问:
Uniqueness in separatioin of variables: how can we assume $f(x(t)) \neq 0$ on a fixed interval?
I know that the answer to the following question should be out somewhere already, but I could not really find it, so if someone has a link I am happy to close this again.

I recently had to look back at the proof of separation of variables for some exercise I did, but I got stock with the proof of uniqueness:

Suppose we have the ODE

$$ x'(t) = f(x(t)), \  x(0) = x_0 $$

with $f(x_0) \neq 0$, $f$ continuos on $\mathbb{R}$. Then there exists some interval $I = (a,b)$ such that on $(a,b)$ the unique solution is given by $g^{-1}(t)$, with $$g(y) = \int_{x_0}^{y} \frac{1}{f(u)}du$$

Now to show that there exists some interval on which $g$ is well defined is easy, and then by the rules of differentiation one also gets that $g^{-1}(t)$ satisfies the ODE. However, for uniqueness the proof that I had learned is basically the same that is written down in this post.

I.e., one starts by observing that any hypothetical solution should satisfy

$$\frac{x'(t)}{f(x(t))} =1$$

and integrates this to show that the solution also has to satisfy $$ t = g(x(t))$$, which one then can invert on the given interval to show that indeed $x$ would have the correct form.

However, to make sense of the above fraction, one would need to know a priori that on some interval we have $f(x(t)) \neq 0$. Now of course by continuity for any solution we will have an interval where this holds , but how do we get an interval $(a,b)$ that works for any solution (which then ofc immediately gives uniqueness on that interval) ?

答:
You are looking at it wrong. You do not need an interval $[a,b]$ that "works for any solution". What you need is for any solution an interval $[a,b]$ that works.

Because $f(x_0) \ne 0$ and $f$ is continuous, there is an interval $[c, d]$, containing $x_0$ in its interior, on which $f$ is non-zero. And by continuity $f$ is single-signed on $[c,d]$, which means that $g$ is strictly monotone and invertible on $[c,d]$, carrying it to some interval $[a, b]$. And inversely $g^{-1}$ carries $[a,b]$ to $[c,d]$.

Now let $x(t)$ be an arbitrary solution, extended to its maximal domain.. As $x$ is continuous, there is some maximal interval $[a', b']$ carried by $x$ into $[c,d]$. By the argument you've indicated, $x = g^{-1}$ on their common domain. But if $[a,b]$ is not a subset of $[a',b']$ then we can extend $x$ to the larger domain by making it equal to $g^{-1}$ there as well. Thus $x$ is defined and equal to $g^{-1}$ on all of $[a,b]$.

Therefore $g^{-1}$ is the unique solution inside $[a,b]$.


问:
Describe the graph locus represented by this equation
I want to know the shape of the region described by
$$ Im(z^2) = 4 $$

so I did the following:
$$ z=x + iy  $$

$$z^2 = x^2 + 2xiy -y^2  $$then
$$Im(z^2) = 2xy
 $$
then the locus is  $$ 2xy = 4 $$
$$ xy = 2 $$

then it's the line  $$ xy = 2 $$

what does this equation describe?

答:
We get the equation $2xy=4$. This is a hyperbola, not a line. You are undoubtedly familiar with its close relative, $y=\dfrac{1}{x}$.

You can put the hyperbola $xy=2$ into more familiar form by letting $x=\frac{u+v}{\sqrt{2}}$, $y=\frac{u-v}{\sqrt{2}}$. This is just a rotation. Then $xy=2$ becomes $u^2-v^2=4$.

ContentLicense: CC BY-SA 3.0


问:
Why does calculating probabilities intuitively result in differences from Bayesian calculations?
As an example, say you're asked to solve for the following problem:

You have three bags, labeled A, B, and C. Bag A contains two red marbles and three blue marbles. Bag B contains five red marbles and one blue marble. Bag C contains three red marbles only. Suppose a bag is randomly selected (again, with equal probability), but you do not know which it is. You randomly draw a marble and observe that it is blue. What is the probability that the bag you selected this marble from is A?

I understand that you can calculate P(A|blue) using Baye's Theorem which would ask me to solve for P(blue|A) * P(A) / P(blue) which comes out to be 0.78. However, if I were to be asked this question outside the context of a class on bayesian statistics, I would think that the probability would be 0.75 or 3/4 since there are 4 blue marbles between the three bags. Since there is an equal chance of drawing from either bag, couldn't we disregard the probability of drawing from a given bag and focus on the distribution of blue marbles between bags? What is the error in my thinking here?

答:
Suppose I pick a random marble from all the marbles, with each marble equally likely, and I show you that it's blue. In that case, your reasoning would make sense: you should figure that there's a 3/4 chance that my marble came from bag A.

However, in the scenario you described, the marbles are not equally likely to be chosen, because the bags have different numbers of marbles. Each blue marble in bag A has a higher probability of being chosen than the one blue marble in bag B has, because bag B has 6 marbles total while bag A only has 5 marbles. That's why the correct answer is a little bigger than 3/4.

ContentLicense: CC BY-SA 4.0


问:
A graph theory problem from mobile games
Example

game interface

This is the question that comes to my mind when I play a game called QuickyRoute,which essentially a Hamilton graph problem,the game will randomly generate a number of points,you need to draw the undirected complete graph with the smallest sum of distances.

As the difficulty of the game increases, the more points, the harder it is to find the correct answer.I want to solve these problems through programming. Obviously there are at most half of n factorial possibilities.

But enumeration is not very convenient, Because this is a famous graph theory problem,I want to ask if there is a better algorithm idea or some excellent papers can refer to this game

答:
The problem is actually equivalent to the traveling salesman problem, except we don't have to return to where we started. The general traveling salesman problem is an NP-hard problem in theoretical computer science. Furthermore, it is well-known that the traveling salesman problem cannot even be approximated within a constant factor.

However, it seems like the problem you have listed is actually a special case of the traveling salesman problem, known as the metric traveling salesman problem (i.e., an instance of the traveling salesman problem in which the triangle inequality holds). Unfortunately, the metric traveling salesman problem is still NP-hard, which means that you will probably not be able to find any efficient algorithms that compute an optimal solution when the number of vertices gets even moderately large. However, there are efficient polynomial-time approximation schemes (EPTAS) that can approximate solutions to the metric traveling salesman within a constant factor.

The best approximation heuristic currently known is Christofides algorithm, which was discovered in 1976.  Christofides's algorithm is a $3/2$-approximation, which means that it guarantees that its solutions will be within a factor of $3/2$ of the optimal solution length.


问:
Complete Cayley table for a field

Let $M=\{0,1,a,b,c\}.$  Can I complete the Cayley tables so that $M$ is a field?

My thought was that $(M,+)$ must be an abelian group. This abelian group must be isomorphic to $(\mathbb Z_5,+).$ Correct? I managed to find that if $a=3, b=4$ and $c=2$ it works for addition.

But this does not work for the multiplication table. Does that mean that I can´t complete the Cayley table so that $M$ is a field?

答:
Although you can use ad-hoc arguments (as the other answer did) for specific cases, the general way to deal with this kind of question is in fact to prove that any two finite fields of the same size are isomorphic! The easiest proof is as follows. Take any finite field $F$. Then $\text{char}(F) = p$ for some prime $p$. Thus $F$ has a subfield isomorphic to$\def\ff{\mathbb{F}}$ $\ff_p$ and hence $F$ is a vector space over $\ff_p$, which implies that $\#(F) = p^k$ for some $k∈ℕ$. Since $(F^*,·)$ is a group where $F^*$ is the set of nonzero elements of $F$, we have $x^{p^k-1} = 1_F$ for every $x∈F^*$ by Lagrange's theorem. Thus every element of $F$ is a root of the polynomial $f = (x↦x^{p^k}-x)$ over $\ff_p$, and hence $f$ splits completely over $F$ since $f$ has at most $p^k$ distinct roots over any field. Therefore $F$ (being nothing more than the roots of $f$ in $F$) is a splitting field for $f$ over $\ff_p$. But splitting fields of the same polynomial over the same base field are unique up to isomorphism. Thus we are done.

Just for example, in this question, the field has size $5$ so it must be isomorphic to $\ff_5$. The first table allows us to easily find that the elements corresponding to $0,1,2,3,4$ are $0,1,c,a,b$ respectively since $\{a,c\} = \{2,3\}$ and $a$ is followed by $b$. But then clearly the second table is wrong because $3·4≠1$ in $\ff_5$. Although it may seem like more trouble here, it is easier in general since you know exactly what the finite field must look like.

Title: Complete Cayley table for a field


问:
Precise meaning of "wellfounded part has been transitivized"
When discussing ultrapowers, one often says something like "we assume the wellfounded part of $M$ has been transitivized", where $M$ is some structure (say an ultrapower). What is the precise meaning of this? I understand that for a structure $(M,E)$, its wellfounded part is the class of all $x\in M$ such that there is no infinite descending $E$-sequence beginning with $x$. I also understand the Mostowski collapse lemma. What I don't understand is what one is doing precisely when one "assumes the wellfounded part of $M$ has been transitivized".

Is it: we divide $M$ into a wellfounded part and the rest, and apply Mostowski collapse to the wellfounded part, and then union the transitive isomorph with the rest?

答:
I'm turning Andreas Blass's comment into an answer to move this question off the "unanswered" queue; I've made this CW to avoid reputation gain, and will delete this answer of Andreas posts one of his own.

You are exactly right - modulo the subtlety that an annoying $M$ might already use one of the sets in the Mostowski-collapse-image of $M$'s well-founded part. So first we "relabel" everything in $M$ to avoid this, and then we appropriately Mostowski collapse.

Basically, all we care about is that at the end of the day we have:

Let $X$ be the set of elements $m\in M$ such that $E\upharpoonright \{x: xEm\}$ is well-founded. Then $X$ is a transitive set and $E\upharpoonright X=\in\upharpoonright X$.


问:
Indefinite integral question with cube roots
Evaluate the following integral:

$$\int\sqrt[3]{x^3+\frac 1 {x^3}} dx$$

For $x^3=u$ and $1+\frac 1 {u^2}=v$, getting $$-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv$$, what is the next step?

答:
I would chose a different substitution:

$$I =\int \big(x^3 +\frac{1}{x^3} \big)^{\frac{1}{3}}dx$$
$$=\int \frac{(x^6 +1)^{\frac{1}{3}}}{x} dx $$
Let: $ u=(x^6+1)^{\frac{1}{3}} $, then:
$x^6=u^3-1 \text{ , and } dx= \frac{(x^6 +1)^{\frac{2}{3}}}{2x^5} du$ .
$$I=\frac{1}{2} \int \frac{u^3}{u^3-1}du$$
$$=\frac{1}{2} \int \frac{1}{u^3-1}du \text{ }+ \text{ } \frac{1}{2}u$$

The first integral can easily be solved now with partial fraction decomposition; and using the derivative of the arctan function.


问:
What kind of distribution should I use to this question?
From every 200 students in High School, 20 students failed. If we were to pick 40 students from the class. Calculate:

(a) the probability of getting 8 failed students.
(b) the expected value for the failed students out of 40 students.

Should I use binomial distribution or hypergeometric?

答:
The sampling procedure is not told thus you can assume if you draw with or without replacement first and then solve the exercise accordingly.

Binomial

$$\mathbb{P}[X=8]=\binom{40}{8}\left(\frac{1}{10}\right)^8\left(\frac{9}{10}\right)^{32}\approx 0.026$$

Hypergeometric

$$\mathbb{P}[X=8]=\frac{\binom{20}{8}\binom{180}{32}}{\binom{200}{40}}\approx 0.018$$

The expectation is the same in the two cases...


问:
For an irreducible Markov Chain let $P^r(x,y)>0$ and $P^l(y,x)>0$, then if m= r+l follows m ∈ T (x)∩T (y) then how T (x) ⊂ T (y) − m holds true?
Lemma- If a Markov chain is irreducible, then the period of all states is equal i.e. $gcd T (x) = gcd T (y)$ for all $x, y ∈ X$

Proof- Fix two states x and y. There exist non-negative integers r and l such
that
$P^r (x, y) > 0$ and $P^l (y, x) > 0$.
Letting $m = r+l$, we have $m ∈ T (x)∩T (y)$ and
$T (x) ⊂ T (y) − m$, whence gcd $T (y)$ divides all elements of $T (x)$. We conclude that
gcd $T (y) ≤ gcd T (x)$. By an entirely parallel argument, $gcd T (x) ≤ gcd T (y)$.

 In this proof can you explain how $T (x) ⊂ T (y) − m$ is derived?

答: Let $n \in T(x)$, so that $P^n(x,x) > 0$.  Then $$0 < P^l(y,x)P^n(x,x)P^r(x,y) \leq P^{l+n+r}(y,y)$$ shows that $$m+n = l+n+r \in T(y).$$  Therefore $n \in T(y) - m$.  Since $n$ was an arbitrary element of $T(x)$, it follows that $T(x) \subset T(y) - m$.


问:
Can't argue with success? Looking for "bad math" that "gets away with it"
I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").

One example would be "cancelling" the 6's in

$$\frac{64}{16}.$$

Another one would be something like

$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$

Yet another one would be

$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$

Note that I am specifically not interested in mathematical fallacies (aka spurious proofs).  Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for all cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

答:
There is a lot of treatment with little rigor around the first infinitesimal calculations. For example consider a circle of radius $r$, we can approximate it's area by filling it with equal triangles.

The area of each triangle is given(approximately) by $lr/2$, where $l$ is the base of the triangle.
If we have $n$ triangles, the sum of their areas is given by $nlr/2$. By increasing
the number of triangles to infinity, the sum of the bases of the triangles
approximates the length of the circumference, i.e, if $n\rightarrow\infty$ then $nl\rightarrow2\pi r$. Thus, the sum of the areas of the triangles
tends to $2\pi r^2/2 = \pi r^2$, which is the area of the circle.

The result is correct but we calculates $\infty\cdot 0$ with almost no rigor at all.


问:
Compute the probability mass function of X
There are two bags A and B. Bag A has 2 green marbles and 3 red marbles. Bag B has 2 green marbles and 3 blue marbles. Two marbles are drawn without replacement from bag A and put in bag B. Then two marbles are drawn without replacement from bag B and put in bag A. Let X be the number of green marbles in bag A after the above operations. Compute the probability mass function of X.

My attempt:

$X=\{0,1,2,3,4\}$

$X=0$: take all green from A and put non-greens back = $\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{7}\cdot\frac{2}{6}=\frac{17}{70}$

$X=1$: take all green from A and put one green back + take 1 green and put non-greens back = $\left(\frac{2}{5}\cdot\frac{1}{4}+\frac{4}{7}\cdot\frac{3}{6}\right)+\left(\frac{2}{5}\cdot\frac{3}{4}+\frac{4}{7}\cdot\frac{3}{7}\right)=\frac{228}{245}$

Then we continue on with $X=2,3,4$. My concern is that if we add up $\mathbb{P}(X=0)$ and $\mathbb{P}(X=1)$, we get a value that is greater than $1$, which should not be the case because everything should sum to $1$. Can someone tell me what I am doing wrong and possibly how I would approach this problem differently to get the right answer? Thank you.

答:
You wrote:

take all green from A and put non-greens back

First sight first error found...

$$\mathbb{P}[X=0]=\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{70}$$

At present I did not check the rest


问:
How many and which unitary transformations in $2d$
I was just wondering, probably totally trivial question, but say I have a $2d$ vector in $\mathbb{R}^2$ or $\mathbb{C}^2$ and the scalar product: Which are the possible unitary groups leaving the scalar product invariant? I suppose one has $\mathrm{U}(1)$, $\mathrm{(S)U}(2)$ and $\mathrm{(S)U}(1,1)$ but it would be nice to know if that is all.

Also, sketchwise, how to proceed to higher dimensions?

Cheers!

答:
Any (nondegenerate) bilinear form is the sum of a symmetric and skew-symmetric one.

The real symmetric bilinear forms are classified by signature $(p,q)$. The standard form on $\mathbb{R}^{p,q}$ is

$$ x\cdot y=(x_1y_1+\cdots+x_py_p)-(x_{p+1}y_{p+1}+\cdots+ x_{p+q}y_{p+q}). $$

Every such form looks like the above in some basis. The corresponding "unitary groups" are called orthogonal groups, $O(p,q)$, with corresponding special orthogonal groups $SO(p,q)$ too.

Over $\mathbb{C}$, we have $-x_jy_j=(ix_j)(iy_j)$, so with a change of basis, every symmetric bilinear form is just the usual dot product $x_1y_1+\cdots+x_ny_n$.

There is a unique skew-symmetric form (up to change of basis) in every even dimension (and none in odd dimension), with corresponding groups $Sp(2n,\mathbb{R})$ and $Sp(2n,\mathbb{C})$ (many, if not a majority, of sources would use $n$ instead of $2n$ in these notations).

Over complex vector spaces (or even quaternionic vector spaces), there are also sesquilinear forms, which are linear in one argument and conjugate-linear in the other. (In physics, conjugate-linearity is in the first argument, but often in math it will be in the second argument.) Every sesquilinear form is the sum of a Hermitian and a skew-Hermitian one.

The Hermitian sesquilinear forms are again classified by signature, the standard form being

$$ \langle x,y\rangle = (\overline{x_1}y_1+\cdots+\overline{x_p}y_p)-(\overline{x_{p+1}}y_{p+1}+\cdots+\overline{x_{p+q}}y_{p+q}) $$

on $\mathbb{C}^{p,q}$ or $\mathbb{H}^{p,q}$. The corresponding unitary groups are called $U(p,q)$ and $Sp(p,q)$. The former have special unitary versions $SU(p,q)$, but the latter don't (there isn't an obvious determinant of a quaternionic matrix, and the non-obvious determinant already only takes values $1$).

You can take a look at Wikipedia's article on Classical groups, or find other resources by googling "classical groups," but be warned there are many complicated and conflicting conventions and definitions.

All of these classifications of forms are essentially proved by generalizing the Gram-Schmidt process to show that "standard bases" for the forms exist.


问:
Is there a unique minimal surface with boundary $S_2$?
Take a loop of wire ($S_2$), twisted into a complicated shape. One can dip the wire in soap to form a minimal surface soap film.

Is there a unique minimal surface or can there be more than one inquivalent surface with minimal area?

答:
There can be more than one with minimal area. Here's an example:

Wrap a piece of wire around two posts stuck in a board, so there's a half-circle, a long straight section, another half-circle, and another long straight piece.

Remove the posts from the board, and lay one of them down across the middle of your loop, so that from overhead you see this:

```<code>           ****
/----------****---------\
|          ****         |
\----------****---------/
           ****
</code>```

where the stars are the post, and the stuff made of up slashes are my best attempt at making semicircles.

Holding down the post, lift up each semicircle until the four "straight" sections are vertical.

The highly-symmetric shape you get has three "extremal" surfaces. Two look like tongue depressors with a 180-degree arcing bend in the middle; the third looks like the "average" of these two. The first two are equal area minimal surfaces (at least if the "straight bits" are long enough; the third is saddle like and is "extremal" in some way I can't precisely remember.

If you shorten the straight-line segments, you get to a point where the two "flatter" surfaces both collapse (if they're soap-films) to the central potato-chip-like one, which is then a unique minimal surface.

Body:
There can be more than one with minimal area. Here's an example:

Wrap a piece of wire around two posts stuck in a board, so there's a half-circle, a long straight section, another half-circle, and another long straight piece.

Remove the posts from the board, and lay one of them down across the middle of your loop, so that from overhead you see this:

           ****
/----------****---------\
|          ****         |
\----------****---------/
           ****

where the stars are the post, and the stuff made of up slashes are my best attempt at making semicircles.

Holding down the post, lift up each semicircle until the four "straight" sections are vertical.

The highly-symmetric shape you get has three "extremal" surfaces. Two look like tongue depressors with a 180-degree arcing bend in the middle; the third looks like the "average" of these two. The first two are equal area minimal surfaces (at least if the "straight bits" are long enough; the third is saddle like and is "extremal" in some way I can't precisely remember.

If you shorten the straight-line segments, you get to a point where the two "flatter" surfaces both collapse (if they're soap-films) to the central potato-chip-like one, which is then a unique minimal surface.


问:
Simplifying radical expressions if possible
How does one go through the process of attempting to simplify $\sqrt[3]{150}$?

What I did was divide the radicand by 2,3,4 etc. and hope that I find a perfect cube.

Is there a better way to simplify this expression?

答: The fastest way is to factor $150$ into primes and see if any prime has an exponent at least $3$.  We have $150=2 \cdot 3 \cdot 5^2$ so there is no factor you can pull out.


问:
every integers from 1 to 121 can be written as 5 powers of 3
We have a two-pan balance and have 5 integer weights with which it is possible to weight
exactly all the weights integers from 1 to 121 Kg.The weights can be placed all on a plate but you can also put some in a dish and others with the goods to be weighted.

It's asked to find the 5 weights that give us this possibility. It also asks you to prove that the group of five weights is the only one that solves the problem.

Easily i found the 5 weights: 1 , 3 , 9 , 27 , 81

but i can' demonstrate that this group of five weights is the only one that solves the problem.

Can you help me ?

Thanks in advance !

答:
Hint: If the first 4 weights sum up to strictly less than 40, then you can not weigh either 41kg or 121 kg.

If the fifth weight is $\leq 81$, then you can't reach 121. If the fifth weight is $\geq 81$, then you can't reach 41.


问:
Value of a generalised probabilty density function with a dirac delta function at a single point
I thought I understood the dirac delta function until I came across a question where I needed to evaluate the probability density function at a single point.

Probability density function:
$$f_X(x) = \frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$

I would like to find the value of the function at $x=1$.

I know that the dirac delta function is used to account for the discontinuities in the cumulative distribution function, and that the dirac delta function is defined as:
$$\delta(x)  = \begin{cases}
    \infty,&  x = 0\\
    0,& otherwise.\end{cases}
$$

I could choose to ignore the discontinuity at $x=1$, but that could result in
$$f_X(x=1) = \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}},$$
$$f_X(x=1) = \frac{1}{2} + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}}$$

or any value in between.

But, according to the definition of the dirac delta function, at point $x=1$ the function could also evaluate to

$$f_X(x=1) = \infty + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}} = \infty$$

This could be reasonable because the dirac delta function is not a traditional function, therefore I could accept a definition, if needed. The problem arises when I try to calculate conditional probabilities in any intervals containing $x=1$. For e.g., if I accept that $f_X(x=1) = \infty$, then

$$P(X=1|X \geq 1) = \frac{P(X=1)}{\int_{1}^{+\infty}(\frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}})} = \frac{P(X=1)}{\frac{1}{2\sqrt{2\pi}} \int_{-\infty}^{1} e^{-\frac{x^2}{2}}} = \infty,$$

which is clearly nonsensical. So then the questions arise, what is the value of $f_X(x=1)$ and why?

答: The Dirac delta is not a function.  It is not defined by its values.  Otherwise there would be no difference between $\delta (x) $ and $5 \delta (x) $.  The Dirac delta is a generalized distribution and does not have values at all.  The definition you give for it is incorrect; there is a way to twist and interpret it into a true statement, but it is very misleading at best.  A generalized distribution is defined by its integrals.  The probability distribution you gave is not a function and does not have a value at 1 at all.  Because of the $\delta$, the entire expression is a generalized distribution and not a function.


问:
System with cubed and squared terms
Find the ordered pair (x,y) such that

$x + y ^3 = 100$

$x^2 - y^2 = 100$

I decided to substitute $x = 100 - y^3$ into the second equation and got

$10000 - 200y^3 + y^6 - y^2 = 100$

Is there any trick or method I can use to solve the 6th degree polynomial above? Any insights would be appreciated.

答:
Beside numerical methods, you could "easily" estimate the solutions.

Consider that you look for the zero's of function
$$f(y)=y^6-200 y^3-y^2+9900$$ for which
$$f'(y)=2 y \left(3 y^4-300 y-1\right)$$ If you neglect the $-1$, you have a solution which is "close" to $y=10^{2/3}$.

Make a series expansion of $f(y)$ around this value
$$f(y)=-10 \left(10+10^{1/3}\right)-2\ 10^{2/3} \left(y-10^{2/3}\right)+\left(900\
   10^{2/3}-1\right) \left(y-10^{2/3}\right)^2+O\left(\left(y-10^{2/3}\right)^3\right)$$ which is a quadratic in $\left(y-10^{2/3}\right)$. Solve it to get
$$y_\pm=10^{2/3}+\frac{10^{2/3}\pm 10 \sqrt{899+900\ 10^{2/3}}}{900\ 10^{2/3}-1}$$ which gives $y_-\sim 4.4721$ and $y_+\sim 4.8133$ while the exact solution would be $4.4656$ and $4.8073$.

Title: System with cubed and squared terms


答:
I would like to find the value of the function at x=1.

Your $f(x)$ is a "mixed density", it is not exactly a density function because your random variable is not absolute continuous. It has a positive probability mass in $X=1$ so you have

$$\mathbb{P}[X=1]=\frac{1}{2}$$

For the rest of the support you have a Standard Gaussian density thus, for example:

$$\mathbb{P}[X\geq 1]=\frac{1}{2}+\frac{1}{2}\left[1-\Phi(1)\right]\approx0.58$$

So for your question you will have

$$\mathbb{P}[X=1|X\geq 1]\approx\frac{50}{58}$$

First observation

Here is the correct (better, here is what your text means as) definition of your Dirac Delta

$$\delta(x-1)=\begin{cases}
1,  & \text{if $x=1$ } \\
0, & \text{elsewhere}
\end{cases}$$

For detail, read here in the paragraph "Generalizations"

Second observation

In these cases it is better to define the random variable with its CDF that always exists rather then a "density" that will not exist.

Anyway I know that in some engineering applications, say telecommunication or Signal's Analysis they often use mixed density with impulse (Dirac delta), Rectangular (uniform) response et cetera et cetera

Title: Value of a generalised probabilty density function with a dirac delta function at a single point

Body:
I thought I understood the dirac delta function until I came across a question where I needed to evaluate the probability density function at a single point.

Probability density function:
$$f_X(x) = \frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$

I would like to find the value of the function at $x=1$.

I know that the dirac delta function is used to account for the discontinuities in the cumulative distribution function, and that the dirac delta function is defined as:
$$\delta(x)  = \begin{cases}
    \infty,&  x = 0\\
    0,& otherwise.\end{cases}
$$

I could choose to ignore the discontinuity at $x=1$, but that could result in
$$f_X(x=1) = \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}},$$
$$f_X(x=1) = \frac{1}{2} + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}}$$

or any value in between.

But, according to the definition of the dirac delta function, at point $x=1$ the function could also evaluate to

$$f_X(x=1) = \infty + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}} = \infty$$

This could be reasonable because the dirac delta function is not a traditional function, therefore I could accept a definition, if needed. The problem arises when I try to calculate conditional probabilities in any intervals containing $x=1$. For e.g., if I accept that $f_X(x=1) = \infty$, then

$$P(X=1|X \geq 1) = \frac{P(X=1)}{\int_{1}^{+\infty}(\frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}})} = \frac{P(X=1)}{\frac{1}{2\sqrt{2\pi}} \int_{-\infty}^{1} e^{-\frac{x^2}{2}}} = \infty,$$

which is clearly nonsensical. So then the questions arise, what is the value of $f_X(x=1)$ and why?


问:
Why $\Psi(H) \le \exp(G)$ for any subgroup $H$ of a finite $p$-group $G$ where $\Psi(H)=\frac{\Sigma_{h\in H}|h|}{|H|}$?
I was reading this paper. I need help to understand a claim.

Let's take $G$ which is a finite $p$-group. For any subgroup $H\le G$ we define $\Psi(H)$ to be the average order of the elements of $H$ that is
$$\Psi(H)=\frac{\Sigma_{h\in H}|h|}{|H|}$$

On page 4 of the paper, it is claimed that

$\Psi(H) \le \exp(G)$ for any subgroup $H$ of $G$.

Can anyone please help me to understand why the claim is true? And is this true for any finite group, not just finite $p$-group?

Thanks in advance.

答:
Hmm, it seems to be true fairly trivially, so I hope I'm not misunderstanding.  For a $p$-group the exponent of $G$ is just the highest order among elements in $G$ by definition.  So $|H| \cdot \Psi(H) = \Sigma_H |h| \leq \Sigma_H \exp(G) = |H| \cdot \exp(G)$ and thus $|H| \cdot \Psi(H) \leq |H| \cdot \exp(G)$.  Canceling gets what you want.  Equality occurs in any elementary $p$-group and a non-trivial cyclic subgroup.

The argument seems to work for any finite group, since in that case $\exp(G)$ is at least as big as $|h|$ for any $h \in H$ (or any $g \in G$).


问:
How does one show that two functors are *not* isomorphic?
Let $C$ be the category of finite-dimensional vector spaces over some field. It is easy to construct pairs of endofunctors $F, G$ of $C$, of the same variance, such that $F(V)$ and $G(V)$ have the same dimension for every $V \in C$, yet are not naturally isomorphic.

For example, we could take $G(V) := (V^*)^{\otimes 2}$, and $F(V) := (V^{\otimes 2})^*$, where $*$ denotes the dual. Or, I'm sure many of us have, at some time, tried showing that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, before realizing that it is false.  A convincing reason why they are not canonically isomorphic is that there is no obvious reason, in terms of universal properties, why they should be. In other words: if you try proving that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, you're going to have a bad time.

However, how does one explicitly show that $F$ and $G$ are not isomorphic? Is there a slick, systematic way of doing this?

Edit: Well, my examples couldn't be worse, as Mariano points out. But the gist of my question still stands (perhaps just get rid of the finite-dimensionality assumption ;))

答:
Here is one way to construct families of endofunctors $F, G$ of $\text{FinVect}$ that can't be distinguished on the basis of dimension counts. First, consider the family of endofunctors $V \mapsto S^n(V)$, and second, consider the family of endofunctors $V \mapsto \Lambda^n(V)$. The first sends vector spaces of dimension $d$ to vector spaces of dimension ${d+n-1 \choose n}$ while the second sends vector spaces of dimension $d$ to vector spaces of dimension ${d \choose n}$. Both of these form bases for the space of integer-valued polynomials in one variable ($d$), hence together there are many linear dependences between them. The smallest one is that

$${d+1 \choose 2} = {d \choose 2} + d$$

which implies that $F(V) = S^2(V)$ and $G(V) = \Lambda^2(V) \oplus V$ can't be distinguished by their dimension counts.

Nevertheless, they are not isomorphic. The reason is that for any endofunctor $F$, the vector space $F(V)$ naturally admits the structure of a $\text{GL}(V)$-representation, and $S^2(V)$ and $\Lambda^2(V) \oplus V$ are not isomorphic as $\text{GL}(V)$-representations. More simply, you can consider the action of a diagonal matrix on $S^2(V)$ and on $\Lambda^2(V) \oplus V$ and verify that the traces (which are symmetric functions) are different.

In fact I think every endofunctor (edit: defining a polynomial function between Hom spaces, see Steve's comment) of $\text{FinVect}$ is equivalent to a direct sum of Schur functors, so they can all be distinguished in this way up to isomorphism. (I need some mild hypothesis on the base field; I think characteristic zero suffices.)


问:
Prove that for all $d,n \in \mathbb{N}$ and all $x \in \mathbb{Z}$, if $d | n$ then $(x \bmod n) \bmod d = x \bmod d$
Prove that for all $d,n \in \mathbb{N}$ and all $x \in \mathbb{Z}$, if $d | n$ then $(x \bmod n) \bmod d = x \bmod d$. I'm confused on where to even start. My initial thought is to perform a proof by contradiction by letting $d \nmid n$, resulting in $0 < n \bmod d < d$. I'm not sure where to go from here or if it's the right approach. Any help is appreciated!

答:
By definition, we have that $x \text{ (mod } n)$ is the least non-negative residue of $x$ modulo $n,$ i.e., it is the remainder (as obtained by the Division Algorithm) of $x$ when divided by $n.$ Consequently, if we write $x = qn + r$ for some integers $q$ and $0 \leq r \leq n - 1,$ then $x \text{ (mod } n)$ is precisely $r.$ Likewise, the integer $[x \text{ (mod } n)] \text{ (mod } d)$ is the remainder (as obtained by the Division Algorithm) of $r = x \text{ (mod } n)$ when divided by $d.$ Put another way, if we write $r = q' d + r'$ for some integers $q'$ and $0 \leq r' \leq d - 1,$ then $[x \text{ (mod } n)] \text{ (mod } d) = r \text{ (mod } d)$ is precisely $r'.$

Let us assume now that $d \,|\, n,$ i.e., assume that there exists an integer $a$ such that $n = ad.$ Can you put this together with the content of the first paragraph to finish the proof?


问:
What is Inverse Fourier Transform of $2^2/(^2−^2)$?
What is the inverse Fourier transform of  $F(w) = \frac{4}{4+(j2\pi f)^2} $?
 I have two suggested solutions:

Assume $j2\pi f = \omega$ and use the standard transform
$e^{-\alpha |\tau |} = \frac{2\alpha}{\alpha ^2 + \omega ^2}$ to obtain $\frac{1}{4}e^{-2|\tau|}$

Calculate $f(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\alpha}{\alpha^2-\omega^2}e^{i\omega\tau}d\omega$

For (2), how can I solve the integral in order to get the inverse Fourier transform $f(\tau)$?

答:
In general Fourier transform (or inverse) take products into convolutions.  Let $F(x)$ and $G(x)$ have transforms $f(t)$ and $g(t)$.  Then the transform of $F(x)G(x)$ is $h(t)=\int f(s)g(t-s)ds$.

For your problem  $f(t)=\frac{1}{16}\int e^{-a|s|-b|t-s|}ds$ where $a=2$ and $b=8$.

(Note:  I may have constant wrong, but you see the point.)


问:
Summation of squared terms?
I am working on a problem involving the Normal distribution. Can the following summation be simplified any further? It is known that $\delta$ and $\mu$ are constants.

\begin{align*}
\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}-\sum_{i=1}^{n}\left(x_{i}-\mu-\delta\right)^{2}
\end{align*}

答:
If $\delta$ is a constant (like $\mu$), you can use the difference of squares identity to simplify this to:

$\displaystyle \delta \sum_{i=1}^{n} (2(x_i-\mu)-\delta) \\=\displaystyle  -n\delta^2 + 2\delta \sum_{i=1}^{n} (x_i-\mu) \\=\displaystyle -n\delta(\delta + 2\mu) + 2\delta \sum_{i=1}^{n} x_i $

Title: Summation of squared terms?


答:
Answers

```<code>f(X=0)=.0143
f(X=1)=.2281
f(X=2)=.5116
f(X=3)=.229
f(X=4)=.0143
</code>```

Method:

In this method, blue and red balls are collectively labeled R because it makes no difference. Create a tree and keep track of number of marbles in each bag at each branch of tree, as well as what you take out of the appropriate bag.

$$\left(\begin{matrix}2G\\3R\end{matrix}\right)\left(\begin{matrix}2G\\3R\end{matrix}\right)\begin{cases}1G1R\left(\begin{matrix}1G\\2R\end{matrix}\right)\left(\begin{matrix}3G\\4R\end{matrix}\right)\begin{cases}1G1R(2)\\2G(3)\\2R(1)\end{cases}\\
2G\left(\begin{matrix}0G\\3R\end{matrix}\right)\left(\begin{matrix}4G\\3R\end{matrix}\right)\begin{cases}1G1R(1)\\2G(2)\\2R(0)\end{cases}\\
2R\left(\begin{matrix}2G\\1R\end{matrix}\right)\left(\begin{matrix}2G\\5R\end{matrix}\right)\begin{cases}1G1R(3)\\2G(4)\\2R(2)\end{cases}\end{cases}$$

This bizarre-looking table tells you

The starting configuration of balls in Box A and Box B, respectively

The combination of balls drawn from Box A

The configuration of balls at this stage in Box A and Box B, respectively

The combination of balls drawn from Box B

The number of green balls in Box A

Simply compute each of the nine probabilities and then add the entries for X=0, X=1, ..., X=5 (these are in the little parentheses at the end of each stem) together to obtain their probabilities, e.g. for X=3 you'll have to add 2 probabilities together. For example, the very first of the nine rows has probability

$$\frac{{2\choose 1}{3 \choose 1}}{{5\choose 2}}*\frac{{3\choose 1}{4 \choose 1}}{{7\choose 2}}=.34\\
\text{or}\\
\frac 2 5 * \frac 3 4 * 2 * \frac 3 7 * \frac 4 6 * 2=.343$$

where you multiply the probability of picking 1 green 1 red from Box A in stage 1 (2G3R pot) with the probability of from picking 1G1R from Box B in stage 2 (3G4R pot).

Body:
There are two bags A and B. Bag A has 2 green marbles and 3 red marbles. Bag B has 2 green marbles and 3 blue marbles. Two marbles are drawn without replacement from bag A and put in bag B. Then two marbles are drawn without replacement from bag B and put in bag A. Let X be the number of green marbles in bag A after the above operations. Compute the probability mass function of X.

My attempt:

$X=\{0,1,2,3,4\}$

$X=0$: take all green from A and put non-greens back = $\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{7}\cdot\frac{2}{6}=\frac{17}{70}$

$X=1$: take all green from A and put one green back + take 1 green and put non-greens back = $\left(\frac{2}{5}\cdot\frac{1}{4}+\frac{4}{7}\cdot\frac{3}{6}\right)+\left(\frac{2}{5}\cdot\frac{3}{4}+\frac{4}{7}\cdot\frac{3}{7}\right)=\frac{228}{245}$

Then we continue on with $X=2,3,4$. My concern is that if we add up $\mathbb{P}(X=0)$ and $\mathbb{P}(X=1)$, we get a value that is greater than $1$, which should not be the case because everything should sum to $1$. Can someone tell me what I am doing wrong and possibly how I would approach this problem differently to get the right answer? Thank you.

Body:
Answers

f(X=0)=.0143
f(X=1)=.2281
f(X=2)=.5116
f(X=3)=.229
f(X=4)=.0143

Method:

In this method, blue and red balls are collectively labeled R because it makes no difference. Create a tree and keep track of number of marbles in each bag at each branch of tree, as well as what you take out of the appropriate bag.

$$\left(\begin{matrix}2G\\3R\end{matrix}\right)\left(\begin{matrix}2G\\3R\end{matrix}\right)\begin{cases}1G1R\left(\begin{matrix}1G\\2R\end{matrix}\right)\left(\begin{matrix}3G\\4R\end{matrix}\right)\begin{cases}1G1R(2)\\2G(3)\\2R(1)\end{cases}\\
2G\left(\begin{matrix}0G\\3R\end{matrix}\right)\left(\begin{matrix}4G\\3R\end{matrix}\right)\begin{cases}1G1R(1)\\2G(2)\\2R(0)\end{cases}\\
2R\left(\begin{matrix}2G\\1R\end{matrix}\right)\left(\begin{matrix}2G\\5R\end{matrix}\right)\begin{cases}1G1R(3)\\2G(4)\\2R(2)\end{cases}\end{cases}$$

This bizarre-looking table tells you

The starting configuration of balls in Box A and Box B, respectively

The combination of balls drawn from Box A

The configuration of balls at this stage in Box A and Box B, respectively

The combination of balls drawn from Box B

The number of green balls in Box A

Simply compute each of the nine probabilities and then add the entries for X=0, X=1, ..., X=5 (these are in the little parentheses at the end of each stem) together to obtain their probabilities, e.g. for X=3 you'll have to add 2 probabilities together. For example, the very first of the nine rows has probability

$$\frac{{2\choose 1}{3 \choose 1}}{{5\choose 2}}*\frac{{3\choose 1}{4 \choose 1}}{{7\choose 2}}=.34\\
\text{or}\\
\frac 2 5 * \frac 3 4 * 2 * \frac 3 7 * \frac 4 6 * 2=.343$$

where you multiply the probability of picking 1 green 1 red from Box A in stage 1 (2G3R pot) with the probability of from picking 1G1R from Box B in stage 2 (3G4R pot).


问:
Expectation of Time Series Process
So I have the following AR(2) process:

$z_t = \delta + \psi_2z_{t-2} + \epsilon_t$

which has the following MA($\infty$) representation:

$\frac{\delta}{1 - \psi_2} + \sum_{n=0}^\infty \psi_2^n \epsilon_{t-2n}$

Now I am looking for $E[z_t|z_1, z_2, . . . ,z_{t-2}, z_{t-1}]$

This is where I am kind of confused.

Would it just be $\frac{\delta}{1 - \psi_2} + \sum_{n=1}^\infty \psi_2^n \epsilon_{t-2n}$

Therefore, the only difference would be the sum begining at $1$ instead of $0$?

答:
Why do you need the $MA(\infty)$ representation here? If you're conditioning on the previous values, they can be taken out of the conditional expectation. So we have

$$\mathbb{E}[z_t|z_1,\dots,z_{t-1}]=\mathbb{E}[\delta +\psi_2 z_{t-2}+\epsilon_t|z_1,\dots,z_{t-1}] = \delta + \psi_2z_{t-2}+\mathbb{E}[\epsilon_t|z_1,\dots,z_{t-1}]$$

You mention the process is AR(2), so presumably the white noise terms are i.i.d. and independent of the history of the process, i.e., $\mathbb{E}[\epsilon_t|z_1,\dots,z_{t-1}]=\mathbb{E}[\epsilon_t]=0$. So we would have

$$\mathbb{E}[z_t|z_1,\dots,z_{t-1}]= \delta + \psi_2z_{t-2}$$

Title: Expectation of Time Series Process


问:
Letters for complex numbers
Suppose that I am writing a proof or some other piece of mathematical writing, and wish to introduce $n$ distinct complex numbers, for some positive integer $n$.  What are the complex numbers called?

If $n=1$, then clearly the (unique) complex number I am interested in is called $z$.

If $n=2$, the two complex numbers are called $z$ and $w$.

It is the case $n\ge3$ that i am concerned about.  There seem to be no other standard letters for complex numbers, and all the other letters around that end of the alphabet ($x,y,u,v,t$ etc.) are reserved for real numbers.  Indeed we often write $z=x+iy$ and $w=u+iv$ where $x,y,u,v$ are stipulated to be real.

What am I missing here?  I can't think of any way to get around this problem.  Is there a theorem in mathematics that states that every problem parametrized by $n$ complex numbers is equivalent (in some precise sense) to a problem involving just two complex numbers?

答:
Subscripts!
$$
z_1 = x_1 + i y_1, \dots, z_n = x_n + i y_n
$$</p>

Title: Letters for complex numbers


问:
Indefinite integral question with cube roots
Evaluate the following integral:

$$\int\sqrt[3]{x^3+\frac 1 {x^3}} dx$$

For $x^3=u$ and $1+\frac 1 {u^2}=v$, getting $$-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv$$, what is the next step?

答:
Continue with

$$I=-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv
= -\frac 1 6 \int \frac1{v-1} \sqrt[3]{\frac v{v-1} } dv
$$

and let $t^3 = \frac v{v-1}$. Then, the integral becomes
\begin{align}
I &= \frac12 \int \frac{t^3}{t^3-1}dt
= \frac12 t +\frac16\ln \frac {t-1}{\sqrt{t^2+t+1}}-\frac1{2\sqrt3}\tan^{-1}\frac{2t+1}{\sqrt3}
\end{align}


问:
Working out efficiently if points are within polygons for large datasets
Perhaps this question should be elsewhere but as its mainly a mathematical problem Im posting here.

I have a dataset of which looks like this:

```<code>[
  [ 242841.86914496083,
  1090.0027001134586,
  11711.344635666988,
  142639.20305005432 ],

  [ 212841.86914496083,
  128007.0027001134586,
  11711.344635666988,
  142639.20305005432 ]
  ...
]
</code>```

Each array is called an "event" and in this case have 4 "parameters". I want to plot these events on multiple scatter charts. For example, on the first scatter chart, i'll plot the first and second parameter of each event, so the (x, y) point for the first event is (242841.86914496083, 1090.0027001134586), for the 2nd event is (212841.86914496083, 128007.0027001134586) etc. On the second chart, i'll plot the second and third parameter of each event, so the (x, y) point for the first event is (11711.344635666988, 142639.20305005432), for the 2nd event is (11711.344635666988, 142639.20305005432) etc.

In the example Im giving the range is 0 to 262144;

The scatter charts are 200px x 200px but can be any pixel size up to 2000px. It's always square.

When plotting, I want to work out which points are within a polygon. Here is a sample polygon:

```<code>[
    [
        200000,
        100000
    ],
    [
        230000,
        180000
    ],
        [
        200000,
        200000
    ],
    [
        200000,
        230000
    ],
    [
        110000,
        220000
    ],
    [
        100000,
        180000
    ],
]
</code>```

The points within the polygon are to be plotted in a different color to the points outside it. The polygon is to be applied to scatter chart 1 and 2, but it's the first and second parameter of each event that is to be used to figure out if the event is within the polygon and what color it should be on both charts. E.g. say the point for event 2 (212841.86914496083, 128007.0027001134586) are within the polygon, this mean that event will appear in red in each chart.

I have a simple algorithm which runs through each point and checks if it's within the polygon. I use https://www.npmjs.com/package/point-in-polygon.

Keep in mind, there are often multiple polygons (n), and these can be applied to different combinations of parameters. So I need to work out if an event is in all n polygons, or if its in n-1 polygons etc.

To sum up, what I need to know is:

percentage of total events within each polygon

number of events within a polygon

number of events within n, n-1 etc polygons

x value mean of event with polygon

y value mean of event with polygon

My algorithm performs well for up to 10,000 events. But then it hits issues. Some of my users have files that contain 1 millions events, and my algorithms take minutes.

How can I work out what I need more efficiently? Binning data is an option but then I lose accuracy in results. E.g. for a 200 by 200 pixel image I could create 200 bins for each axis. Problem is a then lose accuracy on the x and y mean values. Also, users can change to 201 x 201 pixel images, now I need to work out 201 bins etc.

After file upload, these files are on the server and can be pre-processed before graphs are generated. Is there anything that can be done mathematically to improve performance for large number of events?

EDIT:
To be clear - figuring out if a point is in a polygon is not the most performance intensive. The most performance intensive part is looping through 1 million events in order to plot them on a, say, 400px by 400px image. The solution, probably lies somewhere in binning data - but if I make 400 bins, then the user - on the fly - changes the image size to 410px, I now need to make 410 bins from the ORIGINAL dataset. Is there some way around this?

EDIT
So the data is known in advance. This sits on my server and can be manipulated in anyway. The size of the image and the polygons are now knows in advance - users draw the polygons and the image must change on the fly. With, say, 1 million events, this is not possible - the browser usually crashes...I can bin the data, but if I create 200 bins (standard image is 200 x 200), then the user makes the image bigger e.g. 210 x 210, now I have to cycle through 1 million events and create 210 bins, which is obviously a big performance issue

答:
Here is an algorithm that I would generically expect to give you a huge performance boost.  I can't quantify the performance boost without making additional assumptions (such as the probability distribution of the data points or the size/shape of the polygons).  But in realistic cases I would expect it to be much faster.

I'm going to make the following assumptions:

The polygons are not known in advance.

The dataset is known in advance.

The image resolution is not known in advance.  (You mentioned the user changing it suddenly.)

The current problem is that if there are $p$ polygons and $n$ data points, then the naive approach (testing every data point and every polygon independently) scales like $O(pn)$, so it's terrible when $p$ and $n$ are both large.  With the assumptions above, there isn't an obvious solution for slowing down the $p$ factor, but you can do something for the $n$ factor.

My suggestion basically amounts to binning information about the data, but in a resolution-independent way.

Before the user runtime, put the data points into a k-d tree.  This is a binary tree search structure that partitions the data into smaller and smaller geometric regions in a very generic way.  (Here, $k$ is the dimensionality of the space the points are in, so $k=2$.)

For every node in the tree, store two things:

The number of points belonging to that branch of the tree.

The bounding box of all of these points.

At runtime, iterate over all polygons and over the nodes in the tree.  For the node iteration, descend through the tree depth-first.  At each node, check whether the bounding box of the node is completely inside the polygon, completely outside the polygon, or neither.  (See step 4 for a suggested implementation.)  Use the following to respond to each case:

Completely inside the polygon: In this case, count all of the points on this branch of the tree as being inside the polygon and stop descending further into this branch.

Completely outside the polygon: Stop descending further into this branch.

Neither: Keep descending and repeat this check.

Suggested algorithm for checking containment of bounding box and polygon: For every edge $e$ in polygon $p$, check whether $p$ intersects any of the four sides of the bounding box.  If any of them do, immediately put this in the "neither" category (there is partial overlap between the $p$ and the bounding box).

If there are no intersections, then the bounding box must be completely inside or completely outside of the polygon.  To decide which is which, you can check whether any point in the bounding box is inside or outside the polygon (which you already know how to do for a point).

Three things to note about this item:

I am not an expert at computational geometry and this part can probably be improved.

I am assuming that you have a function that can check efficiently whether two line segments intersect.  This is a problem that has been discussed elsewhere (for example, here).

When you get to a leaf node of the tree, then the node corresponds to a point rather than a rectangle, which is much simpler to check.  (You might want to insert special handling for this; I'm not sure if it is worth it.)

The idea here is to prune off whole branches of the tree as soon as you possibly can.  I expect that this will give a huge speedup immediately, but it's hard to prove this.  If you don't see the speedup immediately and want to start investigating/profiling, then here are a few things to keep in mind.

Depth matters.  Anything you can do to classify nodes further up the tree will help.

If the polygons have a lot of edges, then the algorithm I proposed in (4) might be slow, and there might be a better one.

The absolute worst-case scenario here is that you check all of the nodes in the tree.  There are about twice as many nodes as points, and the check that I am proposing is more expensive than the check for a single point, so the behavior would actually get worse.  But I cannot imagine this ever happening in a realistic case.  (You would have to draw a polygon that looks like a really long and winding snake that passes very close to every point.  Otherwise, there would be nodes that are classified as entirely inside or entirely outside of the polygon.)


问:
Having trouble integrating with Gamma function.
I am trying to find an expected value by integration. Not sure if it is hopeless but the statistics part is not that important in this case, rather I want to see if this integral does have a nice anti-derivative. I am sadly not too familiar with gamma function properties besides the very simple properties.

$$\int_{0}^{\infty}\frac{1}{x}\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}x^{-n\alpha-1}e^{-\frac{n\beta}{x}}dx $$

As usual pulling out constants and adding exponents for same base.

$$\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\int_{0}^{\infty}x^{-n\alpha-2}e^{-\frac{n\beta}{x}}dx $$

But this is where I do not know how to continue, I tried following this https://math.stackexchange.com/a/3650717/633922 post, but I am unsure if it is correct or if it even applies in my case. Any help would be appreciated.

答:
Take $n\beta/x=y$ getting$$I=\frac{(n\beta)^{-1}}{\Gamma(n\alpha)}\int_0^\infty y^{n\alpha} e^{-y}dy=\frac{\Gamma(n\alpha+1)}{n\beta\Gamma(n\alpha)}$$What property of the Gamma function can you use to simplify the numerator?

 For $x>0,\Gamma(x+1)=x\Gamma(x)$ giving you the answer $\alpha/\beta$.

Edit: Expanding on my answer, $dx=-n\beta~dy/y^2$ so$$\begin{align*}I&=\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\int_{0}^{\infty}x^{-n\alpha-2}e^{-\frac{n\beta}{x}}dx\\&=\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\int^0_{\infty}\left(\frac{n\beta}y\right)^{-n\alpha-2}e^{-y}\left[\frac{-n\beta}{y^2}\right]dy\\&=\frac{(n\beta)^{-1}}{\Gamma(n\alpha)}\int_0^\infty y^{n\alpha} e^{-y}dy\end{align*}$$


问:
Probability of 3 friends being part of 5 person committee
There are $18$ students that will be forming a committee of $5$ people. The people in the committee will be picked out of a hat randomly. What is the probability that Joe, Joseph, and John will be $3/5$ members of the committee?

I've tried during various things, I know that the total number of combinations of 5 people that can be formed from $18$ is $8568$  $(18C5)$, but I'm not exactly sure what the numerator should be. I would love an explanation as to why the answer is why it is please

答:
You already have @John's 'hidden' answer (+1). Here are
two different methods. Now is the time to consider
how you can confidently get the next such problem correct without help.

Simulation in R: If the three J's are numbered 1,2, and 3 out of 18,
then simulation of a million such draws gives the answer
correct to about 3 places:

```<code>set.seed(2021)
students = 1:18
jjj = replicate(10^6, sum(sample(students, 5)<=3))
mean(jjj == 3)
[1] 0.012278
</code>```

Hypergeometric geometric probability: There are 3 red chips and 15 black chips in an urn
and three are drawn at random without replacement,
what is the probability all three drawn are red?

```<code>dhyper(3, 3,15, 5)
[1] 0.0122549
</code>```

Body:
You already have @John's 'hidden' answer (+1). Here are
two different methods. Now is the time to consider
how you can confidently get the next such problem correct without help.

Simulation in R: If the three J's are numbered 1,2, and 3 out of 18,
then simulation of a million such draws gives the answer
correct to about 3 places:

set.seed(2021)
students = 1:18
jjj = replicate(10^6, sum(sample(students, 5)<=3))
mean(jjj == 3)
[1] 0.012278

Hypergeometric geometric probability: There are 3 red chips and 15 black chips in an urn
and three are drawn at random without replacement,
what is the probability all three drawn are red?

dhyper(3, 3,15, 5)
[1] 0.0122549


问:
Connection between independence number and complete induced subgraphs.
Question:

I'm trying to prove that if the independence number of a graph $G=(V,E)$ is $n$, then $V$ may be partitioned into $n$ sets $S_1, S_2, \dots, S_n$ such that $\bigcup\limits_{i}S_i=V$ and for each $S_i$, the induced subgraph $G[S_i]$ is complete.

My attempt at a constructive proof:

Let $X$ be a maximum independent set in $G$ with $|X|=n$. For each $x_i \in X$, let $S_i$ be the set of vertices of the largest complete subgraph of $G[N(x_i) \cup \{x_i\}]$. From here, I wanted to show that for each $v \in V \setminus X$, $v$ must be in at least one such $S_i$ but I'm not sure how to go about this.

答:
Considering the complementary graph $\overline G$, the statement you are trying to prove is equivalent to the following:

If the clique number of a graph $G=(V,E)$ is $n$, then $V$ be be partitioned into $n$ independent sets, i.e., $\chi(G)\le n$.

This, of course, is falso; the self-complementary graph $C_5$ is the simplest counterexample to this statement and to your statement. In fact, there are triangle-free graphs (clique number $2$) with arbitrarily high chromatic number; see the answers to the question Construction of a triangle-free graph of chromatic number 1526.


问:
Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$

If $x_1,x_2,x_3$ be roots of $x^3+3x+5=0$, find the value of
$$\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$$

To solve this problem I expanded the expression and used Vieta's formula and got the answer $-\frac{29}5$
. but I wonder is there any other approach to solve this problem?

I tried this way as an alternative ansewr:

Consider $x_1,x_2,x_3$ be roots polynomial $P(x)=x^3+3x+5$. then $\frac1{x_1},\frac1{x_2},\frac1{x_3}$ are roots of $P(\frac1x):$

$$P(\frac1x)=0\quad \rightarrow \quad P(\frac1x)=(\frac1x)^3+\frac3x+5=0\rightarrow 5x^3+3x^2+1=0$$
But I don't know whether it helps or not.

Body:
we have to find $$\frac{(x_1^2+1)(x_2^2+1)(x_3^2+1)}{x_1x_2x_3}$$ now $$P(x)=x^3+3x+5=(x-x_1)(x-x_2)(x-x_3)$$ Notice that $$P(i)P(-i)=(x_1^2+1)(x_2^2+1)(x_3^2+1)$$ now use $x_1x_2x_3=-5$ to finish..

$i=\sqrt{-1}$  and $P(i)P(-i)$ can be easily found and is left as an exercise


问:
How is the Prékopa Leindler inequality different from Holder's inequality?
The following is an extract from these notes (Pg. 30), and I'm not able to understand the part towards the end. I would appreciate any help!

Prékopa-Leindler Inequality: Let $f$, $g$ and $m$ be non-negative measurable functions on $\mathbb R^n$, $\lambda\in(0,1)$ and for all $x,y\in\mathbb R^n$,
\begin{equation}
    m(\lambda x + (1-\lambda)y) \geq f(x)^{\lambda}g(y)^{1-\lambda}.
\end{equation}
Then,
$$\int_{\mathbb R^n} m \geq \left(\int_{\mathbb R^n}f\right)^\lambda \left(\int_{\mathbb R^n}g\right)^{1-\lambda}$$
It is perhaps helpful to notice that the Prékopa–Leindler inequality looks like
Hölder’s inequality, backwards. If $f$ and $g$ were given and we set $$m(z) = f(z)^{1-\lambda} g(z)^\lambda$$
(for each $z$), then Hölder’s inequality says that $$\int_{\mathbb R^n} m \leq \left(\int_{\mathbb R^n}f\right)^{1-\lambda} \left(\int_{\mathbb R^n}g\right)^{\lambda}$$
(Hölder’s inequality is often written with $1/p$ instead of $1 − λ$, $1/q$ instead of
$λ$, and $f , g$ replaced by $F^p , G^q$ .) The difference between Prékopa–Leindler and
Hölder is that, in the former, the value $m(z)$ may be much larger since it is a
supremum over many pairs $(x, y)$ satisfying $z = (1 − λ)x + λy$ rather than just
the pair $(z, z)$.

Could someone please explain the part in bold? I don't understand the part where the author talks about the suprema over pairs $(x,y)$ and $(z,z)$ in the two cases. Lastly, what's $F^p, G^q$?

Thank you.

答:
In favour of symmetry, I've switched $\lambda$ and $1-\lambda$ in the Hölder's inequality you have stated.

Hölder's inequality says that given non-negative functions $f$ and $g$, if we define $$m(z)=f(z)^\lambda g(z)^{1-\lambda}$$
for each $z$,
$$\int m \leq \left(\int f\right)^\lambda \left(\int g\right)^{1-\lambda}.$$

In the Prékopa-Leindler Inequality on the other hand, each $m(z)$ is bounded below by some parameter depending on all the $f(x)$ and $g(y)$ for points $x$,$y$ satisfying $z=\lambda x + (1-\lambda)y$.

Since the inequality is a lower bound on the integral of $m$, the worst-case scenario is when $m$ attains the minimum possible value at every point, which is the supremum that is mentioned. That is, if we define
$$m(z)=\sup_{\substack{x,y\in\mathbb{R}^n \\ z=\lambda x + (1-\lambda)y}} f(x)^\lambda g(y)^{1-\lambda}$$
for each $z$ (even for any general $m$ satisfying the condition, $m(z)$ must be at least the quantity above),
$$\int m\geq \left(\int f\right)^\lambda \left(\int g\right)^{1-\lambda}.$$

Comparing the definition of $m$ in the two inequalities, it is seen that we only take the pair $(x,y)=(z,z)$ in Hölder's inequality.

The two inequalities are rather similar in terms of basic structure, but also quite different due to the supremum involved in the Prékopa-Leindler inequality, which causes the direction of the inequality to be reversed.

Lastly, what's $F^p$,$G^q$?

Hölder's inequality usually isn't stated in the format the author (and I) have stated it above. In the Wikipedia article which contains the "usual" statement of the theorem, it is given as

Let $(S, \Sigma, \mu)$ be a measure space and let $p, q \in [1, \infty)$ with $1/p + 1/q = 1$. Then, for all measurable real- or complex-valued functions $F$ and $G$ on $S$,
$$\|FG\|_1 ≤ \|F\|_p \|G\|_q. $$

The equivalence of this and the statement described in the notes is seen on writing $f=F^p$, $g=G^q$, and $1/p=\lambda$.


问:
Convergence of the Laplace Transform
The bilateral Laplace transform is defined as

$$X(s) = \int_{-\infty}^{\infty}x(t)e^{-st}dt$$where $s = \sigma +j\omega$. I'm trying to prove rigorously

The ROC(region of convergence) of $X(s)$ consists of strips parallel to the $j\omega$-axis in the $s$-plane.

In Oppenheim's book it's proved by noting that the ROC of $X(s)$ consists of
those values of $s = \sigma +j\omega$ for which the Fourier transform of $x(t)e^{-\sigma t}$ converges and this means that  the ROC of the Laplace transform of $x(t)$ consists of those values of $s$ for which $x(t)e^{-\sigma t}$ is absolutely integrable:
$$\int_{-\infty}^{\infty}|x(t)|e^{-\sigma t}dt\lt\infty$$Because this condition depends only on $\sigma$, the real part of $s$, we have proved the mentioned statement.

My problem with this proof is that $$\text{convergence of the Fourier transform} \implies \text{absolute integrability }$$In fact the converse of this implication is true and absolute integrability is a sufficient condition for the convergence of the Fourier transform. So I think this proof is wrong. Is my understanding correct? Also is the mentioned statement correct? How can we prove that rigorously?

答:
The first thing to get straight is that convergence of the Fourier transform of $f$ does not imply absolute integrability, i.e., that $f \in L^1(\mathbb{R})$.

For a counterexample, take $f(x) = \sin (x^2)$ where

$$\int_{-\infty}^\infty\sin(x^2)e^{ikx} \, dx = 2\int_0^\infty\sin(x^2) \cos(kx) \, dx = \int_0^\infty(\sin(x^2+kx)+ \sin(x^2-kx)) \, dx$$

Making the change of variables $x = u \mp \frac{k}{2}$ we get

$$\begin{align} \int_0^\infty \sin(x^2\pm kx)\, dx &= \int_0^\infty \sin\left(\left(u\mp \frac{k}{2}\right)^2 \pm k\left(u \mp \frac{k}{2}\right)\right)\, du \\ &= \int_0^\infty\sin\left(u^2 - \frac{k^2}{2}\right)\, du \\ &= \cos \left(\frac{k^2}{4}\right)\int_0^\infty \sin (u^2) \, du - \sin \left(\frac{k^2}{4}\right)\int_0^\infty \cos (u^2) \, du \end{align}$$

The well-known Fresnel integrals on the RHS are $\displaystyle \int_0^\infty \cos (u^2) \, du= \int_0^\infty \sin (u^2) \, du = \sqrt{\frac{\pi}{8}}$.

Substituting above, we get

$$\hat{f}(k) = \int_{-\infty}^\infty\sin(x^2)e^{ikx} \, dx = \sqrt{\frac{\pi}{2}}\left( \cos \left(\frac{k^2}{4}\right)-\sin \left(\frac{k^2}{4}\right)\right), $$

and the Fourier transform converges.  However, $f \not\in L^1(\mathbb{R})$.  This can either be shown directly or inferred from the fact that

$$\lim_{k \to \infty} \left( \cos \left(\frac{k^2}{4}\right)-\sin \left(\frac{k^2}{4}\right)\right)\,\, \text{DNE},$$

If $f \in L^1(\mathbb{R})$ then by the Riemann-Lebesgue lemma we would have $\lim_{k \to \infty}\hat{f}(k) = 0$

ROC Proof

Suppose $x$ is locally integrable and for some $s_0 \in \mathbb{C}$ we have convergence of the unilateral Laplace transform

$$X_+(s_0) = \int_0^\infty x(t) e^{-s_0t} \, dt$$

Let $\alpha(t) = \int_0^t x(u)e^{-s_0 u} \, du$. For any $T>0$ it follows from a basic property of the Riemann-Stieltjes integral (proved here) that

$$\int_0^T x(t) e^{-st}\, dt = \int_0^T e^{-(s-s_0)t} x(t) e^{-s_0t}\, dt = \int_0^T e^{-(s-s_0)t} \, d\alpha(t)$$

Integrating by parts on the RHS, we get

$$\tag{*}\int_0^T x(t) e^{-st}\, dt = \alpha(T)e^{-(s-s_0)T} + (s-s_0)\int_0^T e^{-(s-s_0)t} \alpha(t) \, dt$$

Since $X_+(s_0)$ converges, it follows that $|\alpha(t)|$ is bounded for all $t \geqslant 0$. This implies that for all $s$ such that $\Re(s) > \Re(s_0)$, we have

$$\lim_{T \to \infty}\alpha(T)e^{-(s-s_0)T} = 0,$$

and the integral on the RHS of (*) is absolutely convergent.

Thus,

$$X_+(s) = \lim_{T \to \infty}\int_0^T x(t) e^{-st}\, dt= \lim_{T\to \infty}\int_0^T e^{-(s-s_0)t} \, d\alpha(t) = \int_0^\infty e^{-(s-s_0)t} \, d\alpha(t),$$

where $X_+(s)$ converges for all $s$ such that $\Re(s) > \Re(s_0)$.

In a similar way we can show that if we have convergence of

$$X_-(s_1) = \int_{-\infty}^0 x(t) e^{-s_1t} \, dt,$$

then $X_-(s)$ converges for all $s$ such that $\Re(s) < \Re(s_1)$.

Title: Convergence of the Laplace Transform


问:
Is there a formula of calculating the significance of an increase of a value relative to its size?
This is an idea I had about volume in stock trading (amount of shares traded over a given time interval)

If a stock volume at some point is 5 and the next time unit it's 10, that represents a 100% increase in volume but it only increased by 5, so it's not that significant.

On the other hand, if you had a volume of 10,000 at some point and it went to 11,000 the next time unit, it's only 10% but that's far more significant given the volume's magnitude.

I'm trying to make a program that alerts me based on an increase in volume of a stock, but I wonder if there's a formula that will give me alerts that are actually significant. As the volume goes higher, the increase threshold could be lower for the alert to still make sense (like example above), and vise versa.

Does anybody know of a formula that can determine how much the volume of a stock needs to increase with respect to its initial volume to be "noteworthy" of attention?

Thanks

答:
In order to calculate the significance (usually represented through a "$p$-value") there are two different approaches you can take:

Parametric Approach:

We can assume an underlying distribution of the changes, normally supported by empirical evidence. For example, if you assume that the returns of stock prices, i.e., $(S(t+1)-S(t))/S(t)$, are Normal$(0,\sigma^2)$, and you estimate $\sigma^2$ using past data, then you can calculate a probability of having a move that extreme using the normal distribution. These $p$-values are readily calculated in R, Python, etc., using built in functions. Note that by dividing by $S(t)$ in the above returns we control for the effect of higher priced stocks moving more  - the same reasoning also holds for stock volume.

With regards to financial data, the normal distribution is usually not appropriate as financial data often exhibit heavy tails. Instead, you can use the $t$-distribution, or some other heavy-tailed distribution.

Non-parametric Approach

For this approach, we don't assume an underlying distribution but instead use the previously observed stock volume. For example, let $v_1, v_2, \dots, v_n$  be the daily volumes of a certain stock $i$ over the past $n$ days. Suppose we observe $v_{n+1}$, the volume of current day, and we want to calculate how "significant" the daily volume is. Then we can do this by using the previous days $v_1, v_2, \dots, v_n$. Specifically, calculate the proportion of days with volume larger than the current day, and take this as your $p$-value:

$$\hat{p} = \frac{1}{n}\sum_{i=1}^n \mathbb{1}(v_i > v_{n+1})$$

where $\mathbb{1}(v_i > v_{n+1})$ is the indicator function that is $1$ or $0$ depending on whether the enclosed statement is true or false, respectively. If $\hat{p}$ is small, then the empirical probability of such volume occurring is small, meaning the volume is significant. Conversely, if $\hat{p}$ is large, then such volume occurs frequently, and so it is not significant.

In the above approach we normally use a moving window (using the past $n$ days instead of all previously observed days) since the behavior of financial securities tends to change over time.


问:
Is there always exists a product over a semigroup?
Let $(S,+)$ a semigroup, is there exists a product over $S$, i.e., a binary operation $\cdot$ over $S$ such that $$ x\cdot(y+z)= x \cdot y + x \cdot z,\, (y+z)\cdot x = y\cdot x + z\cdot x,\, x\cdot(y\cdot z) = (x\cdot y)\cdot z$$
for all $x,y,z \in S$ ?

For example, for any finitely abelian group $G$, we know for the fundamental theorem of abelian groups that $G$ is isomorphic to a direct sum $\mathbb{Z}^n \oplus \mathbb{Z}_{p_1} \oplus \cdots \oplus \mathbb{Z}_{p_k}$, where $n,k \geq 0$ and $p_i$ is a prime tower. We can see each one of these groups as a ring under the usual product for $\mathbb{Z}$ and product mod $(p_i)$ for $\mathbb{Z}_{p_i}$, so this direct sum can be seen as a product ring, with $G$ as the underlying abelian group.

答:
Assuming monoid $M$ is commutative with its operation denoted $+$, there is always the trivial product defined by $x \cdot y=0$ for all $x,y \in M$ [where $0$ is the additive identity of $M$ where the commutative monoid $M$ is written in additive notation.] This satisfies the rules you mention, but isn't a very interesting product to say the least. Note this construction works for any abelian monoid, finite or not.

Added later: As I noted in a comment one does not need to assume monoid $M$ is commutative. At the time I wasn't sure and asked for someone to check, which @J.-E. Pin has done.


问:
How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $

$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$

I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.

Body:
We are given $I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)}$

Here we have $7x-10-x^2=(x-2)(5-x)$,
therefore put $\sqrt{7x-10-x^2} =(x-2)t$

on solving and factoring,
we get $x=\frac{\left(2t^2+5\right)}{\left(t^2+1\right)}\ $ therefore

$dx=\frac{-6t}{(t^2+1)^2}$  and $x-2=$ $\frac{(2t^2+5)}{(t^2+1)}-2 =$$\frac{3}{(t^2+1)}$

and $1+\sqrt{7x-10-x^2} = 1+(x-2)t =1+\frac{3t}{t^2+1}$
$=\frac{t^2+3t+1}{t^2+1}$

hence we have $I=\int\frac{t^2+1}{3}.\frac{t^2+1}{t^2+3t+1}.\frac{-6t}{(t^2+1)^2}dt$

$I=-\int\frac{2t}{t^2+3t+1}=-\int\frac{2t+3-3}{t^2+3t+1}dt$

$I=-\int\frac{2t+3}{t^2+3t+1}dt+3\int\frac{dt}{(t+\frac32)^2 -\frac54 }$

$I=ln(t^2+3t+1) -\frac{3}{\sqrt{5}}ln\frac{(2t+3-\sqrt{5})}{(2t+3+\sqrt{5})}$ + C

where $t=\sqrt{\frac{5-x}{x-2}}$.


问:
What is the subsequence of largest points so far called?
Given a sequence $\{x_n\}$, is there a name for the subsequence of largest-points-so-far, that is, $\{x_{n_k} : x_{n_k} > x_i\text{ for all }i<n_k\}$?

答:
I think an official term would be "cumulative maximums".

Sometimes I've seen these called the "champions" of the sequence.


问:
Taking Limits of 2 function
I am taking the log of both numerator and denominator and then trying to find out the limits through L'Hospital's Rule. I am taking log because it makes it easier to differentiate the 2 functions. But I am not sure if this is a valid way to solve the problem. Can anyone clarify this?

$$
\lim_{n\to\infty} \dfrac{f(n)}{g(n)} = \lim_{n\to\infty} \dfrac{\log_2 f(n)}{\log_2 g(n)}
$$

答:
Well, $$\lim_{n \to \infty} \frac{2^n}{3^n} = \lim_{n \to \infty} \Big( \frac23 \Big)^n = 0$$ but
$$\lim_{n \to \infty} \frac{\log_2(2^n)}{\log_2(3^n)} = \lim_{n \to \infty} \frac{n }{n \log_2 3} = \lim_{n \to \infty} \frac{1}{\log_23} = \frac{1}{\log_23}.$$

Title: Taking Limits of 2 function


问:
Algebraic expression in its most simplified form
I am trying to simplify the algebraic expression:

$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$

I am having trouble though. My current thoughts are:

$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$

$$=\bigg(\dfrac{x(x-3)}{1(x-3)}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)}{1(x-3)}+\dfrac{2+6x}{(x-3)}\bigg)$$

$$=\bigg(\dfrac{x(x-3)+(-4)}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)+2+6x}{(x-3)}\bigg)$$

$$=\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x}$$

$$=\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} $$

$$\boxed{=\dfrac{-4(x-3)}{2(1+3x)} }$$ Which does not appear is not the answer. Am I close? Where exactly did I go wrong? I have tried this question multiple times.

Edit: Figured it out!

$\dfrac{x(x-3) - 4}{x(x - 3) + 2(1 + 3x)}\implies\dfrac{x^2-3x-4}{x^2+3x+2}\implies \dfrac{(x-4)(x+1)}{(x+2)(x+1)}$

$(x+1)$'s cancel leaving us with: $\boxed{\dfrac{x-4}{x+2}}$

答:
Comment: After the calculation very well described by AWertheim and amWhy, we end up wit $\dfrac{x^2-3x-4}{x^2+3x+2}$.

Note that the calculation is not over. The top and bottom each factor nicely, and there will be some cancellation.


问:
I can't seem to prove that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology.
Here are my steps:

```<code>(p ∨ q) ∧ (¬p ∨ r) → (q ∨ r)
¬[ (p ∨ q) ∧ (¬p ∨ r) ] ∨ (q ∨ r)   implication to disjunction
¬(p ∨ q) ∨ ¬(¬p ∨ r) ∨ (q ∨ r)      demorgans law
(¬p ∧ ¬q) ∨ (p ∧ ¬r) ∨ (q ∨ r)      demorgans law + double negation
</code>```

I'm stuck on this last step. The only law that seemed hopeful was the distribution law but that won't even work here. I resorted to using a truth table to prove this but I really want to know if it's possible to shrink this proposition to just true to make it a tautology.

Thank you!

答:
You can continue as follows; I think that the steps should all be fairly easy for you to justify.

$$\begin{align*}
&(\neg p\land\neg q)\lor(p\land\neg r)\lor(q\lor r)\\
&\big((\neg p\land\neg q)\lor q\big)\lor\big((p\land\neg r)\lor r\big)\\
&\big((\neg p\lor q)\land(\neg q\lor q)\big)\lor\big((p\lor r)\land(\neg r\lor r)\big)\\
&\big((\neg p\lor q)\land\top\big)\lor\big((p\lor r)\land\top\big)\\
&(\neg p\lor q)\lor(p\lor r)\\
&(\neg p\lor p)\lor(q\lor r)\\
&\top\lor(q\lor r)\\
&\top
\end{align*}$$


问:
How To Negate the Formal Definition of an Injective Function
in my linear algebra class, the professors gave us a formal definition of an injective function, which is

Given a function f,
if $\forall$ x, x' $\in$ $R$ where x $\ne$  x' $\implies$ f(x) $\ne$ f(x'), then f is injective.

I am trying to negate this statement for a proof, but am unsure if I am thinking of this correctly. From what I am able to deduce, this can be broken down into its logical statements. The definition says that we have a P which implies a Q. In this case, the P is [$\forall$ x, x' $\in$ $R$ where x $\ne$  x' $\implies$ f(x) $\ne$ f(x')], and Q is [f is injective.] And if we look at the statement P, we can think about that in terms of R implies T. This is where I get tripped up. The P is [∀ x, x' ∈ R where x ≠ x'] and the Q is [f(x) ≠ f(x')]. Q is clearly a statement. f(x) either does or does not equal f(x'). But, R, which is [∀  x, x' ∈ R where x ≠ x'] does not have a truth value - or the truth value is dependent on x $\ne$ x', which still does not make sense to me.

I say that it does not make sense because I don't understand how to negate this statement. Negating P $\implies$ Q can be written as P $\wedge$ $\neg$Q. This would be equivalent to

[$\forall$ x, x' $\in$ $R$ where x $\ne$  x' $\implies$ f(x) $\ne$ f(x')] $\wedge$ [f is not injective].

While that is logically true, that does not give us any insight into the type of function that a non-injective function is. Intuitively speaking, a function that is not injective can be written as

if $\exists$ x, x' $\in$ $R$ where x $\ne$  x' $\implies$ f(x) = f(x'), then f is not injective.

I cannot figure out how to use logical equivalencies and negations to arrive at that statement. Any help with understanding how to arrive at this definition from the original definition is greatly appreciated.

答:
The reason of your confusion is that the definition you quoted is actually not fully formal, it mixes up English and symbols of formal logic in a bit loosy manner.
 If I was to give a definition of an injective function I would not write
$$ \forall x,\ x' \in R \  where\  x \ne  x' \implies f(x) \ne f(x')$$
but instead I would say that a function $f: R \rightarrow S$ is injective if and only if
$$ \forall x,\ x' \in R: x \neq x' \implies f(x) \ne f(x').$$
And now the negation is easy:
$$ \exists x,\ x' \in R: x \neq x' \land f(x) = f(x').$$

From formal point of view, the literal definition you quoted is incorrect (more strictly speaking it is not even a definition - precisely for the fact that "$x,\ x'\ where\ x \neq x'$" is not a sentence) - so no point in trying to negate it.


问:
Likelihood ratio normal distribution?
Could someone help me fill in the steps to show that
$$
\exp \left\{\frac{1}{2 \sigma^{2}} \sum_{i=1}^{n}\left(x_{i}-\mu_{1}\right)^{2}-\left(x_{i}-\mu_{0}\right)^{2}\right\} = \exp \left\{\frac{n\left(\mu_{0}-\mu_{1}\right)}{\sigma^{2}}\left(\bar{x}-\frac{\mu_{0}+\mu_{1}}{2}\right)\right\}
$$

答:
Expanding the sum at the exponent you get

$$\frac{\Sigma_i\left[x_i^2+\mu_1^2-2\mu_1x_i-x_i^2-\mu_0^2+2\mu_0x_i  \right]}{2\sigma^2}=\frac{n(\mu_1^2-\mu_0^2)-2n(\mu_1-\mu_0)\overline{x}}{2\sigma^2}=$$

$$=\frac{n(\mu_1-\mu_0)}{\sigma^2}\left[\frac{\mu_1+\mu_0}{2}-\overline{x}  \right]$$

....to get exactly your result multiply both factors by $(-1)$

Title: Likelihood ratio normal distribution?


问:
Using the ratio test on $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n^2}$
I already know that I should do the root test on this series due to the $n$th power, but I want to see if I can establish the result using the ratio test first. (Or would I always be stuck with only one kind of test)

I get something like this: $$\lim_{n \to \infty} \left|\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}\right|$$

But then I'm not sure how (or even if I could) simplify this. I've tried on Wolfram Alpha to make it simplify but the best I can get is numerical approximations that seem to converge to $e$

Body:
To find the limit $$ \begin{align}\displaystyle  \dfrac{\left(1+\dfrac{1}{(n+1)}\right)^{(n+1)^2}}{\left(1+\dfrac{1}{n}\right)^{n^{2}}} \cdot \dfrac{\left( 1 + \dfrac{1}{n} \right)^{(n+1)^2}}{\left( 1 + \dfrac{1}{n} \right)^{(n+1)^2}} &= \left( \dfrac{n(n+2)}{(n+1)^2} \right)^{(n+1)^2} \cdot \left( 1 + \dfrac{1}{n} \right)^{2n+1} \\
&= \left( 1 - \dfrac{1}{(n+1)^2} \right)^{(n+1)^2} \cdot \left( 1 + \dfrac{1}{n} \right)^{2n+1} \end{align} $$


问:
What are the common chords formed?

The common chords that I found here were this . (According to me )

C1-C1’ and C1 with C1’-C4’. That equals 4.Then , C2 with C1’-C4’. That equals total of 8.

Then C1 with C1 and similarly til C3 equals 3.

Then , C1’ with C1’ and till C4’.4

Now , my answer is 15 but in book it is 12.

Please tell me where am I wrong ?

答:
Circle $C_1$ has one chord in common with each of the $4$ primed circles.

Same for $C_2$ and $C_3$.

That makes $12$ common chords.

Another way to look at it is to take away all but one circle in each of $G_1$ and $G_2$. In this case it's clear that there's only one common chord. There are $3\times 4$ such pairs of circles.

Title: What are the common chords formed?


问:
Is there a mathematical way to measure complexity?
I was thinking of the mandelbrot set. When you draw a picture of it at some resolution, you are able to get some idea of what it is, but there will be details that are lost.

Is there a way to quantify the amount of detail/complexity of an object? Is there a branch of math that studies this (I am aware of kolmogorov complexity)?

The mandelbrot set would be have a measure of complexity that is somehow absolute (infinite), I suppose.

答:
To the best of my knowledge, there isn't an overarching/unifying theory for this. There are some directions on this within Computational Complexity and Theoretical Computer Science that are worth considering.

As you said, Kolmogorov Complexity measures the length of the shortest program required to generate an object.

Descriptive complexity measures how succinctly we can express certain properties about a given object with respect to an underlying logic. For example, in first-order logic, we are interested in both the number of variables and the quantifier depth of a given expression. The descriptive complexity of an object is closely related to its Weisfeiler--Leman dimension (see for instance, the Cai-Furer-Immerman paper, https://people.cs.umass.edu/~immerman/pub/opt.pdf). Descriptive Complexity also seeks to characterize complexity classes in terms of logical definability. Immerman's text (https://www.springer.com/gp/book/9780387986005) on this is a good starting point. Grohe has a text on descriptive complexity for graphs (https://www.cambridge.org/core/books/descriptive-complexity-canonisation-and-definable-graph-structure-theory/BC758F6004BD96F6995D5F1EF1E29BAD).

One way to approach Automata Theory is to ask the following: given a class $\mathcal{C}$ of languages, what is the minimally powerful type of automaton needed to characterize $\mathcal{C}$? For example, regular languages are precisely captured by finite-state automata.

In circuit complexity, we study the size and depth of circuits, as well as the minimal complexity of circuits needed to compute a given function. One result on the latter is that $\textsf{Parity} \not \in \textsf{AC}^{0}$. This result provides a separation between $\textsf{AC}^{0}$ and $\textsf{ACC}^{0}$ (that is, $\textsf{AC}^{0} \subsetneq \textsf{ACC}^{0}$).

Edinah Gnang has a very recent paper on Partial Differential Encodings of Boolean Functions. This is an algebraic analogue of Kolmogorov Complexity (https://arxiv.org/pdf/2008.06801.pdf).

In Geometric Complexity Theory, one way to try and separate complexity classes $\mathcal{C}_{\text{Easy}}$ and $\mathcal{C}_{\text{Hard}}$ is by finding a module $T$ (think polynomials) that vanishes on every function in $\mathcal{C}_{\text{Easy}}$, but does not vanish on at least one function in $\mathcal{C}_{\text{Hard}}$. The Introduction and Preliminaries of Joshua A. Grochow's paper (https://link.springer.com/article/10.1007%2Fs00037-015-0103-x) serve as a nice introduction.

As was mentioned above, the dimension of a mathematical object is a measure of complexity.

Edit: Since Hausdorff Dimension was mentioned in one of the comments, it is worth pointing out that Hausdorff Dimension has close connections both to fractal geometry and Kolmogorov Complexity (e.g., https://arxiv.org/pdf/cs/0208044.pdf). Jack Lutz is an expert on measures of geometric dimension, including their connections to computational complexity and information theory (e.g., relationships to Kolmogorov Complexity). It would be worthwhile to check out some of his papers (http://web.cs.iastate.edu/~lutz/papers.html).


问:
Considerable Change of Geodesics on Some Surfaces
In $\mathbb{R}^2$, beginning with two points $A$ and $B$ and moving $A$ in a continuous manner, the segment $AB$ changes slightly. Look at the figure below:

However it's not the case on some other surfaces, e.g. the geodesic between $A$
and $B$ compared to the one between $A'$ and $B$ on $S^2$, as is shown in this figure:

What's behind this fact? How can the surfaces of the former kind be distinguished from the latter?

Thanks.

答:
This has to do with the cut locus.  Let $S$ be a surface, and assume that $S$ is geodesically complete, meaning that every geodesic can be defined for all time.  Given a point $B\in S$ and a geodesic $\gamma$ starting at $B$, it follows from general properties of geodesics that there is some $\delta>0$ such that the restriction of $\gamma$ to $[0,\delta]$ is length-minimizing.  If $\gamma$ ever stops being length-minimizing, then there is a time $c>0$ such that $\gamma$ is length-minimizing on $[0,c]$ and not minimizing on any larger interval.  In this case, the point $\gamma(c)$ is called the cut point of $B$ along $\gamma$.   Define the cut locus of $B$ to be the set of all cut points of $B$ along all geodesics.

The important fact is that the complement of the set of cut points of $B$ is a connected open subset $U\subset S$ containing $B$; and for all $A\in U$, the minimizing geodesic from $A$ to $B$ varies continuously with $A$.

There is a related notion (mentioned by @Ted) called conjugate points. Intuitively, a point $A$ is conjugate to $B$ along a geodesic $\gamma$ if there's a nontrivial one-parameter family of geodesics that all start at $B$ and end at $A$.  (The actual definition is more complicated than this, but this will give you the general idea.)  There's a theorem that says the cut point along $\gamma$ (if there is one) comes at or before the first conjugate point (if there is one).

There are a number of powerful theorems that use hypotheses on curvature to draw consequences about conjugate points, and therefore about cut points.  For example, if $S$ is a complete surface with Gaussian curvature everywhere greater than or equal to $1/R^2$ (such as, for example, a sphere of radius $R$ or less), then every geodesic has a conjugate point (and therefore also a cut point) no farther away than $\pi R$.  On the sphere of radius $R$, each geodesic has a cut point  exactly at distance $\pi R$, namely the point antipodal to the starting point.  Thus given a point  $B$ in the sphere, the minimizing geodesic from $A$ to $B$ depends continuously on $A$ as long as $A$ is not equal to the point antipodal to $B$.


问:
Let $A,B,C$ be a complexity classes, are $A \subseteq B \Longrightarrow A^C \subseteq B^C$?
Let $A,B,C$ be a compleixty classes is $A \subseteq B \Longrightarrow A^C \subseteq B^C$.

So $A \subseteq B$ given oracle access to some $L \in C$ complete, follow that $A^C \subseteq B^C$.

is that correct for any complexity classes?

答: Yes- containments relativize. If we can simulate $A$ machines using $B$ machines, then the same simulation goes through for $A^{C}$ machines using $B^{C}$ machines. The only difference is that the $A^{C}$ machines may make oracle queries; but those can be simulated by the $B^{C}$ machines.


问:
Finding the mgf of $X_1 - X_{(1)}$ from Basu's Theorem
Suppose that $X_1, \ldots, X_n$ where $n \geq 2$ are iid random variables with pdf
$$f_X(x) = e^{-(x-\theta)}I(x \geq \theta).$$

I need to find the moment-generating function of the statistic $U = X_1 - X_{(1)}$ where $X_{(1)}$ is the sample minimum. Prior to this, I have found that the distribution function of $T = X_{(1)}$ is $$
f_T(t) = ne^{-n(t-\theta)}I(t\geq\theta).$$

I have also shown that $X_{(1)}$ is complete sufficient and that $U$ is ancillary, and as such, by Basu's theorem, I know that these two statistics are independent.

Now I am supposed to use this information to find the mgf and cdf of $U$. Since $U$ is bounded I know that I can determine the moment-generating function by taking
$$E[U^k] = E[(X_1 - X_{(1)})^k]$$
rather than directly computing the moment generating function (since I do not know the distribution of $U$, just that $U$ is ancillary w.r.t. $\theta$). However, I do not know how to compute this expectation.

I was hoping there would be an algebraic trick that would allow me to use the independence of $T$ and $U$ in computing this expectation, but I cannot find one (I believe that this is how the exercise is intended to be solved, but I am stumped). I did try writing
$$E[U^k] = E\left[\sum_{j=1}^k \binom{k}{j} X_1^k (-X_{(1)})^{k-j}\right] = \sum_{j=1}^k \binom{k}{j} E\left(X_1^j)\right)E\left(-X_{(1)}^{k-j}\right) $$
but this has not helped me compute the expectation, as I do not think that $E(X_1^j)$ has an elementary integral.

If anyone can help me compute this expectation or figure out what trick to use, it would be appreciated.

答: I think you can compute the cdf of $U$ directly. Fix $u \geq 0$. Note that $$F_{U}(u)=P(U\leq u)=\int_{\theta}^{\infty}P(U\leq u|X_{(1)}=t)f_{X_{(1)}}(t)dt$$ Let's take a closer look at $P(U \leq u|X_{(1)}=t)$. We can say $$P(U \leq u|X_{(1)}=t)=P(U \leq u|X_{(1)}=t,X_{(1)}=X_1)P(X_{(1)}=X_1|X_{(1)}=t)+P(U \leq u|X_{(1)}=t,X_{(1)}\neq X_1)P(X_{(1)} \neq X_1|X_{(1)}=t)$$ This simplies to $$P(U \leq u|X_{(1)}=t)=1\cdot \frac{1}{n}+ \frac{\int_t^{t+u}f_{X}(x)dx}{\int_t^{\infty}f_{X}(x)dx}\cdot \frac{n-1}{n}=\frac{1+(n-1)(1-e^{-u})}{n}$$ Plugging this into our formula for $F_{U}$ and integrating yields $$F_{U}(u)=\frac{1+(n-1)(1-e^{-u})}{n}\cdot I(u \geq 0)$$ I'm reluctant to utilize the pdf of $U$ in in our calculation of the moment generating function since $U$ isn't technically a continuous random variable; in fact, the event $\{U=0\}$ occurs with probability $1/n$. To get around this, we can compute $M_{U}(t)=\mathbb{E}(e^{tU})$ via the integral $\int_{0}^{\infty}P(e^{tU}>x)dx$. If $t>0$ is fixed, then $e^{tU}$ is supported on $[1,\infty)$ and so $$M_{U}(t)=1+\int_1^{\infty}\bigg[1-F_{U}\big(\ln(x)/t\big)\bigg]dx=1+\frac{n-1}{n}\int_1^{\infty}\frac{dx}{x^{1/t}}$$ The improper integral on the RHS converges iff $0<t<1$ and evaluates to $\frac{t(n-1)}{n(1-t)}$. In other words, $$M_{U}(t)=1+\frac{t(n-1)}{n(1-t)}$$ for $t\in(0,1)$. Proceeding in this manner for $t<0$ yields the same formula $M_{U}(t)$, and clearly $M_{U}(0)=1$. We conclude $M_{U}(t)=1+\frac{t(n-1)}{n(1-t)}$ is our mgf and has domain $(-\infty,1)$.


问:
Subspace of Vectors orthogonal to every vectors in a given Vector space
In a vector space V(dim-n), prove that the set of all vectors orthogonal to any vector( not equal to 0)  form a subspace V[dim: (n-1)].

I am wondering how the n-1 is coming in the in the picture?

答:
It is possible to prove this without the rank-nullity theorem. Let $V$ be our $n$-dimensional space, and suppose $w \in V$ is non-zero. Let
$$W = \{w\}^\perp = \{v \in V : w \cdot v = 0\}.$$
Since you asked specifically where the $n - 1$ comes from, I'm not going to show that $W$ is a subspace.

The way we shall prove it is to show that, given any basis $B$ for $W$, that $B \cup \{w\}$ is a basis for $V$. $V$ is $n$-dimensional, meaning that $B \cup \{w\}$ must therefore contain $n$ vectors. Note that $w \cdot w = \|w\|^2 > 0$ as $w \neq 0$, hence $w \notin W$. We therefore have $w \notin B$, and so $B$ must only contain $n - 1$ vectors.

Suppose $B = \{u_1, \ldots, u_k\}$ is a basis for $W$. Since $w \notin W = \operatorname{span}\{u_1, \ldots, u_k\}$, we therefore have $B \cup \{w\}$ is linearly independent. So, we simply need to show that $B \cup \{w\}$ is spanning.

Suppose that $v \in V$. We need to write $V$ as a linear combination of vectors from $B \cup \{w\}$. Let
$$u = v - \frac{v \cdot w}{w \cdot w}w.$$
Then $u \in W$, as
$$u \cdot w = \left(v - \frac{v \cdot w}{w \cdot w}w\right) \cdot w = v\cdot w - \frac{v \cdot w}{w \cdot w} w \cdot w = 0.$$
Since $B$ spans $W$, there exist $a_1, \ldots, a_k$ such that
$$a_1 u_1 + \ldots + a_k u_k = u = v - \frac{v \cdot w}{w \cdot w}w.$$
But then, rearranging,
$$a_1 u_1 + \ldots + a_k u_k + \frac{v \cdot w}{w \cdot w}w = v.$$
The left hand side is a linear combination of $B \cup \{w\}$, proving $B \cup \{w\}$ is indeed a basis for $V$ and contains $n$ vectors. Thus, $B$ contains $n - 1$ vectors.


问:
Prove that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$ if and only if $a$ does not divide $b$ in $R$.
Let $R$ be a commutative ring with $1$. Suppose that $a, b \in R$, $b \notin R^\times$ and $\langle a \rangle$ is a maximal ideal of $R$. Prove that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$ if and only if $a$ does not divide $b$ in $R$. [Notice that $R^\times = \{ r \in R ~|~ r \text{ is a unit.} \}$].

$\textbf{My Attempt:}$

Prove of "$\implies$":

Which want to prove that if $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$, then $a$ does not divide $b$ in $R$.

By contra-positive, we can prove that if $a$ divide $b$ in $R$, then $a + \langle b \rangle$ is not a unit in the quotient ring $R/\langle b \rangle$.

Assume $a$ divide $b$ in $R$.

Then, there exist $k \in R$ such that $b = ak$.

Since, $b \in R$ and $b \notin R^\times$.

Then, $b = b \cdot 1 \in \langle b \rangle$.

Since, $a \in R$. So, $ak = b \in \langle b \rangle$.

Then, $(a + \langle b \rangle)(k + \langle b \rangle) = ak + \langle b \rangle = \langle b \rangle = 0 + \langle b \rangle \neq 1 + \langle b \rangle$.

So, $a + \langle b \rangle$ is not a unit in the quotient ring $R/\langle b \rangle$.

So, this proves that if $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$, then $a$ does not divide $b$ in $R$.

Prove of "$\Longleftarrow$":

Which want to prove that if $a$ does not divide $b$ in $R$, then $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$.

Assume $a$ does not divide $b$ in $R$.

Then, there exist $k \in R$ such that $b = ak + r$, where $r \in R$ and $r \neq 0$.

Which I am stuck on how to prove that $r = -1$. Which then I can prove that $ak - 1 \in \langle b \rangle$ and conclude that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$.

Also, is the prove of "$\implies$" correct ?

And, I don't know where to apply the given condition that $\langle a \rangle$ is a maximal ideal of $R$.

答:
Hint: Note that for $ak-1 \in \langle b\rangle$ you don't need to find $k$ with $ak-b=1$ (no such $k$ will exist in general, try $a=7, b=10$ in $R=\mathbb Z$), but only a $k$ with $ak-bl=1$ for some $l\in R$. Maybe now you see where to use the maximality of $\langle a\rangle$?

In your proof for the other direction, you are a bit imprecise at the end: It seems like you want to argue that the existence of some $k$ such that $ak +\langle b\rangle \neq 1 +\langle b\rangle$ shows that $a +\langle b\rangle$ is not a unit. That's not the case. But the fact that there is a $k$ such that $ak +\langle b\rangle = 0 +\langle b\rangle$ indeed shows what we need, namely that there can be no other $k$ with $ak +\langle b\rangle = 1 +\langle b\rangle$. Do you see why?


问:
Taylor approximation $f(x)=e^{-e^x}$ for $\int_0^1 f(x)dx$ with an error less than $10^{-2}$
Taylor approximation of $f(x)=e^{-e^x}$ for $\displaystyle\int_0^1 f(x)dx$ with an error less than $10^{-2}$ from $g(x)=e^x$ using substitution.

This is what I did for getting the Taylor series of $e^{-e^x}$:

$ e^{\displaystyle-e^x}=e^{-\displaystyle\sum_{k=0}^n\frac{x^k}{k!}}=\displaystyle \prod_{k=0}^n\sum_{j=0}^n\frac{(-1)^j}{j!}\bigg(\frac{x^k}{k!}\bigg)^j$.

I'm not sure how to compute the error, using Lagrange Reminder, or something that helps. I'm stuck because of the productory.

答:
I totally understand your difficulties with such an approach.

Let $y=e^x$ and develop $e^{-y}$ as an infinte series around $y=1$ (corresponding to $x=0$). Replace $y$ by $e^x$ to have
$$f(x)=e^{-e^x}=\sum_{n=0}^\infty (-1)^n\frac{ e^{n x}}{n!}$$ and
$$\int_0^1 f(x)\,dx=\sum_{n=0}^\infty (-1)^n\frac{ 1}{n!}\int_0^1 e^{n x}\,dx=\sum_{n=0}^\infty (-1)^n\frac{ 1}{n!}\frac{e^n-1}{n}$$ So, if you make the summation up to $p$, what you want is that
$$R_p=\frac{e^{p+1}-1}{(p+1)^2 \,p!} \leq 10^{-k}$$ Neglecting the $-1$ in numerator, we then want
$$(p+1)^2\,p! \geq e^{p+1}\, 10^k$$ We shall simplify noting that $(p+1)^2\,p!\sim (p+2)!$ and write
$$(p+2)! \geq e^{p+2}\, \frac{10^k} e$$

Looking at this question of mine and adapting @robjohn's magnificent approximation to this problem, we have

$$p\sim e^{2+W(t)}-\frac 52 \qquad \text{where} \qquad t=\frac{2 k\log \left(10\right)-3-\log (2 \pi )}{2 e^2}$$ $W(t)$ being Lambert function.

Trying for $k=2$  gives $p=6.83823$ than $\lceil p \rceil=7$. Notice that, without any approximation, the exact result would be $p=6.93724$.

Checking with the exact $R_p$
$$R_{6}=0.0311 > 10^{-2}\qquad \text{but} \qquad  R_{7}=0.0092<10^{-2}$$


问:
Help me prove $\left(\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$ isn't a perfect square
I need some help in proving

$$\left( \displaystyle\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$$

isn't a perfect square. The only way I can think is repeated integration by parts which is obviously impractical and I still likely wouldn't be able to deduce if the result was a perfect square.

Thanks very much guys! :D

答:
In 1975
P. Erdős and J. L. Selfridge
proved
"The product of consecutive integers is never a power"
in
Illinois J. Math. Volume 19, Issue 2 (1975), 292-301.

One URL:  http://projecteuclid.org/euclid.ijm/1256050816


问:
Sum of squared probability distribution $\sum_c p_c^2$ is a measure of uncertainty?
Can the sum of squared probabilities, i.e. $\sum_c p_c^2$,
be considered to be a measure of uncertainty? If so, does it have a mathematical name or theory?

The form is similar to Shannon's entropy, $-\sum_c p_c \log(p_c)$. It satisfies $\frac{1}{C} \leq \sum_c p_c^2 \leq 1$, partly according to:
Is there some general lower bound for sum of squared probabilities?, which shows that the lower bound is satisfied when $\vec{p}=\frac{1}{C}(1,\cdots,1)$. $\sum_c p_c^2$ never exceeds 1, since $\sum_c p_c=1$ and $p_c^2 \leq p_c$ ($\because 0 \leq p_c \leq 1$). An obvious case for the upper bound is $\vec{p}=(0,\cdots,0,1,0,\cdots,0)$. Taking them together, it seems like that it can be considered to be a measure of uncertainty.

It is appreciated if anyone could provide any names/theories. Best,

答:
It is commonly referred to as the "collision probability," and, by taking the logarithm, you get the collision entropy (which is a Rényi entropy of order 2).

To see why this is called the collision probability (which is the  squared $\ell_2$ norm of the probability mass function), note that if you have two i.i.d. random variables $X,Y$ with probability mass function $p$, then
$$
\mathbb{P}\{X=Y\} = \sum_{c} \mathbb{P}\{X=c, Y=c\}
= \sum_{c} \mathbb{P}\{X=c\}\mathbb{P}\{Y=c\}
= \sum_{c} p_c^2 = \|p\|_2^2
$$

As you mentioned, it is minimized for the uniform distribution on the domain, since, if $p$ has a domain of size $n$ and we denote by $u$ the uniform distribution on this domain, then
$$
0 \leq \|p-u\|_2^2
= \sum_c (p_c-u_c)^2
= \sum_c p_c^2 - \frac{2}{n}\sum_{c} p_c + \frac{1}{n}
= \sum_c p_c^2 - \frac{2}{n} + \frac{1}{n}
= \|p\|_2^2 - \frac{1}{n}
$$
and so $\|p\|_2^2 \geq \frac{1}{n} = \|u\|_2^2$. (Note that as a corollary, it exactly captures the $\ell_2$ distance to the uniform distribution over the domain.)