答:
A cubic b-spline curve should be $C_2$ continuous at a single knot (multiplicity = 1), and $C_1$ continuous at a double knot (multiplicity = 2).

In your example, all knots have multiplicity = 1, so the spline should be $C_2$ everywhere. So, if you see any slope discontinuities in the curve, you’ve done something wrong.

Title: b-spline of degree 3 interpolation formula

Body:
I know that the recursive formula for a bspline is the following:
$$ B_{i}^k (x)= (\frac{x-t_i}{t_{i+1}-t_i})B_{i}^{k-1}(x) + (\frac{t_{i+k+1} - x}{t_{i+k+1}-t_{i+1}})B_{i+1}^{k-1} (x)$$
Where:
$$
B_{i}^0 (x)=
 \begin{cases}
      1 & x \in [t_i,t_{i+1}] \\
      0 & x \in else \\
\end{cases}
$$
I want to find $B_{0}^3 (x)$, then I would recursively find what $B_{i}^3 (x)$ is and then plug in 0 for i.
Here is the overall answer I get for $B_{0}^3 (x)$ and when I graph it using some of my points, it does not look continuous which is wrong... What am I doing wrong here? I have also provided my intermediary steps of my findings $B_{i}^1 (x)$ and $B_{i}^2(x)$.
$$
B_{i}^1 (x)=
 \begin{cases}
      \frac{x-t_i}{t_{i+1}-t_i} & x \in [t_i,t_{i+1}] \\
      \frac{t_{i+2}-x}{t_{i+2}-t_{i+1}} & x \in [t_{i+1},t_{i+2}] \\
      0 & else
\end{cases}
$$
$$
B_{i}^2 (x)=
 \begin{cases}
      \frac{(x-t_i)^2}{(t_{i+2}-t_i)(t_{i+2}-t_i)} & x \in [t_i,t_{i+1}] \\
      \frac{(x-t_i)(t_{i+2}-x)}{(t_{i+2}-t_i)(t_{i+2}-t_{i+1})} + \frac{(x-t_{i+1})(t_{i+3}-x)}{(t_{i+3}-t_{i+1})(t_{i+2}-t_{i+1})} & x \in [t_{i+1},t_{i+2}] \\
      \frac{(t_{i+3} - x)^2}{(t_{i+3}-t_{i+1})(t_{i+3}-t_{i+2})} & x \in [t_{i+2},t_{i+3}] \\
      0 & else
\end{cases}
$$
So, for $B_{i}^3 (x)$ I have the following:
$$
B_{i}^2 (x)=
 \begin{cases}
      \frac{(x-t_i)^3}{(t_{i+3}-t_i)(t_{i+2}-t_{i})(t_{i+1}-t_{i})} & x \in [t_i,t_{i+1}] \\
      \frac{(x-t_{i})^2(t_{i+2}-x)}{(t_{i+3}-t_{i})(t_{i+2}-t_i)(t_{i+2}-t_{i+1})}+\frac{(x-t_{i+1})^2(t_{i+3}-x)}{(t_{i+3}-t_{i})(t_{i+3}-t_{i+1})(t_{i+2}-t_{i+1})}+\frac{(x-t_{i+1})^2(t_{i+4}-x)}{(t_{i+3}-t_{i+1})(t_{i+2}-t_{i+1})(t_{i+4}-t_{i+1})} & x \in [t_{i+1},t_{i+2}] \\
      \frac{(t_{i+3}-x)^2(x-t_i)}{(t_{i+3}-t_{i})(t_{i+3}-t_{i+1})(t_{i+3}-t_{i+2})} + \frac{(x-t_{i+1})(t_{i+3}-x)(t_{i+4}-x)}{(t_{i+3}-t_{i+1})(t_{i+3}-t_{i+2})(t_{i+3}-t_{i+1})} + \frac{(x-t_{i+2})(t_{i+4}-x)^2}{(t_{i+4}-t_{i+2})(t_{i+3}-t_{i+2})(t_{i+4}-t_{i+1})}  & x \in [t_{i+2},t_{i+3}] \\
      \frac{(t_{i+4}-x)^3}{(t_{i+4}-t_{i+1})(t_{i+4}-t_{i+2})(t_{i+4}-t_{i+3})} & x \in [t_{i+3},t_{i+4}] \\
      0 & else
\end{cases}
$$
Plugging in i = 0 to get $B_{i}^3 (x)$ I have the following:
$$
B_{i}^3 (x)=
 \begin{cases}
      \frac{(x-t_0)^3}{(t_3-t_0)(t_2-t_0)(t_1-t_0)} & x \in [t_i,t_{i+1}] \\
      \frac{(x-t_0)^2(t_2-x)}{(t_3-t_0)(t_2-t_0)(t_2-t_1)}+\frac{(x-t_1)^2(t_3-x)}{(t_3-t_0)(t_3-t_1)(t_2-t_1)}+\frac{(x-t_1)^2(t_4-x)}{(t_3-t_1)(t_2-t_1)(t_4-t_1)} & x \in [t_{i+1},t_{i+2}] \\
      \frac{(t_3-x)^2(x-t_0)}{(t_3-t_0)(t_3-t_1)(t_3-t_2)}+\frac{(x-t_1)(t_3-x)(t_4-x)}{(t_3-t_1)(t_3-t_2)(t_4-t_1)}+\frac{(t_4-x)^2(x-t_2)}{(t_4-t_2)(t_3-t_2)(t_4-t_1)}  & x \in [t_{i+2},t_{i+3}] \\
      \frac{(t_4-x)^3}{(t_4-t_1)(t_4-t_2)(t_4-t_3)} & x \in [t_{i+3},t_{i+4}] \\
      0 & else
\end{cases}
$$
Taking the example knots: {-1, -0.75, 0, 0.5, 1} and drawing the graph, we see that there is a discontinuity at point 0. Did I miss an index in the spline or something?  but if i change the knots a bit, the function becomes continuous. Isn't it a characteristic of bsplines to be continuous, or is it dependent on the knots I choose? If so, would this mean that my discontinuous spline is valid?


问:
Problem with composite relations
The composition $\mathcal{R_2}\circ \mathcal{R_1}$ of the relations $\mathcal{R_1}$ and $\mathcal{R_2}$ is defined as follows:

$\mathcal{R_2}\circ \mathcal{R_1}:=\{(x,z)\ |\exists\ y(x\mathcal{R_1}y\land y\mathcal{R_2}z)\}$.

In particular if $\mathcal{R_1}\subset X\times Y $ and $\mathcal{R_2}\subset Y\times Z$, then $\mathcal{R}=\mathcal{R_2}\circ \mathcal{R_1}\subset X\times Z$, and

$x\mathcal{Rz}:=\exists\  y\ ((y\in Y)\land(x\mathcal{R_1y)\land(y\mathcal{R_2z}))}$.

Let $\Delta_X$ be the diagonal of $X^2$ and $\Delta_Y$ be the diagonal of $Y^2$. Show that if the relations $\mathcal{R_1}\subset X\times Y$ and $\mathcal{R_2}\subset Y\times X$ are such that $(\mathcal{R_2}\circ \mathcal{R_1}=\Delta_X)\land(\mathcal{R_1}\circ \mathcal{R_2}=\Delta_Y)$, then both relations are functional and define mutually inverse mappings of $X$ and $Y$.

My working thus far:

$(x,x)\in \Delta_X\Rightarrow (x,x)\in \mathcal{R_2}\circ \mathcal{R_1}\Rightarrow \mathcal{R_2}\circ \mathcal{R_1}\subset X^2$.

Similarly,

$(y,y)\in \Delta_Y\Rightarrow (y,y)\in \mathcal{R_1}\circ \mathcal{R_2}\Rightarrow \mathcal{R_1}\circ \mathcal{R_2}\subset Y^2$.

This is the furthest I got. I am not sure if I have any errors in my working and how to show that the relations are functional.

答:
Interesting to see that this can be also answered in a different way, in which we avoid mentioning elements as far as possible and argue more algebraically.

We are given $\mathcal{R}_1 \subseteq X \times Y$ and that
$\mathcal{R}_2 \subseteq Y \times X$. We are also given that $\mathcal{R}_2 \circ \mathcal{R}_1 = \Delta_X$ and $\mathcal{R}_1 \circ \mathcal{R}_2 = \Delta_Y$. From these we have to show that $\mathcal{R}_1$ and $\mathcal{R}_2$ are functional and inverse to each other.

Note that $\mathcal{R}_1$ being functional is equivalent to
$$\mathcal{R}_1 \circ \mathcal{R}_1^{-1} \subseteq \Delta_Y.$$
Informally: imagine tracing backwards along $\mathcal{R}_1$ from an element $y \in Y$ to some $x \in X$ and then forwards from $x$ to any $y' \in Y$. Being functional means that whenever we do this we have to have $y = y'$, that is, $(y,y') \in \Delta_Y$.

Similarly, $\mathcal{R}_2$ being functional is equivalent to
$$\mathcal{R}_2 \circ \mathcal{R}_2^{-1} \subseteq \Delta_X.$$
We also need to know that the diagonal is the identity relation for composition,
so that
$$\Delta_Y \circ \mathcal{R}_1 \circ \Delta_X = \mathcal{R}_1, \text{ and } \Delta_X \circ \mathcal{R}_2 \circ \Delta_Y = \mathcal{R}_2.$$
We do need this property which holds for an arbitrary relation $\mathcal{R}$:
$$\mathcal{R} \subseteq \mathcal{R} \circ \mathcal{R}^{-1} \circ \mathcal{R}.$$
Now, everything falls out very neatly:
$$\begin{array}{rcl}
\mathcal{R}_1 \subseteq \mathcal{R}_1 \circ \mathcal{R}_1^{-1} \circ \mathcal{R}_1
& \Rightarrow &
\mathcal{R}_1 \circ \mathcal{R}_2 \subseteq
\mathcal{R}_1 \circ \mathcal{R}_1^{-1} \circ \mathcal{R}_1 \circ \mathcal{R}_2\\
& \Rightarrow &
\Delta_Y \subseteq
\mathcal{R}_1 \circ \mathcal{R}_1^{-1} \circ \Delta_Y \\
& \Rightarrow &
\mathcal{R}_2 \circ \Delta_Y \subseteq \mathcal{R}_2 \circ
\mathcal{R}_1 \circ \mathcal{R}_1^{-1} \\
& \Rightarrow &
\mathcal{R}_2 \subseteq \Delta_X \circ \mathcal{R}_1^{-1} \\
& \Rightarrow & \mathcal{R}_2 \subseteq \mathcal{R}_1^{-1}
\end{array}$$
For exactly the same reason $\mathcal{R}_1 \subseteq \mathcal{R}_2^{-1}$, so taking inverses of the relations, $\mathcal{R}_1^{-1} \subseteq \mathcal{R}_2$ and hence (using the fact that $\mathcal{R}_2 \subseteq \mathcal{R}_1^{-1}$) we get $\mathcal{R}_2 = \mathcal{R}_1^{-1}$ and $\mathcal{R}_1 = \mathcal{R}_2^{-1}$.

The property of being functional can be justified explicitly by noting:
$$\mathcal{R}_1 \circ \mathcal{R}_1^{-1} = \mathcal{R}_1 \circ \mathcal{R}_2 = \Delta_Y,
\text{ and } \mathcal{R}_2 \circ \mathcal{R}_2^{-1} = \mathcal{R}_2 \circ \mathcal{R}_1 = \Delta_X.$$

ContentLicense: CC BY-SA 3.0

Title: Problem with composite relations

ContentLicense: CC BY-SA 4.0


问:
Finding side and angle of isosceles triangle inside two circles
I'm having a problem that I'm not sure how to solve (or if it's even possible). It's not homework, just something I'm struggling with for a project. :)

Basically, there are two circles, represented here by the red-ish lines, that have the same center point $E$.
The radii of the circles are not known, these are the distances $AE$ and $CE$.

Distances $AE$ and $BE$ are identical.

Distances $CE$ and $DE$ are identical.

Hence, distances $AC$ and $BD$ are identical.

Known metrics:

$AC = BD = 10$ meters.

$AD = 200$ meters.

From $A$ and $B$ are tangent lines, marked as blue. From the tangent line that is connected by point $A$ to the line $AD$, the angle is $8^\circ$.

The problem here is: How do I find the radius of any of the circles? I'm thinking setting up an isosceles triangle is the way to move forward, but I'm stuck. Can the distance $CD$ and angle $CDE$ be found to solve the sides of the triangle? Is there another way of finding the radii that I haven't thought of?

答:
Consider the figure below:

Let's calculate $CD$ using the Law of cosines in $\triangle CAD$:
$$ CD^2=10^2+200^2-2 \cdot 10 \cdot 200 \cdot \cos 82^{\circ} \Rightarrow$$
$$\Rightarrow CD \approx 198.855 \, \mathrm{m}.$$
Now using the Law of cosines in $\triangle ACD$, let's calculate $\cos \alpha$:
$$ 200^2=10^2+CD^2-2 \cdot 10 \cdot CD \cdot \cos \alpha \Rightarrow$$
$$\Rightarrow \cos \alpha \approx -0.0896857.$$
Note that $\triangle CEF$ is a right-angled triangle, $CE=r$, and $CF=\frac{CD}{2}$, so:
$$\cos(180^{\circ} - \alpha)= \frac{CF}{CE} \Rightarrow$$
$$\Rightarrow -\cos \alpha =\frac{CD}{2r} \Rightarrow$$
$$\Rightarrow r = \frac{CD}{-2\cos \alpha} \Rightarrow$$
$$\Rightarrow r \approx 1108.6 \, \mathrm{m}.$$


问:
Going from a discrete model to a continuous one in wsn
We have a WSN (Wireless sensor network) deployed in an agricultural field for precision agriculture.

The sensors gathers some physical parameters (Temperatures, PH ...) essential for the automated treatment of this field (irrigation and so on).

We have a constraint in the number of sensors deployed as it doesn't make up all the surface ( let's say we have 20 sensors deployed in a 300x300m² surface)

When collecting data I want to be able to process it through some machine learning (or something else) so to transit from a discrete model (20 samples distributed in 300x300) to a continuous one (prediction over all the surface).

Do you have any idea how it could be done ? or is it even something doeable ?

PS1: This is not Time series forecasting, prediction of temperatures between dates of collections isn't the goal.

PS2: The data set I found Data package for automated in situ soil sensor network, provide the data of sensors in 42 instrumented locations in the field, so we can't use regular supervised regression methods, as we don't have data over all the surface to train our model.

答:
So you want to interpolate between sensors.  The simplest method, and there does not seem to be any reason to use anything more complicated, is "linear interepolation".  Any point will be closest to 4 of the sensors.  Let's say there are sensors at (a, a) which gives reading A, (a+b, a) which gives reading G, (a, a+b) which gives reading C, and (a+b, a+b) (b is the distance between sensors) which gives reading D.  Now suppose the point (x,y) lies within those four points.  The distance from (x,y) to (a, a) is $\sqrt{(x-a)^2+ (y-a)^2}= D_1$.  The distance from (x,y) to (a+b, a) is $\sqrt{(x- a-b)^2+ (y-a)^2}= D_2$.  The distance from (x,y) to (a, a+b) is $\sqrt{(x-a)^2+ (y-a-b)^2}= D_3$.  And the distance from (x,y) to (a+b,a+b)  is $\sqrt{(x-a-b)^2+ (y-a-b)^2}= D_4$.

The "weighted average" at (x,y) is $\frac{AD_1+ BD_2+ CD_3+ DD_4}{D_1+ D_2+ D_3+ D_4}$.


问:
Representation of numbers in Cantor set as ternary expansion - in need of a rigorous proof
Background Facts

It is known that every number in the Cantor set $\mathcal{C}$ has a ternary expansion,
where $a_k \in \{0,2\}$:

$$
x=\sum_{k=1}^{\infty} a_{k} 3^{-k}
$$

Cantor set has the representation

$$\mathcal{C}=[0,1] \backslash \bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^{n}-1}\left(\frac{3 k+1}{3^{n+1}}, \frac{3 k+2}{3^{n+1}}\right)$$

Problem

What exactly is the relationship between (2.) and (1.)? That is, how should one go about proving that

$$ x \in \mathcal{C} \iff x=\sum_{k=1}^{\infty} a_{k} 3^{-k}\, \,\,\,\,\,\,,a_k \in \{0,2\}$$

Progress

Following the proof which appears in Representation of elements in Cantor set., I have problem understanding why if $a_{k} \neq 0$ for some $k>n,$ then $x \in\left(\frac{3 k+1}{3^{n}}, \frac{3 k+2}{3^{n}}\right)$ for some $n .$

I can see that $x \geq \frac{1}{3^n} + \frac{1}{3^k}$, but how can one prove $x$ indeed lies in the given interval?

答:
Definition. Let $[x]$ be the largest integer not exceeding $x$ and let {$x$}$=x-[x].$

Def'n. Let $D =[0,1]\setminus C,$ according to your representation 2. of the Cantor set.

Def'n. Let $E$ be the set of all $x\in [0,1]$ that $must$ have a digit $1$ in ternary.

(1). Show that $(1/3,2/3)\subset E.$ (Easy. The digit immediately right of the "decimal" point $must$ be $1.$)

(2). If $n>0$ and $0\le k\le 3^n-1$ and if $(3k+1)/3^{n+1}<x<(3k+2)/3^{n+1}$ then $x\in E.  $ Because if $x\not\in E$ then {$3^nx$}$\not\in E.$ But {$3^nx$} $\in (1/3,2/3)$ and we have $(1/3,2/3)\subset E$  by (1).

(3). By (1) and (2) we have $D\subseteq E.$

(4). If $x\in E$ let $x=\sum_{n=0}^{\infty}x_n/3^{n+1},$ where each $x_n\in \{0,1,2\}.$ Now let $n^*$ be the least $n$ such that $x_n=1.$ It cannot be that $x_n=0$ for every $n>n^*,$ nor that $x_n=2$ for every $n>n^*,$ otherwise $x$ would have another ternary presentation without the digit $1.$ So $$(\bullet)\quad 0<x-\sum_{n=0}^{n^*}x_n/3^{n+1}<1/3^{n^*+1},\text { and also  }x_{n^*}=1$$

Now let $k=3^{n^*}\sum_{n=0}^{n^*-1}x_n/3^{n+1}$ if $n^*>0$, or let $k=0$ if $n^*=0$. By $(\bullet)$ we have $0\le k<3^{n^*}$ and also $0<x-(3k+1)/3^{n^*+1}<1/3^{n^*+1},$ implying $x\in D.$

So $E\subseteq D.$

(5). From (3) and (4) we have $D\subseteq E\subseteq D.$


问:
Problem with system of equations $(x,y,z)$
System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:

$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$

$x^{2}+y^{2}+z^{2}=xy+yz+zx$

$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$

$(x+y+z)^{2}-2*3=3$

$(x+y+z)^{2}=9$

What should I do next?

答: By the calculations you have made, your system is actually eqivalent to $$\begin{cases}x^2+y^2+z^2=3\\(x+y+z)^2=9\end{cases}$$Which is, at the same time, equivalent to the systems $$\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases} \quad \begin{cases}x^2+y^2+z^2=3\\x+y+z=-3\end{cases}$$But you can check using the distance formula between a point and a plane (or in many other ways) that the planes $x+y+z=3$ and $x+y+z=-3$ are both tangent to the sphere $x^2+y^2+z^2=3$, so the only solutions are $x=y=z=\pm1$


问:
Taylor expansion for $1/(1+x^2)$
In the book Short Calculus The author, in order to prove the taylor series for the arctangent, he presented the taylor expansion of $$\frac{1}{1+x^2}, x\in \mathbb R$$

Which is (the one of the author):$$\frac{1}{1+x^2} =1-x^2+x^4-x^6+...+(-1)^{n-1}x^{2n-2}+(-1)^{n}\frac{x^{2n}}{1+x^2}$$

But using the long division,i’ve found a slightly different one:
$$\frac{1}{1+x^2} =1-x^2+x^4-x^6+...+(-1)^{n}x^{2n}+...$$

Which one is correct? This is an important thing because it will change the remainder term of the taylor expansion for the arctangent.

Title: Taylor expansion for $1/(1+x^2)$

Body:
\begin{align}\frac1{1+x^2}&=\sum_{i=0}^\infty (-x^2)^i
\\&=  \sum_{i=0}^{n-1} (-x^2)^i +\sum_{i=n}^\infty (-x^2)^i \\
&=\sum_{i=0}^{n-1} (-x^2)^i +(-x^2)^n\sum_{i=0}^\infty (-x^2)^i \\
&= \sum_{i=0}^{n-1} (-x^2)^i +\frac{(-x^2)^n}{1+x^2}\end{align}

You can also view it as:

$$\frac{1-(-x^2)^n}{1-(-x^2)}=\sum_{i=0}^{n-1}(-x^2)^i$$

where you can replace $-x^2$ by $r$, and get the geometric sum.


问:
G is finite group such that H is normal and P is a Sylow p-subgroup of H
So, suppose that $G$ is a finite group and $H$ is normal subgroup and $P$ is a Sylow $p-$subgroup of $H$. Then prove that there exist $Q$ a Sylow $p-$subgroup of $G$ such that $Q\leq N_G(P)$. Honestly, I don't know how tackle the problem. Just to know that $[G:N_G(P)]=[H:N_H(P)]$ :-(</p>

答: Let $Q$ be a Sylow $p$-subgroup of $N_G(P)$ containing $P$ (using the Sylow containment theorem in $N_G(P)$). We claim that $Q$ is also a Sylow $p$-subgroup of $G$, that is, that $[G:Q]$ is not divisible by $p$. Indeed, $[G:Q] = [G:N_G(P)][N_G(P):Q] = [H:N_H(P)] [N_G(P):Q]$. The first factor $[H:N_H(P)]$ is not divisible by $p$ since $P$ is a Sylow $p$-subgroup of $H$. The second factor $[N_G(P):Q]$ is not divisible by $p$ since $Q$ is a Sylow $p$-subgroup of $N_G(P)$. Since $p$ is prime and divides neither factor, it does not divide their product, $[G:Q]$.</p>


问:
Pool board question about trajectories of balls!

In this problem we are dealing with rectangular pool boards.
Our boards have pointwise holes at the corners and nowhere else. A pointwise pool ball is running on the board and it never stops but falls down at a corner when it arrives there.
If the ball starts its run at the left bottom corner at angle 45o to both of the adjacent sides of a 4×6 board then its path is shown below.

(I.) To which corner the ball arrives?

(II.) What is the total number of cushions until it falls down at the corner?

(III.) What is the total length of its path? (Through how many 1×1 squares does it go through?)

Try to answer these questions in the following cases:

a) 3 × 5;

b) 4 × 8;

c) 15 × 18;

d) 2021 × 235;

e) 2021 × 900.

f) Try to answer questions (I.) , (II.) , (III.) in the case of an m×n board
in terms of m and n where m and n are positive integers.

I have got no idea how to solve any of these!

Would you mind telling me how I can begin to understand and solve this question?

Cheers!

答:
This is not an answer, more like a remark that may or may not be helpful.  It's a way of diagramming the problem.

Consider the case of a billiard table $6$ units long and $4$ units high.  In the following figure every point where the $45^{\circ}$ line crosses one of the dotted lines corresponds to a bounce off a cushion, so the ball bounces off $3$ cushions before dropping into a corner pocket.


问:
If $X_n \to X $ in probability then $f(X_n) \to f(X)$ in probability for a Borel function $f$
$(X_n)_n$ is a sequence of identically distributed random variables, $f:\mathbb{R} \to \mathbb{R}$ a Borel function.

Prove that if $X_n$ converges in probability to $X,$ then $f(X_n)$ converges in probability to $f(X).$

答:
This is an alternative proof for the problem, where we will suppose that $X,X_n$ take values in a separable metric space $(U,d),f:U \to \mathbb{R}$ measurable function.

Consider $\mathcal{K}=\{E \in \mathcal{B}(U),1_E(X_n) \to^P1_E(X)\}.$ We will prove that $\mathcal{K}$ is a $\sigma$-algebra containing $\mathcal{B}(U).$

Clearly $U \in \mathcal{K}$ and $\mathcal{K}$ is a $\pi$-system. Let $E,F \in \
\mathcal{K}$ such that $E \subset F.$ Write $1_{F-E}=1_F-1_E$ to conclude that $F-E \in \mathcal{K}.$ Taking an increasing sequence $(E_n)_n$ in $\mathcal{K}$ and letting $E=\bigcup_nE_n.$

So, $ \lim_k1_{E_k}=1_E.$ Notice that for all $n,k \in \mathbb{N}$ $$P(|1_E(X_n)-1_E(X)|>\epsilon) \leq P(|1_E(X_n)-1_{E_k}(X_n)|>\epsilon/3)+P(|1_{E_k}(X_n)-1_{E_k}(X)|>\epsilon/3)+P(|1_{E_k}(X)-1_E(X)|> \epsilon /3)$$
In other word, using the fact that $X_n$ are identically distributed,
$$P(|1_E(X_n)-1_E(X)|>\epsilon) \leq P(|1_{E_k}(X_n)-1_{E_k}(X)|>\epsilon/3)+2P(|1_{E_k}(X)-1_E(X)|> \epsilon /3)$$
so $\limsup_n P(|1_E(X_n)-1_E(X)|>\epsilon)=0$ and $E\in \mathcal{K}.$

So $\mathcal{K}$ is a $\sigma$-algebra.

If $E$ is a closed of $U,$ then by considering $f_k(x)=\frac{1}{(1+d(x,E))^{k}}$ (which is continuous on $U$ and $\lim_k f_k=1_E$) then $$P(|1_E(X_n)-1_E(X)|>\epsilon) \leq P(|1_E(X_n)-f_k(X_n)|>\epsilon/3)+P(|f_k(X_n)-f_k(X)|>\epsilon/3)+P(|f_k(X)-1_E(X)|> \epsilon /3)$$

so, $\limsup_nP(|1_E(X_n)-1_E(X)|>\epsilon)=0$ this means that $E \in \mathcal{K}$ and $\mathcal{B}(U)=\mathcal{K}.$

Now if $f:U \to \mathbb{R}$ is a measurable function, then it can be approximated by a sequence of simple function $\phi_k$ and from the above for all $k \in \mathbb{N},\phi_k(X_n) \to^P \phi_k(X)$ and again since $$P(|f(X_n)-f(X)|>\epsilon) \leq P(|f(X_n)-\phi_k(X_n)|>\epsilon/3)+P(|\phi_k(X_n)-\phi_k(X)|>\epsilon/3)+P(|\phi_k(X)-f(X)|>\epsilon/3),$$ so $\limsup_nP(|f(X_n)-f(X)|>\epsilon)=0.$

In this proof we didn't need to use the continuity on a compact (uniform continuity).

Separability is supposed just to ensure that $(X,X_n)$ is indeed a random variable $(\mathcal{B}(U^2)=\mathcal{B}(U) \otimes \mathcal{B}(U))$

If $f(X_1) \in L^p$ and $X_n \to ^{L^p} X$ then $f(X_n) \to ^{L^p} f(X).$

The statement is not true for almost sure convergence.


问:
Each integer has one of three colors. Prove that there exists two distinct integers of the same color with the difference between them being a square.

Each integer has one of three colors. Prove that there exists two distinct integers of the same color with the difference between them being a square.

Pretty sure this can be proved using the pigeon hole principle, but I don't know how to prove it?

答:
Just a guess. I guess you need to find 4 distinct integers $a \gt b \gt c \gt d$ such that the difference between every 2 of them is a square. Then since you have just 3 colors, that  would prove the statement. I just don't know yet how easy it is to find such 4 integers. Maybe one can construct something using certain Pythagorean triples, or some squares of numbers, and also the number 0.

But if this problem is about pigeon hole principle, that seems like the most obvious idea.

EDIT:

Yeah, this idea works here. Here are 4 such integers:

$697^2, 185^2, 153^2, 0^2$

The difference between any two of them is a square.

One can show more examples. But one example is enough for this problem.

Here are a few more examples:

$697^2,  185^2,  153^2,  0^2$

$697^2,  680^2,  672^2,  0^2$

$925^2,  533^2,  520^2,  0^2$

$925^2,  765^2,  756^2,  0^2$


问:
Change of basis matrix for inner product space
If $B = (e_1,e_2,\ldots, e_n)$ is a basis for an inner product space $V$ and  $B' = (f_1,f_2,\ldots,f_n)$ is an orthonormal basis of $V$.

Is the change of basis matrix $P$ necessarily orthogonal?

答:
The change basis matrix $P$ is
$$P=\left(f_1 f_2\cdots f_n\right)$$
where $f_i$ is a column vector and since the basis $B'$ is orthonormal we have
$$f_i \ . f_j=\delta_{ij}$$
where $\delta$ is the Kroneker symbol and hence $P$ is orthogonal</p>


答:
For the system $\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases}\quad$ just offset the variables $\begin{cases}x=1+X\\y=1+Y\\z=1+Z\end{cases}$

Then you get $\begin{cases}3+X+Y+Z=3\\3+2X+2Y+2Z+X^2+Y^2+Z^2=3\end{cases}\implies X^2+Y^2+Z^2=0$

Which in the real domain has only the $(0,0,0)$ solution.

For $x+y+z=-3$ proceed similarly with $x=X-1,y=Y-1,z=Z-1$.

Title: Problem with system of equations $(x,y,z)$

Body:
System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:

$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$

$x^{2}+y^{2}+z^{2}=xy+yz+zx$

$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$

$(x+y+z)^{2}-2*3=3$

$(x+y+z)^{2}=9$

What should I do next?


问:
Monotonicity of the Variance
Given a random variable X and two functions that satisfy: $|f(X)| \leq |g(X)| $ and $Var(f(X)) \neq 0 $ , $Var(g(X)) \neq 0 $.

Is $Var(f(X)) \leq Var(g(x)) $ ?

Trying to prove this I noticed that if $E(f(X)) = E(g(X)) = 0$  then the statement is true but I'm not sure about the general case.

Thank you!

答:
Let $g$ be the constant function $g(x)=100$ and let $$f(x)=\begin{cases} 0 &, x = 0 \\ 1 &, x =1\end{cases}$$

Let $X$ follows Bernoulli($0.5)$., then $Var(f(X)) > 0$ but $Var(g(X))=0$.

Edit since the condition that non-zero variance is added:

We have $Var(f(X))=\frac14$

let $$g(x) = \begin{cases} 2+\epsilon &, x = 0 \\ 2-\epsilon&, x=1 \end{cases}$$

$$Var(g(X))=Var(g(X)-(2-\epsilon))=\frac{4\epsilon^2}{4}=\epsilon^2$$

We can choose $\epsilon$ to be arbitrarily small.

Title: Monotonicity of the Variance


答:
The expansion is based on the high-school identity:
$$1-x^n=(1-x)(1+x+x^2+\dots+x^{n-1})$$
which can be rewritten as
$$\frac 1{1-x}=1+x+x^2+\dots+x^{n-1}+\frac{x^n}{1-x}.$$
Now substitute $-x^2$ to $x$ in this formula, and you obtain exactly the formula in the book.

Other than that, you also can obtain the same result, not by long division, but by the division along increasing powers  of $1$ by $1+x^2$. As an example, here is this division up to order $8$:
\begin{array}{rcl}
&&\phantom{-}1-x^2+x^4-x^6 \\
1+x^2&\Bigl(&\phantom{-}\not1 \\
&& -\not1-x^2 \\
&& \phantom{-1}\enspace+x^2 +x^4\\
&& \phantom{-1\enspace+x^2} -x^4- x^6\\
&& \phantom{-1\enspace+x^2 -x^4}+ x^6+x^8\\
&& \phantom{-1\enspace+x^2 -x^4+ x^6}+x^8\\
\end{array}
So the remainder at order $8$ is $x^8$, and this division means that
$$1=(1-x^2+x^4-x^6)(1+x^2)+x^8\implies\frac 1{1+x^2}=1-x^2+x^4-x^6+\frac{x^8}{1+x^2}.$$

Title: Taylor expansion for $1/(1+x^2)$

Body:
In the book Short Calculus The author, in order to prove the taylor series for the arctangent, he presented the taylor expansion of $$\frac{1}{1+x^2}, x\in \mathbb R$$

Which is (the one of the author):$$\frac{1}{1+x^2} =1-x^2+x^4-x^6+...+(-1)^{n-1}x^{2n-2}+(-1)^{n}\frac{x^{2n}}{1+x^2}$$

But using the long division,i’ve found a slightly different one:
$$\frac{1}{1+x^2} =1-x^2+x^4-x^6+...+(-1)^{n}x^{2n}+...$$

Which one is correct? This is an important thing because it will change the remainder term of the taylor expansion for the arctangent.


问:
When $\left(W_1^{\perp}+W_2^{\perp}\right)^{\perp}=W_1+W_2$ is true?
I was trying to figure out a theorem about $\left(W_1^{\perp}+W_2^{\perp}\right)^{\perp}=W_1+W_2$. I know that $(W^{\perp})^{\perp}=W$ and I also know that $W_1^\perp + W_2^\perp = (W_1 \cap W_2 )^\perp$.  But what condition do we need to have so $\left(W_1^{\perp}+W_2^{\perp}\right)^{\perp}=W_1+W_2$ is true?

答: With the mentioned rules we get $W_1\cap W_2=W_1+W_2$ and therefore the equivalent condition $W_1=W_2$.


答:
Slightly different ??

Both Taylor expansions are

$$\sum_n(-1)^nx^{2n} $$ (summation limits omitted on purpose).

Your expression is just missing the remainder term.

Title: Taylor expansion for $1/(1+x^2)$


问:
For a specific $v$ in $V$, let $p(x)$ be the monic polynomial such that $p(T)v=0$. Then $p$ divides the minimal polynomial

Before starting problem, I was wondering why we assume that the degree of $p$ is smaller or equal to the degree of the minimal polynomial. Intuitively, it makes a lot of sense since the minimal polynomial applied to all vectors in $V$ is $0$ where as $p(T)$ was specifically constructed just so $p(T)v=0$, so $p$ does not need to carry as much information as the minimal polynomial, but how do we rigorously show this?

答: Let $f$ be the minimal polynomial of $T$. Then $f$ is monic and $f(T)v = 0$, so, as $p$ is the monic polynomial with minimal degree such that $p(T)v = 0$, $\deg(p) \le \deg(f)$. Note the fact $(*)$ that any non-zero polynomial $g$ such that $g(T)v = 0$ has $\deg(g) \ge \deg(p)$ (because we can turn $g$ into a monic polynomial with the same property by dividing it by its leading coefficient). Using polynomial division, we know that there are polynomials $q$ and $r$, such that $f = qp + r$ and either $r = 0$ or $\deg(r) < \deg(p)$. But $r(T)v = f(T)v - (q(T)p(T))v = 0$, so if $r \neq 0$, this contradicts fact $(*)$.


问:
Relationship between the Reeb field of a contact manifold and the Reeb field of a contact submanifold
Suppose we have a contact submanifold $i:(S,\xi_S) \hookrightarrow (M,\xi)$. This means that if $\xi = \ker \alpha$, then $\xi_S = \ker i^* \alpha$.

What is the Reeb vector field $X_S$ on $S$ and how does it relate to the Reeb vector field $X_M$ of $M$? I thought we might try studying a tubular neighborhood of $S$ viewed as the normal bundle: $\pi:N_S \to S$ and $i:S \hookrightarrow N_S$ is the zero section. Then $X_M \in TN_S = TS \oplus V$ where $V$ are the vertical fibers; we can project $\pi_* X_M \in TS$ and I thought perhaps this should be $X_S$.

However, $i^*\alpha(\pi_*X_M) = (i \circ \pi)^* \alpha(X_M)$. Unfortunately, $\pi \circ i = \text{id}$ but $i \circ \pi \neq \text{id}$.

答:
In general, I believe there exists no easy relation between the Reeb vector field of a contact submanifold and the Reeb vector field of the ambient contact manifold.

Example. Let $(M,g)$ be a Riemannian manifold, then $STM$, which is the unit tangent bundle of $(M,g)$, is canonically a co-oriented contact manifold whose Reeb flow is the geodesic flow on $M$. Similarly, if $N$ is a submanifold of $M$, then $STN$ is a contact submanifold of $STM$ whose Reeb flow is the geodesic flow on $N$. However, unless $N$ is totally geodesic, it is difficult to relate the geodesic flow on $N$ and the geodesic flow on $M$, so that it is illusionary to express the Reeb vector field of $STN$ in terms of the Reeb vector field of $STM$.


问:
How can you prove that the square root of two is irrational?
I have read a few proofs that $\sqrt{2}$ is irrational.

I have never, however, been able to really grasp what they were talking about.

Is there a simplified proof that $\sqrt{2}$ is irrational?

答:
Proof of a more general statement:-

I am going to prove the following statement using proof by contradiction:

Square root of a non-perfect square number is irrational.

So let's start,

Let $N$ be a non-perfect square and let $a,b\in\mathbb{Z}$. So the statement becomes

$\sqrt{N}=\frac{a}{b}$ and assume $gcd(a,b)=1$

$\implies N=\frac{a^2}{b^2}$

$\implies a^2=Nb^2$

Which means that $N$ is a perfect square which contradicts our assumption.

$\therefore \sqrt{N}\neq\frac{a}{b}$.QED

Answer to your question:-

By the proof above, If we let $N=2$

Then $\forall a,b\in\mathbb{Z},\sqrt{2}\neq\frac{a}{b}$

ContentLicense: CC BY-SA 2.5


问:
Simplifying equation of a cone's volume, radius, height when height equals $2\cdot \rm{radius}$
I have a cone shape that has $2\cdot \rm{radius} = \rm{height}$. When taking the formula for a cone and its volume,

$$h = \frac{3V}{π \cdot r^2},$$

I'd like to rewrite this to infer $h$ when $V$ is given. How do I simplify this? Some pointers would be much appreciated!

Body:
Replace $r$ with $\frac h2$:

$$h=\frac{3V}{\pi \frac{h^2}{4}} \\ h^3 =\frac{12V}{\pi} \\ h =\sqrt[3]{\frac{12V}{\pi}} $$


答:
Volume of cone,

$V = \frac13 \pi r^2 h$.

Here, $2r =h $ or $r = \frac h2$. So,

$V = \frac13 \pi \cdot\frac{h^2}{4}\cdot h = \frac{\pi}{12} h^3 \Rightarrow \boxed{h = \left(\frac{12V}{\pi}\right)^{1/3}}$

Title: Simplifying equation of a cone's volume, radius, height when height equals $2\cdot \rm{radius}$

Body:
I have a cone shape that has $2\cdot \rm{radius} = \rm{height}$. When taking the formula for a cone and its volume,

$$h = \frac{3V}{π \cdot r^2},$$

I'd like to rewrite this to infer $h$ when $V$ is given. How do I simplify this? Some pointers would be much appreciated!


问:
Messed up on the inverse of $\cosh(x)$ but I don't know where I went wrong
If someone can tell me where I went wrong, that would be extremely appreciated :)

The function $f:[0,\infty)\mapsto[1,\infty)$ defined by $f(x)=\cosh x$ is bijective. Find a formula for $f^{-1}.$

If $y=\cosh x$, then
\begin{align}
& y = \frac{1}{2}\left(e^x+e^{-x}\right) \\[5pt]
\implies & 2y = e^x+e^{-x} \\[5pt]
\implies & 2ye^x = e^x(e^x+e^{-x}) \\[5pt]
\implies & 2ye^x = e^{2x} + e^0 \\[5pt]
\implies & 2ye^x = e^{2x}+1 \\[5pt]
\implies & -e^{2x}+2ye^x - 1 = 0 \\[5pt]
\implies & e^x = \frac{-2y \pm \sqrt{4y^2-4(-1 \times -1)}}{-2} \\[5pt]
\implies & e^x = y + \sqrt{-2y+2} \text{ as $e^x>0$} \\[5pt]
\implies & x = \ln(y+\sqrt{-2y+2}) \\[5pt]
\implies & \cosh^{-1}x = \ln(y+\sqrt{-2y+2})
\end{align}

答:
We have
\begin{align}
e^x &= \frac{-2y\pm\sqrt{(2y)^2-4}}{-2} \\[5pt]
&= \color{red}{y\pm\frac{\sqrt{4y^2-4}}{-2}} \\[5pt]
&= y\pm\frac{\sqrt{4}\sqrt{y^2-1}}{-2} \\[5pt]
&= y\pm \frac{2\sqrt{y^2-1}}{-2} \\[5pt]
&= y \mp\sqrt{y^2-1} \, .
\end{align}
Since $x$ is nonnegative, we know that $e^x \geq 1$. Hence, $y\mp\sqrt{y^2-1} \geq 1$. Notice that it has to be greater than or equal to $1$, not $0$ as you wrote in your question. Since $0<y-\sqrt{y^2-1}\leq 1$ for $y\geq1$, the negative branch doesn't work. Hence, $e^x=y\color{red}{+}\sqrt{y^2-1}$, and the result follows.


答:
If $y=\cosh x$, then
\begin{align}
& y = \frac{1}{2}\left(e^x+e^{-x}\right) \\[5pt]
\implies & 2y = e^x+e^{-x} \\[5pt]
\implies & 2ye^x = e^x(e^x+e^{-x}) \\[5pt]
\implies & 2ye^x = e^{2x} + e^0 \\[5pt]
\implies & 2ye^x = e^{2x}+1 \\[5pt]
\implies & -e^{2x}+2ye^x - 1 = 0 \\[5pt]
\implies & e^x = \frac{-2y \pm \sqrt{4y^2-4(-1 \times -1)}}{-2} \\[5pt]
\color{red} \implies & \color{red}{e^x = y + \sqrt{-2y+2} \text{ as $e^x>0$}}\\[5pt]
\end{align}

Indeed you should get:

\begin{align}
\implies & e^x = y ±  (-1) \sqrt{y^2 -1} \\[5pt]
\end{align}

Then, following the justification given in the comments by Soheil, you can conclude that since $x \in [0,\infty)$, you end $$\cosh^{-1}(x)= \log(x + \sqrt{x^2-1})$$

Title: Messed up on the inverse of $\cosh(x)$ but I don't know where I went wrong

Body:
If someone can tell me where I went wrong, that would be extremely appreciated :)

The function $f:[0,\infty)\mapsto[1,\infty)$ defined by $f(x)=\cosh x$ is bijective. Find a formula for $f^{-1}.$

If $y=\cosh x$, then
\begin{align}
& y = \frac{1}{2}\left(e^x+e^{-x}\right) \\[5pt]
\implies & 2y = e^x+e^{-x} \\[5pt]
\implies & 2ye^x = e^x(e^x+e^{-x}) \\[5pt]
\implies & 2ye^x = e^{2x} + e^0 \\[5pt]
\implies & 2ye^x = e^{2x}+1 \\[5pt]
\implies & -e^{2x}+2ye^x - 1 = 0 \\[5pt]
\implies & e^x = \frac{-2y \pm \sqrt{4y^2-4(-1 \times -1)}}{-2} \\[5pt]
\implies & e^x = y + \sqrt{-2y+2} \text{ as $e^x>0$} \\[5pt]
\implies & x = \ln(y+\sqrt{-2y+2}) \\[5pt]
\implies & \cosh^{-1}x = \ln(y+\sqrt{-2y+2})
\end{align}


问:
What is the radius of a sphere given in the form $ \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 +z^2 = R \}$?
Let $R > 0$. What is the radius of a sphere given in this format:

$$ \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 +z^2 = R \}$$

I used $R$ indicating the relation to the radius $r$ that I want to find. It might be trivial, but my mind still has difficulties understanding this type of notation hence an explanation would be appreciated. Thank you

答: In general, $$S = \{(x,y,z)\in \mathbb R^3: x^2+y^2+z^2 = r^2\}$$ is the equation of a sphere in $\Bbb R^3$ centered at the origin $(0,0,0)$ with radius $r$. In your case, $r^2 = R$, and so $r = \sqrt R$. $r = -\sqrt R$ is an algebraic possibility, but we throw this away because the radius of a sphere cannot possibly be negative!


答: With some ring theory: Let $k$ be the field of coefficients. Then $k[T]$ is a principal ideal domain. The set of polynomials $f$ defined by $f(T)v=0$ is an ideal and generated by $p$. The minimal polynomial is element of the ideal and therefore divisible by $p$.

Title: For a specific $v$ in $V$, let $p(x)$ be the monic polynomial such that $p(T)v=0$. Then $p$ divides the minimal polynomial

Body: Before starting problem, I was wondering why we assume that the degree of $p$ is smaller or equal to the degree of the minimal polynomial. Intuitively, it makes a lot of sense since the minimal polynomial applied to all vectors in $V$ is $0$ where as $p(T)$ was specifically constructed just so $p(T)v=0$, so $p$ does not need to carry as much information as the minimal polynomial, but how do we rigorously show this?


问:
Solutions of the functional equation $f(x+1)= xf(x)$
I have come across a question of functional equation
$$f(x+1)= xf(x).$$
My question is how can we solve these type of functions. I have tried to use substitutions like $x= x-1$ and some manipulations but could't find the function. I also have tried to use differentiation's basic definition. I am still unable to solve the functional equation. Please try to clear my query.

答:
This can be seen as a first order recurrence relation, which can be recast to a first order difference equation by taking the logarithm.

$$f(x+1)=xf(x)$$

can be written

$$g(x+1)=g(x)+\log x.$$

The solution of such recurrences is made of the general solution of the homogeneous equation (obviously a constant), and a particular solution. In this particular case, we see that for integer $x$,

$$g(n)=\sum_{k=0}^{n-1}\log(k)+g(0)=\log((n-1)!)+g(0).$$

More generally, for real $x$,

$$g(x)=g(\lfloor x\rfloor+\{x\})=\sum_{k=0}^{\lfloor x\rfloor-1}\log(k+\{x\})+g(\{x\})$$

and

$$f(x)=\prod_{k=0}^{\lfloor x\rfloor-1}(k+\{x\})\,e^{g(\{x\})}=\frac{\Gamma(x)}{\Gamma(\{x\})}e^{g(\{x\})}=\frac{\Gamma(x)}{\Gamma(\{x\})}f(\{x\}).$$

Without more specification, $f$ is arbitrary on $[0,1)$ or any other unit interval.


问:
Need to find the limit of a ratio of 2 functions
I am very new to calculus, limits and more complex mathematics in general and I really need some help.

I need to compare the growth rate of 2 functions and because of this I need to find the limit of ratio of these functions $\lim\limits_{n \to \infty} \frac{g(n)}{f(n)}$

The functions are as follows:

$f(n) = \sqrt{n^3log_{2}n}$

$g(n) = 10n\sqrt{n}+100n$

How would I go about doing this? I've seen other people use $k=\sqrt{n}$ as a substitute when doing this.

Thank you in advance!

答:
Observe that $n\sqrt n=\sqrt{n^3}$.
$$\frac{10n\sqrt{n}+100n}{\sqrt{n^3\log_{2}n}}=\frac{10\sqrt{n^3}}{\sqrt{n^3\log_{2}n}}+\frac{100n}{n\sqrt{n\log_{2}n}}=\frac{10}{\sqrt{\log_{2}n}}+\frac{100}{\sqrt{n\log_{2}n}}$$
As you can see in the last expression we only have $n$ in the denominator. So as $n\to\infty$ the limit goes to zero.


答:
As Jean Marie wrote the radius of the sphere is $r = \sqrt{R}$.

Since your sphere is centered at the origin the answer lies really close at hand. The radius is just the length of any of the vectors on the surface and that length is given by the euclidian norm on $\mathbb{R}^3$ which is defined for any $v = (x,y,z) \in \mathbb{R}^3$  exactly as

$$ \left\lVert v\right\rVert =\left\lVert (x,y,z)\right\rVert = \sqrt{x^2+y^2+z^2} = \sqrt{R}$$

where the last equality holds if $v$ lies on the surface of your sphere.

Title: What is the radius of a sphere given in the form $ \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 +z^2 = R \}$?

Body:
Let $R > 0$. What is the radius of a sphere given in this format:

$$ \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 +z^2 = R \}$$

I used $R$ indicating the relation to the radius $r$ that I want to find. It might be trivial, but my mind still has difficulties understanding this type of notation hence an explanation would be appreciated. Thank you


答:
$$\lim_{n \to \infty} \frac{g(n)}{f(n)} = \lim _{n \to \infty} \frac{10n\sqrt{n}+100n}{\sqrt{n^3\log_{2}n}} = \lim _{n \to \infty}\frac{10n^{3/2}+100n}{n^{3/2}(\log_{2}n)^{1/2}} = \lim _{n \to \infty}\frac{10+100n^{-1/2}}{(\log_{2}n)^{1/2}} $$

The numerator tends to $10$ while the denominator tends to $\infty$ so...

Title: Need to find the limit of a ratio of 2 functions

Body:
I am very new to calculus, limits and more complex mathematics in general and I really need some help.

I need to compare the growth rate of 2 functions and because of this I need to find the limit of ratio of these functions $\lim\limits_{n \to \infty} \frac{g(n)}{f(n)}$

The functions are as follows:

$f(n) = \sqrt{n^3log_{2}n}$

$g(n) = 10n\sqrt{n}+100n$

How would I go about doing this? I've seen other people use $k=\sqrt{n}$ as a substitute when doing this.

Thank you in advance!


问:
Find the greatest integer $N$ such that...

Find the greatest integer $N$ such that $N<\dfrac{1}{\sqrt{33+\sqrt{128}}+\sqrt{2}-8}$.

The way I did it is this: first, I rewrote the biggest square root as $\sqrt{1+2*16+8\sqrt{2}}$. Then I made the substitution $x=\sqrt{2}$, so this became $\sqrt{16x^2+8x+1}$, which factors to $\sqrt{(4x+1)^2}$ which is equal to $4x+1$. Then I substituted back: $\dfrac{1}{4\sqrt{2}+1+\sqrt {2}-8}$, or $\dfrac 1{5 \sqrt{2}-7}$. This is where I have a problem. I am able to find the answer because I know that $1.4<\sqrt 2 <1.5$. However, is there any way to find this answer without knowing this? Thanks!

答: Well you can show that $1.4<\sqrt 2<1.5$, just use the following inequalities $49<50$ and $8<9$. (Hint: take square root in both of them)</p>


问:
Closedness v.s. convergent sequence
When are the following statements equivalent?

A subset $A \subset X$ is closed for the given topology on $X$.

Every sequence $\{a_n\}$ in $A$ that converge to $a \in X$, we must have $a\in A$.

答: These are exactly equivalent in so-called sequential spaces. These include all spaces that are first countable (but also more general non first-countable spaces, e.g. all quotient images of metrisable spaces, like $\Bbb R{/}\Bbb Z$).

Title: Closedness v.s. convergent sequence


问:
Arithmetic-quadratic mean and other "means by limits of means"
For $x,y$ positive real numbers, and $p\neq 0$ real, define the Hölder $p$-mean
$$M_p(x,y) := \left(\frac{x^p+y^p}{2}\right)^{1/p}$$
whereas
$$M_0(x,y) := \sqrt{xy}$$
is the limit of $M_p(x,y)$ when $p\to 0$, i.e., the geometric mean.

Furthermore, when $p,q$ are two real numbers, define $L_{p,q}(x,y)$ as the obvious fixed point satisfying the equation
$$L_{p,q}(x,y) = L_{p,q}(M_p(x,y),M_q(x,y))$$
—meaning that $L_{p,q}(x,y)$ is defined as the common limit of the sequences $(x_n)$ and $(y_n)$ such that $(x_0,y_0) = (x,y)$ and $(x_{n+1},y_{n+1}) = (M_p(x,y),M_q(x,y))$.  (Showing convergence is easy, and evidently, $L_{p,q} = L_{q,p}$.)

Thus, $L_{0,1}(x,y)$ is the famous arithmetic-geometric mean.

Among other things, all these functions satisfy $\min(x,y) \leq F(x,y) \leq \max(x,y)$, as well as $F(y,x) = F(x,y)$ and $F(\lambda x, \lambda y) = \lambda F(x,y)$ — so we could just define them from $F(1,x)$, which is continuous and monotonically increasing with value $1$ at $1$.  It is well-known that $M_p(x,y) < M_q(x,y)$ whenever $x\neq y$ and $p<q$, so of course $M_p(x,y) < L_{p,q}(x,y) < M_q(x,y)$ under those circumstances.

Surprisingly, I can find no reference to the $L_{p,q}$ in the literature except for $(p,q)=(0,1)$ (mentioned above), and $(p,q)=(-1,1)$ which this page points out is just $M_0$.  More generally, we have
$$L_{p,-p}(x,y) = M_0(x,y)$$
and
$$L_{0,p}(x,y) = \big(L_{0,1}(x^p,y^p)\big)^{1/p}$$
—both are easy, and $L_{p,p}(x,x) = M_p(x,y)$ is trivial.

I couldn't find anything intelligent to say about $L_{1,2}$ (the "arithmetic-quadratic mean"), however, let alone the other $L_{p,q}$.  Experimentally, it satisfies $L_{1,2}(x,y) \leq M_{3/2}(x,y)$, something which is probably easy to prove and not interesting enough to try, so that's not my question.  Instead, my question will be:

Can $L_{1,2}$, or more generally $L_{p,q}$ (with $0<p<q$), be expressed in closed form using elementary functions and the arithmetic-geometric mean $L_{0,1}$?  Or at least using "standard" special functions (e.g. elliptic integrals)?

(I realize that the answer "no" might be exceedingly difficult to prove, so I'll be liberal in what I accept as an answer: perhaps the question should really be "please say something interesting about $L_{1,2}$ or $L_{p,q}$ in general".)

答:
These type of questions are investigated in "Compromise, consensus, and the iteration of means" and "Markov chains, Gauss soups, and compromise dynamics" by Ulli Krause. These papers also provide more pointers to relevant literature.

An earlier reference is "Pi and the AGM" by Borwein and Borwein and Chapter 8 there introduces the concept of limits of iterated means (only for two different means, as far as I see). A particular results is: If $M$ and $N$ are means, and the iterated mean converges to a function $\Phi$, then this is again a mean and characterized by
$$\Phi(M(a,b),N(a,b)) = \Phi(a,b).$$

So the function you are looking for is simply characterized by the functional equation
$$L_{1,2}(1,x) = L_{1,2}\left(\sqrt{\tfrac{1+x^2}2},\tfrac{1+x}2\right),$$
but you seem to know that already.
The book contains a few examples, but I could not find an example in the direction you are asking…


问:
Prove Positivity of a Function Involving $\log$.
I want to prove

The function $f:(0,\infty)^2 \to \mathbb{R}$ defined by
\begin{equation}\label{key}
   f(x,y) =  \ln \left( \dfrac{1+y}{1+ x}\right) + \dfrac{x}{1+ x} \ln \left( \dfrac{x}{y}\right)
  \end{equation}
is non-negative.

My attempt was to compute   $\dfrac{\partial f}{\partial y}$ and then see where the zeros of the function are.

I think that works fine, but I was wondering if there was a slicker way to do it.

答:
Writing the logarithms as integrals over $1/t$ gives
$$
 (x+1)f(x, y) = (x+1) \int_{1+x}^{1+y} \frac{dt}{t} + x \int_y^x \frac{dt}{t} = \int_x^y \left(\frac{x+1}{t+1} - \frac xt\right) \, dt \\
= \int_x^y \frac{t-x}{t(t+1)} \, dt  = \int_0^{y-x} \frac{s}{(s+x)(s+x+1)} \, ds\, ,
$$
and that is non-negative: The integrand is $\ge 0$ if $x \le y$, and $\le 0$ otherwise.

One can also see that equality holds exactly if $x=y$.


问:
A slight generalization of the Sieve of Sundaram that might shed light on the $6n \pm 1$ phenomenon of sequence A002822.
There's the $n$ such that $6n \pm 1$ is a twin prime pair sequence: https://oeis.org/A002822
It contains all twin prime averages (divided by $6$) other than $4$.

Notice this sequence: Positive numbers $n = 6xy + x + y$ such that $x,y \in \Bbb{Z}$.  That sequence is the complement (all entries found in $\Bbb{N}$ that are not in A002822, by its very definition).

But notice the $6xy + x + y$ part.  That reminded me of the Sieve of Sundaram's $2ij + i + j$ form. Thus:

Start with a list of integers from $81$ to $N$, and cross out all elements of the form $6ij + ki + lj$ such that $k, l \in \{1,3,5\}$ and $i,j \geq 1$.  Then taking the remaining elements, multiplying them by $6$ and adding respectively $lk = 1,3,5,9,15, $ or $25$; and of course we union the result of each case / cross-out form.

We have the following possibilities:

$$
6(6ij + i + j) + 1 = (6i + 1)(6j + 1) \\
6(6i + i + 3j) + 3 = (6i + 3)(6j + 1) \\
6(6ij + i + 5j) + 5 = (6i + 5)(6j + 1) \\
6(6ij + 3i +3j) + 9 = (6i + 3)(6j + 3) \\
6(6ij + 3i + 5j) + 15 = (6i+5)(6j+3) \\
6(6ij + 5i +5j) + 25 = (6i + 5)(6j + 5) \\
$$

The above will cover all odd numbers $\geq  81$ which we find by setting $i=j = 1$ and computing the smallest formula above which is $(6i + 1)(6j + 1) = 81$.

Thus we've filtered out all and only composites $\geq 81$, clearly, and via inspection of the above factorizations.

See: Sieve of Sundaram.

However, all primes $\gt 3$ are of the form $6n + 1$ or $6n + 5$.  Thus we can cross out lines $2,3,5$ in the above 6 formulas, or namely the lines with added $3,9,15$.  But $6n + 25 = 1 \pmod 6$ so it is left and the list of cross-out forms in the generalized Sundaram sieve becomes:

$$
6ij + i + j \\
6ij + i + 5j \\
6ij + 5i + 5j \\
i,j \geq 1
$$

Thus, that is the Sundaram sieve method genralized to a multiple of $6$ instead of just $2$.  You could probably do similarly, with other multiples.

Question. Can we remove even more unecessary cross-out forms so that there would be less than $3$?

答:
First note that $6(0)+5$ is prime, so you would need it in $6(1)-1$ form for the last inequality to hold.  But yes, the Sieve of Sundaram has generalizations to all multiplicative groups. It can be reduced to $6ij\pm i\pm j$ if you need to save space but might want to save that for mod 30 with 36 cases.

In Addition, it can be generalized to find semiprimes with lower bounds on factors etc. Just let $i,j$ not be also representable using some other values. Then it's a product of primes. If exactly 1 is of the form of a semiprime has 3 prime factors ...  But 3 cases exist with 2 possibilities each.


问:
The relationship between Cauchy-Riemann equations and differentiability
What is the relationship between Cauchy-Riemann equations and differentiability of a complex valued function at any point $z=z_0$? If Cauchy-Riemann equations are satisfied at $z=z_0$ then can we say that $f(z)$ is differentiable at $z=z_0$?

答:
The reply is probably too late, but:

According to the theorem, that f(z) satisfies Cauchy-Riemann equation is the necessary condition for the complex function to be complex differentiable. The sufficient conditions also require all first order partial derivatives of u, v (where f(x,y)=u(x,y) + iv(x,y)) are continuous (at z=z_0.)

Hence, complex differentiability implies f satisfying C-R equation, but not the other way around. Here I found an counter example: Show the Cauchy-Riemann equations hold but f is not differentiable.


问:
Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$
It's an inequality based on two found on the website MSE (see the reference):

Let $a,b,c>0$ then we have:

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$

Lemma 1 :

$a,b,c>0$ then we have :

$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$

Proof of lemma 1 :

First we remark that the inequality is homogenous and we can try the substitution $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$ and apply the uvw's method .

We have :
$$a^4+b^4+c^4=(9u^2-6v^2)^2-2(9v^4-6uw^3)$$
$$a^2b^2+b^2c^2+c^2a^2=9v^4-6uw^3$$
$$a^2+b^2+c^2=9u^2-6v^2$$

And :
$\left(\frac{((3u)((3u)^2-4(3v^2))+5w^3)}{3u3v^2-w^3}+2\right)^2\geq \frac{9}{4}+\frac{(9/4)((9u^2-6v^2)^2-2(9v^4-6uw^3)-(9v^4-6uw^3))+(2.25u+0.75w)(9u^2-9v^2)(3u)}{(2.25u+0.75w)(3u3v^2-w^3)}$

it's enough to find an extreme value of our expression for the extreme value of $w^3$ wich happens for an equality case of two variables .

Since the last inequality is homogeneous, we can assume that $b=c=1$.

$$\frac{2}{a+1}+\frac{a}{2}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(a^4+1-2a^2)}{((a+2)\frac{3}{4}+\frac{3}{4}(a)^{\frac{1}{3}})(a+1)^2(2)}+\frac{(a^2+1-2a)(a+2)}{(a+1)^2(2)}}$$

Now it seems to be clear : we get a polynomial  with a root equal to one . See the factorization by Wolfram alpha .

End of the proof of the lemma 1

Remains to show that $ a\geq b \geq c>0$:

$$P(a,b,c)\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$

Wich is not hard I think .

Question :

How to show it ?

Reference :

M. A. Rozenberg, “uvw–Method in Proving Inequalities”, Math. Ed., 2011, no. 3-4(59-60), 6–14

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

Stronger than Nesbitt inequality

Body:
As username says it's not symmetric so here there is a refinement and a symmetric formula :

$a,b,c\in[0.95,1]$:

$$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(0.75(abc)^{\frac{1}{3}}+0.75(a+b+c))(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}+\frac{9}{4}\left(abc\frac{\frac{a^2+b^2+c^2}{ab+bc+ca}-1}{a^3+b^3+c^3}\right)^2}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$


答:
I will give a solution that involve only calculations with 2 digit numbers; all the other contributions will be estimated. I will call my approach the sloppy continued fractions method. Unluckily, if you want to cut down computations you have to be smarter, and in particular lengthy. Sorry for that, but you will enjoy! After the theoretical framework is set up, the problem is easy to understand, but if you prefer you can skip to the "out of the blue" section at the bottom.

Theory

Firstly, notice that the problem can be reformulated as $\log_7(75) < \frac{71}{32}$. The instance of finding good rational approximants to irrational numbers is a classical one, and it is addressed by the continued fractions method. In case of logarithms and inequalities, it has the following simpler form. Say we want to understand whether $\log_a(b) < \frac{p}{q}$. Then:

Take the maximum integer $n_0$ less than $\log_a(b)$, i.e. solve $a^{n_0} < b < a^{n_0+1} $, and let $m_0$ be the maximum integer less than $p/q$.

If $m_0 \ge n_0 +1$ the inequality is true, because
$$\frac{p}{q} \ge m_0 \ge n_0 +1 \ge \log_a(b) $$
Else, if $n_0 \ge m_0 +1$ the inequality is false, because
$$ \frac{p}{q} \le m_0 +1 \le n_0 \le \log_a(b) $$
Otherwise, go to step 3 supposing $n_0 = m_0$.

Since $\log_a(b) - n_0 = \log_a(b/a^{n_0})$ is less than one, we can write
$$ \log_a(b) = n_0+ \frac{1}{\log_{a_1}(b_1) } $$
where $a_1 = b/a^{n_0}, b_1 = a$. Analogously we can write
$$ \frac{p}{q} = n_0 + \frac{1}{ \frac{p_1}{q_1} } $$
where $p_1 = q, q_1 = p-n_0 q$.

Now we reformulate the inequality and we get
$$ n_0 + \frac{1}{\log_{a_1}(b_1) } < n_0 + \frac{p-n_0 q}{q} $$
$$ \log_{a_1}(b_1) > \frac{p_1}{q_1} $$
And we can repeat from step 1 with a similar problem (but sign reversed).

There is a little problem: while the fraction gets simpler at step (5), the numbers involved in logarithms becomes harder. Let me state a little lemma.

Logarithm approximation. Let $\alpha, \beta \in \mathbb{N}$ and $k,j \in \mathbb{Z}$ small with respect to $\alpha, \beta$. Then if $x = k/\alpha, y = j/\beta$:

$$ 1+\frac{1}{\ln(\alpha/\beta)} (x -y -(x^2+y^2)) <  \log_{\frac{\alpha}{\beta}} (\frac{\alpha+k}{\beta+j} )  < 1+\frac{1}{\ln(\alpha/\beta)} (x-y +(x^2+y^2))$$

Proof. This is a simple consequence of $ x-x^2 \le \ln(1+x) \le x+x^2$. Also, depending on the sign of $k,j$ and the inequality that must be used, sometimes the quadratic term is not necessary (indeed $\log(1+x) \le x$ for positive $x$ and $\log(1+x) \ge x$ for negative $x$).

Now let me restate the continued fraction method with a sloppiness parameter. We want to solve inequalities of the form $\log_a (b) < \frac{p}{q} \cdot \alpha$, where $\alpha \simeq 1 $ is a rational number (the approximation parameter). The bad thing is that it can stop at some time because of our sloppiness, but if it doesn't stop it works with few digits!! Also, don't worry if you can't follow the exact formulas, because they will get explicit thereafter.

Take the maximum integer $n_0$ less than $\log_a(b)$, i.e. solve $a^{n_0} < b < a^{n_0+1} $, and let $m_0$ be the maximum integer less than $p/q$.

If If $\alpha m_0 \ge n_0 +1$ the inequality is true, because
$$\alpha\frac{p}{q} \ge \alpha m_0 \ge n_0 +1 \ge \log_a(b) $$
Else, if $n_0 \ge \alpha (m_0 +1)$ the inequality is false, because
$$ \alpha \frac{p}{q} \le \alpha(m_0 +1) \le n_0 \le \log_a(b) $$
Else, if $n_0 = m_0$ go to step 3.
In any other case, return "YOU HAVE BEEN TOO SLOPPY! TRY AGAIN!".

Since $\log_a(b) - n_0 = \log_a(b/a^{n_0})$ is less than one, we can write
$$ \log_a(b) = n_0+ \frac{1}{\log_{a_1}(b_1) } $$
where $a_1 = b/a^{n_0}, b_1 = a$. Analogously we can write
$$ \alpha \frac{p}{q} - n_0 = \alpha(n_0 + \frac{q_1}{p_1} ) - n_0  = (\alpha -1)n_0 + \alpha \frac{q_1}{p_1} = \frac{q_1}{p_1} ( \alpha + \frac{p_1}{q_1} n_0 (\alpha-1) )= \frac{1}{ \alpha_1 \frac{p_1}{q_1} } $$
where $p_1 = q, q_1 = p-n_0 q, \alpha_1 = \left (  \alpha + \frac{p_1}{q_1} n_0 (\alpha-1) \right )^{-1} $.

Now we reformulate the inequality and we get
$$ n_0 + \frac{1}{\log_{a_1}(b_1) } < n_0 + \frac{1}{\alpha_1 \frac{p_1}{q_1} }  $$
$$ \log_{a_1}(b_1) > \alpha_1 \frac{p_1}{q_1} $$
If we want to, we can approximate a bit the logarithm on the left using the logarithm approximation and move this contribution in $\alpha_1$. Then we can repeat from step 1 with a similar problem (but reversed inequality).

Calculations

Let's start! At the beginning $a=7, b= 75, p=71, q=32, \alpha=1$. Also, let me notice that the continued fraction of $71/32$ is $[2,4,1,1,3]$, that is:
$$ \frac{71}{32} = 2+\frac{1}{4+\frac{1}{1+ \frac{1}{1+\frac{1}{3}}}} $$

Since we suspect this approximations is really close, probably the continued fraction of the logarithm will be the same for a lot of time, so that we have a guess of what the "$n_0$" should be.

STEP 1

1.1 Note that $7^2 < 75 < 7^3$, so that $n_0 = 2= m_0$ and we go to point 3.

1.3 We have $a_1 = 75/49, b_1 = 7, p_1 = 32, q_1 = 71 - 64 = 7$. The inequality is now
$$ \log_{75/49} (7) > \frac{32}{7}$$
here it is a trick we will use repeatedly:
$$ \log_{75/49}(7) = \color{red}{\log_{75/49}(75/50)}\log_{75/50}(7) > \color{red}{\left(1- \frac{1}{50} \right)} \log_{3/2}(7) = \frac{49}{50} \log_{3/2}(7) $$

1.5 It is enough to verify that something smaller than LHS is greater than RHS, and we get the sloppy inequality
$$ \log_{3/2}(7)  > \frac{50}{49} \frac{32}{7} $$

STEP 2

2.1 The suggestion from the continued fraction of 32/7 makes us try $n_1 = 4$, and indeed
$$ 3^4 = 81 < 112 = 7 \cdot 16 = 7 \cdot 2^4, \ \ \ 3^5 = 343 > 7 \cdot 32 $$
2.3 Formulas yield $a_2 = 112/81 , b_2 = 3/2, p_2 = 7, q_2 = 4$ and
$$\alpha_2 = \left ( \frac{50}{49} + \frac{7}{4} \cdot 4 \cdot \frac{1}{49}\right)^{-1} = \left ( \frac{57}{49} \right)^{-1} = \frac{49}{57} $$
The inequality is now
$$ \log_{112/81} (3/2) < \frac{7}{4} \frac{49}{57}$$
and the log trick yields
$$ \log_{112/81}(3/2) = \log_{112/81}(110/80) \log_{11/8}(3/2) < \left ( 1-\frac{2}{112} + \frac{1}{81} \right ) \log_{11/8}(3/2) <  \log_{11/8}(3/2)$$

2.5 The sloppy inequality remains
$$ \log_{11/8} (3/2) < \frac{7}{4} \frac{49}{57} $$

STEP 3

3.1 Under the usual suggestion we try $n_2 = 1$ and it works:
$$ \frac{11}{8} = 1+ \frac{3}{11} < 1+ \frac{1}{2} = \frac{3}{2}, \frac{11^2}{8^2} = \left( 1+ \frac{3}{11} \right)^2 > 1+ \frac{6}{11} > \frac{3}{2}  $$
3.3 We get $p_3 =4, q_3 = 3, a_3 = 24/22 = 12/11,  b_3 = 11/8$ and

$$ \alpha_3 =  \left (  \frac{49}{57} -\frac{4}{3} \frac{9}{57} \right )^{-1} = \frac{57}{37} $$

3.5 The fraction $12/11$ is nice, so we don't do the trick at this step. The sloppy inequality is just

$$ \log_{12/11}( 11/8) > \frac{4}{3} \frac{57}{37} $$

STEP 4.

4.1 Here it is the surprise: using the rule for the cube of a sum we have that

$$ \left  ( 1 + \frac{1}{11} \right) ^3 = 1 + \frac{3}{11} + \frac{3}{11^2} + \frac{1}{11^3}  < 1 + \frac{3}{11} + \frac{10}{121} + \frac{11}{11^3} = \frac{15}{11} < \frac{11}{8}$$

because $15 \cdot 8 = 120 < 121 = 11^2$. This translates into $\log_{12/11}(11/8) \ge 3$. On the other side

$$ \frac{4}{3} \frac{57}{37} < \frac{4}{3} \frac{60}{36} = \frac{20}{9} < 3 \le \log_{12/11}(11/8) $$

And we are done. Victory! Yey!

Out of the blue

Let's start from the observation that

$$ (\#1) \ \ \ \ \ \   \left ( \frac{12}{11} \right ) ^3 = \left (1+ \frac{1}{11} \right ) ^3 = $$
$$ =1+ \frac{3}{11} +\frac{3}{11^2} +\frac{1}{11^3} \le 1+ \frac{3}{11} + \frac{10}{11^2} + \frac{11}{11^3} $$
$$  = \frac{15}{11} < \frac{11}{8} $$

So that $\log_{12/11}(11/8) > 3$. Also, note that

$$ (\#2) \ \ \ \ \ \    3 > \frac{20}{9} = \frac{60}{36} \frac{4}{3} > \frac{57}{37} \frac{4}{3} $$
If we combine (1) and (2) we get $ \log_{12/11}(11/8) > \frac{57}{37} \frac{4}{3}$. Taking reciprocal (that swaps base and argument in the logarithm) and adding 1 we get

$$ (\#3) \ \ \ \ \ \    \log_{11/8} \left ( \frac{12}{11} \frac{11}{8} \right ) = \log_{11/8}(3/2) < $$
$$ <1+\frac{37}{57} \frac{3}{4} < \frac{37}{57}+ \frac{21}{57} + \frac{37}{57} \frac{3}{4} =  $$
$$ = \frac{37}{57} \cdot 1 + \frac{12}{57} \frac{7}{4} + \frac{37}{57} \frac{3}{4}  = $$
$$ =\frac{37}{57} \frac{7}{4} + \frac{12}{57} \frac{7}{4} = \frac{49}{57} \frac{7}{4}$$

Using the inequality

$$ (\#4) \ \ \ \ \ \   \log_{112/81} (11/8) = 1+ \log_{112/81}\left (\frac{110 \cdot 81}{112 \cdot 80} \right )  = $$
$$ = 1+ \log_{112/81} \left (1- \frac{2}{122} \right ) - \log_{112/81}\left ( 1-\frac{1}{81} \right ) < $$
$$ < 1+ \frac{1}{\ln(112/81) } \left ( -\frac{2}{122} + \frac{1}{81} \right ) < 1 $$

We can multiply inequality (3) by the (4), and using $\log_x y \log_y z = \log_x z$:

$$ (\#5) \ \ \ \ \ \    \log_{11/8}(3/2) < \frac{49}{57} \frac{7}{4} \Rightarrow \log_{112/81}(3/2) = \log_{112/81}(11/8) \log_{11/8}(3/2) < \frac{49}{57} \frac{7}{4} $$

Taking reciprocals and adding 4 we get

$$ (\#6) \ \ \ \ \ \    \log_{3/2} \left ( \frac{112}{81} \frac{81}{16} \right) = \log_{3/2}(7) > $$
$$ > 4+ \frac{57}{49} \frac{4}{7}  = \frac{57}{49} \cdot 4 - \frac{8}{49} \cdot 4 + \frac{57}{49} \frac{4}{7} = $$
$$ = \frac{57}{49} \frac{32}{7} - \frac{32}{7} \frac{7}{49} = \frac{50}{49} \frac{32}{7} $$

Consider the following inequality
$$ (\#7) \ \ \ \ \ \   \log_{75/49}(3/2) = 1+ \log_{75/49} \left ( \frac{49 \cdot 75}{75 \cdot 50} \right ) = $$
$$ = 1+ \log_{75/49} \left ( 1- \frac{1}{50}  \right ) > 1+ \frac{1}{\ln(75/49) } (- \frac{1}{50} + \frac{1}{50^2}) > $$
$$ > 1- 2 \cdot \frac{1}{51} > 1- \frac{1}{50} =  \frac{49}{50} $$
To show (7) we used, between the second and the third row, that
$$ (75/49)^2 < (75/48)^2 = (25/16)^2 < e \Rightarrow \frac{1}{\ln(75/49) } = \log_{75/49}(e) > 2 $$
Indeed, $625 < 256 \cdot e$.  We are almost finished: combining the inequalities

$$\log_{3/2}(7) > \frac{50}{49} \frac{32}{7}, \ \ \color{red}{\log_{75/49}(3/2) > \frac{49}{50}} $$

we get $ \log_{75/49}(7) > \frac{32}{7}$. Taking inverses and adding 2 we get

$$ 2+ \log_7(75/49) = \log_7(75) < 2+ \frac{7}{32} = \frac{71}{32} $$

Putting this to the exponent of $7$ we get
$$ 75 < 7^{71/32} \Rightarrow 75^{32} < 7^{71} $$

as desired!

Title: Prove $7^{71}>75^{32}$

Body:
My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$

I did the first question: \begin{align}\frac{6^9}{10^7}&=\frac{4}{5}\times\frac{27^3}{25^3}\\&=0.8\times1.08^3\\&>0.8\times(1+3\times0.08+3\times0.08^2)\\&>0.8\times(1+3\times0.086)\\&>0.8\times1.25=1.\end{align}

But I can't work out the second,  I calculated it out on my computer, $\frac{7^{71}}{75^{32}}=1.000000949\cdots$


问:
$\tilde{H}_n(X \cup CA) \simeq H_n(X,A)$
I'd like to prove that $\tilde{H}_n(X \cup CA) \simeq H_n(X,A)$, where $CA$ denotes the cone of $A$, i.e $CA = A \times I / A \times \{0\}$ where $A$ is a subset of $X$ topological space.

The first step in order to do so is to prove the following isomorphism $\tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA,CA)$. Since $CA$ is contractible I thought of taking the long exact sequence of the pair and i get $$\longrightarrow H_n(CA) \longrightarrow H_n(X \cup CA) \longrightarrow H_n(X \cup CA,CA) \longrightarrow H_{n-1}(CA)\longrightarrow$$

So if $n-1 \ne 0$ i.e $n > 1$ since $H_i(CA) = 0$ we have the $\tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA,CA)$ since for those $n, \hspace{0.1cm} \tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA)$.

Here is where I got stuck, since the terms are no longer $0$ I don't know how to continue in the case $n=0,n=1$. Any help would be appreciated.

答:
The long exact sequence has two variants: One for unreduced homology groups ands one for reduced homology groups. The latter is
$$\dots \to \tilde H_n(A) \to \tilde H_n(X)  \to H_n(X,A) \to \tilde H_{n-1}(A) \to  \dots $$
Note that $\tilde H_n(Y) = H_n(Y)$ for $n > 0$. See see Hatcher's "Algebraic Topology"  p.118 and Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$..

Now look at
$$H_1(CA) \to H_1(X \cup CA) \to H_1(X \cup CA, CA) \to \tilde H_0(CA) \to \tilde H_0(X \cup CA) \to H_0(X \cup CA,CA) \to 0 $$
Since $H_1(CA) = \tilde H_0(CA) = 0$, you get the desired result.


问:
Generalization of Bertrand's Postulate
Bertrand's postulate states that there is a prime $p$ between $n$ and $2n-2$ for $n>3$. According to Dirichlet's theorem we have that a sequaence
$$a\cdot n+b$$
has infinite primes iff $a$ and $b$ are relatively prime. So in some sense, Bertrand's postulate gives a maximum of time for encountering a prime in the sequence
$$2\cdot n+1$$
So, the question is: there is a generalization of Bertrand's Postulate for sequences $a\cdot n+b$ that accomplish the Dirichlet's theorem?

EDIT: (For a more concise explanation of the particular generalization.) We know that given
$$a_n=2\cdot n+1$$
we have that for all $m$ there is a prime in the sequence greater than $a_m$ and less than $a_{2m}$. So, the thing is that if there is some generalization of Bertrand's Postulate using the sequence form, for an arbitrary sequence
$$c_n=a\cdot n+b$$
with $a$ and $b$ coprime. Something as, for every relatively prime $a$ and $b$, there is a $k\leq a\cdot b$, such that for all $m$ there is a prime in the sequence between $c_m$ and $c_{k\cdot m}$.

Such kind of thing is what I am looking for.

答:
The answer is that there is such a generalization of Bertrand Postulate. For fizing constants, in the rest of the answer $a$ and $b$ will be coprime numbers, and $c_n$ will be given the sequence by

$$c_n=a+bn$$

We consider the prime number theorem for arithmetic progressions, this say that

$$\pi_{a,b}(x)\sim \frac{1}{\varphi(b)}\frac{x}{\log x}\text{,}$$

where

$$\pi_{a,b}(x)=\text{Card}\{p\leq x\,|\,p\text{ is prime and }p\equiv a\text{ (mod }b\text{)}\}\text{,}$$

$\phi$ is the Euler's totient function and $\sim$ denote that the limit of the quotient of the two functions tends to $1$ as $x$ tends to infinity. Let now $\rho_{a,b}$ be given by

$$\rho_{a,b}(n)=\text{Card}\{k\leq n\,|c_k\text{ is prime}\}\text{,}$$

we can show by straighforward calculation that

$$\rho_{a,b}(n)=\pi_{a,b}(a+bn)-\pi_{a,b}(a)\text{.}$$

From this, we have by the previous theorem that

$$\rho_{a,b}(n)\sim \frac{1}{\varphi(b)}\frac{a+bn}{\log (a+bn)}\text{,}$$

and so, by standard methods of analysis,

$$\rho_{a,b}(n)\sim \frac{b}{\varphi(b)}\frac{n}{\log n}\text{.}$$

Here, we can show, that for all $\varepsilon>0$,

$$\lim_{n\to \infty}\rho_{a,b}((1+\varepsilon)n)-\rho_{a,b}(n)=\infty$$

and thus define $N_\varepsilon$ as the first $n$ for which the previous difference is positive for all natural numbers greater or equal than it, i.e. $\rho_{a,b}((1+\varepsilon)n)-\rho_{a,b}(n)$ is positive for all $n\geq N_\varepsilon$ and negative for $n=N_\varepsilon-1$.

Therefore, we conclude that given $\varepsilon>0$, for all $n\geq N_\varepsilon$ there is a $k$ between $n$ and $(1+\varepsilon)n$ such that $c_k$ is prime. Furthermore, we can make $\varepsilon$ big enough such that $N_\epsilon$ is zero, since

$$\varepsilon\mapsto N_\varepsilon$$

is monotonous decreasing function. In this way, we obtain Bertrand's postulate for arithmetic progressions.

Title: Generalization of Bertrand's Postulate


问:
How to calculate the probability of dice outcomes of a D$30$?
One D$30$ dice is rolled three times, will the difference between the maximum and minimum value be greater than $15$?

So I guess this is the same as just picking three random numbers between $1-30$, and looking at the biggest and smallest number to see if the difference is greater than $15$. But what I am wondering is; How do I calculate the different probabilities here?

答:
First choose which two dice will be min and max. For each combination of min and max on the two chosen dice, the third dice will be between min and max.

There are $3 \cdot 2 = 6$ ways to choose two dice and which will be max and which will be min.

If dice A has min and dice B has max,

For example, if A shows $1$ then B must show between $17$ and $30$ for the difference to be greater than $15$. If B is $17$, C can be any number between $1$ and $17$. So you can count the number of permissible outcomes as,

$\displaystyle 6 \cdot \sum_{i=1}^{14} \sum_{j=16+i}^{30} (j - i - 1)$

This does not include cases when the third number is equal to min or max. There are $ \ 3 \cdot  (15-i) \ $ possibilities for third number to be equal to min and same for third number to be equal to max. So the final expression is,

$\displaystyle 6 \cdot \sum_{i=1}^{14} \sum_{j=16+i}^{30} (j - i) = 12810$

Now divide by the total of $30^3$ unrestricted outcomes for desired probability.


答:
First we will calculate the probability such that the difference of maximum and minimum will be $d$.

There are $30-d$ different possible extreme cases, say $(1,d+1),\, (2,d+2),\, \ldots ,\, (30-d,30)$. Now, if all three dice have different values then there would be $3!=6$ ways to order them, and each pair could have the middle value different from the extremes of $d-1$ different ways, so for each pair there are $6\cdot (d-1)$ different possible triplets of values.

If the middle value repeat then the resulting triplet can be ordered of three different ways, so if it repeat for each pair will be $2\cdot 3=6$ possible different triplets with a repetition (the $2$ there comes because it can repeat the minimum or the maximum value).

Therefore in total you have $(30-d)(6(d-1)+6)=6(30-d)d$ valid cases where the difference of the maximum and the minimum is exactly $d$. Then the probability that you want to calculate is the sum

$$
\frac6{30^3}\sum_{d=16}^{29}(30-d)d=\frac{6}{30^3}\left(29\frac{30\cdot 29-16\cdot 15}{2}-\frac{30\cdot 29\cdot 28-16\cdot 15\cdot 14}{3}\right)\\
=\frac{12810}{30^3}\approx 0.47
$$

Title: How to calculate the probability of dice outcomes of a D$30$?

Body:
One D$30$ dice is rolled three times, will the difference between the maximum and minimum value be greater than $15$?

So I guess this is the same as just picking three random numbers between $1-30$, and looking at the biggest and smallest number to see if the difference is greater than $15$. But what I am wondering is; How do I calculate the different probabilities here?


问:
Using the rounding function to compute $x=0.21045$ should the output be $0.210$ or $0.211$ within a certain set of floating points?
I'm self-learning LA through an online book. And wanted to clear up my confusion regarding the Rounding Function.

I'm finding it hard to describe the problem, so I first started off by describing how this book defines the rounding function for floating-point numbers, then a few obvious examples, then finally my doubts with how this function would result for $x=0.21045$, or any number outside the given precision($p$) value.

The rounding function is defined as:

For any number within the set of floating point numbers $$\mathbb{F}_{p,b,N} =\pm0.d_1d_2\dots d_p\times b^n $$where $p$ is the precision, $b$ is the base, and $n$ is the exponent within the range $\{-N\leq n\leq N\}$.

A couple of examples are given as follows for the values that are not in $\mathbb{F}_{3,10,3}$ but rounded off to a number within the set:
\begin{align}
x = 0.2103 × 10^3 &\rightarrow 0.210\\
x = 0.21051 × 10^3 &\rightarrow 0.211\\
x = 0.21037 × 10^3 &\rightarrow 0.210
\end{align}

Third example is maybe done as follows:
\begin{align}
0.21037 \\
0.2103\tag{since $p=3$, we ignore all numbers after $d_{3+1}$ i.e 3}\\
0.210\tag*{$\because d_{p+1}=3<5$}
\end{align}

Here are my doubts:

I came up with a number similar to the examples above, $x=0.21045$, intuitively its round-off should be $x=0.211$ but by the same methodology as the above example we get $x=0.210$.
\begin{align}
0.21045 \\
0.2104\tag{since $p=3$, we ignore all numbers after $d_{3+1}$ i.e 4}\\
0.210\tag*{$\because d_{p+1}=4<5$}
\end{align}

I am not sure what exactly is wrong here, I think there may be a problem with

either my methodology of implementing the rounding function: ignoring all the values after a certain radix point.

or with the assumption that the answer should be $0.211$ when in fact within the set $\mathbb{F}_{3,10,3}$ the output $0.210$ is correct.

答:
The definition says you just look at the first digit past what you want to round to.  If it is $5$ or greater you increase the last digit you want to keep by $1$.  If it is $0$ to $4$ you leave the last digit the same.  In your example of $21037$ being rounded to three places, the first digit to remove is $3$, so you round down.  You do not even look at the $7$.  Similarly, in $21045$ the first digit you want to remove is $4$, which is less than $5$.  You don't even look at the $5$.  You also round this to $210$.  This makes sense because you are always picking the closer value unless the part to remove is precisely $0.50000\ldots$.  In that case there is no closer value.  Either way you round is the same distance away.

When we round fixed point numbers we worry about an upward bias if we round all the exactly $0.5$ values up to $1$.  Often we round exactly $0.5$ to the even number.  For real numbers getting exactly $0.5$ is not supposed to happen or not enough to matter.


问:
Find the n-th base 2 palindrome
So I want to compute the n-th base 2 palindrome, where n is a number less than or equal to 5000, and we can pick it.

So we know that in base 2 only odd numbers can be palindromes. Okay so that saves us some computation power, but still not enough.

My first thought was, to compute all the possible odd numbers in base 2 and check for every one of them if it is a palindrome. Then I would count in order how many palindromes there are, and when we reach the target $n$ we can print out the n-th palindrome. I think this is not enough, or my method of finding such numbers is bad. My question is, is there any other property of base 2 palindromes, besides that they are odd, that could help my search ? What kind of technique would you pick for search instead of a simple iteration (that's whagt I have done).
(I am coding in python.)

答:
Notice the following:

0 is a palindrome, 1 is a palindrome

11 is palindrome; so is 00, which is relevant to what follows but this is not a different number in base 2 than 0.

101 and 111 are the "unique" palidromes of length 3; but notice that they both wrap two 1s over the palindromes of length 1 (n - 2). 000 and 010 are the "non-unique palindromes", and 010 represents a number (10) which is not a palindrome when "left-trimmed".

1001 and 1111 are the unique palindrome, and wrap 1s over the palindromes of length n-2 (both unique and non-unique); 0000, 0110 are the non-unique (wrapping 0s around the same n-2 palindromes).

Etc, with this pattern (two arrays, one of "all" palindromes, one of "unique" palindromes; and wrapping 0s or 1s around previous palindromes of length n-2), you should have a simpler way to generate all palindromes. Then, when your "unique" palindrome array reaches the right size, just return the $n$-th element.

Title: Find the n-th base 2 palindrome


问:
Proving that Kurtosis is bounded from below by skewness plus $1$
Let $X$ be a random variable that has finite fourth moments, i.e. $X\in L^4$, and that has a non-zero variance, i.e. $\mathsf V(X)\neq0$. I want to prove that the kurtosis of $X$ is bounded from below by its skewness squared plus 1, i.e. that

$$\mathsf E\left(\left(\frac{X-\mathsf E(X)}{\sqrt{\mathsf V(X)}}\right)^4\right)\geq1+\mathsf E\left(\left(\frac{X-\mathsf E(X)}{\sqrt{\mathsf V(X)}}\right)^3\right)^2.$$

My attempt. If we multiply both sides with $V(X)^3$, it boils down to, where $\lVert\cdot\rVert_{L^p}$ denotes the $L^p$-norm,

$$\lVert X-\mathsf E(X)\rVert_{L^4}^4\lVert X-\mathsf E(X)\rVert_{L^2}^2\geq \lVert X-\mathsf E(X)\rVert_{L^2}^6+\lVert X-\mathsf E(X)\rVert_{L^3}^6.$$

The inequality is homogenous since both sides are of "$6$-th power", so I would expect that this is reducible to some known analytic inequality. Can somebody help me out on how to do that?

答:
This inequality is sometimes called Pearson's inequality. See the following paper for proof and further references:

SKEWNESS, KURTOSIS AND NEWTON’S INEQUALITY (SHARMA AND BHANDARI)


问:
Why is the regularized beta function differentiable?
We have the definition of Regularized Incomplete Beta Function:
$$I_x (a,b) = \frac{B(x;a,b)}{B(a,b)},$$ where $B(x;a,b)$ is the incomplete Beta function, $B(a,b)$ is the Beta function and $x \in [0,1], a,b > 0$.

Is the Regularized Incomplete Beta Function differentiable with respect to x ? If yes why ?

I think Regularized Incomplete Beta Function is differentiable since I found differentiation of Regularized Incomplete Beta Function on functions.wolfram.com website but I don't know why.
Thanks in advanced.

$$B(x;a,b) := \int_{0}^{x}{t^{a-1} (1-t)^{b-1} dt}$$
$$B(a,b) := \int_{0}^{1}{t^{a-1} (1-t)^{b-1} dt}.$$

答: By the fundamental theorem of calculus,$$I_x^\prime=\frac{B_x^\prime}{B(a,\,b)}=\frac{x^{a-1}(1-x)^{b-1}}{B(a,\,b)}.$$This at least ensures differentiability on $(0,\,1)$. As $x\to0^+$, $B_x\sim\int_0^xt^{a-1}dt=x^a/a$, so $B_0=0=\lim_{x\to0^+}B_x$. We can treat $x\to1^-$ similarly.


问:
Second partial derivative test question
I'm looking at the solution for finding maximum and minimum of

$f(x,y) = xye^{-x² -y^2}$

To do so, it first finds the critical points using the second derivative, and these points are $(0,0),({\pm {1\over\sqrt[]{2}}},{1\over\sqrt[]{2}}),({\pm {1\over\sqrt[]{2}}},{-{1\over\sqrt[]{2}}}) $

By inspection it finds that

$
f(0,0)=0
$

$
f({\ {1\over\sqrt[]{2}}},{1\over\sqrt[]{2}})=({-{1\over\sqrt[]{2}}},-{1\over\sqrt[]{2}})={e^{-1}\over 2} > 0
$

$
f(-{{1\over\sqrt[]{2}}},{1\over\sqrt[]{2}})=({{1\over\sqrt[]{2}}},-{1\over\sqrt[]{2}})={-{e^{-1}\over 2}} < 0
$

And it reasons that ${e^{-1}\over 2}, -{e^{-1}\over 2}$ are maximum and minimum.

It also uses the Hessian matrix to find out about $(0,0)$.

The determinant of   $det(H(f)(0,0))=1$  but $f_{xx}=0$.

It then claims that because because the matrix is undefined but it also claims that $(0,0)$ is a saddle point.

So I have two questions:

Is it correct that if  $det(H(f)(a,b)) > 0$ and $f_{xx}=0$ the second partial derivative test is inconclusive?

How can we claim that $(0,0)$ is a saddle point if the second derivative test shows that it is inconclusive?

答:
Saddle point for sure. The Hessian of $f$ is:
$$
H_f = \begin{pmatrix}
2x(2x^2-3) y e^{-x^2-y^2} & (2x^2-1)(2y^2-1) e^{-x^2-y^2}
\\
(2x^2-1)(2y^2-1) e^{-x^2-y^2} & 2y(2y^2-3) x e^{-x^2-y^2}
\end{pmatrix},
$$
thus
$$
H_f(0,0) = \begin{pmatrix}
0 & 1
\\
1 & 0
\end{pmatrix},
$$
and $\mathrm{det}(H_f(0,0)) = -1$ not $1$, which shows $(0,0)$ is a saddle point.

Now to answer your first question.

It is impossible for a smooth function that can have $\mathrm{det}(H_f(x_0,y_0)) > 0$ and $f_{xx}(x_0,y_0) = 0$ at $(x_0,y_0)$.

Let's say if $f_{xx}(x_0,y_0) = 0$, then
$$
H_f(x_0,y_0) = \begin{pmatrix}
0 & f_{xy}(x_0,y_0)
\\
f_{yx}(x_0,y_0) & f_{yy}(x_0,y_0)
\end{pmatrix}.
$$
Unless you construct some special smooth functions that bear the property that $f_{xy} \neq f_{yx}$, the determinant $\mathrm{det}(H_f(x_0,y_0))  = - (f_{xy}(x_0,y_0))^2 \leq 0$.

For the second question: when the test is inconclusive,

$f_{xx}$ and $f_{yy}$ have different signs, then we have a saddle point. You can look at the picture to see the geometric meaning of this (though in example they are zero): If our viewponint is somewhere on the $y$-axis and to observe the change in $x$, $f_{xx}>0$ means what we see is a concave up curve near the neighborhood of the point of interest $(x_0,y_0)$;  Moving our viewponint to somewhere on the $x$-axis and to observe the change in $y$, $f_{yy}<0$ means what we see is a concave down curve. Then, clearly at $(x_0,y_0)$ we have a saddle.

$f_{xx}$ and $f_{yy}$ have the same signs, then we have a local maximum/minimum.

Title: Second partial derivative test question


Body: $$\frac{f(n)}{g(n)}=\frac{n^{3/2}\,\sqrt{\log_2(n)}}{10n^{3/2}+100n}=\frac{\sqrt{\log_2(n)}}{10+100n^{-1/2}}\to\infty.$$


问:
Calculus the extreme value of the function $f(x,y)$
Calculus the extreme value of the $f(x,y)=x^{2}+y^{2}+xy+\dfrac{1}{x}+\dfrac{1}{y}$

pleasee help me.

答: You can solve this with the am/gm inequality, by rewriting the function $ x^2 + y^2 + xy + 1/(3x) + 1/(3x) + 1/(3x) + 1/(3y) + 1/(3y) + 1/(3y)$. This will have an absolute minimum for positive x,y when all the terms are equal.</p>


问:
Evaluating differential forms(volume form) on vector field
I was trying to evaluate the volume form $\Omega=dx_1\wedge \cdots \wedge dx_n$ in $\mathbb{R}^n$ on an vector fields $(X_1,\cdots, X_n)$ where each $X_i\in \Gamma(TM)$. Is there a quicker way to compute this instead of using definition of wedge product to write $\Omega$ as a tensor product?

答: Write out each $X_b=\sum_{a=1}^n\xi_{ab}\frac{\partial}{\partial x^a}$, then write $[\xi_{ab}]$ as an $n\times n$ matrix and calculate the determinant. More generally, if you have the differential $k$-form $dx^{i_1}\wedge \dots \wedge dx^{i_k}$, and you want to calculate its value on the vector fields $X_1,\dots, X_k$, then write out each $X_b=\sum_{a=1}^n\xi_{ab}\frac{\partial}{\partial x^a}$, then write out the $n\times k$ matrix $[\xi_{ab}]$, and calculate the determinant of the $k\times k$ submatrix obtained by taking rows $i_1,\dots, i_k$.


问:
If $x + y \in \mathbb Q$ and $x − 3 y \in \mathbb Q$, prove $x$ and $y$ are rational
Consider $,  ∈ ℝ$, such that $ +  ∈ ℚ$ and $ − 3 ∈ ℚ$. Then $,  ∈ ℚ$.

Hey all. I'm recently working through a course-book that's involved with the math course I'm taking next semester and have come across this problem, although it has no solution (this would be taught in the course but I'm just looking over the content to get ahead)

I'm really stumped on how to solve this problem. My initial guess was to subtract $ + $ from $ − 3$ to prove that $-4y$ would be rational provided both equations were rational, and therefore we can assume $y$ is rational, and could potentially do the same for $x$ by multiplying the first equation... However I don't think this would be valid and am not sure.

How would I go about solving this type of problem? Thanks.

答:
Let $u=x + y$ and $v=x − 3 y$. Then
$$
\pmatrix{u \\v} = \pmatrix{1 & \hphantom-1 \\ 1 & -3} \pmatrix{x \\y}
$$
The matrix has nonzero determinant and so is invertible over $\mathbb Q$. Therefore, $u,v \in \mathbb Q \iff x,y \in \mathbb Q$.

Explicitly,
$$
\pmatrix{x \\y} = \pmatrix{3/4 & \hphantom-1/4 \\ 1/4 & -1/4} \pmatrix{u \\v}
$$


问:
Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed
$x^2-|5x-3|-x<2$

For this problem I took two cases where in one case $5x-3 \geq 0$ and in second case $5x-3 \lt 0$.

$1st$ Case  when $5x-3 \geq 0$ :-

For this case I got the solution set as $x \in [0.6,3+2\sqrt{2}]$.

$2nd$ Case  when $5x-3 \lt 0$ :-

For this case I got the solution set as $x \in [-5,0.6]$.

Upon combining both the cases I am getting the solution set as $x \in [-5,3+2\sqrt{2}]$ but in the answer that has been provided for this problem is not including the points and thus the answer given is $x \in (-5,3+2\sqrt{2})$.

What am I doing wrong here that is making my endpoints of the solution inclusive whereas the the solution set provided has got endpoints exclusive. Please help me on this !

Thanks in advance !

Body:
Given $x^2-|5x-3|-x<2$

when  5x−3≥0, $x^2-6x+1<0$,   x $\in$ [0.6,3+2$\sqrt2$)

when  5x−3$\leq$0, $x^2+4x-5<0$, $(x-1)(x+5)<0$,  x $\in$ (-5,0.6] (You should not include 5 in this case.)

So the answer is x $\in$ (-5,3+2$\sqrt2$).


问:
Integrating $x^x$ and getting a graph
I've heard many times of functions that cannot be integrated. For example, $x^x$, which is the most common. But what I don't know is how could you, even if the graph has no equation, plot this integral. If anyone could give me a graph, I would be extremely pleased. Or at least, could you tell me how to graph it? Maybe in Wolfram Mathematica?</p>

答:
Here is the graph of the function with Maple

Plot( $x^x$, $x=0..2$ );

The graph of the integral of $x^x$ using the truncated Taylor series

$$ x^x=1+\ln  \left( x \right) x+\frac{1}{2}\, \left( \ln  \left( x \right)  \right)
^{2}{x}^{2}+\frac{1}{6}\, \left( \ln  \left( x \right)  \right) ^{3}{x}^{3}+\frac{1}{
24}\, \left( \ln  \left( x \right)  \right) ^{4}{x}^{4}+O(x^5). $$

Title: Integrating $x^x$ and getting a graph


问:
Exercise 2.26 (Idempotents and products) from Eisenbud's Commutative Algebra

Exercise 2.26 (Idempotents and products): Show that $R$ can be written as a direct product of two or more (nonzero) rings iff $R$ contains a non-trivial idempotent. Show that if $e$ is an idempotent, then $R = Re\times R(1-e)$ and that $Re$ may be realized as a localization, $Re = R[e^{-1}]$.

I'm most interested in showing that $Re = R[e^{-1}]$. I assume that $R[e^{-1}]$ means localizing at the multiplicative set $U = \{ 1,e \}$ where the fact that $U$ is multiplicative follows from idempotence: $e^2=e$.

After writing it out, I can't seem to picture what $R[e^{-1}]$ looks like. I show below that $r/1=r/e$ in $R[e^{-1}]$ so does that mean $R[e^{-1}]$ have the same number of elements as $R$? If so, why wouldn't the same proof show that $R\cong Re$? What is special about $R[e^{-1}]$ that makes this work? Thanks for your help!

I will use the universal property of localization. Let $\varphi:R\to Re$ be the ring homomorphism $r\mapsto re$. (Here, we are treating $Re$ as a ring in its own right with $e\in Re$ being the identity element.) We see that $\varphi(1)=\varphi(e)=e$ and hence $\varphi$ maps $U$ to units.

By the universal property, there is a  map $\psi:R[e^{-1}]\to Re$ given by the following definition (recall that $s=1,e$):
$$ \psi\left(\frac{r}{s}\right) = \varphi(r)\varphi(s)^{-1} = \varphi(r)e = re^2 = re $$
It is clearly surjective. Notice that $r/1=r/e$ because $e(er-r)=er-er=0$. Therefore, to show injectivity it suffices to show that $re=r'e$ (in $Re$) implies $r/1=r'/1$ (in $R[e^{-1}]$). Suppose $re=r'e$ then $(r-r')e=0$. But this exactly means that $r/1=r'/1$. So $\psi$ is a ring isomorphism as desired.

答:
Question: "Exercise 2.26 (Idempotents and products): Show that R can be written as a direct product of two or more (nonzero) rings iff R contains a non-trivial idempotent. Show that if e is an idempotent, then R=Re×R(1−e) and that Re may be realized as a localization, Re=R[e−1]."

Partial-answer:  Assume $A\cong A_1 \oplus A_2$. It follows $e:=(1,0)\neq 1$ is a non trivial idempotent. Assume conversely that $1\neq e\in A$ is a non trivial idempotent.
Let $I:=(e), J:=(e-1) \subseteq A$ be the ideal generated by $e,e-1$. It follows $I+J=(1)$ hence these ideals are coprime. Moreover $IJ=I\cap J$. Since $IJ:=A(e(e-1))=A(e^2-e)=A(0)=(0)$ is the zero ideal, we get by the CRT an isomorphism

$$ A\cong A/IJ\cong A/I \times A/J$$

where $A/I, A/J \neq 0$. Hence by the CRT we get that $A$ has a non-trivial idempotent iff $A$ may be written as a (non-trivial) direct sum of two rings.


问:
Proving an inequality in number theory.
I am going through Apostol's number theory text and there is a step in the proof of theorem 4.1 that is not clear to me.
It  is this inequality which I am not understanding how to prove.
$$\sum_{ 2 \leq  m\leq \log_{2}x} x^{1/m} \log \left(x^{1/m}\right) \leq \log_{2}x \sqrt{x} \log(\sqrt x)$$

答: The largest term is $\sqrt x\log( \sqrt x)$, and there are fewer than $\log_2 x$ of them.

Title: Proving an inequality in number theory.


问:
Algebraic simplification into difference of squares
I have an equation of the form:

$$
f(x) = x^2 + y^2 - (r+1) xy
$$

where $r > 0$.

I want to write this into the difference of squares form: $(A - B)^2$. I have tried to rewrite this into:

$$
f(x) = (x-y)^2 + (1-r)xy
$$

I am stuck with the term $(1-r)xy$. Any hints would be appreciated! Thanks in advance.

答:
Since you call it $f(x)$, not $f(x,y)$ or $f(x,y,r)$, I am going to assume that $x$ is more important than $y$. I suggest treating $y,r$ as constants and completing the square on $x$:
$$\begin{align}x^2+y^2 - (r+1)xy &= x^2 -2\dfrac{(r+1)y}2x + \dfrac{(r+1)^2y^2}4 +y^2-\dfrac{(r+1)^2y^2}4 \\&= \left(x - \dfrac{(r+1)y}2\right)^2 - \left(y\sqrt{\dfrac{(r+1)^2}4-1}\right)^2\end{align}$$

This still requires $r \ge 1$ to be strictly real.


问:
Expressing the subscript that produces the largest element in a set
If I have a set $D=\{ d_1, d_2, ..., d_n \}$, and I wish to find $k\, (\leq n)$, where $d_k$ is the largest element in $D$, is it understandable that I write

$$
k = \arg\max_i \big( \big\{ d_i | d_i \in D \big\} \big)
$$

, or is there a more rigorous / acceptable / simpler way of expression?

答:
Yes that is understandable. If you want to make it explicit that $n$ is the largest index you are considering, you could write
$$k = \text{argmax}_i \{d_i | 1 \leq i \leq n\}.$$
As noted in a now-deleted comment, it is perfectly rigorous to say something like "let $k$ be such that $d_k$ is the largest element of $D$" which is both rigorous and more readable in a proof.


答:
The question is a little confusing.  Sets don't normally have an order, so it's not clear what the subscripts mean.   If you want the number that is the largest element of $D$ then $$\max D$$ is the simplest way to write it.

If $D$ is actually a sequence, rather than a set, then the subscripts do make sense.  But now you have the possible confusion that there might be more than one largest element.  If $D = \langle 1, 4, 2, 4, 4\rangle$ are you looking for $i=2, i=4$, or $i=5$?

Supposing that you don't care which subscript you get, as long as it identifies one maximum element, probably the simplest thing to say is:

Let $k$ be such that $d_k = \max D$.

This is simple, clear, and also 100% rigorous.

Title: Expressing the subscript that produces the largest element in a set

Body:
If I have a set $D=\{ d_1, d_2, ..., d_n \}$, and I wish to find $k\, (\leq n)$, where $d_k$ is the largest element in $D$, is it understandable that I write

$$
k = \arg\max_i \big( \big\{ d_i | d_i \in D \big\} \big)
$$

, or is there a more rigorous / acceptable / simpler way of expression?


问:
Exercise solved topology
I do not understand the resolution of this exercise

For any pair of integers $a, b \in \mathbb{R}$ , with $b > 0$, let’s write $N_{a,b}= \{ a+kb |  \in \mathbb{Z} \} $.  Prove the following facts:

arithmetic progressions $\beta = \{ N_{a,b} | a, b   \in \mathbb{Z},>0 \} $ form a basis for a topology $\tau$ on $\mathbb{Z}$

every $N_{a,b}$ is both open and closed in $\tau$ ;

call $P = \{ 2, 3, . . . \} \subset \mathbb{N}$ the set of primes. Then

$\mathbb{Z} − \{ −1, 1 \} = \cup \{ N_{0,p} | p \in P \}$

Therefore if $P$ were finite $\{−1, 1 \}$ would be open in $\tau$ .

Solutions

We need to show that the family of progressions $N_{a,b}$ satisfies this Theorem (but
this is an immediate consequence of the formulas) (Why was the theorem not used?)

(Theorem
Let $X$ be a set and $\mathcal{B} \subset \mathcal{P} (X)$ a family of subsets. There exists a
topology on $X$ for which $\mathcal{B}$ is a basis if and only if two conditions hold:

$X = \cup \{ B | B \in \mathcal{B} \}$

for any pair $A, B \in \mathcal{B}$ and any point $x \in A \cap B$ there exists $C \in \mathcal{B}$ such that $x \in C \subset A \cap B$)

$N_{0,1} = \mathbb{Z}$,  $N_{a,b} \cup N_{c,d} = \cup \{ N_{s,bd} | s \in N_{a,b} \cap N_{c,d} \}$ (
(I do not understand this)

As $N_{a,b}$ is the complement in $\mathbb{Z}$of the open union

$N_{a+1,b} \cup N{a+2,b} \cup . . . \cup N_{a+b−1,b}$ , (I do not understand this)

the open set $N_{a,b}$ is also closed. Note that any non-empty open set contains at least
one arithmetic progression, and hence it must be infinite.

Thanks

答:
$N_{0,1} = \Bbb Z$ is clear, and shows that condition 1 of the basis theorem is satisfied, even one basic element already covers the whole space.

To show 2, the statement you gave is not enough. Let $x \in N_{a,b} \cap N_{c,d}$.

So $x= a+k_1b = c + k_2d$ for some $k_1,k_2 \in \Bbb Z$. Now let $e:=(b,d)$ be the gcd of $b$ and $d$, it's then clear that $x \in N_{x,e} \subseteq N_{a,b}\cap N_{c,d}$ so that 2 is satisfied. (this is an improvement on the wrong argument you cited.)

So the $N_{a,b}$ form a base. If $x \in \Bbb Z$ and $b>0$, the clases modulo $b$ divide $\Bbb Z$ into $b$ many disjoint classes which are $N_{0,b}, N_{1,b}, \ldots N_{b-1,b}$, covering $\Bbb Z$, depending on whether the remainder modulo $b$ is $0, \ldots, b-1$.

So all sets $N_{a,b}$ are also closed as the complement is the union of the other $b-2$ sets in this partition, and thus open.

The final remark that $$\Bbb Z\setminus \{-1,1\} = \bigcup_{p \in P} N_{0,p}$$
follows trivially from the fact "any integer unequal to $1$ or $-1$ has a prime divisor".

If $P$ were finite, that union would be closed as a finite union of closed sets and $\{-1,1\}$ would be open (and finite), a contradiction, as all open sets are unions of basic open sets, and these are all infinite. So finite open sets cannot exists (except $\emptyset$)

Title: Exercise solved topology


答:
Here is a quick picture of $\ x\mapsto \int_0^x t^t\;dt$

I added the plot of the real and imaginary part of the integral in the complex plane $\ z\mapsto \int_0^z t^t\;dt\ $ with $z=x+iy$ (note the 'branch jump' on the negative axis $(-\infty,0)$).

Title: Integrating $x^x$ and getting a graph

Body: I've heard many times of functions that cannot be integrated. For example, $x^x$, which is the most common. But what I don't know is how could you, even if the graph has no equation, plot this integral. If anyone could give me a graph, I would be extremely pleased. Or at least, could you tell me how to graph it? Maybe in Wolfram Mathematica?</p>


问:
Space of weak$^*$-weak-continuous operators is closed
In the book Tensor Norms and Operator Ideals by A. Defant and K. Floret, pg. 48, the authors use the fact that the subspace
$$Z = \{ T \in \mathcal{L}(X^*,Y) : T\; \text{is weak}^*\text{-weak-continuous} \}$$
is closed in $\mathcal{L}(X^*,Y)$. Here, $X,Y$ are Banach spaces and $\mathcal{L}(X,Y)$ denotes the space of all bounded linear operators from $X$ to $Y$ with the usual norm.

I've tried to prove this but found no success. Can somebody point me to a proof or reference of this fact? Thanks in advance!

答:
If the sequence $\{T_n\}_n\subseteq Z$ converges to $T\in  \mathcal L(X^*, Y)$, we must to prove that $T$ lies in $Z$.  In order to do
this it is enough to show that, for every $y^*\in Y^*$, the linear functional
$$
  x^*\in  X^*\mapsto \langle T(x^*),y^*\rangle \in {\mathbb R}
  $$
is $w^*$-continuous.  This functional is clearly the norm limit of the functionals
$$
  x^*\in  X^*\mapsto \langle T_n(x^*),y^*\rangle \in {\mathbb R}
  $$
which are $w^*$-continuous since the $T_n$ lie in $Z$ by hypothesis.  The conclusion thus follows from the fact that the
$w^*$-continuous linear functionals on $X^*$ (aka the cannonical image of $X$) is closed in $X^{**}$.


问:
Prove Positivity of a Function Involving $\log$.
I want to prove

The function $f:(0,\infty)^2 \to \mathbb{R}$ defined by
\begin{equation}\label{key}
   f(x,y) =  \ln \left( \dfrac{1+y}{1+ x}\right) + \dfrac{x}{1+ x} \ln \left( \dfrac{x}{y}\right)
  \end{equation}
is non-negative.

My attempt was to compute   $\dfrac{\partial f}{\partial y}$ and then see where the zeros of the function are.

I think that works fine, but I was wondering if there was a slicker way to do it.

答:
Here is an elementary proof without calculus: Combine the logarithms and obtain
$$
f(x,y)=\ln\left[\frac{1+y}{1+x}\cdot\left(\frac{x}{y}\right)^{\frac{x}{1+x}}\right].
$$
We have to show that the argument of the logarithm is always $\geq 1$. This is equivalent to
$$
\frac{1+y}{1+x}\geq \left(\frac{y}{x}\right)^{\frac{x}{1+x}}.
$$
However, using Bernoulli's inequality and keeping in mind that $0<\frac{x}{1+x}\leq 1$ and $\frac{y-x}{x}>-1$, we have
$$
\left(\frac{y}{x}\right)^{\frac{x}{1+x}}=\left(1+\frac{y-x}{x}\right)^{\frac{x}{1+x}}\leq 1+\frac{y-x}{x}\cdot\frac{x}{1+x}=1+\frac{y-x}{1+x}=\frac{1+y}{1+x}
$$
and so we are done!


问:
How to prove $A=\mathcal{O}_n$, use trace of matrix
Let $A=B+C$, and $B=B^T$, $C=-C^T$, $BC=\mathcal{O}_n$. Prove that if $A^2=\mathcal{O}_n$, then $A=\mathcal{O}_n$.
I want to prove $B=C=
\mathcal{O}_n$, and I got that
$$A^2=B^2+CB+BC+C^2\implies0=\mathrm{tr}(A^2)=\mathrm{tr}(B^2)+\mathrm{tr}(C^2)$$
So$$\mathrm{tr}(BB^T)=\mathrm{tr}(CC^T)\geqslant0.$$
But I don't know how to go on.

答:
I was reading your answer and it’s seems fine to me, but there is something that I want to say, just to make it clear. I’m gonna present here two possible resolutions.

I’ll start by a Lemma that will be important.

Lemma. Let $n \in \Bbb{N}$ and $A \in \mathcal{M}_n(\Bbb{K})$ such that $A$ is symmetric matrix. Then, $A = \mathcal{O}_n$ if and only if $A^2$.

Proof. $\implies.$ This is trivial. If $A= \mathcal{O}_n$, then clearly $A^2=\mathcal{O}_n$.

$\Longleftarrow.$ Suppose that $A^2 = \mathcal{O}_n$. Since, $A$ is symmetric, we have that $AA^T = \mathcal{O}_n = A^TA$. Let $i \in \Bbb{N}$ such that $i \leq n$. Then,
\begin{align*}
[A^TA]_{ii} = 0 & \implies \sum_{i = 1}^n [A^T]_{ik}[A]_{ki} = 0 \\
& \implies \sum_{i=1}^n [A]_{ik}^2 = 0 \\
& \implies [A]_{ik} = 0 \qquad \forall k \in \Bbb{N} \colon k \leq n.
\end{align*}
Hence, $A = \mathcal{O}_n$. $\square$

Now, let’s start with the proof of the result.

Proof. Let $n \in \Bbb{N}$ and $A,B,C \in \mathcal{M}(\Bbb{K})$ such that $A = B+C$, $B^T = B$, $-C^T = C$ and $BC = \mathcal{O}_n$. Suppose that $A^2 = \mathcal{O}_n$. Hence,
\begin{align*}
A^2 = \mathcal{O}_n & \iff (B+C)(B+C)=\mathcal{O}_n \\
& \iff B^2 + BC + CB + C^2 = \mathcal{O}_n.
\end{align*}
By hypothesis, we have that
\begin{align*}
BC = \mathcal{O}_n & \iff (BC)^T = (\mathcal{O}_n)^T \\
& \iff C^TB^T = \mathcal{O}_n \\
& \iff -CB = \mathcal{O}_n \\
& \iff CB = \mathcal{O}_n.
\end{align*}
So, we have that $\mathcal{O}_n = B^2+C^2$. Then, it follows that
\begin{align*}
\mathcal{O}_n = B^2 + C^2 & \iff \mathcal{O}_n = B^2(B^2 + C^2) \\
& \iff B^4 + B^2C^2 = \mathcal{O}_n \\
& \iff B^4 = \mathcal{O}_n.
\end{align*}
Then, by the previous Lemma, we have that $B^4 = \mathcal{O}_n \iff B^2 = \mathcal{O}_n \iff B = \mathcal{O}_n$. Then, it follows that $C = \mathcal{O}_n$ (because $C$ is skew symmetric and $C^2 = \mathcal{O}_n \iff CC^T = \mathcal{O}_n$). Therefore, we have that $A = \mathcal{O}_n$. $\square$

My point is be careful when using the trace of a matrix. Note that, in general, $\text{tr}A^2 = 0 \iff A = \mathcal{O}_n$ is false. The $\Longleftarrow$ implication if trivially true, but the $\implies$ implication is false (for example, take $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.) Although, it is true that $\text{tr}A^2 = 0 \iff A = \mathcal{O}_n$ for symmetric matrices. (I’m gonna leave the proof of this result as an exercise).

In short always be careful when justifying this sort of things using the trace of a matrix.

Finally, I would like to suggest another solution.

Proof. Since $A = B + C$, where $B$ is a symmetric matrix and $C$ is a skew symmetric matrix, then $B = \frac{1}{2}(A + A^T)$ and $C = \frac{1}{2}(A - A^T)$ (because every matrix can be expressed in a unique way as a sum of a symmetric and a skew symmetric matrices). Since $BC = \mathcal{O}$, we have that $AA^T=A^TA$. But note that $AA^T=A^TA$ is a symmetric matrix and, since $A^2 = \mathcal{O}_n$, we have that $(AA^T)^2 = \mathcal{O}_n$. Then, by our previous Lemma, we have that $AA^T = \mathcal{O}_n$. So clearly we have that $\mathcal{O}_n$. $\square$

This end up being too long than I expected, but if you have any doubt don’t hesitate, you can always ask. Continuation of a good work. $\ddot\smile$