问:
How can we derive the dot product from the covariance of two random vectors?
I'm trying to understand a geometric interpretation of covariance as explained in this lecture pdf, which states:

If X, Y are two random variables of zero mean, then the covariance Cov[XY ] = E[X · Y ] is the dot product of X and Y. The standard deviation of X is the length of X. The correlation is the cosine of the angle between the two vectors.

I can understand how, given a zero mean, the standard deviation of X is the length of X. What I don't understand is how the the covariance of X and Y is equal to it's dot product.

Specifically, the covariance is defined as:

$Cov(X,Y) = E[(X - \mu_X) (Y - \mu_Y)]$

Which with we can write as:

$Cov(X,Y) = \sum\limits_{i}\sum\limits_{j} p_{ij} (x- \mu_X) (y-\mu_Y)$

And since for two centered random variables the means are zero, and the probability is uniform, we can rewrite the formula as:

$Cov(X,Y) = \dfrac{1}{N} \sum\limits_{i}\sum\limits_{j} x y$ (Eqn 1)

The dot product, in contrast consists of a single summation:

$dot(X,Y) = \sum\limits_{i} x_i y_i$ (Eqn 2)

Note that the $\frac{1}{N}$ factors out of Eqn 1 when I divide by the standard deviations of X and Y. So I almost can derive the dot product from the covariance, but the covariance still contains a double summation and the dot product has a single element-wise summation. What am I missing?

UPDATE

I don't have the answer, but I realize that this formula:

$Cov(X,Y) = \sum\limits_{i}\sum\limits_{j} p_{ij} (x- \mu_X) (y-\mu_Y)$

... where N is the dimension of the vectors, is not equal to Eqn 1:

$Cov(X,Y) = \dfrac{1}{N} \sum\limits_{i}\sum\limits_{j} x y$ (Eqn 1)

Since if $p_{ij}$ equals $\dfrac{1}{N}$, it would sum to be greater then one in the double summation. $p_{ij}$ only equals $\dfrac{1}{N}$ when it's a single summation, which maps easily to the dot product. I've posted a new question here to try and address this confusion in isolation, which I hope in turn will clarify this derivation problem.

答:
The notes seem to be giving a result before stating the interpretation on which the result is based. If you only read the first page of the notes, the statements about vectors don't make sense.

On the second page we start to get an explanation of what is really meant by the use of vector notation on the first page.
We are supposed to consider a probability space $\Omega$ with $n$ elements ($n$ a positive integer), where each element has the same probability measure as any other.
The elements are listed in sequential order from the $1$st element to the $n$th.

We are then supposed to consider a vector in $\mathbb R^n$ as a function that takes the sequential number of an element of the probability space as input,
and gives a real number as output.
This matches the definition of a real random variable over that probability space, which is a function that maps each element of the probability space to a real number.

Now let's consider something like the expected value of the square of a random variable $X$, that is, $E[X^2]$.
The formula $E[X^2] = \sum_i x_i^2 p(x_i)$, where $p$ is the pmf of the random variable $X$ and the values of $X$ are represented by $x_i$, is not the only way to compute this expected value.
For a finite probability space we can also compute the expected value by taking the sum of a different sequence
of products,
$$ E[X^2] = \sum_{\omega \in \Omega} \mu(\omega) (X(\omega))^2, \tag1 $$
where $\omega$ ranges over the elements of the probability space $\Omega$,
$\mu(\omega)$ is the probability measure of the element $\omega,$
and $X(\omega)$ is the value of the random variable $X$ for input $\omega$.

The equation
$$ E[X^2] = \frac1n \sum_{i=1}^n X_i^2 \tag2 $$
is just a rewriting of Equation $(1)$ for a probability space with $n$ elements of equal probability $\frac1n$, where $X_1,\ldots, X_n$ are the values of the random variable $X$ for the $1$st through $n$th elements of the probability space.
The right-hand side of $(2)$ is also
$$ \frac1n (X \cdot X) = \frac1n \lVert X\rVert^2 $$
if $X$ is represented by a vector as suggested on the second page of the notes.

Similarly, for $E[XY],$ the expected value of the product of random variables $X$ and $Y$, we can write
$$ E[XY] = \sum_{\omega \in \Omega} \mu(\omega) X(\omega) Y(\omega). $$

For a probability space of $n$ elements of probability $\frac1n$ each,
using vectors to represent $X$ and $Y,$ this translates to
$$ E[XY] = \frac1n \sum_{i=1}^n X_i Y_i. \tag3$$

Note: this is not a double summation over $i$ and $j$ of terms in $X_i Y_j$,
and there's a good reason for that.
The summation in this formula was really over the elements of the probability space. Each index value $i$ picks out one element. Once you have picked an element of the probability space, each random variable's value is determined.
So when we pick the $i$th element, the value of $X$ is $X_i$, the value of $Y$ is $Y_i$, and the value of $XY$ is $X_i Y_i$.

Now note that if we use vectors for $X$ and $Y$, the right-hand side of $(3)$ can also be written
$$ \frac1n (X \cdot Y). $$

So the notes are literally correct when talking about a finite probability space with a uniform probability measure, with vectors in $\mathbb R^n$ interpreted as random variables over that probability space, except for one detail: they seem to have omitted the factor $\frac1n$.

A further note to explain why we sometimes use a single summation for covariance and sometimes a double summation.
Suppose we have a probability space
$\Omega = \{1,2,3,4,5\}$ with uniform probability measure
(each element has probability measure $\frac15$).
Suppose $X$ and $Y$ are random variables and that
\begin{align}
X(1) &= 11, & Y(1) &= 21, \\
X(2) &= 11, & Y(2) &= 22, \\
X(3) &= 12, & Y(3) &= 21, \\
X(4) &= 12, & Y(4) &= 23, \\
X(5) &= 13, & Y(5) &= 21. \\
\end{align}
So we see that the possible values of $X$ are
$x_1 = 11,$ $x_2 = 12,$ and $x_3 = 13,$
whereas the possible values of $Y$ are
$y_1 = 21,$ $y_2 = 22,$ and $y_3 = 23.$
When we take pairs $(x_i,y_j)$ there are therefore $3\times 3 = 9$ possible pairs, and when we write
$$ \sum_{i=1}^3 \sum_{j=1}^3 x_i y_j P(X=x_i, Y=y_j) $$
we have to add up $9$ terms in total from the double summation.

But actually, according to the definitions of $X$ and $Y$ above, there are only $5$ possible values of $(x_i,y_j)$:

$$ (11,21), (11,22), (12,21), (12,23), (13,21). $$

So the double summation is "wasteful" in that it counts several outcomes that have zero probability.
The single summation
$$ \frac15 \sum_{i=1}^5 X(i) Y(i), $$
on the other hand, counts each of the $5$ pairs of values from $X$ and $Y$ that can occur and does not bother to count any of the zero-probability pairs. So, among the outcomes that actually can occur, each sum counts exactly the same results. They just use different ways of identifying those results.


问:
How many clopen subsets does the subspace [0,1]∪[3,4] have?
Show that the subspace $[0,1]∪[3,4]$ of $\mathbb R$ has at least $4$ clopen subsets. Exactly how many clopen subsets does it have?

Just to clarify, this is the subspace induced by $[0,1]∪[3,4]$ from $\Bbb R$, so the space is just the intersection of all open sets in $\Bbb R$ with $[0,1]∪[3,4].$

Finding $4$ clopen sets is easy. Just form half open intervals with one of the $4$ endpoints, like $[0, 1/2).$ As for how many clopen sets there are, I think the answer is that there are  uncountable many, since the intervals $(0,1)$ and $(3,4)$ have uncountably many elements.

答:
Clopen sets are both closed and open.  For example, the empty set and the entire space are always clopen.  Half-open intervals are not necessarily clopen; for example, $[0,1/2)$ is not closed, because it does not contain $1/2$.

One set that is clopen in the subspace topology is $[0,1]$.  Even though it is not open in $\mathbb R$,  it is open in the subspace topology, because it is the intersection of the subspace $[0,1]\cup[3,4]$ and $(-1,2), $ which is open in $\mathbb R$.

I have given you three clopen sets in the subspace topology, and I hope you can find a fourth.

ContentLicense: CC BY-SA 4.0


问:
How to solve $\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx$

\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx
\end{equation}

This is a MCQ question and there are 5 options to choose which are "A.$2\sqrt{3}\pi+9\ln 3$, B.$2\sqrt{7}\pi+8\ln 3$ , C.$2\sqrt{3}\pi+8\ln 3$ , D.$2\sqrt{2}\pi+2\ln 3$, E.Other solution."

How do you solve this integral?

This appears in MCQ test, so I think there should be a trick to solve this without using too much force. The test have 30 questions and 2 hours, so each question should be finished under 4mn ( I doubted it though). Here is my solution that I spend around 3h to solve it. ( Sorry for not using latex and sorry for the inconveniences )Photo

I appreciate any solution tricks. Thanks!

答:
Tou have done a good work and obtained the correct result.

Trying on my side, using as you did $s=\sin(x)$, we ned with
$$\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx=36 \int_{0}^1  \frac{ds}{ \left(s^4+s^2+1\right)^2}$$

$$s^4+s^2+1=\left(s^2-s+1\right) \left(s^2+s+1\right)$$ Using partial fraction decomposition
$$\frac{1}{ \left(s^4+s^2+1\right)^2}=\frac{1-s}{2 \left(s^2-s+1\right)}+\frac{s+1}{2 \left(s^2+s+1\right)}-$$ $$\frac{s}{4
   \left(s^2-s+1\right)^2}+\frac{s}{4 \left(s^2+s+1\right)^2}$$ The first and the second are not difficult
$$I_1=\int\frac{1-s}{ s^2-s+1}\,ds=-\frac 12\Bigg[\int \frac{2s-1}{ s^2-s+1}ds -\int\frac{ds}{ s^2-s+1}\Bigg]$$
$$I_2=\int\frac{s+1}{s^2+s+1}ds=\frac 12\Bigg[\int \frac{2s+1}{ s^2+s+1}ds -\int\frac{ds}{ s^2+s+1}\Bigg]$$
Now, for the third and fourth antiderivatives which look like
$$J=\int \frac s{(s^2+as+1)^2} ds=\int \frac s{(s-r_1)^2(s-r_2)^2} ds$$ partial fraction decomposition again
$$\frac s{(s-r_1)^2(s-r_2)^2}=\frac{r_1}{(r_2-r_1)^2 (s-r_1)^2}+\frac{r_1+r_2}{(r_2-r_1)^3 (s-r_1)}-$$ $$\frac{r_1+r_2}{(r_2-r_1)^3
   (s-r_2)}+\frac{r_2}{(r_2-r_1)^2 (s-r_2)^2}$$

This is a pure nightmare !

At the end, after recombining evrything, the antiderivative is
$$\frac{6 s\left(1-s^2\right)}{s^4+s^2+1}+4 \sqrt{3} \tan ^{-1}\left(\frac{16 s \left(1-s^2\right)}{3 \sqrt{3}}\right)+9 \log \left(\frac{s^2+s+1}{s^2-s+1}\right)$$ and, for the definite integral, your good result
$$2 \sqrt{3} \pi +9 \log (3)$$

All this work took me more than one hour. Be sure that I have been looking for tricks but ... no one came to my mind.


问:
Proof of Alternating Binomial Coefficient Identity relating
I am looking for a proof of the following binomial identity. I encountered it in an article on Euler’s derivation of the gamma function. Euler begins by evaluating the integral:

$$\int_0^1 x^a(1-x)^n\,dx$$

He performs a binomial expansion on the integrand and makes use of the following identity involving alternating binomial coefficients:

$$\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{1}{a+k+1} = n! \prod_{k=0}^{n}\left( \frac{1}{a+k+1} \right).$$

It is not obvious to me on inspection that this formula is true, but I have found it cited elsewhere in the mathematical literature, without proof. I have verified it by hand for the cases of n=2 and n=3, but I am unable to prove the identity in general. I am looking for a good combinatorial or inductive proof of this very interesting fact.

I am aware that this formula relates to the beta function, but I am trying to derive this relationship without reference to this topic.

Thank you for enlightening me.

答:
I am not quite sure if this counts as a duplicate but I give three proofs of an identity equivalent to this identity here (one of which is the beta function proof). I think it is cleanest to write it as

$$\sum_{k=0}^n \frac{(-1)^k}{z + k} {n \choose k} = \frac{n!}{z(z + 1) \dots (z + n)}$$

(here $z$ is $a+1$) which makes it clearer that for $n$ fixed it is an equality between two rational functions of $z$; in particular it holds for all complex values of $z \neq 0, -1, \dots -n$, and written this way it can be proven by calculating the residues at each pole of the RHS and verifying that they line up with the coefficients of the LHS. I give another proof via thinking of the LHS as the $n^{th}$ finite difference of the sequence $a_k = \frac{1}{z + k}$, which can be computed by induction and verified to line up with the RHS.

If we substitute $z \mapsto -\frac{1}{z}$ and clear denominators we get the equivalent identity

$$\sum_{k=0}^n \frac{(-1)^k}{1 - kz} {n \choose k} = \frac{n! z^n}{(1 - z) \dots (1 - nz)}$$

which is an ordinary generating function identity for the Stirling numbers of the second kind, and which can be proven by proving an exponential generating function identity and then translating it over.


问:
Is sequential criterion of continuity holds for Sequence of continuous functions?
I know that, If $(f_n)$ is Sequence of continuous functions on interval $I$ that converges uniformly on $I$ to function $f$ and If $(x_n)\subset I$ converges to $x_0\in I$ then,

$$lim(f_n(x_n))=f(x_0)$$

My question: Is the other direction holds? I mean, is the following is true?

Let $(f_n)$ be Sequence of continuous functions on $I$ and If for $c\in I$ we have, for every Sequence $(x_n)$ in $I$ that conveges to $c$,  $lim(f_n(x_n))=f(c)$ then, Sequence $(f_n)$ converges uniformly to $f$ on $I$?

Please help...

答:
Not true on the interval $(-\infty,\infty)$: let $$f_0(x) = \begin{cases} 0 & x<0, \\ x & 0\le x<1,\\ 1 & x\ge 1\end{cases}$$
And then set $f_n(x) := f_0(x-n)$. We have $f_n \to 0=:f$ pointwise, but not uniformly.  Yet if $x_n\to c$, then $x_n$ is bounded, so there is some $M$ where $x_n < M$ for all $n$. Then for all $n>M$, $f_n(x_n)=0= f(c)$.

The same example works on $[0,\infty)$.


问:
Proof of the uniqueness of maximal ideal
Let $R$ be a commutative ring with $1$. Let $M$ be a maximal ideal of $R$ such that $M^2 = 0$. Prove that $M$ is the only maximal ideal of $R$.</p>

答:
The other solutions have the best elementary solutions, so I'll try to give a distinct way.

The Jacobson radical $J(R)$ of $R$ contains all nilpotent ideals and is contained in all maximal ideals.

By these conditions, $M\subseteq J(R)\subseteq M$. Thus, the Jacobson radical is a maximal ideal. Thus, there cannot be any different maximal ideals, since there are no ideals between $M$ and $R$.

ContentLicense: CC BY-SA 3.0


问:
What is a good book to study linear algebra?
I'm looking for a book to learn Algebra. The programme is the following. The units marked with a $\star$ are the ones I'm most interested in (in the sense I know nothing about) and those with a $\circ$ are those which I'm mildly comfortable with. The ones that aren't marked shouldn't be of importance. Any important topic inside a unite will be boldfaced.

U1: Vector Algebra.
Points in the $n$-dimensional space. Vectors. Scalar product. Norm. Lines and planes. Vectorial product.

$\circ$ U2: Vector Spaces.
Definition. Subspaces. Linear independence. Linear combination. Generating systems. Basis. Dimesion. Sum and intersection of subspaces. Direct sum. Spaces with inner products.

$\circ$  U3: Matrices and determinants.
Matrix Spaces. Sum and product of matrices. Linear ecuations. Gauss-Jordan elimination. Range. Roché Frobenius Theorem. Determinants. Properties. Determinant of a product. Determinants and inverses.

$\star$ U4: Linear transformations.
Definition. Nucleus and image. Monomorphisms, epimorphisms and isomorphisms. Composition of linear transformations. Inverse linear tranforms.

U5: Complex numbers and polynomials.
Complex numbers. Operations. Binomial and trigonometric form. De Möivre's Theorem.
Solving equations. Polynomials. Degree. Operations. Roots. Remainder theorem. Factorial decomposition. FTA. Lagrange interpolation.

$\star$ U6: Linear transformations and matrices.
Matrix of a linear transformation. Matrix of the composition. Matrix of the inverse. Base changes.

$\star$ U7: Eigen values and eigen vectors
Eigen values and eigen vectors. Characteristc polynomial. Aplications. Invariant subspaces. Diagonalization.

To let you know, I own a copy of Apostol's Calculus $\mathrm I $ which has some of those topics, precisely:

Linear Spaces

Linear Transformations and Matrices.

I also have a copy of Apostol's second book of Calc $\mathrm II$which continues with

Determinants

Eigenvalues and eigenvectors

Eigenvalues of operators in Euclidean spaces.

I was reccommended Linear Algebra by Armando Rojo and have Linear Algebra by Carlos Ivorra, which seems quite a good text.

What do you reccomend?

答: Serge Lang has an Introduction to Linear Algebra, and a more advanced $\textit{Linear Algebra}$. Both of the pdfs can be found easily (on google) for free. Lang is very good at getting the point across and synthesizing abstract concepts with lots of examples. They are both more or less undergraduate level. The introduction book is more-so intended for a gentle engineering course, while the other book focuses on efficiently developing the structure theorems for vector spaces, including the spectral theorem, symmetric diagonalization, triangulation, invariant subspaces, and the Jordan Normal Form. Lang's program in the second book is extremely well laid-out, but he was an extremely experienced and talented writer, so his first book is a good choice also, if that is more of what you want.


问:
In a Rubik's cube what is the probability of randomly getting a face blue?
When we solve a Rubik's cube we make planned moves and easily get the desired results but here I am talking about getting a colour  by randomly moving a Rubik's cube without making an effort to solve it.
I am getting 1/269,438,400 as the probability but I am not sure that it's correct.

I solved the problem this way. I started with the middle block and considered the probability of it being blue as 1/6. Then I considered the blocks located b/w 2 corner blocks. For the first such block the probability is 4/24 that is 1/6 as 4 such blocks are available and a total of 24 blocks of that position are available but for the next such block I considered the probability as 3/22 as in the previous step 2 of the total such blocks because unavailable and one desired (blue) block of that position became unavailable. This way I got the probabilities as 1/10 and 1/18.
I did the same for corner blocks. I then multiplied all the values I got as they are all interdependent.

答: Your problem is unclear, but from your attempt it seems you allow slice turns, and are looking at a fixed face, so that indeed the probability that the centre piece is blue is $1/6$. I have no clear idea what you are trying to do after that, but simply consider that all permutations of the edge pieces are equally likely given fixed centres, so the probability that the 4 relevant edge pieces have blue is $1/\binom{12}{4}$. The probability that those 4 edge pieces have blue facing the right way is $1/2^4$, since all flip combinations are equally likely (note that this does not hold if you want to look at all edge pieces; it holds here only because we are looking at only $4$ of them). This does give the same answer $1/495$ as you obtained, so I suppose you must have reasoned correctly for that. For the corner pieces, similarly probability of having blue is $1/\binom{8}{4}$ and probability of their blue facing the right way is $1/3^4$. Again, it is not trivial that the edge-pieces-blue and corner-pieces-blue events are independent, but is true because the parity restriction has no effect when we are only looking at one face-full of pieces. Final answer is indeed $1/269438400$ as you stated.


问:
Prove the following inequality if a and are reals
Prove inequality $(a^3+b^3) ≤ (a^2 + b^2)(a^4 + b ^4)$  where $a,b$ are reals

They are given reals so cannot apply AM-GM inequality or can we?

Also I am a beginner in inequality so please try to keep it simple....
Thanks in advance

I tried it in the following way:

$a^2 + b^2 ≥ 2ab$

$a^4 + b^4 ≥ 2a^2b^2$

So.. $$(a^2 + b^2)(a^4 + b^4) ≥ 4a^3b^3$$

Am I going in the right direction?

答:
Note that by Cauchy-Schwarz
\begin{align}
a^3+b^3 = a\cdot a^2+b\cdot b^2 \le \sqrt{a^2+b^2}\sqrt{(a^2)^2+(b^2)^2}.
\end{align}

By homogeneity, the inequality as written in the problem is not right. For instance, let $a = b = 1/2$, then we see that
\begin{align}
\frac{1}{4}=\frac{1}{8}+\frac{1}{8}\nleq \left(\frac{1}{4}+\frac{1}{4} \right)\left(\frac{1}{16}+\frac{1}{16} \right) = \frac{1}{2}\frac{1}{8}= \frac{1}{16}
\end{align}


答:
The inequality is false: $a=\frac1 2 , b=0$ isa  counter-example.

However, it becomes correct if you square the left side. To prove this apply Cauchy-Schwarz inequality after splitting $a^{3}$ as $(a) (a^{2}) $ and $b^{3}$ as $(b) (b^{2}) $.

Title: Prove the following inequality if a and are reals

Body:
Prove inequality $(a^3+b^3) ≤ (a^2 + b^2)(a^4 + b ^4)$  where $a,b$ are reals

They are given reals so cannot apply AM-GM inequality or can we?

Also I am a beginner in inequality so please try to keep it simple....
Thanks in advance

I tried it in the following way:

$a^2 + b^2 ≥ 2ab$

$a^4 + b^4 ≥ 2a^2b^2$

So.. $$(a^2 + b^2)(a^4 + b^4) ≥ 4a^3b^3$$

Am I going in the right direction?


问:
Possible description of closed subset of a projective variety
Let $k$ be an algebraically closed field. Let $P\subset \mathbb{P}^n(k)$ be a projective variety, and $X\subset P$ be a subset.

Suppose that $X_i = X\cap U_i$ is an affine closed subset for every affine open $U_i$ of $\mathbb{P}^n$, i.e. that for every $i$ we have a polynomial $\;f_i \in A(U_i \cap P)$ such that $X_i = \mathcal{Z}(f_i)$.

Is it always true that, then, $X$ is a closed subspace of $P$ ? Or do we need some extra conditions?

答:
Thanks to Jared for his suggestion! Below there's a possible easy proof of the following fact (notice there's a slight difference: in my claim $U_i\cap X$ is closed in $U_i$, not in $X$. This is what I need to answer the above question) :

If $P$ is a topological space, $X\subset P$, and ${U_i}$ is an open
  cover of $P$ such that $U_i\cap X$ is closed in $U_i$ for all $i$,
  then $X$ is closed in $P$.

Proof:

Since $X_i = U_i\cap X$ is closed in $U_i$, we have that $U_i \setminus X_i$ is open for every $i$. Then we find that
$$ P\setminus X = \bigcup_i \left( U_i \setminus (\cup_j X_j)\right)
= \bigcup_i U_i \setminus X_i $$
is open being a union of the open sets $U_i \setminus X_i$. Therefore $X$ is closed in $P$.

Hence the answer to the above question is: yes, under those assumptions $X\subset P$ is a closed set.


问:
Conditional probability of sum of random variables
Let $X$ and $Y$ be independent random variables and let $Z = X + Y$. To find $P(Z = z | X = x)$ this is the same as finding $P(Y = z - x)$ (correct me if I'm wrong).

However, if we wish to find $P(X = x | Z = z)$, we cannot simply take $P(Y = z - x)$. Why is that? I know that we can solve for $P(X | Z)$ using Bayes' theorem, but is there an intuitive explanation why $P(Y = z - x)$ won't work?

答:
$P(Z=z|X=x)=P(Y=z-x|X=x)$ and independnec allows us to drop the conditionin on the right side. However, $P(X=x|Z=z)=P(Y=z-x|Z=z)$ and we cannot drop the conidtioning since $Y$ and $Z$ are not independent.

$P(X=x|Z=z)=P\frac {(Y=z-x)} {P(Z=z)}$ so you have to compute $P(Z=z)$ to find this conditinal probability.


问:
A question about impredicative type theory
I am playing with impredicative type theories (CC and UTT). I am not quite familiar with the distinction between $\textsf{Prop}$ and $\textsf{Type}$ as it is not available in MLTT. Here is my question.

Let $\Gamma$ be a valid context and let $A:\textsf{Prop}$, $D:\textsf{Type}$ and $B:D\rightarrow\textsf{Prop}$. My question can be divided into two parts:

By the above definitions, $A\rightarrow\Sigma x:D.B(x)$ is a well-formed type. Is it possible that $A\rightarrow\Sigma x:D.B(x)$ is not a theorem in an empty context but is a theorem in context $\Gamma$?

If $A\rightarrow\Sigma x:D.B(x)$ is not a theorem in an empty context but is a theorem in context $\Gamma$, then what amounts to a proof object of $A\rightarrow\Sigma x:D.B(x)$ in $\Gamma$?

Thanks!

答:
$\newcommand{\Prop}{\mathsf{Prop}}$

If we take $P : \Prop$ to be some independent proposition, then we can take: $$A = ⊤\\D = 2\\B(0) = ¬P\\B(1) = P$$

Then a derivation of $$⊢ ⊤ → Σb:2. B(b)$$ is deciding $P$. If no non-constructive principles are assumed, presumably there is no such derivation. However there is an easy derivation of $$P ⊢ ⊤ → Σb:2.B(b)$$ In the sort of theories you're talking about, constructions in $\Prop$ aren't really very different from those in $\mathsf{Type}$, aside from rules for equality and possible restrictions on eliminators and the like. So a proof term could just look like:

$$p:P ⊢ λt. (1, p) : ⊤ → Σb:2.B(b)$$

I.E. propositional implication and $∀$ are proved by functions, $∧$ and $∃$ by pairs, and $∨$ by terms that look like they belong to a disjoint union, but values of a proposition are considered equal even if they are not syntactically identical, and e.g. you may not be able to do case analysis on an $∨$ value to produce a value of $A : \mathsf{Type}$.

Edit: perhaps I should mention, the bits about how $\Prop$ behaves differently are mostly unrelated to impredicativity, and are instead related to how people think propositions should be different from other types. The possible exception is that impredicative $Σ$ types are inconsistent unless you place restrictions on the way they are eliminated. However, I'm not sure the restrictions I mentioned above are the ones necessary to fix that problem.


问:
Probability concerning Normal Distributions independent of one another
A team comprised of 10 individuals are tasked to put balls inside a box. Suppose that each individual’s contribution is normally distributed and has a mean of 12 balls and standard deviation of  1 ball. The team wins only if the total number of balls they put is less than 100 balls. Assuming that each of their contribution is independent, What is the probability that they win?

Attempt:
Using the formula of z= (x̄- μ) / σ. I find the probability that x is less than 10 which is 0.0275 for z=-2. I then multiply the value by 10 leading to an answer of 0.275. Is this a correct way of approaching the problem or should I be computing it as a whole?

答:
You did not mention that the distribution is gaussian.  Assuming this, the total contribution is gaussian with mean

$$\underbrace{12+12+\dots +12}_{\text{10 times}}=120$$

and variance

$$\underbrace{1^2+1^2+\dots +1^2}_{\text{10 times}}=10$$

If normality is not assumed you have to use another way, i.e. Chebishev's inequality


问:
Taylor's Polynomial of a Function: how we can arrive at it naturally?
Let $f:(a,b)\rightarrow\mathbb{R}$ be a function, which is differentiable, say infinitely many times, in the domain. [We can minimize the hypothesis, but for question, we can assume it.]

Consider the polynomial $P(x)=f(c) + (x-c)f'(c) + \cdots + \frac{f^{k}(c)}{k!}(x-c)^k$.

This polynomial has two properties:

(1) This is a good approximation to $f$ near $c$.

(2) From definition, $P^{(i)}(c)=f^{(i)}(c)$ for $i=1,2,\ldots, k$.

But, I didn't get any "natural motivativation" to arrive at exactly this polynomial.

After discussion with some teachers, they said, it generalizes, mean value theorem: $f(x)-f(x)=f'(\alpha)(x-c)$ i.e. $f(x)=f(c) + f'(\alpha)(x-c)$. But, I didn't get from this mean value theorem, how one naturally arrives at the above $P(x)$, which involves some "factorials", which were not looking in mean value theorem.

In mean value theorem, $f'(\alpha)$ can be written as $f'(\alpha)/1!$ and so one may justify this way for appearance of factorial things; but I would ask then, *we can write $f'(\alpha)$ as $f'(\alpha)/1$', $f'(\alpha)/1^2$' so can't we take the polynomial as  $P^{(i)}(c)=f^{(i)}(c)$ for $i=1,2,\ldots, k$, then can't we take polynomial as $f(c) + (x-c)f'(c) + \cdots + \frac{f^{k}(c)}{k^2}(x-c)^k$?

Why the choice of $P(x)$ above is most natural one and what is "natural way" to arrive at the polynomial $P(x)$?

答:
I think of Taylor series as a trick that lets you extract coefficients.

Let's say you have a polynomial $p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_n x^n$.  However, the coefficients $a_i$ are unknown; but let's say you're allowed to evaluate $p(x)$ at any point, as well as evaluate the derivative $p^n(x)$ for any $n$, at any point $x$.  For example you can compute $p(3) = 5$, $p'(1) = 0$, $p^3(2) = 4.532$, or whatever.  The values here are just examples; the point is that you can use $p(x)$ and its derivatives as black boxes, but you don't know the formula for $p(x)$. Can you still find the coefficients?

Sure.  Finding $a_0$ is easy: Set $x=0$.  $$a_0 = p(0).$$  Thus, we can always extract the constant coefficient.

To find $a_1$, just note: if we differentiate, the constant term $a_0$ vanishes, and then we can extract the constant again:

$$p'(x) = 0 + a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + 5a_5x^4 + \cdots$$
and so $a_1 = p'(0)$.

Continuing to differentiate gives

\begin{align*}
p''(x) &= 2a_2 + 6a_3x + 12a_4x^2 + 20a_5x^3\cdots\\
p'''(x) &= 6a_3 + 24a_4x + 60a_5x^2 + \cdots
\end{align*}

and so $$p''(0) = 2a_2,$$ giving $a_2 = p''(0) / 2$, and similarly $$p'''(0) = 6a_3,$$ so $a_3 = p'''(0) / 6$, and so on.

Hopefully you see the pattern.  The factorials come from the fact that the $n$-th derivative of $x^n$ is the constant $n!$: $$x^n \rightarrow nx^{n-1} \rightarrow n(n-1)x^{n-2} \rightarrow \cdots \rightarrow n! x^0$$ where $\rightarrow$ is differentiation.

This shows that it works for polynomials.  But now if you assume $p(x)$ is a power series $a_0 + a_1x + a_2x^2 + \cdots$, the trick still works.

Now do this a little more carefully, keeping track of the errors in approximating by successive polynomials using some clever integral manipulations, and you get Taylor's Theorem.


问:
Question about Lambert W function
I'm looking for a series for $W_0(x)$ for x $\in [\frac{-1}{e},\infty [$ but every time i found only for $x\in [\frac{-1}{e}, \frac{1}{e}]$

and what about a series for $W_-1(x)$

if it is no series with x$\in [\frac{-1}{e},\infty [$ so how we can have an expression for those function ?

also what about $W_k(x)$ how they evaluate $W_k(x)$ or what is the expression of $W_k(x)$

i mean in the (expression) : series... any thing lead to evaluate any x

please help

thanks for all

答:
What kind of series?  The Taylor series for $W_0(x)$ at $x=0$ has radius of convergence $1/e$, so it cannot represent anything when $x>1/e$.

There is a transseries  you could use, like this
$$
W_0(x) = \log(x) - \log(\log(x)) + \frac{\log(\log(x))}{\log(x)} + \cdots
$$
as $x \to \infty$.

(comput in Maple: http://www.mapleprimes.com/questions/120720-Asymptotics-Of-Lambert-W )

added:  More terms
$$
\ln  \left( x \right) -\ln  \left( \ln  \left( x \right)  \right) +{
\frac {\ln  \left( \ln  \left( x \right)  \right) }{\ln  \left( x
 \right) }}+{\frac {-\ln  \left( \ln  \left( x \right)  \right) +1/2\,
 \left( \ln  \left( \ln  \left( x \right)  \right)  \right) ^{2}}{
 \left( \ln  \left( x \right)  \right) ^{2}}}+{\frac {\ln  \left( \ln
 \left( x \right)  \right) -3/2\, \left( \ln  \left( \ln  \left( x
 \right)  \right)  \right) ^{2}+1/3\, \left( \ln  \left( \ln  \left( x
 \right)  \right)  \right) ^{3}}{ \left( \ln  \left( x \right)
 \right) ^{3}}}+ \left( -\ln  \left( \ln  \left( x \right)  \right) +3
\, \left( \ln  \left( \ln  \left( x \right)  \right)  \right) ^{2}-{
\frac {11}{6}}\, \left( \ln  \left( \ln  \left( x \right)  \right)
 \right) ^{3}+1/4\, \left( \ln  \left( \ln  \left( x \right)  \right)
 \right) ^{4} \right)  \left( \ln  \left( x \right)  \right) ^{-4}+
 \left( \ln  \left( \ln  \left( x \right)  \right) -5\, \left( \ln
 \left( \ln  \left( x \right)  \right)  \right) ^{2}+{\frac {35}{6}}\,
 \left( \ln  \left( \ln  \left( x \right)  \right)  \right) ^{3}-{
\frac {25}{12}}\, \left( \ln  \left( \ln  \left( x \right)  \right)
 \right) ^{4}+1/5\, \left( \ln  \left( \ln  \left( x \right)  \right)
 \right) ^{5} \right)  \left( \ln  \left( x \right)  \right) ^{-5}+O
 \left( {\frac { \left( \ln  \left( \ln  \left( x \right)  \right)
 \right) ^{6}}{ \left( \ln  \left( x \right)  \right) ^{6}}} \right)
$$

I put $x=11$ in there and got $1.80705$, but according to Maple, the true value is $W(11) = 1.80650\dots$.

Title: Question about Lambert W function


问:
How to apply chain rule on matrix

Gradient of $\frac{dL}{dX}$ using chain rule

With the chain rule, $\frac{dL}{dX} = \frac{dY}{dX} \cdot \frac{dL}{dY}$, and $\frac{dY}{dX} = W$ for the product Y = X•W.

Q1

I suppose I need to make $\frac{dY}{dX}$ into  a transpose $W^\intercal$ to match the shape. For instance, if X shape is (, 3) as per numpy, the last axis of shape($\frac{dY}{dX}$) needs to be 3 (so that $\frac{dY}{dX} \cdot dX^\intercal \rightarrow dY$ : (m,3) • (3,n) → (m, n)?)

However, not sure if this is correct and why, hence appreciate any explanations.

Q2

How can I apply the chain rule formula to matrix?

$\frac{dL}{dX} = W^\intercal \cdot \frac{dL}{dY}$

This cannot be calculated because of the shape mismatch where $W^\intercal$ is (4, 3) and $\frac{dL}{dY}$ is (, 4).

Likewise with $\frac{dL}{dW} = X^\intercal \cdot \frac{dL}{dY}$ because $X^\intercal$ is (,3) and $\frac{dL}{dY}$ is (, 4).

What thinking, rational or transformation can I apply to get over this?

There are typos in the diagram. (,4) is (4,) etc. In my
brain, 1D array of 4 element was (,4) but in NumPy, it is
(4,).

For $\frac{dL}{dX}$

I saw the answer is swapping the positions, but no idea where it came from and why.

$\frac{dL}{dX} = \frac{dL}{dY} \cdot W^\intercal$

Instead of:

$\frac{dL}{dX} = W^\intercal \cdot \frac{dL}{dY}$

For $\frac{dL}{dW}$

The shape of $X^\intercal$ (,3) and $\frac{dL}{dY}$ (, 4) need to be transformed into (3, 1) and (1, 4) to match the shapes, but no idea where it came from and why.

Geometry

In my understanding, X•W is extracting the $\vec{\mathbf{W}}$ dimension part of X by truncating the other dimensions of X geometrically. If so,  $\frac{dL}{dX}$ and $\frac{dL}{dW}$ are  restoring the truncated dimensions? Not sure this is correct but if so, would it be possible to visualize it like X•W projection in the diagram?

答:
Thanks to @Reti43 for pointing to the reference. The detail math is provided by the cs231 Justin Jonson (now in Michigan University) as http://cs231n.stanford.edu/handouts/linear-backprop.pdf which is also available as Backpropagation for a Linear Layer.

cs231n lecture 4 explains the idea.

The math calculation from step (5) to (6) seems to be a leap, because dot product would not be resulted from two 2D matrices, and numpy.dot would produce matrix multiplication as np.matmul, hence it would not be a dot product.

The answer in numpy function to use for mathematical dot product to produce scalar addressed a way.

Format of the Weight vector W

Need to note on the weights representation of W by Justin Johnson.

In Coursera ML course, Andrew Ng uses a row vector to capture weights of a node. When the number of features in input to a layer is n, the row vector size is n.

Justin Johnson uses a row vector to represent a layer size, the number of nodes in a layer. Hence if there are m nodes in a layer, the row vector size is m.

Hence the weight matrix for Andrew Ng is m x n meaning m rows of weight vectors, each of which are weights for n features for a specific node.

The weight matrix for Justin Johnson is n x m meaning n rows of weight vectors, each of which are weights for m nodes in a layer, per feature.

I suppose Justin Johnson regards layer is a function whereas Andrew Ng regards node is a function.

As I studied Andrew Ng's ML course first, I am using weight vector per node approach which results in W as m x n matrix. My confusion came from applying W = m x n to Justin Jhonson's paper.

Understanding

My understanding by reading through the Justin Johnson's paper is below.

Dimension analysis

First frame the dimensions/shapes of the gradients.

Derive the gradient

Using the simple one input record X shape(d,), derive the dL/dX and extend it to two dimensional input X shape(n, d) resulting in dL/dY @ W. This is different from  dL/dy @ W.T in the Justin Johnson's paper, because of the difference of the weight matrix W representation.

If something incorrect, very much appreciate any feedbacks.

Title: How to apply chain rule on matrix


问:
Empty set in countability proofs
It is a provable theorem (defining countable as finite or countably infinite) that a nonempty set $S$ is countable if and only if there exists a surjection $f: \mathbb{N} \to S$ if and only if there exists an injection $g: S \to \mathbb{N}$.

Every single time I have seen the result quoted, it is assumed that $S$ is nonempty. Is this necessary, or is just done to eliminate an extra edge case? The result seems fine to me. If $S = \emptyset$, it's finite, hence countable. Then there is only one map $f: \mathbb{N} \to \emptyset$ -- the empty list -- which is vacuously surjective and injective. (It's surjective since there are no $x \in \emptyset$ we need to map to, and vacuously injective because $f(a) \neq f(b)$ for any $a,b \in \mathbb{N}$.) There is also only one map $g: S \to \mathbb{N}$, once again the empty list, which is again vacuously injective.

Is this correct?

答: There is certainly an injection (vacuously) from the empty set to any set, so by the usual definition, it is countable. It really should be anyway, because it is finite (in the sense that there is a bijection from the empty set, to a set with a natural number of elements, i.e. itself). There are no maps to the empty set from a nonempty set though, because you need somewhere to send the elements (you would have an "empty map", but the problem is that definitionally, for $f:A\to B$, you need to have that: for each $a\in A$, $\exists!b\in B$ s.t. $(a,b)\in f$).

Title: Empty set in countability proofs


问:
If $f(0)=f^{\prime}(0)=0$ and $|f''(x)|\leqslant |f(x)|+|f'(x)|$, then $f(x)\equiv 0$
Is there a simpler way to solve this problem with derivatives?

Let $f$ be second-order differentiable on $(-\infty,+\infty)$, and
$$f(0)=f^{\prime}(0)=0, \quad|f''(x)|\leqslant |f(x)|+|f'(x)|$$
Show that $$f(x)\equiv 0$$

I do as follows,for all $x\in\left[-\frac{1}{3},\frac{1}{3}\right]$,let$$M=\max\limits_{x\in\left[-\frac 1 3,\frac 1 3\right]}|f'(x)|$$Using Lagrange mean value theorem I got it$$\begin{align}
|f'(x)| & = |f''(\eta_1)||x|\\
        & \leqslant \frac{|f(\eta_1)|}3+\frac{|f'(\eta_1)|}3 \\
        &  =\frac{|f'(\zeta _1)||\zeta_1|}3+\frac{|f'(\eta_1)} 3 \\
        & \leqslant \frac{2}3|f'(\xi_1)|, \text{ here }|f'(\xi_1)|=\max\{|f'(\zeta _1)|,|f'(\eta_1)|\}
\end{align}$$So we have$$|f'(x)|\leqslant \frac{2^n}{3^n}|f'(\xi_n)| \leqslant \frac{2^n}{3^n}M\to 0(n\to\infty).$$
Hence $$|f'(x)|\equiv 0\equiv f'(x)\implies f(x)\equiv f(0)\equiv 0\text{ for }x\in \left[-\frac 1 3,\frac 1 3\right].$$Repeat the above steps and use induction complete the proof.But
I think my method is too complicated.
Do you have a simpler and refreshing proof? Thank you!

答:
Note that
\begin{align}
g(x):=&\ |f'(x)|+|f(x)| \\
=&\ \left|\int^x_0 f''(y)\ dy \right| +\left|\int^x_0f'(y)\ dy \right|\\
\le&\ \int^x_0 |f''(y)|\ dy + \int^x_0 |f'(y)|\ dy\\
\le&\ \int^x_0 2|f'(y)|+|f(y)|\ dy\\
\le&\ 2\int^x_0 g(y)\ dy.
\end{align}
If we call $G(x) = \int^x_0 g(y)\ dy$ then we have that
\begin{align}
G'(x) \le 2 G(x) \ \ \implies \ \ G'(x)-2G(x) \le 0.
\end{align}
Now, we will use some basic ODE trick, i.e. integrating factor method to get
\begin{align}
\frac{d}{dx}(e^{-2x}G(x))
=e^{-2x}G'(x)-2e^{-2x}G'(x) \le 0
\end{align}
which means
\begin{align}
e^{-2x}G(x)-G(0) = \int^x_0 \frac{d}{dy}(G(y)e^{-2y})\ dy \le 0.
\end{align}
Hence, we arrive at
\begin{align}
\int^x_0 |f'(y)|+|f(y)|\ dy =G(x) \le G(0)e^{2x} = 0.
\end{align}
So, $f(x) = 0$.

The ODE trick is also called Gronwall's inequality.

Edit: As mentioned by @MartinR, if $f''$ is not continuous then we need to be a little more careful.

Fix $x>0$. By the mean-value theorem, we have that

\begin{align}
|f(x)|+|f'(x)| =&\ |f(x)-f(0)|+|f'(x)-f'(0)|\\
\le&\ |f'(\xi_1)||x| + |f''(\xi_2)||x|
\end{align}
where $\xi_1, \xi_2 \in (0, x)$. Using the hypothesis, we have that
\begin{align}
|f(x)|+|f'(x)| \le 2\left(\sup_{\xi \in [0, x]}|f'(\xi)| +\sup_{\xi \in [0, x]}|f(\xi)|\right)|x|.
\end{align}

Choose $x>0$ such that $1-2|x|>0$ then it follows
\begin{align}
(1-2|x|)\left(\sup_{\xi \in [0, x]}|f'(\xi)| +\sup_{\xi \in [0, x]}|f(\xi)|\right)\le 0.
\end{align}
Hence it follows
\begin{align}
\sup_{\xi \in [0, x]}|f'(\xi)| +\sup_{\xi \in [0, x]}|f(\xi)| = 0
\end{align}
for the chosen $x$. Repeat the argument for the interval $[x, 2x]$ so on... This gives you the desired result.


问:
How to compute the integral $\int_0^\infty \arctan\left(\frac 4x\right)\sin(x)\,\mathrm dx$
I want to compute the definite integral $$\int_0^\infty \arctan\left(\frac 4x\right)\sin(x)\,\mathrm dx.$$

My answer is below.

答:
Another solution $$I=\int\frac{e^{ix}}{x^2+16}\, dx=\int\frac{e^{ix}}{(x+4i)(x-4i)}\, dx$$
$$I=\frac i 8\int \Bigg[\frac{e^{ix}}{x+4 i}-\frac{e^{ix}}{x-4 i}\Bigg]\,dx$$
$$I=\frac i {8e^4} \left(e^8 \text{Ei}(i x-4)-\text{Ei}(i x+4)\right)$$

So, for the definite integral
$$\frac i {8e^4}\Big[i \left(e^8-1\right) \pi-\text{Ei}(4)+e^8 (\text{Ei}(-4)+i \pi )\Big]=-\frac{i \left(e^8 \text{Ei}(-4)+\text{Ei}(4)+i \pi \right)}{8 e^4}$$
$$\Re(I)=\frac{\pi }{8 e^4}$$


问:
Why is $\log_{-7}49$ undefined instead of $2$?
When you have $\log_{7}49$, you can tell that it is equal to $2$ because $7^2 = 49$. But, when you have $\log_{-7}49$, isn't this also equal to $2$ because $(-7)^2 = 49$ also? Instead, this is undefined.

Can someone explain why this is?

答:
Log49 to the base -7 is not a real number. However it is a complex number

Log49 to the base -7= ln49/ln(-7)

Ln(-7) is not a real number
Remember exp(ia)= cos(a)+isin(a)

-1= exp(iπ)
ln(-7)= ln(-1)+ln(7)=ln(exp(iπ))+ln(7)
=Ln(7)+ iπ.  (Complex number)

Log49 to the base -7= log(49)/( some complex number)
Hence proved.


问:
Explaining an incorrect solution involving derivative of an inverse.
Suppose $f$ is injective and $f(1)=5$ and $f'(1)=3$. Then a quick application of the inverse function theorem gives $(f^{-1})'(5)=1/3$. However, what is incorrect in the following?:

$$(f^{-1})'(5)=(f^{-1}(5))'=f'(1)=3$$

My question is, how do I explain to a student that the first equality above is incorrect?

答:
The error is in the step

$$(f^{-1})'(5) = (f^{-1}(5))'$$

Let $f^{-1} = g$

$$\implies g'(5) = (g(5))'$$

What you have on the left hand side is the value of the derivative of the inverse evaluated at $x=5$. What you have on the right doesn't make sense - you are taking the derivative of an instance of the function at a point - which makes no sense, as you can only define a derivative of a function

I'm not sure what grade this is being discussed in, but if derivatives have been introduced I'm assuming there is an understanding of the basics of limits and differentiability


问:
Combinatorics and pigeon hole principle
How many subsets of $\{1,2,3,\dots,10\}$, each containing at least 3 and at most 5 elements, must be selected in order to ensure that the selection contains three subsets $A , B , C\subset \{1,2,\dots,10\}$ such that sums of elements of each of the subsets are equal?

35 sums are possible, from minimum sum 6 (for the subset $\{1,2,3\}$) to maximum sum 40 (for the subset $\{6,7,8,9,10\}$). So by the Pigeonhole Principle, at least 71 sets must be chosen to ensure there are three with equal sums. But in the book where I found this question, the answer is given as 67. Where am I wrong?

答:
The question seems to be requiring that the subsets be distinct.  In this case, certain sums can only be attained in one way, and it is thus impossible to put two pigeons in these pigeonholes, reducing the count.

Explicitly:

The sum 6 can only be attained by the set $\{1,2,3\}$.

The sum 7 can only be attained by the set $\{1,2,4\}$.

The sum 40 can only be attained by the set $\{6,7,8,9,10\}$.

The sum 39 can only be attained by the set $\{5,7,8,9,10\}$.

You can only fit one pigeon in each of these pigeonholes.  The other 31 numbers can fit two or more.  Then the maximum number of pigeons without any pigeonholes having three is $4+2*31=66$.  Hence 67 guarantees three pigeons in some hole.

Title: Combinatorics and pigeon hole principle


问:
If we know nullspace of matrix, how to find reduced row echelon form of that matrix?
vectors u = [4 1 0 0]  and v = [1 0 2 1] form a base of nullspace of matrix $$ A\in M_{5,4}(R) $$ Find a reduced row echelon form of Matrix A.

Since $ n-r = dimN(A) $ we know we got two base variables and two free ones. And reduced row echelon form will look like this:
$$
        \begin{bmatrix}
        1 & 0 & a & b \\
        0 & 1 & c & d \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        \end{bmatrix}
$$

I don't know where to go from here. I am also confused about something, if I plug in [4 1 0 0] instead of x1 x2 x3 and x4 in my "potential" reduced row echelon form I would get this:

1*4 + 0*1 + a*0 + b*0 = 0

0*4 + 1*1 + c*0 + d*0 = 0

and I get 4=0 and 1=0 from that. How is that possible? (I know I am making a mistake somewhere, just don't know where.)

答:
I believe that the accepted solution does not have full generality and will note work if, say, the first two columns of the matrix are linearly dependent.

If the columns of matrix $N$ represent the null space, then the RREF can be reconstructed by the following matlab command

```<code>rref(transpose(null(transpose(N))))
</code>```

See this lesson on Lemma for a discussion of the problem (but not the above solution).

Body:
I believe that the accepted solution does not have full generality and will note work if, say, the first two columns of the matrix are linearly dependent.

If the columns of matrix $N$ represent the null space, then the RREF can be reconstructed by the following matlab command

rref(transpose(null(transpose(N))))

See this lesson on Lemma for a discussion of the problem (but not the above solution).


问:
How to derive a formula with interacting variables (with example problem)?
Imagine you are an enterprising child: and instead of starting a lemonade stand, you open a lemonade bank.

People can deposit lemons or sugar. In return, they get credit at your bank which they can used to withdraw either lemons or sugar.

Your mom gives you 1 lemon and your dad gives you 1 ounce of sugar. You issue each of them 1 unit of credit in return. Your bank has been seeded.

You decide you wish to maintain a 1:1 ratio between lemons and sugar. You decide you will do so by issuing more credit for deposits of what you have less of while charging more credits to withdraw what you have less of, and vice versa.

If your mom gives you another lemon and you issue her another full unit of credit then lemons will account for 67% of all credit issued, while lemons would be worth the same amount of credit as sugar (so 50%). If sugar is more valuable to the external market your mom could deposit lemons and extract sugar cheaply.

This will not do. This is why you must have a dynamically adjusting credit issuance which depends on the resulting internal ratio after any transaction.

So your mom says she would like to deposit another lemon into your bank. You currently have 1 lemon, 1 ounce of sugar, and 2 credits issued (in circulation). You must give your mom an amount of credit in return for her lemon, but how much? Giving her a full credit could result in an imbalance, so you must give her less than a full credit, but how much less?

You decide to guess. You think, "if I give a full credit, then all credits in circulation (3) will be backed by 2 lemons and 1 ounce of sugar, so 67% backed by lemons. So instead of 1 full credit, I'll give her .67 credits in return for the second lemon."

You realize this is a good guess because now 2.67 credits will be backed by 2 lemons at .67 credits apiece, totaling 1.34 credits; exactly half of 2.67. You have maintained 1:1 ratio in the value of lemons to sugar regardless of their count. But that makes you realize, now the 1 ounce of sugar in the bank is worth 1.34 credits instead of 1. Sugar's price has increased, in fact, in terms of lemons, it's now twice as valuable.

This dramatic change doesn't quite sit well with you so you tell your mom you'll have to get back to her about the amount of credit you will issue to her for her second lemon deposit.

You decide to do a comparison of scenarios just to recap. If you paid out 1 credit for the lemon then your bank credit would be backed by 67% lemons and 33% sugar, and lemons would cost the same as sugar or you could say lemons would account for 50% of the cost. If instead, you paid out .67 credits for the lemon then your bank credit would be backed by 50% lemons and 50% sugar, but lemons would account for 33% of the cost.

You already ruled out the first scenario, but the second scenario seems to push the imbalance into the price, rather than the backing of your credit. The second scenario seems like an over-correction. So you want to strike a balance somewhere between .67 and 1 credit. But where?

It seems you want to balance the two concerns so that they are mirror images of one another: the backing % for lemons should equal 1 - the unit cost % for lemons. But you don't know the formula to achieve such a balance. So again, you guess. Perhaps smack dab in the middle: .67+((1-.67)/2)~.83334 You give that a try:

Turns out that's really close: you compare 58.8% backing for lemons to 58.3% unit cost for ounces of sugar. It's almost a perfect match. But you're kind of a little psycho and you want the exact formula to balance the two concerns.

Why didn't the middle of 2/3rds and 1 give you the perfect balance you were looking for?

How would you create such a formula? What if your mother wanted to deposit 2 more lemons into your bank or withdraw .5 lemons?

Given this kind of input to the formula, what is the credit cost per lemon you should charge or pay?

EDIT (follow up):

Turns out the exact balance is 0.82842712474619. Which was arrived at iteratively. My question is, can a function be derived that computes this number? If so, how? and what is the function?

EDIT:

How are you calculating the backing?

Backing is calculated by multiplying the cost (or how many credits you're willing to mint) by the number of that asset, for each asset.

Continuing with the above example, imagine you go to your mom and accept her 2nd lemon deposit in exchange for the 0.82842712474619 credits. Once the transaction is carried out you would be in the situation outlined by the last spreadsheet image above; namely, you would have 2 lemons and 1 ounce of sugar in your bank, and you would have issued 2.82842712474619 credits in total.

So 2 lemons * 0.82842712474619 credits = 1.65685424949238 credits are backed by lemons and since there are 2.82842712474619 credits in circulation then we know 2 things: A) 1.65685424949238 / 2.82842712474619 = ~58.6 % so what we really mean is that each unit of credit in circulation is backed 58.6% by lemons and ~41.4% by sugar, B) which means we know the implied updated sugar price even though no sugar transaction has occurred: 2.82842712474619 credits in circulation - 1.65685424949238 credits accounted for by lemons = 1.17157287525381 credits accounted for by sugar (aka backed by sugar). To get the price of sugar we merely divide the total number of credits backed by sugar by the number of ounces of sugar: 1.17157287525381 credits / 1 sugar ounce = 1.17157287525381 credits.

How are you calculating the cost?

That is explained above, but to get the percentage, of course, we simply add up how much each asset costs and divide by the total. Thus 1.17157287525381 sugar cost + 0.82842712474619 lemon cost = 2 total cost so 0.82842712474619 / 2 ~ 41.4 % and 1.17157287525381 / 2 = 58.6 %

Again, notice the cost percentages per asset are the inverse of the backing percentages. That's what I mean by balanced. Finding the right cost for the number of units proposed to be traded that produces this inverse mirrored relationship is the entire goal of the quadratic function we're trying to deduce.

That means if your mom had wanted to deposit 3 or 40 lemons the number of credits you would have issued her per lemon would have been lower than 0.82842712474619. If she had wanted to withdraw the number of credits you would have required of her would have been higher (withdrawing might actually be more nuanced because there's a limit to how many you can withdraw, let's not go down that rabbit hole). I just wanted to say that the relationship between how much someone wants to deposit and the price you will pay per unit has a curve to it, not sure what that curve looks like but I imagine it's something like this:

I'll make additional edits to answer your questions later, but I hope this is sufficient in the meantime to show how backing % and costs % are calculated.

EDIT:

I hope you have enough to define the problem but I'll try to explicitly answer the other questions too:

you say credit can be used to withdraw "either lemons or sugar."
What do you mean by this? Do you mean 1 unit of credit can be used to
withdraw 1 lemon or 1 ounce of sugar? Or are you setting prices
dynamically?

dynamically, in fact, the whole point is to map an amount of any asset someone has to deposit to an amount of issued credit that will balance the two concerns listed above. the amount of credit that is issued must be dynamic.

What do you mean by "lemons will account for 67% of all credit
issued"" in paragraph 5? What implicit relationship between deposits
and credit is being tracked here?

"If your mom gives you another lemon and you issue her another full unit of credit then lemons will account for 67%" because you would have 2 lemons and you would have issued 3 credits and the price of lemons would be 1 credit. so 2 lemons * 1 credit = 2 credits and 2 credits (for lemons) / 3 total credits

What do you mean by "while lemons would be worth the same amount of
credit as sugar" in paragraph 5? This seems to imply that you don't
have the freedom to set prices dynamically; is that right? If so, how
are the prices being set?

"If your mom gives you another lemon and you issue her another full unit of credit then lemons will account for 67% of all credit issued, while lemons would be worth the same amount of credit as sugar" means if you paid 1 credit for another lemon, lemons, and sugar would be the same price. it doesn't mean prices are not dynamic. At this point in the word-problem, I'm trying to show why prices must be dynamic by showing what would happen if you treated them as static.

In paragraph 8 you say "all credits in circulation (3) will be backed by 2
lemons and 1 ounce of sugar, so 67% backed by lemons" which seems to imply that
each deposit backs each credit proportionately. But in paragraph 9 you say "now
2.67 credits will be backed by 2 lemons at .67 credits apiece, totaling 1.34
credits"; why?

If you pay 1 full credit the number of credits in circulation would be 3. if you pay .67 credits the number of credits in circulations would be 2.67. I hope that helps.

cont.: specifically, why does the .67 credit number also apply to
the first lemon, which was deposited for 1 credit? Why isn't the
backing calculation being done as follows: 2.67 credits are backed by
2 lemons and 1 ounce of sugar, so 2.67 / 3 = 0.89 credits apiece for
each of the 2 lemons and the ounce of sugar? Or: each deposit backs
the credit it was exchanged for, so 1 credit for the sugar, 1 credit
for the first lemon, 0.67 for the second lemon?

Because the current price of lemons is the latest price they transacted for. We don't care what lemons went for in the past, they are priced as of now. the system has (or would have, after the transaction is completed) 2 lemons, whatever price is chosen in the transaction is the value that the system thinks lemons are worth - all lemons.

In paragraph 11, what do you mean by "but lemons would account for
33% of the cost"? The cost of what? You seem to have calculated this
number by dividing 0.67 by 2. But what does the 2 represent here?

I'm dividing by 2 because you can add up all the prices of the assets and you come up with 2. so this asset's (lemon's) price represents a certain % of the total of all prices of assets in the system. That what it is, but technically price of sugar which makes up the rest of the number 2 is actually implied by the total credits in circulation as explained in the 3rd paragraph of my previous EDIT.

FINAL EDIT:

Here is the generalized solution (see comments of the answer below):

```<code>e = existing lemons (1)(1) (state)
d = deposit lemons (1)(2) (input variable)
s = existing sugar (1)(1)(state)
c = existing credits (2)(2) (state)
x = cost of lemons (.8284...)(.73...) (output)
</code>```

(the second equation is for a deposit, not sure how to apply the first one, perhaps to withdrawals.)

答:
Here's a neat feature I didn't realize existed; you can ask WolframAlpha if it thinks a constant has a closed form and it will spit out suggestions. For your $0.8284 \dots$ it suggests

$$2 \sqrt{2} - 2 = \frac{2}{\sqrt{2} + 1} \approx 0.82842712474619 \dots $$

which is an exact match to all of your digits.

I don't understand your problem statement (I put a long list of questions into the comments) but fortunately the mathematical content can be reverse-engineered from the spreadsheet screenshots alone, together with the big hint that it should all come down to solving a quadratic equation. It looks like if you pay out $x$ credits for the third lemon then the spreadsheet reads

$$\begin{array}{cccc}
  \text{assets} & \text{units} & \text{backing} & \text{cost} \\
  \text{lemons} & 2 & 2x & x \\
  \text{sugar} & 1 & 2 - x & 2 - x \\
 \text{total} & 3 & 2 + x & 2 \\
\end{array}$$

(I don't understand this, especially the backing calculations) and the condition you want appears to be that the fraction of the cost (I don't understand what this means) made up by the lemons, which is $\frac{x}{2}$, is equal to the fraction of the backing (I don't understand why you're calculating this the way you are) made up by the sugar, which is $\frac{2 - x}{2 + x}$, so

$$\frac{x}{2} = \frac{2 - x}{2 + x}.$$

Clearing denominators gives a quadratic equation

$$x^2 + 4x - 4 = 0$$

which has unique positive real solution $x = 2 \sqrt{2} - 2$. But given the lack of precision in the problem statement I am not at all confident that this is the number you actually wanted.

Edit: Here is my current understanding of how this calculation generalizes to arbitrary numbers. Suppose we have $L$ lemons, $S$ ounces of sugar, and $C$ credits in circulation, and Mom wants to deposit $L'$ new lemons and we want to determine how many credits $x$ to give her. If I've understood everything correctly, the spreadsheet should now read

$$\begin{array}{cccc}
  \text{assets} & \text{units} & \text{backing} & \text{cost} \\
  \text{lemons} & L + L' & (L + L')x & x \\
  \text{sugar} & S & C + (1 - L - L')x & \frac{C + (1 - L - L')x}{S} \\
 \text{total} & L + L' + S & C + x & \frac{C + (1 - L - L' + S)x}{S} \\
\end{array}$$

(this shouldn't be telling you anything you didn't already know because you must already have formulas equivalent to these in your spreadsheet if I've correctly understood the problem). The fraction of the cost made up by lemons is now $\frac{Sx}{C + (1 - L - L' + S)x}$, while the fraction of the backing made up by sugar is now $\frac{C + (1 - L - L')x}{C + x}$, and the condition that these are equal is

$$\frac{Sx}{C + (1 - L - L' + S)x} = \frac{C + (1 - L - L')x}{C + x}$$

which is again, after clearing denominators, a quadratic equation in $x$.


问:
Suppose $G$ is a finite group. What can we say about the representations (matrices) corresponding to elements of $G$?
$G$ is finite and $\rho: G \rightarrow GL_n(\mathbb{C})$ is a group representation of $G$.  </p>

<p>What I would like to know is what do the properties of $G$ (it being finite) tell us about the matrices corresponding to group elements $\rho(g)$?  What can we conclude?  For example, what can we say about diagonalizability and why?  </p>

答:
Here is another approach:

Since every representation $ \rho: G\to GL(n,\mathbb C)$ can be taken to be unitary, i.e., there exists an $M\in GL(n,\mathbb C)$ for all $g\in G$, such that $M^{-1}\rho(g)M\in U_n=\{P\in GL(n,\mathbb C)|P^\ast P=I_n\}$ where $P^\ast$ is the complex conjugate of $P$.

The spectral theorem tells us that every unitary matrix is diagonalizable with all the eigenvalues having absolute value one.

In particular, $\rho(g)$ is diagonalizable.


问:
How to find the limit of $\frac{\frac{1 - 2n}{n}}{5 + 3^{-n}}$
I have an intuition that the limit is $\frac{-2}{5}$, by noting that $\frac{1}{n} \to 0$ and $3^{-n} \to 0$ as $n \to \infty$. But I do not know how to show this formally, i.e. we can always find a starting index $n$ such that $\left| \frac{2}{5} - \frac{\frac{1 - 2n}{n}}{5 + 3^{-n}} \right| < \epsilon$ for a given epsilon. By manipulating the absolute value I managed to get it to form $\left| \frac{-2n3^{-n} - 5}{5n3^{-n} + 25n}\right|$. I suppose the squeeze theorem is one way to go, but as of now I do not know what my two other sequences should be.

答:
Are you allowed to use arithmetical operations with limits? You can use that $\frac{1 - 2n}{n} \to-2$ and $5 + 3^{-n} \to 5$ separately and then divide limits.

If no, then
$$\left|\frac{\frac{1 - 2n}{n}}{5 + 3^{-n}}+\frac{2}{5}\right|=\left|\frac{5+2n3^{-n}}{5n(5 + 3^{-n})}\right|< \frac{6}{5n}$$


问:
Let $(G,*)$ be a finite group and $a \in G$. Show that $\circ(a) \le |G|$.
Let $(G,*)$ be a finite group and
$a \in G$. Show that $\circ(a) \le |G|$.
($\circ(a)$ denote the order of $a$.)

This is the given hint:
Suppose that $|G|=n$. Note that
$\{e_G,a,a^2,\ldots,a^n\} \subseteq G$ and $G$ has $n$ elements.

How to approach $\circ(a) \le |G|$?
Does the given hint means that: "there exists integers $p$ and $q$ with $p>q$ such that $a^p=a^q$.
Then, $a^{p-q} = e_G$. Write $m=p-q$ and hence $a^m = e_G$ for some integer $m$, with $0\le m \le n-1$." ?

答: If all the elements of the set $\{e_G,a,a^2,\ldots, a^n\}$ were distinct, it would contain $n+1$ elements. As $G$ has only $n$ elements, there must be a repeat. Therefore, the argument you propose must follow.


问:
Inversion Formula for Asymptotic Behaviour
Suppose that for some functions $f,g:\mathbb{R} \to \mathbb{R}$, we have the following relationship
$$\sum_{n=1}^{x} f(n) \sim g(x)$$
or in limit notation
$$\lim_{x \to \infty} \frac{\sum_{n=1}^{x} f(n)}{g(x)} =1 $$
Is there a way to retrieve asymptotic behaviour of $f(x)$ in terms of $g(x)$, i.e. can we find a function $h: \mathbb{R} \to \mathbb{R}$ such that
$$f(x) \sim h(g(x))$$

答:
In general the asymptotic behavior of $f$ is not determined by the asymptotic behavior of its partial sums. Loosely speaking the problem is that the summation averages out oscillatory behavior.

A natural source of examples of this behavior is functions coming from number theory; for example, the totient function $\varphi(n)$ has the property that

$$\sum_{k=1}^n \varphi(k) \sim \frac{3n^2}{\pi^2}$$

which is also true of the function $\frac{6n}{\pi^2}$, but it is not true that $\varphi(n) \sim \frac{6n}{\pi^2}$; in fact $\lim_{n \to \infty} \frac{\varphi(n)}{n}$ does not exist, and we have $\limsup_{n \to \infty} \frac{\varphi(n)}{n} = 1$  (by taking $n$ to be prime) but $\liminf_{n \to \infty} \frac{\varphi(n)}{n} = 0$ (by taking $n$ to be the product of the first $k$ primes and letting $k \to \infty$).


问:
Obtaining equation
As I was studying differential equations, I got a question in my mind that we are taking out general solutions for a differential equation.
Like for example:
$$\frac{dy}{dx} = \frac{1}{(1+x³)} - \frac{3x²}{(1 + x²)}y$$
Solution:
$$\Rightarrow \frac{dy}{dx} + \frac{3x²}{(1 + x²)} y = \frac{1}{(1+x³)}$$

Comparing it with $\frac{dy}{dx} + Py = O$, we get

$$P = \frac{3x²}{1+x³}$$

$$Q= \frac{1}{1 + x³}$$

Let’s figure out the integrating factor(I.F.) which is $e^{\int P\,dx}$

$$ \Rightarrow ~\text{I.F}~  =  e^{\int \frac{3x²}{(1+x³)}\,dx}=e^{\ln(1+x³)}$$

$$\Rightarrow~\text{I.F.}~ = 1 + x³$$

Now, we can also rewrite the L.H.S as:

$$\frac{d(y \times I.F)}{dx}, $$

$$ \Rightarrow d(y \times (1 + x³)) dx = \frac{1}{(1 +x³)} \times (1 + x³)$$

Integrating both the sides w. r. t. $x$, we get,

$$\Rightarrow y \times ( 1 + x³) =  x$$

$$\Rightarrow y = \frac{x}{(1 + x³)}$$

$$\Rightarrow y = \frac{x}{(1 + x³)} +  C$$

But how do we get that differential equation (as given in the above question $\rightarrow \frac{dy}{dx} = \frac{1}{(1+x³)} – \frac{3x²}{(1 + x²)}y$) from our real life ?

答:
I am afraid that,  with the $x^2$ and  $x^3$, there is no solution.

For the homogeneous equation, we have
$$y= C \,e^{-3 \left(x-\tan ^{-1}(x)\right)}$$ Variation of the parameter leads to
$$C'=\frac{e^{3 (x- \tan ^{-1}(x))}}{1+x^3}$$ which is impossible to integrate.

Same problem if we had $x^2$ twice.

So, as already said in comments, the problem must be
$$\frac{dy}{dx} = \frac{1}{(1+x^3)} - \frac{3x²}{(1 + x^3)}y$$ which is very simple.

Title: Obtaining equation


问:
What is the probability of spelling out "miss" in "mississippi"?
Suppose that the word "mississippi" is written on a piece of paper. If we cut this piece of paper into 11 smaller pieces of paper, each containing exactly one letter, and then place these letters in a bag, what is the probability that if we select four numbers at random with no replacement, the first letter will be "m", the second letter "i", the third letter "s", and the fourth letter is "s" (so that we form the word "miss")?

At this point, I know that computing ${}_{11}P_4$ will give us the total number of ways we can form $4$-letter words. Would I have to divide ${}_{11}P_4$ by $4!\cdot 4! \cdot 2!$ to get the total number of ways to uniquely spell out "miss". Then, I could find the probability by dividing $1$ by the total number of ways found?

答:
The TOTAL number of ways you could select 4 letters out of 11 would be 11P4.

So, the sample space (denominator) of the probability would be 11P4.

Sample space = $7920$

When we choose miss out of mississippi we could choose from one m and two sss and 4 is.

So there would be 4P2 = $12$ ways where the ss part of miss could be chosen. And there are 4 is to choose from. And only one m to choose from.

So the total number of ways miss ALONE can be selected = $1*4*12 = 48$ ways

So, therefore, the probability to spell out miss would be $\frac{8}{7920}=\frac{1}{165}$

Comment if my answer doesn't satisfy you.


答:
It's quite reasonable to consider the empty set "countable", and if you leave out the surjection bit in the OP, you can easily get rid of the assumption of nonemptiness:

A set $S$ is countable iff there is an injection $g: S \to \Bbb{N}$.

But if you let functions have empty images, then the concept of "injection" becomes utterly trivial, because as you note, maps with empty image would be vacuously injective.

For instance, I could get an injection of $\omega_1$, the first uncountable ordinal, into $3 = \{ 0, 1, 2 \}$, by sending all the uncountably many elements of $\omega_1$ to the empty subset $\emptyset \subset 3$.

A definition of "function" that would make $3$ uncountable is obviously going to create way more technical issues than it solves.

In practice, we handle situations where some potential inputs don't get assigned outputs (think of $f: \Bbb{R} \to \Bbb{R}$, $f(x) := 1/x$, with the potential input $x = 0$) not by saying, for instance, that

"$1/0 = \emptyset,$"

but by saying,

"$x = 0$ is not in the domain of $f(x) = 1/x$"

or,

"$f(x) = 1/x$ is undefined when $x = 0$."

According to this framework, when we write down a function $f: \Bbb{N} \to \emptyset$, none of the numbers $k \in \Bbb{N}$ are actually in the domain of $f$, i.e. the domain of $f$ is empty, and $f: \emptyset \to \emptyset$ is mundanely revealed as the empty function from the empty set to itself.

Title: Empty set in countability proofs

Body:
It is a provable theorem (defining countable as finite or countably infinite) that a nonempty set $S$ is countable if and only if there exists a surjection $f: \mathbb{N} \to S$ if and only if there exists an injection $g: S \to \mathbb{N}$.

Every single time I have seen the result quoted, it is assumed that $S$ is nonempty. Is this necessary, or is just done to eliminate an extra edge case? The result seems fine to me. If $S = \emptyset$, it's finite, hence countable. Then there is only one map $f: \mathbb{N} \to \emptyset$ -- the empty list -- which is vacuously surjective and injective. (It's surjective since there are no $x \in \emptyset$ we need to map to, and vacuously injective because $f(a) \neq f(b)$ for any $a,b \in \mathbb{N}$.) There is also only one map $g: S \to \mathbb{N}$, once again the empty list, which is again vacuously injective.

Is this correct?


答: At each stage calculate the probability of picking the required letter and multiply. The M has probability $\frac1{11}$ of being picked, the I $\frac4{10}$, the first S $\frac49$ and the second S $\frac38$. The final answer is $\frac1{165}$.

Title: What is the probability of spelling out "miss" in "mississippi"?

Body:
Suppose that the word "mississippi" is written on a piece of paper. If we cut this piece of paper into 11 smaller pieces of paper, each containing exactly one letter, and then place these letters in a bag, what is the probability that if we select four numbers at random with no replacement, the first letter will be "m", the second letter "i", the third letter "s", and the fourth letter is "s" (so that we form the word "miss")?

At this point, I know that computing ${}_{11}P_4$ will give us the total number of ways we can form $4$-letter words. Would I have to divide ${}_{11}P_4$ by $4!\cdot 4! \cdot 2!$ to get the total number of ways to uniquely spell out "miss". Then, I could find the probability by dividing $1$ by the total number of ways found?


问:
Find Force from Moment Vector, Given Radius and Unit Vector
Given a known Moment ($\vec{M}$), distance ($\vec{r}$), and unit vector $\hat{f}$  of force ($\vec{F}$), is it possible to find the resultant magnitude of force (using the formulation of the cross product)? Here the unit vector of force $\hat{f}$, $\vec{r}$, and $\vec{M}$ are all known.

I've been trying to think through using a Skew-Symmetric matrix but I'm not sure how to bound the solutions from:

$\vec{M} = \vec{r} \times \vec{F}\\$

$\vec{M} = [\vec{r}]_{x} \vec{F}\\$

$\vec{F} = [\vec{r}]^{-1}_{x} \vec{M}$

with a given unit vector $\hat{r}$

Is this even possible?

答:
Note that you won't get an equation of the type $\vec F=k\vec M$ as that would imply $\vec F,\vec M$ are parallel while the original equation implies they are orthogonal.

Using $\vec F=|\vec F|\hat f$ we get $\vec M=|\vec F|(\vec r\times \hat f)$ so you can find $|\vec F|$ by taking the ratio of the $x$ components $[\vec M]_x$ and $[\vec r\times\hat f]_x$ (or the $y/z$ components) or the ratio of the magnitudes $|\vec M|$ and $|\vec r\times\hat f|$.

I am not sure what you want to do with a skew-symmetric matrix.


答: How about the theorem of Lagrange?  For one number to divide another it must not be greater than it.</p>

Body:
Let $(G,*)$ be a finite group and
$a \in G$. Show that $\circ(a) \le |G|$.
($\circ(a)$ denote the order of $a$.)

This is the given hint:
Suppose that $|G|=n$. Note that
$\{e_G,a,a^2,\ldots,a^n\} \subseteq G$ and $G$ has $n$ elements.

How to approach $\circ(a) \le |G|$?
Does the given hint means that: "there exists integers $p$ and $q$ with $p>q$ such that $a^p=a^q$.
Then, $a^{p-q} = e_G$. Write $m=p-q$ and hence $a^m = e_G$ for some integer $m$, with $0\le m \le n-1$." ?


问:
Taylor's Polynomial of a Function: how we can arrive at it naturally?
Let $f:(a,b)\rightarrow\mathbb{R}$ be a function, which is differentiable, say infinitely many times, in the domain. [We can minimize the hypothesis, but for question, we can assume it.]

Consider the polynomial $P(x)=f(c) + (x-c)f'(c) + \cdots + \frac{f^{k}(c)}{k!}(x-c)^k$.

This polynomial has two properties:

(1) This is a good approximation to $f$ near $c$.

(2) From definition, $P^{(i)}(c)=f^{(i)}(c)$ for $i=1,2,\ldots, k$.

But, I didn't get any "natural motivativation" to arrive at exactly this polynomial.

After discussion with some teachers, they said, it generalizes, mean value theorem: $f(x)-f(x)=f'(\alpha)(x-c)$ i.e. $f(x)=f(c) + f'(\alpha)(x-c)$. But, I didn't get from this mean value theorem, how one naturally arrives at the above $P(x)$, which involves some "factorials", which were not looking in mean value theorem.

In mean value theorem, $f'(\alpha)$ can be written as $f'(\alpha)/1!$ and so one may justify this way for appearance of factorial things; but I would ask then, *we can write $f'(\alpha)$ as $f'(\alpha)/1$', $f'(\alpha)/1^2$' so can't we take the polynomial as  $P^{(i)}(c)=f^{(i)}(c)$ for $i=1,2,\ldots, k$, then can't we take polynomial as $f(c) + (x-c)f'(c) + \cdots + \frac{f^{k}(c)}{k^2}(x-c)^k$?

Why the choice of $P(x)$ above is most natural one and what is "natural way" to arrive at the polynomial $P(x)$?

答:
The usual proof when assuming the most hypothesis on $f$ seems to explain pretty well where the coefficients come from. There are several other proofs if $f$ doesn't satisfy enough hypothesis, giving you different conclusions, but the more hypothesis you assume, the more intuitive is the reasoning. Here is the derivation I'm talking about.

$$\begin{align}f(x)&=f(0)+\int_0^xf'(y)dy\\
&=f(0)+\int_0^x(f'(0)+\int_0^yf''(z)dz)dy\\
&=f(0)+xf'(0)+\int_0^x\int_0^yf''(0)+\int_0^zf'''(w)dwdzdy\\
&=f(0)+xf'(0)+(x^2/2)f''(0)+\int_0^x\int_0^y\int_0^zf'''(w)dwdzdy\\
&=\cdots
\end{align}$$

We see that the coefficient of $f^{(n)}(0)$ is $n$ nested integrals starting by $\int_0^x \cdots dy$ and thus it is the area of a simplex of side $x$ with a right angle. One way to get the coefficients is to use the recursion $\int_0^x y^n/n! dy = x^{n+1}/(n+1)!$.

But there is a lot more to say about this basis $x^n/n!$ that I am not aware of (you find it at a lot of places, not only in Taylor expansion). For instance, a consequence of the things I don't know is that we can think of $x^n/n!$ as "take the product of $n$ copies of $x$ and cancel out the permutations you can make of the factors". Ok, so what I just said probably doesn't make a lot of sense, but one way to get something formal out of it is to say that $x^n$ represents the area of a cube of side $n$
$$ \{(t_1,\dots,t_n)\in\mathbb{R}^n\,|\,0<t_1,\cdots,t_n<x\} $$
while $x^n/n!$ is the area of the simplex
$$ \{(t_1,\dots,t_n)\in\mathbb{R}^n\,|\,0<t_1<\cdots<t_n<x\} $$
since in the cube, we have $n!$ ways to order the coefficients $t_i$ and hence $n!$ copies of the the simplex.


问:
Representation of n digits number
I am confused in the representation of $n$ digits number. If $N$ is a natural number with $n$ digits. Then $a_n, a_{n-1}, \ldots$ represents the  number $N$. But if for example $N$ is the $3$ digits number. What would be the value of $a_0$ digit. Because obviously $3$ digits number has $3$ digits $(n_3, n_2, n_1)$.

I hope u understand my question.

Thanks

答: Obviously an $n$ digit number is represented as $a_{n-1} a_{n-2} \ldots a_0$ not $a_n a_{n-1} \ldots a_1$.

Title: Representation of n digits number


问:
If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism?
$X$ and $Y$ denote Hilbert spaces. If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism? Homeomorphism means continuous map with continuous inverse.

I think the answer is yes, the only thing I am unable to show is what is the inverse of $T^*$ and if it is continuous?

答:
The proof of boundedness of $T^\ast$:

$$\|T^\ast(x)\|^2 = \langle T^\ast x, T^\ast x\rangle= \langle TT^\ast x, x\rangle \leq \|TT^\ast x\|\|x\| \leq \|T\|\|T^\ast x\|\|x\|,$$ where the second to last inequality is Cauchy-Schwarz. If $\|T^\ast x\| = 0$, then there's nothing much to say. Otherwise, we divide and get $$\|T^\ast(x)\| \leq \|T\|\|x\|.$$ So $\|T^\ast\|$ is bounded and $\|T^\ast \| \leq \|T\|.$

The reason that $T^\ast$ is a bijection is given in other answsers. The inverse of $T^\ast$ is bounded since $(T^\ast)^{-1}$ is the adjoint of a continuous function, namely $T^{-1}$. (Or if you like shiny tools, you can use the open mapping theorem!)

Edit: Once you know about reflexivity of Banach spaces, the above will give $\|T^\ast\| \leq \|T^{\ast\ast}\| = \|T\|$ and so $\|T^\ast\| = \|T\|$ follows.


答:
I think you have an indexing problem, if you start indexing from $0$ and in total there are $n$ digits, the last index should be $n-1$.

$$N = \sum_{i=0}^{n-1} a_i \cdot 10^i,    \text{ where } 0 \le a_i \le 9$$

For example, the number $123$ has $a_0=3, a_1=2, a_2=1$.

In contrast, if you start your index from $1$, then the last index should be $n$.

$$N = \sum_{i=1}^{n} b_i \cdot 10^{i-1},     \text{ where } 0 \le b_i \le 9$$

For example, the number $123$ has $b_1=3, b_2=2, b_3=1$.

Title: Representation of n digits number

Body:
I am confused in the representation of $n$ digits number. If $N$ is a natural number with $n$ digits. Then $a_n, a_{n-1}, \ldots$ represents the  number $N$. But if for example $N$ is the $3$ digits number. What would be the value of $a_0$ digit. Because obviously $3$ digits number has $3$ digits $(n_3, n_2, n_1)$.

I hope u understand my question.

Thanks


问:
Step by step solution of an Indefinite Integral
I had this on my test, I just couldn't get it, after solving it a lot.
$$\int\left\{\frac{\log x-1}{1+(\log x)^2}\right\}^2\,dx=\,?$$

I have tried substituting $\log(x)=t$ and then
$$
\int \frac{(t-1)^{2} e^{t} d t}{\left(t^{2}+1\right)^{2}}=\int\left\{\frac{\left(t^{2}+1\right)}{\left(t^{2}+1\right)^{2}}-\frac{2 t}{\left(t^{2}+1\right)^{2}}\right\} e^{t} d t
$$
and then, applying the by parts rule,
$$
=e^{t}\left(\tan ^{-1} t-\frac{1}{t^{2}+1}\right)-\int e^{t}\left(\tan ^{-1} t-\frac{1}{t^{2}+1}\right) d t
$$
after that I can't get ahead.

P.S.- All Curly brackets used here are normal brackets, not the ones used in fractional part. Thanks to @ParclyTaxel for pointing it out.

答:
Since $1+\log^2x$ is squared and appears in the denominator, we might try an antiderivative of something over $1+\log^2x$ because of the quotient rule:
$$\frac d{dx}\frac f{1+\log^2x}=\frac{f'(1+\log^2x)-f((2/x)\log x)}{(1+\log^2x)^2}$$
The numerator contains only $\log$s, so $f$ must cancel the $\frac1x$ of $\frac2x\log x$. So $f$ might be $x$; we check and it works, and the answer is
$$\frac x{1+\log^2x}+K$$


答: we have a result $$\int e^x(f(x)+f'(x))=e^x f(x)+c \tag 1$$ this ie very easy to prove by intgration by parts. You have already done the difficult part of splitting the integral to $$\int e^t\left(\frac{1}{t^2+1}-\frac{2t}{{(t^2+1)}^2}\right)dt $$Now if $$f(t)=\frac{1}{t^2+1} \implies f'(t)= \frac{-2t}{{(t^2+1)}^2}$$  which is of form (1).......

Title: Step by step solution of an Indefinite Integral

Body:
I had this on my test, I just couldn't get it, after solving it a lot.
$$\int\left\{\frac{\log x-1}{1+(\log x)^2}\right\}^2\,dx=\,?$$

I have tried substituting $\log(x)=t$ and then
$$
\int \frac{(t-1)^{2} e^{t} d t}{\left(t^{2}+1\right)^{2}}=\int\left\{\frac{\left(t^{2}+1\right)}{\left(t^{2}+1\right)^{2}}-\frac{2 t}{\left(t^{2}+1\right)^{2}}\right\} e^{t} d t
$$
and then, applying the by parts rule,
$$
=e^{t}\left(\tan ^{-1} t-\frac{1}{t^{2}+1}\right)-\int e^{t}\left(\tan ^{-1} t-\frac{1}{t^{2}+1}\right) d t
$$
after that I can't get ahead.

P.S.- All Curly brackets used here are normal brackets, not the ones used in fractional part. Thanks to @ParclyTaxel for pointing it out.


答:
One way to do this is to find the number of ways to rearrange the letters such that the first four letters are the same (i.e. the number of ways to rearrange the letters of "issippi"), and divide this by the number of ways to rearrange the letters of "mississippi." This can be calculated using multinomial coefficents as follows, where the top number is the number of letters in the word, and the bottom numbers are the counts of each unique letter.

$$\frac{
\begin{pmatrix}
7\\
2,2,3
\end{pmatrix}
}{
\begin{pmatrix}
11\\
1,2,4,4
\end{pmatrix}}
= \frac{210}{34650} = \frac{1}{165}
$$


答:
What about this approach?

Suppose that $|G| = n$. Let $a \in G$
and $\circ(a) = k$. Let $H = \langle a \rangle$. Then, $|H| = |\langle a \rangle| = \circ(a) = k$. Hence, by Langrange's Theorem, $k$ divides $n$ and
thus $k \le n$ as desired.

Title: Let $(G,*)$ be a finite group and $a \in G$. Show that $\circ(a) \le |G|$.


问:
Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$
I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.

I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.

I want to see an actual proof behind this equation.  Does anyone have any ideas?

答:
Consider,

$$ (1+x)(1+x)(1+x)(1+x).. \text{n times}$$

Notice that sequence of each factor is given as:

$$ \{a_i \} = \{ 1,1, 0, 0... \}$$

By $b_i^j$ denote the coefficents of $x^i$ term in the product of $j$ factors, from cauchy product rule:

$$ b_0^0 =1$$
$$ b_k^1=  \sum_{i=0}^k b_i^0 a_{k-i}$$
$$ b_k^2 = \sum_{i=0}^k b_k^1 a_{k-i} $$

$$ \vdots$$
by induction,
$$ b_k^n = \sum_{i=0}^k b_k^{n-1} a_{k-i}= \sum_{i=0}^k b_k^{n-1} a_{k-i}= \sum_{i=0}^k b_{k-i}^{n-1} a_i = b_k^{n-1}  + b_{k-1}^{n-1}$$

Hence,

$$ b_k^n =b_k^{n-1} + b_{k-1}^{n-1}$$

Now use the fact that coefficent of binomial product is just the binomial coefficent:

$$ \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$


问:
In a Locally Compact Hausdorff and second countable space every compact subset is a $G_\delta$ set
Let $X$ be a second countable LCH (Locally Compact Hausdorff) then every compact subset of $X$ is a $G_{\delta}$ set.

A set $E$ is $G_{\delta}$ if is a countable intersection of open sets.

I have two questions,

How to prove this result,

I am also wondering if that would be true for the closed sets in this space, or what would be necessary to be true for closed sets.

Edit:

I posted a resolution, which I found very interesting, of question 1.

答: If $X$ is LCH and C-II, as indicated, $K:=\alpha (X)$, the Aleksandrov (one-point) compactification is compact and metrisable. In a metric space, all closed sets are $G_\delta$ and a compact subset of $X$ is still closed in $K$. Then relativize to $X$ again.


问:
How many clopen subsets does the subspace [0,1]∪[3,4] have?
Show that the subspace $[0,1]∪[3,4]$ of $\mathbb R$ has at least $4$ clopen subsets. Exactly how many clopen subsets does it have?

Just to clarify, this is the subspace induced by $[0,1]∪[3,4]$ from $\Bbb R$, so the space is just the intersection of all open sets in $\Bbb R$ with $[0,1]∪[3,4].$

Finding $4$ clopen sets is easy. Just form half open intervals with one of the $4$ endpoints, like $[0, 1/2).$ As for how many clopen sets there are, I think the answer is that there are  uncountable many, since the intervals $(0,1)$ and $(3,4)$ have uncountably many elements.

答: $[0,1]$ and $[3,4]$ are connected spaces. So they have no non-trivial clopen subsets. So if $C$ is clopen in $X=[0,1]\cup[3,4]$, then $C \cap [0,1]$ is also clopen in $[0,1]$ and so can only be $\emptyset$ or $[0,1]$, and, likewise, $C \cap [3,4]$ is $\emptyset$ or $[3,4]$. These $2 \times 2$ options give you four options for $C$, and it's easy to check that these are all clopen.


问:
Empty set in countability proofs
It is a provable theorem (defining countable as finite or countably infinite) that a nonempty set $S$ is countable if and only if there exists a surjection $f: \mathbb{N} \to S$ if and only if there exists an injection $g: S \to \mathbb{N}$.

Every single time I have seen the result quoted, it is assumed that $S$ is nonempty. Is this necessary, or is just done to eliminate an extra edge case? The result seems fine to me. If $S = \emptyset$, it's finite, hence countable. Then there is only one map $f: \mathbb{N} \to \emptyset$ -- the empty list -- which is vacuously surjective and injective. (It's surjective since there are no $x \in \emptyset$ we need to map to, and vacuously injective because $f(a) \neq f(b)$ for any $a,b \in \mathbb{N}$.) There is also only one map $g: S \to \mathbb{N}$, once again the empty list, which is again vacuously injective.

Is this correct?

答:
Then there is only one map f:N→∅ -- the empty list -- which is vacuously surjective and injective.

That is not the empty list.

The empty function is $g:\emptyset \to \mathbb N$ and that is vacouosly injective and assuredly is NOT  surjective.

It is a function as for every $x \in \emptyset$ (of which there are none) it is vacuously true that there is a distinct $y$ so that $f(x) = y$.  And it is injective for any value of $y$ in the image of the mapping, of which there are none, it is vacuously true that there is only one $x \in \emptyset$ so that $f(x) = y$.  But it is not surjective as for any $y \in \mathbb N$, of which there are infinitely many, there are no $x \in \emptyset$ so that $f(x) = y$.

.....

However a function $f:\mathbb N \to \emptyset$ simply can not exist at all (although it would be vacuously surjective if it existed at all).

To be a function for every $x\in N$, for which there are infinitely many, there must be a distinct $y \in \emptyset$ so that $f(x)=y$ and that is obviously not true.

.....

So the theorem is not true for the empty set; there exists an injective function $f:\emptyset \to \mathbb N$, but there does not exist any function (surjective or otherwise) $g:\mathbb N \to \emptyset$.

Title: Empty set in countability proofs


问:
If $S$ is dense in $X$ and $S\subset T$, then $T$ is dense in $X$
Problem is from Topology without Tears by Morris

If $S$ is dense in $X$ and $S\subset T$, then  $T$ is dense in $X$.

Try

Assume $S$ is dense in $X$. Suppose $S\cap X=\varnothing$. Then every element of $X\setminus S$  is a limit point of $S$. By hypothesis  $S\subset T$  so every element of $X\setminus T$  is a limit point of $T$.

Thus $T$ is dense in $X$.

I am not too good with this topic. I basically tweaked a theorem and hope it works

Help

答: If any non-empty open subset of $X$ intersects $S$, a fortiori it also intersects $T$. So $T$ is dense whenever $S$ is.


问:
If we know nullspace of matrix, how to find reduced row echelon form of that matrix?
vectors u = [4 1 0 0]  and v = [1 0 2 1] form a base of nullspace of matrix $$ A\in M_{5,4}(R) $$ Find a reduced row echelon form of Matrix A.

Since $ n-r = dimN(A) $ we know we got two base variables and two free ones. And reduced row echelon form will look like this:
$$
        \begin{bmatrix}
        1 & 0 & a & b \\
        0 & 1 & c & d \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        \end{bmatrix}
$$

I don't know where to go from here. I am also confused about something, if I plug in [4 1 0 0] instead of x1 x2 x3 and x4 in my "potential" reduced row echelon form I would get this:

1*4 + 0*1 + a*0 + b*0 = 0

0*4 + 1*1 + c*0 + d*0 = 0

and I get 4=0 and 1=0 from that. How is that possible? (I know I am making a mistake somewhere, just don't know where.)

答:
First write your null space basis as a matrix:

$$N = \begin{bmatrix} 4 & 1 \\ (1) & 0 \\ 0 & 2 \\ 0 & (1) \end{bmatrix}$$

Then modify the basis (without changing the nullspace) so that each vector's "last" element is a "1" and in a different row from all other basis vectors.  That is already done in your problem.

Then pad the matrix with zero columns to put the pivots in the right place:

$$N_2 = \begin{bmatrix} 0 & 4 & 0 & 1 \\ 0 & (1) & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & (1) \end{bmatrix}$$

Then subtract the matrix from the identity:

$$I - N_2 = \begin{bmatrix} 1 & -4 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

That's the RRE for the matrix.  The reverse of this can be used to find the null space in a fairly easy way.


问:
Is this limit right, or is there any way to deal with it?
calculate
$$\displaystyle\lim_{n\to+\infty}\displaystyle\int_{0}^{1}\dfrac{nx^{1/2}}{1+n^2x}\sin^5{nx}dx$$
Because$$\left|\dfrac{nx^{1/2}}{1+n^2x}\right|\leqslant\frac{1}{2}.$$
So$$\displaystyle\lim_{n\to+\infty}\displaystyle\int_{0}^{1}\dfrac{nx^{1/2}}{1+n^2x}\sin^5{nx}dx=\displaystyle\int_{0}^{1}\displaystyle\lim_{n\to+\infty}\dfrac{nx^{1/2}}{1+n^2x}\sin^5{nx}dx=0$$

答:
Your answer is totally right but you can also just squeeze it:

\begin{eqnarray*}0
& \leq & \int_{0}^{1}\left|\dfrac{nx^{1/2}}{1+n^2x}\sin^5{nx}\right|dx \\
& \leq & \frac 1n \int_{0}^{1}\dfrac{x^{1/2}}{\frac1{n^2}+x}dx \\
& \leq & \frac 1n \int_{0}^{1}\dfrac{1}{\sqrt x}dx \\
& = & \frac 2n \stackrel{n\to \infty}{\longrightarrow} 0
\end{eqnarray*}


问:
Finding elements of order $8$ in $\mathbb{Z}_{8000000}$.
I want to find elements of order $8$ in $\mathbb{Z}_{8000000}$.

I know that elements of order $n$ in a group $\mathbb{Z}_m$ is $\phi(n)$. But how can I apply this result for a group having such a large number of elements?

Thanks for the help.

答: There are $4$ elements of order $8$. This is because Φ(8)=4, and these elements are $1,3,5,7$. So the generators of $8000000$ of order $8$ are $1000000,3000000,5000000$ and $7000000$.


问:
"smallest" irreducible polynomial over GF(p)
Let $n = 2^{e_0} q_1^{e_1} \cdots q_i^{e_i}$ be the prime factorization of $n$.

Let $p$ be an odd prime.

Is it true that: $\exists k$ such that $x^n+k$ is irreducible over GF($p$) $\iff$ $(2^{\min(e_0,2)}q_1 \cdots q_i) | (p-1)$ ?

Are there similar criteria for "larger" polynomials such as $x^n+k_1x+k_0$ ?

答:
The $\Leftarrow$-direction of the first question is answered in the affirmative with the aid of LTE, the Lifting-the-exponent Lemma.

Let $u$ be a primitive root modulo $p$. That is, an element of order $p-1$ in the multiplicative group of $GF(p)$. I claim that $f(x)=x^n-u$ is irreducible in $GF(p)[x]$.

This is seen as follows. Let $\alpha$ be a zero of $f(x)$ in some extension field $K$ of $GF(p)$. We can assume that $K=GF(p)(\alpha)$. Furthermore, $|K|=p^m$ where $m$ is the degree of the minimal polynomial of $\alpha$. Because every prime factor of $p-1$ is also a factor of $n$, we can immediately conclude that the order of $\alpha$ in $K^*$ is $n(p-1)$. Therefore $n(p-1)$ is a factor of $p^m-1$.

Let $q$ be one of the odd prime factors of $n$ and let $q^e$ be the (highest) power $q$ that appears in the factorization of $n$. We know that $q\mid p-1$. So LTE says that
$$
\nu_q(p^m-1)=\nu_q(p-1)+\nu_q(m).
$$
On the other hand
$$\nu_q(n(p-1))=\nu_q(p-1)+\nu_q(n)=\nu_q(p-1)+e.$$
As $n(p-1)\mid p^m-1$ we can deduce that $\nu_q(m)\ge e$. In other words, $m$ must be divisible by $q^e$.

We still need to deal with the even prime. If $4\mid n$, your condition conveniently implies that we also have $4\mid p-1$. This is just what the doctor ordered for LTE to bite! We can, again,
deduce that
$$
\nu_2(p^m-1)=\nu_2(p-1)+\nu_2(m)
$$
as well as
$$
\nu_2(n(p-1))=\nu_2(p-1)+e_0.
$$
Again, the conclusion is that $2^{e_0}\mid m$. Lastly, if $n\equiv2\pmod 4$ then we need $(p^m-1)/(p-1)=1+p+p^2+\cdots+p^{m-1}$ to be even. This is the case if and only if $m$ is even, so even in this case we have $2^{e_0}\mid m$.

We have proven that $m$ must be divisible $n$. But $m\le\deg f(x)= n$. Therefore $m=n$, and $f(x)$ must be that minimal polynomial. Hence it is irreducible.

The $\Rightarrow$-direction:

Assume that there is an odd prime $q$ such that $q\mid n$, $q\nmid p-1$. As $GF(p)^*$ is cyclic of order $p-1$, the latter assumption means that every element $k\in GF(p)$ is a $q$th power of some element of $GF(p)$, say $k=a^q$.
Then
$$
x^n+k=(x^{n/q})^q+a^q=(x^{n/q}+a)\sum_{j=0}^{q-1}(-1)^jx^{jn/q}a^{q-1-j}
$$
so it cannot be irreducible.

If $4\mid n$ and $4\nmid p-1$, then it is also impossible for $x^n+k$ to be irreducible. If $k$ is a square in $GF(p)$ so is $k/4$. As $\gcd(4,p-1)=2$,
every square is also a fourth power, so $k/4=a^4$ for some $a\in GF(p)$. We get
a Sophie Germain style factorization from
$$
\begin{aligned}
x^n+k=x^n+4a^4&=(x^n+4a^2x^{n/2}+4a^4)-4a^2x^{n/2}\\
&=(x^{n/2}+2a^2)^2-(2ax^{n/4})^2.
\end{aligned}
$$
If $k$ is a non-square, then $-k$ is a square. If $k=-a^2$, we have the obvious
$$
x^n+k=x^n-a^2=(x^{n/2}-a)(x^{n/2}+a).
$$


问:
Can you use Gaussian Elimination for three non linear equations?
The equations are

$na^{n-1}b=12$

$\frac{n\left(n-1\right)}{2}a^{n-2}b^2=60$ and

$\frac{n\left(n-1\right)\left(n-2\right)}{6}a^{n-3}b^3=160$

I know I could brute force it easily by eliminating one variable at a time, but perhaps there is a methodic way for three systems of equations containing exponential functions as well.

答:
If you want something else than brute force, divide $(2)$ by $(1)$ and $(3)$ by $(2)$ to obtain
$$\frac{b (n-1)}{2 a}=5 \tag 4$$
$$\frac{b (n-2)}{3 a}=\frac{8}{3}\tag 5$$ Divide $(5)$ by $(4)$
$$\frac{2 (n-2)}{3 (n-1)}=\frac{8}{15} \implies n=6$$ Go back to $(4)$ and $(5)$ to get $a$ and $b$.


问:
unsure how to expand a LAX pair for KDV equation
This is my first post so im not sure how to make it all mathsy so im going to write it on here, I know that to find the lax equation you find [LM-ML]=0 but im struggling to follow the expansion,

for eg $$L=\left(\frac{d}{dx}\right)^2+u$$
$$M=a\left(\frac{d}{dx}\right)^3+bu\frac{d}{dx}+c\frac{du}{dx}$$

so LM would be $=a(d/dx)^5+$(on the book it says $b(d/dx)(u(d/dx)^2+(du/dx)(d/dx)$) i am unsure how this part comes about, ive tried to expand it and do it all but it doesn't simplify to that

im sorry if it doesnt look clear, again this is my first post.

答:
The composition of $L$ and $M$ involves the application of differential operators in a certain order. It may be helpful to think about each composition $LM$ or $ML$ as an operator acting on a differentiable function $f(x)$, and just carry out the operations systematically by using the usual rules of calculus.

For example, $LM(f)$ would be
$$
\left(\frac{d^2}{dx^2}+u\right)\left(a\frac{d^3f}{dx^3}+bu\frac{df}{dx}+c\frac{du}{dx}f\right).
$$
Assuming $a$, $b$, $c$ are constants, the first part of the result (from applying $d^2/dx^2$) is
$$
a\frac{d^5f}{dx^5}+b\frac{d^2}{dx^2}\left(u\frac{df}{dx}\right)+c \frac{d^2}{dx^2}\left(\frac{du}{dx}f\right).
$$
Note that this expands out to give
$$
a\frac{d^5f}{dx^5}+b
\left(\frac{d^2u}{dx^2}\frac{df}{dx}+2\frac{du}{dx}\frac{d^2f}{dx^2}+u\frac{d^3f}{dx^3}\right)+c\left(\frac{d^3u}{dx^3}f+2\frac{d^2u}{dx^2}\frac{df}{dx}+\frac{du}{dx}\frac{d^2f}{dx^2}\right).
$$
And, the second part is obtained by multiplication alone
$$
  au\frac{d^3f}{dx^3}+bu^2\frac{df}{dx}+cuf\frac{du}{dx}.
$$

Now if you remove the placeholder $f$, these would give the answer:
$$
LM = a\left(\frac{d^5}{dx^5} + u\frac{d^3}{dx^3} \right)
+b\left(\frac{d^2u}{dx^2}\frac{d}{dx}+2\frac{du}{dx}\frac{d^2}{dx^2}+u\frac{d^3}{dx^3} +u^2\frac{d}{dx}\right)
+c\left(\frac{d^3u}{dx^3}+2\frac{d^2u}{dx^2}\frac{d}{dx}+\frac{du}{dx}\frac{d^2}{dx^2}+ u\frac{du}{dx}\right).
$$
And, you can combine some of these terms again to get a more concise expression like the one you are quoting in your post.

But, note that you may not be quoting all the terms in $LM$ in the reference you are reading. For example, there are clearly two parts multiplied by $a$ in $LM$, given by
$$
a\left(\frac{d^5}{dx^5} + u\frac{d^3}{dx^3} \right),
$$
but this is not in your post. Actually, the second term cancels out with a corresponding term in $ML$.


问:
If $K/k$ is a finite Galois extension with a galois abelian group, and $ L/k$ any field extension
any hints would be greatly appreciated to help me prove the following:

If $K/k$ is a finite Galois extension with a galois abelian group, and $ L/k$ any field extension then $KL/L$ is a finite galois extension with an abelian group.

So far I have proved that $KL/L$ is finite, and attempted to prove that its a separable and normal extension by trying show that $L$ is the fixed field of $Aut(KL/L)$ but failed, $L\subset L^{Aut(KL/L)}$ was trivial i and tried doing the other side by contradiction but got nowhere.

i had also previously briefly considered trying to show that  $KL$ is the splitting field of a separable polynomial in $L[x]$ but disregarded that ianidea immediately since i don't think its even possible to think of an explicit polynomial.

答:
Let $\sigma$ be an embedding of $KL$ in a given algebraic closure over $L$. Then $\sigma(KL)=\sigma(K)\sigma(L)=KL$ since $K/k$ is Galois and in particular normal. This shows that $KL/L$ is normal. Since $KL/L$ is finite and $K/k$ is separable, $KL=L(a_1,...a_n)$ where each $a_i\in K$ are separable over $k$ and thus separable over $L$. Then $KL$ is separable over $L$.

Hence, $KL/L$ is finite Galois, and we denote its Galois group by $G(KL/L)$, and similarly for $K/k$. We define a restriction map $f:G(KL/L)\rightarrow G(K/K\cap L)$. $f$ is clearly a homomorphism. The identity automorphism in $G(K/K\cap L)$ fixes $K$ by definition and thus any of its extension to $KL/L$ must fix all of $KL$, i.e. it is the identity in $G(KL/L)$. On the other hand, any $\sigma\in G(K/K\cap L)$ may be extended to an automorphism on $KL$. We contend that the fixed field $K^{G(KL/L)}$ is precisely $K\cap L$.

Let $x\in K\cap L$. Then $x$ is fixed by any automorphism in $G(KL/L)$ and so it lies in $K^{G(KL/L)}$. Conversely, let $x\in K^{G(KL/L)}$. Then $\tau x=x$ for each $\tau\in G(KL/L)$ and thus $x\in L$. But $K^{G(KL/L)}$ is a subfield of $L$, so we must have $x\in L$. So $x\in K\cap L$. This prove that $K^{G(KL/L)}=K\cap L$.

Therefore $f$ is in fact an isomorphism, and we have $G(KL/L)\cong G(K/K\cap L)$. Since any embedding $\tau$ satisfies $\tau(K\cap L)=\tau(K)\cap\tau(L), K\cap L/k$ is Galois, and we have a homomorphism defined by restriction $g: G(K/k)\rightarrow G(K\cap L/k)$. One sees at once it is surjective, and using the same method as before, we see it has kernel $G(K/K\cap L)$, which is isomorphic to $G(KL/L)$. Then it is a subgroup of $G(K/k)$ which is abelian, so it is abelian also. This concludes our proof.


问:
How many partitions of number 25 exist such that the ferrers diagram matches the transposed ferrers diagram?
I need help solving this question. I though about the symmetry about the y=-x line. I had an idea of running through all possible number of points on this line (only even number would work, since the line divides the diagram into two equal parts and the number of total points is 25), but I don't know how to count the possible arrangements of points on one side. Any help would be appreciated. Thanks!</p>

答:
The number of self-conjugate partitions of $n$ is the same as the number of distinct odd partitions of $n$. (This is easy to see by splitting a self-conjugate diagram into "ells" from the corner.) $25$ is an odd number, so may only be split into $1,3,5$ distinct odd parts – and the only partitions with $1$ and $5$ such parts are $25$ and $1+3+5+7+9$ respectively. The number of partitions with $3$ distinct odd parts is nine:
$$21+3+1$$
$$19+5+1$$
$$17+7+1,17+5+3$$
$$15+9+1,15+7+3$$
$$13+11+1,13+9+3,13+7+5$$
So there are $11$ admissible partitions of $25$.