问:
Question regarding dx being infinitely small
In the book "Calculus Made Easy",there is a chapter on smallness of different degrees.There it was said that it was okay to treat high powers of a small fraction negligible and can be ignored.They took $dx$ to mean numeracally $\frac{1}{60}$ or $\frac{1}{1000}$ for example and discarded their high powers as relatively small to the other terms and ignored them.Now why do we need to take $dx$ infinitely small when we can take $dx=0.01$?In the end if we have something like $1000000060$ we discard the $60$.Then what is the need to take $dx$ infinitely small?Same goes for integration.If we can ignore the relatively small area,then we get the perfect area.What is the need to take $dx$ infinitely small?We can for example take $dx=1/64$ and can go with this forever.
I have given a picture for examples.

答:
It may be worthwhile to repeat here what @fleablood wrote in a comment (I hope I'm not mis-representing anything). Calculus is not about neglecting small quantities (relative small, or absolutely small, or any kind of small). Calculus is about computing precisely. Very precisely. It's about computing things that often involve some sort of infinite approximating process. This is often codified in the form of a limit. The language often employed is dynamical (e.g., $x$ vanishes, or tends to infinity). That is so for historical reasons and in order to appeal to a certain intuition that the formalism is trying to capture.

So, calculus is not the art of neglecting small quantities. It is the art very precisely quantifying how a quantity changes in relation to another quantity, with an emphasis on quantifying the effect of 'small' changes. The difference between small and large here is a question of perspective. It guides the questions we ask. For instance, asking if a given function has a local maximum is a question about the behaviour of the function in the very small scale. Asking if it has a global maximum is a large scale question. The tools of calculus are geared toward answering the former, not the latter (of course, we often answer the latter in a calculus setting, but the process of doing so is distinctly different than the classification of local extrema).

Now, this is from the perspective of pure mathematics. One may argue that from an engineering perspective one can (?) set a threshold of smallness and just discard things if they are small enough. Famously, for an engineer, $\sin(x)=x$ for all $x$ small enough. But it takes little effort to improve this and write the correct statement $\lim _{x\to 0}\frac{\sin x}{x}=1$ and then proceed to express $\sin(x)=x+r(x)$ where the remainder term satisfies $\lim _{x\to 0}r(x)=0$. This is the first order approximation of $\sin (x)$ expressed precisely and the remainder can further be analysed to produce more control over the approximation. In other words, we do not neglect $r(x)$. Rather, we replace it under control with something else, easier to compute, but we make sure not to lose track of our approximations. After all in the absence of further specification, $17$ is just a good an approximation of $\pi$ as any other value.


问:
Different homology groups for same space? (cellular homology of the 3-Torus ($T^3$))
I had an exercise to find a CW-structure on $T^3$ = $S^1 \times S^1 \times S^1$. And i thought that i have found one (i thought about $T^3$ as the regular polygon with 12 sides with points identified):

one $0$-cell

six $1$-cells

one $2$-cell (interior of the polygon)

And i got the homology groups:

$\mathbb Z$ (degree $0$)

$\mathbb Z^6$ (degree $1$)

$\mathbb Z$ (degree $2$)

$0$ in all other degrees

In the solutions they got something different. They did the same thing as in Hatcher , page 142, Example 2.39 (https://pi.math.cornell.edu/~hatcher/AT/AT.pdf)

Obviously it's not possible to get different homology groups for the same space, right? But i also can't see how my solution is wrong, does someone see it?

答:
It seems like you are confusing 'product of $3$ circles' with 'surface with $3$ holes'. The former is a $3$-manifold, but your proposed CW-structure has no 3-cell.

Hatcher's cellular structure on $T^3$ in the example you reference is standard.

ContentLicense: CC BY-SA 4.0


问:
Integral of a polynomial fraction: $\int \frac{4x^2-3x+2}{4x^2-4x+3}dx$
I have initially used long division to obtain: $$1+\frac{x-1}{4x^2-4x+3}$$
Now, I'm having difficulty solving $$\int \frac{x-1}{4x^2-4x+3}dx$$
Even if I do the $u$-substitution i.e. $u=4x^2-4x+3$ hence, $du=8x-4$ and $dx=\frac{1}{8x-4}du$, I can't seem to figure out how to deal with the $x-1$ in the numerator.
Any help regarding this will be appreciated. Thank you.

答:
Hints:

$$2\frac{x-1}{4x^2-4x+3}=\frac{2x-2}{(2x-1)^2+2}=\frac{2x-1}{(2x-1)^2+2}-\frac1{(2x-1)^2+2}.$$

In the first term, the numerator is the derivative of the denominator (to a coefficient). In the second, we recognize the derivative of the arc tangent, modulo simple transformations.


问:
Uniform convergence of a sequence of functions to a function (g) but their derivatives converge pointwise to a function which is not (g')
Question is that find a sequence ($f_n$) of continuously differentiable real functions defined on $[0,1]$ converges uniformly to a differentiable function ($g$) and ($f_n'$) converge pointwise to a function that is not ($g'$). I am trying to find this function. I think that $\frac{x}{1+x^2n^2}$ converges uniformly to zero on [0,1] and derivate $\frac{1-n^2x^2}{(1+x^2n^2)^2}$ converges pointwise to 1 at zero and 0 for all other points. Is my logic correct? Any help will be appreciated.Thanks

答: You are correct. Another example: $g_n(x)=x^n$ and $f_n(x)=\int_0^x g_n(t)dt=\frac {x^{n+1}}{n+1}.$ Then $f_n\to f=0$ uniformly on $[0,1]$ so $f'=0.$ But $g_n(x)\to 0$ for $x\in [0,1)$ while $g_n(1)=1$ for all $n$.


问:
Proof explanation on one-to-one analytic mapping with two fixed points being the identity
Prove that if $f:D\to D$ is a one-to-one analytic mapping of the unit disc $|z|<1$ onto itself withtwo distinct fixed points, then $f(z)=z$.

(A complex number $w\in D$ is a fixed point for the map $f:D \to D$ if $f(w)=w$. )

I found an answer here but I have two questions:

(1) Would the extra condition of f being one-to-one/injective simplify @Kavi Rama Murthy's answer?

(2) For his answer and comment, why does $h(z)=z*e^{ic}$, show $h(z)=z$?

答:
It is only needed that $f$ maps the unit disk $D$ into itself and has two fixed points $z_1, z_2$. The fact that $f$ is injective does not simplify the solution.

The same proof works also if $D \subsetneq \Bbb C$ is any simply connected domain, because of the Riemann mapping theorem.

$h=g\circ f \circ g^{-1}$ îs constructed such that is maps the unit disk into itself and has two fixed points $g(z_1) = 0$ and $g(z_2) = w^* \ne 0$. The Schwarz Lemma states that $|h(z)| \le |z|$, and then that $h(z) = e^{ic}z$ (for some $c \in \Bbb R$) because $|w^*| = |g(w^*)|$. Finally, $w^* = g(w^*) = e^{ic} w^*$ shows hat $e^{ic} = 1$, i.e. that $h$ is the identity function.


问:
Find matrices which are fixed by a certain permutation matrix

Let $H$ be an adjacency matrix of a graph. Given a certain permutation matrix $P$ such that $PHP^T=H$, describe the structure of the graph corresponding to $H$ (or the matrix).

This problem is opposite of computing automorphism group of a graph. If $\sigma$ is a permutation associated with the matrix $P$, then we have $H_{\sigma(i), \sigma(j)}=H_{i,j}$. I don't think I can say anything more about the graph. Can you point me towards references where such problem is studied in detail?

答: What you need to do is consider the action of the permutation $P$ on the edges of the complete graph $K_{n}$. These edges will be partitioned into orbits. Then the edges of the graph with adjacency matrix $H$ must be made up of a union of these edge orbits.


问:
Invariant subspace for a given operator
I have studied that given an operator f  that can be either a symmetric, or a skew-symmetric, or an orthogonal operator
in a Euclidean space $V$ and $U ⊂ V$ be its invariant subspace. Then $U^⊥$ is also an invariant for f.

But I came across a question that asks to provide a counterexample for this i.e. give an example of an operator f on some Euclidean (or Hermitian) vector space
such that it has an invariant subspace $U$ and $f(U^⊥)$ does not belong to $U^⊥$

I do not understand this we proved that given any general operator it will have an invariant subspace and we are asking to provide a counterexample to that. It's confusing.

答: `Define $T: \mathbb R^{2} \to \mathbb R^{2}$ by $T(x,y)=(x+y,0)$. Verify that $U=\{(x,y): y=0\}$ is an invariant subspace but $U^{\perp} = \{(x,y): x=0\}$ is not invaraint.

Title: Invariant subspace for a given operator


问:
Intersection form of $\mathbb P^1 \times \mathbb P^1$
This comes from Vakil 20.2 C, self-study.

We have to show the intersection form of $X = \mathbb P^1 \times \mathbb P^1$ is given by $l \cdot l = m \cdot m = 0$ and $l \cdot m = 1$, where $l$ is the curve $\mathbb P^1 \times \{0\}$ and $m$ is the curve $\{0\} \times \mathbb P^1$.

By exercise 20.2 A, we see $l \cdot l = h^0(l, \mathcal O_l)$. I suppose if we could show this was a negative degree line bundle, we would be done by 19.2.3, but I do not see why that should be the case.

We also have just proven Adjunction: given a curve $C$ in a smooth projective surface $X$,  if $K_X$ is the canonical divisor, then

$$C\cdot (K_X + C) = 2g_C - 2$$

where $g_C$ is the genus of $C$. Right after that (and right before this problem), we showed Riemann-Roch for Surfaces:

If $D$ is a Weil divisor on $X$, where $X$ is as above, then

$$\chi(X, \mathcal O_X(D)) = \frac{D \cdot (D - K_X)}{2} + \chi(X, \mathcal O_X)$$

How do I finish?

答:
Your formula is wrong, we don't have $l \cdot l = h^0(l, \mathcal O_l)$. Indeed in this case $l \cdot l = 0$ and $h^0(l, \mathcal O_l) = 1$.

Actually you don't need so much machinery. It's easy to see that $\mathbb P^1 \times \{ p\}$ is linearly equivalent to $\mathbb P^1 \times \{ q\}$ for any $p,q \in \mathbb P^1$. By basic properties of intersection product, you obtain $l \cdot l = 0$.

If you want to use the adjunction formula, then as Mohan suggests you can try to compute the canonical class. To do this, you can notice that $\Omega_X \cong p_1^* \Omega_{X_1} \oplus p_2^* \Omega_{X_2}$ if $X = X_1 \times X_2$ and $p_i : X_i \to X$ is the projection.


问:
Prove that the series converges for $|z| <1$ to a rational function
Suppose that the power series $\sum a_nz^n$ has a recurring sequence of coefficients;
that is, $a_{n+k} = a_n$ for some fixed positive integer $k$ and all $n$. Prove that the series
converges for$ |z| < 1$ to a rational function $p(z)/q(z)$ where $p, q$ are polynomials, and that the roots of $q$ are all on the unit circle.
What happens if $ a_{n+k} = a_{n}/k$ instead?

I was thinking of using the ratio test but didn't really get anywhere with it.

答:
As the coefficients repeat periodically, the first series can be written as

$$\sum_{i=0}^\infty A(z)z^{ki}= A(z)\sum_{i=0}^\infty z^{ki}=\frac{A(z)}{1-z^k}$$ where $A$ denotes a polynomial of degree $k-1$ (made from the $a_n$). It indeed converges absolutely for $|z|<1$.

For the second series, you can observe that the denominators will be increasing powers of $k$, repeated $k$ times and the series is

$$A(z)\sum_{i=0}^\infty\frac{z^{ki}}{k^i}=\frac{A(z)}{1-\dfrac{z^k}{k}}.$$

This is again a rational function and it converges for $|z|<\sqrt[k]k$.


问:
Proper Method for such questions (Algebra)
I came across many questions which generally require manipulation of the terms. $(a,b,c \in \mathbb{R})$

For instance

$a^2+b^2+c^2-ab-bc-ca=0$

In this case, we just multiply the equation with $2$ and obtain $a=b=c$ as a solution of the given equation. I attribute this to pure intuition and observation.

Another example

$a^2+b^2+5c^2+10d^2-6ad-4bc-6c-10d+34=0$

Now, again by observation we try to complete squares

$(a^2-6ad+9d^2)+(d^2-10d+25)+(4c^2-4bc+b^2)+(c^2-6c+9)=0$

$(a-3d)^2+(d-5)^2+(2c-b)^2+(c-3)^2$

Leading us to the solution

$a=15, b=6, c=3, d=5$

My question is Let's say I give the following equation

$a^2+b^2+c^2+d^2-ab-bc-cd-d+2/5=0$

( I can land on the solution by trying to make perfect squares.... but this is not an efficient method to solve )

Is there any proper method to solve such equations ?

I'm not asking for a solution for this specific question rather a viable method which I could incorporate.

Thanks

答:
You can convert the given equation to a quadratic equation with respect to a single variable, and you just need to satisfy the condition $\Delta_x≥0$ for the Polynomial Discriminant for the equation to have real roots. If the polynomial is homogeneous, you can divide all polynomial terms into a particular variable to reduce the number of variables and use the substitution, for example $ z = \dfrac{x}{y} $ (for a polynomial with variables $ x $ and $ y $). Sometimes various inequalities can be helpful. Think of it this way. It is not a good idea to use a general formula for a quartic equation where all roots are rational.  I will go over a simple example to show the "general way":

$$a^2+b^2+c^2-ab-bc-ac=0$$

$$\implies a^2-a(b+c)+(b^2+c^2-bc)=0$$

$$\implies \Delta_a =(b+c)^2-4(b^2+c^2-bc)=-3(b-c)^2≥0$$

$$\implies -3(b-c)^2=0\Longrightarrow b=c$$

$$\implies a=\dfrac {b+c}{2}=\dfrac {2b}{2}=b=c$$

So, we find real solutions as  $$a=b=c=\text {any real number}.$$


问:
How does $b^2=a^2-ab$ lead to $\frac{b}{a}=\frac{-1+\sqrt 5}{2}$?
So basically, I came across this particular solution of a problem that starts with

$$\frac {b}{a-b}=\frac ab \quad\implies\quad b^2=a^2-ab \quad\implies\quad\frac ba=\frac{-1+\sqrt 5}{2}$$

Can someone please explain how the $\frac ba=\frac{-1+\sqrt 5}{2}$ part came about?

The $\frac {b}{a-b}=\frac ab$ part is given as part of the problem.

答:
So we start here:

$$b^2 = a^2 - ab$$

(You can get there just by cross-multiplying in your original equation, the one with the fractions.) Move everything to the left-hand side first:

$$b^2 + ab - a^2 = 0$$

Divide both sides by $a^2$:

$$\frac{b^2}{a^2} + \frac{b}{a}  - 1 = 0$$

To make life easier, define the variable $x=b/a$:

$$x^2 + x - 1 =0$$

This is a quadratic equation in $x$, which is easily solved via the quadratic formula:

$$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$$

But since $x = b/a$, then we have the desired

$$\frac b a = \frac{-1 \pm \sqrt{5}}{2}$$

(I'm not sure why what you were given neglects the negative solution of $(-1-\sqrt 5)/2$, as far as I know it's valid.)


答: $b^{2}=a^{2}-ab$ can be written as $x^{2}+x-1=0$ where $x=\frac b a $. Solve this quadratic. You get $\frac b a =\frac {-1\pm \sqrt 5} 2$.

Title: How does $b^2=a^2-ab$ lead to $\frac{b}{a}=\frac{-1+\sqrt 5}{2}$?

Body:
So basically, I came across this particular solution of a problem that starts with

$$\frac {b}{a-b}=\frac ab \quad\implies\quad b^2=a^2-ab \quad\implies\quad\frac ba=\frac{-1+\sqrt 5}{2}$$

Can someone please explain how the $\frac ba=\frac{-1+\sqrt 5}{2}$ part came about?

The $\frac {b}{a-b}=\frac ab$ part is given as part of the problem.


答:
Dividing $b^2=a^2-ab$ by $a^2$ yields
$$
\left(\frac b a\right)^2 = 1-\frac b a.
$$
Let $z=\tfrac b a$ to get the quadratic equation $z^2=1-z$ or equivalently
$$
z^2+z-1 = 0.
$$
Solve this using your favorite method of solving a quadratic equation to obtain
$$
z = \frac { -1 \pm \sqrt{5} } 2.
$$


答: As you have to find $\dfrac{a}{b}$ let  $a=bt$ and form a quadratic equation in $t$ to get $$ t=\dfrac{1}{t-1}$$


问:
Question Regarding the Axiom of Extensionality
Jech's text on Set Theory states the following:

If X and Y have the same elements, then X = Y :

∀u(u ∈ X ↔ u ∈ Y ) → X = Y.

The converse, namely, if X = Y then u ∈ X ↔ u ∈ Y, is an axiom of
  predicate calculus. Thus we have

X = Y if and only if ∀u(u ∈ X ↔ u ∈ Y).

The axiom expresses the basic idea of a set: A set is determined by
  its elements.

To check my understanding, are the following true?

(1) The axiom of extension states only that if two sets have the same members, then those two sets equal.

(2) It is, on the other hand, an axiom of the language of set theory (i.e., predicate calculus) that if two sets equal, then they have the same members.

A third question:

(3) It seems that the axioms of ZFC are very much independent of the language being worked with.  That is, the axioms of ZFC don't state we must be working in the language predicate calculus.  Then I take it there exist languages we could be working with that don't have the property that if two sets are equal, they have the same members (so that we could only conclude if two sets have the same members, they are the same and not the reverse).  Is this true?  Do mathematicians ever work in languages besides predicate calculus?

答:
(1) and (2) are correct.

As for (3), recall Leibniz's Law -- if $a$ and $b$ are one and the very same thing, then whatever property $a$ has $b$ must have too. Applied to sets, if $a$ and $b$ are one and the very same set, then whatever property $a$ has $b$ must have too. So in particular, if $a$ and $b$ are one and the very same set, if $a$ has the property of having $x$ as a member, $b$ has the property of having $x$ as a member. In symbols, if that helps, if $a = b$ then $x \in a \to x \in b$. And of course, we likewise have if $a = b$ then $x \in b \to x \in a$. So:

if $a$ and $b$ are sets, then if $a = b$ then $x \in a \equiv x \in b$

But $x$ was arbitrary, so there's implicit generalization here which we can make explicit, again borrowing logical notation

if $a$ and $b$ are sets, then if $a = b$ then $\forall x(x \in a \equiv x \in b)$

Now, in getting this far we are just using the informal Leibniz's Law, notation for set-membership, and some handy logical notation. We aren't appealing to a formal system; rather we are appealing to informal mathematical reasoning (the sort of reasoning that logic books aim to formalize using the classical predicate calculus). Because we do want the formal predicate calculus to replicate this informal mathematical reasoning, we will get -- inside the calculus, applied to a language with $\in$ available

$a = b \to \forall x(x \in a \equiv x \in b)$

But the ultimate grounds for this half of the formal version of the extensionality principle are the background informal reasoning using Leibniz's Law which we aim to regiment formally. What's going on here is not an arbitrary appeal to one formal calculus rather than another.

ContentLicense: CC BY-SA 3.0


问:
having a non-vanishing derivative is sufficient for a holomorphic function to be injective?
If a holomorphic function is injective then its derivative is non-zero everywhere but is the reverse true?

Also for real functions, we know positive derivative corresponds to increasing function and negative corresponds to decreasing function, for holomorphic function can we draw conclusions on the behavior of the functions if the real part of the derivative is positive or if the derivative is purely imaginary, etc ...?
thanks

答: The exponential function is periodic and thus not injective, but its derivative, the funcion itself, is vanishing nowhere.</p>


问:
Number of non-crossing and non-partially-overlapping arcs -- Catalan numbers?
Let $n$ be an integer and consider the $n$ points $1,2,\ldots,n$ on a horizontal line. Consider sets of arcs on these points such that

no two arcs cross each other, i.e. there exist no two arcs $(a,b),(c,d)$ with $a<b<c<d$,

no two arcs do partially overlap, i.e. there exist no two arcs $(a,b),(a,c)$ with $a<b<c$ or $(b,a),(c,a)$ with $b<c<a$.

However, note that two arcs are allowed to fully overlap, e.g. $(a,b),(c,d)$ with $c<a<b<d$.

The set for all such graphs for $n=2,3,4$ is

Now it seems that the number of such graphs is
$$2, 5, 14, 42, 132, 429, 1430,\ldots$$
which are simply Catalan numbers.

I know there are many interpretations of Catalan numbers, but I could not directly find the above one. Is the above set of non-crossing and non-partially-overlapping arcs (a) known under a different name, or (b) in a one-to-one correspondence to a known interpretation of Catalan numbers?

答:
I’ll briefly sketch a bijection between these diagrams on $n$ points and valid strings of $n$ pairs of parentheses. The sketch will be a bit rough, as I’m going to have to shut down shortly and will explain it largely by example.

Start with one of your diagrams. Each point represents a matched pair () of parentheses. If $k<\ell$, and there is an arc from point $k$ to point $\ell$, then all of the pairs corresponding to points $k+1,\ldots,\ell$ are inside pair $k$. If this is a maximal arc, all pairs $j$ with $j<k$ are to the left of pair $k$, and all pairs $j$ with $\ell<j$ are to the right of pair $j$. If it is maximal within a larger arc from $p$ to $q$, all pairs $j$ with $p<j<k$ are to the left of pair $k$ within pair $p$, and all pairs $j$ with $\ell<j<q$ are to the right of pair $j$ within pair $p$.

These rules, applied recursively if necessary, suffice to convert your diagrams to valid parenthesis strings. For $n=4$, for instance, your diagrams, in order from the top down, correspond to the following strings, in which each pair of matched parentheses is labelled with the number of the corresponding point:

$$\begin{align*}
&(_1)_1(_2)_2(_3)_3(_4)_4\\
&(_1(_2)_2)_1(_3)_3(_4)_4\\
&(_1(_2)_2(_3)_3)_1(_4)_4\\
&(_1(_2)_2(_3)_3(_4)_4)_1\\
&(_1)_1(_2(_3)_3)_2(_4)_4\\
&(_1)_1(_2(_3)_3(_4)_4)_2\\
&(_1)_1(_2)_2(_3(_4)_4)_3\\
&(_1(_2(_3)_3)_2)_1(_4)_4\\
&(_1(_2(_3)_3(_4)_4)_2)_1\\
&(_1(_2)_2)_1(_3(_4)_4)_3\\
&(_1(_2)_2(_3(_4)_4)_3)_1\\
&(_1(_2(_3)_3)_2(_4)_4)_1\\
&(_1)_1(_2(_3(_4)_4)_3)_2\\
&(_1(_2(_2(_4)_4)_3)_2)_1
\end{align*}$$

With a bit of thought it’s clear that the procedure produces only valid parenthesis strings.

Starting with a valid string of $n$ pairs of parentheses, you simply reverse the process to get a set of arcs on a line of $n$ points. First number the pairs from $1$ to $n$ in the order in which their left parentheses appear. For example the string (()(())) becomes $(_1(_2)_2(_3(_4)_4)_3)_1$. There are parentheses inside pair $1$, so there will be an arc from point $1$. The top level pairs inside pair $1$ are pairs $2$ and $3$, so the arc from point $1$ will be to point $3$, corresponding to the rightmost of these pairs. Pair $4$ is inside pair $3$, so there will be an arc from point $3$ to point $4$. The resulting graph is the fourth one up from the bottom in your list.

Again, a bit of thought shows that no crossing arcs are created by this process.


问:
Theorem about connectedness.
$(X, \mathcal{O})$ : topological space

$\{ M_{\lambda} | \lambda \in \Lambda \}$ : subset system of X

Theorem;

\begin{equation}
\forall \lambda \in \Lambda ; M_{\lambda} \text{ is connected space}, x \in M_{\lambda} \Longrightarrow M:=\cup_{\lambda \in \Lambda} M_{\lambda} \text{ is connected space.}
\end{equation}

I could understand the proof partially, but I couldn't understand the last part.

Proof;

Let $N\subset M$ is open and closed in $(M, \mathcal{O_M})$.

$\exists \, G\text{(open in } (X,\mathcal{O})), F\text{(closed in }(X,\mathcal{O})); N= M\cap G=M\cap F.$

Then, $N\cap M_{\lambda} =M_{\lambda}\cap G=M_{\lambda}\cap F.$

Thus, $N\cap M_{\lambda}$ is open and closed in $(M_{\lambda}, \mathcal{O_{M_{\lambda}}})$

Since $(M_{\lambda},\mathcal{O_{M_{\lambda}}} )$ is connected space, $N\cap M_{\lambda}= M_{\lambda} \text{ or } \emptyset.$

And I have to prove $N=M \text{ or } \emptyset.$

I could prove that $\forall \lambda \in \Lambda; N\cap M_{\lambda} =\phi \Longrightarrow N=\emptyset.$

But I cannot prove that $\exists \lambda \in \Lambda; N\cap M_{\lambda}=M_{\lambda} \Longrightarrow N=M.$

I would like you to give me ideas about why this holds.

答:
I'll assume (as in common in this result) that $x$ is some fixed point of $X$, common to all $M_\lambda$.

Suppose $C$ is a clopen and non-empty subset of $M$. We can assume WLOG that $x \in C$ (or we take $M\setminus C$ instead, which is then also non-empty clopen in $M$ and then contains $x$).

Fix $\lambda \in \Lambda$. Then $C \cap M_\lambda$ is clopen in $M_\lambda$ (as each $M_\lambda$ has the subspace topology w.r.t. $M$). It still contains $x$ so it's non-empty. As $M_\lambda$ is connected, $M_\lambda \cap C = M_\lambda$, or equivalently $M_\lambda \subseteq C$.

As $\lambda$ was arbitrary, $M \subseteq C$ and so $C=M$ and we're done: $M$ is connected.

Title: Theorem about connectedness.


答:
You missed the crucial part of the hypothesis. It is given that there is a point  $x$ which belings to every $M_{\lambda}$. Suppose $x \in N$. Then $x \in N \cap M_{\lambda}$ so the first option does not arise and you get $N \cap M_{\lambda}= M_{\lambda}$ for every $\lambda$ which gives $N=M$.

Now suppose $x \notin N$. Then the second option does not arise so we get $N \cap M_{\lambda}=\emptyset $ for every $\lambda$. This gives $N=\emptyset$.

Title: Theorem about connectedness.

Body:
$(X, \mathcal{O})$ : topological space

$\{ M_{\lambda} | \lambda \in \Lambda \}$ : subset system of X

Theorem;

\begin{equation}
\forall \lambda \in \Lambda ; M_{\lambda} \text{ is connected space}, x \in M_{\lambda} \Longrightarrow M:=\cup_{\lambda \in \Lambda} M_{\lambda} \text{ is connected space.}
\end{equation}

I could understand the proof partially, but I couldn't understand the last part.

Proof;

Let $N\subset M$ is open and closed in $(M, \mathcal{O_M})$.

$\exists \, G\text{(open in } (X,\mathcal{O})), F\text{(closed in }(X,\mathcal{O})); N= M\cap G=M\cap F.$

Then, $N\cap M_{\lambda} =M_{\lambda}\cap G=M_{\lambda}\cap F.$

Thus, $N\cap M_{\lambda}$ is open and closed in $(M_{\lambda}, \mathcal{O_{M_{\lambda}}})$

Since $(M_{\lambda},\mathcal{O_{M_{\lambda}}} )$ is connected space, $N\cap M_{\lambda}= M_{\lambda} \text{ or } \emptyset.$

And I have to prove $N=M \text{ or } \emptyset.$

I could prove that $\forall \lambda \in \Lambda; N\cap M_{\lambda} =\phi \Longrightarrow N=\emptyset.$

But I cannot prove that $\exists \lambda \in \Lambda; N\cap M_{\lambda}=M_{\lambda} \Longrightarrow N=M.$

I would like you to give me ideas about why this holds.


问:
Does the union of a set of sets not contain sets as elements?
I am new to set theory and I saw this problem online.

$A=\{\{1,2\},\{3,4\}\}$

Find $A\cup A$

I think that the answer is$A=\{\{1,2\},\{3,4\}\}$

However the answer is $A=\{1,2,3,4\}$

Is there some rule im not aware about its confusing me a lot.

Thanks in advance.

答:
You're right that $A \cup A = A$.

$\bigcup A$ is equal to $\{1,2,3,4\}$. See wikipedia arbitrary unions.


答:
The first two questions are answered yes. It should be remarked that in older texts equality was omitted completely from the language, and was defined using $\in$ as to make the axiom of extensionality trivial.

To the third question, we work with whatever fits our context best. Set theory is usually thought as a theory in first-order logic, but it can be thought as a theory of second-order logic (including a nice upgrade to the axiom schemata), and it can be thought of in the context of intuitionistic logic as well.

There are several levels to formality, the logic (axioms, inference rules), the language, the axioms, the semantics. We can modify whatever we want for whatever we want to modify it. We just need to wonder whether or not it is going to be useful, possibly consistent, or interesting.

Title: Question Regarding the Axiom of Extensionality

Body:
Jech's text on Set Theory states the following:

If X and Y have the same elements, then X = Y :

∀u(u ∈ X ↔ u ∈ Y ) → X = Y.

The converse, namely, if X = Y then u ∈ X ↔ u ∈ Y, is an axiom of
  predicate calculus. Thus we have

X = Y if and only if ∀u(u ∈ X ↔ u ∈ Y).

The axiom expresses the basic idea of a set: A set is determined by
  its elements.

To check my understanding, are the following true?

(1) The axiom of extension states only that if two sets have the same members, then those two sets equal.

(2) It is, on the other hand, an axiom of the language of set theory (i.e., predicate calculus) that if two sets equal, then they have the same members.

A third question:

(3) It seems that the axioms of ZFC are very much independent of the language being worked with.  That is, the axioms of ZFC don't state we must be working in the language predicate calculus.  Then I take it there exist languages we could be working with that don't have the property that if two sets are equal, they have the same members (so that we could only conclude if two sets have the same members, they are the same and not the reverse).  Is this true?  Do mathematicians ever work in languages besides predicate calculus?


问:
Is $\lnot\lnot\forall x(A(x)\vee\neg A(x))$ a theorem of intuitionistic logic?
I have a short question. I know that $\neg\neg(A\vee\neg A)$ is a theorem of intuitionistic logic. What about $\neg\neg\forall x(A(x)\vee\neg A(x))$?

答:
No, it is not.  If it were, then using the internal language of a topos, the interpretation of the formula $\lnot\lnot \forall p : \Omega, p \lor \lnot p$ would be $\mathrm{true}$ on any topos.

However, consider the topos of sheaves of sets on $\mathbb{R}$.  In this topos, for each $x \in \mathbb{R}$, we have a global section of $\Omega$ corresponding to the open subset $\mathbb{R} \setminus \{ x \} \subseteq \mathbb{R}$.  For this section $p$, we get $\lnot p = \emptyset$, so $p \lor \lnot p = \mathbb{R} \setminus \{ x \}$.  Taking the intersection over all $x$, we see that $\forall p : \Omega, p \lor \lnot p$ must evaluate to $\emptyset$, and so $\lnot\lnot \forall p : \Omega, p \vee \lnot p$ also evaluates to $\emptyset$ instead of to $\mathrm{true} = \mathbb{R}$.

(Actually, in this argument, there is nothing all that special about the topological space $\mathbb{R}$ that we used.  Pretty much the same argument would work on any $T_1$ topological space with no isolated points.)


问:
Questions about locally Euclidean
Here is the definition of locally euclidean I was given:
A topological space $(X,τ)$ is said to be locally euclidean if there exists a positive integer n such that each point $x∈X$ has an open neighbourhood homeomorphic to an open ball about 0 in R^n with the euclidean metric. A Hausdorff locally euclidean space is said to be a topological manifold.

This is the first time I learned what locally euclidean means so I am wondering if I am on the right track with these questions below

(i)Prove that every non-trivial interval $(a,b)$, $a,b ∈ \Bbb{R}$, is locally euclidean.

Every open neighbourhood of elements of open intervals are just open intervals which are homeomorphic to any open interval in $\Bbb{R}$. So they are all homeomorphic to $(0-e,0+e)$.

(ii)  Let $S^1$ be the subspace of $\Bbb{R}^2$ consisting of all $x∈\Bbb{R}^2$ such that $d(x,0) = 1$, where $d$ is the Euclidean metric. Show that the space $S^1$ is locally Euclidean.

Every neighborhood of point of $S^1$ are just connected segment of the circle without the endpoints and we are given that these are homeomorphic to open intervals in $\Bbb{R}$ so they must be homeomorphic to $(0-e, 0+e)$.

(iii) Find an example of a locally Euclidean space which is not a topological manifold?

I am currently stuck on this one. This is the first time I learned what a topological manifold is and it seems like a lot of locally Euclidean space are Hausdorff.

答:
$(i)$ is trivial, we take $n=1$ and note that $X$ itself is homeomorphic to $(-1,1)$, say.

For $(ii)$ we note that $\phi(x)=e^{2i\pi(t + \phi_0)}$ is a homeomorphism between $(-1,1)$ and an open neighbourhood of (ana arbitrary point, chose $\phi_0$ as you please) $e^{2\pi i\phi_0}$ in $S^1$ (seen for convenience as the unit circle in $\Bbb C$).

Problem $(iii)$ is one you know or don't. The standard example is to take a Hausdorff space like $\Bbb R$ and double one of its points, so that we cannot separate it from its "twin". It's called the line with two origins, usually. It shows that just being locally Euclidean (so locally Hausdorff too) doesn't mean it's Hausdorff globally.

Title: Questions about locally Euclidean


问:
A very simple question about an integration problem
I have recently returned to school and am currently brushing up on some old Calculus before the next semester begins. I'm working several problems from the MIT Calculus book online and have come across this one :

9) Draw $y = \sin x$ from $0$ to $\pi$. Three rectangles (base $\dfrac{\pi}{3}$)
and six rectangles (base $\dfrac{\pi}{6}$) contain an arch of the sine function. Find the areas and guess the limit.

My problem is that I can't figure out why there is a discrepancy between my answer and the one provided in the solutions manual.

The solutions manual states that the solution is $\dfrac{\pi}{3} ( 1 + {\sqrt 3})$ ,  $\dfrac{\pi}{6} (3 + {\sqrt 3})$ , $2$.

Now. I'm not really interested in getting the whole answer, I will show you what I did, and I would like somebody to show me why I am looking at this wrong.

We can just stick to the first part, because after that I can try it from there.

I tried to paste a picture, but the picture taken from my phone was too large a file size.

So, I will just say I drew the sin curve from $0$ to $\pi$ : It increases, reaches a maximum at $\dfrac{\pi}{2}$ , and then unfolds its way symmetrically back down to $\pi$ , as sin curves are won't to do.

I moved over to $\dfrac{\pi}{3}$ drew a vertical line to $\sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{2}$ and  drew leftward back $\dfrac{\pi}{3}$ and down for the first rectangle. I repeated that again starting at $\dfrac{\pi}{3}$  and then on the third one $\sin(\pi)=0$ , so I ended up with two rectangles (not three, perhaps this is the start of my difficulty) .

The area of this is equal to $\dfrac{\pi}{3}\left( \dfrac{\sqrt3}{2}  + \dfrac{\sqrt3}{2}\right) = \dfrac{\pi}{3}(\sqrt 3)$ .

Therefore , I am off by one extra $\dfrac{\pi}{3}$. I want to know where this extra $\dfrac{\pi}{3}$ is coming from.

No point in going through the rest of the problem as the process is the same and I get a wrong answer there as well.

Thank you very much for any help you have to offer.

答:
It is advantageous to use a symbol at first and the assign a value later for limits defining the desired interval. Vary $i$ from 1 to 6 in the area integral of each one-sixth slice:

$$ \int_{(i-1) \pi/6}^{i \pi/6} \sin x dx $$


问:
Is learning (theoretical) physics useful/important for a mathematician?
I'm starting to read The Princeton Companion to Mathematics, at the beginning it says:

A proper appreciation of
  pure mathematics requires some knowledge of applied
  mathematics and theoretical physics.

Some of my professors have told me that modern Mathematics require some knowledge about Quantum Mechanics and theoretical Physics.

I attended the second and third day of the José Adem Memorial Lecture Series by Matilde Marcolli at the beginning of this year, they were about Number theory, Quantum statistical Mechanics and Quantum Field theory. I did not understand a single word but somehow the few things I understood had a big impression on me.

Please give me some examples of pure Mathematics that require/use Physics.

答:
My own knowledge of pure mathematics is very limited, but I have observed that ideas from physics at least motivate "pure math" results. For instance, I have heard that Riemann's original proof the Riemann Mapping Theorem was based on intuition about electrostatic forces, as is Dirichlet's principle (http://en.wikipedia.org/wiki/Dirichlet's_principle). Similarly, some properties of harmonic functions stems from the intuition that an initial charge distribution on the boundary must stabilize. </p>

<p>So, in my own experience I have not encountered mathematics where knowledge of theoretical physics is "required" (but I have not dabbled in mathematical physics, where I'm sure this is true), but even at my elementary level I've seen a lot of mathematical ideas which are motivated by physics. Even more vaguely, I've found that some physical principles prove useful in grasping the content of mathematical theorems -- for example, thinking about degrees of freedom involved, special cases, etc. </p>

Body:
My own knowledge of pure mathematics is very limited, but I have observed that ideas from physics at least motivate "pure math" results. For instance, I have heard that Riemann's original proof the Riemann Mapping Theorem was based on intuition about electrostatic forces, as is Dirichlet's principle (http://en.wikipedia.org/wiki/Dirichlet's_principle). Similarly, some properties of harmonic functions stems from the intuition that an initial charge distribution on the boundary must stabilize.

So, in my own experience I have not encountered mathematics where knowledge of theoretical physics is "required" (but I have not dabbled in mathematical physics, where I'm sure this is true), but even at my elementary level I've seen a lot of mathematical ideas which are motivated by physics. Even more vaguely, I've found that some physical principles prove useful in grasping the content of mathematical theorems -- for example, thinking about degrees of freedom involved, special cases, etc.


问:
Reflexivity, Transitivity, Symmertry of the square of an relation
$\def\p{\mathrel p}$If $\p$ is a relation on a set $A$, define $\p^2$ by $a \mathrel{\p^2} b$ if and only if there exists $c$ with $a \p c$ and $c \p b$.

If $p$ is reflexive/symmetric/transitive does $p^2$ have the same properties?

I'm not even sure how to start this, I assume I would need to use the $a$ related to $c$, $c$ related to $b$ somehow?

答:
Let's do reflexivity: Suppose, $p$ is reflexive. Let $a \in A$. We want to show that $a \mathrel{p^2} a$. By the definition of $p^2$, we have to find a $c \in A$ with $a \mathrel p c$, $c \mathrel p a$. But by the reflexivity of $p$, we know that $a \mathrel p a$. So if we let $c = a$, we have $a \mathrel p c$, $c \mathrel p a$, so $a \mathrel{p^2} a$ holds and $p$ is reflexive.

I hope, this will help you to do symmetry and transitivity on your own.


问:
Bound on l1 norm given bound on l2 norm
While doing self-study exercices, I found the following bound without explanation and was not able to see why it is always the case. I found some examples, it seems legitimate but I am unable to produce a formal proof of the point. Here is the property:

Given an arbitrary unit vector (wrt $\ell_2$) $v$ in $\mathbb{R}^d$, then $||v||_1 \leq \sqrt{d}$

答:
Since $v$ is unitary $v_i^2 \leq 1$. Using Jensen's inequality with $\phi(x)=x^2$:
$$\lVert v \rVert_1^2 = \phi\Bigl(\sum_{i=1}^d \lvert v_i \rvert \Bigr) \leq \sum_{i=1}^d \phi(v_i) \leq d$$

Title: Bound on l1 norm given bound on l2 norm


问:
Weak lower semi continuity weakly sequentially closed
Let X be a Banach space and $J:X \rightarrow \mathbb{R}$
a functional. Show that J is weakly lower semi-continuous, if and only if the set $U_\alpha$ is weakly sequentially closed for any $\alpha$ $\in \mathbb R$
provided that it is non-empty.

$U_\alpha := \{w ∈ X : J(w) ≤ \alpha\}$

I am confused on how to use the definitions to proceed.

A functional $f : X \rightarrow \mathbb{R}$ is weakly lower semicontinuous on X if for all $x \in X$ and every sequence $x_n → x$ which
converges weakly to x ∈ X, we have
$$\liminf\limits_{x\rightarrow0} f(x_n) ≥ f(x)$$.

We need to use the following lemma:

Let $C\subset X$ be a closed and convex subset of a normed space $X$. Then C is weakly sequentially closed,
i.e. for a sequence $(u^n )_n$ in $C$ with $u^n\rightharpoonup u$ in X for $n\rightarrow \infty$ we have that $u \in C$.

and also the fact that:

Let $X$ be a Banach space. If a functional $J : X \rightarrow
\mathbb R$ is continuous and convex, then $U_\alpha$
is weakly sequentially closed for any $\alpha \in
\mathbb R$.

答:
Suppose that $J$ is weakly lower semi-continuous on $X$, that is, according to your definition, for all $x \in X$ and every sequence $x_n \rightharpoonup x$  we have $$ \liminf\limits_{n\rightarrow\infty} J(x_n) \geq J(x).$$

Suppose $U_\alpha \neq \emptyset$. Given a sequence $(x_n)\in U_\alpha^\mathbb{N} $, we have $ \alpha \geq J(x_n)$, and by w.l.s.c., $$\alpha \geq  \liminf\limits_{n\rightarrow\infty} J(x_n) \geq J(x),$$ therefore $x\in U_\alpha$ so  $U_\alpha$ is weakly sequentially closed.

Conversely given a sequence  $(x_n)\in X$  with  $x_n \rightharpoonup x$. Consider the sequence $(J(x_n))\in\mathbb{R}^\mathbb{N}$.

By definition, $$\liminf \limits_{n\rightarrow\infty} J(x_n) = \lim \limits_{N\rightarrow\infty} \inf\limits_{n \geq N} J(x_n).$$
Let us suppose first $\liminf \limits_{n\rightarrow\infty} J(x_n)$ is finite,
$$\liminf \limits_{n\rightarrow\infty} J(x_n) =\alpha \in \mathbb{R}.$$

Choose $\epsilon>0$ (thanks @ammath). Then for every $n\in \mathbb{N}$, since the sequence $\inf\limits_{n \geq N} J(x_n)$ is monotone increasing (the sets on which the infimum is taken are nested), a subsequence ($x_{\sigma(n)}$) satisfies $J(x_{\sigma(n)})\leq \alpha +\epsilon$, so $x_n\in U_{\alpha+\epsilon}$ (as the weak limit of the subsequence is the weak limit of the sequence). If $U_{\alpha+\epsilon}$ is weakly sequentially closed then $x\in U_{\alpha+\epsilon}$. , which means $$ \epsilon+\liminf \limits_{n\rightarrow\infty} J(x_n)=\epsilon +  \alpha \geq J(x).$$
Passing to the limit as $\epsilon\to0$, we have obtained
$$
\liminf \limits_{n\rightarrow\infty} J(x_n)\geq J(x).
$$
Finally, let us discuss the $\liminf \limits_{n\rightarrow\infty}J(x_n)=-\infty$ issue (thanks @daw below). By assumption, $J(x)\in \mathbb{R}$. Take $\alpha = J(x)-1$. A subsequence $(x_{\sigma(n)})$ is in $ U_{\alpha}$. Therefore $x\in U_{\alpha}$, that is, $J(x)\leq J(x)-1$ which doesn't happen.


问:
Prove $\lim_{ (x,y) \to (0,0)} \frac{x^3-y^3}{x^2+y^2}=0$
Can some one prove the limit using epsilon delta method to prove that the limit exists
$$
\lim_{(x,y) \to (0,0)} \frac{x^3-y^3}{x^2+y^2}=0
$$

答:
Let $x=r\cos\theta$ and $y=r\sin\theta$.

We have $$\lim_{(x,y) \to (0,0)} \frac {x^3-y^3}{x^2+y^2} = \lim_{r \to 0} \frac {r^3\cos^3\theta-r^3\sin^3\theta}{r^2\cos^2\theta+r^2\sin^2\theta} \\=\lim_{r \to 0} \frac {r^3\cos^3\theta-r^3\sin^3\theta}{r^2} =\lim_{r \to 0} r(\cos^3\theta-\sin^3\theta).$$

Can you complete?


问:
Proving an elementary proposition about modules.
I want to prove the following:

Let $R$ be a commutative ring and $M$ is an abelian group then $M$ is an $R$-module iff there exists a ring homomorphism $f: R \rightarrow End_Z(M).$

My professor said that its proof is elementary but I can not see that, could someone  push me in the right direction of the proof please?

EDIT:

An abelian group $M$ can be given the structure of $R$-module if and only if there exists a ring homomorphism
$$
\varphi\colon R\to\operatorname{End}_{\mathbb{Z}}(M)
$$
If $M$ is given a structure of $R$-module, then the homomorphism $\varphi$ is defined by $\phi(r)\colon x\mapsto rx$. Conversely, given the ring homomorphism $\varphi$, we can define $rx=\varphi(r)(x)$.

Is this a valid proof?

答:
I will give you the corresponding maps/actions and you can then verify that they satisfy the desired properties:

Let $M$ be an $R$-Module. Define $f:R\to \operatorname{End}_{\Bbb Z}(M)$ by $f(r)=g_r$ where $g_r:M\to M$ is given by multiplication by $r$, i.e. $g_r(m)=r\cdot m$. Using the module properties check that this $f$ gives a well-defined homomorphism.

Conversely given such a morphism $f:R\to \operatorname{End}_{\Bbb Z}(M)$, define the $R$-module structure on $M$ by $r\cdot m= f(r)(m)$ for $r\in R,m\in M$. Here as well we can check that this indeed defines an $R$-module structure.

Furthermore these two constructions are inverse to each other.

Edit: For $r,s\in R$ we have: $$(r+s)\cdot m:=f(r+s)(m)=(f(r)+f(s))(m)=f(r)(m)+f(s)(m)=r\cdot m+s\cdot m$$
and $$(rs)\cdot m=f(rs)(m)=(f(r)\circ f(s))(m)=f(r)(f(s)(m))=f(r)(s\cdot m)=r\cdot (s\cdot m)$$
The other properties can be proved similarly.


问:
Continuity for $f(x)=\begin{cases} x, \text{when $x\leqslant0$} \\ \frac{1}{x-1}, \text{when $x>1$} \end{cases}$

Let $f:(-\infty,0) \cup (1,\infty) \to \mathbb{R},$ $$f(x)=\begin{cases} x, \text{when $x\leqslant0$} \\ \frac{1}{x-1}, \text{when $x>1$} \end{cases}$$ Determine if $f$ is continuous? Also determine if $f^{-1} : \mathbb{R} \to (-\infty,0) \cup (1,\infty)$ is continuous.

The problem for continuity with $f$ occurs at $x=1$ and doesn't seem to be continuous when approaching from the positive axis. For the inverse of $f$ I got that $$f^{-1}(x) = \begin{cases} x, \text{when $x\leqslant0$} \\ \frac{x+1}{x}, \text{when $x>1$} \end{cases} $$

for which when approaching $x=1$ from the negative $x$-axis evaluates to $1$ and from the positive axis it evaluates to $2$. So also the inverse also isn't continuous?

答:
There is no problem whatsoever at $1$, since $1$ does not belong to the domain of $f$.

And you got the inverse wrong. Remember that the domain of $f^{-1}$ is the range of $f$, which is $\Bbb R$. Actually, we have$$f^{-1}(x)=\begin{cases}x&\text{ if }x\leqslant0\\1+\frac1x&\text{ if }x>0,\end{cases}$$which is discontinuous at $0$.


答: $1$ is not in the domain  and so continuity at $1$ does not arise.  $f$ is  continuous  but the inverse function is not continuous at $0$.

Title: Continuity for $f(x)=\begin{cases} x, \text{when $x\leqslant0$} \\ \frac{1}{x-1}, \text{when $x>1$} \end{cases}$

Body:
Let $f:(-\infty,0) \cup (1,\infty) \to \mathbb{R},$ $$f(x)=\begin{cases} x, \text{when $x\leqslant0$} \\ \frac{1}{x-1}, \text{when $x>1$} \end{cases}$$ Determine if $f$ is continuous? Also determine if $f^{-1} : \mathbb{R} \to (-\infty,0) \cup (1,\infty)$ is continuous.

The problem for continuity with $f$ occurs at $x=1$ and doesn't seem to be continuous when approaching from the positive axis. For the inverse of $f$ I got that $$f^{-1}(x) = \begin{cases} x, \text{when $x\leqslant0$} \\ \frac{x+1}{x}, \text{when $x>1$} \end{cases} $$

for which when approaching $x=1$ from the negative $x$-axis evaluates to $1$ and from the positive axis it evaluates to $2$. So also the inverse also isn't continuous?


问:
Discriminant of elliptic curves as 1-dimensional abelian variety
We define an abelian variety over a field k as a connected and complete algebraic group over k. And we define an elliptic curve 1-dimensional(as schemes) abelian variety. This should equals a normal elliptic curve, so this should have the discriminant. However the definition of an elliptic curve as an abelian variety don't have Weierstrass forms, hence the discriminant can't be defined as usual. Can we define the discriminant?</p>

答:
I am expanding a little bit on my comment.

Let $E$ be an elliptic curve over $k$, i.e., a $1$-dimensional Abelian variety over $k$. Let us suppose for simplicity that $k$ is algebraically closed. Denote the identity of the group law of $E$ by $P_0$.

Being an Abelian variety, the canonical divisor $K_E$ is trivial. Moreover, the genus of $E$ is $1$. Hence, if $D_n := n[P_0]$ for $n \in \mathbb Z$, Riemann-Roch (cf. Hartshorne, Chapter IV, Theorem 1.3) for the divisor $D_n$ reads as

$$l(D_n) - l(-D_n) = n.$$

Here, $l(D) = \dim L(D)$, where $L(D) = \{ f \in k(E)^* \ | \ \mathrm{div}(f) \geq -D\}$. Since the degree of a principal divisor is $0$, we obtain that $l(D) = 0$ if $\deg(D) < 0$. Thus, the formula obtained by Riemann-Roch yields

$$l(D_n) = n, \ \ \ \ \text{ if } n \geq 0.$$

Having obtained this, we may choose elements

$$y \in L(D_3) \setminus L(D_2), \ \ \ \ x \in L(D_2) \setminus L(D_1).$$

The $7$ elements

$$1, \ x, \ x^2, \ x^3, \ y, \ xy, \ y^2$$

are all contained in $L(D_6)$, which is a $6$-dimensional space. Hence, they are linearly dependent, so that there are $\lambda_0, ..., \lambda_6 \in k$ such that

$$\lambda_0 \cdot 1 + \lambda_1 x + \lambda_2 x^2 + \lambda_3 x^3 + \lambda_4 y + \lambda_5 xy + \lambda_6 y^2 = 0.$$

Note that $x^3$ and $y^2$ have a pole at $P_0$, which has order $6$, and all of the other five remaining elements have poles of order $< 6$ in $P_0$. Hence, in order for the above linear combination to equal $0$, the poles of $x^3$ and $y^2$ have to cancel out. Thus $\lambda_3 \neq 0 \neq \lambda_6$.

Without loss of generality, $\lambda_6 = 1$ and so one gets an equation of the form

$$y^2 = p(x) + yq(x),$$

where $p(x)$ is a cubic polynomial in $x$ and $q(x)$ is a linear polynomial in $x$. By elementary changes of coordinates, this equation can be simplified to an equation in Weierstraß form.


问:
If $M=R/(p^k)$ and $N=p^{k-1}M$ then N is contained in every non-zero submodules of M
I have got this problem where I can do but only the tiny last bit of the proof where I got stuck. So here is the problem:

Let $R$ be a PID, $p\in R$ and irreducible element, $k\ge 1$ and let $M$ be the $R$-module $R/(p^k)$. Let $N=p^{k-1}M$. Show that N is a submodule of M and N is contained in every non-zero submodule of M.

My approach:

M is of the form $\{r+(p^k)|r\in R\}$, so $N=p^{k-1}M=p^{k-1}(r+(p^k))=rp^{k-1}+(p^k)$ where $r\in R$. Let $u=rp^{k-1}+(p^k),v=sp^{k-1}+(p^k)\in N$ where $r,s\in R$. We need to show $u+\rho v\in N$ for $\rho\in R$.

$$u+\rho v=rp^{k-1}+(p^k)+\rho sp^{k-1}+(p^k)$$$$=rp^{k-1}+\rho sp^{k-1}+(p^k)=(r+\rho s)p^{k-1}+(p^k)$$ but since $(r+\rho s)p^{k-1}\in R$, so $(r+\rho s)p^{k-1}+(p^k)\in N$, hence $u+\rho v\in N$.

Now let $\varphi:R\to R/(p^k)$ be natural projection $\varphi(a)=a+(p^k)$. Let $L\le R/(p^k)$ be a non-zero submodule.Then $\varphi^{-1}(N)=(p^{k-1})$ and $L'=\varphi^{-1}(L)$ is a submodule of $R$ that contains $(p^k)$. This is equivalent to $L'$ being an ideal in $R$ that contains $(p^k)$. Then I am stuck.

I know that we need to show $L\supseteq N$ by showing some inclusions in the domain. But how?

Here are some doubts and questions:

1. Is $(p^k)=(p^{k-1})$ ? I think it is since $\frac{1}{p}p^k\in(p^{k-1})$ and $p(p^{k-1})\in (p^k)$. Am I right or wrong?

2. Can $L'$ equal to $(p^k)$? If yes why, if not why not?

3. Do we have to show N is a submodule first before we can show N is contained in every non-zero submodule of M?

Also please feel free to correct my proof on N is a submodule if there are any mistakes.

Thanks very much in advance! I will really appreciate any helps.

答:
You are overcomplicating the problem too much. First recall that an $R$ - submodule of $R/(p^k)$ is just an ideal of this ring. By the correspondence theorem, the ideals of $R/(p^k)$ are in one to one correspondence with the ideals of $R$ that contain $(p)^k$. You should check that the only ideals that contain this are
$$(p)^i, \hspace{5mm} 0 \leq i \leq k$$

with the understanding that $p^0 = (1)$. Now we have the chain of strict inclusions

$$(p)^k \subsetneqq (p)^{k-1}\subsetneqq\ldots  \subsetneqq R$$

and so if we look in the quotient, we have the following strict inclusion of non-zero ideals

$$\Bigg((p)^{k-1}/(p)^k\Bigg) \subseteq \Bigg((p)^{k-2}/(p)^k\Bigg) \subsetneqq \ldots \subsetneqq  R/(p)^k$$

which completes your problem.


问:
Hard problems solving tricks
I'm a math student (last year undergraduate) and I really enjoy solving math problems but sometimes I face problems that I don't see how to solve even after trying every idea I had. When I look up at the solution I understand it perfectly but I can't figure out how the author actually found his solution. Their thinking process is never explained.
Let me take an example:

Let $P \in \Bbb R[X]$. Show that if $\forall x \in [0,1], P(x) \ge 0$ then $\exists A,B,C,D \in \Bbb R[X], P=A^2+XB^2+(1-X)C^2+X(1-X)D^2$.

What I've done:

I tried to look at some special cases ($\deg P = 0,1,2$).

tried to see if there is a structure behind this (by introducing $E= \{A^2+XB^2+(1-X)C^2+X(1-X)D^2 \ | \ A,B,C,D \in \Bbb R[X]  \}$ and looking if $E$ is stable by product) but this was hard.

tried to reinforce the hypothesis by assuming $P \ge 0$, this worked well because I remembered another problem I solved ($P\ge 0 \implies \exists A,B \in \Bbb R[X], P=A^2 + XB^2$). I tried to adapt the proof (what I already tried previously) but it didn't help.

tried to use coefficients (really bad idea here).

tried Lagrange interpolation (not helpful either).

I posted this problem here and @orangeskid advised me to show that, in fact, $P(x)= A^2(x) + x(1-x)D^2(x)$ ie. we could let $B=C=0$, it was a very good hint. Even in a parallel world I'd never have thought of doing so. But why didn't I think of doing this? I don't know, it wasn't obvious for me or maybe I don't see far enough.

Another example:

Let $A, B \in \Bbb Z[X]$ such that $(\gcd(A(n), B(n)))_n$ is periodic. What can we say about $A$ and $B$?

What I tried:

I supposed that $\deg A = \deg B = 1$ even with that $\gcd(A(n),B(n))$ is something that scares me because I don't have a lot of intuition on this sequence. So I applied, Bézout's theorem which says that we can find two sequences $a_n,b_n$ such that $(a_n A(n) + b_n B(n))_n$ is periodic.
And I got stuck here.

I saw a solution here using the resultant of two polynomials, something I never heard of.

There are other examples here, here and here: they all need a little trick, something close to magic (is that cleverness?).

So my questions are how could I find these tricks myself and make them natural? Why the origin of these tricks isn't explained? Should I just learn them "on the job"? Should I be worried (as a student in mathematics) if I don't see them? Thanks for your time!

答:
I think that a typical situation of a working mathematician is described by Nicholas Bourbaki, who said that “the mathematician does not work like a machine, nor as the workingman on a moving belt; we can not over-emphasize the fundamental role played in his research by a special intuition, which is not the popular sense-intuition, but rather a kind of direct divination (ahead of all reasoning) of the normal behavior, which he seems to have the right to expect of mathematical beings, with whom a long acquintance has made him as familiar as with the beings of the real world”.

This vision should be helpful to solve the most of usual problems.

The base of a general method of hard problem solving is sketched in a book “Mathematical discovery: on understanding, learning and teaching” by George Polya.

Should I just learn them "on the job"?

I think this is a usual way.

Why the origin of these tricks isn't explained?

I think there is no need for that. Also there is no need to memorize tons of tricks, because they come naturally, when you have knowledge. I think a trick usually emerges while thinking on the problem. I recommend Chapters 10–12 of Polya’s book for details.

how could I find these tricks myself and make them natural?

I can add to the above a few finding hints. Try to see a problem from different points of view, search familiar elements in the problem. Return to the problem time after time.

Should I be worried (as a student in mathematics) if I don't see them?

I think that although there can be specific mind problems blocking trick vision, usually a student in mathematics should not be worried about that. Leonardo da Vinci said: “There are three classes of people: those who see, those who see when they are shown, those who do not see”. A problem solver sometimes sees some tricks. More advanced see more, less advanced see less. But trick vision often requires a luck and even a very good problem solver can miss a simple trick.

Is there a book that lists some of them

I think a source for the tricks should be not a list but a general mathematical knowledge. Remark that “Mathematical quickies” by Charles Trigg is a collection of problems with tricky solutions.

Title: Hard problems solving tricks


问:
Fisher information of a Binomial distribution
The Fisher information is defined as $\mathbb{E}\Bigg( \frac{d \log f(p,x)}{dp} \Bigg)^2$, where $f(p,x)={{n}\choose{x}} p^x (1-p)^{n-x}$ for a Binomial distribution. The derivative of the log-likelihood function is $L'(p,x) = \frac{x}{p} - \frac{n-x}{1-p}$. Now, to get the Fisher infomation we need to square it and take the expectation.

First, we know, that $\mathbb{E}X^2$ for $X \sim Bin(n,p)$ is $ n^2p^2 +np(1-p)$. Let's first focus on on the content of the paratheses.

$$
\begin{align}
\Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2&=\frac{x^2-2nxp+n^2p^2}{p^2(1-p)^2}
\end{align}
$$

No mistake so far (I hope!).

\begin{align}
\mathbb{E}\Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2 &= \sum_{x=0}^n \Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2 {{n}\choose{x}} p^x (1-p)^{n-x} \\
&=\sum_{x=0}^n \Bigg( \frac{x^2-2nxp+n^2p^2}{p^2(1-p)^2} \Bigg) {{n}\choose{x}} p^x (1-p)^{n-x} \\
&= \frac{n^2p^2+np(1-p)-2n^2p^2+n^2p^2}{p^2(1-p)^2}\\
&=\frac{n}{p(1-p)}
\end{align}

The result should be $\frac{1}{p(1-p)} $ but I've been staring at this for a few hours incapable of getting a different answer. Please let me know whether I'm making any arithmetic mistakes.

答:
So, you have $X$ ~ Binomial($n$, $p$), with pmf $f(x)$:

You seek the Fisher Information on parameter $p$. Here is a quick check using mathStatica's FisherInformation function:

which is what you got :)


问:
Let $f: A\rightarrow B$ and $g: B\rightarrow A$ be 2 surjective Functions. Show that $f\circ g = id \iff g\circ f = id$
I've been sitting on this problem for a bit and trying to solve it using the definition of surjectivity and the definition of the composition, but I feel stuck.

According to surjectivity, for each $b \in B$ exists an $a \in A$ with $f(a) = b$, and for each $a \in A$ exists a $b \in B$ with $f(b) = a$.

For $f\circ g$ we have: $f\circ g: B \rightarrow B$

For $g\circ f$ we have: $g\circ f: A \rightarrow A$

The definition of composition would tell us that: $g(b) \subseteq A$ and $f(a) \subseteq B$

So they seem to be their own inverse functions, but are they that truly? Can we say that g(b) produces a value that will take us back to b if we apply f to it?

Would it be simpler to just prove $f\circ g = g\circ f$?

答: Assume that $f\circ g=\operatorname{id}$. Let $a\in A$; you want to prove that $g\bigl(f(a)\bigr)=a$. Since $g$ is surjective, there is some $b\in B$ such that $a=g(b)$. But$$a=g(b)\implies f(a)=f\bigl(g(b)\bigr)=b$$and therefore$$g\bigl(f(a)\bigr)=g(b)=a.$$


问:
Derivative of the Lagrangian dual: Can I treat the optimal parameter $x^*$ in the dual $g(\lambda) = L(x^*, \lambda)$ as a constant w.r.t $\lambda$?
Suppose I have a standard constrained optimization problem
$$
\min_x f(x) \quad \text{s.t.} h_i(x) \ge 0,
$$
where $f$ and $h_1,...,h_m$ are convex functions $\mathbb{R}^n \to \mathbb{R}$. The dual is given as
$$
g(\lambda) = \min_xL(x, \lambda) = L(x^*, \lambda)
$$
with $L$ the Lagrangian. To obtain the optimal Lagrangian multipliers, I need to solve
$$
\lambda^* = \text{argmax}_\lambda \;g(\lambda).
$$
I would go on and take the derivative w.r.t $\lambda$. My question is now:

Can I treat the optimal parameter $x^*$ in the dual $g(\lambda) = L(x^*, \lambda)$ as a constant w.r.t $\lambda$?

On my first thought: No, since it depends on $\lambda$. (If it didn't we would already have solved the primal problem). But I have found several examples, where we can take the derivative without the substitution of $x^*$ and treat it as a constant. So I was wondering if that is a general rule. If that is somehow true, this would simplify the derivative a lot.

I'll give a concrete example:
Let's solve
$$
\min x^2 \quad \text{s.t.} -x -2 \le 0
$$
The resulting Lagrangian is $ L(x, \lambda) = x^2 + \lambda(-x -2)$.
Taking the derivative w.r.t $x$, and setting it to 0 yields
$ 2x - \lambda = 0 \implies x^* = \lambda / 2$.

The dual is therefore
$$
g(\lambda) = x^{*2} + \lambda(-x^* -2) = \frac 1 4 \lambda^2 + \lambda(-\frac \lambda 2 -2).
$$
Now, when i take the "correct" derivative by taking the derivative of the right side w.r.t. to $\lambda$, I'll get
$$
g'(\lambda) = - \frac \lambda 2 - 2.
$$
This is the same as if I take $x^{*}$ as a constant on the left side and take the derivative, resulting in $$-x^* - 2 = - \frac \lambda 2 - 2.$$

Why does this work?
Any help is much appreciated!

答:
In general, no, $x^*$ is not a constant and you cannot differentiate the Lagrangian as if it were.

Consider the Lagrangian $L(x, \lambda) = x^2 + 1 + \lambda(x-2)(x-4)$. If you take the infimum over $x$ by taking $\partial L /\partial \lambda$ and setting it to zero, you'll find $x^* = 3\lambda/(\lambda + 1)$. Clearly $x^*$ depends on $\lambda$ so you must use the chain rule to differentiate $L$ w.r.t. $\lambda$.


问:
Step 1 in proof of denseness of $\Bbb Q$
I am doing the prof for the denseness of $\Bbb Q$.
I want to find $m,n$ in $\Bbb Z$ ($n>0$) such that $a<\frac mn<b$.

My prof says the first step is to find $n$ in $\Bbb N$ such that $bn-an>1$.

And the second step is to prove the following claim: if [left with] two real numbers $r_1,r_2 \in \Bbb R$ [such that] $r_2-r_1>1$, then there’s always an integer between, so there exists an $m$ in $\Bbb Z$ such that $r_1<m<r_2$.

To be honest, I haven’t even got to why the second step is the way it is, however I am confused when it comes to the first step; it’s like the inequality $bn-an>1$ just pops out of nowhere. I’d appreciate some help. Please and thank you.

答: You have$$bn-an>1\iff(b-a)n>1\iff n>\frac1{b-a}.$$And there is such a $n$ by the Archimedian property.


问:
$\sum_i x_i^n = 0$ for all $n$ implies $x_i = 0$
Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost.

Let $x_1 , x_2, \dots, x_k$ be complex numbers satisfying:

$$x_1 + x_2+ \dots + x_k = 0$$

$$x_1^2 + x_2^2+ \dots + x_k^2 = 0$$

$$x_1^3 + x_2^3+ \dots + x_k^3 = 0$$

$$\dots$$

Then $x_1 = x_2 = \dots = x_k = 0$.

The statement seems obvious because we have more than $k$ constraints (constraints that are in some sense, "independent") on $k$ variables, so they should determine the variables uniquely.  But my attempts so far of formalizing this intuition have failed.  So, how do you prove this statement?  Is there a generalization of my intuition?

答:
For a slightly different method than Potato's second answer (but the idea is mainly the same):

Without loss of generality, the system of equations can be written as: $$\left\{ \begin{array}{lcl} \lambda_1x_1 + \lambda_2x_2+ \dots + \lambda_k x_k &= &0 \\ \lambda_1x_1^2 + \lambda_2x_2^2+ \dots + \lambda_k x_k^2 & = & 0 \\ & \vdots & \\ \lambda_1x_1^k + \lambda_2x_2^k+ \dots + \lambda_k x_k^k & = & 0 \end{array} \right.$$

where $\lambda_i>0$, $x_i \neq 0$ and $x_i \neq x_j$ for $i \neq j$. Indeed, if $x_i=x_j$ replace $x_i+x_j$ with $2x_i$ and if $x_i=0$ just remove it. By contradiction, suppose $k \geq 1$.

Now, the family $\{ (\lambda_1 x_1^j, \dots , \lambda_k x_k^j) \mid 1 \leq j \leq k \}$ cannot be linearly independent since the vector space $\{(y_1,\dots, y_k) \mid y_1+ \dots+ y_k=0 \}$ has dimension $k-1$ (it is a hyperplane). Therefore, the matrix

$$A:=\left( \begin{matrix} \lambda_1x_1 &  \lambda_2x_2 & \dots & \lambda_k x_k  \\ \lambda_1x_1^2 & \lambda_2x_2^2 & \dots & \lambda_k x_k^2  \\ \vdots & \vdots & & \vdots  \\ \lambda_1x_1^k & \lambda_2x_2^k & \dots & \lambda_k x_k^k  \end{matrix} \right)$$

is not invertible. Using Vandermonde formula, $$0= \det(A)=  \prod\limits_{i=1}^k \lambda_i   \cdot \prod\limits_{i=1}^k x_i  \cdot \prod\limits_{i<j} (x_i-x_j) $$

which is nonzero by assumption.


Title: Prove $\lim_{ (x,y) \to (0,0)} \frac{x^3-y^3}{x^2+y^2}=0$

Body:
Can some one prove the limit using epsilon delta method to prove that the limit exists
$$
\lim_{(x,y) \to (0,0)} \frac{x^3-y^3}{x^2+y^2}=0
$$

Body:
Let $(x, y)  \not = (0, 0);$

$\dfrac{|xx ^2-yy^2¦}{x^2+y^2}| \le\dfrac{|x|x^2+|y|y^2}{x^2+y^2}\le|x|+|y|\le 2\sqrt{x^2+y^2}.$

$\delta =? $

P. S. Tito' s comment.


问:
Concern regarding infinitesimal errors
When finding out the area under a curve,we divide it into many thin strips and them sum up.My question is when we write for example $dW=F×ds$ and then integrate in physics terminology,we are referring to a width of $ds$ and a height of $F$.But in a curve we cannot find a perfect rectangle because the two heights won't be the same,though they will be close but not exact.If calculus is the measure of precision,why can we then write $dW=F×ds$ while we know the area of the small region will not exactly be the area of the rectangle and there will be some errors.So when we add all the infinitely thin rectangles,won't there be a small error since we are actually leaving space?I mean even when we have infinitely small width,it still isn't a perfect rectangle. P

答:
Yes that is quite a genuine doubt which arises in the mind if every beginner calculus student. No doubt the error will always be there, but we can make that error as small as WE please..... And if we make the sizes of the rectangles tending towards zero, we will get  the error tending towards zero..... Essentially this is what forms the fundamental theorem of calculus

Theoretically yes, we can get zero error as per Mathematics. And intheoretical physics too, we can tend the small areas towards zero. The infinitesimal is defined as the smallest, as small as possible. And any finite addition or subtraction does not affect it. The too small errors in the calculation of the rectangles area time to zero as does  the Dx element

In the sense of practical physics and computational sciences where we practically cannot achieve 0, we have to take Dx as per our computational capacity and conditions.
And does an error of about 0.001% really matter? The more small Dx you use the more smaller error and more the computational resources required.

Title: Concern regarding infinitesimal errors


问:
How to compare binomial distributions with different number of trial?
I want to know which event has the best score (or luck or odd)

$p$ (prob of success) is equal with all event.
There are givne set of events, that is (number of try, number of success)

$$\mathrm{event1} (n_1, x_2) \sim \mathcal{B}(n_1,P)$$
$$\mathrm{event2} (n_2, x_2) \sim \mathcal{B}(n_2,P)$$
$$\dots$$
$$\mathrm{event99} (n_{99}, x_{99}) \sim \mathcal{B}(n_{99},P)$$

The number of trials,success is all different.

And how compare probability of each cases?
for example,
ranking NO.1 = event33 /
ranking NO.2 = event44 /
... ranking NO.99 = event11

==============================

In the same trial, the more success
In the same success, the less trial

Q1. I try cdf of Binomial distribution. (quantile) But I don't know it is okay. Because Binomial distribution has diffrent distribution with different number of trial.

Q2. so, I try2 Normal distribution approximation. But N is not big enough.

Q3. and I don't know if I should use the negative binomial distribution. In that cases, the prob of binomial dist is bed.. there are too many 1

Thank you

答:
If you want probabilities of specific events, you can use the binomial PDF function. For example, if $X \sim \mathsf{Binom}(5, .4),$ then $P(X = 2) = {5 \choose 2}(.4)^2(.6)^3 = 0.3456.$

```<code>10*.4^2*.6^3
[1] 0.3456
</code>```

You can use the procedure dbinom in R to compute this easily:

```<code>dbinom(2, 5, .4)
[1] 0.3456
</code>```

Notice that this is quite different from getting 4 successes in 10 trials, when $p = 0.4:$

```<code>dbinom(4, 10, .4)
[1] 0.2508227
</code>```

If you want confidence intervals for $p$ given $X = 2$ successes
in $n = 5$ trials, then there are various methods: You can read about them in the Wikipedia article on binomial confidence intervals.
I will show you two of them:

A 95% Agresti-Coull confidence
interval in this case would use $\hat p = (X+2)/(n+4) = 4/9$ to get the interval $\hat p \pm 1.96 \sqrt{\hat p(1-\hat p)/(n+4)}.$
This computes to the interval $(0.119, 0.769),$ which does include $p = 0.4.$

```<code>n = 5;  x = 2
hat.p = (x+2)/(n+4)
hat.p + qnorm(c(.025,.975))*sqrt(hat.p*(1-hat.p)/9)
[1] 0.1198065 0.7690824
</code>```

Another style of confidence interval is the Jeffries interval,
based on a beta distribution: $(0.094, 0.791),$ which also includes $p = 0.4.$

```<code>qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.09439033 0.79058334
</code>```

Note: For larger $n$ the two types of intervals would
tend to be shorter and to agree more closely. For example, consider
these two styles of confidence intervals for 20 successes in 50 trials:

Agresti-Coull

```<code>x = 20; n = 50
hat.p = 22/54
hat.p + qnorm(c(.025,.975))*sqrt(hat.p*(1-hat.p)/54)
[1] 0.2763554 0.5384594
</code>```

Jeffries:

```<code>qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.2729524 0.5382564
</code>```

Body:
If you want probabilities of specific events, you can use the binomial PDF function. For example, if $X \sim \mathsf{Binom}(5, .4),$ then $P(X = 2) = {5 \choose 2}(.4)^2(.6)^3 = 0.3456.$

10*.4^2*.6^3
[1] 0.3456

You can use the procedure dbinom in R to compute this easily:

dbinom(2, 5, .4)
[1] 0.3456

Notice that this is quite different from getting 4 successes in 10 trials, when $p = 0.4:$

dbinom(4, 10, .4)
[1] 0.2508227

If you want confidence intervals for $p$ given $X = 2$ successes
in $n = 5$ trials, then there are various methods: You can read about them in the Wikipedia article on binomial confidence intervals.
I will show you two of them:

A 95% Agresti-Coull confidence
interval in this case would use $\hat p = (X+2)/(n+4) = 4/9$ to get the interval $\hat p \pm 1.96 \sqrt{\hat p(1-\hat p)/(n+4)}.$
This computes to the interval $(0.119, 0.769),$ which does include $p = 0.4.$

n = 5;  x = 2
hat.p = (x+2)/(n+4)
hat.p + qnorm(c(.025,.975))*sqrt(hat.p*(1-hat.p)/9)
[1] 0.1198065 0.7690824

Another style of confidence interval is the Jeffries interval,
based on a beta distribution: $(0.094, 0.791),$ which also includes $p = 0.4.$

qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.09439033 0.79058334

Note: For larger $n$ the two types of intervals would
tend to be shorter and to agree more closely. For example, consider
these two styles of confidence intervals for 20 successes in 50 trials:

Agresti-Coull

x = 20; n = 50
hat.p = 22/54
hat.p + qnorm(c(.025,.975))*sqrt(hat.p*(1-hat.p)/54)
[1] 0.2763554 0.5384594

Jeffries:

qbeta(c(.025,.975), x+.5, n-x+.5)
[1] 0.2729524 0.5382564


问:
A Data Recovery proof on K-means algorithm objective function
There are quite a few materials about the K-means algorithm,
but what I need is proof of how its objective function is derived based on the so-called data recovery approach.
Ref. for this approach: B. Mirkin, Clustering: A Data Recovery Approach, CRC Press (1st Edition, 2005; 2d Edition, 2012). The book can be downloaded freely:
http://www.dcs.bbk.ac.uk/~mirkin/papers/n.pdf

My question:

Given an entity-to-feature matrix $Y = (y_{iv}) \in \mathbb R^{N}$ and given number of clusters K. The goal is to cluster Y into K disjoint sets, that is, $S = \{S_1, S_2, ...., S_K \}$. And the cluster $S_{k}$ is represented by a $N \times 1$ column vector $s_{k} = (s_{ik})$, where $s_{ik}=1$ if $i \in S_{k}$ and  $s_{ik}=0$, otherwise ($i\in I$).

To related the notion of members of the clusters having similar feature values, we consider the standard $V$-dimensional point of $S_{k}$, $c_{k}=(c_{k v})$ ($v=1, ..., V, k=1, ..., K$).

Then, as it is described on page 288:

\begin{equation}
\label{e1}
y_{iv} = c_{kv} + e_{iv}; \  i \in S_{k}   \ \ \ (7.18)
\end{equation}

As the author wrote:
multiplying equations (7.18) by themselves and summing up the results,
it is not difficult to derive the following equation:

\begin{equation}
\label{eqn:y_iv_sqr_sum}
\sum_{i,v} y_{iv}^{2} =  \sum_{k=1}^{K} \sum_{v} c_{kv}^{2} |S_{k}| +  \sum_{i,v} e_{iv}^{2}  \ \ \ \ (7.19)
\end{equation}

Now my question is whether the textbook's proof is correct or not?
Because I obtain something which slightly differs from it, I have the coefficient $K$ multiplied with the LHS and last term of RHS.

Here is my proof

defining the model as:

\begin{equation}
\label{e:2}
y_{iv} = c_{kv} s_{ik} + e_{iv} \ \ (i=1, ..., N, v \in V, k=1, ..., K)
\end{equation}

Rearranging, squaring and summing over all clusters, all data points and all features implies:

\begin{equation}
\sum_{k=1}^{K} \sum_{i,v} e_{i v}^{2} = \sum_{k=1}^{K} \sum_{i, v}(y_{i v} - c_{k v} s_{i k})^{2}
\end{equation}

Furthermore, we have:

\begin{equation}
\begin{split}
K \sum_{i,v} e_{i v}^{2} = \sum_{k=1}^{K} \sum_{i, v} (y_{i v}^{2} - 2 y_{iv} c_{k v} s_{i k} + c_{k v}^{2} s_{i k}^{2})
\end{split}
\end{equation}

Recalling the definition of the mean:

\begin{equation}
    c_{k v} = \frac{\sum_{i} y_{iv} s_{ik}}{|S_{k}|}=\frac{\sum_{i\in S_{k}}y_{iv}}{|S_{k}|}
    \label{eqn:def_ck_same_mdl}
\end{equation}

And by substituting it in its above-mentioned Eqn. and little algebric manipulation we will obtain:

\begin{equation}
      K \sum_{i,v} y_{iv}^{2} =  \sum_{k=1}^{K} \sum_{ v} c_{kv}^{2} |S_{k}| + K \sum_{i,v} e_{iv}^{2}  \ \ \ \ (*)
\end{equation}

The above Eqn. holds because:

$\sum_{i} s_{ik}^2 = \sum_{i} s_{ik} = |S_{k}|$ since $s_k$ is a N-dimensional binary vector;

for a given $k$ recalling the definition of the mean, $ c_{k v} = \frac {1}{|S_{k}|} \sum_{i \in S_{k}} y_{i v} $, we find $\sum_{v} c_{k v} \sum_{i \in S_{k}} y_{i v}= \sum_{v} c_{k v} c_{k v} |S_{k}| = |S_{k}| \sum_{v} c_{k v}^2$.

Am I right?
(Because in the book there is no such a $K$ as I obtained Eqn. (*))

Thanks in advance!

答:
After spending several hours I finally managed to prove it properly so that it is compatible with the proof in the textbook.  (Although I will not close this question  for a while so that probably someone wants to provide another answer)

Here is my proof:

Since we have $K$ non-overlapping clusters therefore, we can define our model as:

\begin{equation}
\label{model_f}
y_{iv} = \sum_{k=1}^{K} c_{kv}s_{ik} + e_{iv},
\end{equation}

Rearranging and squaring implies:

\begin{equation}
\label{model_f_sqr}
e_{iv}^2 =  (y_{iv} - \sum_{k=1}^{K} c_{kv}s_{ik})^2,
\end{equation}

opening the parenthesis yields:

\begin{equation}
\label{model_f_sqr_opn}
e_{iv}^2 =  y_{iv}^2 - 2 y_{iv} \sum_{k=1}^{K} c_{kv}s_{ik} + \sum_{k=1}^{K} c_{kv}^2 s_{ik}^2,
\end{equation}

By summing over all entities, all features and, rearranging a bit:

\begin{equation}
\label{model_f_sqr_opn_sums}
\sum_{i,v} e_{iv}^2 =  \sum_{i,v} y_{iv}^2 - 2 \sum_{i,v} \sum_{k=1}^{K} y_{iv} s_{ik} c_{kv} + \sum_{i,v} \sum_{k=1}^{K} c_{kv}^2 s_{ik}^2.
\end{equation}

Now we should recall two facts:

definition of the average of a cluster, $k$ over a feature, $v$:
$c_{kv} = \frac{\sum_{i}y_{iv}s_{ik}}{N_{k}}$. Where $N_{k}$ is the cluster cardinality. Therefore we have $\sum_{i}y_{iv}s_{ik} = c_{kv} N_{k}$

Since the cluster membership vectors ($s_{k}$) are binary, thus: $\sum_{i} s_{ik}^2 = \sum_{i} s_{ik} = N_{k}$

Therefore,

w.r.t 1) $ -2 \sum_{i,v} \sum_{k=1}^{K} y_{iv} s_{ik} c_{kv} = \sum_{k} \sum_{v} c_{kv}^{2} N_{k}$;

And

w.r.t 2) $ \sum_{k=1}^{K} \sum_{v} \sum_{i} c_{kv}^2 s_{ik}^2 = \sum_{k=1}^{K} \sum_{v} c_{kv}^2 N_{k}$

Therefore, we have:

\begin{equation}
\label{model_proof}
\sum_{i,v} e_{iv}^2 =  \sum_{i,v} y_{iv}^2 - \sum_{k=1}^{K} \sum_{v} c_{kv}^2 N_{k}
\end{equation}

And the proof is complete.


问:
Can we count the number of triangles in the figure with combination methods?
How many triangle can you find in following picture?

Solution: check picture solution,

But can we solve this problem with combination methods?

答: $n$ lines emanating from a common point and crossing one or two other lines define $\binom n2$ triangles. You see such a configuration four times.


问:
Definition of "Topological Equivalence" for metric spaces
Consider a set $A$ on which $2$ metrics are defined: $d_1$ and $d_2$.

Let the two resulting metric spaces $(A, d_1)$ and $(A, d_2)$ be topologically equivalent.

Given the above, is it always the case that the topology induced on $A$ by $d_1$ is necessarily identical to the topology induced on $A$ by $d_2$?

By topological equivalence in the context of metric spaces, I have the definition that $d_1$ and $d_2$ are topologically equivalent iff ($U \subseteq A$ is "$d_1$-open" iff $U \subseteq A$ is "$d_2$-open"). $(1)$

Again, my definition of $d_1$-open is iff it is the neighborhood (under the metric $d_1$) of each of its points.

A neighborhood of a point $x$ is defined as a subset of $A$ which completely contains an open $\epsilon$-ball centered at $x$ for some $\epsilon \in \mathbb R_{>0}$.

The open $\epsilon$-ball of $x$ is the set:

$B_\epsilon (a) := \{x \in A: d_1 (x, a) < \epsilon\}$

... and we are finally down to the bare metal, so to speak.

Now, Bert Mendelson, in his "Introduction to Topology" (3rd ed. 1975, Dover Publications), gives this exercise in his chapter 3, section 2 "Topological Spaces" exercise 3:

"Let $(\mathbb R^n, d)$ and $(\mathbb R^n, d')$ be defined as"

$d(x,y) = \displaystyle \max_{1 \le i \le n} \{|x_i - y_i\}$ (the Chebyshev or chessboard metric)

$d'(x,y) = \left({\displaystyle \sum_{i = 1}^n (x_i - y_i)^2}\right)^{1/2}$ (the usual Euclidean metric)

Prove that the two metric spaces $(\mathbb R^n, d)$ and $(\mathbb R^n, d')$ give rise to the same topological space."

Now this can be straightforwardly addressed (probably, I haven't gone through the motions) of constructing $\tau$ and $\tau'$ corresponding to $d$ and $d'$ respectively, and demonstrating that an arbitrary open set of $\tau$ is in $\tau'$ and an arbitrary open set of $\tau'$ is in $\tau$.

But (as an exercise set in a different work, i.e. W.A. Sutherland's "Introduction to Metric and Topological Spaces") I have already proved that these two spaces are Lipschitz equivalent:

$$d' (x, y) \ge d (x, y) \ge \dfrac 1 2 d' (x, y)$$

and thence, by another result that I have a proof of that Lipschitz equivalent metrics are topologically equivalent in the sense as defined as in $(1)$ above, that they are in fact now proved to be topologically equivalent.

If I could use the result that "two metrics on the same set are topologically equivalent if they give rise to equal topologies (in the sense that the open sets of one are exactly the open sets of the other), then I can answer the question in Mendelson by merely quoting my result above (which I arrived at via that long route via Lipschitz equivalence) that $(\mathbb R^n, d)$ and $(\mathbb R^n, d')$ are proven to be topologically equivalent.

But is it actually the case that "two metrics on the same set are always topologically equivalent if and only if they give rise to equal topologies"?

I have never seen that result stated formally (although I admit I have not read all that widely), and so I am wondering whether, as it has not been so stated, whether it is actually true or not?

答:
I address some of these issues in this older answer.

But in short: yes $(X,d_1)$ and $(X,d_2)$ are exactly topologically equivalent iff $d_1$ and $d_2$ induce the exact same topology on their set $X$. If the metrics are Lipschitz-equivalent (as they will always be on $\Bbb R^n$, if they come from a norm..., as your examples) then they will be topologically equivalent, but the reverse need not need hold. This is related to the so-called uniform structure on a metric space. Two metrics can be non-equivalent for Cauchy sequences and uniform continuity, but still topologically equivalent.

In topology we typically only care about the topology the metric induces. A metric has other uses, but in pure general topology we choose to ignore them.


问:
Equality of expectations in martingale
Let $(X_n)_{n \geq 0}$ be an $\mathcal{F}_n$-martingale that converges to $X_{\infty}$ in $L^p$. Why is it true that for $A \in \mathcal{F}_n$ and $n\leq m$

$$ E[X_{\infty}\mathbb{1}_A]=\lim_{m\to\infty}E[X_m \mathbb{1}_A] = E[X_n\mathbb{1}_A]$$

I know that martingales have constant expectation, but I didn't think

it stayed true for $n \to \infty$

$E[X_n]=E[X_m]$ implied $E[X_m \mathbb{1}_A] = E[X_n\mathbb{1}_A]$ for all $A\in \mathcal{F}_n$

答:
I suppose you are taking $p \geq 1$. $|EX_{\infty}1_A-EX_m1_A| \leq E|X_{\infty}-X_m| \to 0$  as $ m\to \infty$ since convergence in $L^{p}$ implies convergence in $L^{1}$.

Since $EX_m1_A=EX_n1_A$ for all $m \geq n$ when $A \in \mathcal F_n$ the equality of the limit with $EX_n1_A$  follows.

Title: Equality of expectations in martingale


问:
Find Maclaurin series and radius of convergence for $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$
Given $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$, find  the Maclaurin series and radius of convergence.

My attempt:

By considerably simple computation we get that

$f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2} = \ln(1+x) -\frac{1}{(x+1)}$, for $t \ge 0$,

Now what I'm confused about is that the Maclaurin series of $\displaystyle \ln(x+1)=\sum_{n=o}^\infty (-1)^{n+1} \frac{x^n}{n}$, and $\frac{1}{(t+1)} = \left( \ln(t+1) \right) ' $

does that means that: $\displaystyle f(x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} - (\frac{1}{x+1})$ but it seems to me that the convergence radius is $R=1$, but I'm missing the algebraic step to achieve that

答:
There is no need to integrate. $f(0)=0$, obviously, and

$$f'(x)=\frac x{(1+x)^2}=\frac1{1+x}-\frac1{(1+x)^2}.$$

From this,

$$f^{(n+1)}(0)=(-1)^nn!+(-1)^{n+1}(n+1)!$$

So the general term of Taylor is

$$(-1)^n\left(\dfrac1{n+1}-1\right)x^{n+1}$$ and by the ratio test, this converges for $|x|<1$.


问:
Given the matrix $A^3$, find A
Given the matrix $A^3= \begin{bmatrix}83&84\\42&41\end{bmatrix}$, find A

I only know how to use the basic method to find A, which is $A^3= \begin{bmatrix}a&b\\c&d\end{bmatrix}^3$, then solve a,b,c,d.
Is there any other method?

答: Hint : $A^3$ is diagonalizable and its eigenvalues are $5^3$ and $(-1)^3$.

Title: Given the matrix $A^3$, find A


问:
Homotopy-invariance of sheaf cohomology for locally constant sheaves

Suppose we have a homotopy equivalence $f: X \to Y$ (with homotopy
  inverse $g: Y \to X$) and a local system (i.e. a locally constant
  sheaf) $\mathscr{S}$ on $Y$. Is the homomorphism

  $$f^*: H^k(Y,\mathscr{S}) \to H^k(X, f^{-1}\mathscr{S})$$

  induced by the unit $1 \Rightarrow f_*f^{-1}$ of the adjunction then an isomorphism?

I have a feeling that this should be the case. Under the equivalence between locally constant sheaves and representations of the fundamental groupoid (on sufficiently nice spaces), we have that $(f \circ g)^{-1}\mathscr{S} \cong \mathscr{S}$ and $(g \circ f)^{-1}\mathscr{L} \cong \mathscr{L}$ for $\mathscr{L}$ a locally constant sheaf on $X$. I'm pretty sure this part is correct.

Thus, from the unit map $f^*\mathscr{S} \to g_*g^{-1}f^{-1} \mathscr{S} \cong g_*\mathscr{S}$ we get a morphism of sheaf cohomology groups
$$g^*: H^k(X, f^*\mathscr{S}) \to H^k(Y, g^{-1}f^{-1}\mathscr{S}) \cong H^k(Y, \mathscr{S}).$$
I think that this should be the inverse to the morphism $f^*$ above.

The usual way to show this would be first prove the following Lemma:

Lemma Given a locally constant sheaf $\mathscr{S}$ on $Y$ and homotopic maps $f_0: X \to Y$ and $f_1: X \to Y$, then the induced maps in cohomology are equal:  $$f_0^*=f_1^*: H^k(Y,\mathscr{S}) \to H^k(X, f^{-1}\mathscr{S}).$$

Unfortunately, I'm not very experience with sheaves yet, so I'm not sure how to get this result.

For motivation, the particular application I have in mind is for cohomology with local coefficients on a manifold with boundary, where the coefficients are given by the sheaf of parallel sections of a vector bundle with flat connection. I know a manifold with boundary is homotopy equivalent to it's interior, so if we have a vector bundle with flat connection on the manifold with boundary, is the twisted de Rham cohomology the same as the twisted de Rham cohomology when we pull the flat vector bundle back to the interior?

答:
I don't think that your argument is correct. Proving that two functors are isomorphic when restricted to the full subcategory of locally constant sheaves is not enough to deduce an isomorphism between their derived functors, because the definition of derived functor uses the whole category (locally constant sheaves are almost never acyclic).

For a proof (at least in the constant case) see Kashiwara-Shapira, Sheaves on Manifolds, Proposition 2.7.5. Essentially they prove a vanishing theorem for cohomology on $[0,1]$ and use proper base change for the projection $X \times [0,1] \rightarrow X$ to deduce that homotopic maps induce the same morphisms in cohomology for constant sheaves.


答:
Here is a categorical approach. Let us first derive a nice cancelation property of surjective functions which is useful on its own.  Suppose that $f:A\to B$ is surjective, and $h_1,h_2:B\to C$ are arbitrary.  If $h_1\circ f=h_2\circ f$, then $h_1=h_2$.

We will prove the contrapositive, thet if $h_1\neq h_2$, then $h_1\circ f\neq h_2\circ f$. If $h_1\neq h_2$, that means we can find some $b\in B$ such that $h_1(b)\neq h_2(b)$.  Because $f$ is surjective, we can find some $a\in A$ such that $f(a)=b$.  Then $h_1\circ f(a)=h_1(b)\neq h_2(b)=h_2\circ f(a)$, so $h_1\circ f$ disagrees with $h_2\circ f$ on at least one element, so $h_1\circ f\neq h_2\circ f$

N.B. Functions with this right cancelation property are called "epic."  There is also a left cancelation property, and functions that satisfy left cancelation are called "monic".  The proof shows that all surjective functions are epic, and one can show that all injective functions are monic.  One reason we care is that we can talk about "epic" or "monic" things even in contexts where injectivity or surjectivity don't quite make sense.

Proposition:  If $f:A\to B$ and $g:B\to A$ are epic maps, then $f\circ g=\operatorname{id}\Leftrightarrow g\circ f=\operatorname{id}$

Proof: If $f\circ g=\operatorname{id}$, then $g\circ (f\circ g) = g\circ \operatorname{id}=g=\operatorname{id}\circ g.$  However, since $g$ is epic, we can cancel the $g$ on the right: $g\circ f \circ g = \operatorname{id}\circ g$ implies $g\circ f=\operatorname{id}$.  The other direction is exactly the same argument.

Title: Let $f: A\rightarrow B$ and $g: B\rightarrow A$ be 2 surjective Functions. Show that $f\circ g = id \iff g\circ f = id$

Body:
I've been sitting on this problem for a bit and trying to solve it using the definition of surjectivity and the definition of the composition, but I feel stuck.

According to surjectivity, for each $b \in B$ exists an $a \in A$ with $f(a) = b$, and for each $a \in A$ exists a $b \in B$ with $f(b) = a$.

For $f\circ g$ we have: $f\circ g: B \rightarrow B$

For $g\circ f$ we have: $g\circ f: A \rightarrow A$

The definition of composition would tell us that: $g(b) \subseteq A$ and $f(a) \subseteq B$

So they seem to be their own inverse functions, but are they that truly? Can we say that g(b) produces a value that will take us back to b if we apply f to it?

Would it be simpler to just prove $f\circ g = g\circ f$?


Title: Find Maclaurin series and radius of convergence for $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$

Body:
Given $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$, find  the Maclaurin series and radius of convergence.

My attempt:

By considerably simple computation we get that

$f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2} = \ln(1+x) -\frac{1}{(x+1)}$, for $t \ge 0$,

Now what I'm confused about is that the Maclaurin series of $\displaystyle \ln(x+1)=\sum_{n=o}^\infty (-1)^{n+1} \frac{x^n}{n}$, and $\frac{1}{(t+1)} = \left( \ln(t+1) \right) ' $

does that means that: $\displaystyle f(x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} - (\frac{1}{x+1})$ but it seems to me that the convergence radius is $R=1$, but I'm missing the algebraic step to achieve that

Body:
$$\int_0^x \frac{t}{(t+1)^2} \, dt=\left[\frac{1}{t+1}+\log (t+1)\right]_0^x=\log (x+1)-\frac{x}{x+1}$$
$$\log(x+1)=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^n}{n}$$

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

$$\int_0^x \frac{t}{(t+1)^2} \, dt=\sum _{n=1}^{\infty } \frac{(-1)^n (n-1) x^n}{n}$$

radius of convergence is $r=1$.


问:
Is a Cauchy sequence bounded in a metric space if the space is not complete?
I don't think a metric space needs to be complete for a Cauchy sequence to be bounded. Being complete means the sequence converges to a point in the space, but the standard proof for Cauchy sequence being bounded does not require you to know what the sequence converges to. For that reason, I think it doesn't matter whether the metric space is complete or not.</p>

答:
Yes it is.

Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in a metric space $(E,d)$. By definition, there exists $N \in \mathbb{N}$ such that for every $p \geq N$, you have
$$d(x_p,x_N)\leq 1$$

Let
$$M = \max \lbrace d(x_i,x_N), 0 \leq i \leq N-1 \rbrace + 1$$

You have then that for every $i \in \mathbb{N}$,
$$d(x_i,x_N) \leq M$$

so for every $i \in \mathbb{N}$,
$$x_i \in B(x_N, M)$$

(where $B(x_N, M)$ denotes the ball centered at $x_N$ of radius $M$) ; i.e. the sequence is bounded.


答: If $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence in a metric space $(X,d)$, then there is some $N\in\Bbb N$ such that $n\geqslant N\implies d(x_m,x_n)<1$. Therefore$$\{x_n\mid n\geqslant N\}\subset B_1(x_N).$$Now, let $M=\max\{d(x_N,x_k)\mid k<N\}$; it exists, since it is the maximum of a finite set. Then$$\{x_n\mid n<N\}\subset B_{M+1}(x_N)$$and therefore, since $M+1>1$,$$\{x_n\mid n\in\Bbb N\}\subset B_{M+1}(x_N).$$

Title: Is a Cauchy sequence bounded in a metric space if the space is not complete?

Body: I don't think a metric space needs to be complete for a Cauchy sequence to be bounded. Being complete means the sequence converges to a point in the space, but the standard proof for Cauchy sequence being bounded does not require you to know what the sequence converges to. For that reason, I think it doesn't matter whether the metric space is complete or not.</p>


答: Hint:For fixed $y$ the sequence ${d(x_n,y)$ is  a Cauchy sequence of real numbers. Hence it is bounded.


问:
formula of pascal's triangle
I want to make a program for pascal's triangle,I was reading through the details and found something like this:

```<code>0:                                1
1:                             1     1
2:                           1    2    1
3:                         1   3     3   1
4:                       1   4    6    4   1
5:                      1  5   10   10   5   1
</code>```

by looking at row number 5 of the triangle, one can quickly read off that:

$$(x + y)^5 = 1\cdot x^5 + 5\cdot x^4y + 10\cdot x^3y^2 + 10\cdot x^2y^3 + 5\cdot x y^4 + 1\cdot y^5$$

but I haven't been able to understand what are these x and y and what shall be their values.and also how shall I proceed for further rows.
I shall be really thatnkful if someone explained this to me.

答:
The $x,y$ are just formal symbols; the can stand for anything (for instance numbers) that you can add and multiply; in order for the binomial formula you wrote to be valid, one must assume they commute: $xy=yx$ (so they cannot stand for an arbitrary pair of square matrices). For the formula itself the nature of $x$ and $y$ have no importance; thus the formula expresses a very general fact.

To compute these formulas requires just rearranging symbols a bit. Mulitplying the formula for $(x+y)^n$ by $x+y$ and gathering terms, you can see that each new coefficient is obtained as the sum of two previous ones; denoting as is conventional the (binomial) coefficient of $x^ky^{n-k}$ in $(x+y)^n$ by $\binom nk$, you get the relation
$$
  \binom{n+1}k=\binom n{k-1}+\binom nk\qquad\text{for $0<k\leq n$,}
$$
which is known as Pascal's recurrence. Using it allows computing Pascal's triangle more rapidly and safely (if you are using integers of bounded size) than using the explicit multiplicative formula for individual binomial coefficients.

Title: formula of pascal's triangle


问:
I have to show that the sum of this double series is $\frac{1}{2}$
i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right)^{i}\right)\right)$$

$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{(1)^{i}}{k+1}  - \frac{(1)^{i}}{k+2} $$

$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{k+1}  - \frac{1}{k+2} $$

this is a telescoping series. Because of that, the Series is convergence to $\lim_{n\to\infty} \frac{1}{2} - \frac{1}{n+2} = \frac{1}{2}$

Is this solution right and enough?

答:
Given that
$$\color{red}{\sum_{i=1}^{\infty}} \sum_{k=1}^{\infty} \bigg(\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}\bigg) - \bigg(\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)\bigg)$$
let's deal with the black sum first.
$$\begin{align}
&\sum_{k=1}^{\infty} \bigg(\underbrace{\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}}_{a_k}\bigg) - \bigg(\underbrace{\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)}_{a_{k+1}}\bigg)  \\
&= \lim_{n\to\infty}\sum_{k=1}^n(a_k - a_{k+1}) \\
&= \lim_{n\to\infty}(a_1 - a_{n+1}) \\
&= \lim_{n\to\infty}\left(\frac{1}{2}\left(\frac{1}{2}\right)^i - \left(\frac{1}{n+2}\left(\frac{n+1}{n+2}\right)^i\right)\right) \\
&= \left(\frac{1}{2}\right)^{i+1}
\end{align}$$
Now, we will move to the red sum:
$$\color{red}{\sum_{i=1}^\infty} \left(\frac{1}{2}\right)^{i+1} = \frac{\left(\frac{1}{2}\right)^2}{1 - \frac{1}{2}} = \frac{1}{4}\cdot 2 = \frac{1}{2}$$


问:
2-Norm of a Submatrix is $\leq$ 2-Norm of Original Matrix
Say $A$ is a submatrix of $B$. How do I prove that the $\|A\|_2 \leq \|B\|_2$?
I can easily show this for $\|\cdot\|_1, \|\cdot\|_\infty, $ and $\|\cdot\|_F$ and thought maybe the solution lies in relating the inequalities of these other norms to the 2-norm, but this path hasn't proved fruitful.</p>

答:
Without loss of generality, let $B=\pmatrix{A&X\\ Y&Z}$. Then
\begin{aligned}
\|A\|_2
&=\sup_{\|x\|_2=1}\|Ax\|_2\\
&\le\sup_{\|x\|_2=1}\left\|\pmatrix{A\\ Y}x\right\|_2\\
&=\sup_{\|x\|_2=1}\left\|B\pmatrix{x\\ 0}\right\|_2\\
&\le\sup_{\|u\|_2=1}\left\|Bu\right\|_2\\
&=\|B\|_2.
\end{aligned}
In particular, if $Y$ has full column rank, then $Yx\ne0$; hence the first inequality above is strict and $\|A\|_2<\|B\|_2$. Since the induced $2$-norm of a matrix $M$ is also equal to $\sup_{\|y\|_2=1}\|y^\ast M\|_2$, we also have $\|A\|_2<\|B\|_2$ when $X$ has full row rank.


答:
Here is my attempt.

Since all terms are positive, we are allowed to swap the two summations.
$$\sum_{i=1}^\infty \sum_{k=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty }\sum_{k=1}^N \sum_{i=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{1}{k+1} \sum_{i=1}^{\infty} (\frac{k}{k+1})^i - \sum_{k=1}^N \frac{1}{k+2} \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{1}{k+1} \sum_{i=1}^{\infty} (\frac{k}{k+1})^i - \sum_{k=1}^N \frac{1}{k+2} \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{k}{k+1}  - \sum_{k=1}^N \frac{k+1}{k+2}  \right) = $$
$$= \lim_{N \to \infty } \left( \frac{1}{2} - \frac{N+1}{N+2}  \right) = - \frac{1}{2}$$
I don't know where I did a minus sign mistake. :/

Title: I have to show that the sum of this double series is $\frac{1}{2}$

Body:
i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right)^{i}\right)\right)$$

$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{(1)^{i}}{k+1}  - \frac{(1)^{i}}{k+2} $$

$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{k+1}  - \frac{1}{k+2} $$

this is a telescoping series. Because of that, the Series is convergence to $\lim_{n\to\infty} \frac{1}{2} - \frac{1}{n+2} = \frac{1}{2}$

Is this solution right and enough?


问:
How to test whether a series converges or diverges
I've tried every convergence criterion I know. (Dalamber's, Cauchy's and Raabe's tests)

But in all cases, because of $-1^n$, I got indeterminate form.

$$\sum_{n=1}^\infty (-1)^n \sin(\sqrt{n^2-1} - 1)$$

Tell me how to solve this.

答:
Hint: does the general term tend towards zero?

P.S.
BTW Dalamber's, Cauchy's and Raabe's tests are used with non negative series.


问:
Ellipse curvature

Let $a,b>0$ and let $\alpha$ be the ellipse
$\alpha(t)=\big(x(t),y(t)\big)=\big( a\cos t, b\sin t\big)$.

Let $(u(t),0)$ be the point where the line through $\alpha(t)$
perpendicular to $\alpha'(t)$ meets the $x$-axis, and let $d(t)$ be
the distance between $\alpha(t)$ and $(u(t),0)$.
Find a formula for $\kappa(t)$ in terms of $a$, $b$ and $d(t)$.

The curvature of the ellipse has been discussed thoroughly in this question: how to calculate the curvature of an ellipse

However, I can't find a way to use $d(t)$ in expressing curvature.

I tried to sketch the graph and realized that I can inscribe a circle inside the ellipse. But that is as far as I can go.

I also tried to explicitly write out $d(t)$, but it gives me an ugly expression.

答:
Equation of tangent

$$0=T(x,y)\equiv \frac{x\cos t}{a}+\frac{y\sin  t}{b}-1$$

Equation of normal

$$0=N(x,y) \equiv \frac{ax}{\cos t}-\frac{by}{\sin t}-(a^2-b^2)$$

Note that $N(u,0)=0$ and

\begin{align}
  u(t) &= \frac{(a^2-b^2)\cos t}{a} \\
  d(t) &= \frac{|T(u,0)|}{\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}} \\
  &= \frac{\left| -\sin^2 t-\dfrac{b^2\cos^2 t}{a^2} \right|}
          {\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}} \\
  &= b^2\sqrt{\frac{\cos^2 t}{a^2}+\frac{\sin^2 t}{b^2}} \\
  \kappa &= \frac{1}{a^2 b^2
  \left(
    \dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}
  \right)^{3/2}} \\
  &= \frac{b^4}{a^2 d^3}
\end{align}

Title: Ellipse curvature


问:
Proving that if $f'$ has at most $n-1$ zeros, then $f$ has at most $n$ zeros
Is this proof correct?

The problem is the following.

Let $n$ be a natural number. Suppose that the function $f:\mathbb{R}\to\mathbb{R}$ is differentiable and that the following equation has at most $n-1$ solutions:
  $$f'(x)=0, \quad x \in \mathbb{R}.$$
  Prove that the following equation has at most $n$ solutions:
  $$f(x)=0,\quad x \in \mathbb{R}.$$

My proof is:

Let $f'(x)=0$ have solutions $x_1$, $x_2,\ldots ,x_{n-1}$.

since
$$f'(x_1)=\lim\limits_{{x}\to{x_1}}\frac{f(x)-f(x_1)}{x-x_1} = 0$$
and doing so, $f(x)=f(x_{n-1})$
Therefore $f(x)=f(x_1)=f(x_2)=\cdots=f(x_{n-1})$.

Let $x_n$ be one of solutions of $f(x)=0$.
$$0=f(x_n)=f(x_1)=f(x_2)=\cdots=f(x_{n-1}),$$
so $f(x)=0$ has solutions like $x_1$, $x_2,\ldots, x_n$.

If there is wrong part, please let me know.

答:
Use the mean value theorem.

Fix $n=3$. (The argument generalizes for arbitrary $n \in \mathbb{N}$).

Suppose $f(x)=0$ has more than three (distinct) solutions, say $x_1 < x_2 < x_3 < x_4$. Then the mean value theorem (Rolle's Theorem) implies that there exist $c, d, e \in \mathbb{R}$ such that $x_1 < c < x_2 < d < x_3 < e < x_4$ and $f'(c)=f'(d)=f'(e)=0$, contradiction.