问:
How to find image, pre-image, bijection, onto, one-to-one?
The question I am trying to solve is as follows:

Let A = {1, 2, 3, 4, 5} and define the function f : (A) → Z as follows:
For each set X in (A), where (A) is the power set of A, f(X) is the
number of elements in X.

a. What is the image of {1, 2, 3} under f? Justify.

b. What is the pre-image of 4? Justify.

c. What is the pre-image of 10? Justify.

d. Is f one-to-one? Justify.

e. Is f onto? Justify

My professor is not teaching us anything at all and my understanding of what onto, one-to-one, image, and pre-image is extremely weak because of this. Could someone please explain what these are in concrete, simple terms so that I may understand, and also push me in the right direction as far as solving a problem like this?

Thank you!

答:
Consider a function  f from A = { a, b} to B = { c, d, e, m} such that  f(a) = d  and f(b) = e.

The image of a under f is d and the image of b under f is e.

This is equivalent to saying that the preimage of d is a and the preimage of e is b.

The funbction is one-to-one, since no two different elements in A  have the same image in B.

The function is not onto since there is at least one element of B that has no preimage in A under f.


问:
How to evaluate $\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx$
It's a very simple question but it confuses me. How do I evaluate
$$
\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx
$$
without splitting? And why can't I split it?</p>

答:
You can't, because $x\mapsto \frac 1 x$ is not in $L_1$ and thus cannot be integrated over $[1,\infty)$ (as a matter of fact $\int_{[1,\infty)}\frac {dx} x = +\infty$).

However, $\frac 1 x-\frac 1 {x+1}$ can be rewritten (by reducing to the same denominator) into something which is in $L_1$ (and whose integral is easy to compute).

ContentLicense: CC BY-SA 3.0


问:
Subgroups of free groups which avoid conjugacy classes
Let $G = (\mathbb Z/2\mathbb Z)^{\ast m}$ be a free product of some groups of order $2$.  Let $\alpha_1,\ldots,\alpha_m$ be the generators.

Can I find a free, nonabelian subgroup of $G$ that has no nontrivial elements conjugate to any $\alpha_i \alpha_j$? How can I prove it?

答:
I'll answer the stronger version of your question where the set of words $\alpha_i \alpha_j$ is replaced by any finite subset $A \subset G$.

This is not possible if $m=1$ because $G$ is finite in that case and so has no free, nonabelian subgroup.

It is also not possible if $m=2$ because $G$ is the infinite dihedral group which has an index 2 abelian subgroup (in fact cyclic) and therefore has no free nonabelian subgroup.

So we need to assume $m \ge 3$.

Every element of $G$ is expressed uniquely as a "reduced word" meaning a sequence of the form $\alpha_{i_1} .... \alpha_{i_k}$ in which any two consecutive letters $\alpha_{i_j} \alpha_{i_{j+1}}$ are unequal. The identity corresponds to the empty word with $k=0$.

Every conjugacy class in $G$ has a representative which is expressed semi-uniquely as a "cyclically reduced word" meaning that it is reduced and $b_{i_m}, b_{i_1}$ are unequal; by "semi-unique" I mean that such a representative of the conjugacy class is unique up to cyclic permutation of the word.

Okay, so, the first step is to express the conjugacy class of each element of $A$ as a cyclically reduced word, and then take $k$ to be the maximum length of those words.

Here's a particularly simple construction if $m \ge 4$.

Pick distinct reduced words $w,v$ of length $>k$ such that the starting and ending letters of $w$ and $v$ are 4 different letters, for example:
$$w = (\alpha_1 \alpha_2)^k \alpha_3
$$
$$v = (\alpha_2 \alpha_3)^k \alpha_4
$$
It follows that every nontrivial reduced word in the letters $w$ and $v$, after substitution, becomes a cyclically reduced word in the letters $\alpha_1,\ldots,\alpha_4$, and furthermore it has length $\ge k$. For example
$$w^{-1} v = \alpha_3 (\alpha_2 \alpha_1)^k (\alpha_2 \alpha_3)^k \alpha_4
$$
Therefore, the group $\langle w,v \rangle$ is a rank 2 free group and every nontrivial element in it is cyclically reduced of length $> k$, hence is not conjugate to any element of the set $A$.

If $m=3$ it's not possible to choose $w,v$ in such a simplistic manner. But one can choose the $w,v$ to be long reduced words (of length $\ge k + 4$) in the letters $\alpha_1,\ldots,\alpha_3$ so that each of the concatenations $ww$, $vv$, $wv$, $wv^{-1}$, $vw$, $vw^{-1}$ produces a word in $\alpha_1,\alpha_2,\alpha_3$ with short cancellation (at most $2$ letters are cancelled). It then follows that each reduced word in the symbols $w,v$ evaluates to a word in the letters $\alpha_1,\alpha_2,\alpha_3$ whose cyclic reduction has length $\ge k+2$, and so is nontrivial and not conjugate to any element of $A$.

ContentLicense: CC BY-SA 4.0


问:
Norm of difference of repeated limit is zero implies existence of limit point?
Let $(x_n)_{n=0}^\infty$ be a sequence in a normed vector space $X$ (not necessarily complete). Denote the norm by $\|\cdot \|$. Assume that
$$
\lim_{j\to\infty} ( \lim_{k\to\infty} \|x_j -x_k\|) = 0
$$
and
$$
\lim_{k\to\infty} ( \lim_{j\to\infty} \|x_j -x_k\|) = 0
$$

Does this imply that there is a limit point $x\in X$ such that $x= \lim_{n\to \infty} x_n$?

答:
Notice that your condition implies that $(x_n)_n$ is a Cauchy sequence. Namely, for any $\varepsilon > 0$ by definition of $\lim_{j\to\infty}$ there exists $j_0 \in \Bbb{N}$ such that
$$j \ge j_0 \implies \lim_{k\to\infty} \|x_j-x_k\| \le \frac\varepsilon2.$$
Now for any $j \ge j_0$ by definition of $\lim_{k\to\infty}$ there exists $k_0 \in \Bbb{N}$ such that $$k \ge k_0 \implies \|x_j-x_k\| \le \varepsilon$$
so for all $j,k \ge \max\{j_0,k_0\}$ we have $\|x_j-x_k\| \le \varepsilon$.

So we have to find a counterexample in an incomplete space. Consider $c_{00}$, the space of finitely supported sequences, equipped with the norm $\|\cdot\|_1$. Define
$$x_n = \left(\frac12, \frac1{2^2}, \ldots, \frac1{2^n}, 0, 0, \ldots\right).$$
Then $(x_n)_n$ satisfies your condition. Indeed, assuming $k \ge j$ since we first let $k\to\infty$, we have
\begin{align}\|x_j-x_k\|_1 &= \left\|\left(0, \ldots, 0, -\frac1{2^{j+1}}, -\frac1{2^{j+2}}, \ldots, -\frac1{2^k}, 0, 0, \ldots\right)\right\|_1\\
&= \sum_{i=j+1}^k \frac1{2^i} \xrightarrow{k\to\infty} \sum_{i=j+1}^\infty \frac1{2^i} = \frac1{2^j} \xrightarrow{j\to\infty} 0.
\end{align}
However, $(x_n)_n$ doesn't converge in $c_{00}$ because it converges to $\left(\frac1{2^n}\right)_n \in \ell^1\setminus c_{00}$.


问:
Proof by contradiction : $n^3 −n−6 = 0$ means $m^3 − m − 6 ≠ 0$ when $m≠n$
I have to prove by contradiction that if $n$ is a positive integer such as $n^3-n-6 = 0$, so for any positive integer $m$ such as $m ≠ n$, we have $m^3-m-6 ≠ 0$

I am new to proofs, but I tried and here is what I did.

I first assume the opposite of the conclusion, so
$m^3-m-6 = 0$

$(m-2)(m^2+2m+3) = 0$

m=2

If I do the same thing for the n function and find the solution for $n^3-n-6 = 0, n=2$.

This is a contradiction since $m=n$, but we said $m ≠ n$.

I feel like im doing something wrong here and I would like someone to tell me what it is ! (Again, I am a beginner)

Body:
We have

$$n^3-n-6=0\iff$$
$$(n-1)n(n+1)=1.2.3 \iff n=2$$

thus

$$(m-1)m(m+1)\ne 6 \iff m\ne 2$$


问:
How to find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$? I know that it exists and can't find it with Stolz theorem.
How do I find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$?

I thought about Stolz theorem so I got:
$$ \frac{(n+1)^2-n^2}{7^{\sqrt{n+1}}-7^{\sqrt{n}}}, $$
but it doesn't seem to lead anywhere (at least not through $e$ and $\ln$).

And I know for fact that there is some limit because:
$$\lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^2}{7^{\sqrt{n+1}}}}{\frac{n^2}{7^{\sqrt{n}}}} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n+1}}*n^2} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n}\sqrt{\frac{1}{n}+1}}*n^2} \implies \frac{(n+1)^2}{n^2}\implies 1$$

答: Hint: it suffices to consider the subsequence $n_k = k^2$.


问:
Geometric multiplicity for non zero eigen values of matrices $AB$ and $BA$.
As lot of information is given in this site about eigen values of $AB$ and $BA$ for square matrices $A$ and $B$. As characteristics polynomial of $AB$  and $BA$ are same so both have same set of eigen values with multiplicity. Now I want to know about geoemetric multiplicity and as one of $AB$ and $BA$ may become zero and other not even diagonalizable so i can conclude that geometric multiplicity of eigen value $0$ may not equal. Now what about geometric multiplicity of common non zero eigen values? Are they will be same? i.e. if $a\neq 0$ then can we say $$Geo.Mult_a(AB) =Geo.Mult_a(BA)? $$ please explain or give counter example. Thanks.

答:
Here is a somewhat different explanation for the equality of dimensions of the eigenspaces of $AB$ and $BA$ for nonzero eigenvalues than in the other answers (so far); it gives rise to the somewhat stronger result that the Jordan types (lists of sizes of Jordan blocks) are also the same for nonzero eigenvalues. For any linear operator $T$ there is a unique $T$-stable complementary subspace$~W$ to the generalised eigenspace for the eigenvalue$~0$. There are several ways to describe it: over an algebraically closed field, $W$ is the (direct) sum of all other generalised eigenspaces; it is the image of $T^k$ for sufficiently large$~k$ ($k=n$, the dimension of the space, is certainly sufficient); if $Q$ is the quotient of the characteristic polynomial by any factors$~X$ it contains, then $W=\ker(Q[T])$.

Now let $T$ be the linear operator given by $AB$ and let $W_0$ be this subspace$~W$ for it. By construction the restriction of $T$ to $W_0$ is invertible (does not have $0$ as eigenvalue). If $W_1$ is the image of $W_0$ under multiplication by $B$, we have linear maps $b:W_0\to W_1$ (given by multiplication by $B$) and $a:W_1\to W_0$ (given by multiplication by $A$) whose composition $a\circ b$ is that invertible restriction of $T$ to $W_0$, so $a$ and $b$ must each be invertible. Starting with $T'$ given by $BA$ instead of $AB$, one sees that its subspace $W$ is in fact $W_1$. Now the restriction $a\circ b$ of $T$ to $W_0$ is conjugate to the restriction $b\circ a$ of $T'$ to$~W_1$, since $ab=a(ba)a^{-1}$. Since all (generalized) eigenspaces for nonzero eigenvalues of $AB$ respectively of $BA$ are contained in $W_0$ respectively $W_1$, one gets the desired result.


问:
Why: A holomorphic function with constant magnitude must be constant.
How can I prove the following assertion?

Let $f$ be a holomorphic function such that $|f|$ is a constant.  Then $f$ is constant.

Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations.

答:
Another approach uses the fact that if $f$ and $\bar f$ are both holomorph, then $f$ is constant.

If $|f|=0$, then $f=0$.

Let $|f|=c$ with $c>0$. This implies $|f|^2=f\bar f = c^2$.
Now, because $|f|\neq 0 \implies f\neq 0$, we can divide by $f$ and get $\bar f = c^2/f$.
The right side is holomorph, thus $\bar f$ has to be holomorph.


问:
Showing that the inflection points of $x\sin x$ lie on a certain curve

Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$

I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?

答:
Inflection points indeed check OK, as three concurrent plots should, when plotted together.

$$y - x \sin x =0\tag1$$
$$ y^{''}=2 \cos x - x \sin x=0 \tag2 $$
$$ \text{ From(2) solve for x from the equation} (\frac{2}{x}=\tan x ) \tag3 $$

The above is a transcendental equation which has roots that can be solved for by  Newton-Raphson iteration etc. The roots of green curve are directly above or below point of inflection in graphical visual verification below.

Eliminate $x \sin x $ between (1) and (2)
$$ y = 2 \cos x \tag 4 $$

This is a trigonometric equation, not transcendental. ( Not asked for by OP).

Eliminate $ \sin x $ between (1) and (2)

$$ y =\pm  \frac{2 x }{ \sqrt{4+x^2 }}\tag 5 $$

This is an algebraic equation, not transcendental.

In (4) and (5) although we have not eliminated $y$ totally , we however find that they provide valuable plot sources for inflection point loci.

The same is graphically shown below as three plots concur.

In passing btw it should be mentioned that the procedure is so nice. It even  enables finding an envelope through maxima/minima while eliminating between the curve and its first derivative.


问:
Show that $A \otimes B = A \odot B$ if $A$ is a finite-dimensional $C*$-algebra.
Let $A$ and $B$ be $C^*$-algebras. If $A$ is finite-dimensional, then I want to show that the algebraic tensor product $A \odot B$ is complete for the minimal $C^*$-norm on it (in fact, the choice of norm does not matter since $A$ is nuclear).

Attempt:

I know that $$A \cong M_{n_1}(\mathbb{C}) \oplus \dots \oplus M_{n_k}(\mathbb{C})$$

And thus
$$A \odot B \cong (A \odot  M_{n_1}(\mathbb{C}))\oplus \dots \oplus (A\odot M_{n_k}(\mathbb{C})) $$

so I believe it suffices to show that $A \odot M_{n_1}(\mathbb{C})\cong A \odot \mathbb{C}^{n^ 2}$  is complete.

Then I'm stuck.

Maybe I'm overcomplicating this though....

答:
You wrote as if $B$ is the finite-dimensional one, so I'll stick with that.

Let $\gamma$ be a C$^*$-norm on $A\otimes M_n(\mathbb C)$ (one certainly exists, because we can represent $A\subset B(H)$ and then $A\otimes M_n(\mathbb C)$ can be represented in $B(H\otimes\mathbb C^n)$).

For any $k$, the map $a\longmapsto \gamma(a\otimes E_{kk})$ defines a C$^*$-norm on $A$. As a C$^*$-algebra admits a unique C$^*$-norm, we get that $\gamma(a\otimes E_{kk})=\|a\|$. Even if $k\ne j$,
$$
\gamma(a\otimes E_{kj})^2=\gamma(a^*a\otimes E_{jj})=\|a^*a\|=\|a\|^2.
$$
Now suppose that we have a Cauchy sequence  $\{x_n\}\subset A\otimes M_n(\mathbb C)$. We can write $$x_n=\sum_{k,j} a_n(k,j)\otimes E_{k,j}.$$
For any $k,j$,
\begin{align}
\|a_n(k,j)-a_m(k,j)\|^2&=\gamma((a_n(k,j)-a_m(k,j))\otimes E_{11})\\[0.3cm]
&=
\gamma((1\otimes E_{11})(x_n-x_m)(1\otimes E_{11}))\\[0.3cm]
&\leq\gamma(x_n-x_m).
\end{align}
So each sequence $\{a_n(k,j)\}_n$ is Cauchy on $A$. As $A$ is a C$^*$-algebra, there exist $a(k,j)\in A$ with $\lim_na_n(k,j)=a(k,j)$. We want to show that $x_n\to\sum_{k,j} a(k,j)\otimes E_{k,j}$. For this
\begin{align}
\gamma(x_n-\sum_{k,j} a(k,j)\otimes E_{k,j})
&=\gamma(\sum_{k,j} (a_n(k,j)-a(k,j))\otimes E_{k,j})\\[0.3cm]
&\leq \sum_{k,j} \gamma((a_n(k,j)-a(k,j))\otimes E_{k,j})\\[0.3cm]
&=\sum_{k,j} \|a_n(k,j)-a(k,j)\|\\[0.3cm]
&\to0,
\end{align}
where it is essential in this last step that the sum is finite. So $A\otimes M_n(\mathbb C)$ is complete with respect to $\gamma$, and so it is a C$^*$-algebra. This in particular shows that $\gamma$ is necessarily the norm described in the second paragraph.


问:
JEE Circles Doubt
Can you derive a general formula for length of intercept made by a circle $x^2+y^2+2gx+2fy+c=0$ on the line $ax+by+c=0$? If no, then is there any other method to solve these type of questions?

答:
\begin{align}
x^2+y^2+2gx+2fy+d &= 0
\tag{1}\label{1}
,\\
ax+by+c &= 0
\tag{2}\label{2}
.
\end{align}

Expression \eqref{1} represent a circle
with radius $r=\sqrt{g^2+f^2-d}$
centered at $O=(-g,-f)$:

\begin{align}
(x+g)^2+(y+f)^2
&=g^2+f^2-d
\tag{3}\label{3}
.
\end{align}

Intersection points $X_+,\ X_-$
of the circle $(O,R)$
and the line through the point $B$
in direction $D=z_2-z_1$ can be found as follows:

\begin{align}
X_{\pm}&=B+t_{\pm}\cdot D
\tag{4}\label{4}
,\\
t_{\pm}&=\frac{-(B-O)\odot D\pm\sqrt{\left((B-O)\odot D\right)^2+|D|^2(R^2-|B-O|^2)}}{|D|^2}
\tag{5}\label{5}
,
\end{align}
where $\odot$ is
the dot product.

Given \eqref{2},\eqref{3}, we can apply \eqref{4}, \eqref{5}
by setting (assuming that $b\ne0$)

\begin{align}
O&=(-g,-f)
,\quad
z_1=(0,-\tfrac cb)
,\quad
z_2=(1,-\tfrac{a+c}b)
,\quad
B=z_1
,\quad
D=z_2-z_1=(1,-\tfrac ab)
\tag{6}\label{6}
.
\end{align}

The number of intersections is defined by the discriminant
\begin{align}
\delta&=
(a\,f-b\,g)^2-d\,(a^2+b^2)+c\,(2a\,g+2b\,f-c)
\tag{7}\label{7}
.
\end{align}

If $\delta<0$, there are no intersections,
if $\delta>0$, there are two intersection points,

\begin{align}
X_{\pm}&=
\left(
\frac{-b\,\Big(b\,g-a\,f\pm\sqrt{\delta}\Big)-a\,c}{a^2+b^2},\
\frac{ a\,\Big(b\,g-a\,f\pm\sqrt{\delta}\Big)-b\,c}{a^2+b^2}
\right)
\tag{8}\label{8}
,
\end{align}

and if $\delta=0$, then the line is tangent to the circle at the point

\begin{align}
X_+=X_-&=
\left(
\frac{-b\,(b\,g-a\,f)-a\,c}{a^2+b^2},\
\frac{ a\,(b\,g-a\,f)-b\,c}{a^2+b^2}
\right)
\tag{9}\label{9}
.
\end{align}

The length of intercept is just
\begin{align}
|X_+-X_-|&=
2\sqrt{\frac{\delta}{a^2+b^2}}
\tag{10}\label{10}
.
\end{align}

Title: JEE Circles Doubt


问:
Using goal seek in excel.
Hi I don't know if this is an appropriate question for this page but I thought it was worth a shot and if nothing else a learning experience.

So I am studying actuarial mathematics and I am currently working on problem involving calculating the yield of an investment.

I have been working through these questions alongside prerecorded lectures and have been fairly understanding the concept and how to use it however my lecturer constantly says that these answers are not that exact or precise but would be a lot better when worked out using Goal seek in excel.

However my problem is that he has never shown us goal seek or mentioned any further on the topic.

Hence I was wondering if anyone could explain this too me and/or how to use it as I'm getting answers to questions wrong due to the situation.

Thanks in advance.

Adding from comments:

The question I was given was: You invest £10,000 in a project and receive back £700.87 per annum payable half-yearly in arrears for the next 25 years. Calculate the yield on this investment.

So from this we worked out that the money coming back at the end of this question is £17521.75 and the average time for this question is 12.75 $(last time period-first time period) /2$ which is $(25.5)/(2)$

And from this we do the calculation

$(17521.75/10000)^(1/12.75)$

And we get 4.5% which my lecturer then states in his lecture that this isn't a very good answer as the real answer is 4.98% but doesn't proceed to tell us how to obtain this answer.

答:
After $n$ years at an interest rate of $p\,\%$, an initial amount $A$ becomes $A\left(1+\frac p{100}\right)^n$, of course. Can we find the interest rate that would allow us to double our amount in $n$ years? Still easy: Equate the above with $2A$ and transform in straightforward fashion to arrive at $p=100(\sqrt[n]2-1)$.

But things get complicated when we have more complex situations, e.g., annual payments, perhaps even of varying size.  We can still easily compute forward to find the final amount in such a scenario. But the backwards computation, e.g., to find the interest rate or the annual payment that produces a desired final amount, is way more complicated and typically it is not possible to compute it exactly in closed form.

Enter the goal seeker of any modern spreadsheet software. What it does is the following: It repeatedly varies the value of a cell of your choice, computes everything else "forward", and compares another cell with the desired goal value. Once the difference from the goal value is small enough (hopefully about the order of magnitude of rounding errors), it stops.
Of course, they do not try random values until they make a lucky discovery. Instead they use some numerical methods that usually converge nicely to a solution (at least if one starts with an initial guess that is not too far off). For example, if setting the input cell to $6$ produces $2.5$ in the output cell, and $7$ produces $-0.8$, then (at least for a continuous function dependency) we expect a solution between $6$ and $7$ and might do our next trial computation with $6.5$. Actually, the values suggest that the solution might be closer to $7$ than to $6$, so perhaps our next try should rather be $6,75$. If computing with $6.75$ produces output $0.04$, we now know that we should search only between $6.75$ and $7$ and perhaps try $6.77$ next, and so on.

This simplified description leaves aside a lot of problems that are in the fine details. And as a user, you should be aware of possible problems as well.

no good solution might be found (typically detectable by a still large-ish difference from the goal value)

in spite of a small difference from the goal value, the nearest solution is quite far away

or it might even be very far away because the algorithm git "trapped" near some almost-solution

if there are several solutions, you find only one

...

Nevertheless, goal seek can very often be a quick and easy helper in solving such problems with good-enough accuracy.

Title: Using goal seek in excel.


答:
Not an answer, but far too much for a comment.

Computing in gold standard with no interest rates, the question is easy:

You invest 10k£ and earn £700.87 for the forthcoming 25 years. So you have the investment and the yield  (easy to the percentage) and independent if the payments are per year or per half year.

It gets more complicated if interests are of any concern. Money you have access to now is worth more than a promise of money some-when., even when the risk or default is not to be considered.

Please amend your question.

Title: Using goal seek in excel.

Body:
Hi I don't know if this is an appropriate question for this page but I thought it was worth a shot and if nothing else a learning experience.

So I am studying actuarial mathematics and I am currently working on problem involving calculating the yield of an investment.

I have been working through these questions alongside prerecorded lectures and have been fairly understanding the concept and how to use it however my lecturer constantly says that these answers are not that exact or precise but would be a lot better when worked out using Goal seek in excel.

However my problem is that he has never shown us goal seek or mentioned any further on the topic.

Hence I was wondering if anyone could explain this too me and/or how to use it as I'm getting answers to questions wrong due to the situation.

Thanks in advance.

Adding from comments:

The question I was given was: You invest £10,000 in a project and receive back £700.87 per annum payable half-yearly in arrears for the next 25 years. Calculate the yield on this investment.

So from this we worked out that the money coming back at the end of this question is £17521.75 and the average time for this question is 12.75 $(last time period-first time period) /2$ which is $(25.5)/(2)$

And from this we do the calculation

$(17521.75/10000)^(1/12.75)$

And we get 4.5% which my lecturer then states in his lecture that this isn't a very good answer as the real answer is 4.98% but doesn't proceed to tell us how to obtain this answer.


问:
Semidirect product is supersolvable groups
Let $G$ be a finitely generated group and $N$ be a nilpotent group (not necessarily finitely generated) such that $N$ is a subgroup of $G$ and $G / N$ is $\mathbb{Z}$. That is, there is a short exact sequence $$ 1\rightarrow N \rightarrow G \rightarrow \mathbb{Z} \rightarrow 1$$ such that $G$ is finitely generated and $N$ is a nilpotent subgroup.

Is $G$ then supersolvable or at least polycyclic?

If $N$ were a finitely generated group, it is clearly supersolvable. But is it true even though $N$ is infinitely generated ensured that $G$ is finitely generated?

答: Any subgroup of a polycyclic group is finitely generated. By contraposition, if $N$ is not finitely generated, then $G$ is not polycyclic.


答:
If I understand you correctly, you want

$$\begin{align}\int_1^{\infty} dx \left ( \frac{1}{x} - \frac{1}{x+1}\right ) &= \lim_{R \to \infty}\int_1^{R} dx \left ( \frac{1}{x} - \frac{1}{x+1}\right ) \\ &=\lim_{R \to \infty} [\log{R} - (\log{(R+1)} - \log{2})]\\ &= \log{2}  \end{align}$$

Title: How to evaluate $\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx$

Body:
It's a very simple question but it confuses me. How do I evaluate
$$
\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx
$$
without splitting? And why can't I split it?</p>


问:
If $\,\lim_{p\to\infty} \sum_{j=1}^n a_j\cos(p\pi \theta_j) \to 0,\,$ then $\,a_1 = \dots = a_n = 0$.
The exercise

Let $n \geq 1$, $0 < \theta_1 < \dots < \theta_n < 1$ and $a_1, \dots, a_n \in \mathbb{R}^n$. Assume $ \sum_{j=1}^n a_j\cos(\pi p \theta_j) \underset{p \to +\infty}{\longrightarrow} 0$ ($p \in \mathbb{N}$). Show $a_1 = \dots = a_n = 0$.

My try

I tried to prove this by induction over $n$. I succeeded for $n = 1$, but I have no clue to show how the property can be used for $n \geq 2$.

答:
$\left(\cos(\pi p\vartheta_1)\right)_{p\geqslant 0}$ is bounded, thus there exists $\sigma_1:\mathbb{N}\rightarrow\mathbb{N}$ and $\ell_1\in[-1,1]$ such that $\lim\limits_{p\rightarrow +\infty}\cos\left(\pi\sigma_1(p)\vartheta_1\right)=\ell_1$. By the same argument, since $\left(\cos\left(\pi\sigma_1(p)\vartheta_2\right)\right)_{p\geqslant 1}$ is bounded, there exists $\sigma_2:\mathbb{N}\rightarrow\mathbb{N}$ and $\ell_2\in[-1,1]$ such that $\lim\limits_{p\rightarrow +\infty}\cos\left(\pi(\sigma_1\circ\sigma_2)(p)\vartheta_2\right)=\ell_2$, and we also have $\lim\limits_{p\rightarrow +\infty}\cos\left(\pi(\sigma_1\circ\sigma_2)(p)\vartheta_1\right)=\ell_1$. We do the same for $\ell_3,\ldots,\ell_n$. Let $\sigma=\sigma_1\circ\ldots\circ\sigma_n$, then $\lim\limits_{p\rightarrow +\infty}\cos\left(\pi\sigma(p)\vartheta_j\right)=\ell_j$ for all $j$ and, we can suppose without loss of generality that $(\sigma(p+1)-\sigma(p))_{p\geqslant 0}$ is a nondecreasing sequence. Since $\sin\left(\pi\sigma(p)\vartheta_j\right)=\pm\sqrt{1-\cos^2\left(\pi\sigma(p)\vartheta_j\right)}\underset{p\rightarrow+\infty}{\longrightarrow}\pm\sqrt{1-\ell_j^2}$ we have
$$ \cos\left(\pi\left(\sigma(p+1)-\sigma(p)\right)\vartheta_j\right)=\cos\left(\pi\sigma(p+1)\vartheta_j\right)\cos\left(\pi\sigma(p)\vartheta_j\right)+\sin\left(\pi\sigma(p+1)\vartheta_j\right)\sin\left(\pi\sigma(p)\vartheta_j\right)\underset{p\rightarrow +\infty}{\longrightarrow}\ell_j^2+\left(\pm\sqrt{1-\ell_j^2}\right)^2=1 $$
thus we finally get
$$ \lim\limits_{p\rightarrow +\infty}\sum_{j=1}^n a_j\cos\left(\pi\left(\sigma(p+1)-\sigma(p)\right)\vartheta_j\right)={\rm Card}\{j\in[\![1,n]\!],a_j\neq 0 \} $$
But since $\displaystyle\lim\limits_{p\rightarrow +\infty}\sum_{j=1}^n a_j\cos(\pi p\vartheta_j)=0$, we have $a_j=0$ for all $j$.


问:
$L(V,\prod_i W_i) \cong \prod_i L(V,W_i)$ & $L(\oplus W_i,V) \cong \prod_i L(W_i,V)$.
Let V be a vector space, and $\{W_i\}$ a family of vector spaces. Then prove that

a) $L(V,\prod_i W_i) \cong \prod_i L(V,W_i)$

b) $L(\oplus W_i,V) \cong \prod_i L(W_i,V)$.

a)Let $\ T \in L(V,\prod_i W_i)$ i.e $\ T:V \rightarrow \prod_iW_i$ is a linear transformation. Let $x \in V$ then $\ T(x)=(T_i(x))_i \in \prod_iW_i$, where each $T_i:V \rightarrow W_i$.

$T(x+y)=T(x)+T(y) \implies (T_i(x+y))_i=(T_i(x))_i+(T_i(y))_i=(T_i(x)+T_i(y))_i$ ---(1)

$T(ax)=aT(x)\implies (T_i(ax))_i=a(T_i(x))_i=(aT_i(x))_i$ [for some scalar $a$ and vector $x$] ---(2)

Hence each $T_i \in L(V,W_i)$. One has $\Phi:L(V,\prod_i W_i) \rightarrow \prod_i L(V,W_i) \ \ \text{defined as}\ \ T \mapsto (T_i)_i$.
Which can be shown as a linear transformation, by using (1) and (2). Clearly, $\Phi$ is an onto linear transformation.

$\ T \in \operatorname{Ker}\Phi \implies \Phi(T)=(O_i)_i$ [where $O_i:V \rightarrow W_i$ is the zero-transformation] $\implies (T_i)_i=(O_i)_i \implies T(x)=0 \ \  \text{for all x} \in V.$ Hence $\operatorname{Ker} \Phi = \{0\}$.

Hence $L(V,\prod_i W_i) \cong \prod_i L(V,W_i)$.

b) How do I proceed with this one?

What is a counter-example for $L(\prod W_i,V) \cong \prod_i L(W_i,V)$?

答:
I think it is very clear that for any family $\{f_\alpha\}_{\alpha}$ of linear maps $f_\alpha : W_\alpha \to V$ there is a unique linear map $f : \bigoplus_{\alpha} W_\alpha \to V$ such that $f \circ i_\beta = f_\beta$ for every $\beta$. Here, $i_\beta : W_\beta \to \bigoplus_{\alpha} W_\alpha$ sends $w_\beta$ to $(w_\alpha)_{\alpha}$, where $w_\alpha = 0$ if $\alpha \neq \beta$.

 Just define $f : \bigoplus_{\alpha} W_\alpha \to V$ by $(w_\alpha)_{\alpha} \mapsto \sum_{\alpha} f_\alpha(w_\alpha)$. Notice that this is well-defined since $\{\alpha : w_\alpha \neq 0\}$ is a finite set for every $(w_\alpha)_{\alpha} \in \bigoplus_{\alpha} W_\alpha$.

Thus, the rule $(f_\alpha)_{\alpha} \mapsto f$ is a one-to-one function from $\prod_{\alpha} \mathcal L(W_\alpha,V)$ onto $\mathcal L \big(\!\bigoplus_{\alpha} W_\alpha,V \big)$. Now, I will leave you to check that this is a linear map.


答:
The inflection points occur at

$$f''(x)=2\cos x-x\sin x=0$$ or $$y=2\cos x^*.$$

So we indeed have

$$x^2y^2+4y^2=4x^2\cos^2x+4x^2\sin^2x=4x^2.$$

$^*$This is another locus of the inflection points (magenta).

Title: Showing that the inflection points of $x\sin x$ lie on a certain curve

Body:
Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$

I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?


问:
Normal covering spaces - equivalent definitions for connected spaces
A covering space $p: Y \to X$ is normal when for all $ x \in X$ and for all $x_1', x_2' \in p^{-1}(x)$ there is a deck transformation $\phi$ with $x_2' = \phi(x_1')$.

I am asked to show that when $X$ is connected, then this condition is equivalent to saying that there exists one $x_0 \in X$ so that for all $x_1', x_2' \in p^{-1}(x_0)$, there exists a deck transformation taking $x_1'$ to $x_2'$.
One direction of this is easy, the other I am having trouble with.

I have thought about defining the set $A$ to be the set of points in $X$ for which there exist points in its fibre that aren't mapped to teach other by any deck transformation.
If I can show that $A$ is both open and closed, then I will be done.
However I am not sure how to show this.
Is this the correct approach, and if so, how should I continue?

答:
Say that $x\in X$ has property $(\star)$ if whenever $y_1,y_2\in p^{-1}(x)$ there is a deck transformation $\phi:Y\rightarrow Y$ with $\phi(y_1)=y_2$.

Suppose $x_0\in X$ has $(\star)$. Then any point $x$ contained in a neighbourhood $U\subseteq X$ of $x_0$ over which $p$ is trivial also has $(\star)$. If $V\subset X$ is a second open subset of $X$ over which $p$ is trivial and $U\cap V\neq\emptyset$, then there is a point $x\in V\cap U\subseteq V$ with $(\star)$, so by the above all points of $V$ have $(\star)$.

Now suppose that $U_1,\dots, U_n\subseteq X$ is a finite chain of open subsets such that $1)$ $x_0\in U_1$, $2)$ $U_i\cap U_{i+1}\neq\emptyset$ for each $i=1,\dots,{n-1}$, $3)$ $p$ is trivialisable over each $U_i$. By inducting on the previous observation we see that each point of each $U_i$ has $(\star)$, and in particular each point of $U_n$ has $(\star)$.

The basic idea is apparent. To complete we need to show how any two points of $X$ can be joined by a finite chain of trivialising open sets when it is connected.

For the details let $\mathcal{U}$ be any open covering of $X$. For $V\in\mathcal{U}$ put
$$\mathcal{U}(V)=\{W\in\mathcal{U}\mid \exists\, U_1,\dots,U_n\in\mathcal{U},\, V\cap U_1\neq\emptyset,\;W\cap U_n\neq\emptyset,\;U_i\cap U_{i+1}\neq\emptyset,\;\forall i=1,\dots,n-1\}$$
and write $\widetilde V=\bigcup_{U\in\mathcal{U}(V)}U$. Notice that if $V_1,V_2\in\mathcal{U}$, then $\widetilde V_1\cap\widetilde V_2\neq\emptyset$ if and only if $\mathcal{U}(V_1)=\mathcal{U}(V_2)$ if and only if $\widetilde V_1=\widetilde V_2$. Thus $\{\widetilde V\mid V\in\mathcal{U}\}$ is a covering of $X$ by pairwise-disjoint clopen sets.

Finally assume that $X$ is connected. We take $\mathcal{U}$ to be any covering of $X$ by open sets which trivialise $p$. The argument above shows that $\{\widetilde V\mid V\in\mathcal{U}\}$ contains the single set $X$. Thus any two points of $X$ are connected by a finite chain of sets in $\mathcal{U}$. Returning to the open paragraphs we see that if any point $x_0\in X$ has property $(\star)$, then so does every other point.


问:
Why do we use open intervals in most proofs and definitions?
In my class we  usually use intervals and balls in many proofs and definitions, but we almost never use closed intervals (for example, in Stokes Theorem, etc). On the other hand, many books use closed intervals.

Why is this preference? What would happen if we substituted "open" by "closed"?

答:
The truth of many theorems requires that the domain being considered is either open or closed. For example, a continuous function on a closed bounded set is always bounded and obtains its bounds. This is most certainly not true for continuous functions on open sets.

One major difference between open and closed sets is that any point in an open set $U$ is contained in some open neighbourhood which is a subset of $U$. Thus problems on any arbitrary open set can often be reduced to problems on simpler open sets. (In metric spaces, when working with a point in an open set we can always consider a small open ball that contains the point, and open balls are somehow "nice" sets that we know how to deal with.)

Another thing to bear in mind is the idea of closures and interiors. The closure of a set $A$ is the intersection of all closed sets containing $A$ and the interior of $A$ is the union of all open sets contained within $A$. It is fairly easy to see that $ \operatorname{interior}(A)\subset A\subset\operatorname{closure}(A)$. Thus if we work with a closed set $A$, we can see that it will contain some open set (Although this may be the empty set) and anything that was true for all points in $A$ will be true for all points in $\operatorname{interior}(A)$. For example, if we have a continuous function on a closed ball around a point, the function is still continuous if we restrict it to the the interior of the ball (which is an open set) and so we can apply anything we know about continuous functions on open sets. What we can't always do is extend a continuous function on an open set to its closure in a way that keeps it continuous.


问:
Prove that there is a natural injection $U \otimes V \rightarrow E \otimes F $
Hello everyone I found the following statement in my lecture :

Let $U$ a subset of $E$ and $V$ a subset of $F$ , so there is a natural injection $U \otimes V \rightarrow E \otimes F  $

But I don't exactly see why it's true , and what they mean with natural injection ? Any help would be a lot welcomed thanks in advance.

答: Clearly, the map $U \times V \to E \otimes F$ given by $(u,v) \mapsto u \otimes v$ is bilinear, so, by the universal property there is a linear map $U \otimes V \to E \otimes F$ such that $u \otimes v \mapsto u \otimes v$.


问:
Besov spaces: Regularization by convolution
Consider either a Besov space $B_{p,q}^s$ or a Triebel-Lizorkin space $F_{p,q}^s$ with parameters $p,q \in [1,\infty)$ and $s \in \mathbb{R}$. It is known that the Schwartz space $\mathcal{S}(\mathbb{R})$ is dense in $A_{p,q}^s$ for $A \in \{B,F\}$. I was looking for a "concrete" approximation. From my perspective, the most natural idea is to use convolutions: Fix $f \in A_{p,q}^s$ with $A \in \{B,F\}$. For $\chi \in C_c^{\infty}(\mathbb{R}^n)$ with $\chi \geq 0$ and $\int \chi(x) \, dx =1$, define $\chi_k(x) := k^d \chi(kx)$. Then

$$f_k := f* \chi_k$$

should converge to $f$ in $A_{p,q}^s$, right? Moreover, $f_k$ is a Schwartz function, and we are done. Does anybody know a reference for this result? I checked the books by Triebel and also the book by Sawano, but I couldn't find it (probably because it is "obvious").

答: That is the idea. For Besov spaces, this is the content of Proposition 17.12 of Giovanni Leoni's A First Course in Sobolev Spaces (or Proposition 14.5, if you're looking at the 1st edition).


问:
What are terms and functions in the first order logic?
I am trying to learn the syntax of the first order logic. On Wikipedia I see these rules:

Any variable is a term.

Any expression $f(t_1,...,t_n)$ of $n$ arguments (where each argument $t_i$ is a term and $f$ is a function symbol of valence $n$) is a term.

I understand variables as references to specific objects (for example Berlin, USA, John, Marry, Jupiter). I also understand functions as something that get variable as inputs and generate set of variables. For example: ChildrenOf(John, Marry) will returns a set of children of John and Marry and this set, according to the rule 2, is a term by itself. So, it can be used as an argument of a function. But how then the following function should be interpreted:

```<code>ChildrenOf(ChildrenOf(John, Marry), ChildrenOf(Bill, Anna))
</code>```

In other words, how should I evaluate functions if we use sets as their arguments? I can assume that in the above example the function should return all the people for whom the one of the parents is offspring of John and Marry and another parent is offspring of Bill and Anna but I am not sure that this is correct since it is not stated explicitly on Wikipedia.

答:
I also understand functions as something that get variable as inputs and generate set of variables.

That's incorrect: semantically, a function takes in a tuple of values and outputs a single value. This is present in the definition of a structure, where each function symbol is interpreted as a literal single-valued function on that structure. So we don't run into the problem you're worried about at all, and composition of functions is perfectly benign.

It's also not accurate to say that functions output variables: variables are specific types of symbols, and a string like "$f(x_1,x_2)$" is simply not a symbol of that type. But this is really a side issue.

Separately, a bit of intuition which might be helpful: you should think of a term as a noun, possibly a vague noun in case variables are involved (until we pick a variable assignment such a term doesn't pick out a specific element of a given structure) but a noun nonetheless. A function symbol by contrast is part of a noun phrase which begins with the word "the" to reflect its single-valuedness (e.g. "the smallest prime number greater than ---").


问:
How to evaluate $\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx$
It's a very simple question but it confuses me. How do I evaluate
$$
\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx
$$
without splitting? And why can't I split it?</p>

答:
You can easily compute the integral of $\frac{1}{x}$ and $\frac{1}{x + 1}$ by substitution.  Then, you get logarithmic function.  But it is not defined at $\infty$.  Note that:

$$\int\limits_1^\infty \frac{1}{x} dx - \int\limits_1^\infty \frac{1}{x + 1} dx$$
$$= \lim_{t \rightarrow \infty} \int\limits_1^t \frac{1}{x} dx - \int\limits_1^t \frac{1}{x + 1} dx$$

For the second integrand, use substitution, letting $u = x + 1 \rightarrow du = dx$.  Then:

$$\lim_{t \rightarrow \infty} \int\limits_1^t \frac{1}{x} dx - \int\limits_1^t \frac{1}{u} u$$
$$= \lim_{t \rightarrow \infty} \ln{x} - \ln{u}|_{x = 1}^t$$
$$= \lim_{t \rightarrow \infty} \ln{x} - \ln{x + 1}|_{x = 1}^t$$
$$= \lim_{t \rightarrow \infty} \ln{t} - \ln{1} - (\ln(t + 1) - \ln(1 + 1))$$

So if we substitute $t$ with $\infty$, then we obtain the diverging integral.


答:
Somewhere along the timeline you learned that exponential growth wipes out polynomial growth. So if we let $n=x^2,$ we are looking at

$$\lim_{x\to \infty}\frac{x^4}{7^x}.$$

This is polynomial growth divided by exponential growth, and the limit is $0.$

Title: How to find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$? I know that it exists and can't find it with Stolz theorem.

Body:
How do I find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$?

I thought about Stolz theorem so I got:
$$ \frac{(n+1)^2-n^2}{7^{\sqrt{n+1}}-7^{\sqrt{n}}}, $$
but it doesn't seem to lead anywhere (at least not through $e$ and $\ln$).

And I know for fact that there is some limit because:
$$\lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^2}{7^{\sqrt{n+1}}}}{\frac{n^2}{7^{\sqrt{n}}}} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n+1}}*n^2} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n}\sqrt{\frac{1}{n}+1}}*n^2} \implies \frac{(n+1)^2}{n^2}\implies 1$$


问:
About the proof of a graph is not Hamiltonian.
Given the following graph:

https://i.stack.imgur.com/fg2Q9.png

Is this graph Hamiltonian or not?

The answer is no. What I tried to prove is by using the fact that: "if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit." Following that we know, edges 11,6,4,12 must be in any circuit and the edges 17,18,19 and 9,15 must be in any circuit. Knowing that, what should I do to conclude that the graph is not Hamiltonian?

答:
Let G be any graph. If G is Hamiltonian, then κ(G) ≥ 2.

Since κ(G) = 1, with κ-set = {m},  then G must NOT be Hamiltonian.


问:
Translates by infinite subset of finite index subgroup
Given a countable group $\Gamma$ and a finite index subgroup $\Gamma_0$ and an infinite subset $S \subset \Gamma$ such that $S \cap \Gamma_0 = \emptyset$ can one conclude that $\Gamma_0 \cup S \Gamma_0 = \Gamma$?

答: Let $\Gamma = \mathbb{Z}$, $\Gamma_0 = 3\mathbb{Z}$, and $S = 3\mathbb{Z} + 1$.


问:
Prove that $f(x) = x^a$ is an injective function when a is a real number except 0.
The teacher said there's a mistake in my proof:

Let me start by the definition : $f$ would be an injective function if for any $x,y$ with $x\neq y$, $f(x) \neq f(y)$. $x$ and $y$ are in the domain of the function.

Let $x$ and $y$ be two real numbers where  $x \neq y$ and let's show that $x^a \neq y^a$ .

Suppose that $x^a=y^a$.

$\ln(x^a)=\ln(y^a)$

$a\ln(x)=a\ln(y)$.

This means that $x=y$.

End of proof

答:
Hint, not a complete answer.

If that's the end of the proof, then the proof would be correct for $a = 0$. But is the assertion true for $a = 0$? Why or why not? What did you forget?

Post-comment remarks
If a proof of a statement doesn't use the fact that $a \ne 0$, then the proof also proves the statement in the case where $a = 0$. (The proof leads from a set of assumptions to a conclusion; the only thing needed to justify the conclusion is the set of assumptions.)

But your proof goes from
$$
a \ln x = a \ln y
$$
(presumably, although it's omitted) to
$$
\ln x = \ln y
$$
presumably by "dividing through by $a$"; that operation is really "multiplying through by $a^{-1}$, which is valid only if $a^{-1}$ exists, which is true only for $a \ne 0$.

Anyhow, you've now added that assumption, so you can legitimately get to
$$
\ln x = \ln y.
$$
Now the question becomes "How, from knowing this, do you know that $x = y$?" You didn't provide any justification for that step, so your proof is incomplete.

There's an even earlier problem, which is that $x^a$ may not be well-defined when $x < 0$ and $a < 1$. What, for instance, is $(-1)^\frac12$?

And then (assuming that your teacher really meant to ask you this only for $a \ge 1$), there's the further problem that $\ln x$ isn't defined for $x \le 0$, so the very first step, where you take log of both sides...that's not valid either.

Writing complete proofs can be hard work. I strongly recommend two-column proof format until you get really good; it helps to keep you honest.

I suggest that you consider the following claim:

For $a \ne 0$, let
$$
f_a : \Bbb R^{+} \to \Bbb R^{+} : x \mapsto x^a.
$$
Then $f_a$ is injective as a function on the positive real numbers.

That claim is actually true, and a proof very similar to what you wrote will actually prove it.


问:
Expectation of a sufficient statistic, conditional on another sufficient statistic
A random variable $X$ is distributed with pdf $f(x,θ)$. $T(x)$ is a sufficient statistic for $θ,$ and $S(x)$ is a minimal sufficient statistic for $θ.$ The following is stated in my notes without explanation:
$$E_θ(T\mid S) = g(S)$$
for some function $g$ (independent of $θ$). $E_θ$ here refers to the fact that the expectation is a function of $θ.$

Intuitively this makes sense, as we should be able to express any function of $X$ (including $T(x)$) just as a function of $S,$ as it is sufficient.  However, how can I more formally show this is true, using the definition of sufficiency? For example, how could I show this is true by writing the expectation as a integral (or sum) over $X$?

答:
It is not true that every function of the data can be expressed as a function of the minimal sufficient statistic. For example, if $X_1,\ldots,X_n \sim \text{i.i.d.} \operatorname N(\mu,1)$ and $\mu\in\mathbb R$ indexes this family of distributions, then $X_1+\cdots + X_n$ is sufficient for this family of distributions, in the sense that the conditional distribution of $(X_1,\ldots,X_n)$ given $X_1+\cdots+X_n$ does not depend on $\mu,$ but clearly one cannot express $X_1$ as a function of this sufficient statistic.

But the definition implies the conditional distribution of $T$ given $S$ does not depend on $\theta,$ and from that it follows that the conditional expectation of $T$ given $S$ does not depend on $\theta.$


问:
Show that $0, 1, 2, 2^2,...,2^9$ form a complete residue system 11 module.
I have  question about module congruence

Show that $0, 1, 2, 2^2,...,2^9$ form a complete residue system 11 module.

I have no idea or where to begin to resolve this question.

Can anybody help me? Thanks.

答: Hint: $2^2 \not \equiv 1, 2^5 \not \equiv 1 \bmod 11 \implies 1, 2, 2^2,\dots,2^9$ are all different mod $11$.


问:
Find the limit as n approaches infinite
We have the following function:

$$U_n = \sin \dfrac{1}{3} n \pi$$

What is the limit of this function as n approaches infinity?

I first tried to use my calculator as help, for n I chose some arbitrary large numbers, such as 100 and 1000. Then I I just took $n = 10^{50}$ and it gave me an error.

So the correct answer is it doesn't have one, but why? Why does this function have a solution for $n = 10^2, 10^3$ and not for bigger numbers such as $n=10^{50}$?

答:
As David hinted in his comment, try using the fact that
$$
\sin(x + 2\pi k) = \sin(x)
$$
whenever $k$ is an integer.  Or, if you just write out the first few values of $U_n$ (compute these using the unit circle) and you should notice a pattern.</p>

Title: Find the limit as n approaches infinite


问:
Feller textbook question: Show that the (conditional) probability of a triple occupancy of some cells equals $\frac{1}{4}$
The question is already asked here . But there is no direct answer given there. It is a question from Feller's book Volume 1. According to the book, answer to the question is $\frac{1}{4}$. But my calculation never gets me 0.25.

If my calculation is right, answer to the question is

$$
\frac{ 5 \frac{7!}{3!1!1!1!1!} \frac{5!}{4!1!} }{5^7}
$$
which is 0.2688 which is not $\frac{1}{4}$.  Can anyone help me ?

答:
The numerator will be $\dbinom{7}{3,1,1}\times\frac{5!}{4!} = N, say $

The denominator will be $N + \left[\dbinom{7}{2,2,1,1,1}\times \frac{5!}{2!3!}\right]$

[The conditionality is on only 2 cells being empty, and the reference you gave mentioned that none of the remaining 5 cells were to be empty]


问:
Subgroups of free groups which avoid conjugacy classes
Let $G = (\mathbb Z/2\mathbb Z)^{\ast m}$ be a free product of some groups of order $2$.  Let $\alpha_1,\ldots,\alpha_m$ be the generators.

Can I find a free, nonabelian subgroup of $G$ that has no nontrivial elements conjugate to any $\alpha_i \alpha_j$? How can I prove it?

答:
As Lee Mosher says this is not possible if $m \le 2$. If $m \ge 3$ we can argue a bit differently as follows. $G$ is residually finite (proof), so we can find a normal subgroup $N$ of finite index not containing any finite set of non-identity elements, in particular the set $\{ \alpha_i \alpha_j \}$. Since $N$ is normal it doesn't contain any conjugates of those elements either. It remains to show that $N$ contains a free nonabelian subgroup.

By the Kurosh subgroup theorem, $N$ is a free product of finitely many copies of $\mathbb{Z}$ and $\mathbb{Z}/2$. It has a natural map to the direct product of the copies of $\mathbb{Z}/2$ only, whose kernel is a normal subgroup $N'$ of finite index which is free (this follows from some stuff about coverings of graphs of groups, or equivalently a slightly more precise form of the Kurosh subgroup theorem). Since $N'$ has finite index in $G$ it must be nonabelian (this is where we use the hypothesis that $m \ge 3$), e.g. because $G$ is not virtually abelian, or using the fact that the orbifold Euler characteristic $\chi(G) = \frac{m}{2} - (m-1) = 1 - \frac{m}{2}$ is negative.


问:
Additive inverse notation confusion: $-2^2$ vs $(-2)^2$
My textbook says that the additive inverse of a number is represented by $-$ (minus sign).
For example the additive inverse of a number $-1$ is typed like this: $-(-1)$, and the additive inverse of a number $1$ is then typed like this; $-1$.

In the book there are also some example calculations and this is what confuses me:

$$-2^2 = -4\quad\text{and}\quad (-2)^2 = 4$$

I've always thought $-n^2$ simply like $(-n)\times(-n)$. But according to my textbook this is wrong and $-n^2$ actually means $-(n\times n)$

I tried to wrap my head around this and played with it for a while and found out that when i type $-2^2$ to the Microsoft Windows calculator, I'll get the (wrong?) answer of $4$. But when I type this to my calculator (Casio fx-991EX) I'll get the answer of $-4$.

I think many people will just write $-2^2$ when they actually mean $(-2)\times(-2)$ and depending how you'll look at the minus sign gives you two completely different answers.

So which one is right? Is the minus sign really a notation for additive inverse of the operation.

Which is the correct answer?

$$-2^2 = -4\quad\text{or}\quad-2^2 = 4$$

答:
As stated by others, this simply boils down to order of operations (you can use your favourite acronym to help you, such as PEDMAS, or in the UK: BIDMAS).

So with ${-2^2}$, we have no Parenthesis, but we do have an Exponent, so we calculate this part first. ${2^2 = 4}$. Hence ${-2^2=-4}$.

With ${(-2)^2}$, we do have Parenthesis, then an Exponent - so ${(-2)^2 = (-2)(-2)=+4}$.

I'd like to mention this is all just convention. We could change the meaning of ${-2^2}$ to mean ${(-2)(-2)=+4}$, but the universal convention is to follow PEDMAS (or BIDMAS) since that's what everyone agrees upon. Otherwise it'd be very difficult to people to understand each other's calculations. We just need a set of rules such that given any expression - two people that try to evaluate it following the same set of rules will arrive to the same answer (so there is no ambiguity). Overtime - you will become so  used to the order of operations, you won't need to use PEDMAS or BIDMAS any longer, and you will have a natural understanding of an expression when you see it.


问:
Expectation and Variance of a Voting Problem
In a hotly contested presidential election, suppose the outcome is to be determined by the results in 4 battleground states: Arizona, Georgia, Pennsylvania, and Wisconsin, with 11,16, 20, and 10 electoral votes, respectively. One of the candidates, Mr. B, has already gathered 249 electoral votes elsewhere. He needs an additional 21 electoral votes from these 4 battleground states to win the election. Mr. B is equally likely to win or lose each of these 4 battle ground states, and the outcomes in different battleground states are independent. Let N denote the number of states among these 4 that Mr. B will win.

Let M denote the number of electoral votes Mr. B will win from these 4 battleground states. Compute the mean and variance of M.

I know there are 16 possible total votes Mr. B can get (0, 10, 11, 16, 20, ...). Is each total equally likely? I got my mean, E(M) = 456/16 = 28.5 but I'm not sure if this is the right approach.

答: Yes, given that Mr. B is equally likely to win or lose and since the outcomes are independent, it does translate to all 16 possibilities being equally likely.</p>


问:
Two closed-form analytical solutions for multivariate Gaussian entropy: How are they equal?
Entropy of a multivariate Normal distribution is well-known to be

$$H(\boldsymbol{X}) = \frac{N}{2}\ln2\pi e\left(\prod_{i=1}^{N}\sigma_i^2\right)^\frac{1}{N}$$

but here, it is also shown to equal

$$H(\boldsymbol{X})  =\frac{N}{2}\ln\left(2\pi e\right)+\frac{1}{2}\ln\det(\Sigma)$$
where $\Sigma$ is the sample covariance matrix.

How to show that the two formulas equate to each other? and which is more of the complete simplification of multivariate Gaussian entropy?

答:
I assume that $X=AZ$ where $A$ is a constant matrix and $Z$ is standard Normal distributed, $Z\sim\mathcal{N}(0,I)$. In this case, the covariance of $X$ is $\Sigma$. It results: $$\det\Sigma=\sigma_1^2(A) \sigma_2^2(A) \cdots \sigma_n^2(A)=\det(AA^T).$$

The derivation is as follows:

\begin{align} h(X)
& = \frac{N}{2}\log \left(2\pi e\left(\prod_{i=1}^{N}\sigma_i^2\right)^\frac{1}{N}\right)\\
& = \frac{N}{2}\log \left(2\pi e\right) + \frac{N}{2}\log \left(\left(\prod_{i=1}^{N}\sigma_i^2\right)^\frac{1}{N}\right)\\
& = \frac{N}{2}\log \left(2\pi e\right) + \frac{1}{2}\log \left(\prod_{i=1}^{N}\sigma_i^2\right)\\
& = \frac{N}{2}\log \left(2\pi e\right) + \frac{1}{2}\log \det\Sigma.
\end{align}

It is more common to write the second one. It is even more common to put the first term inside the second one as follows:
$$ h(X) = \frac{1}{2}\log\det(2\pi e\Sigma). $$


问:
The function given by the Rao-Blackwell theorem is a statistic
$(X_1,\ldots,X_n)$ is a random sample, $V_n$ is an unbiased estimator of the population parameter $\theta$ and $T_n$ is a sufficient statistic for $\theta$. Then by Rao-Blackwell theorem the rv

$$\varphi(T_n):=\mathbb E_\theta[V_n\mid T_n]$$

is an unbiased estimator of $\theta$ whic is uniformly better than $V_n$.

I know the definitions of conditional expectation and distribution, but I can't get how $\varphi(T_n)$ is an estimator (a function of the random sample that does't depend on $\theta$) using sufficiency

答:
The definition of sufficiency is that the conditional distribution of all of the data given the value of the sufficient statistic does not depend on $\theta,$ i.e. it is the same for all distributions in the family.

If the conditional distribution does not depend on $\theta,$ then the conditional expectation does not depend on $\theta.$


问:
If polynomial $P(x)$ has integer coefficients and at least three integer roots, then $P(x)+5^m$ has no more than one integer root for $m\geq 1$
I've been doing some polynomial excersises lately and in that one I got completly stuck.

Let $m \geqslant 1$ be natural number and $P(x)$ polynomial with integer coefficients which has at least three different integer roots. Prove that $P(x)+5^m$ has no more than one integer root.

At first I considered the easiest case: $(x-x_1)(x-x_2)(x-x_3)+5$, but it did not turned out in anything helpful, so I am seeking for some clues on how to crack that problem.

Also, I'd like to ask for as elementary hint/solution as possible since this question is from (inactive) high school contest.
https://om.mimuw.edu.pl/static/app_main/problems/om48_1.pdf

答:
Let's look at things backwards: Suppose $Q(x)$ has integer coefficients and two integer roots; we need to show that $Q(x)-5^m$ cannot have three integer roots.

By translation, we may assume that one of the roots of $Q$ is positive and the other is at $x=0$. Let's let the positive root be at $x=n$, so that $Q(x)=x(x-n)R(x)$. Now if $Q(r)=5^m$, then both $|r|$ and $|r-n|$ are powers of $5$, say $r=5^a\sigma$ and $n-r=5^b\tau$, with $\sigma,\tau\in\{1,-1\}$. This implies $n=5^a\sigma+5^b\tau$, with a $+1$ for the sign of the larger power (since $n\gt0$). But no positive number $n$ can be written as a sum or difference of two powers of $5$ in more than one way (most cannot be written as a sum or difference of two powers of $5$ at all), so there are at most two possibilities for $r$ (i.e., either the larger power or the smaller power).


答:
Repeated multiplication by $2$ mod $11$ shows that the remainders of those powers of $2$

when divided by $11$
are distinct:  $1, 2,4,8,5,10,9,7,3,6$.  (To get the next number in this series,

simply multiply by $2$ and subtract $11$ if the result is more than $11$.)

Powers of $2$ cycle through all non-zero residues mod $11$; i.e., $2$ is a primitive root modulo $11$.

In contrast, $3$ is not a primitive root modulo $11$; the powers of $3$ mod $11$ are   $1,3,9,5,4$,

and then that cycle repeats.  (To get the next number in that series, multiply by $3$

and then subtract $11$ or $22$ if needed to get a number less than $11$.)

Body:
I have  question about module congruence

Show that $0, 1, 2, 2^2,...,2^9$ form a complete residue system 11 module.

I have no idea or where to begin to resolve this question.

Can anybody help me? Thanks.


问:
problem from PDEs, H. Brezis
today I read book Sobolev space, PDEs of H.Brezis, and when I read chapter 8, I don't know why following remark is easy:

Remark 11 (page 214). Let $I$ be a bounded interval, let $1\leq p\leq \infty$, and et $1\leq q\leq \infty$, then

$$|||u|||=||u'||_p+||u||_q$$

is equivalent to the norm of $W^{1,p}(I)$.

I know $$||u||_{W^{1,p}(I)}=||u||_p+||u'||_p$$
By Sobolev embedding: there exists a constant $C$ such that $$||u||_q=\left(\int_I |u|^q\right)^\frac{1}{q}\leq \left(\int_I ||u||_\infty^q\right)^\frac{1}{q}=I^\frac{1}{q} ||u||_\infty\leq CI^\frac{1}{q} ||u||_{W^{1,p}(I)}$$
and
$$||u'||_p\leq ||u||_{W^{1,p}(I)}$$
so
$$|||u|||=||u'||_p+||u||_q\leq (CI^\frac{1}{q} +1) ||u||_{W^{1,p}(I)}$$

But I can't find $\alpha$ such that
$$|||u|||\geq \alpha ||u||_{W^{1,p}(I)}$$
Can anyone help me?

答:
We can (for example) consider two cases:

Case 1 - $q> p$

We have that $\|u\|_p^p=\int_I |u|^p$. Because $q>p$, we have that $\frac{q}{p}>1$, hence we can apply Holder inequality, to conclude that $\|u\|_p^p\leq C \|u\|_q^{p}$, where $C>0$ is a positive constant (that can change in every line). From it we have that $\|u\|_{1,p}=\|u\|_p+\|u'\|_p\leq C\|u\|_q+\|u'\|_p\leq C\||u\||$

Case 2- $q\leq p$

In this case $W^{1,p}(I)$ is embedded in $W^{1,q}(I)$, hence $\|u\|_p=\Big(\int_I|u|^p\Big)^{1/p}\leq C\|u\|_\infty\leq C\|u\|_{1,q}$. From here you can conclude.

Title: problem from PDEs, H. Brezis


问:
Integral domain $A $ such that $A\to K(A)$ is epimorphism
Sorry for my bad English.
I want to the naming for integral domain which have next property.

$A$ is commutative domain, and $K(A)$ is fraction of $A$. Then, natural inclusion $A\to K(A)$ is epimorphism in category of commutative rings.

For example, ring of integers $\mathbb Z$ have this property.

If this name exists, please tell me thanks.

答: For any commutative ring $R$ and any multiplicatively closed subset $S\subset R$, the morphism $R \to S^{-1}R$ is an epimorphism. In particular, this holds for any integral domain $A$ with $S=A\setminus \{0\}$ and $S^{-1}A=K(A)$


问:
Can we always find $x_n\to\infty$ with $f_n(x_n)\to0$ if the $(f_n)$ are non-decreasing and pointwise convergent to zero?
Suppose we have a sequence of functions $(f_n)$ with each $f_n:[0, \infty) \to \mathbb{R}$, each $f_n$ is monotone non-decreasing and $f_n\to 0$ pointwise. Is it always possible to choose a sequence $(x_n)$ with $x_n\to \infty$ slow enough such that we still have $f_n(x_n) \to 0$? If not, is this possible with additional hypotheses on the $(f_n)$ (e.g. continuity perhaps)?

答:
I believe the answer to be Yes.
Do you want $(x_n)_n$ to be strictly increasing or not necessarily?

If not, it suffices to only try to construct the sequence via the integers, i.e. $\forall n, \space x_n \in \mathbb{N} .$ It is therefore sufficient to look at $f_n|_{\mathbb N}=\hat f_n: \mathbb N \to \mathbb R$.

Let us look at the auxiliary functions $\hat f_n^k(x) = \chi_{x\le k} \hat f_n(x) +  \chi_{x> k}\hat f_n(k)$ where $\chi_{(.)}$ is the indicator function. One can see that for fixed $k$, $\hat f_n^k \to \hat f^k (\equiv 0)$ uniformly in $n$.

Namely given a pair of positive integers $(m,k)$, we can find a positive integer $n(m,k)$ such that for all $n\ge n(m,k)$ we have $\|\hat f_n^k\| \le \frac{1}{m} $. Now we can choose, in an increasing way, a sequence $(t_r)_{r \in \mathbb N} =(r+\max\{ n(1,1),\cdots, n(r,r)\})_r$ so that $t_r \to \infty$ and increases strictily. Finally, choose $x_n = r$ for $n \in [t_r,t_{r+1})$.

Remark: The construction may look difficult, but it's simply "waiting at the k-th step until you are sure that the (k+1)-st step becomes small enough, then stepping on the (k+1)-st one"

Now if you want $(x_n)$ to be strictly increasing, I feel like a key insight can be drawn from

Lusin-Egorov's theorem: If $f_n$ converges almost surely to $f$ (which is the case here with $f=0$) then, on every compact interval $I$ and for any $\varepsilon>0$ there exists a compact subset $I_\varepsilon \subset I$ with $\lambda(I_\varepsilon)\ge \lambda(I) - \varepsilon$ and $f_n|_{I_\varepsilon} \to f|_{I_\varepsilon}$ uniformly.

Choosing $\varepsilon<1$ and approaching the problem as above should yield a solution.


问:
In how many ways can 5 objects be placed in 3 groups?
Consider this question:

In how many ways can $5$ tourists be placed in $3$ different hotels, such that there is at least one tourist in each hotel?

To start working on this, I removed the lower bound on the number of tourists in a hotel. I reasoned that any tourist can be placed in any of the $3$ hotels, resulting in $5\times 3=15$ permutations.

However, the correct answer with the lower bound is $150$, which is more than $15$.

Obviously, I am undercounting by a large amount. How?

答:
You need to decide the breakup of men into 3 groups and which hotel to place them.

You could solve it rather simply using multimonials  as $\dbinom{5}{2,2,1}\dbinom {3}{2,1} + \dbinom{5}{3,1,1}\dbinom{3}{2,1}$


问:
How to prove that eigenfunction of translationally invariant continuous operator $ K(t-t') $ is $ \exp(iwt) $?
I was studying a book in computational neuroscience where I came across the following equations:
$$ \int{W(t,t')e(t')dt'}=\lambda e(t) $$ and read that if $ W(t,t')=K(t-t') $ then the eigenfunction $ e(t) $ is a complex exponential. Could someone explain why?

答:
If $T$ is a linear map, say $C^0(\Bbb{R})\to L^1_{loc}(\Bbb{R})$, such that for all $g,a$, $$T(g(.+a)) = (Tg)(.+a)$$ then every exponential $e^{st}, s\in \Bbb{C}$ is an eigenfunction.

If $\|T(g)\|_2\le C \|g\|_2$ then $T$ is said bounded in $L^2$ norm.

In that case let $H$ such that $T(e^{2i\pi y t}) =H(y)e^{2i\pi yt}$ for $y \in \Bbb{R}$. Then $T(g)= \mathcal{F}^{-1}(H.\mathcal{F}(g))$ (the Fourier transform).

It is hard to make it rigorous (this is the point of functional analysis) but you can think to the Fourier transform as diagonalizing $T$ (in a way similar to the diagonalization of a self-adjoint matrix).

If $H(y_1)=H(y_2)$ then $e^{2i\pi y_1 t}+e^{2i\pi y_2 t}$ is a non-exponential eigenfunction.


问:
Identical distributions
I am wondering if the following holds/can be proven: Let $X_1, \ldots, X_n$ be iid copies of a random variable $X$, and let $Y_1, \ldots, Y_n$ be iid copies of a random variable $Y$. $X$ and $Y$ are also assumed to be independent. Then,
$$A=\Big(\sum_{i=1}^n X_i\Big)\,\cdot\,\Big(\sum_{i=1}^n Y_i\Big)$$
is identically distributed as
$$B=n\sum_{i=1}^n X_i Y_i.$$
Perhaps a look at characteristic functions can be conclusive?

It holds $A=\sum_{i=1}^n \sum_{j=1}^n X_i Y_{i+j\text{ mod }n}$, where $\sum_{i=1}^n X_i Y_{i+j\text{ mod }n}$ for $j=1,\ldots, n$ clearly follow the same distribution, but are not independent from each other. Any suggestions? Thanks!

答:
This is false. The two random variables have the same mean, namely $n^2 \mathbb{E}(X) \mathbb{E}(Y)$, so assume WLOG that $\mathbb{E}(X) = \mathbb{E}(Y) = 0$ and let's compute the variance. Then

$$\mathbb{E}(A^2) = \mathbb{E} \left( \left( \sum X_i \right)^2 \right) \mathbb{E} \left( \left( \sum Y_i \right)^2 \right) = n^2 \text{Var}(X) \text{Var}(Y)$$

but

$$\mathbb{E}(B^2) = n^2 \mathbb{E} \left( \left( \sum X_i Y_i \right)^2 \right) = n^3 \text{Var}(X) \text{Var}(Y).$$

Title: Identical distributions


问:
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$
I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks</p>

答:
HINT:

Using Partial Fraction Decomposition  formula,

$$\frac{x^4+2x+4}{x^4-1}=1+\frac{ax+b}{x^2+1}+\frac c{x+1}+\frac d{x-1}$$ where $a,b,c,d$ are arbitrary constants to determined by equating the coefficients of the different powers of $x$ in

$$x^4+2x+4=x^4-1+(x^2-1)(ax+b)+c(x-1)(x^2+1)+d(x+1)(x^2+1)$$


问:
Roots of a polynomial must be real
Say $P$ is a polynomial of degree $n$ with $n$ real roots. I have to prove that the polynomial $Q(z)=P(z+i)+P(z-i)$ has $n$ real roots.

First it happens that for a real number $t$ and because $P$ is real polynomial we have $$\overline{Q}(t)= \overline{Q(t)} =\overline{P(t+i)}+\overline{P(t-i)} \\
 = P(\overline{t+i})+P(\overline{t-i}) \\
= P(t-i)+P(t+i) = Q(t) $$
The polynomial $Q-\bar{Q}$ has infinite roots (namely all the real numbers) and thus $Q$ is a real polynomial

Truth is I don't know how to proceed further. This exercice was given at the beginning of a complex analysis course so I don't know if I must search for an analytical argument or a simpler one. If you can just give me a hint (I want to solve the problem by myself) that would be really helpful

答:
Hint: If $t$ is real, what is the complex conjugate of $t+i$. What does that tell you about $Q$?

Hint 2: This is a heuristic hint. If you know, or assume, that a polynomial has only real roots, then you can assume the polynomial can be put in the for of a product of monomial factors, each of the form $(x-r_i)$, where r_i is a real root of the polynomial.
$$a_n\prod_{i=1}^n(x-r_i)$$

The heuristic hint is this. You will exclude all of the polynomials that have complex roots simply by using that form rather than the form:
$$a_0 +\sum_{i=1}^n a_nx^n$$

Hint 3: If $Q(z)= 0$, with $z\in\mathbb{C}$, what can one say about $|P(z+i)|$ as it relates to $|P(z-i)|$?

Title: Roots of a polynomial must be real


问:
Prove that if $a \neq b$ then $(a+b)/2 > \sqrt{ab}$
I have to prove by contraposition that if $a \neq b$, then $(a + b)/2 > \sqrt{ab}$, where $a$ and $b$ are both positive real numbers.

So I need to prove that if $(a + b)/2 \leq \sqrt{ab}$, then $a=b$.

\begin{align}
\left(\frac{a+b}{2}\right)^2 &\leq ab \\
(a + b)^2 &\leq 4ab
\end{align}

I don't know if I started correctly, or what I should do next to get $a=b$.

Body:
Hint:

$(a+b)^2-4ab=(a-b)^2$.


答:
$(a+b)^2\le 4ab$

$a^2+2ab+b^2\le 4ab$

$a^2-2ab+b^2\le 0$

$(a-b)^2\le 0$

Can you see it from here?

Title: Prove that if $a \neq b$ then $(a+b)/2 > \sqrt{ab}$

Body:
I have to prove by contraposition that if $a \neq b$, then $(a + b)/2 > \sqrt{ab}$, where $a$ and $b$ are both positive real numbers.

So I need to prove that if $(a + b)/2 \leq \sqrt{ab}$, then $a=b$.

\begin{align}
\left(\frac{a+b}{2}\right)^2 &\leq ab \\
(a + b)^2 &\leq 4ab
\end{align}

I don't know if I started correctly, or what I should do next to get $a=b$.


问:
Is this version of the Gaussian distribution function correct?
I am currently doing an exercise in which the following is said:

Let $f(x, t)$ be the Gaussian distribution function with variance $t$:
$$
          f(x, t) = \frac{1}{\sqrt{2 \pi t}} e^{-x^2/2t} $$

But when I look at https://en.wikipedia.org/wiki/Normal_distribution the function seems to be different :/ For example, $t$(the variance) is not in the square root with $2 \pi$ in the Wiki link :/

答: Write your $t$ as $t=\sigma ^2$ and you will see that both are equivalent


答:
You are implicitly assuming that:

$$t = \sigma^2.$$

Therefore:

$$\frac{1}{\sqrt{2 \pi t}} e^{-x^2/2t} = \frac{1}{\sqrt{2 \pi}~~ \sigma} e^{-x^2/2\sigma^2}.$$

Title: Is this version of the Gaussian distribution function correct?

Body:
I am currently doing an exercise in which the following is said:

Let $f(x, t)$ be the Gaussian distribution function with variance $t$:
$$
          f(x, t) = \frac{1}{\sqrt{2 \pi t}} e^{-x^2/2t} $$

But when I look at https://en.wikipedia.org/wiki/Normal_distribution the function seems to be different :/ For example, $t$(the variance) is not in the square root with $2 \pi$ in the Wiki link :/


问:
If $\,\lim_{p\to\infty} \sum_{j=1}^n a_j\cos(p\pi \theta_j) \to 0,\,$ then $\,a_1 = \dots = a_n = 0$.
The exercise

Let $n \geq 1$, $0 < \theta_1 < \dots < \theta_n < 1$ and $a_1, \dots, a_n \in \mathbb{R}^n$. Assume $ \sum_{j=1}^n a_j\cos(\pi p \theta_j) \underset{p \to +\infty}{\longrightarrow} 0$ ($p \in \mathbb{N}$). Show $a_1 = \dots = a_n = 0$.

My try

I tried to prove this by induction over $n$. I succeeded for $n = 1$, but I have no clue to show how the property can be used for $n \geq 2$.

答:
We shall prove something slightly more general:

If $\,0<\theta_1<\cdots<\theta_n<1$ and
$$
\lim_{p\to\infty}\sum_{j=1}^n \big(a_j\cos(p\pi \theta_j)+b_j\sin(p\pi \theta_j)\big)=0,
$$
then $a_1=b_1=a_2=b_2\cdots=a_n=b_n=0$.

Clearly
$$
\sum_{j=1}^n \big(a_j\cos(p\pi \theta_j)+b_j\sin(p\pi \theta_j)\big)
=\sum_{j=1}^n c_j\mathrm{e}^{ip\pi\theta_j}+
\sum_{j=1}^n \overline{c}_j\mathrm{e}^{-ip\pi\theta_j},
$$
where
$c_j=\frac{a_j-ib_j}{2}$, and hence it suffices to show that $c_1=\cdots=c_n=0$.

Note that the $2n$ complex numbers
$$
\mathrm{e}^{i\pi\theta_j},\mathrm{e}^{-i\pi\theta_j},\quad j=1,\ldots,n,
$$
and distinct. Each of them of absolute value 1.

Hence it suffices to show the following:

If $w_j\in\mathbb C$, $j=1,\ldots,m$, are distinct complex numbers, with $|w_j|=1$, $j=1,\ldots,m$, and
$$
\lim_{p\to\infty}\sum_{j=1}^m d_jw_j^p=0,
$$
then $d_1=\cdots=d_m=0$.

This can be shown inductively on $m$. For $m=1$ it is obvious. Assume that it is true for $m=k-1$ and we have that $\,\lim_{p\to\infty}\sum_{j=1}^k d_jw_j^p=0$. Then
$$
0=w_k\lim_{p\to\infty}\sum_{j=1}^k d_jw_j^p-\lim_{p\to\infty}\sum_{j=1}^k d_jw_j^{p+1}=
\lim_{p\to\infty}\left(w_k\sum_{j=1}^k d_jw_j^p-\sum_{j=1}^k d_jw_j^{p+1}\right) \\
=\lim_{p\to\infty} \sum_{j=1}^{m-1} d_j(w_k-w_j)w_j^p.
$$
Using the induction hypothesis, we obtain that
$$
d_j(w_k-w_j)=0, \quad j=1,\ldots,k-1
$$
and hence $d_j=0$, $j=1,\ldots,k-1$, which implies also that $d_k=0$.


答:
welcome to math.stackexchange this question were answered already.

Here is the link

use polynomial division, we get $$\int \frac{x^4+2x+4}{x^4-1} dx = \int 1 + \frac{2x+5}{(x^2 - 1)(x^2 + 1)}dx
 = \int 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} dx  $$

Expressing this as partial fractions, we need only find $A, B, C$

$$=  \int \left(1 + \frac{A}{x+1} + \frac B{x-1} +\frac{CX+D}{x^2 + 1}\right)\,dx$$

Title: Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$

Body:
I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks</p>


问:
proove a 3 digit block is divisible by 37

Let $a$ be a positive integer.  If $S(a)$ is the sum of 3-digit part of $a$, then prove $a$ is divisible by 37, if and only if $S(a)$ is divisible by 37.

I mention a, is the sum of 3 digit parts of a. For example, $a=5987654321$, then $S(a)=5 + 987 + 654 + 321 =1967$. I think this is very similar to the divisibility by 3, like all the digits sum is divisible by 3 then the number is divisible by 3. I think it can use decimal representation and then prove like that. And use congruences to help proove it. Can someone help judge if this is right?

答:
You're right.

$a_0+a_1\times10+a_2\times10^2+a_3\times10^3+\cdots+a_n\times10^n$

$\equiv a_0+a_1+a_2+a_3+\cdots+a_n\bmod3,$ because $1\equiv10\equiv10^2\equiv10^3\equiv\cdots\equiv10^n\bmod 3$,

and $b_0+b_1\times1000+b_2\times1000^2+b_3\times1000^3+\cdots+b_n\times1000^n$

$\equiv b_0+b_1+b_2+b_3+\cdots+b_n\bmod37$,

because $1\equiv1000\equiv1000^2\equiv1000^3\equiv\cdots\equiv1000^n\bmod37$.

($1000=999+1=27\times37+1\equiv1\bmod 37$.)


问:
Need help finding a bounding function to apply dominated convergence theorem
Let $f,g\in L^1(\mathbb{R})$. Show that  if $f$ is continuous and bounded, then so is $f*g$, the convolution of $f$ with $g$.

I'm showing continuity of $f*g$ using convergent sequences.

Letting $x_n\to x$, it follows that $f(y-x_n)g(y)$ is a function of $y$ that converges to $f(y-x)g(y)$ (by continuity of $f$). Therefore $$\lim_{n\to\infty}(f*g)(x_n)=\lim_{n\to\infty}\int f(y-x_n)g(y) dm.$$

I would like to continue by saying that $$\lim_{n\to\infty}\int f(y-x_n)g(y) dm = \int\lim_{n\to\infty} f(y-x_n)g(y) dm = \int f(y-x)g(y)dm = (f*g)(x),$$

but I need an integrable function that dominates each $f(y-x_n)g(y)$ before I can pass the limit under the integral.

Thanks in advance for sharing any thoughts.

答: $f$ is given to be bounded. If $|f| \leq M$ then $|f(y-x_n)g(y)| \leq M|g(y)|$ and $M|g|$ is integrable.


问:
How to use the Fourier transform for show that $ \int_{0}^{\infty} \frac{sin^2(x)}{x^2} dx = \frac{\pi}{2} $?
Suppose
$$ \int_{0}^{\infty} f(x)\cos(\alpha x)dx = F(\alpha) = \left\{ \begin{array}{lcc}
             1-\alpha &   if  & 0\le \alpha \le 1 \\
             \\ \hspace{0.6cm} 0 &  if & \alpha>1 \\
             \end{array}
   \right.
$$
$$\mathcal{F}(F(\alpha) = \frac{2}{\pi} \int_{0}^{\infty} F(\alpha)\cos(\alpha x) d\alpha = \int_{0}^{1} (1-\alpha)\cos(\alpha x) d\alpha = \frac{1-\cos(x)}{x^2}  $$

How to use this for show that  $ \int_{0}^{\infty} \frac{\sin^2(x)}{x^2} dx = \frac{\pi}{2} $ ?

Body:
$$ \mathcal{F}(F(\alpha) = \frac{1-\cos(x)}{x^2} = \frac{4\sin^2(\frac{x}{2})}{\pi x^2} $$

Let $\alpha = 0$

$$ \int_{0}^{\infty} \frac{4\sin^2(\frac{x}{2})}{\pi x^2} \ cos(0) dx = 1 $$

$$ t=x/2 \Rightarrow x=2t \Rightarrow \frac{dx}{dt} = 2 $$

$$ 2 \int_{0}^{\infty} \frac{4\sin^2(t)}{4 \pi t^2} dt = 1 \iff \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = 1 \iff \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = \frac{\pi}{2} $$


问:
Find a basis for the kernel of $u \otimes v$ and matrix $u \otimes v$
Hello everyone I have the two following questions left without any correction and I'm having a hard time resolving it if anyone could help me  it would be a lot appreciated.

Let $E,F,G,H$ vector space with finite dimension. We have $u \in \mathcal L(E,G)$ and $v \in \mathcal L(F,H)$ , (where $\mathcal L(F,H)$ is the linear mapping $F \rightarrow H$).
We define $e_1 , ... , e_n $ as a basis for $E$ , $f_1 , ... , f_p$ a basis for $F$

We suppose that the basis of their kernel are respectively :  $e_1 , ... e_k$ and $f_1 , ... , f_l$ .

Find the basis of the kernel for :  $u \otimes v$ ?

Let $(a_{ij})$  the matrix of $u$ in the basis $e_j$ and $g_i$  , Let $(b_{kl})$ the matrix of $v$ in the basis $f_l$ and $h_k$

Find the matrix of $u\otimes v$ in the basis $e_j \otimes f_l$ and $g_i\otimes h_k$?

Thanks in advance for your help.

答:
For the first problem: we will have $\ker(u \otimes v) = \ker(u) \otimes F + E \otimes \ker(g)$. One way to produce a basis is as follows:

Extend $e_1,\dots,e_k$ into a basis $e_1,\dots,e_n$ of $E$, and extend $f_1,\dots,f_l$ into a basis $f_1,\dots,f_p$ of $F$. The set $\{e_i \otimes f_j : 1 \leq i \leq k \text{ or } 1 \leq j \leq l\}$ is a basis for $\ker(u \otimes v)$.

For the second problem: keep in mind that the order of the basis is important. Suppose that we order $\{e_j \otimes f_l\}$ so that the tuples $(j,l)$ are taken in lexicographical order. Suppose that $\{g_i \otimes h_k\}$ is ordered in the same way. The matrix of $u \otimes v$ relative to these basis will be the Kronecker product $A \otimes B$.


问:
The map $\frac1{1-z}$
I want to find the image of the unit circle under the map $\frac1{1-z}$. I first rewrote this as
$$
\frac{1+z}{1-z^2}
$$
But on the unit circle, $z^2 = 1$. So I get $\infty$ for the image of all points on the unit circle as the result. But I thought linear fractional always map circles to lines or circles. Can anyone point out what I did wrong?

EDIT: It turns out that I was confusing $z^2$ with $\bar{z}$, so not all points on the unit circle satisfy $z^2 = 1$. I should let $z = x + iy$ and plug that into the map instead.

答:
There are sort of three things going on here:  $1) z\mapsto-z,2)z\mapsto z+1$ and $3)z\mapsto \dfrac1z$.  The first is reflection about the origin, the second translation, and the third inversion.

But personally I prefer to use the method of checking what happens to $3$ points, since generalized circles get sent to generalized circles by Mobius transformations.

Title: The map $\frac1{1-z}$


问:
$(g(Y)+Z)|Y=y \sim g(y)+Z$
I'm self learning probability and I have a trouble with the following problem: let $Y,Z$ be continuous independent real random vectors in $\mathbb{R}^n$, $\mathbb{R}^m$ respectively. Let $g:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be $C^1$. Find the distribution of $X|Y=y$, where $X=g(Y)+Z$.

Context: I'm particularly interested in the case where $Z$ is some gaussian noise, this kind of model is useful in robotic probabilistic state estimation.

Guess: $X|Y=y \sim Z+g(y)$, but how can I prove this?

My try: one should

Prove $X$ is a continuous random vector, i.e., it admits a continuous density

Supposing point 1, we know that the density of $X|Y=y$ can be expressed as $f_{X|Y=y}(x)=f_{X,Y}(x,y)/f_Y(y)$. So I should compute $\mathbb{P}((X,Y)\in C)=\mathbb{P}((g(Y)+Z,Y)\in C)$. Now I would like to invert $(Y,Z) \mapsto (g(Y)+Z,Y)$, but in general I cannot do this. Does it mean that I canno go on to do this?

Can you please help me out/point me to some reference material?

答:
Your definition of "continuous random variable" isn't the usual one, which merely requires the cumulative distribution function of the random variable to be continuous, not that it have a density.  The latter requires the cumulative distribution function to be absolutely continuous, which is a stronger condition.

If $\ X\ $ and $\ Y\ $ are arbitrary random variables, then the distribution $\ P\big(X\in A\,\big|\,Y=y\big)\ $ of $\ X\ $ given $\ Y=y\ $ is defined to be the function of $\ y\ $ which satisfies the equation
$$
P\big(X\in A,Y\in B\big)=\int_B P\big(X\in A\,\big|\,Y=y\big)\,dF_Y(y)
$$
for all measurable $\ B\subseteq\mathbb{R}^n\ $.  This defines $\ P\big(X\in A\,\big|\,Y=y\big)\ $ uniquely for all $\ y\ $ in the support of $\ F_Y\ $.  If $\ F_Y\ $ has a density $\ f_Y\ $, this equation reduces to
$$
P\big(X\in A,Y\in B\big)=\int_B P\big(X\in A\,\big|\,Y=y\big)f_Y(y)\,dy\ ,
$$
but it is not necessary to assume this.

If $\ Y\ $ and $\ Z\ $ are independent, and $\ X=Z+g(Y)\ $ then
\begin{align}
P\big(X\in A,Y\in B\big)&=P\big(Z+g(Y)\in A, Y\in B\big)\\
&=\int_B\int_{\{z\,|\,z+g(y)\in A\}}\,dF_Z(z)\,dF_Y(y)\\
&= \int_BP\big(Z+g(y)\in A\big)\,dF_Y(y)\ .
\end{align}
Comparing this with the definitional equation of $\ P\big(X\in A\,\big|\,Y=y\big)\ $ above, we see that
$$
P\big(X\in A\,\big|\,Y=y\big)= P\big(Z+g(y)\in A\big)
$$
for all $\ y\ $ in the support of $\ F_Y\ $. If $\ F_Z\ $ has a density $\ f_ Z\ $ then
\begin{align}
P\big(X\in A\,\big|\,Y=y\big)&=\int_{\{z\,|\,z+g(y)\in A\}}f_Z(z)dz\\
&=\int_Af_Z(x-g(y))\,dx\ ,
\end{align}
so in this case the conditional distribution $\ P\big(X\in A\,\big|\,Y=y\big)\ $ has the density $\ f_Z(x-g(y))\ $.

Title: $(g(Y)+Z)|Y=y \sim g(y)+Z$


问:
Why is it so that we consider multiplying a vector by a negative scalar as reversing the direction of it.
I know that negative numbers reside in the opposite direction of positive numbers in the number line. I want a solid argument regarding what might the intuition be behind reversing the direction of a vector when multiplied by a negative scalar in any dimension (1, 2 or 3)</p>

答: This happens simply because $-v$ is the additive inverse of $v$: $-v$ is the only vector $∈ V$ such that $v +(-v)=0∈V$


问:
Prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$ with $a_1=2$ is decreasing
I need to prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$, $a_1=2$ is decreasing.

I tried to use induction: I assumed that $a_n<a_n-1$, but couldn't prove that $a_{n+1}<a_n$.

How can I prove it?

答: $a_{n+1} \geq 1$ by AM-GM inequality. Hence $a_n \geq 1$ for all $n$. Now $a_{n+1} \leq a_n$ reduces to $a_n+\frac  1 {a_n} \leq 2a_n$ or $a_n \geq \frac  1{a_n}$ which is true.


答:
This is Heron's method for calculating square roots applied for $S=1$. It's limit is $\sqrt S=1$, as seen:

$$\lambda=\frac12(\lambda+\frac 1\lambda)\implies 2\lambda=\lambda+\frac 1\lambda\implies \lambda=\frac 1 \lambda\implies \lambda=1$$

(I've ignored $\lambda=-1$ because the function is non-negative when $a_1>0$)

Now supposing $a_n=1+t$ for some $t>0$, we notice:

$$a_{n+1}=\frac12(1+t+\frac{1}{1+t})=1+(\frac{t-1}{2}+\frac{1}{1+t})<1+t=a_n$$
and so the sequence is decreasing.

Title: Prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$ with $a_1=2$ is decreasing

Body:
I need to prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$, $a_1=2$ is decreasing.

I tried to use induction: I assumed that $a_n<a_n-1$, but couldn't prove that $a_{n+1}<a_n$.

How can I prove it?


问:
Math fact family number triangles
A buddy of mine sent this picture of his son's homework and asked for ideas.

He said there are no instructions besides what is seen in the image and that his kid is learning "math fact families" which a quick google search shows is based on writing combinations of three numbers in addition/subtraction problems e.g. given the triple $(7,4,11)$ we have the fact family
$$
\begin{aligned}
7+4 &=11\\
4+7 &=11\\
11-4 &=7\\
11-7 &=4
\end{aligned}
$$
Without any further context, I was unable to spot a pattern. I tried looking at relations between the vertices $(E,A,C)$ and then deducing a pattern which proved to not be fruitful. Any ideas on how these are solved?

答:
$A+C+E=68$, and the numbers at the corresponding vertices of the second figure also sum to $68$. Henceforth I’ll use $A'$ for the vertex in the second figure that corresponds to vertex $A$ in the first, so $A'=24$. $B+E=34$, which is exactly half of $68$, and it seems unlikely to be a coincidence that $C'+F'$ is also $34$. We may tentatively conjecture that $C+F$ and $A+D$ are also supposed to be $34$, so that $F=11$ and $D=9$. Similarly, $B+G=F'+H'=17$, exactly half of $34$, so we may guess that $I=17-D=8$ and $H=17-F=6$. The second figure can be completed similarly. Note that $B+D+F=34$ and $G+H+I=17$. Thus, we have a clear pattern.

And on further examination, we find that it is to a considerable extent composed of triangular ‘math fact families’:

$9+11=20$ for the top triangle;

$9+14=23$ for the lower right triangle;

$11+14=25$ for the lower left triangle;

$3+8=11$ for the upper left blue triangle;

$3+6=9$ for the upper right blue triangle; and

$6+8=14$ for the bottom blue triangle.

This is not entirely satisfactory, since the yellow triangle in the middle is not of this type, and I’d also be happier if the number at the bottom of the problem were $136=2\cdot 68$, but I can’t see anything better at the moment.

Title: Math fact family number triangles


问:
A state on a C*-algebra is faithful if and only if it is injective?
A state $\tau$ on a C*-algebra $A$ is called faithful if $\tau(a^{*}a)=0$ implies $a=0$ for all $a\in A$.

Since $a=0$ if and only if $a^{*}a=0$ (by the C*-identity) and since the positive elements of $A$ are characterized by elements of the form $a^{*}a$, I interpreted the defintion of faithful as $$\ker(\tau)^{+}=\ker(\tau)\cap A^{+}=\{0\}.$$

But since the elements of (the closed ideal, hence C*-subalgebra) $\ker(\tau)$ are spanned by the positive elements in $\ker(\tau)$, this implies that $\ker(\tau)=\{0\}$. Clearly, the converse is also true: If $\ker(\tau)=\{0\}$, then $$\{0\}\subset\ker(\tau)^{+}\subset\ker(\tau)=\{0\}.$$
But then $\tau$ is faithful if and only if $\tau$ is injective. This seems very weird, because this implies that if $A$ has a faithful state, then $\dim(A)\leq\dim(\mathbb{C})=1$, i.e. $A=\{0\}$ or $A=\mathbb{C}$.

I must be doing something wrong. Any help will be greatly appreciated!

答:
Something that in my opinion is not encouraged enough, is to try basic examples. You can produce a faithful state on $\mathbb C\oplus\mathbb C$ by $\tau(a,b)=\frac{a+b}2$.

Then you can easily calculate $$\ker\tau=\{(a,-a):\ a\in\mathbb C\},$$
while $(\ker\tau)^+=\{0\}$.

Also, note that linear functionals always have big kernels, as $\dim A/\ker\tau=1$. Indeed, if $a\in A$ is such that $a\not\in\ker\tau$, then for any $b\in A$ we have $b-\frac{\tau(b)}{\tau(a)}\,a\in\ker\tau$, so $b+\ker\tau=\frac{\tau(b)}{\tau(a)}\,a+\ker\tau$. That is, $A/\ker\tau=\mathbb C\,a+\ker\tau$.


问:
Convert a recursive formula to explicit
Given sequence:

$A_{1} = 0$ $A_{2} = 3$ $A_{3} = \frac{3}7$
$A_{4} = \frac{21}{13}$  $A_{5} =  \frac{39}{55}$

And the recursive formula is given as $_{n+1}=\frac{3}{2_{n} + 1}$

How can one find this sequence's explicit formula?

答:
Hint: Solve $x = \frac{3}{2 x + 1}$. Why is this a good approach?

Alternatively, consider the subsequences of $A_n$ for odd $n$ and for even $n$ and prove that they are monotonic and bounded.

Title: Convert a recursive formula to explicit


问:
Exercise with tensor product base change
I'm refreshing my knowledge of Galois theory and thinking about Galois extensions and splitting fields. Suppose that $k$ is a ground field, $P$ is some polynomial of degree $n$ with a splitting field $K$, and $K_j$ are intermediate fields ($K_1=k$, $K_2=k[x]/P$ and so on).

My hypothesis is that $K\otimes_k K \simeq K \otimes_{K_{n-1}} K$.

I think that it can be deduced straightforward from base change property of tensor product ($A\otimes_C B \simeq (A\otimes_C D)\otimes_D B$), but some problem appears.

$$K\otimes_{K_{n-1}}K_n \simeq (K\otimes_{K_{n-1}} K_{n-2})\otimes_{K_{n-2}}K_n$$
and I have no idea how to proceed to $K\otimes_k K$ from this step.

It will be great if anybody would tell me is it really true and, if it is, help me to finish proof.

答:
To answer your question years later and get this off the unanswered queue, this will be the case if and only if $k\to K_{n-1}$ is an isomorphism.

Recall that if $L/K$ is a field extension, and $V$ and $W$ are finite-dimensional $L$-vector spaces, then we have
$$
\dim_L(V\otimes_L W) = \dim_L(V)\dim_L(W)
$$
and
$$
\dim_K(V) = [L:K]\dim_L(V).
$$

Let $[K_{n-1}:k] = d$. We have
$$
\deg_k(K\otimes_{K_{n-1}} K) = d\deg_{K_{n-1}}(K\otimes_{K_{n-1}}K) = d[K:K_{n-1}]^2,
$$
while
$$
\deg_k(K\otimes_k K) = [K:k]^2 = ([K:K_{n-1}][K_{n-1}:k])^2 = d^2[K:K_{n-1}]^2.
$$
These two quantities will be equal if and only if $d = d^2,$ which is true if and only if $d = 1.$


问:
What is the point of computing for $dy$?
I can understand the significance of computing for a derivative $\frac{dy}{dx}$ because you can see the slope (change in y in relation to the change in x) at any given point of a function.

By that logic, computing for $dy$ can find the change in the y axis at a given point x? What is the point of this?

答:
Personally, I find taking the differential instead of the derivative makes it easier to keep my options open, and also makes it more amenable to multivariable equations.

For instance, say you have the equation $y^2 = e^x + z$.  If you take the differential, you get $2y\,dy = e^x\,dx + dz$.  From here, you can solve for the derivative $\frac{dy}{dx}$ if you like.  Or, you can divide both sides by $dt$ and get related rates.  You can set one of the differentials to zero to get partial differentials, etc.

Anyway, there's nothing wrong with the derivative, but, to me, it seems that always forcing it into a form of a derivative is messy.

Additionally, there are a few places where the derivative is misleading.  Take the equation $x = 0$.  The derivative with respect to $x$ would lead you to say $1 = 0$, but the differential is more clear: $dx = 0$.


问:
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$
I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks</p>

答:
Note that this problems cries out for partial fraction decomposition, for which you'll want to factor you denominator: it factors nicely as the difference of squares:

$$(x^4 - 1) = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x +1)(x-1)$$

See also Paul's Online Notes: Partial Fractions for more insight into using this technique for integration, and some nicely worked out examples.


问:
Prove that there's a unique integer n where : $ 2n^2 − 3n − 2 = 0$.
I have to prove that there's a unique integer n where : $2n^2 − 3n − 2 = 0$.

I factorised and got $(2n+1)(n-2) = 0$

I can then find both $n$ solutions : $-1/2$ and $2$. We can therefore see there's only 1 solution that is an integer.

The problem I have is that I think I'm missing something important here and I need confirmation.

答: You have solved the quadratic in the reals, which are a proper superset of the integers. You have shown there are only two real solutions, which immediately establishes there are at most two integer solutions. Furthermore, by inspection, you've concluded that one solution is not an integer, while the other is. This is sufficient to conclude there is exactly one integer solution.


问:
Surjective closed immersion into an integral scheme is an isomorphism?
I want to prove that if $f: X\rightarrow Y$ is a closed immersion, $Y$ is integral and $\dim X=\dim Y$, then $f$ is an isomorphism.

My idea: $Y$ is integral means $Y$ is irreducible, since $f(X)$ is closed in $Y$ and $\dim X=\dim Y$, then $f$ is surjective. But a surjective closed immersion is not necessary an isomorphism.

How to continue to show $f$ is an isomorphism?

Thanks!

答:
The only surjective closed immersion in to a reduced scheme is an isomorphism. It suffices to verify this affine-locally, where every closed immersion is isomorphic to a closed immersion of the form $\operatorname{Spec} A/I \to \operatorname{Spec} A$.

So let $A$ be a ring, $I\subset A$ an ideal, and suppose that $\operatorname{Spec} A/I \to \operatorname{Spec} A$ is a surjective closed immersion. This means that there is a bijection between prime ideals of $A$ and prime ideals of $A/I$. But the prime ideals of $A/I$ are exactly the prime ideals  of $A$ containing $I$, so this means that $I$ must be contained in the intersection of all prime ideals. This is exactly the nilradical of $A$, which is zero iff $R$ is reduced. Thus every surjective closed immersion in to a reduced scheme is an isomorphism.


问:
Good source to learn about surface singularities?
I am looking for something that treats singularities on algebraic surfaces and curves over $\mathbb{C}$, starting from the very basics but not stopping there.

I checked out Miles Reid his lectures on algebraic surfaces and they seem promising (especially the exercises). Also Harris - algebraic geometry a first course seems nice. However, the first gets very technical quickly while the second book's treatment of singularities is pretty short.

Is there also something that treats loads and loads of examples, furnishing a good intuition about the subject?

Thanks!

答: "Lectures on Resolution of Singularities" (Annals of Mathematics Studies, 166) by Kollár also comes to mind. This probably goes a bit further than you need, but the first two chapters (about two-thirds of the length of the book) deal with curve and surface singularities in detail.</p>


问:
Explanations of Lebesgue number lemma
From Planetmath:

Lebesgue number lemma: For every open cover $\mathcal{U}$ of a compact metric space $X$, there exists a real number $\delta > 0$ such
  that every open ball in $X$ of radius $\delta$ is contained in some
  element of $\mathcal{U}$.

Any number $\delta$ satisfying the property above is called a
  Lebesgue number for the covering $\mathcal{U}$ in $X$.

I feel hard to picture and understand the significance of this result. I was wondering if there are some explanation for this lemma? Intuitively,

a number bigger or smaller than a Lebesgue number may not be a
Lebesgue number. So is a Lebesgue number simultaneously measuring
how separated open subsets in an open cover are between each other,
and how big each of them is?

how is a metric space being compact make the existence of a Lebesgue
number possible?

Added: Is the lemma equivalent to say that for any open cover, there
exist a positive number $\delta$, s.t. any open cover consisting of
open balls with radius $\delta$ is always a refinement of the
original open cover?

Thanks and regards!

答:
Here is an intuitive proof :
For any point $ x $ , there is an $ r_x (> 0) $ such that $ B(x, r_x) $ is contained in some element of the cover. Note $ \{ B(x, r_x / 2 ) : x \in X \}
$ is an open cover, so by compactness it has a finite subcover $ \{ B(x_1, r_{x_1} / 2) , …. , B(x_n, r_{x_n} / 2 ) \} $. Now setting $ \delta := min\{ r_{x_1} / 2 , …. , r_{x_n} / 2 \} > 0 $ will do [Because let $ x $ be an arbitrary point. There is a $ B(x_i, r_{x_i} / 2 ) $ containing it. Now note $ B(x, \delta) $ is contained in $ B(x_i, r_{x_i}), $ which in turn is contained in some element of the cover]

Title: Explanations of Lebesgue number lemma


问:
Conditional Joint Distribution when X is a constant
Given the following joint density function:
$$f(x,y)=2(x+y) \\0 \le x \le y \le 1$$
Determine the probability that Y is greater than 1/2 given that X=0.1.

So, I start from the formula below:
$$P(Y \gt \frac{1}{2}|X=0.1)$$

And suddenly, I realized that X is a continuous random variable. So, is P(X=0.1) = 0

Based on the solution, P(X=0.1) is not 0. Can anyone help me to understand why P(X=0.1) is not zero here?

答:
You should use that which you have been provided: the joint probability density function.

$$\begin{align}\mathsf P(Y>0.5\mid X=0.1)&=\int_{y>0.5} f_{\small Y\mid X}(y\mid 0.1)\,\mathrm d y\\[1ex]&= \int_{y>0.5}\dfrac{f_{\small X,Y}(0.1,y)}{f_{\small X}(0.1)}\,\mathrm d y\\[1ex]&=\dfrac{\int_{0.5}^1 2(0.1+y)\,\mathrm d y}{\int_{0.1}^1 2(0.1+y)\,\mathrm d y}\end{align}$$


问:
Universal Generalisation ($\forall$ - I)
With this deduction rule, in the premise of the rule: the term to be substituted for a variable must be arbitrary (refer to an arbitrary d $\in$ D).

What constitutes arbitrary and not arbitrary?

$ P(a) \quad\quad Premise$

$ \forall x P(x) \quad (1), \forall-I: a/x$

Would term john not be considered arbitrary and thus line 2 be incorrect?

$ P(john) \quad Premise$

$ \forall x P(x) \quad (1), \forall-I: john/x$

答:
First of all, I hope you understand the intuition behind this:

Just because some specific object has some property obviously does not mean that all objects from the domain have that property.

However, if an arbitrary object from the domain has some property, then all objects do.

And just to be clear: by 'arbitrary' object we mean: we know and have assumed nothing about this object other than that it is some object from the domain.

Now, how exactly this is being formalized in a specific formal system depends on a lot of formal details. In some systems, variables are used to denote arbitrary objects, but in other systems, 'temporary constants' are being used, typically in combination with certain kinds of subproofs.

So, if you ask me if whether you can apply $\forall \ I$ to infer $\forall x \ P(x)$ from $P(John)$, I really cannot answer that; it all depends on the specifics of the system you are using.


问:
No Rubik's cube can be solved in 13 moves
I've seen a very interesting paper/article few hours ago where it says: there is NO permutation of the Rubiks Cube from where there is a solution in 13 moves, but no solution in less moves.
i.e., there is no position with distance 13 from the identity.
But I cannot find the paper! Can someone help me find or does someone know the paper?

Or are there similar, simpler examples of other groups where all elements can be reached by combining the generators, but there is no element with a certain distance?

David

答: not sure who told you that but according to https://cube20.org/ there are 531,653,418,284,628 positions solvable in exactly 13 moves (and no less).</p>


问:
How to solve $-4xy + x^2 - y^2 + 4x^2 \frac{dy}{dx} = 0$?

We want to find the exact solution of the following ODE:
$$
-4xy + x^2 - y^2 + 4x^2 \frac{dy}{dx} = 0
\quad
\quad
y = 2, x = 1
$$

This is my attempt:
We need to match this form:
$$M(x,y) + N(x,y) \frac{dy}{dx} = 0$$

Then we find $M(x,y)$ and $N(x,y)$:
$$M(x,y) = -4xy + x^2 -y^2$$
$$N(x,y) = 4x^2$$

From this we get two differential equations:
$$\frac{d\Psi(x,y)}{dx} = M(x,y)$$
$$\frac{d\Psi(x,y)}{dy} = N(x,y)$$

It is not important which one we solve by integrating.
Here we integrate the first one.
$$\Psi(x,y) = \int M(x,y) dx  $$
$$= \int -4xy + x^2 - y^2  dx $$
$$= -2x^2y + \frac{x^3}{3} -xy^2  + C(y) = \Psi(x,y)$$

So far so good. From this onward I try to methods and both fail.

First ATTEMPT:

Integrating the other differential equation we get
$$ 4x^2y  + D(x) = \Psi(x,y)$$

comparing the two

$$ 4x^2y  + D(x) = \Psi(x,y) =-2x^2y + \frac{x^3}{3} -xy^2  + C(y)  $$

not sure what to do next since it is not easy to deduce the $C(y)$ and $D(x)$ from here.

Second ATTEMPT:

Now we use the OTHER differential equation:
$$\frac{d\Psi(x,y)}{dy} = N(x,y)= 4x^2$$
and we use the $\Psi(x,y)$ that we found. We now need to find
$C(y)$.

$$\frac{d\Psi(x,y)}{dy} = 4x^2$$
$$ -2x^2 -2xy + \frac{dC(y)}{dy}= 4x^2$$
$$ -6x^2 -2xy + \frac{dC(y)}{dy}= 0$$

integrating everything with respect to y
with constant of integration K(x), we get

$$ -6x^2y -xy^2 + C(y) + K(x) = 0 $$

solve the above for $C(y)$

$$ C(y) = 6x^2y +  xy^2 - K(x) $$

and substitute into $\Psi(x,y)$ to get

$$\Psi(x,y) = -2x^2y + \frac{x^3}{3} -xy^2 + 6x^2y +  xy^2 - K(x)  = 0   $$
$$\Psi(x,y) =  \frac{x^3}{3}  + 4x^2y  -K(x)  = 0   $$

we differentiate with respect to x

$$\frac{d\Psi(x,y)}{dx} = M(x,y)$$

$$ x^2 + 8xy - \frac{dK(x)}{dx}= -4xy + x^2 -y^2 $$
$$ y^2 + 12xy  =  \frac{dK(x)}{dx} $$
$$ xy^2 + 6x^2y + Q  =  K(x) $$

$$\Psi(x,y) =  \frac{x^3}{3}  + 4x^2y - xy^2 - 6x^2y    + Q    = 0   $$
$$\Psi(x,y) =  \frac{x^3}{3}  -2x^2y - xy^2  + Q  = 0   $$

However I think here we get into an infinite loop because $Q$ is actually $Q(y)$ which
needs to be determined using a similiar integration ... so again I don't know what to do.

The final solution $\Psi(x,y)$ does not agree with::
$$\frac{d\Psi(x,y)}{dx} = M(x,y)$$
$$\frac{d\Psi(x,y)}{dy} = N(x,y)$$

What am I doing wrong?

答:
It does not work because the equation is not an exact DE with $M_y=N_x$. You would need to find an integrating factor.

But you can get an easier result by remarking that the terms are all quadratic or contain the derivative, so that you get the form of an homogeneous DE. Thus set $y(x)=xu(x)$ to find that
$$
x^2(−4u+1−u^2+4(xu'+u))=0.
$$
As that is now a separable equation, you get a solution by integrating
$$
\int \frac{4\,du}{u^2-1}=\int\frac{dx}{x}
$$
and making the solution explicit.

You could also treat the original equation as Riccati equation.