问:
The Product of Consecutive Integers is Never a Power: Lemma 1 (Research Paper Study)
P. Erdos and J. L. Selfridge proved in the  paper THE PRODUCT OF CONSECUTIVE INTEGERS IS NEVER A POWER (click here), that the equation $(n + 1) \cdots(n + k)=x^l \cdots (1)$ has no solution in integers with $k > 2, l > 2, n > 0$. There is a lemma $1$, I have $2$ issues (for better searchability for future reader, I asked in single post).

(i) Please read the following paragraph writtten on page 293 -

First observe that by the well-known theorem of Sylvester and Schur, there is always a prime greater than >$k$ which divides $(n + 1)... (n + k)$, since $n > k$. Such a prime divides only one of the $k$ factors,
so $n + k > (k + 1)^l$, whence $n > k^l \cdots (2)$.

If equation $(1)$ is true I understand  $n + k > (k + 1)^l$ since there is a prime greater than $> k$ and that prime must have power $l$. But how do we derive  $$n > k^l?$$

I mean,  $n + k > (k + 1)^l$ is independent of  $n > k^l$, then how this two inequalities are related?

(ii) On page 294 it is written that -

$$ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= \frac{kl^2}{2n-kl}$$

If geometric series is used then it should be $ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= l \cdot \frac{1}{1-\frac{kl}{2n}} = l \cdot \frac{2n}{2n-kl}$ (by multiplying $\frac{2n}{2n}$), not $\frac{kl^2}{2n-kl}$.

So questions are: why $n > k^l$ and $ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= \frac{kl^2}{2n-kl}$?

答:
(i) is evident by an application of binomial theorem. Noting that $l > 2$:

$$n+k > (k+1)^l =\sum_{j=0}^l \binom ljk^j > \binom ll k^l+\binom l1 k^1=k^l+lk$$

Hence $n > k^l+(l-1)k > k^l$.

Now for (ii), it is indeed by sum of geometric series.

However the first term is not $1$, but $\dfrac {kl}{2n}$. This gives

$$ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= l \cdot \frac{\frac{kl}{2n}}{1-\frac{kl}{2n}} = l \cdot \frac{kl}{2n-kl}=\frac{kl^2}{2n-kl}$$

as expected.


问:
Sketching a function $f(z) = (1-i)z + 3 $ on the complex plane. Am I doing it right?
So I have no problem skeching a set defined on the complex plane. But I don't know how to sketch a function. In my example I have: $f(z) = (1-i)z + 3 $ .

I know that multiplying a complex number by 1-i increases its magnitude by $|1-i|=\sqrt{2} $ and rotates it by π/4 (the argument is increased by π/4). Then adding a 3 should "push" the function 3 points to the left side.

Ultimatelly I should have a function of f(z) rotated by π/4 and moved 3 points to the left side. Am I right? Or is there an error in my way of thinking?

答: The function $f(z)=(1+i)z+3$ rotates $-\frac{\pi}{4}$ and shifts $(3,0)$ to the right values $z\in\mathbb{C}$

ContentLicense: CC BY-SA 4.0


问:
Deriving Marginal Distribution from Two Independent Beta Distributions
I have two independent random variables:

$X$ is Beta($a, b$), $x \in [0, 1]$

$Y$ is Beta($a+b, \gamma$), $y \in [0, 1]$

$$f(x,y) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1}(1-x)^{b-1}\frac{\Gamma(a+b+\gamma)}{\Gamma(a+b)\Gamma(\gamma)}y^{a+b-1} (1-y)^{\gamma-1}$$

I want to perform the transformations for $U = XY$, $V=Y$.
$$U=XY \rightarrow U=XY \rightarrow X=U/V, \qquad u \in [0,1]$$
$$V=Y \rightarrow Y=V, \qquad v\in[0,1]$$

$$\det{J}= \frac{1}{v}*1-0=\frac{1}{v}$$
So I have,

$$f(u, v) = \frac{\Gamma(a+b+\gamma)}{\Gamma(a+b)\Gamma(\gamma)} (u/v)^{a-1}(1-(u/v))^{b-1}v^{a+b-1} (1-v)^{\gamma-1} * \frac{1}{v}, \qquad 0 \le u \le v \le 1$$

I want to find the marginal distribution of $u$.

$$f_{U}(u) = \frac{\Gamma(a+b+\gamma)}{\Gamma(a)\Gamma(b)\Gamma(\gamma)} u^{a-1} \int^1_{u} (1-(u/v))^{b-1}v^{a+b-1} (1-v)^{\gamma-1} \frac{1}{v} dv$$

QUESTION: However, I am unsure how to go about this integral. How can I proceed with this integration find the marginal distribution for $U$?

答:
Suppose $X\sim \text{Beta}(a,b)$ and $Y\sim \text{Beta}(c,d)$ are independently distributed with $c=a+b$.

Then $(X,Y)$ has density $$f_{X,Y}(x,y)\propto x^{a-1}(1-x)^{b-1}y^{c-1}(1-y)^{d-1}\mathbf1_{0<x,y<1}\,,$$ where  $a,b,c,d>0$.

Transforming $(X,Y)\mapsto (U,V)=(XY,Y)$, density of $(U,V)$ is

\begin{align}
f_{U,V}(u,v)&=f_{X,Y}\left(\frac{u}{v},v\right)\cdot\frac1v
\\&\propto u^{a-1}(v-u)^{b-1}(1-v)^{d-1}\mathbf1_{0<u<v<1}
\end{align}

Pdf of $U$ is therefore

$$f_U(u)\propto u^{a-1}\int_u^1 (v-u)^{b-1}(1-v)^{d-1}\,dv\,\mathbf1_{0<u<1}$$

Now you can turn this integral into a Beta function if you change variables $v\mapsto t$ such that $t=0$ when $v=u$ and $t=1$ when $v=1$. This yields the relation $v-u=t(1-u)$.

On simplification you have another Beta distribution:

$$f_U(u)\propto u^{a-1}(1-u)^{b+d-1}\,\mathbf1_{0<u<1}$$

I have ignored the normalizing constants here since these are all guaranteed to be valid densities. You can of course verify that you end up with the correct constant.


问:
Relationship between $_nC_2$ and $\sum_{k=1}^{n-1} k$
$$_nC_2 = \frac{n(n-1)}{2}$$

and so is the following sum: $$\sum_{k=1}^{n-1} k = \frac{n(n-1)}{2}$$

I'm curious if there's any ingenious alternative interpretation of the combination $_nC_2$ in terms of the sum of numbers.

Body:
Note that $$\sum_{k=m}^{n}{k \choose m}={n+1 \choose n+1}$$
$$\sum_{k=1}^{n} k=\frac{1}{2} n(n+1)$$
Alternatively
$$\sum_{k=1}^{n}{k \choose 1}={n+1 \choose 2}=\frac{1}{2}n(n+1)$$


问:
Why would a natural isomorphism $A \cong TA \oplus (A / TA)$ imply that $A \twoheadrightarrow A/TA \rightarrowtail TA \oplus (A / TA)$ is natural?
$\newcommand{\abcat}{\text{Ab}_\text{fg}}$
$\newcommand{\tgroup}{TA \oplus (A/TA)}$
$\newcommand{\epi}{\twoheadrightarrow}$
$\newcommand{\mono}{\rightarrowtail}$
Let $A$ be an object in the category $\abcat$ of finitely generated abelian groups. And let $TA$ denote its torsion subgroup.

In Category Theory in Context the author proves that the isomorphisms $A \cong \tgroup$ are not natural in proposition 1.4.4.

The proof starts out by stating the following (note, $\epi$ denotes an epimorphism, and $\mono$ denotes a monomorphism)

Suppose the isomorphisms $A \cong \tgroup$ were natural in $A$. Then the composite

$$A \epi A/TA \mono \tgroup \cong A$$

of the canonical quotient map, the inclusion into the direct sum, and the hypothesized natural isomorphism would define a natural endomorphism of the identity functor on $\abcat$

My main question is simple: "why?".

But I have a few confusions which may relate to why I am having trouble with the main question.

Confusion 1
A natural transformation is between functors $F, G : C \rightrightarrows D$, so what exactly are the functors in the proposed natural isomorphism? I am guessing that one of the functors $F$ is the identity functor on $\abcat$. Then perhaps the other functor $G$ is an endomorphic functor on $\abcat$  where the action on an object of $\abcat$ is $$A \mapsto \tgroup$$ but then how are the morphisms mapped by the functor? Earlier the author does state,

In practice, it is usually most elegant to define a natural transformation by saying that the arrows $X$ are natural, which means that the collection of arrows defines the components of a natural transformation, leaving implicit the correct choices of domain and codomain functors, and source and target categories.

But as I am new to this, I am not sure how these "correct" choices are "implicit". It may also be the case that the author is proving that there are no functors which have a natural isomorphism between them and also map the objects of $\abcat$ in the way described above. Is that what is going on here?

Confusion 2 If the functor $G$ really does map the objects $A$ to the objects $\tgroup$ then I fail to see how the isomorphism $A \cong \tgroup$ has anything to do with a natural epimorphism $A \epi A/TA$ or a natural monomorphism $A /TA \mono \tgroup$ from a categorical perspective. I have a foggy feeling that it is indeed true intuitively. However, I don't understand how a natural transformation could imply this using only category theory formally. (Unless I missed it, the author has not defined $\oplus$ using category theory yet, would this be necessary?)

I am assuming by the way this first part of the proof was worded and also by the previous proofs and examples in the book that we only need a simple categorical argument here, and that neither a group theoretical argument (appealing to the elements of the group) or an extremely complicated categorical argument is necessary. But I fail to see what the argument is.

答:
Let $\mathcal{A}$ be the category of finitely generated abelian groups. The mapping $A \mapsto TA \oplus (A/TA)$ extends to an endofunctor $F: \mathcal{A} \to \mathcal{A}$ as follows: take a morphism $f: A \to B$ of finitely generated abelian groups. Construct the morphism $Ff: TA \oplus (A/TA) \to TB \oplus (B/TB)$ as follows:

$f$ maps torsion elements to torsion elements (if $n \cdot a = 0$, then $n \cdot f(a) = f(n \cdot a) = 0$, so $n \cdot a$ is torsion), so $f$ induces a morphism $f|_{TA}: TA \to TB$ by restriction;

There is a morphism $g: A/TA \to B/TB$ defined by putting $g(a + TA) = f(a) + TB$. This well-defined: if $a + TA = a' + TA$, then $a - a'$ is in $TA$ and $f(a - a') = f(a) - f(a')$ is in $TB$ by the previous observation, so
$$g(a + TA) = f(a) + TB = f(a') + TB = g(a' + TA)
$$
and $g$ is well-defined. Notice that $g$ is also a morphism of groups because $f$ is;

now we put $Ff = f|_{TA} \oplus g$. That is, an element $(a, a' + TA)$ of $TA \oplus (A/TA)$ gets mapped to $(f(a), f(a') + TB)$ in $TB \oplus (B/TB)$ by $Ff$.

Now it is quite easy to prove that $F$ is indeed a functor. What I did might look complicated, but it is indeed quite tautological: “taking torsion” is a functor $T: \mathcal{A} \to \mathcal{A}$, “modding out by torsion” is a functor $(-)/T(-): \mathcal{A} \to \mathcal{A}$ and “taking direct sums” is a functor $\oplus: \mathcal{A} \times \mathcal{A} \to \mathcal{A}$: by appropriately assembling those three functors you get $F$.

Then saying that there are natural isomorphisms $A \cong TA \oplus (A/TA)$ means that there is a natural isomorphism $\eta: F \Rightarrow \mathsf{id}_{\mathcal{A}}$ between $F$ and the identity endofunctor (think about it: this consists of a family of isomorphisms $\eta_A: TA \oplus (A/TA) \cong A$ for each object $A$ of $\mathcal{A}$).

Now there is a natural transformation $\theta: \mathsf{id}_{\mathcal{A}} \Rightarrow F$ such that for a finitely generated abelian group $A$, the morphism $\theta_A: A \to TA \oplus (A/TA)$ is the composition $A \to A/TA \to TA \oplus (A/TA)$ (prove it). If $\eta$ as above exists, you can consider the composition $\eta \circ \theta: \mathsf{id}_{\mathcal{A}} \Rightarrow F \Rightarrow \mathsf{id}_{\mathcal{A}}$, which is explicitly the composition $A \to A/TA \to TA\oplus A/TA \to A$ (where the last morphism is $\eta_A$). Since $\eta$ is supposed natural (by contradiction) and $\theta$ is natural, the composite $\eta \circ \theta$ must be a natural endomorphism of the identity functor, and this is exactly what the author means by “the hypothesized natural isomorphism would define a natural endomorphism of the identity functor”.


问:
Solving a system of separable ODE's
I am attempting to solve the system of first-order ODE's:

$$ \frac{dy_1(x)}{dx} = g(y_1)\frac{y_2(x)}{\int_{a}^b y_2(z) dz} $$
$$ \frac{dy_2(x)}{dx} = \frac{y_2(x)}{f(y_1(x),x)} \frac{\partial f(y_1(x),x)}{\partial x} $$

with boundary conditions $y_1(a) = c$, $y_1(b) = d$ for constants $a,b,c,d$.

My problem is conceptual, and very likely silly: I'm not sure whether I can break apart this problem apart in order to solve it. I would like to exploit separability to obtain a general solution for the second equation:

$$ y_2(x) = e^N f(y_1(x),x) $$

with N a constant, and use this to reduce the first equation to

$$ \frac{dy_1(x)}{dx} = g(y_1)\frac{f(y_1(x),x)}{\int_{a}^b f(y_1(z),z) dz} $$

which I would then solve numerically, presumably by treating the integral as a constant to be determined. I'm not seeing the problem with this approach, but on the other hand it strikes me as "too good to be true." My questions:

Am I mistaken in thinking that $ y_2(x) = e^N f(y_1(x),x) $ is in fact the general solution for $y_2$, irrespective of the dependence of $y_1$ on $y_2$?

When numerically solving a differential equation (say IVP) of the form $y_1'(x) = h\left(y_1,x,\int_{a}^b f(y_1(z),z) dz\right)$ with $a,b$ constant, is it problematic to treat the integral as a constant $A$, use an initial guess for $A$ + e.g Euler's method to solve for $y_1$, and then iterating until the guess and actual value of $A$ converge?

Any input would be much appreciated.

答:
Your system does not define $y_2(x)$ uniquely. If $(y_1(x), y_2(x))$ is a solution then also $(y_1(x), Cy_2(x))$ where $C \neq 0$ is a constant also is a solution.

Thus you may impose an extra condition on $y_2(x)$, the simplest to me is
$$
\int_a^b y_2(x) dx = 1.
$$
After that the system reduces to
$$
y_1'(x) = g(y_1(x)) y_2(x)\\
y_2'(x) = y_2(x) F(x, y_1(x))
$$
with conditions $y_1(a) = c, y_1(b) = d, \int_a^b y_2(x) dx = 1$.
Here I've denoted $F(x, y) = \frac{f_x(y, x)}{f(y, x)}$.

Having a system of two first-order ODE's with three conditions makes me think that the problem does not have a solution in general (but may have for some particular values of $c$ or $d$, like an eigenproblem).

The general solution to the second equation is
$$
y_2(x) = C_1 \exp\left(\int_a^x F(s, y_1(s)) ds\right)
$$
with $C_1$ determined by $\int_a^b y_2(x) dx = 1$ condition. Practically it is easier to put $y_2(a) = C_1 = 1$ and normalize $y_2(x)$ after integration.

Summarizing, the following algorithm may solve the problem:

Take some initial $y_1(x)$, e.g. linear approximation $y_1(x) = c + \frac{x-a}{b-a} (d - c)$.

Having $y_1(x)$ compute the ${\tilde y}_2(x)$ from the following Cauchy problem
$$
\begin{cases}
\tilde y_2'(x) = \tilde y_2(x) F(x, y_1(x)), &\quad x \in [a, b]\\
I'(x) = \tilde y_2(x), &\quad x \in [a, b]\\
\tilde y_2(a) = 1\\
I(a) = 0
\end{cases}
$$
Here $I(x) = \int_a^x y_2(s) ds$, so $I(b) = \int_a^b y_2(x) dx$.

Compute $y_2(x)$ by $$y_2(x) = \frac{\tilde y_2(x)}{I(b)}.$$

Plug obtained $y_2(x)$ into the first equation and solve the new Cauchy problem
$$
\begin{cases}
\tilde y_1'(x) = g(\tilde y_1(x)) y_2(x)\\
\tilde y_1(a) = c\\
\end{cases}
$$

Now, for sure, $\tilde y_1(b) \neq d$, so we need to correct $y_1(x)$ and start from the item 2. And this is quite tricky. As I said, the problem almost certainly won't have a solution. If it was a boundary problem with three conditions and the functions we might have use the residual $\delta = d - \tilde y_1(b)$ to correct some free parameter (see shooting method), but there is no such parameter in this problem, at least as it was stated. You may try combining new approximation $y_1(x)$ as
$$y_1(x) := (1-\omega) y_1(x) + \omega \left[\tilde y_1(x) + \frac{x-a}{b-a}\delta\right]$$
and try to deduce how $\delta$ behaves in iterations depending on different $\omega$ values. The extra term in brackets is needed to correct the right boundary conditions in new $y_1(x)$ before restarting from item 2.

Title: Solving a system of separable ODE's


问:
Infinite series proof
I need to prove this statement using the limit definition.:

$ a_n$ converges $\Rightarrow a_{n+1} converges$  to the same limit.

I started with $\epsilon>0$ $\exists n_0$ $\forall n>n_0$, $|a_n -L|<\epsilon$

What is the right way to finish this proof?

Thank you.

答: Take $N_0=n_0-1$. Then\begin{align}n>N_0&\iff n>n_0-1\\&\iff n+1>n_0\\&\implies|a_{n+1}-L|<\varepsilon.\end{align}

Title: Infinite series proof


问:
Vertices in a graph with the same number of closed walks
A graph is called walk regular if the number of closed walks starting from vertex $u$ of length $k$ does not depend on $u$. If $A$ is the adjacency matrix of the graph, this means that $A^k$ has equal diagonal values for every $k>0$.

I'm interested in a weaker condition, that there exists two vertices $u,v$ such that the number of closed walks of length $k$ is the same for $u$ and $v$. Are there results or references about this?

In other words, when does $[A^k]_{uu} = [A^k]_{vv}$ hold?

For example, it would be nice to have necessary/sufficient conditions for this property to hold. If there is a graph automorphism that maps $u\to v$ then certainly $u$ and $v$ have the same number of closed walks, though the existence of such an automorphism is not necessary (an example is the Folkman graph which is walk-regular but not vertex-transitive).

I'm aware that if the graph has $n$ vertices and $u,v$ have the same number of closed walks for $k=1,2,\ldots n-1$, then by Hamilton-Cayley's theorem it is true also for all $k\geq n$.

I'm asking in the case of a simple undirected graph, but if there are generalizations for directed and/or weighted graphs they're highly welcome.

答:
Continuing my search through various articles, I found the answer: two such vertices are called cospectral vertices.

There is a result, that for an undirected graph $G$ with adjacency matrix $A$, the following are equivalent to $u,v$ cospectral:

For every integer $k>0$, $[A^k]_{u,u}=[A^k]_{v,v}$.

The graphs obtained removing a vertex, $G\setminus \{u\}$ and $G\setminus \{v\}$ have the same spectrum

For every $\lambda$ eigenvalue of $A$, let $P$ be the spectral projector on the $\lambda$-Eigenspace of $A$ (since $A$ is symmetric, $P$ is orthogonal). The condition is then $P_{u,u} = P_{v,v}$ for every spectral projector.

Equivalently, the third condition can be reformulated through graph angles.

References:

Chris Godsil, Jamie Smith, "Strongly Cospectral Vertices" , on arXiv https://arxiv.org/abs/1709.07975

In this articles more equivalent conditions are given, and is studied a generalization of the concept.

Chris  Godsil, Brendan D. McKay, "Feasibility Conditions for the existence of Walk-Regular Graphs" , Linear Algebra Appl., Vol. 30 (1980), pp. 51-61

This paper proves properties for walk-regular graphs, but its results can be applied also to cospectral vertices.

Dragoš Cvetković, Peter Rowlinson, Slobodan Simić, "Eigenspaces of Graphs", Cambridge University Press (1997)

A very good book, where can be found results concerning graph angles.

Allen J. Schwenk,"Almost all trees are cospectral" , New Directions in the Theory of Graphs, Academic Press  (1973), pp. 275–307

I think this was the first paper to use cospectral vertices (although as in the above condition 2, so not explicitly).


问:
Show $(ab)^{1/n}=a^{1/n} b^{1/n}$ for any $a,b >0$ and $n \in \mathbb N$
So, what I have in hand is the definition of the nth root of a real number, that is $y=x^{1/n}$ is the number that satisfies $y^n=x$. And I also have $(ab)^n=a^nb^n$ for $n\in \mathbb N$, $a,b >0$.

How can I go from here to $(ab)^{1/n}=a^{1/n} b^{1/n}$?

I tried to write things like $ab=((ab)^n)^{1/n}$ using the definition. But I can only get $((ab)^n)^{1/n}=(a^n)^{1/n}(b^n)^{1/n}$ and I am stuck.

答:
Given $a\gt 0$, consider $f(x)=x^n-a$. If it has at least two distinct positive real roots, we may denote two of them by $\alpha$ and $\alpha'$. By Rolle's theorem, there is a number between the two roots, say $\beta$, such that the derivative $f'(\beta)=0$, in other words $n\beta^{n-1}=0$, a contradiction.

We take $a^{1/n}$ as the unique positive real root of $x^n-a=0$, $b^{1/n}$ that of $y^n-b=0$, and $(ab)^{1/n}$ that of $z^n-ab=0$. Then $z^n=ab=x^ny^n=(xy)^n$ and thus by the uniqueness of the root we have $(ab)^{1/n}=a^{1/n}b^{1/n}$.


问:
Average of angles
For example, when calculating an average of angle $\pi$ and $-\pi$,
just adding them up and divide with 2 is 0, but since they are the same angle, shouldn't it be $\pi$ or $-\pi$? (180 deg)

Do I always have to use this fomula when adding, subtracting, averaging an angle?

Guys, why everyone is commenting about the equality of plus and minus?

Using the average method in the link, the average of angle $\pi$ and $-\pi$ is $\pi$!

I want to know what is a correct way of adding, subtracting, averaging an angle

答:
You should think of angles as representing travel around the unit circle, starting at $(1,0)$ on the $x$-axis. Then $\pi$ is a counterclockwise rotation half way around and $-\pi$ is a clockwise rotation by the same amount. In this sense they do in fact average to $0$, even though both rotations end at the same point $(-1,0)$.

Edit in response to comment.

The formula in the link calculates (in one sense) the average of the final positions of the rotations corresponding to two angles.

Whether you want the "average angle" to mean
"average amount of rotation" or "average final position" depends on how you plan to use the average in an application. The link points out that on a clock, so for things like time of day, you want the average of the final position.

So the answer to this part of the question:

Do I always have to use this formula when adding, subtracting, averaging an angle?

is no. You use this method sometimes.

Title: Average of angles


问:
Simple summation algebra question: summation of a summation
What does $$\sum_{i=1}^{n}\sum_{i=1}^{n}x_{i}$$ yield?

答:
The problem with $\sum_{i=1}^n \sum_{i=1}^n x_i$ is that the $i$ in $x_i$ could refer to the index of the inner sum or the index of the outer sum. In this case, it wouldn't matter since both have the same range, but more generally $\sum_{i=a}^b\sum_{i=c}^d x_i$ would have the problem of not knowing if $i$ should vary from $a$ to $b$ or from $c$ to $d$. That's why you'll never see such thing in a good book or a serious paper.

If you are interested in knowing what "rule" it violates read about well-formed formulas and free variables.


问:
If $V$ is a finite dimensional vector space, if $N(T) \cap R(T) = \{0\}$, then is it true that $V = N(T)\oplus R(T)$?
This question is already asked before in this site, but they say this is yes and also they have proved it.

But I doubt,this is even true or not.

Take , $T:\mathbb{C}^{2} \to \mathbb{C}^{2} $ such that $T(x,y)=(x+iy,x+iy)$

Clearly , $N(T)=span\{(-i,1)\}$

And, $R(T)=span\{(1,1)\}$

$N(T)\cap R(T)= \{(0,0)\}$

But $N(T) + R(T) \neq \mathbb{C}^{2}$

So, am I correct?

Or, I missed something.

Please help me.

答:
No, you are not correct, since $N(T)\oplus R(T)=\Bbb C^2$. That's so because $\dim\Bbb C^2=2$, and therefore, since the set $\{(-i,1),(1,1)\}$ is linearly independent, it is a basis of $\Bbb C^2$. So, every element of $\Bbb C^2$ is a linear combination of $(-i,1)$ and $(1,1)$.

Or you can show it directly: if $(x,y)\in\Bbb C^2$, then there are complex numbers $\alpha$ and $\beta$ such that $\alpha(-i,1)+\beta(1,1)=(x,y)$; just take$$\alpha=\left(-\frac12+\frac i2\right)(x-y)\quad\text{and}\quad\beta=\left(\frac12+\frac i2\right)(-ix+y).$$


问:
Normal subgroups of a group of order 12
Let $G$ be a group of order 12 and suppose $G$ does not have normal subgroups of order $4$. How to show that $G$ has a subgroup of order $6$?</p>

答:
By Sylow's Theorem, the number of Sylow 2-subgroups $n_2 = |G:N_G(P)| \equiv 1 \pmod 2$ where $P$ is any Sylow 2-subgroup of G and NG denotes the normalizer. Since $P$ cannot be normal in $G$, $|G:N_G(P)|\neq 1$ and must be $3$ (the only other odd divisor of $12$). Assuming the Sylow 3-subgroup is not normal either, results to a number of $4$ subgroups of order $4$ by the same argument. This would add up to one neutral element, 8 elements of order $3$, $3$ elements of one Sylow 2-subgroup and at least another element for another Sylow 2-subgroup,totaling at least 13 group elements of $G$.

Hence $G$ has a normal subgroup $N$ of order $3$. $G$ also has a subgroup $H$ of order $2$ contributed to Cauchy (also part of Sylow's Theorem). Now since $N$ is normal in $G$ and $N\cap H = 1$, we have a semidirect product $NH \le G$ of order $6$.

Added after Steve D's comment:

Let $N\triangleleft G$ with $|N|=3$ and $N = \langle\sigma\rangle$. Let $H\le G$ be a subgroup of order $4$. Assume first that $H$ is cyclic and $\tau\in H$ of order $4$. Then $\sigma^\tau \in \{\sigma, \sigma^2\}$, hence $\sigma^{\tau^2}= \sigma$ for the involution $\tau^2$. If $H$ on the other hand is isomorphic to Klein's four-group, and two element $\tau,\tau'\in H$ of order $2$ both map $\sigma^\tau = \sigma^{\tau'} = \sigma^2$ its product, the third involution $\tau\tau'$ maps $\sigma^{\tau\tau'} = \sigma$.

In both cases there is an element $\rho \in H\le G$ of order $2$ that commutes with $\sigma$, hence the semidirect product is direct and cyclic of order $6$.

ContentLicense: CC BY-SA 3.0

Title: Normal subgroups of a group of order 12


问:
Vector valued functions
What is a vector valued function ? Vectors are just $n-tuples$ , and then how are we able to describe geometrical shapes using them ?

Is a vector function such that it outputs $n-tuples$ using other $n-$tuples and complex numbers and then you get geometrical shapes according to how you interpret these $n-tuples$ , for example ,input vectors are shifts and final output vectors and position vectors . This is only pure interpretation , if you interchanged this interpretation you won't be getting the same shape as before and in the end , it is just $n-tuples$ we are getting , and then shape is how we interpret them and nothing else ?

And if it is true , why don't books mention this ? That shapes are determined according to how we are interpreting certain $n-tuples$ in our space and we have these specific interpretations defined for this geometric space, like from now on we will interpret input vectors are shifts and final output vectors and position vectors/points and this set of points will be our plane but vectors have no role in this , it is just our interpretation in this particular case .

答: A vector-valued function is one where the result is a vector.  In the finite-dimensional case, this is an $n$-tuple.  The domain might be vectors of the same size, or of a different size, or a scalar.  In the first case, one can interpret the function as a vector field.  In the last case, one can interpret the function as a curve in space.  All of these have geometric interpretations and are well-studied.</p>

Title: Vector valued functions


问:
Deriving integral expression with multiple components
I have trouble deriving the following equation with respect to t (that is, $\frac{d}{dt}$): $$ V_{\nu} (t) = \int_{t}^{\infty } e^{-\int_{t}^{s} r(s') ds'} \pi_{\nu}(s) ds$$ (I know basic derivative rules but am not sure how to handle the Euler-function part of the equation). Would be very happy about help! Thank you in advance!

The function is an economic function describing the profit discounted over time (the integral expression). r is the rental rate of capital and s is just some time index other than t. s' is not any derivative and is just independent from s!

答:
Let $V_\nu(t)=\int_t^\infty e^{-\int_t^s r(s')\,ds'}\pi_\nu(s)\,ds$.  Then, we have from Leibniz's Rule

$$\begin{align}
\frac{d}{dt}V_\nu(t)&=\frac{d}{dt}\int_t^\infty e^{-\int_t^s r(s')\,ds'}\pi_\nu(s)\,ds\\\\
&=-\left.\left(e^{-\int_t^s r(s')\,ds'}\pi_\nu(s)\right)\right|_{s=t}+\int_t^\infty \frac{\partial}{\partial t}\left(e^{-\int_t^s r(s')\,ds'}\,\pi_\nu(s)\right)\,ds\\\\
&=\pi_\nu(t)+\int_t^\infty \left(r(t)e^{-\int_t^s r(s')\,ds'}\right)\pi_\nu(s)\,ds\\\\
&=\pi_\nu(t)+r(t)V_\nu(t)
\end{align}$$


问:
MLE for mean of a normal distribution given the range of the mean
If $X \sim \mathcal{N} (\mu,1)$ and $\mu \in (-m,m)$ where $0<m<1$ is a known constant. Then how can I work out the MLE of $\mu$ in this case?

I know in the case where $\mu$ is completely unknown, then the MLE $\hat{\mu} (X)$ is just $X$. However, I really need some insights into how to take into account the given range of $\mu$. Many thanks for any help.

答: The likelihood of $\mu$ outside the range you specified is 0. Accordingly, the MLE is the same as in the usual case except truncated to the range you've specified.


问:
Compute the remainder when $67!$ is divided by $71$.
This is how far I've been able to get.

By using Wilson's Theorem:

$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$

EDIT: Here is how I proceeded using TMM's and  Carl Mummert's hints.

$$\begin{align}
&68 \equiv -3 \pmod{71}\\
and\\
&69 \equiv -2 \pmod{71}\\
\end{align}$$

So:
$$\begin{align}
67!(-3)(-2) &\equiv 1 \pmod{71} \\
67! &\equiv 6^{-1} \pmod{71} \\
\end{align}$$

By using the Euclidean algorithm:
$$71 = 6 \cdot 11 + 5$$
$$6 = 5 \cdot 1 + 1$$
$$5 = 1 \cdot 5 + 0$$

Now, going backwards:
$$\begin{align}
1 &= 6 - 5 \\
&= 6 - (71 - 6 \cdot 11) \\
&= 6 + 6(11) - 71 \\
&= 6(1 + 11) - 71(1) \\
&= 6(12) - 71(1)
\end{align}$$

Therefore, $67! \equiv 6^{-1} \equiv 12 \pmod{71}$.

答:
Hint: $$69 \equiv -2 \pmod{71}, \qquad 68 \equiv -3\pmod{71}.$$
Further hint: $72 = 6 \cdot 12$, so the inverse you need to compute should not be too difficult...


问:
If $\gcd(a, b) = p$, what are possible values of $\gcd(a^3, b^2)$?

If $\gcd(a, b) = p$, where $p$ is a prime. Then what are possible values of $\gcd(a^3, b^2)$?

I have already calculated that $\gcd(a^2, b^2) = p^2$. $\gcd(a^2, b) = p$ or $p^2$, $\gcd(a^3, b) = p$ or $p^2$ or $p^3$. Can anybody give me any idea? how will I find the possible value of $\gcd(a^3, b^2)$?

答:
Since $p$ is prime, each of $a$ and $b$ contain at least one factor of $p$, and one of them contains only one.

Case 1: If $b$ contains only one factor of $p$ regardless of how many factors of $p$ are contained in $a$, then $a^3$ contains at least $3$ factors, and $b^2$ contains only two factors of $p$, so $\gcd (a^3,b^2)=p^2$.

Case 2: If $a$ contains only one factor of $p$ and $b$ contains more than one factor of $p$, then $a^3$ contains $3$ factors of $p$, and $b^2$ contains at least $4$ factors of $p$, so $\gcd (a^3,b^2)=p^3$.


问:
If E is measurable, inner and outer measure could differ?
I got stuck in this question and I can't find a counterexample.  I proved that this is true:

If ${|E|_e}<\infty\Rightarrow$ E is measurable $\Leftrightarrow{|E|_i}={|E|_e}$. Where ${|E|_i}$ and ${|E|_e}$ stand risp. for inner and outer measure.
Well, I cannot find a counterexample in the case when E has infinite outer measure. My book says it is not true but nothing comes to my mind. Everything I think of, has some closed subset with infinite measure so the equivalence still holds. To find one, I should find a set whose closed subsets have all finite measure or such that they are not measurable,right? Any ideas or suggestions?

P.s for those who do not know, the inner measure of E is defined by ${|E|_i}$= sup|F| where F$\subset$E is closed.

答:
Here is a simple counter-exemple: a non-measurable set $E$ such that $|E|_i=|E|_e=+\infty$

Let $A$ be a non-measurable subset of $[0,1]$.  Take $E=(-\infty,-1] \cup A$. Clearly $E$ is non-measurable. We also have that
$$ |E|_i \geqslant |(-\infty,-1] |_i =+\infty$$
and
$$ |E|_e \geqslant |(-\infty,-1] |_e =+\infty$$
So, we have $|E|_i=|E|_e=+\infty$.


问:
Linear shooting method for Second Order BVP
How can we use the linear shooting method to solve the boundary value problem $$y'' = 2y' - y, ~y(0) = 1, \text{and} ~~y(1) = 2?$$

I tried to convert it to a first order system, but didn't get what I needed.

答:
Hints:

We are given:

$$\tag 1 \displaystyle y'' = 2y' - y, ~y(0) = 1, \text{and} ~~y(1) = 2$$

We know that if the linear boundary-value problem:

$y'' = p(x)y' + q(x)y+r(x), a \le x \le b, y(a) = \alpha, y(b) = \beta$, satisfies:

(i) $p(x), q(x)$, and $r(x)$ are continuous on $[a, b]$

(ii) $q(x) \gt 0$ on $[a, b].$

We can get a unique solution.

For this problem, we have:

$y_1(x): y'' = p(x) y' + q(x) y + r(x)$, and

$y_2(x): y'' = p(x) y' + q(x) y$

So,

$p(x) = 2$

$q(x) = -1$

$r(x) = 0$

With the same BCs given in $(1)$.

For comparison purposes, $(1)$ has the exact solution:

$$\displaystyle y(x) = e^{x-1} \left(-e x+2 x+e\right)$$

Now, you just have to apply all of the above to the Linear Shooting Algorithm.

Note, did they provide an "N" for how many steps they wanted because $\displaystyle h = \frac{b-a}{N}$?

Also note, this algorithm is pretty involved and it is easier to code it up rather than manually cranking values.


答:
As mentioned in the comments we will allow $\mu \in [-m,m]$ rather than $(-m,m)$. The likelihood function is
$$L(\mu)=\frac{1}{2\pi}\exp(-\frac{(x-\mu)^2}{2}),$$
so maximizing the likelihood function is equivalent to minimizing the distance
$$|x-\mu|.$$
If $x> m$, then the distance is minimized by $\hat{\mu} = m$ and if $x< -m$, then the distance is minimized by $\hat{\mu} = -m$, thus the maximum likelihood estimator is given by:
$$\hat{\mu} := \begin{cases} -m & X < -m \\ X & X \in [-m,m] \\ m & X > m \end{cases}$$

Remark: If we only allow $\mu \in (-m,m)$, then the MLE is technically undefined for $X \in (-\infty,-m] \cup [m,\infty)$, that is why we should consider $\mu \in [-m,m]$ instead.

Title: MLE for mean of a normal distribution given the range of the mean

Body:
If $X \sim \mathcal{N} (\mu,1)$ and $\mu \in (-m,m)$ where $0<m<1$ is a known constant. Then how can I work out the MLE of $\mu$ in this case?

I know in the case where $\mu$ is completely unknown, then the MLE $\hat{\mu} (X)$ is just $X$. However, I really need some insights into how to take into account the given range of $\mu$. Many thanks for any help.


问:
Evaluate $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}$
$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}$$

I got asked this question in a group, and solved it the following way:

$$\lim\limits_{n\to\infty}\left(
\int\limits_1^{n+1}\frac{x}{x^2+n^2}\space dx\leq \sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}\leq \int\limits_1^n \frac{x}{x^2+n^2}\space dx + f(1)\right)
$$

$$\frac{\ln2}{2}\leq \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}\leq\frac{\ln2}{2}$$

Therefore:

$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2} = \frac{\ln2}{2}$$

The guy who asked me the question said that the answer key said $\frac{\ln\left(\frac{5}{2}\right)}{2}$ but also said that it might be incorrect, so I don't have the answer.

What I want to ask is that how come this sum is not equal to $0$? We were able to use the squeeze theorem because the limit for $f(1)$ goes to zero. Every other value after $f(1)$ also goes to $0$ (even faster?) and that had me thinking about how this summation can be equal to such a value.

Can anyone please explain how this happens? And it would be great if you could also verify or correct my solution.

答:
If you write
$$\frac k {k^2+n^2}=\frac k {(k+i n)(k-in)}=\frac i 2 \left(\frac 1 {k+i n}-\frac 1 {k-i n} \right)$$
$$S_n=\sum_{k=1}^n\frac k {k^2+n^2}=\frac{1}{2} \left(H_{(1-i) n}+H_{(1+i) n}-H_{-i n}-H_{i n}\right)$$ Using asymptotics
$$H_p=\gamma+\log (p)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$
$$S_n=\frac{\log (2)}{2}+\frac{1}{4 n}-\frac{1}{12
   n^2}+O\left(\frac{1}{n^4}\right)$$

Edit

After @achille hui's comment, I consider the most general case of
$$\Sigma=\sum_{k=a n +b}^{c n+d}\frac k {k^2+n^2}$$ Using the same method, I obtained as asymptotics
$$\Sigma=\frac{1}{2} \log \left(\frac{c^2+1}{a^2+1}\right)+\frac{2 \left(a^2+1\right) c d-2 a b \left(c^2+1\right)+(a+c) (a c+1)}{2\left(a^2+1\right) \left(c^2+1\right) n}+O\left(\frac{1}{n^2}\right)$$ and then the limit only depends on $a$ and $c$ (nothing to do with $b$ and $d$ which only define how is approached the limit).


问:
How to transpose eigendecomposition?
When reading Gilbert Strang's Introduction to Linear Algebra (chapter 6.4) I noticed the following transpose of an eigendecomposition:

For eigendecomposition $A=S\Lambda S^{-1}$ $(S\Lambda S^{-1})^{T}=(S^{-1})^T\Lambda S^T$.

Could somebody please explain how is the tranpose above justified, since I would expect it to be $(S\Lambda S^{-1})^{T}=S^T\Lambda (S^{-1})^T$ instead, following that $(AB)^T=A^TB^T$?

答: For transpose we have $(AB)^T=B^TA^T$ and not the other way around.

Title: How to transpose eigendecomposition?


问:
n*2 mod p is not always equal to 0 mod 2?
Edit: as a first-timer here, I sincerely do not understand the downvotes to this legitimate question.

With regular integers in Z, n*2 is always equal to 0 mod 2.

In a finite field Zp of order P, n*2 is equal to 0 mod 2 only if n < (P+1)/2 and 1 mod 2 otherwise.

Is it possible to multiply n by another integer in order to obtain a number equal to 0 mod 2 in Zp for any n in Zp? Would appreciate the proper way to reason about this problem.

答:
To expand on some comments: there's a distinction between the computer science notion of 'modulo $n$' — which is a function mod(-,n) from integers to integers between $0$ and $n-1$ — and the mathematical notion of 'mod $n$', which is a function from integers to equivalence classes of integers. It happens that these always have a representative between $0$ and $n-1$, so we often label them $\langle0\rangle$, $\langle 1\rangle$, etc. to aid with reasoning, and sometimes we just drop the brackets.

The field $C_p$ of residue classes mod $p$ is a mathematical notion; again, we choose to label the things in this field by $\langle0\rangle, \langle1\rangle, \ldots, \langle p-1\rangle$ or similar because those labels help us make sense of the field, but they shouldn't be confused with the integers $0, 1, \ldots, p-1$. The notion of taking $\langle 1\rangle \bmod 2$ doesn't make sense mathematically because there are many different representatives of that residue class; for instance if $p=5$, then the integers $1$, $6$, $1166$, etc. are all members of the equivalence class $\langle1\rangle$.

Now, it happens that the CS notion of modulo shares some of the properties of the mathematical notion; for instance, mod(n+k,n) = mod(k,n). But it doesn't share all of them; we don't even necessarily have that mod(a,n)+mod(b,n)=mod(a+b,n), for instance. It also doesn't behave well with respect to itself, and this is why trying to look at elements of a field $C_p$ mod some other number doesn't work; we don't have an identity mod(mod(a,m),n)=mod(a,n).


问:
Show convolution on a compact group is continuous
Let $G$ be a compact group and $\phi \in C(G)$. Given $f \in L^1(G)$, show that
$$T_\phi f: G \to \mathbb{C}: g \mapsto \int_G \phi(gh^{-1})f(h) dh$$ is continuous.
Here $dh$ is normalized Haar measure on $G$.

Attempt:

Let $g,g'\in G$. Then
$$|\int_G \phi(gh^{-1})f(h) dh- \int_G \phi(g'h^{-1}) f(h) dh| \leq \int_G |\phi(gh^{-1}) - \phi(g'h^{-1})| |f(h)| dh$$
Somehow, I want to be able to estimate the factor $|\phi(gh^{-1})-\phi(g'h^{-1})|$ using continuity of $\phi$ but I don't quite see how this is possible.

If I was allowed to work with sequences, I could just apply DCT but here we need net continuity so that does not work as well.

Any help is appreciated!

答:
First prove the following lemma (which basically says that $\phi$ is uniformly continuous).

For every $\epsilon > 0$ there exists an open neighborhood $V_{\epsilon}$ of $e$ in $G$ such that $|\phi(x_1) - \phi(x_2)| < \epsilon$ for every $x_1,x_2\in G$ satisfying $x_1x_2^{-1}\in V_{\epsilon}$.

Suppose now that $g' \in V_{\epsilon}g$ (note that $V_{\epsilon}g$ is an open neighborhood of $g$). Then $g'g^{-1}\in V_{\epsilon}$. Hence
$$g'h^{-1}(gh^{-1})^{-1} = g'h^{-1}hg^{-1} = g'g^{-1} \in V_{\epsilon}$$
Thus
$$|\phi(g'h^{-1}) - \phi(gh^{-1})| < \epsilon$$
Hence
$$\int_G |\phi(gh^{-1}) - \phi(g'h^{-1})| |f(h)| dh < \epsilon \cdot \int_G |f(h)|dh = \epsilon \cdot ||f||_1$$


问:
The letters A B C D E are arranged randomly in a line. In how many arrangements is the letter A in front of D?
The letters A B C D E are arranged randomly in a line. In how many arrangements is the letter A in front of D?

Any idea how I can do this? I was thinking of drawing out all possible cases, but is there a faster way that can be used to generalize to a large number of unique items?

Right now I have :

A L L L  D = 3!

L A L L D = $^3C_2 \cdot 2$

L L A L D = $^3C_2 \cdot 2$

L L L A D = 3!

So total there are 24 possible arrangements. Am I on the right track?

答:
You can choose $3$ places out of $5$ and place $B, C, E$ in those in $3!$ ways. Order of $A, D$ is fixed.

So total number of desired ways = ${5 \choose 3} \times 3! = 60$


答:
There'll be exactly half of total permutations in which A is before D and half in which A is after D.

Hence desired is $$ \dfrac{5!}{2}=60$$

Body:
The letters A B C D E are arranged randomly in a line. In how many arrangements is the letter A in front of D?

Any idea how I can do this? I was thinking of drawing out all possible cases, but is there a faster way that can be used to generalize to a large number of unique items?

Right now I have :

A L L L  D = 3!

L A L L D = $^3C_2 \cdot 2$

L L A L D = $^3C_2 \cdot 2$

L L L A D = 3!

So total there are 24 possible arrangements. Am I on the right track?


问:
How do I find sum of the series?

I calculated the radius of convergence is $\frac{1}{3}$. Can anyone suggest to me the last part of the question that is the sum of the series.

答:
HINT: Write separately the odd and even terms, as $2n$ and $2n+1$
$$\sum_{k=0}^\infty(2+(-1)^k)^kz^k=\sum_{n=0}^\infty3^{2n}z^{2n}+\sum_{n=0}^\infty1^{2n+1}z^{2n+1}$$

Title: How do I find sum of the series?


问:
Not Quite Invertible... But Close. Property of a Matrix.
Let $b\in \mathbb{R}^d,A\in\mathbb{R}^{d\times d}$ be given.

If $A$ is invertible (with inverse $G$) then $GAw=w$, $w\in \mathbb{R}^d$. What properties of $A,b$ would we need to have to guarantee the existence of a $G$ such that

$$ G(b+Aw)=w ~~~~~~(1)$$

for any $w$? Say we allow $G$ above to be any operator ( not just a matrix ) could one exist for $(1)$ to hold? I mean I can just define the operator like this right?

答:
Given $A$ and $b$, suppose that there exists a matrix $G$ such that, for any $w$:

$$G(b+Aw) = w ~~~~~~(1) .$$

This means that $G$ does not depend on $w$. Rather, it depends on $A$ and $b$ only. We may write:

$$G = G(A, b).$$

Equation (1) is equivalent to:

$$Gb = w - GAw \Rightarrow Gb = (I - GA)w, ~~~~~~(2)$$

where $I$ is the identity matrix.

Suppose that $I-GA$ can be inverted, and let $M$ be the inverse. Again, $M$ depends only on $A$ and $b$, i.e.:

$$M = M(A,b).$$

Now, observe that multiplying both sides of equation (2) by $M$ one gets:

$$MGb = M(I-GA)w \Rightarrow MGb = w. ~~~~~~(3)$$

More precisely, equation (3) reads as:

$$M(A,b)G(A,b)b = w.$$

But this is a contradiction, since all factors on the left hand side does not depend on $w$! Hence, we must conclude that the matrix $G$ does not exist!


问:
Simplifying complex radicals that includes prime numbers
In my mathematics class we learned radicals and how to simplify them. However some of the questions involved prime numbers that were in the radical, and we had to simplify them.

For example:

How on earth can we simplify radicals that have prime numbers in them?

Body:
$$11-6 \sqrt{2}=9+2-6 \sqrt{2}=\left(3-\sqrt{2}\right)^2\to \sqrt{11-6 \sqrt{2}}=3-\sqrt 2$$
$$11+4\sqrt6=8+3+4\sqrt{2\cdot 3}=\left(2\sqrt 2+\sqrt 3\right)^2\to \sqrt{11+4\sqrt6}=2\sqrt 2+\sqrt 3$$


问:
Prove that $\sum_{Y} P(X, Y) = P(X)$
For a homework assignment I need to prove that $\sum_{Y} P(X, Y) = P(X)$, given the following properties (Kolmogorov axioms):

For a given set of events $S$ and a probability mass function $P(\cdot)$, we have

$P(a) \geq 0 $ for all $a \in S$;

$P(S) = \sum_{a \in S}P(a) = 1$;

For $A, B \subseteq S$ that satisfy $A \cap B = \emptyset: P(A \cup B) = P(A) + P(B) $

$P(A|B) = P(A, B) / P(B)$

It is a homework question, but I am really stuck and would like someone to give me a hint. I understand intuitively why $\sum_{Y} P(X, Y) = P(X)$, but I have no idea how to prove it.

Some help would be much appreciated.

答:
Start with what you are given:

$$\sum^Y P(X,Y) $$.

Do you understand the law of total probability?

Title: Prove that $\sum_{Y} P(X, Y) = P(X)$


问:
Computable Distance in a Projective Space?
So this is absolutely not my area of expertise. Still, I wondered if someone could point me in the direction of a computational distance measure between points in a projective space? (ie. an algorithm I could implement to find distance between two points in projective space).

I am trying to find a distance measure for objects which should be scale invariant and can be embedded as points in $\mathbb{R}^n$, naturally, for scale invariance I thought of 1d subspaces and projective spaces. I know the Grassmanian over a real vector space and thus real projective spaces are smooth manifolds and not necessarily endowed with a metric, so this may bear no fruit. But I have seen some mention of metrics on Grassmannians while trying to understand the literature myself.

Any links to papers (preferable readable by a non-specialist) are much appreciated.

答: Well, the representation of points in a projective space is not unique. But you can always find a canonical representative of a point, namely $(0,\ldots,0,1,p_1,\ldots,p_k)$, where $p_1,\ldots,p_k$ are arbitrary. From here you could use the Euclidean distance.


问:
matrix norm formulas for operators between $\ell_1$ and $\ell_\infty$
Let $A=(a_{i,j}):\ell_\infty\to\ell_1$ and $B=(b_{i,j}):\ell_1\to\ell_\infty$ be linear operators.  We can define their operator/matrix norms as follows:
$$
\|A\|_{\infty,1}
=\sup\left\{\sum_{i=1}^\infty\left|\sum_{j=1}^\infty a_{i,j}x_j \right|:\sup|x_j|\leq 1\right\}
$$
and
$$
\|B\|_{1,\infty}
=\sup\left\{\sup\left|\sum_{j=1}^\infty b_{i,j}x_j\right|:\sum_{j=1}^\infty|x_j|\leq 1\right\}
$$
Are there any "nicer" formulas for these norms which can be written only in terms of $(a_{i,j})$ and $(b_{i,j})$, without reference to arbitrary vectors $(x_j)$?  I vaguely remember that there are such formulas, but I can't seem to find them today.

Thanks!

答: Hint: For $|x_j|\leq 1$, $$\left|\sum_{j=1}^\infty a_{i,j}x_j \right|\le\sum_{j=1}^\infty|a_{i,j}|$$ with equality when $x_j=\mathrm{sgn}(a_{i,j})$.


问:
Why leave-one-out residuals (studentized residuals) is t-distribution?
From my text book, there are several types of residuals:

residuals (or ordinary residuals) :

$\hat\varepsilon_i = y_i - \hat y_i = y_i - \boldsymbol{x}_i^T \hat{\beta}$

standardized residuals :

$r_i = \frac{\hat\varepsilon_i}{\hat\sigma\sqrt{1-h_{ii}}}$ ,

where $h_{ii}$ is the i 'th diagonal element of the hat matrix $H = X(X^TX)^{-1}X^T$.

studentized residuals :

$r_i^* = \frac{ \hat\varepsilon_{i} }{ \hat\sigma_{(i)}\sqrt{1-h_{ii}} }$,

where $\hat\sigma_{(i)}$ is the estimator of $\sigma$ where i 'th data are not used.

Also, it said that standardized residuals $r_i$ is not t-distribution because $\hat\sigma$  is a function of $\hat\varepsilon_i$, thus the numerator and denominator of $r_i$ are not independent.

On the contrary, studentized residuals $r_i^*$ is t-distribution because "we did not use the i'th observation $y_i$ when calculating $\hat\sigma_{(i)}$".

I am a little confused about this reason. Numerator of $r_i^*$ is using all data ($y_1$, ..., $y_n$), denominator of $r_i^*$ is using all data except $y_i$. So there are still $n-1$ data are shared. Why are they independent?

答:
Independence does not imply that one statistic is not constructed by other. For example, in normal distribution $S^2 = \frac{1}{n}\sum_{i=1}^n(X_i - \bar{X}_n)^2$ is independent of $\bar{X}_n$, but clearly $S^2$ uses $\bar{X}_n$ in its construction. Indepednece simply imply that
$$
f_{S^2, \bar{X}_n}(s, x) = f_{S^2}(s)f_{\bar{X}_n}(x).
$$

The same idea holds for the studentized residuals.


问:
How many different "Dwayne" "The Rock" "Johnson" orderings can be made for $n$ Dwayne The Rock Johnsons?
Let me explain. Whenever I hear Dwayne Johnson's name, I immediately start mixing it up in weird ways by saying his name and nickname multiple times but in random orders. For example "Dwayne" "The Rock" "Dwayne" "The Rock" "Johnson" "Johnson". But I always follow the same set of rules:

If I say a number of "Dwayne"'s, $n$, I have to say the other two items "The Rock" and "Johnson" $n$ times as well. Basically the full name needs to be said a whole number of times throughout the expression.

If I say "The Rock" it must be because I said a corresponding "Dwayne" sometime before it. For example: "Dwayne" "The Rock" "The Rock" "Dwayne" "Johnson" "Johnson" is not valid. The same is the case for the "Johnson"'s after "The Rock". i.e. They have to appear in order.

So my question is:

For a given number $n$ of "Dwayne", "The Rock", and "Johnson"'s how many different Dwayne The Rock Johnson orderings can I have that follow these rules.

答:
OEIS: A005789

Number of words consisting of n 'x' letters, n 'y' letters and n 'z' letters such that the 'x' count is always greater than or equal to the 'y' count and the 'y' count is always greater than or equal to the 'z' count; e.g., for n=2 we have xxyyzz, xxyzyz, xyxyzz, xyxzyz and xyzxyz.

In this case, 'x' is "Dwayne", 'y' is "The Rock" and 'z' is "Johnson".


问:
To find norm of a bounded linear operator
I have given $S:l^{\infty} \to l^1$ by $S(x_1, x_2, x_3,...)=(\frac{x_1}{2},\frac{x_1+x_2}{2^2},\frac{x_1+x_2+x_3}{2^3},....)$. I showed it is bounded linear operator but I am not getting idea about how to find $\mid\mid S \mid\mid$. Any start or hint. Thanks

答: If $s\in\ell^\infty$ and $\|s\|\leqslant1$, then $s=(x_1,x_2,x_2,\ldots)$ and $(\forall n\in\Bbb N):|x_n|\leqslant1$. So\begin{align}\bigl\|S(s)\bigr\|_1&=\sum_{n=1}^\infty\frac{|x_1+x_2+\cdots+x_n|}{2^n}\\&\leqslant\sum_{n=1}^\infty\frac{|x_1|+|x_2|+\cdots+|x_n|}{2^n}\\&\leqslant\sum_{n=1}^\infty\frac n{2^n}\\&=2.\end{align}So, $\|S\|\leqslant2$. But $\|(1,1,1,\ldots)\|_\infty=1$ and $\bigl\|S\bigl((1,1,1,\ldots)\bigr)\bigr\|_1=2$. So, $\|S\|=2$.


问:
Area of triangle inside a $30-60-90$ triangle

In a right angle triangle ABC with hypotenuse BC and C=60 degrees, M and N are the middles of AB and AC respectively. Draw ND perpendicular to BC (D is point on the side BC). If MD=7cm calculate the area of triangle DNM.

I solved this question in the following way:

$NM//=\frac{CB}{2}$ (Thales theorem)

ABC=30 degrees hence $CA=\frac{CB}{2}$ and hene $NM=CA$

$sinNCD=\frac{ND}{CN}$

$\frac{\sqrt{3}}{2}*CN=ND$

$CN=\frac{2}{\sqrt{3}}*ND$

$AC=\frac{4}{\sqrt{3}}*ND$

$MN=\frac{4}{\sqrt{3}}*ND$

since NM//CB then DNM=90 degrees.

Hence, $DN^2+DN^2*\frac{16}{9}=49$ (Pythagoras), so $DN^2*\frac{25}{9}=49$, so $DN*\frac{5}{3}=7$, so $DN=\frac{21}{5}$

From this we have that $MN=\frac{4}{\sqrt{3}}*\frac{21}{5}=\frac{84}{5\sqrt{3}}$.

Hence $Area=\frac{MN*ND}{2}=\frac{42}{5\sqrt{3}}*\frac{21}{5}=\frac{21*42}{25\sqrt{3}}=\frac{882}{25\sqrt{3}}$

Could you please tell me if I am correct and if I am wrong how much would it be reasonable to assume I would get in an olympiad for my work?

答:
You have done a mistake here

$$ DN^2+DN^2*\frac{16}{\color{red}{3}}=49$$

You wrote $9$.

With correct value, I got $$ [DNM] = \dfrac{98\sqrt{3}}{19}$$


问:
Does the definition of rational numbers already assumes an understanding of them?
I mean, how do we know that the correct equivalence relation is the equality of the crossed products? Does it already assume an intuitive informal understanding of rationals? If yes, what would that intuition be?</p>

答:
What do you mean "correct" equivalence relation?

In a way the answer to your question is "yes" because or goal is to take our intuitive understanding of rations and formalize it in a way so that our formulation fits our intuition.  But in a more important way the answer to your question is "no", because our goal is that are formalization does not rely on any intuition and can apply to other mathematical structures that may prove very different than our intuition.

Our formulation is the $\mathbb Q$ is the set of equivalence classs on the relation on $\mathbb Z\times \mathbb N$ that $(a,b)\sim (c,d) \iff ad=bc$.  And for that formulation all we need to prove is that $\sim$ is an equivalence relation which is fairly straightforward to show.

There is no question that it is the "correct" equivalence relation.  Correct for what?

We can show that if we do have an intuitive idea of what that rationals are that this equivalence fits our intuition.  If we intuitively have some idea that $\frac ab$ is $a\div b$ and $\div$ is the inverse of $\times$ then if $a\div b = c\div d$ then $(a\div b)\times b = (c\div d)\times b$ so $a = (c\div d)\times b$ so $a\times d =(c\div d)\times b \times d = [(c\div d)\times d]\times b= c \times d$.

And alternatively if $ad = bc$ and $d\ne 0;c\ne 0$ then $(ad)\div d = (bc)\div d$ so $a =(bc)\div d$ so $a\div c = [(bc)\div d]\div c= [(bc\div c)]\div d = b\div d$.

So yes, it fits our intuition.


问:
Simple summation algebra question: summation of a summation
What does $$\sum_{i=1}^{n}\sum_{i=1}^{n}x_{i}$$ yield?

答:
Let inner sum be $S$ i.e.,

$$ \sum_{i=1}^n x_i = S$$

Now this is a constant. Hence

$$ \sum_{i=1}^n \Big(\sum_{i=1}^n x_i\Big) =\sum_{i=1}^n S = nS$$


问:
Why this recursion is the same as OEIS sequence A005252
I have the recursion:

$$\left\{ \begin{array}{l}
  f(0) = f(1) = f(2) = f(3) = 1\\
  f(n) = f(n - 1) + 1 + \sum_{i = 1}^{n - 4}f(i) \text{ if $n \geqslant 4$}
\end{array} \right.$$

So we have e.g.

$$\begin{array}{lll}
  f(4) & = & f(3) + 1 = 2\\
  f(5) & = & f(4) + 1 + f(1) = 2 + 1 + 1 = 4\\
  f(6) & = & f(5) + 1 + f(2) + f(1) = 4 + 1 + 1 + 1 = 7\\
  \ldots &  &
\end{array}$$

I noticed this sequence is the same as A005252 - OEIS, which is defined as:

$$a(n) = \sum_{k=0..\lfloor n/4 \rfloor} \binom{n-2k}{2k}$$

I'm not sure how to get there? I tried to use a generating function, but not sure how to manipulate it.

答:
Write the recursion formula $f(n)=\ldots$, and below it $f(n-1)=\ldots\ $. Subtract these two equations, and you obtain the linear difference equation
$$f(n)=2f(n-1)-f(n-2)+f(n-4)\qquad(n\geq4)\ .$$
Its characteristic equation is
$$\lambda^4-2\lambda^3+\lambda^2-1=0$$
with the solutions
$$\lambda_1={1+\sqrt{5}\over2},\quad \lambda_2={1-\sqrt{5}\over2},\quad \lambda_3=e^{i\pi/3},\quad \lambda_4=e^{-i\pi/3}\ .$$
It follows that
$$f(n)=\sum_{k=1}^4C_k \lambda_k^n$$
with certain values for the $C_k$ that can be found using the initial conditions. The result is: Consider the two sequences
$$\eqalign{g(0)&=1,\quad g(1)=1,\quad g(n):=g(n-1)+g(n-2)\ ,\cr
h(0)&=1,\quad h(1)=1,\quad h(n):=h(n-1)-h(n-2)\ .\cr}$$
The first is essentially the sequence of Fibonacci numbers, the second is periodic $(1,1,0,-1,-1,0,\ldots)$ with period $6$. Then
$$f(n)={1\over2}\bigl(g(n)+h(n)\bigr)\qquad(n\geq0)\ .$$


问:
Compute the remainder when $67!$ is divided by $71$.
This is how far I've been able to get.

By using Wilson's Theorem:

$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$

EDIT: Here is how I proceeded using TMM's and  Carl Mummert's hints.

$$\begin{align}
&68 \equiv -3 \pmod{71}\\
and\\
&69 \equiv -2 \pmod{71}\\
\end{align}$$

So:
$$\begin{align}
67!(-3)(-2) &\equiv 1 \pmod{71} \\
67! &\equiv 6^{-1} \pmod{71} \\
\end{align}$$

By using the Euclidean algorithm:
$$71 = 6 \cdot 11 + 5$$
$$6 = 5 \cdot 1 + 1$$
$$5 = 1 \cdot 5 + 0$$

Now, going backwards:
$$\begin{align}
1 &= 6 - 5 \\
&= 6 - (71 - 6 \cdot 11) \\
&= 6 + 6(11) - 71 \\
&= 6(1 + 11) - 71(1) \\
&= 6(12) - 71(1)
\end{align}$$

Therefore, $67! \equiv 6^{-1} \equiv 12 \pmod{71}$.

答:
Recall that if
$$xy \equiv a \pmod{n}$$
then
$$x \equiv ay^{-1} \pmod{n}$$ if $y$ is invertible in $\pmod{n}$. In your case, $n$ is $71$ a prime, which guarantees that any $y \not \equiv 0 \pmod{71}$ is invertible in $\pmod{71}$. Hence,, we have
$$67! \times 68 \times 69 \equiv 1 \pmod{71} \implies 67! \equiv (69)^{-1} (68)^{-1} \pmod{71}$$
We have
\begin{align}
68 \times 47 - 71 \times 45 & = 1\\
69 \times 35 - 71 \times 34 & = 1
\end{align}
Hence,
$$(68)^{-1} \equiv 47 \pmod{71} \text{ and }(69)^{-1} \equiv 35 \pmod{71}$$
Hence,
$$67!\equiv 35 \times 47 \pmod{71} \equiv 12 \pmod{71}$$</p>


问:
The Product of Consecutive Integers is Never a Power: Lemma 2 (Research Paper Study)
P. Erdos and J. L. Selfridge proved in the  paper THE PRODUCT OF CONSECUTIVE INTEGERS IS NEVER A POWER (click here), that the equation $(n + 1) \cdots(n + k)=x^l \cdots (1)$ has no solution in integers with $k > 2, l > 2, n > 0$. There is a lemma $2$ on page 294, 295 -

LEMMA 2. By deleting a suitably chosen subset of $\pi(k-1)$ of the numbers
$a_i (1 \leq i \leq k)$, we have
$$a_{i_l}...a_{i_k'},|(k -1)!\cdots  (9)$$
where $k’=k -\pi(k-1)$.

For each prime $p < k-1 $ we omit an $a_m$ for which $ n + m$ is divisible by $p$
to the highest power. If $1 \leq i \leq k$ and $i \neq m$, the power of $p$ dividing $n +i$ is the same as the power of $p$ dividing $i-m$. Thus $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)!$.

I understand  $p | i-m$ but how does  $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)! \:\:$ ?

Edit:

Here, $n + i= a_ix^l$, where $a_i$ is $l^{th}$-power free and all its prime factors are less than $k$ (supposing Theorem 2 is false, see page 293 of the  paper (click here) for the definition).

答:
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Just write it out.

Let $ p^{k_i} || n+i$.

For $ i  \neq m$, we now show that $p^{k_i} || i-m$. By definition of $m$, $p^{k_i} \mid n+m$.

Case 1: $ p^{k_i + 1} \mid n+m$. Verify that we do indeed have $ p^{k_i} || (n+i)-(n+m) = i-m$.

Case 2: $ p^{k_i} || n+m$. Then $p^{ k_i} \mid (n+i)-(n+m)$ and $ |i-m| \leq  p^{k_i + 1}$ hence $ p^{ k_i} || i-m$.

Let $K=k_i \pmod{l}$ (in reduced modulo class). Then by definition, $ p^{K_i} || a_i$, and $K_i \leq k_i$.

Hence, we have $ p^{\sum k_i } || \prod_{i\neq m}(i-m) = (k-m)!(m-1)! $ and $p^{\sum K_i } || \prod a_i $.

Since $ \sum K_i \leq \sum k_i$, the desired result follows.

Namely, with $\alpha = \sum K_i$, we have that $ p ^\alpha ||  \prod a_i$ and $ \alpha \leq \sum k_i$ so $ p^\alpha \mid p^{\sum k_i} \mid (k-m)!(m-1)!$.


问:
What happen to units when we take Ln?
What happens to units when we take the Ln of the both side of the equation?
For example for an equation $PV^k = C$ (which becomes $\ln P = -k \ln V + \ln C$), how can we find the units of $k$ and $C$?
-->(the unit of $P$ is mmHg and the unit of $V$ is $cm^3$)

Thank you so much..

答:
Well, $k$ must be unitless, because exponentiation isn't defined for anything except numbers.

So the units of $C$ are $mmHg * cm^{3k}$.

Title: What happen to units when we take Ln?


问:
Why is that a Cauchy sequence in $L^2(\mathbb{P})$, by $(1)$?
I quote Øksendal (2003).

Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$. [...]

Starting from a probability space $\left(\Omega,\mathbb{P},\mathcal{E}\right)$ and a Brownian motion $\left(B_t\right)_{t\ge0}$, if $\phi(t,\omega)$ is bounded and elementary, then
$$\mathbb{E}\left[\left(\int_S^T\phi(t,\omega)dB_t(\omega)\right)^2\right]=\mathbb{E}\left[\int_S^T\phi(t,\omega)^2 dt\right]\tag{1}$$ [...]

If $f\in\mathcal{V}$ one can show that it is possible to choose elementary functions $\phi_n\in\mathcal{V}$ such that:
$$\mathbb{E}\left[\int_S^T|f-\phi_n|^2 dt\right]\to0\tag{2}$$
Then, define
$$\mathcal{I}\left[f\right](\omega)=\int_S^T f(t,\omega)dB_t(\omega)=\lim_{n\to\infty}\int_S^T\phi_n(t,\omega)dB_t(\omega)\tag{3}$$
The limit exists as an element of $L^2(\mathbb{P})$, since $\left\{\int_S^T\phi_n(t,\omega)dB_t(\omega)\right\}$ forms a Cauchy sequence in $L^2(\mathbb{P})$, by $(1)$.

What I cannot understand is the statement in bold above. Why is that true?

答:
Note that (2) implies:
$$\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0. \tag{4}$$
Thus
\begin{align}
E \left|\int_S^T \phi_n(t)dB_t-\int_S^T \phi_m(t)dB_t \right|^2 &=E \left|\int_S^T (\phi_n(t)-\phi_m(t))dB_t \right|^2  \\
&=E \int_S^T |\phi_n(t)-\phi_m(t)|^2 dt \rightarrow 0.
\end{align}
The last equality follows from the Ito isometry and the convergence to zero follows follows from (4).
This proves that $\{\int_S^T \phi_n(t)dB_t \}$ is a Cauchy sequence in $L^2$.

Edit: proof that (2) implies (4):
\begin{align}
E\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]&=E\left[\int_S^T|\phi_n(t)-f(t)+f(t)-\phi_m(t)|^2 dt\right] \\
&\leq E\left[\int_S^T 2|\phi_n(t)-f(t)|^2+2|f(t)-\phi_m(t)|^2 dt\right]  \\
&= 2E\left[\int_S^T |\phi_n(t)-f(t)|^2dt\right]+2E\left[\int_S^T|f(t)-\phi_m(t)|^2 dt\right] \rightarrow 0.
\end{align}
The '$\leq$' follows from the inequality $(a+b)^2 \leq 2a^2 + 2b^2$ and the convergence to zero from (2).


问:
uniform continuity of $e^{-\frac{1}{|z+1|}}$
Is function $e^{-\frac{1}{|z+1|}}$ uniform continuous on set of  complex numbers $z \neq -1$.
I tried to prove by definition, or make an example that it is not( tried sequences around -1, such as $x_{n}=-1+\frac{1}{n}$ or $x_{n}=-1+\ln(\frac{1}{n}+1)$ but still nothing.

答:
Consider the function$$\begin{array}{rccc}f\colon&\Bbb C&\longrightarrow&\Bbb R\\&z&\mapsto&\begin{cases}e^{-1/|z+1|}&\text{ if }z\ne-1\\0&\text{ if }z=1.\end{cases}\end{array}\tag1$$Then $f$ is an extension of your function. Therefore, if $f$ is uniformly continuous, then so is your function. Also, note that $f$ is continuous and that therefore the restriction of $f$ to any compact subset of $\Bbb C$ is uniformly continuous.

If $f$ was not uniformly continuous, then there would be a number $\varepsilon>0$ and there would be two sequences $(z_n)_{n\in\Bbb N}$ and $(w_n)_{n\in\Bbb N}$ of complex numbers such that$$(\forall n\in\Bbb N):|z_n-w_n|<\frac1n\quad\text{and that}\quad\bigl|f(z_n)-f(w_n)\bigr|\geqslant\varepsilon.$$ There are wo possibilites now:

The sequence $(z_n)_{n\in\Bbb N}$ is unbounded. Then we can assume WLOG that $\lim_{n\to\infty}|z_n|=\infty$. So, $\lim_{n\to\infty}|w_n|=\infty$ too. But then $\lim_{n\to\infty}f(z_n)=\lim_{n\to\infty}f(w_n)=0$ and therefore we cannot have $(1)$.

The sequence $(z_n)_{n\in\Bbb N}$ is bounded. Then there is some $R>0$ such that every $z_n$ belongs to $\overline{D_R(0)}$ and so $(\forall n\in\Bbb N):z_n,w_n\in\overline{D_{R+1}(0)}$. But $\overline{D_{R+1}(0)}$ is compact. So, the restriction of $f$ to $\overline{D_{R+1}(0)}$ is uniformly continuous and, again, we canot have $(1)$.


问:
differential equation of first order - omitting absolute value operator
i have the equation:
$$y'+2y\:=1$$

and i solve it the regular way for first order differential equation:
$$y'\:=1-2y$$
$$\frac{dy}{dx}=1-2y$$
$$\int \:\frac{1}{1-2y}dy=\int \:1dx$$
$$-\frac{1}{2}\int \:-\frac{2}{1-2y}dy=\int \:1dx$$
and using the integral formula:

$$-\frac{1}{2}\ln\left(\left|1-2y\right|\right)=x+\ln\left(c\right)$$
Why Symbolab omits the absolute value operator and writes:

答:
$$y'+2y\:=1$$
With integrating factor method:
$$(ye^{2x})'=e^{2x}$$
$$ye^{2x}=\dfrac 12 e^{2x}+K$$
$$\boxed {y(x)=\dfrac 12 +Ke^{-2x}}$$
Then we can rewrite this as:
$$\dfrac {2y-1}K=e^{-2x} \geq 0$$
$$\ln \left (\dfrac {2y-1}K\right )=-2x $$
It seems to me that the absolute value is needed.


问:
Length of Triangle BCD

Hey, well I'm doing some higher level revision and I'm stuck...

In the diagram triangle BCD is mathematically similar to triangle ACE.

So what is the length of BD? How do you work it out?

答:
Your problem tells you that the two triangles are similar triangles. Similar triangles always have the same three angles (even if they are rotated).

When we have two sides, the angle between them determines the length of the third side, the side opposite the angle. Because the angles of the two triangles are the same (because they are similar triangles), then the sides must be proportionate to one another.

Because the sides are proportionate to one another, we can work out ratios: let's take one side from triangle A and one side from triangle B, both of which are between the same two angles (or opposite the same angle). Given that we then have the ratio, we can work out the length of the side we are trying to find.

So let's take side DC $=4$ and EC $=(4+6)=10.$ Let's divide the smaller side by the bigger side: $\frac{DC}{EC} = \frac{4}{10}.$ Let's call $\frac{4}{10}$ the ratio.

Let's look at the ratio a bit more:

$\frac{Smaller Side}{Bigger Side}= ratio$

 then $Smaller Side = ratio \cdot BiggerSide$

 and $\frac{SmallerSide}{ratio} = BiggerSide$

Now that we have this, we can use it to work out side BD. In this case, BD is the smaller side, and EA is the bigger side. So we use the ratio and the bigger side to find the smaller side:  $Smaller Side = ratio \cdot BiggerSide$

  i.e. $BD = ratio \cdot EA$

$BD = \frac{4}{10} \cdot 7.5$

$\therefore BD = 3$

Title: Length of Triangle BCD


答:
Addendum to @fleablood 's correct answer.

Throughout mathematics, formal definitions always come after intuitive understanding. The intent of the formal definition is to capture that understanding so we can reason carefully even in places we don't quite yet understand.

You can see this as early as Greek geometry. The elementary theorems there were known long before Euclid's codification.

Newton and Leibniz and their followers developed calculus long before the $\epsilon - \delta$ definition of a limit provided rigor.

In your question about rational numbers you do have the right intuition based on years of elementary school arithmetic, so you can see that the definition captures that intuition.

Students often find new mathematics difficult because they see formal definitions before they have developed much intuition. It's hard to understand the definition of a group before you've actually looked at  different kinds of groups in different contexts. I think courses in abstract algebra usually move to abstraction too soon - but if they didn't then they couldn't "cover as much". That tension is hard to resolve.

Title: Does the definition of rational numbers already assumes an understanding of them?

Body: I mean, how do we know that the correct equivalence relation is the equality of the crossed products? Does it already assume an intuitive informal understanding of rationals? If yes, what would that intuition be?</p>


问:
Measurable space of nonempty compact intervals, $\sigma ( CI_{\mathbb R}) \neq \mathcal I_{\mathbb R}$
If we define

$$CI_{\mathbb R} = \{ [a,b] \subset \mathbb R : (a,b \in \mathbb R \ \text{ and } \ a \leq b ) \}$$

$$ \mathcal I_{\mathbb R} = \sigma_{CI_{\mathbb R}}  \left ( \{ \{ A \in CI_{\mathbb R} : \inf( A) < c \} : c \in \mathbb R \}  \cup \{ A \in CI_{\mathbb R} : \sup( A) < c \} : c \in \mathbb R \} \right )  $$

The idea is to construct a measurable space for confidence intervals. The space $( CI_{\mathbb R}, \mathcal I_{\mathbb R} ) $ is nice. However, I am trying to understand how come these two sets are not equal :

$$\sigma ( CI_{\mathbb R}) \neq \mathcal I_{\mathbb R}$$

LHS should be bigger, do you have an example of an element in LHS that is not in RHS?

答:
An answer given by my lecturer :

On the LHS, we have the power set of $CI_{\mathbb R}$ (which is the largest \sigma-algebra on $CI_{\mathbb R}$).

Let $X$ be an uncountable set (in our case $X = CI_{\mathbb R}$), and let $Y = \sigma_{X} \{ Y_j, j \in J \}$ (in our case $Y = I_{\mathbb R}$),  where $J$ is some (uncountable) set, and $Y_j \subset 2^{X}$ (i.e., power set of $X$).
Note that in our case every $Y_j$ is uncountable, its complement is uncountable, and $Y_j \cap Y_i$ is either uncountable or empty set.

Hence, one can show that every element of $Y$ is uncountable. However, $2^{X}$ also includes countable sets (e.g., singletons). Hence, there are uncountable many sets which are contained in $2^X$ but not in $Y$.


问:
Computable Distance in a Projective Space?
So this is absolutely not my area of expertise. Still, I wondered if someone could point me in the direction of a computational distance measure between points in a projective space? (ie. an algorithm I could implement to find distance between two points in projective space).

I am trying to find a distance measure for objects which should be scale invariant and can be embedded as points in $\mathbb{R}^n$, naturally, for scale invariance I thought of 1d subspaces and projective spaces. I know the Grassmanian over a real vector space and thus real projective spaces are smooth manifolds and not necessarily endowed with a metric, so this may bear no fruit. But I have seen some mention of metrics on Grassmannians while trying to understand the literature myself.

Any links to papers (preferable readable by a non-specialist) are much appreciated.

答:
A detour by Affine Geometry, using barycentric coordinates $(p,q,r)$ (cousin of projective coordinates : see p. 13 here) with respect to a fixed affine basis characterized by triangle $ABC$.

One can imbed metric properties into Affine Geometry with barycentric coordinates. Let us take the example  of the equation of the circumcircle (which is a metric concept, not invariant by a general affine transform, as are usually the properties in this geometry) with classical notations $a=BC,b=CA,c=AB$ is :

$$a^2qr+b^2rp+c^2pq=0$$

In fact, the presence of lengths $a,b,c$  provide the "metrical information"...

Here is the general formula (2) which can be placed at the basis of this metric.

Let us define the

$$\text{Barycentrical coordinates of} \ \vec{MM'}=M'-M : (p',q',r')-(p,q,r)\tag{1}$$

(where $(p,q,r)$ and $(p',q',r')$ are the barycentrical coordinates of $M,M'$ resp.)

The fundamental formula is the way (ordinary Euclidean) dot product between  vectors can be expressed using their barycentrical coordinates:

$$\vec{u} · \vec{v} = 2 {\frak{A}} (\alpha p_1p_2 + \beta q_1q_2 + \gamma r_1r_2)\tag{2}$$

where $\frak{A}$ is the area of triangle $ABC$ with $\alpha=\operatorname{cotan}(\hat{A}), \ \beta=\operatorname{cotan}(\hat{B}), \ \gamma=\operatorname{cotan}(\hat{C}).$

Here is a reference in croatian which is rather understandable (using Google translate...), where formula (2) is established.

I am going looking for an equivalent reference in English.

See as well here.


问:
Derivative and divergence of measures, product of measures and vector field
I am having trouble understanding the following notation: this comes from optimal transport theory and the continuity equation. $\mu: [0,1] \to P_2(\mathbb{R}^d)$ is a curve in space of probability measures having finite second moment in $\mathbb{R}^d$ and $v:[0,1] \times \mathbb{R}^d \to \mathbb{R}^d$ is a vector field. Then the continuity equation is
$$
\partial_t \mu_t + \nabla\cdot (\mu_t v_t) =0
$$
in the sense of distribution.

What does this mean? I do not understand taking derivative of a measure, multiplying a measure and a vector field, and taking the divergence of the result. Any help or reference?
Thank you!

答:
A distribution is a linear functional on $C_c^\infty(\mathbb{R}^d),$ the space of infinitely differentiable functions with compact support. The action of a distribution $u$ on the test function $\varphi \in C_c^\infty(\mathbb{R}^d)$ is often denoted by $\langle u, \varphi \rangle.$

A measure $\mu$ induces a distribution by
$\langle \mu, \varphi \rangle := \int \varphi \, d\mu.$

For $\mu : [0,1] \to P_2(\mathbb{R}^d)$ we define $\partial_t \mu_t$ by
$$
\langle \partial_t \mu_t, \varphi \rangle
:= \frac{d}{dt} \langle \mu_t, \varphi \rangle
= \frac{d}{dt} \int \varphi \, d\mu_t,
$$
define $\mu_t v_t$ by
$$
\langle \mu_t v_t, \varphi \rangle
= \langle \mu_t, v_t \varphi \rangle
= \langle \mu_t, (v_t^1, \ldots, v_t^d) \varphi \rangle
= (\langle \mu_t, v_t^1 \varphi \rangle, \ldots, \langle \mu_t, v_t^d \varphi \rangle),
$$
and
$$
\langle \nabla\cdot(\mu_t v_t), \varphi \rangle
= - \langle \mu_t, \nabla\cdot(v_t \varphi) \rangle
= - \langle \mu_t, (\nabla\cdot v_t) \varphi + v_t \cdot \nabla\varphi \rangle
.
$$

Note: For the above to make sense with respect to the definition of distributions as being linear functionals on $C_c^\infty$ the vector field $v_t$ needs to be $C^\infty.$ But I guess that $\mu_t$ having finite second moment allows us to reduce the requirements on the test functions and on $v_t.$


问:
Show that $f^{-1}(a)$ is a submanifold of $\mathbb{R}$
Let $f:\mathbb{R}\to\mathbb{R}$ be a real analytic function with infinitely many zeros. Let $a\neq 0$ be a real number. Show that  $f^{-1}(a)$ is a submanifold of $\mathbb{R}$.</p>

答: Hint: If an analytic function's zero set has a limit point, the function is constantly zero. Use that fact to show that $f^{-1}[\{a\}]$ is a discrete set.


问:
How do Lagrange multipliers work for functionals? (Motivation: the catenary)
I'm fairly new to the calculus of variations, so please bear with me. This question is very similar to this other question posted six months ago that has as yet not received any answers. If a good discussion is had here I will be sure to send the  findings over to the previous asker. With that out of the way, here is my question:

I want to derive the equation of the catenary using the Euler-Lagrange equation. I know that this has been done many times, but I want to do it myself. Below is a diagram:

Let the path of the wire between the two points be described by some function $y$. It is natural to want to minimize the potential energy of the wire. Assuming the wire has a constant density $\rho$ this boils down to
$$U[y]=\rho g\int_{x_1}^{x_2}y~\mathrm{d}x$$
But of course, there are three constraints:
$$y(x_1)=y_1~;~y(x_2)=y_2$$
$$\int_{x_1}^{x_2}\sqrt{1+{y'}^2}~\mathrm{d}x=L$$
The problem is I don't really know how to apply these constraints properly. I tried using the Euler-Lagrange equation but this only brought me to
$$\rho g(x_2-x_1)=0\implies x_1=x_2$$
Which is obviously nonsense. I looked at Wikiversity's derivation of the catenary and they mention using a Lagrange multiplier. But, the only context in which I'm familiar with using Lagrange multipliers in is when we want to extremize some function $f:\mathbb{R}^n\to\mathbb{R}$ subject to some constraint $g(\mathbf{x})=C$. My main question is: how do we use Lagrange multipliers for extremizing functionals? And why do the same methods work as to extremizing functions?

Thanks for your time,

-K.defaoite

答:
The problem with your formulation is that the integration along the solution curve should be

$$
\mathcal{L}=\rho g\int_{x_1}^{x_2}y \sqrt{1+y'^2}dx+\lambda\left(\int_{x_1}^{x_2}\sqrt{1+y'^2}dx-L\right)
$$

You are using in the first integral $dx$ instead of $ds$.

NOTE

The potential energy is given by

$$
\rho g \int_{x_1}^{x_2}y ds
$$

and the length is given by

$$
L = \int_{x_1}^{x_2}ds
$$

so the lagrangian is

$$
\mathcal{L}=\rho g\int_{x_1}^{x_2}y ds + \lambda\left(\int_{x_1}^{x_2}ds-L\right)
$$

or

$$
\mathcal{L}=\int_{x_1}^{x_2}\left(\left(\rho g y+\lambda\right)\sqrt{1+y'^2}-\frac{\lambda L}{(x_2-x_1)}\right)dx
$$

because $ds = \sqrt{dx^2+dy^2} = \sqrt{1+y'^2}dx$


答:
Let $F(z)=\sum_{n \ge 0} f_n z^n$ be the generating function.  Then the recurrence relation implies that
\begin{align}
F(z)
&= \sum_{n=0}^3 1 z^n + \sum_{n \ge 4} \left(f_{n - 1} + 1 + \sum_{i = 1}^{n - 4}f_i\right)z^n \\
&= \sum_{n\ge 0} z^n + z \sum_{n \ge 4} f_{n - 1}z^{n-1} + \sum_{i \ge 1} f_i  \sum_{n\ge i+4} z^n \\
&= \frac{1}{1-z} + z \left(F(z)-\sum_{n=0}^2 f_n z^n\right) + \sum_{i \ge 1} f_i \frac{z^{i+4}}{1-z} \\
&= \frac{1}{1-z} + z F(z)-z\sum_{n=0}^2 z^n + \frac{z^4}{1-z}\left(F(z)-1\right), \\
\end{align}
so
$$F(z)=\frac{\frac{1}{1-z} -z(1+z+z^2) - \frac{z^4}{1-z}}{1-z- \frac{z^4}{1-z}}
=\frac{1-z}{(1-z+z^2)(1-z-z^2)},$$
which matches the generating function shown in OEIS.

Title: Why this recursion is the same as OEIS sequence A005252

Body:
I have the recursion:

$$\left\{ \begin{array}{l}
  f(0) = f(1) = f(2) = f(3) = 1\\
  f(n) = f(n - 1) + 1 + \sum_{i = 1}^{n - 4}f(i) \text{ if $n \geqslant 4$}
\end{array} \right.$$

So we have e.g.

$$\begin{array}{lll}
  f(4) & = & f(3) + 1 = 2\\
  f(5) & = & f(4) + 1 + f(1) = 2 + 1 + 1 = 4\\
  f(6) & = & f(5) + 1 + f(2) + f(1) = 4 + 1 + 1 + 1 = 7\\
  \ldots &  &
\end{array}$$

I noticed this sequence is the same as A005252 - OEIS, which is defined as:

$$a(n) = \sum_{k=0..\lfloor n/4 \rfloor} \binom{n-2k}{2k}$$

I'm not sure how to get there? I tried to use a generating function, but not sure how to manipulate it.


问:
Discrete Mathematics - Modular arithmethics
I've been trying to solve a task for some while now but I'm currently very stuck.

For the integers $a$ and $b$ it applies that $b \equiv a \pmod{91}$ and $\gcd (a, 91) = 1$. Determine a positive number $k> 1$ such that $b^k \equiv a \pmod{91}.$

I know that  $b^{\phi(91)} = 1\pmod{91}$ and  $b^{\phi(91)} = 72$.

After this I'm pretty empty. Would anyone be so kind to guide me or help me in the right direction?

答:
Since we have $a\equiv b\mod 91$ and also $\varphi(91)=72$ added to $(a,91)=(b,91)=1$ giving $b^{72}\equiv 1\mod 91,$ then we can say $b\cdot b^{72}\equiv b\cdot 1\equiv a\equiv b^{73}\mod 91$.

Thus $b^k\equiv a\mod 91$ using $k=\varphi(91)+1$.


问:
Differential geometry - For any vector in $R^n$ there exists a velocity vector $\gamma\prime(t)$ of closed curve $\gamma$ which is orthogonal to it
I have the following problem.

Suppose that I have smooth (regular) closed curve $\gamma: [a,b]\rightarrow\mathbb R^n$.

I need to prove that for every vector $\phi\in\mathbb R^n$ there exists point $t\in[a,b]$ such as the value of the velocity speed vector at this point ($\gamma\prime(t))$ is orthogonal to this vector.

My approach (Intuition) was to assume that if it isn't true then for any vector $\gamma=(r_1,...,r_n)$ and for any point $t$ the velocity speed vector at this point (i.e. derivative vector) isn't orthogonal to this vector so the curve isn't turning back to the origin point $\gamma(a)$ and therefore it will not be closed eventually (at point $b$) and this will give me a contradiction to the fact that this is a closed curve, But I'm really struggling with how to convert my approach to formal proof.

答:
I assume

$\phi \ne 0, \tag 0$

since the notion of orthogonality is at best trivial and at worst meaningless when (0) fails.

Consider the function

$f(t) = \phi \cdot \gamma(t); \tag 1$

since $\gamma(t)$ is a closed regular curve on $[a, b]$, a compact interval, the function $f(t)$ attains its extremal values on $[a, b]$; at such a

$\tau \in [a, b] \tag 2$

we have

$\phi  \cdot \gamma'(\tau) = f'(\tau) = 0; \tag 3$

but this is precisely the assertion that $\gamma'(\tau)$ is orthogonal to $\phi$. $OE\Delta$

Note Added in Edit, Saturday 7 November 2020 12:47 PM PST:  The geometry of this situation may perhaps be further illuminated bye scrutinizing the meaning of $f(t)$: if we write $\phi$ in polar form:

$\phi = \vert \phi \vert \mathbf e_\phi, \tag 4$

where

$\mathbf e_\phi \in S^{n - 1} \tag 5$

is the unit vector in the direction of $\phi$, then $f(t)$ becomes

$f(t)= \vert \phi \vert \mathbf e_\phi \cdot \gamma(t); \tag 6$

aside from the factor of $\vert \phi \vert$, the quantity $f(t)$ is the component of $\gamma(t) \in \Bbb R^n$ along $\mathbf e_\phi$; $\vert \phi \vert > 0$ may then be seen as an arbitrary (constant) scaling factor, and there is no loss of information by taking $\vert \phi \vert = 1$; then

$f(t) = \mathbf e_\phi \cdot \gamma(t) \tag 7$

is of itself precisely the component of $\gamma(t)$ along $\mathbf e_\phi$, and this component is extremal at $\tau$ such that

$f'(\tau) = \mathbf e_\phi \cdot \gamma'(\tau) = 0, \tag 8$

that is, where $\gamma'(\tau)$ is normal to $\mathbf e_\phi$; hence we may think of $\gamma'(\tau)$ as tangent to or lying in the hyperplane normal to $\mathbf e_\phi$ passing through $\gamma(\tau)$.

We may in fact find the second derivitive of $f(t)$:

$f''(t) = \mathbf e_\phi \cdot \gamma''(t), \tag9$

which of course provides information aas to the nature of an extremum $\tau$; when

$f''(\tau) > 0, \tag{10}$

then $\tau$ is a local minumum, and so forth and so on..

A more geometric approach is had by parametrizing $\gamma(s)$ by its arc-length $s$; then

$\dot \gamma(s) = T(s), \tag{11}$

the unit tangent vector field to $\gamma(s)$.  Then

$f'(s) = \mathbf e_\phi \cdot \dot \gamma(s) = \mathbf e_\phi \cdot T(s); \tag{12}$

proceeding in this direction, we have

$f''(s) = \mathbf e_\phi \cdot \ddot \gamma(s) = \mathbf e_\phi \cdot \dot T(s) = \mathbf e_\phi \cdot \kappa(s) N(s), \tag{13}$

where we have used the Frenet-Serret relation

$\dot T(s) = \kappa(s) N{s}, \tag{14}$

where

$\kappa(s) > 0 \tag{15}$

is the curvature, and $N(s)$ is the unit normal vector to $\gamma(s)$.  In light of (15) we see that the sign of $f''(s)$ is given by $\mathbf e_\phi \cdot N(s)$, that is,

$f''(s) > 0 \tag{16}$

when $\gamma(s)$ is turning toward the positive side of the hyperplane normal to $\mathbf e_\phi$, which side is in fact characterized by the equation

$X \cdot \mathbf e_\phi - Y \cdot \mathbf e_\phi = (X - Y) \cdot \mathbf e_\phi >  0, \tag{17}$

where $Y \in \Bbb R^n$ lies in the hyperplane itself and $X \in \Bbb R^n$ on the positive side.  (Under these hypotheses on $X$ and $Y$,

End of Note.


问:
Determine whether vector system $x^2, x^3, |x^3|$ is linearly independent
How to determine whether vector system $x^2, x^3, |x^3|$ is linearly independent in vector space $\mathbb{R}^{\mathbb{R}}$ or not?

I know that linear independence means that $\lambda_1x^2+\lambda_2x^3+\lambda_3|x^3|=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$.

From that I get $x^2(\lambda_1+\lambda_2x)+\lambda_3|x^3|=0$ and
\begin{cases}{
  x^2(\lambda_1+\lambda_2x) = 0, \\
  \lambda_3|x^3|=0
  }
\end{cases}
But how to finish this using definition?

答:
I would proceed like that. Dividing by $x^2$ for $x \neq 0$ you get
$$\lambda_1 + \lambda_2 x + \lambda_3 \vert x \vert =0$$

Taking the limit at $0$ implies $\lambda_1 =0 $. Dividing again by $x$ you get the equality $\lambda_2 + \lambda_3 = 0$ valid for all $x \gt 0$. And therefore $\lambda(x - \vert x \vert)=0$ for all $x \in \mathbb R$ where $\lambda=\lambda_2=-\lambda_3$.

Taking finally $x=-1$ you get $\lambda=0$ and can finally conclude.


问:
I have difficulty counting
Take this example:

$$84 + 87 + 90 + 93 .. + 180 + 183$$

If we want to use Gauss' way of finding this sum, we have to find the number of elements in this. What I do is just take the difference of the highest and lowest sum (99) and divide by 3, but this yields 33 and that can't be correct because it needs to be an even number in order to make sense. Why does it count 34 elements instead of the 33 my (faulty) arithmetic gives me?

答: Try finding $3+6+\cdots+183$, and then $3+6+\cdots+81$, and subtracting.</p>

Title: I have difficulty counting


问:
Old Math Books Request
I'm looking for books that are written in an old style but are also concise and rigorous. Some examples of what I mean:

Calculus on Manifolds by Spivak

Complex Analysis by Ahlfors

Mechanics by Landau and Lifshitz (not a math book, but consistent with the style I'm looking for)

All of these books are written very concisely (and as a plus, they're written in a very nice old-style font!). Any other recommendations people would have of math books that resemble these? Thanks for the help!

答:
There are so many it's impossible to even know where to begin ! For one thing, one has to define what one considers "old". I'll say, somewhat arbitrarily, pre-1970, but really in my case with an emphasis on 1600 - early 20th century.

Since this is a long post I think I'll say at the outset some of my favorite older but not that much older mathematical writers are Courant (I think his 1934 Calculus book is better than any modern Calculus book I have seen), Coxeter, Burnside, and Casey. Going back further, Euler (Gauß as well but I quickly get lost, would love a tour guide), Serrè, and Möbius, and further still Euclid's «Elements», and Appolonius of Perga, «Conics».

I can try to note some of the titles that I think are better in terms of being concise and rigourous, but for myself, I would say that going to the limit of concise and rigourous is not always primary. An extreme example that comes to mind immediately is one of Abel's original articles, translated into German in 1889 by H. Maser, «Abhandlung über die algebraischen Gleichungen, in welcher die Unmöglichkeit der Auflösung der allgemeinen Gleichung fünften Grades bewiesen wird» (1824) on the impossibility of solving fifth degree equations which is literally 4 pages long. Historically, in this case, Abel is said to have been so poor that he could not afford the paper to write as much as he would have liked to. It qualifies as concise and rigourous, but speaking for myself, I have found it almost impossible to make headway because it is so much so.

I'm still glad I own it, which leads into a variety of other reasons one can be interested in older works. I would list them roughly, unordered, as historical perspective (not least, for me, seeing all the amazing calculation techniques that were used before computers which have now been relegated to the dust-bin), connection with the thoughts of brilliant minds, appreciation of old publishing (which contrary to a comment on the original question is sometimes beautifully typeset and illustrated on very high quality paper), study of other languages besides English, accessiblility (I often find myself understanding older authors better than modern ones, especially in what were at the time textbooks, because often nothing was assumed about the reader's preparation as well as having self-learners in view as target audience) discovering some hidden, lost, or forgotten gems (for example, who has read or even heard of Euler's essay on the theory of music? or books such as Praktisches Zahlenrechnen, which might be thought of as an idiot savant's playbook, or Gauß' review of Babbage's Logarithm Tables where he glows over the typesetting and ink, or for that matter his writings on calculating the date of Easter).

A very incomplete listing is below, sort of a semi-random walk, early printings when I've been able to buy them. Although I've spent a fair amount I usually note to myself that on the average I pay less for a vintage volume in good condition than a university student pays for a brand new required undergraduate or graduate coursebook (eg hundreds of dollars). I should note I cannot claim to have read a high proportion of what I own, although there are certainly a few I have read cover to cover. In some cases just reading slivers here and there has been enough, in others it becomes almost like stamp collecting - but a lot more interesting than stamps.

Anyway, here is a very partial unordered list of some older works I own. In some cases the date is unclear; in a few, date is the date of my volume, not the original work. Several of these are translations from the original, but are themselves still very old - so don't go too hard on me if a date is after the author was dead or the book was written!

Some of these I freely admit are too difficult for me to really understand well, but I have read and understood enough of each of them to get something, and to appreciate them. Others are surprisingly accessible. Others are really meant to be on the easy side by those who wrote them, but are not less worthy because of that. There are a few I am not mentioning, because they are almost too obvious, such as whatever you can understand of Gauß, the works of Hardy and his associates that have been receiving a lot of completely deserved attention, etcetera etcetera. These I'm hoping are more obscure. I'm omitting a lot - works aren't great just because they are old - but I'm hoping that all of these have some merit.

«Differential and Integral Calculus», Courant, 1934. (translation into English)

«Theory of Equations», Burnside and Patton, 1918.

Books by Coxeter - many.

«A Treatise on Conic Sections», Casey, 1885 (several other books by Casey also).

«Praktisches Zahlenrechnen», Werkmeister (gotta love the name with the title), 1945.

«Darstellende Geometrie», Haußner, 1908.

«Astronomie», Möbius, 1910.

«Number Theory», Mathews 1890?.

«Algèbra Supérieure», Serrèt, 1910.

«Geometry and the Imagination (Anschaulich Geometrie)», Hilbert, 1946.

«Essai d'une Nouvelle Théorie de la Musique», Euler, 1839.

«Darstellende Geometrie», Fischer, 1921. (and several other books by Fischer).

«Projective Geometry», Veblen and Young, 1910.

«Theory of Groups», Kurosh, 1956.

«Élemens D'Algebre», Euler, 1774 (Translated by one of the Bernoulli's, plus La Grange).

«Induction and Analogy in Mathematics», Polya, 1954.

«Lineare Algebra», Lorenz, 1989 (so not completely 'old' by my definition).

«Gesammelte Mathematische Abhandlungen», Schläfli, 1956.

«Generatingfunctionology», Wilf, 1994. (again breaking my 'old' definition).

«Introduction to the Theory of Groups», Alexandroff, 1959.

«Theory of Equations», Thomas, 1938.

«Introduction to Inequalities», Beckenbach and Bellman, 1961.

«Gesammelte Werke», Steiner, 1881 (incredible fold-out illustrations).

«Die Ausdehnungslehre», Grassmann, 1862.

«Geometric Transformations», Yaglom, 1962.

«Euclidean and Non-Euclidean Geometries», Greenburg, 1972.

«Irrational Numbers», Niven, 1956. (and other books by Niven - he's a good writer).

«The Theory of Algebraic Numbers», Pollard, 1950.

«The Geometry of the Zeroes», Marden, 1949.

Title: Old Math Books Request


问:
$P(A)=a$ and $X=1_{\{A\}}$ what is Var(X)
$P(A)=a$ and $X=1_{\{A\}}$ what is $Var(X)$

So what I did:

$$Var(X)=E[X^2]-E[X]^2$$

now $E[X]=P(A)=a$ so we have

$$Var(X)=E[X^2]-a^2$$

Does $E[X^2]$ also equal $a^2$?

making the answer $Var(X)=0$?

答: Expectation of $X^2$ is $1^2\cdot a+0^2(1-a)=a$, giving the variance $a-a^2=a(1-a)$. Note that $X$ is a Bernoulli trial so its variance was expected to be $pq=p(1-p)=a(1-a)$.


答:
In the sciences we never take the log of a number with a unit.  For any unitless $k$, $\ln(k)$ is defined. But for a $k$ with a unit, $\ln(k)$ is typically avoided. Its undefined. Its undefined because the answer you get and the relationships you find end up being dependent on the unit, and not on the pure mathematics. Strictly speaking this doesnt really matter. Every equation - in the sciences at least - use constants and ratios to convert units, but the idea is to avoid the need.

The same logic holds with trig functions. The evaluation of $\sin(x)$ will change depending on if youre using radians, grads, degrees, or any other unit.  The radian is a formal standard, and there are a number of mathematical reasons for that.

Often what we do is have a constant that solves the problem. $\ln(kc)$, where $c$ is the inverse units of whatever units $k$ has. Thus $kc$ becomes unitless.  Sometimes that $c$ constant is hidden and presumed, and youre simply told "make sure youre using mks units". Similarly with trig functions, $\sin(kx)$ works for degrees if $k$ is in units radians per degree.

As a matter of fact, we can take this a step further. Beyond the scope...

Inside of trig functions such as $\sin(x)$, $x$ is treated as "unitless" for the same argument. But in fact it does have a unit, the radian.  This is the scientific and mathematical standard.  A radian is sort of a unitless unit; an angle that represents the ratio between the length of an arc to the length of a radius, $\frac{\mathrm{length}}{\mathrm{length}}$, what you might think is unitless is in context just a radian. What comes out of the trig function though is still unitless.

We can do a similar thing with exponentials.  The unit Neper, $\mathrm{Np}$, is the unit on the value $x$ that you stick into an exponential $e^x$ to get a unitless solution.  Naturally it kind of makes sense for the values inside the argument and outside the function to be different. The Neper is a sort of unitless unit that represents the logarithm of a unitless number.

Title: What happen to units when we take Ln?

Body:
What happens to units when we take the Ln of the both side of the equation?
For example for an equation $PV^k = C$ (which becomes $\ln P = -k \ln V + \ln C$), how can we find the units of $k$ and $C$?
-->(the unit of $P$ is mmHg and the unit of $V$ is $cm^3$)

Thank you so much..


问:
Orientation induced from inner and exterior
I'm reading From Calculus to Cohomology, Chapter 11 Index of Vector Fields. In the proof of Theorem 11.22, $U\subset\mathbb{R}^n$ is an open subset, $D\subset U$ is a closed ball, then it says that the orientation of $\partial D$ induced from $U\backslash\mathrm{Int}(D)$, is opposite to the orientation induced from $D$.How do I check this?

答: $D$ is a manifold with boundary, and the orientation on $\partial D$ is induced by a outward vector field, so if $D=B(0,1)$ for example the outward vector field is $v(x)=x$, on other hand, the outward vector field of $\partial D$ on $U-int(D)$ it's $u(x)=-x=-v(x)$ (Draw the situation), so the orientation it's opposite orientation.


问:
What happen to units when we take Ln?
What happens to units when we take the Ln of the both side of the equation?
For example for an equation $PV^k = C$ (which becomes $\ln P = -k \ln V + \ln C$), how can we find the units of $k$ and $C$?
-->(the unit of $P$ is mmHg and the unit of $V$ is $cm^3$)

Thank you so much..

答:
It should be obvious that in the $PV^k=C$ equation the value of $C$ depends on the units of $P$ and $V$. For example, one can measure volume $V$ is in $m^3$, or liters, or $ft^3$. Then I would write the above equation as $$\frac P{Nm^{-2}}\frac{V^k}{m^{3k}}=\frac{C}{Nm^{3k-2}}$$
Then one can use dimensionless quantities $p=\frac P{Nm^{-2}}$, $v=\frac v{m^3}$, $c=\frac{C}{Nm^{3k-2}}$.
With this, you can take the logarithm
$$\ln p=-k\ln v+\ln c$$

Title: What happen to units when we take Ln?


问:
Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$

I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.

Link: http://web.mit.edu/rwbarton/Public/func-eq.pdf

答:
Since $\rm\;p(2x)\;$ has roots being half the roots $\rm\;r_i\;$ of $\rm\;p(x)\;$, and $\rm\;\frac{p(x)}{p(2x)} = \frac{x-16}{16(x-1)}\;$, for the roots to all cancel like that leaving only one term in the numerator and denominator, we must have simply that the map $\rm\;r\to r/2\;$ is a left shift map on the (intermediate) roots. The numerator shows the largest root is 16, and the denomimator shows half the smallest root is 1. The other intermediate roots must obey the left shift $\rm\;r_i/2 = r_{i-1},\;$ so the roots are $\rm\;2,4,8,16,\;$ i.e. $\rm\;p(x) = c(x-2)(x-4)(x-8)(x-16)\;$

EDIT$\:$ I purposely omitted a detail above since I thought it might be a homework problem. Based on the comments, this caused some confusion, so I will elaborate. Namely, notice that each linear linear factor of $\rm\;p(x)\;$ introduces a constant factor of $2\:$ in the denominator of $\rm\: p(x)/p(2x)$. But said denominator has constant factor $16$, so the $4$ factors found above are complete. In particular, $0$ cannot be a root of $\rm\:p(x).$

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Intuitive understanding of why the sum of nth roots of unity is $0$
Wikipedia says that it is intuitively obvious that the sum of $n$th roots of unity is $0$.  To me it seems more obvious when considering the fact that $\displaystyle 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$ and that $\{\alpha \in \mathbb{C}: \alpha^n=1\}$ is a cyclic group.

For me, it is hard to, intuitively, see this in the case that $n$ is an odd composite natural number.

Here is my reasoning thus far:

Let $n=5$. (I understand this is simple when considering the minimal polynomial).

Since the conjugate of an $n$th root of unity is also an $n$th root of unity, it is easy to see that, when written in trigonometric form, the sum (possible reordering here)

$1+\alpha^1+\alpha^2+\alpha^3+\alpha^4$

$=1+( \cos (72)+i \sin (72))+( \cos (144) + i \sin (144))+( \cos (72)-i \sin (72))+( \cos (144)-i \sin (144))$

$=1+2 \cos (72)+2 \cos (144)$.

Somehow $2 \cos (72)+2 \cos (144)=-1$, but I'm not sure this is obvious, or perhaps I'm being silly here.

What do you think?  Silly?

答:
Appealing to physical intuition, it is just the center of mass.

Think of the roots of unity as the coordinates of $n$ identical balls equally spaced around a circle in $\mathbb{R}^2$.  What is the location of the center of mass?  It is the center of the circle.  Consequently, adding the roots of unity gives the center of the unit circle, which is $0$.


答: There are 34 terms starting with $84+0 \times 3=84$ and ending with $84+33\times 3=183$</p>

Title: I have difficulty counting

Body:
Take this example:

$$84 + 87 + 90 + 93 .. + 180 + 183$$

If we want to use Gauss' way of finding this sum, we have to find the number of elements in this. What I do is just take the difference of the highest and lowest sum (99) and divide by 3, but this yields 33 and that can't be correct because it needs to be an even number in order to make sense. Why does it count 34 elements instead of the 33 my (faulty) arithmetic gives me?


问:
Determine how many group homomorphisms from $\Bbb{Z}_3$ (with addition as the group operation) to $S_4$ are there
First of all, is  $\Bbb{Z}_3$ with addition a set that contains ${0,1,2}$ with $\bmod(3)$?
Then, $h(1)=$ identity element is a possible homomorphism.
If $h(1)$ is not the identity element, are we trying to find how many element in S4 has a order of $3$?
If so, given $S_4$ is isomorphic to the symmetry group of a cube, we have the elements included rotation $270^\circ$ are of order $3$. If that's the case the answer equal to $13$ since there are 13 rotational symmetry axis in a cube?

答:
Let $G$ and $H$ be groups, and $h\colon G\to H$ a homomorphism. Then, if $g\in G$ has finite order $o(g)$, we get:

\begin{alignat}{1}
e_H &= h(e_G) \\
&=h(g^{o(g)}) \\
&=(h(g))^{o(g)} \\
\end{alignat}

whence $o(h(g))\mid o(g)$. In our case, $G=\Bbb Z_3=\{0,1,2\}$ and $o(1)=o(2)=3$. Therefore, there are at most as many nontrivial such homomorphisms as the number of $h$'s which send $0$ to $()$, $1$ to some $3$-cycle and $2$ to some $3$-cycle either. Now, $h(1)$ and $h(2)$ can't be the same $3$-cycle, say $\sigma$, because $h(\Bbb Z_3)=\{(),\sigma\}$ is not a subgroup of $S_4$. Hence, there are actually at most as many nontrivial such homomorphisms as the number of $h$'s which send $0$ to $()$, $1$ to some $3$-cycle, say $\sigma$, and $2$ to a different $3$-cycle, say $\tau$ ($\ne\sigma$). Finally, $\{(),\sigma,\tau\}\le S_4\iff \sigma\tau=()$ (closure requirement), which cuts the number of combinations down to $8$, namely:

\begin{alignat}{1}
&h(\Bbb Z_3)=\{\space(),\space h(1)=(1)(234),\space h(2)=(1)(243)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(1)(243),\space h(2)=(1)(234)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(2)(134),\space h(2)=(2)(143)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(2)(143),\space h(2)=(2)(134)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(3)(124),\space h(2)=(3)(142)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(3)(142),\space h(2)=(3)(124)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(4)(123),\space h(2)=(4)(132)\} \\
&h(\Bbb Z_3)=\{\space(),\space h(1)=(4)(132),\space h(2)=(4)(123)\} \\
\end{alignat}

So, with the trivial one ($h(\Bbb Z_3)=\{()\}$), there are $9$ homomorphisms from $\Bbb Z_3$ to $S_4$ altogether.