问:
Cardinality of intersection of sets
Consider the following problem: find $n(A \cap B)$ if $n(A)=10$, $n(B)=13$ and $n(A \cup B) = 15$.

I know if I want to find the union I use the Cardinal Number formula:

$$n(A\cup B) = n(A) + n(B) - n(A\cap B)$$

But how do I do it the other way: to find $n(A\cap B)$?

Would it be $n(A\cap B) = n(A) + n(B) - n(A\cup B)$ ?

答: Yes.  The reason why this works lies in that "n" consists of a function which maps sets to cardinal numbers (which are sets too in set theory, but that doesn't matter here).  So, for n(A), n(B), and so on, we can treat n(A) just like any other sort of number, and thus use variables, and write things n(A)=x, n(B)=y, n(A∪B)=z and so on.  Thus, we can prove that n(A∩B)=n(A)+n(B)−n(A∪B) follows from n(A∪B)=n(A)+n(B)−n(A∩B) using the basic properties of ordinary algebra.</p>

Title: Cardinality of intersection of sets


问:
When do I have to take care if something is a row vector or a column vector?
I understand that it is important in the field of matrix-vector multiplication when vectors can be interpreted as a matrix of one row/column vector.

But I do not really understand when I should use column vectors and when to use row vectors generally - when does it make a difference? Is it more a question on asthetics/style?

Let $v \in \mathbb R^d$, is it a column vector or a row vector? (I think, it depends on interpretation, doesn’t it?) When does one care about it?

答:
Generally speaking, if $x \in \mathbb{R}^d$, then $x$ is conventionally chosen to be a column vector, and perhaps to avoid using extra space, it is written in text as $x = (x_1 , \cdots, x_d)^T$. Since $x$ is a column vector, the order of matrix multiplication with a $d \times d$ matrix $A$ will be $Ax$.

You can consider $x$ to be row vector if you like, then you just need to be careful with matrix multiplication.

ContentLicense: CC BY-SA 4.0


问:
How to find what power of a number is of another number
In the equation, $a=b^x$, Is there an easy way to find the value of $x$ using a calculator (Other than trying multiple $x$ values)?

答:
For $a,b>0$ we have

$$a=b^x \iff \ln a = x \ln b.$$

can you proceed ?


问:
Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $?

I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$

I know my method gets to the same answer as WolframAlpha, namely:

$$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C,\quad (*)$$

but there is one step I can't justify. When I got to
$$\int \tan^2(u)\sec(u)\ du = \frac{1}{2}\left(\ \tan(u)\sec(u) - \ln\left|\sec(u)+\tan(u)\right|\ \right) + C,$$
I then have to substitute stuff back in terms of $x$. Now
$$x=\sec(u) \implies \tan^2(u) = x^2 - 1,$$
but I don't see how this implies $\tan(u) = \sqrt{x^2 - 1}$.

Comparing the graphs of $\sec(u)$ and $\tan(u)$, I don't see why not: $\ \tan(u) = -\sqrt{x^2 - 1}$, which would give:

$$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ -x \sqrt{x^2-1} - \ln\left|x - \sqrt{x^2-1}\right|\ \right) + C,\quad (**)$$

which is a different answer than $(*)$ ?

Now I noticed that $(*) = -(**)\ $ (ignoring the $C \to -C)$.

I can see from the graph of $\sqrt{x^2-1}$ that for $x>1$, the definite integral $\int^x_1\sqrt{t^2-1}\ dt = (*),$ and for $x<-1,\ \int^{-1}_x\sqrt{t^2-1}\ dt = (**)$

So is the indefinite integral sort of poorly defined, or would you say it is:

$$\int\sqrt{x^2-1}\ dx = \pm \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C?$$

答:
Let's differentiate and see by using which sign we get the integrand.
$\begin{align}\frac{d}{dx}\left[\frac{x}{2}\sqrt{x^2-1}-\frac12\ln|x+\sqrt{x^2-1}| + c\right]& = \frac{1}{2}\sqrt{x^2-1}+\frac x2\frac{x}{\sqrt{x^2-1}}-\frac{1+\frac{x}{\sqrt{x^2-1}}}{2(x+\sqrt{x^2-1})}\\
\\& = \frac{2x^2-1}{2\sqrt{x^2-1}}-\frac{1}{2\sqrt{x^2-1}}\\
\\& = \sqrt{x^2-1}\end{align}$

But using the negative sign yields $-\sqrt{x^2-1}$. So the one we've been using is correct.


问:
What is "induction on complexity of formula"
The fundamental theorem of ultraproducts

Given a language $L$ and a family of models $M_i$ of $L$ and ultrafilter $\mathcal U$ on $I$ and $\varphi$ a formula of $L$ then
$$ \prod_{i \in I}A_i/\mathcal U \models \varphi(s_0/_{\mathcal U}, \dots, s_n/_{\mathcal U}) \iff \{i \in I : \varphi (s_0(i), \dots, s_n (i))\}\in \mathcal U$$

Is proved by induction on the complexity of $\varphi$. I expect the proof to do atomic formula case first. Atomic formula is formula of form $\varphi = (t_1 = t_2)$ or of form $\varphi = R(t_1, \dots , t_n)$ where $t_i$ are term. But the proofs for example this prove something about terms not atomic formula. Can someone explain me please what is it we need to show if not atomic formula case first? Why terms?

答: Sometimes, when doing induction on the complexity of formulae, you need to do some sort of induction on the complexity of terms first in order to prove your basis cases for the induction on the complexity of formulae. That's what Chang & Keisler, the book you referenced, do. Their theorem has parts (i), (ii), and (iii); part (i) is just a statement about terms, proven by induction on the complexity of terms, and part (ii) is the part you're concerned about, which is proven by induction on the complexity of formulae using part (i) to get the basis case rolling. So you are doing the atomic case first, but it's just that in order to do the atomic case, you need a separate argument, which happens to be an argument via induction on the complexity of terms.

ContentLicense: CC BY-SA 3.0


问:
Can we prove that $F(x)=\int_0^x \sin(t^2) e^t dt, x\geq 0$ is not bounded on $[0,+\infty)$?
Can we prove that $F(x)=\int_0^x \sin(t^2) e^t dt, x\geq 0$ is not bounded on $[0,+\infty)$? I believe yes. Is there any sequence that we can take to prove this? thanks

答:
As @Sangchul Lee said in his comment, $|F(\sqrt{n\pi})|\to \infty$ as $n \to \infty$.  This can be shown as follows: Put $a_n = \sqrt{n\pi}$. If $F(a_n)$ is bounded, then $|F(a_{n+1})-F(a_n)|$ should also be bounded.  However,

\begin{align}|F(a_{n+1})-F(a_n)| &\geq \int_{a_n}^{a_{n+1}}|\sin(t^2)|e^tdt \\
&\geq \int_{\sqrt{n\pi + \frac \pi 6 }}^{\sqrt{(n+1)\pi -\frac \pi 6  }  } |\sin(t^2)|e^tdt \\ & \geq  \frac 12 \int_{\sqrt{n\pi + \frac \pi 6 }}^{\sqrt{(n+1)\pi -\frac \pi 6}}e^tdt  \\ & \geq \frac12\exp\left(\sqrt{n\pi + \frac \pi 6} \right)\cdot \left( \sqrt{(n+1)\pi -\frac \pi 6} - \sqrt{n\pi + \frac \pi 6} \right) \\
&= \frac12 \exp\left(\sqrt{n\pi + \frac \pi 6} \right)\cdot  \frac {\frac{2}{3}\pi }{ \sqrt{(n+1)\pi -\frac \pi 6} + \sqrt{n\pi + \frac \pi 6} }
 \end{align}

Thus $|F(a_{n+1}) - F(a_n)| \to \infty $ as $n \to \infty$; contradiction.


问:
Why is the Indefinite Integral of a Step Function and continuous?
As I posted before, say I have:\begin{equation}
  S(X)=\left\{
  \begin{array}{@{}ll@{}}
    c_0, & \text{if}\ t_0 \leq x <t_1 \\
    c_1, & \text{if}\ t_1 \leq x <t_2 \\
    c_2, & \text{if}\ t_2 \leq x <t_3 \\
. & \ . \  \\
. & \ . \  \\
. & \ . \  \\
. & \ . \  \\
c_{n-1}, & \text{if}\ t_{n-1} \leq x \leq t_n \\\end{array}\right.
\end{equation} .

Why is the indefinite integral of S(x) piecewise linear and continuous?

I think I understand why it it is piecewise linear (hopefully as I was trying to do in my last post) but why is the indefinite integral of a step function necessarily continuous? It is not clear to me why the integral of $S(x)$, or an arbitrary step function for that matter cannot have jump discontinuities.  Thanks for the help.

Here is a picture stating that it must necessarily be continuous:

Perhaps it would help me if someone could help me understand how they calculated the indefinite integral of this step function:

Essentially they have in this picture:

$T(X) =1$ for $0 \leq x < 2$

$T(X) =-1$ for $2 \leq x \leq 4$.
They get the indefinite integral to be:

$x$ for $0 \leq x < 2$

$4-x$ for $2 \leq x \leq 4$.

How did they compute this?

答: The "new" piece of curve is not "added" as $x$ moves forward. For each $x$ we use the Riemann sums, this is why it is smoother then expected.


问:
How to show two identical integrals $\; 2\int_{-r}^{r}f(x)dx=r\int_{-r}^{r}\sqrt{1+(df/dx)^{2}}dx$
With
$$ \;f(x)=(r^2-x^2)^{1/2}$$

show that$$\; 2\int_{-r}^{r}f(x)dx=r\int_{-r}^{r}\sqrt{1+(df/dx)^{2}}dx$$

by using integration by parts. Then, conclude that the area of a circle of radius r is $\;\pi r^2$ if we define $\;\pi=\;$ circumference$\;\div\;$ diameter

I thought the only method I can use is via trig substitutions. But, the problem asks to solve it with integration-by-parts.

I need help to solve this problem.
Thank you.

答:
Integrate by parts as follows
\begin{align}
LHS= &\> 2\int_{-r}^{r}\sqrt{r^2-x^2} dx
=\int_{-r}^{r}\frac{\sqrt{r^2-x^2}}xd(x^2)\\
= & \>x \sqrt{r^2-x^2}\bigg|_{-r}^{r} + \int_{-r}^{r}\frac{r^2}{\sqrt{r^2-x^2}}dx\\
=&\>0+\>r\int_{-r}^{r}\sqrt{1+\left(\frac{df }{dx}\right)^{2}}dx=RHS
\end{align}


问:
Why a matrix with a low entropy will be sparser?
The entropy of a matrix $\mathbf{P}$:
$$H(\mathbf{P})=-\sum_{i j} \mathbf{P}_{i j} \log \mathbf{P}_{i j}$$

Why a matrix with low entropy will be sparser, and with high entropy will be smoother?

答:
Notice that, for the evaluation of the entropy:

$$0 \log(0) = 1 \log(1) = 0.$$

While in general

$$x \log(x) > 0,$$

for $x \in (0, 1)$.

Then:

${\bf P}_{ij} = 0$ stands for a zero entry of the matrix ${\bf P}$, then the higher is the number of ${\bf P}_{ij} = 0$, the sparser ${\bf P}$ will be;

Since ${\bf P}$ is row-stochastic, ${\bf P}_{ij} = 1$ stands for row composed by one "1" entry, and all others are "0"; again, this is related to the sparsity of the matrix.


问:
Proving an identity for circulant matrices
In my studies of linear algebra, I have encountered this exercise

Let $A$ be a circulant matrix defined as
$$ A_{jl} = \left\{\begin{array}{cc}    r_{l-j} & l\ge j, \\    r_{N+l-j} & l<j\end{array}\right\} $$
and we index our matrices starting from zero. We define \begin{equation*}  U_{jl} = e^{2\pi i j l/N}, \qquad \left(\begin{array}{c} 0\le j\le N-1 \\ 0\le l\le N-1\end{array}\right), \qquad  \Lambda = \begin{pmatrix} \lambda_0 \\ & \ddots \\ & & \lambda_{N-1} \end{pmatrix}\end{equation*} and $\lambda_l = \sum_{j=0}^{N-1} r_j e^{2\pi i jl/N}$. We are asked to show that $$AU=U\Lambda$$

I have no idea how to actually show this. I think this is a diagonalization argument where $U$ are eigenvectors and $\Lambda$ are the eigenvalues. Can someone please show me how to do this? I thank all helpers.

答:
An $N\times N$ circulant matrix $A$ whose first zeroth row is $(r_0,r_1,\ldots,r_{N-1})$ can be written as a polynomial $p(C)=r_0I+r_1C+\cdots+r_{N-1}C^{N-1}$ where $C$ is the cyclic permutation matrix
$$
C=\pmatrix{0&1\\ &\ddots&\ddots\\ &&\ddots&1\\ 1&&&0}.
$$
So, once you obtain a diagonalisation $UDU^{-1}$ of $C$, you automatically obtain a diagonalisation $A=p(C)=p(UDU^{-1})=Up(D)U^{-1}$ of every circulant matrix $A$.

$C$ is the companion matrix of the polynomial $x^N-1$. Its eigenvalues are therefore $1,\omega,\omega^2,\ldots,\omega^{N-1}$ where $\omega=e^{2\pi i/N}$ is the primitive $N$-th root of unity. It is very easy to find an eigenvector $x$ corresponding to the eigenvalue $\omega^l$. Let $x=(x_0,x_1,\ldots,x_{N-1})^\top$. The characteristic equation $Cx=\omega^l x$ implies that
$$
x_1=\omega^lx_0,\ x_2=\omega^lx_1,\,\ldots\tag{1}
$$
and so on. Therefore $x_0$ cannot be zero, or else $x$ is a zero vector. So, we can take $x_0=1$ and the recurrence relation $(1)$ gives $x_j=\omega^{jl}$ and $x=\left(1,\omega^l,\omega^{2l},\ldots,\omega^{(N-1)l}\right)^\top$. Take this as the $l$-th column of $U$, we obtain the $U$ in question and $C=UDU^{-1}$ where $D=\operatorname{diag}(1,\omega,\omega^2,\ldots,\omega^{N-1})$. Now $Up(D)U^{-1}$ is a diagonalisation of $A$ where the $l$-th eigenvalue along the diagonal of $\Lambda=p(D)$ is
$$
\lambda_l=p(\omega^l)=\sum_{j=0}^{N-1}r_j(\omega^l)^j=\sum_{j=0}^{N-1}r_je^{2\pi ijl/N},\ l\in\{0,1,\ldots,N-1\}.
$$


问:
How to solve this minimization problem?
I have a question which asks:

A cylinder shaped can holds $5000cm^3$ of water.  Find the dimensions that will minimise the cost of metal in making the can.

What I did was express the height in terms of the radius as:

$$h=\frac{5000}{(\pi)(R^2)}$$

Then I differentiate it to:

$$\frac{(\pi)(R^2)-2R}{((\pi)(R^2))^2}$$

Now, where should I go from here? should I let it equal zero or should I do something else?

答:
Hint: The amount of metal needed to make the can would consist of the bottom and the lid (both of which are circles) and the cylinder which when unrolled is a rectangle. Thus, you have to formulate your problem as follows:

Objective: Minimize Area of (bottom + lid + cylinder)

Subject to the condition that:

Volume of cylinder =  5000

Note: You have two variables $r$ and $h$. Use the constraint to eliminate one of these variables from the objective function and then use calculus to find the optimum.

Title: How to solve this minimization problem?


问:
Convergence of a recursive sequence (logistic differential equation)
The logistic differential equation $\dot{x}=x(1-x)$ with $x(0)=x_0>0$ can be discretized with the explicit Euler method, i.e. $x_{k+1}=x_k+hx_k(1-x_k)$. Now we only consider $h\in (0,2)$. How can I show that the fixed point $x^*=1$ is asymptotically stable, that is: there is $\delta>0$ such that
$$||x_0-x^*||<\delta \Rightarrow \lim_{k\to\infty} x_k=x^*.
$$
This is a question from numerical analysis but in this context it is pretty much a pure analysis question. What I've tried is the following:
$$|x_{k+1}-1|=|(x_k-1)-hx_k(x_k-1)|=|x_k-1|\cdot|1-hx_k|\le |x_k-1|(|x_k-1|+|h-1|\cdot |x_k|).
$$
And since $|h-1|<1$, I hoped to achieve some kind of linear convergence, but I don't see how that would work. Is there another way to show convergence?

答:
If $x_{k+1} = f(x_{k})$ and the derivative of $f$ at $x_{0}$ is less than 1 in absolute value then you have stability. This follows from the "contraction principle" i.e.
if $I$ is an interval containing $x_{0}$ where derivative is less than $M < 1$  ( well, we need derivative of $f$ to be continuous as well)
$$|x_{k+1} - x_{0}| = |f(x_{k}) - f(x_{0})| = |f^{'}(y)|*|x_{k} - x_{0}| < M*|x_{k}-x_{0}|$$ where the point y is between $x_{k+1}$ and $x_{0}$.
To be exact, one uses the estimation above and induction.
That is we take $x_{1}$ in the interval $I$ and we get that the distance between $x_{2}$ and $x_{0}$ is less than $M*|x_{1}-x_{0}|$ so $x_{2}$ is in the interval I as well and so on.
One gets $$|x_{k+1}-x_{0}| < M^{k}*|x_{1}-x_{0}|$$
That is $x_{k}$ converges to $x_{0}$ as long as $x_{1}$ is in the interval $I$.

In your case $f(x) = x + hx(1-x)$ so $f^{'}(x) = 1+h-2hx$ and $f^{'}(1)=1-h$


问:
Number of Collatz iterations for numbers of the form $2^n-1$
I noticed a peculiar relation while briefly thinking about the Collatz conjecture, which occurs when the iteration is specifically applied to numbers of the form $2^n-1$. I am wondering whether this pattern is known (my expectation is that it is known since long ago...), and whether a proof of this particular pattern exists.

Let me define the function $f(n)$ to be the number of Collatz iterations needed to reach $1$ if starting from $n$. So, for example, $f(2^k) = k$ and $f(27) = 111$ and $f(1)=0$.

The observation is that for $k\geq 2$:
$$f(2^{2k}-1) -1 = f(2^{2k-1}-1)$$

The simplest instance corresponds to $k=2$, when $2^{2k}-1 = 15$ (with $f(15) = 17$) and $2^{2k-1}-1=7$ (with $f(7) = 16$). Although the relation superficially looks very simple, with the "$-1$" just being swapped between two positions, the actual Collatz sequences for these two numbers are of course rather different,

$$7\mapsto 22 \mapsto 11 \mapsto 34 \mapsto 17 \mapsto 52 \mapsto 26 \mapsto 13 \mapsto 40 \mapsto \ldots$$
$$15\mapsto 46 \mapsto 23 \mapsto 70 \mapsto 35 \mapsto 106 \mapsto 53 \mapsto 160 \mapsto 80 \mapsto 40 \mapsto \ldots$$
(while agreeing in the remaining steps $\ldots \mapsto 20 \mapsto 10 \mapsto 5 \mapsto 16 \mapsto 8 \mapsto 4 \mapsto 2 \mapsto 1$.)

Similar pair of sequences for $k=3$ are already rather long, with $f(31) = 106$ and $f(63) = 107$. Quite generally, it seems that the Collatz sequences of $2^{2k}-1$ and $2^{2k-1}-1$ become identical after $4n+2$ steps (for the larger number) resp. after $4n+1$ steps (for the smaller number), which perhaps could be a clue for proving this pattern.

(I numerically checked the above relations for the first ~100 numbers $2^{n}-1$.)

Do you know if this feature has been further investigated somewhere?

答:
Starting with the odd number $3^m2^n - 1$ (where $n \geq 1, m \geq 0$), one iteration of Collatz gives $3^{m+1}2^n - 2$, and the next iteration gives $3^{m+1}2^{n-1} - 1$. This process repeats for $2n$ iterations until the number reaches $3^{m+n} - 1$. In particular, after $4k$ respectively $4k-2$ steps, your numbers reach $3^{2k} - 1$ and $3^{2k-1} - 1$ respectively.

Let us look at $3^{2k-1} - 1$. We can see that modulo 4, it is equal to 2, so that the next two steps will be an "even step" followed by an "odd step", which results in the number
$$
3 \frac{3^{2k-1} -1}{2} + 1 = \frac{3^{2k} - 3 + 2}{2},
$$
but this is precisely the result of doing one more Collatz iteration on the even number $3^{2k} - 1$. Thus after $4k + 1$ respectively $4k$ steps, the numbers agree, and the remainder of their Collatz sequence is identical.


答:
That mostly depends on the context.

If one wants to write a linear transformation on the vector space $\mathbb{R}^n$, like matrix multiplication, then one writes the element in $\mathbb{R}^n$ as a $1\times n$ or $n\times 1$ matrix and call it a row or column vector respectively.

Using column vectors, the linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is written as $Ax$ where $A$ is an $m\times n$ matrix.

Using row vectors, the linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is written as $xA$ where $A$ is an $n\times m$ matrix.

If in general, people only talk about linear transformations on the vector space $V=\mathbb{R}^n$ without thinking about matrix multiplications, they simply refer to an element of $V$ as a vector, without saying whether it is a "row vector" or "column vector".

Title: When do I have to take care if something is a row vector or a column vector?

Body:
I understand that it is important in the field of matrix-vector multiplication when vectors can be interpreted as a matrix of one row/column vector.

But I do not really understand when I should use column vectors and when to use row vectors generally - when does it make a difference? Is it more a question on asthetics/style?

Let $v \in \mathbb R^d$, is it a column vector or a row vector? (I think, it depends on interpretation, doesn’t it?) When does one care about it?


问:
An other total-variation inequality of mine

Given differentiable, continuous $f\left ( x \right )$ on the interval $\left [ 0, 1 \right ]$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x= 0.$ Prove that
$$\left | \int_{0}^{1}xf\left ( x \right ){\rm d}x \right |\leq \frac{1}{12}\max\left | {f}'\left ( x \right ) \right |$$

I think I should transform the constant $1/12$ into an integral like $k\int_{0}^{1}x^{2}{\rm d}x,$ but $k$ is very unusual, I need to your helps, even an example of $f\left ( x \right )$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x= 0$ in order to know what I must do with the constant. Thanks a real lot.

答: $$\int_0^1 xf(x)\,dx=\int_0^1\left(x-\frac12\right)f(x)\,dx=\underbrace{-\frac12 x(1-x)f(x)\Bigg|_0^1}_{=0}+\frac12\int_0^1 x(1-x)f'(x)\,dx.$$ Hence, if $\displaystyle M=\sup_{x\in(0,1)}|f'(x)|$, then $\displaystyle\left|\int_0^1 xf(x)\,dx\right|\leqslant\frac{M}{2}\int_0^1 x(1-x)\,dx=\frac{M}{12}$.


问:
Permutations in lexicographic order
A permutation is a bijection from a set to itself.

$$\pi: \{1,\ldots , n\} \mapsto \{1,\ldots , n\}$$

One way to get permutations in lexicographic order is based on the algorithm successor which finds each time the next permutation.

This procedure works as follows:

We start from the end of the list and we find the first number that is smaller from its next one, say $x$.

Then we start again from the end of the list and we find the first number that is bigger than $x$, say $y$.

We replace $x$ and $y$.

We reverse the sublist that starts from the next position of (the new position) of $y$ till the end.

The algorithm is :

I haven't really understood the idea of that algorithm. Could you explain that further to me? Do we change the order of the elements at the end of the list to get the next permutation in lexicographic order?

$$$$

Other methods to get get permutations in lexicographic order is with rank and unrank.

These algorithms are:

Is the idea of these ones that we write all the permutations in lexicograhic order and enumerate them to get the rank and the unrank is, given a number (rank), that we can give the respective permutation?

Is that correct? Or is the idea of rank or unrank related to permutations elsewise?

$$$$

EDIT :

I am trying to understand the formula that is used in permLexRank(n,L).

The formula is $$\text{rank}(\pi,n)=(\pi(1)-1)(n-1)!+\text{rank}(\pi',n-1) \ \text{ with } \ \text{rank}([1],1)=0$$ where $\pi'$ is the permutation that we get if at $\pi$ we delete the first element and subtract all elements that are greater than that by $1$.

The idea for this formula is :  At the lexicographic order of permutations of $S=\{1,\ldots  n\}$ first there are $(n-1)!$ permutations that start by $1$ then there are $(n-1)!$ permutations that start by $2$, etc.

So it holds that $$(\pi(1)-1)(n-1)!\leq \text{rank}(\pi)\leq \pi(1)(n-1)!-1$$
If $r'$ is the rank of $\pi$ in the set of $(n-1)!$ permutations that start with $\pi(1)$, then $r'$ is the rank of $[\pi(2), \ldots , \pi(n)]$ if we conside it as a permutation of $\{1, \ldots , n\}\setminus \{\pi(1)\}$.

If we subtract each element of $[\pi(2), \ldots , \pi(n)]$ by $1$ we get a permutation $\pi'$ of $\{1, \ldots , n-1\}$ that has again rank $r'$.

I haven't really understood that. Could you explain that to me?

答:
For an outline of the proof/working of the first algorithm (permLexSuccessor), you can refer to this post on cs.stackexchange:

("Correctness proof of the algoritm to generate permutations in lexicographic order")

https://cs.stackexchange.com/a/134600/123596

Title: Permutations in lexicographic order


问:
A tangent to a differentiable function $f$ cannot cross it again given $f'$ is injective
Given $f: \mathbb{R} \to \mathbb{R}$ a differentiable function on $\mathbb{R}$ such that $f'$ is injective and given a tangent at $(x_0,f(x_0))$, prove that the tangent doesn't cross $f$ at any other point.

My reasoning is that $f$ must be convex but not sure how to use it.

答:
The tangent at $(x_0,f(x_0))$ has the equation

$$y=f(x_0)+f'(x_0)(x-x_0).$$

Now suppose that there is $x_1$ such that $x_1 \ne x_0$ and

$$f(x_1)=f(x_0)+f'(x_0)(x_1-x_0).$$

Then we have

$$ \frac{f(x_1)-f(x_0)}{x_1-x_0}=f'(x_0).$$

On the other hand ,the mean value theorem gives some $t$ between $x_1$ and $x_0$ such that $x_1 \ne t \ne x_0$ and

$$ \frac{f(x_1)-f(x_0)}{x_1-x_0}=f'(t).$$

Then we have $f'(x_0)=f'(t).$ But $f'$ is injective, hence $t=x_0,$ a contradiction.


答: Suppose that $f'(a)(b-a)+f(a) = f(b)$ with $a<b$ (meaning that the tangent intersects in the two points $a$ and $b$). By the mean value theorem we know that $f(b)-f(a) = f'(\xi)(b-a)$ for some $\xi$ strictly between $a$ and $b$. What conclusion can you then draw?

Title: A tangent to a differentiable function $f$ cannot cross it again given $f'$ is injective

Body:
Given $f: \mathbb{R} \to \mathbb{R}$ a differentiable function on $\mathbb{R}$ such that $f'$ is injective and given a tangent at $(x_0,f(x_0))$, prove that the tangent doesn't cross $f$ at any other point.

My reasoning is that $f$ must be convex but not sure how to use it.


问:
Reduction of order method - Differential Equations
I need help with this question please.

Find the general solution by using reduction of order method to

$$ty'' -(t+1)y'+y = 0$$

given that $y_1(t) = e^t$ is a solution

答:
$$ty'' -(t+1)y'+y = 0$$
$$(ty''-y')-(ty'-y)=0$$
$$ \left ( \dfrac  {y'}{t} \right)'- \left ( \dfrac  yt \right)'=0$$
$$ \left ( \dfrac  {y'}{t} -\dfrac  y t\right)'=0$$
$$ \left ( \dfrac  {(ye^{-t})'}{te^{-t}} \right )'=0$$
Now substitute $y=y_1v=e^tv$ and integrate.
$$ \left ( \dfrac  {v'}{te^{-t}} \right )'=0$$


问:
Prove that three specific lines in a triangle are concurrent
In a triangle ABC, through circumcenter O we draw three lines parallel to the triangle sides. Call the intersection of lines with the triangle as D,E,F,G,H,K as on the image. Connect the neighboring of these which do not lie on the same side, as on the image. Prove that perpendicular bisectors on these new segments (KD, EG, FH) are concurrent.

I found four properties but cannot progress further. (check image for reference)

A'B'C' has circumcircle also centered at O with radius 1/2 of the (ABC) one. Thus homothety from O with factor 2 into larger circle.

B'C' C'A' A'B' are also parallel with respective sides of the triangle and at half the distance between the respective side and it's parallel that is passing through O.

If you extend KD, EG and HF they form a new triangle A1, B1, C1. Surprisingly these A1, B1, C1 are collinear with CC'O AA'O and BB'O. (but don't know how to prove it)

Call the intersection that needs to be proven as I. Then A1B'IC' are concyclic. Same for B1 and C1.

答: Comment: Triangle ABC is isosceles and lines parallel with sides pass O. As can be seen in figure , GE and DK are two chords of a circle and their perpendicular bisectors pass through I , the center of this circle. Since AB=AC  the parallel line with side BC will have it's midpoint on O. FH is parallel with this line so the perpendicular bisector of FH is also that of GK and BC, hence it is the altitude of ABC from A and it passes through I.


问:
Confusion on the definition of the standard topology on $\mathbb{R}$
So I've been learning a bit about topology recently so I am a complete novice and there are a lot of definitions that confuse me my first confusion regards the topology on $\mathbb{R}$. I think that the standard  topology consists of all open intervals $(a,b)$ and that the topology is generated by the basis that consists of all open balls.
However open balls are used only in metric spaces since we have a distance function so when using these open balls do we no longer have a topological space but a metric space?

My second confusion is that of half open intervals in the standard topology on $\mathbb{R}$ for example I know that singletons are closed when considered as a subset of the reals such as $[0]$ is closed however what is $[0,2)$ considered?

Thanks in advance.

答: A metric space is a topological space (the converse is not always true). The set $ [0,2) $ is neither open nor closed. So are the set $ (0,1) \cup [4,5] $ or the set of rationals $ \mathbb{Q} $.


问:
How do the normal equations *always* have a solution?
My professor says that "the normal equations always have a solution", even when $A$ is not full rank.  HOwever, this does not make sense to me.
 The normal equations are

$$A^\dagger=(A^TA)^{-1}A^T$$

$A^TA$ is invertible IFF $A$ is full column rank.  So, it seems to me like there is only a solution to the normal equations if $A$ is full column rank?

edit:  I wonder if he meant to say "we can always find the Moore Penrose Inverse even when $A$ is not full rank".  that makes more sense to me but I just wantto confirm.

答: I reply to this question by completing the answer by Olittle above that says that "It seems intuitive that there should always exist a (possibly non-unique) value of $x$ that minimizes $\|Ax−b\|^2$". This is true because first observe that solving the normal equation of $Ax=b$ is equivalent to the minimization of $\|Ax−b\|^2$ in the Euclidean space $\mathbb{K}^m$. Now, for $A\in \mathbb{K}^{m\times n}$, let $\mathcal{R}(A)$ denote the range of $A$ which is a closed and convex subset of $\mathbb{K}^m$ because $\mathcal{R}(A)$ is a subspace of a finite-dimensional normed space  (Recall that a finite-dimensional subspace of a normed space is a closed set,  see Every finite-dimension subspace of $\mathcal{X}$ is closed.). Then for any $b\in\mathbb{K}^m$ the problem: minimize $\|y−b\|^2$ for $y\in \mathcal{R}(A)\subseteq\mathbb{K}^m$ has a solution (and is unique) given that it is equivalent to the problem of the minimum distance of the point $b\in\mathbb{K}^m$ to the closed convex subset $\mathcal{R}(A)$ of the Euclidean space  (which is a Hilbert space) $\mathbb{K}^m$. Denote then $\bar y$ such solution. Since $\bar y\in \mathcal{R}(A)$, then $Ax=\bar y$ has solution. This is then a minimizer of $\|Ax−b\|^2$, thus it is a solution of the normal equations of $Ax=b$.


答:
The set $\mathbb{R}$ is a metric space with the metric
$$
d(x,y)=|x-y|
$$

In this metric space, every open interval $(a,b)$ is an open ball:
$$
(a,b)=B(x_0,r),\quad x_0=\frac{b+a}{2},\quad r=\frac{b-a}{2}
$$

The standard topology on $\mathbb{R}$ does not only consist of all open intervals. It is generated by all the open intervals.

In general, for a metric space $(X,d)$, the topology generated by the open balls is sometimes called the metric topology on $X$.

The set $V=[0,2)$ is neither open nor closed in $\mathbb{R}$. But if you consider the subspace topology on $V$, then $V$ is both open and closed.

Title: Confusion on the definition of the standard topology on $\mathbb{R}$

Body:
So I've been learning a bit about topology recently so I am a complete novice and there are a lot of definitions that confuse me my first confusion regards the topology on $\mathbb{R}$. I think that the standard  topology consists of all open intervals $(a,b)$ and that the topology is generated by the basis that consists of all open balls.
However open balls are used only in metric spaces since we have a distance function so when using these open balls do we no longer have a topological space but a metric space?

My second confusion is that of half open intervals in the standard topology on $\mathbb{R}$ for example I know that singletons are closed when considered as a subset of the reals such as $[0]$ is closed however what is $[0,2)$ considered?

Thanks in advance.


问:
Solving modular equation
$$13863x \equiv 12282 \pmod {32394}$$

I need to solve this equation. If I'd found the inverse of 13863 and multiply the equation by this, I'd get the solution. So:

$$13863c \equiv 1 \pmod {32394}$$

And now - how can I find this inverse? The numbers are too big to just look for the multiplier of $13863$ . . .

答: Hint: $\ {\rm mod}\,\ 3m\!:\ 3ax\equiv 3b\iff 3m\mid 3ax-3b \iff  m\mid ax-b\iff {\rm mod}\,\ m\!:\ ax\equiv b $</p>

Title: Solving modular equation


问:
How can we expand partial derivative for complex numbers?

How this equation $$\frac{\partial }{\partial x} + i\frac{\partial }{\partial y}$$ is
expanded in the image given above?

答:
It seems to me that the first line in the RHS is an application of the chain rule for multivariable functions + the Cauchy-Riemann equations.

Namely, we can write $$\frac{\partial}{\partial x} = \frac{\partial u}{\partial x}\frac{\partial}{\partial u} + \frac{\partial v}{\partial x}\frac{\partial}{\partial v},$$ and similarly for $\frac{\partial}{\partial y}$. Combine this observation with the C-R equations, and you should be good to go.


问:
If $\sum_{k=1}^{n-1}\lambda_k^2<K\lambda_n^2$ for some $K$, then $\sum_{k=1}^{n-1}\lambda_k<K'\lambda_n$ for some $K'$
Miklos Schweitzer 1971 Problem 5:

Let $\lambda_1\leq \lambda_2\leq\cdots$ be a positive sequence and let $K$ be a constant such that $$\sum\limits_{k=1}^{n-1} \lambda_k^2 < K\lambda_n^2\qquad(n\in\mathbb{N})$$ Prove that there exists a constant $K'$ such that $$\sum\limits_{k=1}^{n-1} \lambda_k < K'\lambda_n$$

$\textbf{My work}$: We will prove by induction. For $n=2$, we know $\exists K_1: \lambda_1^2< K_1\lambda_2^2\iff \lambda_1 < \sqrt{K_1}\lambda_2$ as the sequence consists of positive terms, and the second inequality holds if we let $K_1'=\sqrt{K_1}$. Now assume $\exists K'_{n-1}: \sum\limits_{k=1}^{n-1} \lambda_k<K_{n-1}'\lambda_n$. Then for the next index, we have $ \sum\limits_{k=1}^n \lambda_k<(K_{n-1}'+1)\lambda_n$. Additionally, we know that $\sum\limits_{k=1}^{n-1}\lambda_k^2<K_{n-1}\lambda_n^2$, so $\sum\limits_{k=1}^n \lambda_k^2<(K_{n-1}+1)\lambda_n^2\leq^* (K_n+K_{n-1})\lambda_{n+1}^2$ where $\leq^*$ follows because $\lambda_n<\lambda_{n+1}$, so $K_{n-1}\lambda_n^2<K_{n-1}\lambda_{n+1}^2$ and so we can add it to both sides without reversing the inequality. Hence we can reduce the above inequality to saying $\lambda_n\leq \sqrt{\frac{K_n+K_{n-1}}{K_{n-1}+1}}\lambda_{n+1}$. Thus $\sum\limits_{k=1}^n \lambda_k<(K_{n-1}'+1)\lambda_n<(K_{n-1}'+1)\sqrt{\frac{K_n+K_{n-1}}{K_{n-1}+1}}\lambda_{n+1}$ and so if we choose $K_n'=(K_{n-1}'+1)\sqrt{\frac{K_n+K_{n-1}}{K_{n-1}+1}}$ we have proven that we can find a $K'$ specific to the choice of $n$ such that $\sum\limits_{k=1}^{n-1}\lambda_k<K'\lambda_n$ if we assume we can find a $K_n: \sum\limits_{k=1}^{n-1} \lambda_k^2<K_n\lambda_n^2$.

My Question: As pointed out by Calvin Lin, this doesn't answer the original question as $K'$ varies based on the choice of $n$. I can't seem to use a similar argument for proving the original question and wondered if anyone could shed some light on a better way to approach the problem.

答:
Let us provide a solution without the assumption that the sequence is nondecreasing: we rewrite the condition as

$$ \lambda_n^2 \ge \frac{1}{K} \left( \sum_{j=1}^{n-1} \lambda_j^2\right).$$

Iterating this condition once (that is, using it for $\lambda_{n-1}^2$), we get

$$ \lambda_n^2 \ge \frac{1}{K}  \left(1+\frac{1}{K}\right) \left(\sum_{j=1}^{n-1} \lambda_j^2 \right).$$

Reiterating this $r-$times, $r < n,$ we have

$$ \lambda_n^2 \ge \frac{1}{K} \left(1 + \frac{1}{K}\right)^r \left(\sum_{j=1}^{n-r} \lambda_j^2 \right) \ge \frac{1}{K} \left(1 + \frac{1}{K}\right)^r \lambda_{n-r}^2,$$

which is equivalent to

$$ \lambda_{n-r} \le \frac{\sqrt{K}}{(1+ (1/K))^{r/2}} \lambda_n, \forall 1 \le r < n.$$

Therefore,

$$\sum_{r=1}^{n-1} \lambda_{n-r} \le \sqrt{K} \lambda_n \left(\sum_{r=1}^{n-1} \frac{1}{(1+(1/K))^{r/2}}\right).$$

As the last sum is $\le \frac{1}{(1+(1/K))^{1/2}-1},$ we conclude the problem.


问:
Is it true that $f(x)=k ,k\in \mathbb{R}$ is a decreasing and increasing and constant function in the same time over it's domain of definition?
It is known that $f(x)=k,k\in \mathbb{R}$ is a constant function  ,one of my friend argued me that he has an example of functions which it is increasing and decreasing and constant in the same time over its domain of definition such that he gave me the following example $f(4)=5 $, I said him that is just a constant function and we can't say that is satisfies monotonicity namely increasing and decreasing and constant in the same time since that is needed  to be compared with other point , Any comment?

答: Your 'friend' is using extremely sloppy notation (or you are sloppy in stating what he said). The proper counter-example to your false claim is that any real function on a singleton set of reals is constant and (strictly) increasing and (strictly) decreasing, and this is a trivial basic mathematical fact.


问:
Could any one tell me how to show the expectation is zero for this random variable?
Could anyone tell me how to show $$\lim_{y\to \infty}yE(\frac{1}{X}I_{X>y})=0$$ Given that $X\ge 0$ but $\frac{1}{X}$ may not be integrable. Thanks for the help. $I$ is an indicator function.

答:
Hint

$$\mathbb E\left[\frac{1}{X}\boldsymbol 1_{X>y} \right]\leq \frac{1}{y}.$$


问:
Questions about Hopf-Rinow theorem
I am reading about Hopf-Rinow theorem using the Jonh M. Lee "Riemannian manifolds: an introduction to curvature",page 108, Theorem 6.13. My doubts are the next ones:

In the book, there is an exercise asking to prove that $\mathbb{H}_{\mathbb{R}}^n$, the hiperbolic space, is complete. How can I do this? I know I could just try to compute the geodesics of the space, but I am trying to deduce this from Hopf-Rinow theorem.

In the statement of the main theorem, one of the hypothesis is for $M$ being connected. I am looking for an example where the completeness fails due to $M$ being not connected, and where is used this hypothesis of being connected in the proof of Hopf-Rinow theorem.

答:
For the hiperbolic space, since every homogeneous  Riemannian connected is complete, if you prove that $\mathbb{H}_{\mathbb{R}}^n$ is homogeneous, it's all done. For that, just check $O_{+}(n,1)$ acts transitively on the set of orthonormal bases on $\mathbb{H}_{\mathbb{R}}^n$.

As they told you, the connectedness question has been developed in Connectedness and Hopf-Rinow Theorem .

Title: Questions about Hopf-Rinow theorem


答: To elaborate slightly on @response's answer, you should minimize $2\pi r^2 + 2\pi r h$ (the area of the cilinder) given that $\pi r^2 h = 5000$ (the volume of the cylinder is 5000). So plugging in the constraint in the objective function, we find that you should minimize $2\pi r^2 + \frac{10000}{r}$. Can you take it from here?</p>

Title: How to solve this minimization problem?

Body:
I have a question which asks:

A cylinder shaped can holds $5000cm^3$ of water.  Find the dimensions that will minimise the cost of metal in making the can.

What I did was express the height in terms of the radius as:

$$h=\frac{5000}{(\pi)(R^2)}$$

Then I differentiate it to:

$$\frac{(\pi)(R^2)-2R}{((\pi)(R^2))^2}$$

Now, where should I go from here? should I let it equal zero or should I do something else?


答:
Your intuition in your first question is close, but we can be a bit more precise.  Remember that a topological space is an ordered pair $(X,\mathcal{T})$, consisting of a set $X$ and a topology $\mathcal{T}$.  A metric space on the other hand, is an ordered pair $(X,d)$, consisting of a set $X$ and a metric $d$.  So, in that sense, a topological space and a metric space are fundamentally different things.  That said, there is a natural way to get from a metric space to a topological space.  Since metrics let us define open balls, we can use a metric to define a topology.  In the case where the metric $d$ induces the same topology $\mathcal{T}$, we say that the topological space $(X,\mathcal{T})$ is metrizable.  This is the case with $\mathbb{R}$ and the usual metric.  The topology generated by open intervals, and the topology generated by open balls with the standard Euclidean metric, are the same, and so $\mathbb{R}$ with the open interval topology is metrizable.  In this case it doesn't really matter whether we think of having a metric space or a topological space, but depending on what we're doing it may be useful to think of it one way or the other.

You may also be interested in reading about metrizable spaces, and examples of topological spaces which are not metrizable, for example $\mathbb{R}$ with the lower limit topology.

As for your second question, it's important to remember that open and closed are not the only types of sets.  We say that a set is closed if its complement is open, but that's not the same as saying that a set which is not open is closed (which is false).  For $[0,2)$, the complement is $(-\infty,0)\cup[2,\infty)$, which is not open, and so we know that $[0,2)$ is not closed.  Because we also know that $[0,2)$ is not open, we conclude that it is neither open nor closed.


问:
index of finite subgroup
Let $H$ be a finite subgroup of a group $G$ (finite or infinite). Is the amount of left cosets (index) of $H$ in $G$ always finite ? I suspect this is true but I cannot seem to find a bijection between $H$ and the set of all cosets of $H$ in $G$.

答: No.  The subgroup $H = \{\pm 1\}$ of the multiplicative group $\mathbb{R}^*$ of non-zero reals has infinitely many cosets.  The cosets of $H$ are pairs of the form $\{-a, a\}$ for non-zero real $a$.  Certainly there can't be finitely many such pairs to exhaust all of $\mathbb{R}^*$.

Title: index of finite subgroup


问:
Derivative of a linear function
I have to find the coefficients for m and n in the following linear function
$() =  + $
the tangent goes through the points $(1,-3)$
then $(1) =  +  = −3$
and then this comes up $′() = 4 − 3$
I understand that this is a derivative, but I don't remember any formula where this $(4x)$ comes from
so my question is what is this $4x$?
The quadratic function is f(x) = 2x ² -3x - 2

答:
Since
$$\frac{\mathrm{d}}{\mathrm{d}x} x^n = n x^{n-1}$$

$\forall n \in \mathbb{R}$ then

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^2-3x-2)= 2(2x)-3=4x-3$$

Title: Derivative of a linear function


问:
Is a pre-Lie algebra homomorphism multiplicative?
A pre-Lie algebra $A$ is a vector space with a binary operation $(x, y) \mapsto xy$
satisfying
$(xy)z − x(yz) = (yx)z − y(xz)$ for all $x, y, z$.

I assume that a homomorphism of pre-Lie algebras is defined as a linear map that preserves the above relation. Am I right in assuming this definition?

Further, does this imply that the homomorphism is multiplicative? I could use help in showing this.

答:
It does not really mean anything that the map "preserves the relation"; what is preserves is the product.

Consider the corresponding property for a morphism of associative algebras: a morphism is a linear map that preserves the product. Asking it to "preserve the associativity" is not really meaningful.


问:
Formula for finding $\cos(\theta)$?
Suppose we have that $\cos(\theta) = \sqrt{3} / 2$. Now I know that I can find the value of $\theta$ easily by referring to a trigonometric table but I want to ask is there any other way to find it's value without the table ? A sort of relationship or formula etc.

答:
https://en.wikipedia.org/wiki/Special_right_triangle

This link basically has everything you are asking about. I can clarify things if you like, if you find anything unclear.

Title: Formula for finding $\cos(\theta)$?


问:
Prove that if $X_n \to X$ in probability and $X_n \to Y$ in probability then $X = Y$ almost surely
I want to prove that if $X_n \to X$ in probability and $X_n \to Y$  in probability then $X = Y$ almost surely.

By definition, for any $\epsilon > 0$ we have:

$$\lim\limits_{n \to +\infty} \mathbb{P}(|X_n - X| \geq \epsilon) = 0$$
$$\lim\limits_{n \to +\infty} \mathbb{P}(|X_n - Y| \geq \epsilon) = 0$$

I should now prove that $$\mathbb{P}(X \neq Y) = 0$$

Maybe I should show that for any $\epsilon > 0$

$$\mathbb{P}(|X - Y| > \epsilon) = 0$$

which is equivalent to what I want to prove, but isn't helpful, because $|X - Y|$ is not greater $|X_n - X| + |X_n - Y|$ anyway (not vise-versa).

Any ideas, how to proceed?

答:
For each $\epsilon>0$,
\begin{align}
\mathsf{P}(|X - Y| > \epsilon) &=\mathsf{P}(|X -X_n+X_n- Y| > \epsilon) \\
&\le \mathsf{P}(|X_n-X|+|X_n- Y| > \epsilon) \\
&\le \mathsf{P}(|X_n-X| > \epsilon/2)+\mathsf{P}(|X_n- Y| > \epsilon/2)\to 0
\end{align}
as $n\to\infty$, which implies that $\mathsf{P}(|X - Y| > \epsilon)=0$. Thus,
$$
\mathsf{P}(X\ne Y)=\mathsf{P}\!\left(\bigcup_{k\ge 1}|X-Y|>k^{-1}\right)=\lim_{k\to\infty}\mathsf{P}(|X-Y|>k^{-1})=0.
$$


问:
Is it correct for $\int {x^x}\, dx $?
Recently, I was working with power towers. I was interested in $x^x$; it is easy to know its derivative, but I wanted to find its integral. Here's what I found. Please let me know if it is right or wrong.

We have $$x^x = e^{x\ln(x)}$$

Now, $e^{x \ln x}$ is same as writing $ e^{\left(\sqrt{x\ln x}\right)^2}$.

We know that,
\begin{align}\int e^{x^2} \,dx &= \frac {\sqrt\pi}{2} \operatorname{erfi}(x) +c\\[5pt]
\implies \int e^{\sqrt {(x \ln x)}^2}\, dx &= \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}} +c
\end{align}
Comparing, we get,$$\int {x^x}\, dx = \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}}+c $$
Simplifying, we get$$\int {x^x}\, dx = \frac {\sqrt{\pi x \ln x} \cdot \operatorname{erfi}\left(\sqrt{x \ln x}\right)}{1+\ln x} + c$$

Please help me know if it is correct/incorrect. Also, if correct, please tell me how to evaluate it by taking upper & lower limits.
 Thanks !

答:
The function $f(x) = x^x$ cannot be integrated in terms of elementary functions.

There is no simple answer to your question, unless you are willing to consider a series approximation, obtained by expanding the exponential as a series:

$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$

Title: Is it correct for $\int {x^x}\, dx $?


问:
Is my proof right? (show that there are no isolated point in the real numbers)
Question: Prove that in the metric space of all real numbers there are no isolated points.

Proof: Showing that for every ball with radius $\epsilon>0$ and center $x_0$, exists $x \in \mathbb{R}$ such that $|x-x_0|< \epsilon$ is equivalent to show that there are no isolated point in the real numbers. Then, take $x =\frac{\epsilon}{2}+x_0$ which implies $|x-x_0|=|\frac{\epsilon}{2}+x_0-x_0|=\frac{\epsilon}{2} < \epsilon$. $\Box$

I would appreciate some suggestions on how to improve or correct.

Thanks in advance!

答: Actually, what you should prove is that there is some $x\in\Bbb R$ such that $\color{red}{0<}|x-x_0|<\varepsilon$; without that $0<|x-x_0|$, you could simply take $x=x_0$, which proves nothing. However, since your $x$ is actually distinct from $x_0$, there is no problem.


问:
How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$

The answer is
 $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

Please help me out; thanks!

答:
As Ross pointed out, and as was previously discussed, we know that

$$
(x+a)(x+b) = x(x+b) + a(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab.
$$

Therefore, to factor a quadratic expression $x^2 + cx + d$, all one has to do is find two numbers that multiply to $d$ and add to $c$.  Let $m$ and $n$ be those two numbers that add to $c$ and multiply $d$.  Then:

$$
x^2 + cx + d = (x+m)(x+n).
$$

Note that it doesn't matter which order we pick the two numbers since $a+b = b+a$ and $ab = ba$.

In your case, to factor $x^2 + x - 2 = x^2 + 1x - 2$, we need to find two numbers that add to $1$ and multiply to $-2$.  Just by picking and trying different numbers, we find the two numbers are $2$ and $-1$.

A good way to go about finding the numbers is to determine all the factors of the constant term, which for us is $-2$.  Since $-2 = 2 \cdot (-1) = (-2) \cdot 1$, we only have to check whether $2 + (-1)$ or $(-2) + 1$ equals one.  Since it is the first pair of numbers with this property, these are the two numbers we were looking for.  Therefore,

$$
x^2 + x - 2 = (x+2)(x+(-1)) = (x+2)(x-1).
$$

The reason this works:  Suppose that we want to factor $x^2 + cx + d$, and we have found the numbers $m$ and $n$ such that $m + n = c$ and $mn = d$.  Then:

$$
\begin{align}
x^2 + cx + d &= x^2 + (m+n)x + mn
\\
&= x^2 + mx + nx + mn
\\
&= x(x+m) + n(x+m)
\\
&= (x+n)(x+m)
\end{align}
$$

Title: How to factor quadratic $ax^2+bx+c$?


问:
Fastest Algorithm for NLP with linear constraints
I have an minimization problem of the following form:

(Im not a mathematician, i come from the programming side, so excuse me if i have not the perfect standard of writing the formulas)

$Z(x) = \sqrt[2]{x_1^2 \cdot a_1+...+x_n^2 \cdot a_n} + x_1 \cdot b_1+...+x_n \cdot b_n$

with the constraints:

(1) $\sum_{i=1}^n{x_i} =1$

(2) $0<x_i<=c_i \ \text{with} \ c = \{c_1,..,c_n\}$

Im interested in an algorithm that finds a local minimum as fast as possible, since i want to implement it.

I have already read a lot of stuff with different approaches for different type of constraints, but they mostly didnt fit or where hard to understand, so i could not be sure if they fit.

I understand that this is a NLP with linear constraints, but since im pretty new to NLP i hope for someone with more experience who can help me.
Thank you!

答:
If the $a_i$ values are negative, all bets are off; it is a non-convex problem. But if the $a_i$ values are nonnegative, this is a convex problem, and therefore it is tractable. In fact, it can easily be converted to a second-order cone program, or SOCP, which is a relatively well known modern standard form for convex optimization.

To convert to a "standard" SOCP, define $\bar{a}_i=\sqrt{a}_i$, and create a new variable $y\in\mathbb{R}$. Then this problem is equivalent to
$$\begin{array}{ll}
\text{minimize}   & \sum_i b_i x + y \\
\text{subject to} & \sqrt{\textstyle \sum_i (\bar{a}_i x_i)^2 } \leq y \\
& \sum_i x_i = 1 \\
& 0 \leq x_i \leq c_i, ~ i=1,2,\dots,n
\end{array}$$
The key here is that the objective is linear, and the constraints are composed of linear equations and inequalities, and convex constraints involving norms. No other types of nonlinearities are permitted in an SOCP other than convex $\ell_2$-norm constraints.

Here is a good online resource on convex optimization, with a section on second-order cone programming. There are a variety of solvers out there that can solve SOCPs. If you don't mind me offering my own software, the MATLAB-based modeling framework CVX can handle this quite readily; you don't even need to do the SOCP conversion. Here would be the model in CVX, that assumes the values of $a_i$, $b_i$, and $c_i$ are stored in MATLAB vectors a, b, and c:

```<code>cvx_begin
    variable x(n)
    minimize( norm(sqrt(a).*x) + b'*x )
    subject to
        sum(x) == 1;
        0 <= x <= c;
cvx_end
</code>```

CVX will certainly not give you the fastest results; it is meant to make problems easy to specify. But you can worry about speed after you're satisfied with the results.

Body:
If the $a_i$ values are negative, all bets are off; it is a non-convex problem. But if the $a_i$ values are nonnegative, this is a convex problem, and therefore it is tractable. In fact, it can easily be converted to a second-order cone program, or SOCP, which is a relatively well known modern standard form for convex optimization.

To convert to a "standard" SOCP, define $\bar{a}_i=\sqrt{a}_i$, and create a new variable $y\in\mathbb{R}$. Then this problem is equivalent to
$$\begin{array}{ll}
\text{minimize}   & \sum_i b_i x + y \\
\text{subject to} & \sqrt{\textstyle \sum_i (\bar{a}_i x_i)^2 } \leq y \\
& \sum_i x_i = 1 \\
& 0 \leq x_i \leq c_i, ~ i=1,2,\dots,n
\end{array}$$
The key here is that the objective is linear, and the constraints are composed of linear equations and inequalities, and convex constraints involving norms. No other types of nonlinearities are permitted in an SOCP other than convex $\ell_2$-norm constraints.

Here is a good online resource on convex optimization, with a section on second-order cone programming. There are a variety of solvers out there that can solve SOCPs. If you don't mind me offering my own software, the MATLAB-based modeling framework CVX can handle this quite readily; you don't even need to do the SOCP conversion. Here would be the model in CVX, that assumes the values of $a_i$, $b_i$, and $c_i$ are stored in MATLAB vectors a, b, and c:

cvx_begin
    variable x(n)
    minimize( norm(sqrt(a).*x) + b'*x )
    subject to
        sum(x) == 1;
        0 <= x <= c;
cvx_end

CVX will certainly not give you the fastest results; it is meant to make problems easy to specify. But you can worry about speed after you're satisfied with the results.


问:
Is there a numeral system for real numbers that is always unique, but still has the usual convenient properties?
For each integer $b\ge 2,$ we know that representations of real numbers are usually unique in the base-$b$ positional notation. The only time that uniqueness fails is if the form ends in a tail of $0$'s or a tail of $(b-1)$'s, in which case it is easy to convert between these dual representations. However, the fact that multiple representations are ever possible forces the mathematician to be additionally careful in writing some proofs. For example, in the standard application of Cantor's diagonal argument to show that the continuum is uncountable, one has to be careful to mention that we  are constructing the rows using only terminating forms when there are dual representations and that the (anti-)diagonal element constructed is not somehow a dual form of one of the those terminating forms.

Question: Can a numeral system be constructed which represents all real numbers uniquely and only real numbers while still admitting some or all of the following convenient properties of the ordinary positional notation, and perhaps additional nice properties of its own:

Being exponentially more efficient than unary, meaning the number of distinct integers represented by at most a certain number of digits is something like the number of distinct symbols in the system to the power of the number of digits.

Admitting convenient pen-and-paper and computer algorithms for performing the arithmetic operations of addition, subtraction, multiplication, division and exponentiation, at least when integers or rationals are involved.

Allowing for the existence of some convenient divisibility rules of integers, though not necessarily the same ones as those admitted by base-$b.$

Having predictable (eg. periodic/cyclic) patterns in the representations of some large classes of real numbers, like the rationals.

If these properties are not possible to fulfill, I would still be interested in a system where there is uniqueness at the cost of losing these features. References to non-standard numeral systems that aim for such a goal (or perhaps other goals of convenience) would be appreciated.

答:
The first thing that came to mind is something like "base $n$ but don't allow tails of the "digit" $(n-1)$ in expansions". But the OP said "the idea of unnecessarily restricting the way that the symbols are used is unattractive to me." If we allow a few more things to be represented together with the real numbers, then there may be another option.

Gonshor's sign expansion

(Part of this section is a modification of my answer to Construction of an infinite number type and other ideas, since I don't have another tidy description of the sign expansion idea to quote.)

The "surreal numbers", or just the "surreals", are a proper class-sized object containing an isomorphic copy of any ordered field. For our purposes, you can just think of them as something that contains all of the reals and also some extra numbers like "infinitesimals" that are positive but less than $\frac1n$ for any positive integer $n$.

There is a way of writing surreal numbers called "Gonshor's sign expansion". Basically, every surreal number is a "string" of $+$s and $-$s (actually a map from an ordinal to $\{+,-\}$). For the finite strings, this matches somewhat closely to tally marks and binary "decimals". $``"=0$, $``+"=1$, $``++"=2$, $``+++"=3$, $``-"=-1$, $``--"=-2$, etc. $``+-"=\frac12=.1_2$,
$``\underline{++}+-\underline{++-+---+-}"=\underline{10}\,\,. \underline{110100010}\,1_2$, etc.

These expansions are nice because the ordering of surreals is obvious ($``\cdots+">``\cdots">``\cdots-"$) and the negative of a surreal is obtained by just switching all of the signs.

Rationals and Reals

Similar to binary, you can represent all reals with a finite string or a string indexed by the natural numbers (thought of as the ordinal $\omega$). For example, since $\frac53=1.101010\ldots_2$, we have $``++-+-+-+-\cdots"=\frac53$. And $\pi=``+++---+--+----+++\cdots"$, etc. And we can identify the rational numbers as those with a finite expansion or an expansion indexed by the naturals that is eventually periodic and not eventually constant.

Extra Numbers

Some expansions indexed by the naturals do not correspond to real numbers, though. If there is no sign change, then we get something beyond the bounds of the reals: we have $``---\cdots"<r$ for any real $r$. If there is a sign change, but the signs are eventually constant, then we get something closer to a rational number than any other real number could be. For example, $``++----\cdots"$ is greater than $1$ (since it starts with $``++"$), but less than $``++"=2$ or $``++-"=3/2$ or $``++--"=5/4$ or $9/8$, etc. Similarly, $``+-+++\cdots"$ is less than $1$ (since it starts with $``+-"$), but greater than $1/2$ or $3/4$ or $7/8$, etc.

Modified sign expansions

The sign expansions that are finite or indexed by naturals have some of the nice properties requested in the OP, but they are not efficient for representing the integers, and hence don't have any nice divisibility rules for integers either.

We can fix this by combining the Gonshor expansions that represent reals with binary expansions. Let's add a dot to mean something like "convert the finitely many symbols to the left to a binary integer" where starting with $+$ or $-$ tells you the sign of the number.

For example, $``+-+.+--"=4+1+``+--"=101.01_2=5+\frac14=``++++++--"$. And $``-+--."=-1011_2=-11$, etc. Then we would keep the efficiency and divisibility rules of binary, without losing anything.

(It's clear what things mean when we only use this new dot when there are only finitely many signs before the dot. I haven't given it much thought, but it may be reasonable to use it for other surreals by exploiting the base 2 variant of Cantor normal form.)


问:
Let $n$ be prime, for which $k \in \mathbb{N}$ is the map $\phi_k$ a group isomorphism.
Problem:

Let $k,n \in \mathbb{N}$ We define the following group homomorphism.

$$\phi_k: (\mathbb{Z}/n\mathbb{Z}) \to (\mathbb{Z}/n\mathbb{Z}), [a] \mapsto [ka]$$

Let $n$ be prime. For which $k \in \mathbb{N}$ is $\phi_k$ a group isomorphism.

Questions:

Right now from what i know is that $\phi_k$ is a homomorphism. So what i need to prove for the group isomorphism, is that $\phi_k$ is either injective or surjective. Though right now i am coming to a little bit of a blockade involving the use of the fact that n is prime. I would require either a comprehensive solution to this relatively probably simple problem. Or some way of helping me get onto the right track of using the properties of n and $\phi_k$.
I thank you for your help in advance.

答:
For a finite set $X$ a function $X\to X$ is injective if and only if it is surjective. So it is enough to check when $\phi_k$ is injective. And for homomorphisms it is enough to check when the kernel is trivial.

So assume that $\phi_k([x])=[0]$ which is if and only if $[kx]=[0]$ which is if and only if $n$ divides $kx$. Now if $gcd(k,n)=1$ then the only possibility is that $n$ divides $x$, i.e. $[x]=[0]$. So we get that when $gcd(k,n)=1$ then $\phi_k$ is an isomorphism.

On the other hand, assume that $g=gcd(k,n)\neq 1$. Put $k=k'g$. Then for $x=n/g$ we get
$$\phi_k([x])=[kx]=[k'g\frac{n}{g}]=[k'n]=[0]$$
Note that when $g\neq 1$ then $1\leq x <n$ and thus $[x]\neq[0]$. Therefore $\phi_k$ is not an isomorphism.

All in all we get: $\phi_k$ is an isomorphism if and only if $gcd(k,n)=1$. Which for prime $n$ means: if and only if $n$ does not divide $k$.


问:
Number of integers $a$ that satisfy $a^n \equiv a \pmod{b}$, where $b = pq$?
Prove that the number of $a$ which satisfy:

$$
a^n \equiv a \pmod{b},
$$where $b = pq$, $\quad$$p,q$ primes.

is $[\text{gcd}(n-1,p-1)+1] \cdot [\text{gcd}(n-1,q-1)+1]$

Attempt:
Since $p,q$ are coprime, then it is equivalent to find the number of $a$ which satisfy:
\begin{align}
a^n &\equiv a \pmod{p} \iff  a^{n-1} \equiv 1 \pmod{p} \tag{1}\\
a^n &\equiv a \pmod{q} \iff  a^{n-1} \equiv 1 \pmod{q}\tag{2}
\end{align}

From here, I can't figure out why the number of $a$ which satisfy $(1)$ for example, is $\text{gcd}(n-1,p-1)$ (the $+1$ part in the result must be that you have to count $0$). Any help would be appreciated.

答: Hint: Suppose that $p \nmid a$. Since $a^{p-1} \equiv 1 \pmod p$, if $a^{n-1} \equiv 1 \pmod p$, then $a^{(n-1)r + (p-1)s} \equiv 1 \pmod p$ for every $r,s \in \mathbb Z$. Take $r,s$ as in Bézout's Thm, so that $a^{\gcd(n-1,p-1)} \equiv 1 \pmod p$. Since $\mathbb Z_p$ is a field, there are exactly $\gcd(n-1,p-1)$ solutions for $a^{n-1} \equiv 1 \pmod p$.


答:
Your mistake is in this step

\begin{align}\int e^{x^2} \,dx &= \frac {\sqrt\pi}{2} \operatorname{erfi}(x) +c\\[5pt]
\implies \int e^{\sqrt {(x \ln x)}^2}\, dx &= \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}} +c
\end{align}

You can't just substitute $x$ by $\sqrt{x \ln x}$ in the integral.

Title: Is it correct for $\int {x^x}\, dx $?

Body:
Recently, I was working with power towers. I was interested in $x^x$; it is easy to know its derivative, but I wanted to find its integral. Here's what I found. Please let me know if it is right or wrong.

We have $$x^x = e^{x\ln(x)}$$

Now, $e^{x \ln x}$ is same as writing $ e^{\left(\sqrt{x\ln x}\right)^2}$.

We know that,
\begin{align}\int e^{x^2} \,dx &= \frac {\sqrt\pi}{2} \operatorname{erfi}(x) +c\\[5pt]
\implies \int e^{\sqrt {(x \ln x)}^2}\, dx &= \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}} +c
\end{align}
Comparing, we get,$$\int {x^x}\, dx = \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}}+c $$
Simplifying, we get$$\int {x^x}\, dx = \frac {\sqrt{\pi x \ln x} \cdot \operatorname{erfi}\left(\sqrt{x \ln x}\right)}{1+\ln x} + c$$

Please help me know if it is correct/incorrect. Also, if correct, please tell me how to evaluate it by taking upper & lower limits.
 Thanks !


问:
Are inverse images of separated presheaves separated?
$\newcommand{\Sh}{\operatorname{Sh}}$
Let $f: X \to Y$ be a continuous map, so that there is an induced geometric morphism $f^* \dashv f_* : \Sh X \to \Sh Y$, where
$$
(f^*G)(U) = (f^\dagger G)^a(U) = (\operatorname{colim}_{V \supseteq f(U)} G(V))^a
$$
where $\cdot^a$ denotes sheafification and $f^\dagger$ denotes the inverse image presheaf.

Question. Is $f^\dagger G$ separated when $G$ is separated?

My guess is yes. Here's why.

First of all, we gotta get an hold on $f^\dagger G$. My intuition is that sections $s \in f^\dagger G(U)$ are the same things as sections of $G$ defined on an open neighbourhood of $f(U)$, up to the equivalence defined by
$$
(s, V_1) \sim (t, V_2) \quad\text{iff}\quad \exists V_1 \cap V_2 \supseteq W \supseteq f(U),\ s\vert_W = t\vert_W.
$$
This is the same intuition behind the definition of stalk, which is indeed an inverse image.

Now, let $U$ be an open of $X$, and $\{U_i\}_{i \in I}$ an open cover of $U$. We want to check whether, given $s, t \in f^\dagger G(U)$, then $s\vert_{U_i} = t\vert_{U_i}$ for all $i \in I$ implies $s = t$.

Unpacking this, it means we are given $s \in G(V_1)$ and $t \in G(V_2)$, where $V_1,V_2 \supseteq f(U)$, and we know that $s\vert_{U_i} = t\vert_{U_i}$ for every $i \in I$. Notice that restriction here is 'trivial', since $V_1,V_2 \supseteq f(U_i)$ so $s$ and $t$ are valid elements of $f^\dagger G(U_i)$ (up to equivalence, of course). What changes is the equivalence relation which is now larger, so when we say $s\vert_{U_i} = t\vert_{U_i}$ we now mean
$$
\exists V_1 \cap V_2 \supseteq W_i \supseteq f(U_i),\ s\vert_{W_i} = t\vert_{W_i}.
$$
Therefore we know $s$ and $t$ coincide on a neighbourhood of every $f(U_i)$. Let $T = \bigcup_{i \in I}W_i$. This is an open neighbourhood of $f(U)$ and it's contained in $V_1 \cap V_2$. We know $G$ is separated: thus from $s\vert_{W_i} = t\vert_{W_i}$ for every $i \in I$ we can conclude $s\vert_T = t\vert_T$, which is enough to conclude $s \sim t$ in $f^\dagger G(U)$, proving the claim.

答:
Here's a higher-tech proof:

By definition, the geometric morphism induced by $f$ makes the following square commute:
$\require{AMScd}$
\begin{CD}
[\mathcal{O}(X)^{\mathrm{op}},\mathbf{Set}] @>{f'}>>
[\mathcal{O}(Y)^{\mathrm{op}},\mathbf{Set}];\\
@A{i}AA @AA{j}A \\
\mathrm{Sh}(X) @>{f}>> \mathrm{Sh}(Y),
\end{CD}
where $f'$ is the geometric morphism whose direct image is composition with (the opposite of) the functor $f^{\#}:\mathcal{O}(Y) \to \mathcal{O}(X)$ and the vertical maps are the inclusions of subtoposes.

Now, a presheaf $G$ on $\mathcal{O}(Y)$ is separated if and only if its unit $\eta_G: G \to j_*j^*(G)$ is monic. So we want to know whether $\delta_{f'^*(G)}: f'^*(G) \to i_*i^*f'^*(G)$ is monic, where $\delta$ is the unit of $i$. Considering the naturality of $\delta$ along $f'^*\eta_G$, we have a commutative square:
\begin{CD}
f'^*(G) @>{\delta_{f'^*(G)}}>>
i_*i^*f'^*(G);\\
@V{f'^*\eta_G}VV @VV{i_*i^*f'^*\eta_G}V \\
f'^*j_*j^*(G) @>{\delta_{f'^*j_*j^*(G)}}>>
i_*i^*f'^*j_*j^*(G),
\end{CD}
The left hand vertical morphism is a mono since $f'^*$ preserves monos. The right hand vertical morphism is equal to $i_*f^*j^*\eta_G$ by commutativity of the diagram above, and so is an isomorphism, since $j^*\eta_G$ is the canonical isomorphism $j^*(G) \cong j^*j_*j^*(G)$. Finally, we use the fact that $f'^*$ preserves sheaves to deduce that $\delta_{f'^*j_*j^*(G)}$ is an isomorphism, whence $\delta_{f'^*(G)}$ is a monomorphism, as required.

Of course, verifying that last step directly is just as much work as your proof, but it's a better known result, since it follows from the fact that $f^{\#}$ is a morphism of sites, and this proof will generalise to any similar situation involving a morphism of sites.


问:
Eigenvalues of $A$ with $A^3 = A$
I just got a quick practice question here that I think should be simple but I can't find a definitive answer.

Let $A$ be a square matrix such that $A^3 = A$. What can you say about the eigenvalues of $A$?

It is multiple choice and all of the answers are combinations of $-1$, $1$, and $0$. I'm pretty sure $0$ and $1$ are possible but I'm not sure how to prove any of them. Thanks in advance.

答: Minimal polynomial of matrix $A$ satisfying $A^3=A$ is given by $x^3-x=0$,so the possible eigenvalues will be $0,1,-1$, depending on the given matrix now, we can select it from MCQ.</p>

Title: Eigenvalues of $A$ with $A^3 = A$


问:
Quick Volume of Revolution Question
So I was given the following prompt:

"Let $R$ be the region in the first quadrant enclosed by the graph of $f(x)=\sqrt(\cos(x))$, the graph of $g(x)=e^x$, and the vertical line $x=\frac {\pi}{2}$, as shown in the figure below. Write, but do not evaluate, an integral expression that gives the area of $R$. "

I'm a bit confused about how I'd go about finding the area for this specific question. I understand that the volume of revolution would be found by integrating the area formula with the bounds from $0$ to the top of the graph on the y-axis, but I'm a bit confused about what the area formula would look like here. Any help would be appreciated!

答:
$$R = \int_{0}^{\frac{\pi}{2}} e^{x}dx- \int_{0}^{\frac{\pi}{2}} \sqrt{\cos(x)}dx,$$

i.e. $R$ is the difference between the area under the highest curve ($e^x$) and the area under the lowest curve ($\sqrt{\cos(x)}$).

Title: Quick Volume of Revolution Question


问:
What is the fifth digit from the left of $12345678987654321\times625$

We have been calculated $$12345678987654321\times625$$

Then what is the fifth digit from the left of this number?

I can rewrite it as  $$12345678987654321\times625={\underbrace{111111111}_{9\text{times}}}^2\times25^2=(111111111\times20+111111111\times5)^2=(2222222220+555555555)^2=2\underbrace{77777777}_{8 \text{times}}5^2$$
But I can't find fifth digit from the left of this square number.

答:
Since $625=5^4$, we have $n\times625={n\times10000\over16}$, so it suffices to do long division of $123456789\ldots$ by $16$. This is perhaps easiest to do by dividing twice by $4$:

$${123456789\ldots\over4}=3086419\ldots$$

and

$${3086419\ldots\over4}=771604\ldots$$

so the fifth digit (from the left) is $0$.

The key observation is that the special nature of $625$ allows us to change an ostensibly messy multiplication problem into a relatively simple long division problem that lets us to ignore all but the first few digits of the number we're dealing with.


答:
The argument of Kevin is right. Presently this integral cannot be expressed in terms of a finite number of available standard functions (including elementary and special functions defined and standardized in the past).

But new special functions can be defined and standardized in the futur. See this paper : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function
$$\int_0^x t^{\alpha\,t} dt=\text{Sphd}(\alpha\:;\:x)$$
Thus
$$\int x^{x} dx=\text{Sphd}(1;x)+\text{constant}$$
About the common use of special functions a paper for the general public : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

Title: Is it correct for $\int {x^x}\, dx $?


答: We know that if $F(x)$ is an antiderivative of $f(x)$, then $F\big(g(x)\big)$ is an antiderivative of $f\big(g(x)\big)g'(x)$. In your question, you seem to think that $f\big(g(x)\big)$ has an antiderivative $F\big(g(x)\big)/g'(x)$. But $$\Big(F\big(g(x)\big)/g'(x)\Big)'=f\big(g(x)\big)\underbrace{{}-F\big(g(x)\big)g''(x)/g'^2(x)}_{\text{your mistake}}.$$

Title: Is it correct for $\int {x^x}\, dx $?


问:
Is $\mathbb{R/Z}$ isomorphic to $\mathbb{Q/Z}$?
I’ve seen a proof that those quotient groups are isomorphic to the circle group, but I don’t know if they’re isomorphic to each other.
By transitivity, they should be, but I cannot prove it directly.</p>

答:
They can't be isomorphic because they have different cardinalities. Indeed, note that
$$\mathbb{R} \to S^1: t \mapsto \exp(2 \pi it)$$
induces an isomorphism $\mathbb{R}/\mathbb{Z}\cong S^1$, so $\mathbb{R}/\mathbb{Z}$ is uncountable. On the other hand, $\mathbb{Q}/\mathbb{Z}$ is countable because $\mathbb{Q}$ is countable.


答:
The issue here is that $$\int e^{\left(\sqrt{x\ln x}\right)^2}\, dx = \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}} +c$$ is not true.  If we differentiate the right-hand side, we should obtain $e^{\left(\sqrt{x\ln x}\right)^2}$, but we don't.  Instead, we get $$\frac{\sqrt{\pi} \operatorname{erfi}\left(\sqrt{x\ln x}\right)}{2\sqrt{x\ln x}} - \frac{\sqrt{\pi x\ln x}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{x\left(\ln x + 1\right)^2} + x^x,
$$ see WolframAlpha.  The issue is that there is no substitution that gives the change of variables $$\int e^{\left(\sqrt{x\ln x}\right)^2}\, dx \longrightarrow \int e^{x^{2}}\,dx.$$  We would need try to use the substitution
\begin{align}
u &= \sqrt{x\ln x}\\
du &= \frac{\ln x + 1}{2\sqrt{x\ln x}}\,dx
\end{align}
which will not give us the intended transformation.

Title: Is it correct for $\int {x^x}\, dx $?


问:
Sum of even and odd naturals
I want to prove that the sum of the first even $k$ natural numbers is $k^2+k$ given that the sum of the first odd $k$ naturals is $k^2$. So, \begin{align*} \underbrace{1+3+5}_{k=3}&=k^2=3^2=9 \\ &\vdots \\ \underbrace{1+3+5+\cdots+(2k-1)}_{k \ terms}&=k^2, n\in\mathbb N \end{align*}
needs to be used to prove that
\begin{align*}
\underbrace{2+4+\cdots+(2k)}_{k \ terms}= k^2+k, n\in\mathbb N
\end{align*}
This would be easier to do for me if I could write this using summation notation, but I can't figure out a way to do this because $k$ isn't an upper limit, but the number of terms. All I can say is that if the given sum is true, then if I add $1$ to every term $k$ times, then every term on the lefthand side would be even and the righthand side would add a $k$ term. Like so, \begin{align*}1+3+\cdots+(2k-1)&=k^2 \\ 1+1+\cdots+1&=k
\\ \implies 2+4+\cdots+(2k)&=k^2+k\end{align*}

But It feels like this explanation is sort of hand-wavy.

Title: Sum of even and odd naturals

Body:
We are given that $$\sum_{i=1}^k (2i-1)=k^2$$

That is $$\sum_{i=1}^k (2i)-\sum_{i=1}^k1=k^2$$

$$\sum_{i=1}^k 2i = k^2+k$$


答: $\Bbb Q/\Bbb Z$ has no infinite cyclic subgroup, whereas $\Bbb R/\Bbb Z$ does.

Title: Is $\mathbb{R/Z}$ isomorphic to $\mathbb{Q/Z}$?

Body:
I’ve seen a proof that those quotient groups are isomorphic to the circle group, but I don’t know if they’re isomorphic to each other.
By transitivity, they should be, but I cannot prove it directly.</p>


问:
Value of $\sin(\pi/18)$ by nested radicals
I recently asked a question about how to find value of $\sin(\pi/18)$ and I understand that there is no expression for $\sin(\pi/18)$ that uses the ordinary arithmetical operations. but I know that we have exact value by nested radicals. I want to know how we get this value ?

http://mathworld.wolfram.com/TrigonometryAnglesPi18.html

答:
Deriving $\sin\frac{\pi}{18}$ will be simpler when we implement Half angle cosine formula
$$2\sin\frac{\theta}{2} = \sqrt{2-2\cos\theta}$$
$$2\cos\frac{\theta}{2} = \sqrt{2+2\cos\theta}$$

and cosine angle identity

$$2\cos(\pi-\theta) = -2\cos\theta$$

We can derive as follows
$2\sin(\frac{\pi}{18})$  = $\sqrt{2-2\cos(\frac{\pi}{9})}$ = $\sqrt{2-\sqrt{2+2\cos(\frac{2\pi}{9}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{4\pi}{9}}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\frac{8\pi}{9}}}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{9}}}})}$

Now you can observe $2\cos(\frac{\pi}{9})$ repeats in cycle infinitely

$$\therefore 2\sin\frac{\pi}{18} = \sqrt{2-\sqrt{2+\sqrt{2+...}}}$$ $[-++]$ is the repeating pattern in infinite nested radical. Refer here. Welcome to the world of infinite nested square roots of 2 as solution to cosine angles!


问:
Two Lebesgue's integrals
I need to calculate these two integrals:
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda$$
I did the second one like this:
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda=\int_0^11\cdot\chi_{(\frac{1}{2},1)}+2\cdot\chi_{(\frac{1}{3},\frac{1}{2})}+\dots d\lambda=$$
$$=\int_0^1\chi_{(\frac{1}{2},1)}d\lambda+2\cdot\int_0^1\chi_{(\frac{1}{3},\frac{1}{2})}d\lambda+3\cdot\int_0^1\chi_{(\frac{1}{4},\frac{1}{3})}d\lambda+\dots$$

First integral if equal to the area of rectangle $(\frac{1}{2},1)\times [0,1]$ which is $\frac{1}{2}$. Every next integral is the area of rectangle $(\frac{1}{n+1},\frac{1}{n})\times [0,1]$ times $n$. From that we get:
$$\frac{1}{2}+2\cdot \frac{1}{6}+3\cdot \frac{1}{12}+\dots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots=\sum_{n=1}^\infty\frac{1}{n+1}=\infty.$$

For the first one I think I can use one of the theorem of Lebesgue which allow me to rewrite it as
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)=\int_\mathbb{R}\lim_{n\rightarrow\infty} \frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
I'm not sure how will that limit behave in that case though. I would be grateful for any hint on how to complete this.

答:
You have for $\vert x \vert \ge 1$

$$\begin{aligned}
\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert &\le \frac{x^2+1}{x^4+n} \le \frac{x^2+1}{x^4+1}
\end{aligned}$$

and for $\vert x \vert \le 1$

$$\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert \le \frac{n+1}{n^2} \le 2.$$

Therefore $f_n(x) = \frac{nx^2+1}{nx^4+n^2} $ is integrable for all $n \ge 1$, and bounded by

$$g(x) = \begin{cases}
\frac{x^2+1}{x^4+1} & \vert x \vert \ge 1\\
2 & \vert x \vert \lt 1
\end{cases}$$ which is integrable. As $\{f_n\}$ converges pointwise to zero, you can apply Lebesgue dominated convergence theorem to conclude that
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x) = 0$$

Title: Two Lebesgue's integrals


答:
A group-theoretic answer why they are not isomorphic is that $\mathbb Q/\mathbb Z$ is a torsion group, i.e. each element has a finite order. (The order of $p/q+\mathbb Z$ divides $q$.)

On the other side, some elements in $\mathbb R/\mathbb Z$ are of infinite order (e.g. $\sqrt{2}+\mathbb Z$).


问:
Find $\oint_S (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS$ where $S$ is the surface bounded by $x^2+y^2=z^2$ and the plane $z=1$.

Find $$\oint_S (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS$$ where $S$ is the surface bounded by $x^2+y^2=z^2$ and the plane $z=1$.

By divergence theorem the integration is $$\begin{align}\oint_S (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS=\underset{V}{\iiint}\operatorname{div(x \hat i+y\hat j+z^2 \hat k)}\,dV=\underset{V}{\iiint}2+2z\,dV\\=\underset{R}{\iint}\,\int_0^1 (2+2z)\,dz\,dA=3\underset{R}{\iint}\,dA=3\pi\end{align}$$
where $R$ is the area inside $x^2+y^2=1$.

But If I try to do this by first finding the integral on the top surface (in the plane $z=1$) and then on the bottom surface (surface of the cone) then I'm getting different answer.

For top surface

$\hat n=\hat k$, so $$\iint_{S_1} (x \hat i+y\hat j+z^2 \hat k)\cdot\hat n \,dS=\iint_{S_1}z^2\,dA$$ since here $z=1$, we get $$=\iint_{S_1}\,dA=\pi$$

For bottom surface

Here normal is towards negative $z-$axis. Since $z^2=x^2+y^2$ so $\frac{\partial z}{\partial x}=\frac x z,\frac{\partial z}{\partial y}=\frac y z$ so we'd have $$\iint_{S_2} (x \hat i+y\hat j+z^2 \hat k)\cdot\left(\frac{\partial z}{\partial x}\hat i+\frac{\partial z}{\partial y}\hat j-\hat k\right) \,dS$$$$=\iint_{S_2}(x \hat i+y\hat j+z^2 \hat k)\cdot\left(\frac{x}{z}\hat i+\frac{y}{z}\hat j-\hat k\right)\,dA$$$$=\iint_{S_2}\frac{x^2+y^2}{z}-z^2\,dA$$$$=\iint_{S_2}\sqrt{x^2+y^2}-x^2-y^2\,dA\\=\int_0^{2\pi}\int_0^1(r-r^2)r\,dr\,d\theta=\frac {\pi}6$$

I think I did something wrong in bottom surface. Please help.Thank you.

答:
When you are applying divergence theorem, your integral should be

$\displaystyle \int_0^{2\pi} \int_0^1 \int_r^1 (2+2z) \ r \ dz \ dr \ d\theta = \frac{7 \pi}{6}$

As you are integrating over the region which is volume of the cone, the $z$ goes from $r$ to $1$ where $z = 1$ is the top surface of the cone. The radius goes from $0$ to $1$.

If you have $0 \leq z \leq 1$ and $0 \leq r \leq 1$, independent of each other, you will get flux through the closed cylinder of radius $1$ and height $1$.


问:
Can two distinct sets of N numbers between -1 and 1 have the same sum and sum of squares?
Is is possible to find two different sets of numbers $\{ a_1, a_2, \dots, a_N\}$ and  $\{ b_1, b_2, \dots, b_N\}$ with $a_i,b_i\in[-1,1]$ such that $\sum a_i = \sum b_i$ and $\sum a_i^2 = \sum b_i^2$ are both true at the same time?

EDIT: Also, no number in each set may be $0$, i.e.   $a_i,b_i \neq 0$. John Omielan has posted an example for $N=3$, before I edited the question.

For $N=1$, it is obvious that this is impossible, because $a_1 = b_1$ (sum over elements in the set) and $a_1 \neq b_1$ (the sets must be different) cannot be true at the same time. Is there a minimum number of $N$ for which it is possible to find such sets, or is it never possible?

答:
The equality condition on the sums and sums of squares is affine invariant, so you can look for solutions in integers and then move them to the required range, which can be any interval.

$$
(1,4,6,7) \text{ and } (2,3,5,8)
$$
do the job for $n=4$. Dividing each by $10$ puts them all in $(0,1)$.

For lots more along these lines see
The Prouhet-Tarry-Escott problem and generalized Thue-Morse sequences at
https://www.intlpress.com/site/pub/files/_fulltext/journals/joc/2016/0007/0001/JOC-2016-0007-0001-a005.pdf

There you will see that for $n=8$ you can make the sums of cubes match too.


问:
Solve log equation $\frac{3}{log_2(10)}-log(x-9)=log(44)$
I am to solve $\frac{3}{log_2(10)}-log(x-9)=log(44)$.

The textbook solution provided is $x=\frac{101}{11}$ but I am unable to arrive at a solution.

Here's as far as I was able to get:

$$\frac{3}{log_2(10)}-log(x-9)=log(44)$$

I want to convert the denominator $log_2$ into the same base (10) as the others:
$$log_2(10)=\frac{log(10)}{log(2)}$$

Now swapping this in I have:

$$\frac{3}{(\frac{log(10)}{log(2)})}-log(x-9)=log(44)$$

Which I think I can rewrite as:

$$\frac{3}{log(\frac{10}{2})}-log(x-9)=log(44)$$

Which can be simplified to:

$$\frac{3}{log(5)}-log(x-9)=log(44)$$

This is as far as I got. I don't know where to go next or if I'm on the right track. The 3 in the numerator is what's stumping me, I don't know how to handle it.

How can I solve this equation to arrive at $x=\frac{101}{11}$?

More granular baby steps much appreciated.

[EDIT]
From a commenter, I'm working through an open online textbook, exercise number 63 at the bottom of this page. The corresponding solution for chapter 66, exercise 63 is on this page.

Body: $\log(10)/\log2=1/\log2$.  Note $\log(10/2)=\log10-\log2$


问:
Question about angles in a triangle having a median line

If PS = SR and x = y, is it true that angle QSR must be 90 degrees?

More generally, if PS = SR, is it true that x+y must equal 90?

I have tried drawing circumcircles, expanding the triangle into a parallelogram, and drawing an altitude from S onto QR.  I can't seem to prove QSR is 90.  Which makes me suspect it doesn't have to be.  On the other hand, I can't seem to visualize a triangle (or draw one) where it is clear that PS=SR and x=y, but angle QSR is not 90.  I also feel like using the laws of sines/cosines doesn't prove helpful.

If angle QSR is 90 and PS=SR, x need not equal y.  If angle PQR is 90 and PS=PR, x need not equal y.  If angle PQR is 90 and PS=SR and x=y, then, yes, angle QSR is 90.  There's a lot of combinations of givens and unknowns we can consider.  But for the particular 2 givens, PS=SR and x=y, and the unknown, angle QSR, I seem to be at a loss.  Any thoughts?

EDIT: I have solved this.  The answer to both questions is "no."  For a fixed value of y and a fixed length of QR, x can be written as a function of the length of PR, call it b.  Then taking the example of the fixed values y=30 and QR=2/sqrt(3), as b increases from 2, x strictly decreases from 60 (to 0), and angle QSR decreases from 90 degrees.  Then at some value of b>2 (approximately b=3.1), x will be equal to 30, while QSR will be less than 90.  Thus, there exists a triangle where PS=SR and x=y, but angle QSR is less than 90 degrees.  The same example shows that x and y can both be equal to 30, and thus x+y need not equal 90.

答:
To answer the first question, lets assume that angle QSR is 90 degrees given that x = y and PS = SR. Angle SQR would then be 90 - y, and angle PQR would be 90 - y + x = 90. This means that PQR is a right triangle with the median from Q being an altitude. This would only be possible if the triangle was also isosceles, so QSR can only be 90 degrees if PQR is an isosceles right triangle with right angle Q.
It is clear that all isosceles right triangles will work, but not all triangles with the original conditions are isosceles right triangles. For example, whenever PR = sqrt2 * PQ, the conditions are satisfied but QSR is not 90. Therefore, it is not always true that angle QSR = 90.

For the second question, consider a right triangle this time with right angle at P. It is clear that angles y and PQR sum to 90. However, since angle x comes from the median from Q, it would always be true that angle x is less that PQR, or x + y < 90. Therefore, it is not always true that x + y = 90.


问:
Is it correct for $\int {x^x}\, dx $?
Recently, I was working with power towers. I was interested in $x^x$; it is easy to know its derivative, but I wanted to find its integral. Here's what I found. Please let me know if it is right or wrong.

We have $$x^x = e^{x\ln(x)}$$

Now, $e^{x \ln x}$ is same as writing $ e^{\left(\sqrt{x\ln x}\right)^2}$.

We know that,
\begin{align}\int e^{x^2} \,dx &= \frac {\sqrt\pi}{2} \operatorname{erfi}(x) +c\\[5pt]
\implies \int e^{\sqrt {(x \ln x)}^2}\, dx &= \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}} +c
\end{align}
Comparing, we get,$$\int {x^x}\, dx = \frac {\frac {\sqrt\pi}{2}\operatorname{erfi}\left(\sqrt{x\ln x}\right)}{\frac {d\left(\sqrt{x \ln x}\right)}{dx}}+c $$
Simplifying, we get$$\int {x^x}\, dx = \frac {\sqrt{\pi x \ln x} \cdot \operatorname{erfi}\left(\sqrt{x \ln x}\right)}{1+\ln x} + c$$

Please help me know if it is correct/incorrect. Also, if correct, please tell me how to evaluate it by taking upper & lower limits.
 Thanks !

答:
What you write is wrong. But anyhow, just plot the integral function
$$
F(t)=\log\left(\int_0^t x^x dx\right), \quad t\geq 0,
$$ and you can see all the properties you are interested in. After that, you can focus on proving just those properties of $F$ you see in the graph and you are interested in. Soon, you will realize that you know about $F$ the same you know about $e^t$.

Title: Is it correct for $\int {x^x}\, dx $?


Title: Solve log equation $\frac{3}{log_2(10)}-log(x-9)=log(44)$

Body:
I am to solve $\frac{3}{log_2(10)}-log(x-9)=log(44)$.

The textbook solution provided is $x=\frac{101}{11}$ but I am unable to arrive at a solution.

Here's as far as I was able to get:

$$\frac{3}{log_2(10)}-log(x-9)=log(44)$$

I want to convert the denominator $log_2$ into the same base (10) as the others:
$$log_2(10)=\frac{log(10)}{log(2)}$$

Now swapping this in I have:

$$\frac{3}{(\frac{log(10)}{log(2)})}-log(x-9)=log(44)$$

Which I think I can rewrite as:

$$\frac{3}{log(\frac{10}{2})}-log(x-9)=log(44)$$

Which can be simplified to:

$$\frac{3}{log(5)}-log(x-9)=log(44)$$

This is as far as I got. I don't know where to go next or if I'm on the right track. The 3 in the numerator is what's stumping me, I don't know how to handle it.

How can I solve this equation to arrive at $x=\frac{101}{11}$?

More granular baby steps much appreciated.

[EDIT]
From a commenter, I'm working through an open online textbook, exercise number 63 at the bottom of this page. The corresponding solution for chapter 66, exercise 63 is on this page.

Body:
There's a mistake $\frac3{\log_2 10 } \ne \frac{3}{\log_{10} 5}$

Using, $\boxed{\log_b(a) = \frac{1}{\log_a(b)}}$

$\frac3{\log_2 10 } = 3\log_{10} 2 = \log_{10}8$

From here, $ \log_{10}(x-9) = \log_{10}(\frac2{11})$ or $x = \frac{101}{11}$


答:
If $\log$ means $\log_{10}$ then your mistake is in this step
$$\frac{3}{(\frac{\log(10)}{\log(2)})} = \frac{3}{(\log(10/2)}$$

You should have $\log_{10}-\log{2}$.

By properties of $\log$

$$\log \left( \frac{a}{b}\right) = \log a - \log b$$


问:
Let $G$ a group show and $x\in G$ show that $x^{n+m}=x^nx^m$
Let $G$ a group show and $x\in G$ show that

$x^{n+m}=x^nx^m$

I take an arbitrary $n\in \mathbb{Z}$ and first I did induction on $m\geq 0$

If $m=0$ then $x^{n+0}=x^n=x^ne=x^nx^0$

If $m>0$ we assume its valid for m, $x^{n+m}=x^nx^m$

then $x^nx^{m+1}=x^n(x^mx)=(x^nx^m)x=x^{n+m}x=x^{n+m+1}$

so its valid for all $m\geq0$

now if $-m<0$
we assume $x^{n-m}=x^nx^{-m}$ then

$x^{x-m-1}=x^{n-m}x^{-1}=x^nx^{-m}x^{-1}=x^nx^{-m-1}$

is right?

答:
You could make your proof more efficient by first proving that $x^ax^{-a}=e$ and then applying induction to the total, $n$, of the two powers.

The base case $n=0$.

This is the preparatory result noted above.

The inductive step for positive $n$

Let $a$ and $b$ be  integers such that $a+b=n+1$. Then
$$x^ax^b=x^a(x^{b-1}x)=(x^ax^{b-1})x=x^{n}x=x^{n+1}.$$

The inductive step for negative $n$

Let $a$ and $b$ be  integers such that $a+b=n-1$. Then
$$x^ax^b=x^a(x^{b+1}x^{-1})=(x^ax^{b+1})x^{-1}=x^{n}x^{-1}=x^{n-1}.$$


问:
Is a pre-Lie algebra homomorphism multiplicative?
A pre-Lie algebra $A$ is a vector space with a binary operation $(x, y) \mapsto xy$
satisfying
$(xy)z − x(yz) = (yx)z − y(xz)$ for all $x, y, z$.

I assume that a homomorphism of pre-Lie algebras is defined as a linear map that preserves the above relation. Am I right in assuming this definition?

Further, does this imply that the homomorphism is multiplicative? I could use help in showing this.

答: A homomorphism of pre-Lie algebra is a linear map $f\colon A\rightarrow B$ with $f(a\cdot b)=f(a)\circ f(b)$, where $(A,\cdot)$ and $(B,\circ)$ are the pre-Lie products. This definition is for all "nonassociative" algebras ("not necessarily associative" algebras).


问:
Shapes with $7$ lines of symmetry
I am trying to find shapes with $7$ lines of symmetry. Regular $7$-gon(Heptagon) has this property. but can you give example of other shapes with $7$ lines of symmetry? (I know I can draw an small circle on every vertices of the Heptagon and still have the symmetry lines but I want to see other examples)

答:
Here you have a shape with 7 axes of symmetry. The axes of symmetry always have to be arranged in this shape. Now you can obtain any shape with 7 axes of symmetry by a simple algorithm:

Draw anything you want into one of the segments. I painted a triangle (marked by red colour).

Paste the same thing into the other segments. (marked by light blue)

You can also do the same thing using two mirrors. Bend two mirrors together in angle $\frac{360}{14}$ degrees and then put them on a paper. Draw anything on the paper and the thing you see in the mirrors will have 7 axes of symmetry.

Title: Shapes with $7$ lines of symmetry


问:
Check for sufficiency
We have the function

$$f(x) = \frac{1}{2x\sqrt{2\pi}}e^{-\frac{(\ln x - \theta)^2}{8}}$$ for the I.I.D (identically independent distributed) sample $X_1,\dots, X_n$.

I have to show that $\sum_{i=1}^n{ln X_i}$ is a sufficient statistic.

My approach was to take the product of the pdf and then to split the product up in two parts, one with the sufficient statistic and $\theta$ and the second part containing $X_1,\dots, X_n$.

But for me it's an impossible task to split the product up and find the sufficient statistic, if someone could help, that would be great!

答:
Your random variable $X$ has pdf $f(x; \theta)$:

(source: tri.org.au)

The joint density of $\overset{\rightharpoonup }{X}=\left(X_1,\ldots ,X_i,\ldots ,X_n\right)$, a random sample of size $n$ on $X$, is given by:

$$f_* \left(\overset{\rightharpoonup }{x};\theta \right)=\prod _{i=1}^n f \left(x_i;\theta \right)$$

The mathStatica function Sufficient constructs the joint density $f_* \left(\overset{\rightharpoonup }{x};\theta \right)$, and then simplifies it. The output from Sufficient can be useful when attempting to identify sufficient statistics for a parameter. For your example, we get:

(source: tri.org.au)

From this, it is easy to see that if we define $S=\sum _{i=1}^n \log  X_i$, we can construct :

$$f_* \left(\overset{\rightharpoonup }{x};\theta \right) = g(s;\theta) h(\vec x) $$

where $g(s;\theta)$  a non-negative function (not necessarily a density function), and $h(\vec x)$ a non-negative function that does not involve $\theta$. In particular, take:

$$g(s;\theta) = e^{\frac{1}{8} \left(2 \theta  s-\theta ^2 n\right)}$$

and

$$h(\vec x)=2^{-\frac{3 n}{2}} \pi ^{-\frac{n}{2}} e^{-\frac{1}{8} \sum _{i=1}^n \log ^2\left(x_i\right)} \prod _{i=1}^n \frac{1}{x_i} $$

Then, by the Factorisation Criterion, $S$ is a sufficient statistic for $\theta$.

Title: Check for sufficiency


问:
Given a curve $\alpha$, show that its osculating plane is the same at any two points.

Given a curve $\alpha:\mathbb{R}\to\mathbb{R}^{3}$ defined by $\alpha=\left(\frac{1}{\sqrt{2}}\cos \sqrt{2}s,\frac{1}{2}\sin \sqrt{2}s,\frac{1}{2}\sin \sqrt{2}s\right)$, show that its osculating plane is the same at any two points $s_{1},s_{2}$.

I was only given the definition of the osculating plane at $s$ to be

$P_{s}=\left\{\lambda_{1}n(s)+\lambda_{2}\alpha'(s)+\alpha(s):\lambda_{1},\lambda_{2}\in\mathbb{R}\right\}$,

where $n(s)$ is the normal unit vector. Is this enough to answer the above question though? Any hint on how to start?

答:
For all $t \in \mathbb R$, you have $\alpha_y(t) = \alpha_z(t)$. Therefore $\alpha$ lies in the plane having for equation $y=z$. Moreover, $\alpha^\prime(t), \alpha^{\prime \prime}(t)$ are linearly independent, meaning that the osculating plane is defined for all points of the curve.

We can conclude that the osculating plane of the curve is the plane $y=z$ at any point of the curve.


Body:
Remember that in your exercise $\log$ means $\log_{10}$. So you have $\log(10)=1$.

Now,

\begin{align}
\frac{3}{\log_2(10)}-\log(x-9)&=\log(44)\\
\log(x-9)&=\frac{3}{\log_2(10)}-\log(44)&\textrm{(put the unknown to one side)}\\
\log(x-9)&=\frac{3\log(2)}{\log(10)}-\log(44)&(\log_2(10)=\frac{\log(10)}{\log(2)})\\
\log(x-9)&=3\log(2)-\log(44)\\
\log(x-9)&=\log\frac{2^3}{44}\\
x-9&=\frac{8}{44}\\
x&=\frac{101}{11}
\end{align}

Notes.

In general,
$$
\log \frac{b}{a}\ne \frac{\log b}{\log a}
$$


问:
Calculating quadratic polynomial, from linear interpolation of its coefficients
I have a polynomial $xt^2 + yt + z$ that I want to evaluate for some value $t$. $t$ is always between 0 and 1.

But I cannot access the values $x$, $y$ and $z$ directly. Instead I can only access linear interpolated values between them:

$A(\alpha) = x\alpha + y(1 - \alpha)$

$B(\beta) = y\beta + z(1 - \beta)$

$C(\gamma) = x\gamma + z(1 - \gamma)$

The arguments $\alpha$, $\beta$, $\gamma$ need to be derived from $t$, and be between 0 and 1.

Is there any way algebraically to calculate $xt^2 + yt + z$, using only two of these values $A(\alpha)$, $B(\beta)$, $C(\gamma)$?

It is for a CUDA program, that may use the GPU's linear texture interpolation, to evaluate this polynomial using only 2 instead of 3 memory lookups.

Body:
Choose $\alpha = 1$, $\beta = 1-1/(t+1)$.  Consider

$$t^2 A(\alpha) + (t+1)B(\beta) = t^2 x + (t+1)[y(1-1/(t+1)) + z(1/(1+t))] = xt^2 + yt + z.$$


问:
Transforming of a nilpotent matrix to a specific form
I wish to build a question in the field of discrete state spaces representation (control theory). The canonical form has a very unique, but not singular, representation. I am focusing on the $A$ matrix of the representation of the form:

$$\begin{bmatrix}
-a_1   & -a_2   & \cdots & -a_{N-1}& -a_N\\
1      & 0      & \cdots & 0       & 0   \\
0      & 1      & \cdots & \cdots  & 0   \\
\vdots & \vdots & \vdots & \ddots  & \vdots   \\
0      & 0      & \cdots & 1       &0   \\
\end{bmatrix}$$

That is a subdiagonal matrix + a matrix of the same size where only the first row has specific values.

Other representations can be generated using a transformation matrix $T$ by $T^{-1}AT$. In my case, $A\in3\times3$.

I want to use a representation with a nilpotentic matrix $B$. Given, $A$ and $B$ which answer these definitions, is there a way to calculate $T$ which transition between them so that  $B=T^{-1}AT$? I do not mind the values of $a_1,\ldots,a_N$ being dictated by $B$. All I care about is the form of the resulting $A$.

For example:

$$B=\begin{bmatrix}
2 & 2 & -2\\
5 & 1 & -3\\
1 & 5 & -3\\
\end{bmatrix},
A=\begin{bmatrix}
-a_1 & -a_2 & -a_3\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{bmatrix}$$

Is it even possible to transform between the two or am I missing a limitation here? If it is possible, is there a close solution for $T$?

答:
This is an extended comment rather than a solution.

Your matrix $A$ is in what is called Rational Canonical Form. Note that the characteristic equation of $A$ is $X^n +a_1 X^{n-1} +\dots + a_n$.

If $B$ is nilpotent then the characteristic polynomial of $B$ is $X^n$.

Similar matrices have the same characteristic polynomial, so if you want $A$ to be in this shape, and want it to be similar to a nilpotent matrix $B$ then we must have that all the $a_j=0$.

Now if you start with a nilpotent $B$ then it is indeed similar to a matrix whose only non-zero elements lie on the subdiagonal, and are all either $1$ or $0$. There are algorithmic ways of finding an appropriate $T$: if you search for "finding a Jordan basis" you'll find them.

In the explicit case of the $3\times 3$ $B$ you give, you can do the following to find the columns of $T$: (i) find an eigenvector and call it $u_3$; (ii) find a solution of $Bx=u_3$ and call it $u_2$; (iii) find a solution of $Bx=u_2$ and call it $u_1$. (These steps just require us to solve linear equations.)


问:
Convexity over the specified domain
I have the following problem:

My attempts:

for i: Hessian matrix I found for 1 is given below.

$$
    \begin{bmatrix}
    \frac {x_2^2}{4(x_1x_2)^\frac 32}+4& \frac {8(x1x2)^{\frac 32}-x1x2}{4( {x1x2)^{\frac 32}}}& 0 \\
    \frac {8(x1x2)^{\frac 32}-x1x2}{4( {x1x2)^{\frac 32}}} & \frac {x_1^2}{4(x_1x_2)^\frac 32}+4 & -2 \\
    0 & -2 & 6 \\
    \end{bmatrix}
$$
for iii: I could not figure out how to approach to this as I know ln function is convex.

for v: Hessian matrix I found for second part of 5 is given below.
$$
    \begin{bmatrix}
    12x_1^2+12x_1+4x_2^2+4x_2+10 & 8x_1x_2+4x_1+4x_2+2 \\
    8x_1x_2+4x_1+4x_2+2 & 12x_2^2+12x_2+4x_1^2+4x_1+10 \\
    \end{bmatrix}
$$

for vi: Hessian matrix I found for 6 is given a

$$
    \begin{bmatrix}
    12x_1^2+\frac {13}3x_2^2 & \frac {26}3x_1x_2 \\
    \frac {26}3x_1x_2 & 12x_2^2+\frac {13}3x_1^2 \\
    \end{bmatrix}
$$

答:
For 1, you cannot separate the function. Try the Hessian.

For 3, try the perspective of the convex function $f(x) = \sum_i x_i \log x_i$

For 4: try relating the function to a norm, writing $Q=UU^T$

For 5: the square root of a nondecreasing and convex function is not necessarily convex, think $\sqrt{x}$. Can you relate this to the result of 4?

For 6: the product of convex functions is not always convex, e.g., multiply $x^2$ with $x$

Title: Convexity over the specified domain


问:
Analysis: Integration (Riemann/Step functions)

Using the definition of the integral of a continuous function, and
  that $\displaystyle\sum_{j=0}^{2^n-1} j = (2^n-1)2^{n-1}$ to show that
  $\int_0^1x \ dx = \frac{1}{2}$

I'm having trouble even starting this question as I'm not sure how to take advantage of the sum given.

Thanks!

答: Since $f$ is continuous, it is integrable. Hence any Darboux sum converges to integral of $f$ when partition's tightness tends to $0$. Now I suggect you to consider the simplest partition $\Delta_j:=[\frac{j}{2^n},\frac{j+1}{2^n}]$ with points $\xi_j=\frac{j}{2^n}$, where $j=0,\ldots, 2^{n}-1$


问:
How to show that the Zermelo hierarchy is really a hierarchy?
In Cameron's Sets, Logic and Categories (p. 48-49), he sets out prove the following fact about the Zermelo hierarchy $V$, namely, that $V_\alpha\subseteq V_{s(\alpha)}$ for all ordinals $\alpha$. His proof proceeds by induction:

Case 1: $\alpha=0$. Then $V_\alpha$ is the empty set, which is a subset of any set.

Case 2: $\alpha=s(\gamma)$ for some $\gamma$. Take $x\in V_\alpha$. Then, $$x\subseteq V_\gamma\subseteq V_{s(\gamma)}=V_\alpha$$ so $x\in\mathcal{P}V_\alpha=V_{s(\alpha)}$.

Case 3: [limit ordinal case...]

I'm quite confused by Case 2. It seems like Cameron is taking $V_\gamma\subseteq V_{s(\gamma)}$ as the (unstated) inductive assumption. But even then, if $x\in V_\alpha$, how do we know immediately that $x\subseteq V_\gamma$?

答: By definition, $V_\alpha=V_{s(\gamma)}$ is the power set of $V_\gamma$.  So, an element of $V_\alpha$ is the same thing as a subset of $V_\gamma$.


问:
Young-like inequality
I'm having trouble with the following statement:

For positive variables $\{x_1,\dots,x_n\}\subset \mathbb{R_+}$ and $k\geq 1$ an integer, there exists $C=C(k)$ a numerical constant such that

\begin{align*}
(\sum_n x_n)^k\leq C \sum_n x_n^k.
\end{align*}

For $k=2$ this might be done by the Young inequality, what about the case $k\geq 3$?

答:
I was able to solve this question using Jensen's inequation:

The above setting is equivalent to
\begin{align*}
n^k\frac{1}{n^k}(\sum_n x_n)^k=n^k(\frac{1}{n}\sum_n x_n)^k\leq n^k\frac{1}{n}\sum_n x_n^k
\end{align*}
where the inequation comes from Jensen's inequality and $n^{k-1}$ is our $C$.

Title: Young-like inequality