问:
Showing a property for a set of rewriting rules
Let $\to$ be a relation over the set of binary strings of 0 and 1. $\to$ is defined by the following rules:

R1. $x10y \to x0001y $

R2. $x01y \to x1y $

R3. $x11y \to x0000y $

R4. $x00y \to x0y$

Where $x,y$ are arbitrary strings. The relation is interpreted as a rewriting system, mapping strings to stirings.

The exercise I am trying to solve ask for a function $f$ mapping strings of 0 and 1 into naturals such that $ x \to y $ implies $f (x) > f(y)$.

If the rewriting system would not have rules R1 and R3 one could simply take the lenght of the input string as the function f.

However it seems hard to find a function that yields the required inequality considering also the rules R1 and R4 that amplify the initial string.

Anyone happens to see such a function?

Edit: After the comment of Peter I would be interested in knowing if there is any easy way to prove that the rewriting system is locally confluent. That is, if a -> b and a -> c then b ⇝ d and c ⇝ d

答:
Too long for a comment.  Community wiki since it doesn't really answer the original question.

Here is a way to view string rewriting systems.  The rules define a relation on the set of strings.  I usually write a→b if one can get from a to b by applying one of the rules and a⇝b if one can get from a to b by applying a bunch of rules in succession (so ⇝ is the reflexive, transitive closure of →).

For example, 10 → 0001 → 01 → 1 first using R1 with x and y empty, next using R4 with x empty and y=01, and finally using R2 with x and y empty.  Once we are comfortable using the rules, we would just write 10 ⇝ 1.

Notice that we must be "done" since none of the four rules can apply to just "1".  We call a string a final form if no rule applies to it.  Group theorists like final forms, and so we want our rewriting systems to at least have the property that every string has a final form, that is, for every string a, there is a final form b, such that a⇝b.  More practically though we want to know that no matter how we choose to apply the rules, we always end up at a final form.  Stated more negatively: there is no infinite sequence of words a1→a2→a3→….  Such a system is called locally noetherian or terminating.  Actual group theorists are very demanding.  They don't want a string to have more than one final form.  More positively, if a,b,c are strings such that a⇝b and a⇝c, then there must be a string d such that b⇝d and c⇝d.  In other words, if I go to b and you go to c, we are not lost, we can both meet at d.  Such a system is called confluent.  In a confluent system, a string has at most one final form.  In a locally noetherian system, a string has at least one final form.  Hence group theorists demand a locally noetherian confluent system so that every string has a unique final form.

How does one check the noetherian condition?  This can be really tricky.  You could have infinite loops, or you could have infinitely growing strings.  I believe in general this is a very hard and open area of computer science.  One great way to do this is to show that the rules "make things simpler."  If the "complexity" of strings can be quantified by f as natural numbers, and if rules decrease the complexity (as measured by f), then of course every sequence of rules must stop:  if f(a) = 37, then you cannot apply more than 37 rules for sure!

Group theorists use a slightly more general (I think) crutch.  Instead of assigning each string a number, we just compare to strings and decide which one is simpler.  We call ≺ a reduction ordering if (1) ≺ is a total order on the set of strings, (2) a ≺ b implies xay ≺ xby for all strings a,b,x,y, and (3)  there is no infinite descending sequence of strings a1 ≻ a2 ≻ a3 ≻ a4 ≻ ….

If every rule xLy → xRy satisfies L ≻ R, then the rewriting system is terminating.  Additionally, under these hypotheses to check confluence under these hypotheses is a little easier: instead of checking all a⇝b and a⇝c, one need only check all a→b and a→c, still only searching for a d such that b⇝d and c⇝d.

A rewriting system is reduced if the right hand sides of the rules are all final forms.

If a⇝b then there is some sense in which a and b are equivalent.  We can define an equivalence relation ↭ as the symmetric, transitive closure of ⇝.  In other words, a↭b means there is a sequence of strings a1, a2, …, an such that a1⇝a2, a3⇝a2, a3⇝a4, … an⇝b, so that by some sort of back-and-forth application of rules we can get from a to b.  If a system is confluent and terminating, then each word is ↭equivalent to exactly one final form.

This gives us a way to go from terminating, confluent rewriting systems to equivalence relations with distinguished representatives.  Notice that if the system is terminating and confluent, then replacing the right hand sides by their final forms does not affect the equivalence relation.

If we have a reduction ordering ≺, then the final form of a string is the unique smallest member of the ↭ equivalence class containing that string.  This lets us go from an equivalence relation with distinguished representative (using ≺) back to a reduced, terminating, confluent rewriting system: The left hand sides are all strings that are not final forms but such that every proper substring is a final form, and the right hand sides are the final form.  The result system is reduced, terminating (using ≺), and confluent (using the equivalence relation).

A standard use case is for finitely presented groups.  The relations are turned into rules, but in group theory one is allowed to use the rules in either order.  This is not good for algorithms, and so one looks for equivalent relations such that rules define a terminating confluent (and hopefully reduced) rewriting system.  The type of ordering suggested in the answers here is used for extensions of groups given by rewriting systems.  One has a group G with normal subgroup N such that both N and G/N are given by rewriting systems with reduction orderings ≺.  One wants a reduction ordering on G that plays well with the reduction orderings on N and G/N.  This is called the wreath product ordering.  Basically you sort the quotient part first by ignoring the normal subgroup completely.  If the two strings are the same, then one compares the intervening normal subgroup parts lexicographically (from the right in our example) using ≺ for N to compare corresponding terms.

In the original question, the "1" corresponds to the quotient group.  We define a ≺ b if b has strictly more 1s than a does.  If they have the same number of 1s, then look at the sequence of intervening runs of 0s.  Compare the two strings right to left. Then a ≺ b if, in the first position where the runs are different lengths, the run in a is smaller than the run in b.  If all the runs are the same length, then by definition the two strings are the same.  Since the orderings on the strings of 1s (counting) and the ordering on the strings of 0s (counting) are both reduction orderings, this construction (their wreath product) is also a reduction ordering.  Hence the system is terminating.

Note that a system might have the property that it is not terminating, while its reduction is terminating.  In group theory this doesn't matter, since they determine the same equivalence relation and we want the terminating version.


问:
Geometrically describing a linear transformation?
Let A = $\begin{bmatrix} 0 & 0 & -1 \\ 1 & 0 & 0 \\ 1 & -1 & 1 \end{bmatrix}$. Describe the linear transformation T(x) = Ax geometrically. If a reflection, what is the plane of reflection? If a rotation, what is the axis/angle of rotation?

I'm not sure what this transformation is. Initially, I thought that since x maps to -z, y maps to x, and z maps to x-y+z that this would be a reflection over the plane x-y+z, but that doesn't seem to be the right answer. Any help appreciated.

答:
Let's see how far we can get without doing any calculations requiring working. A quick expansion along the top row shows that $\det A = 1$. By inspection, it can be seen that $v = (1, 1, -1)$ satisfies $Av = v$ and so the line spanned by $v$ is fixed under the action of $A$. The product of the three eigenvalues is the determinant $\det(A) = 1$, and the sum of the three eigenvalues is the trace $\operatorname{tr}(A) = 1$. Since we have already found that $1$ is an eigenvalue, the product of the other two eigenvalues is $1$ and their sum is $0$, therefore the other two eigenvalues are $\pm i$ and so there must be a plane which is rotated by $90^\circ$.

The plane of rotation can be found by looking at the null-space of $(A - i)(A + i) = A^2 + I$. Here we do some actual work:
$$ A^2 + I = \begin{pmatrix} 0 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & -1 & 1 \end{pmatrix}.$$
This matrix has a null-space spanned by $u = (1, 0, 0)$ and $w = (0, 1, 1)$, and we can check indeed that $Au = w$ and $Aw = -u$, so $A$ indeed acts as a rotation on the plane spanned by $u$ and $w$.

Therefore $A$ acts as a rotation, fixing the axis $\mathbb{R}v$ and rotating the plane $\mathbb{R}\{u, w\}$ by $90^\circ$. As a final point, note that under the usual dot product, the axis is not perpendicular to the plane, and while $u$ and $w$ are perpendicular they do not have the same length, so we might want to call this an "algebraic" rotation, rather than a "geometric" one which we would expect to preserve lengths and angles.

ContentLicense: CC BY-SA 4.0


问:
Tangent Line Intercept
The question is like this: there is a curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$, where $c$ is a real number and $c > 0$. If $L$ is a tangent line, any tangent line, with only 1 x-intercept and 1 y-intercept, prove that the sum of the x, y-intercepts of $L$ is $c$.

First of all, I used implicit differenciation on the equation. $$\sqrt{x}+\sqrt{y}=\sqrt{c}$$

Then I also put a graph here (The curve when c=4)

But I still do not have idea. Thank you for looking at this question.

答: From implicit differentiation you get $y'=-\sqrt{\frac{y}{x}}$. Hence the tangent line at any point $(x_0,y_0)$ is given by $y=-\sqrt{\frac{y_0}{x_0}}(x-x_0)+y_0$. Now find the intercepts and check that there sum is $x_0+\sqrt{x_0y_0}+y_0+\sqrt{x_0y_0}=(\sqrt{x_0}+\sqrt{y_0})^2=c.$

Title: Tangent Line Intercept


问:
Is not first order axiomatizable?
I have the following question: ¿Being a: connected graph, Archimedean field, characteristic field 0,  well order, is not definable first order?, if the answer were true we would have to show that there is no first order $\varphi$  statement (in proper language) such that $\mathfrak{A}⊨\varphi$ if and only if $\mathfrak{A}$ is connected graph, characteristic field 0, etc. I need help to prove or disprove this question.

答:
In addition to the compactness technique mentioned in the other answer, we can use Ultraproducts or Games to show that two structures satisfy all the same first order sentences. Then by arguing that one of the structures is well ordered/connected/characteristic 0/etc. while the other isn't, we get the claim.

Let's start with ultraproducts. The important theorem in this area is that, given some sequence of models $\mathfrak{A}_n$, we can find a model $\mathfrak{A}$ (their ultraproduct) so that $\mathfrak{A} \models \varphi$ if and only if "many" $\mathfrak{A}_n \models \varphi$.

Here "many" is measured by some ultrafilter, but for most cases of interest we get

If all but finitely many $\mathfrak{A}_n \models \varphi$, then the ultraproduct $\mathfrak{A} \models \varphi$ too.

If $\mathfrak{A} \models \varphi$, then infinitely many $\mathfrak{A}_n \models \varphi$.

So, for instance, if we want to show that "characteristic 0" isn't first order in the language of rings. We might proceed as follows:

Let $\mathfrak{A}_p$ be the field $\mathbb{Z}/p$ for every prime $p$. We can write down a sentence $\varphi_q$ which says "I am not characteristic $q$", and all but finitely many of the $\mathfrak{A}_p$ will satisfy $\varphi_q$ (indeed, all but $p = q$ will satisfy it).

Then when we ultraproduct the $\mathfrak{A}_p$ together, the resulting model $\mathfrak{A}$ cannot be of characteristic $p$ for any $p$. It must be a field, though, since all but finitely many (indeed all) of the $\mathfrak{A}_p$ satisfy the field axioms. So $\mathfrak{A}$ is a field of characteristic $0$.

Now if any $\varphi$ individually picked out the fields of characteristic $0$, every $\mathfrak{A}_p$ would model $\lnot \varphi$. This means $\mathfrak{A}$ would have to model $\lnot \varphi$ too. A contradiction.

Another useful technique is that of Ehrenfeucht-Fraisse Games. Loosely, Alyss really likes some model $\mathfrak{A}$. Bob likes Alyss, but only has access to some model $\mathfrak{B}$. He (perhaps misguidedly) has decided to trick Alyss into thinking he has access to model $\mathfrak{A}$. She decides to test him with the following game:

Alyss names a number $n$.

Alyss names either an element of $\mathfrak{A}$ or $\mathfrak{B}$

Bob names an element of whichever model Alyss didn't choose.

Alyss again names an element of either model

Bob responds.

Play continues for $n$ rounds, and Bob wins if and only if the $n$ elements of $\mathfrak{A}$ is the same (finite) structure as the $n$ elements of $\mathfrak{B}$.

If Bob can successfully trick Alyss with this game, then (theorem) $\mathfrak{A}$ and $\mathfrak{B}$ satisfy all the same first order sentences.

Ok, why is this useful? Let's look at two graphs:

$A$ an infinite path (think $\mathbb{Z}$ where $n$ is adjacent to $n \pm 1$)

$B$ the disjoint union of two infinite paths.

I claim Bob can successfully win against Alyss in the above game. Why?

Whenever Alyss chooses some element of $A$, then Bob names the corresponding element in the first component of $B$. Whenever Alyss chooses some element of the first component of $B$, then Bob can again be honest and pick the corresponding element in $A$.

If, however, Alyss chooses some element of the second component of $B$, then Bob has to lie. But Bob knows how long the game will last. So he picks some element of $A$ that's really far away from all of the other elements that have been named so far. In particular he picks a big enough number that Alyss doesn't have enough turns left to connect that element to anything played before.

Since Bob has a winning strategy, $A$ and $B$ are elementary equivalent. But since $A$ is connected and $B$ isn't, we find there can be no first order formula defining connectedness.

I hope this helps ^_^

Title: Is not first order axiomatizable?


问:
Signless Stirling numbers of the first kind with minimum 2 elements
Let $c_2(n, k)$ be the number of permutations of [n] with k cycles in which each
cycle length is at least two. Find a recursive formula for the numbers $c_2(n, k)$.

Thought process:

Case 1 - element n is in a cycle with one other element

Then we can pair n with any of the n-1 elements and permute the rest into $c_2(n-2, k-1)$

Case 2 - element n is in a cycle with more than one other element

Then we can permute the other n-1 elements in $c_2(n-1,k)$ ways and place n after any element in a cycle n-1 ways.

Result

This would result in

$c_2(n, k) = (n-1)c_2(n-2, k-1) + (n-1)c_2(n-1,k)$

Is this answer correct?

答:
Seems correct to me. As a way to check, notice that $$\sum _{k=1}^{n}c_2(n,k)=D_n,$$
the derangement numbers (permutations without fixed points). What you just found is essentially that $D_n=(n-1)(D_{n-1}+D_{n-2}).$

Also, in the literature, you might find them with their generalization as Associated Factorial numbers.


问:
How to find probability of 5 letter word appearing in 1800 character long random string?
So say there is 26 possible characters and you've got 1800 long random string. Now how would you go about finding the chance of there being a specific 5 letter word within?

I already found a related question to this but the given formula in the answer doesn't seem to make sense: Probability of a four letter word from a sequence of n random letters

According to the formula in the accepted answer, the denominator would get really big 26^1800 with 1800 length string and mean very little probability. But shouldn't probability increase when the random string length is increased instead of decrease?

答:
This is a long-winded comment that is being posted as an answer only for legibility.

The only approach that I am aware of is Inclusion-Exclusion which results in very ugly math.

I will make the simplying assumption that the word is composed of 5 distinct characters.  My approach will be

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$

where

$$D = (26)^{(1800)}.$$

There are $(1801 - 5) = 1796$ slots that the word can begin in.  For each slot, there are $(1800 - 5) = 1795$ unconstrained letters, which can be chosen in $(26)^{(1795)}$ ways.

Therefore, I will set

$$N_1 = (1801 - 5) \times (26)^{(1800 - 5)}.$$

The eventual answer will be something like

$$N = N_1 - N_2 + N_3 - N_4 + \cdots.$$

To compute $N_2$, it would be wrong to presume that the computation is

$$N_2 = (1801 - [1 \times 5]) \times (1801 - [2 \times 5]) \times (26)^{(1800 - [2 \times 5])}.$$

The reason that the above approach is wrong is that it presumes that regardless of the placement of one of the words, there will automatically be 1791 possible locations for the start of the 2nd word.

If the first word happens to start in position 1 or position 1796, then yes there will be 1791 possible locations for the 2nd word.  However, as the first letter of the first word changes location from 1 to 2 to 3 to 4 to 5, the possible locations for the start of the second word decrease by 1.

For example, if the first word begins on position 5, then the second word can not begin before position 10.  Note that by the simplifying assumption that the 5 letters in the word are distinct, the two words can not overlap.

If the first letter of the first word is located anywhere between positions 5 and 1792 inclusive, (1788 positions) then there will be (1796 - 9) = 1787 possible starting points for the second word.

If the first word starts in position $i$, for $i \in \{1,2,3,4\}$ then there will be $(1792 - i)$ possible starting positions for the second word.  By symmetry, similar considerations apply if the first word starts on or after position 1793.

Let

$$T_2 = \left[2 \times (1788 + 1789 + 1790 + 1791)\right] ~+~
\left[1788 \times 1787 \right].$$

Then the correct computation is

$$N_2 = T_2 \times (26)^{(1800 - [2 \times 5])}.$$

The math for $N_3$ will get uglier still than the math for $N_2$.  This is because in addition to being concerned about the starting position of the leftmost word and the rightmost word,
you also have to be concerned about the gap between two of the words.  This will also affect how many positions the third word can start in.

Similarly when considering $N_4, N_5, \cdots$ you have to be concerned about each of the gaps between words.

Told you the math was ugly.

If somebody has a better way of doing this that in no way involves a computer algorithm, I'd like to know about it.


问:
Is log x + log y monotonically increasing with respect to x + y?
We know $\log x + \log y = \log(xy)$ which is a monotonically increasing function with respect to the product $xy$.

I am wondering if $\log x + \log y$ is still a monotonically increasing function of the sum $x+y$?

Body:
No.

For example, $\log 2+\log9=\log 18<\log 25=\log5+\log5$,

though $2+9=11>10=5+5$.


问:
Anti-commutative matrices
If $A$ and $B$ are anti-commutative square matrices, so $AB+BA=0$, how do you

a) prove that $\mathrm{tr}(A)=\mathrm{tr}(B)=0$ and

 b) prove that the order of the matrices is even?

答:
If the matrices are non-singular, then writing $A=-BAB^{-1}$ and taking the trace, we get $\mathrm{tr}A=-\mathrm{tr}A$. Hence $\mathrm{tr}A=0$, and the procedure for $B$ is analogous.

Next compute the determinant of both sides of $AB=-BA$: this yields $\mathrm{det}\,A\,\mathrm{det}\,B=(-1)^N\mathrm{det}\,B\,\mathrm{det}\,A$, where $N$ stands for size of matrices. Now since the $A,B$ are non-singular, both sides of the equality are non-zero and the equality is possible only for even $N$.

ContentLicense: CC BY-SA 3.0

Title: Anti-commutative matrices


问:
Limit of $\sqrt[2^{n}]{2^{n}-n}$
I need to find the limit of this expression:

$\sqrt[2^{n}]{2^{n}-n} = e^{\frac{1}{2^{n}}ln(2^{n}-n)} = e^{\frac{1}{2^{n}}[ln(2^{n}) + ln(1-\frac{n}{2^n})]} \implies e^{\frac{1}{2^{n}}ln(2^{n})} = \sqrt[2^{n}]{2^{n}} \implies 1$

Did I do it right?

Title: Limit of $\sqrt[2^{n}]{2^{n}-n}$

Body: Hint: $ 2^{n-1} < 2^n - n < 2^n, n \ge 2$.


Title: How to determine the values of $x$ such that $\frac{1}{x}=[1,1,1,\dots,x]$ with $n$ $1'$s?

Body:
I'm not sure I understood the question correctly. In the hope of finding some pattern, I made a code in Mathematica that expands $[1,1,1,\dots,x]$ with $n$ $1$'s and solve that equation, for $n=2,3,4,\dots,20$ I obtained:

$$\begin{array}{cc}2&\begin{array}{cc}\begin{array}{c}x\to-\frac{1}{\sqrt{2}}\\\end{array}&\begin{array}{c}x\to\frac{1}{\sqrt{2}}\\\end{array}\\\end{array}\\3&\begin{array}{cc}\begin{array}{c}x\to-\frac{1}{\sqrt{3}}\\\end{array}&\begin{array}{c}x\to\frac{1}{\sqrt{3}}\\\end{array}\\\end{array}\\4&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{2}{5}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{2}{5}}\\\end{array}\\\end{array}\\5&\begin{array}{cc}\begin{array}{c}x\to-\frac{\sqrt{\frac{3}{2}}}{2}\\\end{array}&\begin{array}{c}x\to\frac{\sqrt{\frac{3}{2}}}{2}\\\end{array}\\\end{array}\\6&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{5}{13}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{5}{13}}\\\end{array}\\\end{array}\\7&\begin{array}{cc}\begin{array}{c}x\to-2\sqrt{\frac{2}{21}}\\\end{array}&\begin{array}{c}x\to2\sqrt{\frac{2}{21}}\\\end{array}\\\end{array}\\8&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{13}{34}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{13}{34}}\\\end{array}\\\end{array}\\9&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{21}{55}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{21}{55}}\\\end{array}\\\end{array}\\10&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{34}{89}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{34}{89}}\\\end{array}\\\end{array}\\11&\begin{array}{cc}\begin{array}{c}x\to-\frac{\sqrt{55}}{12}\\\end{array}&\begin{array}{c}x\to\frac{\sqrt{55}}{12}\\\end{array}\\\end{array}\\12&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{89}{233}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{89}{233}}\\\end{array}\\\end{array}\\13&\begin{array}{cc}\begin{array}{c}x\to-\frac{12}{\sqrt{377}}\\\end{array}&\begin{array}{c}x\to\frac{12}{\sqrt{377}}\\\end{array}\\\end{array}\\14&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{233}{610}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{233}{610}}\\\end{array}\\\end{array}\\15&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{377}{987}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{377}{987}}\\\end{array}\\\end{array}\\16&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{610}{1597}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{610}{1597}}\\\end{array}\\\end{array}\\17&\begin{array}{cc}\begin{array}{c}x\to-\frac{\sqrt{\frac{987}{646}}}{2}\\\end{array}&\begin{array}{c}x\to\frac{\sqrt{\frac{987}{646}}}{2}\\\end{array}\\\end{array}\\18&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{1597}{4181}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{1597}{4181}}\\\end{array}\\\end{array}\\19&\begin{array}{cc}\begin{array}{c}x\to-2\sqrt{\frac{646}{6765}}\\\end{array}&\begin{array}{c}x\to2\sqrt{\frac{646}{6765}}\\\end{array}\\\end{array}\\20&\begin{array}{cc}\begin{array}{c}x\to-\sqrt{\frac{4181}{10946}}\\\end{array}&\begin{array}{c}x\to\sqrt{\frac{4181}{10946}}\\\end{array}\\\end{array}\\\end{array}$$

But I don't think there is a visible pattern in there. Now I'm clueless again. Can you help?

Body:
Expanding the first few continued fractions:

$$[1,1,x]=1+\frac 1{1+\frac1x} = 1+\frac x{x+1} = \frac{2x+1}{x+1}$$

$$[1,1,1,x]=1+\frac1{[1,1,x]} = 1+\frac{x+1}{2x+1} = \frac{3x+2}{2x+1}$$

$$[1,1,1,1,x] = 1+\frac1{[1,1,1,x]} = 1+\frac{2x+1}{3x+2} = \frac{5x+3}{3x+2}$$

$$[1,1,1,1,1,x] = 1+\frac1{[1,1,1,1,x]} = 1+\frac{3x+2}{5x+3} = \frac{8x+5}{5x+3}$$

By following this pattern (or by induction), one can observe that

$$[\underbrace{1,1,\dots,1}_{n \text{ ones}},x] = \frac {F_{n+1}x+F_n}{F_nx+F_{n-1}}$$

hence we just need to solve

$$\frac1x=\frac {F_{n+1}x+F_n}{F_nx+F_{n-1}} \\ F_n x+F_{n-1}=F_{n+1}x^2+F_nx \\x = \sqrt{\frac{F_{n-1}}{F_{n+1}}}$$


问:
$\mathcal{L} \leq \mathcal{X}$, $\dim \mathcal{L} < \infty$, and $\mathcal{M}\leq \mathcal{X}$, then $\mathcal{L} + \mathcal{M}$ is closed
Taken from Conways A Course In Functional Analysis Chapter 5 section 2 problem 1.

Problem Statement: In relation to Claim 2.4, show that if $\mathcal{L} \leq \mathcal{X}$, $\dim \mathcal{L} < \infty$, and $\mathcal{M}\leq \mathcal{X}$, then $\mathcal{L} + \mathcal{M}$ is closed.

I wanted to see if the solution I had below was correct.

For some $x \in \mathcal{X}\backslash \mathcal{M}$: $\mathcal{L} = \{\alpha x: \alpha \in \mathbb{F}\}$. Since $\mathcal{X}$ is a finite dimensional subspace of $\mathcal{X}$, it follows that $\mathcal{L}+ \mathcal{M}$ is also a finite dimensional subspace (with dimension $\dim \mathcal{M}+1$) of $\mathcal{X}$. Thus, it is closed.

答:
Here is a slightly more direct proof that $\mathbb F v+M$ is closed, provided $M$ is closed and $v\notin M$.

The hypothesis immediately implies that
$$
  d:=\text{dist}(v,M) >0,
  $$
so, if $\lambda $ is any nonzero scalar, and $m\in M$, one has that
$$
  \|\lambda v+m\| = |\lambda |   \left\|v-\left(\frac{-m}\lambda \right)\right\| \geq  |\lambda |d,
  $$
which implies that
$$
  |\lambda |\leq \frac 1d   \|\lambda v+m\|.
  $$
The grand conclusion is that the map
$$
  \lambda v+m\in \mathbb F v+M \quad \longmapsto \quad \lambda \in \mathbb F
  $$
is continuous.
So, if $\{\lambda _kv+m_k\}_k$ is a sequence in $\mathbb F v+M$, converging to some $x\in  X$, we deduce that the $\lambda _k$ converge
to some $\lambda \in \mathbb F$, and hence also that the $m_k$ converge to some $m$, which must be in $M$.  Needless to say
$$
  x=\lim_k\lambda _kv+m_k = \lambda v+m\in \mathbb F v+M.
  $$

One may then complete the argument using induction,  as in @user847970's answer.


问:
Why do we say A given B in Bayes theorem?
I enjoyed this blog on visualizing Bayes theorem but I have a question on the wording for .

$p(A\mid B) = p(A,B)/p(B)$

In words I understand this to be "probability of A given B"

if we are "given B" why do we need to divide by p(B) ?   Surely if I am "given" something then its probability is $1$?

Why do we say it this way?

[Update]

I am seeing that p(B) = 1 (because we are "given" it )

thus p(A,B)/p(B) = p(A,B)

I am looking for the right words to articulate that we still need to divide by p(B) and that it is not 1

If we were saying "given we know the probability of B" instead of "given B" I would not be so confused.

答:
I think I am conflating "given" with GivenWhenThen in programming.

I might be better of reading it as "when". Thus "when" we zoom in to make $B$ the whole universe, "then" we need to adjust $AB$ by the same ratio that we used to blow up B.


问:
Missing something in a simple derivation of the Double Negation rule via the Negation Introduction Rule.
I quote Shapiro's entry on Classical Logic ( in SEP) where (As) refers to a " rule of assumptons" :

(As) If ϕ is a member of Γ, then Γ⊢ϕ.

Our next clauses are for the negation sign, “¬”. The underlying idea
is that a sentence ψ is inconsistent with its negation ¬ψ. They cannot
both be true. We call a pair of sentences ψ,¬ψ contradictory
opposites. If one can deduce such a pair from an assumption θ, then
one can conclude that θ is false, or, in other words, one can conclude
¬θ

(¬I)If Γ1,θ⊢ψ and Γ2,θ⊢¬ψ, then Γ1,Γ2⊢¬θ

By (As), we have that {A,¬A}⊢A and {A,¬A}⊢¬A. So by ¬I we have that
{A}⊢¬¬A. However, we do not have the converse yet. Intuitively, ¬¬θ
corresponds to “it is not the case that it is not the case that” . One
might think that this last is equivalent to θ, and we have a rule to
that effect...

I see easily how both $A$ and $\neg A$ are derivable by the (As) rule from the set $\{A, \neg A\}$ , but I cannot see how it follows from this that $\{A\}⊢¬¬A$.

That is, I do not understands in which way $\{A\}$ plays the role of the set union $\Gamma_1, \Gamma_2$ in the statement of the rule just before . I even do not understand what plays the role of $\Gamma_1$, nor of $ \Gamma_2$, nor of $ \theta$.

What substitutions should be done in order to recognize clearly in
this proof an instantiation of the negation introduction rule?

答:
You seek to have $\{A\}$ be the union of two sets.  At least one must be $\{A\}$, the other may be $\{A\}$ or $\{\}$.

You seek to derive $A$ from $\neg A$ and one of the above sets (call it $\Gamma_1$).  Since, $A$ cannot derive from $\neg A$ and the emptyset, but is trivially derived from $\{A\}$, the later is what that set shall be.$$\{A\}, \neg A\vdash A$$

You also seek to derive $\neg A$ from $\neg A$ and the other of the above sets (call it $\Gamma_2$); and as either $\{A\}$ or $\{\}$ works... we may use either.$$\{\},\neg A\vdash\neg A$$

So there we go $$\begin{split}\Gamma_1,\theta&\vdash \varphi\\\Gamma_2,\theta&\vdash \neg \varphi\\\hline \Gamma_1,\Gamma_2&\vdash \neg\theta\end{split}\qquad\begin{split}A,\neg A&\vdash A\\\neg A&\vdash \neg A\\\hline A&\vdash \neg\neg A\end{split}$$


答:
Just a little to add to the intuition. The other answer is correct.

One thing I'm confused about the question is that the query is not well-defined: log(xy) can take different values even if you know the value of x+y.

You can plot log(xy) on WolframAlpha and imagine "scanning across the graph" using hyperbolas xy=c (y=c/x) where you adjust (increase, for example) c. You can see such hyperbolas are "level sets" of f.

However, trying to do the same with diagonal lines x+y=b doesn't make sense at all.

All this fundamentally comes down to the fact that log turns multiplication into addition

Title: Is log x + log y monotonically increasing with respect to x + y?

Body:
We know $\log x + \log y = \log(xy)$ which is a monotonically increasing function with respect to the product $xy$.

I am wondering if $\log x + \log y$ is still a monotonically increasing function of the sum $x+y$?


问:
Epicycles as precursors of Fourier series
How convincing an argument can be formulated to claim that the Ptolemaic epicycles were actually an early precursor of Fourier series? Ptolemy lived ~200AD, and so well pre-dates Fourier ~1800.</p>

答:
You might want to take a look at the paper Epicycles are Almost as Good as Trigometric Series by Acosta, Smith, Kosheleva & Kreinovich

They report that epicycles were originally proposed by Appollonius of Perga, late 3rd to early 2nd century BCE, developed by several others, and finalised by Ptolomy in the 1st century AD.

They remark that traditional textbooks term it as 'bad science' that was overthrown by the 'genius' of Copernicus.

But that from the mathematical sense a system of epicycles make perfect sense as a trigonometric series.

They don't comment on any direct connection between epicycles and Fourier series. However, it is known that Fourier was interested in physics - and so likely, astronomy.


问:
Can't find Inverse Gaussian derivative w.r.t. to its parameter
I was looking for the derivative of the Inverse Gaussian cdf w.r.t to its parameters $\lambda$ (here called $k$) and $\mu$, however I couldn't find a single source on the internet with a full derivation. (I actually need the second order derivatives and partial derivatives, but for now I'm ok with the first oder derivatives). I tried to derive it myself as follows:

$$
P(t) =Pr(X< t)= \Phi\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)+e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
$$
Where $\Phi$ is the standard normal distribution cdf.
$$
\frac{\partial}{\partial k}P(t)=\\
\Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)
\color{blue}{\frac{\partial}{\partial k}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)}
+\\
\frac{2}{\mu}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
+\\
\Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
\color{purple}{\frac{\partial}{\partial k}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)}
e^{\frac{2k}{\mu}}
\\= \\
\Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)
\color{blue}{\left(\frac{1}{2t\sqrt{\frac{k}{t}}}\left(\frac{t}{\mu}-1\right)\right)}
+\\
\frac{2}{\mu}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
+\\
\Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
\color{purple}{
\left(-\frac{1}{2t
\sqrt{\frac{k}{t}}}\left(\frac{t}{\mu}+1\right)\right)}
e^{\frac{2k}{\mu}}
$$
$$
\frac{\partial}{\partial \mu}P(t)=\\
\Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)
\color{blue}{\frac{\partial}{\partial \mu}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)}
+\\
\frac{2}{\mu}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
+\\
\Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
\color{purple}{\frac{\partial}{\partial \mu}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)}
e^{\frac{2k}{\mu}}
\\=\\
\Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)
\color{blue}{\left(-\sqrt{\frac{k}{t}}\cdot\frac{t}{\mu^2}\right)}
+\\
\frac{2}{\mu}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
+\\
\Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)
\color{purple}{\left(\sqrt{\frac{k}{t}}\frac{t}{\mu^2}\right)}
e^{\frac{2k}{\mu}}
$$

Does it look right? Is there a smarter/faster way to obtain the second order derivatives?

答:
You asked for a "smarter/faster way":  use a program that does symbolic derivatives.  Here's an approach with Mathematica.  First, define your function.

```<code>P[t_] := CDF[NormalDistribution[0, 1], (t/μ - 1) Sqrt[k/t]] +
  Exp[2 k/μ] CDF[NormalDistribution[0, 1], -(t/μ + 1) Sqrt[k/t]]
</code>```

$\frac{\partial}{\partial k}P(t)$:

```<code>(D[P[t], k] // FullSimplify) //. Erfc[z_] -> 1 - Erf[z] //.
  Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // Together
</code>```

$$\frac{e^{-\frac{k (t-\mu )^2}{2 \mu ^2 t}} \left(4 \sqrt{\pi } k e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(-\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)-\sqrt{2} \mu  \sqrt{\frac{k}{t}}\right)}{2 \sqrt{\pi } k \mu }$$

$\frac{\partial^2}{\partial k^2}P(t)$:

```<code>D[P[t], {k, 2}] //. Erfc[z_] -> 1 - Erf[z] //.
  Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // FullSimplify
</code>```

$$\frac{e^{-\frac{k (t-\mu )^2}{2 \mu ^2 t}} \left(-16 \sqrt{\pi } t^3 \left(\frac{k}{t}\right)^{3/2} e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)+16 \sqrt{\pi } t^3 \left(\frac{k}{t}\right)^{3/2} e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}}+\sqrt{2} \left(-k t^2+\mu ^2 (k+t)-4 k \mu  t\right)\right)}{4 \sqrt{\pi } \mu ^2 t^3 \left(\frac{k}{t}\right)^{3/2}}$$

$\frac{\partial}{\partial \mu}P(t)$:

```<code>D[P[t], μ] //. Erfc[z_] -> 1 - Erf[z] //. Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // Together
</code>```

$$\frac{\exp \left(-\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}-\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}\right) \left(4 \sqrt{\pi } k \exp \left(\frac{2 k}{\mu }+\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}+\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}\right) \Phi \left(-\sqrt{\frac{k}{t}} \left(-\frac{t}{\mu }-1\right)\right)-4 \sqrt{\pi } k \exp \left(\frac{2 k}{\mu }+\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}+\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}\right)-\sqrt{2} t \sqrt{\frac{k}{t}} e^{\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}}+\sqrt{2} t \sqrt{\frac{k}{t}} e^{\frac{2 k}{\mu }+\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}}\right)}{2 \sqrt{\pi } \mu ^2}$$

$\frac{\partial^2}{\partial \mu^2}P(t)$:

```<code>(D[P[t], {μ, 2}] // FullSimplify) //. Erfc[z_] -> 1 - Erf[z] //. Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // Together
</code>```

$$\frac{k e^{-\frac{k (t-\mu )^2}{2 \mu ^2 t}} \left(4 k e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(-\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)+4 \mu  e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(-\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)+\left(-\sqrt{\frac{2}{\pi }}\right) t \sqrt{\frac{k}{t}}\right)}{\mu ^4}$$

Body:
You asked for a "smarter/faster way":  use a program that does symbolic derivatives.  Here's an approach with Mathematica.  First, define your function.

P[t_] := CDF[NormalDistribution[0, 1], (t/μ - 1) Sqrt[k/t]] +
  Exp[2 k/μ] CDF[NormalDistribution[0, 1], -(t/μ + 1) Sqrt[k/t]]

$\frac{\partial}{\partial k}P(t)$:

(D[P[t], k] // FullSimplify) //. Erfc[z_] -> 1 - Erf[z] //.
  Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // Together

$$\frac{e^{-\frac{k (t-\mu )^2}{2 \mu ^2 t}} \left(4 \sqrt{\pi } k e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(-\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)-\sqrt{2} \mu  \sqrt{\frac{k}{t}}\right)}{2 \sqrt{\pi } k \mu }$$

$\frac{\partial^2}{\partial k^2}P(t)$:

D[P[t], {k, 2}] //. Erfc[z_] -> 1 - Erf[z] //.
  Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // FullSimplify

$$\frac{e^{-\frac{k (t-\mu )^2}{2 \mu ^2 t}} \left(-16 \sqrt{\pi } t^3 \left(\frac{k}{t}\right)^{3/2} e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)+16 \sqrt{\pi } t^3 \left(\frac{k}{t}\right)^{3/2} e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}}+\sqrt{2} \left(-k t^2+\mu ^2 (k+t)-4 k \mu  t\right)\right)}{4 \sqrt{\pi } \mu ^2 t^3 \left(\frac{k}{t}\right)^{3/2}}$$

$\frac{\partial}{\partial \mu}P(t)$:

D[P[t], μ] //. Erfc[z_] -> 1 - Erf[z] //. Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // Together

$$\frac{\exp \left(-\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}-\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}\right) \left(4 \sqrt{\pi } k \exp \left(\frac{2 k}{\mu }+\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}+\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}\right) \Phi \left(-\sqrt{\frac{k}{t}} \left(-\frac{t}{\mu }-1\right)\right)-4 \sqrt{\pi } k \exp \left(\frac{2 k}{\mu }+\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}+\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}\right)-\sqrt{2} t \sqrt{\frac{k}{t}} e^{\frac{k \left(-\frac{t}{\mu }-1\right)^2}{2 t}}+\sqrt{2} t \sqrt{\frac{k}{t}} e^{\frac{2 k}{\mu }+\frac{k \left(\frac{t}{\mu }-1\right)^2}{2 t}}\right)}{2 \sqrt{\pi } \mu ^2}$$

$\frac{\partial^2}{\partial \mu^2}P(t)$:

(D[P[t], {μ, 2}] // FullSimplify) //. Erfc[z_] -> 1 - Erf[z] //. Erf[z_] -> 1 - 2 Φ[-Sqrt[2] z] // Together

$$\frac{k e^{-\frac{k (t-\mu )^2}{2 \mu ^2 t}} \left(4 k e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(-\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)+4 \mu  e^{\frac{k (\mu +t)^2}{2 \mu ^2 t}} \Phi \left(-\frac{\sqrt{\frac{k}{t}} (\mu +t)}{\mu }\right)+\left(-\sqrt{\frac{2}{\pi }}\right) t \sqrt{\frac{k}{t}}\right)}{\mu ^4}$$


问:
Maximum and Minimum value of Quadratic form
Consider the quadratic form $Q(\vec{x})=-2x^2+2\sqrt{33}xy+6y^2$ where $\vec{x}=\begin{bmatrix}x
\\
y
\end{bmatrix}$

It is easy to figure out the maximum and minimum value of $Q(\vec{x})$ when $\left \| \vec{x} \right \|=1$

I know how to do this, but what if $\left \| \vec{x} \right \|=5$???? then I don't know how to do it,

For example the minimum value of $Q(\vec{x})$ is $-5$ when $\left \| \vec{x} \right \|=1$, but what is the answer if $\left \| \vec{x} \right \|=5?$
Note: the correct answer is $-125$, but I have no idea why is it -125, any suggestion how to get this?

Body:
J.W. Tanner is correct

Suppose $\vec{x}=5\vec{x}_{0}$ where $||\vec{x}_{0}||=1$. Then

$$
\begin{align}
Q(\vec{x})&=-2x^{2}+2\sqrt{33}xy+6y^{2}\\
&=-2(5x_{0})^{2}+2\sqrt{33}(5x_{0})(5y_{0})+6(y_{0})^{2}\\
&=25\left(-2x_{0}^{2}+2\sqrt{33}x_{0}y_{0}+6y_{0}^{2}\right)\\
&=25Q(\vec{x}_{0})\\
\\
\min{\left(Q(\vec{x})\right)}&=25\min{\left(Q(\vec{x}_{0})\right)}
\end{align}
$$


问:
Zorn's lemma and maximal linearly ordered subsets
Let $T$ be a partially ordered set. We say $T$ is a tree if $\forall t\in T$ $\{r\in T\mid r < t\}$ is linearly ordered (such orders can be considered on connected graphs without cycles, i.e. on trees). By a branch we mean a maximal linearly ordered subset in $T$.

It is easy to prove that each tree has a branch using Zorn's lemma. However, the converse is also true (I read it recently in an article). Can anybody give a sketchy proof?

答:
Assume that we have a poset $(X,\leq)$ such that each linearly ordered subset of it has an upper bound. Define the following tree:  </p>

<p>$s\in T$ if $s\subseteq X$ and $(s,\leq)$ is a linear order and let the order relation of the tree be $s<t$ if and only if $s$ is an initial segment of $t$. It's easy to see that this is a tree. Now take a branch of the tree, $B$. If you take $(\bigcup B,\leq)$  you have a linearly ordered set. By the assumption it has an upper bound, $x$. This upper bound is a maximal element of $X$. Indeed if $x<y$ then $\bigcup B\cup\{y\}$ is above (w.r.t. the tree order) every element of $B$, contradicting the maximality of $B$.</p>

Body:
Assume that we have a poset $(X,\leq)$ such that each linearly ordered subset of it has an upper bound. Define the following tree:

$s\in T$ if $s\subseteq X$ and $(s,\leq)$ is a linear order and let the order relation of the tree be $s<t$ if and only if $s$ is an initial segment of $t$. It's easy to see that this is a tree. Now take a branch of the tree, $B$. If you take $(\bigcup B,\leq)$  you have a linearly ordered set. By the assumption it has an upper bound, $x$. This upper bound is a maximal element of $X$. Indeed if $x<y$ then $\bigcup B\cup\{y\}$ is above (w.r.t. the tree order) every element of $B$, contradicting the maximality of $B$.


问:
$f\colon\mathbb{R}\to\mathbb{R}$ is continuous if and only if for every open set $A$ in $\mathbb{R}$ we have $f^{-1}(A)$ open in $\mathbb{R}$
I would like some feedback and corrections to my proof below that $f\colon\mathbb{R}\to\mathbb{R}$ is continuous if and only if for every open set $A$ in $\mathbb{R}$ we have $f^{-1}(A)$ open in $\mathbb{R}$, using $\varepsilon\text{-}\delta$ definition of continuity.

($\Rightarrow$). If $f$ is continuous, then for all $a\in\mathbb{R}$ we have that for all $\varepsilon>0$ there is a $\delta>0$ such that for all $x\in X$ with $\left|x-a\right|<\delta$ this is going to imply that $\left|f(x)-f(a)\right|<\varepsilon$.

Take an open set $A\subset f[\mathbb{R}]\subset \mathbb{R}$ (I'm not sure if it's necessary to take it as subset of $f[\mathbb{R}]$). For any $f(a)\in A$, because $f$ is continuous, we can choose any $\varepsilon>0$, so we do it by choosing $\varepsilon_{f(a)}>0$ such that $(f(a)-\varepsilon,f(a)+\varepsilon)\subset A$, and this is possible because $A$ is open by hypothesis.

As $f$ is continuous, this $\varepsilon_{f(a)}>0$ will gives us a $\delta_{a}>0$ such that all $x\in \mathbb{R}$ that are within a distance $\delta_a$ from $a$ will have their image $f(x)$ within a distance $\varepsilon_{f(a)}$ from $f(a)$. That is, we are going to have an open interval $(a-\delta_a,a+\delta_a)$, and because all of the $x\in\mathbb{R}$ inside of this interval have their image $f(x)\in A$, this interval $(a-\delta_a,a+\delta_a)$ is a subset of $f^{-1}(A)$.

If we follow this for all $f(a)\in A$, we are going to end up with an open interval with center $a$ and radius $\delta_a$ for each $a\in f^{-1}(A)$, so $f^{-1}(A)$ is open.

($\Leftarrow$). We have that for every open set $A$ of $\mathbb{R}$ we have $f^{-1}(A)$ open in $\mathbb{R}$.

As $A$ is open in $\mathbb{R}$, for all elements $f(a)\in A$ there is an $\varepsilon_{f(a)}>0$ such that $(f(a)-\varepsilon_{f(a)},f(a)+\varepsilon_{f(a)})$ is a subset of $A$.

So we can obtain an $a\in f^{-1}[(f(a)-\varepsilon_{f(a)},f(a)+\varepsilon_{f(a)})]$. And because $f^{-1}[(f(a)-\varepsilon_{f(a)},f(a)+\varepsilon_{f(a)})]\subset f^{-1}(A)$ and $f^{-1}(A)$ is open, we choose a $\delta_a>0$ such that $(a-\delta_a,a+\delta_a)\subset f^{-1}[(f(a)-\varepsilon_{f(a)},f(a)+\varepsilon_{f(a)})]$.

Therefore, we have that $f[(a-\delta_a,a+\delta_a)]\subset (f(a)-\varepsilon_{f(a)},f(a)+\varepsilon_{f(a)})$ and we conclude that $f$ is continuous.

答:
Before I get in there and deliver all the negative feedback, I'd like to say that you have a firm grasp on the logic behind this proof, and none of the issues I'm going to point out are serious. If I were marking this as an assignment in undergraduate maths, I would award it full marks.

Take an open set $A \subset f[\Bbb{R}] \subset \Bbb{R}$ (I'm not sure if it's necessary to take it as subset of $f[\Bbb{R}]$).

It's not necessary, and indeed (depending a little on how this is interpreted) problematically restrictive. For example, if $f$ is the indicator function of the rationals, then $f(\Bbb{R}) = \{0, 1\}$. There are no open subsets of $\Bbb{R}$ contained in this set except the empty set, and $f^{-1}(\emptyset) = \emptyset$, which is open! So, according to this condition, $f$ should be continuous, when in reality, it is discontinuous everywhere.

You should only assume $A \subset \Bbb{R}$ is open. It doesn't matter if $A$ contains points that are not in $f(\Bbb{R})$ (though these points will not contribute anything to $f^{-1}(A)$).

The rest of the $\implies$ direction is good, as you don't use the assumption that $A \subseteq f(\Bbb{R})$. I would caution you to be a little more consistent with your $\varepsilon$ notation. Are you denoting $\varepsilon$, or $\varepsilon_{f(a)}$? Stick with one or the other, as either one would be acceptable.

For the $\impliedby$ direction, there's no need to talk about an arbitrary open $A$. Specifically, where you say

As $A$ is open in $\Bbb{R}$...

you haven't actually defined an $A$ to speak of.

All you need to do is show $f$ is continuous at an arbitrary $a \in \Bbb{R}$: fix arbitrary $\varepsilon > 0$, and consider $f^{-1}(f(a) - \varepsilon, f(a) + \varepsilon)$. Use the fact that this is open and contains $a$, in much the same way that you have. That is, there's no need to consider an arbitrary open $A$, when you have a very specific open set $(f(a) - \varepsilon, f(a) + \varepsilon)$.


问:
Is there formula that solves quartic equation $ax^4+bx+c=0$
In general form, a quartic equation is $ax^4+bx^3+cx^2+dx+e=0$. I was thinking of the quartic equation of the form
$$ax^4+bx+c=0$$
which resembles a depressed cubic equation.

Does anyone know a formula that solves this type of depressed quartic equation,  simlilar to the formula

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
for the quadratic equation $ax^2+bx+c=0$, or
$$\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ for the cubic equation $x^3+px +q=0$.

答:
The quartic equation of the form

$$ax^4+bx+c=0$$

has the following formulas for its four roots

$$x= -s \pm \sqrt{ \frac b{4as}-s^2},\>\>\> s \pm \sqrt{ -\frac b{4as}-s^2} $$

where $s^2$ satisfies the cubic equation $s^6 -\frac c{4a} s^2-\left(\frac b{8a}\right)^2=0$. Take the example of
$$x^4+2x+\frac12=0$$

for which $s^6 -\frac 18 s^2-\frac1{16}=0$ or  $s^2=\frac1{2}$.
Plug $s$ into the formulas above to obtain
$$x=\frac{-1\pm\sqrt{\sqrt2-1}}{\sqrt2},\>\>\>
\frac{1\pm i\sqrt{\sqrt2+1}}{\sqrt2} $$


问:
Filippov's Inductive proof for Jordan Canonical form

Let $A$ be an $n×n$ complex matrix. Then there exists an invertible matrix $T$ such that $T^{−1}AT=J$ where $J$ is a Jordan form matrix having the eigenvalues of $A$. Equivalently, the columns of $T$ consist of a set of independent vectors $x_1, . . . , x_n$ such that $Ax_k=λ_kx_k$ , or $Ax_k=λ_kx_k+x_{k−1}$.

There exists an inductive proof by Filippov,

When $n=1$ the Jordan canonical form of the matrix $[a]$ is $[a]$ itself.

Assume the existence of a Jordan canonical form for all $r\times r$ matrices, $r=1,2,...,n-1$, ie. $A_{r\times r}$ is simlar to a Jordan matrix.

Consider an $n\times n$ matrix

Assume that $\lambda=0$ is an eigenvalue, ie., $A$ is singular. So the dimention of $C(A)$ is $r<n$.

Then it says that, by the induction hypothesis A (more precisely the linear operator associated with A) restricted by its range has a Jordan canonical form.

How do I make sense of this statement ?

I think that means,

thanks @ancientmathematician for pointing into the right direction.

If we think of another tansformation, $T:range(A)\to range(A)$ and the corresponding matrix is $B_{r\times r}$ associated with it, then from the induction hypothesis there exists a Jordan canonical basis $(w_1,\cdots,w_r)$ for the $range(A)$ such that $Bw_k=λ_kw_k$ , or $Bw_k=λ_kw_k+w_{k−1}$.

The book seems to use same $A$ for the $B_{r\times r}$, might be the source of confusion here !

Ok.

Even if that is the case, how can one advance the proof from here and find the conditions for the Jordan basis for the original matrix $A_{n\times n}$ ?

How do I get from $Bw_k=λ_kw_k$ , or $Bw_k=λ_kw_k+w_{k−1}$.  to $Aw_k=λ_kw_k$ , or $Aw_k=λ_kw_k+w_{k−1}$ for the original $n\times n$ matrix $A$ ?

Please check A  Primer of Abstract Mathematics By Robert B. Ash.

答:
This is not an answer to your question. Speaking of an inductive proof, I remember the one by Gelfand, which appeared in the second revised Russian edition (1950) of his textbook Lectures on Linear Algebra and predated Filippov's (1971) proof. The original proof of Gelfand was divided into several steps, but some steps can be merged to make the proof shorter. Below is the shortened proof.

Let $A$ be an $n\times n$ matrix over an algebraically closed field $\mathbb F$. To prove that it is similar to a Jordan form, we use mathematical induction on $n$. The base case $n=1$ is trivial. In the inductive step, pick any left eigenpair $(\lambda,\mathbf z)$ of $A$. By considering $A-\lambda I$ instead of $A$, we may assume that $\lambda=0$. Hence $\mathbf z^\top A=0$ and $S=\{\mathbf x\in\mathbb F^n:\mathbf z^\top\mathbf x=0\}$ is an $(n-1)$-dimensional invariant subspace of $A$. By induction assumption, the restriction of $A$ on $S$ is similar to an upper triangular Jordan form. Therefore we may assume that $A$ takes the form of the block matrix that appears on the LHS of the following equality:
$$
\underbrace{\pmatrix{
C&&&&\mathbf u\\
&N_1&&&\mathbf v_1\\
&&\ddots&&\vdots\\
&&&N_k&\mathbf v_k\\
&&&&0}}_{A}
\ \underbrace{\pmatrix{
-C^{-1}\mathbf u\\
-N_1^\top\mathbf v_1\\
\vdots\\
-N_k^\top\mathbf v_k\\
1
}}_{\mathbf x}
=\underbrace{\pmatrix{
\mathbf 0\\
\mathbf w_1\\
\vdots\\
\mathbf w_k\\
0
}}_{\mathbf y}.\tag{1}
$$
Here $C$ is a direct sum of nonsingular Jordan blocks and each $N_i$ is a nilpotent Jordan block. With the vector $\mathbf x$ defined in $(1)$, we obtain
$$
\mathbf w_i:=-N_iN_i^\top\mathbf v_i+\mathbf v_i=(0,\ldots,0,v_i^{\text{last}})^\top,\tag{2}
$$
where $v_i^{\text{last}}$ denotes the last entry of $\mathbf v_i$.

Let $\{\mathbf b_1,\mathbf b_2,\ldots,\mathbf b_n\}$ be the standard basis of $\mathbb F^n$. If $A\mathbf x=0$, we are done because the matrix representation of the linear map $\mathbf t\mapsto A\mathbf t$ with respect to the ordered basis $\{\mathbf b_1,\mathbf b_2,\ldots,\mathbf b_{n-1},\mathbf x\}$ is the Jordan form $C\oplus N_1\oplus\cdots\oplus N_k\oplus0$.

If $A\mathbf x\ne0$, by a suitable permutation we may assume that among all $N_i$s such that $\mathbf w_i\ne0$, $N_k$ has the largest index of nilpotency. Let $N_k$ be $m_k\times m_k$. Denote by $\{\mathbf e_1,\ldots, \mathbf e_{m_k}\}$ the standard basis of $\mathbb F^{m_k}$. Then by $(1)$ and $(2)$, we have
$$
A^r\mathbf x
=\pmatrix{\mathbf 0\\ N_1^{r-1}\mathbf w_1\\ \vdots\\ N_{k-1}^{r-1}\mathbf w_{k-1}\\ N_k^{r-1}\mathbf w_k\\ 0}
=\pmatrix{\mathbf 0\\ \ast\\ \vdots\\ \ast\\ v_k^{\text{last}}\mathbf e_{m_k+1-r}\\ 0}\tag{3}
$$
where the first equality is true for every positive integer $r$ and the second one is true when $1\le r\le m_k$. Since $N_k$ has the largest index of nilpotency ($m_k$) among all $N_i$s such that $\mathbf w_i\ne0$, we have $N_i^{m_k}\mathbf w_i=0$ for every $i$. It follows from the first equality in $(3)$ that $A^{m_k+1}\mathbf x=0$ and from the second equality that $\mathcal B=\{\mathbf b_1,\mathbf b_2,\ldots,\mathbf b_{n-(m+1)},A^{m_k}\mathbf x,A^{m_k-1}\mathbf x,\ldots,A\mathbf x,\mathbf x\}$ is a basis of $\mathbb F^n$. Now we are done, because the matrix representation of $\mathbf t\mapsto A\mathbf t$ with respect to $\mathcal B$ is $C\oplus N_1\oplus\cdots\oplus N_{k-1}\oplus J_{m_k+1}(0)$, where $J_{m_k+1}(0)$ denotes a nilpotent Jordan block of size $m_k+1$.


问:
What is the cycle index for the Dihedral Group acting on a hexagon?
What is the cycle index for the Dihedral Group acting on a hexagon?

So I believe the answer is:
$$
\frac{1}{12}(x_1^6 + 2x_6^1 + 2x_3^3 + 4x_2^3 + 3x_2^1x_2^2)
$$
but I'm not sure if I may have mixed up something. These things seem quite tedious!

答:
I think you're not quite right, so let me try and go through the elements one at a time. If I have this correct $x_n^m$ means there are $m$ cycles of order $n$. Let me know if you think one of these is incorrect, and I'll try and explain my reasoning.

id=$x_1^6$

$f$ contributes to $x_2^2x_1^2$

$r,r^{-1}$ both equal $x_6^1$

$r^2,r^{-2}$ both equal $x_3^2$

$r^3,r^3f,rf,r^{-1}f$ contribute to $x_2^3$

$r^2f,r^{-2}f$ both equal $x_2^2x_2^1$

so in total $x_1^6+4x_2^3+2x_6^1+2x_3^2+3x_2^2x_1^2$


问:
Prove that there are infinitely many pairs of positive integer $(p,q)$ such that given functions have distinct integer roots
$f(x)=x^3 -px^2 + qx$ and $g(x)=3x^2-2px+q$

I want to prove that there are infinitely many pairs of positive integer $p,q$ such that $f(x)$ has three distinct integer roots and $g(x)$ has two distinct integer roots. I can show that if $f,g$ have the required roots, then $p$ is multiple of $3$, $q$ is multiple of $9$, $p^2-3q$ is positive perfect square and $p^2-4q$ is positive perfect square. But i do not know how to show there are infinitely many required pairs. We also require that $gcd(p,q)=3$

答: $f(x)=x(x^2-px+q)=0 \implies x = 0 \text{ or } x^2-px+q=0 \implies D_1 \text{ must be a square.} \implies $ $$ p^2-4q \text{ must be a square}$$  In addition $g(x) = 3x^2-2px+q = 0 \implies D_2 = 4(p^2-3q) \text{ must be a square.} \implies$ $$ p^2-3q \text{ must also be a square}$$  Now, take $p=2$ and you will get $4-4q=4(1-q)$ and $4-3q$. From the first one you get $q = 1-n^2$ for some integer $n$, if you plug it to the second one $m^2=4-3q=4-3(1-n^2)=3n^2+1$. $$m^2-3n^2=1,$$ which is a Pell's equation for $d=3$, which generates infinitely many integer solutions from the base solution of $(m,n)=(2,1)$. Finally, for each generated pair of $(m,n)$ we will get a corresponding pair of $(p,q)=(2,1-n^2)$.


问:
How to show $T$ is diagonalizable?
Let $T\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be linear with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$.  Show that $T$ is diagonalizable.

It seems as if this is a very simple question but I'm a bit confused about how to start!

答: Pick eigenvectors corresponding to each of the three distinct eigenvalues. Then they will be linearly independent since the eigenvalues are distinct (by a theorem). So you can make a basis with these eigenvectors. So the matrix of $T$ with respect to this basis is diagonal.</p>

Title: How to show $T$ is diagonalizable?


问:
What is an infinite set with increment operation called?
What is an infinite set with a unary increment operation (along with the inverse, decrement) called?
For example, the set of integers, $\mathbb{Z}$, along with an increment operation (f: x -> x + 1), without any other operations. The presence of such an operation will make the elements of the set have a strict order (I think).
Is there a name for this particular algebraic structure?

答:
This seems to be called a discrete linear order in the literature. See Decidable Discrete Linear Orders (1988, M. Moses). According to these notes, a discrete linear order can be defined as a strict linear order without endpoints which satisfies a certain discreteness condition, namely that every element has an immediate predecessor and an immediate successor in the ordering:

\begin{align}
& \forall x \exists y (x < y \land \neg \exists z (x < z < y)) & \text{($y$ is a successor of $x$)} \\
& \forall x \exists y (y < x \land \neg \exists z (y < z < x)) & \text{($y$ is a predecessor of $x$)}
\end{align}


问:
Properties of a certain matrix in a certain field

Let $\mathbb{F}$ be a field on two elements, $0$ and $1$ and $G = (V,E)$ be a graph. Let $S$ be a $|V|\times |E|$ matrix in $F$ whose $(i,j)$ entry is $1$ if edge $j$ is incident with vertex $i$ and $0$ otherwise. Determine, with proof, which vectors are in the null space of $S$ in the field $\mathbb{F}$ and which vectors are in the row space of $S$ in terms of $G$.

By definition, the right null space of $S$ is the set of vectors $x\in \mathbb{F}^{|E|}$ so that $S x = 0.$ Each column of $S$ contains $2$ nonzero entries, since each edge is connected to $2$ vertices. In order for $Sx = 0,$ we need each vertex of $G$ to be incident with an even number of edges corresponding to entries of $x$ ($0$ if the edges are included and $1$ otherwise). But I'm not sure what the set of these vectors represents in terms of $G$.

As for the row space, by definition, the row space is the set of vectors in $\mathbb{F}^{|E|}$ of the form $S^T x$ for some $x\in \mathbb{F}^{|V|}.$ The rows of $S^T$ each have exactly two $1$'s and the rest of the entries are $0$. So if row $e$ of $S^T$ is incident with two vertices $v_1$ and $v_2$ corresponding to entries of $x$, then the $e$th entry of $S^T x$ will be $0.$ Similarly, if row $e$ of $S^T$ is incident with no vertices corresponding to entries of $x$, then the $e$th entry of $S^T x$ will be $0.$ Only when row $e$ of $S^T$ is incident with exactly one vertex corresponding to an entry of $x$ will the $e$th entry of $S^T x$ be $1$. But again I'm not sure what these vectors $S^T x$ represent in terms of $G$.

答:
In the product $Sx$, $x\in\mathbf{F}_2^{\lvert E\rvert}$, the vector $x$ corresponds to a set of edges of $G$. As you've observed, $Sx=0$ means that every vertex of $G$ is incident on an even number of edges of the edge set represented by $x$. The set of edges of a cycle of $G$ has this property. More generally, the set of edges of tour of $G$ has this property. (In a tour, vertices may be visited multiple times, but edges are used only once.) More generally still, the set of edges of a disjoint union of tours has this property. Such disjoint unions are equivalent to edge sets of even-degree subgraphs, or, in other words, subgraphs in which every connected component has an Euler tour.

With regard to the row space of $S$: in the product $x^TS$, $x\in\mathbf{F}_2^{\lvert V\rvert}$, the vector $x$ represents a set of vertices of $G$. The elements of the vector $x^TS$ correspond to edges. As you have found, an element of $x^TS$ equals $0$ when either both vertices on the corresponding edge are in the vertex set corresponding to $x$ or neither vertex is in that set; an element of $x^TS$ equals $1$ when exactly one of the vertices on the corresponding edge is in the vertex set corresponding to $x$. Hence the vector $x^TS$ represents the set of edges that join a vertex in the set corresponding to $x$ to a vertex not in that set. There is an induced subgraph corresponding to $x$, and $x^TS$ can be thought of as the set of external edges—edges connecting to vertices outside the induced subgraph.

The correspondence between row-space vectors and sets of external edges of induced subgraphs is not one-to-one for the following reasons: (1) if $X$ is the sets of vertices corresponding to $x$, the the subgraph induced by $X$ has the same external edge set as the subgraph induced by $V\setminus X$; (2) if $G$ is not connected then any connected component or union of connected components has no external edges, just as the empty subgraph does. In other words, if $y_1$, $y_2$, ..., $y_m$ are vectors corresponding to vertex sets of connected components, then $y_j^TS=0$ for all $j\in\{1,2,\ldots,m\}$ and
$$
(x^T+y_1^T+\ldots+y_m^T)S=x^TS
$$
for any $x\in\mathbf{F}_2^{\lvert V\rvert}$.

So there is a one-to-one correspondence between vectors in the row space of $S$ and equivalence classes of partitions of the vertex set of $G$ into two parts, where the partition $\{X,Y\}$ is the same partition as $\{Y,X\}$. The notion of equivalence being used here is the following: let $V$ be partitioned as $\{X,Y,C\}$ where $C$ is the set of vertices of a connected component of $G$. Then $\{X\cup C,Y\}\sim\{X,Y\cup C\}$.


问:
Find the center of mass of a quarter circle given by $\sqrt{r^2−x^2}$, $x\in[0,r]$
Q: The center of mass of the quarter circle given by
$y=\sqrt{r^2−x^2}$, $x\in[0,r]$
is the point $(x, y)$.

My first thought was that the circle was bounded by the points $(0,0)$, $(0,r)$, $(r,0)$ and the graph $\sqrt{r^2−x^2}$ but even with that I'm not sure how to proceed.

答:
HINT

The coordinates of the center of mass are given by $x_{cm}$ and $y_{cm}$, where
\begin{align*}
x_{cm} = \frac{\displaystyle\int_{D}x\mathrm{d}m}{\displaystyle\int_{D}1\mathrm{d}m} = \frac{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}x\mathrm{d}y\mathrm{d}x}{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}1\mathrm{d}y\mathrm{d}x}
\end{align*}
Similarly, one has that

\begin{align*}
y_{cm} = \frac{\displaystyle\int_{D}y\mathrm{d}m}{\displaystyle\int_{D}1\mathrm{d}m} = \frac{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}y\mathrm{d}y\mathrm{d}x}{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}1\mathrm{d}y\mathrm{d}x}
\end{align*}

Can you take it from here?


问:
Example of a Function that is Both Continuous and Uniformly Continuous
Let $S \subseteq \mathbb{R}$ be a countable closed subset and let $f: S \to \mathbb{R}$. Is it always possible to produce such a function $f$ that is both continuous and uniformly continuous?

答:
Answer to original question:

No. If $S$ is also bounded, then it is compact, and a continuous function on a compact set is uniformly continuous.

Answer to new question:

Yes. Just take a constant function on S.


问:
Tangent Line Intercept
The question is like this: there is a curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$, where $c$ is a real number and $c > 0$. If $L$ is a tangent line, any tangent line, with only 1 x-intercept and 1 y-intercept, prove that the sum of the x, y-intercepts of $L$ is $c$.

First of all, I used implicit differenciation on the equation. $$\sqrt{x}+\sqrt{y}=\sqrt{c}$$

Then I also put a graph here (The curve when c=4)

But I still do not have idea. Thank you for looking at this question.

答:
Result: Let $\phi$ be the curve in the $xy$-plane defined by $\sqrt{x}+\sqrt{y}=\sqrt{c}$, where $c\in\mathbb{R}^+$, and let $L$ be a line tangent to $\phi$ with exactly one $x$-intercept $x_i$ and exactly one $y$-intercept $y_i$. Then $x_0+y_0=c$.

Proof: We implicitly differentiate $\phi$ to find the slope of $L$:
\begin{align}
\frac{d}{dx}\left[\sqrt{x}+\sqrt{y}\right] &= \frac{d}{dx}\left[\sqrt{c}\right] \\
\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot\frac{dy}{dx} &= 0 \\
\frac{1}{2\sqrt{y}}\cdot\frac{dy}{dx} &= -\frac{1}{2\sqrt{x}} \\
\frac{dy}{dx} &= -\sqrt{\frac{y}{x}}.
\end{align}
Given any point $(x_0,y_0)$ on $\phi$, the slope of $L$ through $(x_0,y_0)$ is $$\frac{dy}{dx}\bigg|_{(x_0,y_0)}=-\sqrt{\frac{y_0}{x_0}}$$ and so the rectangular equation for $L$ is
\begin{align}
y-y_0 &= -\sqrt{\frac{y_0}{x_0}}\left(x-x_0\right) \\
\frac{y}{\sqrt{y_0}}-\frac{y_0}{\sqrt{y_0}} &= -\frac{x}{\sqrt{x_0}}+\frac{x_0}{\sqrt{x_0}} \\
\frac{y}{\sqrt{y_0}}+\frac{x}{\sqrt{x_0}} &= \sqrt{x_0}+\sqrt{y_0} \\
\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}} &= \sqrt{c}.\tag{Since $(x_0,y_0)$ lies on $\phi$}
\end{align}
We now set $y=0$ and $x=0$ to find the two intercepts of $L$, $x_i$ and $y_i$ respectively. Observe that
\begin{align}
\frac{x_i}{\sqrt{x_0}}+0 &= \sqrt{c} \\
x_i &= \sqrt{c\,x_0},
\end{align}
and
\begin{align}
0+\frac{y_i}{\sqrt{x_0}} &= \sqrt{c} \\
y_i &= \sqrt{c\,y_0}.
\end{align}
Finally, we sum $x_i$ and $y_i$, which yields
\begin{align}
x_i+y_i &= \sqrt{c\,x_0}+\sqrt{c\,y_0} \\
&= \sqrt{c}\left(\sqrt{x_0}+\sqrt{y_0}\right) \\
&= \sqrt{c}\cdot\sqrt{c} \\
&= c.\tag*{$\blacksquare$}
\end{align}

Title: Tangent Line Intercept


问:
Finding the sides of a right triangle given sine and one side length
The Question:
Consider a right triangle with a side of length x opposite angle A, a side of length y opposite angle B,  and a hypotenuse of length z opposite the right angle. If sinB= 1/√3 and x=4, find the length of the other side (y) and the length of the hypotenuse (z).

Explanation:

This is how I sketched the problem
right triangle diagram

It seems to me that the question has already gave us the value of y and z because sinB = opposite/hypotenuse = 1/√3 = y/z

When I enter 1 as the value for y and √3 as the value for z, I get the answer wrong.

What's the correct way to solve this problem? What am I not getting?

答:
From $\sin B = \frac1{\sqrt 3} = \frac yz$ we have $z= \sqrt 3 y$.

By Pythagoreas' Theorem, we have $x^2+y^2=z^2$.

Hence:

$$16 + y^2 = (\sqrt3 y)^2 = 3y^2$$

This gives $y^2 = 8 \leadsto y = 2\sqrt 2, z = 2\sqrt 6$.


问:
Is log x + log y monotonically increasing with respect to x + y?
We know $\log x + \log y = \log(xy)$ which is a monotonically increasing function with respect to the product $xy$.

I am wondering if $\log x + \log y$ is still a monotonically increasing function of the sum $x+y$?

答:
Let $M$ be fixed and let $0 < h \le \frac M2$.  Then $\lim_{h\to 0}(\ln h + \ln(M-h)  = -\infty$ and $\lim_{h\to \frac M2} (\ln h + \ln(M-h)) = 2\ln \frac M2 = 2\ln M - \ln 2$.

$\ln a + \ln b$ with restriction $a + b = M$ has no lower bounded value.


问:
Optimal allocation between loans
I have a two part problem. The first asks whether I can optimally determine how I should split my weekly paycheck towards two loan payments to ensure the shortest total time to payoff my loans. The second, if possible, asks how to allocate funds if I also split funds into an investment account that grows at some rate. I have changed numbers from my real world use case.

Question 1

Given two loans, the first at \$2,000 compounded daily with an annual rate of 6% and the second at \$4,000 compounded daily with an annual rate of 7%, and I save \$10 a day. How can I optimally split my \$10 between the loans each day, to ensure I payoff the loans in the shortest time possible?

I have come up with three equations that I need to solve for.
$$f_\alpha(t) : \mathbb{N} \rightarrow \mathbb{R} $$
The domain is whole numbers representing each day and the co-domain is a range [0,1] that represents the percent of funds that are added to each loan daily. So if $f_\alpha(1) = 0.2$ then on the first day $t_1$, I would add \$2 to loan1 and \$8 to loan2. Then there are the recursive equations representing the daily loan amounts over time.

$$loan_1(t) = loan_1(t-1)*(1 + \frac {0.06}{365}) + 10*f_\alpha(t) \mid \ loan_1(0) = -2000 $$
$$loan_2(t) = loan_2(t-1)*(1 + \frac {0.07}{365}) + 10*(1-f_\alpha(t)) \mid \ loan_2(0) = -4000 $$

The loan amount on each day is the previous day's loan total multiplied by the fixed rate compounded daily plus the dollar amount added to each loan.

Is there a way for me to solve $f_\alpha(t)$ to minimize the time it takes to get $loan_1(t) + loan_2(t) = 0$? Because $f_\alpha(t)$ has a discrete input, that makes this problem non-differentiable. So I don't think I've defined the problem properly to do this with calculus but I'd be happy to change the problem statement so that $f_\alpha(t) : \mathbb{R} \rightarrow \mathbb{R}$ with your suggestion and a modification to $loan_1(t)$ and $loan_2(t)$ that isn't recursive.

Question 2

Bonus points for adding in a fourth function that represents gains in the stock market with an annual percent yield of 7%. The updated functions would look like this:

$$f_\alpha(t) : \mathbb{Z} \rightarrow \mathbb{R^3} $$
mapped from the the integer domain representing days, to a vector $v_\alpha = [v_1,v_2,v_3]$ where $v_1+v_2+v_3=1.0$ and each value is constricted to the range [0,1]. Each value in the vector represents the percent of funds that are added to each loan/investment daily. So $f_\alpha(1) = [0.1,0.3,0.6]$ would have me insert \$1 into loan1, \$3 into loan2, and \$6 into my investment account, namely the function invest3.

$$loan_1(t) = loan_1(t-1)*(1 + \frac {0.06}{365}) + 10*f_\alpha(t)[1] \mid \ loan_1(0) = -2000 $$
$$loan_2(t) = loan_2(t-1)*(1 + \frac {0.07}{365}) + 10*f_\alpha(t)[2] \mid \ loan_2(0) = -4000 $$
$$invest_3(t) = invest_3(t-1)*(1 + \frac {0.07}{365}) + 10*f_\alpha(t)[3] \mid \ invest_3(0) = 1000 $$

I'm also curious to find out if, for varying market gains, if these functions even converge. Is the problem even constrained enough? For example do I need to set a 5 year limit, and simply try to maximize wealth by paying off loans and investing over a 5 year period? I do not know, but I believe answering Question 1 will help in my quest.

Conclusion

I am a first time user so please point out any mistakes in my line of reasoning. Solving this answer will help me in the real world because the true loan amounts are much much larger. I'm an American student who paid on the order of $100,000 for tuition, so solving atleast Question 1 will help me tremendously!

答:
The simple answer is to look at what allocation helps your net worth tomorrow the most.  When looking at loans, pay the high interest one first.  Probably each loan has a minimum payment that if you don't pay it generates a fee that is effectively a higher interest rate.  In that case, make the minimum payment on the $6\%$ loan and put all the rest toward the $7\%$ loan until it is paid off.  You may hear of a "snowball" strategy, which says to pay the maximum on the smallest loan.  The justification is psychological, in that getting a loan paid off is rewarding and will help you stay on course.  The difference in interest is often small and if the encouragement is valuable to you it can be worthwhile.  That is a nonmathematical question.  If you are interested, I suggest making a spreadsheet that shows the balance each month based on how you make your payments.  With these numbers, I think you will find the difference is tiny, so encouragement may be worth the small cost, but that is up to you.

When you look at investments, until taxes intervene you are better off borrowing if the investment returns more than the interest cost of the loan.  If you borrow a dollar at $7\%$ to invest in something that returns $7\%$ after a year you have an investment worth $1.07$ and owe $1.07$ for a wash.  If you can deduct the interest but not pay taxes on the profit (unlikely), you are ahead.  If you pay taxes on the profit but cannot deduct the interest (more likely), you are behind.

Title: Optimal allocation between loans


问:
On the no trivial $3$-tuples $(p, q, \alpha) \in \mathbb{N}^3$ such that $\sum_{k = 1}^{n}k^p =\Big [\sum_{k=1}^{n}k^q\Big]^\alpha $.
It is well known that $\sum_{k = 1}^{n}k^3 =\Big [\sum_{k=1}^{n}k^1\Big]^2$.  My question is very simple.

There are   $3$-tuples $(p, q, \alpha) \in
 \mathbb{N}\times\mathbb{N}\times\mathbb{N}$, in addition to $(3,1,2)$,
   such that $\alpha\geq 2$ and  $$\sum_{k = 1}^{n}k^{\,p} =\Big [\sum_{k=1}^{n}k^{\,q}\Big]^\alpha, \quad \forall n \in \mathbb{N}{\;\Large ?}$$

答:
[The following claims should be obvious, and will not be proven.]

Claim: The degree of $ \sum_{i=1}^n i^r$ is $r+1$.

Claim: The leading coefficient of $ \sum_{i=1}^n i^r$ is $\frac{1}{r+1} $.

The first claim gives us $p+1 = \alpha (q+1) $. The second claim gives us $\frac{1}{p+1} = \left( \frac{ 1}{q+1} \right)^\alpha$.

Multiplying these two equations, we get $(q+1) = \alpha^\frac{1}{\alpha-1}$

Claim: $1 \leq \alpha^{\frac{1}{\alpha-1}} \leq 2$, with equality if and only if $\alpha = 1, 2$.

With this claim, the RHS is never an integer except for $\alpha = 1, 2 $.

If $\alpha = 1$, then $p+1 = q+1, p=q$.

If $\alpha = 2$, this forces $q+1=2$ or $q=1$, and that $p+1 = 2(q+1) $ so $p=3$.

Finally, verify that $(p,p,1) $ and $(3,1,2)$ are solutions.


答:
First notice that every uniformly continuous function is always continuous, so that part is redundant in your question.

HINT Consider a function $f:\mathbb{N}\rightarrow\mathbb{R}.$

ANSWER
Next let $f:\mathbb{N}\rightarrow\mathbb{R}$ be a some function. Let $\epsilon>0$ and $\delta=0.5.$ Then for all $|x-y|<\delta,$ $x,y\in\mathbb{N},$ we have that $|f(x)-f(y)|=0<\epsilon.$ (Why?) So $f$ is uniformly continuous.

Title: Example of a Function that is Both Continuous and Uniformly Continuous

Body: Let $S \subseteq \mathbb{R}$ be a countable closed subset and let $f: S \to \mathbb{R}$. Is it always possible to produce such a function $f$ that is both continuous and uniformly continuous?


问:
Curl Vector Fields
Quite a simple question:

$\textbf{u} = (a\cos(x)\cos(y)\cos(z), b\sin(x)\sin(y)\cos(z), c\sin(z))$ is a curl vector field. Which of the following may correctly be deduced?

A: None of the Above

B: $c = 0$, but $a$ and $b$ are arbitrary;

C: $c = 0$, and $a = b$;

D: $a$, $b$ and $c$ are all arbitrary

E: $a = 1$, $b = 1$ and $c = 0$;

I believed the correct answer to be D, however, I know now that the correct answer is C. But I don't know how this can be deduced?

答:
You need to use the properties of curl vector fields. Let $\textbf{u} = \nabla \times\textbf{f}$

Now, we know that

$$\nabla \cdot\textbf{u} = \nabla\cdot(\nabla \times\textbf{f}) = 0$$

$$\implies -a\sin x \cos y\cos z + b\sin x \cos y \cos z + c\cos z = 0$$

Since this is an identity, it means that this should be true for all $x,y,z$

$$\implies \cos z \left(c + (b-a) \sin x \cos y\right) = 0$$

Now for this to be true for all $x,y,z$, we have

$$c = 0, a=b$$

Title: Curl Vector Fields


问:
Derivative of Quadratic form using chain rule
I want to differentiate
$$
\frac{\partial}{\partial b} \left( [Z^{\top} (y - X b)]^{\top} W [Z^{\top}(y - X b)] \right)
$$

Note that $b : K \times 1$, $y:G \times 1$, $X:G \times K$,$Z : G \times L$,$W:L \times L$.

I know that $\frac{\partial}{\partial b} A^{\top} x A = (A^{\top} + A)x$, but I can't apply that to chain rule.

How to solve that?

(This equation is needed to solve GMM in econometrics.)

答:
$\def\p{\partial}$
Define the vector
$$\eqalign{
p &= Z^T(Xb-y) \quad\implies\quad dp = Z^TX\,db \\
}$$
Write the function in terms of this new vector and calculate its gradient
$$\eqalign{
\phi &= p^TWp \\&= W:pp^T \\\\
d\phi &= W:(dp\,p^T+p\,dp^T) \\&= (W+W^T):dp\,p^T \\
 &= (W+W^T)p:dp \\
 &= (W+W^T)p:Z^TX\,db \\
 &= X^TZ(W+W^T)p:db \\
 &= X^TZ(W+W^T)Z^T(Xb-y):db \\\\
\frac{\p\phi}{\p b}
 &= X^TZ(W+W^T)Z^T(Xb-y) \\
}$$

In the above, a colon has been used to denote the trace/Frobenius product
$$\eqalign{A:B = {\rm Tr}(A^TB) = {\rm Tr}(B^TA) = B:A}$$
The properties of the trace allow terms in such a product to be rearranged in many ways, e.g.
$$\eqalign{A:BC &= B^TA:C = AC^T:B = etc}$$


问:
A problem about the estimations of a parameter
Let $X_{1}, X_{2}, \ldots, X_{n}$ denote a random sample from an exponential  distribution with density $$f(x)=\left\{\begin{aligned} \left(\frac{1}{\theta} \right)e^{-x/\theta}, x>0\\ 0, \quad \text{elsewhere}\end{aligned} \right.$$
Consider the following five estimators of $\theta$ $$\hat{\theta_{1}}=X_{1}, \quad \hat{\theta_{2}}=\frac{X_{1}+X_{2}}{2}, \quad \hat{\theta_{2}}=\frac{X_{1}+2X_{2}}{3}, \quad \hat{\theta_{4}}=\min(X_{1},X_{2},X_{3}), \quad \hat{\theta_{5}}=\overline{X}$$

Which of these estimators are unbiased?

Among the unbiased estimators, which has the smallest variance?

I know that an estimator is unbiased if $\mathbb{E}[\hat{\theta_{i}}]=\theta$ so, by definition we can see that
\begin{eqnarray*}
\mathbb{E}[\hat{\theta_1}]&=&\int_{-\infty}^{\infty}xf(x)dx=\int_{0}^{\infty}x\left(\frac{1}{\theta}\right)e^{-x/\theta}dx=\theta
\end{eqnarray*}
\begin{eqnarray*}
\mathbb{E}[\hat{\theta_2}]=\frac{1}{2}\mathbb{E}[X_1]+\frac{1}{2}\mathbb{E}[X_2]=\theta
\end{eqnarray*}

\begin{eqnarray*}
\mathbb{E}[\hat{\theta_2}]=\frac{1}{3}\mathbb{E}[X_1]+\frac{2}{3}\mathbb{E}[X_2]=\theta
\end{eqnarray*}

\begin{eqnarray*}
\mathbb{E}[\hat{\theta_5}]=\mathbb{E}\left[ \frac{1}{n}\sum_{1\leq i \leq n}X_{i}\right]=\frac{1}{n}\sum_{1\leq i \leq n}\mathbb{E}[X_{i}]=\frac{n}{n}\theta=\theta
\end{eqnarray*}

Question 1: But how can I calculate $\mathbb{E}[\hat{\theta_4}]$?

b. Using the fact $\mathbb{V}[X_{i}]=\mathbb{E}[X^{2}_i]-\mathbb{E}[X_i]$
I obtained $\mathbb{V}[\hat{\theta_1}]=\theta^{2}, \mathbb{V}[\hat{\theta_2}]=\theta^{2}/2, \mathbb{V}[\hat{\theta}_3]=5\theta^{2}/9$.

Question 2: How can I calculate $\mathbb{V}[\hat{\theta}_4]$ and $\mathbb{V}[\hat{\theta}_5]$?

答:
$$\Pr[\min(X_1, X_2, X_3) > x] = \Pr[(X_1 > x) \cap (X_2 > x) \cap (X_3 > x)],$$ because if the smallest of $X_1, X_2, X_3$ is greater than $x$, then each of them must be greater than $x$; conversely, if each is greater than $x$, then the smallest of them is also greater than $x$.

Then, by independence, the right-hand side is $$\Pr[X_1 > x]\Pr[X_2 > x]\Pr[X_3 > x],$$ and since the $X_i$ are identically distributed, this is $$(e^{-x/\theta})^3.$$  Therefore, $$\Pr[\min(X_1, X_2, X_3) \le x] = 1 - e^{-3x/\theta}.$$  Hence $\hat \theta_4$ is exponentially distributed with parameter $\theta/3$; so what is the expectation?

It should also be obvious how to compute the variance of $\hat \theta_4$.  To compute the variance of $\hat \theta_5$, just use the linearity of variance when the $X_i$ are independent.  In other words,

$$\operatorname{Var}\left[\frac{X_1 + X_2 + \cdots + X_n}{n} \right] = \frac{1}{n^2} \left(\operatorname{Var}[X_1] + \operatorname{Var}[X_2] + \cdots + \operatorname{Var}[X_n]\right) = \frac{n \operatorname{Var}[X]}{n^2}.$$  I leave you to compute the rest.  Note that the question of which estimator has the smallest variance depends on the choice of $n$.


问:
The unique root on (1,2)
For any given $n\geq 2$, let $x^n=\sum\limits_{k=0}^{n-1}x^{k}$ be the equation, prove: there is only one real root which in (1,2). </p>

答:
We have $x^n=\frac{1-x^n}{1-x}$, which rearranges to $0=\frac{1-x^n}{1-x}-x^n=\frac{1-x^n-x^n+x^{n+1}}{1-x}$.  For this to be zero (on $x>1$), $f(x)=x^{n+1}-2x^n+1$ must be zero.  We now have the following facts:

$f(2)=2^{n+1}-2^{n+1}+1=1>0$.

$f(1)=1-2+1=0$.

$f'(x)=(n+1)x^{n}-2nx^{n-1}=0$ has two solutions, $x=0$ and $x=\frac{2n}{n+1}\in(1,2)$

$f'(1)=(n+1)-2n=1-n<0$.

Combining (2) and (4) we know $f(x)<0$ immediately to the right of $1$.  By (1) we know $f(x)>0$ at $2$.  By IVT, $f(x)=0$ somewhere in $(1,2)$.  Hence $f(x)$ has two zeroes, at least, in $[1,2)$.  If there were a third one, then $f'(x)=0$ would have at least two solutions in $(1,2)$ by MVT.  But by (3) there is only one such solution, so there is at most one zero of $f(x)$ in $(1,2)$.

Title: The unique root on (1,2)


问:
What is the 5th digit from the end of the number $5^{5^{5^{5^5}}}$?
I thought of using $\text{mod } 10000$ on this but $10000$ and $5$ are not co-prime nor $10000$ is prime so I was not able to apply Fermat or Euler's Theorem . Any Hint would be appreciated.

答:
You want $\bmod 100000$ to get the fifth digit from the end of the number.

$5^5\equiv1\bmod4$, so by Euler's theorem $5^{5^5}\equiv5\bmod8$,

so by Euler's theorem $5^{5^{5^{5}}}\equiv5^5\equiv5\bmod16$,

so by Euler's theorem $5^{5^{5^{5^5}}}\equiv5^5=3125 \bmod32$.

Also,  $5^{5^{5^{5^5}}}\equiv0\equiv3125\bmod5^5 $.

Therefore, since $5^{5^{5^{5^5}}}\equiv3125\bmod 5^5$ and $\bmod2^5$,

by the Chinese remainder theorem $5^{5^{5^{5^5}}}\equiv3125\bmod10^5$,

so the last five digits of $5^{5^{5^{5^5}}}$ are $\color{blue}03125$.


问:
surface integral with normal unit vector
Let  $\vec f(\vec r) = (y,-x, zxy)$ and let $S$ be the surface $x^2+y^2+3z^2=1$ , $z≤0$, with unit normal vector $\vec n$ pointing in the positive $z$-direction. The value of the surface integral $$\iint_S(\nabla \times \vec f) \cdot \vec n\,\mathrm dS$$ is $n\pi$ where $n$ is an integer. What is the value of $n$?

what i know:

I know you need to use stokes theorem when calculating this. I have calculated the curl of the vector field as $(xz, -yz, -2)$. I dont know where to go from this, any help would be appreciated.

答:
You mention you need to use Stokes Theorem but that is not correct. You can calculate it without that too but usually it is simpler to find surface integral use Stokes Theorem.

You already found curl of $\vec{F}$ as $(xz, -yz, -2)$.

Parametrizing our surface $z = -\frac{1}{\sqrt3} \sqrt {1 - x^2 -y^2}$.

$\frac{\partial z}{\partial x} = -\frac{x}{3z}, \frac{\partial z}{\partial y} = -\frac{x}{3z}$

So $d\vec{S} = (-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1) = (\frac{x}{3z},\frac{y}{3z},1)$

So integral becomes $I = \displaystyle \frac{1}{3}\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (xz, -yz, -2).(\frac{x}{z}, \frac{y}{z}, 3) dx \, dy$

$ = \displaystyle \frac{1}{3}\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (x^2 - y^2 - 6) \, dx \, dy = -2 \pi$


问:
why inverse of CDF generates the samples of PDF
I programmed the "inverse transform sampling" according to its wikipage. It sounds amazing:
Given a PDF:
$$
p(x)
$$
we can generate the samples by
$$
s= F^{-1}(r)
$$
where $r\in (0,1)$ is a uniform distribution and F(x) denotes the CDF of p(x).

I understand what PDF and CDF are, but I cannot see why the inverse of CDF generates the samples of PDF. Could someone explain it to me?

If I shot a vertical ray to a steep segment of the CDF, and then make a vertical line to x-axis, the value $p(x_0)$ should has a high value. This makes sense because the steep segment is easy to get horizontal shots.

答:
$P(F^{-1} (U) \leq t) = P(U \leq F(t)) = F(t)$

Where the last equality follows by the cdf of a uniform. Notice that $F(t)$ is the cdf of the random variable you would like to create samples of.


答: Let $n=2$, so $(1+2^p)=(1+2^q)^{\alpha}$, so $(1+2^q)^{\alpha}-2^p=1$. If $p=1$, then $q=\alpha=1$, contradicting $\alpha \geq 2$. Otherwise $1+2^q, \alpha, 2, p>1$, so by Mihailescu's theorem $1+2^q=3, \alpha=2, p=3$. This gives $(3, 1, 2)$, which is indeed a solution. </p>

Title: On the no trivial $3$-tuples $(p, q, \alpha) \in \mathbb{N}^3$ such that $\sum_{k = 1}^{n}k^p =\Big [\sum_{k=1}^{n}k^q\Big]^\alpha $.

Body:
It is well known that $\sum_{k = 1}^{n}k^3 =\Big [\sum_{k=1}^{n}k^1\Big]^2$.  My question is very simple.

There are   $3$-tuples $(p, q, \alpha) \in
 \mathbb{N}\times\mathbb{N}\times\mathbb{N}$, in addition to $(3,1,2)$,
   such that $\alpha\geq 2$ and  $$\sum_{k = 1}^{n}k^{\,p} =\Big [\sum_{k=1}^{n}k^{\,q}\Big]^\alpha, \quad \forall n \in \mathbb{N}{\;\Large ?}$$


答: $F(s)$ is a mapping of the domain of the random variable with that cdf into unit interval.

Title: why inverse of CDF generates the samples of PDF

Body:
I programmed the "inverse transform sampling" according to its wikipage. It sounds amazing:
Given a PDF:
$$
p(x)
$$
we can generate the samples by
$$
s= F^{-1}(r)
$$
where $r\in (0,1)$ is a uniform distribution and F(x) denotes the CDF of p(x).

I understand what PDF and CDF are, but I cannot see why the inverse of CDF generates the samples of PDF. Could someone explain it to me?

If I shot a vertical ray to a steep segment of the CDF, and then make a vertical line to x-axis, the value $p(x_0)$ should has a high value. This makes sense because the steep segment is easy to get horizontal shots.


问:
Angle between 3 vectors
Meta elements are tags used in HTML and XHTML documents to provide structured metadata about a Web page. They are part of a web page's head section. Multiple Meta elements with different attributes can be used on the sam</p>

答:
Three points determine a triangle which has three angles.  The angles of a triangle add up to $180^\circ$, so if one of them is more than $45^\circ$, the others must each be less than $135^\circ.$  I did not actually check your results, but perhaps you should calculate all three angles, and see if your answer is self-consistent and also consistent with the problem given.

Here is a picture of the situation:

As others have pointed out, the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of the length of each of the vectors.

Title: Angle between 3 vectors


问:
How do I proof this in Number-Theoretic Function?
Show that $\sigma(p^n)=\frac{p^{n+1}-1}{p-1}$ , where $p$ is prime.

I know I can't take for granted for just plugging a question and request for the solution.
But I have tried it by using the proof of $\sigma(n)$ but somehow my solution is not convincing, and I think I am not in the right path. Can anyone please guide me? Will appreciate it alot...

This is the formula for $\sigma(n)$ :

答:
You don't need anything to prove it. If $p$ is a prime, the only prime divisor of $p^n$ is $p$. Also, for all natural $k \leqslant n$, we have $p^k | p^n$, and these are the only divisors of $p^n$ since $p$ is a prime.

Thus, $$\sigma(p^n)=1+p+p^2+\cdots+p^n = \frac{p^{n+1} - 1}{p-1}$$


问:
How to evaluate $\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{itx^2}e^{-x^2/2} \, dx$?
I need the solution of this integral:

$$\dfrac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{itx^2}e^{-x^2/2} \, dx$$ with $t
\in \mathbb R$.

I know that $e^{itx^2}=\cos(tx^2)+i\sin(tx^2)$.
Maybe the Fourier transform or the theory of complex numbers can be useful.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{{1 \over \root{2\pi}}\int_{0}^{\infty} \expo{\ic tx^{2}}\expo{-x^{2}/2}\,\dd x}
\\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,&
{1 \over 2\root{2\pi}}\int_{0}^{\infty}
x^{\color{red}{1/2} - 1}\,\,
\expo{-\pars{1/2 - \ic t}x}\,\,\,\dd x
\end{align}
I'll evaluate the integral by means of the
Ramanujan's Master Theorem.
Note that
$$
\expo{-\pars{1/2 - \ic t}x} =
\sum_{k = 0}^{\infty}{\bracks{-\pars{1/2 - \ic t}x}^{\,k} \over k!} =
\sum_{k = 0}^{\infty}\color{red}{\pars{{1 \over 2} - \ic t}^{\,k}}\,
{\pars{-x}^{k} \over k!}
$$
Then,
\begin{align}
&\bbox[5px,#ffd]{{1 \over \root{2\pi}}\int_{0}^{\infty} \expo{\ic tx^{2}}\expo{-x^{2}/2}\,\dd x}
\\[5mm] = &
{1 \over 2\root{2\pi}}\Gamma\pars{1 \over 2}
\pars{{1 \over 2} - \ic t}^{\,-1/2}
\\[5mm] = &\
\bbx{1 \over 2\pars{1 - 2\ic t}^{1/2}} \\ &
\end{align}


Title: Angle between 3 vectors

Body: Meta elements are tags used in HTML and XHTML documents to provide structured metadata about a Web page. They are part of a web page's head section. Multiple Meta elements with different attributes can be used on the sam</p>

Body:
You seem to have done something wrong in your calculations

$\vec{BA} = (-0.5,-0.5,-0.5)$

$|\vec{BA}| = \frac{\sqrt{3}}{2}$

$\vec{BC} = (0.5,0.5,-0.5)$

$|\vec{BC}| = \frac{\sqrt{3}}{2}$

$\vec{BA}\cdot\vec{BC} = -0.5(0.5) + (-0.5)(0.5) + (-0.5)(-0.5) = -0.25$

So the angle is $\arccos(\frac{-0.25}{(\frac{\sqrt{3}}{2})^2})=\arccos(\frac{-1}{3})=109.5^\circ$


问:
$ \lim_{n \to +\infty} \sqrt[n]{\frac{1}{3^n}+\frac{n^5}{5^n}+\frac{1}{2^{nn^{10^{10}}}}}$ Did I do it right?
I need to find the limit of this expression:

$$ \lim_{n \to +\infty} \sqrt[n]{\frac{1}{3^n}+\frac{n^5}{5^n}+\frac{1}{2^{nn^{10^{10}}}}} \implies \sqrt[n]{0 + 0 + 0} \implies 0$$

from squeeze theorem (for every $n \geq 2$ : $ n^5 < 5^n$):
$$ 0 \Longleftarrow \frac{n^5}{6^n} \leq \frac{n^5}{5^n} \leq \frac{(n+1)^5}{5^n} \implies 0$$

Every other part seams obvious. Did I do it right?

答:
You can't let the argument of the $n$-th root go to infinity while leaving the $n$ in the index of the root fixed, it has to go to infinity too (indeed, it would give you a $0^0$ indeterminate form).
The correct way is notice that
$$\sqrt[n]{\frac{1}{3^n}+\frac{n^5}{5^n}+\frac{1}{2^{nn^{10^{10}}}}}=e^{\frac{1}{n}\log\left(\frac{1}{3^n}+\frac{n^5}{5^n}+\frac{1}{2^{nn^{10^{10}}}}\right)}$$
And the exponent tends to $\log(1/3)$ as $n\to\infty$, to see it you can use the property of log after writing $\log\left(\frac{1}{3^n}+\frac{n^5}{5^n}+\frac{1}{2^{nn^{10^{10}}}}\right)=\log\left(\frac{1}{3^n}\left(1+\frac{3^nn^5}{5^n}+\frac{3^n}{2^{nn^{10^{10}}}}\right)\right)$.


问:
How to label a cube
I can't come up with a clever way to label a cube's faces, such that if I turned it in one of four directions, "Up", "Down", "Left", "Right", I'd know the resulting face by applying a function to the current face number, e.g., right(f) = f + 1, left(f) = f - 1. Any ideas?

Please excuse my terrible ASCII cube. I needn't label them 1 - 6.

Cube:

```<code>     ____
    / 5 /|   3 (back)
   /___/_|
4  | 1 |2|
   |___|/

     6 (underside)
</code>```

答: This is a PDF of my attempt to express rotations of the cube in terms of permutations of the faces. Trying to reproduce it here would cost us both some time.</p>

Title: How to label a cube


问:
Finding the subgroup of $(\mathbb Z_{56},+)$ which is isomorphic with $(\mathbb Z_{14},+)$ by GAP
I am sorting some easy questions for the students in Group Theory I. One of them is:

Is $(\mathbb Z_{14},+)$ isomorphic to a subgroup of $(\mathbb Z_{35},+)$? What about $(\mathbb Z_{56},+)$?

I know the first claim is false because if it is true then I have $14\nmid 35$ which is a contradiction. For the second one, I see the function $f:(\mathbb Z_{14},+)\to(\mathbb Z_{56},+),  f(g)=4g$ is a nice one-one homomorphism.

My problem is to find $f(\mathbb Z_{14})$ by using GAP. I know how to define Cyclic Groups or Free Groups in GAP's environment also. Thanks for the ANY help.

答:
```<code>gap> G:=CyclicGroup(56);
<pc group of size 56 with 4 generators>
gap> S:=AllSubgroups(G); # available since GAP 4.5, mainly for teaching purposes
[ Group([ <identity> of ... ]), Group([ f3*f4^3 ]), Group([ f2*f3*f4 ]),
  Group([ f4 ]), Group([ f1*f2*f3 ]), Group([ f3 ]), Group([ f2 ]),
  Group([ f1, f2, f3, f4 ]) ]
gap> H:=Filtered(S,x->Size(x)=14);
[ Group([ f3 ]) ]
</code>```


问:
Proving linear in
$\langle a_a, a_a \rangle =
\begin{cases}
\frac{1}{9},  & \text{if $a$ $\ne$ $a$}\

\end{cases}$

答:
To begin with, let us consider the linear combination
\begin{align*}
\alpha_{1}a_{1} + \alpha_{2}a_{2} + \ldots + \alpha_{k}a_{k} = 0
\end{align*}
If we take the inner product of such linear combination with $a_{1}$ we get the equation
\begin{align*}
2\alpha_{1}b + \alpha_{2} + \alpha_{3} + \ldots + \alpha_{k} = 0\\\\
\end{align*}
If we apply such inner product to $a_{j}$, where $2\leq j\leq k$, we get the following system of equations:
\begin{align*}
\begin{bmatrix}
2b & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{bmatrix}
\begin{bmatrix}
\alpha_{1}\\
\alpha_{2}\\
\vdots\\
\alpha_{k}
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
\vdots\\
0
\end{bmatrix}
\end{align*}
Such system of linear equations has only the trivial solution iff the determinant of the coefficient matrix is different from zero. Based on such result, we need to determine its value in terms of $b$ and $k$:
\begin{align*}
\begin{vmatrix}
2b & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{vmatrix} =
\begin{vmatrix}
2b + k - 1 & 1 & \cdots & 1\\
2b + k - 1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
2b + k - 1 & 1 & \cdots & 2b
\end{vmatrix} & =
(2b + k - 1)
\begin{vmatrix}
1 & 1 & \cdots & 1\\
1 & 2b & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 2b
\end{vmatrix}\\\\
& =(2b + k - 1)
\begin{vmatrix}
1 & 0 & \cdots & 0\\
1 & 2b - 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
1 & 0 & \cdots & 2b - 1
\end{vmatrix}\\\\
& = (2b+k-1)(2b-1)^{k-1} \neq 0
\end{align*}

and we are done.

Hopefully this helps!

Title: Proving linear in


问:
Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$
How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)\;?$$
I came across this Ahmed integral on the site "Art of problem solving", and have found no proof so far. (These two problems seems to be related though). Any help will be appreciated!

Body:
\begin{align}
J&=\int_0^1 \frac{\arctan\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx\\
K&=\int_0^1 \int_0^1 \frac{1}{(x^2+2)(y^2+2)}dxdy \\
&=\int_0^1 \int_0^1 \frac{1}{4+x^2+y^2}\left(\frac{1}{2+x^2}+\frac{1}{2+y^2}\right)dxdy\\
&=2\int_0^1 \int_0^1 \frac{1}{(4+x^2+y^2)(2+x^2)}dxdy\\
&=2 \int_0^1 \left[\frac{\arctan\left(\frac{y}{\sqrt{4+x^2}}\right)}{(2+x^2)\sqrt{4+x^2}}\right]_{y=0}^{y=1} dx\\
&=2\int_0^1 \frac{\arctan\left(\frac{1}{\sqrt{4+x^2}}\right)}{(2+x^2)\sqrt{4+x^2}}dx\\
&=\pi \int_0^1 \frac{1}{(2+x^2)\sqrt{4+x^2}}dx-2J\\
&=\frac{\pi}{2} \left[\arctan\left(\frac{x}{\sqrt{4+x^2}}\right)\right]_0^1-2J\\
&=\frac{\pi}{2}\arctan\left(\frac{1}{\sqrt{5}}\right)-2J\\
\end{align}

On the other hand,

\begin{align}K&=\left(\int_0^1 \frac{1}{2+x^2}dx\right)^2\\
&=\left(\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_0^1\right)^2\\
&=\frac{1}{2}\arctan^2\left(\frac{1}{\sqrt{2}}\right)
\end{align}

Therefore,

$\displaystyle \boxed{J=\frac{\pi}{4}\arctan\left(\frac{1}{\sqrt{5}}\right)-\frac{1}{4}\arctan^2\left(\frac{1}{\sqrt{2}}\right)}$


问:
Ideal $(2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$

Consider the ideal $J = (2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $J \neq (1)$.

I am a bit stuck. If $J = (1)$, then $J = \{1 \cdot r \mid r \in \mathbb{Z}[\sqrt{-5}\}$, so $J = \mathbb{Z}[\sqrt{-5}]$, so I would only need to find one element of $\mathbb{Z}[-5]$. I feel this isn't the correct strategy, and instead I should in some way use the norm function and the fact that the norm function is multiplicative.

Any help would be appreciated.

答:
Suppose $$(1)=(2,1+\sqrt{-5})$$
then $$1 \in (2,1+\sqrt{-5})$$
thus $\exists$ $a+b\sqrt{-5}, c+d\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ such that

\begin{align*}
   1&=2(a+b\sqrt{-5})+(1+\sqrt{-5})(c+d\sqrt{-5})\\
 &=(2a+c-5d)+(2b+c+d)\sqrt{-5}
 \end{align*}

Thus $2a+c-5d=1$ and $2b+c+d=0$. if so then $$(2a+c-5d)+(2b+c+d)=2a+2b+c-4d=1+0=1$$

reducing $\operatorname{modulo}$ $2$ gives you $0=1$ which is a contradiction.


问:
Is limit of composite function applicable when the function depends on $n$?
I know limit of composite function rule from https://teachingcalculus.com/2019/08/26/limit-of-composite-functions/.

However, is it still possible to compute the following like this? The function itself depends on $n$, so I'm not sure function and limit can be interchanged in this case.

$$\lim_{n\to\infty} \left(1+\frac 1n(t^2+t^3/\sqrt n +t^4/\sqrt {n^3}+ ...)\right)^n=
\lim_{n\to\infty} \left(1+\frac 1nt^2\right)^n=\exp(t^2)$$

答:
As mentioned by Greg Martin, generally you cannot but asymptotics can be useful to guess the answer and then prove it rigorously. As is the case here, the limit is indeed $\exp(t^2)$

Notice that,

$$\left(1+\frac 1n(t^2+t^3/\sqrt n +t^4/\sqrt {n^3}+ ...)\right)^n=\exp\left[n\ln\left(1+\frac 1n(t^2+t^3/\sqrt n +t^4/\sqrt {n^3}+ ...)\right)\right]$$

Since $\exp$ is continuous, the limit equals the $\exp$ of the limit of the inside expression as $n\to\infty$. Now,

$$n\ln\left(1+\frac 1n(t^2+t^3/\sqrt n +t^4/\sqrt {n^3}+ ...)\right)=\frac{\ln(1+t^2/n+t^3/n^{3/2}+t^4/n^{5/2}+\ldots)}{1/n}$$

after which you can use L'Hopital once and see this goes to $t^2$ as $n\to\infty$ and hence the original limit is $\exp(t^2)$


问:
Drawing of Quotient Graph
Question: If R is the equivalence relation defined by the partition {{a,f},{e,b,d},{c}}, find the quotient graph, GR. Graph:

Definition: if G has edge set E and vertex set V and R is the equivalence relation induced by the partition, then the quotient graph has vertex set V/R and edge set {([u]R, [v]R) | (u, v) ∈ E(G)}.

Vertex set ={a,b,c,d,e,f}

Now, I am not sure how I find {([u]R, [v]R) | (u, v) ∈ E(G)} i.e. the edge set?

答:
The vertex set of $G^R$ is the set of the equivalence classes: the three vertices are $\{a,f\}$, $\{e,b,d\}$, and $\{c\}$.

For every edge of $G$, put the an edge between the corresponding vertices of $G^R$. For example, there is an edge $ac$ in $G$: this corresponds to an edge between $\{a,f\}$ and $\{c\}$ in $G^R$. Go through all the edges of $G$ in this way, and you will get all the edges of $G^R$.

It is a matter of definitions whether $G^R$ includes loops (for example, coming from the edge $af$) and parallel edges (for example, edges $ac$ and $cf$ in $G$ correspond to the same edge in $G^R$).

Title: Drawing of Quotient Graph


答:
Let $A$ be the $n \times k$ matrix whose columns are $a_i$. Linear independence of the columns of $A$ is equivalent to $\text{rank}(A)=k$.

You are given that
$$A^\top A = \frac{1}{2} J_k + \left(b-\frac{1}{2}\right) I_k,$$
where $I_k$ is the $k \times k$ identity matrix, and $J_k$ is the $k \times k$ "all ones" matrix. (Phrased differently, $A^\top A$ has diagonal entries equal to $b$, and all other entries equal to $1/2$.)

$\text{rank}(A) = \text{rank}(A^\top A)$

Thus, "$\text{rank}(A) = k$" is equivalent to "$A^\top A$ is invertible," which is equivalent to "zero is not an eigenvalue of $A^\top A$."

The eigenvalues of $J_k$ are $0$ and $k$, so the eigenvalues of $A^\top A$ are $b - \frac{1}{2}$ and $b - \frac{1}{2} + \frac{k}{2}$.

Title: Proving linear in

Body:
$\langle a_a, a_a \rangle =
\begin{cases}
\frac{1}{9},  & \text{if $a$ $\ne$ $a$}\

\end{cases}$