问:
Determinant of a permutation matrix plus identity

Let $A$ be a permutation matrix. Calculate $\det(I + A)$.

I tried with the definition of the determinant but couldn't find it. I also tried to use the decomposition of permutations as products of disjoint cycles.

答:
Here's a reasonable approach. First, consider the case in which $A$ is the size $n$ matrix of a single cycle of length $n$. We find that the associated characteristic polynomial is
$$
\det(A - \lambda I) = (-1)^n(\lambda^n - 1).
$$
To calculate $\det(A + I)$, it suffices to plug in $\lambda = -1$. We find that
$$
\det(A + I) = (-1)^n((-1)^n - 1) = \begin{cases}2 & n \text{ is odd,}\\ 0 & n \text{ is even.}\end{cases}
$$
For the general case, let $A_1,\dots,A_k$ denote the matrices associated with each of the disjoint cycles in the cycle decomposition of $A$. We see that $A$ is similar (permutation similar, in fact) to the block diagonal matrix
$$
PAP^{-1} = \pmatrix{A_1\\ & \ddots \\ && A_k}.
$$
It follows that $\det(A + I) = \det(PAP^{-1} + I) = \det(A_1 + I) \cdots \det(A_k + I)$.

Thus, we reach the following conclusion: suppose that the permutation associated with $A$ can be decomposed into a product of $k$ cycles. If one of those cycles has even length, then $\det(A + I) = 0$. Otherwise, we find that $\det(A + I) = 2^k$.


问:
System of equations involving 4 variables

If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$

I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.

答:
From the last three equations, substract the previous multiplied by $x$ to eliminate $a$:

$$b(y-x)y^2=62-24x,\\b(y-x)y=24-10x,\\b(y-x)=10-6x.$$

Now take the ratios to eliminate the common factors:
$$y=\frac{62-24x}{24-10x}=\frac{24-10x}{10-6x}$$

This is equivalent to

$$x^2-3x+1=0$$ and there are two real solutions. From $x$, compute $y$, then $b(y-x)y^3$, which leads to $ax^4+by^4$.

ContentLicense: CC BY-SA 4.0


问:
prooving the harmonic inequality
My real analysis textbook shows this result but gives no proof for it, it would be great if someone could show me why the following result holds.

Show

$$\sum_{n=1}^{\infty} \frac{H_n}{n(n+1)} = \sum_{n=1}^{\infty}
 \frac{1}{n^{2}}$$

where $H_n = \sum_{k=1}^{n} 1/k$.

Title: prooving the harmonic inequality

Body:
$$S=\sum_{n=1}^{\infty} H_n \left ( \frac{1}{n}-\frac{1}{1+n} \right)$$
$H_n=\int_{0}^{1} \frac{1-x^n}{1-x} dx.$
So $$ S=\sum_{n=1}^{\infty} \int_{0}^{1} \int_{0}^{1} \frac{1-x^n}{1-x}  (t^{n-1} -t^{n})~dx~dt=\int_{0}^{1}\int_{0}^{1}\frac{1-t}{1-x} \sum_{n=1}^{\infty} (t^{n-1}-(tx)^{n}t^{-1})$$
$$S=\int_{0}^{1} \int_{0}^{1}\frac{1-t}{1-x}  \frac{1-x)}{(1-x)(1-xt)} dt dx=\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xt} dt dx =\sum_{n=1}^{\infty}\int_{0}^{1} \int_{0}^{1} x^nt^n dx dt.$$
$$\implies S=\sum_{n=1}^{\infty} \frac{1}{n^2}$$


问:
Range of values for $y$ given $-2 \leq x \leq 3$ and $9 \leq x + y \leq 11$
I want to find the range of values of $y$ for which

$-2 \leq x \leq 3$ and $9 \leq x + y \leq 11$

Now initially I thought to subtract the inequalities, but that gives $8 \leq y \leq 11$

However, $y$ cannot be any of these values because then $\max x + \max y = 3 + 11 = 14$ and this is bigger than the maximum value of $x + y$.

Any help is appreciated. Probably being a bit stupid!

答:
A bit of geometry, consider an $x,y$ coordinate system.

1)$-2 \le x \le3.$

Draw lines parallel to the $y-$axis through $x=-2$ and $x=3.$

2)$9 \le x+y \le 11.$

$x+y=C$ is a line with $y-$intercept $C$ and slope $-1.$

Then $9\le C \le11$,

Draw the lines $y=-x +9$ and $y=-x+11.$

3)The lines in 1) and 2) form a parallelogram, refer to Goh's nice drawing.

4)Allowed $x,y$ values are within this parallelogram.

5)Smallest $y:$

Interception of $y=-x +9$ with $x=3$, i.e. the lower right hand corner of the parallelogram:

$y_{min}=6$.

6)Largest $y:$

Interception of $y=-x+11$ with $x=-2$, i.e. the upper left hand corner of the parallelogram:

$y_{max}=13.$

7)Altogether   $6 \le y\le 13.$


问:
$y_1(t) = t^2$ and $y_2(t) = t|t|$ linearly independent on [-1,1]
For $t<0$, we have $t|t|= -t^2$. For $t>0$, we have $t|t|=t^2$.

Hence the Wronskian $$W[y_1,y_2](t) = y_1 y_2' -y_1' y_2  \\
= t^2 \cdot (-2t) - (2t)\cdot(-t^2) =  0 \quad t<0 \\ t^2\cdot (2t)-(2t)\cdot t^2 = 0 \quad t>0\ .
$$
In either case, the Wronskian is zero. How are these functions then linearly independent on [-1,1]?

Context: this is problem 2.1.11 of Braun's Differential Equations and Their Applications, 4th edition.

(a) Show that $y_1$ and $y_2$ are linearly dependent on the interval $0\leq t \leq 1$.

(b) Show that $y_1$ and $y_2$ are linearly independent on the interval $-1\leq t \leq 1$.

答:
You don't need the Wronskian to know these two functions are linearly independent on the interval $[-1,1]$ because two functions are linearly dependent just when one is a constant multiple of the other. That's clearly not the case here. These function are dependent on $[0,1]$ since they are identical. They are also dependent on $[-1,0]$.

At https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Ordinary_Differential_Equations/3%3A_Second_Order_Linear_Differential_Equations/3.6%3A_Linear_Independence_and_the_Wronskian you can read that

Let $f$ and $g$ be differentiable on $[a,b]$. If Wronskian $W(f,g)(t_0)$ is
nonzero for some $t_0$ in $ [a,b]$ then $f$ and $g$ are linearly independent on
$[a,b]$.

Note the implication there: if the Wronskian is nonzero then the functions are independent. You are invoking the converse. But nowhere does it say that an identically zero Wronskian implies dependence.

Your example shows that the converse is in fact false. You can find it in the wikipedia page that @ChristianBlatter notes in his comment.


问:
Relation of eigenvectors and eigenvalues between commuting matrices
Let $A,B\in\mathbb{C}^{n\times n}$ commute, let $B$ be nonsingular and let $\lambda\in\sigma(A)$ have geometric multiplicity 1 and a corresponding eigenvector $\mathbf{v}.$

I have so far shown that $B\mathbf{v}$ is an eigenvector of $A$ corresponding to $\lambda$ and I have to show that

There exists a $\mu\in\sigma(B)$ such that $\mathbf{v}$ is an eigenvector of $B$ corresponding to $\mu.$

If $B=B^{-1},$ then $\mu=\pm 1.$

I am a bit stuck at both of these parts in how to tackle them, I think part 1 requires use of the multiplicity and that the eigenspace will be one dimensional but I am not sure.

答:
Because the geometric multiplicity of $\lambda$ is $1$, the eigenvectors $v$ and $Bv$ must be parallel.

Assuming $B=B^{-1}$, we have $v=B^{-1}Bv=B^2v=\mu^2v$, hence $\mu^2=1$.


问:
What is a condition for two real functions $f,g$ to "commute", so $f(g(x))=g(f(x))$?
Say I'm given two functions $f,g$. Can I tell if they "commute" without actually trying them in the formula
$f(g(x))=g(f(x))$?

And given a function $f$, is there a way to find all functions $g$ such that
$f(g(x))=g(f(x))$?

I've tried using derivatives and the chain rule but haven't got anything interesting yet.

答:
There is, in general, no characterization other than the evaluation of $f(g(x))$ and $g(f(x))$.

Also, for fixed $f$ there is no an easy way to find all the $g$ such that $f \circ g = g \circ f$.

Special cases can be studied. For example, a theorem of J. Lipiński says that if $f$ is a continuous and strictly monotonic function on an interval $I$ such that $f(I) \subseteq I$ then there is an infinity of solutions $g$ (depending on an arbitrary function).


问:
Is it possible to define a bijection from nonnegative to positive numbers?
Let $\mathbb{R}_{\geq 0}$ be the set of nonnegative numbers and $\mathbb{R}_{>0}$ the set of positive numbers, that is

$$
\mathbb{R}_{\geq 0} = \{\,x \geq 0 \mid x \in \mathbb{R} \,\}
$$

and

$$
\mathbb{R}_{> 0} = \{\,x > 0 \mid x \in \mathbb{R} \,\}
$$

Is it possible to define a bijection $f$ between these two sets?

答: Yes, of course.  First map every number that is not a non-negative integer to itself.  Then map every non-negative integer n to n+1.</p>


问:
Good source to learn about surface singularities?
I am looking for something that treats singularities on algebraic surfaces and curves over $\mathbb{C}$, starting from the very basics but not stopping there.

I checked out Miles Reid his lectures on algebraic surfaces and they seem promising (especially the exercises). Also Harris - algebraic geometry a first course seems nice. However, the first gets very technical quickly while the second book's treatment of singularities is pretty short.

Is there also something that treats loads and loads of examples, furnishing a good intuition about the subject?

Thanks!

答: You might enjoy "Introduction to Singularities and Deformations" by Greuel, Lossen, and Shustin. The flavour is complex analytic, but many examples and techniques overlap with the algebraic category. (And it is available on SpringerLink, so it should be easy to find a copy online.)</p>

ContentLicense: CC BY-SA 3.0


问:
Solving the given system of equations

$$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$

I got $x^3 + y^3 = 6xy = (x+y)^2$ which I simplified to $x^2 - xy + y^2 = x + y$ and I'm not sure what to do now. Please help.

答:
In the initial equations, make the substitution $x/y=z$. From here $y=x/z$. Then the second equation is $$z+\frac1z=4$$ and the first equation is $$xz+x\frac1{z^2}=6$$
Calculate first $z$ as $$z_{1,2}=2\pm\sqrt3$$Now calculate $x$ then $y$.

Title: Solving the given system of equations


问:
Given sum of 4 variables as a constant, how to maximize the sum of products?
If variables a, b, c, and d are positive numbers and
a + b + c + d = 1,

Can I find the maximum of the expression below?

ab + cd

and how can I reach that?

答: Hint: $ab+cd \leq (a+c)(b+d)$.

Body: Hint: $ab+cd \leq (a+c)(b+d)$.


问:
Logic - Proof of a resolution rule
I am going to show the following resolution rule:
$$
(p \lor q) \land (\lnot p \lor r) \vdash (q \lor r)
$$
The proof I tried:
\begin{align}
&\quad \ (p \lor q) \land (\lnot p \lor r) \\
&\equiv ((p \lor q) \land \lnot p) \lor ((p \lor q) \land r) \\
&\equiv ((p \land \lnot p) \lor (\lnot p \land q)) \lor ((p \lor q) \land r) \\
&\equiv (\lnot p \land q) \lor ((p \lor q) \land r) \\
&\equiv ((\lnot p \land q) \lor(p \lor q)) \land ((\lnot p \land q) \lor r) \\
&\equiv (((\lnot p \land q) \lor p) \lor q) \land ((\lnot p \land q) \lor r) \\
&\equiv (((\lnot p \lor p) \land (p \lor q)) \lor q) \land ((\lnot p \land q) \lor r) \\
&\equiv (p \lor q) \land ((\lnot p \land q) \lor r) \\
&\equiv (p \lor q) \land (\lnot p \lor r) \land (q \lor r) \\
&\vdash (q \lor r)
\end{align}

It is strange that $(p \lor q) \land (\lnot p \lor r)$ appeared again with $q \lor r$ being connected by conjunction. Is this correct?

答:
Yes, it is correct ... you just derived the dual form of the Consensus Theorem:

$xy + x'z = xy + x'z + yz$

And if you look at a K-map, it's is actually not so strange, because it shows that the $YZ$ term is redundant given the other two.

Title: Logic - Proof of a resolution rule


答:
You can also take every $[n,n+1)$ interval and map it into $(n,n+1]$ by reflecting it.

The idea here being mapping $[0,\infty)$ into $(0,\infty]$ via $\frac{1}{x}$, but we can only do it if the right extreme of the second interval is included. Luckily, we can cover $\mathbb{R}$ with intervals of such form.

Title: Is it possible to define a bijection from nonnegative to positive numbers?

Body:
Let $\mathbb{R}_{\geq 0}$ be the set of nonnegative numbers and $\mathbb{R}_{>0}$ the set of positive numbers, that is

$$
\mathbb{R}_{\geq 0} = \{\,x \geq 0 \mid x \in \mathbb{R} \,\}
$$

and

$$
\mathbb{R}_{> 0} = \{\,x > 0 \mid x \in \mathbb{R} \,\}
$$

Is it possible to define a bijection $f$ between these two sets?


问:
Using implication to argue for all members of a set, something is true
I am trying to express some statements in a formal language that allows universal quantification over domains, but not over sets. I was wondering how to express that something is true for all elements of a given set.

Let us assume $D$ is a domain and we can write $\forall x:D, A(x)$ to mean that "Forall $x$ in the domain $D$, $A(x)$ is true".

Now, let us assume we have a non-empty set $S$ that contains elements of the domain $D$ and
the $\in$ relation is a binary predicate over the domain $D_S$ of all such sets such that $\in : D_S \times D \rightarrow \{true,false\}$.

My goal is to be able to express the statement "For every member of the set $S$, $A(x)$ is true".

However, the language I am using does not allow me to write something like $\forall x \in S: A(x)$.

In this case, would it be correct to write $\forall x:D, ((x \in S) \implies A(x))$?

I realize that if $S$ was allowed to be empty, the statement would be vacuously true, but that is not a concern for the universe I am trying to reason about. My main concern is if $\forall x:D, ((x \in S) \implies A(x))$ means "For every member of the set $S$, $A(x)$ is true".

答:
Assume $x\in S$. If you can then prove $A(x)$, and you have discharged any and all other subsequent assumptions about $x$ in your proof, then you can generalize to obtain:

$~~~~~~\forall x: D, [x \in S \implies A(x)]$

if I understand your notation correctly. You don't really need '$D,$' however. You could write:

$~~~~~~\forall x: [x \in S \implies A(x)]$

or

$~~~~~~\forall x\in S: A(x)$


答:
Hint:

Let $x+y=a, xy=b$

$$\implies 6b=a^3-3ab\  \ \ \  (1)\text{ and }4b=a^2-2b\implies a^2=6b\  \ \ \  (2)$$

Clearly, $ab\ne0$

Replacing the value of $b$ in $(1)$,

$$a^2=a^3-3\cdot a\cdot\dfrac{a^2}6\iff2a^2=a^3\implies a=?\text{ as }a\ne0$$

So, we can find $b=\dfrac{a^2}6$

So, $x,y$ are the roots of $$t^2-at+b=0$$

Title: Solving the given system of equations

Body:
$$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$

I got $x^3 + y^3 = 6xy = (x+y)^2$ which I simplified to $x^2 - xy + y^2 = x + y$ and I'm not sure what to do now. Please help.


Title: Given sum of 4 variables as a constant, how to maximize the sum of products?

Body:
If variables a, b, c, and d are positive numbers and
a + b + c + d = 1,

Can I find the maximum of the expression below?

ab + cd

and how can I reach that?

Body: Hint:$${(a+b+c+d)}^2-4ab-4cd={(a-b)}^2+{(c-d)}^2+2(ac+bd+ad+bc)\ge 0$$


问:
Lottery Competition
Suppose I want to hold lottery competition. I just want two people to win the lottery and there are 100 people buying my lottery. What should the maximum probability (of each player wining the lottery) be to ensure that exactly two people win the lottery?

I obviously think it's $2/100=1/50$. But I think it should not be so trivial. I would appreciate any comments.

答: For the new question, you have a binomial distribution.  If the probability that each player wins is $x$, the probability of exactly $2$ winning is ${100 \choose 2}x^2(1-x)^{98}$.  The first term is the number of ways to choose the winners, the $x^2$ is the probability they both win, and the $(1-x)^{98}$ is the probability that all the others lose.  Now take the derivative and set to zero.  Yes, you should get $1/50$</p>

Title: Lottery Competition


问:
I don't understand the solution to this geometric problem

The figure shows a circle with $AB$ as a diameter.

Given that $\angle{ADE} = \angle{DCA}$ and $\angle{CBA} = 70^\circ$, find

$\angle{FEB}$

$\angle{EFC}$

I could not solve this question and I do not understand the answer:

\begin{align}
\text{Let } \angle{ADE} &= \angle{DCA} = x^\circ \\
\angle{ACB} &= 90^\circ \color{grey}{ \text{(rt. } \angle \text{ in a semicircle)} } \\
\angle{DAE} &= 180^\circ - 90^\circ - x^\circ \color{grey}{(\angle \text{ sum of a } \triangle) } \\
&= (90 - x)^\circ \\
\angle{FEB} &= (90 - x)^\circ + x^\circ \color{grey}{ \text{ext. } \angle = \text{ sum of int. opp. } \angle{s} } \\
&= 90^\circ
\end{align}

\begin{align}
\angle{EFC} &= 360^\circ - 90^\circ - 90^\circ - 70^\circ \\
&= 110^\circ
\end{align}

In 1, I do not understand the line $\angle{DAE} = 180^\circ - 90^\circ - x^\circ$. Where does $90^\circ$ come from?

答:
Since $ABCD$ is a cyclic quadrilateral

$$\angle DAB + \angle DCB =180$$

$$\angle DAE = 180-\angle DCB = 180 - (90+x)= 90 -x$$

(opposite angles of cyclic quadrilateral are supplementary)

$$\angle FEB =\angle DEB = \angle ADE +\angle DAE=x+(90-x)=90$$

(exterior angle of $\triangle DEB$)


问:
What can be Jordan forms of matrix with charcterstic polynomial $(x-3)^2 (x-2)^4$ and minimal polynomial $(x-3)(x-2)^2$
The following question was asked in my linear algebra quiz and I was unable to think which result I should use to solve it.

Question: Let A be a $6 \times 6$ matrix over $\mathbb{R}$ with chacterstic polynomial =$(x-3)^2 (x-2)^4$ and minimal polynomial = $(x-3)(x-2)^2$ .Then Jordan canonical form of A can be? I have to find all the matrix which are possible.

Algebraic multiplicity of 2 and 3 are 4,2 respectively and Geometric multiplicity of 2 and 3 are 2 and 1 respectively.

But i don't know which result should be used to move foreward. I calculated these in case they are useful.

Can you please shed some light on this?

答:
By assumption, $A$'s group of elementary divisors can be one of the following:

$\lambda - 3, \lambda - 3, (\lambda - 2)^2, (\lambda - 2)^2$;

$\lambda - 3, \lambda - 3, (\lambda - 2)^2, \lambda - 2, \lambda - 2$.

Therefore, up to row/column permutations, there are two possibilities of $A$'s Jordan form
$\newcommand{\diag}{\mathrm{diag}}$
\begin{align*}
& \diag(3, 3, J_{(2)}(2), J_{(2)}(2)), \\
& \diag(3, 3, J_{(2)}(2), 2, 2),
\end{align*}
where $J_{(m)}(\lambda)$ denotes the Jordan block of size $m$ associated with eigenvalue $2$.


问:
$\lambda $- geodesic convexity
Let $(X,d)$ be a geodesic space, suppose $g:X \to \mathbb{R}$ is geodesically $\lambda$ convex with $\lambda >0$, that is for all $x,y \in X$ there exists a constant speed geodesic $\gamma$ from $x$ to $y$ such that for all $s\in [0,1]$
$$
g(\gamma_s) \leq s g(x) + (1-s) g(y) -\frac{\lambda}{2}s(1-s)d(x,y)^2
$$
Let $x_0$ be the unique minimizer of $g$. Show that $\forall x$
$$
g(x) \geq g(x_0) + \frac{\lambda}{2}d(x,x_0)^2
$$

EDIT: thanks for the comment.
Honestly I did not do much because I was stuck from the beginning, since I had no idea how to use the geodesic property. I just picked a geodesic between $x$ and $x_0$ and tried to plug in the property but did not see how that could help!
Any hint or suggestions?

答:
Choose your geodesic with $\gamma_{0}=x_{0}$ and $\gamma_{1}=x$. Since $x_{0}$ is minimal we have
\begin{equation}
0=\frac{d}{ds}|_{s=0}g(\gamma_{s}).
\end{equation}
Then, using your first inequality it follows
\begin{equation}
0=\lim_{h\to 0}\frac{1}{h}(g(\gamma_{h})-g(x_{0}))\le\lim_{h\to 0}\frac{1}{h}\left(hg(x)+(1-h)g(x_{0})-\frac{\lambda}{2}h(1-h)d^{2}(x,x_{0})-g(x_{0})\right)
\end{equation}
which yields the claim.

Title: $\lambda $- geodesic convexity


问:
On the proof of Schur's lemma in Fulton & Harris
I'm reading the book on representation theory by Fulton and Harris. I'm stuck with the proof of Schur's Lemma 1.7:

Schur's lemma 1.7 If $V$ and $W$ are irreducible representation of $G$ and $\phi:V\rightarrow W$ is a $G$-module homomorphism, then
  1. Either $\phi$ is an isomorphism, or $\phi = 0$.
  2. [...]

Proof.  The first proof follows from the fact that the kernel of $\phi$ and the image of $ \phi$ are invariant subspaces. [...]

I understand that a $G$-homomorphism is a synonym for a map between two representations.  I figured out why the kernel and the image are invariant (under any $g\in G$), by I have no idea why that fact leads to the conclusion.

I would be most grateful if you could fill in the gap.

答:
Since the kernel is an invariant subspace and $V$ is a simple representation, the kernel is $0$ or $V$.

Since the image of $\phi$ is an invariant subspace of $W$, it is either $0$ or $W$.

If $\phi$ is nonzero, then $\ker(\phi)=0$ and $Im(\phi)=W$.

If $\phi$ is zero, then $\ker(\phi)=V$ and $Im(\phi)=0$.


问:
show that harmonic series is divergent with $ \sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2} $
I'm pretty stuck with this exercise. Hope somebody can help me:

show that
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2}
$$
and use this to show that the harmonic series is divergent.

First I can't figure out how to show the inequality.
Second I'm not quite sure how to use the inequality to show divergence. Because it doesn't really help letting k go to infinity.

Writing out the series gave me this, but where do I go from here:
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}}
$$

Edit:
So I understand how to show the inequality now. Thanks!
But I'm still lost with how to use the inequality for the divergence proof.

Body:
HINT
$$
\begin{split}
\sum_{n=2^k+1}^{2^{k+1}} \frac1n
 \ge \sum_{n=1}^{2^k} \frac{1}{2^k+n}
 \ge \sum_{n=1}^{2^k} \frac{1}{2^{k+1}}
 = \frac{1}{2^{k+1}} \sum_{n=1}^{2^k} 1
\end{split}
$$
Can you finish?


Title: show that harmonic series is divergent with $ \sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2} $

Body:
I'm pretty stuck with this exercise. Hope somebody can help me:

show that
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2}
$$
and use this to show that the harmonic series is divergent.

First I can't figure out how to show the inequality.
Second I'm not quite sure how to use the inequality to show divergence. Because it doesn't really help letting k go to infinity.

Writing out the series gave me this, but where do I go from here:
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}}
$$

Edit:
So I understand how to show the inequality now. Thanks!
But I'm still lost with how to use the inequality for the divergence proof.

Body:
In fact
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}\ge \frac{1}{2^{k}+2^k}+\frac{1}{2^{k}+2^k}+\cdots+\frac{1}{2^{k}+2^k}=(2^{k+1}-2^k)\frac{1}{2^{k+1}}=\frac12.
$$


问:
Given a solution to a system of ODE, prove that A has some pure imaginary eigenvalue (non zero).
$A$ is an $n$-dimensional square matrix:

Let $X(t)$ be the solution of the problem of initial value $\frac{d}{dt}X=AX$ with $X(t_0)=X_0≠0$. Suppose there exists a real $t_1$ such that $X(t_1)=-X_0$. Prove $A$ has some purely imaginary eigenvalue (non zero).

This was an exam problem from last year in my ODE course and I'm trying to solve it to prepare for my exam.

答:
Every square matrix has eigenvalues and we need to demonstrate that one of those  eigenvalues of $A$ has a purely imaginary part.

The solution to $\frac{d}{dt}X=AX$ with $X(t_0)=X_0≠0$ is $e^{At}X_0$.

Plug in the second condition, you get $e^{At_1}X_0=-X_0$ which means $-1$ is an eigenvalue of $e^{t_1A}$.

Let $J$ be a Jordan normal form of $A$ and $A=PJP^{-1}$ where $P$ is some invertible square matrix. Let $J_\lambda$ be the block associated with eigenvalue $\lambda$.

Now $e^{At_1}X_0=-X_0$ is equivalent as $Pe^{t_1J}P^{-1}X_0=-X_0$ and therefore $-1$ is an eigenvalue of $e^{t_1J}$.

Jordon canonical blocks have this nice property: the exponentiation of a Jordon block is an upper triangular matrix with the main diagonal equal to the exponentiation of the eigenvalue.

The same property holds for the whole Jordan form.

In turn, the diagonal elements of upper triangular matrices are the eigenvalues of the matrix.

Therefore, one diagonal element of $e^{t_1J}$ is $-1$. Let the Jordon block where $-1$ is located in be in $e^{t_1J}$ be $J_\lambda$. We then have $e^{t_1\lambda}=-1=e^{i(\frac{\pi}{2}+2n\pi)}$ for $n\in\mathbb{Z}$.

Or, $\lambda=\frac{i\pi}{t_1}(\frac{1}{2}+2n)$, which makes $\lambda$ purely imaginary and non-zero.


问:
Probability density function on a random variable
Problem: The life expectancy $X$ of a lightbulb can be modeled as an exponential distribution with $\lambda > 0$. Every started hour shall be counted fully. Therefore consider $K:=\lceil X \rceil $ and distribution of $K$.

I'm not quite sure I understand the question correctly, so I wanted to ask for some advice concerning the topic. I kind of think of it as a discrete modeled distribution since we are always looking at $\lceil X \rceil $.
So here it would be: $$P(K = t)=P(\lceil X \rceil = t)=P(t-1<X\le t)$$ Is this a correct way of approaching the problem? If yes where do I go from here? Thanks in advance.

答:
I guess you mean $K$ not $T$ in the beginning. Yes, indeed,
$$
\mathbb{P}[K=t]
 = \mathbb{P}[t-1 < X \le t]
 = F_X(t) - F_X(t-1).
$$
Can you find the cdf of the exponential distribution?


问:
Value of surface integral with triangle
Let S be the triangle with vertices at the origin and at a pair of vectors a,b $ \in \mathbf R^3 $ with a × b = ( 0, 3, 4) with unit normal vector n pointing in the direction a × b and let f=(1,2,1). What is the value of the surface integral $ \iint\limits_{S}\mathbf{f\cdot \mathbf{n}}dS $ ?

what I know so far:

The equation is the flux of f through S in the orientation n . I know the definition of the flux but don't know how to use it to compute this as its with a triangle i.e limits.

Any help would be appreciated.

答:
Hint:

$\hat{n} = ( 0, \frac{3}{5}, \frac{4}{5})$ as the unit normal is in direction of $\vec{a} \times \vec{b}$.

$f = (1, 2, 1)$

Instead of a double integral and knowing its bounds, in this case you will use the area of the triangle that is known from the magnitude of the cross product of two vectors of the triangle $(|\vec{a} \times \vec{b}|$).

Title: Value of surface integral with triangle


问:
A binomial identité
I tried to solve this exercise but i failed. If anyone can give me hint or solution i would be grateful.

Let $(n,p,q)\in\mathbb{N}^{3}$ such that $n\leq p$ and $n\leq q$. Show that:$$\sum_{k=0}^{n}\binom{p}{k}\binom{q}{n-k}=\binom{p+q}{n}.$$

答:
Since
$$ (1+x)^p(1+x)^q=(1+x)^{p+q} $$
or
$$ \sum_{k=0}^p\binom{p}{k}x^k\sum_{i=0}^q\binom{q}{i}x^i=\sum_{i=0}^{p+q}\binom{p+q}{i}x^i $$
comparing the coefficients of $x^n$ for both sides, you will have
$$\sum_{k=0}^{n}\binom{p}{k}\binom{q}{n-k}=\binom{p+q}{n}.$$

Title: A binomial identité


问:
Is $-a^2+2ta-1$ and $-b^2+2tb-1$ are also coprime for any positive integer $t>a$ and $t>b$
Let $a,b$ be two coprime positive integers. I am asking if $-a^2+2ta-1$ and $-b^2+2tb-1$ are also coprime for any positive integer $t>a$ and $t>b$.

答:
There are numerous counter examples even for small values. Just go find one.

Rewrite the expression as $\gcd(2(t-1)a - (a-1)^2 , 2(t-1)b -(b-1)^2)$.

Does this immediately suggest how we could make the GCD (a multiple of) 2?

 E.g.  $a, b$ odd coprime numbers, $t$ anything, then clearly both terms are even.


Title: Is $-a^2+2ta-1$ and $-b^2+2tb-1$ are also coprime for any positive integer $t>a$ and $t>b$

Body: Let $a,b$ be two coprime positive integers. I am asking if $-a^2+2ta-1$ and $-b^2+2tb-1$ are also coprime for any positive integer $t>a$ and $t>b$.

Body:
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

In[1]:=ParallelTable[
  If[TrueQ[GCD[a, b] == 1 &&
     GCD[2*t*a - a^2 - 1, 2*t*b - b^2 - 1] == 1 && t > a &&
     t > b], {a, b, t}, Nothing], {a, 2, 10}, {b, 2, 10}, {t, 2,
   10}] //. {} -> Nothing

Running the code gives:

Out[1]={{{{2, 3, 4}, {2, 3, 6}, {2, 3, 7}, {2, 3, 8}, {2, 3, 9}}, {{2, 5,
    6}, {2, 5, 7}, {2, 5, 9}, {2, 5, 10}}, {{2, 7, 8}, {2, 7,
    9}}}, {{{3, 2, 4}, {3, 2, 6}, {3, 2, 7}, {3, 2, 8}, {3, 2,
    9}}, {{3, 4, 5}, {3, 4, 6}, {3, 4, 7}, {3, 4, 8}, {3, 4,
    10}}, {{3, 8, 9}}}, {{{4, 3, 5}, {4, 3, 6}, {4, 3, 7}, {4, 3,
    8}, {4, 3, 10}}, {{4, 5, 6}, {4, 5, 7}, {4, 5, 8}, {4, 5, 9}, {4,
    5, 10}}, {{4, 7, 8}, {4, 7, 9}}}, {{{5, 2, 6}, {5, 2, 7}, {5, 2,
    9}, {5, 2, 10}}, {{5, 4, 6}, {5, 4, 7}, {5, 4, 8}, {5, 4, 9}, {5,
    4, 10}}, {{5, 6, 7}, {5, 6, 8}, {5, 6, 9}, {5, 6, 10}}, {{5, 8,
    9}, {5, 8, 10}}}, {{{6, 5, 7}, {6, 5, 8}, {6, 5, 9}, {6, 5,
    10}}, {{6, 7, 8}, {6, 7, 9}, {6, 7, 10}}}, {{{7, 2, 8}, {7, 2,
    9}}, {{7, 4, 8}, {7, 4, 9}}, {{7, 6, 8}, {7, 6, 9}, {7, 6,
    10}}, {{7, 8, 9}}}, {{{8, 3, 9}}, {{8, 5, 9}, {8, 5, 10}}, {{8, 7,
     9}}, {{8, 9, 10}}}, {{{9, 8, 10}}}}

So, we can see that you're right!


问:
Whats the intuitive difference between ${n \choose 2}$ and taking ${n \choose 1} {{n-1} \choose 1}$?
Example: I am trying to find the number of combinations of full house in poker. In that case I take ${13 \choose 1}{4 \choose 3} {12 \choose 1}  {4 \choose 2}$.

In a different scenario suppose I want to find the number of combinations of two pairs. There I take ${13 \choose 2}  {4\choose 2}  {4 \choose 2}$.

Why shouldn't I take ${13 \choose 1}$ and then ${12 \choose 1}$ like in the previous case?

答: The number $\binom{n}{2}$ is the number of ways to pick a $2$-element subset $\{a,b\}$ from a set $X$ of size $n$, while $\binom{n}{1}\binom{n-1}{1}$ is the number of ways of first picking an element $a$ from $X$ and then choosing an element $b$ from the set $X\setminus \{a\}$. In both cases, you end up with a $2$-element subset of $X$ but when we chose the 2-elements the first way, the order didn't matter. However, in the second way of choosing the elements, the order did matter. For instance, with the first method, you may have chosen the set $\{a,b\}$ while with the second method you may have picked $a$ and then $b$, or $b$ and then $a$. This is why there are in fact twice as many ways to choose $2$-elements in an ordered way versus in an unordered way.


问:
Bounding integrals using Maclaurin series

Prove that $$\int_0^1 \frac{ \sin x}{\sqrt{x} } < \frac23$$

I didn't see any easy way to integrate directly, so I just used the series approximation:

$$ \sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$

Hence,

$$\int_0^1 \frac{ \sin x}{\sqrt{x} }= \sum_{k=0}^{\infty}  \frac{  (-1)^k}{(2k+1)! (2k+ \frac32)}= \frac23- \frac{1}{3! (10.5)} + \text{stuff}$$

The argument I had is that, the increments and decrements added by taking more and more terms keeps shrinking as higher terms are smaller in magnitude for the above series. Hence, the value of integral is no greater than the very first increment which is $2$.

The problem is I feel this argument is not precise/ rigorous enough. Is there any theorems /results I can refer to when stating this?

答: You can use that $\sin x <x$ for $x>0$, hence $\;\dfrac{\sin x}{\sqrt x}<\sqrt x$.


答:
In a full house, there’s an asymmetry between the two values making up the poker hand - one occurs three times, and one occurs twice. In two pair, both values appear the same number of times. This means that if I tell you “I have two pair made of kings and tens,” you know what my hand is, but if I say “I have a full house made of kings and tens” you can’t tell whether I have three kings and two tens or three tens and two kings.

Mathematically, $\binom{13}{2}$ represents “pick two different values, and order doesn’t matter,” which is great for a two pair where the two different values don’t have an associated order. On the other hand, $\binom{13}{1}\binom{12}{1}$ means “pick some value as the first value, then pick some other value as the second value” means that order does matter, since one of them is specifically picked as the three-of-a-kind and one is picked as the two-of-a-kind.

Hope this helps!

Title: Whats the intuitive difference between ${n \choose 2}$ and taking ${n \choose 1} {{n-1} \choose 1}$?

Body:
Example: I am trying to find the number of combinations of full house in poker. In that case I take ${13 \choose 1}{4 \choose 3} {12 \choose 1}  {4 \choose 2}$.

In a different scenario suppose I want to find the number of combinations of two pairs. There I take ${13 \choose 2}  {4\choose 2}  {4 \choose 2}$.

Why shouldn't I take ${13 \choose 1}$ and then ${12 \choose 1}$ like in the previous case?


答: Since $W$ is irreducible under $G$, so the possible submodules of $W$ can only be either trivial module or itself, which are the only possible images of $V$.</p>

Title: On the proof of Schur's lemma in Fulton & Harris

Body:
I'm reading the book on representation theory by Fulton and Harris. I'm stuck with the proof of Schur's Lemma 1.7:

Schur's lemma 1.7 If $V$ and $W$ are irreducible representation of $G$ and $\phi:V\rightarrow W$ is a $G$-module homomorphism, then
  1. Either $\phi$ is an isomorphism, or $\phi = 0$.
  2. [...]

Proof.  The first proof follows from the fact that the kernel of $\phi$ and the image of $ \phi$ are invariant subspaces. [...]

I understand that a $G$-homomorphism is a synonym for a map between two representations.  I figured out why the kernel and the image are invariant (under any $g\in G$), by I have no idea why that fact leads to the conclusion.

I would be most grateful if you could fill in the gap.


问:
About Newton-Cotes numerical integration for degree >8.
I'm studying Newton-Cotes integration for nine points of interpolation, let's say

$$(a_i)_{0 \leq i \leq 8}=\{0,\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}, 1 \}$$

The approximation of $\int_{0}^1 f(t) dt$ is

$$\sum_{i=0}^8 f(a_i) \int_0^1 L_i(t) dt= \sum_{i= 0}^8 f(a_i) w_i$$

where $L_i$ is the lagrange elementary polynôme :

$$L_i(t) = \prod_{j=0, j \neq i}^8 \frac{t-a_j}{a_i-a_j}.$$

Now I am asked to prove that some $w_i=\int_0^1 L_i(t) dt$ are negatives. If we do the whole calculation it is true indeed and can be checked online. However I'm looking for a proof which don't calculate explicitly the weights $w_i$. This proof should be accessible to bachelor student but I can't see it right now...

Any help is welcomed !

答:
There is a beautiful and elegant proof due to Bernstein. Start with
$$P_5^{\prime}(x)=\frac1{2^55!}\frac{d^6}{dx^6}(x^2-1)^5=\frac18(21x^4-14x^2+1)$$
Then for any polynomial $Q_3(x)$ of degree at most $3$,
$$\int_{-1}^1(1-x^2)P_5^{\prime}(x)Q_3(x)dx=0$$
Proof: just integrate by parts $6$ times. It follows that
$$\int_{-1}^1(1-x^2)P_5^{\prime}(x)\frac{P_5^{\prime}(x)}{x-x_4}dx=0$$
Where
$$x_4=\sqrt{\frac{7+2\sqrt7}{21}}$$
is the largest zero of $P_5^{\prime}(x)$. Since the $9$-point Newton-Cotes formula is exact for polynomials of degree at most $9$ it must also give this zero result. However, the integrand is only positive for $x_4\lt x\lt1$ so unless it uses a sample point in this interval or all sample points lie on zeros of the integrand, one of the weights must be negative or the integral would be negative.


问:
Is it possible to define a bijection from nonnegative to positive numbers?
Let $\mathbb{R}_{\geq 0}$ be the set of nonnegative numbers and $\mathbb{R}_{>0}$ the set of positive numbers, that is

$$
\mathbb{R}_{\geq 0} = \{\,x \geq 0 \mid x \in \mathbb{R} \,\}
$$

and

$$
\mathbb{R}_{> 0} = \{\,x > 0 \mid x \in \mathbb{R} \,\}
$$

Is it possible to define a bijection $f$ between these two sets?

答:
Of course. Let f be from the set with 0 to the set without 0:

f(x) = x when x is not integer;

f(0)= 1

f(1) = 2

f(2) = 3

etc.


问:
Can $(\Bbb Z[x], +)$ be isomorphic to $(\Bbb Q, +)$ as groups?

Can $(\Bbb Z[x], +)$ be isomorphic to $(\Bbb Q, +)$ as groups?

My attempt $:$ We know that $(\Bbb Z[x], +)$ is isomorphic to $(\Bbb Q_{\gt 0}, \cdot)$ as groups. Now if there exists a group isomorphism between $(\Bbb Z[x], +)$ and  $(\Bbb Q, +)$ then we have a group isomorphism $\varphi : (\Bbb Q, +) \longrightarrow (\Bbb Q_{\gt 0}, \cdot).$ But then there exists $a \in \Bbb Q$ such that $\varphi (a) = 2 \implies \varphi \left (\frac {a} {2} + \frac {a} {2} \right ) = 2 \implies \varphi \left (\frac {a} {2} \right )^2 = 2 \implies \varphi \left (\frac {a} {2} \right ) = \pm \sqrt {2} \notin \Bbb Q_{\gt 0},\ $ a contradiction. Hence we conclude that there does not exist any group isomorphism (in fact onto group homomorphism) between $(\Bbb Z[x], +)$ and $(\Bbb Q_{\gt 0}, \cdot).$

Am I right? Can anybody please verify my argument above? Thanks for your time.

答:
That's correct, but uses much more than it's needed.

For every $a\in\mathbb{Q}$, there exists $b\in\mathbb{Q}$ such that $a=2b$.

On the other hand, there is no $p(x)\in\mathbb{Z}[x]$ such that $2p(x)=1$.

This is essentially the same argument, but doesn't use the knowledge about $\mathbb{Z}[x]$ being isomorphic to $\mathbb{Q}>0$ which is not relevant.

Any abelian group that's not divisible cannot be isomorphic to $\mathbb{Q}$.


问:
What does $x^\pi$ mean?
I was just wondering, what does $x^\pi$ or for that matter, $x$ raised to any irrational number mean? For example, I want to represent $x^2$ then that would mean $x * x$ or if I want to do $x^\frac{2}{3}$ then that would mean, I first square it then cube root it but what does that mean for irrantional numbers such as $\pi$?

Thanks!

答:
It's better to think of it as rather

$$x^{\pi} = e^{\pi \log{x}}$$

Title: What does $x^\pi$ mean?


问:
Bound for $\|U_1XU_1^\star- U_2XU_2^\star\|_\infty$ for unitary $U_i$
Denote the operator norm by $\|\cdot \|_\infty$. That is $\|X\|_\infty = \max\{\|Xv\|:\|v\| = 1\}$ and $\|\cdot\|$ is the Euclidean 2-norm for vectors.

Let $U_1, U_2$ be unitary matrices (i.e. $U_i^\star U_i = I$ where $\star$ denotes the transpose conjugate) such that

$$\|U_1 - U_2\|_\infty \leq \varepsilon$$

For positive semidefinite $X$, is there a bound on

$$\|U_1 XU_1^\star - U_2X U_2^\star\|_\infty$$

in terms of $\varepsilon$? If yes, how can one show this?

答:
Note that the bound will have to scale with $\|X\|$, because the map $X\mapsto U_1XU_1^*-U_2XU_2^*$ is linear.

Assuming you are using the (Euclidean) operator norm, we have
\begin{align*}
\|U_1XU_1^*-U_2XU_2^*\|&\leq\|U_1XU_1^*-U_1XU_2^*\|+\|U_1XU_2^*-U_2XU_2^*\|\\
&\leq\|U_1X\|\|U_1^*-U_2^*\|+\|U_1-U_2\|\|XU_2^*\|\\
&\leq 2\|X\|\varepsilon,
\end{align*}
Since $\|UX\|=\|X\|$ and $\|X^*\|=\|X\|$ for any unitary $U$ and any matrix $X$.


问:
finding value of X that equates two proportions of true inequalities
Given two sets of numbers of different lengths, how would one find the value of X such that the proportion of numbers in the first set greater than X equates the proportion of numbers in the second set smaller than X?</p>

答:
Order the two sets separately in ascending numbers. Say the first ordered set is now {$a_i$}, containing $A$ numbers, and the second one {$b_i$}, containing $B$ numbers. Then, if I understand the question correctly, you are asking for two proportions, given some number $X$, with the following condition:
$$
\qquad \frac{1}{A} \sum_{a_i > X} 1 -  \frac{1}{B} \sum_{X > b_i} 1  = 0
$$

First observation: you won't always reach equality, since the first and the second term are two fractions with different denominator, which cannot necessarily be made equal.

So a better criterion is given below.

A way to proceed would be the following:

Unite the sets {$a_i$} and  {$b_i$} into a new set and order this set in ascending order, giving {$c_i$}.
Define a set {$x_i$} with $x_1 = c_1 - \epsilon$, for $1 < i \le A+B$: $x_i = \frac12 (c_i + c_{i-1})$, and lastly,  $x_{A+B+1} = c_{A+B} + \epsilon$, where $\epsilon$ is a small positive number. This set will have $A+B+1$ elements.

Then do the following:

initialize $k = 1; X = x_k$.

while $\frac{1}{A} \sum_{a_i > X} 1 -  \frac{1}{B} \sum_{X > b_i} 1  > 0 $

$\qquad k = k+1; X = x_k$

end

Once this is processed, the last assignment of $X$ is the $X$ you are looking for.

The way this works is the following:

For $k=1$, you have   $\frac{1}{A} \sum_{a_i > X} 1 -  \frac{1}{B} \sum_{X > b_i} 1  =1$.

For every transition from $x_k$ to $x_{k+1}$, either one of the two sums will change: either the first sum ($a_i$) looses one element, or the second  ($b_i$) gains one element. So the stop criterion is not exactly matched when the difference equals zero, but when it changes sign. This works for all sizes $A$ and $B$.

The stop criterion will always be reached, since for $k=A+B+1$, you have   $\frac{1}{A} \sum_{a_i > X} 1 -  \frac{1}{B} \sum_{X > b_i} 1  =-1$.


答:
Your proof is correct.

This is how you can make your argument more rigorous.
For any integer $j \geq 0$ let $f(j) = \frac{1}{ (2j+1)! (j+\frac{3}{2})}.$

You have shown
$\int_{0}^1 \frac{\sin t}{\sqrt{t}} = f(0) - (f(1) - f(2)) - (f(3) - f(4)) - \dots$

So it is sufficient to show $f(j) > f(j+1)$ for $j \geq 1$, i.e., $ \frac{f(j)}{f(j+1)} > 1$ for $j \geq 1.$

But $\frac{f(j)}{f(j+1)}  = \frac{(2j+3)! (j+\frac{5}{2})}{(2j+1)! (j+\frac{3}{2})} = \frac{(2j+2)(2j+3)(2j+5)}{2j+3} = (2j+2)(2j+5) > 1$ for $j \geq 1.$

Title: Bounding integrals using Maclaurin series

Body:
Prove that $$\int_0^1 \frac{ \sin x}{\sqrt{x} } < \frac23$$

I didn't see any easy way to integrate directly, so I just used the series approximation:

$$ \sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$

Hence,

$$\int_0^1 \frac{ \sin x}{\sqrt{x} }= \sum_{k=0}^{\infty}  \frac{  (-1)^k}{(2k+1)! (2k+ \frac32)}= \frac23- \frac{1}{3! (10.5)} + \text{stuff}$$

The argument I had is that, the increments and decrements added by taking more and more terms keeps shrinking as higher terms are smaller in magnitude for the above series. Hence, the value of integral is no greater than the very first increment which is $2$.

The problem is I feel this argument is not precise/ rigorous enough. Is there any theorems /results I can refer to when stating this?


问:
On the definition of divisors in Riemann Surfaces
The sum notation for a Divisor $D$ in a Riemann Surface $X$ (as in Miranda's "Algebraic Curves and Riemann Surfaces") is
$$
D=\sum_{p\in X} D(p)\cdot p
$$
That is, $D$ assumes the value $D(p)$ at $p$. For example, a principal divisor of $f$ is the divisor
$$
div(f)=D=\sum_{p\in X} ord_p(f)\cdot p
$$

which, if I understood it correctly, means that this divisior is the function
$$
\begin{array}{rcc}
div(f):&X&\rightarrow&\mathbb{Z}\\
&p&\mapsto &ord_p(f)
\end{array}
$$

Now, defining the divisor of zeroes and the divisor of poles as
$$
div_0(f)=\sum_{p\text{ with }ord_p(f)>0} ord_p(f)\cdot p
$$
and
$$
div_{\infty}(f)=\sum_{p\text{ with }ord_p(f)<0} (-ord_p(f))\cdot p
$$

respectively, how are those functions defined on the points $x\in X$ such that $ord_x(f)\leq 0$ (in the first case) or $ord_x(f)\geq 0$ (on the second)? The question arises since the divisor must be defined in the whole Riemann Surface $X$.

Thanks in advance.

答:
For points $p$ that don't appear in the summation defining the divisor of zeros (resp. the divisor of poles) the value of the divisor is zero. This ensures that $\mathrm{div}(f)=\mathrm{div}_0(f)-\mathrm{div}_\infty(f)$.

In general, when working with a free abelian group on some set $X$ (the Riemann surface in your example), if one writes down an expression which sums over a subset of $X$, it is to be understood that the coefficients corresponding to the other elements of $X$ are all zero.


问:
Dirichlet's test for uniform convergence of improper integrals
Dirichlet's test states that $F(x)=\int_a^\infty f(x,y)g(x,y)dy$ converges uniformly if primitive function of $f(x,y)$ is uniformly bounded for every value of parameter $x$, and $g(x,y)$ is monotonic and for $y\to\infty$ converges uniformly to $0$. My question is: Does $g(x,y)$ have to be monotonic for all $y\in[a,\infty)$ or is it enough if $g(x,y)$ is monotonic on an interval $[b,\infty)$, where $b>a$?
For example, we know that $g(x,y)=\frac{y}{1+y^4}$ converges for $y\to\infty$ uniformly to $0$, but it is not monotonic on $[0,\infty)$. Can be Dirichlet's test used in this case when the lower bound of integration is $0$?

答:
Suppose $f\cdot g$ is Riemann integrable on finite intervals $[a,c]$. There is a Cauchy criterion for uniform convergence of the improper integral over $[a,\infty)$ whereby for every $\epsilon > 0$ there exists $C(\epsilon) > 0$, independent of $x \in D$, such that for all $c_2 > c_1 > C(\epsilon)$ and for all $x\in D$ we have

$$\tag{1}\left|\int_{c_1}^{c_2} f(x,y) g(x,y) \, dy \right| < \epsilon$$

As this  criterion is used to prove Dirichlet's test for uniform convergence, it is enough that $g(x,y)$ is monotone with respect to $y$ eventually.

To see this, assume that $g(x,y)$  is monotone decreasing with respect to $y$ for all $x \in D$ and only for $y \in [b, \infty)$.  With $c_2 > c_1 > b$ and $x$ fixed,  by the second mean value theorem for integrals there exists $\xi(x) \in (c_1,c_2)$ such that

$$\tag{1}\int_{c_1}^{c_2} f(x,y) g(x,y) \, dy = g(x,c_1) \int_{c_1}^{\xi(x)}f(x,y) \, dy$$

By the uniform boundedness condition for $f$, there exists $B > 0$ (independent of $c_1$) such that for all $x \in D$,

$$\tag{3}\left|\int_{c_1}^{\xi(x)}f(x,y) \, dy\right|< B$$

Since $g(x,y) \underset{y \to \infty}\longrightarrow  0$ uniformly for $x \in D$ there exists $C(\epsilon) > b $ such that  for all $c_1> C(\epsilon)$,

$$\tag{4}|g(x,c_1)| < \epsilon/B$$

Altogether, (2), (3), and (4) imply that (1) holds and the improper integral converges uniformly.


问:
Distribution of Y, which is Poisson with mean 1 minus 1. How do we derive the Expectation and Variance?
My question is about the following statement:

Consider a random walk whose step size has distribution Y, which is Poisson with mean 1 minus 1. Asymptotically the critical items are that Y has zero expectation and unit variance.

I am a little confused as to why the expectation and variance differ for this Poisson distribution. I assume this has to do with the statement " Asymptotically the critical items are ". Can someone point me in the right direction how to derive this?

答:
We are subtracting $1$ from the Poisson distribution. We have a random variable $\mathbf X \sim \text{Poisson}(1)$, with mean $1$ and variance $1$; then, we define $\mathbf Y = \mathbf X - 1$, which has mean $0$ and variance $1$.

The next statement is just saying that to understand the asymptotic behavior of the random walk, the mean and variance of $\mathbf Y$ are all that we need to know.


问:
How to find image, pre-image, bijection, onto, one-to-one?
The question I am trying to solve is as follows:

Let A = {1, 2, 3, 4, 5} and define the function f : (A) → Z as follows:
For each set X in (A), where (A) is the power set of A, f(X) is the
number of elements in X.

a. What is the image of {1, 2, 3} under f? Justify.

b. What is the pre-image of 4? Justify.

c. What is the pre-image of 10? Justify.

d. Is f one-to-one? Justify.

e. Is f onto? Justify

My professor is not teaching us anything at all and my understanding of what onto, one-to-one, image, and pre-image is extremely weak because of this. Could someone please explain what these are in concrete, simple terms so that I may understand, and also push me in the right direction as far as solving a problem like this?

Thank you!

答:
The function is defined on the power set of A, that is the set (or collection) of all subsets of A. The function $f$ just assigns to each subset of A the number of elements in the subset. The set {1,2,3} has 3 elements so f maps it to?
To see what the premiage of 4 is, all you need to know is how many subsets of A are mapped to 4 under f. This translates to how many subsets of A have 4 elements in them.
How many subsets of A have 10 elements?
Can two different subsets of A have the same number of elements.
For each integer $n$, is there a subset of A which has$n$ elements? Question 3 might be useful here.


问:
Prove inequality $e-(1+1/t)^t < \frac{e}{2t+1}$
I am trying to prove that
$$e-(1+1/t)^t < \frac{e}{2t+1}$$

I got a lower bound
$$\frac{e}{2t+2} < e-(1+1/t)^t$$ using Hermite-Hadamard inequality.

Any hint is appreciated!

答:
Rewrite the inequality $e-(1+\frac1t)^t < \frac{e}{2t+1}$ as
$(1+\frac1t)^t > \frac{e}{1+\frac1{2t}} $,
which is equivalent to
$$(1+x)^{\frac1x } (1+\frac x2 )> e\tag1$$
with $x =\frac 1t$ over the domain $x>-1,\> x\ne 0$. Observe
$$\frac1x \ln(1+x)
+\ln(1+\frac x2)-1= \int_0^x \frac{g(t)}{t^2}dt \tag2$$
where
$$g(t) = \frac{t(t^2+2t+2)}{(t+1)(t+2)} -\ln(1+t)$$
Note that $g(0)=0$ and
$$g’(t) = \left(\frac t{t+2} \right)^2 \left(\frac1{t+1} +t+4\right)> 0$$
for $t>-1$, which means $ \int_0^x \frac{g(t)}{t^2}dt >0$ and, according to (2)
$$\frac1x \ln(1+x)
+\ln(1+\frac x2)>1$$
Apply the inequality as the power of $e$ to obtain (1).


问:
Why do we need to extend the definition of trigonometric ratios to all angles?
Usually (in India, at least), in 10th grade, trigonometric functions are introduced as ratios of sides of a right triangle. For example, if we have a right triangle $\triangle ABC$ such that $\angle ABC = 90^\circ = \dfrac\pi2^c$ and $\angle C=\theta$. Then, $\sin\theta = \dfrac{\mathrm{perpendicular}}{\mathrm{hypotenuse}} = \dfrac{AB}{AC}$, $\cos\theta = \dfrac{\mathrm{base}}{\mathrm{hypotenuse}} = \dfrac{BC}{AC}$, etc.

We notice that here, if $f$ is a trigonometric function, then $f(\alpha)$, for some angle $\alpha$ is defined only for $\alpha\in(0^\circ,90^\circ)$.

Later on, in 11th grade (again, at least in India), the unit circle definition of trigonometric functions is taught where $\sin\beta$, for example is defined $\forall~\beta\in(-\infty,\infty)$ [all angles, basically].

I want to know why we need to extend the definition of trigonometric functions to all angles. Why can't they just be defined for acute angles? I asked this from a friend of mine and he said that this definition is used multiple times in physics. But as far as I know (and most probably), the unit circle definition was given at a time when physics was not so advanced. So, why exactly did the mathematician(s) who extended the definition do so? Was this extended definition created to solve some problem that could not be solved using the right triangle definition?

Thanks!

答:
Quoting Wikipedia:

Madhava (c. 1400) made early strides in the analysis of trigonometric functions and their infinite series expansions. He developed the concepts of the power series and Taylor series, and produced the power series expansions of sine, cosine, tangent, and arctangent.[24][25] Using the Taylor series approximations of sine and cosine, he produced a sine table to 12 decimal places of accuracy and a cosine table to 9 decimal places of accuracy. He also gave the power series of π and the angle, radius, diameter, and circumference of a circle in terms of trigonometric functions. His works were expanded by his followers at the Kerala School up to the 16th century.

From the power series expansion, it is natural to consider trigonometry functions
as periodic functions defined on all of $\mathbb R$. I would think that this analytic point of view (as opposed to the geometric point of view where it might make more sense to limit to $0 \leq x \leq \frac{\pi}{2}$) is what you're asking about. It seems the motivation for the Taylor series expansion was a way to compute numerically trigonometric values.

Edit:

Power series expansions are particular formulas for expressing some special functions such as $\sin$, $\cos$, but also $\exp$ or $\ln$. They are useful for computing approximate values for instance, or to do other computations such as differentiating or integrating. At the beginning of calculus (with Newton and Leibniz in Europe), analysis was at first mostly concerned with such functions.


答:
If $0< x$ we have:
$$x^\pi=\sup\{x^r\mid r<\pi, r\text { is a rational number}  \}$$</p>

Title: What does $x^\pi$ mean?

Body:
I was just wondering, what does $x^\pi$ or for that matter, $x$ raised to any irrational number mean? For example, I want to represent $x^2$ then that would mean $x * x$ or if I want to do $x^\frac{2}{3}$ then that would mean, I first square it then cube root it but what does that mean for irrantional numbers such as $\pi$?

Thanks!


问:
In how many ways can 5 objects be placed in 3 groups?
Consider this question:

In how many ways can $5$ tourists be placed in $3$ different hotels, such that there is at least one tourist in each hotel?

To start working on this, I removed the lower bound on the number of tourists in a hotel. I reasoned that any tourist can be placed in any of the $3$ hotels, resulting in $5\times 3=15$ permutations.

However, the correct answer with the lower bound is $150$, which is more than $15$.

Obviously, I am undercounting by a large amount. How?

答:
This is not correct. Without the lower bound you can assign one hotel to one by one the tourists. So, if you line up the tourists in a line and record the hotel you will get: $3$ possibilities for each one, giving you $3\times 3\times \cdots \times 3=3^5.$

Notice that this is the same as the number of functions from $[5]$ to $[3]$ (you are making a functional choice, i.e., you dont want to send one tourist to two different hotels).

Now, with the lower bound, you have to take away the times that you missed one of the hotels. The problem becomes in that you will be overcounting so you have to include and exclude. You will get $3^5-3\cdot 2^5+3,$ where the $2^5$ is choosing out of now $2$ hotels(you are missing one of the $3$ hotels). The last $3$ is because when you take out the hotels you are overcounting the choice in which you send all the tourist to the same hotel!

In fancy terms this is the number of surjective functions from $[5]$ to $[3]$ that can be established using the Stirling numbers of second kind as ${5\brace 3}\cdot 3!$


问:
Probability of random line through a point intersecting a circle

Consider a circle of radius $r$ with center $C$ and let $p$ be a point outside the circle ( that is $\operatorname{dist}\left(p,C\right) > r$ ).

What is the probability that a random line through $p$ will intersect the circle $?$.

My intuition tells me it will be something like $r/\operatorname{dist}\left(p,C\right)$ but I would like a formal proof.

答:
Outline: If all lines are equally likely, they the given problem is equivalent to asking what proportion of lines that contain p also intersect with C. Since only one side of the line can intersect with p, it is natural to introduce an axillary problem which is identical, but lines are replaced with rays. The axillary problem is very easy to solve if you already have the formula for angular diameter. The solution for the axillary problem can then be used to solve the given problem because for every ray that intersects with C there is a ray that points in the exact opposite direction and these two rays are associated with the same line.

Axillary Problem: What is the probability P(i) that a ray originating from point p will intersect with circle C, assuming that p is not circumscribed by C.

1: P(i) = (number of rays originating from p that intersect with C) / (number of rays originating from p)

[This comes from the mathematical definition of probability.]

2: P(i) = angular diameter of C from p

[This comes from the definition of angular diameter, although I can not find a source that defines angular diameter in this way.]

3: P(i) = 2 * arctan(d/2D), where d = the diameter of C and D is the distance from p to the center of C.

[Source that this is the formula for angular diameter in radians.]

Given Problem: What is the probability P(I) that a line that passes through point p will intersect with circle C, assuming that p is not circumscribed by C.

Assumption 1: For any line that contains p, there are exactly 2 rays that originate from p and are collinear with that line.

[I am not going to prove this because I don't think anyone here is going to question it. I know that is a bad practice, but I am trying to keep this as concise as possible.]

1: P(I) = (number of lines containing p that intersect with C) / (number of lines containing p)

[This comes from the mathematical definition of probability.]

2: P(I) = (2 * number of rays originating from p that intersect with C) / (2 * number of rays originating from p)

[This follows from assumption 1.]

3: P(I) = (number of rays originating from p that intersect with C) / (number of rays originating from p)

[The 2 in the numerator cancels out with the 2 in the denominator.]

4: P(I) = 2 * arctan(d/2D), where d = the diameter of C and D is the distance from p to the center of C.

[We proved this with the axillary problem.]


问:
Essential self-adjointness of restriction of symmetric operator
Suppose $T$ is a (unbounded) symmetric operator on Hilbert space $\mathscr{H}$, defined on $D$. Let $D_1 \subseteq D$ is a linear subset dense in $D$, $T|_{D_1}$ be the restriction of $T$ on $D_1$. Show that if $T|_{D_1}$ is essentially self-adjoint, then $T$ is essentially self-adjoint, and $\overline{T} = \overline{T|_{D_1}}$.
I have the answer following.

It suffices to show that $\overline{\Gamma (T)} \subseteq \overline{\Gamma (T|_{D_1})}$, then I choose a sequence in $D_1$ converge to x $\in D$, using diagnal technique.Simply let $x_n(\in D) \rightarrow x$, while $T(x_n) \rightarrow y$ and let $x_k^{(n)}(\in D_1)\rightarrow x_n$ then $x^{(n)}_n \rightarrow x$, but how can I show that $T(x^{(n)}_n) \rightarrow y$?
Thanks a lot in advance!

答:
Let $(x, Tx)$ lie in $\Gamma(T)$. Then for all $y\in D_1$ you have:

$$|\langle x, T\lvert_{D_1}y\rangle| = |\langle Tx,y\rangle| ≤\|Tx\|\,\|y\|$$

and $x$ lies in the domain of $(T\lvert_{D_1})^*$ and that $(T\lvert_{D_1})^*x = Tx$.

Since $T\lvert_{D_1}$ is essentially self-adjoint you have that $(T\lvert_{D_1})^*=\overline{T\lvert_{D_1}}$, so we have just seen $(x, Tx)\in\overline{\Gamma(T\lvert_{D_1})}$, ie
$$\Gamma(T)\subseteq \overline{\Gamma(T\lvert_{D_1})}$$


问:
Using implication to argue for all members of a set, something is true
I am trying to express some statements in a formal language that allows universal quantification over domains, but not over sets. I was wondering how to express that something is true for all elements of a given set.

Let us assume $D$ is a domain and we can write $\forall x:D, A(x)$ to mean that "Forall $x$ in the domain $D$, $A(x)$ is true".

Now, let us assume we have a non-empty set $S$ that contains elements of the domain $D$ and
the $\in$ relation is a binary predicate over the domain $D_S$ of all such sets such that $\in : D_S \times D \rightarrow \{true,false\}$.

My goal is to be able to express the statement "For every member of the set $S$, $A(x)$ is true".

However, the language I am using does not allow me to write something like $\forall x \in S: A(x)$.

In this case, would it be correct to write $\forall x:D, ((x \in S) \implies A(x))$?

I realize that if $S$ was allowed to be empty, the statement would be vacuously true, but that is not a concern for the universe I am trying to reason about. My main concern is if $\forall x:D, ((x \in S) \implies A(x))$ means "For every member of the set $S$, $A(x)$ is true".

答:
If $\in$ is part of the formal language, then yes, you can write it this way.

Technically, you would then have to explicitly specify the interpretation for this relation symbol as well as write "$\in(x, S)$" rather than "$x \in S$". However, in practice people will understand (perhaps even better) what you mean if you just write $x \in S$ without further comment.

That being said, unless you are doing set theory itself, there is no real need for the additional complication of having a $\in$ relation symbol: You could instead consider defining the set as a predicate, given that the interpretation of a one-place predicate is a set. That is, instead of writing $x \in S$ where $S$ denotes, say, the set of even numbers, you could also just define $S$ as a one-place predicate $S(x)$ with the interpretation $\mathcal{I}(S) = \{x: x \text{ is even}\}$. That way you can convey the same meaning using just the default inventory of FOL syntax.


问:
$H$ and $K$ are two normal subgroups of group $G$, $H\cap K=\{1\}$, $G=HK$, how to prove $G$ is isomorphic to $H\times K$
If group $G$ has two normal subgroups $H$ and $K$ satisfying $H\cap K=\{1\}$ and $G=HK$, how to prove that $G$ is isomorphic to $H\times K$?

Since the factoring of any $g\in G$ as product $hk$ is unique under given conditions, I tried to prove that the map $\phi:G\ni g\mapsto (h,k)\in H\times K$ is the isomorphism. I can show that $\phi$ is a bijection. Next, I need to show it preserves group operation. Assuming $\forall g_1,g_2\in G$, $g_1=h_1k_1$ and $g_2=h_2k_2$ (the factoring is unique), we have $g_1g_2=h_1k_1h_2k_2$. But $\phi(g_1)=(h_1,k_1),\phi(g_2)=(h_2,k_2)$, so in product group $H\times K$, the operation $\phi(g_1)\phi(g_2)=(h_1h_2,k_1k_2)$. Its pre-image of $\phi$ in $G$ should be the  product $h_1h_2k_1k_2$. So we must establish $h_1k_1h_2k_2=h_1h_2k_1k_2$, which is equivalent to $k_1h_2=h_2k_1$. But I cannot prove this commutative relation based on the given conditions. The normality of $H$ implies $k_1^{-1}h_2k_1\in H$, but it does not necessarily equal $h_2$. This is where I got stuck. Is the above commutative relation true and how do I prove it? Or, was I wrong at the very beginning about what the isomorphism is? Could you please give me some help? Thank you.

答:
This doesn't add much to your excellent effort, other than some details.

For any pair of subgroups $H, K\le G$, the set $H\times K$ can be partitioned into equivalence classes all equicardinal to $H\cap K$, and such that $(H\times K)/\sim$ is equicardinal to $HK$. In fact:

Let's define in $H\times K$ the equivalence relation: $(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$. The equivalence class of $(h,k)$ is given by:

$$[(h,k)]_\sim=\{(h',k')\in H\times K\mid h'k'=hk\} \tag 1$$

Now define the following map from any equivalence class:

\begin{alignat*}{1}
  f_{(h,k)}:[(h,k)]_\sim &\longrightarrow&  H\cap K \\
  (h',k')&\longmapsto& f_{(h,k)}((h',k')):=k'k^{-1} \\
\tag 2
 \end{alignat*}

Note that $k'k^{-1}\in K$ by closure of $K$, and $k'k^{-1}\in H$ because $k'k^{-1}=h'^{-1}h$ (being $(h',k')\in [(h,k)]_\sim$) and by closure of $H$. Therefore, indeed $k'k^{-1}\in H\cap K$.

Lemma 1. $f_{(h,k)}$ is bijective.

Proof.

\begin{alignat}{2}
f_{(h,k)}((h',k'))=f_{(h,k)}((h'',k'')) &\space\space\space\Longrightarrow &&k'k^{-1}=k''k^{-1} \\
&\space\space\space\Longrightarrow &&k'=k'' \\
&\stackrel{h'k'=h''k''}{\Longrightarrow} &&h'=h'' \\
&\space\space\space\Longrightarrow &&(h',k')=(h'',k'') \\
\end{alignat}

and the map is injective. Then, for every $a\in H\cap K$, we  get $ak\in K$ and $a=f_{(h,k)}((h',ak))$, and the map is surjective. $\space\space\Box$

Now define the following map from the quotient set:

\begin{alignat}{1}
  f:(H\times K)/\sim &\longrightarrow&  HK \\
  [(h,k)]_\sim &\longmapsto& f([(h,k)]_\sim):=hk \\
\tag 3
 \end{alignat}

Lemma 2. $f$ is well-defined and bijective.

Proof.

Good definition: $(h',k')\in [(h,k)]_\sim \Rightarrow f([(h',k')]_\sim)=h'k'=hk=f([(h,k)]_\sim)$;

Injectivity: $f([(h',k')]_\sim)=f([(h,k)]_\sim) \Rightarrow h'k'=hk \Rightarrow (h',k')\in [(h,k)]_\sim \Rightarrow [(h',k')]_\sim=[(h,k)]_\sim$;

Surjectivity: for every $ab\in HK$ , we get $ab=f([(a,b)]_\sim)$. $\space\space\Box$

As a corollary, if $|H\cap K|=1$, then all the classes $[(h,k)]_\sim$ are singletons, and hence the map $\tilde f\colon H\times K\to (H\times K)/\sim$, defined by $(h,k)\mapsto [(h,k)]_\sim$, is bijective; this, in turn, implies that the composite map $f\circ\tilde f\colon H\times K\to HK$ is bijective, too. If, in addition, $H\unlhd G$ and $K\unlhd G$, then for every $h\in H, k\in K$ we have that $hkh^{-1}k^{-1}\in H\cap K$, whence $hkh^{-1}k^{-1}=e$ and finally $hk=kh$. This, and the assumption $HK=G$, make of $f\circ\tilde f$ a (bijective) group homomorphism (namely an isomorphism). In fact:

\begin{alignat}{1}
(f\circ\tilde f)((h,k)(h',k')) &= (f\circ\tilde f)(hh',kk') \\
&= f(\tilde f(hh',kk')) \\
&= f([hh',kk']_\sim) \\
&= hh'kk' \\
&= hkh'k' \\
&= (hk)(h'k') \\
&= f([h,k]_\sim)f([h',k']_\sim) \\
&= f(\tilde f(h,k))f(\tilde f(h',k')) \\
&= ((f\circ\tilde f)(h,k))((f\circ\tilde f)(h',k')) \\
\end{alignat}


问:
What is the answer of $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$
I found a limit question in my textbook but i could not obtained the result given by answer key.

Let me introduce the question :  $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$

I tried to separate it into two part such that $ \lim_{n\to \infty}{(-1)}^{(n-1)}\cdot\lim_{n\to \infty}\frac{n}{(n+1)!}$

As you see, this partition gave me "$\text{oscillating (undetermined)}$".$0$

However, the answer is zero. What am I missing? Thank you for your helps ...

答:
Try squeeze.

$0 = \lim_{n \to \infty}\frac{-n}{(n+1)!} \leq \lim_{n \to \infty}\frac{(-1)^{n-1}n}{(n+1)!} \leq \lim_{n \to \infty}\frac{n}{(n+1)!} = 0$


答:
From a very abstract viewpoint, the position $$x^\pi = e^{\pi \log x}$$ is a circular definition: after all $e$ is as much a real number as $x$. There are two ways to break this cirle:

$e^{\pi \log x}$ means $\exp (\pi \log x)$, provided you have defined the exponential function in an indipendent way;

we study the theory real numbers, which includes the construction of $b^x$ for $b>0$ and $x \in \mathbb{R}$. This is not trivial at all, as can be seen in W. Rudin's Principle of mathematical analysis.

Title: What does $x^\pi$ mean?


问:
A $\triangle ADE$ has $\angle AED = 120^\circ$, and $B$ and $C$ are points on $AD$ such that $\Delta BCE$ is equilateral.

A $\triangle ADE$ has $\angle AED = 120^\circ$, and $B$ and $C$ are points on $AD$ such that $\triangle BCE$ is equilateral. If $AB = 4$ cm, $CD = 16$ cm and $BC = x$ cm, find $x$ .

What I Tried: Here is a picture :-

No idea for this problem. I could proceed by angle-chasing, considering $\angle BEA = \theta$ , but I don't think that will help me in finding $BC$ without Trigonometry. I think I need to use Law of Cosines on $\angle AEC$ to solve this, but I am new to it and I prefer simple geometry techniques. There are nothing I can use such as similarity, Pythagoras theorem or areas, etc.

Can anyone help me? Thank You.

答:
Notice $\angle ECD = 120 = \angle ABE$. In $\triangle AED$, one angle is $120$ and it shares one angle each with smaller unshaded triangles.

We conclude

$$ \triangle AED \sim \triangle ABE \sim \triangle ECD$$

by AA similarity.

Thus $\triangle ABE \sim \triangle ECD$ and $$\dfrac{AB}{BE}=\dfrac{EC}{CD}$$

 $$\Rightarrow x = 8 $$


问:
Most ambiguous and inconsistent phrases and notations in maths
What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $\frac{1}{f(x)}$.

$f^2(x)$ can be $(f\circ f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f\:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $\left(f^{(2)}\circ f^{(-3)}\right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

答: It is true that 5 is a divisor of 0, and yet it is true that there are no divisors of 0.</p>


问:
Evaluating $\int_a^b \arccos\left(x\,/\sqrt{(a+b)x-ab\,}\,\right)\,\mathrm {d}x$ assuming $0<a<b$
I have been trying to solve the following integral for a little while now

$$I = \int_a^b \arccos\left( \frac{x}{\sqrt{(a+b)x - ab \,}\,}\right)\mathrm{d}x \quad 0<a<b$$

After doing some testing in maple, I was able to discover that $\displaystyle I=\frac{\pi}{4}\frac{(a-b)^2}{b+a} $

But I want to show this algebraically. (using mathematics)

I have done some effort and one can notice that the denominator equals $x^2-(x-a)(x-b)$. Another thing i tried using was the fact that

$$\arccos(x) = \arcsin\left(\sqrt{1-x^2}\,\right)$$

also if on tries solving it by parts

$$ \left[ \arccos\left( \frac{x}{\sqrt{(a+b)x - ab \,}\,}\right) \right]_a^b = 0 $$

but the remaining integral looks impossible. If one uses the substitution, $u=\text{denominator}$ I end up with

$$ \frac{2}{a+b}\int u^2 \arccos\left( \frac{u^2 + ab}{u(b+a)} \right) \, \mathrm{d}u$$

Also $$I = \int_a^b \text{arccsc}\left( \sqrt{\frac{(a-x)(b-x)}{x^2}}\right)\mathrm{d}x$$

not sure what this buys me though. I also tried a variety of clever things such at differentiating under the integral sign, but alas nothing worked. Anyone mind helping me ? =)

答:
I don’t understand why a should be strictly positive.

Indeed a = 0 works giving I(0,a) = Pi/4 . a

Also, considering that I(a,b) = I(0,b) - I(0,a) and applying my above result we have : I(a,b) = Pi/4 (b-a)

I am probably missing something or wrong somewhere because my result is different from the one provided by AMM and others.

fjaclot


Title: What is the answer of $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$

Body:
I found a limit question in my textbook but i could not obtained the result given by answer key.

Let me introduce the question :  $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$

I tried to separate it into two part such that $ \lim_{n\to \infty}{(-1)}^{(n-1)}\cdot\lim_{n\to \infty}\frac{n}{(n+1)!}$

As you see, this partition gave me "$\text{oscillating (undetermined)}$".$0$

However, the answer is zero. What am I missing? Thank you for your helps ...

Body: $\lim_{n\to\infty}{n\over(n+1)!}=\lim_{n\to\infty} \frac{1}{(n+1)(n-1)!}=0\Longrightarrow\lim_{n\to\infty}{(-1)^{n-1}n\over(n+1)!}=0 $


问:
Prove that 4 is not a 3-adic cube
I'm trying to show that 4 is not a 3-adic cube.

On the Wikipedia page for Hensel's lemma (https://en.wikipedia.org/wiki/Hensel%27s_lemma) I read that: "4 is not a 3-adic cube since it is not a cube mod 9"

However, I'm not sure why a number not being a cube mod 9 implies it is not a 3-adic cube. Is there a nice proof of this fact?

答:
If $4 \in \Bbb Z_3$ was a cube $4=a^3$, then $|a|_3=1$, so $a \in \Bbb Z_3$ and we can reduce the equation $4=a^3$ mod $9$, using that $\Bbb Z_3/9\Bbb Z_3 \cong \Bbb Z/9\Bbb Z$.

To elaborate on the last point, note that any $x \in \Bbb Z_3$ can be written as $x=\sum_{n=0}^\infty a_n 3^n$ for $a_n \in \{0,1,2\}$, pulling out the first two summands gives $x=a_0+3a_1+9 \sum_{n=2}^\infty a_n 3^{n-2}$. If $a_0$ and $a_1$ vary over $\{0,1,2\}$, then $a_0+3a_1$ varies between $\{0, \dots, 8\}$. If we reduce mod $9$, we just get $x \equiv a_0+3a_1 \pmod{9}$. This shows that $\Bbb Z_3/9\Bbb Z_3 \cong \Bbb Z/9\Bbb Z$. The same reasoning can be used to show that $\Bbb Z_p/p^n\Bbb Z_p \cong \Bbb Z/p^n\Bbb Z$ for all primes $p$ and $n \in \Bbb N$.

Title: Prove that 4 is not a 3-adic cube


问:
Prove that $AM = MF$ where $M$ is the intersection between the perpendicular of circle $ω$ and line $AF$

Given is circle $ω$ and line $k$ disjoint with circle $ω$. Point $M$ is
the foot of the perpendicular from the center of the circle onto line $k$. A line
intersects line $k$ at point $A$ and circle $ω$ at points $B$ and $C$. Lines $\overline{BM}$ and
$\overline{CM}$ intersect circle $ω$ for the second time at points $D$ and $E$. Line $\overline{DE}$ meets
line $k$ at point $F$. Prove that $\overline{AM} = \overline{MF}$.

Once again, I've been assigned another set of geometry problems, and this is one of the harder ones that I haven't been able to figure out yet. I've asked my instructor what I could use for this problem, and he told me that Power of a Point would be useful here. But so far, I've tried applying Power of a Point, but I'm not getting anywhere with the few points that currently exist. Can anyone help out?

答:
COMMENT.-  The asked property is verified with any good function plotter taking arbitrary values and a suitable proof is geometric in nature. On the other hand, an analytical proof involves the consideration of seven unknowns (six by the coordinates of points C, D and E and one by the first coordinate of point F in the attached figure). We want to do it but this is a challenge that without trickery like the one we will use can be too annoying.

$\hspace{9 cm}$****************

Line $ABC\hspace{1cm}\dfrac{y-d}{x-a}=\dfrac{d-b_2}{a-b_1}\iff y=\dfrac{d-b_2}{a-b_1}x+\dfrac{ab_2-db_1}{a-b_1}=mx+h$ where $m=\dfrac{d-b_2}{a-b_1}$

Line $MBD\hspace{1cm}\dfrac{y-d}{x}=\dfrac{d-b_2}{-b_1}\iff y=\dfrac{b_2-d}{b_1}x+d=m_1x+d$ where $m_1=\dfrac{b_2-d}{b_1}$

Line $MEC\hspace{1cm}\dfrac{y-d}{x}=\dfrac{d-y_1}{-x_1}\iff y=\dfrac{y_1-d}{x_1}x+d=m_2x+d$ where $m_2=\dfrac{y_1-d}{x_1}$

►Point $C=(x_1,y_1)\in w$ in line $ABC$

$x_1^2+y_1^2=r^2\space\space$ then $\begin{cases}(m^2+1)x_1^2+2hmx_1+(h^2-r^2)=0\\(m^2+1)y_1^2-2hy_1+(h^2-m^2r^2)=0\end{cases}$

It follows (Vieta) $b_1x_1=\dfrac{h^2-r^2}{m^2+1}$ and $b_2y_1=\dfrac{h^2-m^2r^2}{m^2+1}$.

Similarly we have

►Point $D=(x_2,y_2)\in w$ in line $MBD$

$b_1x_2=\dfrac{d^2-r^2}{m_1^2+1}$ and $b_2y_2=\dfrac{d^2-m_1^2r^2}{m_1^2+1}$

►Point $E=(x_3,y_3)\in w$ in line $MEC$

$x_1x_3=\dfrac{d^2-r^2}{m_2^2+1}$ and $y_1y_3=\dfrac{d^2-m_2^2r^2}{m_2^2+1}$

$\hspace{9 cm}$****************

Now line $DEF$ is given by $\dfrac{y-y_2}{x-x_2}=\dfrac{y_3-y_2}{x_3-x_2}$ and because point $F$ belongs to this line one has
$\dfrac{d-y_2}{\alpha-x_2}=\dfrac{y_3-y_2}{x_3-x_2}$ so making $\alpha=-a\iff\overline{AM}=\overline{MF}$, after substitutions we obtain an equality $$f(a,d,b_1,b_2)=0$$ whose verification completes the solution. We leave this exercise for whoever wants to do it.


问:
Good source for Triebel-Lizorkin spaces?
I'm trying to look into different types of function spaces.  At the moment, at least for function spaces involving integration, I only have $L^p$ and $W^{k,p}$.  The next function spaces I thought I'd read about are Besov spaces and Triebel-Lizorkin spaces.  Admittedly I usually look at wikipedia and Wolfram.  However, wikipedia has next to nothing on Triebel-Lizorkin spaces, Wolfram has absolutely nothing on Triebel-Lizorkin and very little on Besov spaces.

Does anyone know of a good source for Triebel-Lizorkin spaces?  And maybe Besov spaces too.  Google just gives me these crazy articles which are way above my head.

答: There's the books by Triebel: Theory of Function Spaces, II, III.</p>


问:
Find an increasing function in $\mathbb{R}$ such that its set of discontinuity is $\mathbb{Q}$.
Question: Find an increasing function in $\mathbb{R}$ such that its set of discontinuity is $\mathbb{Q}$.

Attempt: Suppose $(q_n)_{n \in \mathbb{N}}$ be an enumeration of the rationals, and define
$$f(x) := \sum_{k=1}^\infty 2^{-k} \chi_{(q_k, \infty)}(x).$$

Then the function is increasing by construction.

Now, I am stuck at showing that $f$ is discontinuous at $x \in \mathbb{Q}$.

In particular, I am having trouble showing that if for some $n \in \mathbb{N}$ such that $$x=q_n < y < \max_{q_i > q_n,i =1 \ldots, n} q_i\, ,$$ then we have
$f(x) + 2^{-n} \leq f(y)$.

Any help would be appreciated.

答: Just look at the left-hand and right-hand limits. For any rational $x = q_k$, we have by construction $$\lim_{x \to q_k^-} f(x) = f(q_k)$$ but $$\lim_{x \to q_k^+} f(x) = f(q_k) + 2^{-k}.$$


答:
The rule that $\lim_n (a_n b_n) = \left( \lim_n a_n\right)\left(\lim_n b_n\right)$ is valid only when both of the latter limits exist, meaning both are actual finite numbers (not $\pm\infty$ and not something that oscillates). And $(-1)^n$ oscillates.

But the limit you started with exists.
$$
\frac{-n}{(n+1)!} \le \frac{(-1)^n \cdot n}{(n+1)!} \le \frac n {(n+1)!}
$$
Now observe that
$$
\frac n {(n+1)!} = \frac n {n+1} \cdot \frac 1 {n!}.
$$
And this one you can partition, finding limits of the two factors separately. The first one approaches $1$ and the second approaches $0.$ Thus the sequence whose limit you seek is squeezed between two sequences whose limit is $0.$


问:
What does $x^\pi$ mean?
I was just wondering, what does $x^\pi$ or for that matter, $x$ raised to any irrational number mean? For example, I want to represent $x^2$ then that would mean $x * x$ or if I want to do $x^\frac{2}{3}$ then that would mean, I first square it then cube root it but what does that mean for irrantional numbers such as $\pi$?

Thanks!

答:
Ultimately, it's nothing near so nice as the rational case. If we're given an irrational number $\alpha$, then $$x^\alpha:=\underset{r\to\alpha}{\lim_{r\in\Bbb Q}}x^r,$$ so long as this limit is defined (for example, it doesn't work when $x$ is negative, and it doesn't work when $x=0$ and $\alpha$ is negative).

Alternately, it is at times more beneficial to define it as $$x^\alpha:=e^{\alpha\ln x}.$$ These are equivalent definitions for $x>0$. We can even do a bit better and define $$x^\alpha:=\lim_{t\to x^+}e^{\alpha\ln t}.$$ That definition works precisely when the original definition does--namely, whenever $x>0$, and whenever $x=0$ and $\alpha>0$. Of course, both of the latter two definitions do require some independent definition of $e^w$, such as $$e^w:=\lim_{n\to\infty}\left(1+\frac wn\right)^n.$$

Title: What does $x^\pi$ mean?


问:
Is $\mathbb R$ completely normal space?

Is $\mathbb R$ completely normal ( $T_5$ ) space?

I've seen this Wiki Page about Urysohn's lemma, which states that a topological space is normal if and only if any two disjoint closed sets can be separated by a continuous function.

Two plain subsets $A$ and $B$ are said to be separated by a function if there exists a continuous function $f : X \to[0,1]$ such that $f(A) = 0$ and $f(B) = 1$. Any such function is called a Urysohn function for $A$ and $B$.

So, now I need to construct continuous function $f : \mathbb R \to [0, 1]$ such that for all closed $A, B \subseteq \mathbb R: A \cap B = \varnothing,$ we have $f(A)=0$ and $f(B)=1$.

However, I wonder if one could construct such continuous function. I tried some thoughts by brute force but no success. Is there any algorithm to solve this kind of problems?

Thanks in advance!

答:
In fact, $\Bbb{R}$ is not only $T_5$, it's  $T_6$.

If you surf from the Urysohn's lemma Wikipedia page over to the Wikipedia page for "Normal space," you learn:

All metric spaces (and hence all metrizable spaces) are perfectly normal Hausdorff

Since $\Bbb{R}$ is a metric space, and being perfectly normal is stronger than being completely normal, $\Bbb{R}$ is completely normal.

Title: Is $\mathbb R$ completely normal space?


问:
Doubts in a textbook lemma
There is a lemma in the book saying:

"If the primal basic solution is an optimal solution of a linear
program (P), B is not necessarily an optimal basis."

I don't understand because, by the dual theory, if the primal has a feasible solution and so does the dual, then B has to be optimal? I tried to find an example that matches this but I can't think of any. Can someone give me an example so that I can think this through?

答:
It depends on how you define "optimal basis".  In the presence of degeneracy, you can have a basis where the basic solution of the dual is not feasible, but the basic solution of the primal turns out to be optimal (and after some pivots you get an optimal basis with the same basic solution of the primal.

For an extreme example, consider the problem

$$ \eqalign{\text{maximize }\ x_1 &\cr
  \text{subject to }\ x_1 + x_2 \le 0\cr
      x_1, x_2 \ge 0\cr} $$
There is only one basic feasible solution: $x_1 = 0$, $x_2 = 0$, slack variable $s_1 = 0$.  But not every basis is optimal.

Title: Doubts in a textbook lemma


问:
Calculus 1 - Differential Equation

$$\pi  \left(6 h(t)-h(t)^2\right) h'(t)=-\frac{h^{1/2}(t)}{\sqrt{3}},h(0)=3$$

Im supposed to solve this differential equation when $h(0) = 3$, and im not sure how it's supposed to be solved. And im not sure what kind of differential equation it is.

So i would really appriciate if i could get some help solving this equation.

答:
Set $y=h(t)$ you get
$$dx=\sqrt{3} \pi  (y-6) \sqrt{y}\,dy$$

Title: Calculus 1 - Differential Equation


问:
Identity up to isomorphism treated as identity in proof
In the following corollary to the inverse mapping theorem by Serge Lang, Fundamentals of Differential Geometry, 1999, p.17-18, there are two things in the proof which I don't understand, the first step and the last one:

If there is an identity up to isomorphism between E and $ F_1 $ as established by $ f'(x_0) $, why can we limit our consideration in the proof to the actual identity? This I have seen several times in proofs, but I don't understand why it can be done here and what the precise circumstances have to be in a proof to allow for this.

I dont't see why the local inverse $ \big( \varphi'(0,0) \big)^{-1} $, which is called g at the end of the proof, satisfies the two requirements defined in the corollary for the map g used there.

Thanks for any help.

Notes: $E, F_1, F_2 $ are Banach spaces. "Morphism" means a $ C^p $-map with $ p \geq 1 $. "Local isomorphism" means a local $ C^p $-isomorphism (dt.: lokaler $ C^p $-Diffeomorphismus). "Toplinear isomorphism" means an isomorphism between topological vector spaces.

Maybe the following drawing is helpful:

答:
The proof is indeed worded poorly wrt $\varphi'(0,0)$. It is also not necessary to do the identification $E=F_1$; the complication you get from considering them to be different is very manageable. Additionally using the word toplinear is a crime. Let me rewrite the proof with your two questions in mind:

Define
$$\varphi: U\times F_2\to F_1\times F_2, \qquad (x,y)\mapsto f(x)+(0,y)$$
Then the derivative of $\varphi$ at $(x_0,0)$ is equal to $f'(x_0)+(0,\mathrm{id}_{F_2})$. Since $f'(x_0)$ is valued in $F_1$ and is a linear isomorphism you have that $\varphi'(x_0,0)$ is also a linear isomorphism.

By the inverse function theorem it follows that there is a neighbourhood $V\subseteq U\times F_2$ of $(x_0,0)$ so that $\varphi\lvert_V$ is a diffeomorphism (understood to imply $p$-times differentiable). Let $h$ denote the inverse of $\varphi\lvert_V$ and define $g:=(f'(x_0),\mathrm{id}_{F_2})\circ h$. As a composition of diffeomorphisms $g$ is a diffeomorphism. Let $U_1$ be the projection of $V$ onto the $E$ component.

Then for $x\in U_1$:
$$g(f(x)) = (f'(x_0), \mathrm{id}_{F_2})\left[h(\varphi(x,0))\right]= (f'(x_0),\mathrm{id}_{F_2}) [(x,0)] = (f'(x_0)[x],0)$$
and $g\circ f$ is valued in $F_1\times\{0\}\subseteq F_1\times F_2$.


问:
Find all triples of positive integers (x, y, z)

I've no idea how to attempt such a question. I tried eliminating z from the equations to get a quadratic equation in terms of x, so I could use the fact that the discriminant would be a perfect square. Please help

答:
$$(x,y,z) = (-3,6,z)\quad {\rm or}\quad (12,13,z)$$

for any integer value of $z$.

So there are an infinite number of solutions.


答:
Ron already gave the canonical answer: $x^\alpha = e^{\alpha \log{x}}$ by definition, and then you may refer to the very nice properties of $e^x$. If you prefer, think at two approximating series of rational numbers $\{a_n\}$ and $\{b_n\}$ converging from above and from below to $\pi$, compute $\{x^{a_n}\}$ and $\{x^{b_n}\}$ and see that they actually converge to a single value.

At least as far as I am concerned, there is not a lot of difference asking what $x^\alpha$ is and asking what an irrational number $\alpha$ is.

Title: What does $x^\pi$ mean?


问:
Stability of 2nd order linear vector differential equations?
Given the system $\ddot{x} +B \dot{x}+ C x =0$ where $x \in R^{2}$ and $B$ is positive definite symmetric matrix and I know that the eigenvalues of $C$ are either $\lambda_1<0<\lambda_2$ or $\lambda=\pm \omega j$.

My question is: Can I conclude from these givens that the system is unstable and can the type the equilibrium point be determined?

答:
Let us write $B = \begin{bmatrix} b_1 & b_2 \\ b_2 & b_3\end{bmatrix}$ and $C = \begin{bmatrix} c_1 & c_2 \\ c_3 & c_4\end{bmatrix}$. Since $B$ is postive definite, we know that $b_1>0$, $b_3>0$, and $\det B = b_1b_3 - b_2^2 >0$. For $C$ we know that either

(pure imaginary poles) $\operatorname{trace} C = c_1+c_4 = 0$ and $\det C = c_1c_4-c_2c_3>0$,

or $\det C = c_1c_4-c_2c_3<0$ (the $\lambda_1 \lambda_2<0$ case).

Let us define $\eta := [x_1, x_2, \dot{x}_1, \dot{x}_2]^\top$. Then your system is given by
$\dot{\eta} = F\eta$, where
$$F := \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -c_1 & -c_2 & -b_1 & -b_2 \\ -c_3& -c_4& -b_2& -b_3\end{bmatrix}.$$

The characteristic polynomial is given by
$$\det(sI-F) = s^4+\left(b_{1}+b_{3}\right)\,s^3+\left(-{b_{2}}^2+c_{1}+c_{4}+b_{1}\,b_{3}\right)\,s^2+\left(b_{3}\,c_{1}-b_{2}\,c_{2}+b_{1}\,c_{4}-b_{2}\,c_{3}\right)\,s+c_{1}\,c_{4}-c_{2}\,c_{3}.$$

So, in the $\lambda_1 \lambda_2<0$ case we have that the free term of the polynomial is negative, and the system is unstable.

For the second part, the answer is negative. Try (a random search result)
$$B = \begin{bmatrix} 3.3 & -1 \\ -1 & 8.4 \end{bmatrix}, \quad C = \begin{bmatrix} 4.8 & 3.3 \\ -9.68 & -4.8 \end{bmatrix}.$$ The matrix $B$ is positive-definite, the matrix $C$ has eigenvalues on the imaginary axes, and $F$ is Hurwitz.


问:
Why do we not measure angles like this?
An angle is a simple geometric figure that is obtained by taking a ray and rotating it about a point to form another ray, both of which have their starting point in common. The initial rays is called the initial side, the final one is called the terminal side and the common point is called the vertex.

The point of measuring angles is to describe, in a way to what degree (pun intended) the initial line is rotated about the vertex to obtain the terminal line. The three usual measures of angles i.e. degree, radian and grads are dimensionless since they are all based on measuring an angle as a ratio of the arc length to the radius (I might be wrong, please let me know if I am).

Now, let's say that we have an angle $AOB$, as shown here :

Here, $OP = OQ = 1~\mathrm{unit}$. $L$ is the length of the arc $PQ$.

So, why can't we call say that $L$ is the angle measure of $\angle AOB$ (in a new angle measurement system). For such a system, the measure of any given angle will be the same as it's measure in radians, only the dimensionality will be different. The dimensionality of angle measure in this system becomes the same as of lengths. I don't see how that's a problem, though.

Also, I find thinking about measuring angles like this much simpler than as a ratio of arc lengths.

So, can this be called a valid way of measuring angles. If not, why?

Thanks!

答:
The units on a measurement tell you, among other things, how the number will change if you decide to change the units you use.

Suppose that "$1$ unit" in your case is $1$ meter, and you measure the arc length $L$ to be $0.7$ meters. Reporting the angle as "$0.7$ meters" suggests that if we were using centimeters instead, the angle would be $70$ centimeters.

But that's not what actually happens. If we make "$1$ unit" be $1$ centimeter, then the arc will become $100$ times smaller, and the arc length $L$ will change to $0.7$ centimeters.

So it make sense to report the angle simply as "$0.7$" in both cases, because if we change the units we're using, the angle does not change.

Title: Why do we not measure angles like this?


问:
Minimise $|f''(x)|$ on an interval when you know the values of the function and the values of the derivative at the endpoints of the interval only.
Given $a, b,c \in \mathbb{R}$, let $A_{abc}$ be the set of twice-differentiable real functions $f:[0,1] \to \mathbb{R}$ such that $f(0) = 0$, $f(1) = a, f'(0) = b$ and $f'(1) = c.\ $ Find the infimum of {$\max_{x \in [0,1]}$$|f''(x)|: f \in A_{abc}$ }$\subset \mathbb{R}$ in terms of $a, b$ and $c$.

I'm not sure I've written that correctly: I'm sure I'll have to re-write that more clearly. Anyway, the title should be somewhat clear as to what I'm after.

Example 1:

$$$$
Example 2:

I think we should also add the constraint that $ f'(0) \geq 0$, because if we then reflect everything the functions in the x-axis, we get all the cases where $f'(0) < 0$, so we don't lose any generality by assuming $ f'(0) \geq 0$.

Anyway, I'd be very surprised if this wasn't some sort of duplicate as it's such a natural optimisation question: I just think I'm not using the right lingo.

答:
In this answer I show that the exact answer is $sign((b-a)(c-a))(b+c-2a)+\sqrt{2((b-a)^2+(c-a)^2)}$ (formula (5) below).

As noted in AlexRavsky's and MartinR's answer, we may assume without loss that $a=0$, so that $f\in B_{b,c}=A_{0,b,c}$.

Next, if $f\in B_{b,c}$ with $b\neq 0$, then $g(x)=\frac{f(x)}{b}$ satisfies $g\in B_{1,\frac{c}{b}}$. So, (assuming $b\neq 0$, which we do since this limit case will turn out to be similar and simpler than our generic case) we may assume without loss
that $b=1$, so that $f\in C_{c}=A_{0,1,c}$.

Further, if $f \in C_c$, then $h(x)=-\frac{f(1-x)}{c}$ satisfies $h \in C_{\frac{1}{c}}$. So we may assume without loss that $|c| \geq 1$.

We assume $|c| \gt 1$ (the limit case $|c|=1$ will turn out to be similar and simpler than our generic case). Let $\varepsilon =\frac{c}{|c|}$ be the sign of $c$.

Let $m=||f''||_{\infty}$, let $z\in (0,1)$ be a constant to be fixed later, and let

$$
I_1 = \int_0^z (m+\varepsilon f''(t))(z-t)dt, \ I_2 = \int_z^1 (m-\varepsilon f''(t))(t-z)dt \tag{1}
$$

Both $I_1$ and $I_2$ can be computed by two successive integration by parts, and we find

$$
I_1 = m\frac{z^2}{2} - \varepsilon z + \varepsilon f(z), \ I_2 = m\frac{(1-z)^2}{2}+\varepsilon c(z-1)- \varepsilon f(z)\tag{2}
$$

Now both $I_1$ and $I_2$ are nonnegative by construction, so $I_1+I_2\geq 0$. This means that $m\geq m_0$ where

$$
m_0=\frac{2\varepsilon (c+z-cz)}{z^2+(1-z)^2} \tag{3}
$$

The inequality $m\geq m_0$ will be an equality iff the integrands in $I_1$ and $I_2$ are zero a.e., so that $f''$ is $- \varepsilon m$ on $[0,z]$ and $+ \varepsilon m$ on $[z,1]$. Integrating twice from $0$, it follows that $f(x)=-\varepsilon m\frac{x^2}{2}+x$ for $x\in [0,z]$ and $f(x)=\varepsilon m\frac{x^2}{2}+(1-2\varepsilon mz)x+\varepsilon mz^2$ for $x\in [z,1]$. The two conditions $f'(1)=c$ and $f(1)=0$ give us a (nonlinear) system of two equations in $m$ and $z$, and it turns out that this system has the unique solution

$$
m=\varepsilon (c+1)+\sqrt{2(c^2+1)}, z = \frac{c-\varepsilon \sqrt{\frac{c^2+1}{2}}}{c-1}\tag{4}
$$

Returning to the general case and unrolling the symmetries of the initial problem, we find the fully general bound for $f\in A_{a,b,c}$ :

$$
m=sign((b-a)(c-a))(b+c-2a)+\sqrt{2((b-a)^2+(c-a)^2)} \tag{5}
$$

Note 1. To make the formula work when of $b-a,c-a$ is zero and the other is not, we must use the convention $sign(0)=+1$.

Note 2. The unique solution we found does not strictly satisfy the requirements in the OP (the second derivative is discontinuous at $z$), but this is not a problem ; by well known density results, the non-$C^2$ optimal solution can be approximated by $C^2$ solutions and the infimum stays the same (although it is never attained by $C^2$ solutions).


问:
Let $Z\sim N(0,1)$, $X=Z^2$, $Y=e^{-X}$. Compute the expectations to confirm change of variables.
Let $Z\sim N(0,1)$, $X=Z^2$, $Y=e^{-X}$.

There's a couple of parts to this question; the first part I believe is setting up to the part that I posted.

The first part says to find the distributions $P_X$ and $P_Y$ on $(\mathbb{R},\mathcal{B}(\mathbb{R})).$

Letting $\phi(x)$ be the density for the normal distribution, I calculated the following;

$$P_X(X(\omega)\leq x)=P(Z(\omega)^2\leq x) = P(-\sqrt{x}\leq Z(\omega)\leq \sqrt{x}) = \int_{-\sqrt{x}}^\sqrt{x}\phi(t)dt,$$

And I could similarly find $$P_Y(Y(\omega)\leq y)=\int_{-\sqrt{\ln(y)}}^\sqrt{\ln(y)}\phi(t)dt.$$

After this, I'm then asked to calculate $$\int_\mathbb{R}e^{-z^2}\phi(z)dz, \int_\mathbb{R}e^{-x}dP_X, \text{ and } \int_{\mathbb{R}}ydP_Y,$$ and verify that all integrals agree.

Getting the first integral wasn't too hard.

Here's where I'm having trouble:  even though I think what I have for the first part is technically correct, it's not very useful in actually calculating the integrals. For example, The only thing I can think to do to calculate $\int_\mathbb{R}e^{-x}dP_X$ would be to find a sequence of simple functions converging to $e^{-x}$ and then taking the limit of the lebesgue integrals w.r.t. $P_X$, which I'm pretty sure is not what is expected in this problem.

More consequentially, this problem is prefaced with the statement that, when using change of variables, one of the three integrals is usually easier to calculate than the others, and tells us ahead of time that the first integral will be easier to get than the others. This leads me to believe that we're expected to calculate the integrals independently, without assuming validity of the CoV formula, in order to demonstrate the usefulness of it.

Is there a non-ridiculous way to get these integrals without assuming change of variables, or do you think I'm misunderstanding something about what this problem is asking of me?

Thanks in advance for any thoughts.

答:
Going on with your calculations (derivating the CDF you found) you will easy realize that

$$X\sim \chi_{(1)}^2$$

$$f_X(x)=\frac{1}{2\sqrt{x}}\phi(\sqrt{x})+\frac{1}{2\sqrt{x}}\phi(-\sqrt{x}) =\dots=\frac{\Big(\frac{1}{2}\Big)^\frac{1}{2}}{\Gamma\Big(\frac{1}{2}\Big)}x^{-\frac{1}{2}}e^{-\frac{x}{2}}$$

For $Y=e^X$,

In this case with monotonic transformation it is faster  to get immediately the density with the fundamental transformation Theorem

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

Getting

$$f_Y(y)=\frac{1}{y\sqrt{2\pi logy}}e^{-\frac{logy}{2}}\cdot\mathbb{1}_{(1;\infty)}(y)$$


答:
Units of measurement are attached to values to represent their behavior under change of scale, or change of unit.

Consider a stick, whose length is measured in units of length. Let's say it is $x$ metres long. If we scale the object up by a factor of $c$, the stick will be $c\cdot x$ metres long. If we choose to measure in centimetres instead, the stick will be $100 \cdot x$ centimetres long.

Consider a carpet, whose area is measured in units of length squared. Let's say it has an area of $y$ square metres. If we scale the object up by a factor of $c$, the carpet will have an area of $c^2 \cdot y$ square metres. If we choose to measure in centimetres instead, the carpet will have an area of $100^2 \cdot y$ square centimetres.

In general, if a value in some system is measured in $m^n$, then scaling the system, or our unit of measurement, will cause that value to be scaled with the $n$-th power of the scale factor.

Now let's look at your angle example, and find the appropriate $n$ such that an angle should be measured in $m^n$. Let's say our initial unit of measurement is $m$ and the size of our angle is $1$ radian large. If we scale our angle up by a factor of $c$, it is still $1$ radian large, its value has multiplied by $c^0$. If we measure in $cm$ instead, the angle will still be $1$ radian large, it has been scaled by a factor of $100^0$. So it only makes sense for our angle to have units of length to the power of $0$, which means the angle is unitless.

Title: Why do we not measure angles like this?

Body:
An angle is a simple geometric figure that is obtained by taking a ray and rotating it about a point to form another ray, both of which have their starting point in common. The initial rays is called the initial side, the final one is called the terminal side and the common point is called the vertex.

The point of measuring angles is to describe, in a way to what degree (pun intended) the initial line is rotated about the vertex to obtain the terminal line. The three usual measures of angles i.e. degree, radian and grads are dimensionless since they are all based on measuring an angle as a ratio of the arc length to the radius (I might be wrong, please let me know if I am).

Now, let's say that we have an angle $AOB$, as shown here :

Here, $OP = OQ = 1~\mathrm{unit}$. $L$ is the length of the arc $PQ$.

So, why can't we call say that $L$ is the angle measure of $\angle AOB$ (in a new angle measurement system). For such a system, the measure of any given angle will be the same as it's measure in radians, only the dimensionality will be different. The dimensionality of angle measure in this system becomes the same as of lengths. I don't see how that's a problem, though.

Also, I find thinking about measuring angles like this much simpler than as a ratio of arc lengths.

So, can this be called a valid way of measuring angles. If not, why?

Thanks!


答:
Every metric space is normal. Since every subspace of a metric space is a metric space, it thus follows that every metric space is completely normal.

You seem to be asking for a proof that every metric space is perfectly normal. Let's just go ahead and write something down.

Fix a metric space $(X,d)$. If $A\subset X$ is closed, then $$f_A:x\mapsto d(x,A)=\inf\{d(x,a)\mid a\in A\}$$
is a continuous function $X\rightarrow[0,\infty)$ with $f^{-1}_A(0)=A$. To see that this function is continuous just use the triangle law to show that it is even Lipschitz.

Suppose $B\subseteq X$ is closed and disjoint from $A$. Form $f_B$ as above. Now consider
$$h=\frac{f_A}{f_A+f_B}.$$
Since $A\cap B =\emptyset$ the function is well-defined. Clearly $h^{-1}(0)=A$ and $h^{-1}(1)=B$.

(Edit: To put my open comment in contex note that a space is completely normal if and only if it is hereditarily normal. This is a standard definition of completely normality. The equivalence of the two conditions is stated on the wikipedia page. I've shown that each metric space is perfectly normal. Every perfectly normal space is completely normal. For istance with the notation above consider the disjoint neighbourhoods $U_A=h^{-1}[0,1/3)$ and $U_B=h^{-1}(2/3,1]$.)

Title: Is $\mathbb R$ completely normal space?

Body:
Is $\mathbb R$ completely normal ( $T_5$ ) space?

I've seen this Wiki Page about Urysohn's lemma, which states that a topological space is normal if and only if any two disjoint closed sets can be separated by a continuous function.

Two plain subsets $A$ and $B$ are said to be separated by a function if there exists a continuous function $f : X \to[0,1]$ such that $f(A) = 0$ and $f(B) = 1$. Any such function is called a Urysohn function for $A$ and $B$.

So, now I need to construct continuous function $f : \mathbb R \to [0, 1]$ such that for all closed $A, B \subseteq \mathbb R: A \cap B = \varnothing,$ we have $f(A)=0$ and $f(B)=1$.

However, I wonder if one could construct such continuous function. I tried some thoughts by brute force but no success. Is there any algorithm to solve this kind of problems?

Thanks in advance!