答:
Note that $(m+1)^3<2m^3\iff\left(\frac{m+1}m\right)^3<2\iff\left(1+\frac1m\right)^3<2$. This is easy to prove if $m=4$ and, if $m\geqslant4$, then $\left(1+\frac1m\right)^3\leqslant\left(1+\frac14\right)^3<2$.

The author is assuming that $k^3\leqslant2^k$. It follows from this that $2k^3\leqslant2^{k+1}$. So, in order to prove that $(k+1)^3\leqslant2^{k+1}$, it will be enough to prove that $(k+1)^3\leqslant2k^3$, which is equivalent to the assertion $3k^2+3k+1\leqslant k^3$.

Title: Alternative proof for $(m+1)^3 \leq2m^3$

Body:
This is the proof on the textbook:

Until here, I was able to get to the same conclusion. What I tried to do was to prove by induction that $(m+1)^3 \leq2m^3$. I started from $(m + 2)^3$, but got nowhere.

The author then does the following:

I don't understand the first three lines of the previous picture (before the proof starts) nor how he uses it in the proof.

Is it possible to prove $(m+1)^3 \leq2m^3$ by induction? Or is there any other simpler proof that I'm not seeing?

Could someone please throw some light on the proof used in the textbook? I find it highly unintuitive.


问:
Finding the limit at positive and negative infinity
$$y = (3-x)(1+x)^2(1-x)^4$$

Hello,

how can I find the limit when $x \to \infty$ and when $x \to - \infty$?

Thanks.

答:
Well for $x\rightarrow +\infty$ you can say that
$$y=(3-x)(1+x)^2(1-x)4 \sim (-x)(x)^2(-x)^4=-x^7\longrightarrow -\infty\mbox,$$
similarli for $x\rightarrow -\infty$
$$y\sim -x^7 \longrightarrow +\infty\mbox.$$

ContentLicense: CC BY-SA 4.0


问:
ANR is locally contractible
Recall that a space $X$ is contractible if there exists a homotopy $h:X\times [0,1]\to X$ such that $h$ is equal to the identity map on $X\times\{0\}$ and $h$ is constant on $X\times\{1\}$.

Please help me out since I'm stuck with this question, namely, why each ANR is locally contractible. It seems that I can prove that each AR is contractible this way: Consider $A = X\times \{0\} \cup X\times \{1\} \cup \{x_0\}\times [0,1] \subset X\times [0,1]$, where $x_0$ is any point from $X$. This set is closed in $X\times [0,1]$. Define $h: A\to X$ by $h|_{X\times \{0\}} = id_X$, $h|_{X\times \{1\}} = x_0$, $h|_{\{x_0\}\times [0,1]} = x_0$. This function is continuous, we can extend it (since being an AR is equivalent to being an AE) to the whole space $X\times [0,1]$. And this will be the required homotopy that squeezes $X$ into a point ($x_0$).

But similar approach seems not to work in case of an ANR. Because here I get a neighborhood $V$ of $A$ in $X\times [0,1]$ and an extension of $h$ onto this neighborhood. I can of course find an open $U\subset X$ such that $x_0\times [0,1] \subset U\times [0,1]\subset V$ (where $x_0$ is the point for which I want to find a contractible neighborhood) and consider the restriction of my extension onto $U\times [0,1]$, but the problem is that this map is not necessarily a map from $U\times [0,1]$ to $U$ (the range of my restricted extension can be larger than $U$).

Can anybody please give a hint on what's going on? (a hint would be even better than a full answer).

答:
Just to clarify the terminological confusion, continuing with Paul Frost's notation:

Definition 2 should be regarded as the standard one, it is the one introduced by Borsuk and used to prove that each ANR is locally contractible.

Definition 1 is not equivalent to Definition 2, see two examples here: These are ANRs (even ARs) which fail Definition 1.

I am not sure who is responsible for renaming, but Hatcher in his "Algebraic Topology" refers to Definition 2 as "locally contractible in weak sense." But at least he spells out both definitions. In contrast, nlab here pretends that the standard definition (Definition 2) does not even exist. Ditto wikipedia here.

ContentLicense: CC BY-SA 3.0

Title: ANR is locally contractible


问:
Show that rank($A^{n+1}$) = rank($A^n$)
Suppose $A$ is a $n \times n$ matrix i.e. $A \in \mathbb{C}^{n \times n}$, prove that rank($A^{n+1}$) = rank($A^n$). In other words, I need to prove that their range spaces or null spaces are equal. If it helps, $A$ is a singular matrix.

Note that, I don't want to use Jordan blocks to prove this. Is it possible to prove this without using Jordan form? I can use Schur's triangularization theorem. Also, it's not known if A is diagonalizable.

答:
$\newcommand{\rg}{\operatorname{range}}$
Obviously for every $m$, $\rg A^{m+1}\subset\rg A^{m}$, so if $d_m=\dim\rg A^m$, $d_{m+1}\le d_m$. If $d_{m+1}=d_m$ for some $m$, then $\rg A^{m+1}=\rg A^{m}$ and therefore $\rg A^m=\rg A^{m+1}=\rg A^{m+2}=\dotsb{}$. That is, the sequence $d_0,d_1,\dots$ becomes constant once it stops descending.

Because $d_0= n$, the sequence must stop descending within $n$ terms.

Edit: For the problem you phrased in the comment, $\rg A^{m+1}=\{AA^{m}y:y\in \mathbb C^n\}=\{Ax:x=A^my\in\rg A^m\}=\{Ax:x\in\rg A^m\}$,

Therefore  $\rg A^{m}=\rg A^{m+1}\implies$

$\rg A^{m+1}=\{Ax:x\in\rg A^m\}=\{Ax:x\in\rg A^{m+1}\}=\rg A^{m+2}$.

Title: Show that rank($A^{n+1}$) = rank($A^n$)


问:
Absurd result in inequality
I have this inequality

$(x-2)^2 \ge 0 $

So I solve it like this

$(x-2)^2 \ge 0 \implies (x-2)(x-2)\ge0$

$\implies x(x-2)-2(x-2)\ge0 \implies x(x-2)\ge2(x-2)\implies x\ge2$

But obviously $x\ge2$ is false in R

What did I miss?

答:
You couldn’t have divided both sides by $x-2$ As it can be negative.
$$-1>-2 \not \Rightarrow 1>2$$

Title: Absurd result in inequality


答:
Hint

You can prove that for $k \ge 0$ $$\mathrm{rank}(A^{k+2}) - \mathrm{rank}(A^{k+1}) \le \mathrm{rank}(A^{k+1}) - \mathrm{rank}(A^{k})$$

Therefore,  $$\mathrm{rank}(A^{n+1}) < \mathrm{rank}(A^{n})$$ would imply the contradiction $\mathrm{rank}(A) \gt n$.

Title: Show that rank($A^{n+1}$) = rank($A^n$)

Body:
Suppose $A$ is a $n \times n$ matrix i.e. $A \in \mathbb{C}^{n \times n}$, prove that rank($A^{n+1}$) = rank($A^n$). In other words, I need to prove that their range spaces or null spaces are equal. If it helps, $A$ is a singular matrix.

Note that, I don't want to use Jordan blocks to prove this. Is it possible to prove this without using Jordan form? I can use Schur's triangularization theorem. Also, it's not known if A is diagonalizable.


问:
A question about regularity of Borel measures
From Wikipedia:

"Any Borel probability measure on any metric space is a regular measure."

I've been thinking about how to prove this. And now I'm wondering whether being Borel isn't enough. Is $\mu(X) = 1$ (or $\mu(X) < \infty$) necessary and does $X$ have to be a metric space? I assume the answer to this is yes but I don't see why. Maybe someone could give me a hint for a proof or a proof?

Thanks for your help.

答:
It's a reply to this comment. I'm writing it as an answer since there is a bit too much $\LaTeX$ in it.

Well, for each $\varepsilon$ you need $n, G_i, F_i$ such that: $\mu\Big(\bigcup_{i\in \mathbb N} G_i \setminus \bigcup_{i<n} F_i\Big) < \varepsilon$.

We can equivalently require:
$\mu \Big(\bigcup_{i\in \mathbb N} G_i \setminus \bigcup_{i\in \mathbb N} A_i \Big) < \varepsilon$ and $\mu \Big(\bigcup_{i\in \mathbb N} A_i \setminus \bigcup_{i<n} F_i\Big) < \varepsilon$. So actually the problem is the second assertion.

We have $F_i$ such that $\mu(A_i\setminus F_i)< \frac{\varepsilon}{2^{i+2}}$, so
$\mu \Big(\bigcup_{i\in \mathbb N} A_i \setminus \bigcup_{i\in\mathbb N} F_i\Big) \leq
\sum_{i \in \mathbb N} \frac{\varepsilon}{2^{i+2}} = \frac{\varepsilon}{2}$. Thus we know:
$$\mu \Big(\bigcup_{i\in\mathbb N} F_i\Big) \geq \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \frac{\varepsilon}{2}.$$
Finally, we use continuity of measure to conclude that there exists $n$ such that
$$\mu \Big(\bigcup_{i<n} F_i\Big) > \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \varepsilon.$$


问:
How to calculate wedge product of differential forms for vectors?
I'm working on an exercise where I have to calculate the wedge product of differential forms with vectors and I feel like I'm missing something very basic.

The question is asking to calculate $(\phi_1\wedge dx_2+\phi_2\wedge\phi_3)(u,v)$. (All of these are given of course.)

So far I've developed $(\phi_1\wedge dx_2+\phi_2\wedge\phi_3)$ using the properties of wedge product that we saw in class, but what does it mean to calculate it for $u$ and $v$? What exactly am I calculating here?

(For reference $u=(1,0,2,-1)$, $v=(3,2,1,0)$ and I developed $(\phi_1\wedge dx_2+\phi_2\wedge\phi_3)=dx_2\wedge dx_1+3dx_4\wedge dx_1 +9dx_4\wedge dx_3$.)

答:
I'm assuming, since you have a $dx_1,\dots$ basis, that $u=(1,0,2,-1)$ is written in the corresponding $\partial_{x_1}$ basis. So $u=1\partial_{x_1} + 2\partial_{x_3} -1\partial_{x_4}$, and similarly for $v$.

Do you know the following rule for how a 2-form acts on 2 vectors?
$$(\alpha\wedge\beta)(X,Y) = \alpha(X)\beta(Y)-\beta(X)\alpha(Y)$$
i.e. $\alpha\wedge\beta = \alpha\otimes\beta - \beta\otimes\alpha$, if that makes any sense to you (I don't know how exactly your class chose to define the wedge product).

Now you can use $dx_1(\partial_{x_1})= 1$ and $dx_1(\partial_{x_j})= 0$ if $j\neq 1$, and use the linearity of everything.


问:
Induced connection on Whitney sum of vector bundles
Let $(E, B, V)$ and $(E',B,V)$ be two vector bundles with connections $\nabla$ and $\nabla'$ respectively. If I do their Whitney sum, do these connections induce a connection in $(E \oplus E', B, V)$? If so, how do they relate? Can you recommend any literature that deals with this?

Appreciate.

答: Suppose $E_1 \to B$ and $E_2 \to B$ are vector bundles with connections $\nabla^1$ and $\nabla^2$ respectively. Note that a section $\sigma$ of $E_1\oplus E_2$ takes the form $\sigma = \sigma_1 + \sigma_2$ where $\sigma_i$ is a section of $E_i$. Therefore, one defines a connection $\nabla$ on $E_1\oplus E_2$ by $\nabla\sigma := \nabla^1\sigma_1 + \nabla^2\sigma_2$.


问:
Find the cumulative distribution function of $M_{n} - m_{n}$
Again, in this exercise how can I find the cumulative distribution function of $M_{n} - m_{n}$ and how I can calculate the limit in distribution of the sequence $(n (1 - M_{n} + m_{n})) \ n \in \mathbb{N}$?

Let $(U_{n})_{n\in\mathbb{N}}$ be a sequence of random variables i.i.d. with uniform distribution in the interval $[0, 1]$.
We define for each $n\in \mathbb{N}$ the random variables:
$$M_{n} = \max\{U_{1} ,...,U_{n}\}$$
$$m_{n} = \min\{U_{1} ,...,U_{n}\}$$

Show that $M_{n} -m_{n}\ \underrightarrow{d}\ 1$

Now I have to:

$$P(\min\{U_{1} ,...,U_{n}\}>a)=P(U_{1}>a)P(U_{2}>a)\dotsi P(U_{n}>a)=(1-a)^n$$
$$\rightarrow F_{m_{n}}(a)=1-(1-a)^n \rightarrow f_{m_{n}}(a)=n(1-a)^{n-1}$$

$$P(\max\{U_{1} ,...,U_{n}\}\leq a)=P(U_{1} \leq a)P(U_{2} \leq a)\dotsi P(U_{n} \leq a)=a^n$$
$$\rightarrow f_{m_{n}}(a)=na^{n-1}$$

I don't know how to continue, could you help me?

答:
Use continuous mapping theorem:

$$m_n\xrightarrow{d}0$$

$$M_n\xrightarrow{d}1$$

Thus

$$M_n-m_n\xrightarrow{d}1$$


问:
Solve in rational numbers, the equation, $x\lfloor x\rfloor\{x\}=58$
Solve in rational numbers, the equation,
$$x\lfloor x\rfloor\{x\}=58$$
where $\lfloor x\rfloor$ and $\{x\}$ are the greatest integer less than or equal to $x$ and the fractional part of $x$ respectively.

I tried to make an equation like:
$$(a+b)ab=58$$
where $a=\lfloor x\rfloor$ and $b=\{x\}$ and giving the restrictions that $a$ is an integer and $b$ is a rational number such that $0≤b<1$.

Then I made a quadratic equation over $b$ and used the famous quadratic formula. But everything became complex (not the mathematical 'complex') and so please help me out. Thanks in advance!

答:
Let $b=p/q$ where $\gcd(p,q)=1$ and $0 \le p<q$. The equation becomes $$(qa+p)ap=58q^2$$

Since $\gcd((qa+p)p,q)=\gcd(p^2,q)=1$ we have $q^2|a$. Write $a=kq^2(k \in \Bbb{Z})$(apparently $k \ne 0$) and we get $$(kq^3+p)kp=58$$

Then $p|58$. Moreover, $|kq^3+p| \ge |k|q^3-p \ge (p+1)^3-p>p$, thus $p$ is the smaller positive factor of $58 \Rightarrow p=1,2$. A simple discussion will conclude that $p=2,q=3,k=1,a=9$, so the solution is $x=29/3$.


答: In your last step you divide both sides of the inequality by $(x-2)$ which is negative for $x < 2$, and this would require reversing the inequality sign.  Instead you could simply split your solution into two cases: $x\geq 2$ and $x< 2$, and derive the entire solution space by considering each case individually.

Title: Absurd result in inequality

Body:
I have this inequality

$(x-2)^2 \ge 0 $

So I solve it like this

$(x-2)^2 \ge 0 \implies (x-2)(x-2)\ge0$

$\implies x(x-2)-2(x-2)\ge0 \implies x(x-2)\ge2(x-2)\implies x\ge2$

But obviously $x\ge2$ is false in R

What did I miss?


问:
Proof that a set $A$ is nowhere dense set in $R$ if and only if every sphere $S \subset R$ contains another set $S'$ such that $S' \cap A = \emptyset$
I am having problems trying to understand and prove this assertion. For instance to prove the first part:

Let $A$ be a nowhere dense set and let $S$ be any set in $R$. Then if $A \cap S = \emptyset$ of course any set $S' \subset S$ has an empty intersection with $A$. Now suppose $A \cap S \neq \emptyset$. This implies that $S - A \neq \emptyset$ or $S=A$. If it is the first case, just let $S' = S - A$ and it follows that $S' \cap A = \emptyset$. Finally if $S=A$ then select $S' = \emptyset$ (which is a subset of every set) and thus again $S' \cap A = \emptyset$.

What is wrong with this line of reasoning? And if nothing is wrong, it seems to me that we don't require $A$ to be a nowhere dense set. Could you provide a proof and/or comment on this one?

Thanks a lot and happy saturday!

答:
The problem most likely assumes you are not allowed to take $S'= \emptyset$.

Following what you write, observe that $S=A \implies \exists$ a point in the interior of $A$ (which follows directly from the fact that $S$ is a sphere) which isn't allowed since $A$ is nowhere dense set.


问:
probability of N same type pairs from a heterogeneous group
There are 16 people: 8 males and 8 females. The people are randomly divided to pairs, each pair get the exact same mission. Let $X$ be the number of male-male pairs.

calculate $P\{X=3\}$

calculate $P\{X=i\}$ where $i$ is each value $X$ can take

find $E[X]$

for (1) there are $\binom{8}{2,2,2,1,1}$ ways to divide the males to 3 all male pairs and  $\binom{8}{2,2,2,1,1}$ ways to divide the females to 3 all female pairs and there are 4 ways to pair the remaining males and females so I expected the probability to be $$P\{X=3\}=\frac{\binom{8}{2,2,2,1,1}\cdot \binom{8}{2,2,2,1,1} \cdot 4}{\binom{16}{2,2,2,2,2,2,2,2}} = \frac{\frac{8!}{8}\cdot \frac{8!}{8}\cdot 4}{\frac{16!}{2^8}}=\frac{8!^2\cdot 4\cdot 64 \cdot 4}{16! \cdot 64}=\frac{8!^2}{15!} \approx 0.0012$$ but it is incorrect so instead I tried, I choose 6 males and pair them in $\binom{8}{6}\binom{6}{2,2,2}$ and the 2 remaining males I have $\binom{2}{1}\binom{8}{1}$ and $\binom{7}{1}$ ways to pair with a female, and the remaining 6 females I have $\binom{6}{2,2,2}$ ways to pair so $$\frac{\binom{8}{6}\binom{6}{2,2,2}\binom{2}{1}\binom{8}{1}\binom{7}{1}\binom{6}{2,2,2}}{\binom{16}{2,2,2,2,2,2,2,2}}=\frac{\frac{8!}{16}\cdot 16 \cdot 7\cdot\frac{6!}{8}}{\frac{16!}{2^8}}=\frac{7!\cdot 7! \cdot 2^8}{16!}\approx 0.0003$$ which is also incorrect. I'm not sure how I can calculate $P\{X=3\}$ or any other $P\{X=i\}$

after reading @saulspatz's answer I found a possible solution:
all possible pairing is given by $\binom{16}{2,2,2,2,2,2,2,2}\cdot \frac{1}{8!}$ because all the pairs have the same role but the same works for each of the 3 male-male or female-female pairs, hence $\binom{6}{2,2,2}\cdot \frac{1}{3!}$ and because the 2 male-female pairs also have the same role I get $$P\{X=3\}=\frac{\frac{\binom{8}{6}\binom{6}{2,2,2}}{3!}\frac{\binom{2}{1}\binom{8}{1}\binom{7}{1}}{2!}\frac{\binom{6}{2,2,2}}{3!}}{\frac{\binom{16}{2,2,2,2,2,2,2,2}}{8!}}=\frac{\frac{8!}{16\cdot 3!}\cdot \frac{16\cdot 7}{2!}\cdot\frac{6!}{8\cdot 3!}}{\frac{16!}{2^8 \cdot 8!}}=\frac{8!\cdot 7! \cdot 7!\cdot 2^8}{16!\cdot 3!\cdot 3!\cdot 2!}\approx 0.1740$$ and from here calculating $P\{X=i\}$ is the same and and there is no problem finding $E[X]$

答:
The number of ways to divide the $16$ people is not pairs is not $$\binom{16}{2,2,2,2,2,2,2,2}$$ but $$\frac1{8!}\binom{16}{2,2,2,2,2,2,2,2}$$  To divide the people into pairs, arrange them into a line, and pair the first and second, the third and fourth, and so on.  The order of the tow people in the pairs doesn't matter, so we have $$\frac{16!}{(2!)^8}$$ but also the order of the pairs themselves doesn't matter, so we must divide by $8!$.

You make a similar mistake at other points in the calculation.

I'm not sure that this is the only problem.


问:
Example of application of stochastic calculus to discrete probability
Integrals extend the concept of sum in the continuous domain, and the usefulness of such extension with respect to the generalized concepts (i.e. discrete mathematics) can be easily illustrated with some examples, such as providing bounds on the value of a series by upper and lower bounding it with appropriate integrals.

Stochastic calculus extends the concept of sum of random variables in a similar fashion. Can we also provide a corresponding motivating example of an application of stochastic calculus in a discrete-probability setting? In other words, can we prove something about a discrete process by the means of stochastic calculus, that we could not easily prove otherwise?

答:
I guess one of the mightiest tools in order to infer from the contiuous setting to the discrete setting is Skorokhod's second embedding theorem, which will tell you that given a random walk $S_{n}$ and a Brownian motion $B_{t}$ there exist stopping times $\tau_{n}$ such that $S_{n}$ and $B_{\tau_{n}}$ are identically distributed. This can be used for instance to prove the law of the iterated logarithm for random walks using the corresponding statement for Brownian motion. See e.g. https://www.math.upenn.edu/~pemantle/papers/Student-theses/Masters/Choi120424.pdf .
I would expect that other asymtotic statements can be proved in a similar fashion, such as arc-sine laws.


问:
Is there (or can there be) a general algorithm to solve Rubik's cubes of any dimension?
I love solving Rubik's cube (the usual 3D one). But, a lecture by Matt Parker at the Royal Institute (YouTube Link) led me to an app that can simulate a four dimensional rubik's cube. But unfortunately it was so complex, that I soon got bored as I failed to solve it.

The website to the 4D cube : http://superliminal.com/cube/cube.htm

Following which today, I also found an app that can simulate a 5D cube!!

The website to the 5D cube: http://www.gravitation3d.com/magiccube5d/

So, my two questions are :

Is there a general (non brute-force) algorithm that can be used to solve a well-scrambled cube of any dimension (even though it may not be very efficient, but yet is not a simple search over all the available space of move sequences) ? [Point modified after reading @RavenclawPrefect's answer :D ]

Mathematically, what is common in all these cubes, and "hypercubes"?

NOTE: By dimensions I mean physical dimension and not the number of rows of pieces on a face of the cube. So a four-dimension cube is a cube in x,y,z,$\delta$ dimensions where $\hat x,\hat y,\hat z,\hat \delta$ are mutually orthogonal unit vectors in the respective dimensions.

For example, here is how the aforementioned 4D cube moves:
https://miro.medium.com/max/2552/1*ga32DoV_Hc6e8t6PC1hFHw.gif

Addendum:

List of resources that may help finding an answer to this question:

Solving Rubik's cube and other permutation puzzles

答:
The algorithmic problem is a special case of ``constructive membership test in permutation groups'' -- given a permutation (you label every face of the cube with a number, then every slice rotation corresponds to a permutation of the numbers, as does the mixed state), write it as a product of generators.
This does not care for the dimension of the cube you work in, you just get more and more labels.

The generic solution to the problem is called the Schreier-Sims algorithm (https://en.wikipedia.org/wiki/Schreier–Sims_algorithm). However in its naive form, it produces very long solutions (length >100k for the $3\times3\times 3$ cube).

One can add heuristics (basically to pre-seed with short products as "extra" generators), see for example:
https://doi.org/10.1006/jsco.1998.0202 .
With such heuristics, the $3\times3\times 3$ cube gets solutions of length $\sim 40-100$, so a magnitude too large, but not astronomically so. One can tune heuristics (use more memory) to get better solutions, though the only (known) way to find a shortest solution is brute-force.


问:
If $X\sim\text{Bin}(3,2)$, find $\Pr(X=k|X\ge 1)$
I'm having some trouble with this. I wrote

$$\Pr(X=k|X\ge 1)=\frac{\Pr(X=k\cap X\ge1)}{\Pr(X\ge1)},$$

but I'm stuck in calculating the quantity $\Pr(X=k\cap X\ge1)$.

答:
What about rewriting it as

$$Pr(X = k | X \geq 1) = \frac{Pr(X \geq 1 | X = k) P(X = k)}{Pr(X \geq 1)}$$

That should make it clear, if you think about what is $Pr(X \geq 1|X = k)$.

You should consider $k = 0$ and $k > 0$ separately maybe.


问:
If the union of $A$ and $B$ is linearly independent then the intersection of the spans $= \{0\}$

$\newcommand{\sp}{\operatorname{sp}}$ Let $V$ be a vector space over $F$ field, and let $A,B$ be two different, disjoint, non-empty sets of vectors from $V$.

Prove or disprove the following:
If $A \cup B$ are linearly independent, $\sp(A) \cap \sp(B) = \{0\}$.

It's an easy one to example over $\mathbb{R}^2$, because the $\text{span}$ of all $\mathbb{R}^2$ vectors has $(0,0)$ in it, but something keeps telling me that it might be disproved for other vector spaces.

答:
Let $v$ be a vector
$$v\in\operatorname{sp}(A)\cap\operatorname{sp}(B)$$
Since $v\in\operatorname{sp}(A)$ you can write
$$v=\sum_jf_ja_j$$
for some $f_j\in F,a_j\in A$. Similarly, since $v\in\operatorname{sp}(B)$, you can also write
$$v=\displaystyle\sum_ig_ib_i$$
for some $g_i\in F,b_i\in B$.
Then you get
$$\displaystyle\sum_jf_ja_j-\displaystyle\sum_ig_ib_i=0$$
but LHS is a linear combination of vectors in $A\cup B$, hence by assumptions linearly indipendent, thus all coefficients are zero, implying that $v=0$</p>


问:
$\sum \dfrac{n^k}{2^n} (k\in \mathbb{N})$
I think I solve it but I want some verify.
First I take $k=n$ and rewrote the series.
$\sum \dfrac{n^{n}}{2^{n}}$ and I apply Root Test. And in the end I got:
$\lim _{n\rightarrow \infty }\dfrac{n}{2}=\infty >1$ So that, series is divergent for $k=n$
And then, I assume that $k<n$
I applied Ratio Test and I got:
$\lim _{n\rightarrow \infty }\dfrac{1}{2}\left( 1+\dfrac{1}{n}\right) ^{k}$ So I thought that, k<n in this respect I wrote this inequality:
$\dfrac{1}{2}\lim _{n\rightarrow \infty }1< \lim _{n-\infty }\left( 1+\dfrac{1}{n}\right) ^{k} <\lim _{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}$ So, I conclude that limit of second term must be between $\dfrac{1}{2}$ and $\dfrac{e}{2}$. My question is was it true? Or can I bound more clearly because $e/2$ is like $1.35...$ which means the inequality does not always give us convergent series.

答:
You cannot take $k=n$ because $k$ is an up-front given, costant number from $\mathbb N$.

You can apply the ratio test, because

$$\frac{(n+1)^k/2^{n+1}}{n^k/2^n}=\frac{1}{2}\left(1+\frac{1}{n}\right)^k\to\frac{1}{2}<1\text{ as }n\to\infty$$

(note $\frac{1}{n}\to 0$ so $\left(1+\frac{1}{n}\right)^k\to 1^k=$ as $k$ is fixed.

As a way how to convince yourself that this is a correct view at the things, try to solve the same problem for some concrete $k$, e.g. $k=2$, $k=3$, $k=4$ ... and then notice that the logic of the solution doesn't really depend on $k$.


问:
Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$.

Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$. Given that $AE = 169$ cm and $EC = 119$ cm, find $ED$.

What I Tried: Here is a picture :-

We have $AE = 169, EC = 119$.

Let $AB = r$ , $DE = x$ , $AD = y$ , $BC = z$ , $BE = a$.

Now, applying Pythagoras Theorem gives all these equations one by one :-
$$(i) x^2 + y^2 = 169^2$$
$$(ii) z^2 + 119^2 = a^2$$
$$(iii) (x+a)^2 + y^2 = R^2$$
$$(iv) 288^2 + z^2 = R^2$$
We also have $\Delta CEB \sim \Delta DAB \sim \Delta DAE$ . They give information as :-
$$(i) \frac{CE}{DA} = \frac{CB}{DB} = \frac{EB}{AB}$$
$$(ii) \frac{DE}{EC} = \frac{DA}{CB} = \frac{AE}{EB}$$

The Problem is that all these don't really make me find the value of $ED$, or I am unable to find it by using them appropriately in some way.

Can anyone help me? Thank You.

答:
By angle bisector rule $$\frac{CB}{AB}=\frac{119}{169}$$ also by Pythagoras $$AB^2+CB^2={(169+119)}^2$$ $$AB=?,CB=?$$ Now in $\Delta CEB $ $$\tan x=\frac{119}{CB}=?$$ From here you can find
$\sin x$ but $$\sin x=\frac{DE}{169}$$

We are done!


问:
use power series expansion to prove $e^{z_1+z_2}=e^{z_1}e^{z_2}$
prove $e^{z_1+z_2}=e^{z_1}e^{z_2}$

hint: If $\sum_{n=0}^{\infty} a_n \sum_{n=0}^{\infty} b_n = \sum_{n=0}^{\infty} c_n$ then

$ c_n=\sum_{k=0}^{n} a_k b_{n-k} $ provided that $ \sum_{n=0}^{\infty} a_n , \sum_{n=0}^{\infty} b_n$  are convergent

using power series expansion: $e^{z}= \sum_{n=0}^{\infty} {z^n\over n!}$

$e^{z_1}= \sum_{n=0}^{\infty} {z_1^n\over n!}$ and $e^{z_2}= \sum_{n=0}^{\infty} {z_2^n\over n!}$

$e^{z_1}e^{z_2}= \sum_{n=0}^{\infty} {z_1^n\over n!} \sum_{n=0}^{\infty} {z_2^n\over n!}=\sum_{k=0}^{n} {z_1^k\over k!}  {z_2^{n-k}\over {n-k}!} = e^{z_1+z_2}= \sum_{n=0}^{\infty} {(z_1+z_2)^n\over n!}$ [Using binomial thm.]

Now, $e^{z_1+z_2}= \sum_{n=0}^{\infty} {(z_1+z_2)^n\over n!}=e^{z_1}e^{z_2}$
but iam not sure of the binomial thm part

答:
Let $S_n(x)=\sum_{j=0}^n x^j/j!.$

Then $S_n(x)S_n(y)-S_n(x+y)=R_n(x,y)$ where $$R_n(x,y)=\sum_{j\le n\land k\le n\land j+k>n}x^jy^k/(j!k!).$$ The number of terms in the RHS above cannot be more than $(n+1)^2.$ And each term cannot have a modulus greater than $r^{2n}/m!$ where $r=\max (1,|x|,|y|),$ and (i) $m=1+n/2$ if $n$ is even,  and (ii) $m=(n+1)/2$ if $n$ is odd.

So $|R_n(x,y)|\le (n+1)^2r^{2n}/m!,$ which $\to 0$ as $n\to \infty$ because $ m> n/2.$


问:
Use power series to solve the initial-value problem: $y''-2xy'-2y=0$ where $y(0)=1$ and $y'(0)=0$.

Use power series to solve the initial-value problem:
$$y''-2xy'-2y=0; y(0)=1, y'(0)=0$$

How to we solve this in terms of summation sign?

I got this :

$a_{n+2}=2\frac{a_n}{n+2}$

$y(x)=c_0(1-x^2-\frac12x^4)+c_1(x+\frac23x^3+\frac4{15}x^5)$

I dont know what will it be in terms of odd and even summations.

Body:
$$a_{n+2}=2\frac{a_n}{n+2};\;a_0=1;\;a_1=0$$
$$\begin{cases}
a_{2n+1}=0\\
a_{2n}=\frac{1}{n!}\\
\end{cases}
$$
$$y=\sum _{n=0}^{\infty } \frac{x^{2 n}}{n!}=e^{x^2}$$


答:
It's true. Suppose $A \cup B$ is linearly independent. Assume by contradiction that the intersection of the spans contains a vector $x \neq 0$. Suppose $A = \{v_1,\ldots,v_m\}$ and $B = \{w_1,\ldots,w_n\}$ so that
$$
x = \alpha_1 v_1 + \cdots + \alpha_m v_m = \beta_1 w_1 + \cdots + \beta_n w_n.
$$
However, subtracting the RHS from the LHS gives
$$
\alpha_1 v_1 + \cdots + \alpha_m v_m - \beta_1 w_1 - \dots - \beta_n w_n = 0
$$
and since $A \cup B$ is linearly independent, that implies $\alpha_i = \beta_j = 0$ for all $1 \leq i \leq m$ and $1\leq j \leq n$. This is a contradiction since this would imply $x = 0$. Thus, the intersection of the spans must only contain the zero vector.</p>

Title: If the union of $A$ and $B$ is linearly independent then the intersection of the spans $= \{0\}$

Body:
$\newcommand{\sp}{\operatorname{sp}}$ Let $V$ be a vector space over $F$ field, and let $A,B$ be two different, disjoint, non-empty sets of vectors from $V$.

Prove or disprove the following:
If $A \cup B$ are linearly independent, $\sp(A) \cap \sp(B) = \{0\}$.

It's an easy one to example over $\mathbb{R}^2$, because the $\text{span}$ of all $\mathbb{R}^2$ vectors has $(0,0)$ in it, but something keeps telling me that it might be disproved for other vector spaces.


问:
Show that rank($A^{n+1}$) = rank($A^n$)
Suppose $A$ is a $n \times n$ matrix i.e. $A \in \mathbb{C}^{n \times n}$, prove that rank($A^{n+1}$) = rank($A^n$). In other words, I need to prove that their range spaces or null spaces are equal. If it helps, $A$ is a singular matrix.

Note that, I don't want to use Jordan blocks to prove this. Is it possible to prove this without using Jordan form? I can use Schur's triangularization theorem. Also, it's not known if A is diagonalizable.

答:
Every matrix corresponds to a linear map. Suppose that $A=\mathcal{M}(T)$, where $T\in \mathcal{L}(V)$ and $dim (V)=n$. Using Theorem 8.4 in Axler,
$$Null(T^n)=Null(T^{n+1}).$$

According to the Fundamental Theorem of Linear Maps,
$$n=dim Null(T^n)+ dim Range(T^n)$$
$$n=dim Null(T^{n+1})+dim Range(T^{n+1})$$

$dim Range(T^n)=dim Range(T^{n+1})$ implies that $rank(A^n)=rank(A^{n+1})$.

Title: Show that rank($A^{n+1}$) = rank($A^n$)


问:
Bundles of relative scalars
Let $M$ be a smooth manifold with an atlas $\{(U_i,\psi_i)\}$. If $x\in U_i\cap U_j$, let $a_{ji}(x)$ denote the Jacobian matrix at $\psi_i(x)$ of the smooth map $\psi_j^{-1}\circ \psi_i:\psi_i(U_i\cap U_j)\to \psi_j(U_i\cap U_j)$. Then $a_{ji}$ is a smooth map $U_i\cap U_j\to GL(n,\Bbb R)$ and satisfies the cocycle condition $a_{ki}=a_{kj}a_{ji}$.

Now define $g_{ji}=a_{ji}^m$, where $m$ is a fixed positive integer, and let $E\to M$ be a smooth vector bundle that has the transition functions $\{g_{ji}\}$. Then $E$ is called the the bundle of relative scalars over $M$ of weight $m$, according to Steenrod's The Topology of Fibre Bundles, section 6.

I see that if $m=1$, then $E\to M$ is the tangent bundle of $M$. But for $m>1$, what is $E$? I mean, I haven't seen these kind of definitions in textbooks about smooth manifolds, for example, Lee's Introduction to Smooth Manifolds. But maybe these bundles have another familiar names nowadays, I hope.

答:
I think you're misinterpreting Steenrod's definition. (Admittedly, his exposition is hard to follow on this point.) What he calls the bundle of relative scalars of weight $\boldsymbol w$ is the rank-$1$ vector bundle with transition functions $\det(a_{ij})^w$, where $(a_{ij})$ are the transition functions for the tangent bundle as you defined them above.

Modern mathematicians usually call this the bundle of densities of weight $\boldsymbol w$. See my answer to this question for a detailed explanation.

Title: Bundles of relative scalars


问:
Find the solution for Cauchy problem using transformation
I'm learning differential equations for the first semester and I'm not sure how to even begin with the following problem:

Find the solution for Cauchy problem using transformation:

$\left\{\begin{matrix}
(1-t^2)x''-tx'+n^2x=0 & n \in \mathbb{Z}\\
y(0)=1 & y'(0)=0
\end{matrix}\right.$

Could you please help me?

答:
In fact this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0220.pdf

Let $s=\int\dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}t$ ,

Then $\dfrac{dx}{dt}=\dfrac{dx}{ds}\dfrac{ds}{dt}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}$

$\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left(\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}\right)=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d}{dt}\left(\dfrac{dx}{ds}\right)+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d}{ds}\left(\dfrac{dx}{ds}\right)\dfrac{ds}{dt}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d^2x}{ds^2}\dfrac{1}{\sqrt{1-t^2}}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{1-t^2}\dfrac{d^2x}{ds^2}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}$

$\therefore(1-t^2)\left(\dfrac{1}{1-t^2}\dfrac{d^2x}{ds^2}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}\right)-\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}+n^2x=0$

$\dfrac{d^2x}{ds^2}+\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}-\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}+n^2x=0$

$\dfrac{d^2x}{ds^2}+n^2x=0$

$x=C_1\sin ns+C_2\cos ns$

$x=C_1\sin(n\sin^{-1}t)+C_2\cos(n\sin^{-1}t)$

$x(0)=1$ :

$C_2=1$

$\therefore x=C_1\sin(n\sin^{-1}t)+\cos(n\sin^{-1}t)$

$x'=\dfrac{C_1n\cos(n\sin^{-1}t)}{\sqrt{1-t^2}}-\dfrac{n\sin(n\sin^{-1}t)}{\sqrt{1-t^2}}$

$x'(0)=0$ :

$C_1n=0$

$C_1=0$

$\therefore x=\cos(n\sin^{-1}t)$


答:
Hint:

$$x\in Sp(A)\cap Sp(B)\implies \exists\,a_1,...,a_k\in A\;,\;b_1,...,b_m\in B\,,\,c_1,...,c_{k+m}\in \Bbb F\;\;s.t.$$

$$x=\sum_{i=1}^kc_ia_i=\sum_{i=1}^mc_{k+1}b_i\implies c_1a_1+\ldots c_ka_k-c_{k+1}b_1-\ldots c_mb_m=0\implies$$

$$\implies c_r=0\;\;\forall r=1,\ldots,m\;,\;\text{since}\;a_i,b_k\in A\cup B\ldots$$


问:
Evaluate $\lim_{x\to\ 0}\left(\frac{\cot(x)(1-\cos(3x)^{2})}{x^{2} + 5x}\right)$
I tried find a this limit:

$$\lim_{x\to\ 0}\left(\frac{\cot(x)(1-\cos(3x)^{2})}{x^{2} + 5x}\right)$$

I came to this:

$$\lim_{x\to\ 0}\frac{\frac{\cos(1-\cos(3x)^{2})}{\sin(1-\cos(3x)^{2})}}{(x^{2} + 5x)} = \frac{\frac{\cos(1)}{1-\cos(3x)^{2}}}{(x^{2} + 5x)} =  \lim_{x\to\ 0}\frac{1}{\sin(3x)^{2}(x^{2} + 5x)} = \infty $$

But book says that the correct answer is $\frac{9}{5}$. How should I solve it?

答:
Originally the numerator was "${ctgx(1-cos(3x)^{2})}$," i.e. $\cot(x)\cdot (1-\cos^2(3x))$. Then the limit is indeed $9/5$, as can be seen by invoking $\lim_{\theta\to 0}\sin(k\theta)/\theta=k$ for constant $k$:
$$
\lim_{x\to 0}\frac{\cot(x)\sin^2(3x)}{x^2+5x}
$$
$$
=\lim_{x\to 0}\frac{x}{\tan(x)}\cdot \frac{\sin^2(3x)}{x^2}\cdot \frac{1}{x+5}=1\cdot 9\cdot \frac{1}{5}
$$


问:
How do I solve a first order differential equation of the following form
If I had a differential equation $x'(t) + \frac{x(t)}{t} = e^{t^2}$

And I have a question saying solve the following first ODE for $x(t)$ how would I go about doing this?

答:
This is a linear differential equation so first let's find the solution to the homogenous equation. We see that $x(t)=A\exp(-\ln(t))=\frac{A}{t}$ where $A$ is a constant. Using a well-known method, we suppose that $A=A(t)$, so we have $x(t)=\frac{A(t)}{t}$. Using the equation, we get$$
A'(t)=t \exp(t^2)
$$
so $A(t) = \frac{\exp(t^2)}{2}+C$ where $C$ is a constant. Thus
$$
x(t)=\frac{\exp(t^2)}{2t}+\frac{C}{t}.
$$


问:
How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$?
I need to prove something like that:

For $a,b,c>0$  prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.

I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.

Can anyone help me?

答:
Hint: Using CS inequality,

$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$


答:
$$x'(t) + \frac{x(t)}{t} = e^{t^2}\tag1$$
It is a first order linear ODE. Integrating factor $~=e^{\int(1/t)dt}=e^{\ln t}=t~.$

Multiplying both side of $(1)$ with the integrating factor, we have
$$tx'(t) + x(t)= te^{t^2}\tag2$$Integrating,
$$\implies d\left(tx\right)=\int te^{t^2}~dt$$
$$\implies tx=\dfrac 12\int e^{t^2}~d(t^2)$$
$$\implies tx=\dfrac 12 e^{t^2}+c$$
$$\implies x(t)=\dfrac {e^{t^2}}{2t}+\dfrac{c}{t}$$where $c$ is integrating constant.

Title: How do I solve a first order differential equation of the following form

Body:
If I had a differential equation $x'(t) + \frac{x(t)}{t} = e^{t^2}$

And I have a question saying solve the following first ODE for $x(t)$ how would I go about doing this?


答:
If $k \ge 1$, then $P(X=k \cap X \ge 1) = P(X=k)$.

If $k = 0$, then $P(X=k \cap X \ge 1) = 0$.

Title: If $X\sim\text{Bin}(3,2)$, find $\Pr(X=k|X\ge 1)$

Body:
I'm having some trouble with this. I wrote

$$\Pr(X=k|X\ge 1)=\frac{\Pr(X=k\cap X\ge1)}{\Pr(X\ge1)},$$

but I'm stuck in calculating the quantity $\Pr(X=k\cap X\ge1)$.


问:
Prove that $\;0\le-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\le\ln n$
Prove the following property.

If $\;n\in\mathbb{N}\;,\;p_1,p_2,\ldots,p_n>0\;$ and $\;p_1+p_2+\ldots+p_n=1\;,\;$ then
$0\le-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\le\ln n\;.$

Could you tell me if my proof is correct?

Is it possible to prove it in a simpler way?

My proof:

Since $\;f(x)=\ln x:\left]0,+\infty\right[\to\mathbb{R}\;$ is a concave function, it follows that

$f(t_1x_1+t_2x_2+\ldots+t_nx_n)\ge t_1f(x_1)+t_2f(x_2)+\ldots+t_nf(x_n)$

for all $\;x_1,x_2,\ldots,x_n\in\left]0,+\infty\right[\;$ and for all $\;t_1,t_2,\ldots,t_n\ge0\;$ such that $\;t_1+t_2+\ldots+t_n=1\;.$

By letting $\;x_i=\dfrac{1}{p_i}\in\left]0,+\infty\right[\;,\;t_i=p_i>0\;$ for any $\;1\le i\le n\;,\;$ we get that

$f\left(p_1\dfrac{1}{p_1}+p_2\dfrac{1}{p_2}+\ldots+p_n\dfrac{1}{p_n}\right)\ge p_1f\left(\dfrac{1}{p_1}\right)+p_2f\left(\dfrac{1}{p_2}\right)+\ldots+p_nf\left(\dfrac{1}{p_n}\right)\;,$

$f\left(n\right)\ge p_1f\left(\dfrac{1}{p_1}\right)+p_2f\left(\dfrac{1}{p_2}\right)+\ldots+p_nf\left(\dfrac{1}{p_n}\right)\;.$

Consequently,

$\ln n\ge p_1\ln\left(\dfrac{1}{p_1}\right)+p_2\ln\left(\dfrac{1}{p_2}\right)+\ldots+p_n\ln\left(\dfrac{1}{p_n}\right)\;,$

$\ln n\ge-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\;,\;$

$\color{brown}{-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\le\ln n}\;.$

Since $\;p_1,p_2,\ldots,p_n>0\;$ and $\;p_1+p_2+\ldots+p_n=1\;,\;$ it results that $\;p_1,p_2,\ldots,p_n\in\left]0,1\right]\;,\;$ hence

$\color{brown}{-p_1\ln p_1-p_2\ln p_2-\ldots-p_n\ln p_n\ge0}\;.$

答: Your proof is correct. I don't see a simpler way than this to prove that.</p>


答:
Let $AB = 169y$, $BC=119y$.

$$AC^2=AB^2-BC^2\Rightarrow y^2=\ldots$$

$$BE^2=CE^2+BC^2\Rightarrow BE^2=\ldots$$

$$DE\cdot BE = AE\cdot CE \Rightarrow DE=\ldots$$

Body:
Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$. Given that $AE = 169$ cm and $EC = 119$ cm, find $ED$.

What I Tried: Here is a picture :-

We have $AE = 169, EC = 119$.

Let $AB = r$ , $DE = x$ , $AD = y$ , $BC = z$ , $BE = a$.

Now, applying Pythagoras Theorem gives all these equations one by one :-
$$(i) x^2 + y^2 = 169^2$$
$$(ii) z^2 + 119^2 = a^2$$
$$(iii) (x+a)^2 + y^2 = R^2$$
$$(iv) 288^2 + z^2 = R^2$$
We also have $\Delta CEB \sim \Delta DAB \sim \Delta DAE$ . They give information as :-
$$(i) \frac{CE}{DA} = \frac{CB}{DB} = \frac{EB}{AB}$$
$$(ii) \frac{DE}{EC} = \frac{DA}{CB} = \frac{AE}{EB}$$

The Problem is that all these don't really make me find the value of $ED$, or I am unable to find it by using them appropriately in some way.

Can anyone help me? Thank You.


问:
Prove that $(n! +n) \leq (n+1)!$
How can I prove that $(n! +n) \leq (n+1)!$ given that $n \geq 0$?

答:
If $n \ge 1$, this is equivalent to
$$
1 \le n(n-1)!
$$
which is clear.

Title: Prove that $(n! +n) \leq (n+1)!$


Title: How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$?

Body:
I need to prove something like that:

For $a,b,c>0$  prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.

I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.

Can anyone help me?

Body:
Hint:

$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$


答: let $Q$ be the foot of the perpendicular from $E$ to $AB$ then  $EC=EQ$ ,  $AQ^2 = AE^2-EQ^2 $ and $$\frac{AQ} {EQ} =\frac{AC} {CB}$$  You can calculate $EB$ by pythagoras from $\triangle EBC$ , and finally $$AE \cdot EC= EB \cdot DE$$

Title: Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$.


问:
Finding a limit at negative infinity
$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$

Hello,

I know that the answer to this is $- \infty$, but I'm having trouble showing it. Can anyone please point me in the right direction?

Thanks!

答:
$$\lim_{x \to -\infty}(x^2 + 2 x^7)=-\infty$$
For any $M>0$ there exists an $N>0$ such that if $x<-N$ then $f(x)<-M$

That is $x^2 + 2 x^7<-M$.

For $x<0$ we have $2x^7<x^2+2x^7<-M$ therefore taking $N=\sqrt[7]{\frac{M}{2}}$ if $x<-N$ we have $f(x)<-M$

Title: Finding a limit at negative infinity


问:
Figuring out function composition
If I know $f(x)$ and $f(g(x))$,

how do I figure out what is $g(x)$?

I can solve already some simple cases, but I'm looking for a general strategy.

Particularly in harder cases when just trying to figure it out is hard.

答:
Set $h$ so that $f(g(x))=h(x)$.

To get $g(x)$, $f$ is disturbing. $g$ is composited with $f$ from the left. You need the inverse operation. The inverse operation of composition from the left is the left inverse.

Let $f^{-1}$ denote the left inverse of $f$:

$$g(x)=f^{-1}(h(x))$$
$\ $

Let's presuppose that $f$ and $g$ are functions.

If $f$ is bijective, you can set the inverse function (compositional inverse / inverse) of $f$ for $f^{-1}$.

If $f$ is not bijective, you have to choose suitable partial inverses of $f$ for $f^{-1}$. The equation above can give you $g$ as a piecewise given function then.

Title: Figuring out function composition


问:
How to decide on the standard error for difference of standard deviations

In this problem the null hypothesis is that $\sigma_1 = \sigma_2$ then why in the standard error a common sigma value is not used? It can he found as $\sigma = ({n_{1}s_1^{2}+ n_{2}s_{2}^2})/({n_{1}+n_{2}})$
And also while testing for difference of sample means i.e $H_0: \mu_1=\mu_2$, does it always mean that $\sigma_1=\sigma_2$? Or does that happen only when the samples are drawn from the same population? Basically what’s the difference from the null hypothesis that $H_0: \mu_1=\mu_2$ and the case when samples are drawn from same population? Is the testing done for the significant difference of population mean or sample mean? also in difference of proportions, under the hypothesis $P_1=P_1$ the standard error also has one term $P_1=P_2=P$ why doesn’t that happen here for sigma? Or let me put it this way, under the null hypothesis $P_1=P_1$ the standard error also has one term $H_0: P_1=P_2=P$ Is this $P$ same as the one calculated from $P=(n_1p_1+n_2p_2)/(n_1+n_2)$ where $p_i$ is the sample proportion.

EDIT: I referred another text and here it says, under null hypothesis $H_0: \sigma_1=\sigma_2$ and continues to say that, i.e sample standard deviations don’t differ significantly. But are not $\sigma_1$ and $\sigma_2$ population standard deviations?

答:
The answer to your first question appears to be that a "common sigma value" was in fact used in  the original definition of the $\ z\ $ statistic, given as
$$
z=\frac{s_1-s_2}{\sigma\sqrt{\frac{1}{2n_1}+\frac{1}{2n_2}}}\ ,
$$
on page $20$ of Statistics for Engineers, in which the problem described under Example $2$ of your first illustration appears as Problem $6$ on page $24$.  However, when you replace $\ \sigma\ $ with its estimate $\ \sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}}\ $ you get the following estimate $\ z_e\ $ for the value of $\ z\ $:
\begin{align}
z_e&=\frac{s_1-s_2}{\sqrt{\left(\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}\right)\left(\frac{1}{2n_1}+\frac{1}{2n_2}\right)}}\\
&=\frac{s_1-s_2}{\sqrt{\frac{s_1^2}{2n_2}+\frac{s_2^2}{2n_1}}}
\end{align}
Please note the following:

In your question, you've omitted the square root sign from $\ \sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}}\ $, which is a random quantity whose value is only an estimate of $\ \sigma\ $, and not equal to it (except by a very lucky coincidence).

In the above expression for $\ z_e\ $  the indices on $\ s\ $ and $\ n\ $ in the fractions in the denominator
do not match.  In the examples given in the above-cited book, this expression for $\ z_e\ $ gets replaced without explanation by $\ \frac{s_1-s_2}{\sqrt{\frac{s_1^2}{2n_1}+\frac{s_2^2}{2n_2}}}\ $, which corresponds to taking $\ \sqrt{\big(n_2s_1^2+n_1s_2^2\big)\big/\big(n_1+n_2\big)}\ $ as an estimate for $\ \sigma\ $ instead of the much more natural $\ \sqrt{\big(n_1s_1^2+n_2s_2^2\big)\big/\big(n_1+n_2\big)}\ $.

As heropup alludes to in this answer (which I fully agree with), none of $\ s_1, s_2, s_1-s_2, z\ $, or either version of $\ z_e\ $, have an approximately normal distribution, even for very large samples from a normally distributed population—contrary to the presumptions of the above-cited book.  Thus, any calculation of significance levels based on the presumption that $\ z_e\ $ is normally distributed is completely unjustified and will almost certainly be wrong.

Update

If the samples are independently drawn from two homogeneous, normally distributed populations with the same variance (but possibly different means), then the true  distribution of the absolute value of the statistic $\ z\ $ defined in Example $2$ of the OP's first illustration can be expressed in terms of the $F$-distribution with $\ n_1-1\ $ and $\ n_2-1\ $ degrees of freedom.  This makes it easy to construct examples showing how the assumption that this statistic has a standard normal distribution can give completely erroneous results.

Take the case when $\ n_1=150,$ $n_2=250,$$\ s_1=14.0\ $, and $\ s_2=11.3\ $. Then
$$
z=\frac{14.0-11.3}{\sqrt{\frac{(14.0)^2}{300}+\frac{(11.3)^2}{500}}}\approx2.83\ .
$$
If $\ z\ $ had a standard normal distribution, this would give a $p$-value of
$$
P\big(|z|>2.83\big)=2{\cal N}(0,1;-2.83)\approx0.00465\ ,
$$
implying that the null hypothesis could be rejected at the $0.5\%$ level of significance.

To obtain the true distribution of $\ |z|\ $, rewrite $\ z\ $ as
$$
z=\frac{\frac{s_1}{s_2}-1}{\sqrt{\frac{1}{2n_1}\left(\frac{s_1}{s_2}\right)^2+\frac{1}{2n_2}}}\ .
$$
A little algebraic manipulation then shows that
\begin{align}
|z|\le x&\iff \left(1-\frac{x^2}{2n_1}\right)\left(\frac{s_1}{s_2}\right)^2-2\left(\frac{s_1}{s_2}\right)+1-\frac{x^2}{2n_2}\le0\\
&\iff\ell_\ell\le \frac{s_1}{s_2}\le\ell_u\ \ \ (\text{ provided }\ \ x^2<2n_1),
\end{align}
where
\begin{align}
\ell_\ell&=\frac{1-\sqrt{1-\left(1-\frac{x^2}{2n_1}\right)\left(1-\frac{x^2}{2n_2}\right)}}{1-\frac{x^2}{2n_1}}\ \ \text{ and}\\
\ell_u&=\frac{1+\sqrt{1-\left(1-\frac{x^2}{2n_1}\right)\left(1-\frac{x^2}{2n_2}\right)}}{1-\frac{x^2}{2n_1}}\ .
\end{align}
Under the hypotheses stated in the opening paragraph of this update, the statistic $\ \frac{n_1\big(1-n_2\big)}{n_2\big(1-n_1\big)}\left(\frac{s_1}{s_2}\right)^2\ $ has an $F$-distribution with $\ n_1-1\ $ and $\ n_2-1\ $ degrees of freedom, and so
\begin{align}
P\big(|z|\le x\big)=&\,P\left(\frac{n_1\big(1-n_2\big)\ell_\ell^2}{n_2\big(1-n_1\big)}\le\frac{n_1\big(1-n_2\big)}{n_2\big(1-n_1\big)}\left(\frac{s_1}{s_2}\right)^2\le\frac{n_1\big(1-n_2\big)\ell_u^2}{n_2\big(1-n_1\big)}\right)\\
=&\,F_{n_1-1,n_2-1}\left(\frac{n_1\big(1-n_2\big)\ell_u^2}{n_2\big(1-n_1\big)}\right)-F_{n_1-1,n_2-1}\left(\frac{n_1\big(1-n_2\big)\ell_\ell^2}{n_2\big(1-n_1\big)}\right)\ .
\end{align}
For the example immediately above $\ \frac{n_1\big(1-n_2\big)\ell_\ell^2}{n_2\big(1-n_1\big)}\ $ and $\ \frac{n_1\big(1-n_2\big)\ell_u^2}{n_2\big(1-n_1\big)}\ $ evaluate approximately to $\ 0.668\ $ and $\ 1.539\ $ respectively, giving
\begin{align}
P\big(|z|\le2.83\big)&\approx F_{149,249}\big(1.539\big)-F_{149,249}\big(0.668\big)\\
&\approx0.930\ ,
\end{align}
or $\ P\big(|z|>2.83\big)\approx0.070\ $. Thus, the true $p$-value is more than an order of magnitude larger than the erroneous value obtained by assuming $\ z\ $ has a standard normal distribution, and implies that the null hypothesis cannot be rejected even at a $5\%$ level of significance.


答:
Also if $n=0$ it is obvious ($1\le 1$),

for $n\ge 1$:
$$n!+n\le n!+n!=n!2\le (n+1)!$$

Title: Prove that $(n! +n) \leq (n+1)!$

Body: How can I prove that $(n! +n) \leq (n+1)!$ given that $n \geq 0$?


问:
Proof that a $2^n . 2^n$ Chessboard with a Missing Square can be covered by L-Shaped Tiles - Missing Step
I understand the recursive solution to tiling a $2^n. 2^n$ chessboard with a missing square using L-shaped tiles such as these:

using the following method to divide the original problem into subproblems:

However the proof depends on the fact that a chessboard containing $2^n . 2^n - 1$ squares can be covered by the L-shaped tiles. Whereas I can see that the number of squares that in such a defective chessboard is a multiple of $3$, I can't see why it is necessarily the case that all these squares can be covered by the L-shaped tiles, regardless of the position of the missing square.

Can anyone please help me to understand what this is the case?

答:
This is an inductive proof. The base cases are $n = 0$ which is the first image and $n = 1$ which is the next three images (the case where the missing tile is in the bottom right corner is missing - ah well.) The key to the $n \to n + 1$ step is the last image, but that can be formalized.

In particular, the $n + 1$ case can be reduced to the $n$ case by dividing the board into four quadrants, and placing a single L-shaped tile in the pictured location so as to remove a square from those three quadrants that are not already missing a square.

Each of the four quadrants can then be tiled using the inductive hypothesis.


问:
$\int \frac{dx}{(x^4 + 1)^2}$
What would be a relatively simple method for computing the indefinite integral below?

$\displaystyle \int \frac{dx}{(x^4+1)^2}$

Furthermore, how would one evaluate the following, possibly by detouring the computation of the indefinite integral?

$\displaystyle\int\nolimits_{-\infty}^\infty \frac{dx}{(x^4+1)^2}$

Note: The resolved value of the definite integral (according to wolfram alpha) is $\displaystyle \frac{3\pi}{4\sqrt{2}}$

答: The definite integral can be done using residues.  Note that the zeros of $(z^4+1)^2$ in the upper half plane are at $\frac{\pm 1 + i}{\sqrt{2}}$, both with multiplicity 2.  The residues are $\frac{3 \sqrt{2}}{32}(1-i)$ at $\frac{-1+i}{\sqrt{2}}$ and $\frac{3 \sqrt{2}}{32}(-1-i)$ at $\frac{1+i}{\sqrt{2}}$ (note that the residue of $\frac{1}{((x-r) f(x))^2}$ at $x=r$, where $f$ is analytic at $r$ with $f(r) \ne 0$, is $\frac{-2 f'(r)}{f(r)^3}$).   So the integral is $2 \pi i \frac{3 \sqrt{2}}{32} (1-i -1 - i) = \frac{3 \pi \sqrt{2}}{8}$.  </p>

Title: $\int \frac{dx}{(x^4 + 1)^2}$


答:
All does depend on $n$. So this is a nice case for complete induction over n.

n=1: A=a real or complex and nonzero. $Rank(A)=1=Rank(A^{n=1})=n=1=rank(a^2)=rank(A^2)=rank(A^{n+1})$

For $n$ natural the hyptheses is $true$.

For $n+1$ A change exactly in one row or column to the case $n$. This row or column can be either but not linear dependent on the other making up the A for $n$. The implies implicitly the at least one element in the column or row is nonzero exactly in the dimension added to A for $n$.

We can now uses some of the equivalent definitions for the $rank$ of a square matrix. With restrict to generality the added row or column has only one nonzero element. This acts as a factor for example in the determinate development or is a new eigenvalue or the matrix A for $n+1$. So the determinante is nonzero at least in that development because we have a nonzero value and the knowledge that the determinant of our A for $n$ is nonzero and $rank(A)=n$.

The main idea for the induction step is ring of matrice rank or Rankconserse by multiplication of a nonsingular matrix in general a $A$ itself specially.
Matrices with nonzero $rank$ preserve the rank under multiplication. The multiplication under consideration is commutative because we only multiply A. That is another indicator for our hypotheses for $n+1$. Eigenvalues and Schur decomposition are closely related. One of the matrices in the Schur decomposition is an upper triangle matrice. So increasing the dimension from $n$ to $n+1$ simply adds another last if the last row and column in a unite vector with only a value in the new dimension.

The Schur decomposition is equivalent to that the matrix $∈ℂ^{+1×+1}$ has the property relying on the matrix $∈ℂ^{×}$. The matrice of $rank$ from a group and can transformed into each other under conservation of the $rank$. And the prove is done.

Title: Show that rank($A^{n+1}$) = rank($A^n$)


问:
Find a counter example for the convergence of a sequence that has a property close from cauchy
I want to find a counter example for this statement : Let ($x_n$) be a sequence such as $\forall m > 0,
lim_{n \rightarrow \infty} (x_{n+m} -x_n) = 0$, also ($x_n$) is bounded. Then the sequence $(x_n)$ converges.

I think that the statement is false so I tried to find a counter example.I tried with cosinus , sinus , tangent but each time there is a problem.  Maybe something with the logarithmic function can work ? But I assume this function is nothing usual and need a complex construction.

Can you help me find it please ?

答: Let $x_n=\sin(2\pi \ln(n))$. Recall that $|\sin(a)-\sin(b)| \le |a-b|$, so $\forall m \in \Bbb{N}$ $$\lim_{n \rightarrow \infty}|x_{n+m}-x_n| \le \lim_{n \rightarrow \infty}|2\pi \ln(n+m)-2\pi \ln(n)|=2\pi \lim_{n \rightarrow \infty}|\ln(1+\frac{m}{n})|=0$$ and $\{x_n\}$ does not converge(this derives from the fact that $y_n=\{\ln(n)\}$ is dense in $[0,1]$).


问:
Find Taylor polynomial of an integral function
I ran into this problem and I'm absolutely clueless. The problem is:

Let $f$ be: $$f(x)=\int_0^{\ln(1+2x)} \frac{e^t}{1+t^2}\;dt$$ Find the Taylor polynomial of order $1$ around $x=0$.

The only idea I've had is that you could write $f(x)= F(ln(1+2x))-F(0)$, and then I could use that $f'(x)=F'(ln(1+2x))-F'(0)$, but whenever I try that I always just get $0$.

Any help is greatly appriciated.

答:
The taylor polynomial formula (around $x=0$) is:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n$$
if we only want the first order polynomial, then we will have the first two terms $x^0,x^1$ and the rest of the terms are neglected and will form our error. This gives:
$$f(x)\approx\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x=f(0)+f'(0)x\tag{1}$$
now we need to work out these values (the first is given by showing the integrand tends to 1 for $t\to0)$:
$$f(0)=\int_0^{\ln(1)}\frac{e^t}{1+t^2}dt=0$$
$$f'(x)=\frac d{dx}\left[\ln(1+2x)\right]\frac{e^{\ln(1+2x)}}{1+\ln^2(1+2x)}=\frac{2(1+2x)}{(1+2x)(1+\ln^2(1+2x))}=\frac{2}{1+\ln^2(1+2x)}$$
$$\therefore f'(0)=\frac{2}{1+\ln^2(1)}=2$$
and so putting these values into our first equation we get:
$$f(x)\approx 2x$$


答:
$\frac{1}{n+1} + \frac{n}{(n+1)!} \leq \frac{1}{n+1} + \frac{n}{(n+1)} = 1$

Therefore since $\frac{1}{n+1} + \frac{n}{(n+1)!}= \frac{n!}{(n+1)!} + \frac{n}{(n+1)!} $ your inequality follows immediately.

Title: Prove that $(n! +n) \leq (n+1)!$


问:
Number of ways to write $2n$ as sum of two primes is unbounded
For all $k>0$, we are to show that

there exists an even number $2n > 0$ that can be written as sum of two primes in at least $k$ ways.
($8=3+5=5+3$ counts as one way)

I tried to use pigeonhole principle to "fill" odd numbers in the lower half of $2n$ with primes, i.e. to show $\pi(2n)-[\frac{n}{2}]$ can be arbitrarily large, but it turns out to diverge to negative infinity instead.

Of course, requiring the entire half to be primes is an overkill, but I don't know how to rephrase the problem into a manageable form.

It probably has something to do with Prime Number Theorem, but I am not sure.

Thank you for any help.

答:
There are $\pi(n)-1 \approx \pi(n)$ odd primes less than $n$. Summing them two at a time, we obtain $\frac{(\pi(n))!}{(\pi(n)-2)!2!}=\frac{(\pi(n))\cdot (\pi(n)-1)}{2}$ even sums, all of which are smaller than $2n$. In addition, there are $\pi(n)$ occurrences of sums of the form $p_i+p_i$ not generated by the pairwise summing, and including those with the previous result, we obtain $\frac{(\pi(n))(\pi(n)+1)}{2}\approx \frac{(\pi(n))^2}{2}$ even sums to consider. There are $n$ even numbers smaller than $2n$

If for a chosen $n$ and any $k$, the number of sums exceeds $kn$, then by the pigeon hole principle, at least one even number $\le 2n$ is expressible as a sum of two odd prime numbers in at least $k$ different ways. This occurs with certainty when $\frac{(\pi(n))^2}{2}>kn$ or $(\pi(n))^2>2kn$

According to PNT, $\pi(n) \approx \frac{n}{\ln(n)}$. Substituting, the necessary condition becomes $\frac{n}{\ln^2(n)}>2k$

Since $\frac{n}{\ln^2(n)}$ grows without bound, it is always possible to choose an $n$ large enough to ensure that some even number smaller than $2n$ can be expressed as the sum of two odd primes in at least $k$ different ways.


答:
With this kind of problems, the standard approach is to start with finding a solution to the homogenous differential equation, and then add what's called a 'particular solution' for the inhomogenous part. If you're not familiar with this solving method, you should learn more about it very soon, it's quite common :)

For the homogenous equation:
$x'(t) = -\frac{x(t)}{t}$
I used the so called 'separation of the variables' method:
$\frac{dx}{x(t)} = -\frac{dt}{t}$

And came to the homogenous solution:
$x(t) = \frac{c}{t}$
for some constant $c \in \mathbb{R}$.

Then, to find a particular solution (you only need one of them), I used the method of the variation of constants:
$x(t) = \frac{1}{t}\left(c + \int_{0}^{t}x\cdot e^{x^2}dx\right)$

And finally found the following general solution to your ODE problem:
$x(t) = \frac{1}{2}\cdot\frac{e^{t^2}}{t} + \frac{c}{t}$

So, I hope this will help you to understand the steps to follow! And I wrote my results so that you can check on your way :)


问:
Equation of a hyperplane in n- dimensions.
I know that the equation of a hyperplane in n-dimensions is given by:
$$w^Tx+w_0=0$$
Where $w$ is a vector that is perpendicular to the surface of the hyperplane and $w_0$ is a constant.
I also know that if $w_0=0$, then the plane passes through the origin.
My question is, what is the physical significance of the constant term $w_0$. Is it the distance of the plane to the origin? If not then what is its physical significance?

答:
Assume that the length of $w$ is $1$.

Since $w$ is perpendicular to the hyperplane, the orthogonal projection of the origin to the hyperplane is $\lambda w$ for a unique real number $\lambda$.

The length of $\lambda w$ is $|\lambda|$, this is the distance of the origin to the hyperplane, and since $\lambda w$ is assumed to be on the hyperplane, it satisfies
$$w^T(\lambda w)+w_0=0\implies \lambda=-w_0$$
since $w^Tw=\|w\|^2=1$ by assumption.


答:
Another argument for not convergence of sin(ln(n)) would be to note that ln(n) is dense on [0,1] mod 1. This is easy to see since ln(n+1) - ln(n) tends to 0 monotonically.
Lastly, since sin(x) is continuous sin(ln(n)) is dense in [0,sin1].</p>

Title: Find a counter example for the convergence of a sequence that has a property close from cauchy

Body:
I want to find a counter example for this statement : Let ($x_n$) be a sequence such as $\forall m > 0,
lim_{n \rightarrow \infty} (x_{n+m} -x_n) = 0$, also ($x_n$) is bounded. Then the sequence $(x_n)$ converges.

I think that the statement is false so I tried to find a counter example.I tried with cosinus , sinus , tangent but each time there is a problem.  Maybe something with the logarithmic function can work ? But I assume this function is nothing usual and need a complex construction.

Can you help me find it please ?


问:
$f ( n ) = n ^ { 2 } + \sqrt { n }$. Whether $2 ^ { 2 ^ { f ( n ) } } = \Omega ( g )$

Let $f : \mathbb{N} \rightarrow \mathbb{R}$ be the function $f ( n ) = n ^ { 2 } + \sqrt { n }$. Determine whether the following statement is true or false, providing a proof for your answer.

$2 ^ { 2 ^ { f ( n ) } } = \Omega ( g )$, where $g(1) = 1$, $g(2) = 2$, and $g(n) = [g(n - 1)] ^2 + [g(n - 2)] ^2$ for $n \ge 3$.

I think the statement is false as $g(n)$ seems to grow faster, but how can I formulate a proof for that?

Body:
The first seven values are
$$
\begin{array}{r|r|r}
n  & g(n) &2^{2^{f(n)}}\\
\hline
 1 & 1. & 16. \\
 2 & 2. & 6.8\times 10^{12} \\
 3 & 5. & 1.02\times 10^{512} \\
 4 & 29. & 1.6\times 10^{78913} \\
 5 & 866. & 9.6\times 10^{47586437} \\
 6 & 750797. & 2.9\times 10^{112995071916} \\
 7 & 5.63697\times 10^{11} & 2.\times 10^{1060547355989709} \\
\end{array}
$$
unless you meant $2^{f(n)}$ there are no doubts. Even in this case $2^{f(n)}$ grows faster


问:
Solution set of $\frac x{x+2}>0\land\frac{x+1}{x+2}<1$
Okay people, $\Im$ stupid. I have been working on this for literal 2 days but still I can't understand.

Here is the question:
$$\frac x{x+2}>0\land\frac{x+1}{x+2}<1$$
What's the solution set? [Answer is $(0,\infty)$]

I found first inequality means $x > 0$ and second one says $x > -2$ but that doesn't make the solution set given by my lecturer which is $(0,\infty)$.

答:
ok so lets first consider the first inequality:
$$\frac{x}{x+2}>0\tag{1}$$
for this to be true either $x>0$ so the top and bottom are both positive, or we can have $x<-2$ and so the solution for this inequality would be $x\in(-\infty,-2)\wedge(0,\infty)$.

Now for the second:
$$\frac{x+1}{x+2}<1\tag{2}$$
$$1-\frac{1}{x+2}<1$$
$$-\frac 1{x+2}<0$$
$$\frac{1}{x+2}>0$$
and it is clear from this that the solution is $x>-2$ and so: $x\in(-2,\infty)$. Now for both to be simultaneously true we need to find where these domains overlap, which would be $x\in(0,\infty)$


问:
Continuous function that is only differentiable on irrationals
Can you help me finding a function $f : \mathbb{R} \rightarrow \mathbb{R}$ that is continuous in $\mathbb{R}$ and differentiable at $x$ iff $x \notin \mathbb{Q}$  ?

Thank you very much !

答:
Interpreting the question as asking for a continuous function that is differentiable only at the irrationals, we might perturb one of Weierstrass's functions and look at

$$f(x) = \sum_n \frac{|\sin (n \pi x)|}{n^3}$$

It's easy to see it's continuous, but a bit more challenging to see about its derivative.


问:
Prove that $x^2> \frac1e-1 $ and $e \gt 0$ $\implies x\lt -\sqrt{\lvert \frac 1e-1\rvert} $ or $ x\gt \sqrt{\lvert \frac 1e-1\rvert} $

$x^2> \frac1e-1 $ and $e \gt 0$ $\implies x\lt -\sqrt{\lvert \frac 1e-1\rvert} $ or $    x\gt \sqrt{\lvert \frac 1e-1\rvert} $

In my text book I have come across this implication. I am not able to justify its validity. In my opinion It should be as below

Given : $x^2> \frac1e-1 $ and $e \gt 0$

Case 1: $(\frac 1e-1) \gt0$ then $x^2> \frac1e-1 \implies \lvert x\rvert^{2} \gt$ $(\sqrt{\lvert \frac 1e-1\rvert})^2 \implies \lvert x\rvert \gt (\sqrt{\lvert \frac 1e-1\rvert})$

Case 2: $(\frac 1e-1) \lt0$ then $x^2> \frac1e-1 \implies  -x^2\lt -(\frac1e-1) \implies -x^2\lt \lvert \frac 1e-1\rvert$

I am not able to reach the given implication.

Any help towards this will be appreciated.

答:
I think that this statement is false: Take for example $e=4/3$, so the statement is
$$
x^2>-\frac{1}{4}\ \Rightarrow\ x<-\frac{1}{2}\ \text{or}\ x>\frac{1}{2}
$$

$x=0$, for instance, is a counterexample.

Regarding the reference: The proof is not quite accurate. Given $\epsilon>0$ you need to prove that there is $K$ such that if
$x<K$, then $|\frac{1}{1+x^2}-0|<\epsilon$, that is
$$
x<K\Rightarrow \frac{1}{1+x^2}<\epsilon.
$$
It should be separated into two cases: if $1<\epsilon$, then you may take $K=0$, since $\frac{1}{1+x^2}\leq 1<\epsilon$ for every $x$. If $0<\epsilon\leq 1$, then as indicate in the reference, you may take $K=-\sqrt{\frac{1}{e}-1}$.


问:
Need help with a double integral in a triangular region
I'm trying to learn some maths in my own time. I am 13 and trying to learn some more maths on my own.

Can someone help me with question? My teacher could not :(

$\iint_{D}-sin(y+x) dx dy$

Where D is the triangle with vertices (0,0), (1,2) and (2,1).

From where I read online it says I should draw a triangle, but I don;tt know how this helps me...

答: One way to do it is to draw the rectangle $0\leq x,y\leq2$.  This is dissected into four triangles.  It's easy to integrate over the rectangle and also over each of the three right triangles, and subtraction gives the integral over the triangle you want.


问:
n points sampled uniformly from $[-1,1]^2$, how many lie within the unit circle?
I was looking at the expected output of an algorithm and translated it into this equivalent problem:

Consider $n$ points sampled uniformly from $[-1,1]^2$ and let $A=\frac{4Z}{n}$ where $Z$ is how many of these $n$ points lie within the interior and boundary of the unit circle. Find $\mathbb{E}[A]$.

I tried running the algorithm but I fail to recognize if the output value is approximating a familiar number. How does one compute for this expected value?

答:
The probability that a single point lies in the unit circle is $p := \frac{\text{area of unit circle}}{\text{area of $[-1,1]^2$}}$.

$E[Z] = np$ (why?)

$E[A] = \frac{4}{n} E[Z]$ (why?)


问:
Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$.

Let a circle with diameter $AB$ have points $C$ and $D$ on same side so that $BD$ bisects $\angle CBA$. Chords $AC$ and $BD$ intersect at $E$. Given that $AE = 169$ cm and $EC = 119$ cm, find $ED$.

What I Tried: Here is a picture :-

We have $AE = 169, EC = 119$.

Let $AB = r$ , $DE = x$ , $AD = y$ , $BC = z$ , $BE = a$.

Now, applying Pythagoras Theorem gives all these equations one by one :-
$$(i) x^2 + y^2 = 169^2$$
$$(ii) z^2 + 119^2 = a^2$$
$$(iii) (x+a)^2 + y^2 = R^2$$
$$(iv) 288^2 + z^2 = R^2$$
We also have $\Delta CEB \sim \Delta DAB \sim \Delta DAE$ . They give information as :-
$$(i) \frac{CE}{DA} = \frac{CB}{DB} = \frac{EB}{AB}$$
$$(ii) \frac{DE}{EC} = \frac{DA}{CB} = \frac{AE}{EB}$$

The Problem is that all these don't really make me find the value of $ED$, or I am unable to find it by using them appropriately in some way.

Can anyone help me? Thank You.

答:
Note $ DE \cdot EB = AE \cdot EC =169\cdot 119 $, or
$$DE \cdot (DB -DE)  =169\cdot 119$$
Substitute  $$DB = AB \cos x= AB \cdot\frac{\sin 2x}{2\sin x}=AB \cdot\frac{\frac{AC}{AB}}{2\frac{DE}{AE}}= \frac{144\cdot169}{DE}$$
to get
$$DE^2= 169\cdot(144-119)=169\cdot 25=(13\cdot5)^2$$
Thus, $DE=65$.


问:
Prove that $(n! +n) \leq (n+1)!$
How can I prove that $(n! +n) \leq (n+1)!$ given that $n \geq 0$?

答:
$n!+n\le n!+n\cdot n!=n!(1+n)=(n+1)!\;$

for all $\;n\in\mathbb{N}\cup\left\{0\right\}.$

Title: Prove that $(n! +n) \leq (n+1)!$


问:
Bayes Theorem containing 3 events
"Billy goes to his grandparents house to help out with chores, he's always running late though.

The probability he is late when he walks is 30%
The probability he is late when he runs is 15%
The probability he is late when he bikes is 5%
The probability he walks is 0.2
The probability he runs is 0.1
The probability he bikes is 0.7

What is the probability he will be late?

Billy ended up being late once again, what is the probability he walked, what is the probability he ran?"

I am stuck on how to set this up using Bayes theorem
Would it be: 0.2 * 0.1 * 0.7 for being late
Then,
0.3 * 0.2 / 0.3 * 0.2 + 0.2 (late that he didn't walk, which is 0.15 + 0.5) * 0.8 ( this is 0.7 + 0.1, where he ran or biked, not walked)

答:
The probability to be late is

$$\mathbb{P}[L]=0.3\times0.2+0.15\times0.1+0.05\times0.7=0.11$$

Thus, simply dividing, you get

$$\mathbb{P}[W|L]=\frac{0.3\times0.2}{0.11}\approx 54.55\%$$

$$\mathbb{P}[R|L]=\frac{0.15\times0.1}{0.11}\approx 13.64\%$$

Title: Bayes Theorem containing 3 events


问:
Using a symmetric matrix to solve a second order DE
Suppose I have the second order DE

$\frac{d^2}{dt^2}\underline{x}=-S\underline{x}(t)$

How could I use the knowledge that S is a symmetric matrix, $S=O^TDO$, to show that a general solution to the above is:

$\underline{x}(t) = [A\sin(\omega_it)+B\cos(\omega_it)]\underline{v_i}$

Any help would be greatly appreciated

答:
:) Hi d3V!

If your matrix is symmetric, it means, as you already mentioned, that you can diagonalize it, which is in such a case very useful.

For an $x(t) \in \mathbb{R}^n$, you have the following problem:
$$\dfrac{d^2x(t) }{dt^2}= -S x(t) \tag 1$$

And for such a problem in one dimension, we already know that a common solution is given by $x(t) = A'e^{ikx} + B'e^{-ikx}$, or equivalently $A \sin(kx) + B \cos(kx)$ for some $k \in \mathbb{R}$.

For our case, the diagonalisation is interesting, because otherwise, computing the exponential of matrices is really not funny. In fact, it holds:
$e^{-S x} = e^{-O^TDO x} = O^T e^{-D x} O$

Now, the space of solution to  $( 1)$, our initial problem, is given by:
$O^T e^{-D x} O = im(O^T e^{-D x})$

So you can come to the expression that you claimed, if you consider the vectors $v_i$ to be the columns of O^T, and the real values $\omega_i$ to be the diagonal elements of D.

With this elements, I hope you'll be able to finish your problem! :)

EDIT: the $\omega_i$ are the square root of the eigenvalues, not the eigenvalues themselves! sorry for the confusion ;)


问:
What is the meaning of this sequence of random variables?
Let consider a sequence of independent identically distributed random variables $(X_i)_{i\ge 1}$ which takes value in $\{1,...d\}$ with $p_k=\mathbb{P}(X_1=k)$, for $k=1,...,d.$ Let $N \sim Poisson(\lambda)$ which is independent from the $(X_i)_{i\ge 1}$. Then we define for $k=1,...,d$ : $N_k = \sum \limits_{i=1}^{N} \mathbb{1}_{\{X_i=k\}}$.

I have to find the probability distribution of the vector $(N_1,...,N_d)$.

But the fact is that I do not know how to interpret the random variable $N_k$. Indeed the last term of the sum is $\mathbb{1}_{\{X_N = k\}}$ contains the index $N$ which is a random variable.

For instance if we consider $k=1$ then $N_1 =\mathbb{1}_{\{X_1 = 1\}}+...+\mathbb{1}_{\{X_N = 1\}}$.

Thanks in advance !

答: Here is a concrete example which may help to understand the meaning of $N_k$: you have customers which come to a shop, the $i$th of them spends a certain amount of money $X_i$ between $1$ and $d$. The random variable $N$ may be the number of customers entering in the shop in a certain interval of time, say one hour. Then $N_k$ is the number of customers who have spent exactly $k$ say euros.


问:
Operators commuting with tensor product representations of SU(2)
I am currently investigating $SU(2)$ symmetric qubit systems. In the course of this work I proved the following theorem:

Let $S_n$ denote the permutation group of $n$ elements. For $\sigma\in S_n$ define the tensor product permutation operator $P_\sigma\in \operatorname{End}((\mathbb{C^2})^{\otimes n})$ as
$$P_\sigma: v_1 \otimes \cdots \otimes v_n \mapsto v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)}$$

Let $T \in \operatorname{End}((\mathbb{C^2})^{\otimes n})$ be a linear operator. Then

$$\left(\exists \{c_\sigma\}_{\sigma \in S_n} \subset \mathbb{C}: T=\sum_{\sigma\in S_n} c_\sigma P_\sigma\right)\iff \forall U \in SU(2): [T, U^{\otimes n}]=0$$

In words: $T$ commutes with the tensor product of the fundamental representation of $SU(2)$ if and only if $T$ is a linear combination of permutation operators.

The implication “$\implies$” is trivially true. However I also proved the other direction. Now my question is whether this theorem is already known? If yes, I would be grateful if someone could point me to some references.

答:
Congrats! You have rediscovered on your own a cornerstone of representation theory. This is the First Fundamental Theorem of Classical Invariant Theory for the group $SU(2)$ or $SL(2)$. The easy part $\Rightarrow$ was discovered by Arthur Cayley in:

A. Cayley, On linear transformations, Cambridge Dublin Math. J. 1 (1846) 104–122.

The hard part $\Leftarrow$ was first proved by Alfred Clebsch in:

A. Clebsch, Ueber symbolische Darstellung algebraischer Formen, J. Reine Angew.
Math. 59 (1861) 1–62.

You can find a proof in my two MathOverflow answers:

https://mathoverflow.net/questions/255492/how-to-constructively-combinatorially-prove-schur-weyl-duality/255853#255853

and

https://mathoverflow.net/questions/255492/how-to-constructively-combinatorially-prove-schur-weyl-duality/255938#255938


问:
How do I find condition on $f$?
Find the necessary and sufficient conditions for $f \in L^2(U)$ so that the equation

$-\sum_{i,j=1}^n (a^{ij}u_{x_i})_{x_j}-7u=f$ in $U$ and $u=0$ on $\partial U$.

has a weak solution $u \in H_{0}^1(U)$. Here the matrix $a^{ij}$ is symmetric, bounded, and uniformly elliptic.

I tried to solve this question by solving the bilinear form and Lax Milgram theorem but not able to reach the conclusion. Anyone can suggest some hints?

答:
First of all stating that $u = 0$ on $\partial U$ and stating that a solution has to lie in $H_0^1(U)$ is superfluous (you could either remover the first or state $u \in H^1(U)$). It does not really matter, just a small remark.
Second, I will assume that $\partial U$ is (piecewise) Lipschitz/smooth enough that we can use integration by parts.

Then, the bilinear form corresponding to this PDE is equal to
$$
B: H_1^0(U) \times H_1^0(U) \rightarrow \mathbb{R}; \quad B[u,v] = \int_U (\nabla u)^TA (\nabla v) - 7 u^2,
$$
where $(A)_{ij} = a_{ij}$.
Ignoring this $7$ for a moment we would have $B[u,u] \geq \theta  \left \lVert  u \right \rVert_{H^1(U)}$ (where $\theta$ is the constant from the uniform ellipticity) and we could use Lax-Milgram (even just Riesz representation Theorem as the bilinear form is symmetric). So we would have an answer, any $f$ suffices.

However, due to the $-7u^2$ term we are in a tougher situation.
Then the trick I know comes from Evans' book on PDE.
In particular, in Evans' PDE book he uses the Fredholm alternative for compact operators to answer this question:
$Lu = f $ has a (unique) solution for every $f \in L^2(U)$ if and only if $\langle f, v \rangle_{L^2(U)} = 0$ for all $v \in H_0^1(U)$ such that $Lu = 0$ (This is the second existence theorem, Theorem 4 of $\S$ 6.2.3 in my version, where I used that your A is symmetric).
In other words, when $f \perp Ker(\tilde{B}^*) = Ker(\tilde{B})$, where $\tilde{B}:H_0^1(U) \rightarrow (H_0^1(U))^*; u \mapsto B[u,\cdot]$ is the 'weak version of L'.

This is not a very explicit condition of $f$ since it is just a slight rearrangement of the statment that $f \in ran(\tilde{B})$ together with the usual theorem from functional analysis that $ran(A)^\perp = ker(A^*)$ for a bounded linear operator and the fact that $\tilde{B}$ has closed range, and $\tilde{B}$ is symmetric.

A different view of the above can be found using spectral theory.
Notice that $ker(\tilde{B}) = \{0\}$ if and only if $0$ is not an element of the spectrum of the function $\tilde{B}$ i.e. if 7 is not an eigenvalue of the function
$$
H_0^1(U) \ni \, u \mapsto \int_u (\nabla u)^T A(\nabla \cdot) \, \in (H_0^1(u))^*
$$
but this condition would, again, imply that there is no condition necessary for $f$ to lie in the range (i.e. $\tilde{B}$ would be surjective).

My intuition would be that, to describe $ran(\tilde{B})$ (i.e. by explicitely describing $ker(\tilde{B})$), we would need some more information of $B$ so, as daw mentioned in his comment, some more info on the a_{ij}. But there may be some other explicit characterizations of $ker(\tilde{B})$ or $ran(\tilde{B})$ that work in general (for example, possibly using harnack's inequality, that $u \in ker{L}$ implies $\inf u \leq C \sup u$).

Title: How do I find condition on $f$?


问:
If $\,a^2+b^2\,$ is a prime, how often is $\,(ab)^2+(a+b)^2\,$ a prime too?
For each prime $\;p\equiv 1\mod 4\;$ a unique decomposition exists as a sum of two squares $\;p=a^2+b^2$.

I ask how often is $\,(ab)^2+(a+b)^2\,$ a prime too.

If we obtain another prime we can repeat the procedure and, eventually, generate a chain of such primes. Up to the first $10^7$ primes, 4999454 primes are congruent to 1 (mod 4), and some computations have given the following results:

just $1$ chain of lenght $7$ (starting with $\,p=2\,$ that, even if not congruent to $1$ (mod $4$), is equal to $\,1^2+1^2$)

$5$ chains of lenght $6$ (starting with: $5, 23434781, 65715821, 165664753, 176884153$)

$159$ chains of lenght $5$ ($0.0032$ %)

$3750$ chains of lenght $4$ ($0.075$ %)

$57989$ chains of lenght $3$ ($1.16$ %)

$592927$ chains of lenght $2$ ($11.86$ %)

We may conjecture that:

the lenght of the chains is upper bounded by $7$

the only chain of lenght $7$ is the one starting from $\,p=2$ $(2, 5, 13, 61, 1021, 110581, 183198541)$

there are infinitely many chains of lenght less than 7

Many thanks for any suggestion, comment or answer.

答: The 'odds' that a random odd number $\lt 10^7$ is prime are roughly $2/\ln(10^7)\approx1/8$, which comes pretty close to your 11.8% figure. I suspect there's nothing known about this but I would expect them to follow the heuristics. If I'm doing my pre-coffee math correctly, then since abstractly the number you're hoping to be prime is of size $\approx p^2$, heuristically it seems as though chains of arbitrary fixed length should exist: for all $p$, the heuristic probability of a chain of length $n$ starting from $p$ is $\displaystyle\approx \frac1{\ln p}\cdot\frac1{\ln(p^2)}\cdots\frac1{\ln(p^{2^n})} \approx \frac1{2^{n^2/2}\ln^np}$. The constant term is smaller, but this is broadly akin to the probability for Cunningham Chains of length $n$.


问:
Find all pairs of primes $p, q$ such that 3$p^2q+16pq^2$ equals to square of natural number
Find all pairs of prime numbers $p, q$ such that 3$p^2q+16pq^2$ equals to square of natural number

My attempt: I've been trying to calculate equation through square root but now I'm stuck, please help

答:
Assume $3p^2q+16pq^2=k^2$

Since $q$ divides the LHS, it must divide $k^2$, and since $q$ is prime, it must be true that $q\mid k^2 \Rightarrow q^2\mid k^2$. Since $q^2\mid 16pq^2$, it follows that $q^2\mid 3p^2q \Rightarrow q\mid 3p^2$. But $q\not \mid p^2$, so $q\mid 3 \Rightarrow q=3$

Since $p$ divides the LHS, it must divide $k^2$, and since $p$ is prime, it must be true that $p\mid k^2 \Rightarrow p^2\mid k^2$. Since $p^2\mid 3p^2q$, it follows that $p^2\mid 16pq^2 \Rightarrow p\mid 16q^2$. But $p\not \mid q^2$, so $p\mid 16 \Rightarrow p=2$

This checks out, as $3\cdot 4\cdot 3+16\cdot 2\cdot 9=324=18^2$


问:
Calculating residue of a line integral
My book has a confusing residue calculation:

$$\int_c \frac{4-3z}{z^2-z}dz$$

$$\int_c \frac{4-3z}{z(z-1)}dz$$

So when evaluating the residue at z = 0
$$Res_{z=0} [ \frac{4-3z}{z(z-1)}]$$

I don't understand how subbing z=0 gives us this? Shouldn't the denominator just be 0 since when z=0 the whole denom = 0?

$$= [\frac{4-3z}{z-1}]_{z=0} = -4$$

答: The notation $\operatorname{res}_{z=a}f(z)$ means “residue of $f$ at $a$”. So, $\operatorname{res}_{z=0}\frac{4-3z}{z(z-1)}$ stands for the residue of $\frac{4-3z}{z(z-1)}$ at $0$ (which is indeed $-4$).

Title: Calculating residue of a line integral


问:
Schwarz's inequality integration - Probability distribution
Hi I am trying to solve the problem below and somehow I do not have the right idea.
A particle is located between x and x + dx with probability P(x)dx. If

$\displaystyle \langle |x|^n \rangle \equiv \displaystyle \displaystyle\int_{-\infty}^{\infty} |x|^nP(x)\ dx$

show, using Schwarz's inequality, that $\langle |x| \rangle^2 \leqslant \langle |x|^2 \rangle $ .

答:
This is a special case of Jensen's inequality. The proof of this special case can be obtained from Cauchy-Schwarz's inequality.

Note that \begin{align}
\langle |x|\rangle &=\int_{-\infty}^{\infty}|x|P(x)dx \le \left(\int_{-\infty}^{\infty}|x|^2P(x)dx\right)^{1/2}\left(\int_{-\infty}^{\infty} 1^2P(x)dx\right)^{1/2}\\
&=\left(\int_{-\infty}^{\infty}|x|^2P(x)dx\right)^{1/2} = \langle |x|^2\rangle^{1/2}.
\end{align}

Squaring both sides gives you the desired inequality. The inequality in the first line uses Cauchy-Schwarz and the $\int_{-\infty}^{\infty} 1^2P(x)dx=1$ because $P(x)$ is a probability density.

As an aside, one can prove a general version of this result known as Jensen's inequality. For any positive convex function $\phi,$ we have $\phi(\langle |x|\rangle)\le \langle \phi(|x|)\rangle.$


问:
Putting negations before predicates using quantifiers
I'm trying to transform this logical statement $\neg\forall x\exists y(P(x, y)\lor Q(x, y))$ to have
negation before predicates and I'm asking for a little explanation. Are these steps
I performed earlier correct or am I wrong, and how to do that the proper way?

$\neg\forall x\exists y(P(x, y)\lor Q(x, y))$

My steps:

$\exists x\neg(\exists y(P(x, y)\lor Q(x, y)))$

$\exists x\forall y\neg(P(x, y)\lor Q(x, y))$

$\exists x\forall y(\neg P(x, y)\land\neg Q(x, y))$

So this is the final statement I get :

$$\exists x\forall y(\neg P(x, y)\land\neg Q(x, y))$$

Is this correct? You guys are the best :)

答:
Yes, you seem to have a clear understanding of bringing the negation into what is being quantified.

For you're first step, you could have started with $$\exists x \lnot\exists y\Big((P(x, y) \lor Q(x, y)\Big).$$

But either way, the steps you demonstrate are correct.


答:
First, we need to establish the boundaries of D, ie the equations of the three lines that make up the triangle.

The line $L_1$ through (0,0) and (1,2) has equation y = 2x.

The line $L_2$ through (0,0) and (2,1) has equation y = x/2.

The line $L_3$ through (1,2) and (2,1) has equation y = -x + 3.

If you sketch these on a diagram, it becomes clear that we need to consider all x between 0 and 2, and for each x we consider, we need to consider points above $L_2$. For $x\le 1$ we need to ensure the point is below $L_1$ and for $x> 1$ we need to ensure the point is below $L_3$.

This leads us to break up the integral into two parts with limits as follows:

$\int^1_{0} \int_{x/2}^{2x}-sin(y+x) dy dx +\int^2_{1} \int_{x/2}^{3-x}-sin(y+x) dy dx$

Note that I have switched dy and dx so that y becomes the "inner" variable.

To finish the question, the integral is now:

$\int^1_{0} [cos(y+x)]_{y=x/2}^{2x} dx +\int^2_{1}[cos(y+x)]_{y=x/2}^{3-x} dx$

$=\int^1_{0} cos(3x)-cos(3x/2) dx +\int^2_{1}cos(3)-cos(3x/2) dx$

$=\left [ \frac{\sin(3x)}{3}-\frac{2\sin(3x/2)}{3} \right ]_0^1 + \left [ x \cos(3)-\frac{2\sin(3x/2)}{3} \right ]_1^2 $

$=\frac13\sin(3) + \cos(3)$

Title: Need help with a double integral in a triangular region

Body:
I'm trying to learn some maths in my own time. I am 13 and trying to learn some more maths on my own.

Can someone help me with question? My teacher could not :(

$\iint_{D}-sin(y+x) dx dy$

Where D is the triangle with vertices (0,0), (1,2) and (2,1).

From where I read online it says I should draw a triangle, but I don;tt know how this helps me...


问:
Roots of a cubic equation
I have the following equation:

$s^3+as+b=0$

Now I want the values for a and b for which the given equation has the following complex roots:

$c \pm di$

I don't really care about the remaining root.

Any ideas? I have Matlab available, but I don't know how I can use it to solve this problem.

答: plug both solutions for $s$ in the equation and solve for $b$ and $a$ resp. </p>

Title: Roots of a cubic equation


问:
Let $f(z) = u(x,y) + iv(x,y)$ be an holomorphic function and $2u(x,y) + v(x,y) = 5$. Show that $f$ is constant
I understand that it suffices to show that $f'(z)=0$

What I don't understand though, is how to take a partial derivative from $2u(x,y) + v(x,y) = 5$ ...

Another solution I thought of is,

If $2u(x,y) + v(x,y) = 5$, and 5 is a real number. Then 2 $y1$ and $y2$ is equal to $0$, and 2 $x1$ and $x2$ is equal to $5$.

meaning that if I take the derivative of 2 $x1$, its always going to be zero because 5 is a constant.  $\frac{du}{dx} = 0$ and since it's analytic, by CRE it implies that $\frac{dv}{dy}$ is also zero.

Likewise if I take the derivative of $x2$, $\frac{dv}{dx}$ is also going to be $0$, which again by CRE implies that $-\frac{dv}{dx} = \frac{du}{dy}$.

So it's all $0$, meaning that $f$ is a constant.

Any hints? or suggestions in regards to this solution?

答: The image of such an analytic function would be a subset of a line in the plane.  I don't know where you are in your study of complex variable, but such a function would not be an open mapping [if it is nonconstant], so it cannot be analytic.</p>


答: Taking partial derivatives of $$2u+v=5$$ gives $$2u_x+v_x=0\\2u_y+v_y=0$$  Use the Cauchy-Riemann equations to express the partials with respect to $y$ in terms of the partials with respect to $x$, and conclude that all fo them are $0$.

Title: Let $f(z) = u(x,y) + iv(x,y)$ be an holomorphic function and $2u(x,y) + v(x,y) = 5$. Show that $f$ is constant

Body:
I understand that it suffices to show that $f'(z)=0$

What I don't understand though, is how to take a partial derivative from $2u(x,y) + v(x,y) = 5$ ...

Another solution I thought of is,

If $2u(x,y) + v(x,y) = 5$, and 5 is a real number. Then 2 $y1$ and $y2$ is equal to $0$, and 2 $x1$ and $x2$ is equal to $5$.

meaning that if I take the derivative of 2 $x1$, its always going to be zero because 5 is a constant.  $\frac{du}{dx} = 0$ and since it's analytic, by CRE it implies that $\frac{dv}{dy}$ is also zero.

Likewise if I take the derivative of $x2$, $\frac{dv}{dx}$ is also going to be $0$, which again by CRE implies that $-\frac{dv}{dx} = \frac{du}{dy}$.

So it's all $0$, meaning that $f$ is a constant.

Any hints? or suggestions in regards to this solution?


答:
"I understand that it suffices to show that $f′(z)=0$".
This is true only for domains in the complex plane. For a counterexample, take $f \equiv 1$ in $B_1 (0)$ and $f \equiv 0$ in $B_1 (3)$. $f$ is only locally constant. Other than that, @saulspatz already gave the answer I was about to give you.


问:
Conjecture about the Jordan-Polya numbers
The Jordan-Polya numbers are the products of factorial numbers A001013:

```<code>1,2,4,6,8,12,16,24,32,36,48,64,72,96,120,128,144,192,216,240,256,288,384,432,480,512,576,720,768,864,960,1024,1152,1296,1440,1536,1728,1920,2048,2304,2592,2880,3072,3456,3840,4096,4320,4608,5040,5184,5760,...
</code>```

It seems that the difference between two consecutive Jordan-Polya numbers itself is a Jordan-Polya number. Is this true and if so, are there a proof of this somewhere?

答:
This is false.  Using the data in your list...

$$  4320 - 4096 = 224  \text{.}  $$

The next few violations of the conjecture are:
\begin{align*}
8192 - 7776 &= 416  \text{,}  \\
8640 - 8192 &= 448  \text{,}  \\
16384 - 15552 &= 832  \text{,}  \\
17280 - 16384 &= 896  \text{, and}  \\
25920 - 24576 &= 1344  \text{.}
\end{align*}
So, while it is common in the first few instances that violations include a power of $2$, this is not true for all violations.

There are $6853$ Jordan-Polya numbers up to $20!$.  Computing the gaps and splitting up the first $6800$ gaps into blocks of $100$ gaps ("centuries").  We plot the number of gaps conforming with the conjecture.

After initial high levels of conformance, by the tenth century (thousandth gap), the likelihood of a Jordan-Polya number gap being a Jordan-Polya number seems to be around $50\%$ and is, at least for the range covered, fairly stably so.

Repeating up to $30!$, there are $91\,802$ Jordan-Polya numbers and $53\,065$ gaps are violating.  The plot by centuries ...

... suggests that conforming instances become rarer as one proceeds.


问:
Deduce dice configuration knowing 2 adjacent faces
Let's say I have a die. I know it's up face and right face. I'd like to know what's on other faces, in this configuration.

I know the dice configuration:

```<code>  0
3 1 2
  5
</code>```

where $1$ is on the front face and $4$ is at the back (not visible).

On $2$ opposite faces, the sum of values is $5$ ($2-3$, $0-5$, $1-4$).

Basically, I would like to find a function f(x, y):

```<code>f(0, 1) = 3    f(1, 0) = 2
f(0, 2) = 1    f(2, 0) = 4
f(0, 3) = 4    f(3, 0) = 1
f(0, 4) = 2    f(4, 0) = 3

f(1, 2) = 5    f(2, 1) = 0
f(1, 3) = 0    f(3, 1) = 5
f(1, 5) = 3    f(5, 1) = 2

f(2, 4) = 5    f(4, 2) = 0
f(2, 5) = 1    f(5, 2) = 4

f(3, 4) = 0    f(4, 3) = 5
f(3, 5) = 4    f(5, 3) = 1

f(4, 5) = 2    f(5, 4) = 3
</code>```

Is there a simple and elegant math formula for this function?

答:
Let $g$ be the identification
\begin{align}
  g(0) = (1,0,0) &\quad g(5) = (-1,0,0) \\
  g(1) = (0,1,0) &\quad g(4) = (0,-1,0) \\
  g(2) = (0,0,1) &\quad g(3) = (0,0,-1).
\end{align}
Then $f$ is given by
$$ f(m,n) = g^{-1}(  g(n)\times g(m) ) ,$$
so under the identifications given by $g$ your map is just the vector product (in reverse order because of an unfortunate difference in convention between the vector product and what you want your map to be).
Is this elegant enough?</p>


问:
Closed form to find minimum last index of a summation so that it's greater than a given value
I'd like to find the minimum integer value of $t$ for which the following inequations 1 and 2 are true (one value of $t$ by inequation).

These equations represent someone contributing monthly an amount $M$ to an investment portfolio and the investor is wondering when he will achieve a certain amount $C$ given the compounded interest rate. In the equation 2, the person adapts its contribution to the inflation rate (he contributes more as the inflation rise).

$t$ is a positive integer. $M$ and $C$ are positive real numbers. $\mathit{interest\_rate}$ and $\mathit{inflation\_rate}$ are real numbers strictly greater than 1.

\begin{equation}
C < \sum_{i=0}^{t} M \cdot interest\_rate^i
\end{equation}
\begin{equation}
C < \sum_{i=0}^{t} M \cdot inflation\_rate^i \cdot interest\_rate^i
\end{equation}

I can solve this problem using code (example for second equation, Python 3.8):

```<code>interest_rate = 1.08 ** (1 / 12)

inflation_rate = 1.02 ** (1 / 12)

portfolio = 0
M = 1000
C = 1_750_000
t = 0
while C > portfolio:
    t += 1

    portfolio += M * inflation_rate ** t * interest_rate ** t

print(f"{t // 12} years and {t % 12} months")
</code>```

But I'm a complete loss to find a closed form.

答:
Let $x=t$, $r$ be the interest rate (or the product of interest rate and inflation rate), and thus you want to know when:

$$\sum_{i=0}^x{r^i} = \frac{r^{x+1}-1}{r-1}\gt \frac{C}{M}$$

i.e.

$$r^{x+1}>\frac{C}{M}(r-1)+1$$

$$(x+1)\log{r}>\log{\left(\frac{C}{M}(r-1)+1\right)}$$

$$x>\frac{\log{\left(\frac{C}{M}(r-1)+1\right)}}{\log{r}}-1$$

So the result is:

$$\Big\lceil{\frac{\log{\left(\frac{C}{M}(r-1)+1\right)}}{\log{r}}-1}\Big\rceil$$

Note that when $r \approx 1$, $\log{\left(\frac{C}{M}(r-1)+1\right)} \approx \frac{C}{M}(r-1)$ and $\log{r} = \log{(r-1+1)} \approx (r-1)$, and so the value is about $\frac{C}{M}-1$ as one would expect.


问:
Finding the points which have equal area under the curve for Gausian distribution?
I have a normal distribution N(1,0) and have to split it in n-pieces where the area under the curve is equal i.e. 1/n

Now I need to find the specific points/values where this happen.

```<code>  f(1 .. n) =
</code>```

How do I do that ? If you can provide pseudo code it will be good too i.e. i need to make it python code.

PS> I'm trying to implement this :

https://www.cs.ucr.edu/~eamonn/SAX.pdf

page 5 : Discretization

I hope I'm describing the problem OK!
I'm abit confused by why they need table, instead of just list of numbers i.e. points/values.

答:
Let $\Phi^{-1}$ be the quantile function of the standard normal distribution, then the points
$$\Phi^{-1}(0),\Phi^{-1}(\frac1n), \Phi^{-1}(\frac2n),\dots\Phi^{-1}(\frac{n-1}{n}),\Phi^{-1}(1),$$
where $\Phi^{-1}(0)=-\infty$ and $\Phi^{-1}(1)=\infty$, will divide the real line into intervals, which each have probability $\frac{1}{n}$ with respect to a $N(0,1)$ distribution.

Let's verify this. Let $Z\sim N(0,1)$, then
\begin{align*}
P(\Phi^{-1}(\frac{k}{n}) \leq Z \leq \Phi^{-1}(\frac{k+1}{n}))  &= \Phi(\Phi^{-1}(\frac{k+1}{n}))-\Phi(\Phi^{-1}(\frac{k}{n})) \\ &= \frac{k+1}{n}-\frac{k}{n} \\ &= \frac{1}{n}
\end{align*}
The quantile function $\Phi^{-1}$ can be computed in Python with the norm.ppf() function from the scipy.stats module https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.norm.html .


问:
Two circles $C_1$ and $C_2$ of radii $10$ cm and $8$ cm are respectively are tangent to each other internally at a point $A$.

Two circles $C_1$ and $C_2$ of radii $10$ cm and $8$ cm are respectively are tangent to each other internally at a point $A$. $AD$ is the diameter of $C_1$ and $P$ and $M$ are points on $C_1$,$C_2$ respectively such that $PM$ is tangent to $C_2$ . If $PM = \sqrt{20}$  and $\angle PAD = x^\circ$, find $x$.

What I Tried: Here is a picture :-

I absolutely have no good ideas to start this. Let $O$ and $K$ be the centers of the $2$ circles. The only thing which I could do was to find $PO$ using Pythagoras Theorem, which will come out to be $\sqrt{84}$.

I also did, for example, checked this in Geogebra and I found $\angle PAD$ to be $60^\circ$, so I suppose that is the answer, but why is it coming so? I didn't find any ideas using Geogenra either.

Can anyone help me? Thank You.

答:
$OK = 10 - 8 = 2$, $KP = 10, OP = \sqrt{84}$

Using cosine law,

$84 = 2^2 + 10^2 - 40 \cos \angle AKP \implies cos \angle AKP = \frac{1}{2}$

$\angle AKP = 60^0$

$\angle PAD = \frac{1}{2}(180^0 - 60^0) = 60^0$

EDIT: Just using Pythagoras,

Drop perp from $P$ to $AK$ which meets $AK$ at say, $X$ and $KX = x$.

Then in $\triangle POX$ and $\triangle PKX$,

$PX^2 = 10^2 - x^2$

$PX^2 = (\sqrt{84})^2 - (x-2)^2$

Equating both, $x = 5 = \frac{AK}{2}$.

Which means perpendicular from $P$ to $AK$ meets it at midpoint, hence $AP = PK = KA = 10$. So, $\triangle APK$ is equilateral.


问:
How to interpret $|x|^{-1}$ as a distribution
Currently I'm reading a little about regularization of distributions which usually have annoying singularities like $x^{-1}$ or $|x|^{-1}$. To the best of my understanding,
$$
\frac{1}{x}
$$
is interpreted as a distribution by considering it's Cauchy principal value instead of doing the bruteforce integration. This way, it's possible to recover a meaningful result by carefully cancelling contributions to the singularity. So if $g$ is, say, a function from Schwartz space, then
$$
\left\langle \text{p.v.}\, \frac{1}{x}, \, g(x) \right\rangle = \int _{0}^{\infty} \frac{g(x) - g(-x)}{x} \, dx.
$$
Intuitively, this makes total sense to me. However, when reading about $|x|^{-1}$ I get slightly confused. The definition states that $|x|^{-1}$ (or, rather, it's regularized cousin $\mathcal{P}|x|^{-1}$) acts on test functions as per
$$
\left\langle \mathcal{P}\frac{1}{|x|}, \, g(x) \right\rangle = \int _{|x|<1} \frac{g(x)-g(0)}{|x|} \, dx + \int_{|x|>1} \frac{g(x)}{|x|} \, dx
$$
The second integral is pretty straightforward; no singularities to sneak around. But the first one...
Does it solve the singularity problem? Yes, I can see that, since
$$
\frac{g(x)-g(0)}{|x|}
$$
remains bounded as $x \rightarrow0$. But how do we get to it? What's the reasoning behind it? I've been struggling to see it for a little while. Any and all help will be appreciated :).

答:
Let $U,V$ be two open sets with $U\subset V$. One has a restriction map for distributions
$$
r:\mathscr{D}'(V)\rightarrow\mathscr{D}'(U)
$$
which is defined by $r(T)(f)=T(\tilde{f})$ where $T$ is a distribution on $V$, $f$ is a smooth compactly supported function on $U$ and $\tilde{f}$ is its extension by zero to all of $V$. Now given a distribution $S\in \mathscr{D}'(U)$ one can ask for a characterization of all extensions $T$ to $V$, i.e., find $T$'s such that $S=r(T)$.

Now take $V=\mathbb{R}$ and $U=(-\infty,0)\cup(0,\infty)$, and $S$ given by integration times $\frac{1}{|x|}$. The construction in the question is typically how you produce a $T$ which works. To get all other, you just add a distribution supported at the origin like the Dirac Delta or a linear combination of its derivatives.

To single out the construction in the question one can also bring into the game the notion of homogeneity. Thinking of $T$ as "function" $T(x)$, homogeneity of exponent $\alpha$ means
$$
T(\lambda x)=\lambda^{\alpha}T(x)
$$
for all nonzero $\lambda$ (the usual definition takes $\lambda>0$ but never mind).
In other words, for every $\lambda$, $T$ is an eigenvector with eigenvalue $\lambda^{\alpha}$ for the rescaling operator $T(x)\rightarrow T(\lambda x)$.
The $T$ in the question is not homogeneous, since up to scale the only homogeneous distribution of exponent $-1$ is the Dirac Delta at the origin. But $T$ is the next best thing, i.e., a generalized eigenvector, in the sense of Jordan reduction. The key word for this theory is associated homogeneous distributions.


问:
how to solve $\lim\limits_{x \to +\infty} f(x) \over \lim\limits_{x \to +\infty} g(x)$?
how to solve $\lim\limits_{x \to +\infty} f(x) \over \lim\limits_{x \to +\infty} g(x) $ when $\lim\limits_{x \to +\infty} f(x) = +\infty$ and $\lim\limits_{x \to +\infty} g(x) = +\infty$?

for example what are the solutions of $\lim\limits_{x \to +\infty} x \over \lim\limits_{x \to +\infty} 2x $ or $\lim\limits_{x \to +\infty} x \over \lim\limits_{x \to +\infty} x^2 $?

I think we can't use the L'hopital rule because ${\lim\limits_{x \to +\infty} f(x) \over \lim\limits_{x \to +\infty} g(x)} \neq \lim\limits_{x \to +\infty} {f(x) \over  g(x)} $ . (not sure)

So please tell me how to solve these kinds of problems and which theorems we use must use.

Thanks

答: $$\frac{\lim_{x\to\infty} 2x}{\lim_{x\to\infty} x}$$ doesn't make sense, because neither the limit in the numerator nor that in the denominator exists.  The rule that the limit of a quotient is the quotient of the limits only holds when the latter limits exist.


答:
If $r$ is the other root. we want $2c+r=0$ and $r(2c)+c^2+d^2=a$ and $(c^2+d^2)r=-b$. The first equation tells us what $r$ is and now it's over.

Remark: We used the fact that if $p$, $q$, and $r$ are the roots, then the coefficient of $s^2$ is $-(p+q+r)$, the coefficient of $s$ is $pq+qr+rp$, and the constant term is $-pqr$.

Title: Roots of a cubic equation

Body:
I have the following equation:

$s^3+as+b=0$

Now I want the values for a and b for which the given equation has the following complex roots:

$c \pm di$

I don't really care about the remaining root.

Any ideas? I have Matlab available, but I don't know how I can use it to solve this problem.