问:
$\sup_{\Gamma}\sum |\phi(x_i)-\phi(x_{i-1})|=\infty$ but $\lim_{|\Gamma|\to 0}\sum (\phi(x_i)-\phi(x_{i-1}))$ exists
Let $\Gamma=\{a=x_0,\cdots,b=x_m\}$ a partition of $[a,b]$, is it possible to find a function $\phi$ (any) such that $$\sup_{\Gamma}=\sum_{i=1}^m |\phi(x_i)-\phi(x_{i-1})|=\infty$$ but such that  the following limit exists $$\lim_{|\Gamma|\to 0}\sum_{i=1}^m  (\phi(x_i)-\phi(x_{i-1}))\ \ \ \ ?$$
I think the answer is yes, I just could not think of the $\phi$ that works.

答:
Note that $\phi(x) = \begin{cases} x \sin \frac{1}{x}, & 0 < x \leqslant 1 \\0, &x = 0\end{cases}$ has unbounded variation but is continuous, and for any partition of $[0,1]$,

$$\sum_{i=1}^m  (\phi(x_i)-\phi(x_{i-1})) = \phi(1) - \phi(0) = \sin 1$$


问:
Decreasing functions $f:\mathbb{N} \rightarrow \mathbb{N}$
Let $P=\{f:\mathbb{N} \rightarrow \mathbb{N}; f(i+1)\leq f(i) \ \forall \ i \in \mathbb{N}\}.$

I need to show 2 elements of $P$ and prove that $P$ is countable (or not).

I think that $P$ only have constant functions, and if this is right, i can prove that $P$ is countable. There is some counter-example for this?

答:
This answer builds on a nice principle outlined in a post by @Riemann'sPointyNose that has been deleted because of an argumentation gap, which I believe is closed in the following.

The idea is to assign to any function $f: \mathbb{N}\to\mathbb{N}$ a real number in such a way that every $f\in P$ is mapped to a rational number.

Let for example $f(1) = 245$, $f(2) = 123$, $f(3) = 10$, $f(4) = 4$, $f(5) = 4$, ...

Then concatenate the decimal representations to get an infinite string
$$\color{red}{2451231044\ldots}$$
However we cannot uniquely reconstruct the function $f$ from this string because we don't know where the boundaries of the numbers are.
To fix this, we interleave it with another sequence consisting of 0's and 1's, where we put a 1 whenever a "new number begins". That is:
$$\color{red}{2}\color{blue}{0}\color{red}{4}\color{blue}{0}\color{red}{5}\color{blue}{1}\color{red}{1}\color{blue}{0}\color{red}{2}\color{blue}{0}\color{red}{3}\color{blue}{1}\color{red}{1}\color{blue}{0}\color{red}{0}\color{blue}{1}\color{red}{4}\color{blue}{1}\color{red}{4}\color{blue}{1}\ldots$$
This enables us to uniquely reconstruct the function $f$.

Now note that if $f$ becomes constant eventually, the resulting string will be periodic eventually.
Hence, if we regard it as a decimal expansion of a number in $[0,1]$, the number will be rational.

Since every function in $P$ becomes constant eventually, this gives us an injective mapping
$$P \to \mathbb{Q}$$
and $P$ must therefore be countable.

ContentLicense: CC BY-SA 4.0


问:
Existence of $C$ such that $p(x) \leq p(y) + C$ for all $x \leq y$
Suppose I have a polynomial $p$ of order $n$, which we can assume is positive, i.e. if $p(x) = \sum_{i=0}^n a_ix^i$ then $a_n > 0$. I am interested in the existence or non-existence of a constant $C \geq 0$ such that $p(x) \leq p(y) + C$ for all $0 \leq x^n \leq y^n + K$ and with $x, y \geq 0$.

Here are some thoughts when $K = 0$ and hence $0 \leq x \leq y$. We let $A = \{0 \leq x \leq y\} \subset \mathbb R^2$, and I think I can reformulate by question as whether or not there exists a $C$ such that $f(x, y) = p(y) - p(x) + C \geq 0$ for all $(x, y) \in A$.

I can find a $C$ if the region is compact by continuity, say if $y \leq L$ for any $L$, and for $x \geq L$ where $L$ is sufficiently large, eventually $p(x)$ is increasing so $p(y) - p(x)$ for $y \geq x$. On the region $x \leq L$ and $y \geq L$, and for $L$ sufficiently large again, we have that $p(x)$ is bounded and $p(y)$ is positive so a $C$ exists here as well.

What if $K \neq 0$? In this case I can see the analogous argument with $0 \leq x \leq y + K'$ is not strong enough. (This we can obtain from the fact if $x^n \leq y^n + K$ then $x \leq (y^n + K)^{1/n} \leq y + K^{1/n}$.) For instance if $p(x) = x^2$ and $y = x - K'$ then $y^2 - x^2 = -2xK' + (K')^2$ which can be arbitrarily large and negative by taking $x \rightarrow \infty$. So it seems to me that control over what happens at top order is necessary (thus we need $x^n \leq y^n + K$ and not just $x \leq y + K'$), but I don't know how to put this argument correctly/nicely.

EDIT: Thanks to a comment below, an extra condition $x, y \geq 0$ has been imposed.

答:
One can find an upper bound for $p(x)-p(y)$ as follows:

First consider a decomposition of the polynomial into it's negative and positive coefficients

$$p(x)=p_+(x)+p_-(x)=\sum_{i=1}^{A}P_{i}x^{p_i}+\sum_{i=1}^{n-A}Q_{i}x^{q_i}$$

where $P_i>0, Q_i<0$. Next, one can show that on each curve $x^n-y^n=K$, the function $p(x)-p(y)=p((y^n+K)^{1/n})-p(y)$ is bounded. Consider the function

$$f(y)=(y^n+K)^{m/n}-y^m$$

This function is monotonically decreasing for $y\geq 0$ and also $\lim_{y\to\infty} f(y)=0$ and it is bounded below and above by

$$0<f(y)\leq K^{m/n}$$

We can now write

$$p(x=(y^n+K)^{1/n})-p(y)=\sum_{i=1}^{A}P_{i}(x^{p_i}-y^{p_i})+\sum_{i=1}^{n-A}Q_{i}(x^{q_i}-y^{q_i})<\sum_{i=1}^AP_iK^{p_i/n}=p_+(K^{1/n})$$
and we conclude that in the interval $K\in(0,K_{\max})$, which spans the region $0\leq x^n-y^n\leq K_{max}~,~ x\geq0, y\geq 0$, the quantity of interest is bounded, and since $p_+$ is an increasing function for $K\geq 0$ the universal bound is

$$p(x)-p(y)<C:=p_+(K_{\max}^{1/n})$$


问:
Categorical products are associative
Let $\mathbf{C}$ be a category with products, and let $A, B, C \in \mathbf{C}$. I wish to show that there exists a morphism $h: (A \times B) \times C \to A \times (B \times C)$ which is an isomorphism.

I believe I can get that we have morphisms $g_1:(A \times B) \times C \to A \times (B \times C)$ and $g_2:A \times (B \times C) \to (A \times B) \times C$ and that they are the unique morphisms satisfying the universal property for products.

However, I don't see how to show that one/either of them is an isomorphism. We need to prove that their compositions are identity morphisms on those product objects. This amounts to showing that for any objects $U, V, W, X$ and arrows
$$
\begin{aligned}
f_U&: U \to (A \times B) \times C \\
f_V&: (A \times B) \times C \to V \\
f_W&: W \to A \times (B \times C) \\
f_X&: A \times (B \times C) \to X \\
\end{aligned}
$$
we have
$$
\begin{aligned}
(g_2 \circ g_1) \circ f_U &= f_U \\
f_V \circ (g_2 \circ g_1) &= f_V \\
(g_1 \circ g_2) \circ f_W &= f_W \\
f_X \circ (g_1 \circ g_2) &= f_X.
\end{aligned}
$$
The first two equalities above assert that $g_2 \circ g_1 = \mathrm{id}_{(A \times B) \times C}$, and the second pair asserts that $g_1 \circ g_2 = \mathrm{id}_{A \times (B \times C)}$. From here, though, I am stuck.

答: Hint: As in the comments, prove that both $(A\times B)\times C$ and $A\times (B\times C)$ are limits for the discrete 3-point diagram $A,B,C$ (ternary product).

Title: Categorical products are associative


问:
I need help with a stupid webassign problem. It's about partial fraction
The question is following:

Write out the form of the partial fraction decomposition of the function (as in this example). Do not determine the numerical values of the coefficients.
$$\frac{1}{x^2+x^4}$$

I know how to solve this. but the problem is I have to do it like the example and I can't figure out what the "proper way" the Webassign is looking for.

$$\frac{x^3+x^2+1}{x(x-1)(x^2+x+1)(x^2+1)^3}=\frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+x+1}+\frac{Ex+F}{x^2+1}+\frac{Gx+H}{(x^2+1)^2}+\frac{Ix+J}{(x^2+1)^3}$$

Body:
Is it $$\frac A {x^2}+\frac B{x^2+1}$$?

No, it's

$$\frac A{x}+\frac B {x^2}+\frac C{x^2+1}$$


问:
Conditional Probability Multivariate Question
Two life insurance policies, each with a death benefit of 10,000 and a one-time premium of 500, are sold to a couple, one for each person. The policies will expire at the end of the 10th year. The probability that only the wife will survive at least 10 years is 0.025, the probability that only the husband will survive at least 10 years is 0.01, and the probability that both of them will survive at least 10 years is 0.96. what is the expected excess of premiums over claims, given that the husband survives at least 10 years?

I understood that it was a conditional probability problem but I tried to solve it a different way and I'm not sure where my thinking is wrong. Here's what I tried:

X: wife survives 10 years, Y: husband survives 10 years

Since we are assuming the husband survives the 10 years, I disregarded the probability that only the wife survives the 10 years.

-(Premium for both policies) + (Death Benefit For One) * (Probability Only Husband Survives) + (No Death Benefit) * (Probability Husband and Wife Both Survive)
$$-1000 + 10000\times P(X' \cap Y) +0\times P(X \cap Y)$$

I have the solution so I know how to "correctly" solve it, I'm just confused about why this method I tried won't work.
Thanks in advance!

答:
You want the conditional expectation: $-1000+10000\,\mathsf P(X'\mid Y)+0\,\mathsf P(X\mid Y)$.

You are given $\mathsf P(X'\cap Y)=0.01$ and $\mathsf P(X\cap Y)=0.96$.

Recall the definition of conditional probability: $$\begin{align}\mathsf P(X\mid Y)&=\dfrac{\mathsf P(X\cap Y)}{\mathsf P(Y)}\\[2ex]\mathsf P(X'\mid Y)&=\dfrac{\mathsf P(X'\cap Y)}{\mathsf P(Y)}\end{align}$$

Also recall the law of total probability : $$\mathsf P(Y)=\mathsf P(X\cap Y)+\mathsf P(X'\cap Y)$$


问:
homogeneous ideal in $k[x_0,...,x_n]$
Yesterday I did some self study about projective varieties in algebraic geometry. I found homogeneous ideal is a concept that is rather hard to understand. I googled it but most of the material online introduce concept via graded ring but the lecture notes I read introduced this from scratch.

Definition An ideal $I\subset k[x_0,...,x_n]$ is homogeneous if for every $f\in I$, all the homogeneous components $f^{(i)}$ are in $I$

Proposition An ideal $I\subset k[x_0,...,x_n]$ is homogeneous if and only if it is generated by homogeneous elements

My question is how to prove the only if part(Well, the writer claims it is trivial). By Hilbert's Basis theorem, all the ideals in $k[x_0,...,x_n]$ can be finitely generated since any field is a Noetherian ring(the only ideals are 0 and k so the chain will become stationary). So let ${f_1,...,f_r}$ be a set of generators, then by definition $f_j = \sum_{i=0}^{a_j}f_j^{(i)}$, where all $f_j^{(i)}$ are in $I$. But how could I select a set of generators so they are homogeneous i.e they have the same $(i)$?

Or maybe even more awarkward, my english is bad and did not understand this concept correctly.

答: In what you've written, $\{f_j^{(i)}\}$ is the homogeneous set of generators of $I$: each element is homogeneous, contained in $I$, and the ideal generated by them contains the $f_j$, which generate $I$. There's no requirement that all of the $(i)$ need to be the same, and in fact you cannot do this for all homogeneous ideals: for instance, $(x,y^2)\subset k[x,y]$ cannot be generated by any set of elements all of the same degree.

Title: homogeneous ideal in $k[x_0,...,x_n]$


问:
Why differentiate between continuous on $[a,b]$ and differentiable on $(a,b)$?
Before starting I want to say that my main language is not english so you may find some grammar mistakes. I will try to avoid making them as much as I can, so please forgive me.

A few months ago I decided that I wanted to learn Calculus so I got some books and started to study, during that time I faced a statement that is pretty common among math definitions related to it, and that statement is:

$f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$

What I was thinking was that a limit needs a closed interval to compute its value, but after making some research I found that the function $f(x) = \sqrt{x}$ is continuous at $0$, so I started to make myself this question.

If we consider, for example $f(x) = x^2$. and the interval $[1,3]$. We know that the function is differentiable and continuous at its whole domain but we do we say that it has to be continuous at $[1,3]$ and differentiable at $(1,3)$, this of course, considering the function over the interval mentioned before, when in fact it is differentiable at $1$ and $3$.

Also the function $\sqrt{x}$ is considered continuous at its whole domain, even also at $x = 0$ but it is not differentiable at 0. Why?

Why do we do this difference between $[a,b]$ and $(a,b)?$

Sorry if I did not express myself the best way possible, again, English is not my main language.

答:
Consider the following function. It's continuous on $[0,1]$, differentiable on $(0,1)$ but neither at $0$ nor $1$. Yet Rolle's theorem still applies and guarantees there is an $x$ in $(0,1)$ such that $f'(x)=0$. It's for this kind of situation that it's fundamental to distinguish between continuity on $[a,b]$ and differentiability on $(a,b)$. Had Roll's theorem been proved for $f$ differentiable on $[a,b]$, we could not draw any conclusion for this function.

A bit of nostalgia: in the equivalent of first year undergraduate studies, in 1998, our teacher asked the class to try and prove Rolle's theorem. He inadvertently assumed differentiability on $[a,b]$, and I inadvertently drew the picture above to find inspiration. Until he noticed my mistake, which led him to notice his - that differentiability at the bounds is unnecessary. It's during those years I learned the importance of the slightest details in assumptions. I hope you will see too how important this is.


问:
For $W=\cup_{U\in\mathcal U} U$ show that there exists $U_1,\dots,U_n: \ \sum_{i=1}^n \lambda(U_i) > \frac{1 - \epsilon}{3^d}\lambda(W)$
I'm trying to solve the following, which is Exercise 13.15 in the book Probability Theory by A. Klenke.

Let $C \subset \mathbb R^d $ be an open, bounded and convex set and assume that
$$
\mathcal{U} \subset \left \{ x+rC: x \in \mathbb R^d, r>0 \right \}
$$
is such that
$$
W:= \bigcup_{U \in \mathcal{U}}U
$$
has finite Lebesgue measure $\lambda(W)$. Show that for any $\epsilon > 0$ there exists finitely many pairwise disjoint sets $U_1, \dots, U_n$ such that
$$
\sum_{i=1}^n \lambda(U_i) > \frac{1 - \epsilon}{3^d}\lambda(W). \tag 1
$$
Show by a counterexample that the condition of similarity of the open sets in $\mathcal{U} $ is essential.

This is my approach: Given that the Lebesgue measure $\lambda$ is inner regular, pick a compact set $K \subset W$ such that
$$
\lambda(W) - \epsilon < \lambda(K).
$$
Since the open sets\footnote{We take it as given that $C$ is open implies that $x + rC$ is open.} in $\mathcal{U}$ covers $K$ and $K$ is compact there exists a finite number $m$ of them such that $U_i, \dots, U_m$ covers $K$. Order the $U_i$:s such that, if
$$
U_i = x_i + r_iC, \qquad i=1, \dots m
$$
then $r_1 \ge r_1 \ge \dots \ge r_m$.

Now, I have seen a similar Lemma in Rudin's Real and complex analysis (Lemma 7.3) where the sets $U_i$ are open balls $U_i = B(x_i,r_i)$. In that case one may do as follows: to get a disjoint collection of sets we let $i_1=1$, and then we discard every $U_j$ that intersects $U_{i_1}$. Let $U_{i_2}$ be the first remaining $U_j$ (if any exists) and discard the remaining $U_j$ that intersects $U_{i_2}$. Continuing this process gives a a collection of $n$ disjoint sets. Then one may claim that,
$$
\bigcup_{i=1}^m x_i+ r_iC \subset \bigcup_{k=1}^n x_{i_k} + 3r_{i_k}C.
$$
The conclusion then follows (for that Lemma) from
$$
\lambda \left(B(x, 3r)\right) =3^d \lambda \left(B(x, r)\right)
$$
and subadditivity.

I don't think the exact same argument works here but is there a similar approach to get at least pairwise disjoint sets in the case of this exercise? And further such that (1) holds?

Secondly, what is the meaning of the last sentence in the exercise, "Show by a counterexample that the condition of similarity of the open sets in $\mathcal{U} $ is essential."?

Much grateful for any help provided!

答:
For your last question regarding the statement:

Show by a counterexample that the condition of similarity of the open sets in $\mathcal U$ is essential.

This means (as I understand it) that if $\mathcal U$ simply is a collection of convex open sets, and not necessarily sets of the form $x + rC$ for a fixed convex open set $C$, then this result fails.

Here is one counterexample in $\mathbb R^2$. Let $q \in (0,1)$, and let $\tilde C_0$ be the sector given in polar coordinates by $\{(r,\theta) : 0 < r < 1, \, 0 < \theta < 2\pi q\}$. Let $A_\theta : \mathbb R^2 \to \mathbb R^2$ denote a rotation by $\theta$, let $\alpha > 0$, let $a = -\alpha(\cos\pi q, \sin\pi q) \in \mathbb R^2$, and let $C_\theta = A_\theta\left(a+\tilde C_0\right)$. Then each $C_\theta$ is an open sector with radius $1$ and angle $2\pi q$ containing the origin. If $\mathcal U = \left\{ C_\theta : 0 \leq \theta <2\pi\right\}$, $W = \bigcup_{\theta \in [0,2\pi)} C_\theta$, then we find that $W$ is the open disc of radius $1-\alpha$ centered at the origin. So $\lambda^2(W) = \pi(1-\alpha)^2$, and $\lambda^2(C_\theta) = \pi q$ for every $\theta$. Since no two sets in $\mathcal U$ are disjoint, the statement of the result implies:
$$
\forall \epsilon > 0 : \quad \pi q = \lambda^2(C_\theta) > \frac{1-\epsilon}{9}\lambda^2(W) = \frac{1-\epsilon}9 \pi(1-\alpha)^2.
$$
Since this must hold for every $\epsilon > 0$, this means in particular that $q \geq \frac 1 9 (1-\alpha)^2$. This is false given appropriate choices of $q$ and $\alpha$, e.g. if $q \in (0,\frac 1 9)$ and if $\alpha < 1-3\sqrt q$.


问:
Confusion on Spivak's proof chapter 7-12
So I've been trying to work through Spivak's Calculus and I keep getting stuck.
I am terribly confused on Spivak's proof on  chapter 8 theorem 7-2 that a continuous function $f$ on an interval $[a,b]$ is bounded above the proof goes as follows:

They are quite a few things that confuse me in this proof for example why can we say that the set $A$ is not empty because it is bounded above by b i know that if a  function is continuous at a point $a$ then there is a number  $\delta > 0$ such that $f$ is bounded on the interval $(a-\delta,a+\delta)$ however how can we know that the delta interval is contained in the interval $[a,b]$ and say that then b must be an upper bound?.

My 2nd confusion has to do with why  he assumes the supremum $\alpha=b$ this really confuses me like what reason is there for this to be true and in his proof by contradiction he assumes $\alpha < b$ and then states that it must be the case that $\alpha$=$b$ however why is it not true that $\alpha \leq b$?.

Thanks in advance.

答:
They are quite a few things that confuse me in this proof for example why can we say that the set A is not empty because it is bounded above by b

No, that's the justification that $A$ is bounded. The justification that $A$ is non-empty is written immediately after that is claimed: $a \in A$. This isn't hard to see: $f$ is clearly bounded on $[a,a]$ since that is just one point.

however how can we know that the delta interval is contained in the interval [a,b] and say that then b must be an upper bound?

It doesn't need to be contained in $[a,b]$ but if say, the interval extends outside $[a,b]$ then you can always replace it by a smaller interval which is contained inside $[a,b]$. For instance. Suppose $[a,b] = [0,1]$ and you tell me that $f$ is bounded on $(0.5, 1.1)$, well then $f$ must also be bounded on $(0.6, 1)$.

However, this has nothing to do with $b$ being an upper bound. $b$ is an upper bound because by definition, $A = \{x : a \le x \le b \text{ and ...}\}$ so automatically, if $x \in A$ then $a \le x \le b$: every element of $A$ is less than or equal to $b$.

My 2nd confusion has to do with why he assumes the supremum α=b this really confuses me like what reason is there for this to be true and in his proof by contradiction he assumes α<b and then states that it must be the case that α=b however why is it not true that α≤b?.

$b$ is an upper bound of $A$. Therefore $\sup A \le b$ meaning either $\sup A < b$ or $\sup A = b$. If we can rule out that $\sup A < b$ then it must be the case that $\sup A = b$. This is what happens in the first paragraph.


问:
Solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$?
I am trying to solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$.

I can see that the solution to the homogenous $h_n = 3h_{n-1}$ is clearly $h_n = 3^n$. That being said, I have no idea how to solve $h_n - 3h_{n-1} = 3(2)^n$: I assume it's some solution of the sort $(c_1n + c_2)(2)^n$, but I don't know where we'd go from here! Can anyone give advice?

Body: Let $ n\in\mathbb{N}^{*} $, note that for all $ k\in\mathbb{N} $, we have : \begin{aligned}h_{n-k}&=3h_{n-k-1}+3\times 2^{n-k}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3^{k}h_{n-k}&=3^{k+1}h_{n-k-1}+3^{k+1}\times 2^{n-k}\\ \Longrightarrow\sum_{k=0}^{n-1}{\left(3^{k}h_{n-k}-3^{k+1}h_{n-k-1}\right)}&=3\times 2^{n}\sum_{k=0}^{n-1}{\left(\frac{3}{2}\right)^{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h_{n}-3^{n}h_{0}&=3\times 2^{n}\times\frac{\left(\frac{3}{2}\right)^{n}-1}{\frac{3}{2}-1}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  h_{n}&=3^{n}+6\left(3^{n}-2^{n}\right)\end{aligned}


问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答: Actually, the French system was designed regularly, i.e. with "septante" for 70, "octante" for 80 and "nonante" for 90. But this system was not accepted at the time (roughly the 16th century), because people were used to the old system ("quatre-vingt", etc.). But the old system remained in Belgium (for "septante" and "nonante") and Switzerland (for "septante, the slightly modified "huitante" in Vaud canton, and "nonante").</p>

ContentLicense: CC BY-SA 3.0

Title: Why do the French count so strangely?


问:
Constant in Second Order PDE
I am trying to solve this PDE
$$
u_{xy} = x - y
$$

$$ \implies \iint u_{xy}\,dy\,dx = \iint ( x-y )\,dy\,dx$$
$$ = \int (xy - \frac{y^2}{2} + k_0(x))\,dx$$

My question is: How should I proceed? $k(x)$ is a constant with respect to $y$, however it depends on $x$ and it could be any function, therefore I don't know how to integrate it. Should I leave $k_0(x)$ just as a real number or ignore it?

If I ignore it I get:

$$
u(x,y) = \frac{x^2y - xy^2}{2} + k_0(x) + k_1(y)
$$

If I take this $u(x,y)$ and differentiate with respect to $x$ and the with respect to $y$ I would get the original PDE but I don't know how to justify it.

答:
You're on the right track.

If $u$ is a sufficiently smooth solution (let's say $C^2(\mathbb R^2)$), then from the condition $u_{xy}(x,y) = x - y$ it follows by integrating with respect to $y$, that $$ \int u_{xy}(x,y)dy = \int (x-y)dy $$
Hence for some constant $C$ (constant with respect to $y$) we have $$u_{x}(x,y) = \frac{x^2}{2} - xy + C$$
The point is, $C$ is constant with respect to $y$, but not a constant (necessarily) with respect to $x$!. So in fact, $C$ is a function of $x$, and it should be written $$ u_x(x,y) = \frac{x^2}{2} - xy + C(x) $$
Note that due to $u_x \in C^1(\mathbb R^2)$ then $C \in C^1(\mathbb R)$, too.
Now, what's left, is to integrate one more time, with respect to $x$ (note we need to integrate function $C$, but it's in particular continuous, hence we can, and let $F$ be such function, that $F'(x)=C(x)$.
Then
$$ \int u_x(x,y)dx = \int (\frac{x^2}{2} - xy +C(x))dx $$
And for some constant $A$ (constant with respect to $x$), but again, not necessarily constant with respect to $y$, so in fact for some function (continuous) $A$ of variable $y$ we have

$$ u(x,y) = \frac{x^2y - y^2x}{2} + F(x) + A(y) $$
Where $F'=C$ and $C \in C^1(\mathbb R)$ (so $F \in C^2(\mathbb R)$ and $A \in C^2(\mathbb R)$
In other words, if $u$ is a solution to your equation (of the class $C^2(\mathbb R^2)$) then there exists two functions of one variable, sufficiently smooth, that is $f,g \in C^2(\mathbb R)$, such that $$ u(x,y) = \frac{x^2y - xy^2}{2} + f(x) + g(y) $$

Title: Constant in Second Order PDE


问:
What is the probability that a string of $100$ of tosses of a fair coin doesn't have an unlikely substring?
This question relates to the YouTube video "How Lucky is Too Lucky?" by Matt Parker.  In it, he poses the following question.  Parker has published a list of $100$ tosses of a fair coin.  A "malicious actor" wants to claim that in fact, Parker made the tosses up.  He searches through the published tosses, and finds $12$ consecutive tosses comprising $2$ tails and $10$ heads.  The probability of $2$ or fewer tails in $12$ tosses of a fair coin is less than $2\%$, so the actor claims that the tosses cannot possibly be legitimate.

Now Parker addresses the probability that such an anomalous string occurs.  There are $5050=\binom{100}2+100$ sub-runs of length $1$ to $100$.  What is the probability that at least one of them is unlikely?  If we have $n$ tosses comprising $t$ heads and $h$ tails, then the run is unlikely if the probability of getting $t$ or fewer tails is $\leq.019$ or the probability of getting $h$ or fewer heads is $\leq.019$.

I should stress that the malicious actor does not choose $n$ in advance.  He looks for an unlikely string of any length.

At about $16:25$ in the video, Parker says that the exact probability is $88.3\%$, and gives no indication how this number is arrived at.  The problem of course, is that substrings overlap, so we don't have independent events.

Of course, he says throughout that he won't go into the technical details of the math, but I haven't figured out how this number was arrived at.  It's easy to confirm by simulation, but I don't think Parker would have used the phrase "exact value" without a theoretical calculation to back it up.

$2^{100}$ is on the order of $10^{29}$ so generating and counting the admissible runs is infeasible.  I've thought about trying to write a recurrence relation, but it seems hopeless, because there are too many possibilities.  Usually, in this sort of problem, if $a_n$ is the number of admissible strings of length $n$, we have to break the recurrence up depending on the last characters of an admissible string of length $n-1$, $n-2$, and so on.  There seem to be too many possibilities in this case.

An approximate calculation with a small error bound would be fine.  Can you point me in the right direction?

Just in case I haven't described the problem comprehensibly, I append my simulation script:

```<code>from math import factorial
from random import choices

def choose(n,m):
    return factorial(n)//(factorial(m)*factorial(n-m))

epsilon = .019

critical = { }
for n in range(6,101):
    prob = 0
    mu = 2**(-n)
    for m in range(n+1):
        prob += choose(n,m)*mu
        if prob > epsilon:
            critical[n] = m-1
            break

def test(trials):
    success = 0
    for _ in range(trials):
        flips = choices(range(2), k=100)
        success += anomalous(flips)
    return success/trials

def anomalous(flips):
    for m in range(6, 101):
        for s in range(101-m):
            run = flips[s:s+m]
            tails = run.count(0)
            if min(tails, m-tails) <= critical[m]:
                return True
    return False

</code>```

答:
My own attempt at simulation in R suggests a probability of about $0.41$ for the one-tailed version of $2$ or fewer tails in $12$ consecutive flips and about $0.68$ for the two tailed version, rather smaller than  $88.3\%$.  Testing it on $6$ flips, runs of $4$ and extremes of $1$ or fewer tails/heads, it come close enough to the exact $\frac{31}{64}$ one-tailed and $\frac{58}{64}$ two-tailed to suggest the larger simulation was reasonable.

```<code>simlookingforextreme <- function(totalflips, runlength, extreme){
  flips <- sample(c(0,1), totalflips, replace=TRUE)
  runs <- diff(c(0,cumsum(flips)), runlength)
  c(min(runs) <= extreme , max(runs) >= runlength - extreme)
  }
</code>```

giving

```<code>set.seed(2021)
sims <- replicate(10^6, simlookingforextreme(100, 12, 2))
c(low=mean(sims[1,]), high=mean(sims[2,]), two=mean(sims[1,] | sims[2,]))
#      low     high      two
# 0.414857 0.414524 0.677665

set.seed(1)
sims <- replicate(10^6, simlookingforextreme(6, 4, 1))
c(low=mean(sims[1,]), high=mean(sims[2,]), two=mean(sims[1,] | sims[2,]))
#      low     high      two
# 0.484576 0.484273 0.906638
</code>```

This encouraged me to try to check my simulations.  There is an exact approach which does not count $2^{100}$ binary strings, but instead $2^{12}=4096$ such strings, spotting which are extreme, lopping off an end digit and sticking a new digit at the other end, and cycling through the remaining flips.  Again in R

```<code>exactprobextreme <- function(totalflips, runlength, extreme,
                             extremetail){ # "low", "high", "two"
  runscores <- 0
  for (i in 1:runlength){
    runscores <- c(runscores, runscores+1)
    }
  scoreprobs <- rep(1/2^runlength, 2^runlength)
  extremeprob <- 0
  for (n in runlength:totalflips){
    if(extremetail == "low" | extremetail == "two"){
      extremeprob <- extremeprob +
                     sum(scoreprobs[runscores <= extreme])
      scoreprobs[runscores <= extreme] <- 0
      }
    if(extremetail == "high" | extremetail == "two"){
      extremeprob <- extremeprob +
                     sum(scoreprobs[runscores >= runlength - extreme])
      scoreprobs[runscores >= runlength - extreme] <- 0
      }
    scoreprobs[1:(2^(runlength-1))] <- (
       scoreprobs[2*(1:(2^(runlength-1)))] +
       scoreprobs[2*(1:(2^(runlength-1)))-1] )/2
    scoreprobs[(2^(runlength-1)+1):(2^runlength)] <-
        scoreprobs[1:(2^(runlength-1))]
    }
    extremeprob
  }
</code>```

giving (faster than the simulation)

```<code>exactprobextreme(totalflips=100,runlength=12,extreme=2, extremetail="low")
# 0.4145669
exactprobextreme(totalflips=100,runlength=12,extreme=2, extremetail="high")
# 0.4145669
exactprobextreme(totalflips=100,runlength=12,extreme=2, extremetail="two")
# 0.6770409
exactprobextreme(totalflips=6,  runlength=4, extreme=1, extremetail="low")
# 0.484375
exactprobextreme(totalflips=6,  runlength=4, extreme=1, extremetail="high")
# 0.484375
exactprobextreme(totalflips=6,  runlength=4, extreme=1, extremetail="two")
# 0.90625
</code>```

strengthening my opinion that the $88.3\%$ should be $67.7\%$.

Or maybe I have misunderstood.  Looking at the video after writing this and then rereading your question, it seems that there did not to have been run lengths of $12$, and any unlikely runs of any length could have been considered.

Body:
My own attempt at simulation in R suggests a probability of about $0.41$ for the one-tailed version of $2$ or fewer tails in $12$ consecutive flips and about $0.68$ for the two tailed version, rather smaller than  $88.3\%$.  Testing it on $6$ flips, runs of $4$ and extremes of $1$ or fewer tails/heads, it come close enough to the exact $\frac{31}{64}$ one-tailed and $\frac{58}{64}$ two-tailed to suggest the larger simulation was reasonable.

simlookingforextreme <- function(totalflips, runlength, extreme){
  flips <- sample(c(0,1), totalflips, replace=TRUE)
  runs <- diff(c(0,cumsum(flips)), runlength)
  c(min(runs) <= extreme , max(runs) >= runlength - extreme)
  }

giving

set.seed(2021)
sims <- replicate(10^6, simlookingforextreme(100, 12, 2))
c(low=mean(sims[1,]), high=mean(sims[2,]), two=mean(sims[1,] | sims[2,]))
#      low     high      two
# 0.414857 0.414524 0.677665

set.seed(1)
sims <- replicate(10^6, simlookingforextreme(6, 4, 1))
c(low=mean(sims[1,]), high=mean(sims[2,]), two=mean(sims[1,] | sims[2,]))
#      low     high      two
# 0.484576 0.484273 0.906638

This encouraged me to try to check my simulations.  There is an exact approach which does not count $2^{100}$ binary strings, but instead $2^{12}=4096$ such strings, spotting which are extreme, lopping off an end digit and sticking a new digit at the other end, and cycling through the remaining flips.  Again in R

exactprobextreme <- function(totalflips, runlength, extreme,
                             extremetail){ # "low", "high", "two"
  runscores <- 0
  for (i in 1:runlength){
    runscores <- c(runscores, runscores+1)
    }
  scoreprobs <- rep(1/2^runlength, 2^runlength)
  extremeprob <- 0
  for (n in runlength:totalflips){
    if(extremetail == "low" | extremetail == "two"){
      extremeprob <- extremeprob +
                     sum(scoreprobs[runscores <= extreme])
      scoreprobs[runscores <= extreme] <- 0
      }
    if(extremetail == "high" | extremetail == "two"){
      extremeprob <- extremeprob +
                     sum(scoreprobs[runscores >= runlength - extreme])
      scoreprobs[runscores >= runlength - extreme] <- 0
      }
    scoreprobs[1:(2^(runlength-1))] <- (
       scoreprobs[2*(1:(2^(runlength-1)))] +
       scoreprobs[2*(1:(2^(runlength-1)))-1] )/2
    scoreprobs[(2^(runlength-1)+1):(2^runlength)] <-
        scoreprobs[1:(2^(runlength-1))]
    }
    extremeprob
  }

giving (faster than the simulation)

exactprobextreme(totalflips=100,runlength=12,extreme=2, extremetail="low")
# 0.4145669
exactprobextreme(totalflips=100,runlength=12,extreme=2, extremetail="high")
# 0.4145669
exactprobextreme(totalflips=100,runlength=12,extreme=2, extremetail="two")
# 0.6770409
exactprobextreme(totalflips=6,  runlength=4, extreme=1, extremetail="low")
# 0.484375
exactprobextreme(totalflips=6,  runlength=4, extreme=1, extremetail="high")
# 0.484375
exactprobextreme(totalflips=6,  runlength=4, extreme=1, extremetail="two")
# 0.90625

strengthening my opinion that the $88.3\%$ should be $67.7\%$.

Or maybe I have misunderstood.  Looking at the video after writing this and then rereading your question, it seems that there did not to have been run lengths of $12$, and any unlikely runs of any length could have been considered.


问:
How to find the maximal interval of existence for the following IVP?
The IVP is:   $\frac{du}{dt} = u^2 (u-1)$ ; $u(0)=2$ .

Effort: I solved by using separation of variables method. And I got:  $$\ \ln(|u-1|) - \ln(|u|) + \frac{1}{u} = t + 1/2 -\ln(2)$$

But I am not able to obtain $u$ as an explicit function of $t$. The graph given by the above implicit equation supports the same.

So how do I find the maximal interval of existence? Any hint would be helpful.

答: Your right hand side is a continuous function with continuous partial (w.r.t. $u$) on the whole plane. Your (unique) solution must exit any rectangle in the plane. Next observe that, as $u\to\infty$, $t\to\ln 2-1/2$ and, as $u\to 1$, $t\to -\infty$. Therefore your solution has domain $(-\infty, \ln 2-1/2)$.


问:
Taylor series of $\sin\ln x$ at $x = 1$
I tried to discover the Taylor series for and get the answer from Wolfram Alpha, but I couldn't prove the expression provided.

The expression is:

$$\sin\ln x=\sum_{n=0}^{\infty}\frac{1}{2}i(x-1)^n\Bigg(\binom{-i}{n}-\binom{i}{n}\Bigg)$$

How can I prove this relation?

答:
A solution based on differential equations seems made to order.

Render $y=\sin(\ln x)$.  Differentiate using the chain and quotient rules:

$\dfrac{dy}{dx}=\dfrac{\cos(\ln x)}{x}$

$\dfrac{d^2y}{dx^2}=\dfrac{[x(-\sin(\ln x)/x)]-[\cos(\ln x)]}{x^2}=-\dfrac{\sin(\ln x)+\cos(\ln x)}{x^2}$

Now take the linear combination $x(\dfrac{d^2y}{dx^2})+\dfrac{dy}{dx}$ which eliminates the cosine term, and put in $y=\sin(\ln x)$ to get rid of the sine function.  After clearing fractions and collecting terms you have the Cauchy-Euler equation

$x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}+y=0$

whose methods of solution are described in the link.  You basically assume solutions having the form $x^\lambda$ and derive an algebraic equation for $\lambda$ that fits the differential equation, which in this case turns out to be $\lambda^2+1=0$.  Thus we get a general solution

$y=Ax^i+Bx^{-i}$

To render $A$ and $B$ that fit the specific function $y=\sin(\ln x)$, use the function value, and the first derivative given above, evaluated at $x=1$ to render

$A+B=0, A-B=-i$

Thereby $A=-i/2,B=i/2$ and then

$\sin(\ln x)=(-i/2)x^i+(i/2)x^{-i}$

Now render $x^i=(1+y)^i$ with $y=x-1$ and expand with the Binomial Theorem.  Do the same with $x^{-i}$ and combine the results to get the claimed expansion.

Title: Taylor series of $\sin\ln x$ at $x = 1$


问:
Why this vector field $f$ belongs to $C^1({\bf R}^2\times {\bf R})$?
The following system is an example in a book of dynamical system(in the section about Hopf Bifurcation).

$$
\begin{align}
\dot{x}=\mu x- y-x\sqrt{x^2+y^2} \\

\dot{y}=x + \mu y-y\sqrt{x^2+y^2}
\end{align}
$$

But I don't understand why it can be called a $C^1$-system, i.e., the vector field $f:R^2\times R\to R^2$ defined by this systemm belongs to $C^1(R^2\times R)$, i.e., all of the first partial derivatives with respect to $x,y$ and $\mu$ are continuous for all $x,y$ and $\mu$.

I set it as an exercise, and found that
$$
\frac{\partial f}{\partial x}=
\left(
  \begin{array}{c}
     \mu-(\sqrt{x^2+y^2}+\frac{x^2}{\sqrt{x^2+y^2}})  \\
     1-\frac{xy}{\sqrt{x^2+y^2}}  \\
  \end{array}
\right)
$$
This is not even defined at $(0,0,\mu)$. Then why $f\in C^1(R^2\times R)$?

Edit: The original question finally boils down to another question:

What exactly is the definition of
  $C^1$ functions, or more generally
  $C^k$ functions? Are the $k$-th derivatives allowed to
  have removable-discontinuity points?

答:
You have a function $f:(x,y) \mapsto f(x,y)=(u(x,y),v(x,y))$ defined by expressions containing square roots (or maybe denominators). At all points $(x_0,y_0)$ where the radicand (the denominator) is nonzero you may calculate the partial derivative ${\partial f\over \partial x}$ using the rules of calculus. At points $(x_0,y_0)$ where the radicand is zero the function $f$ may fail to be differentiable because the function $t\mapsto \sqrt{t}$ is not differentiable at zero. It follows that we have to check such points separately (hoping that a "special effect" comes to our help). In the example at hand the only special point is the origin. Here we use the actual definition of partial derivative and obtain, e.g.,
$${\partial u \over\partial x}(0,0)=\lim_{x\to 0}{u(x,0)-u(0,0)\over x} = \lim_{x\to 0}{\mu x -x|x| - 0 \over x} =\mu\ .$$
This shows that ${\partial u\over\partial x}$ is also defined at $(0,0)$, so it is defined everywhere. We still have to check that it is continuous at $(0,0)$. Now for $(x,y)\ne(0,0)$ you have obtained
$${\partial u\over\partial x}=\mu-\sqrt{x^2+y^2}\Bigl(1+{x^2\over x^2+y^2}\Bigl )\ .$$
Here the last factor has a value between $1$ and $2$. As $\lim_{(x,y)\to(0,0)}\sqrt{x^2+y^2}=0$ it follows that $$\lim_{(x,y)\to(0,0)}\ {\partial u\over\partial x}=\mu={\partial u\over\partial x}(0,0)\ .$$
This proves that $u(\cdot,\cdot)$ is indeed $C^1$ on all of ${\mathbb R}^2$.</p>


问:
Am looking to prove following conjecture:
Am looking to prove that the tangent to the cubic $y = x^3 + 7x^2 + 4x + 28$ at any point midway between any two roots will itself cut the x-axis at the remaining root. The cubic factorizes as $y = (x+2i)(x-2i)(x+7)$. And here's my problem: complex roots. How can I solve the above?</p>

答:
The claim is true for any cubic, whether the roots are real or not. Look at this:

The above is Mathematica output. Here is what it does:

We begin with an arbitrary cubic polynomial $p(x):=(x-a)(x-b)(x-c)$ with given real or complex zeros $a$, $b$, $c$ and put
$$y_0:=p\left({a+b\over2}\right)=-{1\over8}(a-b)^2(a+b-2c),\quad y_1:=p'\left({a+b\over2}\right)=-{1\over4}(a-b)^2\ .$$
In the case where the roots are real the equation
$$y:=q(x):=y_0 + y_1 \left(x-{a+b\over 2}\right)$$
describes the tangent to the graph of $p$ at $x_0:={a+b\over2}$. The last command ("Solve") in the above Mathematica dialog computes the zero of the linear function $q$, and sure enough it found $x=c$.


问:
First countability of $[0,1]^\mathbb{R}$
The proofs I know for the fact that the space $[0,1]^\mathbb{R}$ is not first countable, use the product topology in some step of the demonstration. (Reference)

So, I would like to know if this space is also not first countable in the box topology and also if there is any topology that makes this space first countable.

Edit:
As Brian commented in his answer, I was looking for a topology that was different from the trivial and the discrete.

答:
It’s even easier to show that it’s not first countable in the box topology than it is to show that it’s not first countable in the product topology: even $[0,1]^{\Bbb N}$ isn’t first countable in the box topology.

There are many first countable topologies that can be put on the set $[0,1]^{\Bbb R}$, including the trivial and discrete topologies, but I expect that you want one that is directly related to the structure of the product. Any metrizable topology will work, and there is a natural metric on it: for $x,y\in[0,1]^{\Bbb R}$ let

$$d(x,y)=\sup\{|x_r-y_r|:r\in\Bbb R\}\,.$$


问:
Proving that none of these elements 11, 111, 1111, 11111...can be a perfect square
How can i prove that no number in set S
S = {11, 111, 1111, 11111...}
Is a perfect square.

I have absolutely no idea how to tackle this problem i tried rewriting it in powers of 10 but that didn't really get me anywhere...

Thanks in advance.

答:
First notice that each element in S can be rewritten as:

$4(25m+2) +3$ where m is an integer, the next step is to realise that all squares leave a remainder of either 1 or 0 when divided by 4.

As $(2n)^2 = 4n^2$ which clearly leaves a remainder of 0 when divided by 4

And for odd squares: $(2n-1)^2 = 4n^2 -4n +1$ which clearly leaves a remainder of 1 upon division by 4.

But by writing each element of S as $4(25m+2) +3$ we see that it leaves a remainder of 3 when divided by 4, therefore it can never be a perfect square.


问:
Given an arbitrary infinite set $S$, how to construct an injection $\mathbb{N} \to S$ with the Axiom of Choice?
Let $S$ be an arbitrary infinite set. As a part of our homework, I have the following question:

Question: How can I construct an injection $\mathbb{N} \to S$?

Intuitively, I understand that such an injection should exist but describing it explicitly is probably not possible. Therefore, I think we need to use the

Axiom of Choice: If $X$ is a set of non-empty sets, then there exists a choice function $F$ on $X$, i.e. a function on $X$ such that for all $M \in X$, we have $F(M) \in M$.

Our class was given the following hint: Construct a family of $A_n \subseteq S$ with $|A_n| = 2^n$ for all $n \in \mathbb{N}$. Next, construct a family $B_n \subseteq S$ of non-empty and pairwise disjoint subsets.

However, I do not really know what to do with this hint. Maybe the condition $|A_n| = 2^n$ indicates that we should use the power set of some set with $n$ elements but that is just a vague idea.

I would really appreciate it if you could help me advancing with this problem.

答:
I am assuming that your definition of infinite is such that $S$ is infinite if for each $n\in\Bbb Z^+$ there is an injection $f_n$ of $\{1,\ldots,n\}$ into $S$ (or something from which that can easily be inferred). For each $n\in\Bbb N$ let $A_n=f_{2^n}\left[\left\{1,\ldots,2^n\right\}\right]$.

Show that $A_n\setminus\bigcup_{k<n}A_k\ne\varnothing$ for each $n\in\Bbb N$. HINT: What is the largest possible value of $\sum_{k<n}|A_k|$?

Now let $B_n=A_n\setminus\bigcup_{k<n}A_k$ for each $n\in\Bbb N$ and apply the axiom of choice to $\{B_n:n\in\Bbb N\}$ to construct an injection from $\Bbb N$ to $S$.


问:
Generalized delta method
Suppose $f\in C^{k+1}$ satisfies $f^{(j)}(\mu)=0$ for all $1\le j\le k$ and $f^{(j)}(\mu)\ne 0$ where $\mu$ is fixed. Furthermore, assume $a_n\to \infty$ and $a_n(X_n-\mu)$ converges in distribution to some random variable $X$.

I am trying to see if the following two claims are true, which are somewhat generalizations of the Delta method. That is,

$a_n^j(f(X_n)-f(\mu))$ converges in distribution to 0 for all $j$ and that $a_n^{k+1}(f(X_n)-f(\mu))$ converges in distribution to a non-degenerate random variable. However I cannot just mimic the standard proof of the Delta method as it hinges on assuming $a_n=n$.

答:
Let $f \in C^{k+1}$ and let $j \in \{1,...,k\}$ be fixed.
Note that $f(X_n) = f(\mu) + (X_n-\mu)f'(\mu) + ... + \frac{1}{j!}(X_n-\mu)^j f^{(j)}(\mu) + h(X_n-\mu)(X_n-\mu)^j $ where $h$ is such a function, that $h(x) \to 0$ as $x \to 0$.

Note that all derivatives disappear, hence $f(X_n)-f(\mu) = h(X_n-\mu) (X_n-\mu)^j $

Now, the point is, since $a_n(X_n-\mu) \to X$ in distribution, and $a_n \to \infty$, then $(X_n-\mu) \to 0$ in distribution (since $(X_n-\mu) = \frac{1}{a_n} \cdot (a_n(X_n-\mu))$), so in probability, too.

From this follows $h(X_n-\mu) \to 0$ in probability (easiest argument would be to argue by characterisation $Y_n \to Y$ in probability, iff for all $(n_k)$ there exists $(n_{k_m})$ such that $Y_{n_{k_m}} \to Y$ almost surely). By convergence in probability, we have convergence in distribution, too.

Hence $a_n^j(f(X_n)-f(\mu)) = (a_n (X_n-\mu))^j \cdot h(X_n-\mu) \to X^j \cdot 0 = 0$ in distribution (we used that continuous functions preserve convergence in distribution and used slutsky lemma, which says that $Y_n \to Y$ in distribution, and $Z_n \to c$ in distribution, where $c$ is a constant, implies $Y_nZ_n \to cY$ in distribution).

For the last part, do the same writting with remainder, but now for $j=k+1$, we'll get
$f(X_n)-f(\mu) = \frac{1}{(k+1)!}(X_n-\mu)^{k+1}f^{(k+1)}(\mu) + h(X_n-\mu)(X_n-\mu)^{k+1}$

We show as before, that $a_n^{k+1}(X_n-\mu)^{k+1}h(X_n-\mu) \to 0$ in distribution.
And the first part, similarly $(a_n (X_n-\mu))^{k+1} \frac{f^{(k+1)}(\mu)}{(k+1)!} \to X^{k+1} \frac{f^{(k+1)}(\mu)}{(k+1)!}$ in distribution and again slutsky lemma (but now version $Y_n \to Y$ in distribution, $Z_n \to c$ in distribution ($c$ constant) implies $Y_n + Z_n \to Y + c$) and we finally get

$$ a_n^{k+1}(f(X_n) - f(\mu)) \to X^{k+1} \frac{f^{(k+1)}(\mu)}{(k+1)!} $$ in distribution

Title: Generalized delta method


答:
How would you naively argue that an infinite set has an injection from $\Bbb N$ into the set?

Well, pick some $x_0$, then suppose that $\{x_0,\dots,x_{n-1}\}$ were chosen, since the set is infinite, there is some $x_n$ which is different from those $n$ elements. So we defined $\{x_n\mid n\in\Bbb N\}$, which is clearly the range of an injection from $\Bbb N$ into the set.

Where did we use the axiom of choice? Well, the choice of $x_0$ was arbitrary, but the choice of $x_1$—while arbitrary—depends on the choice of $x_0$, since $x_1\neq x_0$ by design. And the choice of $x_{216}$ depends on the $216$ previous choices. So the choice here is literally in the ability to consolidate the real proof that an infinite set has arbitrarily large finite subsets, into a proof that it has a countably infinite subset.

Where did we use choice? We chose the $x_n$. So by fixing a choice function, $F$, and applying the recursion theorem to choose $F(n)$ from $X\setminus\operatorname{rng}(F\restriction n)$ we can get our infinite sequence.

Fun exercise. The above proof relies on the principle of dependent choice, but we can still prove that every infinite set has a countably infinite subset from the weaker "axiom of choice for countable families of sets".

Title: Given an arbitrary infinite set $S$, how to construct an injection $\mathbb{N} \to S$ with the Axiom of Choice?

Body:
Let $S$ be an arbitrary infinite set. As a part of our homework, I have the following question:

Question: How can I construct an injection $\mathbb{N} \to S$?

Intuitively, I understand that such an injection should exist but describing it explicitly is probably not possible. Therefore, I think we need to use the

Axiom of Choice: If $X$ is a set of non-empty sets, then there exists a choice function $F$ on $X$, i.e. a function on $X$ such that for all $M \in X$, we have $F(M) \in M$.

Our class was given the following hint: Construct a family of $A_n \subseteq S$ with $|A_n| = 2^n$ for all $n \in \mathbb{N}$. Next, construct a family $B_n \subseteq S$ of non-empty and pairwise disjoint subsets.

However, I do not really know what to do with this hint. Maybe the condition $|A_n| = 2^n$ indicates that we should use the power set of some set with $n$ elements but that is just a vague idea.

I would really appreciate it if you could help me advancing with this problem.


问:
Entropy of a non-central chi-squared distribution with one degree of freedom
Let $Y=X^2$ where $X$ is a non-central Gaussian distributed random variable. So $Y$ has a non-central chi-squared distribution with one degree of freedom. What is the entropy of $Y$? I write the integral formula of entropy, but I cannot simplify the integral at all. </p>

答:
The answer should be in this recent paper:

Stefan Moser (2020): "Expected Logarithm and Negative Integer Moments of a Noncentral χ2-Distributed Random Variable", Entropy, vol. 22, no. 9, https://doi.org/10.3390/e22091048

The result is provided for any odd or even integer number of degrees of freedom and is pretty complicated. It seems to be a novel result, not derived before.

Set the function $g_n(\xi)$ for an arbitrary $n$, defined for $\xi>0$ by

$$g_n(\xi) = \ln(\xi) - \mathrm{Ei}(-\xi) + \sum_{i=1}^{n-1}(-1)^i\left( e^{-\xi} (i-1)! - \frac{(n-1)!}{i(n-1-i)!} \right)\left(\frac{1}{\xi}\right)^i $$

Set the function $h_n(\xi)$ for an odd $n$, defined for $\xi>0$ by

$$h_n(\xi) = -\gamma - 2\ln(2)+2 \xi {}_2 F_2\left(1,1,\frac{3}{2},2,-\xi\right) +\sum_{i=1}^{(n-1)/2} (-1)^{i-1}\Gamma\left(i-\frac{1}{2}\right)\left(\sqrt{\xi}e^{-\xi}\mathrm{erfi(\sqrt{\xi})}+\sum_{j=1}^{i-1}\frac{(-1)^j\xi^j}{\Gamma\left(j+\frac{1}{2}\right)} \right)\left(\frac{1}{\xi}\right)^i$$

where $\mathrm{Ei}$ is the exponential integral function, $\gamma$ is the Euler constant, ${}_2 F_2$ is a generalised hypergeometric function and $\mathrm{erfi}$ is the imaginary error function.

Then the result for $X\sim\mathcal{X}'^2_n(\lambda)$ is, for $n$ odd,
$$ \mathbb{E}\left[-\ln(X)\right]=\ln(2)+h_n\left(\frac{\lambda}{2}\right)$$
and, for  $n$ even,
$$ \mathbb{E}\left[-\ln(X)\right]=\ln(2)+g_{n/2}\left(\frac{\lambda}{2}\right)$$


问:
The greatest integer common divisor of the three polynomials is the same as the $gcd$ of their combined coefficients
I have three polynomials $f(x,y,z), g(x,y,z),h(x,y,z)$ in three integer variables $x,y,z$ and with integer coefficients.

I am searching for conditions in which the three polynomials have a common prime divisor. I came across with this problem when solving a system of algebraic equations involving the equilibrium points of a dynamical system. By "prime divisor" I mean an integer prime divisor not an irreducible integer polynomial divisor.  I read that the greatest integer common divisor of the three polynomials is the same as the $gcd$ of their combined coefficients. But I am not understanding the meaning of this phrase, i.e., combined coefficients.

答:
If $\,f,g,h\in\Bbb Z[x]\,$ with coef's $\,f_i,g_i,h_i,\,$ and $\,d\in \Bbb Z,\,$ then we have $$d\mid f\ \,{\rm in}\,\ \Bbb Z[x]\iff f/d\in \Bbb Z[x]\iff \{f_i/d\}\subset \Bbb Z\iff \forall\,i\!:\, d\mid f_i \iff d\mid \gcd\{f_i\}$$
So $\,d\mid f,g,h\!\iff\! d\mid \gcd\{f_i\},\gcd\{g_i\},\gcd\{h_i\}\! \!\iff\! d\mid \gcd\{f_i, g_i, h_i\}\,$ by gcd is associative.


问:
$\sum_{n=1}^{\infty} |a_n|<\infty \implies \exists N$ such that $\sum_{i=N}^ \infty |a_i|<\epsilon$ for any $\epsilon>0$
$\displaystyle\sum_{n=1}^{\infty} |a_n|<\infty \implies \exists m$ such that $\displaystyle\sum_{i=m}^ \infty |a_i|<\epsilon$ for any $\epsilon>0$.

Proof) Let $S_n=\displaystyle\sum_{k=1}^{n} |a_k|$. Since $S_n$ converges, it is Cauchy, so

for any $\epsilon>0$, $\exists N$ such that $|S_n-S_m|=\left|\displaystyle\sum_{k=m+1}^n|a_k|\right|<\epsilon$ for all $n,m>N$.

Taking the limit as $n\to\infty$ of both sides, we have $\displaystyle\sum_{k=m+1}^{\infty}|a_k|<\epsilon$.

I am wondering if this proof is correct.

答:
First of all, your statement is not true. You should move the "for all $\epsilon>0$" part to right after "$\Rightarrow$".

$\displaystyle\sum_{n=1}^{\infty} |a_n|<\infty \implies \forall\epsilon>0\ \exists N$ such that $\displaystyle\sum_{i=N}^ \infty |a_i|<\epsilon$

In your proof, when taking the limit $n\to\infty$, you should conclude that
$$
\sum_{k=m+1}^\infty |a_k|\le \epsilon\tag{1}
$$
not the strict inequality.

But since $\epsilon>0$ is arbitrary, you can eventually replace $\le$ with the $<$. To be precise, the following two statements are equivalent

for every $\epsilon>0$, $A\le \epsilon$;

for every $\epsilon>0$, $A<\epsilon$.


问:
Why differentiate between continuous on $[a,b]$ and differentiable on $(a,b)$?
Before starting I want to say that my main language is not english so you may find some grammar mistakes. I will try to avoid making them as much as I can, so please forgive me.

A few months ago I decided that I wanted to learn Calculus so I got some books and started to study, during that time I faced a statement that is pretty common among math definitions related to it, and that statement is:

$f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$

What I was thinking was that a limit needs a closed interval to compute its value, but after making some research I found that the function $f(x) = \sqrt{x}$ is continuous at $0$, so I started to make myself this question.

If we consider, for example $f(x) = x^2$. and the interval $[1,3]$. We know that the function is differentiable and continuous at its whole domain but we do we say that it has to be continuous at $[1,3]$ and differentiable at $(1,3)$, this of course, considering the function over the interval mentioned before, when in fact it is differentiable at $1$ and $3$.

Also the function $\sqrt{x}$ is considered continuous at its whole domain, even also at $x = 0$ but it is not differentiable at 0. Why?

Why do we do this difference between $[a,b]$ and $(a,b)?$

Sorry if I did not express myself the best way possible, again, English is not my main language.

答:
The short answer is: we often desire for the assumptions of a theorem to be least restrictive possible, so that the theorem is most general and widely applicable. It turns out that in many theorems you require continuity on $[a, b]$, but you do not actually need differentiability at the endpoints, only on $(a, b)$.

For example, in the proof of Rolle's theorem, we have a function $f$ defined on $[a, b]$ such that $f(a)=f(b)$ and we wish to establish the existence of a point $c \in (a, b)$ such that $f'(c) = 0$. We know that $f$ must have the maximum on $[a, b]$ and we define $d$ to be an argument for which the maximum is attained. Now there are two cases. Either $d$ is an endpoint of $[a, b]$ or an interior point. In the first case $f$ is the constant function and we are done without explicitly invoking differentiability of $f$ (even though every constant function on a suitable domain is differentiable). In the second case we invoke differentiability in order to say that $f'(d)$ exists before we observe that $f'(d)=0$.

This example shows that sometimes differentiability of a function in the interior $(a, b)$ of an interval $[a, b]$ may simply be all that is needed to establish a conclusion.

There is also the subtle point that consideration of differentiability and derivative of a function defined on a closed interval at an endpoint of the interval requires that we appropriately interpret the definition as referring to one-sided limit. The subtlety disappears if we only consider differentiability and derivatives on open sets.

Regarding the reason that the function $\sqrt{x}$ is not differentiable at $x=0$ it is because the tangent to its plot becomes vertical at this point. Equivalently, the limit $$\lim_{x\to 0_+}\frac{\sqrt{x}}{x} = \lim_{x\to 0_+}\frac{1}{\sqrt{x}} $$ does not exist.


问:
What is the most general definition of a graph?
I would like to give the most general definition of a graph. Based on what I read about graphs, I came up with the following definition:

Let us introduce some symbols. They are

(a) $V$:: a non-empty set of vertices;

(b) $E$:: a set of edges $E$;

(c) $\mathcal{P}(X)$:: the power set of a set $X$;

(d) $\mathit{\Pi}(X)=(\mathcal{P}(X)\cup
\bigcup \limits_{i=2}^{\infty}X^i)\setminus \varnothing$:: the alternative power set (with ordered and unordered tuples);

(e) $F$:: a class of functions defined over any subset of $\mathit{\Pi}(V \cup E) \cup (V \cup E)$.

The structure $G=(V,E,F)$ is a (general) graph if there is an $I \in F$ with $I: E \rightarrow \mathit{\Pi}(V \cup E)$, which is called incidence function.

Is this a specification which includes all versions of graphs known in the literature (or at least the most important ones)? Can it happen that an even more general definition is needed for some very special graphs?

It would be also interesting if this formalism could be reduced.

答:
Common definition is that a graph $G = (V, E)$ is a finite set of vertices $V$ and a set of edges $E$, where an edge $e \in E$ is a set of exactly two vertices.
Some allow an infinite set of vertices.

The other stuff (neighbors of $v$, and so on) can be deduced from this.

Note that digraphs (directed graphs) $D = (V, E)$ add a significant difference: The set of edges is a set of pairs of vertices, $E \subseteq V \times V$. They are very different beasts.


问:
4 cards are shuffled and placed face down. Hidden faces display 4 suits: spades, clubs, diamonds, hearts. You turn over cards until win or lose
Problem: 4 cards are shuffled and placed face down. Hidden faces display 4 suits: spades, clubs, diamonds, hearts. You turn over cards until win or lose. You win if you turn over spades AND clubs. You lose if you turn over hearts. Find P(win)

I can't seem to find what's wrong with my approach.

Law of total probability:

$P(lose) = P(lose | H_1) * P(H_1) + P(lose | H_2) * P(H_2) + P(lose | H_3) * P(H_3) + P(lose | H_4) * P(H_4)$
where $Hx$ denotes the Hearts card being in the $x^{th}$ position

$P(lose) = 1/4 * 1 + 1/4 * 1 + 1/4 * (1/4)^2 + 1/4 * 0 = 0.515625$

If Hearts is in either 1st or 2nd position, we always lose. If Hearts is in 3rd position, we lose if Spades & Clubs are exactly in 1st and 2nd position, which happens with $P = (1/4)^2$, in the last scenario we cannot lose

$P(win) = 1-P(lose) = 0.484375$
But the official answer is $2/3$, so I don't know where I went wrong. Please also explain the correct answer like I'm 5.

Thanks for your help!

答:
I think your problem is that your tree of possibilities needs pruning.  You also are missing a conditional probability: The probability of spades and clubs being the first two given the heart being 3rd is not 1/16 but rather 1/3.
This problem is more easily done by drawing the tree.


问:
Least squares by tolerance?
In the documentation for MicroStation (https://docs.bentley.com/LiveContent/web/OpenBuildings%20Designer%20Help-v3/en/GUID-2B6B090B-CEC2-C255-8A0B-425F1075A9EE.html), there exists a method for creating B-splines curves using a method called "Least squares by tolerance" where the user supplies a "tolerance" i.e. a maximum deviation from the input polygon that the B-Spline may lie.

How would this be implemented? I've done a lot of research and have only found a method on Least squares where the number of control points is specified, but not a tolerance.

Not necessarily specfically, but what would a least squares by tolerance algorithm look like?

答:
The "By tolerance" LS fitting actually works like this:

Perform LS fitting with a B-spline of single segment. Denoting the output curve as $C(t)$.

Compute the error of each data point. $Err_i= ||P_i - C(t_i)||$ where $P_i$ are the data points and $t_i$ the parameters, then compute the maximum error.

If the maximum error is smaller than the tolerance, you are done. If not, increase the number of segments (and therefore the number of control points) and go back to step 1.

Several issues to note:

Make sure that the number of control points is always smaller than the number of data points during the iterations. If they become equal, then you will be doing interpolation, instead of LS fitting.

How do you increase the number of segments is a matter of choice. You can increase the number of segments one by one, which will make the whole iteration converge slowly but will never produce a B-spline curve with more control points than necessary. Alternatively, you can also double the number segments each time for a faster convergence and accept the fact that the final B-spline curve might have too many control points.

Also, the error vector computed from step 2 is typically not orthogonal to the curve produced by the LS fitting. So, the maximum error obtained this way is typically bigger than the "true" maximum deviation between the data points and the B-spline curve. There is a technique that will adjust the parametrization to make the error vector become more orthogonal to the output curve. This technique is described in "Intrinsic parametrization for approximation, by J. Hoschek, published in CAGD Vol5, 1988.

Title: Least squares by tolerance?


问:
Smoothly connecting PDEs with finite differences

A PDE with non-smooth inhomogeneity

Let $\mathcal{L}$ be a second-order, linear, elliptic differential operator acting on $\mathcal{C}^2([0,2]^2)$.

I'm numerically solving the inhomogeneous PDE
\begin{align*}
\mathcal{L}u(x,y)+(x-1)^+=0,
\end{align*}
where $(\cdot)^+$ denotes the positive part.

Put differently, I solve two PDEs which need to be connected at $x=1$:
\begin{align*}
\begin{cases}
\mathcal{L}u(x,y)=0 & \text{for }(x,y)\in[0,1]\times[0,2], \\
\mathcal{L}u(x,y)+x-1=0& \text{for }(x,y)\in(1,2]\times[0,2].
\end{cases}
\end{align*}

Approximating  all partial derivatives by central differences, I get the nine-point stencil
\begin{align*}
c_1 u_{i-1,j-1} + c_2 u_{i,j-1} + c_3 u_{i+1,j-1} + c_4 u_{i-1,j} + c_5 u_{i,j} &\\
  + c_6 u_{i+1,j}+ c_7 u_{i-1,j+1} + c_8 u_{i,j+1} + c_9 u_{i+1,j+1} + (x_i-1)^+ &=0.
\end{align*}
Thus, $u$ is the solution to a system of linear equations.

Problem

Plotting the solution $u$ for a fixed $y$, it all looks fine and perfect. However, a plot of $\frac{\partial u}{\partial x}$ as a function of $x$ shows that the derivative is not smooth at $x=1$. The above FD scheme works fine with value-matching (the solution $u$ is perfectly continuous) but struggles with smooth-pasting at $x=1$ (the derivative is not smooth).

Question

How do I ensure smooth-pasting with a finite difference scheme at $x=1$?

Some of my failed attempts include

Impose that forward and backward differences at $x=1$ equal each other (= 2nd order central difference is zero).

Use higher order approximations around $x=1$ such as $u_{xx}\approx \frac{-u(-2h)+16u(-h)-30u(0)+16u(h)-u(2h)}{12h^2}$ and $u_{x}\approx \frac{u(-2h)-8u(-h)+8u(h)-u(2h)}{12h}$ and central differences everywhere else.

Approximating partial derivatives using points only from one side of $x=1$ (i.e. only using either $u(0), u(h), u(2h)$ or instead $u(0),u(-h),u(-2h)$).

Imposing that $u_x\approx\frac{u_{i+1,j}-u_{i-1,j}}{2h}$ equals an average of partial derivatives at $1+h$ and $1-h$.

Note: This problem arises as part of a larger system of free boundary problems. Thus, it's necessary to solve the PDE numerically. This question is also posted here.

答:
I think you're running into theoretical limitations, but you're seeing the solution. The elliptic regularity theorem guarantees only so many continuous derivatives in the solution when the nonhomogeneous part is not completely smooth.

Take for example the similar but stripped-down problem $u_{xx}(x) = (x)_+$ on $(-1, 1)$ subject to $u(-1) = u(1) = 0$. For the left region ($-1 \leq x \leq 0$), $u^l(x) = a + ax$ for some constant $a$ satisfies the
differential equation and the boundary condition at $x = -1$. And for the right region ($0 \leq x \leq 1$),
$u^r(x) = A + Bx + x^3/6$ satisfies the differential equation.
To ensure $u$ and $u_x$ are continuous at $x=0$ you need $A = a$ and $B = a$ so
\begin{align*}
        u^l(x) &= a + a x, \\
        u^r(x) &= a + a x + x^3/6.
\end{align*}
The right boundary condition requires $a + a + 1/6 = 0$, so $a = -1/12$.

You can see $u$, $u_{x}$, $u_{xx}$ all are continuous at $x = 0$, but $u_{xxx}$ isn't.


答: Each number is of the form $(10^k - 1)/9$. Thus it is a perfect square if and only if $10^k - 1$ is a perfect square. But notice that for $k \ge 2$ we have $10^k - 1 \equiv 3 {\pmod 4}$, and every square is either $0$ or $1$ modulo $4$.</p>

Title: Proving that none of these elements 11, 111, 1111, 11111...can be a perfect square

Body:
How can i prove that no number in set S
S = {11, 111, 1111, 11111...}
Is a perfect square.

I have absolutely no idea how to tackle this problem i tried rewriting it in powers of 10 but that didn't really get me anywhere...

Thanks in advance.


答:
If Hearts is in either 1st or 2nd position, we always lose.

Correct. This happens with probability $\frac{1}{2}$.

If Hearts is in 3rd position, we lose if Spades & Clubs are exactly in 1st and 2nd position.

This is not correct. You lose if Diamonds is in either 1st or 2nd position. This happens with probability $\frac{1}{4}\times \frac{2}{3}=\frac{1}{6}$.

The probability of losing is therefore $\frac{1}{2}+\frac{1}{6}=\frac{2}{3}.$

Body:
Problem: 4 cards are shuffled and placed face down. Hidden faces display 4 suits: spades, clubs, diamonds, hearts. You turn over cards until win or lose. You win if you turn over spades AND clubs. You lose if you turn over hearts. Find P(win)

I can't seem to find what's wrong with my approach.

Law of total probability:

$P(lose) = P(lose | H_1) * P(H_1) + P(lose | H_2) * P(H_2) + P(lose | H_3) * P(H_3) + P(lose | H_4) * P(H_4)$
where $Hx$ denotes the Hearts card being in the $x^{th}$ position

$P(lose) = 1/4 * 1 + 1/4 * 1 + 1/4 * (1/4)^2 + 1/4 * 0 = 0.515625$

If Hearts is in either 1st or 2nd position, we always lose. If Hearts is in 3rd position, we lose if Spades & Clubs are exactly in 1st and 2nd position, which happens with $P = (1/4)^2$, in the last scenario we cannot lose

$P(win) = 1-P(lose) = 0.484375$
But the official answer is $2/3$, so I don't know where I went wrong. Please also explain the correct answer like I'm 5.

Thanks for your help!


答:
Not if one polynomial is the negative of the
sum of the other two.</p>

Title: The greatest integer common divisor of the three polynomials is the same as the $gcd$ of their combined coefficients

Body:
I have three polynomials $f(x,y,z), g(x,y,z),h(x,y,z)$ in three integer variables $x,y,z$ and with integer coefficients.

I am searching for conditions in which the three polynomials have a common prime divisor. I came across with this problem when solving a system of algebraic equations involving the equilibrium points of a dynamical system. By "prime divisor" I mean an integer prime divisor not an irreducible integer polynomial divisor.  I read that the greatest integer common divisor of the three polynomials is the same as the $gcd$ of their combined coefficients. But I am not understanding the meaning of this phrase, i.e., combined coefficients.


答:
Think about where the diamond may be when given that the heart is somewhere.

We lose if the heart is in position 1 or 2, or if the heart is in position 3 and the diamond is not in position 4.

$$\begin{align}\mathsf P(L)&=\mathsf P(H_1\cup H_2\cup(H_3\cap D_4^{\small\complement}))\\[1ex]&=\mathsf P(H_1)+\mathsf P(H_1)+\mathsf P(H_3)\;\mathsf P(D_4^{\small\complement}\mid H_3)\end{align}$$

When given that the heart is in position 3, the probability that the diamond is not in position 4 is: $2/3$.

$$\begin{align}\mathsf P(L)&=\tfrac 14+\tfrac 14+\tfrac 14\cdot\tfrac 23\\[1ex]&=\tfrac{2}{3}\end{align}$$

Title: 4 cards are shuffled and placed face down. Hidden faces display 4 suits: spades, clubs, diamonds, hearts. You turn over cards until win or lose


问:
Out-of-print textbooks that should be reprinted
One of the things I'm looking into doing is beginning a Dover-like publishing company for putting out-of-print textbooks back into print in nice, cheap editions for students. So I'd like to ask the mathematicians here who are aware of such books:

What books that are out of print would you like to see republished in such editions—books for which you think it's a tragedy that they are out of print?

I think it's a good question for professional mathematicians and would form the basis for the beginnings of my book line. No limit on how many you'd like to suggest.

I have a ton of such books, but I'll kick it off with the two books I think it's tragic are out of print:

Functional Analysis, 2nd. ed. by L.V.Kantorovich and G.P. Akilov.

Notes On Differential Geometry by Noel J.Hicks. (Yes, I know there's a PDF of it floating around online, but a lot of us would like an actual physical book.)

答:
Jon Barwise, Handbook of Mathematical Logic

W.A.J. Luxemburg, Lecture Notes on Nonstandard Analysis

W.A.J. Luxemburg, Introduction to Nonstandard Analysis


答:
The function $f(x)=\sin\log x$ is $C^{\infty}$ on $(0,\infty)$ and analytic in a neighborhood of $1$, as the composition of two analytic functions. We will compute $f^{(n)}(x)$ to derive the Taylor series at $x=1$.

The derivative of $a_n\dfrac{\cos\log x}{x^n}+b_n\dfrac{\sin\log x}{x^n}$ is

$$-na_n\dfrac{\cos\log x}{x^{n+1}}-a_n\dfrac{\sin\log x}{x^{n+1}}-nb_n\dfrac{\sin\log x}{x^{n+1}}+b_n\dfrac{\cos\log x}{x^{n+1}}\\=(-na_n+b_n)\dfrac{\cos\log x}{x^{n+1}}+(-a_n-nb_n)\dfrac{\sin\log x}{x^{n+1}}$$

And it's easy to verify this formula is still valid for $n=0$. So, with $a_0=0, b_0=1$, the $n$th derivative of $f(x)=\sin\log x$ (the $0$th derivative is simply $f$) is:

$$f^{(n)}(x)=a_n\dfrac{\cos\log x}{x^n}+b_n\dfrac{\sin\log x}{x^n}$$

With

$$\left\{\begin{eqnarray}
a_{n+1}&=-na_n+b_n \\
b_{n+1}&=-a_n-nb_n
\end{eqnarray}\right.$$

We are only interested in $a_n$ since the $b_n$ part will contribute $0$ to $f^{(n)}(1)$. Rewrite $b_n=a_{n+1}+na_n$ from the first relation, then the second can be written:

$$a_{n+2}+(n+1)a_{n+1}=-a_n-n(a_{n+1}+na_n)$$

That is

$$a_{n+2}+(2n+1)a_{n+1}+(n^2+1)a_n=0$$

With $a_0=0$ and $a_1=1$.

Then the Taylor series at $x=1$ is:

$$f(x)=\sum_{n=0}^\infty a_n\frac{(x-1)^n}{n!}$$

We need two solutions of the recurrence equation. Let $a_n=(s)_n=s(s-1)\cdots(s-n+1)$ for some $s$. Then

$$a_{n+2}+(2n+1)a_{n+1}+(n^2+1)a_n=(s)_n(s-n)(s-n-1)+(s)_n(s-n)(2n+1)+(n^2+1)(s)_n\\
=(s)_n\left[(s-n)(s-n-1)+(s-n)(2n+1)+(n^2+1)\right]$$

We want

$$(s-n)(s-n-1)+(s-n)(2n+1)+(n^2+1)=0$$
$$s^2+(-2n-1+2n+1)s+[n(n+1)-n(2n+1)+n^2+1]=0$$
$$s^2+1=0$$

With $s=\pm i$, we get two linearly independent solutions, hence the general solution of the recurrence equation is:

$$a_n=\alpha (i)_n+\beta (-i)_n$$

Now use the initial conditions:

$$\left\{\begin{eqnarray}
a_0&=&0=\alpha+\beta \\
a_1&=&1=i(\alpha-\beta)\end{eqnarray}\right.$$

Hence $\alpha=-i/2$ and $\beta=i/2$.

Finally:

$$\sin\log x=\sum_{n=0}^\infty \frac{i}{2}\frac{(x-1)^n}{n!}\left[(-i)_n-(i)_n\right]$$
$$\sin\log x=\sum_{n=0}^\infty \frac{i}{2}(x-1)^n\left[\binom{-i}{n}-\binom{i}{n}\right]$$

Now to complete the proof, we still need the radius of convergence. It's easy to see that $\binom{\pm i}{n}$ is bounded, so the radius of convergence must be at least $1$. It can't be larger because the function has a singularity at $x=0$.

Write:

$$\binom{i}{n}=\frac{i}{1}\cdot\frac{i-1}{2}\cdots\frac{i-n+1}{n}$$

And for $k\ge1$,

$$\left|\frac{i-k+1}{k}\right|^2=\frac{1+(k-1)^2}{k^2}=1-2\frac{k-1}{k^2}\le1$$

Hence

$$\left|\binom{i}{n}\right|\le1$$

Likewise,

$$\left|\binom{-i}{n}\right|\le1$$

Title: Taylor series of $\sin\ln x$ at $x = 1$

Body:
I tried to discover the Taylor series for and get the answer from Wolfram Alpha, but I couldn't prove the expression provided.

The expression is:

$$\sin\ln x=\sum_{n=0}^{\infty}\frac{1}{2}i(x-1)^n\Bigg(\binom{-i}{n}-\binom{i}{n}\Bigg)$$

How can I prove this relation?


问:
Converting propositional logic formula to CNF
I have been trying to do this all day but I am not getting anywhere with it, could anyone help me?

My formula is

```<code>r → (p ↔ ¬q)
</code>```

And I want to convert it into CNF

When I do it, I get this answer, which does not seem correct:

```<code>(Q ∨ P ∨ ¬R) ∧ (¬P ∨ ¬Q ∨ ¬R)
</code>```

答:
When I do it, I get this answer, which does not seem correct:

(Q ∨ P ∨ ¬R) ∧ (¬P ∨ ¬Q ∨ ¬R)

That is correct.

Suppose $(Q ∨ P ∨ ¬R) ∧ (¬P ∨ ¬Q ∨ ¬R)$ and assume $R$.  The assumption and supposition can both be satisfied when $P$ and $\lnot Q$, or when $\lnot P$ and $Q$ are satisfied: thus $R\to(P\leftrightarrow\lnot Q)$

Suppose $R\to(P\leftrightarrow\lnot Q)$.  This is satisfied when $R$ is falsified or $P\leftrightarrow\lnot Q$ is satisfied.  The later is satisfied at least one from $P,Q$ is satisfied and at least one is falsified.   This: $(\lnot R\lor P\lor Q)\land(\lnot R\lor\lnot P\lor\lnot Q).$


问:
Conjugacy class matrices of SL_2(K)
I've these 2 matrices, with $\lambda, \mu \in \mathbb{K} $
\begin{array}{cc} A=\begin{pmatrix} 1 & \lambda \\ 0 & 1 \end{pmatrix} \qquad B=\begin{pmatrix} 1 & \mu \\ 0 & 1 \end{pmatrix} \end{array}

Proof that these are conjugate iff $\frac{\lambda}{\mu}$ is a square in $\mathbb{K}$

Deduce the number of conjugacy classes of such matrices when $\mathbb{K} =\mathbb{C}, \mathbb{R}, \mathbb{Q}$

I did the first question, but stuck on the second one, "of such matrices" what does that mean ? I don't understand what we are looking for.
I know being on the same class means being on each others orbit, but no clue on what I got asked there

答:
For $\mathbb{C}$, all numbers are squares. So, by your first result, all matrices of this form are conjugate to either

$$\begin{array}{cc} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ and } \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{array}.$$

Similarly, for $\mathbb{R}$, representatives of the conjugacy classes are
$$\begin{array}{cc}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \text{ and } \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \end{array}.$$

Finally, for $\mathbb{Q}$, there are infinitely many classes since every square-free rational number is in a different class. E.g. $1,2,3,5,6,7,10,...$ and these are only the whole numbers.

Title: Conjugacy class matrices of SL_2(K)


问:
Worst Time Complexity of Talks Scheduling
Question is to find the worst case time complexity of a brute-force algorithm for scheduling the talks by examining all possible subsets of talks.

Suppose we have $n$ talks that we want to schedule using greedy algorithm. There is a theorem which says that $n$ elements could give $2^n$ subsets. So, we have a combination of $2^n$ talks. Then we an compare $n\times(n-1)\times2^n$ pairs of talks and thus the worst-case complexity is $\mathcal{O}(n\times(n-1)\times2^n)$.

Each pair of talks in a subset have to be compared against each other to check for overlapping. There are $n$ talks and thus there are $n\times(n-1)$ pairs of (distinct) talks.

Question: Could you please explain why there are $n\times(n-1)$ pairs of (distinct) talks?

答:
Grind through it.

There are $\binom{n}{k}$ subsets of exactly $k$ talks to consider, to check a $k$ set for overlaps means considering $k (k - 1) / 2$ pairs (pick first element in $k$ ways, there are $k - 1$ options for the second one; but the order $a, b$ is the same a $b, a$, thus the $2$). Or just $\binom{k}{2} = k (k - 1) / 2$ pairs.

So in all there are $W$ comparisons:

$\begin{align*}
   W
     &= \sum_{0 \le k \le n} \binom{n}{k} \frac{k (k - 1)}{2}
\end{align*}$

Quite a mouthful.

Consider:

$\begin{align*}
    f_m(x)
      &= (1 + x)^m \\
      &= \sum_{0 \le k \le m} \binom{m}{k} x^k \\
    f_m''(x)
      &= m (m - 1) (1 + x)^{m - 2} \\
      &= m (m - 1) \sum_{0 \le k \le m - 2} \binom{m - 2}{k} k (k - 1) x^{k - 2}
\end{align*}$

Time to relate to your sum:

$\begin{align*}
   \sum_{0 \le k \le n} \binom{n}{k} \frac{k (k - 1)}{2}
     &= \frac{1}{2} \sum_{0 \le k \le n} \binom{n}{k} k (k - 1) \\
     &= \frac{1}{2} f_{n + 2}''(1) \\
     &= \frac{1}{2} (n + 2) (n + 1) \cdot 2^{n + 2} \\
     &= (n + 2) (n + 1) \cdot 2^{n + 1}
\end{align*}$

Now for the $O(\cdot)$ part. It is easy to see that:

$\begin{align*}
   (n + 2) (n + 1) \cdot 2^{n + 1}
      &= O(n^2 \cdot 2^n) \\
      &= O(n (n - 1) \cdot 2^n)
\end{align*}$


答: Hint: Notice that all numbers are $-1 \pmod 8$ except $11$.


问:
How to extend scalars of a MeatAxe $kG$-module $M$ given in GAP?
Let $G$ be a finite group. Suppose $k:=\mathbb{F}_5$, let char$(k)\mid |G|$, and let $M$ be a $kG$-module given in GAP, as in the following example:

```<code>G:=Group((1,2),(1,2,3,4,5));
M:=RegularModule(G,GF(5))[2];
</code>```

Question:

How can one define a new module $N$ in the MeatAxe of GAP such that $N=\mathbb{F}_{625}\otimes_{k} M$ ?

The module above is only an example. I would be interested in doing this with any finite-dimensional $kG$-module in GAP.

Is this already implemented somewhere (via Conway-polynomials) ?

I am using GAP 4.11.0 and have also installed the Shared C MeatAxe (see https://users.fmi.uni-jena.de/~king/SharedMeatAxe/).

Thank you very much for the help.

答:
A MeatAxe module in GAP is defined by the matrices describing an algebra action (the component .generators) and the field. You can simply take the existing matrices of a module and define a new module over a larger field:

```<code>gap> N:=GModuleByMats(M.generators, GF(625));;
</code>```

(In your example all irreducible modules are absolutely irreducible, so you will not see much difference of results.)

Body:
A MeatAxe module in GAP is defined by the matrices describing an algebra action (the component .generators) and the field. You can simply take the existing matrices of a module and define a new module over a larger field:

gap> N:=GModuleByMats(M.generators, GF(625));;

(In your example all irreducible modules are absolutely irreducible, so you will not see much difference of results.)


问:
Prove that there is a continuous $f$ not Riemann-Stieltjes integrable with respect to $\phi$
Let $V[\phi;a,b]=\infty$, show there is a continuous function $f$ such that $$\int_a^b f\ d\phi$$ not exists.

Info: This result seems easy to follow from the results already proved. But I couldn't answered yet.  $V$ represents the total variation of $\phi$, $V=\sup_{\Gamma}\sum_{i}|\phi(x_i)-\phi(x_{i-1})|=\infty$, $\Gamma$ is a partition of $[a,b]$. The hint says use the following two results:

If $V[\phi; a,b]=\infty$. Show there exists a point $z\in[a,b]$ and a monotone sequence $(z_n)_{n\geq 1}$ such that $z_n\to z$ and such that $\sum_{i=1}^{\infty} |\phi(z_i)-\phi(z_{i-1})|=\infty$

Which I have already proved here: There exists $(z_n)_{n\geq 1}$ monotone $z_n\to x_0$ such that $\sum |\phi(z_i)-\phi(z_{i-1})|=\infty$

If $\{a_k\}$ is a sequence of positive terms with $\sum a_k=\infty$, then there is another sequence of positive terms $\{\xi_k\}$ such that $\xi_k\to 0$ and $\sum\xi_k a_k=\infty$.

Any help is appreciated. The complicated hint should be there for a reason, I can't see yet. I tried using for example $f\equiv 1$ but then $\int_a^b fd\phi=
\lim_{m\to\infty}\sum_{i=1}^m (\phi(x_i)-\phi(x_{i-1}))$ this is not precisely $\infty$, at least not immediatly, because here $\phi$ could be any function (including discontinuous).

答:
One solution would be to use Riesz representation theorem, i.e. if the integral is finite for any continuous function then since it is linear it is represented by a function of a bounded variation say $\psi$. Then one would have to prove that if $\int fd\phi = \int f d\psi$ for every continuous function $f$ then $\phi = \psi + constant$.
(Which probably is also part of Riesz's theorem i.e. the function we integrate against is unique).

But going back to the hints in the problem.
You have to construct continuous function $f$ such that
$\sum f(y_i)(\phi(x_{i+1})-\phi(x_i))$ tends to $\infty$.
Ideally, $f(y_i) = sign(\phi(x_{i+1}) - \phi(x_i)$ but such function is not continuous of course.

Here come your hints, take $\xi_k\to 0$ such that $\sum_{i=1}^{\infty} \xi_k|\phi(z_i)-\phi(z_{i-1})|=\infty$ and define $f$ to be $f(z_i) = \xi_k * sign (\phi(x_{i+1}) - \phi(x_i))$. That is, at this moment, $f$ is defined on a convergent sequence only but extend it linearly to continuous function on $[0,1]$.


问:
Conditional probability on a multiple choice test
A student is taking a multiple-choice test. Each question on the test has five possible answers and only one choice is correct. The probability this student knows the correct answer is 70%. If the student does not know the answer, they select an answer at random with each answer having an equal probability of being picked. Calculate the conditional probability that the student knew the answer given they answered the question correctly.

I started off like this:
Let B denote the event that the student knew the answer. Let A denote the event that the student answered the question correctly.

I was able to work out that $P(A)=\frac{3}{10}*\frac{1}{5} + \frac{7}{10} = \frac{19}{25}$

And I know the formula $P(B|A)=\frac{P(A \cap B)}{P(A)}$

I am unsure on how to work out $P(A \cap B)$. Any advice would be greatly apprectiated.

答:
It is easiest to draw a tree for these.

```<code>
know        70%   -   correct   1       .7

                  /   correct   1/5     .06
don't know  30%
                  \   incorrect 4/5     .24
</code>```

Thus $P(A\cap B)=.7$ and your desired probability is $\frac{P(A\cap B)}{P(A)}=\frac{.7}{.76}= \frac{35}{38}$. It is very likely that the student knew the answer if they got it correct.

Body:
It is easiest to draw a tree for these.

know        70%   -   correct   1       .7

                  /   correct   1/5     .06
don't know  30%
                  \   incorrect 4/5     .24

Thus $P(A\cap B)=.7$ and your desired probability is $\frac{P(A\cap B)}{P(A)}=\frac{.7}{.76}= \frac{35}{38}$. It is very likely that the student knew the answer if they got it correct.


问:
Correspondence subring of algebraic integers - subfield of algebraic numbers
Let $K=\overline{\mathbb{Q}}$ be the field of algebraic numbers over $\mathbb{Q}$, and $R=\mathcal{O}_K$ be the ring of integers of $K$.

Let $R' \subset R$ be a subring of $R$, and $K' = \langle R' \rangle \subset K$ be the smallest subfield of $K$ containing $R$.

Question: Is it true than $R' = \mathcal{O}_{K'}$?

答:
Not necessarily. Take $K=\mathbb{Q}(\sqrt{5})$, so $R=\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$. Let $R’=\mathbb{Z}[\sqrt{5}]$. Then $K’=K$, but $R’\neq R$.

Your $K’$ is the field of fractions of $R’$, so the questions is really equivalent to asking whether $R’$ is integrally closed. Not every subring of an integrally closed ring is integrally closed, though.


问:
How would the mathematics relevant to physical theory be different if it didn't use real numbers?
Real numbers assume we can have infinite precision and some of the theory behind them uses infinite processes to establish certain proofs. A small band of mathematicians–eg ultrafinitists–disagree with this and deny the possibility of infinite processes or the logical structure of proofs that require them.

I always assumed that this was an interesting but faddish idea.

But I was recently reading a book by the very influential mathematician and computer scientist Donald Knuth (incidentally one of the popularisers and inventors of surreal numbers, a totally different way of establishing a theory of numbers) and though he didn't doubt the theory of real numbers he made the argument that it is wrong to assume they apply to the real world.

He argues (in chapter 6 of Things a Computer Scientists rarely Talks About, bold highlight mine, italics Knuth):

When I say that the question "finite or infinite?" is a red herring, I don't mean simply that philosophers and theologians have often been arguing about an unimportant issues. I also mean that physicists and scientists fail to realise this. For example, take the literature of chaos theory: Hundreds of papers have been written about the behaviour of solutions to unstable recurrences, by people who assume that real numbers are real.

...But it is a tremendous leap of faith to assume that real numbers apply perfectly to the real world...

...It seems to me that a new branch of physics is needed, called maybe "discrete physics" or something like that, to study the effects of the assumption that parameters can be infinitely precise and to consider instead that the universe probably has only a finite–but extremely large–number of states.

I'm not interested in the philosophical implication or speculations, or whether ultrafinitism has a point. But the question of whether we misunderstand some physics because we assume real numbers apply to the mathematics underpinning physical theory sounds about as important as realising that the universe is not based on flat Euclidian geometry.

So what parts of mathematics relevant to physics would be different if the concept of real numbers doesn't work for physical theory? And what are the implications for physics?

答:
I will try an address the question in a couple ways although as you pointed out, this line of questioning inevitably leads down the Philosophy of Physics rabbit hole, which I am not sure members of this SE would appreciate much.

Firstly, what is physics in the most raw sense? It is the generation of mathematical abstractions that can be used to describe patterns that we see in nature. Over the last century there has been tremendous work done in the area of mathematical abstraction; so the empirical basis of physics is often overlooked(String Theory is guilty of this). But in so far as Physical theories go, they must be consistent with our observations and explain future ones sufficiently. So let's look at paradigm shifting discoveries in Physics; did Einstein say Newton was wrong? Well sort of, but not quite, Newton was an approximation. He couldn't have been completely wrong since objects still fall at constant acceleration, and last I checked we're still in an Elliptic Orbit around the sun. Einstein's work changed our view of the world not because of his cool math tricks, but because he developed a framework that could explain two inconsistent physical theories in a unified way, using cool math tricks.

Now let's look at the alternate formulations of Classical Mechanics done by Hamilton, Euler and Lagrange. Their work was at its core math. They did not explain any new phenomena, or expand the laws of physics in a meaningful way; they developed deeper techniques of mathematical abstraction to express Newton's laws with a further level of generality. Is this really useful for Physics? Yes of course! Is it new Physics? ...sadly no (at least I would say so).

I don't want to reduce the role that discoveries in Math plays in Physics, in fact Einstein and the founders of QM made tremendous use of the Math from their predecessors to develop their new theories of reality. But the Math itself was not enough.

Now that we got all of that out of the way. Where does the existence(or lack thereof) of real numbers come into play? Well it doesn't...unless someone gives us a reason to believe it does. Such a question on the impacts of the relationship of the truth of mathematics to our universe is equivalent to asking "does our world obey Math, or do we use Math to describe it", which is a question that dates back to Aristotle and Plato, and we haven't made much progress since. As far as Physics is concerned Math is a tool, and in so far as Calculus (an inherently continuous field of study) and other tools are useful, let's use them to describe our world.

As to their inherent truth, we don't really care (remember Newton didn't even have a rigorous theory of Calculus, he just kinda made it up as he went along). You may have noticed I said "unless someone gives us a reason to" earlier, and what I mean by that is someone has to make use of the Math to make accurate predictions about our world. It may in fact be that real numbers don't exist, Calculus is flawed, and the ultrafinitists have been wrong all along, they can get a Fields Medal for that! But until someone demonstrates a new theory with new predictions about the world using this ultrafinitism, nobody will be getting a Nobel Prize for it.

Note: In Quantum theory the universe actually does have a minimal length, this is because making smaller and smaller measurements requires more and more energy, and at a certain point it will take so much energy that the act of measurement will create a black hole. This finite minimal length shows us that Quantum Theory inherently implies that the Universe is Discrete, but that does not mean we are banned form using irrational numbers that don't 'technically' exist in the real world. This video describes it quite well https://www.youtube.com/watch?v=nyPdIBnWOCM


问:
Construction of the projective plane over $\mathbb{F}_3$
I have a question about constructing  projective plane over $\mathbb{F}_3$. We first establish seven equivalence classes $P= \{ [0,0,1], [0,1,0], [1,0,0], [0,1,1], [1,1,0], [1,0,1], [1,1,1] \}$.

Given a triple $(a_0, a_1, a_2) \in \mathbb{F}^3_3 \setminus (0, 0, 0)$ we define the line
$L(a_0, a_1, a_2)$ as follows: $L(a_0, a_1, a_2) := \{ [x_0; x_1; x_2] \in P : a_0x_0 + a_1x_1 + a_2x_2 = 0 \}$.

It is quite easy for $L(0,0,1), L(0,1,0), L(1,0,0)$, because here we take the points which have zero on the first, second and third coordinate.

It gets more difficult for me for $L(0,1,1)$, because here we need to have $x_1+x_2=0$. Do we treat the coordinates of the points as elements of $\mathbb{F}_2$ or $\mathbb{F}_3$?

There are $26$ nonzero triples in $\mathbb{F}^3_3 $.

Do we check all 26 $L(x_0, x_1, x_2) $ sets?

Please help, because I really want to understand it.

Thank you.

答:
You are describing a method for getting the Fano plane, which by definition is the projective plane over $\mathbb F_2$.

For getting the projective plane over $\mathbb F_3$ in a similar way, do the following:

Select a set of projective representatives of the nonzero elements of $\mathbb F_3^3$. One way to do this is to select only the vectors whose first nonzero entry is a $1$.
Since $\mathbb F_3$ has $2$ units, you end up with $(3^3 - 1) / 2 = 13$ vectors. Those vectors are called coordinate vectors and give you the $13$ points of the projective plane.

You can also use the coordinate vectors for the description of the lines: Each coordinate vector $v$ corresponds to the line containing all the points $w$ such that the scalar product of $v$ and $w$ equals $0$ (so $\langle v,w\rangle = 0$).
In this way, there are $13$ lines containing $4$ points each.

Typically, the projective plane over a field $K$ is defined in this way:

The subspaces of $K^3$ of dimension $1$ are the points, and the subspaces of $K^3$ of dimension $2$ are the lines.
It may be worth to convince yourself that the above construction is compatible with this description.


答:
Let us denote the events: $A$: answered correctly, $B$: knew the answer.

We are provided that: $\mathsf P(B)=0.70$, $\mathsf P(A\mid B')=0.20$, and $\mathsf P(A\mid B)=1.00$.

NB: Thus ${\mathsf P(A\cap B)~{=\mathsf P(A\mid B)\,\mathsf P(B) \\= \mathsf P(B)\\=0.70}\\\mathsf P(A)\qquad~{=\mathsf P(B)+\mathsf P(A\mid B')\;\mathsf P(B')\\=0.70+0.20\cdot0.30\\=0.76}}$

We seek $\mathsf P(B\mid A)$.   Just use Bayes' Rule and the Law of Total Probability. $$\begin{align}\mathsf P(B\mid A) &=\dfrac{\mathsf P(A\mid B)\,\mathsf P(B)}{\mathsf P(A)} \\[1ex]&=\dfrac{\mathsf P(A\mid B)\,\mathsf P(B)}{\mathsf P(A\mid B)\;\mathsf P(B)+\mathsf P(A\mid B')\;\mathsf P(B')}\\&=\dfrac{35}{38}\end{align}$$

Title: Conditional probability on a multiple choice test

Body:
A student is taking a multiple-choice test. Each question on the test has five possible answers and only one choice is correct. The probability this student knows the correct answer is 70%. If the student does not know the answer, they select an answer at random with each answer having an equal probability of being picked. Calculate the conditional probability that the student knew the answer given they answered the question correctly.

I started off like this:
Let B denote the event that the student knew the answer. Let A denote the event that the student answered the question correctly.

I was able to work out that $P(A)=\frac{3}{10}*\frac{1}{5} + \frac{7}{10} = \frac{19}{25}$

And I know the formula $P(B|A)=\frac{P(A \cap B)}{P(A)}$

I am unsure on how to work out $P(A \cap B)$. Any advice would be greatly apprectiated.


问:
Can I multiply equations of a system of equations by a function without affecting the result?
I have a system of equations as the following:

$$
\left\{\begin{array}{rcl}
f_1(x,y)=0\\
\frac{f_2(x,y)}{g_1(x)}=0\\
f_3(x,y)=0\\
\frac{f_4(x,y)}{g_2(x)}=0
\end{array}
\right.
$$

where $g_1(x)$ and $g_2(x)$ are functions that contain trigonometric functions like $\cos x$ or $\cos^2 x$.

I know that I could multiply each line by a different scalar without affecting the solution. However, how will the solution be affected if I multiply the second line by $g_1(x)$ and the third line by $g_2(x)$ in order to eliminate these terms from the denominator?

答:
Short answer - yes, if you're careful.

Longer answer:

A fraction is $0$ if and only if its numerator is $0$, so you can multiply your fractions by the functions in their denominators as long as those denominators are not $0$. When the functions in the denominator are $0$ the fractions are ambiguous - they may make sense if the numerator vanishes too in such a way as to cause the fraction to have a limit.


问:
Consecutive Percentage Increase-Decrease Problem
If the price of sugar increases by 10% on the first day and decreases by 20% on the second day and increases by 30% on the third day and decreases by 40% on the fourth day and like this it continues on. On which day the price of the sugar will become 1.2 times of the original price?

I came across this Arithmetic question which I was not able to solve mathematically i.e. I was not able to prove mathematically that this is not possible. However I deduced this logically that this is not possible as the percentage by which the price falls is always greater than the percentage by which the price of the sugar increases and hence there will be no such day the price of the sugar will become 1.2 times of the original price?

Can someone please tell me that if there is a mathematical way possible to prove that there could be no such day possible?

Also what will happen if the question is flipped like on the first day the price decreases by 10% and the second day the price increases by 20% and on the third day the price decreases by 30% and on the fourth day it increases by 40% and like this it continues. Can we find the day when the price could be 'x' times of the original price?

Thanks!

答:
$\delta = 10\% = \dfrac {1}{10}$

Then we need to find $n$ such that

$(1+\delta)(1-2\delta)(1+3\delta)\cdots (1\pm n\delta) = 1.2$

Note

$$(1+\delta)(1-2\delta) = 1 - \delta -2\delta^2 < 1$$

$$(1+3\delta)(1-4\delta) = 1 - \delta - 12\delta^2 < 1$$

$$(1+5\delta)(1-6\delta) = 1 - \delta - 30\delta^2 < 1$$

$$\vdots$$

$$(1+(2n-1)\delta)(1-2n\delta) = 1 - \delta - 2n(2n-1)\delta^2 < 1$$

Hence the price of sugar is, overall, decreasing. So it can never get to $1.2$.