问:
Trying to understand correlation and independence geometrically
I am trying to understand correlation and independence of two random variables geometrically, but found it difficult to grasp the intuition and to explain it rigorously.

First, given two uniformly distributed RVs X,Y, and given the distribution of (X,Y) and (-X,Y) are the same, how can we show and explain $\mathbb{E}XY=\mathbb{E}(-X)Y=-\mathbb{E}XY$. From this equality, is there any way to show whether X,Y are independent or not, uncorrelated or not?

Suppose consider a point with cartesian coordinates (X,Y), on the unit circle with uniformly distributed X and Y. Are X,Y uncorrelated and why? Are X,Y independent and why?

My attempt to reason with the unit circle is that if we pick a point outside the circle then $0\not=\mathbb{E}X\mathbb{E}Y\not=\mathbb{E}XY=0$, am I right?

How about on the unit square?

Sorry for so many questions at once, I am trying to understand the concept as a whole.

Thanks.

答:
I will answer your question about the unit circle. I think what you meant to say is that the point with coordinates $(X,Y)$ is chosen uniformly from the unit circle. This doesn't mean that $X$ and $Y$ are themselves uniform. In fact, they're not.

$X$ and $Y$ are not independent. Intuitively, two variables are independent if knowing something about one of the variables doesn't affect the probability distribution of the other variable (more precisely, the conditional density of $Y$ given $X=x$ is the same as the density of $Y$). But here, it does. For example, if we know $X=0$, then the conditional distribution of $Y$ is uniform between $-1$ and $1$ (think of a vertical cross-section at $X=0$). But if we know $X=3/5$, then the conditional distribution of $Y$ is uniform between $-4/5$ and $4/5$ (vertical cross-section at $X=3/5$). Thus, knowing things about $X$ changes the distribution of $Y$, so $X$ and $Y$ are not independent.

$X$ and $Y$ are uncorrelated. Here it's best to use the formula for correlation: $\rho_{X,Y}=E((X-\mu_x)(Y-\mu_y))/\sigma_x\sigma_y$. In the case of the circle, $X$ and $Y$ are symmetric about the origin, so this expectation evaluates to $0$.

Intuitively, two variables are positively correlated if when you know one of them is relatively large you also expect the other to be relatively large. Two variables are negatively correlated if when you know one of them is relatively large then you expect the other one to be relatively small. Two variables are uncorrelated if when you know something about how small or large one of them is, it doesn't change what you expect about how small or large the other one is.

According to this intuition, we might have guessed that $X$ and $Y$ are indeed uncorrelated since no matter what $x$ is, the expected value of $Y$ given that $X=x$ is always $0$ (since $Y$ is symmetric about $0$). Caution: this intuition shouldn't always be trusted. At the end of the day, you should back up your intuition with calculations.

Note: independent always implies uncorrelated, but not the other way around (as this example shows)


问:
How to calculate $(n+1)$'th derivative of $\prod_{i=0}^{n}(x-a_i)$ with constants $a_i \in \mathbb{R}$?
Letting $f(x):=\prod_{i=0}^{n}(x-a_i)$, with a real variable $x \in \mathbb{R}$ and $n\in \mathbb{N}$, I want to calculate the $(n+1)$'th derivative of $f$ with respect to $x$. $f(x)$ is a product of $(n+1)$ linear functions in $x$, each of which are $(n+1)$ times differentiable. In this regard I was thinking of maybe applying the general leibniz rule to calculate the derivative, but I don't fully understand the basis of it yet. Another approach that comes to mind is a calculation involving induction on $n$. What would be a desirable approach?

答: It may be more clear to write it out. By definition, $$\prod_{i=0}^n (x - a_i) = (x-a_0)(x-a_1)(x-a_2)\cdots(x-a_n).$$ We can expand the right hand side to get $x^{n+1} -(a_0 + a_1 + a_2 + \dots + a_n)x^{n} + \dots + a_0a_1a_2\cdots a_n.$ Taking the derivative, it is clear that we have $(n+1)!.$

ContentLicense: CC BY-SA 4.0


问:
Conservative field in $~\mathbb R ^2$ \ $\{(0,0)\}~?$
Sorry if I make some mistakes with terminology but I'm studying that in Italian, any way...

I have a lot of exam question very similar each other. One of example:

$$\Omega = \mathbb{R}^2\setminus\left \{(0,0)  \right \}$$
$$F:\Omega \rightarrow \mathbb{R}^2$$
$$F(x,y) = (\frac{x}{\sqrt[4]{x^2+y^2}}, \frac{y}{\sqrt[4]{x^2+y^2}})$$
$$\gamma :[0,2\pi]\rightarrow \mathbb{R}^2 $$
$$\gamma(t)=(\cos t, \sin t)$$

They gave me some answers to choose and the correct answer is: $$\int_{\gamma} F dl  = 0$$ because the field is conservative.

My question is: why is conservative if the field has a hole and so it's not a simply connected space?

(I know the integral result is 0... it's a theory question)

答: Well, if it's written like that, I presume the expectation was that you should have recognized immediately $F$ as being the gradient of a function $G:\Bbb R^2\setminus\{0\}\to\Bbb R$. I suggest $G(x,y)=\frac23\left(x^2+y^2\right)^{3/4}$.


问:
Cosine of complex variable

Show that $|\cos z|\leq2$  for all $z \in B(0,r)\subset C$ for some $r>0$ . Can we have $|\cos z|\leq 1$  for all $z \in B(0,r)\subset C$ for some $r>0$ ?

I have found the answer for the first part but can't think of a solution for the second question. For the first part

$$|\cos z|=\biggl|\frac{e^{iz}+e^{-iz}}{2}\biggr|$$
$$ =\biggl|\frac{e^{ix}*e^{-y}+e^{-ix}*e^{y}}{2}\biggr|$$
$$ \leq\biggl|\frac{e^{ix}*e^{-y}}{2}\biggr|+ \biggl|\frac{e^{-ix}*e^{y}}{2}\biggr| $$
$$ \text{But }\:|e^{-ix}|=|e^{ix}|=1$$
$$ =\biggl|\frac{e^{-y}}{2}\biggr|+ \biggl|\frac{e^{y}}{2}\biggr|$$
$$= \cosh y$$
$$\text{So }\: |\cos z|\leq \cosh  y$$
and since $\cosh 0=1$ we can find a neighborhood around the point $0$ such that $|\cos z|\leq 2$.

Since $\cosh x$ is a strictly increasing function I can't use it to prove that $|\cos z|\leq 1 $ for some neighborhood. Another method I can think of is taking partial derivative of $|\cos z|$ along $x$ and $y$-axes to check the direction in which it increases or decreases. But is there a simpler way for checking whether such a neighborhood exists ?

答:
Note that
$$\cos ix=\cosh x$$
and so for any disc of radius $r>0$ centred on the origin, $\cos\frac r2i>1$ and the task is impossible.

Title: Cosine of complex variable


答:
Since $\cos(0)=1$ and since $\cos$ is continuous, there is some $r>0$ such that$$|z|<r\implies|\cos(z)-1|<1\implies|\cos(z)|\leqslant2.$$

But it follows from the maximum principle that there is no $r>0$ such that$$|z|<r\implies|\cos(z)|\leqslant1.$$

Title: Cosine of complex variable

Body:
Show that $|\cos z|\leq2$  for all $z \in B(0,r)\subset C$ for some $r>0$ . Can we have $|\cos z|\leq 1$  for all $z \in B(0,r)\subset C$ for some $r>0$ ?

I have found the answer for the first part but can't think of a solution for the second question. For the first part

$$|\cos z|=\biggl|\frac{e^{iz}+e^{-iz}}{2}\biggr|$$
$$ =\biggl|\frac{e^{ix}*e^{-y}+e^{-ix}*e^{y}}{2}\biggr|$$
$$ \leq\biggl|\frac{e^{ix}*e^{-y}}{2}\biggr|+ \biggl|\frac{e^{-ix}*e^{y}}{2}\biggr| $$
$$ \text{But }\:|e^{-ix}|=|e^{ix}|=1$$
$$ =\biggl|\frac{e^{-y}}{2}\biggr|+ \biggl|\frac{e^{y}}{2}\biggr|$$
$$= \cosh y$$
$$\text{So }\: |\cos z|\leq \cosh  y$$
and since $\cosh 0=1$ we can find a neighborhood around the point $0$ such that $|\cos z|\leq 2$.

Since $\cosh x$ is a strictly increasing function I can't use it to prove that $|\cos z|\leq 1 $ for some neighborhood. Another method I can think of is taking partial derivative of $|\cos z|$ along $x$ and $y$-axes to check the direction in which it increases or decreases. But is there a simpler way for checking whether such a neighborhood exists ?


问:
How to think of affine maps composition.
Say one has two affine maps in the form $f_1=A_1x+b_1$ and $f_2=A_2x+b_2$ with $A_1, A_2 \in M_{n\times n}(\mathbb{R})$, $b_1,b_2\in\mathbb{R}^n$. Keeping to this notation - i.e. without using extended matrix notation - how would one write $f_1
\circ f_2$? I think it should be equal to $A_1A_2 +(A_2b_1+b_2)$ i.e. the linear part and the translation stay "separated" but I'm rusty in linear algebra and would like someone to confirm my guess.

Body: $$(f_1\circ f_2)(x):=f_1(f_2(x))=f_1(A_2x+b_2)=A_1A_2x+A_1b_2+b_1=\\=(A_1A_2)x+(A_1b_2+b_1)$$


问:
On two different definitons of a module
I'm reading the book "Symmetry and the Standard Model: Mathematics and Particle Physics" by Matthew Robinson. On page $70$ of the book the author gives the following characterisation of a module of a set $M$.

Previously, we discussed how elements of a group act on each other, and we also talked about how elements of a group act on some other object or set of objects (like three painted eggs). We now generalize this notion to a set of $q$ abstract objects a group can act on, denoted $M=\{m_0,m_1,m_2,\dots,m_{q-1}\}$. Just as before, we build a vector space, similar to the one above used in building an algebra. The orthonormal vectors here will be

\begin{equation}
m_0\mapsto|m_0\rangle,\quad m_1\mapsto|m_1\rangle,\quad\dots\quad m_{q-1}\mapsto|m_{q-1}\rangle\tag{3.48}
\end{equation}

This allows us to understand the following definition. The set

\begin{equation}
\mathbb{R}\mathbf{M}=\left\{\left.\sum_{i=0}^{q-1}a_i|m_i\rangle\,\right|\,a_i\in\mathbb{R}\,\forall i\right\}\tag{3.49}
\end{equation}

is called Module of $M$ (we don't use square brackets here to distinguish modules from algebras).

But then I looked at the definition of a module on Wikipedia and it seems to be something very different.

Can someone explain the difference/similarity between the two definitons?

答:
A more general note first: modules over rings are a natural generalization of vector spaces over fields. In particular, a $R$-module with $R$ a field is nothing else than a $R$-vector space. So one can call vector spaces modules but I think this is more confusing than everything else.

What is constructed here is the so-called free module over $M$. This is the most general module structure one can construct given a set $M$ and some base ring $R$. Formally, this may be introduced by means of universal properties but I do not think this is beneficial right now. Intuitively, the construction is clear: we need to somehow close our set under linear combinations, that is allow addition of its elements and scalar multiplication by elements of our ground ring (an entirely analogous construction exists for vector spaces, given a ground field yielding the free vector space over $M$) as the module axioms dictate us.

To elaborate on this (as I have seen your question in the comments): a module $M$ over a ring $R$ is an abelian group $M$ with an external scalar multiplication $\cdot\colon R\times M\to M$ satisfying some compatibility axiom (i.e. making the whole thing well-behaved). As it turns out it suffices to just expect every (finite) linear combination of elements of $M$ with coefficients in $R$ to be in it.

Indeed, the trivial linear combination is our neutral element in $M$ and we have $(-1)\cdot m=-m$ as additive inverse. So this will gives us an abelian group. The compability axioms are only concerned with the scalar multiplication and its interaction with addition. The situation is much like for vector spaces and the axioms all follow at once if we consider the set of (finite) linear combinations.

In the given case we actually have a basis (something modules do not have in general and why their general definition is different; modules with a basis are called free) given by the $|m_i\rangle$ which simplifies the discussion significantly. In fact, it is only of importance that $M$ is finite in your case which makes everything so easy.

The map $m_i\mapsto|m_i\rangle$ is a "vectorfication" of the elements of $M$ to distinguish between their role as elements of $M$ and as base vectors for $\mathbb R\mathbf M$. The definition of the Module of $M$ then does exactly what I mentioned above: include all linear combinations in the $|m_i\rangle$ manually. You can check that this construction gives in a fact a $\mathbb R$-vector space (i.e. a $\mathbb R$-module but $\mathbb R$ happens to be a field; so here is your connection to the general notion of modules).

Title: On two different definitons of a module


问:
On the product of Bernoulli random variables
Let $X$ and $Y$ be two independent random variables which take only the values $0$ and $1$ with identical positive probabilities. Can $X$ and $Y$ be chosen such that $P(XY=0)=1$?

My attempt:

Let $P(X=0)=P(Y=0)=p$.

Then,$P(XY=0)=2p-p^2$.

Therefore $P(XY=0)=1 \Rightarrow p=1$.

Therefore not possible as $1-p=0$.

答:
Since $X,Y$ have the same Bernoulli distribution, we can write $P(X=0)=P(Y=0)=p_1$ and $P(X=1)=P(Y=1)=p_2$ where $p_1,p_2\in[0,1]$.

Now, the probability $P(XY=0)=1-P(X\ne0 , Y\ne 0)=1-P(X=1,Y=1)$.

And since $X,Y$ are independent, $P(X=1,Y=1)=P(X=1)P(Y=1)=p_2^2$.

So, yes, $X,Y$ can be chosen such that $P(XY=0)=1$, by choosing $p_2=0$.


问:
Proving a given function is convex function
I need help to prove the following question.
Prove that function $f$ is convex on set $E \in R^n$ if and only if
$$f(y + \beta(y - x)) \geq f(y) + \beta(f(y)-f(x))$$ for any $x, y \in E, \beta \geq 0$ and $y + \beta(y -x) \in E.$

答:
See that you can re-write $y=\big(\frac{1+\beta}{1+\beta}\big)y+\big(\frac{\beta}{1+\beta}\big)x-\big(\frac{\beta}{1+\beta}\big)x=\big(\frac{\beta x}{1+\beta}+\frac{y}{1+\beta}+\frac{\beta}{1+\beta}(y-x)\big) $.

Then, $f(y)\leq \frac{\beta}{1+\beta}f(x)+\frac{1}{1+\beta}f(y+\beta(y-x))$ (by definition of convexity).

$\implies (1+\beta)f(y)-\beta f(x)\leq f(y+\beta(y-x))$. Hence, proved.


问:
Expectation of minimum exponentialy distrubited variables
Let's consider two independent variables $X \sim \mathrm{Exp}(4), Y \sim \mathrm{Exp}(12)$. I want to find $$E[\min(4X,Y)]$$

I have my approach but I'm not sure if I'm correct with my way of thinking:

First observation is that :
$$
\min(X,Y) =
\begin{cases} 4X, & \mbox{if } 4X<Y \\ Y, & \mbox{if } 4X \ge Y \end{cases}
$$
So I'm not sure if I'm correct with statement that :

$$E[\min(4X,Y)] = \iint_{\{4x < y\}}4x \cdot \rho_x \,\mathrm{d}x \,\mathrm{d}y + \iint_{\{4x \geq y\}} y \cdot \rho_y \,\mathrm{d}x \,\mathrm{d}y$$
$$= \int_0^\infty \int_0^{\frac{y}{4}} 4x \cdot \rho_x \,\mathrm{d}x \,\mathrm{d} y + \int_0^\infty \int_0^{4x} y \cdot \rho_y \,\mathrm{d} y \,\mathrm{d} x ,
$$

where $\rho_x, \rho_y$ are the probability density functions of $X$ and $Y$ respectively.

Is my justification correct so far ?

答:
Your approach looks reasonable. The integrals you have look right if you evaluate them. But you should be careful with using the proper notation. For example, the first integral $\iint_{\{4x < y\}} 4x \cdot \rho_{x,y} \,\mathrm{d} x \,\mathrm{d} y$ should still be a double integral with respect to the joint pdf.

To offer a neat alternative approach, we can also use the properties of the Exponential distribution.

$4 X \sim \mathrm{Exp}(1)$ by closure under scaling by positive factor.

$\min(4X, Y) \sim \mathrm{Exp}(1 + 12)$ by distribution of the minimum of independent Exponential random variables.


问:
Prove that if $k^2+pk+q=0$ and $l^2-pl-q=0$ then between $k$ and $l$ there is a solution to $x^2-2px-2q=0$.
Real number $k$ satisfies $x^2+px+q=0$ and $l$ satisfies $x^2-px-q=0$. Prove that between $k$ and $l$ there is a solution to the equation $x^2-2px-2q=0$.

I started the solution this way:
$$ k = \frac{-p\pm\sqrt{p^2-4q}}{2} $$
$$ l = \frac{p\pm\sqrt{p^2+4q}}{2} $$
$$ x = p\pm\sqrt{p^2+2q} $$
But I failed to compare these solutions. What should be done next?

答:
HINT

Let $f_1 = x^2 + px + q$, $f_2 = x^2 -px-q$, $f = x^2-2px-2q$

$$f_1(-q/p) = \frac{q^2}{p^2} > 0 $$

$$f_2(-q/p) = \frac{q^2}{p^2} > 0$$

$$f(-q/p) = \frac{q^2}{p^2} > 0$$

Since all of these quadratics have positive leading coefficient, that means that all of them are upward facing

Now if you examine the given function in the neighbourhood of this point, can you make any useful observations?


问:
The number of integers 'a' ($1\leq a\leq 200$) such that a^a is a perfect square are:$ A). 105$ $ B) 103$ $C)107$ $D)109?$

The number of integers $a$ ($1\leq a\leq 200$) such that $a^a$ is a perfect square are is:

$A)\;105$

$B)\;103$

$C)\;107$

$D)\;109$

My work

$a^a$ is a perfect square when $a$ is even.
$2^2 = 4. 4^4 = 256$ and so on.

$a = 2,4,6.....198$

$198 = 2 + (n-1)2$

$196/2 + 1 = 99$

Is my work correct?  However there is no option matching this.

答:
First of all, $a^a$ being a perfect square number does not necessarily indicate that $a$ is an even number. If required conditions meet, $a$ might take on odd values as well. Below you may see the required conditions.

$$a=2n+1, \space \space \space a^a=(a^n)^2 a \space \space \land \space \space a=b^2, \space \space b\in \Bbb Z $$

The number of even numbers that makes $a^a$ a perfect square number is $\frac{200-2}{2}+1=100$. On the other hand, there are odd perfect square numbers such that $$1^2, 3^2, 5^2,... 13^2$$
The number of odd numbers that makes $a^a$ a perfect square number is $\frac{13-1}{2}+1=7$. The correct answer is $107$.

ContentLicense: CC BY-SA 3.0


问:
Two hyperbolas $ y_1 = \dfrac{k_1}{x} $ and $ y_2 = \dfrac{ k_2}{x} $ $ (k_1, k_2> 0) $ intersets a circle which touch the X-axis

The circle whose center is located in the first coordinate quarter
touches the $ Ox $ axis at the $ M $ point, intersects two hyperbolas
$ y_1 = \dfrac{k_1}{x} $ and $ y_2 = \dfrac{ k_2}{x} $ $ (k_1, k_2> 0) $
at the points $ A $ and $ B $ such that the line $ AB $ passes through
the origin $ O $. It is known that $ \dfrac {4} {k_1} + \dfrac {1}{k_2} = 4 $. Find the smallest possible length of the $ OM $ segment.
In response, write down the square of the length of the segment $ OM$.

I know if a circle equation is $(x - h)^2 + (y - k)^2 = a^2$ , when it touches x-axis then, $k = a$ and so the equation will be $(x - h)^2 + (y - a)^2 = a^2 => x^2 + y^2 - 2hx - 2ay + h^2 = 0$.
The length of $OM$ will be $h$. The equation of $AB$ will be in $y = mx$ form.

Now, from $ y_1 = \dfrac{k_1}{x_1} $ and $ y_2 = \dfrac{ k_2}{x_2} $ $ (k_1, k_2> 0) $ , if we will $\dfrac{1}{y_1} + \dfrac{1}{y_2} = \dfrac{x_1}{k_1} + \dfrac{x_2}{k_2} = p  $ and comparing it with $ \dfrac {4} {k_1} + \dfrac {1}{k_2} = 4 $ we get $x_1 = 4; x_2 = 1; p = 4 $.

I can't get the idea to proceed further. Please help me with the solution.

答:
Consider $y=mx$ as the equation of the line $OA$ and find the coordinates of $A$ and $B$

$$B=\left\{\frac{k_1}{\sqrt{k_1 m}},\sqrt{k_1 m}\right\};A=\left\{\frac{k_2}{\sqrt{k_2 m}},\sqrt{k_2 m}\right\}$$

We have
$$OA^2=k_2 m+\frac{k_2}{m};\;OB^2=k_1 m+\frac{k_1}{m}$$

Now apply the theorem which states that $OM^2=OB\cdot OA$

$$OM^2=f(m)=\sqrt{k_1 m+\frac{k_1}{m}} \sqrt{k_2 m+\frac{k_2}{m}}$$

and set derivative equal to zero to find extrema.

$$f'(m)=\frac{k_1 k_2 \left(m^4-1\right)}{m^3 \sqrt{k_1 \left(m+\frac{1}{m}\right)} \sqrt{k_2 \left(m+\frac{1}{m}\right)}}$$

$f'(m)=0\to m=1$ since $m=-1$ gives a line crossing the second and fourth quadrant.

Since $f'(m)>0$ for $m>1$ we can conclude that $m=1$ is a minimum

$$OM^2=2 \sqrt{k_1k_2}$$

knowing that $\frac{4}{k_1}+\frac{1}{k_2}=4$ we can conclude that the minimum for $OM^2$ is
$$OM^2=\frac{k_1}{\sqrt{k_1-1}}$$

$$...$$


问:
Calculating the number of conjugate subgroups
What is the general formulae for calculating the number of conjugate subgroups in Sn for a particular cycle?

And what is the relation between number of commuting elements with a particular cycle of Sn and conjugating elements? Please explain with an example.

Any help would be greatly appreciated.

答:
suppose you have a cycle $c=(a_1,\dots,a_k)$. For any permutation $\sigma$ of $S_n$, you´ll have that $\sigma^{-1} c\sigma$ is the cycle $(\sigma(a_1),\dots,\sigma(a_n))$ of $S_n$. Thus, the conjugates of a cycle of length $k$ in $S_n$ are all the cycles of length $k$. And the conjugate subgroups of the one generated by the cycle are the subgroups generated by cycles of length $k$.

How many of them are there? Well, there are $\frac{n!}{(n-k)!}$ ways of picking $k$ ordered elements out of $n$, and each cycle can be represented in $k$ different ways (for example $(123)=(231)=(312)$), you have $\frac{n!}{k\cdot(n-k)!}$ conjugate elements of the cycle.

However, in the subgroup generated by a cycle of order $k$ there are $\phi(k)$ different $k$-cycles which generate it (where $\phi(k)$ is the Euler Phi function). That means there will be just $\frac{n!}{k\phi(k)(n-k)!}$ conjugate subgroups.

As for the second question, the elements of the group which commute with a given elements for a subgroup called the centralizer. To work out its order, you can consider the action of $S_n$ on itself by conjugation. Then the orbit-stabilizer theorem tells us that the order of the group is the order of the centralizer of an element times the size of its conjugation class.

For example, if you consider the element $(123)$ in $S_5$, it has $\frac{5!}{3(5-3)!}=20$ conjugate elements (all the $3$-cycles), so its centralizer will have order $\frac{|S_5|}{20}=6$. (In this case it will be the subgroup generated by $(123)$ and $(45)$)


Title: Prove that if $k^2+pk+q=0$ and $l^2-pl-q=0$ then between $k$ and $l$ there is a solution to $x^2-2px-2q=0$.

Body:
Real number $k$ satisfies $x^2+px+q=0$ and $l$ satisfies $x^2-px-q=0$. Prove that between $k$ and $l$ there is a solution to the equation $x^2-2px-2q=0$.

I started the solution this way:
$$ k = \frac{-p\pm\sqrt{p^2-4q}}{2} $$
$$ l = \frac{p\pm\sqrt{p^2+4q}}{2} $$
$$ x = p\pm\sqrt{p^2+2q} $$
But I failed to compare these solutions. What should be done next?

Body:
Let $k^2+pk+q=0$, $l^2-pl-q=0$ and $h(x)=x^2-2px-2q$.

Then
$$ h(k)=k^2-2pk-2q =-pk-q-2pk-2q=-3pk-3q=-3(pk+q)=3k^2 $$
and
$$ h(l)=l^2-2pl-2q=pl+q-2pl-2q=-pl-q=-l^2 $$

So we have $h(k)>0$ and $h(l)<0.$ So by IVT $h(x)=0$ has one root between $k$ and $l$.


问:
Let $S$ be a finite set. Show that $\text{Sym}(S)$ is a group.
Wondering if someone could check I have answered this correctly, I’ve just started a course on Group theory and don’t feel confident in ability so far. The Question asks: “Let $S$ be a finite set. Show that $\operatorname{Sym}(S)$ is a group.”

Much appreciated,
L

Answer:

Let $X,Y,W,Z\subseteq S$ be groups such that
$$\mu\colon X\to Y,\quad\Phi\colon Y\to W,\quad p\colon W\to Z$$
Then we have

$G0)$ Closure: Let $x\in X$ be arbitrary. Then
$$\Phi\circ\mu(x)=\Phi(\mu(x))=\Phi(y)\in\operatorname{Sym}(S)$$
for some $y\in Y$.

$G1)$ Associativity:
$$\rho(\Phi\circ\mu(x))=\rho(\mu\circ\Phi(x))=\rho(\Phi(\mu(x)))\in\operatorname{Sym}(S)$$
and
$$(\rho\circ\Phi)\circ\mu(x))=(\Phi\circ\mu)\circ\mu(x))=\rho(\Phi(\mu(x)))\in\operatorname{Sym}(S)$$
Thus, $\operatorname{Sym}(S)$ is closed under associativity.

$G2)$ Identity: Let $\sigma\colon S\to S,\,s\mapsto es$, then $\sigma$ is identity map $\operatorname{Id}\colon S\to S$
$$\sigma(s)=es=s$$
where $e$ is the identity element of $S$.

$G3)$ Inverse: Let $f\colon X\to Y$ and $g\colon Y\to X$ be maps where $X,Y\subseteq S$ and $f,g\in\operatorname{Sym}(S)$. Then,
$$f\circ g (y)=f(g(y))=e\quad\text{and}\quad g\circ f(x)=g(f(x))=e$$
$\Rightarrow g(y)=f^{-1}(y)$

Thus, $\operatorname{Sym}(S)$ is a group under composition.

答:
Besides a few mistakes you made (for example, you swapped the order of composition in your proof of $G1$ which is not possible in general) you seem to not have the right definition of $\operatorname{Sym}(S)$ at hand and do not completely understand what you have to show.

So let's start there. We define $\operatorname{Sym}(S)$ for a finite set (at some point you seem to assume that $S$ is a group, which is not the case) $S$ as the set of all its permutations which are nothing else than bijective maps $S\to S$. Indeed, permuting the elements of $S$ is nothing else than swapping them around one-to-one and onto. Hence,
$$\operatorname{Sym}(S)=\{f\colon S\to S\,\mid\,f\ \text{bijective}\}$$
Now, we want to show that this set  is a group under composition. For example, we have to show that there is an element $e\in\operatorname{Sym}(S)$ such that $f\circ e=f=e\circ f$ for all $f\in\operatorname{Sym}(S)$. So you have to explicitely construct a bijective map $S\to S$ which behaves as identity under composition. I will include the solution under a spoiler:

 A good try would be the identity on $S$ (the name is chosen for a reason). This map is defined by
$$\operatorname{id}\colon S\to S\,s\mapsto s$$
So, take some $f\in\operatorname{Sym}(S)$. We have to show that $f\circ\operatorname{id}=f=\operatorname{id}\circ f$, that is for $x$ arbitrary we have to show that $(\operatorname{id}\circ f)(x)=f(x)=(f\circ\operatorname{id})(x)$. By definition, $(\operatorname{id}\circ f)(x)=\operatorname{id}(f(x))=f(x)$ and similarily $(f\circ\operatorname{id})(x)=f(x)$ as desired.

The other group axioms follow using similar ideas.

There is a particular no need for introducing some subsets $X,Y,W,Z\subseteq S$ and maps between them. I think I understand what lead you to this discussion (namely, image considerations of maps $S\to S$) but it is unnecessary at best, highly confusing at worst.


问:
Can Isotopic Arcs be Isotoped So That They Don't Intersect Until $t = 1$?
Suppose $A, B$ are two disjoint arcs in $X = \mathbb{R}^3$.  An 'arc' is a homeomorphic image of $[0,1]$.  Let $F$ be an ambient isotopy on $X$ carrying $A$ to $B$ - that means there is a homeomorphism $f_1$ on $X$ with $f_1(A) = B$ and $F = \lbrace f_t$ $|$ $0\leq t \leq 1 \rbrace$ is a homotopy from $f_0 = \text{id}_X$ to $f_1$ such that $f_t$ is a homeomorphism for every $t$.

In other words, $A$ can be carried onto $B$ via a continuous family of homeomorphisms of $X$.

The question is: Does this imply the existence of an ambient isotopy $G$ such that $G_t(A) \cap B = \varnothing$ for all $t \neq 1$?  I think this is not true; any of the standard examples of wild arcs seem to be counterexamples, but I'm having trouble showing how to prove it explicitly.  It's not my normal sort of stuff.

Anyone wanna have a go at it, or know a reference?  Surely it's been considered before.

答:
Definition. Given a subset $E\subset R^n$ we say that $A$ can be "pushed-off itself instantly" if there exists an isotopy
$$
F: E\times I\to R^n, f_t=F(\cdot, t),
$$
such that $f_0=id$  and $f_t(E)\cap E=\emptyset$ for all $t>0$. David Wright in

Wright, David G., Pushing a Cantor set off itself, Houston J. Math. 2, 439-447 (1976). ZBL0332.57002.

proved that a totally disconnected compact subset $E$ in $R^n$ satisfies this property if and only if it is tame, i.e. there is a homeomorphism $R^n\to R^n$ sending $E$ to a subset of a straight line. Since $R^3$ contains wild subsets homeomorphic to the Cantor set (e.g. Antoine discontinuum), the latter cannot be pushed off itself instantly. Now, as I explained here, each totally disconnected compact subset in $R^n$ is contained in an (embedded) topological arc $B$. Taking such $B$  and an arbitrary topological arc $A\subset R^3$ disjoint from $B$ and ambient isotopic to $B$ (say, obtained by a suitable translation of $B$), you obtain the required example.


问:
Definition of $\sin()$, $\cos()$ and $\tan()$
This is just a clear up of what I just learnt.

Imagine a right triangle where $a$ and $b$ are the legs and $c$ is the hypotenuse. $\alpha$ is the angle between $a$ (the short side) and $c$. Is it always true that:
\begin{align*}
\sin(\alpha) & = \frac{b}{c}\\[2mm]
\cos(\alpha) & = \frac{a}{c}\\[2mm]
\tan(\alpha) & = \frac{b}{a}
\end{align*}

Because it seems to me that it is awfully specific to have that rule, and is there any proof that it is true? I just want to clear it up so I can move on. (I'm self-taught so there is no specific lesson, take your time)

Thanks in advance,
Alex

答:
These are the definition for sin, cos and tan in the context of right triangles. Definitions are, by definition, always true. You can think it as a way to assign the same thing by a shorter name. So instead of write b/c all the time, you can write sin(alpha) instead, and vice versa.

It seems superfluous, but once the ideas of sine and cosine extends to all real numbers, it becomes very beautiful and powerful.


问:
Prove that the algebraic expression is greater than zero
Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.

It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.

My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...

答:
Let's complete the alternate sign polynomial with  $\begin{cases}a=x^{8}-x^{5}+x^{2}-x+1\\b=-x^{7}+x^{6}+x^{4}-x^{3}\end{cases}$

When $x\le 0$ then $(x^5+x)\le 0$ and since $(x^8+x^2+1)\ge 1$ we conclude easily that $a>0$.

So let's focus on the case $x>0$.

Notice that $a+b=\dfrac{x^9+1}{x+1}>0$

But since $b=(1-x)(x^6-x^3)=-x^3(1-x)^2(x^2+x+1)\le 0$

We can conclude that $a=(a+b)+(-b)>0$ in this interval.


问:
Differential Equations Reference Request
Currently I'm taking the Differential Equations course at college, however the problem is the book used. I'll try to make my point clear, but sorry if this question is silly or anything like that: the textbook used (William Boyce's book) seems to assume that the reader doesn't familiarity with abstract math, so it lacks that structure of presenting motivations, then definitions, then theorems and corolaries as we see in books like Spivak's Calculus or Apostol's Calculus.

I've already seem a question like this on Math Overflow, however unfortunatelly some people felt offended somehow and said: "are you saying that Boyce is easy? It doesn't matter if you know how to prove things, you must learn to compute", and my point is not that: Apostol and Spivak teaches how to compute also, however since their books are aimed to mathematicians they take care to build everything very fine and their main preocupation is indeed in the theoretical aspects.

I really don't like the approach of: well, in some strange way we found that this equation works, so memorize it, know how to compute things and everything is fine. I really want to understand what's going on, and until now I didn't find this possible with Boyce's book (certainly there are people that find this possible, but I'm used to books like Spivak's and Apostol's, so I don't really do well with books like Boyce's).

I've already seem Arnold's book on Differential Equations, but the prerequisites for reading it are bigger: he uses diffeomorphisms many times and although I'm currently also studying differential geometry, I don't feel yet comfortable on reading a book like this one.

Can someone recommend a book that covers ordinary differential equations, systems of differential equations, partial differential equations, and so on, but that can be read without much prerequisite, and that still has the structure of a book of mathematics? And when I say "structure of book of mathematics" is being like Spivak's and Apostol's books: not mixing up definitions, theorems and examples inside histories of how the theory developed. Since I'm a student of Mathematical Physics, of course motivations and examples from Physics are welcome, but not all mixed up in the text.

In truth I don't really believe that a book like the one I described exists (if a book on this topic is good in my point of view, I believe it'll have a lot of prerequisites). Anyway I hope that I don't get misunderstood in what's my doubt, and really sorry if this question is silly in someway.

答: Try Birkhoff/Rota and Hirsch/Smale (I actually don't know the latest edition with Devaney).</p>


问:
How do you show gcd( p+qi, p-qi) | 2 in this case?
Consider the ring of numbers $\Bbb{Z}$ [i] = {a + bi | a, b $\in$ $\Bbb{Z}$, $i^{2}$ = 1}

Prove that if p and q are coprime integers, then gcd(p + qi, p - qi) | 2.

so gcd(p, q) = 1, you can say there exists a,b $\in$ $\Bbb{Z}$[i] where ap + bq = 1

Any help would be appreciated thanks

答: We can see that $\,\gcd(p+qi,p-qi)$ divides both $(p+qi)+(p-qi)=2p$ and $(p+qi)-(p-qi)=2qi$. Since $\gcd(2p,2qi)=2$, we conclude that $\gcd(p+qi,p-qi) \mid 2$.


问:
Help to know what kind of differential equation is this
I've to solve this differential equation:

$
(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0
$

But I even don't know what kind of differential equation is it.

Please, any help?

Thanks in advance.

答:
As I've mentioned in a comment above, this is a differential equation reducible to an exact differential equation.

A differential equation of the form $M(x,y)dx +N(x,y)dy=0$ is said to be an exact D.E. if it's left hand member is the exact differential of some function $U(x,y)$

Or, $$du=M(x,y)dx +N(x,y)dy=0 $$Thus, it's solution is $U(x,y)=c$
Ex. Consider the D.E.  $xdy +ydx=0$

Note that, $d(xy)= xdy + ydx$

Therefore, the solution of the given D.E. is $xy=c$

A necessary condition for a D.E. $M(x,y)dx +N(x,y)dy=0$ to be exact is $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

The solution of an exact differential equation $M(x,y)dx +N(x,y)dy=0$ is given by:
$$\int_{y=constant} Mdx + \int N(y)dy=c$$ where in the first part of LHS, we are integrating the expression $M(x,y)$ treating y as a constant, and $N(y)$ are the terms of $N(x,y)$ which contain only y.

I will not be providing the proof for the points 1 and 2, but you can check  it out here

Even if it may be a bit hard to follow, it will definitely make life easier with respect to solving most differential equations.

Sometimes a differential equation which is not exact may become so, on multiplication by a suitable function known as the integrating factor ().

The IF can be obtained as follows:

1.If  $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} }{N}=g(x)$ (function of x alone), then IF=$e^{\int g(x)dx}$

2.If  $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} }{M}=h(y)$ (function of y alone), then IF=$e^{\int -h(y)dy}$

There are more ways to obtain the IF, but for your question, the first one will do.

SOLUTION:
$$(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0$$
$$M(x,y)=2y^3-2y^2+3x$$
$$\frac{\partial M}{\partial y}= 6y^2-4y$$
$$N(x,y)=3xy^2 -2xy$$
$$\frac{\partial N}{\partial x}= 3y^2-2y$$
$$\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}= 3y^2 - 2y$$
$$\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{N}= \frac{3y^2 - 2y}{3xy^2-2xy}= \frac{1}{x}$$
As$\frac{1}{x}$ is a function of x alone, the IF would be:
$$I.F.= e^{\int \frac{1}{x}dx}=x$$

Multiplying the equation throughout by x, we get:

$$(2xy^3-2xy^2+3x^2)dx + (3x^2 y^2 -2x^2y)dy=0$$

This will definitely satisfy the first condition I've mentioned(in the beginning)

Thus, it's solution will be:
$$\int_{y=constant} (2xy^3-2xy^2+3x^2)dx=c$$
As you can see, I haven't taken the second integral, as there is no term of y, independent of x.
$$x^2y^3-x^2y^2+x^3=c$$

This is the general solution of the given differential equation.


问:
Show that this compound Poisson process is a Levy process
Consider a Poisson Process $(N_{t})_{t\geq 0}$ with parameter $\lambda > 0$ and a sequence of iid random variables $(U_{i})_{i \in \mathbb N}$ with distribution $Q$, now define a process $X_{t}:= \sum\limits_{k=1}^{N_{t}}U_{k}$ which we shall call the compound Poisson process.

Show that this compound Poisson process is a Levy process, i.e. Show that it satisfies the following four conditions:

$(i) \; X_{0}=0$

$(ii) \; \text{there exists a family of measures }(\mu_{t})_{t\geq 0}\text{ such that for }t>s\geq 0\ \; X_{t}-X_{s}\sim \mu_{t-s}$

$(iii)$ for $n \in \mathbb N$, $0=t_{0}< t_{1}<...<t_{n}$, we have $X_{t_{i+1}
}-X_{t_{i}}\; \forall i =0,...,n-1$ are  independent.

$(iv)$ $(X_{t})_{t\geq 0}$ has a.s. cadlag paths.

My attempt:

$(i):$ is trivial

$(ii):$ I have computed the characteristic functions, so for

$$ E[\exp(iuX_{t})]=E[(\int_{\mathbb R}\exp(iux)Q(dx))^{N_{t}}]=\sum\limits_{n \in \mathbb N_{0}}(\int_{\mathbb R}\exp(iux)Q(dx))^{n}P(N_{t}=n)\\=\exp(-\lambda t)\sum\limits_{n \in \mathbb N_{0}}(\int_{\mathbb R}\exp(iux)Q(dx))^{n}\frac{(\lambda t)^{n}}{n!}=\exp(\lambda t(\int_{\mathbb R}\exp(iux)Q(dx)-1))$$

Now by the characteristic function of $X_{t}$, following the same steps we can show that $X_{t+h}-X_{h}$ has the same characteristic function. And thus $X_{t+h}-X_{t}=X_{t}$ in distribution. Therefore this should suffice for (ii), although I would be rather interested to know what form $\mu_{t}$ takes on.

To show (iii), I do not know if there is a nice way to prove it.

For (iv): Consider a sequence from the right to $t \geq 0$ and call this sequence $(t_{n})_{n \in \mathbb N}$ since $N$ is a.s. cadlag, we have that $\lim\limits_{n \to \infty} X_{t_{n}}=\sum\limits_{k=1}^{N_{t}}U_{k}=X_{t}$ and from the left we obtain the same.

I would appreciate any help particularly on the points  (ii) and (iii).

答:
Stationary and independent increments would follow if we can show, for any $n\in\mathbb{N}$, $\alpha_{1},\ldots ,\alpha_{n}$ and $0 = t_{0} < \cdots < t_{n}$, that

$$\mathbb{E}\left[ \exp\left(\sum_{j=1}^{n}\text{i}\langle \alpha_{j}, Y_{t_{j}}-Y_{t_{j-1}}\rangle \right) \right]  = \prod_{j=1}^{n}\exp\left( -\lambda (t_{j}-t_{j-1})(1-\varphi (\alpha_{j})) \right)$$
where $\varphi_{j}$ is the characteristic function of $U_{j}$. The characteristic function of the process $N$ itself is
$$\mathbb{E}\left[ e^{\text{i}\langle \alpha , N_{t}\rangle} \right] = \exp \left( -\lambda t(1-\varphi (\alpha)) \right)$$
as you also calculated.

Below I have shown the desired formula in the case $n=2$. I suppose there are no trouble doing the same for a general $n\in\mathbb{N}$, just with more notation. We have
\begin{align}
\mathbb{E}\left[ \exp(\text{i}(\langle \alpha_{1},X_{t_1} - X_{t_0}\rangle + \langle \alpha_{2},X_{t_2} - X_{t_1}\rangle) \mid N_{t_1} = n_{1}, N_{t_2} - N_{t_1} = n_{2}\right] &= \varphi (\alpha_{1})^{n_1}\varphi (\alpha_{2})^{n_2}
\end{align}
So we can consider
\begin{align}
\mathbb{E}\left[ \varphi (\alpha_2)^{N_{t_2}-N_{t_1}} \right] &= \sum_{j=0}^{\infty}\varphi (\alpha_{2})^{j}\mathbb{P}(N_{t_2} - N_{t_1} = j)\\
&= \sum_{j=0}^{\infty}\frac{\varphi(\alpha_2)^{j}\lambda(t_{2}-t_{1})^{j}}{j!}e^{-\lambda (t_{2} - t_{1})} \\
&= \exp(-\lambda (t_{2} - t_{1})(1-\varphi (\alpha_{2})))
\end{align}
You can then do the same for $\varphi (\alpha_{1})^{N_{t_1}}$ and use independence of the increments of $N$ to obtain that
\begin{align}
\mathbb{E}\left[ \exp(\text{i}(\langle \alpha_{1},X_{t_1} - X_{t_0}\rangle + \langle \alpha_{2},X_{t_2} - X_{t_1}\rangle) \right] &= \mathbb{E} \left[ \varphi (\alpha_{1})^{N_{t_1}}\varphi(\alpha_2)^{N_{t_2}-N_{t_1}} \right] \\
&= \mathbb{E} \left[ \varphi (\alpha_{1})^{N_{t_1}}\right]\mathbb{E}\left[\varphi(\alpha_2)^{N_{t_2}-N_{t_1}} \right] \\
&= \exp(-\lambda t_{1}(1-\varphi (\alpha_{2})))\exp(-\lambda (t_{2} - t_{1})(1-\varphi (\alpha_{2})))
\end{align}
which proves the desired result for $n=2$.


问:
How to prove that $\lim_{x \to \infty}\int^{x+1}_x\frac {t^2+1}{t^2 + 20t + 8}dt=1$?
$$\lim_{x \to \infty}\int^{x+1}_x\frac {t^2+1}{t^2 + 20t + 8}dt=1$$

This was in a multiple choice question, I managed to disprove all choices except this one which was the answer, I wonder how can I prove it formally (or atleast without relying on intuition that on $x \to \infty$ the integrand goes to $1$ , so the area of the rectangle is somehwat $1\times 1=1$).

Appreciate any help, thanks in advance.

Body:
Hint:
$$
\frac{{t^2  + 1}}{{t^2  + 20t + 8}} < \frac{{t^2  + 1}}{{t^2  + 1}} = 1
$$
and
$$
\frac{{t^2  + 1}}{{t^2  + 20t + 8}} = 1 - \frac{{20t + 7}}{{t^2  + 20t + 8}} > 1 - \frac{{20}}{t}
$$
for all $t>0$.


答:
The function
$$f:t\mapsto \frac{t^2+1}{t^2+20t+8}$$ is increasing on $[2,\infty )$.

Therefore $$f(x)\leq f(t)\leq f(x+1),$$
for all $x\leq t\leq x+1$. Hence $$f(x)\leq \int_{x}^{x+1}f(t)\,\mathrm d t\leq f(x+1).$$
Taking $x\to \infty $ gives the wished result.

Title: How to prove that $\lim_{x \to \infty}\int^{x+1}_x\frac {t^2+1}{t^2 + 20t + 8}dt=1$?

Body:
$$\lim_{x \to \infty}\int^{x+1}_x\frac {t^2+1}{t^2 + 20t + 8}dt=1$$

This was in a multiple choice question, I managed to disprove all choices except this one which was the answer, I wonder how can I prove it formally (or atleast without relying on intuition that on $x \to \infty$ the integrand goes to $1$ , so the area of the rectangle is somehwat $1\times 1=1$).

Appreciate any help, thanks in advance.


问:
Is it true that if $f$ is strictly convex and $d$ a metric, then $f\circ d$ is not a metric?
As I was studying functional analysis, this question came to my mind. Let $X$ be a metric space with metric $d$. Is it true that for any function $f$ strictly convex, then $f(d(x,y))$ is not a metric? What about crescent and convex and $f(0)=0$?

I tend to think it’s true, since I can easily show that for $f(x)=x^2$ then $f\circ d$ is not a metric.

答: This is not true. Take $X=\{0,1\}$, and set $d(0,1):=1$. Then each function  $f$ with $f(0)=0$ and $f(1)>0$ will create another metric $f\circ d$ on $X$.


问:
How can I solve this non-linear differential equation?
I'm trying to solve the equation $$y' = 1 - y^2$$
Here is my attempt:
$$y' = 1 - y^2$$
Divide by (1-y^2)
$$\frac{y'}{1-y^2} = 1$$
Integrate both sides:
$$\frac{1}{2}\log|\frac{y+1}{y-1}|=t+c$$
Rearrange
$$y = \frac{ke^{2t}+1}{ke^{2t}-1}$$
I'd have thought that solution was right, but we have to figure out a specific solution with y(0) = 0. But this isn't possible with the above equation.</p>

答:
Reducing from what you have a little more, we get that equal to Tanh[x-k].

Tanh[-k] == 0 //Seting x to zero

Therefore k = 0, leaving Tanh[x] as your function.


问:
Tangent Line Error Bound with Taylor Series
I have an equation, $e^x$, based at 0 (b=0).

I am supposed to us the Tangent Line Error Bound to bound the error $|f(x)-T1(x)|$ on the interval I=[-1,1].

(Aside: I have already computed the first Taylor Polynomial $T1(x)$ for $e^x$ at b=0 to be $1+x$)

To find the error, I have done the following:

Found the second derivative of $f(x)$ to be $f''(x)=e^x$.

Found $M$, the maximum of $f''(x)$, to be $e^1$ on the interval [-1,1].

Now, the notes I have on finding the error are slightly confusing in that they indicate the method to find the error is one of the following:

$\frac{M}{(n+1)!} | x - b |^{n+1}$

$\frac{M}{(n+1)!} | d - b |^{n+1}$

However, using either of these in the homework submission website indicate wrong answers. What am I doing wrong?

答: Same exact problem here. You must be using webassign. It will work if you put in a good decimal approximation according to the Lagrange estimate explained in the other answer. I put in 1.359 and it finally marked my answer correct.</p>


问:
Asymptotic growth of recurrence sequences
A sequence $(a_n)_{n≥0}$ in $R$ satisfies $6_ = a_{n-2} + a_{n-1},$ where $ _0 =-1$ $ _1 = -1$

Determine the asymptotic growth of $(a_n)_{n≥0}$.

My question is: Should we first solve this recurrence and then find the asymptotic growth?
I am familiar with the asymptotic growth functions and recurrences but I am not sure how to start this problem..

Any hint or idea would be much appreciated.

答:
This recurrence relation can be written as
$$\begin{pmatrix} a_{n-1} \\ a_n \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ \frac{1}{6} & \frac{1}{6}\end{pmatrix}\begin{pmatrix} a_{n-2} \\ a_{n-1}\end{pmatrix}.$$
Hence, by induction,
$$\begin{pmatrix} a_{n-1} \\ a_n \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ \frac{1}{6} & \frac{1}{6}\end{pmatrix}^{n-1}\begin{pmatrix} a_0 \\ a_1\end{pmatrix}.$$
Computing the eigendecomposition of $A := \begin{pmatrix} 0 & 1 \\ \frac{1}{6} & \frac{1}{6}\end{pmatrix}$ will give you the asymptotic growth of $(a_n)_{n \in \mathbb{N}}$. Usually it will be determined by the largest eigenvalue of $A$ unless you exactly hit the eigenvector of the lower eigenvalue or the eigenvalues are equal in size.


问:
Proof of Probability Measure Decomposition
I want to prove that all probability measures $\mathbb{P}$ on the measurable space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ σ-algebra Borel, can be decomposed as

$$\mathbb{P}=\mu+\nu$$

where $\mu$ is a discrete measure and $\nu$ a diffuse measure.

Can you provide some guidance or steps that I have to follow in order to prove that ?

答:
Lebesgue's decomposition theorem

For every two σ-finite signed measures $\mu$  and $\nu$ on a measurable space $(\Omega,\Sigma)$, there exist two σ-finite signed measures $\nu_0$ and $\nu_1$, such that:

$\nu=\nu_0+\nu_1$

$\nu_0\ll \mu$ (absolutely continuous)

$\nu_1 \perp \mu$ (singular)

Moreover, $\nu_0$ and $\nu_1$ are uniquely determined by $\nu$ and $\mu$.

Since you have a probability space, any measure is $\sigma$-finite and signed.
Choose $\nu=\mathbb{P}$, let $\mu$ be the point masses of $\mathbb{P}$.
The Lebesgue decomposition theorem gives you $\nu_0$ and $\nu_1$, such that $\nu_0$ only has point masses (discrete measure) and $\nu_1$ has no point masses (diffuse measure, for $\mathcal{B}(\mathbb{R})$).


答:
Recurrence sequences have a lot in common with linear differential equations. Here we have the analogous differential equation
$$
6y'' - y' - y = 0,
$$
$y'(0) = -1$, $y(0) = -1$. Characteristic equation is $6\lambda^2 - \lambda - 1 = 0$, $\lambda = \frac{1}{2}, -\frac{1}{3}$. So the solution is $y = C_1 e^{\frac{1}{2}t} + C_2 e^{-\frac{1}{3}t}$. From the condition we obtain $C_1 = -\frac{8}{5}$ and $C_2 = \frac{3}{5}$. Now it's obvious that the asymptotic behaviour is close to $-\frac{8}{5}e^{\frac{1}{2}t}$ (or $-\frac{8}{5}e^{\frac{1}{2}n}$) since the second summand tends to zero.

Title: Asymptotic growth of recurrence sequences

Body:
A sequence $(a_n)_{n≥0}$ in $R$ satisfies $6_ = a_{n-2} + a_{n-1},$ where $ _0 =-1$ $ _1 = -1$

Determine the asymptotic growth of $(a_n)_{n≥0}$.

My question is: Should we first solve this recurrence and then find the asymptotic growth?
I am familiar with the asymptotic growth functions and recurrences but I am not sure how to start this problem..

Any hint or idea would be much appreciated.


答:
Background: Recall that if $\mu$ is a measure on a measurable space $(X,\mathcal{A})$, then $A \in \mathcal{A}$ is an atom (with respect to $\mu$) if $\mu(A) > 0$ and, for each $B \in \mathcal{A}$ with $B \subseteq A$, we have $\mu(B) \in \{0,\mu(A)\}$.  We say that a measure is atomic if every measurable set of positive measure contains an atom.
According to Wikipedia, a measure $\mu$ is discrete if it equals a (countable) weighted sum of Dirac measures; while a measure $\nu$ is diffuse if it has no atoms.

Using the metric properties of $\mathbb{R}$, one can show that if $\mu$ is atomic, then $\mu$ is discrete.  This holds more generally whenever $\mu$ is a Borel measure on a Polish (= separable, metric) space (e.g. if we replace $\mathbb{R}$ by $\mathbb{R}^{d}$).  (Actually, each atom is "equivalent" to a point.)  We won't need that here, but it's worth thinking about.

Hint: To start with, let $\mathcal{A}$ be the set of atoms of $\mu$ in $\mathscr{B}(\mathbb{R})$.  Show that if $A$ is an atom, then there is an $x \in \mathbb{R}$ such that $\mathbb{P}(\{x\}) = \mathbb{P}(A)$.  (Use the ordering of $\mathbb{R}$.)  Let $\{x_{n} \, \mid \, n \in J\}$ be the set of all points obtained from atoms in this way, where $J \subseteq \mathbb{N}$ (why?).

Define $\mu = \sum_{n \in J} \mathbb{P}(\{x_{j}\}) \delta_{x_{j}}$ and $\nu = \mathbb{P} - \mu$.  Check that $\mu$ is discrete and $\nu$ is diffuse, and notice that $\mathbb{P} = \mu + \nu$.

Title: Proof of Probability Measure Decomposition

Body:
I want to prove that all probability measures $\mathbb{P}$ on the measurable space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ σ-algebra Borel, can be decomposed as

$$\mathbb{P}=\mu+\nu$$

where $\mu$ is a discrete measure and $\nu$ a diffuse measure.

Can you provide some guidance or steps that I have to follow in order to prove that ?


问:
"Large" metric spaces
I'm studying Functional Analysis from Joseph Muscat's book. The author says the non separable spaces are very large. This makes sense too as finite metric spaces and countable spaces are trivially separable and so the only metric spaces which can be non-separable must be uncountable. This is what he says to justify to say that non-separable spaces are too large.

Can someone explain to me what he means by computing distances "precisely", "approximately" and "to any accuracy"? Given two points, we know distance between those points, so, where does the impreciseness occur here?

答:
I emailed the author and he saw this question and his response was the following (which I decided to add as an answer and which I believe might be helpful):

The answer given by CyclotomicField is accurate for the first two
examples, but is out of point for the non-separable spaces because
he/she is using a different concept of separability.

The main issue is that an algorithm can only work with finite data. So
it can only handle a countable number of points, each of which is
described by a finite amount of data. Let's take some examples to
illustrate what's written:

The rationals $\mathbb{Q}$ are an example of a countable metric space:
each number requires just two integers to describe it, and the
distance between two rationals can be worked out precisely, i.e.,
exactly. Rationals with extremely large numerators and denominators
will take longer to work out because additions and multiplications of
them take longer.

The reals $\mathbb{R}$ are an example of a separable metric space:
most real numbers cannot be specified by a finite amount of data, but
that doesn't mean we cannot work out, say, $|\pi - e|$ as an
approximation. First find rational numbers close to each of them, say,
$3.141592654$ and $2.718281828$, then subtract. If one wants a better
approximation, take better rational approximations of them. So it is
not possible in general to find the exact value of the distance
between two reals, but one can approximate it as closely as one
wishes.

Finally take an example of a non-separable metric space, the set of
bounded integer sequences with the infinity norm; that is,
$d(\mathbf{x},\mathbf{y})=\sup_n|x_n-y_n|$. Then the algorithm needs
to calculate the differences $|x_n-y_n|$ (where $x_n$ and $y_n$ are
both integers) but it can never decide that it has reached the
supremum, not even approximately. Even if it calculates the
differences for $n=1,\ldots ,10^6$, and then take the maximum, there
are still an infinite number of terms that follow them to check. For
example, suppose the differences happen to be $0$ for $n$ less than a
million; the algorithm cannot decide that $d(x,y)=0$ (even
approximately), because there could be a large difference at the $2$
millionth position. This argument does not apply to the reals example
above; if two reals have agreeing decimals up to the millionth place,
then $x$ is very close to $y$ and the algorithm can conclude that the
distance is less than $10^{-1000000}$.

Title: "Large" metric spaces


问:
Is there a minus thickening operator on a metric space?
Let $S$ be a metric space and $A$ a subset. For some $\varepsilon>0$ define the $\varepsilon$-thickening of $A$ as

$$A^{\varepsilon} = \left\{p \in S \mid \exists q \in A \;\;\text{with}\;\; d(p,q) < \epsilon\right\} = \bigcup_{q\in A} B_{\varepsilon}(q)$$

Does this thickening operation have an inverse for $\operatorname{diam}A>\varepsilon$, $A$ open? In what conditions would $((A^c)^\varepsilon)^c$ work for $A$ clopen?

答:
My take on the matter would be to define first an open and a closed thickening operator.
Let $B(x,r)$ denote the open ball of radius $r$ around $x$. Define
$$
A^{r)} = \bigcup_{x \in A} B(x,r)
$$
for $r>0$ and for $r\geq 0$
$$
A^{r]} = \bigcap_{r' > r} A^{r')}
$$
Note $A^{r]}$ closed since $\bigcap_{r' > r} A^{r')} \supset \bigcap_{r' > r} \overline{A^{r'/2)}} = \bigcap_{r' > r} \overline{A^{r')}}  $. Further
\begin{align*}
A^{0)} &= A^{\circ}
\\
    A^{-r)}
&=
    A^{\mathrm{C}\; r]\; \mathrm{C}}
=
    \left(\bigcap_{r' > r} A^{\mathrm{C}\; r']}\right)^{\mathrm{C}}
=
    \bigcup_{r' > r} A^{\mathrm{C}\; r']\;\mathrm{C}}
\\
    A^{-r]}
&=
    A^{\mathrm{C}\; r)\; \mathrm{C}}
=
    \left(\bigcup_{x \in A} B(x,r) \right)^{\mathrm{C}}
=
    \bigcup_{x \in A} B(x,r)^{\mathrm{C}}
\end{align*}
Note that negative closed/open thickenings are open/closed.

For $ r \geq 0 $ and $s > 0$ note the inclusions
\begin{equation}
\label{eq:openThickeningIncl}
    A^{r)\;s)}, A^{r]\;s)} \subset A^{r+s)}
\end{equation}
that hold since
\begin{align*}
        A^{r)\;s)} \subset A^{r]\;s)}
    &=
        \left( \bigcap_{r' > r} A^{r')} \right)^{s)}
\\
    &=
        \left\{ y\in X \mid \exists x \in \bigcap\nolimits_{r' > r} A^{r')} \colon |x\;y| < s\right\}
\\
    &=
        \left\{ y\in X \mid
          \exists z \in A \colon |x\;z| < s\right\}
\\
    &=
        A^{r+s)}
\text{.}
\end{align*}
Further note the inclusion
\begin{equation}
    A^{r]\;s]}, A^{r]\;s]} \subset A^{r+s]}
\end{equation}
for $r,s \geq 0$ that holds due to
\begin{equation*}
        A^{r)\;s]} \subset A^{r]\;s]}
    =
        \bigcap_{s' > s} A^{r]\; s')}
%\\
    \subset
        \bigcap_{s' > s + r} A^{s')}
%\\
    =
        A^{r]}
\text{.}
\end{equation*}
The inclusions also hold when $s$ is negative with a reverse inclusion sign. Does anyone know if these definitions have shown up in the literature somewhere?


问:
Is $\mathbb{N} \cup \{ \mathbb{N}\}$, $\{n \in \mathbb{N}: 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2} \}$ or $ \{ \emptyset \}$ an inductive family?
I know what is an inductive set. I know that $\mathbb{N}$ is an inductive set and I know that $\emptyset$ is an inductive set. But what is an inductive family? Is $\mathbb{N}$ an inductive family?

What about $\mathbb{N} \cup \{ \mathbb{N}\}$, $\{n \in \mathbb{N}: 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2} \}$ or $\{ \emptyset  \}$?

答:
The (usual) definition of inductive set is:

We call a set, $A$, an inductive set when $∅∈A$ and whenever $x∈A$ then $Sx=x∪\{x\}∈A$

So $∅$ is not an inductive set.

Note that if there exists $B$ such that $SB=A$ then $A$ is not inductive:

If $A=SB$ then $B∈A$ but $SB=A∉A$

So, $Sℕ=ℕ∪\{ℕ\}$ and $S∅=∅∪\{∅\}=\{∅\}$ are both not inductive.

$\{n \in \mathbb{N}: 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2} \}$ is just $ℕ$, so it is inductive.


问:
With reference to Figure Q3, write parametric functions $x (u), y (u), u \in [0, 1]$ defining this spiral curve.
I am having a hard time trying to figure this out:
U domain is [0,1]
[]1

答:
The shape is an Archimedean spiral, where the radius is continuously increasing to $0.5$ linearly.

Think about the simple spiral shown below, defined as $x=Ru\cos(2k\pi u), y=Ru\sin(2k\pi u)$, where $R$ is the final radius and $k$ is the total revolutions.

Now compare the shape of the question and the shape above, the former is rotated by $\pi/2$ and the initial radius starts from $0.1$ instead of $0$.

Hence the hint would be $x=f(u)\cos(\phi(u))$, where $f(u)$, a linear function, represents the radius, and $\phi(u)$ affects the revolution, rotate direction (clockwise and the rotation of the whole shape).


问:
Envelope of oscillator signal
How do we find envelope ODE of a modulated oscillator obeying

$$y^{''}(t)+y (t)\left(\dfrac{2 \pi}{t-c}\right)^2 = 0 $$

after obtaining solution say for initial conditions $y(0) = 1, y'(0) = 0?$. Does it oscillate indefinitely?

Could not readily apply $\; c-, p-$  discriminant methods. Please help.

Numerical solution with $ c=t _{max} =6$

Envelope sketched by hand.

答:
If you set $x=t-c$, then $$x^2y''(x+c)+4\pi^2y(x+c)=0$$ is an Euler-Cauchy equation with characteristic polynomial
$$
0=r(r-1)+4\pi^2=(r-\tfrac12)^2+4\pi^2-\tfrac14
$$
which has for the domain $x<0$ basis solutions $(-x)^{\frac12\pm iω}=(-x)^{\frac12}e^{\pm iω\ln(-x)}$ which combine to the real basis solutions
$$
(-x)^{\frac12}\cos(\omega\ln(-x)+\phi_0)
~\text{ and }~
(-x)^{\frac12}\sin(\omega\ln(-x)+\phi_0),
$$
where $ω=\frac12\sqrt{16\pi^2-1}$. The initial conditions and setting $\phi_0=-\ln(c)$ then lead to the formula in the other answer.

Title: Envelope of oscillator signal


问:
continuous mapping from a topology to a product topology

Let $(Y,\tau)$ and $(X_i,\tau_i)$, $i= 1,2,\ldots,n$ be topological spaces. Further for each $i$, let $f_i$ be a mapping of $(Y,\tau)$ into $(X_i, \tau_i)$. Prove that the mapping
$$f:(Y,\tau)\longrightarrow \prod(X_i,\tau_i),$$
given by $f(y) =\langle f_1(y),f_2(y),\ldots,f_n(y)\rangle$,is continuous if and only if every $f_i$ is continuous. Hint: $f_i=p_i(f)$ where $p_i$ is the projection mapping of the $i$th component.

This seems trivial for both directions so I'm not sure if I'm missing something.

Assume each $f_i$ is continuous. The take an open set $A$ in the product topology where its basis are open sets from each $X_i$. Since $f$ is continuous, the preimage of $A$ in the product topology is an $n$-tuple of open sets in $Y$ which means each component $f_i$ is continuous.

For the other direction, we assume $f$ is open and show each $f_i$ is open. Using the hint, each $f_i=p_i(f)$ which is a composition of two continuous functions since the projection mapping is continuous. Then each $f_i$ has to be continuous.

答:
Your proof is almost correct. If $f = (f_1, \ldots, f_n)$ and all $f_i$ are continuous, then for a basic open subset $O=\prod_{i=1}^n O_i$, where $O_i \subseteq X_i$ is open for all $i$, we have that $$f^{-1}[O] = \{y \in Y\mid f(y) \in \prod_{i=1}^n O_i\}= \{y \in Y\mid \forall i \in \{1, \ldots n\}: f_i(y) \in O_i \} = \bigcap_{i=1}^n f_i^{-1}[O_i]$$

which is a finite intersection (not product or $n$ tuple!) of open subsets by the continuity of the component mappings $f_i$, and hence open. As inverse images under $f$ of basic open subsets are open, the same holds for all open subsets of the product topology and so $f$ is continuous.

The reverse indeed simply follows from $f_i = \pi_i \circ f$ for all $i$ and the fact that compositions of continuous maps are continuous.

It's easy to see how this generalises to infinite products (if you've treated these as well), but not to the box topology on such a product.


问:
Show that Frac($R$) is algebraic over $K(\beta_{1},...,\beta_{d})$
Let $K$ be a field and $R$ be a finitely generated $K$-algebra without zero divisors. Let $\beta_{1},...,\beta_{d}\in R$. If $R$ is integral over $K[\beta_{1},...\beta_{d}]$, then I want to show Frac($R$) is algebraic over $K(\beta_{1},...,\beta_{d})$.

I'm stuck on defining the polynomial that $f(\frac{a}{b})=0$ for $a,b\in R$, can anybody help?

答:
You want a polynomial $f= a_0 + a_1 x + \ldots + a_{n-1}x^{n-1} + x^n$ with $a_i = p_i/q_i$ and $p_i,q_i \in K[\beta_1, \ldots, \beta_n]$ for which

$$
a_0 + a_1 (a/b) + \ldots + a_{n-1}(a/b)^{n-1} + (a/b)^n = 0.
$$

Clearing denominators - i.e. multiplying by $b^n \prod_i q_i$ -  this means precisely that

$$
b^np_0 + p_1b^{n-1}a + \ldots + p_{n-1}b a^{n-1} + a^n = 0.
$$

The "precisely" uses that we can go back, since $R$ has no zero divisors. We have an integral equation

$$
c_0 + \ldots + c_{n-1}a^{n-1}+a^n = 0
$$

with each $c_j \in K[\beta_1,\ldots,\beta_r]$ by hypothesis, so we would like to define $p_i = c_i/b^{n-i}$. This in turn shows that $a/b$ is algebraic over $K(\beta_1,\ldots,\beta_n,b) = K(\beta_1,\ldots,\beta_n)(b)$. But $b$ is integral over $K[\beta_1, \ldots, \beta_n]$ hence algebraic over $K(\beta_1, \ldots, \beta_n)$ and so the result follows by transitivity.


问:
What is the order of $\wedge^2(\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z)$?
I am learning wedge product and face a question to find out what is the wedge square of the group  $\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z$, $$\wedge^2(\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z)=?$$
Actually I have a solution saying it has order $p$. But I am wondering how to compute this. Any hint is appreciated.

答:
If the vector space setting is anything to go by, just expand in terms of a basis:

Write $a,b$ for the generators of $\mathbb Z/p \mathbb Z$.

Then, you can certainly expand any individual product, by linearity (coefficients all integers):

$$\begin{align}
x \wedge y &= (λ\, a + ν\, b)\wedge(\eta\,a + ρ\, b)
%
\\
&= λ\, a \wedge η \, a + λ\,a \wedge ρ \,b + ν \,β \wedge \eta\,a + ν\, b \wedge ρ\,b
\\
&=λη\, (a \wedge a) + λρ(a \wedge b) + νη(β \wedge a) + νρ(b \wedge b).
%
\end{align} $$

Now, if you apply the antisymmetry relation: the first and the last product must equal zero, and the third wedge is  $(-1)$ the second wedge. I.e., your product is really a multiple of $a \wedge b$.

Therefore (since this applies to sums of $x\wedge y$'s as well) $a \wedge b$ is a generator of the group.

As for its order:

$$n (a \wedge b) = (na) \wedge b = 0 \iff p |n,$$

as the order of $a$ is $p$. I.e., the order of the generator, hence the group, is $p$.


答:
I'm assuming this $\wedge^2 ( \mathbb{Z}/p \oplus \mathbb{Z}/p)$ is the $k$th exterior power of a vector space, $\Lambda^k V^*$ being the alternating $k$-tensors on the vector space $V$.

Then $V = \mathbb{Z}/p \oplus \mathbb{Z}/p$ is a two dimensional $\mathbb{Z}/p$ vector space, so $\Lambda^2 V^*$ is $1$-dimensional, and is isomorphic to the ground field, $\mathbb{Z}/p$, which has order p.

We can see this by general considerations: when dim$V = n$, $\Lambda^k V^*$ has dimension $\binom{n}{k}$. Proving this isn't too hard. Giving $V^*$ basis $\epsilon^1 , \cdots, \epsilon^n$, the exterior power has a basis consisting of all wedge products $\epsilon^{i_1} \wedge \cdots \wedge \epsilon^{i_k}$, where the indices $i_1 < \cdots < i_k$ are increasing.

Title: What is the order of $\wedge^2(\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z)$?

Body:
I am learning wedge product and face a question to find out what is the wedge square of the group  $\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z$, $$\wedge^2(\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z)=?$$
Actually I have a solution saying it has order $p$. But I am wondering how to compute this. Any hint is appreciated.


答:
In general, if $V$ is a finite dimensional $k$-vector space with $n = \dim_k V$, one can show that the wedges $v_{i_1} \wedge v_{i_2} \wedge \cdots \wedge v_{i_k}$ for each $i_1 < \cdots i_k$ form a basis for $\bigwedge^kV$, hence $\dim \bigwedge^kV = {n \choose k}$.

In your example, you are dealing with a vector space over $\Bbb Z_p$ of dimension $2$ and $k = 2$, hence

$$
\dim_{\Bbb Z_p} \bigwedge^2 \Bbb Z_p \oplus \Bbb Z_p = {2 \choose 2} = 1
$$

and this implies $|\bigwedge^2 \Bbb Z_p \oplus \Bbb Z_p| = p$.


问:
Finding planar curve $\beta(s)$ given curvature $k(s) = \frac{1}{s}$, $\beta(1) = (0, 0)$, $\beta'(1) = (1, 0)$
Find the curve $\beta(s) = (x(s),y(s))$ with $|\beta'(s)|= 1$ for all $s >0$ such that $\beta(1) = (0,0)$, $\beta'(1) = (1,0)$ , and curvature $k(s) =\frac{1}{s}$ for all $s >0$.

I am guessing some sort of integral is required since we are given 'initial' conditions, but I do not see how to begin.

答:
Let's write everything down.
$$\left|\beta^{\prime}(s)\right|=1 \implies \left(x'(s)\right)^2+\left(y'(s)\right)^2=1$$
An arc length paramtrization of $\beta$ is going to be given by
$$\left(\int_1^s \cos(\theta(t))dt,\int_1^s \sin(\theta(t))dt \right)  $$
for some $\theta$. In the above equation we used the first initial condition.
The curvature is given by
$$\beta''(s)=\kappa(s) e_2$$
where $e_2$ is orthonormal to $\beta'$ so we have that
$$|\beta''(s)|^2=\frac{1}{s^2}$$
Thus we get
$$\sin^2(\theta(s))\theta'(s)^2+\cos^2(\theta(s))\theta'(s)^2=\frac{1}{s^2}$$
This implies
$\theta'(s)=\frac{1}{s}$. Thus we get $\theta(s)=\ln(s)$ by the initial condition for $\beta'$.


问:
First derivative of a quadratic function
I have the following question about the derivative.

Find out the quadratic function, which takes the value $41$ at $x=-2$ and the value $20$ at $x=5$ and is minimized at $x=2$ . Calculate the minimum value of this function.

If the function is minimized at $x=2$, then $f'(2) = 0$.

My question is: the derivative of the quadratic is a line tangent to the function or the slope of the function for a given $x$? Then the derivative equals $0$ because it is minimized at the vertex and at the vertex there is no slope?

This is not really about the question above but i just want to be sure if that is the case.

答:
You don't need derivatives to solve this problem; it is known from high school that a quadratic polynomial $q(x)=ax^2+bx+1$ has an extremum at $\xi=-\frac b{2a}$, which is a minimum if $a>0$, a maximum if $a<0$, and that this extremum is equal to
$$q(\xi)=\frac{4ac-b^2}{4a}=-\frac{\Delta(q)}{4a}.$$
To determine the coefficients of $q(x)$, you simply have to solve the system of equations:
$$\begin{cases}\begin{aligned}
q(-2)&=\phantom{2}4a-2b+c=41 \\
q(5)&= 25a+5b+c=20 \\
-\frac b{2a}&= 2
\end{aligned}\end{cases}
\implies
\begin{cases}\begin{aligned}
  21a+7b &=20 \\
b&= -4a
\end{aligned}\end{cases}
$$
Can you end the calculations?

Added: To answer the question of why the derivative is $0$ at a vertex $S(x_0,f(x_0))$ on the curve, suppose $S$ corresponds, say, to a local minimum. The intuitive explanation is that the slope of a chord $(SM)$ is negative is $M$ is before $M(x,f(x))$, i.e. since $x<x_0$ and $f(x)>f(x_0)$ the rate of variation is negative. Passing to the limit on left, we obtain that $f'(x_0)\le 0$.

On the other hand, if $M$ is  *after $s$, we have $x>x_0$ and still $f(x)>f(x_0)$, so the rate of variation is positive, and wwe can conclude <ith the same argument that $f'(x_0)\ge 0$.

Therefore we have both $f'(x_0)\le 0$ and  $f'(x_0)\ge 0$. Conclusion?

Note: This is only a necessary condition  to have a local extremum, certainly not a sufficient condition.


问:
How can I find this distribution function?
The random variable $X$ has the probability density function $f_X(x)$ represented
in the figure

How can I find the $F_X(x)$ (distribution function)? And how can I find the value of $C$ when $P(|X|\le  0.5) \ge C$?

答:
In general, to find the CDF $F_X(x)$ given the PDF $f_X(x)$, you should integrate:
$$
   F_X(x) = \int_{-\infty}^x f_X(t) \,dt.
$$
In this case, $f_X$ is a piecewise function defined separately for $x<-1$, $-1 \le x < 0$, $0 \le x < 1$, and $x \ge -1$. So you have to integrate separately on those intervals:

For $x < -1$, $F_X(x) = \int_{-\infty}^x f_X(t) \,dt = \int_{-\infty}^x 0 \,dt = 0$.

For $-1 \le x < 0$, $F_X(x)$ can be broken up as $\int_{-\infty}^{-1} f_X(t)\,dt + \int_{-1}^x f_X(t)\,dt$, which becomes $0 + \int_{-1}^x (t+1)\,dt$.

For $0 \le x < 1$, $F_X(x)$ can be broken up as $\int_{-\infty}^{-1} f_X(t)\,dt + \int_{-1}^0 f_X(t)\,dt + \int_0^x f_X(t)\,dt$, which becomes $0 + \int_{-1}^0 (t+1)\,dt + \int_0^x (1-t)\,dt$.

Finally, for $x \ge 1$, we have to integrate over all four parts, but we expect that at this point $F_X(x) = 1$.

Another way to phrase it is that if we have intervals $(-\infty,-1) \cup [-1,0) \cup [0,1) \cup [1,\infty)$ on which $f_X$ has different definitions, we:

Set $F_X(t) = \int_{-\infty}^x 0\,dt$ for $x \in (-\infty,-1)$ and use this to compute $F_X(-1)$. (We'll get $0$.)

Set $F_X(t) = F_X(-1) + \int_{-1}^x (t+1)\,dt$ for $x \in [-1,0)$ and use this to compute $F_X(0)$.

Set $F_X(t) = F_X(0) + \int_0^x (1-t)\,dt$ for $x \in [0,1)$ and use this to compute $F_X(1)$. (We'll get $1$.)

Set $F_X(t) = F_X(1) + \int_1^x 0\,dt$ for $x \in (1,\infty)$.

The idea is that on every interval, $F_X(x)$ should be the whole probability of the previous intervals, plus some portion of the probability of the current interval.

Finally, to find $\Pr[|X| \le 0.5]$ or $\Pr[-0.5 \le X \le 0.5]$, we can take $F_X(0.5) - F_X(-0.5)$.


答:
Integrating the triangular density you will easy find

$$  F_X(x) =
\begin{cases}
0,  & \text{if $x<-1$} \\
x+\frac{x^2}{2}+\frac{1}{2},  & \text{if $-1\leq x <0$} \\
x-\frac{x^2}{2}+\frac{1}{2},  & \text{if $0 \leq x <1$} \\
1,  & \text{if $x \geq 1$}
\end{cases}$$

To calculate the requested probability

$$\mathbb{P}[|X|>0.5]=0.25$$

it is immediate to calculate the area of the two triangles in the tails (you will do it by mind, without any complicate calculation)

Title: How can I find this distribution function?

Body:
The random variable $X$ has the probability density function $f_X(x)$ represented
in the figure

How can I find the $F_X(x)$ (distribution function)? And how can I find the value of $C$ when $P(|X|\le  0.5) \ge C$?


问:
Meaning of average mark
What does the average mark tell you about an exam, for example??

For example, if the average mark is 70, does this mean that most of yhe students got a mark around 70?

Thank you in advance.

答: An average test score is the sum of all the scores on an assessment divided by the number of test-takers. For example, if three students took a test and received scores of 69, 87, and 92, these numbers would be added together and then divided by three to get an average of 82.6.</p>

Title: Meaning of average mark


问:
How are these two equivalent?
$$\frac{\ln(e^x+x)}{x}=\frac{e^x+1}{e^x+x}$$

I see that they did something to get rid of the natural log. I couldn't find any properties that would allow me to do this. I also think that they raised both the numerator and denominator by $e$. I have tried it and I did not get the same result. Does anyone know how the solution manual got this? I am supposed to use L'Hopital's rule to find the limit as x approaches $0$.

答:
$$\frac{\ln(e^x+x)}{x} \; \neq \;\;\frac{e^x+1}{e^x+x}$$

It's the application of l'Hospital's rule to the left hand side: take the derivative of the LHS's numerator and of the denominator.

$$\large \frac{e^x+1}{e^x+x}=\frac{\frac{e^x+1}{e^x+x}}{1} = \frac{\frac d{dx}\left({\ln(e^x+x)}\right)}{\frac d{dx}(x)}$$

So $$\lim_{x\to 0} \frac{\ln(e^x+x)}{x} = \lim_{x\to 0} \frac{e^x+1}{e^x+x}$$

Title: How are these two equivalent?


问:
Question about the percentages of sets in other sets
I am puzzling over the question:
If there are three sets, $A, B, $ and $C$, where:

$A \cap B \ne \emptyset$

$A \cap C \ne \emptyset$

$B \cap C \ne \emptyset$

i.e. $A, B$ and $C$ have to intersect in some way.

From there:

a) what % of set $D$ is contained inside set $A$ if $D=A \cap B$ ?

b) what % of set $E$ is contained inside set $A$ if $E=D \cap B$  (i.e. $A \cap B \cap C)$ ?

c) what % of set $F$ is contained inside set $A$ if $F$ is the triangle in the diagram below ?

State any variables that will have to be included to solve these problems.

How should I start??

EDIT:
I meant what percentage of all items in $D, E$ or $F$ are contained in all of the items in $A$

答:
For any sets $A, B$ such that $A \cap B \neq \emptyset$, $A \cap B$ represents the elements in both $A$ and $B$, therefore, by definition,

$$A\cap B \subset A \textrm{, and } A \cap B \subset B$$
Hence, 100% of $D = A \cap B$ is contained in A, and similarly, 100% of $E = A \cap B \cap C$ is contained in $A$.


问:
Right-continuous for every decreasing sequence
Let $f \colon I \to \overline{\mathbb{R}}$ a function. I want to show that $f(x_n) \to f(x)$ for every sequence $(x_n)$ with $x\leq x_n$ and $x_n \to x$ iff $f(x_n) \to f(x)$ for every strictly decreasing sequence $(x_n)$ with $x\leq x_n$ and $x_n \to x$. But I don't know how i can do this. Can I maybe use the sequence $a_k= \sup_{n \geq k} x_n$?

答:
One direction ($\implies$) of the equivalence is trivial. For the other direction ($\impliedby$) assume that $(x_n)$ is a sequence with $x_n \ge x$, $x_n \to x$, but $f(x_n) \to f(x)$ does not hold. Then you can extract a subsequence $(x_{n_k})$ such that $(f(x_{n_k}))$ is convergent, with $f(x_{n_k}) \to c \ne f(x)$. Note that $x_{n_k} > x$ must hold for all but finitely many $k$. Finally extract a strictly decreasing subsequence $(x_{n_{k_j}})$ to get a contradiction.

This is an example of the following principle:

Let $(a_n)$ be a sequence with the property that every subsequence of $(a_n)$ has a subsequence that converges to $A$. Then $(a_n)$ converges to $A$.


问:
How can I find the difference between weakly converge in Sobolev spaces and all orders weak derivative weakly converge in $L^p$
$U$ is a open subset in $\mathbb{R}^n$, $u_k$ is a sequence in $H^1(U)$ (Sobolev space).

$u_k$  weakly converges to $u$ in $H^1(U)$.

$u_k$ and $Du_k$ weakly converge to $v$ and $Dv$ in $L^2(U)$.
How to show that (1)  implies (2), and (2)  implies (1)?

I can't find the way to prove them. In other words, what is the difference between weakly converge in Sobolev spaces  and the all  order weak derivative weak converge in  $L^p(U)$?
Could someone give me some details, thank you!

答:
If $u_k \rightharpoonup u$ in $H^1(U)$, then $u_k\rightharpoonup u$ and $Du_k\rightharpoonup Du$ in $L^2(U)$ as $id:H^1\to L^2$ and $D:H^1\to L^2$ are linear and continuous.

Now let $u_k\rightharpoonup u$ and $Du_k\rightharpoonup Du$ in $L^2(U)$. Let $f\in H^1(U)^*$. Since $H^1(U)$ is a Hilbert space there is $w\in H^1(U)$ such that
$$\int_\Omega \nabla w\cdot \nabla v +wv\ dx = f(v)$$ for all $v\in H^1(U)$. Using this representation it is easy to prove $f(u_k)\to f(u)$.


问:
Quadratic-exponential integral divided by $x^2$. How to solve?
Is it possible to solve the following integral?

$$\int_a^b \frac{e^{-x^2-c x}}{x^2} dx$$

(where $c$ is a positive number)

I tried both with Mathematica and looking on an integral table (Gradshtein's big book) and I wasn't able to find an answer.

EDIT:

I've been asked to add a bit of context to this question.
I'm trying to calculate the solution of the 1D diffusion equation for arbitrary initial condition and inhomogeneous term independent on time:

$$\rho_t=D\rho_{xx}+v\rho_x+\phi\rho+\epsilon(x)$$
with a change of variable you reduce it to
$$f_t=D f_{xx}+\eta(x)$$
where $\eta(x,t)=\epsilon(x) e^{-\phi t}$

if $\phi$ is set to zero, $\eta$ would be independent on time.

The integral I've presented you is the time-integral of the one in Eq. 1.17 of
http://www-m6.ma.tum.de/~kuttler/script_reaktdiff.pdf
the coefficient $c$ in the integral is proportional to $\phi$, and I need it to impose degradation of the particles of my random walk.

I hope now the frame of the problem is a bit more clear .

答:
The integral has no elementary solution, and it's a problem also because to the unknown nature of the limits $a$ and $b$.
One way to attack the problem, under some assumptions, is the following:

First we complete the square by supposing there does exist $d$ and $p$ such that
$$e^{-x^2 - cx} = e^{-(x^2+cx)} = e^{-(x^2 + cx + d - d)} = e^{d}\ e^{-(x+p)^2}$$

Where of course $c = 2p$ and $d = p^2$.

This being said, the integral reads

$$e^{d}\int_a^b \dfrac{e^{-(x+p)^2}}{x^2}\ \text{d}x$$

Due to the unknown nature of the limit, we shouldn't proceed.
Yet with some poker face, assuming we can, we develop a Taylor Series for the exponential term, which brings us to

$$e^{d}\int_a^b \dfrac{1}{x^2}\sum_{k = 0}^{+\infty} \dfrac{(-1)^k}{k!} (x + p)^{2k}\ \text{d}x = $$

$$ = e^d \sum_{k = 0}^{+\infty} \dfrac{(-1)^k}{k!} \int_a^b \dfrac{(x+p)^{2k}}{x^2}\ \text{d}x$$

The general integral can be calculated in terms of Hypergeometric Functions:

$$\int \dfrac{(x+p)^{2k}}{x^2}\ \text{d}x = \frac{(p+x)^{2 k+1} \, _2F_1\left(2,2 k+1;2 k+2;\frac{x}{p}+1\right)}{(2 k+1) p^2} = G(k,p,x)$$

Adding the limits $a$, $b$ means just to take $G(k,p, b) - G(k, p,a)$.

Eventually the result is expressed in terms of a Series of Hypergeometric Functions.

Not something nice for sure, but for the lovers of numerical methods it might work!

$$e^d \sum _{k=0}^{\infty } \frac{(-1)^k \left((b+p)^{2 k+1} \, _2F_1\left(2,2 k+1;2 k+2;\frac{b+p}{p}\right)-(a+p)^{2 k+1} \, _2F_1\left(2,2 k+1;2 k+2;\frac{a+p}{p}\right)\right)}{(2 k+1) p^2 k!}$$

The first term is

$$\frac{1}{a}-\frac{1}{b}$$

The second term is

$$-\frac{p^2}{a}+2 p \log \left(-\frac{a}{p}\right)+a+\frac{p^2}{b}-2 p \log \left(-\frac{b}{p}\right)-b$$

The third term is

$$\frac{1}{6} \left(-a^3-6 a^2 p+\frac{3 p^4}{a}-12 p^3 \log \left(-\frac{a}{p}\right)-18 a p^2+b^3+6 b^2 p-\frac{3 p^4}{b}+12 p^3 \log \left(-\frac{b}{p}\right)+18 b p^2\right)$$

Putting them together so far you can write down the first three terms of the series as

$$-\frac{a^3}{6}-a^2 p+\frac{p^4}{2 a}-3 a p^2-\frac{p^2}{a}-2 \left(p^2-1\right) p \log \left(-\frac{a}{p}\right)+a+\frac{1}{a}+\frac{b^3}{6}+b^2 p-\frac{p^4}{2 b}+3 b p^2+\frac{p^2}{b}+2 \left(p^2-1\right) p \log \left(-\frac{b}{p}\right)-b-\frac{1}{b}$$

And so on. As I specified: this solution couldn't work due to the unknownness of $a$ and $b$ and $c$.


问:
What is the analogy to logic when denoting independence of random variables as $p\models X\perp Y$?
I'm reading Nir Friedman and Daphne Koller's "Probabilistic Graphical Models: Principles and Techniques". The authors occasionally use the notation
$$p\models X\perp Y \mid Z$$

to indicate that the set of random variables $X$ is independent of $Y$ given $Z$, under $p$ (which is a joint distribution over the values of all of these variables).

I've never studied mathematical logic, but I read a little about models, semantic and syntactic consequences and am still unclear as to the precise meaning of the symbol $\models$ in the expression above. How is this analogous to formal logic systems? Are the conditional independence assertions analogous to formulae? Axioms?

答:
As Mauro ALLEGRANZA explains in the above, double turnstile is a model operator. To get into a bit more detail: double turnstile is the validity operator, which expresses validity of a proposition in a model. So, given a model p, A is valid in p is expressed by p|=A. It is worth noting that this is a little bit different from truth: a model is composed of worlds (if we are using Kripke semantics, which are quite popular, and works for this case if we reinterpret things). So, let p be a set of worlds. A may be true at some worlds in p, or not. If A is true at all worlds in p, then we say A is valid in p, and express this using p|=A (read: "A is valid in p").

In your case, the author apparently uses p|=A to mean "across the whole joint distribution p, A is the case." This is more or less explaining a hieroglyph, but if you want to look into the nuances of |=, looking up kripke semantics for modal logic is a great way of getting into it without necessarily dealing with very heavy logic.


问:
Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$
Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?

答:
 Edited: thanks to several comments below, I find that the original answer made a (serious) mistake in counting the degrees of relevant polynomials and thus a mistaken use of the fundamental theorem of algebra. I apologize for the errors. A not-so-elegant way to fix the argument is added.

Systems of nonlinear equations are in general difficult; solutions are found by numerical methods.

Nevertheless, for this particular problem,
by inspection$\dagger$, there are at least four solutions:
$$
(-2,-2),\quad (-2,2),\quad (2,-2),\quad (2,2)\tag{1}
$$

We claim that the system has at most four solutions.

On the other hand, by symmetries, it suffices to show that $(2,2)$ is the unique solution in the first quadrant: any other solution not in (1) would give you one more solution in the first quadrant.

If you introduce the new variables $u=x^2$ and $v=y^2$:
$$
v=20-u^2,\quad u=20-v^2,\quad u,v>0
$$

The pair $(u,v)$ is the intersection of two parabolas on the $u$-$v$ plane.

If one draws a picture$\dagger\dagger$, one can see that one has only one intersection on the first quadrant, which implies that the first-quadrant solution to the original system is unique.

$\dagger$Notes. The observation $20=4+16$ gives an obvious hint to get a solution.

$\dagger\dagger$ Yes, this is just a hand-waving geometric "proof".


答:
The equation that you wrote is false. The correct statement is an application of l’Hospital’s rule:

$$\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{e^x+1}{e^x+x}\;.$$

Since both $\ln(e^x+x)$ and $x$ approach $0$ as $x\to 0$, you can apply l’Hospital’s rule to evaluate the limit. To do this you must differentiate the numerator and the denominator separately.

$$\frac{d}{dx}\ln(e^x+x)=\frac{\frac{d}{dx}(e^x+x)}{e^x+x}=\frac{e^x+1}{e^x+x}\;,$$

and of course $\frac{dx}{dx}=1$, so

$$\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}\;.$$

It is not true without the limits.

Title: How are these two equivalent?

Body:
$$\frac{\ln(e^x+x)}{x}=\frac{e^x+1}{e^x+x}$$

I see that they did something to get rid of the natural log. I couldn't find any properties that would allow me to do this. I also think that they raised both the numerator and denominator by $e$. I have tried it and I did not get the same result. Does anyone know how the solution manual got this? I am supposed to use L'Hopital's rule to find the limit as x approaches $0$.


问:
How to derive this equality involving kets and momentum?
$p_{x} | \psi \rangle = \frac{\hbar}{i} \int dx' |x' \rangle \frac{\partial \langle x'| \psi \rangle}{\partial x'}$

$p_{x} = \frac{i}{\hbar} \frac{\partial}{\partial x}$

I can't understand how does this equality hold:

$p_{x} | \psi \rangle = \int dx' p_{x} |x'\rangle \langle x'|  \psi \rangle= \frac{\hbar}{i}\int dx' \frac{\partial}{\partial x'} |x'\rangle \langle x'|  \psi \rangle $

Which is near, but not equal to the equality :/

答:
Integration by parts gets you everything but the sign. The sign error seems to arise in the inconsistent definition and application of the momentum operator (the  $\hbar / i$ factor).

Quick edit:

It would be more precise to say that the momentum operator acts as a derivative in the position space basis which arises due to the matrix elements
$$\langle x \vert \hat{p}_{x} \vert x'\rangle = -i\hbar \frac{\partial}{\partial x}\delta(x - x')$$
which could be introduced if we put another resolution of the identity to the left of the operator.


问:
How to calculate a pdf of a random variable
I'm reading a paper about kernel estimation and struggling with this part, which constructs the pdf of logZ^2 (Z has Standard normal distribution)
I don't understand how they compute the function like this. Can someone provide the detailed computation from step 1 to step 2?
Many thanks in advance.

Let $f_Z$ denote the density of $\log Z_i^2$. We obtain,
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\dx}{\mathrm{dx}}$
$\newcommand{\deriv} [2]   {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$
\begin{align*}
  f_Z(x)
    &= \deriv{}{x}\brp{\frac{2}{\sqrt{2\pi}}\int_{-\infty}^{e^{x/2}} e^{-s^2/2}ds-1}
  \\&= \frac{2}{\sqrt{2\pi}}\brs{\brp{\deriv{}{x}e^{x/2}}e^{-e^x/2}}
  \\&= \frac{2}{\sqrt{2\pi}}\brs{\frac{1}{2}e^{x/2}e^{-e^x/2}}
  \\&= \frac{1}{\sqrt{2\pi}}e^{x/2}e^{-e^x/2}
\end{align*}

[1]function: https://i.stack.imgur.com/oaKMB.jpg

答:
Sorry but reading it with my cell phone I cannot see your picture. That is one of the reason why it is not a good idea to substitute formulas with pictures.

Anyway, just using the Fundamental Transformation theorem, Here in the paragraph "Functions of Random Variables, you will easy find that

$$y=\log z^2$$

$$z^2=e^y$$

$$|z|=e^{y/2}$$

and

$$\Bigg|\frac{d}{dy}z\Bigg|=\frac{1}{2}e^{y/2}$$

thus

$$f_Y(y)=\frac{2}{\sqrt{2\pi}}e^{-e^{y}/2}\frac{1}{2}e^{y/2}=\frac{1}{\sqrt{2\pi}}e^{-e^{y}/2}e^{y/2}$$

At present I cannot check if my result matches with yours but I think it does.


答:
By substracting the two equations you get
$$ x^2(1-x^2) + y^2(y^2-1) = 0$$
$$ -(x^2-\frac12)^2 + (y^2-\frac12)^2 = 0 $$
so
$$ x^2 - \frac12 = \pm (y^2-\frac12) $$
that is
$$ x^2 = y^2 $$
or $$ x^2 = 1- y^2 $$
Analysing theses two cases it's easy to solve the equations.

Title: Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$

Body: Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?


问:
Is there a difference in continuity between a function having range $\subseteq \mathbb{R}-\mathbb{Q}$ and codomain $\subseteq \mathbb{R}-\mathbb{Q}$?
What I'm about to ask and/or say may be horribly wrong, but I feel there some subtlety to this topic that I don't really get right now.

Suppose you wonder whether there is a nonconstant continuous function $f:[a,b]\to\mathbb{R}-\mathbb{Q}$ and the answer is quickly decided to be no: the Intermediate Value Theorem would forbid this because if $f(x_0) < f(x_1)$ for $x_0\neq x_1$ and  $x_0,x_1\in [a,b]$ then $f$ must obtain every output between $f(x_0)$ and $f(x_1)$ for some input and thus output some rational numbers.

However, this argument seems to me to essentially suppose that $f$'s codomain is $\mathbb{R}$ and so IVT shows that being nonconstant violates a fundamental property of continuous functions $\mathbb{R}\to\mathbb{R}$ but perhaps there could be a nonconstant function which is continuous and its codomain is $\mathbb{R}-\mathbb{Q}$.

To illustrate clearer what I mean:

One way of seeing the above result is to recall the fact that the image of any compact subset of $\mathbb{R}$ under a continuous map is also compact. If the image of $[a,b]$ had been, say, $[\sqrt{2},\sqrt{3}]\cap (\mathbb{R}-\mathbb{Q})$, then we'd have a contradiction as this image is not closed and thus not compact. However, this set is both closed and bounded in $\mathbb{R}-\mathbb{Q}$ and so if $f$'s codomain really is $\mathbb{R}-\mathbb{Q}$ (and not simply $\mathbb{R}$ with its range being a subset of $\mathbb{R}-\mathbb{Q}$) then this argument on its own doesn't work.

Question: Does any of this make sense? Can continuity essentially depend on what you consider to be the codomain?

答:
To answer the question in bold: no, continuity does not depend of the codomain (in a sense which is formalized in the next paragraph). But this made sense, it was a valid question to ask even if it was not exactly formal.

If $f \colon A \to B$ is continuous and $B \subset C$ then so is the composition $f \colon A \to B \xrightarrow{\mathbf{inc}} C$, if $B$ is a subspace of $C$. Likewise, if $f \colon A \to C$ is continuous, so is its corestriction. I can sketch some proofs if needed.

Also, as a side remark: the intermediate value theorem is actually a theorem about connectedness, not compactness. It is a special case of the fact that continuous functions send connected sets to connected sets (in $\Bbb R$ these are always intervals). This bypasses your issue: since the connected components of $\Bbb R \setminus \Bbb Q$ are singletons and $[a,b]$ is connected, we have $f([a,b]) = \{c\}$ for some $c$.


问:
How Can the Vector Space $\mathbb{R}^n$ be identified with the Column space
Hi i am reading An introduction to manifolds by Loring and have some doubts in section 2.1 and 2.3. First i am writing some statements from the sections from which i have doubts and then i'll state my doubts.So the first statement that i have doubt in is:

By the correspondence between arrows and column vectors, the space $\mathbb {R}^n$ can be identified with this column space.

The second statement that i have doubt in is:

We usually denote the standard basis for $\mathbb {R}^n$ or $T_p(\mathbb R^n)$ by $e_1,...e_n$.

The third statement i have doubt in is:

This proves that $\mathcal D=\mathcal D_v$ for $v=\langle \mathcal Dx^1,...\mathcal xD^n \rangle$.This theorem shows that one may identify the tangent vectors at $p$ with the derivations at $p$. Under the vector space isomorphism $T_p\mathbb R^n\simeq\mathcal D_p\mathbb R^n$ the standard basis $e_1,...e_n$ for $T_p(\mathbb R^n)$ corresponds to the set $\left.\frac{\partial}{\partial x^1}\right|_p,...,\left.\frac{\partial}{\partial x^n}\right|_p$.

My doubts are as follows:

How can we identify the space $\mathbb R^n$ with the column space? Is this an isomorphism ?

What does it mean that we have identified the space $R^n$ with column space? Does this mean we can generate the space $R^n$ as linear combinations of the vectors from the column space? Also it is written in the book that "we visualize the tangent space $T_p \mathbb R^n$ at $p$ in $R^n$ as the vector  space of all vectors emanating from $p$". So will this vector space include all the possible vectors that have $p$ as starting point,even the normal vectors? Or only the vectos that pass tangentially through $p$?

In the second statement(doubt) why we are denoting the standard basis $e_1,...,e_n$ both for $R^n$ and $T_p \mathbb R^n$? Aren't they different vector spaces? How can they have the same basis. For example, in $\mathbb R^3$ we have $i,j,k$ as basis vectors.Are these basis vectors in $R^3$ different from $\left.\frac{\partial}{\partial x^1}\right|_p,\left.\frac{\partial}{\partial x^2}\right|_p,\left.\frac{\partial}{\partial x^3}\right|_p$?

In the third(doubt) how does $\mathcal D=\mathcal D_v$ for $v=\langle \mathcal Dx^1,...\mathcal xD^n \rangle$ mean allow us to identify the tangent vectors at $p$ with the derivations at $p$? I mean i know that the tangent vector is $v$ , $\mathcal D$ is the derivation at $p$ and $\mathcal D_v=v^i \frac{\partial}{\partial x^i}$ is the directional derivative at $p$.

For reference i am attaching the screenshots of the statement where i have highlighted different statements from the book.

答:
Parts of your question have been answered here.

$\mathbb R^n$ is the vector space of all $n$-tuples of real numbers. Usually one writes $x \in \mathbb R^n$ as $x = (x_1,\ldots,x_n)$ in row form, but you can use also the column form
$$x = \left[ \begin{array}{rrr}
x_1  \\
. \phantom{.} \\
. \phantom{.} \\
. \phantom{.} \\
x_n  \\
\end{array}\right]$$
This is just a notational issue. If you like, you can regard column vectors as $n \times 1$-matrices. Tu also prefers to use for coordinates upper indices instead of lower (for example $x^i$  instead of $x_i$).

Tu introduces the tangent space $T_p(\mathbb R^n)$ at $p$ in $\mathbb R^n$ as the vector space of all arrows emanating from $p$. He illustrates this for $n = 3$:

This generalizes of course to arbitrary $n$. Thus, formally we should regard such an arrow as a pair $a = (p,v)$ with $v \in \mathbb R^n$, that is we have $T_p(\mathbb R^n) = \{p\} \times \mathbb R^n$. The vector space structure on $\{p\} \times \mathbb R^n$ is of course given by $(p,v) + (p,w) = (p,v+w)$ and $r(p,v)= (p,rv)$. This achieves that the canonical bijection
$$\beta : \{p\} \times \mathbb R^n \to  \mathbb R^n, \phi((p,v)) = v $$
becomes a linear isomorphism of vector spaces. Note that an arrow emanating at $p$ can geometrically be viewed as the directed line segment from $p$ to $p + v$, in other words as the pair of points $(p,p+v)$.

Tu uses the convention to write the starting point $p$ of an arrow $a = (p,v)$ in row form and the "vectorial part" $v$ of $a$ in column form. In Tu's own words, he does so to distinguish between points and vectors. Moreover, by an abuse of notation he drops $p$ from $a = (p,v)$, i.e. identifies $a$ with the column vector $v$. Thus in the sense of Tu there is difference between the column vector with components $v^i$ (which is an element of $T_p(\mathbb R^n))$ and the row vector $v = (v^1,\ldots,v^n) \in \mathbb R^n$. Personally I do not like this way of notational distinction of $T_p(\mathbb R^n)$ from  $\mathbb R^n$, it does not elucidate anything and does not have any advantage over properly writing $a = (p,v)$ for the elements of $T_p(\mathbb R^n)$. In fact, Tu introduces an oversubtle and unnecessary distinction between row and columns vectors -  but perhaps it is a matter of taste.

Anyway, this should answer your questions 1. and 2. In 2. you also ask "So will this vector space include all the possible vectors that have $p$ as starting point, even the normal vectors? Or only the vectors that pass tangentially through $p$?" Here you are misled by Fig. 2.2. A tangent vector $v$ to a surface at $p$. In fact, $T_p(\mathbb R^3)$ is not the set of tangent vectors to some surface $S \subset \mathbb R^3$ at the point $p \in S$, it is the set of all column vectors in the above sense. The set of tangent vectors to $S$ at $p$ is a two-dimensional linear subspace of $T_p(\mathbb R^3)$, as you shall see later when studying the book. You can write $T_p(S)$ for it.

Concerning your question 3.: Yes, it is an abuse of notation. Tu uses the same symbols for basis elements of $\mathbb R^n$ and of $T_p(\mathbb R^n)$. If $e_1, \ldots,e_n$ denote the standard basis vectors of $\mathbb R^n$, he should properly write $(p, e_i)$ for the standard basis vectors of $T_p(\mathbb R^n)$. But due to Tu's notational conventions we may omit $p$ and regard $e_i$ as a column vector. I think the danger of confusion is very little.

Of course $e_i$ does not live in $\mathcal D_p(\mathbb R^n)$ and thus is different from the derivation $D_{e_i} = \dfrac{\partial}{\partial x_i} \mid_p$. But for each $v \in \mathbb R^n$ we have the usual directional derivative $D_v = D_v \mid_p$ at $p$. This is the map which associates to any smooth $f : U \to \mathbb R$ defined on an open  neigborhood $U$ of $p$

$$D_v \mid_p(f) = \lim_{t \to 0}\frac{f(p +tv)-f(p)}{t} .$$
Tu formalizes this and gets a map
$$D_v = D_v \mid_p : C^\infty_p \to \mathbb R$$
defined on the set $C^\infty_p$ of germs of smooth functions at $p$. This set is an $\mathbb R$-algebra (i.e. a vector space with a compatible multiplication of its elements). Tu shows that each $D_v$ is a derivation and thus gets his linear map
$$\phi : T_p(\mathbb R^n) \to \mathcal D_p(\mathbb R^n), \phi(v) = D_v.$$
In this formula $v \in  T_p(\mathbb R^n)$ is a column vector in the sense of Tu's convention. It would be more precise to understand it as $\phi((p,v)) = D_v \mid_p$.

The formula $D_v =  \sum v^i \dfrac{\partial}{\partial x_i} \mid_p$ follows from the fact that $D_{e_i} = \dfrac{\partial}{\partial x^i} \mid_p$: We have $v = \sum v^i e_i$ and therefore $D_v = \phi(v) = \phi(\sum v^i e_i) =  \sum v^i \phi(e_i) = \sum v^i  \dfrac{\partial}{\partial x^i} \mid_p$.

It is by no means trivial that $\phi$ is an isomorphism of vector spaces. This is proved in Theorem 2.2. It answers your question 4.: As Tu writes

This theorem shows that one may identify the tangent vectors at $p$ with the derivations at $p$. Under the vector space isomorphism $T_p(\mathbb R^) \simeq \mathcal D_p(\mathbb R^n)$, the standard basis $e_1,\ldots,e_n$ for $T_p(\mathbb R^n)$ corresponds to the set $\dfrac{\partial}{\partial x^1} \mid_p, \ldots, \dfrac{\partial}{\partial x^n} \mid_p$  of partial derivatives.


问:
Why does this gamma function equality hold?
I am trying to understand a 'given' in Titchmarsh's third proof of the functional equation for Riemann's zeta function (E. C. Titchmarsh (1986). The Theory of the Riemann Zeta-function (2nd ed.). Oxford: Oxford Science Publications. Section 2.4, pp. 21-22, also quoted on the Wikipedia page for the zeta function here). The proof begins with the observation that for $\Re(s)>0$

$$\int_0^{\infty } x^{\frac{1}{2}s-1} e^{-n^2  \pi  x} \, dx = \frac{\Gamma  \left(\frac{s}{2}\right)}{n^s \pi ^{s/2}}$$

I realise it probably only requires basic algebra, but I don't understand how one gets to this statement from the definition of the gamma function

$$\Gamma (s) = \int_0^{\infty } x^{s-1} e^{-x} \, dx, \Re(s)>0$$

I reason as follows:

$$\begin{aligned}
\Gamma \left(\frac{s}{2}\right) &= \int_0^{\infty } x^{\frac{1}{2}s-1} e^{-x} \, dx
\\
\\ \frac{\Gamma \left(\frac{s}{2}\right)}{n^s \pi^{s/2}} &= \frac{1}{n^s \pi ^{s/2}} \int_0^{\infty } x^{\frac{1}{2}s-1} e^{-x} \, dx
\\
\\ &= \int_0^{\infty } x^{\frac{1}{2}s-1} \frac{e^{-x}}{n^s \pi ^{s/2}} \, dx
\end{aligned}$$

with the last step being justified because the integral is over $x$ and $\frac{1}{n^s \pi ^{s/2}}$ is therefore effectively a constant in relation to it.

But then I get lost, because it's obvious that $ n^{-s} \pi^{-s/2} e^{-x} \neq e^{-n^2  \pi  x}$.

Could someone explain how the integral expression above for $\frac{\Gamma \left(\frac{s}{2}\right)}{n^s \pi^{s/2}}$ is derived?

答:
Let $u=n^2 \pi x$, so the integral becomes

$$ \int_0^\infty dx \ x^{-1+s/2} e^{-n^2 \pi x} = \int_0^\infty \frac{du}{n^2 \pi} \left( \frac{u}{n^2\pi} \right)^{-1+s/2} e^{-u}=\frac{1}{(n^2\pi)^{s/2}} \int_0^\infty du \ u^{-1+s/2} e^{-u}$$

The integral on the right is $\Gamma(s/2)$, so we are left with

$$ \int_0^\infty dx \ x^{-1+s/2} e^{-n^2 \pi x} =\frac{1}{n^s \pi^{s/2}}\Gamma\left(\frac{s}{2}\right) $$$$\tag*{$\blacksquare$}$$


问:
If $p \mid a^{n}$ Prove that $p^{n} \mid a^{n}$
$p$ Is a prime number.
I’ve found this problem in a math book.

My Proof :

We know that :$$p\mid n_{1}n_{2}…n_{i} \implies \exists j (1\le j \le i)$$such that $p/n_j $

So:
$$p \mid a^{n} \implies p\mid a \implies p^{n} \mid a^{n}$$
Please tell whether there is any mistake in my proof.

答:
To prove the implication we need to prove that $a$ multiple of $p$ ,and $p$ is a prime number meaning

$PGCD(a.p)\neq 1$

Assume.  $PGCD(a.p)=1$

$(1)$...  $p\mid a^n=aa^{n-1} \wedge PGCD(a.p)=1 \Rightarrow p\mid a^{n-1} $ ....

$(2)$....   $p\mid a^{n-1}=aa^{n-2}  \wedge PGCD(a.p)=1 \Rightarrow p\mid a^{n-2} $

.

.

$(n-1)$..  $p\mid a^2=a.a \wedge PGCD(a.p)=1 \Rightarrow p\mid a $

This is a contradiction to the assumption above

So $PGCD(a.p)\neq 1 \wedge p$ is a prime number we find

$ (a=kp) ,   k\in\mathbb N^{*} $

Finally :
$ a^n= k^n.p^n \Leftrightarrow p^n\mid a^n $


答:
First, you can guess the answer, which might be $(u, v) = (2, 2)$. Since there might exist other solutions, we need to think more.

The next thing is: I believe that you can draw the graph of $u = 20 - v^{2}$ and $v = 20 -u^{2}$ on $(u, y)$-plane, which gives two parabolas that are reflections each other with respect to the line $u = v$. And, you have another guess: The other solutions do not satisfy $u\geq 0$ and $v\geq 0$, which should happen since $u = x^{2}$ and $v = y^{2}$. (Note that there are 3 more intersection points) So we can say confidently that the only solution is $(u, v) = (2, 2)$, but this is not a proof. There's another solution that you can guess that corresponds to another intersection point with $u = v$ other than $(4, 4)$. If you set $u = v$, then you obtain $u = -5, 4$ by solving a quadratic equation.

At last, to prove that this is the only solution, note that the degree 4 polynomial should be divisible by $(v-4)$ and $(v+5)$ since $4, -5$ are roots of it. We have
$$
v^{4} - 40v^{2} +v + 380 = (v-4)(v+5)(v^{2} - v - 19)
$$
and $v^{2} - v - 19  = 0 \Leftrightarrow v = \frac{1 \pm \sqrt{77}}{2}$, where both of them fail to satisfy $(u \geq 0) \wedge (v \geq 0)$. So the only answer is $(u, v) = (2, 2) \Leftrightarrow (x, y) = (\pm 2, \pm 2)$.