问:
Normal vector of an implicit curve
We consider the equation $y^2-4y+4+x^2y+\arctan(x)=0$, the exercice of my book say that the normal vector $n$ of the cartesian curve implicitally defined by the previous equation is $n=l(1,0)$ for $l \neq 0$. I don't understand the reason of this affermation, because I think that if $y(x)$ is the implicit function obtained form the equation, the parametric equation of the curve is $(t, y(t))$, so the first coordinate of the normal vector is zero.
I would therefore expect that the normal vector is $n=l(0,1)$. Is it right?

答:
$$f(x,y)=x^2 y+\arctan x+y^2-4 y+4=0\tag{1}$$
The points where the normal is parallel to $(1,0)$ are those where the tangent is vertical.

This happens when looking for intersection with a vertical line $x=k$ the equation gives a double root.Plug $x=k$ into equation $(1)$

$$k^2 y+\arctan k+y^2-4 y+4=0$$
discriminant is zero when
$$k^4-8 k^2-4 \arctan k=0$$
which gives $4$ values  for $x$ as can be seen in the image below

$$x=0,x=-2.71,x=-0.48,x=2.93$$

Points are $$(0,2),(-2.71,-1.67),(-0.48,1.89),(2.93,-2.29)$$

Title: Normal vector of an implicit curve


问:
How to sum the series $\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}$
I have been tinkering with telescoping sum and I've encountered by chance this series:

$$\sum_{n=4}^{+\infty}\dfrac{1}{(2n-7)(2n-6)}=\log(2)$$

I use Wolfram to sum this series because I am not sure if it is a telescoping sum. I wonder how can you sum this series? By converting the infinite sum to an improper integral?

I integrate this function:

$$\int_{4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}dx=\dfrac{\log(2)}{2}$$

I am new to improper integration and numerical infinite series. I thought the sum of the series and the result of the integral should match, but they don't.

It seems that I miss something very basic and crucial. I am learning myself Cal 3 so I may miss some crucial theorems. What are conditions for the sum of the series and the limit of the integral to be the same?

Come to think of it, summation is just a discrete version of integration, so the values naturally cannot be the same.

Is there a way to sum this series and obtain its sum? This doesn't seem to be a telescopic series in my humble opinion.

答:
$S=\displaystyle \sum_{n=4}^\infty \frac{1}{(2n-7)(2n-6)}=\sum_{n=4}^\infty\left(\int_{0}^1x^{2n-8}-x(x^{2n-8}) \mathrm{d}x\right)$

$=\displaystyle\int_{0}^1 \left(\sum_{n=4}^\infty x^{2n-8}-x(x^{2n-8}) \right)\mathrm{d}x$

$\text{(Note: the interchange of the summation is legitimate since the limits of integration are from 0 to 1)}$

$$S = \int_{0}^{1} \frac{1-x}{1-x^2} \mathrm{d}x=\log2 $$

ContentLicense: CC BY-SA 4.0


问:
A mapping of an infinite set into itself has an invariant subset
Let $E$ be an infinite set and $f:E\rightarrow E$. Then there exists a subset $S$ of $E$ such that $\emptyset\neq S\neq E$ and $f(S)\subset S$. </p>

<p>This seems easy, but I don't have an idea. A hint would be nice. </p>

答:
Thanks to a hint of user6312 and on his suggestion:

Let $a$ be an element of $E$. If there exists $n\geq 1$ such that $f^n(a)=a$, then define $S=\{a,f(a),f^2(a),\ldots,f^{n-1}(a)\}$. Since $S$ is finite, $S\neq E$. Obviously, $f(S)\subset S$.

If $f^n(a)\neq a$ for all $n\geq 1$, then define $S=\{f^n(a)|n\geq 1\}$. Obviously, $f(S)\subset S$, and since $a\notin S$, $S\neq E$.

ContentLicense: CC BY-SA 3.0


问:
How to find remainder?
$$a=r\mod (r+1) \ \ \forall r\in\{2,3,4,\dots,9\}$$

Then how do we find $'x'$ if $$a=x\mod 11$$

I get $$2a=9\mod11$$ but that does not help.

Please keep solution simple , i don't now number theory.

The above is the crux what I got from the question:

Let $n_1,n_2,... $ be an increasing sequence of natural numbers each of which leaves remainder $r+1$ when divided with $r\in\{2,3,....9\}$. Find the remainder when $n_{2008}$ is divided with $11$.

答: Find the least common multiple $L$ of the numbers $3,4,\dots,10$. Then $$a=L-1+kL$$ where $k$ is arbitrary. There is no restriction on $a$ modulo $11$ (unless there's more information on $a$, such as if you take the smallest possible $a$, which is $L-1$). </p>

Title: How to find remainder?


问:
Why do we care about the Hilbert Cube?
The Hilbert Cube is defined to be the countable infinite Cartesian products of the interval $[0,1]$ or anything homeomorphic to $[0,1]$. Why do we care about this object?

答: In addition to Henno's answer and many comments, here is my take, as a geometric topologist. Below is a quote from the introduction to Chapman's book "Lectures on Hilbert cube manifolds" (Q-manifolds, which are analogues of topological manifolds, except that the local structure is modeled on the Hilbert cube):

Title: Why do we care about the Hilbert Cube?


问:
Spanning tree created with cycles breaking different form the Maximal Spanning Tree.
Let $G$ be a connected graph with $N$ nodes, $B$ the set of the nodes and $V$ the set of the edges of $G$.

For all edges $e \in V$, there is an associate positive weight $\nu_e > 0$ on this edge ($\nu_e=0$ if there is no connection). We suppose that there are no self-connecting edges (an edge that connects a node to himself) and that all the edges have different weigth.

It is very easy, using Kruskal or Prim's algorithm, to find a maximal spanning tree $T_{max} \subseteq V$ that is the spanning tree that maximizes the graph sum $$\sum_{e \in T_{max}} \nu_e.$$

Now, I want to operate on $G$ following this instruction :

For every cycle $c$, I remove the edge of the cycle that have the most minimal weight (and I do nothing if this edge has already been removed before).

At the end of this simple algorithm, we get a spanning tree $T$ that covers every edges, and by definition of the maximal spanning tree we have :

$$\sum_{e \in T} \nu_e \leq \sum_{e \in T_{max}} \nu_e.$$

However, on all the examples I've tried, I always end up with both sums being equal... (I end up with $T=T_{max}$)

Is their a counter-example that would verify the following
$$\sum_{e \in T} \nu_e <\sum_{e \in T_{max}} \nu_e.$$

Or maybe the tree $T$ constructed using this method has to be a maximal spanning tree ?

Any help or ideas are welcomed.

答:
The tree you get this way is guaranteed to be a maximal spanning tree.

If $T_\max$ is a maximal spanning tree, and $C$ is any cycle, then $T_\max$ can't possibly include the minimal-weight edge $e$ of $C$. Suppose it did. Deleting $e$ leaves $T_\max$ in two components, and since $C$ contains at least one edge crossing from one component to the other (the edge $e$) it must contain at least two. So there's some other edge $e'$ lying on $C$ which crosses from one component to the other, and which we can add to $T_\max - e$ to make it connected again.

But this is an improvement: we've deleted $e$, and added $e'$ which has higher weight than $e$. This contradicts our assumption that $T_\max$ was a maximal spanning tree.

As a result, $T_\max$ already can't include any minimal-weight edge of any cycle. If we go through the cycles and delete their minimal-weight edges, we never lose any edges of $T_\max$, so the remaining subgraph must contain $T_\max$. However, the remaining subgraph is a tree (it doesn't have any cycles, we ensured that) and so it must be $T_\max$.


问:
"Introduction to Smooth Manifolds" - differential of a function query
I'm trying to understand the differential $df$ as an approximation to $\Delta f$, in Lee's Introduction to Smooth Manifolds (p282-283 - see image below). He says “let $p$ be a point on $M$” but then, in the diagram, labels $p$ in $U$ but not on $M$. Why does he do that?

答: Paul Frost's answer is completely correct. But I also want to add that what's really going on here is an example of using a coordinate chart to "identify" an open subset of $M$ with the corresponding open subset of $\mathbb R^n$. I discuss this at some length starting at the bottom of page 15.


答:
Let $e_1, e_2,\dots,e_m$ be an ordering of the edges such that $\nu_{e_1} \leq \nu_{e_2} \leq \dots \leq \nu_{e_m}$. If you run Kruskal's algorithm according to this ordering, the edges of the tree you get will be precisely those $e_i$ which are not in a cycle in the subgraph $G_i = (B, \{e_j, j \geq i\})$ of $G$. (Equivalently, those $e_i$ whose endpoints lie in different connected components of $G_{i+1}$.) On the other hand, your algorithm will delete those edges $e_i$ which are in a cycle in $G_i$. So the two algorithms actually give the same spanning tree!

Although this might not be useful to you right now, I can't resist adding a note about this question from the point of view of matroid theory. The above observations actually work in any matroid to show that your algorithm gives a maximum weight basis. There is actually a deeper underlying reason for this. The complement of a maximum weight basis is a minimum weight basis in the dual matroid. So running the greedy algorithm in the dual matroid to find a minimum weight basis and then taking the complement gives a maximum weight basis in the original matroid, and this is exactly what your algorithm does!

Title: Spanning tree created with cycles breaking different form the Maximal Spanning Tree.

Body:
Let $G$ be a connected graph with $N$ nodes, $B$ the set of the nodes and $V$ the set of the edges of $G$.

For all edges $e \in V$, there is an associate positive weight $\nu_e > 0$ on this edge ($\nu_e=0$ if there is no connection). We suppose that there are no self-connecting edges (an edge that connects a node to himself) and that all the edges have different weigth.

It is very easy, using Kruskal or Prim's algorithm, to find a maximal spanning tree $T_{max} \subseteq V$ that is the spanning tree that maximizes the graph sum $$\sum_{e \in T_{max}} \nu_e.$$

Now, I want to operate on $G$ following this instruction :

For every cycle $c$, I remove the edge of the cycle that have the most minimal weight (and I do nothing if this edge has already been removed before).

At the end of this simple algorithm, we get a spanning tree $T$ that covers every edges, and by definition of the maximal spanning tree we have :

$$\sum_{e \in T} \nu_e \leq \sum_{e \in T_{max}} \nu_e.$$

However, on all the examples I've tried, I always end up with both sums being equal... (I end up with $T=T_{max}$)

Is their a counter-example that would verify the following
$$\sum_{e \in T} \nu_e <\sum_{e \in T_{max}} \nu_e.$$

Or maybe the tree $T$ constructed using this method has to be a maximal spanning tree ?

Any help or ideas are welcomed.


问:
Stable and Unstable Manifolds Computation
I want to compute (approximately) the stable and unstable manifolds around the origin of the following system.

$x' = 6x + 8y + (sinx)^2 - y^2$

$y' = 8x - 6y + x^2 + xy$

How can I solve it? Thank you very much.

答:
Making $y=h(x) = \sum_{k=1}^n a_k x^k$ over the manifolds we have

$$
\dot y = h_x(x)\dot x
$$

or

$$
8x-6h(x)+x^2 + x h(x) = h_x(x)\left(6x+8h(x)+\left(x^2-\frac{x^4}{3}\right)-h^2(x)\right)
$$

considering $n = 3$ we have after reducing to a null polynomial

$$
\left\{
\begin{array}{l}
 4 \left(a_1+2\right) \left(2 a_1-1\right) =0\\
 -a_1^3+6 \left(4 a_1+3\right) a_2-1 =0\\
 16 a_2^2+\left(1-4 a_1^2\right) a_2+8 \left(4 a_1+3\right) a_3 =0\\
\end{array}
\right.
$$

and after solving we have

$$
\cases{
h_1(x) = -\frac{1183 x^3}{18000}+\frac{7 x^2}{30}-2 x\\
h_2(x) = -\frac{9 x^3}{16000}+\frac{3 x^2}{80}+\frac{x}{2}
}
$$

Those are two representations for unstable manifolds (in red) depicted in the following figure

NOTE

$\left(x^2-\frac{x^4}{3}\right)$ represents the first two term series representation for $(\sin x)^2$


答:
Use the Chinese reminder Theorem. It looks to me that you have the relations
$$\begin{align}
a&\equiv4\mod{5}\\
a&\equiv6\mod{7}\\
a&\equiv7\mod{8}\\
a&\equiv8\mod{9}\\
\end{align}$$
and all of the other relations give redundant information. There is a unique solution modulo $5\cdot 7\cdot8\cdot9=2520$. Since each of the moduli divide this number, $2519$ is a solution for $a$. So $a\equiv 2519\mod{2520}$. Now what would $a$ be modulo $11$? Anything is possible since $11$ does not divide $2520$. Are you sure this is the question that was asked? If $a$ is supposed to be the smallest positive solution, then $x\equiv2519\equiv0\mod{11}$. If $a$ is the smallest solution in absolute value, then $x\equiv-1\mod{11}$.

Now with the full question posted, do you see how $n_1=2519$? Continue to find $n_{2008}$.

Title: How to find remainder?

Body:
$$a=r\mod (r+1) \ \ \forall r\in\{2,3,4,\dots,9\}$$

Then how do we find $'x'$ if $$a=x\mod 11$$

I get $$2a=9\mod11$$ but that does not help.

Please keep solution simple , i don't now number theory.

The above is the crux what I got from the question:

Let $n_1,n_2,... $ be an increasing sequence of natural numbers each of which leaves remainder $r+1$ when divided with $r\in\{2,3,....9\}$. Find the remainder when $n_{2008}$ is divided with $11$.


问:
Find curve using midpoint theorem

Determine a curve such that, at every point, the tangent exists and is
not horizontal. The segment of this tangent between the axis has as
middle point, the contact point of the curve, and the curve touches
the points $$(2,4), (-2,4)$$

I'm having trouble with this problem. I have tried to use the midpoint theorem, but I don't see it.

答:
Tangent $AB$ has equation
$$Y-y=y'(X-x)$$
we have axis intersection points $A\left(\frac{xy'-y}{y'},0\right)$ and $B\left(0,y-xy'\right)$

so the midpoint is
$$P\left(\frac{xy'-y}{2y'},\frac{y-xy'}{2}\right)$$
as $P$ is on the curve we must have
$$\begin{cases}
x=\frac{xy'-y}{2y'}\to y+xy'=0\\
y=\frac{y-xy'}{2}\\
\end{cases}
$$
Consider the second equation, which is equivalent to the first
$$2y=y-xy'\to \frac{y'}{y}=-\frac{1}{x}\to \log y =-\log x+C\to y=\frac{k}{x}$$
as the curve must pass through $(2,4)$ we have $k=8$ so the solution is
$$y=\frac{8}{x}$$

Title: Find curve using midpoint theorem


问:
What significance is there if a theorem can be proved in the same system in more than one way?
Doesn't that imply some sort of redundancy or point to some inner connection between the axioms? Or is it the case that when all theorems used in the proofs are broken down to their axioms that the two proofs must in fact be equivalent?</p>

答:
Some proofs are of a very different style from others; geometrical vs algebraic vs formal vs generalized, etc. Technically everything known in mathematics can be proven from the axioms; that doesn't mean those theorems are all superfluous. One gauge of a mathematician's breadth of knowledge is how many totally different proofs of the Pythagorean theorem they know.  (I still only know two).
Mathematics is highly interconnected; we usually have several ways to say the same thing, and it is often much easier to prove something one way than another.  That is the whole reason different coordinate systems exist, for example.
We can write down the axioms in a small space; but the endless consequences of those axioms, and the chains of thought to get there, are interwoven and form a rich structure.</p>


问:
How to solve cross and scalar product equation knowing length and angle
We have two vectors $a$ and vector $b$. they are both non zero vectors and the angle between them is $\frac{\pi}{6}$. Their lenghts are connected with equation $|b| = \sqrt{3}|a|$

We need to solve this equation (*...scalar product, $\times$... cross product):

I find it very hard since I know that scalar product is not associative.

My try:
$$ (x*a+x*b)*a + x \times b = 2a\times b + 3|a|^{2}b\\
(x*a)*a+(x*b)*a + x \times b = 2a\times b + 3|a|^{2}b$$
If I solve the distance of $2a\times b$, I get nowhere. I do not know hot to continue.

The solution is $x = 2a+ \frac{2}{9}b+ 2a\times b $

答:
Use the fact that $a$, $b$, and $2a\times b$ is a basis.  Let
$$x=\alpha a+\beta b+\gamma (2a\times b).$$
Multiply out and compare coefficients; use that $3\|a\|^2=\|b\|^2$ and $2\langle a,b\rangle=\|b\|^2$.


问:
Need help to resolve equation with delta
Hello I need to solve this equation, I have the result but I do not understand how we solve it:

$\Delta n$ is the difference between the expected and observed number in a $\chi^2$, we try to find out how many individuals would have been needed to reject $H_0$.

I do not understand the part where $\Delta n \ge 14.919$ becomes $n \ge 246.669$ or $n \le 216.831$.

答: The average of $246.669$ and $216.831$ is $231.75,$ which is $14.919$ away from either bound.


问:
Does the path component functor admit a right adjoint?
Does $\pi_0:Top\to Set$, sending a topological space to its path-connected components, admit a right adjoint?

If not, does it admit a right adjoint if we restrict it to certain topological spaces, say CW complexes?

答:
As someone mentioned, this can't be done in general, but if we restrict ourselves to locally path connected spaces (in which path components and connected components coincide) $\pi_0$ has a right adjoint.

Let $Top$ denote the category of locally path connected spaces,
define $F: Set \rightarrow Top$ by taking a set $A$ to $A$ with the discrete topology. Since continuous maps preserve connected components, the value of a function
$$f: X \rightarrow A$$
in a connected component is determined by a single point in the component. Thus
$$\mathrm{Hom}_{Top}(X,F(A)) \cong \mathrm{Hom}_{Set}(\pi_0(X),A).$$


问:
positive semidefinite, positive definite?
Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$. Consider the $n \times n$ matrix $A=(a_{ij})$.
Then

It is possible to choose $a_1.\dots,a_n$ such that $A$ is non-singular

matrix $A$ is positive definite if $(a_1,\dots,a_n)$ is nonzero vector

matrix $A$ is positive semi definite for all $(a_1,\dots,a_n)$

for all $(a_1,\dots,a_n)$, $0$ is an eigen value of $A$

Please help. I can't even make a guess.

答:
If $v=\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix}$ then $A=(a_{ij})=vv^T$

(1) is false for such a matrix always has rank no more than 1

(2) is false for positive definite matrices better be invertible!

(3) is true, for $x^TAx=x^Tvv^Tx=(v^Tx)^Tv^Tx$ which is a non-negative real number for all vectors $x$ as it is just self inner product.

(4) is true if $n>1$ for then $A$ is singular.


问:
Understanding the solution of #14.5.1 in Artin
Here is the question:

Determine a presentation matrix $A$ as an $R-$module for the ideal $ I = (2, 1 + \sqrt{-5})$ where $R = \mathbb Z[\sqrt{-5}].$

Here is a solution of the question:

Exercise 14.5.1. Let $R=\mathbb{Z}[\delta]$ where $\delta=\sqrt{-5}$. Determine a presentation matrix as an $R$-module for the ideal $(2,1+\delta).$

Proof. Let $\varphi:R^2\longrightarrow I$ that sends $(x,y)\rightsquigarrow2x+(1+\delta)y$. Now
$$\ker\varphi=\left\{(x,y)\in R^2\,\middle|\,x=-\frac{1+\delta}{2}y\right\}\leftrightarrow\left\{y\in R^2\,\middle|\,\frac{1+\delta}{2}y\in R\right\}$$
where $\leftrightarrow$ denotes bijection. But $y=2,1-\delta$ are the elements in $R$ with smallest norm satisfying this condition, and moreover the corresponding vectors $(-3,1-\delta),(-1-\delta,2)\in R$ are independent over $R$ since $2r\ne1-\delta$ for any $r\in R$. Thus, $\ker\varphi$ is generated by these two vectors, and by pp. 424-425 we have a presentation matrix
$$A=\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).\qquad\square$$

My questions are:

Here is what I know:

If $I = (2, 1 + \delta )$ where $\delta = \sqrt{-5}.$ Define $\varphi: R^2 \rightarrow I$ by $(r,s) \mapsto 2r + (1+ \delta)s.$ Then $\operatorname{ker}\varphi =
\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)s\\s\end{matrix}\right)=\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right)s.$ And hence a generator for $\operatorname{ker}\varphi$  is $\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right).$ But above in a solution of excercise 14.5.1 in Artin book "Algebra"(second edition) the author did not use this generator when the author needed to use a set of generators of the module of relations to be expressed as a linear combination of the set of generators of $I$ to find a presentation matrix as an $R$− module for the ideal $I$, he did not use that generator.
1- Does  $\mathbb Z[\sqrt{-5}]$ make any difference in the set of generators I should find?

2- Also, why we need the $s$ satisfying the condition of  $\operatorname{ker}\varphi$  to be of minimum norm, could someone clarify this to me please?

3-Also, is the solution in the picture above correct?

Note that this question here Determining the presentation matrix for a module helped me a lot in understanding the idea of the presentation matrix.

Note there is no need to answer all questions below as they were already  answered in the comments

1- why in the second braces $y \in R^2$?

2- How do the author get that that $y =2, 1 - \delta $? And how did we know that they have the smallest norm? and what is the importance of having the smallest the norm in the solution?

3- How did the author get the corresponding vectors $(-3, 1- \delta)$ and $(-1-\delta , 2 )$? how these 2 dimensional vectors belong to $R$? Also, I did not uderstand the argument the author used for showing that they are linearly independent, and why we have to show that they are linearly independent?

4- It seems for me until this point the author is trying to find the coefficients of the set of relations because the columns of the presentation matrix are built from them, is that correct?

5- Finally, is that solution correct?

Could anyone clarify these points to me please?

Here is also another solution:

The surjective map is considered as shown below;
\begin{align*}
R^2   & \longrightarrow\varphi(2,1+\delta),\\
(x,y) & \longrightarrow2x+(1+\delta)y.
\end{align*}
As $\ker\varphi$ has two generators. Then, there is the relation that $2(-1-\delta)+(1+\delta)2=0$ and also the relation $2(-3)+(1+\delta)(1-\delta)=0$.

Therefore, two relations cannot be derived from each other. Then $\ker\varphi$ is find by finding $(x,y)\in\mathbb{R}$ such that $2x+(1 +\delta)y=0$.

It means that
$$\ker\varphi=\left\{y:\ y\in\mathbb{R}\ \text{and}\ \left(\frac{1+\delta}{2}\right)y\ \text{is also in}\ \mathbb{R}\right\}.$$

Therefore, the presentation matrix is
$$\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).$$

答:
Observe that the ring $R = \mathbb Z[\sqrt{-5}]$ is equipped with a norm $\upsilon : R \to \mathbb Z$ defined by $\upsilon(a + b \sqrt{-5}) = a^2 + 5b^2,$ hence $\upsilon(R)$ is a subset of the non-negative integers. Consequently, $$V = \left \{\upsilon(y) \,\, \bigg| \,\, y \in R - \{0\} \text{ and } \dfrac{1 + \sqrt{-5}} 2 y \in R \right\}$$ is well-ordered, i.e., there is a minimum element of $V.$ Considering that the norm of an element $a + b \sqrt{-5}$ of $R$ is of the form $a^2 + 5b^2,$ the possible norms $\upsilon(y)$ are $1, 4, 5, 6,$ etc. But the only elements of norm $1$ are $\pm 1 = \pm 1 + 0 \sqrt{-5},$ and it is straightforward to show that $y$ cannot be $\pm 1$ (by the second condition defining $V$). Likewise, the only elements of norm $5$ are $\pm \sqrt{-5} = 0 + \pm \sqrt{-5},$ and by the same rationale as before, one can show that $y$ cannot be $\pm \sqrt{-5}.$ Clearly, we have that $2$ satisfies both conditions of $V,$ hence $2$ is an element of least norm $\upsilon(2) = 4$ in $V.$ Further, one can show that the only elements of $R$ of norm $4$ are $\pm 2.$ Last, the element $1 - \sqrt{-5}$ with norm $\upsilon(1 - \sqrt{-5}) = 6$ satisfies both conditions of $V.$ Once you know that $\ker \varphi$ is generated by two elements, it suffices to show that $2$ and $1 - \sqrt{-5}$ do not divide each other in $R$ to conclude that $\ker \varphi$ is generated by $(-1 - \sqrt{-5}, 2)$ and $(-3, 1 - \sqrt{-5}).$*

Ultimately, in order to determine a presentation of a finitely generated ideal $I = (x_1, \dots, x_n)$ of a ring $R,$ it suffices to find a finite generating set of the kernel of the map $\varphi : R^n \to I$ defined by $\varphi(r_1, \dots, r_n) = r_1 x_1 + \cdots + r_n x_n.$ I don't see how the structure of $R$ changes this goal; however, if your ring $R$ is "nice" (i.e., if $R$ is a field or a Euclidean domain or a principal ideal domain), then this will simplify the problem of finding the generators of $\varphi.$ Unfortunately, the ring $R = \mathbb Z[\sqrt{-5}]$ is not even a unique factorization domain, so the argument is a bit tougher.

*Towards this end, observe that every ideal of a Dedekind domain is either principal or generated by two elements. Once you have shown that $\ker \varphi$ is isomorphic (as an $R$-module) to an ideal of $R,$ it follows that $\ker \varphi$ is generated by two elements (because it contains at least two non-associate elements, as computed in the first paragraph above).


问:
A confusion in an IMO geometry problem.
So, I am solving EGMO and had a confusion on this problem;

A circle has center on the side AB of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.

My question is ;

Is it necessary that the center of circle is the midpoint of $AB$? Does it matter?

As I was able to solve most of the other problems, and from what I have heard, this is one of the easier problems so I shouldn't be having much trouble with it but I don't know why before starting the problem (at 2 in the night), I had this confusion in my mind so now I can't move past.

I am sorry if this is a dumb question, but please please clear my confusion.

答: Attempting a sketch with Geogebra, the answer seems to be no, $O$ does not have to be the centre of $AB$.

Title: A confusion in an IMO geometry problem.


答:
Answer: No, it's not necessary that the center of the circle is, at the same time, the midpoint of $AB$.

For the proof, consider a point $T$ on $AB$, such that $BT=BC$, and let us employ directed angles. We will proceed to show that $AT=AD$. In fact, let $O$ be the center of the circle, and observe that $DOTC$ is a cyclic quadrilateral too: $$\angle ODC=\frac{\angle ADC}2=\frac{180^\circ-\angle CBA}2=\angle BTC$$ But then $$\angle DTA=\angle DCO=\frac{\angle DCB}2=\frac{180^\circ -\angle BAD}2\implies AT=AD$$

 Notice: We did not require $O$ to be the midpoint of $AB$. And playing around (say: with geogebra or desmos) will show that such a configuration is possible.

Title: A confusion in an IMO geometry problem.

Body:
So, I am solving EGMO and had a confusion on this problem;

A circle has center on the side AB of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.

My question is ;

Is it necessary that the center of circle is the midpoint of $AB$? Does it matter?

As I was able to solve most of the other problems, and from what I have heard, this is one of the easier problems so I shouldn't be having much trouble with it but I don't know why before starting the problem (at 2 in the night), I had this confusion in my mind so now I can't move past.

I am sorry if this is a dumb question, but please please clear my confusion.


问:
The centroid problem of a plane geometry (using integral)
Let D be a thin slice with a radius of R, and L be a tangent to D. The point density of any point P on D is proportional to the square of the distance d from P to L, and the proportional coefficient K>0. Find the position of the center of mass of the slice.

I am confused about the steps to find the center of mass.

答:
The object, of which the center of mass is sought, is assumed to be a thin circular sector having a radius $R$ and a thickness $t$. The given tangent, which controlled the point-densities, touches the outer edge of the sector at $C$ (see $\mathrm{Fig.\space 1}$). The arc $AB$ subtends an angle $\phi$ at its center $O$. The radius $OC$ make an angle $\omega$ with the $x$-axis. Selected arbitrary point $P$ lies on this sector a distance of $r$ from $O$ and $\measuredangle POA$ is equal to $\alpha$. Therefore, $P$ has the coordinates $\left(r\cos\alpha, r\sin\alpha\right)$. Let the density at $P$ be $\rho$.

$\underline{\mathbf{\large{1}}.}\space\mathrm{Finding\space Density\space at\space an\space Arbitrary\space Point}$

First of all, we need to express $\rho$ in terms of $R, K,$ and $r$. We drop two perpendiculars $PH$ and $PQ$ to $OC$ and the tangent respectively. Now, we have,
$$OH = r\cos \left(\omega - \alpha\right)\space\rightarrow\space b = CH = R- r\cos \left(\omega - \alpha\right).$$

$$\therefore\quad \rho = Kb^2 = K\bigl[R-r\cos\left(\omega - \alpha \right)\bigr]^2. \tag{1}$$

$\underline{\mathbf{\large{2}}.}\space\mathrm{Finding\space Weight\space and\space Moment\space of\space a\space Small\space Section}$

Next, we are going to determine the volume $\Delta V$ of the small section $DEFG$ of the sector shown in $\mathrm{Fig.\space 2}$.
$$\Delta  V = \left[\frac{1}{2} \left(r+\frac{1}{2}\Delta r\right)^2\Delta\alpha - \frac{1}{2} \left(r-\frac{1}{2}\Delta r\right)^2\Delta\alpha\right]t \approx r\cdot t\cdot \Delta r\cdot \Delta\alpha. \tag{2}$$

As shown below, using (1) and (2), we can express the weight $\Delta W$ of the section $DEFG$ and its moment $\Delta M_x$ about the $x$-axis.
$$\Delta W = \rho\Delta V \approx K\cdot t\cdot r \bigl[R-r\cos\left(\omega - \alpha \right)\bigr]^2 \Delta r\cdot \Delta\alpha \tag{3}$$
$$\Delta M_x = \Delta W\times PN \approx K\cdot t\cdot r^2 \bigl[R-r\cos\left(\omega - \alpha \right)\bigr]^2\sin \left(\alpha\right) \Delta r\cdot \Delta\alpha \tag{4}$$

$\underline{\mathbf{\large{3}}.}\space\mathrm{Finding\space Total\space Weight\space of\space the\space Sector}$

The total weight $W$ of the thin circular sector can be found by carrying out the following integration.
$$W= \int\Delta W = \int^{\phi}_{0}\int^{R}_{0} K\cdot t\cdot r \Bigl[R-r\cos\left(\omega - \alpha \right)\Bigr]^2 dr\cdot d\alpha$$

Since $r$ and $\alpha$ do not depend on each other, we shall expand the expression to be integrated and write it as two independent integrations, one inside the other, as shown below
$$W = K\cdot t \int^{\phi}_{0}\Biggl[\int^{R}_{0} \Bigl\{R^2r^2-2Rr^2\cos\left(\omega - \alpha \right)+r^3\cos^2\left(\omega - \alpha \right)\Bigr\} dr\Biggr] d\alpha.$$

Once we do the inside integration, it reduces to
$$W =K\cdot t\int^{\phi}_{0}\left[\frac{1}{2}R^2 r^2-\frac{2}{3}Rr^3\cos\left(\omega - \alpha \right)+\frac{1}{4}\cos^2\left(\omega - \alpha \right) \right]^{R}_{0} d\alpha,$$
and, then, to
$$W = \frac{1}{12}K\cdot R^4\cdot  t \int^{\phi}_{0} \Bigl\{6-8\cos\left(\omega - \alpha \right)+3\cos^2\left(\omega - \alpha \right)\Bigr\}d\alpha.$$

We need to expand the $3^{rd}$ term to obtain
$$W = \frac{1}{24}K\cdot R^4\cdot t \int^{\phi}_{0} \Bigl\{15-16\cos\left(\omega - \alpha \right)+3\cos\left(2\omega - 2\alpha \right)\Bigr\}d\alpha.$$

When we integrate this, we get
$$W = \frac{1}{48}K\cdot R^4\cdot t \biggl[30\alpha+32\sin\left(\omega - \alpha \right)-3\sin\left(2\omega - 2\alpha \right)\biggr]^{\phi}_{0}.$$

$$\therefore\quad W=\frac{1}{48}K\cdot R^4\cdot t \Bigl\{30\phi-32\sin\left(\omega - \phi \right) - 3\sin\left(2\omega – 2\phi\right)+32\sin\left(\omega\right)\\ - 3\sin \left(2\omega\right)\Bigr\}. \tag{5}$$

$\underline{\mathbf{\large{4}}.}\space\mathrm{Finding\space Total\space Moment\space of\space the\space Sector\space about\space the\space \it{x-}Axis}$

The total moment about the $x$-axis, i.e. $M_x$, of the thin circular sector can be found by carrying out the integration given below.
$$M_x = \int\Delta M_x = K\cdot t\int^{\phi}_{0}\Biggl[\int^{R}_{0} \Bigl(R-r\cos\left(\omega - \alpha \right)\Bigr)^2r^2  dr\Biggr]\sin\left(\alpha\right)  d\alpha$$

This yields the following expression for $M_x$.
$$M_x = \frac{1}{12}K\cdot R^5\cdot t\biggl[ 52\bigl\{1-\cos\left(\phi\right)\bigr\} -30\phi\sin\left(\omega\right)+15\cos\left(\omega - 2\phi\right)\\ + 6\cos\left(2\omega - \phi\right) - 2\cos\left(\omega - 3\phi\right) - 13\cos\left(\omega\right) -6\cos\left(2\omega\right)\biggl] \tag{6}$$

$\underline{\mathbf{\large{5}}.}\space\mathrm{Finding\space Coordinates \space of\space the\space Sector’s\space center\space of\space Gravity}$

If we denote the $y$-coordinate of the sector’s center of gravity as $y_G$, then, it can be calculated using (5}, (6) and the following relationship.
$$y_G = \frac{M_x}{W} \tag{7}$$
Now, it is up to the OP, that is you by the way, to find an expression for $M_y$ - the total moment about the $y$-axis – and, then, calculate the $x$-coordinate of the sector’s center of gravity using a relationship similar to (7).

We are still in want of the $z$-coordinate of the sector’s center of gravity $z_G$. However, finding it does not pose any difficulty. Since we have assumed that the thin sector has a uniform thickness $t$,
$$z_G = \frac{1}{2}t. \tag{8}$$

$\underline{\mathbf{\large{6}}.}\space\mathrm{A\space Test\space Case}$

It is always advisable to verify the validity of mathematically derived results using  known cases. The selected case here is the thin disk, where we have $\phi = 2\pi$ (see $\mathrm{Fig.\space 3}$). It is obvious that the point-density distribution is symmetrical about the diameter $CD$, whatever value is chosen for $\omega$. Therefore, disk’s center of gravity is located on that particular diameter. This becomes even more evident if we choose $\omega = 0^O$, because, then, the CG lies on the $x$-axis, making $y_G = 0$. Let us check our formulae (5), (6), and (7).

$$W = \frac{5}{4}KtR^4\pi \qquad\mathrm{and}\qquad M_x = 0 \quad\rightarrow\quad y_G = \frac{M_x}{W} = \frac{0}{\frac{5}{4}KtR^4\pi}=0 $$
You are welcome to perform the same test for any $\omega$, once you obtain an expression for $x_G$.

$\mathrm{Fig.\space 4}$ is the 3D-plot of the point-density distribution of a disk when $\omega = 0^{o}$. It has the shape of a parabolic dish antenna. Interestingly enough, when we add the disk to the picture, the resulting object resembles a chopped-off cylinder with a flat base and the mentioned dish at top. We think that the $x$- and $y$-coordinates of this solid cylinder is the same as that of our thin disk.


问:
If $\sigma$-algebra of product space is product of $\sigma$-algebras then at least one of the algebras is trivial
I want to show that if $(X, \mathcal{A}),(Y, \mathcal{B})$ are measurable spaces and
$$
\mathcal{A} \otimes \mathcal{B} = \{ A\times B : A \in\mathcal{A}, B\in\mathcal{B} \}
$$
then $\mathcal{A} =\{\emptyset, X\}$ or $\mathcal{B} = \{\emptyset, Y\}$.

答:
The idea is to use contra position and to find a set in $\mathcal A \otimes \mathcal B$ which is not "rectangular", i.e. not of the form $A\times B$ for $A\in\mathcal A, B\in\mathcal B$.

Suppose $\{\emptyset, N, N^c, X\} \subseteq \mathcal A$ and $\{\emptyset, M, M^c, Y\} \subseteq \mathcal B$. Then
$$
K = (N\times M) \cup (N^c \times M^c) \in \mathcal A \otimes \mathcal B
$$
but
$$
K \notin \{A\times B : A \in\mathcal{A}, B\in\mathcal{B} \}.
$$


问:
Find a basis to represent a transformation
I've been trying to solve the following problem:

Let $T:\mathbb R^2 \rightarrow \mathbb R^2 $ be a linear transformation defined as $T(x,y)=(2x+y,x+3y)$. Prove that a basis $B$ of $\mathbb R^2 $ so that the representing matrix is:

$[T]_B = \begin{bmatrix}1&1\\-1&4\end{bmatrix}$

I'm just not sure where to begin, as I can tell the representing matrix by the standard basis is:

$[T]_E = \begin{bmatrix}2&1\\1&3\end{bmatrix}$

Am I supposed to find where $(2,1) = (1,1)$ and $(1,3) = (1,4)$ ?

Thank you!

答:
Let the transformation matrix under the standard basis and the standard basis be $$A=\begin{pmatrix}2&1\\1&3\end{pmatrix}, E=\{e_1,e_2\}=\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\}$$

and the transformation matrix under the new basis be

$$B=\begin{pmatrix}1&1\\-1&4\end{pmatrix}, \beta=\{u_1,u_2\}=\left\{\begin{pmatrix}a\\b\end{pmatrix},\begin{pmatrix}c\\d\end{pmatrix}\right\}$$

It is known that the columns of B are

$$\begin{split}B&=\begin{pmatrix}\begin{bmatrix}Au_1\end{bmatrix}_\beta&\begin{bmatrix}Au_2\end{bmatrix}_\beta\end{pmatrix}\\
&=\begin{pmatrix}\begin{bmatrix}2a+b\\a+3b\end{bmatrix}_\beta&\begin{bmatrix}2c+d\\c+3d\end{bmatrix}_\beta\end{pmatrix}\end{split}$$

where the columns are the tranformation in the standard basis applied to the basis vectors under the new basis, represented in the new basis's coordinates. Thus

$$\begin{split}\begin{pmatrix}2a+b\\a+3b\end{pmatrix}&=1\cdot u_1-1\cdot u_2=\begin{pmatrix}a-c\\b-d\end{pmatrix}\begin{matrix}(1)\\(2)\end{matrix}\\
\begin{pmatrix}2c+d\\c+3d\end{pmatrix}&=1\cdot u_1+4\cdot u_2=\begin{pmatrix}a+4c\\b+4d\end{pmatrix}\begin{matrix}(3)\\(4)\end{matrix}\end{split}$$

From equations (1) and (2) you find that

$$\begin{split}c&=-a-b \text{ (1)}\\d&=-2-2b\text{ (2)}\end{split}$$

Plugging these into (3) and (4) you only get identities back

$$\begin{split}-3a-4b&=-3a-4b\text{ (3)}\\
-7b-4a&=-7b-4a\text{ (4)}\end{split}$$

Thus you can choose any a, b, c, d such that equations (1) and (2) are satisfied and that $u_1,u_2$ would be linearly independent. One such choice could be a=1, b=2, thus c=-3, d=-5.

$$\beta=\left\{\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}-3\\-5\end{pmatrix}\right\}$$


问:
absolute convergence of the series $\sum\limits_{n=1}^\infty (-1)^n \left(e - \left(1 + \frac{1}{n}\right)^n\right)$
I have to check whether this alternating series $\sum\limits_{n=1}^\infty (-1)^n \left(e - \left(1 + \frac{1}{n}\right)^n\right)$ is convergent and absolute convergent. Using the Leibniz criteria I could show convergence. But I got stuck with the absolute convergence ...

I can show $\left|e - \left(1 + \frac{1}{n}\right)^n\right| \leq \frac{e}{n}$ but that would give a divergent upper bound. Anyway my guess is that it is not absolute convergent. But using $\left(1 + \frac{1}{n}\right)^n < \sum\limits_{k=0}^n \frac{1}{k!}$ and then $e - \left(1 + \frac{1}{n}\right)^n \geq \sum\limits_{k=n+1}^\infty \frac{1}{k!}$ leads (I think) to a convergent lower bound.

Can anyone point me in the right direction?

答:
You can use the fact that
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^n & = \exp \left( {n\log \left( {1 + \frac{1}{n}} \right)} \right) = \exp \left( {n\left( {\frac{1}{n} - \frac{1}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)} \right) \\ & = e\exp \left( { - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e - \frac{e}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
\end{align*}


问:
Numbers of the kind $0.aaa\ldots =\frac{1}{aaa\ldots a}$
What are the integers $a$ such that
$$
0.\underbrace{aaa\ldots}_{\infty\text{ times}} =\frac{1}{\underbrace{aaa\ldots}_{k\text{ times}}}
$$

eg.
$$
0.333\ldots=1/3\\
$$
while
$$
0.1616\ldots\ne1/16\ne1/1616\ne1/\underbrace{161616\ldots}_{\text{any }k}
$$

I was able to reduce $a$ to
$$
a=\frac{10^d-1}{\sqrt{10^{k\,d}-1}}
$$

where
$d$ is the number of digits in $a$ in base $10$. If this is correct, is $k=1,d=1$ the only solution for integral $a$?

答:
Let $a$ have $d$ digits in base $b$. Then
$$
\begin{align}
a\ge b^{d-1}=1{\underbrace{00..0}_{d-1\text{ times}}}{}_{b}\\
\end{align}
$$
Therefore
$$
\begin{align}
\underbrace{aaa\ldots a}_{k\text{ times}}&\ge1000\ldots0100\ldots0100\ldots0_b\\
&>1000\ldots0000\ldots0000\ldots0_b\\
&=b^{k\, d -1}
\end{align}
$$
So
$$
\begin{align}
\frac{1}{aaa\ldots a}&<b^{1-k\,d}\\
\end{align}
$$
If $k\,d>1$,then
$$
\begin{align}
\frac{1}{aaa\ldots a}&<b^{-(k\,d-1)}\\
&=0.\underbrace{000\ldots 0}_{k\,d-2\text{ times}}1\\
\therefore 0.aaa\ldots&<0.000\ldots 01\\
\therefore 0.a_1a_2a_3\ldots a_d a_1a_2a_3\ldots a_d \ldots&<0.000\ldots 01\tag{1}
\end{align}
$$

where $a_1a_2a_3\ldots a_d$ is written showing its digits in base $b$. If $k\,d>2$, there is a non-zero number of $0$s after the decimal in the RHS of eqn. $1$ implying $a_1=0$. If $k\,d=2$, then the LHS $<0.1$ which again implies $a_1=0$. In either case, this contradicts the assumption that $a$ has $d$ digits in base $b$. Therefore, for postive integral $k,d$
$$
k\,d=1\implies k =1,d=1
$$
So the required numbers are of the form
$$
\begin{align}
0.a_1a_1a_1\ldots&=\frac{1}{a_1}\\
a_1.a_1...&=b/a_1\\
a_1+\frac{1}{a_1}&=\frac{b}{a_1}\\
a_1&=+ \sqrt{b-1}=a
\end{align}
$$

same as @player3236's result above. For e.g. $0.\bar{2}_5=\left(\frac{1}{2}\right)_{5}$. Note that $b>2$ since $a=0,1$ don't satisfy the condition in the question.


问:
How do I minimize $||L||_* + \lambda ||S||_1$
Assume that we got a $n * m$ picture called $X$ and $X$ contains the noise picture $S$.
$X - S = L$ where $L$ is the clean filtered ideal picture and $X = L + S$ is the real picture taken by camera.

I want to minimize
$$||L||_* + \lambda ||S||_1$$
With the subject to: $$X = L + S$$

Where $||L||_*$ is the Nuclear norm (Sum of singular values) and $||S||_1$ is the L1-norm (Sum of absolute values) and $\lambda$ is just a tuning parameter e.g a number.

Questions:

What method should I use to solve this optimization problem?

What algorithm should I use with that method?

Can I solve this using least squares?

Can I solve this with linear programming e.g simplex method?

Can I solve this with dynamic programming?

I assume that this is a quadratic programming problem because we got two matrices to work with, $L$ and $S$.

答:
What method should I use to solve this optimization problem? What
algorithm should I use with that method?

Your problem is a non-smooth convex optimization problem that would typically be solved using a first-order method such as ADMM.

Since the objective function involves the nuclear norm of a nonsymmetric matrix you will have to compute SVD's during the solution of the problem.  The subgradient and prox operator for the nuclear norm term can both be computed from the SVD.

Can I solve this using least squares?

No.

Can I solve this with linear programming e.g simplex method?

No.

Can I solve this with dynamic programming?

I don't believe so.


问:
Condition for $\det(A^{T}A)=0$
Is it always true that

$\det(A^{T}A)=0$, $\hspace{0.5mm}$ for $A=n \times  m$ matrix with $n<m$?

From some notes I am reading on Regression analysis, and from some trials, it would appear this is true.

It is not a result I have seen, surprisingly.

Can anyone provide a proof?

Thanks.

答:
A possible short-cut to prove this:

Let
$$
\tilde{A}:=\begin{pmatrix}A\\0\end{pmatrix}
$$
such that $A\in\mathbb{R}^{m\times m}$, then immediately you have $A^TA=\tilde{A}^T\tilde{A}$, but the latter clearly has determinant 0.

For a general result, consult:
http://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula

Title: Condition for $\det(A^{T}A)=0$


问:
Rationals: dense ordering question
I'm reading about construction of $\mathbb{Q}$ on Wiki and dense ordering on Wiki.

In the former:

$(\frac a b < \frac c d) \implies (\frac a b < \frac {a + c} {b + d} < \frac c d)$

Is a theorem that establishes this? How adding numerators and denominators yields a fraction in between.

答: You can do this just by using arithmetic. For example, $a(b+d) < b(a+c)$ because $ab+ad < ab+bc$ because $ad<bc$.

Title: Rationals: dense ordering question


问:
What am I doing wrong with my probability equation?
I have these 3 values given to me

```<code>f(0) = P(X = 0) = 0.592704
f(2) = P(X = 2) = 0.064512
f(3) = P(X = 3) = 0.004096
</code>```

I found f(1) by 1- sum of all other decimal numbers (f(1) = .338, I then need to then find E(x) and Var(x), for E(X) I know you take sum of xf(x) for x=0 to 3, I do .592 + .338 + .064 + .004 and it says I am wrong? and then for the Var(x) I am unsure why I am getting incorrect results, I am doing the deviation of the numbers. I get .479, .521, 1.521 and 2.521 and then the deviation squared to get .229, .271, 2.31, 6.35 and then I multiply by the probabilities and add everything up (to get the variance) and I get .401 but it says I am wrong, I dont know whether I am missing something important? But i am almost positive i am solving it correctly, any help on that would be great!

答:
Probability that X equals 1, as you correctly noted:

```<code>> 1-.593-.0645-.0041
[1] 0.3384
</code>```

The expectation is not just the sum of the probabilities. It's the sum of the probabilities times the values of X at each probability:

```<code>> mu=0*.593+1*.338+2*.0645+3*.0041
> mu
[1] 0.4793
</code>```

Variance is the expected value of X-mu, squared. So subtract mu from each X, square it, and then sum over the probabilities.

```<code>> (0-mu)^2*.593+(1-mu)^2*.338+(2-mu)^2*.0645+(3-mu)^2*.0041
[1] 0.4030796
</code>```

Body:
Probability that X equals 1, as you correctly noted:

> 1-.593-.0645-.0041
[1] 0.3384

The expectation is not just the sum of the probabilities. It's the sum of the probabilities times the values of X at each probability:

> mu=0*.593+1*.338+2*.0645+3*.0041
> mu
[1] 0.4793

Variance is the expected value of X-mu, squared. So subtract mu from each X, square it, and then sum over the probabilities.

> (0-mu)^2*.593+(1-mu)^2*.338+(2-mu)^2*.0645+(3-mu)^2*.0041
[1] 0.4030796


问:
Linear map such that $\ker f \cap \operatorname{im} f = \{0\}$. Show that $V = \ker f \oplus \operatorname{im} f$.

Let $V$ be a finite dimensional vector space and $f \colon V\to V$ be a linear map such that $\ker f \cap \operatorname{im} f = \{0\}$. Show that $V = \ker f \oplus \operatorname{im} f$.

I have that $\{v \in V \mid f(v) =0 \} \cap \{f(v) \mid v\in V\}= \{0\}$. If I understand this correctly it means that there's no such element that $f(v) = 0$ and is in the image of $f$? From the problem statement it seems that this would somehow imply that $V$ is the sum of these two sets?

答: $\ker \ f , \operatorname{im} \ f \subset V$ subspace of $V$. $\ker \ f \cap \operatorname{im} \ f=\{0_v\}$ then there is no non zero element $x$ in $V$ that $x$ in $\ker f$ and $\operatorname{im} f$. So we can say $\ker f \oplus \operatorname{im} f$ exists Let $\dim V=n$  $$\dim V=\dim{ (\ker \ f)} +\dim{ (\operatorname{im} \ f) }-\dim( \ker \ f \cap \ \operatorname{im} \ f) $$ Also $\dim ( \ \ker \ f \cap \ \operatorname{im}\ f) =\dim\{ 0_v \}=0$ so $\dim V= \dim( {\ \ker \ f}) + \dim( { \ \operatorname{im} \ f })$ we can get $V=\ker \ f \oplus \operatorname{im} \ f$


问:
Which function can be used for Substitution
Find the value of $$I=\int_{0}^{\frac{\pi}{2}}\left(\sin(x)-\cos(x)\right)\,\log(\sin(x))dx$$

Method $(1)$. I split up the Integral into two Integrals as

$$I=I_1-I_2$$ where $I_1=\int_{0}^{\frac{\pi}{2}}\sin(x)\,\log(\sin(x))dx$ and $I_2=\int_{0}^{\frac{\pi}{2}}\cos(x)\,\log(\sin(x))dx$

For $I_1$, using Substitution $\cos(x)=t$ we get $$I_1=\int_{1}^{0}-\log(\sqrt{1-t^2})dt=\frac{1}{2}\int_{0}^{1}\log(1-t)+\log(1+t)dt=\frac{1}{2}\int_{0}^{2}\log(x)dx=\log(2)-1$$

Similarly in $I_2$ Use substitution $\sin(x)=t$ we get

$$I_2=\int_{0}^{1}\log(t)dt=-1$$ Hence

$$I=I_1-I_2=\log(2)$$

Method $(2)$. If for $I$, I use Substitution $\sin(x)+\cos(x)=t$ ,then $\left(\cos(x)-\sin(x)\right)dx=dt$ and $\sin(x)$ will be some function of $t$. But since limits are getting $1$ and $1$, Can't we conclude that the Integral is Zero?

答:
However, in the integral$$ \int_{0}^{\frac{\pi}{2}}{\log{(\sin x)}}d(\sin x+\cos x) $$
when our t changes from $1$ and back to $1$, the $ \sin x $ doesn't remain symmetric.
I think this is clear enough to answer the question.


问:
Spivak Prologue, Chapter 2, problem 7: prove $\sum_{i=1}^n k^p$ can always be written as sum of powers of n
This is problem 7 of Spivak's Calculus, 4th Edition, Prologue Chapter 2:

Show that $$\sum_{i=1}^n k^p$$ can always be written in the form $$\frac{n^{p+1}}{p+1} + An^{p} + Bn^{p-1} + Cn^{p-2} + ...$$

There is also a hint to use the method from problem 6.

I have a few questions on some specific details about the solution in the solution book, which I reproduce in what follows:

Let A be the set of all natural numbers for which $$\sum_{i=1}^n k^p$$ can be written in the form $$\frac{n^{p+1}}{p+1} + An^{p} + Bn^{p-1} + Cn^{p-2} + ...$$.

Using induction on $p$, for $p=1$ we have

$$\sum_{k=1}^n k = \frac{n(n+1)}{2} = \frac{n^2}{2} + n$$

so $p=1$ is in $A$.

Suppose the statement is true for all natural numbers smaller than $p$ (actually the solution book has smaller than or equals; so here is the first question: is it $<$ or $\leq$ here?).

The Binomial Theorem yields: $$(k+1)^{p+1}-k^{p+1} = (p+1)k^p + terms\ involving \sum_{k=1}^n k^r for\ r < p\tag{1}$$

By our induction assumption, we can write each $\sum_{k=1}^n k^r$ as an expression involving powers $n^s$ with $s\leq p$.

Writing out $(1)$ for $k=1,...,n$ we obtain:
$$2^{p+1}-1^{p+1}=(p+1)1^p +\ ...$$
$$3^{p+1}-2^{p+1}=(p+1)2^p +\ ...$$
$$(...)$$
$$(n+1)^{p+1}-n^{p+1}=(p+1)n^p +\ ...$$

Adding the equations, we obtain:

$$(n+1)^{p+1} - 1 = (p+1)\sum_{k=1}^n k^p +\ terms\ involving\ \sum_{k=1}^n k^r,\ for\ r<p\tag{2}$$

Now at this point, the solution actually says we obtain something else: $$\frac{(n+1)^{p+1}}{p+1}=\sum_{k=1}^n k^p+\ terms\ involving\ \sum_{k=1}^n k^r,\ for\ r<p\tag{3}$$

So the second question is: what assumptions are involved in obtaining $(3)$? Ie, in (2) does the term 1 on the lefthand side cancel with one of the terms on the righthand side? It is also true that each term on the righthand side has a (p+1) term, so is that term the one dividing the lefthand side?

Finally, as we noted above, because of the induction assumption for all natural numbers less than $p$, we can write each $\sum_{k=1}^n k^r$ as an expression involving powers of $n^s$ with $s\leq p$. Therefore, isolating $\sum_{k=1}^n k^p$ we get: $$\sum_{k=1}^n k^p = \frac{(n+1)^{p+1}}{p+1}+\ terms\ involving\ powers\ of\ n\ less\ than\ p+1$$.

Therefore, if $s$ is in $A$ for $s<p$ then $p$ is in $A$; therefore by induction $A=N$.

Final question: is this considered a rigorous proof?

答:
Let $s_p(n)=1^p+2^p+\cdots+ n^p$. We know that $s_1(n) = {1 \over 2} n^2+ q_1(n)$, where $q_1$ is a polynomial with degree $\partial q_1 \le 1$. So suppose we know that $s_r(n) = {1 \over r+1}n^{r+1}+ q_r(n)$, with $\partial q_r \le r$ for $r=1,...,p-1$.

Then
$(k+1)^{p+1} - k^{p+1} = \sum_{i=0}^{p} \binom{p+1}{i} k^i = (p+1)k^p +  \sum_{i=0}^{p-1} \binom{p+1}{i} k^i $.
Rearranging gives
$k^p = {1 \over p+1}((k+1)^{p+1} - k^{p+1} - \sum_{i=0}^{p-1} \binom{p+1}{i} k^i )$.

Summing both sides with $\sum_{k=1}^n$ gives
$s_{p}(n) = {1 \over p+1}((n+1)^{p+1} - 1 - \sum_{i=0}^{p-1} \binom{p+1}{i} s_i(n) )$ and since
$\partial \sum_{i=0}^{p-1} \binom{p+1}{i} s_i \le p-1$, we have the desired result.


问:
How to solve cross and scalar product equation knowing length and angle
We have two vectors $a$ and vector $b$. they are both non zero vectors and the angle between them is $\frac{\pi}{6}$. Their lenghts are connected with equation $|b| = \sqrt{3}|a|$

We need to solve this equation (*...scalar product, $\times$... cross product):

I find it very hard since I know that scalar product is not associative.

My try:
$$ (x*a+x*b)*a + x \times b = 2a\times b + 3|a|^{2}b\\
(x*a)*a+(x*b)*a + x \times b = 2a\times b + 3|a|^{2}b$$
If I solve the distance of $2a\times b$, I get nowhere. I do not know hot to continue.

The solution is $x = 2a+ \frac{2}{9}b+ 2a\times b $

答:
The given data gives $\vec a. \vec b=a^2$.
Let $\vec x= u\vec a+v \vec b+ w (\vec a \times \vec  b).$
where $\vec a$ and $\vec b$ are non-collinear. Inserting it in the equation
$$[\vec x.(\vec a+\vec b)]\vec a+\vec x \times \vec b= 2\vec a\times \vec b + 3a^2\vec b.$$
$$ [\{u \vec a + v \vec b+ w(\vec a\times \vec b)\}.(\vec a + \vec b)].\vec a+ [u\vec a+v \vec b+ w (\vec a \times \vec  b)]\times \vec b=2 \vec a \times \vec b +3a^2 \vec b$$
In this Eq. One has to compare the co-efficients of the bases $\vec a$, $\vec b$ and $\vec a \times \vec b$ to get the scalars $u,v,w$. One can make use of $(\vec a \times \vec b)\times \vec b= (\vec a. \vec b)\vec b-b^2 \vec a.$ and $\vec a.(\vec a \times \vec b)=0.$

I hope to come back.


问:
If $f_n\to f$ uniformly on each ball, does it follow that $\left|\frac{f_n(tx)}{f_n(sx)}-\frac{f(tx)}{f(sx)}\right|\to0$ for all $s,t\in[0,1]$?
Let $E$ be a normed vector space and $f_n,f:E\to\mathbb C\setminus\{0\}$ be uniformly continuous with $$\min\left(\inf_{x\in\overline B_r(0)}|f_n(x)|,\inf_{x\in\overline B_r(0)}|f(x)|\right)>0\;\;\;\text{for all }r>0\tag1$$ for $n\in\mathbb N$ and $$\sup_{x\in\overline B_r(0)}|(f_n-f)(x)|\xrightarrow{n\to\infty}0\;\;\;\text{for all }r>0.\tag2$$

Given $s,t\in[0,1]$, are we able to show $$\sup_{x\in\overline B_r(0)}\left|\frac{f_n(tx)}{f_n(sx)}-\frac{f(tx)}{f(sx)}\right|\xrightarrow{n\to\infty}0\tag3$$ for all $r>0$?

Intuitively, $(3)$ should be true and actually I think that $(1)$ is not important for that. But how can we prove it rigorously?

答:
I think that (2) is enough, together with the fact that $f_n(x), f(x)\neq 0$ for every $x$.
Specifically, let us fix $r>0$. By (2), the sequence is uniformly bounded on $\overline{B}_r$, hence there exists $M_r > 0$ such that
$$
|f(x)|, |f_n(x)| \leq M_r
\qquad \forall x\in \overline{B}_r,\ \forall n\in\mathbb{N}.
$$
Moreover, since $|f(x)|\neq 0$ for every $x$, again by (2) there exists $m_r > 0$ and $N\in \mathbb{N}$ such that
$$
m_r \leq |f(x)|, |f_n(x)|
\qquad\forall x\in \overline{B}_r, \ \forall n\geq N.
$$
Hence, for $n\geq N$, $t,s\in [0,1]$, $x\in \overline{B}_r$,
$$
\begin{split}
\left|\frac{f_n(tx)}{f_n(sx)} - \frac{f(tx)}{f(sx)}\right|
 &\leq
\frac{1}{m_r^2} \left|f_n(tx) f(sx) - f(tx) f(sx) + f(tx) f(sx) - f_n(sx) f(tx)\right|
\\ & \leq
\frac{M_r}{m_r^2} \left( |f_n(tx) - f(tx)| + |f_n(sx) - f(sx)| \right)\,,
\end{split}
$$
and the claim follows again from (2).


问:
Understanding a step in normal distribution function derivation
I am a HS student with minimal knowledge of statistics and calculus (Calculus 1 level) . Could someone explain in detail what happens in between these two steps in this proof:https://web.sonoma.edu/users/w/wilsonst/papers/Normal/default.html

答:
They integrated it.

In general,

$$\int \frac 1 x dx=ln |x| + c$$

and

$$\int x dx=\frac{x^2}{2}+c$$

but if you have something like $x-\mu$, you do a u-substitution. Let $u=x-\mu$, then $du=dx$ and

$$\begin{split}\int (x-\mu)dx&=\int udu\\
&=\frac{u^2}{2}+c\\
&=\frac{(x-\mu)^2}{2}+c\end{split}$$

Now you can see the step they took was:

$$\begin{split}\int \frac{df}{f}&=-k\int (x-\mu)dx\\
\ln|f|+\ln c_1&=-k\frac{(x-\mu)^2}{2}+\ln c_2\end{split}$$

Now you know that $\log a-\log b=\log \frac a b$. Thus

$$\begin{split}\ln|f|&=-k\frac{(x-\mu)^2}{2}+\ln c_2-\ln c_1\\
&=-k\frac{(x-\mu)^2}{2}+\ln \frac{c_2}{c_1}\\
&=-k\frac{(x-\mu)^2}{2}+\ln C\end{split}$$

with $C=\frac {c_1}{c_2}$. But since $c_1,c_2$ were arbitrary, C is as well.


问:
Defining the branch of $z^{1/n}$
If $Log_{\gamma}(z) :=  Log|z| + iarg_{\gamma}(z)$, where $arg_{\gamma}(z)$ is the angle of $z$ as lying on$ [\gamma, \gamma + 2\pi)$, how can I similarly define $z^{1/n}$?

答:
We have $$z^{1/n} = \exp\big(\tfrac 1n \log z\big) = \exp\big(\tfrac 1n (\log|z| + i\arg(z))\big).$$
In other words, $z^{\frac 1n}$ is multivalued in general.

Title: Defining the branch of $z^{1/n}$


答:
As indefinite integrals

$\int \frac1{f} \, df = \ln(f)+A$ for some constant $A$

$-k\int (x-\mu) \, dx = -k\frac{(x-\mu)^2}{2}+B$ for some constant $B$

If you let these be equal, you get  $\ln(f)+A = -k\frac{(x-\mu)^2}{2}+B$.  Now let $C=\exp(B-A)$, i.e. $\ln(C)=B-A$, you can rewrite this as $$\ln(f) = -k\frac{(x-\mu)^2}{2}+\ln(C)$$ for some constant $C>0$.  You can go further and rewrite this as $$f=C\exp\left(-k\frac{(x-\mu)^2}{2} \right)$$ which is starting to look related to the density function of a normal distribution with mean $\mu$ and variance $\frac1k$.

Title: Understanding a step in normal distribution function derivation

Body: I am a HS student with minimal knowledge of statistics and calculus (Calculus 1 level) . Could someone explain in detail what happens in between these two steps in this proof:https://web.sonoma.edu/users/w/wilsonst/papers/Normal/default.html


问:
Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$.
Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$. Consider the $n \times n$ matrix $A=(a_{ij})$.
Then

It is possible to choose $a_1.\dots,a_n$ such that $A$ is non-singular

matrix $A$ is positive definite if $(a_1,\dots,a_n)$ is nonzero vector

matrix $A$ is positive semi definite for all $(a_1,\dots,a_n)$

for all $(a_1,\dots,a_n)$, $0$ is an eigen value of $A$

I have calculated upto $3\times 3$ that determimant is $0$ but I have no idea how to conclude rigoriously. please help?

答:
Here are a few comments:
First, 4. implies that $A$ is always singular, answering 1. Follow Gerry Myerson's advice to prove 4 (basically, you are trying to show that the first two columns are linearly dependent). Now 2. cannot be true since 0 is an eigenvalue.

Also, it may be helpful to think of $A$ as the product $a a^T$ where $a$ is the column vector of the $a_i's$. Edit: you can use this to show 3.


问:
Show there exists isomorphism $\phi\colon V\to V$ s.t. $\phi(\phi(v))=-v$ for all $v\in V$
Problem: Let $ V $ be a vector space over $ \mathbb{F} $ and suppose its dimension is even, dim$V = 2k $. Show there exists an isomorphism $ \phi \colon V \rightarrow V $ s.t. $ \phi(\phi(v)) = -v $ for all $ v \in V$

My idea: Most importantly, I know there exists a basis $ \{ v_1,...,v_{2k} \} $ of $ V $ and on the basis I have to define $ \phi(v_i) $ for every $ 1\leq i \leq 2k $ ( and then show it's a bijection and a linear map ). But I'm completely clueless as to how actually define $ \phi(v_i)$ and the fact that we want $ \ker\phi=\{ \vec0\} $ and $\operatorname{im}\phi = V $ ( since we want an isomorphism ) doesn't help me much. Please help.

答:
It is enough to define $\phi$ on a basis $\{v_1,\ldots, v_{2k}\}$ of $V$ and extend it by linearity.
Let $\phi:V\to V$ be given by
$$\begin{array}{l}
\phi(v_{2i-1})= v_{2i}\\
\phi(v_{2i})= -v_{2i-1} \end{array} \qquad i=1,\ldots, k.$$

We then get $\phi(\phi(v_{2i-1}))=\phi(v_{2i})=-v_{2i-1}$ and
$\phi(\phi(v_{2i}))=\phi(-v_{2i-1})=-\phi(v_{2i-1})= -v_{2i}$. Therefore, $\phi \circ \phi= -id_V$.


问:
Show that a set $ S_m $ is equal to $ \mathbb{Z_m} $
Problem:

Let $$ {S_m} = \left\{ \left[ \frac{k(k+1)}{2} \right]_{\pmod{m}} \mid k \in \mathbb{N} \right\}.$$ Show that $ S_m = \mathbb{Z}_m $ if and only if $ m = 2^s $ for some $ s \in \mathbb{N}.$

Attempted solution:

First we observe that for each prime $ p > 2 $ there exists a $ b \in \mathbb{Z_p} $ such that $ x^2 = b \pmod{p}$ has no solutions.

So $m$ can't be a prime, which is excluded if $ m = 2^s $. Now I'm stuck, I don't know how to approach this problem. Any hints?

Muchos gracias!

答:
You can easily show that if $S_m=\mathbb Z_m$ and $n\mid m$ then $S_n=\mathbb Z_n.$

Now, if $n$ is odd, then $8$ is invertible modulo $n$ and $$8^{-1}\left((2k+1)^2-1\right)\equiv \frac{k(k+1)}2\pmod n$$

The squares modulo $n$ take at most $(n+1)/2$ distinct values, and thus $\frac{k(k+1)}2$ modulo $n$ can take at most $(n+1)/2$ distinct values. Thus, when $n>1$ is odd, though can’t have $S_n=\mathbb Z_n.$

Now, if $m$ has an odd prime factor, we again get that $S_m\neq \mathbb Z_m.$

So if $\mathbb S_m=\mathbb Z_m$ we must have that $m$ is a power of $2.$

I’ll leave the other direction to you. Hint: do it by induction on $s.$


问:
How many words/strings of length 5 can we make using the first 10 letters of the alphabet with at least one repeated letter?
How would you approach a problem like this?
If I were to make words of length 5 from the first 10 letters it would be 10^5 or 10x10x10x10x10, right?
But how do I account for the repetition part? repeated does not mean that they have to be next to each other. It just means that the same letter exists more than once.</p>

答:
First 10 alphabets of English language are :

A,B,C,D,E,F,G,H,I,J.

So, if you want to write the 5 letter string without any repetition, then, the ways should be : $$\binom{10}{5}*5!$$
Also, atleat 1 means : TOTAL WAYS(can repeat) - NONE REPEATING.

Thus, Total ways = $10*10*10*10*10 = 10^5$

Thus, required number of ways is:
$$10^5 - \binom{10}{5}*5!$$

And that's your answer!


问:
Find limit definitions manually (without WolframAlpha)
There was a question where
$\quad0\le \big(3-\sin(2\pi x)\big)\sin(\pi x - \frac{\pi}{4})-  \sin(3\pi x +\frac{\pi}{4})\le 2\sqrt{2}\quad $
and the user wanted to know what values of $x$ would be solutions.

I reasoned algebraically how both components involved
$\quad  (3-0)\sin(180 x^\circ - 45^\circ)-  \sin(180x^\circ + 45^\circ) \quad$
and how how $\quad2n-1\quad $ was a low limit of one range and
$\quad \frac{4n+1}{4}\quad $ was a high limit for another. I still had to use
WolrframAlpha here to find the in-between limits

$$
(2 n - 1)<x\le \frac{2 \big(\pi n - \frac{3 π}{8}\big)}{\pi} \land n \in\mathbb{ Z}\quad
\land \quad\frac{2 \big(\pi n + \frac{\pi}{8}\big)}{\pi}\le x\le\frac{(4 n + 1)}{2} \land n\in\mathbb{Z} $$

With these limits defined, they can be calculated manually because the 2n-and-4n values are easy and the $\pi$-values cancel. The problem is that I have not been able to reason how to define the two limits I did not find but that WolframAlpha did find.

I tried incrementing up or down from the "limits I found" and I did find that the high-low limits varied the $x$-values varied by no more that $\quad0.25\quad$ before $f(x)$ went out of range. The problem remained, I did not know how to define the other limits. I could just specify them as $\pm 0.25$ of the limits I found but that would be heuristic and not something I could prove.

Does anyone have an idea how WolframAlpha came up with the limits that involve $\space\pi?$

答:
You are considering the function
$
f = \big(3-\sin(2\pi x)\big)\sin(\pi x - \frac{\pi}{4})-  \sin(3\pi x +\frac{\pi}{4})
$.

After some manipulations, I got the identity
$f =  2\cos(\pi(x+1/4))^3 - 5\cos(\pi(x+1/4))$. So it is clear that the period length is $2$ and we only need to consider $x \in [0 \quad 2]$.

Setting $y = \cos(\pi(x+1/4))$ your inequality becomes $0\le y (2y^2 -5)\le 2\sqrt{2}$ and we need to look at the range $-1 < y < 1$.

So $0\le y (2y^2 -5)$ holds for $-1 < y < 0$ which is $\pi/2 < \pi(x+1/4) < 3 \pi/2$ or $1/4 < x < 5/4$.

Likewise, $y (2y^2 -5)\le 2\sqrt{2}$ holds for $-\sqrt{0.5} < y < 1$ which corresponds to two areas:  $3 \pi/4 > \pi(x+1/4) $ or $0 < x < 0.5$; and $5 \pi/4 < \pi(x+1/4)$ or $x > 1$.

Combining lower and upper bound gives that the inequality holds for  $0.25 < x < 0.5$ and for $1 < x < 1.25$, and of course for shifts of these areas by integer multiples of $2$. $\qquad \Box$


问:
I am trying to come up with the "Lebesgue-derivative"
so I discovered that the lebesgue integral exists I'm still on chapter 3 of baby Rudin, chapter 2 of Munkres. Is it possible to make something similar that applies to derivatives, if yes, what is it? If no, why, any proof? Basically a derivative for functions which are not continuous. As an example, can you use it on the function f(x) = x! (I know it can be done by turning it into the gamma function but can it be done with this weird derivative I'm talking about which might not even exist?)

答:
Since this is a broad question, I'll treat it more as a reference request. You may be interested in the notion of weak derivative (https://en.wikipedia.org/wiki/Weak_derivative), which are used at great length in the study of Sobolev spaces (https://en.wikipedia.org/wiki/Sobolev_space), harmonic analysis, and partial differential equations. These are technical and delicate topics (though the weak derivative page is quite understandable) usually taught at the graduate level.

Essentially, you recall that if $f$ is differentiable, then for any sufficiently smooth test function $\phi$ which is zero outside a big ball, one can integrate by parts $\int f \phi' = -\int f' \phi$. If $f$ is not differentiable, sometimes you can still find a function $g$ such that for any $\phi$ as above one has $\int f \phi' = -\int g \phi$.
In this case $g$ is called a weak derivative of $f$. Such a weak derivative has many nice properties like being unique (up to Lebesgue measure 0).

You can also see other common generalizations of the derivative (of which there are many) on the wiki page: https://en.wikipedia.org/wiki/Generalizations_of_the_derivative.

Additionally, the study of $f(x) = x!$ is a separate issue. The issue with this one is not that it is not differentiable, it is that $x!$ is usually only defined for natural numbers.
Once you pick the gamma function as a continuation, there is no issue of differentiability. But you could have picked infinitely many other real valued differentiable functions that agree with the factorial on integer values, so it is not like the gamma function is "special" in the sense that it agrees with the factorial and is differentiable. What makes the gamma function special (what uniquely determines it) is that in addition it is logarithmically convex (https://en.wikipedia.org/wiki/Gamma_function#19th%E2%80%9320th_centuries:_characterizing_the_gamma_function).


问:
find a plane perpendicular to a line and a certain distance from the origin
We have a line

$$ \frac{x}{2} = \frac{y+4}{2} = z-3$$

We need to find a plane that is perpendicular to this line and it's distance from the origin is exactly 3.

My try:

The normal of the plane is the same as the vector $s = (2,2,1)$

Then we need to position it: $$x^2+y^2+z^2 = \pm 9$$

from the line we get:
$$x = 2t$$
$$y = 2t-4$$
$$z = t+3$$

We input into the equation:$x^2+y^2+z^2 = \pm 9$
and we get a complex solution for $9t^2-2t+25 = \pm 9$.

Where have I done the mistake?

The official solution is: $2x + 2y + z = \pm 9$

答:
From your parametric equation of the line, you get the vector ditector which is
$$\vec{v}=(a,b,c)=(2,2,1)$$

this vector must be normal to the plane.

So, the cartesian equation of the plane will be as
$$ax+by+cz-d=0$$
or
$$2x+2y+z-d=0$$

its distance to the origine is
$$\frac{|2.0+2.0+1.0-d|}{\sqrt{2^2+2^2+1^2}}=3$$

thus $$d=\pm 9$$


问:
Convert an OGF to an EGF
If I have an ordinary generating function (OGF) as a rational polynomial:

$$
Q(y) = \frac{f_1(y)}{g_1(y)}
$$

Which has a power series representation

$$
Q(y) = \sum_{n \geq 0} a_n y^n
$$

How can I compute the exponential generating function (EGF)
$$
Z(y) = \sum_{n \geq 0} b_n y^n / n!
$$
as a rational polynomial?
$$
Z(y) = \frac{f_2(y)}{g_2(y)}
$$

As an explicit solved example, consider the case where
$$ Q(y) = \frac{yv -1}{y(2+v) -1} $$
The intermediate terms are
$$ a_n  = 2 (v+2) ^ {n-1} $$
Giving as a final answer
$$ Z(y) = \frac{2 e^{(v+2)y}}{v+2} $$

... ideally, I like to be able to do this by going from one rational polynomial to another without computing $a_n$. We can assume that I can factor the polynomial $g_1(y)$.

答: What you want is the Borel transform.  This is related to the inverse Laplace transform.  Specifically, $Z(s)$ is the inverse Laplace transform of $Q(1/t)/t$.</p>

Title: Convert an OGF to an EGF


问:
How do you use generating functions in this problem?
Let's say that I would like to find all the ways I could make a dollar using pennies, nickels, dimes, quarters, half-dollars, and dollar coins, assuming that all coins of a type are indistinguishable. I would only need to use $100$ pennies or $20$ nickels or $10$ dimes or $4$ quarters or $2$ half-dollars or $1$ dollar coin. I can represent each possible choice of the number of pennies as $(x^0+x^1+x^2...+x^{100})$ from $0$ to $100$ pennies. Likewise, I can represent choices for each possible choice for the number of the other coins as $(x^0+x^5+x^{20}...+x^{100})$ for nickels $(x^0+x^{10}+x^{20}...+x^{100})$ for dimes $(x^0+x^{25}+x^{50}+x^{100})$ for quarters $(x^0+x^{50}+x^{100})$ for half dollars and $(x^0+x^{100})$. As you can see, the exponents represent how much value that each number of coins has. For example, the $2$nd term in the expression that represents quarters represents that $1$ quarter had $25$ value $x^{25}$. Let's see what happens when I multiply all of these factors together.

$(x^0+x^1+x^2...+x^{100})(x^0+x^5+x^{20}...+x^{100})(x^0+x^{10}+x^{20}...+x^{100})(x^0+x^{25}+x^{50}+x^{100})(x^0+x^{50}+x^{100})(x^0+x^{100})$. When I multiply the factors together, the exponents add, and I will eventually end up with a term with $x^{100}$, which represents a value of $100$, or one dollar, which is what we are looking for. However, note that each term in our factors represent the possible values for a different number of coins. So, the coefficient of $x^{100}$ represents all the ways to make a dollar by differing the number of each coin you use. Now, I just need to find the coefficient of $x^{100}$. I can simplify the formula I found by using the formula for geometric series to get $\frac{x^{101}-1}{x-1}*\frac{x^{105}-1}{x^5-1}*\frac{x^{110}-1}{x^{10}-1}*\frac{x^{125}-1}{x^{25}-1}*\frac{x^{150}-1}{x^{50}-1}*\frac{x^{200}-1}{x^{100}-1}$. Now, this is where I am stuck, even though I simplified it, it still looks complicated. This isn't some random thing I came up with, the expressions I used are called "generating fucntions", so I know that this method would be useful in solving problems like these. However, I do not see how to effectively use these generating functions to solve this problem. Could somebody show me the way?

答:
This is maybe not what you're looking for as an answer, but generating functions are cool in part because they're so computable. In particular, computers are very very good at computing taylor series (that is, at getting the terms from a generating function).

In practice, we are usually interested in proving theorems, and here we can use facts about rational functions, complex analysis, etc. to say something about how (often combinatorial) series behave. In this case we will frequently do manipulations with the series by hand.

If we are ever interested in a concrete example, though, there is no reason to work by hand. Once you've computed the generating function (as you have) then every subsequent question (except theoretical ones) can and should be bashed out by a computer.

Your function does look gross if your measure of "gross" involves differentiating it by hand. However, if you modify your definition of "gross" to take into account "something I can type into a computer", then this function isn't gross at all!

So, with that in mind, say you want the $x^{100}$ term from this series:

Open up sage or something similar.

You'll use code like

```<code>sage: f(x) = (x^101 - 1)/(x-1) * (x^105 - 1)/(x^5 - 1) * (x^110 - 1)/(x^10 - 1) * (x^125 - 1)/(x^25 - 1) * (x^150 - 1)/(x^50 - 1) * (x^200 - 1)/(x^100 - 1)

sage: f.series(x,101)
</code>```

Then you can simply look at the 100th coefficient. Sage will tell you the answer is $293$ in a matter of microseconds!

Edit:

You can see how to compute this by hand in Grahm, Knuth, and Patashnik's "Concrete Mathematics". There's an example computation on pages $327$ to $330$ of my edition, and a closed form is computed on pages $344$ - $346$.

I hope this helps ^_^

Body:
This is maybe not what you're looking for as an answer, but generating functions are cool in part because they're so computable. In particular, computers are very very good at computing taylor series (that is, at getting the terms from a generating function).

In practice, we are usually interested in proving theorems, and here we can use facts about rational functions, complex analysis, etc. to say something about how (often combinatorial) series behave. In this case we will frequently do manipulations with the series by hand.

If we are ever interested in a concrete example, though, there is no reason to work by hand. Once you've computed the generating function (as you have) then every subsequent question (except theoretical ones) can and should be bashed out by a computer.

Your function does look gross if your measure of "gross" involves differentiating it by hand. However, if you modify your definition of "gross" to take into account "something I can type into a computer", then this function isn't gross at all!

So, with that in mind, say you want the $x^{100}$ term from this series:

Open up sage or something similar.

You'll use code like

sage: f(x) = (x^101 - 1)/(x-1) * (x^105 - 1)/(x^5 - 1) * (x^110 - 1)/(x^10 - 1) * (x^125 - 1)/(x^25 - 1) * (x^150 - 1)/(x^50 - 1) * (x^200 - 1)/(x^100 - 1)

sage: f.series(x,101)

Then you can simply look at the 100th coefficient. Sage will tell you the answer is $293$ in a matter of microseconds!

Edit:

You can see how to compute this by hand in Grahm, Knuth, and Patashnik's "Concrete Mathematics". There's an example computation on pages $327$ to $330$ of my edition, and a closed form is computed on pages $344$ - $346$.

I hope this helps ^_^


答:
The key idea is that obtaining the count outright might be hard, but we might be able to more easily write down a function $f(x)$ such that in the Taylor expansion of $f(x)$, the coefficient of $x^{n}$ provides the count when our object is of size $n$.

For the making change problem, computing the number of ways to make change for a fixed $n$ might be hard, but we can write down an (ordinary) generating function where the coefficient of $x^{n}$ is our desired count. This is the next best thing.

As was mentioned above, a computer algebra system is useful in ascertaining specific coefficients. In some cases, it may be feasible to compute the coefficient outright by hand (e.g., Computing problems and generating functions).

Title: How do you use generating functions in this problem?

Body:
Let's say that I would like to find all the ways I could make a dollar using pennies, nickels, dimes, quarters, half-dollars, and dollar coins, assuming that all coins of a type are indistinguishable. I would only need to use $100$ pennies or $20$ nickels or $10$ dimes or $4$ quarters or $2$ half-dollars or $1$ dollar coin. I can represent each possible choice of the number of pennies as $(x^0+x^1+x^2...+x^{100})$ from $0$ to $100$ pennies. Likewise, I can represent choices for each possible choice for the number of the other coins as $(x^0+x^5+x^{20}...+x^{100})$ for nickels $(x^0+x^{10}+x^{20}...+x^{100})$ for dimes $(x^0+x^{25}+x^{50}+x^{100})$ for quarters $(x^0+x^{50}+x^{100})$ for half dollars and $(x^0+x^{100})$. As you can see, the exponents represent how much value that each number of coins has. For example, the $2$nd term in the expression that represents quarters represents that $1$ quarter had $25$ value $x^{25}$. Let's see what happens when I multiply all of these factors together.

$(x^0+x^1+x^2...+x^{100})(x^0+x^5+x^{20}...+x^{100})(x^0+x^{10}+x^{20}...+x^{100})(x^0+x^{25}+x^{50}+x^{100})(x^0+x^{50}+x^{100})(x^0+x^{100})$. When I multiply the factors together, the exponents add, and I will eventually end up with a term with $x^{100}$, which represents a value of $100$, or one dollar, which is what we are looking for. However, note that each term in our factors represent the possible values for a different number of coins. So, the coefficient of $x^{100}$ represents all the ways to make a dollar by differing the number of each coin you use. Now, I just need to find the coefficient of $x^{100}$. I can simplify the formula I found by using the formula for geometric series to get $\frac{x^{101}-1}{x-1}*\frac{x^{105}-1}{x^5-1}*\frac{x^{110}-1}{x^{10}-1}*\frac{x^{125}-1}{x^{25}-1}*\frac{x^{150}-1}{x^{50}-1}*\frac{x^{200}-1}{x^{100}-1}$. Now, this is where I am stuck, even though I simplified it, it still looks complicated. This isn't some random thing I came up with, the expressions I used are called "generating fucntions", so I know that this method would be useful in solving problems like these. However, I do not see how to effectively use these generating functions to solve this problem. Could somebody show me the way?


问:
Find $P(B^2 \ge 4 A C)$ where $A,B,C$ are uniform random variables
Suppose $A,B,C$ are independent random variables uniformly distributed over $(0,1)$. Then, I want to find the probability that $B^2 \ge 4 ~A\cdot C$

Attempt: The following argument is likely to have some error but I am not able find out, what exactly:

Because $A,B,C$ are independent random variables, thus, $f_{A,B,C} (A = a,B = b, C = c) = f_A(a)~ f_B(b)~ f_C(c)$. Thus:

$f_{A,B,C} (A = a,B = b, C = c) =  \begin{cases}
      1 & 0\leq a,b,c \le1 \\
      0 & \text{otherwise}
   \end{cases}
$

Now, the required probability is $$\int \int_{\dfrac{B^2}{4} \ge AC} \int~f_{A,B,C} (a,b,c)~ da ~db ~dc$$

$$ = \int_{a=0}^{a=1} \int_{b=0}^{b=1} \int_{c=0}^{c={\frac{b^2}{4a}}} 1 \cdot dc~ db~da$$

Solving this integral gives an unexpected result.

Could someone please tell me where could I be makingg a mistake?

答:
The upper bound $c = \frac{b^2}{4a}$ could be bigger than $1$; however, $c$ should only range from $0$ to $1$. So the integral should be
$$
   \int_{a=0}^{a=1} \int_{b=0}^{b=1} \int_{c=0}^{c=\min\{\frac{b^2}{4a},1\}}1\,dc\,db\,da
$$
which gives the right answer, eventually.


问:
Intuition behind why continuous random variables cannot take a particular value?
It was iterated over and over again in our probability class that continuous random variables CANNOT have a probability mass function, instead they NEED TO have a density function. I am trying to find some intuition behind this. Any help would be greatly appreciated.

Additionally the following facts were mentioned:

The probability distribution function of some continuous random variable may take a value larger than 1. Why/How?

In the case of a continuous random variable, $P(X=x) \ne p_{x}(x)$. Why?

答:
We know from the probability axioms that the total probability over a sample space is $1$. For a continuous random variable, this corresponds to the area to the left of a point under the curve. So, the probability $P(X=x)$ is the area of an infinitely small strip with some height, and so the area is $0$.

The way to formalise this is to consider the probability $P(X-\epsilon<x<X+\epsilon)$.

As for your first question, can you elaborate further with a specific example?


问:
Divergence of $\sum_{n=1}^{\infty}\frac{a \cos(b \ln n)+b \sin(b \ln n)}{n^a}$

Divergence of $$\sum_{n=1}^{\infty} \frac{a \cos(b \ln n) + b\sin(b \ln n)}{n^a}$$ where $0<a<1, b>0$ are constants.

My try:
$$\cos(x)\geq 1-\frac{x^2}{2} , \forall x$$
$$\sin (x)\geq -x , \ \forall x$$
$$a\cos(b \ln n)+b\sin(b \ln n)\geq a\left(1-\frac{(b\ln \ n)^2}{2}\right)-b^2 \ln\  n$$

How to bound the given summation?

答:
Hint: Let's try something simpler: $\sum_{n=1}^{\infty}\dfrac{\cos(\ln n)}{n^a}.$ For each positive integer $m,$ let $E_m$ be the set of integers $n$ such that

$$2\pi m <\ln n < \pi/3+2\pi m.$$

Then

$$\sum_{n\in E_m}\frac{\cos(\ln n)}{n^a} > |E_m|\cdot\frac{1}{2}\cdot \frac{1}{(e^{\pi/3+2\pi m})^a}.$$


问:
Giving an example to the following statement regarding set theory.
Write down in roster notation a set  of cardinality 3, of which all elements are sets, and which
satisfies the following property:
∀,  ∈  ( ⊆  ⇒  ⊆ ).
No justification is needed. Do not use ellipses (“…”) in your answer.

What I understands from this question is that all the 3 sets have to be the same to fulfill the statement.Example of an answer would be { {1,2,3},{1,2,3},{1,2,3}} am i right?

答:
$U$ is a set of sets.  The inner sets have to have three members each but nothing is specified about the size of $U$.  Your example has two problems.  First, the elements of a set have to be distinct.  Second, you need a set of outer braces for the set.

It turns out any set of sets of cardinality three  where the sets are disjoint works.  For any finite sets $A,B$ of the same size $A \subseteq B \implies B \subseteq A$.  This is because the only way one finite set can be a subset of another finite set of the same size is for the two sets to be equal.

The simplest $U$ that satisfies the requirement is the empty set.  Since there are no sets that are elements of $U$ the requirement is vacuously satisfied.


问:
Given $a, b$ positive integers, such that $\gcd(a, b) = 2$, $\gcd(a, 5) = 1$ and $a|5b$, show $a = 2$

Given $a, b$ positive integers, such that $\gcd(a, b) = 2$, $\gcd(a, 5) = 1$ and $a|5b$, show $a = 2$.

My solution:

Once for $d = gcd(a, b) \implies d|a$ and $d|b$, therefore for $\gcd(a, b) = 2$, $2|a$ and $2|b$, this implies $a, b$ are even numbers. Once the greatest common factor of $a$ and $b$ is $2$, therefore $a$ or $b$ must be equal to $2$.

According to Bézout's Theorem, for $a, b \in\mathbb{Z}$, such that  $d = gcd(a, b)$. Then there exist $x, y \in\mathbb{Z} : ax + by = d$. Therefore, $2 = ax+by$ and $1 = ax + 5y$. So,

$$2-by = 1 - 5y \implies 1 = by - 5y \implies 1 = y(b-5)$$

Considering $b = 2$,  so $1 = y(2-5) \implies 1 = -3y \implies y = -\dfrac{1}{3}$, but  $-\dfrac{1}{3}\not\in \mathbb{Z}$, therefore $a = 2$.

I am not sure if this solution is correct. I even not considered $a|5b$, how can I use it in the proof?

I would like to know a better solution or what would make my attempt better.

答:
Once the greatest common factor of a and b is 2, therefore a or b must be equal to 2.

Wrong : counter example is $a = 4, b = 6.$

From Number Theory, if $r,s$ relatively prime and $r|(ks)$, then $r|k$.

$\gcd(a, b) = 2$, $\gcd(a, 5) = 1$ and $a|5b$, show $a = 2$

From the premises, you therefore have that $a|b$.

Therefore, $\gcd(a,b) = a = 2.$


Title: Show there exists isomorphism $\phi\colon V\to V$ s.t. $\phi(\phi(v))=-v$ for all $v\in V$

Body:
Problem: Let $ V $ be a vector space over $ \mathbb{F} $ and suppose its dimension is even, dim$V = 2k $. Show there exists an isomorphism $ \phi \colon V \rightarrow V $ s.t. $ \phi(\phi(v)) = -v $ for all $ v \in V$

My idea: Most importantly, I know there exists a basis $ \{ v_1,...,v_{2k} \} $ of $ V $ and on the basis I have to define $ \phi(v_i) $ for every $ 1\leq i \leq 2k $ ( and then show it's a bijection and a linear map ). But I'm completely clueless as to how actually define $ \phi(v_i)$ and the fact that we want $ \ker\phi=\{ \vec0\} $ and $\operatorname{im}\phi = V $ ( since we want an isomorphism ) doesn't help me much. Please help.

Body:
$$
M =
\left(
\begin{array}{rr|rr|rr}
0&1&0&0&0&0 \\
-1&0&0&0&0&0 \\ \hline
0&0&0&1&0&0 \\
0&0&-1&0&0&0 \\ \hline
0&0&0&0&0&1 \\
0&0&0&0&-1&0 \\
\end{array}
\right)
$$
gives
$$ M^2 = -I$$


问:
Proving from Hilbert axioms in propositional logic
I am using axioms similar to the Hilbert system of logical axioms:

\begin{align}
&P\rightarrow (Q\rightarrow P)  \tag{A1}\\[5pt]
&(P\rightarrow (Q\rightarrow R))\rightarrow ((P\rightarrow Q)\rightarrow (P \rightarrow R)) \tag{A2}\\[5pt]
&(\neg Q \rightarrow \neg P)\rightarrow (P\rightarrow Q) \tag{A3}
\end{align}

I may also use Modus Ponens and the deduction theorem. I am trying to prove two things:

$(a)\;\;\; \vdash \neg P \rightarrow (\neg Q\rightarrow (P\rightarrow R))$

$(b)\;\;\; \vdash (P \rightarrow ¬P) → (\neg\neg P \rightarrow \neg P)$

I know what I am supposed to do but I just can't quite get there with either one. I tried use deduction theorem in both so I am trying to prove in $(a)$ that $\neg P \vdash (\neg Q\rightarrow (P\rightarrow R))$ and similarly in $(b)$. Here is my effort for $(a)$:

\begin{align}
&1. \neg P  &&\text{premise}\\[5pt]
&2. \neg P \rightarrow (\neg Q \rightarrow \neg P) &&(A1)\\[5pt]
&3. (\neg Q \rightarrow \neg P) &&MP:1+2\\[5pt]
&4. (\neg Q \rightarrow \neg P) \rightarrow (P \rightarrow Q) &&(A3)\\[5pt]
&5. P \rightarrow Q  &&MP:3+4
\end{align}

I don't know how to proceed. Feels like I have all the needed pieces but I can't put them together. I have quite a similar situation with $(b)$. Help would be appreciated!

答:
A)Use deduction theorem twice. So then you have to show that $\neg P,\neg Q\vdash_{ax}P\rightarrow R$. You are almost done, you only need to show that $\vdash_{ax}P\rightarrow(Q \rightarrow R)$ and apply MP.

B) Use deduction theorem once and notice that by axiom $A2$ $(\neg\neg P\rightarrow(P \rightarrow \neg P)) \rightarrow ((\neg\neg P \rightarrow P) \rightarrow (\neg\neg P \rightarrow \neg P))$. Then by axiom $A1$ $(P \rightarrow \neg P) \rightarrow (\neg\neg P\rightarrow (P \rightarrow \neg P))$. I think you are able to finish from here.


问:
Prove that any Subspace of Hausdorff is Hausdorff

Prove that any Subspace of Hausdorff is Hausdorff

Attempt

$\mathcal{T}_Y$={O$\cap$ Y:O$\in\tau$}

Let X be any topology where Y$\subset X$
where (X,$\tau$)

Let a,b $\in Y$.

Since X is Hausdorff then for a$\ne$b,
there exists U,V $\in$ X, s.t a$\in $U and b$\in$V and
U$\cap V=\emptyset$

As a$\in $U we have W= Y$\cap U$ for W$\in\tau$

But for b$\in $ V. we get Y=Y$\cap$V then

( Y$\cap U$ )$\cap$ (Y$\cap V$)=Y$\cap$(U$\cap V)=\emptyset$

Thus ( Y$\cap U$ ) and ( Y$\cap V$ ) are
separate ,thus Hausdorff.

Any help would be appreciated.

Did l miss anything?

Hope l am getting better?

答:
There are indeed several \ errors in your attempt although you started in the
correct way. Let $(X, \mathcal{T})$ be a Hausdorff topological space, $Y
\subseteq X$ using the subspace topology $\mathcal{T}_Y = \left\{ O \bigcap
Y|O \in \mathcal{T} \right\}$. To prove that $(Y, \mathcal{T}_Y)$ is
Hausdorff. Take $a, b \in Y$ such that $a \neq b$ then as $\mathcal{T}$ is
Hausdorff there exists $U, V \in \mathcal{T}$ so that $a \in U, b \in V$ and
$U \bigcap V = \emptyset$. Now as $a, b \in Y$ we have that $a \in U \bigcap
Y$ and $b \in V \bigcap Y$. Further $\left( U \bigcap Y \right) \bigcap \left(
V \bigcap Y \right) = \left( U \bigcap V \right) \bigcap Y = \emptyset$,
finally by definition of $\mathcal{T}_Y$ we have that $U \bigcap Y, V \bigcap
Y \in \mathcal{T}_Y$. Thus proving that $(Y, \mathcal{T}_Y)$ is Hausdorff.

Your assertion that as $a \in U$ we have that $U = Y \bigcap U$ is incorrect
and [it is only true if $U \subseteq Y$] and not needed for the proof,
likewise your assertion that for $b \in V$ we get $Y = Y \bigcap V$ is
incorrect and also does not help with the proof.

Hope this clarify the proof for you

$E$ is a spelling error, it should be $Y$ so that $\mathcal{T}_Y$ is the
subspace topology of $Y$. You do not have to construct $U \bigcap Y$ or $V
\bigcap Y$ only show that it is in the subspace topology of $Y$


问:
To prove or disprove that for each $y\in N_\epsilon(A)\setminus A$, $B(y,\epsilon)\cap [X\setminus N_\epsilon(A)]\neq \emptyset$.
Let $A$ be a subset in a normed space $(X,\|\cdot\|)$, and $N_\epsilon(A)=\bigcup\limits_{a\in A}B(a,\epsilon)$, where $B(a,\epsilon)$ is open ball centered at $a$ with radius $\epsilon>0$. To prove or disprove that for each $y\in N_\epsilon(A)\setminus A$, $B(y,\epsilon)\cap [X\setminus N_\epsilon(A)]\neq \emptyset$, if $X\setminus N_\epsilon(A)\neq\emptyset$.

What I tried is:

I wanted to prove that if $B(y,\epsilon)\cap [X\setminus N_\epsilon(A)]= \emptyset$, then $y$ has to be in $A$, but could not succeed. We have, for two distinct points $u,v\in X$, there is an isometry $$I:[0,\|u-v\|]\to \{\alpha u+(1-\alpha)v:0\leq \alpha \leq 1\}$$ such that $I(0)=u, I(\|u-v\|)=v, I([0,\|u-v\|])=\{\alpha u+(1-\alpha)v:0\leq \alpha \leq 1\}$. I took a $z\in X\setminus N_\epsilon(A)$, so $[0,\|y-z\|]$ is isometric to $\{\alpha y+(1-\alpha)z:0\leq \alpha \leq 1\} $. Then, if $\epsilon\geq\|y-z\|$, then done; but if $0<\epsilon<\|y-v\|$, there is $w\in \{\alpha y+(1-\alpha)v:0\leq \alpha \leq 1\}$, such that $I(\epsilon)=w$, then how to prove $w\in X\setminus N_\epsilon(A)$.

答: This is not true: let $\varepsilon=1$ and in the complex plane take $A$ to be a disk minus a disk, $A=\{z\in\mathbb{C}: 0.9<|z|<3\}$. Then $0\not\in A$, but $0\in N_\varepsilon(A)$. On the other hand $N_\varepsilon(A)=\{z\in\mathbb{C}:|z|<4\} \neq\mathbb{C}$, so obviously $\mathbb{C}\setminus N_\varepsilon(A)\neq\emptyset$, and $D(0,\varepsilon)\cap (\mathbb{C}\setminus N_\varepsilon(A)=\emptyset$.


答:
The probability distribution function of some continuous random variable may take a value larger than 1. Why/How?

For the same reason that you can drive somewhere one mile away while going at speeds faster than 1 mph.

The same general idea applies to the rest of the questions.

Title: Intuition behind why continuous random variables cannot take a particular value?

Body:
It was iterated over and over again in our probability class that continuous random variables CANNOT have a probability mass function, instead they NEED TO have a density function. I am trying to find some intuition behind this. Any help would be greatly appreciated.

Additionally the following facts were mentioned:

The probability distribution function of some continuous random variable may take a value larger than 1. Why/How?

In the case of a continuous random variable, $P(X=x) \ne p_{x}(x)$. Why?


问:
Prove that a given meromorphic function is rational
I'm doing some exercises in complex analysis, and I've reached one I simply can't figure out on my own, which is why I'm hoping for some help.

The exercise:

We assume that $h:\Bbb C\to \Bbb C \cup \{\infty\}$ is meromorphic with finitely many poles $z_1,...z_n$ and assume that there exist $k\gt0$, $N\in\Bbb N$ and $R\gt0$ such that $|h(z)|\le k|z|^N$ for $|z|\gt R$. Prove that h is a rational function.

What I've been thinking so far:

The definition of a rational function is, that you have to be able to write it on the form $f(z)=p(z)/q(z)$, where $p,q\in\Bbb C[z], q\neq 0$. So I guess I have to show that my function $h$ can be written this way too? I have, however not yet been able to figure out a way to do this which makes sense.

Earlier we've done an exercise, in which we've proven that if we let f be an entire function and we assume that $|f(z)|\le A+B|z|^n$ for $z\in\Bbb C$, where $A,B\ge0$ and $n\in\Bbb N$, then f is a polynomial of degree $\le n$. I've been thinking this might be useful, although a meromorphic function isn't entire.

Thank you for your time.

答:
Since $h$ has a finite number of poles, $h(z)\cdot (z-z_1)\cdots (z-z_n)$ is an entire function. Let $w_1,\ldots,w_n$ (this could be an empty list) be the zeros of $h(z)\cdot (z-z_1)\cdots (z-z_n)$. There must be a non-zero entire function $f$ such that $$ h(z)\cdot (z-z_1)\cdots (z-z_n)=(w-w_1)\cdots (w-w_n)\cdot f(z) .$$

We show that $f$ must be a constant and the problem will be solved.

By assumption $$|(w-w_1)\cdots (w-w_n)\cdot f(z)|\le k|z|^N\cdot |(z-z_1)\cdots (z-z_n)|\le j|z|^M$$ for some $j$ and $M$.

Thus $(w-w_1)\cdots (w-w_n)\cdot f(z)$ must be a polynomial and consequently $f$ is a non-zero polynomial (i.e. a constant).


答:
Maybe this will help clarify things:

Suppose you tried to use a mass function for a continuous r.v.  This is a function that assigns nonzero probability to some basic, atomic outcomes.  There is an uncountably infinite number of outcomes that have values, though, given a continuous distribution.

How could you write a summation to define the expectation?  There are not enough numbers in the infinite series of integers that would index a summation.

OK, but suppose, magically (doesn't really make sense, but let's try it out for the sake of a reductio) you could write the expectation as a sum.  Since there is an infinite number of values, each with equal probability (otherwise the distribution will not be continuous), all values have zero probability, in which case the total probability is zero, or else the probabilities are nonzero, in which case the total probability is infinite.  But the sum must be equal to 1.

With a density function, you can "sum" using an integral, though. Each possible value of the r.v. has zero probability, but regions of values have nonzero probability.  That's why the expectation of a continuous r.v. can be written as an integral over a density function.


问:
Solving integral of a function of derivative
I'm having problems in solving the following integral:

$$\int_0^z s(1-s)\left[\frac{A'(s)}{A(s)}\right]^2 ds - \int_0^z s(1-2s)\left[\frac{A'(s)}{A(s)}\right]ds$$

Where $A'(s)$ denote the first derivative of $A(s)$ with respect to s.Here $A(s)$ is an extreme value generator, for extreme value copulas, but that shouldn't be relevant for the purpose.

This integral should be simplified as follows:

$$\int_0^z \frac{s(1-s)}{A(s)}dA'(s) -z(1-z)\frac{A'(z)}{A(z)}$$

I actually tried several times, changing measure for logs and so on, but I don't see such solution!
Does anybody have any idea? Probably I'm forgetting some rule from calculus, but I did not find anything on integral of function of derivatives.
As always, any help would be very much appreciated. Many thanks!

答:
As a test, let $A(z)=e^z$. Then $A'(z)=e^z$ and $A'(z)/A(z)=1$, so the expressions in the OP would become

\begin{align}
\int_0^z s(1-s)\,ds-\int_0^z s(1-2s)\,ds
&=\int_0^z s\,ds\\&=\frac12 z^2,\\\\
\int_0^z \frac{s(1-s)}{e^s}d(e^s)-z(1-z)
&=\int_0^z s(1-s)\,ds-z(1-z)\\
&=\frac{z^2}{2}-\frac{z^3}{3}-(z-z^2)\\
&=-z+\frac{1}{2}z^2-\frac{1}{3}z^3.
\end{align}
Since these don't agree, the two expressions are not equal in general.


答:
If there is some $c\in \mathbb{R}$ such that $\Pr [X=c]>0$ then the distribution function of $X$ at $c$ is discontinuous. To see this we will show the contrapositive statement: if $F_X$ will be continuous at $c$ then we will show that $\Pr [X=c]=0$.

First, as by assumption $F_X$ is continuous at $c$, then

$$
\lim_{x\to c}F_X(x)=F_X(c)\tag1
$$

Also we have that

$$
\Pr [X=c]= \Pr [X\leqslant c]-\Pr [X<c]\tag2
$$

But as a probability is a measure then

$$
\begin{align*}
\Pr [X<c]&=\Pr [X^{-1}((-\infty ,c))]\\
&=\Pr \left[\bigcup_{n\in \mathbb N }X^{-1}((-\infty ,c-1/n])\right]\\
&=\lim_{n\to\infty}\Pr [X^{-1}((-\infty ,c-1/n])]\\
&=\lim_{n\to\infty}F_X(c-1/n)\tag3
\end{align*}
$$

Thus from (1) and (3) we find that if $F_X$ is continuous at $c$ then $\Pr [X<c]=\Pr [X\leqslant c]$, and consequently from (2) we have that $\Pr [X=c]=0$.∎


问:
Formula to determine amount before percentage deductions?
For some reason, i'm having trouble working out what should be a rather simple algebraic formula.

I have amount (X), which is the amount added to an account after tax (T) and fees (F) are deducted from the original payment (P). I know the value of X, T and F and need to determine the value of P. The part i'm getting hung up on is that T and F are percentages of P (ie: T = 20%, F = 3%).

Example:

```<code>P = 100
T = .20
F = .03
X = 77

P - (P * (T + F)) = X
100 - (100 * (.20 + .03) = 77
</code>```

How can I calculate P, if I only know T, F and X ?

答:
The terms $P$ and $P\times(T+F)$ have a common factor of $P$.

Use distribution of multiplication over addition to get a single factor $P$:

$$  xa + xb = x(a+b).$$


问:
Fitting a set of points in a plane to a smooth curve obtained by joining a half-line and an arc of a circle
I have a set of points in the plane and I want to find a curve that best fits these points (e.g., in a least squares manner, or using some other convenient "measure"). I want that the curve be either:

a line

a circle

a circle whose center lies on a given line

a curve that looks like a hook, that is obtained from:

(a) an arc of a circle whose center lies on a given line

(b) a half-line tangent to one of the endpoints of the arc circle (in this way I can get a curve that is smooth at the point where the arc of the circle and the half-line meet).

I managed to get an answer for all but the last problem. Any suggestions are welcomed.

答:
Such curves have $5$ real parameters:

$(s,t)$, the point where the arc meets the line

$r$, the radius of the arc

$\theta$, the angle of the arc

$d$, the length of the line segment

And one discrete parameter that chooses the orientation of the hook.

You don't lose anything by letting the line extend forever and the arc complete its circle. It seems to me that doing so only increases your ability to obtain a smaller distance from the data to the curve. So then you have three real parameters and one discrete parameter.

Theoretically, if you sum the squared distances from your data points to the curve with these parameters, and use calculus to minimize, you will have your result.

But there are questions: what will be your measurement of distance? In least-squares regression, the distance is measured as the vertical separation. Do you instead want to minimize direct distance from data points to the curve?


答:
The original equation:
$$P-(P(T+F)) = X$$

Factoring P:
$$P\cdot [1-(T+F)] = X$$

Dividing both sides of the equation by $1-(T+F)$:
$$P = \frac{X}{1-(T+F)}$$

This should be clear enough, if you still don't understand something, feel free to ask!

Title: Formula to determine amount before percentage deductions?

Body:
For some reason, i'm having trouble working out what should be a rather simple algebraic formula.

I have amount (X), which is the amount added to an account after tax (T) and fees (F) are deducted from the original payment (P). I know the value of X, T and F and need to determine the value of P. The part i'm getting hung up on is that T and F are percentages of P (ie: T = 20%, F = 3%).

Example:

```<code>P = 100
T = .20
F = .03
X = 77

P - (P * (T + F)) = X
100 - (100 * (.20 + .03) = 77
</code>```

How can I calculate P, if I only know T, F and X ?


问:
PDF of a rectangle
If I want to find the $c$ of a $PDF$ when it's given:

$f_{X,Y}\left(x,y\right)=c\:\:\:\:\left(The\:area\:in\:blue\right),\:otherwise:\:0$

I try to do that:

$$\int _{\frac{1}{2}}^1\:\int _{-x+\frac{3}{2}}^1\:c\,dy\,dx+\int _0^1\:\int _{-x+\frac{1}{2}}^{-x+1}\:c\,dy\,dx=1$$
But I got $c=\frac{8}{5}$ and the result is $c=2$.

(I know we can calculate it with by calculating the area of the triangles but I am wondering to know why my way is not working)

Title: PDF of a rectangle

Body:
You need to calculate like this:

$$\int _0^{\frac{1}{2}}\:\int _{\frac{1}{2}-x}^{-x+1}\:cdydx+\int _{\frac{1}{2}}^1\int _0^{-x+1}\:\:cdydx+\int _{\frac{1}{2}}^1\int _{-x+\frac{3}{2}}^1\:\:cdydx=1$$

And then you'll get: $c=2$


问:
Prove that $\angle AEF =90^\circ$ given a square $ABCD$
Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.

My attempt:

Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish to prove that $AE^2 +EF^2 = AF^2$. I’ve tried a lot of methods to reach this point but none of them worked.

答:
Triangles EDA FCE  are similar. Scale factor =2.

$$ FC= EC/2,\;DE= AD/2 \;$$

Since the triangles are similar corresponding angles are same. (Given square contains a right angle).