问:
Solving parametric in rationals
I want to find all values of a such that the equation

$$\sqrt{a - x} = a - x^2$$

has at least one real root and none of its roots are irrational. I made some decent progress (I think) but stuck at the latest stages.

Here's what I deduced myself so far:

First, we must have $a \ge 0$ because on the left hand side we have a square root which should be non-negative and if $a < 0$ then the right hand side will be strictly negative since $x^2$ is non-negative. Now, for the special case when $a=0$ we have $\sqrt{-x} = -x^2$ which has one real solution $x=0$ and it satisfies the condition. So, $a=0$ works.

Let's now have $a>0$. I solve the equation directly this way:

Square both sides. We get $a-x = (a-x^2)^2$, so $a-x = a^2-2ax^2+x^4$ and this is quadratic in a :

$a^2-(2x^2+1)a+(x^4+x)=0$ for which we have $D=(2x^2+1)^2-4(x^4+x)=4x^2-4x+1=(2x-1)^2$ so it's a perfect square and we get the solutions in a :

$a_{1,2} =\frac{2x^2+1\pm\sqrt{D}}{2} = \frac{2x^2+1\pm(2x-1)}{2}$ (we don't have to use absolute value here because we got $\pm$ anyways). Now, considering both values separately we arrive at two different equations:

$x^2-x+(1-a)=0$ and $x^2-x-a=0$ and we're off to the final stage here

The final stage is to ensure that these equations have solutions and if they do, all of those are not irrational. First, the easier case is - when only one of this equations has a solution. We have:

$$D_{1} = 1 - 4(1-a) = 4a - 3$$
$$D_{2} = 1 - 4(-a) = 4a + 1$$
($D_{1,2}$ stand for the respective discriminants of these equations). We have a strict inequality $4a-3 \lt4a+1$ therefore if it is the case of only one equation having a real root, it must be the second one. This means we need to solve the following system of inequalities:
$$4a-3\lt0$$
$$4a+1\ge0$$
which yields $a\in(0,\frac{3}{4})$ - (as established above, a must be non-negative and we exclude $a=0$ since we handled that case earlier) for these values of a, so long as 4a+1 is a square of a rational number, things will work. We can set $4a+1=\frac{p^2}{q^2}$ where p and q have no common divisors and get $a=\frac{p^2-q^2}{4q^2}$ ; Now, substituting our restriction for a we get: $0\lt a\lt\frac{3}{4}$, so $0\lt 4a\lt3$, so $0\lt\frac{p^2-q^2}{q^2}\lt3$ and finally $q^2\lt p^2 \lt 4q^2$

Finally, the most interesting is the case when both of the equations will have a solution. Starting with the value of $a=\frac{3}{4}$ (the second discriminant is equal to 0). We can check that in this case the first discriminant is equal to $4\times\frac{3}{4}+1 = \frac{12}{4}+1 = \frac{16}{4} = 4$ which is a perfect square, hence the solutions will be rational numbers. Thus $a=\frac{3}{4}$ works. Now, let's consider $a\gt\frac{3}{4}$ In this case our goal is to make sure that both discriminants are perfect squares.

It is sort of "obvious" that there can be no solutions to this when a is a natural number. This is because $4a-3$ and $4a+1$ are separated by 4, but no perfect squares of natural numbers have such gap: the smallest one is $4-1=3$ followed by $9-4=5$ and then it will only be increasing, thus the difference of 4 between perfect squares of two natural numbers is an impossibility. (We can also prove that more rigorously if we set $D_{1}=n^2$ and $D_{2}=m^2$, then subtracting one from another and factoring it) ; This leaves us with the only possibility - that a is a rational number. Then a is expressible as $a=\frac{p}{q}$ in lowest terms (so p and q are co-primes). Then we have $4a-3=4\times\frac{p}{q}-3=\frac{4p-3q}{q}$ but since p and q are co-primes, then so are 4p-3q and q and since we want a perfect square of a rational number, the denominator must be itself a perfect square. Thus for $q=k^2$ for some integer k. Rewriting these both using k we get
$$4a-3=\frac{4p-3k^2}{k^2}$$
$$4a+1=\frac{4p+k^2}{k^2}$$
Since our denominators are perfect squares, then the numerators must be too. And we can also notice that the first numerator differs from the second one by $4k^2$ which itself is $(2k)^2$ , hence - a perfect square. In the end we have a Pythagorean triple of numbers $4p-3k^2, (2k)^2, 4p+k^2$ as a solution here and... that's it. I don't really know how to arrive at some closed form for a in terms of p and q . I tried with Euclid Formula, but I didn't go far with it. I know for sure that such solutions exist. For instance $\frac{p}{q}=\frac{21}{16}$ works, but I want to find all of the values of a in some closed form which make this work in the case when both equations have real roots.

Any ideas on how to finish it off? You can, of course, offer a completely different way of solving it, but I would also like to know how to complete my take on the problem.

答:
I came up with a general idea for a working solution, I think. Will post the idea here, there are several cases to consider carefully.

Ok. If we continue from where I left off, we can say that the solution, whatever it might be, should satisfy:

$$4p-3k^2=n^2$$
$$4p+k^2=m^2$$

For some integers n and m . Now, it is obvious that $m\gt n$ because $k^2$ is strictly positive (we can't have 0 as a denominator). Also we can safely say $m>n>0$ because otherwise we have at least one discriminant as 0 and that is the case we approached before and now we are under assumption that both of the discriminants are positive. Continuing from here we get:

$$4p-3k^2=n^2$$
$$12p+3k^2=3m^2$$

therefore, $16p = n^2+3m^2$ and if we subtract $4m^2$ from both sides, we get $16p-4m^2=n^2-m^2$ or $4\times(4p-m^2) = (n-m)\times(n+m)$

As we noticed before, $m\gt n$ but let's also note that $4p\lt m^2$ (because $4p=m^2-k^2$). Hence we're actually dealing with both negative sides of the equation and we can flip the order:

$$4\times(m^2-4p)=(m-n)\times(m+n)$$

or, if we write is as factors of 4:

$$4=\frac{(m-n)\times(m+n)}{m^2-4p}$$

And this is where we start considering different factorization of 4 and its distribution among all these members on the right hand side. I'll give several examples.

First : $m-n=1$ , then we have $m=n+1$ and $\frac{m+n}{m^2-4p}=4$ this leads us to $\frac{2n+1}{m^2-4p}=4$, thus $2n+1=4(m^2-4p)$ which is clearly an impossibility because one side of this is odd while another one is even

Second : $m+n=1$ - immediately an impossibility because we defined $m>n>0$

Third : $m-n=4$ and $\frac{m+n}{m^2-4p}=1$. Here we have $m=n+4$, so $2n+4=n^2+8n+16-4p$ and $4p=n^2+6n+16$ . Since left hand side is divisible by 4, we need $n=2b$ for some integer b (in which case both sides are divisible by 4).

So far we have $n=2b$ , thus $m=2b+4$ , now returning to our k from the second original equation:

$$k^2=m^2-4p = (2b+4)^2-(n^2+6n+12)=4b+4=4(b+1)$$

As we know, if we have a perfect square as a product of "something" and another perfect square, then that "something" must itself be a perfect square. The "something" we have is b+1 , therefore for some integer c the equality $b+1=c^2$ must hold. This means $b=c^2-1$ and substituting it further we get:

$$n=2c^2-2$$
$$m=2c^2+2$$

which leads to the following solution:

$$k^2 = m^2-4p = 4c^4+8c^2+4-4c^4-4c^2-4 = 4c^2 = (2c)^2$$
$$4p-3k^2 = 4c^4+4c^2+4-12c^2 = (2c^2-2)^2$$
$$4p+k^2 = 4c^4+4c^2+4+4c^2=(2c^2+2)^2$$

for any arbitrary integer c . In the end, getting back to the p and q notation, we get:

$$D_{1} = 4a-3 = \frac{4p-3q}{q} = \frac{4p-3k^2}{k^2}=\frac{(2c^2-2)^2}{(2c)^2}$$
$$D_{2} = 4a+3 = \frac{4p+q}{q} = \frac{4p+k^2}{k^2}=\frac{(2c^2+2)^2}{(2c)^2}$$

or in other words,

$$\frac{p}{q} = \frac{c^4+c^2+1}{4c^2}$$

An example that I found randomly before (the $\frac{p}{q}=\frac{21}{16}$) is an outcome of this closed form when $c=2$ . I think many others posted in the comment above come from this too. Didn't check all of them, but some definitely are (like for $c=3$)

Fourth : $m+n=4$ . Also an easy case, since we have $m>n>0$ the only possibility it leaves us with is $m=3, n=1$ but in that case we arrive at $3k^2=5$ so no solutions here either

Fifth : $m+n=2$ . An impossibility as much as in the second case since $m>n>0$

Sixth : $m-n=2$ Similarly to the third case, this means $\frac{m+n}{m^2-4p} = 2$ or $4p = n^2+3n+3 = n(n+3) + 3$ . We can now notice that if we have two numbers $n$ and $n+3$ when one of them is guaranteed to be even meaning that our right hand side is a sum of an even and an odd numbers. This is an impossibility because the sum of an even and an odd numbers is odd while our left hand side is even. Hence we have no solutions here either.

And that's it! These are all the cases of how we can factor 4! This means that our $a=\frac{p}{q}=\frac{c^4+c^2+1}{4c^2}$ is indeed the only family of solutions to this problem in case $a>\frac{3}{4}$ . A cool thing is that the border value of $a=\frac{3}{4}$ is also included into this family of solutions when $c=1$!

Final Answer

The complete answer therefore is the combination of all the special values for a and the two general cases, the first is when only one equation has a solution and the second one when both have:

$$a =
\begin{cases}
\frac{p^2-q^2}{4q^2},  & \text{for $0<q^2\le p^2 \lt 4q^2$} \\
\frac{c^4+c^2+1}{4c^2}, & \text{for any $c\in \Bbb N$}
\end{cases}$$

($a=0$ is included into the first one when $p=q$ and $a=\frac{3}{4}$ is included into second one when $c=1$)

EDIT

I'd still like to know if the approach with Pythagorean triples is a viable one and if so, how is it possible to get that to a working form. If you know how - please post an answer

Title: Solving parametric in rationals


问:
Is it possible to make a regular 3-polygon by selecting $3$ points in $S$
I would appreciate if somebody could help me with the following problem:

Let $\mathbb{Z}$ be the set of all integers and let $ S = \mathbb{Z} \times \mathbb{Z} $.

Question:

1). Is it possible to make a regular 3-polygon by selecting $3$ points in $S$ ?

2). Is it possible to make a regular 5-polygon by selecting $3$ points in $S$ ?

3). Is it possible to make a regular $n$-polygon by selecting $n$ points in $S$ ?

答: The sides of lattice triangles have rational slopes, and using the addition formula for $\tan$ one deduces that all three $\tan\alpha_i$ are rational. In particular $\alpha=60^\circ$ cannot be realized.</p>

ContentLicense: CC BY-SA 3.0


问:
Infinite subset of $\mathbb{N}$ is countable. Proof by math induction.
Proposition: Infinite subset of $\mathbb{N}$ is countable.

Proof: Suppose that $E$ is infinite subset of $\mathbb{N}$ and our goal to show that $E$ is countable, i.e. there is $f:\mathbb{N}\to E$ which is bijection.

Since $E\neq \varnothing$ then $\exists! e_1:=\min E$, then $E \backslash \{e_1\}\neq \varnothing$ then $\exists! e_2:=\min E \backslash \{e_1\}$. Also $\exists! e_3:=\min E \backslash \{e_1,e_2\}$ and we can continue this process by induction. The we define function $f:\mathbb{N}\to E$ in the following way: $f(n)=e_n$ for each $n\in \mathbb{N}$.

Intuitively I understand this very well. But how induction works here in a rigorous way?

Principle of mathematical induction is the following: If $S\subset \mathbb{N}$ with $1\in S$ and for each $n\in S$ we have $n+1\in S$ then it follows that $S=\mathbb{N}$.

How to write this set $S$ in our case in a rigorous way?

Would be very thankful for help!

答:
HINT: Say that a finite $f\subseteq\Bbb N\times E$ is good if

$f$ is a function,

$\operatorname{dom}f=\{1,\ldots,n\}$ for some $n\in\Bbb N$, and

$f(k)=\min(E\setminus\{f(i):i<k\})$ for each $k\in\operatorname{dom}f$.

Step $\mathbf1$. Show first that if $f$ and $g$ are good and have the same domain $\{1,\ldots,n\}$, then $f=g$. You can do this by induction by considering the set $$\{k\in\Bbb N:f(k)=g(k)\text{ or }k>n\}\,.$$

Step $\mathbf2$. Show by induction that for each $n\in\Bbb N$ there is a good function $f_n$ with domain $\{1,\ldots,n\}$; the first step ensures that $f_n$ is unique.

Step $\mathbf3$. Show that if $1\le k\le m\le n$, then $f_m(k)=f_n(k)$; this is similar to the first step.

Define a function $h:\Bbb N\to E:n\mapsto f_n(n)$. Use the result of Step $3$ to show that $h$ is injective, and use the fact that $\Bbb N$ is well-ordered to show that $h$ is surjective. (If $h$ is not surjective, consider the least element of $E\setminus h[\Bbb N]$ and get a contradiction.)

ContentLicense: CC BY-SA 4.0


问:
How to multiply 2 Binary Numbers?
Suppose I have two regular numbers $a$ and $b$ in base $10$ like this: (where $N$ is even)

$$a=a_{\frac N2}a_{\frac N2-1}\ldots a_1,\qquad b=b_{\frac N2}b_{\frac N2-1}\ldots b_1$$

So the result of their multiplication is:

$$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)10^{i+j-2}$$

Now suppose $a=a_Na_{N-1}...a_1$ and $b=b_Nb_{N-1}...b_1$ are (unsigned) binary numbers, such that each digit is actually 16 bits by its own. For example: $a_1$ can be: $1111010101000000$

So if $a$ has $N$ digits it's actually: $16N$ long binary number.

Given this fact how can I multiply them in a similar way to what is shown above?

Why I need this?

I want to multiply big numbers using the fact that I know only to multiply 16 bits binary numbers.

Small But Different Example:

if $a_M, b_M, a_L, b_L$ are all 16 bits binary numbers then I could use this fact to multiply 32 bits numbers like this:

答:
In base $10$, the digit $a_1$ can take values from $0$ to $9$. Any numbers higher than that can only displayed if you also use $a_2$.
In your setting, $a_1$ can take values from $0$ to $65535$ (which is $1111111111111111$ in binary). Any numbers higher than that can only displayed if you also use $a_2$.
So, in summary, we should consider base $65536$ (instead of base $10$).

The formula becomes
$$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)65536^{i+j-2},$$
where instead of $65536^{i+j-2}$
we can also write $(2^{16})^{i+j-2}$, which is the same as $2^{16 (i+j-2)}$.
The number $16 (i+j-2)$ in the exponent of the last expressions
is the number of bits that you need to shift by. to do (because if you multiply by $2^k$ this corresponds to a shift by $k$ bits).

Title: How to multiply 2 Binary Numbers?


问:
Equivalent definition of Riemann integral.
We can define Riemann integral using Darboux sums. In particular, we partition a segment in this way: $a = x_0 < x_1 < ... < x_{n-1} < x_n = b$ (1).
But can we take any partition of $[a; b]$ such that $[a; b] = \sqcup_{k=1}^nE_k$, $E_k$ is Jordan measurable,  and replace $\Delta x_i$ with $\mu_J(E_k)$ (2).

In our lecture we prove that $(2) \implies (1)$, and our teacher omitted prove $(1) \implies (2)$ because it's "a tricky exercise". I'm confused a bit...

We can replace $[x_i; x_{i+1}]$ with $[x_i; x_{i + 1})$ to get disjoint union and with a couple of equations we're done. Or it's not so simple?

The question is: does (1) implies (2) and is it difficult to prove?

答:
Suppose $f$ is Riemann integrable under definition (1).  Given any Jordan measureable set $E \subseteq [a,b]$, we define the integral of $f$ over $E$ as

$$\tag{*}\int_E f = \int_a^b f(x) \chi_{E}(x) \, dx,$$

The integral on the RHS of (*) exists since the boundary of $E$ has zero content and if $f$ is Riemann integrable on $[a,b]$, then $f \chi_{E}$ is discontinuous on a set of at most measure zero.  We also have the usual property that

$$\inf_{x \in E} f(x) \cdot\mu(E) \leqslant \int_E f \leqslant\sup_{x \in E} f(x) \cdot \mu(E)$$

Let $\mathcal{S} = \{P=\{E_k\}_{k=1}^n |\, n \in \mathbb{N}, \,\cup_{k=1}^nE_k = [a,b], \, E_k  \text{ Jordan measureable}   \}$, the collection of Jordan measureable partitions of $[a,b]$ and define upper and lower sums as

$$U_{\mathcal{S}}(P,f) = \sum_{k=1}^n \sup_{x \in E_k}f(x) \cdot \mu(E_k), \,\,\, L_{\mathcal{S}}(P,f) = \sum_{k=1}^n \sup_{x \in E_k}f(x) \cdot \mu(E_k)$$

Definition (2) of Riemann integrability is

$$\tag{**}\int_a^b f(x) \, dx = \sup_{P \in \mathcal{S}}\, L_{\mathcal{S}}(P,f) = \inf_{P \in \mathcal{S}}\, U_{\mathcal{S}}(P,f)$$

To prove that (1) implies (2), note that for any partition $P = \{E_k\}_{k=1}^n$

$$\int_a^b f(x) \, dx = \sum_{k=1}^n\int_{E_k}f \geqslant \sum_{k=1}^n \inf_{x \in E_k}f(x) \cdot \mu(E_k) = L_{\mathcal{S}}(P,f),$$

and, thus, $\int_a^b f(x) \, dx) \geqslant \sup_{P \in \mathcal{S}} L_{\mathcal{S}}(P,f)$.  To prove that the first equality in (**) holds, we must show that for any $\epsilon > 0$ there exists a partition $P \in \mathcal{S}$ such that

$$\int_a^b f(x) \, dx - \epsilon \leqslant L_{\mathcal{S}}(P,f)$$

As you suggested this is obvious by taking $P = \{[a,x_1), [x_1,x_2), \ldots, [x_{n-1},b]\}$and using Riemnann integrability of $f$ under definition (1).

In a similar way, the second equality of (**) can be proved.


问:
Hausdorffness of $\mathbb{R}$ with cocountable topology
I know that $(\mathbb{R}, \tau_{cocountable})$ is a $T_1$ space, and that it is not $T_2$. It is easy for me to prove that it is $T_1$ (let $x, y \in \mathbb{R}$, take $\mathbb{R}\backslash \{x\}$ and $\mathbb{R}\backslash \{y\}$ as the open sets). However, I can't come up with a solid proof for the fact that this space is not $T_2$. Could someone help me with this?

答:
You need only to see how open sets look like. Let $U$ be a nonempty open set $\tau_{\text{co}}$; then $$U=\mathbb R\setminus(\text{countable set})$$

Now, let $x,y\in\mathbb R$, then you can not find two disjoint nonempty open sets $U,V$such that $x\in U$ and $y\in V$


问:
$f_i^N(uv_\lambda)=\sum_{k=0}^N {{N}\choose{k}}((ad\ f_i)^ku)(f_i^{N-k}v_\lambda)$
Carter states that the following holds
$$f_i^N(uv_\lambda)=\sum_{k=0}^N {{N}\choose{k}}((ad\ f_i)^ku)(f_i^{N-k}v_\lambda)$$
where $u\in\mathfrak{g}$ the Kac-Moody algebra and $v_\lambda$ is the highest-weight vector of the module $L(\lambda)$ (the quotient of the Verma module). However I don't think these conditions are actually necessary. I'm having trouble seeing why this holds.

My stratetegy would be to prove it by induction. The induction start is pretty clear, however I don't seem to get anywhere with the inductive step.

Or maybe there is another way of seeing why this holds?

答:
Let $x, y \in \mathfrak{g}$ be any two elements of a Lie algebra $\mathfrak{g}$, then in $U(\mathfrak{g})$ we have the relation $xy = [x, y] + yx$, where $[w, z] \in \mathfrak{g}$ is the Lie bracket (rather than the commutator bracket). Considering $x^2 y$ and applying this relation three times:

$$ \begin{aligned}
x^2 y
&= x([x, y] + yx) \\
&= x[x, y] + (xy) x \\
&= ([x, [x, y]] + [x, y]x) + ([x, y] + yx)x \\
&= [x, [x, y]] + 2 [x, y]x + yx^2.
\end{aligned}$$

From here we can do induction: suppose that for $x, y \in \mathfrak{g}$ and for some $n \geq 1$ we have
$$ x^n y = \sum_{k = 0}^n \binom{n}{k} (\operatorname{ad}_x^k y) x^{n - k}.$$
Then we have
$$\begin{aligned}
x^{n + 1} y
&= x \sum_{k = 0}^n \binom{n}{k} (\operatorname{ad}_x^k y) x^{n - k} \\
&= \sum_{k = 0}^n \binom{n}{k} \left( [x, \operatorname{ad}_x^k y] + (\operatorname{ad}_x^k y)x \right) x^{n - k} \\
&= \sum_{k = 0}^n \binom{n}{k} (\operatorname{ad}_x^{k + 1} y)x^{n-k} + (\operatorname{ad}_x^k y) x^{n + 1 - k} \\
&= \sum_{j = 1}^{n + 1} \binom{n}{j - 1} (\operatorname{ad}_x^j y)x^{n + 1 - j} + \sum_{j = 0}^{n} \binom{n}{j} (\operatorname{ad}_x^j y)x^{n + 1 - j} \\
&= (\operatorname{ad}_x^{n+1} y) + \sum_{j = 1}^n \left( \binom{n}{j - 1} + \binom{n}{j}\right)(\operatorname{ad}_x^j y)x^{n + 1 - j} + x^{n + 1} \\
&= (\operatorname{ad}_x^{n+1} y) + \sum_{j = 1}^n \binom{n + 1}{j}(\operatorname{ad}_x^j y)x^{n + 1 - j} + x^{n + 1} \\
&=\sum_{k = 0}^{n+1} \binom{n+1}{k} (\operatorname{ad}_x^k y) x^{n + 1 - k}.
\end{aligned}$$

As you can see, the induction is long and tedious, and it would be nice to be able to directly use the binomial theorem.

Otherwise, we can argue more abstractly. On the tensor algebra $T(\mathfrak{g})$, given an element $x \in \mathfrak{g}$ we have three linear operators: $l_x(\tau) = x \otimes \tau$ (left multiplication by $x$), $r_x(\tau) = \tau \otimes x$ (right multiplication by $x$), and the derivation $\operatorname{ad}_x$, defined on $\mathfrak{g}$ by $\operatorname{ad}_x(y) = [x, y]$, and extended as a derivation, so that $\operatorname{ad}_x(\tau \otimes \omega) = \operatorname{ad}_x(\tau) \otimes \omega + \tau \otimes \operatorname{ad}_x(\omega)$.

These operators pairwise commute, which is easy to see for $l_x$ and $r_x$. For $\operatorname{ad}_x$ and $l_x$ we check
$$ \operatorname{ad}_x l_x \tau = \operatorname{ad}_x(x \otimes \tau) = \operatorname{ad}_x(x) \otimes \tau + x \otimes \operatorname{ad}_x(\tau) = 0 + x \otimes \operatorname{ad}_x(\tau) = l_x \operatorname{ad}_x \tau,$$
and the proof for $\operatorname{ad}_x$ and $r_x$ is similar. From the relation $xy = [x, y] + yx$, we find that in the quotient $U(\mathfrak{g})$ we have the equality of operators $l_x = \operatorname{ad}_x + r_x$. Now the proof is very easy:

$$ l_x^n y = (\operatorname{ad}_x + r_x)^n y = \sum_{k = 0}^n \binom{n}{k} r_x^{n - k} \operatorname{ad}_x^k y = \sum_{k = 0}^n \binom{n}{k} (\operatorname{ad}_x^k y) x^{n - k}.$$


问:
Why doesn't integrating the area of the square give the volume of the cube?
I had a calculus course this semester in which I was taught that the integration of the area gives the size (volume):

$$V = \int\limits_a^b {A(x)dx}$$

But this doesn't seem to work with the square. Since the size of the area of the square is $x^2$ then $A(x) = {x^2}$, then:

$$V = \int\limits_{ - r}^r {{x^2}dx}  = \left[ {\frac{{{x^3}}}{3}} \right]_{ - r}^r = \frac{{{r^3}}}{3} - \frac{{ - {r^3}}}{3} = \frac{2}{3}{r^3}$$

It's clear that this is not the volume of the cube. Why is this the case? Am I misunderstanding something?

答:
It should be:

$$V = \int_0^a a^2 dz$$

where $a$ is the length of one of the sides of the square.

Or using your notation:

$$V = \int_0^x x^2 dz$$

where $z$ is the dimension over which you are integrating.


问:
A non-negative solution $f$ that satisfies $Qf = \alpha f$ is indeed also smaller than $1$ where $Q$ is conservative $Q$ matrix
Let $\alpha>0$ and $Q=(q(x,y))_{x,y \in E}$ such that $E$ is a countable state space and $Q$ is conservative $Q$-matrix with transition matrix $P=(p(x,y))_{x,y \in E}$ where $p(x,x)=0$ for any $x \in E$. Further define $c(x) = \lvert q(x,x)\rvert$ such that $p(x,y)c(x)=q(x,y)$ for any $y \neq x$

Show that a non-negative solution $f$ on $E$ that satisfies $Qf = \alpha f$ is indeed also $\leq  1$

Approach thus far:

Let $f\geq 0$ be non-negative solution on $E$, then we have that for any $x\in E$:

$ \sum\limits_{y\in E}q(x,y)f(y)=\alpha f(x)\implies f(x)= \sum\limits_{y\in E\setminus \{x\}}\frac{c(x)}{\alpha+c(x)}f(y)p(x,y)\leq \sum\limits_{y\in E\setminus \{x\}}f(y)p(x,y)$

I believe there must be some kind of contradiction here, if I assume that $f\geq 0$ and that there exists $x^*$ such that $f(x^*)> 1$ but I cannot seem to find it. Any ideas?

Edit: A conservative Q-matrix is one such that all entries are finite, the off-diagonals elements are positive and the row sums are equal to zero, i.e. $\sum\limits_{s\in E}q(r,s)=0$ for any $s \in E$

答:
Since I already did quite a bit to squeeze into a comment, here is the proof for a special case, and some examples to argue why some more assumptions may be needed.

Something much more general holds for rate matrices for CTMCs, namely, their eigenvalues, if they exist, should be all non-positive in a reasonable domain (for instance, rate bounded reversible rate matrices). Does $f$ in the question belong to a reasonable class of functions? (say, bounded functions or at least summable functions?)

Otherwise, consider $q(i, j) = 4^{-(j-i)^+}$ and $q(i, i) = -1/3$, and consider the functions $f(i) = 2^i$. Then, we have that $\alpha = 2/3$ is an eigenvalue, as $\sum q(i, j) f(j) = (-1/3) 2^i + 2^i (1/2 + 1/4 + ...) = (2/3) 2^i$.

Thus, let us assume that the conjectured $f$ is bounded.

You already have the main idea in your attempt. Let $\epsilon > 0$, and let $x^*$ be such that $f(x^*) > \sup f - \epsilon > 0$. For notational convenience, let $-q(x^*, x^*) := \lambda > 0$.

We now have

$ \sum\limits_{y\in E}q(x^*,y)f(y)=\alpha f(x) \implies \sum_{y \in E \setminus \{x^*\}} q(x^*,y) f(y) = (\alpha + \lambda) f(x^*)$

However, the left hand side is a non-negative quantity upper bounded by $\lambda \sup_{x \in E} f(x)$ (which one can observe by bounding each $f(y)$ by the sup, and then using the definition of a  conservative matrix), so we have $\lambda \sup f \geq (\alpha + \lambda) (\sup f - \epsilon)$. Now, simplifying, we have:

$\lambda \epsilon \geq \alpha (\sup f - \epsilon)$

which, if there is a uniform bound on $\lambda$s, gives you a nice contradiction, by picking $\epsilon$ small enough.

If there is no uniform bound on $\lambda$s, similar to the counterexample before, let $-q(i, i) = q(i, i+1) = \alpha_i$ where $\alpha_i$ solves the equation below, and let $f(i) = 2 - 1/i$, which gives

$\alpha_i ((2 - 1/(i+1)) - (2- 1/i)) = \alpha (2 - 1/i)$.

Note that this answer does not use the $p(x,x)= 0$, or anything about $p$, since as written, the question does not seem to really use it, so there could be some slackness there.


问:
Divided difference and polynomial.
Theorem: If $f$ is a polynomial of degree $\leq n$, then all of the divided differences $f[x_0,x_1,\cdots, x_i]$ is zero for $i\geq n+1$.

This can be proved by using the result that $f[x_0,x_1,\cdots, x_i]=\frac{f^{(n)}(c)}{n!}$ for some $c\in (a,b)$, for $n$th continuously differentiable function $f$ on $[a,b]$.

Now I want to know the converse of above theorem . Can I say that if $f$ is $n$-times continuously differentiable on $[a,b]$ having all of the divided differences $f[x_0,x_1,\cdots, x_i]$ is zero for $i\geq n+1$ then $f$ must be polynomial of degree $\leq n$?

According me converse is true as $f[x_0,x_1,\cdots, x_i]=\frac{f^{(n)}(c)}{n!}=0$ for some $c$ will give $n$th derivative of $f$ as zero as $c$ varies according $x_i$ and hence $f$ must be a polynomial. Am I right ? Please comment. Thank you.

答:
I think the answer is yes. Suppose $f \colon [a, b] \to \mathbb{R}$ and that
$$f[x_0, x_1, \dots, x_{n + 1}] = 0$$
for any set of distinct points $\{x_0, x_1, \dots, x_{n + 1}\} \subseteq [a, b]$.
Choose $n$ distinct points $x_0, x_1, \dots, x_n$ in $[a, b]$. Let
$$p_n(x) = \sum_{i = 0}^{n}f[x_0, \dots, x_i](x - x_0) \dots (x - x_{i - 1})$$
be the interpolation polynomial for $f$ through $x_0, \dots, x_n$.
Now fix an arbitrary $c \in [a, b]$. We can add $c$ to our list of interpolation points to get a polynomial $p_{n + 1}$ that matches $f$ at $x_0, x_1, \dots, c$. Thus
$$f(c) = p_{n + 1}(c) = p_n(c) + f[x_0, x_1, \dots, x_n, c](x - x_0)(x - x_1) \dots (x - x_n) = p_n(c).$$
Thus $f(x) = p_n(x)$ for all $x \in [a, b]$.

Title: Divided difference and polynomial.


问:
if $V=\mathbb{R^3}$ and $S=\{(x,y,0)\in\mathbb{R}^3\}$ why the coset v +S represents parallels planes to xy plane?
if $V=\mathbb{R^3}$ and $S=\{(x,y,0)\in\mathbb{R}^3\}$ why the coset v +S represents parallels planes to xy plane?

I know a plane $h$ is parallel to other plane  $z$ if $h=\lambda z$ how can I prove it?

I don't have any idea in how to justify it any hint?

答:
Your definition of $S$ is understood as:
$$
S=\{(x,y,0)\in\mathbb{R}^3\mid x,y\in\mathbb{R}\}.
$$
This is a subspace of $\mathbb{R}^3$. Given any vector $v=(v_1,v_2,v_3)$, the coset
$$
v+S=\{v+s\mid s\in S\}.
$$

So basically you are shifting every vector in $S$ by the same vector $v$, which is the same as shifting the $x$-$y$ plane by the vector $v$. So what you get is a plane that is parallel to the $x$-$y$ plane.

To see a simple example, consider $v=(0,0,1)$. Then $v+S$ is the plane $z=1$, which is parallel to the plane $z=0$ (i.e. S).


答:
The $(x,y)$ plane $S\,$ has equation $z=0$, and and any plane parallel to $S$ has equation $z=k,\enspace k\in\mathbf R$.

Now if $v=(a,b,c)$, the coset $v+S$ comprises all points with coordinates $(x+a,y+b, c)$, so it has equation $z=c$.

Another explanation:
any two points in the coset have difference in $S$, which corresponds to the definition of an affine subspace with direction $S$.

Body:
if $V=\mathbb{R^3}$ and $S=\{(x,y,0)\in\mathbb{R}^3\}$ why the coset v +S represents parallels planes to xy plane?

I know a plane $h$ is parallel to other plane  $z$ if $h=\lambda z$ how can I prove it?

I don't have any idea in how to justify it any hint?


答: Let's say we can take it as given that $v+S$ is a plane, and we define parallel planes to be ones that are either equal or disjoint. If $v\in S$ then it's easy to show $v+S=S.$ If $v\not\in S$ suppose there is a vector $w$ such that $w\in S$ and $w\in v+S$. Then for $s\in S,$ $w=v+s$. Then because $S$ is a subspace, $v=w-s\in S,$ a contradiction.

Title: if $V=\mathbb{R^3}$ and $S=\{(x,y,0)\in\mathbb{R}^3\}$ why the coset v +S represents parallels planes to xy plane?


问:
Prove a well-known inequality using Cauchy-Schwarz or AM-GM
For $a;b>0$ and $ab \geq 1$ we have a well-known inequality:

$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2} \geq \dfrac{2}{1+ab}$

Which is equivalent to: $\dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)} \geq 0$ (true)

But is there a solution for this inequality by Cauchy-Schwarz or AM-GM? Thank you.

答:
Edit: Came up with another solution that actually uses the AM-GM inequality.

We start with the initial inequality, and multiply the first two factions by $\frac{1}{2}b^2$ and $\frac{1}{2}a^2$ on the top and bottom respectively:
$$\frac{\frac{1}{2}b^2}{\frac{b^2+(ab)^2}{2}}+\frac{\frac{1}{2}a^2}{\frac{a^2+(ab)^2}{2}}\geq\frac{2}{1+ab}$$

This sets us up to use the AM-GM inequality on the denominators of both (keeping in mind since $a$ and $b$ are always greater than zero this works):
$$\frac{1}{2a}+\frac{1}{2b}\geq\frac{2}{1+ab}$$

We can tidy that up - rinse and repeat with the AM-GM inequality by dividing by 2:
$$\frac{a+b}{2}(1+ab)\geq 2ab$$
$$(1+ab)\geq 2\sqrt{ab}$$

And again, to finally arrive that:
$$\sqrt{ab}\geq\sqrt{ab}$$

Which is always true for $a,b>0$ and $ab\geq 1$.
(Not a very technical answer, there are gaps, should really apply AM-GM to the smaller inequality, but hopefully that gets you started)

Probably kinda convoluted, but I've done this more algebraically.

Let $a=u+v$ and $b=u-v$, where $u>v$ by the restrictions given.

Expanding the expression produced we get the following:
$$\frac{1}{1+u^2+2uv+v^2}+\frac{1}{1+u^2-2uv+v^2}\geq\frac{2}{1+u^2-v^2}$$

And then simplifying using difference of two squares:
$$\frac{2(1+u^2+v^2)}{(1+u^2+v^2)^2-4(uv)^2}\geq\frac{2}{1+u^2-v^2}$$

Cross multiplying and simplifying again gets:
$$(1+u^2)^2-(1+u^2+v^2)^2\geq v^4-4(uv)^2$$

And we expand and simplify:
$$u^2-v^2\geq1$$

By difference of two squares, we can substitute $a$ and $b$ back in and we're left with $ab\geq1$, which is one of the restrictions given.

Edit - so I realised you wanted it via one of the inequalities directly, but this solution doesn't give it that way. However, I did take inspiration from the AM-GM inequality - if you notice, $u=\frac{a+b}{2}$ if you solve for $u$ independently, and $ab$ winds up being $u^2-v^2$, which ends up being a fairly helpful expression. It works out to be slightly more intuitive if you do the substitution for $a$ and $b$ rather than applying AM-GM directly (using this solution anyway). Hope this helps.


问:
For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable?
I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.

My idea was this. Since $G$ is solvable, there exists a normal abelian tower
$$
G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}.
$$
I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence
$$
G/H\supset K_1\supset\cdots\supset K_r=\{e\}.
$$
My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.

Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If $g'H_i\in K_i$, then I want to show $g'H_igH_{i+1}g'^{-1}H_i$ is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you.

答:
I can't quite show that $G$ is solvable implies $G/H$ is solvable.

$G$ is solvable if and only if the process $G_0 = G, G_i = [G_{i-1}, G_{i-1}]$ of repeatedly taking commutator subgroups eventually terminates in the trivial group. If it terminates after $n$ steps, this is equivalent to saying that a certain word made of $n$ levels of nested commutators vanishes identically for every choice of elements of $G$, and this property is automatically preserved by homomorphisms $G \to G/H$ (as well as taking subgroups).


问:
Find all of the distinct non-planar graphs with 6 vertices.
Find all of the distinct non-planar graphs with 6 vertices.  Draw a picture of each such graph, and explain why the graphs you have found are the only possibilities.

HINT: Proceed by cases, considering the number of edges in each possible graph.

答:
By Kuratowski's Theorem you are looking for the graph that have either the subdivisions of the $K_{3,3}$ or the $K_5$ as a subgraph.

The $K_{3,3}$ is a nonplanar graph with 6 vertices. For the other graphs, start with a $K_5$ and add a vertex. You have now different possibilities how to connect (or not connect) the additional vertex to the $K_5$. Try to go through all the cases by iterating over the degree of the additional vertex. Finally, you have the $K_5$ with a subdivided edge.


问:
Same number of generators and coincidence of integral closures of ideals ensures equality?
Let $(R,\mathfrak{m})$ be a commutative, Noetherian regular local ring with infinite residue field $k=R/\mathfrak{m}$. When $I$ is an ideal of $R$ there is the well-known integral closure $\bar{I}$ of $I$ consisting of elements $s\in R$ satisfying monic equations $s^n+a_1 s^{n-1}+\ldots+a_n=0$ with $a_j\in I^j$, $j=1,\ldots,n$.

A subideal $J\subset I$ is said to be a reduction of $I$ if there exists $t\in\mathbb{N}$ such that $JI^t=I^{t+1}$. It is also well-known that $J\subset I$ is a reduction of $I$ if and only if $\bar{I}=\bar{J}$ (equality of integral closures). Reductions can be proper: for instance, if $R=k[[x,y]]$ then $(x^2,y^2)$ is a proper (non-trivial) reduction of $(x^2,xy,y^2)$.

My question could be phrased as (EDITED): Are there any general conditions under which $\bar{I}=\bar{J}$ ensures equality of the original ideals $J\subset I$, when the ideals $I$ and $J$ have the same number of generators? Suppose it is known that the ideal $J\,\,$ I'm interested in is not a minimal (regarding the $\subseteq$ order...) reduction of $I$. As far as I understand this last condition could be concisely put in terms of anaytic spread $\ell(I)$ of $I$: that is $\mu(J)=\mu(I)>\ell(I)$.

To avoid trivial answers assume also that the ideals are not integrally closed. If it matters, the $R=k[[x_1,\ldots,x_n]]$ would be enough to me.

Thanks.

答:
We have a positive answer in the case where the residue field is infinite and $J$ is a minimal reduction.

In this case, $J$ is a minimal reduction of $I$ if and only if the number of generators of $J$ is equal to $\ell$, the analytic spread of $I$. Furthermore, the minimal generating set of $J$ can be extended to that of $I$, See Prop 8.3.3 in Integral closures by Swanson-Huneke. The example following the proposition shows that in general, the equality on the number of generators does not imply equality.


问:
Dividing both sides by 5, when solving $x = 5(\sin y) + 2$ for $y$?
How do I know which operations to do when solving an equation?
For example,
$$x = 5(\sin y) + 2 \tag1$$
If I wanted to make $y$ a function of $x$,
$$\begin{align}
x &= 5(\sin y) + 2 \tag2 \\[4pt]
x - 2 &= 5(\sin y) \tag3 \\[4pt]
\frac{x - 2}{5} &= \sin y \tag4 \\[4pt]
y &= \sin^{-1}\left(\frac{x - 2}{5}\right) \tag5
\end{align}$$
I believe this is the correct answer, but I was wondering why it isn't this way:

$$\begin{align}
x &= 5(\sin y) + 2 \tag6 \\
\frac{x}{5} &= (\sin y) + 2 \tag7 \\
\frac{x}{5} - 2 &= \sin y \tag8 \\
y &= \sin^{-1}\left(\frac{x}{5} - 2\right) \tag9
\end{align}$$

Why is equation $(9)$ wrong?

答:
A lot of algebra boils down to doing the same thing to both sides of an equation. For instance, if I started with the equation:
$$x=y+1$$
I could subtract $1$ from each side to get
$$x-1=y+1-1$$
which doesn't look that useful - except that using what I know about addition and subtraction, I can figure out that $y+1-1$ must equal $y$ to deduce a more useful expression:
$$x-1=y.$$
It's easy to learn this sort of manipulation by rote, but what we're doing is just cleverly using a fundamental property of equality: if two things are equal, I can do the same operation to both, and they will remain equal.

I could do other correct manipulations, if I wished - I could double each side to get
$$2\cdot x = 2\cdot (y+1)$$
which would simplify to
$$2\cdot x = 2\cdot y + 2.$$
Would this be useful? Probably not - but it's certainly legal. The trick with solving algebraic equations is that you choose your manipulations cleverly - you try to sort of "unwrap" the variable you wish to solve for, usually by undoing whatever operations were applied to it. I mention this because it's important to remember that we're not just shuffling terms about from side to side - to get from $x=y+1$ to $x-1=y$, I didn't magically move a $1$ from one side to the other - I subtracted from both sides in a way that conveniently made $y+1-1$ and then cancelled.

In the given example, removing the $\sin$ since it isn't at fault here, if I start with:
$$x=5y+2$$
It's absolutely true that there are multiple ways to start: One way is to start by subtracting $2$ from both sides giving
$$x-2=5y+2-2$$
$$x-2=5y$$
and then dividing both sides by $5$ to get
$$\frac{x-2}5=\frac{5y}5$$
$$\frac{x-2}5=y.$$
This is a nice sequence to go about it, because $5y+2$ means "take $y$, multiply it by $5$, then add $2$" and if I want to undo that sequence of operations, I just undo each operation in reverse order - that is, "subtract $2$, then divide by $5$".

But, you're right that you can start in other ways: if I divide both sides by $5$, I get
$$\frac{x}5=\frac{5y+2}5$$
which would then simplify as
$$\frac{x}5=y+\frac{2}5.$$
This step - from (6) to (7) in your reasoning - appears to be where the issue arises. From there, I could subtract $\frac{2}5$ from both sides as
$$\frac{x}5-\frac{2}5=y+\frac{2}5-\frac{2}5$$
$$\frac{x}5-\frac{2}5=y$$
which is equivalent.

As long as you take both sides and do the exact same thing to each of them, you'll arrive at a correct statement at the end*. Succinctly put, the property we're using is that if $L=R$ then $f(L)=f(R)$ for any function $f$ - where that function could be "add $2$" or "divide by $5$" or anything else we might dream of. Algebra is just knowing the correct thing to do so that at the end your result is useful.

(*Caveat: Sometimes students will assume this goes both ways, but if you do something like "square both sides" you can end up with extraneous solutions. It's still true that if $L=R$ then $L^2=R^2$, but it's not true that if $L^2=R^2$ then $L=R$. This doesn't come up with adding and subtracting stuff from both sides or multiplying by non-zero coefficients since you can undo those operations)


问:
About general Homology with negative index
Let $0 > n\in \mathbb{Z}$. I know, that in Singular Homology $ H_n(X)=0 $ for a toplogical space $ X \neq \emptyset $. If we now use a general Homology which satisfies the Eilenberg–Steenrod axioms does this property also hold? If yes, how to prove? And if no, is there an example of a Homology theory which is non zero for negative $n$?

Furthermore if it is not zero, how does this affect the reduced homology?

答:
No.

The "classical" Eilenberg-Steenrod axioms are homotopy invariance, exactness, excision and dimension. These describe ordinary homology theories. It is well-known that for finite CW-pairs the homology groups $H_n(X,A)$ are (up to natural isomorphism) uniquely determined by the coefficient group $G = H_0(*)$, where $*$ is a one-point space. In fact, they agree with the singular homology groups of $(X,A)$ with coefficients in $G$.
In particular $H_n(X,A) = 0$ for $n < 0$.

Beyond finite CW-pairs things are more sophisticated. In

I.M. James and J.H.C. Whitehead, "Homology with zero coefficients", The Quarterly Journal of Mathematics, Volume 9, Issue 1, 1958, Pages 317–320

one can find examples of non-trivial ordinary homology theories with zero coefficient group. There are infinite CW-complexes $X$ with nonvanishing homology groups. Making a dimension shift ($H'_n = H_{n+k}$ for some $k \in \mathbb N$) we can achieve that $H'_n(X) \ne 0$ for negative $n$. The theory $H'_*$ has coefficient group zero, if you do not like that consider $H'_* \oplus H^{sing}_*$, where $H^{sing}_*$ is singular homology with $\mathbb Z$-coefficients.


问:
If $\alpha,\beta,\gamma$ are the roots of equation $x^3-7x^2+12x-11=0$, find the last four digits of $\alpha^{21007}+\beta^{21007}+\gamma^{21007}$

If $\alpha,\beta,\gamma$ are roots of the equation $x^3-7x^2+12x-11=0$, then find the last four digits of $\alpha^{21007}+\beta^{21007}+\gamma^{21007}$.

I know that $\alpha+\beta+\gamma=7, \alpha\beta+\beta\gamma+\gamma\alpha=12,\alpha\beta\gamma=11$. From this how I can proceed. please give some idea

答:
Let $z_i = \alpha^i + \beta^i + \gamma^i$

It's easy to check that $z_0 = 3, z_1 = 7, z_2 = 25$ etc.

Our objective is to compute $z_{21007} \bmod 10000$

Consider the sequence of rational integers $3, 7, 25, \ldots$

and the corresponding recurrence relation $$z_{n+3} - 7z_{n+2} + 12z_{n+1} - 11z_n = 0$$

We'll find the periodicity of the above sequence modulo $10000 = 2^4 5^4$

We'll do it in $3$ steps and need $3$ theorems (viz., Theorems $2, 8, 10$) from On sequences defined by linear recurrence relations  by H. T. Engstrom

$1.$ Find the periodicity modulo $2$ and modulo $5$ (See Theorem $2$)

Consider $$F(z) = z^3 - 7z^2 + 12z - 11 = 0$$

Or, $$F(z) \bmod 2 = (z^3 + z^2 + 1) \bmod 2$$
and  $$F(z) \bmod 5 = (z^3 + 3z^2 + 2z + 4) \bmod 5 = (z+4)(z^2+4z+1) \bmod 5$$

The period of the sequence modulo $2$ is $2^3-1 = 7$

and the period of the sequence modulo $5$ is $LCM(5^1-1, 5^2-1) = 24$

$2.$ Find the periodicity modulo $2^4$ and modulo $5^4$ (See Theorem $8$)

The period of the sequence modulo $2^4$ is $2^{4-1}.7 = 56$

and the period of the sequence modulo $5^4$ is $5^{4-1}.24 = 3000$

$3.$ Find the periodicity modulo $2^4 5^4$ (See Theorem $10$)

Finally the period of the sequence modulo $2^4 5^4$ is $LCM(56, 3000) = 21000$

The final step is to observe that $$z_{21007} \bmod 10000 = z_{21007 \bmod 21000} \bmod 10000 = z_7 \bmod 10000 = 4637$$


问:
What's the formal definition of an indefinite integral?
I understand that definite integrals can be defined as the limit of Riemann sums but I have yet to see a similar definition for the indefinite integral. I suppose
$$
\int f(x) \, dx = \{F(x):F'=f \}
$$
might do the trick but I'm unsure if this really a satisfactory definition. Alternatively, with the knowledge of the fundamental theorem of calculus, we might define
$$
\int f(x) \, dx = \int_{a}^{x} f(t) \, dt + C
$$
but this definition seems odd in that uses quite a deep result to define something which is usually introduced very early on in calculus classes. So is there a formal definition of an indefinite integral, and if so, what is it?

答:
You should stick to the definition of an indefinite integral.

Given a function $f:I\to\mathbb{R}$ defined on an open interval $I$, if $F:I\to\mathbb{R}$ is a function such that $F'(x)=f(x)$ for every $x\in I$, then we call $F$ an antiderivative of the function $f$. It is easy to check the two following two facts:

If $F$ is an antiderivative $f$ on the interval $I$, then so is $F+C$ for any constant $C$;

If $F$ and $G$ are two antiderivatives of $f$, i.e, $F'(x)=G'(x)=f(x)$ for all $x\in I$, then there exists (exercise!) a constant $C$ such that $F=G+C$.

We define the notion of the "indefinite integral" of the function $f$ (on the interval $I$) as a family of functions:

$$
\textrm{indefinite integral of  } f=\{F:I\to\mathbb{R}\mid F'(x)=f(x)\ \textrm{for all } x\in I\}
$$

Because of these two facts above, we write
$$
\int f(x)\,dx=F(x)+C
$$
for any antiderivative $F$ of $f$ (on the interval $I$).


问:
Proof regarding a colored tree
Let $T$ be a tree with $n$ vertices. Vertices of $T$ are colored in white and black such that the number of black vertices is always less than white ones and white and black vertices are not adjacent. Prove that the white vertices can be numbered with { 1 , 0 , -1 } such that the sum of numbers of the adjacent white vertices to every black vertex is zero considering that we have at least one none-zero vertex.

It's easy to prove that there is at least one white leaf. And also if there is a black vertex connected to more that one white leaf then those leafs can be numbered in the way described in the question and the rest of the white vertices can be zero. But I don't know how to prove this for trees that are not like that.

答:
Suppose there is a black leaf. Then its white neighbor must be colored $0$. We can delete the black leaf and its white neighbor, leaving one or more tree components. At least one of those components has more white vertices than black vertices, and we can solve the problem there (and label everything $0$ in the other components). We can keep reducing the problem to a smaller problem until we get down to a subproblem with no black leaves at all.

In a tree with no black leaves, we can just find a labeling greedily. For concreteness, give the tree a root, which will be some white vertex. Each other vertex will have a parent closer to the root, and possibly some children. There are no black leaves, so each black vertex has at least one child.

Label the root $+1$. Then, for each black vertex whose parent has a $\pm 1$ label, give one of its children a $\mp 1$ label to cancel it, and give any other children it has a $0$ label. Repeat until there are no unlabeled vertices.

Title: Proof regarding a colored tree


问:
Can a ring without unity have a non-trivial subring with unity?
The original problem was to prove that for any commutative ring R, there is a canonical (unique) ring homomorphism from Z to R. The problem would be simple, if R has $1_R$ and we can let the homomorphism map $\phi$ sends $0$ to $0_R$ and $1$ to $1_R$. But if R doesn't have an identity, and $\phi(Z)$ forms a subring of $R$ with unity $\phi(1)$, then the Rng $R$ has a subring with unity. If ring without unity can only have a trivial subring with unity, $\phi$ must be the trivial map, and the proof would be complete. But I'm not sure of this part. Any help would be appreciated!

答: Consider the ring $\Bbb Z \oplus 2\Bbb Z$, with componentwise addition and multiplication; it has no unit, but the subring $\Bbb Z$ (first component) is a unital ring.


答:
Many "commutative rings" are assumed from context to contain the identity, and homomorphisms of these rings must take the identity to the identity. If that is the case, define $\phi:\mathbb{Z}\to R$ so that $\phi(n)=1_R+...+1_R$ (n times).

If this is not assumed, it is false that for any commutative ring $R$ there exists a unique ring homomorphism $\phi:\mathbb{Z}\to R.$ For example if $R=\mathbb{Z}$ either the trivial map or the identity map is a homomorphism in the non-unital ring sense.

Title: Can a ring without unity have a non-trivial subring with unity?

Body: The original problem was to prove that for any commutative ring R, there is a canonical (unique) ring homomorphism from Z to R. The problem would be simple, if R has $1_R$ and we can let the homomorphism map $\phi$ sends $0$ to $0_R$ and $1$ to $1_R$. But if R doesn't have an identity, and $\phi(Z)$ forms a subring of $R$ with unity $\phi(1)$, then the Rng $R$ has a subring with unity. If ring without unity can only have a trivial subring with unity, $\phi$ must be the trivial map, and the proof would be complete. But I'm not sure of this part. Any help would be appreciated!


问:
maximum eigenvalue of rank 1 update matrix
I am stuck with the following problem. The ingredients are:

$A=diag(a_1,\dots,a_n)$, $a_i \in (0,1)$ and $a_1+\dots+a_n=1$,

$B=diag(b_1,\dots,b_n)$, $b_i \in (0,1)$ and $b_1<b_2<\dots<b_n$,

the entries of the matrices $A$ and $B$ are related by $\sum\limits_{i=1}^n a_i {b_i \over 1- b_i}<1$.

Denote ${\bf b}^T=(\sqrt{a_1} b_1,\dots,\sqrt{a_n}b_n)$, $c={1 \over \sum_{i=1}^n a_i (1-b_i)}$. The problem is to prove that the maximum eigenvalue of the matrix $D=B+ c {\bf b} {\bf b}^{T}$ is less than 1.

I've been studying the steady-state distribution of the two-dimensional Markov chain. Its steady-state vector turned out to be ${\vec p}_k={\vec p}_1 D^{k-1}$, $k \ge 1$. As usual, ${\vec p}_1$ is found from the normalization condition, which requires evaluation of $\sum_{k=0}^\infty D^k$. To make it $(I-D)^{-1}$, the aforementioned proof is required. My numerous numerical expreiments show that the maximum eigenvalue is $<1$. But I cannot prove it.

What I've tried and failed:

interative methods (like power iteration) don't help since the power $D^k$ does not have a nice expression.

Gershgorin theorem does not help, since some entries of $D {\vec 1}$ may be $>1$.

the matrix $c {\bf b} {\bf b}^{T} $ is of rank 1 and has one eigenvalue $\rho=c \sum_{i=1}^n {a_ib_i^2}$. This matrix is symmetric.

the eigenvalues of $D$ are all different and do satisfy the well-known secular equation $f(\lambda)=0$, where:
$$
f(\lambda)=1+ c \sum_{i=1}^n {a_i b_i^2 \over b_i - \lambda}.
$$

denote the roots of $f(\lambda)=0$ by $\lambda_1,\dots,\lambda_n$. According to the theory, the roots satisfy the inequalities:
$$
b_1<\lambda_1<b_2<\lambda_2<\dots<b_n<\lambda_n.
$$
This does not help since the upper bound for the maximum eigenvalue $\lambda_n$ remains unknown. The theory says that $\lambda_i=b_i+m_i \rho$, where $m_i \in (0,1)$ and $\sum_{i=1}^n m_i=1$. But I can't prove that $\lambda_n=b_n+m_n\rho<1$.

$\sum\limits_{i=1}^n a_i {b_i \over 1- b_i}<1$ is equivalent to $1<\sum\limits_{i=1}^n {a_i \over 1-b_i}<2$. Since $\sum_i (u_i/v_i) \sum_i (u_iv_i) \ge 1$ we have $\sum\limits_{i=1}^n a_i (1-b_i)>{1 \over2}$ and thus $1<c<2$.

one of the classical inequalities gives
$$
\sum_{i=1}^n a_i (1-b_i)+ (1-b_1)(1-b_r) \sum_{i=1}^n {a_i \over 1-b_i} \le 2- b_1-b_r,
$$
and thus
$$
\sum_{i=1}^n a_i (1-b_i) \le 2- b_1-b_n - (1-b_1)(1-b_n) = 1-b_1 b_n <1,
$$
and thus, instead of $1<c<2$ we have ${1 \over 1-b_1 b_n}<c<2$. Moreover, since $\sum_{i=1}^n a_i (1-b_i)>{1\over 2}$, from the last inequality we get
${1\over 2} \le 1-b_1 b_n$ or $b_1 b_n < {1\over 2}$. Since $b_1<b_2<\dots<b_n$ we have
$$
b_1 b_i < {1\over 2}.
$$

Unfortunately, all this did not help me to make a step towards the proof and I would be grateful for any useful suggestions.

答:
what you want to do is show

$D-I=\big(B-I)+ c {\bf b} {\bf b}^{T}\prec \mathbf 0$

via congruence transform, this is the same as showing

$-I + c \big(I-B\big)^\frac{-1}{2}{\bf b} {\bf b}^{T}\big(I-B\big)^\frac{-1}{2}\prec \mathbf 0$

or that the trace of that rank 1 matrix is $\lt 1$.  Rescaling by $c^{-1}$ gives the criterion

$\sum_{i=1}^n a_i\frac{ b_i^2}{1-b_i}  =\text{trace}\Big( \big(I-B\big)^\frac{-1}{2}{\bf b} {\bf b}^{T}\big(I-B\big)^\frac{-1}{2}\Big)\lt c^{-1}=\sum_{i=1}^n a_i(1-b_i) $

if you subtract the LHS from the RHS this is equivalent to proving

$0\lt \sum_{i=1}^n a_i\frac{(1-b_i)^2- b_i^2}{1-b_i}$  which holds because

$0$

$\lt 1 -\sum_{i=1}^n a_i\frac{b_i}{1-b_i}$

$= \big(\sum_{i=1}^n a_i\frac{1-b_i}{1-b_i}\big) - \sum_{i=1}^n a_i\frac{b_i}{1-b_i} $

$= \sum_{i=1}^n a_i\frac{1-2b_i}{1-b_i} $

$=\sum_{i=1}^n a_i\frac{(1-b_i)^2- b_i^2}{1-b_i}$

where the inequality is bullet point 3 in the original post


问:
Proof for $I \in \binom {[r]}{k}$ with $k \in \mathbb{N_0}$ there exist exactly $(r-k)^n$ r-tuples $(X_1, ..., X_r)$
For $n, r \in \mathbb{N}$,  $(X_1, ...,X_r)$ is an ordered partition of $[n]$ in $r$ non-empty sets, if $X_1,..., X_r$ are disjoint and non-empty subsets  of $[n]$ for which it holds that $X_1 \cup  ... \cup X_r = [n]$.

How can one prove that

for $I \in \binom {[r]}{k}$ with $k \in \mathbb{N_0}$ there exists exactly $(r-k)^n$ r-tuples $(X_1, ..., X_r)$ where

$X_1, ..., X_r$ are disjoint subsets of $[n]$

for all $i \in I$ the set $X_i$ is empty and

$X_1 \cup  ... \cup X_r = [n]$ holds

I found this here, but it answers a different question and I don't know if we can apply the same here. This proof used to give 3 points in our exam, so it shouldn't require a detailed answer to this extent.
Any help is appreciated!

答: A tuple $(X_1,\dots, X_r)$ where $X_i$ is empty for all $i\in I$, corresponds exactly to a function $f$ from $[n]$ to $[r]\setminus I$. Namely, for any $x\in [n]$, $f(x)$ is the unique element $j$ of $[r]\setminus I$ for which $x\in X_j$, which exists because $(X_1,\dots, X_r)$ is a partition. Clearly, the number of such functions is $(|[r]-I|)^{|[n]|}=(r-k)^n$.


问:
How to calculate the norm of an ideal?
Would someone please help explain how to calculate the norm of an ideal? I can't find a source that explains this clearly.
For example, I know that the norm $N_\mathbb{Q(\sqrt10)}(\langle2,\sqrt10\rangle)=2$, but not clear on how to get it myself.
Many thanks</p>

答:
In general, given $I=(s_1,\dots,s_n)\lhd\mathcal{O}_K$ in the ring of integers of some number field $K$, we can factor it into primes to try and work out its norm.

You can see the answer here which explains the rational behind it, but the important points are that:

If a prime $\mathfrak{p}\mid I$, then $s_1,\dots,s_n\in\mathfrak{p}$. Let $(p)=\mathfrak{p}\cap\mathbb{Z}$. Then by the multiplicativity of the norm, we know that $\text{N}(\mathfrak{p})\mid\text{N}(s_i)$, and $\text{N}(\mathfrak{p})$ is always a prime power. This means that the primes in $\mathcal{O}_K$ dividing $I$ are above the rational primes dividing $\gcd(\text{N}(s_1),\dots,\text{N}(s_n))$: this gives us a finite list to check!

If $\mathcal{O}_K=\mathbb{Z}[\alpha]$, then find the possible primes $\mathfrak{p}=(p,g(\alpha))$. Then $\mathfrak{p}\mid I$ if and only if $s_1,\dots,s_n\in(\overline{g})\lhd(\mathbb{Z}/p\mathbb{Z})[\alpha]$.

In your example, $\text{N}(a+b\sqrt{10})=a^2-10b^2$, so $\text{N}(2)=4$, and $\text{N}(\sqrt{10})=-10$. So the only primes dividing $I=(2,\sqrt{10})$ are of norm $2$. In $\mathcal{O}_K=\mathbb{Z}[\sqrt{10}]$, there is in fact only one prime ideal of norm $2$, namely $\mathfrak{p}_2=(2,\sqrt{10})=I$.

Alternatively, you can use the definition that $\text{N}(I)=|\mathcal{O}_K/I|$, and this method may well be quicker with simpler examples such as these. It should be quite clear that
$$
\mathcal{O}_K/I=\mathbb{Z}[\sqrt{10}]/(2,\sqrt{10})\cong\mathbb{Z}/2\mathbb{Z}
$$

Title: How to calculate the norm of an ideal?


问:
Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices?
My question might be very silly as I've only recently started to learn about Lie algebras.

Given a Lie algebra $\mathfrak{g}$, and a vector space basis $\{e_i|i=1 \ldots \dim \mathfrak{g}\}$, we can write the adjoint action $\text{Ad}_x: \mathfrak{g} \to \mathfrak{g}: y \mapsto [x,y]$ in matrix form, where the $i,j$th entry of the matrix $[\text{Ad}_x]$ equals $\lambda_i$ if $[x,e_j] = \sum_{k}\lambda_k e_k$. I am looking for necessary and sufficient conditions on our Lie algebra $\mathfrak{g}$ such that there exists a basis in which all of these matrices $\{ [\text{Ad}_x]| x \in \mathfrak{g} \}$ are skew-symmetric.

If the condition turns out to be either simplicity of compactness of the associated simply connected Lie group, I would be very happy.

答:
The problem has haunted me for the last couple of weeks, and I want to add another answer:

Let's assume our Lie algebra $\mathfrak g$ is of finite dimension $n$ over a field $k$. The computation in Malkoun's answer shows that the question is equivalent to:

Does there exist a (symmetric, nondegenerate) $ad$-invariant bilinear form on $\mathfrak g$ that in some basis is represented by the identity matrix $Id_n$, i.e. such that the corresponding quadratic form is equivalent to $q_{st} (x_1, ..., x_n) =x_1^2 + x_2^2 + ... + x_n^2$?

Now let $\mathfrak g$ be absolutely simple over a field $k$ of characteristic $0$. Then it's well-known that if $b: \mathfrak g \times \mathfrak g \rightarrow k$ is any $ad$-invariant, nondegenerate bilinear form, then there is $c \in k^*$ such that $b(x,y) = c \kappa(x,y)$ for all $x,y \in \mathfrak g$, where $\kappa(\cdot, \cdot)$ is the Killing form. So our question becomes whether or not the Killing form, up to scalar multiple, is represented by $Id_n$.

(NB. For symmetric BLFs/ quadratic forms one usually looks at the equivalence relation on $M_n(k)$ given by
$$ A \simeq B :\Leftrightarrow \exists P \in M_n(k): A= P^TBP$$
and says A, B are congruent. Here, we are looking at the "coarser" equivalence relation $$ A \sim B :\Leftrightarrow \exists c \in k^*, P \in M_n(k): A=cP^TBP$$

This comes up a lot in the study of classical Lie algebras. I think Jacobson called two such matrices "cogredient", but I don't think that has become standard nomenclature. Elman/Karpenko/Merkurjev in The Algebraic and Geometric Theory of Quadratic Forms call it "similar". For now I will just say "congruent up to scalars".)

Now to decide whether this is the case for a given absolutely simple Lie algebra over a specific field can become quite intricate, and can depend a lot on both the Lie algebra and arithmetic of the field. The answers given for $\mathbb R$ and $\mathbb C$ are, in way, very good cases.

First off, the assertion is true if $k$ is quadratically closed (in particular, if $k$ is algebraically closed), because then, $q_{st}$ is the only quadratic form up to congruence (we don't even need scalar multiples there).

Over $\mathbb R$, via Sylvester's law of inertia, the only scalar multiples of $q_{st}$ up to congruence are $\pm q_{st}$, which are up to congruence the only definite quadratic forms there; i.e. our question is equivalent to deciding whether the Killing form is definite, and it is well-known that it is iff $\mathfrak g$ is a compact form.

For other fields, however, things might get tough. Let's just look at the

Easiest example $\mathfrak g:= \mathfrak{sl}_2(k)$

Since this Lie algebra (of dimension $n=3$) contains non-trivial $ad$-nilpotent elements, the Killing form contains isotropic vectors. A necessary condition for our assertion is therefore that there are $(0,0,0) \neq (x,y,z) \in k^3$ such that $x^2+y^2+z^2=0$. Actually, the theory of quadratic forms also shows that this is sufficient. Indeed, one can do this with explicit matrix computations: W.l.o.g. assume there are $x \in k, y \in k^*$ with $x^2+y^2+1=0$. Then

$$e_1 := \pmatrix{-\frac{1}{y}&\frac{x}{y}\\\frac{x}{y}&\frac{1}{y}} , e_2 := \pmatrix{-\frac{x}{y}&-\frac{1}{y}\\-\frac{1}{y}&\frac{x}{y}} , e_3 := \pmatrix{0&1\\-1&0}$$

form a basis of $\mathfrak{sl}_2(k)$ which is "orthonormal" with respect to $-\frac{1}{8}$ times the (isotropic) Killing form. (Note e.g. the special case that $k$ contains a square root of $-1$ called $i$: Then we can set $x=0$, $y=i$ and get $$e_1=\pmatrix{i&0\\0&-i}, e_2 = \pmatrix{0&i\\i&0}, e_3 = \pmatrix{0&1\\-1&0}$$
a standard basis of the compact real $\mathfrak{su}_2$ as well as $\mathfrak{sl}_2(\mathbb C)$, matching previous result for those fields.)

But for which fields there exist such $x,y (,z)$ is a non-trivial problem in general! E.g. while all extensions of some $p$-adic field $\mathbb Q_p$ for odd primes $p$ do, and also all seven quadratic extensions of $\mathbb Q_2$, the $2$-adic numbers $\mathbb Q_2$ themselves do not, and for other extensions of $\mathbb Q_2$ the answer seems to be hard to find.

Meaning that e.g. all $\mathfrak{sl}_2(\mathbb Q_p)$ for odd $p$ have bases for which all matrices representing the adjoint action are skew-symmetric, but there is no such basis for $\mathfrak{sl}_2(\mathbb Q_2)$.

Further, we can easily see that for any number field $k$ which can be embedded into $\mathbb R$, $\mathfrak{sl}_2(k)$ has no such basis (this generalises to any non-anisotropic form over any formally real field); certainly over many other number fields $k$ (trivial example $k=\mathbb Q(i)$), $\mathfrak{sl}_2(k)$ does have such a basis; but for example $\mathfrak{sl}_2(\mathbb Q(\sqrt{-7}))$ does not, as it embeds into $\mathbb Q_2$.

Now $\mathfrak{sl}_2$ can have other forms over fields, the classification of which is equivalent to that of division quaternion algebras over those fields. If $k$ is a finite extensions of some $\mathbb Q_p$, up to isomorphism there is only one, just like over $\mathbb R$. We call it

$\mathfrak{su}_2$ (over a $p$-adic field $k$)

Its elements are of the form $$\pmatrix{r\sqrt{a}&s+t\sqrt{a}\\b(s-t\sqrt{a})&-r\sqrt{a}}$$ where $r,s,t \in k$, and $a,b$ are chosen such that the quadratic form $X_0^2-aX_1^2-bX_2^2+abX_3^2$ is anisotropic (up to congruence, there is exactly one anisotropic quadratic form in four variables over $k$). One can check that the Killing form of $\mathfrak{su}_2$ is anisotropic again, and up to scaling is congruent to the unique (up to scaling) anisotropic quadratic form in three variables over $k$, which is $X_0^2-aX_1^2-bX^2$.

But our previous results showed that for all $p \neq 2$ (and also for any quadratic extension of $\mathbb Q_2$), the form $q_{st}$ is isotropic over $k$, so it cannot be congruent up to scaling to that Killing form! (Whereas conversely for odd degree extensions of $\mathbb Q_2$, it is the unique-to-scaling anisotropic form with $a=b=-1$). This means that for all $\mathbb Q_p$ ($p$ odd) the forms of $\mathfrak{sl}_2$ behave exactly the other way than over $\mathbb R$: The compact one cannot be written as skew-symmetric matrices, but the split one can!

Finally, note that although some of the above arguments generalise, some of them do not even generalise to $\mathfrak{sl}_3$: At some steps above I implicitly used that quadratic forms in three variables are still relatively easy to handle (in particular, up to scaling there is exactly one isotropic one). Another way to think of the $\mathfrak{sl}_2$-forms is in terms of isos of split or non-split forms of $\mathfrak{sl}_2$ and $\mathfrak{so}_3$, respectively. Any non-zero map between these is an isomorphism, for dimension reasons, so one is compact iff the other is. This line of argument would also break down already for forms of $\mathfrak{sl}_3$. For higher dimensions, one needs to know more about the Witt groups of specific fields than I do.


问:
Does $\,n\mid a_1b_1,\,a_2b_2, a_1b_2+a_2b_1\Rightarrow n\mid a_1b_2$?
Imagine that we have two pairs of integers $(a_1,b_1)$ and $(a_2, b_2)$ where

$$ a_1b_1\equiv 0,\,\ a_2b_2\equiv 0,\,\ a_1b_2+a_2b_1\equiv 0\pmod n$$

Does this imply that
$$ a_1 b_2 \equiv 0\pmod n $$

I assume that $a_1 b_1 \equiv 0\pmod n$ is only true if $a_1$ and $b_1$ are divisors of $n$. Using that, I've checked $a_1b_2 = 0$ mod $n$ numerically for a large number of values of $n$ and it seems to hold, but I am not sure how to prove it.

答:
The result is true.

Let $x=a_1b_2$ and $y=a_2b_1$ and let $p^m$ divide $n$ for some prime $p$.

Then $xy = 0 \mod p^{2m}$ and so $p^m$ divides at least one of $x$ and $y$.

However  $x+y = 0 \mod p^m$ and so $p^m$ divides both $x$ and $y$.

Since this is true for all prime power divisors of $n$ it is true for $n$. Thus $n$ is a factor of $x=a_1b_2$, as required.


问:
What do we call an exponential object in a concrete category that basically equals the one in Set
So suppose you've got a concrete category $\mathcal C$ and two objects $X, Y$ of $\mathcal C$. For instance, $\mathcal C$ could be the category of topological spaces and we could also suppose that $Y$ is locally compact, so that $X^Y$, if endowed with the compact-open topology, is an exponential object of $X$ to the power of $Y$.

Now in this scenario, the space $X^Y$ is nothing but a subset of the set $X^Y$ (that is, the continuous functions from $Y$ to $X$) with a topology on it, so that the underlying set happens to be a subset of the exponential object in the category $\mathbf{Set}$. Moreover, all the functions associated to this exponential object are merely restrictions of those given in the category $\mathbf{Set}$.

Now my question is:

Is there a special name for an exponential object of $X$ to the power of $Y$ which, when projected to $\mathbf{Set}$, is nothing but a subset of the set $X^Y$, together with the canonical evaluation map and the canonical assignment of the $\lambda$ function as explained in the Wikipedia article "Exponential object"?

答:
Yes. If the underlying set of your exponential object is just an exponential in $\mathsf{Set}$ and the evaluation maps get sent to evaluation maps in $\mathsf{Set}$, then we call the category Concretely Cartesian Closed (over $\mathsf{Set}$, if we want to be precise).

More generally, from The Joy of Cats (Chapter 27, Definition 11):

A concrete category $(\mathbb{A}, U)$ over $\mathbb{X}$ is called Concretely Cartesian Closed provided that the following hold:

$\mathbb{A}$ and $\mathbb{X}$ are cartesian closed.

$U$ preserves finite products, power objects, and evaluation; in particular, whenever $A \times B^A \overset{\text{ev}}{\longrightarrow} B$ is an evaluation in $\mathbb{A}$, then
$$U \left ( A \times B^A \overset{\text{ev}}{\longrightarrow} B \right ) = UA \times UB^{UA} \overset{\text{ev}}{\longrightarrow} UB$$
is an evaluation in $\mathbb{X}$.

Edit:

As commenters have noted, this definition is too restrictive. In fact, it excludes the example that you gave.
The authors of Joy of Cats actually mention that every concretely cartesian closed category over $\mathsf{Set}$ with a discrete terminal object is, up to concrete isomorphism, a full subcategory of $\mathsf{Set}$. So this is a very restrictive definition.

They give another definition (Chapter 27, Definition 17), which tries to
be more accommodating:

A construct $(\mathbb{A}, U)$ is said to have function spaces provided the following hold:

$(\mathbb{A}, U)$ has finite concrete products

$\mathbb{A}$ is cartesian closed and the evaluation morphisms $A \times B^A \overset{\text{ev}}{\longrightarrow} B$ can be chosen in such a way that $U \left ( B^A \right ) = \text{hom}_\mathbb{A}(A,B)$ and $\text{ev}$ is the restriction of the canonical evaluation map in $\mathsf{Set}$

This is still too restrictive for your example, though, since it requires all exponentials to exist. However, if we restrict attention to a "nice" subcategory of $\mathsf{Top}$ (for instance, compact hausdorff spaces) then we do have function spaces, and they agree with your example.

I haven't seen a definition which allows for only some exponential objects, but based on the language of "has function spaces", I think calling such an exponential a function space when it exists would be reasonable.

I hope this helps ^_^


问:
Complexity class of comparison of power towers
Consider the following decision problem: given two lists of positive integers $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_m$ the task is to decide if $a_1^{a_2^{\cdot^{\cdot^{\cdot^{a_n}}}}} < b_1^{b_2^{\cdot^{\cdot^{\cdot^{b_m}}}}}$.

Is this problem in the class $P$?

If yes, then what is the algorithm solving it in polynomial time?

Otherwise, what is the fastest algorithm that can solve it?

Update:

I mean polynomial type with respect to the size of the input, i.e. total number of digits in all $a_i, b_i$.

$p^{q^{r^s}}=p^{(q^{(r^s)})}$, not $((p^q)^r)^s$.

答:
My approach is similarly numeric to Leonid's, but more precise and perhaps easier to analyse. It supports real exponents > 0.

The idea is to represent power towers as a single floating point number with $n$ exponentiations: $(x\mid n) := exp^n(x)$. Normalizing $x\in[0,1)$, this format allows easy comparison between numbers.

What remains is a way to calculate $a^{(x\mid n)}$ for any real, positive $a$. My Python code below is an attempt to do so, while being as numerically stable as possibly, e.g. by using the log-sum trick. My code runs in time proportional to the height of the tower (number of apow calls) and the iterated-log of it's value (number of recursive calls).

I haven't been able to find two towers with values close enough to case my method to fail. At least for integer exponents. With fractional exponents it is possible to create very towers too close for my representation to handle. E.g. $$ 2^{2^{2^{2^0}}} \\
< \\
2^{2^{2^{2^{(1/2)^{2^{2^{2^2}}}}}}} $$

I would be interested in suggestions to other types of counter examples, especially integer ones.

It seems to me that for the problem to be in P, we need to non-numerical methods. It doesn't seem unlikely at all, that certain analytical cases are harder than P.

```<code>from math import log, exp

def normalize((x,n)):
  """ Adjusts n to put x in the range [0,1) """
  if x >= 1: return normalize((log(x),n+1))
  if x < 0:  return normalize((exp(x),n-1))
  return (x,n)
normdec = lambda f: lambda *a: normalize(f(*a))

@normdec
def apow(a,(x,n)):
  """ Calculates a^(x|n) """
  if a == 1: return (1,0)
  if a < 1: return (rpow(x,n)*log(a), 1)
  if n == 0: return (x*log(a), 1)
  if n >= 1:
    y, k = cpow(log(log(a)), x, n-1)
    return (y, k+2)

@normdec
def cpow(c,x,n):
  """ Calculates (x|n) + c """
  if c == 0: return (x, n)
  if n == 0: return (x + c, 0)
  z = rpow(x,n-1)
  if z <= log(abs(c)):
    return (exp(z)+c, 0)
  if c < 0: y, k = cpow(log(1-exp(log(-c)-z)), x, n-1)
  if c > 0: y, k = cpow(log(1+exp(log(c)-z)), x, n-1)
  return (y, k+1)

def rpow(x,n):
  """ Calculates (x|n) as a float
      Returs Infinty if the value is out of range"""
  try:
    for _ in range(n):
      x = exp(x)
  except OverflowError:
    # We get into this case in two situations
    # 1) We are calculating log((x|n) + c) and c contributes very little
    # 2) We are calculating a^(x|n) for a < 1 and (x|n) is so small it doesn't
    #    fit into a float
    return float('inf')
  return x

def powtow(bs):
  """ Calculates b[0]**b[1]**b[2]**...**b[m-1] in the (x|n) form.
      Equivalent to `foldr apow (1,0) bs'
      e.g. apow(b[0], apow(b[1], apow(b[2], (1,0)))) """
  if not bs: return (1,0)
  return apow(bs[0], powtow(bs[1:]))

if __name__ == "__main__":
  print powtow([1,2,3,4,5])
  print powtow([2,3,4,5])
  print powtow([5,4,3,2])
  print powtow([4,4,3,3,3])
  print powtow([3,3,3,3,3])
  print powtow([4,6,8,8,9])
  print powtow([2,2,5,2,7,4,9,3,7,6,9,9,9,9,3,2])
  print powtow([3,3,6,3,9,4,2,3,2,2,2,2,2,3,3,3])
  print powtow([2,3,2,3,5,8])
  print powtow([3,2,2,7,6,7])
  print powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2])
  print powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,16])
  print powtow([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9])
  print powtow([1.54090919967, 1.46228204461, 1.78555495826, 1.75545819035, 2.21730941808, 1.0797564499, 7.90630125423, 0.881978093585, 1.75085618709, 2.23911325176, 1.39697337886, 1.16053659586, 1.5939192079, 6.11401961748, 0.844860266481, 1.92758094038, 4.64573316954, 0.870819420274, 1.49026447511, 1.77839910981, 1.46208378213, 2.29956158055, 1.00884903003, 1.77521724246, 2])
  print powtow([2.32185478602, 1.88198918762, 2.27614315145, 1.77518235487, 1.4841479727, 0.563158971798, 0.732132856919, 0.669957968262, 2.16345101714, 2.23185963501, 0.824885385628, 0.873101580546, 1.45714899023, 2.3973000247, 0.507154709525, 1.94022843601, 1.29982267606, 0.578058713016, 1.58207655843, 1.79417433851, 1.18630377782, 1.37314328673, 0.655551609076, 1.57569812897, 1])
  print powtow([2, 2, 2, 1])
  print powtow([2, 2, 2, 2, .5, 2, 2, 2, 2])

  flip = lambda (a,b): (b,a)
  snd = lambda (a,b): b
  import itertools
  f = lambda xs: [1./x for x in xs]
  # Print all power towers of permutations [2..6], sorted
  print list(reversed(sorted((flip(powtow(p)),p) for p in itertools.permutations(range(2,7)))))
  # Print all power towers of permutations [1/2..1/6] sorted
  print list(reversed(sorted((flip(powtow(f(p))),p) for p in itertools.permutations(range(2,7)))))
</code>```

Examples:

```<code>powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2]) = (0.1184590219613409, 18)
powtow([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9]) = (0.10111176550354063, 18)

powtow([2,2,5,2,7,4,9,3,7,6,9,9,9,9,3,2]) = (0.10111176550354042, 17)
powtow([3,3,6,3,9,4,2,3,2,2,2,2,2,3,3,3]) = (0.19648862015624008, 17)
</code>```

Counter examples:

```<code>powtow([2,2,2,2,2,2,2]) = (0.8639310719129168, 6)
powtow([3,2,2,2,2,2,2]) = (0.8639310719129168, 6)
</code>```

Body:
My approach is similarly numeric to Leonid's, but more precise and perhaps easier to analyse. It supports real exponents > 0.

The idea is to represent power towers as a single floating point number with $n$ exponentiations: $(x\mid n) := exp^n(x)$. Normalizing $x\in[0,1)$, this format allows easy comparison between numbers.

What remains is a way to calculate $a^{(x\mid n)}$ for any real, positive $a$. My Python code below is an attempt to do so, while being as numerically stable as possibly, e.g. by using the log-sum trick. My code runs in time proportional to the height of the tower (number of apow calls) and the iterated-log of it's value (number of recursive calls).

I haven't been able to find two towers with values close enough to case my method to fail. At least for integer exponents. With fractional exponents it is possible to create very towers too close for my representation to handle. E.g. $$ 2^{2^{2^{2^0}}} \\
< \\
2^{2^{2^{2^{(1/2)^{2^{2^{2^2}}}}}}} $$

I would be interested in suggestions to other types of counter examples, especially integer ones.

It seems to me that for the problem to be in P, we need to non-numerical methods. It doesn't seem unlikely at all, that certain analytical cases are harder than P.

from math import log, exp

def normalize((x,n)):
  """ Adjusts n to put x in the range [0,1) """
  if x >= 1: return normalize((log(x),n+1))
  if x < 0:  return normalize((exp(x),n-1))
  return (x,n)
normdec = lambda f: lambda *a: normalize(f(*a))

@normdec
def apow(a,(x,n)):
  """ Calculates a^(x|n) """
  if a == 1: return (1,0)
  if a < 1: return (rpow(x,n)*log(a), 1)
  if n == 0: return (x*log(a), 1)
  if n >= 1:
    y, k = cpow(log(log(a)), x, n-1)
    return (y, k+2)

@normdec
def cpow(c,x,n):
  """ Calculates (x|n) + c """
  if c == 0: return (x, n)
  if n == 0: return (x + c, 0)
  z = rpow(x,n-1)
  if z <= log(abs(c)):
    return (exp(z)+c, 0)
  if c < 0: y, k = cpow(log(1-exp(log(-c)-z)), x, n-1)
  if c > 0: y, k = cpow(log(1+exp(log(c)-z)), x, n-1)
  return (y, k+1)

def rpow(x,n):
  """ Calculates (x|n) as a float
      Returs Infinty if the value is out of range"""
  try:
    for _ in range(n):
      x = exp(x)
  except OverflowError:
    # We get into this case in two situations
    # 1) We are calculating log((x|n) + c) and c contributes very little
    # 2) We are calculating a^(x|n) for a < 1 and (x|n) is so small it doesn't
    #    fit into a float
    return float('inf')
  return x

def powtow(bs):
  """ Calculates b[0]**b[1]**b[2]**...**b[m-1] in the (x|n) form.
      Equivalent to `foldr apow (1,0) bs'
      e.g. apow(b[0], apow(b[1], apow(b[2], (1,0)))) """
  if not bs: return (1,0)
  return apow(bs[0], powtow(bs[1:]))

if __name__ == "__main__":
  print powtow([1,2,3,4,5])
  print powtow([2,3,4,5])
  print powtow([5,4,3,2])
  print powtow([4,4,3,3,3])
  print powtow([3,3,3,3,3])
  print powtow([4,6,8,8,9])
  print powtow([2,2,5,2,7,4,9,3,7,6,9,9,9,9,3,2])
  print powtow([3,3,6,3,9,4,2,3,2,2,2,2,2,3,3,3])
  print powtow([2,3,2,3,5,8])
  print powtow([3,2,2,7,6,7])
  print powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2])
  print powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,16])
  print powtow([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9])
  print powtow([1.54090919967, 1.46228204461, 1.78555495826, 1.75545819035, 2.21730941808, 1.0797564499, 7.90630125423, 0.881978093585, 1.75085618709, 2.23911325176, 1.39697337886, 1.16053659586, 1.5939192079, 6.11401961748, 0.844860266481, 1.92758094038, 4.64573316954, 0.870819420274, 1.49026447511, 1.77839910981, 1.46208378213, 2.29956158055, 1.00884903003, 1.77521724246, 2])
  print powtow([2.32185478602, 1.88198918762, 2.27614315145, 1.77518235487, 1.4841479727, 0.563158971798, 0.732132856919, 0.669957968262, 2.16345101714, 2.23185963501, 0.824885385628, 0.873101580546, 1.45714899023, 2.3973000247, 0.507154709525, 1.94022843601, 1.29982267606, 0.578058713016, 1.58207655843, 1.79417433851, 1.18630377782, 1.37314328673, 0.655551609076, 1.57569812897, 1])
  print powtow([2, 2, 2, 1])
  print powtow([2, 2, 2, 2, .5, 2, 2, 2, 2])

  flip = lambda (a,b): (b,a)
  snd = lambda (a,b): b
  import itertools
  f = lambda xs: [1./x for x in xs]
  # Print all power towers of permutations [2..6], sorted
  print list(reversed(sorted((flip(powtow(p)),p) for p in itertools.permutations(range(2,7)))))
  # Print all power towers of permutations [1/2..1/6] sorted
  print list(reversed(sorted((flip(powtow(f(p))),p) for p in itertools.permutations(range(2,7)))))

Examples:

powtow([2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2]) = (0.1184590219613409, 18)
powtow([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9]) = (0.10111176550354063, 18)

powtow([2,2,5,2,7,4,9,3,7,6,9,9,9,9,3,2]) = (0.10111176550354042, 17)
powtow([3,3,6,3,9,4,2,3,2,2,2,2,2,3,3,3]) = (0.19648862015624008, 17)

Counter examples:

powtow([2,2,2,2,2,2,2]) = (0.8639310719129168, 6)
powtow([3,2,2,2,2,2,2]) = (0.8639310719129168, 6)


问:
Proving $\phi$ is continuous
In this video the lecturer gave an example, that is:

$\phi : M\to N$, where $M$ is equipped with arbitrary topology $\mathcal O_M$, and $N$ is equipped with the chaotic topology $\{\varnothing, N\}$. Then any $\phi$ is continuous.

It's clear that the preimage of $\varnothing$ must be $\varnothing$ and thus a open set in $M$, but why the preimage of $N$ is an open set in $M$?

答: The preimage $\phi^{-1}(N) = \{x \in M : \phi(x) \in N\}$ is equal to $M$ and $M$ is open by definition of a topology.

Title: Proving $\phi$ is continuous


问:
Is there a function that is monotonic increasing but not cyclically monotonic increasing?
Could someone provide an example of a function that is monotonic increasing but not cyclically monotonic increasing?

This question comes from reading the statement on page 2 just before Prop 3 here

答: Consider an $N\times N$ matrix $A$ and the function $f(x)=Ax$. For $f$ to be cyclically monotone, you need $f$ to be the gradient of a convex function, and so $A$ must be symmetric and positive semi-definite. For $f$ to be monotone, it is enough that $A+A^T$ be positive semi-definite.


问:
Conditional independence.
I am trying to understand the sentence, "X and Z are independent given Y". I think this means that $X|Y$ is independent of $Z|Y$. Is this the same thing as saying $X|(Y,Z)$ only depends on $Y$? This is basically saying that once $(Y,Z)$ takes on a value, we don't care about what $Z$ takes on since $X$ doesn't depend on $Z$ once $Y$ occurs. How do I prove this rigorously?

答:
If $X \perp \!\!\! \perp Z$ given $Y$ then yes that means $(X \mid Y) \perp \!\!\! \perp (Z \mid Y)$

(read $A \perp \!\!\! \perp B $ as 'A is independent of B'.)

Moreover we have $P(X \cap Z \mid Y) = P(X \mid Y) \cdot P(Z \mid Y)$

So if we additionally condition on $Z$ : $\displaystyle P(X \mid Z,Y) = \frac{P(X \cap Z \mid Y)}{P(Z)} = \frac{P(X \cap Z \cap Y)}{P(Z)P(Y)} = \frac{P(X \cap Y \mid Z)}{P(Y)}$

Note: this of course this requires that $P(Y), P(Z) > 0$

Interestingly enough in the case where we condition $X$ upon $Z$ and $Y$, it turns out it does not matter the order in which we condition.

Title: Conditional independence.


问:
Find the limit $\lim\limits_{x\to \infty}(e^{-x}+2\cos3x)$, or show that it does not exist.
I have calculus homework question that to be quite frank I don't even know how to begin to solve. Here is my attempt on how to to solve the question; however I'm not sure if I'm tackling this correctly. I'm hoping that someone here can tell me if I'm doing this correctly and if my answer would be correct.

Here is the question.

Find the limit or show that it does not exist:
  $$\lim_{x\to \infty}(e^{-x}+2\cos3x)$$

My attempt to the solution is DNE or does not exist simply because cosine oscillates between -1 and 1 and it never approaches any one number to reach a horizontal asymptote. Is this answer correct? Or how should I approach this problem if it's not? Or should it be
$$-\infty$$ since that would be the the value of $$e^{-x}\quad ?$$Please forgive my stupidity if I'm incorrect in all fronts.

答:
The accepted answer is incorrect.

You have $$\lim_{x\to\infty}f(x)+g(x)=\lim_{x\to\infty}f(x)+\lim_{x\to\infty}g(x)$$ when both $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ exist!

To PROVE that the limit does not exist you can do it as follows.

Assume that the limit exists and
$$
\displaystyle\lim_{x\to\infty}(e^{-x}+2\cos(3x))=L.
$$

Then, since $\lim_{x\to\infty}e^{-x}=0$, we conclude that
$$
L=\lim_{x\to\infty}(e^{-x}+2\cos(3x))-\lim_{x\to\infty}e^{-x}
=\lim_{x\to\infty}2\cos(3x)\tag{1}
$$
where we use the fact that whenever $\lim_{x\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ exist, so does $\lim_{x\to\infty}[f(x)+g(x)]$.

But (1) cannot be true by considering to different sequences $a_n=(2n+1)\pi/3$ and $b_n=2n(\pi)/3$.


问:
Why doesn't integrating the area of the square give the volume of the cube?
I had a calculus course this semester in which I was taught that the integration of the area gives the size (volume):

$$V = \int\limits_a^b {A(x)dx}$$

But this doesn't seem to work with the square. Since the size of the area of the square is $x^2$ then $A(x) = {x^2}$, then:

$$V = \int\limits_{ - r}^r {{x^2}dx}  = \left[ {\frac{{{x^3}}}{3}} \right]_{ - r}^r = \frac{{{r^3}}}{3} - \frac{{ - {r^3}}}{3} = \frac{2}{3}{r^3}$$

It's clear that this is not the volume of the cube. Why is this the case? Am I misunderstanding something?

答: The real problem here is not the endpoints of your integral, it's that the function you are integrating is not constant with respect to the variable of integration.  A cube has the same cross section everywhere, while in your original integral the cross section is bigger at the ends than in the middle.  See @response's solution for the right way to set this up.</p>


问:
Hitting time on a closed set of a continuous stochastic process is a stopping time.
Let $X$ be a stochastic process with continuous sample paths and the canonical filtration. $\Gamma\in {\cal B} (\mathbb{R}^n)$ be a closed set and let $T \triangleq \inf\{t\ge 0:X_t \in \Gamma\}$. Show that $T$ is a stopping time.

I understand the proof using $\Gamma_n = \{x:$dist$(x,\Gamma)<\frac{1}{n}\}$ and the fact $T_n = \inf\{t\ge 0: X_t \in \Gamma_n\}$ is an optional time. My question is why is the following method wrong or not obvious?

Using continuity, if $T(\omega) \le t$, then given any $\epsilon > 0$, $A =\{X_s(\omega):s \in [T(\omega), T(\omega) +\epsilon)\} \cap \Gamma\neq \emptyset$. Then there exists $\{s_n\}\subset [T(\omega), T(\omega)+\epsilon)$ non-increasing (not necessarily distinct) such that $s_n \rightarrow T(\omega)$ with $X_{s_n}(\omega) \in \Gamma$, and since $\Gamma$ is closed, we know $\lim_{n\rightarrow \infty}X_{s_n}(\omega) = X_{T(\omega)}(\omega)\in \Gamma$. Therefore, $\{T \le t\} = \bigcup_{0 \le s \le t}\{X_s \in \Gamma\}\in \bigcup_{0\le s \le t}{\cal F}_s^X = {\cal F}_t^X$.

The reason I am asking is that I think this method is more straight forward than the one I mentioned before, and I don't see this method anywhere online for this problem, so there must be some problem with it.

So, please let me know if this is correct. Thanks in advance.

答:
The issue with your argument is that
$$\bigcup_{0 \le s \le t}\{X_s \in \Gamma\}$$ is an uncountable union and a priori may not be in $\mathcal{F}_t$. If you tried replacing the union with one over a countable subset of $[0,t]$ you may find that $X$ hits your set only at a single point $s$ that is not in your countable set, so the argument cannot be fixed this way either. Indeed, consider $X(t) = t$ deterministic and $\Gamma = \{x\}$ to be a single point.


问:
What does it mean when there is an equals line through an angle?

What does this line through the angle mean?

答:
That kind of stroke at an angle might be a way to say that angle is the same as some other angle (not shown). It would be analogous to the single and double strokes through the segments HE and HY. If there were another segment elsewhere with two strokes it would be the same length as HE.

To know whether this is what the stroke on the angle means we would need to see much more context.


问:
Differentiability of $f(x)$ at $0$ provided that $\lim_{x\to0} [f(2x)-f(-x)]/x$ exists?
Assume that $f:\mathbb{R}\to\mathbb{R}$ is continuous at $x=0$, and the limit
\begin{equation}
\lim_{x\to0}\frac{f(2x)-f(-x)}{x}
\end{equation}
exists. Is $f$ necessarily differentiable at 0?

I was able to deduce that if $f$ is continuous in a neighbordhood of $0$ and has an isolated zero at $x=0$, it must be differentiable (unless I made an error). I believe that this result holds if there exists some $\delta>0$ such that $f(x)$ does not change sign in the neiborhoods $-\delta<x<0$ and $0<x<\delta$.

But for the life of me I can't figure out a general proof (or a counterexample).

Any help would be greatly appreicated!

答:
Let $ \varepsilon >0 $, there exists some $ \eta >0 $ such that $ \left(\forall x\in\left(-\eta,\eta\right)\right),\ \left|f\left(2x\right)-f\left(-x\right)-\ell x\right|\leq\varepsilon\left|x\right| $.

Let $ n\in\mathbb{N}^{*} $, we have, for all $ x\in\left(-\eta,\eta\right) $ : \begin{aligned}\small\left|f\left(x\right)-f\left(0\right)-\frac{\ell}{3} x\right|&\small=\left|\sum_{k=0}^{n-1}{\left(f\left(\frac{\left(-1\right)^{k}}{2^{k}}x\right)-f\left(\frac{\left(-1\right)^{k+1}}{2^{k+1}}x\right)-\ell\frac{\left(-1\right)^{k}}{2^{k+1}}x\right)}+f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\sum_{k=0}^{n-1}{\left|f\left(\frac{\left(-1\right)^{k}}{2^{k}}x\right)-f\left(\frac{\left(-1\right)^{k+1}}{2^{k+1}}x\right)-\ell\frac{\left(-1\right)^{k}}{2^{k+1}}x\right|}+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\sum_{k=0}^{n-1}{\varepsilon\left|\frac{\left(-1\right)^{k}}{2^{k+1}}x\right|}+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\varepsilon\left(1-\frac{1}{2^{n}}\right)\left|x\right|+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\end{aligned}

Giving first some clarifications about what we did, we introduced first a telescopic sum combined with a geometric sum, then we applyied our hypothesis for $ y=\frac{\left(-1\right)^{k}}{2^{k+1}}x $, because whatever $ k $ is, $ y $ will always be in $ \left(-\eta,\eta\right) $, after that we simplified the geometric sum that we obtained to end up with the result.

How we've assumed that $ f'\left(0\right) $, if exists, must be $ \frac{\ell}{3} $ ? Well to get it, we must take look at the following, for $ x\in\left(-\eta,\eta\right) $ : $$ \frac{f\left(2x\right)-f\left(-x\right)}{x}=2\times\frac{f\left(2x\right)-f\left(0\right)}{2x}+\frac{f\left(-x\right)-f\left(0\right)}{-x} $$

Taking $ n $ to $ +\infty $, we end with : $$ \left|f\left(x\right)-f\left(0\right)-\frac{\ell}{3}x\right|\leq\varepsilon\left|x\right| $$

Which means $ f $ is differentiable at $ 0 $.


问:
Find all of the distinct non-planar graphs with 6 vertices.
Find all of the distinct non-planar graphs with 6 vertices.  Draw a picture of each such graph, and explain why the graphs you have found are the only possibilities.

HINT: Proceed by cases, considering the number of edges in each possible graph.

答:
Hint In any planar graph with 6 vertices you have

$$e \leq 3v-6=12 .$$

This proves that any graph with at least 13 edges is non-planar.

All you need to do now, is check the graphs with 12 or less edges.

Since $K_5$ has $10$ edges and $K_{3,3}$ has $9$ edges, any non-planar graph has at least 9 edges.

9 edges: Your graph must be $K_{3,3,}$ (Why?)

10 edges: Your graph must be $K_{3,3,}$ with an extra edge or $K_5$ with an isolated vertex. (Why?)

11 edges and 12 edges: If your graph contains $K_{3,3}$, it is easy: it must be $K_{3,3}$ with enough extra edges.

For $K_5$ you need to look at two different situations: your graph contains $K_5$ as a subgraph, or the sixth vertex is the vertex of degree $2$ in a subdivision of $K_5$.


问:
How to calculate "undo" rotation except for parallel rotation
Let's say we start with a vector $u = [0, 0, 1]$ which represents the positive z-axis.

Now, let's say we arbitrarily transform this vector via a rotation $z' = R \begin{bmatrix}0 \\ 0\\ 1 \end{bmatrix}$.

There are arbitrarily many rotations that can lead to the same $z'$.

I want to calculate a new rotation from $R$, $R_{undo}$, which inverts $R$ except it does not invert any portion of the rotation corresponding to a planar rotation around $z'$ (e.g. $z'$ is the normal vector of the plane of rotation).

For example, let's say in Euler angles with order zyx, $R = [90, 90, 0]$. Then, using the same zyx convention, $R_{undo} = [0, -90, 0]$, not $[-90, -90, 0]$.

How do we calculate $R_{undo}$ in general, for an arbitrary axis u?

答:
It seems to me that your $R_{\text{undo}}$ is not very well defined, and that we should assume everything is in $\mathbb{R}^3$.
From your example, if $z'=Ru$ for some rotation $R$, it looks like you want $R_{\text{undo}}$ to be the unique rotation that maps $z'\mapsto u$ and fixes the plane spanned by the vectors $z'$ and $u$ (take a moment to think about what this means geometrically). But note that if $z'=-u$, it is not at all clear what $R_{\text{undo}}$ should do -- the vectors won't span a plane, and there is no "prefered direction" to bring $z'$ back to $u$ anymore.

So let's put this problem aside, and assume $z'\neq-u$. Then you can calculate $R_{\text{undo}}$ following the procedures in the answers to this question, which looks very much like yours:

Calculate Rotation Matrix to align Vector A to Vector B in 3d?


问:
Is there a one-to-one corrospondence between integer functions and the reals?
Assuming $\Bbb{Z}^\Bbb{Z}$ to mean the set of all functions from $\Bbb{Z}$ to $\Bbb{Z}$, are $\Bbb{Z}^\Bbb{Z}\cong\Bbb{R}$?

Another way to put it, assuming a set of functions $Z:=\{f:\Bbb{Z}\rightarrow\Bbb{Z}\}$, does there exist a bijective function $g:\Bbb{Z}^\Bbb{Z}\rightarrow\Bbb{R}$ that assigns a real number for every element of $Z$?

In essence, is the following statement true:

"for any function $h\in Z$, there exists one, and only one, real number $r$, such that $g(h)=r$ and $g^{-1}(r)=h$."

答:
To follow GEdgar's hint, consider $r\in\mathbb{R}$. Then $r$ can be uniquely written as
$$
r=\pm a.a_1a_2a_3\cdots
$$
where $a$ is a positive integer and $a_k\in\{0,1,2,\cdots,9\}$ such that $a_k$ is not "eventually" $9$. This gives an injective map from $\mathbb{R}$ to $\mathbb{Z}^{Z}$.

For the other direction, consider instead $\mathbb{N}^{\mathbb{N}}$. Given any sequence of natural numbers $q_n$, it represents a "unique" real number
$$
r=a_0.a_1a_2a_3\cdots
$$
except for countably many sequences that are "eventually" $9$.

Now you only needs to show $\mathbb{N}^{\mathbb{N}}\cong \mathbb{Z}^{\mathbb{Z}}$.


问:
Fraction field commutes with extension of scalars?
Let $K$ be a field, $A$ a $K$-algebra, and $L$ an extension of $K$. Denote $B:= A \otimes_K L$, the extension of scalars of $A$, and assume that $B$ is an integral domain (and therefore $A$ too). Under what conditions do we have
$$B_{(0)}\cong A_{(0)} \otimes_K L,$$
where the index ${(0)}$ denotes taking the fraction field? Or, more in general, what is the relation between $B_{(0)}$ and $A_{(0)} \otimes_K L$? It seems to be quite a specific situation that relies on $L$ being a field and on $B$ being an integral domain, but at the same time it looks very natural to consider this question.

答:
Assuming as KReiser points out in the comments that $A, K, L$ have been chosen so that $B$ remains integral, which is not automatic, this is still very rarely true. A simple example is $A = K[x], B = L[x]$, which is about as nice as it gets, with fraction fields $K(x), L(x)$. There's a natural injection

$$K(x) \otimes_K L \to L(x)$$

and it is never a surjection unless $L = K$. We can see this as follows: if $L \neq K$ then there is some $\alpha \in L$ not in $K$, and $\frac{1}{x - \alpha}$ is not in the image of $K(x) \otimes_K L$ because it cannot be written as an element of $L$ times a quotient $\frac{p(x)}{q(x)}, p, q \in K[x]$. (It is perhaps easier to split into the cases that $\alpha$ is algebraic and $\alpha$ is transcendental: in the transcendental case $x - \alpha$ does not divide any polynomial in $K[x]$, and in the algebraic case $\alpha$ has some minimal polynomial with other roots besides $\alpha$, so the partial fraction decomposition of $\frac{p(x)}{q(x)}$ over $\overline{K}$ necessarily contains $\frac{1}{x - \alpha'}$ where $\alpha'$ is one of these other roots and hence cannot contain $\frac{1}{x - \alpha}$ alone.)

Edit, 10/6/22: This argument is incorrect as Eoin points out in the comments. The image of $K(x) \otimes_K L$ can be described exactly as $\frac{L[x]}{K[x]}$, by which I mean the set of rational functions that can be expressed as the quotient of an element of $L[x]$ by an element of $K[x]$. If $L$ is algebraic over $K$ then every polynomial in $L[x]$ divides a polynomial in $K[x]$ so this is all of $L(x)$ and the above map is an isomorphism.

To get a counterexample we need to assume that $L$ is transcendental over $K$ and then if $\alpha \in L$ is transcendental over $K$ then $\frac{1}{x - \alpha}$ is not in the image of the above map since $x - \alpha$ does not divide any polynomial in $K[x]$ by hypothesis.


问:
How to find the period of this sinusoid?
I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.

$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$

I would like to turn the product into a sum and do something about that $2$ in the exponent of $\sin$

$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3}) = \cos(\frac{4\pi t}{5})\left[ 1 - \cos(\frac{16 \pi t}{3}) \right]\frac{1}{2} $$

$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \cos(\frac{4\pi t}{5}) \cos(\frac{16 \pi t}{3})\right]$$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \frac{1}{2}\cos(\frac{4\pi t}{5} + \frac{16 \pi t}{3}) \cos(\frac{16 \pi t}{3} - \frac{4\pi t}{5})\right]$$

So the sum of cosines will have a period that is the LCM of of the periods of the individuals terms we added. The angular frequencies are $\frac{4\pi}{5}, \frac{92\pi}{15}, \frac{68\pi}{15}$.

Converting this to periods -> we get $\frac{5}{2}, \frac{15}{46}, \frac{15}{34}$, but the answer is $\frac{30}{4}$

答: You get 7.5 if you simply apply the LCM to $\frac{4\pi t}{5} = 2\pi$ (period = 2.5) and $\frac{8\pi t}{3} = 2\pi$ (period = 0.75).


问:
Prove a well-known inequality using Cauchy-Schwarz or AM-GM
For $a;b>0$ and $ab \geq 1$ we have a well-known inequality:

$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2} \geq \dfrac{2}{1+ab}$

Which is equivalent to: $\dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)} \geq 0$ (true)

But is there a solution for this inequality by Cauchy-Schwarz or AM-GM? Thank you.

答:
Again not AM-GM or C-S (unless you count $a^2+b^2\geq2ab$) but I thought this was kinda sweet:
\begin{align*}
\frac{1}{1+a^2}+\frac{1}{1+b^2}=1-\frac{(ab)^2-1}{1+a^2+b^2+(ab)^2}&\geq1-\frac{(ab)^2-1}{1+2ab+(ab)^2}\\
&=1-\frac{ab-1}{ab+1}\\
&=\frac{2}{1+ab}.
\end{align*}


问:
Quick Area of Cross Section Question
"The base of a solid is the region in the first quadrant between the graph of $y=2x$ and the $x$-axis for $0\leq1\leq2$. For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the $x$-axis and one radius in the $xy$-plane. What is the volume of the solid?"

I guess I'm a bit confused about both the wording and where I'd have to go in this question, I understand how to find the area of a cross-section when these cross-sections might be a square or a rectangle, but I'm a bit lost when it comes to circles as well. Any help would be appreciated!

答: Is this the volume you're defining?

Title: Quick Area of Cross Section Question

Body: Is this the volume you're defining?


问:
Splitting a single integral
After performing a dot product inside the integral this result was obtained. $P_1$ and $P_2$ are points in 3D space.

$$\int_{P_1}^{P_2} \textbf{A} \cdot \textbf{dl} = \int_{P_1}^{P_2} (xy \ dx + (3x-y^2) dy) = \int_{P_1}^{P_2} xy \ dx + \int_{P_1}^{P_2} 3x-y^2 \ dy  $$

What rule was used to split this integral into two integrals in the last step? What confuses me is that there are two variables that produce two differential elements dx and dy that I dont know how to deal with during this split of the integral. In normal case I would use this:

$$ \int_a^bf(x)+g(x)dx = \int_a^bf(x) \ dx + \int_a^bg(x) \ dx  $$

答:
It does not make sense to write $\int_a^b$ for a line integral (of a vector field) unless you are considering specifically the points $(a,0)$ and $(b,0)$ on the $x$-axis and the line segment between them. In this case, you simply have $\int_a^b (3x-y^2)\,dy=0$.

Another possible interpretation is that you consider specific parameterization $\gamma:[a,b]\to\mathbb{R}^2$ and interpret your line integral as
$$
\int_\gamma p(x,y)dx+q(x,y)dy=\int_a^b p(\gamma(t)),q(\gamma(t)))\cdot \gamma'(t)\,dt
$$

Okay, according to your edit, you mean the first case. Now you simply have
$$
\int_P^Q F(r)\cdot dr=\int_P^Q F_1(x,y,z)dx+F_2(x,y,z)dy+F_3(x,y,z)dz.
$$
where $F=(F_1,F_2,F_3)$.

In your example you have
$$
F_1(x,y,z)=xy,\quad
F_2(x,y,z)=3x-y^2,\quad
F_3(x,y,z)=0
$$

Now note that
$$
F=F_1e_1+F_2e_2+F_3e_3
$$
and
$$
\begin{align}
\int_P^Q F\cdot dr
&= \int_P^Q (F_1e_1+F_2e_2+F_3e_3 )\cdot dr\\
&=\int_P^Q F_1 e_1\cdot dr+
\int_P^Q F_2 e_2\cdot dr+
\int_P^Q F_3 e_3\cdot dr\\
&=\int_P^Q F_1(x,y,z) dx+
\int_P^Q F_2(x,y,z) dy+
\int_P^Q F_3(x,y,z) dz
\end{align}
$$

Title: Splitting a single integral