问:
Geometric series of complex numbers
Let $ z $ be a complex number. I want to find the radius of convergence of $$ \sum_{n=0}^{\infty}z^{n} $$

My intuition is that this series converge for $ z\in D\left(0,1\right) $ (open unit disk).

Basicly, my intuition is based on the formula for real numbers $$ \sum_{n=0}^{\infty}x^{k}=\frac{1}{1-x} $$

Where $|x|<1 $.

Is this formula holds for complex $ |z|< 1 $? if so, how can I prove it?
That would prove that for $ |z| < 1 $ this series converge.

Now what about $ |z|=1 $ ? obviously for $ z=1 $ this series diverge, but what about $ |z|=1, z\neq1 $ ? My intuition is that the series would converge in this case, since both the imaginary part and the real part of $ z $ are smaller then $ 1 $ (also except for z=i, z=-i etc..).

How can I prove in a formal way that this series diverge for $ z=i, z=-i $ ?

Im not sure what tools from real series we can use in the complex case, so a few clarifications would be very helpful.

Thanks in advance.

答:
You are correct that the series converges for $|z|<1$. It diverges for all $z$ with $|z|=1$ because $|z^n|=1$ (i.e. does not converge to $0$).

EDIT:
For the sum of the geometric progression we have:
$$
s_n=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}.
$$
Therefore for $|z|<1$:
$$
\lim_{n\to\infty}s_n=\frac{1}{1-z}
$$
since $\lim_{n\to\infty} z^{n+1}=0$.

Title: Geometric series of complex numbers


问:
A quick question about addition step when proving vector subspaces
$$U = \{X\in\text{Real}^{2\times2};\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}X = X\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}\}$$

My question is, when I wanted to show that it is a vector subspace, can I do the following:

$$\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}X_1 + \begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}X_2 = X_1\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}+X_2\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}$$
I written this equation, and then factored out the matrices (distributive property) and seen that it holds. Is this enough for addition property when proving that it is a subspace. (The alternative that I could think of is, that I would actually compute X, but I want to see your thoughts on the upper approach)

答:
Looks good. You should definitely not try to actually compute all possible $X$ for this problem. But to write it properly, you should not start with
$$\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}X_1 + \begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}X_2 = X_1\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}+X_2\begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}.$$

Instead, I would write it like this:

Let $A = \begin{bmatrix}
0 &1 \\
2 & 1\\
\end{bmatrix}.$
Take $X_1, X_2 \in U$. We will show $X_1 + X_2 \in U:$
$$A(X_1 + X_2) = AX_1 + AX_2 = X_1A + X_2A = (X_1 + X_2)A.$$
Therefore $X_1 + X_2 \in U$.

ContentLicense: CC BY-SA 4.0


问:
What is the symmetry between the definitions of the bounded universal/existential quantifiers?
What is the symmetry between the definitions of the bounded universal/existential quantifiers?

$\forall x \in A, B(x)$ means $\forall x (x \in A \rightarrow B(x))$

$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$

These make intuitive sense, but I would expect there to be some kind of symmetry between how the definitions of the bounded quantifiers work, and I can't see one. $A \rightarrow B$ means $\lnot A \lor B$ which doesn't seem to have a direct relationship with $A \land B$. What am I missing?

答:
Is this more symmetrical?

$\forall x \in A, B(x)$ means $\forall x (x \notin A \lor B(x))$

$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$

ContentLicense: CC BY-SA 3.0


问:
The minimum of $f(x)=1+x+\cdots+x^{2n}$
For the function $$
f(x)=\sum_{k=0}^{2n}{x^k}
$$
I think this function has no zeros, and if ${f(x_0)}$ is the minimum point, then ${x_0\in \left[-1,0 \right] }$.

For $1+x+x^2+x^3+x^4$ and $1+x+x^2$, I can get the minimum by calculating $f'(x)$ and then using the root fomula. But this method doesn't work for $n\geqslant 3$ since there is no root fomula for equations higher than the fifth degree.

I have no idea for the situation where  $n\geqslant 3$, so any advice could be helpful

答:
We can establish the minimum $x_0\in(-1,0)$ without explicitly finding it. This can be done by first re-writing, $$f(x)=\frac{x^{2n+1}-1}{x-1}$$ and then differentiating, $$f'(x)=\frac{2nx^{2n+1}-(2n+1)x^{2n}+1}{(x-1)^2}$$ Descarte's Rule of Signs reveals that $f'(x)$ has only one negative real root, call it $\alpha$. In particular, we see that $f'(-1)=-n$ and $f'(0)=1$, so we know $\alpha\in(-1,0)$, thus we know that $f(x)$ is decreasing (negative derivative) from $(-\infty,\alpha)$ and is increasing (positive derivative) from $(\alpha,0)$. Thus, since $f(x)>1$ for $x>0$ and $f(0)=1$ and $f(-1)=0$, then we are assured that $f(\alpha)<0$ and will be the absolute minimum of the whole function $f(x)$.

Finding $\alpha$ will most likely be done with an approximation (instead of closed form). There are a number of ways to approximate $\alpha$. One particularly straight forward approach, is to half the intervals testing if you are positive or negative at each step using the derivative. Since you know $\alpha\in[-1,0]$ this method will reach a really good approximation very quickly.

Title: The minimum of $f(x)=1+x+\cdots+x^{2n}$


答:
By Cauchy-Hadamard, the radius of convergence is $r=\limsup_{n\to\infty}1/\sqrt [n]{|a_n|}=1/\sqrt [n]1=1$.

In general,  you need to investigate convergence on $|z|=r $ independently.  But in this case,  when  $|z|=1$, we have that $z^n\not\to0$.  Hence the series diverges.

Title: Geometric series of complex numbers

Body:
Let $ z $ be a complex number. I want to find the radius of convergence of $$ \sum_{n=0}^{\infty}z^{n} $$

My intuition is that this series converge for $ z\in D\left(0,1\right) $ (open unit disk).

Basicly, my intuition is based on the formula for real numbers $$ \sum_{n=0}^{\infty}x^{k}=\frac{1}{1-x} $$

Where $|x|<1 $.

Is this formula holds for complex $ |z|< 1 $? if so, how can I prove it?
That would prove that for $ |z| < 1 $ this series converge.

Now what about $ |z|=1 $ ? obviously for $ z=1 $ this series diverge, but what about $ |z|=1, z\neq1 $ ? My intuition is that the series would converge in this case, since both the imaginary part and the real part of $ z $ are smaller then $ 1 $ (also except for z=i, z=-i etc..).

How can I prove in a formal way that this series diverge for $ z=i, z=-i $ ?

Im not sure what tools from real series we can use in the complex case, so a few clarifications would be very helpful.

Thanks in advance.


问:
How many set will result from sorting $n$ objects into sets of $m$ elements such that each set overlaps with each other set by $p$ elements?
I have $n$ items and I would like to sort them into set of $m$ items. I would like to create these sets such that every pair of sets shares exactly $p$ elements. Additionally, all elements should be used the same number of times. Assuming this is possible with a given triplet $(n, m, p)$, what is the maximum number of sets one may construct?

A few examples: $(a, b, 0)$ yields $\frac{a}{b}$, $(q, q-1, q-2)$ yields $q$, $(n, 2, 1)$ yields $\frac{n!}{2(n-2)!}$, and $(7, 3, 1)$ yields $7$.

Thank you for any help you may provide.

答:
The question is not so clear. Anyway, I hope this can help.

Let us fix the number of sets $N$, each of which will have $m$ items and assume that each pair of sets shares exactly $p$ items.

How many items $n$ will there be?
I will say
$$ n = N * m - \binom{N}{2} * p . $$
In fact, we can count $m$ items for each set and then subtract $p$ items for each pair of sets.

Furthermore, each element will be exactly in $k$ sets, where $k$ is given by
$$ n * k = N * m , $$
which means
$$ k = \frac{N * m}{n} = \frac{N * m}{N * m - \binom{N}{2} * p} .$$

Now, let us assume that a triplet $(n,m,p)$ is given. We can re-write the question as follows: Find $(k,N)$ such that the equations are satisfied and $N$ is maximum.


问:
Feasible set of linear optimization problem
Consider the set

$$\{ x \in \mathbb{R}^n \mid x_1 = \cdots = x_{n-1}=0, 0 \le x_n \le 1 \}$$

Could this be the feasible set of a problem in standard form?

I'm leaning towards no, but I am unsure how to prove this.

答:
I'm assuming that the standard form of a constraint set is a collection of inequalities $a_1^Tx \le c_1, ..., a_m^Tx \le c_m$. In this case, yes, this set can be written in standard form.

The constraint $x_i = 0$ can be written as the equation $e_i^Tx = 0$ where $e_i$ is the $i^{th}$ standard basis vector, $e_1 = (0, ..., 1, ..., 0)^T$ where the $1$ is in the $i^{th}$ position. An equation can always be written as two inequalities, $e_i^Tx \ge 0$, $e_i^Tx \le 0$, or equivalently $-e_i^Tx \le 0, e_i^Tx \le 0.$

For the last inequalities: they are $e_n^T \ge 0$ and $e_n^T \le 1$, or equivalently $-e_n^T \le 0$ and $e_n^T \le 1.$


问:
If $\mathcal{A}$ is an abelian category then $\text{Hom}_{\mathcal{A}}(M,-)$ is left exact functor
Let $0\to A\xrightarrow{i} B \xrightarrow{j}C$ be an exact sequence in an abelian category $\mathcal{A}$. For an object $M\in \mathcal{A}$, I want to show that the sequence $0\to \text{Hom}(M,A)\xrightarrow{i_*} \text{Hom}(M,B) \xrightarrow{j_*}\text{Hom}(M,C)$ is exact. I know the proof of this when $\mathcal{A}$ is the category of $R$-modules for some ring $R$, and the same proof applies to show that $i_*$ is injective and $j_*i_*=0$, but I can't see how to show $\ker(j_*)\subset \text{image}(i_*)$. In the case $\mathcal{A}=R$-mod, given $f\in \ker(j_*)$ I may take $i^{-1}f\in \text{Hom}(M,A)$ to get $i_*(i^{-1}f)=f$. How can we show this in a general abelian category?

答: By exactness of the original sequence, $i$ is a kernel of $j$.  So by definition of a kernel, a morphism $f:M\to B$ factors through $i$ iff $jf=0$.  This says exactly that if $f\in\ker(j_*)$ then $f\in\operatorname{im}(i_*)$.


问:
When specifying the universal property that defines $V \otimes W$, what category are we working in?
What category does $V \otimes W$ satisfy its universal property in? I'm having a hard time figuring out what the arrows in the category in question are supposed to be. I don't think it's just the category of vector spaces with arrows being linear maps because bilinear maps are not linear maps and I'm not sure what collection of arrows contains bilinear and linear maps without simply containing all functions between vector spaces.

All vector spaces in this question are over a fixed field $F$.

The universal property of definition of the tensor product is given here on Wikipedia.

I've reproduced a similar diagram to this one below, but added in an extra identity arrow from $Z$ because I can't figure out how to draw a diagonal arrow, also the arrow from $V \otimes W$ to $Z$ should be dashed. I also changed the notation for $h'$ to use a prime instead of a superscript tilde.

In the diagram below, $h$ is bilinear, $\varphi$ is bilinear, and $h'$ is linear. Additionally $h'$ is uniquely determined.

$$
\require{AMScd}
\begin{CD}
V \times W @>{\varphi}>> V \otimes W\\
@V{h}VV @VV{h'}V \\
Z @>{\text{id}}>> Z
\end{CD}
$$

I don't understand which category specifically this diagram is in.

$V$, $W$, $V \times W$, $Z$ and $V \otimes W$ are all vector spaces. Additionally, $h$ and $\varphi$ are bilinear, but $h'$ is linear, so it's not clear to me what the collection of arrows as a whole should be.

What follows is my understanding of how to build $V \otimes W$ by imposing equalities on the free vector space over $V \times W$.

This is my attempt to understand how the tensor product of vector spaces is defined using different machinery.

The concrete construction for the tensor product makes intuitive sense and I think I understand the intuition for why a bilinear map from $V \times W$ to an arbitrary $Z$ would necessarily be completely determined by what it does to the basis of $V \otimes W$. I don't have a proof of this fact at the moment though.

Here's my understanding of how to produce a specific concrete vector space given two vector spaces.

I'll use ${v \choose w}$ to nonstandardly denote the formal tensor product of $v$ and $w$, which I'm thinking of as a fancy kind of ordered pair.

Let $V$ and $W$ be vector spaces. I will now define $T$, the tensor product of $V$ and $W$.

Let $T^1$ be defined as $\{ {v \choose w} \mathop| v \in V \land w \in W \} $.

Let $T^2$ be the set of formal finite linear combinations of elements of $T_1$, with coefficients taken from $F$, our fixed field of scalars.

Let $T^3$ be defined by taking $T^2$ and modding out by the congruence relation $\simeq$.

Let $\simeq$ be the congruence relation generated by $(\simeq_0)$ and respecting the operations $+$ and scalar multiplication by $k$. $(\simeq_0)$ is defined below.

$$ {v_1 \choose w} + {v_2 \choose w} \;(\simeq_0)\; {v_1 + v_2 \choose w} $$
$$ {v \choose w_1} + {v \choose w_2} \;(\simeq_0)\; {v \choose w_1 + w_2} $$
$$ (k){v \choose w} \;(\simeq_0)\; {kv \choose w} $$
$$ (k){v \choose w} \;(\simeq_0)\; {v \choose kw} $$

Let $T$ be equal to $T^3$.

答:
The term "universal property" does not have a universally agreed upon definition, as far as I know. But most universal properties can be understood in terms of representable functors.

Let $C$ be a category. A functor $F\colon C\to \mathsf{Set}$ is representable if there is some object $x\in C$ such that $F$ is naturally isomorphic to the Hom  functor $h^x\colon  C\to \mathsf{Set}$, which is defined on objects by $h^x(y) = \mathrm{Hom}(x,y)$ and on arrows by $h^x(f) = f\circ -\colon \mathrm{Hom}(x,y)\to \mathrm{Hom}(x,z)$ for  $f\colon y\to z$.

If $F$ is representable by $x$, then the set $\mathrm{Hom}(x,y)$ is in natural bijection with the set $F(y)$. That is, arrows out of $x$ to $y$ are classified by elements of $F(y)$.

In fact, the situation is even better than that! The identity arrow $\mathrm{id}_x\in \mathrm{Hom}(x,x)$ corresponds to an element $u\in F(x)$, and $u$ is "universal" in the following sense: For every object $y$ in $C$ and every element $a\in F(y)$, there is a unique arrow $f\colon x\to y$ in $C$ such that $F(f)(u) = a$.

The existence of $u\in F(x)$ with  this universal property is equivalent to the statement that $F$ is representable by $x$ - the proof is essentially the proof of the Yoneda lemma. Of course this whole discussion has  a dual version for universal properties in which maps into $x$ from $y$ are classified  by  elements of $G(y)$ for  a contravariant functor $G$ to $\mathrm{Set}$.

Ok, what about tensor products? Fixing vector spaces $V$ and $W$ over a field $k$, there is a functor $B_{V,W}\colon \mathrm{Vect}_k\to \mathrm{Set}$ defined by $B_{V,W}(U) = \{f\colon V\times W\to U\mid f\text{ is bilinear}\}$. If $T\colon U\to U'$ is a morphism of vector spaces (a linear transformation) and $f\colon V\times W\to U$ is bilinear, then $T\circ f\colon V\times W\to U'$ is bilinear. This defines the action of $B_{V,W}$ on morphisms: $B_{V,W}(T)\colon B_{V,W}(U)\to B_{V,W}(U')$ is defined by $B_{V,W}(T)(f) = T\circ f$.

Now what does it mean to say that this functor $B_{V,W}$ is represented by the tensor product $V\otimes W$? By the discussion above, there is a universal $b\in B_{V,W}(V\otimes W)$ such that for any vector space $U$ and any element $f\in B_{V,W}(U)$, there is a unique morphism of vector spaces $\tilde{f}\colon V\otimes W\to U$ such that $f = B_{V,W}(\tilde{f})(b)$. Unpacking, we have a distinguished bilinear map $b\colon V\times W\to V\otimes W$, and for any bilinear map $f\colon V\times W\to U$, there is a unique morphism of vector spaces $\tilde{f}\colon V\otimes W\to U$ such that $f = \tilde{f}\circ b$. This is exactly the familiar universal property of the tensor product.

Note that there was no need to consider any categories in the above discussion other than the category $\mathrm{Vect}_k$ of vector spaces over $k$ and the category $\mathrm{Set}$ of sets. To define the functor $B_{V,W}$, we did have to consider composing two different kinds of maps, namely bilinear maps and linear maps. But of course these are special kinds of functions, and we know how to compose functions - to make sense of this, we are not required to consider a category where the objects are vector spaces and the arrows include both linear maps and bilinear maps.

Nevertheless, it is reasonable to wonder whether such a category exists. It turns out that the right type of structure to consider here is not a category, but a multicategory. Multicategories are generalizations of categories in which every arrow has a finite sequence of objects as its domain. The main motivating example is the multicategory where the objects are vector spaces and the arrows $(V_1,\dots,V_n)\to W$ are the multilinear maps $V_1\times\dots\times V_n\to W$.

In this multicategory, the tensor product $V_1\otimes\dots\otimes V_n$ has the universal property that any arrow $(V_1,\dots,V_n)\to W$ factors uniquely as the canonical arrow $(V_1,\dots,V_n)\to V_1\otimes \dots\otimes V_n$ followed by an arrow $V_1\otimes \dots\otimes V_n\to W$.

Since I know that you (the OP) are interested in logic, I'll mention that other examples of multicategories come from logic. For example, take objects to be sentences and arrows $(P_1,\dots,P_n)\to Q$ to be sequents $P_1,\dots,P_n\vdash Q$ (so there is an arrow $(P_1,\dots,P_n)\to Q$ if and only if $Q$ follows from assumptions $P_1,\dots,P_n$). In classical propositional logic, the conjunction $P_1\land \dots \land P_n$ has the universal property of the tensor product.


问:
The correspondence between independence systems and basis systems of a matroid
Let $E$ be a finite set.

$1)$ If $I$ is an independence system on $E$, then the family of maximal elements of $I$ is a basis system

$2)$ If $B$ is a basis system, then $I_{b \in B} 2^b$ is an independence system

And finally, I would like to understand why the above two constructions are mutual inverses

Thank you. I used to study mathematics when I was young, and now I'm just dabbling in random texts. This is a proposition on a text on algebraic combinatorics that I've been reading, and the proof is given as an exercise, and I have not been able to crack it. Thank you!

答:
Suppose that $\mathscr{I}$ is an independence system on $E$, and let $\mathscr{M}$ be the family of maximal elements of $\mathscr{I}$. Plainly $\mathscr{M}\ne\varnothing$, so $\mathscr{M}$ will be a basis if and only if it has the following property: whenever $A,B\in\mathscr{M}$, and $a\in A\setminus B$, there is a $b\in B\setminus A$ such that $(A\setminus\{a\})\cup\{b\}\in\mathscr{M}$.

Suppose that $A,B\in\mathscr{M}$; I claim that $|A|=|B|$. If not, without loss of generality suppose that $|A|>|B|$; $\mathscr{I}$ has the augmentation property, so there is an $a\in A\setminus B$ such that $B\cup\{a\}\in\mathscr{I}$, contradicting the maximality of $B$, and the claim follows. Now suppose that $A,B\in\mathscr{M}$, and $a\in A\setminus B$. Let $A'=A\setminus\{a\}$; $A'\in\mathscr{I}$, and $|B|>|A'|$, so there is a $b\in B\setminus A'$ such that $(A\setminus\{a\})\cup\{b\}=A'\cup\{b\}\in\mathscr{I}$. Let $M\in\mathscr{I}$ be maximal such that $(A\setminus\{a\})\cup\{b\}\subseteq M$; then $M\in\mathscr{M}$, so $|M|=|A|=|(A\setminus\{a\})\cup\{b\}|$, so $(A\setminus\{a\})\cup\{b\}=M\in\mathscr{M}$, and $\mathscr{M}$ is a basis.

Now suppose that $\mathscr{B}$ is a matroid basis on $E$, and let $\mathscr{I}=\bigcup_{B\in\mathscr{B}}\wp(B)$; we want to show that $\mathscr{I}$ is an independence system on $E$. (Here I am guessing at the meaning of your $I_{b\in B}2^b$: the $I$ is rather obscure.) Clearly $\varnothing\in\mathscr{I}$, and $\mathscr{I}$ is downward-closed (meaning that if $J\subseteq I\in\mathscr{I}$, then $J\in\mathscr{I}$), so it only remains to show that $\mathscr{I}$ has the augmentation property.

It is very helpful to prove first that all members of $\mathscr{B}$ have the same cardinality; I’ll sketch the argument. Suppose that $M,N\in\mathscr{B}$, and $|M|<|N|$. Clearly there is an $a\in N\setminus M$, so by the exchange property there is a $b\in M\setminus N$ such that $(N\setminus\{a\})\cup\{b\}\in\mathscr{B}$. Let $N_1=(N\setminus\{a\})\cup\{b\}$; $|N_1|=|N|>|M|$, so you can invoke the exchange property again to get an $N_2\in\mathscr{B}$ such that $|N_2|=|N|$, and $N_2$ contains two elements of $M\setminus N$. Continue invoking the exchange property between $N_k$ and $M$; eventually you get $N_{|M\setminus N|}\in\mathscr{B}$, and at that point $M\subseteq N_{|M\setminus N|}$, so that it is impossible to invoke the exchange property one more time, even though $|N_{|M\setminus N|}|=|N|>|M|$. This is contrary to the definition of a basis, so we conclude that all members of $\mathscr{B}$ must have the same cardinality.

Now let $A,B\in\mathscr{I}$ be such that $|A|>|B|$; we want to show that there is an $a\in A\setminus B$ such that $B\cup\{a\}\in\mathscr{I}$. By definition there are $M,N\in\mathscr{M}$ such that $A\subseteq M$ and $B\subseteq N$. We can use the base exchange property repeatedly as in the last paragraph to replace all elements of $N\setminus B$ with elements of $M$, so we may as well assume that $N\setminus B\subseteq M$. $|N|=|M|$ by the preceding paragraph, and $|B|<|A|$, so $|N\setminus B|>|M\setminus A|$, and therefore $N\setminus B\nsubseteq M\setminus A$. Thus, $(N\setminus B)\cap A\ne\varnothing$, and we may choose $a\in(N\setminus B)\cap A$. But then $a\in A\setminus B$, and $B\cup\{a\}\subseteq N$, so $B\cup\{a\}\in\mathscr{I}$, as desired.


问:
matrix and eigenvalues
Please help me solve this question. This is not a homework assignment. If $\lambda_M$ is the eigenvalue of the matrix M, please prove that $max|\lambda_M|<1$
$$M_{2n\times2n}=\begin{vmatrix} A_{n\times n} & I-A\\I_{n\times n} & 0_{n\times n}\end{vmatrix}$$
NOTE: $max|\lambda_A|<2$

I easily solved this problem in scalar mode but I have no idea about its matrix mode

答:
The block matrix in question is similar to
$$
\pmatrix{I&-I\\ 0&I}\pmatrix{A&I-A\\ I&0}\pmatrix{I&I\\ 0&I}
=\pmatrix{A-I&0\\ I&I}.
$$
Its spectrum is therefore the union of the spectrum of $A-I$ and $n$ copies of $1$.

Title: matrix and eigenvalues


问:
H is normal in K iff K is a subgroup of the normalizer of H - is this false?
I think I found a mistake in my book.. It says that if $G$ is a group and we define the normalizer of $U\in P(G)$ as $N_G(U) = \{ g\in G \mid gU = Ug\}$, then if $H,K\leq G$, we have:
$$ H \unlhd K  \Leftrightarrow K\leq N_G(H)$$ .

The $\Rightarrow$ part of the implication is true and I was able to prove it. However, the reverse does guarantee that $H$ is normal in $K$ only if we can somehow prove that $H$ is actually a subgroup of $K$. I actually believe there is a counterexample without the addition of $H$ being a subset of $K$.

NOTE:  The book actually has an inverted subgroup sign for the right hand side of the equivalene, but I found out it should definitely be the other way around. They even say $N_G(H)$ is the smallest subgroup of $G$ containing $H$ as a normal subgroup, but I found this to also be false (it's actually the largest).. they seem to have messed up the ordering.

答:
I'm sure your counterexample is right. Consider any element $g$ of the group which normalizes $H$, but $g \not\in H$. Then $\langle g\rangle$, the subgroup generated by $g$, is a subgroup of $N_G(H)$, but we likely do not have $H \subseteq\langle g\rangle$.

To be more concrete, you could consider the alternating group $A_n$. Let $\tau$ be any transposition. Then we have $\langle\tau\rangle \leq N_{S_n}(A_n) = S_n$, but we certainly do not have $A_n \leq \langle\tau\rangle$.

(Or cyclic groups of the dihedral group could give you a similar counterexample.)

The authors were probably thinking of the LHS of your displayed equation as simply the "normalizing" property, without the subset property.


问:
On the number of ways to draw kissing circles
So I was watching Numberphile with Neil Sloane of OEIS fame on the number of ways to make circles intersect. During the intro, he explicitly forbid kissing (touching, tangent) circles. This made me think about the number of ways to draw circles that must all kiss, where intersections are not allowed. Kind of the dual problem, if you will. So in a nutshell:

Circles in the affine plane can be of any finite size

Circles must kiss at least one other (no separate circles, all must be connected)

Solutions that can be transformed into each other through mirroring, rotation, stretching, moving circles without making new tangent points, are considered the same

For $n=2$ circles there are two ways: touching on the outside, and touching on the inside of the larger circle.

For $n=3$ I believe the answer is 9. Try it, should be easy.

For $n=4$ I started drawing and came up with, so far, and with help from you guys, 38, 39, 42, 43, 44, 45, 47, 44 ways. The colored dots represent the number of kissing circles at the tangent points.

Questions:

Did I miss any? I'm sure I did.

Do I have inadvertent duplicates?

Does an integer sequence exist, or even a formula mapping $n = 2, 3, 4, ...$ to $2, 9, 44, ...$?

答:
Let's be a little systemic about drawing these.

For $n=4$, there are 9 hierarchies of circles (I think this is from OEIS A000081).

(((()))) the circles are all nested inside each other.

There are four of these: the differences come from whether the points of tangency coincide or not.  You have them labelled as #26, #28, #29, and #30 in the diagram when I started writing this.

((()())) two innermost circles surrounded by a nested pair.

There are six of these: the two innermost circles can either 1. each touch only their parent, 2. touch each other but only one touch the parent, or 3. touch each other and both touch the parent, and then the point of tangency with the grandparent can be involved or not.  Of these, only one is included in your diagram, as #34.

((())()) two children, one grandchild.

I'll call the child with its own child A, and the child without its own child B.  The children can touch just the parent (and the grandchild can touch or not the outer circle); A can touch only B and not parent (and the grandchild can touch or not B); B can touch only A and not parent (and the grandchild can touch B, the outer circle, or neither); A and B can touch both each other and the parent (and the grandchild can touch B, the outer circle, or neither), for a total of 10.  Of these, you have five: #22, #23, #31, #32, and #33.

((()))() one with child and grandchild, one empty.

There are four of these: the grandchild can touch or not the grandparent, and the child can touch or not its uncle.  Of these you have two, #27 and #45.

(()()()) one parent with three children.

The three children can be in four layouts: they can not touch at all; there can be one pair and one lonely one; there can be three in a line; there can be all three in a triangle.  This leads to eleven in the whole class: the three independent siblings can all touch parent #16, the pair can both touch #13 or only one can #35, the line can touch once on the end or in the middle (you have neither) or twice on both ends #12 or end and middle #25 or three times #19, the triangle can touch once (not present) twice (double counted #15 and #24) or three times (#11)

(()())() one with two children, one empty.

This lineup is basically the same as type 2; the only difference is that the "outermost circle" in 2 has become an uncle instead of a grandparent.  You have five of these #14 #17 #18 #39 #43, and have double counted two as #20 and #21 - you're missing the uncle-tangent counterpart to #14.

(())(()) two with one child each.

There's only three of these: the question is how many children touch their uncles.  It could be zero #38, one #37, or two #36.

(())()() one with one child, two empty.

Geometry finally comes into play!  The three outer circles can be in a line or triangle.  If they're in a line, the child can be in one of the ends, or in the middle, and it can touch or not an uncle #6, #7, #8, #44; if they're in a triangle, the child can touch one of the uncles #10, or outside the triangle formed by the uncles #9, or inside that triangle (not present)

()()()() four empties.

I think you actually have all of these: there's line #5, triangle plus tail #1, rhombus #2, square #3, star #40, and then there are three that are a triangle with circle in the middle that touches one #42 two #41 or three #4 of the outer circles.

So in total you appear to have (in the diagram when I started writing this answer double counted three (#20 is the same as #18, #21 is the same as #17, and #24 is the same as #15), and missed sixteen, for a grand total of 58 that I can count.


问:
Meaning of "defined"
What are the precise meanings of terms "defined", "well defined" and "undefined", etc.? We can't define what "defined" means since then we would run into circular definitions. (If definitiveness is established in terms of something else like "existence", then again we face the same problem.)

答:
"Well defined" usually means unambiguously defined, which is necessary for the definition to be any good.  For example, suppose you're defining multiplication modulo $11$.  Observe that $6$ and $17$ are equivalent to each other modulo $11$.  Similarly $4$ and $37$ are equivalent.  So if I find $6\cdot4$ and also $17\cdot 37$, will those be equivalent?  Notice that the first is $24$ and that reduces to $2$ and the second is $629$ and that reduces to $2$.  If they didn't reduce to the same thing, then multiplication modulo $11$ would not be well defined.

"Undefined" refers to an expression, not to a value of an expression.  Thus "$20/4$" is an expression, and "$2+3$" is an expression, and they're two different expression, but both have the same value.  The expression "$0/0$" is also an expression, but it doesn't have a value.  Hence it is undefined.

What "defined" means in general is a tougher problem and I won't attempt to deal with it here unless the Muse get more emphatic.

Title: Meaning of "defined"


问:
Understanding why the lines in $\mathbb{P}^2$ can be seen as planes passing through the origin of $\mathbb{R}^3$
I am reading the text Multiple-View Geometry in Computer Vision where the author is introducing the projective plane $\mathbb{P}^2$ as the augmentation of the homogeneous co-ordinates of $\mathbb{R}^2$ of the form $(x_1, x_2, x_3)$ with $x_3=0$ for points at infinity, and the line at infinity defined as $(1,0,0$), and regular euclidean points in $\mathbb{R}^2$ as $(x_1/x_3, x_2/x_3)$ for $x_3 \neq 0$.

The author writes on page $29$:

A fruitful way of thinking of $\mathbb{P}^2$ is as a set of rays in $\mathbb{R}^3$. The set of all vectors $k(x_1, x_2, x_3)^T$ as $k$ varies forms a ray through the origin. Such a ray may be thought of as representing a single point in $\mathbb{P}^2$. In this model the lines in $\mathbb{P}^2$ are planes passing through the origin.

This last sentence is giving me difficulty. Take two rays that are not identical say $k(x_1, x_2, x_3)$ and $t(x'_1, x'_2, x'_3)$, then since they both contain the origin, they span a plane. I'm not sure what the corresponding "line" is in $\mathbb{P}^2$, I'm having a bit of difficulty understanding the terminology/concepts. Any insights appreciated.

答:
First, "ray through the origin" is (imo) a bad term to use. Indeed a point in $\mathbb{P}^2$ corresponds to a set of the form $\{k (x_1, x_2, x_3) : k \in \mathbb{R}\}$ for some nonzero vector $(x_1, x_2, x_3)$, but this is a line, not a ray! The distinction is very important, since $(x_1, x_2, x_3)$ and $(-x_1, -x_2, -x_3)$ trace out the same line (and thus correspond to the same point in $\mathbb{P}^2$).

Anyway, this is really a way to define "line in $\mathbb{P}^2$". A line in $\mathbb{P}^2$, by this definition, is the set of points in $\mathbb{P}^2$ corresponding to lines through the origin in $\mathbb{R}^3$ which are contained in a given plane through the origin.


答:
Hartley & Zisserman define $\mathbb{P}^2$ as the quotient space $(\mathbb{R}^3 \setminus \{0\}) / \sim$, where two points are equivalent iff the one dimensional subspace spanned by the points is the same.
Let $[x]$ denote the equivalence classes.

So, not only is it fruitful to think of points as 'rays' (they seem to mean $k \neq 0$ as opposed to the more usual meaning of ray, $k\ge 0$) but that is essentially how they define them. (This interpretation is consistent with other texts.)

As far as I can determine, the authors use the sentence "In this model..." to define what they mean by a line in $\mathbb{P}^2$. That is, a line in $\mathbb{P}^2$ corresponds to a plane (passing through the origin) in $\mathbb{R}^3$.

If $[a],[b]$ are two distinct elements of $\mathbb{P}^2$, then $a,b$ are linearly independent elements of $\mathbb{R}^3$ and define a unique plane $\{ta+sb| s,t \in \mathbb{R} \} \subset \mathbb{R}^3$ which is, by definition, a line in $\mathbb{P}^2$.

Similarly, if you are given two distinct lines in $\mathbb{P}^2$, these correspond to two distinct planes passing through the origin. Their intersection defines a line (in $\mathbb{R}^3$) passing through the origin which is a point in $\mathbb{P}^2$.

(Frankly I find that particular section of the book to be a notational morass.)

Title: Understanding why the lines in $\mathbb{P}^2$ can be seen as planes passing through the origin of $\mathbb{R}^3$

Body:
I am reading the text Multiple-View Geometry in Computer Vision where the author is introducing the projective plane $\mathbb{P}^2$ as the augmentation of the homogeneous co-ordinates of $\mathbb{R}^2$ of the form $(x_1, x_2, x_3)$ with $x_3=0$ for points at infinity, and the line at infinity defined as $(1,0,0$), and regular euclidean points in $\mathbb{R}^2$ as $(x_1/x_3, x_2/x_3)$ for $x_3 \neq 0$.

The author writes on page $29$:

A fruitful way of thinking of $\mathbb{P}^2$ is as a set of rays in $\mathbb{R}^3$. The set of all vectors $k(x_1, x_2, x_3)^T$ as $k$ varies forms a ray through the origin. Such a ray may be thought of as representing a single point in $\mathbb{P}^2$. In this model the lines in $\mathbb{P}^2$ are planes passing through the origin.

This last sentence is giving me difficulty. Take two rays that are not identical say $k(x_1, x_2, x_3)$ and $t(x'_1, x'_2, x'_3)$, then since they both contain the origin, they span a plane. I'm not sure what the corresponding "line" is in $\mathbb{P}^2$, I'm having a bit of difficulty understanding the terminology/concepts. Any insights appreciated.


问:
How do I construct polynomials with complex numbers?
Can I have a simplified version of how to construct polynomials with complex numbers? Can I have an example on how to do it?</p>

答:
A complex polyonmial of degree $n\ge0$ looks like this
$$\sum_{k=0}^{n}a_kx^k=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...,$$
where $a_k,x\in\mathbb{C}$. Here are a few examples:

$i+3x^2-(1+4i)x^7$

$2x^4+\sqrt{2}x^5$

$3/4ix+\sqrt{i}x^9$

As the comments already made clear it is not clear what your asking. If my answer gives you not a solution you wanted, I will delete my answer. You should then specify your question.


问:
Find the index of the equilibrium points of the system (Question on solution)
I have the following system:
$$\dot{x} = 2xy$$
$$\dot{y} = 3x^2-y^2$$

I have the following solution: The system has one equilibrium point at the origin. Let the curve $\Gamma$ surrounding the origin be the ellipse $x = \cos\theta$, $y = \sqrt{3}\sin\theta$ for $0\leq\theta\leq 2\pi$. Then
$$\dot{x} = 2xy = \sqrt{3}\sin2\theta$$
$$\dot{y} = 3\cos2\theta$$

Then on $\Gamma$
$$\tan\phi = \sqrt{3}\tan[\frac{\pi}{2}-2\theta].$$
As $\theta$ increases from $0$ to $2\pi$, $\phi$ decreases $0$ to $-4\pi$. Hence the index is 2.

My question about this solution is how do I see that as $\theta$ increases from $0$ to $2\pi$, $\phi$ decreases $0$ to $-4\pi$? I feel like this is a typo because at $\theta = 0$ isn't $\phi = \frac{\pi}{2}$?

Any help and comments would be greatly appreciated. Thank you.

答: I'm 7 years late, here it is anyways. The following picture allows you to visualize why the index of the fixed point should be 2 (edit: the index is -2!).


问:
Is it possible to form a $\mathbb{R}^{n \times n}$ basis only with symmetric matrices?
It is a simple question but I don’t know if it is possible.

Can only symmetric matrices form a $\mathbb{R}^{n \times n}$ basis?

答: No. The set $V$ of symmetric matrices forms a proper subspace of $\mathbb{R}^{n \times n}$, so the span of any set of symmetric matrices must be a subspace of $V$.


问:
Lebesgue outer measure - Trouble understanding proof that outer Lebesgue measure of an interval is its length
I am stuck understanding Royden's version of the proof that an interval's outer Lebesgue measure is equal to its length, specifically just the first part where we consider a closed, bounded interval $[a,b]$, and want to prove that $m^*([a,b]) = b-a$. (full proof is here)

I understand the first step, which we just consider an arbitrary open interval $(a-\varepsilon, b+\varepsilon)$ and arrive at $m^*([a,b]) \leq b-a$. From here, I agree that if we can show $m^*([a,b]) \geq b-a$, then we have $m^*([a,b]) = b-a$, which is what we want for closed bounded intervals.

For the converse, I understand the application of Heine-Borel on an arbitrary open cover of $[a,b]$ and obtaining the subsequent sum of lengths of the intervals in a finite subcover. We get from these calculations that for any arbitrary finite subcover $\{I_k\}_{k=1}^N$, we have $\sum_{k=1}^N l(I_k) > b-a$.

The connection I'm not making here is why $\sum_{k=1}^N l(I_k) > b-a$ implies $m^*([a,b]) \geq b-a = l([a,b])$. The former is a statement about the sum of lengths over a finite subcover of $[a,b]$, so it seems like if anything, I want to say that $m^*([a,b]) \leq b-a$ since $m^*(A) = \inf\{\sum_k l(I_k)|A\subseteq I_k\}$.

EDIT: for reference, Royden defines Lebesgue outer measure of a set $A$ of real numbers as $m^*(A) = \inf\{\sum_{k=1}^\infty l(I_k) | A \subseteq \bigcup_{k=1}^\infty I_k\}$, where $l$ of an interval is defined to be the difference of its endpoints if the interval is bounded, and $\infty$ otherwise.

答:
Let $E$ be the set of families of open intervals of open intervals that cover $[a,b].$ Let $F$ be the set of finite families of open intervals of open intervals that cover $[a,b].$

For any $e\in E$ there is an $f\in F$ with $f\subseteq e.$  So for any $e\in E$ there is an $f\in F$ with $$\sum_{j\in e}l(j)\ge \sum_{j\in f}l(j).$$ So we have $$\inf \, \{\sum_{j\in e}l(j):e\in E\}\ge \inf \, \{\sum_{j\in f}l(j): f\in F\}.$$


问:
Proving that a relation is symmetric
I need to show that $x \sim y$ if $x=2y$. I started off by assuming that $x,y \in R$. Since $x=2y$, $y=x/2$. Since $y \neq 2x$, $\sim$ is not symmetric. Could I approach the problem is this way?

答:
Yes, you are correct. The relation is not symmetric.

Hope this helps. Ask anything if not clear :)

Title: Proving that a relation is symmetric


答: Yes, you can. Note that all you have to do is to find a pair of real numbers $x,y$ such that $x \sim y$ but $y \nsim x$. Take $x = 2, y = 1$.

Title: Proving that a relation is symmetric

Body: I need to show that $x \sim y$ if $x=2y$. I started off by assuming that $x,y \in R$. Since $x=2y$, $y=x/2$. Since $y \neq 2x$, $\sim$ is not symmetric. Could I approach the problem is this way?


问:
Probability that XOR of an arbitrary number of random bits is 1
How can we say than the XOR between $n$ (uniform, independent) random bits is $1$ with probability $1/2$?
For example if we have 4 random bits, we know that the XOR of them will be $1$ with half probability. The same happens if we have 3, or 4 random bits. But if we have $n$ random bits?

答:
It's One if number of 1s is Odd. So ignore the first n-1 bits. (their parity is completely random*) Now the last bit determine the final answer (based on number of 1s in the first n-1 bits)

*: Probability that number of 1s in the first n-1 bits is Odd, is 1/2. Try to prove this part yourself.

hint: You can find a 1-to-1 correspondence between even and odd parities. for example for every sequence with even parity there is sequence of bits with odd parity (by switching the last bit) and vice versa.


问:
Why is $(\frac{x}{y})^x=n^x$ linear when graphed.
I am currently learning the end of Algebra $1$. I was experimenting in demos and found that $(\frac{x}{y})^x=n^x$ is a linear equation. Furthermore, $n$ functions as $m$ would in $y=mx+b$ with. Why is this?

答:
$1^x$ is always $1$, so you're just graphing $(\frac xy)^x=1$.

A number to any power is $1$ if and only if the number itself is $1$, so you're really just graphing $\frac xy=1$.

And that's just $x=y$.


问:
Minkowski sum of the intersection of a closed and an open set with a compact set
Consider $\mathbb R^n$ with the usual topology and the Borel sigma-algebra. Let $A$ be open and $B$ be closed sets, respectively, in $\mathbb R^n$. Let $C$ be a compact set. Is the set $(A \cap B) \oplus C$ always Borel measurable?

Comments: The set $A \oplus C$ is open (and therefore measurable), $B \oplus C$ is closed (and measurable). In general, the Minkowski sum of a Borel measurable set with a compact set need not be Borel measurable.

答:
$(A \cap B) \oplus C$ must be $F_\sigma$ (and therefore Borel measurable).

Indeed, since $A$ is open it is $F_\sigma$, and say
$A=\cup_{n<\omega} F_n$ where each $F_n$ is closed.

Then $F_n\cap B$ is closed and hence $(F_n\cap B) \oplus C$ must be closed (where we use that $C$ is compact). Clearly $(A \cap B) \oplus C=\cup_{n<\omega}\big((F_n\cap B) \oplus C\big)$, showing that
$(A \cap B) \oplus C$ is $F_\sigma.$


答:
A much simpler way is to just consider the equation
$$x^2=2^2$$
To solve for $x$ we take the square root of both sides. One way of thinking of this equation is
$$\left(\frac{x}{y}\right)^x=n^x$$
and in a similar fashion, we take the $x$ root of both sides
$$\sqrt[x]{\left(\frac{x}{y}\right)^x}=\sqrt[x]{n^x}$$
which then cancels the $x$. giving
$$\frac{x}{y}=n\implies y=\frac{x}{n}$$
Which is the equation of a straight line $y=mx+b$ with $b=0$ and $m=1/n$

Title: Why is $(\frac{x}{y})^x=n^x$ linear when graphed.

Body: I am currently learning the end of Algebra $1$. I was experimenting in demos and found that $(\frac{x}{y})^x=n^x$ is a linear equation. Furthermore, $n$ functions as $m$ would in $y=mx+b$ with. Why is this?


答: Let us assume that $x \not =0$. Then there is a nonzero value of $y$ that solves this equation. Multiplying both sides by $y^x$ then gives $$x^x = 1^xy^x =y^x,$$ and so this equation is solved if the equation $x=y$ holds.


问:
Ideal generated by $v\otimes v - \Phi(v)1$
I'm trying to understand the proof of the existence of a Clifford algebra. Let $V$ be a vector space over $\mathbb{K}$, $\varphi: V\times V \to \mathbb{K}$ a bilinear form and $\Phi(v) = \varphi(v,v)$ its associated quadratic form. The prove existence of a Clifford algebra, one takes the quotient of the tensor algebra $T(V)$ with the ideal generated by elements of the form $v\otimes v - \Phi(v)1$, $v \in V$ and $1 \in \mathbb{K}$.

I don't have background in algebra, so I'd like to know what exactly is "the ideal generated by elements of the form $v \otimes v - \Phi(v)$". Also, is this the same as the ideal generated by $u\otimes v + v\otimes u - \varphi(u,v)1$ (which is also found in some references)?

答:
In general, if $A$ is an algebra, and $S$ is a subset, then the ideal generated by $S$, is the smallest ideal $I_{S} \subseteq A$ that contains $S$.

In this specific case, the set $S$ is
\begin{equation*}
 S = \{ v \otimes v - \Phi(v)1 \in T(V) \mid v \in V \}.
\end{equation*}
One way to get your hands on $I_{S}$ in this case is to simply set
\begin{equation*}
 I_{S} = \{ a \in T(V) \mid \exists (b_{i},b_{i}' \in T(V), s_{i} \in S): a = \sum_{i} b_{i}s_{i}b_{i}' \}.
\end{equation*}

To answer your question about the relation to the other ideal, we'd have to consider
\begin{equation*}
 S' = \{ u \otimes v + v \otimes u - 2\varphi(u,v)1 \mid u,v, \in V \}
\end{equation*}
and show that $I_{S} = I_{S'}$.
(You definitely need the factor 2, just consider the case $v=u$.)

So, let $V$ and $\varphi$ be as you stated, and let's define $\mathrm{Cl}(V)$ to be algebra $T(V)/I_{S}$.
Now we'd like to prove that this satisfies the universal property.
To that end, let $A$ be an associative, unital algebra, and let $f: V \rightarrow A$ be an arbitrary map.
Then, the map $f: V \rightarrow A$ extends to an algebra homomorphism $F: T(V) \rightarrow A$.
Now, suppose that $f$ was in fact a Clifford map, i.e.,
\begin{equation*}
 f(v)^{2} = \Phi(v) 1_{A},
\end{equation*}
for all $v \in V$.
One can use this equation that the map $F: T(V) \rightarrow A$ vanishes on the ideal $I_{S}$, and thus descends to an algebra homomorphism $T(V)/I_{S} = \mathrm{Cl}(V) \rightarrow A$.

If you want to know more about Clifford algebras, I strongly recommend the first chapter of Lawson & Michelsohn's Spin Geometry.


答: It's actually not a function. It's a straight line $y=x$, except $x=2t$, $y=\pm 2t$ are solutions for all $t \in \mathbb{Z}$. Think of it like the line $y=-x$ is also a solution but only when $x$ is an even integer and $y$ is negative $x$.


答: No, since the set of symmetric matrices has dimension $\frac{n(n+1)}{2}<n^2$ whatever set of $\big(\frac{n(n+1)}{2}+1\big)$ you take will be linearly dependent and by definition won't be a basis for $\mathbb{R}^{n\times n}$

Title: Is it possible to form a $\mathbb{R}^{n \times n}$ basis only with symmetric matrices?

Body:
It is a simple question but I don’t know if it is possible.

Can only symmetric matrices form a $\mathbb{R}^{n \times n}$ basis?


问:
The L ' Hopitals rule serves for some kinds of limits?
I was with this doubt whether the L 'Hopitals rule only serve for $\sin x$, $\arccos x$, $\tan x$, $\ln x$, $e^x$, these crazy trigonometrys there, because on the internet I just found explanations of the theorem of L'Hopital just in these examples. Could enlighten me if there is the possibility of using this theorem for any expression, in terms which seem must have for apply L'Hopital:

Ratio of two functions: $\frac{f(x)}{g(x)}$, when the limits are in indetermination, $f(x)$ and $g(x)$ differentiable and $g'(x) \neq 0$ for $x \to c$ (except possibly at $c$) and using polynomial equations of varying degrees, with or without root, in simple limits.

Please, the more examples to post better!

答:
L'Hospital's rule does work for any two functions $f(x)$ and $g(x)$, to find the limit
$$\lim_{x \to c} \frac {f(x)}{g(x)} = \lim_{x \to c} \frac {f'(x)}{g'(x)},$$ as long as $f(x) = g(x) = 0$ and both functions are differentiable. If $f$ and $g$ are in fact continuously differentiable (meaning that their derivatives are also continuous) and $g'(x) \ne 0$, there is a relatively easy proof of this:
$$\lim_{x \to c} \frac {f(x)}{g(x)} = \lim_{x \to c} \frac {f(x)-0}{g(x)-0} = \lim_{x \to c} \frac {f(x) - f(c)}{g(x) - g(c)},$$
since $f(c) = g(c) = 0$, and
$$ \lim_{x \to c} \frac {f(x) - f(c)}{g(x) - g(c)} = \lim_{x \to c} \frac {\frac{f(x) - f(c)} {x-c}}{\frac{g(x) - g(c)}{x-c}} = \frac {\lim_{x \to c} \frac{f(x) - f(c)} {x-c}}{\lim_{x \to c} \frac{g(x) - g(c)}{x-c}} = \frac {f'(x)}{g'(x)} = \lim_{x \to c} \frac {f'(x)}{g'(x)},$$
using properites of limits and the definition of derivative.
For a proof of the more general case, see the Wikipedia page.</p>


问:
How to move a single vector on Riemannian manifold?
On Riemannian manifold $(M,g)$, we know
$$
u^{\mu}\nabla_{\mu}v^{\nu}=0
$$
determinates the curve on which we can move $v^{\nu}$ parallel. But here $v^{\nu}$ is a vector field. If we are in Euclidean space, we can move a single vector freely, so my question is how to define the movement of a single vector on a curve in Riemannian manifold?

I think because I only have a single vector on one point, I can't use the covariant derivative. But Lie derivative is still helpful because it only depends on the map between two manifolds. But I don't know whether we can read a map between two manifolds from a curve on manifold.

I would be very grateful for any hints or references.

PS: This problem comes from physics. If a man has his own coordinate system on a point, how to know the corresponding coordinate system after his movement. I think it is a solved question, but I can't find related references.

答: Given a (piecewise) smooth curve $\alpha\colon [0,1] \to (M,g)$, and $v \in T_{\alpha(0)}M$ there is a unique solution to the initial value problem $$\frac{{\rm D}V}{{\rm d}t}(t) = 0, \quad V(0) = v$$defined on the interval $[0,1]$, since ${\rm D}V/{\rm d}t = 0$ is a first order linear ODE for the field $V$ along $\alpha$. Then $P\colon T_{\alpha(0)}M \to T_{\alpha(1)}M$ given by $P(v) = V(1)$ is what you want. In short, how to do the parallel translation of a vector along a curve? Take the unique parallel extension $V$ of the single vector $v$ to the curve $\alpha$, and look at the final value of this field $V$.


答:
If $n=1$, then yes. Otherwise, no.

Any linear combination $A$ of symmetric matrices $A_k=A^T_k$ is itself symmetric

$$
A^T = \left(\sum_k a_k A_k\right)^T = \sum_k a_k A_k^T = \sum_k a_k A_k = A
$$

but if $n>1$ then $\mathbb{R}^{n\times n}$ includes matrices which are not symmetric. Therefore, there is no basis of $\mathbb{R}^{n\times n}$ which consists of symmetric matrices only.


答:
Maybe using the log makes it more obvious, (for $y\ne 0$, $\left(\dfrac{x}{y}\right)^x > 0$, $n^x > 0$ )
$$\log\left(\left(\dfrac{x}{y}\right)^x\right) = \log\left(n^x\right) $$
since $\log(a^b) = b\cdot \log(a)$
$$x\cdot \log\left(\dfrac{x}{y}\right) = x\cdot \log(n)$$
simplify the $x$, and since $e^{\log(x)}=x$
$$\dfrac{x}{y} = n$$
we get the usual linear $y = ax + b$ form
$$y = \dfrac{1}{n}\cdot x + 0$$
which is linear.


答:
@Peter's answer is great for your specific case, but some insight on the more general algebraic construct you're dealing with should help, I think, because it helped me a lot back in the day.

I think the concept you are lacking is the idea of "an algebraic quotient by an equivalence class", and how this is linked to ideals of rings.

I'll take it that you're familiar with the very basics of set theory/abstract algebra/category theory: ie, that you at least know what an equivalence relation, a partial order, a binary operator, closure, associativity, identity, invertibility, commutativity, a monoid, a group, a ring, a field, a vector space, and a $\mathbb{K}$-algebra are. [If you're not acquainted with each of these concepts, I've written a concise introduction to this subject that might interest you, and that's freely available online; but such introductions abound on the web in any case.]

An ideal is a particular subset of a ring (or a $\mathbb{K}$-algebra, which is very similar to a ring). Ideals are special because they are subsets of a ring that are closed (stable) under both addition and multiplication. This gives them some important properties. Ideals of a ring $R$ are generally expressed as a set $eR = \{ e * r \; | \; r \in R \}$ (note that in non-commutative rings, you have to distinguish left-ideals from right-ideals). Thus, an ideal can generally be understood as "the set of multiples of $e$ in $R$". For example $4\mathbb{Z} = \{\dots, -8, -4, 0, 4, 8, \dots\}$, and $(x^2 + 1)\mathbb{R}[X]$ is the set of all polynomials which you can factor by $(x^2 + 1)$ (all polynomials that have $i$ and $-i$ amongst their list of roots).

One very important property of ideals is that they separate a ring into a "subspace" (the ideal itself; which is not exactly a ring, because it lacks the multiplicative identity in general) and a list of "affine subspaces" (the ideal plus some element of the ring) which are parallel to the "subspace".  For example,

$4\mathbb{Z} = \{\dots, -8, -4, 0, 4, 8, \dots\}$

$4\mathbb{Z} + 1 = \{\dots, -7, -3, 1, 5, 9, \dots\}$

$4\mathbb{Z} + 2 = \{\dots, -6, -2, 2, 6, 10, \dots\}$

$4\mathbb{Z} + 3 = \{\dots, -5, -1, 3, 7, 11, \dots\}$

$4\mathbb{Z} + 4 = 4\mathbb{Z}$

With this, we can define an equivalence relation over elements of the base ring (here $\mathbb{Z}$), based on any ideal: we consider any element to be equivalent if they are in the same affine space for that ideal. This is expressed algebraically as $e \sim f \Leftrightarrow e - f \in I$, where I is the ideal. The "algebraic quotient of a ring by an ideal" is then the homomorphism mapping an element of the input ring to its equivalence class. The output set, generally written $R/e$ or $R/eR$, has the same number of elements as there were "subspaces". $\mathbb{Z}/4\mathbb{Z}$ is obtained by mapping $\mathbb{Z}$ via the "modulo $4$" operator (the remainder of the Euclidean division by four); it is a space which functions like a clock with $4$ hours. $\mathbb{R}[X]/(x^2 + 1)$ is isomorphic to $\mathbb{C}$. The kernel of this homomorphism is ALWAYS the ideal itself.

Algebraic quotients come up all over mathematics; everywhere you see an equivalence relation. They're not limited to rings. Parallelism is an equivalence relation, and for example, taking the set of lines $\mathbb{R}^2$, and quotienting it by parallelism, you'll get a "looping half-circle" isomorphic to $S^1$. You can define the $u - v \in W$ equivalence relation for any subspace $W$ of any vector space $V$. Most spaces you'll see have their "most elegant construction" done through algebraic quotients. Generally, you use a free functor to "get an abstract block of marble" from some other structure (for example, the free functor from $Vect_\mathbb{K}$ to $Alg_\mathbb{K}$, which turns a vector space into its tensor algebra, which is the "least specific/most general way" of building an algebra from a vector space) and then you use algebraic quotients to "inject desired mathematical properties" into your quotient space (for example, the ideal generated by the elements of the form $a \otimes b + b \otimes a = 0$ injects anticommutavity into your tensor algebra over $V$; turning into the exterior algebra over $V$).

In the case of $\mathbb{R}[X]/(x^2 + 1)$, there's "only one element of the form $(x^2 + 1)$", the polynomial $I = x^2 + 1$; but this also means every polynomial which is a multiple of $I$ is part of the kernel; so in $\mathbb{R}[X]/(x^2 + 1)$, any $Q = (x^2 + 1)P = 0$. This makes the property $x^2 + 1 = 0$ be injected into $\mathbb{R}[X]$. Consequently, there exists at least one element $i$ such that $i^2 + 1 = 0$.

Now, in your case, you have an ideal generated by multiple elements, but the principle is the same. For every vector $v$ in $V$, you're going to inject the property that the "abstract, propertyless (safe for bilinearity)" multiplication of the tensor algebra must respect $v \otimes v = \Phi(v)$ (ie, the product of two vectors is a globally defined quadratic, euclidean (pseudo)norm). It is equivalent, @Peter shows, to defining the (pseudo)norm via the use of a (generalized) inner product (ie, $u \otimes v + v \otimes u = \phi(u,v)$).

Title: Ideal generated by $v\otimes v - \Phi(v)1$

Body:
I'm trying to understand the proof of the existence of a Clifford algebra. Let $V$ be a vector space over $\mathbb{K}$, $\varphi: V\times V \to \mathbb{K}$ a bilinear form and $\Phi(v) = \varphi(v,v)$ its associated quadratic form. The prove existence of a Clifford algebra, one takes the quotient of the tensor algebra $T(V)$ with the ideal generated by elements of the form $v\otimes v - \Phi(v)1$, $v \in V$ and $1 \in \mathbb{K}$.

I don't have background in algebra, so I'd like to know what exactly is "the ideal generated by elements of the form $v \otimes v - \Phi(v)$". Also, is this the same as the ideal generated by $u\otimes v + v\otimes u - \varphi(u,v)1$ (which is also found in some references)?


问:
What property results in $Pr(Y_1 < 0.5n^{0.75} - \sqrt{n}) \leq Pr(|{Y_1 - E[Y_1]}| > \sqrt{n})$?
I'm looking through a probability derivation, and I see the following inequality.

$$Pr(Y_1 < 0.5n^{0.75} - \sqrt{n}) \leq Pr(|{Y_1 - E[Y_1]}| > \sqrt{n})$$

Is there sufficient information provided here to know where this inequality comes from? $Y_1$ is a binomial random variable. $n^{0.75}$ is the number of trials.

So $E[Y_1] = n^{0.75} p_{success}$.

It is also given that $p_{success} = 0.5 + 0.5/n$

答: It's easy to show that if $y<0.5n^{0.75}-\sqrt n$, then $|y-(0.5n^{0.75}+0.5n^{-0.25})| > \sqrt n$. This means that the event $Y_1 < 0.5n^{0.75}-\sqrt n$ is contained in the event $|Y_1-E[Y_1]| > \sqrt n$.


问:
Finding minima and maxima of the map $f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}, x \in [0,1]$
Suppose that $c>0$ and define $f : [0,1] \to \mathbb R,$
$$
f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}
$$
I want to show that $x=1/2$ is a minimizer of this map and the maxima are attained at $x=0$ and $x=1$. The above map seems to be a Jacobi theta function of third kind. I'm not very familiar with special functions so any help or reference would be appreciated!

答:
Let's call $g(t)=e^{-ck^2}$ then the fourier transform is $\hat{g}(\xi)=e^{-c\xi^2}$ which means means $\hat{g}(k)=g(k)$ and now let use the poisson summation formula (PSF) which says that

$\sum_{k \in \mathbb Z}h(yk)=\frac{1}{y}\sum_{k \in \mathbb Z} \hat{h}(\frac{k}{y})\qquad$ (multiplicative PSF)

$\sum_{k \in \mathbb Z} h(k+y)=\sum_{k \in \mathbb Z} \hat{h}(k)e^{2\pi iky}\qquad $ (additive PSF)

The first case is interesting when $f(x) = \sum_{k \in \mathbb Z} e^{-cxk^2}$, because then we have that $f(x)=\sum_{k \in \mathbb Z}g(\sqrt{x}k)=\frac{1}{\sqrt{x}}\sum_{k \in \mathbb Z}\hat{g}(\frac{k}{\sqrt{x}})=\frac{1}{\sqrt{x}}\sum_{k \in \mathbb Z}g(\frac{k}{\sqrt{x}})=\frac{1}{\sqrt{x}}f(x^{-1})$

when $c=\pi$ we have $f(x)=\vartheta(0;-ix)$ the theta function which is important in number theory.

Finally the second case is more important for our question $f(x) = \sum_{k \in \mathbb Z} e^{-c(k+x)^2}$, here we have that
$f(x)=\sum_{k \in \mathbb Z} g(x+k)=\sum_{k \in \mathbb Z} \hat{g}(k)e^{2\pi ikx}=\sum_{k \in \mathbb Z} g(k)e^{2\pi ikx}$

$\Rightarrow f(1)=f(0)=\sum_{k \in \mathbb Z}g(k)$ and $f(1/2)=\sum_{k \in \mathbb Z}(-1)^kg(k)$


问:
Possible Choice Function for Collection of Non-Empty Subsets of Real Numbers?
If a step doesn’t choose a number, proceed to the next.

Note: the phrase a’s distance to b or $d(a, b)$ in this post means $|a - b|$, where $|x|$ is the absolute value function.
Note 2: as used in this post, the phrase A is dense around b means that the set of the distances of elements of A to B has no least element.

Note: as used in this post, the phrase $a \bmod b$ means the distance from the largest integer multiple of B smaller than a to a, or if there exists $n \in \mathbb Z$ such that $a = nb$, then 0.

Step 1: If there exists $n \in \mathbb E$ such that $|n| \leq |w|$ for any other $w \in \mathbb E$, choose $n$. (if we are to proceed, $\mathbb E$ has no smallest element with respect to distance to $0$)

Step 2: If there exists $n \in \mathbb E$ such that $|1 - n| \leq |1 - w|$ for any other $w \in \mathbb E$, choose $n$. (if we are to proceed, $\mathbb E$ has no smallest element with respect to distance to $1$)

Step 3: Set $a = 0$, $b = 1$. Call the interval $[a, b]$ $I$. If $P = \mathbb Z \cap I \cap E$, set $a = a - 1$, $b = b + 1$. Repeat until $I$ is non-empty.
Call the set of all real numbers contained in I that are also members of $\mathbb E$ as $P$. If $|P| = 1$, choose its only element. (if we are to proceed, $|P| = 2$)

Step 4: If $P$ has a maximum, choose it. (if we are to proceed, $P$ is dense around its supremum)

Step 5: If P has a minimum, choose it. (if we are to proceed, $P$ is dense around its infimum)

Step 6: Define a function $λ(s, x, m) = x \bmod m^{-s}$. Repeat the following increasing s until you choose a number:

a. Set m = e.

b. For all numbers n that are elements of P, plot $(λ(s, n, m), n)$, and if multiple n cause the same $λ(s, n, m)$, then choose the greatest as the point to plot. An infinite conflict cannot occur since the interval is finite and $λ(s, x, m)$ is periodic with a finite period. —> (P is dense around every number $k$ where $k mod m^{-s} = 0$.)

c. If there is a point with the greatest distance from 0 or the most extreme x-value, choose the height of that point, which will be in E because it’s also in P.

d. If m = e, set m = 2, and go to step b. Otherwise, continue.

Incomplete and non-rigorous proof that Step 6 will always terminate:

($m$ can be $e$ or $2$)

If $P$ contains any intervals then $m^{-s}$ will eventually be smaller than the size of the interval and the endpoints of the plot of (λ(s, n, m), n) will be in P.

Call the set of all the numbers that $P$ is dense around $D$. Now, for every $M \in D$, there must be infinitely many points in $P$ around $M$, otherwise it would be possible to loop over each point and determine its distance to $M$ and find the minimum, which would violate the definition of dense around. For Step 6 to not terminate for some $s$, the following are required:

For some $M$, $M \bmod m^{-s} = 0$, so that there can be no least or greatest point $n \in P$ with respect to $λ(s, n, m)$.

There must be no $n \in P$ such that $n\bmod m^{-s} = 0$, otherwise it would be the least in the interval.

These conditions become untenable as the $m^{-s}$ becomes smaller and smaller and each successive value is linearly independent with the alternate value of $m$ (and the previous value of $s$, in the case of $m = e$).
(This needs a rigorous proof).

答:
Your algorithm seems convolution and unreadable. But without even reading it, I can happily tell you that it won't work.

If you can describe a choice function for arbitrary sets of reals, then you can prove that the real numbers can be well-ordered without using the Axiom of Choice. We know that this is not the case. Case in point, Paul Cohen proved that if $\sf ZF$ is consistent, then $\sf ZF+\Bbb R$ cannot be well-ordered is consistent as well.

But wait, there's more! It gets worse! You cannot even prove that there is a choice function from countable sets of reals.


问:
Is $P(Y > X + c) \leq P(Y > X)$ for any random variables $X,Y$ and non-negative constant $c$?
Does this hold in the general case? It seems intuitive that this should always be the case since adding a non-negative constant to a random variable will thus make it the same or larger, and therefore the probability of $Y$ being greater should never increase.

How do you actually prove this formally?

答: Because holds $\{\omega \colon Y(\omega) \gt X(\omega)+c  \} \subset \{\omega \colon Y(\omega) \gt X(\omega)  \}$ for $c \gt 0$, then we can say $P(Y > X + c) \leqslant P(Y > X)$.


问:
How do you prove that a quadrilateral is a parallelogram using vectors?

PQRS is a quadrilateral where A, B, C, and D are the midpoints of SP, PQ, QR, and RS, respectively. Prove using vector methods that ABCD is a parallelogram.

I am unsure how I can prove it is a parallelogram with only this information. Thanks.

答:
Outline.

Assign position vectors (relative to some arbitrary origin $O$ which you don't have to characterise further) to $P, Q, R, S$. For example for $P$, you can call it $\mathbb{p}$.

Note that the midpoint of points $M$ (position vector $\mathbb{m}$) and $N$ (position vector $\mathbb{n}$) has the position vector $\frac 12 (\mathbb{m} + \mathbb{n})$.

Use that to find the vectors for the four sides of $ABCD$ by vector subtraction.

Now use the property that in a parallelogram, the parallel sides have vectors of the same magnitude and either the same or the exact opposite direction (depending on how you take the order of the points). In vector algebra terms, this would be represented by $\mathbb{u} = \pm \mathbb{v}$. Show this for both sets of parallel sides and hence complete the proof.


问:
Interpreting Game Theory Questions
I have come across two game theory questions, and got stuck in a similar stage in both.

Problem 1:

There are two boxes, correctly labeled as 4 candies and 6 candies according to their contents. You and your partner are famously perfectly rational. The game proceeds by each of you deciding independently which box to choose. If you both pick the same box, you split its' contents (2 or 3 candies each, depending on the box chosen). If you pick different boxes, you split the total (i.e. 5 candies each). What is the optimal strategy for you?

Solution of Problem 1:
I got stuck with too many undetermined probability values:

$$p_{4}^\text{me}, (p_{6}^\text{me} = 1-p_{4}^\text{me}),p_{4}^\text{partner},(p_{6}^\text{partner}=1-p_{4}^\text{partner}).$$

I couldn't see how to decide on my probabilities, as I don't know my partner's.
Apparently it is ok to assume that my partner's probabilities equal mine. From there, it is simple to get the expected payoff and maximize it as a single variable function.

Problem 2: (Mark Joshi et al, Quant Job Interview Questions and Answers)

We play a game: I pick a number $n$ from 1 to 100. If you guess correctly, I pay you \$$n$ and zero otherwise. How much would you pay to play this game?

My Attempt at Problem 2:
Let the choice of $n$ by the dealer be given by the probabilities
$$p_n^\text{delaer},\quad n \in \{1,2,\dots,100\}. $$
Here is where I found myself stuck again. How can I decide on a strategy when I don't know the dealer's probabilities?

The solution in the book reads

Note that this is really a game theory question. Each player has to pick a strategy. In this case, an optimal strategy will be one in which the expected winnings is independent of the strategy the opponent picks....
The solution is to pick $k$ with probability proportional to $1/k$. We therefore pick $k$ with probability $$\frac{1}{k} \left( \sum_{j=1}^{100} \frac{1}{j} \right)^{-1}$$.

I can't see how the author got around the unknown dealer probabilities, and found this strategy.

My questions are:

In problem 1, is it supposed to be obvious that my probabilities are equal to my partner's? Does that follow from our perfect rationality?

How does one end up with the $C/k$ probabilities given in the solution? How do they follow from the question? It seems to me that more wording is necessary for that, isn't it?

Are there some general rules when interpreting game theory problems as equations which I am missing?

Thank you!

答:
For problem 1, when using mixed strategies we must use one player’s payoffs to solve for the player’s strategy. This is because the point of my mixed strategy is to make you indifferent, and vice versa. Suppose that Row is going to randomize. Then Row’s payoffs must be equal for all strategies that Row plays with positive probability. But that equality in Row’s payoffs doesn’t determine the probabilities with which Row plays the various rows. Instead, that equality in Row’s payoffs will determine the probabilities with which Column plays the various columns. The reason is that it is Column’s probabilities that determine the expected payoffs for Row; if Row is going to randomize, then Column’s probabilities must be such that Row is willing to randomize.

Problem 1 gives the game
\begin{array}{c|lcr}
I/II & \text{Box4} & \text{Box6} \\
\hline
\text{Box4} & (2,2) & (5,5)  \\
\text{Box6} & (5,5) & (3,3)  \\
\end{array}

If player I best responds with a mixed strategy player II must make him indifferent between the choice of box. So we want the expected payoff between choosing Box4 and Box6 to be equal for player I. This gives

$$2p_2 + 5(1 - p_2) = 5p_2 + 3(1 - p_2)$$
$$p_2 = \frac{2}{5}$$

This means the only way player I can be indifferent to choosing either box is if player II chooses Box4 with probability $\frac{2}{5}$ and Box6 with probability $1 - \frac{2}{5} = \frac{3}{5}$. To find the probabilities for player I we use the same process.

It ends up being that in mixed strategies the Nash Equilibrium is player I: $(\frac{2}{5}, \frac{3}{5})$ and player II: $(\frac{2}{5},\frac{3}{5})$. In addition there are two pure equilibriums (Box4, Box6) and (Box6, Box4).

Title: Interpreting Game Theory Questions


问:
Is it possible to simplify $\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$?
Is it possible to simplify this expression?
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$
Is there a systematic way to check ratios of Gamma-functions like this for simplification possibility?</p>

答:
Amazingly, this can be greatly simplified.  I'll state the result first:

$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\sqrt{5}+1}{3^{1/10} 2^{6/5} \sqrt{\pi}}$$

The result follows first from a version of Gauss's multiplication formula:

$$\displaystyle\Gamma(3 z) = \frac{1}{2 \pi} 3^{3 z-1/2} \Gamma(z) \Gamma\left(z+\frac13\right) \Gamma\left(z+\frac{2}{3}\right)$$

or, with $z=2/15$:

$$\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right) = 2 \pi \,3^{1/10} \frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)}$$

Now use the duplication formula

$$\Gamma(2 z) = \frac{1}{\sqrt{\pi}}\, 2^{2 z-1}  \Gamma(z) \Gamma\left(z+\frac12\right)$$

or, with $z=2/5$:

$$\frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)} = \frac{\sqrt{\pi} \, 2^{1/5}}{\displaystyle\Gamma\left(\frac{9}{10}\right)}$$

Putting this all together, we get

$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\displaystyle\Gamma\left(\frac{1}{10}\right) \Gamma\left(\frac{9}{10}\right)}{\sqrt{\pi^3} \, 2^{6/5} \, 3^{1/10}}$$

And now, we may use the reflection formula:

$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$

With $z=1/10$, and noting that

$$\sin{\left(\frac{\pi}{10}\right)} = \frac{\sqrt{5}-1}{4} = \frac{1}{\sqrt{5}+1}$$

the stated result follows.  This has been verified numerically in Wolfram|Alpha.


问:
Find the pattern of a series
I am given the series $$\frac{1}{6} + \frac{1\cdot 8}{6\cdot 10} + \frac{1\cdot 8\cdot 15}{6\cdot 10\cdot 14}$$

I am asked to find a formula for this series $a_n$

So I have found that the last fraction multiplied at $nth$ term is $\frac{(1+7(n-1))}{(6+4(n-1))}$

But I am stuck, unsure how to proceed after this.

答:
Try recursion.

You know that each term (after the first term, of course) is simply the previous term multiplied by a varying constant that relates to $n$.

So try something like:

$$a_1 = \frac{1}{6} $$ and $$a_n = a_{n-1} \cdot \frac{1+7(n-1)}{6+4(n-1)}.$$

Title: Find the pattern of a series


问:
What is the fastest way to multiply and divide polynomials?
I want to get faster at multiplying and dividing polynomials. Is there a way I can do it faster? For multiplying I just do it normally. But for division I have to use the box method.</p>

答:
Here is a way I like to do division of polynomials that has the advantage that it doesn't require writing anything down except the final answer; I find that this is convenient and faster than setting everything up. Here's an example:

Suppose I want to divide $2x^4+3x^3+x^2+1$ by $x^2+x+1$. Maybe I start by writing (though you can omit all the writing except the answer if you want) something like $2x^4+3x^3+x^2+1 = (x^2+x+1)($, where I'm going to start writing my answer after that parenthesis. I think, what must the highest order term be? Well, whatever I multiply $x^2+x+1$ by, the highest degree term will be a multiple of that $x^2$. We need to end up with $2x^4$, so, I see I'll want to multiply the $x^2$ by $2x^2$, and so I write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2$. Now, I know whatever lower terms I add on to my answer, I've gotten the fourth power of the end result correct.

So, now I move on to the third power. We want our product $2x^4+3x^3+x^2+1$  to have a $3x^3$ term in it, and we know we want to multiply our $x^2+x+1$ by some multiple of $x$ to make this happen. You need to be careful though: looking at our $(x^2+x+1)(2x^2$, we see that we have already contributed to the $x^3$ term in the answer. In particular, when we multiply what we have so far, the squared $2x^2$ and the linear $x$ in $x^2+x+1$ multiply to the cubic term $2x^3$. So, we already have $2x^3$ of the $3x^3$ that we want, so we only need to get a $1x^3$ more: so we write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x$, as that new $x$ term combines with the $x^2$ to make the $x^3$.

Now we know that we've secured the correct fourth and third power of our final product, so it's time to work on the $x^2$ part. $2x^4+3x^3+x^2+1$ just has $x^2$ for that term, but we need to see how our previous terms we've written so far have affected that term. So looking at $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x$, we see two multiplications ended up making a contribution to that: the $1 * 2x^2$ and $x*x$, which make $3x^2$. So we want an $x^2$ in the end, but already have contributed $3x^2$, so we actually need to remove $2x^2$. So we write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2$, as that $-2$ will multiply by that $x^2$ to contribute what we want. (note in this step, we had to check two contributions that our previous terms did to the $x^2$ term. For longer polynomials, this step gets harder and harder, involving summing three, four, etc terms; the nice thing is that when doing all this adding, you don't need to keep much in your head at one time, as you can figure out what one previous thing you wrote contributed, and move on, keeping a running total. So even for large polynomials, if your arithmetic is fast and accurate, you can use this)

Now we've found the constant term, so next we need to find the part that didn't divide in nicely. Perhaps we write $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2 + \frac{}{x^2+x+1}$, and we know on the top of that fraction we have a linear term, to deal with the linear and constant term of the product. We see $2x^4+3x^3+x^2+1$ has no linear term, but our $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2$ contributed to the linear term $x*1$ and $-2*x$, for a net of $-x$; so we need to add $x$ back in to cancel that. So we write $(x^2+x+1)(2x^2+x-2 + \frac{x}{x^2+x+1}$. Finally, the constant term of $2x^4+3x^3+x^2+1$ is $1$, but in $2x^4+3x^3+x^2+1 = (x^2+x+1)(2x^2+x-2$ we already contributed $-2$ to the constant term, so we need to use the fraction to add a $3$, so we write $(x^2+x+1)(2x^2+x-2 + \frac{-x+3}{x^2+x+1})$.

For multiplication, you can similarly work out the product term by term by keeping running totals: it's easy to determine the constant term, two products will contribute to the linear term of the answer, three to the quadratic term, and so on. It's an easier process than the case of division, so I'll leave the details for you to ponder.


问:
In the history of mathematics, has there ever been a mistake?
I was just wondering whether or not there have been mistakes in mathematics. Not a conjecture that ended up being false, but a theorem which had a proof that was accepted for a nontrivial amount of time before someone found a hole in the argument. Does this happen anymore now that we have computers? I imagine not. But it seems totally possible that this could have happened back in the Enlightenment.

Feel free to interpret this how you wish!

答:
I am the proud coauthor of a 2018 paper in which we point out a wrong proof (and a wrong statement) in someone else's 2003 paper, which in turn was supposed to answer a question from yet someone else's 1993 paper. The funny thing is that one of the authors of the 1993 paper had written a Zentralblatt review for the 2003 paper without realizing that the example claimed in the latter was incorrect, and that the original 1993 question remained open. (The question itself is not something of a great importance, so I will only provide a link to the latest of these three papers below, in case someone might want to take a look: Section 4 deals with that error. I also tend to believe that the referee of the 2003 paper had not done a good job. The error in the proof boiled down to the claim that if a family $\mathcal A$ refines a family $\mathcal B$ (i.e. for every $A\in\mathcal A$ there is $B\in\mathcal B$ with $A\subseteq B$) then $\cap\mathcal A\subseteq\cap\mathcal B.$ My coauthor and I realized something was wrong since one of our papers together with those 2003 and 1993 papers produced a contradiction, so we had to dig to see exactly where it came from.)

ON MONOTONE ORTHOCOMPACTNESS

Strashimir G. Popvassilev, John E. Porter

Serdica Math. J. 44 (2018), 177–186

paper link pdf file


问:
Conditional Monte Carlo simulation
Suppose random variables $Z_i$ ~ $p(z)$ and $X_i$ ~ $p(x|z)$. $\frac{\sum_{i=1}^n X_i}{n}$ converges to $ E(X)$?

答: If you generate $Z$ from $p_Z(z)$, then $X$ from $p(x\, \vert Z=z)$ you're simply generating $(X, Z)$ samples from $p_{XZ}(x, z)$ since $p_{XZ}(x, z) = p(x\, \vert Z=z) p_Z(z)$. So the expected value of each $X_i$ is $\text{E}X$. Take the sample average of $n$ of them and the answer stays the same.

Title: Conditional Monte Carlo simulation


问:
Is it possible to form a $\mathbb{R}^{n \times n}$ basis only with symmetric matrices?
It is a simple question but I don’t know if it is possible.

Can only symmetric matrices form a $\mathbb{R}^{n \times n}$ basis?

答: A symmetric matrix is determined by the entries on and above  (or below) the diagonal.   There are $1+2+\dots+n=n (n+1)/2$ such entries.   It is easy to see that they form a subspace  whose dimension is $n (n+1)/2$.  Thus the answer in general is no.


问:
What's wrong with the following argument showing that $L^p$ is Banach.
I want to show that $L^p$ is complete (assume I have everything else done). Here's my approach which is slightly different from the usual approach.

If $f_n$ is Cauchy then let $g_k$ be subsequence such that $\|g_{k+1}-g_k\|_p<\dfrac{1}{2^k}$. Then let $$f=g_1+\sum_{k=1}^{\infty}g_{k+1}-g_k$$

As the name suggest, I want to show that $f_n\rightarrow f$. Now suppose that I already know that $f$ is in $L^p$, then I only need to show that $f_n$ converges to $f$.

To show that I say, for $g_n$ we have

$$\| f-g_n\|=\|(g_1+\sum_{k=1}^{\infty}g_{k+1}-g_k)-g_n\|$$

But the sum is telescoping and it is equal to $g_n+\sum_{k=n}^{\infty}g_{k+1}-g_k$

and so
$$\| f-g_n\|=\|\sum_{k=n}^{\infty}g_{k+1}-g_k)\|\le \sum_{k=n}^{\infty}\|g_{k+1}-g_k)\|\le \dfrac{1}{2^n}$$

the second to last inequality is the Minkowski's inequality and continuity of the norm. So $g_n\rightarrow f$. But since $f_n$ is Cauchy, we conclude that $f_n\rightarrow f$.
I can't tell what's wrong with the argument, but I have my suspicions. First, the book does it differently and the difference is significant. Second, I can apply this argument to conclude that any functional space over any space (not necessary banach) is Banach, which sounds dubious. I would appreciate your feed back. Even more dubiously, the same argument tells me that $\|f\|= 0$

答:
Let $(X,\mathcal{F},\mu)$ be a measure space. Let $p\in[1,\infty)$.
We go to show that $L^{p}(X,\mathcal{F},\mu)$ is complete. We assume
that $L^{p}$ is a normed space and that Minkowski's inequality holds.
Hence, we are going to prove that the $L^{p}$-norm is complete.

Let $([f_{n}])_{n}\in L^{p}$ be a Cauchy sequence. For each equivalence
class $[f_{n}]$, pick a representative $f_{n}$ in an arbitrary way.
(Note that $f_{n}:X\rightarrow\mathbb{R}$ is a measurable function)
Choose integers $1\leq n_{1}<n_{2}<\ldots$ such that $||f_{n_{k+1}}-f_{n_{k}}||_{p}\leq\frac{1}{2^{k}}$.
For simplicity, denote $g_{k}=f_{n_{k}}$.

Claim 1: There exists $A\in\mathcal{F}$ with $\mu(A^{c})=0$ such
that $\sum_{k=1}^{\infty}|g_{k+1}(x)-g_{k}(x)|<\infty$ for each $x\in A$.

Proof of Claim 1: For each $K\in\mathbb{N}$, define $S_{K}:X\rightarrow\mathbb{R}$
by $S_{K}(x)=\sum_{k=1}^{K}|g_{k+1}(x)-g_{k}(x)|$. Clearly, $S_{K}$
is non-negative and measurable. Moreover, $S_{1}\leq S_{2}\leq\ldots$.
Let $S=\lim_{K\rightarrow\infty}S_{K}$ be defined pointwisely. Note
that $S$ may take value $+\infty$. By Minkowski's inequality, $||S_{K}||_{p}\leq\sum_{k=1}^{K}||g_{k+1}-g_{k}||_{p}\leq\sum_{k=1}^{K}\frac{1}{2^{k}}\leq1$.
That is, $\int S_{K}^{p}d\mu\leq1$. Letting $K\rightarrow\infty$
and by Monotone Convergence Theorem, we have $\int S^{p}d\mu\leq1$.
In particular, $S$ is $\mathbb{R}$-valued $\mu$-a.e. Let $A=\{x\in X\mid S(x)\in\mathbb{R}\}$,
then $\mu(A^{c})=0$ and $\sum_{k=1}^{\infty}|g_{k+1}(x)-g_{k}(x)|<\infty$
for each $x\in A$.

Claim 2: Let $f=g_{1}1_{A}+1_{A}\sum_{k=1}^{\infty}(g_{k+1}-g_{k})$
be defined pointwisely. Then $f$ is a $\mathbb{R}$-valued measurable function. Moreover, $\int|f|^{p}d\mu<\infty$.

Proof of Claim 2: If $x\in A^{c}$, then $[1_{A}\sum_{k=1}^{\infty}(g_{k+1}-g_{k})](x)=0$.
If $x\in A$, then $\sum_{k=1}^{\infty}|g_{k+1}(x)-g_{k}(x)|<\infty\Rightarrow\sum_{k=1}^{\infty}[g_{k+1}(x)-g_{k}(x)]$
converges to an element in $\mathbb{R}$ (here, we have invoked the
completeness of $\mathbb{R}$ implicitly). Therefore, $\left(1_{A}\sum_{k=1}^{K}(g_{k+1}-g_{k})\right)_{K}$
is a sequence of measurable functions which converges pointwisely
and hence the pointwise limit is $\mathbb{R}$-valued and measurable.
That is, $f$ is $\mathbb{R}$-valued and measurable. Denote $h_{K}=g_{1}1_{A}+1_{A}\sum_{k=1}^{K}(g_{k+1}-g_{k})$.
Note that $|h_{K}|^{p}\rightarrow|f|^{p}$ pointwisely. By Minkowski's
inequality, we have that $||h_{K}||_{p}\leq||g_{1}1_{A}||_{p}+||1_{A}\sum_{k=1}^{K}(g_{k+1}-g_{k})||_{p}\leq||g_{1}||_{p}+1$.
That is, $\int|h_{K}|^{p}d\mu\leq\left(1+||g_{1}||_{p}\right)^{p}$.
Letting $K\rightarrow\infty$ and by Fatou's lemma, we have that $\int|f|^{p}d\mu\leq\left(1+||g_{1}||_{p}\right)^{p}<\infty$.

Claim 3: $g_{k}\rightarrow f$ in $L^{p}$-norm.

Proof of Claim 3: For $K>k$, we have that $h_{K}-g_{k}1_{A}=1_{A}g_{K+1}-1_{A}g_{k}=1_{A}\sum_{i=k}^{K}(g_{i+1}-g_{i}).$
By Minkowski's inequality, we have that
\begin{eqnarray*}
 &  & \int|h_{K}-g_{k}1_{A}|^{p}d\mu\\
 & \leq & \left(\sum_{i=k}^{K}\frac{1}{2^{i}}\right)^{p}\\
 & \leq & (\frac{1}{2^{k-1}})^{p}.
\end{eqnarray*}
Since $h_{K}\rightarrow f$ pointwisely, by Fatou's lemma, we have
$\int|f-g_{k}|^{p}d\mu=\int|f-g_{k}1_{A}|^{p}d\mu\leq\liminf_{K}\int|h_{K}-g_{k}1_{A}|^{p}d\mu\leq(\frac{1}{2^{k-1}})^{p}.$
That is, $||f-g_{k}||_{p}\leq\frac{1}{2^{k-1}}$ and hence $g_{k}\rightarrow f$
in $L^{p}$.

Claim 4: $f_{n}\rightarrow f$ in $L^{p}$.

Proof of Claim 4: Let $\varepsilon>0$ be given. Choose $N$ such
that $||f_{m}-f_{n}||_{p}<\varepsilon$ whenever $m,n\geq N$. Choose
$k'$ such that $n_{k}>N$ whenever $k\geq k'$. By claim 3, there
exists $k''>k'$ such that $||g_{k}-f||_{p}<\varepsilon$ whenever
$k\geq k''$. Let $n\geq N$ be arbitrary, then $||f_{n}-f||_{p}\leq||f_{n}-f_{n_{k''}}||_{p}+||f_{n_{k''}}-f||_{p}<2\varepsilon.$
Therefore $f_{n}\rightarrow f$ in $L^{p}$-norm.


问:
Laplace Transformations
can someone kindly help me with these few questions? :)

Find $L{e^tf(t)}$ in terms of $f*(s)$ and state a range of $s$ which this is defined.

I couldn't figure this out. You use the definition but i then get $e^t(1-s)f(t)$ and then i don't know what to do....

Using the Convolution Theorem, find the function f(t) satisfying the equation

$$f(t) = \int_0^t e^u f(t-u)\mathrm du + e^t$$

I know how to take the LT of both sides. I'm not sure how to figure out the LT of the integral though. Ive for f(s) = something + 1/(s-1) at the moment.

答:
The solution to the first question follows easily from the definition: Can you bring $\int_0^\infty  {e^{ - st} e^t f(t)\,dt}$ into the form $\int_0^\infty  {e^{ - s't} f(t) \,dt}$? (Consider the hint you were already given above.)

The solution to the second question follows easily from the Convolution Theorem. You want to find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation
$$
f(t) = \int_0^t {e^u f(t - u)\,du}  + e^t, \;\; t \geq 0.
$$
Your approach is right.
Taking Laplace transform on both sides gives
$$
\hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1,
$$
where $\hat \varphi$ is the Laplace transform of the convolution
$$
(f*e^t )(t) = \int_0^t {e^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)e^{t - u}\,du}\bigg).
$$
(Note that the upper limit of the integral is $t$ in order to satisfy $t-u \geq 0$.) The Convolution Theorem states that the Laplace transform of a convolution is the product of the Laplace transforms of the individual functions. Hence,
$$
\hat \varphi(s) = \hat f(s) \frac{1}{{s - 1}}.
$$
Now you can solve for $\hat f(s)$, and, in turn by inversion (assuming that $s > 2$), find $f(t)$.

Note: After you find $f(t)$, verify that it satisfies the original equation. This is very easily done; indeed $f(t)$ is only slightly different from $e^t$ (in form).

EDIT (completing the solution; see Remark below):

Solving for $\hat f(s)$ gives
$$
\hat f(s) = \frac{1}{{s - 2}}.
$$
Assuming that $s > 2$, it follows by inversion that $f(t)=e^{2t}$. Indeed, this $f$ satisfies the original equation; that is,
$$
e^{2t}  = \int_0^t {e^u e^{2(t - u)} \,du}  + e^t ,\;\; t \geq 0.
$$

Remark. In view of the OP's comments below, it appears that the factor $u$ was forgotten in the convolution term of the original equation. A solution to the modified problem has now been posted.

Title: Laplace Transformations


问:
Count users with big space between pings logic
please try to be lenient with me because I really have forgotten most of the stuff, so I will probably be making incorrect assumptions, word the problem incorrectly, etc.

Context

I'm trying to calculate the amount of daily users an application has. Currently the application sends a ping to the servers if at least 24 hours have passed since the last ping.

It needs to be taken into account that the application might not send a signal if it's not switched on. Examples:

The application is opened for the first time, ping is sent immediately

The application is opened for 24h and 1 second, a ping is sent at the beginning and at the 24th hour

The application is opened and closed, and opened again after 8 hours, a ping is sent only the first time

The application is opened and closed, and opened again after 30 hours, a ping is sent the first time, and at hour 30

The pings are summed on a daily basis and presented as the daily users the application has. I will refer to this as daily signal.

Questions

My objective is to understand:

How the original signal (users using the application) is transformed by the different design decisions to arrive to the daily users number.

If and where does Nyquist–Shannon sampling theorem apply.

I have failed to understand what happens when a signal has higher frequency than a the sampling its being measured with, and what does this mean to when reconstructing the signal. From what I understand, a lot of aliasing is introduced, however I don't see how it can't be dealt with in this scenarios.

How to calculate the error of the final output VS the desired output (users that pinged in a day VS users active in a day)

答:
So, if I am understanding correctly,

There is a system where users enter and, I suppose, leave after using it for some time (this is a, so called, queue)

The number of users in the system is sampled with some (maybe non-uniform) frequency

You want to find the number of users that entered the system within a 24h interval given the samples

I don't see how the sampling theorem is relevant here. As you pointed out, the process you are trying to reconstruct is discontinuous (infinite bandwidth) and the sampling theorem does not apply here.

I think it is impossible to give a definite answer for the number of users with gaps between measurements. You can only hope to provide a probabilistic answer, i.e., something of the form "given the measurements, the number of users within 24h was 10 with probability p(10), 11 with probability p(11), and so on."

The only approach I can think of is to consider a stochastic model for the arrival and departure of users in the system, which  will provide a probabilistic characterization of the number of users entering the system in any given interval, without any measurements. Given the measurements, you would consider the conditional probabilities (i.e., if you sampled, say, 8 users at some instant during the interval of interest, the probability of the number of users being less than 8 is zero, although there might be a non-zero probability that this will be the case if no measurements are available.)

You could maybe start by the simplest mathematical model for the arrival times in a queue, the Poisson point process. This however, does not model how long the users stay in the system (i.e., when they depart). You should look into queueing theory for such models.

P.S.: Apologies if I came off as offensive in my comment, that was not my intention.


问:
Dimensional coincidences between Lie groups
There are several exceptional isomorphisms between classical Lie groups that occur in low dimensions; of course it is necessary for the dimensions of the groups to coincide for this to happen! I'm curious if coincidences with pairs of higher dimensional Lie groups having the same dimension have meaning of some sort too, even if it isn't from a full-blown isomorphism. I'll relate some examples of what I'm looking for below.

The dimensions of $SU(n)$ and $SO(m)$ are $n^2-1$ and $m(m-1)/2$, respectively. Small cases where these are equal include $n=2, m=3$ and $n=4,m=6$, which relate to the exceptional isomorphisms $SU(2) \cong Spin(3)$ and $SU(4) \cong Spin(6)$; these and other low dimensional isomorphisms (which can be found here) make it natural to be curious if anything similar occurs in higher dimensions; the Diophantine equations involved are cute (involving a square being one more than a triangular number), which would make positive results in this vein all the more satisfying.

There are infinitely many pairs $(n,m)$ where these dimensions are equal; for example, take $n=11, m=16$. This might make one hope there is a something interesting happening with $SU(11)$ and $SO(16)$. There isn't an exceptional isomorphism with $Spin(16)$ (it's not on the lists of exceptional isomorphisms I've seen, which I assume are complete), but could there still be some relationship between these groups, or some interesting (possibly geometric) consequence to these? Of course, that value of $(n,m)$ is just an example, I'm interested in this phenomenon in general; any Lie groups are fair game.

(as one last example, a fun Diophantine equation comes from asking when a square number can be a triangular number: an example is $dim(U(6)) = 36 = 6^2 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = dim(SO(10))$; it would be very nice if this "meant" something)

答:
There are no special isomorphisms among different simple Lie groups of dimension $> 15$. Still, there are a few things that can be said:

Different real forms of the same complex simple Lie group have, of course, the same dimension. For example, for any $n > 1$, $\dim SL(n, \Bbb R) = SU(p, n - p)$ for any $p \in \{0, \ldots, n\}$ (both groups are real forms of the complex simple Lie group $SL(n, \Bbb C)$.)

If you're familiar with the $X_n$ nomenclature for complex simple Lie groups, we have $$\dim_{\Bbb C} B_n = \dim_{\Bbb C} C_n = n (2 n + 1) = n (2 n + 1),$$ that is,
$$\dim_{\Bbb C} SO(2 n + 1, \Bbb C) = \dim_{\Bbb C} Sp(2 n, \Bbb C) .$$ One consequence of this coincidence for real groups is that $$\dim SO(2 n + 1) = \dim PSp(n) = n (2 n + 1)$$ for all $n \geq 2$. In both cases the groups are isomorphic only for $n = 2$.

I don't know of anything that can be said about the relationship between generic pairs $G, G'$ of simple Lie groups whose dimensions coincide.


问:
Function whose support lies in neighborhood of point, then a function exists and is well defined
I am self studying the book Analysis on Manifolds, and saw this problem in chapter 16. I was wondering how one would do this.

Let $S$ be an arbitrary subset of $\mathbf{R}^{n} ;$ let $x_{0} \in S .$ We say that the function $f: S \rightarrow \mathbf{R}$ is differentiable at $\mathbf{x}_{0},$ of class $C^{r},$ provided there is a $C^{r}$ function $g: U \rightarrow R$ defined in a neighborhood $U$ of $x_{0}$ in $\mathbf{R}^{n},$ such that $g$ agrees with $f$ on the set $U \cap S .$ In this case, show that if $\phi: \mathbf{R}^{n} \rightarrow \mathbf{R}$ is a $C^{r}$ function whose support lies in $U,$ then the function
$$
h(\mathbf{x})=\left\{\begin{array}{ll}
\phi(\mathbf{x}) g(\mathbf{x}) & \text { for } \mathbf{x} \in U \\
0 & \text { for } \mathbf{x} \notin \text { Support } \phi
\end{array}\right.
$$
is well-defined and of class $C^{r}$ on $\mathbf{R}^{n}$.

答:
Hint:   This'll work because the product of  $C^r $ functions on an open set $U$ is $C^r $ on $U $.  So use $\hat g:=\phi\cdot g:U\to\mathbb R $ in the definition.

The complement of the support of  $\phi $ is open,  so for $x\notin U $, there's an open set  $V $ on which  $h\equiv0$.


问:
How to get a parametrically representation from a line
I have this question:

The equations
$$
x-1 = \frac{1}{2}(y-2) = \frac{1}{3}(z-2)
$$

describe a line. The same path is given parametrically by $x=1+t$, what are the parametric representations of $y$ and $z$?

I am a bit lost, can anybody help me?

David

答:
Just set each "side" of your equation equal to $t$.  For $x$, we obtain $x-1=t$ which gives the answer you mention.  For $y$, we have:

$$\frac{1}{2}(y-2)=t\quad \Longrightarrow\quad y=2t+2$$

I'll let you try to find the parametric equation for $z$.


问:
A vector along a trapezoid's diagonal

In the trapezoid TXYZ, $\vec{TX} = 2 \vec{ZY},$ and the diagonals meet at $O.$ Find an expression for the vector $\vec{TO}$ in terms of the vectors $\vec{TX}$ and $\vec{TZ}.$

I do not know how to answer this question. It involves the properties of vectors.

What I have done so far:

Let $\vec{TX}$ = unit vector $\mathbf{a},$ then $\vec{ZY}=\frac12$ unit vector $\mathbf{a}.$

Could you please give me a hint?

答:
For a hint without reading the rest of the following answer, you could refer to this different-looking example, which utilises the same solution technique.
$$$$

We need to determine the position of point $O.$ Start by constructing a diagram, labelling it as helpfully as possible. As such, let
$$\vec{ZY}=\mathbf{a},\\
\vec{TZ}=\mathbf{b},\\
\vec{TO}=m\,\vec{TY},\\
\vec{ZO}=n\,\vec{ZX}.$$

Now we form an equation relating $m, n, \mathbf{a},$ and $\mathbf{b}:$

 $$\vec{TO}=\vec{TZ}+\vec{ZO}\\ m\,\vec{TY}=\mathbf{b}+n\,\vec{ZX}\\ m(\mathbf{a}+\mathbf{b})=\mathbf{b}+n(2\mathbf{a}- \mathbf{b})\\ (m-2n)\mathbf{a}+(m+n-1)\mathbf{b}=\mathbf{0}.$$

Since $\mathbf{a}$ and $\mathbf{b}$ are not collinear, the coefficients of both vectors are $0.$

 Therefore, $m=\frac23.$ So
 $$\vec{TO}=\frac23\vec{TY}\\ =\frac23(\mathbf{a}+\mathbf{b})\\ =\frac13(2\mathbf{a})+\frac23(\mathbf{b})\\ =\frac13\vec{TX}+\frac23\vec{TZ}.$$

Alternatively, to determine the position of point $O$ without relying just on vector properties, we can just use triangle similarity.

 Triangle $OTX$ is similar to triangle $OYZ.$ So
 $$\frac{OT}{OY}=\frac{TX}{YZ}\\=2.$$
 Thus $$TO=\frac23TY\\ \vec{TO}=\frac23\vec{TY}\\ =\frac23(\vec{TZ}+\vec{ZY})\\ =\frac23(\vec{TZ}+\frac12\vec{TX})\\ =\frac13\vec{TX}+\frac23\vec{TZ}.$$

Title: A vector along a trapezoid's diagonal


问:
What if we take step functions instead of simple functions in the Lebesgue integral
When we define the Lebesgue integral, we first define it for simple functions $s(x) = \sum\limits_{j=1}^n c_j\chi_{A_j}(x)$ (where $A_j$ are measurable) as $\int sd\mu = \sum\limits_{i=j}^n c_j \mu(A_j)$ and then for $f\ge 0$ as $\int fd\mu = \sup\{\int sd\mu$ : s simple and $0\le s\le f\}$. But I was wondering what could go wrong if instead of taking simple functions in the supremum, we would take step functions, i.e. $s(x)=\sum\limits_{j=1}^nc_i\chi_{I_j}(x)$ where $I_j$ are intervals (any type, like $(a,b), (a,b], [a,b), [a,b])$).

答:
Your question: What could go wrong if instead of taking simple functions in the supremum, we would take step functions, i.e. $s(x)=\sum\limits_{j=1}^nc_i\chi_{I_j}(x)$ where $I_j$ are intervals (any type, like $(a,b), (a,b], [a,b), [a,b])$)?

Answer: If the Integral was defined using step functions, it would not be additive. Note:

$$ \int \chi_{ \mathbb{Q}\cap[0,1]}  =
\sup\{\int sd\mu : \textrm{ s step function and } 0\le s\le \chi_{\mathbb{Q})\cap[0,1]}\}=0 $$

$$\int \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]}  =
\sup\{\int sd\mu : \textrm{ s step function and } 0\le s\le \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]}\}=0 $$

$$ \int \chi_{ [0,1]}  =
\sup\{\int sd\mu : \textrm{ s step function and } 0\le s\le \chi_{[0,1]}\}=1 $$

So $\chi_{ [0,1]} =   \chi_{ \mathbb{Q})\cap[0,1]} + \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]}$ , but
$$ \int \chi_{ [0,1]}  = 1 \neq 0 + 0 = \int \chi_{ \mathbb{Q}\cap[0,1]}  +  \int \chi_{(\mathbb{R} \setminus \mathbb{Q})\cap[0,1]} $$

The only way to recover additivity would be to reduce the class of integrable functions. So either Integrals would not be additive or we would have less integrable functions.