答:
Let DB and EA meet at X. From the triangle median theorem
$$DB^2= \frac12( AB^2+ BC^2)-\frac14 AC^2, \>\>\>
EA^2= \frac12( AB^2+ AC^2)-\frac14 BC^2
$$
Combine them, along with the given $BC^2+AC^2=5AB^2$, to get
$$DB^2 +EA^2 = AB^2 +\frac14(BC^2+AC^2)= \frac94 AB^2$$

and substitute it into the cosine rule below
\begin{align}
\cos\angle DXE= \frac{DX^2+EX^2- DE^2}{2 DX\cdot EX}= \frac{\frac19 (DB^2+EA^2)- \frac14 AB^2}{2DX\cdot EX}=0
\end{align}Thus, $\angle DXE =90^\circ$ and the two medians are perpendicular.

Body:
The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$. Prove that the medians drawn from $A$ and $B$ are perpendicular.

What I Tried: Here is a picture :-

I would normally solve these kind of problems using elementary geometry, but I wasn't able to do it this time, simply because $(a,b,c)$ can take many values and there is probably some kind of construction or some clever proof which I am missing. One idea I thought of is that the medians divide the side lengths into $2$ equal parts, and the centroid itself divides the medians in $2:1$ ratio, and I probably have to show somehow, that Pythagoras Theorem on either of the $3$ triangles work. I took an example in Geogebra by drawing an accurate diagram with $a = 12$ , $b = 18.66$ nearly, and then $c = 10$ . This actually showed that the medians from $A$ and $B$ are really perpendicular, and I just drew a picture of that taken from Geogebra.

Now given $(a,b,c)$ , is there any way to prove this? I did not find any elementary methods like angle-chasing , pythagoras theorem , similarity and areas , etc. , but I did not find any clever use of them till now. Maybe some extra construction is required?

Can anyone help me? Thank You.


问:
Does the number/expression above the sigma notation mean the last n-value, or the number of terms?
The way I've understood it, is that the number or expression above the sigma function gives the number of terms. Here's an example:

$$\sum_{k=2}^4 k^2 = 2^2 + 3^2 + 4^2 + 5^2$$

However, I have now seen another way of using the sigma function:

$$\sum_{k=2}^4 k^2 = 2^2 + 3^2 + 4^2$$

The former example illustrates how I've learned to use it, the value above the sigma function denoting the number of terms. The latter example illustrates the other definition, where the value above denotes the last value. Which of these definitions are correct? And if both are correct, how does one tell the reader which one is being used?

答:
The number or expression on top of $\sum$ in a summation indicates the highest value of the index.

For example, $\sum\limits_{k=2}^4 k^2=2^2+3^2+4^2$.

If you want the number of terms to be $b$, write $\sum\limits_{k=a}^{a+b-1}k^2.$

Note that, when $a=1$ (but not otherwise!), this is simply $\sum\limits_{k=1}^b k^2.$

ContentLicense: CC BY-SA 4.0


问:
Ambiguity in orbit space notation
For an action of a group $G$ on a topological space $Y$ where each element of $G$ corresponds to a homeomorphism of $Y$, Hatcher defines the orbit space $Y/G$ of the action to be the quotient of $Y$ by the equivalence relation of being in the same orbit of the action. This confuses me. The notation $Y/G$ mentions the space and the group, but not the action. From visualising, I’m pretty sure two $Y/G$’s would not even have to be homeomorphic. Consider the subspace $X$ of $\mathbb{R}^2$ given by the union of all lines $x=z$ and $y=z$ with $z\in\mathbb{Z}$. Here are two actions of $\mathbb{Z}$ on $X$. In the first, a generator maps to the function $x+(1,0)$, in the second it maps to $x+(1,1)$. The quotient by the first action is a line  with a bunch of circles glued to it at a discrete set of points, whereas the quotient by the second is a bunch of circles indexed by the integers with the North pole of one glued to the South pole of the next. These are homotopy equivalent but not homeomorphic. So why the notation?

答:
Just to augment the answer of @PaulFrost a bit, there is a alternate method for denoting orbit spaces that sidesteps this abuse of notation.

As you probably know, an action of $G$ on a topological space $Y$ is the same as a homomorphism
$$\alpha : G \to \text{Homeo}(Y)
$$
where $\text{Homeo}(Y)$ denotes the group of self-homeomorphisms of $Y$ under the operation of function composition. If we let $\Gamma = \text{image}(\alpha) < \text{Homeo}(Y)$ then $\alpha$ becomes an isomorphism between $G$ and $\Gamma$, and we then write $Y/\Gamma$ for the quotient. This removes all ambiguity of the quotient space itself. It nonetheless loses a bit in that the original group $G$ has been lost, as has its homomorphism to $\Gamma$.

This alternate notation is particularly common in geometric situations. For example, if $X$ is some kind of geometric space such as the Euclidean plane or the hyperbolic plane, one is interested in isometric actions on $X$, which means actions that preserve the geometric structure (i.e. the Riemannian metric). The subgroup of $\text{Homeo}(X)$ consisting of isometries is denoted $\text{Isom}(X)$. It is then very common to denote (and study) quotient spaces $X / \Gamma$ of subgroups $\Gamma < \text{Isom}(X)$.

Title: Ambiguity in orbit space notation


问:
Min value of $a\sec \theta +b\cos \theta$
Min value of$$a\sec \theta +b\cos \theta$$ for $$\theta \in (-90^\circ,90^\circ) $$

I have trouble in approaching this problem .Could only a method/hint be given ? I would to try it myself.

答:
The exact solution as a function of $a$ and $b$ is very difficult to calculate, even with numerical values. Indeed, the derivative equal to zero gives the equation $$b\cos^3(x)=a\sin(x)$$
which leads to a polynomial equation of sixth degree in $\cos(x)$ quite difficult to solve.

The real solutions of this equation given by Wolfram calculator is showed in the attached figure.


问:
The difference between $\mathbb{Z}$ and $\mathbb{Z}^2$
I know that $\mathbb{Z}$ is the set of integers. But, what does $\mathbb{Z}^2$ mean? How is it different from $\mathbb{Z}$?

Thanks.

答:
Instead of $\mathbb{Z}$ and $\mathbb{Z}^2$ I will draw for you $\mathbb{N},\mathbb{N}^2$.
The set $\mathbb{N}$ may be drawn as
$$\begin{array}{ccc}
*&*&*&*&*&*&*  &* &*&\cdots\end{array}$$, where the left most $*$ is the number zero. Now if you look at $\mathbb{N}^2$, you are looking at ordered pairs of numbers. We may visualize this as the following

$$\begin{array}{ccc}
.&.&.&.&.&.&.  &. &.\\
.&.&.&.&.&.&.  &. &.\\
.&.&.&.&.&.&.  &. &.\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\
*&*&*&*&*&*&*  &* &*&\cdots\\   \end{array}$$

Where the lower left hand $*$ is the point $(0,0)$ and the point to the right of $(0,0)$ is $(1,0)$, and the point above the origin is $(0,1)$. Note that this is similar to the way we picture $\mathbb{R}$ as a line and $\mathbb{R}^2$ as  the $xy$-plane. What we did here was to look at only those points whose co-ordinates are positive integers. The picture for $\mathbb{Z}$ and $\mathbb{Z}^2$ is similar except that it is infinite in all four directions. As an exercise, how would one visulaize $[0,1]^2$ where $[0,1]$ is the set of reals between zero and one inclusive.

ContentLicense: CC BY-SA 3.0


问:
Identify the equivalence relation on a set
Let $A$ be the set of all points in the $x - y$ plane. Define a
relation $\sim$ on A as follows:

For $a, b\in A$, $a\sim b$ if and only if $a$ and $b$ have the same distance from the origin. Is $\sim$ an equivalence relation on $A$?

答:
Yes, this is an equivalence relation, because it satisfies

reflexivity: a point $a \in A$ has the same distance from the origin as itself;

symmetry: if $a$ and $b$ have the same distance from the origin, so do $b$ and $a$.

transitivity: if $a$ and $b$ have the same distance from the origin and $b$ and $c$ have the same distance from the origin, then $a$ and $c$ have the same distance from the origin.


问:
Projective norm for Banach spaces
The projective norm of tensors from (algebraic) tensor product of Banach spaces $X,Y$ is defined as
$$\|t\|_\wedge = \inf\left\{ \sum\limits_{j=1}^N \|x_j\|\|y_j\| \, : \, t=\sum\limits_{j=1}^N x_j\otimes y_j\right\}.$$

In one place I found slightly different definition:
$$\|t\|_\wedge '=\inf\left\{\left(\sum\limits_{j=1}^N \|x_j\|^2\right)^{1/2}\left(\sum\limits_{j=1}^N \|y_j\|^2\right)^{1/2} \, : \, t=\sum\limits_{j=1}^N x_j\otimes y_j\right\}.$$

Are they indeed equal? It is easy to see one inequality, but why we have the opposite one?

答:
The two norms are indeed equal. On the one hand, by the Cauchy-Schwarz inequality we have
$$\sum_{j = 1}^{N} \lVert x_j\rVert \lVert y_j\rVert \leqslant \Biggl(\sum_{j = 1}^{N} \lVert x_j\rVert^2\Biggr)^{1/2} \Biggl(\sum_{j = 1}^{N} \lVert y_j\rVert^2\Biggr)^{1/2}$$
which yields $\lVert t\rVert_{\wedge} \leqslant \lVert t\rVert'_{\wedge}$. I suppose that's the inequality you saw.

On the other hand, consider when we have equality in the Cauchy-Schwarz inequality. A particular case where we have equality is when $\lVert x_j\rVert = \lVert y_j\rVert$ for $1 \leqslant j \leqslant N$.

We can always achieve that by moving a scalar factor from $x_j$ to $y_j$, unless one of $x_j$ and $y_j$ is zero. But excluding all representations
$$t = \sum_{j = 1}^{N} x_j \otimes y_j \tag{1}$$
where at least one of the $x_j$ or $y_j$ is zero doesn't change either norm. For the first, including or excluding such terms doesn't change $\sum \lVert x_j\rVert \lVert y_j\rVert$ at all, for the second, the value for the sum without zero factors is smaller (or equal, if $x_j$ and $y_j$ are both zero for such terms), so the terms we exclude can't make the infimum, i.e. $\lVert t\rVert'_{\wedge}$ smaller.

Thus for a representation $(1)$ where no $x_j$ or $y_j$ vanishes, define
$$c_j = \sqrt{\frac{\lVert x_j\rVert}{\lVert y_j\rVert}}$$
and $x_j' = c_j^{-1}\cdot x_j$, $y_j' = c_j \cdot y_j$. Then we have
$$t = \sum_{j = 1}^{N} x_j' \otimes y_j'$$
and
$$\sum_{j = 1}^{N} \lVert x_j\rVert \lVert y_j\rVert = \sum_{j = 1}^{N} \lVert x_j'\rVert \lVert y_j'\rVert = \Biggl(\sum_{j = 1}^{N} \lVert x_j'\rVert^2\Biggr)^{1/2} \Biggl(\sum_{j = 1}^{N} \lVert y_j'\rVert^2\Biggr)^{1/2}$$
which shows
$$\lVert t\rVert'_{\wedge} \leqslant \lVert t\rVert_{\wedge}\,.$$

Title: Projective norm for Banach spaces


问:
Find the length of base of a triangle without using Pythagorean Theorem
I'm curious whether it is possible to find the length of base of the triangle without using Pythagorean Theorem

No Pythagorean Theorem mean:

=> No trigonometric because trigonometric is built on top of Pythagorean Theorem.
etc $\sin{\theta} = \frac{a}{r}$

=> No Integration on line or curve because the integration is built on top of Pythagorean Theorem.
etc: $s(x) = \int \sqrt{f'(x)^2 + 1}$

答:
Since $AB^2=a^2+a_1^2$, any derivation of the length of
$AB$, if it does not use the Pythagorean theorem must essentially be a proof of the Pythagorean theorem.

There are so many proofs of the theorem that there seems to be an infinite number since no matter how many you find there is always one more.


问:
Evaluating $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$
How to calculate that integral? I have no idea what substitution to use. I think polar coordinates will be needed here.

$$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$$

Body:
You can even do it with $(x,y)$. Complete the square first

$$\int{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}} \text{erf}\left(\frac{4 x-y}{2 \sqrt{2}}\right)}{4
   \sqrt{2 \pi }}$$
$$\int_{-\infty}^\infty{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}$$
$$\int \frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}\,dy=\frac{\text{erf}\left(\frac{1}{2} \sqrt{\frac{11}{2}} y\right)}{2 \sqrt{11}}$$
$$\int_{-\infty}^\infty \frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}\,dy=\frac{1}{\sqrt{11}}$$

In fact, for the most general case
$$I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac 1{2\pi} e^{-(ax^2+bxy+by^2)}\,dx\,dy=\frac{1}{\sqrt{4 a c-b^2}}$$ if
$$\Re(a)>0\land \Re\left(\frac{b^2}{a}\right)<4 \Re(c)$$


答:
Assume $a,b>0$, then
$$a\sec \theta +b\cos \theta=(\sqrt{a\sec\theta } - \sqrt{b\cos\theta })^2 +2\sqrt{ab }\ge 2 \sqrt{ab }
$$
Others cases of $a,b$ can be examined similarly.

Title: Min value of $a\sec \theta +b\cos \theta$

Body:
Min value of$$a\sec \theta +b\cos \theta$$ for $$\theta \in (-90^\circ,90^\circ) $$

I have trouble in approaching this problem .Could only a method/hint be given ? I would to try it myself.


问:
Computing a double integral over $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$

Compute the integral $\int_{D} f \ dx \ dy$, when $f : D \to \mathbf{R}, f(x,y)= y^2x+x^3$ and $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$

It seems that $D$ defines a semicircle with radius $5$ in the $xy$-plane. So converting to polar coordinates I have $x=r\cos\theta, y=r\sin\theta$ and the limits for integration would be $r \in [0,5]$ and $\theta \in [0, \pi]$, thus the integral would become $$\int_{0}^{\pi} \int_{0}^{5} r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta \ dr \ d\theta.$$

However I doubt that this is correct since seems to evaluate to zero?

答:
Note that
$$
\begin{split}
 I
 &= \int_0^\pi \int_0^5
      \left(r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta\right) r\ dr \ d\theta \\
 &= \int_0^5 r^4 dr
    \int_0^\pi \left(\sin^2\theta +\cos^2 \theta\right) \cos\theta  d\theta \\
 &= \frac{5^5}{5} \int_0^\pi \cos\theta  d\theta
\end{split}
$$
which does evaluate to zero since cosine is symmetric around $\pi/2$...

A different way to see it without going polar is the boundary curve (since $y>0$) is given by $y < \sqrt{25-x^2}$ and $f(x,y) = x(x^2+y^2)$, which is odd in $x$, and since we will integrate an odd function over a symmetric interval, the integral will be zero.


问:
Adjoint question in the Category

If $\langle F,G,φ \rangle : X \to A$ is adjunction with $G$ full and every unit $η_x$ a monic, then every $η_x$ is also epi.

Some similar questions like this, maybe I do not catch the key, who can help me?

答:
The general strategy is to postcompose and precompose all possible arrows with units and counits of the adjunction, until the universal property of the adjunction can be applied sufficently many times.

Let $f, g \colon GF(A) \rightarrow B$ be two parallel morphisms and assume that $f \circ \eta_A = g \circ \eta_A = r$. We have to show that $f = g$.

Postcompose $r$ with $\eta_B$ to obtain a morphism $A \rightarrow GF(B)$. By universality of $\eta_A$, we know that there exists exactly one morphism $h \colon F(A) \rightarrow F(B)$ such that $G(h) \circ \eta_A = \eta_B \circ r$. But $G$ is full, so every morphism $GF(A) \rightarrow GF(B)$ is of the form of $G(k)$ and thus there exists exactly one morphism $GF(A) \rightarrow GF(B)$ such that the above condition is satisfied. Obviously, both $\eta_B \circ f$ and $\eta_B \circ g$ make the triangle commute. So,  $\eta_B \circ f = \eta_B \circ g$, what gives $f = g$ beacuse $\eta_B$ is mono.

Title: Adjoint question in the Category


问:
Some Results in $\mathbb{Z} [\sqrt{10}]$
This is a question from an old Oxford undergrad paper on calculations in $\mathbb{Z} [\sqrt{10}]$. We equip this ring with the Eucliden function $d(a+b\sqrt{10})=|a^2-10b^2|$. I want to prove the following results:

If $d(x)=1$, then $\frac{1}{x} \in \mathbb{Z} [\sqrt{10}]$

Any non-zero element of $\mathbb{Z} [\sqrt{10}]$ which is not a unit can be expressed as a product of finitely many irreducibles in $\mathbb{Z} [\sqrt{10}]$

The ideal generated by $2$ and $\sqrt{10}$ is not principal in $\mathbb{Z} [\sqrt{10}]$

Thought so far

Suppose $x=a+b\sqrt{10}$. Clearly if $x$ is a unit then $d(x)=1$, though I'm not sure if this helps. Are we OK simply to note that $\frac{1}{x}=\frac{a-b\sqrt{10}}{a^2-10b^2}$ and since $d(x)=1$ then the deonminator is either $1$ or $-1$.

I know this is true in general in a principal ideal domain and every Euclidean ring is a principal ideal domain, but this proof is lengthy. Is there any calculation one can perform in $\mathbb{Z} [\sqrt{10}]$ to demonstrate this property more quickly.

Any help would be appreciated; I'm not actually too sure what this ideal looks set. Could someone put it in a set notation for me?

Many thanks.

答:
It seems to me that (1) and (3) are dealt with adequately in the comments (but I would be happy to incorporate that here if you need). For (2), the key point is that the ring is Noetherian, which is effectively a consequence of the Hilbert basis theorem, implying that a polynomial ring $\mathbb{Z}[x]$ is Noetherian, and the fact that a quotient ring of a Noetherian ring is Noetherian. Given this, every element can be written as a product of irreducibles, since otherwise you obtain an infinite ascending chain of ideals.

Here are some more details for (3): First, observe that $d(xy)=d(x)d(y)$ for all $x,y$ and that $d(2)=4$ and $d(\sqrt{10})=10$. Thus any common divisor $x$ has $d(x)|2$. If $x$ is a non-unit this implies $d(x)=2$. Thus assuming a non-unit common divisor $x$ exists, there are integers $a,b$ with
$$\pm 2=a^2-10b^2.$$ Consider this equation modulo $10$. It implies that either $2$ or $8$ is a square mod $10$, but the squares modulo $10$ are just $0,1,4,9,6,5$. Hence there is no non-unit common divisor $x$. On the other hand, the ideal is proper since its elements are all of the form $a+b\sqrt{10}$ with $a$ even. Therefore it is not principal.


问:
Is total variation additive?
Definition. The total variation of $F$ on $[a,x]$ (where $a\leq x\leq b$) is defined by $$T_F(a,x) = \sup\sum_{j=1}^N|F(t_j)-F(t_{j-1})|$$
where the sup is over all partitions of $[a,x]$.

here, we assume $F$ is real-valued function.

Under this definition, if $a\leq c\leq b$, then $T_F(a,c)+T_F(c,b) = T_F(a,b)$?

答: Hint for the part you're missing: Take any partition $(t_0,t_1,\ldots,t_N)$ of $[a,b]$ and consider the sum $$\sum_{j=1}^N|F(t_j)-F(t_{j-1})|$$ for that particular partition. Make a new partition by inserting $c$ at the suitable place, and see what happens to the sum.

Title: Is total variation additive?


问:
Standard notation for the the collection of all *minimal* elements of the set of all upper bounds?
Let $(P,\leq)$ denote a poset and suppose $X \subseteq P$. Then the minimum element of the set of all upper bounds of $X$ can be denoted $\operatorname{sup} X$, or $\bigvee X$. Is there a similar notation for the collection of all minimal elements of the set of all upper bounds?

答:
There is none, as far as I know.

However, the set of all upper bound of a subset $X$ is sometimes denoted by
$X^\uparrow$ and the set of all minimal elements of a subset $Y$
is sometimes denoted by $\min(Y)$; this gives us the notation $\min(X^\uparrow)$.

I think this is a usable notation, because if one wants to prove something about the set in question, the proof will probably use the set $X^\uparrow$. So it probably did not make sense to invent an "irreducible" notation.


问:
$x$ positive implies $\text{Tr}(x-)$ positive.
Let $H$ be a complex Hilbert space and $x \in T(H)$ be a positive trace class operator. I want to show that
$$\text{Tr}(x-): B(H) \to \mathbb{C}: y \mapsto \text{Tr}(xy)$$
is a positive functional.

Attempt:

Let $y \geq 0$ a positive operator in $B(H)$. I need to show that
$$\text{Tr}(xy) \geq 0$$

Of course, we can write $x = a^* a, y =b^ *b$ for some operators $a,b$. And then we need to show that

$$\text{Tr}(a^* a b^*b) \geq 0$$

It seems natural to try to use the cyclicity of the trace:
$$\text{Tr}(a^*ab^*b) = \text{Tr}(ba^* ab^*) = \text{Tr}((ab^*)^*(ab^*))$$

and then we can conclude since the trace of a positive operator is positive.

However, I only know that $\text{Tr}(uv) = \text{Tr}(vu)$ when either $u$ or $v$ is trace class, so what I need for the above to work is that $a$ or $b$ is trace class?

答:
Since $x$ is trace-class, so is $xy$. And $xy^{1/2}$. As you know that you can commute inside the trace if one of the two operators is trace-class,
$$
\operatorname{Tr}(xy)=\operatorname{Tr}(xy^{1/2}y^{1/2})=\operatorname{Tr}(y^{1/2}xy^{1/2})\geq0,
$$
as this last operator is trace-class and positive.


问:
Maximal unramified subextension at one prime
Let $L/K$ be a finite Galois extension of algebraic number fields with Galois group $G$, $\mathfrak{p}$ is a prime of $K$ and $\mathfrak{q}_1,...,\mathfrak{q}_m$ be the primes of $L$ lying over $K$. Then we have the interia subgroups $I_{\mathfrak{q}_i/\mathfrak{p}}$ of $G$, then we can define the subgroup $H$ of $G$ generated by these $I_{\mathfrak{q}_i/q}$.

Then I guess that $L^H$ is the maximal unramified subextension of $L$ at $\mathfrak{p}$, where maximality means that if $M$ is any subextension which is unramified at $\mathfrak{p}$, then $M$ is contained in $L^H$.

Is my guess right? If not, I hope someone could explain how we can construct the maximal unramified subextension at one prime. Thank you.

答:
Yes,

Let $F$ an intermediate field then the exponent $e((q_j\cap O_F)/p)$ of $q_j\cap O_F$ in $pO_F$ is $1$ iff $$|I(q_j/p)|=e(q_j/p)=e(q_j/(q_j\cap O_F))=|I(q_j/(q_j\cap O_F))|$$ iff $I(q_j/p)=I(q_j/(q_j\cap O_F))$ iff $F$ is fixed by $I(q_j/p)$.

Thus $p$ is unramified in $O_F$ iff all the $e((q_j\cap O_F)/p)=1$ iff $F$ is fixed by all the $I(q_j/p)$.


问:
If $rank(S) \leq nullity(T)$ then is $rank(S^*) \geq nullity(T^*)$?
Here $T:V \rightarrow W$ and $S:W \rightarrow V$, where V and W are finite dimensional inner-product spaces over the same field. It's also given that $TS=0$.

The question is to prove that $rank(S^*) \geq nullity(T^*)$.

My attempt: Using the fact that $TS=0$, I derived that $rank(S) \leq nullity(T)$. Does the inequality asked automatically follow from this? If not, please give alternate ways to prove the same.

答:
It is not true that $\operatorname{Rank}(S) \leq \operatorname{Nullity}(T)$ implies that $\operatorname{Rank}(S^*) \geq \operatorname{Nullity}(T^*)$. For a counterexample, let $V = W$ be vector spaces of positive dimension and let $S = T = 0$. Then $\operatorname{Rank}(S) = 0$ and $\operatorname{Nullity}(T) = \operatorname{dim}(V) > 0$, but $S^* = T^* = 0$, so also $\operatorname{Rank}(S^*) = 0 < \operatorname{dim}(V) = \operatorname{Nullity}(T^*)$.

The same counterexample actually shows that it cannot follow from $TS = 0$ that $\operatorname{Rank}(S^*) \geq \operatorname{Nullity}(T^*)$, so it appears that the question you are trying to prove is incorrect.

Something that you can prove though is that $\operatorname{Nullity}(S^*) \geq \operatorname{Rank}(T^*)$. This follows from what you did in your attempt by observing that $S^*T^* = (TS)^* = 0$.


问:
Finding the area not by integral

Original problem. Find the area under $y= \sqrt{x}$ in the range $\left [ 0, 1 \right ]$

My friend, she wants to use total area instead of calculating the integeral. So I tried something:

Dividing the area by $n$ parts with each part is $\Delta x= \frac{1}{n}$.

The part $k$ is $\left [ \left ( k- 1 \right )\frac{1}{n}, \frac{k}{n} \right ]$, we calculate the total area by formula:

$$S_{n}= \sum_{k= 1}^{n}\frac{1}{n}\sqrt{\frac{k}{n}}= \frac{1}{n^{3/2}}\sum_{k= 1}^{n}k^{1/2}$$

How should I do next ?? Thank you....

答:
You have to select the partition points more carefully. They need not be equidistant, just near to each other. Choose an $N\gg1$ and put
$$x_k:={k^2\over N^2}\qquad(0\leq k\leq N)\ .$$
You then have
$$\sqrt{x_k}={k\over N},\qquad x_k-x_{k-1}={2k-1\over N^2}<{2\over N}\ .$$
This leads to the Riemann sums
$$R_N:=\sum_{k=1}^N f(x_k)\,(x_k-x_{k-1})=\sum_{k=1}^N{k\over N}\,{2k-1\over N^2}=\ldots\quad.$$
It is easy to compute the last sum explicitly, and you will then see that $$\lim_{N\to\infty}R_N={2\over3}=\int_0^1\sqrt{x}\>dx\ .$$

Title: Finding the area not by integral


问:
The path of complex structure.
$J = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)$. If $I(t)$ is a path in $\rm M(2n, \mathbb R)$ ($2n \times 2n $ real matrix) such that $I(0)=0$ and $I(t)J+JI(t)=0$, then for sufficiently small $t$ can we find a path $J(t) \in \rm M(2n, \mathbb R)$ such that (1) $J(0)=J$, (2) $J^2(t)+I^2(t)=-I_{2n}$, (3) $I(t)J(t)+J(t)I(t)=0$?</p>

答:
OK last go I promise...

$[A,B]=AB-BA$,

$\{ A,B \} =AB+BA$.

Define the following matrix under the assumption the power series converges :

$\sqrt{\mathbb{I}_{2n}+I(t)^2}=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k(2k)!}{(1-2k)(k!)^2(4^k)}I(t)^{2k}\in M(2n,\mathbb{R})$,

...I think it converges for "sufficiently small t", because if $I(t)\rightarrow0$ as $t\rightarrow0$ then the elements can be made arbitrarily small at some $t$; the equivalent 1d series converges for $I(t)<1$.

Note the following properties:

$\left[I(t),\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$,

because:

$\left[I(t),I(t)^{2n}\right]=0$,

and:

$\left[J,\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$,

because:

$\{J,I(t)\}=0 \Rightarrow J.I(t)^n=(-1)^nI(t)^n.J \Rightarrow [J,I(t)^{2n}]=0$.

With that said, let us define the following path:

$J(t)=J.\sqrt{\mathbb{I}_{2n}+I(t)^2}\in \rm M(2n, \mathbb{ R})$,

$I(0)=0,J(0)=J: J,I(t)\in \rm M(2n, \mathbb R)$.

Then note:

$\{J(t),I(t)\}=J.\sqrt{\mathbb{I}_{2n}+I(t)^2}.I(t)+I(t).J.\sqrt{\mathbb{I}_{2n}+I(t)^2}=\{J,I(t)\}.\sqrt{\mathbb{I}_{2n}+I(t)^2}=0$,

because:

$\left[I(t),\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$ ...and... $\{J,I(t)\}=0$,

and:

$J(t)^2+I(t)^2=J.\sqrt{\mathbb{I}_{2n}+I(t)^2}.J.\sqrt{\mathbb{I}_{2n}+I(t)^2}+I(t)^2=J^2.\left(\mathbb{I}_{2n}+I(t)^2\right)+I(t)^2=-\mathbb{I}_{2n}$,

because:

$\left[J,\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$ ...and... $J^2 = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right).\left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)=-\mathbb{I}_{2n}$.

NOTE:
If we were forbidden to use the power series to prove the commutation relations, perhaps diagonalization may work. For instance if $I(t)$ is diaganolized by $S$ to leave a diagonal matrix $D$, with $D_{i,i}=e_i$ the eigenvalues of $S$, then we could define:

$I(t)=SDS^{-1},I(t)^2=SD^2S^{-1},\sqrt{\mathbb{I}_{2n}+I(t)^2}=S\sqrt{\mathbb{I}_{2n}+D^2}S^{-1}$,

this would certainly prove:

$\left[I(t),\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$,

but it’s not yet obvious how it might prove:

$\left[J,\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$.

Title: The path of complex structure.


问:
One-dimensional Integral
I have seen an integral \begin{equation}\frac{1}{(2 \pi)^{\frac{n_z}{2}}} \int_{\varepsilon_{v}} e^{-\frac{1}{2}|| \boldsymbol{v} \|^{2}} \,\mathrm{d}v\,,\end{equation} where $\mathcal{E}_{v}:=\left\{v:\|v\|^{2} \leq r^{2}\right\}.$

It results in the integration $$I=\frac{1}{2^{n_{z} / 2} \Gamma\left(n_{z} / 2\right)} \int_{0}^{r^{2}} \chi^{\frac{n_{z}}{2}-1} e^{-\frac{\chi}{2}} d \chi$$

I am wondering why the result of the first integral is not the same as $\int e^{-\frac{x^2}{2}}$, which leads to $\sqrt{2}\Gamma{(\frac{1}{2})}$. And how should I evaluate the integral over $0$ to $r^2$ in the second equation.

Any hints are appreciated. Thanks.

答:
To solve your integral in terms of $\chi$, change variables to $y = \chi / 2$. This removes the factor of $1/2$ in the exponential and then you'll be able to see that it's an incomplete gamma function integral.

The reason that the integral is different from
$\int_{-\infty}^\infty e^{-\frac{x^2}{2}}dx$  is that your original integral is over an n-dimensional space. Your integral is spherically symmetric, so to get to your final form you need to first change coordinates to spherical polars. The factor of $\chi^{\frac{n_{z}}{2}-1}$ comes from the volume element in spherical polars, and the $\Gamma\left(n_{z} / 2\right)$ in the answer comes from the integration over the angular coordinates, it's related to the volume of the n-ball.

Title: One-dimensional Integral


问:
Expected number of coin flips until number of tails equals number of heads
Here is a simple probability question: An unfair coin with probability of heads equals $p$ is flipped once and lands heads. Compute the expected number of flips until the number of heads equals the number of tails.

The expected number of coin flips to get a tail is 1/(1-p) since it follows a geometric distribution. How to use this to answer the question? Does it have anything to do with catalan numbers?

Thanks.

答:
Using linearity of expectation,

$1 + np = n(1-p) \implies n = \frac{1}{1-2p}$

Where $n$ is additional flips after the first.

And given you have the first landing as a head, we need $p \lt \frac{1}{2}$.

Another way to look at it -

I am starting with a difference of $1$ ($1$ head more than tail). So if it takes me $n$ flips to bridge that gap,

$n = 1 + p \times (n + n) \implies n = \frac{1}{1-2p}$ (I either bridge that gap with probability $(1-p)$ or I increase the gap by $1$ with probability $p$).


问:
Two ways of evaluating $\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}}$, giving discrepancy of a factor of $2$.
This is related to my previous question here. Here I am evaluating (in the sense of distributions) for a vector $\mathbf{r} = (x,y,z)$
$$
\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \delta(x) \delta(y) \delta(z) \ .
$$
At the same time, you can evaluate the above integral as
$$
\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \frac{1}{2 \pi^2 |\mathbf{r}|} \int_0^\infty dP \; P \sin(|\mathbf{r}| P) = - \frac{\delta'(|\mathbf{r}|)}{2\pi|\mathbf{r}|}
$$
where the last line uses the representation $\int_0^\infty \frac{dP}{\pi} P \sin( \alpha P) = - \delta'(\alpha)$ (which follows from $\int_0^\infty \frac{dP}{\pi} \cos( \alpha P) = \delta(\alpha)$). There is additionally the identity $\alpha \delta'(\alpha) = - \delta(\alpha)$ would imply that
$$
\cdots = \frac{\delta(|\mathbf{r}|)}{2\pi|\mathbf{r}|^2} = 2 \delta(x)\delta(y)\delta(z)
$$
where in the last line I have used the identity $\delta(x) \delta(y) \delta(z) = \frac{\delta(|\mathbf{r}|)}{4\pi|\mathbf{r}|^2}$ which is equation (3.1.30) from Kanwal's book on distributions.

The above seems to say that $\delta(x)\delta(y) \delta(z) = 2 \delta(x)\delta(y) \delta(z)$, which is obviously wrong. Where am I making a mistake in my second way of computing?

EDIT: Note that the $3D$ Fourier transform of a rotationally invariant function $f(|\mathbf{p}|)$ is also rotationally invariant $F(|\mathbf{x}|)$,
$$
F(|\mathbf{x}|) = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} f(|\mathbf{p}|) e^{i \mathbf{r} \cdot \mathbf{p}} =  \int \frac{d^3 \mathbf{p}}{(2\pi)^3} f(|\mathbf{p}|) e^{i |\mathbf{r}| |\mathbf{p}| \cos\theta }
$$
where I have taken $z \to |\mathbf{r}|$ (free to do so by rotationally symmetry) and used spherical coordinates. From here the $\phi$ integral evaluates to $2\pi$, then doing the $\theta$ integral gives the $\propto \sin(|\mathbf{r}|P)/|\mathbf{r}|$ function in my formula above. I then use $f(|\mathbf{p}|) =1$ in the above.

答:
Your computation of the integral using spherical coordinates $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}}=\frac1{(2\pi)^2}\int_{\Bbb R^+}\int_{[0,\pi]} r^2\sin\theta\cdot e^{i|{\bf r}|P\cos\theta}\,d\theta\,dP=\frac{\delta(|{\bf r}|)}{2\pi|{\bf r}|^2}$$ is correct as are the identities you have used up to this point. However, it is important to note the lack of consensus in the proportionality constant of $|{\bf r}|^{-2}\delta(|{\bf r}|)$. For example,

Kanwal (2004) defines $\delta(x) \delta(y) \delta(z) = \dfrac{\delta(|\mathbf{r}|)}{4\pi|\mathbf{r}|^2}$ as you have stated, whereas

Bracewell (1999) defines $^3\delta(x,y,z)=\dfrac{\delta(\rho)}{2\pi\rho^2}$ where $\rho=|{\bf r}|$ (see equation $(51)$ of this link).

This primarily stems from the ill-defined nature of $x^{-\kappa}\delta(x)$ when $\kappa\ge1$ as discussed in Delta function at the origin in polar coordinates and Delta function of the Euclidean norm $\delta(|\mathbf x|)$ / in polar coordinates at origin $\delta(r)$ for polar coordinates representations.


问:
How can you show that $I=J=\phi^{-2}$
How can you show that $I=J=\phi^{-2}$

Where $\phi=\frac{\sqrt{5}+1}{2}$
$$I=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{4x^4+5x^2+1}\mathrm dx$$

$$J=\int_{-\infty}^{+\infty}\frac{\sin^2(4x\ln\phi)}{4x^4+5x^2+1}\mathrm dx$$

I don't think there are such formulas to represent $\sin^2(a)+\sin^2(b)=\sin^2(c)$

$$I=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{4x^4+5x^2+1}\mathrm dx=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{(4x^2+1)(x^2+1)}\mathrm dx$$

$$I=\int_{-\infty}^{+\infty}\frac{4\sin^2(x\ln\phi)+4\sin^2(3x\ln\phi)}{3(4x^2+1)}\mathrm dx-\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(x\ln\phi)}{3(x^2+1)}\mathrm dx$$

In general we got evalaute,
$$K=\int_{-\infty}^{+\infty}\frac{\sin^2(ax\ln\phi)}{bx^2+c}\mathrm dx$$

$\sin^2(a)=\frac{1-\cos(2a)}{2}$

$$K=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1-\cos(2ax\ln\phi)}{bx^2+c}\mathrm dx=\frac{\pi}{2b}\sqrt{\frac{b}{c}}-\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\cos(2ax\ln\phi)}{bx^2+c}\mathrm dx$$

答:
The key integral given at the start of @xpaul's answer can be obtained using algebra
$$\int\frac{\cos (x)}{x^2+a^2}\,dx=\int\frac{\cos (x)}{(x+ia)(x-ia)}\,dx$$
$$I=\int \frac{i\,\cos(x)}{2 a (x+i a)}\,dx-\int\frac{i\,\cos(x)}{2 a (x-i a)}\, dx$$
$$I=\frac{i \cosh (a) \text{Ci}(i a+x)-\sinh (a) \text{Si}(i a+x)}{2 a}+$$
$$\frac{\sinh (a) \text{Si}(i a-x)-i \cosh (a) \text{Ci}(x-i a)}{2 a}$$

Assuming $a>0$
$$\int_{-\infty}^\infty \frac{\cos x}{x^2+a^2}dx=\frac{\pi}{a} e^{-a}$$

Title: How can you show that $I=J=\phi^{-2}$


问:
Strong characterization of $\mathbb C$ with respect to $\mathbb R$

$\mathbb R^2$ is not a field, but $2$-tuple arithmetic rules like $(a,b)(c,d)=(ac-bd,ad+bc)$ coupled with $\mathbb R^2$ make it a field, but are there other rules than $(a,b)(c,d)=(ac-db,ad+bc)$ combined with $\mathbb R^2$ that make it a field?

Is $\mathbb C$ the only field that contains $\mathbb R$ as a subfield or are there other fields that contain $\mathbb R$ as subfields?

Is $\mathbb C$ the only that can be made by coupling $\mathbb R^2$ and rules like $(a,b)(c,d)$ $=$ $(ac-db,ad+bc)$? Are there other rules that can be coupled with $\mathbb R^2$ to make it a field?

PS: Just trying to pin down some properties that make $\mathbb C$ special/unique from other fields that can/may be constructed using some specific algebraic structure of 2 tuples in $\mathbb R$ e.g. $(\sqrt 2,\pi)$. At this point this is just a wide net as I lack the terminology to make this question specific. But any hints on how to make this question specific to listing properties that make $\mathbb C$ unique or gives strong characterisation of relationship between  $\mathbb C$ and $\mathbb R$ are welcome.

答:
Just because I don't think the very first question is explicitly addressed in other answers:

Are there other rules than $(a,b)\times(c,d)=(ac−db,ad+bc)$ that make (the $\Bbb R$-vector space) $\Bbb R^2$ a field?

Yes there are, but the field defined is always isomorphic to $\Bbb C$ (by transfer through some linear isomorphism of $\Bbb R^2$, usually different from the identity). To limit trivial variations a bit we may insist that the element $1$ of the field will be $(1,0)$, so that in any case $(a,0)\times(c,0)=(ac,0)$. If you call $x$ the element $(0,1)$, then $(a,b)=a+bx$, so the multiplication is completely determined by knowing $x^2=(s,t)$, namely $(a,b)\times(c,d)=(ac+sbd,ad+bc+tbd)$. Not all choices will result in a field though. The element $x$ will satisfy the equation $x^2=s+tx$, so if we put $P=X^2-tX-s$ then the structure will be that of the quotient ring $\Bbb R[X]/(P)$. This will be a field if and only if $P$ has no real roots, which happens whenever it has negative discriminant: $D=t^2+4s<0$. In this case the linear map to the complex numbers, i.e., to $\Bbb R^2$ with the usual product, which sends $(1,0)$ to $(1,0)$, and $x=(0,1)$ to the complex root $(\frac t2,\frac12\sqrt{-D})$ of the polynomial $P$, is easily shown to be an isomorphism of fields.


问:
Problem about extending $S\subseteq V$ to a Jordan basis of a nilpotent operator $T:V\to V$

Let $V$ be a finite-dimensional vector space over $\mathbb{F}$, on which $T\in\mathcal{L}(V)$ is a nilpotent linear operator. Assume that $k$ is the smallest integer such that $T^k=0$. Show that any linearly independent $S\subseteq V$ such that $L(S)\cap\ker T^{k-1} =\{0\}$ can be extended to a Jordan basis for $T$.

I am trying to prove this using induction, but I would appreciate any other ideas (possibly simpler or more direct). I chose to do induction on $\dim V = n$.

If $n=1$, and $T$ is nilpotent, then $T$ must be the zero operator. Any linearly independent non-empty subset $S$ of $V$ contains only one vector, say $0\neq v\in V$ since $\dim V = 1$. Here, $L(S) = V$ and $\ker T^{k-1} = \ker I = \{0\}$. $L(S)\cap\ker T^{k-1} =\{0\}$ holds, and $S$ itself is a (Jordan) basis for $T$. If $S=\emptyset$, then $L(S)=\{0\}$ and $L(S)\cap\ker T^{k-1} =\{0\}$ holds. We can add any $0\neq v\in V$ to $S$ to construct a basis of the one-dimensional space $V$. The base case ends here.

Now, assume that the statement holds for all $\dim V \in \{1,2,...,n-1\}$. We must show that it holds for $\dim V=n$.

Consider a linearly independent subset $S\subseteq V$ such that $\dim L(S) = r$ where $0\leq r\leq n$, and $L(S)\cap\ker T^{k-1} =\{0\}$. We want to construct a Jordan basis $B$ of $V$ for $T: V\to V$ by adding $n-r$ linearly independent vectors to $S$ from $V\backslash L(S)$. How should I choose these vectors so that the resulting basis is a Jordan basis? I am stuck here.

Clearly, I am unable to make use of the condition that $L(S)\cap\ker T^{k-1} =\{0\}$.

I would appreciate any hints on how to take this solution forward, and also ideas for any other ways of approaching this problem. Thank you!

答:
I believe it is simpler to prove by induction on $k$ rather than on $n$.
The main observation is that, for Jordan basis of a nilpotent operator, if $v\in B$ then either $Tv\in B$ or $Tv=0$, meaning that if $S \subset B$ we have $\{Tv \; | \; v\in S, Tv\ne 0\} \subset B$.

The induction basis of $k=1$ is identical to the solution you presented for $n=1$, so I skip it.
Suppose now that $k>1$ and that you have proved the assertion for linear transformations with nilpotency index smaller than $k$.
Consider the set $T(S):=\{Tv \; | \; v\in S\} \subset \operatorname{Im}(T)$. I assert it is linearly independent. Indeed, if $\sum_{v\in S}\alpha_v Tv=0$, then $\sum_{v\in S}\alpha_vv \in \ker T \subset \ker T^{k-1}$ so $\sum_{v\in S}\alpha_vv=0$ since $L(S)\cap\ker T^{k-1}=\{0\}$.

By the induction hypothesis, exist a Jordan basis $B_0$ of $\operatorname{Im}(T)$ containing $T(S)$, we are left to understand what vectors from $V\setminus \operatorname{Im}(T)$ we need to add to it to get a Jordan basis of $V$.
Obviously $S$ has to go there since it was our original goal.
For any other $v\in B_0$ s.t. $v\notin T(S)$ and $v\notin T(B_0)$ we also need to pick a preimage $w_v\in V$ s.t. $Tw_v=v$ and add it to our Jordan basis (this is always possible since $v\in \operatorname{Im}(T)$).
Finally, we take $B_0\cap \ker T$ and add to our Jordan basis its completion to a basis of $\ker T$, to get all the length $1$ chains.
So the Jordan basis is
$$ B:= B_0 \cup S \cup \{ w_v \; | \; v\in B_0\setminus (S\cup T(B_0)) \} \cup B'\setminus (\ker T\cap B_0) $$
where $B'$ is a basis of $\ker T$ containing $B_0\cap \ker T$.

Until here is the construction, let prove it works.
If we were to prove the resulting set is indeed a basis, it is immediate that this basis is a Jordan basis for $T$ since by construction every $v\in B$ satisfies that either $Tv=0$ or $Tv\in B$.
Suppose that $\sum_{v\in B}\alpha_v v=0$ is a linear dependency.
Applying $T$ to it we get $\sum_{v\in B\setminus \ker T}\alpha_v Tv = 0$ which means that $\alpha_v=0$ for all $v\in B\setminus \ker T$, since for all such $v$, $Tv\in B_0$.
So any linear dependency includes only $B\cap \ker T$ but this was chosen as to be a basis of $\ker T$ so can not consist of non trivial linear dependencies.

We are left with showing that $B$ spans $V$.
Let $u\in V$.
Since $B$ contains a basis for $\operatorname{Im}(T)$ and since every $v\in B\cap \operatorname{Im}(T)$ is an image of an element of $B$, $T(u)=\sum_{v\in B\setminus \ker T}\alpha_vT(v)$.
Consider now the vector $w = \sum_{v\in B\setminus \ker T}\alpha_v v$. The difference $u-w$ is clearly in the kernel of $T$ so we can span it by $B$ and thus get that $u\in Span(B)$.


问:
Relation between closure of a set and the set of limit points of the interior of the set
Suppose $S \subseteq \mathbb{R}^{n}$. It seems intuitive that $\overline{S}$ is equal to the set of limit points of $S^{0}$, where $\overline{S}$ is the closure of $S$ and $S^{0}$ is the interior of $S$. Is this intuition true? Is there a proof for it?

答:
No they may be different. Consider $S$, a subset of $\mathbb{R}^2$, as the union of the closed unit disk and the interval $[1,2]$.

Then $\overline{S}=S$ (as it is closed), $S^0$ is the open unit disk but the closure of $S^0$ is just the closed unit disk.

The result always holds if $S$ is open. In this case $S^0=S$ so equality of the closures is obvious.


问:
A problem on Division and Distribution of objects.
Find Number of ways to distribute $4$ identical red balls , $4$ identical  green balls, $4$ identical white balls, $4$ identical  black balls among four people so that each person receives at least one ball but no one gets $4$ red balls.

There is a really similar question here ; Number of ways of distributing $4$ identical red balls,$1$ green ball,$1$ black ball among $4$ persons . But in my book, it was written : Number of ways of distributing $4$ identical red, green, black, white balls among $4$ persons so that each person receives at least one ball but no one gets $4$ red balls.

So i interpreted the question as the question i am posting now and i just couldn't solve it. So i looked it up on MSE and found the question (link attached). So i am wondering if it is even possible to solve that question because I can not stop thinking about it but i just can't solve it.

Thanks!

I THINK THIS PROBLEM IS WRONG

答:
Note [2020-12-25]: This is a completely revised answer.

This answer consists of three parts. In the first part we give a solution to the problem. The second part is a generalisation which can be derived easily and we find a nice identity of Stirling numbers of the second kind. In the third part we calculate a small example by hand as plausibility check.

A solution: This solution is based upon PIE, the principle of inclusion-exclusion.

The number of ways to distribute four identical balls to four persons so that each of them has zero or more balls is using stars and bars
\begin{align*}
\binom{4+4-1}{4-1}=\binom{7}{3}\tag{1}
\end{align*}
We start with a formula for distributing zero or more balls and not at least one ball, since balls with other colors will also be used for distribution to the people.

We have four different colors with four balls each. The number of ways to distribute these balls, so that each of the persons has zero or more balls is according to (1)
\begin{align*}
\binom{7}{3}^4\tag{2}
\end{align*}

Now we subtract from (2) according to PIE, the principle of inclusion-exclusion, the cases where at least one of the four persons has no ball. Since one person can be selected in $\binom{4}{1}$ ways, we get as number to subtract
\begin{align*}
\binom{4}{1}\binom{4+3-1}{3-1}=\binom{4}{1}\binom{6}{2}
\end{align*}
But here we did some overcounting, since we counted the cases where two of the four persons has no ball, twice. To compensate this we add $\binom{4}{2}$ times the number of ways to distribute the balls to two persons and so on. We obtain this way
\begin{align*}
\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^4\tag{3}
\end{align*}

Finally we have to respect that four red balls are not allowed to be distributed to one person only. In order to do so we calculate the number of ways to distribute three times four balls of the other colors to four persons. Since we have $\binom{4}{1}$ ways to give the four red balls to a person we have to subtract:
\begin{align*}
\binom{4}{1}\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^3\tag{4}
\end{align*}
One final aspect we missed so far is that we additionally have to subtract the $\binom{4}{1}=4$ cases where four red balls are given to a person and this person has no other balls. These $4$ cases also have to be subtracted.

We obtain from (3) and (4) the number of wanted ways as
\begin{align*}
&\color{blue}{\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^4-\binom{4}{1}\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^3-\binom{4}{1}}\\
&\quad=\left[\binom{4}{0}\binom{7}{3}^4-\binom{4}{1}\binom{6}{2}^4+\binom{4}{2}\binom{5}{1}^4-\binom{4}{3}\binom{4}{0}^4\right]\\
&\quad\quad-\binom{4}{1}\left[\binom{4}{0}\binom{7}{3}^3-\binom{4}{1}\binom{6}{2}^3
+\binom{4}{2}\binom{5}{1}^3-\binom{4}{3}\binom{4}{0}^3\right]-\binom{4}{1}\\
&\quad=\left[1\cdot1\,500\,625-4\cdot50\,625+6\cdot625-4\cdot1\right]\\
&\quad\quad-4\left[1\cdot 42\,875-4\cdot 3\,375+6\cdot125-4\cdot1\right]-4\\
&\quad=1\,301\,871-120\,484-4\\
&\,\,\quad\color{blue}{=1\,181\,383}\tag{5}
\end{align*}

A generalisation:

We can easily derive the following generalisation of (3) from the approach above: We consider $n$ pairwise distinct colors $C_1, C_2, \ldots, C_n$ and  $b_1+b_2+\cdots+b_n$ balls where $b_j$ balls have color $C_j, 1\leq j\leq n$. The number of ways to distribute $b_1+\cdots+b_n$ balls to $k$ persons is
\begin{align*}
\color{blue}{\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}\prod_{j=1}^n\binom{b_j+k-i-1}{k-i-1}}\tag{6}
\end{align*}

We consider the special case $b_1=b_2=\cdots=b_n=1$. Here we have $n$ different(ly colored) balls and we can reformulate the wanted number as $k!$ times the number of ways to partition $n$ numbers into $k$ sets. This is ${n \brace k}k!$, with $n\brace k$ the well-known Stirling numbers of the second kind.

We derive from (6) the identity
\begin{align*}
\color{blue}{{n\brace k}k!}&=\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}\prod_{j=1}^n\binom{1+k-i-1}{k-i-1}\\
&=\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}\binom{k-i}{k-i-1}^n\\
&\,\,\color{blue}{=\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}(k-i)^n}\\
\end{align*}

An example:

Here we consider as application and plausibility check of the formulas (5) and (6) the following problem.

Find the number of ways to distribute $2$ identical red balls , $2$ identical  green balls, $2$ identical white balls among two people so that each person receives at least one ball but no one gets $2$ red balls.

We have the situation $n=3, b_1=b_2=b_3=2$ and we have two persons, which means $k=2$. The wanted number is following (5) and (6)

\begin{align*}
&\sum_{i=0}^{2-1}(-1)^i\binom{2}{i}\binom{2+2-i-1}{2-i-1}^3\\
&\qquad-\binom{2}{1}\sum_{i=0}^{2-1}(-1)^i\binom{2}{i}\binom{2+2-i-1}{2-i-1}^2-2\\
&=\sum_{i=0}^{1}(-1)^i\binom{2}{i}\binom{3-i}{1-i}^3
-2\sum_{i=0}^{1}(-1)^i\binom{2}{i}\binom{3-i}{1-i}^2-2\\
&=\left[\binom{2}{0}\binom{3}{1}^3-\binom{2}{1}\binom{2}{0}^3\right]
-2\left[\binom{2}{0}\binom{3}{1}^2-\binom{2}{1}\binom{2}{0}^2\right]-2\\
&=25-2\cdot7-2\\
&\,\,\color{blue}{=9}
\end{align*}

We list all $25$ cases, give the $9$ valid cases in boldface and mark the $2\cdot7$ and $2$ cases which are not valid blue and red. We have
\begin{align*}
\begin{array}{cccccc}
\mathbb{g|g\color{blue}{rr}ww}&\mathbb{gg|\color{blue}{rr}ww}&\mathbf{ggr|rww}&\mathbb{gg\color{blue}{rr}|ww}&\mathbb{gg\color{blue}{rr}w|w}&\mathbf{ggrw|rw}\\
\mathbf{ggrww|r}&\mathbb{ggw|\color{blue}{rr}w}&\mathbb{ggww|\color{red}{rr}}&\mathbf{ggr|grww}&\mathbb{g\color{blue}{rr}|gww}&\mathbb{g\color{blue}{rr}w|gw}\\
\mathbb{g\color{blue}{rr}ww|g}&\mathbf{grw|grw}&\mathbf{grww|gr}&\mathbb{gw|g\color{blue}{rr}w}&\mathbb{gww|g\color{blue}{rr}}&\mathbf{r|ggrww}\\
\mathbb{\color{red}{rr}|ggww}&\mathbb{\color{blue}{rr}w|ggw}&\mathbb{\color{blue}{rr}ww|gg}&\mathbf{rw|ggrw}&\mathbf{rww|ggr}&\mathbb{w|gg\color{blue}{rr}w}\\
\mathbb{ww|gg\color{blue}{rr}}&&&&&
\end{array}
\end{align*}


问:
For what a will this series converge?
$\sum_{n=1}^{\infty}n^a\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$

I tried to use the D'Alembert ratio test and it didn't quite work out. The series inside is telescoping but I don't know if that information will be useful. Can someone help?

答:
We have
$$
\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}} =  \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} + \sqrt{n})} \sim \frac{1}{2n^{3/2}}.
$$
Hence $a_n = n^{\alpha} \Bigl( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \Bigr) \sim \frac{1}{2n^{3/2-a}}$. The sum $\sum_n a_n$ is convergent iff $\sum_n \frac{1}{n^{3/2-a}}$ is convergent. Using integral test for convergence we get that $\sum_n \frac{1}{n^{3/2-a}}$ is convergent iff $3/2-a > 1$.
So, the sum $\sum_n a_n$ is convergent iff $a<\frac12$.

Title: For what a will this series converge?


问:
Expression of a smooth multivariable function around a point
In Robert Wald's General Relativity, he states that

If $F: \mathbb{R}^n \to \mathbb{R}$ is $C^\infty$, then for each $a=(a^1, \dots, a^n) \in \mathbb{R}^n$ there exist $C^\infty$ functions $H_\mu$ such that for all $x \in \mathbb{R}^n$ we have
$$ F(x) = F(a) + \sum_{\mu = 1}^n (x^\mu - a^\mu)H_\mu(x) $$
Furthermore, we have
$$ H_\mu(a) = \frac{\partial F}{\partial x^\mu}\Bigr|_{x=a}$$

Where the superscripts $x^\mu$ indicate indices, not exponents.
Problem 2 at the end of the chapter then gives a hint to prove this by induction:

(Hint: For $n=1$, use the identity
$$ F(x) - F(a) = (x-a) \int_0^1 F'[t(x-a) + a]dt$$
then prove it for general $n$ by induction.)

I've seen other ways of proving the statement, but for the life of me I can't see how to prove it by induction. Any insight would be greatly appreciated!

答:
The base case proceeds as follows. For a smooth function $F: \mathbb{R} \to \mathbb{R}$ and some $a \in \mathbb{R}$, we can construct $H(x) = \frac{F(x) - F(a)}{x-a}$ so that $F(x) - F(a) = (x-a)H(x)$ and $H(a) = \frac{d}{dx}F(x) \Bigr|_{x=a}$. Such an $H$ is smooth and well-defined on $\mathbb{R}$ because $F$ is smooth.

Now suppose the proposition is true for $n<k-1$. Given a smooth function $F: \mathbb{R}^k \to \mathbb{R}$ and some $a \in \mathbb{R}^k$, we can consider fixing the last coordinate $x^k = a^k$, letting us apply the inductive proposition

$$ F(x^1, \dots, x^{k-1}, a^k) = F(a) + \sum_{\mu=1}^{k-1} (x^\mu - a^\mu) H_\mu(x^1, \dots, x^{k-1})$$

Now if we define

$$ H_k(x) \equiv \frac{F(x) - F(x^1, \dots, x^{k-1}, a^k)}{x^k - a^k} $$

We find that $H_k$ is smooth and well-defined on $\mathbb{R}^k$ again because $F$ is smooth, and we recover the intended proposition

$$ F(x) - F(a) = \sum_{\mu=1}^{k-1} (x^\mu - a^\mu) H_\mu(x^1, \dots, x^{k-1}) + (x^k - a^k)H_k(x) = \sum_{\mu=1}^{k} (x^\mu - a^\mu) H_\mu(x)$$

and

$$ H_k(x) = \frac{\partial}{\partial x^k} F(x) \Bigr|_{x=a} $$


问:
Is the statistical standard deviation of elements of x=(x1,x2,...,xn) a convex function?
For $x\in R^n$, is the statistical standard deviation of its elements, formulated as below, a convex function?

$(\frac{1}{n} \sum_{i=1}^n x_i^2 - (\frac{1}{n}\sum_{i=1}^n x_i)^2)^{\frac{1}{2}}$

答:
Using matrix-vector notation:
$$
\begin{aligned}
f(x) &= \left(\frac{1}{n}x^Tx - \left(\frac{1}{n}x^T 1\right)^2 \right)^{1/2}
\\
&= \left(\frac{1}{n}\left(x- \frac{1}{n}1^Tx \ 1\right)^T \left(x- \frac{1}{n}1^Tx \ 1\right) \right)^{1/2}
\\
&= \frac{1}{\sqrt{n}} \left\|\left(I- \frac{1}{n}11^T\right)x \right\|_2
\\
&= \frac{1}{\sqrt{n}} \left\| (I-P) x \right\|_2
\end{aligned}
$$
Where $P=\frac{1}{n}11^T$ is the idempotent projector to the ones vector. So $f$ is a composition of a linear map with the $\ell_2$ norm, so it's convex by the triangle inequality. In words, $f$ is proportional to the length of the component of $x$ which is orthogonal to the ones vector.


问:
Find the length of base of a triangle without using Pythagorean Theorem
I'm curious whether it is possible to find the length of base of the triangle without using Pythagorean Theorem

No Pythagorean Theorem mean:

=> No trigonometric because trigonometric is built on top of Pythagorean Theorem.
etc $\sin{\theta} = \frac{a}{r}$

=> No Integration on line or curve because the integration is built on top of Pythagorean Theorem.
etc: $s(x) = \int \sqrt{f'(x)^2 + 1}$

答:
It is perfectly possible to construct this figure in a way such that
$a_1 \neq a.$

But in the case where $a_1 = a,$ we can observe that the perpendicular line labeled $g$ divides the large triangle into two congruent triangles,
hence the triangle with hypotenuse $AB$ is twice the area of the triangle with hypotenuse $a.$

The larger triangle also is similar to the smaller triangle.
In order to have twice the area, all its dimensions, including the length of the hypotenuse, must therefore be greater by a factor of $\sqrt2.$

A somewhat more complicated version of this argument can be applied in the case where $a_1 \neq a,$ resulting in a proof of the Pythagorean Theorem.
I would argue that proving the Pythagorean Theorem is technically not the same thing as using it but would still be a violation of the spirit of the question. Proving only the special case $a_1 = a,$ however, does not rise to the level of a proof of the Pythagorean Theorem (which is more general).


问:
Prove that $h_n(x) = n^2 (e^{x/n} -1 - x/n)$ does not converge uniformly on $\mathbb{R}$
$h_n(x) = n^2 (e^{x/n} -1 - x/n)$

So I know that the pointwise limit is $x^2/x.$ I found it by using L'hopital's rule (is there a way to show it without using l'hopital's rule?).

I know to show that the convergence is not uniform I need to:

Find $\epsilon >0$ st $\forall N \in \mathbb{N}, \exists \geq N, x \in \mathbb{R}$ such that $|h_n(x)-x^2/n| = |n^2 (e^{x/n} -1 - x/n) - x^2/2| \geq \epsilon$.

Is the negation above correct?

From there I said let $\epsilon =1/10$ then $\forall N \in \mathbb{N},$ let $n=N$ and $x=0$ then $|n^2 (e^{x/n} -1 - x/n) - x^2/2| = |n^2 (1-1-0) - 1/2| = |-1/2| = |1/2| \geq 1/10.$

edit: that dosent work.

答: Since$$n^2(e^{x/n}-1-x/n)=x^2/2+x^3/(6n)+o(x^3/n),$$the error term $\sim x^3/(6n)$ exceeds $\epsilon$ for $x\gtrsim\sqrt[3]{6n\epsilon}$.


问:
Inner Products why is $\langle \langle x, v_i \rangle v_i, y \rangle=\langle x,v_i \rangle\langle v_i,y \rangle$
I have seen in a few different proofs $\langle \langle x, v_i \rangle v_i, y \rangle=\langle x,v_i \rangle\langle v_i,y \rangle$ by linearity but I can't seem to wrap my head around why. Could someone please explain this

答: $\langle x, v_i \rangle$ is a scalar, so it just comes from the linearity axiom of $\langle a\vec v, \vec w \rangle=a\langle \vec v, \vec w \rangle$


答:
Let us notice (with the aid of Taylor series) that $h_n(x) \to \frac{x^2}2$ for any fixed $x$. So, $h_n(x)$ converges uniformly iff $\sup_{x} |h_n(x) - \frac{x^2}2| \to 0$, $n \to \infty$.

But if $x = n$ then $|h_n(x) - \frac{x^2}2| = |n^2(e - 1 -1 ) - \frac{n^2}2|$ doesn't converge to $0$. Hence, $h_n(x)$ doesn't converges uniformly.

Title: Prove that $h_n(x) = n^2 (e^{x/n} -1 - x/n)$ does not converge uniformly on $\mathbb{R}$

Body:
$h_n(x) = n^2 (e^{x/n} -1 - x/n)$

So I know that the pointwise limit is $x^2/x.$ I found it by using L'hopital's rule (is there a way to show it without using l'hopital's rule?).

I know to show that the convergence is not uniform I need to:

Find $\epsilon >0$ st $\forall N \in \mathbb{N}, \exists \geq N, x \in \mathbb{R}$ such that $|h_n(x)-x^2/n| = |n^2 (e^{x/n} -1 - x/n) - x^2/2| \geq \epsilon$.

Is the negation above correct?

From there I said let $\epsilon =1/10$ then $\forall N \in \mathbb{N},$ let $n=N$ and $x=0$ then $|n^2 (e^{x/n} -1 - x/n) - x^2/2| = |n^2 (1-1-0) - 1/2| = |-1/2| = |1/2| \geq 1/10.$

edit: that dosent work.


问:
If $f \geq 0$ is RI on $[a,b]$ show that its integral is also $\geq 0$
$f \geq 0 \ on \ [a,b] \ RI.$

Then, $$\int_{a}^{b} f(x) dx \geq 0$$

I have seen this

Is the Riemann integral of a strictly positive function positive?

However, I'm not fully satisfied with the answers there as the conditions are slightly different. In the accepted answer, they say that since $f$ is continuous and positive on I (to make it similar to my problem let $I = [a,b]$, then $\int_{I} f > 0$. However, this is the conclusion. To me, it seems like they used the final answer to prove the result.

In my problem, however, I am not told $f$ is continuous in my problem. So that is a major assumption I am missing because while every continuous function is RI, being RI does not guarantee continuity.

I did look at a second answer on that same problem that has more upvotes. However, they prove it by assuming that $\int_{0}^{1} f = 0$. This works for that problem because $f > 0$ strictly. In my problem, however, it could be equal to $0$. So showing this case won't really help me here.

There was one more answer, but that uses Lebesgue integration, which I have not covered in this class.

I am really not so sure how to start it myself. I am convinced it's true after trying to come up with counterexamples because I wasn't convinced at first, but I really have no idea how to show it. Can anyone help me with this one? Thanks!

答:
The definition of the Riemann integral looks something like

$$ \sup_{\mathcal{P}} \sum_{i=1}^n \left(\inf_{[x_{i},x_{i+1}]} f(x) \right) (x_i - x_{i-1}) = \int_a^b f = \inf_{\mathcal{P}} \sum_{i=1}^n \left(\sup_{[x_{i},x_{i+1}]} f(x) \right) (x_i - x_{i-1}). $$

You might write this somewhat differently but the point is that every piece of this is $\ge 0$ by assumption. Specifically, if $f \ge 0$ then $\inf f \ge 0$ on any interval.


问:
Proof of an equality in Landau Notation
Definition 1: $\ g(h) = o(h) \quad {\rm iff} \quad g(h) = h \epsilon(h)$

where $\epsilon(h) \to 0$ as $h \to 0$

Definition 2: $\ g(h) = O(h) \quad {\rm iff} \quad \exists M > 0 \ {\rm such\ that} \ |g(h)| < Mh$

Show that if $g(h) = O(h^2)$ then $g(h) = o(h)$

I think my main difficulty comes from the fact that this notation is brand new to me. I've never worked with it before, and my professor only mentioned it briefly in his lectures. He thought we would have seen it before at some point, but I had not.

In terms of how to start this one, I really don't have much intuition here. I'm assuming that $O(h^2)$ would be $|g(h)| < Mh^2$, but I can't be sure because I'm not sure if the $h$ does something different when it is squared in that notation.

To add to my confusion here, the relation for $O(h)$ is an inequality, but it would have to become an equal sign in $o(h)$ to make this true. So I'm really not sure how these two would relate. I'm not even convinced it's true right now. Can anyone shed some light on this? Thank you!

答:
Your definitions don't cover the full thing.

Assume that we are dealing with a limit $x\to\xi\ $ ($\xi=\infty$ allowed), and that there is a well understood function $p(x)>0$, defined in the neighborhood of $\xi$.

We furthermore have a not so well understood function $x\mapsto g(x)$, and we want to express that $g$ behaves asymptotically in a certain way when $x\to\xi$. Landau's symbols allow the following two declarations:

$$g(x)=o\bigl(p(x)\bigr)\quad(x\to\xi)\qquad:\Leftrightarrow\qquad \lim_{x\to\xi}{g(x)\over p(x)}=0\ .$$
$$g(x)=O\bigl(p(x)\bigr)\quad(x\to\xi)\qquad:\Leftrightarrow\qquad \exists M>0 \quad {\rm such \ that}\quad |g(x)|\leq M p(x)\quad\bigl(x\ {\rm near}\ \xi\bigr)\ .$$
Now, when $$g(h)=O(h^2)\quad(h\to0)\ ,$$
then there is an $M>0$ and a $h_0>0$ such that
$$|g(h)|<M h^2\quad\bigl(|h|<h_0\bigr)\ .$$
This implies
$${|g(h)|\over|h|}<M\,|h|\qquad(|h|<h_0)\ ,$$
and therefore $\ \lim_{h\to0}{|g(h)|\over|h|}=0$, and this is $\>g(h)=o(h)$ $(h\to0)$.

Title: Proof of an equality in Landau Notation


问:
Is this integral answer 1 ?? However wolfram alpha tells me 1/3 ...
Is the answer "1" right ?? But the answers on the integral calculators tell $\frac{1}{3}$. Please help me.
$$\int_1^\infty\frac1{(x+2)^2}\,dx$$

答:
If you rendered $u=x+2$ to get $\int(1/u^2)du=-(1/u)+C$, you may have forgotten to do either one of two things:

If you keep the antiderivative in terms of $u$, the limits on the integrand must be values of $u=x+2$, thus $x=1$ means $u=3$ for the lower limit.

If you use back substitution to render the antiderivative as $-(1/(x+2))+C$, then the value of $-1/(x+2)$ at $x=1$ is $-1/3$ rather than $-1/1$.

Either way, $1/3$ is the correct value of the definite integral.


问:
Solve equation $\sqrt{4t + 1} = 3 - 3t$
Solve equation $\sqrt{4t + 1} = 3-3t$

→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$

→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$

I'm stuck at that point. Can someone tell me what I am doing wrong?

Title: Solve equation $\sqrt{4t + 1} = 3 - 3t$

Body:
Given: $\sqrt{4t+1}=3-3t$

$$\bigg(\sqrt{4t+1}\bigg)^2=\bigg(3-3t\bigg)^2$$

$$4t+1=\bigg(3-3t\bigg)\bigg(3-3t\bigg)$$

$$4t+1=9-9t-9t+9t^2$$

$$4t+1=9-18t+9t^2$$

$$0=9-1-18t-4t+9t^2$$

$$0=9t^2-22t+8$$

Which is a quadratic in the form:$$0 =ax^2+bx+c$$
Use the quadratic equation:
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

$$x = \frac{ -(-22)\pm \sqrt{(-22)^2-4(9)(8)}}{2(9)}$$

$$x = \frac{22\pm\sqrt{484-288}}{18}$$

$$x = \frac{22\pm\sqrt{196}}{18}$$

$$x = \frac{22\pm{14}}{18}$$

$$x = 2 $$ $$or$$ $$x =\frac{4}{9} $$

Understand that it still be checked that these are indeed solutions to your initial equation.


问:
Finding the area not by integral

Original problem. Find the area under $y= \sqrt{x}$ in the range $\left [ 0, 1 \right ]$

My friend, she wants to use total area instead of calculating the integeral. So I tried something:

Dividing the area by $n$ parts with each part is $\Delta x= \frac{1}{n}$.

The part $k$ is $\left [ \left ( k- 1 \right )\frac{1}{n}, \frac{k}{n} \right ]$, we calculate the total area by formula:

$$S_{n}= \sum_{k= 1}^{n}\frac{1}{n}\sqrt{\frac{k}{n}}= \frac{1}{n^{3/2}}\sum_{k= 1}^{n}k^{1/2}$$

How should I do next ?? Thank you....

答:
Hint

Find the $S$ area of $y=x^2$ in $[0,1]$. The wanted area is $1-S$

Title: Finding the area not by integral


问:
Taylor series for complex functions $f(z)=\sqrt [n] {g(z)}$ and $f(z)=\log(g(z))$
I'm looking for a good way of computing Taylor series for complex functions $f(z)=\sqrt [n] {g(z)}$ and $f(z)=\log(g(z))$ ($f(z_0)$ is fixed for $z_0$ and there is a suitable plane cut). May be one can find them using Taylor series for real  $f(z)=\sqrt [n] {g(z)}$ and $f(z)=\log(g(z))$?

答:
Let $f(z)$ be a holomorphic branch of $\log(g(z))$ in a disk $B_r(z_0)$. By decreasing $r$ if necessary we can assume that
$$
 \left| \frac{g(z)}{g(z_0)} - 1\right| < 1
$$
for all $z \in B_r(z_0)$. Then
$$
 f(z) = f(z_0) + \operatorname{Log} \left(\frac{g(z)}{g(z_0)}\right)
$$
where $\operatorname{Log}$ is the principal branch of the logarithm (i.e. the one satisfying $\operatorname{Log}(1) = 0$). The Taylor series
$$
 \operatorname{Log}(1+w) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} w^n
$$
is valid for $|w| < 1$ so that substituting $w = \frac{g(z)}{g(z_0)} - 1$ allows to compute the Taylor series of $f$.

Similarly, if $f$ is a holomorphic branch of $(g(z))^{1/n}$ in a neighborhood of $z=z_0$ then
$$
 f(z) = f(z_0) \left(\frac{g(z)}{g(z_0)}\right)^{1/n}
$$
allows to compute the Taylor series of $g$ via the binomial series for $(1+w)^{1/n}$.


问:
What is the common usage for "let" for multiple variables?
Should I use

Let $x$ be (...), let $y$ be (...), ...

or

Let $x$ be (...). Let $y$ be (...). ...

or

Let $x$ be (...), $y$ be (...), ...

or even something different? What variant is commonly used in English?

答: All three are common.  The second variant is preferred if the descriptions are long.</p>


答:
Since $\lambda=\langle x, v_i\rangle$ is a scalar. then
$$
\langle\langle x,v_i\rangle v_i,y\rangle=\langle\lambda v_i,y\rangle=\lambda\langle v_i,y\rangle=\langle x,v_i\rangle\langle v_i,y\rangle.
$$

Title: Inner Products why is $\langle \langle x, v_i \rangle v_i, y \rangle=\langle x,v_i \rangle\langle v_i,y \rangle$

Body: I have seen in a few different proofs $\langle \langle x, v_i \rangle v_i, y \rangle=\langle x,v_i \rangle\langle v_i,y \rangle$ by linearity but I can't seem to wrap my head around why. Could someone please explain this


问:
Help to clarify proof of Euler's Theorem on homogenous equations

Why is the last step (setting $\lambda = 1$ allowed?

I have trouble accepting this because if I set $\lambda =1 $ at the very start, then:

$f(\lambda x , \lambda y)=\lambda^r f(x,y)$

becomes

$f(x,y) = f(x,y)$

and

so I can't prove the theorem.

Why does setting $\lambda = 1$ at the end of the proof work?

EDIT

For example, let $f(x,y)=x^2y^2$

Therefore, $f(x,y)$ is homogeneous of degree 4.

Therefore, the second to last step will be:

$4x^2y^2=4\lambda^3x^2y^2$

The above equation is only true if $\lambda=1$

Why doesn't the theorem make a qualification that $\lambda$ must be equal to 1?

It seems to me that this theorem is saying that there is a special relationship between the derivatives of a homogenous function and its degree but this relationship holds only when $\lambda=1$. Please correct me if my observation is wrong.

答:
You can derive Euler theorem without imposing $\lambda=1$.

Starting from $f(\lambda x, \lambda y) = \lambda^n \times f(x,y)$, one can write the differentials of the LHS and RHS of this equation:

LHS

$df(\lambda x, \lambda y) = \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} \times d(\lambda x) + \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} \times d(\lambda y)$

One can then expand and collect the $d(\lambda x)$ as $xd\lambda + \lambda dx$ and $d(\lambda y)$ as $yd\lambda + \lambda dy$ and achieve the following relation:

$df(\lambda x, \lambda y) = \lambda \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} \times dx + \lambda \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} \times dy + \left[x\left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x}\right] \times d\lambda$

RHS

$d\left[\lambda^n f(x,y)\right] = n\lambda^{n-1}f(x,y) \times d\lambda + \lambda^n \left[\left(\frac{\partial f}{\partial x}\right)_y \times dx + \left(\frac{\partial f}{\partial y}\right)_x \times dy\right]$

RHS = LHS

Both differentials must be equal. Thus, terms in front of $dx$, $dy$ and $d\lambda$ must be equal in both expressions. This leads to the following system:

$\begin{cases}
\lambda \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} = \lambda^n \times \left(\frac{\partial f}{\partial x}\right)_y\\
\lambda \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} = \lambda^n \times \left(\frac{\partial f}{\partial y}\right)_x\\
x\left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} = n \lambda^{n-1} \times f(x,y)
\end{cases}$

From this point, the classical next step is to take the third equation, set $\lambda=1$ (although you want the identity to be true for all $\lambda$) and conclude that the identity is proven. Clearly this assumption is not required as you can inject the first and second equations into the third and obtain the required identity:

$\lambda^{n-1} x\left(\frac{\partial f}{\partial \lambda x}\right)_{y} + \lambda^{n-1} y\left(\frac{\partial f}{\partial \lambda y}\right)_{x} = n \lambda^{n-1} \times f(x,y)$

You can see that all $\lambda^{n-1}$ cancel and lead to the expected relation:

$x\left(\frac{\partial f}{\partial \lambda x}\right)_{y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{x} = n \times f(x,y)$


问:
Represent a function as a difference of two convex function
I know that if $f:\mathbb{R}\to\mathbb{R}$ is $C^2$ it can be rewrite as a difference of two convex function. I ask to you if there is someone who know if it's possible extend this property to the function which have $\mathbb{R}^n$ as domain.

答:
It is possible, for example:

If $f \in C^2(D,\mathbb{R})$ and $D$ a non-empty, convex and compact subset of $\mathbb{R}^n$, then $f$ is a dc-function, i.e. the difference of two convex functions.

The proof can't be done the same way as in $\mathbb{R}$. Note that in this statement you need a compact subset.

Here is a reference (Sanjo's answer): https://math.stackexchange.com/a/843020/797553

For a proof use $g(x)=f(x) + \rho/2 \cdot x^Tx$ and $h(x)=\rho/2 \cdot x^Tx$, where $\rho=\left| ~min ~\{\lambda_{min}(Hessf(x)): x\in D\} ~\right|$. Then $f=g-h$. Note that $\rho$ only exists as $D$ is compact.

EDIT: For $D=\mathbb{R}^n$ the statement is still true. (see [Konno H., Thach P.T., Tuy H. (1997) D.C. Functions and D.C. Sets. In: Optimization on Low Rank Nonconvex Structures. Nonconvex Optimization and Its Applications, vol 15. Springer, Boston, MA]). In this book you'll also find a proof for the statement above.


问:
Simplify $\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$
How do I evaluate the product:

$$\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$$

I know that $$\log_ba=\frac{\log\ a}{\log\ b}$$

How can I apply it?

Thanks!

答: Just apply the rule you stated and write the terms down , you will notice that all the terms cancel out except the denominator of the first and the nominator of the last one.</p>


问:
The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$.

The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$. Prove that the medians drawn from $A$ and $B$ are perpendicular.

What I Tried: Here is a picture :-

I would normally solve these kind of problems using elementary geometry, but I wasn't able to do it this time, simply because $(a,b,c)$ can take many values and there is probably some kind of construction or some clever proof which I am missing. One idea I thought of is that the medians divide the side lengths into $2$ equal parts, and the centroid itself divides the medians in $2:1$ ratio, and I probably have to show somehow, that Pythagoras Theorem on either of the $3$ triangles work. I took an example in Geogebra by drawing an accurate diagram with $a = 12$ , $b = 18.66$ nearly, and then $c = 10$ . This actually showed that the medians from $A$ and $B$ are really perpendicular, and I just drew a picture of that taken from Geogebra.

Now given $(a,b,c)$ , is there any way to prove this? I did not find any elementary methods like angle-chasing , pythagoras theorem , similarity and areas , etc. , but I did not find any clever use of them till now. Maybe some extra construction is required?

Can anyone help me? Thank You.

答:
Note $P$ the intersection point and consider the triangle $\triangle{ABP}$ so we have by the well known formulas for the medians
$$AP=\frac23\left(\frac{\sqrt{2(b^2+c^2)-a^2}}{2}\right)\\PB=\frac23\left(\frac{\sqrt{2(a^2+c^2)-b^2}}{2}\right)\\AB=c$$
It is verified directly and easily that $AP^2+PB^2=AB^2$ because of $a^2+b^2=5c^2$


问:
Injective image of quotient ring
I was checking the 2nd answer of this
How to deal with polynomial quotient rings.
There is an isomorphism $F_2[X]/(X+1)^4 \cong F_2[Y]/Y^4$. How to explicitly construct this isomorphism? Could this be generalized to fields like rationals or reals?

答:
I seem to have seen that it could be generalized because any element of $F_2[x]$ can be expressed as a polynomial with variable $x+1$ by long division. And it does generalize, if I am correct?

The shift map $\,s(x)= x+c\,$ is a ring hom by the universal mapping property of polynomial rings (it implies there is a unique ring hom $\,s\,$ with $\,s(x) = x+c).\,$  $\,s\,$ is invertible $\,(s^{-1}(x) = x-c)\,$ thus it follows that $\,s\,$ is a ring automorphism. Ditto for $\,s(x) = u\:\! x + c\,$ if $\,u\,$ is a unit (invertible).

That the automorphism extends to the stated isomorphism of quotient rings follows by

Lemma $ $ If $\ s: R\to S$ is an injective ring homomorphism then $\,R/I\,\cong\, s(R)/s(I),$

Proof $\ $ See e.g. this answer.

Title: Injective image of quotient ring


问:
Problem about martingale, bounded stopping time
Let $F_n$, $n\geq0$, be a filtration, and let ${X_n}$ be a sequence of r.v.'s such that $X_n\in F_n$ and $E\lvert X_n \rvert < \infty$. Show that if $EX_T = EX_0$ for all bounded stopping times T, $X_n$ is a martingale.

This is my problem. Actually, I can show that If T is bounded stopping time then $EX_T =EX_0$. But, this isn't problem.

How can I solve above problem? or any hints please.

答:
From the definition of a martingale, we want to show that $\mathbb{E}[X_t | \mathcal F_s] = X_s$ for all $s < t$, which is equivalent to showing that for all $A \in \mathcal F_s$ we have $\mathbb{E}[X_t 1_A] = \mathbb{E}[X_s 1_A]$.  We define a stopping time $\tau := t 1_A + s 1_{A^c}$ (check this is a stopping time!), and from the hypotheses we have

\begin{align*}
\mathbb{E}[X_s] = \mathbb{E}[X_0] &= \mathbb{E}[X_\tau] = \mathbb{E}[X_t 1_A + X_s 1_{A^c}]
\end{align*}

so $\mathbb{E}[X_s 1_A] = \mathbb{E}[X_t 1_A]$.  We conclude $\mathbb{E}[X_t | \mathcal F_s] = X_s$ and therefore $(X_n)$ is a martingale since we already had that it was adapted and integrable.


答: Hint: Its $\frac{log 3}{log 2}\cdots \frac{log 8}{log 7}$</p>

Title: Simplify $\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$

Body:
How do I evaluate the product:

$$\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$$

I know that $$\log_ba=\frac{\log\ a}{\log\ b}$$

How can I apply it?

Thanks!


问:
Christmas Tree Problem: String around Cone
This problem came up when I was trying to figure out how long of a string of lights I need for my Christmas tree. It was easy to estimate and obviously this isn't for use in the real world, but I found it a lot harder than I thought it would be at first.

A cone has a base radius $3$ ft and height of $8$ ft. A line ("string") is wrapped around the cone so that if you go straight up from any point on the string along the surface of the cone, the vertical distance (not distance along the cone's surface) to the next intersection of the string is $1$ ft.

The string is wrapped counterclockwise, looking from the top. As a more stringent restriction, a section of the string that encompasses some angle $\theta$ around the surface of the cone, the change in height from the bottom point to the top point is given by:
$$h = \frac{\theta}{2\pi}$$

What is the length of the string?

My first try at this problem assumed that the angle of the string relative to the horizontal would always be a constant. I quickly realized the string would have to get steeper as it goes up the cone ("tree"), in order to keep up with the shrinking circumferennce. I pretty much banged my head into a wall from there because I have no clue how to factor that in.

答:
Please see this link to my calculation of Christmas tree light string length.

The length of Christmas Tree Lights wrapping a conical tree from bottom to top with uniform vertical spacing.
(Reproduced below.)

Our $20$-year old artificial Christmas tree has built-in lights, but the top and bottom thirds are burnt out. So I decided to buy several (much cheaper) new "tree light" strings (during a sale after the end of last Christmas) the majority of which would continue to work even if a light were to burn out (wired in parallel, I assume).

The bottom tree boughs are $4$-feet in diameter, $r_0 = 2$-feet, the top tree boughs are $1$-foot in diameter, $r_h = 0.5$-foot, and the vertical height distance between the top and bottom boughs is $h = 5.5$-feet. The top center spike rises, but will not include lights.

If I want to wrap the tree from bottom to top, with $1$-foot vertical spacing between wraps, what length of light string is required?

This is how I attacked the problem.

I related all variables in terms of the number of turns wrapping the tree from bottom to top.

The height (of light string wrapping) is $5.5$-feet with $1$-foot vertical spacing, thus there are $5.5$-turns = $11 \cdot \pi$ radians, of rotation as you progress from bottom to top.

I let $\theta$ (the angle) be the variable of integration from $0$ to $11 \cdot \pi$.

I, imaginatively, affixed a sector of cylindrical shell element at the end of a bough at the middle height of the tree. The light strand would wrap diagonally through the middle of this element. The edges of the element represent: the differential arc length component, $ds$, the differential height component, $dh$, and the differential radius component, $dr$.

The differential diagonal length, $dl$, in terms of $d\theta$ would be $$\frac{dl}{d\theta} = \sqrt{(ds_\theta)^2 + (dh_\theta)^2 + (dr_\theta)^2}$$

$r$ as a function of theta: $$r_\theta = r_0 - \frac{( r_0 - r_h ) \cdot \theta}{11 \cdot \pi}$$
$\frac{dr_\theta}{d\theta}:$ $$dr_\theta = - \frac{ r_0 - r_h }{11 \cdot \pi} \cdot d\theta$$
$h$ as a function of theta:$$h_\theta = \frac{5.5 \cdot \theta}{11 \cdot \pi}$$
$\frac{dh_\theta}{d\theta}:$ $$dh_\theta = \frac{5.5}{11 \cdot \pi} \cdot d\theta$$
$s$, arc length, as a function of theta:$$s = r \cdot \theta$$
$\frac{ds_\theta}{d\theta}:$ $$ds_\theta = r_\theta \cdot d\theta$$
$$ds_\theta = (r_0 - \frac{( r_0 - r_h ) \cdot \theta}{11 \cdot \pi}) \cdot d\theta$$
Plugging the above into an HP-$42$S simulator on my phone and integrating $dl$ as $\theta$ revolves from $0$ to $11 \cdot \pi$: $$L=\int_0^{11 \cdot \pi} \frac{dl}{d\theta} \cdot d\theta$$yields, $L\approx43.6$-feet. By varying the vertical spacing this allows one to evenly accommodate a modularly assembled light string length. (Note this also alters the total number of turns.)

Before I had a chance to test the modification, my wife said, "No.", and we returned the new "tree light" sets and got a new Christmas Tree (last year's model) on sale after last Christmas instead.$$\prod_0^{n=2}(Ho_n) = Ho \cdot Ho \cdot Ho$$


问:
What are the values of $(a,b)$ so that this weakly stationary process is well-defined?
I'm solving this exercise from previous final exam of Time Series. Because tomorrow is my exam, it's too late for me to send the professor an email.

Could you explain on what does it mean for $X$ to be well-defined in this case? I think the question have mentioned that $X$ is weakly stationary process and thus already well-defined.

答:
This is an ARMA(1,1) process. I think the question b) requires you to find the coefficient values consistent with stationarity. You can write this as

$$(1-aB)X_t=(1-bB)\epsilon_t$$

As explained e.g. here this is stationary if the roots of $P(z)=1-az$ lie outside the unit circle. The roots solve

$$P(z)=0$$

or

$$1-az=0$$

or

$$z=\frac{1}{a}$$

So you need $|\frac{1}{a}| > 1$ or $-1<a<1$.


问:
Meaning of negative angle
If I consider $\alpha=-\frac{\pi}{3}$ this is exactly equal to $2\pi-\frac{\pi}{3}$?
So the consider blu angle is equal to consider the red one?

答:
If we are talking about your special question:

When considering angles on the circle, you should see them modulo $2\pi$.
Let $\alpha = -\frac{\pi}{3} $ and $\beta = 2\pi - \frac{\pi}{3}$. Since $\alpha \equiv \beta \pmod {2\pi}$ they will coincide on the circle.

If we are talking about the meaning of the negative angle in general:

These all are just an agreement of mathematicians (plus everyone else working in related fields).

For convenience, we say "rotate $-30^\circ$" instead of saying "rotate $30^\circ$ clockwise" or "rotate $300^\circ$"

So, whenever you see a negative angle, you should think of its direction.

Title: Meaning of negative angle


问:
Determine for which α ∈ R the series is absolutely convergent.
Determine for which α ∈ R the series:

$$\sum_{n=1}^{\infty}\frac{\sin\big(e^{\sqrt{n+1}-\sqrt{n}} + \pi n -1\big)}{\ln(\arctan\frac{1}{n})}n^\alpha$$

is:
a) aboslutely convergent,
b) convergent,
c) divergent.

I've tried an approach where I considered the nominator and the denominator separately using direct comparison test for both and:

Taylor's theorem to approximate  $e^{\sqrt{n+1}-\sqrt{n}}$ as $e^x = x + 1 + o(x):$

$e^{\sqrt{n+1}-\sqrt{n}} = {\sqrt{n+1}-\sqrt{n}} + 1 + o(\sqrt{n+1}-\sqrt{n}) $

The fact that $\arctan\frac{1}{n} = \frac{π}{2} -\arctan{n} $

Neither of these helped me solve this problem. Any suggestions? Thanks for the help.

答:
Notice that

$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}\sim \frac{1}{2\sqrt{n}}$$

and since $\sin(x+n\pi) = (-1)^n\sin x$, we have an alternating series as well. That's it for the numerator, now looking closely at the denominator we have that

$$\log\left(\arctan\frac{1}{n}\right) \sim \log\left(\frac{1}{n}\right) = - \log n$$

meaning that (calling the above series $\sum a_n$) :

$$|a_n| \sim \frac{n^{\alpha - \frac{1}{2}}}{2\log n}$$

For $\alpha > \frac{1}{2}$ this will always diverge. At the very least for $\alpha < -\frac{1}{2}$ the series will converge absolutely. I will leave it to you to prove that statement, and figure out where else the series will absolutely or conditionally converge.


Body:
You have everything that you need:

$$\log_n(n+1)=\frac{\log_2(n+1)}{\log_2n}\;,$$

so

$$\begin{align*}
\log_23\log_34\log_45\log_56\log_67\log_78&=\log_23\cdot\prod_{n=3}^7\left(\frac{\log_2(n+1)}{\log_2n}\right)\\
&=\frac{\prod_{n=3}^8\log_2n}{\prod_{n=3}^7\log_2n}\\
&=\log_28\\
&=3\;.
\end{align*}$$


问:
Why does rotation $r^2$ by $\pi$ give an order $2$ subgroup in $D_8?$
Find a subgroup of $D_8$  of order $2.$

My attempt :

A presentation of $D_{2n}$ is $\langle r,s\mid r^n=s^2=\mathrm{id},srs=r^{-1}\rangle$, the reflections are $s,sr,sr^2,\ldots,sr^{n-1}$

Put $ n=4$ then $D_{8}$ is $\langle r,s\mid r^4=s^2=\mathrm{id},srs=r^{-1}\rangle.$

The element in $D_8$ of order $2$ are the reflection  $s,sr,s^2$ and $sr^3$ because they all are  identity symmetry

But there are $5$ distinct subgroup  of order $2$

Rotation $r^2$ by $\pi$ also give order $2$

Here im not getting why rotation $r^2$ by $\pi$  give order $2$ subgroup  in $D_8?$

答: Doing it twice gets you back to where you started (a rotation of $2\pi$), so the element has order $2$, and hence generates a subgroup of order $2$.


问:
How to understand Urysohn’s lemma?
I am studying general topology. And I have problem in understanding Urysohn's lemma.

Urysohn's Lemma : If $X$ is normal and $C$, $D$ are disjoint subsets of $X$, there exists continuous map $f$ : $X \rightarrow [0, 1]$  s.t. $f(C)=0$, $f(D)=1$.

How can $f$ be defined  like that? For example, if $X=R$, $C=[0, 1]$, $D=[2, 3]$, how can we define such $f$?

$f([0, 1])$ and $f([2, 3])$ can be defined on the 'set of subsets of $R$', not on $R$.

'$f$ defined on $R$' means $f(x)$, where $x$ is a real number. Not the interval.

How can I understand it? Is there something I misunderstood?

答:
If $A$ and $B$ are sets, $f$ is a function from $A$ to $B$ and $S$ is a subset of $A$, then$$f(S)=\{f(s)\mid s\in S\}.$$

So, the statement of Urysohn's Lemma actually is:

If $X$ is normal and $C$ and $D$ are disjoint closed subsets of $X$, then there exists a continuous map $f\colon X\longrightarrow[0,1]$ such that $f(C)=\{0\}$ and that $f(D)=\{1\}$.

Title: How to understand Urysohn’s lemma?


问:
Understanding the following definition of torsion free modules
So I'm having some trouble seeing why the following is true immediately. So according to wikipedia;
https://en.wikipedia.org/wiki/Torsion-free_module.

The first paragraph reads:

"In integral domains the regular elements of the ring are its nonzero elements, so in this case a torsion-free module is one such that zero is the only element annihilated by some non-zero element of the ring. Some authors work only over integral domains and use this condition as the definition of a torsion-free module, but this does not work well over more general rings, for if the ring contains zero-divisors then the only module satisfying this condition is the zero module"

I'm definitely missing something trivially obvious; but I cant see immediately why having a zero-divisor immediately leads to the conclusion that the module MUST be the zero module ?

Any help or insights is appreciated.
Cheers !

答:
Let $R$ be an $M$-module and suppose $R$ is not an integral domain so $ab = 0$ for some $a,b \in R\setminus\{0\}$. Let $m \in M \setminus\{0\}$.

If $bm = 0$ we are done. Otherwise, $a(bm) = (ab)m = 0 m = 0$ so it is not torsion free. Hence, the only torsion free module over $R$ would be $0$.


问:
$\omega$-saturation of $(\mathbb{R},<)$
Could anyone of you explain me why $(\mathbb{R},<)$ is $\omega$-saturated?

EDIT: do you know also why the theory of Boole algebras without atoms is $\omega$-categoric?

Added: The added question about Boolean algebras is at Boolean algebras without atoms

答: Your structure is an endless dense linear ordering, and this is a theory that admits elimination of quantifiers. Thus, every assertion in this language is equivalent to a quantifier-free assertion. If one has finitely many parameters, then the only possible consistent types are the assertions about how those parameters are ordered, and how the new variable $x$ fits into the resulting intervals. That is, the variable $x$ is either equal to one of them, or between two successive ones, or above all or below all of them. All such types are already realized in $\mathbb{R}$, since it has such points already for each of these possible patterns, and so the structure is $\omega$-saturated.</p>

Title: $\omega$-saturation of $(\mathbb{R},<)$


问:
Trace of matrices
Well, I was solving exercises and here is one I cant solve.

$tr(A^TA)tr(B^TB)\geq tr(A^TB)^2$

I have tried it and I come with proof that $tr(A^TA)>0$ I dont know how it helps.

Thank you for your help.

答:
For any square or rectangular matrx $C,$  we find $$  \operatorname{trace} C^T C \geq 0$$
because it is simply the sum of the squares of all elements.

Next, for any square matrix $M$ we have
$$  \operatorname{trace} M^T =  \operatorname{trace} M$$
Detail, $$(G^T H )^T = H^T G $$

Finally,
$$   \operatorname{trace} \; (A-B)^T (A-B) \geq 0.    $$
Expand this, there is a factor of $2$ you are missing.

Title: Trace of matrices


答: Suppose that $A$ has a zero divisor $a$ (so that $ba = 0$ for some nonzero $b$) and $M$ is an $A$-module. If $M \neq 0$ then let $m \in M$ be nonzero. Then it is the case that $am = 0$ or we have $am \neq 0$ and $b(am) = 0$ - in both cases something non-zero is killed by a nonzero element of $A$.

Title: Understanding the following definition of torsion free modules

Body:
So I'm having some trouble seeing why the following is true immediately. So according to wikipedia;
https://en.wikipedia.org/wiki/Torsion-free_module.

The first paragraph reads:

"In integral domains the regular elements of the ring are its nonzero elements, so in this case a torsion-free module is one such that zero is the only element annihilated by some non-zero element of the ring. Some authors work only over integral domains and use this condition as the definition of a torsion-free module, but this does not work well over more general rings, for if the ring contains zero-divisors then the only module satisfying this condition is the zero module"

I'm definitely missing something trivially obvious; but I cant see immediately why having a zero-divisor immediately leads to the conclusion that the module MUST be the zero module ?

Any help or insights is appreciated.
Cheers !


问:
Prove that Cantor distribution is continuous
For a sequence $ a_{n}\in\left\{ 0,1,2\right\}  $ we define $ [a]_{3}=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}} $

(Thats just Base 3 expansion).

Now let $ A=A_n $ be a sequence of i.i.d random variables, each uniformly distributed over {0,2}

Define $ X=[A]_{3}=\sum_{n=1}^{\infty}\frac{A_{n}}{3^{n}} $

Such $ X $ is said to have to Cantor distribution.

That is all the given details, the way I see it, $ X $ is a random variable on some $ \varOmega $ such that for any $ \omega \in \varOmega$  we have:

$ X\left(\omega\right)=\sum_{n=1}^{\infty}\frac{A_{n}\left(\omega\right)}{3^{n}} $

And the value of $ A_n(\omega) $ can be $ 0 $ or $ 2 $, So that $ X(\omega) $ is in $ [0,1] $ for any $ \omega $.

Now based on the way I understood the details, I want to prove that X is continuous RV. Thus, its sufficies to show that

$ \mathbb{P}\left(X=r\right)=0 $

For any $ r \in \mathbb{R} $.

fix some $ r \in \mathbb{R} $. If there is no sequence $ (a_n)_n $ such that $ r=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}} $ and  $ a_{n}\in\left\{ 0,2\right\}  $ for any $ n $,

Then $ \mathbb{P}\left(X=r\right)=0 $ because $ X^{-1}\left\{ r\right\} =\emptyset $

So let $ R $ be the set such that for any $ r \in R $ we have a sequence $ \left(a_{n}^{r}\right)_{n} $ such that

$ \sum_{n=1}^{\infty}\frac{a_{n}^{r}}{3^{n}}=r $

and $ a_{n}^{r}\in\left\{ 0,2\right\}  $ for any $ n $.

CLAIM 1:
I calim that since we can only use the digits $ 0 $ and $ 2 $, there is only one unique seuquence $ a_n^r $ for any $ r \in R $ such that all the demands holds, and further more for $ r_1,r_2 $ we have $ r_{1}\neq r_{2}\iff\left(a_{n}^{r_{1}}\right)_{n}\neq\left(a_{n}^{r_{2}}\right)_{n} $.

If this claim is ture, then:

Note that for any $ \omega \in \varOmega $ we have:

$ X\left(\omega\right)=r\iff\sum_{n=1}^{\infty}\frac{A_{n}\left(\omega\right)}{3^{n}}=r\iff\forall n:A_{n}\left(\omega\right)=a_{n}^{r} $

And this is true because of claim 1, the uniquness of the sequence that defines a number $ r \in R $ , there is no other way for the sum $ \sum_{n=1}^{\infty}\frac{A_{n}\left(\omega\right)}{3^{n}} $ to converge to $ r $ unless $ A_{n}\left(\omega\right)=a_{n}^{r} $ because if we'll have just one term such that $ A_{j}\left(\omega\right)\neq a_{j}^{r} $, we'll get other number than $ r $.
(as long as claim 1 is correct).

Now if we agree up to this point, then for any $ r \in R $ :

$ \mathbb{P}\left(X=r\right)=\mathbb{P}\left(\omega\in\varOmega:A_{n}\left(\omega\right)=a_{n}^{r},\forall n\in\mathbb{N}\right)=\mathbb{P}\left(A_{1}=a_{1}^{r}\cap A_{2}=a_{2}^{r}...\right) $

$ =\mathbb{P}\left(\bigcap_{j=1}^{\infty}A_{j}=a_{j}^{r}\right) $

Since $ (A_n)_n $ is a sequence of independent RVs, we have that all those events in the intersecntion are mutually independent and so :

$ \mathbb{P}\left(\bigcap_{j=1}^{\infty}A_{j}=a_{j}^{r}\right)=\prod_{j=1}^{\infty}\mathbb{P}\left(A_{j}=a_{j}^{r}\right) $

And since for any $ j $ we have that $ A_j $ is uniformly distributed on $ \left\{ 0,2\right\}  $ and $ a_j^r \in$  {0,2} $ then

$ \mathbb{P}\left(A_{j}=a_{j}^{r}\right)=\frac{1}{2} $

So we have:

$ \mathbb{P}\left(X=r\right)=\prod_{j=1}^{\infty}\mathbb{P}\left(A_{j}=a_{j}^{r}\right)=\prod_{j=1}^{\infty}\frac{1}{2}=\lim_{k\to\infty}\prod_{j=1}^{k}\frac{1}{2}=\lim_{k\to\infty}\frac{1}{2^{k}}=0 $

And thus for any $ r \in \mathbb{R} $,  $ \mathbb{P}(X=r) =0 $

And so $ X $ is continious.

So questions:

Is claim 1 correct?

2.Is this proof holds?

答:
There are few things which I would like to say here. The first claim is true, and the idea seems to be true as well. I want to give out some facts I hope will let you feel better about that:

We need to pick the expansion of the interval [0,1], where for every number the expansion is unique.

It turns out that there is , at least for the "most" of points, unique expansion. The problem of uniqueness arise where we try to expand numbers of the form $\frac{p}{3^k} $. In these cases, there are two optional expansions- One s.t $ \forall j>k
 $ ,$a_j =0$. The other is of the form $a_j=2 $ ($\forall j>k $). If we pick the second expansion, then it is unique on the whole unit interval.

For example, $ \frac{1}{3}=0.10000...$

And the second expansion, which we prefer, is given by  $ \frac{1}{3}=0.02222...$\

I hope it makes you feel more confident about that proof.


问:
Find convex hull or simplex on hypersphere that enclose all vectors
Assume that we have x1,x2,...xn vectors in d-dimensional (d>=64) hypersphere, How can we find a convex hull or simplex for all of them that encloses all the vectors? can we describe that shape with limited number of vectors?</p>

答:
For any finite set of vectors $V = \{v_1, \ldots, v_n\}$, the convex hull is given by $$\mathcal H(V)=\left\{\sum_{k=1}^n a_kv_k\;\middle|\; \forall k, a_k > 0; \sum_{k=1}^n a_k = 1\right\}$$

Note that this is the entire hull, not just the boundary hypersurface. Also, a point in the hull may be given by many different coefficient sequences $(a_1, a_2, ..., a_n)$.

Added: to show that the set defined above is indeed the convex hull of a finite set.

Convex hulls $C(V)$ and the sets
$$\mathcal H(V) := \left\{\sum_{k=0}^n a_kv_k\;\middle|\; n \in \Bbb N; \forall k, a_k > 0\text{ and }v_k \in V; \sum_{k=0}^n a_k = 1\right\}$$
satisfy a similar property:

For any set of vectors $V$ and any vector $x, C(V \cup \{x\})$ is the union of all line segments connecting $x$ to elements of $C(V)$.

For any set of vectors $V$ and any vector $x, \mathcal H(V\cup \{x\})$ is the union of all line segments connecting $x$ to elements of $\mathcal H(V)$.

To prove the first, Let $\displaystyle C' = \bigcup_{w \in C(V)} \overline{xw}$. Note that $C(V \cup \{x\})$ is a convex set containing $V$, so as the intersection of all such sets, $C(V) \subseteq C(V \cup \{x\})$. And since $C(V \cup \{x\})$ is convex, it must therefore contain all line segments between $x$ and points of $C(V)$. That is, $C' \subseteq C(V\cup \{x\})$.

Conversely, given any two points $a,b \in C'$, there must be $p,q \in C(V)$ such that $a$ lies on $\overline{xp}$ and $b$ lies on $\overline{xq}$. Now $\triangle xpq$ has its base $\overline{pq}$ in $C(V)$, and every point in its interior lines on a segment from $x$ to a point on $\overline{pq}$, and is therefore in $C'$. In particular, the line segmqent $\overline{ab}$ is interior to the triangle, and so is in $C'$. Since $a, b$ where arbitrary, $C'$ is convex. And so $C(V\cup \{x\}) \subseteq C'$.

Hence $\displaystyle C(V\cup \{x\}) = C' = \bigcup_{w \in C(V)} \overline{xw}$ as required.

To prove the second, let $\displaystyle H = \bigcup_{w \in \mathcal H(V)} \overline{xw}$. Note that the elements of $\mathcal H(V\cup\{x\})$ that do not include $x$ as one of the $v_k$ in the sum are exactly the elements of $\mathcal H(V)$, so $\mathcal H(V)\subseteq \mathcal H(V\cup\{x\})$. Further, if $v = \sum_{k=0}^n a_kv_k \in \mathcal H(V\cup\{x\})$ with $v_0 = x$, $v_k \in \mathcal H(V)$ for all $k > 0$, and $a_0 \ne 1$, we can rewrite the sum as
$$v = a_0x + (1 - a_0)\sum_{k=1}^n \dfrac{a_k}{1-a_0}v_k$$
and note that
$$\sum_{k=1}^n \dfrac{a_k}{1-a_0} = \dfrac 1{1-a_0}\sum_{k=1}^n a_n = \dfrac 1{1-a_0}(1-a_0) = 1$$
since $\sum_{k=0}^n a_n = 1$. So $w = \sum_{k=1}^n \frac{a_k}{1-a_0}v_k \in \mathcal H(V)$ and $v = a_0x + (1-a_0)w$ is a point on the line segment connecting $x$ and $w$. Thus $\mathcal H(V\cup\{x\}) \subseteq H$.

Conversely, if $v\in H$, then there exists a $t \in [0,1]$ and a $w \in \mathcal H(V)$ such that $v = tx + (1-t)w$. Now by definition, $w = \sum_{k=1}^n a_kv_k$ with all $v_k \in V$ and $\sum_{k=1}^n a_k = 1$. Let $v_0 = x, b_0 = t$ and for all $k > 0, b_k = (1-t)a_k$. Then

$$\begin{align}v &= tx + (1-t)w\\&= tx + \sum_{k=1}^n (1-t)a_kv_k\\&=b_0v_0 + \sum_{k=1}^n b_kv_k=\sum_{k=0}^n b_kv_k\end{align}$$
And $$\sum_{k=0}^n b_k = b_0 + \sum_{k=1}^n b_k = t + (1-t)\sum_{k=1}^n a_k = t + (1-t)1 = 1$$
Hence $v \in \mathcal H(V\cup\{x\})$. Since $\mathcal H(V\cup\{x\}) \subseteq H$ and $H \subseteq \mathcal H(V\cup\{x\})$, the two are equal, as required.

Now given any finite set $V = \{v_0, \ldots, v_N\}$, we can set $V_n = \{v_0, \ldots v_n\}$ for all $n \le N$. Since $V_0$ consists of a single point, it is trivially convex, so $C(V_0) = V_0$. But also by the definition, $\mathcal H(V_0) = V_0$. In particular, $C(V_0) = \mathcal H(V_0)$.

But then $C(V_1)$ and $\mathcal H(V_1)$ are both the union of all line segments connecting $v_1$ to elements of $V_0$, and thus are equal as well. Continuing inductively $C(V_n) = \mathcal H(V_n)$ for every $n$.


问:
Show that there exist infinitely many positive integers $x$ such that $2a$ divide $2bx+(a^2-b^2)$
Let $a,b$ be two positive integers. Then my question is:

Show that there exist infinitely many positive integers $x$ such that $2a$ divide $2bx+(a^2-b^2)$

答: This is not true. Say $b=a-1$ then your expression is $$2bx +(a-b)(a+b) =2ax-2x+2a-1$$ so $2a$ should divide $2x+1$ which is not true since $2x+1$ is odd and $2a$ even.


Body: $$log_23\ log_34\ log_4 5\ log_5 6\ log_6 7\ log_7 8=\frac{\log 3}{\log 2}\frac{\log 4}{\log 3}\frac{\log 5}{\log 4}\frac{\log 6}{\log 5}\frac{\log 7}{\log 6}\frac{\log 8}{\log 7}=\frac{\log 8}{\log 2}=log_28=3$$