问:
Let n be a positive integer. Find the number of positive integers whose base n representation has some other properties (USAMO 1990)
Let $n$ be a positive integer. Find the number of positive integers whose base $n$ representation consists of distinct digits with the property that except for the leftmost digit, every digit differs by $\pm1$ from some digit further to the left.

Through a lot of guesswork I managed to conclude that if $a_n$ denotes the number of suitable base $n$ representations, then $a_{n+1}=a_n+2^{n+1}-2$ and since $a_1=0$ then $a_{n+1}=2^{n+2}-4-2n$ which equates to $a_n=2^{n+1}-4-2(n-1)=2^{n+1}-2n-2$.

Could you please explain to me how I could have mathematically come up with the recursive relation $a_{n+1}=a_n+2^{n+1}-2$?

答:
Establishing the recurrence for base $n+1$ (which uses the digits from $0$ to $n$).

(Wishful thinking, will revisit if this doesn't work out) $a_n$ corresponds to those strings which do not use $n$.

It remains to count the number of strings have the digit $n$ in them. We do so by creating a (almost) bijection.

Given any string of length $k+1$, we can associate it to a new string of length $k$ of "Up" and "Down" as follows. Starting from the left, ignore the first digit,

If the $i+1$th digit is higher than all of the previous digits, call the new string "Up" in the $i$th position from the left.

If the $i+1$th digit is lower than all of the previous digits, call the new string "Down" in the $i$th position from the left.

Conversely, given any string of length $k$ of "Up" and "Down", we can associate it to a string of length $k+1$ that contains $n$.

Fill in $n, n-1, \ldots$ in the following order.

Focus on "Up", work from right to left.

Fill in the starting digit.

Focus on "Down", work from left to right.

This is almost a bijection, except in 1 specific instance (Which one? Why is this unique?)

This establishes that there are $1 + 2 + 2^2 + \ldots 2^n -1 = 2^{n+1} - 2$ strings with $n$ in them.

Thus $ a_{n+1} = a_n + 2^{n+1} - 2 $.

As an explicit example of the almost bijection, with $n = 2$,

$\{\} \leftrightarrow 2$

$ \{ \uparrow \}  \leftrightarrow 12$

$ \{ \downarrow \}  \leftrightarrow 21$

$ \{ \uparrow, \uparrow \}  \leftrightarrow 012$

$ \{ \uparrow, \downarrow \}  \leftrightarrow 120$

$ \{ \downarrow, \downarrow \}  \leftrightarrow 210$

$ \{ \downarrow, \uparrow \}  \leftrightarrow 102$

Which number is duplicated?


问:
Which values of $a$ and $b$ will give $x^{2}+2$ as a factor of $x^{17}+ax+b$ over $\mathbb{Z}_{3}$?
The question:

Let $x^{17}+ax+b$ be a polynomial over $\mathbb{Z}_{3}$. For which values of $a$ and $b$ will $x^{2}+2$ be a factor?

I know that I can find a solution by a brute force technique, but is there a more efficient way to solve this problem? Thanks.

答: If $x^2+2$ is a factor of that polynomial p, then every root of $x^2+2$ will be a root of p. $\mathbb{Z}_{3}$ only has 3 elements. Check which ones are roots of $x^2+2$. Then those same roots need to be roots of the polynomial. So $p(x_1)=0$ and $p(x_2)=0$. You get two equations with two unknowns, so you can solve for a and b.

ContentLicense: CC BY-SA 4.0


问:
Solving recurrence equation using generating functions
$$a_{0} = 0, a_{1} = 1, a_{2} = 2, a_{3} = 6$$
$$a_{n} = a_{n+3} - a_{n+2}$$

$\sum = a_{n}x^{n}$

$A(x) = \frac{A(X) - x - 2x^{2}}{x^{3}} - \frac{A(x) - x}{x^{2}}$

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

Is this the correct solution? The answer is different as far as I know, why is that so? Is it needed to replace the recurrence equation with minuses after n's? (well, all examples I've seen are with minuses there) And if so, how can I do this?

答:
If we make a blanket assumption that $a_n=0$ for all $n<0$, we can write the recurrence as

$$a_n=a_{n-1}+a_{n-3}+[n=1]+[n=2]+4[n=3]\;,\tag{1}$$

where the last three terms are Iverson brackets, and it will be correct for all $n\ge 0$.

If you’ve not seen this idea before, try evaluating $(1)$ for $n=0,1,2,3$ on the assumption mentioned at the beginning: $$\begin{align*}&a_0=a_{-1}+a_{-3}+0+0+4\cdot0=0\\&a_1=a_0+a_{-2}+1+0+4\cdot0=1\\&a_2=a_1+a_{-1}+0+1+4\cdot0=2\,\text{ and}\\&a_3=a_2+a_0+0+0+4\cdot1=6\;.\end{align*}$$

Multiplying through by $x^n$ and summing over $n\ge 0$, we get the generating function:

$$\begin{align*}
A(x)&=\sum_{n\ge 0}a_nx^n\\
&=\sum_{n\ge 0}\Big(a_{n-1}+a_{n-3}+[n=1]+[n=2]+4[n=3]\Big)x^n\\
&=\sum_{n\ge 0}a_{n-1}x^n+\sum_{n\ge 0}a_{n-3}x^n+x+x^2+4x^3\\
&=xA(x)+x^3A(x)+x+x^2+4x^3\;,
\end{align*}$$

so

$$A(x)=\frac{x+x^2+4x^3}{1-x-x^3}\;.$$

I find this method the easiest way to account for the initial values correctly.

ContentLicense: CC BY-SA 3.0


问:
How to tell when two advertising panels showing an ad will synchronize?
The problem is as follows:

A billboard turns on every $5$ minutes after turning off and displays
a colored ad for $45$ seconds and then turns off. Another similar
panel turns on every $8$ minutes after turning off and displays
another colored advertisement for $10$ seconds and then turns off. If
both panels are turned on at $\textrm{3:00 a.m.}$, what time will they
turn on simultaneously for third time?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{9:48 p.m}\\
2.&\textrm{9:46 p.m}\\
3.&\textrm{9:45 p.m}\\
4.&\textrm{9:47 p.m}\\
5.&\textrm{9:44 p.m}\\
\end{array}$

How exactly should this problem be assessed?. I do remember there's a formula which related the number of single events to avoid falling in telephone post error

$$\textrm{number of individual events}=\frac{\textrm{total elapsed time}}{\textrm{time elapsed between each event}}+1$$

This is assuming that the time between each event is the same and by rearranging the above equation into:

$$(\textrm{number of individual events}-1)t=\textrm{total elapsed time}$$

From the looks of this situation to find when both panels will synchronize again will require to use the least common multiple. Is this part correct?.

But from reading at the problem it makes me feel confused the part of turns on every $5$ minutes after turning off. I think what it is intended to say is the panel will turn on every 5 minutes and display a 45 second ad and so on and on.

While the other will do the same but instead for $8$ minutes while its ad will take $10$ seconds.

But the part which makes me feel confused is what to find the lcm to?

Would it be to $5$ and $10$? Can someone guide me in the right path for this? Since I feel lost in this question, it would really help me a lot that an answer would include a detailed explanation for this problem and step by step using the approach mentioned. Is this the right adequate approach?.

答:
The question is asking at what time will they simultaneously turn on for the third time. The key term to determine the inter-advertisement times is "after turning off". The first billboard turns on every 5 minutes and 45 seconds = 345 seconds. The second billboard turns on every 8 minutes and ten seconds = 490 seconds.

We need to find the least common multiple because we need to find when both billboards will be turned on. For example, if the first one turned on every 2 seconds and the second one every 3 seconds, we could enumerate the times they will turn on to find the common times.
0, 2, 4, 6, 8...
0, 3, 6, 9...
We can see that 6, the least common multiple of 2 and 3, is the first time they turn on at the same time.

Knowing this, we can apply a similar principle to the given problem and find the least common multiple between 345 and 490 which is 33,810. We divide by 60 to get the number of minutes between each simultaneous turn-on event, which is 536.5 minutes, or 9 hours and 23.5 minutes.

Knowing this, we simply add to get the third time. The first time occurred at 3 am.

3 am -> first time

3:00 am + 9:23.5 =
12:23.5 pm -> second time

12:23.5 pm + 9:23.5 =
9:47 pm -> third time

Therefore we can see the correct answer is #4, 9:47 pm.


问:
Prove the identity $\frac{1}{1+x^{1}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$
Show that for all non-negatives integers $n$, it is true that
$$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$$
For $x \ne 1$, $x \ne -1$
I tried to solved it by appliying geometric series but I got the following
after doing the $2f\left(x\right)-f\left(x\right)$ step, where $f\left(x\right)=\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}$

$$f\left(x\right)=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n}}}$$
Not sure if Fermat's theorem is needed in order to solve it

答:
A proof by induction is elementary and not difficult.  Let $$S_n(x) = \sum_{k=0}^n \frac{2^k}{1 + x^{2^k}}, \quad P_n(x) = \frac{1}{x-1} + \frac{2^{n+1}}{\color{red}{1 - x^{2^{n+1}}}}.$$  Note that you have a typographical error for $P_n(x)$; the correct expression is shown in red text.  The claim is that for all nonnegative integers $n$, we have $S_n = P_n$.  The proof of the inductive step is simply

$$\begin{align}
S_{n+1}(x) &= S_n(x) + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\
&= P_n(x) + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\
&= \frac{1}{x-1} + \frac{2^{n+1}}{1 - x^{2^{n+1}}} + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\
&= \frac{1}{x-1} + 2^{n+1} \left( \frac{(1 + x^{2^{n+1}}) + (1 - x^{2^{n+1}})}{(1 - x^{2^{n+1}})(1 + x^{2^{n+1}})} \right) \\
&= \frac{1}{x-1} + 2^{n+1} \left( \frac{2}{1 - x^{2^{n+2}}} \right) \\
&= P_{n+1}(x).
\end{align}$$

The way that this simplifies also suggests that a direct proof is possible via algebraic simplification of a telescoping sum, e.g., $$\frac{2^{k+1}}{1 - x^{2^{k+1}}} = \frac{2^k}{1 - x^{2^k}} + \frac{2^k}{1 + x^{2^k}}$$ suggests defining $$a_k (x) = \frac{2^k}{1 + x^{2^k}}, \quad b_k(x) = \frac{2^k}{1 - x^{2^k}} $$ hence $$a_k(x) = b_{k+1}(x) - b_k(x)$$ and $$S_n(x) = \sum_{k=0}^n a_k(x) = \sum_{k=0}^n b_{k+1}(x) - b_k(x) = b_{n+1}(x) - b_0(x) = P_n(x).$$


问:
$\int_{-1}^{+1} \sqrt{1-x^2 + g (1-x^4)} \: dx, \quad g > 0,$
I need to compute the following integral analytically

\begin{equation}
\int_{-1}^{+1} \sqrt{1-x^2 + g (1-x^4)} \: dx, \quad g > 0,
\end{equation}

Here $g$ is a real parameter. I have tried to compute it in Mathematica but it fails to give a result. So I am wondering if the integral can be computed analytically. Thanks.

答:
$$1-x^2+g(1-x^4)=-gx^4-x^2+(g+1)=-g\left[x^4+\frac{x^2}g-\frac{(g+1)}g\right]$$
$$x^4+\frac{x^2}{g}-\left(1+\frac1g\right)=\left(x^2+\frac{1}{2g}\right)^2-\left(1+\frac1g+\frac{1}{4g^2}\right)$$
now if for ease we say: $$\alpha=-g,\beta=\frac{1}{2g},\gamma=1+\frac1g+\frac1{4g^2}$$
now we have the integral in the form:
$$I=\sqrt{\alpha}\int_{-1}^1\sqrt{(x^2+\beta)^2-\gamma}\,dx$$
and now maybe try a trig substitution here


问:
$d\mid p\!-\!1\Rightarrow x^d-1\pmod{\!p^2}$ has exactly $d$ roots for odd prime $p$
I'm trying to figure out the number of solutions to the congruence equation $x^d \equiv1 \pmod{p^2}$ where $p$ is prime and $d\mid p-1$.

For the congruence equation ${x^d}\equiv1 \pmod p$ where $p$ is prime and $d\mid p-1$ I've shown that there are exactly $d$ solutions modulo $p$.

I'm trying to use the above result to extend it to the higher power. I'm aware of something called Hensel's Lemma which says if a polynomial has a simple root modulo a prime  $p$, then this root corresponds to a unique root of the same equation modulo any higher power of $p$. We can 'lift' the root iteratively up to higher powers.

I'm unsure exactly how this process works. All I have to start with is that I assume a solution of the form $m+np$ solves the 'base' congruence and I'm trying to somehow extend it to $p^2$.

Any help would be appreciated.
Thanks.

答: Set $f(x)=x^d-1$.  Each solution $m$ mod $p^k$ lifts to a unique solution mod $p^{k+1}$ exactly when $f'(m)\neq 0\pmod{p}$.  Fortunately, $f'(m)=dm^{d-1}$.  Since $d\mid p-1$, $p\nmid d$.  Since $m^d\equiv 1\pmod{p}$, $p\nmid m$.  Hence each solution mod $p$ lifts to a unique solution mod $p^2, p^3, \ldots$.  There are plenty of resources to read more about Hensel's lemma, such as Wikipedia, or here on mse.</p>


问:
Probability of choosing a cell in a $ N\times M$ grid with specific conditions
I have this problem :
Given a $N\times M$ grid.
You cover $2$ cells by a $1\times2$ tile.
If you pick a cell,what is the probability for it to be covered by the tile?

It's not hard to evaluate the number of the tile possible positions, which is $N(M-1)$.
But how could I find the answer ?

答:
Let :

T the random variable in {0,1} that determines if we choose a tile (C=1).

I is a random variable in $\{(i,j)/ (i,j) \in [1,N] \times [1,M-1]\}$ that determines the Initial position of the left tile in the board.

$$p(T=1) = \sum_{i=1, j=1}^{i=N, j=M-1} p(T=1 | I = (i,j))p(I=(i,j))$$

With :

Given the possible initial positions : $p(I=(i,j)) = \frac{1}{N(M-1)}$

Because we have 2 tiles covered, then : $p(T=1 | I = (i,j)) = \frac{2}{N \cdot M}$

So, $p(T=1) = \frac{2}{N \cdot M}$


问:
Solve cubic equation $x^3 - 15x - 4 = 0$ with Cardano's method
Suppose that we wanted to solve
$$x^3 - 15x - 4 = 0$$
by using Cardano's method. In doing so, I end up with
$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$
However, by checking my answer, it seems there are two other answers I haven't accounted for: $\sqrt{3} - 2$ and $-2 - \sqrt{3}$. Is there a way to generate these solutions from this method, or would I need to find them in other ways, e.g., by factoring?

答:
The Cardano’s formula is normally used to calculate the real root of the a cubic equation with one real root and a pair of complex roots. In the case of three real roots, e.g. $x^3 - 15x - 4 = 0$, the formula results in the expression

$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$

which is of multiple complex values. However, it can still be used to obtain the three real roots if their real parts are taken, i.e.

$$x=Re(\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i})= 2Re (\sqrt[3]{2 + 11i})\tag1 $$
Write the cubic radicant as
$$2+11i=\sqrt{125}e^{i \cos^{-1}\frac2{5\sqrt5}}
= \sqrt{125}e^{i \theta}$$
where
$$\cos\theta = \frac2{5\sqrt5}=4\cos^3\frac{\theta}3-3\cos\frac{\theta}3
$$
which factorizes as
$$\left(\cos\frac{\theta}3- \frac2{\sqrt5}\right)\left(\cos\frac{\theta}3+\frac{\sqrt3+2}{\sqrt5}\right)\left(\cos\frac{\theta}3- \frac{\sqrt3-2}{\sqrt5}\right)=0
$$
Then
$$\cos\frac{\theta}3=\frac2{\sqrt5},\>\frac{\sqrt3-2}{2\sqrt5},\>
\frac{-\sqrt3-2}{2\sqrt5}
$$
and the corresponding roots (1) are
\begin{align}
x&
=2Re \left( \sqrt{125}e^{i \theta}\right)^{\frac13} = 2\sqrt5 \cos\frac{\theta}3=4,\>\sqrt3-2,\>-\sqrt3-2
\end{align}


问:
Is the finitely additive integral linear in probability measures?
Let $(X, \mathcal X, \mu)$ be a finitely additive probability space. That is, $\mu$ is assumed to be finitely, but not necessarily, countably additive.

We can define integration for non-negative extended-real functions with respect to $\mu$ in the usual way. If $f: X \to [0, \infty]$ is an $\mathcal X$ measurable step function of the form
$$f = \sum_{i=1}^n \alpha_i 1_{A_i},$$
where $\alpha_i \in \mathbb R$ and $A_i \in \mathcal X$, let
$$\int f d\mu = \sum_{i=1}^n \alpha_i \mu(A_i).$$
Then for any non-negative extended-real $\mathcal X$-measurable $f$, let
$$\int f d\mu = \sup\Big\{ \int g d \mu: 0 \leq g \leq f, g \ \text{is a step function} \Big \}.$$
The resulting integral is linear in the sense that
$$\int \alpha(f+g) d\mu = \alpha \int f d\mu + \alpha \int g d\mu$$
for all non-negative extended-real $\mathcal X$-measurable $f$ and $g$, and all $\alpha \in \mathbb R^+$.

I'm wondering whether the dual form of linearity holds as well. That is: Is it true that for all non-negative extended-real $\mathcal X$-measurable $f$, all $\alpha \in [0,1]$, and all finitely additive probability measures $\mu_1$ and $mu_2$
$$\int f d(\alpha \mu_1 + (1 - \alpha) \mu_2) = \alpha \int f d\mu_1 + (1-\alpha) \int f d\mu_2?\tag{1}$$

I can show that (1) is true for all bounded $f$, and for general $f$, I can show that
$$\int f d(\alpha \mu_1 + (1 - \alpha) \mu_2) \leq \alpha \int f d\mu_1 + (1-\alpha) \int f d\mu_2.$$
So all I need is the reverse inequality for general $f$.

答:
Sorry. I can't really see what is your problem.

I assume $\mu_1$ and $\mu_2$ are positive.Otherwise, your finitely additive integral would behave weird.

Let $g,h$ be two positive step functions. So $u=max(g,h)$ is also a positive step function. Furthermore, if $g,h \le f$ then $u$ is also not bigger than $f$.

Hence,

$\alpha \int g d\mu_1+ (1-\alpha) \int h d\mu_2 \le \alpha \int u d\mu_1+ (1-\alpha) \int u d\mu_2 = \int u d( \alpha \mu_1+(1-\alpha)\mu_2)\le \int fd( \alpha \mu_1+(1-\alpha)\mu_2) $

So $\alpha \int f d\mu_1+ (1-\alpha) \int f d\mu_2 \le \int fd( \alpha \mu_1+(1-\alpha)\mu_2) $

On the other hand,

For all step function $0 \le g \le f$

$\int g d( \alpha \mu_1+(1-\alpha)\mu_2) = \alpha \int g d\mu_1+ (1-\alpha) \int g d\mu_2 \le  \alpha \int f d\mu_1+ (1-\alpha) \int f d\mu_2 $

So, $\alpha \int f d\mu_1+ (1-\alpha) \int f d\mu_2 \ge \int fd( \alpha \mu_1+(1-\alpha)\mu_2) $

Hence, conclusion.

Am I wrong somewhere?


问:
Logic proving that $A$ is a subset of $B$
Let $A = \{x \in \mathbb{Z} \mid x \equiv 5 \pmod{15}\}$ and $B = \{ x \in \mathbb{Z} \mid x \equiv 2 \pmod{3}\}.$

Give an outline of a proof that $A \subseteq B,$ being as detailed as possible. Then prove the statement.

To create a proof of this would I be correct in starting off by:

Let $A = \{ x \in \mathbb{Z} \mid x \equiv 5 \pmod{15}\},$ and let $B = \{x \in \mathbb{Z} \mid x \equiv 2 \pmod{3}\}.$

Let $x \in \mathbb{Z}$ be arbitrary,

I am a bit stuck on where to go from here though in this outline.

答:
Outline: Note that you want to prove that given two subsets $A$ and $B$ we have that $A \subseteq B.$ By definition, this means that, given an object $x$

$$x \in A \implies x \in B.$$

So this is our way: we pick an element from $A$ and then we show that it have to belong to $B$ as well.

Proof: Let $x \in A.$ By definition of $A,$ we know that

$$x \equiv 5 \pmod{15}.$$

This means that

$$x = 15k + 5 \quad \text{for some } k \in \mathbb{Z}.$$

But we also have that

$$\begin{align}
x & = 15k + 5\\
& = 3(5k) + 3 + 2\\
& = 3(5k + 1) + 2
\end{align}$$

Since $k \in \mathbb{Z}$ then $5k + 1 \in \mathbb{Z}.$ Then we conclude that $x \equiv 2 \pmod{3}.$ Hence $x \in B.$ Since $x \in A \implies x \in B$ we have that $A \subseteq B.$ $\square$


问:
Does Viviani's theorem hold on higher dimensions?
Viviani's theorem claims that given a point inside an equilateral triangle, the sum of the distances from the point to the sides of the triangle is constant an equal to the height of the triangle. Also, it can be proven that the only 2D shape that satisfies the property that the sum the three distances is constant is an equilateral triangle.

Generalisations to 3D have been made, and for a regular tetrahedron, if an inner point is chosen, the sum of the distances from that point to the sides of the tetrahedron remains constant. In this case, nevertheless, there are more 3D polyhedrons that satisfy this property, as long as the faces have equal area.

And here it is my question: does Viviani's theorem hold for n-simplex in dimensions higher than 3? My intuition says that this is the case, but I cannot find any proof of such claim on the web.

Many thanks in advance!

答:
The "volume" (area, volume, hypervolume, etc) of an $n$-simplex in $\mathbb{R}^n$ is given by $$\tfrac1n\cdot(\text{distance from a vertex to opposite facet})\cdot|\text{facet}|$$ (where "$|\;\;|$" indicates "volume", and a "facet" is an ($n-1$)-dimensional side of the simplex). Joining an interior point $P$ to the vertices determines $n+1$ subsimplices such that $$|\text{simplex}|=\sum_k |\text{subsimplex $k$}|=\sum_k\tfrac1n\cdot(\text{distance from $P$ to facet $k$})\cdot|\text{facet $k$}|$$ (We can allow $P$ to be exterior if we use signed distances to the facets.)

If the facets have equal volume, then we can divide-through by that volume (and $1/n$) to get that the sum of the distances from $P$ to the facets is a constant (namely, the "height" from a vertex), thus generalizing Viviani.


答:
I also find that Mathematica does not evaluate the integral, even when informed of the condition on the radicand.

```<code>Assuming[{g >= -1/(x^2 + 1)},
    Integrate[Sqrt[1 - x^2 + g (1 - x^4)], {x, -1, 1}]
]
</code>```

$$  \int_{-1}^1 \; \sqrt{1-x^2 + g \cdot (1-x^4)} \,\mathrm{d}x  $$

However, the fundamental theorem of calculus allows us to use any antiderivative to compute the definite integral.  Since we've established Mathematica as a suitable method of integration,

```<code>Assuming[{g >= -1/(x^2 + 1)},
    Integrate[Sqrt[1 - x^2 + g (1 - x^4)], x]
]
</code>```

$$  \frac{\sqrt{-g x^4+g-x^2+1} \left(g x \left(x^2-1\right) \left(g
   x^2+g+1\right)+\sqrt{-(g+1) \left(x^2-1\right) \left(g x^2+g+1\right)}
   \left(E\left(\sin ^{-1}(x)|\frac{1}{g+1}-1\right)-(2 g+1) F\left(\sin
   ^{-1}(x)|\frac{1}{g+1}-1\right)\right)\right)}{3 g \left(x^2-1\right) \left(g
   x^2+g+1\right)}  \text{,}  $$
where $E(\varphi|k^2)$ is the incomplete elliptic integral of the second kind and $F(\varphi|k^2)$ is the incomplete elliptic integral of the first kind.  If we try to directly evaluate this at $-1$ or $1$, the result is undefined, so as usual for an improper integral, we take limits.

```<code>Simplify[(Limit[#, x -> 1, Direction -> "FromBelow"]
    - Limit[#, x -> -1, Direction -> "FromAbove"])] & [(
    Sqrt[1 + g - x^2 - x^4] (
        g x (-1 + x^2) (1 + g + g x^2) +
        Sqrt[-(1 + g) (-1 + x^2) (1 + g + g x^2)] (EllipticE[
      ArcSin[x], -1 + 1/(1 + g)] - (1 + 2 g) EllipticF[
       ArcSin[x], -1 + 1/(1 + g)]))
    )/(3 g (-1 + x^2) (1 + g + g x^2))
]
</code>```

$$  \frac{2 \sqrt{2 g^2+3 g+1} \left((2 g+1)
   K\left(-\frac{g}{g+1}\right)-E\left(-\frac{g}{g+1}\right)\right)}{3 g \sqrt{2
   g+1}}  \text{,}  $$
where $K(k)$ is the complete elliptic integral of the first kind and $E(k)$ is the complete elliptic integral of the second kind.

This is real-valued if $g \geq -1/2$.

Title: $\int_{-1}^{+1} \sqrt{1-x^2 + g (1-x^4)} \: dx, \quad g > 0,$

Body:
I need to compute the following integral analytically

\begin{equation}
\int_{-1}^{+1} \sqrt{1-x^2 + g (1-x^4)} \: dx, \quad g > 0,
\end{equation}

Here $g$ is a real parameter. I have tried to compute it in Mathematica but it fails to give a result. So I am wondering if the integral can be computed analytically. Thanks.

Body:
I also find that Mathematica does not evaluate the integral, even when informed of the condition on the radicand.

Assuming[{g >= -1/(x^2 + 1)},
    Integrate[Sqrt[1 - x^2 + g (1 - x^4)], {x, -1, 1}]
]

$$  \int_{-1}^1 \; \sqrt{1-x^2 + g \cdot (1-x^4)} \,\mathrm{d}x  $$

However, the fundamental theorem of calculus allows us to use any antiderivative to compute the definite integral.  Since we've established Mathematica as a suitable method of integration,

Assuming[{g >= -1/(x^2 + 1)},
    Integrate[Sqrt[1 - x^2 + g (1 - x^4)], x]
]

$$  \frac{\sqrt{-g x^4+g-x^2+1} \left(g x \left(x^2-1\right) \left(g
   x^2+g+1\right)+\sqrt{-(g+1) \left(x^2-1\right) \left(g x^2+g+1\right)}
   \left(E\left(\sin ^{-1}(x)|\frac{1}{g+1}-1\right)-(2 g+1) F\left(\sin
   ^{-1}(x)|\frac{1}{g+1}-1\right)\right)\right)}{3 g \left(x^2-1\right) \left(g
   x^2+g+1\right)}  \text{,}  $$
where $E(\varphi|k^2)$ is the incomplete elliptic integral of the second kind and $F(\varphi|k^2)$ is the incomplete elliptic integral of the first kind.  If we try to directly evaluate this at $-1$ or $1$, the result is undefined, so as usual for an improper integral, we take limits.

Simplify[(Limit[#, x -> 1, Direction -> "FromBelow"]
    - Limit[#, x -> -1, Direction -> "FromAbove"])] & [(
    Sqrt[1 + g - x^2 - x^4] (
        g x (-1 + x^2) (1 + g + g x^2) +
        Sqrt[-(1 + g) (-1 + x^2) (1 + g + g x^2)] (EllipticE[
      ArcSin[x], -1 + 1/(1 + g)] - (1 + 2 g) EllipticF[
       ArcSin[x], -1 + 1/(1 + g)]))
    )/(3 g (-1 + x^2) (1 + g + g x^2))
]

$$  \frac{2 \sqrt{2 g^2+3 g+1} \left((2 g+1)
   K\left(-\frac{g}{g+1}\right)-E\left(-\frac{g}{g+1}\right)\right)}{3 g \sqrt{2
   g+1}}  \text{,}  $$
where $K(k)$ is the complete elliptic integral of the first kind and $E(k)$ is the complete elliptic integral of the second kind.

This is real-valued if $g \geq -1/2$.


问:
How to prove $(\frac{n+1}{e})^n<n!<e(\frac{n+1}{e})^{n+1}$ without integrating method?
How to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that
$$i<x<i+1\Rightarrow \ln i<\ln x<\ln (i+1),$$ then integrating it.</p>

答:
The left hand inequality is immediate from the power series for $e^x$. Plugging in $x = n+1$ gives
$$e^{n+1} = \sum_{m=0}^{\infty} {(n+1)^m \over m!} > {(n+1)^{n+1} \over (n+1)!}$$
$$= {(n+1)^n \over n!}$$
This rearranges into the left inequality.</p>


问:
Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$
So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$

So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$

答:
The sequence is

$$1=1 \  , \ 3+5 = 8 \ , \ 7+9+11 = 27 \ , \ \dots$$

So you can see that

$$a_n = \sum_{k=1}^{1+2+...+n} (2k-1) - \sum_{k=1}^{n-1} a_k =$$
$$\sum_{k=0}^{n(n+1)/2} (2k-1) - a_{n-1} =\frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} a_k$$

because you can view the $n$-th term as the sum of $1+2+\dots+n$ odd numbers minus the sum of all the other previous terms ($27 = 1+3+5+7+9+11 - (1)-(3+5)$).

So the base step is clear, now for the inductive one we use strong induction (we say it works from $1$ to $n-1$) then

$$a_{n} = \frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} k^3 = \frac{1}{4}n^2(n+1)^2 -\frac{1}{4}(n-1)^2n^2 = n^3$$

so the formula is proven.


问:
Functions of functions in calculus
I have just finished working through a chapter on calculus and have coped OK, answering most of the questions correctly. The chapter dealt with composite functions, along with other topics.

But I just cannot understand this question. I can see we are dealing with functions of functions but they are not set out in a form I can understand. Any help would be appreciated.

In each of the following cases, express $f'$, the derivative of $f$ (with respect to $x$) in terms of $g'$. The number $a$ is constant:

$\begin{align} i) \quad & f(x) = g(x + g(a)) \\
ii) \quad & f(x) = g(a + g(x)) \\
iii) \quad & f(x) = g(x^2)
\end{align}$

答:
It is the use of the chain rule. For example

$ii) \quad f(x)=g(x+g(x))$

$$\frac{df}{dx} = \frac{d\big(g(x+g(x))\big)}{d\big(x+g(x)\big)} \frac{d\big(x+g(x)\big)}{dx} = g’(x + g(a))(1 + g’(x))$$

(the big parenthesis are only there to avoid confusion)

Title: Functions of functions in calculus


问:
If not associative, then what?
Consider a binary operation $*$ acting from a set $X$ to itself. It's useful and standard to work with operations which are associative, such that $(a*b)*c = a*(b*c)$. What about operations which are not associative?

Is there any way to characterize all different possible types of such binary operations $*$ which are not associative? Eg. Can we say that if $*$ is not associative,  then it must instead satisfy one of set of other possible properties, depending on any other additional operations that we have on our set $X$?

If we also add some additional structure to our set $X$ so that we can add elements together and multiply by scalars, it's standard to quantify the amount that two elements of $X$ commute with each other under $*$ by calculating the commutator $[a,b] = a*b - b*a$.
Is it ever useful to consider an 'associative commutator' $[abc] = (a*b)*c - a*(b*c)$, for a given non-associative $*$?

Finally, I know from Lie algebras that if $*$ anticommutes then it can be natural to consider a Jacobi identity

$(a*b)*c = a*(b*c) - b*(a*c)$

Are there other natural extensions of associativity in different settings?
Why do Lie algebras use this Jacobi identity and not for example

$(a*b)*c = a*(b*c) + k b*(a*c)$

Where k is a scalar?

答:
Is there any way to characterize all different possible types of operation which are not associative?

This is too broad and subjective to answer I think. What exactly is a "type" of operation? I assume you're already talking about binary operations, so presumably a "type" of operation is one which satisfies certain identities, like the associative identity. Certain specific examples come to mind:

Jacobi identity for Lie algberas,

Jordan identity for Jordan algebras,

Moufang identites for loops,

Self-distributive laws for racks and quandles,

and certainly others (I am no an expert in nonassociative algebra). Many of the identities above are not three-variable identities, but still. Generally, interesting algebras and their identities are not chosen randomly but rather follow from certain canonical examples whose properties are generalized. The algebras are meant to represent certain structures, and the identities ensure that. For instance, Lie algberas linearize Lie groups, and similarly Jordan algebras linearize projective spaces, Moufang identities generalize octonions' alternativity, racks and quandles represent how groups act on themselves by conjugation, etc.

Ultimately there is a "type" of operation for every possible set of "words" you can pick from the free magma (or if you allow addition, free nonassociative algebra) on so many generators. (There is going to be redundancy in this - different sets of words can yield the same class of algebras.)

Can we say that if ∗ is not associative, then it must instead satisfy one of set of other possible properties, depending on any other additional operations that we have on our set $X$?

Probably not. For instance the free nonassociative algebra on some generating set strikes me as a candidate for not having any "properties" (i.e. identities).

Is it ever useful to consider an 'associative commutator' for a given non-associative ∗?

Yes. The associator is useful for instance in (efficiently) proving the octonions are an alternative algebra (which is like halfway to being associative), which is in turn useful for many things like simplifying octonion expressions and classifying subalgebras and reasoning about automorphisms of $\mathbb{O}$. The octonion associator also gives rise to the exceptional ternary 8D cross product.

There's probably a lot more you can do with it in general nonassociative algebras but I wouldn't know.

Why do Lie algebras use this Jacobi identity

Consider where Lie algebras come from. Start with a Lie group $G$. The tangent space $\mathfrak{g}$ tells you all the directions that one-parameter subgroups can point in. The addition operation on $\mathfrak{g}$ corresponds to the group operation on $G$. Indeed, the exponential $\exp:\mathfrak{g}\to G$ is approximately linear in a neighborhood of $0$ with quadratic error term. As $G$ acts on itself by conjugation (and there are many sources listing example after example to show conjugation in a group is very important), so too it acts on $\mathfrak{g}$ by conjugation. Define $\mathrm{Ad}_A(X)=AYA^{-1}$ for $A\in G,Y\in\mathfrak{g}$. If we differentiate this at $A=I$ with tangent vector $X$ we get $\mathrm{ad}_X(Y)=XY-YX=[X,Y]$, the "commutator bracket." Note the adjoint action preserves this operation, and if we differentiate $\mathrm{Ad}_A[Y,Z]=[\mathrm{Ad}_AY,\mathrm{Ad}_AZ]$ at $A=I$ again with the product rule we get the identity $\mathrm{ad}_X[Y,Z]=[\mathrm{ad}_XY,Z]+[Y,\mathrm{ad}_XZ]$, which says $\mathrm{ad}_X$ is a "derivation" (i.e. satisfies the "product rule" like a derivative, but with the commutator bracket instead of multiplication). This identity may be rearranged to the more cyclically symmetric form you know as the Jordan identity.

All of the other identities I listed above have similar stories of where they come from. The Jordan identity comes from an algebraic investigation of spaces of Hermitian matrices (which are the span of projection operators, which correspond to points in projective spaces). Apparently the Jordan identity also has an interpretation in terms of the inversion symmetry of a Riemannian symmetric space, but I don't know how that story goes. The Moufang identity comes from investigating real normed division algebras, which leads to the octonions, which leads to the alternative identities, and then the simplest four-term identities one can check are where one term is repeated. The self-distributive law for racks and quandles comes from the fact conjugation is an automorphism in a group.

Title: If not associative, then what?


问:
How to calculate $\cos$(number) or $\sin$(number) without use of a scientific calculator?
Can somebody please help me how to calculate numerical value for $f(x)=-18\cos (3x)$ where $x=5$ so $f(5)=-18\cos(15)$ I am allowed to use only simple calculator and I don't know how to find the numerical value for this function. How much is $\cos(15)$ and how can I find how much cosine (of a number)is without using calculator with special functions?

I also don't know what to do with $\sin $(number) example $f(x)=-6\sin (3x)$ where $x=5 - f(5)=-6\sin (15)$
How much is $\sin (15)$ and how can I find how much is sine (of a number) without using calculator with special functions?

Thank you so much for your help!

答:
For $\alpha = 15°$ you can do it without calculator.

But it's not so easy for any $\alpha$, i.e. it's not always possible

to get a nice solution (say in radicals) with paper and pencil only.

First of all $$\sin(30°) = 1/2$$ this is well-known to you, I assume.

Then probably you also know that

$$\cos(30°) = \frac{\sqrt{3}}{2}$$

But then using the double-angle formula gives us

$$\frac{\sqrt{3}}{2} = \cos(30°) = 2 \cos^2{15°} - 1 $$

$$\frac{\sqrt{3}}{2} = \cos(30°) = 1 - 2 \sin^2{15°}$$

Solving the last 2 equations

(while keeping in mind that $\sin{15°} \gt 0$ and $\cos{15°} \gt 0$ )

gives us the exact values for $\sin{15°}$ and $\cos{15°}$

$$\cos(15°) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$$
$$\sin(15°) = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$$


答:
The integral is even in $x$.  So in Mathematica:

$$2 \int\limits_{x=0}^1 \sqrt{1 - x^2 + g (1 - x^4)}\ dx = \frac{2 \sqrt{g+1} \left((2 g+1)
   K\left(\frac{1}{g+1}-1\right)-E\left(\frac{1}{g+1}-1\right)\right)}{3 g}$$


答:
First, I think it will help to point out that these are not functions of functions. A composition of functions isn't a function of functions,$^*$ it's just an ordinary function which you might think about as being "built in two (or more) steps."

OK, now on to the problem itself. Consider a simpler example. Suppose $$f(x)=g(2x).$$Both $f$ and $g$ are ordinary functions; maybe $f(x)=4x^2$ and $g(x)=2x$. We don't know enough about $f$ to compute $f'(x)$, but we do know enough about how $f$ relates to $g$ to compute a relationship between $f'$ and $g'$, via the chain rule:

First, we explicitly write $f$ as a composition of two functions: the outer one is just $g$ itself, and the inner one is $h:x\mapsto 2x$. So $f=g\circ h$.

That's much more confusing to write than just "$f(x)=g(2x)$," but it makes things clearer for applying the chain rule to get $$f'(x)=g'(h(x))\cdot h'(x)=g'(2x)\cdot 2=2g'(2x).$$ So we've shown $f'(x)=2g'(2x)$; again, this doesn't amount to a computation of $f'$ itself, but it does tell us how to compute $f'$ in terms of $g'$.

Now your cases are a bit more complicated, but they're using the same principle. That first bulletpoint is going to be crucial, so focusing on the first problem for simplicity:

Suppose $f(x)=g(x+g(a))$. What's a function $h$ such that $$f=g\circ h$$ (if you can find this $h$ you'll then be able to just apply the chain rule directly)?

$^*$That said, there is a genuine "function-of-functions" floating around here - namely, composition itself! Composition (of functions $\mathbb{R}\rightarrow\mathbb{R}$, for simplicity) is the operation which takes a pair of functions $h_1,h_2$ and spits out the function $h_1\circ h_2$. So composition is a binary function-of-functions.

Title: Functions of functions in calculus

Body:
I have just finished working through a chapter on calculus and have coped OK, answering most of the questions correctly. The chapter dealt with composite functions, along with other topics.

But I just cannot understand this question. I can see we are dealing with functions of functions but they are not set out in a form I can understand. Any help would be appreciated.

In each of the following cases, express $f'$, the derivative of $f$ (with respect to $x$) in terms of $g'$. The number $a$ is constant:

$\begin{align} i) \quad & f(x) = g(x + g(a)) \\
ii) \quad & f(x) = g(a + g(x)) \\
iii) \quad & f(x) = g(x^2)
\end{align}$


问:
partial derivatives of order $n$
Be $f : \Bbb R^m \rightarrow R$  a function with continuous partial derivatives up to order $n$. How many derivatives of order $n$ does the function have?

I proved it for $n=1$
and I got that the number of derivatives was $m$

then I proved it for $n=2$
(second-order partial derivatives)
and I got $m^2$

but I'm stuck, because for proving by induction, I need a hypothesis right?

to get $m^n$?

答:
Supposing  the order of differentiation is not taken into account, this is a purely combinatorial problem: it is the number of $m$ objects taken $n$ at a time with repetitions:
$$\Gamma_m^n=\binom{m+n-1}n.$$

Title: partial derivatives of order $n$


问:
Which values of $a$ and $b$ will give $x^{2}+2$ as a factor of $x^{17}+ax+b$ over $\mathbb{Z}_{3}$?
The question:

Let $x^{17}+ax+b$ be a polynomial over $\mathbb{Z}_{3}$. For which values of $a$ and $b$ will $x^{2}+2$ be a factor?

I know that I can find a solution by a brute force technique, but is there a more efficient way to solve this problem? Thanks.

Body:
If $x^2 + 2$ is a factor of

$p(x) = x^{17} + ax + b, \tag 0$

any zero of $x^2 + 2$ must also satisfy $p(x)$;. the zeroes of $x^2 + 2$ in $\Bbb Z_3$ are $1$ and $2$:

$1^2 + 2 = 3 \equiv 0 \mod 3, \tag 1$

$2^2+ 2 = 4 + 2 = 6 \equiv 0 \mod 3; \tag 2$

indeed

$(x - 1)(x - 2) = x^2 - 3x + 2 = x^2 + 2 \mod 3; \tag 3$

if $1$ and $2$ are roots of $p(x)$, then

$1 + a + b = 1^{17} + a \cdot 1 + b = p(1) = 0 \mod 3, \tag 4$

$2^{17} + a \cdot 2 + b = p(2) = 0 \mod 3; \tag 5$

now,

$2^{17} = 2(2^4)^4 = 2(16)^4 = 2(1)^4 = 2 \mod 3; \tag 6$

thus (5) becomes

$2 + 2a + b = p(1) = 0; \mod 3; \tag 7$

thus (4) and (7) form the system

$a + b = -1 = 2 \mod 3, \tag 8$

$2a + b = -2 = 1 \mod 3; \tag 9$

we subtract (8) from (9):

$a = -1 = 2 \mod 3, \tag{10}$

and then substitute this into (8):

$b + 2 = 2 \mod 3, \tag{11}$

whence

$b = 0 \mod 3; \tag{12}$

therefore

$p(x) = x^{17} + 2x. \tag{13}$

Note that

$p(1) = 1 + 2 \cdot 1 = 3 = 0 \mod 3 \checkmark \tag{14}$

and

$p(2) = 2^{17} + 2 \cdot 2 = 2 + 4 = 6 = 0 \mod 3 \checkmark  \tag{15}$


问:
prove that the unit circle $x^2+y^2=1$ is a closed set
I need to prove that the unit circle $x^2+y^2=1$ is a closed set in $\mathbb{R}^2$ is closed using convergent sequence method.</p>

答:
Another way is to show that the complement is open. The complement of the your set is
$$S = \{(x,y) \in \mathbb{R}^2: x^2 + y^2 \neq 1\}$$
Consider $(x,y) \in S$. For this $(x,y)$, the $\epsilon$-ball centered at $(x,y)$, where $\epsilon = \left\vert \sqrt{x^2+y^2} - 1 \right\vert$, lies completely in $S$. Hence, $S$ is open.

EDIT:
Here is a pictorial description of finding the open sets. The blue colored unit circle if your set $x^2+y^2 = 1$. The remaining white space is the complement of the unit circle. You want to show that this complement is open.

Now pick a point, say $(0.25,0.8)$. This is a point "inside" the blue unit circle circle. For this point, you can find a open ball of radius $1-\sqrt{0.25^2 + 0.8^2}$, which lies completely "inside" the blue unit circle and therefore lies completely in the complement of the unit circle.

Similarly, pick a point, say $(1.125,0.6)$. This is a point "outside" the blue unit circle circle. For this point, you can find a open ball of radius $\sqrt{1.125^2 + 0.6^2}-1$, which lies completely "outside" the blue unit circle and therefore lies completely in the complement of the unit circle.

In general, you can do this for any point not lying on the unit circle. This shows that the complement of the unit circle is open.


问:
A question on a $6 \times 6 $ table(Olympiad question)
There are some mines in some cells of a $6 \times 6$ table. Among all $2 \times 2 $ squares, exactly $n$ squares contain an odd number of mines and the rest contain an even number of mines. What is the possible values of $n$ ?

Here is my try to solve this question :
For convenience I did it for a $ 4 \times 4 $ table which contains nine, $2\times 2$ squares. In this case suppose the following coordinates are mines : $(1,1) , (1,3) , (3,1) , (3,3)$, in this case all $9$ squares contain a mine, now we could reduce their number one by one forexample at first we add a mine in the $(4,4)$ square, so the number of $2\times 2$ square in which there is the number of odd mines is $8$. In the next step we'll put a mine in the square with $(4,3)$ coordinate in this case two of squares contain the even number of mines, and seven squares contain the number of odd mines.

I think we could continue above solution for solving this question in general, but I'm a little confused. I hope you can help me.

答:
We make a much stronger claim.

On a $ (k+1) \times (k+1)$ table, we can fill it up with mines such that any chosen subset of $2 \times 2$ squares contain an odd number of mines, and the rest contain an even number of mines. Hence, all $k^2 $  possibilities (unordered arrangement of squares) are possible.

In addition, all $2^{k^2}$ possibilities (ordered arrangements of squares) are possible.

In addition, there are $ 2^{2k+1}$ ways of doing so for each selection of odd/even squares.

Proof:

Let the $ 2 \times 2 $ squares be indexed by the lower left square. IE The $(1,2)$  $2 \times 2$ square comprises of $(1,2), (1,3), (2,2), (2,3)$.

Fill in the $2k+1$ squares of the form $(i, k+1)$ and $(k+1, i)$ (where $i$ is a variable) with mines or not. This gives us $ 2^{2k+1}$ ways. We will show that way will yield chosen subsets of odd/even mines as desired.

Consider the $(k, k)$ $2 \times 2 $ square. This uniquely determines the $(k, k)$ square (based on what we chose for $(k, k+1), (k+1, k), (k+1, n+1)$.

Consider the $(k-1, k)$ $2 \times 2$ square. This uniquely determines the $(k-1, k)$ squares (based on what we have for $(k, k), (k-1, k+1), (k, n+1)$.

Continue by inducting on $(k-i, k)$ to fill out the top row.

Likewise, we can then induct on $(k-i, k-1)$ to fill out the next row, then induct on $(k-i, k-2)$ to fill out the next row, all the way till we induct on $(k-i, 1)$ to fill out the bottom most row.


问:
MAE(mean absolute error)
For objects $x_1,..., x_n$ with correct answers $y_1,...,y_n$ from R, construct a constant model $a(x)=c$ for the loss function.
$$MAE=\frac{1}{N}\sum_{i=1}^{n}|y_i-c|$$

As I understand, I need to take a derivativе of $c$ and find minimum, but the equation of derivative doesn't have solutions.
$$(MAE)'= \frac{1}{N}\sum_{i=1}^{n}\frac{-y_i+c}{|y_i-c|}$$
but I did this task by using inequality $$\sum_{i=1}^{n}|y_i-c|\geq|\sum_{i=1}^{n}y_i-n  \cdot c|$$ and I get the answer $$c=\frac{\sum_{i=1}^{n}y_i}{n}$$

Does it correct?
Tanks for the help

答:
Relabel your $y_i$ in increasing order: $y_1 \le y_2 \le ... \le y_n$. When $c < y_1$, $$MAE = \frac 1n \sum_{i=1}^n y_i - c$$ and $MAE' = -n$. That means that the $MAE$ is going down as $c$ is increasing.

For $y_k < c < y_{k+1}$, $$MAE = \sum_{i=1}^k c-y_i + \sum_{i=k+1}^n y_i - c$$ and $MAE' = k - (n-k)= 2k - n$. When $2k < n$, this is still negative, meaning that increasing $c$ still decreases $MAE$. But when $n > 2k$, the derivative is positive, at which point the $MAE$ is increasing as $c$ increases.

So where do you think the minimum will be?

Title: MAE(mean absolute error)


问:
Finding expectation of minimum of $(X,Y)$ where $(X,Y)$ is bivariate normal distribution.
Let $(X,Z)$ be bivariate normal with parameters
$\mu_X := E(X) = 1, \mu_Z := E(Z ) = 1, \sigma_X^2 := Var(X) = 1$,

$ \sigma_Z^2
:= Var(Z ) = 1$,
and the correlation coefficient of (X, Z ) is $\rho$ with density

$$f(x,z)=\dfrac{1}{2\pi\sqrt{1-\rho^2}}\exp(-\dfrac{1}{2(1-\rho^2)}((x-1)^2-2\rho (x-1)(z-1)+(z-1)^2)).$$

Let $Y=\min(X,Z)$. Then what is $E[Y]$

For the above what I have so far is

$\textbf{MY Attempt}$

we first transform the variables so that $X=1+X_1$ and $Z=1+X_2$. Then
\begin{align*}
\mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 - \mathbb{P}\left\{ X_1 > y, X_2 > y \right\} \\
&= 1 - \int_{y}^{\infty} \int_y^{\infty} f_{X_1,X_2}(s,t)dsdt \\
&= 1 -\int_y^{\infty} f_{X_2}(t)\int_y^{\infty} f_{X_1|X_2}(s,t) ds dt
\end{align*}
where
$$
f_{X_2} (t)=\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}, \qquad f_{X_1|X_2}(s,t)=\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(s-\rho t)^2}{2(1-\rho^2)}},
$$
and so
\begin{align*}
\mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 - \int_{y}^{\infty} \varphi(t) \left(1 - \Phi\left( \frac{y -\rho t}{\sqrt{1-\rho^2}}\right) \right) dt. \\
&= 1 - \int_{y}^{\infty} \varphi(t) \Phi\left( \frac{\rho t - y }{\sqrt{1-\rho^2}} \right) dt
\end{align*}
To get the density we differentiate with respect to $y$ giving
\begin{align*}
f_{Y}(y) &= -\frac{\partial}{\partial y}\int_{y}^{\infty}\varphi(t)\Phi\left(\frac{\rho t - y}{\sqrt{1-\rho^2}}\right) dt \\
&= \varphi(y)\Phi\left(\frac{\rho y - y}{\sqrt{1-\rho^2}}\right) + \int_{y}^{\infty}\varphi(t) \frac{1}{\sqrt{1-\rho^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\rho t - y)^2}{2(1-\rho^2)}}dt \tag{1}
\end{align*}
Completing the square of the last term in $(1)$ we have
\begin{align*}
\varphi(t)\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(\rho t-y)^2}{2(1-\rho^2)}} &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left((1-\rho^2)t^2 +(\rho t -y)^2  \right)} \\
&=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left(t^2 - 2 t \rho y + \rho^2 y^2 + (1-\rho^2)y^2  \right)} \\
&= \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}.
\end{align*}
So putting this back in to $(1)$ we get
\begin{align*}
f_{Y}(y) &= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\int_y^{\infty} \frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}dt \\
&= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) +  \varphi(y)\left( 1 - \Phi\left(\frac{y - \rho y}{\sqrt{1-\rho^2}}\right)\right) \\
&= 2\varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right).
\end{align*}

$\textbf{Hence I think I have the density of random variable Y. From here its difficult for me to find the expectation of Y.}$

答:
You can use the following
$$\min(X,Z) = \frac{(X+Z)}{2}-\frac{|X-Z|}{2} $$

the distribution of $W =(X -Z)$ is normal with mean $0$ and variance $2(1-\rho)$
Now this is a random variable and we can take out the distribution of $|W|$ as
$$\mathbb{P}(|W|\leq t) = \mathbb{P}(W\leq t)-\mathbb{P}( W \leq -t)=F_W(t) - F_W(-t)$$
Hence we will have $$f_{|W|}(t) = f_{W}(t)+f_{W}(-t) = 2f_{W}(t) $$
there we can get the $\mathbb{E}(|W|) = \frac{2}{\sigma\sqrt{2\pi}}\displaystyle\int_0^{\infty}z\exp{ \frac{-z^2}{2\sigma^2} }dz = \sigma\sqrt{\frac{2}{\pi}}$

Hence we will have the answer using $\mathbb{E}\Big[\min(X,Z)\Big] = .5\mathbb{E}\Big[X\Big]+.5\mathbb{E}\Big[Z\Big]-.5\mathbb{E}\Big[|W|\Big]$.


答:
For a general function $f$ (that has $n$-th order partial derivatives), you may have up to $m^n$ many different $n$-th order partial derivatives. The logic that you used to make this claim is that you can take the derivative with respect to any variable. Then do it again. And again, until you get to the $n$th derivative. This is all correct in general.

However, if these $n$-th order partial derivatives are continuous, then many of these derivatives are in fact the same. By Clairaut's Theorem, $f_{xy}=f_{yx}$ for the continuous function $f(x,y)$. Similarly, $f_{xxxyy}=f_{xyxyx}$, etc.

So for the question that you asked, you have to handle this idea of repetition. As Bernard mentioned, you are effectively asking how many ways are there to choose $n$ objects out of the $m$ choices, where repetition is allowed. This is indeed given by the formula $$\binom{n+m-1}{n}.$$

Title: partial derivatives of order $n$

Body:
Be $f : \Bbb R^m \rightarrow R$  a function with continuous partial derivatives up to order $n$. How many derivatives of order $n$ does the function have?

I proved it for $n=1$
and I got that the number of derivatives was $m$

then I proved it for $n=2$
(second-order partial derivatives)
and I got $m^2$

but I'm stuck, because for proving by induction, I need a hypothesis right?

to get $m^n$?


问:
Showing that, if $\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$, then $\sqrt2\sin\beta=\sin\alpha-\cos\alpha$

If
$$\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$$
then prove that
$$\sqrt2\sin\beta=\sin\alpha-\cos\alpha$$

I have been trying to solve this exercise but I don't get it. I need help. Thanks.

答:
Writing $\alpha=x,\beta=y,$

As $(\sin x-\cos x)^2+(\sin x+\cos x)^2=2$

Using the given condition,
$2=(\sin x-\cos x)^2(1+\cot^2y)$

$\implies(\sin x-\cos x)^2=2\sin^2y$

As we have taken square, we have introduced
When do we get extraneous roots?

For example, $$a=b\implies a^2=b^2\implies a=\pm b$$

But actually $a\ne-b$


问:
Proof of divisibility with combinatorics
Prove that, for every positive integer $\,k, {2k\choose k}\,$ is divisible by $k+1$. Hint: $\frac{2k+1}{k+1}{2k\choose k}$

What I have done so far

${2k\choose k} = 2k!/k!k!=((k+k)(k+k+1)(k+k+2)\cdots k!)/k!k!= ((k+1)(2k!/(k+1))/k!k!$

Is this proof correct or is there a better way to prove it?

答:
It is the special case $\,n=2k\,$ below, by $\,\color{#c00}{\gcd(k\!+\!1,2k\!+\!1)= 1},\,$ by Euclid.

Lemma $\ \ $ Both $\ \  k\!+\!1\mid {n\choose k}\ \ $ and $\ \ n\!+\!1\mid {n+1\choose k+1}\  $ when $ \ \color{#c00}{\gcd(k\!+\!1,n\!+\!1)=1}$

$\begin{align} {\bf Proof}\qquad  \dfrac{1}{k\!+\!1}\:\! \dfrac{n!}{k!(n\!-\!k)!} &\,=\, \dfrac{1}{n\!+\!1}\:\!\dfrac{(n\!+\!1)!}{(k\!+\!1)!(n\!-\!k)!}\\[.6em]
\Rightarrow\ \ \ q\, :=\, \dfrac{1}{\color{#c00}{k\!+\!1}}\ \ \ {n \choose k}\ \ \ =\!\!\!  &\qquad\!\dfrac{1}{\color{#c00}{n\!+\!1}}\,{n\!+\!1\choose k\!+\!1}
\end{align}$

Therefore  $\,q\,$ is an integer, being a fraction writable with $\rm\color{#c00}{coprime}$ denominators.

See here for the straightforward generalization to the non-$\rm\color{#c00}{coprime}$ case.

Remark $ $ Conceptually, the order of $\,q\,$ (= least denominator) divides coprimes so it must be $1,\,$ where the order is interpreted in the group $\,\Bbb Q/\Bbb Z = $ rationals $\!\bmod 1.\,$ Equivalently, said more simply: it is well known since grade school that the least denominator of a fraction divides every other denominator (though this is usually not proved till much later, e.g. by Euclid's Lemma or unique prime factorization); thus since the least denom divides coprimes it must be $1$.  The prior link gives proof(s) of this property of fractions, and elaborates on the various conceptual viewpoints emphasized above.


问:
Countable intersection of open dense subsets of $\mathbb{R}$
This particular question was asked in my analysis assignment and I was unable to solve it. Kindly help.

Let A be a subset of $\mathbb{R}$ satisfying $A=\bigcap_{n\geq 1} V_n$, where for each n $\geq1$, $V_n$ is an open dense subset of $\mathbb{R}$ . Which 1 of following is correct?

A.  A is non -empty

B A is countable

C A is uncountable

D A is dense in $\mathbb{R}$.

I marked B which is wrong . A is correct and I marked it.For C,D i don't know how to attempt them. SO, Kindly tell me how to reason b,c,d,

Waiting for your help.

答:
A space in which the intersection of countably many dense open sets is dense is called a Baire space, and the tool that you want is the Baire category theorem, specifically, the version that says that every complete metric space is a Baire space: $\Bbb R$ is a complete metric space, so it is a Baire space.

This immediately tells you that $A$ is dense in $\Bbb R$ and hence that (D) is true. Of course the empty set is not dense in $\Bbb R$, so (A) is also true. To dispose of (B) and (C), suppose that $A$ is countable, say $A=\{a_n:n\in\Bbb Z^+\}$. For each $n\in\Bbb Z^+$ let $U_n=\Bbb R\setminus\{a_n\}$. Let

$$\mathscr{W}=\{U_n:n\in\Bbb Z^+\}\cup\{V_n:n\in\Bbb Z^+\}\,;$$

each $U_n$ is dense and open in $\Bbb R$, so $\mathscr{W}$ is a countable family of dense open subsets of $\Bbb R$, and it’s easy to check that $\bigcap\mathscr{W}=\varnothing$, contradicting the Baire category theorem. Thus, (B) must be false, and (C) must therefore be true.


问:
Why isn't $\sum_{n=1}^{\infty}(-1)^n$ a telescoping series?
I read that the reason $\sum_{n=0}^\infty(-1)^n$ does not converge is because the terms alternate between -1 and 0. So I realize that this sentence does answer my question, but what I'm having difficulty understanding is that why wouldn't it collapse to 0 at infinity in the same way that another alternating series, even if it doesn't collapse to 0 before infinity, does?

答:
Consider the sequence $(a_n)_n$ defined as $a_n = (-1)^n.$

Now you can form the sequence of the partial sums, $(s_n)_n$, as

$$s_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n.$$

By definition, $\sum_{k=1}^{\infty} a_n = \lim s_n.$

Let’s study the convergence of $(s_n)_{n}.$ Consider this two subsequences of $(s_{n})_{n}.$

$$s_{2n} = (-1)^{2n} = 1 \quad \quad s_{2n-1} = (-1)^{2n-1} = -1$$

As you can easily check

$$\lim s_{2n} = 1 \neq -1 = \lim s_{2n-1}.$$

So there are two subsequences of $s_n$ that converge to different values. Therefore $s_n$ does not converge.

If a sequence converges, then every subsequence converges for the same limit.

Check this for more information on this result

Prove: If a sequence converges, then every subsequence converges to the same limit.

Therefore $\lim s_n$ does not exist. Hence $\sum_{k=1}^{\infty}$ diverges.

Note that in this case the cancellation of the terms in $s_n$ wouldn’t work because the result would depend on what $n$ you are deciding. That’s why this is not a telescoping series, you can’t express it as a certain amount of terms after cancellation and still have the same result


答:
Avoid squaring whenever possible as it immediately introduces
When do we get extraneous roots?

$$\tan\beta=\dfrac{1-\tan\alpha}{1+\tan\alpha}=\tan(\pi/4-\alpha)$$

$$\beta=n\pi+\pi/4-\alpha$$ where $n$ is any integer

$$\implies\sin\beta=\sin(n\pi+\pi/4-\alpha)$$

Now as $\sin(n\pi+x)=(-1)^n\sin x,$

$$\sin\beta=(-1)^n\sin(\pi/4-\alpha)=?$$

Title: Showing that, if $\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$, then $\sqrt2\sin\beta=\sin\alpha-\cos\alpha$

Body:
If
$$\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$$
then prove that
$$\sqrt2\sin\beta=\sin\alpha-\cos\alpha$$

I have been trying to solve this exercise but I don't get it. I need help. Thanks.


问:
Integration of a continuously increasing function and its inverse
If $f$ is a function that is continuous strictly increasing and differentiable on $[0, a]$ with $f(0)=0$ and g is the inverse of f, show that for any $x \in [0,a],  \int_{0}^{x} f(t) \,dt +  \int_{0}^{f(x)} g(t) \,dt  - xf(x) = 0$.

Looking at some graphs, I know this is true because the first integral gives the area under the curve and the second gives the area above the curve but below $y=f(x)$ which when added together equals $xf(x)$ the area of the rectangle, but I'm having a hard time turning this intuition into a proof. I'm pretty much stuck after "pick an arbitrary $x \in [0, a].$"

答: Verify that the derivative of the left side is $0$ and that it vanishes when $x=0$. This implies that it is identically $0$.


问:
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$</p>

答:
I got $\quad \displaystyle 8\sqrt{\pi}\,e^{-2}$

Now let's search a proof...

I'll use the classical 'duplication formula' for $\Gamma$ :
$$\Gamma(z)\Gamma\left(z+\frac 12\right)=\sqrt{2\pi}\ 2^{1/2-2z}\,\Gamma(2z) $$
so that we have :
$$\frac{\Gamma(2^n)\Gamma\left(2^n+\frac 12\right)}{2^{2^{n+1}}\Gamma\left(2^{n+1}\right)}=2\sqrt{\pi}\ $$
and I thought at using 'kind of multiplicative telescoping' but it seems that considering telescoping of the logarithms as nicely done by O.L. is less confusing!


问:
Is it true that $\mathbb{E}|X-\mathbb{E}(X)|=0$ where $\mathbb{E}(X) $ is finite?
I'm not sure

whether this is always true? and

If it is true, whether it is a trivial assertion to make?

I've tried integrating using the definition of expectation, but I'm not getting anywhere with it.

答: It is true if and only if $X$ is a constant random  variable. Reason: $|X-\mathbb E X|$  is  non-negative random variable and its expectation can be $0$ only when it is zero almost surely.


问:
For $f(x) = a x^2 + bx + c$, find $ d +e(i) $ that will generate a real value

This is the challenge question, I don't understand how because $i$ must be squared in order for it to be $-1$

答:
Hint $1$:

you just have to make the sum of the terms with $i$ involved $0$, the only condition will be $d=-\frac{b}{2a}$ when $e \neq 0 $ otherwise it always happens. (assuming $a,b,c,d,e \in \mathbb{R}$)

Hint $2$ :

$$f(d+ei) = a(d+ei)^2+b(d+ei)+c = ad^2-ae^2+2daei +bd+bei+c = $$
$$ad^2-ae^2+bd+c +(2dae+be)i$$


问:
inequality with binomial coefficients
I'd like to show that if $s \le n/4$, then we have
$$ {n \choose s-1} + {n \choose s-2}+ \dots {n \choose 0} \le {n \choose s}.$$
I'm happy with any short proof of this, but a "combinatorial proof" would be especially nice.

Comment: this is a special case of something listed as an "easy exercise" in Chapter 5 of Babai and Frankl's book. (See Proposition 5.13): http://people.cs.uchicago.edu/~laci/CLASS/HANDOUTS-COMB/BaFrNew.pdf

答:
Prove Proposition 5.13 in the book by induction.

For $ n\geq 2s$, we have
$$ { n \choose s } + {n \choose s-1 } + \ldots + { n \choose 0 } < { n \choose s } \times ( 1 + \frac{s}{n-2s+1} ). $$

It is indeed an easy exercise (when you know this is what you have to prove).

Then, follow the book for the next 2 lines:

If $ s \leq n/ l $ then the right hand side is less than $ { n \choose s} \cdot  ( 1 + \frac{1}{ l - 2 } )$. For example, with $s \leq n/4$, we obtain the inequality

$$ { n \choose s } + {n \choose s-1 } + \ldots + { n \choose 0 } < 2 \cdot { n \choose s } . $$

I'm not quite sure why he didn't pick $l = 3$ though. That is almost certainly a typo.

Title: inequality with binomial coefficients


问:
If $S^1 \hookrightarrow X$ induces an injection of $H_1$, then $X$ retracts onto $S^1$
This is an exercise in Hatcher, section 4.3, exercise 3, page 419, on which I'm struggling.

Suppose that a CW complex X contains a subcomplex $S^1$ such that the inclusion
  $S^1 \hookrightarrow X$ induces an injection $H_1(S^1;\mathbb Z) \to H_1(X; \mathbb Z)$ with image a direct summand of $H_1(X;\mathbb Z)$. Show that $S^1$ is a retract of $X$.

I've been trying to use obstruction theory to solve this, but I'm stuck. Showing that $S^1$ is a retract is equivalent to showing there is a map $r : X \to S^1$ extending the identity $id : S^1 \to S^1$. By obstruction theory ($S^1$ is an abelian space, we can assume $X$ is connected), we need to consider groups of the form $H^n(X, S^1; \pi_{n-1}(S^1))$. The only nonzero such group is $H^2(X, S^1)$ (with integer coefficients).

Let $H_1(X) = H_1(S^1) \oplus M$. The long exact sequence in homology gives an exact sequence:

$0 \to H_2(X) \to H_2(X, S^1) \to H_1(S^1) \to H_1(X) \to H_1(X, S^1) \to 0$

Thanks to the hypothesis on $H_1$, we therefore have that $H_2(X) \simeq H_2(X, S^1)$ and $H_1(X, S^1) \simeq M$. By universal coefficients:

$$\begin{align}
H^2(X, S^1) & \simeq \mathrm{Hom}(H_2(X,S^1), \mathbb Z) \oplus \mathrm{Ext}(H_1(X,S^1), \mathbb Z) \\
& \simeq \mathrm{Hom}(H_2(X), \mathbb Z) \oplus \mathrm{Ext}(M, \mathbb Z) \oplus \mathrm{Ext}(\mathbb Z, \mathbb Z)\\
& \simeq \mathrm{Hom}(H_2(X), \mathbb Z) \oplus \mathrm{Ext}(H_1(X), \mathbb Z) \\
& \simeq H^2(X)
\end{align}$$

And this is where I'm stuck. There's an obstruction $\omega_2(id) \in H^2(X, S^1; \pi_1(S^1)) \simeq H^2(X)$, and I'd like to show that it's zero, but I don't know how. Could you give me some insight?

答:
EDIT: Sorry, I forgot the initial step of the argument.  Your calculation shows that the map $H^2(X,S^1;\pi_1 S^1) \to H^2(X;\pi_1 S^1)$ is an isomorphism.  This means that the obstruction cocycle for constructing a map extending the one on $S^1$ is zero precisely when the obstruction cocycle for an unconstrained map, and it suffices to show that the latter is always zero.

Roughly, this particular obstruction is always zero, because this obstruction cocycle only depends on an existing choice of map from the $0$-skeleton.

More in detail.  Obstruction cocycles answer particular questions.  Given a map from the $s$-skeleton of $X$ to $Y$, we have such an obstruction cocycle measuring whether it's possible (given a current choice of map on the $(s-1)$-skeleton) to make some choice of maps on the $s$-cells to get a map which can be extended to the $(s+1)$-skeleton.  This is the obstruction that lives in $H^{s+1}(X, \pi_s Y)$, roughly because it's asking about whether the attaching maps for the $(s+1)$-cells can be made nullhomotopic in $Y$.

In your case, $s=1$.  You can take whatever map $X^{(0)} \to S^1$ that you are given and lift it up to a map $X^{(0)} \to \mathbb{R}$, and all such maps automatically have an extension to a map $X^{(2)} \to \mathbb{R} \to S^1$.

This pattern continues for Eilenberg-Mac Lane spaces: the obstruction cocycle for a map $X \to K(G,n)$ depends only on a map from the $(n-1)$-skeleton, and all the maps from the $(n-1)$-skeleton to a $K(G,n)$ are nullhomotopic.  The obstruction cocycle is then the same as that for a constant map.  You only tend to get interesting obstructions when you've had the opportunity to make interesting choices on the lower skeleta.


问:
Limit of the form 0 $\cdot$ inf
I have been trying for hours to solve the following limit without using De L'Hospital theorem, but I got stuck. Can you help me? Thanks in advance! (The solution should be $-\sqrt2$)

$\lim_{x\rightarrow\pi/4}(2\sin(x)-\sqrt2)\tan(x-3\pi/4)$

答:
$$2\sin x-\sqrt2=2\left(\sin x-\sin\dfrac\pi4\right)$$

Use https://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$\tan(x-3\pi/4)=\tan(x+\pi/4-\pi)=\cdots=\dfrac{\sin(x+\pi/4)}{\cos(x+\pi/4)}$$

Now $\cos(x+\pi/4)=\sin(\pi/4-x)=-2\sin(x-\pi/4)$

Finally use $\sin2y=2\sin y\cos y$

As $x\to\pi/4,x\ne\pi/4,$

$\sin\dfrac{4x-\pi}8\ne0$ so can be cancelled out safely

Title: Limit of the form 0 $\cdot$ inf


答: Here's a diagrammatic verification for acute $\alpha$ and $\beta$:


问:
Find the generating function for $ a_n = n^4$ .
Find the generating function for $a_n = n^4$.

How should I approach this question? I have tried to go $1,16,81\dots$ and generate the function with the series but no idea how to do so.

Body:
\begin{align}
\sum_{n \ge 0} n^4 z^n
&=\sum_{n \ge 0} \left(\binom{n}{1}+14\binom{n}{2}+36\binom{n}{3}+24\binom{n}{4}\right) z^n \\
&=\sum_{n \ge 1} \binom{n}{1} z^n + 14\sum_{n \ge 2}\binom{n}{2} z^n + 36\sum_{n \ge 3}\binom{n}{3} z^n + 24\sum_{n \ge 4}\binom{n}{4} z^n \\
&=z \sum_{n \ge 0} \binom{n+1}{1} z^n + 14z^2\sum_{n \ge 0}\binom{n+2}{2} z^n + 36z^3\sum_{n \ge 0}\binom{n+3}{3} z^n + 24z^4\sum_{n \ge 0}\binom{n+4}{4} z^n \\
&=\frac{z}{(1-z)^2} + \frac{14z^2}{(1-z)^3} + \frac{36z^3}{(1-z)^4} + \frac{24z^4}{(1-z)^5} \\
&= \frac{z + 11 z^2 + 11 z^3 + z^4}{(1-z)^5}
\end{align}


问:
How to prove that a real solution to this algebraic equation exists ? (roots of a given multivariate polynomial)
The equation is

$$(y_0z_1-y_1^2) \sum\limits_{i=0}^2 c_i x_i=x_0x_1z_1+y_0\left(\sum\limits_{i=0}^2 c_i y_i\right)\left(c_0y_1+\sum\limits_{i=1}^2 c_i z_i\right) -y_1x_0\left(c_0y_1+\sum\limits_{i=1}^2 c_i z_i\right) -y_1x_2\left(\sum\limits_{i=0}^2 c_i y_i\right)$$

The unknown are $c_0,c_1,c_2$, the other variables are fixed on the real line under the following assumptions: $y_0\geq 0, z_1 \geq 0, y_0z_1-y_1^2 \neq 0$.

Is there a real solution to this equation under those general assumptions ? It looks like the equation corresponds to finding the real root(s) of a polynomial of degree $2$ in $c_0,c_1,c_2$. Is this possible ? I only need to know that a real solution exists, not necessarily compute it.

The challenge is hidden in the square roots we take to arrive at a solution: for example if we fix $c_0=0=c_1$, we get a quadratic equation for $c_2$, but then nothing guarantees that $c_2$ is real.

If there are no real solutions under those assumptions, can you find what extra (minimal) assumptions should the variables satisfy to get a real solution ?

答:
Under the given assumption, the equation has no real solutions iff $$y_0=x_0z_1-x_1 y_1+x_2 y_1=x_0z_2- x_2 y_1+x_2y_2=0\mbox{ and }x_0x_1z_1\ne 0.$$

Let us show this. I did routine calculations with Mathcad, so we skip them.

Impose first a restriction $c_2=0$. Then the initial equation becomes a quadratic equation with respect to $c_1$ whose discriminant is a quadratic polynomial with respect to $c_0$ with the highest coefficient $y_0^2(y_0z_1-y_1^2)^2$. So if $y_0\ne 0$ then for sufficiently big $c_0$ the discriminant is non-negative and so the initial equation has a real solution.

If $y_0=0$ then $y_1\ne 0$ and the initial equation reduces to a linear equation

$$(x_0z_1-x_1 y_1+x_2 y_1)c_1+(x_0z_2- x_2 y_1+x_2y_2)c_2= x_0x_1z_1/y_1.$$


答: A telescoping series (sum) is usually defined as a finite series in which consecutive terms cancel, so that you are only left with the initial and final terms. The series you're considering is an infinite series. In general, you can't treat infinite series like finite series.

Title: Why isn't $\sum_{n=1}^{\infty}(-1)^n$ a telescoping series?

Body: I read that the reason $\sum_{n=0}^\infty(-1)^n$ does not converge is because the terms alternate between -1 and 0. So I realize that this sentence does answer my question, but what I'm having difficulty understanding is that why wouldn't it collapse to 0 at infinity in the same way that another alternating series, even if it doesn't collapse to 0 before infinity, does?


问:
Sketching a set of $Re(z^2(1+i))$ on the Complex plane.
So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.

I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies  Re(z^2) = a^2 - b^2$$

Similarly:

$$Re(z^2(1+i)) \implies  (a^2 + 2abi - b^2)(1+i) =  a^2 + 2abi - b^2 + a^2i - 2ab - b^2i \implies Re(z^2(1+i)) =  a^2 - 2abi + b^2 = (a-b)^2 $$

So how am I supposed to sketch a set of complex numbers $(a-b)^2 $ on a 2D plane? Or did I do something wrong?

答:
When dealing with complex number, polar form is almost always more useful than breaking it into real and imaginary part.

$$
Re((1+i)z^{2})<0\implies \frac{\pi}{2}<\arg{((1+i)z^{2})}<\frac{3\pi}{2}
$$

Now we use the following
$$
\begin{align}
\arg{((1+i)z^{2})}&=\arg{(1+i)}+ 2\arg{(z)}\\
&=\frac{\pi}{4}+2\arg{(z)}
\end{align}
$$
Substitute this to the first equation to obtain
$$
\frac{\pi}{8}<\arg{(z)}<\frac{5\pi}{8}
$$


答:
Here's a short direct proof of your inequality for $ s \leq n/3$.

Lemma: For $ s \leq n/3$, $ {n \choose s } = \frac{n-s+1}{s} { n \choose s-1}  > 2 { n \choose s-1} $.

Then, it follows that

$$\begin{array} {r l l } { n \choose s } & > 2 { n \choose s-1} &  = { n \choose s-1} + { n \choose s-1} \\
& > { n \choose s-1 } + 2 {n \choose s-2}  & = { n \choose s-1} + { n \choose s-2} + { n \choose s-2}\\
& > {n\choose s-1 } + {n \choose s-2} + 2 { n \choose s-3} \\
& > \ldots \\
& > \sum { n \choose s-i}. \end{array} $$

One way to motivate the idea is via the induction part of proposition 5.13.

Title: inequality with binomial coefficients

Body:
I'd like to show that if $s \le n/4$, then we have
$$ {n \choose s-1} + {n \choose s-2}+ \dots {n \choose 0} \le {n \choose s}.$$
I'm happy with any short proof of this, but a "combinatorial proof" would be especially nice.

Comment: this is a special case of something listed as an "easy exercise" in Chapter 5 of Babai and Frankl's book. (See Proposition 5.13): http://people.cs.uchicago.edu/~laci/CLASS/HANDOUTS-COMB/BaFrNew.pdf


问:
Rescaling a function to equalize impact of input variables
Suppose $f:\mathbb{R}^2\to\mathbb{R}$ is a smooth function. Suppose that disturbing one variables changes the function value much more than disturbing the other variable,
$$
|f(x+\varepsilon,y)-f(x,y)| \gg |f(x,y+\varepsilon)-f(x,y)| \,.
$$
This poor scaling property challenges computer solvers to find the roots of $f$, see here for Matlab. The matlab team suggests to rescale the function such that

''each coordinate has about the same effect on the objective function''.

Example (from Mathwork). A poorly scaled function is $f(x,y)= 10^6 x^2 + 10^{-6}y^2$. Instead, we can try to solve $g(x,y)=0$ where $g(x,y)= f\left(\begin{pmatrix} 10^{-3} & 0 \\ 0 & 10^3 \end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix}\right)=f(10^{-3}x,10^3y)$.

Question: Is there a more general approach to rescale a function? In Mathwork's example, $\begin{pmatrix} 10^{-3} & 0 \\ 0 & 10^3 \end{pmatrix}$ is pretty obvious but how could one come up with a suitable rescaling if the problem isn't that obvious? Is there a general hint on how I can equalize the effect of input variables on the function value?

While my question is a bit more general, it originates from my problem of finding the roots of
\begin{align*}
f(x,y) = \begin{pmatrix} \left(\frac{c_1c_3 }{c_1-c_2}\left(\frac{c_2 c_3}{(y-c_6)(c_1-c_2)}\right)^{-\frac{c_2}{c_1}} +c_6 \right) \frac{c_2}{c_1}x^{c_2-c_1}+\frac{x^{1-c_1}}{c_1c_5}-y \\
\left(\frac{c_1c_3 }{c_1-c_2}\left(\frac{c_2 c_3}{(y-c_6)(c_1-c_2)}\right)^{-\frac{c_2}{c_1}} +c_6\right) x^{c_2}+ c_5x - c_7-yx^{c_1}
\end{pmatrix},
\end{align*}
where all the $c_i$ are a bunch of constants. I asked Matlab to minimise $\|f\|$ but that turned out to be poorly scaled.

答:
Here's an ad-hoc idea for if you function is Lipschitz-smooth.

In a sense, the issue of scaling stems from the fact that you've got very different Lipschitz constants in each component of your function. In particular, I mean that if we look at the restriction of $f$ in its first component and guesstimate its Lipschitz constant $L_x$, that number may be very different than if we looked at the restriction of $f$ in its second component $L_y$. To alleviate this issue, if we rescale each input component by the inverse of the respective Lipschitz constants, we should get an improvement. In math, this means we rescale by $(x,y)\mapsto f(\frac{x}{L_x},\frac{y}{L_y})$.

Now, how do we get $L_x$ and $L_y$? The key fact in this scheme is that any $\gamma\in[\sup_{z\in\mathbb{R}}|g'(z)|,\infty)$ is a Lipschitz constant of $g$. Using this fact as inspiration, we can guesstimate choices for $L_x$ and $L_y$ in their respective ranges. Note that if our current point is at $(x_0,y_0)$, the "$g$" above will be either $y\mapsto f(x_0,y)$ for guesstimating $L_y$ or $x\mapsto f(x,y_0)$ for guesstimating $L_x$. The hope is that, at the very least, our estimate of $L_x$ and $L_y$ will work locally. While $\sup_{z\in\mathbb{R}}|f'(z)|$ might not always be computable,  a numerical approximation of the directional derivative is easy to compute (e.g. using a numerical differentiation scheme). To be safe and make sure we've really got an upper bound, we could scale up the magnitude of the numerical derivative by rescaling (e.g. $2\times$, if need be).

If you're actually going to implement this, then the idea above does dictate that you'll want to approximate $L_x$ and $L_y$ locally at every iteration. That said, you could probably recycle those gradient evaluations in your minimization method.


问:
Einstein notation • Cartesian reference frame • Cartan, theory of spinors
I’m having a problem understanding Cartan’s notation in theory of spinors §2. Here are the definitions:

Set x to $3 \\ 4$ so the length is 5, then set as a simple test basis $\eta_1$ to $1 \\ 1$ and $\eta_2$ to $0 \\ 1$.

$u^1$ I set to 3 and $u^2$ to 1. The scalar product of $\eta_1 . \eta_2$ is 1. If I run over the $u^i u^j$ successively and independently, that gives me $$3 \quad 3 \\ 3 \quad 1 \\ 1 \quad 3 \\ 1 \quad 1$$, which sum to 16, not 25.

If changing basis is supposed to change length, then I don’t understand why Cartan’s $$u^i u^j \eta_1 . \eta_2$$ is a good definition of fundamental form.

答:
Let $x = (3, 4) = 3e_1 + 4 e_2$, with square length $3^2 + 4^2 = 25$. Now express $x$ in the new basis of $\eta_1 = (1, 1)$ and $\eta_2 = (0, 1)$, and we get that
$$ x = (3, 4) = (3, 3) + (0, 1) = 3 \eta_1 + \eta_2.$$
Therefore in the expression $x = u^i\eta_i$ we have $u^1 = 3$ and $u^2 = 1$, as you found.

Now consider the fundamental form $g_{ij} = \eta_i \cdot \eta_j$, which we can write as a matrix
$$ g = \begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}.$$
Evaluating $\Phi$ we get
$$ \begin{aligned}
\Phi
  &= g_{ij} u^i u^j\\
  &= g_{11} u^1 u^1 + g_{12} u^1 u^2 + g_{21} u^2 u^1 + g_{22} u^2 u^2 \\
  &= 2 \cdot 3 \cdot 3 + 1 \cdot 3 \cdot 1 + 1 \cdot 1 \cdot 3 + 1 \cdot 1 \cdot 1 \\
  &= 25.
\end{aligned}
$$


答:
Some facts:

We require the real part to be less than zero, which means $(1 + i)z^2$ must be located on the left half of the complex plane.  In terms of the argument, this means $(1+i) z^2$ must have an argument between $\pi/2$ and $3\pi/2$, plus some integer multiple of $2\pi$.

Squaring a complex number doubles its argument.

Multiplying a complex number by $1+i$ increases its magnitude by $|1+i| = \sqrt{2}$ and rotates it by $\pi/4$; i.e., the argument is increased by $\pi/4$.

Therefore, $$\arg((1+i)z^2) = \frac{\pi}{4} + 2\arg (z)$$ and we need this to be between $\pi/2$ and $3\pi/2$, or between $5\pi/2$ and $7\pi/2$.  This gives us $$\frac{\pi}{8} < \arg (z) < \frac{5\pi}{8} \quad \text{or} \quad \frac{9\pi}{8} < \arg (z) < \frac{13\pi}{8}.$$  Using interval notation, we can also write this as $$\arg (z) \in (\tfrac{\pi}{8}, \tfrac{5\pi}{8}) \cup (\tfrac{9\pi}{8}, \tfrac{13\pi}{8}).$$

We can visualize this region in the complex plane by coloring quadrants I and III (i.e., those regions where $\Re(z)\Im(z) = 1$, and then rotating those regions by $\pi/8$ counterclockwise.

Title: Sketching a set of $Re(z^2(1+i))$ on the Complex plane.

Body:
So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.

I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies  Re(z^2) = a^2 - b^2$$

Similarly:

$$Re(z^2(1+i)) \implies  (a^2 + 2abi - b^2)(1+i) =  a^2 + 2abi - b^2 + a^2i - 2ab - b^2i \implies Re(z^2(1+i)) =  a^2 - 2abi + b^2 = (a-b)^2 $$

So how am I supposed to sketch a set of complex numbers $(a-b)^2 $ on a 2D plane? Or did I do something wrong?


问:
Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$
So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$

So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$

答:
Let us find the general term of $1,3,7,13,\cdots$

As the next consecutive differences are $2,4,6,\cdots$

$$T_n=an(n-1)+b(n-1)(n-2)+cn(n-2)$$

$n=1\implies1=c(1-2),c =-1$

$n=2\implies3=2a$

$n=3,7=6a+2b+3c=3(3)+2b+3(-1),2b=1$

$$\implies T_n=\cdots=n^2-n+1$$

We need  the sum of $n$ terms from $k=n^2-n+1$ to $(n+1)^2-(n+1)+1-1$ with the common difference $=2$

which is $$n(n^2-n+1)+\sum_{k=1}^n(2k)=n^3-n^2+n+\dfrac n2(2+(n-1)2)=?$$


答:
There are several ways to prove this. Here is an algebraic one.
Start with the difference of two integers (binomial coefficients)
and end up with the Catalan number quotient:

$$ {2n \choose n}  -  {2n \choose n+1} = \tag{1}$$
$$ \frac{ (2n)! }{n!\, n!} - \frac{(2n)!}{(n+1)!\,(n-1)!} = \tag{2}$$
$$ \frac{ (2n)!\,(n+1)}{n!\,(n+1)\,(n-1)!\,n} -
   \frac{ (2n)!\,n}{(n+1)!\,(n-1)!\,n} = \tag{3}$$
$$ \frac{ (2n)!}{ (n+1)!\,(n-1)!\,n} \,( (n+1)-n ) = \tag{4}$$
$$ \frac{ (2n)! }{ n!\,(n+1)\,n!}\,(1) = {2n \choose n}\frac1{n+1}. \tag{5}$$

Title: Proof of divisibility with combinatorics

Body:
Prove that, for every positive integer $\,k, {2k\choose k}\,$ is divisible by $k+1$. Hint: $\frac{2k+1}{k+1}{2k\choose k}$

What I have done so far

${2k\choose k} = 2k!/k!k!=((k+k)(k+k+1)(k+k+2)\cdots k!)/k!k!= ((k+1)(2k!/(k+1))/k!k!$

Is this proof correct or is there a better way to prove it?


问:
Find a closed curve where $F(x,y,z)=(y,x,x)$ closed line integral isn't 0.
Given the function $F(x,y,z)=(y,x,x)$, I have to show it isn't conservative and find a close curve where it isn't equal to 0.

It was easy to show that $D_1F_3\neq D_3F_1$, but I can't find an easy curve that isn't 0. I tried unit circle but it equals to zero.

Watching the plot I can't figure it out. In general, I don't know how to solve this when I'm in $\mathbb{R^3}.$

答:
Consider the circle $C : t \mapsto (\cos t, 0, \sin t)$.

You have $C^\prime(t)=(-\sin t, 0, \cos t)$ and

$$\int_C F.dr =\int_0^{2\pi} \cos^2 t \ dt >0$$


问:
Standardized residuals
I am having a hard time understanding the concept of standardizing residuals and how the variance of a residual is decomposed. In a linear model, we defined residuals as:

$e = y - \hat{y} = (I-H)y$ where H is the hat matrix $X(X^TX)^{-1}X^T$

and we defined standardized residuals as:

$r_i = \frac{e_i}{s\sqrt{1-h_{ii}}}$,  $i = 1,...,n$

where $s^2$ is the usual estimate of $\sigma^2$, $var(e_i) = \sigma^2h_{ii}$, and $h_{ii}$ is the diagonal entry of H at the $i^{th}$ row and $i^{th}$ column

However, I am not sure why $r_i$ and $e_i$ are functions of $h_{ii}$ rather than the whole row $h_i$. Basically I am confused about what $h_{ii}$ stands for as opposed to row $h_{i}$.

答:
Let's look at the variance of the residuals vector $e$,
$$
Var(e) = Var( (I-H)y ) = (I-H)^TVar(y) (I-H) = \sigma^2(I-H).
$$
The main diagonal of $\sigma^2(I -H)$ are the variance terms of $e$. Particularly, the variance of $e_i$ is the $i$th diagonal element of $\sigma^2(I-H)$, that is $\sigma^2(1 - h_{ii})$. Since you don't know $\sigma^2$ and estimate it with $s^2$, thus the estimated variance of $e_i$ is $s^2(1-h_{ii})$.

Title: Standardized residuals


问:
What do we know about $Z$ if $Z = U - V$ where $U$ and $V$ are indep. and both exponentially distributed
Assume now that Ali and Berta independently assign exponentially distributed scores, with rates $a$ and $b$, respectively. But they only record the difference between the two scores (Ali’s score minus Berta’s score, say.) What is the maximum likelihood estimator for $a$ and $b$, using only the score differences?

In a previous part I calculated the MLE of the RV $X$ that was the sum of their scores that was exponentially dist with parameter $\lambda$ as:

$\hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i}$

Note that in that part we only observed $X$ and nothing about their scores or score rates individually.

Can I just consider this same result for my new RVs $U$ and $V$ with their parameter's $a$ and $b$? How would you suggest approaching this new RV $Z$ that is their difference?

I see here Find the distribution of $Z=X+Y$ where both $X$ and $Y$ are exponentially distributed. that there might be something to the new variable having the gamma dist. Is this a correct path to take?

I'm just looking for a helpful push or maybe a resource that shows me the right direction.

Thanks for your time.

Edit: $U$, $V$ $\geq 0$ sorry forgot to add this before.

答:
I don't know if you can find anything interesting.

Based on my calculation, the density of $Z$ would be:

$ f(v) = \frac{ ab}{a+b} e^{ -a|v|}$ if $v \ge 0$

and $f(v)= \frac{ ab}{a+b} e^{ -b|v|}$ if $v \le 0$

So the standard procedure of finding MLE here seems to be not complicated. Though the final result might not be as explicit as when there is only one exponential variable.


问:
How can I prove this cosine equation?
How to prove that $\cos(90)\cos(\theta)+\sin(90)\sin(\theta)=\sin(\theta)$ ?</p>

答:
Note that:

$$\cos(90)\cos(\theta) + \sin(90)\sin(\theta) = \cos(90 - \theta)$$

Then, by the angle sum trig form, we have:

$$\cos(90 - \theta) = \sin(\theta)$$

To show why this is true, consider the usual graph $\cos(\theta)$.  Since it's the even function, $\cos(\theta) = \cos(-\theta)$.  Geometrically, the graphs of $\cos(-\theta)$ and $\cos(\theta)$ are congruent.  Now, if you shift the graph of the function by $90$ degrees or $\frac{\pi}{2}$, then you move the graph of $\cos(-\theta)$ that much units to the right.  Finally, you get the sine graph.

Compare both graphs here

OR

You can simplify $\cos(90)$ and $\sin(90)$ and show that equality holds.  You get the same results.

$$\cos(90)\cos(\theta) + \sin(90)\sin(\theta) = 0 \cdot \cos(\theta) + 1 \cdot \sin(\theta)$$
$$= \sin(\theta)$$

Title: How can I prove this cosine equation?


问:
Re-allocating cells in rows in columns to equal given row and column sums
Suppose I have a matrix which looks like:
$$
\left[\begin{array}{cccc}
a_{11} & a_{12} & ... & a_{1n}\\
a_{21}\\
.\\
a_{n1} &  &  & a_{nn}
\end{array}\right]
$$
where $a_{ij}$ denotes the value of $a$ in row $i$ and column $j.$
For each row $i,$ I have a given constraint on its sum. So, the sum of the first row should be:
$$
\sum_{j}x_{1j}=R_{1}
$$
Similarly, we have it for all other $n-1$ rows. Moreover, I have
another constraint, which for the first column is:
$$
\sum_{i}x_{i1}=C_{1}
$$
These also make up a total of $n$ column constraints. So, I have
$n$ row constraints and $n$ column constraints in this example.
As of now, there is no guarantee that these constraints are being
met. Is there any well known algorithm ( a least squares type?) to re-allocate cells in a manner which doesnt change original data much, but still meets the constraints? This could  take the form of, say:
$$
\text{min}\sum_{ij}\left(a_{ij}-\epsilon\right)^{2}
$$
s.t. the $2N$ row and column constraints? Is there a closed form
solution for this? Many thanks!

答:
This is called matrix balancing and can be solved via iterative proportional fitting.

See Schneider and Zenios, A Comparative Study of Algorithms for Matrix Balancing, Operations Research 38 (1990).

Here are two informative blog posts:

https://blogs.sas.com/content/iml/2020/09/08/matrix-balancing.html

https://blogs.sas.com/content/iml/2020/09/10/iterative-proportional-fitting-sas.html


问:
Let $f,g:[a,b]\rightarrow \mathbb{R}$ continuous and $f(x)<g(x)$ for all $x\in [a,b]$, then exists a $\delta >0$ s.t $f+\delta \leq g$
My attempt.

We know that $f(x)<g(x)$ for all $x\in [a,b]$. We define $h(x)=g(x)-f(x)$, then $h(x)>0$. By the archimedean property exists a $n_{x}$ such that $0<\frac{1}{n_{x}}<h(x)$. Then we choose $n=max\{n_{x}:x\in [a,b]\}$ and define $\delta=\frac{1}{n}$.

We conclude that $f+\delta\leq g$.

答:
$g-f$ is a continuous map on the compact $[a,b]$. It therefore attains its minimum on that interval at let say $c \in [a,b]$ according to the extreme value theorem. We have for all $x \in [a,b]$

$$g(x)-f(x) \ge g(c)-f(c) \gt0$$

as desired.


问:
Properties of the horocycles of a tree
I am willing to fill in the gaps I found in the introduction of the following two papers :

W. Woess, Lamplighters, Diestel-Leader graphs, random walks, and harmonic functions, (link).

A. Terras and D. Wallace, Selberg's trace formula on the k-regular tree and applications (link).

$\newcommand{\bbN}{\mathbb{N}}$
By $T_p$ I mean the regular $(p+1)$-tree (the tree where each vertex has degree $p+1$). Define a ray of $T_p$ as a sequence $(e_n)_{n\in\bbN}$ of vertices such that $$d(e_i,e_j)=|i-j|$$

for all $i,j\in\bbN$. Define an end of $T_p$ to be an equivalence class of rays, and denote the set of such ends as $\partial T_p$, where two rays $(e_n)_{n\in\bbN}$ and $(f_n)_{n\in\bbN}$ are equivalent if the sets $\{e_n,\;n\in\bbN\}$ and $\{f_n,\;n\in\bbN\}$  have infinite intersection.

Now, fix an end $\omega\in\partial T_p$. The situation can be described picturally :

The line in bold represents a ray leading to an end $\xi\neq\omega$.

I have proven that for any vertex $x\in T_p$, there is a unique ray $\overline{xw}$ starting at $x$ and leading to $\omega$. Now, define the confluent of two vertices $x,y\in T_p$ to be the unique $z=x\curlywedge y$ such that $$\overline{x\omega}\cap\overline{y\omega}=\overline{z\omega}.$$

Finally, by also fixing a vertex $o\in T_p$, one can define a Busemann function $\mathfrak{h}$ on $T_p$ by : $$\mathfrak{h}(x)=d(x,x\curlywedge o)-d(o,x\curlywedge o),$$

with $d$ the graph metric on $T_p$. The horocycle of level $k$ is $$H_k=\{x\in T_p\;/\;\mathfrak{h}(x)=k\}.$$

Picturally, we have :

It is clear that horocycles partition $T_p$. What I am concerned about are the other claims that are made in the papers :

Horocycles are infinite.

Each $x\in H_k$ has exactly one neighboor in $H_{k-1}$ and $p$ neighboors in $H_{k+1}$.

If those two properties are very intuitive and visual once we picture the situation, I have no idea how to prove them formally.

How would one proceed to prove these claims ?

So far, for 2., by writing $y_0,...,y_p$ the $p+1$ neighboors of $x\in H_k$, I tried computing some relations between $y_i\curlywedge o$ and $x\curlywedge o$, but without success. Considering a path $\gamma:e_0=x\to e_1\to...\to e_{n-1}\to e_n=x\curlywedge o$ lead me to two cases : either $y_i=e_1$, or $y_i\notin\gamma$, but in both cases, I wasn't able to prove anything useful.

For 1., if I knew that some $H_k$ was infinite, then by 2., I would also get that all the $H_{k-\ell}$, $\ell\geqslant0$ are infinite, by induction, by considering the function $H_k\to H_{k-1}$ sending a vertex $x\in H_k$ to its only neighboor in $H_{k-1}$ (since pre-images of vertices $y\in H_{k-1}$ have at most $p+1$ elements, because we are in $T_p$). But this has two issues : we don't know for sure how to prove that one horocycle is infinite (otherwise, by mimicing the argument, 1. would be solved already), and this would only tell us that somewhat "half" of the horocycles would be infinite...

答:
To get arbitrarily many elements of $H_k$, first take $n$ steps from $o$ along $\overline{o\omega}$ (reaching a vertex we'll call $y$). Then, take $n+k$ more steps that don't follow $\overline{o\omega}$ and don't backtrack. (This requires $n \ge -k$, but otherwise works for arbitrary $n$.) This gives us a vertex $x$ and I claim $x \in H_k$.

To see this, note that following the last $n+k$ steps backward from $x$ to $y$, then following the remainder of $\overline{o\omega}$, gives a ray leading to $\omega$; you've proven that it must be unique, so it is $\overline{x \omega}$. Similarly, starting at $y$ and following the remainder of $\overline{o\omega}$ (without the first $n$ edges) must give $\overline{y\omega}$. This tells us $\overline{x \omega} \cap \overline{o \omega} = \overline{y \omega}$, so $y = x \curlywedge o$, and therefore $$\mathfrak{h}(x) = d(x,y) - d(o,y) = (n+k) - n = k.$$

To prove 2, let's first be more specific. If $x \in H_k$, let $x_0$ be $x$'s neighbor along $\overline{x \omega}$, and $x_1, \dots, x_p$ be $x$'s other neighbors. Then $x_0 \in H_{k-1}$ and $x_1, \dots, x_p \in H_{k+1}$.

Taking $\overline{x\omega}$ and deleting $x$ from the beginning gives a ray starting at $x_0$ and leading to $\omega$; this must be $\overline{x_0 \omega}$. Similarly, for $1 \le i \le p$, taking $\overline{x\omega}$ and prepending $x_i$ must give $\overline{x_i \omega}$.

In most cases, we have $x \curlywedge o = x_i \curlywedge o$ for $0 \le i \le p$, and we get from $x_i$ to $x \curlywedge o$ by walking along $\overline{x_i \omega}$. This tells us that $d(x_0, x \curlywedge o) = d(x, x\curlywedge o) - 1$ and $d(x_i, x \curlywedge o) = d(x, x\curlywedge o) + 1$ for $1 \le i \le p$. From here, we get $x_0 \in H_{k-1}$ and $x_1, \dots, x_p \in H_{k+1}$.

This doesn't quite happen if $x$ lies on $\overline{o\omega}$, in which case two of $x$'s neighbors are also on $\overline{o\omega}$, and you'll have to handle those separately.

Title: Properties of the horocycles of a tree


问:
Verify Functional Limit of $\lim_{h \to 0} \frac{\int_{1-h}^{1+h} f(x) \,dx}{h}$
I want to find $$\lim_{h \to 0} \dfrac{\displaystyle\int_{1-h}^{1+h} f(x) \,dx}{h}$$. Where $f$ is a continuous function defined on $\mathbb{R}$

What I have so far:

Since $$\lim_{h \to 0} \int_{1-h}^{1+h} f(x) \,dx = 0 \text{ and } \lim_{h \to 0} h = 0$$ we can use L'Hopital's rule.

$\dfrac{d}{dh}h =1$ and $\frac{d}{dh} \displaystyle\int_{1-h}^{1+h} f(x) \,dx = f(1+h)+f(1-h)$. (Is this part right?).

Then from that we apply L'Hopital's rule:

$$\lim_{h \to 0} \frac{\displaystyle \int_{1-h}^{1+h} f(x) \,dx}{h} = \lim_{h \to 0} f(1+h)+f(1-h) = f(1) +f(1)=2f(1)$$

Does this look right? Thanks!

答:
By mean value theorem, for all $h>0$, there is always a number $x_h \in (1-h,1+h)$ such that:

$\int^{1+h}_{1-h} f(x)dx= 2h f(x_h)$

And as $h$ converges to $0$, clearly, $x_h$ converges to one. Hence by continuity, the desired limit is $2f(1)$.


问:
What happens to the singular values if we take absolute values of all matrix entries?
$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
Consider $|A|:=(|a_{ij}|)$. I have to prove or disprove that $\norm{|A|}_2=\norm{A}_2$, where $\norm{A}_2=\sqrt{\rho(A^*A)}$ and $\rho(M)$ denotes the spectral radius of a matrix $M$.

First I tried to find counterexpmples using any random matrix I could think of. Than I constructed some, where taking the abs of every entry causes maximal changes to the matrix. But nothing worked.

Then I tried to proof it. First I defined $A:=\pmatrix{a & b \\ c & d}$ and formally calculated all the other matrix, hoping something nice will show up. Then I tried to do some algebra, using defintions and norm properties. But I always stuck at $|A|^*|A|$ since I do not think I can give any statements about its eigenvalues depending from $A$.

Can anyone tell me if this claim is true or not and give me a little hint how to solve it?

答:
If you want to generate a counter-example, what you want are simple examples -- so select real symmetric matrices.

$A:=\begin{bmatrix}
-2 & -1\\
-1 & 2\\
\end{bmatrix}$

$\det\big(A\big) = -4- 1 = -5$ and A is traceless so
$\lambda_1 = \sqrt{5}$, $\lambda_2 = -\sqrt{5}\implies \Big \Vert A\Big \Vert_2 = \sqrt{5}\lt 3$

$B:=\begin{bmatrix}
2 & 1\\
1 & 2\\
\end{bmatrix} = \mathbf {11}^T + I$

$\lambda_1 = 3, \lambda_2= 1\implies \Big\Vert B\Big\Vert_2 =3$

Thus  $\Big \Vert {|A|}\Big\Vert_2\neq \Big\Vert{A}_2\Big\Vert$


问:
Prove that $\prod_{i=1}^n(1+x_i)\leq \sum_{i=0}^n\frac{S^i}{i!}$, where $x_i\in\mathbb{R^+}$.

Let $x_1$, $x_2$, $\ldots$, $x_n$ be positive real numbers, and let $$S=x_1+x_2+\cdots+x_n.$$ Prove that $$(1+x_1)(1+x_2)\cdots(1+x_n)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots+\frac{S^n}{n!}.$$

My first thought is about using induction. For $n=1$, $LHS=1+x_1\leq1+S=RHS$.

Now I suppose that this inequality holds true for $n=k$; that is, $$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$

And here I am stuck. I can't really see how I can find the relation between when $n=k$ and $n=k+1$.

Perhaps induction may not be the way?  Any hints or suggestions will be much appreciated. Thanks.

答:
Show that

$$ i! \sum_{1 \leq a_1 < a_2 \ldots < a_i \leq n } x_{a_1} x_{a_2} \ldots x_{a_i}  \leq S^i.$$

This is almost immediately obvious by expanding the RHS and noticing that we have a ton of extra terms.

In fact, we have strict inequality when $n \geq 2$.

Then, sum from $i = 0$ to $n$ to obtain the desired inequality.


Title: Prove that $\prod_{i=1}^n(1+x_i)\leq \sum_{i=0}^n\frac{S^i}{i!}$, where $x_i\in\mathbb{R^+}$.

Body:
Let $x_1$, $x_2$, $\ldots$, $x_n$ be positive real numbers, and let $$S=x_1+x_2+\cdots+x_n.$$ Prove that $$(1+x_1)(1+x_2)\cdots(1+x_n)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots+\frac{S^n}{n!}.$$

My first thought is about using induction. For $n=1$, $LHS=1+x_1\leq1+S=RHS$.

Now I suppose that this inequality holds true for $n=k$; that is, $$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$

And here I am stuck. I can't really see how I can find the relation between when $n=k$ and $n=k+1$.

Perhaps induction may not be the way?  Any hints or suggestions will be much appreciated. Thanks.

Body:
Induction, as you requested, works too. Just do it. Where are you stuck?

$$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$

Let $ S' = \sum_{i=1}^{k+1} x_i = S + x_{k+1}.$

$\prod_{i=1}^{k+1} (1 + x_i) \leq (1+x_{i+1}) (1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}) \\
= 1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!}     $

Claim $ S^i ( S + (i+1) x_{i+1} ) \leq S' ^{i+1}$.

This is true because the sum of the terms on both sides are equal to $(i+1)S'$, and the terms are evenly distributed on the RHS.

Corollary $x_{i+1} \frac{S^i}{i!} + \frac{S^{i+1}}{(i+1)!} \leq \frac{S'^{i+1}}{{i+1}!}$

Now, do a sum of

$ 1 \leq 1 $

$x_{i+1} \frac{S^i}{i!} + \frac{S^{i+1}}{(i+1)!} \leq \frac{S'^{i+1}}{{i+1}!}$ for $ i = 1$ to $k$

$x_{i+1} \frac{ S^k}{k!} \leq \frac{ S'^{k+1} } { (k+1)!}$.

to conclude that

$1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!}    \\
\leq 1 + S' + \frac{ S'^2}{2!} + \ldots + \frac{ S' ^{k+1} } { (k+1)!} $


答:
I think you guys are trying to overkill these problems. Clearly
$${\rm LHS} \le \left(1 + \frac{S}{n}\right)^n = \sum_{k \ge 0}^n \frac{n!}{(n-k)!\cdot n^k} \frac{S^k}{k!} \le {\rm RHS}$$


问:
Derivation of Quantifier Negation: $-(\forall x)(Px \rightarrow -Qx) \vdash (\exists x)(Px \wedge Qx)$
I am working on a series of derivations demonstrating the equivalence of negated quantifiers, etc.

I've worked through a bunch, but am stuck on one that seems to have Demorgan's law in it when I get my proof going.

$-(\forall x)(Px \rightarrow -Qx) \vdash (\exists x)(Px \wedge Qx)$

From there, I started with the premise, and then assumed:

$-(\exists x)(Px\wedge Qx)$

And then to:

$-(Pa\wedge Qa)$

From there, I dont really know what steps I should take. I think I would like to get to contradict on each parts of the conjunction and end with showing a contradiction between $-(\forall x)(Px \rightarrow -Qx) \wedge (\forall x)(Px \rightarrow -Qx)$, but without the availability to freely use Demorgan's law, I feel like it would be more work to try to insert a sub-proof to break it apart.

Does anyone have any suggestions or know if I am going about this the right way at all?

答:
My answer assumes that you are able to either assume or to prove that in general

$$(A \implies B) \iff [(-A) \vee B].$$

Using this property, you have

$$-(\forall x)(Px \implies -Qx) \iff -(\forall x)[(-Px) \vee (-Qx)] \iff$$

$$(\exists x)(Px \wedge Qx).$$


问:
Radius of convergence of a series which shows convergence for x>-a.
We need to find the radius of convergence of  $\sum_{i=1}^n a_n$ where
$a_n$=$(-4)^n*(x+2)^(2n)$/$n(n+1)$ .
By ratio test the series converges for |x+2|<1/4 i.e.
-9/4<x<-7/4 then what would be the radius of convergence? I know that if series converges for x=a then radius of convergence is R=0 but is it 1/4 here?

答: By Cauchy-Hadamard, the radius of convergence is: $r=\limsup_{n\to\infty}\dfrac1{\sqrt[2n]{\dfrac{4^n}{n(n+1)}}}=\dfrac12$.


答:
Just use that
$$\begin{align}
\sin(90) &= 1\\
\cos(90) &= 0
\end{align}
$$
(I assume that by $90$ you mean $90$ degrees).</p>

Title: How can I prove this cosine equation?

Body: How to prove that $\cos(90)\cos(\theta)+\sin(90)\sin(\theta)=\sin(\theta)$ ?</p>


问:
MLE for rates of exponential distributions
I have two sequences of exponentially distributed independent variables, $X_1, \ldots, X_n$ and $Y_1, \ldots, Y_n$ $$X_{i}\sim \exp(a)\\ Y_i \sim \exp(b)$$ However,  I only have the difference of the two sequences $X_1-Y_1, \ldots, X_n-Y_n$ recorded. I was wondering if there is a way to create MLE estimators for the parameters $a$ and $b$ using only the differences between the variables?

答:
Sometimes "exponentially distributed with parameter $a$" means the distribution is $e^{-ax}(a\,dx) \text{ for } x>0,$ and sometimes it means $e^{-x/a} (dx/a) \text{ for } x>0.$ For now I will assume the former. I will further assume the sequences are of independent random variables. I would have stated that in the problem.

Robert Israel's answer to a related question tells us that the density of $X-Y$ is
$$
\frac{ab}{a+b} \begin{cases} e^{-ax}\,dx & \text{if }x>0, \\ e^{-bx} \,dx & \text{if } x<0. \end{cases}
$$
The the likelihood function is
\begin{align}
L(a,b) & = \left( \frac{ab}{a+b} \right)^n \left( \prod_{i\,:\,x_i \,>\,0} e^{-ax_i} \right) \left( \prod_{i\,:\,x_i\,<\,0} e^{-bx_i} \right) \\[8pt]
& = \left( \frac{ab}{a+b} \right)^n \exp\left( -a \sum_{i\,:\,x_i \,>\,0} x_i - b\sum_{i\,:\,x_i \,<\,0} x_i \right).
\end{align}
Therefore
\begin{align}
& \ell(a,b) = \log L(a,b) \\[8pt]
= {} & n(\log a + \log b - \log(a+b)) -a \sum_{i\,:\,x_i \,>\,0} x_i - b\sum_{i\,:\,x_i \,<\,0} x_i \\[8pt]
\text{and so } & \frac{\partial\ell}{\partial a} = \frac n a - \frac n{a+b} - \sum_{i\,:\,x_i \,>\,0} x_i, \\[8pt]
\text{and } &  \frac{\partial\ell}{\partial b} = \frac n b - \frac n{a+b} - \sum_{i\,:\,x_i \,<\,0} x_i. \\[8pt]
\end{align}
These are both $0$ when
$$\frac b {a(a+b)} = \overline{x}_{>0} \quad\text{and}\quad \frac a{b(a+b)} = \overline{x}_{<0}
$$
where $\overline{x}_{>0}$ and $\overline{x}_{<0}$ are respectively the means of the positive and negative $x$-values. Where there are no positive $x$-values and where there are no negative $x$-values, the MLEs for $a,b$ respectively are undefined.

Possibly to be continued$\,\ldots$