问:
Definition of 'product states' for $C^*$-algebras
I am slightly confused by the definition of 'product states' for $C^*$-algebras. So far I have the following:

Let $A$ and $B$ be $C^*$-algebras, $\varphi\in\mathcal{S}(A)$, $\psi\in\mathcal{S}(B)$. Then $\varphi\odot\psi$ is a positive linear functional on $A\odot B$. But at this point there is no $C^*$-norm defined on the $*$-algebra $A\odot B$, so I am at a loss how to establish that the 'product state' is indeed a state on $A\odot B$ i.e. that its norm is $1$.

Also I have seen in a proof reference to a function from 'product states of $A\odot B$ to product states of $C\odot D$'. Is every state on $A\odot B$ (for a given $C^*$-norm) a product of states on $A$ and $B$?

Also is there is any good reference where I can read more about this?

答:
1. Tensor product of $C^*$-algebras

A  $C^*$-algebra $A$ is called nuclear if, for every other $C^*$-algebra $B$, there is a unique $C^*$-norm on
$A\odot B$, whose completion is denoted   $A\otimes B$, and called the $C^*$-tensor product of $A$ and $B$.

Every commutative $C^*$-algebra is nuclear, and so are AF-algebras, and a large number of $C^*$-algebras one encounters
in applications.  A notable exception  is the group $C^*$-algebra of the free group on two generators.

If neither $A$ or $B$ are nuclear, there are two important $C^*$-norms on   $A\odot B$, respectively denoted
$\|\cdot\|_{\text{min}}$ and $\|\cdot\|_{\text{max}}$, and any other $C^*$-norm sits in between these two.

2. Product states

If $A_1$ and $A_2$ are $C^*$-algebras equipped with states $\varphi _1$ and $\varphi _2$, then one may look at the GNS representation
$(\pi _i,H_i,\xi _i)$ for each $\varphi _i$.  The algebraic tensor product   $A_1\odot A_2$ may then be represented on the Hilbert
space $H_1\otimes H_2$ via a $^*$-representation  $\pi $ such that
$$
  \pi (a\odot b) = \pi _1(a)\otimes \pi _2(b).
  $$

The vector state $\varphi $ associated to the vector $\xi _1\otimes \xi _2$, namely
$$
  \varphi (c) =   \langle \pi (c)(\xi _1\otimes \xi _2),\xi _1\otimes \xi _2\rangle , \quad \forall c\in   A_1\odot A_2,
  $$
then satisfies
$$
  \varphi (a\odot b) =
  \langle \pi (a\odot b)(\xi _1\otimes \xi _2),\xi _1\otimes \xi _2\rangle  = $$$$ =
  \langle \pi _1(a)\xi _1,\xi _1\rangle    \langle \pi _2(b)\xi _2,\xi _2\rangle  =
  \varphi _1(a)\varphi (b),
  $$
so we see that $\varphi $ is the product state of $\varphi _1$ and $\varphi _2$.

It can be shown that $\pi $ is continuous wrt $\|\cdot\|_{\text{min}}$ and hence also with respect to any $C^*$-norm on
$A_1\odot A_2$,  because,  as already mentioned,  $\|\cdot\|_{\text{min}}$ is the smallest of all $C^*$-norms.

As a consequence, the product state is also continuous with respect to any $C^*$-norm on
$A_1\odot A_2$.

Every vector state associated to a vector of norm 1 (such as $\xi _1\otimes \xi _2$) is a state (i.e. has morm 1), so the product state is indeed a state.

Regarding references, I'd say Murphy's book on C*-algebras is a good starting point.


问:
Are some irrational numbers closer to rational numbers than others?
I'm not even sure this question is meaningful because any irrational number can be approximated by a rational as close as you want.  But given that there are infinitely many irrationals between any two rationals, surely some must be closer to a rational number than others?

Is there a metric by which irrationals can be ranked, or can we compare two irrationals, based on how far they are from a rational?

I'm having trouble wrapping my head around this, can anyone assist?

答:
This notion can be formalized in terms of distance between sets. In a metric space $(M,d)$ with $X,Y \subseteq M$, we can formalize the Hausdorff distance in particular by

$$d_{\mathrm H}(X,Y) := \max\left\{\,\sup_{x \in X} \left( \inf_{y \in Y} d(x,y)\right), \sup_{y \in Y} \left( \inf_{x \in X} d(x,y)\right) \right\}$$

In particular, if $\overline X = \overline Y$, then $d_{\rm H} = 0.$ This is the case for $\Bbb Q$ and $\Bbb R \setminus \Bbb Q$, which have closure $\Bbb R$, and so the distance between the rationals and irrationals is zero. In particular what this, more intuitively, claims is that for any irrational $x$, you can pick a rational $y$ that is within any distance you want from $x$.

For instance, take $y_n$ to be $x$, up to $n$ places after the decimal, i.e.

$$y_n := \frac{ \lfloor 10^n x \rfloor}{10^n}$$

Then $y_n \in \Bbb Q$ $\forall n \in \Bbb N$ while $x \in \Bbb R \setminus \Bbb Q$, but $y_n \to x$ as $n \to \infty$. Thus, the sequence $\{y_n\}_{n \in \Bbb N}$ gets "arbitrarily close" to $x$ as $n$ grows, at least within the Euclidean distance.

Which touches on an important point -- the question is somewhat ill-defined, in the sense of "how do we say two points or two sets are close? how do we quantify closeness?" Absent of any other definition, we can see that the rationals and irrationals are a distance zero from each other, but only under a particular definition. The Hausdorff distance is a metric motivated by the simpler distance $d(x,Y)$ between a point and a set, and a distance $d(X,Y)$ between sets, defined by

\begin{align*}
d(x,Y)&:=\inf \{ d(x,y) \mid y \in Y \}\\
d(X,Y)&:=\sup \{ d(x,Y) \mid x \in X \}
\end{align*}

Another interesting take, as given in the comments by Robert Israel, is the irrationality measure. If $x$ has irrationality measure $\mu$, then $\mu$ is the smallest number where $\exists Q \in \Bbb Z^+$ such that, $\forall \varepsilon > 0$ and $\forall q \ge Q$,

$$\left| x - \frac p q \right| > \frac{1}{q^{\mu + \varepsilon}}$$

which helps to encode how poorly one can approximate $x$ by rational numbers $p/q$, in the sense that higher $\mu$ implies it is easier to approximate. (Specifically, approximate $x$ by rational numbers which are not $x$ itself, since we may define $\mu$ for all $x \in \Bbb R$.)

$\pi$ is well-known for having somewhat decent approximations through its continued fraction. It's $\mu$ is known to be between $2$ and about $7.10320$.

Liouville numbers have $\mu = \infty$, implying they're well-approximated. (You can also see a section on the irrationality measure on Wikipedia via that link.) Pretty obvious considering how many zeroes they have.

$\varphi := (1+\sqrt 5)/2$ was claimed as "the most irrational number" in a video by Mathologer, because its continued fraction expansion converges slower than any other one to its corresponding constant -- which is part of why $\mu = 2$ for it, as it is for all other algebraic numbers. (Note that some transcendental numbers, like $e$, also have $\mu = 2$.) So I suppose one could say $\varphi$ is "tied" for the "most irrational number."

(Which I find quite interesting -- the definition of $\mu$ and the values it takes on $\Bbb Q$, the algebraic numbers, and the transcendental numbers imply that it's easier to approximate most transcendental numbers than it is algebraic numbers, or to approximate irrational numbers than it is to approximate rational numbers by other rational numbers.)

ContentLicense: CC BY-SA 4.0


问:
Question regarding big O notation.
Let $z\in \mathbb{C}$. I want to show that $\frac{1}{z+n}+\frac{1}{z-n}=O(\frac{1}{n^2})$.

By definition, I need to show that for some constant $M$ and $N$, we have $\mid\frac{1}{z+n}+\frac{1}{z-n}\mid=\mid\frac{2z}{z^2-n^2}\mid\leq \frac{M}{n^2}$ for every $n\geq N$. I need to change this inequality and treat $z$ as a constant. Is this the right approach?

答: You have started the right way. Note that $|\frac {2z} {z^{2}-n^{2}}| \leq \frac {2|z|} {n^{2}-|z|^{2}}\leq \frac M {n^{2}}$ if $M( n^{2}-|z|^{2}) \geq 2|z|n^{2}$. This can be written as $n^{2} (M-2|z|) \geq M|z|^{2}$. As long as $M >2|z|$ we can always find $N$ such that this inequality holds for all $n \geq N$. Can you finish?

Title: Question regarding big O notation.


问:
Solving an integral equation using the Fourier transform
I have to solve the equation

$\int_0^{\infty} f(x) \cos{(\alpha x)}\, dx=\frac{\sin{\alpha }}{\alpha}$

Using fourier transform. I know this is half of the usual fourier cosine transform, and so that I would get back $f(x)$ using $\frac{2}{\pi} \int_0^{\infty} \frac{\sin{\alpha} \cos{\alpha x}}{\alpha}d{\alpha}$

Is this correct? How do I do this integral?

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

\begin{align}
&\bbox[5px,#ffd]{{2 \over \pi}\int_{0}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha} =
{1 \over \pi}\int_{-\infty}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha
\\[5mm] = &\
{1 \over \pi}\,\Re\int_{-\infty}^{\infty}\expo{-\ic \alpha x}\,\,{\sin\pars{\alpha}  \over \alpha}\,\dd\alpha \\[5mm] = &\
{1 \over \pi}\,\Re\int_{-\infty}^{\infty}\expo{-\ic \alpha x}
\pars{{1 \over 2}\int_{-1}^{1}\expo{\ic k\alpha}\dd k}\dd\alpha
\\[5mm] = &\
\Re\int_{-1}^{1}\ \underbrace{
\int_{-\infty}^{\infty}\expo{\ic\pars{k - x}\alpha}\,\,\,
{\dd\alpha \over 2\pi}}_{\ds{\delta\pars{k - x}}}\,\dd k =
\bracks{\verts{x} < 1\vphantom{\Large A}}
\end{align}
The case $\ds{x = \pm 1}$ yields $\ds{\color{red}{1/2}}$ which can be directly derived from the proposed  integral. Then,
\begin{align}
&\bbox[5px,#ffd]{{2 \over \pi}\int_{0}^{\infty}{\sin\pars{\alpha} \cos\pars{\alpha x} \over \alpha}\,\dd\alpha} =
\left\{\begin{array}{lcl}
\ds{1} & \mbox{if} & \ds{-1 < x < 1}
\\[1mm]
\ds{1 \over 2} & \mbox{if} & \ds{x = \pm 1}
\\[1mm]
\ds{0} &           & \mbox{otherwise}
\end{array}\right.
\end{align}

ContentLicense: CC BY-SA 3.0


问:
ZF Extensionality axiom
To familiarize myself with axiomatic set theory, I am reading Kenneth Kunen's The Foundations of Mathematics that presents ZF set theory. I haven't gotten really far since I am stuck at the axiom of Extensionality, stated as follows:

$$\forall x,y \; [\forall z(z \in x \leftrightarrow z \in y) \rightarrow x = y]$$

As far as I understood it, the purpose of this axiom is to state that every two sets that have exactly the same members are the same set. In terms of the above formula, when antecedent is true, the consequent needs to be true. What confuses me here is the case when the antecedent is false. Then, the consequent can be false or true, i.e., we can't say much about the consequent. But, don't we really want to say that in that case those two sets are not equal? More precisely, shouldn't we use equivalence instead of implication? If not, what is the reason?

答: The reason the converse is not part of the axiom is that it already follows from the axioms of first-order logic: $x = y \implies (\varphi(x) \iff \varphi(y))$ for any formula $\varphi$.  This is known as the substitution property of equality.</p>

Title: ZF Extensionality axiom


问:
Short routes in vehicle routing problem
Consider the vehicle routing problem (VRP), which poses the question of what the optimal set of routes for a fleet of vehicles to traverse in order to deliver to a given set of customers is.

In the classical problem, we seek to minimize the total distance traveled across all vehicles. Instead, I am considering a variant of the problem in which we primarily seek to minimize the number of vehicles and secondarily seek to minimize the total distance.

In this variant, we have a few break time constraints. In particular, these constraints depend on how long a route takes. If a route takes less than $4$ hours, then no break is required. If a route takes between $4$ and $8$ hours, then there will be a $30$ minute break required. Finally, if a vehicle's time exceeds $6$ hours, then a $60$ minute break is required (just to be clear: if a route takes $7$ hours, we'd require only a break of $60$ minutes and not $60 + 30 = 90$ minutes). No route can exceed $8$ hours.

My question is the following: Is it ever possible to do better (i.e. use fewer vehicles) with shorter routes as opposed to longer routes? I am pretty sure the answer is no. I can't come up with a counterexample. If the answer is no, is it possible to show that it's never possible to do better? I would like the triangle inequality to be satisfied.

答:
I think it is possible, particularly if driving time is not strictly proportional to distance between customers (which is normal). Consider the following two diagrams, which have the same depot (A) and customers (B through G). Returns to depot are not shown.

In the first diagram, there are four routes (black, red, green and blue). The black and blue routes are assumed to take long enough that there is only enough time after nodes C or E to make a beeline back to the depot. Similarly, the time between F and G is enough that it is not possible for one vehicle to cover both.

In the second diagram, only three routes are used. The black and blue routes are shorter than in the first diagram. The red route is longer in geometric (distance) terms than either the red or green routes in the first diagram (but perhaps not longer than their combined lengths). It may involve less driving time than either of those routes (for instance, if the red route in the second diagram runs along a highway and the red and green routes in the first diagram use congested local roads with traffic lights).

Title: Short routes in vehicle routing problem


Title: Question regarding big O notation.

Body:
Let $z\in \mathbb{C}$. I want to show that $\frac{1}{z+n}+\frac{1}{z-n}=O(\frac{1}{n^2})$.

By definition, I need to show that for some constant $M$ and $N$, we have $\mid\frac{1}{z+n}+\frac{1}{z-n}\mid=\mid\frac{2z}{z^2-n^2}\mid\leq \frac{M}{n^2}$ for every $n\geq N$. I need to change this inequality and treat $z$ as a constant. Is this the right approach?

Body:
$$\frac1{z+n}+\frac1{z-n}=\frac{z+n+z-n}{(z+n)(z-n)}=\frac{2z}{z^2-n^2}$$
$$\left|\frac{2z}{z^2-n^2}\right|=2|z|\frac{1}{|z^2-n^2|}$$
now we can pick apart this second term:
$$|z^2-n^2|=|n^2[(z/n)^2-1]|=n^2|(z/n)^2-1|$$
so:
$$\left|\frac{2z}{z^2-n^2}\right|=\frac{2}{n^2}\left|\frac{z}{(z/n)^2-1}\right|$$


问:
How often should I compound my Algorand (or other proof of stake crypto assets)?
Note to editors/mods: Please help me tag this appropriately! I'm not really a math wiz... is it multivariable calculus?

See this note from the Algorand FAQ about how collecting rewards on staked assets works with compounding.

The participation rewards are calculated automatically, however the compounding effect is not automatic. This is because the rewards are calculated from the last recorded balance on the blockchain, the easiest way to force rewards compounding is to send a zero Algo payment transaction to the target address on a frequent, recurring basis. This transaction will trigger the commit of all accrued rewards and record them to the on-chain balance of the account.

As of this question, the network fee of a 0.00 ALGO transaction to myself is .001 ALGO, and the APR for rewards is ~5.827% (converted from ~6% APY). Let's say I have some amount of ALGO principal, P, and I decide to claim my rewards every d days.

It stands to reason then that every time I claim my rewards, the resulting interest, I is:

I = P * ( d * .05827 /365)-.001

Every d days, P goes to P+I.

How do we find, for a given value P, which value d is the optimal compound interval such that we maximize the rate that P increases?

My suspicion is that as P increases sufficiently high, it will make sense to compound much more frequently. It's been 11 years since I took Calculus 1 and 2, and this seems like an interesting problem. It seems like there's some recursive nature to this that makes it challenging.

答:
Instead of writing d for the number of days between collecting interest, let's write n for the number of times a year you do it.  Also write r for the annualized interest rate and f for the fee per transaction.  At subsequent collections the balance is

$P$,

$P(1 + r / n) - f$,

$(P(1 + r / n) - f)(1 + r / n) - f
= P(1 + r / n)^2 - f[(1 + r/ n) + 1]$,

$P(1 + r / n)^3 - f[(1 + r/ n)^2 + (1+r/n) + 1]$,

and so on.

At the end of the year, then, we have

$$P_{new} = P\left(1 + \frac{r}{n}\right)^n - f\left[\left(1 + \frac{r}{n}\right)^{n-1} + \ldots + \left(1+\frac{r}{n}\right) + 1\right]$$

and of course we want to maximize this quantity.

Summing the geometric series on the right, we can write this as

$$P_{new} = P\left(1 + \frac{r}{n}\right)^n - \frac{1 - (1 + r/ n)^n }{1 - (1+r/n)} f$$

which we can simplify to

$$P_{new} = P\left(1 + \frac{r}{n}\right)^n - \left[\left(1 + \frac{r}{n}\right)^n - 1\right]\frac{f}{r} n$$

or, equivalently, to

$$P_{new} = \left(P - \frac{f}{r} n\right)\left(1 + \frac{r}{n}\right)^n + \frac{f}{r} n$$

It's tedious but not difficult to write out an expression for the derivative with respect to n, but unfortunately the roots of the derivative are not given by an elementary function of $P$, $r$, and $f$.

However, let's take a look at what would happen for some of the values you've given.  You say that $f$ is 0.001 and $r$ is 5.827%.  Let's say you have the equivalent of \$1 million, which is apparently around 2.95 million of these.

Calculating by brute force, the optimal number of compounding periods per year is 33, which corresponds to something like once every eleven days.  This will result in a balance of ALGO 3,126,343 or roughly \$1,061,875 assuming a constant exchange rate.

On the other hand, let's say that instead of 33 times a year you do it six times a year.  This will result in a final balance of ALGO 3,126,124, or roughly \$1,061,631, which is \$244 less than you made compounding 33 times per year.

In fact, let's say you just do it three times a year.  Then you'll end up with a final balance of ALGO 3,125,256.92 or roughly $1,061,337.25, about \$538 less than you would have made with 33 compounding periods.  That represents less than a 1% difference in your total profit, or about 5 basis points of your initial investment.  In other words, these are amounts that would be swamped by small fluctuations in exchange rates.

Of course nobody said that these need to be equally spaced throughout the year like we've been assuming.  But that's even more work for what I hope is clear is a fairly paltry reward.

By the way, they could do away with all this nonsense if they simply accounted for compounding when they added in the interest.  All they'd have to do is give you an interest payment of $P(e^{r \Delta t} - 1)$ rather than an interest payment of $P r \Delta t$.  I guess this way they collect more 0.001 ALGO fees, though...


问:
n-sphere and n-torus
Given the fact that $\mathbb{R}^n = \mathbb{R}\times\mathbb{R}\times\cdots\times\mathbb{R}$, do we have $\mathbb{S}^n = \mathbb{S}\times\mathbb{S}\times\cdots\times\mathbb{S}$ ($\mathbb{S}$ is the unit circle)? If not, what makes them different?

From Chris Custer's answer, the n-sphere is simply connected while the n-torus is not.

The $\mathbb{R}^n$ is always simply connected while $\mathbb{T}^n = \mathbb{S}\times\mathbb{S}\times\cdots\times\mathbb{S}$ is not simply connected. However, we have $\mathbb{S} = \mathbb{R}/2\pi$, is there any theorem telling the simply-connectedness of a direct-product group?

答: We don't.  Take $S^2$, the standard  sphere, and $T=S^1\times S^1$, the torus, for starters.  The first is simply connected, whereas the second is not.  Thus they are not homeomorphic.

Title: n-sphere and n-torus


问:
Prove the inner product on C^1[0,1] is bigger than its supreme
prove that $ \forall f$ in $C^{1}[0,1]$, if $(f,g)=\int_0^1fg+f'g' dx$, then there exist $C>0$ , s.t. $C*(f,f)^{\frac{1}{2}}\geq sup(|f|)$
Assume that f(r)=max(|f|), I have proven that if   min(|f|)=0, the result can be got easily. But how to prove the result when the function is always bigger than 0 on [0,1]?

答: I think I get the answer by myself. When $f>0, \forall x \in [0,1]$, then assume f(s)=min(f)Since there is no zero point on [0,1], we just need to prove the case of f>0). Then $(\int_0^1 f^2 dx)^{\frac{1}{2}}\geq min(f)=f(s)$ and f(r)-f(s)=$\int_s^r f’dx \leq (\int_0^1 (f’)^2 dx)^ {\frac{1}{2}}$ .That lead to the conclusion.


问:
$z^2-1$ has infinitely many roots
I'm thinking about the fundamental theorem of algebra. It says: Let $F$ any field and $p(x)\in F[x]$ then there is an extension $K$ such that $p(x)$ has $n$ roots over $K$ ¿Does it mean has maximum $n$ roots ? Think in the polynomial $z^2−1$ over complex numbers, it has infinitely many roots because of $z=\cos(\theta)+i\sin(\theta)$ is a root but $\theta$ could be any real number except $0$. I mean $z$ could be any complex number over the circle with center $0$ and radius $1$. I'm asking about this because of I've found in many sites that fundamental theorem of algebra guarantees there are at most n roots and if a polynomial has infinitely many roots then it is the zero polynomial. Thanks

答:
$$z^2-1=0$$
$$z=re^{ix}$$
$$r^2e^{2ix}=1$$
then since $|e^{2ix}|=1$ we must have $r^2=1$ and so $r=1$ since $r>0$
and so now we are solving:
$$e^{2ix}=1$$
notice that for this we want $2x=2n\pi$ and so $x=n\pi$ for $n\in\mathbb{Z}$ so yes there are infinitely many roots in the polar form but if we view the complex solutions as vector numbers then there are two $\pm1$

Title: $z^2-1$ has infinitely many roots


问:
A slight variation on the Pythagorean theorem
Are there any solution to $$a^2+b^2=c^2+1 $$ where $a \not=0$ and $b\not=0$

This is a follow-on from a previous question For what $n$ and $m$ is this number a perfect square?, which ultimately boils down to the above. Any help would be greatly appreciated.

答:
My brother built a garage.  Its horizontal cross-section measured 45 feet by 30 feet.  In inches, that's 540 inches by 30 inches.  In order to make sure the right angles at the corners were accurately measured, he wanted to measure the diagonals and make sure they were equally long.  By the Pythagorean theorem, their lengths should be
$$
\sqrt{540^2+360^2} \approx 648.999229584\text{ inches}.
$$
Those three $9$s are striking.  Why is it so close to an integer?  Here is an identity:
$$
540^2+360^2 = 649^2 - 1.
$$
Not quite what you want, but it superficially seems somewhat similar.


问:
Reference Needed: $m$-faced dice with $n$ options
I would like to investigate the following question and I am not sure if similar ideas have been investigated. I cannot really find a detail discussion about this matter on the Internet. It will be great if you can tell me what domain I should look into, or you may also give your view on the following questions.

Suppose we have one fair dice. We would like to choose an option randomly from 6 possible options. We roll the dice once then we can decide which option we should choose.

Suppose we now have 8 options. To choose an option randomly, different algorithms can be used.
For example, roll the dice once first.

If the result is odd, we focus on option 1 to 4. Then roll the dice again, if we get 1 to 4, we know which option to use. If it is a 5 or 6, repeat the second step again.

If the result is even, we focus on option 5 to 8. Then roll the dice again, if we get $n$, which $1\leq n\leq 4$, we choose option $n+4$. If it is a 5 or 6, repeat the second step again.

One can check the the probability of choosing each option is exactly $\frac{1}{8}$.

My question now is to investigate if it is possible that for whatever number of options we have, we can always use a dice to choose an option randomly. And is there a general algorithm to follow in order to choose an option?

My second question is more advance. Suppose we have a fair $m$-faced dice and we have $n$ total options. Can we choose an option randomly by using the dice for any $m$ and $n$?

Once again, you don't have to answer the above two questions. I am looking for hints or domain that I should look into. Thank you.

答:
Here's another method based on similar ideas to those used in jlammy's nice answer.

Let $\ b_i=\frac{i}{n}\ $ for $\ i=0,1,\dots,n\ $, $\ D_j\ $ the face observed on the $\ j^\text{th}\ $ toss of your $\ m$-die (with faces labelled $\ 0,1,\dots m-1\ $) for $\ j=1,2,\dots\ $, and $\ B_t=\sum_\limits{j=1}^tD_jm^{-j}\ $. Keep tossing the die until there is an integer $\ k=0,1,\dots,n-1\ $ such that $\ b_k\le B_t\le b_{k+1}-m^{-t}\ $.  With probability $1$ this will occur after only a finite number of tosses, and the integer $\ k\ $ is then equally likely to be any of $\ 0,1,\dots,n-1\ $.

The integer $\ k\ $ will not exist after $\ t\ $ tosses if and only if $\ D_1, D_2, \dots, D_t\ $ are the first $\ t\ $ digits in the base-$m$ expansion of some  $\ b_i\ $  with $\ 0\le i\le n-1\ $. The probability of this being the case is $\ \frac{n_t}{m^t}\ $, where $\ n_t\ $ is the number of distinct such  leading $\ t$-digit  sequences among the base-$m$ expansions of the numbers $\ b_0,b_1, \dots, b_{n-1}\ $.  The probability that $\ k\ $ will be determined on the $\ t^\text{th}\ $ toss is thus $\ \frac{n_{t-1}}{m^{t-1}}-\frac{n_t}{m^t}\ $ for $\ t>1\ $, or $\ 1-\frac{n_1}{m}\ $ for $\ t=1\ $, and the expected number of tosses it will take to determine $\ k\ $ is
$$
\sum_{t=1}^\infty\frac{n_t}{m^t}\le n\sum_{t=1}^\infty m^{-t}=\frac{n}{m-1}\ .
$$
One advantage of this method is that it generalises easily to the case where the $\ n\ $ options are not necessarily equally likely.  If $\ p_i\ $ is the probability of option $\ i\ $, and $\ b_i\ $ is defined by
$$
b_i=\cases{0&if $ i=0$\\
        \sum_\limits{j=0}^{i-1}p_j&if $\ 1\le i\le n\ $,}
$$
then the same method will generate the integer $\ k\ $ with probability $\ p_k\ $.

Example (as requested by OP)

If $\ n=7 $ and $\ m=3\ $, then $\ b_i=\frac{i}{7}\ $ for $\ i=0,1,\dots7\ $.

First toss the die and calculate $\ B_1\ $.  Suppose, for example that the first toss results in $\ D_1=1\ $. Then $\ B_1=\frac{D_1}{3}=\frac{1}{3}\ $.

Next, check to see whether $\ B_1\ $ satisfies the condition $\ b_k\le B_1\le b_{k+1}-\frac{1}{3}\ $ for any integer $\ k\ $. If there is such an integer, then it will be unique, and you can stop with that as your randomly selected choice.  For $\ B_1=\frac{1}{3}\ $ we have $\ \frac{2}{7}=b_2<B_1<b_3=\frac{3}{7}\ $, but $\ B_1\not\le b_3-\frac{1}{3}=\frac{2}{21}\ $, so there is no integer $\ k\ $ satisfying the stopping condition $\ b_k\le B_1\le b_{k+1}-\frac{1}{3}\ $, and we have to toss the die again.  Suppose the result of the second toss is $\ D_2=0\ $.  Then $\ B_2=\frac{1}{3}+\frac{0}{9}=\frac{1}{3}\ $.

Now check to see whether $\ B_2\ $ satisfies the condition $\ b_k\le B_2\le b_{k+1}-\frac{1}{9}\ $ for any integer $\ k\ $.  For $\ B_2=\frac{1}{3}\ $, as before, we have $\ \frac{2}{7}=b_2<B_2<b_3=\frac{3}{7}\ $, but $\ B_2\not\le b_3-\frac{1}{9}=\frac{19}{63}\ $, so we have to toss the die again.  Suppose the result of the third toss is $\ D_3=1\ $. Then $\ B_3=\frac{1}{3}+\frac{0}{9}+\frac{1}{27}=\frac{10}{27}\ $.

On now checking the stopping condition, we find that $\ b_2\le B_3\le b_3-\frac{1}{27}=\frac{74}{189}\ $, so it is satisfied by $\ k=2\ $ and this becomes our randomly chosen option.

The procedure becomes easier to understand and execute if we use ternary numerals to represent the numbers $\ b_i\ $:
$$
  \begin{align}
     b_1=\frac{1}{7}&=0.\overline{010212}_3\\
     b_2=\frac{2}{7}&=0.\overline{021201}_3\\
     b_3=\frac{3}{7}&=0.\overline{102120}_3\\
     b_4=\frac{4}{7}&=0.\overline{120102}_3\\
     b_5=\frac{5}{7}&=0.\overline{201021}_3\\
     b_6=\frac{6}{7}&=0.\overline{212010}_3\ .
  \end{align}
 $$
As long as the results, $\ D_1,D_2,\dots,D_t\ $, of the first $\ t\ $ tosses of the die duplicate the first $\ t\ $ ternary digits of any of the numbers $\ b_i\ $ for $\ 1\le i\le 6\ $, the stopping condition will not be satisfied and we have to toss again.  If, however, $\ D_1,D_2,\dots,D_t\ $ duplicate the first $\ t\ $ ternary digits of $\ b_k\ $ (say) but $\ D_1,D_2,\dots,D_{t+1}\ $ do not duplicate the first $\ t+1\ $ ternary digits of any of $\ b_{k-1}, b_k\ $ or $\ b_{k+1}\ $ then we must have either $\ b_{k-1}\le B_{t+1}\le b_k-3^{-(t+1)}\ $ (if $\ D_{t+1}\ $ is less than the $\ (t+1)^\text{th}\ $ ternary digit of $\ b_k\ $), in which case $\ k-1\ $ will be our randomly chosen option, or $\ b_k\le B_{t+1}\le b_{k+1}-3^{-(t+1)}\ $ (if $\ D_{t+1}\ $ is greater than the $\ (t+1)^\text{th}\ $ ternary digit of $\ b_k\ $), in which case $\ k\ $ will be our randomly chosen option.

Here are a few further examples to illustrate the procedure:

Tosses $\ 0,0\ $ give $\ k=0\ $.

Tosses $\ 1,2,0,2\ $ give $\ k=4\ $.

Tosses $\ 1,2,0,0\ $ give $\ k=3\ $.

Tosses $\ 2,0,1,0,1\ $ give $\ k=4\ $.

Tosses $\ 2,1,2,1\ $ give $\ k=6\ $.

Why does it work?

First note that the sequence $\ \left\{B_t\right\}_{t=1}^T\ $ is non-decreasing, so if $\ b_k\le B_\tau\ $ for some $\ \tau<T\ $, then $\ b_k\le B_t\ $ for all $\ t\ge\tau\ $.

Next, if $\ B_\tau\le b_{k+1}-m^{-\tau}\ $, then $\ B_t<b_{k+1}\ $ for all $\ t\ge\tau\ $, because $\ \sum_\limits{s=\tau+1}^tD_sm^{-s}<m^{-\tau}\ $.

If $\ b_{k+1}-m^{-\tau}<B_\tau\le b_{k+1}\ $ then $\ D_1, D_2, \dots, D_\tau\ $ must be the first $\ \tau\ $ digits of the base-$m$ expansion of $\ b_{k+1}\ $. Thus, if you toss again, and the result, $\ D_{\tau+1}\ $, is greater than the $\ (\tau+1)^\text{th}\ $ digit of the base-$m$ expansion of $\ b_{k+1}\ $, then you will have $\ B_{\tau+1}>b_{k+1}\ $

The upshot of these observations is that if the stopping rule is satisfied at $\ t=\tau\ $, then you can be sure that $\ B_t\ $ will lie in the interval $\ \left[b_k, b_{k+1}\right)\ $ for all $\ t\ge\tau\ $, no matter how large you make $\ t\ $. If it is not satisfied, however, then even if $\ b_k \le $$B_\tau<$$b_{k+1}\ $, it is still possible that if you continue tossing you will get $\ B_t\ge b_{k+1}\ $ for some $\ t>\tau\ $.

Now if $\ \tau $ is the time determined by the stopping rule, and $\ c=$$\sum_\limits{i=1}^\infty c_im^{-i}\ $, with $\ c_i\in\{0,1,\dots,m-1\}\ $ for all $\ i\ $, is any of the numbers $\ b_0,b_1,\dots,b_n\ $, then
$$
\begin{align}
\left\{B_\tau<c\right\}&=\left\{D_1<c_1\right\}\cup\\
&\hspace{2em}\bigcup_\limits{i=2}^\infty\left\{D_i<c_i, D_{i-1}=c_{i-1},\dots,D_1=c_1-1\right\}\ .
\end{align}
$$
The events $\ \left\{D_i<c_i, D_{i-1}=c_{i-1},\dots,D_1=c_1-1\right\}\ $ and $\ \left\{D_1<c_1\right\}\ $ are mutually exclusive, and
$$
\begin{align}
P\big(D_1<c_1\big)&=\frac{c_1}{m}\\
P\big(D_i<c_i, D_{i-1}=c_{i-1},\dots,D_1=c_1-1\big)&=\left(\frac{1}{m^{i-1}}\right)\left(\frac{c_i}{m}\right)\\
&=\frac{c_i}{m^i}\ \text{ for }\ i\ge2\ .
\end{align}
$$
Thus
$$
P\big(B_\tau<c\big)=\sum_{i=1}^\infty\frac{c_i}{m^i}=c\ ,
$$
and therefore
$$
\begin{align}
P\big(b_k\le B_\tau<b_{k+1}\big)&=b_{k+1}-b_k\\
&=p_k\ ,
\end{align}
$$
if $\ b_k\ $ are defined as above, because $\ \big\{B_\tau<b_k\big\}\subseteq$$\big\{B_\tau<b_{k+1}\big\}\ $.


答:
I think you are confusing "a" rational number meaning a specific rational number with "a" rational number meaning any rational number.

Given the rational numb $3$ and the irrational numbers $e$ and $\pi$.  $\pi$ is clearly closer to $3$ than to $e$ is. And there are irrational numbers closer.  $\pi =0.14 = 3.0015926535897932384626433832795....$ is even closer to $3$.

And we can make a norm where metrics are ranked on how far they are from $3$. Via $|x|_n = |x-3|$.

That's easy but fairly pointless.

But what it sounds like what asking whether there are some irrational numbers that are closer to their closest rationals than other irrational to their closest rationals.

The answer to that is no.  For any irrational we can find rationals arbitrarily close to it so there is not closest rational and there is no measurable distance between an irrational number and all the potential (infinitely many of them) rationals that can be arbitrarily close to it.

For any $\epsilon > 0$ there will always be (infinitely many) rationals within $\epsilon$ of the irrational.

Title: Are some irrational numbers closer to rational numbers than others?

Body:
I'm not even sure this question is meaningful because any irrational number can be approximated by a rational as close as you want.  But given that there are infinitely many irrationals between any two rationals, surely some must be closer to a rational number than others?

Is there a metric by which irrationals can be ranked, or can we compare two irrationals, based on how far they are from a rational?

I'm having trouble wrapping my head around this, can anyone assist?


问:
Proving a unique solution with the norm of a matrix in a linear system
Let $B\in\mathcal{M}_n$ with $\left \| B \right \|<1$ and $c$ a vector in $\mathbb{R}^n$. Prove that the linear system:

$u=Bu+c$

Has a unique solution $u$

I know how to prove a unique solution, but I don't know how to use the norm of the matrix to prove it.

答: Elaborating on Kavi Rama Murthy's comment: the eigenvalues of $I-B$ are of the form $1-\lambda$ where $\lambda$ is an eigenvalue of $B$. If $\|B\| < 1$, then $|\lambda| < 1$ for all eigenvalues of $B$. This implies $0$ is not an eigenvalue of $I-B$.


问:
Wolfram Alpha's Complex Integration involving Log in the antideriviative, gives a different result
I need to compute:

$$ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{i}{9 e^{i 2 \theta} + 1}d\theta $$

The indefinite integral above is:

$$ F(\theta)= i \theta - \frac{1}{2} \log(1 + 9 e^{2 i \theta}) $$

According to Wolfram Alpha and my own computation.

The problem is that if I assign the values of the limit myself to it I get:

$$ F(\frac{\pi}{2}) - F(-\frac{\pi}{2}) = \\
i (\frac{\pi}{2} - (-\frac{\pi}{2}))-\frac{1}{2} \left (\log(1+9 e^{2 i\frac{\pi}{2}}) -\log(1+9e^{2 i \cdot (-\frac{\pi}{2})}\right ) = \\
i \pi -\frac{1}{2}\left ( \log(\frac{1-9}{1-9})\right ) = i\pi
$$

And wolfram's solution for the definite integral is $0$ (link).

Why? Is it related to the principal value of the natural logarithm?

Body:
$$\frac{z^3}{z^2+1}=z\frac{z^2}{z^2+1}=z\left(1-\frac{1}{z^2+1}\right)=z-\frac{z}{z^2+1}$$
$$\int zdz=\int_{-\pi/2}^{\pi/2}3e^{i\theta}d\theta$$
$$\int\frac{z}{z^2+1}dz=\frac12\int\frac{1}{u}du$$
$u=z^2+1=9e^{2i\theta}+1$


问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

答:
These conjectures are correct and can be proved by reducing the determinants to suitable triangular forms to determine the exact values.

Setup of the problem

Consider two $n \times n$ matrices $A_n$ and $B_n$ defined by the 2 dimensional sequences defined on positive integers $\mathbb{Z}^+$, or natural numbers where $\mathbb{Z}^+ = \mathbb{N} = \{1,2,3,\cdots\}$. The sequences are  $a_{i,j}^{(n)}$ and $b_{i,j}^{(n)}$ defined by the following relations;

$$
a_{i,j}^{(n)} := \min\{i,j\}
\quad \quad
b_{i,j}^{(n)} := \max\{i,j\}
\quad \quad
\forall i,j \leq n \in \mathbb{Z}^+
$$

The generalized matrices are thus defined by the following templates;

$$
A_n:=
\begin{pmatrix}
a_{i,j}^{(n)}
\end{pmatrix}_
{i,j = 1, 2, ..., n}
=
\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
1 & 2 & 3 & \cdots & 3 & 3\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 2 & 3 & \cdots & n-1 & n-1\\
1 & 2 & 3 & \cdots & n-1 & n\\
\end{bmatrix}
$$

$$
B_n:=
\begin{pmatrix}
b_{i,j}^{(n)}
\end{pmatrix}_
{i,j = 1, 2, ..., n}
=
\begin{bmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
2 & 2 & 3 & \cdots & n-1 & n\\
3 & 3 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
n-1 & n-1 & n-1 & \cdots & n-1 & n\\
n & n & n & \cdots & n & n\\
\end{bmatrix}
$$

Proof for $\det{A_n}$

Consider the determinant of the matrix $A_n$;

$$
\det{A_n}
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
1 & 2 & 3 & \cdots & 3 & 3\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 2 & 3 & \cdots & n-1 & n-1\\
1 & 2 & 3 & \cdots & n-1 & n\\
\end{vmatrix}
$$

Subtracting the $(n-1)^{th}$ row from the $n^{th}$ row. $\mathcal{R}_n := \mathcal{R}_n - \mathcal{R}_{n-1}$

$$
\det{A_n}
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
1 & 2 & 3 & \cdots & 3 & 3\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 2 & 3 & \cdots & n-1 & n-1\\
0 & 0 & 0 & \cdots & 0 & 1\\
\end{vmatrix}
$$

Next subtract the $(n-2)^{th}$ row from the $(n-1)^{th}$ row. $\mathcal{R}_{n-1} := \mathcal{R}_{n-1} - \mathcal{R}_{n-2}$

$$
\det{A_n}
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
1 & 2 & 3 & \cdots & 3 & 3\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & 1\\
0 & 0 & 0 & \cdots & 0 & 1\\
\end{vmatrix}
$$

Similarly, following this pattern upward of removing the previous row from the current row, upto the second row. $\mathcal{R}_{i} := \mathcal{R}_{i} - \mathcal{R}_{i-1} \quad \forall i=2,3,\cdots,n$

$$
\det{A_n}
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
1 & 2 & 3 & \cdots & 3 & 3\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & 1\\
0 & 0 & 0 & \cdots & 0 & 1\\
\end{vmatrix}
=
\cdots
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
0 & 0 & 1 & \cdots & 1 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & 1\\
0 & 0 & 0 & \cdots & 0 & 1\\
\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
0 & 1 & 1 & \cdots & 1 & 1\\
0 & 0 & 1 & \cdots & 1 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & 1\\
0 & 0 & 0 & \cdots & 0 & 1\\
\end{vmatrix}
$$

Finally an upper triangular matrix is left after performing these $(n-1)$ row operations of subtracting the previous from each row. The sequence in which these operations are performed is important starting from the last row upwards to the second row. The upper triangular matrix has all $1's$ in the upper triangle and all $0's$ below the diagonal. The determinant of a triangular matrix is the product of the entries of the diagonal. Thus;

$$
\det{A_n}=\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
0 & 1 & 1 & \cdots & 1 & 1\\
0 & 0 & 1 & \cdots & 1 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & 1\\
0 & 0 & 0 & \cdots & 0 & 1\\
\end{vmatrix}
=
(1)^n=1
\quad \blacksquare
$$

Proof for $\det{B_n}$

For the proof of the determinant of $B_n$ we will follow a similar approach as for the previous case. We will try to convert the determinant into a triangular form through elementary row operations.

$$
\det{B_n}=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
2 & 2 & 3 & \cdots & n-1 & n\\
3 & 3 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
n-1 & n-1 & n-1 & \cdots & n-1 & n\\
n & n & n & \cdots & n & n\\
\end{vmatrix}
$$

Subtracting the $(n-1)^{th}$ row from the $n^{th}$ row. $\mathcal{R}_n := \mathcal{R}_n - \mathcal{R}_{n-1}$

$$
\det{B_n}=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
2 & 2 & 3 & \cdots & n-1 & n\\
3 & 3 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
n-1 & n-1 & n-1 & \cdots & n-1 & n\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
$$

Next subtract the $(n-2)^{th}$ row from the $(n-1)^{th}$ row. $\mathcal{R}_{n-1} := \mathcal{R}_{n-1} - \mathcal{R}_{n-2}$

$$
\det{B_n}=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
2 & 2 & 3 & \cdots & n-1 & n\\
3 & 3 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
$$

Similarly, following this pattern upward of removing the previous row from the current row, upto the second row. $\mathcal{R}_{i} := \mathcal{R}_{i} - \mathcal{R}_{i-1} \quad \forall i=2,3,\cdots,n$

$$
\det{B_n}
=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
2 & 2 & 3 & \cdots & n-1 & n\\
3 & 3 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
=
\cdots
=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
2 & 2 & 3 & \cdots & n-1 & n\\
1 & 1 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
$$

In order to convert this into a triangular form, the first row must be bought all the way to the bottom, while shifting every other row up by $1$. This is performed by consecutive swapping of 2 rows. Recall that every row swap flips the sign of the determinant of the matrix, thus;

$$
\det{B_n}
=
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-1 & n\\
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
$$

Swapping $1^{st}$ and $2^{nd}$ rows gives the following. $\mathcal{R}_{1} <=> \mathcal{R}_{2}$

$$
\det{B_n}
=
(-1)
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 2 & 3 & \cdots & n-1 & n\\
1 & 1 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
$$

Next swap the $2^{nd}$ and $3^{rd}$ rows. $\mathcal{R}_{2} <=> \mathcal{R}_{3}$

$$
\det{B_n}
=
(-1)^{2}
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
1 & 2 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
$$

Similarly continue this pattern of swapping the rows $(n-1)$ times in total till the last two rows. $\mathcal{R}_{i} <=> \mathcal{R}_{i+1} \quad \forall i=1,2,\cdots,n-1$

$$
\det{B_n}
=
(-1)^{2}
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
1 & 2 & 3 & \cdots & n-1 & n\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
=
(-1)^{3}
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
= \cdots =
(-1)^{n-2}
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 2 & 3 & \cdots & n-1 & n\\
1 & 1 & 1 & \cdots & 1 & 0\\
\end{vmatrix}
=
(-1)^{n-1}
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 & 0\\
1 & 1 & 0 & \cdots & 0 & 0\\
1 & 1 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \cdots & 1 & 0\\
1 & 2 & 3 & \cdots & n-1 & n\\
\end{vmatrix}
$$

This determinant is finally in a lower triangular form. The sequence of swaps between two consecutive rows starting from the first two rows down to the last two rows is important to bring this matrix into the triangular form as shown above. The lower triangular matrix contains all $0’s$ above the diagonal. It contains all $1’s$ in the lower triangle except for the last row. The last row contains entries from $1$ to $n$ in column numbers $1$ to $n$ respectively. Since the determinant of a triangular matrix is the product of the diagonal entries, thus;

$$
\det{B_n} = (-1)^{n-1} \times (1)^{n-1} \times n
= (-1)^{n-1}n
\quad
\blacksquare
$$

But these hold true for any positive integer value of $n$ and thus the conjectures hold true for any $n\in \mathbb{Z}^+$.


答:
Equality is often a part of the logical symbols, and if $x=y$ then for every relation $R$, and every $z$ we have $R(x,z)\leftrightarrow R(y,z)$ and $R(z,y)\leftrightarrow R(z,x)$.

In particular $\in$ is such relation.

In older texts, you can sometimes find that people don't assume that $=$ is part of the language and then we simply define $x=y$ as the formula $\forall w(w\in x\leftrightarrow w\in y)$, from which extensionality follows as a theorem. Then the language containing only the binary relation symbol $\in$ (without equality) is called the Language of Set Theory (often abbreviated as LST).

Title: ZF Extensionality axiom

Body:
To familiarize myself with axiomatic set theory, I am reading Kenneth Kunen's The Foundations of Mathematics that presents ZF set theory. I haven't gotten really far since I am stuck at the axiom of Extensionality, stated as follows:

$$\forall x,y \; [\forall z(z \in x \leftrightarrow z \in y) \rightarrow x = y]$$

As far as I understood it, the purpose of this axiom is to state that every two sets that have exactly the same members are the same set. In terms of the above formula, when antecedent is true, the consequent needs to be true. What confuses me here is the case when the antecedent is false. Then, the consequent can be false or true, i.e., we can't say much about the consequent. But, don't we really want to say that in that case those two sets are not equal? More precisely, shouldn't we use equivalence instead of implication? If not, what is the reason?


问:
Come up with some fun "equation Limericks"
We were discussing "Limericks" in my Calculus class. Specifically, "equation Limericks".

A Limerick is a poem with five lines.

The first, second, and fifth lines should have nine syllables each and rhyme with each other,
 and the third and fourth should have six syllables each and rhyme with each other.

An obscure subtype of the limerick is the "equation Limerick", which states an equation.

Here are some examples given in class:

A dozen, a gross, plus a score

Plus three times the square root of four

Divided by seven

Plus five times eleven

Is nine squared (and not a bit more).

The integral tee squared dee tee

From one to the cube root of three

Times the cosine

Of three pi over nine

Is the log of the cube root of e.

The log of e to the four

Times the square root of ten twenty-four

Adding six dozen please

Minus eight twenty-three's

Is sixteen, case is closed, shut the door.

I was able to come up with a couple of my own Limericks, but they are a bit simple compared to the ones above.

Surprisingly, there are not many resources online regarding equation Limericks. Can anyone come up with their own that they would like to share?

答:
You start out with seventy-four,
take eleven away from a score.
If then you divide
and set eight aside,
you'll have half of point infinite fours

74/(20-11)-8 = 0.444.../2


问:
dimension of infinite dimensional vector space

Let $V$ be a vector space over a field $\mathbb{K}$ (either $\mathbb{C}$ or $\mathbb{R}$) that has an infinite linearly independent subset. Prove that if $B$ and $B'$ are two bases for $V,$ then $B$ and $B'$ have the same cardinality. Denote the unique cardinality of a basis for $V$ as $\dim_{\mathbb{K}}V,$ the dimension of $V$ over the field $\mathbb{K}.$ Determine without appealing to the continuum hypothesis, $\dim_{\mathbb{Q}}\mathbb{R}.$

For any set $S,$ let $\mathcal{F}(S)$ denote the set of finite subsets of $S.$ Let $S$ be an infinite linearly independent subset of $V.$ Then for any basis $B$ for $V,$ since $B$ is maximally linearly independent, $|S|\leq |B|. $ Since $S$ is infinite, $\aleph_0\leq |S|.$ Let $B$ and $B'$ be two bases for $V.$ I know that $|\mathcal{F}(B)| = |B|$ and $|\mathcal{F}(B')| = |B'|,$ so it suffices to show that $|\mathcal{F}(B)| = |\mathcal{F}(B')|,$ but I'm not sure how to show this.

In the vector space $\mathbb{R}$ over $\mathbb{Q},$ I think, but I'm not sure how to show, that $\{\sqrt{2},\sqrt{2}^\sqrt{2}, \sqrt{2}^{\sqrt{2}^\sqrt{2}},\cdots \}$ is an infinite linearly independent subset.  I'm not sure how to determine the cardinality of this vector space. Is it $\aleph_1 = 2^{\aleph_0},$ and if so, is there some proof for this?

答:
$\aleph_1$ is irrelevant: the statement that $\aleph_1=2^{\aleph_0}$ is the continuum hypothesis, which is independent of $\mathsf{ZFC}$ and which you are not supposed to use.

What you need to show is that $\dim_{\Bbb Q}\Bbb R=2^{\aleph_0}$. Suppose that $B$ is a base for $\Bbb R$ over $\Bbb Q$. $B$ is infinite, so it has $|B|$ finite subsets, and $\Bbb Q$ has $\aleph_0$ finite subsets, so there are at most $|B|\cdot\aleph_0$ linear combinations of elements of $B$ with rational coefficients. $B$ spans $\Bbb R$, so $$2^{\aleph_0}=|\Bbb R|\le\aleph_0\cdot|B|=|B|\le|\Bbb R|=2^{\aleph_0}\,,$$ and therefore $|B|=2^{\aleph_0}$.


问:
derive using the chain rule
Given the polinomyal $f(x)=\frac{x^3}{(4-x^2)^3}$ find $f'(x)$

So, If I try to derive this, first I must to apply the chain rule in the denominator and then derive of the division (...)

$$f'(x)=\frac{x^3}{3(4-x^2)^2(-2)} = \frac{x^3}{-6(4-x^2)^2}$$

(...)?

Or there is another way to do this?

答:
You can probably use quotient rule,

$$h(x)=\dfrac{f(x)}{g(x)}$$

$$h'(x)=\dfrac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$$

Title: derive using the chain rule


答:
$z = \cos \theta + i \sin \theta$ is a solution to $|z^2| = 1$.  But it is not a solution to $z^2 =1$.

I think you are confused because if $x$ is real then $|x^2| = x^2$.  But that is because if $x$ is real then we must have either $x = 1\cdot |x|$ or $x =-1\cdot |x|$.  And either way $x^2 = (\pm 1)^2|x|^2$ and $|\pm 1|^2 = 1$.

In complex numbers that just isn't true.

In complex numbers $z$ may equal $|z|(\cos \theta + i\sin \theta)$ for any angle $\theta$ (including $0$... why did you leave $0$ out?).  ANd so $z^2 = |z|^2(\cos \theta + i\sin \theta)^2$.  But $(\cos\theta + i\sin)^2 = (\cos^2 \theta - \sin^2\theta) + 2i\cos\theta \sin\theta$ and that does not have to equal $1$.  (It could equal $1$ if $\theta = 0$ or $\pi$; and its absolute value is $1$; but it, itself, doesn't equal $1$ if $\theta \ne 0$ or $\pi$.)

The only two solutions to $z^2 -1 = 0$ are $z^2 =1$ so $z =\pm 1$.

There are infinitely many solutions to $|z|^2 -1 = 0$.  But there are only !$2$! solutions to $z^2 -1 = 0$.

Title: $z^2-1$ has infinitely many roots

Body: I'm thinking about the fundamental theorem of algebra. It says: Let $F$ any field and $p(x)\in F[x]$ then there is an extension $K$ such that $p(x)$ has $n$ roots over $K$ ¿Does it mean has maximum $n$ roots ? Think in the polynomial $z^2−1$ over complex numbers, it has infinitely many roots because of $z=\cos(\theta)+i\sin(\theta)$ is a root but $\theta$ could be any real number except $0$. I mean $z$ could be any complex number over the circle with center $0$ and radius $1$. I'm asking about this because of I've found in many sites that fundamental theorem of algebra guarantees there are at most n roots and if a polynomial has infinitely many roots then it is the zero polynomial. Thanks


问:
How to compute the trace of an exponential raised to a power of bras and kets? (bra-ket notation)
So, I am working on the following equation:

$$\frac{|A^2|}{|B^2|+|A^2|} = \frac{e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}}{Tr[e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}]} \tag{1}$$

Which I am told is expressible in the form:

$$|A^2| = C|B^2| \tag{2}$$

Where C is a value that is a function of $\omega$ and $k_b T$ (both constants).

With some basic rearranging I have obtained:

$$|A^2| = \frac{e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}}{Tr[e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}] - e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}} }|B^2| \tag{3}$$

But one issue that I have with my solution is the $Tr[e^{...}]$ term; I cannot further simplify it. I am familiar with computing the trace of a function involving kets and bras, but no so much when taking the trace of an exponential to the power of bras and kets.

My question is: how does one compute such a trace? I have attempted to use the Macluaurin expansion of $e$ and compute the trace of each individual term but didn't have much luck. Thank you for your time.

For clarification $\langle e\vert e\rangle=1$ and $\langle e\vert k\rangle=0$ for $k\ne e$.

答:
What you have written is a little vague but let's go with some reasonable assumptions: $\langle e\vert e\rangle=1$ and $\langle e\vert k\rangle=0$ for $k\ne e$.  Then your operator
\begin{align}
\hat E=\vert e\rangle\langle e\vert
\end{align}
is diagonal and idempotent: $\hat E^2=\hat E$, with eigenvalue $1$) (or at there there is a basis where it's diagonal). Thus
\begin{align}
e^{\alpha \hat E}= \sum_n \frac{\alpha^n}{n!} \hat E^n = \sum_n \frac{\alpha^n}{n!}  \hat E = e^{\alpha}\vert e\rangle\langle e\vert.
\end{align}
You can easily deal with the rest.

Note that your denominator in (3) must be handled with care (or is downright incorrect) unless you have a way of defining "division by an operator".  It could be defined by the power series of $(1+\hat A)^{-1}$ and you won't have to worry about ordering since $\hat E$ will commute with powers of $\hat E$ in the numerator.  You better start with (1) which is free of this issue.


问:
How to prove $\frac{6n}{n-6}$ can not be integer when $n >42$?
This was a junior math contest question. However, it's not obvious for me.
Could anyone offer a clue?</p>

答:
By the Euclidean algorithm,
$$\gcd(6n,n-6)=\gcd(n-6,6n-6(n-6))=\gcd(n-6,36).$$
For $n>42$, this is strictly less than $n-6$. So $n-6$ doesn't divide $6n$.


问:
Finite Factors of Refactorable Numbers
This was just a question that I came up with while learning about refactorable numbers.

While looking through the sequence of refactorable numbers (1, 2, 8, 9, 12, 18....), I decided to look at the number of factors that each refactorable number had. This gave me the sequence (1, 2, 4, 3, 6, 6....). I saw that while refactorable numbers were all distinct, my numbers repeated. My question is that will all the numbers in my sequence repeat infinitely, or will some of these stop showing up after a finite value? Either way how do I prove this?

Since this is just a problem that I came up with I don´t have much work on it, but here is the wikipedia page that I  reading about refactorable numbers from:

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

答:
For an odd prime $p$ the number $8p$ has $4 \cdot 2 = 8$  divisors, which divides $8p.$  This is repeated infinitely often

For an odd prime $q > 3$ the number $12q$ has $6 \cdot 2 = 12$  divisors, which divides $12q.$

For an odd prime $r > 5$ the number $80r$ has $10 \cdot 2 = 20$  divisors, which divides $80r.$

For any prime $s \neq 3$ the number $9s^2$ has $3 \cdot 3 = 9$  divisors, which divides $9s^2.$

For any prime $t \neq 5$ the number $625 t^4$ has $5 \cdot 5 = 25$  divisors, which divides $625 t^4.$

For any prime $u \neq 7$ the number $117649 u^6$ has $7 \cdot 7 = 49$  divisors, which divides $117649 u^6$

There are also slowly growing families. For any prime $v > 2$ take
$$   v^{v-1} $$ which has exactly $v$ divisors.

Any $w $  take $$  2^{\left( 2^w -1\right)}  $$
$$  3^{\left( 3^w -1\right)}  $$
$$  5^{\left( 5^w -1\right)}  $$
$$  7^{\left( 7^w -1\right)}  $$
or
$$ \color{red}{ p^{\left( p^w -1\right)} } $$

Title: Finite Factors of Refactorable Numbers


问:
Base change of purely inseparable extension
If $K$ is a purely inseparable extension  of $k$ and $A$ is a $k$-algebra, since $A\otimes_kK$ may be not a field, hence I want to ask the following question:

If $\mathfrak{p}$ is a prime of $A\otimes_kK$ and $f^{-1}(\mathfrak{p})$ be the preimage of $\mathfrak{p}$ under the natural map $f:A\rightarrow A\otimes_kK$, then how to prove that $\operatorname{Frac}(A/f^{-1}(\mathfrak{p}))\rightarrow\text{Frac}(A\otimes_kK/\mathfrak{p})$ is purely inseparable?

The motivation is that for a scheme over $k$, I want to prove that the induced residue field map of $X_K\rightarrow X$ at any point is purely inseparable.

Thanks in advance.

答:
For completeness, and to get this question off the unanswered queue, here's a fully written out solution. Let $p=\operatorname{char} k$ and let $\overline{x}\big/\overline{y}\in\text{Frac}([A\otimes_k K]\big/\mathfrak{p})$, where $x,y\in (A\otimes_k K)\setminus\mathfrak{p}$. We want to find $n\in\mathbb{N}$ such that $\left(\overline{x}\big/\overline{y}\right)^n$ lies in the image of $\text{Frac}(A\big/f^{-1}(\mathfrak{p}))$, which consists of all elements of the form $\overline{a_1\otimes 1}\big/\overline{a_2\otimes 1}$ for $a_1\in A$ and $a_2\in A\setminus f^{-1}(\mathfrak{p})$.

Now, $x$ and ${y}$ are sums of simple tensors. Suppose ${x}=\sum_{i=1}^na_i\otimes t_i$. Since $K\supseteq k$ is purely inseparable, for each $i$ there is some $m_i$ such that $t_i^{p^{m_i}}\in k$. Thus, letting $m=\sum_{i=1}^nm_i$, we have
\begin{align}
{x}^{p^m}&=(a_1\otimes t_1+\dots+a_n\otimes t_n)^{p^m} \\
&= (a_1\otimes t_1)^{p^m}+\dots+(a_n\otimes t_n)^{p^m} \\
&=a_1^{p^{m}}\otimes t_1^{p^m}+\dots+a_n^{p^m}\otimes t_n^{p^m},
\end{align}
where the omitted terms vanish because we are working in characteristic $p$. Note that $t_i^{p^m}=\left(t_i^{p^{m_i}}\right)^{p^{m-m_i}}\in k$, so each $a_i^{p^m}\otimes t_i^{p^m}=(a_i^{p^m}\big/t_i^{p^m})\otimes 1\in f(A)$ and hence ${x}^{p^m}\in f(A)$; for convenience say $x^{p^m}=a_1\otimes 1$. Similarly, we can find $m'$ such that $y^{p^{m'}}=a_2\otimes 1\in f(A)$; note in particular that, since $y\notin\mathfrak{p}$ and $\mathfrak{p}$ is prime, we have $a_2^{p^{m}}\notin f^{-1}(\mathfrak{p})$. Hence, letting $n=m+m'$, we have $$\left(\frac{\overline{x}}{\overline{y}}\right)^{p^{n}}=\frac{\overline{x^{p^{m+m'}}}}{\overline{y^{p^{m+m'}}}}=\frac{\overline{a_1^{p^{m'}}\otimes 1}}{\overline{a_2^{p^m}\otimes 1}},$$ and so we are done.


Title: derive using the chain rule

Body:
Given the polinomyal $f(x)=\frac{x^3}{(4-x^2)^3}$ find $f'(x)$

So, If I try to derive this, first I must to apply the chain rule in the denominator and then derive of the division (...)

$$f'(x)=\frac{x^3}{3(4-x^2)^2(-2)} = \frac{x^3}{-6(4-x^2)^2}$$

(...)?

Or there is another way to do this?

Body:
You want to use the quotient rule then the chain rule.

$$\frac{d}{dx}\frac{f(x)}{g(h(x))}=\frac{g(h(x))f'(x)-f(x)g'(h(x))h'(x)}{g(h(x))^2} $$

so:

$$\frac{d}{dx}\frac{x^3}{(4-x^2)^3} = \frac{(4-x^2)^3(3x^2)-x^3(3(4-x^2)^3)(-2x)}{(4-x^2)^6} $$


答:
We have
$$
\frac{6n}{n-6} = 6 + \frac{36}{n-6}
$$
This is an integer iff the last term is an integer iff $n-6$ divides $36$. In particular, $n-6 \le 36$, which gives $n \le 42$.

We can continue and conclude that the fraction is an integer iff $$n \in \{-30,-12,-6,-3,0,2,3,4,5,7,8,9,10,12,15,18,24,42\}$$

Title: How to prove $\frac{6n}{n-6}$ can not be integer when $n >42$?

Body:
This was a junior math contest question. However, it's not obvious for me.
Could anyone offer a clue?</p>


问:
Are some irrational numbers closer to rational numbers than others?
I'm not even sure this question is meaningful because any irrational number can be approximated by a rational as close as you want.  But given that there are infinitely many irrationals between any two rationals, surely some must be closer to a rational number than others?

Is there a metric by which irrationals can be ranked, or can we compare two irrationals, based on how far they are from a rational?

I'm having trouble wrapping my head around this, can anyone assist?

答:
Aside from a set of measure zero there is no much difference.

On the one hand, classical result due to Dirichlet asserts that for
any irrational x (say x between 0 and 1), there are infinite many integers p,q such that  $${|x - p/q|}  < 1/q^{2}$$
On the other hand, for fixed $\epsilon$ > 0 the set of those numbers which can be approximated by order 2 + $\epsilon$ (denoted below by A) has measure zero.
To see this define
$$A_{k,N}^{\epsilon} = \{x; x\in [0,1], |x-k/N| < 1/N^{2 + \epsilon}, 1 \leqq k\leqq N\}$$
One has
$$A = \bigcap_{n=1}^{^\infty}\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}$$
Since
$$|\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}| \leqq \sum_{k,N\geq n}^{}|A_{k,N}^{\epsilon}|$$
and
$$|A_{k,N}^{\epsilon}|\leqq 1/N^{1+\epsilon}$$
we get
$$|\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}| \leqq \sum_{k,N\geq n}^{}|A_{k,N}^{\epsilon}| \leqq \sum_{N=n}^{\infty}1/N^{1+\epsilon}$$
In words, A has really measure zero as advertised.


问:
Constructing a measure space given the definition of a measurable function
Suppose:

$(X,\mathcal{S},\mu)$ is a (complete) measure space,

$(\mathbb{R},\mathcal{B}_\mathbb{R})$ is the Borel space associated to $\mathbb{R}$, and

$f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B}_\mathbb{R})$ is a measurable function.

Given this information, is it possible to construct the measure space $(Y,\mathcal{T},\lambda)$ defined by:

$Y=f(X):=\{f(x):x\in X\}\subset\mathbb{R}$,

$\mathcal{T}=f(\mathcal{S}):=\{f(A):A\in\mathcal{S}\}\subset\mathcal{B}_\mathbb{R}$,

$\lambda=f_*(\mu)$ (i.e. the pushforward of $\mu$ by $f$).

If possible, do we need to consider the completion of $(X,\mathcal{S},\mu)$ for the above statement to hold?

Thank you,
Frederick

答:
In the general case, you can not expect that $f(\mathcal{S}):=\{f(A):A\in\mathcal{S}\}\subseteq\mathcal{B}_\mathbb{R}$.

Here is a simple counter-example:

Consider $X=\mathbb{R}$, $\mathcal{S}=2^\mathbb{R}$ and $\mu$ the counting measure. Then
$(X,\mathcal{S},\mu)$ is a complete measure space. Consider the function $f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B}_\mathbb{R})$ defined by $f(x)=x$.  It is easy to see that $f$ is a measurable function. Now, take any $E\subseteq \mathbb{R}$ such that $E \notin \mathcal{B}_\mathbb{R}$. Then $E\in 2^\mathbb{R}$ and $f(E)=E \notin \mathcal{B}_\mathbb{R}$. So
$f(\mathcal{S})\nsubseteq\mathcal{B}_\mathbb{R}$.

So to do what you want to do, you need to define the $\sigma$-algebra in different way. Let us see it.

Let us construct the measure space $(Y,\mathcal{T},\lambda)$ defined by:

$Y=f(X):=\{f(x):x\in X\}\subset\mathbb{R}$,

$\mathcal{T}:=\{A \cap Y:A\in\mathcal{B}_\mathbb{R}\}$,

$\lambda=f_*(\mu)$ (i.e. the pushforward of $\mu$ by $f$).

Note that, for any $E\in \mathcal{T}$ there is $A\in\mathcal{B}_\mathbb{R}$ such that $E=A \cap Y$ and $\lambda(E) = \mu(f^{-1}(A))$.

Note that $\lambda$ is well defined. In fact, given $E\in \mathcal{T}$ and  $A_1, A_2\in\mathcal{B}_\mathbb{R}$ such that $E=A_1 \cap Y=A_2 \cap Y$, then
$f^{-1}(A_1)=f^{-1}(A_2)$ and so $\mu(f^{-1}(A_1))=\mu(f^{-1}(A_2))$.


问:
Derivative of trace of matrix expression with respect to a matrix
In exercise 4.2 of Pattern Recognition and Machine Learning the official solution states that the  derivative of $$E_D(\mathbf{\tilde W}) = \frac{1}{2}Tr\{(\mathbf{XW}+\mathbf{1}w_0^T - \mathbf{T})^T(\mathbf{XW}+\mathbf{1}w_0^T - \mathbf{T}) \}$$  with respect to $w_0$ (where $w_0$ is a column vector of bias weights and $\mathbf{1}$ is a column vector of N ones) is $$ 2Nw_0+2(\mathbf{XW-T})^T\mathbf{1} $$ but I do not know how they got to this result. Can you please explain how to get to this result as well as a link to a resource which explains this.

答:
For typing convenience, define the auxiliary matrix $A$ as

$$\eqalign{
A= {\tt1}w_0^T + (XW -T) \\
A^T= w_0{\tt1}^T + (XW -T)^T
}$$
and use a colon as an infix product notation for the trace function, i.e.
$$A:B = {\rm Tr}(A^TB) = {\rm Tr}(B^TA) = B:A$$
Rewrite the objective function in a form which makes it easy to calculate the gradient
$$\eqalign{
\phi &= \tfrac 12A:A = \tfrac 12A^T:A^T \\
d\phi &= A^T:dA^T = A^T:dw_0\,{\tt1}^T = A^T{\tt1}:dw_0 \\
\frac{\partial\phi}{\partial w_0} &= A^T{\tt1} = Nw_0 + (XW-T)^T{\tt1} \\
}$$
This is the same result as your notes except for the factor of ${\tt2}$, which I suspect is a typo.

The standard reference for this subject is Matrix Differential Calculus by Magnus and Neudecker.


问:
On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity.
Trying to solve similar type equation to this.

On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity. According to the diagram, vinegar is how many times as acidic as pure water?

答:
pH represents the negative logrithm of H+ ions.  So the number ic $ln(c)/ln(0.1)$.

Acidity could then be written as $7-pH$ as a logrithm, or $10^{7-pH}$ as a number.

If you are using a different number system, the units are gram-moles per litre, and the adjustment to water is to add 3 to the pH, eg $pH_{12}=(pH_{10}+3)/log_{10}(12)$.  This would reference it to a 'water-mole', eg gram-mole / cc, where 'water' = 'gram/cc'.


问:
Triple of primes satisfying 3 equations
Let $P_1$, $P_2$, $P_3$ be $3$ distinct odd primes, and $n>1$, such that:

$$nP_1 \equiv P3 \pmod{P_2},\ nP_2 \equiv P_1 \pmod{P_3},\  nP_3 \equiv P_2 \pmod {P_1},$$
Take $n = 2$.

Question: Is there a triple $(P_1,P_2,P_3)$ satisfying the above $3$ equations?

Comment : I noticed that there are solutions $(P_1,P_2,P_3)$ if $n>2$ in general. But I could not find one triple with $n = 2$. Any comment will be helpful. Thank you.

答:
There are no triples with $n = 2$. First, consider if $P_2$ is the largest prime. Your first congruence relation gives that, for some $k \in \mathbb{Z}$,

$$2P_1 \equiv P_3 \pmod{P_2} \implies 2P_1 - P_3 = kP_2 \tag{1}\label{eq1A}$$

Since $P_1 \lt P_2 \implies 2P_1 \lt 2P_2$, then $k \lt 2$. Also, $P_3 \lt P_2 \implies -P_3 \gt -P_2$, so $k \gt -1$. Thus, $k \in \{0, 1\}$. With $k = 0$, this gives $2P_1 = P_3$, which is not allowed. Thus, we must have $k = 1$ which gives

$$2P_1 - P_3 = P_2 \implies 4P_1 - 2P_2 = 2P_3 \implies -2P_2 \equiv 2P_3 \pmod{P_1} \tag{2}\label{eq2A}$$

Your third congruence equation then gives

$$2P_3 \equiv P_2 \pmod{P_1} \implies -2P_2 \equiv P_2 \pmod{P_1} \implies 3P_2 \equiv 0 \pmod{P_1} \tag{3}\label{eq3A}$$

Since the primes are distinct, this means $P_1 = 3$. The left side of \eqref{eq2A} gives

$$2P_1 = 6 = P_3 + P_2 \tag{4}\label{eq4A}$$

However, since $3$ is the smallest odd prime and the primes are distinct, then $P_2 \gt 3$ and $P_3 \gt 3$, so \eqref{eq4A} is not possible.

With $P_3$ being the largest prime instead, using similar arguments as before, your second congruence equation gives

$$2P_2 - P_1 = P_3 \implies 4P_2 - 2P_1 = 2P_3 \implies -2P_1 \equiv 2P_3 \pmod{P_2} \tag{5}\label{eq5A}$$

Thus, your first congruence equation gives

$$2P_1 \equiv P_3 \pmod{P_2} \implies -2P_3 \equiv P_3 \pmod{P_2} \implies 3P_3 \equiv 0 \pmod{P_2} \tag{6}\label{eq6A}$$

Thus, as done before, this means $P_2 = 3$, with the first part of \eqref{eq5A} giving a contradiction.

Finally, consider the case where $P_1$ is your largest prime. As done before, your third congruence equation results in

$$2P_3 - P_2 = P_1 \implies 4P_3 - 2P_2 = 2P_1 \implies -2P_2 \equiv 2P_1 \pmod{P_3} \tag{7}\label{eq7A}$$

Your second congruence equation now becomes

$$2P_2 \equiv P_1 \pmod{P_3} \implies -2P_1 \equiv P_1 \pmod{P_3} \implies 3P_1 \equiv 0 \pmod{P_3} \tag{8}\label{eq8A}$$

As was done earlier, this results in $P_3 = 3$, so the first part of \eqref{eq7A} gives a contradiction.

This covers all possibilities, showing there are no solutions for $n = 2$.

Title: Triple of primes satisfying 3 equations


问:
$a,b,c > 0$ and $a+b+c=1$. Prove that $\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $
$a,b,c > 0$ and $a+b+c=1$. Prove that

$$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geqslant \frac{13}{2(1-abc)} $$

This inequality can be solved by computer ( essentially, remove the constrain through homogenizing, clear denominator, expanding and proving positive terms). Since this mechanical strategy is very simple to do but impractical during math contest, I am looking forward to solution using classical inequalities approach.

答:
I would not call this an answer, but perhaps could be helpful. Observe that:
\begin{align*}
(1-a)(1-b)(1-c)&=\color{red}{1-a-b-c}+ab+ac+bc-abc\\
&=ab+ac+bc-abc \\
&=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}-abc \\
&=\frac{1-(a^2+b^2+c^2)}{2}-abc \\
&=1-abc-\frac{1+a^2+b^2+c^2}{2} \\
&=1-abc-\frac{a+b+c+a^2+b^2+c^2}{2}.
\end{align*}
Thus, $$a+b+c+a^2+b^2+c^2=2[(1-abc)-(1-a)(1-b)(1-c)].$$
By Cauchy Schwartz inequality, $$(a+b+c+a^2+b^2+c^2)\left(\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\right)\geq3^2=9.$$
As a result, $$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geq\frac{9}{2[(1-abc)-(1-a)(1-b)(1-c)]}.$$
If we can show $$\frac{9}{2[(1-abc)-(1-a)(1-b)(1-c)]}\geq\frac{13}{2(1-abc)},$$ then we are done. Suppose this is true. It gives $$13[(1-abc)-(1-a)(1-b)(1-c)]\leq 9(1-abc),$$ or equivalently, $$(1-abc)\leq\frac{13}{4}(1-a)(1-b)(1-c).$$
If this inequality is true under the constraint $a+b+c=1$ and $a,b,c>0$, then we are done. I checked that the equality precisely holds when $a=b=c=1/3$. Hope anyone can finish it.


问:
An exercise of infinite sets
I have to proof that if A is an infinite set, then $P(P(A))$ has a countable set. They ask me to do it without using the Axiom of Choice

答:
I assume that by countable you actually mean countably infinite.

HINT: For $n\in\Bbb N$ consider the set $\{F\in\wp(A):|F|=n\}$.

Title: An exercise of infinite sets


答:
You're doing wrong. A good way to do this when still not expert in derivatives, is to write
$$
f(x)=\frac{g(x)}{h(x)}
$$
where
$$
g(x)=x^3,\qquad h(x)=(4-x^2)^3.
$$

Then, first of all, you have to differentiate the quotient:
$$
f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}
$$

So you need to compute $g'(x)$ and $h'(x)$. The first one is easy:
$$
g'(x)=3x^2.
$$
The second one needs the chain rule:
$$
h'(x)=3(4-x^2)^2(-2x)=-6x(4-x^2)^2.
$$
Now plug in the formula above:
\begin{align}
f'(x)&=
\frac{3x^2\cdot(4-x^2)^3 - x^3\cdot(-6x(4-x^2)^2)}{(4-x^2)^6}\\[2ex]
&=\frac{3x^2(4-x^2)^3+6x^4(4-x^2)^2}{(4-x^2)^6}\\[2ex]
&=\frac{3x^2(4-x^2)+6x^4}{(4-x^2)^4}\\[2ex]
&=\frac{12x^2+3x^4}{(4-x^2)^4}=\frac{3x^2(4+x^2)}{(4-x^2)^4}
\end{align}

A different strategy may be noting that
$$
f(x)=\left(\frac{x}{4-x^2}\right)^3
$$
so, letting
$$g(x)=\frac{x}{4-x^2}$$
you have, by the chain rule,
$$
f'(x)=3g(x)^2g'(x)
$$
The derivative of $g(x)$ can be computed again with the quotient rule:
$$
g'(x)=\frac{1\cdot(4-x^2)-x\cdot(-2x)}{(4-x^2)^2}=
\frac{4-x^2+2x^2}{(4-x^2)^2}=\frac{4+x^2}{(4-x^2)^2}
$$
and so
$$
f'(x)=3\left(\frac{x}{(4-x^2)}\right)^2\frac{4+x^2}{(4-x^2)^2}
$$
which gives the same result as before.

Title: derive using the chain rule


问:
How do we show that ${\sum}_{w\in\wedge}\frac{1}{(z+w)^2}$ is not absolutely convergent?
This is from Complex Analysis by Shakarchi and Stein in the chapter of Elliptic functions.

How do we show that ${\sum}_{w\in\wedge}\frac{1}{(z+w)^2}$ is not absolutely convergent, where $\wedge$ is a double lattice in complex plane? It seems difficult for me to show the absolute convergence of the complex infinite sums. Maybe I can try to bound this series and use the fact from real series, but I don't know how.

答:
By a rotation we can assume the lattice is $m+n\tau, \tau=a+ib, b>0$ and wlog we can assume $a \ge 0$ as otherwise we use $n <0$ in what follows.

Fix $z=x+iy$, so $|z+m+na+inb|^2=(m+na+x)^2+(nb+y)^2$.

Then if $Nb>|y|$, we get $(nb+y)^2<4b^2n^2, n \ge N$

and similarly $M>0, M+Na >|x|$ implies $(m+na+x)^2<4(m+na)^2, m \ge M, n \ge N$

This means that $\frac{1}{|z+m+na|^2} \ge \frac{1}{4b^2n^2+4(m+na)^2}, m \ge M, n \ge N$

But now summing only those terms and calling that sum $S$ we get that:

$S \ge \sum_{m \ge M, n \ge N}\frac{1}{4b^2n^2+4(m+na)^2}$

Using that a double series of positive numbers can be interchanged at will (with the same result either finite or infinite) we immediately get (as summand is decreasing in $m$) that for fixed $n \ge N$:

$\sum_{m \ge M}\frac{1}{b^2n^2+(m+na)^2} \ge \int_{M+1}^{\infty}\frac{dt}{b^2n^2+(t+na)^2}=$

$=\frac{1}{bn} \tan^{-1}(\frac{t+na}{nb})|_{t=M+1}^{t=\infty}=\frac{1}{bn}(\pi/2-\tan^{-1}(\frac{M+1+na}{nb})) \ge \frac{1}{bn}(\pi/2-c) =A/n, n \ge N$

where $c=\tan^{-1}(\frac{M+1+Na}{Nb})$ as $\frac{M+1+na}{nb} \le \frac{M+1+Na}{Nb}, n \ge N$ and the arctangent is increasing

But this shows that $S \ge \sum_{n \ge N}\frac{A}{4n}=\infty$ so the double series of absolute values on a lattice subset is already infinite and we are done!


问:
Induction of measure by a measurable function
Suppose:

$(X,\Sigma,\mu)$ is a measure space,

$(\mathbb{R},\mathcal{B}_\mathbb{R})$ is the Borel space associated to $\mathbb{R}$ (as topological space),

$f:(X,\Sigma)\to(\mathbb{R},\mathcal{B}_\mathbb{R})$ is a measurable function.

Then, it is clear that one can induce a measure, $\nu$, on $(\mathbb{R},\mathcal{B}_\mathbb{R})$ by pushing forward the measure $\mu$ by $f$ as follows:
$$\nu:=f_*\mu=\mu\circ f^{-1}:\mathcal{B}_\mathbb{R}\to[0,\infty]\\
B\mapsto\nu(B)=\mu\circ f^{-1}(B),$$
yielding therefore the induced measure space $(\mathbb{R},\mathcal{B}_\mathbb{R},\nu)$.

My question is: if we had to map every element of $\Sigma$ by $f$ (symbolically defined here as $f(\Sigma):=\{f(A):A\in\Sigma\}$), then:

Do we get as a result of doing this a $\sigma$-algebra on $\mathbb{R}$?

If so, is this $f(\Sigma)\subseteq\mathcal{B}_\mathbb{R}$?

If not, what about if $X$ is a topological space and $\Sigma$ the associated Borel $\sigma$-algebra. Does point 1. or 2. still not hold?

Thank you, Frederick.

答:
If $X$ and $Y$ are sets,  and $\mathscr P(X)$ and $\mathscr P(Y)$ are the corresponding power sets,  then the
correspondence
$$
  B\in \mathscr P(Y) \mapsto  f^{-1}(B) \in  \mathscr P(X)
  $$
has much beter properties if compared to the correspondence
$$
  A\in \mathscr P(X) \mapsto  f(A) \in  \mathscr P(Y).
  $$
The former respects unions, intersections and complementation, while the latter only respects unions.

For that reason things often work much better when sets are moved backwards, as opposed to forward, through  functions.  For example, a function between
topological spaces is continuois iff the inverse image of open sets is open.  Open functions, namely  functions that map open sets to open
sets, have a much more limited use in Math if compared to continuous functions.

This said, and adopting the terminology in the question, $f(\Sigma )$ is not always  a $\sigma $-algebra, since $\mathbb R$ is not
necessarily in $f(\Sigma )$, unless $f$ is surjective.   Moreover, it might happen that $f(A)\cap f(B)$ is not in
$f(\Sigma )$ even if $A$ and $B$ are in $Σ$ (try to
find an example to illustrate this).

The second question also has a negative answer.  The reason is that there are continuous functions $f:[0,1]\to \mathbb R$, such as the Devil staircase,
sending the Cantor set $C$ to a set of  positive Lebesgue measure.    If we take $\Sigma $ to be the set of all Lebesgue
measurable sets of $[0,1]$, then every subset $A$ of $C$  is in $\Sigma $,  but $f(A)$ cannot all be Borel measurable simply
because there are $2^{2^{\aleph_0}}$ subsets of $C$ and only $2^{\aleph_0}$ Borel measurable subsets of $\mathbb R$.

Even if $X$ is a topological space and $\Sigma $ is its Borel $\sigma $-algebra, question (2) may have a negative answer.  For a
trivial example let $X$ be any non-Borel subset of $\mathbb R$, with its induced topology, and let $f$ be the inclusion
of $X$ in $\mathbb R$.  Obviously $X$ is Borel in itself, but $f(X)$ is not Borel in $\mathbb R$.

For another perspective I suggest  reading about analytic sets,  where
emphasis is put  on direct image of Borel  sets.


问:
Rod partitioning in n pieces
A straight rod of length n units (obviously n is an integer) is to be partitioned into n pieces of random length. All pieces are to be used to form triangular constructions, so we are looking for the boundaries for the length of the largest piece, to ensure that this will be possible with any 3.

If the largest piece is $n_i$ units long, and its largest and smallest possible lengths are $x_i$ and $y_i$, then $y_i \leq n_i \leq x_i$ and then for all other pieces, it will be:
$n_j \leq y_i$.

Also $\sum n_i \leq \sum y_i = n*y_i$.

Therefore for any 3 pieces, we will have:

$y_i \leq y_1 \leq n_1 \leq x_1 \leq x_i$ and same for $n_2$ and $n_3$.
Wlg we consider $n_1 \leq n_2 \leq n_3$.

In order for the 3 pieces to form a triangle, it must be $n_1+n_2 \leq n_3$, $n_1+n_3 \leq n_2$ and $n_2+n_3 \leq n_1$. Also $n_3-n_2 \geq n_1$, $n_2-n_1 \geq n_3$ and $n_3-n_1 \geq n_2$.

An obvious solution would be to have all segments equal to 1 unit, so we could form $\frac {n}{3}$ equilateral triangles but obviously this is not the general case.

I have seen several solutions to similar problems but with very advanced math (integrals etc) and I can't follow the solutions.

I wonder if there is any solution with basic calculus (11th or 12th grade) because I don't have any maths background. FYI this is not homework or assignment.

Thank you very much.

答: $n\geq 3$ and let $l$ be the longest rod piece length, then $l\geq 1$ otherwise we would have $n$ pieces of lengths less than $1$ and so the total sum less than $<n$ contradicting that the sum is $=n$, so $1$ is the lower bound for $l$ and we want to find the upper bound of the interval for values for the longest length, if $l$ the maximum length then $ \frac{n-l}{n-1}$ is the average length of the remaining $n-1$ pieces and so there must be a pair such that there combined length is $ \leq 2 \frac{n-l}{n-1}$ otherwise all pairs lengths are bigger $ > 2\frac{n-l}{n-1}$ which means that the sum of all lengths is bigger than $>n$ which is contradiction and so we need that $2 \frac{n-l}{n-1} \geq l $ which means that $ 2n \geq (n+1)l $ and so $l \leq \frac{2n}{n+1}$ and so the interval of the longest rod piece is $[1,\frac{2n}{n+1}]$.

Title: Rod partitioning in n pieces


问:
|matrix A| = 1 , ||vector B|| = 1 ➜ ||A x B|| always = 1?
Matrix A is a 3x3 matrix.

Vector B is a tuple of (x,y,z).

In my real application, A is a normal matrix, and B is a normal vector.

I know that :-

Det of A = 1

Size of B = ||B|| = sqrt(B.x ✖ B.x + B.y ✖ B.y + B.z ✖ B.z) = 1

Is ||A x B||  always 1?

Please prove or contradict it by an example.

If this is true, I will be able to optimize a 3D model loader for Opengl.

If it is true, is it possible to extend the theory to any sizes of matrix/vector?

I think I should change the vector B to a 3x3 matrix to make it easier to prove.

I am new to Math.

答: No. See for instance $A=\begin{pmatrix}\frac12&0&0\\ 0&6&0\\ 0&0&\frac13\end{pmatrix}$ and $B=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$.


问:
Computing the exponential matrix with complex eigenvalues
Let $a,b\in\mathbb{R}$ two fixed parameters, and consider the following matrix $$
A:=\left(\begin{matrix} 0 & -a & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 &0&0&b \\ 0&0&b&0\end{matrix}\right).
$$
I am trying  to compute the exponential matrix $e^{Ax}$ associated with $A$ above. In order to do it, I started by computing its eigenvalues, from where I obtained $\{\pm b,\pm ia\}$, and hence $A$ is diagonalizable. Here is where everything became dark for me, specifically, the fact that I have two complex eigenvalues confuses me. For example, in order to obtain the eigenspace associated with $\pm b$, I know that I just have to solved the following system $$
(A\mp bI)\vec{v}=\vec{0},
$$
from where I obtained (respectively): $$
\left(\begin{matrix} 0 \\ 0 \\ 1 \\ 1\end{matrix}\right) \qquad \hbox{and}\qquad \left( \begin{matrix} 0 \\ 0 \\ 1 \\ -1\end{matrix}\right).
$$
However, as I said before, when trying to mimic the previous computations with the eigenvalues $\pm ia$, I got confused and I blocked. For example, if I pick $+ia$, the first two equations associated with $(A-iaI)\vec{v}=\vec{0}$ writes $$
ia v_1+av_2=0 \quad \hbox{ and } \quad av_1-iav_2=0,
$$
from where I cannot solve for $v_1$ or $v_2$. Does anyone knows how to follow?

答:
You actually have two paths to solution.

The first is as you suggest go to the diagonal form.  If $U$ is the matrix obtained from the eigenvectors of $A$ arranged in columns, then
\begin{align}
U^{-1}AU=D\, ,\qquad
D=\left(\begin{array}{cccc}
b&&&\\
&-b&&\\
&&ia&\\
&&&-ia\end{array}\right)
\end{align}
so that
\begin{align}
e^A &= 1+UDU^{-1}+\frac{1}{2}UD^2U^{-1}+\ldots \, ,\\
&= U \left(\begin{array}{cccc}
e^b&&&\\
&e^{-b}&&\\
&&e^{ia}&\\
&&&e^{-ia}\end{array}\right) U^{-1}\,
\end{align}

The second path is to observe that your $A$ is actually block diagonal and that it's
easy to take powers of each block since (for instance)
\begin{align}
\tilde a&= a\sigma\, ,\qquad \sigma=\left(\begin{array}{cc}
0&-1\\ 1&0\end{array}\right)\, ,\\
\tilde a^2 &= - a^2 1\, ,\qquad \tilde a^3=- a^3\sigma \ldots
\end{align}
so that, for instance, the exponentiation of this submatix gives
\begin{align}
e^{\tilde a}&= \left(1-\frac{a^2}{2}+\frac{a^4}{4!}+\ldots\right)
+ \sigma \left(a - \frac{a^3}{3}+\frac{a^5}{5!}+\ldots\right)\,  \\
\end{align}
and you can recompose the sums into trigonometric functions.

The same reasoning applied to the submatrix containing $b$ entries except that
this time $\Sigma=\left(\begin{array}{cc}
0&1\\ 1&0\end{array}\right)$ is such that $\Sigma^2=1, \Sigma^3=\Sigma$ etc so that $e^{\tilde b}$ will recompose in terms of hyperbolic sine and cosine.


问:
Evenly distributing $n^2$ ball between $n$ pockets by chosing $2$ of them at a time.
We have $n^2$ balls distributed among $n$ pockets (some of them may be empty). We can choose any two pockets of which sum of balls is even and then distribute the balls equally among these two pockets. We can repeat it as many times as we wish. For every natural, even number $n$ such that $2 ≤ n ≤ 10$ decide if by using this method we can always distribute the balls so that at the end every bag contains n balls, no matter how many balls there are in each pocket at the beginning.

I think and I am informed it is true that we can do it for every n number of pockets which is a power of two. I suppose, we split them in two until we are left with $n/2$ pairs of pockets. Then we make a move on each pocket and make our way out merging the subsets of pairs. For $n=4$ we have two pairs which we first want to even out and then make the move on the first elements from the each pair an then on the second elements from each pair. However the number of ball combined in the selected pockets has to be even. That feels to be the case as if $n$ is a power of two, $n^2$ is even so we have an even number of odd numbers of balls. How would a proper proof of that look? How to prove we cannot do it if $n$ is not a power of two?

答:
First, call a distribution of balls "dead" when some pockets have an even number $E$ of balls and the rest have an odd number $O$ of balls.  When we start with such a dead distribution, it's impossible to change the distribution of balls, so we cannot evenly spread the balls among the pockets.

We will be able to generate a special class of dead distributions as follows: Given any $n>1$,

Choose $k$ such that $0 \leq k < n$.  Put $k$ balls into each pocket.  At the end of this step, we have $n^2 - nk = n(n-k)$ balls remaining to distribute.

Write $n(n-k) = 2^i\cdot m$ where $m$ is odd.  If $2^i < n$, then pick any $2^i$ pockets and put $m$ balls in each, so these pockets will have $m+k$ balls in total; otherwise, do nothing.

With this procedure, we have two claims:

This procedure can test whether a given $n$ has a dead distribution.

When the procedure finishes, some pockets have $k$ balls and others have $m+k$ balls and $(m+k) - k = m$ is odd, so the resulting distribution is dead.

On the other hand, when a dead distribution exists with $E$ and $O$ as in my first paragraph, we can set $k = \min(E,O)$ and take $|E-O|$ balls from each of the $P$ pockets currently holding $\max(E,O)$ balls.  Every pocket now has $k$ balls.  Also, you now hold $$n(n-k) = P\cdot |E-O| = 2^i\cdot m$$ balls in your hand, and since $|E-O|$ is odd, $2^i \leq P < n$, so the procedure also yields a dead distribution.

The existence or non-existence of a dead distribution is the key question we must investigate.

In short, this is because the existence of a dead distribution provides an initial distribution from which you cannot equally distribute the balls, and the non-existence of a dead distribution, combined with the fact any distribution $(b_1, b_2, \ldots, b_n)$ of balls can move at most $\sum b_i^2$ times before you cannot make any more moves, which only happens when the final distribution is equally split among all pockets or is a dead distribution.

Now consider the cases:

$n$ is not a power of two:  If $n$ is odd, choose $k=0$ in the above procedure; otherwise choose $k=1$.  Then $n-k$ is odd and $2^i$ divides $n(n-k)$, so $2^i$ divides $n$.  Since $n$ is not a power of $2$, $2^i < n$ so the above procedure successfully yields a dead distribution.  Therefore, when $n$ is not a power of two, it is not always possible to evenly distribute the balls.

$n$ is a power of two:  In this case, regardless of the $0 \leq k < n$ chosen, we will always have $n$ dividing $2^i$, so $n \leq 2^i$.  Therefore the above procedure can never complete, and in particular no dead distribution of balls exists.  Therefore, when $n$ is a power of two, it is always possible to evenly distribute the balls.


问:
Find sufficient condition over the coefficients of a quartic polynomial to have only positive roots
For a stability question I'm looking for conditions under the coefficients $a,b,c,d\in\mathbb{R}$ of a monic quartic polynomial $$f(x)=x^4+ax^3+bx^2+cx+d$$ that guarantee that all the roots of this polynomial are positive. Also, I need to find sufficient conditions under the coefficients that guarantee at least one positive root (like $d<0$), or conditions that guarantee that there isn't any positive root. I'm aware of approaches like the one of the following article: https://www.researchgate.net/publication/266543070_Absence_of_positive_roots_of_complex_cubic_and_quartic_polynomials, but I need simpler relations between the coefficients, since my polynomial of interest has coefficients in terms of a lot of variables. Thank you in advance for your comments.

答:
More than a comment:

The necessary and sufficient conditions that a polynomial has all the roots real is that the Hermite matrix
$$(p_{i+j})_{0\le i,j \le n-1}$$
is positive semi-definite, where
$$p_i = \sum_{k=1}^n x_k^i$$
are the power sums of the roots of the equation. Thus we get a condition on the coefficients of the equation, using Newton-Girard identities.

This is rather cumbersome.

Let's consider the case of a polynomial $P$ of degree $4$. We are looking for conditions that the polynomial has all roots real. A necessary condition is that $P'$ has all roots real. But we know that this is equivalent to its discriminant $\operatorname{Disc} P'$ being $\ge 0$.

Assume that the roots $t_i$ of $P'$ are real and distinct ($ \operatorname{Disc} P'> 0$). Now, $P$ will have all roots real if two of the number $P(t_i)$ are negative and one positive.  A necessary condition for this is $\prod P(t_i) > 0$. This product equals the resultant of $P$ and $P'$ ( up to a positive constant), or, the discriminant of $P$, $\operatorname{Disc}P$.

Let's sum up: the condition
$\operatorname{Disc} P, \operatorname{Disc} P'> 0$ is equivalent to the roots $t_i$ of $P'$ are real and distinct and $\prod P(t_i) > 0$. We have $2$ negatives among the $P(t_i)$  if and only if : at least one of the inequalities
$$P(t_1)+ P(t_2) + P(t_3)<0\\
P(t_1) P(t_2) + P(t_1)P(t_3)+ P(t_2) P(t_3)<0$$
is satisfied.

Let $P(x) = x^4 + a x^3 + b x^2 + c x + d$, $t_1$, $t_2$, $t_3$ the roots of $P'(x)$. Then $P(t_i)$, $i=1,3$ are the roots of the equation
(obtained with WA)
$$-a^2 b^2 c^2 + 4 b^3 c^2 + 4 a^3 c^3 - 18 a b c^3 + 27 c^4 + 4 a^2 b^3 d - 16 b^4 d - 18 a^3 b c d + 80 a b^2 c d + 6 a^2 c^2 d - 144 b c^2 d + 27 a^4 d^2 - 144 a^2 b d^2 + 128 b^2 d^2 + 192 a c d^2 - 256 d^3+\\
+(- 4 a^2 b^3  + 16 b^4  + 18 a^3 b c  - 80 a b^2 c  - 6 a^2 c^2  + 144 b c^2  - 54 a^4 d  + \\+ 288 a^2 b d  - 256 b^2 d  - 384 a c d  + 768 d^2 )y +\\
 + (27 a^4  - 144 a^2 b  + 128 b^2  + 192 a c  - 768 d )y^2 + 256 y^3=0$$

Now, the free term of this equation is $- \operatorname{Disc} P$, and this must be negative, like we mentioned above. Now, if either
$$-(27 a^4  - 144 a^2 b  + 128 b^2  + 192 a c  - 768 d )<0$$
or
$$(- 4 a^2 b^3  + 16 b^4  + 18 a^3 b c  - 80 a b^2 c  - 6 a^2 c^2  + 144 b c^2  - 54 a^4 d  + \\+ 288 a^2 b d  - 256 b^2 d  - 384 a c d  + 768 d^2 )<0$$

( coupled with $\operatorname{Disc} P , \operatorname{Disc} P>0$), then all of the roots of $P$ are real. If we also want them positive, then we need also $a<0$, $b>0$, $c<0$, $d>0$.


问:
evaluate $\int\ln x\tan x\, \mathrm dx$
How to evaluate  $$\int\ln x\tan x\, \mathrm dx \:?$$
I've tried to do integration by parts but after calculations it cancel out the main question.

答:
Please note that the no-closed-form integrals in fact can be further classified as four types:

Type $1$: Can be expressed as infinite series whose its radius of convergence covers on $\mathbb{C}$ or $\mathbb{R}$

Type $2$: Can only be expressed as infinite series whose its radius of convergence covers on finite ranges, to get the results suitable on $\mathbb{C}$ or $\mathbb{R}$ should have cases divisions

Type $3$: Other than Type $1$ and Type $2$ but the software can express them as known special functions, we can only follow the expressions from the software

Type $4$: Other than Type $1$ and Type $2$ and even the software cannot express them as known special functions, if we really want to force to solve them we should only use the formula e.g. this one.

Unfortunately, $\int\ln x\tan x~dx$ is in fact belongs to Type $4$ because of the following reasons:

$1.$ Wolfram fails to solve this integral.

$2.$ The radius of convergence of the power series of $\ln x$ is only $1$ , and $\int x^n\tan x~dx$ still have no-closed-form

$3.$ The radius of convergence of the power series of $\tan x$ is only $\dfrac{\pi}{2}$ and its coefficients have no-closed-form, and we should have infinitely many times of cases divisions in order to get the results suitable on $\mathbb{C}$ or $\mathbb{R}$

Title: evaluate $\int\ln x\tan x\, \mathrm dx$


问:
Directional vs Total Derivative
The total derivative appears similar to the directional derivative.  I have a question on the similarity.  I provide some up front definitions and then questions.

Directional Derivative Wiki of $f(x) = f(x_1, ..., x_n)$ is:

$\triangledown_v f(x) = \triangledown f(x) \cdot v = \lim_{h \rightarrow 0} \frac {f(x + hv) - f(x)} h$

where RHS is the dot product of the gradient and directional vector $v = (v_1, ..., v_n)$

Total Derivative Wiki as a differential form
where $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $\Delta x = (\Delta x_1, ..., \Delta x_n)^T$, the total derivative of $f$ at $a$ is $d_{f_a} = (\frac {\partial f} {\partial x_1} (a), ..., \frac {\partial f} {\partial x_n} (a)) = \triangledown f(a)^T$.  Then Wiki later continues by stating $f(a + \Delta x) - f(x) \approx df_a \cdot \Delta x$, then goes on by redefining $d_{f_a} = \sum_{i=1}^n \frac {\partial f} {\partial x_i} (a) d x_i$.

Questions

It appears from these definitions the total derivative is reduced to the gradient.  Is this true?

$d_{f_a}$ is later redefined as the dot product of $\triangledown f(x)^T \cdot \Delta x$.  This appears to be the directional derivative.  Is this correct or an error?  Is the direction vector a scaling of a unit vector in every direction?

答:
It appears from these definitions the total derivative is reduced to the gradient. Is this true?

The gradient is a vector. The total derivative, on the other hand, is the linear map $df_a \colon \Bbb{R}^n \to \Bbb{R}$ defined by taking the dot product with the gradient:
$$ df_a(v) = \nabla f(a) \cdot v = \nabla_v f(a) $$
So while $\nabla f$ is a vector, $df$ is a function (which happens to use $\nabla f$ in its definition). It just happens that the matrix representation of the function $df$ has the same components as the gradient vector, since matrix multiplication is the same as taking the dot product.

I think this also answers your question #2.

Title: Directional vs Total Derivative


问:
What rational number is between these two real numbers?
According to several texts and professors, there exists a rational number between any two real numbers. But suppose you had two real numbers which had the same digits in the same places up to some place, where they differed by one digit - say the smaller one had a $4$ and the bigger one had a $5$. The smaller number's digits following the four are all $9$s, and the bigger number's digits after the five are all $0$s. For example:
$$ \beta=1.235\overline{0} \\ \alpha=1.234\overline{9}$$

What rational number is between $\alpha$ and $\beta$?

答:
Whenever you see that 'bar stuff' at the end of a decimal expansion, you know that is shorthand for a repeating decimal and that the number is rational.

So the two numbers

$\beta=1.235\overline{0} \\ \alpha=1.234\overline{9}$

are both rational numbers.

The average of two rational numbers is also a rational number, and unless you are taking the average of two equal numbers, you'll be getting a new number between them.

So, just take the average and see what happens,

$\quad (.5) \times \bigr(1.235\overline{0} + 1.234\overline{9}\bigr) = $

$\quad (.5) (2.46\overline{9}) =$

$\quad 1.23 + (5) (0.\overline{9})\cdot 10^{-3} =$

$\quad 1.23 + 5 \cdot 10^{-3} =$

$\quad 1.235 = \beta$

So we are starting with two equal numbers.


问:
Doubt in open dense sets of a metric space.
Let $B_d(x,r) \cap G_1$ where $G_1$ is an open dense set. Now this intersection is non empty. Let $x_1 \in B_d(x,r) \cap G_1$.Then, $B_d(x_1,r_1) \subset G_1$ .Now my doubt is that will there exists a $r_2 > 0$ such that $B_d(x_1,r_2) \subset G_1$ and $B_d(x_1,r_2) \subset B_d(x,r)$.

I think it will be true. However I was wondering if there is a contradicting example.

答: Whether $B_d(x_1,r_1)\subseteq G_1$ depends on $r_1$. What is true is that there is an $r_1>0$ such that $B_d(x_1,r_1)\subseteq G_1$. As for your question, $B_d(x,r)\cap G_1$ is an open set containing $x_1$, and the open balls are a base for the topology, so necessarily there is an $r_2>0$ such that $B_d(x_1,r_2)\subseteq B_d(x,r)\cap G_1$, which of course implies that $B_d(x_1,r_2)\subseteq B_d(x,r)$ and $B_d(x_1,r_2)\subseteq G_1$.


问:
Finite Factors of Refactorable Numbers
This was just a question that I came up with while learning about refactorable numbers.

While looking through the sequence of refactorable numbers (1, 2, 8, 9, 12, 18....), I decided to look at the number of factors that each refactorable number had. This gave me the sequence (1, 2, 4, 3, 6, 6....). I saw that while refactorable numbers were all distinct, my numbers repeated. My question is that will all the numbers in my sequence repeat infinitely, or will some of these stop showing up after a finite value? Either way how do I prove this?

Since this is just a problem that I came up with I don´t have much work on it, but here is the wikipedia page that I  reading about refactorable numbers from:

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

答: Each prime $p$ appears exactly once in your list.  Proof:  $p^{p-1}$ is a refactorable number with $p$ divisors, and it is the only one because every number with exactly $p$ divisors has the form $q^{p-1}$ for some prime $q$.

Title: Finite Factors of Refactorable Numbers