问:
Is Robinsonian Semantics a special case of Tarskian Semantics?
Like the question says,

Putting all the nitty-gritty aside, Tarksi semantics boils down to an assignment function that maps a sequence of elements in the universe to a set of variables in a term.

Robinson takes another approach:

consider a language $L$ and structure $\scr S$ of $L$ and universe $S$. Instead of including an assignment function, we take an expansion of $L$, $L^+$, which is just $\scr {S}$ $\cup$ $C$ where $C$ includes a constant $c_i$ for each $s_i$ in $S$.

I can't prove this, but I think every Robinsonian term is a Tarksian term with the only difference being: if we consider a Robinsonian term within $\scr S$ it just has a particular assignment-namely- $\bar c$ which is from $C$.

答:
There isn't a substantive difference between the two (and I've never heard the term "Robinson semantics" before - I've seen both approaches subsumed by the term "Tarskian semantics"). Basically, in order to recursively define the satisfaction relation for normal sentences, it turns out that we have to consider some kind of "generalized sentence" along the way due to the behavior of quantifiers. There are two reasonable ways to get such a generalization, namely variable assignments or parameters in the syntax. These do exactly the same thing, but look different on the surface.

Using the terminology in the OP, here's a possibly-overly-detailed description of the situation. Tarskian semantics recursively defines a class $\mathfrak{C}_{Tarski}$ of triples of the form $(\mathcal{A},\varphi,\nu)$ where $\mathcal{A}$ is a structure in some language $L$, $\varphi$ is an $L$-formula, and $\nu$ is an assignment of elements of $\mathfrak{A}$ to the free variables in $\varphi$: we have $(\mathcal{A},\varphi,\nu)\in\mathfrak{C}_{Tarsk}$ iff $\mathcal{A},\nu\models\varphi$ in the Tarskian sense. Similarly, Robinsonian semantics recursively defines a class $\mathfrak{C}_{Robinson}$ of pairs of the form $(\mathcal{A},\hat{\varphi})$ where $\mathcal{A}$ is a structure in some language $L$ and $\hat{\varphi}$ is a sentence in the language $L_\mathcal{A}$ gotten by adding to $L$ a constant symbol for each element of $\mathcal{A}$: we have $(\mathcal{A},\hat{\varphi})\in\mathfrak{C}_{Tarsk}$ iff $\mathcal{A}\models\hat{\varphi}$ in the Robinsonian sense.

The point is that $\mathfrak{C}_{Tarsk}$ and $\mathfrak{C}_{Robinson}$ are really equivalent. For example, fix a structure $\mathcal{A}$ in a language $L$ and an $L_\mathcal{A}$-sentence $\hat{\varphi}$. Let $c_{a_1},...,c_{a_n}$ be the "$\mathcal{A}$-constants" appearing in $\hat{\varphi}$, let $x_1,...,x_n$ be variables not occurring in $\hat{\varphi}$, and let $(\varphi,\nu)$ be the following pair:

$\varphi$ is the $L$-formula gotten by replacing each "$c_{a_i}$" in $\hat{\varphi}$ by "$x_i$," and

$\nu:\{x_1,...,x_n\}\rightarrow\mathcal{A}: x_i\mapsto a_i$ is the variable assignment "remembering" what we've replaced.

We have $(\mathcal{A},\hat{\varphi})\in\mathfrak{C}_{Robinson}$ iff $(\mathcal{A},\varphi,\nu)\in \mathfrak{C}_{Tarski}$. And we can similarly translate the other way. So (up to annoying variable/constant symbol naming issues) Tarski and Robinson provide exactly the same picture. In particular, we have:

Suppose $\varphi$ is an $L$-sentence and $\mathcal{A}$ is an $L$-structure. Then $$(\mathcal{A},\varphi,\emptyset)\in\mathfrak{C}_{Tarski}\quad\iff\quad(\mathcal{A},\varphi)\in\mathfrak{C}_{Robinson}.$$ (Note that $\emptyset$ is indeed a variable assignment for $\varphi$ since $\varphi$, being a sentence, has no free variables whatsoever.)


问:
Confusion applying the shifting property of z-transform
I'd like to calculate the z-transform of $Z=(-3)^{-n}u[n-2]=\left(-\dfrac{1}{3}\right)^nu[n-2]$. Also, I'd like to use shifting property, which states:
$$x(n-k)\rightarrow z^{-k}\left(X(z)+\sum_{n=1}^kx(-n)z^k\right)$$
For the function in question, that would be:
$$\left(-\dfrac{1}{3}\right)^nu[n-2] \rightarrow z^{-2}\left(\dfrac{1}{1+\frac{z^{-1}}{3}}-3z+9z^2\right)$$
After some simplification $Z=\dfrac{3}{3z^2+z}-\dfrac{3}{z}+9$. After further simplification verified here, it becomes $Z=\dfrac{27z}{3z+1}$. However, wolfram alpha gives a different transform, which is $\dfrac{1}{9z^2+3z}$. So, I am confused. How should one apply the shifting property? what if the function were $\left(-\dfrac{1}{3}\right)^{n-2}u[n-2]$, or more generally $a^{n-k}u[n-j]$.

答:
Note that there are two versions of $\mathcal{Z}$-transform. The bilateral $\mathcal{Z}$-transform is defined as $$X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$And the unilateral $\mathcal{Z}$-transform is defined as $$X(z) = \sum_{n=0}^{+\infty}x[n]z^{-n}$$Adopting the bilateral transform, yields $$\mathcal{Z}\{\alpha^nu[n]\} = \frac{1}{1-\alpha z^{-1}} \\ \mathcal{Z}\{x[n-n_0]\} = z^{-n_0}X(z)$$
Combining these facts, we have $$\mathcal{Z}\{(-\dfrac{1}{3})^{n-2}u[n-2] \} =  \frac{z^{-2}}{1+\frac{z^{-1}}{3} } \implies \\ \mathcal{Z}\{(-\dfrac{1}{3})^{n}u[n-2] \} = \frac{z^{-2}}{9+3z^{-1} } = \frac{1}{9z^2 +3z} = \frac{1}{3z(3z+1)}$$ This can be easily verified by WA. Also read this documentation which mentions WA computes the unilateral transform by default. The unilateral transform gives $$\mathcal{Z}\{\alpha^nu[n]\} = \frac{1}{1-\alpha z^{-1}} \tag{1}$$ $$\mathcal{Z}\{x[n-1]\} = z^{-1}X(z) + x[-1] \implies \mathcal{Z}\{x[n-2]\} = x[-2] + x[-1]z^{-1} + z^{-2}X(z) \tag{2}$$
By choosing $\alpha=\frac{-1}{3}$ in equation $(1)$, we have $$\mathcal{Z}\{(-\frac{1}{3})^nu[n]\} = \frac{1}{1+\frac{z^{-1}}{3}}$$Now let $x[n] =  (-\frac{1}{3})^nu[n]$, so we have $x[-1] = x[-2] = 0$ which implies, by equation $(2)$, $$\mathcal{Z}\{(-\frac{1}{3})^{n-2}u[n-2]\} = z^{-2}\frac{1}{1+\frac{z^{-1}}{3}}$$
Finally, by linearity property we have $$\mathcal{Z}\{(-\frac{1}{3})^{n}u[n-2]\} = \frac{z^{-2}}{9}\frac{1}{1+\frac{z^{-1}}{3}}$$
In this case both transforms give the same result since $(-\dfrac{1}{3})^{n}u[n] = 0$ for $n\lt0$.

ContentLicense: CC BY-SA 4.0


问:
I want an operation that keeps digits in order and result in a year number
I remember a math teacher wishing us a good year with a mail title that was something like:

101 = (1 + 2 + 3)* 4 - 5*(6 - 7) + 8*9

Except the result was 2006.

Is there a website that can find a number this way, or is there some kind of script that does this already?

I'm curious about how this could be called. Number targeting game? I guess I could find it by brute force with a python eval() but placing parenthesis looks like it's quite a bit of work.

I guess it's still a fun curiosity to share. Is there a method other than brute force?

2021 is 43 x 47, so I'm not sure it's doable by hand.

What makes it beautiful is that the number are in order. Someone suggested:

1 * 2 * 3 * 4 * 7 * 8 * 9 / 6 + 5

but it's not ordered.

答:
$$-1\times 2+345\times 6-7\times 8+9=2021$$

And, for next year,

$$1- 2+345\times 6-7\times 8+9=2022.$$

However, I don't have a general method! Noticing that $345\times 6$ is reasonably close means one then has lots of options to adjust the answer using the other numbers.

OwnerDisplayName: jokoon


问:
Confused about one Concept
Today in my differential equation lecture, the teacher begin to write some examples in which differential equations appear, first begin with the classic example of a rocket, the example of the mass and the resort, and then he request to us give to he a example of a family of lines, and I tell to he

$y=mx+b$ where $m,b$ are parameters and then he put
$\frac{dy}{dx}=m$ and then he remplaze it in the original equation
$y=\frac{dy}{dx}x+b$
and put that this are the "Equivalent differential form of the family of curves.

Then he  put an exercise and exactlly say the following

Consider $y=A \sin x+ B \cos x$ familly of curves with two parameters $A,B$.Find it´s equivalent differential form.

I Never watch these kind of substitution in my life, and since he say that is a differential form I´m confused because the differential form of one equation is an expression like a $dz=f(x,y)dx+g(x,y)dy$

Is correct the teacher or it´s wrong, in the second case, can someone explain me or put the name of the concept.

I think that is wrong with the word differential form because if I don´t make a mistake
the differential of the equation $y=mx+b$ is $dy=mdx$

答:
\begin{align*}
y&=A\sin(x)+B\cos(x)\\
\frac{d}{dx}(y)&=\frac{d}{dx}(A\sin(x)+B\cos(x))\\
\frac{dy}{dx}&=A\frac{d\sin(x)}{dx}+B\frac{d\cos(x)}{dx}\\
\frac{dy}{dx}&=A\cos(x)-B\sin(x)\\
dy&=(A\cos(x)-B\sin(x))dx
\end{align*}
Therefore :
$$
dy-(A\cos(x)-B\sin(x))dx=0
$$
Is the equivalent differential form of $y$

Title: Confused about one Concept


问:
Derivative of linear map?
Given $f: \mathbb{R}^{n\times n}\to \mathbb{R}^{n\times n}$, $f(X)=X^TX$, how do I know that $df(A)H=A^TH+H^TA$ for $A\in O(n,\mathbb{R})$ and $H\in \mathbb{R}^{n\times n}$ where $O(n,\mathbb{R})$ is the orthogonal group.

答:
Recall that a function $\phi : \mathbb R^m \to \mathbb R^k$ is differentiable at $x$ if there exists a linear map $L : \mathbb R^m \to \mathbb R^k$ such that
$$\lim_{h \to 0} \frac{\lVert \phi(x+h) - \phi(x) - L(h)\rVert}{\lVert h \rVert} = 0 .$$
In that case $L$ is uniquely determined and is written as $d\phi(x)$.

Given $X$, the map $L_X(H) = X^TH + H^TX$ is linear and we have
$$f(X + H) - f(X) - L_X(H) = X^TX + X^TH + H^TX + H^TH - X^TX - (X^TH + H^TX) = H^TH .$$
Thus
$$\frac{\lVert f(X + H) - f(X) - L_X(H) \rVert}{\lVert H \rVert} = \frac{\lVert H^TH \rVert}{\lVert H \rVert}$$
Since all norms on finite-dimensional Euclidean spaces are equivalent, we may assume that $\lVert - \rVert$ is a submultiplicative matrix norm (for square matrices) which means that $\lVert A B \rVert \le \lVert A \rVert \lvert B \rVert$. See for example Proof of matrix norm property: submultiplicativity. Hence
$$ \frac{\lVert H^TH \rVert}{\lVert H \rVert} \le \lVert H^T \rVert$$
which goes to $0$ as $H \to 0$. This shows that $df(X) = L_X$.

Title: Derivative of linear map?


问:
Evaluating $\sum\limits_{i=0}^n{n\choose i}2^i$
I'm being asked to "evaluate the sum"

$$\sum\limits_{i=0}^n{n\choose i}2^i$$

Writing out the first few values of $n$ it is clear that the sum is equal to $3^n$, however I'm not sure how to prove this. I know that it is straight forward to show that $\sum\limits_{i=0}^n{n\choose i}=2^n$, we just set $x=y=1$ in the binomial formula, but the $2^i$ seems to make things much less obvious.

The wording of the question seems ambiguous. I'm not sure if "evaluate" means that I don't need to prove the result? But just evaluating a few values of $n$ and calling it a day seems insufficient.

If anybody could give me a hint to maybe set me on the right track that would be great. Thanks in advance!

答:
HINT

Rewrite the given expression as
\begin{align*}
\sum_{i=0}^{n}{n\choose i}2^{i} = \sum_{i=0}^{n}{n\choose i}2^{i}1^{n-i}
\end{align*}

and apply the binomial theorem.

Can you take it from here?


问:
Does this proof of the sequential completeness of a reflexive space require the Eberlein–Šmulian theorem?
I'm self-teaching functional analysis through Conway's book at the moment, and can't seem to get one of the steps in this proof:

The statement that $\{ x\in X : ||x||<M\}$ implies the sequence $\{x_n\}$ has a cluster point is throwing me off. I can't seem to find a way to justify this without invoking the Eberlein–Šmulian theorem - which Conway doesn't mention in his book until a few chapters later. My question is - am I missing something? The definition Conway gives for a Cauchy sequence in this situation is given below:

Is there a way to see thus statement without using such heavy machinery? Or is this just a small oversight in his book?

Edit: the problem is that weak compactness implies weak sequential compactness here. I should’ve specified that fact that ballX is weakly compact isn’t the problem

答: Weak sequential compactness isn’t actually required here. K a compact space $\implies$ every net in K has a cluster point.


问:
If $P_n = span\{p_1(x),p_2(x),...,p_k(x)\}$ and $a \in \mathbb{R}$ show that $p_i(a) \neq 0$ for some $i$.
If $P_n = span\{p_1(x),p_2(x),...,p_k(x)\}$ and $a \in \mathbb{R}$ show that $p_i(a) \neq 0$ for some $i$.

I don't understand how to approach this problem. If I make $k=n$ the set becomes a minimal spanning set. If I have a $f(a) = a_1p_1(x) + ... + a_ip_i(a) + ... + a_np_n(a)$ and $p_i(a)=0$, then my intuition tells me that $f(a)$ can not be represented by that spanning set. Am I on the right track? can someone help formalize this proof?

Added Attempt:

Show contrapositive: if $p_i(a) =0$ for all $1\leq i \leq k$ than $P_n \neq span\{p_1(x),p_2(x),...,p_k(x)\}$

Let $f(x) = a \in P_n$ ($a\neq0$) then $f(x) = a_1p_1(x) +...+a_ip_i(x) + ... + p_k(x)$. However, $f(a) = 0$ which is a contradiction.

Is this right?

答: Suppose there is $a\in\Bbb R$ such that every $p_i(a)=0$.  Set $f\equiv 1$. Then $f\in\rm{span}\{p_1,\dots,p_k\}\implies 1=0$.


问:
How to show that $A$ is increasing?

Suppose that $f$ is a twice differentiable real function such that
$f''(x)>0$ for all $x\in[a,b]$. Find all numbers $c\in[a,b]$ at which
the area between the graph $y=f(x)$, the tangent to the graph at
$(c,f(c))$, and the lines $x=a$, $x=b$, attains its minimum value.

I understand that the area mentioned above is

$$A(c)=\int\limits_a^b (f(x)-f(c)-f'(c)\cdot(x-c))dx,$$

and using this formula I can get that $c= \frac{a+b}{2}$ is a critical point of $A$, but I'm falling to see why this has to be where $A$ attains, specifically, a minimum. If anyone could enlighten me, that would be wonderful.

答:
We want to find the minimum of the function $A(c)$, which will be located at a point where $A'(c)=0$ so:
$$A'(c)=-f''(c)\int_a^b(x-c)dx$$
now since we know that $f''>0$ this cannot equal zero so the integral must:
$$\int_a^b(x-c)dx=\left[\frac {x^2}{2}-cx\right]_a^b=\frac{b^2-a^2-2c(b-a)}{2}=0$$
now we need to solve for c, so:
$$2c(b-a)=b^2-a^2$$
$$2c=b+a$$
$$c=\frac{a+b}{2}$$
and there is your answer

Title: How to show that $A$ is increasing?


问:
Primary ideals are primal
Let $R$ be a commutative ring with unity.

Recall an ideal $I \subseteq R$ is primary if $ab \in I$ implies $a \in I$ or $b^n \in I$ for some $n$.

$I$ is primal if its adjoint $I^{\not\perp}\ := \{ a \in R: \exists\, b \notin I \text{ such that } ab \in I \}$ is an ideal.

I am struggling to prove that every primary ideal is primal. It is easily shown that the adjoint of any ideal contains $0$, is closed under negation and absorbent under multiplication. So, somehow I need to use the primary-ness of $I$ to show $I^{\not\perp}\ $ is closed under addition, but I can't see how.

答: Let $a_1,a_2\in I^{\not\perp}$. Then there are some $b_1,b_2\notin I$ such that $a_1b_1, a_2b_2\in I$. Since $I$ is primary it follows that we must have $a_1,a_2\in\sqrt{I}$, and so $c:=a_1+a_2\in\sqrt{I}$ as well. Take the minimal $n\in\mathbb{N}$ such that $c^n\in I$. Then $c\cdot c^{n-1}=c^n\in I$, where $c^{n-1}\notin I$. Hence $c\in I^{\not\perp}$.

Title: Primary ideals are primal


问:
Does This Concept for Defining a Nearest Neighbor in High Dimensions Make any Practical Sense?
I'm not particularly well-educated in mathematics. I've been reading about the, 'curse of dimensionality,' and how distance metrics become meaningless in high-dimensional spaces.

I then thought: What if you treat each dimension in a high-dimensional vector as a 1D scalar, and find the vector with the nearest scalar value for the same dimension. Then, you add the index for this, 'single-dimension nearest neighbor,' to a new vector. You iterate through this process for every dimension in the vector and in the end, you define the nearest neighbor as the vector that appears most often in the vector of single-dimensional nearest neighbors.

Is this utter nonsense from a mathematical perspective? I have no idea how to even begin analyzing the efficacy of this mathematically.

答:
When you make a definition like this there are two criteria of interest.  The first is whether it is well defined-is it clear what vector it will choose as the nearest neighbor.  The second is whether it is useful.  I think well defined fails here.  If you choose one vector out of a group of $10,000$ in $100$ dimensional space, it sounds like you will find the vectors that are closest in each coordinate direction to the chosen vector and take a plurality vote among them.  If you have many more vectors than dimensions, the nearest vector in each dimension may be different, so each of them gets one vote.  Which is the nearest?  Maybe you just want to make a list of the several nearest vectors.  That will clearly be well defined.

Whether it is interesting depends on whether the vector or vectors you select as nearest neighbor(s) are similar to the original one in the sense of interest to you and your community.  I can't comment on that.


问:
Compact hermitian operator and eigenvalues
I would like to show that all non-zero eigenvalues of a (possibly unbounded) compact and hermitian operator $T:\mathcal{D}(T)\to \mathcal{H}$ on a seperable, infinite-dimensional Hilbert space have finite multiplicity.

My idea so far: I would like to use the characterisation:

$\lambda\in\sigma (T)$ is eigenvalue with finite multiplicity if and only if $\lambda\in\sigma (T)$ and for all sequences $(x_{n})$ in $\mathcal{D}(T)$ with $\Vert x_{n}\Vert=1$ and $\Vert (T-\lambda )x_{n}\Vert\to 0$ there exists a convergent subsequence.

So let $\lambda\in\sigma (T)$ be non-zero and $(x_{n})$ be such a sequence. Since this sequence is by assumption bounded, there exists a weakly-convergent subsequence by Banach-Alaoglu, so lets write $x_{n_{k}}\rightharpoonup x_{0}\in\mathcal{H}$. By compactness of $T$, it follows that $Tx_{n_{k}}\to Tx_{0}$. But now I don't know how to proceed. Maybe I can show that $\Vert x_{n_{k}}\Vert\to \Vert x_{0}\Vert$, because than it would follow that $x_{n_{k}}\to x_{0}$....I haven't use the assumptions $\Vert (T-\lambda )x_{n}\Vert\to 0$ and $\lambda\neq 0$ yet....

答: If $N=\mathcal{N}(T-\lambda I)$ is infinite-dimensional for some $\lambda\ne 0$, then there exists a sequence $\{ x_n \}$ of unit vectors that are mutually orthogonal and satisfy $(T-\lambda I)x_n=0$. Then $T$ maps the bounded sequence $\{ x_n \}$ to a sequence $\{ \lambda x_n \}$ with no convergent subsequence. So $\mathcal{N}(T-\lambda I)$ must be finite-dimensional for all $\lambda\ne 0$.


答:
Given that your teacher referred to $y=\frac{dy}{dx}X + b$ as the "differential form" of $y=mx + b$, and given that you're just encountering some simple differential equations, as a way of introducing differential equations, I think you're hearing a false similarity.  The teacher should have more correctly said the "differential equation form", and either wasn't thinking of differential forms or didn't know about them.  All the teacher is looking to do (and we can tell from the beginning and result) is to show $y=mx+b$ as a differential equation.  We do that by differentiating, in order to find an equivalent expression for $\frac{dy}{dx}$ and somehow work that $\frac{dy}{dx}$ back into the original equation.

Thus, what your teacher would be expecting (if I'm right) is that you express

$$
y=A\sin x + B\cos x
$$

as a differential equation.  Differentiate once or twice to get an expression  in terms of a derivative that you can substitute back into this equation.

Title: Confused about one Concept

Body:
Today in my differential equation lecture, the teacher begin to write some examples in which differential equations appear, first begin with the classic example of a rocket, the example of the mass and the resort, and then he request to us give to he a example of a family of lines, and I tell to he

$y=mx+b$ where $m,b$ are parameters and then he put
$\frac{dy}{dx}=m$ and then he remplaze it in the original equation
$y=\frac{dy}{dx}x+b$
and put that this are the "Equivalent differential form of the family of curves.

Then he  put an exercise and exactlly say the following

Consider $y=A \sin x+ B \cos x$ familly of curves with two parameters $A,B$.Find it´s equivalent differential form.

I Never watch these kind of substitution in my life, and since he say that is a differential form I´m confused because the differential form of one equation is an expression like a $dz=f(x,y)dx+g(x,y)dy$

Is correct the teacher or it´s wrong, in the second case, can someone explain me or put the name of the concept.

I think that is wrong with the word differential form because if I don´t make a mistake
the differential of the equation $y=mx+b$ is $dy=mdx$


问:
Differences behind different methods of fminunc in MATLAB?
Assume I have some .m file with a function (and it's gradient) to be used by fminunc() in MATLAB for some unconstrained optimization problem.

To solve the problem in the most simple way, I do this:

```<code>clear all
[x,fval] = fminunc(@fun, [1;1])
</code>```

This will minimize fval and return the optimized values of x. For a more accurate optimization, I do this:

```<code>clear all
op = optimset('GradObj', 'on', 'LargeScale', 'off');
[x,fval] = fminunc(@fun, [1;1], op)
</code>```

Both fval and x values still are the solution to the problem only that now they are more accurate, because of the supplied gradient. Correct?

Both of the above methods use the line-search algorithm but I can also use the trust-region algorithm, like this:

```<code>clear all
op = optimset('GradObj', 'on');
[x,fval] = fminunc(@fun, [1;1], op)
</code>```

Both fval and x values are different from the previous ones. What does this mean? Is this algorithm better or worse? Or maybe it's different in a way that it's not better nor worse. What does it mean than?

答:
The main benefit of providing a gradient is that convergence should be quicker. In general, even without gradient, you should be able to converge to a solution to within a desired accuracy, although at the expense of more iterations. A more direct control over accuracy is given by options 'TolX' and 'TolFun'.

If the objective functions have local minima, the solution you find will naturally depend on initial conditions. Differences in algorithms cause them to follow different paths in their way to the solution, so when applied to functions having many local minima, they may end up in different solutions.

In minimization problems, it's actually easy to choose the best among different solutions: just choose the one with the lower value! :-) Their meaning depends, of course, on the actual problem being solved...

ContentLicense: CC BY-SA 3.0


问:
What books should I read before books of Advanced Calculus?
I need books to get the right understanding of calculus to read books like Spivak, Apostol and Courant Calculus, I'm a complete beginner at it, since I've had heard those are good books, and I've look at them and they are really interesting books which I'd I'd love to study, but as I've said I don't know much (almost any) about calculus.

I want to know if there are some good references to study before that can help me to don't get stuck (easily) when I am studying or should I go directly to those books, if that is ghe case which one should I read first: Courant, Apostol or Spivak

答: I love love loved my calculus book from when I took AP in high school.  It was the two volumes by Best and Penner.  The red and the green.  Highly recommend them.</p>


问:
Probability that two randomly chosen people out of 10 have birthday in the same month

There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?

My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility

答: I'm sure that your answer of 1/12 is correct. When you randomly choose 2 people, the probability of person 2 having a birthday in the same month as person 1 would just be 1/12, assuming that the 12 months have an equal chance at being someone's birth month.</p>


问:
In a CCC with the initial object, can I have a morphism A -> 0?
I'm in a cartesian-closed category with the initial object $0$. What are the consequences of having a morphism from some object $A$ to $0$? Does it mean that there is also a global element of $0$, that is $1\to 0$ ?  Does it mean that we have a zero object, $1 \cong 0$ ? Does it mean that all objects are isomorphic?

Assume we don't have coproducts.

答:
Since the asker is not satisfied with Qiaochu's answer I'll try to expand on it and show more of the intermediate steps.

First we show that $X \times 0 \cong 0$, by showing that $X \times 0$ is an initial object.
$$Hom(X \times 0, A) \cong Hom(0, A^X) \implies |Hom(X \times 0, A)| = 1$$

This means that the projection $\pi_0: X \times 0 \to 0$ is an isomorphism. We use the $Hom(A, -)$ functor to get the isomorphism (functors send isomorphisms to isomorphisms).
$$Hom(A, \pi_0) : Hom(A, X \times 0) \to Hom(A, 0)$$
which is just post-composition by $\pi_0$.

The universal property of the product gives us another isomorphism, which we'll call $f$:
$$ f: Hom(A, X)\times Hom(A,0) \to Hom(A, X \times 0) $$

We can compose these two isomorphisms to get a third isomorphism:
$$ Hom(A, \pi_0) \circ f : Hom(A, X)\times Hom(A,0) \to Hom(A, 0)$$

It turns out that $ Hom(A, \pi_0) \circ f = \pi_{Hom(A,0)} $. This is because
$$ (Hom(A, \pi_0) \circ f)(g,h) = \pi_0 \circ f(g,h) = h $$

The last equality is due to the universal property of $A \times 0$. Hence the projection:
$$ \pi_{Hom(A,0)}:  Hom(A, X)\times Hom(A,0) \to Hom(A, 0)$$
is an isomorphism. If $Hom(A, 0)$ is non-empty then $Hom(A, X)$ also has to be non-empty. If $Hom(A, X)$ had more than one element then $ \pi_{Hom(A,0)} $ would not be injective which means $ Hom(A, X)$ has exactly one element and $A$ is an initial object.


问:
On theorem 11.8 Rudin Analysis
Here is a copy of theorem 11.8 Rudin Analysis. How the inequality just after (20) holds?
$$\sum_{n=1}^\infty\mu(A_n)\leq\mu^*(A)+\varepsilon$$

I checked every previous definitions and equations but I can't derive the mentioned inequality.

My second question is Rudin has defined subaddivity for $\mu*$ not $\mu$, so how the third inequality in $\mu(A)\leq \mu(F)+\epsilon \leq \mu\left(A_1\bigcup\cdots\bigcup A_N\right)+\epsilon\leq \sum_{1}^N \mu(A_n)+\epsilon \leq \mu^*(A)+2\epsilon$ holds?

Please help!

答:
Definition 11.7 provides the answer you're looking for: since $\mu^*(E)$ is the greatest lower bound of $\sum_{n=1}^{\infty} \mu\left(A_{n}\right)$ over all possible families of open elementary sets $\{A_n\}$ covering $E$, for every $\varepsilon>0$ you can find one such family such that your inequality holds.

As for your second question, that inequality holds because the collection of sets $\{A_n\}$ is finite and the measure $\mu$ is finitely additive: see equations (6) and (11) in that chapter.

Title: On theorem 11.8 Rudin Analysis


问:
Finding the intersection of a plane and a sphere.
Given a sphere with the equation of $x^2+y^2+z^2=144$ and the plane $x=5$, find the equation of the resulting intersection.

I know that the intersection will be a circle. I tried to plug x in as five. However, when graphing this online, I realized that this was a cylinder. I am not sure how to limit this region to only the intersection.

答:
Here $y^2+z^2=119$ would indeed be a cylinder, if $x$ can take any value,

but that is not an expression for this intersection, since we know $x=5$,

so the answer is "$y^2+z^2=119$ and $x=5$", which is a circle.

You could make this a messy single equation such as $(y^2+z^2-119)^2 +(x-5)^2=0$ though it would not be particularly informative and not so obviously be a circle


问:
Let $M = K + L$ and let $f: M \to N$ be an epimorphism. Prove that $N = f(K) \oplus f(L)$ if $K \cap L = \operatorname{Ker}f$
Let $M = K + L$ and let $f: M \to N$ be an epimorphism. Prove that
$N = f(K) \oplus f(L)$ if $K \cap L =  \operatorname{Ker}f$

I think we should start with take an element from $K + L$ and then we can use $f(M)=f(K + L)=N$ but I don't know how to do the rest.

The book is rings and categories of modules. Writers Anderson and Fuller

Section 5 exercise 5

答:
Clearly, $f(K)+f(L) = f(K+L)= N$.

Let $x\in f(K)\cap f(L)$. Then there exist $k\in K$ and $l\in L$ such that $f(k)=x=f(l)$, and hence $f(k-l)=0$. So $k-l\in\ker(f) = K\cap L\subset L$, implying that $k = (k-l) + l \in L+L=L$. This means that $k\in K\cap L = \ker(f)$ and hence $x=f(k)=0$. So $f(K) \cap f(L) = \{0\}$.


问:
Show that an element of the factor group $\mathbb{R}/\mathbb{Z}$ has finite order if and only if it is in $\mathbb{Q}/\mathbb{Z}$

Show that an element of the factor group $\mathbb{R}/\mathbb{Z}$ has finite order if and only if it is in  $\mathbb{Q}/\mathbb{Z}$.

What I have done: So far I know that if $a$ is an element in $\mathbb{Q}/\mathbb{Z}$, then $a$ has the form $\frac{b}{c}+\mathbb{Z}$, with $b,c \in \mathbb{Z},c\neq0$. Without loss of generality suppose that $c>0$, then $ca=c\left ( \frac{b}{c}+\mathbb{Z} \right )=b+\mathbb{Z}=\mathbb{Z}$ wich is the identity element in $\mathbb{Q}/\mathbb{Z}$. Thus $ a$ has finite order. I think this part proves ($\leftarrow $)

I have trouble establishing the other implication and verifying the theorem. I mean...

If an element of the factor group $\mathbb{R}/\mathbb{Z}$ has finite order, then this element lies in   $\mathbb{Q}/\mathbb{Z}$

How can I prove this?

答: Hint: suppose $x+\mathbb{Z}$ has finite order where $x\in\mathbb{R}$. That means that $nx\in\mathbb{Z}$. Can you take it from there?


问:
Is the following definition of an elliptic curve correct?
Im new to algebraic geometry so I want to make sure im getting my definitions right. I know there are a few ways to state what an elliptic curve is (ex a smooth projective curve of genus one with distinguished $K$-rational point). But I am wondering if the following is equivalent. For simplicity lets just work over $\mathbb{C}$.:

$\textbf{Definition:}$ An elliptic curve $E$ is a non-singular projective curve in $\mathbb{P}^2$ of the form
$$E: Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3 $$

I am wondering if this is sufficient to define an elliptic curve?

答:
Yes, this will do. The abstract definition of an elliptic curve is that it is nonsingular of genus $1$ as you mentioned. If your curve is cut out by a degree $d$ equation in $\Bbb{P}^2$, the degree-genus formula gives
$$
g=\frac{(d-1)(d-2)}{2}.
$$
Hence, since you have a cubic $C$, $g(C)=\frac{2}{2}=1$. Conversely, you can show that any abstract genus $1$ curve over an algebraically closed field $k$ has an embedding into projective space $\Bbb{P}^2$ which realizes it as a degree $3$ curve.


问:
$f(z)=\frac{1}{z}$ is not uniformly continuous in $|z|<1$, $z \ne 0$.
Show that $f(z)=\frac{1}{z}$ is not uniformly continuous in $|z|<1$, $z \ne 0$. Suppose by the sake of contradiction that it is, then for all $\epsilon>0$ exists $\delta(\epsilon)>0$ such that for all $z_1,z_2$ such that $|z_1|<1$ and $|z_2|<1$,  $|z_1-z_2|<\delta(\epsilon) \implies \left|\frac{1}{z_1}-\frac{1}{z_2}\right|<\epsilon$.

Suppose without loss of generality that it is $0<\delta(\epsilon)<1$, then since the definition of uniform continuity holds for all $z_1,z_2$ such that $|z_1|<1$ and $|z_2|<1$ I can choose $z_1=\delta(\epsilon)$ and (since $\frac{\delta(\epsilon)}{1+\epsilon} <\delta(\epsilon)<1$) I can choose $z_2=\frac{\delta(\epsilon)}{1+\epsilon}$ for it follows
$$\left|\frac{1}{\delta(\epsilon)}-\frac{1}{\frac{\delta(\epsilon)}{1+\epsilon}}\right|<\epsilon$$
But $\left|\frac{1}{\delta(\epsilon)}-\frac{1}{\frac{\delta(\epsilon)}{1+\epsilon}}\right|=\frac{\epsilon}{\delta(\epsilon)}$, so it is $\frac{\epsilon}{\delta(\epsilon)}<\epsilon \iff \delta(\epsilon)>1$ but this contradicts the fact that $\delta(\epsilon)<1$. So $f$ is not uniformly continuous in $|z|<1$. Is this correct? I am not sure because I'm not sure if all my supposition are correct, in particular: I'm not sure that I can suppose $\delta(\epsilon)<1$, because this is a proof by contradiction so I suppose that $f$ is uniformly continuous and I don't know if the $\delta(\epsilon)$ I get is actually $<1$ (but I think this is ok because if $\delta(\epsilon) \geq 1$ is the one that works for the uniform continuity then all the $\delta^*(\epsilon)<1$ work as well) and I'm not sure if the fact that when I get $\delta(\epsilon)>1$ I actually get a contradiction, since it was a supposition of mine that $\delta(\epsilon)<1$.

答:
If $f$ is uniformly continuous in $(a,b)$ then $f$ is bounded in $(a,b)$. Since your function is not bounded couldn't be uniformly continuous.

Check this: If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$.


问:
How to "invent" a function?
I'm currently making a game. I was given some numbers the leader wants me to stick close to. They are used for leveling; level 1, for example, needs 100 experience points, level 2 needs 250 experience points, etc.

Is there any way, maybe with a computer program, to calculate a function that will do something like this, instead of having to hard-code it?

Thanks in advance!

Simon.

答: It depends upon how accurately you want to reproduce the numbers.  For example, you could use $40*2.5^{level}$.  This gives 100, 250,625,1562.5,3906.25 and so on.  The first two match exactly.  Exponential growth seems to be what many games want.  Does this meet your needs?</p>

Title: How to "invent" a function?


问:
A combinatoric identity involving Eulerian and Stirling numbers
Let
$$
A_{n}(x)=\sum_{k=0}^{n}\left\langle\begin{array}{l}
n \\
k
\end{array}\right\rangle x^{n-k}
$$
Eulerian Polynomials
and $$
\left\langle\begin{array}{l}
n \\
k
\end{array}\right\rangle
$$ are Eulerian numbers.

I want to ask whats the proof of this interesting equation?
$$
\sum_{k=1}^{n}\left\langle\begin{array}{l}
n \\
k
\end{array}\right\rangle x^{n-k}=(1-x)^{n} \sum_{k=0}^{n}\left\{\begin{array}{l}
n \\
k
\end{array}\right\} k !\left(\frac{x}{1-x}\right)^{k}
$$

I have a reference https://www.mat.uniroma2.it/~tauraso/AMM/AMM11007.pdf

In this paper(American Mathematical Monthly Problem)  there are a similar approach. It solves 2 seperate equation with induction.You can look at it.

When i solved this equation in same way ,i couldnt reach a recurence relation about eulerian and stirling numbers. Maybe i go wrong in somewhere.
So can you help me ?

答:
Here is another proof for the curious. We seek to verify that
(with $n\ge 1$, $n=0$ holds by inspection)

$$\sum_{k=0}^n \left\langle n \atop k \right\rangle x^{n-k}
= (1-x)^n \sum_{k=0}^n {n\brace k} k!
\left(\frac{x}{1-x}\right)^k.$$

We get using standard EGFs for the RHS

$$n! [z^n] (1-x)^n
\sum_{k=0}^n \frac{(\exp(z)-1)^k}{k!} k!
\left(\frac{x}{1-x}\right)^k
\\ = n! [z^n] (1-x)^n
\sum_{k=0}^n (\exp(z)-1)^k
\left(\frac{x}{1-x}\right)^k.$$

Now because $\exp(z)-1 = z+\cdots$ we have $(\exp(z)-1)^k = z^k +
\cdots$ so when $k\gt n$ there is no contribution to the coefficient
extractor and we get

$$n! [z^n] (1-x)^n
\sum_{k\ge 0} (\exp(z)-1)^k
\left(\frac{x}{1-x}\right)^k
\\ = n! [z^n] (1-x)^n
\frac{1}{1-(\exp(z)-1)x/(1-x)}
\\ = n! [z^n] (1-x)^n
\frac{1-x}{1-x-(\exp(z)-1)x}
\\ = n! [z^n] (1-x)^n
\frac{1-x}{1-x \exp(z)}
\\ = n! [z^n] \frac{1-x}{1-x\exp(z(1-x))}.$$

On the other hand we have for the LHS by the mixed GF of the Eulerian
numbers

$$n! [z^n] \sum_{k=0}^n x^{n-k}
[w^k] \frac{w-1}{w-\exp((w-1)z)}$$

Now we have $\left\langle n \atop k \right\rangle = 0$ when $k\ge n$ so
this is

$$n! [z^n] x^n \sum_{k\ge 0} x^{-k}
[w^k] \frac{w-1}{w-\exp((w-1)z)}
\\ = n! [z^n] x^n \frac{1/x-1}{1/x-\exp((1/x-1)z)}
\\ = n! [z^n] x^n \frac{1-x}{1-x\exp((1/x-1)z)}
\\ = n! [z^n] \frac{1-x}{1-x\exp((1/x-1)zx)}
\\ = n! [z^n] \frac{1-x}{1-x\exp((1-x)z)}.$$

The LHS is the same as the RHS and we have the claim.

Addendum. We have

$$n! [z^n] [w^k] \frac{w-1}{w-\exp((w-1)z)}
\\ = n! [z^n] [w^{k+1}] \frac{w-1}{1-\exp((w-1)z)/w}
\\ = n! [z^n] [w^{k+1}] (w-1)
\sum_{q\ge 0} \frac{1}{w^q} \exp(q(w-1)z)
\\ = [w^{k+1}]
\sum_{q\ge 0} \frac{1}{w^q} q^n (w-1)^{n+1}
= \sum_{q\ge 0} [w^{k+1+q}] q^n (w-1)^{n+1}
\\ = (-1)^{n-k} \sum_{q=1}^{n-k} (-1)^q q^n {n+1\choose k+1+q}.$$

This justifies that $\left\langle n \atop k \right\rangle = 0$ when $k\ge
n$ and hence the two coefficient extractors combined return  zero in that
case as  claimed.


问:
What's a more general nullstellensatz?
$\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\mr}{\mathscr}$ $\newcommand{\mc}{\mathcal}$

Let $k$ be an algebraically closed field.
For an ideal $I$ in $k[x_1,\ldots, x_n]$ we define $\mc V(I)$ as the set of all the points in $k^n$ on which each element of $I$ vanishes.
For a subset $S$ of $k^n$ we define $\mc I(S)$ as the set of all the functions $f\in k[x_1,\ldots, x_n]$ which vanish on each point of $S$.
The nullstellensatz states that $\mc I(\mc V(I)) = \sqrt{I}$ for any ideal $I$ in $k[x_1,\ldots, x_n]$.

Now suppose we were working over the ring $R =k[x_1^{\pm},\ldots, x_n^{\pm}]$, that is, the localization of $k[x_1,\ldots, x_n]$ with respect to the multiplicatively closed set $\set{x_1^{m_1} \cdots x_n^{m_n}:\ m_i\geq 0}$.
Here we define, for an ideal $I$ in $R$, $\mc V(I)$ as
$$
\mc V(I) = \set{p\in (k^\times)^n:\ f(p) = 0 \text{ for all } f\in I}
$$
For any subset $S$ of $(k^\times)^n$ we define $\mc I(S)$ as the set of all the elements of $R$ which vanish at every point of $S$.
Then again we have
$$
\mc I(\mc V(I)) = \sqrt{I}
$$
and this can be proved using the nullstellensatz stated before.

So we recover the nullstellensatz by replacing $k$ with $k^\times$.
I believe the `reason' for this is that $k[x_1^{\pm},\ldots, x_n^{\pm}]$ is the coordinate ring of $(k^\times)^n$.

Question.
Is there a more general version of the nullstellensatz which captures both the statements simultaneously (and also is there is a unified proof?)

答:
Here are many such statements. We will assume $R$ is Noetherian. Let $X=\operatorname{Spec} R$ the set of all prime ideals of $R$ and let $X_m\subset X$ be the set of maximal ideals.

Let $I\subset R$ be any ideal and let $V(I)\subset
 X$ be the the set of all prime ideals containing $I$. Let $J=\cap_{P\in V(I)} P$. Then $\sqrt{I}=J$.

Assume $R$ is Jacobson. This means for any prime ideal $P$, $P=\cap_{P\subset M\in X_m} M$. Then, $J=\cap_{M\in X_m,\, I\subset M} M$.

If $R$ is a finitely generated algebra over a field $k$, then $R$ is Jacobson.

Finally, if $k$ is algebraically closed field, $R=k[x_1,\ldots, x_n]$ then any maximal ideal is of the form $(x_1-a_1,\ldots, x_n-a_n)$ for $a_i\in k$.

These  give both your statements (and many more).

Title: What's a more general nullstellensatz?


问:
Young's generalizated inequality
If $a_{1},a_{2},\dots,a_{n}$ are non-negative numbers, $p_{1},p_{2},\dots,p_{n}, \ p_{i}>1,$ for all $i=1,\dots,n$ and  $\sum_{i=1}^{n}\dfrac{1}{p_{i}}=1,$ then $$\prod_{i=1}^{n} a_{i} \leq \sum_{i=1}^{n} \frac{a_{i}^{p_{i}}}{p_{i}}.$$

Could anyone help me? For case $n=2$ is classic and I know. But I couldn't generalize.

答:
$\log$ is concave, therefore using Jensen's inequality we have
$$ \log\left(\prod_{i=1}^n a_i\right)=\sum_{i=1}^n\log(a_i)=\sum_{i=1}^n\frac{\log(a_i^{p_i})}{p_i}\leqslant \log\left(\sum_{i=1}^{n}\frac{a_i^{p_i}}{p_i}\right)$$
Thus $\displaystyle\prod_{i=1}^n a_i\leqslant\sum_{i=1}^n\frac{a_i^{p_i}}{p_i}$

Title: Young's generalizated inequality


答: Calculus by Michael Spivak is one of the most celebrated books in calculus & mathematical analysis.it is the best choice for self study for a beginner. It starts from basic definition and slowly slowly devlopes given all prerequisite it keeps you engaged.Would definitely recommend it. Go for it</p>

Title: What books should I read before books of Advanced Calculus?

Body:
I need books to get the right understanding of calculus to read books like Spivak, Apostol and Courant Calculus, I'm a complete beginner at it, since I've had heard those are good books, and I've look at them and they are really interesting books which I'd I'd love to study, but as I've said I don't know much (almost any) about calculus.

I want to know if there are some good references to study before that can help me to don't get stuck (easily) when I am studying or should I go directly to those books, if that is ghe case which one should I read first: Courant, Apostol or Spivak


答: Well, to fully understand the concepts of advanced calculus, you should first understand the formal language, so logic, set theory, induction. Then you could start with some basic book on calculus, I have Spivak calculus 4th edition, which I have used a few years ago in University, it's a really nice book, it has a lot of exercises, explain the theory, it has a lot of images, and provides intuition on was going on, so I recommend you this one. Then you can go one step further with numerical series, series of functions, integration (Riemann, Darboux, Stieltjes) and you will be prepared for advanced calculus, Apostol make integrals before derivatives, which is not the usual way to learn, but it's fine. I used Buck's book advanced calculus to learn this course.</p>


问:
Variance from a joint probability distribution

The joint distribution for the claims amount X and Y is:
$$\begin{array}{c|c|c|}
 & X=10 & X=50 & X=100\\ \hline
Y=25 & 0.08 & 0.14 & 0.13\\ \hline
Y=75 & 0.15 & 0.3 & 0.2 \\ \hline
\end{array}$$
Find the variance of $X – 3Y$.

My attempt:

Let $Z=X-3Y$. The probability distribution of $Z$ is given by

$$\begin{array}{c|c|c|}
Z & P(Z=z) & Z^2 \\ \hline
-65 & 0.08 & 4225\\ \hline
-215 & 0.15 & 46225\\ \hline
-25 & 0.14 & 625\\ \hline
-175 & 0.30 & 30625\\ \hline
25 & 0.13 & 625\\ \hline
-125 & 0.20 & 15625\\ \hline
\end{array}$$

Now I computed $Var(Z)$ using $E(Z^2)-[E(Z)]^2$.
$$Var(Z) = 19753-(121.7)^2$$

This did not lead to the correct answer. Is there something conceptually wrong with my approach (I've double-checked my computations many times so they should be correct)? Note: I do not want to use the formula $Var(Z) = Var(X)+9Var(Y)-6Cov(X, Y)$ for this question unless I have to.

Body:
Your table is correct, but your calculation of the expectation is not.  To complete your table, we have:

$$\begin{array}{|c|c|c|c|c|}
\hline
Z & P(Z=z) & Z^2 & Z\Pr[Z = z] & Z^2 \Pr[Z = z] \\ \hline
-65 & 0.08 & 4225 & -5.2 & 338. \\ \hline
 -215 & 0.15 & 46225 & -32.25 & 6933.75 \\ \hline
 -25 & 0.14 & 625 & -3.5 & 87.5 \\ \hline
 -175 & 0.3 & 30625 & -52.5 & 9187.5 \\ \hline
 25 & 0.13 & 625 & 3.25 & 81.25 \\ \hline
 -125 & 0.2 & 15625 & -25. & 3125. \\ \hline
& & & \color{red}{-115.2} & 19753 \\
\hline
\end{array}$$


问:
Can a limit be thought of as an operation that makes a discontinuous function continuous, and evaluates it?
Here are some examples. If you take the limit of $x^2/x$ as it approaches $0$, you get $0$, because the limit just plugs in the hole of the graph to make $x$ and evaluates it for $x=0$. Another example would be $|x|/x$. When you make the function continuous, you get a vertical bar of infinite values at $x=0$, so since there is no one value, this limit is undefined.

答:
A general statement is as follows.

If you have a function $f:\mathbb{R}\setminus\{a\}\to\mathbb{R}$ such that
$$
\lim_{x\to a}f(x)=L,
$$
then the function $g:\mathbb{R}\to\mathbb{R}$ with
$$
g(x)=\begin{cases}
f(x),&x\ne a\\
L,&x=a
\end{cases}
$$
Then $g$ is continuous at $x=a$.


问:
Derivative of the Fundamental Solution to the Laplace Equation
The fundamental solution of Laplace equation is given as $\Phi(x) = \frac{1}{n(n-2)\alpha(n)}\frac{1}{|x|^{n-2}}$ for $n \geq 3$. and $\alpha(n)$ is a constant

We want to check $$u(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)dy$$ solves the poisson’s equation.

A naive attempt will be to move the Laplacian to inside the Integral:  $$\Delta u(x)=\int_{\mathbb{R}^n}\Delta \Phi(x-y)f(y)dy$$

But one cannot do it because $\Phi(x-y)$ is not summable near the singularity at $y=x$. My task is to show why.

A hint from my tutor is to show that $|D^2\Phi| \leq \frac{C}{|x|^n} \notin L^1(B_{\epsilon}(0))$

I don’t understand the $\notin$ part which means given an n-dimensional ball with radius $\epsilon$ centered at $0$ we should show:

$$\int_{B_{\epsilon}(0)}\frac{1}{|x|^n} = \infty$$ where $x \in \mathbb{R}^n$

答:
The function $x\mapsto 1/|x|^n$ is the prime (the only) example of an $n$-th order pole: integrating it over any domain "containing" the origin will lead to an explosion. This is because that function is not locally integrable (and thus it does not define a distribution). Indeed, if we specialize to (punctured) balls centered at the origin,
$$\int_{B_\varepsilon(0)} \frac{dx}{|x|^n} = \int_0^\varepsilon \frac{1}{r^n} r^{n-1} dr \int_{\mathbb S^{n-1}} d\Omega_{n-1} = \operatorname{vol}(\mathbb S^{n-1}) \cdot \ln r\Bigg|_{0}^\varepsilon = \infty,$$
where $\mathbb S^k$ is the unit $k$-sphere, $d\Omega_k$ is the $k$-dimensional solid angle element, and $r^{n-1}$ is the radial part of the hyperspherical jacobian in $n$ dimensions.


问:
Fundamental Theorem of Calculus.
Fundamental Theorem of Calculus says

Let $f$ be an integrable function on $[a,b].$ For $x$ in $[a,b],$  let $F(x)=\int_{a}^{x}f(t)dt.$ Then $F$ is continuous on $[a,b].$ If $f$ is continuous at $x_{0}$ in $(a,b),$ then $F$ is differentiable at $x_{0}$ and $F^{'}(x_{0})=f(x_{0}).$

Now my question is why we are saying that this is "Fundamental Theorem of whole Calculus".

It is clear that it connects intergal and differential calculus. But i am not getting the exact concept, why this is so important as its name is fundamental theorem of calculus. Why we says that it is fundamental theorem of calculus?  Please suggest me. Thanks in advance.

答:
As folks have already noted, the FTC states that derivation and taking accumulation functions are inverse processes, much the same way that multiplying and diving by 5 are inverse processes.
From another point of view, if you think about derivation and antidifferentiation, it's not clear at all: 1) which functions have antiderivatives, 2) how on earth to construct those antiderivatives. The (first) FTC answers both questions.</p>

Title: Fundamental Theorem of Calculus.


问:
Surfaces: behavior of Pluricanonical maps
Let $S$ be an algebraic smooth surface (over $\Bbb{C}$).

Suppose we have some fibration $p\colon S\rightarrow C$ onto a smooth curve. Let $f$ be a fiber, a curve, say.

Let $K$ be a canonical divisor on $S$ and assume $K\cdot f=0$. My question is: can we conclude that each $m$-canonical map $\varphi_{mK}\colon S\rightarrow\Bbb{P}^N$, when defined, will contract $f$ to a point ?

答:
One slick way to explain why the answer is yes uses the projection formula. For our purposes the following formulation is enough: if $f: X \rightarrow Y$ is a morphism between smooth projective varieties, $L$ is a line bundle on $Y$, and $C$ is a curve on $X$, then $$f^*L \cdot C = L \cdot f_*C$$ (where $f_*$ denotes the push-forward of cycles, and $\cdot$ is intersection of cycles).

By definition of the morphism $\varphi_L: X \rightarrow \mathbf{P}^n$ associated to a basepoint-free line bundle on a variety $X$, we have $\varphi_L^* \mathcal{O}(1) =L$. So if $C$ is any curve with $L \cdot C =0$, the projection formula gives $(\varphi_L)_*C \cdot \mathcal{O}(1)=0$. But $\mathcal{O}(1)$ has strictly positive degree on any curve in $\mathbf{P}^n$, so the 1-cycle $(\varphi_L)_*(C)$ must be zero: that is, $C$ must be contracted by $\varphi$.

Now just apply this in your case with $L=mK$. (Note that for each integer $m$, the conditions $K \cdot C=0$ and $mK \cdot C=0$ are equivalent.)


问:
Proving a solution to ODE exists without solving it
I am analysing the ODE

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y(x)+\frac{c}{y(x)}\right)=1,\quad y(1)=1.$$

I would like to show, without attempting to solve the ODE, that $\exists$ a solution $y(x)>c$ if $\ln c-c<-2$. I am wondering what techniques I should employ to do so? I attempted to stuff it into a form involving Grönwall's inequality, but have not had much progress with that. What should I do? Thanks!

答:
Your ODE is of first order:

$\frac{dy}{dx}=\frac{y^2}{y-c}, y(1)=1$

Can you take from here using Picard's existence theorem?


问:
If a winning strategy does not exist for player 2, does it exist for player 1?
I have been studying some instances of strategy-stealing. The issue is that I am not sure whether proving the non-existence of a winning strategy for player 2 necessarily proves a winning (or at least tying) strategy for player 1. Wikipedia says

In combinatorial game theory, the strategy-stealing argument is a general argument that shows, for many two-player games, that the second player cannot have a guaranteed winning strategy.

The arguments that I have seen make sense, in terms of proving that player 2 cannot have a winning strategy. For example, the Wikipedia article on Chomp makes the modest claim

The second player therefore cannot have a winning strategy.

However, in Engel's Problem-Solving Strategies, Double Chess is analyzed in problem #18 of the games chapter and the following seemingly grander claim is made:

The rules of chess are changed as follows: Black and White make alternately two legal moves. Show that there exists a strategy for white which guarantees him at least a tie.

The proof is

Suppose $B$ can win no matter what $A$ does. On his first move, $A$ moves one of his knights to any one of the two possible squares and then back to its original position. Now all the pieces are in their original position, but $A$ has becmoe the second player and must win. Contradiction!

When does the negation of the existence of a winning strategy for player 2 imply the existence of a winning/tying strategy for player 1?

I might be confused because I am actually not sure of how to formally describe a winning strategy, so a rigorous explanation of that might be helpful as a preface. (I considered using alternating existential and universal quantifiers to describe winning strategies, but didn't know how to terminate the sequence.)

P.S. I have included set theory as a tag because, according to the Wikipedia article on determinacy, the subject is a subfield of set theory. However, I am not familiar with this part of set theory and I did not really understand the article.

答:
The Simplest Case

Simple Conditions

For now, let's consider two-player games of perfect information where there is a pre-known bound on the number of moves a game might take, and one of the two players wins in the end, and the first player is fixed (e.g. the player who controls the white pieces must move first).

Strategies

As mentioned in the comments, there is a theorem/collection of related theorems called Zermelo's theorem* (in game theory), that is relevant here. Summarized, it says that in a game of this type, either the first player to move or the second player to move must have a "winning strategy" — a function that provides a move for every situation that player could possibly be in (a strategy) which guarantees a win no matter how the opponent plays.

(*As an aside, most discussions of "Zermelo's theorem" aren't quite in line with what Zermelo wrote about and how his proof went. See Zermelo and the Early History of Game Theory by Ulrich Schwalbe and Paul Walker for some clarification about the history.)

The Argument

A full formal setup and proof would make this post much longer, and might not add much more conceptual understanding. Briefly, the idea is to use induction on, say, the maximum length of a game. A game with only one turn either has a winning move for the first player, in which case the winning strategy for them is "play one of the winning moves". Or there isn't, in which case the winning strategy for the second player is an empty function since they have no choices to make. For the induction step: if there are any moves to a "non-next player has a winning strategy" situation, the first player has a winning strategy to make one of those moves and then follow the existing strategy. If there are no such moves, then all moves are to a "next player has a winning strategy" position, and the second player can follow that strategy no matter which move the first player makes. Another, less brief, account can be found in Lecture 15 from the Open Yale Course ECON 159: Game Theory, which has accompanying lecture notes and transcript, etc.

Variants

There are many variants on the conditions that go beyond the simplest case, and I won't cover them all here, but I wanted to address the comments about Chess and give a picture of how this stuff can be extended.

Ties

Some games, like "Tic Tac Toe"/"Noughts and Crosses" don't satisfy the condition where someone must win in the end. The game can end in a "tie" (also known colloquially as a "draw", but we make a distinction in combinatorial game theory).

From a game $G$ like this, we can build a new game $G_1$ by breaking ties in favor for first player, and another game $G_2$ by breaking ties in favor of the second player. For $G_1$, Zermelo's Theorem tells us that either the first player or the second player has a winning strategy in $G_1$. This means that the first player can force at least a tie in $G$ or the second player can force a win in $G$. Analogously, Zermelo's Theorem applied to $G_2$ tells us that either the second player can force at least a tie in $G$ or the first player can force a win.

Now it's just a matter of listing the possibilities:

If the first player can force a win, then the second player can't even force "at least a tie".

If the second player can force a win, then the first player can't force "at least a tie".

If neither player can force a win, then by the observations about $G_1$ and $G_2$ above, it must be that case that both player can force "at least a tie". And since neither can force a win, it must be the case they can simply force a tie.

Repeated positions (like Chess)

Some games like chess have things like repeated board positions, which would seem to throw a wrench in the induction proof of Zermelo's Theorem. However, this is sort of an illusion because of things like the Chess's repetition rules. Chess doesn't allow real repetitions if you include a list of all previous positions with their frequency counts as part of the "position". With this clarified, Chess and similar games are basically just games with ties (though you could complicate the earlier story if you want to distinguish stalemates from repetition draws, etc.)

Repeated positions (for real)

Some other games (much less likely to be actually played) really can repeat positions, and hence can go on forever. For example, maybe Nim-but-you-can-return-objects-too, or "Fair Shares and Varied Pairs" (a description can be found in this mathstrek blog post or Winning Ways). For these so-called "loopy" games, we must consider "drawn" games which last forever, in addition to the ones that end.

It takes some extra work (see IV.4 of Combinatorial Game Theory, for instance), but it turns out there's a tidy description of things with just one more category in addition to first or second player having a winning strategy: those positions where both players can force the game to be drawn. So it's not obvious, but these drawn games basically work out the same as ties.

Partizan Games

If you study more combinatorial game theory, you'll encounter partizan games where players have an extra distinction aside from who moves first. For example, maybe you could consider a position on a chessboard and the "moves for Black" even if it would be White's turn in the original position. For the purposes of things like Zermelo's Theorem (ignoring a lot of good theory), this doesn't really matter. Given a partizan game, you could basically consider $G_1$ where Black moves first (and the color whose turn it is is part of the data of the position) and $G_2$ where White moves first and then apply a previous theorem to each of those.


答:
Your equation is equivalent to $\log y+\frac{c}{y}=x+\lambda$ for some $\lambda\in\mathbb{R}$. Plugging $x=1$ in the above equality gives $\lambda=c-1$. Thus your differential system is equivalent to $-\frac{c}{y}e^{-\frac{c}{y}}=-ce^{1-c-x}$. Let $\varphi(x)=xe^x$, then the above equation can be written as $\varphi\left(-\frac{c}{y(x)}\right)=-ce^{1-c-x}$.  But a quick study of $\varphi'$ shows that $\varphi$ is a bijection from $(-1,+\infty)$ to $\left(-\frac{1}{e},+\infty\right)$, therefore there exists a solution to your initial equation such that $y>c$ iff $-ce^{1-c-x}<-\frac{1}{e}$ for all $x\geqslant 0$ (is suppose that $y$ is defined on $[0,+\infty)$), which is equivalent to $\log c-c<x-2$ for all $x\geqslant 0$, and this is equivalent to $\log c-c<-2$. Therefore if $\log c-c<-2$, let $y(x)=-\frac{c}{\varphi^{-1}(-ce^{1-c-x})}$, then $y$ is well defined because of what said above and since $\varphi^{-1}>-1$ and $\varphi^{-1}(-ce^{-c})=\varphi^{-1}\circ\varphi(-c)=-c$, we have $y>c$ and $y(1)=1$ and $y$ is solution of the differentiel equation.

I don't know if this solution suits you since I found the solution of the differential equation, but we can still prove the existence without writing what $y$ is by ignoring what I wrote from "Therefore if...".

Title: Proving a solution to ODE exists without solving it

Body:
I am analysing the ODE

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y(x)+\frac{c}{y(x)}\right)=1,\quad y(1)=1.$$

I would like to show, without attempting to solve the ODE, that $\exists$ a solution $y(x)>c$ if $\ln c-c<-2$. I am wondering what techniques I should employ to do so? I attempted to stuff it into a form involving Grönwall's inequality, but have not had much progress with that. What should I do? Thanks!


问:
For $f : \mathbb{R} \stackrel{\text{cont.}}{\to} \mathbb{R}$, show that $\int_{0}^{1} f(x)x^2\,dx = \frac{1}{3}f(c)$ for some $c \in [0,1]$.

For $f : \mathbb{R} \stackrel{\text{cont.}}{\to} \mathbb{R}$, show that $$\int_{0}^{1} f(x)x^2 \,dx= \frac{1}{3}f(c)$$ for some $c \in [0,1]$.

I have seen similar problems that Rolle's theorem or LMVT can solve neatly. This one seems to have a scalar of $\frac{1}{3}$ attached with it, which screams of $x^2$ being integrated somewhere.

答:
Hint: Use the following fact: if $f$ is continuous on $[a,b]$ and $g$ is integrable and nonnegative on $[a,b]$, then there exists $c\in[a,b]$ such that:
$$\int_a^bf(x)g(x)\,dx=f(c)\int_a^bg(x)\,dx.$$
(I can give the proof, but it's not too difficult if you want to try by yourself first.)


问:
Solve non-homogeneous recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$
I'm trying to solve this recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$, but the solution of the homogeneous is $a_n=b+cn$, and I don't know how to change it to find the non-homogeneous solution.

答:
Another way is applying unilateral $\mathcal{Z}$-transform which is defined as $$X(z) = \sum_{n=0}^{+\infty}x[n]z^{-n}$$The equation is $$x[n+2] -2x[n+1] +x[n] = 8u[n]$$We can easily show that the following properties holds:$$\mathcal{Z}\{\alpha^nu[n]\} = \frac{1}{1-\alpha z^{-1}} \\ \mathcal{Z}\{x[n+1]\} = zX(z) - zx[0]\implies\mathcal{Z}\{x[n+2]\} = z^2X(z) - z^2x[0] - zx[1] \\ \mathcal{Z}\{nx[n]\} = -z\frac{dX(z)}{dz}\\ \mathcal{Z}\{x[n-1]\} = z^{-1}X(z) + x[-1] \\ \mathcal{Z}\{\delta[n]\}  = 1$$Applying the transform to the equation yields $$z^2X(z) - z^2x[0] - zx[1] - 2(zX(z) - zx[0]) +X(z) = \frac{8}{1-z^{-1}}$$For convenience, let $x[0] = a$ and $x[1] = b$. Partial fraction decomposition gives $$X(z) = a + \frac{8}{(z-1)^3} + \frac{8 - a + b}{(z-1)^2} + \frac{b}{z-1}$$Taking the inverse transform gives $$x[n] = a\delta[n] + bu[n-1] + (8 - a + b)(n-1)u[n-1] + 8(\frac{1}{2}(n-2)(n-1)u[n-1])$$Which can be written as $$x[n] = a - 4n - an + bn + 4n^2 = 4n^2 + n(b-a-4) + a$$
It can be verified by WA. Note that the counterpart of the Laplace transform for discrete-time signals is the $\mathcal{Z}$-transform.


问:
Probability that two randomly chosen people out of 10 have birthday in the same month

There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?

My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility

答: The first person $A$(say) is free to have any month as his/her birthday month so $P(A)=\frac{12}{12}$ . But the second person $B$(say) is compelled to possess the same birthday month as that of $A$'s so $P(B)=\frac{1}{12}.$ So required probability is $P(A).P(B)=\frac{1}{12}$


问:
Entropy of a Fair Coin Toss
The problem: A fair coin is tossed until a a heads is reached for the first time. What is the entropy $H\left(x\right)$ in bits?

My solution:

Because $P\{X=i\}=\left(\frac{1}{2}\right)^i$,

$H\left(x\right)=  -\sum_{i}^{ \infty}p(x)\log_{2}\left(p(x)\right)$

$\qquad = -\sum_{i}^{ \infty}\left(\frac{1}{2}\right)^i\log_{2}\left(\left(\frac{1}{2}\right)^i\right)$

$\qquad=-\sum_{i}^{\infty}i\left(\frac{1}{2}\right)^i\log_{2}\left(\frac{1}{2}\right)$

$\qquad = \frac{\frac{1}{2}}{\left(1-\frac{1}{2}\right)^{2}}$

$\qquad = 2$.

So 2 bits.

However, the text provides this solution:

Setting $p=\frac{1}{2}$ where $P\left(X=n\right)=pq^{n-1},n \in\{1,2,...\}$ the entropy is

$H\left(X\right)=-\sum_{n=1}^{\infty}pq^{n-1}\log_{2}\left(pq^{n-1}\right)$

$\qquad=-\left[\sum_{n=0}^{\infty}pq^n\log_2\left(p\right)+\sum_{n=0}^{\infty}npq^{n}\log_{2}\left(q\right) \right]$

$\qquad =\frac{-p\log_{2}\left(p\right)}{1-q}-\frac{pq log_{2}\left(q\right)}{p^{2}}$

$\qquad=\frac{-p\log_{2}\left(p\right)-q\log_{2}\left(q\right)}{p}$

$\qquad = \frac{H\left(p\right)}{p}$
and setting $p=\frac{1}{2}$, we have 2 bits.

My questions:

Is my solution valid? Also, if someone could explain in detail how the text provided solution went from this step

$-\sum_{n=1}^{\infty}pq^{n-1}\log_{2}\left(pq^{n-1}\right)$ to this step $-\left[\sum_{n=0}^{\infty}pq^n\log_2\left(p\right)+\sum_{n=0}^{\infty}npq^{n}\log_{2}\left(q\right) \right]$,

as I am a little lost at how to bridge these two parts of the solution. Thanks in advance.

答:
You have the right idea, they just worked in a more abstract setting. Here $p$ is your probability of hitting heads on the next flip and $q$ is the probability of flipping tails.  Since there are only two outcomes on a given flip, then $q=1-p$. In this particular case $p=\frac{1}{2}$ so $q=1-p=\frac{1}{2}$. So the probability they give is
$$P(X=n)=pq^{n-1}=\Big(\frac{1}{2}\Big)\Big(\frac{1}{2}\Big)^{n-1}=\Big(\frac{1}{2}\Big)^{n} $$
Which is exactly what you found for this particular case.

The other step you mentioned is using the properties of $\log$ and then reindexes the series.

$\log(ab)=\log(a)+\log(b)$

$\log(a^n)=n\log(a)$

$\sum_{n=1}^\infty a_{n-1}=\sum_{n=0}^\infty a_n$

Title: Entropy of a Fair Coin Toss


问:
a series of functions that converges uniformly, converges to a harmonic function and so its derivative
let $u_{n}:G\rightarrow\mathbb{R}$ be a series of harmonic on a domain G, and $u_{n}\rightarrow u$ uniformly.
Then

u must be also harmonic. it was written here, but not formalize:Does a uniformly sequence of harmonic function converge to a harmonic function?

$\frac{\partial u_{n}}{\partial y}\rightarrow\frac{\partial u}{\partial y}\land\frac{\partial v_{n}}{\partial x}\rightarrow\frac{\partial v}{\partial x}$

答:
Let $z \in G$ then since $u_n$ are harmonic they satisfy the mean value property.

That is, $u_n(z) = \frac{1}{2 \pi}\int_0^{2 \pi} u_n(z+re^{i \theta})d\theta$ for some $r > 0$.  Now use uniform convergence to show that $|u_n (z) - \frac{1}{2 \pi}\int_0^{2 \pi} u(z+re^{i \theta})d\theta |\to 0$ and conclude that $u$ is harmonic because it satisfies the mean value property and because the limit is unique.


问:
What book is better for learning calculus Courant, Apostol or Spivak?
I want fo learn (Real) calculus and I've seem those books are very recommended, now I know why, I've given them a look and I think they are worth their reputation.

I'm talking about

Calculus by Michael Spivak

Calculus by Tom Apostol

Introduction to Calculus and Analysis by Fritz John and Richard Courant

I want to read them all, but I have to chose:
what should I pick and why?
are there any sustactial difference?
if I have to read them all, in what order should I read them?

答:
I'm about halfway through Spivak and it's great. I haven't gone through Courant or Apostol. Spivak is known for uniquely interesting and difficult problems, and a tone that's widely considered more playful than is typical for a math textbook (though subtly so; it's not Mad Magazine).

Spivak also includes answers to about half of the exercises, and an Answer Book for the others is available for purchase.

I strongly encourage you to buy the Answer Book if you're going the Spivak route, as the problems are difficult.

You should expect to get stuck, often. You will likely be unable to do many of the problems without at least some hints from the book. There have been occasions in which it took me a few days to even understand the published solution to a problem I looked up.

Don't be discouraged when you get stuck. Take some time. Do something else. Maybe try thinking about it away from the book and your paper. Go for a walk. This is how learning math works. Not everything is immediately comprehensible. Some things take a little while to crystalize in you mind. Once they do its wildly gratifying.

Another Spivak tip: do every problem. Spivak often puts important results in the problems, and they typically build off each other in a very purposeful way.

Some disadvantages of Spivak are

It's single variable only. Apostol and Courant each cover single variable calculus, multivariable calculus, and some linear algebra and other material.

Spivak doesn't have many easy problems. You're not going to get practice doing a thousand rote calculations the way you might with a more standard (and mediocre) text like Stewart.

Spivak doesn't focus much on applications. IIRC Courant and Apostol have more examples from Physics and other fields, and those things are cool and can help with your intuition and ability to picture what's going on.

It should be noted, Spivak is also a single volume. Those others are each two volume collections IIRC. I imagine (hope!) it's feasible to use Spivak for single variable, and then hop over to part II of either Apostol or Courant to tackle multivariable stuff.

Whatever you decide, don't be put off by "getting stuck"!


问:
Solving modular equation
$$13863x \equiv 12282 \pmod {32394}$$

I need to solve this equation. If I'd found the inverse of 13863 and multiply the equation by this, I'd get the solution. So:

$$13863c \equiv 1 \pmod {32394}$$

And now - how can I find this inverse? The numbers are too big to just look for the multiplier of $13863$ . . .

答:
The $\gcd$ if $32394$ and $13863$ is $3$, so there is no inverse. What you need to do is divide out any common factors in the original equation: everything here is divisible by $3$.
$$13863x\equiv 12282\bmod 32394$$
In general, if you want to solve $$dx\equiv da\bmod dm,$$
then for any solution $y$ of the congruence $x\equiv a\bmod m$, the numbers $$y,\quad y+m,\quad\ldots,\quad y+(d-1)m$$
will all be solutions to $dx\equiv da\bmod dm$, because
$$d(y+km)=dy+k(dm)\equiv dy\equiv da\bmod dm.$$
When faced with a congruence where the $\gcd$ of all the terms, together with the modulus, is not $1$, divide through by their $\gcd$, solve that congruence, and then "move back up" to obtain the solutions to your original congruence.

So, let's solve
$$4621x\equiv 4094\bmod 10798$$
(which is what we get after dividing everything by their $\gcd$ of $3$). As you've noted in your question, we want to find the inverse for $4621$ modulo $10798$.

Run the Euclidean algorithm for computing $\gcd(4621,10798)$:
$$\begin{align*}
10798 &= 2\cdot 4621 + 1556\\
4621&= 2\cdot 1556 + 1509\\
4668&=1\cdot 1509 + 47\\
1509 & = 32\cdot 47+ 5\\
47&= 9\cdot 5+2\\
5&=2\cdot 2+\fbox{1}\\
2&=2\cdot 1+0
\end{align*}$$
then "run it backwards":
$$\begin{align*}
\fbox{1}&=\mathbf{5-2\cdot2}\\
&\quad =5-2\cdot (47-9\cdot 5)\\
&\quad =5-2\cdot 47+ 18\cdot 5\\\\
&=-2\cdot 47+ 19\cdot 5\\
&\quad=-2\cdot 47+19\cdot(1509- 32\cdot 47)\\
&\quad= -2\cdot 47+19\cdot 1509-608\cdot 47)\\\\
&=19\cdot 1509-610\cdot 47\\
&\\
&\vdots\\
&\\
&=u\cdot 4621+v\cdot 10798
\end{align*}$$
And now we have our inverse: $u$, because
$$u\cdot 4621=1-v\cdot 10798$$
implies that
$$u\cdot 4621\equiv 1\bmod 10798.$$

Title: Solving modular equation


问:
Why is this integral diverging? $\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$
$$\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$$

I can easily prove that this integral is diverging by taking the limit over the following proper integral,

$$\lim_{R_1,R_2\to \infty} \int\limits^{R_2}_{-R_1} \,\frac{x}{x^2+1} dx$$

Mathematically, this makes sense to me, but intuitively I am not able to absorb this.

If we observe the function $y=\frac{x}{x^2+1}$, it is clearly an odd function. And since, integrals return signed areas, an integral to an odd function having limits that are the negatives of each other, should evaluate to zero.

As clearly seen by the following plot.

Can someone please clarify this?

Thank you.

答:
From the graph of $\frac{x}{1+x^2}$;

For $x>0$, $I_1=\lim_{R_1\rightarrow\infty}\int_0^{R_1}\frac{x}{1+x^2}dx=\lim_{R_1\rightarrow\infty}\frac{1}{2}In(1+R_1^2)$

For $x<0$, $I_2=\lim_{R_2\rightarrow\infty}\int_0^{R_2}\frac{x}{1+x^2}dx=\lim_{R_2\rightarrow\infty}-\frac{1}{2}In(1+R_2^2)$

Signed area$=I_1+I_2=\lim_{R_1\rightarrow\infty}\frac{1}{2}In(1+R_1^2)-\lim_{R_2\rightarrow\infty}\frac{1}{2}In(1+R_2^2)$

The problem is that $RHS=\infty-\infty\ne 0$


问:
How to describe the specific circle that is formed by an intersection a plane and sphere.
Given sphere with equation $x^2+y^2+z^2=100$, and plane $x=5$, describe the intersection of these two surfaces.

I understand that I could plugin $5$ for $x$. However, this yields a cylinder that does not describe the specific circle. How do I write the equation of this circle?

答:
The graph in $\mathbb{R}^3$ of a single equation in the variables $x,y,z$ is typically a surface, not a curve.

The circle in question can be represented algebraically using the system of two equations
$$
y^2+z^2=75,\;\;\;x=5
$$
or alternatively, in parametric form
\begin{align*}
x&=5\\[4pt]
y&=5\sqrt{3}\cos(t)\\[4pt]
z&=5\sqrt{3}\sin(t)\\[4pt]
\end{align*}
for $0\le t < 2\pi$.

It's possible to represent the circle in question as the graph in $\mathbb{R}^3$ of the single equation
$$
(x-5)^2+(y^2+z^2-75)^2=0
$$
but such a representation is not very useful.


问:
For a surjective function $f:G\rightarrow H$, if $f(a)=f(b)$, does it follow that $a=b$?
I am wondering the question in the title. What I understand about surjectivity is that if a $f$ is surjective (onto), then for all $h\in H$, there exists some $g\in G$ such that $f(g)=h$. I also know that $H$ is "bigger" than $G$ provided that $f$ is surjective.

Can someone please clarify this?

Thanks a lot.

答: No. Consider a function $f:A \rightarrow B$ with $B = \{b\}$ a singleton and $A$ containing more than one element. This function is constant so $f(a_1) = f(a_2)=b$ for any $a_1, a_2 \in A$ even if $a_1 \neq a_2$ and is clearly surjective.


问:
Norm of a smooth function is smooth?
My lecture notes seem to use this, although I can't shake the feeling that it's wrong. Let $\phi: [a,b] \to \mathbb{R}^n$ for $n \ge 2$ be smooth (infinitely differentiable) satisfying $|\phi '(t)| \neq 0$ for all $t \in [a,b]$. Then, $|\phi'(t)|$ is also a smooth function.

Does anyone have a nice proof for this? (My first 'counterexample' was $\log$, but it doesn't satisfy the condition $|\phi '(t)| \neq 0$.) Also, is $|\phi(t)|$ a smooth function too?

Although this is from a course in elementary differential geometry, my question is essentially a calculus question...

答:
Let $g(v) = \langle v, v\rangle = \lVert v\rVert^2$ and $f(t) = \sqrt{t}$.
Then $\lVert \phi'(t)\rVert = \left(f\circ g\circ\phi'\right)(t)$.

You know that $\phi'$ is smooth and nonvanishing.

It's easy to see that $\nabla_g(v) = 2v$; in other words, $\nabla_g$ is twice the identity, so $g$ is smooth.
It follows that the composition $g\circ \phi'$ is smooth.

Moreover, we have that $g(v) = 0 \iff v = 0$, from which we see that the composition $g\circ\phi'$ is also nonvanishing.

Finally, it's a simple univariate calculus exercise to show that $f:[0,+\infty)\longrightarrow \Bbb R$ is smooth away from the origin.
Hence, the composition $f\circ (g\circ \phi')$ is smooth.

Title: Norm of a smooth function is smooth?


问:
Can you factor a second order polynomial with negative leading coefficient with only two factors?
Suppose we have a polynomial $ax^2 + bx + c$ with $a\in \left(-\infty, 0\right)$ and $b, c\in \mathbb{R}\setminus \left\lbrace 0 \right\rbrace$. Is it possible to find $p, q, r, s \in \mathbb{R}$ such that $\left(px + q\right)\left(rx + s\right) = ax^2 + bx + c$? And if not, why not? I was trying to factor $-y^2 + y - h$, with $h>0$ and I couldn't find any way to factor this into only two factors. I welcome answers involving abstract algebra if that is relevant to this question.

答:
Regardless of the sign of $a \ne 0$, the roots are $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Hence we have

\begin{align}ax^2+bx+c &= a\left(x- \frac{-b + \sqrt{b^2-4ac}}{2a}\right) \left(x- \frac{-b - \sqrt{b^2-4ac}}{2a}\right) \\&= \left(ax- \frac{-b + \sqrt{b^2-4ac}}{2}\right) \left(x- \frac{-b - \sqrt{b^2-4ac}}{2a}\right)\end{align}

Note that it is possible that $b^2-4ac$ to be negative though.


问:
Error in evaluating this limit?
No matter how I look at this limit, it looks like it should evaluate to $0$. Am I missing some rule about limits with differentials?

The limit is from page 19 of the third edition of Viscous Fluid Flow by Frank White.

$$d\alpha = \lim_{dt->0}\left(\arctan\frac{\frac{\partial v}{\partial x}dxdt}{dx+\frac{\partial u}{\partial x}dxdt}\right) = \frac{\partial v}{\partial x}dt$$

My best guess is that the author is assuming $dx >> \frac{\partial u}{\partial x}$ and then using the small-angle approximation, but that is not mentioned in the text.

Any help is appreciated.

答:
There's not so much a typo as there is a seeming abuse of notation.  It seems that White is using $$\lim_{dt->0}\left(\arctan\frac{\frac{\partial v}{\partial x}dx\,dt}{dx+\frac{\partial u}{\partial x}dx\,dt}\right) = \frac{\partial v}{\partial x}dt$$ to mean that $$\arctan\left(\frac{\frac{\partial v}{\partial x}dx\,dt}{dx+\frac{\partial u}{\partial x}dx\,dt}\right)\sim \frac{\partial v}{\partial x}dt\quad\text{ as } dt\to 0.$$  This is relatively straightforward to show.  First, rewrite
\begin{align}
\arctan\left(\frac{\frac{\partial v}{\partial x}dx\,dt}{dx+\frac{\partial u}{\partial x}dx\,dt}\right) = \arctan\left(\frac{\frac{\partial v}{\partial x}dt}{1+\frac{\partial u}{\partial x}dt}\right).
\end{align}

Then, using $$\tan(\theta) \sim \theta\quad\text{ as }\theta \to 0 \implies \arctan(\theta) \sim \theta\quad\text{ as }\theta \to 0,$$  as well as

$$\frac{ax}{1+bx}\sim ax\quad\text{ as }x\to 0,
$$
we have that

\begin{alignat}{2}
\arctan\left(\frac{\frac{\partial v}{\partial x}dt}{1+\frac{\partial u}{\partial x}dt}\right) &\sim \frac{\frac{\partial v}{\partial x}dt}{1+\frac{\partial u}{\partial x}dt} &&\quad\text{ as }dt\to 0\\
&\sim \frac{\partial v}{\partial x}dt&&\quad\text{ as } dt\to 0.
\end{alignat}

Title: Error in evaluating this limit?


问:
Formal proof for first-order logic question using natural deduction
I'm new to first-order logic and need a little bit of help with proving the following:

∀x∃yA(x,y) ⊢ ¬∃x∀y¬A(x,y)

It seemed straightforward but I have been stuck at it for hours. This is what I have come up with so far:

attempt.jpg

Any help is greatly appreciated!!

答:
$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}$

Here's the Fitch style layout for this proof.

$$\fitch{~~1.~\forall x~\exists y~A(x,y)}{\fitch{~~2.~\exists x~\forall y~\lnot A(x,y)}{\fitch{~~3.~[b]~\forall y~\lnot A(b,y)}{~~4.~\exists y~A(b,y)\hspace{8ex}{\forall}\mathsf E~1\\\fitch{~~5.~[c] A(b,c)}{~~6.~\lnot A(b,c)\hspace{8ex}{\forall}\mathsf E~4\\~~7.~\bot\hspace{14ex}{\lnot}\mathsf E~5,6}\\~~8.~\bot\hspace{16ex}\exists\mathsf E~4,5{-}7}\\~~9.~\bot\hspace{18ex}\exists\mathsf E~2,3{-}8}\\10.~\lnot\exists x~\forall y~\lnot A(x,y)\hspace{6ex}{\lnot}\mathsf I~2{-}9}$$

As you can see, there are three assumptions raised - one for the negation introduction, and two for the existential eliminations.  The two universal eliminations are made to the free variables raised in the assumption for the existential eliminations (called the witness variables).

Note: Some systems simply reuse the variable bound to the existential statement but, when allowed, you should use a local variable for clarity.

Now for your tree system: An existential elimination takes the following form: $$\dfrac{\lower{1.5ex}{\exists z~P(z)}~~{\small[a]}\dfrac{\begin{array}{c}[P(a)]^n\\[-1ex]\vdots\end{array}}{Q}{\tiny\textsf{somehow}}}{Q}{\small\exists\mathsf E^n}$$

Remember, the witness variable must not occur free within the derived consequent ($Q$). However, it may occur free in other statements within that derivation--such as for example, generated by a universal elimination.

Relevantly here we have the proof for $\forall w~\lnot P(w), \exists z~P(z)\vdash\bot$

$$\dfrac{\lower{1.5ex}{\exists z~P(z)}~~{\small[a]}\dfrac{\lower{1.5ex}{[P(a)]^n}\dfrac{\forall w~\lnot P(w)}{\lnot P(a)}{\small\forall\mathsf E}}{\bot}{\small\lnot\mathsf E}}{\bot}{\small\exists\mathsf E^n}$$

Well, the proof you seek is a little more complicated, but that is the basis.

$$\phantom{\dfrac{\dfrac{\lower{1ex}{[\exists x~\forall y~\lnot A(x,y)]^1}~~{\small[b]}\dfrac{\dfrac{\forall x~\exists y~A(x,y)}{\exists y~A(b,y)}{\small\forall\mathsf E}~~{\small[c]}\dfrac{\dfrac{[\forall y~\lnot A(b,y)]^2}{\lnot A(b,c)}{\small\forall\mathsf E}~\lower{1.5ex}{[A(b,c)]^3}}{\bot}{\small\lnot\mathsf E}}{\bot}{\small\exists\mathsf E^3}}{\bot}{\small\exists\mathsf E^2}}{\lnot\exists x~\forall y~\lnot A(x,y)}{\small\lnot\mathsf I^1}}$$


问:
Direct sum of two vector spaces
Can someone point out where is the fallacy in my proof below?

Claim: If $U_1,U_2,W$ are subspaces of a vector space $V$ and  $V = U_1\bigoplus W = U_2 \bigoplus W $, then $U_1 = U_2$

My counter example: Let $V = R^2$, the 2-dimensional space.

$W$: the x-axis, $(y=0)$.

$U_1$: the y-axis, $(x=0)$.

$U_2$: the straight line $x=y$.

Then

$(1): V = W + U_1$, because coordinate wise sum of $(x,0)$ and $(0,y)$ will span whole of $R^2, (x,y)$,

$(2):$ any $v\in V$ can be uniquely represented by a unique element in $W$ and an unique element in $U_1$. $(x,y)=(x,0)+(0,y)$

So $V=W \bigoplus U_1$

Also,

$(1): V = W + U_2$, because coordinate wise sum of $(x,0)$ and $(y,y)$ will span whole of $R^2, (x,y)$,

$(2):$ any $v\in V$ can be uniquely represented by a unique element in $W$ and an unique element in $U_2$. $(x,y)=((x-y)+y,y)=((x-y),0)+(y,y)$

So $V=W \bigoplus U_2$

But, as we see, $U_1 \neq U_2$

My proof: $V=U_1\bigoplus W\Rightarrow \forall v\in V, \exists$ a unique $u\in U_1$ and $w\in W$, such that $v=u+w$.

Let us consider a specific $v_1=u_1+w_1......(1)$

$V=U_2\bigoplus W\Rightarrow \forall v\in V, \exists $ unique $u\in U_2, w\in W$, such that $v=u+w$.

With respect to the specific $v_1$ and $w_1$ in equation $(1), \exists$ a unique $u_2\in U_2$, satisfying $v_1=u_2+w_1.............(2)$

Comparing equations $(1)$ and $(2)$, we get $$u_1+w_1=u_2+w_1$$

Adding $(-1)w$ to both sides, we get
$$u_1 + w_1 -w_1= u_2+w_1-w_1$$
$$\Rightarrow u_1+0 = u_2 +0$$
$$\Rightarrow u_1 = u_2.........(3)$$

Now, for any $u_1\in U_1, \exists $ unique $v_1, w_1$ and $u_2$, satisfying equation $(1)$ and $(2)$. And $u_1=u_2\in U_2$. In other words, $u_1\in U_1\Rightarrow u_1\in U_2$. Therefore $U_1\subseteq U_2.......(4)$

Similarly, for any $u_2\in U_2$, from equations $(1),(2) $ and $(3)$, we can show $u_2\in U_2 \Rightarrow u_2\in U_1$. Therefore $U_2\subseteq U_1.......(5)$.

Hence, from equations $(4)$ and $(5)$, we conclude $U_1=U_2$.

答:
Comparing equations (1) and (2), we get u1+w1=u2+w1

There's no reason to say that the $w$ you get from $U_1 \oplus W$ is the same as the one you get from $U_2 \oplus W$.

Title: Direct sum of two vector spaces


问:
weird finding function from the graph

The graph is on the interval [-3,3], defined on R

This function maintain the pattern on all of R

How to find the $f(x)$?

See how the graph goes parallel at certain intervals. I am thinking adding one more variables to indicate one of the direction. However, I cannot merge the two together.

Any thoughts?

答:
You can use a triangle wave written in terms of sine and arcsine:

$$f\left(x\right)\ =\ \frac{1}{\pi}\arcsin\left(\sin\left(\pi\left(x+\frac{1}{2}\right)\right)\right)+\frac{1}{2}.$$  This gives you the following plot:

Title: weird finding function from the graph


答:
You can define the function in part-wise intervals. For example,
$$y=1-x, ~~\text{if}~ x \in [0,1] $$
Similarly, define the function for other intervals.

Title: weird finding function from the graph

Body:
The graph is on the interval [-3,3], defined on R

This function maintain the pattern on all of R

How to find the $f(x)$?

See how the graph goes parallel at certain intervals. I am thinking adding one more variables to indicate one of the direction. However, I cannot merge the two together.

Any thoughts?


问:
Complete geometry theory other than Tarski?
It is well known that Tarski's axioms for Euclidean geometry is recursively axiomatizable and complete, and hence also decidable. Basically it is because the theory of real closed field has these properties. Does there exist other Euclidean or non-Euclidean geometry theory, that extends Hilbert's axioms and share these properties?

Regarding Euclidean geometry, since models of Hilbert's axioms plus parallel axiom are exactly cartesian products over ordered Pythagorean fields, the question is, as I understand, the same as asking for complete extension of theory of ordered Pythagorean field. I know very little about model theory but it seems there are not many theories known to be decidable.

I know even less about models of non-Euclidean geometry. I just learned that (to my surprise) the theory of hyperbolic geometry is decidable. I could not find the reference. Does this refer to the finite theory consisting of Hilbert's axioms plus a suitable hyperbolic axiom, or some infinite theory that resembles Tarski's axioms?

答:
Simone Ramello is correct in identifying Schwabhäuser's Metamathematische Methoden in Der Geometrie as the proper reference for this subject, but as he suggests this book has not been translated into English.

That said, you only need translations from two pages. I omit a discussion of Tarski's axioms, which can be found on Wikipedia or, in a more digestible form, from section 4 of this paper by Beeson et al. When looking at Beeson's list, note that the circle axiom CA is actually a restricted special case of the continuity axiom A11 and does not need to be counted separately. (The distinction here is that the more general continuity axiom A11 allows you to find a point between any pair of sets, whereas the circle axiom only allows you to locate the point of intersection between a circle and a straight line. On an intuitive level, it's something like the difference between extending the rational numbers to the reals by either providing the fully-general least upper bound axiom, or just asserting the existence of square roots.) In Beeson's list, A10 is equivalent to Euclid's parallel axiom.

Schwäbhauser in Part II, section 2.1(i)(1) and 2.1(iii) defines $n$-dimensional absolute geometry as absolute geometry with continuity essentially by taking axioms A1-A7 from Beeson's list and calling them "dimension-free absolute geometry," then replaces A8 and A9 with lower- and upper-dimensional axioms (though you can keep A8 and A9 as written if you are interested in 2-dimensional geometry) to get $n$-dimensional absolute geometry. In 2.1(i)(5) he defines "full n-dimensional absolute geometry" by adding axiom A11.

So for 2-dimensional geometry we basically have all of Beeson's axioms (A1 through A11) except A10, which is the Euclidean parallel postulate. In section 2.4 Schwabhäuser replaces it with Hilbert's Axiom of Hyperbolic Parallels to get what he would call "full n-dimensional hyperbolic geometry." The formula for this new axiom (called HP in the book) is a symbolic rendition of this large expression, which can be expressed entirely in terms of collinearity and betweenness (where collinearity can itself be defined in terms of betweenness:

If points $a$, $c$, and $d$ are not collinear, then there exist points $b_1$ and b_2$ such that the following three things are true:

$a$, $b_1$, and $b_2$ are not collinear,

For all $u$, if $u$ is between $b_1$ and $b_2$ and the three points $u$, $b_1$, and $b_2$, are distinct (unequal to each other), then there exists an $x$ collinear with $c$ and $d$ such that $u$ is between $a$ and $x$.

There is no $x$ collinear with $c$ and $d$ for which either $b_1$ or $b_2$ is between $a$ and $x$.

(End of the HP axiom.)

He goes on to explain the axiom intuitively: "For any non-collinear points $a$, $c$, $d$ there are always two rays $\overrightarrow{ab_1}$, $\overrightarrow{ab_2}$ starting from $a$, which are not contained in a line [that is, they're not exactly opposite to each other] and which do not intersect the line $\overleftrightarrow{cd}$, while every other ray emitted from $a$ within the angle $\angle b_1ab_2$ and starting from $a$ intersects the line $L(ad)$" and he provides a figure that alas, I cannot reproduce here.

So the axiomatization you seek would comprise:

Axioms A1 through A7 from Beeson's list (dimension-free absolute geometry)

A lower- and upper-dimensional axiom (A8 and A9 from Beeson's list will suffice if you are working in two dimensions; otherwise you have to use the axioms from PlanetMath)

Hilbert's Axiom of Hyperbolic Parallels. This replaces axiom A10, which is explicitly Euclidean.

The continuity axiom A11.

Schwabhäuser calls that collection of axioms $H^2_n$, where the $2$ means that the full second-order axiom of continuity is being used, and the $n$ tells the dimensionality of the axioms you're using to replace A8 and A9.

In section 2.5, Schwabhäuser asserts that a structure is a model of axioms $H_n$ if and only if it's isomorphic to the $n$-dimensional Klein space over the real numbers. The Klein space is essentially the Klein-Beltrami model from Wikipedia: the points are  $n$-dimensional vectors of real numbers whose lengths [the usual vector norm] are less than 1, where $y$ is between $x$ and $z$ iff there's a real number $\lambda$ between $0$ and $1$ for which ${\bf y}-{\bf x}=\lambda({\bf z}-{\bf x})$ [this is the same definition of "betweenness" that you would use for ordinary vectors in ordinary models], and $xy\equiv uv$ iff the distance $(1-{\bf x}\cdot{\bf y})^2\over(1-|{\bf x}|^2)(1-|{\bf y}|^2)$ equals the similarly-defined distance for $u$ and $v$.

What he means in 2.5 is that any formula about points in $n$-dimensional hyperbolic geometry, written in terms of Tarski's betweenness and equidistance operators, can be converted into an equivalent formula about $n$-dimensional vectors of real numbers, where distances are represented in terms of that funky Klein-Beltrami metric but betweenness is the same as it always was. But the equivalent formula involving real numbers uses only field operations (addition, subtraction, multiplication, and division) so it's decidable if and only if your theory of the real numbers is decidable.

Tarski proved that the first-order theory of the real numbers (the theory of "real closed fields") is complete and decidable in 1933. I'm pretty sure the full second-order theory of the real numbers is not complete because set theories generally aren't complete. Schwabhäuser noted this distinction in section 2.5 and I glossed over it; basically he said that if you replace the second-order continuity axiom A11 with an axiom scheme consisting of all of the first-order axioms (that is, the only sets you're allowed to use in axiom A11 are those that are defined in terms of relationships between points---no funny business with sets defined in terms with other set quantifiers) then the resulting geometry is equivalent to the Klein-Beltrami model over a real-closed field (that is, the real numbers with a first-order completeness axiom instead of the full second-order least upper bound axiom), which is what Tarski proved is decidable.

So to sum things up: take Beeson's axioms, replace A10 with AP, optionally replace A8 and A9 with corresponding axioms for a higher-dimensional space, and restrict A11 to a first-order axiom schema, and you have a complete Tarski-style axiomatization of $n$-dimensional hyperbolic geometry.


问:
How to use substitution to evaluate $ \int \frac{3}{\sin^2{x}} dx~ $?
I've learned a couple of methods of integrating, but I'm still not sure when to use which one.

$$ \int \frac{3}{\sin^2{x}} dx $$

I tried using a method where I set something to $u$ and $dv$ and go from there, but I don't end up anywhere with this problem.
 I know you can use substitution method and then integrate by parts, but I'm not sure which part of the integral I should begin substituting.

答: Just let $u = \frac{\cos x}{\sin x}$ and find $\frac{du}{dx}$ then proceed further.


答:
Generally speaking, this is a standard integral

$$I = \int \frac{3}{\sin^2x}dx = 3\int\frac{1}{\sin^2x}dx = 3\int\frac{\sin^2x + \cos^2x}{(\sin x)^2}dx$$

Now if $v = \frac{\cos x}{\sin x} \implies \frac{dv}{dx} = ?$

Title: How to use substitution to evaluate $ \int \frac{3}{\sin^2{x}} dx~ $?

Body:
I've learned a couple of methods of integrating, but I'm still not sure when to use which one.

$$ \int \frac{3}{\sin^2{x}} dx $$

I tried using a method where I set something to $u$ and $dv$ and go from there, but I don't end up anywhere with this problem.
 I know you can use substitution method and then integrate by parts, but I'm not sure which part of the integral I should begin substituting.


问:
Big Greeks and commutation
Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering?

Clearly if $\mathbf{x}_i$ is a matrix then:

$$\prod_{i=0}^{n} \mathbf{x}_i$$

depends on the order of the multiplication. But, even if one accepts that it has a sequence, it is not clear if it should mean $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ or $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$.

A similar question, is there a "big" wedge product convention?

$$\overset{n}{\underset{i=0}{\Huge\wedge}} \;{}^{\Large{\mathbf{x}_i} \;=\; \mathbf{x}_0 \wedge \mathbf{x}_1 \;\cdots \mathbf{x}_{n-1}\; \wedge \mathbf{x}_{n}} $$

答:
I think that even if it's not written explicitly anywhere, the $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ convention is the most predictable and sensible.

I've never seen the distinction made explicit, since in most circumstances the operation involved is commutative.

I did see somewhere on m.SE someone suggest $\mathbf{x}_i\prod_{i=1}^n$ to denote $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$, but that may have been with tongue in cheek...

Title: Big Greeks and commutation


问:
Isomorphism between fractional ideals
Let the restricted direct product
$$V_N = (0, \infty) \times \prod_{p \mid N} \{x \in \mathbb Z_p \ : \ x \equiv 1 \mod N\} \times \prod_{p \nmid N} \mathbb Q_p^\times$$

Let $I_N$ be the group al fractional ideals of $\mathbb Q$ prime to $N$. Let $P_N$ be the subgroup of principal fractional ideals $\alpha \mathbb Z$ for $\alpha \in \mathbb Q^\times \cap V_N$. I would like to understand why (and how to make inttuitive sense of)
$$I_N /P_N \simeq (\mathbb Z/N \mathbb Z)^\times.$$

I see that there is a natural map $(\mathbb Z/N \mathbb Z)^\times \to I_N$, but why the quotient?

答:
$$I_N = \{ \frac{a}b \Bbb{Z}, \gcd(N,ab)=1\}, \qquad P_N = \{ \frac{a}b \Bbb{Z}, \gcd(N,ab)=1, a\equiv b\bmod N\}$$

$$\phi:I_N\to \Bbb{Z}/N\Bbb{Z}^\times/\pm 1, \qquad \phi(\frac{a}b\Bbb{Z})=\pm \frac{a}b $$ is surjective and its kernel is $P_N$, therefore $$I_N/P_N\cong \Bbb{Z}/N\Bbb{Z}^\times/\color{red}{\pm 1}$$

Title: Isomorphism between fractional ideals


答:
Using the absolute value function together with the floor function, you can write
$$
f(x)=1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
$$
Verification:

With $f$ as defined above, we get
\begin{align*}
f(x+2)
&=
1-\left|(x+2)-\left(2\left\lfloor\frac{(x+2)+1}{2}\right\rfloor\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left\lfloor\frac{x+1}{2}+1\right\rfloor\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left(\left\lfloor\frac{x+1}{2}\right\rfloor+1\right)\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left(\left\lfloor\frac{x+1}{2}\right\rfloor+1\right)\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left(\left\lfloor\frac{x+1}{2}\right\rfloor+1\right)\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)-2\right|
\\[4pt]
&=
1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
\\[4pt]
&=
f(x)
\\[4pt]
\end{align*}
so $f$ is periodic with period $2$.

It remains to verify that $f$ works correctly on the interval $[-1,1)$.

Thus suppose $f$ is restricted to the interval $[-1,1)$.
\begin{align*}
\text{Then}\;\;&
-1\le x < 1
\\[4pt]
\implies\;&
0\le x+1 < 2
\\[4pt]
\implies\;&
0\le \frac{x+1}{2} < 1
\\[4pt]
\implies\;&
\left\lfloor\frac{x+1}{2}\right\rfloor=0
\\[4pt]
\end{align*}
hence for $-1\le x < 1$ we get
$$
f(x)
=
1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
=
1-|x|
$$
which matches the graph for the interval $[-1,1)$.

Title: weird finding function from the graph


答: Hint: recall that $(\cot x)'=-\frac{1}{sin^2 x}$


答:
Question $\bf1$: I've learned a couple of methods of integrating, but I'm still not sure when to use which one.

Answer: Yes, there are so many methods to solve the integrations. And if you want to use the most effective one, you have to watch the integrand function carefully and think about the rules you learn. Basically the more you practice, more you learn.

Solution of the given integral :
$$I=\int \frac{3}{\sin^2{x}} dx=3\int \dfrac{1/\cos^2(x)}{\sin^2(x)/\cos^2(x)} dx=3\int \dfrac{\sec^2(x)}{\tan^2(x)} dx$$Putting  $~u=\tan(x)~\implies du=\sec^2(x)~ dx$,

$$I=3\int \dfrac{\sec^2(x)}{\tan^2(x)} dx=3\int \dfrac{du}{u^2}=-\dfrac 3u+c=-\dfrac{3}{\tan(x)}+c=-3\cot(x)+c$$where $~c~$ is arbitrary independent constant.


问:
Expressing a polynomial in terms of basis
Suppose $\alpha$ is a root of the irreducible polynomial $x^3 -2x-2 \in \mathbb{Q}[x]$.

PROBLEM:

Express $(\alpha +1)(\alpha^2+\alpha+1)^{-1}$ in terms of $\{1, \alpha, \alpha^2\}$ for $\mathbb{Q} (\alpha)$ over $\mathbb Q$.

My attempt: I tried to find $(\alpha^2+\alpha+1)^{-1}$, $(a+b\alpha+c\alpha^2)(\alpha^2+\alpha+1)=1$. Implies that $a=1,b=0$ and $c=0$ . From here I dont know what to do or did I do something wrong here?

Any help and hints would be much appreciated!

答: Note that $\{1, \alpha, \alpha^2\}$ is a basis of $\Bbb Q(\alpha)$ over $\Bbb Q$. Thus you should try to determine $a, b  c$ from $(a+b \alpha+c\alpha^2)(1+\alpha+\alpha^2)=1$. When expanding this expression you should use $\alpha^3-2\alpha-2 =0$. This results for example in $\alpha^4=\alpha \alpha^3=\alpha(2\alpha+2)=2\alpha^2+2\alpha$.


答:
You did not do something correct there, but your intention seems to be along the right path.

Since $\alpha$ is a root of the polynomial, you know that $\alpha^3 - 2\alpha - 2 = 0$, or in other words, $\alpha^3 = 2\alpha + 2$.

This means you can express any polynomial in $\alpha$ in terms of $\{1, \alpha, \alpha^2\}$, by substituting (repeatedly, if necessary) $\alpha^3$ in powers of $\alpha$ greater than $2$.

Now, if you can express $(\alpha^2+\alpha+1)^{-1}$ as a polynomial (not a rational function) in $\alpha$, then by the observation above you are done.
Of course, by the observation itself, if such a polynomial exists, it is expressible in terms of $\{1, \alpha, \alpha^2\}$.
Hence, we're looking for $a$, $b$ and $c$ such that

$$(\alpha^2+\alpha+1)^{-1} = a + b\alpha + c\alpha^2 \iff 1 = (\alpha^2+\alpha+1)(a + b\alpha + c\alpha^2)$$

We can expand that to find that

$$\begin{align}
1
&= a + (a+b)\alpha + (a+b+c)\alpha^2+(b+c)\alpha^3 + c\alpha^4
\\&= a + (a+b)\alpha + (a+b+c)\alpha^2+(b+c)(2\alpha+2) + c\alpha(2\alpha+2)
\\&= (a+2b+2c) + (a+3b+4c)\alpha + (a+b+3c)\alpha^2
\end{align}$$

This yields the following system of linear equations:

$$\begin{align}
a+2b+2c &= 1
\\a+3b+4c &= 0
\\a+b+3c &= 0
\end{align}$$

Do you think you can take it from here?

Title: Expressing a polynomial in terms of basis

Body:
Suppose $\alpha$ is a root of the irreducible polynomial $x^3 -2x-2 \in \mathbb{Q}[x]$.

PROBLEM:

Express $(\alpha +1)(\alpha^2+\alpha+1)^{-1}$ in terms of $\{1, \alpha, \alpha^2\}$ for $\mathbb{Q} (\alpha)$ over $\mathbb Q$.

My attempt: I tried to find $(\alpha^2+\alpha+1)^{-1}$, $(a+b\alpha+c\alpha^2)(\alpha^2+\alpha+1)=1$. Implies that $a=1,b=0$ and $c=0$ . From here I dont know what to do or did I do something wrong here?

Any help and hints would be much appreciated!


问:
Intuition behind the exponential loss function
I'm reading about AdaBoost from the book The Elements of Statistical Learning.
The book mentions that, to train the model, the exponential loss function is used:
$$L(y, f (x)) = e^{−y f (x)},$$
where $y$ is the expected output and $f(x)$ is the model output given the feature $x$.
The problem I have understanding this loss function is that "same errors" give different loss.

For example,

$y=1$ and $f(x)=0.8$ then $L(y, f (x)) = e^{−0.8}$

$y=0$ and $f(x)=.2$ then $L(y, f (x)) = e^{0.2} \neq e^{−0.8}$

In both cases, $y-f(x)= 0.2$, but the loss functions give different values.

Could anyone tell me what I'm missing here?
Thank you!

答:
Note the multiplicative form, exponential loss function is meant for classification.

$y$ should be either $1$ or $-1$.

If $y=1$, $f(x)=0.8$, $L(y, f(x))=e^{-0.8}$.

If $y=-1$, $f(x)=-0.8$, $L(y, f(x))=e^{-0.8}$.

If you share the same sign as the truth as have large magnitude, we want the value to be small.


问:
Prove Weierstrass theorem for a general $[a,b]$ follows from the result on $[0,1]$.
Give a detailed proof of the assertion that the Weierstrass theorem for a general $[a,b]$ follows from the result on $[0,1]$ (by using lemma 11.1)

It's been two days since  I am trying to figure out the way of proving this problem in (N.L. Carothers), any hints please?

答:
Hint: Start with $g \in C[a,b]$, from Lemma 11.1, there is $f \in C[0,1]$ such that $g(x) = f(t(x))$ where $t(x) = \frac{x-a}{b-a} \in [0,1]$.

Do the logical steps till you get:

$$\lVert g(x) - p_n (x) \rVert = \max_{a\le x \le b} |f(t(x)) - P_n(t(x))|=\max_{0\le t \le 1} |f(t) - P_n(t)| < \epsilon.$$

That is because we have already know that the space of polynomials is dense in $C[0,1]$.