问:
Multiset identity.
How does one get to the following combinatorial identity: $$\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{i}=\binom{n+k-1}{k}$$

I'm well aware of the definition of a multiset, as well as of the derivation of cardinalities of such. The closest one I came across is a special case of the Chu-Vandermonde identity, namely for any integers $i,k$ and $n$ satisfying $0\leq i\leq k \leq n$ the following holds true $$\sum_{m=0}^n\binom{m}{i}\binom{n-m}{k-i}=\binom{n+1}{k+1}$$

Any help is appreciated. Thanks in advance.

Title: Multiset identity.

Body:
$$
\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{i}
=\sum_{i=0}^{k-1}\binom{n}{n-i-1}\binom{k-1}{i}
=\binom{n+k-1}{n-1}
=\binom{n+k-1}{k}
$$


问:
Densely-defined unbounded operators in Hilbert spaces
I've found the following exercise about densely-defined unbounded operators in Hilbert spaces quite challenging. Here's the text:

Let T : D(T) $\subset$ H $\rightarrow$ $\mathbb{C}$ be a linear operator such that its closure $\overline T$ is densely defined in H. Show if T is densely defined in H.

I tried to prove this by absurd but I'm not sure my thinking is correct.

Let's assume that T and its closure are densely defined in H (and therefore T is closable). Then, if $T \subset$ $\overline T$ it follows that $(\overline T)^{*} \subset T^{*}$. By hypothesis, T is dense in H so I can use the following theorem which states that:

If T : D(T) $\subset$ H $\rightarrow$ H is densely defined in H then: 1) $T^{*} = \overline{T^*}$ 2) T is closable if and only if $D(T^{*})$ is dense in H and $T \subset \overline T = T^{**}$

to prove that $T^{*}$ is dense in H. Furthermore, being $\overline T$ densely defined in H, we can conclude that $T^{**}$ is densely defined in H as well.

But then by a corollary of this theorem we have that if $T, T^*$ and $T^{**}$ are densely defined in H, then $T^* = {\overline{T}}^{*} = \overline{T^*} = T^{***}$.

However, this implies that $(\overline T)^{*} = T^{*}$.
I reasoned as follows: if T is extended (but not equal, by hypothesis we do not know if T is closed) by its closure, then $(\overline T)^*$ is extended by $T^*$ which equals $\overline {T^*}$. However, what I conclude is that: $(\overline T)^* = \overline{T^*}$ which means that simultaneously $(\overline T)^* \subset \overline{T^*}$ and $\overline {T^*} \subset( \overline T)^*$. But then, $\overline T \subset T$ which implies T and its closure must be equal which is absurd since by hypothesis $T \subset \overline T$.

Note: the notation I used $T \subset \overline T$ means that $\overline T$ extends T. $T^*$ does mean the adjoint of T.

I believe that the question here is subtle: I've just proved that the if and only if implication doesn't hold in this case. It is clear to me that T being densely defined in H does not imply that its extension $\overline T$ is dense in H (roughly saying: I don't know if the closure behaves well). In spite of this, I've encountered some difficulties in proving (or disproving) the vice-versa. Thus, if anyone could tell me whether the reasoning above is correct (or at least whether I'm on the right path towards the solution) it would be extremely appreciated.

答:
$\overline{D(T)}$  is dense in $H$

Let $x \in D(\overline{T}) \subset H$. Since, by hypothesis, $T$ is closable $\overline{G(T)}$ is the graph of a closed operator ($\overline T$): $\overline{G(T)} = G(\overline{T})$.

It follows that: $\exists$ {${x_n}$} $_{n \in \mathbb{N}}$ such that ${x}_n \in D(T)$ and $\lim_{n \rightarrow \infty}$ $(x_n, Tx_n) = (x,y) \in \overline{G(T)} = G(\overline{T})$. Therefore, $y = \overline{T}x$.

As a consequence, $\lim_{n \rightarrow \infty}$ $x_n = x \in D(\overline{T})$. Since - by definition of closure - $\overline{D(T)}$ contains the limit points of all succession in $D(T)$ we can conclude that: $x \in \overline{D(T)}$.
Thus, $D(\overline{T}) \subset \overline{D(T)}$ and being $\overline{T}$ densely defined in $H$, it follows that $\overline{D(T)}$ is dense in $H$
as well.

$D(T)$ is dense in $H$

$D(T)$ dense in $H$ means that: $\forall \phi \in H$ and $\forall \epsilon > 0$ $\exists \tilde{\phi} \in D(T)$ such that: $|\phi -\tilde{\phi}| < \epsilon$

$\overline{D(T)}$ dense in $H$: $\forall \phi \in H$ $\forall \epsilon > 0$ $\exists x \in \overline{D(T)}$ such that: $|\phi -x| < \epsilon /2$

$\overline{D(T)}$ is the closure of $D(T)$: $x \in \overline{D(T)} \subset H$ then $\forall \epsilon > 0$ $\exists \tilde{\phi} \in D(T)$ such that: $|x - \tilde{\phi}| < \epsilon /2$

Therefore:
$|\phi - \tilde{\phi}| = |\phi - x + x - \tilde{\phi}| \le |\phi - x| + |x - \tilde{\phi}|< \epsilon$
which proves the thesis.

ContentLicense: CC BY-SA 4.0


问:
Isomorphism between the Tangent Space and Euclidean Space
I attempt to understand the invariance of dimension of topological manifolds from Loring Tu's An Introduction to Manifolds (Second Edition, page no. 89). The corresponding section of the book is given below.

My Question

In the proof of Corollary 8.7, it is claimed that the tangent space $T_p(U)$ at point $p \in U \subset \mathbb{R}^n$ is isomorphic to the Euclidean $n$-space $\mathbb{R}^n$. How do we know that this is the case? In other words, what is the proof of this claim?

答:
Let $A$ be an affine space with translation space $V$. That is, $A$ is a set whose elements are points, and we have an operation $+:A\times V \to A$ such that $p+0=p$ for all $p \in A$, we have $(p+v)+w=p+(v+w)$ (where the last $+$ is addition in $V$) and given $p,q \in A$, there is a unique $v\in V$ such that $q=p+v$ (people write $v = q-p$).

Once $O \in A$ has been chosen, we have a bijection $V \to A$ given by $v\mapsto O+v$, and so every structure $V$ has (vector space structure, topology, manifold structure, etc) is transferred to $A$. A different choice of $O$ gives an isomorphic vector space structure, same topology, and same manifold structure.

For every $p \in A$, we have $T_pA =V$ in a natural way. For example, if $$T_pA = \{\alpha'(0)\mid \alpha:(-\epsilon,\epsilon)\to A \mbox{ with }\alpha(0)=p\},$$one can write $$v = \frac{\rm d}{{\rm d}t}\bigg|_{t=0}p+tv.$$If you define $T_pA$ as the space of derivations of germs of smooth functions at $p$, then the isomorphism is $$v\mapsto \left([f]_p \mapsto \frac{\rm d}{{\rm d}t}\bigg|_{t=0}f(p+tv)\right),$$and so on.


问:
$\frac{d}{dx}(\sin(x^{\frac{1}{3}}))$ from first principle
The question contains a hint: at the appropriate point use the result $a^{3} - b^{3} =
\left(a - b\right)\left(a^{2} + b^{2} + ab\right)$.
$$
\frac{{\rm d}}{{\rm d}x}\sin\left(x^{1/3}\,\right)
$$

My attempt:

I did not use the hint as it was not immediately obvious to me what the simplification was. Instead I wrote with binomial expansion that $(x+h)^{\frac{1}{3}}=x^{\frac{1}{3}}(1+\frac{h}{x})^{\frac{1}{3}} \approx x^{\frac{1}{3}}(1+(\frac{1}{3})\frac{h}{x})$, used $\sin(()+())=\sin()\cos()+\sin()\cos()$ and applied $\cos() \approx 1$, $\sin() \approx ()$ for small $h$ which gave the correct answer.

But I do not see the point of using the cubes difference formula given in the question? Could this be a typo because it would make more sense to apply the identity to something like e.g. $\sin({x})^{\frac{1}{3}} (\text{not }\sin(x^{\frac{1}{3}}))\to \sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}} \iff (\sin(x+h)-\sin(x)=(\sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}})(\sin(x+h)^2+\sin(x)^2+\sin(x+h)\sin(x))$

Or am I missing something here?

Body:
You have

$$\begin{aligned}
\frac{\sin(x^{\frac{1}{3}}) - \sin(c^{\frac{1}{3}}) }{x-c} &= \frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{x-c}\\
&=\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{(x^{\frac{1}{3}} - c^{\frac{1}{3}})(x^{\frac{2}{3}} + c^{\frac{2}{3}} + x^{\frac{1}{3}}c^{\frac{1}{3}})}
\end{aligned}$$

Now as $$\lim\limits_{x \to c }\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)}{x^{\frac{1}{3}} - c^{\frac{1}{3}}} = 1$$

we get $$\frac{d}{dx}(\sin(x^{\frac{1}{3}})) = \frac{\cos(x^{\frac{1}{3}})}{3 x^{\frac{2}{3}}}$$


问:
Does the MIP*=RE presentation imply that the Turing Halting Problem is solvable?
I noticed an article where the author seemed to imply the halting problem was solved so I found the paper he was referencing but it is over my level of knowledge.

Was hoping someone in the community would be so kind to make a brief comment. The first link below is the article in quanta magazine and the second link is the article that is under discussion.

https://www.quantamagazine.org/landmark-computer-science-proof-cascades-through-physics-and-math-20200304/

https://arxiv.org/abs/2001.04383

答:
No, it doesn't.

What it does do is reduce the halting problem to a different type of problem, which on the face of it doesn't seem particularly related. But this doesn't mean that the halting problem is solvable, only that it's solvable relative to a different problem (which consequently is itself unsolvable).

There is in fact a rich structure of "degrees of unsolvability," more commonly called the Turing degrees. It's worth noting that in contrast with complexity theory where many-one reductions are standard, the default notion of reducibility in the context of computability theory is Turing reducibility. To see the difference, note e.g. that we trivially have a polynomial-time Turing equivalence between $\mathsf{NP}$ and $\mathsf{coNP}$ while it's open whether they are polynomial-time many-one equivalent.


问:
Möbius inversion formula for two functions f(x) and g(x)
Given the 2 functions $$ g(x)= \sum_{n=1}^{\infty}f\left(\frac{x}n\right)\log(n)\;, $$ how can I use Möbius inversion to recover $f$ from $g$?? I believe that

$$ f(x)=  \sum_{n=1}^{\infty}\mu (n)g\left(\frac{x}n\right)\log(n)\;. $$ Here 'mu' is the Möbius function.

答:
I was re-reading through some old posts on this site. I want to respond to Gerry's concern given in the comment from above in more complete detail. For the series Gerry cites using $g(x) = x^{10}$, we have (by uniformly convergent series analysis, and differentiation of this formal series) that
$$g(x) = -\zeta^{\prime}(10) \cdot x^{10}.$$

Fair enough, here the exact $|\zeta^{\prime}(10)| \approx 0.000697033$, indeed a small constant factor. Let's take $C_0 := -\zeta^{\prime}(10)$. By similar reasoning on the second series expansion that results for $f(x)$, we see that

$$f(x) = \sum_{n \geq 1} \mu(n) g(x/n) \log n = C_0 \cdot x^{10} \times \frac{d}{ds}\left[\frac{1}{\zeta(s)}\right]\Biggr\rvert_{s=10}.$$

By formally computing the last derivative of the reciprocal of the Riemann zeta function with respect to $s$, we have that $f(x) = x^{10}$, e.g., the former notation does in fact yield $CD = 1$ in this case.

Note that we have used the property that the Dirichlet series over the Moebius function satisfies $$\sum_{n \geq 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}, \forall \Re(s) > 1.$$

ContentLicense: CC BY-SA 3.0


问:
generate any random integer
I apologize in advance as I am not very experienced with any formal notion of randomness.

The title says most of it: I want to generate a random integer within a reasonable time, where every integer can appear, whether with equal frequency or not isn't important. As an add on, computer memory is not an issue, as even with infinite memory space to store these generated numbers it isn't obvious how one could do this. I haven't made any progress in actually figuring out a proper algorithm but here are my observations.

If you can generate any real number randomly then you could use functions like the floor function to generate any integer. If you could randomly generate any real number between any interval $[a,b]$, then you could use asymptotic functions like $\tan$ to generate any real number.

In general if I have a set S which has a larger or equal cardinality to the integers, and I can randomly generate an element within S, then I can randomly generate any integer by mapping members of S onto the integers.

I know that there are sequences, such as the prime gap sequence, which are random and contain arbitrarily large integers, but are not computable easily.

However that is about it with regards to what I can think of. I would not be surprised if there was no easy solution to the problem, but if anyone has a reason as to why this is not possible I would like to hear as well.

答:
The arbitrary size has no meaning since the computation cannot halt!

Consider that you toss a coin for every bit of the random integer, then you can see that the coin tossing is never-ending.

One should be careful when playing with the arbitrary size. Mathematically you can say that let $x$ be a random integer, i.e. $x \stackrel{R}{\leftarrow} \mathbb Z$ , however, when you try to find a value of this you will face the generation of it. If you want a uniform random integer then obviously it will fail!

Now assume that you have a limit like $0\color{red}{<} x \leq 2^L$ then you can use LFSR's to generate random numbers in the range. If an LFSR with size $L$ is maximal then it is periodic and it has a period of $2^L-1$. In this period it visits all possible $L$-bit numbers except the all-zero state. You can get a seed from the time and start using it.

Note that LFSR is far from away from being a Cryptographically Secure Random Pseudo Generator (CSPRNG). Having just $2L$ bit output from the LFSR is enough to determine the next bits due to the Berlakamp-Massay algorithm - and actually, Gaussian elimination is enough, however, BM is a lot faster-.

Title: generate any random integer


问:
Showing that degree of polynomial reduced modulo $p$ divides degree of splitting field
Let $f$ be a polynomial of degree $n$ with integer coefficients and $p$ a prime for which $f$, considered modulo $p$, is a degree-$k$ irreducible polynomial over $\mathbb{F}_p$. Show that $k$ divides the degree of the splitting field of $f$ over $\mathbb{Q}$.

(this is Miklos Schweitzer 2020, Problem 10)

My miscellaneous thoughts: We know that for a polynomial of degree $n$, the degree of its splitting field is at most $n!$ from induction on the Ring Tower Theorem. Additionally, since $f(x)$ is irreducible over $\mathbb{F}_p$, then the field $\mathbb{F}_p[x]/(f(x))$ has a root of $f(x)$. Now suppose $k = 2$. Then $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in \mathbb{Z}[x]$ reduces to $f(x) \equiv b_2 x^2 + b_1x + b_0 \in \mathbb{F}_p [x]$ where $b_i \equiv a_i \pmod{p}$ where applicable. I'm really not sure if these things are particularly useful or how to combine them.

This is really hard!

答:
If $p\mid f(0)$ then $\overline f(x)=cx$ (using the notation from the previous posts) for some $c\in\mathbb F_p^\times$, and thus $k=1$. So let's assume $p\nmid f(0)$ from now on. Let $x_1,\ldots,x_n$ be the roots of $f(x)$ in $K$ and let * denotes the reciprocal polynomial. Then $f^*(x)$ has degree $n$ (since $f(0)\ne0$) and roots $1/x_1,\ldots,1/x_n$. Furthermore, we have $f(0)/x_1,\ldots,f(0)/x_n\in\mathcal O_K$ since they are roots of the monic polynomial $g(x):=f(0)^{n-1}\cdot f^*(x/f(0))\in\mathbb Z[x]$. Let $\mathfrak P$ be a prime ideal of $\mathcal O_K$ above $p$, then since $g(x)$ splits in $\mathcal O_K$, $\overline g(x)$ splits in $\mathcal O_K/\mathfrak P$. We have $\overline{f^*}(x)=x^{n-k}\cdot\overline f^*(x)$, and since $p\nmid f(0)$, $\overline f^*(x)$ has degree $k$ and is irreducible over $\mathbb F_p$, and $\overline g(x)=\overline{f(0)}^{k-1}x^{n-k}\cdot\overline f^*\big(x/\overline{f(0)}\big)$. The polynomial $\overline f^*\big(x/\overline{f(0)}\big)$ is also irreducible over $\mathbb F_p$ of degree $k$, and thus the splitting field of $\overline g(x)$ is just $\mathbb F_p[x]\big/\big(\overline f^*\big(x/\overline{f(0)}\big)\big)$ and
$$k=\big[\mathbb F_p[x]\big/\big(\overline f^*\big(x/\overline{f(0)}\big)\big):\mathbb F_p\big]\mid\big[\mathcal O_K/\mathfrak P:\mathbb F_p\big]\mid\big[K:\mathbb Q\big].$$


问:
Equation using $f(x)=\frac{\sin\pi x}{x^2} $
If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$

Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.

Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.

Then which of the options is(are) correct.

(A) $x_1<y_1$

(B) $x_{n+1}-x_n>2$ for all $n$

(C) $x_n \in (2n,2n+\frac{1}{2})$ for every $n$

(D) $|x_n-y_n|>1$ for every $n$
The official answer is $B,C,D$

My approach is as follow

$$y' = \frac{x^2 \times \pi  \times \cos \pi x - 2x\sin \pi x}{x^4}$$

$$y' = \frac{x^2 \times \pi  \times \cos \pi x - 2x\sin \pi x}{x^4} = \frac{x \times \pi  \times \cos \pi x - 2\sin \pi x}{x^3} = \frac{2\cos \pi x \left( \frac{\pi x}{2} - \tan \pi x \right)}{x^3}$$

We should consider $x\in\frac{1}{2},\frac{3}{2},\ldots.$ as inflection point even though $\tan\pi x$ is not defined because we need to consider the numerator part $\pi x\cos \pi x - 2\sin \pi x$

答:
$$f'(x)=\frac{\pi  x \cos (\pi  x)-2 \sin (\pi  x)}{x^3}$$
$$f'(x)=0\to \sin(\pi x)=\pi x\cos(\pi x)\tag{1}$$
Second derivative is
$$f''(x)=-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi  x)+4 \pi  x \cos (\pi  x)}{x^4}$$
To understand which are max/min we use second derivative test using relation $(1)$ in the second derivative
$$f''(x^*)-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi  x)+4 \sin(\pi x)}{x^4}=-\frac{\left(\pi ^2 x^2+4 \pi -6\right) \sin (\pi  x)}{x^4}$$
As the factor $\left(\pi ^2 x^2+4 \pi -6\right) $ is positive for any $x$ we have that second derivative is negative when $\sin(\pi x)>0$ that is $2k\pi <\pi x<(2k+1)\pi$ $\forall k\in\mathbb{N}$ and finally $2k<x<2k+1$.

In other words the solutions of $(1)$
are max when
$ 2k<x<2k+1$ and are minimum when $k+1<x<2k$

So we can see that statement (A) is false.


问:
Is $2 + 2 + 2 + 2 + ... = -\frac12$ or $-1$?
Using zeta function regularization, the divergent series $1+1+1+1+...$ can be evaluated to yield $$1+1+1+1+1+...=\sum_{n=1}^\infty\frac1{n^0}=\zeta(0)=-\frac12.$$

But what is $2+2+2+2+...$ then? On the one hand, it should be twice as much, but on the other hand each $2=1+1$ so it could also be $-\frac12$ again. My gut feeling is that factoring out the $2$ of this divergent series is formally "less" valid than expanding the twos into sums of ones, but is the a sensible non-ambiguous answer?

答:
This claim is wrong:

$$\sum_{n=0}^\infty\frac1{n^0}=-\frac12$$

These equalities are correct:

$$\operatorname{reg}\sum_{n=0}^\infty1=\frac12$$

$$\operatorname{reg}\sum_{n=1}^\infty1=-\frac12$$

You can check them in Wolfram Mathematica.

```<code>Sum[1,{n,0,Infinity},Regularization->Dirichlet]
Sum[1,{n,1,Infinity},Regularization->Dirichlet]
</code>```

So,

$$\operatorname{reg}\sum_{n=0}^\infty2=2\operatorname{reg}\sum_{n=0}^\infty1=1$$

Body:
This claim is wrong:

$$\sum_{n=0}^\infty\frac1{n^0}=-\frac12$$

These equalities are correct:

$$\operatorname{reg}\sum_{n=0}^\infty1=\frac12$$

$$\operatorname{reg}\sum_{n=1}^\infty1=-\frac12$$

You can check them in Wolfram Mathematica.

Sum[1,{n,0,Infinity},Regularization->Dirichlet]
Sum[1,{n,1,Infinity},Regularization->Dirichlet]

So,

$$\operatorname{reg}\sum_{n=0}^\infty2=2\operatorname{reg}\sum_{n=0}^\infty1=1$$


问:
Can you "construct" a category from another category?
It's perfectly possible that I have no idea what I'm talking about here, as I know little about category theory, but my understanding is that a category is a space of some objects equipped with some "arrow" (functions in a loose sense) on those objects. Since it seems that these categories are defined typically quite independently from each other, and in a standard undergrad curriculum you might see the "bottom-up" construction of certain spaces from more elementary spaces, I am wondering if you can somehow "construct" one category from the other (and so in a sense they are not independent).

As an example, it is my understanding that you can (in principle) construct much of mathematics starting from nothing more than the ZFC axioms of set theory (the first few lectures in this series does this nicely; in particular, he constructs topological spaces.). This makes it look like (to my naive eyes) you can construct Top from Set. However, reading on the Axiom of Choice, we know that it "fails" on Top, but "works" in Set, so presumably there is some contradiction. See for example, this answer on Math Overflow (incidentally, I don't totally understand this answer; topologists use the Axiom of Choice in the form of Tychonoff's Theorem or Zorn's Lemma often by my understanding, yet the category they work in forbids it?).

Perhaps my misunderstanding lies in some difference between the category Set and ZFC set theory, or the space of topological objects that topologists work in and Top?

Edit: the comments gave some examples of categories constructed from other categories, a trivial one being the opposite category. Refining my question, I am wondering if you can specifically construct Top from Set, and if construction of objects (ignoring category theory) from ZFC implies that you can construct the category of those objects from Set. Also, is the axiom of choice (as in the linked MO question) problematic, because it "fails" in some categories but not others?

答:
The statement "Choice fails in Top" has a very specific technical meaning, and it's not "topologists can't use Choice." Rather, the point is this: many statements about sets can be reformulated in terms of maps alone. Such reformulations then make sense in the context of arbitrary categories - although they may only hold in some.

Here are a couple examples:

Take the theorem "There is a set $x$ such that for any set $y$ there is exactly one function from $x$ to $y$" (namely $x=\emptyset$). In categorical terms, this is the statement that Set has an initial object. Some categories have initial objects while others don't, so we see here an example of a set-theoretic principle holding in some but not all categories.

A more interesting example is provided by the Cantor-Bernstein theorem. Classically this says that whenever we have a pair of injections from each of two sets to the other, there is a bijetion between those two sets; its categorical formulation is "Given monomorphisms $f:A\rightarrow B$ and $g:B\rightarrow A$, there is an isomorphism $h:A\cong B$." This fails in Top for example since e.g. $(0,1)$ and $[0,1]$ with the usual topologies are not homeomorphic but continuously embed into each other. See e.g. here for a discussion of which categories satisfy this.

In the case of the axiom of choice, the relevant statement is "Surjections split," or "For every surjective $f:X\rightarrow Y$ there is a map $g:Y\rightarrow X$ such that $f\circ g=id_Y$." This can be reformulated in categorical terms (replacing "surjection" with "epimorphism"), and holds in some but not all categories. In particular, we have for example:

$\mathsf{ZFC}$ proves that choice holds in Set but fails in Top.

So there is no obstacle to Top being "buildable" from Set. This can in fact be done in a precise way: in the model-theoretic sense, Top is interpretable in Set. This is a rather tedious construction, and certainly more complicated than e.g. the construction of Set$^{op}$ from Set, but it's perfectly sensible.

A good first step is to familiarize yourself with the definition of "topological space" within the $\mathsf{ZFC}$ context. This will help motivate the categorical construction. Note that it's not quite true that we can recover the universe of sets $V$ from the category Set, but there is also plenty of information we do not lose in the passage $V\leadsto$ Set, and this is one such case.


问:
Splitting of prime ideals in algebraic extensions
I'm reading Algebraic Number Theory by Jurgen Neukirch. I have some problems with some of the exercises in Section 9 of Chapter 1.

They are:

1) If $L / K$ is a Galois extension of algebraic number fields with non-cyclic
Galois group, then there are at most finitely many non-split prime ideals of $K$.

2) Let $L / K$ be a finite (not necessarily Galois) extension of algebraic number
fields and $N / K$ the normal closure of $L / K$. Show that a prime ideal $p$ of $K$ is totally
split in $L$ if and only if it is totally split in $N$.

I have worked on them for a long time but couldn't get any idea of it.

Can you please help? Thank you!

答:
For (2), I think the key is the discussion Neukirch has earlier in that section (top of page 55), namely:

Let $\mathfrak{p}$ be a prime in $K$, and let $P_{\mathfrak{p}}$ be the set of primes of $L$ lying over $\mathfrak{p}$. Let $G=\mathrm{Gal}(N/K)$, and let $H=\mathrm{Gal}(N/L)$ be the subgroup corresponding to $H$. If $\mathfrak{P}$ is a prime of $N$ lying over $\mathfrak{p}$, then $G_{\mathfrak{P}} = \{\sigma\in G\mid \sigma\mathfrak{P}=\mathfrak{P}\}$, the decomposition group of $\mathfrak{P}$, is also a subgroup of $G$.

Neukirch states (and leaves as an exercise) that the map from the double cosets of $G$ modulo $H$ and $G_{\mathfrak{P}}$, $H\setminus G/G_{\mathfrak{P}}$ to $P_{\mathfrak{p}}$ given by
$$H\sigma G_{\mathfrak{P}} \longmapsto \sigma \mathfrak{P}\cap L$$
gives a bijection between the double cosets and $P_{\mathfrak{p}}$.

Assume that this is indeed the case (that is, the bijection is as given; I'll prove it below).

Showing that $\mathfrak{p}$ splits completely is equivalent to showing that $G_{\mathfrak{P}}$ is trivial (middle of page 54). So we prove that $G_{\mathfrak{P}}$ is trivial if and only if $\mathfrak{p}$ splits completely in $L$.

If $G_{\mathfrak{P}}$ is trivial (that is, if $\mathfrak{p}$ splits completely in $N$), then the double cosets are just the cosets of $H$ in $G$, and there are $[G:H] = [L:K]$ cosets (by the Fundamental Theorem of Galois Theory); that means that there are $[L:K]$ primes of $L$ lying over $\mathfrak{p}$, so $\mathfrak{p}$ splits completely in $L$ (this part can also be done simply by looking at the ramification and decomposition indices, which are multiplicative in towers).

Conversely, if $\mathfrak{p}$ splits completely in $L$, then the number of double cosets $H\setminus G/G_{\mathfrak{P}}$ equals $[L:K] = [G:H]$; this is the same as the number of right cosets of $H$; since each double coset decomposes as a disjoint union of right cosets of $H$, it follows that $H\sigma G_{\mathfrak{P}}  = H\sigma$ for all $\sigma\in G$, and in particular all conjugates of $G_{\mathfrak{P}}$ are contained in $H$. That is, the normal subgroup generated by $G_{\mathfrak{P}}$ is contained in $H$.

But since $N$ is the normal closure of $L$, and $H$ corresponds to $L$, then by the Fundamental Theorem of Galois Theory we know that $H$ is core-free: the largest normal subgroup of $G$ contained in $H$ is the trivial group. That means that the normal subgroup generated by the decomposition group is trivial, hence the decomposition group $G_{\mathfrak{P}}$ itself is trivial. And this proves that $\mathfrak{p}$ splits completely in $N$, as desired.

So it all comes down to establishing the bijection mentioned above: the map takes the double coset $H\sigma G_{\mathfrak{P}}$ to $\sigma\mathfrak{P}\cap L$.

First, the map is well defined: if $\tau\in G_{\mathfrak{P}}$, then $\tau\mathfrak{P}=\mathfrak{P}$, so $\sigma\mathfrak{P}\cap L = \sigma\tau\mathfrak{P}\cap L$. And if $\rho\in H$, then $\rho$ fixes $L$ pointwise, so $\rho\sigma\mathfrak{P}\cap L = \rho(\sigma\mathfrak{P}\cap L) = \sigma\mathfrak{P}\cap L$. So $\rho\sigma\tau$ corresponds to the same prime of $L$ as $\sigma, and the map is well-defined.

To see that the map is onto, given any prime $\mathfrak{q}$ of $L$ lying above $\mathfrak{p}$, there is a prime $\mathfrak{Q}$ of $N$ lying above $\mathfrak{q}$, and the transitive action of the Galois group guarantees the existence of $\sigma\in G$ such that $\sigma\mathfrak{P}=\mathfrak{Q}$. Thus, $H\sigma G_{\mathfrak{P}}$ maps to $\sigma\mathfrak{P}\cap L = \mathfrak{Q}\cap L = \mathfrak{q}$ (since $\mathfrak{Q}$ lies above $\mathfrak{q}$). This proves that the map is onto.

Finally, to show that the map is one to one, suppose that $\sigma\mathfrak{P}\cap L = \phi\mathfrak{P}\cap L = \mathfrak{q}$. Then $\sigma\mathfrak{P}$ and $\phi\mathfrak{P}$ both lie above $\mathfrak{q}$, so there exists $\rho\in \mathrm{Gal}(N/L) = H$ such that $\rho\sigma\mathfrak{P} = \phi\mathfrak{P}$. Therefore, $\phi^{-1}\rho\sigma\mathfrak{P} = \mathfrak{P}$, so $\phi^{-1}\tau\sigma\in G_{\mathfrak{P}}$, hence there exists $\rho\in G_{\mathfrak{P}}$ such that $\tau\sigma\rho^{-1} = \phi$. Hence, $\phi$ lies in the double coset $H\sigma G_{\mathfrak{P}}$, so $H\phi G_{\mathfrak{P}}=H\sigma G_{\mathfrak{P}}$, showing that the correspondence is one-to-one.


问:
Confusion Regarding Comparison Theorem
Determine if the integral converges or diverges using the Comparison Theorem (CT).

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

I saw solutions online arrive at the following inequality

$$\frac{x+1}{\sqrt{x^4 -x}} \ge \frac{1}{x}$$

But this inequality only holds for  $x > 1$  and not for $x \ge 1$ since at $x=1$ the LHS of the inequality, $\frac{x+1}{\sqrt{x^4 -x}}$, is undefined.

I thought the inequality had to be true over the limits of integration, $x \ge 1$ or $[1, \infty]$ in this case, in order to use the CT.

Furthermore, the CT requires the functions being compared to be continuous over a closed interval, but at $x = 1$ function $\frac{x+1}{\sqrt{x^4 -x}}$ is infinitely discontinuous (vetical asymptote) which I'm assuming means we can't apply the CT over $[1, \infty]$.

But I have seen solutions online show that

$$\int_{1}^{\infty} \frac{1}{x}dx$$

diverges and thus $$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$ diverges by the CT. Are these solutions correct?

As a work around I was thinking the inequality is true for $x \ge 2$ so the integral could be "split up" into the following

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx = \int_{1}^{2} \frac{x+1}{\sqrt{x^4 -x}}dx + \int_{2}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

The CT could then be applied to the second integral. The first integral, however, still has an infinite discontinuity at $x = 1$ and thus is an "improper" integral. This would require evaluating the integral directly which defeats the purpose of using the CT in the first place to determine convergence or divergence.

My questions are:

Does it matter if the inequality is not true over the interval or limits of integration $[a, \infty)$ or $x \ge a$ in order to apply the Comparison Theorem?

Were the solutions online correct?

答:
I cannot tell you whether or not the online solutions are correct, since I did not sse them. But I can tell you that your idea of doing the decomposition$$\int_1^\infty \frac{x+1}{\sqrt{x^4 -x}}\,\mathrm dx=\int_1^2\frac{x+1}{\sqrt{x^4 -x}}\,\mathrm dx+\int_2^\infty\frac{x+1}{\sqrt{x^4-x}}\,\mathrm dx$$is fine. Then, you can indeed deduce from the inequality$$\frac{x+1}{\sqrt{x^4-x}}\geqslant\frac1x$$that the second integral diverges. On the other hand$$\lim_{x\to1}\frac{\frac{x+1}{\sqrt{x^4-x}}}{\frac1{\sqrt{1-x}}}=2$$and therefore, since the integral$$\int_1^2\frac{\mathrm dx}{\sqrt{1-x}}$$converges, then so does$$\int_1^2\frac{x+1}{\sqrt{x^4 -x}}\,\mathrm dx.\tag1$$The final conclusion is, of course, that the original integral diverges.

Note that it would still diverge if the integral $(1)$ was divergent. So, in fact, you do not have to analyze the convergence of $(1)$.

Title: Confusion Regarding Comparison Theorem


问:
What is the opposite category of the category of Presheaves?
I know that the category of presheaves on $\textbf{C}$ is defined as $\textbf{Psh}(C):= \textbf{Func}(\textbf{C}^{op}, \textbf{Set})$.

I have been wondering what the category $(\textbf{Psh}(\textbf{C}))^{op}$ would be like.

I have a hunch that $(\textbf{Psh}(\textbf{C}))^{op}$ would be isomorphic to $\textbf{Func}(\textbf{C}, \textbf{Set})$.

Am I correct in this guess? I have tried to work on this but I do not even know whether the hunch is correct. Any help with this would be great.

Thank you.

答:
Presheaf categories are particularly nice examples of locally presentable categories, and you can show that the opposite of a locally presentable category is never locally presentable unless it is a preorder. So $\text{Psh}(C)^{op}$ is not isomorphic to any presheaf category. Many other arguments are possible; for example, presheaf categories are also Grothendieck topoi and so have, for example, exponential objects, while their opposites don't.

$\text{Psh}(C^{op})^{op}$ has a universal property dual to the universal property of presheaves: in the same way that $\text{Psh}(C)$ is the free cocompletion of $C$ (at least if $C$ is essentially small), $\text{Psh}(C^{op})^{op}$ is the free completion of $C$. That is, it's the category obtained from $C$ by "freely adjoining (small) limits." Mostly this is not too useful except as a convenient place to embed the category of pro-objects $\text{Pro}(C)$.


问:
Integration and periodic divergent series
Consider the divergent infinite series

$$ \sum_{n=1}^\infty 5^n \cos(nt) $$

where $t \in \mathbb{R}$.  I don't believe it has a closed form.  However, if I take the integral with respect to $t$:

$$ \int_{-\pi}^\pi \sum_{n=1}^\infty 5^n \cos(nt) \, dt = \sum_{n=1}^\infty 5^n \int_{-\pi}^\pi \cos(nt) \, dt$$
$$ = \sum_{n=1}^\infty 5^n \cdot 0 = 0$$

So even though the series is divergent it still has a finite integral.  This seems to be a consequence of the fact that it's periodic over the integration domain: the infinities sort of cancel each other out.

This can extend also to the product of series:
$$ \int_{-\pi}^\pi \left( \sum_{n=1}^\infty 5^n \cos(nt) \right) \left(\sum_{n=1}^\infty \frac 1 {7^n} \cos(nt) \right) \,dt$$
$$ = \pi \sum_{n=1}^\infty \left( \frac 5 7 \right)^n = \frac 5 2 \pi$$

I'm trying to understand the underlying properties at work here.  Specifically, is there a way to come up with a "closed form" for a periodic infinite divergent series which behaves the same under integration (for instance, something I could use with integration by parts) without actually being the limit of the partial sums of $f(t)$?  That is, something like a proxy function with the infinities sort of preemptively normalized out?

答:
You are asking about asymptotics$^\dagger$. Specifically assigning a meaning to divergent series. The strategy is to first construct something that does converge out of your series. Using Borel summation, we write

$$ \phi(x)=\sum_{n=0}^{\infty} c_n \frac{x^n}{n!}$$

Where the $c_n$ are given by your series

$$ c_n =5^n \cos(nt)$$

Now let

$$ B(x)=\int_0^\infty  dz \ \phi(xz)\exp(-z)$$

We will define your (divergent) sum as $B(1)$, if that exists. It's not obvious, but the above things are defined so that $B(1)=\sum_n c_n$, which follows after a few manipulations. In any case, with the assistance of Mathematica, I find

$$\phi(x)=\cos(5x\sin(t)) \left( \cosh(5x\cos(t))+\sinh(5x\cos(t))\right)$$

$$B(x)=\frac{1}{2}\left( 1+\frac{1}{1-5xe^{it}}+\frac{5x}{-5x+e^{it}} \right)$$

$$ B(1) = \frac{1}{2}+\frac{6}{-13+5\cos(t)}$$

This is nice, but the manipulations are obscure and the integrals annoying. Let us apply Euler summation to get the same answer with much less fuss$^\ddagger$. Let

$$ I(x)=\sum_n c_n x^n $$

With the $c_n$ given by your series.  This is more intuitive than Borel summation because it is obvious that when $x=1$, we have $\sum_n c_n$, which is your original sum. We will define $\lim_{x\rightarrow 1}I(x)$ as that sum, if the limit exists. Substituting your $c_n$

$$ I(x)=\sum_n (5x)^n \cos(nt)=\frac{1}{2} \sum_n (5x)^n \left(e^{int}+e^{-int} \right)$$

$$ I(x)=\frac{1}{2} \sum_{n=0}^\infty \left( 5xe^{it} \right)^n +\frac{1}{2} \sum_{n=0}^\infty \left( 5xe^{-it} \right)^n$$

We now have two geometric series, and importantly, they converge for $|x|<1/5$.

$$ I(x)=\frac{1}{2} \frac{1}{1-5xe^{it}}+\frac{1}{2} \frac{1}{1-5xe^{-it}}$$

Having performed the sum, we take the limit $x \rightarrow 1$ to find

$$ I(1)=\frac{1}{2}+\frac{6}{-13+5\cos(t)}$$

Which we may define as the value of your sum. As promised, $I(1)=B(1)$. Why does this (sometimes) work? Because for small enough $x$, perhaps $I(x)$ converges, even though the original sum didn't. Similar logic applies to Borel summation, with that terrific $1/n!$ damping of terms.

$\dagger$ Too large a topic to explain here. I'll just refer you to Carl Bender's excellent book: Advanced mathematical methods.

$\ddagger$ There are other types of summation, too. For a given series, some may work while others fail.

EDIT: Using a method inspired by the other answer:

For real $t$, the sum must diverge because $|\cos|\leq1$ but $5^n$ grows exponentially, so $\lim_{n\rightarrow \infty}c_n \neq0$. However, if we allow $t \in \mathbb{C}$, and writing $\cos$ in exponential form, the sums converge for $t$ with sufficiently large imaginary part. After performing the sum we can define the answer to be $\lim_{\Im(t)\rightarrow 0}$, which turns out to be exactly $B(1)$.


问:
Why is it true that $\int_0^\infty \sin(qx) dx = 1/q$.
In a Quantum Mechanics class the professor pointed out a calculation from Zetilli's textbook on Quantum Mechanics (page 630).

There he says that:

$$\int_0^\infty \sin(qx) dx = \lim_{\lambda \to 0} \int_0^\infty e^{-\lambda x} \sin(qx) dx$$

Then he writes the $\sin$ as complex exponentials and reaches the result that the integral is $1/q$. WHAT? I was taught at my calculus course that this integral does not converge. For me this result is plain wrong, and I do not understand it at all. I even don't see why the limit can be imposed like that. Any thoughts on this "trick"?

答: The mean value of the integral as the upper bound goes to infinity is $1/q$. The Abel regularization in your source is essentially finding the mean value.


问:
Method to solve this indefinite integral $ \int \frac{x^{4}}{\sqrt{x^{2}+9}}dx $
I'm not able to solve this integral. Can anybody help me?
I tried doing it by parts noticing that
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .
$$
where we know that $$ D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9} $$

so that we can view the integral as

$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx  .
$$
which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{\bbox[5px,#ffd]{x \equiv {9 - t^{2} \over 2t} \implies
t = \root{x^{2} + 9} - x}}$:
\begin{align}
&\bbox[5px,#ffd]{\int{x^{4} \over \root{x^{2} + 9}}
\,\dd x}
\\[5mm] = &\
\int\pars{-\,{t^{3} \over 16} + {9t \over 4} - {243 \over 8t} + {729 \over 4t^{3}} - {6561 \over 16t^{5}}}\dd t
\\[5mm] = &\
-\,{t^{4} \over 64} + {9t^{2} \over 8} - {243\ln\pars{t} \over 8} - {729 \over 8t^{2}} + {6561 \over 64t^{4}} +
\mbox{a constant}\label{1}\tag{1}
\end{align}
Replace
$\ds{\quad t = \root{x^{2} + 9} - x\quad}$ in (\ref{1}).


问:
What is an analytical definition or $\epsilon-\Bbb N$ definition for a sequence in real numbers which doesn't have any convergence subsequence.
I have one conceptual question: what is an analytical definition or $\epsilon-\Bbb N$ definition for a sequence in real numbers which doesn't have any convergence subsequence.

I was thinking in this way that $\not \exists$ any subsequence $(x_{n_i})$ s.t $x_{n_i} \not \to L$ so for any increasing sequence $(n_i)$ there exist $\epsilon >0$ s.t for all $N \in \Bbb N$ there is a $m,n>N$ s.t $|x_m-x_n|>\epsilon$.

My question is to make the definition smoother. So that I can see this more easily what I am observing is that the sequence has to be unbounded. Can I say that there exists $\epsilon>0$ s.t $|x_n-x_m|>\epsilon$ for all but finitely many $m,n \in \Bbb N$? Please help with your idea.

I also invite to have your answer about how do you write the definition of a sequence in real numbers which doesn't have any convergence subsequence.

Further edit:

Another idea that came to my mind is that there is an $\epsilon>0$
s.t there is $N \in \Bbb N$ so that for all $k \in \Bbb N$ we have
$|x_N-x_{N+k}|\geq \epsilon$.

If statement 1 holds then the sequence can't have a convergent subsequence.
If statement 1 doesn't hold then for any $\epsilon>0$ and for any $N \in \Bbb N$ there exist $k \in \Bbb N$ s.t $|x_{N+k}-x_N|<\epsilon$ and then we will start with $x_1$ and $ \epsilon=1$ then we will get $x_{1+k_1}$ then $x_{1+k_1+k_2}$ for $\epsilon=\frac 12$ and proceed in this way to get a convergent subsequence.

答:
Suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ has no convergent subsequence; then $\{n\in\Bbb N:x_n=x\}$ is finite for each $x\in\Bbb R$, and $A=\{x_n:n\in\Bbb N\}$ is infinite. Suppose that $x$ is an accumulation point of $A$. Then there is a sequence $\tau$ in $A\setminus\{x\}$ converging to $x$, and it’s not hard to show that $\tau$ has a subsequence $\tau'$ that is also a subsequence of $\sigma$. This is impossible, so $A$ has no accumulation points and is therefore a closed, discrete subset of $\Bbb R$.

It would be nice if this meant that there were an $\epsilon>0$ such that $|x_n-x_m|\ge\epsilon$ for all $n,m\in\Bbb N$ such that $x_n\ne x_m$, but this is not necessarily the case. Let $x_0=0$, and for $n>0$ let $x_n=\sum_{k=1}^n\frac1k$, the $n$-th harmonic number $H_n$. Then $\sigma=\langle x_n:n\in\Bbb N\rangle$ is strictly increasing, and $\lim\limits_{n\to\infty}x_n=\infty$, so $\sigma$ has no convergent subsequence. However, for any $\epsilon>0$ there is an $n\in\Bbb N$ such that $\frac1n<\epsilon$, so that

$$|x_{k+1}-x_k|=\frac1{k+1}<\frac1n<\epsilon$$

whenever $k\ge n$.


答:
Let $\epsilon > 0$ (something very small perhaps) and then

\begin{align} \int_{1}^{n} \frac{x+1}{\sqrt{x^4 -x}}dx  & = \int_{1+\epsilon}^{n} \frac{x+1}{\sqrt{x^4 -x}} + \int_{1}^{1 + \epsilon} \frac{x+1}{\sqrt{x^4 -x}} \\
\end{align}

You could then show that (a sentence or two)

$$\int_{1}^{1 + \epsilon} \frac{x+1}{\sqrt{x^4 -x}} \geq 0$$

and that

$$\int_{1+\epsilon}^{n} \frac{x+1}{\sqrt{x^4 -x}} \geq \int_{1 + \epsilon}^n\frac{1}{x}dx $$

diverges (which you have already done). A way to think about it (this certainly isn't rigorous) might be that

$$\frac{x+1}{\sqrt{x^4 - x}} \geq \frac{1}{x}$$

holds apart from at a single point, and given that a single point isn't going to add anything to the integral we can simply ignore it.

Title: Confusion Regarding Comparison Theorem

Body:
Determine if the integral converges or diverges using the Comparison Theorem (CT).

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

I saw solutions online arrive at the following inequality

$$\frac{x+1}{\sqrt{x^4 -x}} \ge \frac{1}{x}$$

But this inequality only holds for  $x > 1$  and not for $x \ge 1$ since at $x=1$ the LHS of the inequality, $\frac{x+1}{\sqrt{x^4 -x}}$, is undefined.

I thought the inequality had to be true over the limits of integration, $x \ge 1$ or $[1, \infty]$ in this case, in order to use the CT.

Furthermore, the CT requires the functions being compared to be continuous over a closed interval, but at $x = 1$ function $\frac{x+1}{\sqrt{x^4 -x}}$ is infinitely discontinuous (vetical asymptote) which I'm assuming means we can't apply the CT over $[1, \infty]$.

But I have seen solutions online show that

$$\int_{1}^{\infty} \frac{1}{x}dx$$

diverges and thus $$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$ diverges by the CT. Are these solutions correct?

As a work around I was thinking the inequality is true for $x \ge 2$ so the integral could be "split up" into the following

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx = \int_{1}^{2} \frac{x+1}{\sqrt{x^4 -x}}dx + \int_{2}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

The CT could then be applied to the second integral. The first integral, however, still has an infinite discontinuity at $x = 1$ and thus is an "improper" integral. This would require evaluating the integral directly which defeats the purpose of using the CT in the first place to determine convergence or divergence.

My questions are:

Does it matter if the inequality is not true over the interval or limits of integration $[a, \infty)$ or $x \ge a$ in order to apply the Comparison Theorem?

Were the solutions online correct?


问:
a lemma for Lebesgue outer measure
An interval in $\mathbb{R}$ can be either open, closed, or half closed, $l(I)= b-a$ where $b, a,\ (a<b)$ are exactly the boundaries of $I$.

The following lemma is important in order to prove the countable subadditivity of the Lebesgue outer measure;

Let $(I_m)$ and $(J_n)$ be sequences of intervals such that the $I_m$'s are pairwise disjoint, prove for $k,\ p$ arbitrary in $\mathbb{N}$, that $\bigcup\limits_{m=1}^k I_m\subseteq \bigcup\limits_{n=1}^p J_n$ implies $\sum\limits_{m=1}^kl(I_m)\leq \sum\limits_{n=1}^pl(J_n)$

答:
We prove this by mathematical induction on $k$, considering the proposition :

$P(k):$ $\quad$ for any $p$ in $\mathbb{N}$, if $I_m$'s are pairwise disjoint intervals such that $\bigcup\limits_{m=1}^k I_m\subseteq \bigcup\limits_{n=1}^p J_n$ then $\sum\limits_{m=1}^kl(I_m)\leq \sum\limits_{n=1}^pl(J_n)$

Clearly $P(1)$ holds; if $I_1\subseteq \bigcup\limits_{n=1}^p J_n$ then we can show by induction on $p$ that $l(I_1)\leq \sum\limits_{n=1}^pl(J_n)$

Suppose $P(k)$ holds, we want to prove $P(k+1)$, we will use induction on $p$;

Clearly for $p=1$, if $I_m$'s are pairwise disjoint intervals such that $\bigcup\limits_{m=1}^{k+1} I_m\subseteq J_1$ then $\sum\limits_{m=1}^{k+1}l(I_m)\leq l(J_1)$

Suppose now that the implication holds for a number $p$ of intervals $J_n$'s, and let $\{I_m\}_{m=1}^{k+1}$ be pairwise disjoint intervals such that $\bigcup\limits_{m=1}^{k+1} I_m\subseteq \bigcup\limits_{n=1}^{p+1} J_n$

We want to prove that $\sum\limits_{m=1}^{k+1}l(I_m)\leq \sum\limits_{n=1}^{p+1}l(J_n)$ so to show that $P(k+1)$ holds and we will be done.

If one of the $p+1$ intervals $J_n$'s doesn't contain any point from any interval $I_m$, name it $J_{n_0}$, then obviously $\bigcup\limits_{m=1}^{k+1} I_m\subseteq \bigcup\limits_{n=1, n\neq n_0}^{p+1} J_n$ and by the hypothesis $\sum\limits_{m=1}^{k+1}l(I_m)\leq \sum\limits_{n=1, n\neq n_0}^{p+1}l(J_n) \leq \sum\limits_{n=1}^{p+1}l(J_n)$,

So we can assume that each interval of the $J_n$'s intersect with at least one of the $I_m$'s,

we can discuss two cases here:

In the case where there exists one $J_{n_1}$ that contains a whole interval $I_{m_1}$, we can have $k$ intervals $\{I_m\}_{m=1, m\neq m_1}^{k+1}$ contained in $\bigcup\limits_{n=1}^{p+1} J_n$ so by the first induction hypothesis $$\sum\limits_{m=1, m\neq m_1}^{k+1}l(I_m)\leq \sum\limits_{n=1}^{p+1}l(J_n)$$ but also we know that this covering contain the interval $J_{n_1}$ with length exceeding the total length of the $I_m$'s by a quantity greater or equal to $l(I_{m_1})$, recall that the $I_m$'s are pairwise disjoint, so we can add this length without losing the inequality :

$l(I_{m_1})+ \sum\limits_{m=1, m\neq m_1}^{k+1}l(I_m)\leq \sum\limits_{n=1}^{p+1}l(J_n)$, in that case then we have verified the needed.

In the other case, when there is no interval from $(J_n)$ that contains a whole interval $I_m$, this means each interval $I_m$ is covered by at least two intervals from $(J_n)$, recall again that the $I_m$'s are pairwise disjoint, so necessarily in this case $p+1> k+1$, we can consider now the interval $I_{m_{LE}}$ that is the most 'at the left' (considering simply the interval that contain a real number lesser than all numbers in the other intervals), Notice that this interval is covered partially by $J_{n_{LE}}$ such that also this $J_{n_{LE}}$ cover no other interval, we can then have $k$ intervals $\{I_m\}_{m=1, m\neq m_{LE}}^{k+1}$ contained in $\bigcup\limits_{n=1, n\neq n_{LE}}^{p+1} J_n$ so by the first induction hypothesis $$\sum\limits_{m=1, m\neq m_{LE}}^{k+1}l(I_m)\leq \sum\limits_{n=1, n\neq n_{LE}}^{p+1}l(J_n)$$
but also here we have for one part of the interval $I_{m_{LE}}$: $l(I_{m_{LE}}\cap J_{n_{LE}})\leq l(J_{n_{LE}})$,
the other remaining part has length that is already in the 'excess' of the total length $\sum\limits_{n=1, n\neq n_{LE}}^{p+1}l(J_n)$, Thus $\sum\limits_{m=1}^{k+1}l(I_m)\leq \sum\limits_{n=1}^{p+1}l(J_n)$. QED

Title: a lemma for Lebesgue outer measure


问:
Help evaluating $\int_0^\infty \frac{1}{x^{1/2}(x+1)}dx$
I began solving this with U sub and partial fractions...first for $x^{1/2}$ and then for $x+1$  but neither of those methods got me the answer of $\pi$.

I know the indefinite integral should be $2\tan ^{-1} x^{1/2} + C$.

答:
HINT:

Put $x=\tan^2\theta$ so that $dx=2\tan\theta\sec^2\theta d\theta$

If $x\to0, \tan^2\theta\to0\implies \tan\theta\to0\implies \theta\to0$ (taking the principal value)

and if $x\to\infty ,\tan^2\theta\to\infty\implies \tan\theta\to\pm\infty$

If we take $\tan\theta\to+\infty$ i.e., $\theta\to\frac\pi2$ (taking the principal value) $,\sqrt x=+\tan\theta$

Then the integral becomes $$\int_0^\frac\pi2 2d\theta=2\left(\frac\pi2-0\right)$$

If we take $\tan\theta\to -\infty$ i.e., $\theta\to -\frac\pi2$ (taking the principal value) $,\sqrt x=-\tan\theta$

Then the integral becomes  $$\int_0^{-\frac\pi2} (-2)d\theta=-2\left(-\frac\pi2-0\right)$$


问:
Finding a nice solution to $\sum_{a,b,c\ge 1, a+b+c=9}\frac{9!}{a!b!c!}=18150$
I am trying to find a nice way to solve for
$$\sum_{a,b,c\ge 1, a+b+c=9}\frac{9!}{a!b!c!}=18150$$
I managed to solved it (and  verified by computer) by doing manually (on paper) on $7$ cases and got a "nice" answer of $216+1512+3024+2268+1890+7560+1680=18150=2\cdot 3\cdot5^2 \cdot11^2$.

答: This is the number of surjections from a 9-set onto a 3-set.  More generally, the number of surjections from an $n$-set onto a $k$-set involves Stirling numbers of the second kind: https://oeis.org/A019538


问:
Maximum number of right-angled triangles

Let $S$ be a set of $n$ points in the plane, no $3$ collinear. Determine the maximum number of right-angled triangles with all three vertices as points in $S$.

This is a slightly more difficult and precise question than IMO 1970. For $n=3$, clearly the maximum is one. For $n=4$, we can have four triangles, etc. However, I don't know how to continue.

答:
OK, for record's sake here's a solution.

The answer is $\frac{n(n-2)}{2}$ for even $n$ and $\frac{(n-1)(n-2)}{2}$ for odd $n$. The construction is to draw a circle, and pair up your points to be diametrically opposite (so when $n$ is odd, we have one lonely point not in a pair).

To show that we can't do better, for each point $P$, we count the number of triangles for which $P$ is the right angle. The key insight is that if $X,Y$ form a right angle at $P$, then this is the only right triangle at $P$ involving $X$ or $Y$. Otherwise, if (say) $\angle YPZ=90^{\circ}$, then $X$, $P$, $Z$ are collinear, contradiction. So there are at most $\lfloor\frac{n-1}{2}\rfloor$ triangles that their right angle at $P$.

Applying this to all $n$ choices of $P$ implies the result for even $n$. Some more fiddling is required when $n$ is odd.


问:
Derive shoulder position from head and hands (Virtual Reality)
The minimal set of positional data common to current VR headsets is the position and orientation of the headset and also of a controller held in each hand. Direction of travel is typically a choice between moving where the headset is looking (gaze) or the direction the offhand controller is facing (point).

More ideal I think would be to travel in the direction the player's torso is facing - being more intuitive/immersive than pointing while permitting the player to look around at the same time. To get the torso orientation I think I would need to derive the position of both shoulders from the position and orientation of each controller.

If I assume the player is standing upright and that wrists are not bent I think I can count on the following data-points; scalene triangle between neck/spine, shoulder and elbow

N(xyz) The top of the spine (neck), derived from headset position and orientation.

E(xyz) The elbow, derived from the hand position and orientation (projected back a pre-calculated distance which is proportional to the player height).

E-N The distance measured between the elbow and neck/spine.

S-N A pre-calculated horizontal distance between the shoulder and neck/spine (proportional to player height).

E-S A pre-calculated length for the upper arm (proportional to player height).

The math is beyond me but it's clear S (shoulder) could ordinarily exist anywhere on a circle perpendicular to E-N but for the fact that its vertical component (z) is known. Can anybody tell me how I might go about finding the x/y for S such that I have its full 3D coordinate? If it's not already clear I don't have a lot of math so baby steps would be much appreciated. Ultimately I'd need to code this (to run every frame).

EXAMPLE VALUES

In-game metres, x/y are horizontal coordinates, z is vertical (from ground at 0).

N(xyz): -0.02,  0.12, 1.39

S(xyz): ?.??, ?.??, 1.39

E(xyz):  0.03, -0.30, 1.41

E-N: 0.42

S-N: 0.18

E-S: 0.30

screenshot

Body:
Consider this specific example. You can immediately form the following two equations based on the length of $SE$ and $NS$:

$$(x-0.03)^2 + (y+0.30)^2 + (z-1.41)^2 = (0.30)^2$$

$$(x+0.02)^2 + (y-0.12)^2 + (z-1.39)^2 = (0.18)^2$$

Since $z = 1.39$, the above system of equations simplifies to

$$(x-0.03)^2 + (y+0.30)^2 = 0.0896$$

$$(x+0.02)^2 + (y-0.12)^2 = 0.0324$$

Solving, we get the following two solutions:

$(0.104428, -0.0100681)$ and $(-0.110414, -0.0356446)$

Most probably only one of the above solutions is acceptable in this case, right?

For the general case, I found a solution for the following system of equations:

$$(x-A)^2+ (y-B)^2 + (z-C)^2= G^2$$
$$(x-D)^2 + (y-E)^2 + (z-F)^2 = H^2$$
$$z = F$$
$$A \neq D, B \neq E$$

The lengthy solution follows. I'll check the solution once again.

$$x = \frac{A^3 - A^2 D - \sqrt{-(B - E)^2 (A^4 - 4 A^3 D + 2 A^2 B^2 - 4 A^2 B E + 2 A^2 C^2 - 4 A^2 C F + 6 A^2 D^2 + 2 A^2 E^2 + 2 A^2 F^2 - 2 A^2 G^2 - 2 A^2 H^2 - 4 A B^2 D + 8 A B D E - 4 A C^2 D + 8 A C D F - 4 A D^3 - 4 A D E^2 - 4 A D F^2 + 4 A D G^2 + 4 A D H^2 + B^4 - 4 B^3 E + 2 B^2 C^2 - 4 B^2 C F + 2 B^2 D^2 + 6 B^2 E^2 + 2 B^2 F^2 - 2 B^2 G^2 - 2 B^2 H^2 - 4 B C^2 E + 8 B C E F - 4 B D^2 E - 4 B E^3 - 4 B E F^2 + 4 B E G^2 + 4 B E H^2 + C^4 - 4 C^3 F + 2 C^2 D^2 + 2 C^2 E^2 + 6 C^2 F^2 - 2 C^2 G^2 + 2 C^2 H^2 - 4 C D^2 F - 4 C E^2 F - 4 C F^3 + 4 C F G^2 - 4 C F H^2 + D^4 + 2 D^2 E^2 + 2 D^2 F^2 - 2 D^2 G^2 - 2 D^2 H^2 + E^4 + 2 E^2 F^2 - 2 E^2 G^2 - 2 E^2 H^2 + F^4 - 2 F^2 G^2 + 2 F^2 H^2 + G^4 - 2 G^2 H^2 + H^4)} + A B^2 - 2 A B E + A C^2 - 2 A C F - A D^2 + A E^2 + A F^2 - A G^2 + A H^2 + B^2 D - 2 B D E - C^2 D + 2 C D F + D^3 + D E^2 - D F^2 + D G^2 - D H^2}{2 (A^2 - 2 A D + B^2 - 2 B E + D^2 + E^2)}$$

$$y = \frac{B^4 - 2 E B^3 + A^2 B^2 + C^2 B^2 + D^2 B^2 + F^2 B^2 - G^2 B^2 + H^2 B^2 - 2 A D B^2 - 2 C F B^2 + 2 E^3 B - 2 E F^2 B + 2 E G^2 B - 2 E H^2 B - 2 C^2 E B + 4 C E F B - E^4 - A^2 E^2 + C^2 E^2 - D^2 E^2 + 2 A D E^2 + E^2 F^2 - E^2 G^2 + E^2 H^2 - 2 C E^2 F + A \sqrt{-(B - E)^2 (A^4 - 4 D A^3 + 2 B^2 A^2 + 2 C^2 A^2 + 6 D^2 A^2 + 2 E^2 A^2 + 2 F^2 A^2 - 2 G^2 A^2 - 2 H^2 A^2 - 4 B E A^2 - 4 C F A^2 - 4 D^3 A - 4 D E^2 A - 4 D F^2 A + 4 D G^2 A + 4 D H^2 A - 4 B^2 D A - 4 C^2 D A + 8 B D E A + 8 C D F A + B^4 + C^4 + D^4 + E^4 + F^4 + G^4 + H^4 - 4 B E^3 - 4 C F^3 + 2 B^2 C^2 + 2 B^2 D^2 + 2 C^2 D^2 + 6 B^2 E^2 + 2 C^2 E^2 + 2 D^2 E^2 + 2 B^2 F^2 + 6 C^2 F^2 + 2 D^2 F^2 + 2 E^2 F^2 - 4 B E F^2 - 2 B^2 G^2 - 2 C^2 G^2 - 2 D^2 G^2 - 2 E^2 G^2 - 2 F^2 G^2 + 4 B E G^2 + 4 C F G^2 - 2 B^2 H^2 + 2 C^2 H^2 - 2 D^2 H^2 - 2 E^2 H^2 + 2 F^2 H^2 - 2 G^2 H^2 + 4 B E H^2 - 4 C F H^2 - 4 B^3 E - 4 B C^2 E - 4 B D^2 E - 4 C^3 F - 4 C D^2 F - 4 C E^2 F - 4 B^2 C F + 8 B C E F)} - D \sqrt{-(B - E)^2 (A^4 - 4 D A^3 + 2 B^2 A^2 + 2 C^2 A^2 + 6 D^2 A^2 + 2 E^2 A^2 + 2 F^2 A^2 - 2 G^2 A^2 - 2 H^2 A^2 - 4 B E A^2 - 4 C F A^2 - 4 D^3 A - 4 D E^2 A - 4 D F^2 A + 4 D G^2 A + 4 D H^2 A - 4 B^2 D A - 4 C^2 D A + 8 B D E A + 8 C D F A + B^4 + C^4 + D^4 + E^4 + F^4 + G^4 + H^4 - 4 B E^3 - 4 C F^3 + 2 B^2 C^2 + 2 B^2 D^2 + 2 C^2 D^2 + 6 B^2 E^2 + 2 C^2 E^2 + 2 D^2 E^2 + 2 B^2 F^2 + 6 C^2 F^2 + 2 D^2 F^2 + 2 E^2 F^2 - 4 B E F^2 - 2 B^2 G^2 - 2 C^2 G^2 - 2 D^2 G^2 - 2 E^2 G^2 - 2 F^2 G^2 + 4 B E G^2 + 4 C F G^2 - 2 B^2 H^2 + 2 C^2 H^2 - 2 D^2 H^2 - 2 E^2 H^2 + 2 F^2 H^2 - 2 G^2 H^2 + 4 B E H^2 - 4 C F H^2 - 4 B^3 E - 4 B C^2 E - 4 B D^2 E - 4 C^3 F - 4 C D^2 F - 4 C E^2 F - 4 B^2 C F + 8 B C E F)}}{2 (B - E) (A^2 - 2 D A + B^2 + D^2 + E^2 - 2 B E)}$$

$$z = F$$

and

$$x = \frac{A^3 - A^2 D + \sqrt{-(B - E)^2 (A^4 - 4 A^3 D + 2 A^2 B^2 - 4 A^2 B E + 2 A^2 C^2 - 4 A^2 C F + 6 A^2 D^2 + 2 A^2 E^2 + 2 A^2 F^2 - 2 A^2 G^2 - 2 A^2 H^2 - 4 A B^2 D + 8 A B D E - 4 A C^2 D + 8 A C D F - 4 A D^3 - 4 A D E^2 - 4 A D F^2 + 4 A D G^2 + 4 A D H^2 + B^4 - 4 B^3 E + 2 B^2 C^2 - 4 B^2 C F + 2 B^2 D^2 + 6 B^2 E^2 + 2 B^2 F^2 - 2 B^2 G^2 - 2 B^2 H^2 - 4 B C^2 E + 8 B C E F - 4 B D^2 E - 4 B E^3 - 4 B E F^2 + 4 B E G^2 + 4 B E H^2 + C^4 - 4 C^3 F + 2 C^2 D^2 + 2 C^2 E^2 + 6 C^2 F^2 - 2 C^2 G^2 + 2 C^2 H^2 - 4 C D^2 F - 4 C E^2 F - 4 C F^3 + 4 C F G^2 - 4 C F H^2 + D^4 + 2 D^2 E^2 + 2 D^2 F^2 - 2 D^2 G^2 - 2 D^2 H^2 + E^4 + 2 E^2 F^2 - 2 E^2 G^2 - 2 E^2 H^2 + F^4 - 2 F^2 G^2 + 2 F^2 H^2 + G^4 - 2 G^2 H^2 + H^4)} + A B^2 - 2 A B E + A C^2 - 2 A C F - A D^2 + A E^2 + A F^2 - A G^2 + A H^2 + B^2 D - 2 B D E - C^2 D + 2 C D F + D^3 + D E^2 - D F^2 + D G^2 - D H^2}{2 (A^2 - 2 A D + B^2 - 2 B E + D^2 + E^2)}$$

$$y = \frac{B^4 - 2 E B^3 + A^2 B^2 + C^2 B^2 + D^2 B^2 + F^2 B^2 - G^2 B^2 + H^2 B^2 - 2 A D B^2 - 2 C F B^2 + 2 E^3 B - 2 E F^2 B + 2 E G^2 B - 2 E H^2 B - 2 C^2 E B + 4 C E F B - E^4 - A^2 E^2 + C^2 E^2 - D^2 E^2 + 2 A D E^2 + E^2 F^2 - E^2 G^2 + E^2 H^2 - 2 C E^2 F - A \sqrt{(-(B - E)^2 (A^4 - 4 D A^3 + 2 B^2 A^2 + 2 C^2 A^2 + 6 D^2 A^2 + 2 E^2 A^2 + 2 F^2 A^2 - 2 G^2 A^2 - 2 H^2 A^2 - 4 B E A^2 - 4 C F A^2 - 4 D^3 A - 4 D E^2 A - 4 D F^2 A + 4 D G^2 A + 4 D H^2 A - 4 B^2 D A - 4 C^2 D A + 8 B D E A + 8 C D F A + B^4 + C^4 + D^4 + E^4 + F^4 + G^4 + H^4 - 4 B E^3 - 4 C F^3 + 2 B^2 C^2 + 2 B^2 D^2 + 2 C^2 D^2 + 6 B^2 E^2 + 2 C^2 E^2 + 2 D^2 E^2 + 2 B^2 F^2 + 6 C^2 F^2 + 2 D^2 F^2 + 2 E^2 F^2 - 4 B E F^2 - 2 B^2 G^2 - 2 C^2 G^2 - 2 D^2 G^2 - 2 E^2 G^2 - 2 F^2 G^2 + 4 B E G^2 + 4 C F G^2 - 2 B^2 H^2 + 2 C^2 H^2 - 2 D^2 H^2 - 2 E^2 H^2 + 2 F^2 H^2 - 2 G^2 H^2 + 4 B E H^2 - 4 C F H^2 - 4 B^3 E - 4 B C^2 E - 4 B D^2 E - 4 C^3 F - 4 C D^2 F - 4 C E^2 F - 4 B^2 C F + 8 B C E F)} + D \sqrt{-(B - E)^2 (A^4 - 4 D A^3 + 2 B^2 A^2 + 2 C^2 A^2 + 6 D^2 A^2 + 2 E^2 A^2 + 2 F^2 A^2 - 2 G^2 A^2 - 2 H^2 A^2 - 4 B E A^2 - 4 C F A^2 - 4 D^3 A - 4 D E^2 A - 4 D F^2 A + 4 D G^2 A + 4 D H^2 A - 4 B^2 D A - 4 C^2 D A + 8 B D E A + 8 C D F A + B^4 + C^4 + D^4 + E^4 + F^4 + G^4 + H^4 - 4 B E^3 - 4 C F^3 + 2 B^2 C^2 + 2 B^2 D^2 + 2 C^2 D^2 + 6 B^2 E^2 + 2 C^2 E^2 + 2 D^2 E^2 + 2 B^2 F^2 + 6 C^2 F^2 + 2 D^2 F^2 + 2 E^2 F^2 - 4 B E F^2 - 2 B^2 G^2 - 2 C^2 G^2 - 2 D^2 G^2 - 2 E^2 G^2 - 2 F^2 G^2 + 4 B E G^2 + 4 C F G^2 - 2 B^2 H^2 + 2 C^2 H^2 - 2 D^2 H^2 - 2 E^2 H^2 + 2 F^2 H^2 - 2 G^2 H^2 + 4 B E H^2 - 4 C F H^2 - 4 B^3 E - 4 B C^2 E - 4 B D^2 E - 4 C^3 F - 4 C D^2 F - 4 C E^2 F - 4 B^2 C F + 8 B C E F)}}{2 (B - E) (A^2 - 2 D A + B^2 + D^2 + E^2 - 2 B E)}$$

$$z = F$$


问:
Why is this axiom missing?
My course book for mathematical structures uses a different set of axioms than Peano in order to define the natural numbers, and the operations on those:

We take a set $\mathbb{N}$, two elements $0,1 \in \mathbb{N}$, and two functions $+: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$, and $\cdot: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ such that the following statements hold $\forall a,b,c \in \mathbb{N}$:

$a + b = b+a$

$a + (b+c) = (a+b) + c$

$a + 0 = 0 + a = a$

$a + b = a + c \implies b = c$

$a \cdot b = b\cdot a$

$a \cdot(b\cdot c) = (a\cdot b)\cdot c$

$a \cdot1 = 1\cdot a = a$

$a \cdot(b + c) = a \cdot b + a \cdot c$

$0 \neq 1$

$1 + a \neq 0$

If $A \subset \mathbb{N}$ such that $0 \in A$ and $a \in A \implies a+1 \in A$, then $A = \mathbb{N}$.

Their claim is now, that this uniquely defines the natural numbers, and I believe that. I did, however, notice that the addition axioms 1-3 were repeated as 5-7 for multiplication, but the 4th axiom was not. I have checked an older (by 10 years) version of the course book, and this axiom was there at the time, meaning it was for some reason removed somewhere between then and now. This seems to imply that "$a \cdot b = a\cdot c \implies b = c$" can be constructed from the other axioms, but I don't see how that would be the case. The only similar statement would be 4, but it is not clear to me how a multiplication can be written as only additions, which would be necessary to use that axiom. Does anyone have any hint on why this should be the case?

I have tried googling this question, but not knowing what the property this axiom describes is called in English didn't help (I only know it's "schrapwet" in dutch).

答: If I am right, the property $a\ne0,a\cdot b=a\cdot c\implies b=c$ can be proven by induction, using 8., 4., 5. and 11.

Title: Why is this axiom missing?


答:
A broad family of categories - aka algebraic structures - can be constructed from sets with a unified kind of operation.

There is a generalization of categories that are called "operads": here you prescribe not only arrows from $x$ to $y$, but also " multiarrows" from $x_1, \ldots, x_n$ to $y$. You can imagine these as corollas with root $y$ and leaves $x_1, \ldots, x_n$. If you then have $n$ new corollas with roots $x_1, \ldots, x_n$, you can compose them through the old corolla and get a new one based at $y$. Beside the details, this is a way of generalizing composition to multimaps.

An operad is now associated to a certain kind of structures; for example, there will be the operad $Ass$ that encodes the axioms for a monoid (an associative multiplication with unity). A functor of operads generalized exactly a function of categories: it sends objects to objects, multiarrows to multiarrows; it preserves identity and composition.

Also, given an ordinary category with a "tensor product" (like the product in set and the tensor product in vector spaces), it has a natural operad structure with multimaps given by

$$Mul(x_1, \ldots, x_n ; y) := Hom(  x_1 \otimes \ldots \otimes x_n, y) $$

It turns out that if you want to define the category of, say, all monoids, it coincides with the category of functors from $Ass$ to $Set$, where the latter is equipped with the operad structure given by the ordinary product of sets.

What is the advantage of such reformulation? We actually had to define an operad $Ass$ (say "the axioms") to have a well defined category $Fun(Ass,Set)$ of $Ass$-algebras (the algebras defined by the axioms). Now say you want to define the category of $\mathbb{C}$-algebras, i.e. associative rings over complex numbers. This is the same as a monoid where the ground set is a vector space, and the multiplication is bilinear; in other the words, the functors should be $Vec_{\mathbb{C}}$ valued. We get for free that $\mathbb{C}-Alg=Fun(Ass, Vec_{\mathbb{C}}) $.

In this way you get for free something: you separated the degree of freedom about the axioms, and about where the axioms take place. But remember: if you want to create something new, you always have to put new information somewhere. I also recommend you look up the ind-category construction.

Title: Can you "construct" a category from another category?

Body:
It's perfectly possible that I have no idea what I'm talking about here, as I know little about category theory, but my understanding is that a category is a space of some objects equipped with some "arrow" (functions in a loose sense) on those objects. Since it seems that these categories are defined typically quite independently from each other, and in a standard undergrad curriculum you might see the "bottom-up" construction of certain spaces from more elementary spaces, I am wondering if you can somehow "construct" one category from the other (and so in a sense they are not independent).

As an example, it is my understanding that you can (in principle) construct much of mathematics starting from nothing more than the ZFC axioms of set theory (the first few lectures in this series does this nicely; in particular, he constructs topological spaces.). This makes it look like (to my naive eyes) you can construct Top from Set. However, reading on the Axiom of Choice, we know that it "fails" on Top, but "works" in Set, so presumably there is some contradiction. See for example, this answer on Math Overflow (incidentally, I don't totally understand this answer; topologists use the Axiom of Choice in the form of Tychonoff's Theorem or Zorn's Lemma often by my understanding, yet the category they work in forbids it?).

Perhaps my misunderstanding lies in some difference between the category Set and ZFC set theory, or the space of topological objects that topologists work in and Top?

Edit: the comments gave some examples of categories constructed from other categories, a trivial one being the opposite category. Refining my question, I am wondering if you can specifically construct Top from Set, and if construction of objects (ignoring category theory) from ZFC implies that you can construct the category of those objects from Set. Also, is the axiom of choice (as in the linked MO question) problematic, because it "fails" in some categories but not others?


问:
Equivalence of categories preserves colimits
Is it true that if $F: \cal{C} \rightarrow \cal{D}$ is an equivalence of categories (I mean, is full, faithfull and essentially surjective over objects), then $F$ preserve colimits?

答:
Yes.

In such a case $F$ has an inverse functor $G:\mathcal D\to\mathcal C$, such that $FG\simeq{\mathrm id}_{\mathcal D}$ and $GF\simeq{\mathrm id}_{\mathcal C}$, and it's easy to see that $G$ is both a left and right adjoint of $F$.

Consequently $F$ preserves all limits and colimits, and so does $G$.


问:
How to find the integral of $\tan x - \sec x\tan x$?
$$\int(\tan x - \sec x\tan x)dx$$

I rewrote the integral like this:

$$\int\left(\frac{\sin x}{\cos x}- \sec x\tan x\right)dx$$

and using $u$-substitution for the first term I got a final answer of:

$$-\ln|\cos x| - \sec x + c$$

Where was my mistake?

答:
You made no mistake.

Going to explain a bit to be clearer.

The first part is correct for
$$\int \tan(x)\ \text{d}x = -\ln|\cos(x)| + c_1$$
and this is easy (you just see the tangent as sine/cosine recognising it has the form $f'(x)/f(x)$ (with a minus sign) which is nothing but a logarithmic derivative.

About the second term:

$$\int \sec(x)\tan(x)\ \text{d}x = \sec(x) + c_2$$

Indeed by the definition: $\sec(x) = \dfrac{1}{\cos(x)}$

We have

$$\dfrac{\text{d}}{\text{d}x} \sec(x) = \dfrac{\sin(x)}{\cos^2(x)} \equiv \dfrac{1}{\cos(x)}\dfrac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)$$

And of course the constant vanished in the derivative.

You made no mistake.


问:
Prove that $z_1z_2z_3 = 1$ for complex numbers $z_1$, $z_2$, $z_3$.
Let $z_1$, $z_2$, $z_3$ be distinct complex numbers such that $|z_1| = |z_2| = |z_3| > 0$. If $z_1 + z_2z_3$, $z_2 + z_3z_1$, and $z_3 + z_1z_2$ are real numbers, prove that $z_1z_2z_3 = 1$.

I believe that this problem comes from 1979 Romanian Math Olympiad (State Competition, 10th grade).

I tried several approaches: The first was to substitute $a + ib$, $c + id$, and $f + ig$ for $z_1$, $z_2$, and $z_3$, respectively, and bash. I used $(1+z_1)(1+z_2)(1+z_3)$, which didn't really get me anywhere.

After a couple more attempts at substitutions and such, I hit a wall and didn't really know what to do.

答:
My solution relies heavily on the fact that the two sides of a parallelogram are equal if and only if the diagonals of the said parallelogram are orthogonal (when the parallelogram is a rhombus). On the complex plane this can be reformulated as follows: two complex numbers share the same magnitude if and only if their sum and difference vectors are orthogonal to each other.

I will first prove that the common length of the three complex numbers is $=1$. I will show that $|z_3|=1$, the others are done in the same way.

Consider the known real numbers $u=z_1+z_2z_3$ and $v=z_2+z_1z_3$. We have that
$$
u+v=(z_1+z_2)(1+z_3)
$$
and
$$
u-v=(z_1-z_2)(1-z_3)
$$
are also both real. Let us denote $\phi_k=\arg z_k, k=1,2,3$. Because $|z_1|=|z_2|$, the parallelogram with sides $z_1$ and $z_2$ is a rhombus. From this (or by a number of other means) it follows that modulo an integer multiple of $\pi$ we have
$$
\arg(z_1+z_2)\equiv\frac{\phi_1+\phi_2}2
$$
and
$$
\arg(z_1-z_2)\equiv\frac{\pi+\phi_1+\phi_2}2.
$$
In other words, the arguments of these two numbers differ from each other by $\pm\pi/2$.

As $\arg(u+v)\equiv \arg(u-v)\equiv0\pmod\pi$, the arguments of
$1-z_3$ and $1+z_3$ must also differ from each other by $\pm\pi/2$. Therefore the diagonals $1\pm z_3$ of the parallelogram with sides $1$ and $z_3$ are orthogonal, making the said parallelogram a rhombus. Therefore $|z_3|=1$ as claimed.

Another consequence of $|z_3|=1$ is that
$$
\arg(1-z_3)=\frac{\pi+\phi_3}2.
$$
Putting these pieces together we get (modulo $\pi$)
$$
0\equiv\arg(u-v)=\arg(z_1-z_2)+\arg(1-z_3)=\frac{2\pi+\phi_1+\phi_2+\phi_3}2\equiv\frac{\phi_1+\phi_2+\phi_3}2.
$$
So $\phi_1+\phi_2+\phi_3$ is an integer multiple of $2\pi$ and the claim follows.


问:
Equivalence Relations and functions on partitions of Sets
let $f$ be a function on $A$ onto $B$.  Define an equivalence relation $E$ in $A$ by: $aEb$ if and only if $f(a)=f(b)$.

Define a function $\phi$ on $A/E$ by $\phi([a]_{E})=f(a)$.

Hint: Verify that $\phi([a]_{E})=\phi([a']_{E}) $ if $[a]_{E} = [a']_{E}$

I know that A/E is a partition of A.

I know that a binary relation $F$ is called a function if $aFb$ and $aFc$ implies $b=c$ for any $a, b, \text {  and  } c$

答:
Hint: Suppose that $\langle [a]_E,x\rangle\in \phi$ and $[a]_E=[a']_E$, then $x=f(a)=f(a')$, conclude the wanted conclusion.

Bonus points: Show that $\phi$ is injective!

More bonus points: Show that every equivalence relation is actually induced like this by some function $f$.

(If you do both the bonus points thingies (they are quite short actually), then you have completely understood the topic.)


问:
How to evaluate Riemann Zeta function for argument $1-2^{1-s}=0$
I would like to code Riemann Zeta function in Scilab. Even if I would only need computation for real arguments, I tried to code a fully operational function for complex argument (hoping it can help anyone…). I based my code on previous exchanges referring to this paper indicating P.Borwein process.

$$\zeta(s)=\frac{1}{d_0(1-2^{1-s})}\sum_{k=1}^{n}\frac{(-1)^{k-1}d_k}{k^s}+\gamma_n(s)$$
with
$$d_k=n\sum_{j=k}^{n}\frac{(n+j-1)!4^j}{(n-j)!(2j)!}$$
and by noting $s=\sigma+it$
$$\left | \gamma_n(s) \right |\leq \frac{3}{(2+\sqrt8)^n} \frac{(1+2\left | t \right |)e^{\left | t \right |\pi/2}}{\left | 1-2^{1-s} \right |}$$

This computation is really efficient but the amount of terms according to residual error is only valid for $\Re(s)\geq \frac{1}{2}$.

Another computation is also described in the paper with the residual error maximization available for $\Re(s)>0$ or $\Re(s)<0$.

However, when $s\approx 1+p \cdot \frac{2\pi i}{log(2)}$, the functional equation cannot be used to reflect to a computable argument: real part is $0$ or to close to ensure accurate computation... Well it seems I cannot use this fast computation process for these arguments.

I found on Wikipedia (French) the orientation to compute $\frac{1}{\zeta(s)}$, and also the following more complexe alternating series, that I also read in this document or D.V. Widder, Laplace Transform book.

$$\zeta(s)=\frac{1}{1-3^{1-s}}\sum_{n=1}^{\infty}\left ( \frac{1}{(3n-2)^s} + \frac{1}{(3n-1)^s} -\frac{2}{(3n)^s}\right )$$

It looks like promising but no orientation as P.Borwein with Chebychev polynomials. I did not succeed in identifying any paper, so if anyone can help…

Does a similar fast method with error evaluation exist for this expression?
If no, I guess that Euler Maclaurin summation method will work but I will need to investigate it to understand.

Thanks in advance!

答:
If $1-2^{1-s}=0$ and $s\ne 1$ then
$$\zeta(s)\log 2= (\zeta(s)(1-2^{1-s}))'=\frac{1}{d_0}\sum_{k=1}^{n}\frac{(-1)^{k}d_k}{k^s}\log k+\gamma_n(s)\log 2$$

For $z$ near such a $s$ you get that $(\zeta(z)(1-2^{1-z}))'$ is close to $\zeta(z)\log 2$, where near and close relies on a bound for $|\zeta'(1+it)|$ obtained from a bound for $|\zeta(\sigma+it)|$.


问:
Application of Induction in the analyze of the convergence a sequence defined recursive.

Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad  n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.

My approach: Note that  $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly  I would like to find $a_{n}$. Now, I was trying to find a  pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.

How can find $a_{n}$?

Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.

答:
Define the function
$$
f(a)=\frac1{4-3a}\tag1
$$
Note that
$$
\begin{align}
f(a)-a
&=\frac{(3a-1)(a-1)}{4-3a}\tag{2a}\\
&\left\{\begin{array}{}
\lt0&\text{if }a\in\left(\frac13,1\right)\cup\left(\frac43,\infty\right)\\
\gt0&\text{if }a\in\left(-\infty,\frac13\right)\cup\left(1,\frac43\right)
\end{array}\tag{2b}
\right.
\end{align}
$$
Consider the two sequences for $n\in\mathbb{Z}$,
$$
\begin{align}
p_n
&=\frac{3^{n-1}+1}{3^n+1}\tag{3a}\\
&=\frac13\left(1+\frac2{3^n+1}\right)\tag{3b}
\end{align}
$$
and
$$
\begin{align}
q_n
&=\frac{3^{n-1}-1}{3^n-1}\tag{4a}\\
&=\frac13\left(1-\frac2{3^n-1}\right)\tag{4b}
\end{align}
$$
where $q_0=\pm\infty$.

Note that
$$
\begin{align}
f(p_n)&=p_{n+1}\tag{5a}\\
f(q_n)&=q_{n+1}\tag{5b}
\end{align}
$$
where, in the case of $q_0$,
$$
\begin{align}
f(q_{-1})&=f\!\left(\tfrac43\right)=\infty=q_0\tag{6a}\\
f(q_0)&=f(\infty)=0=q_1\tag{6b}
\end{align}
$$
Define the intervals
$$
\begin{align}
P_n&=(p_{n+1},p_n)\tag{7a}\\
Q_n&=(q_n,q_{n+1})\tag{7b}
\end{align}
$$
where $Q_{-1}=\left(\frac43,\infty\right)$ and $Q_0=\left(-\infty,0\right)$:

In the animation above, the solid red and green lines are the $P_n$ and $Q_n$. The arrows point to the dotted intervals $P_{n+1}$ and $Q_{n+1}$. The intervals are red if $f(a)\lt a$ on that interval and green if $f(a)\gt a$; these intervals are described in $(2)$.

Since $f'(a)\gt0$ except at $q_{-1}=\frac43$ (which is between $Q_{-2  }$ and $Q_{-1}$), we have the bijections
$$
\begin{align}
f&:P_n\to P_{n+1}\tag{8a}\\
f&:Q_n\to Q_{n+1}\tag{8b}
\end{align}
$$
Since
$$
\bigcup_{n\in\mathbb{Z}}P_n\cup\bigcup_{n\in\mathbb{Z}}Q_n\cup\left\{p_n:n\in\mathbb{Z}\right\}\cup\left\{q_n:n\in\mathbb{Z}^{\ne0}\right\}=\mathbb{R}\tag9
$$
$(5)$ and $(8)$ show that for all points except $\left\{q_n:n\le0\right\}\cup\{1\}$, iterating $f$ will produce a seqence converging to $\frac13$ (one might even say that $q_{-\infty}=1$).


问:
Problem about absolute value of percentage error

$(a)$ Bound the error in the approximation $\sin(x)\approx x$ for
$-\frac{\pi}{4}\le x \le \frac{\pi}4$.

$(b)$ Since this is a good approximation for small values of $x$, also
consider the "percentage error"

$$\frac{\sin(x)-x}{\sin(x)}\approx\frac{\sin(x)-x}{x}$$ Bound the
absolute value of the latter quantity for $-\delta\le x\le\delta$. Pick
$\delta$ to make the absolute value of the percentage error less than
$1$%.

I've successfully solved part $a$. for part $b$ I think I supposed to solve the problem with Taylor Remainder theorem (like first part):

$$|\frac{\sin(x)-x}{x}|\le0.01$$
For $\sin(x)$ we have: $$R_{2n}=\frac{x^{2n+1}}{(2n+1)!}\times\cos(c),\quad 0\le c\le x$$
Substitute it in the inequality:

$$0.99\le\frac{x^{2n}}{(2n+1)!}\times\cos(c)\le1.01,\quad0\le c \le x$$
I don't know how to find $\delta$ in original question from the above inequality.

答:
If you have the Taylor series $\sin(x)= x -\frac1{3!}x^3+\frac1{5!}x^5-\cdots$

then $\left|\frac{\sin(x)-x}{x}\right|\le \frac1{3!}x^2$

so $|x|<\sqrt{0.06}\approx 0.24$ will give $\left|\frac{\sin(x)-x}{x}\right|\le 0.01$


问:
What is the difference between, a "square" and a "perfect-square", number?
Is, "36", a perfect square?

I know that, "4" is a perfect square.

Similarly, "1","9","25", are "perfect-square"s.

答:
It's interesting how a discipline that is so reliant on the precision of its language can still have terms that are fuzzy.

Anyhow, for what it's worth, my interpretation of "square number" is any number that is formed by squaring a natural number (or maybe an integer at a push).

Many people refer to $a^2-b^2$ as a difference of perfect squares and that informs my interpretation of "perfect square number" as one whose square root is a rational number (remembering that this includes integers).


问:
How do I evaluate the following expression?
How to evaluate the following expression:  </p>

<p>$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}$</p>

答:
HINT:

$$\frac1{\sqrt{r+1}+\sqrt r}=\frac{\sqrt{r+1}-\sqrt r}{r+1-r}=\sqrt{r+1}-\sqrt r$$

Can you see the Telescoping Series?


问:
confusion in finding $I=\frac{1}{2\pi j} \oint_c Re\{ z\} \, dz$

If C denotes the counterclockwise unit circle, then the value of the contour integral
$$\frac{1}{2\pi j} \oint_c Re\{ z\} \, dz \quad \text{is _____ }?$$

Method $1$:

$$Re\{ z\} = \frac{z + \bar{z}}{2} \quad \dots (1)$$
$$\implies \frac{1}{2\pi j} \oint_c Re\{ z\} \, dz = \frac{1}{2\pi j} \oint_c \frac{z + \bar{z}}{2} \, dz$$
$$=\frac{1}{2\pi j} \oint_c \frac{z }{2} \, dz + \frac{1}{2\pi j} \oint_c \frac{\bar{z}}{2} \, dz \quad \dots (2)$$
Now using Cauchy-Integral theorem, we get:
$$\frac{1}{2\pi j} \oint_c \frac{z }{2} \, dz=0 \quad \dots (3)$$
as $f(z)=z$ is differentiable at every point inside C ($|z|=1$)

and $\bar{z}=\frac{1}{z}$ for C( $|z|=1$),

so, $eq(2)$ can be rewritten as:
$$\frac{1}{2\pi j} \oint_c \frac{z }{2} \, dz + \frac{1}{2\pi j} \oint_c \frac{\bar{z}}{2} \, dz = 0 + \frac{1}{2}\frac{1}{2\pi j} \oint_c \frac{1}{z} \, dz$$
$$\frac{1}{2} .1=\frac{1}{2}$$

Method $2$:

Let $z=re^{j \theta}$

$$\implies \frac{1}{2\pi j} \oint_c Re\{ z\} \, dz = \frac{1}{2\pi j} \oint_c r cos (\theta) \, d(re^{j \theta}) \quad \dots (4)$$
Now, for C : $r=1 $ and $  \theta \to 0 $ to $2\pi $
$$=\frac{1}{2\pi j} \int_0^{2\pi} cos (\theta) e^{j \theta}j \, d\theta$$
$$=\frac{1}{2\pi } \int_0^{2\pi} cos (\theta) e^{j \theta} \, d\theta \quad \dots (5)$$
Now,
$$\int e^{ax}cos(bx) dx = \frac{e^{ax}}{a^2 +b^2}\{ a cos(bx) + b sin (bx)\} +C \quad \dots  (6)$$
$$\implies \int_0^{2\pi} cos (\theta) e^{j \theta} \, d\theta = [\frac{e^{jx}}{j^2 +1^2}\{ j cos(x) +  sin (x)\}]_0^{2\pi}$$
Here, denominator is coming to $0$ , so we can't use the formula described in $eq(6)$;

Hence how to evaluate $eq(5)$ ?

Any help or suggestions please...

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{{1 \over 2\pi\ic}\oint_{\verts{z} = 1}
\Re\pars{z}\dd z} =
{1 \over 2\pi\ic}\oint_{\verts{z} = 1}
{z + \overline{z} \over 2}\,\dd z
\\[5mm] = &\
{1 \over 2\pi\ic}\oint_{\verts{z} = 1}
{1 \over 2}\pars{z + {1 \over z}}\,\dd z
\\[5mm] = &\
{1 \over 2\pi\ic}\,{1 \over 2}\,\pars{2\pi\ic \times 1} =
\bbx{1 \over 2} \\ &
\end{align}


问:
Finding a probability when there is a biased die
Let W
be the number shown on a biased die. It is known that ℙ(=)=0.2 for
w
even, and also that the probability the die shows a prime number is 0.5. What is ℙ(=1)?

[Note: You can assume the die has six sides.]

How do i approach this question as generally i would draw a pmf table however doing this gives me more than 1 which indicates that it is incorrect, any tips?

My working out -
P(w=2,4,6)=.2
P(w=2,3,5)=0.5
As 2 is common i multiplied give p(w=2)=0.10

Body:
We know that $P(W=2) = P(W=4) = P(W=6) = 0.2$, and that $0.5 = P(W\in\{2,3,5\}) = P(W=2)+ P(W=3) + P(W=5)$. This implies that $P(W=3) + P(W=5) = 0.3$.

As we also know that
$\begin{align*}
1 &= P(W=1) + P(W=2) + P(W=3) + P(W=4) + P(W=5) + P(W=6)\\
 &= P(W=1) + 0.2 + 0.2 + 0.2 + (P(W=3) + P(W=5))\\
&= P(W=1) + 0.6 + 0.3
\end{align*}$

we can calculate the value of $P(W=1)$.


答:
Here are two formulas for the Riemann zeta-function:
$$
\zeta(s) = \frac{\zeta_2(s)}{1 - 2/2^s} = \frac{\zeta_3(s)}{1 - 3/3^s},
$$
where
$$
\zeta_2(s) =  1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \frac{1}{5^s} - \frac{1}{6^s} + \dots
$$
and
$$
\zeta_3(s) = 1 + \frac{1}{2^s} - \frac{2}{3^s} + \frac{1}{4^s} +
\frac{1}{5^s} - \frac{2}{6^s} +
\frac{1}{7^s} + \frac{1}{8^s} - \frac{2}{9^s} + \dots,
$$
are series that converge for ${\rm Re}(s) > 0$.
It appears that you are using the first formula with some accelerated version of the series for $\zeta_2(s)$ and you also mention awareness of the second formula. Have you tried to accelerate the convergence of $\zeta_3(s)$? Maybe Section 3 here could be helpful.

The denominator $1 - 2/2^s$ vanishes when $s = 1 + 2\pi ik/\log 2$ and the denominator $1 - 3/3^s$ vanishes when $s = 1 + 2\pi i\ell/\log 3$, for integers $k$ and $\ell$. No number has both forms except $s = 1$, which is a pole of the zeta-function.  So away from $s = 1$ you can use the formula for $\zeta(s)$ in terms of $\zeta_3(s)$ is $s$ is close to a zero of $1 - 2/2^s$.

Title: How to evaluate Riemann Zeta function for argument $1-2^{1-s}=0$

Body:
I would like to code Riemann Zeta function in Scilab. Even if I would only need computation for real arguments, I tried to code a fully operational function for complex argument (hoping it can help anyone…). I based my code on previous exchanges referring to this paper indicating P.Borwein process.

$$\zeta(s)=\frac{1}{d_0(1-2^{1-s})}\sum_{k=1}^{n}\frac{(-1)^{k-1}d_k}{k^s}+\gamma_n(s)$$
with
$$d_k=n\sum_{j=k}^{n}\frac{(n+j-1)!4^j}{(n-j)!(2j)!}$$
and by noting $s=\sigma+it$
$$\left | \gamma_n(s) \right |\leq \frac{3}{(2+\sqrt8)^n} \frac{(1+2\left | t \right |)e^{\left | t \right |\pi/2}}{\left | 1-2^{1-s} \right |}$$

This computation is really efficient but the amount of terms according to residual error is only valid for $\Re(s)\geq \frac{1}{2}$.

Another computation is also described in the paper with the residual error maximization available for $\Re(s)>0$ or $\Re(s)<0$.

However, when $s\approx 1+p \cdot \frac{2\pi i}{log(2)}$, the functional equation cannot be used to reflect to a computable argument: real part is $0$ or to close to ensure accurate computation... Well it seems I cannot use this fast computation process for these arguments.

I found on Wikipedia (French) the orientation to compute $\frac{1}{\zeta(s)}$, and also the following more complexe alternating series, that I also read in this document or D.V. Widder, Laplace Transform book.

$$\zeta(s)=\frac{1}{1-3^{1-s}}\sum_{n=1}^{\infty}\left ( \frac{1}{(3n-2)^s} + \frac{1}{(3n-1)^s} -\frac{2}{(3n)^s}\right )$$

It looks like promising but no orientation as P.Borwein with Chebychev polynomials. I did not succeed in identifying any paper, so if anyone can help…

Does a similar fast method with error evaluation exist for this expression?
If no, I guess that Euler Maclaurin summation method will work but I will need to investigate it to understand.

Thanks in advance!


问:
A problem of "guessing number"
The problem statement:

I have two people A and B, and think of a natural number n. Then I
give the number n to A and the number n + 1 to B. I tell them that
they have both been given natural numbers, and further that they are
consecutive natural numbers. However, I don't tell A what B's number
is and vie versa. I start by asking A if he knows B's number. He says
\no". Then I ask B if he knows A's number, and he says \no" too. I go
bak to A and ask, and so on. A and B an both hear each other's
responses. Do I ever get a \yes" in response? If so, who responds first
with \yes" and how many times does he say \no" before this? Assume
that both A and B are very intelligent and logical.

I understand we should start from a simple case where $A = 1$, and $B = 2$ ---- obviously number of "no" from $A$ is $0$.

Then for $A = 2, B = 3$ case, $A$ responds with "no" because $B$ could be $1$ or $3$, then $B$ responds with "no", so $A$ knows $B$ is not $1$, then in the second respond he says "yes".

So on and so forth.

However, it's not clear to me how should I extend the case to $n$ and $n+1$? Maybe better to have a recursive relation between $(n,n+1)$ and $(n-1,n)$ case?

答: Your examples suggest that if $A$'s number is $n$, then $A$ can guess the number within $n$ guesses.  You've already proved the $n=1$ case.  Suppose that $A$'s number is $n>1$ and that the statement is true for $n-1$.  $A$ knows that $B$'s number is $n-1$ or $n+1$.  Suppose no one has guessed the number after  $n-1$ turns.  Then $A$ can reason as follows.  Reversing the roles of $A$ and $B$, if $B$'s number were $n-1$ he would have known it by turn number $n-1$, by the induction hypothesis.  ($B$ can just ignore $A$'s first statement.)  Therefore, on his $n$th turn, $A$ knows that $B$'s number is $n+1$.

Title: A problem of "guessing number"


问:
List of interesting integrals for early calculus students
I am teaching Calc 1 right now and I want to give my students more interesting examples of integrals.  By interesting, I mean ones that are challenging, not as straightforward (though not extremely challenging like Putnam problems or anything).  For example, they have to do a $u$-substitution, but what to pick for $u$ isn't as easy to figure out as it is usually.  Or, several options for $u$ work so maybe they can pick one that works but they learn that there's not just one way to do everything.

So far we have covered trig functions, logarithmic functions, and exponential functions, but not inverse trig functions (though we will get to this soon so those would be fine too).  We have covered $u$-substitution.  Thinks like integration by parts, trig substitution, and partial fractions and all that are covered in Calc 2 where I teach.  So, I really don't care much about those right now.  I welcome integrals over those topics as answers, as they may be useful to others looking at this question, but I am hoping for integrals that are of interest to my students this semester.

答:
I remember one, rather good:

$$\int \dfrac{\text{d}x}{x^4+1}$$

Contrarily to a very similar one where there is a minus sign, this is rather tough but challenging. It starts with partial fractions and substitutions. Give yourself a try, the result is rather elegant:

$$\frac{\log \left(x^2+\sqrt{2} x+1\right)}{4 \sqrt{2}}-\frac{\log \left(x^2-\sqrt{2} x+1\right)}{4 \sqrt{2}}+\frac{\tan ^{-1}\left(\sqrt{2} x+1\right)}{2 \sqrt{2}}-\frac{\tan ^{-1}\left(1-\sqrt{2} x\right)}{2 \sqrt{2}} + c$$


答: To show that $\phi$ is a function, you must follow the hint. If $[a]_E=[a']_E$, then what can we say about $f(a)$ and $f(a')$? What then can be said about $\phi\left([a]_E\right)$ and $\phi\left([a']_E\right)$? You'll need to keep the definitions of $E,\phi$ in mind.</p>

Title: Equivalence Relations and functions on partitions of Sets

Body:
let $f$ be a function on $A$ onto $B$.  Define an equivalence relation $E$ in $A$ by: $aEb$ if and only if $f(a)=f(b)$.

Define a function $\phi$ on $A/E$ by $\phi([a]_{E})=f(a)$.

Hint: Verify that $\phi([a]_{E})=\phi([a']_{E}) $ if $[a]_{E} = [a']_{E}$

I know that A/E is a partition of A.

I know that a binary relation $F$ is called a function if $aFb$ and $aFc$ implies $b=c$ for any $a, b, \text {  and  } c$


问:
If $ax^3+bx+c=0$, $bx^3+cx+a=0$, $cx^3+ax+b=0$ have a common root for distinct non-zero $a$, $b$, $c$, then at least one has three real roots

Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations
$$\begin{align}
E_1: ax^3+bx+c=0 \\
E_2: bx^3+cx+a=0 \\
E_3: cx^3+ax+b=0
\end{align}$$
have a common root, prove that at least one of these equations has three real roots (not necessarily distinct).

This question was a math olympiad problem.Below was my attempt but I've got confused:

I know that the discriminant of the equation $x^3+Ax+B=0$ is given by $-4A^3-27B^2$, so it suffices to prove that the discriminant of at one least of the equations is non-negative.

It's easy to prove that the common root $r$ must be real. Now assume on the contrary that all three equations had a pair of complex conjugate roots. Let $\alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma, \bar{\gamma}$ be the complex conjugate roots of $E_1, E_2$ and $E_3$ respectively.

Then by viete theorem, we have $$\alpha \bar{\alpha} r=|\alpha|^2r=\frac{-c}{a},$$
$$|\beta|^2r=\frac{-a}{b},$$
$$|\gamma|^2r=\frac{-b}{c}.$$
Multiplying the three equations gives $$|\alpha \beta \gamma|^2r^3=-1.$$
Hence $r<0.$ Hence $-c/a<0$ or $c/a>0.$ Similarly $a/b>0$ and $b/c>0.$ This implies that $a,b,c$ all all positive or or negative.

By symmetry, we can assume $a>b>c>0$, but then none of the discriminant could be non-negative. So what's wrong with the argument and how can I complete the proof? Thanks in advance!

答:
Suppose all three have only one solution.

Then $3ax^2+b$ is always positive or negative i.e. $ab<0$. The same way we get $ac<0$ and $bc<0$. Impossible.


问:
How do I show that $\sum^{\infty}_{n=1}\frac{x}{n^2}$ is not uniformly convergent on $\mathbb{R}$?
How do I show something is not uniformly convergent? I was thinking maybe showing that the property of uniform convergence(continuity/boundedness preserved, integrity commutes etc) doesn't hold, but the problem is I haven't learned how to calculate $\sum^{\infty}_{n=0}1/n^2$

答: With $f : x \in \mathbb{R} \mapsto \displaystyle \sum_{k=1}^{+\infty} \dfrac{x}{k^2}$ and for all $n \in \mathbb{N}^*$, $f_n: x \in \mathbb{R} \mapsto \displaystyle \sum_{k=1}^{n} \dfrac{x}{k^2}$. We know that $(f_n)_n$ converges simply to $f$. But for $n \in \mathbb{N}^*$, we have $ f-f_n : x \in \mathbb{R} \mapsto \displaystyle \sum_{k=n+1}^{+\infty} \dfrac{x} {k^2} = x \sum_{k=n+1}^{+\infty} \dfrac{1} {k^2}$ that is not bounded because $\sum_{k=n+1}^{+\infty} \dfrac{1} {k^2} \neq 0$ therefore $(f_n)_n$ doesn't uniformly converge on $\mathbb{R}$.


问:
Showing that a function in a measure space does not converge in measure
Let $(X, \mathbb{X}, \mu)$ be a measure space. A sequence of real-valued measurable functions $(f_n)$ is said to converge in measure to a real-value measurable function $f$ if

$$\lim_{n\to \infty}\mu(\{x\in X:|f_n(x) - f(x)|\ge \alpha\})=0$$

for all $\alpha>0$.

I have to prove that $f_n = \chi_{[n,n+1]}$ converges pointwise to $f=0$ and that it does not converge in measure.

I have shown that it does converge pointwise but I don't know how to show that it doesn't converge in measure to any function.

If it converges in measure then it must converge in measure to $f=0$? If so, then assuming that it does converge in measure, for $\alpha=1$ and $n\in\mathbb{N}$, $\mu(\{x\in X:|f_n(x) - f(x)|\ge \alpha\}) = \mu(\{x\in X:\chi_{[n,n+1]}(x)\ge 1\})=\mu([n,n+1])=1$ (I'm assuming Lebesgue measure), a contradiction.

But I'm struggling to prove this assumption.

答:
First, start by saying, we assume that $f_n$ converges in measure to some function $f$.

Fix some $R>0$, then we have for $n>R$ and every $\alpha >0$
$$ \mu(\{ x \in X  \ : \ \vert f_n(x) - f(x) \vert \geq \alpha \})
\geq \mu(\{ x \in [-R,R]  \ : \ \vert f_n(x) - f(x) \vert \geq \alpha \})
= \mu(\{ x\in [-R,R] \ : \ \vert f(x) \vert \geq \alpha \}). $$
However, the RHS is independend of $n$ and the LHS goes to zero, thus, we get
$$ \mu(\{ x\in [-R,R] \ : \ \vert f(x) \vert \geq \alpha \}) =0 $$
for every $\alpha >0$. This implies that $f=0$ a.e. on $[-R,R]$ as we have
$$ \mu(\{ x\in [-R,R] \ : \ \vert f(x) \vert \neq 0 \})
= \mu( \bigcup_{m\geq \mathbb{N}_{\geq 1}}\{ x\in [-R,R] \ : \ \vert f(x) \vert \geq 1/m \})
= \lim_{m\rightarrow \infty} \mu( \{ x\in [-R,R] \ : \ \vert f(x) \vert \geq 1/m \})
= \lim_{m\rightarrow \infty} 0 =0.
$$
However, $R>0$ was arbitrary, so we get that $f=0$ a.e.


问:
Calculating expected value of non-empty chambers. What is wrong in my solution?
Instering $5$ different balls to $12$  different chambers, all the options are equally probabable. Let $X$ be the number of non-empty chambers, calculate $\mathbb{E}[X]$

I know that $\mathbb{E}(X)=\sum_{t \in R_{X}} t \cdot \mathbb{P}(X=t)$ so I just need to calculate $\mathbb{P}(X=t)$ for every $1\leq t \leq 5$
$$\mathbb{P}(X=1)= \frac{1}{12^4}$$
$$\mathbb{P}(X=2)= \frac{11 \cdot 2^3}{12^4}$$
$$\mathbb{P}(X=3)= \frac{11\cdot 10\cdot 3^2}{12^4}$$
$$\mathbb{P}(X=4)= \frac{11\cdot 10 \cdot 9 \cdot4}{12^4}$$
$$\mathbb{P}(X=5)= \frac{11\cdot 10 \cdot 9 \cdot8}{12^4}$$

now I just substitute this values in the expression $$\mathbb{E}(X)=\sum_{t \in R_{X}} t \cdot \mathbb{P}(X=t)$$ but I am not getting the right answer. I assume I have mistake in develop the expressions above. The options are:

A. $\displaystyle \frac{161051}{20736}$

B. $\displaystyle \frac{161051}{248832}$

C. $\displaystyle \frac{87781}{20736}$

D. $\displaystyle \frac{87781}{248832}$

答:
As have been already noted the probabilities you wrote down are not correct. The correct probability of $k$ from $n$ bins being non-empty after $m$ balls are placed into the bins is
$$
\mathsf P(X=k)=\frac{\binom nk {m \brace k}k!}{n^m}.
$$

Here the factor $\binom nk$ counts the number of ways to choose the $k$ non-empty bins, and the factor ${m \brace k}k!$ (where ${m \brace k}$ refers to Stirling number of the second kind) counts the number of ways to distribute $m$ distinct balls between these $k$ bins.

It is the factor ${m \brace k}$ which is missing in your solution. The origin of the error was (as explained in a previous answer) unduly fixed order of placing the balls into the bins.


问:
Code that corrects a length $s$ burst error
Show that exists an $(n,k)=(155,135)$ binary code that corrects burst errors of length $6$.

P.S. A  Code with length $n$  fixes burst error (pattern) of length $s$ if the error set of this code consists of sets of length $n$ whose non-zero components are located in $s$  consecutive bits, that is, in each such set, the locations of the first and last non-zero components differ by no more than $s - 1$.

答:
Basically if you want to correct a burst of length $s,$ your code must have the property that no two codewords differ by the sum of two (non overlapping) bursts of length $s.$

This leads to the Rieger bound, that $n-k\geq 2s$. So your parameters satisfy this. However, you probably need to look up Fire codes which would actually achieve a so called burst error correction efficiency ratio given by
$$
\frac{2s}{n-k} \geq 2/3.
$$
Good luck.


问:
Why is it true that $\int_0^\infty \sin(qx) dx = 1/q$.
In a Quantum Mechanics class the professor pointed out a calculation from Zetilli's textbook on Quantum Mechanics (page 630).

There he says that:

$$\int_0^\infty \sin(qx) dx = \lim_{\lambda \to 0} \int_0^\infty e^{-\lambda x} \sin(qx) dx$$

Then he writes the $\sin$ as complex exponentials and reaches the result that the integral is $1/q$. WHAT? I was taught at my calculus course that this integral does not converge. For me this result is plain wrong, and I do not understand it at all. I even don't see why the limit can be imposed like that. Any thoughts on this "trick"?

答:
Not an answer, just a long comment

If we consider the sum of a series in a different way (Cesàro summation), then
$$\sum_{n=1}^{\infty} \sin n x=\frac{1}{2} \cot \left(\frac{x}{2}\right)$$
Indeed
$$S_k=\sum _{i=1}^k \sin (i x)\csc \left(\frac{x}{2}\right) \sin \left(\frac{k x}{2}\right) \sin \left(\frac{1}{2} (k+1) x\right)$$
and
$$\frac{1}{n}\sum _{k=1}^n \csc \left(\frac{x}{2}\right) \sin \left(\frac{k x}{2}\right) \sin \left(\frac{1}{2} (k+1) x\right)=\\=\frac{1}{4n}\csc \left(\frac{x}{2}\right) \left(2 n \cos \left(\frac{x}{2}\right)-\csc \left(\frac{x}{2}\right) \sin ((n+1) x)+\sin (x) \csc \left(\frac{x}{2}\right)\right)$$

and finally
$$\underset{n\to \infty }{\text{lim}}\frac{1}{4n}\csc \left(\frac{x}{2}\right) \left(2 n \cos \left(\frac{x}{2}\right)-\csc \left(\frac{x}{2}\right) \sin ((n+1) x)+\sin (x) \csc \left(\frac{x}{2}\right)\right)=\frac{1}{2} \cot \left(\frac{x}{2}\right)$$


问:
Differentiating 2 sides of an equation
I'm self-learning the course Multivariable Calculus of MIT and i'm stuck at understanding the concept of "Constrained Differentials".

In the course's note, there is something like:

Suppose $x^2 + y^2 = 1$, it is claimed that that we can differentiate both sides of this constraint to get $2x dx + 2y dy = 0$

Could you please explain me why the operation like that valid ?

Thank you very much for your help!

答:
When you have a constraint it means that the values of $x$ and $y$ are not independent, but linked by $f(x,y)=0$. You are going to consider increments $dx$ and $dy$ that keep you on the same curve to the first approximation, that is you need $f(x+dx,y+dy)=0$. But $f(x+dx,y+dy)=f(x,y)+df((x,y),(dx,dy))$ up to higher order infinitesimals, so you need $df=f_xdx+f_ydy=0$.

For the second part, you get $dx=0$.

Title: Differentiating 2 sides of an equation