问:
What base is Roman Numerals?
What is the base for Roman Numerals? It starts off with unary then goes back and forth between multiples of 5 and 10.</p>

答:
Roman numbers represent the two-on-five abacus, so the columns process by tens, but it's a cooperation between a count of 5 and a division of 2.

The actual numbers are token-weights (like money).  You don't need a zero-cent coin to separate the $1 coin and the 1c coin.   The idea of representing four and nine as one-before-five and one-before-ten is a later invention, but the sumerians had a one-before-ten rune for nine too.  You don't write though, 49 as IL, but XL IX, which proves that there is token-weighted decimal digits in use.

```<code>       D    L     V       Decimal tokens.
  -------------------        is base 10, because any symbol is 10 times
  M    C    X     I       the value to the right.

 £50    £5   50p    5p     Decimal currency.
 £20    £2   20p    2p        I don't need to have any 10p, 20p or 50p
 £10    £1   10p    1p     coins to fill out the 'empty column'

1/4  £100   £5   5s   3d   1/4d    Base 20 abacus, for money.
    ---------------------------       the last column is broken into 3
1     £20   £1   1s   1d              rather than five, without fault.
</code>```

One can see that there is indeed a notion of place in the roman column.  One writes a unit-stone as C if it's in the second column, or 'M' if it's in the third column, or I if it's in the first column.  It's probably like saying that you have no notion of place of money, because the £100 note in your pocket isn't supported by zero-coins to fill the £10, the £1 and 10p and 1p columns.

Alternating Bases

By Neugebauer 'the exact sciences in antiquity'

The Sumerian number system for ordinary numbers, is a variety of hodgepodge of decimal, sixty and twelfty symbols.  A number like dec 192, might be written as 100 + 60 + 30 + 2, might be represented in the matching tables as 3A2 (ie 3 10 2).

The correct style to represent sumerian numbers, is not the manner of our base 60 time, but to use rune for rune: ie A = 10, B=20, C=30, D=40, E=50, and numbers 1-9 as such.

The runes A to E start off life as fractions, ie 1/6 to 5/6.  So one might write 1 1/2, as 1C.  It's kind of like having numbers fro 0/12 to 11/12 for fractions of a dozen, or uncia, like the romans do.   But where the romans ultimately use weight-fractions, the sumerian identification of A mina = 10 shekels, (ie 1/6 mina = 10 shekels), and that one can write numbers as eg 1C mina 2A shekels ie 1 3/6 mina, 2 1/6 shekels.  (where we would write 1:32:10 mina).

In any case, the implementation of the abacus everywhere is that the units are in the low column, and the divisions start in the high column, so

```<code>                  C = 3/6 mina      A = 1/6 shekel.
    ----------------------------------------------
     1 = 1 mina   2 = 2 shekels
</code>```

Sumerian numbers are written as 'alternating divisions and multiples', rather than a tight sixty-number.  The number we write as 1.24 51 10, is written as 1B4E1A  without any spacing, and its recriprocal (also given) is .42 25 35 as D2B5C5, again without any spacing.  One does not really need spacing when the high and low digits are different.

It's also a division system.  That is 1B4E1A is not to be read as 1 25 51 10 = dec 212570, but as 1:25 51 10, = 1.41421356.  It's like our time and angle.  A three-number angle is degrees, a five-number angle is degrees and minutes, and a seven-figure angle is degrees, minutes and seconds.  Also 1500 hours is not 62 1/2 days, but 15 o'clock. denoted as HHMM.  The hundred is not 5 score but three score.

The sumerian numbers were constructed as a division system to avoid division.  We note this from the appearence of tables of pairs of numbers whose product is 60, eg 1C D  (1.5 * 40 = 60), sorted in increasing size.  So if one wanted to divide by 72, one sees in the table 1A2   E.  The second table to consult is a set of ready-reckners that gives multiples of E for 1 to 20, and then 40.  Such tables were still in commercial use until the 1970s, and might still be found in book-sales.

There is reference in Neugebauer, of 'the problem of seven brothers', since 7 does not appear in the recriprocal tables.  The thing is referred to as the equal of a scientific paper, that eg  8C4A6 < '7 < 8C4A8   (where '7 is the egyptian style of writing 1/7).  The actual result is to repeat 8C4A7.

There is also examples of e v e n l y   s p a c e d numbers, like writing 8 C 4 A 6.  This is little pussle to me, since i often do this to effect criss-cross multiplication in base 120.

Alternating bases demand alternating arithmetic.  This is not too hard.  It's just that you have to detangle this nonsence that a base requires digits from 0 to b-1, and that the multiplication table is of the order of $O(b^2)$.  I don't use much more than 12 * 12 = 1.24 in base 120.  You just don't need to multiply 77 * 88 = 56.56, for example.

Title: What base is Roman Numerals?

Body:
Roman numbers represent the two-on-five abacus, so the columns process by tens, but it's a cooperation between a count of 5 and a division of 2.

The actual numbers are token-weights (like money).  You don't need a zero-cent coin to separate the $1 coin and the 1c coin.   The idea of representing four and nine as one-before-five and one-before-ten is a later invention, but the sumerians had a one-before-ten rune for nine too.  You don't write though, 49 as IL, but XL IX, which proves that there is token-weighted decimal digits in use.

       D    L     V       Decimal tokens.
  -------------------        is base 10, because any symbol is 10 times
  M    C    X     I       the value to the right.

 £50    £5   50p    5p     Decimal currency.
 £20    £2   20p    2p        I don't need to have any 10p, 20p or 50p
 £10    £1   10p    1p     coins to fill out the 'empty column'

1/4  £100   £5   5s   3d   1/4d    Base 20 abacus, for money.
    ---------------------------       the last column is broken into 3
1     £20   £1   1s   1d              rather than five, without fault.

One can see that there is indeed a notion of place in the roman column.  One writes a unit-stone as C if it's in the second column, or 'M' if it's in the third column, or I if it's in the first column.  It's probably like saying that you have no notion of place of money, because the £100 note in your pocket isn't supported by zero-coins to fill the £10, the £1 and 10p and 1p columns.

Alternating Bases

By Neugebauer 'the exact sciences in antiquity'

The Sumerian number system for ordinary numbers, is a variety of hodgepodge of decimal, sixty and twelfty symbols.  A number like dec 192, might be written as 100 + 60 + 30 + 2, might be represented in the matching tables as 3A2 (ie 3 10 2).

The correct style to represent sumerian numbers, is not the manner of our base 60 time, but to use rune for rune: ie A = 10, B=20, C=30, D=40, E=50, and numbers 1-9 as such.

The runes A to E start off life as fractions, ie 1/6 to 5/6.  So one might write 1 1/2, as 1C.  It's kind of like having numbers fro 0/12 to 11/12 for fractions of a dozen, or uncia, like the romans do.   But where the romans ultimately use weight-fractions, the sumerian identification of A mina = 10 shekels, (ie 1/6 mina = 10 shekels), and that one can write numbers as eg 1C mina 2A shekels ie 1 3/6 mina, 2 1/6 shekels.  (where we would write 1:32:10 mina).

In any case, the implementation of the abacus everywhere is that the units are in the low column, and the divisions start in the high column, so

                  C = 3/6 mina      A = 1/6 shekel.
    ----------------------------------------------
     1 = 1 mina   2 = 2 shekels

Sumerian numbers are written as 'alternating divisions and multiples', rather than a tight sixty-number.  The number we write as 1.24 51 10, is written as 1B4E1A  without any spacing, and its recriprocal (also given) is .42 25 35 as D2B5C5, again without any spacing.  One does not really need spacing when the high and low digits are different.

It's also a division system.  That is 1B4E1A is not to be read as 1 25 51 10 = dec 212570, but as 1:25 51 10, = 1.41421356.  It's like our time and angle.  A three-number angle is degrees, a five-number angle is degrees and minutes, and a seven-figure angle is degrees, minutes and seconds.  Also 1500 hours is not 62 1/2 days, but 15 o'clock. denoted as HHMM.  The hundred is not 5 score but three score.

The sumerian numbers were constructed as a division system to avoid division.  We note this from the appearence of tables of pairs of numbers whose product is 60, eg 1C D  (1.5 * 40 = 60), sorted in increasing size.  So if one wanted to divide by 72, one sees in the table 1A2   E.  The second table to consult is a set of ready-reckners that gives multiples of E for 1 to 20, and then 40.  Such tables were still in commercial use until the 1970s, and might still be found in book-sales.

There is reference in Neugebauer, of 'the problem of seven brothers', since 7 does not appear in the recriprocal tables.  The thing is referred to as the equal of a scientific paper, that eg  8C4A6 < '7 < 8C4A8   (where '7 is the egyptian style of writing 1/7).  The actual result is to repeat 8C4A7.

There is also examples of e v e n l y   s p a c e d numbers, like writing 8 C 4 A 6.  This is little pussle to me, since i often do this to effect criss-cross multiplication in base 120.

Alternating bases demand alternating arithmetic.  This is not too hard.  It's just that you have to detangle this nonsence that a base requires digits from 0 to b-1, and that the multiplication table is of the order of $O(b^2)$.  I don't use much more than 12 * 12 = 1.24 in base 120.  You just don't need to multiply 77 * 88 = 56.56, for example.


问:
Sum of $\aleph_0$ numbers
I tried to search online for this and look in some textbooks, but didn't seem to find anything. My question is:

What is the sum of all positive rational numbers? Can it be defined? What type of infinity is it?

Of course, we can ask ourselves the same question in a more general way:

What is the sum of $\aleph_0$ numbers? Again, can it be defined? What does it depend on? What about $\aleph_i$ numbers?

For example, the sum $$\sum_{i=1}^{\infty}\frac{1}{2^i}=1$$

so in this case, the sum of $\aleph_0$ numbers is $1$. However, we can find a series of rational numbers that converges to any irrational numer too, thus, sometimes the sum of $\aleph_0$ numbers is irrational.

I know this is a bit vague, but, to sum up again:

What is the sum of all positive rational numbers? Can we tell anything about how big the sum of $\aleph_0$ or even $\aleph_i$ numbers can get?

Thanks

答:
What is the sum of all positive rational numbers? Can it be defined? What type of infinity is it?

The question is, quite simply, incredibly ill-defined.

$
\newcommand{\N}{\mathbb N}
\newcommand{\Q}{\mathbb Q}
\newcommand{\R}{\mathbb R}
\newcommand{\eqdef}{\stackrel{\text{def}}{=}}
\newcommand{\set}[1]{\left\{ #1 \right\} }
$

Infinite Limits:

When we say a summation approaches or diverges to infinity, this is not the "infinity" of the aleph numbers that you're thinking of. The infinity in this context is more of a notion or concept, rather than an actual number itself.

To see what I mean, recall: an infinite summation can be defined as the limit of its partial sums, i.e. for a sequence $(a_n)_{n \in \Bbb N}$,

$$\sum_{n=0}^\infty a_n \eqdef \lim_{N \to \infty} \sum_{n=0}^N a_n$$

Now, to further this discussion, we have to appeal to the formal, rigorous definition of a sequence approaching infinity as its index approaches infinity: a sequence $(s_n)_{n \in \N}$ is said to approach infinity if

$$(\forall R \in \R)(\exists N \in \N)(\forall n \ge N)(s_n > R)$$

In words: $s_n \to \infty$ if and only if...

...for every real number $R$...

...there exists some natural number $N$ such that...

...for every integer $n$ greater than or equal to $N$...

...it holds that $s_n > R$

(Of course, $s_n$ in our case is your sequence of partial sums.)

This is quite a mouthful, but essentially it says that, for any real number, the sequence is eventually always greater than that real number. It encodes the notion and intuition of divergence to infinity.

Notice in particular: this definition does not, in any way, actually reference infinity itself. This is what I mean when I say this infinity is more conceptual than an actual thing - it is not inherently a number itself.

Infinite Cardinals:

What about the infinities that, say, describe how many numbers there are in a set? You have presumably seen, for instance, we say that

the cardinality of the naturals, denoted $|\N|$, is equal to $|\aleph_0|$

moreover, $|\Bbb Q| = |\Bbb Z| = \aleph_0$

however, $|\R| = |\Bbb C| = |(0,1)| = c$ (or $\aleph_1$, depending on the continuum hypothesis)

the set $\R^\R$ of all functions $f : \R \to \R$ has cardinality $2^c$

...and so on. But what does this mean?

We call these "cardinal numbers." Particularly, these are what we call "infinite cardinals." Cardinality is a means of measuring how many elements are in a set. For a finite set, it is easy enough:

$$|\set{0,1,2}| = 3 \qquad |\set{1}| = 1 \qquad |\set{1,2,\cdots,100}| = 100 \qquad |\varnothing|=0$$

and so on. To generalize things to infinite sets, however, makes the definition a bit murkier. Assuming the axiom of choice, we say that the cardinality of a set $S$ is given by the least ordinal number $\alpha$ such that there exists a bijective (i.e. invertible) function $f : S \to \alpha$. Ordinal numbers are a bit more complicated to get into; the two are intuitively similar but formally not at all the same, albeit intimately linked.

That said, essentially, we then can say that two sets have equal cardinality if there exists a bijection between them - and this works for infinite sets as well as finite sets. Moreover, we can say $|S| \le |T|$ if there is an injective function $f : S \to T$, and $|S| \ge |T|$ if there is a surjective function $g: T \to S$.

Uncountable Sums:

The main takeaway from this is that the infinities used to describe cardinality are precisely that: they are used to describe how many elements a set has. On the other hand, the infinity from limits is a purely conceptual construct, and is not inherently representative of any such infinity.

However, there is slight detour I want to go on: sums on arbitrary sets.

In a loose sense, you can imagine the usual infinite sum as being a sum over $\N$ or another countable set. (A set is countable if its cardinality is at most $\aleph_0$.) Perhaps we can generalize the concept? You can this talked about here on MSE, but essentially the takeaway is this: we can sum over any set $S$ whose members are nonnegative by

$$\sum_{i \in S} a_i \eqdef \sup_{F \subseteq S} \set{ \sum_{i \in F} a_i \; \middle| \; F \subseteq S \text{ is finite} }$$

Ultimately, then, what this means is that the sum over a set is the least upper bound on the value of any finite sum we can get. However this ultimately means there are at most countably many nonzero summands $a_i$  and thus can be reduced to some finite or (typical) infinite sum - so it's not particularly useful or noteworthy in general. Figured it'd be neat to bring up though.

What is the sum of $\aleph_0$ numbers? Again, can it be defined? What does it depend on? What about $\aleph_i$ numbers?

If by this you mean "what is the sum of a set of $\aleph_i$-many numbers?", I guess the above explains it.

Of course, we can take the sums of these numbers, we just need at most countably many nonzero such numbers. And of course, by merit of how I started this discussion, the sum ultimately still has to be defined.

In the general case, I'm not sure if there is a known necessary and sufficient condition for an arbitrary infinite series $\sum a_n$ (of rationals or anything else) to exist. One of the main necessary conditions is that $a_n \to 0$, but it is not sufficient (consider $a_n = 1/n$). Some particular types of series do have convergence tests that you can look at, at least.

What is the sum of all positive rational numbers?

I suppose we can use the spirit of the definition I gave earlier.

Define the set $\N_n \eqdef \set{1,2,\cdots,n}$. Observe that $\N_n \subseteq \Q^+$ and it is finite. Thus, from the definition of the sum and of supremum,

\begin{align*}
\sum_{r \in \Q^+} r
&= \sup_{F \subseteq \Q^+} \set{ \sum_{r \in F} r \; \middle| \; F \subseteq \Q^+ \text{ is finite} }\\
&\ge \sum_{r \in \N_n} r
\end{align*}

and this holds for every $n \in \N$. Observe that, moreover,

$$\sum_{r \in \N_n} r < \sum_{r \in \N_{n+1}} r \qquad \& \qquad \sum_{r \in \N_n} r = \frac{n(n+1)}{2}$$

Therefore, taking the limit to infinity,

$$\sum_{r \in \Q^+} r \ge \lim_{n \to \infty} \sum_{r \in \N_n} r = \lim_{n \to \infty} \frac{n(n+1)}{2}$$

which is clearly infinite, and thus the sum of the positive rationals does not converge. (Again, we do not say "what kind" of infinity it is, simply because that question is not really well-defined, or is nonsensical really, per the earlier discussion.)

Can we tell anything about how big the sum of $\aleph_0$ or even $\aleph_i$ numbers can get?

Since every summation (within the definitions I've given) are limited to essentially countable sums at most, bear that in mind going forward.

As you noted, we can essentially make a sequence of rational numbers approach any real number. (This is because $\R$ forms what we call a "complete metric space" and in particular is what we call the "completion" of the metric space for $\Q$.) In particular, if you want a sequence of rational numbers that approach $r \in \R$, simply define

$$r_n \eqdef \frac{ \lfloor 10^n r \rfloor}{10^n}$$

If you want a summation that does so, take

$$s_n \eqdef \frac{ \lfloor 10^n r \rfloor - 10^n \sum_{i=0}^{n-1} s_i }{10^n} , s_0 = \lfloor r \rfloor$$

and then $\sum_{n=0}^\infty s_n = r$. (Essentially just a formal way of summing up the digits of the number, relative to their base-$10$ representation.)

Thus, within the real numbers, there isn't really a "limit" as to how big a sum can get: it can be any finite number you want. Of course, there is also infinity (which is often defined as being greater than every real number), but if you include that then $\infty$ is the absolute most you can achieve (because once again, there aren't different "kinds" of this particular conceptualization of infinity).

ContentLicense: CC BY-SA 4.0

Title: Sum of $\aleph_0$ numbers


问:
Calculate $I=\int_0^{2\pi}dx \int_0^{\pi} e^{\sin y(\cos x-\sin x)}\sin y\,dy$
I'm not quite sure how to calculate this integral:

$$I=\int_0^{2\pi}dx \int_0^{\pi}  e^{\sin y(\cos x-\sin x)}\sin y\,dy$$

With Mathematica I see the result is $2\pi \sqrt 2 \sinh \sqrt 2$.

Is there any special technique for calculating this integral?

Any help will be appreciated.

Thank you for these brilliant solutions using Bessel function or complex analysis.

After looking into this integral for a long time I found another "primary" solution:

Rewrite it as

$$I=\int_0^{2\pi}d\theta \int_0^{\pi}  e^{\sin \varphi(\cos \theta-\sin \theta)}\sin \varphi\,d\varphi$$

By polar coordinates it becomes

$$I=\int_{u^2+v^2+w^2=1} e^{u-v}\,dS$$

since $dS=\sin\varphi d\theta d\varphi$. Then choose an orthogonal transform such that $z=\frac{u-v}{\sqrt 2}$, we get

$$I=\int_{x^2+y^2+z^2=1} e^{\sqrt 2 z}\,dS$$

Going back to polar coordinates it becomes

$$I=\int_0^{2\pi}d\theta \int_0^{\pi}  e^{\sqrt 2 \cos\varphi}\sin \varphi\,d\varphi$$

And finally we get

$$I=2\pi \int_{-1}^1 e^{\sqrt 2 t}\,dt$$

答:
Denoting : $ f : \mathbb{C}\rightarrow\mathbb{C},\ z\mapsto\frac{\mathrm{e}^{\frac{a\left(1+\mathrm{i}\right)z}{2}}\,\mathrm{e}^{\frac{a\left(1-\mathrm{i}\right)}{2z}}}{\mathrm{i}\,z} $, where $ a\in\mathbb{R} \cdot $

$ \left(\forall n\in\mathbb{Z}\right),\ a_{n}=\left\lbrace\begin{matrix}\frac{a^{n}\left(1+\mathrm{i}\right)^{n}}{2^{n}n!},\ \text{If }n\geq 0\\ \ \ \ \ \ 0, \ \ \ \ \ \ \text{If }n<0\end{matrix}\right. $, and $ \left(\forall n\in\mathbb{Z}\right),\ b_{n}=\left\lbrace\begin{matrix}\frac{a^{-n}\left(1-\mathrm{i}\right)^{-n}}{2^{-n}\left(-n\right)!},\ \text{If }n\leq 0\\ \ \ \ \ \ 0, \ \ \ \ \ \ \text{If }n>0\end{matrix}\right. $

$$ \oint_{\left|z\right|=1}{\frac{\mathrm{e}^{\frac{a}{2}\left(z+\frac{1}{z}\right)-\frac{a}{2\,\mathrm{i}}\left(z-\frac{1}{z}\right)}}{\mathrm{i}\,z}\,\mathrm{d}z}=\oint_{\left|z\right|=1}{\frac{\mathrm{e}^{\frac{a\left(1+\mathrm{i}\right)z}{2}}\,\mathrm{e}^{\frac{a\left(1-\mathrm{i}\right)}{2z}}}{\mathrm{i}\,z}\,\mathrm{d}z} $$

\begin{aligned} f\left(z\right)&=\frac{1}{\mathrm{i}\,z}\left(\sum_{n=0}^{+\infty}{\frac{a^{n}\left(1+\mathrm{i}\right)^{n}}{2^{n}n!}z^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{a^{n}\left(1-\mathrm{i}\right)^{n}}{2^{n}n!}z^{-n}}\right)\\ &=\frac{1}{\mathrm{i}\,z}\left(\sum_{n=-\infty}^{+\infty}{a_{n}z^{n}}\right)\left(\sum_{n=-\infty}^{+\infty}{b_{n}z^{n}}\right)\\ f\left(z\right) &=\frac{1}{\mathrm{i}\,z}\sum_{n=-\infty}^{+\infty}{c_{n}z^{n}} \end{aligned}

Where $$ \left(\forall n\in\mathbb{Z}\right),\ c_{n}=\sum_{k=-\infty}^{+\infty}{a_{k}b_{n-k}}=\sum_{k=\max\left(n,0\right)}^{+\infty}{\frac{a^{2k-n}\left(1+\mathrm{i}\right)^{k}\left(1-\mathrm{i}\right)^{k-n}}{2^{2k-n}k!\left(k-n\right)!}} $$

Setting $ n $ to $ 0 $, we'll get : $$ \sum_{k=\max\left(n,0\right)}^{+\infty}{\frac{a^{2k-n}\left(1+\mathrm{i}\right)^{k}\left(1-\mathrm{i}\right)^{k-n}}{2^{2k-n}k!\left(k-n\right)!}}=\sum_{n=0}^{+\infty}{\frac{a^{2n}}{2^{n}\left(n!\right)^{2}}} $$

Thus, applying the residue theorem : $$ \int_{0}^{2\pi}{\mathrm{e}^{a\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}=2\pi \sum_{n=0}^{+\infty}{\frac{a^{2n}}{2^{n}\left(n!\right)^{2}}} $$

For $ y \in\mathbb{R} $, setting $ a=\sin{y} $, then integrating with respect to $ y $, we get : $$ \int_{0}^{\pi}{\sin{y}\int_{0}^{2\pi}{\mathrm{e}^{\sin{y}\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}\,\mathrm{d}y}=4\pi\sum_{n=0}^{+\infty}{\frac{W_{2n+1}}{2^{n}\left(n!\right)^{2}}} $$

Where $ W_{n} $ for $ n\in\mathbb{N} $, is the Wallis integral $ W_{n}=\int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\,\mathrm{d}x} \cdot $

Note that since we're integrating on a segment, switching the integral and the infinite sum, which we've just done, is in deed possible.

Since $ \left(\forall n\in\mathbb{N}\right),\ W_{2n+1}=\frac{2^{2n}\left(n!\right)^{2}}{\left(2n+1\right)!} $, we have : \begin{aligned} \int_{0}^{\pi}{\sin{y}\int_{0}^{2\pi}{\mathrm{e}^{\sin{y}\left(\cos{x}-\sin{x}\right)}\,\mathrm{d}x}\,\mathrm{d}y}&=4\pi\sum_{n=0}^{+\infty}{\frac{2^{n}}{\left(2n+1\right)!}} \\ &=2\pi\sqrt{2}\sinh{\sqrt{2}}\end{aligned}


问:
In what sense is logical entailment a set-theoretic relation?
In set theory a relation is a subset of a Cartesian product. I suppose that in logics this product is the Cartesian square of the powerset of all propositions(?).

Semantic/model-theoretic entailment between two sets holds (is "true"?) when there is no interpretation that makes elements left of the double turnstile $\vDash$ all true and elements on the right all false.

Syntactic/proof-theoretic entailment $\vdash $ holds if you can derive the RHS from the LHS.

In what formal sense are these set theoretic relations, and is this at all important?

答:
Under the most common definition of entailment, it is a relation between a (possibly empty or singleton) set of premise propositions and a single conclusion proposition, i.e. a subset of $\wp{(\text{PROP})} \times \text{PROP}$ with $\langle \Gamma, \phi \rangle \in R \text{ iff } \Gamma \vDash \phi$.

If one permits, rather than a single proposition, a set of conclusions which are read as disjoined (entailment holding if at least one of the conclusions is true), one gets $R \subseteq \wp{(\text{PROP})} \times \wp{(\text{PROP})}$ wiht $\langle \Gamma, \Delta \rangle \in R \text{ iff } \Gamma \vDash \Delta$ by the definition you cited.

While the relation holds between sets of propositions, a set-theoretic definition can be given via the propositions' models: By definition, entailment holds (not: "is true") iff all interpretations which are models of all premises are also models of at least one conclusion. This means that the set of models of the premises is a subset of the set of models of the conclusions. The models of the premises are those interpretations which are models of all of the premises; this is the intersection between the models of the premises. The models of the conclusions are those interpretations which are models of at least one of the conclusions; this is the union between the models of the conclusions. Thus, we have:

$$\psi_1, \ldots, \psi_m \vDash \phi_1, \ldots, \phi_n\\
\text{ iff } \bigcap_{i=1}^{m} \{\mathcal{M}: [\![\psi_i]\!]^\mathcal{M} = 1\} \subseteq \bigcup_{j=1}^{n} \{\mathcal{M}: [\![\phi_j]\!]^\mathcal{M} = 1\}$$

In order to make the limit cases behave as desired, one needs to define that the set of models for an empty set of premises (empty intersection of models) is the set of all structures, and the models for an empty set of conclusions (empty union of models) is the empty set.

For the variant with a singular conclusion, it is the same without disjunction of truth/union of models on the right-hand side:

$$\psi_1, \ldots, \psi_m \vDash \phi\\
\text{ iff } \bigcap_{i=1}^{m} \{\mathcal{M}: [\![\psi_i]\!]^\mathcal{M} = 1\} \subseteq \{\mathcal{M}: [\![\phi]\!]^\mathcal{M} = 1\}$$

As for whether this is relevant: It does not really add new information because it is pretty much a direct consequence or even just a paraphrase of the definition, but perhaps it helps to better understand some of the properties of the entailment relation (contradictory premises entail arbitrary conclusions because the empty set is a subset of every set; an argument with no premises is valid iff the conclusion is a tautology because the conclusion's models then must be the set of all structures; entailment is reflexive and transitive but not symmetric because so is the subsethood relation, etc.)

A similar idea of accounting for relations between propositions via their models is also useful in the field of cognition, to model an agent's beliefs and changes in those beliefs through the addition of new information, though this is usually done with the related concept of possible worlds rather than structures.


问:
Prove $\sqrt{2}$ is irrational using prime number properties
I found the following argument in a textbook that uses a number theory approach. However I do not understand the last two sentences that seems to use some number theory properties, can someone please help explain them?

Let
$$
a^2 = 2b^2, \quad (a,b)=1, \quad a,b\in \mathbb{N}
$$

Rearranging,
$$
b^2 = (a+b)(a-b)
$$

Let $p$ be a prime factor of $b$, then
$$
p \mid (a+b)\quad or\quad p\mid (a-b)
$$

If $p$ divides any of the above, then $p$ divides both of them, hence $p$ must divide $a$. Then $p$ would be a common divisor of $a$ and $b$, a contradiction.

答:
The conclusion that $p\mid(a+b)$ or $p\mid(a-b)$ comes from the property of primes that if $p\mid(rs)$ then $p\mid r$ or $p\mid s$. This, in fact, is the defining property of prime elements in a ring.

For the next sentence: if $p$ divides $a+b$ then it divides $(a+b)-2b=a-b$; if $p$ divides $a-b$ then it divides $(a-b)+2b=a+b$. The author didn't need this step to conclude that $p$ divides $a$, however; they could have simply proceeded as in Malkoun's comment: if $p$ divides $a+b$ then $p$ divides $(a+b)-b=a$; if $p$ divides $a-b$ then $p$ divides $(a-b)+b=a$.

The final sentence of the proof just uses the assumption that $p\mid b$ and the conclusion that $p\mid a$ to infer that $p$ is a common divisor of $a$ and $b$, which violates the earlier assumption $(a,b)=1$.


答:
You can fix a bijection $a:\Bbb N \to \Bbb Q^+$ and then talk about the series $\sum_{n = 1}^\infty a(n).$

Since all terms are positive, it does not matter which bijection you fix. Of course, this series will then evaluate to $\infty$ since $a(n)$ is unbounded.

Title: Sum of $\aleph_0$ numbers

Body:
I tried to search online for this and look in some textbooks, but didn't seem to find anything. My question is:

What is the sum of all positive rational numbers? Can it be defined? What type of infinity is it?

Of course, we can ask ourselves the same question in a more general way:

What is the sum of $\aleph_0$ numbers? Again, can it be defined? What does it depend on? What about $\aleph_i$ numbers?

For example, the sum $$\sum_{i=1}^{\infty}\frac{1}{2^i}=1$$

so in this case, the sum of $\aleph_0$ numbers is $1$. However, we can find a series of rational numbers that converges to any irrational numer too, thus, sometimes the sum of $\aleph_0$ numbers is irrational.

I know this is a bit vague, but, to sum up again:

What is the sum of all positive rational numbers? Can we tell anything about how big the sum of $\aleph_0$ or even $\aleph_i$ numbers can get?

Thanks


问:
How to find out when this expression is greater than 0
For what value of $a$ is this expression always greater than zero?

$$ax_2^2+10x_1^2+2x_2x_1+6x_3x_1+x_3^2+2x_2x_3$$

I know that the answer is 5 but I have no idea of how to calculate that by hand. How do one even solve such a problem by hand?

答:
No value of $a$ guarantees that the expression is always greater than zero, since $x_1=x_2=x_3=0$ will yield the value zero, regardless of the value of $a$. If you meant greater or equal than zero, the associated symmetric matrix can be either positive definite or positive semi-definite.

$$
ax_2^2+10x_1^2+2x_2x_1+6x_3x_1+x_3^2+2x_2x_3= x^T \begin{pmatrix}10 & 1 & 3 \\
1 & a & 1\\3 & 1 & 1 \end{pmatrix}x.
$$

If you study this matrix (using the determinants of the principal minors or the eigenvalues) you'll see that:

If $a < 5$ the matrix is either indefinite or negative definite and so it does not meet the requirements.

If $a > 5$, the matrix is positive definite and so the expression takes positive values for non-zero vectors (and zero for the zero vector).

If $a=5$ the matrix is positive semi-definite. The expression takes the value zero for eigenvectors associated with $\lambda = 0$ and positive values for all other vectors.

The final answer should be $a \ge 5$.


问:
Prove that ${a_{1} +2a_{2} +3a_{3}+\cdots+np\ a_{n}}_{p} \ =\frac{np}{2}( 1+p)^{n}$
In expansion of ${\ \left( 1+x+x^{2} +\cdots+x^{p}\right)^{n} =a_{0} +a_{1} x+a_{2} x^{2} +\cdots+a_{np}} x^{np}$,
prove that ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p} \ =\frac{np}{2}( 1+p)^{n}$.

My attempt:
${\ \left( 1+x+x^{2} +\cdots+x^{p}\right)^{n} =a_{0} +a_{1} x+a_{2} x^{2} +\cdots+a_{np}} x^{np}$
for $x=1$, $(1+p)^n=a_0+a_1+a_2+a_3+\cdots+a_{np}$  if we multiply $(a_o+a_1x+a_2x^2+\cdots+a_{np}x^{np})(1+2\frac{1}{x}+3\frac1{x^2}+4\frac{1}{x^3}+\cdots)$ we will get terms independent of $x$, $a_0+2a_1+3a_2+\cdots+(np+1)a_{np}$ then if we subtract $(1+p)^n$ from it, then we will get ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p}$ so we need to find term independent of $x$ in $(1+x+x^2+\cdots+x^{p})^n (1-\frac{1}{x})^{-2}$ simplifying further  $(1+x+x^2+\cdots+x^{p})^n x^2 (1-x)^{-2}$ here no terms will be free of $x$ hence term will be $0$, subtracting $(1+p)^n$ from it we have ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p}=-(1+p)^n$ .

Body:
$$(1+x+x^2+...+x^p)^n=(a_0+a_1x+a_2x^2+....+a_{np}x^{np})$$
D.w.r.t. $x$ on both sides
$$n(1+x+x^2+...+x^p)^{n-1}(1+2x+3x^2+4x^3+...+px^{p-1})=a_1+2a_2x+a_3x^2+a_4x^3+...+np a_{np} x^{np-1}$$
Put$x=1$ to get
$$n (1+p)^{n-1}\frac{p(p+1)}{2}=a_1+2a_2+3a_3+...+np a_{np}=\frac{1}{2}np(1+p)^n.$$


问:
Dominated Convergence and Monotone Convergence
Assuming that the measure space is $\sigma$-finite, how can I show that the Dominated convergence theorem implies the monotone convergence theorem?

答:
This is a nice exercise. Since the monotone convergence theorem is often the very first theorem proved in measure theory, the only tools we can safely use from measure theory are the basic properties of measures, the definition of Lebesgue integral and the dominated convergence theorem (DCT). Once you realize this, you're almost there.

Let $(X,\mu)$ be a $\sigma$-finite measure space. Take an increasing sequence $\{f_{n}\}$ of integrable functions with $f_{n} \geq 0$. Let $f = \lim_{n} f_{n}$. We want to show $\lim_{n} \int f_{n} d\mu = \int f d\mu$. (I assume that showing the measurability of $f$ is not part of the exercise.)

If $\int f d\mu < \infty$, then $\lim_{n} \int f_{n} d\mu = \int f d\mu$ by DCT.

Suppose $\int f d\mu = \infty$. If $\lim_{n} \int f_{n} d\mu= \infty$, we're done. Suppose this limit is finite, say $B$. So for the rest of the proof, we assume
$$
\int f d\mu = \infty, \quad \lim_{n} \int f_{n} d\mu = B < \infty.
$$
Lemma: If $\mu(X) < \infty$, then  $\lim_{n} \int f_{n} d\mu = \int f d\mu$ by DCT.

Proof: Since $\int f d\mu = \infty$, by the definition of Lebesgue integral for nonnegative functions, there's a simple function $s \leq f$ such that
$$
B < \int s d\mu < \infty.
$$
The finiteness of $\mu$ is needed for the second inequality above. Let $g_{n} = \min \{f_{n}, s\}$.  Since $f_{n}$ increases to $f \geq s$, $g_{n}$ increases to $s$. So by DCT,
$$
\lim_{n} \int g_{n} d\mu = \int \lim_{n} g_{n} d\mu = \int s d\mu > B.
$$
But
$$
\lim_{n} \int g_{n} d\mu \leq \lim_{n} \int f_{n} d\mu = B,
$$
a contradiction.  QED

Back to the case $(X,\mu)$ is $\sigma$-finite.  There's an increasing sequence $\{A_{j}\}$ of measurable sets such that $\cup_{j} A_{j}=X$ and $\mu(A_{j}) < \infty$. By the lemma above, for each $j$,

$$
\int_{A{j}} f d\mu = \lim_{n} \int_{A_{j}} f_{n} d\mu  \leq \lim_{n} \int f_{n} d\mu = B.
$$
Letting $j \rightarrow \infty$,
$$
\lim_{j} \int_{A_{j}} f d\mu \leq B.
$$
Since $\int f d\mu = \infty$, by the definition of Lebesgue integral for nonnegative functions, there's a simple function
$$
s = \sum_{k=1}^{m} a_{k} 1_{C_{k}} \leq f
$$
such that
$$
B < \int s d\mu = \int \sum_{k=1}^{m} a_{k} 1_{C_{k}} d\mu = \sum_{k=1}^{m} a_{k} \mu(C_{k}).
$$
(The difference here from the proof of the lemma above is that the rightmost side may not be finite.) We have
$$
\int_{A_{j}} f d\mu \geq \int_{A_{j}} \sum_{k=1}^{m} a_{k} 1_{C_{k}} d\mu = \sum_{k=1}^{m} a_{k} \mu(C_{k} \cup A_{j}).
$$
Letting $j \rightarrow \infty$,
$$
B \geq \lim_{j} \int_{A_{j}} f d\mu \geq \lim_{j} \sum_{k=1}^{m} a_{k} \mu(C_{k}\cup A_{j}) = \sum_{k=1}^{m} a_{k} \lim_{j} \mu(C_{k} \cup A_{j}) = \sum_{k=1}^{m} a_{k} \mu(C_{k}) > B.
$$
The last equality above is valid because any measure is continuous from below. The first equality is valid because the limit of the finite sum of convergent sequences is the sum of the limits of the sequences. Now we have a contradiction.


问:
Definite integration of a trigonometric function

How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$

my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$

I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.

答:
Essentially, we want to integrate
$$I = \int_0^{\pi/2} \dfrac{\sin^2(t)dt}{1+c \sin^2(t)}$$
where $c<1$.
We have
$$I = \sum_{k=0}^{\infty} (-c)^k\int_0^{\pi/2}\sin^{2k+2}(t)dt = \dfrac{\pi}8 \cdot\sum_{k=0}^{\infty}\left(\dfrac{-c}4 \right)^k \dbinom{2k+2}{k+1} = \color{red}{\dfrac{\pi}{2c} \left(\dfrac{\sqrt{1+c}-1}{\sqrt{1+c}}\right)}$$

$\dfrac1{1+r} = \displaystyle \sum_{k=0}^{\infty}(-r)^k$; $\displaystyle \int_0^{2\pi} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{2\pi}{2^{2k}}$; $ \dfrac1{\sqrt{1-4x}} = \displaystyle\sum_{k=0}^{\infty} \dbinom{2k}k x^k \,\, \forall x \in \left[-\dfrac14,\dfrac14 \right)$

The integral you have is
$$J = \dfrac{2a}{b^2} \int_0^{\pi/2} \dfrac{\sin^2(x)}{1+ \underbrace{\left(\dfrac{a^2-b^2}{b^2}\right)}_{c} \sin^2(x)}dx = \dfrac{2a}{b^2} \cdot \frac{\pi}{2c} \left(\dfrac{\vert a/b \vert-1}{\vert a/b \vert}\right) = \dfrac{a \pi}{a^2-b^2}\dfrac{\vert a \vert - \vert b \vert}{\vert a \vert}$$
This gives you
$$\boxed{\color{blue}{J = \dfrac{\text{sign}(a)}{\vert a \vert + \vert b \vert} \pi}}$$

ContentLicense: CC BY-SA 3.0


问:
What is the rationale behind the method behind generation of normally distributed vectors?
For a long time, I thought that a normally distributed vector $x \in \mathbb{R}^n$ was generated in the following way

Sample $x_i$ from a Gaussian distribution, $x_i = \mu +
 \sigma^2\mathcal{N}(0,1)$, where $\mu \in \mathbb{R}$ and $\mathcal{N}(0,1)$ is a normally distributed value. Then form a vector $x = (x_i)$.

However, I recently discovered that in fact the approach is,

Let $C$ be a covariance matrix. Factorize covariance matrix $C$ using
Cholesky decomposition, $C = AA^\top$, where $A$ is lower triangular.
Then $x = m + A  \mathcal{N}(\mathbf{0}, I)$, where $m \in \mathbb{R}^n$ is the mean vector, $\mathcal{N}(\mathbf{0}, I)$ is a normal vector

Are these approach one and the same? I am guessing not because the
first way is assuming a uniform variance.

Which is more correct?

And what is the rationale behind factorizing $C$ using Cholesky
decomposition.

答:
In your original approach, if you want $\ \sigma\ $ to be the standard deviation of $\ x_i\ $ you should have $\ x_i=\mu+\sigma\mathcal{N}(0,1)\ $, rather than $\ x_i=\mu+\sigma^2\mathcal{N}(0,1)\ $.  If you do that, your original approach is just a special case of the second, in which your generated $\ x_i\ $ will be independent normal variates with mean $\ \mu\ $ and variance $\ \sigma^2\ $—that is, $\ m=(\mu,\mu,\dots,\mu)\ $ and $\ C=\sigma^2I_{n\times n}\ $.

Which approach is "correct" depends entirely on what you want to do.  If all you want to do is generate a sequence of independent, normally distributed , random variables, all with mean $\ \mu\ $ and variance $\ \sigma^\ $, then your original approach is perfectly fine, but if you want to generate a random vector which is multivariate normally distributed with mean $\ \big(m_1,m_2,\dots,m_n\big)\ $ and a covariance matrix $\ C\ $ that is not just a multiple of the identity, then your original approach cannot do that, and you'd have to resort to something like the second.

It's not necessary for the $\ A\ $ you use in the second approach to be obtained from the Cholesky decomposition, merely that it satisfy the equation $\ AA^\top=C\ $. However, since the Cholesky decomposition gives you an $\ A\ $ that's lower triangular, it will then require less work to calculate the product $\ A\mathcal{N}(\mathbf{0},I)\ $ than if $\ A\ $ were filled entirely with non-zero entries.


问:
Prove that $2^{n-1}(a^n+b^n)\geq(a+b)^n$
I'm trying to prove for algorithm-purposes that given $a,b,n$ positive integers:
$$2^{n-1}(a^n+b^n)\geq(a+b)^n$$
I tried by induction, and got the step as follows:
$$2^{n}(a^{n+1}+b^{n+1})\geq^?(a+b)^{n+1}$$
I tried using the binomial expansion $(a+b)^n=\sum^n_{k=0} {{n}\choose{k}}a^kb^{n-k}$ and then excluding the last element
$$(a+b)^{n+1}=\sum^{n+1}_{k=0} {{n+1}\choose{k}}a^kb^{n-k+1}=\sum^{n}_{k=0} {{n+1}\choose{k}}a^kb^{n-k}b+{{n+1}\choose{n+1}}a^{n+1}b^0$$$$=\sum^{n}_{k=0} (n+1){{n}\choose{k}}a^kb^{n-k}b+{{n+1}\choose{n+1}}a^{n+1}b^0=[(n+1)b]\sum^{n}_{k=0}{{n}\choose{k}}a^kb^{n-k}+{{n+1}\choose{n+1}}a^{n+1}b^0$$$$=[(n+1)b](a+b)^n+a^{n+1}\leq[(n+1)b]\times2^{n-1}(a^n+b^n)+a^{n+1}$$
Assuming that everything is correct so far, I don't how to proceed from there to get $\leq 2^n(a^{a+1}+b^{n+1})$

My second attempt was by doing the step as follows:
$$2^{n-1}(a^n+b^n)\geq(a+b)^n \setminus\cdot(a+b)$$
$$2^{n-1}(a^n+b^n)(a+b)\geq(a+b)^{n+1}$$
$$2^{n-1}(a^{n+1}+b^{n+1}+a^nb+b^na)\geq(a+b)^{n+1}$$
Now I don't know how to eliminate $a^nb+b^na$, and to proceed to $2^n$

Is there another way to prove this? Or any hints to continue my step?

答:
From the rearrangement inequality, we have the following: $$a^n+b^n\geq a^i\cdot b^{n-i}+a^{n-i}\cdot b^i$$

$$(a+b)^n=\sum_{i=0}^{n}\binom{n}{i}a^ib^{n-i}\leq\frac{a^n+b^n}{2}\sum_{i=1}^{n}\binom{n}{i}$$

which is exactly what you want


问:
Trouble Understanding Seemingly Simple Property of the Cantor Ternary Space
In Cullen's Introduction to General Topology the very first result involving the Cantor Ternary Space (CTS) is that no single point is open. I'll detail the proof below, but first would like to state my innate disbelief of this result: the product topology defines open sets as those products which are of all open sets... Is every sequence of the Cantor set not just the infinite product of open sets from the discrete space N2={0,1}? This seems to completely contradict the definition of the product topology as far as I can see, but clearly I'm missing something. Here's the proof as stated in the text:

By definition of the product topology any base element $U^*$ is $\Pi_{i=1}^\infty U_i$, where $U_i$ is open in $Y_i=\{0,1\}$ in $U_i=Y_i$ for $i>k\in\mathbb{N}$. Hence for any point in the CTS denoted by $(x_1,x_2,x_3,...,x_i,...)$, no set containing this one point contains a base element, since $\{x_{k+1}\}$ does not contain $Y_{k+1}$. Hence, $\Pi_{i=1}^\infty \{x_i\}$ does not contain $U^*$, so no one point set is open.

Perhaps it's the notation that's confusing me? As far as I can tell from searching for the past thirty minutes, this is the only source that addresses this specific result... I simply can't find an explanation that makes sense.

答: Maybe what you are missing is that, in the product topology, open sets are rarer than you think. (See: https://en.wikipedia.org/wiki/Product_topology , the base consists of $\prod_{i\in\mathbb N}U_i$ where $U_i$ is open in $\{0,1\}$ and only finitely many $U_i$'s are $\ne\{0,1\}$). Without the last condition, any subset of CTS would be open - i.e. the topology would be discrete. Another way to look at this is that the CTS is homeomorphic to Cantor set (https://en.wikipedia.org/wiki/Cantor_set), which you can visualise on the real line


问:
$X$ and $Y$ are random variables. If $E(X) = E(Y)\;$ and $P(X \geq Y) =1\;$ then $X = Y$ almost surely.
I came across the statment that if $X$ and $Y$ are random variables. If $E(X) = E(Y)\;$ and $P(X \geq Y) =1\;$ then $X = Y$ almost surely.

I know that it is sufficient to show that $P(Y \geq X) =1\;$, I tried by using the linearity of the expectation. I know that:

$E(X) - E(Y) = E(X-Y) = \sum _{\omega \in \Omega} (X-Y)(\omega)P(\left \{ \omega \right \}) = 0$

However, I don't really know how to proceed from here.

I'd love to get some help!

答:
I would try to argue by contradiction.

If $X$ isn't equal to $Y$ almost surely, then there is a positive probability that $X \neq Y$, say $\mathbb{P}(X \neq Y) \geq \delta > 0$. Now, together with the fact that $\mathbb{P}(X \geq Y) =1$ it follows that $\mathbb{P}(X > Y) \geq \delta$.

Now, if as indicated in your question we're working in a finite probability space this is already sufficient to find a contradiction, since $\min \{ X(\omega) - Y(\omega) \colon X(\omega) \neq Y(\omega)\} = \alpha$ exists and then it's not hard to show that $\mathbb{E}(X) > \mathbb{E}(Y) + \delta \alpha$.

Otherwise one has to be a bit more careful and consider the events
$$
A_n = \{ \omega \colon \frac{1}{n-1} > X(\omega) - Y(\omega) > \frac{1}{n}\}
$$
We know that $\mathbb{P}\left( \bigcup A_n) \right) > \delta$ and so there must be some $m$ and $\epsilon$ such that $\mathbb{P}(A_m) > \epsilon$, and then we can argue as before.


问:
Understanding roots of complicated equations
This is a very broad question apparently, but is there any systematic way to obtain (sort of) sharp estimates for roots of complicated equations? Or maybe bounds for the roots?

For example, consider the equation $$x = A - B\dfrac{e^x}{1+e^x}$$where $A,B$ are large positive integers with $B>A$. It is easy to see that this equation has a unique root, by observing that the function $f(x)=x - A + B\dfrac{e^x}{1+e^x}$ is increasing in $x$. But I want to get some pretty sharp estimates of where the root $x^*$ lies.

I have looked at several numerical examples and it seems that $x^*$ is very close to $\log(A/(B-A))$. But I cannot really prove it.

So I am asking, given a complicated equation, what is a good way to get a good estimate on where the root must lie?

答:
In fact, there is a formal solution to this equation. Rewrite it as
$$e^{-x}=-\frac{x-(A-B)}{x-A}$$ and the solution is given in terms of the generalized Lambert function (have a look at equation $(4)$). The bad news is that, from a practical point of view, it does not bring much.

What we could try is to use an high order iterative method which would give for example
$$x=-\frac{4 (2 A-B) \left(-B \left(-2 A^2+2 A B+(B+6) (B+12)\right)-96\right)}{(B+4)
   \left(B \left(-8 A^2+8 A B+B (B+36)+144\right)+192\right)}$$

Trying for $A=1234$ and $B=5678$, this would give as an estimate
$x=-1.28244$ while the solution is $x=-1.27997$. Pure luck ?

In this case, your guess is impressive since $-\log \left(\frac{2222}{617}\right)\sim -1.28129$ (close to mine - which is not !)

Trying two orders above (formula not typed - just too long), the guess is $x=-1.27994$.


问:
Hypothesis of dominated convergence theorem
The dominated convergence theorem says :

Suppose $(f_n)$ is a sequence of measurable function s.t. $f_n(x)\longrightarrow f(x)$ a.e. as $n\to \infty $. If $|f_n(x)|\leq g(x)$ and $g$ is integrable, then $$\lim_{n\to\infty }\int f_n=\int f.$$

Quest 1) If $|f|$ is integrable, do we directly have $\lim_{n\to\infty }\int f_n=\int f$ or we need to have $|f_n(x)|\leq |f(x)|$ ?

Quest 2) If no, I have that $g$ is continuous with compact support (and thus integrable). Why the dominated convergence theorem allows me to conclude that $$\lim_{\delta\to 1}\int g(\delta x)dx=\int g(x)dx\ \ ?$$

Don't talks about change the variable $u=\delta x$ to prove it, it's not my question (I know how to prove this). My question is precisely : Why can we conclude using dominated convergence ? Because we don't necessarily have $|g(\delta x)|\leq g(x)$, and thus I'm a little annoyed by this argument.

答:
For question 1, to apply the dominated convergence theorem, you definitely need $|f_{n}| \leq |f|$ even if $f$ is integrable (as the first comment points out).

Here's a simple example to show why the integrability of $f$ is not enough. Let $f_{n}(x) = 1$ if $x \in [n,n+1)$, and $0$ otherwise. Here the domain of $f_{n}$ is $\mathbb{R}$. Then $f$ is identically equal to zero and obviously integrable, but
$$
1=\lim_{n} \int f_{n} > \int f = 0.
$$

On the other hand, to show $\lim_{n} \int f_{n} = \int \lim_{n} f_{n}$, you don't have to use the dominated convergence theorem. There're some extensions of the theorem that do not require the existence of a dominating function. There's also a recent paper that gives a necessary and sufficient condition for $\lim_{n} \int f_{n} = \int \lim_{n} f_{n}$.


问:
Proof of Dedekind cut

If there exist a rational number whose square = D, then there exist
two positive integers $t, u$, that satisfy the equation $t^2 − Du^2 = 0$, and we may assume that $u$ is the least positive integer.
In the end of the proof, he proved that there exists a $u'$ less than
$u$ and a $t'$ such that $t'^2 − Du'^2 = 0$. So it's contrary to the
assumption of u. Then, he concluded that the square of every rational
number x is either < D or > D.

答: Let $\frac{t}{u}$ be the rational number whose square is $d$. Then $\left(\frac{t}{u}\right)^2 = d \Rightarrow t^2 = du^2 \Rightarrow t^2-du^2=0$, as stated. Having shown by contradiction that $u$ cannot exist, it follows that the supposedly rational number $\frac{t}{u}$ cannot exist, i.e. there does not exist a rational number whose square is $d$.

Title: Proof of Dedekind cut


问:
How to find number of ways to put $n$ different balls in 3 different boxes?
$n$ is an even number. I have $n$ different balls and I need to put them into $3$ different boxes satisfying conditions:

$\frac{n}{2}$ balls are in the first box;

Number of balls in the second box is in the form $3k+1$.

It seems simple, but somehow I struggle to see the solution.

答:
The number of ways to distribute the balls is:
${n\choose n/2}\left({n/2\choose 1}+{n/2\choose 4}+{n/2\choose 7}+\ldots\right)$ (simple combinatorial argument). The trouble is, of course, to calculate the sum in the brackets.

One way of calculating this sum is this: label $\epsilon=e^{\frac{2i\pi}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$. We know that $\epsilon^3=1$ and $\overline{\epsilon}=\epsilon^2$. Now write the following powers of sums:

$$A=(1+1)^{n/2}=1+{n/2\choose 1}+{n/2\choose 2}+\ldots$$
$$B=(1+\epsilon)^{n/2}=1+{n/2\choose 1}\epsilon+{n/2\choose 2}\epsilon^2+\ldots$$
$$C=(1+\overline{\epsilon})^{n/2}=1+{n/2\choose 1}\overline{\epsilon}+{n/2\choose 2}\overline{\epsilon}^2+\ldots$$

What is $A+\overline{\epsilon}B+\epsilon C$? When you expand this, you will end up with terms of the form $(1+\overline{\epsilon}+\epsilon){n/2\choose 3k}$ (which are all zero as $1+\epsilon+\overline{\epsilon}=0$), with terms of the form $(1+\overline{\epsilon}\epsilon+\epsilon\overline{\epsilon}){n/2\choose 3k+1}$ (the value in the first bracket is actually precisely $3$ as $\epsilon\overline{\epsilon}=1$) ... and with the terms of the form $(1+\overline{\epsilon}^2+\epsilon^2){n/2\choose 3k+2}$ - which are again all zero.

The conclusion is that $A+\overline{\epsilon}B+\epsilon C=3\left({n/2\choose 1}+{n/2\choose 4}+{n/2\choose 7}+\ldots\right)$. Thus, the total number of distributions of the balls is:

$${n\choose n/2}\frac{2^{n/2}+\overline{\epsilon}(1+\epsilon)^{n/2}+\epsilon(1+\overline{\epsilon})^{n/2}}{3}$$

This can be further compactified by noticing that $1+\epsilon=e^\frac{i\pi}{3}, 1+\overline{\epsilon}=e^{-\frac{i\pi}{3}}$, but this would require distinguishing cases for $n/2\pmod 6$. For example, if $n/2\equiv 0\pmod 6$, then $(1+\epsilon)^{n/2}=(1+\overline{\epsilon})^{n/2}=1$, so the formula boils down to:

$${n\choose n/2}\frac{2^{n/2}-1}{3}\text{ when }n\equiv 0\pmod{12}$$

I would leave it to you to finish the calculation for the other five cases.


问:
What is the name of Fibonacci variation when $F(n) = a\cdot F(n-1) + b\cdot F(n-2) + c$, were $c$ is a constant, and $a >0, \ b >0, \ c>0$
I am trying to write $\log(n)$ algorithm for the above. I don't know if there is a specific name for the Fibonacci variation when:

$$F(n) = a\cdot F(n-1) + b\cdot F(n-2) + c$$

where: $a >0, \ b >0, \ c>0$

Could someone help me with the name of this variation?

答:
For general coefficients $a,b,c$ what you have is a linear second order difference equation with constant coefficients and constant RHS. I'm not sure what is your algorithm, but you can get a closed form expression for $F_n$ by following these steps:

Compute the general solution of the homogeneous equation $F_n = aF_{n-1}+bF_{n-2}$. This is accomplished knowing the roots of the characteristic equation $p(\lambda) = \lambda^2 - a\lambda -b$. In this case ($a,b>0$) the solution is
$$
F^h_n = c_1 \left(\frac{a+\sqrt{a^2+4b}}{2}\right)^n + c_2 \left(\frac{a-\sqrt{a^2+4b}}{2}\right)^n.
$$
This solution is valid, more generally, if $a^2 + 4b>0$. If $a^2+4b = 0$, $p(\lambda)$ has a real root with multiplicity 2 and the solution would be
$$
F_n^h = (c_1 + c_ 2 n)\left(\frac a2 \right)^n
$$
Finally, if $a^2 + 4b < 0$, say $a^2+ 4b = -\beta^2 (\beta > 0)$, there are two complex conjugate roots of $p(\lambda)$ and
$$
F_n^h = \left(\frac a2\right)^n \left(c_1 \cos \left(\frac{\beta n}{2}\right) + c_2 \sin\left(\frac{\beta n}{2}\right)\right)
$$

Obtain a particular solution of the equation, $F_n^*$. If $a+b \ne 1$ you can take $F_n^* = \frac{c}{1-a-b}$. If $a+b = 1$, you can use $F_n^* = \frac{2c n}{a}$

The general solution to the recurrence is
$$
F_n = F_n^h + F_n^*
$$

If you are given some extra conditions, for instance the values of $F_1$ and $F_2$, you can compute $c_1, c_2$.

In conclusion, in this particular situation,

If $a+b \ne 1$
$$
F_n = c_1 \left(\frac{a+\sqrt{a^2+4b}}{2}\right)^n + c_2 \left(\frac{a-\sqrt{a^2+4b}}{2}\right)^n + \frac{c}{1-a-b}
$$

If $a+b=1$,
$$
F_n = c_1 \left(\frac{a+\sqrt{a^2+4b}}{2}\right)^n + c_2 \left(\frac{a-\sqrt{a^2+4b}}{2}\right)^n + \frac{2c n}{a}
$$


问:
Integral and semi-local ring
Let $R$ be a ring and let $S$ be a subring of R. If $R$ is a semi-local ring and $R$ is integral over $S$, why $S$ is semi-local as well?</p>

答: A ring is semi-local when it has finitely many maximal ideals. By Corollary 5.8 of Atiyah-Macdonald, for every prime ideal $\frak{q}$ of $R$, the ideal ${\frak{p}}={\frak{q}}\cap S$ is maximal if and only if $\frak{q}$ is maximal. By Theorem 5.10 of Atiyah-Macdonald, for every prime $\frak{p}$ of $S$ there is some prime $\frak{q}$ of $R$ with ${\frak{q}}\cap S=\frak{p}$. Thus, because $R$ has finitely many maximal ideals, so will $S$.</p>

Title: Integral and semi-local ring


问:
Poisson process- write down $p_{n+m}$ in terms of $p_n$ and $p_m$
Reading the list of exercises about Poisson Process I came along to such task:

Let $(Sn; n \geq 1)$ be a partial sum of iid sequence with exponential distribution ($EX=\frac{1}{\lambda}$) and let $p_n$ be a  density of variable $S_n$.
How can we write down $p_{n+m}$ using $p_n$ and $p_m$. Discuss the importance of assumption that fucntion $x \mapsto p_n(x) $ is continous $(x \in R )$

I appreciate any hints, because I have no idea what should I do.

答:
For any $n\in N$, $S_n \sim Erlang(n,\lambda)$

where $Erlang(k,\lambda)$ is the Erlang distribution, with pdf $ f(x;n,\lambda)= \frac{\lambda^nx^{n-1}e{-\lambda x}}{(n-1)!}$

The problem then just reduces to the property that if $X \sim Erlang(n,\lambda)$ and $Y \sim Erlang(m,\lambda)$ then, $X+Y \sim Erlang(n+m,\lambda)$. So all you need to do is to write the pdf of $S_n+S_m$ where $S_n$ and $S_m$ are independent(as they occur at different times), which is the convolution between them.


问:
Logic operations between independent random variables
I have one queue with two servers $S_1$ and $S_2$. The serving times are modelled $\sim\exp(\mu_1)$ and  $\sim\exp(\mu_2)$ respectively.

The first server is free while the second has two clients, $A$ whose service is in progress and $B$ that is waiting.
Then another client $X$ arrives.

Now, I need to calculate:

$P_b$, the probability that customer B is still in the system when $S_1$ finishes to serve $X$.

So, intuitively that's simple: let $T_1$ be the time used by $S_1$ to finish one task, and let $T_2$ the time consumed by $S_2$. Saying that $B$ is still in the system means that:

2$T_2$ > $T_1$

In other words, if I change the variable names in $T_A$ and $T_B$, respectively the time $S_2$ needs to serve $A$ and the time for $B$ what I need is:

$P(T_A + T_B > T_1)$

The problem is that I don't know how to step further. The only thing I know is that the sum of two exponential r.v has a density function equals to $\lambda^2 x \ e^{-\lambda x}$.

答:
First of all, the relation $2T_2 > T_1$ is not what you're after. That would implicitly assume that the waiting times for $A$ and $B$ are exactly the same, while they really are independent samples from the same distribution. Also note that the "memorylessness" of the exponential distribution is important here: when customer $X$ arrives, customer $A$'s service is already in progress for some time, let's call it $t$. If we were dealing with a continuous variable other than the exponential distribution, we'd have to take this information into acccount, however the memorylessness says that

$$ \mathbb{P}(T_A > t+s | T_A > t) = \mathbb{P}(T_A > s), \quad s \geq 0.$$

So, you are correct in saying that the probability we are looking for is $\mathbb{P}(T_A + T_B > T_1)$. Here, $T_A,T_B$ are independent exponential random variables with parameter $\mu_2$, and thus $T_A + T_B$ is exponentially distributed with parameter $2\mu_2$. If $f$ is the density of $T_1$, we can now continue by conditioning (aka law of total probability), using the independence of $T_A +T_B$ and $T_1$:

\begin{align*}\mathbb{P}(T_A + T_B > T_1) &= \int_{0}^{\infty} \mathbb{P}(T_A + T_B > T_1 | T_1 = x)f(x)~dx =\int_{0}^{\infty} \mathbb{P}(T_A + T_B > x)f(x)~dx \\
&=\int_{0}^{\infty} e^{-2\mu_2x}\mu_1 e^{-\mu_1x}~ dx = \mu_1 \int_{0}^{\infty}e^{-(2\mu_2+\mu_1)x}~dx \\
&=\frac{\mu_1}{\mu_1 + 2\mu_2}. \end{align*}


问:
Proving limit using definition
I need to prove $$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$ using epsilon-delta definition.
I know I need to show that for every $\epsilon >0$ there exist a $\delta >0$ such that if $|x|<\delta$, then $| {\frac{x^2-8}{x-8}}-1|<\epsilon$

WHat I did:

$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$, but Im having troubles now deciding what my delta should be.

I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.

答:
Hint:

$$\left|\frac{x^2-x}{x-8}\right| = |x|\cdot \left|\frac{x-1}{x-8}\right|.$$

Now, if $\delta<1$, then you can further estimate that:

$|x-1| < 3$

$|x-8| > 6$.

Title: Proving limit using definition


答:
Note that$$\frac{x^2-x}{x-8}=x\frac{x-1}{x-8}.$$Now, if $|x|<1$, then $-1<x<1$ and therefore you have two things:

$-9<x-8<-7$, which implies that $|x-8|>7$;

$-2<x-1<0$, which implies that $|x-1|<2$.

So,$$|x|<1\implies\left|x\frac{x-1}{x-8}\right|<\frac27|x|.$$Therefore, given $\varepsilon>0$, take $\delta=\min\left\{\frac72\varepsilon,1\right\}$and then\begin{align}|x|<\delta&\implies|x|<\frac72\varepsilon\quad\text{and}\quad|x|<1\\&\implies\left|x\frac{x-1}{x-8}\right|<\varepsilon.\end{align}

Title: Proving limit using definition

Body:
I need to prove $$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$ using epsilon-delta definition.
I know I need to show that for every $\epsilon >0$ there exist a $\delta >0$ such that if $|x|<\delta$, then $| {\frac{x^2-8}{x-8}}-1|<\epsilon$

WHat I did:

$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$, but Im having troubles now deciding what my delta should be.

I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.


问:
what base number system is the hebrew language?
In the hebrew bible, there number system is based off of hebrew letters. There is a single digit used going all the way up to 10, then it uses two digits... Untill it gets to 20, which is a single digit again. The actual hebrew numbers go in sequence of: 1,2,3,4,5,6,7,8,9,10,20,30,40,50,60, 70, 80, 90, 100, 200, 300, 400. What base of numbers would that be+</p>

答:
Hebrew uses a decimal system, with heiratic tokens.

The decimal nature corresponds to that there are symbols for 1, 10, 100, etc.

The heiratic nature corresponds to using separate symbols for 1, 2, 3, 4, 5, ..., 9 in each place.  This was the style of the heiratic system in Egypt, and used in alphabetic systems throughout the ancient world.

The modern form might be seen in a series of stamps for 1c, 2c, ... 9c, then 10c, 20c, 30c, .. 90c, and then £1, £2, ..., £9, ... etc, as formerly used on railway small freight.  One makes up £1.63 with a £1 stamp, a 60c and a 3c stamp.

Of course, one notes that the choice of letter implies a notion of place, even if this is not so presented.  Consider the abacus. This was the mainstay of calculation in ancient times, and ultimately the reference of numbers.

The western abacus consists of columns and rows, where the column value is the product of the row values.  Carry depends on if there were a change of column.

The notation of the chinese and mayan is to attach a column weight and position, like saying 3c 2x 4i, meaning 3 in the hundreds, 2 in the tens, and 4 in the units.

Other notations use different ways of showing this.  The tokens of the romans and kinderd folk, use the same scale we use of coins: each token means 1 or 5 in a specific column.  So 2x is written as XX.

The alphabetic codes like the hebrew, greek and gothic, use a large number of symbols for "n in column C", the order of the alphabet provides a ready-made collection.

```<code>       D    L    V
  M    C    X    I
</code>```

Since the notion of base and position rests in the abacus, and not the notation, it is fair to say that the base is decimal, and the selection from 7i vs 7x or 7c suggests that position was perfectly understood.

The modern notation derives from the greek alphabetic one, with 1x drafted to mean 0i.  This is why, for example, 0 is after 9, rather than before 1.  The use of the symbols for the i column, and a later substition of say 3i 2i as 3x 2i is an unfolding of the results.

Place was understood: it is what makes a base a base.  Position is a function of notation, and does not destroy the place values.

Body:
Hebrew uses a decimal system, with heiratic tokens.

The decimal nature corresponds to that there are symbols for 1, 10, 100, etc.

The heiratic nature corresponds to using separate symbols for 1, 2, 3, 4, 5, ..., 9 in each place.  This was the style of the heiratic system in Egypt, and used in alphabetic systems throughout the ancient world.

The modern form might be seen in a series of stamps for 1c, 2c, ... 9c, then 10c, 20c, 30c, .. 90c, and then £1, £2, ..., £9, ... etc, as formerly used on railway small freight.  One makes up £1.63 with a £1 stamp, a 60c and a 3c stamp.

Of course, one notes that the choice of letter implies a notion of place, even if this is not so presented.  Consider the abacus. This was the mainstay of calculation in ancient times, and ultimately the reference of numbers.

The western abacus consists of columns and rows, where the column value is the product of the row values.  Carry depends on if there were a change of column.

The notation of the chinese and mayan is to attach a column weight and position, like saying 3c 2x 4i, meaning 3 in the hundreds, 2 in the tens, and 4 in the units.

Other notations use different ways of showing this.  The tokens of the romans and kinderd folk, use the same scale we use of coins: each token means 1 or 5 in a specific column.  So 2x is written as XX.

The alphabetic codes like the hebrew, greek and gothic, use a large number of symbols for "n in column C", the order of the alphabet provides a ready-made collection.

       D    L    V
  M    C    X    I

Since the notion of base and position rests in the abacus, and not the notation, it is fair to say that the base is decimal, and the selection from 7i vs 7x or 7c suggests that position was perfectly understood.

The modern notation derives from the greek alphabetic one, with 1x drafted to mean 0i.  This is why, for example, 0 is after 9, rather than before 1.  The use of the symbols for the i column, and a later substition of say 3i 2i as 3x 2i is an unfolding of the results.

Place was understood: it is what makes a base a base.  Position is a function of notation, and does not destroy the place values.


问:
Dominated Convergence Theorem assumptions
I'm trying to find an example to show that it is impossible to weaken the hypothesis of the dominated convergence theorem that $|f_n|\leq g$ for all $n \ge 1$ with g satisfying   $\int_{X} gd\mu \lt \infty$,
$X$
even if we assume that $μ(X) \lt \infty$.

I'm not actually sure what the question is specifically asking me to do...

答:
It's actually possible to weaken the hypothesis of the dominated convergence theorem.

For example, let $X=(0,1]$ and $\mu$ be the Lebesgue measure. Let
$$
f_{n}(x) =
\begin{cases}
[1/n - 1/(n+1)]/n & \text{if $x \in (1/(n+1),1/n]$,}
\\ 0 & \text{otherwise.}
\end{cases}
$$
Then $\int f_{n} = 1/n$ and $\lim_{n} f_{n} = 0$. In this case, there's no dominating function that is integrable, but
$$
\lim_{n} \int f_{n} = \int \lim_{n} f_{n} = 0.
$$

There're various extensions of the theorem, and there's a recent paper that gives a necessary and sufficient condition for the first equality above to be valid. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability, which is also somewhat complicated and in turn weaker than the hypothesis of the dominated convergence theorem.


问:
A counter example which shows that $\Omega(f|_{\Omega}) \neq \Omega(f)$
I'm looking for an example :

A counter example which shows that $\Omega(f|_{\Omega}) \neq \Omega(f)$

$(X,f)$ is a Dynamical System if $f:X \to X$ is a homeomorphism and $X$ is a compact space. $\Omega(f)$ is the set of non wandering points, i.e. all $x$ such that $\forall$ U open containing $x$ and $\forall$ $N>0$ there exists some $n>N$ such that $f^n(U) \cap U \ne \emptyset$.

I think we have to take $f$ on a circle but I don't know how to compute the non wandering set.

答:
[DISCLAIMER]
This answer is wrong, I keep it to show an exemple for a non continuous function. See the second answer below for a good one

Take $X= \ [0;1] \cup \{i ,-i\} \cup \{ i(1-\frac{1}{n}),n \in \mathbb{N}\} \cup \{ -i(1-\frac{1}{n}),n \in \mathbb{N}\}$ with the induce topology of $\mathbb{C}$. This is a compact set made of the segment $[0,1]$ and two sequences of point accumulating at $i$ and $-i$

Now take $f:x \mapsto x^2$ if $x \in ]0;1]$

$f(i)=i$

$f(-i)=-i$

$f(0)=\frac{i}{2}$

$f(i(1-\frac{1}{n}))=i(1-\frac{1}{n+1})$

$f(-i(1-\frac{1}{n}))=-i(1-\frac{1}{n-1})$, if $n \geq 2$

This function is indeed continuous and $\Omega(f)=\{1,0,i,-i\}$ but $\Omega(f_{| \Omega})= \{1,i,-i \}$

You can see The nonwandering set for an example of a counter-example on a connected space but with only a continuous map, not a homeomorphism.

The question is there a counter-example on connected space with an homeo is a good one. I will think about it and keep you informed.

I add an edit to discuss the discret dynamic case as ask in commentary.

In a discret space every singleton in open. So if $x \in \Omega(f)$, taking $U = \{ x \}$ we have that there exist $n$ such that $f^n(x)=x$. $x$ is a periodic point !

Of course periodic point for $f$ are also periodic for $f_{| \Omega}$ and therefor $\Omega(f) = \Omega(f_{| \Omega})$ in the discret case.


问:
Definite integration of a trigonometric function

How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$

my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$

I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.

答: Actually $$\int_0^{\pi/2}\!\dfrac{2a}{a^2 +b^2 cot^2 x}\, dx $$ is also the way. Then $u = \cot x $ and $du = - \csc^2x = -(1+u^2)$ which will lead you to partial fractions.</p>


问:
Functions on quotient topologies
I'm having an issue of (seemingly) conflicting information from textbooks involving mappings between topological spaces. If $ \pi : S \rightarrow S/\sim $ is the projection of a topology S into a quotient over the relation $ \sim $, the topology of $ S $ is transferred to the quotient by requiring that all sets $ V \in S / \sim \, $ are open if $ \pi^{-1} (V) $ are open in $ S $. I'm assuming that this mapping is probably defined such that $ \pi^{-1}(V) := \{ x \in S \, | \, \pi(x) = [x] \; \;  \forall [x] \in V \subset S/\sim \}$.

In Tu's manifolds (page 72), this condition constitutes a topology because:

$$ \pi^{-1} \Big( \bigcup_\alpha V_\alpha \Big) = \bigcup_\alpha \pi^{-1} (V_\alpha) \\
\pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) = \bigcap_\alpha \pi^{-1} (V_\alpha) $$

for an indexed collection of sets $ V_\alpha \in S/\sim $. However, Baker's Introduction to Topology (page 22) states that for a mapping between sets $ f:X \rightarrow Y $ and a collection of indexed sets $ U_\alpha \in X $,

$$ f \Big( \bigcap_\alpha U_\alpha \Big) \subseteq \bigcap_\alpha f(U_\alpha ) $$

and that equality of these two sets is satisfied only when $ f $ is one-to-one. However, when this is applied to the mapping $ \pi^{-1} $ above, it is not only clear that $ \pi^{-1} $ isn't one-to-one, it isn't even well defined as a function, because it's very probable that many elements of $ S $ will map to a single equivalence class in $ S/\sim $, making the inverse multi-valued. Additionally, I can't seem to find any flaw in the following proof that the missing $ \subseteq $ relationship is true all the time:

$$ \text{Let } x \in \bigcap_\alpha \pi^{-1} (V_\alpha) \implies x \in \pi^{-1}(V_\alpha) \quad \forall \alpha \\
\implies \exists [x] \in V_\alpha \ni \pi(x) = [x]  \quad \forall \alpha \implies [x] \in \bigcap_\alpha V_\alpha \\
\implies x \in \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) \implies \bigcap_\alpha \pi^{-1} (V_\alpha) \subseteq \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) $$

So is my proof valid, and Baker is simply wrong? I highly doubt that, it seems I'm missing something critical on why Tu is still correct, and Baker's theorem doesn't apply to this situation? I've seen a similar proof to the above posted here before, I guess my main question is why Baker's one-to-one restriction on the equality isn't killing my proof and as a result Tu's definition.

答:
If we have a function $\phi : A \to B$ between two sets $A,B$, then we can define for each subset $M \subset A$ its image
$$\phi(M) = \{ \phi(a) \mid a \in M \} \subset B$$
and for each subset $N \subset B$ its preimage
$$\phi^{-1}(N) = \{ a \in A \mid \phi(a) \in N \} \subset A.$$
$\phi^{-1}$ is not a function from $B$ to $A$. It is actually a function
$$\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A), $$
where $\mathfrak P (X)$ denotes the powerset of the set $X$. Alternatively we could interpret $\phi^{-1}$ as "multivalued function from $B$ to $A$".  Note that for a bijection $\phi : A \to B$ one usually writes $\phi^{-1} : B \to A$ for the inverse function. Thus we have the same notation for two different things:

$\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A)$ which is defined for all functions $\phi : A \to B$

$\phi^{-1} : B \to A$ which is defined only for bijections $\phi : A \to B$

This may be confusing, but if you are used to it you will not regard it as a problem. The interpretation of $\phi^{-1}$ is normally clear from the context.

Similarly one writes
$$\phi : \mathfrak P (A) \to \mathfrak P (B)$$
for the "image function" determined by $\phi$ which is also a little ambiguous.

Concerning unions the functions $\phi : \mathfrak P (A) \to \mathfrak P (B)$ and $\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A)$ have the same behavior:
$$\phi(\bigcup_\alpha M_\alpha) = \bigcup_\alpha \phi(M_\alpha) \phantom{x}, \phantom{x} \phi^{-1}(\bigcup_\alpha N_\alpha) = \bigcup_\alpha \phi^{-1}(N_\alpha)  .$$
However, concerning intersections their behavior is different:
$$\phi(\bigcap_\alpha M_\alpha) \subset \bigcap_\alpha \phi(M_\alpha) \text{ but in general } \phi(\bigcap_\alpha M_\alpha) \ne \bigcap_\alpha \phi(M_\alpha)  \phantom{x}, \phantom{x} \phi^{-1}(\bigcap_\alpha N_\alpha) = \bigcap_\alpha \phi^{-1}(N_\alpha) .$$
Baker is right,  the injectivity of $\phi : A \to B$ is a necessary and sufficent condition for
$$(*) \phantom x \phi(\bigcap_\alpha M_\alpha) = \bigcap_\alpha \phi(M_\alpha). $$
To see that it is necessary, consider a non-injective function $\phi :A \to B$. There are $a, a' \in A$ such that $a \ne a'$ and $\phi(a) = \phi(a') = b$. Then $\emptyset = \phi(\emptyset) = \phi(\{a\} \cap \{a'\}) \ne \{b\} = \phi(\{a\}) \cap \phi(\{a'\})$. To see that it is sufficient, let $b \in  \bigcap_\alpha \phi(M_\alpha)$, i.e. $b \in \phi(M_\alpha)$ for all $\alpha$. Thus $b = \phi(a_\alpha)$ for suitable $a_\alpha \in M_\alpha$. But since $\phi$ is injective we must have $a_\alpha = a_{\alpha'}$ for all $\alpha,\alpha'$. This means $a_\alpha = a$ for all $\alpha$. Clearly $a \in \bigcap_\alpha M_\alpha$, thus $b \in \phi(\bigcap_\alpha M_\alpha)$.

Your proof for $\phi^{-1}$ is correct and you do not need the injectivity of $\phi : A \to B$. But there is no contradiction: It only means that different functions (like $\phi : \mathfrak P (A) \to \mathfrak P (B)$ and $\phi^{-1} : \mathfrak P (B) \to \mathfrak P (A)$) may behave in a different manner.

Title: Functions on quotient topologies


问:
Show that $2^n-1 \neq k^y$ for odd $y$
For $n\in \mathbb N$, $n>1$ prove that $$2^n-1 \neq k^y$$ for all $k,y \in \mathbb N_{\geq 2}.$

Assuming for contradiction that there exists $(k,y)$ such that $2^n-1 = k^y$, I succeeded to prove that the pair does not exist for an even k, and for an even y.

I need to prove that it does also not exist for an odd y.

I need to use in this proof that

$$\frac{x^{2k+1}+1}{x+1} = x^{2k} -x^{2k-1}+\cdots+1.$$

Thank you!

答:
If $y$ is odd (e.g. $y=2z+1$), then:

$$2^n=k^y+1=(k+1)(k^{2z}-k^{2z-1}+\ldots+ 1)$$

This means that the sum in the second brackets on the right has $2z+1$ terms, all being odd, so the whole sum is odd.

This in turn means that $2^n\mid k+1$ as all occurrences of the prime factor $2$ must be present in the first factor $k+1$.

However, as we also have $k+1\mid 2^n$, this means that $k+1=2^n$, i.e. $k=2^n-1=k^y$. So either $k=1$ and so $2^n=2$, i.e. $n=1$ (contradiction), or $k>1$, which implies $y=1$ (contradiction).

Title: Show that $2^n-1 \neq k^y$ for odd $y$


问:
$Y$ have the pdf $g ( x ) = \frac { S ( x ) } { \mu } \mathbb{1} _ { \{ x > 0 \} }$. Find the expected value of $Y$.

Let $X$ be a positive continuous random variable with mean $\mu$,
variance $\sigma ^ { 2 }$ and survival function $S(x)$ for $x > 0$.

$g ( x ) = \frac { S ( x ) } { \mu } \mathbb{1} _ { \{ x > 0 \} }$

Suppose $Y$ is a continuous random variable having the probability
density function $g$. Find the expected value of $Y$ in terms of $\mu$
and $\sigma ^ { 2 }$.

I'm only able to show $g$ is a valid probability density function. How can I compute $E(Y)$?

答: $$EX^{p}=\int_0^{\infty} pt^{p-1} P(X>t)dt$$ for any $p\geq 1$ and any positive random variable $X$. [This is a standard application of Fubini's Theorem]. In  particular $$EX^{2}=\int_0^{\infty} 2t P(X>t)dt.$$ Now you see that $$EY=\int_0^{\infty} xg(x)dx=\frac {EX^{2}} {2\mu}=\frac {\mu^{2}+\sigma^{2}} {2\mu}.$$


问:
Area of a Region Bounded by a Circle and Vertical Line
What is the area of the right region bounded by $x^2 + y^2 =25$ and $x = -3$?

My attempt solution: I solved first for the area of the left region and subtracted it from the area of the circle, which is

$$ 25 \pi - \int_{-4}^4 -3 + \sqrt {25-y^2} \,dy $$
$$ 78.5 - 11.18 = 67.32 $$
So, the area of the right region is 67.32. But what is the solution wherein I will not use the equation of the area of the circle, and just use integration alone?

答:
"But is there a solution wherein I will not use the equation of the area of the circle, and just use integration alone?"

Yes, there are a number of ways, but they will not be simpler than the method you used.

One way, is to divide the region of the circle to the right of $x=-3$ into two halves, the region above $y=0$ and the region below. By symmetry, they have the same area. You can then use integration to find one of those areas and multiply by 2. If you had a similar problem, but without symmetry, you could simply integrate to find the top part, integrate to find the bottom part, and add the two parts.

A second way, would be to convert your problem to polar coordinates. This would be ugly, because of the line $x=-3$, but would give you the correct answer without subtracting from the area of a circle, and using integration alone.


问:
Curvature of a curve whose unit tangent creates a constant angle with z-axis
I have this question:
Consider a curve $\gamma$(s) and its projection to the plane $\beta$(s), i.e
$$\gamma(t) = \begin{bmatrix} f(s)\\ g(s) \\ h(s) \end{bmatrix},  \beta(t) = \begin{bmatrix} f(s) \\ g(s) \\ 0 \end{bmatrix} $$
Where s is the arc length parameter for $\gamma$.

Is s also the arc length parameter for $\beta$?

Let $T$ be the unit tangent vector of $\gamma$. Prove that the angle between $T(s)$ and the z-axis is at least $\frac{\pi}{4}$ iff $|h'(s)|\leq \frac{1}{\sqrt 2}$.

Assume $|h'(s)|\leq \frac{1}{\sqrt 2}$ holds for all $s$. Assume there is a constant C so that $$\kappa_{\gamma}\leq C$$ is satisfied everywhere. Prove that the curvature $$\kappa_{\beta}\leq 100C.$$

It would be great if someone could help me with this problem because I have been working on it for quite some time and have no idea how to arrive at the conclusion for part 3. Thank you in advance!

答:
An alternative answer: $$k\gamma=\frac{\mid g'h''-h'g''h'f''-f'h''f'g''-g'f''\mid}{((f')^{2}+(g')^{2}+(h')^{2})^{3/2}}$$ greater and equal to C #.

Thanks to Hieu Hoang who wrote this short equation.


问:
Identify $S = 0$ for any set $S$ and form a non-boolean ring out of $\Delta$ on sets?
Let $S$ be a set and define for each $A \subset S$ the negative of $A$ to be $-A = S \setminus A$.  So that $A + S\setminus A = S$.  Thus we identify $0 = S$.  The $+$ operation is still $\Delta$ so if $-A$ were more than $S \setminus A$ it would be less than $S$ when $\Delta$'d with $A$, similarly it can't be smaller than $S \setminus A$.  Therefore for $|S| \geq 2$, $R$ the ring formed is not boolean, as is the usual case for such $\Delta$ (symmetric difference) rings.  Is this then just $\mathcal{P}(S)/(S)$ where $(S)$ is the ideal generated by the whole set $S \in \mathcal{P}(S)$ the power set of $S$?

Therefore, you can quotient a boolean ring and achieve a non-boolean ring?  That doesn't seem possible, you can't have a surjective $h: A \to B$ such that $A$ is boolean while $B$ is non-boolean.

So where did I make an error?

答:
You do not have any freedom to redefine $-A$ while keeping $+$ the same as it was before.  The zero for $+$ has to satisfy $0+0=0$, and $A\Delta A=A$ implies $A=(A\setminus A)\cup (A\setminus A)=\emptyset$.  So the additive identity is necessarily $\emptyset$, and the additive inverses are necessarily $-A=A$.  Given an operation that makes an abelian group, you are already stuck with the inverses it provides you.

You would have to find a new operation for which $-A$ really behaves as the inverse, in order to succeed.  Unfortunately, this is impossible: since the neutral element is always its own inverse, you would need $Z=Z^c$, which is of course impossible if $S$ is nonempty.  Therefore you don't have an abelian group, much less a ring.

It is true that the quotient of a boolean ring is always a boolean ring.  It is as simple as noting that $x^2=x$ implies $(x+I)^2=x^2+I=x+I$ for all $x$ in the ring.


问:
upper bound for $\frac{x+y}{ax+y}$
I need an upper bound for
$$\frac{x+y}{ax+y}$$

I know that $$1\leq a< 2$$
$x\geq 0 $ and $y\geq 0 $ . This upper bound can include just $a$ and constant numbers not $x$ or $y$.

thanks a lot.

答:
Here is how you advance, since $x,y>0$, then

$$ 1\leq a < 2\implies  x\leq ax < 2x \implies x+y\leq ax+y < 2x+y   $$

$$ \implies \frac{1}{2x+y}<\frac{1}{ax+y}\leq \frac{1}{x+y} \implies \frac{x+y}{ax+y}\leq \frac{x+y}{x+y}=1 $$

$$ \implies \frac{x+y}{ax+y}\leq 1. $$

Title: upper bound for $\frac{x+y}{ax+y}$


问:
Find the residue of ζ(2s) and ζ(6s)
I know that the residue of the Riemann zeta function is 1. I'm wondering how one would calculate the residue if you have ζ(2s) and ζ(6s) as the pole has shifted to 1/2 and 1/6 respectively.
Also do the trivial zeros change? So ζ(2s) has trivial zeros at s=-1,-2,-3,...?</p>

答:
You are correct about the placement of the trivial zeros of $\zeta(2s).$ This could be seen, for example, from the functional equation (i.e. reflection formula) of $\zeta(2s)$:

$$\zeta(2s)=\zeta(1-2s)\Gamma(1-2s)\sin(πs)π^{2(s-1)}4^s$$ which follows from the functional equation of $\zeta$ after some algebraic massage. This thing goes to 0 precisely when $s=-n\quad\land\quad 1\le n\in\mathbb{N}.$ So yes, they do in fact change values.

The poles of $\zeta(2s)$ and $\zeta(6s)$ are, in fact, simple, and as you have stated are $\frac{1}{2}$ and $\frac{1}{6}$ respectively, so we have the residue of $f$ at a simple pole $x_0$: $\lim_{z\to x_0}(z-x_0)f(z)$ and as to your question, the residues could be found by:

$\lim_{z\to\frac{1}{2}}(z-\frac{1}{2})\zeta(2z)$

$\lim_{z\to\frac{1}{6}}(z-\frac{1}{6})\zeta(6z)$

respectively.

Title: Find the residue of ζ(2s) and ζ(6s)


答:
Your structure is not a ring, and not even an Abelian group (since you didn't even specify multiplication).

The only element $x$ in $P(S)$ which satisfies $x\,\Delta\,A=A$ for all $A$ is $x=\emptyset$, so this is the additive identity. (Just plug in $A=\emptyset$.)

Therefore $-A$ must be an element which satisfies $(-A)\,\Delta\,A=\emptyset$, and then $-A=A$ follows.

As an Abelian group, or if you wish, as a vector space over $\Bbb F_2$, your quotient construction can make sense: every $A$ is identified with $S\setminus A=S\,\Delta\,A$, and using associativity you can verify that symmetric difference will work fine in the quotient.

We can even identify it with the Abelian group $(P(S\setminus\{a_0\}),\Delta)$ for a fixed element $a_0\in S$, using the representation $A$ or $S\setminus A$ of a set $A\in P(S)$ which doesn't contain $a_0$.

But, the quotient as rings is the trivial ring $\{0\}$, because $S=1$ is the multiplicative identity (multiplication is intersection), and therefore $1=0$ implies $x=0$ for every element $x$.

Title: Identify $S = 0$ for any set $S$ and form a non-boolean ring out of $\Delta$ on sets?

Body:
Let $S$ be a set and define for each $A \subset S$ the negative of $A$ to be $-A = S \setminus A$.  So that $A + S\setminus A = S$.  Thus we identify $0 = S$.  The $+$ operation is still $\Delta$ so if $-A$ were more than $S \setminus A$ it would be less than $S$ when $\Delta$'d with $A$, similarly it can't be smaller than $S \setminus A$.  Therefore for $|S| \geq 2$, $R$ the ring formed is not boolean, as is the usual case for such $\Delta$ (symmetric difference) rings.  Is this then just $\mathcal{P}(S)/(S)$ where $(S)$ is the ideal generated by the whole set $S \in \mathcal{P}(S)$ the power set of $S$?

Therefore, you can quotient a boolean ring and achieve a non-boolean ring?  That doesn't seem possible, you can't have a surjective $h: A \to B$ such that $A$ is boolean while $B$ is non-boolean.

So where did I make an error?


问:
Understanding roots of complicated equations
This is a very broad question apparently, but is there any systematic way to obtain (sort of) sharp estimates for roots of complicated equations? Or maybe bounds for the roots?

For example, consider the equation $$x = A - B\dfrac{e^x}{1+e^x}$$where $A,B$ are large positive integers with $B>A$. It is easy to see that this equation has a unique root, by observing that the function $f(x)=x - A + B\dfrac{e^x}{1+e^x}$ is increasing in $x$. But I want to get some pretty sharp estimates of where the root $x^*$ lies.

I have looked at several numerical examples and it seems that $x^*$ is very close to $\log(A/(B-A))$. But I cannot really prove it.

So I am asking, given a complicated equation, what is a good way to get a good estimate on where the root must lie?

答:
Consider the function $$x-\left(A-\frac{B e^x}{e^x+1}\right)$$
plugging $x=\log \left(\frac{A}{B-A}\right)$ we get
$$\log \left(\frac{A}{B-A}\right)+\frac{A B}{(B-A) \left(\frac{A}{B-A}+1\right)}-A=\log \left(\frac{A}{B-A}\right)$$
thus when $B\sim 2A$  we have $\log \left(\frac{A}{B-A}\right)\sim 0$ and $x^*=\log \left(\frac{A}{B-A}\right)$ is a good approximation of the solution.

If we take $A=1;\;B=10000$ the approximation gives $-9.2$ while the root is $-7.1$

For $A=60000;\;B=100000$ we get respectively $0.405465;\;0.405448$ with an error less than $10^{-4}$.

Hope this helps


问:
Prove that there exists $n \in \Bbb N$ such that $p|(n^n-7^p)$

Fix an arbitrary prime $p \in \Bbb N$. Prove that there exists $n \in \Bbb N$ such that $p|(n^n-7^p)$

Clearly if $p=7$, can choose $n=7$

For $p=2$, any odd number would work for $n$, but for a general prime, how to proceed? I would like to state that I have been working on a problem and this is where I could reduce it.

Thanks in advance for help.

答:
We will try to find $n$ of the form $n=mp+7$ for some $m\in\mathbb N$. Then, $n^n=(mp+7)^n=\sum_{k=0}^n{n\choose k}m^kp^k7^{n-k}$. The last term here is $7^n$, while all the others are divisible by $p$. So:

$$n^n-7^p\equiv 7^n-7^p\pmod{p}$$

Now, $7^n=7^{mp+7}-7^p$. Assuming $p\ne 7$, we have $7^{p-1}\equiv 1\pmod{p}$, so we are looking for $m$ such that $p-1\mid mp+7-p$. This is possible as $p$ is coprime with $p-1$.


问:
Quadratic formula in differential equations

$$(y')^2 + y' =\frac{y}{x} \tag{0}$$

The solution of this differential equation involves using the quadratic formula for a quadratic in terms of $y'$ but I'm a bit bothered that we get a $ \pm$ when we do that:

$$y'  =- \frac12 \pm \sqrt{\frac{4y}{x} +1} \tag{1}$$

And then we could do $y=xt$ and solve but how exactly do we understand the plus or minus quantity which we get in step-1? It seems that the procedure of completing the quadratic formula generates two differential equation which solves the one in (0). So, should I solve both ones and the actual solution for (0) is a linear combination of both?

答:
Incomplete solution:

Let $y'=p$, then the ODE is
$$p^2+p=\frac{y}{x} \implies x=\frac{y}{p^2+p}~~~~~(1)$$
D. (1) w.r.t. $y$ and write $\frac{dx}{dy}=\frac{1}{p}$ to get
$$p^3+p^2=-y(2p+1)\frac{dp}{dy} \implies \int \frac{dy}{y}=-\int\frac{2p+1}{p^3+p^2}dp$$
$$\ln Cy=\frac{1}{p}-\ln p+\ln(1+p) \implies y=\frac{(1+p)e^{1/p}}{Cp}~~~~(2)$$
Putting this in (1), we get $$x=\frac{ e^{1/p}}{Cp^2}~~~~(3)$$
(2) and (3) give the complete solution of the first order ODE (1) in terms of one undetermined constant $C$, here $p$ acts as a real parameter. One may eliminate $p$ between (2) and (3) to get the cartesian solution with one constant $C$. Se the solution $y(x)$ for $C=1,2,3$


问:
Visual understanding of $\frac{\mathbb R} {\mathbb Q} $
Consider the quotient set $\frac{\mathbb R} {\mathbb Q} $ obtained by the equivalence relation $$ x \equiv y \mod \mathbb Q \quad \text{iff} \quad x-y \in \mathbb Q  $$
I was wondering if there exists a way to get a visual intuition of it.
I mean, $\frac{\mathbb R} {\mathbb Z} $ can be thought of as the interval $[0,1) $.
To this extent, what does $\frac{\mathbb R} {\mathbb Q} $ look like?
And $\frac{\mathbb [0,1] } {\mathbb Q} $ (if defined in the same way)?

I know the example I provided is a bit different from what I asked, and there may not be an "obvious" or simple answer. I don't have any problem with the "abstract" definition, I was looking for a more "tangible" way to understand it.
Thank you for your help, sorry if my answer seems useless or silly.

答:
A way to represent $\mathbb R / \mathbb Q$ would be using the axiom of choice to pick up one element per class and to obtain a subset $A$ of the reals.

This is possible. Unfortunately, $A$ has a « bad behavior »! In particular $A \cap [0,1]$ is not Lebesgue measurable. See here for a proof.

So, representing $A$, for example graphically on the real line is not a simple mission...


问:
Finding solutions of a system of linear equations
I was asked the following:

$$\begin{align}2x_1 - 4x_2 + 8x_4 &= 10\\
3x_1 + 2x_2 &= 3\\
3x_1 + 2x_3 + 6x_4 &= 11\\
8x_1 + 6x_2 + 2x_3 + 14x_4 &= 24\end{align}$$
  Find the solution set $W$. Is $W$ a vector space?

I know I could express one variable from one of the equations and substitute into the remaining ones. For example, from the second equation I get
$x_1=1-\frac23x_2$.

After substituting this into the remaining equations I have
$$
\begin{align*}
-\frac{16}3x_2+8x_4&=8\\
-2x_2+2x_3+6x_4&=7\\
\frac23x_2+2x_3+14x_4&=16\\
\end{align*}
$$
I could continue by expressing another variable and plug into other equations. But this approach seems to be rather cumbersome and if I have more than three equations I am prone to make mistakes. Is there a more elegant way to do this?

I also do not know what to say about the part of the assignment asking about vector space.

Body:
I am posting my solution - mainly for the purpose of showing you how can you write down the solution; in case you have a mistake in it and want to ask others whether your solution is correct.

$\begin{pmatrix}
 2 & 4 & 0 & 8 &|&10\\
 3 & 2 & 0 & 0 &|& 3\\
 3 & 0 & 2 & 6 &|&11\\
 8 & 6 & 2 &14 &|& 24
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-2 & 2 & 6 &|& 8\\
 4 & 3 & 1 & 7 &|&12
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 4 & 3 & 1 & 7 &|&12
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 1 & 1 & 1 & 7 &|& 9
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 0 &-1 & 1 & 3 &|& 4
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 2 & 0 & 0 &-4 &|&-2\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 1 & 0 & 0 &-2 &|&-1\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 2 & 0 & 6 &|& 6\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 1 & 0 & 3 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 1 & 0 & 3 &|& 3\\
 0 & 0 & 1 & 6 &|& 7\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}$

If I did not made a mistake, the rows of the first matrix should be linear combinations of the rows of the result. I can use this to make a sanity check. When I look on the first three columns (because the last matrix has pivots in the first three columns) I get the coefficients 2,4,0. So I can check whether I get $(2,4,0,8,10)=2(1,0,0,-2,-1)+4(0,1,0,3,3)+0(0,0,1,6,7)$; i.e., whether the first row is linear combination with this coefficients. (If not, I must have made mistake somewhere.) I can do the same thing with other rows.

Perhaps it is worth mentioning that even if this verification works, I cannot be entirely sure that everything is correct (that I have all solutions of the system).

This can be used to find where is a mistake, as illustrated below.

I see that you asked repeatedly in comments how to express $W$.

Notice that the last matrix corresponds to the system
$$
\begin{align}
x_1-2x_4&=-1\\
x_2+3x_4&=3\\
x_3+6x_4&=7
\end{align}
$$
Clearly, for any choice of $x_4$ I can find $x_1$, $x_2$, $x_3$ which fulfill those equations. Namely $x_1=-1+2x_4$, $x_2=3-3x_4$, $x_3=7-6x_4$.
So the set of all solutions is
$$W=\{(-1+2x_4, 3-3x_4, 7-6x_4, x_4); x_4\in F\}$$
where $F$ is the field you are working with. (For example, $F=\mathbb R$ if you are solving this system over real numbers.)

The same method works in general. You can choose the variables corresponding to columns without pivot and you can calculate the remaining values.

Now you can also see that $W$ is not a subspace of $F^4$ since $(0,0,0,0) \notin W$ and every subspace contains the zero vector.

This is an incorrect solution, I've added it here solely for the purpose of illustrating how to find the place where I made a mistake

Suppose I made computations like this:
$\begin{pmatrix}
 2 & 4 & 0 & 8 &|&10\\
 3 & 2 & 0 & 0 &|& 3\\
 3 & 0 & 2 & 6 &|&11\\
 8 & 6 & 2 &14 &|& 24
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-2 & 2 & 6 &|& 8\\
 4 & 3 & 1 & 7 &|&12
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 4 & 3 & 1 & 7 &|&12
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 1 & 1 & 1 & 7 &|& 9
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 0 &-1 & 1 & 3 &|& 4
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 3 & 2 & 0 & 0 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 2 & 0 & 0 &-4 &|&-2\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 1 & 0 & 0 &-2 &|&-1\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 2 & 0 & 2 &|& 6\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 1 & 0 & 1 &|& 3\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 1 & 0 & 1 &|& 3\\
 0 & 0 & 1 & 4 &|& 7\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}$

When I try to check my solution, I see that
$2(1,0,0,-2,-1)+4(0,1,0,1,3)=(2,4,0,0,10)\ne (2,4,0,8,10)$. So I know that there is a mistake somewhere.

I can do the same verification for each matrix I obtained in the process.

Now when I look at these two matrices

$\begin{pmatrix}
 1 & 2 & 0 & 4 &|& 5\\
 1 & 0 & 0 &-2 &|&-1\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}\sim
\begin{pmatrix}
 1 & 0 & 0 &-2 &|&-1\\
 0 & 2 & 0 & 2 &|& 6\\
 0 &-1 & 1 & 3 &|& 4\\
 0 & 0 & 0 & 0 &|& 0
\end{pmatrix}$

I found out that the verification does not work for the first one, but it works for the second. This means that in this step there must be a mistake. And the second row in the later matrix is indeed incorrect.

In general, if I want to find where I made a mistake, I can look for two consecutive matrices such that for one of them the verification works and for the other it does not. (This works whenever I am trying to find a matrix in reduced row echelon form, it is not specific to solving linear systems.)

I should probably add the caveat that in this way we find the mistake which is closest to the end of the computation. So if I made several mistakes in my computations, this will be rather cumbersome. (I would have to do the same thing several times.)


答:
Let $g$ be a primitive root $\pmod p$, with $g^i\equiv 7\pmod p$.

It suffices to solve $$n\equiv g\pmod p \quad \&\quad n\equiv i\pmod {p-1}$$

As $\gcd(p,p-1)=1$ there is always a solution.

Note:  if you prefer, you can skip the primitive root part and simply solve
$$n\equiv 7\pmod p \quad \&\quad n\equiv p\equiv 1\pmod {p-1}$$

Body:
Fix an arbitrary prime $p \in \Bbb N$. Prove that there exists $n \in \Bbb N$ such that $p|(n^n-7^p)$

Clearly if $p=7$, can choose $n=7$

For $p=2$, any odd number would work for $n$, but for a general prime, how to proceed? I would like to state that I have been working on a problem and this is where I could reduce it.

Thanks in advance for help.


问:
Galton Watson and generating function
In the treatment of the Galton-Watson process one usually introduces the "generating function" of the random variable $Z_n$ as $\psi_{Z_n}(s)=E[s^{Z_n}]$. Why would $\psi_{Z_n}(0)=P(Z_n=0)$? It seems to me that this identity being true would hinge on $0^0=1$ and $0^k=0$ for $k\geq 1$. Is it usually assumed that $0^0=1$ in this context? Is there another way to see this?

答: $Es^{X}$ is always interpreted as $P(X=0)+sP(X=1)+s^{2}P(X=2)+\cdots$. This is the correct definition of generating function and it avoids the indeterminate form $0^{0}$. When $s=0$ the sum becomes $P(X=0)$.

Title: Galton Watson and generating function


答:
One of the ways would be to change the order of your integral and rewrite as -

$\displaystyle \int_{-3}^5 \int_{-{\sqrt {25-x^2}}}^{\sqrt {25-x^2}} dy \, dx$

Title: Area of a Region Bounded by a Circle and Vertical Line

Body:
What is the area of the right region bounded by $x^2 + y^2 =25$ and $x = -3$?

My attempt solution: I solved first for the area of the left region and subtracted it from the area of the circle, which is

$$ 25 \pi - \int_{-4}^4 -3 + \sqrt {25-y^2} \,dy $$
$$ 78.5 - 11.18 = 67.32 $$
So, the area of the right region is 67.32. But what is the solution wherein I will not use the equation of the area of the circle, and just use integration alone?


答: In this context, $s^0$ uses $0$ as a monomial index and hence a non-negative integer, and the $s=0$ case is just the empty product $1$. (You can think of this as the hypervolume of a $0$-dimensional cube of side length $s$; in $0$ dimensions, hypervolume just counts the single point present.) The "$0^0$ is an indeterminate form" concern is a separate issue that results when the exponent can vary continuously, e.g. $\lim_{x\to0^+}x^x\ne\lim_{x\to0^+}(e^{-1/x^2})^{x^2}$.

Title: Galton Watson and generating function

Body: In the treatment of the Galton-Watson process one usually introduces the "generating function" of the random variable $Z_n$ as $\psi_{Z_n}(s)=E[s^{Z_n}]$. Why would $\psi_{Z_n}(0)=P(Z_n=0)$? It seems to me that this identity being true would hinge on $0^0=1$ and $0^k=0$ for $k\geq 1$. Is it usually assumed that $0^0=1$ in this context? Is there another way to see this?


答:
Note that by Fermat's little theorem, $7^p \equiv 7 \mod p$.
Assume $p \neq 7,2$ Therefore we want to prove that there exists a solution to $n^n \equiv 7 \mod p$.
 Take $n = (2p-6)(p-1) + 1 \rightarrow n \equiv 7 \mod p$ so $n$ is not divisible by $p$.
 By Fermat's little theorem $n^{p-1} \equiv 1$ (mod $p$) So we get:
$$n^n \equiv 7^n \equiv 7 \cdot (7^{(p-1)})^{2p-6} \equiv7 \mod p $$
As desired.

Title: Prove that there exists $n \in \Bbb N$ such that $p|(n^n-7^p)$


问:
Cannot solve quadratic equation through product sum product when the the first product is fraction
I have the folowing equation

$$\frac{16}{3}x^2 - 2x -45 = 0$$
I am trying to solve it for about 40 minutes. I am starting to think I did something wrong. Maybe the equation insn't right, I derived it.

I got as far as this

```<code>(16/3)x2 - 2x -45 = 0  (-45 the signs will be different)
               1 . 45
16/3  1        3 . 15
4/3   4        9 . 5
2/3   8

</code>```

The pairs below the equation are the products. I cannot find the pairs.

答:
$$\frac{16}3x^2 - 2x -45 = 0\\
16x^2 - 6x -135 = 0\\
\left(4x-\frac34\right)^2 -\frac{3^2}{4^2} -135 = 0\\
\left(4x-\frac34\right)^2 = 135 + \frac{9}{16} = \frac{2169}{16}$$
You can finish the rest yourself. The right hand side is not a nice square, so the solutions are not rational numbers, and it is therefore not surprising that they cannot be found by trying to factorise the original quadratic. Directly applying the quadratic formula is easier than completing the square in a case like this.


问:
I can't understand Intermediate Value Theorem Proof
Intermediate Value Theorem. If ƒ is a continuous function on a closed interval [a, b] and if u is any value between ƒ(a) and ƒ(b), then u = ƒ(c) for some c in [a, b].

I'm trying to understand this theorem from Wikipedia(https://en.wikipedia.org/wiki/Intermediate_value_theorem#Proof), it says that if $A:= \{x \in [a,b]:f(x)≤u\}$ and $s:=supA$ then $f(s)=u$.

I can't understand it. Please help me to prove this theorem.

Edited:I don't understand this part on Wikipedia:

答:
About the Wikipedia proof quoted in revision 3:

By the $(\varepsilon, \delta)$-definition of limit and continuity, the proof one line above writes

$$f(x) - \varepsilon < f(c) < f(x)+\varepsilon \tag 1$$

for all $x \in (c-\delta, c+\delta)$.

$c$ is chosen to be the smallest number that is greater than or equal to every member of $S$ ($c=\sup S$).

So there are some $a^* \in (c-\delta, c]$ that's "just smaller" than $c$ and also $a^*\in S$. So $f(a^*) \le u$.

By the right hand side of inequality $(1)$, $f(c) < f(a^*) + \epsilon$, and so

$$f(c) < u + \varepsilon$$

Similarly, since $c = \sup S$, for $a^{**}\in (c, c+\delta)$ that's "just larger" than $c$, $a^{**}\not\in S$ and so $f(a^{**}) > u$.

By the left hand side of inequality $(1)$, $f(a^{**}) - \varepsilon < f(c)$, and so

$$u - \varepsilon < f(c)$$