答:
Hint. If we have that
$$\prod_{i=0}^n \frac{1}{(x+i)}=\sum_{i=0}^n \frac{A_i}{(x+i)}.$$
then, for $k=0,\dots,n$,
$$\lim_{x\to -k}(x+k)\prod_{i=0}^n \frac{1}{(x+i)}=A_k.$$
The general formula of $A_k$ is related to a binomial coefficient.

Once that the partial fraction decomposition is ready, the integration is straightforward.

Title: Integrate $\int \prod_{i=0}^n \frac{1}{(x+i)}dx$

Body:
I initially tried to find a pattern using partial fraction decomposition:
$$
n= 1 \, : \, \, \frac{1}{x(x+1)} = \frac{1}{x}+\frac{-1}{x+1}
$$

$$
n=2 \, : \, \, \frac{1}{x(x+1)(x+2)} = \frac{1/2}{x}+\frac{-1}{x+1}+\frac{1/2}{x+2}
$$

$$
n=3 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)} = \frac{1/6}{x}+\frac{-1/2}{x+1}+\frac{1/2}{x+2} + \frac{-1/6}{x+3}
$$

$$
n=4 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)(x+4)} = \frac{1/24}{x}+\frac{-1/6}{x+1}+\frac{1/4}{x+2} + \frac{-1/6}{x+3} + \frac{1/24}{x+4}
$$
But I don't seem to find a pattern. Is there another method to resolve this? Thank you.


问:
A question on the image of a Möbius function
Let $a\in\mathbb C$ such that $|a|<1$ and consider the Möbius function

$$
f(z)=\frac{z-a}{1-\bar{a}z}
$$
I ask to show that if $|z|=1$, then $|f(z)|=1$ and if $|z|<1$, then $|f(z)|<1$ using the maximum modulus theorem.

I noticed that
$$
|1-\bar{a}z|^2-|z-\bar{a}|^2=
1+|a|^2|z|^2-2Re(\bar{a}z)-(|z|^2-2Re(z\bar{a})+|a|^2)
=(1-|a|^2)(1-|z|^2)
$$

Consequently, if $|z|=1$, then $|f(z)|=1$ and if $|z|<1$, then $|f(z)|<1$. This is quite straightforward. Is there an easy way to prove that using the maximum modulus theorem ?

答: Since you are specifically asked to use MMP you can do the following: Let $|z|=1$. Then $|z-a|=|\overline {z-a}|=|\frac1 z-\overline a|=|1-\overline a z|$ so $|f(z)|=1$. MMP implies that $|f(z) | <1$ for $|z|<1$ since $f$ is analytic and not a constant.

ContentLicense: CC BY-SA 4.0


问:
Property of transpose Tr in category of modules (Auslander-Reiten theory)
I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 , and I have a question on its proof of main properties of transpose $Tr$, appearing in Auslander-Reiten theory.

Here we assume $A$ is a finite dimensional algebra over an algebraically closed field $K$, and $mod A$ is a category of finitely generated modules of $A$. Also, consider a functor $(−)^t = Hom_A(−,A) : mod A → mod A^{op}$.

For an object $M$ in $mod A$, we define $Tr M$ as follows:

If $P_1→P_0→M→0$ is a minimal projective presentation, then $Tr M=Coker(P_0^t→P_1^t)$. In other words, $P_0^t→P_1^t→TrM→0$ is exact.

Proposition 2.1.(b) in the textbook p.107 is about showing that

If M is not projective, then the exact sequence $P_0^t→P_1^t→TrM→0$ is a minimal projective presentation of an $A^{op}$ module $Tr M$.

In the proof, author assumes $P_0^t→P_1^t→TrM→0$ is not a minimal projective presentation. Then there exist non-trivial decompositions (by projective modules) $P_0^t=E_0'⊕E_0''$, $P_1^t=E_1'⊕E_1''$ and an isomorphism $v:E_0''→E_1''$ and a homomorphism $u:E'_0→E'_1$ such that the previous exact sequence is expressed as
$E_0'⊕E_0''\overset{\begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}}{\rightarrow}E_1'⊕E_1''→Tr M→0$.

(I think the author assumed $E'_0\overset{u}{\rightarrow}E'_1→Tr M→0$ is a minimal projective presentation.)

But I couldn't quite understand the next step. It is written as,

But then applying $(−)^t$ yields a projective presentation of M of the form ${E'_1}^t\overset{u^t}{\rightarrow}{E'_0}^t→M→0$

But I couldn't figure out why we get $M$ as a cokernel of $u^t$. Any helps will be appreciated. Thank you for reading.

答:
For finitely generated projective modules $X$, the evaluation map $X\to X^{tt}$ (defined by $x\mapsto[\varphi\mapsto\varphi(m)]$) is a natural isomorphism.

So $P_1\to P_0$ is isomorphic to $P_1^{tt}\to P_0^{tt}$, which in turn is isomorphic to the direct sum of $E_1'^t\xrightarrow{u^t}E_0'^t$ and $E_1''^t\xrightarrow{v^t}E_0''^t$.

But $v$ is an isomorphism, so $v^t$ is an isomorphism and so its cokernel is zero. So $M$, which is the cokernel of $P_1\to P_0$, is also the cokernel of the first summand $E_1'^t\xrightarrow{u^t}E_0'^t$.


问:
Canonical transformation, Lagrangian submanifold and generating function
If we know a generating function $\phi$ for a canonical transformation and also we know a generating function W for a Lagrangian submanifold $\Lambda$, what is a generating function for $\phi(\Lambda)$?

答:
You can find an answer in the paper by Brunella $\textit{On a Theorem of Sikorav}$,
Enseign. Math. (2) 37, No. 1--2, 83--87 (1991; Zbl 0739.58015). Lemma 4 is the result you are looking for, and you can find the paper here. Note that however there is the hypothesis that the symplectomorphism is compactly supported, and I do not know a reference where it is removed (nor if in fact the result is still true).


问:
Extending a function to obtain continuity
I am working on a two part problem for studying. Not homework. Not exam prep.

The problem is

Let $f\colon(0,1]\rightarrow\mathbb{R}$ be a uniformly continuous function on $(0,1].$

a.  State the definition of uniform continuity and use it to show that $g(x)=\ln x$ is not uniformly continuous on $(0,1]$,

b.  Prove that $f$ can be extended uniquely to a continuous function on $[0,1]$.

I know how to do part a but don't know how to prove the extenstion portion in part b.

答:
Let $(x_n) \in (0,1]^{\mathbb N}$ be such that $x_n \to 0$. We will prove that $(f(x_n))$ converges, as $\mathbb R$ is complete, it suffices to prove that $(f(x_n))$ is Cauchy. Let $\epsilon > 0$, by uniform continuity, there is an $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)|< \epsilon$. As $x_n \to 0$, $(x_n)$ is Cauchy, hence there is an $N$ such that $|x_n - x_m| < \delta$ for $n,m \ge N$. So $|f(x_n) - f(x_m)| <\epsilon$ for $n,m \ge N$.

As for every sequence $x_n \to 0$, $\lim_{n\to\infty} f(x_n)$ exists, the value of the limit does not depend on $(x_n)$. So $\lim_{x\to 0^+} f(x)$ exists, and
$$ \bar f(x) = \begin{cases} f(x) & x > 0\\ \lim_{x\to 0^+} f(x) \end{cases} $$
is the unique continuous extension of $f$.

ContentLicense: CC BY-SA 3.0


问:
What is the modular representation of an integer?
What is the modular representation of an integer using a set of primes?

More specifically, a problem on my homework asks to convert 49 to a modular representation using primes 7,11,13,17.

Would appreciate a general solution.

答:
I saw this in the book, Applied Algebra- Codes, Ciphers and Discrete Logarithms by Darel W.Hardy. It's on page 138 of the second edition, in the section about the Chinese Remainder Theorem and Extended Precision Arithmetic. The accepted answer gives a wonderful summary of the book's content.

In the general case, the author states, "Let $m_1,m_2,...,m_r$" be pairwise relatively prime positive integers with the property that $m_i^2$ can be computed exactly in machine arithmetic. Then any positive integer $n$ less than the product $m_1m_2...m_r$ is determined by the numbers $$ a_1  = n\: mod\:m_1, a_2  = n\: mod\:m_2,...,a_r  = n\: mod\:m_r$$ The modular representation of n relative to the bases $m_1,m_2,...,m_r$ takes the form $$ n = [a_1, a_2,...,a_r]$$

Knowing this, we can represent 49 using the primes 7,11,13,17.
$$49\:mod\:7 = 0$$ $$49\:mod\:11 = 5$$ $$49\:mod\:13 = 10$$ $$49\:mod\:17 = 15$$ We get $$ 49 = [0, 5,10,15]$$

We can get 49 from its modular representation by solving the system of congruences
$$x \equiv 0\;mod\: 7$$ $$x \equiv 5\;mod\: 11$$ $$x \equiv 10\;mod\: 13$$ $$x \equiv 15\;mod\: 17$$


问:
Length of a triangle side proportional to angle size?
I'm trying to find the values of a and b in the following situation: https://i.stack.imgur.com/YxavX.jpg

I can get the correct answer using trigonometry, but I was wondering why the following solution does not give a correct result:

$$ a = 15 * \frac{35}{38+35}$$
$$ b = 15 * \frac{38}{38+35}$$

Is it because the length of the side of a triangle (a, b) isn't directly proportional to the angle (35°, 38°)?

答:
Yes, that's correct. Considering an extreme case, if one angle were $90º$, $a$ would be infinitely long whereas the incorrect method would be a fraction of $15$.

$a$ and $b$ are correctly related to each other by the extended angle bisector theorem, which works even if the triangles are not right-angled. In your case, this would be:

$$\frac{a}{b} = \frac{x \sin 35º}{y \sin 38º}$$

but this is not the right method unless you know $x,y$ beforehand.

For this question, you would need to equate the common side as follows: $\frac{a}{\tan 35º} = \frac{b}{\tan 38º}$, which I presume is what you did. Finding one variable in terms of the other and using the fact that $a+b=15$ gives you $a,b$.


Body:
You may use:

$$\prod_{k=0}^{n} \frac{1}{x+k}=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k-1} \frac{k{n \choose k}}{x(x+k)}=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k} [\frac{1}{x}-\frac{1}{x+k}]$$
So $$\int \prod_{k=0}^{n} \frac{1}{x+k} dx=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k}[\ln x- \ln (x+k)]$$


问:
Sum of possible values of $PH$ when altitude intersects another median

The median $ AM $ and the height $ BH $ of the $ ABC $ triangle ($ H $
on the $ AC $ side) intersect at the $ P $ point. Find $ PH $ if $ AM = BH = 196 $, $ MN = 61 $ where $ N $ is the intersection point of
the continuation $AM $ with the circle circumscribed around the
triangle $ ABC $. In response, write down the sum of possible values $ PH $.

I can't come out with the solution, I tried. Please anyone help me with the solution. I think I need to form an equation or inequality to find all the possible values.

答:
Since $BH=AM$ and both are opposite $\angle C$, triangles $BHC$ and $AMC$ have same circumradius namely, $R_{\triangle AMC}=MC$.

By sine-rule in $AMC$,

$$\frac{AM}{\sin C}=2R_{\triangle AMC}=\frac{MC}{1/2}=\frac{MC}{\sin 30}$$
$$\therefore \angle MAC=30 \Rightarrow \angle BPM=60$$

If we let $PH=x$, then $AP=2x$, $PM=196-2x$, $BP=196-x$. Also $BM^2=AM\cdot MN=196\cdot 61$

By cosine-rule in $\triangle BPM$,

$$BM^2=BP^2+PM^2-BP\cdot PM$$
$$196\cdot 61=(196-x)^2+(196-2x)^2-(196-2x)(196-x)$$

Sum of values of $PH$ is $(-\,\text{coeff of}\, x/\text{coeff of} \, x^2)$ of this quadratic.


问:
$\sum_{n=1}^{\infty} X_n$ converges almost surely
question:

$\{ X_n, n \in N\}$ be sequence of rrvs and suppose that $E(X_n) = 3^{-n}$ and $var (X_N)=2^{-n}$. Prove the series $\sum_{n=1}^{\infty} X_n$ converges almost surely (a.s.)

My solution:

By Chebyshev inequality

$P(|X_n - 3^{-n} | > e) \le 2^{-n} /e^2$ for small enough e>0.

Define $A_n = \{ w: |X_n - 3^{-n} | > e\}$ such that

$lim P(A_n) = lim P(|X_n - 3^{-n} | > e) \le lim (2^{-n} /e^2) \to 0$ as $n \to 0$.

i.e. $lim P(A_n) \le 0 $

since probability is between 0 and 1, $lim P(A_n) = 0$

At this point, I am stack.

$\sum P(A_n) = 0 $ Then,

$\sum |X_n - 3^{-n}  = 0 $

....

And I cannot continue to find the convergence a.s.

答: $EX_n^{2}=\frac  1{2^{n}}+(\frac 1 {3^{n}})^{2}$. Using the inequality $E|X_n| \leq \sqrt {EX_n^{2}}$ you can easily check that $\sum_n E|X_n| <\infty$. But this im plies that $E(\sum_n |X_n|) <\infty$, so $\sum_n |X_n| <\infty$ almost surely. Hence, the series $\sum X_n$ converges absolutely with probability $1$.


问:
Show that a ball in Euclidean space is convex.
Let $B(x,r)$ be the ball cantered at $x$ with radius $r$, in the Euclidean space,  prove that it is convex.

The notes I have give the following proof:

Take $y,y' \in B(x,r)$ and set $z=ay +(1-a)y'$ for $0 \leq a \leq 1$
So $$d(x,z) =d(x,ay +(1-a)y') \leq d(x,ay)+d(x,(1-a)y')=ad(x,y)+(1-a)d(x,y')$$

I don't understand why it's the case that $d(x,ay)=ad(x,y)$, for example, if the metric was the absolute value I can't see how $|x-ay|$ can be equal to $a|x-y|$.

Can you explain why $d(x,ay)=ad(x,y)$ holds?

答:
The proof you have given only holds true if $x=0$, as then we have:

$$d(x,ay)=d(0,ay)=ad(0,y)$$

using the Euclidean metric.

It is false for all other values of $x$.


问:
Is it a known result that the Collatz Tree repeats on itself indefinitely?
With reference to my visual pattern of the Collatz problem, I recognize that the Collatz tree for numbers of the form $6x-1$ (for $x\in\mathbb{Z}$ & $x>0$) and $1-6x$ (for $x\in\mathbb{Z}$ & $x\leq0$) has layers (a known fact) but also that the first layer alone repeats on itself indefinitely.
My question is whether the repetition of the first layer on itself is a known result and what would be it's implication.  My research has come up empty-handed.

Details.

The binary tree below shows the first layer up to depth 4 and its repetitive nature. Notice we would have infinite subtrees with root 0 should we extend the tree indefinitely.

What do I mean by ‘layer’? Each number, x, has two children on each layer given by the formulas below where $n\in\mathbb{N}$ and $n=1$ produces children on the first layer, $n = 2$ produces children on the second layer, and so on.

To make my post brief, I have omitted a proof of my claim. However, if a proof is needed, I can supply it upon request. My thinking is that this may be a known fact due to the trivial 1,4,2,1 cycle, but I have not come across any Collatz tree with this repetitive nature.

答:
This is actually a necessary consequence of the function you describe always mapping 2 to 1.

Given any function $f:\mathbb{Z}\rightarrow\mathbb{Z}$, such that $f(0)=0$ and $\#f^{-1}(n) = 2$ for each $n$. Then we can describe 2 functions $g$ and $h$ where $g<h$ and $\{g(n), h(n)\} = f^{-1}(n)$.

In this way the sequence of finite sequences described as $X_0 = (0)$ and $$X_{n+1} = \big(g(X_n[0]), h(X_n[0]), g(X_n[1]), h(X_n[1]), \ldots, g(X_n[2^n-1]), h(X_n[2^n-1])\big).$$ This is exactly what you defined in your graph, but without the edges.

We can also see that finite sequence $X_{n+1}$ contain a copy of $X_n$. You can prove this by induction. The basestep consists of recognizing that $X_1 = \big(g(0), h(0)\big)$.
Because $0$ is an element of the pre-image of $f$ either $g(0)$ or $h(0)$ must equal $0$. So $X_1$ contains the sequence $(0)$.

The induction step goes as follows. Let the statement hold for $X_n$ meaning that $X_n$ contains a subsequence equal to $X_{n-1}$ without loss of generality say the first half. Then the first half of $X_{n+1}$ is defined as
$$\big(g(X_n[0]), h(X_n[0]), g(X_n[1]), h(X_n[1]), \ldots, g(X_n[2^{n-1}-1]), h(X_n[2^{n-1}-1])\big),$$ but we assumed that $X_n$ was equal to $X_{n-1}$ on the first half of its terms so this is equal to
$$\big(g(X_{n-1}[0]), h(X_{n-1}[0]), g(X_{n-1}[1]), h(X_{n-1}[1]), \ldots, g(X_{n-1}[2^{n-1}-1]), h(X_{n-1}[2^{n-1}-1])\big).$$
This by definition is the sequence $X_n$ concluding the induction step.

So the pattern you discovered is indeed real, and much stronger even as every layer repeats infinitely many times!


问:
A name/use for this general decomposion of any function?
Any function $\{(x_1,y_1), (x_2,y_2),... \}$ can be viewed as the union of "disjoint singleton" functions (I mean functions with distinct singleton co-domains, i.e. $\{x_1 \} \neq \{ x_2 \}$).

This reduction seems pleasing to me. Does it have a conventional uses or names?

Perhaps in the category of sets, where singleton sets are encoded as functions from the terminal element? Or elsewhere?

答: The categorical notions of presheaf and sheaf are an abstraction of the idea of constructing a function by gluing together smaller functions that are compatible in the sense that they agree whenever they are both defined. Your example arranges for the domains of the functions to be disjoint, so that they are automatically compatible. So an appropriate technical term would be "gluing": you are observing that any set-theoretic function can be obtained by gluing together singletons. See the Wikipedia page on the gluing axiom for some more information, in particular, how the presheaf and sheaf ideas pan out in situations where topological and algebraic considerations have to be taken into account.


问:
How to prove that this map is homotopic to a constant map?
I recently did an exam of Basic Algebraic Topology and could not get my head around this question.

I have this map $f:\Bbb S^1\to\Bbb S^1\times\Bbb S^1$, $f(z)=(z,z^q)$, for $q\ge1$. I could calculate the fundamental group for $X = (\Bbb S^1\times\Bbb S^1)\cup_f e^2$, and if I'm not mistaken it is $\Bbb Z$.

But then they asked me to prove that any map $g:X\to\Bbb S^1\times\Bbb S^1$ is homotopic to a constant map. I tried to use the lifting properties of the covering maps, and tried to check $\operatorname{im}(g_*)$, where $g_*:\pi_1(X)=\Bbb Z\to\pi_1(\Bbb S^1\times\Bbb S^1) = \Bbb Z \times \Bbb Z$, in order to check if $\operatorname{im}(g_*)\subset \operatorname{im}(p_*)$, for $p_*$ the induced map of a covering space. Is it true that $\operatorname{im}(g_*) = \Bbb Z$ or must it be a subset of $\Bbb Z \times \Bbb Z$? Also, which covering map could I choose to do this approach?

The last question asked me to get all covering spaces of $X$, for $q=4$ up to homotopy equivalences. I don't get why did they need to fix $q$, if we could easily do it with the classification of covering spaces with a base point. But then which covering spaces do I choose for the subsets of $\Bbb Z$ (again, if I'm not mistaken, they are $m\Bbb Z$, for $m\in \Bbb Z$ and $\{0\}$)?

Edit: I know that the question in hand is wrong. Thanks to user881318 for the answer and to Кряжев Арсений for the comment!

I also wanted to correct that the last question did not ask of computing the covering spaces of $X$, but rather say how many covering spaces $X$ had. This is much simpler, since it has as many covering spaces as subgroups $\pi(X,x_0)$ has (which is infinitely many, considering that $\pi(X,x_0) = \Bbb Z$, and the subgroups are $m\Bbb Z$, for $m\in \Bbb Z$ and $\{0\}$).

答:
Every map $X \to T^2$ arises as a pair of a map $g: T^2 \to T^2$ and a null-homotopy of $fg$.

The map $g: T^2 \to T^2$ is determined up to homotopy by its induced map on $\Bbb Z^2$, which is given by some 2 x 2 integer matrix $A$.

The assumption that $fg$ is null-homotopic, then, is the assumption that $$A \begin{pmatrix} 1 \\ q \end{pmatrix} = 0;$$ this is precisely the necessary condition to extend $g$ to a map $\overline g$ from $X$.

This condition is simple enough. It means that $$A = \begin{pmatrix} nq & -n \\ mq & - m \end{pmatrix}$$ for some pair of integers $n,m$.

So take such a nonzero matrix $A$ and write $\overline g_A: X \to T^2$ for a choice of extension of the corresponding $g_A$ to $X$. Then $\overline g_A$ is not null-homotopic, because its restriction to $T^2$ --- $g_A$ --- is not null-homotopic, because its induced map on fundamental groups --- $A$ --- is nonzero.

So the premise of your exam question is either wrong or misstated.


问:
Regression of a time series approaching a power law
I have a time series that has a power-law $A \cdot t^{b}$ behaviour for $t\xrightarrow{}\infty$, while for low $t$ it is far from that. Something like this:

How can I make a linear fit and find an estimation for $A$ and $b$? How many points shall be discarded?

Edit:
as you can see from the picture, taken from 3 independent runs, for $t < t_0$ (let's say, 10,000) the behaviour is quite random, and unrelated, while for $t > t_0$ the 3 series are almost parallel, meaning that they have the same exponent.
A simple linear fit will give the value of this exponent (about -0.0093). But the problem is: how to identify $t_0$, in order to make the regression only in the range $[t_0, \infty]$ where it follows the model.

答:
The best way should be to start from a physical model which allows to derive a convenient kind of function.

of course the function includes some parameters. With numerical data a non-linear regression would give approximate values of the parameters.

This is the best way because the parameters are likely to have physical significance, which is useful in practice.

If no physical model is available one have to guess some function susceptible to fit more or less correctly the data.

In the pesent case a linear piecewise function appears convenient (next figure).

The equation of the fitted curve drawn in red is :
$$y(x)=(a_1x+b_1)H(\alpha-x)+(a_2x+b_2)\big(H(x-\alpha)-H(x-\beta)\big)+(a_3x+b_3)H(x-\beta))$$
$$\alpha=\frac{b_2-b_1}{a_1-a_2}\quad;\quad \beta=\frac{b_3-b_2}{a_2-a_3}$$
$H$ is the Heaviside function.

One can smooth the angular transition between the linear segments : One replace the Heaviside function by a approximate function of the logistic kind :
$$H(\xi)\simeq \frac{1}{1+e^{-\gamma\:\xi}}\qquad \gamma>0$$
The more $\gamma$ is large the more this function is close to the Heaviside function.

Thus a smooth function to be fitted to the data can be :
$$y(x)=\frac{a_1x+b_1}{1+e^{-\gamma_1\:(\alpha-x)}}+\frac{a_2x+b_2}{1+e^{-\gamma_1\:(x-\alpha)}}-\frac{a_2x+b_2}{1+e^{-\gamma_2\:(x-\beta)}}+\frac{a_3x+b_3}{1+e^{-\gamma_2\:(x-\beta)}}$$

The parameters to be found thanks to non-linear regression are : $a_1,b_1,a_2,b_2,a_3,b_3,\gamma_1,\gamma_2$. The two above equations relate $\alpha,\gamma$ to $a_1,b_1,a_2,b_2,a_3,b_3$.

From the graph provided in the question without numerical data one cannot proceed. Scannig the graph gives the coordinates of the pixels. But the numerical data obtained is not accurate enough and too scrambled.

On the figure below the red curve isn't fitted to data. This is only to show how the angular transitions are smoothed, for example with $\gamma_1=3$ and $\gamma_2=5$. These values must not be trusted.


问:
$f$ be an entire function, that is also injective. Then $f$ take the form $f(z) = az+b$ with $a, b \in \Bbb C$ and $a \neq 0$
I am looking at the proof of the following statement: Let $f$ be an entire function, that is also injective. Then $f$ take the form $f(z) = az+b$ with $a, b \in \Bbb C$ and $a \neq 0$.

The proof is given as follow: Assume $f$ is an entire injective function. Then $f$ is non-constant, so $g(z) :=f(\frac{1}{z})$ has either a pole or an essential singularity at $z = 0$.

1) Why we have that I can't be a removable zero too?

We will show first that the singularity at $0$ cannot be an essential singularity. If it were an
essential singularity, then the Cazorati-Weierstrass theorem would imply that
the set $g(B(0, 1) \setminus {0})$ is dense in $\Bbb C$. However, $g(B(2,\frac{1}{2}))$ is an open set by the open mapping theorem. Therefore these two sets intersect, which shows that
$g(z)$ and hence $f(z)$ is not injective.

2) I don't see why we need to dense set to conclude that $g$ and hence $f$ is not injective.

Therefore, the singularity at $z = 0$ must be a pole, implying that $f(z)$ is a polynomial. Suppose $f(z)$ is a polynomial of degree $m$. Then $f$ has $m$ roots, counting multiplicity. Evidently, if $f$ has two distinct roots, then $f$ is not injective. Thus $f(z) = c(z − z_0)^m$ for some complex numbers $c$ and $z_0$.
However, for $m \geq 2$ such functions are also non-injective: $f(z_0 + 1) = c =
f(z_0 + e^{
2πi/m})$. Thus $m = 1$ and $f(z)$ is a linear polynomial (evidently $c \neq 0$
since $f$ is non-constant).

I am pointed out my 2 questions. Thanks for some help!

答:
If it were removable $g$ would be bounded in a neighborhood of $0$, hence $f$ would be bounded and therefore constant.

A dense set intersects any non-empty open set non-trivially, hence there are $x\in B(0,1)\setminus \{0\}, y\in B(2,1/2)$ with $g(x)=g(y)$, i.e. $f(1/x)=f(1/y)$. As these two sets are disjoint we have $x\ne y$. If $g(B(0,1)\setminus 0)$ were not dense there would be no reason at all why it should intersect with $g(B(2,1/2))$


问:
How to prove that the dimension of a hyperplane is n-1
The hyperplane $H$ defined by $$H:=\{x\in\mathbb {R}^n:a^Tx=b\}$$ is the set that has dimension $n-1$, my question is why or how can we prove that its dimension is $n-1$? Thank you to every one who provide any help or if possible the proof for that.</p>

答:
First of all, I think the way you frame the question is confusing. A hyperplane is n-1 dimensional by definition. You should probably be asking "How to prove that this set- Definition of the set H goes here-  is a hyperplane, specifically, how to prove it's n-1 dimensional"

With that being said. The proof can be separated in two parts:

-First part (easy): Prove that H is a "Linear Variety"

Second Part: Prove that this linear variety H is n-1 dimensional.

I hope the answer from this question would be sufficient

Help with proof: Hyperplane is an $(n-1)$-dimensional linear variety


答:
First question: If it is removable singularity then $f$ is bounded and hence constant by Liouville's Theorem. That contradicts injectivity.

Second question: $B(0,1)\setminus \{0\}$ and $B(2,\frac 1 2 )$ are disjoint. Since $g$ is injective the images of these two sets under $g$ have no point in common. But the dense set $g(B(0,1)\setminus \{0\})$ intersects every non-empty open set, in particular $g(B(2,\frac 1 2))$. So we have  a contradiction.

Body:
I am looking at the proof of the following statement: Let $f$ be an entire function, that is also injective. Then $f$ take the form $f(z) = az+b$ with $a, b \in \Bbb C$ and $a \neq 0$.

The proof is given as follow: Assume $f$ is an entire injective function. Then $f$ is non-constant, so $g(z) :=f(\frac{1}{z})$ has either a pole or an essential singularity at $z = 0$.

1) Why we have that I can't be a removable zero too?

We will show first that the singularity at $0$ cannot be an essential singularity. If it were an
essential singularity, then the Cazorati-Weierstrass theorem would imply that
the set $g(B(0, 1) \setminus {0})$ is dense in $\Bbb C$. However, $g(B(2,\frac{1}{2}))$ is an open set by the open mapping theorem. Therefore these two sets intersect, which shows that
$g(z)$ and hence $f(z)$ is not injective.

2) I don't see why we need to dense set to conclude that $g$ and hence $f$ is not injective.

Therefore, the singularity at $z = 0$ must be a pole, implying that $f(z)$ is a polynomial. Suppose $f(z)$ is a polynomial of degree $m$. Then $f$ has $m$ roots, counting multiplicity. Evidently, if $f$ has two distinct roots, then $f$ is not injective. Thus $f(z) = c(z − z_0)^m$ for some complex numbers $c$ and $z_0$.
However, for $m \geq 2$ such functions are also non-injective: $f(z_0 + 1) = c =
f(z_0 + e^{
2πi/m})$. Thus $m = 1$ and $f(z)$ is a linear polynomial (evidently $c \neq 0$
since $f$ is non-constant).

I am pointed out my 2 questions. Thanks for some help!


问:
Number of all possible ways of arranging N balls, M at a time in M boxes in cyclic fashion
I have a set of $N$ types of balls out of which I want to pick $M$ at a time and arrange in $M$ boxes.

The conditions I have are
1. I am allowed to pick all $M$ balls of the same type.
2. The boxes are arranged in a cycle.
3. An additional constraint is that clockwise and anti-clockwise arrangements are considered to be equivalent.

In how many distinct ways this can be done? Any help is much appreciated.

I give some examples for simple cases.

$M$ = 5, $N$ = 1

A  A  A  A  A

$M$ = 5, $N$ = 2

A  A  A  A  A

A  A  A  A  A

A  A  A  A  B

A  A  A  B  B

A  A  B  A  B

A  A  B  B  B

A  B  A  B  B

B  B  B  B  A

B  B  B  B  B

Note that in the second example the arrangements A A A B B and B B A A A are considered equivalent hence only A A A B B is given. Also other cyclic permutations are ignored.

On the other hand A A A B B and A A B A B are considered distinct because they are not related by cyclic permutations.

答:
If you only have $M$ boxes without the cycle, this is given by the set of words of length $M$ on $N$ letters. There are $N^M$ such words.

Now, let's assume the equivalence up to cycles. In the combinatorics literature, these are called "necklaces," and you can find out more about them here.

If you also want the equivalence up to reflection, see "bracelets" or "turnover necklaces" on the same Wikipedia page.


问:
How to find the first term of a geometric series, given the sum to n is $8 - 2^{3 - 2n}$.
I study maths as a hobby.

I have come across this question:

The sum of the first n terms of a geometric series is $8 - 2^{3 - 2n}$. Find the first term of the series, its common ratio and its sum to infinity.

Now I know that the formula for a geometric series is

$S_n = \frac{a(1 - r^n)}{1 - r}$ and that

$8 - 2^{3 - 2n} = 2^3 - \frac{2^3}{2^{2n}}$ but I cannot see how that leads me to the first term.

答:
You need to put in values, starting from 1,2,3 and so on.
Since it asks for first term, use $n=1$.

Also for finding the sum to infinity, you need to use another formula:

$S$$\infty$ = $a/(1-r)$

where $a$ is the first term of the geometric progression
and $r$ is the common ratio.

Or, like @Rob Arthan mentioned, you can also calculate the sum to infinity by using :

$\lim_{n\to\infty} 8 - 2^{3-2n}$

This would be more useful as you wouldn't have to calculate the common ratio at all.


问:
Generalisation of an inequality to infinite dimensional normed linear spaces
Lemma 6.3 in Rudin's Real and Complex Analysis states that:

For every $n$ and for every $n$-tuple of complex number $z_1,...,z_n$, there exists $S\subseteq \{1,...,n\}$ such that

$$\left| \sum_{k \in S} z_k \right| \geqslant \frac1\pi\sum_{i=1}^{n} | z_i |$$

This can be generalised to $\mathbb R^d$ for all $d$ (see, for example, here) and, exploiting the fact that all norms on a finite dimensional vector space are equivalent, we arrive at the following generalisation

Let $(X,||\cdot||)$ be a finite dimensional normed space. There exists a constant $c>0$, that depends only on $X$ and $||\cdot||$, such that the following holds: for any $n$ and any $n$-tuple $(v_1,v_2,…,v_n)$ of elements of $X$, there exists a subset $J⊆\{1,2,…,n\}$ such that $$\left\| \sum_{j \in J} v_j \right\| \geqslant c \sum_{i=1}^{n} \| v_i \|.$$

This is not always the case for infinite dimensional normed spaces. For example, if $(X,\langle\cdot,\cdot\rangle)$ is an infinite dimensional Hilbert space, pick a countable orthonormal subset $\{e_n\}_{n\in\mathbb N}$. Then for all $n$ and for all subsets $J\subseteq\{0,...,n\}$, we have that $\sum_{i=0}^{n} \| e_i \|=n$ and $\left\| \sum_{j \in J} e_j \right\|=\sqrt{|J|+1}\leqslant\sqrt{n+1}$, but there is no constant $c>0$ such that $\forall n\in\mathbb N:\sqrt{n+1}\geqslant cn$. A similar argument also works for $\ell^p$ for all $p\in(1,\infty]$, however, I wasn't able to apply the same argument for $\ell^1.$

My question is: does there exist an infinite dimensional normed space for which the theorem above holds? I couldn't come up with an example, so i tried to prove that is is never the case. My idea was to use Riesz's lemma, but I got stuck.

答:
One argument would be based on Dvoretzky's theorem, i.e. given Banach space $(X,||.||)$ then there are constants $c_{1}$ and $c_{2}$ such that for any $n$ one can find a a subspace $X_{n}$ of $X$ such that
$$c_{1}*(\sum a_{k}^{2})^{1/2} <= ||\sum a_{k}*x_{k}|| <= c_{2}*(\sum a_{k}^{2})^{1/2}$$
for $x_{k} \in X$ and $a_{k}$ scalars.
In words, $X$ admits uniformly copies of $l_{n}^{2}$.
Then one would use your argument for Hilbert space above.

Now, this is too complex argument and there must be a simpler one but I do not see it ... .


问:
Given $f(x) = x^3 − 3x^2 − 1$, $x \geq 2$, find $f^{−1} {'} (−1)$

Given $f(x) = x^3 - 3x^2 -1, x\ge2$, find $f^{-1}{'}(-1)$

I know I should show work I have done on this question when I ask on stackexchange but I literally get stucked at the beginning. I cannot find the inverse of this function and needless to say the derivative of it. Any help would be appreciated!

答:
You do not need to find the algebraic expression of $f^{-1}(x)$

You just need the derivative of the inverse at $x=-1$

So you calculate where $f(x)=-1;\;x\ge 2$ solving

$$x^3-3 x^2-1=-1;\;x\ge 2\to x=3$$
And apply the rule of the derivative of the inverse
$$f^{{-1}'}(-1)=1/f'(3)$$
As $f'(x)=3x^2-6x$ we have $f'(3)=9$ and finally
$$f^{{-1}'}(-1)=1/9$$


问:
Triangle inequality for cheapest simple paths
Let $G=(V,E,I,W)$ is a directed graph, where $V$ is the set of vertices, $E$ is the set of edges, $I: E \rightarrow V^2$ is an incidence function, and $W: E \rightarrow \mathbb{R}$ is a weight function.

A simple path of length $L$ from $v_1 \in V$ to $v_{L+1} \in V$ through $v_2, v_3, \ldots, v_L \in V$
is a subset of $E$ with edges $e_1, e_2, \ldots, e_L \in E$, where $I(e_i)=(v_i,v_{i+1})$,
$e_i \ne e_j$, and $v_k \ne v_l$ for all  $1 \le i,j \le L$ with $k \ne l$ and $1 \le k,l \le L+1$
with $k \ne l$.
$P=\{e_1,e_2,...,e_L\}$ is a cheapest simple path from $v_1$ to $v_{L+1}$ if
$c(P)=\sum\limits_{i=1}^L W(e_i)$ is minimal, where $c(P)$ is the cost of $P$.

The simple cycle $C=\{e_1,e_2,...,e_{L+1}\}$ is the union of $P$ and $e_{L+1}$
with incidence $I(e_{L+1})=(v_{L+1},v_1)$.
The cost of cycle $C$ is defined as $c(C)=\sum\limits_{i=1}^{L+1} W(e_i)$.

Here only simple paths and simple cycles are considered.

My question is how to prove the following lemma for a graph $G$ without negative-cost simple cycles:

If $Q$ and $R$ are cheapest simple paths from $s \in V$ to $u \in V$
and from $s \in V$ to $v \in V$, respectively, and $\epsilon \in E$
with incidence $I(\epsilon)=(u,v)$, then $c(Q) + W(\epsilon) \ge c(R)$.

答:
There are two cases. If $Q$ does not pass through $v$, then the edge set $Q'$, consisting of the edges of $Q$ followed by $\epsilon$ is a simple path from $s$ to $v$, so by the assumption that $R$ is a cheapest $s-v$ path we have $c(Q) + W(\epsilon) = c(Q') \geq c(R)$.

On the other hand, if $Q$ does pass through $v$, then $Q'$ is made up of a simple $s-v$ path $P$, followed by a simple cycle $C$. Then $$c(Q) + W(\epsilon) = c(Q') = c(P) + c(C) \geq c(R) + 0 = c(R).$$


问:
prove that $Z_n = \sum_{k=1}^{n} (X_k - \mu_k)$ converges a.s.
Let  $\{ X_n, n\in N\}$ be a sequence of rrvs in $L_2 (\omega, A, P)$ with $\mu_n = E(X_n)$ and $\sigma_n^2 = \sigma^2 (X_n) $

Part - 1:

suppose that $\sum_{k=1}^{\infty} \sigma_n^2 < \infty$. Prove that $\{ X_n - \mu_n, n\ge 1\}$ converges to 0 a.s.

Part - 2:

suppose that $\sum_{k=1}^{\infty} n^4 \sigma_n^2 < \infty$. Then prove that $Z_n = \sum_{k=1}^ n (X_k - \mu_k)$ converges a.s.

My solution for part - 1:

By Cheybshev inequality,

$P(|X_n - \mu_n|>e ) \le \sigma^2_n / e^2$ for small enough e>0

Define $A_n = \{ w: |X_n - \mu_n|>e \}$ such that

$P(A_n) = P(|X_n - \mu_n|>e ) \le \sigma^2_n / e^2$

$\sum P(A_n) \le \sum \sigma^2_n / e^2 < \infty$

By Borel Contelli lemma

if $\sum P(A_n) < \infty$, $P(limsup A_n) = P( U \{ X_n- \mu_n\})=0$

Then, $\{ X_n- \mu_n\} \to 0$ a.s.

So far, I hope it is true. (since I am not sure) And I am stack with doing part 2.How can I deal with the part - 2?

答:
You have done the first part correctly.

$E|X_k-\mu_k| \leq \sqrt {E(X_k-\mu_k)^{2}}=\sigma_k$. Note that $\sum \sigma_k =\sum (k^{2}\sigma_k) (\frac  1{k^{2}})$. By C-S/Holder's inequlity we see that $E\sum |X_k-\mu_k|=\sum \sigma_k <\infty$. Hence $\sum (X_k-\mu_k)$ converges absoluetly with probaility $1$.


问:
Finding the number of pairs (a,b) such that gcd(a,b,n)=1
This is question 6E from the 2019 Cambridge Mathematics Tripos paper 1A (which can be accessed at https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2019/paperia_4_2019.pdf).

"Let $n\geq 2$ be an integer. Let $X = \{0,1,2,...,n-1\}$ and $Y=\{(a,b) \in X^2\mid \operatorname{gcd}(a,b,n)=1\}$.

Show that $$|Y|=n^2 \prod_{p \mid n}(1-\frac{1}{p^2})$$ where the product is over all primes $p$ dividing $n$."

The previous part of the question asks respondents to state the inclusion-exclusion principle, so I believe that this is supposed to be a hint.

My thoughts on the solution include the following:

If we have $\operatorname{gcd}(a,n)=k$ and $\operatorname{gcd}(b,n)=j$, then the condition that $\operatorname{gcd}(a,b,n)=1$ seems to imply that:

$k$ and $j$ must be relatively prime

$\operatorname{gcd}(a,b)=m$, where $\operatorname{gcd}(m,k) = \operatorname{gcd}(m,j)=1$.

In fact, these two conditions collectively seem to imply that $\operatorname{gcd}(a,b,n)=1$. But I could not get anywhere with this. I am also aware of the totient function $\phi(n)$, and I've tried to extend the proof that $$\phi(n)=\prod_{p\mid n} n(1-\frac{1}{p})$$ but was unable to. I noticed that the desired form in the question is actually equivalent to $\phi(n^2)$, but was unable to use this to generate a proof.

Any hints / solutions would be appreciated.

答:
Consider first the case of $n$ having just one prime factor $p$, so $n = p^e$ for some integer $e \ge 1$. As you surmised, the inclusion-exclusion principle can be used. To get the elements $(a,b) \in Y$, first include those where $\gcd(a, p) = 1$, with this being $n\left(\frac{p - 1}{p}\right)$ integers, and with $b$ being any integer in $X$, i.e., $n$ of them, for a combined total of $\left(n\left(\frac{p - 1}{p}\right)\right)n$ ordered pairs. Next, include the $(a,b)$ where the roles of $a$ and $b$ are switched, i.e., $a$ is any integer in $X$ and $\gcd(b, p) = 1$, with this resulting in the same product. However, we've double-counted where both $a$ and $b$ are relatively prime to $p$, with this being $n\left(\frac{p - 1}{p}\right)n\left(\frac{p - 1}{p}\right)$ pairs, so these need to be excluded from the sum. Putting this together gives

$$\begin{equation}\begin{aligned}
\left| Y \right| & = 2n^2\left(\frac{p - 1}{p}\right) - n^2\left(\frac{p - 1}{p}\right)^2 \\
& = n^2\left(\frac{2p}{p}\left(\frac{p - 1}{p}\right) - \frac{p^2 - 2p + 1}{p^2}\right) \\
& = \frac{n^2}{p^2}\left(2p^2 - 2p - p^2 + 2p - 1\right) \\
& = \frac{n^2}{p^2}\left(p^2 - 1\right) \\
& = n^2\left(1 - \frac{1}{p^2}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This proves your statement for the case where $n$ has only one prime factor. Next, let

$$f(n) = \left| Y \right| \tag{2}\label{eq2A}$$

Since you've written you're willing to accept just hints, I'll leave it to you to extend \eqref{eq1A} to get the result you're asking to prove where $n$ has any number of distinct prime factors, such as by proving & using that $f(n)$ is a multiplicative function (e.g., by using a similar argument outlined in Euler's totient function of proving a bijection between sets using the Chinese remainder theorem).


问:
A question regarding the Poisson distribution
The number of chocolate chips in a biscuit follows a Poisson distribution with and average of $5$ chocolate chips per biscuit. Assume that the numbers of chocolate chips in different biscuits are independent of each other. What is the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips?

Let $X$ be the number of chocolate chips in a biscuit. We know that $\lambda = E[X]=5$.

Then the probability that each biscuit has more than $7$ chocolate chips is
$$\Pr(X \gt 7) = 1 - \Pr(X \le 6) =p_1,$$

where $p_1$ is a value to be found.

Let $Y$ be the number of biscuits in a box of $20$ that has at more than $7$ chocolate chips. Then the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips is

$$\Pr(Y \le 1) = \Pr(Y=0) + \Pr(Y=1) = \frac{{p_1}^0 e^{-p_1}}{0!} + \frac{{p_1}^1 e^{-p_1}}{1!}=e^{-p_1}(1+p_1).$$

Is this correct? I appreciate your help.

答:
The probability that at least one has more than $7$ is, in the notation of your post,
$$1 -\left(\Pr(X\le 7)\right)^{20}.$$

This because the event "at least one has more than $7$" is the complement of the event "all $20$ have $\le 7$." By independence, the probability that $20$ cookies in a row have $\le 7$ is  $\left(\Pr(X\le 7)\right)^{20}$.

Remark: In the post, you were looking at $\Pr(X\le 6)$. That's not quite the relevant probability, since the problem says more than $7$. Similarly, in the attempted calculation of the probability of at least one, the wrong index was being talked about. The probability of at least one is $1$ minus the probability of none.

In the answer, I assumed you know how to find $\Pr(X\le 7)$. We have
$$\Pr(X\le 7)=\sum_{k=0}^7 e^{-\lambda}\frac{\lambda^k}{k!},$$
where $\lambda=5$. A somewhat tedious calculation!


问:
Reference for Hyperbolic Geometry related with Complex Analysis?
I'm a first year Graduate student.In my first semester complex analysis course there was a topic hyperbolic geometry but due to time limit unfortunately this topic was not touched during the course.Now i am trying to read this topic on my own from the point of view of complex analysis.I tried to find out some books but could not find much,the only book which i found was Gamelin's complex analysis.This book also does not contain,it discuss some hyperbolic geometry on disc only.

What are the interesting result in complex analysis related with hyperbolic geometry? Are there any good notes/book which contains the same material?

Thank you

答: Plane hyperbolic geometry can be represented in the complex half-plane(Poincare half-plane model). This produces certain connexions between complex functions, complex domains and hyperbolic geometry.</p>


问:
Looking for a trick to solve $2\sqrt {2x}+\sqrt {2x+3}=\sqrt {3x+2}+\sqrt {6x+20}$
Consider the equation:

$$2\sqrt {2x}+\sqrt {2x+3}=\sqrt {3x+2}+\sqrt {6x+20}.$$

Find a trick ( if exists ) which allows to solve it elegantly i.e. with avoiding  the  systematic squaring.

(The systematic squaring inevitably leads to a fourth-degree equation:

$$
\begin{align}
0
&=
207x^4-12564x^3+27738x^2+231084x-40401\\[6pt]
&=9\left( 23x^2-1258x-4489\right) \left( x^2-6x+1\right)\;,
\end{align}
$$
so the answer is
$$x=\dfrac {629+\sqrt {498888}} {23}.$$

答:
This equasion is defined for all $x>0$.

Function on the left hand side is strictly increasing (i.e always increasing).

Function on the right hand side is strictly increasing (i.e always increasing) too.

We can draw a simple graph on large scale, but with low prescision, to find out where too look for this root. We'll see on the graph that there is 1 intersection. Beacuse both functions strictly increasing there will be no other roots for $x>0$.

You can use this or any other tool to draw two graphs for left and right hand side functions.

From that graph we'll see that $x \in (58.02 , 58.08)$. You can get the decimal answer depending on the prescision required by incrasing prescision of interval ends.


问:
Does an Abelian group under multiplication modulo $n$ contain $0$?
I read the following definition in a notebook.

The set $\mathbb{Z}^{*}_{n}$ that consists of all integers $i = 0, 1, ..., n-1$ for which the $\operatorname{gcd}(i, n) = 1$ forms an abelian group under the binary operation multiplicaiton modulo $n$. The identity element, of course, is $1$.

But I think it should be,

The set $\mathbb{Z}^{*}_{n}$ that consists of all integers $i = 1, ..., n-1$ for which the $\operatorname{gcd}(i, n) = 1$ forms an abelian group under the binary operation multiplicaiton modulo $n$. The identity element, of course, is $1$.

Or could it be that both the definition are valid?

答: Both are valid. Note that $\operatorname{gcd}(0,n)=n\ne1$ (for $n>1$; but this is not serious restriction). Hence $0$ is excluded anyways.


问:
Norm of functional on $C[-1,1]$ defined as $P(f) = f(1)+f(-1)-2f(0)$
I am studying for my final exam and kinda struggling with the following:

Calculate the norm of functional on $C[-1,1]$ defined as $P(f) = f(1)+f(-1)-2f(0)$

This seems like one-liner but I am clueless anyway. I appreciate your time.

答:
Clearly

$$\vert P(f) \vert \le 4 \Vert f \Vert_\infty.$$ So the norm of $P$ is less or equal to $4$. Now $f(x) = -\cos \pi x$ is such that

$$f(0) = -f(1)=-f(-1)=-1$$ and $ \Vert f \Vert_\infty =1$. So $\vert P(f) \vert = 4$ and $\Vert P \Vert = 4$.


问:
Show that $\lambda\in\sigma_p(T)$ iff $\overline\lambda\in\sigma_r(T')$
Let $X$ be a complex Banach space, $T\in\mathfrak L(X)$ and $\lambda\in\mathbb C$.

Are we able to show $\lambda\in\sigma_p(T)$ (the point spectrum of $T$) if and only if $\overline\lambda\in\sigma_r(T')$ (the residual spectrum of $T':X'\to X'$)?

I'm only able to (partially) prove this claim under the assumption that $X$ is a Hilbert space and $T'$ is replaced by the Hilbert space adjoint $T^\ast:X\to X$:

In that situation, if $S\in\mathfrak L(X)$, then $$\mathcal N(S)={\mathcal R(S^\ast)}^\perp\tag1$$ from which we infer that $S$ is injective if and only if $\mathcal R(S^\ast)$ is dense. Substituting $S=\lambda-T$ and noting that $(\lambda-T)^\ast=\overline\lambda-T^\ast$, it is only left to show that if $\lambda-T$ is not injective (hence $\mathcal R(\overline\lambda-T^\ast)$ is not dense), then $\overline\lambda-T^\ast$ is injective.

How can we show this? And are we able to prove the desired equivalence in the general Banach space case well?

答:
No.  If $X_1=X_2$, and $T$ is given by $T(x)=\lambda x$, with $\lambda\in \mathbb C \setminus \mathbb R$, then
$$
  T'(\phi)=\lambda \phi, \quad\forall \phi\in  X_2',
  $$
so the spectra of both operators coincide with their point spectra and
$$
  \sigma (T) = \sigma (T')=\{\lambda \} \neq  \{\bar \lambda \}.
  $$


问:
Show that if $R$ is reflexive, then $S$ is a subset of $R$ composition $S$
I know how to show that a specific relation is reflexive, but I have problems with more abstract though still basic examples.

Let $R$, $S$ be binary relations on the $X \neq \emptyset$.

Show that:

$R$ is reflexive $\implies$ $S \subset R \circ S$.

答:
By definition, showing that $S \subseteq R \circ S$ amounts to show that if $(x, y) \in S$ then $(x,y) \in R \circ S$.

How can we show it, under the hypothesis that $R$ is reflexive?

Let $(x,y ) \in S$. As $R$ is reflexive, $(x,x) \in R$. By definition of composition, from $(x,x) \in R$ and $(x,y ) \in S$ it follows that $(x,y) \in R \circ S$.
Since we are talking about a generic $(x,y ) \in S$, we have showed that $S \subseteq R \circ S$.

The crucial point in the proof above is the definition of composition for relations: $(x,y) \in R \circ S$ means that there exists $z \in X$ such that $(x,z) \in R$ and $(z,y) \in S$.

In the case above, since $R$ is reflexive, it is easy to find the "intermediate term" $z$ for the composition: just take $x$.


问:
Positiveness of Limit Superior
Let $f:(0,1] \to \mathbb R$ be a bounded non negative integrable function. Assume the limit $$\lim_{x \to 0+} \frac{1}{|\ln(x)|} \int_x^1 \frac{f(t)}{t} \mathrm dt$$ exists and has a positive value. How to show (or give a counterexample) that $\limsup_{t \to 0+} f(t)$ also has a positive value?</p>

答:
Since $f$ is non-negative, then $\limsup\limits_{x\to0+}f(x)\geq\liminf\limits_{x\to0+}f(x)\geq 0$. Assume $\limsup\limits_{x\to0+}f(x)\leq 0$. Then $\lim\limits_{x\to0+}f(x)=\liminf\limits_{x\to0+}f(x)=\limsup\limits_{x\to0+}f(x)=0$. Fix $\varepsilon>0$, then there exist $\delta>0$ such that $|f(x)|<\varepsilon$ for all $x\in(0,\delta)$. In this case
$$
\left|\int_x^1\frac{f(t)}{t}\right|
\leq\left|\int_x^\delta\frac{f(t)}{t}\right|+\left|\int_\delta^1\frac{f(t)}{t}\right|
\leq\int_x^\delta\frac{|f(t)|}{t}+\left|\int_\delta^1\frac{f(t)}{t}\right|
\leq\int_x^\delta\frac{\varepsilon}{t}+\left|\int_\delta^1\frac{f(t)}{t}\right|
\leq\varepsilon(\ln(\delta)-\ln(x))+\left|\int_\delta^1\frac{f(t)}{t}\right|
$$
Hence
$$
\left|\frac{1}{|\ln(x)|}\int_x^1\frac{f(t)}{t}\right|\leq\varepsilon+\frac{1}{|\ln(x)|}\left(\left|\int_\delta^1\frac{f(t)}{t}\right|+\varepsilon\ln(\delta)\right)
$$
and we have the bound
$$
\limsup\limits_{x\to0^+}\left|\frac{1}{|\ln(x)|}\int_x^1\frac{f(t)}{t}\right|\leq\varepsilon
$$
Since $\varepsilon>0$ is arbitrary we conclude
$$
\lim\limits_{x\to0^+}\frac{1}{|\ln(x)|}\int_x^1\frac{f(t)}{t}=0
$$
Сontradiction, so $\limsup\limits_{x\to0+}f(x)>0$</p>

Title: Positiveness of Limit Superior


问:
Taking square roots in equations
When are we allowed to take square roots in an equation?

For example, $x^2 = (x+2)^2$

Are we allowed to take square roots to solve it?

答:
You can take roots, just be aware of the fact that $\sqrt{a^2} = |a|$. So solving $x^2 = (x+2)^2$ is equivalent to solving $|x| = |x+2|$.

UPDATE

Next step is to remember that

if $x\le-2$ then the equation becomes $-x = -(x+2)$, which simplifies to $0=-2$ and so has no solutions;

if $-2<x<0$ then we get $-x=x+2$, which implies $x=-1$

if $x\ge 0$ then we get $x=x+2$, which again has no solutions.

So there is only one solution to the equation at $x=-1$.

Title: Taking square roots in equations


问:
Deriving Vieta formula for polynomials
So I just learned the Vieta formula for both the quadratic and cubic form. I am wondering if there is a form of the Vieta formula for polynomials with more than just 3 terms, and if there is one, is there any way to quickly derive it for a polynomial with many roots so that I can use it?</p>

答:
Take a polynomial of the form $f(x)=a_nx^n+\cdots+a_1x+a_0$. We can w.l.o.g. assume that $a_n=1$ as this does not change anything about the roots and makes the following a bit easier.

Suppose you know all the roots, let's call them $\alpha_1,\dots,\alpha_n$ (there are exactly $n$ counted with multiplicity by the fundamental theorem of algebra). Then:

$$f(x)=(x-\alpha_1)\cdots(x-\alpha_n)$$

And here the fun begins. We note that $(-1)^n\alpha_1\cdots\alpha_n=a_0$ as this is the only term not containing any $x$. We also note that $-(\alpha_1+\cdots+\alpha_n)=a_{n-1}$ as these are all terms of with power $x^{n-1}$. Similarily you can obtain all the Vieta relations (but they are a bit messy in between). For example:

$$\alpha_1\alpha_2+\alpha_1\alpha_3+\cdots+\alpha_{n-1}\alpha_n=\sum_{1\le i<j\le n} \alpha_i\alpha_j=a_{n-2}$$

In general, the coefficients will be so-called elementary symmetric polynomials (as stated on the Wikipedia page as well) in the roots.

Title: Deriving Vieta formula for polynomials


答: Let's work through equivalences:$$\begin{align}x^2=(x+2)^2&\iff x=x+2\lor -x=x+2\\&\iff-x=x+2\\&\iff x=-1.\end{align}$$

Title: Taking square roots in equations

Body:
When are we allowed to take square roots in an equation?

For example, $x^2 = (x+2)^2$

Are we allowed to take square roots to solve it?


问:
Egoroff's Theorem-related
Let $f_n: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue measurable for $n \geq 1$.
Suppose $f(x)=\lim_{n \rightarrow } f_n(x)$ exists almost everywhere.

Show that if $\epsilon>0$, then there exists a measurable set $E \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$ for every $n \geq 1$.

This is sort of like Egoroff's Theorem. But $m(\mathbb{R})=\infty$ in this case. So we can't use Egoroff's theorem.

I tried to follow the prove of Egoroff's Theorem. But not sure how to proceed, because the measure of $\mathbb{R}$ is not finite.

Thanks in advance!

答:
First let us clarify the notation in the question.

The question is:

Let $f_i: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue measurable for $i \geq 1$.
Suppose $f(x)=\lim_{i \rightarrow \infty} f_i(x)$ exists almost everywhere.

Show that, for every $n \geq 1$, if $\epsilon>0$, then there exists a measurable set $E \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$.

Given any $n \geq 1$, we have that $m([-n,n]) <+\infty$. So we can apply Egoroff's Theorem to $f_i$ and $f$ restricted to $[-n,n]$.

By apply Egoroff's Theorem to to $f_i$ and $f$ restricted to $[-n,n]$, we have that:
if $\epsilon>0$, then there exists a measurable set $F \subset [-n,n]$ with $m([-n,n]-F)< \epsilon$ and the convergence is uniform on $F$.

Now, take $E= [-n,n]-F$. Then, since $F \subset [-n,n]$, we have $F= [-n,n]-E$.

So, we have that: if $\epsilon>0$, then there exists a measurable set $E \subset [-n,n] \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$.

So, we have proved that: for every $n \geq 1$, if $\epsilon>0$, then there exists a measurable set $E \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$.

Title: Egoroff's Theorem-related


答: If |a|= |b| then either a= b or a= -b.  So if |x|= |x+ 2| either x= x+ 2 or x= -(x+ 2)= -x- 2.  If x= x+ 2 then, subtracting x from both sides, 0- 2 which, of course, is not true.  x is not equal to x+ 2.    If x= -x- 2, adding x to  both sides, 2x= -2 so x= -1. You can (and should) check back in the original problem:  if x= -1, x^2= (-1)^2= 1 while (x+ 2)^2= (-1+ 2)^2= 1^2= 1.  Yes, they are equal!</p>

Title: Taking square roots in equations


答:
This can be solved instead as $$(x+2)^2-x^2=0=(x+2+x)(x+2-x)=4(x+1)$$

Using the difference of two squares in this way saves the taking of square roots, and can be less error-prone as well as making it clear what is going on.

The "missing solution" from the quadratic is identified by setting $x=\frac 1y$ so that $$\frac {1}{y^2}=\left(\frac 1y+2\right)^2$$ or $$1=\left(1+2y\right)^2$$ which has the solution $y=0$ and indicates a "missing solution at infinity". $y=-1$ corresponds to the original solution for $x$.

Title: Taking square roots in equations


问:
infinitely differentiable function $\exp(-1/(x^2(1-x)^2))$
Given the function :
$f:\mathbb{R}\rightarrow\mathbb{R}$ defined as

$f(x) = \begin{cases}\exp(-1/(x^2(1-x)^2)) &\text{for $x\in (0,1)$} \\0 &\text{otherwise} \end{cases}$

How can I formally show that this function is infinitely differentiable?
This function is clearly quite difficult only on $(0,1)$ and given that $\exp(x)' = \exp(x) \cdot x'$ we have that $(-\frac{1}{x^2(1-x)^2})' = \frac{2\left(2x-1\right)}{x^3\left(x-1\right)^3}$ and this tells me that the derivative will always be of the form "exponential times fraction of polynomials". Should I use induction? Are there any cases I should watch out for?

答:
Here's a solution to show that $f$ is infinitely differentiable at $x = 0$.

By definition
$$ f'(0) = \lim_{h \to 0} \frac{f(h)-0}{h}$$

the limit from the left is clearly $0$ so consider $h \in (0,1)$.

Notice that forall all $x \in (0,1)$ we have

$$ \frac{-1}{x^2(1-x)^2}<\frac{-1}{x^2} < \frac{-1}{x}.$$

Therefore $$0<  \frac{1}{x}\exp\frac{-1}{x^2(1-x)^2} < \frac{1}{x} e^{-1/x}$$

and we have that $f$ is differentiable at $0$ with $f'(0) = 0$  since $1/xe^{-1/x} \to 0.$

We now show that higher order derivatives at $0$ exists and are equal to $0$. That is  $f^{(k)}(0) = 0.$ We will rely on induction.

We've already shown the base case $f'(0) = 0.$

Assume that $f^{(k)}(0) = 0$ then

$$ f^{(k+1)}(0) = \lim_{h \to 0} \frac{f^{(k)}(h)- f^{(k)}(0)}{h} = \lim_{h \to 0} \frac{f^{(k)}(h)}{h}.$$

The limit from the left is clearly $0$ so consider what happens for  $h \in (0,1).$

I make the following claim which can easily be proven by induction

Claim: Let $g(h) = \frac{-1}{h^2(1-h)^2}.$  $$(e^{g})^{(k)} = Q_ke^{g}$$ for some rationnal function $Q_k$

It follows that

$$\frac{f^{(k)}(h)}{h} = \tilde Q_k(h) \exp {g(h)}$$

for some rationnal function $\tilde Q_k = Q_k/h$. By what was done previously we have

$$ \vert \tilde Q_k(h) \exp g(h) \vert \leq \vert \tilde Q_k (h)\exp \frac{-1}{x}\vert$$

by definition we can find $a_i, b_i$ such that

$$ \tilde Q_k(h ) = \frac{b_nh^n + ... + b_1h + b_0 }{a_mh^m +... + a_1h + a_0}$$

If $a_0  \neq 0$ then $\tilde Q_k \exp{-1/x}$ converges to $0$ and we are done.

If $a_0 = 0$ then we can factor out a power of $h$, $h^l$ in the denominator. By taking $l$ as large as possible for this property we have

$$ \lim_{h \to 0} h^{l+1} \tilde Q_k(h) = 0$$

Hence for $h$ small enough we have $\vert \tilde Q_k(h)\vert < \vert \frac{1}{h^{l+1}}\vert$

Therefore for small values of $h$ we have

$$ 0 \leq \left\vert \frac{f^{(k)}(h)}{h}\right \vert \leq \left\vert \frac{1}{h^{l+1}}e^{-1/h}\right\vert.$$

Since the RHS converges to $0$ as $h \to 0$ we deduce that

$$ f^{(k+1)}(0) = 0.$$

To show smoothness at $x = 1$ I suspect (haven't checked) you can use similar arguments but by using the inequality

$$ \exp \frac{-1}{x^2(1-x)^2} \leq \exp \frac{-1}{1-x}$$

and factoring out powers of $(1-h)$ instead of powers of $h$ in the latter part.


问:
Showing that the set of isomorphism class of simple left module is finite
Let $R$ be an Artinian Ring, then prove that the isomorphism classes of simple left $R$-modules is finite.

Morever show that the isomorphims class of simple left modules and the isomorphism classes of simple right modules has the same cardinality.

What am I thinking is to use descending chain condition, but I am unable to use it. Suppose that there are infinitely many isomorphism classes, then I am trying to use Artinian property of $R$, but it doesn't help

答:
The simple right modules of $R$ correspond exactly with those of $R/J(R)$ where $J(R)$ is the Jacobson radical (exercise.)

Suppose $R$ is right Artinian.  Then $R/J(R)$ is a semisimple ring, i.e. a finite product of matrix rings over division rings.

Since all right ideals are summands in such a ring, each simple right module is isomorphic to a minimal right ideal. (A simple module $S$ is isomorphic to $R/M$ for some maximal ideal, but since $M$ is a summand it’s complement is a minimal right ideal isomorphic to $S$.)

Finally it is also easy to show the minimal right ideals of a finite product of rings is as one would expect: the right ideals which are nonzero only on one fixed coordinate, and in that coordinate its values are all from a minimal right ideal in that ring.

Since matrix rings over division rings each have exactly one simple right ideal each, you only have finitely many minimal right ideals in the product, and hence only finitely many isoclasses of simple right modules.


问:
What does it mean by "number of permutations on n elements that are involutions"
https://en.wikipedia.org/wiki/Telephone_number_(mathematics)

This is another question on telephone numbers.
As mentioned in this wikipedia article, telephone numbers are here defined as "the number of permutations on n elements that are involutions"
Can someone please explain what this means?
The article later says this:

"Every pattern of pairwise connections between n subscribers defines an involution, a permutation of the subscribers that is its own inverse, in which two subscribers who are making a call to each other are swapped with each other and all remaining subscribers stay in place. Conversely, every possible involution has the form of a set of pairwise swaps of this type. Therefore, the telephone numbers also count involutions."

I know the concept of an involution, and I know what permutations are, but I can't seem to understand this explanation.

答:
One involution on the numbers 1 to 6 is $(12)(35)$.

This represents 1 phoning 2, 3 phoning 5, and 4 and 6 not talking to anyone


答: Hint: please note that the box has 20 biscuits and they are independent, and the number of chocolate chips in (one*) biscuit follows a Poisson distribution with $\lambda=5$, so first you need to compute the probability that one biscuit has more than 7 chocolate chips and then find the desired probability.</p>

Title: A question regarding the Poisson distribution

Body:
The number of chocolate chips in a biscuit follows a Poisson distribution with and average of $5$ chocolate chips per biscuit. Assume that the numbers of chocolate chips in different biscuits are independent of each other. What is the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips?

Let $X$ be the number of chocolate chips in a biscuit. We know that $\lambda = E[X]=5$.

Then the probability that each biscuit has more than $7$ chocolate chips is
$$\Pr(X \gt 7) = 1 - \Pr(X \le 6) =p_1,$$

where $p_1$ is a value to be found.

Let $Y$ be the number of biscuits in a box of $20$ that has at more than $7$ chocolate chips. Then the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips is

$$\Pr(Y \le 1) = \Pr(Y=0) + \Pr(Y=1) = \frac{{p_1}^0 e^{-p_1}}{0!} + \frac{{p_1}^1 e^{-p_1}}{1!}=e^{-p_1}(1+p_1).$$

Is this correct? I appreciate your help.


问:
Estimate $ \int_{\mathbb R} \left|\frac{d}{dx}f(g(x))\right| dx$, where $f \in BV(\mathbb R)$ and $g$ is invertible
Let $f \in BV(\mathbb R)$ and $g: \mathbb R \to \mathbb R$ be an invertible function. How can we estimate
$$ \int_{\mathbb R} \left|\frac{d}{dx}f(g(x))\right| dx$$
that is, the total variation of the composition? It is bounded by $TV(f) = \int_{\mathbb R} |\frac{d}{d y} f(y)| dy$ (maybe times a constant depending on the derivative of $g$ or of $g^{-1}$)?

答:
First we should note that the total variation is not always the integral of the absolute value of the derivative: $f=\chi_{(0,\infty)}$.

I actually doubt that what you ask is true. But if we assume in addition that $g$ is continuous then it's trivial from the definitions that $TV(f\circ g)= TV(f)$.

Note that $g$ must be monotone; assume wlog that $g$ is increasing. Suppose $x_1\le x_2\le\dots\le x_n$. Then $$\sum|f\circ g(x_{j+1})-f\circ g(x_j)|=\sum|f(g(x_{j+1}))-f(g(x_j))|\le TV(f)$$by the definition of $TV(f)$, since $g(x_1)\le\dots\le g(x_n)$. Now taking the sup of the lhs shows $TV(f\circ g)\le TV(f)$.

Say  $h=f\circ g$. Then $g=h\circ g^{-1}$, so it follows that $TV(g)\le TV(h)$, since $g^{-1}$ is also a bijection.


问:
please help me to solve the integral of equation F as depicted .$F(\mu)=1-\int_{x-(-2\alpha^2\ln\mu)^{1/2}}^{x+(-2\alpha^2\ln\mu)^{1/2}}p(c)dc$
I am a civil engineer and for one part of my thesis I studied an article. In that article a parametric equation p(c) is defined as below:
$$p(c)=
\frac{1}{(2\pi)^{1/2}\sigma}\exp\left(-\frac{(c-u)^{2}}{2\sigma^2}\right)$$   eq.1

and $F(\mu)$ is probability distribution function (CDF)and it is defined as:

$$F(\mu)=1-\int_{x-(-2\alpha^2\ln\mu)^{1/2}}^{x+(-2\alpha^2\ln\mu)^{1/2}}p(c)dc$$   eq.2

in that article the probability density function (PDF) is defined as:

$$p(\mu)=(F(\mu))'=-(\frac{1}{(2\pi)^{1/2}\sigma}\exp(-\frac{(x+(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2})\times(x+(-2\alpha^2\ln\mu)^{1/2})'-\frac{1}{(2\pi)^{1/2}\sigma}\exp(-\frac{(x-(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2})\times(x-(-2\alpha^2\ln\mu)^{1/2})')=\frac{1}{2(2\pi)^{1/2}\mu\sigma}(\frac{-2\alpha^2}{\ln\mu})^{1/2}\times(\exp(-\frac{(x+(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2})+\exp(-\frac{(x-(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2}))$$         eq.3

In my thesis, I need to write these equations in the Simulink function (Matlab). So, I need to know what is the simplified form of equation $F(\mu)$ when we put $p(c)$ in eq.2.

I think it is very complicated and I would be appreciated if anybody can solve the parametric equation $F(\mu)$.

答:
There is no closed-form for the integral. The function $p$ is the density function of the normal distribution.

You can write the result in terms of CDF of standard normal distribution though.

In MATLAB, you can use the Normal cumulative distribution function normcdf to represent $F$.


答:
$$\int \frac{e^{-\frac{(c-u)^2}{2 \sigma ^2}}}{\sqrt{\pi } \sigma } \, dc=\frac{\text{erf}\left(\frac{c-u}{\sqrt{2} \sigma }\right)}{\sqrt{2}}+C$$

$$F(\mu)=1-\int_{x-\sqrt{-2 \alpha ^2 \log (\mu )}}^{x+\sqrt{-2 \alpha ^2 \log (\mu )}} \left(\frac{e^{-\frac{(c-u)^2}{2 \sigma ^2}}}{\sqrt{\pi } \sigma }\right) \, dc=$$
$$=1-\frac{1}{\sqrt{2}}\left[\text{erf}\left(\frac{\sqrt{-\alpha ^2 \log (\mu )}+\frac{u}{\sqrt{2}}-\frac{x}{\sqrt{2}}}{\sigma }\right)+\text{erf}\left(\frac{\sqrt{2} \sqrt{-\alpha ^2 \log (\mu )}-u+x}{\sqrt{2} \sigma }\right)\right]$$

Title: please help me to solve the integral of equation F as depicted .$F(\mu)=1-\int_{x-(-2\alpha^2\ln\mu)^{1/2}}^{x+(-2\alpha^2\ln\mu)^{1/2}}p(c)dc$

Body:
I am a civil engineer and for one part of my thesis I studied an article. In that article a parametric equation p(c) is defined as below:
$$p(c)=
\frac{1}{(2\pi)^{1/2}\sigma}\exp\left(-\frac{(c-u)^{2}}{2\sigma^2}\right)$$   eq.1

and $F(\mu)$ is probability distribution function (CDF)and it is defined as:

$$F(\mu)=1-\int_{x-(-2\alpha^2\ln\mu)^{1/2}}^{x+(-2\alpha^2\ln\mu)^{1/2}}p(c)dc$$   eq.2

in that article the probability density function (PDF) is defined as:

$$p(\mu)=(F(\mu))'=-(\frac{1}{(2\pi)^{1/2}\sigma}\exp(-\frac{(x+(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2})\times(x+(-2\alpha^2\ln\mu)^{1/2})'-\frac{1}{(2\pi)^{1/2}\sigma}\exp(-\frac{(x-(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2})\times(x-(-2\alpha^2\ln\mu)^{1/2})')=\frac{1}{2(2\pi)^{1/2}\mu\sigma}(\frac{-2\alpha^2}{\ln\mu})^{1/2}\times(\exp(-\frac{(x+(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2})+\exp(-\frac{(x-(-2\alpha^2\ln\mu)^{1/2}-u)^2}{2\sigma^2}))$$         eq.3

In my thesis, I need to write these equations in the Simulink function (Matlab). So, I need to know what is the simplified form of equation $F(\mu)$ when we put $p(c)$ in eq.2.

I think it is very complicated and I would be appreciated if anybody can solve the parametric equation $F(\mu)$.


问:
Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?
Let $X$ be a continuous random variable with pdf,
$$f(x;\theta)=\frac {\theta^m.x^{m-1}e^{-\theta x}} {(m-1)!} ; x\geq0, \theta>0$$
Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?</p>

答:
For $m \geq 2$,
$$
\begin{align}
\newcommand{\e}{\mathbb{E}}\newcommand{\rd}{\,\mathrm{d}}
\e \frac{m-1}{X}
= \int_0^\infty \frac{m-1}{x} \frac{\theta^m x^{m-1}}{(m-1)!} e^{-\theta x} \rd x
= \theta \int_0^\infty \frac{\theta^{m-1} x^{m-2}}{(m-2)!} e^{-\theta x} \rd x
= \theta
\end{align}
$$
since the integrand after the second equality is simply a gamma pdf with parameters $m-1$ and $\theta$ and so must integrate to one.

This is closely related to your previous question, by the way.

This "trick" pops up quite frequently in introductory mathematical statistics, so it's worth learning and being able to recognize when it can be applied.


问:
Minimum values of multiple modulus function
Find the minimum value of the equation $|x-1|+|x-2|+|x+1|+|x+2|$.
I checked the values at x=1,x=2,x=-1 and x=-2 and getting the answer 6. Is there any other way to find the minimum values.

答: For an even number of data their median minimizes the sum of the absolute deviations.  As in our case the median equals $0$, the minimum is $6$.


问:
How to solve 4 equation 4 unknown when some there are some product of unknowns
recently I want to solve these four equations

$a(b+c) = 1$

$c^2+cd = 0$

$a^2+bc = 1$

$c(a+d) = 0$

Right now, I can get

$a = \frac{1}{b+c} $

$c = -d$

$a = -d$

However, I cannot actually solve one equation. How can I solve it? Thanks a lot

答:
Your conclusions are only true if $c\ne 0$. In that case, we have $a = c = -d$, and hence $a^2 + ab - 1=0$, which is a quadratic in $a$.

By the quadratic formula, we have $a = \dfrac {-b\pm \sqrt {b^2 + 4}}2 = c = -d$ as the solutions. Every value of $b$ will give a different solution.

For the case $c = 0$, the second and fourth equations are satisfied; we now have $a^2 = ab = 1$, which gives $a =b = \pm1$ and no restriction on the value of $d$.


问:
On maximizing $\operatorname{Trace}(AX)$ subject to $\|X\|=1$
Let $\Bbb S_n (\Bbb R)$ denote the set of $n \times n$ real symmetric matrices. Given $A \in \Bbb S_n (\Bbb R)$, let

$$ X^* := \arg\max_{X \in\Bbb S_n (\Bbb R)} \operatorname{Trace}(AX) \quad\mbox{s.t.}\quad \|X\| = 1 $$

For what set of matrix norms (e.g., Frobenius, spectral, $p$-Schatten, etc) in the problem above, would the maximizer have the following form?

$$ X^* = \frac{1}{\|A\|} A $$

答:
Sometimes the simplest thing to do is draw a picture

Note: This shows the entry-wise norms $L_{p,p}$, i.e. treating $X$ as a $n^2$ vector.


问:
Solving $\sqrt{x+5} = x - 1$
I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:

$$\sqrt{x+5} = x - 1$$

So I used this logic:

$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$

Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.

But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:

$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$

So using 4 works as expected, but when using $-1$:

$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$

At what stage am I going wrong?

And according the WolframAlpha, the solution is $x = 4$.

答:
Function on the left hand side is strictly increasing (i.e always increasing).

Function on the right hand side is strictly increasing (i.e always increasing) too.

We draw a graph to find out where too look for the roots. You can use this or any other tool to draw two graphs for left and right hand side functions. We'll see that there is one interception at $x=4$ and there will be no other interceptions because functions striclty increase.

Title: Solving $\sqrt{x+5} = x - 1$


问:
Construction of characteristics polynomial
Suppose we have a matrix $A$ of order $n$. Let $\lambda$ be an eigenvalue of $A$ and $y$ be a corresponding eigenvector. Then from the relation $Ay = \lambda y$ we get some relation about the coordinates of the eigenvector.

From those relations can we construct the characteristics polynomial?

答:
$Ay = \lambda y \Rightarrow Ay- \lambda y=0$

$Ay- \lambda y=(A-\lambda I)y=0 \Rightarrow Ker((A- \lambda I))\neq \{0\}$

$Ker((A- \lambda I))\neq \{0\} \Rightarrow Det((A- \lambda I))=0$


答:
$$
\begin{cases}
a (b+c)=1\\
c^2+c d=0\\
a^2+b c=1\\
c (a+d)=0\\
\end{cases}
$$
if $c=0$ then $ab=1,a^2=1$ give infinite solutions $(1,1,c,d);\;(-1,-1,c,d)$.

Suppose $c\ne 0$. The system becomes
$$
\begin{cases}
a (b+c)=1\\
c+ d=0\\
a^2+b c=1\\
a+d=0\\
\end{cases}
$$
that is
$$
\begin{cases}
a b+ac=1\\
d=-c\\
a^2+b c=1\\
d=-a\\
\end{cases}\to a=c
$$
$$
\begin{cases}
a b+a^2=1\\
a^2+ab=1\\
c=a\\
d=-a\\
\end{cases}
$$
The system is indeterminate

solving $a^2+ab-1=0$ we get
$$a=\frac{-b\pm\sqrt{b^2+4}}{2}$$
So we have two sets of solutions, with $b$ as parameter
$$a=\frac{1}{2} \left(-\sqrt{b^2+4}-b\right),c=\frac{1}{2} \left(-\sqrt{b^2+4}-b\right),d=\frac{1}{2} \left(\sqrt{b^2+4}+b\right)$$
$$a=\frac{1}{2} \left(\sqrt{b^2+4}-b\right),c=\frac{1}{2} \left(\sqrt{b^2+4}-b\right),d=\frac{1}{2} \left(b-\sqrt{b^2+4}\right)$$

Title: How to solve 4 equation 4 unknown when some there are some product of unknowns

Body:
recently I want to solve these four equations

$a(b+c) = 1$

$c^2+cd = 0$

$a^2+bc = 1$

$c(a+d) = 0$

Right now, I can get

$a = \frac{1}{b+c} $

$c = -d$

$a = -d$

However, I cannot actually solve one equation. How can I solve it? Thanks a lot


问:
Necessary and sufficient conditions for the existence of a derivative at one point
Let $f:[-1, 1] \rightarrow \mathbb{R}$ be differentiable in $[-1, 1] \setminus \{0\}$. Consider the following condition: The left and right limits of $f'(x)$ at $0$ exist and are finite and equal.

a. Show, through examples, that the condition is not sufficient nor necessary for the existence of $f'(0)$.

b. Add the condition.

This exercise reminded me of the definition of a point of discontinuity of the third kind but I don't know how to proceed.

答:
$a.$ Take $f(x) =x \mbox{ for } x>0$ and $f(x) =x+1 \mbox{ for } x\leq 0$

$b.$ If we add that $f$ is continuous then $f$ is differentiable at $0$.


问:
Aggregate Relationships
Which of the following represents an aggregate relationship (has-a)?

Parent and child.

Animal and dog.

teacher and computer

phone and fax machine

All of the bove

The correct answer is 3. But why is not 1 also an aggregate relationship?

答:
It depends on your application. Using UML specification:

Sometimes a Property is used to model circumstances in which one instance is used to group together a set of instances;
this is called aggregation.

whereby there are different possibilities:

none: Indicates that the Property has no aggregation semantics.

shared: Indicates that the Property has shared aggregation semantics. Precise semantics of shared aggregation
varies by application area and modeler.

composite: Indicates that the Property is aggregated compositely, i.e., the composite object has responsibility for
the existence and storage of the composed objects.

Now, the key difference between composition and aggregation is that "in composition, when the owning object is destroyed, so are the contained objects", while aggregation "does not imply ownership" (see wikipedia).

The child can not survive without its parent in this scenario. Whether this makes sense or not depends on the used model.

Title: Aggregate Relationships


问:
Is $\lim_{n \to \infty} n((1-\frac{1}{n})^n - \frac{1}{e}) = 0$?
Intuitively I think this statement is true, but I am unable to proof it. Can someone help me? If possible, I would like bound $\vert(1-\frac{1}{n})^n - \frac{1}{\mathrm{e}}\vert$ (or even $\vert{(1+\frac{1}{n})^n - \mathrm{e}}\vert$) because I think this is a good think to know in general.

答:
$$\left(1-\frac{1}{n}\right)^n\sim \frac{1}{e}-\frac{1}{2 e n};\;n\to\infty$$
$$\lim_{n \to \infty}n\left(\left(1-\frac{1}{n}\right)^n - \frac{1}{e}\right) = \lim_{n \to \infty}n\left(\left(\frac{1}{e}-\frac{1}{2 e n}-\frac{1}{e}\right)\right)=$$
$$=-\frac{1}{2e}$$


问:
Finding the number of Partitions in a set into two subsets
Here is the problem:

Let A be a set with n elements. Find an expression for S(n, 2), the number of partitions of A into exactly two subsets. You can either start with the general recurrence for S(n, k), or count S(n, 2) directly.

I'm having trouble understanding what exactly it wants me to do. So far I know that it wants me to find an expression that gives the number of combinations of A into sets like (x, y). I got the equation: (n - 1) * 2 for number of partitions of A into two parts, but I don't know understand what the problem means by creating two subsets. What am I doing wrong?

答:
I think that you need S(n, 2)=2^(n-1)+1, assuming that both sub-sets are required to not be empty.

For example n=3 gives 3 subsets (i.e {{1,2}, {3}}, {{1,3}, {2}}, and {{2,3}, {1}}).

There is a recurrence relation: S(n+1,2)=2*S(n,2)+1

This can be interpreted as saying that when you and a new element, then this element can be put in any of the previous sub-sets, or in a sub-set by itself. So, for S(4,2) you would have:
{{1,2,4}, {3}}, {{1,3,4}, {2}}, and {{2,3,4}, {1}}) as well as
{{1,2}, {3,4}}, {{1,3}, {2,4}}, and {{2,3}, {1,4}}), and as well as
{{1,2,3}, {4}}

So S(4,2)=2*3+1=7 etc.

In general, there is a recurrence relation: S(n+1,k)=kS(n,k)+S(n,k-1) for k between 1 and n. This can be interpreted in the same way as the example where k=2. [and the sum over all k for any fixed n is the number of partitions, which is known as Bell's number]


问:
Determining the kernel and image of a Linear Map
Hello to all math people,

maybe there is someone who could help me with this task?

$V$ is a $K$-vector space and $U$ is a subspace of $V$. The Linear Map to consider is:

$$f : V → V/U, v ↦ v + U$$

Problem: Determine the Kernel and Image of this Linear Map.

Thought on this: Do I get it right, that $\ker f= U$; because only for $u∈U$ we have $u+U=U = 0$ in $V/U$?

Thanks for any help!

答:
You are correct, the kernel is equal to $U$ since $v+U=U$ (which is the zero in $V/U$) if and only if $v\in U$.

Since $V/U$ consist of all cosets $v+U$ with $v\in V$ by definition, the image of $f$ is all of $V/U$.


问:
Disproving asymptotic lower bound of central binomial coefficient
This is sort of a cross-post from here, since I'm looking for a more mathematical proof.

I was recently learning about binomial coefficients and was wondering about how to disprove 2nCn (or the central binomial coefficient) not being lower-bounded by 4^n; in other words:

$$\binom{2n}{n} \notin \Omega (4^n)$$

Some extremely generous bounds can be easily constructed, such as the following:

$$\frac{4^n}{2n+1} \leq \binom{2n}{n} \leq 4^n \text{ for all } n \geq 1$$

I sought to prove by contradiction, so to assume:

$$\exists c_1 \text{ or } c_2 : 0 \lt c_14^n \leq \frac{4^n}{2n+1} \leq c_24^n \leq \binom{2n}{n} \leq 4^n$$

Clearly, $c_1$ cannot exist, since $\lim_{n \rightarrow \infty}{\frac{1}{2n + 1}} = 0$. It can also be seen that $c_2 \in (0,1]$. And... I'm stuck. Intuitively, it seems rather obvious that $c_2$ cannot exist.

I am aware a similar question has been asked here, but there wasn't really a proof provided. I'm also aware that you could prove $\lim_{n \rightarrow \infty}\frac{\binom{2n}{n}}{4^n} = 0$, but I was wondering if there was another way to do so - particularly, by proving that $c_2$ cannot exist.

答: You can prove by induction on $n\geq 1$ that $\binom{2n}{n} \leq \frac{4^n}{\sqrt{n}}.$


问:
Compatibility of a metric for a given topological space.
It is indeed a very easy question but I am not %100 sure about it.

Assume you are given a topological space $(X,\tau_x)$ with base $\mathcal B$.

And you are given a metric on $X$. We know that $d$ induces a topological space with base $\mathcal B_d=\{B(x,\epsilon)|x\in X, \text{for some}\quad  \epsilon>0 \}$

Is the problem showing $d$ is compatible with given $(X,\tau_x)$, actually showing the given topological space is equivalent to the induced by $d$?

答:
Yes, $\mathcal{B}$ induces the  topology $\mathcal{T}_X$ (the original topology on $X$),by taking unions, and a metric $d$ induces a base $\mathcal{B}_d = \{B(x,r): x \in X; r>0\}$.

Your task is then indeed to show that the topology induced by $\mathcal{B}_d$ is also exactly $\mathcal{T}_X$, or any $B \in \mathcal{B}$ is metric open and vice versa. So you need

$$\forall B \in \mathcal{B}: \forall x \in B: \exists r>0: B(x,r) \subseteq B\tag{1}$$

and also

$$\forall x \in X: \forall r>0: \exists B \in \mathcal{B}: x \in B \subseteq B(x,r)\tag{2}$$


问:
How can $B(a)$ be utilized in the attached question in graph theory
Let $G := (V,E)$ be a graph. Show that there is a function $α : V → \{0,1\}$
such that, for each $v ∈ V$, at least half of the neighbours of $v$ have a different
$α$-value than $v$.
Here's the question. It is part of my Graph Theory homework and I've done every other problem except this one... I've tried an approach utilizing the fact that every graph $G$ contains a bipartite subgraph with at least half the edges, but that didn't work. Can somebody tell me how this hint is related to other theorems in Graph Theory? Because I'm really a little stuck around here.

答:
As you write in the comments, let $B(\alpha)$ denote the number of edges of $G$ that have different $\alpha$-values on the endpoints.

Hint: choose $\alpha$ so that $B(\alpha)$ is maximal and show that it has the desired property.

Solution:

 If there was a vertex $v$ such that less than half of its neighbours have $\alpha$-values different from $\alpha(v)$, then changing $\alpha(v)$ would increase $B(\alpha)$.


问:
Quadratic Equations Mathematical Method
The question is :

If $a,b$ are integers such that all the roots of the equation $ ( x^2+ax+20)(x^2+17x+b)=0 $ are negative integers. What are the smallest possible vaues of $a,b$?

My approach to this goes like this:

Either $( x^2+ax+20)=0$             $\rightarrow$   $(1)$

or  $(x^2+17x+b)=0 $     $\rightarrow$   $(2)$

Let the roots of $(1)$ be $\alpha $ , $\beta $

so, $\alpha $ + $\beta $ = $- a $

and $\alpha \beta $ = $20$

Now, I just put in possible values and subsequently got $-4,-5$ as the numbers which lead to the smallest possible sum, $i.e  $     $  9$.

Let the roots of $(2)$ be $\gamma $ , $\delta $

so, $\gamma $ + $\delta $ = $- a $

and $\gamma \delta $ = $20$

Again, I just put in the possible values and got $-1,-16$ as the numbers which lead to the smallest possible product, $i.e  $     $  16$.

Of course the value of $a+b=25$ but I want to know the proper mathematical method to reach the answer. Please help me with the same. Thanks!

答:
Minimize $a=(-\alpha)+(-\beta)$ subject to $(-\alpha)(-\beta)=20$ using $a^2=80+((-\alpha)-(-\beta))^2$, implying $-\alpha,\,-\beta$ need to be as close together as possible, namely $4,\,5$ in some order.

Minimize $b=(-\gamma)(-\delta)$ subject to $(-\gamma)+(-\delta)=17$ using $4b=17^2-((-\gamma)-(-\delta))^2$, implying $-\gamma,\,-\delta$ need to be as far apart as possible, namely $1,\,16$ in some order.

Title: Quadratic Equations Mathematical Method


问:
Converting Propositional Statements into Logics
Suppose

A= Ali has passed the exam

B= Bilal has passed the exam

U= Umar has passed the exam

S= Sana has passed the exam

The statement to perform is:

If Ali has passed the exam then Bilal and Umar have also passed the exam.

Ali can only pass the exam if Bilal and Umar has not passed, but Sana passed the exam.

Ali, Bilal and Umar has passed the exam if and only if Sana has not passed the exam, but if
neither Ali nor Bilal pass the exam then Sana passed only if Umar passed.

This is What I have done so far

$A \to (B\land U)$

$A \to (\neg B\land \neg U) \land S$

$[(A\land B\land U) \leftrightarrow \neg S)] \land  (\neg A\land \neg B) \to (S \to U)$

I just wanted to confirm that, What I did is correct? or If it is not correct what changes I needed to perform... Thanks.

答:
Basically correct (good job!) ... though you need a few extra parentheses:

$A \to (B\land U)$

$\color{red}[ A \to (\neg B\land \neg U)\color{red}] \land S$

$[(A\land B\land U) \leftrightarrow \neg S)] \land \color{red}[ (\neg A\land \neg B) \to (S \to U)\color{red}]$


问:
If $p(x)\leq 1$ for $x\in [-1,1]$ find the greatest value of $\int_{-1}^{1}p(x)dx$
Let p be a polynomial of degree at most 4 such that $p(-1)=p(1)=0$ and $p(0)=1$. If $p(x)\leq 1$ for $x\in [-1,1]$ find the greatest value of $\int_{-1}^{1}p(x)dx$

My Attempt:
Using the given information I was able to obtain a 4 degree polynomial as $f(x)=ax^4+bx^3-(a+1)x^2-bx+1$.

Thus $$\int_{-1}^1p(x)dx=2\int_{0}^{1}\left(ax^4-(a+1)x^2+1\right)dx=\frac{4}{3}-\frac{4a}{15}$$

Answer given is $\frac{8}{5}$.

I have not been able to apply the information that $p(x)\leq 1$ for $x\in [-1,1]$

How to use it

答: Since $f(0) = 1$, if $f$ does not exceed $1$ then $0$ must be a local maximum. Therefore $f'(0) = 0$ and $f''(0) \le 0$. This implies $b=0$ and $a \ge -1$. By your calculation, the integral is maximised with $a$ as negative as possible. Therefore $a=-1$.


问:
A question regarding the Poisson distribution
The number of chocolate chips in a biscuit follows a Poisson distribution with and average of $5$ chocolate chips per biscuit. Assume that the numbers of chocolate chips in different biscuits are independent of each other. What is the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips?

Let $X$ be the number of chocolate chips in a biscuit. We know that $\lambda = E[X]=5$.

Then the probability that each biscuit has more than $7$ chocolate chips is
$$\Pr(X \gt 7) = 1 - \Pr(X \le 6) =p_1,$$

where $p_1$ is a value to be found.

Let $Y$ be the number of biscuits in a box of $20$ that has at more than $7$ chocolate chips. Then the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips is

$$\Pr(Y \le 1) = \Pr(Y=0) + \Pr(Y=1) = \frac{{p_1}^0 e^{-p_1}}{0!} + \frac{{p_1}^1 e^{-p_1}}{1!}=e^{-p_1}(1+p_1).$$

Is this correct? I appreciate your help.

答:
Yes, you are on the correct track, but this might be a bit easier to reason about.

Change the problem into a Bernoulli random variable problem by considering this event: a biscuit has more than 7 chips or not. Let $$p = P(X > 7) = \sum_{i = 8}^{\infty}\frac{\lambda^ie^{-\lambda}}{i!},$$ where $\lambda = 5$.

Then, our problem becomes what is the probability that out of 20 biased coin flips with probability $p$ of success, at least one will succeed?

This would be found using the reasoning: what is the probability that NONE of them succeed? $(1 - p)^{20}$. We then subtract this from one to get our desired value: $$1 - (1 - p)^{20}.$$

$p$ can be calculated using a computer, by the way. It is simply a matter of routine calculations.


问:
Three unit vectors whose sum is zero
Let $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ be unit vector such that $\overrightarrow a  + \overrightarrow b  + \overrightarrow c  = 0$. Which of the following is correct ?

(A) $\overrightarrow a  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c  = \overrightarrow c  \times \overrightarrow a  = \overrightarrow 0 $

(B) $\overrightarrow a  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c  = \overrightarrow c  \times \overrightarrow a  \ne \overrightarrow 0 $

(C) $\overrightarrow a  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c  = \overrightarrow a  \times \overrightarrow c  \ne \overrightarrow 0 $

(D) $\overrightarrow a  \times \overrightarrow b ,\overrightarrow b  \times \overrightarrow c ,\overrightarrow c  \times \overrightarrow a $ are mutually perpendicular

The official answer is (B).

My approach is s follow

$\overrightarrow a  + \overrightarrow b  + \overrightarrow c  = 0$

$\overrightarrow a  \times \overrightarrow b  + \overrightarrow b  \times \overrightarrow b  + \overrightarrow c  \times \overrightarrow b  = 0$

$\overrightarrow a  \times \overrightarrow b  =  - \overrightarrow c  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c $

$\overrightarrow a  \times \overrightarrow c  + \overrightarrow b  \times \overrightarrow c  + \overrightarrow c  \times \overrightarrow c  = 0 \Rightarrow \overrightarrow b  \times \overrightarrow c  = \overrightarrow c  \times \overrightarrow a $.

Now I am confused between option (A) and (B) if $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are parallel to each other then option (A) is correct or option (B)

Vector being a unit vector plays any vital role in deciding between (A) and (B)

答: (A) would imply the vectors are pairwise parallel or antiparallel, so their sum is of length $1$ or $3$.

Title: Three unit vectors whose sum is zero


问:
Discontinuity points in linearly ordered space
It is a simple well known result that given a monotonic function $f:\mathbb{R}\to \mathbb{R}$, its discontinuity points can only be jump discontinuities.  My question is probably a simple one. Given an order preserving map $f:(X,\mathcal{T}_X)\to (Y,\mathcal{T}_Y)$, where these space are linearly topological ordered spaces, can we say that a discontinuity point $x_0\in X$ of $f$ is a jump discontinuity in general?
i.e., can we say that
$$ \underset{x<x_0}{\sup} f(x)< \underset{x>x_0}{\inf} f(x)? $$

Do we need to assume further that $Y$ is a complete lattice, or is this true in the general case?

答:
Hint/idea : for any order preserving function $f$ we already know, for any $x_0$ (which is not the minimal or maximal element of $X$) that

$$\sup_{x < x_0} f(x) \le f(x_0) \le \inf_{x > x_0} f(x)\tag{1}$$

So it suffices to show that if equality holds between the leftmost and rightmost term, $f$ is in fact continuous at $x_0$. I think this is not too hard using the order topology definitions and order preservingness.


答:
If $\vec a, \vec b$ are parallel, they have the same direction.

Since they are both unit vectors, this implies $\vec a = \vec b$ and $\vec c = -2\vec a$.

But now $|\vec c| = |(-2\vec a)| = 2$, contradicting the fact that $\vec c$ is also a unit vector.

Title: Three unit vectors whose sum is zero

Body:
Let $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ be unit vector such that $\overrightarrow a  + \overrightarrow b  + \overrightarrow c  = 0$. Which of the following is correct ?

(A) $\overrightarrow a  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c  = \overrightarrow c  \times \overrightarrow a  = \overrightarrow 0 $

(B) $\overrightarrow a  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c  = \overrightarrow c  \times \overrightarrow a  \ne \overrightarrow 0 $

(C) $\overrightarrow a  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c  = \overrightarrow a  \times \overrightarrow c  \ne \overrightarrow 0 $

(D) $\overrightarrow a  \times \overrightarrow b ,\overrightarrow b  \times \overrightarrow c ,\overrightarrow c  \times \overrightarrow a $ are mutually perpendicular

The official answer is (B).

My approach is s follow

$\overrightarrow a  + \overrightarrow b  + \overrightarrow c  = 0$

$\overrightarrow a  \times \overrightarrow b  + \overrightarrow b  \times \overrightarrow b  + \overrightarrow c  \times \overrightarrow b  = 0$

$\overrightarrow a  \times \overrightarrow b  =  - \overrightarrow c  \times \overrightarrow b  = \overrightarrow b  \times \overrightarrow c $

$\overrightarrow a  \times \overrightarrow c  + \overrightarrow b  \times \overrightarrow c  + \overrightarrow c  \times \overrightarrow c  = 0 \Rightarrow \overrightarrow b  \times \overrightarrow c  = \overrightarrow c  \times \overrightarrow a $.

Now I am confused between option (A) and (B) if $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are parallel to each other then option (A) is correct or option (B)

Vector being a unit vector plays any vital role in deciding between (A) and (B)


答:
If they are all parallel to each other then $\vec{b} = k\vec{a} $, $\ \vec{c} = m\vec{a} $ which means that $(1+k+m)\vec{a} = \vec{0} $

We can't have $\vec{a} = \vec{0}$ since $\vec{a}$ is a unit vector. Similarly   we notice that since all three are unit vectors then $k$ and $m$ must be either 1 or -1 so $k + m + 1 \neq 0$ which proves that you can't have three linearly dependent unit vectors that are parallel to each other (try to think of this geometrically - what would the sum of two parallel unit vectors on a plane be?)

Title: Three unit vectors whose sum is zero


答:
$$x^2+a x+20=0\to x=\frac{-a\pm\sqrt{a^2-80}}{2}$$
as $a$ is the sum of the solutions, if we want them both negative we must have

$a>0;\;a^2-80\ge 0$ that is $a\ge 4\sqrt 5$.

As is given $a$ integer, then must be $a\ge 9$. If $a=9$ then $x_1=-4;\;x_2=-5$.

In a similar way

$$x^2+17 x+b=0$$ has both solutions negative if $b\ge 1$. If $b=16$ then $x_1=-1;\;x_2=-16$

Thus the minimum values are $a=9;\;b=16$

Title: Quadratic Equations Mathematical Method

Body:
The question is :

If $a,b$ are integers such that all the roots of the equation $ ( x^2+ax+20)(x^2+17x+b)=0 $ are negative integers. What are the smallest possible vaues of $a,b$?

My approach to this goes like this:

Either $( x^2+ax+20)=0$             $\rightarrow$   $(1)$

or  $(x^2+17x+b)=0 $     $\rightarrow$   $(2)$

Let the roots of $(1)$ be $\alpha $ , $\beta $

so, $\alpha $ + $\beta $ = $- a $

and $\alpha \beta $ = $20$

Now, I just put in possible values and subsequently got $-4,-5$ as the numbers which lead to the smallest possible sum, $i.e  $     $  9$.

Let the roots of $(2)$ be $\gamma $ , $\delta $

so, $\gamma $ + $\delta $ = $- a $

and $\gamma \delta $ = $20$

Again, I just put in the possible values and got $-1,-16$ as the numbers which lead to the smallest possible product, $i.e  $     $  16$.

Of course the value of $a+b=25$ but I want to know the proper mathematical method to reach the answer. Please help me with the same. Thanks!


问:
How to change variables in surface integral
Now I know how to change variable of multiple integral, and I'd like to find out how to do this on surface integral. Like this
$$\iint_\Sigma x^2 \,dy\,dz+y^2\,dz\,dx+z^2\,dx\,dy,\,where \Sigma:(x-1)^2+(y-1)^2+\frac{z^2}{4}=1(y>1)
$$
I think it would be convenient if let
$$u=x-1,v=y-1,w=\frac{z}{2}
$$
It's just an example, you don't have to solve this.

答:
$\displaystyle \iint_S x^2 \,dy\,dz+y^2\,dz\,dx+z^2\,dx\,dy,\ $

where $S:(x-1)^2+(y-1)^2+\frac{z^2}{4}=1(y>1)$

Please note,

$\displaystyle \iint_S x^2 \,dy\,dz+y^2\,dz\,dx+z^2\,dx\,dy = \iint_S \vec{F} \cdot \hat{n} \ dS , $ where $\vec{F} = (x^2,y^2,z^2)$.

If you parametrize the surface in spherical coordinates as,

$r(\theta, \phi) = (1 + \cos\theta \sin\phi, y = 1 + \sin\theta \sin \phi, z = 2\cos\phi)$, the normal vector is $(2\cos\theta \sin^2\phi, 2\sin\theta \sin^2\phi, \sin\phi \cos\phi); 0 \leq \phi \leq \pi, 0 \leq \theta \leq \pi$ (Note $y \geq 1)$

$\vec{F} = ((1+\cos\theta\sin\phi)^2,(1+\sin\theta\sin\phi)^2,4\cos^2\phi)$

Now you do the dot product and integrate.

Or you could parametrize in cylindrical coordinates as $(1+\rho\cos\theta, 1+\rho\sin\theta, 2 \sqrt{1-\rho^2})$.

Or in cartesian, $u = x-1, v = y-1, w = \frac{z}{2}$.

But it may be a bit easier using divergence theorem,

To apply divergence theorem, we close the surface by adding an elliptic disc at $y=1$. We can subtract flux through it later.

$\nabla \cdot \vec{F} = 2(x+y+z)$

Using $u = x-1, v = y-1, w = \frac{z}{2}, |J| = 2$, the region has transformed to $R: u^2+v^2+w^2 = 1, v \geq 0$ which is a hemisphere of unit radius.

Divergence is rewritten as $2(x+y+z) = 2(2+u+v+2w)$

So the volume integral to find flux through closed surface is,

$\displaystyle \iiint_R 2(2+u+v+2w) \ |J| \ du \ dv \ dw$

Now we note that $u$ and $w$ are both odd function and due to symmetry wrt $VW$ plane and $UV$ plane respectively, the integral will be zero. The reason the integral is not zero for $v$ as our transformed surface is hemisphere with $v \geq 0$. So it boils down to,

$8 \displaystyle \iiint_R \ du \ dv \ dw + 4 \displaystyle \iiint_R v \ du \ dv \ dw$

First part is simply $8$ times volume of the hemisphere of radius $1$. Now you can find the second part and add.

Then subtract flux through elliptic disc that we added. Normal vector for the disc is $(0,-1,0)$ and $y = 1$.

$\vec{F}\cdot\hat{n} = (x^2,y^2,z^2) \cdot (0,-1,0) = - 1 \ (y = 1)$.

Ellipse has semi axes $a = 1, b = 2$ and its area is $2\pi \ $ (using $\pi ab)$. The flux through the surface is $-2\pi$.


问:
Visualizing the domain of the square root
I would like to show someone the domain of the complex square root function (the 2-sheeted riemann surface).  Is there a good interactive visualization software for this?

I would like some sort of GeoGebra style app where there is a forbidden point or disk, but outside of that you can drag a point around.  As you drag around the pole, it changes from red to blue smoothly, so that on the bottom sheet it is red, and on the top sheet it is blue (or some reasonable periodic color scheme where 50% apart is always very distinct).

It would be doubly nice if one could have simple geometric shapes do the same.  Basically I want a nice double covering of a plane symmetry group that is still very geometric.

It would be n-tuply nice if one could handle nth roots and the n-sheeted Riemann surface, but n=2 suffices for me, I think.

I've used other people's geogebra apps, but never made my own, and have no idea how to keep track of what sheet the point is on (or honestly how to animate the color, though I assume once I have a 0…4π valued argument function, I should be fine).

Some pretty images from wikipedia:

答:
Here is a visualization I have coded for moving points around on the Riemann surface for $f(z) = z^{1/n}$; points are colored according to the argument of $f(z)$ on the branch currently shown, fading to white with decreasing radius. Click to toggle a view with $z$ shown in black and $f(z)$ shown in white.

By default, the program displays $z^{1/3}$, but this can be changed by clicking on the code icon at the top and altering the line n = 3.


问:
Studying for the Putnam Exam
This is a question about studying for the Putnam examination (and, secondarily, other high-difficulty proof-based math competitions like the IMO). It is not about the history of the competition, the advisability of participating, the career trajectories of former participants, or other such things. This is a question about how one can most effectively prepare to do well.

Many such questions have been asked before on this site. The answers, while helpful, all seem to boil down to recommendations of the same standard canon of books with the encouragement to do more problems. This is very good advice, but I want to approach the topic from a different angle.

We have a wealth of talent on this website, and in particular many users who have done quite well at these competitions. I would like to hear their personal stories. In particular, how did you prepare, and what kind of time commitment did you put forth? And advice, especially practical study tips, is always welcome. :)

edit: Also, it seems to me that the majority of people who do very well at the Putnam had developed the majority of their skill in high school, and focused mainly on their classes in college (while attending whatever Putnam seminar their college offered). I would appreciate comments on this matter too.

(Motivation: It seems to me like the role of talent is vastly overestimated in mathematics, and in mathematics competitions in particular, to the point where the Putnam exam gets used as a sort of pseudo-IQ test. Of course the people who do well have gifts, but it also seems that, without fail, they all have a history of doing many hours of mathematics a day for years on end. For example, I recall reading an interview with Tao where he admitted his childhood consisted of nothing but math and computer games. I am trying to gather some evidence on this matter [and advice for Putnam preparation!]. Please keep the answers focused on the actual question, though.)

答:
I took the Putnam four times, $1965$-$1968$. I never knew my scores, but my ranks, if I remember correctly, were in the $80$s my first year, abysmal my second year — in the $700$s, I think — and somewhere between $100$ and $150$ my last two years. (My background and interests just weren’t well-suited to that $1966$ exam; the other three exams suited me much better.) That’s hardly a spectacular Putnam career, but it’s certainly respectable.

I never did anything specifically to prepare for for the exam: I was simply taking it for fun.

However, I’d known since before I was ten that I was going to be a mathematician, teaching mathematics at the college/university level. By the time I actually went off to college, I’d completed standard college courses in calculus, advanced calculus, differential equations, and complex variables. I’d also had a course in elementary number theory and another combining bits of group theory and linear algebra in the Ross Mathematics Program, then at Notre Dame, the summer after my first year of high school. It’s perfectly true, therefore, that I’d had a lot of exposure to mathematics, including writing it, before I first took the Putnam. At the same time I should point out that I had quite a few other interests and by no means spent all my time on mathematics and closely related subjects.

A couple of points come to mind that seem worth mentioning. That was a different era: there was nothing like the number of mathematics competitions that are available today, and there was less that one could easily do to prepare for the Putnam. Nowadays it’s much easier to gather competition experience, if one has the talent and interest, and I’ve a strong suspicion that such experience confers a significant advantage. It’s also much more common now for a mathematics department to run problem-solving seminars, or even courses. On the other hand, I would not have been much interested in developing specifically competition-oriented skills even if I’d had the opportunity: I enjoy solving problems that I find interesting, and a lot of the standard competition tricks and topics simply don’t much interest me and never did. I’d have considered my time much better spent learning more ‘real’ mathematics.

Title: Studying for the Putnam Exam