答:
Continuing from where you left off, let $p=\log x$ and $q=\log y$. Then:
$$ \begin{align*}
(\log xy)^2 &= 4 \log x \log y \\
(\log x + \log y)^2&= 4 \log x \log y \\
(p+q)^2&= 4 pq \\
p^2 + 2pq + q^2 &= 4 pq \\
p^2 - 2pq + q^2 &= 0 \\
(p-q)^2 &= 0 \\
p-q &= 0 \\
p &= q \\
\log x &= \log y\\
x &= y
\end{align*} $$
as desired.</p>

Title: if $\log_xa+\log_ya=4\log_{xy}a$ prove that $x=y$

Body:
Let $x,y$ be numbers in the interval $(0,1)$ with the property that there exists a positive number $a\ne1$ such that
$$\log_xa+\log_ya=4\log_{xy}a$$
I have used the property
$$\log _ba=\frac{\log a}{\log b}$$

and I have
$$\log_xa+\log_ya=\frac{\log a}{\log x}+\frac{\log a}{\log y}$$

$$\log a\left(\frac{\log x+\log y}{\log x \log y}\right)=4\frac{\log a}{\log(xy)} \implies (\log(xy))^2=4 \log x \log y$$

I don't see how this implies $x=y$


问:
Nth term of recursively defined sequence
I have the following sequence defined recursively: $s_{n+1}=1+\frac{s_{n}}{2}$. I understand that the nth term of this sequence is given by $f(n)=2-(2-t)2^{1-n}$. However, I am not sure how this conclusion is reached (in particular, how this general term is derived from the recursive definition). Could someone provide a hint (if not demonstrate) how this is done? Thank you.

答:
One general way uses generating functions. Define $S(z) = \sum_{n \ge 0}
 s_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$, recognize the resulting sums:

$\begin{align*}
  s_{n + 1}
    &= \frac{1}{2} s_n + 1 \\
  \sum_{n \ge 0} s_{n + 1} z^n
    &= \frac{1}{2} \sum_{n \ge 0} s_n z^n + \sum_{n \ge 0} z^n \\
  \frac{S(z) - s_0}{z}
    &= \frac{1}{2} S(z) + \frac{1}{1 - z}
\end{align*}$

You don't give the initial value, just call it $s_0$. Solve for $S(z)$, write as partial fractions (a computer algebra system is nice to have...):

$\begin{align*}
   S(z)
     &= \frac{2 s_0 - 2 (s_0 - 1) z}{2 - 3 z + z^2} \\
     &= \frac{2}{1 - z} + \frac{s_0 - 2}{1 - z / 2}
\end{align*}$

We want the coefficient of $z^n$ in this. Just two geometric series:

$\begin{align*}
   [z^n] S(z)
      &= 2 + (s_0 - 2) \cdot 2^{-n}
\end{align*}$

ContentLicense: CC BY-SA 4.0


问:
What do you call a graph that has vertex groups?
I was wondering what to call a graph that can have a group of vertices as one vertex while at the same time allowing individual vertices in the group to be linked to other vertices in the graph. Here is a figure to illustrate:

In the example figure, note the following:

we cannot have 2 edges from a and b to c. The semantics here is: the (a,b) group has to be linked to C and not a and b individually.

we can have the group act as one vertex ((d,e) -> f) and we can have a subset of the group to act as one vertex as well (c -> e)

let's assume we are only dealing with a directed graph

let's assume the group cannot have an inward edge (the destination of every edge is a single vertex)

This cannot be represented as a hypergraph because the vertices inside a group are never linked together.

What I have in mind: we can look at this graph as a directed graph that can have vertex groups (including those that with 1 vertex). So every vertex becomes a vertex group, and then the notation would be: G(V, S, E). V is the set of individual vertices, S is the set of vertex groups defined over V, and E is the set of edges connecting v -> w s.t. v and w are in S. Does this seem sound?

Any pointers would be appreciated.

答:
This is called a directed hypergraph.

Each edge of a directed hypergraph is a pair of subsets of the vertex sets: a "tail subset" and a "head subset". We think of the hyperedge as going from the tail subset to the head subset.

In the example you drew, the three edges are

an edge with tail $\{a,b\}$ and head $\{c\}$

an edge with tail $\{c\}$ and head $\{d\}$

an edge with tail $\{d,e\}$ and head $\{f\}$.


问:
Changing the interval for alternating series test
Say I am doing an alternating series test, on a series from one to infinity, but after finding the first derivative of my sequence and drawing a number line, I find that it is only decreasing on the interval 2 to infinity. Would it be acceptable to say that the series has the same end behavior as a series from 2 to infinity, and continue with checking if the limit as n approaches infinity of that is 0? If I find that the second one is convergent, does that mean the first one must be?</p>

答:
Not taking anything away from the other, correct answer.  It can help to see this:
$$  \sum_{i=1}^\infty f(i) = f(1) + \sum_{i=2}^\infty f(i)  $$
Let $(S_i)_{i=1}^\infty$ be the sequence of partial sums on the left and suppose $\lim_{i \rightarrow \infty} S_i = S$ exists.  The sum on the right has the sequence of partial sums $(S_i - f(1))_{i=2}^\infty$ and $\lim_{i \rightarrow \infty} (S_i - f(1)) = S-f(1)$, so the right-hand side gives $f(1) + (S - f(1)) = S$, just like the left-hand side.


问:
A typo on page 42 of Robinson's book on group theory.
There's a typo on page 42 of Robinson's,  "A Course in the Theory of Groups (Second Edition)", ISBN 987-1-4612-6442-9.

Here's a picture:

It reads as follows:

If $H$ and $K$ are permutation groups on finite sets $X$ and $Y$, show that the order of $\color{red}H\quad \color{red}K$ is $\lvert H\rvert^{\lvert Y\rvert}\lvert K\rvert$.

The typo is the gap between the red $H$ and $K$.

Based on the context, I believe it should be $\color{red}{H\wr K}$. In previous instances of the Wreath product in the book, the symbol $\wr$ is slanted backward a bit, so it takes up about the same amount of space as the gap in question; besides, the result seems to match the properties of the Wreath product listed on Wikipedia.

Is this correct?

Please help :)

答:
When $K$ acts on a set $\Omega$, the (unrestricted) wreath product $H\wr_{\Omega}K$ is the semidirect product of $\prod_{\omega\in\Omega}H$ by $K$, letting $K$ act on the product by permuting the coordinates according to its action. When no set is specified, $H\wr K$ is usually considered to be the (unrestricted) regular wreath product with $K$ acting on itself by translation (depending on how you write your wreath products, it can be the left or the right action...). Rotman uses $\wr_r$  to denote the regular wreath product.

(The restricted wreath product uses the restricted direct product instead, i.e. the subgroup of the cartesian product of almost null elements; Hanna Neumann's book uses $H\mathop{\mathrm{Wr}}_{\Omega}K$ for the unrestricted product, and $H\mathop{\mathrm{wr}}_{\Omega}K$ for the restricted one.)

In the case at hand, it is clear that Robinson means $H\wr_Y K$, given the cardinality.


问:
Floor function inequality proof by contradiction
I've got this exercise to prove :

if $x \leq y$ then $\lfloor x \rfloor \leq \lfloor y \rfloor$

And I want to know if logically my approach is fine:

Suppose for the sake of contradiction that $x \leq y$ and $\lfloor x \rfloor \gt \lfloor y \rfloor$
then if we assume $x = y$ then $\lfloor x \rfloor = \lfloor y \rfloor$ which is contradiction.

is this legal proof logically? Do I need to look for contradiction for the second case as well (where $ x < y$) or it's enough like that?

Thanks alot !

答:
No, this is not sufficient as written. You need to handle the other case.

A recommendation: If $x<y$ then either $x< \left \lfloor{y}\right \rfloor $ or $\left \lfloor{y}\right \rfloor \leq x< y$. This splits your second case into two sub-cases. In my opinion, this is the easiest approach to finishing off your proof.

Edit: Just to be 100% clear, I did not finish the proof here. I am just recommending what's written be your next step.

Edit 2: I changed my two cases slightly. The earlier version was not wrong, but this way is better.


问:
Prove that an analytic function is zero
Suppose that I have an analytic function in four variables $$f = f(z_1,z_2,z_3,z_4)$$
such that I know the following facts:
$$\Re [f(z_1,z_2,0,0)] = 0$$
and that:
$$f(0,0,z_2,z_3) = 0$$
I was wondering whether these constraints are enough to say that the function is zero everywhere in $\mathbb{C}^4$ or not. I would be interested in knowing which tools one might use in understanding this.

答:
Here is a counterexample
$$f(z_1, z_2, z_3, z_4) = z_1 z_2 z_3 z_4$$

Title: Prove that an analytic function is zero


问:
if $\log_xa+\log_ya=4\log_{xy}a$ prove that $x=y$
Let $x,y$ be numbers in the interval $(0,1)$ with the property that there exists a positive number $a\ne1$ such that
$$\log_xa+\log_ya=4\log_{xy}a$$
I have used the property
$$\log _ba=\frac{\log a}{\log b}$$

and I have
$$\log_xa+\log_ya=\frac{\log a}{\log x}+\frac{\log a}{\log y}$$

$$\log a\left(\frac{\log x+\log y}{\log x \log y}\right)=4\frac{\log a}{\log(xy)} \implies (\log(xy))^2=4 \log x \log y$$

I don't see how this implies $x=y$

答:
Continuing your work...

$$(\log{x y})^2 = (\log{x}+\log{y})^2 = (\log{x})^2 + 2 \log{x} \, \log{y} + (\log{y})^2 = 4 \log{x} \, \log{y}$$

which means that

$$(\log{x})^2 - 2 \log{x} \, \log{y} + (\log{y})^2 = (\log{x} - \log{y})^2 = 0$$

which means that

$$\log{x} - \log{y} = 0 \implies x=y$$

ContentLicense: CC BY-SA 3.0


问:
How do I find the equation of the tangent line to the curve?
We have a curve described by the system of equations:
$$\begin{cases}
    x^2z+2xy=-8\\
   y^2z^2+xy=-4
  \end{cases}$$
We are asked to find a parameterization of the tangent line to the curve at the point $(-2,2,0)$

I've solved similar problems before but I've never seen the curve defined with a system of equations. I need help figuring out how to approach the problem!

My attempt:

$$\text{Let }
  \begin{cases}
   f(x,y,z) = x^2z+2xy+8 = 0 \\
   g(x,y,z) = y^2z^2+xy=-4
  \end{cases}$$
The common normal vector to $\nabla f \;\text{ and }\; \nabla g$ is tangent to the original curve, i.e their cross product.
$$\begin{cases}
   \nabla{f} = (2xz+2y,2x,x^2) \\
   \nabla{g} = (y,2yz^2+x,2zy^2)
  \end{cases}$$
$$\nabla{f}(-2,2,0) = (4,-4,4) \\
  \nabla{g}(-2,2,0) = (2,-2,0)
  $$
$$\nabla{f}\times\nabla{g} = (8,8,0) || (1,1,0)$$
Ultimately, the equation of the tangent is
$$\begin{cases}
   x = -2 +t \\
   y = 2 + t \\
   z = 0
  \end{cases}$$

答:
Hint: Let $(u,v,w)$ be a directing vector of the tangent.

Expand

$$\begin{cases}
    (x+\lambda u)^2(z+\lambda w)+2(x+\lambda u)(y+\lambda v)&=&-8\\
   (y+\lambda v)^2(z+\lambda w)^2+(x+\lambda u)(y+\lambda v)&=&-4
  \end{cases}$$
You will obtain 2 polynomials equations in variable $\lambda$.

Annihilate the first order terms in $\lambda$ (of course replacing $(x,y,z)$ by $(-2,2,0)$).

This will give a system of 2 equations in unknowns $u,v,w$ :

$$\begin{cases}
    4u-4v+4w&=&0\\
    2u-2v+0w&=&0
\end{cases}$$

which is enough for determining a direction.

Do you see why ? Intuitively, this annihilation is  the way to express that for an infinitesimal "step" in the direction of vector $(u,v,w)$, one remains on each surface, therefore on their intersection curve.

Remark: We assume that the way the surfaces are meeting at the considered point is not "pathological"...


问:
How do I interpret a log-plot that looks exponential?
When looking at an x,y plot such that the y-axis is the logarithm of $f(x)$, generally the plot looks like a line. How am I supposed to interpret a plot that still looks exponential after applying a log on the output?

答:
Your observation means that the function is growing faster than exponentially. You might consider taking logarithms again.

This image from wikipedia shows that the logarithm of the number of digits of the largest known prime $p$ is increasing linearly with time. But the number of digits of $p$  is the logarithm of $p$, so if you just graphed the logarithm of $p$ over time it would grow exponentially.


问:
Is it permissible to use a partial order to perform a generalized intersection or union?
Suppose I have a partially ordered index set $(I, \geq_b)$,
which indexes a family of sets $\{S_i\}_{i \in I}$.

If we let $i \in I$, is it permissible to
perform a generalized intersection as follows?
$$
\bigcap_{j \geq_bi} S_j
$$And, also for a generalized union?

答:
Yes. This is just another case of an intersection indexed by a set - here, fixing $i\in I$ the indexing set is $$J=\{j\in I: j\ge_bi\}$$ so that we're looking at $\bigcap_{j\in J}S_j$ or $\bigcup_{j\in J}S_j$ respectively. This really has nothing to do with the partial order.

I suspect the point at issue is that we initially think of unions/intersections as happening in some order. While this framework is satisfactory at first, once we move into more advanced topics it's really essential to forget the idea of ordering our sets entirely and just think about intersections/unions of arbitrary families of sets. (That said, note that things are a bit weird if we try to take the intersection or union of no sets, but this is an edge case you can ignore at first.)

EDIT: to address the comment below, yes, this is fine. We often abbreviate something like $$\bigcup_{j\in\{x: \varphi(x)\}}A_j$$ with $$\bigcup_{\varphi(j)}A_j,$$ so e.g. $$\bigcup_{j\ge_bi}S_j=\bigcup_{j\in\{x: x\ge_bi\}}S_j.$$ This is completely unambiguous and quite common; it's analogous to something like $$\sum_{1\le i<j\le n}{i\choose j}.$$


问:
What structure is preserved in a vector space isomorphism?
The definition of an isomorphism is "a structure-preserving mapping between two structures of the same type that can be reversed by an inverse mapping".

In my linear algebra class, an isomorphism is defined as "a bijective linear transformation." I understand some of the structure that is preserved between the spaces, but is there some exact fixed set of things that are preserved by a bijective linear transformation? (or an isomorphism in general).  Further, before we can say that a bijective linear transformation is an isomorphism, do we not need to prove that all of the structure is actually preserved? I ask this because we seem to just assume that a lot of things are preserved between different vector spaces, but never have we proven that these things are preserved between spaces. Below are a few things that I am a little confused about is they need to be proven or follow from the fact that $T$ is a bijective linear transformation.

Let $T: V \to W$ be an isomorphism.  I understand that since $T$ is linear, scalar multiplication and addition will be preserved. Must it be shown that if there is a linear transformation $A$ in $V$ such that for all $x \in V$, there exists a unique transformation $B$ in $W$ such that $A(x) = T^{-1}(B(T(x))$, or that if $S$ is a basis of $W$ of where $S = (S_1, \dots, S_n)$, then $(T^{-1}(S_1), \dots, T^{-1}(S_n))$ is a basis of $V$?

I'm not exactly sure where this preservation of structure stops, and when I am no longer using the fact that we have an isomorphism, and rather sort of just saying things that are not proven true. Maybe I am not entirely sure what constitutes as structure that needs to be preserved in terms of vector spaces.

Any help clearing up this idea of isomorphism is greatly appreciated.

答:
(This is a response to a question OP asks in the comments which is too long to keep in the comments, but is still relevant as an answer to the original question.)

Model Theory and/or Universal Algebra provide a very general way of approaching the question you're asking about what 'structure' is and what it means for 'isomorphisms' to preserve 'properties' of that structure. The other questions have provided more concrete explanations of this for the specific case of vector spaces, so I'll give a more abstract answer. I will also be approaching $\mathbb{F}$-vector spaces as a single-sorted structure, in contrast to user21820's approach.

A (first-order) language is a triple $\langle \mathcal{F}, \mathcal{R}, \operatorname{ar}\rangle$ consisting of disjoint sets $\mathcal{F}$ and $\mathcal{R}$ and a function $\operatorname{ar} \colon \mathcal{F} \sqcup \mathcal{R} \to \mathbb{N}$. Our interpretation of this triple is that $\mathcal{F}$ denotes a set of function symbols, $\mathcal{R}$ denotes a set of relation symbols, and $\operatorname{ar}$ associates to each function and relation symbol an arity.

In our particular example of $\mathbb{F}$-vector spaces (where $\mathbb{F}$ is any field, e.g., $\mathbb{R}$ or $\mathbb{C}$), we can take $\mathcal{F} = \{+\} \cup \{s_a \mid a \in \mathbb{F}\}$, $\mathcal{R} = \emptyset$, and $\operatorname{ar}(+) = 2$ and $\operatorname{ar}(s_\alpha) = 1$ for each $\alpha \in \mathbb{F}$. $+$ has its usual interpretation as vector addition, while the function symbols $s_a$ correspond to scalar multiplication by $a$ (this is how we get around the 'two-sorted' nature of vector spaces after fixing a field). Optionally you can include a $0$-ary function (i.e., constant) symbol $0$ for the zero vector and/or a unary function symbol $-$ for vector negation, but these are 'definable' in the sense I'll indicate below.

Given a language $\langle \mathcal{F}, \mathcal{R}, \operatorname{ar}\rangle$, a structure over that language is a triple $\mathbf{A} = \langle A, \langle f^\mathbf{A}\rangle_{f \in \mathcal{F}}, \langle R^\mathbf{A} \rangle_{R \in \mathcal{R}}\rangle$ such that for each $f \in \mathcal{F}$, $f^\mathbf{A}$ is a function $A^{\operatorname{ar} f} \to A$ and for each $R \in \mathcal{R}$, $R^\mathbf{A}$ is a $(\operatorname{ar} R)$-ary relation over $A$. In other words, a structure over a language is a set and instantiations of all the function and relation symbols (such that the arities line up appropriately). Note that at this point we haven't introduced any 'axioms' or the like.

If $\mathbf{A}$ and $\mathbf{B}$ are two structures over a language $\langle \mathcal{F}, \mathcal{R}, \operatorname{ar}\rangle$, an isomorphism between $\mathbf{A}$ and $\mathbf{B}$ is a bijection $\varphi \colon A \to B$ such that for function symbol $f \in \mathcal{F}$ and relation symbol $R \in \mathcal{R}$ (say, where $\operatorname{ar} f = \operatorname{ar} R = n$) and $a_1,a_2,\ldots,a_n$,
$$\varphi(f^\mathbf{A}(a_1,a_2,\ldots,a_n)) = f^\mathbf{B}(\varphi(a_1),\varphi(a_2),\ldots,\varphi(a_n))$$
and
$$\langle a_1,a_2,\ldots,a_n\rangle \in R^\mathbf{A} \iff \langle \varphi(a_1),\varphi(a_2),\ldots,\varphi(a_n)\rangle \in R^\mathbf{B}.$$
In other words, an isomorphism is a bijection which is compatible with the operations (or commutes with the operations) and both preserves and reflects each relation.

In our particular example of $\mathbb{F}$-vector spaces with the language I gave above, a structure over the language of $\mathbb{F}$-vector spaces is just a set $V$, a binary operation $+\colon V^2 \to V$, and unary operations $s_\alpha \colon V \to V$. An isomorphism $\varphi$ between such structures asks $\varphi$ to be a bijection and for $\varphi(v+w) = \varphi(v)+\varphi(w)$ and $\varphi(s_\alpha(v)) = s_\alpha(\varphi(v))$ for all $v,w$ in the domain of our first structure and $\alpha \in \mathbb{F}$. So far, this doesn't look quite like the typical definition of an isomorphism of $(\mathbb{F})$-vector spaces which would ask that $\varphi$ satisfy $\varphi(s_\alpha(v)+w) = s_\alpha(\varphi(w)) + \varphi(w)$ for all $v,w \in V$ and $\alpha \in \mathbb{F}$. We'll address that now.

We like to deal with structures that exhibit particular properties, like groups, rings, posets, etc. One way to realize this in the framework above is to define a first-order theory $\Phi$ over a language $\sigma = \langle \mathcal{F},\mathcal{R},\operatorname{ar}\rangle$ to be a set of well-formed formula in that language. To give the shortest definition I can of what is meant by 'well-formed formula', fix a countably infinite set of variables $X$ and consider strings over the set $\Sigma = X \cup \mathcal{F} \cup \mathcal{R} \cup \{ \text{`('},\text{`)'},\text{`,'},=,\vee,\wedge,\to,\neg,\leftrightarrow,\forall,\exists\}$ with the usual interpretation of the logical symbols and assuming that everything is distinct. $\mathrm{Term}_\sigma$ is the smallest subset of $\Sigma^\ast$ (the set of strings built up over the alphabet $\Sigma$) such that:

$X \subseteq \mathrm{Term}_\sigma$,

if $f \in \mathcal{F}$ is $n$-ary and $t_1,t_2,\ldots,t_n \in \mathrm{Term}_\sigma$, then $f(t_1,t_2,\ldots,t_n) \in \mathrm{Term}_\sigma$ (where if $n=0$ we interpret this as just '$f$').

In other words, $\mathrm{Term}_\sigma$ is the set of expressions built up from the variables and function symbols. Then define $\mathrm{Form}_\sigma$ to be the smallest subset of $\Sigma^\ast$ such that:

For any terms $t,s \in \mathrm{Term}_\sigma$, $(t = s) \in \mathrm{Form}_\sigma$.

If $\varphi,\psi \in \mathrm{Form}_\sigma$, then $(\varphi \wedge \psi)$, $(\varphi \vee \psi)$, $(\varphi \to \psi)$, $\neg \varphi$, and $(\varphi \leftrightarrow \psi)$ are in $\mathrm{Form}_\sigma$.

If $\varphi \in \mathrm{Form}_\sigma$ and $x \in X$, then $\exists x \varphi$ and $\forall x \varphi$ are in $\mathrm{Form}_\sigma$.

If $R \in \mathcal{R}$ is $n$-ary and $t_1,t_2,\ldots,t_n \in \mathrm{Term}_\sigma$, then $R(t_1,t_2,\ldots,t_n) \in \mathrm{Form}_\sigma$.

I'll call elements of $\mathrm{Form}_\sigma$ wffs (well-formed formulas). A sentence is a wff $\varphi$ such that if $x$ is a variable appearing in $\varphi$, then it appears in a subformula of $\varphi$ of the form $\exists x \psi$ or $\forall x \psi$ -- in other words, they're wffs without free variables, or equivalently wffs all of whose variables are bound. The point of sentences is that they correspond to wffs whose 'truth' in any particular structure shouldn't depend on providing an additional assignment of values to free variables.

Finally we can say that a first-order theory is nothing more than a set $\Phi$ of sentences; in other words, we're picking out sentences which will be our 'axioms'.

In our particular example of $\mathbb{F}$-vector spaces, the theory we take consists of the following sentences/axioms (diverging a bit in notation by using infix notation):

$\forall v \forall w (v+w = w+v)$ (associativity of vector addition)

$\exists 0 \forall v ((v+0 = v) \wedge \exists w (v+w = 0))$ (existence of a zero vector and additive inverses)

For each $\alpha \in \mathbb{F}$, $\forall v \forall w (s_\alpha(v+w) = s_\alpha(v) + s_\alpha(w))$ (scalar multiplication distributes over vector addition)

For each $\alpha,\beta \in \mathbb{F}$, $\forall v ( s_\alpha(s_\beta(v)) = s_{\alpha \cdot \beta}(v))$ (compatibility of scalar multiplication with field muliplication)

For each $\alpha,\beta \in \mathbb{F}$, $\forall v (s_{\alpha+\beta}(v) = s_\alpha(v)+s_\beta(v))$ (compatibility of scalar multiplication with vector addition)

$\forall v (s_1(v) = v)$ (identity element of scalar multiplication; here $1$ is the multiplicative identity of the field)

A $\mathbb{F}$-vector space is then a structure over the aforementioned language which satisfies the above first-order theory. Giving a precise definition of satisfaction would be too long, but an intuitive idea probably suffices. It can then be easily shown that for any map $\varphi$ between vector spaces, having $\varphi(\alpha v + w) = \alpha \varphi(v)+\varphi(w)$ is equiavlent to asking for $\varphi(\alpha v) = \alpha \varphi(v)$ and $\varphi(v+w) = \varphi(v)+\varphi(w)$.

Last but not least, we double back to what all this means in terms of isomorphisms. The general result is the following:

Theorem. Given a language $\sigma = \langle \mathcal{F},\mathcal{R},\operatorname{ar}\rangle$ and two structures $\mathbf{A}$ and $\mathbf{B}$ over that language, if there exists an isomorphism from $\mathbf{A}$ to $\mathbf{B}$, then for every sentence $\varphi$ in the language $\mathbf{A}$ satisfies $\varphi$ if and only if $\mathbf{B}$ satisfies $\varphi$.

So isomorphic structures agree with each other on any property/identity/etc that can be written down as a first-order sentence.

Disclaimer: The specific way I defined what a wff is isn't entirely universal; some approaches use Polish notation, some move $=$ from being a logical symbol to being a relation symbol and then add appropriate axioms characterizing it, some use a single variable symbol and include a way to generate infinitely many variables out of that (e.g., given $x$, we add another logical unary operation symbol $'$ such that $x'$, $x''$, $x'''$, etc. are all independent logical variables), etc. The specific ways I wrote the axioms for $\mathbb{F}$-vector spaces isn't standard (though pretty typical), nor is the language I used (as mentioned above, a constant ($0$-ary function) symbol $0$ and/or a unary funciton symbol $-$ might be used). Some folks like to distinguish constant symbols from function symbols. Etc. etc.


问:
Sequential characterization of integrability
I'm having some difficulties trying to understand a step from the following proof, which is about the sequential characterization of integrability. Here's the theorem:

And here's the part of the proof I cannot totally understand:

I understand why squeeze theorem can be applied, and I agree with the fact that it implies:

$$
\lim_{n\rightarrow \infty}{[U(f;P_n)-U(f)]} = 0
$$

Yet still I don't see why the above limit can be 'split' into two different limits, since, in general, the fact that:
$$
\lim_{n\rightarrow \infty}{(a_n-b_n)} = 0
$$
does not imply the existence of the following limits:
$$
\lim_{n\rightarrow \infty}{a_n}, \,\,\,\,\,\,\,\,
\lim_{n\rightarrow \infty}{b_n}.
$$

Thanks beforehand!!

答:
Since $U(f) = L(f) = \int_a^b f(x) \, dx$, we have

$$L(P_n,f) \leqslant \int_a^bf(x) \, dx \leqslant U(P_n,f),$$

and

$$0 \leqslant U(P_n,f) - \int_a^bf(x) \, dx \leqslant U(P_n,f) - L(P_n,f)\underset{n \to \infty}\longrightarrow 0,$$

which implies  $U(P_n,f) \to \int_a^b f(x) \,dx$ as $n \to \infty$.

We can show in a simlar way that  $L(P_n,f) \to \int_a^b f(x) \,dx$.


问:
Long Division of polynomials
How can we prove that $x^n$  isn't divisible by $g(x)=x^4+x+1$ without remainders?

I understand why for all $n<4$ it's working, but how with $n \ge  4$? How we can prove it formally?

Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$

答:
By Euclid: $\,\gcd(g,\color{#c00}x) = \gcd(g\bmod x,x) = \gcd(1,x)=1\,$ thus $\,\gcd(g,\color{#c00}{x^n}) = 1$

Or prime $\,x\nmid g\,$ so $\,g\,$ is coprime to $\,x\,$ so also to prime power $\,x^n$ (by unique factorization)

Title: Long Division of polynomials


问:
Prove that there exists a unitary matrix $U$ such that $U^H A U = B$

Given a non-null  $3 \times 3$ matrix $A$ such that $A^2 = 0$, then prove that there exists a unitary matrix $U$ such that $$U^H A U = \begin{bmatrix} 0 & 0 & u \\ 0 & 0 & v \\ 0 & 0 & 0\end{bmatrix}$$

I tried taking a generic $A$ as:
$$A=[v_1\:v_2\:av_1+bv_2]$$ where $v_1,v_2,av_1+bv_2$ are linearly dependent column vectors of $A$ since $Det(A)=0$.
Now let the unitary matrix be:
$$U=[u_1\:u_2\:u_3]$$
But the matrix multiplication is too heavy.

答:
the result trivially holds when $A=\mathbf 0$ so we can assume WLOG that $A\neq \mathbf 0$.

1.) By Sylvester's Rank Inequality we have

$\text{rank}\big(A\big)+\text{rank}\big(A\big)\leq \text{rank}\big(A^2\big)+n = 0 +3\implies \text{rank}\big(A\big) = 1$

2.) By rank-nullity $\dim \ker \big(A\big) = 3-1=2$.    So select 2 linearly independent vectors $\mathbf v_1, \mathbf v_2$ from the kernel.  Rescale so that $\mathbf v_1$ has length (2 norm) of 1 and call it $\mathbf u_1$.  Now run Gram Schmidt on $\mathbf v_2$ to get a second orthogonal vector that is re-scaled to have length one and call it $\mathbf u_2$.

3.)
now find a 3rd basis vector, $\mathbf v_3$, that is linearly independent from the other $\mathbf v_k$'s.  There are many ways to do this.  An easy way is to note that at least one standard basis vector $\mathbf e_k$ must work. (Why? a simple dimension argument will do it.)  Run Gram Schmidt on $\mathbf v_3$ as before to get $\mathbf u_3$.

Alternatively: consider the matrix $B:=\begin{bmatrix}\mathbf u_1^*\\\mathbf u_2^* \end{bmatrix}$

$2=\text{row rank}\big(B\big)=\text{col rank}\big(B\big)\implies \dim \ker\big( B\big) = 1$. So select a non-zero vector in that kernel and rescale so that it has length one, and call it $\mathbf u_3$.

4.)

$U  :=
\bigg[\begin{array}{c|c|c}
\mathbf u_1 & \mathbf u_2 & \mathbf u_{3}
\end{array}\bigg]$

where $U$ is unitary. Then

$AU = U\begin{bmatrix}0 &0 & x_1\\ 0 &0 & x_2\\0 &0 & x_3 \end{bmatrix}= U\begin{bmatrix}0 &0 & x_1\\ 0 &0 & x_2\\0 &0 & 0 \end{bmatrix}= UR$

where we know $x_3=0$ because $A$ is nilpotent and hence has $n=3$ eigenvalues equal to zero (i.e. $A=URU^*$ and the eigenvalues of a triangular matrix are on the diagonal).


问:
Fubini Theorem for measure zero
I know Fubini Theorem in calculus, but the measure zero version does not make sense to me:

$n=k+1$, and $V_c$ is the "vertical slice" {c}$\times R_l$.
Let $A$ be a closed subset of $R^n$ such that $A \cap V_c$ has measure zero in $V_c$ for all $c \in R^k$. Then $A$ has measure zero in $R^n$.

Could someone help me explain what does this mean, and how it relates to the Fubini Theorem in calculus?

Thank you very much!

答:
how it relates to the Fubini Theorem?

Let $\chi_A$ be the characteristic function of $A$: that is, $\chi_A(x)=1$ when $x\in A$ and $0$ otherwise. The integral of $\chi_A$ with respect to whatever measure is equal to the measure of $A$. This is why the characteristic function is used to make the transition from measures to integrals and back.

Fubini's theorem is about integrals. Applied to $\chi_A$, it says that the integral of $\chi_A$ can be found by integrating over each vertical slice first, and then integrating over $c$. Well, if every slice has measure zero, then the integral over every slice is zero. Then $\int 0\,dc =0$, and the conclusion is that $A$ has measure zero.

what does this mean?

Think of the calculus exercise about finding volume by integrating the area of cross-sections. If the area of each cross-section happened to be zero, the volume is zero. That's all there is to it, but a rigorous proof takes some effort, mostly to make sure that the concepts are properly defined.

Title: Fubini Theorem for measure zero


问:
Prove that if $f_n \rightarrow f$ in measure and $\phi$ is continous, then $\phi \circ f_n \rightarrow \phi \circ f$ in measure.
Let $f_n,f: E \rightarrow \mathbb{R}$ where $E$ has finite measure, suppose that $\phi:\mathbb{R} \rightarrow \mathbb{R}$ is continuous. Then $\phi \circ f_n \rightarrow \phi \circ f$ in measure. Show that this can fail when $E$ has infinite measure.

Here is my work so far: Since $f_n \rightarrow f$ in measure there exists a subsequence $f_{n_k}$ such that $f_{n_k} \rightarrow f$ pointwise a.e. on $E$. Then by continuity we have $\phi\circ f_{n_k} \rightarrow \phi \circ f$ pointwise a.e. on E. Of course this implies that $\phi \circ f_{n_k} \rightarrow \phi \circ f$ in measure. My trouble is that this only shows a subsequence converges in measure and not the full sequence itself. I feel like I almost there but I am just confused about how to end the proof.

Any help is appreciated!

答:
As mentioned in the comments, the following theorem trivializes the problem.

Theorem. Let $\mu$ be a finite measure. Then $(f_n)$ converges to $f$ in measure if and only if for every subsequence of $(f_n)$ there is a further subsequence that converges almost everywhere.

Proof. Suppose $(f_n)$ converges to $f$ in measure, i.e. $\mu(|f_n - f| > \epsilon) \to 0$ for all $\epsilon > 0$, and let a subsequence be given. For convenience just write it as $(f_n)$ again to avoid subscripts. Take a sequence $\epsilon_n \to 0$ decreasing and $\delta_n$ decreasing with $\sum_n \delta_n < \infty$ and choose $N(0) = 0$ and $N(n) > N(n-1)$ such that $\mu(|f_{N(n)} - f| > \epsilon_n) < \delta_n$ for all $i \geq N(n)$. Since $\sum_n \delta_n <\infty$, by the Borel-Cantelli lemma, except on a set of measure $0$, one has $|f_{N(n)}-f| \leq \epsilon_n$ for all but finitely many $n$. Since the $N(n)$ are increasing and $\epsilon_n$ decreasing, $|f_{N(i)}-f| \leq \epsilon_i \leq \epsilon_n$ for all $i \geq n$ except finitely many exceptions. Thus $f_{N(n)} \to f$ as  $n \to \infty$ except on a set of $0$ measure.

Next suppose that for each subsequence there is a further subsequence converging almost surely. Assume for contradiction that $\limsup_n \mu(|f_n - f| > \epsilon) = \delta > 0$ for all $n$. Then choose a subsequence (which to avoid subscripts we assume is the whole sequence) such that $\mu(|f_n - f| > \epsilon) \geq \delta/2$ for all $n$. By the assumed property, choose a further subsequence such that $f_{N(n)}  \to f$ except on a set of measure $0$. Then
$$\delta/2 \leq \mu(|f_{N(n)}-f| > \epsilon) = \int 1_{|f_{N(n)}-f|>\epsilon} d\mu \to 0$$
by the dominated convergence theorem (using here that $\mu$ is finite), a contradiction. Thus it must be that $f_n \to f$ in measure.


答:
In your case you should denote $a_n:=L(f;P_n)$, $c:=L(f)$, $d:=U(f)$ and $b_n:=U(f;P_n)$, where

$$ a_n \leq c\leq d\leq b_n. $$

You know that $a_n-b_n\to 0$, which as you said implies that

$$ a_n-d\to0 \quad \text{and} \quad b_n-c\to 0. $$

Title: Sequential characterization of integrability

Body:
I'm having some difficulties trying to understand a step from the following proof, which is about the sequential characterization of integrability. Here's the theorem:

And here's the part of the proof I cannot totally understand:

I understand why squeeze theorem can be applied, and I agree with the fact that it implies:

$$
\lim_{n\rightarrow \infty}{[U(f;P_n)-U(f)]} = 0
$$

Yet still I don't see why the above limit can be 'split' into two different limits, since, in general, the fact that:
$$
\lim_{n\rightarrow \infty}{(a_n-b_n)} = 0
$$
does not imply the existence of the following limits:
$$
\lim_{n\rightarrow \infty}{a_n}, \,\,\,\,\,\,\,\,
\lim_{n\rightarrow \infty}{b_n}.
$$

Thanks beforehand!!


问:
Is there a name for $(x, f(x))$?
Given a function $f$, and an $x$ from its domain, is there a name for the pair $(x, f(x))$?

Is there a defined terminology for one (any) such pair? I think the set of all $(x, f(x))$ is called the graph of the function, but I am asking is there's a name for one (or any) point of the graph.

答:
It's called an ordered pair and it is usually written as $(x,y)$ instead of $(x,f(x))$ because $y=f(x)$. It's called an ordered pair because it's a pair of numbers, and the order matter. For example, $(2,3)$ is different from $(3,2)$.

You may read more about it here

https://en.wikipedia.org/wiki/Ordered_pair

Or watch some videos on this topic here

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-negative-number-topic/cc-6th-coordinate-plane/v/plot-ordered-pairs

https://www.youtube.com/watch?v=YlT726odQcM

Title: Is there a name for $(x, f(x))$?


问:
Creating sets of $4$ elements such that no two sets share $3$ elements

The city of Mathlandia has $10$ politicians. Let $S$ denote the set containing all $\dbinom{10}{4}$ possible groups of $4$ of politicians.

Denote a subset, $M$, of $S$, as special if no two elements of $4$ politicians in $M$ share $3$ politicians. For example, if we label the politicians, $1, 2, 3, \cdots, 10$, then the subset: $\{ (1, 2, 3, 4), (2, 3, 5, 6) \}$ is special but the subset $\{(1, 3, 4, 10),  (1, 3, 6, 9), (2, 3, 4, 10) \}$ is not because elements $1$ and $3$ share $3, 4, 10$.

What is the maximal size of a special subset?

My Understanding

This problem is similar to this: Largest set of 4 digit codes that do not share 3 digits

However, what makes it different is that the digit $0$ is not allowed, order does not matter, and we cannot put the same number twice in a set of $4$ politicians (we need $4$ distinct politicians).

Due to these distinctions, I was unable to use Mike Earnest's solution to the linked problem. Does anyone know how to solve it?

How could we generalize for $n$ politicians? Is there some clean combinatorial method?

My Current Idea for Solution: Why Is this Wrong?

There are $\dbinom{10}{3} = 120$ sets of $3$ politicians.

Each set of $4$ politicians in $M$ contains $4$ different of $3$ politicians. So, the total number of sets of $4$ must be $\frac{120}{4} = 30$.

I checked this via a Java code and saw that the answer of $30$ is incorrect: it should actually be $18$.

Furthermore, if we try another number instead of $10$, for example $6$, this solution gives $5$ sets of $4$. However, the maximal amount of sets is $3$ : $$(1234), (1256), (3456)$$

For $n = 5$, we don't even get an integer. We get $2.5$. What incorrect assuption is this solution making?

答:
You observed that the four subsets of size $3$ derived from each subset of size four must be distinct from one another. This means that if $30$ sets were attainable, they would have to consist of all $4\cdot 30=120$ subsets of size $3$ exactly once. Therefore, you would need what is called a Steiner quadruple system (SQS) of order $10$. Fortunately, Steiner quadruple systems of order $n$ exist precisely when $n\equiv 2$ or $4\pmod 6$, so you can achieve $30$ sets.

It seems that constructing SQS's is a complicated matter, based on the abstract of this paper: A new existence proof for Steiner quadruple systems. However, in your case,  you can find the $30$ sets of size four in the La Jolla covering repository: Covering design C(10, 4, 3). A covering design is more general then a Steiner system, but in this case they are the same.

What you are looking for is also known as a constant weight binary code with length $10$, weight $4$, and distance $4$. To translate this to the language of coding theory, you view subsets of $\{1,\dots,10\}$ of size $4$ as binary strings of length $10$ with $4$ ones. The condition that no two subsets share $3$ elements is equivalent to saying any two strings differ in at least four places, hence the distance of $4$.

In general, the number of strings in a code of length $n$, weight $w$, and distance $d=2e$ is at most
$$
\left\lfloor \frac{n}{w}\left\lfloor \frac{n-1}{w-1}\cdots \left\lfloor \frac{n-w+e}{e}\right\rfloor\cdots\right\rfloor\right\rfloor
$$
This is the Johnson bound. For length $n$, size $4$ and weight $4$, this bound is certainly achievable when $n\equiv 2,4$ by a Steiner system of order $n$. There are lots of ad hoc ways to construct codes like this, you can find more codes and known upper bounds on Andries Brouwer's website.


问:
Upper bound for Poisson distribution
For a Poisson random variable $Z$ with the parameter $\lambda,\,$ what would be a good upper bound (sub-exponential type perhaps?) for $P(Z \geq \frac{\lambda}{2})?$

The issue here is that I can't use the large deviation bound for Poisson. What would be an alternative argument?

答:
$P(Z \ge \frac \lambda 2) \to 1$ as $\lambda$ increases, with jumps every time $\lambda$ is an even integer.

So $1$ is a trivial upper bound.

$1-e^{-\lambda}$ is slightly better, but is $P(Z \gt 0)$, so exact when $\lambda \le 2$ but not so brilliant for larger $\lambda$.

Empirically something like $1-0.1e^{-0.16 \lambda}$ looks better as an upper bound for $\lambda>2$.

For an empirical lower bound at least for $\lambda>1$, it seems $1-0.6e^{-0.15 \lambda}$ also seems to work reasonably.

This chart compares these bounds:

and the following chart shows the same bounds for the logarithm of the complementary probability with larger $\lambda$

Title: Upper bound for Poisson distribution


问:
Convolution of a compactly supported function with a uniform continuous function
If we have a compactly supported, integrable sequence of functions $g_i$ over $\mathbb{R}^n$ with their integrals $\int_{\mathbb{R}^n}g_i=1\ \forall i
$. Then, if $f$ be a uniform continuous function, can we say that $g_i*f\to f$ uniformly over $\mathbb{R}^n$? When can it be possible if it is not possible with the given conditions?

I have a theorem that says if $g_i$ satisfy the conditions as above and $f$ be bounded and integrable over $\mathbb{R}^n$, then $g_i*f\to f$ over $\mathbb{R}^n$ with the convergence being uniform over all compact subsets of $\mathbb{R}^n$. Now, if we make $f$ uniform continuous, how can we ensure uniform convergence throughout $\mathbb{R}^n$? Any hints? Thanks beforehand.

答:
What is $g_i$ ? Did you mean $g_i(x)=i^n g_1(i x)$ ? Otherwise you'll need that the supports of the $g_i$ "tend" to $\{0\}$ and $A=\sup_i \|g_i\|_{L^1}<\infty$.

The proof is that we have a sequence $c_i\to 0$ such that the support of $g_i$ is in $|x|<c_i$ so that $$g_i\ast f(y)-f(y)=\int_{\Bbb{R}^n} g_i(x ) (f(y-x )-f(y))dx\le A \|f(y-x )-f(y)\|_{L^\infty(|x|<c_i)}$$


问:
Continuity and uniform continuity on $[0,\infty)$
I have a problem proving the following statement:

Let $f$ be a function which is continuous on $[0,\infty)$ and there exists $c$ for some $c>0$ such that $f$ is uniformly continuous on $[c,\infty)$. Prove that $f$ is uniformly continuous on $[0,\infty)$

Maybe I can claim that $[0,c+1]$ has an overlap of length 1 with $[c,\infty)$. Then I can argue that two points $x,y$ such that $|x-y|<1$ is contained in one of $[0,c+1]$ and $[c,\infty)$. But how would I argue $f$ is uniformly continuous on the whole $[0,\infty)$?

And, especially how would I argue that $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$?

答:
Suppose $f$ is uniformly continuous on $[c, \infty)$, $c > 0$, and is continuous on $[0, \infty)$. Then $f$ is uniformly continuous on $[0, c + 1]$ since $f$ is continous and $[0, c + 1]$ is compact.

Consider that for all $[0, c + 1] \cup [c, \infty) = [0, \infty)$. Consider further that for all $x \in [0, \infty)$, the open ball $B_{1/3}(x) \cap [0, \infty)$ is a subset either of $[0, c + 1]$ or of $[c, \infty)$.

Now consider some $\epsilon > 0$. Take $\delta_1$ such that for all $x, y \in [0, c + 1]$, if $|x - y| < \delta_1$ then $|f(x) - f(y)| < \epsilon$. Take $\delta_2$ similarly for $[c, \infty)$.

Define $\delta = \min(\delta_1, \delta_2, 1/3)$. Consider $x, y \in [0, \infty)$ such that $|x - y| < \delta$. Suppose WLOG that $B_{1/3}(x) \cap [0, \infty) \subseteq [0, c + 1]$. And $y \in B_{1/3}(x) \cap [0, \infty)$. Then $x, y \in [0, c + 1]$. And $|x - y| < \delta \leq \delta_1$, so $|f(x) - f(y)| < \epsilon$. The case where $B_{1/3}(x) \cap [0, \infty) \subseteq [c, \infty)$ is analogous.

So $f$ is uniformly continuous on $[0, \infty)$.

To prove this for the function $f(x) = \sqrt{x}$, note that $f'$ is bounded on the interval $[1, \infty)$ and therefore $f$ is uniformly continous on $[1, \infty)$. Since $f$ is continuous on $[0, \infty)$, the uniform continuity of $f$ follows.


问:
Basic question on logic substitution
Wikipedia explains that substitution is a function $\sigma : V \rightarrow T$ that maps all variables $x_i$ to the corresponding term $t_i$ for some natural number $i$. Additionaly, according to the article, some authors state that $\sigma(x)=x$. What does that mean? That the substitution function $\sigma$ is an identity function, in which every replaced variable $x$ is equal to the replacing term $\sigma(x)$? That interpretation does not seem quite right for me, since some paragraphs later, the text talks about invertible and not invertible substitution functions. If the substitution function $\sigma$ is an identity function, then, if $card(T) \leq card(V)$, $\sigma$ is surjective and injective and, hence, always invertible.

答:
Here's the full quote from that paragraph of Wikipedia for reference

In first-order logic, a substitution is a total mapping $\sigma : V \to T$ from variables to terms; many, but not all, authors
additionally require that $\sigma(x) = x$ for all but finitely many
variables $x$.

This means that $\sigma$ leaves all but finitely many variables the same when transforming one well-formed formula into another. "$\sigma(x)=x$" is not meant to imply that all variables $x$ are sent to themselves by $\sigma$, but rather that some of them (specifically cofinitely many) are.

The idea here is that a substitution, when written in the set of pairs notation shown below with fixed points omitted, should be finite.

$$ \{ (x_1, t_1), (x_2, t_2), \cdots ,(x_7, t_7) \} \;\; \text{completely describes $\sigma$ and is finite} $$

We could have chosen to define $\sigma$ differently, with a domain of $V'$, a finite subset of $V$. In that case, $\sigma$ when applied to a formula would only replace variables in $V'$. Variables in $V\setminus V'$ would be untouched.

Title: Basic question on logic substitution


问:
The motivation behind the properties of height functions
I'm reading Hindry & Silverman's book on Diophantine Geometry, specifically the section on height functions.

At the beginning they speak of two important properties that a height function $h:V\rightarrow [0,\infty)$ should satisfy: first, the set of points in the complex projective variety $V$ of bounded height should be finite. Second, it should translate geometric properties of points in $V$ to arithmetic properties.

The second property is a natural thing to strive for, if somewhat difficult to quantify. But I don't see any particular reason to ask for the first property to be true. I understand that such a property is fundamental in many finiteness proofs - to show that $V(\mathbb{Q})$ is finite, for example, one can simply show that its points have finite, bounded height.

So, what exactly is the motivation for seeking this property in height functions? Or was the theory actually developed the other way around - geometers knew that they needed functions with this property in order to tackle certain problems?

答:
It might help if you substitute the word "complexity of an object" for the "height of an object." It's very natural, when studying infinite sets such as the set of points in $\mathbb P^N(K)$ where $K$ is a number field, to try to (roughly) order them by complexity. For any sort of counting results, one wants only finitely many elements whose complexity is smaller than any given bound $B$. In principle, and I think this is the correct intuition, is that the height (complexity) of an object $\mathcal{O}$ is equal to (again, somewhat roughly)
$$ \operatorname{height}(\mathcal{O}) = \operatorname{Complexity}(\mathcal{O}) = \text{Number of bits it takes to describe $\mathcal{O}$.} $$
E.g.,  for a an integer $n\in\mathbb Z$, we have $h(n)=\log(n)$. (Use $\log_2$ if you want, but it won't matter much.) Here's a typical problem: How many primes are there? The way people approach the problem is to ask how fast the following function grows:
$$ \pi(X) = \#\{ \text{primes $p$} : \operatorname{Complexity}(p) \le X \}. $$
Similarly for $\mathbb P^N(K)$. So heights are not used only  to prove finiteness theorems by showing that the elements in a  set have bounded height, but also for infinite sets they are used to give an arithmetic (complexity theoretic) measure of how large the set is by measuring the growth of the height (complexity) counting function.

Then, when trying to solve problems and prove theorems, one is naturally lead to analyze how the complexity of objects change when one has a map between sets of objects. It would be nice if the complexity always varied in some regular way, but generally that's only true if the maps are nice enough. So it's a balancing act.

For an illustration of this, Weil heights in arithmetic geometry have nice functorial transformation properties for morphisms between smooth (or at least not too singular) projective varieties. But for dominant rational maps, and indeed even for birational automorphisms, heights can behave in ways that are very hard to analyze.


问:
Variance definition notation question
I have a question on the Wiki definition of Variance.  Does this form represent abuse of notation or am I interpreting it incorrectly?

The definition:
$$
Var(X) = E[(X - E[X])^2]
$$
and
$$
E[X] = \sum_{i=1}^k c_i p_x(c_i) = \int_{\mathbb{R}} c p_x(c) dc
$$

It's my understanding that the parameter of $Var$ and $E$ is identified by a random variable $X$ and the result is a value of the random variable, typically $\in \mathbb{R}$.  It seems that the outermost $E$ is given some number other than the symbol for the random variable.  I'm also confused by $X - E[X]$, where $X$ appears to be a random variable subtracted by a value $\in \mathbb{R}$.

Appreciate your guidance.

答:
$E[X] = \int x f_X(x) \ dx$ is the expected value of $X$.

You have to imagine that $X$ is a function (it is indeed a function) so what $X-E[X]$ means is just a function minus a real value (which is a function).

For example, if $f(x) = x^2$ you understand what is $f(x) - 2$.

But what you have inside the variance, is taking the expectation of that "function":

\begin{equation*}
\begin{split}
Var[X] & =E[(X-E[X])^2]=E[X^2 - 2 X E[X] - E[X]^2] \\
& = E[X^2] - 2E[X]^2+E[X]^2 = E[X^2]-E[X]^2.
\end{split}
\end{equation*}
$$$$

If you want to understand everything, you first need to know what a random variable is.

It is a function $$X: \Omega \to E$$ (where usually $E = \mathbb{R}$) with some properties.

So for example, if you consider $E = \mathbb{R}$, when you write $P(X =2)$, this is an abuse of notation for
$$P(\{\omega \in \Omega : X(\omega) = 2\})$$
where $P$ is a probability measure and in that case the expectation would be written as
$$E[X] = \int_{\Omega} X(w) P(d \omega) = \int_{\Omega} X(w) dP$$

If you want to understand this terms better, I suggest for example that you see the definition of random variable in wikipedia.

Title: Variance definition notation question


问:
Proving that if $f^{-1}((a, \infty]) \in\mathcal{S} \; \forall a \in \mathbb{R}$, then $f$ is an $\mathcal{S}$-measurable function.
I am trying to prove:

Suppose $(X, S)$ is a measurable space and $f:X\mapsto[-\infty, \infty]$ is a function such that
\begin{align}
     f^{-1}((a, \infty]) \in\mathcal{S}
 \end{align}
for all $a\in\mathbb{R}$. Then $f$ is an $\mathcal{S}$-measurable function.

This is how I began this proof:
Let $\mathcal{T}=\{A\subset\mathbb{R}: f^{-1}(A)\in\mathcal{S}\}$. Then it suffices to show that every Borel subset of $\mathbb{R}$ is in $\mathcal{T}$. To do this, I showed that $\mathcal{T}$ is a $\sigma$-algebra on $\mathbb{R}$ (for the sake of this question, let's assume that I did this correctly). By hypothesis, $\mathcal{T}$ contains $\{(a, \infty]: a \in \mathbb{R}\}$.

This is where I got stuck. I don't know how to complete the proof after this step. I was thinking of writing $(a, \infty] = \cap_{k=1}^\infty\left(a, \infty-\frac{1}{k}\right)$, but I am not sure how to use this to complete the proof. How can I complete the proof?

答:
To finish the proof, we must show that every basic open set is an element of $\mathcal{T}$. This follows from the fact that $\mathcal{T}$ is a $\sigma$-algebra and contains all opens of the form $(a, \infty]$.

This is because we have all sets of the form $[-\infty, b]$ as elements of $\mathcal{T}$ by complementing the sets $[a, \infty)$. And $\bigcup\limits_{q \in \mathbb{Q}, q < a} [-\infty, q] = [-\infty, a)$.

To get the set $(a, b)$, just take $[-\infty, b) \cap (a, \infty]$.

Then we have the "basic open sets" of $[-\infty, \infty]$, which are intervals $[-\infty, a)$, $(a, b)$, and $(a, \infty]$. And any open set of $[-\infty, \infty]$ can be written as a countable union of these basic open sets. So we see that $\mathcal{T}$ contains all open sets. Hence, the Borel algebra is a subset of $\mathcal{T}$.


问:
Normalization of an affine toric variety is toric
In the book "Toric Varieties" by Cox-Little-Schenck, Proposition 1.3.8 is left as an exercise. It essentially characterizes what the normalization of an affine toric variety is. The construction is as follows:

Pick $V=\operatorname{Spec}(\mathbb{C}[\mathsf{S}])$ where $\mathsf{S}$ is an affine semigroup. Let $\mathscr{A} = \{m_1,\ldots,m_s\}\subseteq \mathsf{S}$ be such that $\mathbb{N}\mathscr{A} = \mathsf{S}$, and let $M = \mathbb{Z}\mathscr{A}$ be the lattice of characters of the torus of $V$. They pick the rational cone given by $\sigma := \operatorname{Cone}(\mathscr{A})\subseteq N_{\mathbb{R}}$, and the affine semigroup given by $\mathsf{S}_\sigma := \sigma^\vee \cap M$. They claim that the obvious inclusion $\mathbb{C}[\mathsf{S}]\hookrightarrow \mathbb{C}[\mathsf{S}_\sigma]$ induces a morphism at the level of varieties, $\operatorname{Spec}(\mathbb{C}[\mathsf{S}_\sigma]) \to \operatorname{Spec}(\mathbb{C}[\mathsf{S}])$ which is a normalization map.

What I tried to do so far is to prove using the plain definition of normalization map, that if $\alpha$ is in the fraction field $\operatorname{Frac}(\mathbb{C}[\mathsf{S}])$ and satisfies a monic polynomial equation with coefficients in $\mathbb{C}[\mathsf{S}]$ then it has to be an element of $\mathbb{C}[\mathsf{S}_{\sigma}]$.

I suspect, however, that if one could prove first that the normalization of an affine toric variety is again an affine toric variety (without necessarily being precise of who the lattice, torus and action are), then the characterization of normal toric varieties given by Theorem 1.3.5 could be used somehow. Any help would be highly appreciated.

答: Hint: use the fact that for a variety over a field, the normalization is the maximal finite birational morphism $W \to V$; it seems you have found a normal variety with a finite birational morphism $ V' \to V$, so...


问:
Show that the language wb^|w| (any string w over {a,b} followed by as many b’s as the size of w) is not regular.
Question : Show that the language wb^|w| (any string w over {a,b} followed by as many b’s as the size of w) is not regular.

My proof (as per my opinion) is that: We suppose that w is a regular language so (w^2)b^|w| would also be regular and we will use pumping lemma that this is not regular by contradiction.

So, our language would be:
If

W=1: ab

W=2: aabb

W=3: aaabbb (That's where I am confused at If I am doing if rightly)

then, X = XYZ where, |XY|<=n and |Y|>0,

Let's take X = a, Y=ab, Z=b, and then for XY^iZ, we will take i = 3.

Then our X = aabababb. Which wouldn't be available in our language. Hence proved by contradiction.

That's what I think The answer is but I am bit doubtful. Any comments would be helpful.

答:
Let $L=\left\{wb^{|w|}:w\in\{a,b\}^*\right\}$; you want to prove that $L$ is not regular. This can easily be done with the pumping lemma, but what you’ve done doesn’t make much sense.

First, there is no reason to bring in the language $\left\{w^2b^{|w|}:w\in\{a,b\}^*\right\}$, and it is not at all clear that this language would be regular if $L$ were. Next, it is not at all clear what you are doing with $W$: you’ve given us no explanation of the bulleted lines. Finally, if $X=XYZ$, then $Y=Z=\epsilon$, the empty word, which I’m sure is not what you intended.

To use the pumping lemma properly, suppose that $L$ is regular, let $p$ be the pumping length, and let $u=a^pb^p$; setting $w=a^p$, we can see that $u\in L$. The pumping lemma now tells us that $u$ can be decomposed as $u=xyz$, where $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for each integer $k\ge 0$.

Since $|xy|\le p$, $xy$ must be contained in the $a^p$ part of $u$. Thus, there are integers $r\ge 0$ and $s\ge 1$ such that $x=a^r$, $y=a^s$, and $r+s\le p$. Of course this means that $z=a^{p-r-s}b^p$. Then for each $k\ge 0$ we have

$$xy^kz=a^ra^{ks}a^{p-r-s}b^p=a^{p+(k-1)s}b^p\in L\,.$$

But $p+(k-1)s>p$ whenever $k>1$, and in particular, $xy^2z=a^{p+s}b^p$. If we can show that this word is not in $L$, we’ll have a contradiction proving that $L$ is not regular after all.

To show that $a^{p+s}b^p\notin L$, we have to show that there is no $w\in\{a,b\}^*$ such that $a^{p+s}b^p=wb^{|w|}$. Suppose that $a^{p+s}b^p=wb^{|w|}$; there are only $p$ $b$s in $a^{p+s}b^p$, so $|w|\le p$. But then

$$\left|wb^{|w|}\right|=2|w|\le 2p<2p+s=\left|a^{p+s}b^p\right|\,,$$

since $s\ge 1$, so in fact $wb^{|w|}$ cannot be equal to $a^{p+s}b^p$, and $a^{p+s}b^p$ cannot be in $L$. This is the contradiction that we wanted, showing that $L$ is not regular.


问:
Interesting natural rearrangements of conditionally convergent sums?
In my first course in analysis, we briefly covered the proof of the Riemann Rearrangement theorem as an exercise, from which we are assured that we could rearrange the alternating harmonic series to converge to, for example, $\pi$.

However, the rearrangement required to achieve $\pi$ in this way would be quite contrived - I'm assuming that there would not be any obvious pattern to this rearrangement. I'm curious to know whether or not there are many known "natural" rearrangements with more sensible patterns.

The linked Wikipedia article gives one quite nice natural rearrangement:

Suppose that two positive integers $a$ and $b$ are given, and that a rearrangement of the alternating harmonic series is formed by taking, in order, $a$ positive terms from the alternating harmonic series, followed by $b$ negative terms, and repeating this pattern at infinity [...]

It follows that the sum of this rearranged series is: $$\ln\bigg({2\sqrt\frac{a}{b}}\bigg)$$

Are there any other such natural rearrangements (for any conditionally convergent series) that give nice results?

答:
Let me get you started, since I judge that your comment means you don't know how to begin.  We rearrange the series $$1-\frac13+\frac15-\frac17+\cdots$$ by adding two positive terms and then one negative term, ad infinitum.  $$S=1+\frac15-\frac13+\frac19+\frac1{13}-\frac17$$
Separate the sum into groups of three:
$$S=\left(1+\frac15-\frac13\right)+\left(\frac19+\frac1{13}-\frac17\right)+\cdots+\left(\frac1{8n+1}+\frac1{8n+5}-\frac1{4n+3}\right)+\cdots$$

A moment's thought shows that if this series converges, so does the original series, to the same sum.  The partial sums are the subsequence $s_{3n}$ of the sequence $s_n$ of partial sums of the original series.  Since the terms go to $0$ if $s_{3n}\to S$ then also $s_{3n+1}\to S$ and $s_{3n+2}\to S$ and then $s_n\to S$.

We have $$\begin{align}s_{3n}
&=\sum_{k=0}^n\left(\frac1{8k+1}+\frac1{8k+5}\right)-\sum_{k=0}^n\frac1{4k+3}\\
&=\sum_{k=0}^{2n+1}\frac1{4k+1}-
\sum_{k=0}^n\frac1{4k+3}\\
&=\sum_{k=0}^n\left(\frac1{4k+1}-\frac1{4k+3}\right)
+\sum_{k=n+1}^{2n+1}\frac1{4k+1}
\end{align}$$

As $n\to\infty$, we know that the first sum $\to\frac\pi4$, so that we need to evaluate $$t:=\lim_{n\to\infty}\sum_{k=n+1}^{2n+1}\frac1{4k+1}$$  It's easy to see that the limit exists and that $\frac18\leq t\leq\frac14$.

This is as far as I've gone, and I'm going to turn it over to you now, with the suggestion that you might try to relate this to the harmonic numbers. (I'm going to guess $t=\frac\gamma4$, where $\gamma$ is Euler's constant, $\gamma\approx.5772$, but that may be wildly wrong.)

I think if you can do this case, you'll be able to generalize it to other rearrangements of the same sort.


问:
Definite integral $\int_0^{\pi} \sin^{n} x \ln(\sin x) dx$?
Is the integral
$$I(n) = \int_0^{\pi} \sin^{n}(x) \ln(\sin(x)) dx$$
analytically tractable for $n \in \mathbb{N}$? If not, are there good upper and lower bounds?

To clarify, although this has a fairly standard-looking form, I haven't been able to find an answer anywhere and I've tried a few of the usual tricks (although I'm a tad rusty).

I have tried integration by parts to reduce to a simpler or recursive form but this seems to make the expression more complex, and likewise with the substitution $u=\sin(x)$.

Equally, it is relatively easy to see that the expression under the integral is non-positive. Taking a derivative
$$ \frac{d}{dx} \sin^{n}(x) \ln(\sin(x)) = \sin^{n-1}(x) \cos(x) (1 + n \ln(\sin(x))) = 0$$
gives a has minimum at $-\frac{1}{en}$.

Together these give the reasonably simple bounds: $$-\frac{\pi}{en} < I(n) < 0.$$

But numerically these don't appear to be particularly tight, so I'm still hopeful of an exact solution or an improvement.

答:
Integrate by parts to obtain a recursive formula as follows

\begin{align}
I_n&=\int_0^{\pi} \sin^{n}x \ln(\sin x) dx\\
&= -\int_0^{\pi} \sin^{n-1}x \ln(\sin x)\> d(\cos x)\>\\
& =\int_0^{\pi}((n-1) \sin^{n-2}x \cos^2x\ln(\sin x)+ \sin^{n-2}x\cos^2x)dx\\
&= (n-1) (I_{n-2}-I_n)+ \frac1{n-1}\int_0^{\pi}\sin^{n}x\>dx
\end{align}
Thus
$$I_n = \frac{n-1}n I_{n-2} +\frac1{n(n-1)} \int_0^{\pi}\sin^{n}x\>dx
$$
with $I_0 = -\pi\ln2$ and $I_1= \ln2 -1$. (See here for evaluating $\int_0^{\pi/2}\sin^{n}x\>dx$.)


问:
Given $3$ lines, prove that $3$ points cannot be collinear
Given $3$ distinct lines, $A_1A_2$, $A_3A_4$, and $A_5A_6$, which are concurrent at point $P$, prove that the lines $A_2A_5$ and $A_1A_6$ cannot both pass through point $A_4$.

When I drew out this diagram, it was quite evident to me that this statement is true, however, I have no idea how to actually formally prove this.

Any help would be greatly appreciated!

NOTE: All $7$ of these points are distinct

Update

I'm not sure if I should make a new thread for this, so I'm just putting it here.

I thought of an interesting generalization to the problem, which I'm not too sure how to prove:

Given $n$ distinct lines, $$A_1A_2, A_3A_4, \cdots A_{2n-1}A_{2n}$$ which are all concurrent at point $P$, prove/disprove that the following $n-1$ lines can pass through $A_{n+1}$: $$A_1A_{2n}, A_2A_{2n-1}, A_3A_{2n-2} \cdots A_{n-1}A_{n+2}.$$

How would one go about solving this now?

答: I think it's possible. Let me know if I'm missing something.


问:
Defining product topology in terms of basis elements?
In the text that I have, we define the product topology $X \times Y$ on two topological spaces $(X, \mathcal{T}), (Y, \mathcal{U})$ as the topology generated by the basis $\{U \times V: U \in \mathcal{T}, V \in \mathcal{U}\}$.

Now, I am wondering:

Does anything go wrong if we define the product topology as the topology generated by the basis $\{B_1 \times B_2: B_1 \in \mathcal{B}_1, B_2 \in \mathcal{B}_2\}$, where $\mathcal{B}_1$ and $\mathcal{B}_2$ are bases for $X$ and $Y$, respectively?

答: No, because if $U\in\mathcal T$ and $V\in\mathcal U$, then $U$ can be written as an union of elements of $\mathcal B_1$ and $V$ can be written as an union of elements of $\mathcal B_2$. Therefore, $U\times V$ can be written as an union of elements of that set that you have mentioned.


答: Here's a counter-example to your "answer":

Title: Given $3$ lines, prove that $3$ points cannot be collinear

Body:
Given $3$ distinct lines, $A_1A_2$, $A_3A_4$, and $A_5A_6$, which are concurrent at point $P$, prove that the lines $A_2A_5$ and $A_1A_6$ cannot both pass through point $A_4$.

When I drew out this diagram, it was quite evident to me that this statement is true, however, I have no idea how to actually formally prove this.

Any help would be greatly appreciated!

NOTE: All $7$ of these points are distinct

Update

I'm not sure if I should make a new thread for this, so I'm just putting it here.

I thought of an interesting generalization to the problem, which I'm not too sure how to prove:

Given $n$ distinct lines, $$A_1A_2, A_3A_4, \cdots A_{2n-1}A_{2n}$$ which are all concurrent at point $P$, prove/disprove that the following $n-1$ lines can pass through $A_{n+1}$: $$A_1A_{2n}, A_2A_{2n-1}, A_3A_{2n-2} \cdots A_{n-1}A_{n+2}.$$

How would one go about solving this now?


问:
I need to find out how many zeros the function $f(z) = z^5 +2+e^z$ has in the left half plane by considering a D-shaped contour.
I need to consider the D-shaped contour from -iR to iR of radius R. I am thinking, that this contour covers the whole LHplane in the limit $R \rightarrow \infty$ so by considering the argument principle, if I can calculate
$$ \int_\gamma \frac{f^{'}(z)}{f(z)}dz$$ I should be able to get a value for the number of zeros, since f(z) has no poles.
But this is where I am struggling to get anywhere, so if I can get any pointers that would be great!

答: Hint: for $z$ in the left half plane, $|e^z| \le 1$. Use Rouché.


问:
Antipodal map - is there a different homotopy with Lie group structure?
The antipodal map $a: S^k \rightarrow S^k$ sends $x \in S^k \subset R^{k+1}$ to $-x$.

It is apparent that $a: S^1 \rightarrow S^1$ is homotopic to the identity map by

$$f_t = \left(\begin{matrix}
\cos \pi t & -\sin \pi t\\
\sin \pi t & \cos \pi t
\end{matrix}\right).$$

And this can be easily generalized to $2k+1$ dimensional.

However, for $a: S^3 \rightarrow S^3$, is there a different homotopy with Lie group structure?

答:
Yes, there is. First, recall that the group structure of $S^1$ comes from it being the set of unit complex numbers. The multiplication by $e^{it}$, $0\le t\le \pi$, furnishes the desired homotopy.

The group structure of $S^3$ comes from thinking it being the set of unit quaternions. The multiplication by $e^{it}$, $0\le t\le \pi$, furnishes the desired homotopy again. The field of complex numbers naturally sits inside of the skewed field of quaternions, since we can identify the complex $i$ with the quaternionic $i$ (or $j$, or $k$).


问:
$\int_{0}^\pi \frac{\sin(nx)}{\sin x} dx$
How do I integrate :$\int_{0}^\pi \frac{\sin(n\theta)}{\sin \theta} d\theta $

I did the following:
$\int_{0}^\pi \frac{\sin(n \theta)}{\sin \theta}d\theta = \mbox{Im} \int_{0}^{\pi} \frac{e^{i n \theta}}{\sin \theta} d\theta = \mbox{Im} \int_{0}^{\pi} 2i \frac{e^{i n \theta}}{e^{i \theta}-e^{- i \theta}} d\theta$

Put $ z=e^{i \theta}$ then

$ \mbox{Im} \int_{|z|=1} \frac{z^n}{(z^2-1)} dz$

Now I have problem , because the function $f(z)= \frac{z^n}{(z^2-1)}$ is not analytic on the boundary of the unit disk. Now I can not use the Cauchy Residue Theorem.

The answer is $0$ if $n$ is even and $ \pi$ if $n$ is odd.

I stuck:

Any suggestions ?

答:
$$I=\int_{0}^\pi \frac{\sin(n\theta)}{\sin \theta} d\theta =\frac12 \int_{-\pi}^\pi \frac{\sin(n\theta)}{\sin \theta} d\theta $$
$$= \frac12 \int_{-\pi}^\pi \frac{e^{in\theta}-e^{-in\theta}}{e^{i\theta}-e^{-i\theta}}  d\theta  $$
$$= \frac12 \int_{-\pi}^\pi \sum_{k=0}^{n-1} e^{i\theta (2k+1-n)}  d\theta  $$
$$= \frac12  \sum_{k=0}^{n-1} \int_{-\pi}^\pi e^{i\theta (2k+1-n)}  d\theta  $$

If $2k+1-n \ne 0$ then this integral is zero (because $e^{i\pi m}-e^{-i\pi m}=0$ for any integer $m$).
Therefore we get
$$I=0 \;\;\text{if n is even}$$
$$I=\pi \;\;\text{if n is odd}$$


问:
Applying mean value theorem to function of two variables
If I am given a function $f$ of two variables, where partial derivatives exist on all of $\mathbb{R}^2$. Can I apply the mean value theorem on an interval $[x_1,x_2]$ to $f(x,y)$ by fixing a $y_0$ and then apply the mean value theorem for functions of one variable to choose a $c \in (x_1,x_2)$ with $f(x_2,y_0)-f(x_1,y_0)=f'(c)(x_2-x_1)$?

答:
In general for a differentiable function from $\mathbb{R}^2$ to $\mathbb{R}$, there exists a point $(c,d)$ on the line segment from $(x_1,y_1)$ to $(x_2,y_2)$ such that

$$f(x_2,y_2)-f(x_1,y_1)=\frac{\partial f}{\partial x}(c,d)(x_2-x_1)+ \frac{\partial f}{\partial y}(c,d)(y_2-y_1)$$

Taking $y_2 = y_1 = y_0$ the above expression simplifies to

$$f(x_2,y_0)-f(x_1,y_0)=\frac{\partial f}{\partial x}(c,y_0)(x_2-x_1)$$


答: Yes, for a fixed $y_0$, $f(x,y_0)$ can now be thought of as a function of just $x$. However, your $f′(c)$ doesn't quite make sense, since $f$ is a function of two variables. The line would be $$f(x_2,y_0)−f(x_1,y_0)=f_x(c,y_0)(x_2−x_1).$$

Title: Applying mean value theorem to function of two variables

Body: If I am given a function $f$ of two variables, where partial derivatives exist on all of $\mathbb{R}^2$. Can I apply the mean value theorem on an interval $[x_1,x_2]$ to $f(x,y)$ by fixing a $y_0$ and then apply the mean value theorem for functions of one variable to choose a $c \in (x_1,x_2)$ with $f(x_2,y_0)-f(x_1,y_0)=f'(c)(x_2-x_1)$?


答:
Yes, but note that we are in 2 dimensions, so:

$$
f(x_2,y_0) - f(x_1,y_0) = \left. \frac{\partial f(x,y)}{\partial x} \right|_{(x,y)=(c,y_0)} (x_2-x_1)
$$


问:
Is there a method to solve the following diophantine equation: $2x^2+2y^2+2z^2 +u^2 +v^2 = N$?
Just to give a simple numerical example but in general the variables $x,y,z,u,v$ are not equal.

$113= 2*4^2 + 2*4^2 +2*4^2 + 4^2 +1^2$

I am looking for a general method to solve this type of equation or a piece of software to do the same. I already looked in this site for methods that could help but could not find anything dealing with this ind of case.

Question 2 It is also useful to know if there is a test that can tell if the equation does not have a solution.

答:
Here is some brutal code (understand not optimized at all) to get all solutions as a verification.

Since $x,y,z$ and $u,v$ are symmetric, I did not print the shuffles.

There is always at least one solution if we allow some of the variables to be zero:

for $n$ odd set $x=y=0$ then $n=2z^2+u^2+v^2$ has a solution, see this  representing all odd naturals as the sum of four squares, two of them equal

for $n=2p$ even set $u=v$ and we get $x^2+y^2+z^2+u^2=p$ which has a solution by Lagrange four square theorem

It seems to me with $5$ unknowns there is plenty of room to have solutions, so it might be difficult to express them all unless using brute force.


问:
What is another, perhaps quicker and nicer way of solving this question?
In a calculus book, a question reads:

A car is traveling at night along a highway shaped like a parabola with its vertex at the origin. The car starts at a point 100 m west and 100 m north of the origin and travels in an easterly direction. There is a statue located 100 m east and 50 m north of the origin. At what point on the highway will the car's headlights illuminate the statue?

I want to know how you would solve this problem, because the method I am using is very round-about. There must be a much more logical way of solving this problem, and I hope you will share it with me. Thank you.

This is my solution..

The function appears to be of the form $$y = f(x) = ax^2$$

$$100 = a(100)^2$$
$$a = 1/100$$
$$y = \frac{x^2}{100}$$

Then I decided (by solving a few tangent equations and generalizing) that the tangent line expression at a point $(p, f(x))$ is given by the function:

$$T(p) = T_m(p)*x + T_b(p)$$

Where $T_m(p)$ represents the slope for a point, $p$ on $f(x)$ and $T_b(p)$ represents the y-intercept value.

$$T_m(p) = \frac{df(p)}{dx} =\frac{p}{50}$$

$$T_b(p) = f(p) - \frac{df(p)}{dx} * p = -\frac{p^2}{100}$$

So given that the statue rests on $(100, 50)$ we know the equation of this tangent line outputs $50$ when $100$ is input.

$$T(p) = \frac{p}{50}*x - \frac{p^2}{100}$$

$$50 = \frac{p}{50}*100 - \frac{p^2}{100}$$

the $x$ position of the car is at a point $p$ such that $p$ satisfies the equation.

I end up getting the positive result $x = 100 - 50\sqrt{2}$

答:
A coworker came to me with this problem today, and I found the solution a slightly different way. My approach wasn’t any simpler, but I still thought I would share:

As both of your approaches demonstrated, $f(x)=\frac{x^2}{100}$ and $f'(x)=\frac{x}{50}$. We also know that the tangent line intersects $(100,50)$. So the tangent line can be defined as:
$$y=mx+b$$
$$50=\frac{x}{50}100+b$$
$$b=50-2x$$
$$y=\frac{x}{50}x+50-2x$$
$$y=\frac{x^2}{50}-2x+50$$
Solve the system: $y=\frac{x^2}{100}$ and $y=\frac{x^2}{50}-2x+50$.
$$\frac{x^2}{100}=\frac{x^2}{50}-2x+50$$
$$\frac{x^2}{100}-2x+50=0$$
$$x^2-200x+5000=0$$
$$x=100±50\sqrt{2}$$
I ignored that value where $x>100$, so the car's location at which its headlight's illuminate the statue is:$$(100-50\sqrt{2}, 150-100\sqrt{2})$$


问:
How do I find the interior, closure and boundary of an arbitrary set?
I'd like to find the interior, boundary, closure, etc. of an arbitrary set. I mean, if I'm working on a subset of $\mathbb{R}, \mathbb{R}^2,$ or $\mathbb{R}^3$ it could be quite easy using Geogebra, but in other cases or sets that are complex to graph, it turns out to be very difficult.

So I'm asking about an algebraic method (most specifically, an example to understand) to find those sets.

Like $A=\{(x,y)\in \mathbb{R}^2:0<(x-2)^2+y^2<1\}\cup\left\{\left(\dfrac{1}{n}+2,0\right):n\in\mathbb{N}\right\}$

答:
You have to develop intuition about what it means to be a limit point, on the boundary, etc. Sometimes looking at a graph is helpful. Here's a sketch of the set. Note that all but one of the sequence of points converging to $(2, 0)$ are already in the open punctured disk, so the set could be described more simply as:
$$
A = \{ (x,y) \in \mathbb{R}^2: 0 <(x-2)^2 + y^2 < 1 \} \cup \{ (3, 0) \}.
$$
Here's a sketch, where I've indicated the sequence of points so you can see them:

[![A set of points in the plane][1]][1]

To answer your questions, the algebraic inequalities used in defining the sets allow you to rigorously confirm the following conclusions:

The interior of the set consists of all points who have an open neighborhood (e.g. a disk of arbitrarily small size) around them in $A$. This is just the punctured disk:
$$
A^\circ = \{ (x,y) \in \mathbb{R}^2: 0 <(x-2)^2 + y^2 < 1 \}
$$

The closure includes the set and all of its limit points. The limit points are not necessarily in the set, but any neighborhood around a limit point intersects the set. You get the center of the disk and the circle of points at the boundary of the disk:
$$
\overline{A\,} = \{ (x,y) \in \mathbb{R}^2: (x-2)^2 + y^2 \leq 1 \}
$$

The boundary consists of all points who have a neighborhood that intersects both the set and its complement, or equivalently the closure with the interior removed. This consists of the center of the disk and the circle:
$$
\partial A = \{ (2,0) \} \cup \{ (x,y) \in \mathbb{R}^2: (x-2)^2 + y^2 = 1 \}
$$
[1]: https://i.stack.imgur.com/vLWXH.png


问:
who were the amateur mathematicians had won the field medal?
Who were the   amateur mathematicians  had  won the  field medal?

I search the field medal  winners. I  didn't find any single amateur mathematician who won the field medal. All are from academics background (mostly  from Princeton University.)

I think no amateur mathematicians had won the field medal.

答: One who is not a professional mathematician is Edward Witten, who is a physicist.</p>


答:
$$ u^2 + v^2 +2 x^2+ 2 y^2 $$ is positive universal. Thus, take
$$ z = \left\lfloor \sqrt \frac{N}{2} \right\rfloor   $$

after which $N- 2 z^2$ is fairly small, search any way you like in finishing
$$   N - 2 z^2 =  u^2 + v^2 +2 x^2+ 2 y^2  $$

Title: Is there a method to solve the following diophantine equation: $2x^2+2y^2+2z^2 +u^2 +v^2 = N$?

Body:
Just to give a simple numerical example but in general the variables $x,y,z,u,v$ are not equal.

$113= 2*4^2 + 2*4^2 +2*4^2 + 4^2 +1^2$

I am looking for a general method to solve this type of equation or a piece of software to do the same. I already looked in this site for methods that could help but could not find anything dealing with this ind of case.

Question 2 It is also useful to know if there is a test that can tell if the equation does not have a solution.


问:
Is there a name for $(x, f(x))$?
Given a function $f$, and an $x$ from its domain, is there a name for the pair $(x, f(x))$?

Is there a defined terminology for one (any) such pair? I think the set of all $(x, f(x))$ is called the graph of the function, but I am asking is there's a name for one (or any) point of the graph.

答: I call this the graph of a function and I use $\Gamma = (x,f(x))$ as the notation.

Title: Is there a name for $(x, f(x))$?


问:
What is the result of (λx.λy.x + ( λx.x+1) (x+y)) ( λz.z-4 5) 10?
Could you explain what should I do about

λx.λy.x

part?
Thanks.

答:
The parentheses are pretty bad, here is my guess on what the expression should be in order to evaluate to a number in the end:

$(\lambda x.\lambda y.(x + ( \lambda x.x+1) (x+y)))~( (\lambda z.(z-4))~5)~10$

Then you just have to reduce everything.
Here is a step-by-step reduction:

$$\begin{array}{l}
(\lambda x.\lambda y.(x + ( \lambda x.x+1) (x+y)))~( (\lambda z.(z-4))~5)~10\\
\to (\lambda x.\lambda y.(x + (x+y+1)))~(5-4)~10\\
\to (\lambda x.\lambda y.(2x+y+1))~1~10\\
\to (\lambda y.(2*1+y+1))~10\\
\to (2*1+10+1)\\
\to 13
\end{array}$$


问:
Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$
So i was given this question. Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$

Here is my solution:

Given $2n$ objects, split them into $2$ groups of $n$, $A$ and $B$. $2$-combinations can either be assembled both from $A$, both from $B$ or one from each. There are ${n \choose 2}$ from $A$ and ${n \choose 2}$ from $B$. For the mixed pair, each choice from $A$ can be coupled with $n$ choices from $B$ so the total is $n^2$.

Therefore ${2n\choose 2} = 2{n \choose 2} + n^2$

Is this correct. Is there any other combinatoric proof to solve this?

答:
Theorem: C(2n, 2) = 2C(n, 2) + n^2

Proof.

Suppose we have 2 groups, both of size n, which combine to make a class of size 2n, and we wish to know the number of ways we can choose 2 members from the class.  This is represented by C(2n, 2).

There are C(n, 2) ways we can choose 2 students from a single group. Since there are 2 groups, there are 2C(n, 2) ways we can choose 2 students from a single group.  Finally, there are n^2 ways to choose to members from the class where each member is from a different group, so there are 2C(n, 2) + n^2 ways to choose 2 members from the class of size 2n.

Therefore C(2n, 2) = 2C(n, 2) + n^2∎


问:
Does the number of permutations in $S_n$ with major index equal to $k$, satisfy a degree $k$ polynomial?
I'm interested in fixing $k$ and finding a formula, $M_k(n)$, for the number of permutations $\pi \in S_n$ such that $\operatorname{maj}(\pi) = k$, where $\mathrm{maj} \colon S_n \rightarrow \mathbb N_{\geq0}$ denotes the major index of the permutation. (It does not change the question if you replace the major index with the inversion number or any other Mahonian statistic.)

Table

Here's the table of values, given by OEIS sequence A008302, where $T(n,k) = M_k(n)$:

```<code>n\k| 0    1    2     3     4      5      6      7       8
---+------------------------------------------------------
 1 | 1
 2 | 1,   1
 3 | 1,   2,   2,    1
 4 | 1,   3,   5,    6,    5,     3,     1
 5 | 1,   4,   9,   15,   20,    22,    20,    15,      9, ...
 6 | 1,   5,  14,   29,   49,    71,    90,   101,    101, ...
 7 | 1,   6,  20,   49,   98,   169,   259,   359,    455, ...
 8 | 1,   7,  27,   76,  174,   343,   602,   961,   1415, ...
 9 | 1,   8,  35,  111,  285,   628,  1230,  2191,   3606, ...
10 | 1,   9,  44,  155,  440,  1068,  2298,  4489,   8095, ...
11 | 1,  10,  54,  209,  649,  1717,  4015,  8504,  16599, ...
</code>```

The table satisfies the following recurrence $$\begin{align}
T(1, 0) &= 1, \\
T(n, k) &= 0 \text { for } n < 0, k < 0 \text{ or } k > \frac{n(n-1)}{2}, \\
T(n, k) &= \sum_{j=0}^{n-1} T(n-1, k-j), \text{ or alternatively}, \\
T(n, k) &= T(n, k-1) + T(n-1, k) - T(n-1, k-n).
\end{align}$$

Conjecture

It appears that for $n \geq k$, the $k$-th column satisfies:

$M_0(n) = 1$.

$M_1(n) = -1 + n$.

$\displaystyle M_2(n) = \frac{-2 - n + n^2}{2!}$.

$\displaystyle M_3(n) = \frac{-7 n + n^3}{3!}$.

$\displaystyle M_4(n) = \frac{-14 n - 13 n^2 + 2 n^3 + n^4}{4!}$.

$\displaystyle M_5(n) = \frac{120 - 46 n - 65 n^2 - 15 n^3 + 5 n^4 + n^5}{5!}$.

$\displaystyle M_6(n) = \frac{516 n - 356 n^2 - 165 n^3 - 5 n^4 + 9 n^5 + n^6}{6!}$.

I'm curious to know more about these in approximately the following order:

Is it true that the $k$-th column, $M_k(n)$, always satisfies a degree $k$ polynomial for $n \geq k$?

Is the constant part of $M_k(n)$ nonzero if and only if $n$ is a generalized pentagonal number?

If so, is the coefficient of $x^k$ is always $\displaystyle \frac{1}{k!}$?

Is the coefficient of $x^{k-1}$ always $\displaystyle \frac{-3n + n^2}{2k!}$?

Is the coefficient of $x^{k-2}$ always $\displaystyle \frac{-2n + 21n^2 - 22n^3 + 3n^4}{24k!}$?

Is the coefficient of $x^{k-m}$ always a polynomial of degree $2m$?

Is the coefficient of $x^{k-m}$ given by a polynomial whose numerator is divisible by $\displaystyle \prod_{i=0}^{m-1} (x-i)$?

Do these polynomials $M_k(n)$ have a combinatorial explanation in terms of (finite) exponential generating functions?

答:
The generating function for the inversion numbers is well known:
$$
T_{n,k}=[q^k]\frac{(1-q)(1-q^2)\cdots(1-q^{n})}{(1-q)^n}
$$
We can safely remove the factors of $(1-q^{k+1})(1-q^{k+2})\cdots(1-q^n)$ from the numerator without changing the $q^k$ coefficient, so
$$
T_{n,k}=[q^k](1-q)(1-q^2)\cdots(1-q^k)(1-q)^{-n}
$$
By the pentagonal number theorem,
$$(1-q)(1-q^2)(1-q^3)\cdots=\sum_{j\in \mathbb Z}(-1)^jq^{j(3j-1)/2}.\tag{*}$$
For this purpose, we only need the power of $q$ up to $q^k$ in $(1-q)(1-q^2)\cdots(1-q^k)$, which indeed agree with the terms of $q^k$ up to $k$ in $(*)$. Finally, $[q^i](1-q)^{-n}= \binom{n+i-1}{i}$. Therefore,
$$
\begin{align}
T_{n,k}
&=[q^k]\left(\sum_{j\in \mathbb Z}(-1)^jq^{j(3j-1)/2}\right)\left(\sum_i q^i\binom{n+i-1}{i}\right)
\\&=\sum_{j\in \mathbb Z} (-1)^j\binom{n+k-j(3j-1)/2-1}{k-j(3j-1)/2}
\end{align}
$$
Even though this is an infinite sum, there are only finitely many $j$ for which the lower index $k-j(3j-1)/2$ is nonnegative, and only those contribute positively. This explains your observation about the pentagonal numbers.

Each binomial coefficient can be expanded into a polynomial in $n$, but I do not know how to find the coefficients of that nicely. I think the above expansion is quite pretty, since this shows that $T_{n,k}$ is a linear combination of $O(\sqrt{k})$ binomial coefficients, instead of $O(k)$ powers of $n$.


问:
Calculate cosh 1 correct to 6 decimal places.
Can anyone help me with this problem?

Calculate $\cosh 1$ correct to 6 decimal places.

答:
If you are aware of the Taylor series, try using the taylor series for $\cosh$ which is:

$$\sum^\infty_{n=0} \frac{x^{2n}}{(2n)!}$$

Then you can expand the summation and you will have something like the following:

$$\sum^\infty_{n=0} \frac{x^{2n}}{(2n)!} = 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!} \ldots$$

Then you can simply sub in "$1$" for $x$ and you will have your approximation.


问:
Almost-isometric embedding is almost surjective
I'm looking for a statement, if it exists, of the following sort:

Let $X$ be a compact metric space. Let $f:X\to X$ be a function, and suppose that for some (fixed) $\epsilon>0$, for all $x,y\in X$,
$$ | d(f(x) , f(y) ) - d(x,y) | < \epsilon . $$
Then $f$ is $\epsilon$-surjective, meaning that for every $y\in X$,
$$ \inf_{x\in X} d(y, f(x)) < \epsilon .$$

Can anyone help me, and if possible, provide a reference? Possibly replacing $\epsilon$-surjectivity with $2\epsilon$, or any other similar function of $\epsilon$.

答:
What you want to prove is false.

Let $X = \{x_1,\dots,x_n\}$ with $d(x_2,x_1) = \epsilon$, $d(x_3,x_1),d(x_3,x_2) = 2\epsilon$, $d(x_4,x_1),d(x_4,x_2),d(x_4,x_3) = 3\epsilon$, etc.. Define $f: X \to X$ by $f(x_1) = x_1$ and $f(x_i) = x_{i-1}$ for $2 \le i \le n$. Then $f$ is $\epsilon$-close to being an isometry, but is $(n-1)\epsilon$ away from being surjective.


答:
I would call this an element of the graph of $f$, or an element of $\Delta(f)$ where $\Delta$ sends symbols to their denotations, or an element of $f$ as a set.

This amounts to the same thing as calling it an element of $f$, as suggested in the comments, except that it makes explicit the act of sending a function to its corresponding set. This operation is a no-op if you think of a function as just a set of pairs. It is not a no-op if you think of it as an ordered triple with its domain, codomain, and graph as elements, or if you think of a function as some kind of non-set entity.

Title: Is there a name for $(x, f(x))$?

Body:
Given a function $f$, and an $x$ from its domain, is there a name for the pair $(x, f(x))$?

Is there a defined terminology for one (any) such pair? I think the set of all $(x, f(x))$ is called the graph of the function, but I am asking is there's a name for one (or any) point of the graph.


问:
Existence of partials for all vectors $p$ at origin implies continuity
If the partial derivative $\frac{\partial f}{\partial p}(0)=0$ for all vectors $p$ is $f$ necessarily continuous at the origin. I would say no, but it is difficult to come up with an example. Any suggestions?

答: No, indeed. Take, for instance,$$f(x,y)=\begin{cases}0&\text{ if }xy=0\\1&\text{ otherwise.}\end{cases}$$I suppose that you can check that $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$ and that $f$ is discontinuous at $(0,0)$.


问:
Finite Field Question
I'm having trouble with the Concept of finite fields. Say $\mathbb{F}_2 = \{0,1\}$. How many linear maps from $\mathbb{F}_2^5 \to \mathbb{F}_2^5$ (mapped from five tuples to five tuples) are injective? I know that the null has to be zero, but I'm thinking there might be an infinite number of maps.

I also had another follow up question. So now suppose $U = \{(x,x,y,y,y)\}$ where $x,y\in \mathbb{F}_2$. Using the same mapping, how many linear maps have their null space contain $U$ and equal $U$?

答:
Each linear map $A: \mathbb{F}_2^5 \to \mathbb{F}_2^5$ corresponds to a unique $5\times 5$ matrix with entries in $\mathbb{F}_2$. There are $2^{25}=33554432$ such matrices. This immediately rules out the possibility that there are an infinite number of maps $A: \mathbb{F}_2^5 \to \mathbb{F}_2^5$ (injective or otherwise).

By the rank-nullity theorem, each injective linear map $A: \mathbb{F}_2^5 \to \mathbb{F}_2^5$ corresponds to a unique matrix in the general linear group $GL(5, \mathbb{F}_2)$. The
order of the group is

$$
\begin{align}
\prod_{k=0}^{n-1}(2^n - 2^k) &= (2^5 - 1)(2^5 - 2)(2^5 - 4)(2^5 - 8)(2^5 - 16) \\ &=31 \cdot 30 \cdot 28 \cdot 24 \cdot 16 = 9999360
\end{align}\tag1
$$

which means that $\frac{9999360}{2^{25}} \approx 29.8\%$ of the matrices in $\mathbb{F}_2^{5\times 5}$ correspond to injective maps.

The formula $(1)$ captures the fact that the first column of an invertible matrix in $\mathbb{F}_2^{5\times 5}$ can be anything but the zero vector, the second can be anything but the zero vector and the first column, the third anything but any of the four linear combinations of the first two columns and so on.

Title: Finite Field Question


问:
Finding all rational values of $x$ for which $256(x + 15\{x\})\{x\} = 2021$
Find all rational numbers $x$ that satisfy $$256(x + 15\{x\})\{x\} = 2021$$ where $\{x\}$ is the fractional part of $x.$

I first substituted in $x - \lfloor x \rfloor = \{x\},$ which gave me $$256(16x - 15\lfloor x \rfloor)(x - \lfloor x \rfloor) = 2021$$ after simplifying. However, from here, I got stuck.

答:
Hint: Let $\{x\} = a/b$ in lowest terms, $x = y + a/b$ where $y$ is an integer.

Then $x + 15 \{x\} = y + 16 a/b$ and the equation becomes
$$ 256 \left( b y + 16 a\right) \frac{a}{b^2} = 2021 $$
Conclude that $b$ must be a power of $2$...


问:
Mathcounts National 2014 Sprint #15

This problem is from the 2014 Mathcounts National Sprint test. The correct answer is 13/1024, but I keep getting 17/2048s.

My attempt at this problem was to use casework. Obviously, the denominator of this fraction will come out to 161616 = 2^12 = 4096. For the numerator, I came up with 4 cases for how the sum of the squares could be over 15.

7 7 7  :There is one order that this can be thrown in and all three of them must be from the same 1 square so this combination gives us 1.

7 7 5 : There is 3 ways this can be ordered * 3 squares the five can be picked from to give 9.

7 7 3: 3 orders possible * 5 squares for the three so 15.

7 5 5: 3 orders possible * (3 squares choose 2 fives) = 3 * 3 = 9.

In total, from these cases, we have 9 + 9 + 15 + 1 = 34.

34/2^12 = 17/2^11 = 17/2048.

Am I missing cases? Or did I calculate the number of ways incorrectly. Please let me know of any edits I should make to the question before you downvote it.

EDIT: I just realized that the way I calculated case 4 was incorrect, as I assumed that the five squares needed to be 2 different squares. In reality, there are 331 total ways to choose squares times the 3 orders = 27. This makes the final result 52/2^12  = 13/1024.

答: The mistake I had made was that for case #4 "7 5 5", I used 3 choose 2 to determine how many ways the squares could be chosen for the fives. However, I did not take into account that the squares could be the same. This would mean that rather than 3 choose 2, it would be 3 * 3, as there are 3 squares for the first 5 and 3 squares for the second 5. 9 * the 3 ways to order the case give 27, not 9.</p>

Title: Mathcounts National 2014 Sprint #15


问:
On surjectivity of exponential map for Lie groups
A recent question made me realize I didn't know any proof that exponential of a Lie algebra $\mathfrak g$ of a compact connected Lie group $G$ is surjective. After a bit of thinking I've come up with two proofs. First is based on relation between exponential curves and geodesics. This is rather technical but also gives us other useful information. This is not what I want to talk about here though.

The second proof (which I find slicker) is based only on topology and goes like this:
Since $\exp$ is a local homeomorphism, it is both open and closed. Therefore $\exp(\mathfrak g)$ is clopen and so equal to $G$.

The trouble with this "proof" is that it also proves the statement for $G$ non-compact (which is false). So I wonder (and this is my question) what precisely went wrong.

Can the above mentioned "proof" be made into a real proof?

My thoughts on this are that $\exp$ is closed and open only when $G$ is compact because then we can pick a bounded open subset $C \subset \mathfrak g$ such that $\exp(C) = G$  and we can use the relation $\exp(A+\epsilon B) \approx \exp(A)\exp(\epsilon B)$ to conclude that $\exp$ is a local homeomorphism everywhere in $C$ (not just around $0$). This implies that $\exp$ is open (since it is locally open) in $C$. Also, since any closed subset of $C$ is compact, it's image is also compact and therefore closed in $G$.

Where exactly does this argument break down when $G$ is not compact.

答:
I think a local diffeomorphism need not be a closed map?

For example, fix some irrational $a \in \mathbb{R}\setminus \mathbb{Q}$, and consider an irrational winding of the plane around the torus:
$$
f : \mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2 : (x,y) \mapsto (ay+x,y)+\mathbb{Z}^2 .
$$

Then $f$ is a local diffeomorphism, but $f$ maps the closed subset $\{0\} \times \mathbb{R} \subset \mathbb{R}^2$ to the non-closed $\{(at,t)+\mathbb{Z}^2 : t\in\mathbb{R}\} \subset \mathbb{R}^2/\mathbb{Z}^2$.


问:
Set of linearly independent $n$-tuples in infinite dimensional Banach space is open
is the following result true?

Let $X$ be an infinite dimensional Banach space, and
$$U=\{(x_1, \cdots, x_n) \in X^n: \{x_1, \cdots, x_n\} \text{ is linearly independent}\}$$
then $U$ is open in $X^n$ with the product topology.

I think it is true, at least if $X$ is separable because I've encountered this result being used in a paper I'm reading but I cannot prove it.

Any ideas?

答:
I also think that this is true, and I think it amounts to intersecting finitely many (at most $2^n$) open neighbourhoods of any point that remain in $U$.

Saying $x_k$ is linearly independent of $\{x_1,...,x_n\}\setminus \{x_k \}$, is to say that

$$ x_k\neq \sum_{j\neq k}a_j \cdot x_j \quad \text{for all} \quad \{ a_j \}_{j\neq k }\subseteq \mathbb{F} . $$

Then the finite linear space spanned by $\{ x_j \}_{j\neq k}$ can be separated from $x_k$, and we can find neighbourhoods $V'_n, V_f''\subseteq X$ for all $j\neq k$ such that
if $y_k\in V_k'$ and $y_j\in V_j''$, then

$$ y_k \neq \sum_{j\neq k}a_j \cdot y_j \quad \text{for all} \quad \{ a_j \}_{j\neq k}\subseteq \mathbb{F} .  $$

This condition corresponds to the neighbourhood

$$ W_k:= \prod_{j\neq k}V_j'' \times Vk'.  $$

We can construct such a neighbourhood $W_k$ for all $1\leq k \leq n$, and define

$$W:= \bigcap_{k=1}^n W_k. $$

If

$$ x_k\neq \sum_{j\neq k}a_j \cdot x_j \quad \text{for all} \quad \{ a_j \}_{j\neq k }\subseteq \mathbb{F}  $$

for all $1\leq k\leq n$, then $\{x_1,...,x_n\}$ is linearly independent. Hence, every element in $(y_1,...,y_n)\in W$ is linearly independent, and we found a neighbourhood $W\subseteq U$ of $(x_1,...,x_n)$.


问:
Kan Extensions and Coends
I am lost in a bit of a haze of abstraction and am wondering if someone can set me straight.

I'm reading the chapter in Categories for the Working Mathematician, and Mac Lane points out on page 238 that the argument for vanilla categories (that the Kan extension can be given by a coend formula) may fail for Abelian categories and additive functors unless one uses the proper notion of "cotensor."  So I am trying to make sure I understand at what point the usual argument fails. This is my attempt to work through the proof:

Let $T: M\to A$, let $K: M\to C$ be additive functors between Abelian categories.  We want to define
$R = \int_m T(m)^{Hom_C(c, K(m))} : C\to A$ and demonstrate a correspondence, for functors $S: C\to A$,
$Nat(S,R)\cong Nat(SK,T)$ natural in $S$.

Now

$Nat(S, R)\cong \int_c Hom_A(S(c), R(c))$

$\cong\int_c Hom_A(S(c), \int_m T(m)^{Hom_C(c, K(m))})$

$\cong\int_c\int_m Hom_A(S(c),T(m)^{Hom_C(c, K(m))})$

And here is where the divergence arises, as far as I understand it, in the proofs. If we interpret $T(m)^{Hom_C(c,K(m))}$ as being given by $\prod_{f\in Hom_C(c,K(m))}T(m)$, then there is a natural correpondence between morphisms $S(c)\to \prod T(m)$ and $|Hom_C(c,K(m))$|-indexed tuples of morphisms $S(c)\to T(m)$; that is, the next line should be

$\cong\int_c\int_m Hom_{\mathbf{Sets}}(Hom_C(c, K(m)),Hom_A(S(c),T(m)))$
and indeed this is the next line in Mac Lane.

But working in an Abelian category, i.e. take $R$-mod, it perhaps makes more sense to define $T(m)^{Hom_C(c,K(m))}$ as the Abelian group $Hom_{R-mod}(Hom_C(c,K(m)),T(m))$ as one can then, in the next line, get the  isomorphism

$\cong\int_c\int_m Hom_{\mathbf{Ab}}(Hom_C(c, K(m)),Hom_A(S(c),T(m)))$ by simple manipulations with tensor products, etc.

Now I am trying to understand what the difference is between these when we take the end with respect to $c$. As I understand it, the end of the second version should be $\int_c Hom_{\mathbf{Ab}}(Hom_C(c,K(m)),Hom_A(S(c),T(m)))$ should be $Nat(Hom_C(-,K(m)),Hom_A(S(-),Tm))$ as taken in $Ab$; whereas the end of the first one should be more like $\int_c Hom_{\mathbf{Sets}}(...)=Nat(U(Hom_C(-,K(m))),U(Hom_A(S(-),Tm))$ where $U$ is the forgetful functor $Ab\to Sets$.

But aren't these sets the same? In both cases the Yoneda lemma - abelian cat version or normal version - allows us to pull back the natural transformations to morphisms between the original objects, i.e. a map in $A$ from $SK(m)\to T(m)$. So why does it matter which version of the "cotensor" we use?

答:
It's not true that the sets are the same for an arbitrary functor $S$. This is what Mac Lane is trying to point out when he stresses that the universal property of the constructed $R$ holds for additive functors $S$.

It's obviously true that the additive natural transformations $\operatorname{Hom}_C(-,Km) \to \operatorname{Hom}_C(S(-),Tm)$ are a subset of all natural transformations. Let $\sigma : S(K(M))\to Tm$ be a morphism which by Yoneda represents a natural transformation; the natural transformation it is associated to gives for each $c$ a map $\operatorname{Hom}_C(c,Km) \to \operatorname{Hom}_C(Sc,Tm)$ sending $g : c\to Km$ to $\sigma\circ S(g)$.
This is not additive in $g$ unless $S$ is itself an additive functor, and the proof gets stuck. So we shouldn't expect the cotensor formula for the right Kan extension to give the right Kan extension where all functors are regarded as enriched over Sets.

Title: Kan Extensions and Coends


问:
Estimate error using Taylor Series
I was told to use Taylor Series to find the error of this but I am not confident in doing this.

Estimate the error in approximating $\cos x$ by $1-(x^2/2)$ on the interval $[-0.5, 0.5]$.

答:
The taylor development of cos x around 0 is just
$$
1-x^2/2+x^4/24-x^6/720+\dots
$$

When you subtract your given term, you get get an error of magnitude
$$
O(x^4)
$$

Title: Estimate error using Taylor Series


问:
How many equivalence classes are there in $J$.?
Let $G$ be the symmetric group $S_5$ of permutations of five symbols. Consider the set $J$ of subgroups of $G$
that are isomorphic to the non-cyclic group of order $4$. Let us call two subgroups $H$ and $K$ belonging to $J$
as equivalent if they are conjugate (that is,  there exists $g $ such that $gHg^{-1}=K$). How many equivalence
classes are there in $J$.?

I have tried a lot but i found only one. But correct answer is two. But how ?

答:
EDIT: Originally, this answer made the mistaken assumption that the involutions that generate $V$ were transpositions in $S_5$.

The Klein four-group is generated by a pair of commuting <s>transpositions</s> involutions $s$ and $t$:
$$
V = \langle s, t \rangle = \{ 1, s, t, st \}.
$$

There are two cases to consider.

Class 1: pair of commuting transpositions and a double transposition.
An embedding of $V \hookrightarrow S_5$ is determined by selecting two non-overlapping pairs $\{a,b\}$, $\{c,d\}$. The subgroup consists of
$$
1, \quad (a\,b), \quad (c\,d), \quad (a\,b)(c\,d).
$$
But this double-counts the possible embeddings since the order you choose generating transpositions doesn't matter. Thus, the total number is
$$
\frac{1}{2}\binom{5}{2} \binom{3}{2} = \frac{1}{2} \cdot 10 \cdot 3 = 15.
$$

Class 2: three double transpositions.
An embedding of $V \hookrightarrow S_5$ is determined by selecting a set of four elements $\{a,b,c,d\}$. The subgroup consists of
$$
1, \quad (a\,b)(c\,d), \quad (a\,c)(b\,d), \quad (a\,d)(b\,c).
$$
Since all possible pairs of pairs are in the subgroup and any two can be generators, the count is exact and the total number is
$$
\binom{5}{4} = 5.
$$

All $15$ embedded copies of $V$ in Case 1 are conjugate, and all $5$ embedded copies of $V$ in Case 2 are conjugate, but embedded copies of $V$ from opposite classes are not conjugate. Why? Any two permutations are conjugate by some $g \in S_5$ if and only if they are of the same cycle shape. Explicitly, using cycle notation,
$$
g \, (a\,b) \, g^{-1} = (a'\,b')
\qquad \text{where }
g(a) = a'
\text{ and }
g(b) = b'
$$
for Class 1, and
$$
g \, (a\,b)(c\,d) \, g^{-1} = (a'\,b')(c'\,d')
\qquad \text{where }
g(a) = a', g(b) = b', g(c) = c',
\text{ and }
g(d) = d'
$$
for Class 2.


问:
How union and intersection implies complement?
Defining the family of Borel sets to be the smallest σ–algebra that contains all the closed sets in a metric space X, I want to show that it is equivalent to the following definition :

The only part remained to show is that if conditions 1. and 2. of the Theorem 3.3 holds then for any set $E\in B$, also $X\backslash E\in B$. I know being closed under union and complement implies being closed under intersection, but how to prove being closed under union and intersection implies being closed under complement?

Other way to finish my proof is to show that in a metric space every open set is an $F_σ$ ; this is easy for $\mathbb{R}$ but how to prove that for a general metric space?

答:
You can prove instead that each closed set is a $G_\delta$. Let $\langle X,d\rangle$ be a metric space, and let $F\subseteq X$ be closed. For each $n\in\Bbb N$ let $U_n=\bigcup_{x\in F}B(x,2^{-n})$; clearly each $U_n$ is open, and $F\subseteq\bigcap_{n\in\Bbb N}U_n$. Suppose that $x\in\bigcap_{n\in\Bbb N}U_n$; then for each $n\in\Bbb N$ there is an $x_n\in F$ such that $d(x,x_n)<2^{-n}$. Clearly the sequence $\langle x_n:n\in\Bbb N\rangle$ converges to $x$, and $F$ is closed, so $x\in F$. Thus, $F=\bigcap_{n\in\Bbb N}U_n$, and $F$ is therefore a $G_\delta$.

Since every closed set in $X$ is a $G_\delta$, every open set must be an $F_\sigma$.