问:
Cubic Diophantine
I cannot seem to find a solution to the following Diophantine Equation: $x^2-y^3=2$, where $x,y \in \mathbb{Z}.$ I thought that I could maybe reduce it to a simpler equation , maybe check for the extension $\mathbb{Q}\left[\sqrt{2}\right],$ but nothing that I have tried seems to work. Perhaps it is a known equation/curve I am not aware of? I was first introduced to the problem when my friend pointed out that he could not solve it. I would appreciate your help. Thanks, in advance.

答:
COMMENT.- The discriminant $\Delta = - (4a ^ 3 + 27b ^ 2)$ of the curve $x ^ 2 = y ^ 3 + 2$ considering it as a case of $y ^ 2 = x ^ 3 + ax + b$ (Weirstrass form) in which they have reversed the usual roles of the coordinates, it is different from zero so the curve is not singular or elliptic. There is an obvious solution $A = (x, y) = (1, -1)$.

It is doubtful that the tangent to the curve in this integer point (that is, $2A$ in the arithmetic of the elliptic curve) produces another integer point (in fact graphically it does not give it) but if this point $A$ is not torsion then it will produce an infinity of rational points on the curve.

We do not apply the operations that define an additive group law on the curve because this is not the case here.

Regarding solving the equation $x^2-y ^ 3 = 2$ by other means, this has the savor of the Catalan's conjecture that lasted almost two centuries to be proven by Mihăilescu's  so we better stop here this comment.

Title: Cubic Diophantine


问:
Reverse Characteristic Function of the Cantor Set
Let $f: [0,1] \to \mathbb{R}$

$f(x) = 0$ if $x\in C$

$f(x) = 1$ otherwise

Show that $f$ is Riemann Integrable and compute
$$\int_{0}^{1} f(x) dx$$

I called this the "reverse" characteristic function since normally the $1$ and $0$ have their places switched. When I did this problem, I thought the answer was the same as the normal characteristic function.
We want to show $U(f) = L(f) = \int_{0}^{1} f(x) dx$
Let $P$ be a partition
First, notice that $L(f,P) = 0$ because there are points in $[0,1]$ that are in the Cantor set.
This means $L(f) = 0$

Now, $U(f,P) = \frac{2^n}{3^n}$ since this is the length of the nth subinterval and the supremum is $1$ due to their being points in $[0,1]$ that are not in the Cantor set.
Taking the infimum on both sides,
$U(f) = inf (U(f,P)) = inf \frac{2^n}{3^n} = 0$ because as $n \to \infty$, this expression goes to $0$.
Therefore, the function is RI since $U(f) = L(f)$ and the value is $0$.

However, my classmates have told me that the answer is $1$. My professor neither confirmed nor denied this, but he did say that my answer was wrong. Considering that the normal characteristic function has a value of $0$ for this problem, it would make sense that this version has $1$ as an answer. However, I do not see how.

Can anyone explain this and show the steps?

Thank you!

答: If $g$ is the characteristic function of $C$ the $f(x)=1-g(x)$ for all $x$. Hence, $\int_0^{1} f(x)dx=\int_0^{1} 1dx-\int_0^{1} g(x)dx=1-0=1$,

ContentLicense: CC BY-SA 4.0


问:
Constructing simply connected CW complex with prescribed homology groups
I'm stuck on a past qualifying exam problem from my university:

Construct two homotopically not equivalent simply connected CW complexes such that $H_0=\mathbb{Z}$, $H_2=\mathbb{Z}$, $H_3=\mathbb{Z}_2$ with all other homology groups zero.

I was able to construct one such CW complex: let $Y$ be $S^3$ with a four cell attached via a degree 2 map, then the wedge $Y\vee S^2$ satisfies the requirements.

Can someone please help me constructing another one? I came up with a lot of constructions that turned out to be homotopy equivalent or the same. Thank you!

答:
This is a good question, but rather hard for someone who has just learned algebraic topology. Here is my thought process (and an answer):

There is a theory of minimal cell complexes that apply to simply connected finite CW complexes, this can be found in Hatcher. It states that there is a cell structure such that for each copy of $\mathbb{Z}$ in the homology, we may find a cell structure with one cell in that dimension, and for each copy of $\mathbb{Z}/p$ we may have a cell in that dimension and one higher (one to supply the nontriviality and one to make it torsion).

This implies that every homotopy type with the homology you ask for has a representative with one 0-cell, one 2-cell, one 3-cell, and one 4-cell. Further, the differential takes the 4-cell to two times the 3-cell. So how on earth could we have two different homotopy types represented? The 2-cell can only attach one way to the 0-cell; the 3-cell is attached via a map $\pi_2(S^2 ) = \mathbb{Z}$ which we know from the differential must be trivial. And we also know the differential from dimension 4 to dimension 3!

Well, the answer is due to the fact that homology can only see adjacent dimensions. In fact, $\pi_3(S^2 \vee S^3) \neq \mathbb{Z}$, it contains things like $\eta + 2$, i.e. the Hopf map + twice the identity. Because the Hopf map interacts only with the 2-skeleton, attaching a 4-cell via this map will result in the same differential as just attaching via multiplication by 2.

Then by analyzing the third homotopy group of the resulting complexes, you can see that this results in a different homotopy type. I believe also that the $\mathbb{Z}/2$ cohomology ring of this complex will be nontrivial.


问:
one point sets are open as a topology
Let $\mathbb{R}^2$ be our set and let the topology $\tau$ be given by the smallest topology such that the intersection of two lines is open.

So I take this to mean singletons are open so for any $\vec{x} \in \mathbb{R}^2$, $\{\vec{x}\} \in \tau$.

So the first question I have is that, do we still have for any $\epsilon>0$ and $x \in \mathbb{R}^2$ that $B_\epsilon(x) \in \tau$? (i.e., is it standard TOGETHER with one point sets being open or?)

Then I need to tell if $(\mathbb{R}^2,\tau)$ is 1st countable? And 2nd countable?

Does the first one mean having countable basis at each point? And the second one just says we have a countable basis?

Any hints would help.

答:
You're right, it implies that all sets $\{x\}$, where $x \in \Bbb R^2$, is open in $\tau$.

But recall that all unions of opens sets are open too, so if $A \subseteq \Bbb R^2$ is any subset, $A = \bigcup\{\{x\}\mid x \in A\}$ is also open in $\tau$ so that $\tau$ is the powerset of $\Bbb R^2$; the topology is discrete.

So $\Bbb R^2$ is first countable trivially, every point $x$ has a local base $\{\{x\}\}$, which is finite, so countable.

For every base for $\tau$ we must have $\{x\}$ in the base for any $X$, so no countable base exists for $\tau$ and the space is not second countable.

Title: one point sets are open as a topology


问:
Name for three-valued sign $+, -, 0$
Is there an accepted term (an adjective or prefix) like strict, trichotomous, strong or definite sign to indicate the three-valued sign whose values are $+$, $-$, and $0$?

Are there words reserved for this use in mathematical languages besides English?

An example is functions or differential equation solutions that do not change "sgn".  So to speak.

Edit.  The phrase should retain the word "sign" and add to it an adjective, prefix or (less likely) a suffix.  So, "super-sign (it)" is hypothetically a candidate, but not necessarily signito, sssignnnn, signulus, signifier, signiance, signplusplus.

答: The signum function meets your criteria.</p>

ContentLicense: CC BY-SA 3.0

Title: Name for three-valued sign $+, -, 0$


问:
Why is the stationary distribution of a Markov Chain one over the mean sojourn time?
I am reading the book "Markov Chains and Mixing Times" by Levin et al. and i have a problem with proposition 1.14, page 12.

There a stationary distribution is constructed by defining for fixed initial state z

$$\tilde \pi(y):=\mathbf E_y(\text{number of visits to y before returning to z})=\sum_{t=0}\mathbf P(X_t=y,\tau_z^+>t)$$

$\tau_z^+$ is the first return time to z.

The proof is all well, but in the end a stationary distribution $\pi$ is constructedby normalizing such that
$$\pi(x)=\frac{\tilde \pi(x)}{\mathbf E_z(\tau_z^+)}$$

Then, without further comment it is stated that this means in particular that
$$\pi(x)=\frac{1}{\mathbf E_x(\tau_x^+)}$$

How do i see this?

答:
The statement is true, but I think the way it is presented is a mistake by the authors. What is obvious is that

$$\pi(z) = \frac{1}{\mathbb{E}_z (\tau_z^+)}.$$

Indeed, starting from $z$, the Markov chain visits $z$ exactly once (at time $t=0$)  before returning to $z$, so $\tilde{\pi} (z) = 1$.

However, the stationary measure we got depends a priori on the state $z$ chosen to construct it. If we use another state $z'$, then we get another stationary measure $\pi'$ such that $\pi'(z') = \frac{1}{\mathbb{E}_{z'} (\tau_{z'}^+)}$, but there is no guarantee that $\pi'(z') = \pi(z')$! This is actually false for certain examples of Markov chains (take e.g. a Markov chain on two states with the identity as the transition matrix).

What saves us is that, if the Markov chain is irreducible, there is indeed a unique stationary probability measure, which implies that $\pi = \pi'$. Anyway, at this point, the proof must somehow use that fact that the Markov chain is irreducible.

I think that this is a mistake by the authors since they do not explicitely mention the irreducibility at this point of the proof, and the uniqueness is only proved in the following section.


答:
This is an elliptic curve, and can be solved by a computer algebra system.

I use the following code on sage cell server, and you can also try it yourself.

```<code>E = EllipticCurve([0, 2])
print(E.integral_points())
</code>```

The output:

```<code>[(-1 : 1 : 1)]
</code>```

Therefore this is (up to sign) the only non-trivial solution.

An interesting fact is that this curve actually has Mordell-Weil rank $1$, meaning that it has infinitely many rational points.

Examples:

```<code>(17/4 : -71/8 : 1)
(127/441 : 13175/9261 : 1)
(66113/80656 : -36583777/22906304 : 1)
</code>```

etc.

All these calculations are done with Sage.

How does Sage do that, and how can Sage be sure that the results are correct? This is a much deeper subject. A good starting point is the GTM book "The Arithmetic of Elliptic Curves" by Joe Silverman.

Title: Cubic Diophantine

Body: I cannot seem to find a solution to the following Diophantine Equation: $x^2-y^3=2$, where $x,y \in \mathbb{Z}.$ I thought that I could maybe reduce it to a simpler equation , maybe check for the extension $\mathbb{Q}\left[\sqrt{2}\right],$ but nothing that I have tried seems to work. Perhaps it is a known equation/curve I am not aware of? I was first introduced to the problem when my friend pointed out that he could not solve it. I would appreciate your help. Thanks, in advance.

Body:
This is an elliptic curve, and can be solved by a computer algebra system.

I use the following code on sage cell server, and you can also try it yourself.

E = EllipticCurve([0, 2])
print(E.integral_points())

The output:

[(-1 : 1 : 1)]

Therefore this is (up to sign) the only non-trivial solution.

An interesting fact is that this curve actually has Mordell-Weil rank $1$, meaning that it has infinitely many rational points.

Examples:

(17/4 : -71/8 : 1)
(127/441 : 13175/9261 : 1)
(66113/80656 : -36583777/22906304 : 1)

etc.

All these calculations are done with Sage.

How does Sage do that, and how can Sage be sure that the results are correct? This is a much deeper subject. A good starting point is the GTM book "The Arithmetic of Elliptic Curves" by Joe Silverman.


问:
Find the condition in which $\angle ADB = 3\angle BAC$
See below in acute triangle $ABC$, $D$ is on $AC$ such that $AD=BC$. $CF$ is the angle bisector of $\angle ACB$. $DE \parallel CF$. $E$ is on $AB$. $AE = CD$.

Prove that $\angle ABC = 2 \angle BAC$. In addition, find the condition in which $\angle ADB = 3 \angle BAC$

I struggle to find ways to use two conditions $AD = BC$ and $AE = CD$

答:
Not an elegant solution, because I'm using trigonometry, but just to show that the correct implication is $∠=2∠$.

Set $a=AE$, $b=AD$ and $x=a/b$.
Note first of all that from similarity of triangles $AED$, $AFC$ one gets:
$$
EF={a^2\over b}.
$$
Moreover, from the angle bisector theorem in triangle $ABC$ we also get:
$$
BF=a.
$$
If $\alpha=∠$ from the cosine rule we obtain then:
$$
\cos\alpha={2+5x+4x^2+x^3\over4+6x+2x^2}={x+1\over2}.
$$
And if $\beta=∠$ from the same rule we get:
$$
\cos\beta={-2+3x+4x^2+x^3\over4+2x}={x^2+2x-1\over2}
=2\cos^2\alpha-1=\cos2\alpha.
$$

EDIT.

See diagram below to get convinced that $\angle ADB$ is NOT in general equal to $3\alpha$.


问:
How many ways to arrange 3 elephants, 1 lion, 2 rabbits, 1 bird

How many ways to arrange 3 elephants, 1 lion, 2 rabbits, 1 bird that
no elephant sit next to a lion? (assuming all the animals from same
type are identical

I thought of approach the way of, All possibilities  - (The possibility we don't want).

I took 2 cases where

Case 1 : elephant, lion ,elephant

Case 2: x, lion, elephant , x is different than an elephant

since it doesnt matter where the 3rd elephant in 2 cases

then take the elephant and the lion as one object

finally $\frac{7!}{3!2!}- (case 1 + 2*case2)$

multiply case 2 by two since one elephant can sit on both sides on the lion

答:
Problem deserves exposition.  First of all, I agree with N.F.Taussig: "strategy is sound".

However, there is a great deal to discuss on this problem:

Next, the initial term of $420 = \frac{7!}{(3!)(2!)}$ indicates that the seating is in a row, not a circle.

This indicates that you actually have 3 Inclusion-Exclusion cases to deal with, rather than 2.

Case-1

The lion is at one of the two ends, and has an elephant for a neighbor.

There are two end-seats (left or right) to choose from.  Further, there are then $\binom{5}{2} \times \binom{3}{2}$ ways of distributing the 5 remaining animals.

Therefore, the enumeration for case 1 is:

$$2 \times 10 \times 3 = 60.$$

Case 2

There are 5 middle seats that the lion might be in, and the lion has an elephant on both sides.  The rabbits can therefore be distributed in $\binom{4}{2}$ ways around the remaining seats, and there are then 2 choices for the remaining elephant.

Therefore, the enumeration for case 2 is:

$$5 \times 6 \times 2 = 60.$$

Case 3

There are 5 middle seats that the lion might be in, and the lion has an elephant on one side only.  There are 5 seats for the lion to choose, and then 2 ways for determining whether the elephant is on the lion's left or right.

If the lion's other neighbor is the bear, then there are $\binom{4}{2}$ ways of distributing the remaining two elephants and two rabbits.

If the lion's other neighbor is a rabbit, then there are $\binom{4}{2} \times 2$ ways of distributing the remaining 4 animals.

Therefore, the enumation for Case 3 is

$$5 \times 2 \times [6 + 12] = 180.$$

Consequently the Inclusion-Exclusion approach yields the answer of

$$420 - 60 - 60 - 180 = 120.$$

Now the fun starts.  Is this (indirect - Inclusion Exclusion) approach the best way of calculating the answer.  A case can be made either way.  In general, Inclusion Exclusion approaches generalize well, which means that for more complicated problems, the approach is best.

This means that one does well to master this approach.  However, in this case, the direct approach, which is given below, is easier.

Case A

The lion is on one of the two end seats, and has either a rabbit or a bear for a neighbor.  The factor of $2$ indicates that the lion could be on either end.

If the neighbor is the bear, then the enumeration is $\binom{5}{3}$, re placing the elephants.  If instead, the neighbor is a rabbit, then the enumeration is $\binom{5}{3} \times 2$.

Therefore, the enumeration for Case A is

$$2 \times 10 \times (1 + 2) = 60.$$

Case B

The lion could be in any of the 5 middle seats, which yields a first factor of $5$.  There are then 3 choices for the lion's left-right neighbors : rlr, rlb, or blr.  Regardless of which of these 3 choices pertain, 3 of the 4 remaining animals will be the elephants, so the possible distribution will then be $\binom{4}{3}.$

Therefore, the enumeration for Case B is

$$5 \times 3 \times 4 = 60.$$

Therefore, the direct approach, which merely adds Case A and Case B gives

$$60 + 60 = 120.$$


问:
Subalgebras of certain C*-algebras
Let $A$ be a C*-subalgebra of $C(X, M_{n}(\mathbb{C}))$ where $X$ is a compact Hausdorff space, does it follow that $A$ is isomorphic to  $C(Y, M_{m}(\mathbb{C}))$ for some $Y\subseteqq X$ and $m<=n$ ?</p>

答: This is not even even true for a singleton $X=\{\ast\}$. In this case $C(X,M_n(\mathbb{C}))\simeq M_n(\mathbb{C})$. Consider, for instance, $A$ the subalgebra of diagonal matrices. Then $A\simeq \mathbb{C}^n$. So for $n\geq 2$, this is not isomorphic to any $C(Y,M_m(\mathbb{C}))$ for $Y\subseteq X$, as the latter are isomorphic to $M_m(\mathbb{C})\not\simeq \mathbb{C}^n$.</p>

Title: Subalgebras of certain C*-algebras


问:
Simplify conditional probability tree
There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".

This question is similar to my last question except the desired outcome is rolling one of each number, the probability being:

$$\frac{6!}{6^6} = \frac{720}{6^6}$$

An example of this kind of roll would be "123456" or "352461".

I need to determine the breakdown of the amount of rolls by number of gold sides. Using the method of the accepted answer I could get the number of rolls with 6, 4, and 3 gold sides (exactly 5 gold sides is impossible as the last side must also be gold in order for the roll to contain one of each number).

$$G_6 = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 1$$

$$G_4 = {6 \choose 4} \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 15$$

$$G_3 = {6 \choose 3} \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 40$$

Now for $G_2$ the probability of the 4th dice being gold became conditional on the position of the 3rd dice, and the 5th conditional on the 3rd and 4th. I drew a probability tree to find there are 9 positions out of 24 for the last 4 dice not to be the number of their gold side.

And then plug that in to find the answer:

$$G_2 = {6 \choose 2} \times \frac{1}{6} \times \frac{1}{5} \times \frac{9}{24} \times 6! = 135$$

But I am wondering what is the better way to find the answer (without having to draw the tree, as a tree for $G_1$ and $G_0$ would be even larger and more tedious).

答:
There is no easy way to do it with a probability tree. For any $k\in \{0,\dots,6\}$, the number of rolls with $k$ gold faces is equal to the number of permutations of $\{1,\dots,6\}$ with $k$ fixed points, which is in turn equal to $\binom{6}k$ times the number of derangements of a set of size $6-k$. The number of derangements of a set of size $n$ is well known to be
$$
n!\sum_{i=0}^n\frac{(-1)^i}{i!}
$$
which can be found using the principle of inclusion-exclusion.

Title: Simplify conditional probability tree


问:
How to express this axiom in first-order logic
Reading in a book on logic I found some examples of structures and languages in first order logic:

The language $L_{geom}$ of basic plane geometry has two 1-place relation symbols $P$
and $L$ for “point” and “line”, and a 2-place relation symbol $I$ for “point $x$ lies on
line $y$”. Examples of axioms are:

$\forall x(P(x) ∨ L(x))$

$\forall x¬(P (x) ∧ L(x))$

these axioms mean: everything is either a point or a line; and
not both.

how to express this axiom in $L_{geom}$: for every two points there is a unique line they lie on.

答:
$\forall x,y (P(x) \land P(y) \implies \exists z ( L(z) \land I(x,z) \land I(y,z)))$

is the part of the axiom concerning the existence of a line that joins every two points. I’ll let you figure out how to formalize the uniqueness of that line (Hint: what must happen if we pick two lines joining the same pair of points?)


问:
Combinatorics, number theory, rhombic rectangular prism
$$
1\cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n\cdot(n+1)=\frac{n(n+1)(n+2)}{3},
$$
right???

So will $\dfrac{n(n+1)(n+2)}{3}+1$ form a rhombic rectangular pyramid?

答:
Not sure what you mean.

For $1\cdot2+2\cdot3+\cdots+n\cdot(n+1)$ (hence without the extra $+1$), you can get the following, for $n=5$:

However, it doesn't really look like a pyramid.

Of course, with $1\cdot1+2\cdots2+\cdots+n\cdot n$, it's easy to get a pyramid:

You should explain a little bit what you mean by a pyramid in this context, and why, with $\frac13n(n+1)(n+2)+1$, you expect to get a pyramid.


问:
A problem about the limit of a sequence having to do with the partial sums of another sequence
Let us suppose that  $\{a_n\}_{n \in \mathbb{N} \cup \{0\}}$ is a sequence of positive real numbers and that the sequence $s_{0}, s_{1}, s_{2}, \ldots$ is defined by $s_{k}=a_{0}+\ldots+a_{k}$ for every $k \in \mathbb{N}\cup \{0\}$.

Let us suppose that $\lim_{k \to \infty} \ln \left(\frac{s_{k+1}}{s_{k-1}}\right)=0$ and that $\lim_{k \to \infty} \ln(s_{k+1}s_{k})=\ell$ for a certain real number $\ell$.

Do the previous conditions are sufficient to guarantee that the sequence $\{\ln (s_{k})\}$ has a limit as $k \to \infty$?

Thanks in advance for your comments and answers.

答: $(s_k)$ is an increasing sequence and so is $(\ln (s_k))$. So we only have to check if this sequence is bounded. Suppose $\ln (s_k) \to \infty$. Then $\ln (s_{k+1}s_k) =\ln (s_{k+1})+\ln (s_k) \to \infty$ contradicting the hypothesis. Hence $(\ln (s_k))$ is convergent.


问:
3D plotting of real life object to find Volume of Revolution
I have an issue in a project where I am trying to find volume of a water pot through the use of double and triple integration. for that, I have to first plot the pot on a 3d graph but I am not really sure how it works or how it goes. Please help out. Moreover, performing integrals through the graph so I need to obtain equation of graph and also limits. Please helpHere is the picture of the water pot

With the help of the answer I plotted a circle and hyperbola reffereing to the points that I already had. I am not able to graph it on 3D platform, can someone help:2D graph:

2D graph

答:
I don't really know if this is what you want but I tried modeling that pot using simple functions (a sphere and a hyperboloid) and it went something like this:
I made a model in 2D and one in 3D, the 2D one is a slice of the 3D one, the equations are the ones seen in the pictures.
Those are just a sketch, I didn't actually check if the sphere intersects with the hyperboloid or not, and the values I used were chosen by eye, if this is what you're looking for I can help you finding good values for those constants afterwards.
Also, the pot on the picture is a bit more flat than a sphere, I could also change that on our model if you'd like, using something like a parabola or so.
What are your toughts? Does this help or is this not what you're looking for?
(If this doesn't help, you can also try out fourier series, I don't actually know how to model that with those but I know its possible, its just a matter of research).


问:
What's the reasoning behind the parameterization in this line integral?
I could not quite get why we have parameterized r(t) as in the picture below.

In fact, parameterizing curves whenever I try to solve a problem remains to be the only step I'm having trouble with in line integrals even after reading many things on the topic and thinking that I've understood them. It makes sense after calculating, but it seems pretty arbitrary and without any well-defined method at the beginning.

答:
Points given on the curve: $r_0 = (0,0), r_1 = (3,3a), r_2 = (2,4a)$

It is asking you to choose a parametrization for the curve between points $r_0$ and $r_1$ and then between $r_1$ and $r_2$.

It is not that the only curve between the first two points is a straight line. It could be any curve, for example, a parabola $y = \frac{a}{3}x^2$ with parametrization $(t, \frac{at^2}{3}), 0 \leq t \leq 3$. I do not think you have written down the original question but as it says mountain hike, I am assuming there is something in the question why solution has chosen a straight line. Note the line integral may vary depending on which curve you choose as the vector field is not conservative.

Now to your point on confusion in parametrization, see if the below helps in how to parametrize a straight line,

The first step to parametrize a straight line is to subtract position vector of start point from the end point so you get direction vector from start point to end point. Then the equation of oriented line is

$(0,0) + (3-0,3a-0)t = (3t, 3at)$. Now as you can see substituting $t = 0$  gives you the start point $(0,0)$ and $t = 1$ gives you the end point. So $0 \leq t \leq 1$. If we parametrized as $(t, at)$, then $0 \leq t \leq 3$.

$x'_t = 3, y'_t = 3a$ for parametrization $(3t, 3at)$ (which is nothing but the directional vector we found above by subtracting start point from the end point).

From here, the line integral follows with either the limits $0 \leq t \leq 1$ or $0 \leq t \leq 3$ depending on which parametrization we chose. Let me know if you have any questions.


问:
Probability that order statistic lies within interval
I am having trouble understanding how to begin with this question:

If ten points are picked independently and at random on the interval (0,1):

(a) What is the probability that point nearest 1 (i.e., the largest of the 10 numbers selected) will lie between 0.9 and 1.0.

(b) The probability is 1/2 that the point nearest 0 will exceed what number?

I'm assuming I need to apply the formula below somehow with $n = 10$ and $k = 10$ (since this is the largest value in the sample) and assuming $X$ follows a uniform distribution.

$$
g(x_{(k)}) = nF^{n-1}(x_{(n)})f(x_{(n)}) = \left\{\begin{array}{cl} nx^{n-1}_{(n)} & \text{if}\  0 < x_{(n)} < 1;\\ 0 & \text{otherwise}.
\end{array}\right.
$$

Body:
Call $X_1,\dots,X_n$ the random sample, and $A=\min_i(X_i), B=\max_i(X_i)$.

Then, for $t\in[0,1]$,

$$P(B\le t)=P(\cap \{X_i\le t\})=\prod_iP(X_i\le t)=t^{10}$$

Hence

$$P(0.9\le B\le 1)=1-0.9^{10}\simeq0.65$$

Likewise,

$$P(A\ge t)=P(\cap \{X_i\ge t\})=\prod_iP(X_i\ge t)=(1-t)^{10}$$

Therefore $P(A\ge t)=\frac12$ iff $(1-t)^{10}=\frac12$, iff $t=1-2^{-1/10}\simeq 0.067$


问:
Line graph of square lattice
Is it possible to create a line graph of an n by n square lattice? If so what would it look like and does it have a specific name? I've tried drawing it manually and ended with sideways squares tiled with triangles, looking almost like the $(3^2,4,3,4)$ lattice however I believe this is incorrect.

答:
I drew it out, and the best way I can describe it is the following:

You take the lattice and rotate it 45 degrees. Then for alternating tiles, you draw both diagonals to make it a complete graph.

The whole thing looks almost like a grid of $K_4$ complete graphs stitched together.

EDIT: You can find it on page 2 of this PDF.

Title: Line graph of square lattice


问:
Limit of function involving natural logarithm
I was wondering why the limit of the following function is:

$$ \lim_{x\rightarrow 0^{+}}{\dfrac{\ln x}{x}} = -\infty $$

When we try to solve the limit by substituting in 0, one would expect the numerator to go to minus infinity and the denominator to 0. Thus, having an indeterminate form as a quotient, we could apply L'Hôpitals Rule. But then we would get:

$$ \lim_{x\rightarrow 0^{+}}{\dfrac{1}{x}} = +\infty $$
What's wrong with what I've written above? Thanks beforehand.

答: In the interval (0 1), log x is negative and x is positive. Hence the negative ratio tends to - $\infty$


问:
Can you check to make sure I am completely understanding the Fundamental Theorem of Calculus?
I am not sure that I am completely understanding FTC in terms of how the derivative and definite integral are connected. I drew/wrote this depiction of how I currently think they are connected in terms of velocity and position. I was wondering if anyone could point out any potential gaps in my understanding.

答:
I'd say your geometric picture is a good start for understanding the fundamental theorems of calculus. (I say theorems because really there are two of them.) Say the velocity of a car that has been travelling for one second can be modelled by the equation
$$
v(t)=\sqrt{t+1} \quad{1\leq t} \, ,
$$
The velocity-time graph looks like this:

Since the horizontal axis represents time, and the vertical axis velocity, the area under the curve represents the $\text{velocity $\times$ time $=$ displacement}$ of the car. This means that the displacement of the car (from its position at $t=1$) after $T$ seconds is given by
$$
\int_{1}^{T}v(t) \, dt = \int_{1}^{T} \sqrt{t+1} \, dt \, .
$$
But we also know that the derivative with respect to time of displacement is the velocity, meaning that the antiderivative of velocity is displacement. That last sentence was a little sloppy: every function has infinitely many antiderivatives which differ by a constant. But we can use the information given in this example to narrow things down to one antiderivative:
$$
s(t) = \int v(t)\, dt \, = \int \sqrt{t+1} \, dt = \frac{2}{3}\left(t+1\right)^{3/2}+C
$$
When $t=1$, the displacement of the car is zero. Hence,
$$
s(1) = 0 = \frac{2}{3}\left(t+1\right)^{3/2}+C \implies C = -\frac{2}{3}2^{3/2} \, ,
$$
meaning that
$$
s(t) = \frac{2}{3}\left(t+1\right)^{3/2} - \frac{2}{3}2^{3/2} \, .
$$
Hence, after $T$ seconds, the displacement is given by
$$
s(T) = \frac{2}{3}\left(T+1\right)^{3/2} - \frac{2}{3}2^{3/2}
$$
But we also know that
$$
s(T) = \int_{1}^{T} \sqrt{t+1} \, dt \,
$$
which gives us a clue as to how integrals and antiderivatives are related. In more general terms,
$$
s(T) = \int_{1}^{T} v(t) \, dt = \int v(t) \, dt \bigg|_{t=T}
$$
(Here, I'm using the notation
$$
\int v(t) \, dt \bigg|_{t=T}
$$
in a slightly unorthodox way, to refer to one particular antiderivative rather than many.) We are almost there: if we differentiate both sides of
$$
\int_{1}^{T} v(t) \, dt = \int v(t) \, dt \bigg|_{t=T} \, ,
$$
we get
$$
\frac{d}{dt}\int_{1}^{T} v(t) \, dt = v(T) \, ,
$$
which is the first fundamental theorem of calculus. We have at last discovered a concrete connection between integrals and antiderivatives. Again, there is a nice geometric interpretation. Recall that by definition,
\begin{align}
\frac{d}{dt}\int_{1}^{T} v(t) &= \lim_{\Delta t \to 0}\frac{\int_{1}^{T+\Delta t} v(t) \, dt-\int_{1}^{T} v(t) \, dt}{\Delta t} \\[6pt]
&= \lim_{\Delta t \to 0}\frac{\int_{T}^{T+\Delta t}v(t) \, dt}{\Delta t} \, .
\end{align}
As $\Delta t$ shrinks towards zero, the area (marked in green below)
$$
\int_{T}^{T+\Delta t}v(t) \, dt
$$
is better and better approximated by the rectangle with width $\Delta t$ and height $v(T)$.

Since
$$
\lim_{\Delta t \to 0}\frac{\int_{T}^{T+\Delta t}v(t) \, dt}{\Delta t}
$$
is, roughly speaking, the ratio of the area of the rectangle to its width, this limit must be equal to the height of the rectangle $v(T)$, which is what we wanted. If I find the time, I'll try to update this post with more information about the second fundamental theorem of calculus.


问:
Showing by definition that a curve is a 1 dimensional differentiable manifold

Let $\gamma :(a,b) \to \mathbb{R}^d$ be a $C^1$ curve and $S = \left \{ x  |  \exists t\in(a,b) : \gamma(t)=x  \right \}$ its image.

Show that if $\gamma$ is injective and $\gamma' (t) \neq \vec{0}, \forall t \in (a,b)$, then $S$ is a one dimensional differentiable manifold in $\mathbb{R}^d$.

Just to clarify the definition we were taught: I need to show that $\forall x \in S$, there exists an open neighborhood $V \subset \mathbb{R}^d$ of $x$ s.t $S \cap V$ is the graph of a differentiable function $f: U \subset \mathbb{R} \to \mathbb{R}^{d-1}$.

I believe this should be a pretty basic exercise but I'm a little lost. Help appreciated.

*EDIT: The above question is false as mentioned, modifying the question this way:

Let $\gamma :(a,b) \to \mathbb{R}^d$ be a $C^1$ curve and $S = \left \{ x  |  \exists t\in(a+\varepsilon ,b-\varepsilon) : \gamma(t)=x  \right \}$ for some $\varepsilon>0$, the image of $\gamma |_{(a+\varepsilon ,b-\varepsilon)}$.

Show that if $\gamma$ is injective and $\gamma' (t) \neq \vec{0}, \forall t \in (a,b)$, then $S$ is a one dimensional differentiable manifold in $\mathbb{R}^d$.

how would we prove it?

答:
NOTE: we'll assume the function $\gamma$ proper, in order to avoid pathological cases (i.e. the curve getting infinitely close to itself - though without self-intersections). I thank Ted Shifrin for his observations and corrections on this proof.

I'll use the notation $\gamma(t)=(\gamma_1(t),...,\gamma_d(t))$.

For a fixed $t_0$ you can take one of the $d$ axis, say $\hat{x_1}$ (if not, $\hat{x_2}$...) such that $\gamma '(t_0)$ is not perpendicular to $\hat{x_1}$.

Then $\gamma_1'(t_0)\neq 0$ and I can take an interval on $\hat{x_1}$ where it remains $\neq0$. Over this interval, I reparametrise $Im\gamma$ as:

$\alpha(s)=(\gamma \circ \gamma_1^{-1})(s) = (\gamma_1 \circ \gamma_1^{-1},\gamma_2 \circ \gamma_1^{-1},...,\gamma_d \circ \gamma_1^{-1})(s) = (s, \alpha_2(s),...,\alpha_d(s)) $

by using that $\gamma_1' \neq 0 \Rightarrow \gamma_1$ bijective.

And there we have a local graph (if we restrict on $V\cap S$ with $V\subset\mathbb{R}^d$ a suitable open set)


问:
Determining all formal power series satisfying the equation $Q^m=(1+x)^n$ for $n,m\in\mathbb{Z}\setminus\{0\}$
The question:

Let $m$ and $n$ be integers $\neq 0$. Determine all solutions $Q=1+b_1x+\cdots$ of the equation $$Q^m=(1+x)^n.$$

My current attempt at a solution:

Let $Q\in\mathbb{F}[[x]]$. The formal power series $1+x$ is equivalent to the binomial series $B_1$. Hence, we can rewrite the given equation in the following form
\begin{align*}
    1+\sum_{k=1}^\infty b_k^{(m)}x^k=Q^m=(1+x)^n=B_1^n=B_n=\sum_{k=0}^\infty\binom{n}{k}x^k.
\end{align*}
Applying the J.C.P Miller formula to $Q^m$ and comparing coefficients we construct the following linear system of equations
\begin{gather*}
    b_0^{(m)}=b_0^m=1=1=\binom{1}{0}=[B_n]_0\\
    b_k^{(m)}=\frac{1}{k}\sum_{l=1}^k[l(m+1)-k]b_lb_{k-l}^{(m)}=\binom{n}{k}=[B_n]_k \mbox{ for } k=1,2,\ldots.
\end{gather*}
Solving the second equation for $k=1$ we have,
\begin{align*}
    \sum_{l=1}^1[l(m+1)-1]b_lb_{1-l}^{(m)}=mb_1b_0^{(m)}=mb_1=n=\binom{n}{1}.
\end{align*}
Hence, $b_1=\frac{n}{m}$. Solving the second equation for $k=2$ we have,
\begin{align*}
    \frac{1}{2}\sum_{l=1}^2[l(m+1)-2]b_lb_{2-l}^{(m)}=\frac{1}{2}\left[(m-1)b_1b_1^{(m)}+2mb_2b_0^{(m)}\right]\\=\frac{1}{2}\left[\frac{(m-1)n^2}{m}+2mb_2\right]=\frac{n(n-1)}{2}=\binom{n}{2} \Rightarrow{} b_2=\frac{n(n-m)}{2m^2}.
\end{align*}
Hence, $b_2=\frac{n(n-m)}{2m^2}$. Suppose that the $k$th term of $Q$ satisfying the given equation is given by $b_k=\frac{\prod_{l=0}^{k-1}(n-lm)}{k!m^k}$ for $k\geq 1$. Then for $k+1$ it follows,
\begin{align*}
b_{k+1}=\frac{\prod_{l=0}^{k+1-1}(n-lm)}{(k+1)!m^{k+1}}=\frac{\prod_{l=1}^k(n-lm)}{(k+1)k!mm^k}=b_k\frac{n-km}{(k+1)m}
\end{align*}

I can't seem to find a way to finish the induction step and have been unable to find an alternative method. Note that I am using $b_k^{(m)}$ to represent the $k$th coefficient of the $m$th power of the formal power series $Q$.

Edit: The field is considered to be an arbitrary field of characteristic zero.

答:
You didn't specify what $\Bbb F$ was to be. Let me tell you what the answer is in case $\Bbb F$ is a field of characteristic $p>0$.

The easy case is the one where we’re asking for $m$-th roots of $(1+x)^n$ when $m$ is prime to $p$. Note that it’s enough to take the case when $n=1$. So we’re just asking for the existence of an $m$-th root of $1+x$, in $\Bbb F[[x]]$.

You could point out that the binomial coefficients in $(1+x)^{1/m}$ have denominators prime to $p$, but proving this seems a pain. Easier to observe that $1/m$ is a $p$-adic integer, i.e. is $p$-adic limit of a sequence of positive integers. In the $p$-adic topology, the powers of $p$ go to zero; and since the $p^k$ powers of $x$ go to zero in $\Bbb F[[x]]$ as $k\to\infty$, it follows that the $p^k$ powers of $1+x$, namely $(1+x)^{p^k}=1+x^{p^k}$ go to $1$ as $k\to\infty$.

Explicitly, then, let $\{\mu_0,\mu_1,\mu_2,\cdots\}$ be a sequence of positive integers with $1/m$ as $p$-adic limit. Then
$$
(1+x)^{m_0},(1+x)^{m_1},(1+x)^{m_2},\cdots
$$
is a convergent sequence in $\Bbb F[[x]]$, whose limit is an $m$-th root of $1+x$. Don’t see that right away? Well, what makes the sequence $\{m_i\}$ $p$-adically convergent is that $m_{i+1}\equiv m_i\pmod{p^{k_i}}$ for an unceasing sequence of integers $\{k_i\}$. That is, $m_{i+1}-m_i=p^{k_i}r_i$ for suitable integers $r_i$. Thus
$$
\frac{(1+x)^{m_{i+1}}}{(1+x)^{m_i}}=(1+x)^{p^{k_i}r_i}=(1+x^{p^k})^{r_i}\longrightarrow1
$$

I’ll leave it to you to worry about the case where $p|m$.


问:
Showing that a general Hessian matrix is positive semidefinite

Given vector $a \in \mathbb{R}^n$, show that the scalar field $g : \mathbb{R}^n \to \mathbb{R}$ defined by $$g(\mathbf{x}) = -\log ( f(\mathbf{x}))$$ where $$f(\mathbf{x}) = \dfrac{1}{1+\exp(-a^T\mathbf{x})}$$ is convex.

To show that, we need to show that the Hessian matrix is positive semidefinite, i.e., $\nabla \nabla g(\mathbf{x})\succcurlyeq 0$. I calculated the Hessian as follows:

\begin{equation*}
    \mathbf{H} = \nabla \nabla g(\textbf{x}) = \left[\dfrac{\partial^2 f(\mathbf{x})}{\partial x_i \partial x_j} \right] =  \begin{cases}
      a_i^2 f^2(\mathbf{x})\exp{(\mathbf{-a}^T\mathbf{x})}  & , \ \ \text{if} \ \ \  i = j \\
      a_ia_j f^2(\mathbf{x})\exp{(\mathbf{-a}^T\mathbf{x})} & , \ \  \text{if} \ \ \  i \neq j
   \end{cases}
\end{equation*}

I am not sure what is the easiest way to show the positive semidefiniteness of such a functional form of the Hessian. I cannot see how we could show $z^T\mathbf{H}z \geq 0$ for all nonzero $z$ or that the eigenvalues of $\mathbf{H}$ are all non-negative. Any suggestions?

答: You do not need the Hessian. The scalar function $h(s)=\log (1+e^s)$ is convex since its second derivative is non-negative. Your function is the composition of $h$ with the linear function $m(\mathbf{x})=-a^T\cdot \mathbf{x}$ so it is convex as well (compositions of convex and linear functions are convex).


问:
Vectors in Physics vs. Vectors in Abstract Linear Algebra
In physics class, students are often shown vectors in $\mathbb{R}^n$ (for $n=2$ or $n=3$ of course) via arrow diagrams and they are told that a vector is an entity with magnitude and direction.

But then in abstract linear algebra, we learn that yes, $\mathbb{R}^n$ is one example of a vector space but there are others like $C^\infty$ functions for example. However, if $C^\infty$ functions are to have magnitude and direction, we must have a norm and an inner product defined on the space as well. It doesn't come automatically with one.

Therefore, we have two definitions of vector:

(1) An entity with magnitude and direction
(2) An entity which satisfies the vector space axioms

These are very different because a vector space can't be assumed as endowed with something that gives its members magnitude and direction. Doesn't this mean the physics lessons are technically giving a wrong definition?

答: A (normed) vector space is filled with objects that have "magnitude and direction" that is, they have a length(metric, derived from inner product) and we know which way they're pointing (there is a set of basis objects that span the space). These rules make up the basis for vector space axioms and this is how we are able to explore objects that permit a reasonable notion of "magnitude and direction" such as functions. This abstraction has been a critical component of much of modern math.</p>


问:
How to find $\overline{abc}_{7}-\overline{cba}_{7}=\overline{xy4}_{7}$?
The problem is as follows:

Let:

$\overline{abc}_{7}-\overline{cba}_{7}=\overline{xy4}_{7}$

Assuming $a=b-c$

Find: $cba_{7}$ in terms of decimal system.

How exactly should I do this?. Can someone help me here?.

The thing is I don't know how to deal with carry numerals.

So far the only thing which I spotted is that $c-a=4$

For this the only way to get this is assuming $c=6$ and $a=2$ thus it will not carry. Isn't it?.

But this is somewhat contradictory by looking on the third digit because it states $a-c=2-6=-4$

Which it cannot happen. Thus what am I doing wrong?. Can someone help me here?. My only requirement is a method which doesn't really uses much sophisticated math such as modular arithmetics or too many algebraic manipulations. Does it exist a way to do this more intuitively?

答:
BIG HINT

It is important to see that $c-a$ is not $4$. From the leading digit one has that $a$ is at least as big as $c$. So the only possibility is that $7+c-a=4$ i.e. there has been a carry. We therefore obtain $a=c+3$.

You should now be able to see that $\overline{abc}$ must be treated as $$\overline{(a-1)(b+6)(c+7)}$$.

The rest should be easy but I will add that there seems to be a convention that no leading digits are zero.


问:
Interpretation using First-order logic
The following formula:
$$\forall x(\forall y (\forall z((x=y)\lor(y=z)\lor(z=y) )))$$
is interpreted as "there are at most two elements ". how can we write a formula for "there are exactly two elements"?

I have the idea that this is done as follows:
we can find two elements that are different, but if we take more than two elements then the third is equal to one of the other two.

However, I don't have a very clear idea how to write this as a formula

答: $$\exists x \exists y \big(x\neq y \land \forall z(z=x \lor z=y)\big)$$

Title: Interpretation using First-order logic


答:
Personally, I'd tackle this by using a variation on your sentence for "there are at most 2 elements" and then also requiring there to be at least 2. This gives exactly 2.

$$
\forall x(\exists y ((y\neq x) \land \forall z((y=z)\lor(x=z))))
$$

In English: For all $x$ there's a $y$ that is different to $x$ (at least two elements), and any third element is equal to one of the other two (at most two elements).

Title: Interpretation using First-order logic

Body:
The following formula:
$$\forall x(\forall y (\forall z((x=y)\lor(y=z)\lor(z=y) )))$$
is interpreted as "there are at most two elements ". how can we write a formula for "there are exactly two elements"?

I have the idea that this is done as follows:
we can find two elements that are different, but if we take more than two elements then the third is equal to one of the other two.

However, I don't have a very clear idea how to write this as a formula


问:
Proof technique for between any two real numbers is an irrational number
I'm proving that between any two real numbers is an irrational, and my first thought is to prove that between any two real numbers is a rational number, and then show that between any two rational numbers is an irrational. However, I don't think this would complete the proof because I will not have shown that between any two real numbers there are two rational numbers. Would the first part of my proof imply that there are an infinite number of rationals between any two reals, or do I need to show that separately?

Also, I'm really trying to figure out this problem on my own, so I ask not for the answer, but a pointer in the right direction.

Thanks!

答:
Try this one:

First we show that the sum of a rational number $r$ and an irrational number $i$ is irrational. This can be proved by the fact that $\mathbb{Q}$ is closed under addition. (If $i+r$ was rational, then $(i+r) + (-r) = i$ would become rational).

So our task becomes proving the existence of an arbitrarily small irrational number $s$. If $0<s<y-x$, $x+s$ then becomes an irrtional number between $x$ and y.

We require the existence of any positive irrational number. We may use $\sqrt{2}$. Then consider the sequence $\frac{\sqrt{2}}{n}$, which gets arbitrarily small as $n \to \infty$.

The only thing left now is to prove that $\frac{\sqrt{2}}{n}$ is irrational. The proof can be done by using a process similar to how we proved the sum of a ration number and an irrational number is irrational ($\mathbb{Q}$ is also closed under multiplication).


问:
Does $(n+1)^x-n^x$ go to $\infty$ as $n\rightarrow\infty$ for $x>1$?
Does $(n+1)^x-n^x$ go to $\infty$ as $n\rightarrow\infty$ for $x>1$? I suspect it does from plots, but I want to use the ratio test to prove that
\begin{align}
\sum_{n=0}^\infty \exp(cn-vn^x)
\end{align}
converges for $c,v>0$ and $x>1$. Thanks.

答:
I am not seeing how the ratio test would help per se.

We can see however, that $d(n^x)/dn = (x-1)n^{x-1}$. This goes to $\infty$ with $n$ for $x>1$, and furthermore, is an increasing function. So

$$(n+1)x-n^x \ = \ \int^{n+1}_n(x-1)m^{x-1}dm \ \ge \ 1 \times (x-1)n^{x-1}$$

and as the term on the right goes to $\infty$ as $n$ does, then so does $(n+1)^x-n^x$.


问:
How is the dual space generated from the linear forms?
Been reading about dual spaces without knowing much about it (not a mathematician). Now, after stumbled upon linear forms (function mapping vector space to number), that Wikipedia page reads:

The set of all linear functionals from $\mathbf{V}$ to $k$ , denoted by $\mathrm{hom}(V,k)$ is itself a vector space over $k$ with the operations of addition and scalar multiplication defined pointwise. This space is called the dual space of $\mathrm{V}$, or sometimes the algebraic dual space, to distinguish it from the continuous dual space.

The linear functional it talks about it's just
$$
\begin{align}
f(\vec{x})  &= \vec{w}\cdot{}\vec{x}\\
 &= a_1 x_1 + a_2 x_2 + \ldots + a_n x_n
\end{align}
$$

how does this produce a space of vectors?

As I said before, not a mathematician so don't go over a million definitions (please).

答:
Given an element $w$ of a vector space $V$, there is a linear map $T_w:V\to{\bf R}$ given by $T_w(v)=w\cdot v$. As $w$ runs through $V$, the set of these linear maps takes on the structure of a vector space, and it's called the dual space of $V$.

I'm now going to rewrite this in words of one syllable, as it were.

For each $w=(a_1,a_2,\dots,a_n)$ in ${\bf R}^n$ we can define a function $f_w$ from ${\bf R}^n$ to the real numbers by $f_w(x)=w\cdot x=a_1x_1+a_2x_2+\cdots+a_nx_n$ (where $x=(x_1,x_2,\dots,x_n)$).

If we take two of these functions, say, $f_w$ and $f_z$, we can add them to get a new function from ${\bf R}^n$ to the real numbers; if $g=f_w+f_z$ then $g(x)=f_w(x)+f_z(x)=w\cdot x+z\cdot x=(w+z)\cdot x=f_{w+z}(x)$. That is, it turns out that when you add two of these functions, you get another one of these functions, as $f_w+f_z=f_{w+z}$.

If we take one of these functions, $f_w$, and multiply it by a real number, we get a new function from ${\bf R}^n$ to the real numbers; if we call this new function $h=cf_w$, then $h(x)=cf_w(x)=c(w\cdot x)=(cw)\cdot x=f_{cw}(x)$. So, if you multiply one of these functions by a real number, you get another one of these functions, $cf_w=f_{cw}$.

So now if we consider the collection of all of these functions $f_w$, we have a set of functions with the properties that if you add two functions that are in the set, you get another function that's in the set, and if you multiply a function that's in the set by a real number, you get another function that's in the set. Now that's the beginning of the definition of a vector space – a vector space is a set where you can do addition, and where you can do multiplication by a real number, and where moreover these two operations have a whole bunch of other properties, like associativity and commutativity and zero element and additive inverses and distributive laws. And one can show that the collection of all these functions $f_w$ has all of these properties, so the collection is a vector space.


问:
Square of sum of matrices
I'm trying to follow these lecture notes on Linear Discriminant Analysis (LDA) but I can't seem to figure out how the author gets from:

$$ \Sigma_{x\epsilon\omega_{i}} (w^{T}x - w^{T}\mu_{i})^2$$

to

$$ \Sigma_{x\epsilon\omega_{i}} w^{T}(x-\mu_{i})(x-\mu_{i})^Tw$$

Title: Square of sum of matrices

Body:
Since, $w^{T}x - w^{T}\mu_{i}$ is a scalar we can write it as:

$w^{T}x - w^{T}\mu_{i} = (w^{T}x - w^{T}\mu_{i})^T$

Thus,

$(w^{T}x - w^{T}\mu_{i})^2 = (w^{T}x - w^{T}\mu_{i}) (w^{T}x - w^{T}\mu_{i})^T$

Now, we know that:

$(AB)^T = B^TA^T$

Thus, we have:

$(w^{T}x - w^{T}\mu_{i}) (w^{T}x - w^{T}\mu_{i})^T = w^{T} (x - \mu_{i}) (w^{T} (x - \mu_{i}))^T$$

Or in other words:

$(w^{T}x - w^{T}\mu_{i}) (w^{T}x - w^{T}\mu_{i})^T = w^{T} (x - \mu_{i}) (x - \mu_{i})^T w$


问:
Composition $B \to B/J \to B$ equals identity implies that the ideal $J$ is zero
Let $A$ be a commutative ring with $1$ and $B$ an $A$-algebra. Let $J$ be an ideal of $B$ and we consider two arbitrary ring maps preserving $A$-module structure
$f_1: B \to B/J$, $f_2: B/J \to B$ which compose to identity $\mathrm{id}_B = f_2 \circ f_1$.

Question is when the existence of such two maps imply
that $J=0$?

One neccessary condition is to require that $B$ is
finitely generated $A$-algebra, otherwise here is a conterexample:

Take as example
$B= K[X_1, X_2,... ]$, with $K$ arbitry field, $J := (X_1)$ and
consider two $K$-morphisms $f_1: B \to B/J, X_i \mapsto X_{i+1}$
and $f_2: B/J \to B, X_{j+1} \mapsto X_j$. Observe that
$B/J = K[X_2,X_3,...]$ and the composition $f_2 \circ f_1$ is
identity on $B$.

Therefore assume that moreover $B$ is finitely generated $A$-algebra
and there exist $f_1, f_2$ as above with $\mathrm{id}_B = f_2 \circ f_1$.
Does this imply $J=0$? If not, are there any non trivial
conditions one should impose for $B$ to make the claim true?

Let me also note that this problem is closely related
to this question.

答:
Not sure about finitely generated algebra. However, if $B$ is finitely generated as an $A$-module then yes, $J$ must be trivial. Your condition clearly implies that $f_2$ is surjective. So if $\pi:B\to B/J$ is the canonical projection then $f_2\circ\pi:B\to B$ is a surjective homomorphism of $A$-modules. Since $B$ is finitely generated, we conclude that $f_2\circ\pi$ is an isomorphism. (a standard result obtained from Nakayama's lemma). In particular $\pi$ is injective, and so $J=0$.

Edit: We'll show why if $B$ is a finitely generated module and $T:B\to B$ is a surjective homomorphism of modules then $T$ is injective. We can define an $A[x]$-module structure on $B$ by $p(x).b=p(T)(b)$. In particular, $x.b=T(b)$ for all $b\in B$. Now let $I\subseteq A[x]$ be the ideal generated by the polynomial $x$. Since $T$ is surjective we have $IB=B$. From one of the versions of Nakayama's lemma (just noticed that in Atiyah-Macdonald it appears right before Nakayama's lemma, it is corollary 2.5 in chapter 2) there exists some element $p(x)\in A[x]$ such that $p(x)\equiv 1$ (mod $I$) and $p(x)B=0$. Thus $q(x):=1-p(x)$ is an element of $I$ such that $q(x).b=b$ for all $b\in B$.

Finally, since $I$ is generated by $x$ we know there is some polynomial $u(x)\in A[x]$ such that $q(x)=u(x)x$. And now define $S:B\to B$ by $S(b)=u(x).b$. This is clearly a homomorphism of $A$-modules, and:

$ST(b)=u(x).T(b)=u(x).(x.b)=u(x)x.b=q(x).b=b$

So we found a left inverse of $T$, which means it is injective.


答:
A particular linear function $f:V\to k$ is not producing a 'space of vectors', instead it is an element of a vector space, namely of the dual space of $V$.

$\hom(V,k)$ is the set of all linear functionals, and addition and scalar multiplication are done pointwise, and will satisfy the vector space axioms.

Additionally, if $V$ is finite dimensional with a basis $e_1,\dots,e_n$, then elements of $V$ are usually represented by column vectors (of the coordinates with respect the basis), and any linear map $f:V\to k$ can be represented by the row vector $w=(f(e_1),\dots,f(e_n))$, so that $f(v)=w\cdot v$ for all $v$ (coordinated in $e_i$), where $\cdot$ is actually matrix product, but it's the same as the dot product $w^T\cdot v$.

Title: How is the dual space generated from the linear forms?

Body:
Been reading about dual spaces without knowing much about it (not a mathematician). Now, after stumbled upon linear forms (function mapping vector space to number), that Wikipedia page reads:

The set of all linear functionals from $\mathbf{V}$ to $k$ , denoted by $\mathrm{hom}(V,k)$ is itself a vector space over $k$ with the operations of addition and scalar multiplication defined pointwise. This space is called the dual space of $\mathrm{V}$, or sometimes the algebraic dual space, to distinguish it from the continuous dual space.

The linear functional it talks about it's just
$$
\begin{align}
f(\vec{x})  &= \vec{w}\cdot{}\vec{x}\\
 &= a_1 x_1 + a_2 x_2 + \ldots + a_n x_n
\end{align}
$$

how does this produce a space of vectors?

As I said before, not a mathematician so don't go over a million definitions (please).


Body:
Does $(n+1)^x-n^x$ go to $\infty$ as $n\rightarrow\infty$ for $x>1$? I suspect it does from plots, but I want to use the ratio test to prove that
\begin{align}
\sum_{n=0}^\infty \exp(cn-vn^x)
\end{align}
converges for $c,v>0$ and $x>1$. Thanks.

Body:
Hint

$$(n+1)^x-n^x=n^{x-1} \frac{(1+\frac{1}{n})^x-1}{\frac{1}{n}}$$

Now
$$
\lim_{n \to \infty} n^{n-1}= \infty
$$
and
$$
\lim_{n \to \infty}\frac{(1+\frac{1}{n})^x-1}{\frac{1}{n}} = \lim_{z \to 0} \frac{(1+z)^x-1}{z}= \frac{d}{dz} (1+z)^x  |_{z=0}=x
$$


问:
Birthday Paradox with Leap Year
I looked online, and found more than one and inconsistent answers to the Birthday Paradox when we throw the leap year into the mix. None of the answers I saw match with my own. I am posting my solution to see if it is correct, or if I am missing something.

Question:

Assume that the leap year occurs every four years. (i.e. ignore the 100 and 400 year rule). Also assume that the number of people born each day is the same. What is the probability that in a group of $n$ people (each one selected randomly), no two people share the same birthday?

My Solution:

Let $\mathcal{D}$ be the set of all possible dates in an year. (Thus $\mathcal{D}$ contains 366 elements. Note that these possibilities are not equally likely, since a person is four times as likely to be born on (say) Jan 1 than on Feb 29. This is true for any given day other than Feb 29, and it is encoded in the probability assignments given below.)

Now probability that a randomly selected person is born on Feb 29 is $\frac{1}{1 + 4*365} = \frac{0.25}{365.25}$.

Also, the probability that a randomly selected person is born on a given day other than Feb 29 is $\frac{1}{365.25}$.

Now, for a group of $n$ randomly selected people, the Sample Space of birthdays is $\mathcal{D}^n$. Let $\mathcal{A} \subset \mathcal{D}^n$ be the subset such that no two people share the same birthday.

Divide $\mathcal{A}$ into two disjoint sets $\mathcal{A}_1$ and $\mathcal{A}_2$ such that

$\mathcal{A}_1 = \{\xi: \xi \in \mathcal{D}^n \text{ and no two people have same birthday, and none is born on Feb 29} \}$, and

$\mathcal{A}_2 = \{\eta: \eta \in \mathcal{D}^n \text{ and no two people have same birthday, and exactly one is born on Feb 29} \}$

Now, $\mathbb{P}(\xi) = \frac{1}{(365.25)^n}$ for each $\xi \in \mathcal{A}_1$, and $\mathbb{P}(\eta) = \frac{0.25}{(365.25)^n}$ for each $\eta \in \mathcal{A}_2$.

Also, $|\mathcal{A}_1| = \; ^{365}P_{n}$, and $|\mathcal{A}_2| = \; n \; \cdot \; ^{365}P_{n-1}$

Finally, $\mathcal{A}_1$ and $\mathcal{A}_2$ being disjoint, it follows that

$$\mathbb{P}(\mathcal{A}) = \mathbb{P}(\mathcal{A}_1) + \mathbb{P}(\mathcal{A}_2) = \frac{^{365}P_{n}}{(365.25)^n} + \frac{0.25 \; \cdot \; n \; \cdot \; ^{365}P_{n-1}}{(365.25)^n}$$

PS:

As usual, if you want to find out the probability that at least two people share the same birthday then you would calculate $1 - \mathbb{P}(\mathcal{A})$. Interestingly, the number of people required so that this probability is more than 0.5 is still $n = 23$, same as the birthday paradox without leap year.

Please let me know if the solution above looks accurate.

答:
I conclude that the result in the question is accurate, but the method by which it is obtained is questionable.

The questionable aspect is the use of
$\lvert\mathcal{A}_1\rvert$ and $\lvert\mathcal{A}_2\rvert$,
the cardinalities of the sets  $\mathcal{A}_1$ and $\mathcal{A}_2$.
The usual way to use cardinalities of sets of outcomes in order to compute the probability of a desirable event ("success") is to partition the probability space into some number $D$ of equally likely events such that each event is either completely within the favorable outcomes or completely within the unfavorable outcome.
The favorable outcomes are then represented by a set of events
$\mathcal{A}$,
and the probability of success is then simply $\lvert\mathcal{A}\rvert/D.$

When working on the birthday paradox under the usual assumptions ($365$ possible birthdays, any person equally likely to be born on any of those dates),
for a group of $n$ persons the probability space can be partitioned into
$365^n$ equally-likely events, one for each possible combination of birthdays of the $n$ distinct persons in the group.
You can count up the number of events in which no two persons share a birthday,
which is $^{365}P_n,$ and the probability of no match is therefore
$^{365}P_n / 365^n.$

In the question, however, the probability space has not been partitioned into a set of equally likely events.
Instead, the probability is computed with a denominator $365.25^n$ which is not even an integer.

There are, however, at least three ways to fix this. One way (which is detailed in another answer) is to consider each person to be equally likely to have a birthday on every day in a four-year period.
I present two more methods below.

The basis of each of these methods is the assumption that each day other than February $29$ is equally likely, whereas February $29$ is $\frac14$ as likely as any other particular day, and these $366$ possibilities cover the entire probability space. The total probability of all events is therefore $365.25$ times the probability of a birthday on January $1$ (which is $1/365.25$),
while the probability of a birthday on February $29$ is one quarter of that, or $0.25/365.25.$ The probability space is therefore partitioned into $366$ events whose probabilities add up to $1$ as required.

First alternative method

One typical way to approach the usual ($365$-day) problem via conditional probabilities. The probability of no match is the product of conditional probabilities $P_k$ that the $k$th person will have a birthday distinct from the previous $k-1$ persons,
given that the first $k-1$ persons all have birthdays distinct from each other.
If we define $B_0$ equal to the entire probability space,
and for $k > 0$ we define $B_k$ as the event that the first $k$ persons all have distinct birthdays,
then $P_k = \mathbb P(B_k \mid B_{k-1})$ and the probability of no match in the entire group is
$$ \prod_{k=1}^n \mathbb P(B_k \mid B_{k-1}). $$

Adapting this approach to the leap-year version of the problem,
let $A_0$ be the event that all $n$ persons have distinct birthdays not including February $29.$
If $B_0$ is the entire probability space,
and for $k > 0$ we define $B_k$ as the event that the first $k$ persons all have distinct birthdays not including February $29,$
then
$$ \mathbb P(A_0) = \prod_{k=1}^n \mathbb P(B_k \mid B_{k-1}). $$

Now let $A_m$ be the event that all persons in the group were born on distinct days and the $m$th person was born on February $29$,
Let $C_0$ be the entire probability space, let $C_1$ be the event that the first person is born on February $29,$ and for $k > 1$ define $C_k$ as the event that the first person was born on February $29$ and the next $k-1$ persons all have distinct birthdays not including February $29.$
Then
$$ \mathbb P(A_1) = \prod_{k=1}^n \mathbb P(C_k \mid C_{k-1}). $$

By symmetry, $\mathbb P(A_1) = \mathbb P(A_2) = \cdots = \mathbb P(A_n).$
So the probability all $n$ persons in the group are born on distinct days is

$$ \mathbb P(A_0) + \mathbb P(A_1) + \cdots + \mathbb P(A_n)
 = \mathbb P(A_0) + n \mathbb P(A_1). $$

Now let's compute $\mathbb P(A_0).$
We suppose that a person has a $1/365.25$ probability to be born on any particular day other than February $29,$ and a $0.25/365.25$ probability to be born on February $29.$
The probability that the first person's birthday is not February $29$ is therefore
$\mathbb P(B_1 \mid B_0) = 365/365.25.$
More generally, for $k > 0,$ given that the first $k-1$ persons have all distinct birthdays,
the probability that the $k$th person has a birthday distinct from any of the previous $k-1$ and not on February $29$ (that is, the probability that the first $k$ persons have distinct birthdays not including February $29$) is
$\mathbb P(B_k \mid B_{k-1}) = (365 - (k - 1))/365.25.$
Therefore

\begin{align}
\mathbb P(A_0)
  &= \prod_{k=1}^n \frac{365 - (k - 1)}{365.25} \\
  &= \frac{365 \cdot 364 \cdot 363 \cdot \cdots \cdot (365 - (n - 1))}{365.25^n} \\
  &= \frac{^{365}P_n}{365.25^n}.
\end{align}

Next let's compute $\mathbb P(A_0).$
The probability that the first person's birthday is February $29$ is
$\mathbb P(C_1 \mid C_0) = 0.25/365.25.$
For $k > 1,$ given that the first person is born on February $29$ and the first $k-1$ persons have all distinct birthdays,
the probability that the $k$th person has a birthday distinct from any of the previous $k-1$ and not on February $29$ (that is, the probability that the first $k$ persons have distinct birthdays and the first one was born on February $29$) is
$\mathbb P(C_k \mid C_{k-1}) = (365 - (k - 2))/365.25.$
Therefore

\begin{align}
\mathbb P(A_1)
  &= \frac{0.25}{365.25} \prod_{k=2}^n \frac{365 - (k - 2)}{365.25} \\
  &= \frac{0.25}{365.25} \cdot
       \frac{365 \cdot 364 \cdot \cdots \cdot (365 - (n - 2))}{365.25^{n-1}} \\
  &= \frac{0.25 \cdot {}^{365}P_{n-1}}{365.25^n}.
\end{align}

In conclusion,

$$ \mathbb P(A_0) + n \mathbb P(A_1)
 = \frac{^{365}P_n}{365.25^n} + \frac{n \cdot 0.25 \cdot {}^{365}P_{n-1}}{365.25^n},
$$
the same result that was given in the question.
But the fact that we can use the permutation symbol here seems to be coincidental,
since permutations if $n$ persons or $n-1$ persons were not involved in any part of the derivation of this result.

Second alternative method

A second alternative method also uses conditional probabilities, but not so many of them, and also uses a counting argument involving permutations.

Again we consider two possible kinds of event:
$A_0,$ which occurs when all $n$ persons have distinct birthdays from each other and none was born on February $29$;
and $A_k$ for $k\geq 1,$ which occurs when the $k$th person is born on February $29$ and the other $k-1$ have have distinct birthdays from each other, none of which is February $29.$

The probability of $A_0$ is the probability that none of the $n$ persons was born on February $29,$ which is $(365/365.25)^n,$ times the conditional probability that no two of them share a birthday, given that none was was born on February $29.$

The conditional probability is simply the probability of no matching birthdays among $n$ persons in the ordinary $365$-day birthday problem,
which we already know is $^{365}P_n / 365^n$
by counting the number of favorable combinations of birthdays among the $365^n$ possible combinations.
So
$$ \mathbb P(A_0)
 = \left(\frac{365}{365.25}\right)^n \times \frac{^{365}P_n}{365^n}
 = \frac{^{365}P_n}{365.25^n}.
$$

For $A_1$, the probability is the probability that the first person was born on February $29,$ which is $0.25/365.25,$ times the probability that none of the other $n-1$ persons was born on February $29,$ times the conditional probability that the remaining $n-1$ persons were all born on different days,
given that none of them was born on February $29.$

Similarly to the previous case, the conditional probability is simply the probability of $n-1$ distinct birthdays in the  ordinary $365$-day birthday problem, which is $^{365}P_{n-1} / 365^{n-1}.$
So
$$ \mathbb P(A_1)
 = \frac{0.25}{365.25}
   \left(\frac{365}{365.25}\right)^{n-1} \times \frac{^{365}P_{n-1}}{365^n}
 = \frac{0.25 \cdot {}^{365}P_{n-1}}{365.25^n}.
$$

Therefore the final answer is

$$ \mathbb P(A_0) + n \mathbb P(A_1)
 = \frac{^{365}P_n}{365.25^n} + \frac{n \cdot 0.25 \cdot {}^{365}P_{n-1}}{365.25^n}.
$$

In this method, the fact that we can write parts of the formula using permutations is no coincidence; we actually derived those parts of the formula by counting permutations.

Title: Birthday Paradox with Leap Year


问:
one and two dimensional Gaussian integral

a) Evaluate the one-dimensional Gaussian integral

$I(a)$ = $\int_R exp(-ax^2)dx$,  $a>0$

b) evaluate the two-dimensional Gaussian integral using a)

$I_2(a,b)$ = $\int_{R^2} exp(-ax^2  -by^2)dxdy, a,b>0$

For a) I have done the following:

$\int_{-\infty}^{+\infty}$ $e^{-ax^2}dx = 2\int_0^{\infty}e^{-ax^2}dx $

$I^2$ = 4$\int_0^\infty$$\int_0^\infty$$e^{-a(x^2 + y^2)}dydx$

...

$I^2$ = $\sqrt{\pi a}$

$I$ = $\pi$$\sqrt a$

I am having difficulties understanding how to solve b) any help or guide will be appreciated.

答:
Note the suggested answer in the original question is wrong. Should be:
$$I = \sqrt{\frac{\pi}{a}}$$


答:
Trivial in Mathematica:

```<code>g = GridGraph[{3, 4}]
</code>```

```<code>LineGraph[g]
</code>```

Here is the line graph for a $10 \times 10$ grid graph:

No specific name that I know of.

Title: Line graph of square lattice

Body: Is it possible to create a line graph of an n by n square lattice? If so what would it look like and does it have a specific name? I've tried drawing it manually and ended with sideways squares tiled with triangles, looking almost like the $(3^2,4,3,4)$ lattice however I believe this is incorrect.

Body:
Trivial in Mathematica:

g = GridGraph[{3, 4}]

LineGraph[g]

Here is the line graph for a $10 \times 10$ grid graph:

No specific name that I know of.


问:
A well founded relation $R$ on a class $A$ does not need to be set-like
The theorem that justifies transfinite induction on a class states the following:

Assume that $R$ is well-founded and set-like on $A$, and that $X$ is a non-empty sub-class of $A$. Then $X$ has an $R$-minimal element.

However, with the Axiom of Foundation, we can eliminate the requirement that $R$ must be set-like.  My question is regarding a certain step in the proof of such.  Here is what I have.

For $x \in X$, let $f(x) = \min\{\text{rank}(z): z\in X \wedge zRx\}.$  (Here $\text{rank}(z)$ is $\text{rank}_{V,\in}(z)$, this is where we need Foundation.)  This function essentially corresponds each element $x \in X$ to the lowest rank of a set which is below $x$ in the relation $R$.  This function exists because any class of ordinals has a minimal element.  The goal is to find a subset of $X$ whose minimal element is also minimal in $X$.  Suppose we can find a $\gamma$ such that $X \cap V_\gamma$ is non-empty and $\forall x \in X \cap V_\gamma [f(x) < \gamma]$.

Let $y$ be minimal in $X \cap V_\gamma.$  Then $f(y) = \emptyset$, for assume that $f(y) = \alpha$.  Then by our definition of $f$, there is a $z \in X$ such that $zRy$ and $\text{rank}(z) = \alpha$.  Then, because $\alpha < \gamma$ (by our choice of $\gamma$) we get $z \in X \cap V_\gamma$.  But then $y$ is not minimal in $X \cap V_\gamma$ and thus $f(y) = \emptyset$.  Therefore, there is no $z \in X \wedge xRy$ and $y$ is minimal in $X$.

My question is: how do we produce $\gamma$?

答: Start with any ordinal $\gamma_0$ such that $X\cap V_{\gamma_0}$ is nonempty.  Recursively define $\gamma_{n+1}=\sup \{f(x):x\in X\cap V_{\gamma_n}\}+1$.  This gives a sequence of ordinals $\gamma_n$ for each $n\in\omega$; let $\gamma$ be their supremum.  Then any $x\in X\cap V_\gamma$ is in $V_{\gamma_n}$ for some $n$, and so $f(x)<\gamma_{n+1}\leq\gamma$ by definition.


答:
How about: as $n \to \infty$, we have
\begin{align}
(n+1)^x &= n^x\left(1+\frac{1}{n}\right)^x \ge n^x\left(1+\frac{x}{n}\right)
\qquad\text{[Bernoulli's inequality]}
\cr
(n+1)^x - n^x &\ge n^x+xn^{x-1}-n^x = xn^{x-1}
\cr
(n+1)^x - n^x & \to +\infty\qquad\text{[$x > 1$]}
\end{align}

Title: Does $(n+1)^x-n^x$ go to $\infty$ as $n\rightarrow\infty$ for $x>1$?


问:
Any proof (in English) that Generalised Peano Continua are continuous images of the real line?
By generalised Peano continuum I mean a metric, connected, locally connected, locally compact space. I found a paper called A Hahn-Mazurkiewicz Theorem for generalized Peano continua, which says that this result was proved by Mazurkiewicz in his seminal paper Sur les lignes de Jordan. However it is a more than 40 pages long document in French and I don´t understand french, is there a more recent book in English that deals with this topic? I haven´t found any.

答:
After a few days, I was able to find a way to prove the result. So I thought I should write an answer. The proof I wrote for my thesis is highly analogous to the section 31 on Peano Spaces of Willard´s General Topology. This answer can´t be understood without previously reading that section. The proof took me about two pages by latex with wide margins, and it is in spanish, so I´ll just sum up the steps here. Either way I´ll add the proof in Spanish in two images at the end of the answer.

You begin by proving the analogous to theorem 31.2:

If $X$ is a Generalised Peano Space and $U$ is an open connected subset of $X$ such that $\overline{U}$ is compact, then $U$ is path connected.

The proof is can be done the same way as the one of 31.2, thanks to the fact that $U$ is connected and locally connected and $\overline{U}$ is compact, so the sets $C_n$ of that proof will also be compact. By the way, if you try to understand it in detail, you´ll see that at the very least the proof of 31.2 in Willard´s General Topology is incomplete. More info about this proof can be found in this article. The proof can be fixed, although it´s a tedious task. May be worth it to use proofs from other books.

Anyways, with that lemma we know that Generalised Peano spaces are locally path connected, and thus are path connected. The next result you need is this one:

Generalised Peano spaces are $\sigma$-compact (that is, a countable union of compact subspaces).

To prove it you just have use that Generalised Peano spaces are metric, so they are paracompact, and that connected, locally compact, paracompact spaces are $\sigma$-compact (this last fact is proved in the second page of Appendix A of Spivak´s 'A Comprehensive Introduction To Differential Geometry. Vol1').

In Willard General Topology´s section about Peano spaces, they prove that Peano spaces are unformly locally arcwise connected. However, the same can´t be said of generalised Peano spaces. However, it will be enough to prove this weaker property:

If $X$ is a generalised Peano space and $C$ is a compact subspace of $X$, then $C$ is uniformly locally arcwise connected respect to $X$, that is, given $\varepsilon>0$ there is $\delta>0$ such that, if $a$ and $b$ are points of $C$ and $d(a,b)<\delta$, then there is an arc in $X$ from $a$ to $b$ with a diameter $<\varepsilon$.

The proof is essentially the same as Willard´s proof that Peano spaces are uniformly locally arcwise connected.

After this, you are prepared to prove that any Generalised Peano Space  $X$ is a continuous image of $\mathbb{R}$. The proof is analogous to Willard´s proof of 31.5. In that proof, he begins by defining a surjective function from the Cantor set to a Peano space. In our case, we use that we can fit infinitely many Cantor sets into the real line, for example fitting a Cantor set in every interval $[2n,2n+1]$ with integer $n$. Now, as $X$ is $\sigma$-compact, we can define a surjective function $g:C\to X$, where $C$ is the union of infinite Cantor sets.

The second, harder problem comes when we have to extend the continuous function $g:C\to X$ to a continuous function $f:\mathbb{R}\to X$. We have to define the image of $f$ in the countable open intervals that form $\mathbb{R}\setminus C$, in such a way that the function $f$ is continuous. In his proof of 31.5, Willard uses uniform local arc connectedness at this point. What I did at this point was the following:

Given points $a,b\in X$, define $\varrho(a,b)$ to be the infimum of the diameters of the paths from $a$ to $b$. Clearly $\varrho(a,b)>d(a,b)$.

Then, I defined the extension $f$ in each interval $(p_n,q_n)$ to be a path from $g(p_n)$ to $g(q_n)$ with diameter less than $2\varrho(p_n,q_n)$. Defined this way, $f$ will be continuous. However, to prove it is not trivial and requires an epsilon delta argument similar (maybe a bit more complex) than the one in the proof of Willard´s 31.5. This is the hardest part of the proof, in my opinion.

Now the two images with the proof in Spanish:


问:
Proving that unary predicates theory has no $T$-complete formulas
As a follow up to my previous question, as suggested in the comments, I would like to know a way how to prove that the theory of infinitely many unary predicates, is axiomatized by the sentences $\sigma_{F, G}$ (for $F, G$ disjoint finite subsets of $\mathbb{N}$) given by $\exists x\left(\bigwedge_{j \in F} U_{j}(x) \wedge \wedge_{j \in G} \neg U_{j}(x)\right)$ has no principal types. I know that the way to show it is through showing that no formula in $\mathcal{L}$ is $T$-complete.

To clarify, by theory of infinitely many unary predicates, I mean the language with unary predicate symbols $\left(U_{n}: n \geq 1\right)$. and $T$ a theory $\operatorname{Th}(\mathcal{M})$ where $\mathcal{M}$ is the following structure: take $\mathcal{M}=\mathcal{P}(\mathbb{N})$ and define the interpretation of each $U_{n}$ by taking $U_{n}^{\mathcal{M}}(\alpha) \leftrightarrow n \in \alpha$ for each $\alpha \subseteq \mathbb{N}$.

I am using defnitions as in thew original question:

Definitions

Take a satisfiable set of $L$-sentences $\Sigma$ and variables $x = x_1, \ldots, x_n$. Denote by $S_x(\Sigma)$ the set of all $\Sigma$-realizable $x$-types in $L$.

A type  $p(x) \in S_x(\Sigma)$ is principal if it contains a $\Sigma$-complete formula. Equivalently, the singleton $\{p(x)\}$
is an open set in $S_x(\Sigma)$, or principal $x$-types are exactly the isolated
points of $S_x(\Sigma)$.

The source of this example is lecture notes by Henson.

答:
Think about reducts. For each finite set $X\subseteq\mathbb{N}$ (really, every finite sublanguage of $L$) and each $a,b\in\mathcal{M}$, write $a\sim_Xb$ iff $$\{n\in X: U_n(a)\}=\{n\in X: U_n(b)\}.$$ Now you can show that the following holds:

Suppose $\varphi$ is an $L$-formula only using symbols $U_i$ for $i\in X$, and $a\sim_Xb$. Then $\mathcal{M}\models \varphi(a)\leftrightarrow\varphi(b)$.

This has nothing to do with unary predicates, it's a more general feature of arbitrary "filterings" of languages.

Now here's the key point about $\mathcal{M}$: for every finite $X\subseteq\mathbb{N}$ there are $a,b\in\mathcal{M}$ with $tp(a)\not=tp(b)$ but $a\sim_Xb$. For example, pick any $a$ whatsoever and let $b=a\Delta\{\max(X)+1\}.$ This lets us wrap things up as follows:

 For each formula $\varphi$, let $a,b$ be elements of $\mathcal{M}$ with distinct types but which are $\sim_{X_\varphi}$-equivalent.


问:
Derivative of Trig Function with exponent and chain rule logic
I'm new to calculus and self-learning it, I am having trouble grasping why $\sin^2 5x$ can be rewritten as $(\sin 5x)^2$

I fumble when I try to understand the logic behind it. For example, if I plug in a number for 'sin' I would get a different answer when I plugged it in the original equation and when I plugged it into the rewritten equation. Just trying to understand the logic and the 'why', so I don't go through calculus mindlessly solving equations. Thanks.

答:
Welcome to Math SE!

The question you have is just one of notation. It's pretty standard to write, generally, $[\sin(x)]^n$ as $\sin^n(x)$. The latter notation is probably more common than the former, in fact.

This notation holds broadly for integer values of $n$ except in the case of $n=-1$, when $\sin^{-1}(x)$ denotes $\arcsin(x)$. If you want to write $[\sin(x)]^{-1}$, you could write it as $[\sin(x)]^{-1}$, $\frac{1}{\sin(x)}$, or $\csc(x)$.

So, when you need to take the derivative of $\sin^2(5x)$, it may be a bit easier to think of it as $[\sin(5x)]^2$, then apply the chain rule. They both mean the same thing, it's just easier to recognize the ordering of the function composition: $5x$ is the input to $\sin(\cdot)$, and $\sin(\cdot)$ is the input to $(\cdot)^2$.


问:
Finding the self intersection point of two parametric equations.
Find the values of t where the graph of the parametric equations crosses itself:
$x = t^3-4t^2+t+7$ and $y = t^2-t$.

I am asking this problem on behalf of me and my Calculus 3 class. We understand the idea of $t$ being two different times, so making them say $s$ and $t$, then setting $x(s) = x(t)$ and $y(s) = y(t)$.

So we get:

$$x(t) = t^3-4t^2+t+7 = s^3-4s^2+s+7 = x(s)$$

$$y(t) = t^2-t = s^2-s = y(s)$$

However, past that we are completely lost. Our textbook goes over this type of problem for all of one paragraph, summarizing the process heavily. We asked our professor to go over this type of problem again but it still makes next to no sense. We also looked on YouTube and here on StackExchange for examples but they are either not the same type of problem, or the answers given are heavily summarized with no clear process. We are looking for a concrete process or set of ideas to follow in order to solve a problem like this.

Any help to this is greatly appreciated!

答:
By ploting the path we can get a pretty big hint of where it should cross itself.
By looking at the graph it becomes easy to see that the path crosses itself with something like x=1 and y=2. But we wanna do this algebraically, not graphically. So, lets try another method.
We can easily see that the path crosses itself if 2 or more points with different values of t, have the same vallues of both x and y (s and t as you were calling them).
We know that s and t must be different, but x and y must be iqual, lets try once again to visualize this.
We know that x²-x = y²-y, so we get the following plot:

The places where the functions intersect are the places where x²-x=y²-y, so, lets try to find the equation to those intersections: y²-y+(x-x²)=0, using something as simple as the bhaskara's formula, we get that y=0.5±0.5√(4x²-4x+1).
This can be simplified to y=0.5±|x-0.5|.
If you do the same for the equation x³-4x²+x+7=y³-4y²+y+7, we should end up with something like this:
We only want the points where s≠t, so, we're only interested in the points where (s,t) = (-1,2) and (2,-1).
This gives us our solution, with t=-1 and t=2 the coordinates (x,y) of your path are the same, so, the path intersects itself.


问:
Ring of integer-valued polynomials Int(X) is not isomorphic to Z[X]
Write $\operatorname{Int}(\mathbb{Z})$ for the set of polynomials $p$ mapping integers to integers, i.e., $p(\mathbb{Z}) \subseteq \mathbb{Z}$. I have shown that $\operatorname{Int}(\mathbb{Z})$ is a subring of $\mathbb{Q}[X]$. I am now asked to prove that $\operatorname{Int}(\mathbb{Z})$ is not isomorphic to $\mathbb{Z}[X]$.

By dabbling around online I found that $\operatorname{Int}(\mathbb{Z})$ is non-Noetherian, whilst $\mathbb{Z}$ as a PID is Noetherian and so $\mathbb{Z}[X]$ is Noetherian, hence the two rings in question cannot be isomorphic.

However, I would like to have a more elementary proof. We have $\mathbb{Z}[X] \subsetneq \operatorname{Int}(\mathbb{Z})$ since the latter contains polynomials  such as $\frac{1}{2} X^2 + \frac{1}{2} X$. I thought I could get a contradiction by considering the 'extra elements', but since both rings are infinite and a ring automorphism needs not be the identity I don't know how to proceed. What are some other properties of ring isomorphisms that I can consider to reach a contradiction?

答:
$\mathbb{Z}[x]$ is generated by a single element, namely $x$, so you can show that $\text{Int}(\mathbb{Z})$ is not generated by a single element, or equivalently that no homomorphism $\mathbb{Z}[x] \to \text{Int}(\mathbb{Z})$ can be surjective.

In fact $\text{Int}(\mathbb{Z})$ is not even finitely generated. The reason is that any finite set of polynomials in $\text{Int}(\mathbb{Z})$ generates a subring of $\mathbb{Q}[x]$ with the property that the denominators of all coefficients that appear are divisible by finitely many primes. But $\text{Int}(\mathbb{Z})$ contains polynomials ${x \choose n}$ (which form a $\mathbb{Z}$-linear basis) with coefficients whose denominators are divisible by arbitrarily large primes.


问:
Atyiah Macdonald Ex 1.2(iii)

$f\in A[x]$ is a zero divisor if and only if there is $a\neq 0\in A$ with $a\cdot f=0$

One direction is trivial. The other direction I find the hint confusing. It says we should choose a zerodivisor $g$ of $f$ with a minimal degree. We write $f=\sum^n a_ix^i$ and $g=\sum^m b_j x^j$. Then all the coefficients of $fg$ are zero. In particular the highest degree coeeficent is zero, $a_nb_m=0$. By assumption then $a_ng$ must be zero since it has a degree less than $g$ and it annihilates $g$.

Then the hint says to use induction on $a_{n-r}$. Are we inducting on $r$? It is also not clear to me how to result follows from this?

This question is not a duplicate of Zero divisor in $R[x]$ as I am specifically asking how to use the hint and do this problem with the induction.

答:
You computed the coefficient of $x^{n+m}$ in $fg$ and got that $a_ng=0$. Now let's compute the coefficient of $x^{n+m-1}$ in $fg$. We get:

$a_{n-1}b_m+a_nb_{m-1}=0$

Since $a_ng=0$ we know that in particular $a_nb_{m-1}=0$, and thus it follows that $a_{n-1}b_m=0$. So again, $a_{n-1}g$ is a polynomial of degree less than $m$ which annihilates $f$, and so $a_{n-1}g=0$.

Continue like this by induction and show that $a_{n-r}g=0$ for $r=0,1,...,n$. In particular this implies:

$a_0b_m=a_1b_m=...=a_nb_m=0$

And so $b_mf=0$.

Title: Atyiah Macdonald Ex 1.2(iii)


问:
How to find the intersection of two lines on a plane in 3D space.

My Question is: How do I math the intersection($P5$) of $\overline{P3,P0}$ and $\overline{P7,P8}$ ?

I'm trying to find the intersection of two lines in a 3d space, (XYZ) on a plane.

I have a plane, it is formed by 3 points

$$
\begin{bmatrix}
P1\\\hline
1\\
1\\
1\\
\end{bmatrix},
\begin{bmatrix}
P0\\\hline
-1\\
.345\\
1\\
\end{bmatrix},
\begin{bmatrix}
P3\\\hline
1\\
.275\\
-1\\
\end{bmatrix}
$$

I also have 2 other points on that plane

$$
\begin{bmatrix}
P7\\\hline
.420\\
.687\\
.661\\
\end{bmatrix}
\begin{bmatrix}
P8\\\hline
.295\\
.555\\
.410\\
\end{bmatrix}
$$

Now if I draw a line though $\overline{P7,P8}$, it should be intersecting the line $\overline{P0,P3}$, at

$$
\begin{bmatrix}
P5\\\hline
.060\\
.308\\
-.060\\
\end{bmatrix}
$$

(consequently, it should also intersect line $\overline{P1,P0}$ somewhere (.589,.866,1) )

My Question is: How do I math the intersection($P5$) of $\overline{P3,P0}$ and $\overline{P7,P8}$ ?

答:
The vector for a general point on the line $P_7P_8$ is
$$\lambda(P_7-P_8)+P_8=\lambda\begin{pmatrix}0.125\\0.132\\0.251\end{pmatrix}+\begin{pmatrix}0.295\\0.555\\0.410\end{pmatrix},$$
where $\lambda$ is any real number. You can think of this formula as starting at the point $P_8$, then shifting your position by a scalar multiple of the vector $\overrightarrow{P_8P_7}$.

Likewise the equation for a general point on line $P_0P_3$  is
$$\mu(P_3-P_0)+P_0=\mu\begin{pmatrix}2\\-0.07\\-2\end{pmatrix}+\begin{pmatrix}1\\0.275\\-1\end{pmatrix}.$$
Equating these two expressions should give you three equations (one for each component) for $\lambda$ and $\mu$, which you can then solve.


问:
On Lévy collapsing the reals
Consider the Lévy forcing notion. Let $M$ be a transitive standard model of $\mathsf{ZFC}$. Let $\aleph_n$ be the cardinality of the real numbers $2^\omega$ in $M$. Now collapse $\aleph_n$ to $\omega$. The resulting model $M[G]$ is again a model of $\mathsf{ZFC}$ but it is provable in $\mathsf{ZF}$ that $2^\omega$ is uncountable.

Could someone help me resolve this seeming contradiction? I'd be most grateful.

And I have a second question: Where can I access the original document containing the forcing notion explained? Or if that is not available: are there any other resources available? (I have a feeling that I might have been able to sort out my confusion on my own with more documentation available but Wiki and Jech are rather too concise for me.)

答:
Note that for every $\alpha<\aleph_{n+1}$ we add a bijection between $\omega$ and $\alpha$ to $M$. So in $M[G]$ all those ordinals are countable ordinals. So we added subsets of $\omega$ which encode these bijections (or the order type, if you prefer to think about that). So there are $\aleph_{n+1}$ new subsets to $\omega$ in $M[G]$.

Therefore we added $(\aleph_{n+1})^M$ real numbers to $M$, and changed $(\aleph_{n+1})^M$ to be $(\aleph_1)^{M[G]}$, because now whenever $\alpha<(\aleph_{n+1})^M$ we have that $M[G]\models|\alpha|=\aleph_0$. Therefore in $M[G]$ the least ordinal not in bijection with $\omega$ is $(\aleph_{n+1})^M$, which makes it $(\aleph_1)^{M[G]}$.

So we now have that $M[G]\models |2^\omega|=(\aleph_{n+1})^M=\aleph_1$.

As for reference request, I think that Kanamori's The Higher Infinite gives a nice exposition on the Levy collapse, as well one of its famous uses (constructing the Solovay model where all sets of reals are measurable).

Title: On Lévy collapsing the reals


问:
Does ``Converge to'' and ``Strict Equality'' Always Mean the Same Thing? If Not, is This a Paradox?
Consider the geometric series with a = 1 and r = 1/10.

Then, we have

$$
\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = \sum_{n=0}^{\infty}9\left(\frac{1}{10}\right)^{n} - 9 = 9\left[ \lim_{k\to\infty}\left(\frac{ar^{k}}{r - 1}\right) + \frac{a}{1-r}  - 1 \right] = 9\left[ \lim_{k\to\infty}\left(\frac{1\cdot(1/10)^{k}}{1/10 - 1}\right) + \frac{1}{1-1/10}  - 1 \right] = 9\left(0 + \frac{10}{9} - 1 \right) = 9 \left( \frac{1}{9} \right) = 1.
$$

Hence, we say that the geometric series with a = 1 and r = 1/10 converges to 1.

Now, (correct me if I am wrong) the given series will never achieve the sum of 2 (rather, approach arbitrarily close) to 2. I conclude such because the limit of the quotient after the second equals sign never achieves its limit of 0.

Furthermore, there are many generally accepted ways to show that $0.\overline{9} = 1$; that is identically equal to 1.

One conventional way, of course, is to let $x=0.\overline{9}$ and subtract it from $10x$ giving $9x = 9$, and thus $x = 1$. And I've heard nobody complain about that---even though we must subtract an infinite number of terms in the exact same manner as we do a finite number of terms.

But, that means

$$
\underbrace{1 = 0.\overline{9} = \sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n}}_{\text{exactly equal to}} = \overbrace{\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = 1}^{\text{arbitrarily close to}}.
$$

Doesn't it?

It seems to me that the only way the above equalities can hold is if converges to always means exactly equal to.

Otherwise, do we have a paradox here?

Please explain.

Thank you.

答:
Now, (correct me if I am wrong) the given series will never achieve the sum of 2 (rather, approach arbitrarily close) to $1$.

You are .... not correct.

The series will never achieve $1$ within a finite number of terms and we reach a precise value that is close to $2$ depending upon the precise number of steps and we can achieve a result that is as arbitrarily close to $1$ as we like by performing a specific finite number of steps required to get it that close to $1$.

But if we do the series an infinite number of steps.....  well, that's a different story altogether.

... Okay, we can debate what it means to do an infinite number of terms and if that even makes sense and if it does make sense is it possible.

But if doing it an infinite number of steps is meaningful (in my opinion, it isn't) and if figuring out the infinite sum is possible (in my opinion as it isn't meaningful it's not meaningful to talk of it being done) then, if so, then the sum is exactly $1$ and we do get $1$.  Not arbitrarily close to $1$ (we aren't doing this a finite number of times and not getting to the end; we are doing this an infinite number of times and we ARE getting to the end.

More formally we have a sequence of terms:

$a_n = \sum_{k=1}^n 9\cdot \frac1{10^k}=0.\underbrace{999....9}_{n\text {times}}$

This sequence $\{a_i\} = \{0.9,0.99,0.999,0.9999,......\}$ converges to $1$ but none of the terms are equal to one but it ISN'T any of the terms we are talking about when we say the limit is exactly equal to $1$. (It is.)  It's the limit of the terms, not any of the terms itself.

And we define an INFINITE sequence, not equal to any of the finite sequences $\sum_{k=1}^\infty 9\cdot \frac1{10^k} =\lim_{n\to \infty} \sum_{k=1}^n 9\cdot \frac1{10^k}$ and it is the limit that is the infinite series, that is not any of the finite terms that make the sequence, which is equal to exactly $1$.


问:
Derivative of submatrix with respect to the whole block matrix
I am reading a research paper and getting stucked with how they derived a formula. Suppose that we have the following block matrix
$$\underset{d \times d}{\boldsymbol{A}} = \begin{bmatrix} \underset{q \times q}{\boldsymbol{A}_{11}} & \underset{q \times (d - q)}{\boldsymbol{A}_{12}} \\ \underset{(d - q) \times q}{\boldsymbol{A}_{21}} & \underset{(d - q) \times (d - q)}{\boldsymbol{A}_{22}} \end{bmatrix}$$
The formula involves taking the derivative of a quantity that involves $\boldsymbol{A}_{22}$ with respect to $\boldsymbol{A}$. In particular, that quantity is
$$\text{trace} (\boldsymbol{A}_{22} \boldsymbol{B}),$$
where $\boldsymbol{B}$ is matrix with dimension $(d - q) \times (d - q)$. We need to take the following derivative
$$\frac{\partial }{\partial \boldsymbol{A}} \text{trace} (\boldsymbol{A}_{22} \boldsymbol{B}).$$
In the context of my paper, $\boldsymbol{A}$ is a symmetric matrix, but if possible, I would assume $\boldsymbol{A}$ is a square matrix.

Please help me if you have an idea. Thank you so much.

答:
$\def\m#1{\left[\begin{array}{l|l}#1\end{array}\right]}\def\mm#1#2{\left[\begin{array}{l|l}#1\\\hline #2\end{array}\right]}$
Let $\,p=(d-q)$ and write the $d\times d$ identity matrix in two different block forms
$$\eqalign{
I_d &= \m{E_1&E_2}{} = \mm{I_q&0}{0&I_p} \\
E_1 &\doteq \mm{I_q}{0} \qquad\;\; E_2 \doteq \mm{0}{I_p} \\
}$$
The $(E_i,\;E_k)$ matrices can be used to extract the corresponding block of $A,\,$ e.g.
$$\eqalign{
A_{ik} &= E_i^TAE_k \\
}$$
Now we can write the cost function and calculate its gradient
$$\eqalign{
f &= {\rm Tr}(A_{22}B) \\
  &= {\rm Tr}(E_2^TAE_2B) \\
  &= {\rm Tr}(E_2BE_2^TA) \\
\frac{\partial f}{\partial A} &= E_2BE_2^T = \mm{0&0}{0&B} \\
}$$


问:
Determining all formal power series satisfying the equation $Q^m=(1+x)^n$ for $n,m\in\mathbb{Z}\setminus\{0\}$
The question:

Let $m$ and $n$ be integers $\neq 0$. Determine all solutions $Q=1+b_1x+\cdots$ of the equation $$Q^m=(1+x)^n.$$

My current attempt at a solution:

Let $Q\in\mathbb{F}[[x]]$. The formal power series $1+x$ is equivalent to the binomial series $B_1$. Hence, we can rewrite the given equation in the following form
\begin{align*}
    1+\sum_{k=1}^\infty b_k^{(m)}x^k=Q^m=(1+x)^n=B_1^n=B_n=\sum_{k=0}^\infty\binom{n}{k}x^k.
\end{align*}
Applying the J.C.P Miller formula to $Q^m$ and comparing coefficients we construct the following linear system of equations
\begin{gather*}
    b_0^{(m)}=b_0^m=1=1=\binom{1}{0}=[B_n]_0\\
    b_k^{(m)}=\frac{1}{k}\sum_{l=1}^k[l(m+1)-k]b_lb_{k-l}^{(m)}=\binom{n}{k}=[B_n]_k \mbox{ for } k=1,2,\ldots.
\end{gather*}
Solving the second equation for $k=1$ we have,
\begin{align*}
    \sum_{l=1}^1[l(m+1)-1]b_lb_{1-l}^{(m)}=mb_1b_0^{(m)}=mb_1=n=\binom{n}{1}.
\end{align*}
Hence, $b_1=\frac{n}{m}$. Solving the second equation for $k=2$ we have,
\begin{align*}
    \frac{1}{2}\sum_{l=1}^2[l(m+1)-2]b_lb_{2-l}^{(m)}=\frac{1}{2}\left[(m-1)b_1b_1^{(m)}+2mb_2b_0^{(m)}\right]\\=\frac{1}{2}\left[\frac{(m-1)n^2}{m}+2mb_2\right]=\frac{n(n-1)}{2}=\binom{n}{2} \Rightarrow{} b_2=\frac{n(n-m)}{2m^2}.
\end{align*}
Hence, $b_2=\frac{n(n-m)}{2m^2}$. Suppose that the $k$th term of $Q$ satisfying the given equation is given by $b_k=\frac{\prod_{l=0}^{k-1}(n-lm)}{k!m^k}$ for $k\geq 1$. Then for $k+1$ it follows,
\begin{align*}
b_{k+1}=\frac{\prod_{l=0}^{k+1-1}(n-lm)}{(k+1)!m^{k+1}}=\frac{\prod_{l=1}^k(n-lm)}{(k+1)k!mm^k}=b_k\frac{n-km}{(k+1)m}
\end{align*}

I can't seem to find a way to finish the induction step and have been unable to find an alternative method. Note that I am using $b_k^{(m)}$ to represent the $k$th coefficient of the $m$th power of the formal power series $Q$.

Edit: The field is considered to be an arbitrary field of characteristic zero.

答:
I stepped away for a bit to see Lubin gave a much nicer answer. I can't guarantee there are no mistakes but my answer if probably an effectively more cumbersome version with more details left to fill in. Hopefully it does some good to post regardless!

We can use nonarchimedean analysis (ultrametric calculus) to put an absolute value on the elements of $\mathbb{F}[[X]]$ (and so get a metric).

$$|R|:= \begin{cases}a^{\mathrm{ord}(R)}& R \ne 0 \\ 0 & R = 0\end{cases}$$

Here $a$ is an element of $(0,1) \subset \mathbb{R}$ and $\mathrm{ord}$ is the order of the series $R$, the power on the lowest term (you can think of it as the opposite of the degree).

Here Hensel's lemma general form applies, for $f(Q) = Q^m-(1+X)^n$ we'd like to find a root. There is a solution if we can find some guess $Q_0$ such that,

$$|f(Q_0)|<|f'(Q_0)|^2$$

in our case we can simply pick $Q_0=1$ and we have (assuming the characteristic of our field does not divide $m$),

$$|1^m - (1+X)^n| \le |X| = a < 1 = |m*1^{m-1}|^2$$

Further the generalized Hensel lemma tells us this is the unique solution in the open ball of radius $|m*1^{m-1}|=1$ centered at $Q_0=1$, which means there is only one solution in $1 + X\mathbb{F}[[X]]$.

At this point we can compute it by Newton's method, but we can do better and simply write the convergent series, as series converge nicely in ultrametric spaces - all you need is for their term to tend to $0$.

$$Q(z)=(1+X)^z = \sum_{k=0}^\infty \binom{z}{k} X^k$$

Here the binomial coefficient is defined as the image of the rational number in our field $\mathbb{F}$, $\binom{z}{k} = \frac{1}{k!} \prod_{i=0}^{n-1}(z-i)$

Because we have $Q(z) = (1+X)^z$ with $Q: \mathbb{N} \to \mathbb{F}[[X]]$ this infinite sum actually is finite, and so just a polynomial. We can extend it to rationals by putting a p-adic topology on $\mathbb{N}$ and slightly altering Mahler's theorem so the codomain is $\mathbb{F}[[X]]$ instead of $\mathbb{C}_p$. This gets us a unique continuous extension from $\mathbb{N}$ to the rational numbers without $p$ in the denominator. We simply pick $p$ so that it doesn't divide $m$ or the order of the field to avoid any issue.

Now let $z=\frac{n}{m}$,
$$Q=(1+X)^{n/m} = \sum_{k=0}^\infty \binom{n/m}{k} X^k$$

For more detail I recommend looking at W. H. Schikhof's Ultrametric Calculus and Alain M. Robert's A Course in p-adic Analysis.


问:
Prove $X \not\approx \boldsymbol{2}^{X}$ using contradiction
I am proving the theorem $X \not\approx \boldsymbol{2}^{X}$. It is required that contradiction be used in the proof. My thought is presented below.

We can assume there is a function $f: X \to \boldsymbol{2}^{X}$. Intuitively, we understand that $X$ has to be "smaller" than $\boldsymbol{2}^{X}$. So if $f$ is one-one, it cannot be onto. So we should assume the function is one-one and try to argue that it is not onto. The condition of being one-one gives me the following argument:

Since $f$ is one-one, if $x_{1}, x_{2} \in X$ and $x_{1} \neq x_{2}$, then $f\left(x_{1}\right) \neq f\left(x_{2}\right)$. Then without loss of generality, there exists $y \in X$ such that $\left[f\left(x_{1}\right)\right]\left(y\right) = \boldsymbol{0}$ and $\left[f\left(x_{2}\right)\right]\left(y\right) = \boldsymbol{1}$.

I guess I have fully utilized the condition of $f$ being one-one. But how can I construct a contradiction assuming that $f$ is also onto?

答:
Theorem: If X is a set, then X is not equivalent to its power set.

Proof: Suppose for a contradiction that $f:X\to P(X)$ is a bijection. Define $B:=\{x \in X| x\not\in f(x)\}$. Because $f$ is surjective (onto), there is an element $g$ in $X$ with $B=f(g)$.

Since $B$ is a subset of $X$ and $g$ is an element of $X$, it has to be that either $g\in B$ or $g\not\in B$.

If $g\in B$, then $g\in f(g)$. Then by definition of $B$, $g\not\in B$. So this can't be.

If $g\not\in B$, then $g\not\in f(g)$. Then by definition of $B$, $g\in B$. Again, this cannot be.
So it is impossible that there is some $g\in X$ such that $f(g)=\{x\in X∣x\not\in f(x)\}$.


问:
How to find $\overline{abc}_{7}-\overline{cba}_{7}=\overline{xy4}_{7}$?
The problem is as follows:

Let:

$\overline{abc}_{7}-\overline{cba}_{7}=\overline{xy4}_{7}$

Assuming $a=b-c$

Find: $cba_{7}$ in terms of decimal system.

How exactly should I do this?. Can someone help me here?.

The thing is I don't know how to deal with carry numerals.

So far the only thing which I spotted is that $c-a=4$

For this the only way to get this is assuming $c=6$ and $a=2$ thus it will not carry. Isn't it?.

But this is somewhat contradictory by looking on the third digit because it states $a-c=2-6=-4$

Which it cannot happen. Thus what am I doing wrong?. Can someone help me here?. My only requirement is a method which doesn't really uses much sophisticated math such as modular arithmetics or too many algebraic manipulations. Does it exist a way to do this more intuitively?

答:
$$a,b,c,x,y \in \{0,1,2,3,4,5,6\}$$

$$abc-cba7 = xy4$$

$$abc = cba + xy4$$

So we want to end up with this addition result (in base 7).

\begin{array}{c}
     & c & b & a \\
   + & x & y & 4 \\
\hline
     & a & b & c
\end{array}

Let's start with the right column

\begin{array}{c}
     & c & b & \color{red}a \\
   + & x & y & \color{red}4 \\
\hline
     & a & b & \color{red}c
\end{array}

Either {$a+4=c$} or {$a+4 = 7+c$, which implies  $c=a-3$}.

From the last column, we conclude that $c < a$.  So $c = a-3$ and, because $a+4 = 7+c$,
we must carry a $1$.

\begin{array}{c}
     & c & \color{red}{\overset{\displaystyle 1}{b}} & a \\
   + & x & \color{red}y & 4 \\
\hline
     & a & \color{red}b & c
\end{array}

So, either $(b+1)+y = b$ (which implies $y=-1$)
or $(b+1)+y = 7+b$ (which implies $y=6$.

We conclude that $y=6$ and we must carry a $1$.

\begin{array}{c}
     & \color{red}{\overset{\displaystyle 1}{c}}
     & \overset{\displaystyle 1}{b} & a \\
   + & \color{red}x & 6 & 4 \\
\hline
     & \color{red}a & b & c
\end{array}

So it must be true that $(c+1)+x = a$. Substituting $c=a-3$, we conclude $x=2$.

\begin{array}{c}
     & \overset{\displaystyle 1}{c} & \overset{\displaystyle 1}{b} & a \\
   + & 3 & 6 & 4 \\
\hline
     & a & b & c
\end{array}

So, if you want unique answers

$a=3 \qquad b =5 \qquad c=0 \qquad x = 2 \qquad y=6$

\begin{array}{c}
     & 0 & 5 & 3 \\
   + & 2 & 6 & 4 \\
\hline
     & 3 & 5 & 0
\end{array}

$$cba_7 = 053_7 = 38$$

$$xy4_7 = 264_7 = 144$$

$$abc_7 = 350_7 = 182$$