a level curve and traveled in a direction normal to the gradient at that point, will I land on a level curve with the same value as when I began?</p>

答:
Generic Case

If a function $f$ is nice enough that it is differentiable at a point $\mathbf{x}$, the directional derivative in the direction of $\mathbf{v}$ (at least if $\mathbf{v}$ is a unit vector), is given by the dot product of the gradient and $\mathbf{v}$: $\nabla_{\mathbf{v}}f(\mathbf{x})=\nabla f(\mathbf{x})\bullet \mathbf{v}$. If you are working in at least two dimensions, you can do as jlammy suggested and take $\mathbf{v}$ in a direction perpendicular to $\nabla f(\mathbf{x})$ to force $\nabla f(\mathbf{x})\bullet \mathbf{v}=0$, so that the directional derivative is zero and there is no instantaneous increase or decrease.

Exception

However, not every function is differentiable, and it is possible for every directional derivative to be nonzero (including the direction perpendicular to the gradient). See my answer to "Is there always a direction in which the directional derivative of a function is zero?" for an example.

Caveat

I.e., if I were standing on a level curve and traveled in a direction normal to the gradient at that point, will I land on a level curve with the same value as when I began?

However, just because the directional derivative is $0$ doesn't mean you can end up at the same value. For instance, the derivative of $f(x)=x^3$ is $0$ at $x=0$, but $f$ is increasing through the point $(0,0)$, and so walking on that graph can't land at height $0$ at more than one spot.

Title: Vector normal to the gradient


问:
Finding an alternate solution to a plane-geometry problem.
I'm having some trouble finding an alternate solution to the following problem:

Let us have a right-angled triangle $ABC$ ($A$ being the right angle vertex). We draw an altitude to the hypotenuse creating point $D$ and an angle bisector of the vertex angle C which intersected with $AB$ gives us another point $E$. Prove that $|AE| \cdot|AC|=|BE| \cdot |CD|. $

I do know how to solve the problem with the use of analytic geometry, however I'm curious wether there's some less computationally demanding way. Any help is appreciated.

答: $\frac{AE}{BE}=\frac{AC}{BC}$ by the angle bisector theorem, which is true in any triangle. Then $\frac{AC}{BC}=\cos C=\frac{CD}{AC}$.


答:
Here is a straightforward proof using two theorems from Euclidean geometry.

Theorem 1:  If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle.  (You can find this in Euclid's Elements, Book VI, Proposition 3.) In this case, we conclude that $$\frac{|AE|}{|BE|} = \frac{|AC|}{|BC|}$$
or, equivalently,
$$|AE| \cdot |BC| = |BE| \cdot |AC|$$

Theorem 2: The altitude drawn to the hypotenuse of a right triangle divides the triangle into two similar triangles.  In this case, we conclude that $\triangle ADC \sim \triangle BDA$, and hence
$$\frac{|AC|}{|BC|} = \frac{|CD|}{|AC|} $$

Multiply these two equations together and you get the result you want.

Title: Finding an alternate solution to a plane-geometry problem.

Body:
I'm having some trouble finding an alternate solution to the following problem:

Let us have a right-angled triangle $ABC$ ($A$ being the right angle vertex). We draw an altitude to the hypotenuse creating point $D$ and an angle bisector of the vertex angle C which intersected with $AB$ gives us another point $E$. Prove that $|AE| \cdot|AC|=|BE| \cdot |CD|. $

I do know how to solve the problem with the use of analytic geometry, however I'm curious wether there's some less computationally demanding way. Any help is appreciated.


问:
Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct?
I've found this limit by this way. Am I correct?

Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$

Let's see that:

\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1
\end{align}

Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!

答:
Hint: Rewrite as follows:
$$\begin{align*}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n+1-1}{n+1}-1-\frac{1}{n} \\
&=1-\frac{1}{n+1}-1-\frac{1}{n} \\
&=-\left(\frac{1}{n+1}+\frac{1}{n}\right).
\end{align*}$$
As $n\to\infty$, what do $1/n$ and $1/(n+1)$ approach?


答:
You have a mistake in:

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

It should be:

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

However, it would be better to take a factor of $n$:

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

Hence, you will find that the limit will be zero.

Title: Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct?

Body:
I've found this limit by this way. Am I correct?

Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$

Let's see that:

\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1
\end{align}

Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!


答: $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n} \implies \lim_{n\rightarrow \infty}  \dfrac{1}{1+\frac{1}{n}}-1-\dfrac{1}{n} $ equals?


问:
Inequiality with ceil function
I'm having trouble with the following inequality

$$\lceil 2n/3\rceil-1<n-\lceil2n/3\rceil$$

I put it on wolframalpha and got that $1<n \leq 3/2$ (https://www.wolframalpha.com/input/?i=ceil%282n%2F3%29-1%3Cn-ceil%282n%2F3%29). But I don't see how to get that result.

I tried doing this

$\lceil 2n/3\rceil < \frac{n+1}{2}$ and also $2n/3\leq\lceil 2n/3\rceil$ because $\lceil x\rceil = k$ iff $k-1<x\leq k$, so $2n/3 < \frac{n+1}{2} \implies n<3$, but its not the same answer as wolframalpha. Any help would be appreciated.

答:
This sort of equation will generally be a mess of special cases.

Rearrange to $$\left\lceil\frac{2n}{3}\right\rceil<\frac{n+1}{2}$$
first.

Next, look at asymptotic behaviour (for $n$ very large). LHS has $\frac{2}{3}n$, RHS has $\frac{1}{2}$n. Clearly, for large enough positive $n$ there can be no solution, whereas for large enough negative $n$ it is always satisfied.

Since ceiling changes behaviour at integers, you need to look at those breakpoints. Essentially, keep the LHS as an integer (which places a constraint on $n$), then solve for $n$ within that range.

Suppose
$$\left\lceil\frac{2n}{3}\right\rceil=0\Rightarrow n\in\left(-\frac{3}{2},0\right]\text{,}$$
then
$$0<\frac{n+1}{2}\Rightarrow n>-1\Rightarrow n\in(-1,0]$$
works in this range.

Next,
$$n\in\left(0,\frac{3}{2}\right]\Rightarrow1<\frac{n+1}{2}\Rightarrow1<n\Rightarrow n\in\left(1,\frac{3}{2}\right]$$
works.

$$n\in\left(\frac{3}{2},3\right]\Rightarrow2<\frac{n+1}{2}\Rightarrow3<n$$
is impossible, and clearly there are no further solutions above $\frac{3}{2}$.

Going in the other direction,
$$n\in\left(-3,-\frac{3}{2}\right]\Rightarrow-1<\frac{n+1}{2}\Rightarrow-3<n$$
means the entire range works. So, anything below $n<-\frac{3}{2}$ works.

Title: Inequiality with ceil function


问:
Prove the order statistic is a minimal sufficient statistic for the logistic pdf $f(x|\theta)=\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}$
I'm going through Statistical Inference by Casella and Berger, and I'm currently on Chapter 6. In particular, I'm doing exercise 6.9 (c), and I'm trying to prove that if $X_1, ..., X_n$ is a random sample from a population with pdf $f(x|\theta)=\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}$, then the order statistic $T(X) = (X_{(1)},...,X_{(n)})$ is a minimal sufficient statistic for $\theta$.

The joint pdf of the sample is: $$f_X(\boldsymbol{x}|\theta)=e^{n(\theta-\bar{x})}\prod_{i=1}^n \frac{1}{(1+e^{-(x_{(i)}-\theta)})^2}.$$

I want to apply Theorem 6.2.13, so I want to verify that for every two sample points $\boldsymbol{x}$ and $\boldsymbol{y}$, the ratio $\frac{f(\boldsymbol{x}|\theta)}{f(\boldsymbol{y}|\theta)}$ is independent of $\theta$ if and only if $T(\boldsymbol{x}) = T(\boldsymbol{y})$. This ratio is:

$$e^{n(\bar{y}-\bar{x})} \prod_{i=1}^n \bigg(\frac{1+e^{-(y_{(i)}-\theta)}}{1+e^{-(x_{(i)}-\theta)}}\bigg)^2.$$

Now the first part is independent of $\theta$, so we just need to verify (dropping the square) that $$\prod_{i=1}^n \frac{1+e^{-(y_{(i)}-\theta)}}{1+e^{-(x_{(i)}-\theta)}}$$ is independent of $\theta$ if and only iff the $T(\boldsymbol{x})=T(\boldsymbol{y})$ are equal. In the solutions and everywhere else I've looked online this is just stated as true, and while the "if" direction is obvious, the "only if" one isn't. Or rather, sure, it looks like this should only be independent of $\theta$ if the order statistics are equal and I can't find a counterexample, but it doesn't seem necessarily obvious to me. I do believe it's true, but I can't prove it, and I can't even prove this for the case $n=2$, where the relevant part is

$$\frac{1+e^{\theta}(e^{-y_{(1)}}+e^{-y_{(2)}}) + e^{2\theta}e^{-y_{(1)}-y_{(2)}}}{1+e^{\theta}(e^{-x_{(1)}}+e^{-x_{(2)}}) + e^{2\theta}e^{-x_{(1)}-x_{(2)}}}.$$

I'm not sure if I'm missing something obvious, but any help with rigorously proving this would be appreciated (even if you can just answer the question for the special case $n=2$).

答:
Claim: The order statistic $(X_{(1)},\cdots,X_{(n)})$ is equivalent with $\frac{f(x|\theta_j)}{f(x|\theta_0)},j=1\cdots,n+1$.

Now prove $\frac{f(x|\theta_j)}{f(x|\theta_0)},j=1\cdots,n+1$ is minimal sufficient:

Since $\frac{f(x|\theta_j)}{f(x|\theta_0)},j=1\cdots,n+1$ is a minimal sufficient statistic of $\mathcal{P}_0=\{f(x|\theta_1),\cdots,f(x|\theta_{n+1})\}\subset\mathcal{P}=\{f(x|\theta),\theta\in\mathbb{R}\}$, and sufficient for $\mathcal{P}$ (because order statistic is sufficient), thus $\frac{f(x|\theta_j)}{f(x|\theta_0)},j=1\cdots,n+1$ is a minimal sufficient statistic of $\mathcal{P}$ by lemma:

$T(X)$ minimal sufficient for $\mathcal{P}_0\subset \mathcal{P}$ and
sufficient for $\mathcal{P}$, then $T(x)$ is minimal sufficient for
$\mathcal{P}$

Proof of lemma:
For any $U(X)$ sufficient for $\mathcal{P}$, $U(X)$ is sufficient for $\mathcal{P}_0$. Since $T(X)$ is minimal sufficient for $\mathcal{P}_0$, $T(X)$ is a function of $U(X)$. Thus sufficient statistic (w.r.t $\mathcal{P}$) $T(X)$ is a function of any sufficient statistic of $\mathcal{P}$, thus minimal sufficient.

Proof of claim:
\begin{align*}
(X_{(1)},\cdots,X_{(n)}) &=  (X'_{(1)},\cdots,X'_{(n)})\\
\overset{u=e^{x}}{\iff} (u_{(1)},\cdots,u_{(n)}) &= (u_{(1)},\cdots,u_{(n)})\\
\overset{}{\iff} \prod_{i=1}^n\frac{1 + u_{(i)}\xi}{1 + u_{(i)}} &= \prod_{i=1}^n\frac{1 + u'_{(i)}\xi}{1 + u'_{(i)}}, \forall\xi\in\mathbb{R}^+\\
\overset{n-\text{polynomial agrees on n+1 values}}{\iff} \prod_{i=1}^n\frac{1 + u_{(i)}\xi_j}{1 + u_{(i)}} &= \prod_{i=1}^n\frac{1 + u'_{(i)}\xi_j}{1 + u'_{(i)}}, \forall j\in\{1,\cdots,n+1\}\\
\overset{\xi=e^{\theta}}{\iff} e^{n\theta_j}\prod_{i=1}^{n}\frac{1+e^{-x_{(i)}+\theta_j}}{1+e^{-x_{(i)}}} &= e^{n\theta_j}\prod_{i=1}^{n}\frac{1+e^{-x'_{(i)}+\theta_j}}{1+e^{-x'_{(i)}}}, \forall j\in\{1,\cdots,n+1\}\\\iff
\frac{f(x|\theta_j)}{f(x|\theta_0)} &=\frac{f(x'|\theta_j)}{f(x'|\theta_0)},\forall j\in\{1,\cdots,n+1\},
\end{align*}

where $\theta_0=0$ and $\theta_j\text{ (distanct)}\in \mathbb{R},j=1\cdots,n+1$.


问:
Definition of R-Algebra
There are two different definition of algebra:

The first one is given as follows :

Let $R$  be a commutative ring,$R-$algebra is a ring $A$ with ring homomorphism $f:R\to A$ (that mapping identity to identity), such that $f(R)$ is contained in the center of $A$.

Then we can make the $R$-algebra into $R$-module by defining scalar product $r\cdot a = f(r)a$.

If $R$ is a field then the above definition makes it a vector space.

There is another definition for $K$-algebra ($K$ is a field):

Let $K$ be a field, and let $A$ be a vector space over $K$ equipped with an additional binary operation from $A \times A$ to $A,$ denoted here by $\cdot$ (i.e. if $\mathbf{x}$ and $\mathbf{y}$ are any two elements of $A, \mathbf{x} \cdot \mathbf{y}$ is the product of $\mathbf{x}$ and $\mathbf{y})$. Then $A$ is an algebra over $K$ if the following identities hold for all elements $\mathbf{x}, \mathbf{y}, \mathbf{z} \in A,$ and all elements (often called scalars) $a$ and $b$ of $K$ :

Right distributivity: $(\mathbf{x}+\mathbf{y}) \cdot \mathbf{z}=\mathbf{x} \cdot \mathbf{z}+\mathbf{y} \cdot \mathbf{z}$

Left distributivity: $\mathbf{z} \cdot(\mathbf{x}+\mathbf{y})=\mathbf{z} \cdot \mathbf{x}+\mathbf{z} \cdot \mathbf{y}$

Compatibility with scalars: $(a \mathbf{x}) \cdot(b \mathbf{y})=(a b)(\mathbf{x} \cdot \mathbf{y})$.
which is given in wiki here

Are these two definition conside?

I can show the first definition satisfy the second one,the bilinear map defined in the second one with three axiom ,such that first two is given by Ring structure $A$,and the third one is given by $K=R$ lies in the center of $A$.

But it seems the second definition even need not to make $A$ ring ,due to associative and identity is not given

答:
The second one is more general, as it allows one to go on to define Lie algebras, Jordan algebras, Poisson algebras and other non-associative algebras in addition to associative algebras.

The first one determines only associative algebras. It defines a subclass of the class defined by the second definition.

(Overlooked the field/ring difference too.  You can extend the second to be over commutative rings as well, which is what I was thinking of while writing above. A great deal of work on algebras focuses on the scalar ring being a field, but using more general rings has also been extremely useful.)

Title: Definition of R-Algebra


Body:
$$\frac{n}{n+1}-\frac{n+1}{n}=\frac{n2-(n^2+2n+1)}{n(n+1)}=-\frac{n+(n+1)}{n(n+1)}=-\bigg(\frac{1}{n+1}+\frac{1}{n}\bigg)$$

$$\therefore lim_{n\rightarrow\infty} \frac{n}{n+1}-\frac{n+1}{n}=0$$


答:
Some curiosities:

(a) the polynomial in $u=\sqrt x$ is

$$ ( 2u - 5 ) ( 32u^6 + 80u^5 - 180u^4 - 418u^3 - 301u^2 + 2812u - 1805 ) $$

(b) the polynomial in $v=\sqrt y$ is

$$ ( 2v - 3 ) ( 32v^6 + 48v^5 - 300v^4 - 418v^3 + 133v^2 + 3968v - 2883 ) $$

(c) the polynomial in $w=\sqrt{xy}$ is

$$ ( 4w - 15 ) ( 1024 w^6 + 768 w^5 + 5952 w^4 - 120064 w^3 - 591600 w^2 + 1387684 w + 5203815 ) $$

I'm quite convinced there is no elegant way to find them. But I hope someone here will prove I'm wrong.

[EDIT]

John, it's not difficult (for a computer) to solve the system

$$u^2+u\,v^2-95/8$$
$$v^2+v\,u^2-93/8$$
$$w-u\,v$$

using elimination technics based on resultants or Gröbner basis. But if you want to know exactly what I did:

(a) is the Sylvester's Resultant of

$$(u) \, v^2 + (0)  \,v + (u^2-95/8)$$
$$(1) \, v^2 + (u^2)\,v + (-93/8)$$

(b) is the Sylvester's Resultant of

$$(1) \, u^2 + (v^2)\,u + (-95/8)$$
$$(v) \, u^2 + (0)  \,u + (v^2-93/8)$$

(c) is the Sylvester's Resultant of

$$(w) \, v^3 + (-95/8) \,v^2 + (0) \,v + (w^2)$$
$$(1) \, v^3 + (0)\,v^2 + (-93/8)\,v + (w^2)$$

I don't consider those as valid answers because I think everyone (including me) is expecting a solution that only relies on elementary algebra. I mean we are dealing with a system that is supposedly simple.

Title: Solve the system of equations for $\sqrt{xy}$

Body:
$$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$

I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.

$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$

$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$

This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.

I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.


问:
Marginal pdf from joint pdf?

I have a hard time finding the marginal pdf, can someone help?

Best regards.

答:
To illustrate:

$$\displaystyle f_{Y}(y)=\int_{x=y^2}^{x=4}f(x,y)dx=\int_{x=y^2}^{x=4}\frac 1{4x}dx=\frac 1 4\log\frac{4}{y^2}$$ with support from $y=0$ to $y=2.$

$$\displaystyle f_{X}(x)=\int_{y=0}^{y=\sqrt{x}}f(x,y)dy=\int_{y=0}^{y=\sqrt{x}}\frac 1 {4x}dy=\frac{1}{4\sqrt x}$$

with support from $x=0$ to $x=4.$

Title: Marginal pdf from joint pdf?


问:
Measuring the Shannon entropy of an ordered sequence
I have 927 unique sequences of the numbers 1, 2 and 3, all of which sum to 12 and represent every possible one-octave scale on the piano, with the numbers representing the intervals between notes in half-steps (i.e., adjacent keys). For example, the Major Scale is { 2,2,1,2,2,2,1 }, while the hexatonic (6-note) Blues Scale is { 3, 2, 1, 1, 3, 2 }. It can help mentally to add a zero at the beginning to represent the root note.

I'm trying to decide how to categorize them, and believe Shannon entropy is a natural place to begin, with one problem: The default Shannon Equation returns the same figure agnostic of the order of the sequence (as we would expect), whereas this is a case where the order is supremely relevant. (Music theory and information theory have a lot in common, though this appears to be largely unexplored.)

Let's take a simple example: Both of these are valid scales, though neither has a name, to my knowledge:

```<code>{ 1,2,1,2,1,2,1,2 }
{ 2,2,1,2,1,1,1,2 }
</code>```

Since there are equal numbers of 1s and 2s in each set, we don't need a calculator to see that the Shannon Entropy for each is 4:

$$-\sum_{i=1}^8 \frac{1}{2} \log_{2}\frac{1}{2}$$

However, I believe--I may be very wrong here--that there is significantly more information contained in the second sequence, given that the first can be communicated as "{1,2}, four times" while the second is considerably more arbitrary. (I believe this is debate in bioinformatics, where one can locate long spans of simple repeating sequences of nucleotides that some consider "junk DNA," though others dispute that adjective.)

How does one account for order when measuring surprisal? If I were to take, say, The Great Gatsby and sort the ~261,000 characters (as in letters, etc., not fictional people!) in ASCII order, I don't think it would contain the same amount of information as the original text.

One strategy I tried was a rolling measure of probability, where the algorithm is initially unaware of the total frequency of each number but accumulates the odds as it goes. (I suppose this is like measuring the accumulating surprisal of a die you cannot see and don't know how many sides it has.) This did not produce very intuitive or useful results.

Another thought I had was to divide each sequence into "words" that are repeating sub-sequences of at least two digits, then measure the entropy of each word and ... I'm not sure what to do with them. A sum of $H[{1,2}]$ four times still adds to 4. I'm a bit out of my depth here both in reducing sequences to the most efficient subsequences and knowing what to do with them.

I've read Shannon's 1948 paper twice, including the discussion of the relative odds of letters in a given language, which seems relevant but is, in fact, not useful in this case, where every possible ordered combination is present and the intervals are added independently of those preceding them, with the edge case that they cannot exceed a sum of 12.

Which is all to say, how does one quantify the entropy of an ordered sequence, when the order is of extreme relevance but each item is independent of the others? Below I'll include how I generated the 927 sequences for anyone interested, but it's ancillary to the question, I think.

***

First, I wrote a simple recursive algorithm that begins with an empty set and calls itself three times after adding 1, 2 or 3, terminating when the sum is 12 and tossing any that go over. Happy to provide code or pseudo-code, but I think it's intuitive.

To check myself, I generated the all the combinations of [1,2,3] that add to 12--there are 19--using a Coin-Change algorithm, and then calculated the unique, distinct permutations of each of them (treating identical digits as indistinguishable). The sum of the permutations is also 927, which range from 1 for the sequence of twelve 1s (the Chromatic Scale) or four 3s (a sparse diminished scale) to 140 possible scales for the intervals { 1, 1, 1, 2, 2, 2, 3 }, which almost subsumes the Harmonic Minor Scale.

One more thought

Regarding the above example of four 1s and four 2s: If I was flipping an unbiased coin 12 times and coding Heads as "1" and Tails as "2", I would be more surprised, in a general sense of the word, to get alternating results than to get the second sequence with small clusters. So it's possible that "surprisal" is the wrong frame of mind here versus information.

答:
You're starting to think of Kolmogorov complexity, which is a (almost uncomputable) measure of "how hard it is to describe" the sequence. It is completely dependent on "what is allowed to be used to describe" sequences (as computer programs, actually).

Shannon entropy fundamentally describes how much information per character there is when a stream of the specified probability distribution arrives. Serial correlations and so on are specifically not accounted for (you may of course consider characters to be multi-interval, and then your second sequence does have higher entropy).

Another point: surprisal in terms of "the intervals being in an unexpected mathematical sequence" is drastically different from surprisal per "this is a dissonant interval relative to the previous ones", but you're into atonality so that may not matter.

This is also heavy math, but in the direction of the lattice of rationals  (abstract algebra), not information theory.


问:
A System of Matrix Equations with a Surprising Property
Setup

Let $\mathbf a$ be an arbitrary $m\times 1$ vector, and $\mathbf B$ be an arbitrary $n\times m$ matrix, with $m>n$. Let $\mathbf K$ be the $m\times u$ given by
$$\mathbf K = \left[\begin{array}{c}\mathbf I_u\\ \mathbf{0} \end{array} \right],$$
where $m>u$.

The $n\times 1$ vector $\mathbf h$ satisfies the following fixed point problem:
\begin{align*}
\mathbf h =  & \;\left(\mathbf B {\mathbf \Omega}\mathbf B'\right)^{-1}\mathbf B{\mathbf\Omega}\mathbf{KK'a}  ,\\[2ex]
\mathbf {\Omega} =&\;\mathbf I_m   +\mathbf{K } \mathbf W\mathbf K' ,\\[2ex]
\mathbf W  = &\;\left\{  \mathbf I_u-   \mathbf K'\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right)\left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf K\right\}^{-1}.
\end{align*}

Conjecture

There exists a scalar $\gamma$, such that $\mathbf h$ satisfies
$$
\mathbf h =   \left(\mathbf B \mathbf\Gamma \mathbf B'\right)^{-1}\mathbf B\mathbf{ KK'a} ,
$$
where
$$ \mathbf\Gamma  =  \left[\begin{array}{cc}\mathbf I_u &\mathbf{0} \\
\mathbf{0}&\gamma\mathbf I_{m-u}\end{array} \right].$$

Updates:

We found a formula for $\gamma$:
$$\gamma=\frac{1-r}{2-r},\qquad\text{where } r=\left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf {KK'}\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right).$$
I also updated the code below, since it is more stable with this formula.

It follows from the Sherman-Morrison formula that
$$\mathbf W  = \mathbf I_u +   \frac{\mathbf K'\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right)\left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf K}{1-r}.$$

We have tested this conjecture numerically thoroughly and are pretty convinced it is true, but have not been able to prove it. Any help would be much appreciated.

As background, the fixed point problem comes from a variation on a linear forecasting problem with ambiguity aversion.

Here is a simple Matlab code that allows you to test the result:

```<code>clear

% Parameters:

n = 3;
u = 4;
m = 6;

a = randn(m,1);
B = randn(n,m);
K = [eye(u);zeros(m-u,u)];

% Find fixed point for h:

ite    = 0;
d      = 1;
maxite = 1000;
tol    = 1e-10;
h0 =  (a'*(K*K')*B'/(B*B'))';
while (d>tol && ite<maxite)
    ite = ite+1;
    W   = eye(u)/(eye(u)-K'*(B'*h0-a)*(h0'*B-a')*K);
    Om  = eye(m)+K*W*K';
    h   = (a'*(K*K')*Om*B'/(B*Om*B'))';
    d   = max(abs(h-h0));
    h0  = h;
end

% Solve for gamma:

r = ((h0'*B-a')*(K*K')*(B'*h0-a));
gamma = (1-r)/(2-r);

h_gamma = (a'*(K*K')*B'/(B*[eye(u),zeros(u,m-u);zeros(m-u,u),gamma*eye(m-u)]*B'))';

disp(gamma)
disp([h,h_gamma])
</code>```

答:
We start with the relation $\mathbf h =   \left(\mathbf B {\mathbf \Omega}\mathbf B'\right)^{-1}\mathbf B{\mathbf\Omega}\mathbf{KK'a}$ which gives
$ \mathbf B {\mathbf \Omega}\mathbf B'\mathbf h =   \mathbf B{\mathbf\Omega}\mathbf{KK'a}.$ Substituting $\mathbf \Omega=\mathbf{I}+\mathbf{KWK'}$, we get
\begin{align}
\mathbf B \left(\mathbf{I}+\mathbf{KWK}'\right)\mathbf B'\mathbf h &=   \mathbf B\left(\mathbf{I}+\mathbf{KWK}'\right)\mathbf{KK'a}\\
\implies \mathbf B\mathbf{KK'a} &=\mathbf {B B}' \mathbf h +\mathbf{B}\mathbf{KWK}'\left(\mathbf B'\mathbf h -\mathbf{a}\right),\tag{1}
\end{align}
since $\mathbf{K'KK'}=\mathbf{K'}.$ Using the relation $\mathbf W  = \left(  \mathbf I-   \mathbf K'\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right)\left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf K\right)^{-1}$ and the Sherman-Morrison formula, we obtain
\begin{align}
\mathbf{KWK}'\left(\mathbf B'\mathbf h -\mathbf{a}\right)&=\mathbf{K}\left(  \mathbf I-   \mathbf K'\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right)\left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf K\right)^{-1}\mathbf{K}'\left(\mathbf B'\mathbf h -\mathbf{a}\right)\\
&=\mathbf{K}\left(  \mathbf I +   \frac{\mathbf K'\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right)\left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf K}{1-r}\right)\mathbf{K}'\left(\mathbf B'\mathbf h -\mathbf{a}\right)\\
&= \left(  1 +   \frac{r}{1-r}\right)\mathbf{K}\mathbf{K}'\left(\mathbf B'\mathbf h -\mathbf{a}\right)= \frac{1}{1-r}\mathbf{K}\mathbf{K}'\left(\mathbf B'\mathbf h -\mathbf{a}\right)\tag{2},
\end{align}
where $r\triangleq \left(\mathbf h'\mathbf B-\mathbf a' \right)\mathbf K\mathbf K'\left(\mathbf B^{\prime}\mathbf h -\mathbf a \right)\neq 1$. Combining $(1)$ and $(2)$, we arrive at
\begin{align}
\mathbf B\mathbf{KK'a} &=\mathbf {B B}' \mathbf h +\frac{1}{1-r}\mathbf{B}\mathbf{K}\mathbf{K}'\left(\mathbf B'\mathbf h -\mathbf{a}\right)\\
\implies \frac{2-r}{1-r}\mathbf B\mathbf{KK'a} &= \mathbf {B}\left(\mathbf I+\frac{1}{1-r}\mathbf{K}\mathbf{K}'\right)\mathbf{ B}' \mathbf h\\
\implies \mathbf B\mathbf{KK'a} &= \mathbf {B}\left(\frac{1-r}{2-r}\mathbf I+\frac{1}{2-r}\mathbf{K}\mathbf{K}'\right)\mathbf{ B}' \mathbf h.\tag{3}
\end{align}
Since $\mathbf{K}=\begin{bmatrix}\mathbf{I} \\\mathbf 0\end{bmatrix}$, we get that
$$ \frac{1-r}{2-r}\mathbf I+\frac{1}{2-r}\mathbf{K}\mathbf{K}' = \frac{1-r}{2-r}\begin{bmatrix}\mathbf I &\mathbf 0\\ \mathbf 0&\mathbf I\end{bmatrix}+\frac{1}{2-r}\begin{bmatrix}\mathbf{I} \\\mathbf 0\end{bmatrix}\begin{bmatrix}\mathbf{I} &\mathbf 0\end{bmatrix} = \begin{bmatrix}\mathbf I &\mathbf 0\\ \mathbf 0& \frac{1-r}{2-r}\mathbf I\end{bmatrix}\triangleq \mathbf\Gamma.$$
Thus, $(3)$ yields
$ \mathbf B\mathbf{KK'a} = \mathbf {B}\mathbf\Gamma\mathbf{ B}' \mathbf h\implies \mathbf h=\left(\mathbf {B}\mathbf\Gamma\mathbf{ B}'\right)^{-1}\mathbf B\mathbf{KK'a}.$ Here, $\mathbf {B}\mathbf\Gamma\mathbf{ B}'$ is invertible as $r\neq 1$ due to the Sherman-Morrison formula, and from the fact that $\mathbf B {\mathbf \Omega}\mathbf B'$ is invertible (i.e., $n=\mathrm{rank}\{\mathbf B\}=\mathrm{rank}\{\mathbf B\mathbf B'\} $).

Thus, the proof is complete.


问:
How to solve for $ x $: $ ae^{bx}+ce^{dx}+x = f $
I don't know how to use The Lambert W function in this form, and I didn't find anything about it. Can someone help me? I'm trying to make a multimeter, and, to measure the voltage between the terminals, I need to calculate this equation, which referees two diodes in parallel, both in series with a resistor. I can resolve it with graphs, but I want to decrease the work that my microcontroller will do. Again, can someone help me? Thank you very much for your attention :)</p>

答:
Waiting for more information about the parameters, I shall assume that $a$ and $c$ are positive.

You are looking for the zero of function
$$f(x)=ae^{bx}+ce^{dx}+x - f$$ which will not show any analytical solution and for which a numerical method will be required. This implies a "reasonable" starting guess or at least a renge for the solution.

What we can consider is that $f(x)$ is somewhere between
$$g(x)=ae^{bx}+ce^{bx}+x - f \quad \text{and} \quad h(x)=ae^{dx}+ce^{dx}+x - f$$

$$g(x)=0 \implies x_g=f-\frac 1b W\left(b (a+c) e^{b f}\right)$$
$$h(x)=0 \implies x_h=f-\frac 1d W\left(d (a+c) e^{d f}\right) $$

Let us try for $a=2$, $b=\frac 12$, $c=3$, $d=\frac 15$ and $f=23.456$. This would give
$$x_g \sim 2.83386 \quad \text{and} \quad x_h\sim 6.19505$$ Using the average as $x_0$, Newton iterates would be
$$\left(
\begin{array}{cc}
 0 & 4.514450 \\
 1 & 3.885394 \\
 2 & 3.787382 \\
 3 & 3.785395 \\
 4 & 3.785394
\end{array}
\right)$$

Using @Semiclassical's example
$$f(x)=2e^{x}+e^{2x}+x-1$$ we should have
$$x_g \sim -0.617642 \quad \text{and} \quad x_h\sim -0.38607$$ Newton iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & -0.5018562 \\
 1 & -0.5274934 \\
 2 & -0.5277952
\end{array}
\right)$$


问:
Is a subset of a relation that is a partial order also a partial order?
I've sat on this question for a while now:

Question:

Suppose that $R$ is a partial order. If we take some subset $S$ of $R$, will it also
be a partial order? Prove or disprove this.

Now I know the following. For a partial order, the set must follow these criteria.

reflective

anti-symmetric

transitive

So if the relation on a set is a partial order. If we collected any values from that set $R$ and put in set $S$ will that be a partial order?

I believe that the subset is always reflective because a set is always a subset of itself. However I am struggling to prove the other $2$ criteria.

Example:

$R = \{(1,1),(1,2),(2,2),(1,3),(2,3),(3,3)\}$

$S = \{(1,1),(2,3)\}$

Subset $S$ is not a partial order from the set of relation $R$? I could use multiple counter examples and proof by cases?

Is this the right way of looking at the question?

Thanks.

答:
A subset of a partial order need not always be one. For a simple example, take

$$R = \{(1,1),(2,2),(3,3),(1,2),(2,3),(3,1)\}$$

This relation is a partial order. However, take

$$S = R \setminus \{(1,1)\}$$

The relation is no longer reflexive, and thus not a partial order. In general, if $R$ is a partial order over a nonempty set $A$, then $S_a = R\setminus \{(a,a)\}$ for any fixed $a \in A$ is a subset of $R$ which is not a partial order. (And we know such an $(a,a) \in R$ exists since $A \ne \varnothing$ and $R$ is reflexive.)

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Convergence of infinite products flaw in this derivation of Euler product formula?
I am trying to write a derivation of the Euler product formula in a way that is accessible to younger students and those not mathematically trained at university.

Because of this I am starting from the simplest starting point that I can, and use the simplest possible steps in the derivation.

I arrive at the following which I know is wrong as it is missing the exponent $s>1$ to ensure convergence.

$$ \prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n} $$

My question is to ask where I am introducing the error in the following simple derivation.

Step 1

Start with a familiar series, valid for $|x| < 1$.

$$ \frac{1}{(1-x)}=1+x+x^{2}+x^{3}+\ldots $$

We're interested in primes, and might be tempted to set $x$ to a prime $p$, but we can't because $|x|$ needs to be $<1$. Let's try $\frac{1}{p}$ which is always $<1$.

$$ \frac{1}{(1-\frac{1}{p})}=1+\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}+\ldots $$

This series converges to a finite value.

Step 2

Let's now pick two different primes $p_1$ and $p_2$, and multiply the analogous series for each.

$$ \frac{1}{(1-\frac{1}{p_{1}})}\cdot\frac{1}{(1-\frac{1}{p_{2}})}=\left(1+\frac{1}{p_{1}}+\frac{1}{p_{1}^{2}}+\ldots\right)\cdot\left(1+\frac{1}{p_{2}}+\frac{1}{p_{2}^{2}}+\ldots\right) $$

I don't think there is a problem here.

We are multiplying two infinite series, each one known to converge to a finite value. Am I wrong?

Step 3

Now we extend from two primes, $p_1$ and $p_2$, to all primes $p_i$.

$$ \prod_{p_{i}}\frac{1}{(1-\frac{1}{p_{i}})}=\prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) $$

I think this is the flaw in my derivation. We have an infinite product of finite values.

Each factor $\frac{1}{(1-\frac{1}{p_{i}})}$ is $>1$ so the infinite product looks like it might not converge.

However each factor tends to $\rightarrow 1$ as $i \rightarrow \infty$. This suggests the infinite product is ok. Am I wrong?

Step 4

If we multiply out those brackets, each of the form $\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right)$, we will obtain terms which are all combinations of primes, in all the combinations of powers for each prime.

$$ \prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) = 1 + \frac{1}{p_1} + \frac{1}{p_2} + \ldots + \frac{1}{p_1 p_2} + \frac{1}{p_1 p_3} + \ldots + \frac{1}{p_1^2} + \frac{1}{p_2^2} + \ldots  $$

Importantly, each combination occurs only once.

Step 5

The fundamental theorem of arithmetic tells us that each integer can be written as a unique product of primes. This means that each integer is represented in the terms we arrived at in the previous step.

Because each combination of primes occurs only once, each integer is represented only once.

That leads us to the Euler product formula.

$$ \boxed{\prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n}} $$

Discussion

If my suspicion is correct that the error is in the infinite product because it doesn't converge, how does changing $p$ for $p^s$ for $s>1$ help?

答:
Your intuition is right that there is a flaw in step 3. You propose that any infinite product in which the terms tend to $1$ must converge; but this is false. The easiest example is the telescoping product
$$
\prod_{j=1}^J \bigg( \frac{j+1}j \bigg) = J+1,
$$
which show that $\prod_{j=1}^\infty \frac{j+1}j$ diverges to $\infty$. More generally, the convergence of the product $\prod_j (1+x_j)$ is by definition equivalent to the convergence of its logarithm $\sum_j \log(1+x_j)$, and in many circumstances this is further equivalent to the convergence of the logarithm's linear approximation $\sum_j x_j$; the example above is the case $x_j = \frac1j$. This is where we need the assumption $s>1$.

There is also a flaw (a very common one, in my experience) in step 4. You are correct that each of the terms on the right-hand side appears when we expand the left-hand side. However, there are literally uncountably many other terms as well. What about, for example, the term in the expansion where we select $\frac1{p_i^1}$ from every factor? What happens to a term like $\frac1{p_1p_2\cdots}$? And all the others where infinitely many of the factors don't equal $1$? In my opinion, it's not very easy to repair this flaw without starting over and using a more rigorous real-analysis proof.


问:
Let $R$ be a relation on a set $A$.
Define $T(R)=R\cup R^{-1} \cup \{(x,x)\mid x \in A \}$. Show that $T(R)$ is reflexive and symmetric.
Let $R$ be a relation on a set $A$.
Define $T(R)=R\cup R^{-1} \cup \{(x,x)\mid x \in A \}$. Show that $T(R)$ is reflexive and symmetric.

Let $\triangle = \{(x,x) \mid x\in A\}$

I only know that $R \cup \triangle$ is the reflexive closure and $R \cup R^{-1}$ is the symmetric closure.

Any idea?

答:
You almost have it! Let me properly write the idea:

$T(R)$ is reflexive since if $x \in A$, then $(x,x) \in \Delta$ and $\Delta \subseteq T(R)$ by definition of $T(R)$.

$T(R)$ is symmetric since if $(x,y) \in T(R)$, then $(x,y) \in R$ or $(x,y) \in R^{-1}$ or $(x,y) \in \Delta$. In the case that $(x,y) \in R$ we have $(y,x) \in R^{-1} \subseteq T(R)$; in the case that $(x,y) \in R^{-1}$ we have $(x,y) \in R \subseteq T(R)$; and in the case that $(x,y) \in \Delta$ we have $(y,x) = (x,y) \in T(R)$. Thus, $(x,y) \in T(R)$ always implies that $(y,x) \in T(R)$.


答:
You are correct that step 3 is a problem.  (This is not to say that steps >3 are not also problems.)

You suggested that if the factors tend to 1 then it will be okay, but that is not valid.  For instance,
$$
\prod e^{1/n}=e^{\sum 1/n}=e^\infty=\infty
$$

Title: Convergence of infinite products flaw in this derivation of Euler product formula?

Body:
I am trying to write a derivation of the Euler product formula in a way that is accessible to younger students and those not mathematically trained at university.

Because of this I am starting from the simplest starting point that I can, and use the simplest possible steps in the derivation.

I arrive at the following which I know is wrong as it is missing the exponent $s>1$ to ensure convergence.

$$ \prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n} $$

My question is to ask where I am introducing the error in the following simple derivation.

Step 1

Start with a familiar series, valid for $|x| < 1$.

$$ \frac{1}{(1-x)}=1+x+x^{2}+x^{3}+\ldots $$

We're interested in primes, and might be tempted to set $x$ to a prime $p$, but we can't because $|x|$ needs to be $<1$. Let's try $\frac{1}{p}$ which is always $<1$.

$$ \frac{1}{(1-\frac{1}{p})}=1+\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}+\ldots $$

This series converges to a finite value.

Step 2

Let's now pick two different primes $p_1$ and $p_2$, and multiply the analogous series for each.

$$ \frac{1}{(1-\frac{1}{p_{1}})}\cdot\frac{1}{(1-\frac{1}{p_{2}})}=\left(1+\frac{1}{p_{1}}+\frac{1}{p_{1}^{2}}+\ldots\right)\cdot\left(1+\frac{1}{p_{2}}+\frac{1}{p_{2}^{2}}+\ldots\right) $$

I don't think there is a problem here.

We are multiplying two infinite series, each one known to converge to a finite value. Am I wrong?

Step 3

Now we extend from two primes, $p_1$ and $p_2$, to all primes $p_i$.

$$ \prod_{p_{i}}\frac{1}{(1-\frac{1}{p_{i}})}=\prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) $$

I think this is the flaw in my derivation. We have an infinite product of finite values.

Each factor $\frac{1}{(1-\frac{1}{p_{i}})}$ is $>1$ so the infinite product looks like it might not converge.

However each factor tends to $\rightarrow 1$ as $i \rightarrow \infty$. This suggests the infinite product is ok. Am I wrong?

Step 4

If we multiply out those brackets, each of the form $\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right)$, we will obtain terms which are all combinations of primes, in all the combinations of powers for each prime.

$$ \prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) = 1 + \frac{1}{p_1} + \frac{1}{p_2} + \ldots + \frac{1}{p_1 p_2} + \frac{1}{p_1 p_3} + \ldots + \frac{1}{p_1^2} + \frac{1}{p_2^2} + \ldots  $$

Importantly, each combination occurs only once.

Step 5

The fundamental theorem of arithmetic tells us that each integer can be written as a unique product of primes. This means that each integer is represented in the terms we arrived at in the previous step.

Because each combination of primes occurs only once, each integer is represented only once.

That leads us to the Euler product formula.

$$ \boxed{\prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n}} $$

Discussion

If my suspicion is correct that the error is in the infinite product because it doesn't converge, how does changing $p$ for $p^s$ for $s>1$ help?


答:
Let $x \in A$. Then, $(x,x) \in \triangle \subseteq T(R)$. Hence, $T(R)$ is reflexive.

Let $x,y \in A$ and let $(x,y) \in T(R)$.
Then, $(x,y) \in R$ or $(x,y) \in R^{-1}$ or $(x,y) \in \triangle$.

If $(x,y) \in R$, then by definition, we have $(y,x) \in R^{-1} \subseteq T(R)$.
Hence, $T(R)$ is symmetric.

If $(x,y) \in R^{-1}$, then by definition, we have $(y,x) \in R \subseteq T(R)$. Hence, $T(R)$ is symmetric.

If $(x,y) \in \triangle$, then we must have $y = x$. Thus, $(x,y)=(y,x)$. Hence, $T(R)$ is symmetric.

That's it, $T(R)$ is reflexive and symmetric as desired.

Body:
Let $R$ be a relation on a set $A$.
Define $T(R)=R\cup R^{-1} \cup \{(x,x)\mid x \in A \}$. Show that $T(R)$ is reflexive and symmetric.

Let $\triangle = \{(x,x) \mid x\in A\}$

I only know that $R \cup \triangle$ is the reflexive closure and $R \cup R^{-1}$ is the symmetric closure.

Any idea?


问:
What is the point of model theory?
Wikipedia defines model theory as ...

... the study of the relationship between formal theories (a collection of sentences in a formal language expressing statements about a mathematical structure), and their models, taken as interpretations that satisfy the sentences of that theory.

Why does this field need to exist? If we have proof theory, then we can discuss any formal system $\Gamma$ and statements within $\Gamma$ like, say, some arbitrarily chosen sentences $\phi$ and $\psi$, and even theorems like $\Gamma \vdash \phi \to \psi$, assuming that $\psi$ is derivable from $\phi$ in $\Gamma$, of course. Why should anyone care whether $\Gamma \models \phi$ if one already knows that $\Gamma \vdash \phi$?

Thanks for any advice.

答:
If you wish to understand your own question, you need to learn some history of how mathematics became the mathematics of today.

Some accounts of the priority debate between Newton and Leibniz suggest that British mathematics pursued a logical calculus in the spirit of Newton's achievements. Meanwhile, Continental mathematics developed the calculus along Leibnizian methods. One may understand the work of Augustus de Morgan with regard to symbolic algebras in this context. Arguably, explicit accounts of formalism begin with his categories "peculiar symbols," "rules of operation," and "meanings." He refers to algebras with meanings as "significant algebras." He points out that an individual who discovers a symbolic algebra with no instruction of intended meanings might invent meanings which are no less significant than those pragmatically conveyed from teacher to student.

It is important to understand that the significance of Skolem's criticism of Zermelo falls under this approach to studying mathematics.  And, logicism as introduced by Frege rejected this formalist view. And, while Russell's logicism differed from Frege's, it also rejected this formalist view.

Chang and Keisler describe model theory as "logic + universal algebra." Of course, the symbolic algebras studied by de Morgan fall under universal algebra today. But, as logic developed in the late nineteenth century, rules of inference became axiomatized--- that is, logic now had "rules of operation" in the sense described by de Morgan. However, these rules of operation act upon formulas. For formalists, Skolem's criticism of Zermelo had been a first step toward understanding the "definite properties" of mathematical statements in the form of first-order formulas.

Your questions about proof theory speak to the influence of David Hilbert. His interest in formalism seems to have been motivated by the desire to ensure that mathematics could not be responsible for contradictions that might arise in applications of mathematics. A secondary motivation had been to sever mathematics from its association with spatiotemporal intuition. Hilbert did not protest Kant's reduction to a priori intuition nearly as much as others in his day. He simply maintained that Kant attributed too much to a priori intuition. His book, "Foundations of Geometry" had been significant for eradicating temporal verbs from geometric reasoning. And, because the nineteenth century produced a plurality of geometries, he demonstrated that "peculiar symbols" and "rules of operation" could, indeed, support different "meanings."

When Hilbert turned his attention to mathematical logic some time later, Russell had already published several works. Hilbert had been aware of Russell's work. But, he also retained his concerns about consistency. His first attempts to ground mathematical induction attracted criticism from Poincare over circularity. His response to this had been a limited invocation of a priori intuition.  In Hilbert's case it would be the immediate sensible impression made by symbols on a page. You could not have proof theory without this. Formalists maintain that Hilbert's metamathematical induction is weaker than the mathematical induction in the object language. Metamathematical enumeration consists of alternating strokes and plus signs with only strokes as initial and terminal symbols. Formulas are concatenations of the "peculiar symbols." Proofs are the concatenations of formulas according to the "rules of operation" given by the logical calculus.

Of course, there is no reason for platonists to respect the subtleties Hilbert needed to circumvent Poincare's objections. And, there is no reason to actually distinguish formalism from platonism with regard to Goedel's work. Hilbert needed inductive structure for the purpose of verification. His metamathematics specifically assumes an inseparable relationship between arithmetic and logic.  Goedel's approach to recursion is clearly platonistic. And Hilbert's presuppositions are vulnerable to the arithmetization of syntax.

In so far as model theory is understood to arise from universal algebra, "meanings" in semantics ought to relate to one another in much the same way that algebraic realizations relate to one another through homomorphisms and isomorphisms. In an appendix to his book on mathematical logic, Kreisel divides foundational studies into semantical foundations and combinatorial foundations. His usage may be compared with the study of groups as is done in combinatorial group theory. In combinatorial group theory one studies presentations (de Morgan's rules of operations).  In combinatorial foundations one begins with signatures (de Morgan's peculiar symbols).

This is important for understanding Skolem's objections to Zermelo. Above all, Zermelo's set theory had failed to be categorical. The signature of Zermelo's theory admitted more than just a single interpretation (de Morgan's meanings). This is certainly a problem for a theory claiming to be a foundation for mathematics.

There are two particularly important features of first-order logic relating to infinities. One is the compactness theorem. The other is the Lowenheim-Skolem theorem. First-order logic is the only logic closed under conjunction, unary negation, and existential quantification which satisfies both of these theorems. And, while one might not require a complete logic, the completeness of first-order logic speaks to the artifact of Hilbert's concern for the study of  consistency.

Not every logic is complete. Your willingness to set aside semantics in favor of proof theory is only possible by virtue of a history in which Hilbert emphasized the need to consider the possibility that mathematics might contain hidden inconsistencies. That is, it is only possible because first-order logic is complete.

Title: What is the point of model theory?


问:
How to do this counting problem?
I know that the pigeonhole principle is to be applied here but I can’t see yet how.

On a certain planet in the solar system Tau Cetus, more than half the surface of the planet is dry land. Show that the Tau Cetans can dig a tunnel straight through the centre of the planet, beginning and ending on dry land.

答:
If dry land is always opposite of non dry land, then the area of wet land must be at least equal to area of dry land. But the area of dry land itself is already more than half the area of the planet. If you don't like paraphrasing, check out the following alternative:

Let $S$ be the whole planet, $G$ be all dry land and $G'$ be all land opposite of dry land. Suppose that $G\cap G'=\varnothing$
$$
\begin{align}
A(S)&\geq A(G\cup G')\\
&=A(G)+A(G')-A(G\cap G')\\
&=2A(G)
\end{align}
$$

This contradict $2A(G)>A(S)$

Title: How to do this counting problem?


答: In this case there are uncountably many pigeonholes corresponding to all possible pairs of antipodes of the planet (endpoints of all tunnels) and a pigeon is an infinitesimally small patch of land. Since taking one end of each tunnel covers exactly half the planet's area and more than half the planet's surface is land, there must be a pigeonhole with two pigeons, i.e. suitable endpoints for a land-to-land tunnel.</p>

Title: How to do this counting problem?

Body:
I know that the pigeonhole principle is to be applied here but I can’t see yet how.

On a certain planet in the solar system Tau Cetus, more than half the surface of the planet is dry land. Show that the Tau Cetans can dig a tunnel straight through the centre of the planet, beginning and ending on dry land.


答:
Step 3 is where a "big leap" first occurs in the logic.  Putting the variable $s$ back into the exponent, it is an interesting issue that the geometric series expansion
$$
\frac{1}{1-1/p^s} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \cdots = \sum_{k \geq 0} \frac{1}{p^{ks}}
$$
is completely valid when $s > 0$ (or ${\rm Re}(s) > 0$ if you want to use complex variables) and also when multiplying together finitely many of these equations, but multiplying such equations together over all $p$ is invalid when $0 < s \leq 1$.

It's hard to pin down an intuitive reason why something goes wrong in the passage from the finite to the infinite when $0 < s \leq 1$ but not when  $s > 1$. You can't say "well, after some definite finite number of steps there's a problem".  The issue is entirely about what's going on with the full infinite product. It's sort of like the number $\pi = 3.1415926535\ldots$ being irrational but it is a limit of truncated decimals $3$, $3.1$, $3.14$, and so on and at each finite stage the number is rational but the limit is not. There is no definite point at which the terms stop being rational.  That property is true at the limit but not at an earlier step.  Or an infinite series of continuous functions can be discontinuous (Fourier series). Limits are subtle.

The only mathematically acceptable way to "explain" why we need $s > 1$ is to bring in a serious theorem that justifies when infinite products can be expanded and rearranged arbitrarily to get a legal infinite series. Here is such a theorem.

Theorem. Let $\{z_n\}$ be real or complex numbers such that (i) $|z_n| \leq 1 - \varepsilon$ for
some positive $\varepsilon$ (which is independent of $n$)
and (ii) $\sum |z_n|$ converges.

a$)$ The infinite product
$\prod_{n \geq 1} \frac{1}{1-z_n} := \lim_{N \rightarrow \infty}
\prod_{n=1}^{N} \frac{1}{1-z_n}$ converges to a nonzero number.

b$)$ Every rearrangement of factors of the product converges to the same value.

c$)$ The product $\prod_{n \geq 1} \frac{1}{1-z_n} =
\prod_{n \geq 1} (1 + z_n + z_n^2 + \dots)$ has a series expansion
by collecting terms in the expected manner:
$$
\prod_{n \geq 1} \frac{1}{1 - z_n} =
1 + \sum_{r \geq 1}\sum_{\substack{k_1,\dots,k_r \geq 1 \\ 1 < i_1 < \dots < i_r}}
z_{i_1}^{k_1}\cdots z_{i_r}^{k_r},
$$
and this series is absolutely convergent.

Applying this with $z_n = 1/p_n^s$, where $p_n$ is the $n$th prime and $s > 0$, what we need in this theorem to justify the expansion of the Euler product into the Dirichlet series for the zeta-function is (i) $1/p_n^s \leq 1-\varepsilon$ for all $n$ and (ii) $\sum_{n \geq 1} 1/p_n^s < \infty$. There is no issue with the first condition regardless of the value of $s$, since $1/p_n^s \leq 1/2^s < 1$ for each $s > 0$: we can use $1-\varepsilon = 1/2^s$. But for the second condition we need $\sum_{n \geq 1} 1/p_n^s < \infty$. That is true for $s > 1$ but not for $0 < s \leq 1$, and this is how the theorem uses $s > 1$. But that is not intuitive and I do not think such convergence criteria from analysis are  intuitive for inexperienced students.  It is the actual nuts and bolts of the proof of the theorem that shows how the hypotheses lead to the conclusion here.

My advice is to point out there is a subtlety at $s = 1$, which makes it plausible that for $s < 1$ things are also bad (because $1/p^s > 1/p$ when $0 < s < 1$) and then just say mathematicians can justify that the the result of informal handwavy calculations is okay for $s > 1$.  Make some reference to the integral test if the students have seen that already: $\int_1^\infty dx/x^s$ is convergent for $s > 1$ but not for $0 < s \leq 1$ even though $\int_1^b dx/x^s$ is meaningful for all $s > 0$ when $b$ is finite.


问:
Amy has a cuboid with volume at most 8

Amy has a clump of iron with volume $8$. She melts it and make a cuboid (she might have iron left over). Her boyfriend Ben increased the length, width and height of the cuboid by $1$ to form a new cuboid. Show that the surface area of the new cuboid is at least two times of the volume of the new cuboid.

I made this problem, so I know a solution (using inequality). Any solution is accepted as long as it makes sense. There might be a geometric solution that I don't know! I will post my solution if no one posts it.

答:
We let the sides be $a,b,c$, so $abc\le8$. Then we need to prove
$$2\big((a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)\big)\ge2(a+1)(b+1)(c+1)$$
Which is equivalent to
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\ge1$$
We let $a'=\frac8{bc}$. As $abc\le8$, $a\le\frac8{bc}=a'$, so $a+1\le a'+1$, therefore $\frac1{a+1}\ge\frac1{a'+1}$. Also, we have $a'bc=\frac8{bc}\cdot bc=8$, so we can let $a'=\frac{2x}y,b=\frac{2y}z,c=\frac{2z}x$. Then, with above facts and the CS inequality,
$$\begin{split}\frac1{a+1}+\frac1{b+1}+\frac1{c+1}&\ge\frac1{a'+1}+\frac1{b+1}+\frac1{c+1}\\&=\frac1{\frac{2x}y+1}+\frac1{\frac{2y}z+1}+\frac1{\frac{2z}x+1}\\&=\frac y{2x+y}+\frac z{2y+z}+\frac x{2z+x}\\&=\frac{\big(\frac y{2x+y}+\frac z{2y+z}+\frac x{2z+x}\big)(x+y+z)^2}{(x+y+z)^{2}}\\&=\frac{\big(\frac y{2x+y}+\frac z{2y+z}+\frac x{2z+x}\big)\big(y(2x+y)+z(2y+z)+x(2z+x)\big)}{(x+y+z)^{2}}\\&\ge\frac{(y+z+x)^2}{(x+y+z)^2}\\&=1\end{split}$$

Title: Amy has a cuboid with volume at most 8


答:
Proof by contradiction: let's suppose that there was no way to dig a tunnel through the center of the planet starting on dry land which also ends on dry land. This means that for each point on dry land, the point on the opposite side of the planet cannot be dry.

Let's define a function $O(p)$ which acts on the domain of points on the planet which lie on dry land. By our reasoning before, the codomain must consist of entirely wet land. It should be clear that $O$ is injective because each point can only be opposite to exactly one point on the planet.

However, the pigeonhole principle states that there cannot exist an injective function whose domain is larger than its codomain, and the problem tells us that there is more dry land than wet land. Therefore, we have reached a contradiction and there must be some way to dig a tunnel through the center of the planet which starts and ends on dry land.

Hope this helps!

Title: How to do this counting problem?


问:
Let be $T \in L (V) $ a nilpotent operator, $T^{ p }= 0.$ Show that the I-T operator is invertible
I need some help with this exercise, I did the version with matrices but I don't know how to extrapolate it to linear operators, a hint is enough for me</p>

Body: $(I-T)(I+T+T^2+...+T^{p-1})=I-T^p=I$. So $(I-T)^{-1}=I+T+T^2+...+T^{p-1}$.


问:
How to find the length of a segment in a quadrilateral made by two triangles?
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{3 cm}\\
2.&\textrm{4 cm}\\
3.&\textrm{5 cm}\\
4.&\textrm{6 cm}\\
\end{array}$

I'm stuck at trying to find the proper relationships in this figure based on the triangle congruence. Can someone help me here?

The thing is that I'm stuck on how to find $BE$?.

The only relationships which I could spot were:

$AC=AD$

and triangle $BCE$ is isosceles. But nothing more than that?. What else can be done here?.

Please include a drawing in your answer I don't know exactly how to relate a triangle congruence here, which seems to be the intended method of solution?. Can this be approached relying only in euclidean geometry?.

答:
The trick about this problem is that the given angles and lengths show $ABCD$ is inscribed in a circle. This is because
$$AD=DE \Longrightarrow \widehat{DAC}=\widehat{DEA}$$
$$\widehat{DEA} = \widehat{CEB}$$
$$BC=CE \Longrightarrow \widehat{CEB}=\widehat{CBD}$$
$$\Longrightarrow \widehat{DAC} = \widehat{CBD}$$
This is enough to show that the four points $A,B,C \& D$ are on the same circle.

What follows is interesting. From the given figure:
$$\widehat{BCA} = \widehat{ACD} =\alpha$$
And now that we know the points are on the same circle, we can see that
$$ \widehat{ABD} = \widehat{ACD} \quad \& \quad \widehat{ADB} = \widehat{ACB}$$
So:
$$ \widehat{ABD} = \widehat{ADB} \Longrightarrow AD=AB$$
Now, the problem tells us $AB=6$ and $AD=DE$ , so we have $DE=6$, and subtracting from the given length of $BD$ we see that $BE=5$.


问:
Solve limit using definite integration
I have to solve this limit using definite integral

$${\lim_{n\to \infty} \frac{1}{n} \cdot \sum_{i=1}^n \frac{1}{1+(\frac{i}{n})^i}}$$

Well, my development was:

I tried to give the Riemann integral definition form, using a regular partition such that:

$$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{i=1}f\left(a+i\frac{b-a}{n}\right)$$

So, i set $b=1, a=0$ for simplicity, then we have $f(\frac{i}{n})$ must be equal to ${\frac{1}{1+(\frac{i}{n})^i}}$.

Let ${x=\frac{i}{n}}$, then $${f(x)=\frac{1}{1+x^{nx}}}$$

The problem is that $n$ is a dummy variable that does not make sense outside the limit, that is, it does not make sense for the function $f$

So, I need to do some kind of algebraic transformation or variable change for $n$, in such a way that the function $f$ remains only in terms of $x$ and thus I can use it in the definite integral. However, I have not been able to find such a magical algebraic transformation or variable change for $n$.

答:
As it is really shown in OP, for $\;x=\dfrac in\in(0,1]\;$
$$f(x) =\lim\limits_{n\to\infty}\;\dfrac 1{1+x^{nx}}=\genfrac{\{}.0{}{1,\text{ if }x\in(0,1)}{\;^1/_2,\text{ if } \;x=1,\;}$$
i.e. the function $\;f(x)\;$ has a removable discontinuity at the point $\;x=1\;$ and is equaled to $1$ at the other points of the domain.

Therefore,
$$\color{brown}{\mathbf{\lim\limits_{n\to\infty}\,\dfrac1n\,\dfrac1{1+\left(\large\frac in\right)^i} = \int\limits_0^1\, 1\,\text dx =1.}}$$

Title: Solve limit using definite integration


问:
Find the minimum values of $a,b,c,d,e,f$ that satisfy following equations
${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c \right)  }^{ 2 }={ d }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d \right)  }^{ 2 }={ e }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d+e \right)  }^{ 2 }={ f }^{ 2 }$

where $a,b,c,d,e,f$ positive integers. Moreover, prove that we can always find positive integer solutions when the number of equations goes to infinity, i.e.

${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c \right)  }^{ 2 }={ d }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d \right)  }^{ 2 }={ e }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d+e \right)  }^{ 2 }={ f }^{ 2 }\\ \vdots $

Note: You probably see this problem for the first time here.

Edit: A small example:

${ 24 }^{ 2 }+{ 7 }^{ 2 }={ 25 }^{ 2 }\\ { 24 }^{ 2 }+{ \left( 25+7 \right)  }^{ 2 }={ 40 }^{ 2 }$

Body:
We consider the rational solutions first,and then multiply by an integer to make them be integers.

Assume $a=1,$ then $1+b^2=c^2,(c-b)(c+b)=1,$ let $c+b=t_1,c-b=\dfrac{1}{t_1},$ then we get
$$c=\frac{1}{2}(t_1+\frac{1}{t_1}),b=\frac{1}{2}(t_1-\frac{1}{t_1}),$$
we need $1+(b+c)^2=1+t_1^2=d^2,$in the same way,we get
$$d=\frac{1}{2}(t_2+\frac{1}{t_2}),t_1=\frac{1}{2}(t_2-\frac{1}{t_2}),$$
$$e=\frac{1}{2}(t_3+\frac{1}{t_3}),t_2=\frac{1}{2}(t_3-\frac{1}{t_3}),$$
$$f=\frac{1}{2}(t_4+\frac{1}{t_4}),t_3=\frac{1}{2}(t_4-\frac{1}{t_4}),$$
$$\cdots$$

For example, let $t_4=11,$ then $t_3=\frac{60}{11},t_2=\frac{3479}{1320},t_1=\frac{10361041}{9184560},b=\frac{22995028210081}{190323205453920}.$

Then multiply by $190323205453920,$ we get
$a=190323205453920,b=22995028210081,c=191707312997281,d=286914652560962,e=536509581434876,f=1055428684789920.$

You must choose $t_4$ so that $t_3,t_2,t_1,b>0$ .


问:
Probability of disease transmission to future generations
Assume disease D is transmitted through a father (male) to his children. Let us assume that every family in a society have $C\geq1$ children. Let the father of a family have the disease D. What is the probability that disease D is transmitted to at least one person in the $N$-th generation after him $(N\geq1)$? $N=1$ represents the children of the father, $N=2$ represents the grandchildren, etc.

One can calculate the probability for a given $N$ and $C$, but I was wondering if it possible to find a general closed-form solution.

P.S.: The problem is equivalent to the problem of passing the family-name. In a society where the family-name is passed from father to children, what is the probability of having the family-name of a father in the $N$-th generation after him.

答:
What you're referring to is sometimes called a Branching Markov Chain, or simply Branching Process.

To solve we'll introduce some notation.

Let $X_n$ be the number of diseased people at time(generation) $n$.
$p_{d}$ be the probability of having $d$ offspring from a person with disease $D$. $u_n = Pr(X_n = 0)$. Our goal is to find $Pr(X_n > 0) = 1 - u_n$.

Using the law of total probability we can recursively find:

$u_n = \sum\limits_{d=0}^{C^n}p_d(u_{n-1})^d$.

Or more generally

$u_n = \sum\limits_{d=0}^\infty p_d(u_{n-1})^d$.

Allowing us to find the extinction probability at time $n$. From there we simply take it's complement to find the probability of $D$ existing at time $n$

Reference: An Introduction to Stochastic Modeling, Pinsky and Karlin

EDIT:

I've made an error and edited above. $p_d$ is the probability of having $d$ children who could inherit the disease. At $N = 1$ this is simply $C$, but at greater $N$ this some multiple of $C$ depending on previous generations.

$u_n$ is found recursively at each generation using the previous generations $u$ and the total law of probability.


问:
Finding Height In Gerstner Wave Function In World space.
Using the GPU Gems Article Effective Water Simulation From Physical Models I have implemented Gerstner Waves into UE4, I have built the function both on GPU for the tessellated mesh displacement and in code for the purpose of sampling height of the waves, but the issue I have run into is that the points by nature of the formula move away from their original points, so when using the gerstner function to get the height of a point, it is not necessarily at the point you are sampling so provides an incorrect height for that coordinate.

So somehow I need to solve for the point in the gerstner function that is actually lines up with the world position.  I've taught myself enough to implement the waves, but i'm stuck on this solution, though I suspect there is some sort of matrix or way to take the inputs, apply the cos values and use it as a lookup for the actual point to sample.  Any ideas and help is greatly appreciated.

For Reference the Gerstner Wave Formula

Formula Image

Q = Slope

L = WaveLength

A = Amplitude

D = Vector2 Direction

x = x world coordinate

y = y world coordinate

t = time

Where Qi = Q/wi * A * NumWaves

Where wi = 2pi/L

Where phase = Speed * 2/L

答:
Note I'm ignoring a lot of the variables for brevity
Gerstner Waves vaguely simplify to y=sin(x) x = cos(x)

'Sample a vertical offset and send it somewhere that way horizontally' but we want the vertical offset of the point that landed here. We can guess where that point comes from by going backwards by the horizontal offset of our current position.

y = sin(x-cos(x))

This isn't perfectly accurate, but we can keep doing this to make it more accurate.
But if we 'guess' again based on the position we've arrived at we can continue to get closer.

y = sin(x-cos(x-cos(x-cos(x)) or y = sin(x-cos(x-cos(x-cos(x-cos(x-cos(x)))) and so on

By the fifth time we've gotten a pretty accurate guess and I think five steps is about the right amount of times. This solution doesn't work if you step back an even number of times.

Obviously, this isn't very performant because it's about 5 times more complex than sampling a single point. But it's accurate enough for buoyancy.

Proof of concept: https://www.desmos.com/calculator/28taueduzs
It didn't occur to me to step back multiple times until after I watched . this video at timestamp
It's also a good idea to swap from Gerstner Waves to FFT as sampling a texture multiple times over is probably a lot cheaper than calculating the offsets multiple times over. Though FFTs are more complicated concepts to wrap your head around.