问:
Solving $\tau^4 +1=u^2 +v^2$ over the integers
I'm trying to show, that the solutions $t_i$ from Pell's equation $t^2-2s^2=\pm 1$ are not squares (only for the trivial cases). For the even solutions $t_{2i}$ (solutions for the + sign) this is easy, but for the odd solutions $t_{2i+1}$ (solutions for the - sign) this is a little bit harder.

This question leads to the diophantine equation $\tau^4-2s^2=-1$ and the new question, whether it has nontrivial solutions or not.

I do the following approach. First of all I take the equation
$\tau^4+1=u^2+v^2$ (*),
try to find solutions $(u, v, \tau)$ and set in the end $s=u=v$. With

$\begin{align}
u&=u_2 a^2 +u_1 a+u_0 \\
v&=v_2 a^2 +v_1 a+v_0 \\
\tau&=\tau_1 a+\tau_0
\end{align}$

and comparison of the coefficients one gets

$\begin{align}
u&=2a^2 b^2 (b^4 +1)+4ab^3 +1 \\
v&=a^2 (b^8-1)+2ab(b^4 -1)+b^2 \\
\tau&=a(b^4 +1)+b
\end{align}$

with natural numbers $a$ and $b$.

Now the question: are there any other solutions for (*) and if not, why?
Maybe there is also an easy way showing, that the $t_{2i+1}$ are not squares, then I would prefer this proof.

thank you

答:
First, we can find not-all-solutions but  gain insights by re-arranging terms.
$$\tau^4 +1=u^2 +v^2\quad \implies
\tau^4-u^2 =v^2-1\quad \text{or}\quad  \tau^4 -v^2=u^2-1$$

$$\implies (\tau^2)^2-u^2 =v^2-1\quad \text{or}\quad (\tau^2)^2 -v^2=u^2-1\\
\implies u=v=\pm\sqrt{\tau} \quad\lor\quad \tau=u=v=\pm1$$

but this is only a tiny subset of solutions and empirical data might offer more insights. Let

$$\tau = \sqrt[4]{u^2 +v^2-1}$$

$$\big\{(0,1,0)\big\} \\
\big\{(0,0,1), \space  (1,1,1), \space (2,4,1), \space (3,9,1), \space (4,16,1), \space (5,25,1), \space (6,36,1),\space  \cdots\big\}\\
\qquad\qquad\qquad\quad\big\{(2,1,4),\space (3,1,9),\space (4,1,16),\space (5,1,25),\space  (6,1,36)\quad \cdots\big\} $$

$$\big\{(8,56,31), \space (8,31,56)\big\}\quad
\big\{(9,71,39),\space (9,39,71)\big\} \quad
 \big\{(14,191,44), \space (18,321,44)\big\}\quad
 \big\{(10,76,65), \space (10,65,76)\big\} \quad\cdots $$

We see a pattern in many/most solutions but $u,v$ values of $31, 39, 44, 56, 65,71,76,\cdots $ make "patterns" more elusive, except that $u,v$ are interchangeable in many cases. These numbers were derive from tests of $u \le 3000\land v\le 100$. Beyond that, we do have the complimentary cases of
$(14,44,191), \space (18,44,321).$ What remains is the puzzle of the non-squares.


问:
Why does the set of algebraic elements of $K/F$ contain $F$?
Let $K/F$ be a field extension and $\tilde{F}$ the set of elements of $K$ that are algebraic over $F$, i.e.

$$\tilde{F}=\{\alpha \in K \mid \alpha \ \text{algebraic over} \ F \}$$
in my lecture notes for university there is a corollary which states that $\tilde{F}$ is a subfield of $K$ which contains $F$. I'm interested in the last part, i.e. why does $\tilde{F}$ contain $F$? In other words, why is every $x \in F$ algebraic over $F$?

答: Because $x$ is a root of $T-x\in F[T]$.

ContentLicense: CC BY-SA 4.0

Body: Because $x$ is a root of $T-x\in F[T]$.


答: Because for all $\;a\in\Bbb F\;,\;\;P_a(x)=x-a\in\Bbb F[x]\;$ is a polynomial with coefficients in $\;\Bbb F\;$  with $\;a\;$ as a root....

Title: Why does the set of algebraic elements of $K/F$ contain $F$?

Body:
Let $K/F$ be a field extension and $\tilde{F}$ the set of elements of $K$ that are algebraic over $F$, i.e.

$$\tilde{F}=\{\alpha \in K \mid \alpha \ \text{algebraic over} \ F \}$$
in my lecture notes for university there is a corollary which states that $\tilde{F}$ is a subfield of $K$ which contains $F$. I'm interested in the last part, i.e. why does $\tilde{F}$ contain $F$? In other words, why is every $x \in F$ algebraic over $F$?


问:
Finite equivalence class same cardinality
For an equivalence relation $\sim$, if each partition has a finite number of elements, and $X$ is an infinite set, then is it true that $|X/\sim|=|X|$?

I can prove injectivity one way by defining the map $f:X/\sim\rightarrow X$ by picking an element from the equivalence class, but not sure how I would go about the other way.

答:
Assuming the axiom of choice (at least for families of finite sets) then the answer is yes.

The reason that you find it somewhat difficult to continue is exactly this, the axiom of choice is needed and we cannot really do this in an explicit manner. Let me give one of the classical examples.

We say that $S$ is a Russell set if $S$ can be written as a disjoint union of pairs $P_n$ where $n\in\Bbb N$, but $S$ does not have a countably infinite subset. This is equivalent to saying that the product of any infinite set of the pairs is empty, or in simple words: there is no choice function from infinite families of pairs into $S$.

Clearly $\mathscr P=\{P_n\mid n\in\Bbb N\}$ is a partition of $S$ into pairs, which are finite sets, but it is not true that $|\mathscr P|=|S|$. In fact it is not even true that $|\mathscr P|\leq|S|$, which is an even more disturbing fact!

So we really do need the axiom of choice. And this example easily generalize to replace the number of pairs by larger cardinals, in which case we can even save some other limited forms of choice (countable choice, for example), and still have these bizarre sets in our universe.

ContentLicense: CC BY-SA 3.0


问:
There exists a differentiable function for which $f'\left(x\right)=\begin{cases} \frac{\cos x-1}{x^{2}} & x\ne0\\ 1 & x=0 \end{cases}$
Prove or disprove: There exists a differentiable function for which $$f'\left(x\right)=\begin{cases} \frac{\cos x-1}{x^{2}} & x\ne0\\ 1 & x=0 \end{cases}$$

My way:

False. Assume towards a contradiction that there exists such function.

Then we have:

$$f'\left(0\right)=1$$ $$f'\left(2\pi\right)=0$$

By Darboux's theorem, there exists $$0<c<2\pi$$ such that:
$$f'\left(c\right)=\frac{1}{2}$$
Thus:
$$f'\left(c\right)=\frac{1}{2}=\frac{\cos c-1}{c^{2}}\Rightarrow$$ $$\cos c=1+\frac{c^{2}}{2}>1$$
Contradiction!

.

My question is - would it been easier to disprove by integrating and trying to find the original function?

Even if not, could I see the shortest method on this case please?

Thank you.

答:
I think another way of doing it would be showing that the function you have stated has a discontinuinty at $x=0$ and show this by evaluating:
$$\lim_{x\to0}\frac{\cos x-1}{x^2}$$
from above and below if you want to be thorough and then use the fact that no differentiable function exists where its differential has a discontinuity


问:
Null measure implies empty interior
I tried writing a proof for an exercise i found in my Real Analysis book which claims the following:

Problem:

If a set $X \subset \mathbb{R}$ has null measure, then it must have empty interior.

Context/Relevant definitions/Notation:

-> Null measure: One says that $X \subset \mathbb{R}$ has null measure, which we will denote by writing $m(X) = 0$ when, given $\epsilon > 0$ one can obtain a countable collection of open intervals $I_{1}, I_{2}, \ldots, I_{n}, \ldots$ such that:

$$X \subset I_{1} \cup \cdots \cup I_{n} \cup \ldots$$ and $$\sum_{n=1}^{\infty}\left|I_{n}\right|<\varepsilon$$

-> Interior point:

Given $X \subset \mathbb{R}$ a point $x \in X$ is called an interior point when there exists $\epsilon > 0$ such that:

$$
(x-\varepsilon, x+\varepsilon) \subset X
$$

We will denote by $int(X)$ the set of all interior points of $X$.

Attempt:

Given $X \subset \mathbb{R}$, suppose by contradiction that $int(X) \neq \varnothing$. Then, there exists $\alpha \in int(X)$ such that $(x-\bar{\varepsilon}, \alpha+\bar{\varepsilon}) \subset X$ for some $\bar{\varepsilon} > 0$

On the other hand, since $m(X) = 0$, given $\epsilon > 0$ one can obtain a countable collection of open intervals $I_{1}, I_{2}, \ldots, I_{n}, \ldots$ such that:

$$X \subset I_{1} \cup \cdots \cup I_{n} \cup \ldots$$ and $$\sum_{n=1}^{\infty}\left|I_{n}\right|<\varepsilon$$

However, by taking $\epsilon=\frac{\bar{\varepsilon}}{4}>0$ we see that:

$$
(\alpha-\varepsilon, \alpha+\varepsilon) \subset(\alpha-\bar{\varepsilon}, \alpha+\bar{\varepsilon}) \subset X \subset \bigcup_{n \in \mathbb{N}} I_{n}
$$

which means that $\alpha$ is such that: $$\alpha \in
I_{j} \doteq\left(a_j, b_{j}\right)
$$

for some $j \in \mathbb{N}$

Since $\alpha$ is an interior point, it follows that: $$
(\alpha-\bar{\varepsilon}, \alpha+\bar{\varepsilon}) \subset I_{j} = (a_{j}, b_{j})
$$

Finally, since
$\left|I_{j}\right|=b_{j}-a_{j}>(\alpha+\bar{\varepsilon})-(\alpha-\bar{\varepsilon}) = 2\bar{\varepsilon}$ we get a contradiction, as $
\sum_{n \in N}\left|I_{n}\right|>\varepsilon
$, which is impossible, because $m(X) = 0$

I would appreciate if someone could criticize my work and check if the solution is indeed correct!

Thanks in advance, Lucas

答:
If $X$ has non-empty interior in $\Bbb R$ it contains some $x$ and for some $r>0$, $(x-r, x+r) \subseteq X$ from which it follows that

$$\mu X \ge \mu((x-r,x+r)) = 2r > 0$$

Title: Null measure implies empty interior


问:
Show that $f(a+h)=f(a) + hf'(a+t_{h}h)$ and $\lim_{h \to 0} t_{h} = 1/2$
Let $f$ be a function of $C^{2}$ in an interval of the form $[a-h,a+h]$ where $h>0$.

Show that if $f''(a)\ne0$ then there exists $t_{h} \in [-1,1]$ such that $$f(a+h)=f(a) + hf'(a+t_{h}h) \quad, \quad
 \lim_{h \to 0} t_{h} = 1/2$$
The first equality is easy using generalized Rolle's theorem, the limit however is confusing : it seems like it should tend to $0$ since that would give us the definition of the derivative at $a$ ie : $\lim_{h \to 0} \dfrac{f(a+h)-f(a)}{h}= f'(a)$,

Any help would be appreciated!

答:
Since $f'$ is $C^1$, we have the Taylor expansion for $0 \le x \le h$:

$$f'(a+x)=f'(a)+xf''(a)  +o(x)\,.$$
Taking $x=t_h h$ we infer
$$f'(a+t_h h)=f'(a)+t_h hf''(a)+ o(h)\,.$$
Using this in the formula that defined $t_h$ gives
$$f(a+h)=f(a) + hf'(a+t_{h}h)= f(a)+hf'(a)+t_h h^2f''(a)+o(h^2)\,.$$
Now comparing this to the second order Taylor expansion of $f$:
$$f(a+h)=f(a)+hf'(a)+h^2f''(a)/2+o(h^2)$$
yields (since $f''(a) \ne 0$) that $t_h h^2=h^2/2+o(h^2)$, so  $t_h-1/2 \to 0$ as $ h \to 0$.

[1] https://en.wikipedia.org/wiki/Taylor%27s_theorem


问:
How many ways are there to divide 9 distinct candies into groups of 2 or 3?
Recently, I found a question involving dividing distinct candies into groups of specific sizes, and the specific one I found goes as follows:

There are $9$ distinct candies in a large bucket. There are also multiple smaller bags that may contain either $2$ or $3$ candies each. How many ways can the candies from the bucket be divided into the smaller bags? All candies must be divided into a bag, and the smaller bags are all identical to each other.

My initial thoughts was to find all the possible bag combinations. So there would either be $3$ bags each containing $3$ candies, or $3$ bags of $2$ candies and $1$ bag of $3$ candies. But I'm not exactly sure how to proceed in finding the amount of possibilities for either case. Additionally, I figured that this method might take a long time for more complex scenarios, and was wondering if there was any more efficient method or a formula for this kind of a problem. Any ideas?

答: For three bags of three, you choose three candies out of nine for the first bag, three out of six for the second bag, and three out of three for the third bag, giving ${9 \choose 3}{6 \choose 3}{3 \choose 3}=\frac {9!}{3!3!3!}$.  Then as the bags are identical, we don't care what order they were filled in so we have to divide by another $3!$ for that.  The other is attacked similarly but the bag with three candies is distinct now.


问:
Hermitian of A Determinant??
I'm a self-learner working through Gil Strangs Linear Algebra course. In the Chapter on Hermitian, Unitary Matrices, I came across this question:

"  How are the eigenvalues of $A^H$
related to the eigenvalues of the square matrix A?

Answer:

"The eigenvalues of $A^H$ are complex conjugates of the eigenvalues of A: det(A−λI)= 0
gives det($A^H$ − $\overline{λ}I$) = 0

I can't really see why this is the case. I'm assuming he is applying the conjugate transpose to
det(A−λI) ?

If so how, I'm confused as to how the complex conjugate   $[det(A−λI)]^H$ works its way inside the determinant.

I hope the question made sense and appreciate any insight someone might have! -Thanks!

答:
The determinant of the transpose is the transpose of the determinant. So it's all about the conjugate coefficients.

The determinant of a matrix can be written as
$$
\det A=\sum_\sigma \operatorname{sgn}(\sigma)\prod_{j=1}^n a_{j,\sigma(j)}.
$$
Then
$$
\det \overline A=\sum_\sigma \operatorname{sgn}(\sigma)\prod_{j=1}^n \overline{a_{j,\sigma(j)}}
=\overline{\sum_\sigma \operatorname{sgn}(\sigma)\prod_{j=1}^n a_{j,\sigma(j)}}=\overline{\det A}.
$$
It's just the fact that the conjugate preserves sums, products, and real numbers.

Title: Hermitian of A Determinant??


问:
Let $G$ be a group with center $Z<G$. Prove that if every element in $G\setminus Z$ has finite order, then $G$ is periodic.
Here we say that a group $G$ is periodic if every element of $G$ has finite order. $G$ needs not be abelian.

Problem. Let $G$ be a group with center $Z<G$. Prove that if every element in $G\setminus Z$ has finite order, then $G$ is periodic.

My proof is as follows: Assume $z\in Z$ has infinite order.

Select an element $a\in G\setminus Z$. By construction, there exists some $b\in G\setminus Z$ such that $ab\neq ba$.

On the one hand, the element $za$ is not in the center as $$(za)b=z(ab)\neq z(ba)=(zb)a=(bz)a=b(za).$$

On the other hand, the element $za$ also has infinite order: Suppose $(za)^m=(za)^n$. By assumption, $a$ has finite order, say $k$, then we have $$(za)^{mk}=(za)^{nk}\implies z^{mk}a^{mk}=z^{nk}a^{nk}\implies z^{mk}=z^{nk},$$ contradicting the assumption that $z$ has infinite order.

However, by assumption, since $za\in G\setminus Z$, it should have finite order, giving a contradiction.

Therefore, every element in the center $Z$ also has finite order and hence $G$ is periodic. $\square$

Hope anyone can help to check my proof. Different approaches are highly welcomed.

答:
The argument is fine, but perhaps overly complicated; and you don’t actually need to cast it as a proof by contradiction: you can do a direct proof. Note that you never use the assumption that $z$ has infinite order until you are done proving that it has finite order: that’s typical of what I call “fake proofs by contradiction” (fake because they aren’t really proofs by contradiction, they are direct proofs or proofs by contrapositive that are dressed up to appear to be proofs by contradiction).

So, let me point some places where it can be improved:

The argument about $za$ can be done simpler: in general, if $H$ is a subgroup of $G$, $h\in H$, and $g\in G$ with $g\notin H$, then $gh,hg\notin H$. So if $z\in Z$ and $a\in G\setminus Z$, then $za\notin Z$.

Once you know that, we prove that $z$ has finite order directly, essentially how you do it: we know $a$ and $za$ both have finite order, say $n\gt 1$ and $m\gt 1$. Then, since $za=az$, we have:
$$1 = ((za)^m)^n = (za)^{nm} = z^{nm}a^{nm} = z^{nm}(a^n)^m = z^{nm}1^m = z^{nm}.$$
Therefore, $z$ has finite order dividing $nm$.


问:
L'hopital's rule in deriving Cobb-Douglas function from CES production function

$$ \ln(Y) = \ln(A) + \frac{\ln[\alpha K^\gamma + (1-\alpha) L^\gamma]}{\gamma}$$

can be taken to the limit by applying l'Hôpital's rule:

$$\lim_{\gamma\rightarrow 0} \ln(Y) = \ln(A) + \alpha \ln(K) + (1-\alpha) \ln(L).$$

I am not sure how l'Hopital's rule was used - differentiating by $\gamma$ produces some weird results.

答:
Note that $$\frac{d}{d\gamma}\ln[\alpha K^\gamma + (1-\alpha) L^\gamma] = \frac{\alpha K^{\gamma}\ln (K) -(\alpha -1)L^{\gamma}\ln (L)}{\alpha K^{\gamma}-(\alpha-1)L^{\gamma}} $$
 where that fact that $\frac{d}{dx} a^x = a^x \ln(a)$ is used.

Then by L'Hospital, $$\lim_{\gamma\to 0} \ln Y = \ln (A)+ \lim_{\gamma\to 0}\frac{\frac{d}{d\gamma}\ln[\alpha K^\gamma + (1-\alpha) L^\gamma]}{\frac{d}{d\gamma} \gamma}.$$


问:
If $a_n=100a_{n−1}+134$ , find least value of $n$ for which $a_n$ is divisible by $99$

Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$

So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$

But after that what i did is to multiply both sides by $9$ in the congruence

$35n - 11 \equiv 0 \pmod{99}$

$315n \equiv 0 \pmod{99}$

$18n \equiv 0 \pmod{99}$

dividing by $18$ ,we get
$n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?

thankyou

答:
$35n\equiv11\bmod99$ could also be solved with the Chinese remainder theorem.

It's equivalent to $-n\equiv2\bmod9$ and $n\equiv0\bmod11$.

Can you find a multiple of $11$ between $0$ and $99$ that is $2$ less than a multiple of $9$?


问:
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$.

I drew the picture but I can't seem to figure out the way to start.. If someone could give me a hint on how to start it would mean a lot..

答:
With cylindrical coordinates $x= r\cos \theta$ and $z=r \sin\theta$, the surface between the two planes can be integrated along the cross-section circle of radius $r=1$
$$A= \int_0^{2\pi} (y_2-y_1)\cdot 1\cdot d\theta=
 \int_0^{2\pi} ((4-\cos\theta )-(\sin\theta -4))d\theta=16\pi
$$


问:
How many 10 letter 'words' can be made with no repeated letters and at least 8 consonants?
Need a little bit of help with this one.

How many 10 letter 'words' can be constructed which have no repeated letters and at least 8 consonants?

where 'word' is defined as any combination of the 26 letters of the english alphabet, and by consonant, (obviously) a letter of the english alphabet that are not either a,i,u,e, or o.

This is where I am at:

We need to choose the positions for
the 8 consonants which is $c(10,8)$

Then we need to choose the 8
consonants which I think is
$p(21,8)$? (not sure about this one)

Choose the remaining 2 letters: We have already used 8 letters so we only have 18 letters to choose from so $p(18, 2)$

What else needs to be calculated? Do we need to choose the position of the remaining 2 letters?

答:
Your argument has a couple of problems:

You don't specify where the last two letters selected go; you pick the two letters, but you never place them; placing them in different order gives you different solution.

Likewise, you never say how you will distribute the 8 consonants you pick; you select the 8 locations, but you don't say which letter goes where.

Finally, your solution counts some words too many times. Say you first select the eight consonants b,c,d,f,g,h,j,k and the first eight positions for them (in that order). Then you select two more letters, say m,n, and place them in order in positions nine and ten. You will "count" this word again if your eight selected consonants are b,c,d,f,g,h,m,n, the eight positions selected are positions one through six, nine, and ten (in that order); and then the other two letters you select are j and k and you place them in positions seven and eight. You will count this word several times through your process.

So: perhaps it's better if we start by counting how many 10 letter strings have exactly eight consonants and no repeated letters. My recommendation:

First select the eight positions that will contain the consonants.

Then select the consonants and place them in those positions in order (that is, which consonant you select first matters, because that will go in the leftmost selected position).

Finally, select the two vowels that will fill the remaining two positions, again in order.

Then count how many of your strings have exactly 9 consonants in a similar manner; then how many have all letters consonants (that will be the simplest one).


问:
Known Equality in Normal Distribution?
Can someone explain to me the following equality in general normal distribution?

The left part is equal to $\phi(a)=\phi((a-\mu)/\sigma)$

The original version:

My lecturer just wrote it and said it's easy to concluded that...

答:
If $X \sim N(\mu, \sigma^2)$ then $\displaystyle  \frac{X - \mu}{\sigma} = Z \sim N(0,1)$

Or

If $Z \sim N(0,1)$ then $X = \sigma Z + \mu \sim N(\mu, \sigma^2)$

So $\displaystyle \Phi (x) = P(X \le x) = P(\sigma Z + \mu \le x) = P(Z \le \frac{x - \mu}{\sigma}) = \Phi(\frac{x - \mu}{\sigma})$

Title: Known Equality in Normal Distribution?


问:
Find the function p(x)
Solve for $p(x)$:

$$(x-1)\cdot p(x)=x^3+x-2$$

I made this expression:

$$p(x)=\frac{x^3+x-2}{x-1}$$

But no matter what I tried I couldn't simplify this further. I sat with this problem for over 2 hours and could not find a solution. I looked online for an explanation but could not find it. Can you please explain  in detail how and why I can simplify this further. I found the solution but it didn't explain why that was the solution. Please note that this is not my math homework, I just want to know how to solve this problem.

答:
Of course, you can simply use a polynomial division algorithm. Alternatively, by the phrasing of the question, it is obvious it is likely able to be simplified:

$(x^{3}+x-2)=(x-1)(x^{2}+...)$ start this way to handle the $x^{3}$.

Now we want to cancel out the $x^{2}$ term since the $LHS$ does not have one. The only way is to add $x$ which gives us $-x^{2}$ and a $-x$ to take care of. The constant that will add back $2x$ and coincidentally a $-2$ in the final expression to give $+x$ and $-2$ in the final expression.

In questions like this, one can repeat a similar process as the question likely implies a proper divisor.

Title: Find the function p(x)


问:
If a polynomial $P$ has only real roots, so does $P'-2xP$
Let $P$ be a polynomial such that all of the roots of $P$ are real. Prove that all the roots of $P'-2xP$ are also real. I know a proof for this fact but it is very computational and messy. Is there an elegant solution?

答:
May assume that $P$ has $n$ distinct roots $a_1< \ldots < a_n$, otherwise the multiple roots are also roots of the new equation.  The equation is equivalent to $\frac{P'(x)}{P(x)} = 2 x$, that is
$$\sum_{i=1}^n \frac{1}{x-a_i} = 2x$$
On each interval $(a_i, a_{i+1})$ the LHS decreases from $+\infty$ to $-\infty$, while the  RHS increases between two finite values. Therefore, the equality is achieved exactly once in each $(a_i, a_{i+1})$. Now, on the interval $(a_{n}, \infty)$, LHS decreases from $+\infty$ to $0$, while RHS increases from a finite value to $+ \infty$. So we have exactly one root in $(a_{n}, \infty)$. Similarly in $(-\infty, a_1)$. We conclude that the equation has exactly $n+1$ solutions.  Note that we can substitute $2 x$ with any continuous increasing function with infinite limits. The case of a constant is also interesting.
The picture for  $P(x)=(x+1)(x-1)(x-2)$ is below.

Note: @Hagen von Eitzen: 's method works for any functions $P(x)$, since
$(e^{-L(x)}P(x))' = e^{-L(x)}( P'(x) - l(x) P(x))$, so we do get a root ( an odd number) between two (consecutive) roots of $P(x)$. If $e^{-L(x)}P(x)$ is $0$ at $\pm \infty$ $  we do get extra roots.

Does our method work for functions $P(x)$ that are not polynomial? Say $a$, $b$ are consecutive simple roots of $P(x)$. Then $P'(a)$, $P'(b)$ are of opposite value so on the interval $(a,b)$ the function $\frac{P'(x)}{P(x)}$ ranges from $+\infty$ to $-\infty$.  At $\pm \infty$ it's about the behaviour of $\frac{P'(x)}{P(x)}$. The method seems to still work.

$\bf{Added:}$ Say we are want to show that $(x-\alpha) P'(x) + \lambda P(x)=0$ has all roots real ( $\lambda> 0$ or $\lambda < -n$) (Laguerre).  Consider the derivative of $|x-\alpha|^{\lambda} P(x)$.  Or we can do the analysis with $\frac{P'(x)}{P(x)}$.


问:
A Prime $\mathcal P$-filter is contained in a unique $\mathcal P$-ultrafilter?
Some backround:

Let $\mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $\mathcal P$ then $P_1\cap P_2$ and $P_1\cup P_2$ belong to $\mathcal P$. A $\mathcal P$-filter $\mathcal F$  is a collection of nonempty  elements of $\mathcal P$ closed for finite intersections and such that for any $P_1\in \mathcal F$ and $P_1\subseteq P_2\in \mathcal P$ we have $P_2\in \mathcal F$.

A $\mathcal P$-filter $\mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $\mathcal P$ and $P_1\cup P_2\in \mathcal F$, then $P_1\in \mathcal F$ or $P_2\in \mathcal F$. A $\mathcal P$-ultrafilter is just a maximal $\mathcal P$-filter.

My question is, is every prime $\mathcal P$-filter contained in a unique $\mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.

I have proved that if $\mathcal F$ is a $\mathcal P$-ultrafilter and $P\in \mathcal P$ is such that $P\cap F\neq \emptyset$ for all $F\in \mathcal F$, then $P\in \mathcal F$. I think this must be used in the proof but I don't know how.

All hints are appreciated.

答:
In general it's false, I think, even for finite collections.

The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:

Let $X = \mathbb{R}$, say, and $\mathcal{P} = \{[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]\}$ (the lattice diagram is a "double diamond").

Then $\mathcal{F} = \{[0,9], [3,9]\}$ is a prime filter, but both $\mathcal{U} = \mathcal{P}\setminus \{[3,5],[4,5]\}$ and $\mathcal{U}' = \mathcal{P} \setminus \{[4,6],[4,5]\}$ are ultrafilters extending $\mathcal{F}$.


问:
Retrieve variable from modulo equation
I do have the following equation

$x \mod C = A-B$

where $x$ is a variable, while $A$, $B$ and $C$ are parameters. This equation seems quite simple, but I cannot find a way to solve it and get $x$. Any suggestions?

答:
First let's clarify the difference between the two notions I was asking about in my comment. Given integers $a,b,c$ we can ask whether $a$ is congruent to $b$ modulo $c$, often denoted
$$
a \equiv b \mod c
$$
This is equivalent to saying that there exists some integer $k$ such that
$$
a-b = c\cdot k
$$
Note that I am using an equal sign here, and I used the congruence (three lines) in the previous equation.

The other notion, which you may be more familiar with especially if you're coming from a computer science background is the reduction operation, sometimes denoted with $\mathbin\%$ i.e.
$$
a\mathbin\%c = b
$$
Which we take to mean that there exists an integer $k$ such that $0 \leq a+c\cdot k < c$ and that
$$
a+ck = b
$$
These are very interrelated notions, but they are not the same. The you can think of the first idea, congruence modulo $c$ as a kind of loosening of the definition of equality and the second idea, the reduction operation as an operation on the integers. To see how they can be different, note that if $a = 2, b =5,$ and $ c =3$ then $a \equiv b \mod c$ since
$$
2 - 5 = (-1) \cdot 3
$$
However,
$$
2 \mathbin\% 3 = 2 \neq 5
$$
This is a distinction well worth understanding. Based on how you phrased your original question and how you answered my question in the comments, I take it to mean that you mean the reduction operation, i.e. you want to find solutions to
$$
(x\mathbin\%c) = a-b
$$
If $a-b<0$ or if $a-b\geq c$ then there are no solutions since $0\leq x\mathbin\%c < c$. If $0\leq a-b < c$, then you have an infinitely set of solutions namely the equation holds for any $x$ of the form $a-b + k\cdot c$ for all integer values of $k$.

Hope that answers your question.

Title: Retrieve variable from modulo equation


Title: Find the function p(x)

Body:
Solve for $p(x)$:

$$(x-1)\cdot p(x)=x^3+x-2$$

I made this expression:

$$p(x)=\frac{x^3+x-2}{x-1}$$

But no matter what I tried I couldn't simplify this further. I sat with this problem for over 2 hours and could not find a solution. I looked online for an explanation but could not find it. Can you please explain  in detail how and why I can simplify this further. I found the solution but it didn't explain why that was the solution. Please note that this is not my math homework, I just want to know how to solve this problem.

Body:
Consider $p(x+1) = \frac{(x+1)^3+(x+1)-2}{x} = x^2+3x+4$

So $p(x) = (x-1)^2 +3(x-1)+4 = x^2+x+2$


Title: Find the function p(x)

Body:
if x-1=0 then x=1 thus $x^3$+x-2=1+1-2=0 thus (x-1)|($x^3$+x-2)

Now since we know it's divisible by (x-1) we do the division:

$x^3$+x-2 - $x^2$ (x-1) = -$x^2$+x-2

-x^2+x-2 + x (x-1) = 2x-2

2x-2 - 2 (x-1) = 0

thus $\frac{x^3+x-2}{x-1} = x^2-x+2$


问:
K Theory: Book Recommendations
Good people!

So I've been hoping to get into K Theory for a while now, and the book that I have been trying to use (and failing) has been Charles Weibel's book by that very title.

The book itself isn't bad, but it's not for the faint-hearted, and can appear quite dense and challenging to a relative newbie like me. As such, I was wondering if someone might be able to recommend a book or two or three or $n$ to help "ease" my way into Weibel. Something that touches upon the constructions he introduces in the first chapter--Picard groups, topological and algebraic vector bundles, Chern classes, etc.--but being more, well, easy for a newbie to process.

I look forward to your suggestions.

答:
Other than the above recommendations, I would also recommend this paper by Grayson, which gives a big picture overview of Quillen's $K$-theory.

If and when you can understand Grayson's papers above, I would highly recommend Quillen's original paper Higher algebraic K-theory I. While it is certainly not for beginners, I think it still is something which is quite readable with the correct background.

Title: K Theory: Book Recommendations


问:
Show that the number of ways of splitting a set with $n$ elements into $k$ non-empty subsets is equal to $x^n$'s coefficient in a series
The series evaluates to (could not put it in the title): $$\frac{x^{k}}{(1-x)(1-2x)...(1-kx)}$$ I tried to come up with something. A solution would probably require a use of the geometric series formula as well as Cauchy's product. The task also demands one to deduce the following formula for the Stirling numbers of the second kind:

$$S(n, 3) = \frac{3^{n-1} - 2^{n} + 1}{2}$$ which does show up as a coefficient after multiplying: $$({\sum_{n = 0}^{\infty}x^{n}})({\sum_{n = 0}^{\infty}(2x)^{n}})({\sum_{n = 0}^{\infty}(3x)^{n}}).$$ I guess I should be able to follow from there, yet I am stuck.

Edit: Thank you for the link as well as the proof. Is it possible to proceed from my suggestion about using the geometric series formula?

答:
Let $S(n,k)$  be the number of partitions requested.  Let $R(n,k)$ denote the coefficient of $x^n$ in the given series $F_k(x)$. Clearly $S(0,0)=1= R(0,0)$ and for $n,k>0$ we have $S(n,0)=0=R(n,0)$ and $S(0,k)=0=R(0,k)$. One way to generate a partition of
$\{1,...,n+1\}$ into $k$ nonempty sets is to make $\{n+1\}$ a singleton and   partition   $\{1,...,n\}$ into $k-1$ nonempty sets. The only other way is to start with partition of $\{1,...,n\}$ into $k$ nonempty sets, and adjoin $n+1$ to one of these $k$ sets.  This gives the recurrence formula $$S(n+1,k)=S(n,k-1)+kS(n,k) \,. \quad (*)$$
On the other hand
$$F_k(x)=\sum_{j=1}^{\infty} k^{j-1}x^j F_{k-1}(x)$$
implies that
$$R(n,k)=\sum_{j=1}^{n+1-k} k^{j-1}R(n-j,k-1) \,. \; \quad (**)$$
Writing $i=j+1$ we have
$$R(n,k-1)+kR(n,k)=R(n,k-1)+\sum_{i=2}^{n+2-k} k^{i-1}R(n+1-i,k-1) =R(n+1,k)\,,$$
where the rightmost equality follows from (**) with $n$ replaced by $n+1$.
Comparing this to (*) we conclude by induction on $n$ that $R(n,k)=S(n,k)$ for all $n,k$.


问:
A probability problem in discrete times.

Assume that $X_1,X_2....$ are random variables uniformly distributed in $(0,1)$ and we write $N(x)=\min\{k \in \Bbb{N}:X_1+...+X_k \geq x\}$ and $G_n(x)=P(N(x) \geq n)$.

Compute $P(N(x) \geq x),E[N(x)],Var[N(x)]$ and $P(N(x)=n)$

I managed to prove the relation $G_n(x)=\int_0^x G_{n-1}(x-y)dy$ using the convolution of distribution functions and i believe that this relation will help me   find $P(N(x) \geq x)$ but i cannot continue any further!

How can i compute these quantities?

Can someone help me with this?

Thank you in advance.

答:
For $t\leq1$, note that $N(t)>n$ iff $X_1+\dots+X_n<t$. So $\mathbb P(N(t)\geq n)$ is the volume of an $n$-dimensional simplex with side length $t$, which is $\frac{t^n}{n!}$. So
$$\mathbb E[N(t)]=\sum_{n=0}^\infty\mathbb P(N(t)>n)=\sum_{n=0}^\infty\frac{t^n}{n!}=e^t.$$
Also $\mathbb P(N(t)=n)=\mathbb P(N(t)>n-1)-\mathbb P(N(t)>n)=\frac{t^{n-1}(n-t)}{n!}$.

This is all for $t\leq1$. For $t>1$, things get messy fast.

An alternative way to find $\mathbb E[N(t)]$. Assume $X_1=x$. If $x>t$, then $N(t)=1$. If $x<t$, then $N(t)=1+N'$, where $N'$ has the distribution of $N(t-x)$. So
$$m(t)=\mathbb E[N(t)]=1+\int_0^tm(t-x)\mathop{}\!\mathrm{d}x=1+\int_0^tm(x)\mathop{}\!\mathrm{d}x.$$
So $m'=m$, and as $m(0)=1$, we again get $m(t)=e^t$, for $t\leq1$.


问:
How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

I started out by arranging the 5 different dogs which is $5!$, and from here basically I got stuck, I tried to do it by cases, like if I arranged the 5 dogs, I have 6 places to put the cat in,starting from the right, the rat also can be put in 6 places (can be near the cat from the left), and moving one to the left each case, so I got: $6*6+5*5+4*4+3*3+2*2+1*1=91$ and this multiplied by the number of ways to arrange the dogs gave me the answer: $91*5!=10920$, now this seems clearly wrong and I messed up since $7!=5040$, which is the number of ways to arrange all seven, I am trying to understand my mistakes and how to deal with this question.

Thanks in advance to any help.

答:
I have 6 places to put the cat in, starting from the right, the rat also can be put in 6 places (can be near the cat from the left), and moving one to the left each case, so I got: 6∗6+5∗5+4∗4+3∗3+2∗2+1∗1=91

Multiplying is correct only when the two choices are independent. You have six options for where to put the cat, and six options for where to put the rat, but you don't have both choices at the same time. If you put the cat all the way to the right, there are six ways to put the rat to the left of the cat. However, there is only one way to put the cat all the way to the right. So the total number of configurations where the cat is all the way to the right is $1*6$, not $6*6$. So you end up with $6+5+4+3+2+1 = 21$. When you multiply that by $5!$, you get $2520$, which matches the other answers.


问:
How to get the maximum perimeter in a set of straight squares?
The problem is as follows:

Rudy has a toy set which consists of $11$ squares. All of them are equal
in size. The lenght of each edge is equal to $\textrm{1 cm.}$ Assuming
he's been given the instruction to make with those, plane figures that
are made up of squares joined by a complete side. If at the end of the
day he succeeds and builds one such figures whose perimeter is
maximum. What is the value of such perimeter?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{24 cm}\\
2.&\textrm{26 cm}\\
3.&\textrm{28 cm}\\
4.&\textrm{22 cm}\\
\end{array}$

I'm not very sure on this question. The situation is that the kid has been given the task to make different figures using 11 squares, given that these must share a full side or in other words two squares must be glued together not counting corners then he succeeds into making those figures. But it just happens that he makes one peculiar figure which has the largest possible perimeter.

But how to get that?. My first guess is that the only possible way to make a figure given those conditions is to put the squares one next to each other.

Given this it would follow that:

$1*11*2+2=24\,cm$

But I don't know if this should be the right answer?. Is it possible to make a larger figure with those requirements?. Is there any way to ensure that's the only way to maximize the perimeter?.

Can someone help me here?. Please try to include a figure in the answer I don't know but it seems that this may need it to justify the situation.

答:
You would think the greatest perimeter would be when they are all in a row. That way, there would be $11$ on each side plus $2$ ends for a total of $24$.

If you arrange them as in the picture below where each internal corner counts as $2$ sides, you can see that we have a total of $9$ corners, $2$ tops, $1$ bottom, and $3$ ends for a total of $24$

I believe that, no matter how you arrange them, the greatest perimeter is $24$ because the total number of sides is $44$, each $9$ of them must have $2$ side hidden for $-18$ and $2$ will have only $1$ side hidden for $-2$ giving $\quad 44-18-2=24.\quad $ Other arrangements yield the same MAX.


问:
Conditional probability - dice problem
Two dice are rolled. Let X be the lower of the two scores and Y  be the larger of the scores. What is the conditional probability
P(Y=6|X=4)?</p>

答:
I am taking the assumption that $Y > X$ always based off of wording in OP's statement: 'Let X be the lower of the two scores and Y be the larger of the scores.'

We are given the information that $Y > X$ and now we want $P(Y = 6 \mid X=4)$.

This takes us to a subset of the sample space: $\{ 5,6\} \subset \{1,2,3,4,5,6\}$. Either value is just as likely; namely, it reduces to $Y$ is uniform over $\{5,6\}$ given this new information thus $\displaystyle P(Y=6 \mid X=4) = \frac{1}{2}$.

More generally,

If we are given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ if we have a $B \in \mathcal{F}$ such that $\mathbb{P}(B) > 0$.

Then we may define a new probability space: $(\Omega, \mathcal{F}, \mathbb{Q})$ where $\forall A \in \mathcal{F}, \mathbb{Q}(A) = \mathbb{P}(A \mid B)$.

Since we know $Y > X$ and $\mathbb{P}(X=4) = \frac{1}{6} > 0 $ we can do the above transformation to a new space.

Additional note:
Sometimes trying to solve by $\displaystyle P(A\mid B) = \frac{P(A,B)}{P(B)} $ doesn't work so well when there's a dependence. So it is better to consider what does $P(A \mid B)$ mean given some information that relates $A$ and $B$.

How this relates to the above is let $A : = Y = 6$ and $B : = X = 4$ then we should be using $P(A,B) = P(A\mid B)P(B)$ form of the above since we have a dependency here; namely, that $Y > X$.

Title: Conditional probability - dice problem


问:
Sufficient condition for the sum of two closed subspaces of a Banach space to be closed.
Let $X$ be a Banach space and $Y$ and $Z$ its closed subspaces such that $Y \cap Z = \{0\},$ and $$k = \inf \{||y-z|| \ | \ y\in Y, z \in Z, ||y||= ||z||= 1 \}.$$

I need to show that if $k > 0$ then $Y+Z$ is closed. I managed to show the other way around, however here I am stuck. If anyone has any idea how to proceed I would be really greatfull.

答:
Let us first prove that there is a constant $C\geq 0$, such that
$$
  \|y\|\leq C\|y-z\|,  \quad\forall y\in  Y, \quad\forall z\in  Z.
  \tag 1
  $$

Arguing by contradiction we could otherwise find sequences $\{y_n\}_n$ and $\{z_n\}_n$ in $Y$ and $Z$,  respectively,  such that
$\|y_n\|=1$,  while $\|y_n-z_n\|\to 0$.
Observe  that
$$
  \big|\|z_n\|-1\big| =    \big|\|z_n\|-\|y_n\|\big| \leq   \|z_n-y_n\| \to 0,
  $$
so, defining $z'_n = z_n/\|z_n\|$, one has that $\|z_n-z'_n\|\to 0$.

Since $\|y_n\|=\|z'_n\|=1$, we have by hypothesis that
$$
  k\leq \|y_n-z'_n\|\leq   \|y_n-z_n\|+   \|z_n-z'_n\| \to 0,
  $$
a contradiction, hence proving (1)

Replacing $z$ by $-z$ in (1) one also gets
$$
  \|y\|\leq C\|y+z\|,  \quad\forall y\in  Y, \quad\forall z\in  Z.
  \tag 2
  $$

Similarly one proves the existence of a constant $D\geq 0$, such that
$$
  \|z\|\leq D\|y+z\|,  \quad\forall y\in  Y, \quad\forall z\in  Z.
  \tag 3
  $$
Now, to prove that $Y+Z$ is closed, let $x$ be a point in the closure, so we may write
$$
  x=\lim_{n\to \infty }y_n+z_n,
  $$
with $\{y_n\}_n\subseteq Y$, and $\{z_n\}_n\subseteq Z$.
Clearly $\{y_n+z_n\}_n$ is a Cauchy sequence,  whence so are $\{y_n\}_n$ and $\{z_n\}_n$  by (2) and (3),  so we may define
$$
  y=\lim_{n\to \infty }y_n,  \quad\text{and}\quad z=\lim_{n\to \infty }z_n.
  $$
It then follows that
$$
  x = \lim_{n\to \infty }y_n+z_n = \lim_{n\to \infty }y_n + \lim_{n\to \infty }z_n =  y+z \in  Y+Z.
  $$


问:
Problem Solving a Differntial Inequality.
I am having problems solving the following differential inequality.
$$\begin{cases}\alpha r^2 f(r) - \frac{r^2 - \ell^2 M}{\ell^2}\left(f'(r)\right)^2 > 0\\
f(l \sqrt{M}) = 0
 \end{cases}, \qquad r> \ell\sqrt{M}>0$$
I tried to solve first for the equality obtaining,
$$ f(r) = \frac{\ell^2 \alpha}{4} \left( r^2 - \ell^2 M\right).$$
Then, I try to said that $$f(r) > \frac{\ell^2 \alpha}{4}(r^2 - \ell^2 M).$$
However, I am not sure if this is correct or if it exist another way to solve this kind of problems.

答:
In fact, you can do this way. The inequality is
$$ \frac{r^2 - \ell^2 M}{\ell^2}\left(f'(r)\right)^2 <\alpha r^2 f(r) $$
from which, one has $f(r)>0$ for $r>\ell\sqrt M$. So
$$ -\frac{\alpha r\ell}{\sqrt{r^2-\ell^2}}<\frac{f'(r)}{\sqrt{f(r)}}<\frac{\alpha r\ell}{\sqrt{r^2-\ell^2}}. $$
Integrating from $\ell\sqrt M$ to $r$ gives
$$ -\alpha\ell\sqrt{r^2-\ell^2}\bigg|^{r}_{s=\ell\sqrt M}\le2\sqrt{f(s)}\bigg|^{r}_{s=\ell\sqrt M}\le\alpha\ell\sqrt{r^2-\ell^2}\bigg|^{r}_{s=\ell\sqrt M} $$
or
$$ 0\le \sqrt{f(r)}\le \frac12\alpha\ell(\sqrt{r^2-\ell^2}-\ell\sqrt{M-1}). $$
So
$$ 0\le f(r) \le \frac14\alpha^2\ell^2(\sqrt{r^2-\ell^2}-\ell\sqrt{M-1})^2. $$


问:
If $Z = X + Y$ where $X$ and $Y$ are two independent, uniform random variables, then what is the conditional distribution of $X | Z$?
So I have been struggling with this question for a while. Suppose $X$ is uniformly distributed over an interval $(a, b)$ and $Y$ is uniformly distributed over $(-\sigma, \sigma)$, $X$ and $Y$ are independent. Consider a random variable $Z = X + Y$. What would be the conditional distribution of $X | Z$, i.e. $F_{X|Z}(x)$?

I know that $F_X(x) = \frac{x - a}{b-a}$ and $F_y(y) = \frac{y + \sigma}{2\sigma}$ but I am having trouble figuring out $F_{X|Z}(x)$. Could anyone help me? My gut says that $X|Z$ should be uniformly distributed over $(Z - \sigma, Z + \sigma)$ but I can't figure out how to prove this.

答:
$X|Z$ can't, in general, be uniformly distributed over $(z-\sigma,z+\sigma)$ because there are values of $z$ such that $(z-\sigma,z+\sigma)$ is not a subset of $(a,b)$. (Note that $Z$ is a random variable, $z$ is real number drawn from its distribution. $Z-\sigma$ is not a real number, because $Z$ is a random variable, not a real number. A random variable over the real numbers is different from a real number.)

Draw a rectangle with horizontal dimension $(a,b)$ and vertical $(-\sigma,\sigma)$. Then draw diagonal lines with slope $-1$ in the rectangle. Each diagonal line represents a constant value of $z$. For a particular $z_0$, $X$ will be uniformly distributed along the horizontal extension of the corresponding diagonal line. To find $X|Z$, you have to take several cases depending on whether $z<a$, $z>b$, $z<-\sigma$, $z>\sigma$.


问:
Binomial Probability employees problem
A company has $500$ employees, and $60\%$ of them have children.

Suppose that we randomly select 4 of these employees.

What is the probability that exactly 3 of the 4 employees selected have children?

We know that 300 of these employees have children.

I tried to figure it out, how to work with the binomial equation by the given's

答:
The probability that the first person picked has kids is $\frac{300}{500}$. Conditional on that, the probability that the second person picked has kids is $\frac{299}{499}$. Conditional on all that, the probability the third person has kids is $\frac{298}{498}$. Conditional on all that, the probability the fourth person doesn't have kids is $\frac{200}{497}$. So multiply these together and scale by $4$ (since the person without kids could have been any of the four).

An alternative perspective: the number of quadruples where exactly three have kids is $\binom{300}{3}\cdot\binom{200}{1}$. There are $\binom{500}{4}$ quadruples in total, so divide the two to get the probability.

Title: Binomial Probability employees problem


问:
How to evaluate $\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1}, ~~f_k=k/n$
Very often the summations at pre-degree level are done by using differencing-telescoping and by some  symmetry properties. The following summation is fabricated keeping one thing in mind, I may tell it later. $$\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1},~~ f_k=k/n.$$

The question is: How will you evaluate this sum?

答:
I ended up with a similar solution to that of Albus Dumbledore but I wonder why Ablus didn't make it more obvious.

Notice that the numerator invites for a third order binomial manipulation:

$$3(f^2_k-f_k)+1 = f_k^3 + (1-f_k)^3 = f_k^3 + f_{n-k}^3$$

Therefore
$$S = \sum_{k=0}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1} = \sum_{k=0}^{n} \frac{f^3_k}{f_k^3 + f_{n-k}^3} \\
= \frac 12 \sum_{k=0}^{n} \left(\frac{f^3_k}{f_k^3 + f_{n-k}^3} + \frac{f^3_{n-k}}{f_k^3 + f_{n-k}^3} \right)= \frac 12 (n+1).\blacksquare$$


问:
Ring Inside an Algebraic Field Extension
Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?</p>

答:
A more general argument can be given: let $A\subset B$ be two integral domains. An element $b\in B$  is said to be integral over $A$ if there exists a monic polynomial $p(t)\in A[t]$ with $p(b)=0$. The ring $B$ is said to be integral over $A$ is every element in $B$ is integral over $A$.

Theorem: Let $A\subset B$ be I.D.'s , with $B$ integral over $A$. Then, $A$ is a field iff $B$ is a field.


问:
Sum of the series $\sum {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$
Evaluate the sum of the series:
$\displaystyle \sum_{n=0}^\infty {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$

Ratio test confirms convergence of the series. But what's the exact value of the series ?
Computer programming approximates it somewhere near $\frac 12$.

答:
$\displaystyle \sum_{n=0}^\infty \textstyle{(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}=\displaystyle {-1\over 2\pi}\sum_{n=0}^\infty \textstyle {(-1)^n2n\over (2n+1)!}\cdot \pi^{2n+1}$
$\pi \cos \pi-\sin \pi=\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n(2n+1) \over (2n+1)!}\cdot \pi^{2n+1}-\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n \over (2n+1)!}\cdot \pi^{2n+1}$
$=\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n2n\over (2n+1)!}\cdot \pi^{2n+1}$

Hence $\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}={-1\over 2\pi}(\pi \cos \pi-\sin \pi) =\frac 12$


问:
On the Hahn-Banach Theorem
I just started a self-study of functional Analysis from "Functional analysis, Sobolev spaces and partial differential equations" by Haim Brezis. I struggle to understand the underlying assumptions for the proof of Hahn-Banach theorem on the extension of function. In brief, the theorem is written as follows

Let $p : E \rightarrow \mathbb{R} $ be a Minkowski function.

Let $G \subset E$, $g : G \rightarrow \mathbb{R} $ such that $g \leq p $ is a linear function.

Then there exists a linear function $f$ such that $f=g,  \forall x \in G $ and $f \leq p, \forall x \in E$.

The definition of a Minkowski is

$1. p(\lambda x) = \lambda p(x) \forall \lambda > 0$,

$2. p(x+y) \leq p(x)+p(y)$.

My question is why we need the first strict condition for this theorem. Why any convex function does not work with the condition $p(\lambda x) \leq \lambda p(x), \forall \lambda \geq 1$.

答:
First, for the case that $p(0)=0$ holds,
one can show that your conditions on $p$ imply that $p$ is a Minkowski function.

But (perhaps) more important is that Hahn-Banach only
compares the function $p$ to a linear function ($f$ or $g$).
For comparing a function $p$ with a linear function in this context,
it suffices to only consider the ``worst'' values of $p$
along each ray $\{t x : t>0\}$.

To formalize this, suppose $p$ is convex.
We define $q$ via $q(x):=\inf \{t p(x/t) : t>0 \}$.
Then one can show that $q$ is a Minkowski function and that
$$
 g \leq p \iff g \leq q
$$
holds for all linear functions $g$ which are defined on a linear subspace
(I will not give the detailed proof here for these claims).

From this one can conclude that the Hahn-Banach theorem
also holds for convex functions $p$, and not just Minkowski functions.

Title: On the Hahn-Banach Theorem


问:
Does the MLE estimator of the covariance of a multivariate Gaussian have a non-zero determinant?
The MLE estimator of the covariance of a multivariate Gaussian is

$$
\mathbf{\Sigma} = \frac{1}{N} \sum_{n=1}^N \left( (\mathbf{x}_n-\boldsymbol{\mu}) (\mathbf{x}_n-\boldsymbol{\mu})^\intercal \right).
$$

Can we prove that its determinant is non-zero?

答:
Determinant of $\Sigma$ may be zero.

Let $k$ be the dimension of the space to which the sample vectors $x_n$ belong. If $N < k$ then $\Sigma$ is not full rank and therefore $\det \Sigma = 0$.

Moreover, even when $N \ge k$ it is still possible that $\det \Sigma = 0$. For a simple example, suppose that $k=2$ and that the sample vectors $x_n \in\mathbb{R}^2$ originate from the distribution $(X, X)$ where $X \sim \mathcal{N}(0, 1)$, i.e. the two components are perfectly correlated. In this case
$x_n = (a_n, a_n) = a_n (1, 1)^T$ for some real numbers $a_n$ for all $n=1,\dots,N$. Therefore,

$$
\det \Sigma = \det \left(\frac{1}{N} \sum_{n=1}^N x_n x_n^T\right) = \det \left(\frac{1}{N} \sum_{n=1}^N a_n^2 \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} \right) = \det \begin{pmatrix}a & a \\ a & a\end{pmatrix} = 0
$$

where $a = \frac{1}{N} \sum_{n=1}^N a_n^2$.

On the other hand, note that the set

$$
\mathcal{X}_0 = \{(x_1, x_2, \dots, x_N)\,|\, \det \Sigma(x_1, x_2, \dots, x_N) = 0\}
$$

of $N$-tuples of samples $x_n$ for which $\det \Sigma = 0$ has dimension one less than the dimension of the set of all $N$-tuples of samples $x_n$. Therefore, if $N \ge k$ and $\det \Sigma' > 0$ where $\Sigma'$ denotes the covariance matrix of the underlying distribution, then the set $\mathcal{X}_0$ has measure zero and consequently $\det \Sigma > 0$ with probability $1$.

Remark about practical applications: Note that in a practical application one would probably not care about strict equality $\det \Sigma = 0$, but about whether $\Sigma$ is nearly singular (in fact one may care about a stronger condition that $\Sigma$ is well-conditioned).

By the multi-dimensional variant of the Central Limit Theorem as $N\to \infty$, the sample covariance $\Sigma$ converges to $\Sigma'$. Therefore, if the eigenvalues of $\Sigma'$ are large then for large $N$ it becomes increasingly unlikely that $\det \Sigma$ is near zero.

On the other hand, if $\Sigma'$ does have small eigenvalues then one can use Principal Component Analysis to identify a subspace $V$ on which the eigenvalues of the restriction $\Sigma'|_V$ are large and therefore $\det \Sigma|_V > 0$ with high probability.

Lower bounds: There are lower bounds on the eigenvalues of $\Sigma$. For example, for isotropic distributions, i.e. ones with $\Sigma' = \alpha I$ lower bounds on the smallest eigenvalue of $\Sigma$ are derived in this paper.


问:
Showing elements of an ultrapower have the form $j(f)([id]_{\mathcal{U}})$
I am in a set theory class and we just started talking about ultrapowers. My professor mentioned that, if $\kappa$ is a measurable cardinal and $\mathcal{U}$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$, then it is an easy consequence of Łoś's theorem that, for any function $f : \kappa \rightarrow V$, the equivalence class $[f]_{\mathcal{U}}$ is the same as $j(f)([id]_{\mathcal{U}})$, where $j$ is the canonical embedding. But when I tried to prove this, it wasn't easy for me! I know $j(f)([id]_{\mathcal{U}})$ must be an element of the ultrapower, and so I set $j(f)([id]_{\mathcal{U}})$ equal to $[g]$ for some $g : \kappa \rightarrow V$. Then I tried to show $\left\{\xi < \kappa : f(\xi) = g(\xi) \right\} \in \mathcal{U}$. If this was true, then by Los's theorem we would get $[f]_{\mathcal{U}} = [g]_{\mathcal{U}} = j(f)([id]_{\mathcal{U}})$. But I don't really understand the relationship between $g$ and $f$ and kept going in circles (one thing I tried to show was that $[id]_{\mathcal{U}}$ was in $j(\left\{\xi < \kappa : f(\xi) = g(\xi) \right\})$, and that definitely didn't seem to be going anywhere). Where am I going wrong?

答:
As a newbie for ultrapower, handling elements of an ultrapower is confusing and not easy. I also experienced trouble to evaluate $j(f)([\mathsf{id}])$.

For me, viewing $j(f)([\mathsf{id}])$ componentwise was helpful. Imagine the function $g$ representing $j(f)([\mathsf{id}])$. What is the value of that function if we put $\alpha$ into $g$? Well, we may view $j(f)$ as a function from $\kappa$ to $V$ that always returns $f$. $\mathsf{Id}$ is just the identity function.
By assumption, $[g]=j(f)([\mathsf{Id}])$. Now let us see the $\alpha$th section of this equality, then it would be $$g(\alpha)=(j(f)(\alpha))(\mathsf{Id}(\alpha))=f(\alpha).$$
Thus we may guess $g=f$.

Of course, the above argument is not an actual proof. The point is that functions in set theory are binary relations technically. Hence the assertion $f(x)=y$ is syntactic sugar for $(x,y)\in f$.

Thus what we have to prove is $([\mathsf{Id}],[f])\in j(f)$, which is equivalent to
$$\{\alpha<\kappa \mid (\alpha,f(\alpha))\in f\}\in\mathcal{U},$$
but you can easily see that the above sentence holds.


Title: Sum of the series $\sum {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$

Body:
Evaluate the sum of the series:
$\displaystyle \sum_{n=0}^\infty {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$

Ratio test confirms convergence of the series. But what's the exact value of the series ?
Computer programming approximates it somewhere near $\frac 12$.

Body:
$\dfrac{\sin x}{2x}-\dfrac{\cos x}2=\dfrac1{2x}\sum\limits_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}-\dfrac12\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}$

$=\dfrac12\sum\limits_{n=0}^\infty(-1)^nx^{2n}\left(\dfrac{1}{(2n+1)!}-\dfrac1{(2n)!}\right)$

$=\sum\limits_{n=0}^\infty\dfrac12(-1)^{n+1}x^{2n}\left(\dfrac{2n+1}{(2n+1)!}-\dfrac1{(2n+1)!}\right)$

$=\sum\limits_{n=0}^\infty(-1)^{n+1}x^{2n}\dfrac n{(2n+1)!}$.

Now take $x=\pi$.


问:
Does $y = f(x) = ax+b$ actually have two mappings inside it?
I’m just a high school student, so I may be somewhat logically flawed in understanding this.

According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.

But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?

My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.

These two definitions are not quite the same.

On 2022.6.29: The picture below had solved my confusion.

答: The notation $f(1)=0$ means that $f$ maps $1 \in X$ to $0 \in Y$. This agrees with the familiar notation $y=f(x)$ form that most people encounter in high school such as $y=ax^2+bx+c$. This means that $f(x) \in Y$ is always true. There is no other assignment operation occurring. Alternatively if you define functions as a kind of relation $f \subset X \times Y$ then $f(1)=0$ means $(1,0) \in f$. In both cases it's merely two names of the same thing.


问:
How to calculate this multi-integral?
Please calculate $$I=\int_0^1 dx \int_0^x dy \int_0^y \frac{\sin z}{(1-z)^2}dz$$

Any hints? Thank you!

Title: How to calculate this multi-integral?

Body:
The integral $$I=\int_0^1 \frac{\sin(z)}{(1-z)^2}dz\int_z^1 dy\int_y^1 dx\\=\int_0^1 \frac{\sin(z)}{(1-z)^2}dz\int_z^1(1-y)dy\\=
\int_0^1 \frac{1}{2}\sin(z)dz=\frac{1}{2}-\frac{1}{2}\cos(1)$$


问:
Graphing an elliptic integral of the first kind
Is there any way to graph an elliptic integral of the first kind on Excel or some other graphing software that is free?</p>

答:
In Mathematica:

```<code>Plot[EllipticF[\[Phi], 0.5], {\[Phi], 0, Pi/2}]
</code>```

which can also be done in Wolfram|Alpha (for free).

Body:
In Mathematica:

Plot[EllipticF[\[Phi], 0.5], {\[Phi], 0, Pi/2}]

which can also be done in Wolfram|Alpha (for free).


问:
What am I doing wrong on this question about "varies directly" and "varies inversely"?
I don't get this problem:

$R$ varies directly as $S$ and inversely as $T$. When $R = \frac{4}{3}, T = \frac{9}{14}$, $S = \frac{3}{7}$. Find $S$ when $R = \sqrt{48}$ and $T = \sqrt{75}$.

My attempt:

We know

$$ R = c_1 S$$

and

$$ R = \frac{c_2}{T}$$

Plugging things in, we get $c_1 = \frac{28}{9}$ and $c_2 = \frac{6}{7}$. Now, we just solve for $S$.

$$ \sqrt{48} = \frac{6}{7} S$$

$$ 4 \sqrt{3} = \frac{6}{7} S$$

$$ 2 \sqrt{3} = \frac{3}{7} S$$

$$S = \frac{14}{3} \sqrt{3} $$

But, as far as I know, the correct answer appears to be $30$. What am I doing wrong? Am I getting wrong the definition of "directly" and "inversely"?

答:
You can’t treat the relationship between $R$ and $S$ and the relationship between $R$ and $T$ as if they were independent: the hypothesis actually means that

$$R=c\cdot\frac{S}T$$

for some constant $c$.

Thus, the relationship $R=c_1T$ holds only when $T$ is held constant, and the relationship $R=\frac{c_2}T$ holds only when $S$ is held constant.

You know that

$$\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=\frac23\,,$$

so

$$c=\frac{4/3}{2/3}=2\,.$$

When $R=\sqrt{48}$ and $T=\sqrt{75}$ we have

$$\sqrt{48}=\frac{2S}{\sqrt{75}}\,,$$

so

$$S=\frac12\sqrt{48\cdot75}=30\,.$$


问:
17 definitions of algebraic K-theory of a ring. Which should I take?
There are multiple definitions of algebraic K-theory, but I have trouble differentiating between them. Could someone help me out? Let $R$ be a commutative ring. I would like to define $K_n(R)$, and for simplicity $n \geq 0$ is OK for me now.

The plus construction gives us $B\operatorname{GL}(R)^+$, whose homotopy groups give me $K_n(R)$ for all $n \geq 0$.

The Q-construction takes as input an exact category $\mathcal{C}$. At this point we can pick the finitely generated $R$-modules, or the projective finitely generated $R$-modules, or the finitely generated free $R$-modules, or the coherent $R$-modules. This gives us four K-theory groups $K_n(R)$ for $n \geq 0$.

All the exact categories give Waldhausen categories, to which we can apply the S-construction, yielding four more K-groups.

Segal's construction involving $\Gamma$-spaces takes in a symmetric monoidal category $\mathcal{C}$. We can take any of the above categories, along with two choices of monoidal structure, namely $\otimes$ or $\oplus$, yielding another eight K-theory groups $K_n(R)$ for $n \geq 0$.

Ideally, I'd like to see these definitions compared to modern ones. I've been told that one should think of algebraic K-theory as an $\infty$-group completion procedure, but I have yet to find a readable reference on this.

答:
Let me make a few comments relating the various classical constructions you mentioned.

The $BGL(R)^+$ construction produces an infinite loop space. It's constructed by taking $BGL(R)=\varinjlim BGL_n(R)$ and then attaching $2$-cells to kill the commutator subgroup of $\pi_1$ and attaching $3$-cells to preserve  the original homology. It might seem ad-hoc but there was good reason to come up with this construction. You can read more about the motivation in Segal's contribution to the AMS tribute to Quillen. The original source is Quillen's seminal paper.

The Quillen $Q$-construction is far more general and applies to a notion of exact category (i.e. an additive category with a class of exact sequences satisfying some properties), for example coherent sheaves or vector bundles over schemes. Essentially you take a an additive category $\mathcal{C}$, give it an exact structure, make a $Q$ consturction out of it i.e. a category $Q\mathcal{C}$ and then take its classifying space $BQ\mathcal{C}.$ The  $K$-groups are the homotopy groups of the loop space $\Omega BQ\mathcal{C}.$ In fact, $\Omega BQ\mathcal{C}$ is an infinite loop space.

In the $Q$-construction for a scheme you get two groups related to the category of coherent sheaves and the category of vector bundles. However there are theorems in the $Q$ construction of $K$-theory which allows you relate these two groups in nice cases. For example if you are over a smooth projective variety, then every coherent sheaf has a finite resolution by locally free ones, and by Quillen's Resolution theorem, the $K$-theories are the same. You can translate this to the language of commutative algebra if you want.

In the case of a ring $R$ and $P(R)$ the category of finitely generated proejctive modules, both constructions the $Q$ construction and the plus construction are equivalent i.e. $K_{0}(R)\times BGL(R)^+\simeq \Omega BQP(R).$

This is the work of Grayson, which he attributes to Quillen.

Then you have the Waldhausen $S$-construction, which is the most general of all which you mentioned. This is a functorial way to construct connective spectra as opposed to constructing just the infinite loop space. More precisely, for a Waldhausen Category $\mathcal{C}$, the Waldhausen $S$-construction gives you not only an infinite loop space $\Omega |wS_\bullet\mathcal{C}|$  but also infinite deloopings $|wS_{\bullet}S_{\bullet}\mathcal{C}|,|w S_{\bullet}^3\mathcal{C}|\ldots,$ and where $$|wS_{\bullet}^{n}\mathcal{C}|\simeq \Omega|wS_{\bullet}^{n+1}\mathcal{C}|.$$

Every exact category is (canonically) a Waldhausen categoy. In this case the $Q$-construction and the Waldhausen construction are related via a simplicial subdivision method called Segal's subdivision. This is proved in Waldhausen's original paper. Functionally this means that there is an equivalence $$\Omega BQ\mathcal{C}\simeq \Omega |wS_\bullet\mathcal{C}|.$$

Segal's $\Gamma$-machinery is an infinite loop space machine. It takes in an infinite loop space and outputs connective spectra. So for example you can feed it the $Q$-construction of Quillen and you will get connective spectra out. Note that the $Q$-construction already knows about the additive structure and in particular about $\oplus$ (the finite bi-poduct). If you also have a symmetric monoidal structure given by $\otimes$, over which $\oplus$ distributes , then your spectra inherits a coherent multiplication $KC\wedge KC\to KC$, where $KC$ is the $K$-theory spectrum of $C$ (in other words it is an $E_\infty$-ring spectrum).

Finally, there are several other infinite loop space machines such as May's delooping via the bar construction, but these are all equivalent. This is a theorem of May and Thomason.

Edit: A sketch of the $\infty$-categorical construction of the K-theory spectrum of a commutative ring (more generally a symmetric monoidal $\infty$-category) is given in the appendix of the Bhatt-Scholze paper on the Witt vector affine Grassmanian.


问:
Is $f(x) = \frac{\sin(x)}{x}e^{-xy}$ integrable for $x\in ]0,\infty[$ and $y \geq 0$?

Is $\left]0,\infty\right[ \to \mathbb{R},\,\,\, \operatorname{f}\left(x\right) =
\frac{\sin\left(x\right)}{x}\,\,{\rm e}^{-xy}\,\,\,$ Lebesgue-integrable for $\,\,\, y \geq 0\ ?$.

I tried $\,\,\,\left\vert\operatorname{f}\left(x\right)\right\vert \leq \left\vert\,{\frac{\sin\left(x\right)}{x}}\,\right\vert,\,\,\,$ but I don't know whether $\,\,\,\left\vert\,{\frac{\sin\left(x\right)}{x}}\,\right\vert\,\,\,$ is integrable.

Maybe I am taking a wrong direction..

答:
Just for curiosity. Observe that
\begin{align}
\int^\infty_0 \frac{\sin (\alpha x)}{x} e^{-yx}\ dx =&\ \int^\infty_0 \frac{e^{i\alpha x}-e^{-i\alpha x}}{2i x}e^{-xy}\ dx\\
=&\ \int^\infty_0 \frac{e^{i\alpha x}}{2i x}e^{-xy}\ dx- \int^\infty_0 \frac{e^{-ix}}{2i x}e^{-xy}\ dx\\
=&\ \int^\infty_0 \frac{e^{i\alpha x}}{2i x}e^{-xy}\ dx+\int^0_{-\infty} \frac{e^{ix}}{2ix}e^{xy}\ dx\\
 =&\ \frac{1}{2}\int^\infty_{-\infty} e^{i\alpha x}\frac{e^{-y|x|}}{ix}\ dx.
\end{align}
Notice that
\begin{align}
\frac{d}{d\alpha}\int^\infty_0 \frac{\sin (\alpha x)}{x} e^{-yx}\ dx = \frac{1}{2}\int^\infty_{-\infty} e^{i\alpha x}e^{-y|x|}\ dx = \frac{y}{y^2+\alpha^2}.
\end{align}
Finally, we see that
\begin{align}
\int^\infty_0 \frac{\sin (\alpha x)}{x} e^{-yx}\ dx = \tan^{-1}\left(\frac{\alpha}{y} \right).
\end{align}
Set $\alpha = 1$ gives us the desired result.


答:
I see what you are saying, but there are ways to think about this without using more than one mapping, as the other answers have pointed out.

But here is what I think you are getting at (which I agree with):

$x \mapsto f(x) \mapsto ax+b$. where we have a general function of $x$ and we send it to some expression in terms of the variable $x$.

Then $x$ maps to $f(x)$ as normal by using the domain and codomain.

Title: Does $y = f(x) = ax+b$ actually have two mappings inside it?

Body:
I’m just a high school student, so I may be somewhat logically flawed in understanding this.

According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.

But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?

My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.

These two definitions are not quite the same.

On 2022.6.29: The picture below had solved my confusion.


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
The issue largely vanishes if we just consider $\lim$ and $\log$ explicitly as partial functions. A partial function can be seen as a function whose codomain contains one extra (distinguishable!) element, basically the “error value”.
$$\begin{align}
   \log :&& \mathbb{R} \not\to \mathbb{R}
\\ \lim_0 :&& ((\mathbb{R}\setminus\{0\})\to\mathbb{R}) \not\to \mathbb{R}
\end{align}$$
where we have for example
$$\begin{align}
   \log(1) =& \text{OK}(0)
\\ \log(0) =& \text{ERR}
\\ \lim_0( h\mapsto \tfrac{\sin h}{h}) =& \text{OK}(1)
\\ \lim_0( h\mapsto \tfrac1{h}) =& \text{ERR}
\end{align}$$

Now, the logarithm law
$$
  \log(a\cdot b) = \log a + \log b
$$
is to be understood with a “lifted” $+$ operator, that just passes failure on either side on. But that means that for this operator, we can't infer from $p+q=p$ that $q=0$, because $\text{ERR}+q$ is always $\text{ERR}$ regardless! Instead, only from $\text{OK}(p)+q = \text{OK}(p)$ we can infer $q = \text{OK}(0)$. Thus we don't get to the wrong conclusion about $\log(0)$, because that is not an $\text{OK}$ value.

Applied to the limits in the differentiation, we can immediately write
$$
  f'(x) = \lim_0\left(h\mapsto\frac{f(x+h)-f(x)}{h}\right)
$$
just noting that the result might be $\text{ERR}$. What we can also do without any problem is rewrite the expression inside the limit with anything that – as a function $h\mapsto\ldots$ – really is (extensionally) the same. This is in particular no problem for
$$\begin{align}
  f'(x) =& \lim_0\left(h\mapsto\frac{(x+h)^2-x^2}{h}\right)
\\ =& \lim_0\left(h\mapsto\frac{2\cdot h\cdot x+h^2}{h}\right)
\end{align}$$
because $h\mapsto\frac{(x+h)^2-x^2}{h}$ and $h\mapsto\frac{2\cdot h\cdot x+h^2}{h}$ really are the same for all $h\in\mathbb{R}$. Still, at this point we don't know if either of the limits actually exist – they might be both $\text{ERR}$, or both $\text{OK}$, but at any rate equal.

For the next step we need the fact that the limit considers its argument as only a function with nonzero numbers as the domain, because only considered as a function on that domain is $h\mapsto\frac{2\cdot h\cdot x+h^2}{h}$ the same function as $h\mapsto 2\cdot x+h$.

And that's it, at this point we can read off that the limit is indeed $\text{OK}(2\cdot x)$ and going back we see that the other limits must also have been $\text{OK}$ with that same value.


问:
How to calculate overall conversion percentage?
Let's a say there's an exam which has a 10% passing percentage from Prelims to Main then 15% passing percentage from Main to Interview and 2% passing percentage from interview to final selection. Then how will we calculate the overall conversion percentage for a successful candidate with their name in the final selection ?</p>

答:
Am I missing something? It would appear that each percentage applies to the number of people not culled in the exam before. In that case, the overall percentage is just as you imagined.

$$10\%\times15\% \times2\%=0.0003=0.03\%$$


问:
Show $ (2 - b^2c^2) (1 + b^2) \ge (b^2 + c^2)(b + c - 3)^2 $ in some range.
Show $ F = (2 - b^2c^2) (1 + b^2) - (b^2 + c^2)(b + c - 3)^2 \ge 0$ in the following two ranges of $(b,c)$:

$b \in [0 \quad 1]$ and $0 \le c\le b$,

$b \in [1 \quad 1.2]$ and $0 \le c\le 3-2b$.

Equality occurs only at $(b,c)= (1,0)$.

This can be solved by treating $F$ as a polynomial in one of the variables (the other a parameter), and then using Budan's theorem. This has been performed here, but this is a very lengthy business. Does someone have a more direct method? Thanks in advance!

答:
Proof of (1): Clearly, if $b = 0$, the inequality is true. Also, if $c = 0$, the inequality
becomes $(-b^2 + 4b + 2)(b-1)^2 \ge 0$ which is true.

If $b\in (0, 1]$ and $c\in (0, b]$, let $b = \frac{1}{1+s}, c = b \frac{1}{1+t}$ for $s, t \ge 0$.
The inequality becomes
$$\frac{1}{(1+s)^6(1+t)^4}[f(s,t) + 14s^4 - 3s^2 + 2s] \ge 0$$
where $f(s, t)$ is a polynomial with non-negative coefficients. Thus, the inequality is true.

We are done.

$\phantom{2}$

Proof of (2): If $b = 1$, the inequality becomes $c(1-c)(c^2 - 3c + 4) \ge 0$ which is true.
Also, if $c = 0$, the inequality
becomes $(-b^2 + 4b + 2)(b-1)^2 \ge 0$ which is true.

If $b\in (1, 6/5]$ and $c\in (0, 3-2b]$, let $b = 1 + \frac{1}{5}\cdot \frac{1}{1+s}$
and $c = (3-2b)\frac{1}{1+t}$ for $s, t\ge 0$. The inequality becomes
$$\frac{1}{15625(1+s)^6(1+t)^4}g(s, t) \ge 0$$
where $g(s, t)$ is a polynomial with non-negative coefficients. Thus, the inequality is true.

We are done.


问:
Conditional probability - dice problem
Two dice are rolled. Let X be the lower of the two scores and Y  be the larger of the scores. What is the conditional probability
P(Y=6|X=4)?</p>

答:
The other answers and comments are incorrect as they falsely assume that $Y$ must be strictly greater than $X$ in every occasion, however that is invalid.  It is perfectly acceptable for $Y$ to equal $X$ in the event that both dice show the same value.

To clarify and formalize, I am working with the assumption that $A$ and $B$ are independent random variables, each of which uniformly distributed over $\{1,2,3,4,5,6\}$ and $X=\min(A,B)$ and $Y=\max(A,B)$... that is $A$ is the result of the "first die" and $B$ the result of the "second die", that $X$ is colloquially the result of the "smaller die" and $Y$ the result of the "larger die" (yes, even in the event that $A=B$)

Let us imagine that the dice are both differently colored or that they are rolled in sequence, or similar... e.g. one blue and one red die.  There are $36$ equally likely different outcomes.

$\Pr(Y=6\mid X=4) = \dfrac{\Pr(Y=6\cap X=4)}{\Pr(X=4)}$

The probability that the larger die is a $6$ and the smaller die is a $4$, well that corresponds to the two outcomes where the red die is $6$ and the blue die is $4$ or to the outcome that the red die is $4$ and the blue die is $6$.  We have then that $\Pr(Y=6\cap X=4)=\dfrac{2}{36}$

The probability that the smaller die is a $4$ is going to be the probability that both dice are greater than or equal to $4$ while not being the case that both dice are simultaneously greater than or equal to $5$ (as then $4$ would not have been the minimum).  There are $3\times 3$ outcomes corresponding to both dice being greater than or equal to $4$ and $2\times 2$ outcomes corresponding to both dice being greater than or equal to $5$.  As such, there are $3\times 3-2\times 2=9-4=5$ outcomes corresponding to the minimum of the dice rolls being a $4$ and so $\Pr(X=4)=\dfrac{5}{36}$

This used the general result that $\Pr(X= k) = \Pr(X\geq k)-\Pr(X>k)$

Combining the results, the probability is then:

$$\Pr(Y=6\mid X=4)=\dfrac{\frac{2}{36}}{\frac{5}{36}}=\dfrac{2}{5}=0.4$$

Title: Conditional probability - dice problem


问:
Convergence of sequence in p-adic integers
Let $a\in Z_p^{*}$. I need to prove the sequence $\{a^{p^n}\}_{n\geq 0}$ converges in $\mathbb{Z}_p$ to $w$, where $$a=wb$$ for $w$ is the $p-1$ root of unity in $\mathbb{Z}_p^*$ and $b\in 1+p\mathbb{Z}_p$.

I feel like I need to use Hensel's Lemma for this question but I do not know how to start it. Is there any particular hint for this question?

答:
$\Bbb{Z/p^k Z}^\times$ has $p^{k-1}(p-1)$ elements. Thus $order(a\bmod p^k)$ divides $p^{k-1}(p-1)$,

For $n\ge k$, $order(a^{(p-1)p^n}\bmod p^k)= 1$ ie. $$a^{p^{n+1}}\equiv a^{p^n}\bmod p^k$$

This implies that the limit converges $$w=\lim_{n\to \infty} a^{p^n}\in \Bbb{Z}_p$$

Also $a\equiv w\bmod p$ and
$w^p = w$ so that $w$ is the unique $p-1$-th root of unity such that $w^{-1}a\in 1+p\Bbb{Z}_p$.


问:
John and Gary are playing a game. John spins a spinner numbered with integers from 1 to 20.
John and Gary are playing a game. John spins a spinner numbered with integers from 1 to 20. Gary then writes a list of all of the positive factors of the number spun except for the number itself. Gary then creates a new spinner with all of the numbers on his list. John then spins this spinner, and the process continues. The game is over when the spinner has no numbers on it. If John spins a 20 on his first spin, what is the maximum number of total spins (including the one he already made) that John can make before the game is over?</p>

答:
If John spins a 20, then Gary's list contains the numbers 1, 2, 4, 5, 10. Thus, these are the numbers on the second spinner.

If John spins a 1, then Gary's list will be empty because there are no positive factors of 1 besides itself. Thus, the game will be over. This yields a maximum of 1 additional spin.

If John spins a 2, then Gary's list will only contain the number 1. Then on John's next spin, we will have the same scenario as above. This yields a maximum of 2 additional spins.

If John spins a 4, then Gary's list will contain the numbers 1 and 2. As we have already found above, spinning a 2 yields more additional spins than a 1, so the maximum additional spins in this case is 3 spins.

If John spins a 5, then Gary's list will only contain the number 1. As above, this will yield a maximum of 2 additional spins.

Finally, if John spins a 10, then Gary's list will contain the numbers 1, 2, and 5. Of these numbers, 2 and 5 have the highest maximum number of additional spins, so this case has a maximum of 3 additional spins.

Thus, of all of the possibilities, spinning a 4 or 10 next could result in 3 additional spins, so the maximum total number of spins is $\boxed{4}$. These would be achieved by spinning 20, 10, 2, 1 or 20, 10, 5, 1 or 20, 4, 2, 1.


问:
Decay rate of $P(\left\lvert S_n/n - \mu \right\rvert > \epsilon)$ when higher moments exist.
Suppose that $X_1,X_2\ldots$ are iid with $E|X_1| < \infty$. Define $\mu = EX_1$ and $S_n = X_1 + X_2 + \ldots + X_n$. If $E X_1^2 < \infty$, we can use chebyshev's to conclude that:
$$P\left(\left\lvert \frac{S_n}{n} - \mu \right\rvert > \epsilon\right) \leq \frac{\textrm{Var} S_n}{n^2\epsilon^2} = \frac{n \textrm{Var}X_1}{n^2\epsilon^2} \leq \frac{EX_1^2}{n\epsilon^2}.$$

Therefore, we get that $P(\left\lvert S_n/n - \mu \right\rvert > \epsilon)$ decays at least like $\frac{1}{n}$. Now, assume that $E\left\lvert X^k \right\rvert < \infty$. Chebyshev's gives us that:
$$P\left(\left\lvert \frac{S_n}{n} - \mu \right\rvert > \epsilon\right) \leq \frac{E\left\lvert S_n - n\mu \right\rvert^k }{n^k\epsilon^k}$$
I don't know how to handle the expression $E\left\lvert S_n - n\mu \right\rvert^k$.  Can i get a similar result like above e.g. $P(\left\lvert S_n/n - \mu \right\rvert > \epsilon)$ decays at least like $\frac{1}{n^{k/2}}$?

答:
From the central limit Theorem and uniform integrability, or from the Marcinkiewicz–Zygmund inequality and the triangle inequality (see e.g. (1.2) in [R]) we have $$E[|S_n-n\mu|^k]=O(n^{k/2}) ,$$ so this yields the bound
$$P(|S_n/n - \mu |> \epsilon) =O(n^{-k/2}) \,.$$

Also relevant is the following:

Theorem   [K]
Let $X_1,X_2,\ldots$ be i.i.d. random variables of mean $\mu$ satisfying
$E{|X_1|^k}<\infty$ with $k\ge 1$. If $r>k$, then, for all $\epsilon >0$,
$$
\sum_{n=1}^\infty n^{r-2} P\bigl(|S_n-n\mu|\ge n^{r/k} \epsilon\bigr) <\infty.
$$

[R] Rio, E. Moment Inequalities for Sums of Dependent Random Variables under Projective Conditions. J Theor Probab 22, 146–163 (2009). https://link.springer.com/article/10.1007%2Fs10959-008-0155-9

[K] Katz, M., The probability in the tail of a distribution,
Ann. Math. Statist.    34, 1963, 312--318.


问:
Incomplete space
I must show that $C_0(\mathbb{R})$ space of all continuous real functions $f:\mathbb{R} \longrightarrow \mathbb{R}$ with compact support is not a complete space endowed by norm
$\|f\|=\sup\limits_{t \in \mathbb{R}}|f(t)|$.

Thank for any help.

答: Hint: find a sequence whose limit does not have compact support.</p>

Title: Incomplete space


问:
Proof of inequality in Weierstrass approximation theorem proof through probability
I found this exercise in my probability theory book. The problem guides you through a proof of the Weierstrass Approximation Theorem through probability theory. My question is only about part b, so any help or hint is much appreciated. I can make the exercise, except for part $b$, and I show all my workings below. I understand that you might want to see my workings for $b$, but I have really no idea. First I quote Chebyshev's inequality, because apparently we need this.

Thus: my question is only about part b, my work for the other parts can be found below.

$\bf{Chebyshev's \ inequality}$ If $Y$ is a random variable and $\mathbb{E}(Y^2)<\infty$, then $$\mathbb{P}(|Y|\geq t)\leq \frac{\mathbb{E}(Y^2)}{t^2}, \quad \mathrm{for \ } t>0.$$

I quote the exercise:

Let ($X_i$) be a sequence of i.i.d. Bernoulli$(p)$ variables, thus $\mathbb{P}(X_i=0)=1-p$ and $\mathbb{P}(X_i=1)=p$, for all $i\in\mathbb{N}$.

a) Let $f$ be a continuous function of $[0,1]$ and prove that $$B_n(p)=\mathbb{E}\left(f\left(\frac{\sum_{i=1}^nX_i}{n}\right)\right)$$ is a polynomial in $p$ of degree at most $n$.

b) Use Chebyshev's inequality to prove that for all $p$ such that $0\leq p\leq1$ and for any $\epsilon>0$, $$\sum_{k\in K}\binom{n}{k}p^k(1-p)^{n-k}\leq\frac{1}{4n\epsilon^2},$$ where $K=\{k:0\leq k\leq n, |k/n-p|>\epsilon\}$.

c) Using this and the fact that $f$ is bounded and uniformly continuous on $[0,1]$, prove the following version of the Weierstrass approximation theorem: $$\lim_{n\to\infty}\sup_{0\leq p\leq1}|f(p)-B_n(p)|=0.$$

For part $a$, if we let $Y_n:=\sum_{i=1}^nX_i$, then $Y_n\sim \mathrm{Bin}(n,p)$. Thus $$B_n(p)=\mathbb{E}\left(f\left(\frac{\sum_{i=1}^nX_i}{n}\right)\right)=\mathbb{E}\left(f\left(\frac{Y_n}{n}\right)\right)=\sum_{k=0}^n f\left(\frac{k}{n}\right)\binom{n}{k}p^k(1-p)^k$$ is a polynomial in $p$ of degree at most $n$. This follows directly from just writing out the expectation of the binomial distribution.

For part $c$, $f$ is continuous on the closed interval $[0,1]$, so $f$ is uniformly continuous and bounded on $[0,1]$, and assumes it's minimum and maximum on this interval. These are all basic real analysis results. Since $f$ is bounded on $[0,1]$, there exists an $M\in\mathbb{R}$ such that $|f(p)|\leq M$, for all $p\in[0,1]$.

From part $b$, we know that $0\leq p\leq1$ and for any $\epsilon>0$, $$\sum_{k\in K}\binom{n}{k}p^k(1-p)^{n-k}=\mathbb{P}(|k/n-p|\geq \epsilon)\leq\frac{1}{4n\epsilon^2},$$ where $K=\{k:0\leq k\leq n, |k/n-p|>\epsilon\}$. Thus we obtain the following: In equation 1, we take the limit of the result of part b; equation 2 follows because of uniform continuity of $f$; equation 4 follows because of linearity of expectation, part $a$ and because $f(p)$ is fixed; equation 5 follows from the partition theorem for expectations:  $$\begin{align}
&\lim_{n\to\infty}\mathbb{P}(|\frac{k}{n}-p|\geq\epsilon) = 0 & (1)\\
&\lim_{n\to\infty}\mathbb{P}\left(|f\left(\frac{k}{n}\right)-f(p)|\geq\epsilon\right) = 0 & (2)\\
&\lim_{n\to\infty}\mathbb{E}\left(|f\left(\frac{k}{n}\right)-f(p)|\right)  = & (3)\\
&\lim_{n\to\infty}|B_n(p)-f(p)|=& (4)\\
&\lim_{n\to\infty}\mathbb{E}\left(|f\left(k/n\right)-f(p)|\Bigg|k/n-p|\geq\epsilon\right)\cdot\mathbb{P}(|k/n-p|\geq\epsilon)+& \\
&\lim_{n\to\infty}\mathbb{E}\left(|f\left(k/n\right)-f(p)|\Bigg|\frac{k}{n}-p|<\epsilon\right)\cdot\mathbb{P}(|k/n-p|<\epsilon)=& (5)\\
& 0+0=0.& (6)\\
\end{align}$$

The first term is zero because $\mathbb{P}(|\frac{k}{n}-p|\geq\epsilon)$ goes to zero as $n$ goes to $\infty$, as shown in part $b$, and because the boundedness of $f$. The second term goes to zero because: (i) $f(k/n)-f(p)$ goes to zero as $n$ goes to infinity because $f$ is uniformly continuous, because $\lim_{n\to\infty}k/n=p$ because of the law of large numbers (remember that $k$ denotes the number of successes in a $\mathrm{Bin}(n,p)$ experiment); and (ii) because $\mathbb{P}(|k/n-p|<\epsilon)$ is bounded by definition of a probability measure.

Thus $\lim_{n\to\infty}|B_n(p)-f(p)|=0$, for any $p\in[0,1]$, which is the interval where $f$ is uniformly continuous, and thus $\lim_{n\to\infty}\sup_{0\leq p\leq1}|f(p)-B_n(p)|=0$. Thus any continuous function on $[0,1]$ can be uniformly approximated by polyinomials on $[0,1]$. Obvious rescaling proves the Weierstrass Approximation Theorem.

EDIT: Should I use for part $b$ that $\sum_{k\in K}\binom{n}{k}p^k(1-p)^{n-k}$, where $K=\{k:0\leq k\leq n, |k/n-p|>\epsilon\}$ equals $$\begin{eqnarray}\mathbb{P}(|k/n-p|\geq \epsilon) & = & \mathbb{P}(|k-np|\geq \epsilon n)\\&=& \mathbb{P}(|k-\mathbb{E}(Y_n)\tag{with $Y_n\sim\mathrm{Bin}(n,p)$}|\geq \epsilon n)\\ & \leq & \frac{1}{\epsilon^2n^2}\mathbb{E}((k-np)^2)\tag{Chebyshev's inequality}\\ &\leq& \frac{1}{\epsilon^2n^2}(\frac12\cdot\frac12\cdot n) \tag{variance is maximized if $p=\frac12$}\\&=& \frac{1}{4n\epsilon^2}\end{eqnarray}$$

I think this is correct, after posting I came a lot further through reading on wikipedia and the like.

答:
I will present a full proof. Write $S_{n}:=\operatorname{Bin}(n, x)$. Note that $|f|$ is uniformly bounded by some $\mathrm{M}$ and uniformly continuous $-$ fix $\epsilon>0,$ and let $\delta$ be such that $|f(x)-f(y)|<\frac{\epsilon}{2}, \forall|x-y|<\delta.$
Then,
$$
\begin{aligned}
\mid B_{n}(x, f)-f(x) &|=| \mathbb{E}\left(f\left(\frac{S_{n}}{n}\right)-f(x)\right) \mid \\
&=\left|\mathbb{E}\left(f\left(\frac{S_{n}}{n}\right)-f(x)\right) \mathbb{I}\left(\left|\frac{S_{n}}{n}-x\right|<\delta\right)\right| \\
&+\left|\mathbb{E}\left(f\left(\frac{S_{n}}{n}\right)-f(x)\right) \mathbb{I}\left(\left|\frac{S_{n}}{n}-x\right| \geq \delta\right)\right| \leq \epsilon+2 M \cdot P\left(\left|\frac{S_{n}}{n}-x\right| \geq \delta\right) \\
& \leq \frac{\epsilon}{2}+2 M \frac{x(1-x)}{n \delta^{2}} \leq \frac{\epsilon}{2}+\frac{M}{2 n \delta^{2}}
\end{aligned}
$$
The theorem is proved


问:
Maximizing multinomial distribution with constraints using Lagrange multipliers

Use methods of Lagrange multipliers to maximize $$W (N_{1}, \dots, N_{M}) = \frac{N!}{\prod_{j=1}^{M} N_{j}!}$$ under the constraints $\sum N_{j} = N =$ constant, $\sum E_{j} N_{j} = \mathcal{E} =$ constant. Assume $N_{j}$'s are continuous and $N$ large enough to use Stirling's approximation.

After turning the problem into the maximization of $$\ln W = N \ln N - N - \sum (N_{j} \ln N_{j} - N_{j}),$$ I proceeded to setting up equations with Lagrange multipliers $\lambda_{1}, \lambda_{2}$
$$
\frac{\partial}{\partial N_{i}} \left[{\ln W - \lambda_{1} ({\sum N_{j} - N}) - \lambda_{2} ({\sum E_{j} N_{j} - \mathcal{E}}}) \right] = 0 .
$$

However, I encountered a problem when calculating the partial derivative of $\ln W$:
\begin{align*}
 \frac{\partial}{\partial N_{i}} \ln W &= \frac{\partial}{\partial N_{i}} \left[ N \ln N - N - \sum ({ N_{j} \ln N_{j} - N_{j}}) \right] \\
 &= \frac{\partial}{\partial N_{i}} ({N \ln N - N}) - \frac{\partial}{\partial N_{i}} \sum ({ N_{j} \ln N_{j} - N_{j}}) & (N \textrm{ constant}) \\
 &= 0 - ({\ln N_{i} + N_{i} \cdot \frac{1}{N_{i}} - 1}) \\
 &= - \ln N_{i}
\end{align*}
Here, I imposed the constraint that $N$ is constant. However, it occurred to me that, depending on when the constraint is used, there can be different results:
\begin{align*}
 \frac{\partial}{\partial N_{i}} \ln W &= \frac{\partial}{\partial N_{i}} \left[N \ln N - N - {\sum ( N_{j} \ln N_{j} - N_{j}) }\right] \\
&= \frac{\partial}{\partial N_{i}} [N \ln N - N - \sum N_{j} \ln N_{j} + \sum N_{j}] \\
&= \frac{\partial}{\partial N_{i}} [N \ln N - N - \sum N_{j} \ln N_{j} + N] \\
&= \frac{\partial}{\partial N_{i}} (N \ln N) - \frac{\partial}{\partial N} \sum N_{j} \ln N_{j} \\
 &= 0 - ({\ln N_{j} + N_{j} \cdot \frac{1}{N_{j}}}) \\
 &= - \ln N_{j} - 1
\end{align*}
or (treating $N = \sum N_{j}$ as a function of $N_{j}$'s)
\begin{align*}
 \frac{\partial}{\partial N_{i}} \ln W &= \frac{\partial}{\partial N_{i}} \left[N \ln N - N - {\sum N_{j} (\ln N_{j} - N_{j}) }\right] \\
 &= \frac{\partial}{\partial N_{i}} ({N \ln N - N}) - \frac{\partial}{\partial N_{i}} {\sum ( N_{j} \ln N_{j} - N_{j}}) \\
 &= {\frac{\partial}{\partial N_{i}} ({\sum N_{j} \ln \sum N_{j} - \sum N_{j}}}) - ({\ln N_{j} + N_{j} \cdot \frac{1}{N_{j}} - 1}) \\
 &= \ln \sum N_{j} + \sum N_{i} \cdot \frac{1}{\sum N_{j}} - 1 - ({\ln N_{j} + N_{j} \cdot \frac{1}{N_{j}} - 1}) \\
 &= \ln \sum N_{j} - \ln N_{j}
\end{align*}
When should I use the constraint $N = \sum N_{j} =$ constant?

答:
Maximization actually does come from the expression :

$$ d(\ln(W))=0 $$
Giving

Lagrange multiplier on :

$$ \mathcal{L}(W,\lambda_1,\lambda_2)=\ln(W)-\lambda_1(\sum N_j-N)-\lambda_2(\sum e_iN_j-\epsilon)$$

$$d\mathcal{L}=0$$

Factorize by $dN_i$ in the sum, the factors (coordinates of the $dN_i$ in the sum is  null because $dN_i$ is a differential)

You will get the result.


问:
Sum of the series $\sum {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$
Evaluate the sum of the series:
$\displaystyle \sum_{n=0}^\infty {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$

Ratio test confirms convergence of the series. But what's the exact value of the series ?
Computer programming approximates it somewhere near $\frac 12$.

答:
The summation with $x=i\pi$

$$S=\dfrac i\pi\sum_{n=0}^\infty\dfrac{nx^{2n+1}}{(2n+1)! }$$

Writing $2n=2n+1-1, $
$$S=\dfrac i{2\pi}\left(x\sum\dfrac{x^{2n}}{(2n)!}-\sum\dfrac{x^{2n+1}}{(2n+1)!}\right)$$

The first infinite sum $=\dfrac{e^x+e^{-x}}2$

and the second $=\dfrac{e^x-e^{-x}}2$

Use How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?


答:
There is serious confusion about a basic condition of the problem.  The OP is explicitly interested in this case:  "Let X be the lower of the two scores and Y be the larger of the scores."  This a fortiori excludes cases where neither of the values is higher, nor lower, e.g., cases of the form $(n,n)$.

What could be clearer?

Imagine the OP wrote:  "Two dice are rolled.  Let X be the odd score and Y be the even score."

You simply cannot come back and say:  "What about the case $(4,4)$?"  or "What about the case $(3,3)$?"  "What about the case $(4,6)$?"

Those cases are explicitly excluded from consideration.  Hence any calculation of the probability that involves those excluded cases would of course be in error.

Imagine the OP wrote:  "Two dice are rolled.  Let $X$ be a roll above 3.5 and $Y$ be a roll below 3.5.  What is the average difference between the rolls?"

You cannot come back and say "But what about $(5,5)$?"  "What about $(2,2)$?"  Those cases are explicitly excluded.

Nobody is saying that you will never roll a $(4,4)$ or $(4,6)$, etc., just that this is not part of the question.  Just as a question about the rolls being relatively prime, or perfect numbers, or...

Improperly including cases such as $(4,4)$ leads to such confusion.  Suppose you roll a red die and green die and get $(4,4)$ and hear:
"Tell me the color of the single die that is larger than the other."  A perfectly reasonable request.  (It would be a deliberate mis-representation to change this clear question to a different one, such as "Tell the color of the largest."  It would be to deliberately exclude the key word "than.")

Back to the explicit case asked by the OP:  We are told explicitly that one of the rolls is "larger" and the other "smaller."  This is of course not the same as "maximum of the set."  It is of course not correct to say that $4$ is "larger of" the other element of the set $(4,4)$ and that $4$ is "smaller of"...  (um... which?)... just that $4$ is the maximum value.  The author was clear in using "larger of" and any attempt at an answer that deliberately tries to negate those clear words is confused and in error.

Big difference.

Title: Conditional probability - dice problem

Body:
Two dice are rolled. Let X be the lower of the two scores and Y  be the larger of the scores. What is the conditional probability
P(Y=6|X=4)?</p>


问:
Does $y = f(x) = ax+b$ actually have two mappings inside it?
I’m just a high school student, so I may be somewhat logically flawed in understanding this.

According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.

But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?

My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.

These two definitions are not quite the same.

On 2022.6.29: The picture below had solved my confusion.

答:
Equality is not just a mapping. It means two things are one and the same.

We wrote $f(x) = y$ not to indicate that we are mapping $f(x)$ to $y$;
it means that $f(x)$ is $y$.

Consider a function $f$ that maps any value $x$ in the domain $\mathbb Z$ (the integers) to a value $y$ in the codomain $\mathbb Z$.
(Codomain rather than domain is the correct word here.)
In particular, $f$ is the function that maps each integer to the previous integer, so $f$ maps $3$ to $2$, $f$ maps $2$ to $1,$ and $f$ maps $1$ to $0.$

Then the notation $f(1)$ is simply a way of saying
"the number that $f$ maps $1$ to."
So if you ask, "What is $f(1)$," the answer is $0$,
because $f$ maps $1$ to $0$.

Again: $f$ maps $1$ to $0$, therefore the number $f$ maps $1$ to is $0$;
that is, $f(1)$ is $0$. We express this in the equation, $f(1) = 0$.

It is certainly possible to set up a situation in which a member of one set is mapped to something which is then mapped to something else.
But that is not how a single function is defined.
A function defines exactly one transition from one set to another,
and the notation $f(x)$ is one way to express the final result of that transition.
If $f$ maps from a domain $X$ to a codomain $Y$ then $f(x)$ is already a member of $Y$ and needs no further "transitioning" to get to $Y$.

Notice that in the previous paragraph I never mentioned any variable named $y$. Indeed I too take issue with the part of the Wikipedia article that says, "this relation is denoted by $y = f (x)$ ... where the element $x$ is the argument or input of the function, and $y$ is the value of the function ... ." The symbol $y$ is completely superfluous here; if we use $x$ to represent the input of the function, the usual denotation for the value of the function is $f(x)$, and a denotation that actually shows the relation of an element $x$ in the domain to an element in the codomain is
$x \mapsto f(x).$
Here $f(x)$ is the element of the codomain and $\mapsto$ is the relation.


问:
Prove that $f$ is differentiable on $[0,∞)$
Assuming that $e^x$ is differentiable on $\mathbb{R}$, prove that

$$f(x)=\begin{cases}\dfrac{x}{1+e^{1/x}},&x \neq 0\\ 0, &x=0\end{cases}$$

is differentiable on $[0,∞)$. Is $f$ differentiable at $0$?

Here is my textbook problem. when I asked some help from our madam she said that this is not differentiable on $[0,∞)$ it should be $(0,\infty)$ and she said  $[0,∞)$ would be typo

But I think she is wrong but I don't know exactly whether I am correct or not I know $|x|$ is not differentiable at $0$ but it is differentiable on $[0,\infty)$ or$(-\infty,0]$

Am I wrong?

答:
$f'(0)=\displaystyle \lim_{x\to 0} \textstyle {f(x)-f(0)\over x-0}=\displaystyle \lim_{x\to 0}\textstyle {f(x) \over x}$
$\qquad \; =\displaystyle \lim_{x\to 0}\textstyle {1\over 1+e^{1/x}}$

Now $Lf'(0)=\displaystyle \lim_{x\to 0^-}\textstyle {1\over 1+e^{1/x}} =1$
but $Rf'(0)=\displaystyle \lim_{x\to 0^+}\textstyle {1\over 1+e^{1/x}}=0$

Thus $f'(0)$ does not exist.
Your madam is correct.


问:
Tiling the plane with consecutive squares
For which $n$ is it possible to find a region $R$ made of non-overlapping squares of side length $1,2,\ldots,n$ which tiles the plane?

$n=1$ is trivial, and $n=2$ works as well. However, for $n\geq3,$ I am unable to find $R$ that work. Obviously, we can try every possible combination for smaller values, but I want to know for arbitrary $n$ what the conditions are for $R$ to exist.

答:
Here's another way of constructing these. I start with the area of the first N squares, eg for 1 to 9 the area is 285 which is 3x5x19. Then I try to tile a torus of 15x19, then klein bottles of 15x19 and 19x15, finally a manifold (real projective plane) of those dimensions. This all fails for 9, so I use two sets and try torus, klein bottles, rpp for whichever rectangle sizes are convenient. Sooner or later one will work. Here are two sets for 1 to 9 in a 19x30 klein bottle, doubled and flipped so it forms a patch which tiles the plane by translation only. The resulting four sets of squares 1 to 9 are coloured differently.

For 1 to 10 I had to use three sets on a 33x35 torus so I get a translation-only patch with three sets. I told the tiling program to make sure each of the three sets was linked into one self touching area, but if such a tiling did not exist we could use disjoint sets too.

It seems that when we allow multiple sets like this, the probability of a tiling increases with N but a proof is not in my skill set. I might try a few larger N to see whether my program can manage the tiling in reasonable time. One thing I noticed is that the area always seems to be nicely composite, rather than a prime or near prime.

Here is another triple set for 1 to 11, on a cylinder. Note that the sets are contiguous, blue and orange join when wrapped top and bottom.

And here's a two-set tilings for 1 to 11 using a 44x23 Klein bottle. This took my program 86 hours to find.

Just found a 1 to 15 with two sets on a cylinder.

Here's a triple 1 to 12 on a toroid