问:
Does Lagrange interpolation yield a unique polynomial over finite fields?
I understand that part of this question has been addressed before (here), but I am confused about the uniqueness of a Lagrange interpolant.

Given any field $F$, I am trying to prove the existence and uniqueness of a Lagrange interpolant in $F[x]$ for any function $f$ from $F$ to itself over a set of distinct $n$ points :$(x_1,f(x_1)) \dots (x_n,f(x_n))$, where each $x_1 \dots x_n$ is unique. Following the proof outlined here, I see that it says the proof for both the existence and uniqueness of a Lagrange interpolant is satisfied for any given field, which implies that it also satisfied for finite fields. However, there are finitely many functions from $F$ to itself and there are infinitely many polynomials in $F[x]$. I also see that from this post, a polynomial function in a finite field can be given by multiple polynomials. Would this not imply that the uniqueness of a Lagrange interpolant need not hold over a finite field?

Any clarification about how the proof of uniqueness for a Lagrange interpolant over a finite field would be much appreciated!

答:
You are missing the issue that the interpolation polynomial for $n$ points has degree less than $n$ (or is the zero polynomial, but I won't keep repeating this). That makes the polynomial unique even if the field $F$ is finite.  What keeps things unique given the restriction on the degree being less than $n$ is that polynomials of the form $c_0 + c_1x + \cdots + c_{n-1}x^{n-1}$  and $c_0' + c_1'x + \cdots + c_{n-1}'x^{n-1}$  in $F[x]$ that are equal at $n$ different numbers in $F$ must have $c_i = c_i'$ for all $i$.  This is true for all fields, and notice that when $F$ is finite such a result only makes sense when $n \leq |F|$ since $F$ has at most $|F|$ distinct elements.  For instance, you can't find a polynomial in $\mathbf F_3[x]$ taking specified values at $5$ different elements of  $\mathbf F_3$ since there aren't $5$ different elements of $\mathbf F_3$. (You could look for a polynomial in $\mathbf F_9[x]$ taking specified values at 5 different elements of $\mathbf F_9$, but not at 14 different elements of $\mathbf F_9$, and so on.)

You have concern about losing uniqueness over a finite field, but
even over an infinite field you would lose uniqueness without restricting the degree of the interpolation polynomial to have degree less than  $n$: if the distinct points in $F$ are $a_1, \ldots, a_n$ then you can add to an interpolating polynomial an arbitrary polynomial multiple of $(x-a_1)\cdots(x-a_n)$ without changing the values of the polynomial at $a_1,\ldots,a_n$.  You don't lose uniqueness as long as you require the interpolation polynomial at $n$ different numbers in $F$ to have degree less than $n$.

Here is a correct theorem: for a field $F$, distinct $a_1, \ldots, a_n$ in $F$, and arbitrary $b_1, \ldots, b_n$ in $F$, there is a unique polynomial $f(x) \in F[x]$ of degree less than $n$ such that $f(a_i) = b_i$ for $i = 1, \ldots, n$.


问:
When are elements of a p-group independent mod a power of p?
Suppose that an Abelian $p$-group $G$ can be written as a direct sum of countably many finite cyclic groups.  Let $a_1,...,a_k$ be generators of $k$ distinct summands of this direct sum, and let these elements have orders $p^{n_1},...,p^{n_k}$ respectively.  Then for what values of $m$ are $a_1,...,a_k$ linearly independent over $\mathbb{Z}_{p^m}$?  I mean that in the sense that if $0\leq c_1,...,c_k<p^m$ and $c_1a_1+...c_ka_k=0$ then $c_1=...=c_k=0$

I’m particularly interested in the case when not all of the $n_i$’s are equal.

答:
In general, if you have $G=\bigoplus_{i\in I} A_i$, and $x_j\in A_{i_j}$, $j=1,\ldots,k$, and the $i_j$ are pairwise distinct, then (using additive notation)
$$\alpha_1x_1+\cdots+\alpha_kx_k = 0\iff \forall j(\alpha_jx_j=0).$$
This simply because that’s how the direct sum works.

In your case, the $a_i$ are in pairwise distinct summands, so we definitely have that $c_1a_1+\cdots +c_ka_k=0$ if and only if $c_ia_i=0$ for each $i=1,\ldots,k$. This occurs if and only if $p^{n_i}|c_i$ for $i=1,\ldots,k$.

In particular, if $m\gt n_i$ for some $i$, then setting $c_j=0$ for $j\neq i$ and $c_i=p^{n_i}\lt p^m$ gives $0=c_1a_1+\cdots+c_ka_k$, yet $c_i\neq 0$. Thus, we must have $m\leq n_i$ for all $i$; that is, $m\leq \min(n_1,\ldots,n_k)$.

Conversely, if $m\leq\min(n_1,\ldots,n_k$, and $c_1a_1+\cdots+c_ka_k=0$, then $p^{n_i}|c_i$ for each $i$; since $0\leq c_i\lt p^m\leq p^{n_i}$, this forces $c_i=0$ for all $i$.

So the values of $m$ that give you the desired implication are precisely $1\leq m\leq \min(n_1,\ldots,n_k)$.

ContentLicense: CC BY-SA 4.0


问:
Jacobian Determinant of frenet transformation
If anybody can help with this.
Given a point $\boldsymbol x = (x,y)$, it can be represented as $\boldsymbol x=\boldsymbol p(s)+r \boldsymbol u(s) $. p is a curve parametrized with arc length s. $r$ is the distance between $\boldsymbol p$ and $\boldsymbol x$. $\theta$ is the angle between the tangent vector of $\boldsymbol p$ and the x-axis. $\boldsymbol u(s)$ is the unit normal vector of $\boldsymbol p$ at s. How can I prove that

$$\begin{vmatrix}
\frac{\partial x}{\partial s} & \frac{\partial x}{\partial r} \\
\frac{\partial y}{\partial s} & \frac{\partial y}{\partial r}
\end{vmatrix} = 1 + r \dfrac{d\theta}{ds}
$$
I think that relates somehow to the determinant of jacobian of frenet frame but I could not find any resource that has the proof.

答:
Writing the curvature $\kappa = d\theta/ds$ and suppressing the vector nature of the problem just makes it difficult. You are right that the Frenet equations are relevant.

We have $\dfrac{\partial x}{\partial s} = T + r(-\kappa T)$ (where $T,N$ are the Frenet frame of the curve $p$) and $\dfrac{\partial x}{\partial r} = N$. So the determinant should be $1-r\kappa$, not the $1+r\kappa$ you have.


问:
Understanding the internal direct product of a group.
I have come across the statement:

$G$ is the direct product of its subgroups $N_1$ and $N_2$ if the following conditions hold:

1) $N_1,N_2$ are normal subgroups.

2) $N_1\cap N_2=\{e\}$

3) They generate the group, i.e. $G=N_1N_2$

The book had previously defined the external direct product of two groups as consisting of ordered pairs of elements, one from each group. I felt that this wouldn't make much sense in this context, so I assumed this meant you could write each element of $G$ uniquely as a product of $n_1n_2$ where $n_1\in N_1,n_2\in N_2$. Then to prove this I wrote if $n_1n_2=n_1'n_2'$ then we have $n_1'^{-1}n_1=n_2'n_2^{-1}$, and since $N_1\cap N_2=\{e\}$ we have $n_1=n_1',n_2=n_2'$. This didn't however use the fact that these are normal subgroups, and I thought that maybe this is necessary because suppose $n_1n_2\ne e$, then $n_1n_2=n_2(n_2^{-1}n_1n_2)$. As $n_2\in N_2$ the product in the brackets must be in $N_1$, since $n_1n_2\ne e$ and the subsets are trivially disjoint.

I think this is correct, but the book doesn't say anything more on the subject so I can't be sure. Is this chain of reasoning correct?

答: I answered this question (starting from another one and generalizing it) in my post here. I hope it helps. </p>

ContentLicense: CC BY-SA 3.0


问:
Numerical integration of Newtons Second Law
I am having trouble wrapping my head around this concept. I am trying to numerically solve the following equation in python (using ODEint) for a n-body simulation:

$$ m_2\frac{d^2\textbf{x}}{dt^2} = \frac{Gm_1m_2}{r^2}\hat{r} \ $$

Then I chose the substation:

$$ \textbf{v} = \frac{d \textbf{x}}{dt} \longrightarrow \frac{d\textbf{v}}{dt} = \frac{d^2\textbf{x}}{dt^2}$$

with some basic manipulation this leads to:

$$ \frac{d\textbf{v}}{dt} = \frac{Gm_1}{r^2}\hat{r} \ $$

However I'm stuck as to what todo now.

I am unsure how r hat, and how it is inplamentent when solving this equation

Have I made a mistake in cancelling out both of the smaller masses (m2), is this true only if m2<<m1?

I've seen some answers along the lines of $ \frac{d\textbf{v}}{dt} = - \frac{Gm_1m_2}{r^3}\textbf{x} \ $ and I'm wondering if this is the same as what I have derived?

Any help would be much approached in understanding this topic

答:
First of all, I believe you have dropped a minus sign from your equation since Newton's law is normally written $$F=-\frac{Gm_1m_2}{r^2}\hat{r}$$ meaning that $F=m_2\frac{d^2\mathbf{r}}{dt^2}$ gives us $$m_2\frac{d^2\mathbf{r}}{dt^2}=-\frac{Gm_1m_2}{r^2}\hat{r}$$

$$\hat{r}:=\frac{1}{\sqrt{x^2+y^2+z^2}}(x\hat{x}+y\hat{y}+z\hat{z}),$$ This is the position unit vector. The way you have your equation set up it makes more sense to have $$m_2 \frac{d^2 \mathbf{x}}{dt^2}=-\frac{Gm_1m_2}{|\mathbf{x}|^2} \hat{\mathbf{x}}$$ with $\mathbf{x}=\mathbf{r}$

Your starting equation is really only correct when $m_1$ is so large that it doesn't move relative to its starting condition. If you want to be more you have to consider how each body might move in relation to each other. To do this you can write out the forces between the two objects relative to a stationary frame (very tedious and difficult) or you can use the Lagrange equation to solve for the equations of motion and remove a couple of the restraints and utilize the symmetries of the problem.

The equation you have written here is quite similar to the one you started with, though I think it might be slightly wrong. We can write $\mathbf{x}=|\mathbf{x}|\hat{\mathbf{x}}$ which we can convert our equation to $$m_2 \frac{d^2 \mathbf{x}}{dt^2}=-\frac{Gm_1m_2}{|\mathbf{x}|^3}\mathbf{x}$$ It doesn't really make much of a difference.


问:
Is this explanation of normal subgroups and quotient groups correct?
I apologize for the long post, but I'm currently a student finishing up his first semester in group theory. My introduction was pretty definition-heavy so I've found I can internalize concepts (such as quotient groups, normal subgroups, etc.) myself by forming my own way of motivating and teaching them intuitively. I'd like to know if my current presentation/understanding is correct.

I think after learning about subgroups and Lagrange's theorem, a natural question is then if we can break down a group $G$ to better understand its parts and hopefully the whole $G$ (as is tradition in any analytical endeavor). But if we want to pull back any useful understanding of $G$ from this smaller group, it ought to preserve some structure of $G$. That structure is exactly how the operation acts on elements of $G$ (since groups are just elements with an operation relating them).

So for the sake of exploration, we pretend to have the magic function $\phi : G \rightarrow H$ that does exactly this for us–maps $(G, *_G)$ to some smaller part $(H, *_H)$, then ask what we can say about $\phi$. Our original goal was for $\phi$ to preserve the operation, i.e. for all $a, b \in G$ that $\phi(a *_G b) = \phi(a) *_H \phi(b)$.

The next thing I would observe is that since $H$ has smaller order, $\phi$ necessarily maps a multiple elements, say $a, b$, in $G$ to the same element in $H$. In this sense, $a$ and $b$ are "equivalent" under $\phi$. Given that we have "willed" $\phi$ to operation-preserving, we can see that a natural way this arises by letting $ak = b$ for some $k \in G$:

$$ak = b \implies \phi(b) = \phi(a *_G k) = \phi(a) *_H \phi(k).$$

If we want $\phi(b) = \phi(a) *_H \phi(k) = \phi(a)$ then $\phi(k) = e_H$. I think leads naturally to the definition of the kernel: it's a set of elements that maps to the identity, and makes $a$ equivalent to $b$ mod $\ker \phi$. And in fact, as I've learned from this answer, we naturally get equivalence classes of elements that partition the group into cosets analogous to modular arithmetic. So, $\phi$ takes elements and puts them neatly into these equivalence classes (abstracting away some of the details in $G$ that look "the same" in $H$, leading–at least for me–directly to the First Isomorphism Theorem). Then it makes sense to propose the map $\phi : g \mapsto g \ker \phi$.

The next question becomes what the operation of this $H$ looks like. We've established that elements of $H$ are cosets (and equivalence classes), so for two elements in cosets $g_1 \ker\phi$ and $g_2 \ker\phi$, once combined by $*_H$ we'd want for the result to be in $g_1g_2 \ker\phi$ (modular arithmetic analogy works here as well). Set-wise, we might write

$$g_1\ker\phi \cdot g_2\ker\phi = g_1g_2\ker\phi.$$

But does this come for free? For an element $g_1k_1g_2k_2 \in g_1\ker\phi \cdot g_2\ker\phi$ to look like $g_1g_2k$ for some $k$, it must be that $k_1g_2 = g_2k_3$ for some $k_3$. Set-wise this can be written as $g\ker\phi = \ker\phi g$, i.e. left-costs = right-cosets and it turns out it indeed satisfies this condition and we are safe to proceed.

So, in the end, we have designed $H$, a broken-down version of $G$. And how did we do it? By "dividing out" or "quotienting out" the information that looks the same under $\phi$ in $H$–$\ker \phi$. Thus we write $H$ as $G/\ker\phi$, aptly called a quotient group.

Although, you could flip this presentation, and instead of viewing from the kernel perspective, suppose $K$ is some arbitrary group. Then it must satisfy the condition of left-cosets = right-cosets (which we name normality because it is a nontrivial property that gives us a usable quotient) for $G/K$ to be a group, as $\ker\phi$ already does, and through satisfying normality automatically becomes the kernel of some homomorphism (namely the natural, which I've presented).

My questions are:

Is this presentation correct (on an intuitive level, I know there are lots of places for concrete proofs)? It feels right to me, but I also feel like I may have gotten definitions vs. implications mixed up.

If so, does any textbook follows this approach that I can dig into?

I think homomorphisms also can fit into this framework, given I suggest $\phi$ pretty early on, but how would non-surjective homomorphisms be explained?

答:
It seems like you have understood the story pretty well.

To address your specific questions:

Yes, seems good.  I agree that you could do some work polishing this to make everything fully rigorous, including making some decisions about what you want to be a definition and what you want to be a theorem.

I think this is pretty close to how most textbooks present the quotient groups and the first isomorphism theorem.

Non-surjective homomorphisms are surjective onto their image.  The first isomorphism theorem says that if $\phi: A \to B$ is a group homomorphism, then $\textrm{Im}(\phi) \cong A/\ker(\phi)$.  So $\phi$ is picking out a subgroup of $B$ which is isomorphic to the quotient group $A/\ker(\phi)$.


问:
Cancellation laws and group
Maybe all that I am writing below is already exists at forum but I would like to know is my reasoning correct? So please do not duplicate my question.
I am in the begining of group theory and it is very important to me to know that my reasoning is correct.

We know the following fact from group theory: Suppose a finite set $G$ is closed under an associative product and that both cancellation laws hold in $G$ $\Rightarrow$ $G$ must be a group.

However, if suppose that only one of the cancellation laws hold then $G$ need not to be a group.

Let $G$ be the set with three elements, i.e. $G=\{a,b,e\}$ and we define the product $x\cdot y=x$ for any $x,y\in G$. We see that this product is binary and associative since $(x\cdot y)\cdot z=x\cdot y=x$ and $x\cdot (y\cdot z)=x$ and we see that indeed associativity holds in $G$.

If $x\cdot z=y\cdot z$ then $x=y$  by the definition of product $\cdot$ hence right cancellation law holds.

However, from $a\cdot a=a\cdot b=a$ does not follow that $a=b$. So left cancellation law fails in $G$.

It's easy to verify that the set $G$ is not group since if we suppose that $G$ is group then $\exists e'\in G$ such that $e'\cdot a=a\cdot e'=a$ and we conclude that $e'=a$. If we apply this identity element for $b\in G$ we get $b\cdot e'=e'\cdot b=b$ and we get that $e'=b$ thus $a=b$ which is contradiction.

Would be grateful for verification.

答: You have proved via using left and right cancellation in given equalities. By equality, it means a common term in a table. So, $ae'=e'a, be'=e'b$ leads to contradiction as $a,b$ are distinct. Although, your interpretation is correct as mean $e'=a$, hence $aa=a$; also $e'=b$, hence $bb=b$. But, still there seems flaw to me, as identity $e'$ should, by defn., mean $e'a =a$ & $e'b =b$. So, $ae'=e'a\implies ae' = a, be'=e'b\implies be'=b$. So, a better way to use left & right cancellation should be used, for the purpose of the proof. But, am unsure if can find one.

Title: Cancellation laws and group


问:
If you choose two different sets of results to match inside a coin flipping record, does the odds of finding a match depend on what you pick on each?
Example, if you flip a coin 10 times and record the results, and you pick 2 different sets of 5 length (e.g HHTHT and TTHTT where H=heads and T=tails) and the goal is to match any of your two sets anywhere inside the coin results (e.g TTTTHHTHTH contains the example set HHTHT).

The question is, can the combined odds of finding a match on either set be affected depending on the way the sets are chosen?

Also bonus in case the odds vary, explain which strategies could be used to improve and/or worsen the chances of finding a match (if it doesn't matter at all, bonus doesn't apply obviously).

My hypothesis is that if they overlap (e.g set_a = HHHHH and set_b = THHHH) you end up with less odds of finding a match, but usually these kind of problems are counter-intuitive and I couldn't tackle this problem myself, at least not in a non-inductive way.

答:
Yes, your choice of sequence can make a difference. Let's look at one of the very simplest cases: HH vs. TH in a sequence of three flips. Then the sequences that contain HH are HHH, HHT and THH; therefore, your chances are $3/8$ of a match. OTOH, the sequences that contain TH are HTH, THH, THT, and TTH, so your chances of matching there are $1/2$. This gap only gets longer with longer sequences; there are $F_{n+1}$ sequences of coin flips of length $n$ with no HH in them (try to prove this yourself; it's a pretty classic appearance of the Fibonacci numbers), so the number of 'hits' on HH is $2^n-F_{n+1}$, and your hit probability is $1-\frac{F_{n+1}}{2^n}\approx 1-\phi\cdot\left(\frac\phi2\right)^n$. OTOH, the only sequences with no TH in them are those of the form H$^*$T$^*$ and there are only $n+1$ of these, so your hit probability is $1-\frac{n+1}{2^n}$; your chances of a miss are exponentially smaller than they are on HH.

Abstractly, this is a question of how many words of a given length have a given-length subword; my first instinct would be to use Fibonacci words but you 'clearly' want a word of density 1/2 if you can. It's possible that Thue-Morse sequences are ideal, but that's something that will take quite a bit more digging...


问:
Basic questions about fields and ordered fields
I have some basic questions about fields.

Let $(\mathbb K,\oplus,\odot)$ be a field.

Are there any fields which don't have $0$ as additive identity or $1$ as multiplicative identity?
Are there any fields not covering $0$ or $1$?

In some literature they write $0_{\mathbb K}$ instead of $0$ and $1_{\mathbb K}$ instead of $1$.
So what does $n_{\mathbb{K}}$ mean and what is the difference between $n_{\mathbb K}$ and $n$?

Are the operations $\oplus$ and $\odot$ always addition and multiplication?

Assume you have an ordered field $\mathbb F$. If I read about ordered fields I always read about positives and negatives. But does an ordered field have to contain negatives? Is this implication necessary: $a\in \mathbb F \Rightarrow -a\in \mathbb F$?

答:
These questions can be answered in the more general setting of (unital) rings - a field is simply a commutative ring whose elements are all invertible.

A ring has to have identities for addition and multiplication (it is an abelian group under addition, monoid under multiplication). By convention, these are 0 and 1. A field is a ring, so to answer your first question, almost-yes. There are conceptual ideas like the field with one element but these are not fields in the classical sense.

I would guess $n_\mathbb{K}$ simply refers to $1_\mathbb{K} + \dots + 1_\mathbb{K}$ with $n$ summands. The difference is that there may not be an isomorphic copy of $\mathbb{Z}$ in your field (let alone $\mathbb{Z}/n\mathbb{Z}$) - for example, take $\mathbb{K} = \mathbb{Z}/2\mathbb{Z}$ and $n = 3$. Under the above notation $3_{\mathbb{Z}/2\mathbb{Z}} = 1_{\mathbb{Z}/2\mathbb{Z}}$.

The operations are always called 'addition' and 'multiplication', but the terminology for this comes from the fact that we come up with the abstract definition of a 'ring' to generalise the arithmetic properties of e.g. $\mathbb{Z}$.

Finally, if $a \in R$ then under the abelian group structure of $(R,\oplus)$, there must exist an additive inverse $b$ for which $a + b = 0_R$. We can relabel $b$ by $-a$.


问:
Could someone explain to me why this solution is correct/ how you get to this solution?

$$\left|\{x\in\mathbb{Z}:|x|<10\}\right| = 19.$$

Find the following cardinalities.

答: The set $\{x\in\mathbb{Z}:|x|<10\}$ means all the integers whose absolute value is less than $10$.  Writing $\left|\{x\in\mathbb{Z}:|x|<10\}\right|$ means the size of that set, i.e. how many integers are there whose absolute value is less than $10$?  Well, we can list them out: $-9, -8, -7, -6, -5, -4, -3, -2, -1, 0,1,2,3,4,5,6,7,8,9$ and count them, there are $19$.  Alternatively, we want the number of integers between $-9$ and $9$, inclusive, which we can calculate by $9 - (-9) + 1 = 19$.


问:
If a subrectangle $R$ lies in both one of the rectangles $Q_i$ and one of the rectangles $Q_j^{\prime}$, $\cdots$ (Munkres Analysis on Manifolds)
I am reading "Analysis on Manifolds" by James R. Munkres.

On p.94, Munkres wrote as follows:

Divide the collection of all subrectangles $R$ determined by $P$ into two disjoint subcollections $\mathcal{R}$ and $\mathcal{R^{\prime}}$, so that each rectangle $R\in\mathcal{R}$ lies in one of the rectangles $Q_i$, and each rectangle $R\in\mathcal{R^{\prime}}$ lies in one of the rectangles $Q_j^{\prime}$.

I think there is a possibility that a subrectangle $R$ determined by $P$ lies in both one of the rectangles $Q_i$ and one of the rectangles $Q_j^{\prime}$.

答:
It is possible that a subrectangle $R$ lies in both rectangles $Q_i$ and $Q_j'$, but it does not alter the outcome.  In this case we have

$$U(f,P) - L(f,P) < \sum_{R \in \mathcal{R}}(M_R(f) - m_R(f))v(R) + \sum_{R \in \mathcal{R'}}(M_R(f) - m_R(f))v(R)\\ \leqslant 2M \epsilon'+2\epsilon'v(Q), $$

with strict inequality as the two sums may include common terms.

Nevertheless, the bounds on those sums imply that $U(f,P) - L(f,P) < \epsilon$ which proves integrability


问:
If $X$ is distributed $\mathcal{N}\sim(\mu,1)$ then what is the PDF of $X^2$?
If $X$ is distributed $\mathcal{N}\sim(\mu,1)$ then what is the distribution of $X^2$?
My try:
\begin{align*}
    P(X^2<x)\\
    P(-\sqrt{x}<X<\sqrt{x})\\
    \int_{-\sqrt{x}}^{\sqrt{x}} f(x) \,dx
\end{align*}
Then:
\begin{align*}
    f_{X^{2}}(x)=\frac{d}{dx}\int_{\sqrt{-x}}^{\sqrt{x}} f(x) \,dx\\
    =f_X(\sqrt{x})\frac{1}{2\sqrt{x}}+f_X(-\sqrt{x})\frac{1}{2\sqrt{x}}\\
    =\frac{1}{\sqrt{2\pi x}} e^{\frac{-(\sqrt{x}-\mu)^2}{2}}
\end{align*}
How should I continue?

答:
$X^2$ is a member of the family of noncentral chi-squared distributions.  There are a few problems with your calculations as others have pointed out.  Notably, you should not use the same variable to represent both the limits of integration and the variable of integration.  You should write instead
$$\begin{align}
f_{X^2}(x) &= \frac{d}{dx} \int_{t=-\sqrt{x}}^\sqrt{x} f_X(t) \, dt \\
&= \frac{f_X(\sqrt{x})}{2\sqrt{x}} + \frac{f_X(-\sqrt{x})}{2\sqrt{x}} \\
&= \frac{1}{2 \sqrt{2 \pi x}} \left( e^{-(\sqrt{x}-\mu)^2/2} + e^{-(\sqrt{x} + \mu)^2/2} \right).
\end{align}$$
Your main computational error is in assuming that $f_X(\sqrt{x}) = f_X(-\sqrt{x})$.  This is not true unless $\mu = 0$.

We can stop here, or we can perform additional algebraic manipulations:  note
$$(\sqrt{x} \pm \mu)^2 = x \pm 2\mu \sqrt{x} + \mu^2,$$ therefore
$$e^{-(\sqrt{x} - \mu)^2/2} + e^{-(\sqrt{x} + \mu)^2/2} = e^{-x/2} e^{\mu \sqrt{x}} e^{-\mu^2/2} + e^{-x/2} e^{-\mu \sqrt{x}} e^{-\mu^2/2} = e^{-(x + \mu^2)/2} (e^{\mu \sqrt{x}} + e^{-\mu \sqrt{x}}).$$  Since $e^z + e^{-z} = 2 \cosh z$, we can write $$f_{X^2}(x) = \frac{e^{-(x+\mu^2)/2} \cosh (\mu \sqrt{x})}{\sqrt{2\pi x}}, \quad x \ge 0.$$

This density corresponds to a noncentral chi-squared distribution with $1$ degree of freedom and noncentrality parameter $\mu^2$.


问:
True or False: The line y=x is always at 45 degrees to the x-axis?
I have a doubt.

My teacher has posted a question that Is it always/sometimes/never true that the line y=x is at 45 degrees to the x-axis?

I know that the slope of the line y=x is always 1, and the line must make 45 degrees with x-axis.

I am a bit confused when I am plotting the line y=x at different X-scales (see Figures 1 and 2 attached).
I do not visually see the line at 45 degrees to the x-axis in Figure 2.

What am I missing?

Which answer would be correct and why?
(a) It is always true
(b) It is sometimes true because when we change the scale of x or y axis, graph will distort and not be at 45 degrees.

See images attached.
Fig 1 with equal x and y axis
Fig 2 with different axis

答:
A more advanced approach: we can show $y = x$ has a $45^{\circ}$ angle by using polar coordinates.

Let $x = r \cos \theta$ and $y = r \sin \theta.$  As $y = x$, $r \sin \theta = r \cos \theta$, but $r$ cancels out and we have $\cos \theta = \sin \theta$.  Divide by $\cos \theta$ to get $\tan \theta = 1$, and thus $\theta = \arctan (1)$ or $45^{\circ}$.


问:
$x_1 + x_2 + x_3 + x_4 = 28$ number of solutions
How many solutions are there to the equation in the title with the following constraints:
$0 ≤ x_i, x_1 ≤ 6, x_2 ≤ 10, x_3 ≤ 15, x_4 ≤ 21?$

So, to do this I tried the following:

(A) Total number is ${31 \choose 28}$ ways.

(B) $x_1$ having $7$+ is ${24 \choose 21}$ ways.

(C) $x_1$ having $7$+ and $x_2$ having $11$+ is ${13 \choose 10}$ ways.

There is no way that $x_3$ can have 15+ since $7 + 11 + 15 > 28$.

Answer is $A - B + C = 2,757$ ways.

Is this correct?

答:
This is how I would calculate it.

$T_1 = \binom{31}{3}$

ignores all upper constraints.

$T_2 = \binom{24}{3}$

$x_1 \geq 7.$

$T_3 = \binom{20}{3}$

$x_2 \geq 11.$

$T_4 = \binom{15}{3}$

$x_3 \geq 16.$

$T_5 = \binom{9}{3}$

$x_4 \geq 22.$

$L_1 = T_2 + T_3 + T_4 + T_5.$

$L_1$ represents # of ways that at at least one constraint violated.

Running total so far is $T_1 - L_1$.

$T_6 = \binom{13}{3}$

$x_1 \geq 7, x_2 \geq 11.$

$T_7 = \binom{8}{3}$

$x_1 \geq 7, x_3 \geq 16.$

$T_8 = 0$

$x_1 \geq 7, x_4 \geq 22.$ : impossible

$T_9 = \binom{4}{3}$

$x_2 \geq 11, x_3 \geq 16.$

$T_{10} = 0$

$x_2 \geq 11, x_4 \geq 22.$  : impossible

$T_{11} = 0$

$x_3 \geq 16, x_4 \geq 22.$  : impossible

$L_2 = T_6 + T_7 + T_8 + T_9 + T_{10} + T_{11}.$

$L_2$ represents # of ways that at at least two constraints violated.

Running total so far is $T_1 - L_1 + L_2$.

The above is the final total, because you can not have more than two constraints violated and still have the sum $\leq 28.$


问:
Convergence of $\prod\limits_{n=0}^{\infty} (1+x^{2^{n}})$
I never covered infinite products in my real analysis class, so I have no idea. I was wondering about the convergence of the above product. Thanks in advance.</p>

答:
$$\prod\limits_{n=0}^{\infty}( 1+x^{2^{n}}) $$
is such that every term is $1+x^{2^{n}}$ and since the exponent of $x$ is always even, you have that $x^{2^n}\ge 0$. This means that we have 3 cases:

If $x=0$ then the product is $1$,

Else, if $|x|\ge 1$ we have that the general term is $\ge 2$ so the infinite product is bigger than "$2^{+\infty}$", so it diverges.

Finally, if $0<|x|<1$, we use this criteria:
$$\prod\limits_{n=0}^{\infty}( 1+x^{2^{n}}) \mbox{ converges} \iff \sum_{n=0}^{\infty}\log(1+x^{n^2}) \mbox{ converges.} $$
That series is such that
$$\sum_{n=0}^{\infty}\log(1+x^{n^2}) \le \sum_{n=0}^{\infty}x^{n^2}$$
and for the values of $x$ considered it converges.


问:
Polynomial Set Notation
How do you read out the notation in the screenshot below. I know that the polynomials f(x) and g(x) are in the range of the real numbers but what is the significance of the [x] after the range.

Screenshot

答:
I personally dislike this notation, but its pretty standard. Given a polynomial
$$f(x)=\sum_{k=0}^n a_kx^k$$
If
$$a_0,...,a_n\in S$$
For some arbitrary set $S$ (usually a ring or field) then we write
$$f\in S[x]$$
There are two main reasons I think this notation is abhorrent:

$S[x]$ looks like a functional that takes in some function $x:t\to x(t)$

It seems to suggest the variable used when writing out the function must be $x$, when of course this should be completely arbitrary.

Why we can't use something better like $\text{Poly}(S)$ baffles me.

Title: Polynomial Set Notation


答:
Yes, the two sets does affect the probability of finding one match.

Indeed, given a set $A$ of length 5, call $p(A) $ the probability that a random set of length 10 contains $A$. Given two sets $A,B$ we have

$$ p(A \text{ or } B) = p(A) + p(B) - p(A \text{ and } B) $$

The probability $p(A) $ depends on the "shift-symmetries" that $A$ has: the more it is symmetric, the less is probable to find a match.

Example 1: HHHHH can appear in first position in $2^5$ possible combinations, all the ways one can choose the last letters. It can also appear in second position in $2^5$ possible combinations. Continuing like this, it can appear in position up to six, yielding $6*2^5$ possible combinations. But wait..! Actually the first and second block of combinations overlap when the word starts by HHHHHH, so we have counted this combinations twice!
Taking this complication into account, one obtains

$$6* 2^5 - 5* 2^4 + 4* 2^3 ... -1 = 57 < 192 = 6* 2^5 $$

Example 2: if you take THTTT, the first block overlap with the fifth, and the second overlap with the sixth, and that's all. So you will have $6*2^5- 2* 2^4 = 160$, about three times the odds of the word above!

To conclude, the mixed term roughly gets bigger the smaller word you can find containing both of your $A$ and $B$. Let's call the length of the smallest word containing both the "combined rank". This is only a first approximation of how big is the term. Since it has a minus sign, the combined probability gets bigger if the combined rank is high.

Example: $THTTT$ and $HTHHH$ are cool words, because the shortest word containg both is 10 long, and singularly the words have few shift symmetries. Explicitly you get that the possible combinations are $160+160-2 = 318$.

On the other side, if you take $HHHHH$ as first word and $HHHHHH$ as second, this has a very small probability, because they have a lot if shift symmetries and the combined rank is 5. You get $57$ possibilities, way less than above.

Title: If you choose two different sets of results to match inside a coin flipping record, does the odds of finding a match depend on what you pick on each?

Body:
Example, if you flip a coin 10 times and record the results, and you pick 2 different sets of 5 length (e.g HHTHT and TTHTT where H=heads and T=tails) and the goal is to match any of your two sets anywhere inside the coin results (e.g TTTTHHTHTH contains the example set HHTHT).

The question is, can the combined odds of finding a match on either set be affected depending on the way the sets are chosen?

Also bonus in case the odds vary, explain which strategies could be used to improve and/or worsen the chances of finding a match (if it doesn't matter at all, bonus doesn't apply obviously).

My hypothesis is that if they overlap (e.g set_a = HHHHH and set_b = THHHH) you end up with less odds of finding a match, but usually these kind of problems are counter-intuitive and I couldn't tackle this problem myself, at least not in a non-inductive way.


问:
Why is the distance matrix in optimal transport always the same, and doesn't actually measure distances between samples?
Why does the distance matrix computed in the optimal transport linear programming model always look the same (asides from its size changing with the number of histogram bins) no matter how the source and target distributions might change?

If we really are supposed to be interested in the distances between samples in the source and target distributions, why does optimal transport dumb down this step to a distance comparison of cardinality, that is, the distances between the integer-indices of the source and target distributions, without any imputation of actual sample observations or probabilities?

Matrix D in the image below is the distance matrix, which always shows the same pattern regardless which sample data is thrown in the model:

答:
why does optimal transport dumb down this step to a distance comparison of cardinality, that is, the distances between the integer-indices of the source and target distributions, without any imputation of actual sample observations or probabilities?

The matrix $D$ is completely independent from the input distribution, and element $D_{ij}$ defines how hard it is to move one unit of whatever you are moving from location $i$ to location $j$.

The source and target distributions indicate how much should be moved from and to each location. If you decide to move $0.2$ from location $i$ to location $j$, the contribution to the objective function is $0.2D_{ij}$. Let $s_i$ be the amount available at source location $i$, then the contribution of that amount to the objective function is $\sum_j D_{ij} x_{ij}$ where $x_{ij}$ satisfies $\sum_j x_{ij} = s_i$. Therefore, if $s_i$ is large, row $i$ of matrix $D$ is multiplied with (at least some) large coefficients, so this is how cardinality comes into play.


问:
How to find the group presentation of $\mathbb{Z}^n$
How to find the group presentation of $\mathbb{Z}^n$, where $\mathbb{Z}^n$ is direct product of groups.

答: $\langle a_1,\dots,a_n|a_ia_j=a_ja_i, i,j=1\dots,n\rangle$.  Presentations are not unique, btw.


答:
Let
$$U=\{(x_1,x_2,x_3,x_4)\in \mathbb Z^4\mid x_1+x_2+x_3+x_4=28,\  x_i \geq 0,\  i=1,2,3,4 \},$$
$$A_k=\{(x_1,x_2,x_3,x_4)\in \mathbb Z^4\mid x_1+x_2+x_3+x_4=28,\ x_k\leq a_k,\  x_i \geq 0,\  i=1,2,3,4 \},$$ where $a_1=6$, $a_2 = 10$, $a_3 = 15$, $a_4 = 21$.

The number of solutions you want is
\begin{align}
|A_1\cap A_2 \cap A_3 \cap A_4| &= |U| -|(A_1\cap A_2 \cap A_3 \cap A_4)^c| \\
&= |U| - |A_1^c \cup A_2^c \cup A_3^c\cup A_4^c|\\
&= |U| - (|A_1^c|+|A_2^c|+|A_3^c|+|A_4^c|)\\
&+(|A_1^c\cap A_2^c|+|A_1^c\cap A_3^c|+|A_1^c\cap A_4^c|+|A_2^c\cap A_3^c|+|A_2^c\cap A_4^c|+|A_3^c\cap A_4^c|)\\ &-(|A_1^c\cap A_2^c\cap A_3^c|+|A_1^c\cap A_2^c\cap A_4^c|+|A_1^c\cap A_3^c\cap A_4^c|+|A_2^c\cap A_3^c\cap A_4^c|)\\ &+(|A_1^c\cap A_2^c\cap A_3^c\cap A_4^c|)
\end{align}

which follows from the inclusion-exclusion principle.

Note that $A_k^c =\{(x_1,x_2,x_3,x_4)\in \mathbb Z^4\mid x_1+x_2+x_3+x_4=28,\ x_k\geq a_k+1,\  x_i \geq 0,\  i=1,2,3,4 \}.$

I'll show how to calculate these on an example and leave the rest to you (note that lots of the above sets will be empty). Let's look at $|A_1^c\cap A_2^c|$. This is the number of nonnegative integer solutions of $$x_1+x_2+x_3+x_4 = 28,\ x_1 \geq 7,\ x_2\geq 11.$$

Now, $$x_1+x_2+x_3+x_4 = 28 \implies (x_1 - 7) + (x_2 -11) + x_3 + x_4 = 10,$$
so the number of solutions is $\binom {10+4-1}{4-1}$.

The rest can be calculated analogously.

Title: $x_1 + x_2 + x_3 + x_4 = 28$ number of solutions

Body:
How many solutions are there to the equation in the title with the following constraints:
$0 ≤ x_i, x_1 ≤ 6, x_2 ≤ 10, x_3 ≤ 15, x_4 ≤ 21?$

So, to do this I tried the following:

(A) Total number is ${31 \choose 28}$ ways.

(B) $x_1$ having $7$+ is ${24 \choose 21}$ ways.

(C) $x_1$ having $7$+ and $x_2$ having $11$+ is ${13 \choose 10}$ ways.

There is no way that $x_3$ can have 15+ since $7 + 11 + 15 > 28$.

Answer is $A - B + C = 2,757$ ways.

Is this correct?


答:
By uniqueness of binary expansion,
$$\prod\limits_{n=0}^{\infty} (1+x^{2^{n}})=\sum_{k=0}^\infty x^k,$$
which is $$\frac{1}{1-x}$$ if $|x|<1$.

Title: Convergence of $\prod\limits_{n=0}^{\infty} (1+x^{2^{n}})$

Body: I never covered infinite products in my real analysis class, so I have no idea. I was wondering about the convergence of the above product. Thanks in advance.</p>


答:
First of all, if $\sum_{n=0}^{\infty}\ln(1+x^{2^n})$ converges then so does the product in question.

Now $x$ must satisfy $|x|<1$ for this to converge.
However, for each such $y \le 1$; $y$ positive note that $0\le \ln(1+y)  \le 2y$. So $\ln(1+x^{2^n})$ is no larger than $2x^{2^n} < 2x^n$ for each $x$ satisfying $|x| < 1$.

Thus

$$0 \le \sum_{n=0}^{\infty}\ln(1+x^{2^n}) \le \sum_{n=0}^{\infty} x^n $$

and $\sum_{n=0}^{\infty} x^n$ converges.


问:
Curvature of a non-compact complete surface
Assume $\Sigma$ is a non compact, complete surface. Assume the integral $$\int_{\Sigma}K$$ is convergent, where K is the Gauss curvature of $\Sigma$. Is it always true that $$\frac{1}{2\pi}\int_{\Sigma}K$$ is an integer?</p>

答: I believe that Alexander horned sphere is a good example for as a visual representation for as an answer.</p>


问:
Unify $P(x, y)$ and $P(x, y)$
For the atoms $\phi = P(x, y)$ and $\gamma = P(x, y)$, provide a unifier, if it exists.

Obviously a unifer is not needed for these two atoms, but does one exist?

Can a substitution $S = \varnothing $? If so, would that substitution unify the two atoms? $S(\phi) = P(x,y) = S(\gamma) $

What about $ S = \{(z/x)\}$? $S(\phi) = P(x,y) = S(\gamma) $

答:
A substitution $\sigma$ is a unifier for an expression $\phi$ and an expression $\gamma$ if and only if $\phi \sigma=\gamma\sigma$.

A substitution is a finite mapping of variables to terms. I will write substitutions as sets of replacement rules, like the one shown below. In each rule, the variable to which the arrow is pointing is to be replaced by the term from which the arrow is pointing. In this case, $x$ is to be replaced by $z$, and $y$ is to be replaced by $w$.

\begin{align}
\sigma=\{x\leftarrow z,y\leftarrow w\}
\end{align}

Consider the substitution $\sigma$, trivially, this makes the two expressions $\phi$ and $\gamma$ the same.

\begin{align}
\phi\sigma&=P(x,y)\{x\leftarrow z,y\leftarrow w\}\\
&=P(z,w)\\
\gamma\sigma&=P(x,y)\{x\leftarrow z,y\leftarrow w\}\\
&=P(z,w)
\end{align}

Your substitution $\{x\leftarrow z\}$ is a unifier for the expressions $\phi$ and $\gamma$. Application of the empty substitution will also unify the two expressions.

Title: Unify $P(x, y)$ and $P(x, y)$


问:
Orthogonality of vectors
Which of the following sets of vectors is orthogonal?

$(a)v_{1}=(2,4,-2), v_{2}=(1,4)\\ (b)v_{1}=(14,5)^{T}, v_{2}=(10,-2)^{T}\\(c)v_{1=}(5,-2,3)^{T}, v_{2}=(-2,4,6)\\(d)v_{1=}(2,3)^{T}, v_{2}=(-1,4)^{T}$

Trail: I calculated the dot products and got results of $18,130,0,10$ simultaneously. So $(c)$
is the correct answer. Here I wonder can we calculate the dot product of 3 dim and 2 dim vectors as in $(a)$. similarly, is the calculation of dot product of row and column vector correct in case of $(c)$.

答:
No, you cannot dot two vectors of different sizes, unless you pad one vector (or trim the other).

The second question is a bit tricky. Given that with two vectors $a \cdot b = a^\top b$ and $a \odot b= a b^\top$, the first being the inner product and the second being the outer product. So the dot product is always the matrix product of a row vector with a column vector. But I like the separate the meaning of the dot product from the details of how to calculate it. So I vote that the dot product does not care if the vectors are row or column vectors, but matrix products do care.

Title: Orthogonality of vectors


问:
Can 3 Vectors ever span $\mathbb{R}^2$?
If I had: $(0,1)$, $(1,2)$, and $(1,3)$, couldn’t I always set the constant that multiplies the third vector equal to 0, and then the other two vectors could successfully span all of $\mathbb{R}^2$ since they aren’t scalar multiples of each other?

That makes sense to me when I visualize that in my head, but is that wrong? Could that work?

答: Of course, because it only takes two vectors to span $\Bbb R^2$.

Title: Can 3 Vectors ever span $\mathbb{R}^2$?


问:
What is the connection between metric space and topological space?which come first?
I'm trying to say that what is the hierarchical order for these two spaces and what is the definition of topological space?</p>

答:
A topological space is a pair (X, A) where X is a set and A is a topology.
A topology is a subset of $P(X) $ such that:

$\emptyset, X \in A$

If $\{U_i\} \subset A \Rightarrow \cup U_i \in A$

If $U, V\in A \Rightarrow U\cap V\in A$

$A$ is called the collection of open sets.

Now if you a a metric d, you can define a topology on X in a standard way.

So a metric space is automatically a topological space, but not every topogical space is a metric space: the trivial topology ($A=\{\emptyset, X\} $) is non metric on $X=\mathbb{R} $.


问:
Submodules of homology modules
I have been dealing with certain subgroups of group cohomology, and the following general question comes to my mind. Suppose $C$ is a chain complex of $R$-modules and $H_n(C)$ its $n$-th homology module. Now let M be an $R$-submodule of $H_n(C)$. Does there exist a chain complex $D$ of $R$-modules such that $M=H_n(D)$?

Edit: Suppose that there is a sequence of $R$-submodules $M_n$ of $H_n(C)$. When can we find a chain complex $D$ such that $H_n(D)=M_n$ for all $n$?

答:
For any sequence of $R$-modules $M_n$ whatsoever (regardless of whether you obtained them as submodules of $H_n(C)$ for some $C$), the chain complex $D$ defined by
$$\cdots\xrightarrow{\;d_{n+2}\;} M_{n+1}\xrightarrow{\;d_{n+1}\;}M_n\xrightarrow{\;d_{n}\;} M_{n-1}\xrightarrow{\;d_{n-1}\;}\cdots$$
where each $M_n$ is in the $n$th spot, and the maps $d_n$ are all the trivial homomorphisms (sending all of $M_n$ to the zero element of $M_{n-1}$) will have
$$H_n(D)=\ker(d_n)/\mathrm{im}(d_{n+1})=M_n/0\cong M_n.$$</p>

Title: Submodules of homology modules


问:
Notations for Random Variables
Can someone explain what it means when we use random variables of the form $X_{n,m}$ given $1\le m \le n$. I understand what it means for random variables of the form $X_1,X_2, ..., X_n $ which is that some event X is being repeated n times (I am assuming that these are all i.i.d) but I can't quite wrap my head around the other notation.

答:
This notation represents a doubly-indexed array of random variables. The notation $1\le m\le n$ suggests that index $n$ is the "main" index while the index $m$ is the "subsidiary" index within $n$. Writing this out, you can arrange the variables $X_{n,m}$ into a triangular array as $n$ varies from $1,2,\ldots$:
$$
(n=1):\qquad X_{1,1}\\
(n=2):\qquad X_{2,1}, X_{2,2}\\
(n=3):\qquad X_{3,1}, X_{3,2}, X_{3,3}\\
(n=4):\qquad X_{4,1}, X_{4,2}, X_{4,3}, X_{4,4}\\
\vdots\\\
(n=k):\qquad X_{k,1}, X_{k,2}, \ldots, X_{k,k-1}, X_{k,k}
$$
and so on. If you are studying advanced probability theory, the most general version of the Central Limit Theorem is stated in terms of a triangular array, with the notion being that your observed sample of random variables, with a fixed value for $n$, is one particular row of a triangular array. You might then invoke the CLT to justify why the sum of your random variables has approximately a normal distribution.

The reason why there are two indices is that this allows the distribution of the $X$'s to differ from one row to the next, for example in row $n$ the $X_{n,1},\ldots,X_{n,n}$ are an IID sample from the Bernoulli($p_n$) distribution -- you're tossing a coin $n$ times where the coin has probability $p_n$ of landing heads. $X_{n,m}$ is then the 0/1 result of the $m$th toss.

Title: Notations for Random Variables


问:
Orthogonal basis for a subspace of $R^3$
My apologies if the question sounds silly, but I couldn't really wrap my head around this.

I came across this multiple choice question that asked me for the set(s) of vectors that could form an orthogonal basis for a subspace of $R^3$. I got most of the question correct, but there is one set of vectors that I didn't select, and it turned out that it acts as an orthogonal basis for the subspace of $R^3$ too.

$$\{\begin{bmatrix}-3\\1\\2\end{bmatrix}, \begin{bmatrix}1\\1\\1\end{bmatrix}\}$$

I understand why this set of vectors forms an orthogonal basis but I'm quite confused about how this set acts as a basis for the subspace in $R^3$. I haven't done linear algebra in a while so please correct me if I'm wrong, but I remember that dim(A) needs to be at least 3 for a set of vectors to even qualify for the possibility of spanning a $R^3$ subspace. This set could only exist in a subspace of $R^2$ right?

答:
If a subspace of $R^{3}$ had dimension 3, then that subspace would have to be $R^{3}$.

A subspace of $R^{3}$ doesn't need to have dimension 3, for example, a plane in $R^{3}$ that goes through the origin is a subspace of $R^{3}$ with dimension 2

For your specific set, just check that the span of those vectors meet the conditions for a subspace


问:
Test of convergence of $\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$
I am having some trouble with this problem and don't know if I am doing it right:

$$\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$$

so the steps I have taken so far are, I split it into

$$\int_0^\infty \dfrac{x^6+6}{x^8+8} + \int_{-\infty}^0 \dfrac{x^6+6}{x^8+8}$$

for the second integral I made a change of variable $x = -u$

so it would look the same as the first integral and then use $$ \lim_{x\to \infty} \dfrac{f(x)}{g(x)}$$ where  $g(x) = \dfrac1{x^2}$.

Just wondering if my approach is correct.

答:
Your approach is fine, as long as you compare with $1/x^2$ when $|x|\gt1$ and then use the continuity of $\frac{x^6+6}{x^8+8}$ for $|x|\le1$. We can illustrate this with a slightly different comparison function.

First, note that
$$
\lim_{|x|\to\infty}\frac{(x^6+6)(x^2+1)}{x^8+8}=1\tag{1}
$$
Thus, $(1)$ says that there is an $m$ so that if $|x|\ge m$, then $\frac{(x^6+6)(x^2+1)}{x^8+8}\le2$. Since $\frac{(x^6+6)(x^2+1)}{x^8+8}$ is continuous on the compact set $[-m,m]$, it is bounded there. Therefore, there is an $M$ so that
$$
\frac{(x^6+6)(x^2+1)}{x^8+8}\le M\tag{2}
$$
Therefore,
$$
\frac{x^6+6}{x^8+8}\le\frac{M}{1+x^2}\tag{3}
$$
Now, simply use
$$
\int_{-\infty}^\infty\frac{\mathrm{d}x}{1+x^2}=\pi\tag{4}
$$
to show that
$$
\int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x\tag{5}
$$
converges by comparison.

Although this answer uses contour integration, it does give the value for
$$
\int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x=\frac{\pi}{8}\csc\left(\frac{\pi}{8}\right)\left(2^{5/8}+3\cdot2^{-5/8}\right)\tag{6}
$$


问:
Apostol proof of subset openness in metric spaces
The theorem goes as follows:

Let $(S, d)$ be a metric subspace of $(M, d)$, and let $Y \subseteq S$. Then Y is closed if and only if $Y = B \cap S$ for some set $B$ which is closed in $M$.

The Apostol proof of the "if" direction goes as follows:

If $Y = B \cap S$, where $B$ is closed in $M$, then $B = M - A$, where $A$ is open in $M$. Then:

$$Y = S \cap B = S \cap (M - A) = S - A$$

from where Apostol says: "hence Y is closed in S". Page 62 of Mathematical Analysis Edition 2.

How does Apostol jump from "$S - A$" to "$Y$ is closed in $S$"?

IMHO, $A$ is open in $M$, not in $S$. Thus, the implication of open/closed-flipping ("S" is closed in $M$, if $M - S$ is open) does not work for $S - A$ from the outset. So, seems like a big leap of logic here, or incorrect proof, or I am missing something.

Can someone clarify?

答:
Given that Apostol has provided the following theorem, which I'll refer to as Theorem 1:

If $(S,d)$ is a metric subspace of $(M,d)$, then $X \subseteq S$ is open in $S$ if and only if $X = A \cap S$, where $A$ is an open subset of $M$.

Let's prove the closed set version:

If $(S,d)$ is a metric subspace of $(M,d)$, then $Y \subseteq S$ is closed in $S$ if and only if $Y = B \cap S$, where $B$ is a closed subset of $M$.

First, the "if" part that you asked about in your question:

Suppose $Y = B \cap S$, where $B$ is a closed subset of $M$. Thus we can write $B = M - A$, where $A$ is an open subset of $M$. Then

$$Y = B \cap S = (M-A) \cap S = S - A$$

Consequently,
$$S - Y = S - (S - A) = S \cap A$$
so by Theorem 1, $S - Y$ is open in $S$. Hence $Y$ is closed in $S$.

Now, for the "only if" part:

Suppose that $Y$ is a closed subset of $S$. If we define $X = S - Y$, then $X$ is an open subset of $S$. Hence by Theorem 1, we can express $X = S \cap A$ where $A$ is an open subset of $M$. Thus
$$Y = S - X = S - (S \cap A) = S \cap (M - A)$$
Since $M-A$ is a closed subset of $M$, we may set $B = M-A$.


问:
Closed-form solution to an infinite series?
I have the following series that I've confirmed on matlab to have some closed-form solution, but I can't find it through trial-and-error, and I definitely don't have the math background to just solve it. Here it is:
$$ \sum_{t=0}^\infty \beta^t(1-\delta)^t $$
Where $\beta,\delta\in(0,1)$. Any help would save me oodles of time that I need to spend doing the rest of the problem set (that isn't math problems; this is ancillary to the subject).

答:
This is a geometric series, and
$$
\sum_{t=0}^\infty \beta^t(1-\delta)^t
=\sum_{t=0}^\infty (\beta(1-\delta))^t
=\frac{1}{1-\beta(1-\delta)}
$$


问:
Q is a field, extension fields
What's the difference between Q[√ 3,√ 2] and Q(√ 3,√ 2)?

Q is the rational numbers
Please help

答:
They are the same. As for why the different notations, the square brackets denote a polynomial ring, while the round brackets denote a field extension. So
$$
Q[\sqrt{3},\sqrt{2}]=\left\{p(\sqrt{3},\sqrt{2}):\, p(x,y)\in \mathbb{Q}[x,y]\right\}=\{a+b\sqrt{3}+c\sqrt{2}+d\sqrt{6}:\,a,b,c,d\in\mathbb{Q}\},
$$
but since $1/\sqrt{3}=\sqrt{3}/3$ (and ditto for $\sqrt{2}$), it does not differ from the field extension.

You would get a difference if you choose transcendental elements (=polynomial ring variables). Here $Q[x]$ is the polynomial ring in $x$, while $Q(x)$ is its fraction field, containing, for example $\frac{x^2+3x+1}{x^4+2x+10}$.

Title: Q is a field, extension fields


答:
This sum converge becouse it is
$$=\sum_{i=0}^\infty [\beta(1-\delta)] ^t$$
With argument $\beta(1-\delta) <1$.

If converges to
$$\frac{1}{1-\beta(1-\delta)}$$

Title: Closed-form solution to an infinite series?

Body:
I have the following series that I've confirmed on matlab to have some closed-form solution, but I can't find it through trial-and-error, and I definitely don't have the math background to just solve it. Here it is:
$$ \sum_{t=0}^\infty \beta^t(1-\delta)^t $$
Where $\beta,\delta\in(0,1)$. Any help would save me oodles of time that I need to spend doing the rest of the problem set (that isn't math problems; this is ancillary to the subject).


问:
I don't understand why the partial derivative of $ F $ with respect to $v'$ is like this, does $y'$ depend on $v'$?
I don't understand why the partial derivative of $ F $ with respect to $v'$ is like this, does $y'$ depend on $v'$?

Image attached:

答:
You have that
$$
y=y(u,v) , x=x(u,v)
$$
and according to your image you are supposing too that $v$ depends on $u$.
So, using chain rule

$$
dy=y_{u}+y_{v} v'(u)
$$
Similarly,
$$
dx=x_{u}+x_{v}v'(u)
$$
Which implies that
$$
y'(x)= \displaystyle \frac{dy}{dx} = \displaystyle \frac{y_{u}+y_{v} v'(u)}{x_{u}+x_{v}v'(u)}
$$
Is like when you consider the change of variables in polar coordinates : $x=r(\theta)cos(\theta)$, $y=r(\theta)sin(\theta)$.

Best regards.


问:
Find a basis given...
Let $a,b,c\in \mathbb{R}$.
Given $$V:=\left\{ \begin{pmatrix} u+v+w \\ au+bv+cw \\ a^2y+b^2v+c^2w \end{pmatrix} : u,v,w\in\mathbb{R} \right\}$$

and $V$ is said to be a subspace of $M_{3\times1}
(\mathbb{R})$.

So, should the basis of $V$ should be,
$W=u\begin{pmatrix}1\\a\\a^2\end{pmatrix}+v\begin{pmatrix}1\\b\\b^2\end{pmatrix}+w\begin{pmatrix}1\\c\\c^2\end{pmatrix}$

or am I wrong and if so what should the correct basis of $V$ be?

答:
Given that every element of $S$ can be written as $$x\begin{pmatrix}1\\a\\a^2\end{pmatrix}+y\begin{pmatrix}1\\b\\b^2\end{pmatrix}+z\begin{pmatrix}1\\c\\c^2\end{pmatrix}$$
for some $x,y,z \in \mathbb R$, the set $$B := \left\{ \begin{pmatrix}1\\a\\a^2\end{pmatrix}, \begin{pmatrix}1\\b\\b^2\end{pmatrix}, \begin{pmatrix}1\\c\\c^2\end{pmatrix} \right\}$$
is a spanning set (or a generator) for $S$. So, assuming that $a,b$ and $c$ make $B$ also a linearly independent set, then $B$ is basis for $S$.

Title: Find a basis given...


问:
Product of non-trivial group subsets is not (necessarily) a group
If $G$ is a non-trivial group then the product of its non-empty subsets does not (necessarily) define a group structure on the set $\wp \left( G \right)-\left\{ \varnothing  \right\}$.
So far, I've proven that this product defines a monoid structure on the set $\wp \left( G \right)-\left\{ \varnothing  \right\}$. Obviously, there must be an element within $\wp \left( G \right)-\left\{ \varnothing  \right\}$  which has no inverse element. Any suggestion how to construct a non-trivial group $G$ with such a property?

答: As $G$ is non-trivial, it is a non-identity element of $\mathcal{P}(G) \setminus \{\emptyset\}$ (the identity in $\mathcal{P}(G) \setminus \{\emptyset\}$ is $\{e\}$). But $G$ is idempotent in $\mathcal{P}(G) \setminus \{\emptyset\}$, $G \cdot  G = G$, as $x \cdot y \in G, \forall x, y \in G$. Thus, $G$ can't have an inverse in $\mathcal{P}(G) \setminus \{\emptyset\}$.


问:
Binary subsequence of size three, graph with euler cycle
Alan Tucker's Applied Combinatorics state, "A set of eight binary digits (0 or 1) are equally spaced about the edge of a disk. We want to choose the digits so that they form a circular sequence in which every subsequence of length three is different. Model this problem with a graph with four vertices, one for each different subsequence of two binary digits. Make a directed edge for each subsequence of three digits whose origin is the vertex with the first two digits of the edge’s subsequence and whose terminus is the vertex with the last two digits of the edge’s subsequence."

I have to a) build this graph
b) Show how an Euler cycle (which exists for this graph) will correspond to the
desired 8-digit circular sequence.
c)Find such an 8-digit circular sequence with this graph model.

I'm not sure how to build the graph, more explicitly, should the vertex essentially have a coordinate with binary numbers? Such as, (1,0) (0,1), (1,1), (0,0) and would these be in a circle, or some type of cycle?

Anything to get me started please!

答:
This is very pretty.

Hints to get you started:

You have found the four vertices of your graph. Each is a subsequence of two digits

Each vertex will have two directed edges leaving it, pointing at vertices whose first digits match its last digit

For example $(0,1)$ ends with $1$ so will have edges pointing at $(1,0)$ and $(1,1)$, both of which start with $1$

Each edge is in effect a triplet.  For example the edge from $(0,1)$ to $(1,1)$ is the triplet $(0,1,1)$ taking account of the overlap

There are $4 \times 2=8$ edges in total corresponding to the $2^3=8$ different subsequences of three digits

So a Eulerian cycle (there are in fact two) using each edge once will give you what you want

Not that the question asks you to do so, but you can make the triplets vertices with directed quadruplet edges and look for a Hamilonian cycle


问:
How to find E[X] of a multivariable joint PDF f(x,y)?
Question: X and Y  have the following joint probability density function. Find E[X]

$$f(x, y)= \frac{3}{620}y$$    $$0  <  x  <  7,  y  >  0,  x − 4  <  y  <  x + 4$$

My Approach:

To find the expectation of X, I first find the marginal pdf $f(x)$ by integrating y. Originally, my lower bound was $x-4$ but I changed it because if $x<4$, then $y<0$ which violates one of the constraints so:

$$\frac{3}{620}\int_{0}^{x+4}ydy = \frac{3}{620} \frac{(x+4)^2}{2} =\frac{3(x+4)^2}{1240}$$

Then, I find the expectation. I thought that the lower bound would be 4 instead of 0 as $x<4$ would not be possible for y:

$$E[X] = \int_{4}^{7}x*\frac{3(x+4)^2}{1240} dx = \frac{819}{1240} $$

However, this answer is incorrect and I'm having trouble figuring out why. If anyone can explain where I went wrong I would greatly appreciate it. Thank you!

答:
You are correct that the marginal density for $X$ is found by applying the formula
$$f(x)=\int_y f(x,y) dy\tag1$$
but your limits of integration are wrong. If you plot the region where $f(x,y)$ is defined and nonzero, it's a truncated parallelogram in the $x,y$-plane. Specifically, it can be broken into two sub-regions (draw the picture!):
$$
0 < y < x+4\qquad\text{when $0< x< 4$}
$$
and
$$
x-4 < y < x+4\qquad\text{when $4< x< 7$}.
$$
This means you need to break up the integral (1) into two pieces, corresponding to those two regions:
$$
f(x) = \begin{cases}
\int_0^{x+4} \frac3{620}y\,dy& \text{if $0< x<4$}\\
\int_{x-4}^{x+4} \frac3{620}y\,dy& \text{if $4< x<7$}
\end{cases}\tag2
$$
In the end you'll have a marginal density for $X$ that's defined in two pieces. (Your calculation is just the first piece, for the case $0<x<4$.) Having obtained both pieces of (2) you then proceed to compute $E[X]=\int xf(x)\,dx$, again in two pieces:
$$E[X]=\int_0^4 xf(x)\,dx + \int_4^7xf(x)\,dx$$


问:
Gompertz distribution
I need to prove that the Gompertz distribution function is indeed of distribution, I'm certain this function would never be negative, the problem comes when integrating the function with parameters: $b,\eta > 0$

$$f(x)  = \begin{cases}
b\eta\;\exp(\eta+bx-\eta e^{bx})&\text{for $x\geq0$}\\
0&\text{otherwise}\\
\end{cases}$$

So far I have:
$$\int_{0}^{\infty} b\eta\;\exp(\eta+bx-\eta e^{bx})  \,dx  = 1$$

$$\Rightarrow \int_{0}^{\infty} b\eta\;\exp(\eta+bx-\eta e^{bx}) \,dx = b\eta\int_{0}^{\infty} \;\exp(\eta+bx-\eta e^{bx}) dx$$

I don't know how to treat the $exp$, can someone please give me a hint?

答: Write the density as $$f(x) = e^\eta (b \eta e^{bx}) \exp(-\eta e^{bx}).$$  Then with the substitution $$u = \eta e^{bx}, \quad du = b \eta e^{bx} \, dx,$$ we see $$\int f(x) \, dx = e^\eta \int e^{-u} \, du.$$  I leave the rest of the computation as an exercise.

Title: Gompertz distribution


问:
Can we win the lotto twice with the same combination?
Hello everyone : here's Jack's little story

So let's say the lotto Jack's playing is: $5$ numbers between $1$ and $49$

And let's say he likes to play : $ 28 , 13 , 35 , 41 , 8 $

I know that to win we have : $49 \times 48  \times 47 \times 46 \times 45 $ and we have $5!$ permutation ...

Boom , he WON .

Jack says that he will not play with this combination anymore because he already won with it.

It's like winning with $1,2,3,4,5$  or $2,4,6,8.10$ , or also $5,10,15,20,25$ Jack says that we may have more chance to win with this combination because we never saw a draw like that.

I don't think he's right but I see what he meant by that

Are all these statements true?

答: Both of Jack's assertions are examples of the so-called gambler's fallacy, that an event which has happened more often in the past is less likely to happen in the future, or one which has happened less often in the past is more likely to happen in the future.  This is a fallacy whenever (as with lotto draws) the event in question has the same probability of occurring regardless of what has happened in the past.   For lotto draws which choose $5$ numbers out of $49$ without replacement there are $\ {49\choose5}=$$\frac{49!}{5!\,44!}\ $ equally likely possible combinations of $5$ numbers that can be drawn, so, if you have just one entry, your probability of winning is always $\ \frac{5!\,44!}{49!}\approx5.2\times10^{-7}\ $, regardless of what numbers you choose, or what numbers have ever been drawn in the past.


问:
Proving a function has bounded derivatives
The function in question is:

$ \mathrm{e}^\frac{1}{\left(4x+3\right)^2-1}  $   where $ x∈ (-1, -1/2) $

I've tried graphing the derivatives and it seems that they explode as $ x \rightarrow -1/2 $

答: The function actually goes to $ 0$ when $x$ approaches the boundary of the interval, because the exponent shoots down to $-\infty$. This also happens for every derivative, and in particular they are all bounded in that interval.


问:
What could be better than base 10?
Most people use base 10; it's obviously the common notation in the modern world.

However, if we could change what became the common notation, would there be a better choice?

I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?

I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.

Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:

First and foremost, 12 is a highly composite number — the smallest
  number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
  and 12). As noted, 10 has only two. Consequently, 12 is much more
  practical when using fractions — it's easier to divide units of
  weights and measures into 12 parts, namely halves, thirds, and
  quarters.

And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.

So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?

答:
It is not necessarily true that the multiplication is of the order of $O(b^2)$, since this is a particular implementation of the base, rather than the base itself.  The mayans divided their score into four sticks of five dots, and had a true zero in the dot-position (eg fifteen is "3-fives-zero").

One should remember that counting (multiples) and division are separate operations, and that it is possible to use different number-systems for them.  Historically, the sixty-wise system is one of divisions: the first column is that of units, and later places are divisions by sixty.  Likewise, the romans multiplied by 10's, and divided into 12's.

An alternating base like 60 or 120, supposes two rows on each column of the abacus, where the unit (in the bottom row), is counted to 10 to make one carry up, but divides into 12 to be borrowed into the top row of the lower column.  Since one can start in either the top row or bottom row (for counting), the use of tens by twelves or twelves by tens, automatically produces an alternating base.

Using alternating arithmetic then reduces the size of the tables to the order of $O(b)$.

It should be noted that the sumerian system is a division system to avoid division.  We see this from recriprocal tables (eg 2  <=> 30 ), and tables of reckoners of multiples of the recriprocals (eg multiples of 4.26.40).  Even in their reckoners, multiples are supplied for 1 to 20, and 40.  Neugebauer even gives reference to a paper on the seven brothers, ie what is 1/7.  It is concluded it lies between 0.8.34.16 and 0.8.34.18.

But i wrangle base 120 for nearly 30 years, and never felt the need to go past 12*12.

Title: What could be better than base 10?


问:
Prove $ \lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a). $

Suppose $f(x)$ is continuous on $[a, b]$. Prove that $\forall x\in (a,b)$, we have
$$
\lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a).
$$

If $f$ id diffentiable, the conclution can be obtained by Newton-Leibniz formula easily. When $f$ is just continuous, I tried to use knowledge about integral with parameters. Write
$$
F(h,x):=\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t,
$$
but it seems we still need to compute $\lim_{h\rightarrow 0}\frac{f(t+h)-f(t)}{h}$, and prove uniform convergence. I have no idea how to go on.

Is my thoughts workable? If not, please suggest your way. Appreciate any help!

答:
Two things:

This is only a partial answer (you'll see why at the end), but it might still help.

It seems like the limit should be regarded as a right-hand limit.
Please let me know if I'm wrong.

The limit $\lim_{h\to 0^{+}}\frac{1}{h}\int_{a}^{x}\left[f(t+h)-f(t)\right]dt$ looks suspiciously similar to the limit that appears in the definition of the derivative. Given our knowledge of the first part of the Fundamental Theorem of calculus, this observation motivates defining a function $F$ with an integral like the one that FTC1 is concerned with.

$$F(x)=\int_{a}^{x}f(t)\text{ }dt$$

By FTC1, $F$ is continuous over $[a,b]$ and differentiable over $(a,b)$, and its derivative is $F'(x)=f(x)$ for all $x\in (a,b)$. Let's try to express $\frac{1}{h}\int_{a}^{x}\left[f(t+h)-f(t)\right]dt$ in terms of $F$.

If $h$ is small enough, then the integral $\int_{a}^{x}f(t+h)dt$ will be well defined. Applying the substitution rule and basic integral properties then gives

\begin{align*}
\int_{a}^{x}f(t+h)dt &= \int_{a+h}^{x+h}f(t)dt\\
&= \int_{a}^{x+h}f(t)dt-\int_{a}^{a+h}f(t)dt\\
&= F(x+h)-F(a+h)
\end{align*}

Perhaps you see why I think the limit should be a right-hand limit. If $h$ were allowed to be negative, then $a+h<a$, so the domain of integration $[a+h,x+h]$ would contain points outside the “domain” of $f$. Anyway, we may now write

\begin{align*}
\frac{1}{h}\int_{a}^{x}\left[f(t+h)-f(t)\right]dt &= \frac{1}{h}\left(\int_{a}^{x}f(t+h)\text{ }dt-\int_{a}^{x}f(t)\text{ }dt\right)\\
&= \frac{F(x+h)-F(a+h)-F(x)}{h}\\
&= \frac{F(x+h)-F(x)}{h}-\frac{F(a+h)-0}{h}\\
&= \frac{F(x+h)-F(x)}{h}-\frac{F(a+h)-F(a)}{h}
\end{align*}

because $F(a)=\int_{a}^{a}f(t)dt=0$

If one can prove that $\lim_{h\to 0^{+}}\frac{F(a+h)-F(a)}{h}=f(a)$ (this seems intuitively obvious to me, but I’m not sure how to prove it), then the desired result immediately follows from taking the limit and applying FTC1 to the first term. Hope this is helpful!


问:
Every linear operator in irreducible $G$-module can be represented as element of $\mathbb{C}[G]$
Suppose that $G$ is a finite group and $V$ is a (complex) irreducible representation of $G$, that is, for any $g\in G$ we have $gV=V$ (here we use notation $gv$ for the action of element $g$ on vector $v$ instead of, for example, $\rho(g)v$, where $\rho\colon G\to\operatorname{End} V$).

There is a natural embedding of the group algebra $\Phi\colon\mathbb{C}[G]\hookrightarrow\operatorname{End} V$ which is defined by the formula
$$
\Phi\left(\sum_{g\in G}c_g\cdot g\right)(v)=\sum_{g\in G}c_g\cdot gv.
$$

I found somewhere (can't remember now) that $V$ is irreducible iff $\Phi$ is surjective, i. e. every linear operator on $V$ can be realised as an element of $\mathbb{C}[G]$, but I don't know how to prove that.

The "if" part is easy. Indeed, if $V$ is reducible, then there is a $G$-invariant subspace $W\subsetneq V$ and we can take projection $P\in\operatorname{End} M$ onto $U$, where $U\in W$ such that $U\oplus W=V$ (and define $P|_{U}=\operatorname{id}_{U}$ and $P|_{W}=0$). It's clear that for any $a\in\mathbb{C}[G]$ we have $\operatorname{Im}\Phi(a)\subset W$, so there is no $a\in\mathbb{C}[G]$ such that $\Phi(a)=P$.

Thus, it remains to prove the "only if" part. Any ideas? I am a beginner in the representation theory, so  any help would be appreciated.

答:
From Schur's orthogonality relations we have
$$\frac{d}{|G|}\cdot \sum_{g\in G} a_{ji}(g^{-1})\, g = E_{ij}$$
with $1$ at  $(i,j)$ and $0$ in rest, and so, any matrix in $M_{d}(\mathbb{C})$ is a combination of the $g$'s.

$\bf{Added:}$ For every $A\in M_{d}(\mathbb{C})$ we have
$$ \frac{d}{|G|}\cdot \sum_{g\in G} Tr (g^{-1}A) g = A$$


问:
probability to reach a spesific point before other
a frog can jump each time forward or backward (one step) with an equal probability of 0.5,
the frog stands in point 0,
what is the probability that the frog will reach to point 100 before point -1?
tried to look at this answer but not sure that it is the same
The probability of reaching a certain amount of money before going bust in a casino game.</p>

答: Your situation is closer to this version of gambler's ruin, where the random walk is symmetric. If you shift each position one unit to the right, then the frog starts at position ("state") $1$ and you want the probability that the frog hits state $101$ before hitting state $0$. This is the $p=\frac12$ version, so the desired probability equals $\frac{1}{101}$, since state $1$ is $\frac1{101}$ of the way between $0$ and $101$.


答:
Taking what was said

$$ \int e^{-u} du =  -e^{-u} $$

Substitute back in $u =\eta e^{bx}$

$$ - e^{\eta}\cdot e^{-\eta e^{bx}} \Big|_{0}^{\infty} $$

The negative will reverse it and we pull out $e^{\eta}$

$$ = e^{\eta} (e^{-\eta}  - \displaystyle \lim_{x \to \infty} e^{-\eta e^{bx}}) $$

The second term will go to $0$ and we get

$$ e^{\eta} \cdot e^{-\eta} = 1$$

Title: Gompertz distribution

Body:
I need to prove that the Gompertz distribution function is indeed of distribution, I'm certain this function would never be negative, the problem comes when integrating the function with parameters: $b,\eta > 0$

$$f(x)  = \begin{cases}
b\eta\;\exp(\eta+bx-\eta e^{bx})&\text{for $x\geq0$}\\
0&\text{otherwise}\\
\end{cases}$$

So far I have:
$$\int_{0}^{\infty} b\eta\;\exp(\eta+bx-\eta e^{bx})  \,dx  = 1$$

$$\Rightarrow \int_{0}^{\infty} b\eta\;\exp(\eta+bx-\eta e^{bx}) \,dx = b\eta\int_{0}^{\infty} \;\exp(\eta+bx-\eta e^{bx}) dx$$

I don't know how to treat the $exp$, can someone please give me a hint?


问:
The relationship of ${\frak m+m=m}$ to AC
Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.)

Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$?

If not, what is the relationship of this statement for all cardinals ${\frak m}$ to AC?

The weaker principle $\aleph_0\le{\frak m\Rightarrow m}+1={\frak m}$ can be derived in ZF, and it is a well-known result of Tarski that $\forall{\frak m}\ge\aleph_0,{\frak m\times m=m}$ implies the axiom of choice, but I am not sure about the intermediate result with ${\frak m+m=m}$.

Edit: A related question: is it consistent with ZF that there exists a cardinal ${\frak m}\ge\aleph_0$ such that ${\frak m=n+p}$ with ${\frak n<m}$ and ${\frak p<m}$? It's hard to say whether ${\frak m+m=m}$ precludes this scenario.

答:
Suppose that $A$ is infinite and Dedekind finite. Then $\mathfrak m=|A\cup\mathbb N|$ satisfies that $|A|<\mathfrak m$, $\aleph_0<\mathfrak m$, and $\mathfrak m+\mathfrak m>\mathfrak m$.

To see the last inequality, note that if $\mathfrak m+\mathfrak m=\mathfrak m$ then $A\times 2$ embeds into $A\cup\mathbb N$, say via $f$, but only a finite subset of it embeds into $\mathbb N$, so a set strictly larger than $A$ must embed into $A$.

Note that being Dedekind infinite is the same as embedding $\omega$, so if we require $\mathfrak m+\mathfrak m=\mathfrak m$ for all infinite cardinals $\mathfrak m$, or even for all Dedekind-infinite cardinals, then there are no Dedekind-finite sets.  But no, Countable Choice is strictly stronger than the lack of infinite Dedekind finite sets. This is due to Pincus, see this MO question.

As for whether $\mathfrak m+\mathfrak m=\mathfrak m$ (the idemmultiple hypothesis) gives us Countable Choice, the answer is again no, as proved by

Gershon Sageev. An independence result concerning the axiom of choice, Ann. Math. Logic 8, (1975), 1–184. MR0366668 (51 #2915).


问:
Galois group of the compositum of a non-totally-ramified and a the unramified extension of $\mathbb{Q}_p$ of prime power degree
Let $K=\mathbb{Q}_p$ and $L,F$ be extensions of $K$ such that

$[L:K] = [F:K]$ is a prime power,

$L/K$ is not totally ramified,

$F/K$ is unramified,

$L/K$ is cyclic (and $F/K$ too which is implied by the fact that it's unramified).

Question: Can one limit the possibilites for $\operatorname{Gal}(LF/K)$?

Thoughts and Remarks

Since $L/K$ and $F/K$ are both cyclic of the same prime power degree and since both are not totally ramified, the image of a Frobenius element (i.e. a lift of $x \mapsto x^p$ over the residue field of $K$) is a generator in $\operatorname{Gal}(L/K)$ resp. $\operatorname{Gal}(F/K)$.

The degree of $LF/K$ lies between $e(L/K) \cdot n$ (where $n$ is the mentioned prime power degree of $L/K$ and $F/K$, and $e(L/K)$ is the ramification index of $L/K$) and $n^2$.

The possible choices for $\operatorname{Gal}(LF/K)$ should "lie between" $C_e \times C_n$  and $C_n \times C_n$.

Could you help me advancing with my line of thought? Thank you!

答:
$q$ is prime, $L/K$ is cyclic of degree $q^r$, $L'/K$ is unramified of degree $q^{r-s}>1$ and $L/L'$ is totally ramified of degree $q^s>1$, then any automorphism $\sigma\in Gal(L/K)$ such that $\sigma|_{L'}$ is the Frobenius will be a generator of $Gal(L/K)$.

Since $F/K$ is unramified of degree $q^r$ then
$[LF:L]=[F:L'] = q^s$ and $[LF:K] = q^r q^s$

$$Gal(L/L')\times Gal(F/L') \cong Gal(LF/L') = \langle a\rangle \times \langle b\rangle = C_{q^s}\times C_{q^s}$$
where $b$ is the Frobenius of $LF/L$ and $a$ is a generator of $Gal(LF/F)$.

Let $\phi$ be an extension of the Frobenius of $F/K$ to $LF$, then $$\phi^{q^{r-s}}|_F=b|_F$$ so $\phi^{q^{r-s}} = a^n b$, the order of $\phi$ is $q^r$ so that $\langle a\rangle \cap \langle \phi \rangle = \{1\}$ and $$
 Gal(LF/K) = \langle a\rangle \times \langle \phi \rangle = C_{q^s} \times C_{q^ r}$$


答:
If $f(x) =\mathrm{e}^\frac{1}{\left(4x+3\right)^2-1}$, then
$$f^{\prime} (x) =\frac{-8(4x+3)} {[(4x+3)^2-1]^2} \mathrm{e}^\frac{1}{\left(4x+3\right)^2-1}$$ which is zero in $x=-3/4$ so that it is its maximum (since $\lim_{x\to - 1^+}f(x)=\lim_{x\to - (1/2)^-}f(x)=0$).

So $f(x) \le f(-3/4)$

Now if you take the derivative of $f^{\prime} $, which is $f^{\prime \prime} $, you should get the max point, so you can obtain: $f^{\prime} (x) \le f^{\prime} (x_{max}) $.

You should then be able to go on.

Title: Proving a function has bounded derivatives

Body:
The function in question is:

$ \mathrm{e}^\frac{1}{\left(4x+3\right)^2-1}  $   where $ x∈ (-1, -1/2) $

I've tried graphing the derivatives and it seems that they explode as $ x \rightarrow -1/2 $


问:
Fundamental group of the connected sum of $n$ tori with $m$ points removed
I have read through this answer: Fundamental Group of Connected Sum of $n$ Tori with point removed, which deals with the special case of removing a single point. I wish to generalize this. Would anyone be able to nudge me in the right direction? Thanks.</p>

答:
Draw the polygonal representation of your n-holed torus.

Put your $k$ "missing" points in a small disk around the center of your $4n$-gon.

Draw a circle $S_1$ around the $k$ deleted points; draw a circle $S_2$ outside that.

Then the portion of the figure outside $S_1$ is topologically an $n$-holed torus with $1$ point removed, so you can use the previous computation to figure out generators and relations for its fundamental group.

The portion INSIDE $S_2$ is a disk with $k$ holes; you can compute $\pi_1$ for that as well. And then you can use Seifert-van Kampen to compute $\pi_1$ of the union of the two.


答:
The key is that the supported $y$ must be such that for all $0\lt x\lt 7$ both these conditions hold: $0\lt y$ and $x-4\lt y\lt x+4$ .   This indicates that we should partition the domain at the $x = 4$ divide.

$$\begin{align}f(x,y) &= \dfrac{3y}{620}\mathbf 1_{\large 0\lt x\lt 7, \max\{0,x-4\}\lt y\lt x+4 }\\&=\dfrac{3y}{620}\left(\mathbf 1_{\large 0\lt x\lt 4, 0\lt y\lt x+4 }+\mathbf 1_{\large 4\leqslant x\lt 7,x-4\lt y\lt x+4 }\right)\\[4ex] \mathsf E(X)&=\iint_{\Bbb R^2} x\,f(x,y)\,\mathrm d\langle x,y\rangle\\&=\dfrac{3}{620}\left(\int_0^4 x\int_0^{x+4}y\,\mathrm d y\,\mathrm d x+\int_4^7 x\int_{x-4}^{x+4} y\,\mathrm d y\,\mathrm d x\right)\end{align} $$

Title: How to find E[X] of a multivariable joint PDF f(x,y)?

Body:
Question: X and Y  have the following joint probability density function. Find E[X]

$$f(x, y)= \frac{3}{620}y$$    $$0  <  x  <  7,  y  >  0,  x − 4  <  y  <  x + 4$$

My Approach:

To find the expectation of X, I first find the marginal pdf $f(x)$ by integrating y. Originally, my lower bound was $x-4$ but I changed it because if $x<4$, then $y<0$ which violates one of the constraints so:

$$\frac{3}{620}\int_{0}^{x+4}ydy = \frac{3}{620} \frac{(x+4)^2}{2} =\frac{3(x+4)^2}{1240}$$

Then, I find the expectation. I thought that the lower bound would be 4 instead of 0 as $x<4$ would not be possible for y:

$$E[X] = \int_{4}^{7}x*\frac{3(x+4)^2}{1240} dx = \frac{819}{1240} $$

However, this answer is incorrect and I'm having trouble figuring out why. If anyone can explain where I went wrong I would greatly appreciate it. Thank you!


问:
A ring extension of $\mathbb Z$ in which $x^4+y^4+1$ is reducible
My first instinct was the ring of two variable polynomials modulo $a^4+b^4+1$.($\mathbb Z[a,b]/(a^4+b^4+1)$)

But I haven’t managed to reduce it to linear factors.

Can any of you help?

答:
Your question is very unclear. From the title, it looks as if you’re looking for a ring $A$ containing $\Bbb Z$ such that the polynomial $x^4+y^4+1\in\Bbb Z[x,y]$ becomes reducible when viewed as a polynomial in $A[x,y]$.

Perhaps you were thinking of your question as an analog to the one-variable situation, where the $\Bbb Z$-polynomial $x^4+1$ becomes reducible, first over the Gaussian numbers, where it’s $(x^2-i)(x^2+i)$, and better over the ring gotten by adjoining a primitive eighth root of unity $\zeta=\zeta_8=\frac{1+i}{\sqrt2}$, as $x^4+1=(x-\zeta)(x-\zeta^7)(x-\zeta^3)(x-\zeta^5)$.

Now, lamentably, if you’re restricting your rings $A$ to subrings of the complex numbers, $x^4+y^4+1$ remains irreducible no matter what $A$ is. The reason is that as soon as you get into multivariable polynomials, you’re suddenly no longer doing basic number theory nor abstract algebra, but Algebraic Geometry.

Geometrically, one thinks of $x^4+y^4+1=0$ as describing a “curve” in the (affine) plane over the complex numbers; topologically it’s a surface. And if you look at it in the way you should, as a curve in the projective plane over $\Bbb C$, it becomes a compact Riemann surface, of genus three. It’s only when, as a curve, it consists of two (possibly intersecting) curves in the plane that the original polynomial would have been reducible.

Since the algebraic geometers’ curve, topologists’ Riemann surface, defined by your polynomial equation $x^4+y^4+1=0$, does not break up into two separate, perhaps intersecting, curves/surfaces, your polynomial is “absolutely irreducible”, never becomes reducible no matter what field you consider its coefficients to lie in.


答:
Define $F:\Bbb{R}^2\to \Bbb{R}$ as $F(\alpha,\beta):=\int_{\alpha}^{\beta}f(t)\, dt$. Then, by a simple change of variables $u=t+h$ in $\int_a^xf(t+h)\, dt$, it is easy to see that (for $h\neq 0$)
\begin{align}
\dfrac{1}{h}\int_a^xf(t+h)-f(t)\, dt = \dfrac{F(a+h,x+h)-F(a,x)}{h}
\end{align}
In other words, if we fix $a$ and $x$, and define $\gamma:\Bbb{R}\to\Bbb{R}^2$ as $\gamma(h)=(a+h,x+h)$, then the above expression equals $\dfrac{(F\circ\gamma)(h)-(F\circ\gamma)(0)}{h}$, which is precisely the difference quotient for $F\circ \gamma$. Now, since $f$ is continuous, it is easily seen that $F$ is continuously differentiable (the FTC tells you how to calculate the partial derivatives of $F$ in terms of $f$), and $\gamma$ is clearly differentiable. So, by the chain rule, the composition is also differentiable. Now, calculating your limit is a simple exercise in chain rule:
\begin{align}
\lim_{h\to 0}\dfrac{1}{h}\int_a^xf(t+h)-f(t)\, dt &= (F\circ \gamma)'(0)\\
&=\dfrac{\partial F}{\partial \alpha}(\gamma(0))\cdot \gamma_1'(0)+\dfrac{\partial F}{\partial \beta}(\gamma(0))\cdot \gamma_2'(0)\\
&= \left[-f(a)\right]\cdot \left[1\right] + \left[f(x)\right]\cdot \left[1\right]\\
&= f(x)-f(a)
\end{align}
Hopefully the evaluation of partial derivatives of $F$ is clear (this is just the first/second idk which fundamental theorem of calculus).

Title: Prove $ \lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a). $

Body:
Suppose $f(x)$ is continuous on $[a, b]$. Prove that $\forall x\in (a,b)$, we have
$$
\lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a).
$$

If $f$ id diffentiable, the conclution can be obtained by Newton-Leibniz formula easily. When $f$ is just continuous, I tried to use knowledge about integral with parameters. Write
$$
F(h,x):=\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t,
$$
but it seems we still need to compute $\lim_{h\rightarrow 0}\frac{f(t+h)-f(t)}{h}$, and prove uniform convergence. I have no idea how to go on.

Is my thoughts workable? If not, please suggest your way. Appreciate any help!


问:
Explicit formulae for $\binom{n}{1} \cdot 1^y+\binom{n}{2} \cdot (2)^y+\dots +\binom{n}{n}\cdot n^y $
Is there any explicit formulae for

$\binom{n}{1} \cdot 1^y+\binom{n}{2} \cdot (2)^y+\dots +\binom{n}{n}\cdot n^y $

$\binom{n}{1}\cdot 1 \cdot 2^y+\binom{n}{2}\cdot 2\cdot 3^y+\dots+\binom{n}{n}\cdot n \cdot (n+1)^y $

I thought about it, also this is my own question, so I am not sure if there is an explicit formula or not. Thankyou!

答:
You already received answers in comments.

If you like hypergeometric functions, you have a compact form. Writing a few of the
$$f_y=\sum_{k=1}^n \binom{n}{k}k^y $$
$$f_4=n \, _4F_3(2,2,2,1-n;1,1,1;-1)$$
$$f_5=n \, _5F_4(2,2,2,2,1-n;1,1,1,1;-1)$$
$$f_6=n \, _6F_5(2,2,2,2,2,1-n;1,1,1,1,1;-1)$$ and the pattern is quite clear.


问:
What is the connection between metric space and topological space?which come first?
I'm trying to say that what is the hierarchical order for these two spaces and what is the definition of topological space?</p>

答:
As NaveMozza says above, every metric induces a topology: given a metric space $(X,d)$, we get a topology $\tau_d$ on $X$ generated by the open balls according to $d$. Meanwhile, many topologies are not induced by any metric (they are "non-metrizable"). For example, non-Hausdorff topologies are non-metrizable. We can in fact characterize the metrizable spaces, although it's a bit technical.

However, there's a crucial nuance here: different metrics on the same set can induce the same topology. For example, both the Euclidean metric $d_{Euc}((a,b),(c,d))=\sqrt{(a-c)^2+(b-d)^2}$ and the taxicab metric $d_{taxi}((a,b),(c,d))=\vert a-c\vert+\vert b-d\vert$ induce the same topology on $\mathbb{R}^2$.

So the passage from metric spaces to topological spaces is more complicated than just "pure generalization" (compare e.g. the passage from groups to monoids). We gain variety but lose specificity.


问:
Find the basis of $W$ in $\mathbb R^3$ and use the answer to find a basis for $W^⊥$
Question: Find a basis for the subspace $W$ of $\mathbb R^3$, then use it to find a basis for $W^⊥$.
$$W = \{\begin{bmatrix}x\\y\\z\end{bmatrix}, 2x-y+3z=0 \}$$

I'm kinda confused about what the $W$ set above contains. Is it simply referring to the two vectors ($[1, 1, 1]$ and $[2, -1, 3]$) that span the subspace of $W$? In that case, how would we compute the basis for $W^⊥$?

I'd much appreciate if someone could show me how this is done. Thanks in advance!

答:
I think intended was what would usually be written with a $\mid$ or colon instead of a comma:

$W$ is the set of all $\pmatrix{x\\y\\z}$ such that $2x-y+3z=0$.  $W$ is a plane.

We can eyeball that $W$ is spanned by $(1,2,0)$ and $(0,3,1)$.

The orthogonal space is spanned by $(1,2,0)\times(0,3,1)=(2,-1,3)$, a vector perpendicular to the plane and the vectors we gave spanning it.  This normal vector can also be obtained from the equation for the plane, which is $(2,-1,3)\cdot(x,y,z)=0$.


答:
$W$ is a plane through the origin.  You just need two linearly independent solutions.  Say $(2,1,-1)$  and $(0,3,1)$.

You could take their cross product to get the normal.  Or, if you know anything about equations of planes, you can read off $(2,-1,3)$ as the normal direction.  And it's a basis for $W^{\perp}$.

Title: Find the basis of $W$ in $\mathbb R^3$ and use the answer to find a basis for $W^⊥$

Body:
Question: Find a basis for the subspace $W$ of $\mathbb R^3$, then use it to find a basis for $W^⊥$.
$$W = \{\begin{bmatrix}x\\y\\z\end{bmatrix}, 2x-y+3z=0 \}$$

I'm kinda confused about what the $W$ set above contains. Is it simply referring to the two vectors ($[1, 1, 1]$ and $[2, -1, 3]$) that span the subspace of $W$? In that case, how would we compute the basis for $W^⊥$?

I'd much appreciate if someone could show me how this is done. Thanks in advance!


问:
poisson rate and uniform distributions
Question : Customers arrive at a bank at a Poisson rate $\lambda$. Suppose that we already know that exactly two customers arrive during the first hour.

What is the expected arrival time of the first customer ?

I think that the two customers are uniformly dist on $(0,1)$ so the ans should be 20 mins?
Since there is a conditional, I think the distribution changes and is not Poisson anymore.

答:
Your reasoning is correct.

The joint conditional distribution for the arrival times in the interval, under the condition that there are exactly two, is that for the order statistics for two identical and independent uniformly distributed samples over the interval.

Title: poisson rate and uniform distributions


问:
The Gluing of Schemes Along Closed Subschemes Preserves Universal Closedness
This is Vakil 16.4 O a), self-study.

We are to show that if $X_1$ and $X_2$ are universally closed $S$-schemes (meaning their structure maps are closed maps topologically and any base change map is closed topologically), that their gluing along closed subschemes $Z_1$ and $Z_2$ is also universally closed.

The gluing along closed subschemes has underlying topological space the topological gluing of the underlying topological spaces of the individual schemes. Its stalks are the usual stalks outside the gluing locus. Let $Z$ be in the image of either closed subscheme in the gluing. Then if $p \in X_1 \cap X_2$, the stalk at $p$ in the gluing is given by those members of $\mathcal O_{X_1, p} \times \mathcal O_{X_2, p}$ which agree in $\mathcal O_{Z_1, p} \simeq \mathcal O_{Z_2, p}$.

I can show that that structure map for the gluing is closed, but showing that base change preserves this closedness eludes me. Vakil says that in fact not only is this gluing a fibered coproduct, which is evident from the construction, but that it also is a fibered diagram. In particular, it commutes with base change. I am unable to show this is true. If pullback in the category of schemes had a right adjoint that we knew about, then I could prove the result since I know then that base change would commute with colimits, of which this gluing is an example. However, I do not believe we have seen that there is such a right adjoint.

Questions: is there such a right-adjoint to pullback in $Sch$? If not, what is the best way to show that gluing commutes with base change?

答:
This can be shown directly without too much trouble. Consider the following diagram, where each of the three rectangles is a pullback square:

$$\require{AMScd}
\begin{CD}
(X_1\coprod X_2)' @>>> X_1 \coprod X_2\\
@VVV @VVV \\
(X_1\coprod_Z X_2)' @>>> X_1\coprod_Z X_2 \\
@VVV @VVV \\
S' @>>> S
\end{CD}$$

where we write $A'=S'\times_S A$ for an $S$-scheme $A\to S$. Since $X_1\coprod X_2\to S$ is universally closed, $(X_1\coprod X_2)'\to S'$ is universally closed as well. As $X_1\coprod X_2\to X_1\coprod_Z X_2$ is surjective and surjectivity is stable under base change, we have that $(X_1\coprod X_2)'\to (X_1\coprod_Z X_2)'$ is surjective. So every closed subset $Y$ of $(X_1\coprod_Z X_2)'$ is the surjective image of a closed subset $\widehat{Y}$ in $(X_1\coprod X_2)'$. Thus the images of $Y$ and $\widehat{Y}$ in $S'$ are the same and therefore closed, as $(X_1\coprod X_2)'\to S'$ is universally closed. Hence $(X_1\coprod_Z X_2)'\to S'$ is closed for any $S'\to S$, and $X_1\coprod_Z X_2\to S$ is universally closed.

As Daniel Hast remarks in the comments, a right adjoint for base change is in general not representable. This can fail even in cases which are not "so bad" - see MO for an example.


问:
Nonnormal covering space of Klein bottle by Torus
I've been trying to construct a non-normal covering space for a Klein bottle $K$ by some torus $T$.
I've found some non-normal subgroups of  $\pi_{1}(K)= \langle a,b \mid a b a b^{-1}=e \rangle$ that should correspond to Torus covering spaces, e.g: $$, but I'm not sure how to describe the covering map $p \colon T \to K$, especially how do I set a partition of the fundamental polygon of $T$ into the squares representing $K$.

答:
Ok, I got it.

Let $a$, $b$ span a lattice $\mathbb{Z}^{2}$ in $\mathbb{R}^{2}$. If we are considering the covering space corresponding to $\langle a^{i}b^{j},a^{k}b^{l} \rangle \le \pi_{1}(K)$, we can choose the parallelogram $P$ made by vectors $ia+jb$ and $ka+lb$, then cut it into squares along the lattice. An edge of $P$ within each square can be mapped homotopically onto the boundary of the square, and the composition of these boundaries should give back $a^{i}b^{j}$ or $a^{k}b^{l}$ that corresponds to the edge.

Take the case $\langle a^{3},a^{2}b^{2} \rangle$ as an example:

In the parallelogram, the edge $c_{1}$, $c_{2}$ are homotopic to $ab$, $a^{-1}b$, so the edge $c_{1}c_{2}$ is presented by $aba^{-1}b=a^{2}b^{2}$.


问:
Why the ideal sheaf of $\operatorname{Spec}k[x]$ generated by $(x)$ is free but vanishes at $0$?
Consider the affine scheme $\operatorname{Spec}k[x]$ where $k$ is a field. Let $\mathcal{I}$ be the ideal sheaf of Spec$k[x]$ corresponding to the ideal $(x)$. Since $(x)$ is a free $k[x]$-module, $\mathcal{I}$ is a free $\mathcal{O}_{\operatorname{Spec}k[x]}$-modules of sheaf of rank $1$.

I feel that it is strange because all sections of $\mathcal{I}$ vanish at $0$ so it does not look like "free". Do I misunderstand something?

答: This is a source of constant confusion when dealing with locally free sheaves of rank one -- they are subsheaves of the (constant, incoherent) sheaf of rational functions, and it can happen that a section of an invertible sheaf $\mathcal{L}$ vanishes at a point when viewed as a rational function, but this does not mean that it maps to $0$ under a (local) isomorphism of $\mathcal{O}_X$ modules $\mathcal{L} \to \mathcal{O}_X$. Indeed in your case, that isomorphism is division by $x$.


问:
How do I find the volume of $\cos x$ about the $y$-axis?
Evaluate the volume formed by rotating the region bounded by $y=0$ and $y=\cos x$ for $\frac{\pi}{2}\le x\le\frac{3\pi}{2}$ about the y-axis.

I tried making $x=\cos^{-1}y$ then solving for $$\int_{{\frac{\pi}{2}}}^{\frac{3\pi}{2}}\cos^{-1}y\ dy$$ but clearly that is undefined.
How do I solve this?

答:
HINT

\begin{align*}
V = \int_{\pi/2}^{3\pi/2}2\pi x|\cos(x)|\mathrm{d}x = -2\pi\int_{\pi/2}^{3\pi/2}x\cos(x)\mathrm{d}x
\end{align*}

This might be helpful in order to understand such formula.


问:
Iterative solver for linear least squares, involving a large sparse/structured r.h.s.
Let $B$ be a $n$ by $n$ matrix, and $A$ a $n$ by $m$ matrix. I'm looking for the variable $X$, of size $m$ by $n$, that approximately solves $A X = B$; more precisely, the matrix $X$ that minimizes the Frobenius norm $||A X - B||_\text{fro}$.

While $n$ is large, the dimension $m$ is relatively small, and $A$ is dense. The matrix $B$ is either sparse, or products $B Y$ and $Z B$ can be computed efficiently. Thus, I'd like to use an iterative solver for the problem, but wasn't able to find anything.

$B$ is not necessarily symmetric, but I could work with the symmetric case first to benchmark different possibilities.

Is there a name for such a problem already in the community? Are there ways to solve that problem efficiently?

答:
Let $A = Q R$ be the QR decomposition of $A$, where $Q$ is a $n$ by $m$ matrix and $R$ is a $m$ by $m$ matrix. We then minimize $|| Q R X - B ||_\text{fro}$. Let $\overline{Q}$ represent the orthogonal complement to the range of $Q$ (and thus of $A$), so that the horizontal concatenation $(Q, \overline{Q})$ is a unitary matrix.

Then $|| Q R X - B ||_\text{fro} = || R X - Q^+ B ||_\text{fro} + || \overline{Q}^+ B ||_\text{fro}$: we use the fact that the Frobenius norm is invariant under a unitary change of basis ($||U A||_\text{fro} = ||A||_\text{fro}$ for unitary $U$), and split the range of $B$ on the range of $Q$ and $\overline{Q}$.

Obviously only the first term needs to be minimized. If $R$ is non-singular (equivalent to $A$ being full rank), then $X = R^{-1} Q^+ B$. Otherwise, a dense algorithm can be used as the matrices involved are not as big as before.


答:
We break down the area trapped for $y=\cos x$ in between $\frac{\pi}{2} < x < \frac{3\pi}{2}$ into strips of each length $dx$

Now, the differential volume of the strip on revolution is $dV = 2\pi x |\cos(x)|dx$

Hence, the total volume

$$V = \int_{\pi/2}^{3\pi/2}2\pi x |\cos x| dx$$

Title: How do I find the volume of $\cos x$ about the $y$-axis?

Body:
Evaluate the volume formed by rotating the region bounded by $y=0$ and $y=\cos x$ for $\frac{\pi}{2}\le x\le\frac{3\pi}{2}$ about the y-axis.

I tried making $x=\cos^{-1}y$ then solving for $$\int_{{\frac{\pi}{2}}}^{\frac{3\pi}{2}}\cos^{-1}y\ dy$$ but clearly that is undefined.
How do I solve this?


问:
How can you calculate $\int_{-\infty}^\infty\frac{dx}{x^2+1}$ using the residue at $-i$?
Instead of using traditional upper-half semicircle, I have tried to calculate this integral using lower-half and have obtained $-\pi$, that is the opposite to the actual value.
Can anyone show me how to calculate it?

答:
Almost certainly Robert Lee is right about your error. The residue is the result of integrating around a closed curve in the counter-clockwise direction. If you traverse the curve in the clockwise direction, you get the opposite value.

If you use the upper half-circle and travel counter-clockwise, then the part of the curve on the real line is integrated from negative to positive, and taking the limit gives $\int_{-\infty}^\infty \frac{dx}{x^2 + 1}$. But if you use the lower half-circle, you can either go counter-clockwise, in which case the portion on the real line is integrated from positive to negative, so taking the limit gives $\int_{\infty}^{-\infty} \frac{dx}{x^2 + 1}$, or you can travel clockwise, in which case you do get $\int_{-\infty}^\infty$, but it is equal to the opposite of the residue, instead of the residue itself.

Either way, there is an additional negation you missed performing.


问:
How to solve this complicated integration?
The PDF  of random variable $h$ is $$f_h(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/2\sigma^2}$$
I want to integrate PDF $f_\rho(x)$ i.e., $$\int_0^{h}f_\rho(x)\ dx$$
where $$f_\rho(x)=\frac{f_h(\sqrt{x/\rho_t})}{\sqrt{\rho_tx}}$$
I know the final answer includes error function. But having difficulty in reaching to final step. Any help in this is highly appreciated.

答:
If $$f_h=\frac{1}{\sqrt{2 \pi \sigma ^2}}e^{-\frac{x^2}{2 \sigma ^2}}\qquad \text{and}\qquad f_\rho=\frac{f_h\left(\sqrt{\frac{x}{\rho }}\right)}{\sqrt{\rho  x}}$$ then
$$f_\rho=\frac{e^{-\frac{x}{2 \rho  \sigma ^2}}}{\sqrt{2 \pi \rho \sigma ^2} \sqrt{
   x}}$$ Let
$$t^2=\frac{x}{2 \rho  \sigma ^2}\implies x=2 \rho  \sigma ^2 t^2\implies dx=4 \rho  \sigma ^2 t\, dt$$

$$\int f_\rho \,dx=\frac{2 }{\sqrt{\pi }}\int e^{-t^2}\,dt$$


问:
$\mathrm{Hom}$-sets in categories with finite biproducts have the structure of a commutative monoid: a reference for a proof
It seems it is a folk result that $\mathrm{Hom}$-sets in categories with finite biproducts (including the empty biproduct) have the structure of a commutative monoid. I know an analogous result for abelian categories from Boceux' book, but the proof presented there doesn't appear to be generalizable for this case. Does anybody have a reference? Or, perhaps, if someone knows the proof, they can help providing it?

The idea is that, for morphisms $f,g\colon X\to Y$, take $f + g$ to be $(f, g)\circ \Delta$ or $\Delta\circ h$ where $h\colon X\to Y\oplus Y$ is the morphism induced by the universal property of a product. But I struggle to make sense of details.

答:
You can find a fairly detailed proof in my blog post A meditation on semiadditive categories. The sketch is that

In any category with finite coproducts, every object is uniquely and hence canonically a commutative monoid with respect to the coproduct, with multiplication given by the codiagonal $\nabla : X + X \to X$; dually, in any category with finite products, every object is uniquely and hence canonically a cocommutative comonoid with respect to the product, with comultiplication given by the diagonal $\Delta : X \to X \times X$, and

In a category with finite biproducts, finite products and finite coproducts agree, so every object is uniquely and hence canonically a commutative monoid with  respect to the product, and this commutative monoid structure passes to homsets.


问:
Question about direct sum of function space
I am reading Applied linear algebra: the decoupling principle by Lorenzo Adlai Sadun (btw very recommendable!)

On page 30 about direct sums on vector spaces it says:

Let $V$ be the space of continuous functions on a domain $U \subset \mathbb R^3$. Then $V \oplus V \oplus V$ is the space of continuous $\mathbb R^3$-valued functions on $U$.

What I don't understand is how this threefold direct sum of $V$ can lead to a statement about the codomain ("-valued" functions). But I just guess that I don't get it altogether - please enlighten me. Thank you!

EDIT

I think what confuses me most is that we start with $\mathbb R^3$ and end up with it. It would come more natural if we started with $\mathbb R$ and after taking the direct sum three times would end in $\mathbb R^3$.

答:
I know that this question is 9 years old, but the other answers didn't help my tiny brain. If someone comes across like I did, I hope this helps!

Let $C(\mathbb{R}^3)$ denote the space of continuous functions with domain $\mathbb{R}^3$.  Let $V \subset C(\mathbb{R}^3)$ be defined by $V := \{f \in C(\mathbb{R}^3) : f:U\subset \mathbb{R}^3 \rightarrow \mathbb{R}\}$. Then $V$ is simply the subspace of  $C(\mathbb{R}^3)$ where each function is restricted to $U$.

The direct sum $$V\oplus V \oplus V := \{ (f_1,0,0)+ (0,f_2,0) + (0,0,f_3) : f_i \in V\} = \{(f_1,f_2,f_3) : f_i \in V\}$$
is simply the tuple of continuous functions on the domain $V$. In otherwords, you give me an $(x_1,x_2,x_3) \in U \subset \mathbb{R}^3$ and I give you back $(f_1(x_1,x_2,x_3),f_2(x_1,x_2,x_3),f_3(x_1,x_2,x_3))$. Hence, elements of the direct sum are the maps

$$[ (x_1,x_2,x_3) \mapsto(f_1(x_1,x_2,x_3),f_2(x_1,x_2,x_3),f_3(x_1,x_2,x_3))] \in V \oplus V \oplus V$$
This is what is meant by $(f_1,f_2,f_3)$.