问:
Can an essential supreme/infimum norm be defined in terms of supreume and infimum?
From Wikipedia:

the essential supremum of $f$, ess sup $f,$ is defined by
  $$
    \mathrm{ess } \sup f=\inf \{a \in \mathbb{R}: \mu(\{x: f(x) > a\}) = 0\}\,
$$
  if the set $\{a \in \mathbb{R}: \mu(\{x: f(x) > a\}) = 0\}$ of essential upper bounds is not empty, and ess sup $f = +∞$ otherwise.

Exactly in the same way one defines the essential infimum as the largest essential lower bound, that is,
  $$
    \mathrm{ess } \inf f=\sup \{b \in \mathbb{R}: \mu(\{x: f(x) < b\}) = 0\}\,
$$
  if the set of essential lower bounds is not empty, and as $−∞$ otherwise.

I was wondering if they can be defined equivalently in terms of $\sup$ and $\inf$ respectively?

For example,
$$
    \mathrm{ess } \sup f=\inf \{\sup(f(A)): \forall \text{ measurable subset } A, \mu(A^c) = 0\}\,
$$
and
$$
    \mathrm{ess } \inf f=\sup \{\inf(f(A)): \forall \text{ measurable subset } A, \mu(A^c) = 0\}\, ?
$$

Thanks and regards!

答:
In fact, the two definitions you show are the same.

We assume that $f:\mathbb{R}\to\mathbb{R}$ is measurable. Let $A=\{\sup f(S): S\subset \mathbb{R} \mbox{ measurable}, \mu(S^c)=0\}$ and $B=\{a: \mu(\{x: f(x)>a\})=0\}$.

Claim: $\inf A =\inf B$.

Proof: Suppose $p\in A$ and $p=\sup f(S)$ for some measurable set $S$ such that $\mu(S^c)=0$. Then $\{x: f(x)>p\}\subset S^c$, so
$\mu(\{x: f(x)>p\})=0$ and $p\in B$. Thus $A\subset B$ and $\inf A\ge \inf B$.

On the other hand, suppose $a\in B$. Then $\mu(\{x: f(x)>a\})=0$ so the set $S=\{x: f(x)\le a\}$ is one of the defining subset in the definition of $A$. Clearly $\sup f(S)\le a$. This proves that $\inf A\le \inf B$.

The proof for the essential inf is similar.


问:
Parallel lines in 3D and their distance?
Show that the two lines are parallel. Determine the distance between them and the plane containing both lines.

$$L_{1}\left\{\begin{matrix}
x=3-2t\\y=4+t
\\z=1-t
\end{matrix}\right.\quad
L_{2}\left\{\begin{matrix}
x=5+2t\\y=1-t
\\z=7+t
\end{matrix}\right.$$

I'm feeling lost.

EDIT:
They are parallel because the direction vectors are the same. $(2,-1,1)$

To calculate the normal of the plane i take $\begin{pmatrix}-2&1&-1\end{pmatrix}\times \begin{pmatrix}2&-1&1\end{pmatrix}=\begin{pmatrix}0&0&0\end{pmatrix}.$ That is wrong. The normal shall be $(3,10,4)$. How do i calculate this?

Arbitrary point on $L_{1}(3,4,1)$.
Equation of the plane: $(x,y,z)=(3,4,1)+t(-2,1,-1)+s(2,-1,1)$

答:
Hint:This is analytic geometry problem.You have equation of lines in parametric form.  Write these equations in symmetric forms:

$L_1:$, $$\frac {x-3}{-2}=\frac{y-4}1=\frac{z-1}{-1}$$

$L_2:$  $$\frac {x-5}{2}=\frac{y-1}{-1}=\frac{z-7}{1}$$

Now two lines are parallel if the ratio of their corresponding gradients(the denominators of fractions) are equal. We have:

$\frac{-2}2=\frac1{-1}=\frac{-1}1=-1$

So $L_1$ and $L_2$ are parallel.

ContentLicense: CC BY-SA 4.0


问:
Interesting properties of Green's function
I am trying to improve my understanding of the Green's Function and its use to solve second-order linear ODEs.

Is it correct to assume that the Green's functions can be used to solve all second order linear ODEs?

Is there any examples when using the Green’s functions to solve second-order linear ODEs that gives an interesting or special result?

Is there any properties of Green's function that has a special case or is special?

答:
In general, Greens functions do not tend to be well-behaved at all points in space, which can lead to problems.

For example consider finding a solution to electromagnetic radiation problem, using the Green function, but ask for solution at the point where the source, i.e. some oscillating charge distribution is located. See, for example, W. C. Chew "Waves and Fields in Inhomogeneous Media". You then need to think of ways of isolating your observer from the source etc.

Another interesting area is the null-space of Green functions, i.e. non-trivial sources that do not lead to excitation. Again, in electromagnetism, there is a lot of info on this in A. J. Devany, "Mathematical foundations of Imaging, Tomography and Wavefield Inversion". There such things are called the non-radiating configurations.


问:
Range of square root of rational function
How can I determine the range of the function $$f(x)=\sqrt{\frac{-x^2+2x+3}{18x-3x^3}}$$ without using limits or derivatives? I have factorised numerator and denominator, but nothing simplifies. I tried solving $y=f(x)$ for $x$, because the domain of the inverse function is the range of the initial function, but I am somehow stuck. My guess is that the range is all non-negative real numbers. How can I be sure that we reach all of them? If we have it for the fraction, we will also have it for the whole root. Can someone help me here? Thanks!

答:
$$\frac{(x+1)(3-x)}{3x(\sqrt6-x)(\sqrt6+x)}$$ must be non-negative. Notice that for $x$ in

$$(-\sqrt6,-1]$$ this fraction takes all possible non-negative values, as the signs are

$$\frac{-+}{-++}$$ and the denominator tends to $0$ on the left, and the numerator to $0$ on the right.

Hence, $$\mathbb R^+.$$


问:
Sum with two indices, how to handle the condition $i\neq j$?
I'm given
$$
\sum_{i,j=1, i \neq j}^{N} ij \tag 1
$$
I assume this is a short hand for

$$
\sum_{i=1}^{N}
\Bigg (
\sum_{j=1}^{N} ij
\Bigg )
\qquad? \tag 2
$$
But where does $i\neq j$ belong, to the inner or outer sum?

Example with $N=2$ and $i\neq j$ in the inner sum:
\begin{align}
\sum_{i,j=1 \\i\neq j}^{2} ij
&=
\sum_{i=1}^{2}
\Bigg (
\sum_{j=1, i \neq j}^{2} ij
\Bigg )
\tag 3
\\
&=\sum_{i=1}^{2}
\Bigg (
i1+i2
\Bigg )
\tag 4
\\
&=
(11+12 + 21 + 22)
\tag 5
\\
&\text{\{Now discard 11 och 22\}}\\
&=23
\end{align}

I guess the end result is correct, but I don't think my approach is correct ("Now discard"). How should I handle the condition $i\neq j$?

Body:
A simple way is to develop the square of $\sum_{i=1}^N{i}$ and to separate the two cases, $i=j$ and $i \ne j$:

$$\left(\sum_{i=1}^N {i}\right)^2 = \sum_{i,j=1}^N{ij} = \sum_{i,j=1 ; i \ne j}^N{ij} + \sum_{i,j=1; i=j}^N{ij}$$

$$ = \sum_{i,j=1; \,i \ne j}^N{ij} + \sum_{i=1}^N{i^2}$$

As
$$\sum_{i=1}^N {i} = \frac{N(N+1)}{2} $$

And
$$\sum_{i=1}^N {i^2} = \frac{N(N+1)(2N+1)}{6} $$

We get

$$ \sum_{i,j=1 ;\,i \ne j}^N{ij} = \frac{N^2(N+1)^2}{4} - \frac{N(N+1)(2N+1)}{6} = \frac{3N^4+2N^3-3N^2-2N}{12}$$


问:
How many nonnegative integer solutions of the inequality $x_1+x_2+x_3+x_4+x_5<23$ are there with the constraints $x_1 > 2$, $x_2 \geq 3$?
$$x_1+x_2+x_3+x_4+x_5<23$$

$$x_1>2,x_2\ge 3,x_3,x_4,x_5\ge0$$

How can I solve this?

答:
Since $x_i$ are integral, we want to solve for $x_1+x_2+x_3+x_4+x_5+x_6=22$ with $x_{1,2}\ge3,x_{3,4,5}\ge0$ and $x_6=22-x_1-x_2-x_3-x_4-x_5\ge0$.

Replace $x_1=z_1+3$ and $x_2=z_2+3$ where $z_{1,2}\ge0$. Then you will have$$z_1+z_2+x_3+x_4+x_5+x_6= 22-6=16$$

Using the stars and bars approach, the required answer is$$\binom{21}5=20349.$$


问:
How to find the coefficient of $x^{l}$ in the expansion of $(1+x+x^{2}+...+x^{n})^{m}$ for some given $l$?
Suppose I'd like to find the coefficient of $x^{l}$ in the expansion of  $(1+x+x^{2}+...+x^{n})^{m}$, where $n$ and $m$ are given positive integers, for some given integer $l$ such that $n < l < mn$. Is there a straightforward formula (in terms of some multinomial coefficient, or similar)?

答:
For n=1 the answer is obvious so the obvious assumptions are n>1,m>1. For sufficiently small m or n you could likely write a reasonable formula but for general m,n I am not sure of a 'straightforward' or 'closed form' answer, depending upon exactly what one would classify as being in one of those 2 categories. Though one may express the answer perhaps in the form:

$\sum_{m_1=0}^{min(m,L)}\sum_{m_2=0}^{min(m,L)}\dots \sum_{m_{min(L,n)}=0}^{min(m,L)}C(m_1,m_2,\dots,m_{min(L,n)})\prod_{j=0}^{min(L,n)}f_{j}^{(m-m_j)}\;$

where $f_j$ is $\frac{d^j}{d\:x^j}\sum_{i=0}^{n}x^i\;$ , the superscripts $(m-m_j)$ are the powers or exponents
and the $C(m_1,m_2,\dots,m_{min(L,n)})$ are constants which are the result of differentiating $\left(\sum_{i=0}^{n}x^i\right)^m \;$ L  times where have written the capital  L to stand for your $l$. Then we can set $x=0$ and divide by $L!$ to obtain the answer for desired coefficient which would then be

$\frac1{L!}\left(\sum_{m_1=0}^{min(m,L)}\sum_{m_2=0}^{min(m,L)}\dots \sum_{m_{min(L,n)}=0}^{min(m,L)}C(m_1,m_2,\dots,m_{min(L,n)})\prod_{j=0}^{min(L,n)}({j!})^{(m-m_j)}\right)\;$

Though finding and exressing the $C(m_1,m_2,\dots,m_{min(L,n)})$ coeff's may be quite unwieldly and have not done so as yet so

In order that the length of writing the formula or answer does not increase proportional to m or n I choose to solve it by a recursive routine which means it calls itself and is as follows. Have tried to write it in a kind of universal hopefully simple enough language that you could easily transfer it to fortran(my favorite) or basic or most any other symbolic math programs etc. The final answer, that is coeff. of $x^L$, is cf and have named the recursive routine nicf(ni,nl,nm,nn)
with it's 4 arguments in parenthesis which follows where I have written capital L for your $l$.

nicf(ni,nl,nm,nn): for i=1 thru $min(\frac{nl}{nn},nm)$ do( if $nl\!-\!i\:nn\!<\!nm\!-\!i\!+\!1$ then $cf\!=\!cf\!+\!ni\binom{nm}{i}\binom{nm-i}{nl-i\:nn}$
,if $nm\!-\!i>0$ then for j=2 thru $min(nn\!-\!1,nl\!-\!i\:nn)$ do nicf($ni\binom{nm}{i},nl\!-\!i\:nn,nm\!-\!i,j$) )

All quantities are integers and all terms in the routine are local except the accumulated answer cf which is global. Note $ni\binom{nm}{i}...$ means $ni$ times $\binom{nm}{i}...\;$ and $\;i\:nn$ means $i$ times $nn$, the perhaps small figure '1' and the $1$ both mean the same which is one, and where have written
...do cf(...) it means same as call cf(...) which may be more common in other languages. The $\binom{a}{b}$ means $\frac{a!}{b!(a\!-\!b)!}$. Now assuming you understand that routine it is used as follows in a simple program.

if L<m+1 then cf=$\binom{m}{L}$ else cf=$0$, for k=max(ceiling($\frac{L}m$),2) thru n do nicf($1$,L,m,k)

And the final answer will be cf=coeff. of $x^L\;$ Have tested it in symbolic program maxima for sufficiently large enough m,n,L to be reasonably sure it is correct. To try and write a better general formula etc. it seems it may be good to study partitions and perhaps generating functions such as in Knuth 'Art of Computer Programming' vol I fundamental algorithms.


问:
Is it valid to apply L'hopital rule to evaluate the limit?
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{x^2+x}\times\left(\sqrt{x^2+x+1}+1\right)$$

Then it used equivalence and wrote:

$$\lim_{x\to 0^+}\frac{2x\left(\frac12x^2\right)}{x^2+x}\times\left(\sqrt{x^2+x+1}\right)=\lim_{x\to 0^+}\frac{x^3}{x(x+1)}.\left(\sqrt{x^2+x+1}\right)=0$$

I wonder why we should do all these calculation? Is it possible to use L'hopital rule and get $\frac0{\tfrac12}=0$ ?

答:
Yes since the expression is in the form $\frac 0 0$ we are allowed to use l'Hospital to obtain

$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=\lim_{x\to 0^+}\frac{2 \sec^2 x - 2 \cos x}{\frac{2 x + 1}{2 \sqrt{x^2 + x + 1}}}=\frac{0}{\frac12}=0$$

Note that often use l'Hospital's rule is esplicitely not allowed because when we use it blindly we can't really understand what it is going to determine the limit. For these reason is good to solve limits in more than one way, using standard limits when possible.

In this case as an alternative we can proceed as follows

$$\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=2\frac{\tan x}{x}\frac{1-\cos x}{x^2}\frac{x^3}{\sqrt{x^2+x+1}-1}\frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}$$

which allows to conclude by standard limits.

I also exhort not expert users to do not proceed blindly by asymptotic equivalence since it can leads to wrong results if we use this way without the necessary attention. For example as $x \to 0$ for $\frac{1- \cos x}{x^2}$ we could wrongly state that limit is zero since $\cos x \sim 1$ or with $\frac{x- \sin x}{x^3}$ also since $\sin x \sim x$.


问:
How to calculate the distance from a point on ellipse arc to the chord
Now I have an ellipse-arc starting at $A$ and ending at $B$, so $AB$ is the chord of the ellipse-arc. I want to know how to calculate the maximum distance from the points on the ellipise-arc to the chord.

答:
Presumably you know  something about the ellipse, perhaps $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

and presumably you know something about $A$ and $B$, perhaps that they are where $y=mx+c$ intersects the ellipse

Then you can say that the tangents to the ellipse parallel to $AB$ can be expressed as $y = mx \pm \sqrt{a^2m^2+b^2}$

and the maximum distance from the the ellipise-arc to the chord is $\dfrac{|c\mp \sqrt{a^2m^2+b^2}|}{\sqrt{1+m^2}}$

with the $\pm$ and $\mp$ determined by which of the two arcs resulting from the chord you are referring to.


答:
Short answer:

$$\tan x\sim x$$

and

$$\sqrt{x^2+x+1}-1\sim\frac x2.$$

Hence the factor $1-\cos x$ makes the limit tend to zero.

Title: Is it valid to apply L'hopital rule to evaluate the limit?

Body:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{x^2+x}\times\left(\sqrt{x^2+x+1}+1\right)$$

Then it used equivalence and wrote:

$$\lim_{x\to 0^+}\frac{2x\left(\frac12x^2\right)}{x^2+x}\times\left(\sqrt{x^2+x+1}\right)=\lim_{x\to 0^+}\frac{x^3}{x(x+1)}.\left(\sqrt{x^2+x+1}\right)=0$$

I wonder why we should do all these calculation? Is it possible to use L'hopital rule and get $\frac0{\tfrac12}=0$ ?


答:
The lines can be written in vector form as:
$\vec r=(3\vec i+ 4 \vec j+ \vec k)+t(2\vec i+ \vec j- \vec k)=\vec a+ t \vec b$ and other as $\vec r=(5 \vec i+\vec j+7\vec k)+ s(2\vec i+ \vec j- \vec k)=\vec c+ s \vec b.$ The distance between them is given by  the perpendicular projection of $\vec{AC}$ abour $\vec b$
$$d=\left|\frac{(a-\vec c)\times \vec b}{b}\right|=\sqrt{\frac{269}{6}}.$$

The direction vector of both the lines is the same $\vec b$ so they are parallel.

Title: Parallel lines in 3D and their distance?

Body:
Show that the two lines are parallel. Determine the distance between them and the plane containing both lines.

$$L_{1}\left\{\begin{matrix}
x=3-2t\\y=4+t
\\z=1-t
\end{matrix}\right.\quad
L_{2}\left\{\begin{matrix}
x=5+2t\\y=1-t
\\z=7+t
\end{matrix}\right.$$

I'm feeling lost.

EDIT:
They are parallel because the direction vectors are the same. $(2,-1,1)$

To calculate the normal of the plane i take $\begin{pmatrix}-2&1&-1\end{pmatrix}\times \begin{pmatrix}2&-1&1\end{pmatrix}=\begin{pmatrix}0&0&0\end{pmatrix}.$ That is wrong. The normal shall be $(3,10,4)$. How do i calculate this?

Arbitrary point on $L_{1}(3,4,1)$.
Equation of the plane: $(x,y,z)=(3,4,1)+t(-2,1,-1)+s(2,-1,1)$


问:
Position of a point on the boundary of a moving circle
Consider a circular wheel with radius $r$ moving to the right along the ground at a constant velocity. The ground is the x-axis, and the wheel moves in the positive direction. Assume that $Q_0$ is at the origin, so that the position vector of $Q_0$ is the zero vector. Let $Q_t$ be the position of the point after t seconds ($Q_t$ always stays on the boundary of the circle for all $t$ and the point moves in clockwise-direction), and let $P_t$ be the
position of the centre of the wheel after t seconds, for all $t ≥ 0.$ I want to find

i) $\sigma \in \text {Bij}(\mathbb R^2) $ such that $P_{t+1} = \sigma (P_{t}) \,\,\, \forall t \geq 0.$

ii) $\tau \in \text {Bij}(\mathbb R^2) $ such that $Q_{t+1} = \tau (Q_{t}) \,\,\, \forall t \geq 0.$

Here Bij($\mathbb R^2$) is the set of all transformations on $\mathbb R^2.$

My answers:

i) $\sigma$ is the map $T_{A,\mathbb b} : \mathbb R^2 \to \mathbb R^2$, $\mathbb x \mapsto A\mathbb x \,\,  + \dfrac{2\pi r}{T}t  $ where $A$ is the clockwise rotation matrix and $\mathbb b = \dfrac{2\pi r}{T}t $.

ii) I want to express $\tau$ as a product of transformations, defined geometrically. The definition of $\tau$ will certainly involve $P_t.$ But I don't know how to do it.

Any help appreciated.

答:
The wheel is moving at a constant velocity, let's call it $v$. In one unit of time it changes $x$ by $v\cdot 1$, while keeping $y$ fixed, $y=r$. So $$(x_{t+1},y_{t+1})=(x_t+v,y_t)=(x_t,y_t)+(v,0)$$
The point of the wheel rotates around the axis at a constant angular speed. To describe the point in the fixed reference frame, you need to account for the fact that the axis is moving as well, so you would need to add the motion of the center. In the reference frame at the center of the axis $$x_t=-r\sin\omega t\\y_t=r(1-\cos \omega t)$$
When you add the motion of the center, $$x_t=vt-r\sin\omega t\\y_t=r(1-\cos \omega t)$$.
Here $\omega$ is the angular speed. If the wheel rotates without slipping, $$v=\omega r$$

Can you finish from here to write $\tau$? Write $x_{t+1}$ and $y_{t+1}$ in terms of the quantities at $t$.

Note that you are describing the cycloid motion, in case you want to look it up.


问:
Sum over binomial coefficients
I want to find out if the following sum has a closed form:
$$\sum_{j=0}^N{j+k-1\choose k-1}{N-j+k-1\choose k-1}$$

I tried to rewrite the sum such that I could use the following form of the Vandermonde identity:
$$\sum_{m=0}^n{m\choose j}{n-m\choose k-j}={n+1\choose k+1}$$
but I couldn't arrive at any result.

答: Hint: You are doing good! Using what you put there you will need $n-(j+k-1)=N-j+k-1$ implies $n=N+2k-2$ and you also have that $2k-2-(k-1)=k-1$ So take the $k$ in the second equation to be $2k-2.$

Title: Sum over binomial coefficients


问:
Rest with congruences
Find the remainder of the division of $619^{1367}$ by $15$.

I cannot see what the initial congruence associated with the problem would be. Any tips?

答:
Firstly, note that $15$ divides $615$. So, $619 \equiv 615+4 \equiv 4 \pmod {15}$. This means that $619^{1367} \equiv 4^{1367} \pmod {15}$.

Next, note that $4^2 \equiv 1 \pmod {15}$. Therefore, $(4^2)^683 \equiv 4^{1366} \equiv 1 \pmod {15}$, and multiplying by $4$ gives $619^{1367} \equiv \boxed{4} \pmod {15}$.

Title: Rest with congruences


问:
Determine an orthonormal base
Determine an orthonormal base $({\overrightarrow{e_{1}}}, {\overrightarrow{e_{2}}},{\overrightarrow{e_{3}}})$ in $\mathbb{R}^3$ so that ${\overrightarrow{e_{1}}}$ is parallel to the line $\left\{\begin{matrix}
x=1+t\\y=1-t
\\z=1+t
\end{matrix}\right.$ and ${\overrightarrow{e_{2}}}$ is parallel to the plane $x+y+z=1$.

My work:
${\overrightarrow{e_{1}}}=(1,-1,1)$

EDIT: Thanks for all of your comments. Did i understood this correct?

Normalize ${\overrightarrow{e_{1}}}=(1,-1,1)$.
$${\overrightarrow{e_{1}}}=\frac{(1,-1,1)}{\sqrt{(1)^2+(-1)^2+(1)^2}}=(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$

Normal of the plane $(1,1,1)$.
$$\begin{pmatrix}\frac{1}{\sqrt{3}}&-\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\end{pmatrix}\times \begin{pmatrix}1&1&1\end{pmatrix}=\begin{pmatrix}-\frac{2}{\sqrt{3}}&0&\frac{2}{\sqrt{3}}\end{pmatrix}$$

$${\overrightarrow{e_{2}}}=\frac{\begin{pmatrix}-\frac{2}{\sqrt{3}}&0&\frac{2}{\sqrt{3}}\end{pmatrix}}{\sqrt{\left(-\frac{2}{\sqrt{3}}\right)^2+\left(0\right)^2+\left(\frac{2}{\sqrt{3}}\right)^2}}=\begin{pmatrix}-\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\end{pmatrix}$$

$${\overrightarrow{e_{3}}}=\begin{pmatrix}\frac{1}{\sqrt{3}}&-\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\end{pmatrix}\times \begin{pmatrix}-\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\end{pmatrix}=\begin{pmatrix}-\frac{1}{\sqrt{6}}&-\sqrt{\frac{2}{3}}&-\frac{1}{\sqrt{6}}\end{pmatrix}$$

答:
You know that you have an orthonormal system.

You need to normalize $\vec e_1$.

$\vec e_2$ should be perpendicular to both the normal to the plane (because it is parallel to the plane) and $\vec e_1$ (orthonormal system). You can use the cross product. Don't forget to normalize $\vec e_2$.

If $\vec e_1$ and $\vec e_2$ are perpendicular and of length $1$, $$\vec e_3=\vec e_1\times\vec e_2$$
Note that $\vec e_3$ is already normalized.

Title: Determine an orthonormal base


问:
Difference between convention for matrix multiplication of vector in algebra and computer science
In algebra, we typically write $Ax=b$ when we wish to multiply a $n$ dimensional $x$ vector by matrix $A$, requiring $A$ of course to have $n$ columns.

Yet, in computer science, I typically see $xA=b$, with $A$ having $n$ rows.

Of course, any "math" matrix can be transposed to give you its "computer science" matrix, so this is a matter of convention.  Why? Is this for computational efficiency? For clarity? Tradition?

Similarly, the scalar product of two vectors is typically expressed $x \cdot y$ in math, but $x^Ty$ in computer science.  Why?

答:
I don't associate the "row vector" convention to Computer Science more than to other fields and I believe that it is globally much less frequent. It probably arose just because the two conventions are possible.

Computational efficiency cannot be an explanation (or more precisely cache-friendliness), as there exist languages used in scientific applications that store matrices in row-major as well as column-major order.

$x^Ty$ is a kind of notational abuse, as it corresponds to a $1\times1$ matrix, and not exactly a scalar. The dot is more appropriate.


问:
Determine all three-digit positive integers *abc*
Determine all three-digit positive integers abc (a is hundreds b tens and c ones) such that 8abc = 3cba.

Here is my how I have started:

8 (100a + 10b + c) = 3(100c + 10b + a)

800a + 80b + 8c = 300c + 30b + 2a

797a + 50b - 292c = 0

50b = 292c - 797a

b = (292c - 797a)/50

Since abc and cba both need to be even we know that a must be even but not c because factor 292 makes the product even.

Then I started to test different values for a and c, it didn't take so long because a can be larger than 2, so the only values I tested were 1 and 2 for a. For c I tested all values up to 7 (then I found the answer) and it was kind of easy because you don't need to do the whole calculation since if the unit digit of the product of 292 multiplied by c minus 4 and 7 is not 0 we know that the difference is not divisible with 50.

So the only answer I found is c = 7  and a = 2 so b = 9

But I can't prove it algebraically or write a better solution for this problem, I don't think testing is the best method for solving it...

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

I wrote and ran some Mathematica-code:

```<code>In[1]:=ParallelTable[
  If[TrueQ[8*(a*100 + b*10 + c*1) == 3*(100*c + 10*b + 1*a)], {a, b,
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]={{{{2, 9, 7}}}}
</code>```

So, we can see that you're right!

When we expand this, with different values of the constants before the numbers we get:

```<code>In[2]:=ParallelTable[
  If[TrueQ[4*(a*100 + b*10 + c*1) == 7*(100*c + 10*b + 1*a)], {a, b,
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

Out[2]={{{{2, 3, 1}}}, {{{4, 6, 2}}}, {{{6, 9, 3}}}}
</code>```

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

I wrote and ran some Mathematica-code:

In[1]:=ParallelTable[
  If[TrueQ[8*(a*100 + b*10 + c*1) == 3*(100*c + 10*b + 1*a)], {a, b,
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

Running the code gives:

Out[1]={{{{2, 9, 7}}}}

So, we can see that you're right!

When we expand this, with different values of the constants before the numbers we get:

In[2]:=ParallelTable[
  If[TrueQ[4*(a*100 + b*10 + c*1) == 7*(100*c + 10*b + 1*a)], {a, b,
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

Out[2]={{{{2, 3, 1}}}, {{{4, 6, 2}}}, {{{6, 9, 3}}}}


问:
Show if $x^2$ is divisible by $5$ then $x$ is divisible by $5$ as well

Show if $x^2$ is divisible by $5$ then $x$ is divisible by $5$ as well

I tried to do this with contrapositive but I feel like it doesn't work. I start with if $x$ is not divisible by $5$ then $x^2$ is not divisible by $5$.

答:
Note that
$$x^2 \equiv \begin{cases} 0 \pmod5 & \text{if $x \equiv 0 \pmod5$}\\1 \pmod5 & \text{if $x \equiv \pm 1 \pmod 5$}\\-1 \pmod5 & \text{if $x \equiv \pm 2 \pmod 5$}\end{cases}$$
Now conclude what you want.</p>

ContentLicense: CC BY-SA 3.0


问:
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$
I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$

Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.

答:
Let $f\left( x \right) = \sum\limits_{n = 0}^\infty  {a_n x^n } $ be the generating function of the solution of the recurrence.
$$
\sum\limits_{n = 0}^\infty  {na_n x^n }  = \sum\limits_{n = 0}^\infty  {\left( {n + 1} \right)a_n x^{n + 1}  - 4\sum\limits_{n = 0}^\infty  {a_{n - 1} x^n } }  +
12\sum\limits_{n = 5}^\infty  {na_n x^n } H_n
$$
which gives the following DE
$$
xf'\left( x \right) = x^2 f'\left( x \right) - 3xf\left( x \right)-\frac{2 x \left(50 x^5-67 x^4+2 x^3+3 x^2+6 x+6 \log (1-x)\right)}{(x-1)^2}$$
Whose solution is
$$f(x)=\frac{2}{75 (x-1)^2} \left(-7553 x^5+3750 x^5 \log (1-x)+\\+24040 x^4-18750 x^4 \log (1-x)-29405 x^3+37500 x^3 \log (1-x)+\\+16830 x^2-37500 x^2 \log (1-x)-3840 x+18750 x \log (1-x)-3840 \log (1-x)\right)$$

Coefficients of the MacLaurin expansion of $f(x)$ are
$$0,0,0,0,0,\frac{137}{5},\frac{578}{15},\frac{1667}{35},\frac{395}{7},\frac{41137}{630},\frac{13007}{175},\frac{964849}{11550},\ldots$$

There is no closed form for $a_n$, though.


问:
Calculating the Galois group of an (irreducible) quintic
On my homework I have been asked to compute the Galois group of a quintic.  I have no idea how to do this, except

(a) I calculated that it was irreducible (brute-force)

(b) Since it is irreducible, its splitting field must have degree divisible by $5$

(c) The Galois group must be a subgroup of $S_5$.

There is also a fact that looks helpful, about $S_n$ only having one normal subgroup for $n \geq 5$.  Does this mean there's only one possible Galois extension, or is that only if $S_n$ was already some sort of Galois group?

Aside from that, I have no idea how to do the problem, so can you help me?

Thanks!

P.S. The quintic in question is $x^5 + x - 1 \in \mathbb{Q}[x]$

答:
Here are some general facts about Galois groups of irreducible quintics.

There are five transitive subgroups of $S_5$ up to conjugacy: $S_5, A_5, D_5, C_5, F_{20}$. All should be familiar to you except possibly the last one, which is the Frobenius group of invertible affine linear transformations $x \mapsto ax + b$ on $\mathbb{F}_5$.

Of these five, only $S_5$ and $F_{20}$ lie outside $A_5$. Thus if the discriminant is a square, the Galois group must be one of $A_5, D_5, C_5$, and otherwise the Galois group must be one of $S_5, F_{20}$.

$S_5$ is the only one of these groups containing a transposition. Indeed, a more general statement is true: if a transitive subgroup of $S_n$ contains a transposition and a $p$-cycle for some prime $p > \frac{n}{2}$, then it must be $S_n$. See, for example, these notes by Keith Conrad.

Edit: Ah, I see in the comments that the polynomial is reducible. That explains it, then.


答: Look at the prime factors of $ x $ and $ x^2 $.</p>

Body:
Show if $x^2$ is divisible by $5$ then $x$ is divisible by $5$ as well

I tried to do this with contrapositive but I feel like it doesn't work. I start with if $x$ is not divisible by $5$ then $x^2$ is not divisible by $5$.


问:
Show that the tangent line at ($x_{0}$, $y_{0}$) has y-intercept -$y_{0}$
I've been wrapping my head around this problem for days and I'm not able to solve it. Could someone explain to me how to solve it?

Let p be a positive constant and consider the parabola $x^2=4py$ with vertex at the origin and focus at the point (0, p).
(a) Show that the tangent line at ($x_{0}$, $y_{0}$) has y-intercept -$y_{0}$.

What I did this far is:

I used the slope formula (i) $y - y_{0} = m(x - x_{0})$

m = $\frac{dy}{dx} \frac{x^2}{4p} = \frac{x}{2p}$

Evaluated m when $x = x_{0}$

m = $\frac{x_{0}}{2p}$

Replaced m in (i)

$y = \frac{x_{0}}{2p}(x-x_{0}) + y_{0}$

In order to get the y-intercept, I evaluated $x = 0$

$y = \frac{x_{0}^2}{2p}+y_{0}$

What am I missing?

Body: Since the tangent line is$$y=\frac{x_0}{2p}(x-x_0)+y_0,$$if you put $x=0$, then you get\begin{align}y&=-\frac{{x_0}^2}{2p}+y_0\\&=-\frac{{x_0}^2}{2p}+\frac{{x_0}^2}{4p}\\&=-\frac{{x_0}^2}{4p}\\&=-y_0.\end{align}


答:
When $x = 0$, you'll get $\frac{-x_0^2}{2p} + y_0$. Since $(x_0,y_0)$ is a point on $x² = 4py$ we have $x_0^2 = 4py_0$.

The y-intercept is then $$\frac{-x_0^2}{2p} + y_0 = -2y_0 + y_0 = -y_0$$ as desired.

Body:
I've been wrapping my head around this problem for days and I'm not able to solve it. Could someone explain to me how to solve it?

Let p be a positive constant and consider the parabola $x^2=4py$ with vertex at the origin and focus at the point (0, p).
(a) Show that the tangent line at ($x_{0}$, $y_{0}$) has y-intercept -$y_{0}$.

What I did this far is:

I used the slope formula (i) $y - y_{0} = m(x - x_{0})$

m = $\frac{dy}{dx} \frac{x^2}{4p} = \frac{x}{2p}$

Evaluated m when $x = x_{0}$

m = $\frac{x_{0}}{2p}$

Replaced m in (i)

$y = \frac{x_{0}}{2p}(x-x_{0}) + y_{0}$

In order to get the y-intercept, I evaluated $x = 0$

$y = \frac{x_{0}^2}{2p}+y_{0}$

What am I missing?


问:
How to calculate the angle between matrices?
I understand the formula:

$cos\left(θ\right)=\frac{tr\left(AB^T\right)}{\sqrt{tr\left(AA^T\right)\cdot tr\left(BB^T\right)}}$

Only one thing is not clear to me,the number I get is the angle itself?

Or the number should be obtained by the cos function?

I will give an example with simple numbers to make the question clearer.

$A=\begin{pmatrix}1&2\\ 0&1\end{pmatrix} B=\begin{pmatrix}2&0\\ 0&2\end{pmatrix}$

$\frac{tr\left(\begin{pmatrix}1&2\\ 0&1\end{pmatrix}\cdot \begin{pmatrix}2&0\\ 0&2\end{pmatrix}^t\right)}{\sqrt{tr\left(\begin{pmatrix}1&2\\ 0&1\end{pmatrix}\cdot \begin{pmatrix}1&2\\ 0&1\end{pmatrix}^t\right)\cdot tr\left(\begin{pmatrix}2&0\\ 0&2\end{pmatrix}\cdot \begin{pmatrix}2&0\\ 0&2\end{pmatrix}^t\right)}}=\frac{1}{\sqrt{3}}$

So the angle is $\frac{1}{\sqrt{3}}(\approx 0.577)$ or $cos (\frac{1}{\sqrt{3}})\approx 0.837$?

答: Neither. With $\theta$ your angle, you have $\cos \theta = \frac{1}{\sqrt{3}}$, and so $\theta = 2 \pi n \pm \arccos\left(\frac{1}{\sqrt{3}}\right)$, where $\arccos\left(\frac{1}{\sqrt{3}}\right) \approx 0.955$ (radians).


问:
Find different intergrating factor

Find two diffenrent integrating factor of $$3x(y^2-1)dx+2y(x^2+1)dy=0$$

I can only find the $\frac{1}{(y^2-1)(x^2+1)}$, Is it any other integrating factor?

答:
$$3x(y^2-1)dx+2y(x^2+1)dy=0$$
Your integrating factor looks good to me. You can multiply it by any constant it will also work.
$$3(y^2-1)d(x^2+1)+2(x^2+1)d(y^2-1)=0$$
Substitute $u=x^2+1$ and $v=y^2-1$:
$$3vdu+2udv=0$$
Divide by $uv$:
$$\dfrac {3du}{u}+\dfrac {2dv}{v}=0$$
Integrate.

Another integrating factor

$$3vdu+2udv=0$$
Multiply by $u^2v$:
$$3u^2v^2du+2u^3vdv=0$$
$$v^2du^3+u^3dv^2=0$$
$$d(u^3v^2)=0$$
Integrate.
$$u^3v^2=C$$
$$(x^2+1)^3(y^2-1)^2=C$$

Title: Find different intergrating factor


问:
Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer?
I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve the problem.

I know that the First and Second way should be correct because it's using more expressions of the Taylor series around 0.
What I can't figure out is WHY using fewer expressions of the Taylor series in the Third way doesn't give 0/0 but gives a wrong answer?

First way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x-2 \sin x \cos x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x(x-\sin x)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(x-x+\frac{x^3}{6}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(\frac{x^3}{6}\right)}{x^3}=\frac{1}{3}$$
Second way:
$$\lim _{x \rightarrow 0} \frac{2x(1-\frac{x^2}{2})-(2x-\frac{8x^3}{6})}{x^3}=\lim _{x \rightarrow 0} \frac{2x-x^3-2x+\frac{8x^3}{6}}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$
Third way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} =\lim _{x \rightarrow 0} \frac{2 x \cos x-2x}{x^3}=\lim _{x \rightarrow 0} \frac{2x(\cos x -1)}{x^3}=\lim _{x \rightarrow 0} \frac{2x(-\frac{x^2}{2})}{x^3}=-1$$

答:
In all the cases we should use remainder to proceed properly as follows

$$\frac{2 x \cos x- \sin 2x}{x^3}=\frac{2 x \left(1-\frac12 x^2+O(x^3)\right)-  \left(2x-\frac16 (2x)^3+O(x^4)\right)}{x^3}=$$

$$=\frac{2x-x^3-2x+\frac43x^3+O(x^4)}{x^3}=\frac13+O(x) \to \frac13$$

without remainder we can easily get wrong with the solution.


问:
Prove that a subgroup which contains half of all elements is a normal subgroup.
I have been going through some material on elementary group theory (Group Theory for Physicists, Christoph Ludeling) and in one place it states:

A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.

Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$.
How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.

答: The two left cosets of $H$ are $H$ and $G\setminus H$, and the two right cosets of $H$ are also $H$ and $G\setminus H$. Clearly $xH=Hx$ for every $x\in H$, and if $x\in G\setminus H$, then $xH=G\setminus H=Hx$ as well. Thus, $xH=Hx$ for each $x\in G$, and $H$ is normal in $G$.</p>


答:
You are dividing by $x^3$ at the end, so you need all possible terms at least of degree $3$ in the numerator to be present, otherwise you're basically guaranteed to change the value of the limit.

Let's keep the error term in the third way, and see what happens. I will do that the following way: we have
$$
\sin(2x) = 2x + x^3\cdot g(x)\\
\cos(x) = 1-\frac{x^2}2 + x^4\cdot h(x)
$$
for some functions $g$ and $h$ where $g(x)$ and $h(x)$ are bounded as $x\to 0$. (It is more common to use $O(x^3)$ instead of $x^3\cdot g(x)$ and $O(x^4)$ rather than $x^4\cdot h(x)$. But the $O$ terms can be a bit unintuitive to arithmetize with, so if you are unaccustomed to working with error terms, I think that my approach here is closer to what you are already used to.)

Then we follow the steps in your third way and see what we get:
$$
\frac{2 x \cos x- \sin 2x}{x^3} =\frac{2 x \cos x-(2x + x^3\cdot g(x))}{x^3}\\
=\frac{2x(\cos x -1) - x^3\cdot g(x)}{x^3}\\
=\frac{2x(-\frac{x^2}{2} + x^4\cdot h(x)) + x^3\cdot g(x)}{x^3}\\
=-1 + x\cdot h(x) - g(x)
$$
and we see that in order to assess the limit as $x\to 0$, we don't need to know more about $h$ (it is bounded, so $x\cdot h(x)\to 0$), but we do need to know more about $g(x)$. Of course, it is easy to go back and check that $g(x) = -\frac8{3!} + x^2\cdot g_1(x)$ for some function $g_1$ that is bounded for $x\to 0$. Which is enough to conclude that the limit is indeed $\frac13$.

Title: Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer?

Body:
I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve the problem.

I know that the First and Second way should be correct because it's using more expressions of the Taylor series around 0.
What I can't figure out is WHY using fewer expressions of the Taylor series in the Third way doesn't give 0/0 but gives a wrong answer?

First way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x-2 \sin x \cos x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x(x-\sin x)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(x-x+\frac{x^3}{6}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(\frac{x^3}{6}\right)}{x^3}=\frac{1}{3}$$
Second way:
$$\lim _{x \rightarrow 0} \frac{2x(1-\frac{x^2}{2})-(2x-\frac{8x^3}{6})}{x^3}=\lim _{x \rightarrow 0} \frac{2x-x^3-2x+\frac{8x^3}{6}}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$
Third way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} =\lim _{x \rightarrow 0} \frac{2 x \cos x-2x}{x^3}=\lim _{x \rightarrow 0} \frac{2x(\cos x -1)}{x^3}=\lim _{x \rightarrow 0} \frac{2x(-\frac{x^2}{2})}{x^3}=-1$$


问:
How was the following Integral identity calculated?
$$\frac{1}{2\pi} \int_{0}^{2\cdot\pi}\left |1+e^{j\cdot k\cdot d\cdot \cos(\theta )}  \right |^{2}d\theta =2(1+ J_{0}\left ( k d \right ))$$
Where:
$$j=\sqrt{-1}$$
and $J_{0}$
is the Bessel function of the first kind of order zero.
k,d are constants.

答:
Here is my idea despite the fact that I didn't finish the calculation (yet)... I also don't promise to do so. I hope this helps though somehow ... Maybe you can continue from where I stopped my reasoning.

Let $ I = (\frac{1}{2\cdot\pi})\cdot \int_{0}^{2\cdot\pi}\left (1+e^{j\cdot k\cdot d\cdot cos(\theta )}  \right )^{2}d\theta $

Prove that $ I = 2 (1+ J_{0} ( 2 k d  ))$

Let $x=cos \theta $ which means that $d\theta = \frac{-dx}{sin x}$

Therefore $I = -\frac{1}{2 \pi} \int_{0}^{2 \pi} (1 + e^{jkdx})². \frac{1}{sin x}dx$ : (1)

On the other hand (source: https://en.wikipedia.org/wiki/Bessel_function)

$J_0(2kd) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} e^{2jkdsin \alpha}d \alpha = \frac{1}{\pi} \int_{0}^{\pi} cos(-2kdsin(\beta)) d \beta$ :(2)

Now back to (1): $I = -\frac{1}{2 \pi}( \int_{0}^{2 \pi} \frac{1}{sin x} dx + 2 \int_{0}^{2 \pi} e^{jkdx}. \frac{1}{sin x}dx + \int_{0}^{2 \pi}e^{2jkdx}. \frac{1}{sin x} dx)$

Can you continue from here? Or did you do this kind of reasoning already?


问:
Parametric form of a line from two points
Let $L$ be the line in $\mathbb R^3$ that fits through the points $P = (1, 2, 3)$ and $Q = (-1, 1, -1)$. Determine a parametric form of $L$.

So I'm assuming that the task is to determine a scalar $t$ and a vector $v$ so that I can express the line as such:

$(x, y, z,) = (x_{0}, y_{0}, z_{0}) + t \times v$

Where $(x, y, z)$ is $P$ and $(x_{0}, y_{0}, z_{0})$ is $Q$ or vice versa.
But I'm unsure of how I would go about doing this.

答:
You are mostly correct, but there's no need to determine $t$: we'll be leaving that free as our parameter (also, your $(x,y,z)$ will be an arbitrary point on the line, rather than just $P$).

Thus, what we need is a vector going from $P$ in the direction of $Q$ (or vice versa). Shifting everything by $P$ (that is: subtracting $P$ from everything), we want a vector from the origin in the direction of $Q - P$. Well, there's a very obvious such vector: it's $Q - P$ itself.

Now, the parametric equation $(x(t),y(t),z(t)) = t(Q - P)$ doesn't quite work: it's in the right direction, but it goes through the origin, instead of wherever our line is. To fix that, we just need to pick any point on our line to shift it by: to keep things simple, we'll pick $P$. Thus, our parametric equation is exactly

$$(x(t),y(t),z(t)) = P + t(Q - P).$$

I'll leave you to substitute in your particular values for yourself.


问:
Solving $|y-\sqrt{2}| \leq (\sqrt{2} - 1) |x-\sqrt{2}|$
I am an undergraduate student and currently I am working on inequalities and the properties of absolute values.
I am trying to solve the following problem, but I do not know how to approach it:

Being $x >= 0$. Suppose that $y = \frac{x+2}{x+1}$.
Calculate $\frac{y-\sqrt{2}}{x-\sqrt{2}}$. Deduce the following inequality:

$$|y-\sqrt{2}| \leq (\sqrt{2} - 1) |x-\sqrt{2}|$$

答:
Show that

$$\frac{y-\sqrt{2}}{x-\sqrt{2}} = - \frac{ \sqrt{2} - 1 } { x + 1 }. $$

(This is the numerical answer to the intermediate step. The proof is just by expanding the algebra and factoring.)

Hence, conclude that

$$ \left| \frac{y-\sqrt{2}}{x-\sqrt{2}} \right| = \frac{ \sqrt{2} - 1 } { x + 1 } \leq \sqrt{2} - 1 . $$


答:
If a prime $p$ divides a product $ab$, then it divides one of the two factors $a$ or $b$.

This is one of the most important properties of prime numbers.

Now take $a=b=x$.

Title: Show if $x^2$ is divisible by $5$ then $x$ is divisible by $5$ as well


问:
How to find a centroid of a quarter circle?

Find the centroid of the quarter circle $x^2+y^2 \le  10$, $y\ge |x|$ assuming the density $ {\displaystyle \delta }(x,y) = 1$. give your answer as point coordinates in the form (X,Y)

So I first convert to polar coordinates:

$$rsin(\theta)^2+rcos(\theta)^2=10$$

Then I know it's a quarter circle that's bounded by $y=|x|$

So $\theta = \frac{\pi}{4}$ to $\frac{3\pi}{4}$

So now I integrate:

$$\int _{\frac{\pi }{4}}^{\frac{3\pi }{4}}\int _0^{2.5}\:\:\:rsin\left(\theta \right)^2+rcos\left(\theta \right)^2drd\theta $$

But this is just one integral. How do I get two? What am I missing?

Thank you.

edit:

答:
To do this in polar coordinates as the question asks, please consider the surface as a triangle with two sides as $r$ and the third side as $r \mathrm{d}\theta$.

Area of triangle, $dA = \frac{1}{2} r^2 \mathrm{d}\theta$. As we know, the centroid of the triangle will be at $\frac{2}{3}r$.

So,

$\overline{x} = \displaystyle \dfrac{2}{3A}\int_{\pi/4}^{3\pi/4} r\cos\theta \,  \mathrm{d}A = \dfrac{1}{3A}\int_{\pi/4}^{3\pi/4} r^3 \cos\theta \,  \mathrm{d} \theta$

$\overline{y} = \displaystyle \dfrac{2}{3A}\int_{\pi/4}^{3\pi/4} r\sin\theta \, \mathrm{d}A = \dfrac{1}{3A}\int_{\pi/4}^{3\pi/4} r^3 \sin\theta \,  \mathrm{d} \theta$

$A = \displaystyle \dfrac{1}{2}\int_{\pi/4}^{3\pi/4}r^2 \,\mathrm{d}\theta = \frac{\pi r^2}{4}$

So, $\overline {x} =\displaystyle  \frac{4r}{3\pi} [\sin \theta]_{\pi / 4}^{3 \pi / 4} = 0$ (which is obvious due to symmetry around $y$ axis).

$\overline {y} =\displaystyle  \frac{4r}{3\pi} [-\cos \theta]_{\pi / 4}^{3 \pi / 4} = \frac{4 \sqrt2 \, r}{3\pi} = \frac {8 \sqrt5}{3 \pi} \,$ (for $r = \sqrt {10}$).


问:
Equivalent condition of completeness of Boolean algebras

A Boolean algebra $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is said to be complete if every non-empty subset of $B$ has a greatest lower bound (g.l.b). Show that for $\mathcal{B}$ to be complete, it is necessary and sufficient that every non-empty subset have a least upper bound (l.u.b).

Here's my work, albeit incomplete:

Consider $\phi \neq X \subseteq B$, and let $\phi \neq Y = \{x\in B:x^c \in X\}$. We start by assuming that $\mathcal{B}$ is complete, so $Y$ has a g.l.b, say $b$.

Claim: If $a = b^c$, then $a$ is the l.u.b of $X$.

Proof of Claim: $\forall x \in X$, $x^c \in Y$ so $b\leq x^c$ which means $x\leq b^c = a$. So $a$ is an upper bound for $X$. If $m$ is also an upper bound for $x$, then $\forall x \in X$, $x\leq m$ which means $m^c\leq x^c$. By definition of $Y$, if $x\in X$ then $x^c \in Y$, so $m^c$ is a lower bound for $Y$. Since $b$ = g.l.b($Y$), $m^c\leq b$ and $a = b^c\leq m$ which implies that $a$ is the l.u.b of $X$.

In conclusion, what I have shown is that for every subset $X \subseteq B$, we can find a subset $Y\subseteq B$ which has a g.l.b (follows from completeness). Using the claim proved above, we can find a l.u.b for $X$, and this works for all $X\subseteq B$. Hence, all non-empty subsets $X$ of $B$ have a least upper bound.

I think this only proves the necessary part of the argument. How do I show that this is also sufficient?

答:
You proved that if $\bigwedge Y$ exists, for all $\varnothing \neq Y \subseteq B$, then $\bigvee Y$ exists too.

Suppose now that $\bigvee Y$ exists for all $\varnothing \neq Y \subseteq B$.

Define $X = \{y^c : y \in Y\}$ and prove that $\bigwedge Y = \bigvee X$.

(It's exactly the same reasoning you used above.)


问:
Show that $ \sqrt[3]{(z-1)(z-2)(z-3)} $ can be defined on the entire plane minus [1, 3].
My professor gave me the above problem as test preparation, and I am feeling a bit lost at how to solve it. I looked up the answer in the back of my book, and I didn't really understand the logic behind it. I was hoping someone could shed some insight into the problem solving process since I will have a similar, but not identical, question on my test. Without further ado, here is the solution:

Let $$ f(z) = \sqrt[3]{(z-1)(z-2)(z-3)} = \text{exp}(\frac{1}{3}\text{log}[(z-1)(z-2)(z-3)]) $$

where

$$ \text{log}((z-1)(z-2)(z-3)) = \int_4^z{\frac{[(z-1)(z-2)(z-3)]'}{(z-1)(z-2)(z-3)}dz} + \text{log} (6) \text{, for } z \in \mathbb{C} \backslash (-\infty, 3]$$

Then, it can be shown that $ f(z) $ defines a function that is continuous at all points $ x \in \mathbb{R} \text{ such that } x \lt 1 $.

There are many things I don't understand about this, mostly related to the choice of logarithm. Why is the integral from 4 to z? Why do you add log(6)? Why is it defined on $ \mathbb{C} \backslash (-\infty, 3] $? How can I figure out how to solve problems like this in similar cases?

Furthermore, what is the easiest way to show continuity on the real axis where x < 1? Thanks for the help.

答: The constant 4 is an arbitrary choice on $\mathbb{C}\backslash(-\infty,3]$ and $\log 6$ is to match $f(4)=\sqrt[3]{6}$. Your integrand function is holomorphic on $\mathbb{C}\backslash(-\infty,3]$, so the integration is well defined  $\mathbb{C}\backslash(-\infty,3]$. To show $x<1$, $f(x)$ is continuous, you need to show that integration does not depend on path up to $6\pi i$, hence $f(x)$ is well-defined on $(-\infty, 1]$.</p>


问:
Find $\sum_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$
This question comes from "adjusting the order of an infinite series".

We are talking about whether a series' value would change or not by changing the position of some terms.
For example, we have

$$\ln{2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum\limits_{n=1}^{\infty}\left({\frac{1}{2n-1}-\frac{1}{2n}}\right)$$

By changing the order, we get

$$H=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots$$

and
$$H=\sum\limits_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$$

And to generalize, we construct a series which has firstly, k positive terms and then 1 negative term, and the same story repeats. We write it as $H_{k}$.

But how to calculate the values of all these $H_{k}$ becomes a problem, in fact, I'm stuck in $H_{2}$.

Thank you for your attention.

答:
Write
$${1\over4n-3}+{1\over4n-1}-{1\over2n}={3\over 4n(4n-3)}+{1\over4n(4n-1)}\ .$$
Mathematica can sum both terms on the RHS, and obtains
$${1\over8}(\pi+6\log2)+{1\over8}(-\pi+2\log8)={3\over2}\log 2\ .$$


问:
Derivative of Matrix elements w.r.t. to scalar
First of all, excuse the stupid question. Unfortunately, I couldn't find something similar already asked.

Consider the matrix

$$
A(r_1, r_2) = \begin{pmatrix} r_1^2 & 3 \\ 2 & r_2 \end{pmatrix}.
$$

I'd like to know how I can get the Jacobian of this matrix. I'd suggest it's something like:

$$
\begin{align*}
\frac{\partial A(r_1, r_2)}{\partial r_1} &= \begin{pmatrix} 2r_1 & 0 \\ 0 & 0 \end{pmatrix} \\
\frac{\partial A(r_1, r_2)}{\partial r_2} &= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\end{align*}
$$

But what mathematical object is the Jacobian of the Matrix $A$? Is it a matrix with three dimensions? Any help or hints for useful links are really appreciated.

答:
The problem is that you do not define your "space" of matrices. The easiest way to think of $2\times 2$ matrices is as $\mathbb{R}^4$ vectors. Then, the Jacobian of $A:\mathbb{R}^2\to\mathbb{R}^4$ is a $4\times 2$ matrix:
$$
J_A=\begin{pmatrix}
2r_1 & 0\\
0 & 0\\
0 & 0\\
0 & 1\\
\end{pmatrix}
$$

If you want to work on these derivatives as matrices, you can think of operations such as the reshape of Matlab, but it is not clear e.g. how the chain rule would work then.


问:
Arranging word FACETIOUS but vowels must be in order
The word FACETIOUS is the most common word in the English language which has each of the five vowels in order from left to right. How many ways can the letters in the word FACETIOUS be arranged such that all five vowels are in order from left to right?

I am wondering if there is an easier way than bashing and counting all the possiblities.

答: We start with $9!$ and divide by $5!$ to account for the vowels. This is because there are $9!$ permutations without the extra restriction, and then one in every $5!$ of them have the vowels in the desired order. The answer is $3024.$


答:
Hint.\begin{align}
\left|1+e^{j\cdot k\cdot d\cdot \cos(\theta )}  \right|^{2}&=
\big(1+e^{j\cdot k\cdot d\cdot \cos(\theta )}\big)
\big(1+e^{-j\cdot k\cdot d\cdot \cos(\theta )}\big)
\\
&=  e^{-j\cdot k\cdot d\cdot \cos(\theta )} + 2 + e^{j\cdot k\cdot d\cdot \cos(\theta )}
\end{align}
$$=2+2\cdot \cos(k\cdot d\cdot \cos(\theta))=2\cdot(1+\cos(k\cdot d\cdot \cos(\theta))$$

Title: How was the following Integral identity calculated?

Body:
$$\frac{1}{2\pi} \int_{0}^{2\cdot\pi}\left |1+e^{j\cdot k\cdot d\cdot \cos(\theta )}  \right |^{2}d\theta =2(1+ J_{0}\left ( k d \right ))$$
Where:
$$j=\sqrt{-1}$$
and $J_{0}$
is the Bessel function of the first kind of order zero.
k,d are constants.


问:
The union of a locally finite family of functionally closed- RYSZARD ENGELKING
In the book RYSZARD ENGELKING General Topology he states

Show that the union of a locally finite family of functionally closed sets is not necessarily functionally closed. Note that in a perfectly normal space the union of a locally finite family of functionally closed sets is functionally closed.

A topological space $X$ is called a perfectly normal space if $X$ is a normal space and every closed subset of $X$ is $G_{\delta}$-set.

A subset $A$ of a topological space $X$ is called functionally closed if $A=f^{-1}(0)$ for some continous function $f:X \to [0,1]$

Every closed subset of $X$ is $G_{\delta}$-set means that it countable intersections of open sets

On the other hand,

observation 1: I have proven that a subset $A$ of a normal space is closed $G_{\delta}$ if and only if there exists a continuous function $f:X \to [0,1]$ such that $A=f^{-1}(0)$. i.e, in a normal space functionally closed sets coincide with closed $G_{\delta}$.

A family $\{A_s\}_{s \in S}$ of subsets of a topological space $X$ is locally finite if for every point $x \in X$ there exists a neighbourhood $U$ such that the set $\{s \in S : U \cap A_s \neq \emptyset \}$ is finite.

Any ideas:
Suposse that $X$ is perfectly normal espace.
Let $A=\{A_s\}_{s \in S}$ a locally finite family of functionally closed sets, for observation 1, the $A_s$ are $G_{\delta}$-sets.  We have to find a continuous function $f:X \to [0,1]$ such that $\cup_{s \in S} A_s= f^{-1}(0)$. For each $A_s$ there is a continuous function $f_s: X \to [0,1]$ such that $f_s^{-1}(0)=A_s$.I don't know how to use my hypotheses to build my continuous function f? some help?
On the other hand, I can't think of a counterexample for the first part.

答:
Let $N=\Bbb R\times [0,\infty)$ with the topology of the Niemitzky plane (a.k.a the  Niemitzky-Moore plane). For brevity let $E=\Bbb R\times [0,\infty).$ Then $E$ is a closed discrete subspace of $N$ so every $A\subset E$ is closed in $N.$ We show that every $A\subset E$ is the intersection of a functionally-closed family in $N,$ and that some subsets of $E$ are not functionally-closed in $N.$

For $(r,0)=p\in E$ and $s\in [0,\infty$ let $B(p,s)=\{p\}\cup \{(x,y)\in N:(x-r)^2+(y-s)^2<s^2\}.$ Then $B(p,s)$ is open, and $\overline {B(p,s)}= B(p,s)\cup \{(x,y)\in N:(x-r)^2+(y-s)^2=s^2\}.$

For $p\in E$ and $p\ne q\in B(p,1)$ there is a unique circle $C$ containing $q$ and tangent to $E$ at $p.$ Let $t$ be the radius of $C$  and let $f_p(q)=1-t.$ For $ q\in N$ \ $B(p,1)$ let $f_p(q)=0.$ Let $f_p(p)=1.$ We can confirm that $f_p:N\to [0,1]$ is continuous by observing that if $0<z<1$ then $f_p^{-1}(z,1]=B(p,1-z)$ is open, and $f_p^{-1}[0,z)=N$ \ $\overline {B(p,1-z)}$ is open.

The "point" is that $f_p^{-1}(0,1]$ is functionlly open, and its intersection with $E$ is $\{p\}.$

Now  $N$ \ $E$ is functionally open. So for any $A\subset E,$ the set $F_A=\{N \backslash E\}\cup \{f_p^{-1}(0,1]:p\in E\backslash A\}$ is a functionally open family. If $\cup F_A$ is functionally open then $A=N$ \ $\cup F_A$ is functionally closed.

Let $C(N, \Bbb R)$ be the set of continuous $f:N\to \Bbb R.$ Since $\Bbb R$ is Hausdorff and $N$ is separable, we have $|C(N,\Bbb R)|\le |\Bbb R|^{\aleph_0}=2^{\aleph_0}.$ But if every $A\subset E$ were functionally closed in $N,$ we would have $|C(N,\Bbb R)|\ge |\{A:A\subset E\}|=2^{2^{\aleph_0}}.$


问:
Equivalence of two radical equations without certain conditions - correctness of method
I have question related to the following example: $\sqrt{22-x} - \sqrt{10-x}=2$.

First question:

Do I first need conditions $22-x \geq 0 $ and $ 10-x \geq 0$ to obtain the equivalent equation: $\sqrt{22-x}= \sqrt{10-x}+2$, then take square and so on
or  I don't need these conditions

i.e.$\sqrt{22-x} - \sqrt{10-x}=2$  is already equivalent to $\sqrt{22-x}= \sqrt{10-x}+2$, so we need only $10-x \geq 0$. The right side is non-negative, we take square and so on.

I wonder which method is correct.

Second question:

I know from theory that $\sqrt{f(x)}= g(x) \iff g(x) \geq 0 $ and $  f(x)= (g(x))^2 \geq 0$, so condition $f(x)\geq 0$ is automatically fulfilled. Can we do same here i.e. since right side is positive, so is left side in $\sqrt{22-x} - \sqrt{10-x}=2$ and we just take square of both sides?

Please help with this, I am learning for exams and I am confused about this. Thanks a lot in advance.

答:
ANSWER TO 1ST QUESTION

I belive you can just mess around with the terms to find the solutions in $x$ but then you have to check if all of them are valid, ie you do need to use the conditions that $22-x\ge0$ and $10-x\ge0$. Is that helpful?


问:
10 to the power of 3.5: $10^{3.5}$
So $10^3 = 10\times 10\times 10 = 1000$, this is really easy to understand.

But what about: $\,10^{3.5}\,?\,$ My logic would suggest this was

$10\times 10 \times 10\times 5 = 5000,\;$ but the calculator says it's 3162.27...

Can someone illustrate how the calculator calculates the power of when the number is with decimals?

Please keep in mind that I'm a newbie to mathematics!

答:
Your mistake is computing $10^\frac{6}{2} \cdot \frac{10}{2}$ instead of $10^\frac{6}{2} \cdot 10^\frac{1}{2}$. Although 5 is halfway to ten when working with addition, it is past halfway to ten when working with multiplication.

$10^{3.5}$ is equivalent to $10^\frac{7}{2}$. You can perform the '7' and 'over 2' parts of the exponentiation separately. That is to say: $10^\frac{7}{2} = (10^7)^\frac{1}{2}$. That's just $\sqrt{10^7}$, which can be reduced to $10^3 \sqrt{10}$, which is approximately $3163.3$.

Title: 10 to the power of 3.5: $10^{3.5}$


答:
Your argument will be of the form $P_0(x) \implies P_1(x) \implies P_2(x) \implies \cdots$ and, hopefully, you will end up with a set of possible values for $x$, say $x \in D$. Essentially you can now say

$$\text{If $x$ solves $P_0(x)$, then $x \in D$.}$$

This is when you worry about what subset of $D$ will make $P_0(x)$ true.

There is an interesting "trick" that can be used to solve this equation.

\begin{align}
   \sqrt{22-x} - \sqrt{10-x} &= 2 \\
   \sqrt{22-x} + \sqrt{10-x} &= y \\
\hline
   (\sqrt{22-x} - \sqrt{10-x})(\sqrt{22-x} + \sqrt{10-x}) &= 2y \\
   (22-x)-(10-x) &= 2y \\
   12 &= 2y \\
   y &= 6 \\
\hline
   (\sqrt{22-x} - \sqrt{10-x})+(\sqrt{22-x} + \sqrt{10-x}) &= 2+y \\
   2\sqrt{22-x} &= 8\\
   \sqrt{22-x} &= 4 \\
   22-x &= 16 \\
   x &= 6
\end{align}

And a quick check shows that $x=6$ is indeed a solution.

Title: Equivalence of two radical equations without certain conditions - correctness of method

Body:
I have question related to the following example: $\sqrt{22-x} - \sqrt{10-x}=2$.

First question:

Do I first need conditions $22-x \geq 0 $ and $ 10-x \geq 0$ to obtain the equivalent equation: $\sqrt{22-x}= \sqrt{10-x}+2$, then take square and so on
or  I don't need these conditions

i.e.$\sqrt{22-x} - \sqrt{10-x}=2$  is already equivalent to $\sqrt{22-x}= \sqrt{10-x}+2$, so we need only $10-x \geq 0$. The right side is non-negative, we take square and so on.

I wonder which method is correct.

Second question:

I know from theory that $\sqrt{f(x)}= g(x) \iff g(x) \geq 0 $ and $  f(x)= (g(x))^2 \geq 0$, so condition $f(x)\geq 0$ is automatically fulfilled. Can we do same here i.e. since right side is positive, so is left side in $\sqrt{22-x} - \sqrt{10-x}=2$ and we just take square of both sides?

Please help with this, I am learning for exams and I am confused about this. Thanks a lot in advance.


问:
Existential Notation: Is "$\exists mn$" the same as "$\exists m\exists n$"?
I'm doing an exercise for Discrete Mathematics, chapter Logic, topic Quantifiers. However, I never seen this kind of notation: "$\exists mn$".

Is "$\exists mn$" same as "$\exists m \exists n$"?

答: Yes, this is because $\exists m, \exists n, P(m,n)$  is true if and only if $\exists n, \exists m, P(m,n).$  So you can think of $\{m,n\}$ as a block variable where order does not matter and instead write $\exists \{m,n\}, P(m,n)$ or informally $\exists m,n, P(m,n).$


问:
What interpretation shows that $\forall y \;\exists x(Fxy)$ does not imply $\exists x \;\exists y(Fxy \wedge Fyx)$?
So far, I have used the domain, $\{1, 2, 3\}$, and have tried to determine an interpretation for $F$ in which $\forall y\; \exists x(Fxy)$ is true and $∃x\;∃y(Fxy ∧ Fyx)$ is false.

I began by trying to make $∃x\;∃y(Fxy ∧ Fyx)$ false. I reasoned that since this is an existential statement, for it to be false, every case of it, $∃y\;(Fxy ∧ Fyx)$, needs to be false. And since each of these cases, are themselves existentials, for them to be false each of their cases need to be false.

Here are the cases:

$∃y\;(F1y ∧ Fy1)$: $F11 ∧ F11$, $F12 ∧ F21$, $F13 ∧ F31$

$∃y\;(F2y ∧ Fy2)$: $F21 ∧ F12$, $F22 ∧ F22$, $F23 ∧ F32$

$∃y\;(F3y ∧ Fy3)$: $F31 ∧ F13$, $F32 ∧ F23$, $F33 ∧ F33$

Since each of these existentials' cases are conjunctions, for them each to be false at least one of the conjuncts need to be false, i.e. one of the ordered-pairs in the each conjunctions needs to be excluded from the extension of F.

Now, $∀y\;∃x(Fxy)$ is a universal, so to make it true, I need to make all of its instances true. And these instances, $∃x\;(Fxy)$, are themselves existentials, so to make them true, at least one of their instances need to be true. Here are the cases:

$∃x\;(Fx1)$: $F11$, $F21$, $F31$

$∃x\;(Fx2)$: $F12$, $F22$, $F32$

$∃x\;(Fx3)$: $F13$, $F23$, $F33$

So my approach was to rule out one of the ordered pairs from the conjunctions in order to falsify each of the first set of cases, while leaving enough of them not ruled out to make at least one of the ordered pairs in the second set true.

In the first place, I think I have to exclude from the extension of F each of the ordered pairs $<1, 1>$, $<2, 2>$, $<3, 3>$.

From there, I then decided to rule out $<1, 2>$, $<1, 3>$, $<1, 2>$, and $<2, 3>$.

But here's where I run into problems: If I rule out $<2, 3>$, then $'∃x\;(Fx3)$: $F13$, $F23$, $F33'$ turns false. If though, I include $<2, 3>$, then $'∃y\;(F2y ∧ Fy2)$: $F21 ∧ F12$, $F22 ∧ F22$, $F23 ∧ F32'$ turns true.

Note, I think that this extension for $F$ does the trick: $\{<1, 2>, <2, 3>, <3, 1>\}$. But I trying to more methodically work out how to produce this solution, if this is, in fact, a solution.

Thanks for any help. I really appreciate it!

This is IIIB2a from Goldfarb's Deductive Logic

答: Your solution is fine.  In order to make $\exists x \exists y (Fxy \wedge Fyx)$ false, you cannot have any of $F11,F22,F33$.  You also can have only one of each pair $(F12,F21), (F13,F31),$ and $(F23,F32)$.  To make $\forall y \exists x (Fxy)$ you need one instance with each element in the second position, which requires three instances to be true.  You therefore need one of each pair.  If you start with $F12$, you then need $F31$ as that is the only one available with $1$ in the second position.  This forces $F23$ and your solution.  Alternately you could start with $F21$ and get the corresponding solution $(F21,F32,F13)$


答:
Is "$\exists mn$" same as "$\exists m \exists n$"?

Yes, exactly. This is just a shorthand / abbreviation. If you want to be completely formal about it, you would write out $\exists m \; \exists n$.

Title: Existential Notation: Is "$\exists mn$" the same as "$\exists m\exists n$"?

Body:
I'm doing an exercise for Discrete Mathematics, chapter Logic, topic Quantifiers. However, I never seen this kind of notation: "$\exists mn$".

Is "$\exists mn$" same as "$\exists m \exists n$"?


问:
How to calculate triangle's angle
Following this image, I have C1, C2 and H values. How to calculate the angle?</p>

答:
Let the angle you're looking for, be $\theta$. Have you come across trigonometry yet (especially inverse trigonometric functions)? You're looking for $\theta$ such that $$\tan\theta = \frac{C_2}{C_1}$$
$$\implies \theta = \arctan\frac{C_2}{C_1}$$
where $\arctan$ is just $\tan^{-1}(x) \in [-\frac{\pi}{2},\frac{\pi}{2}]$, i.e. the inverse function of $f(x) = \tan x$.

Title: How to calculate triangle's angle


答:
There are two possible answers. One is what you already know and the other is $\{(1,3),(2,1),(3,2)\}$.

Since the domain is small it will help to draw a $3\times3$ table with $y$ values and $x$ values along the rows and columns. Place $1$ in the cell which will be contained in $F$ and cross out the rest.

The second condition crosses out all diagonal cells.

In each column i.e. $\forall y$, we have to place atleast one $1$ in the column due to the first condition.

When you place a $1$, the cell diagonally across gets automatically crossed out due to second condition.

For $y=1$, we may select $x=2$. So $(2,1)$ is selected and $(1,2)$ is crossed out. The only choice for $y=2$ is $x=3$ so $(3,2)$ is selected and $(2,3)$ is crossed. Finally $(1,3)$ is selected and we cross out $(3,1)$.

\begin{matrix}&y=1&y=2&y=3\\x=1&X&\color{red}X&1\\x=2&1&X&\color{red}X\\x=3&\color{red}X&1&X\end{matrix}

So we get $F=\{(1,3),(2,1),(3,2)\}$. In red are the non-diagonal $X$s. You will see that exchanging the $1$s and $X$s will also satisfy our requirements since each red $X$ was placed diagonally across a $1$ and there is a red $X$ in each column. Selecting the red $X$s gives us the $F$ you have mentioned which corresponds to selecting $x=3$ for $y=1$.  Also note that this property is preserved for larger domains: flipping the member pairs of $F$ gives us another valid $F'$.

Title: What interpretation shows that $\forall y \;\exists x(Fxy)$ does not imply $\exists x \;\exists y(Fxy \wedge Fyx)$?

Body:
So far, I have used the domain, $\{1, 2, 3\}$, and have tried to determine an interpretation for $F$ in which $\forall y\; \exists x(Fxy)$ is true and $∃x\;∃y(Fxy ∧ Fyx)$ is false.

I began by trying to make $∃x\;∃y(Fxy ∧ Fyx)$ false. I reasoned that since this is an existential statement, for it to be false, every case of it, $∃y\;(Fxy ∧ Fyx)$, needs to be false. And since each of these cases, are themselves existentials, for them to be false each of their cases need to be false.

Here are the cases:

$∃y\;(F1y ∧ Fy1)$: $F11 ∧ F11$, $F12 ∧ F21$, $F13 ∧ F31$

$∃y\;(F2y ∧ Fy2)$: $F21 ∧ F12$, $F22 ∧ F22$, $F23 ∧ F32$

$∃y\;(F3y ∧ Fy3)$: $F31 ∧ F13$, $F32 ∧ F23$, $F33 ∧ F33$

Since each of these existentials' cases are conjunctions, for them each to be false at least one of the conjuncts need to be false, i.e. one of the ordered-pairs in the each conjunctions needs to be excluded from the extension of F.

Now, $∀y\;∃x(Fxy)$ is a universal, so to make it true, I need to make all of its instances true. And these instances, $∃x\;(Fxy)$, are themselves existentials, so to make them true, at least one of their instances need to be true. Here are the cases:

$∃x\;(Fx1)$: $F11$, $F21$, $F31$

$∃x\;(Fx2)$: $F12$, $F22$, $F32$

$∃x\;(Fx3)$: $F13$, $F23$, $F33$

So my approach was to rule out one of the ordered pairs from the conjunctions in order to falsify each of the first set of cases, while leaving enough of them not ruled out to make at least one of the ordered pairs in the second set true.

In the first place, I think I have to exclude from the extension of F each of the ordered pairs $<1, 1>$, $<2, 2>$, $<3, 3>$.

From there, I then decided to rule out $<1, 2>$, $<1, 3>$, $<1, 2>$, and $<2, 3>$.

But here's where I run into problems: If I rule out $<2, 3>$, then $'∃x\;(Fx3)$: $F13$, $F23$, $F33'$ turns false. If though, I include $<2, 3>$, then $'∃y\;(F2y ∧ Fy2)$: $F21 ∧ F12$, $F22 ∧ F22$, $F23 ∧ F32'$ turns true.

Note, I think that this extension for $F$ does the trick: $\{<1, 2>, <2, 3>, <3, 1>\}$. But I trying to more methodically work out how to produce this solution, if this is, in fact, a solution.

Thanks for any help. I really appreciate it!

This is IIIB2a from Goldfarb's Deductive Logic


问:
Find the minimum number of operations to make a binary number alternating
Let's say there is a binary number with length n. n is even. It is guaranteed that the number will contain n/2 numbers of 0's and n/2 number of 1's. In one operation it is only allowed to reverse a sub portion of that binary number.

How can we find the number of operations to make the binary number alternating(No two 0 or two 1 side by side. Example - 101010 or 010101)

For example,

If the binary number is 0110, we can reverse the bold portion once and we'll get 0101. So, the required number of operation is one.

If the binary number is 11101000, we need two operations:

11101000, reverse the bold portion to get 10101100

10101100, reverse the bold portion to get 10101010

So, the required number of operations is two.

Actually, I found out the problem here

I cannot understand the solution they provided.

Solution

Count the number of 00 or 11. the answer is the maximum number of 00 or 11.
So, for 0110. the number of 11 is 1 (0110). the number of 00 is 0. So, the answer is max(1,0) = 1

So, for 11101000, the number of 11 is 2, One for 11101000, another for 11101000. Similarly, the number of 00 is 2. So, the answer is 2.

Question

Why the above solution works? Could someone please explain?

答:
(Fill in the gaps yourself. They should be almost immediately obvious, )

Claim 1: Given any binary string of length $n$, let the number of consecutive 00 be $a$, and the number of consecutive 11 be $b$, then the number of "reverse a sub portion of that binary string" that we need to convert it into an alternating string is $ \lceil \frac{ a+b}{2} \rceil $.

 Hint: Every 00 or 11 will need to be the start or the end of a sub-portion, in order to become an alternating string.

Claim 2: For a binary string of length $n = 2k$ with $k$ 0's and $k$ 1's, then $ |a-b| \leq 1$.

 Hint: The number of 00's is equal to the number of 0's minus the number of 0-blocks.

 Hint: The number of 0-blocks and 1-blocks is at most 1 apart.

Corollary: For a binary string of length $n = 2k$ with $k$ 0's and $k$ 1's, the number of "reverse a sub portion of that binary string" that we need to convert it into an alternating string is $ \lceil \frac{ a+b}{2} \rceil  = \max (a,b)$.

 Hint: Use $|a-b| \leq 1$.


问:
Showing the inequality holds (binomial coefficient)
I'm struggling to understand how to show that the inequality holds for $ m,n \in \mathbb{N} \; \text{with} \; m<n $
$$\frac{1}{m^k} {m \choose k} < \frac{1}{n^k} {n \choose k} \;\; \text{for all} \; k=2,...,m$$

Any pointers of how to approach this would be very helpful.

答:
$\frac{1}{m^k} {m \choose k} < \frac{1}{n^k} {n \choose k} \;\; \text{for all} \; k=2,...,m$

$\displaystyle {m \choose k} = \frac{m!}{(m-k)!k!} = \frac{1}{k!}\prod_{i=0}^{k-1} (m-i)$

So, $ \displaystyle \frac{1}{m^k} {m \choose k} = \frac{1}{k!}\prod_{i=0}^{k-1} \frac{m-i}{m} = \frac{1}{k!}\prod_{i=0}^{k-1} (1-\frac{i}{m})$

Similarly $ \displaystyle \frac{1}{n^k} {n \choose k} = \frac{1}{k!}\prod_{i=0}^{k-1} \frac{n-i}{n} = \frac{1}{k!}\prod_{i=0}^{k-1} (1-\frac{i}{n})$

As $n \gt m, (1-\frac{i}{n}) \gt (1-\frac{i}{m})$.

This should lead to the proof.


问:
How can we derive the standard form of the linear equation: $Ax+By+C$?
How can we derive the standard form of the linear equation: $Ax+By+C=0$? What do "$A$", "$B$" and "$C$" in the standard form of the linear equation mean? As in the point-slope form of the linear equation: $y-y_0=m(x-x_0)$, $m$ is the slope, and $x$, $y$ are the coordinates of any point. Likewise, $x_0$, $y_0$ are the coordinates of the given point. </p>

答:
Let's say we start with the point-slope form $$y-y_0=m(x-x_0),$$ where $(x, y)$ represents any point on the line, and $(x_0, y_0)$ represents a given point on the line, and $m$ is the given slope of the line. Now,
$$
\begin{align} y-y_0 =m(x-x_0)
& \iff  y - y_0 = mx - mx_0 \\ \\
& \iff y - y_0 - mx - mx_0 = 0 \\ \\
& \iff (-m)x + y + (-y_0 - mx_0) = 0
\end{align}
$$

Now, we can simply let $A = -m,\; B = 1,\; C = (- mx_0 - y_0)$, which means that $$y + (-m)x + (-y_0 - mx_0) = 0 \iff Ax + By + C = 0$$


答:
I will just write a brief outline of the proof.

You only need to show the inequality holds for $n=m+1$.

$$
\frac{1}{m^k} {m \choose k} < \frac{1}{(m+1)^k} {n \choose k} \iff (m-k+1)(m+1)^k < (m+1) m^k
$$

$$
\iff (m-k+1)(m+1)^{k-1} < m^k
$$

Experiment with $k=2,3$ for example, you will see you can apply some famous inequality involving two different kind of averages.

Title: Showing the inequality holds (binomial coefficient)

Body:
I'm struggling to understand how to show that the inequality holds for $ m,n \in \mathbb{N} \; \text{with} \; m<n $
$$\frac{1}{m^k} {m \choose k} < \frac{1}{n^k} {n \choose k} \;\; \text{for all} \; k=2,...,m$$

Any pointers of how to approach this would be very helpful.


问:
Division of polynomials. For which $m$ one polynomial can be divided to other.
Let $p(x) = (x + 1)^m - x^m - 1$ and $q(x) = (x^2+x+1)^2$.
Coefficients if these polynomials are in $\mathbb{R}$

So, the question is: "For which $m$ p(x) can be divided by $q(x)$ without remainder?".

I tried to use this idea: let's simulate GCD's algorithm for $p(x)$ and $q(x)$ and see for which $m$ their gcd is dividable to $q(x)$, but didn't got any intresting results.

So, maybe exists a better solution. I'll be very grateful.

答:
Hint:

As the roots of $q(x)$ are the complex cubic  roots of unity $\;\omega=\mathrm e^{\tfrac{2i\pi}3}$ and $\omega^2=\bar\omega=\mathrm e^{\tfrac{4i\pi}3}$, and $p(x)$ has to be divisible by the square of $q(x)$, you have to check that $\omega$ and $\omega^2$ are double roots of $p(x)$, i.e. that they're roots of $p(x)$ and $p'(x)$.

Some more details:

As $p'(x)=m(x+1)^{m-1}-mx^{m-1}$, we have the conditions

$(\omega+1)^m-\omega^m-1=(-\omega^2)^m-\omega^m-1=0$,

$(-\omega^2)^{m-1}-\omega^{m-1}=0$.


问:
Decomposing a permutation into 3-cycles
I have two exercises I don't know how to solve:

Write the permutation $α = (1 2)(3 4)$ as a product of 3-cycles

Write the permutation $α = (1 2 8 3 7)(4 5 6)$ as a product of 3-cycles

I know how to convert a permutation into a composition of 2-cycles, but not of 3-cycles. I imagine there is a rule for when this is possible, depending on odd/even properties, but since I'm am using the exercises mentioned to help me to learn about permutation parity, I can't yet follow arguments based on that understanding.

Also, I am working with left to right composition such that $(123) = (12)(13)$.

Thanks in advance for any help.

答:
First, notice that, since 3-cycles are even, and products of even permutations are even themselves (if you know what it means, this comes from the fact that the map sending each permutation to its sign is a homomorphism), for a permutation to be written as a product of 3-cycles you need it to be even. Luckily, your $\alpha$'s are.

Now, let me give you the general argument here, that hopefully you will be able to specialize to the cases above. Take an even permutation $\sigma$  and write it as a product of transpositions, say $\tau_1\cdots\tau_k$, where $k$ must be even. Looking at $\tau_1\tau_2$, you can suppose $\tau_1\neq\tau_2$, otherwise their product is the identical permutation. Then, you have two cases:

$\tau_1=(ab),\tau_2=(ac)$, i.e. $\tau_1,\tau_2$ permute a common number, say $a$. In this situation, as you noted for $(123)$, $(ab)(ac)=(abc)$, so $\tau_1\tau_2$ is a 3-cycle;

$\tau_1=(ab),\tau_2=(cd)$, with $a,b,c,d$ different, i.e. they are disjoint. In this case, write $(ab)(cd)=(ab)(bc)(bc)(cd)$ and you get a product of two 3-cycles by point 1.

Iterating and using the fact that $k$ is even, you get a full decomposition of $\sigma$ in 3-cycles. Incidentally, this actually proves that any even permutation can be written as a product of 3-cycles (again, provided you are familiar with the terminology, $A_n$ is generated by 3-cycles).

Title: Decomposing a permutation into 3-cycles


问:
Is quantifiers are dichotomy?
If for example this is false:

$$ \exists x \forall y(P(x,y)$$

Does it mean that

$$ \sim( \exists x \forall y(P(x,y))  $$

Is guaranteed to be true? so is there a dichotomy?

And does it flip all the quantifiers to:

$$ \forall x \exists y( (P(x,y))$$

I know that $ \exists x = \sim( \forall x(\sim P))$ but all I get is:

$$ \sim \exists x \forall y(P) = \sim \sim \forall x \forall y( \sim P) = \forall x \forall y(\sim P)$$

Where is my mistake? Thanks for the help!

答:
Your mistake lies in taking the negation.

The negation of $\exists x\forall y(P(x,y))$ is $\forall x\neg[\forall y(P(x,y))]$ since negation converts $\exists x(P)\to\forall x(\neg P)$. Now $\neg[\forall y(P(x,y))]$ becomes $\exists y(\neg P(x,y))$, giving $\forall x\exists y(\neg P(x,y))$.

Title: Is quantifiers are dichotomy?


答:
There are nine spaces, $5$ vowels that must be in order, and $4$ are consonants.  You can pick any letter first and choose where to place it so lets choose to place the $4$ consonants first.

We can put the $F$ in any of the $9$ places.  We can put the $C$ in any of the remaining $8$.  We can place the $T$ and the $S$ in any of the remaining $7$ and $6$ places.  That's $9\times 8\times 7 \times 6$ ways to place the consonants.

The remaining $5$ spots must be filled with the vowels in order.  There is only one way left to do that.

So there are $9\times 8\times 7 \times 6$ ways.

....

Alternatively.  There are ${9\choose 4}$ ways to pick spots for the consonants, or ${9\choose 5}$ ways to pick spots of the vowels.  For any one of the vowel/consonants placements we can arrange-- in the four spots set for the consonants-- the consonants in $4!$ ways.  So there are ${9\choose 4}\times 4!$ or ${9\choose 5} \times 4!$ ways.

......

Or there are $9!$ ways total to do arrange the nine letters.  However the vowels can be ordered in $5!$ ways and we can only accept $1$ of the the $5!$ orders.  So of all the $9!$ ways to arrange the nine letters we can only accept $1$ out of $5!$ of them.  SO there are $\frac{9!}{5!}$ ways.

===

Method 1: gives us $9\cdot 8 \cdot 7\cdot 6=3024$ ways.

Method 2a: gives us ${9\choose 4}\cdot 4! = \frac {9!}{5!4!}4!= \frac {9!}{5!} = \frac {1\cdot 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4\cdot5}=6\cdot7\cdot8\cdot9 =3024$ ways.  Method 2b: and ${9\choose 5} = \frac {9!}{4!5!} = {9\choose 4}$ gives us the same answer.

Method 3: gives us $\frac {9!}{5!} = \frac {1\cdot 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4\cdot5}=6\cdot7\cdot8\cdot9 =3024$ ways.

Title: Arranging word FACETIOUS but vowels must be in order

Body:
The word FACETIOUS is the most common word in the English language which has each of the five vowels in order from left to right. How many ways can the letters in the word FACETIOUS be arranged such that all five vowels are in order from left to right?

I am wondering if there is an easier way than bashing and counting all the possiblities.


问:
Power series solution for differential equations
Is the solution to differential equations using power series applicable only to homogeneous differential equations?
I mean equations of the form: $$a_2 \phi ''(x)+a_1 \phi '(x)+ \phi(x) = 0$$

答:
If you have a source term $f$ that admits a power series decomposition, you can apply the power series method to find a solution of your equation.

Also, this method can be extended to differential equations of any degrees. It also works if your coefficient $a_i$ are in the form $a_i x_k$ for any $k$.

Finaly, I would add that you need to be careful with this method if you have a non-linear O.D.E.


问:
Mathematic problem
I'm stuck on a problem that maybe most of you will be able to solve instantly. Here is the problem:

A publisher needs to bind $4500$ books. One print shop can bind these books in $30$ days, another shop can do it in $45$ days. How many days are necessary to bind all the books if both shops work in parallel?

The problem's answer is $18$ days, but I got $13$ days. Here is how I calculated this:
First, I found the books per day for each shop:
Shop 1 = $\frac{4500}{30} $ = $150$ books/day
Shop 2 = $\frac{4500}{45}$ = $100$ books/day

Let $x$ be the number of days:
$$150x + 100x = 4500$$
$$x = 4500 / 350 = 12.8$$ or $13$ days

What am I missing here? Thanks

答: A common way of solving these types of problems is figuring out the fraction of books bound in a day. One shop completes ${1}/{30}$ of the total books, and the other shop completes ${1}/{45}$ of the total books in a day. Add these fractions together, and you get ${1}/{30} + {1}/{45} = {3}/{90} + {2}/{90} = {5}/{90} = {1}/{18}$. That means ${1}/{18}$ of the books are bound in a day, so it takes $18$ days to complete them all. This method can be used for other problems of this form.

Title: Mathematic problem


问:
How to improve the precision of this approximation
I have to approximate $\pi$ by using

$\pi=\lim_{n\to\infty}z_n$

with
$z_2=2, z_{n+1}=2^{n-1/2}\sqrt{1-\sqrt{1-4^{1-n}z_{n}^2}}, n=2,3,...$

How can I improve the accuracy of this approximation when calculating the iterations in Matlab? $z_{n+1}=0$ for $n>30$, I assume its because of the squareroot getting too close to $0$, is that correct? How can I modify this equation to reach better results when using a computer?
Any help is very much appreciated (and sorry for my bad english).

答:
As you state, the issue is the cancellation which occurs within the first radical. This may be handled by rationalizing it:

$$\sqrt{1-\sqrt{1-4^{1-n}z_n^2}}=\frac{\sqrt{1-(1-4^{1-n}z_n^2)}}{\sqrt{1+\sqrt{1-4^{1-n}z_n^2}}}=\frac{2^{1-n}|z_n|}{\sqrt{1+\sqrt{1-4^{1-n}z_n^2}}}$$

Multiplying by $2^{n-1/2}$ and noting $z_n\ge0$ leaves

$$z_{n+1}=\frac{z_n\sqrt2}{\sqrt{1+\sqrt{1-4^{1-n}z_n^2}}}$$


问:
Integration by parts help
Integration by parts

$$x\tan^{-1}x~dx=\frac12\left(x^2+1\right)\tan^{-1}x-\frac12x+c$$

and

$$x^2e^{-3x}dx =-\frac13e^{-3x}\left(x^2+\frac23x+\frac29\right)+c$$

I have

$$d(uv)=u~dv+v~du$$

Can anyone please help me?

答:
First one:

Recall that $d(\arctan(x)) = \dfrac{dx}{1+x^2}$.
$$\int x \arctan(x) dx = \underbrace{\int \arctan(x) d\left(\dfrac{x^2}2 \right)}_{\displaystyle \int v du} = \overbrace{\dfrac{x^2 \arctan(x)}2}^{\displaystyle uv} - \underbrace{\int \dfrac{x^2}{2(1+x^2)}dx}_{\displaystyle \int u dv}$$
Now
$$\int \dfrac{x^2}{2(1+x^2)}dx = \int \dfrac{1+x^2-1}{2(1+x^2)}dx = \int \dfrac{dx}2 - \int \dfrac{dx}{2(1+x^2)} = \dfrac{x}2 - \dfrac{\arctan(x)}2$$
Hence, we get that
$$\int x \arctan(x) dx = \dfrac{x^2 \arctan(x)}2 - \dfrac{x}2 + \dfrac{\arctan(x)}2 + c = \dfrac{(1+x^2) \arctan(x)}2 - \dfrac{x}2 + c$$

Second one:

$$\int x^2 e^{-3x} dx = \underbrace{\int x^2 d \left(\dfrac{e^{-3x}}{-3}\right)}_{\displaystyle \int v du} = \overbrace{\dfrac{e^{-3x}}{-3} x^2}^{\displaystyle uv} - \underbrace{\int \dfrac{e^{-3x}}{-3} (2x) dx}_{\displaystyle \int u dv} = \dfrac{e^{-3x}}{-3} x^2 + \dfrac23 \int xe^{-3x} dx$$
Now let us do integration by parts again to evaluate $\displaystyle \int xe^{-3x} dx$.
$$\int xe^{-3x} dx = \underbrace{\int x d \left(\dfrac{e^{-3x}}{-3}\right)}_{\displaystyle \int v du} = \overbrace{x \dfrac{e^{-3x}}{-3}}^{\displaystyle uv} - \underbrace{\int \dfrac{e^{-3x}}{-3} dx}_{\displaystyle \int u dv} = -\dfrac{x e^{-3x}}3 - \dfrac{e^{-3x}}{27}$$
Putting this back, we get that
$$\int x^2 e^{-3x} dx = -\dfrac13 e^{-3x}x^2 - \dfrac29 e^{-3x}x - \dfrac2{27} e^{-3x} + c = - \dfrac{e^{-3x}}3 \left(x^2 + \dfrac{2x}3 + \dfrac2{27}\right) + c$$

Title: Integration by parts help


问:
How to find the dot product using the law of cosines
I'm working with the following problem:

We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$

Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.

$$AC^2=AB^2 +BC^2 -2AB\cdot BC\cos x$$
$$\implies AC^2=3^2 +2^2 -2\cdot2\cdot3\cos 60^\circ$$
$$\implies AC^2=9 +4 -12\cdot0.5$$
$$\implies AC^2=13-6=7$$
$$\implies AC=\sqrt 7.$$

My next step is to calculate the angle $BAC$; we should be able to use the law of cosines here as well:

$$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cos x$$
$$\implies4=9+(\sqrt 7)^2 -2\cdot3\sqrt 7\cos x$$
$$\implies4=9+7 -6\sqrt 7\cos x$$
$$\implies-2=-6\sqrt 7\cos x$$
$$\implies\frac{1}{3}=\sqrt 7\cos x$$
$$\implies\frac{1}{3\sqrt 7}=\cos x.$$

If we want to calculate our dot product using only the vector lengths, we would use the fact that $A\cdotp B=|A||B|\cos x$, which in this case would mean that:

$$AB \cdot AC=3\sqrt7\frac{1}{3\sqrt 7}=1,$$

which is quite wrong since the answer is supposed to be $12$.

Can someone please tell me where I've made a mistake?

In the proposed solution another method seem to be used. The textbook claims that $AB \cdotp AC = AB \cdotp (AB + BC)=AB \cdotp AB + AB \cdotp BC = |AB|^2 + |AB||BC|cos(60)=9+6*\frac{1}{2}=12$

答:
$$\cos\measuredangle BAC=\frac{3^2+(\sqrt7)^2-2^2}{2\cdot3\cdot\sqrt7}=\frac{2}{\sqrt7},$$
which gives $$\vec{AC}\cdot\vec{AB}=3\cdot\sqrt7\cdot\frac{2}{\sqrt7}$$

In your solution this statement is wrong: $-2=-6\sqrt7\cos{x}$.

It should be $$-12=-6\sqrt7\cos{x}.$$
About you last adding.

It should be
$$\vec{AB}\cdotp\vec{AC}=\vec{AB}\cdotp (\vec{AB} + \vec{BC})=\vec{AB}\cdotp\vec{AB}+\vec{AB}\cdotp\vec{BC} =$$
$$=|AB|^2+ |\vec{AB}||\vec{BC}|\cos120^{\circ}=9-6\cdot\frac{1}{2}=6$$


问:
if limit of $a_n$ is $L>0$ then $a_n>0$
I'm trying to prove the following theorem:

Let $a_n$ be a sequence of real numbers. We know that the limit of $a_n$ is $L>0$. Prove: $a_n>0$.

I know that if $\lim\limits_{n\to\infty} a_n=L>0$, so $a_n>\frac{L}{2}>0$ for all $n>n_1$. But it does not help me because, as I understand from the question, I need to show it for all $n\in\mathbb{N}$. How should I prove it?

答: You can't show this result for all $n\in\Bbb N$. Consider $a_1=-1,a_n=2\forall n\ge2$. Then $L=2$ but $a_1\le0$. In fact we can always have finitely many non-positive terms.

Title: if limit of $a_n$ is $L>0$ then $a_n>0$


问:
How to compare the solutions of Jacobi's equations?
If $f$ is the solution of Jacobi's equation
$$f^{\prime\prime}(t) + K(t) f(t) =0,$$ $$  f(0)=0,\quad f^{\prime}(0)=1,$$
where $K$ is a real-valued continous function of $t$, and $f_1$ is the solution of
$$f_1^{\prime\prime}(t) + Cf_1(t)=0 ,$$ $$ f_1(0)=0,\quad f_1^{\prime}(0)=1,$$
where $C$ is a constant, and $K\geq C$.
Then how can I show that $f\leq f_1$?

Body:
Since
$$ f(0)=f_1(0),f'(0)=f'_1(0)=1$$
there is $t^*>0$ such that $f(t)>0, f_1(t)>0$ for $t\in(0,t^*)$. Multiplying the first eqn by $f_1$ and then integrating from $0$ to $t$ give
$$ \int_0^tf''(s)f_1(s)ds+\int_0^tK(s)f(s)f_1(s)ds=0 $$
or
$$ f'(s)f_1(s)\bigg|_0^t-\int_0^tf'(s)f_1'(s)ds+\int_0^tK(s)f(s)f_1(s)ds=0 $$
or
$$ f'(t)f_1(t)-\int_0^tf'(s)f_1'(s)ds+\int_0^tK(s)f(s)f_1(s)ds=0. \tag{1}$$
Similarly
$$ f_1'(t)f(t)-\int_0^tf'(s)f_1'(s)ds+\int_0^tCf(s)f_1(s)ds=0. \tag{2}$$
(1)-(2) gives
$$ f'(t)f_1(t)-f(t)f_1'(t)=\int_0^t(C-K(t))f(s)f_1(s)ds. $$
So for $t\in(0,t^*)$,
$$ \bigg(\frac{f(t)}{f_1(t)}\bigg)'=\frac{1}{f_1^2(t)} \int_0^t(C-K(t))f(s)f_1(s)ds\le0. $$
Namely, $\frac{f(t)}{f_1(t)}$ is decreasing for $t\in(0,t^*)$. Thus for $0<s<t<t^*)$,
$$ \frac{f(t)}{f_1(t)}\le\frac{f(s)}{f_1(s)} $$
Letting $s\to0^+$ gives
$$ \frac{f(t)}{f_1(t)}\le\frac{f'(0)}{f_1'(0)}=1$$
which implies
$$ f(t)\le f_1(t), \text{ for }t\in(0,t^*$). $$


问:
Riemannian connection on product manifolds vanishes
Let $X$ and $Y$ be Riemannian manifolds with metric $g$ and $h$, respectively. Let $(x_1, ..., x_n)$ and $(y_1, ..., y_m)$ be local coordinates on $X$ and $Y$.

Then in local coordinates form $$g = \sum_{i,j = 1}^n g_{(i,j)}(x)dx_i \otimes dx_j \ \mbox{and} \ h = \sum_{s,t=1}^m h_{(s,t)}(y) dy_s \otimes dy_t.$$

Forming a Riemannian manifold $X \times Y$ with the metric $G := g \times h = \sum_{i,j = 1}^n g_{(i,j)}(x)dx_i \otimes dx_j + \sum_{s,t=1}^m h_{(s,t)}(y) dy_s \otimes dy_t.$

Let $F_1 = \sum_{i=1}^n a_i(x)\frac{\partial}{\partial x_i}$ and $F_2 = \sum_{j=1}^m b_j(y)\frac{\partial}{\partial y_j}$ be two vector fields on $X \times Y$.

Then I want to show that $\nabla_{F_1}F_2 = 0, \nabla :=$ the Riemannian connection on $X \times Y.$

Attempt Write $\nabla_{F_1}F_2$ in local coordinate form as $$\nabla_{F_1}F_2 = \sum_{i=1}^n\sum_{j=1}^ma_ib_j\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial y_j} + \sum_{i=1}^n\sum_{j=1}^ma_i\frac{\partial}{\partial x_i}b_j\frac{\partial}{\partial y_j}.$$

Since $b_j = b_j(y) = b_j(y_1, y_2, ..., y_m)$, $\frac{\partial}{\partial x_i}b_j(y) = 0.$

so what left is the first sum, which I am not sure why it is zero ? I guess it might due to symmetry of $\nabla$, but the terms $\nabla_{\frac{\partial}{\partial y_j}}\frac{\partial}{\partial x_i}$ do not present in the sum. I am not sure how to proceed !

答:
You need to calculate the Levi-Civita connection with respect to the product metric. You have not done it yet.

Since
$$ \nabla_{\partial x_i} \partial_ {y_s} = \Gamma^{k}_{is} \partial _{x_k} + \Gamma_{is}^t \partial_{y_t},$$

we need to show
$$\Gamma^{k}_{is} = \Gamma_{is}^t  = 0.$$
This is easy: let $G$ be the product metric, then

\begin{align}
\Gamma^{k}_{is} &= \sum_{l} G^{kl} (-\partial_l G_{is} + \partial_i G_{sl} + \partial_s G_{li}) + \sum_{t} G^{kt} (-\partial_t G_{is} + \partial_i G_{st} + \partial_s G_{ti}) \\
&= \sum_{l} G^{kl} (-\partial_l G_{is} + \partial_i G_{sl} + \partial_s G_{li})  \qquad (\text{since } G^{kt} = 0) \\
&= 0
\end{align}
since $G_{is} = G_{sl} = 0$ and $G_{li}= g_{li}$ is independent of $y$. Similarly you can show that $\Gamma_{is}^t =0$.


问:
Haar measure of $\operatorname{SL}_n(\mathbb{R})$ in terms of the product $K\times A\times N$
I'm misunderstanding a standard argument about the interpretation of the Haar measure on $K\times A\times N$. (See Lemma 2.5 on page 145 of Bekka-Mayer for example.)

Let $G=\operatorname{SL}_n(\mathbb{R})$, $K=\operatorname{SO}(n)$ and let $B\subset G$ denote the upper triangular subgroup.

We have a diffeomorphism $\phi:K\times B \to G$ given by multiplication. Write $dg$ for the (bi-invariant) Haar measure on $G$ and let $dx$ denote the measure on $K\times B$ which pushes forward to $dg$. Alternatively, one may think of them as top forms, $dg$ pulling back to $dx$.
We immediately have that $dx$ is left invariant under the families of maps $(k,b) \mapsto (k_0k,b)$ and $(k,b) \mapsto (k,bb_0)$.

Question: If $dk$ and $db$ denote left and right Haar measures on $K$ and $B$ respectively. Is it then clear that $dx$ must be the product measure $dk\times db$?

$K\times B$ viewed as a product group will certainly have a unique (upto scalar) measure invariant under the maps $(k,b)\mapsto (kk_0,bb_0)$. But does $dx$ satisfy this property?
Or should I be looking at it differently?

答:
Here be the comment of 'reuns' (see above):

From that $dg$ is bi-invariant then $\mu(U\times V)= \int_{U^{-1}V}dg$ is a right Haar measure on $K\times B$?


问:
Need a counterexample of limit of average integral
Assume that $f(x)\geq 0$ is integrable in $[0,1]$ and

$$\lim\limits_{h\to 0^+}\frac{1}{h}\int_{0}^{h}f(x)dx=0$$

I feel that $\lim_{x\to 0}f(x)=0$ is incorrect, but I don’t know the example. I could only think about $1+\sin\frac{1}{x}$, but it seems that it’s not work.

In general, what do we know of $f$ as $x\to 0$ ?

答:
For a continuous function, in the limit $h\to 0$, $$f(h)\approx f(0)$$and you can approximate the integral $$\int_0^h f(x)dx\approx f(0)h$$
Then, $$\lim\limits_{h\to 0}\frac{1}{h}\int_{0}^{h}f(x)dx\approx f(0)$$
So if your function is continuous then $f(0)=0$.


问:
What does it mean when a function is set to zero?
I am reading an article about the Reproduction numbers and I found the next phrase:

If $\mathcal{F}(x)$ is set to zero, then all eigenvalues of $Df (x)$ have
negative real parts.

and my question is what does it mean that a function is set to zero.

答:
In various areas of mathematics, the zero set of a function is the set of all its zeros.

More precisely, if ${\displaystyle f:X\to \mathbb {R} } $ is a real-valued function (or, more generally, a function taking values in some additive group), its zero set is $ f^{-1}(0)$, the inverse image of $\{0\}$ in $X$.


问:
Conformal mapping to upper half plane
I need to find conformal mapping of $U = \{z \in \mathbb{C}:Im(z) >0\}\setminus\{ it:t\in [1;\infty)\}$ to upper half plane. I tried to square $U$ and then use inverse of $w = \frac{1}{2}(z + \frac{1}{z})$ and on paper it seems like i am on the right way, but cannot understand what's wrong. Any hints?

答:
It would be a good idea to embed $U$ into the Riemann sphere $\overline{\mathbb C}$, since it lends itself better to geometrical intuition. See this picture:

The blue line is the real axis, green is the imaginary axis, and red is the unit circle. On this picture, $U$ is the backside of the sphere (with $\mathrm i$ at the center), but with the upper part of the dashed green line removed. Squaring yields $U'=\mathbb C\backslash{(-\infty,-1]\cup[0,\infty)}$. That's the entire sphere, but with 3/4 of the blue circle missing (the arc starting at $0$ and going counterclockwise to $-1$. Which is not far removed from the slashed complex plane ($\mathbb C^-:=\mathbb C\backslash(-\infty,0]$), which is the sphere with a different part of the blue circle missing (the clockwise arc from $\infty$ to $1$). If you can find a conformal mapping from $U'$ to $\mathbb C^-$, and a conformal mapping from $\mathbb C^-$ to the upper half plane, you're done (just compose all of the conformal mappings).

Title: Conformal mapping to upper half plane


问:
Two distinct reduced words are not equivalent
Given a set $X$ and two words on $X$, $w=x_1^{\varepsilon_1} \cdots x_n^{\varepsilon_n}$ and $w'=y_1^{\delta_1} \cdots y_m^{\delta_m}$, we define that $w$ is equivalent to $w'$, $w \sim w'$, if there exists a finite sequence $w_0, w_1, \cdots, w_k$ of words such that $w_0=w$, $w_k=w'$ and for every $i=0, \cdots, k-1$, $w_{i+1}$ is obtained from $w_i$ by applying one of the following operations:

Introduce at the beggining, at the end or between two letters of $w_i$ the empty word or a word of the form $xx^{-1}$ or $x^{-1}x$

Delete from $w_i$ the empty word or a word of the form $xx^{-1}$ or $x^{-1}x$

Now I've proven by induction on the length of the word that every word is equivalent to a reduced word (that's one with no empty word or blocks of the form $xx^{-1}$ or $x^{-1}x$ as subwords) but I'm not able to rigurouly prove that this word is unique, that is, that two distinct reduced words cannot be equivalent.

Could someone give a rigurous proof of this fact?

Thank you.

答:
Suppose that $w_1$ and $w_2$ are equivalent reduced words. We want to prove that $w_1=w_2$.

So there is a path from $w_1$ to $w_2$, where each move in the path consists of either inserting or deleting a word $xx^{-1}$ or $x^{-1}x$.

Let $l$ be the maximum length of any word in this path, and suppose that $k$ words in this path have length $l$. Choose a path with $kl$ minimal. We will derive a contradiction by replacing the path by a different path with a smaller value of $kl$.

If the path has length $0$ then there is nothing to prove, and otherwise there must be successive words $v_1$, $v_2$, $v_3$ in the path where $v_2$ has length $l$ and $v_1$ and $v_3$ have length $l-1$.

So we get $v_2$ from $v_1$ by inserting $xx^{-1}$ somewhere and $v_3$ from $v_2$ by deleting $yy^{-1}$, where $x$ and $y$ are generators or inverses of generators.

If the substrings $xx^{-1}$ of $v_2$ do not overlap, then we can get to $v_3$ by first deleting $yy^{-1}$ and then inserting $xx^{-1}$ thereby reducing $kl$.

Otherwise they overlap. If they are equal then we can just eliminate them both. Otherwise they overlap in a single generator, and $v_2$ contains the substring $xx^{-1}y$ with $y=x$, or $yxx^{-1}$ with $y=x^{-1}$, and in both cases we have $v_1=v_3$, so we can again eliminate both moves, reducing $kl$.

There may be quicker proofs than this, but I like it because it extends to a much more general result in the theory of string rewriting syaytems, which says that local confluence implies confluence.


答:
By this answer the map
$$z\mapsto\sqrt{\frac{z^2}{z^2+1}}$$
achieves the desired objective. It is the successive application of

$z^2$: maps $U$ to $U_1=\mathbb C\setminus((-\infty,-1]\cup[0,\infty))$

$\frac z{z+1}$: maps $U_1$ to $U_2=\mathbb C\setminus[0,\infty)$

$\sqrt z$: maps $U_2$ to upper half-plane

Title: Conformal mapping to upper half plane

Body: I need to find conformal mapping of $U = \{z \in \mathbb{C}:Im(z) >0\}\setminus\{ it:t\in [1;\infty)\}$ to upper half plane. I tried to square $U$ and then use inverse of $w = \frac{1}{2}(z + \frac{1}{z})$ and on paper it seems like i am on the right way, but cannot understand what's wrong. Any hints?


问:
Prove $(A^T)^{-1}$ = $(A^{-1})^T$ whenever $A$ is invertible.
Prove $(A^T)^{-1}$ = $(A^{-1})^T$ for any invertible matrix $A.$

I actually don't know where to start. I do not think I can just apply index laws.

Any help is cool! Thanks.

答:
Try taking the transpose of the equation
$$AA^{-1}=I.$$</p>


问:
Is the limit of a sequence of continuous functions bounded?
Let $(f_n)_{n \in\Bbb{N}}$ be a sequence of continuous functions on $[0,1]$. Suppose $f_n \rightarrow f$ pointwise. Is $f$ bounded? If yes, prove it, if not, give a counter example.

My attempt:

No $f$ is not bounded. Consider $f(x) = \frac{1}{x}$ and $f_n(x) = \frac{1}{x}+\frac{1}{n}$. Clearly $\lim_{n\rightarrow \infty}f_n=f$ and $\lim_{x\rightarrow 0} \frac{1}{x}=\infty$, thus, $f$ is not bounded.

There are some flaws in my counter-example though. The functions should be defined on [0,1] but clearly $f$ is not defined at $0$. I tried to fix this by the following:

$f(x) =
\begin{cases}
1/x,  & x \in(0,1]\\
\infty, &x=0
\end{cases}$

But I don't think this is valid (Would this still be continuous?).

If someone could help fix my counter-example, or show me why $f$ is bounded, that would be greatly appreciated!

答: You can take, for instance,$$f_n(x)=\begin{cases}n^2x&\text{ if }x\in\left[0,\frac1n\right]\\\frac1x&\text{ otherwise.}\end{cases}$$Each $f_n$ is continuous, but $(f_n)_{n\in\Bbb N}$ converges pointwise to$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x=0\\\frac1x&\text{ otherwise.}\end{cases}\end{array}$$


问:
How to calculate the distance from a point on ellipse arc to the chord
Now I have an ellipse-arc starting at $A$ and ending at $B$, so $AB$ is the chord of the ellipse-arc. I want to know how to calculate the maximum distance from the points on the ellipise-arc to the chord.

答: Let $O$ be the centre of the ellipse, $M$ the midpoint of $AB$ and $P$ the intersection between line $OM$ and arc $AB$. The tangent through $P$ is parallel to $AB$, hence you just need to compute the distance from $P$ to line $AB$.


问:
Decomposing a permutation into 3-cycles
I have two exercises I don't know how to solve:

Write the permutation $α = (1 2)(3 4)$ as a product of 3-cycles

Write the permutation $α = (1 2 8 3 7)(4 5 6)$ as a product of 3-cycles

I know how to convert a permutation into a composition of 2-cycles, but not of 3-cycles. I imagine there is a rule for when this is possible, depending on odd/even properties, but since I'm am using the exercises mentioned to help me to learn about permutation parity, I can't yet follow arguments based on that understanding.

Also, I am working with left to right composition such that $(123) = (12)(13)$.

Thanks in advance for any help.

答:
Cristofer already answered your question, so I'll just add the general case since you were wondering if there's a "rule" for when things could be decomposed. This is something I learned from Beardon Algebra and Geometry but surely there are more references.

Theorem. Let $\sigma\in S_n$ be a permutation, $2 \leq m \leq n.$ Then $\sigma$ can be decomposed as a product of $m$-cycles $\iff$ either $m$ is even or $\sigma$ is an even permutation.

Proof. $\implies$ should be easy, as if $\sigma$ is an odd permutation then all the $m$-cycles needs to be odd permutations, so $m$ is even.

$\impliedby$ Case I. If $\sigma$ is an even permutation, then notice $\left(a_{1} a_{2}\right)\left(a_{1} a_{3}\right)=\left(a_{1} a_{2} a_{3} a_{4} \cdots a_{m}\right)\left(a_{m} \cdots a_{4} a_{3} a_{1} a_{2}\right)$ (or, the other way since you are using left-to-right composition) and Cristofer's trick in (2) that $(ab)(cd)=(ab)(bc)(bc)(cd)$ would allow you to write $\sigma$ as a product of $m$-cycles.

Case II. If $\sigma$ is an odd permutation but $m$ is even, then apply Case I to decompose $\sigma (1 2 3 ... m)$ into $m$-cycles which is an even permutation, and multiply $(m (m-1) ... 2 1)$ on the right when you finish gives you $m$-cycle decomposition of $\sigma$.

Title: Decomposing a permutation into 3-cycles


问:
Stronger than Nesbitt's inequality using convexity and functions
Hi it's a refinement of Nesbitt's inequality  and for that, we introduce the function :

$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$

With $a,b,x>0$

Due to homogeneity we assume $a+b=1$ and we introduce the function :

$$g(a)=\frac{a}{1-a+x}$$

Showing that $g(a)$ is convex on $(0,1)$ is not hard so we have :

$$\frac{b}{x+a}+\frac{a}{b+x}\geq 2\frac{a+b}{2(\frac{a+b}{2}+x)}=h(x)$$

So we have :

$$f(x)\geq h(x)+\frac{x}{a+b}$$

Now we put $u=\frac{x}{a+b}$ and we want to show :

$$h(x)+\frac{x}{a+b}=u+\frac{1}{0.5+u}\geq \frac{3}{2}$$

The last inequality is obvious.

My question :

It is correct?

Do you know other refinements?

Thanks in advance!

Ps: I add the tag reference request for the last question.

答:
There are very many refinements of the Nesbitt's inequality.

For example. For positives $a$, $b$ and $c$ we have:

1.$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2};$$
2.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)};$$
3.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3\sqrt{5(a^2+b^2+c^2)-ab-ac-bc}}{4(a+b+c)};$$
4. For any reals $a$, $b$ and $c$ such that $ab+ac+bc>0$ prove that:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$$