问:
Prove the locus of $M$ is a conic passing though $I$ and $J$.
Problem: In the perspective plane $\mathbb{P}^2$, given a conic $(S)$ and a line $d$ having no intersection with $(S)$. Fixed two points $I$ and $J$ on $d$. Take a variety point $M$ such that the tangents from $M$ to $(S)$ intersect $d$ at two points $R$ and $S$ such that $(IJRS)=-1$. Prove the locus of $M$ is a conic passing through $I$ and $J$.

Here is what I've discovered so far.

Let the tangents from $I$ and $J$ touch $(S)$ at $I_1,I_2$ and $J_1,J_2$ respectively. Then the locus of $M$ is the conic passing through $I,I_1,I_2,J,J_1,J_2$. I also proved those 6 points lie on a conic (let's call it $(S')$).

Take any point on $(S')$, call $A$. Then the intersection $A_1, A_2$ of two tangents from $A$ to $(S)$ and $d$ satisfy $(IJ,A_1A_2)=-1$. Now I'm stucking at proving this claim.

2'. Something that might be useful for proving 2: Let $P$ the polar of $d$ wrt $(S)$. Then $PI$ and $PJ$ tangent to $(S')$.

Can I have some hints for the next steps? Any ideas would be appreciated. Thanks for reading.

答:
The other answer here uses a proof by Milne that claims "range (c)=range(d)".  What does that mean?

On page 29 the notations "range (a)" and "pencil P(a)" are defined.
For our purposes, a "range" is any collection of collinear points and a "pencil" is any collection of concurrent lines.  Often ranges and pencils contain just four elements so that we can take cross ratios, but in general they contain an arbitrary number of elements.

For shortness we shall often denote a range (abcde ...) by (a), and a
pencil P{abcde ...) by P(a)

(Chapter IV is worth a read, or at least a skim)

Unfortunately, I cannot find any explicit definition by Milne of what is meant by "range(a)=range(b)" or "pencil P(a)= pencil P'(a')".  From context is seems to be that "range(a)=range(b)" means the two are homographic.  In fact, on page 187 pencils and ranges are said to be equal (homographic) as in:

the pencil $\rho(m)$ = the range of poles $(m)$ = the pencil of polars
$P(\mu)$

Does this help?  Getting back to your question, I interpret

And the range (c) = the range (d), therefore the pencil A(P) = the
pencil B(P),

as

"Because range(c) is homographic to range(d) the pencil A(P) is homographic to B(P)"

In all of the above, ranges and pencils are held to be homographic because they preserve cross ratios. In particular, you should verify that the conjugacy $c \to d$ preserves cross ratio and thus is homographic.


问:
There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color?
A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?

Here's what I tried:

Case 1: 2 black balls and 1 white ball

8/15 * 7/14 * 7/13 = 392/2730 = 28/195

Case 2: 2 white balls, 1 black ball

7/15 * 6/14 * 8/13 = 336/2730 = 24/195

28/195 + 24/195 = 52/195 = 4/15

My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!

答:
Case $1$: $2$ black balls and $1$ white ball. The correct probability will be

$P(2B,1W) = \displaystyle \frac{ 3 \cdot8 \cdot 7 \cdot 7}{15 \cdot 14 \cdot 13} = \frac{28}{65}$.

$\big[\text{It is basically } \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{7}{15} \cdot \frac{8}{14} \cdot \frac{7}{13}\big]$

I would suggest think as selecting $2$ black balls out of $8$ and $1$ white ball out of $7$ vs. selecting any $3$ balls out of $15$ which can be written as -

$P(2B,1W) = \displaystyle {8 \choose 2}{7 \choose 1} / {15 \choose 3}$

Case $2$: $2$ white balls, $1$ black ball

$P(2W,1B) = \displaystyle {8 \choose 1}{7 \choose 2} / {15 \choose 3} = \frac{24}{65}$.

Adding them, we get probability as $\displaystyle \frac{4}{5}$.

ContentLicense: CC BY-SA 4.0


问:
Complete analytic function for $\sqrt{1+\sqrt{z}}$.
Let $U=\mathbb{C}\setminus(-\infty,0]$.

There is an analytic function $\phi$ on $U$, such that $\phi(z)^2=z$. $\phi$ is the inverse to $z\mapsto z^2$ on the right half plane. The image of $U$ under $z\mapsto 1+\phi(z)$ is inside $U$ so we can define $f: U\rightarrow \mathbb{C}$ such that $z \mapsto \phi(1+\phi(z))$.

The complete analytic function for $(U,f)$ is the set of all function elements $(V,g)$ where $V\subset \mathbb{C}$ is connected and open and $g:V\rightarrow \mathbb{C}$ is analytic, such that $(V,g)$ is an analytic continuation of $(U,f)$. I need to find all the germs in the complete analytic function at $1/2$ and at $1$.

Here is my attempt: Let $(V,g)$ be an analytic continuation of $(U,f)$ such that $1\in V$.

Since $(f(z)^2-1)^2=z$ on $U$, then $(V,(g^2-1)^2)$ is an analytic continuation of $(U,\mathrm {id}_U)$. Then since the identity map is an entire function, $(g^2-1)^2=\mathrm{id}_V$. This means that $g(1)=\pm \sqrt2, 0$. Since $h(z)=(z^2-1)^2$ is a left inverse to $g$, $h'(g(1))g'(1)=1$. $h'(0)=0$ so $g(1)\neq 0$. I am left with the possibilities $g(1)=\pm \sqrt2$. Since $h$ is an analytic left inverse to $g$, local behavior of $h$ at $g(1)$ determines $g$ near $1$, so there is at most one germ corresponding to each of $g(1)=\pm \sqrt2$. I know already that $f(1)=\sqrt 2$ so it suffices to prove that $(U,-f)$ is an analytic continuation of $(U,f)$. I feel this is true after drawing some loops in the complex plane.

Similarly for $1/2$. Let $(V,g)$ be an analytic continuation of $(U,f)$ such that $1/2\in V$. Then $f(1/2)=\pm\sqrt{1\pm\sqrt{1/2}}$. I want to show that $\pm \phi\circ(1\pm \phi)$ (well defined near $1/2$) on some small domain around $1/2$ are all analytic continuations of $f$.

Is there a neat and rigorous way to see these? Can anyone offer some ideas?

答:
Some ideas:

You have the composition of $f(z)=\sqrt{1+z}$ and $g(z)=\sqrt{z}$.

$g$ has two germs at any nonzero point. It is not analytic at $0$.

$f$ has two germs at any point other than $-1$. It is not analytic at $-1$.

The composition $f\circ g$ has four germs at $1/2$, whose values at $1/2$ correspond to the choices of $\pm  $ in $\pm \sqrt{1\pm \sqrt{1/2}}$

The composition $f\circ g$ has two germs at $1$, whose values at $1$ correspond to the choices of $\pm  $ in $\pm \sqrt{1+ \sqrt{1}}$. The germ of $g$ that has $g(1)=-1$ cannot be inserted in $f$, per item 2.

ContentLicense: CC BY-SA 3.0


问:
Mathematical way of proving an equation does not produce a prime number for all numbers in a set?
Given $k = n^2 + 9n + 1$

Prove that the statement “$k$ will be a prime number for all integer values of $n$ from $1$ to $9$”, is wrong.

You can try the numbers $1$ to $9$ in turn, and determine that $6$ proves the statement wrong.

But is there an elegant mathematical way of proving the statement wrong?

答: One can look at the polynomial $n^2+9n+1$ modulo small primes $p$. It is never equal to $0$ modulo $2$, $3$, or $5$ (we only need to evaluate $2/3/5$ consecutive values to prove that). However, modulo $7$ the polynomial is congruent to $n^2+2n+1=(n+1)^2$, which is clearly $0\pmod7$ when $n\equiv-1\pmod7$. This shows that $n=6$ will be a counterexample (and indeed that every set of $7$ consecutive integers produces a counterexample).


问:
What is the relationship between $(u\times v)\times w$ and $u\times(v\times w)$?
Given three vectors $u$, $v$, and $w$, $(u\times v)\times w\neq u\times(v\times w)$. This has been a stated fact in my recent class. But what is the ultimate relationship between them? I would presume that one is a scalar multiple of the other?</p>

答:
As others have mentioned, the ultimate relationship between $(u\times v)\times w$ and $u\times(v\times w)$ is given by the Jacobi identity: $u \times (v \times w) + w \times (u \times v) + v \times (w \times u) = 0$.

Hence, $(u\times v)\times w = u\times(v\times w)$ iff $v \times (w \times u) = 0$, and this happens iff $u$ and $w$ are parallel or $v$ is perpendicular to both $u$ and $w$.


问:
A problem with Weirstrass substitution
I hope you all are doing well.

Yesterday, I was doing some undefined integral exercises when I faced an exercise that was something like:
$$\int \frac{F(\sin(x),\cos(x))}{G(\sin(x), \cos(x))}dx$$
I found that a substitution with $t =\tan(\frac{x}{2})$ would allow me to transform this integral in some $\int \frac{P(t)}{Q(t)}\frac{2}{1+t^{2}}dt$ using the following identities:

What I was wondering a few days ago, was that this identities had a problem when for example I express $\sin(x)$ as a $f(t)$ and $x = \pi$ (in this case $f(t)$ is not defined.). Also the fact that I make a $t =\tan(\frac{x}{2})$ restricts $\pi$ from the domain of the resulting $F(u)$ such that $F'(t) = f(t)$ so... What happens when I make a substitution $u = s(x)$ such that the domain of that function excludes some elements of the function I'm integrating? And also, what happens with the identities if $x = \pi$?

I hope to have expressed myself correctly. I do not speak english so feel free to improve my writing and sorry if there are grammar mistakes.

Thanks.

答: Suppose this substitution achieves $f(x)dx=g(t)dt$. For $a<\pi$,$$\int_a^\pi f(x)dx=\lim_{y\to\pi^-}\int_a^yf(x)dx=\lim_{y\to\pi^-}\int_{\tan(a/2)}^{\tan(y/2)}g(t)dt=\int_{\tan(a/2)}^\infty g(t)dt.$$You can treat limits of integration above $\pi$ with one-sided limits in a similar way: if $a>\pi$,$$\int_\pi^af(x)dx=\lim_{y\to\pi^+}\int_y^af(x)dx=\lim_{y\to\pi^+}\int_{\tan(y/2)}^{\tan(a/2)}g(t)dt=\int_{-\infty}^{\tan(a/2)}g(t)dt.$$

Title: A problem with Weirstrass substitution


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
Not sure they saw $\log$ yet, but to "visualize" and get the numbers out of the exponentiation, reduce the scale using the $\log$

$$\log(2^x) = x \cdot \log(2)$$
$$\log(x^2) = 2 \cdot \log(x)$$

Removing the constants, we compare now $x$ and $\log(x)$, and
$$\dfrac{x}{\log(x)}$$grows quickly.


问:
Shimura Varieties actual varieties for any level?
Since I have not been answered on my other thread Shimura varieties/Double coset for non-neat level so far and I can't work with this uncertainty, I'll just put down the most important questions here. I would still be grateful if someone would take time for the other thread. Let $(G,X)$ be a Shimura datum. Are the level $K$ varieties $\text{Sh}_K(G,X)$ also (quasi-projective) varieties if $K$ is not sufficiently small (in which case they are normal)? Do the $X_K=\text{Sh}_K(G,X)$ also have models over a number field if $K$ is not sufficiently small? Does the tower of Shimura varieties $\{X_K\}$ have a model over $E$ even if $K$ runs through all open compact $K$? Now let $D$ be a general Hermitian symmetric domain and $\Gamma \subset \text{Is}^+(D,g)$ a discrete subgroup of the holomorphic isometry group of $D$. Bailey-Borel says $\Gamma\backslash D$ is a normal quasi-projective variety over $\mathbb{C}$ for any $\Gamma$ (not necessarily torsionfree). Does it have a model over a number field?

答:
First as you said Bailey-Borel theorem says that $D/\Gamma$ is analyticification of a quasi-projective smooth variety for any torison free $\Gamma$ and a quasi-projective normal variety for any $\Gamma$.

you can understand all $Sh(G,X)_K$ as varieties, and they all have canonical models over reflex field.(although sometimes it is better to look at the $Sh(G,X)_K$ for big $K$ as a stack because the variety is only a coarse solution to moduli problem you want to consider).

all the connected components have a model over some number field so $D/\Gamma$ always have a model over some number field for any $\Gamma$ that comes from a congurance subgroup of some $G$ such that $G^{ad}=Hol(D)$.maybe for any $\Gamma$ you can find a model over a number field but this model is not in any way canonical and after all the interesting point about canonical models is that you have a model over a fixed number field otherwise you only want a model over $\bar{\mathbb{Q}}$.

*just to clarify what I mean by look at big level shimura variety as a stack: you can write $Sh(G,X)_K=Sh(G,X)_{K'}/\frac{K}{K'}$ for a neat group $K'$, this quotient is represented by a normal quasi-projective variety in the category of schemes(because $\frac{K}{K'}$ is a finite group) but sometimes it is better to look at the quotient stack which could be equal to that normal variety or not.


答:
Assuming the exponential is $c^x$:

Then: $c^x = ((\sqrt{c})^2)^x = (\sqrt{c})^{2x} = ((\sqrt{c})^x)^2$

Our original question is now changed into: let's compare $x^2$ with $((\sqrt{c})^x)^2$. Which has the same result as comparing $x$ directly with $(\sqrt{c})^x$.

As a result, we have reduced our argument to: why does an exponential function eventually get bigger than a linear one? Which likely has been answered already by the teacher.

Title: Why does an exponential function eventually get bigger than a quadratic

Body:
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.


问:
Show that the trivial representation of G occurs exactly once in $\pi_X$
I am looking at the following problem in relation to representation theory:

Let $G$ be a finite group acting on a set $X$ (we denote this action by $(g,x) \mapsto g\cdot x$). We assume that $X$ has at least two elements. Let $\pi_X$ denote the representation attached to this action.
Suppose the action of $G$ on $X$ is transitive, that is for every $x,y \in X$ there exist $g \in G$ such that $g \cdot x=y$. Show that the trivial representation of G occurs exactly once in $\pi_X$

I know that the trivial representation of G is meant as the unique 1-dimensional irreducible trivial representation of G so if I denote it by $\sigma$ then $\sigma (g) = 1 $ for all g in G, but I am not quite sure what it means to "show that the trivial representation of G occurs exactly once in $\pi_X$" thus I am having some difficulties to finding a way of solving it.

EDIT:

I have found out that if $\pi$ is a representation and $\sigma$ is isomorphic to a subrepresentation of $\pi$, then we say that $\sigma$ appears exactly once in $\pi$ if $\sigma \oplus \sigma$ is not isomorphic to a subrepresentation of $\pi$.

However, I am still having difficulties proving this by using the definition.

答:
To get a representation we first need some sort of a vector space for $G$ to act on, to do this we consider the space of formal sums $\displaystyle \sum_{x \in X} c_x e_x $ where each $c_x \in \mathbb{C}$, and $e_x$ are the basis vectors.

This space is usually denoted $\mathbb{C}[X]$, now the action of $G$ on this space is given by:

$$g \cdot \sum_{x \in X} c_x e_x = \sum_{x \in X} c_x e_{gx} =\sum_{y \in X} c_{g^{-1}y} e_y. $$

This representation is the so-called $\pi_X$ you are after.

Finally, to show the trivial rep occurs only once I would suggest using character theory, although this can be done directly too if you'd like.


问:
Density of random variable max{ξ1, ξ2}
Let $\xi_1, \xi_2$ be independent and equally distributed absolutely continuous random variables with density $p$.

How can I find the density of the random variable $\max(\xi_1, \xi_2)$.

答:
Let $\eta=\max\{\xi_1,\xi_2\}$, then this has cdf $F(a)=P(\eta<a)=P(\max\{{\xi_1,\xi_2\}<a})=P(\xi_1<a,\xi_2<a)=P(\xi_1<a)P(\xi_2<a)=\left[G(a)\right]^2$ with G being the cdf of $\xi$.

The pdf of $\eta$ can then be found by taking the derivative wrt a: $\frac{d}{da}[G(a)]^2=2G(a)g(a)$. In this case, the density is $p$, giving us $2G(a)p(a)$. And G can be found by integrating... $G(a)=\int_{-\infty}^a p(t)dt$.

Title: Density of random variable max{ξ1, ξ2}


问:
Evaluating $\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1}$

I am seeking to evaluate the sum $$S=\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1} \\ =(n-1)^2 \binom{n}{1}+(n-3)^2 \binom{n}{3}+(n-5)^2\binom{n}{5}+\cdots$$

I re-wrote the sum as $$S=(n+1)^2\sum_{r \in \mathbb{N}} \binom{n}{2r-1}-4(n+1)\sum_{r\in\mathbb{N}}r\binom{n}{2r-1}+4\sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$$

It is clear from the binomial expansion of $(1+x)^n$, (by plugging $x=\pm1$ ) we get the value of the sum $\displaystyle \sum_{r\in\mathbb{N}}\binom{n}{2r-1}$ as $2^{n-1}$.

Now, after this I can find the value of the sums $\displaystyle \sum_{r\in\mathbb{N}}r\binom{n}{2r-1}$ and  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$ by considering the binomial expansion of $\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$ and substituting $x$ for $\sqrt{x}$ and then differentiating, and by manipulations, I will be able to find the desired sums.

However, this method gets very lengthy, especially when doing differentiation for finding  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$.

I am hoping to see some other different approaches which don't involve so much calculation and are easy to understand.

The final closed form is $n(n+1)2^{n-3}$.

答:
For a combinatorial proof, consider choosing from $n$ people a committee with an odd number of regular members and a secretary and treasurer (which might be the same person).  The LHS conditions on the number $2r-1$ of regular members.  The RHS first selects the secretary and treasurer and then selects an odd number of remaining members.  If one person is both secretary and treasurer, then half of the subsets of the remaining $n-1$ people are odd.  If the secretary and treasurer are different people, then half of the subsets of the remaining $n-2$ people are odd.  The sum of the two counts is:
$$n\cdot\frac{2^{n-1}}{2} + n(n-1) \frac{2^{n-2}}{2} = (2n+(n^2-n))2^{n-3} = n(n+1)2^{n-3}$$


问:
Is $f$ holomorphic in $\mathbb C$?
Let $f:\mathbb C\to \mathbb C$ such that functions $z\mapsto \sin (f(z))$ and $z\mapsto \cos (f(z))$ are holomorphic on the entire complex plane.
 a) Is $f$ also holomorphic in $\mathbb C$?
 b) If we additionally assume that $f$ is continuous, is $f$ holomorphic in $\mathbb C$?

Logic suggests that in point a) the correct answer is NO, and in b) YES, but I do not know how to prove it.

答:
To show a is "no", just let $f$ be a non-constant function taking values only in $2\pi\mathbb{N}$, so that $\sin(f) =0$ and $\cos(f) = 1$, which are holomorphic, but since $f$ is not even continuous it is not holomorphic.

For b, not that if $f$ is continuous, then around some $z_0 \in \mathbb{C}$ we have by continuity that $f(z)$ can be made to stay in a small neighborhood of $f(z_0)$ by continuity. At least one of $\cos$ and $\sin$ has nonzero derivative at $f(z_0)$, so by the inverse function theorem either $\sin$ or $\cos$ is locally invertible around $f(z_0)$ with holomorphic inverse. (If you are not comfortable with inverse function theorem, then just note that all I am saying here is you can pick a branch of $\arcsin$ or $\arccos$ which are analytic near $\sin(f(z_0))$). Then we have either $f(z) = \sin^{-1}(\sin(f(z))$ or $f(z) = \cos^{-1}(\cos(f(z))$ is, being the composition of two holomorphic functions, holomorphic in a neighborhood of $z_0$. Since $z_0$ is arbitrary, $f$ is holomorphic.

Title: Is $f$ holomorphic in $\mathbb C$?


问:
Prove that if Triangles ABC = DEF in a metric geometry, then line AB contains exactly two of the points D, E, and F.
Prove that if Triangles ABC = DEF in a metric geometry, then line AB contains exactly two of the points D, E, and F.

We are not allowed to use the facts:
In a metric geometry, if triangles ABC=DEF, then {A,B,C} = {D,E,F}

or

In a metric Geometry, if A, B, and C are not collinear, then A is an extreme point of triangle ABC.

Naturally, I tried to do this with a proof by contradiction by assuming line AB does NOT contain exactly two of the points D,E, or F. I set up three cases for this.

Line AB contains none of D,E, or F;

Line AB contains one of D, E, or F. (Where I'd imagine the same contradiction will arise no matter which point I choose);

Line AB contains all three of D, E, and F.

In the first two cases, I can't find where a contradiction will arise. Please Help.

Some Definitions

The definition of a triangle that I am given is
If {A,B,C} are not collinear points in a metric geometry, then the triangle ABC is the set $\bigtriangleup$ABC = $\overleftrightarrow{AB}$ $\cup$ $\overleftrightarrow{BC}$ $\cup$  $\overleftrightarrow{CA}$

The definition of a line segment I am given is $\overleftrightarrow{AB}$ = {C $\in$ $\mathscr{P}$ | A-C-B or C=A or C=B}

The definition of between I am given is A-C-B iff A, B, C are collinear and d(A,B) = d(A,C) + d(C,B); where $d$ is the distance function for a metric geometry.

答:
I believe this is a valid proof, however, I used a proposition that is not obvious (I think). ** will denote when I used this assumption. I will prove it at the end.

Prove that if $\bigtriangleup$ABC=$\bigtriangleup$DEF in a metric geometry, then $\overleftrightarrow{AB}$ contains two of the points D, E, or F.

Proof
I will prove this by contradiction. This can be split into three cases. $\overleftrightarrow{AB}$ contains none of D, E, and F; one of D, E, or F; all three of D, E, and F.

Suppose $\overleftrightarrow{AB}$ contains one of D, E or F. Clearly $\overleftrightarrow{AB}$ cannot intersect any other side at any other point. If $\overleftrightarrow{AB}$ intersects an adjacent side to the vertice that it contains, then the $\overleftrightarrow{AB}$ and the adjacent side are equal because they intersect at two points. If $\overleftrightarrow{AB}$ intersects the opposite side of the contained vertice, then there is a point on either side of the intersection that cannot belong to any other side or else $\overleftrightarrow{AB}$ would be equal to that side. In any case, $\overleftrightarrow{AB}$ contains more than one of D, E, or F.

Now suppose $\overleftrightarrow{AB}$ contains all three. Then D, E, and F are collinear and $\triangle$DEF is not a triangle.

Now suppose $\overleftrightarrow{AB}$ doesnt contain any of D, E or F. This means either $\overleftrightarrow{AB}$ doesnt intersect $\triangle$DEF which would mean that $\triangle$ ABC $\neq$ $\triangle$DEF since A, B are not in $\triangle$DEF. If $\overleftrightarrow{AB}$ intersects any of the sides between the vertices then we could at least 3 points on either side of the intersection that couldn't possibly be contained by the remaining sides of $\triangle$ DEF since one side would have to contain two, thus equating the lines, and thus $\overleftrightarrow{AB}$ would contain at least two of D, E or F. This is a contradiction.


问:
Show that $pq$ ($pq\neq6$) is deficient.
Show that $pq$ ($pq \ne6$) is deficient.

This means I need to show that $\sigma(n)-n<n$ or that $\sigma(pq)-pq<pq$. Since $\sigma(pq)=\sigma(p)\sigma(q)=(p+1)(q+1)=pq+p+q+1$ and so $\sigma(pq)-pq=pq+p+q+1-pq=p+q+1$. We must then show that $p+q+1<pq$, but I'm not sure how to proceed from here. Any hints or solutions are greatly appreciated.

答:
Hint: $p+q+1 < pq$ is equivalent to $(p-1)(q-1) > 2$.

WLOG suppose $2 \le p < q$. Then we just need to prove $q-1 > 2$.


答:
You have already been told in the comments that the answer to the first question is negative.

Now, suppose that $f$ is continuous that both of those functions are holomorphic. Then so is$$z\mapsto\cos\bigl(f(z)\bigr)+i\sin\bigl(f(z)\bigr)=e^{if(z)}.$$Now, let $z_0\in\Bbb C$. Let $g(z)=e^{iz}$. Since $g$ is locally invertible (this follows from the fact that $g'$ has no zeros), there is a neighborhood $V$ of $f(z_0)$ such that its restriction to $V$ has an analytic inverse $h$. And, since $f$ is continuous, there is a neighborhood $W$ of $z_0$ such that $f(W)\subset V$. So, $f|_W$ is analytic, since it is equal to $h\circ e^{if}$ there.

Title: Is $f$ holomorphic in $\mathbb C$?

Body:
Let $f:\mathbb C\to \mathbb C$ such that functions $z\mapsto \sin (f(z))$ and $z\mapsto \cos (f(z))$ are holomorphic on the entire complex plane.
 a) Is $f$ also holomorphic in $\mathbb C$?
 b) If we additionally assume that $f$ is continuous, is $f$ holomorphic in $\mathbb C$?

Logic suggests that in point a) the correct answer is NO, and in b) YES, but I do not know how to prove it.


问:
Convolution of two random varaibles
Let $X\sim U[0,20]$ and $Y\sim U[5,15]$ two independent random variables.

Calaculate $P(X+Y\leq t)$.

Can anyone help me in this question, I get here:

$\left\{ X+Y\le t\right\} \Rightarrow\left\{ Y\le t-X\right\} \Rightarrow\left\{ X\in\mathbb{R}\right\} \cap\left\{ Y\le t-X\right\} \Rightarrow\\\int\limits _{0}^{20}\frac{1}{20}\left(\int\limits _{5}^{t-x}\frac{1}{10}dy\right)dx=\\\frac{1}{20}\cdot\frac{1}{10}\int\limits _{0}^{20}1\left(\int\limits _{5}^{t-x}1dy\right)dx=\\\frac{1}{200}\int\limits _{0}^{20}1\left(\int\limits _{5}^{t-x}1dy\right)dx$

But I don't know how to continue now.
Thank you!

答:
Do a drawing of the domain (a rectangle) and pass through the line $X+Y=t$.

The corresponding CDF is the area under the line multiplied by the joint density

$$F_T(t)=\begin{cases}
0,  & \text{if $t<5$} \\
\frac{1}{200}\cdot\frac{(t-5)^2}{2},  & \text{if $5\leq t<15$} \\
\frac{1}{4}+\cdot\frac{(t-15)\cdot10}{200},  & \text{if $15\leq t<25$} \\
1-\frac{(35-t)^2}{2\cdot200},  & \text{if $25\leq t<35$} \\
1, & \text{if $t\geq 35$}
\end{cases}$$

The 3 areas are obviously

a triangle

the triangle with area $\frac{1}{4}$ plus a parallelogram

1 minus the remaining triangle

Hint: if you find too difficult this exercise try first with the same exercise where $X,Y$ are independent uniform on $[0;1]$ then solve your exercise with this rectangle

Here is the drawing

Title: Convolution of two random varaibles


问:
Can one sample uniformly from the surface of an $n$-sphere of non-unit radius using normal r.v.'s?
One can sample coordinates of the surface of a unit radius $n$-dimensional sphere uniformly using the following method: independently generate a vector of $n$ standard normal random variables $\mathbf{x}=[X_1,\ldots,X_n]$, compute the Euclidean norm of this vector $\|\mathbf{x}\|_2=\sqrt{X_1^2+\ldots+X_n^2}$, then vector $\mathbf{x}/\|\mathbf{x}\|_2$ will contain the uniform sample of a coordinate from the surface of a unit-radius $n$-sphere.

I am wondering if that method can be extended to sampling from an $n$-sphere with arbitrary radius $r>0$.  My intuition tells me that multiplying the sample from unit-radius $n$-sphere by $r$ might work, but I haven't seen that mentioned anywhere, which makes me think that either my intuition is wrong or this may be a question with an obvious answer.

So, can someone clarify: is vector $r\mathbf{x}/\|\mathbf{x}\|_2$ a uniform sample from the surface of an $n$-sphere of radius $r$?  If it is not, is there a way to use i.i.d. normal variates to obtain such a sample?

答:
Answer: "Yes." – Rahul Narain Apr 22 at 8:35

Additional remark: if you manage to take uniform samples from the unit sphere by whatever method, multiplication by $r$ will do the same for the sphere by radius $r$. This is because the pushforward of the normalized surface measure on the unit sphere by the map $x\mapsto rx$ is the  normalized surface measure on the  sphere of radius $r$.


问:
Difficult Quadratic Simultaneous Equations
I tried solving this system of equations multiple ways: $3x+2y=8, \, x^2-2y^2+xy=4$

I tried substituting $x$, substituting $y$, equating the two by making the second equation equal to 8, but the closest I got was by breaking down the $-2y^2$ to $-y^2-y^2$:

From first equation, $y=4-\frac{3x}{2}$.
$$x^2-y^2-y^2+xy=4$$
$$(x+y)(x-y)-y(x-y)=4$$
$$(x-y)(x+y-y)=4$$
Then substitute y for the value found in first equation. This gave me $x=4$ or $x=16/5$.

The actual answer is $x=2, y=1$ or $x=18/5, y=-7/5$

How do I solve it step by step?

答:
hint

The second equation can be put in the form

$$(x+\frac y2)^2-\frac{y^2}{4}-2y^2=4$$

or

$$(2x+y)^2-9y^2=16$$

the first will be

$$6x+4y=16=3(2x+y)+y$$
If we put $ X=2x+y $, the system becomes
$$X^2-9y^2=16$$
$$3X+y=16$$
then

$$X^2-9(16-3X)^2=16$$


问:
Proposition 3.6 part 1 - Do Carmo's Riemannian Geometry

Proposition 3.6 Let $p \in \mathcal{M}$, $U$ a normal neighborhood of $p$, $B \subset U$ a normal ball of center $p$. Let $\gamma : [0,1] \to B$ be a geodesic segment with $\gamma(0) = p$. If $c : [0,1]\to \mathcal{M}$ is any piecewise differentiable curve joining $\gamma(0)$ to $\gamma(1)$ then $l(\gamma) \leq l(c)$ and if equality holds then $\gamma([0,1]) = c([0,1])$.

For the proof

Suppose initially that $c([0,1]) \subset B$. Since $\exp_p$ is a diffeomorphism on $U$, the curve $c(t)$, for $t \neq 0$, can be written uniquely as $\exp_p (r(t) \cdot v(t)) = f(r(t),t)$ where $t \to v(t)$ is a curve in $T_p\mathcal{M}$ with $\left| v(t) \right| = 1$ and $r : (0,1] \to \mathbb{R}$ is a positive piecewise differentiable function.

The first question is why can $c(t)$ be written in such a way?
My first guess is because in $B$ by definition of exponential map there's a unique geodesic such that

$$
\exp_p(v) = \alpha(1,p,v)
$$

And I can write $v$ in the form stated in the theorem and the equation

$$
c(t) = \exp_p(v)
$$

Is well defined.

Carrying on with the prof

It follows that, except for a finite number of points,
  $$
\frac{dc}{dt} = \frac{\partial f}{\partial r} r'(t) + \frac{\partial f}{\partial t}
$$

I think it's clear the rule applied is essentially the chain rule. However I do struggle to derive the formula using the definition of differentials in manifolds (rigorously).

My attempt was to decompose $f$ as

$$
t \to (r(t),t) \to f(r(t),t)
$$

So $f = f_2 \circ f_1$, where $f_1 : \mathbb{R} \to \mathbb{R}^2$ and $f_2 : \mathbb{R}^2 \to \mathcal{M}$, for $f_1$ the differential is simply a derivative wrt $t$ componentwise. For $f_2$ I would assume I can use

$$
d f_{2_{(r,t)}} = \left[ d f_{2_{(r,t)}} \left( \frac{\partial}{\partial r} \right) \; d f_{2_{(r,t)}} \left(\frac{\partial}{\partial t} \right) \right]
$$

And using the author notation I'll end up with the same expression, the question is whether my derivation is correct.

Finally

From the Gauss lemma, $\left\langle \frac{\partial f}{\partial r}, \frac{\partial f}{\partial t} \right\rangle = 0$

I'd assume from the Gauss Lemma more specifically it follows that

$$\left\langle \frac{\partial f}{\partial r}, \frac{\partial f}{\partial t} \right\rangle = \left\langle \frac{\partial}{\partial r}, \frac{\partial}{\partial t} \right\rangle$$

And the two vectors $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial t}$ and these are orthogonal.

Again, is this correct?
Very last bit

Since $\left| \frac{\partial f}{\partial r} \right| = 1$, ...
  I'm not quite sure I understand why is 1, could you explain?

I'm still reading through the rest of the proof it seems ok, but I'll potentially ask a different question.

Also just to clarify shouldn't the expression
$$
\frac{dc}{dt} = \frac{\partial f}{\partial r} r'(t) + \frac{\partial f}{\partial t}
$$
actually be
$$
\frac{dc}{dt} = r'(t) \frac{\partial f}{\partial r} + \frac{\partial f}{\partial t}
$$

Mostly because $r'(t)$ is considered an element of a scalar field, while $\frac{\partial f}{\partial r}$ is a vector.

Thank you

答:
In the following, we will consider $c$ as smooth as possible everywhere, so everything is well defined.

As $\exp_p : U_p \to \exp_p(U_p) \subset M$ is a diffeomorphism, if $c : [0,1] \to \exp_p(U_p)$ is a smooth curve, there exist a unique curve $\tilde{c} : [0,1] \to U_p$ such that
$$
\forall t \in [0,1],~ c(t) = \exp_p(\tilde{c}(t)).
$$
Now, as $\tilde{c}$ is a path in $U_p \subset T_pM$, by polar decomposition, one can write $\tilde{c}(t) = r(t) \cdot v(t)$ where $r(t) \geqslant 0$ and $\|v(t)\|_{g_p} = 1$. The author seems to chose a little-bit complicated notation, but it is clear that he uses polar decomposition.

Now, write $c(t) = \exp_p(r(t)\cdot v(t)) = f(r(t),t)$, where $f$ is the function $f(r,t) = \exp_p(rv(t))$ when $r$ is considered as a variable. Then
$$
c'(t) = \left.\frac{\mathrm{d}}{\mathrm{d}s}\right|_{s=0}f(r(t+s),t+s),$$
and using the right-way the chain-rule says
$$
c'(t) = \mathrm{d}f(r(t),t)\cdot( r'(t),1),
$$
where $(r'(t),1)$ is the tangent vector to the curve $s\mapsto (r(s),s)$ at $s=t$. Hence, by the very definition of partial derivatives in differential geometry:
$$
c'(t) = \dfrac{\partial f}{\partial r}(r(t),t) \cdot \left(r'(t),0 \right) + \frac{\partial f}{\partial t}(r(t),t) \cdot (0,1),
$$
which we write for the sake of compactness
$c'(t) = \frac{\partial f}{\partial r}\cdot r'(t) + \frac{\partial f}{\partial t} $. We put the $r'(t)$ at the right because we think of $r'(t)$ as a vector, tangent to the curve $s \mapsto r(s)$. Indeed, $r'(t)$ is just a scalar because $r$ takes value in an interval, so you can move it to the left. Basically, you can do whatever you want with these kind of expressions if you know exactly what are the objects.

Now, recall the general Gauss lemma:

If $(M,g)$ is a Riemannian manifold, if $p\in M$, if $t\geqslant 0$ and if $u,v \in T_pM$, then
$$
g_{\exp_p(tu)}\left(\mathrm{d}\exp_p(tu)\cdot u, \mathrm{d}\exp_p(tu)\cdot v\right)=g_p\left(u,v\right).
$$

Looking closely to $f(r,t) = \exp_p(r\times v(t))$, we find that
\begin{align}
\frac{\partial f}{\partial r}(r,t) &= \mathrm{d}\exp_p(r\times v(t))\cdot v(t) ,\\
\frac{\partial f}{\partial t}(r,t) &= \mathrm{d}\exp_p(r\times v(t))\cdot r\times v'(t).
\end{align}
Hence, by the Gauss lemma, we have
$$
g_{\exp_p(rv(t))}\left(\frac{\partial f}{\partial r}(r,t),\frac{\partial f}{\partial t}(r,t) \right) = g_p(v(t),r\times v'(t)) = rg_p(v(t),v'(t)).
$$
As $t \mapsto v(t)$ is a curve that lies on the unit sphere, $v'(t)$ is orthogonal to $v(t)$, and then, the above scalar product is equal to zero.

Finally, note that the function $f(\cdot,t) : r \mapsto \exp_p(r\times v(t))$ is actually a geodesic. Hence, all its tangent vectors have the same length. At $r=0$, the tengent vector is $v(t)$, which is of length $1$ by definition of polar decomposition. At $r >0$, the tangent vector to the curve is $\partial f /\partial r$. We can conclude that $\left\|\frac{\partial f}{\partial r}\right\| = 1$.

Comment. The function $f$ is defined so that we can write the equation $c'(t) = \frac{\partial f}{\partial r}\cdot r'(t) + \frac{\partial f}{\partial t} $, which tells us, together with the fact that the partial derivatives of $f$ are orthogonal, that we have a natural decomposition of $c'(t)$ into a radial component ($\partial_rf \cdot r'$) and a component orthogonal to geodesic spheres ($\partial_t f)$.


问:
Alternative notation for maximum of a vector.
Do you know of a one symbol notation for $\max(\mathbf{x}), \mathbf{x}\in\mathbb{R}^n$ ?

For instance the mean is often denoted with an overline: $\overline{\mathbf{x}}$.

Alternatively, is the maximum associated to a letter or symbol, like the letter $\mu$ that often denotes the mean ?

答:
There is the $\ell_\infty$ norm, which coincides with the maximum if the vector has nonnegative entries:
$$
\| x\|_\infty = \max_{1\leq k\leq n} |x_k|
$$

For the maximum of a statistical sample (of size $n$) specifically, you are looking for the $n$-th order statistic:
$$
X_{(n)} = \max\{X_1,\dots,X_n\}
$$


问:
Proof of $ \arg (G(j \omega))=-\omega\left(t_{2}-t_{1}\right) $
We have this formula:
$$
\displaystyle \arg (G(j \omega))=-\omega\left(t_{2}-t_{1}\right)
$$

(We understand the amplitude of course, please help us to find a proof for the phase shift)

without a proof (control theory course).
Can someone remind us how it is calculated?

答:
It was not that hard in the end!

We consider the time points when the two curves passes 0:

$$
\begin{aligned}
\sin \left(\omega t_{1}\right) &=0 \\
K \sin \left(\omega t_{2}+\phi\right) &=0
\end{aligned}
$$

$$
\Longrightarrow \omega t_{1} = \omega t_{2}+\phi \\
\phi = \omega t_{1} - \omega t_{2} \\
\phi = -\omega(t_2-t_1)
$$


问:
Determinator of matrix operator
find the determinant of the operator $K[x]_{\leq n} \to K[x]_{\leq n}$ specified by the formula:
$f(x) \rightarrow f(x) + x \cdot f'(x+1)$

I've tried to solve it, but the binomial coefficients came out and it made me scared.

答:
Apply the map on the canonical basis $\;\{1,x,...,x^n\}\;$ of $\;K_n[x]\;$ to find out a nice matrix representation of the linear map $\;Tp(x):=p(x)+xp'(x+1)\;$:

$$\begin{cases}T1=1+x\cdot0=1=&1\cdot1+&0\cdot x+\ldots+\ldots+0\cdot x^n\\{}\\
Tx=x+x\cdot1=2x=&0\cdot1+&2x+\ldots+\ldots+0\cdot x^n\\{}\\
Tx^2=x^2+2x(x+1)=&0\cdot1+&2\cdot x+3\cdot x^2+\ldots+0\cdot x^n\\{}\\
..................................\\{}\\
Tx^n=x^n+nx(x+1)^{n-1}=&0\cdot1+&nx+n(n-1)x^2+\ldots+(1+n)x^n\end{cases}$$

The above is a lower triangular matrix $\;(n+1)\times(n+1)\;$ , so its determinant is easy to evaluate...(though the actual matrix is the transpose of the above, but since $\;\det A=\det(A^t)\;$ for any square matrix $\;A\;$ , we've no problem).

Of course, this is in case you meant "Determinant" ...

Title: Determinator of matrix operator


问:
Question to find all strictly increasing $f:\mathbb{N}\to\mathbb{N}$ mapping only primes to primes and satisfying $f(2n)=f(n)+n$ for all $n$

Let $f:\mathbb{N}\to\mathbb{N}$ be a function such that

$f(m)<f(n)$ whenever $m<n$;

$f(2n)=f(n)+n$ for all natural numbers $n$;

$n$ is a prime number whenever $f(n)$ is a prime number.

Find all functions $f$ satisfying above conditions with proof.

I have noticed that the identity function is a solution. I am however, unable to prove this result. To start off I am unable to prove $f(1)=1$, so I'm unable to proceed by induction. I am also not able to effectively use the third condition of the question. I request somebody to please provide a solution for this. Any help is much appreciated. Thanks a lot :)

答:
We will prove $f(1)=1$, hence by induction you can take it further to show that $f(n)=n ~\forall n \in \mathbb{N}$

Observe that $$f(n)=f(1)+n-1$$ satisfies the conditions of the functional equation. Now, suppose $f(1)=k>1$.

Next observe that all of the numbers $(k+1)!+2,(k+1)!+3,\cdots,(k+1)!+(k+1)$ are composite numbers (why?). Now, if $p$ is the smallest prime greater than $(k+1)!+(k+1)$, then plugging $n=p-k+1$ yields $f(n)=p$. Therefore, $n$ is a prime number. Now because $k>1$, we have $n=p-k+1<p$ and we conclude $(k+1)!+2<n<p$.

Now note that $p$ was defined to be the smallest prime exceeding $(k+1)! + (k+1)$, and since $(k+1)!+2,(k+1)!+3,\cdots,(k+1)!+(k+1)$ are all composite, means it is also the smallest prime exceeding $(k+1)!+1$. But $n$ was supposed to also be a prime less than $p$ but more than $(k+1)! + 1$. This contradicts that $p$ is the smallest prime exceeding $(k+1)!+(k+1)$ which implies it is impossible to have $f(1)>1$.


问:
File process calculation by time and amount processed
I am trying to get a good formula to calculate the time left for file processing. In 1 hour 10.3 GB of files can be processed. How many hours would it take to process the 45 GB files?</p>

答:
$$45\text{ GB} \times \frac{1\text{ hour}}{10.3\text{ GB}} \;\;\approx\;\; 4.369\text{ hours} $$

Assuming the amount of time it takes to process $n$ GB of files in linear in $n$, then $\frac{1\text{ hour}}{10.3\text{ GB}}$ is our "unit" of converting between GB and hours.


问:
Combinatorics and logic equivalences
I've seen the following question and I wondered about whether or not the change I did is valid, the question is:

"For how many numbers $i$ when $1 \leq i \leq 120$, the following statement holds:

'if $i$ is not divisible by at least one of the numbers 3,5 then is divisible by 7' "

Can I use the logical equivalence that says : $p \rightarrow q \equiv \lnot p \lor q$ ?

And to substitute the statement into "$i$ is divisible by 3 and 5 or divisible by 7" ?.

I'll appreciate any help !

答:
"If $i$ is not divisible by at least one of the numbers 3,5 then is divisible by 7"

Can I use the logical equivalence [...] to substitute the statement into "$i$ is divisible by 3 and 5 or divisible by 7"?

No, this is not logically equivalent.

"Not divisible by at least one of 3 or 5" means "(not divisible by 3) and (not divisible by 5)", and negates to "divisible by 3 or divisible by 5".

Title: Combinatorics and logic equivalences


问:
$\sum_{j=1}^{n} f\left(\frac{j}{n}\right) \cdot \chi _{\left[\frac{j-1}{n},\frac{j}{n}\right)} \longrightarrow f$ converges uniformly
Let $f:[0,1]\longrightarrow \mathbb{R}$ be a continuous function

and  $\displaystyle f_n=\sum_{j=1}^{n} f\left(\frac{j}{n}\right) \cdot \chi _{\left[\frac{j-1}{n},\frac{j}{n}\right)}$

where $\chi _{\left[\frac{j-1}{n},\frac{j}{n}\right)}$ is the characteristic function of ${\left[\frac{j-1}{n},\frac{j}{n}\right)}$

How can we prove that $f_n \longrightarrow  f $ uniformly ?

Any hints would be appreciated.

答:
Sure. $f$ is uniformly continuous on $[0,1]$, so, given $\epsilon>0$, there is ... (text hidden)

 $\delta>0$ so that whenever $|x-x'|<\delta$, we have $|f(x)-f(x')|<\epsilon$. Just pick $N$ so that $1/N<\delta$. Then it should follow that whenever $n\ge N$, we have $|f_n(x)-f(x)|<\epsilon$ for every $x\in [0,1]$. The point is that $f_n(x)=f(j/n)$ where $x\in [\frac{j-1}n,\frac jn)$, and $|x-j/n|<\delta$.


问:
$f: M \to \mathbb R$, smooth, $\varphi_t$ 1 parameter group of diffeomorphisms, then $df(\varphi_t(x))/dt = 1$
I will first introduce some preliminaries and notation. The question is highlighted below.

$M$ smooth manifold of dimension $N$, $f:M \to \mathbb R$ smooth.

Fact: a given vector field $\underline v$ on $M$ which vanishes outside a compact subset of $M$ generates a $1$-parameter group of diffeomorphisms $\varphi_t(x)$ which is the solution of
$$
\frac{d \eta}{dt} = \underline v(\eta(t)), \quad \eta(0) = x.
$$

We can construct a Riemannian metric on $M$, which gives a notion of inner product in the tangent spaces and allows for the definition of the gradient vector:
$$
\langle v, \text{grad} f \rangle = df(v).
$$

Consider $f^{-1}([a, b])$ a compact set without critical points of $f$. We can define a map $\rho: M \to \mathbb R$ given by
$$
\rho = \frac{1}{\langle \text{grad} f, \text{grad} f \rangle}
$$
in $f^{-1}([a, b])$ and $0$ outside a compact neighborhood of $f^{-1}([a, b])$. Then $\underline v = \rho \text{grad}f$ generates a $1$-parameter group of diffeomorphisms.

The book presents the following computation:
$$
\frac{df(\varphi_t(x))}{dt} = \left\langle \frac{d\varphi_t(x)}{dt}, \text{grad}f \right \rangle = \langle x, \text{grad}f \rangle = 1.
$$
Clearly there is a typo in the third member of the chain of inequalities. In order for us to get $1$ in the right-hand side, the third member should be $\rho\langle \text{grad}f, \text{grad}f \rangle$. However, as I understand, we only have
$$
\frac{d\varphi_t(x)}{dt} = \rho \text{grad}f(\varphi_t(x)),
$$
and we would obtain
$$
\rho \langle \text{grad}f(\varphi_t(x)), \text{grad}f \rangle.
$$
Please, what am I missing?

This is in pages 64/65 of Damascelli and Pacella's Morse Index of Solutions of Nonlinear Elliptic Equations, and appears in the proof that we can deform a sublevel into another if there is not critical level between them.

Thanks in advance.

答:
The problem is that the author has chosen not to write the variables in the equality. This is more pleasant to read but we can be confused while theses equalities occure. Conversely, writting everything is sometimes harsh.

I will denote by $g$ the Riemannian metric of $M$: if $u$ and $v$ are tangent to $M$ and $p$, then their scalar product is denoted by $g_p(u,v)$ instead of $\langle u, v\rangle$, so we keep the information on where do these vectors live. The gradient of a function $h$ is then defined thanks to the relation
$$
\forall v \in T_pM, \mathrm{d}h(p)\cdot v = g_p\left(\mathrm{grad}h(p),v \right)
$$

If $f : M \to \mathbb{R}$ is smooth, and if $v = \rho \mathrm{grad}f$ is the vector field you defined above, then
\begin{align}
\frac{\mathrm{d} f\left(\varphi_t(x)\right)}{\mathrm{d}t} &= \mathrm{d}f\left(\varphi_t(x)\right)\cdot \left(\frac{\mathrm{d}}{\mathrm{dt}}\varphi_t(x)\right) \\
&=g_{\varphi_t(x)}\left(\mathrm{grad}f(\varphi_t(x)), \frac{\mathrm{d}}{\mathrm{d}t}\varphi_t(x) \right) \\
&= g_{\varphi_t(x)}\left(\mathrm{grad}f(\varphi_t(x),\rho(\varphi_t(x))\mathrm{grad}f(\varphi_t(x)) \right) \\
&= \rho(\varphi_t(x)) g_{\varphi_t(x)}\left(\mathrm{grad}f(\varphi_t(x)),\mathrm{grad}f(\varphi_t(x)) \right) \\
&=1.
\end{align}
As you can see, the notations here are a little bit unpleasant, but we find the right proof.


问:
Question to find all functions satisfying the functional equation $ f \left( f ( m ) ^ 2 + f ( n ) ^ 2 \right) = m ^ 2 + n ^ 2 $
Find all $f:\mathbb{N_{0}}\to\mathbb{N_{0}}$ satisfying
$$f(f(m)^{2}+f(n)^{2})=m^{2}+n^{2}\text{, for all }m,n\in\mathbb{N_{0}}$$

Just to clarify, the set $\mathbb{N_{0}}$ refers to the set of non negative integers. Getting back to the question, I was able to prove that $f$ is injective and I also noticed that the identity function is a solution. However, I am not able to complete my proof. I am also unable to find the initial values at 0,1 to try and proceed by induction. I request somebody to provide a solution for this. Any help is much appreciated :)

 Here is how I proved injectivity of the function. If $f(m)=f(n)$, then
$$f(f(m)^{2}+f(n)^{2})=f(f(m)^{2}+f(m)^{2})$$
$$\Rightarrow m^{2}+n^{2}=m^{2}+m^{2}$$
$$\Rightarrow m=n$$
We are ignoring negative value due to range of the function.

答:
$f$ is injective:
$f(n)=f(m) \to m^2+n^2=f(f(m)^2+f(n)^2)=f(f(m)^2+f(m)^2)=m^2+m^2 \to m=n$

$f$ is not constant: suppose $f(m)=c$ for all $m\in\mathbb{N}_0$. Then $f(2c^2)=m^2+n^2=c$ for all $m,n$. Absurd.

Now, the condition implies $f[f[f(m)^2+f(n)^2)]]=f(m^2+n^2)$.

Putting $m'=f(m)$ and $n'=f(n)$ we have

$f[f[f(m')^2+f(n')^2)]]=f(m'^2+n'^2)=f(f(m)^2+f(n)^2)=m^2+n^2$

Because $f$ is injective, and we have 2 values with image $m^2+n^2$ then:

$f[f(m')^2+f(n')^2)]=f(f(m)^2+f(n)^2)$

Again, because $f$ is injective

$f(m')^2+f(n')^2)=f(m)^2+f(n)^2$

$f(m')^2-f(m)^2=f(n')^2-f(n)^2$

$[f(m')-f(m)][f(m')+f(m)]=[f(n)-f(n')][f(n)+f(n')]$

Both sides of the equality depend on different variables, so it must be a constant. Suppose this constant is not 0. But the equality is in $\in\mathbb{N}_0$, so we know that $f(m')+f(m)\geq f(m)$ would be bounded, absurd since $f$ is injective.That is, the product $[f(m')-f(m)][f(m')+f(m)]$ must be $0$.

Now, $f(m')+f(m)=0 \to f(m)=0 \text{ and } f(m')=0 \to m'=m$. If $f(m')-f(m)=0 \to f(m')=f(m) \to m'=m$. So, $m'=m$ for all $m$, ie $f(m)=m$.


问:
Find the flux for the vector field $F =(x^2,y^2,z^2)$
Find the flux for the vector field $\vec{F}=(x^2,y^2,z^2)$ across the boundary to the ball given as $$(x-1)^2 +(y+1)^2 +(z-2)^2 \le 4$$

Edit: How much I've done: The center of the sphere is $(1,−1,2)$ (I've sketched it), thus $r(θ,ϕ)=(1+cosθsinϕ, −1+sinθsinϕ, 2+cosϕ)$ . Then to find the normal vector $\vec{N}$ , I've calculated $\frac{∂r(θ,ϕ)}{∂ϕ}=(cosθcosϕ,sinθcosϕ,−sinϕ)$ and $\frac {∂r(θ,ϕ)}{∂θ}=(−sinθsinϕ,cosθsinϕ,0)$ then $\frac{∂r(θ,ϕ)}{∂ϕ} \times \frac{∂r(θ,ϕ)}{∂θ}$, and finally got $(sin2ϕ,sinθsin2ϕ,sinϕcosϕ)$. Am I on the right track?

答:
The Divergence Theorem is absolutely the right approach, as all the hypotheses of the theorem hold. You should always prefer to do a volume integral rather than a surface integral. In this case, you might also recognize that
\begin{multline*}
\iiint_K x\,dV = \bar x\, \text{volume}(K), \quad \iiint_K y\,dV = \bar y\, \text{volume}(K), \quad\text{and} \\ \iiint_K z\,dV = \bar z\, \text{volume}(K),
\end{multline*}
where $(\bar x,\bar y,\bar z)$ is the centroid (center of mass) of $K$.
When $K$ is a ball (with uniform density), we all know that its centroid is at its center. Indeed, I would comment that multivariable calculus students should always be on the lookout to apply symmetry arguments.


答:
Since $y=4-\frac{3x}{2}$, you can express the second equation in terms of $x$ only by substituting $y$ and then solving the resulting quadratic. So we have

$$x^2-2y^2+xy=x^2-2(4-\frac{3x}{2})^2+x(4-\frac{3x}{2})$$
$$=x^2-2(16-12x+\frac{9x^2}{4})+4x-\frac{3x^2}{2}=x^2-\frac{9x^2}{2}-\frac{3x^2}{2}+24x+4x-32$$
$$=-5x^2+28x-32=4$$

and rearranging gives

$$5x^2-28x+36=(5x-18)(x-2)=0$$

etc.

Title: Difficult Quadratic Simultaneous Equations

Body:
I tried solving this system of equations multiple ways: $3x+2y=8, \, x^2-2y^2+xy=4$

I tried substituting $x$, substituting $y$, equating the two by making the second equation equal to 8, but the closest I got was by breaking down the $-2y^2$ to $-y^2-y^2$:

From first equation, $y=4-\frac{3x}{2}$.
$$x^2-y^2-y^2+xy=4$$
$$(x+y)(x-y)-y(x-y)=4$$
$$(x-y)(x+y-y)=4$$
Then substitute y for the value found in first equation. This gave me $x=4$ or $x=16/5$.

The actual answer is $x=2, y=1$ or $x=18/5, y=-7/5$

How do I solve it step by step?


问:
Why is height of a complete binary tree O(log n)?
A complete binary tree has $n$ nodes. Then for height $h$, $n=2^h-1$. At every “layer” of height, the number of nodes gets doubled. So we have $1+2^{(2-1)}+2^{(3-1)}+...+2^{(h-1)}=2^h-1$ nodes, for height $h$.
Thus $h=log_2(n+1)$. I understand this.
But how does inserting a new element into this take time $O(log_2n)$ and not $O(log_2(n+1))$?

答:
Because
$f(n)
\in O(\log_2(n))
\iff
f(n)
\in O(\log_2(n+1))
$,
and it is convenient
to use the simpler form.


问:
If $X \mapsto X^tG$ is an isomorphism, then $G$ is an invertible matrix. (Bilinear forms)
Let $f : E \times F \to R$ be a bilinear form that is non-singular on the left, where $E, F$ are free $R$-modules of dimension $n$ (both of them).

Then if $X, Y$ are column vectors for general elements $x,y \in E, F$ respectively.  Then the value $f(x,y)$ can be written as:

$$
f(x,y) = X^t G Y
$$

Lang's Algebra on page 529 says that $X \mapsto X^t G$ is an isomoprhism (by non-singularity), and that isomorphic here implies that $G$ is invertible.  How do we prove that $G$ is invertible, since things are unfamiliar from the usual way of matrix multiplying, i.e. $a(x) = AX$?

$G$ is an $n\times n$ matrix with entries in $R$.

答:
You can revert back to a familiar way of matrix multiplication by exploiting the transpose operator as follows.

Define $\phi:R^{n \times 1} \rightarrow R^{1\times n}$ and $l:R^{n \times 1} \rightarrow R^{n \times 1}$ by $$\phi(X)=X^tG$$ $$l(X)=G^tX$$ Now $\phi(X)^t=l(X)$ for all $X\in R^{n\times 1}$ so $l$ is also an isomorphism. For each $1 \leq i \leq n$ find a unique $X_i\in R^{n\times 1}$ such that $$l(X_i)=G^tX_i=e_i$$ Such an $X_i$ exists by isomorphism. If $M$ is the matrix whose $i^{\text{th}}$ column is $X_i$ then its easy to see that $(G^t)^{-1}=M$ hence $G^t$ is invertible and so is $G$.


问:
Proof by contradiction and invalid reasoning
I am very aware of the several posts that are in this forum, but to be honest I am not quite satisfied with the provided answers. Also, I can tell that the best post is the following:

Proof by Contradiction, Circular Reasoning?

I would love to continue the discussion in that post, however my reputation level does not allow me to add any comments, and I still have some questions. Therefore, I want to continue discussing more about proof by contradiction and circular reasoning.

Suppose we wish to prove $(P \wedge Q) \rightarrow R$. We assume P and Q. Proof by contradiction begins by assuming $\neg R$, and from these two assumptions, a "contradiction" is derived, and sometimes that contradiction is that $\neg P$ or $\neg Q$ (our initial valid assumptions). Therefore, since there is a contradiction with our assumptions, we conclude that it cannot happen that $\neg R$, so it must be the case that $R$.

This proof strategy bothers me, because I have the feeling that there is some kind of circular (or invalid) reasoning going on there i.e. $(P \wedge Q) \rightarrow R$ is equivalent to $\neg R \rightarrow \neg(P \wedge Q)$, so if we already know that the consequent is false, in order for the statement to be true, $\neg R$ has to be false, so R is true, but R is part of the consequent part of the conditional statement (isn't doing this like assuming there is a biconditional i.e. because R is true, then $P \wedge Q$ is true?). I probably don't know how to explain it better, but how/why can we say that this is indeed a proof? Hence, my main question is why it is valid to assume $\neg R$ if our conclusion involves R? I have one explanation that kind of makes sense to me (still a little iffy though):

In more simple terms, where we want to prove that $P \rightarrow Q$, we are trying to do the following: $[(P \wedge \neg Q) \rightarrow Contradiction] \rightarrow (P \rightarrow Q)$ if we can prove that $(P \wedge \neg Q)$ leads to a Contradiction, then we are indeed proving that $P \rightarrow Q$, which was our original goal. Given that this is a tautology, whenever we follow this proof strategy structure, we are always proving $P \rightarrow Q$. However, saying that this is a tautology kind of bothers me for some reason.

I would sincerely appreciate your insights, and if there is anything you would improve in my post, please feel free to do so. I'm starting my formal journey in the proof realm (currently reading How To Prove It), and I have always heard that proofs by contradiction were the easiest strategy. However, I have learned that this strategy seems to be difficult at the beginning for all those who are trying to learn proof strategies.

答:
Circular reasoning is when you assume what you want to prove, not when you assume what you want to refute and derive a contradiction. The difference is crucial.

If we allow circular reasoning as I've defined it, we can "prove" anything as concisely as "$p$, therefore $p$". In particular, we can just as easily "prove" $p$ is true this way as prove it's false.

By contrast, a proof by contradiction can only succeed if what we assume turns out to imply a contradiction, at which point the proof is able to finish. You can quickly derive a contradiction by assuming $\sqrt{2}$ is rational. If you ever manage to derive one by assuming it isn't, let us know, because you'll have found an unknown inconsistency in whatever standard mathematical axioms you've started from.


问:
Probability that a limit exists is 0
Let $(X_{n})_{n>0}$ be i.i.d. real valued random variables and suppose $X_{1}$ is not constant with probability 1.

Show that: P($\lim_{n \to \infty} X_{n}$ exists)=0

My first idea was to let A be an arbitrary real number and to show that P($\lim_{n \to \infty} X_{n}=A$)=0.In order to do so I wanted to rewrite it to a form where I could apply Borel Cantelli however I can't seem to figure out how to rewrite it to such a form. Could I get help with this question? Thanks in advance!

答:
By the assumption on $X_1$, we can find   a real $s$ such that $0<P(X_1 \le s)<1$. Then there must exist $t>s$ such that $P(X_1>t)>0$.
Apply Borel-Cantelli to the events $\{ X_n \le s\}$ to conclude that they occur infinitely often almost surely.
Apply Borel-Cantelli to the events $\{ X_n >t\}$ to conclude that they also occur infinitely often almost surely.

Title: Probability that a limit exists is 0


答:
Just because they say it's $O(\log_2(n))$, that doesn't mean it can't also be $O(\log_2(n+1))$. In fact, we have
$$
O(\log_2(n)) = O(\log_2(n+1))
$$
and the two complexity categories are entirely equal. We see this because on one hand we have
$$
f(n)\in O(\log_2(n))\implies f(n)\leq c\cdot \log_2(n)\leq c\cdot\log_2(n+1)\\
\implies f(n)\in O(\log_2(n+1))
$$
(where $c$ is some constant), while on the other hand we have
$$
f(n)\in O(\log_2(n+1))\implies f(n)\leq d\cdot\log_2(n+1)\leq d\cdot 2\log_2(n)\\
\implies f(n)\in O(\log_2(n))
$$
(where $d$ is some constant). Since the two are equal, the simpler alternative is often preferred over the more transparent option.

Title: Why is height of a complete binary tree O(log n)?

Body:
A complete binary tree has $n$ nodes. Then for height $h$, $n=2^h-1$. At every “layer” of height, the number of nodes gets doubled. So we have $1+2^{(2-1)}+2^{(3-1)}+...+2^{(h-1)}=2^h-1$ nodes, for height $h$.
Thus $h=log_2(n+1)$. I understand this.
But how does inserting a new element into this take time $O(log_2n)$ and not $O(log_2(n+1))$?


问:
Is this kind of mistake I made in my bachelor thesis a serious thing?
A couple of days ago I handed in my bachelor thesis. Unfortunately I found a mistake in two of my definitions. This definitions were recursive ones so I forgot to write $n\in\mathbb{N}\setminus\{1\}$ instead of $n\in\mathbb{N}$. The definitions don't make sense for $n=1$. Would this be a big thing for my professor ? I just can't understand why some mistakes kind of seem impossible to find until you hand in your bachelor thesis... it drives me crazy.

答: Don't worry about it. Your thoughts and ideas are the substance of your thesis. Spelling, grammar, and off-by-one errors are unavoidable side-effects of recording your thoughts and ideas in writing. We only try to minimize those side-effects as courtesy to readers anyways, and we (or editors or reviewers) help stamp those errors out if/when the writing gets published.</p>


问:
Understanding why path $p*p^{-1}$ is homotopic to the identity/trivial loop
I saw this question, but the top answer has the same picture that is confusing me currently, so I was wondering if somebody could explain it to me. I'm reading Concepts of Modern Mathematics by Ian Stewart, and in it he is trying to define an inverse for paths so we can construct a group. He does this as follows:

If $p^{-1}$ is the path $p$ in reverse, then although $p*p^{−1}$ is not equal to the trivial loop, it is homotopic. As in Figure 139, we can shrink $p*p^{−1}$ gradually back towards the base point, at the same time going round it faster and faster.

Here's what I don't understand. Suppose $q = p*p^{-1}$:

Are there only inverses for paths that start and end at the same point? What about a path that starts at point $a$ and ends at point $b$ such that $a \neq b$, this won't be homotopic to the trivial loop

As $q$ is "shrinking", it is not covering the full distance, and instead of traveling from the initial point to some point that is not the initial point before returning. Doesn't this mean we're not dealing with another path, e.g. $r*r^{-1}$, in the second drawing, and another path in the third, etc. as it shrinks?

EDIT: Upon rereading the text, I realized that the author was trying to get to the fundamental group, which is why he works with loops. He explained homotopy by working with general paths, not just loops, and this is what got me confused since I didn't realize the general paths were only mentioned for this brief explanation, and most of the rest of the discussion involved loops. This document has a good explanation of why loops are used, saying that:

By restricting attention to loops with a fixed basepoint $x_0 \in X$, we guarantee that the product $fg$ of any two such loops is defined.

答:
You're correct that you can talk about inverses without assuming that your paths are loops. Let me introduce some notation to clarify this. For $a, b \in X$ I will denote $p: a \longrightarrow b$ to mean that $p$ is a path starting at $a$ and ending at $b$. Given a path $p: a \longrightarrow b$, I will also denote its homotopy class as $[p]$. More specifically, to say $[p] = [q]$ I mean that there is a homotopy $H: I^2 \longrightarrow X$ such that $H(0, s) = p(s)$, $H(1, s) = q(s)$ for all $s$ and such that $H$ fixes the endpoints, so $H(t, 0) = a$ and $H(t, 1) = b$ for all $t$. This fixed endpoint condition means that for all $t$, the path $s \mapsto H(t, s)$ is a path $a \longrightarrow b$. Thus, it is well defined to say that $[p] : a \longrightarrow b$, as the fixed endpoint condition means that if $[p] = [q]$ then $q$ has the same endpoints $a \longrightarrow b$.

With that jargon aside, the point is that if we have paths $p: a \longrightarrow b$, $q: b \longrightarrow c$ then we can concatenate them to get $p*q: a \longrightarrow c$. We can also define the reverse path $p^{-1}: b \longrightarrow a$. Notably, concatenation is well defined on homotopy classes! That is, we can define the product $[p] * [q]$ as $[p * q]$ and have it be well defined. For $a \in X$, we will also denote $c_a: a \longrightarrow a$ as the constant path at $a$.

Now, with all of these definitions set up, we can prove that $[p] * [p^{-1}] = [c_a]$ for $p: a \longrightarrow b$. And we need not insist $a = b$ at all for this statement to make sense. The construction here is exactly as you mentioned for loops. As $[p] * [p^{-1}] = [p * p^{-1}]$ we are trying to show that $p * p^{-1}$ and $c_a$ are homotopic with fixed endpoints. Here, of course, $p * p^{-1} : a \longrightarrow b \longrightarrow a$ and $c_a: a \longrightarrow a$ so $a \longrightarrow a$ are our endpoints. For $t \in [0, 1]$, we will denote $p_t = p|_{[0, t]}$. Then $p_t: a \longrightarrow p(t)$. We also have its reverse $p_t^{-1} : p(t) \longrightarrow a$. Then we are claiming that $H(t, s) = (p_t * p_t^{-1})(s)$ is a homotopy between $p * p^{-1}$ and $c_a$ which fixes the endpoints $a \longrightarrow a$. Certainly, $H(0, s) = p_0 * p_0^{-1}(s) = c_a(s)$ and $H(1, s) = p_1 * p_1^{-1}(s) = p * p^{-1}(s)$. Furthermore, for any $t$, the path $s \mapsto H(t, s)$ is the path $p_t * p_t^{-1}: a \longrightarrow a$. Thus, $H$ does indeed fix endpoints so $H$ witnesses $[p * p^{-1}] = [c_a]$. Again, note that we never needed the fact that $a = b$. If you're interested more in this idea of paths as "maps" $a \longrightarrow b$, you could look into the fundamental groupoid. It is essentially the object I defined here consisting of things like $[p]: a \longrightarrow b$. Restricting this "fundamental groupoid" to only consider loops $[p]: a \longrightarrow a$ yields the fundamental group.

To answer your second question, certainly it is the case that the "shrunken" $p * p^{-1}$ are all completely different maps. Here, I wrote them as $p_t * p_t^{-1}$. However, they all have the same endpoints $a \longrightarrow a$ and essentially, the map $t \mapsto p_t * p_t^{-1}$ is a continuous family of loops at $a$, i.e. a homotopy.


问:
Results on convergence in $r$th mean convergence
In the work I am interested in (primarily statistics), most of the results I have seen about stochastic convergence deal primarily with convergence almost surely, in probability and in distribution. In contrast, I have seen much less work aiming for convergence in $r$th mean. Of course, by Markov's inequality, convergence in $r$th mean implies convergence in probability. Since convergence in $r$th mean does not imply (nor is implied by) convergence almost surely, I'm trying to understand when I'd be interested in showing convergence in $r$th mean.

My understanding of convergence almost surely is that the set of events on which the convergence does not occur has measure $0$; I can't imagine a context in which I would seek out a different (and not weaker) notion of convergence.

I'm mostly looking for examples where convergence in $r$th mean is more relevant than convergence almost surely. By "more relevant" I mean that convergence in $r$th mean is the specific result of interest.

答:
Convergence in the $r$-th mean is often more convenient to prove mathematically than almost sure convergence.
Moreover, if you use Markov's inequality, you may get a rate of convergence for convergence in probability, if you can prove a rate for the convergence in $r$-th mean.

A simple illustrative example is the weak law of large numbers:
$$
\mathbb{P}\left[\left|\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]\right|>\varepsilon\right]
\leq \frac{1}{\varepsilon^2}\mathbb{E}\left[\left|\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]\right|^2\right]$$
The rest of the proof aims to show that $\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]$ converges in the second mean to $0$. This is done as follows:
$$
\leq \frac{1}{\varepsilon^2}\mathbb{E}\left[\frac{1}{n^2}\sum_{i,j=1}^n\left| X_i-E[X_1]\right|\left| X_j-E[X_1]\right|\right]
$$
Now we can use linearity of the expectation, which we couldn't do when working with $\mathbb{P}$ directly:
$$
=\frac{1}{\varepsilon^2}\frac{1}{n^2}\sum_{i,j=1}^n\mathbb{E}\left[\left| X_i-E[X_1]\right|\left| X_j-E[X_1]\right|\right]=\frac{1}{\varepsilon^2}\frac{1}{n}\mathbb{E}\left[\left| X_1-E[X_1]\right|^2\right]=:\frac{\sigma^2}{\varepsilon^2n}
$$

Some concluding remarks:

Proving the strong law of large numbers (almost sure convergence) is much harder.

Moreover, some proofs of the strong law use the weak law as a Lemma (e.g. the martingale argument).
So, implicitly, they use that $\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]$ converges in the second mean to $0$. Indeed, convergence in $r$-th mean can be used as a stepping stone to prove almost sure convergence! (even if the implication is not direct, and of course not always true).

The previous derivation gives you a speed of convergence, which the "almost surely" result doesn't.


问:
What does it mean when one says "$B$ has no limit point in $A$"?
If $B$ is a subset of a set $A$, what does the sentence "$B$ has no limit points in $A$" mean? I am aware that $x$ is a limit point of $A$, if for every neighbourhood $U$ of $x$, $(U-\{x\})\cap A$ is non-empty.

Please let me know. Thank you.

答:
It means that if $x$ is a limit point of $B$, then $x$ is not an element of $A$.  or conversely, if $x$ is an element of $A$, then it is not a limit point of $B$.

This would include the case where $B$ has no limit points at all.

Without more context it's hard to say why this might be important.


问:
The jump operator is well defined on the Turing degrees
This is in a way a continuation of my previous question: Basic questions on Turing degrees and the jump operator

There's this statement:

If $A\leq_T B$, then $K^A\leq_1 K^B$. It follows that the jump operator is well defined on the Turing degrees.

First, what does the statement "the jump operator is well defined on the Turing degrees" mean exactly? The jump operator was defined on sets: for a set $A$, the jump of $A$ is $A'=K^A=\{n\in\mathbb N: \phi_n^A(n)\downarrow\}$. There are no equivalence classes in this case to check if something is well defined.

There are equivalence classes in the definition of $0^{(k)}$. I would say that well definiteness of $0^{(k)} $ means the following: if $A\equiv_T B$ and $A$ lies in $0^{(k)}$, then $B$ also lies in   $0^{(k)}$. Is this what is meant? But even if so, I don't see how the implication  $A\leq_T B\implies K^A\leq_1 K^B$ proves this.

And second, how to prove that  $A\leq_T B\implies K^A\leq_1 K^B$? I suppose it's enough to construct a partial computable function $$V:N\times N\to N$$ such that if $n\in K^A$ then $V(n,n)$ is computable in $B$ and if $n\notin K$, then $V(n,n)$ is not computable in $B$. (After that one can apply the s-m-n theorem or something like that to get a total computable $N\to N$ that $m$-reduces $K^A$ to $K^B$). But I'm not sure how to do that.

答:
There are no equivalence classes in this case to check if something is well defined

Ah, but there are!

The statement "The jump operator is well defined on Turing degrees" is just an example of a more general type of statement: when we have a set $X$, a function $f:X\rightarrow X$, and an equivalence relation $E$ on $X$, we say $f$ is well defined on $X/E$ if we have $$xEy\implies f(x)Ef(y)$$ for all $x,y\in X$. Here, $X=\mathcal{P}(\mathbb{N})$, $E$ is Turing equivalence so that $X/E$ is the set of Turing degrees, and $f$ is the jump function.

The point is that when $f$ is well defined on $X/E$, we get an "analogue" of $f$ given by $$f_E: X/E\rightarrow X/E: [x]_E\mapsto [f(x)]_E$$ (note that well-definedness on $X/E$ is crucial here). Often we conflate $f$ and $f_E$; so e.g. we'll freely talk about the jump of a Turing degree even though that's an abuse of terminology.

Now as to how well-definedness on degrees is proved from the claimed implication, I suspect you're just overthinking this.

We know that $A\le_T B\rightarrow A'\le_1B'$ (my "$C'$" is your text's "$K^C$"). A fortiori then, we have $A\le_T B\rightarrow A'\le_TB'$ and so in particular we get $$A\equiv_TB\rightarrow A'\equiv_TB'$$ as desired (just use the previous implication in both directions).

Note that $\le_1$ here was overkill: we really only needed $\le_T$ in the conclusion of the result. But noting $1$-reductions when they exist is generally worthwhile if it doesn't take much additional work.

Finally, we're left with the task of proving that claimed implication, $A\le_TB\rightarrow A'\le_1B'$. I suggest first trying to show $A\le_TB\rightarrow A'\le_mB'$. The padding lemma will then let you convert that $\le_m$ into an $\le_1$.

So suppose I want to tell whether $\Phi_e^A(e)\downarrow$ (that is, whether $e\in A'$) and I have access to $B'$ as well as a method for getting $A$ from $B$. Consider a machine which, with oracle $B$, first constructs $A$ from $B$ to use as an oracle going forwards ...


问:
Can this system of congruences be solved?
\begin{equation}
  \left\{\begin{array}{@{}l@{}}
    2x\equiv7\mod9 \\
    5x\equiv2\mod6
  \end{array}\right.\,.
\end{equation}
Can this system of congruences be solved?
I notice that $(9,6) = 3 \ne 1$ so I can't apply the Chinese theorem of remainders, but this doesn't imply that it can't be solved, so I thought to rewrite the second equation in two different equations, like this:

\begin{equation}
  \left\{\begin{array}{@{}l@{}}
    2x\equiv7\mod9 \\
    x\equiv0\mod2 \\
    5x\equiv2\mod3
  \end{array}\right.\,.
\end{equation}

But the problem still here: $(9,3) = 3 \ne 1$, so it can be solved or is it impossible?

答:
Hint:

The system of congruences  $\begin{cases}
x\equiv \alpha\mod a,\\ x\equiv \beta \mod b,
\end{cases}$ has a solution if & only if  $\alpha\equiv \beta\mod\gcd(a,b)$.

So, solve each congruence independently from the other, and check whether the condition is satisfied.


问:
Measure of Intersection of Translate of Sets of Infinite Measure
Let $A,B\subset \mathbb{R}$ with $\mu(A)=\mu(B)=\infty$ and $\mu(A\cap B)=0$. Does there necessarily exist some $x\in \mathbb{R}$ for which $\mu(A\cap(B+x))=\infty$?

This seems likely as, if it did not hold, there would exist two sets of infinite measure so that the intersection of any translate has finite measure, which is admittedly counterintuitive, but not unthinkable... However, I'm not quite sure how to prove this.

答:
No. Take $A=(-\infty,0)$ and $B=(0,\infty)$. Then $B+x = (x,\infty)$ and $A\cap B+x = (x,0)$ if $x<0$ and empty otherwise. This set is always of measure $x<\infty$.

If instead you are looking for an example of $A$ and $B$ for which there exists such an $x$. Then take

$B=(0,\frac{1}{2})\cdot\mathbb{N}$ - all the numbers with fractional part smaller than half, $A=(\frac{1}{2},1)\cdot \mathbb{N}$ and $x=\frac{1}{2}$. Then $B+x=A$ and $A\cap B = \emptyset$.


问:
Teacher claims this proof for $\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$ is wrong. Why?
My son's high school teacher says his solution to this proof is wrong because it is not "the right way" and that you have to "start with one side of the equation and prove it is equal to the other". After reviewing it, I disagree. I believe his solution is correct, even if not "the right way", whatever that means. I asked my son how he did it: he cross-multiplied the given identity, simplified it to a known/obvious equality, and then reversed the steps for the proof. This was a graded exam, and the teacher gave him a zero for this problem.

What do you think about my son's solution? Thanks!

Problem: prove the following trigonometric identity
\begin{align*}
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}\ .\\
\end{align*}
Solution:  for all real $\theta$ not equal to an integer multiple of $\pi/2$, we have
\begin{align*}
\cot^2(\theta)&=\cot^2(\theta)\\[8pt]
\frac{\cos^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\frac{1-\sin^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\csc^2(\theta)-1&=\cot^2(\theta)\\[8pt]
\frac{\csc^2(\theta)-1}{\cot(\theta)}&=\cot(\theta)\\[8pt]
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}
\end{align*}

答:
There are multiple motivations for writing proofs: perhaps to pass an exam (from a regular student's viewpoint), or perhaps to test understanding (from a teacher's perspective). Here, I'd like to highlight another important motivation: to enjoy the process of discovery.

Here's a geometric proof of the trigonometric identity. It is not practical to use in an exam, but it makes the abstract symbols more concrete. Coming up with multiple proofs of the same result can help one develop intuition and allow one to appreciate the connections between different areas of mathematics.

Problem statement: $$\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$$

In this picture involving the unit circle, we construct some lengths which correspond to expressions in the problem statement. We start with $\theta$ and $\phi = \frac{\pi}{2} - \theta$.

Basic trigonometric definitions lead to the blue and orange lengths. By some angle-chasing (or a circle theorem), we have the $\color{red}{\text{red}}$ angle. Then, more angle-chasing gives:

The $\color{red}{\text{red}}$ angles are the same $\implies\triangle ADB\sim \triangle BDE \implies \frac{AD}{DB}=\frac{DB}{ED}\iff$ the problem statement.


问:
Lebesgue measure and continuity of a function.

Let $A,B$ be Lebesgue-measurable sets on $\mathbb{R}$ and $m$ be Lebesgue measure and $m(A)<\infty.$

Prove that $f(x):=m((A+x)\cap B)$ is continuous.

$\Big( A+x=\{ a+x | a\in A\}\Big)$.

I can use the following Fact.

Fact

If $E$ is Lebesgue measurable on $\mathbb{R}$ and $m(E)<\infty$, then $\lim_{x\to 0} m((E+x)-E)=0.$

I tried but couldn't prove.

Set $\forall a \in \mathbb{R}.$
I'll prove $|f(x)-f(a)|\to 0 \quad (x\to a)$.

If I could prove $|f(x)-f(a)|<m((A+x-a)-A)$, let $x\to a$ and I could prove that $f(x)$ is continuous, using the Fact.

But I cannot prove
$|f(x)-f(a)|<m((A+x-a)-A)$.

I would like you to give me some ideas.

答:
$\newcommand\e\varepsilon$There might be a smarter way to go about this, but here is a naive approach. Let $\e>0$. Then there exists $V$ open with $V\supset A$ and $m(V\setminus A)<\e/4$. Note that this implies that $m(V)<\infty$. Now
$$
m((V+x)\cap B)-m((A+x)\cap B)=m((V\setminus A)+x)\cap B)\leq m(V\setminus A).
$$
Then
\begin{align}
|f(y)-f(x)|
&\leq \e/2+m((V+y)\cap B)-m((V+x)\cap B).
\end{align}
Now since $V$ is open, we may write it as a disjoint union of open intervals, $V=\displaystyle\bigcup_n(a_n,b_n)$. The finite measure of $V$ gives $\sum_n(b_m-a_n)<\infty$. We have $$m(E)-m(F)=m(E\setminus F)-m(F\setminus E).$$ Then, for $x<y$,
\begin{align}
m((a_n,b_n)+y)\cap B)-m((a_n,b_n)+x)\cap B)&\leq m((a_n+x,a_n+y))+m((b_n+x,b_n+y))\\[0.3cm]
&\leq2(y-x).
\end{align}
So, we choose $n_0$ such that $\sum_{n>n_0}(b_n-a_n)<\e/4$. Then if $|y-x|<\e/(2n_0)$,
\begin{align}
|f(y)-f(x)|&\leq \frac\e2+\sum_nm((a_n,b_n)+x)\cap B)\\[0.3cm]
&\leq\frac\e2+\frac\e4+\sum_{n>n_0}m((a_n,b_n)+x)\cap B)\\[0.3cm]
&\leq\frac\e2+\frac\e4+\frac{2n_0\e|y-x|}{2n_0}=\e.
\end{align}


答:
$9$ and $6$ are both multiples of $3$.

$2x\equiv7\pmod9\implies 2x\equiv1\pmod3$

$5x\equiv2\pmod6\implies 2x\equiv2\pmod3$

$1\equiv2\pmod3$ is a contradiction, so the system cannot be solved.

Body:
\begin{equation}
  \left\{\begin{array}{@{}l@{}}
    2x\equiv7\mod9 \\
    5x\equiv2\mod6
  \end{array}\right.\,.
\end{equation}
Can this system of congruences be solved?
I notice that $(9,6) = 3 \ne 1$ so I can't apply the Chinese theorem of remainders, but this doesn't imply that it can't be solved, so I thought to rewrite the second equation in two different equations, like this:

\begin{equation}
  \left\{\begin{array}{@{}l@{}}
    2x\equiv7\mod9 \\
    x\equiv0\mod2 \\
    5x\equiv2\mod3
  \end{array}\right.\,.
\end{equation}

But the problem still here: $(9,3) = 3 \ne 1$, so it can be solved or is it impossible?


问:
For a cadlag submartingale $X$ the function $t \mapsto E[X_{t}]$ is right continuous.
I want to show that for a cadlag submartingale $(X_{t})_{t \geq 0}$ the function $t \mapsto E[X_{t}]$ is right continuous. So far I've only noticed, that this function is monotonously increasing. I would be really grateful for hints.

答:
Let $\{t_n\}_{n=1}^\infty$ be a sequence of rational numbers in $(t, \infty)$, monotonically decreasing to $t\ge 0$ as $n\to\infty.$ We want to show that $\lim_{n\to\infty} \mathbb{E} [X_{t_n}]=\mathbb{E} [X_t].$ To this end, we use the theory of backward martingales:

Definition: Let $\{\mathcal{F}_n\}_{n=1}^\infty$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$ (i.e., $\mathcal{F}_{n+1}\subset\mathcal{F}_n\subset \mathcal{F}$, $\forall n\ge 1$). Then $\{Y_n,\mathcal{F}_n\}$ is a backward submartingale if $\mathbb{E} [Y_n]<\infty$, $Y_n$ is $\mathcal{F}_n$-measurable, and $\mathbb{E} [Y_n\mid \mathcal{F_{n+1}}]\ge Y_{n+1}$ a.s., for all $n$.

Notice that if $\{X_t,\mathcal{F}_t\}$ is a submartingale, then $\{X_{t_n},\mathcal{F}_{t_n}\}$ is a backward submartingale (for $\{t_n\}$ defined above). Moreover, backward submartingales are uniformly integrable. Hence,  we deduce $$\lim_{n\to\infty} \mathbb{E} [X_{t_n}]=\mathbb{E} [X_t],$$
using the fact that $\lim_{n\to\infty} X_{t_n}=X_t$ a.s. (from right-continuity of $X_t)$ and the uniform integrability of $\{X_{t_n}\}_n$ (see, for example, Theorem 0.2 here). The right-continuity of the function $t\mapsto \mathbb{E} [X_t]$ follows.


问:
How to boil down $n*k + (n+1)*k + (n+2)*k + (n+3)*k + \ldots$
How do I express this in a singel function to be used for exampke in Excel?

$$n*k + (n+1)*k + (n+2)*k + (n+3)*k \ldots$$

$k$ is a constant and $n$ is counting from $1$ to infinity. For example: $k=5$, $n$ starts at $1$ for $4$ iterations.

$$1*5 + 2*5 + 3*5 + 4*5$$

答:
Unless $k = 0$, the infinite sum will not converge.  But if you want the sum of a finite number of terms, say from $n = 1$ to $n = m$, where in your example you chose $m = 4$, then the sum has a convenient closed form:
$$\sum_{n=1}^m kn = \frac{km(m+1)}{2}.$$  So with your choice $k = 5$, $m = 4$, your sum is $$1(5) + 2(5) + 3(5) + 4(5) = \frac{5(4)(4+1)}{2} = 50.$$  In Excel, if A1 contains the value of $k$ and A2 contains the value of $m$, then your sum could be entered in A3 as =A1*A2*(A2+1)/2.


答:
We have, letting $m$ be the number of terms:

$$nk + (n+1)k + (n+2)k + \dots + (n+m-1)k =nmk + (1 + 2 + \dots + (m-1))k$$

Using $1 + 2 + \dots + a = \frac{a(a+1)}{2}$:

$$=\boxed{nmk + \frac{m(m-1)k}{2}}$$

Title: How to boil down $n*k + (n+1)*k + (n+2)*k + (n+3)*k + \ldots$

Body:
How do I express this in a singel function to be used for exampke in Excel?

$$n*k + (n+1)*k + (n+2)*k + (n+3)*k \ldots$$

$k$ is a constant and $n$ is counting from $1$ to infinity. For example: $k=5$, $n$ starts at $1$ for $4$ iterations.

$$1*5 + 2*5 + 3*5 + 4*5$$


问:
Does $\mathbb{E}[(1-\varepsilon_n)^{X_n}]\rightarrow 1$ imply that $\mathbb{P}(X_n\geq A_n)\rightarrow 0$?
Let $X_n$ be a sequence of random variables taking values on the non negative integers which is finite almost surely (that is, for all $n\in\mathbb{N}$, there exists $k=k(n)$ such that $\mathbb{P}(X_n>k)=0$), $\varepsilon_n\in(0,1)$ and $\varepsilon_n A_n\geq O(1)$.

Question: Does $\mathbb{E}[(1-\varepsilon_n)^{X_n}]\rightarrow 1$ imply that $\mathbb{P}(X_n\geq A_n)\rightarrow 0$?

The main tool i've been using to relate these expressions is Markov Inequality, but in this case i get an inequality in the wrong direction. To illustrate this, here is my reasoning on why the converse is true if $\varepsilon_nA_n=o(1)$. By the Markov Inequality,
$$
\mathbb{P}(X_n\geq A_n)=1-\mathbb{P}(X_n<A_n)\geq 1- \frac{\mathbb{E}[(1-\varepsilon_n)^{X_n}]}{(1-\varepsilon_n)^{A_n}}.
$$
Now, as $(1-\varepsilon_n)^{A_n}=[(1-\varepsilon_n)^{1/\varepsilon_n}]^{A_n\varepsilon_n}$ and $\varepsilon_n A_n=o(1)$, we have $(1-\varepsilon_n)^{A_n}\rightarrow 1$, so $\mathbb{P}(X_n\geq A_n)\rightarrow 0$ implies $\mathbb{E}[(1-\varepsilon_n)^{X_n}]\rightarrow 1$.

I also think that the answer to my question could follow from some tail arguments in the expected value, but i'm not been able to formalize that.

答:
Let $c>0$ such that $\varepsilon_n A_n \geq c$ for all $n$. Then
$$\begin{align}
\mathbb{E}[(1-\varepsilon_n)^{X_n}]
&\leq \mathbb{E}[(1-\varepsilon_n)^{X_n}\mathbf{1}_{X_n \geq A_n}] + \mathbb{P}\{X_n < A_n\} \\
&\leq (1-\varepsilon_n)^{A_n}\mathbb{E}[\mathbf{1}_{X_n \geq A_n}] + \mathbb{P}\{X_n < A_n\} \\
&= 1- \left(1- (1-\varepsilon_n)^{A_n}\right)\cdot \mathbb{P}\{X_n \geq A_n\}
\end{align}$$
Now,
$
(1-\varepsilon_n)^{A_n} \leq e^{-\varepsilon_n A_n} \leq e^{-c} < 1
$
so
$$
1-(1-\varepsilon_n)^{A_n} \geq 1-e^{-c} > 0
$$
and so
$$
\mathbb{P}\{X_n \geq A_n\} \leq (1-e^{-c})^{-1}\left(1-\mathbb{E}[(1-\varepsilon_n)^{X_n}]\right) \xrightarrow[n\to\infty]{} 0
$$
showing the result.


问:
Why is such function continuous?
Can anyone help me see why the function $f: \bar{\mathbb{Q}} \to \mathbb{Q} $ where $\bar{\mathbb{Q}}$ denotes the set of algebraic numbers,
$$f(x)=\begin{cases}1,&\ \text{if the real part of $x$ is greater than $\pi$, and}\\
0,&\ \text{otherwise.}\end{cases}$$

is continuous?

答:
Note that $z \in \mathbb C$ is algebraic over $\mathbb Q$ if and only if $Re(z), Im(z)$ are algebraic over $\mathbb Q$. Now, to show that $f$ is continuous we need only show that $f^{-1}[0]$ and $f^{-1}[1]$ are closed. Indeed, $f^{-1}[1]$ is precisely those $z \in \overline{\mathbb Q}$ such that $Re(z) > \pi$. Since $Re(z)$ is algebraic and $\pi$ is transcendental, $Re(z) > \pi$ iff $Re(z) \geq \pi$. Now, let $S \subseteq \mathbb C$ be defined as $\{z \in \mathbb C : Re(z) \geq \pi\}$. The map $Re: \mathbb C \longrightarrow \mathbb R$ is just projection, so it is continuous. Thus, $S = Re^{-1}[[\pi, \infty)]$ is closed. Furthermore, as discussed, $S \cap \overline{\mathbb Q} = f^{-1}[1]$. Then as $S$ is closed in $\mathbb C$, $S \cap \overline{\mathbb Q} = f^{-1}[1]$ is closed in the subspace topology.

$f^{-1}[0]$ consists of $Re(z) \leq \pi$ so essentially the same argument shows $f^{-1}[0]$ is closed. Hence, $f$ is continuous.

Title: Why is such function continuous?


问:
Find the spectrum of self-adjoint operator $T$ that satisfies $T^3+T^2+T+I=0$
Let $H$ be a Hilbert space, and let $T:H \to H$ be a bounded operator such that $T$ is self-adjoint and also it satisfies the equation:
$$T^3+T^2+T+I=0$$

I want to find the spectrum $\sigma(T)$

By multiplying the equation with $T-I$, I got that $||T||=1$ (and also $||T^2||=1$, $||T^4||=1$).

Moreover, from the equation I also get: $0 = (T+I)(T^2+I)$

But I just don't see how from here I get to the spectrum of $T$ apart from $\lambda \not \in \sigma(T)$ if $|\lambda| > 1$.

Help would be appreciated.

答: This is simply the spectral mapping theorem for self-adjoint operators. Define $f:\mathbb C\to \mathbb C$ by $f(z)=z^3+z^2+z+1$. Then $f(\sigma(T))=\sigma(f(T))=\{0\}$. Thus you simply need to find the roots of $f$ (remember that self-adjoint operators can only have a non-empty real spectrum).


问:
Proving $f(x)=\cos x\quad\forall x \in \mathbb{R}$

Suppose the function $f(x)$ is continuous on $\mathbb{R}$, and $f(r)=\cos r$ for every rational number $r$. Prove that $f (x) = \cos x$ for all $x ∈ \mathbb{R}.$

I would very much appreciate a hint.

答:
Let $\;x\;$ be any irrational number and prove that $\;f(x)=\cos x\;.$

Let $\;\left\{x_n\right\}_{n\in\mathbb{N}}\;$ be the sequence of rational numbers defined as

$x_n=\dfrac{\lfloor 10^nx\rfloor}{10^n}\;$ for all $\;n\in\mathbb{N}\;.$

It results that $\;\lim\limits_{n\to\infty}x_n=x\;$ and, since $\;f(x)\;$ and $\;\cos x\;$ are functions continuous at the point $\;x\;,$ we get that

$f(x)=\lim\limits_{n\to\infty}f(x_n)=\lim\limits_{n\to\infty}\cos x_n=\cos x\;,\;$ hence

$f(x)=\cos x\;.$


答:
Hint: start by showing that if $T$ is self-adjoint then $T^2+I$ is injective (in fact it is invertible).  As a start, suppose $(T^2+I)x = 0$ and look at $\langle (T^2+I)x, x \rangle=0$.  You should be able to manipulate this to conclude $x=0$.

Now as you have shown, $(T^2+I)(T+I) = 0$.  Given the injectivity, what does that say about $(T+I)x$ for arbitrary $x$?

This will tell you exactly what operator $T$ must be, and in particular it will be clear what its spectrum is.

Title: Find the spectrum of self-adjoint operator $T$ that satisfies $T^3+T^2+T+I=0$

Body:
Let $H$ be a Hilbert space, and let $T:H \to H$ be a bounded operator such that $T$ is self-adjoint and also it satisfies the equation:
$$T^3+T^2+T+I=0$$

I want to find the spectrum $\sigma(T)$

By multiplying the equation with $T-I$, I got that $||T||=1$ (and also $||T^2||=1$, $||T^4||=1$).

Moreover, from the equation I also get: $0 = (T+I)(T^2+I)$

But I just don't see how from here I get to the spectrum of $T$ apart from $\lambda \not \in \sigma(T)$ if $|\lambda| > 1$.

Help would be appreciated.


问:
Restriction of function respects tangent space of sub manifolds?
Might be a silly question, but I feel like the following should be true but I can't justify it.

If $f:M\rightarrow N$ is a differentiable function between two smooth manifolds, and there is some sub manifold $A$ of $M$ such that $f(A)=B$ where $B$ is some submanifold of $N$. Is it true that given $p\in A$, and the map $df_p:T_p M\rightarrow T_{f(p)}N$, we have $df_p(T_p A)\subset T_{f(p)}B$? If not, is there a situation where this does hold?

答: The answer is yes. Take $v \in T_pA$: then there exists a smooth path $\gamma : (-\varepsilon, \varepsilon) \to A$ with $\gamma(0) = 1$ and $\gamma'(0) = v$. Now, as $\mathrm{d}f(p)\cdot v = \left.\frac{\mathrm{d} f\circ \gamma(t)}{\mathrm{d}t}\right|_{t=0}$, and as $f\circ \gamma : (-\varepsilon,\varepsilon) \to B$ is a smooth path with $f\circ\gamma(0) = f(p)$ and $(f\circ\gamma)'(0) = \mathrm{d}f(p)v$, it follows that $\mathrm{d}f(p)v \in T_{f(p)}B$.


问:
Is there a standard terminology for $|f^{-1}(y)|=1$ for almost every $y$?
Let $\Omega \subseteq \mathbb R^n$ be a nice open, connected, bounded subset (say with Lipschitz boundary) and let $f:\Omega \to \mathbb{R}^n$ be a Lipschitz map.

Is there a standard terminology for the following property?

$|f^{-1}(y)| \le 1$ for almost every $y \in \mathbb R^n$, where $|f^{-1}(y)|=1$ refers to the cardinality of the preimage.

I guess that reasonable suggestions could be "injective a.e. in the image" or "almost injective" or something like that. However, I wonder whether there is an accepted terminology I could use, and I don't want to "overload" or get into conflict with other terminologies. (I can guess that "almost injective" could be referring to a map which is injective after you exclude a set of measure zero from your domain).

(The mathematical context is the Area/Coarea formula.)

答: At least in the context of elasticity theory, this is known as "injectivity almost everywhere", see, e.g., Ciarlet's "Mathematical Elasticity vol. 1", Problem 5.7 (following the work of Ball) and Section 7.9 (following Ciarlet-Nečas).</p>


问:
How to find the value of $x^3(x^3-18)$?
If $x(x-3)=-1$ then the value of $x^3(x^3-18)$ will be ?

$A)1\quad\quad\quad\quad B)-1\quad\quad\quad\quad C)0 \quad\quad\quad\quad D)|-1|$

I tried solving $x(x-3)=-1$:

$x^2-3x+1=0\quad\text{hence}\quad x_{1,2}=\cfrac{3\pm\sqrt{5}}{2}$. But it seems that doesn't help a lot to find the value of $x^6-18x^3$. as an alternative way, from the quadratic we have $x^3=3x^2-x$. if we plug in it $x^3(x^3-18)$ we obtain $(3x^2-x)(3x^2-x-18)$ but I can't continue from here.

Also the options are strange because $|-1|=1$ so it seems the answer should be  $B$ or $C$.

Title: How to find the value of $x^3(x^3-18)$?

Body: Since $x^2=3x-1$, $x^3=3x^2-x=8x-3$, so$$x^3(x^3-18)=(8x-3)(8x-21)=64x^2-192x+63=192x-64-192x+63=-1,$$so $B$) is correct. Alternatively, use$$x^6-18x^3+1=(x^2-3x+1)(x^4+3x^3+8x^2+3x+1).$$


问:
Reasoning about products of reals
Take 2 sets of real numbers:

$x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_m$ such that $\prod\limits_{1 \le i \le n} x_i > \prod\limits_{1 \le j \le m} y_j$.

Let $k$ be any positive real number.

Does it necessarily follow that $\prod\limits_{1 \le i \le n} (x_i+k) > \prod\limits_{1 \le j \le m} (y_j +k)$

If the question were related to addition, then the generalization would apply:
$$\sum\limits_{i=1}^n\left(x_i + k\right) > \sum\limits_{j=1}^m \left(y_j + k\right)$$

Intuitively, adding a positive to each real $x_i$ and each real $y_j$ should increase the product so the question relates to how much it increases each product.

For example, if I choose $x_1 = 10, x_2, = 11$ and $y_1 =1, y_2 = 2$ and $k=1$ it is clear that $110 > 2$ and $132 > 6$.

It seems to me that the answer is yes.  Am I correct?  If yes, how does one prove this?  If no, what is the argument against?

Title: Reasoning about products of reals

Body: A counterexample is given by $m=n=2,\; x_1=x_2=-1,\; y_1=y_2=0,\; k=1.$


问:
Find the Relation between $a,b,c$

In the figure shown find the relation between $a,b,c$.

My try:
When two circles of radii $r_1,r_2$ touch externally, the length of their direct common tangent is $2\sqrt{r_1r_2}$
Let the radius of the bigger circle is $r_1$ and that of smaller one is $r_2$. We have:

$AM=a$

$MK=2\sqrt{r_1r_2}$

By Pythagoras theorem we get:
$$AK^2+4r_1^2=(a+b+c)^2$$

Also by secant tangent theorem we have:
$$AK^2=AC \times AD$$ So we get:
$$AK^2=(a+b)(a+c)$$
So we get:
$$4r_1^2=b^2+c^2+ab+bc+ca$$

Any hint from here?

答:
If you let angle KAD = 2p, then sin(2p) = 2(r1)/(a+b+c) from triangle AKD

from a circle theorem angle KCD is a right angle and angle CKD=2p, so c=2(r1)sin(2p)
and so c=4r(1)^2/(a+b+c)

then perhaps you can use what you did to eliminate r1

...after trying it, it leads to an inconsistency, there might be a mistake in your AK^2 line, second bracket should be (a+b+c), then it's the same result...would then recommend using triangle AB(O2), where O2 is the centre of circle 2, tan(p)=(r2)/a and cos(2p)=AK/(a+b+c) from triangle AKD and then use trig identities to try and eliminate unwanted letters.

...further results:

$$4r_1^2=c(a+b+c) $$

$$AK = \sqrt{(a+b)(a+b+c)}$$

$$sin(2p)=\sqrt{\frac{c}{a+b+c}}$$

$$cos(2p)=\sqrt{\frac{a+b}{a+b+c}}$$

$$tan(2p)=\sqrt{\frac{c}{a+b}}$$

so perhaps they can be related using inverse trig functions, all equalling 2p and each other.

Title: Find the Relation between $a,b,c$


问:
Find the flux for the vector field $F =(x^2,y^2,z^2)$
Find the flux for the vector field $\vec{F}=(x^2,y^2,z^2)$ across the boundary to the ball given as $$(x-1)^2 +(y+1)^2 +(z-2)^2 \le 4$$

Edit: How much I've done: The center of the sphere is $(1,−1,2)$ (I've sketched it), thus $r(θ,ϕ)=(1+cosθsinϕ, −1+sinθsinϕ, 2+cosϕ)$ . Then to find the normal vector $\vec{N}$ , I've calculated $\frac{∂r(θ,ϕ)}{∂ϕ}=(cosθcosϕ,sinθcosϕ,−sinϕ)$ and $\frac {∂r(θ,ϕ)}{∂θ}=(−sinθsinϕ,cosθsinϕ,0)$ then $\frac{∂r(θ,ϕ)}{∂ϕ} \times \frac{∂r(θ,ϕ)}{∂θ}$, and finally got $(sin2ϕ,sinθsin2ϕ,sinϕcosϕ)$. Am I on the right track?

答:
Your region is $R: (x-1)^2+(y+1)^2+(z-2)^2 \leq  4$.

If you are applying divergence theorem, you need to evaluate $\displaystyle \iiint_R 2(x+y+z) \ dV$

Just to simplify your working, consider the below substitution -

$u = x-1, v = y+1, w = z-2$ and region $R$ transforms to $E: u^2+v^2+w^2 \leq 4$ which is a sphere of radius $2$ centered at the origin. Jacobian for the transformation is $1$.

So the integral becomes $\displaystyle \iiint_E 2(2 + u + v + w) \ dV$. Note that $u, v, w$ are odd functions and due to symmetry, their integral over a sphere centered at the origin will be zero. For example, integral of $u$ over half sphere (for $u \leq 0$) will cancel out with the integral over the other half ($u \geq 0)$.

So the problem reduces to finding $\displaystyle \iiint_E 4 \ dV$ which you know even without the integral.


问:
How do we know if a problem is hardest in NP
I read that the definition of
NP-complete is :
    These are the hardest problems in NP. Such a problem is NP-hard and in NP

How do we know if a problem is hardest in NP, and no harder problem exists.
I understand that let's assume that somehow magically we know that a problem L is hardest in NP and then we can find out more hardest problems H if we can reduce H to L and vice versa.But my question is how does it all begin? How do we know 1 hardest problem to begin with?

Also, to be able to say that something is hardest (or any extreme), we need to know all possible problems in NP and then argue about the hardest.. How do we know all possible NP problems? Is this where turing machine comes useful and by using representation of output string in form of 1 and 0 in output tape, we can theoretically talk about all possible NP problems.

I understand that I may not have been able to articulate my question well - due to confusion.

Thanks,

答:
If every problem, $p$, in a complexity class, $C$, can be reduced to another problem $q$, then $q$ is said to be Hard. (it is not required that $q$ belong to $C$)

You can conclude that if you are able to solve $q$, then you can solve every problem $p$ in $C$ and it must therefore be harder to solve than each problem $p$.

If $q$ is Hard for $C$ and $q$ belongs to $C$, then $q$ is said to be Complete.

If $q$ is Complete, then it means it is one of the most difficult to solve in $C$.


问:
What is the complexity class for each one of the following functions
What is the complexity class for each one of the following functions:

$a) (n^3+n^2 \log n)(\log n+1) + (10 \log n+7)(n^3+3)$

$b) (2n + n^2)(4n^3 + 4n)$

$c) (n^n + n2^n + 3n)(n! + 6n)$

答:
If your question mean is "what is the complexity of these functions":

a) $O(n^3\log n) \rightarrow a\in P$

b) $O(n^5) \rightarrow b\in P$

c) $O(n^nn!) \rightarrow c\in EXP$


问:
Best Intro to Number Theory books for non mathematicians?
I want to know if there are any books that can teach me or introduce me to number theory requiring no prerequisites. This is coming from someone who wants to broaden their mathematical horizons and investigate number theory and cryptology.

Any suggestions are welcome.

答: I would absolutely recommend these two books: Number Theory-G.H Hardy, and Introduction to Number Theory - Ivan Niven. The first book sounds a bit outdated at times and has rather strange notation and terminology. I recommend the second book if you want to learn a bit more about quadratic residues and so forth. It is more clear, but it uses less notation and has more "words".</p>