问:
Some horrid integrals I am confused with
$$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$

Speaking as an example on (1).

The only thing I could do from here is only to do a u-sub:

$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$

And so this becomes:

$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$

And from here it's basically a deadend.. I would appreciate if you could give some insights..

Interestingly, these similar integrals have interesting solutions:

$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx =  \sqrt{2 e \pi}$$

$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$

(2) - where and why does the pi comes into play? this looks like an interesting solution..

(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?

Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)

Thank you :-)

Body:
For the first integral, write $$I=\int_{1/e}^\infty\sqrt{\frac{\log x+1}{x^3+1}}\,dx=\int_0^\infty\frac{u^{1/2}e^{u-1}}{(1+e^{3(u-1)})^{1/2}}\,du=I_1+I_2$$ where $$I_1=\int_0^1\frac{u^{1/2}e^{u-1}}{(1+e^{3(u-1)})^{1/2}}\,du=\sum_{n\ge0}\frac{(2n-1)!!(-1)^n}{2^nn!}\int_0^1u^{1/2}e^{(3n+1)(u-1)}\,du$$ and $$I_2=\int_1^\infty\frac{u^{1/2}e^{u-1}}{(1+e^{3(u-1)})^{1/2}}\,du=\sum_{n\ge0}\frac{(2n-1)!!(-1)^n}{2^nn!}\int_1^\infty u^{1/2}e^{-(3n+1/2)(u-1)}\,du$$ so $$I=\sum_{n\ge0}\frac{(2n-1)!!(-1)^n}{2^nn!}\left[\frac1{(3n+1)^{3/2}e^{3n+1}}\int_0^1v^{1/2}e^v\,dv+\frac{e^{3n+1/2}}{(3n+1/2)^{3/2}}\int_1^\infty v^{1/2}e^{-v}\,dv\right]$$ where the integrals are incomplete gamma functions.

For the third integral, another way to see that it is not real is to note that $$\Im\int_{1/e}^\infty\sqrt{\frac{\log x}{x^3+1}}\,dx=\Im\int_{1/e}^1\sqrt{\frac{\log x}{x^3+1}}\,dx=\Im\int_1^e\sqrt{\frac{-\log u}{u^4+u}}\,du\ne0$$ as $\log u/(u^4+u)$ is positive on $(1,\infty)$.


问:
A question about $\varepsilon$-$\delta$ definition of continuity
How can I prove the $f(x)=x^{\frac{2}{3}}-1$ is continuous on $[ -1, 1 ]$. First I wrote the expression for continuity on $\varepsilon$-$\delta$ language. So, $\forall \varepsilon>0$, $\exists \delta>0$ if $x_{1}\in [-1, 1]$ and $|x_{1}-x|<\delta$, then $|f(x_{1})-f(x)|<\varepsilon$. I tried so much but I ultimately get cubic deltas so I couldn't write delta as a function of epsilon. Thanks

答:
First of all, continuity is about one point. What you are trying to prove seems to me like uniform continuity, which is a different thing. Luckily, continuous functions on compact sets are uniformly continuous, so you can do both. So let's prove uniform continuity (which implies continuity).

Fix $\epsilon >0$ arbitrarily small. I say that $\exists \delta$ such that if $x_1 , x_2 \in [-1,1]$ with $|x_1 - x_2 | < \delta$ then
$|f(x_1)- f(x_2 )|< \epsilon$.

We have $$|f(x_1)- f(x_2)|= |(x_1)^{2/3} - (x_2)^{2/3}|=|(x_1)^{1/3} - (x_2)^{1/3}||(x_1)^{1/3} + (x_2)^{1/3}| \leq 2 |(x_1)^{1/3} - (x_2)^{1/3}|$$

Now you can use the identity $a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$ and it should do the trick. I mean that you multiply and divide by $(x_1)^{2/3} + (x_2)^{2/3} + (x_1)^{1/3}(x_2)^{1/3}$ so that on the numerator you have $x_1 - x_2$ and on the denminator something which you can easily estimate.

ContentLicense: CC BY-SA 4.0


问:
Suppose f is a one-one function. Show that $f^{-1} \circ f(x) = x$ for all x in D(f)
I am not sure if my answer below is complete, my math writing is pretty bad so I am not sure if I am covering all the necessary points. Please tell me if you think that the format/style is bad too.

2.F.Suppose f is a one-one function. Show that $f^{-1} \circ f(x) = x$ for all $x$ in $D(f)$.

Let's first remember our  definitions:

Definition Inverse Function - Let f be a one-one function with domain D(f) in A and range R(f) in B. If $g = \{(b, a) \in B \times A: (a, b) \in f \}$ then $g$ is a one-one function with domain $D(g) = R(f)$ in $B$ with range $R(g) = D(f)$ in $A$. The function $g$ is called the function inverse to $f$ and we ordinarily denote $g$ by $f^{-1}$\

Definition Composition of Function - Let $f$ be a  function with domain $D(f)$ in $A$ and range $R(f)$ in $B$ and let $g$ be a function with domain $D(g)$ in $B$ and range $R(g)$ in $C$. The composition $g \circ f$ is the function from $A$ to $C$ given by\

$g \circ f = \{(a, c) \in A \times C: \text{there exists an element }b \in B \text{ such that }(a,b) \in f \text{ and } (b,c) \in g \}$

(Proof) By the definition 12, we need to prove that if $(x, z) \in f^{-1} \circ f$ then $z = x$. Now let $(x, z)$ be any ordered pair in $f^{-1} \circ f$, by the definition 12 of  composition of function, we know that the pair $(x, y)  \in f$ and the pair $(y, z) \in f^{-1}$ exist. By the definition of inverse, the function $f^-1$ is the inverse of  $f$ if and only if considering the pair $(x, y) \in f$ and $(y, z) \in f^{-1}$, $z = x$,  so we proved that for all $(x, z) \in f^{-1} \circ f, z = x.$

答:
Well, suppose $gf(x)=x$ for all $x\in D_f$.

Then $g$ defined on $ran(f)$ is one-to-one:

Let $a,b\in ran(f)$ with $g(a)=g(b)$. There are $u,v$ such that $f(u)=a$ and $f(v)=b$.

Then $u=gf(u) = g(a)= g(b) = gf(v)=v$ and so $a=b$.


问:
What is the largest number such that the number formed by the first $n$ digits is divisible by $n$?
What is the largest number such that the number formed by the first $n$ digits is divisible by $n$?

For example, if we have a number $$abcdefghijklm,$$ and all of these leters stand for digits,
then $a$ is divisible by $1$, $ab$ is divisible by $2$, $abc$ is divisible by $3$, and so on. Also, what is allowed is (besides $a$) the digits can be $0$ and digits can repeat.

答:
By copying and pasting the question, it turns out that these numbers are called polydivisible numbers, and there are finitely many of them, with the largest being:

$3608528850368400786036725$

Source: http://en.wikipedia.org/wiki/Polydivisible_number

ContentLicense: CC BY-SA 3.0


问:
Proving two subgroups of the same order are both normal
Let $X$ be the set of all subgroups of $G$, and let the group action be defined by conjugation, that is for $a\in X$:
$$g.A = gAg^{-1}$$
Now let $A$ and $B$ be the only subgroups of $G$ of order $n$ (N.B. that $G$ may have other subgroups, but they will all have order $\neq n$). Suppose $|G|$ is odd, prove that $A$ and $B$ are both normal in $G$. We are given the hint to use the second form of the orbit-stabiliser theorem given to us by: $$|\rm{Orb}_G(x)| = {|G|\over{|\rm{Stab}_G(x)|}}$$
In a previous part of the question we have proven that if we have $K\in \rm{Orb}_G(A)$ that $|K| = |A|$.

My immediate thoughts for this is that if $A,B$ are normal then $\rm{Stab}_G(A)=G$ but have no idea how to use the fact that $|G|$ is odd or how that could factor in, is there something really obvious that I'm missing?

答: The orbit of $A$ can't have size greater than $2$, because its points are subgroups conjugate to $A$ (and by assumption there are just $2$ subgroups of order $|A|$). But it can't have size $2$ either, by the orbit-stabilizer theorem and the assumption on $|G|$. Therefore, $O(A)$ is a singleton, and hence  $gAg^{-1}=A$ for every $g\in G$. Same conclusion holds for $B$.


问:
Solving $T(n) = 2^n T(n-1) + 2^n$
Is there an exact (and if not, an asymptotic) solution to the following recurrence relation?
$$
T(n) = 2^n T(n-1) + 2^n, \text{for } n > 0
$$

If yes, I'd also like to know whether this relation is an instance of a more general class of relations for which you can apply the same technique for solving it.

Title: Solving $T(n) = 2^n T(n-1) + 2^n$

Body:
A sketch:

Let $u_n$ denote the value of $T(n)$ obtained if $T(0)=0$. Prove by induction $T(n)=2^{n(n+1)/2}T(0)+u_n$.

Define $v_n:=u_n/2^{n(n+1)/2}$ so $T(n)=2^{n(n+1)/2}(T(0)+v_n)$. Prove by induction $v_n=v_{n-1}+2^{-n(n-1)/2}$.

Since $v_0=0$, $T(n)=2^{n(n+1)/2}(T(0)+\sum_{k=1}^n2^{-k(k-1)/2})$.


问:
Let $A$ be a connected subspace of $X$.If $A\subset B\subset \bar A$,then $B$ is also connected.
Hi i am reading Topology by Munkres and i have 2 doubt in the proof of theorem 23.4.The part that i have problem in is:

$B$ cannot intersect D. This contradicts the fact $D$ is a nonempty subset of $B$.

My doubts are as follows:

Why $B$ cannot intersect $D$? That is how does $\bar C\cap D=\phi$ imply B cannot intersect D?

How does the statement "B cannot intersect D" contradicts the fact that $D$ is a nonempty subset of $B$?

For reference i am attaching the screenshot of the proof where i have highlighted the part.

答: We have $B \subset \bar A \subset \bar C$. Since $\bar C \cap D = \emptyset$, also $B \cap D = \emptyset$. But we assumed $B  = C \cup D$, thus $B \cap D = D$.


问:
Asymptotics for $y'(t)=y(t-1)$
Let $y : \mathbb{R}\to\mathbb{R}$ be a differentiable function satisfying $y'(t)=y(t-1)$ for all $t$.

Is it possible to give an asymptotics for $y(t)$ as $t\to \infty$?

It is clear to me that there should be an asymptotics of the form $y(t) \sim C\frac{t^{\lfloor t \rfloor +1}}{(\lfloor t\rfloor +1)! }$, inducting on intervals of lenghts 1. How is it possible to prove it?

答:
This is a delayed DE. Let $y(t)=ce^{\lambda t}$ and then the characteristic equation is
$$ \lambda=e^{-\lambda} $$
which has a unique solution $\lambda>0$. Therefore $y(t)\to\infty$ as $t\to\infty$.

Title: Asymptotics for $y'(t)=y(t-1)$


问:
determining the series solution from given ODE
Given this equation, I'm trying to find the series solution however I'm not sure how to proceed from finding the identity principle to remove $x^-1$ to find the solutions.

$xy'' + (1-x)y' + \lambda y = 0$

Given that:

$y=\sum_{n=0}a_{n}x^{n+a}$ Then differentiating this as $y' =\sum_{n=0}a_{n}(a+n)x^{n+a-1}$ and $y'' =\sum_{n=0}a_{n}(a+n)(a+n-1)x^{a+n-2}$

Then plugging this into the first equation, heres what I've tried:

$x\sum_{n=0}a_{n}(a+n)(a+n-1)x^{a+n-2}+(1-x)\sum_{n=0}a_{n}(a+n)x^{a+n-1}+\lambda \sum_{n=0}a_{n}x^{a+n}=0$

$x^{a}(\sum_{n=0}a_{n}(a+n)(a+n-1)x^{n-1}+\sum_{n=0}a_{n}(a+n)x^{n-1}-\sum_{n=0}a_{n}(a+n)x^{n}+\lambda \sum_{n=0}a_{n}x^{n})$

first and second sum we get $k = n-1$ and $n = k+1$ whilst the third and fourth we get $k=n$

$\sum_{k-1}a_{k+1}(a+k+1)(a+k)x^{k}+\sum_{k=-1}a_{k+1}(a+k+1)x^{k}-\sum_{k=0}a_{k}(a+k)x^{k}+\lambda \sum_{k=0}a_{k}x^{k}=0$

Then separating off the k = -1 and k=0 terms we get:

$(a_{0}a(a-1)+a_{0}a)x^{-1}+(-(a_{0}a)+\lambda a_{0})=0$

Where do I go from here? I understand that I must apply the identity principle to find the solutions.

Do i divide the first part by $x^{-1}$ to get rid of $x^{a}$, then rearrange the equation to make it equal to 0?

such as :

$a_{0}a^{2}+(-a_{0}a+a_{o}a)=a_{0}a^{2}=0?$

答:
You did everything correct. That $a=0$ means that the solution will be an ordinary power series, but this is an exception for expansions at regular singularities. It would be better if coefficients and exponent had a different letter as variable name. The coefficient equation at the degree $a+n-1$ is thus
$$
(a+n)^2a_n+(λ-a-n+1)a_{n-1}=0.
$$
With $a=0$ this reduces to
$$
a_n=\frac{n-1-λ}{n^2}a_{n-1},~~n\ge 1.
$$
If $λ$ is a positive integer, this recursion will give a finite sequence (only a finite number of non-zero coefficients).


问:
Question about a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$
This is from Ross's elementary analysis book. The statement is if $a,b \in \mathbb{R}$ such that $a<b$ then there exists a rational $r \in \mathbb{Q}$ such that $a<r0$ then by the archimidean property there exists a natural number call it $n$ such that $n(b-a)>1$. Now we must prove there exists an integer $m$ such that $an<m<bn$. By the archimidean property again there exists an integer $k$ such that $k>\max(|an|, |bn|)$ so that $-k<an<bn<k$. Then the set $$\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$$ is finite and nonempty and we can set $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$. Then $an < m$ but $m - 1 \leq an$. Also, we have $m = (m - 1) + 1 \leq an + 1 < an + (bn - an) = bn$.

Comment: I get up to the point how the author sets$ k$ to be the integer that is larger than both $an$ and $bn$. And since $a<b$ we get the bounded inequality $-k<an<bn<k$. From here on this is where I get confused.  He creates a set  $\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which he calls finite which I see and nonempty (I'm guessing since this set has a least upper bound $k$ thus it is nonempty). But then he lets  $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which follows $an<m$ but $m - 1 \leq an$. I don't get how the author gets to that point?

答: $S_k := \{j\in Z:−k<j≤k \text{ and }an<j\}$ is finite clearly and it is non-empty because $an \lt k$.  Then you can find the smallest integer $m \in S_k$ such that $an \lt m$.  Once you have this $m$, you want to show that $an < m < bn$.  To do this, he says $m = (m-1)+1 \le an + 1$ because by definition $m-1 \le an$.  And $an + 1 \lt an + (bn -an) = bn$.  Hence $an < m < bn$. </p>


问:
If $f(a;f(b;c)) = f(a;b) + c$ and $f(a;a) = 0$ , then find $f($1.1$; -5)$
Function $f(x;y)$defined on pairs of real numbers satisfies the conditions $f(a;a) = 0$, $f(a;f(b;c)) = f(a;b) + c$ for any $a, b, c$... Find$f($1.1$; -5)$

Let me think first , $f(a;a) = 0 $ for any real numbers. Now it can be said that:

=> $f(1.1;-5)+c = f(1.1;f(-5;c))$

=> $f(1.1;-5)= f(1.1;f(-5;c))-c$

I lost my path waht to do next!! Please help me with the explanation and solution.

答:
Hint: The path to victory is always to try and manipulate variables so that you can get something you know. Since we know something about $f(b;b)$, consider $f(a; 0) = f(a; f(b; b)) = f(a; b) + b$, which has to be constant as $b$ varies. This ultimately lets you find a closed form for $f(0; c)$.

Once you've got that, consider $f(0; f(x; y))$.


问:
Find the relationship between $x$ and $y$ so that $y:=0\rightarrow\frac{\pi}{2}\Leftrightarrow x:=y\rightarrow\frac{\pi}{2}$

Find the relationship between $x$ and $y$ so that
$$y:=0\rightarrow\frac{\pi}{2}\Leftrightarrow x:=y\rightarrow\frac{\pi}{2}$$

I'm having trouble solving the multivariable calculus if I change the order between ${\rm d}x, {\rm d}y,$ so I try a new approach like finding function $y=\!f_{1}\left ( x \right )$ and $x=\!f_{2}\left ( y \right )$ but unsuccessfully . I need to the help

答:
I can give an example of an integral such that changing order of the integration changes limits like you've said in the title of your question. Consider the following double integral:
\begin{equation}
\iint\limits_{0<x<y<\frac\pi2}\mathrm dx\mathrm dy
\end{equation}
That, which I've written below the integral sign is the region of integration, and this integration will find out area of the region given by
\begin{equation}
A = \left\{(x,y)\Bigg|0<x<y<\frac\pi2\right\}
\end{equation}
in the plane.

Separating the limits, and doing integration in this order, we get the required integral to be
\begin{align}
I&=\iint\limits_A\mathrm dx\mathrm dy\\
&=\int_0^\frac\pi2\int_0^y\mathrm dx\mathrm dy\\
&=\int_0^\frac\pi2y\mathrm dy\\
&=\dfrac{\pi^2}{8}
\end{align}

Changing the order of integration, we get,
\begin{align}
I&=\iint\limits_A\mathrm dy\mathrm dx\\
&=\int_0^\frac\pi2\int_x^\frac\pi2\mathrm dy\mathrm dx\\
&=\int_0^\frac\pi2\left(\dfrac\pi2-x\right)\mathrm dx\\
&=\dfrac{\pi^2}{8}
\end{align}

Since the problem is not entirely clear to me, please comment if it doesn't answer your question.


答: By definition of $m$ you have $an<m$. Note that $-k<m-1$ because $-k<an<m$ and $m-1<m<k$. Assume $an<m-1$, then $m$ is not the minimal number of the set $\{j:-k<j\leq k,\; an<j\}$. Contradiction, so $m-1\leq an$.</p>

Title: Question about a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$

Body:
This is from Ross's elementary analysis book. The statement is if $a,b \in \mathbb{R}$ such that $a<b$ then there exists a rational $r \in \mathbb{Q}$ such that $a<r0$ then by the archimidean property there exists a natural number call it $n$ such that $n(b-a)>1$. Now we must prove there exists an integer $m$ such that $an<m<bn$. By the archimidean property again there exists an integer $k$ such that $k>\max(|an|, |bn|)$ so that $-k<an<bn<k$. Then the set $$\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$$ is finite and nonempty and we can set $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$. Then $an < m$ but $m - 1 \leq an$. Also, we have $m = (m - 1) + 1 \leq an + 1 < an + (bn - an) = bn$.

Comment: I get up to the point how the author sets$ k$ to be the integer that is larger than both $an$ and $bn$. And since $a<b$ we get the bounded inequality $-k<an<bn<k$. From here on this is where I get confused.  He creates a set  $\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which he calls finite which I see and nonempty (I'm guessing since this set has a least upper bound $k$ thus it is nonempty). But then he lets  $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which follows $an<m$ but $m - 1 \leq an$. I don't get how the author gets to that point?


问:
Boundedness of a linear functional with a parameter
I'm working on this question:

Let consider the subspace $Z=\{x\in C[0,1];x(0)=0\}$. For which values of $r>0$ the linear functional
$$
f_r(x):=\int_0^1\frac{x(t)}{t^r}dt
$$
is bounded? For those $r$, find $\|f_r\|$.

Well, unitl now, I get

$$
|f_r(x)|\leqslant\|x\|_\infty\int_0^1t^{-r}dr,
$$

and then the values of $r$ that makes sense in the integral are $0<r<1$. So, we obtain

$$
|f_r(x)|\leqslant\frac{1}{1-r}\|x\|_\infty\quad\Longrightarrow\quad\|f_r\|\leqslant\frac{1}{1-r}.
$$

If my idea is right, then candidate to be the norm of $f_r$ is $1/(1-r)$. But I can't find the $x\in Z$ that does the work to the other inequality.

However, there is a big chance that what I'm doing is not correct...

Thanks for any help!

答:
to get norm equality you can test your functional by $\phi_n(t)=t^{1/n}$, and $n$ large enough. By doing the computations you get a maximizing sequence: $$||f_r||=\sup_{||x||_\infty=1, x \in Z} |f_r(x)| \geq |f_r(\phi_n)|$$ for all $n$, and the latter term tends exactly to $1/(1-r)$.

the same sequence shows that for $r=1$ the functional is unbounded


问:
Is every element in an open set an interior point?
So I've just started reading topology without tears and I was a bit confused at this question

Given a set  $X = \{a,b,c,\}$. topology

$\tau$ =$\{X, \emptyset, \{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\}$.

Is the point a an interior point of the set $\{a\}$

I think that it is because all points in an open set are automatically interior points but also because the set

$\{a\}$ is by definition a neighbourhood of the point as there is an open set contained inside of it namely itself is my reasoning correct?
And is it true that all points in an open set are interior points?

I'm not quite sure what the answer is.

Thanks in advance.

答: A point $x\in A$ is an interior point of $A$ if $A$ is a neighborhood of $x$. In other words, $x$ is an interior point of $A$ if there is some open set $S$ such that $x\in S$ and $S\subset A$. But then, if $A$ is open, you can just take $S=A$. So, yes, all points in an open set are interior points of that set. And so, yes, your second argument is correct. And so is the first one, by the way.


问:
True or false? About unitary operation.
Let $V$ be a finite inner product space. Let $T:V\to V$ be a linear transformation. Suppose that $v_1,...,v_n$ is an orthonormal basis of $V$ such that $(Tv_i,Tv_i)=1$ for every $1\leq i\leq n$. Then $T$ is unitary operation.

Is that statement true?

答:
Unitary operators are invertible and therefore surjective.  There are non-surjective maps which satisfy the condition above, so not all operators satisfying that condition are unitary.

Consider the map $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by $(x,y) \to (x+y,0)$.  Taking the standard inner product on $\mathbb{R}^2$, we find that $\langle T(1,0), T(1,0) \rangle = 1$ and $\langle T(0,1), T(0,1) \rangle = 1$.  Therefore $T$ satisfies the above condition, but it is not invertible and therefore not unitary.

Title: True or false? About unitary operation.


问:
Questions About Base and Hausdorff in Topology
I am really confused about below questions. I am new both in this platform and below topics. I hope, I was clearly explain the questions.

For ∀n∈$\mathbb{Z}$+

Based on $B_n$ = {n, n+1, n+2...}, B = {$B_n$|n ∈ $\mathbb{Z}$+} family of sets are given. According to this,

(i) Family B is the base for a topology on $\mathbb{Z}$+. Show it.
(ii) Show that there is no Hausdorff by writing the topology on $\mathbb{Z}$+ produced by B.
(iii) Show that the sequence (2,4,6,8,...) in $\mathbb{Z}$+ converges to each point according to topology produced by the family B

Thank you...

答:
(i) to show that $\mathcal{B}=\{B_n\}$ is a base for some topology on some space $X$, one should shows that $$\forall A, B \in \mathcal{B}\quad\exists C_1,\ldots,C_n\in\mathcal{B}\quad A\cap B = \bigcup_{i=1}^{n}C_n$$
and that $\bigcup_{B\in\mathcal{B}}\ B = X$

This conditions easily verified in your case $X=\mathbb{Z}^+$ and $B=\{\{n, n+1, n+2, \ldots\}| n \in \mathbb{Z}^+\}$.

The topology will be same as $\mathcal{B}$ because $\mathcal{B}$ is closed under union too. So all open sets will be the form $\{n, n+1, n+2, \ldots\}$.

(ii) For all open $U$ if $n \in U$ then $n + 1 \in U$ use this to show that point $n$ and $n+1$ don't satisfy Hausdorff condition.

(iii) Sequence $\{x_n\}$ converges to $x$ $\iff$ $\forall$ open $U$ s.t $x\in U$ $\exists N\ \forall n \ge N\ x_n \in U$

in out case $x_n = 2\cdot n$.
By the property that for all open $U$ if $n \in U$ then $n + 1 \in U$ if $x_n\in U$ then $\forall m\ge n\ x_m \in U$

So you only need to verify that
$\forall x\in\mathbb{Z}^+\ \forall$ open $U$ s.t $x\in U$ $\exists n$ such that $x_n \in U$. That you can derive from the form of open sets in our case.


问:
If $A$ is the adjacency matrix of a graph $G$, and $A^3$ has a $1$ on the diagonal, then $G$ has a triangle

If we have a graph $G$, and $A$ the adjacency matrix that represents $G$, how can we proof the diagonal of $A^3$ having a $1$ implies that there is a triangle in $G$?

I try to use the theorem that $A$ is the adjacency matrix of a graph then $A^k_{ij}=1$ if and only if there is a path of length $k$ from $i$ to $j$, but I how we can link it with the diagonal?

答:
Let $G$ be a graph and $A$ the adjacency matrix for $G$.

If the equation $A^k_{ij} =1$ for any vertices $i$ and $j$ [now $i$ and $j$ may be the same], then there is a walk $W$ from $i$ to $j$ of length precisely $k$, and for the case where $i=j$ then there is a closed walk $W$ of length precisely $k$. Now, a priori, $W$ is not guaranteed to be a path for the case where $i \not = j$, and $W$ is not guaranteed to be a cycle for the case where $i=j$,

For example, if $A$ is the adjacency matrix of a matching on 2 vertices $\{1,2\}$ then $A^{2k}_{ii}$ is positive for $i=1,2$ and all even $k$, but there are clearly no cycles of length $2k$ for $k \ge 2$.

However, for $k=3$ and also a graph with no loops i.e., every diagonal element of $A$ is 0, the following holds: A walk $W$ of length $k=3$ that starts and ends at the same vertex is indeed a 3-cycle which is indeed a triangle. You can see this directly for yourself, try drawing a closed walk of length 3 that is not a triangle and where the same vertex does not appear consecutive times.

However, even for $k=3$, it is crucial that $W$ starts and ends at the same vertex i.e., $W$ be a closed walk. Otherwise, what about if $W$ starts and ends at different vertices? Then you need more information to guarantee that $W$ is indeed a path of length precisely 3. For example, getting back to the instance where $A$ is the adjacency matrix of a matching on 2 vertices $\{1,2\}$ then $A^{k}_{12}$ is positive for all odd $k$, but there are clearly no paths of length precisely $k$ for $k \ge 3$.


问:
find the antiderivative for a function

Hi!
I'm a little confused about a calculus exercise.
Can someone please check if my logic is right?

The task:
Find the antiderivative for a function $f$ if $f$ is a continuous function and if $F'(x)=f(x)$ for all real numbers $x$, of $\int_{1}^{4}{f(2x)\,dx}$.

As far as I understand the answer is going to be $2F(4)-2F(1)$?
Any help appreciated

答:
Hint.

There is no "the" antiderivative of a function $f$. If $F$ is an antiderivative of $f$, then so is $F+C$ for any constant $C$.

The fundamental theorem of calculus tells you that if $F$ is an antiderivative of $f$, then
$$
\int_p^q f(x)dx=F(q)-F(p)
$$

But now you have $
\int_a^b f(2x)dx$. By a change of variable
$$
\int_a^b f(2x)dx=\frac12\int_{2a}^{2b}f(u)du\tag{1}
$$
You can now apply the fundamental theorem of calculus to the right-hand side of (1).

For your problem, $a=1$, $b=4$.

Title: find the antiderivative for a function


问:
Finding $\int_0^{\infty } \frac{\tan ^{-1}(a x)-\tan ^{-1}(b x)}{x} \, dx$

Prove that
$$\int_0^{\infty } \frac{\tan ^{-1}(a x)-\tan ^{-1}(b x)}{x} \, dx=\frac{1}{2} \pi  \log \left(\frac{a}{b}\right);\;0<a<b$$

I do not have a clue on how to solve this. In class I have only been taught how to find converging diverging and I do not know how I could apply it here. I can understand how the numerator could give $\frac {\pi}{2}$ but then there are two arc tans so I do not how how I would solve it. And I do not understand why it is $log(a/b)$. And as usual the solution provided by the professor is inadequate as all it tells is to find the limit of arc tan tending to 0 once and tending infinity once and the next step he did says hence proved. Any help is appreciated.

答:
Defne $$F(a,b)=\int_{0}^{\infty}\left(\frac{\tan^{-1}(ax)-\tan^{-1}(bx)}{x}\right)dx$$
use the formula:
$$g(\alpha)=\int_{0}^{\infty}f(\alpha,t)dt \Longrightarrow g^{\prime}(\alpha)=\int_{0}^{\infty}\frac{\partial f}{\partial\alpha}(\alpha,t)dt$$
then show that:
$$\frac{\partial F}{\partial a}=\frac{\pi}{2a}$$
$$\frac{\partial F}{\partial b}=-\frac{\pi}{2b}$$

$$\frac{\partial F}{\partial a}=\int_{0}^{\infty}{\frac{\frac{1}{1+(ax)^{2}}.x}{x}}dx=\int_{0}^{\infty}{\frac{1}{1+(ax)^{2}}}dx=\frac{1}{a}\int_{0}^{\infty}{\frac{d(ax)}{1+(ax)^{2}}}dx=\frac{1}{a}\arctan(a\cdot \infty)=\frac{\pi}{2a}.$$ because $a>0.$

Finally,
$$F(a,b)=\frac{\pi}{2}\ln(a)+M(b)$$
and
$$-\frac{\pi}{2b}=\frac{\partial F}{\partial b}=M^{\prime}(b).$$
$$M(b)=-\int{\frac{\pi}{2b}}db=-\frac{\pi}{2}\ln(b).$$


问:
A quick Poisson distribution problem
A plane drops 535 bombs over 576 fields. How many fields were hit at least twice?

I have a feeling this is related to the Poisson distribution, but I'm having trouble identifying the variable which is Poisson distributed here. Is it simply the number of fields hit at least twice? But then what would the parameter be? Or is the the number of bombs dropped on any single field, then the number of fields hit at least twice is the probability of a single field being hit at least twice times the number of fields?

答:
We use a Poisson model and cross our fingers. If $X$ is the number of bombs hitting a "field," the distribution of the random variable $X$ can, if aim is poor, be approximately modeled by a Poisson with parameter $\lambda=\frac{535}{576}$. The probability that $X\ge 2$ is then $p=1-e^{-\lambda}-e^{-\lambda}\frac{\lambda^1}{1!}$
The mean number of fields with at least $2$ hits is then $576p$.

Remark: During the Second World War, central London was divided into a grid of equal-sized regions. A selection of $576$ of these was made. A study was done of the distribution of bomb hits (I think it was the V1 or V2, not aircraft-launched bombs).  The empirical distribution of the $537$ "hits" was reasonably consistent with the Poisson model.

Title: A quick Poisson distribution problem


问:
Alternative approaches to prove the following inequality
For $a,b,c \in \mathbb{R^+},$ prove that
$$\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3.$$
I managed to prove this problem using the technique of isolated fudging. In particular, one can prove that $\dfrac{1}{3}\left(\frac{2a}{b+c} \right)^{\frac{2}{3}} \geq  \dfrac{a}{a+b+c} $ (motivation is explained below) using AM-GM, as follows:
\begin{align*}
\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq  \dfrac{a}{a+b+c} & \iff \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq  \dfrac{3a}{a+b+c} \\
& \iff  \left(\dfrac{2a}{b+c} \right)^{2} \geq  \left(\dfrac{3a}{a+b+c}\right)^3  \\
& \iff 4a^2(a+b+c)^3 \geq 27a^3(b+c)^2.
\end{align*}
The preceding inequality is homogeneous, hence W.L.O.G. set $a+b+c=3$. It thus suffices for us to prove that $4 \geq a(3-a)^2 \iff 8 \geq 2a(3-a)^2$. But this is obvious from AM-GM: $2a(3-a)^2 \leq \left(\dfrac{2a+(3-a)+(3-a)}{3}\right)^3=8.$ Similarly, we have  $\dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} \geq  \dfrac{b}{a+b+c}  $ and $\dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq  \dfrac{c}{a+b+c} $. Thus, summing cyclically, we obtain:
$$\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq 1 \Rightarrow \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3 .$$
And we are done.

Some motivation:

Let $f(a,b,c)=\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} - \dfrac{a^r}{a^r+b^r+c^r}$, for some $r \in \mathbb{R}$. Ideally, we want $f(a,b,c) \geq 0$ for all  values of $a,b,c$. Note that $f(1,1,1)=0$, which suggests that we should set $(1,1,1)$ as a local minimum of $f$. Hence, take the partial derivative of $f$ with respect to $a$, and set it to zero at $(1,1,1)$. By solving the resulting equation, we find a corresponding value of $r$:

$$\dfrac{\partial f}{\partial a}=  \dfrac{1}{3} \cdot \sqrt[^3]{4} \cdot \left(\dfrac{1}{b+c} \right)^{\frac{2}{3}} \cdot \dfrac{2}{3} \cdot a^{\frac{-1}{3}} - \dfrac{ra^{r-1}(a^r+b^r+c^r)-a^r(ra^{r-1})}{(a^r+b^r+c^r)^2}$$
Hence,
$$\frac{\partial f}{\partial a }\Bigr|_{(1,1,1)} =0 \Rightarrow \dfrac{2}{9}- \dfrac{2r}{9} =0 \Rightarrow r=1.$$

My question is, is there any other way to solve this inequality besides isolated fudging? I.e. would methods such as Cauchy / Holder / Jensen also work here? I would love to see any other alternative approach.

答: We can assume that $a+b+c = 1$ and so with $$ f(x) = \left(\frac{2x}{1-x}\right)^{2/3}$$ the inequality becomes $$ f(a) + f(b) + f(c) \geq 3.$$ Now observe that $f(x) \geq 3x$ for $x \in [0,1]$.


问:
Where does surjectivity fail in a mapping from the unit interval to the Cantor Set?
I'm asked to state why there cannot be a surjective and continuous function $$g: [0,1]\longrightarrow
\text{Cantor Set} $$
I know that $g^{-1}$ exists and is continuous & surjective since the previous exercise was for me to show exactly that.
Now I know that if $g$ were a bijection, then if $g$ were to be continuous, then it would be a homeomorphism and would have to preserve connectedness of $[0,1]$, which it doesn't, implying that $g$ cannot be continuous.

But this requires me to have a bijectivity of $g^{-1}$, where does surjectivity alone enter in this case?

答: If $g: [0,1] \to C$ is continuous then $g[[0,1]]$ is connected. The only connected non-empty subsets of $C$ are of the form $\{c\}$ ($C$ is totally disconnected), so $g$ is constant.


问:
How do I prove that the following expected values are equivalent?
I have been having difficulties with the following problem:

Show that if $f(x)$ is bounded and $X\sim\mathrm{Poiss}(\lambda)$ then $\mathbb{E}[\lambda f(X+1)]=\mathbb{E}[Xf(X)]$. Here $\mathrm{Poiss(\lambda)}$ denotes the usual Poisson distribution with pmf $p(k)=e^{-\lambda}\lambda^k/k!$ for $k\geq0$.

From what I could find online, it seems it has something to do with exchangeable pairs and/or Poisson approximation, but I am struggling to get started with the problem. How do I prove that the following expected values are equivalent? Thank you in advance!

Body:
Be aware that:

$$\sum_{k=0}^{\infty}\lambda f\left(k+1\right)\frac{\lambda^{k}}{k!}=\sum_{k=1}^{\infty}f\left(k\right)\frac{\lambda^{k}}{\left(k-1\right)!}=\sum_{k=1}^{\infty}kf\left(k\right)\frac{\lambda^{k}}{k!}=\sum_{k=0}^{\infty}kf\left(k\right)\frac{\lambda^{k}}{k!}$$


问:
Probability of finding a non-repeating combination
This is a real life problem, which I'm turning into balls just to make it easier to understand.

Suppose balls can be numbered from 1 to 12.

Suppose I have 10 boxes. Each box has 3 distinct balls, distributed at random.

A valid combination is if I can find, from those 10 boxes and taking 1 ball from each, a combination where all balls have distinct numbers (i.e. there are (12,10) valid combinations).

Here are some examples, to make it more concrete:

Box 1 has (1,2,3), Box 2 has (2,3,4), Box 3 has (3,4,5)... Box 10 has (10,11,12). This is the best case scenario, I can easily make valid combinations such as (1,2,3,4,5,6,7,8,9,10); (1,2,3,..,9,11); and so on

Box 1 has (1,2,3), Box 2 has (1,2,3), Box 3 has (1,2,3), Box 4 has (1,2,3). This is the worst case scenario, as I can't find any valid combination. I will always have to repeat 1, 2 or 3.

I would like to know what is the probability of finding scenarios that yield valid combinations from such a setup. In other words - given a scenario of 10 box with 3 distinct balls in each, what is the probability that from those 10 boxes I can find at least one valid combination (it doesn't matter if there is more than 1 valid combination).

Here's a potential approach:

The universe of potential combinations: A = C(12,3)^10

The cases where numbers repeat and I get an invalid scenario:

B = C(12,9) * C(9,3)^10

Probability is then 1 - B/A.

答:
Consider a simpler case: 4 boxes and balls numbered from 1 to 6. We still put 3 balls in each box. I have chosen this case, since it is possible to write out all combinations and check corresponding probabilities.

For each box, we now have $6\choose3$ possible combinations.

In the first box, we can select any ball, yielding a probability of 1.
For the second box, we have two options to select a valid ball (i.e. a ball not selected previously):

A box without the ball that we already selected. The total number of possibilities for this case are ${1\choose0}{5\choose3}$. ${1\choose0}$ specifies the number of balls we select from the previously selected balls. ${5\choose3}$ specifies the number of combinations of choosing three balls from the remaining five that have not been selected previously. The probability of selecting a valid ball is 1, since the previously selected ball is not in the box.

A box with the ball that we already have selected. Out of the $6\choose3$ possibilities, ${1\choose1}{5\choose2}$ have the already selected ball. Selecting a valid ball has a probability of $\frac{2}{3}$.

Thus, selecting a valid ball from the second box is given by $\frac{1\times {1\choose0}{5\choose3} + \frac{2}{3}\times{1\choose1}{5\choose2}}{6\choose3} = \frac{5}{6}$.

For the third box, we have the following cases:

A box without the previously selected balls: ${2\choose0}{4\choose3}$ possibilities. The probability of selecting a valid ball is 1.

A box with one previously selected ball: ${2\choose1} {4\choose2}$. The probability of selecting a valid ball is $\frac{2}{3}$.

A box with two previously selected balls: ${2\choose2}{4\choose1}$ possibilities, with a probability of $\frac{1}{3}$ of not selecting a previously selected ball.

For the third box, the probability of selecting a valid ball is consequently:
$\frac{1 \times {2\choose0}{4\choose3} + \frac{2}{3} \times {2\choose1} {4\choose2} + \frac{1}{3}\times {2\choose2}{4\choose1}}{6\choose3} = \frac{4}{6}$.

For the fourth box, we have the following cases:

No previously selected balls: ${3\choose0}{3\choose3}$ possibilities. We have a probability of 1 of selecting a valid ball.

One previously selected ball: ${3\choose1}{3\choose2}$. Corresponding probability is $\frac{2}{3}$.

Two previously selected balls: ${3\choose2}{3\choose1}$. Corresponding probability is $\frac{1}{3}$.

Three previously selected balls: ${3\choose3}{3\choose0}$. Corresponding probability is 0.

For the fourth box, the probability of selecting a valid ball is thus given by:
$\frac{1 \times {3\choose0}{3\choose3} + \frac{2}{3}\times {3\choose1}{3\choose2} + \frac{1}{3} \times {3\choose2}{3\choose1} + 0 \times {3\choose3}{3\choose0}}{6\choose3} = \frac{3}{6}$.

We thus find the probability of valid combinations for this scenario as follows:
$\text{P(selected balls are distinct)} = \frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}\times\frac{3}{6} = \frac{\frac{6!}{(6-4)!}}{6^4}$.

This pattern also arises in your scenario. In your scenario, we find a probability of:
$\text{P(Selected balls are all distinct)} = \frac{\frac{12!}{(12-10)!}}{12^{10}}\approx 0.003868$.


问:
Question About Compact Space in Topology
I am really confused about below questions. Actually, I did some inferences but I am not sure. Thank you

(i) Show that T = {A ⊂ R | 1 ∉ A } ∪ {$\mathbb{R}$} is a topology on $\mathbb{R}$ and show that the space ($\mathbb{R}$, t) is a compact space according to this topology.

Thank you for now.

答:
Let $\mathcal{U}$ be an open cover for $\Bbb R$ in this topology. In particular there is an $U \in \mathcal{U}$ such that $1 \in U$.

By the definition of the topology $\mathcal{T}$ the only open set that contains $1$ is $\Bbb R$ itself. So $U = \Bbb R$ and so $\{U\}$ is a (very) finite subcover of $\mathcal{U}$. So the space is compact : every open cover has a finite subcover (always the same boring one in this trivial case).


问:
Prove that $X \in \mathfrak g.$

Let $E$ be a Banach space. Let $G \subseteq GL(E)$ be a closed subgroup i.e. $G$ is a linear Banach Lie group. Define $$\mathfrak g : = \left \{X \in \mathcal L(E)\ \big |\ \forall t \in \Bbb R, e^{tX} \in G \right \}.$$ Let $\gamma : (-\varepsilon, \varepsilon) \longrightarrow G$ be a $C^1$-map with $\gamma (0) = I.$ Show that $X = \dot {\gamma} (0) \in \mathfrak g.$

By Taylor's theorem I get for all $t \in (-\varepsilon, \varepsilon)$ $$\gamma (t) = I  + t X + o (t).$$

From here how do I conclude that $e^{tX} \in G,$ for all $t \in \Bbb R\ $? Would anybody please help me in this regard?

Thanks in advance.

EDIT $:$ If $G = GL(E)$ then we are through since $e^T \in GL(E)$ if $T \in \mathcal L (E).$

答:
Lemma $:$ Let $\{A_n \}_{n \geq 1}$ be a sequence in $\mathcal L(E)$ with $\lim\limits_{n \to \infty} A_n = A.$ Then $$\lim\limits_{n \to \infty} \left (I + \frac {A_n} {n} \right )^n = e^A.$$

Let us prove the required result with the help of the above lemma.

Let $t \in \Bbb R.$ Then there exists $N \in \Bbb N$ such that for all $n \geq N$ we have $\left \lvert \frac {t} {n} \right \rvert \lt \varepsilon$ i.e. $\frac {t} {n} \in (-\varepsilon,\varepsilon).$ Hence it follows that $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N.$ Therefore for all $n \geq N$ we have $\gamma \left (\frac {t} {n} \right )^n \in G,$ since $G$ is a group. Now by Taylor's theorem we have $$\begin{align*} \gamma \left (\frac {t} {n} \right ) & = \gamma (0) + \frac {t} {n} \dot {\gamma} (0) + o \left (\frac {t} {n} \right ) \\ & = I + \frac {t} {n} X + o \left (\frac {t} {n} \right )\ \ (\because \gamma (0) = I\ \text {and}\ \dot{\gamma} (0) = X\ ) \end{align*}$$ Hence from the above lemma it follows that $$\lim\limits_{n \to \infty} \gamma \left (\frac {t} {n} \right )^n = \lim\limits_{n \to \infty} \left (I + \frac {tX + o(1)} {n} \right )^n = e^{tX}\ \ (\because o(1) \to 0\ \text {as}\ n \to \infty\ )$$

Since $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N$ and $G$ is closed it follows that $e^{tX} \in G.$ Since $t \in \Bbb R$ was arbitrarily taken it follows that $e^{tX} \in G,$ for all $t \in \Bbb R.$ Hence $X \in \mathfrak g,$ as required.

QED

Title: Prove that $X \in \mathfrak g.$


答:
Let $T=\{A\subseteq\mathbb R:1\notin  A\}\cup \{\mathbb R\}$ be a topology on $\mathbb R$. First we show that it's indeed a topology.

(i) Observe that $\emptyset,\mathbb R \in T$ (because $1\notin \emptyset\implies \emptyset\in\{A\subseteq\mathbb R:1\notin  A\}$).

(ii) Let $\mathcal A=\{A_i\}_{i\in I}$ be a family of open sets in $T$. Since $\forall i\in I$, $A_i\in T\implies 1\notin A_i\forall i\in I$.

Therefore we can state that $1\notin \underset{A_i\in\mathcal A}{\bigcup}A_i\implies \underset{A_i\in\mathcal A}{\bigcup}A_i\in T$.

(iii) If we now consider $A_1,A_2\in\mathcal A\implies 1\notin A_1,A_2\implies 1\notin A_1\cap A_2\implies A_1\cap A_2\in T$, hence $T$ defines a topology on $\mathbb R$.

Let $\mathcal F$ be a family of open sets that covers $\mathbb R$, then $1$ will be neccesairily in $\mathcal F$, so $\mathcal F$ must contain all the real line.

In other words, if $1\in \mathcal F$, then $\exists A\in\mathcal F:1\in A$, but the unique open set that contains the point $\{1\}$ is $\mathbb R\implies A=\mathbb R$. Since $\mathcal F$ was a generic family of open sets that covers $X=(\mathbb R,T)$, it follows that $X$ is compact because we can always find the open sub-cover $\{A\}$.

Title: Question About Compact Space in Topology

Body:
I am really confused about below questions. Actually, I did some inferences but I am not sure. Thank you

(i) Show that T = {A ⊂ R | 1 ∉ A } ∪ {$\mathbb{R}$} is a topology on $\mathbb{R}$ and show that the space ($\mathbb{R}$, t) is a compact space according to this topology.

Thank you for now.


问:
Number of ways to arrange three 1s, two 2s, two 3s and one 4?
Number of ways to arrange three 1s, two 2s, two 3s and one 4? The order doesn't matter.

The direct answer from the book is $\frac{8!}{3!2!2!1!}$, but is there another way to do it?

答:
Here is an  iterative approach to find the number of ways to arrange three $1$s, two $2$s, two $3$s and one $4$:

Three $1$s: We start with three $1$s:
\begin{align*}
1\ 1\ 1\qquad\rightarrow\quad 1\ \mathrm{way}\quad
\end{align*}

Two $2$s: We have now $4$ positions where we can place two $2$s. We have $\binom{4}{1}$ ways to place  two $2\mathrm{s}$ at one position and $\binom{4}{2}$ ways to place two $2$s at different positions:
\begin{align*}
\bullet\ 1\ \bullet\ 1\ \bullet\ 1\ \bullet \qquad\rightarrow\quad \binom{4}{1}+\binom{4}{2}\ \mathrm{ways}
\end{align*}

Two $3$s: To each of the valid arrangements consisting of three $1$s and two $2$s we have now $6$ positions where we can place two $3$s. We habe $\binom{6}{1}$ ways to place two $3$s at one position and $\binom{6}{2}$ ways to place two $3\mathrm{s}$ at different positions:
\begin{align*}
\bullet\ 1\ \bullet\ 1\ \bullet\ 2\ \bullet\ 1\ \bullet\ 2\ \bullet \qquad\rightarrow\quad \binom{6}{1}+\binom{6}{2}\ \mathrm{ways}\quad\qquad\quad
\end{align*}

One $4$: To each of the valid arrangements consisting of three $1$s, two $2$s and two $3$s we have now $\binom{8}{1}$ positions to place the number $4$.
\begin{align*}
\bullet\ 1\ \bullet\ 1 \bullet\ 3 \bullet\ 3\ \bullet\ 2\ \bullet\ 1\ \bullet\ 2\ \bullet \qquad\rightarrow\quad \binom{8}{1}\ \mathrm{ways}\qquad\qquad\qquad\qquad\qquad\quad
\end{align*}

We conclude: The number of valid arrangements is
\begin{align*}
&1\cdot \left[\binom{4}{1}+\binom{4}{2}\right]\cdot\left[ \binom{6}{1}+\binom{6}{2}\right]\cdot\binom{8}{1}\\
&\qquad=\binom{5}{2}\binom{7}{2}8=10\cdot21\cdot 8\\
&\qquad\,\,\color{blue}{=1\,680}
\end{align*}
in accordance with $\frac{8!}{3!2!2!1!}=1\,680$.

Here we use the binomial identity $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$.


问:
Is it equally correct to state of square root of i is plus or minus -1 to the power 1/4?
I understand that the two square roots of $i$ are

$$\pm {1 \over \sqrt{2}} (1+i)$$

But when asked what is the square root of $i$, is it equally correct to give the answer as

$$\pm {(-1) ^ {1/4}}$$

? And actually, since $\sqrt 2$ is defined to be only the positive root, so square root of $i$ can be just

$${(-1) ^ {1/4}}$$

?

答: This is easiest to see in polar form.  $i$ is at an angle of $\frac \pi 2$.  Its square roots are at angles of $\frac \pi 4$ and $\frac {5\pi}4$.  Its fourth roots are at $\frac \pi 8 + k\frac \pi 2$, which do not match at all.


问:
question about percentages
At the start of the new year, the price for a gallon of milk at a local grocery store increased by 2.4% as a result of inflation. If a gallon of milk cost $2.92 last year, about how much does it cost this year?</p>

```
A.  $2.84
B.  $2.99
C.  $3.62
D.  $5.32
```

答:
So the price of my favorite bike was $\$100$  last year. They increased the price by $\%5$ this year. What would be the new price?

It was $\$100$  last year. What is $\%5$ of $\$100$? Note that $\%5$ is equivalent to saying $\frac{5}{100}$, so $\%5$ of $\$100$ is $\frac{5}{100} \times \$100 = \$5$. Hence, the new price is $\$100+\$5 = \$105$.

Now, you do the same thing for your question, and let me know if you have any problems.

Title: question about percentages


问:
Limit of ratio goes to zero
I  am trying to show, if a sequence $\{a_n\}$ satisfies
$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=0$$
then
$$\lim_{n\to \infty}a_{n}=0.$$

I figured there exists an $\epsilon_N$ such that the following holds

$$|a_{N+1}|< \epsilon_N |a_N|$$

and $\lim_{n\to \infty}\epsilon_{N}=0.$

Since $\lim_{n\to \infty}a_{n}\ne  \infty$, then $\lim_{N\to \infty}\epsilon_{N}*|a_{N}|=0.$

One can conclude $\lim_{N\to \infty}|a_{N+1}|=0$. Is this correct? Is there another way to show this?

答: This is more like killing a mosquito with a sledge hammer but I will give it a try: As $\vert\lim_{n\to \infty}\frac{a_{n+1}}{a_n}\vert=0$, then $\sum \vert a_n\vert<\infty$, so it should also be: $\lim \vert a_n\vert =\lim a_n = 0$.

Title: Limit of ratio goes to zero


问:
Is an isolated singularity of a complex-valued function an isolated point of the domain of the function?
Let $\Omega\subseteq\mathbb C$ be open and $f:\Omega\to\mathbb C$. We say that $z_0\in\mathbb C$ is an isolated singularity of $f$ if $f$ is holomorphic on a punctured neighborhood $N$ of $z_0$.

On the other hand, there is the following paragraph in the Wikipedia article of this notion:

Formally, and within the general scope of functional analysis, an isolated singularity for a function $f$ is any topologically isolated point within an open set where the function is defined.

I don't get this. In a general topological space $E$, we say that $x\in E$ is an isolated point of $B\subseteq E$ if there is an open subset $U\subseteq E$ with $B\cap U=\{x\}$.

But why is $z_0$ an isolated point of $N$?

答:
In general, one should remember that anybody can edit a Wikipedia page, regardless of their qualification. In this case, by looking through the history of edits, this particular offending sentence (or, rather, some version of it) was initially written by an engineer, who clearly did not understand the definition; then another engineer tried to correct it but made it even worse. The correct sentence should be something like:

"In the language of topology, an isolated singularity of $f$ is an isolated point of the set $\partial \Omega$."

or

"In the language of topology, an isolated singularity of $f$ is an isolated point of the set ${\mathbb C}\setminus \Omega$."

(There is some disagreement if $\infty$ is allowed as an isolated singularity; if it is, then $\partial \Omega$ is understood as the boundary of $\Omega$ in the Riemann sphere.)


问:
Solve the limit $\lim_{x\to 0+} x\int_x^1 \frac{f(t)}{t}dt$
Solve the limit $\displaystyle \lim_{x\to 0+} ~ x\int_x^1 \frac{f(t)}{t}dt$.

$f$ is continuos on $[0,1]$.

I tried using mean value theorem (there exists $c\in[0,1] s.t ...) $to see where $\displaystyle\int_x^1 \frac{f(t)}{t}dt$ as $x\to 0+$ so I get the expression $\displaystyle\int_0^1 \frac{f(t)}{t}dt=f(c)\int_0^1 \frac{dt}{t}$ and then solve the improper integral, which diverges.

I'm not sure if I can apply that theorem, nor how to approach this initial limit.

答:
Since $f$ is continuous, it is a bounded function on $[0;1]$. Let
$$m \le f(x) \le M$$
Then
$$x \int_x^1 \frac{m}{t} \mathrm dt \le x \int_x^1 \frac{f(t)}{t} \mathrm dt \le x \int_x^1 \frac{M}{t} \mathrm dt$$
But now
$$x \int_x^1 \frac{m}{t} \mathrm dt = - m x \log x \to 0$$ as $x \to 0$. Similarly you can prove that
$$x \int_x^1 \frac{M}{t} \mathrm dt = - M x \log x \to 0$$
Hence by the squeeze theorem the limit is zero.


问:
Application Church-Turing thesis
I would like to give examples of problems which are solvable with an algorithm, for example the function $f$ which maps the tuple $(n,m)$ to the greatest common divisor. This map is recursive. I would do it in the following way: 1st step: If $m$ equals $0$, then $\mathrm{gcd}(n,m)=n$. 2nd step: otherwise $\mathrm{gcd}(n,m)=gcd(m,n \mod m)$.

What about the set of polynomials in one variable with integer coefficients? I want to argue that the subset of the polynomials which have an integer root is recursive.

Moreover, why is the addition and multiplication of rational numbers a recursive function? What about the exponentiation of two rational numbers?

答:
What about the exponentiation of two rational numbers?

I assume you mean positive rational numbers with a rational exponent.

As a formal symbol $q^r$ with related rules of calculation, it all boils down to prime factorization and rules for integer exponentiation, which are recursive in any reasonable discrete representation.   However, if you mean calculation of real numbers equal to the exponentiated rationals, you will need infinite series (or more complicated things) to define a recursive sequence of rational approximations, and basic operations on the real number representation (like checking which of two expressions with rational exponentials is larger) are either not recursive or it can be difficult to prove correctness or termination of the algorithms.

The other questions are simpler. Test for an integer root of a polynomial with integer coefficients can be done by searching through divisors of the first and last coefficients, which reduces the problem to integer factorization.  Addition and multiplication are done by polynomial formulas, so if you know integer addition and multiplication are recursive, this is enough to get the recursiveness of rational fractions.

Title: Application Church-Turing thesis


问:
When L'Hopital's rule fails (again)
I just realized that there may be a case where L'Hopital's rule fails, specifically

$$ \lim_{x \to \infty} \frac{e^x}{e^x}$$

which evaluates to an indeterminate form, specifically $\frac{\infty}{\infty}$. Sure, we can cancel the $e^x$s, but when we use L'Hopital's, we get

$$ \lim_{x \to \infty} \frac{(e^x)^\prime}{(e^x)^\prime}$$

Since the derivative of $e^x$ is $e^x$, we have

$$ \lim_{x \to \infty} \frac{e^x}{e^x}$$

which is our original limit. Therefore, L'Hopital's fails to work in this example.

Question: Does L'Hopital's rule actually fail in this example, or am I understanding it wrong?

Edit: I mean "fails" in which it does not make progress toward a determinate result.

答: L'Hôpital's Rule says that if $\lim_{x\to a}{\frac{f'(x)}{g'(x)}}=L $, then $\lim_{x\to a}{\frac{f(x)}{g(x)}}=L$. With $f'(x)=e^x$ and $ g'(x)=e^x$ is obviously true. L'Hôpital's Rule is also valid when $x\to \infty $.

Title: When L'Hopital's rule fails (again)


答:
In the first place, L'Hopital's Rule does not fail, in the usual sense of “fail”, only in the sense that you've added at the end of your question: you can apply it as many times as you wish without ever getting an answer.

And there are other cases:$$\lim_{x\to\infty}\frac{x^{-1}}{x^{-1}}=\lim_{x\to\infty}\frac{-x^{-2}}{-x^{-2}}=\lim_{x\to\infty}\frac{2x^{-3}}{2x^{-3}}=\cdots$$

Title: When L'Hopital's rule fails (again)

Body:
I just realized that there may be a case where L'Hopital's rule fails, specifically

$$ \lim_{x \to \infty} \frac{e^x}{e^x}$$

which evaluates to an indeterminate form, specifically $\frac{\infty}{\infty}$. Sure, we can cancel the $e^x$s, but when we use L'Hopital's, we get

$$ \lim_{x \to \infty} \frac{(e^x)^\prime}{(e^x)^\prime}$$

Since the derivative of $e^x$ is $e^x$, we have

$$ \lim_{x \to \infty} \frac{e^x}{e^x}$$

which is our original limit. Therefore, L'Hopital's fails to work in this example.

Question: Does L'Hopital's rule actually fail in this example, or am I understanding it wrong?

Edit: I mean "fails" in which it does not make progress toward a determinate result.


答:
No. L'Hopital does not fail.

The L'Hopital rule says that (under some assumptions) if $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=L$ then
$$
\lim_{x\to\infty}\frac{f(x)}{g(x)}=L
$$

In your example,
$$
\lim_{x\to\infty}\frac{e^x}{e^x}=1
$$

One should not blame L'Hopital if one does not know how to calculate $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$ when it does exist.

Title: When L'Hopital's rule fails (again)


答:
Do remember that anyone can do any changes in wikipedia so don't completely rely on wikipedia.

An isolated singularity of f is an isolated point of the set ∂Ω.

Title: Is an isolated singularity of a complex-valued function an isolated point of the domain of the function?

Body:
Let $\Omega\subseteq\mathbb C$ be open and $f:\Omega\to\mathbb C$. We say that $z_0\in\mathbb C$ is an isolated singularity of $f$ if $f$ is holomorphic on a punctured neighborhood $N$ of $z_0$.

On the other hand, there is the following paragraph in the Wikipedia article of this notion:

Formally, and within the general scope of functional analysis, an isolated singularity for a function $f$ is any topologically isolated point within an open set where the function is defined.

I don't get this. In a general topological space $E$, we say that $x\in E$ is an isolated point of $B\subseteq E$ if there is an open subset $U\subseteq E$ with $B\cap U=\{x\}$.

But why is $z_0$ an isolated point of $N$?


问:
Can a calculator find any limit?
I am using CASIO fx-991ex and I noticed through my homework that when x approaches infinity or zero it's almost like using big or small numbers in calculator. For example $e^x$ while $x$ approaches $\infty$ is a very big number like using $e^9$ (+-) in the calculator, or $1/x$ while x approaches $0$ with $0.0001$ in the calculator (instead of $x$) it returns also a big number which I can assume it's $\infty$. So my question is when this thing doesnt work? I know that knowing how to find it out by myself is another important thing but im asking mainly for verifying purposes.

答:
As a general principle it's fine to use a calculator as a tool to reinforce your intuition or guess an answer or confirm a theoretical calculation.

It's not a substitute for logical analysis.

There will be times when it "doesn't work" for several reasons. The calculator works with limited precision (just a finite number of decimal places) and finite range (there is a largest possible value). It may also fail when the limiting value is approached with oscillations rather than monotonically as in your two examples.

Title: Can a calculator find any limit?


答:
The Ackermann function is defined as:

$$A(0, n) = n+1$$
$$A(m, 0) = A(m-1, 1)$$
$$A(m, n) = A(m-1, A(m, n-1))$$

Define $A(n) = A(n, n)$.

The function $A$ grows extremely quickly. $A(3) = 61$. $A(4)$ is bigger than $10^{1355718576299609}$ - likely much bigger, but that is just the easiest lower bound estimate I got from Mathematica from the fact that computation overflows.

Define the inverse Ackermann function, $a(n)$, as the inverse of this function. So $a(61) = 3$.
For any number your calculator can represent, the value of $a$ is less than $4$. Moreover, for any number Mathematica can represent on my machine, the value of $a$ is less than $4$. And yet the limit of $a(n)$ as $n \to \infty$ is $\infty$.

Or consider $\sum_{n=1}^{\infty} \frac{1}{n}$. This is well-known to be infinite (by the Cauchy condensation test, for example). And yet $\sum_{n=1}^{1,000,000} \frac{1}{n}$ is approximately $14.3927$. If you use your calculator to estimate the value of the limit, you will get it wrong.

Title: Can a calculator find any limit?

Body: I am using CASIO fx-991ex and I noticed through my homework that when x approaches infinity or zero it's almost like using big or small numbers in calculator. For example $e^x$ while $x$ approaches $\infty$ is a very big number like using $e^9$ (+-) in the calculator, or $1/x$ while x approaches $0$ with $0.0001$ in the calculator (instead of $x$) it returns also a big number which I can assume it's $\infty$. So my question is when this thing doesnt work? I know that knowing how to find it out by myself is another important thing but im asking mainly for verifying purposes.


问:
The number of matches in a competition
In a competition there were 10 players , each player plays 4 matches , how many matches in the whole competition ?

My attempt is : $C^{10}_{3}$

Is it correct ?

答: A simple way of solving this problem without knowing anything about combinations or graph theory is to suppose that each player wins half their matches. Then the total number of wins is $10\times 2 = 20$ and so there are precisely $20$ matches.

Title: The number of matches in a competition


答: The process that you are using there tends to work all the time. It's exactly what I did in highschool when I knew nothing about limits. Like taking $x\to\infty$ I just typed $x=999999$ into my calculator and it gave me the correct limit. This was especially applicable when we were looking at the logistic equation for population growth. It worked similarly with taking $x\to 0$ as you said.

Title: Can a calculator find any limit?


答:
Hint: If I have a number $z$, and I

increase $z$ by $p$ percent

that means

take $z$, and then add to it $p$ percent of $z$

Remember, the quantity "$p$ percent of $z$" is defined to be the expression
$$\frac{p}{100}\times z,$$
so that for example, $15$ percent of $200$ is
$$\frac{15}{100}\times 250=37.5.$$

Now, let $x$ be the amount that milk cost last year.

Write down the expression for $2.4$ percent of $x$:

$$\fbox{$\strut\quad\qquad^\strut_{\strut}$}$$

and add that to $x$:

$$x+\;\fbox{$\strut\quad\qquad^\strut_{\strut}$}$$

We are told that this quantity is equal to the price of milk this year: $\$2.92$. Now solve for $x$.

Title: question about percentages

Body:
At the start of the new year, the price for a gallon of milk at a local grocery store increased by 2.4% as a result of inflation. If a gallon of milk cost $2.92 last year, about how much does it cost this year?</p>

```
A.  $2.84
B.  $2.99
C.  $3.62
D.  $5.32
```


问:
linear space W which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$
Give an example of a linear space W of finite dimension $d \geq 4$ which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$.

Is that possible? I mean the only possibility of $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ that I can think of is $V_1=V_2=V_3$ but then $V_1⊕V_2=V_1⊕V_3=V_2⊕V_3 = \emptyset$. It's probably something basic and I wasted way to much time thinking of it.

答:
Consider W =M2×3(R) with subspace V1,V2,V3 such that

V1={A€W:a11,a12&a13 are non zero and all other entries are 0s}

V2={A€W:a21,a22 & a23 are non-zero and all other entries are 0s}

V3={A€W: a23 ,a22& a13 are nonzero and all other entries are 0s}

We get dimV1=dimV2=dimV3=3

And their mutual intersection will give zero space.


答:
Just to add the precise statement to explain clearly what is going on:

$\textbf{Theorem - L'Hospital's Rule:}$ Let $a,b\in [-\infty,\infty]$. Suppose that f and g are differentiable real valued maps on $(a,b)$, with $g'(x)\neq 0$ on $(a,b)$., and the following hold:

(i) $\lim_{x\to b}f(x) = \lim_{x\to b}g(x) = +\infty$

and (ii) $\lim_{x\to b}\frac{f'(x)}{g'(x)} = L\in [-\infty,\infty]$,

then we can finally say that $\lim_{x\to b}\frac{f(x)}{g(x)} = L$.

Certainly $e^x$ is $C^{\infty}$, and  the conditions (i) and (ii) hold with $L = 1$, so our original limit is 1 also (not that we needed L'Hospital's rule for that, but there it is, it works here).

Title: When L'Hopital's rule fails (again)


答:
The rule lets us write $\lim_{x\to a}\frac{f}{g}=\lim_{x\to a}\frac{hf}{hg}=\lim_{x\to a}\frac{(hf)^\prime}{(hg)^\prime}$, so sometimes you need a non-constant $h$ to make progress.

With examples of the form $f/f$, we can take $h=x^k/f$, where the choice of $k$ is a context-dependent exercise.

More interesing is needing to transform. For example, consider $\lim_{x\to0^+}\frac{e^{-1/x}}{x}$, which after $n$ iterations gives $\lim_{x\to0^+}\frac{e^{-1/x}}{n!x^{n+1}}$, not helpful at all. But $y:=\frac1x$ converts it to $\lim_{y\to\infty}\frac{y}{e^y}=\lim_{y\to\infty}\frac{1}{e^y}=0$.

Since such a transformation changes the variable with respect to which we differentiate, by the chain rule it uses the more general $\lim_{x\to a}\frac{f}{g}=\lim_{x\to a}\frac{j(hf)^\prime}{j(hg)^\prime}$.

Title: When L'Hopital's rule fails (again)


问:
Is $(\mathbb{C},+)$ isomorphic to a matrix Lie group?
I'm wondering if $(\mathbb{C},+)$ is isomorphic to a matrix Lie group; I mean, if it is isomorphic to a closed subgroup of $GL(n,\mathbb{C})$. If it is, what is the isomorphism? And if it is not, why?

答: Yes, it is isomorphic to the subgroup of $GL(2,\Bbb C)$ which consists of the matrices of the form$$\begin{bmatrix}1&z\\0&1\end{bmatrix},$$because$$(\forall z_1,z_2\in\Bbb C):\begin{bmatrix}1&z_1\\0&1\end{bmatrix}.\begin{bmatrix}1&z_2\\0&1\end{bmatrix}=\begin{bmatrix}1&z_1+z_2\\0&1\end{bmatrix}.$$


问:
How to calculate limit $\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$ without using L'Hospital rule
I have to calculate next limit without using L'Hospital rule:
$$\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$$
I came this far with that $$1+\frac{1}{e}\lim_{t\rightarrow1}\frac{\ln(t)}{t-1}$$
I know that result for this limit using the L'Hospital rule is $1+\frac{1}{e}$
Any suggestions how to finish this?

答:
Using series expansions in
$$y=\frac{x+\log(e+x)-1}{x}$$
$$\log(e+x)=1-\sum_{n=1}^\infty\frac{\left(-\frac{1}{e}\right)^n}{n} x^n=1+\frac{x}{e}-\frac{x^2}{2 e^2}+\frac{x^3}{3 e^3}+O\left(x^4\right)$$
$$y=\left(1+\frac{1}{e}\right)-\frac{x}{2 e^2}+\frac{x^2}{3 e^3}+O\left(x^3\right)$$ Try it with $x=\frac e{10}$ (very far away from $0$). The exact value is
$$1+\frac{10 \log \left(\frac{11}{10}\right)}{e}\sim 1.35063$$ while the truncated  series gives
$$1+\frac{143}{150 e}\sim 1.35071$$


问:
be $A,B$sets, show that $(A\setminus B) \sim (B\setminus A)$ then $A \sim B $
Be $\sim $ an equivalence relation.

Suppose that $A \cap B = \emptyset$ then $A \setminus B = A$,  same to $B \setminus A$

Then, $A \setminus B \sim B \setminus A = A \sim B$

How can i show it when $A \cap B \neq \emptyset $, also the reciprocal is correct, i think is not but i can´t think in a good counterexample.

答:
Let $A\sim B$ when either $|A|=|B|=1$ or $A=B$. This is an equivalence relation.

Now, letting $A=\{1,2\}$ and $B=\{2,3\}$, we see that $A\setminus B\sim B\setminus A$ but $A\not\sim B$.


问:
Finding the vertical asymptote of $f(x)=x^2e^{-1/x}$
I'm having a problem finding a vertical asymptote for this function.
$$f(x)=x^2e^{-1/x}$$

I also looked for horizontal asymptotes, but there seems to be none of those.

答:
$$f(x)=x^2e^{-1/x}$$ is the same as $$f(x)=\frac{x^2}{e^{1/x}}$$

Observe that $$\lim_{x \to 0^-}{\frac{x^2}{e^{1/x}}}=+\infty$$ and so $x=0$ is a vertical asymptote (see here for the precise definition).

There is no horizontal asymptote because if you consider rewriting as $$f(x)=\frac{x^2}{x^0e^{1/x}}$$ then the leading exponent in the numerator $2$ is larger than the leading exponent in the denominator $0$.

Alternatively, you could determine if there is a horizontal asymptote by computing $$\lim_{x \to \infty}{\frac{x^2}{e^{1/x}}}$$ and $$\lim_{x \to \mbox{-}\infty}{\frac{x^2}{e^{1/x}}}$$


问:
Power Series with $a_{n}$ and $a_{k}$

If $P(x)= \sum_{n=0}^{\infty} a_{n} x^{n}$. It is known that $P$ satisfies: $$P^{\prime}(x)=2xP(x)$$ $\\$ for all $ x\in \mathbb{R}$ and $$P(0)= 1$$
\begin{split}\end{split} (1) Prove that $a_{2k+1}=0$ for every $ k\in \mathbb{\left \{1,2,3,4,...  \right \}}$

Its true that if two power series have the same coefficients then they are equal?

I can demonstrate the first equation transforming the series to an integral, but i cant prove (1) and (2) because $k$ confuses me, I was thinking that $k$ is for $P'(x)$, just as $n$ is for $P(x)$. Also i need orientation with the proof requested.

Thanks.

答:
If $P(x)$ is of degree $n$, then the lhs is of degree $(n-1)$ and the rhs is of degree $(n+1)$.

Since we compare the coefficients of same power, we  have only odd or only even powers.

Now, use the fact that for power $0$, $P(0)\neq 0$ and you have the answer.

Title: Power Series with $a_{n}$ and $a_{k}$


问:
Find the eigen value of a matrix without using the characteristic eqution
For a 3x3 matrix, how to find the eigenvalues without using its characteristic equation, if one of the eigenvalues is given.
Suppose 2 is an eigenvalue of the matrix A, and find the others without using its characteristic equation
$$A = \begin{bmatrix}3&-1&1\\-1&5&-1\\1&-1&3\end{bmatrix}$$

答:
Hint- Let $\alpha$ and $\beta$ are the other two eigenvalues. Then

Sum of eigenvalues=trace(A)=(3+5+3)=11

$\implies \alpha+\beta=9$

Ptoduct of eigenvalues=det(A)=$36$

$\implies \alpha\beta=18$


答:
It's isomorphic to the subgroup of $\operatorname{GL}(2,\mathbb C)$ consisting of matrices of the form

$$\begin{pmatrix}1&0\\z&1\end{pmatrix}$$

via the isomorphism

$$\varphi:z\mapsto\begin{pmatrix}1&0\\z&1\end{pmatrix},$$

since $\varphi(0)=E_2$ and

$$\varphi(a)\varphi(b)=\begin{pmatrix}1&0\\a&1\end{pmatrix}\begin{pmatrix}1&0\\b&1\end{pmatrix}=\begin{pmatrix}1&0\\a+b&1\end{pmatrix}=\varphi(a+b).$$

Injectivity and surjectivity should be clear.

Title: Is $(\mathbb{C},+)$ isomorphic to a matrix Lie group?

Body: I'm wondering if $(\mathbb{C},+)$ is isomorphic to a matrix Lie group; I mean, if it is isomorphic to a closed subgroup of $GL(n,\mathbb{C})$. If it is, what is the isomorphism? And if it is not, why?


问:
Summing a (partial) harmonic series
Given that a is a real number greater than 1, what is the value of

$$S = \lim\limits_{n\rightarrow\infty} S_n$$ where $S_n = \displaystyle \sum\limits_{i=1}^n \dfrac{1}{an-i}$, i.e.,
$$S_n = \frac{1}{an - 1} + \frac{1}{an - 2} + \ldots + \frac{1}{an - n}$$

I have chosen that if $a > 1$, S converges to a finite number, but I can't seem to arrive at an exact answer (which is what I need).

Thanks for any insights you can give.

Ron

Title: Summing a (partial) harmonic series

Body:
$S_n = \sum\limits_{i=1}^n \frac{1}{an-i}
= \sum\limits_{i=an-n}^{an-1} \frac{1}{i}
= H_{an-1}-H_{an-n-2}
$
where
$H_n = \sum\limits_{i=1}^n \frac1{i}$.

We now use the result that
$H_n = \ln(n)+\gamma+O(1/n)$
where $\gamma\approx 0.577$ is Euler's constant.

$\begin{align}
H_{an-1}-H_{an-n-2}
&= \ln(an-1)-\ln(an-n-2)+O(1/n)\\
&= \ln\big(\frac{an-1}{an-n-2}\big)+O(1/n)\\
&= \ln\big(\frac{an(1-1/(an))}{n(a-1)-2}\big)+O(1/n)\\
&= \ln\big(\frac{an(1-1/(an))}{n(a-1)(1-2/(n(a-1)))}\big)+O(1/n)\\
&= \ln\big(\frac{an}{n(a-1)}\big)
+\ln\big(\frac{1-1/(an)}{1-2/(n(a-1))}\big)+O(1/n)\\
&= \ln\big(\frac{a}{a-1}\big)
+\ln\big(\frac{1-1/(an)}{1-2/(n(a-1))}\big)+O(1/n)\\
&= \ln\big(\frac{a}{a-1}\big)+O(1/n)\\
\end{align}
$

for
$n > \max(2/a, 4/(a-1))$,
since,
for $0 < z < 1/2$,
$0 < -\ln(1-z)
= \sum_{k=1}^{\infty} \frac{z^k}{k}
< \sum_{k=1}^{\infty} z^k
= \frac{z}{1-z}
< 2z
$.


问:
What maths concepts should you know to model infectious diseases?
I'm a high school student and for my math research paper, I've to model and predict spread of an infectious disease (e.g. covid) in a country of my choice. Which math concepts should I know to do this activity? I've heard that I should learn differential equations, but is there anything else that I must know?

Could you also suggest some good books to learn how to do this?

Thank you,

答:
For pandemic-like modeling, usually logistic functions are the right approach. You can research a bit about them here (Wikipedia page). The are those "s" form curves(they are called sigmoid curves if I'm not wrong), and they are useful not only in pandemic but in lots of fields of study.

Some real functional analysis might be enough for you to understand this (derivatives, etc.), and maybe some differential equations.


问:
Questions About Base and Hausdorff in Topology
I am really confused about below questions. I am new both in this platform and below topics. I hope, I was clearly explain the questions.

For ∀n∈$\mathbb{Z}$+

Based on $B_n$ = {n, n+1, n+2...}, B = {$B_n$|n ∈ $\mathbb{Z}$+} family of sets are given. According to this,

(i) Family B is the base for a topology on $\mathbb{Z}$+. Show it.
(ii) Show that there is no Hausdorff by writing the topology on $\mathbb{Z}$+ produced by B.
(iii) Show that the sequence (2,4,6,8,...) in $\mathbb{Z}$+ converges to each point according to topology produced by the family B

Thank you...

答:
$(B_n), n \in \Bbb Z^+$ is a base for a topology, because their union is $\Bbb Z^+$ and the collection is closed under finite intersections.

The space is not Hausdorff, because there is no basic element that contains $1$, but does not contain $2$. So $1$ and $2$ cannot be separated by open sets. Writing out the whole topology is not needed (though not hard).

Let $n \in \Bbb Z$. If $B_k$ is any basic element that contains $n$, then $k \le n$ and then all members of the sequence that are $>n$ (all but finitely many one, thus) are in $B_k$. So the sequence converges to $n$.


问:
If $u$ is a solution of $F(x) \cdot \nabla u = h(u) $ and $F(x) \cdot v(x)>0 \forall x \in \partial U$ then $u\equiv 0$

Let $U\subseteq \mathbb{R}^n$ open and bounded; $0 \in U $ and $\partial U$ is smooth. Let $u \in C^1(\bar{U})$ be a solution of the equation
$$F(x) \cdot \nabla u = h(u),$$
with $F:\bar{U} \to \mathbb{R}^n$ continuous satisfying $F(x)\cdot v(x)>0, \forall x \in \partial U$, where $v(x)$ is the normal unit vector exterior to $\partial U$. If $h$ is a decreasing function and $h(0) = 0$ then $u \equiv 0$.

I have to solve this (yes, it is my homework, I signed up to some classes and I don't have the prerequisites and so I'm lost in almost everything I have to do. Please help me learn. I might need to know something very basic but I'm missing that piece of information).

Now, given thet $\partial U$ is smooth I can use the divergence theorem on this. So far I have this:

$$\int_U \nabla\cdot F(x) dx = \int_{\partial U} F(x) \cdot v(x) dx >0,$$
As $h$ is decreasing and $h(0)=0$ I know $h(x)\leq 0 \forall x \in \mathbb{R}$ and so
$$0\geq\int_U h(u(x))dx = \int_U F(x)\cdot \nabla u(x) dx = \int_U \nabla\cdot [F(x)u(x)]dx - \int_U [\nabla\cdot F(x)]u(x)dx$$

So $$\int_U \nabla \cdot [F(x)u(x)]dx \leq \int_U[\nabla\cdot F(x)]u(x)dx$$

How do I carry on? I get that I should find something on the lines of: "on one hand, on $\partial U$ we have that $F(x)\cdot v(x)$ is always positive, on the other $h(u(x))$ is never positive, so $u$ must be $0$ everyhere".

But I don't know how to do so. Thanks in advance.

Tips provided by the professor today (haven't tried it yet, but I'll post it here already in case someone wants to give it a try as well).

This is just my recollection of what he said in class, it is in no way formally written:

The function u is $C^1(\bar{U})$ and therefore has a maximum (and minimum) value somewhere in $\bar{U}$. Now we have two cases, either the maximum (and minimum) is inside $U$ or is in the boundary $\partial U$.

We'll talk about the maximum but the result follows similarly for the minimum.

If u has it's maximum in $x \in U$ then we have that, at the maximum the gradient of $u$ is $0$. and so
$$h(u(x)) = F(x)\cdot \nabla u(x)  = F(x) \cdot 0 =0$$

as the function $h$ is decreasing, and $h(0) = 0$, we have that $u(x)$ must be $0$, as it is the only place where $h$ assumes the value $0$.

and so the maximum of the function $u(x)=0$. Similarly we find that it's minimum is also $0$, so $u \equiv 0$.

If on the other hand the maximum(and minimum) is on the boundary, then we have a problem. The professor sugested using "Lagrange multiplier" to solve this part.

I have no clue what a lagrange multiplier is, but I'll read and after I learn it I believe I'll be able to solve the rest.

答:
I think I have a solution. Fix me if I am wrong.

There are two different options:

$u$ has it's maximum in $x_{\max} \in U$ we obtain that $u(x_{\max}) = 0$

$u$ has it's maximum in $x_{\max} \in \partial U$.

Consider the second case:

As far as $\partial U$ is smooth we can choose such function $g(x) \in C^1(\bar{U})$ so that $\partial U = \{x \, |\, g(x) = 0\}$ and the normal vector $\nabla g(x)$ to $\partial U$ will be external hence $\nabla g(x) = \alpha v(x)$ where $\alpha > 0$. So now we can apply Lagrange necessary conditions for a constrained extrema:
$$\exists \lambda: \,  \nabla u(x_{\max}) = \lambda \nabla g(x_{\max})$$
By multiplying left and right part of the last equation by $F(x_{\max})$ we will get:

$$ F(x_{\max}) \nabla u(x_{\max}) = \lambda F(x_{\max}) \nabla g(x_{\max})$$
$$ h(u(x_{\max})) = \lambda \alpha F(x_{\max}) v(x_{\max})$$

By the condition of the problem $F(x_{\max}) v(x_{\max}) > 0$

Let $\lambda \geq 0$ then: $h(u(x_{\max})) > 0 $ which is impossible because $h$ is decreasing and $h(0) = 0$ hence $h(x) < 0$.

So $\lambda < 0$. Which actually mean that $\nabla u(x_{\max})$ is interior normal of the $\partial U$.

Hence if $h > 0$ we obtain that vector ($x_{\max} + h \nabla u(x_{\max})) \in U$. Since $x_{\max}$ is maximum of $u$:

$$ \frac{u(x_{\max} + h \nabla u(x_{\max})) - u(x_{\max})}{h} \leq 0$$

Going to the limit we will get the directional derivative of $u$ at $x_{\max}$ along the vector $\nabla u(x_{\max})$

$$ 0 \leq \nabla u(x_{\max}) \cdot \nabla u(x_{\max}) = \lim_{x \to 0+} \frac{u(x_{\max} + h \nabla u(x_{\max})) - u(x_{\max})}{h} \leq 0$$

Which means that $\nabla u(x_{\max}) = 0$ which is impossible. Thus we got that $x_{\max} \in U \implies u(x_{\max}) = 0$.

Doing similarity for $x_{\min}$ we conclude that:

if $x_{\min} \in U$ we obtain that $u(x_{\min}) = 0$

if $x_{\min} \in \partial U$ we obtain that $h(u(x_{\min})) = \lambda \alpha F(x_{\min}) v(x_{\min}) < 0$. But since $h$ is decreasing we get $u(x_{\min}) \geq 0 = u(x_{\max})$.

It both cases we get $u(x_{\min}) = u(x_{\max}) = 0$ which implies $u \equiv 0$.


问:
Is the description $\bigwedge^{k} V_{\mathbb{C}}^{*} = \bigwedge^{k} V^{*} \oplus i\bigwedge^{k} V^{*}$ right?
I am trying to learn about complex differential forms and I'm having some problem understanding the equality presented in the title, which I have seen in Volker Scheidemann's book "Introduction to Complex Analysis in Several Variables". I will leave below a screenshot of the statement in the book.

As he says, each $\mu \in \bigwedge^{k} V_{\mathbb{C}}^{*}$ is a real valued alternating multivariable function on $V_{\mathbb{C}}^{k}$, so to me, the equality does not makes sense as an element on the right seems to be a complex valued function right? What am I missing?

I appreciate all the help in advance.

答: Here Scheidemann writes $\bigwedge^k V_{\mathbb C}^*=\bigwedge^k V^*\oplus i\bigwedge^k V^*.$ Notice that there are no $\mathbb C$ subscripts on the right hand side. One way to think about it is like this: if you fix a basis $\{v_j\}_{j=1}^n$ for $V,$ then $V_\mathbb C$ has $\mathbb R$-basis $\{v_j, iv_j\}_{j=1}^n.$ Then the real dual space is spanned by $\{\varphi_j,\psi_j\}_{j=1}^n$ where $$\varphi_j(v_k)=\delta_{jk}\\ \varphi_j(iv_k)=0\\ \psi(v_k)=0\\ \psi_j(iv_k)=\delta_{jk}$$ and the $\varphi_j$'s and the $\psi_j$'s are real linear.


问:
What is the terminology for the non-repeating portion of a rational decimal?
Given a number co-prime with 10, such as thirteen, we can construct a repeating decimal from its reciprocal: $\frac{1}{13}$ = 0.(076923). If we successively divide this number by a factor of 10 (i.e., 2 or 5) we get a sequence of numbers which have a fixed (non-repeating) and repeating portion (shown in parentheses):

$\frac{1}{26} = 0.0(384615)$

$\frac{1}{52} = 0.01(923076)$

$\frac{1}{104} = 0.009(615384)$

$\frac{1}{208} = 0.0048(076923)$

$\frac{1}{416} = 0.00240(384615)$

$\frac{1}{832} = 0.001201(923076)$

What is the correct terminology for the non-repeating and repeating portions of the decimals?

答: Given an eventually-periodic sequence, the piece occurring before the periodicity starts is called the preperiod (at least, this is the term I've always heard for it; note, however, that it might also be used to refer to the length of this piece). I don't know if there's a more specialized term for when we are talking about digits of a decimal expansion, but I think it's a good name for it regardless.


问:
How to find probability of a sum of reciprocals of random variables?

The probability density functions of two independent random variables $X$ & $Y$ is known.

To be more specific $X$ & $Y$ have Rayleigh and Rician distribution respectively.

I want to find the following probability:
$P\left(\frac{XY}{X + Y + 1} < c\right)$, where $c$ is a constant.

After slight reframing the problem, I have put it as
$$
P\left[\left(\frac{1}{X} + \frac{1}{Y} + \frac{1}{XY}\right) >
\frac{1}{c}\right]
$$
Can anyone suggest a simple way to arrive at the final solution $?$.

答:
Sketch for a solution: let $Z:= \frac{XY}{X+Y+1}$ where $X$ and $Y$ are independent and have density, then you have that

$$
\Pr [Z\leqslant c]=\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(x,y)f_X(x)f_Y(y)\mathop{}\!d (x,y)
$$

for $A_c:=\{(x,y)\in \mathbb{R}^2:xy/(1+x+y)\leqslant c\}$. Then when $(1+x+y)(c-x)>0$ we have that
$$
\frac{xy}{1+x+y}\leqslant c \iff y\geqslant c\frac{1+x}{x-c}\\
\text{ and }(1+x+y)(c-x)>0 \iff c+(c-1-y)x-x^2>0
$$
Then, solving the last inequality with the roots, we can divide $A_c$ in a finite number of disjoint sets with the form $\{(x,y)\in \mathbb{R}^2: y\geqslant f(x)\}$ for some functions $f$, then we can apply Fubini's theorem easily in each of these sets to compute each integral (the set where $(1+x+y)(c-x)=0$ have Lebesgue measure zero so we can ignore it).


问:
Evaluating $\lim_{x\to 0}\frac1{x^2}({\arctan(1+x^2)-\arcsin\frac{\cos x}{\sqrt{2}}})$
I've been struggling for a while with the following limit:
$$L = \lim_{x\to 0}\frac{\arctan(1+x^2) - \arcsin\left(\frac{\cos x}{\sqrt{2}}\right)}{x^2}$$

I know that $L = 1$, since it seems obvious from the graph and WolframAlpha confirms it, yet I simply can't find a way to theoretically evaluate it. I've tried all sorts of trig substitutions, two of which seem pretty intuitive:

$\cos x = \sin\alpha+\cos\alpha = \sqrt{2}\sin\left(\alpha + \frac{\pi}{4}\right)$ (suitable for the right part)

$1+x^2 = \tan\left(\theta + \frac{\pi}{4}\right)$ (suitable for the left part)

I thought of those because the limit kind of hints at them and both new variables also tend to $0$ and I feel I'm really close, but I couldn't finish anything and also started doubtingthis might not be the way to go. Any ideas are really welcome and greatly appreciated!

Edit: I forgot the mention, this limit is somehow supposed te be evaluable without using L'Hopital's rule, that's what all the fuzz is about! :D I also grew suspicious if it's even possible, but it's an interesting challenge nonetheless!

答:
Is it permitted to use Taylor expansion? Like

$$ \arcsin(x) = \int^x_0 \frac{1}{\sqrt{1-t^2}}\,dt = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc $$


问:
Finding a quadratic equation given range and vertex to origin distance
The question :

"Find the equation of quadratic function that value positive for $-7 < x < 1$ and the distance of vertex and origin is 5"

I've just tried to let $-7$ and $1$ are the $x$-intercepts, however, nothing works. Could someone help me, please?

答:
You can use vertex form of the quadratic function which is given by,

$y - k = a(x-h)^2$.

Given $y$ is positive for $-7 \lt x \lt 1$, we take these points where the function $y$ turns from positive to negative.

Equating $y$ to zero, we have $x = h \pm \sqrt{-\frac {k}{a}}$

Solving, $h - \sqrt{-\frac {k}{a}} = - 7$ and $h + \sqrt{-\frac {k}{a}} = 1$

$h = -3, -\frac{k}{a} = 16$

As $h^2 + k^2 = 25, k = \pm4$ We know the function is positive for $x \in (-7,1)$ so $k = 4$.

$a = -\frac{1}{4}$.

The equation of quadratic is $y = -\frac{1}{4} (x+3)^2 + 4$


问:
$\lim_{x\to 3}((x+1)(x-3)\tan(\frac{x\pi}{2}))$
Been wondering on how to compute this limit without:

L'hôpital

Any Taylor Series

Only with trigonometric identities:
$\lim_{x\to 3}((x+1)(x-3)\tan(\frac{x\pi}{2}))$

And yeah I've tried some stuff: $$=\lim_{x\to 3}\frac{(x+1)(x-3)2\tan\left(\frac{xπ}{4}\right)}{1-\tan^2\left(\frac{xπ}{4}\right)}$$ by the double angle identity of tan. Then, I expressed tan as sin and cos and cancelled one $\cos(\frac{x\pi}{4})$ $$=\lim_{x\to 3}\frac{\frac{(x+1)(x-3)2\sin(\frac{xπ}{4})}{\cos(\frac{xπ}{4})}}{\frac{\cos^2(\frac{xπ}{4})-\sin^2(\frac{xπ}{4})}{\cos^2(\frac{xπ}{4})}}$$ $$\lim_{x\to 3}\frac{(x-1)(x-3)2\cos(\frac{x\pi}{4})\sin(\frac{xπ}{4})}{\cos^2(\frac{xπ}{4})-\sin^2(\frac{xπ}{4})}$$ then I used $\cos^2(\alpha)=1-\sin^2(\alpha)$ and $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ and simplified to $$\lim_{x\to 3}\frac{(x+1)(x-3)\sin(\frac{x\pi}{2})}{1-2\sin^2(\frac{x\pi}{4})}$$ but I feel like I'm stuck here.

答:
Hint:
As $\lim_{x\to3}(x+1)$ is non-zero finite

$$\lim_{x\to3}(x+1)(x-3)\tan\dfrac{\pi x}2=\lim_{x\to3}(x+1)\cdot\lim_{x\to3}(x-3)\tan\dfrac{\pi x}2$$

Set $x-3=h,$ then $$\tan\dfrac{\pi x}2=\tan\left(\dfrac{\pi(3+h)}2\right)=\tan\left(\dfrac{3\pi}2+\dfrac{\pi h}2\right)=-\cot\dfrac{\pi h}2$$

Use $\lim_{h\to0}\dfrac{\sin h}h=1.$


答:
So a quadratic with roots at $1$ and $\mbox{-}7$ will be $c(x+7)(x-1)$, where $c$ is a real number. We just have to figure out what $c$ should be. We need $c$ to be negative if you want the function to be positive on only $(\mbox{-}7,1)$. Also, here the vertex is the point $\left(\mbox{-}\dfrac{b}{2a},f\left(\mbox{-}\dfrac{b}{2a}\right)\right)$. Let's write $\mbox{-}t$ for our number $c$ (so t will be positive), so our polynomial will be:

$$f(x)=c(x^2+6x-7)=cx^2+6cx-7c$$

To find what c is, we assume that the distance between the vertex and the origin is $5$ as desired. Compute the vertex of $f(x)$:

$$-\frac{b}{2a} = -\frac{6c}{2c} = -3$$

$$f(-\frac{b}{2a}) = f(-3) = c((-3)^2 + 6(-3) - 7) = c(9 -18-7) = -16c$$

which is what it means for the vertex to be:

$$v = \bigg(-\frac{b}{2a},f(-\frac{b}{2a})\bigg) = (-3,-16c)$$

The distance between the vertex and the origin being 5 means that:

$$5 = \sqrt{(-3-0)^2+(-16c-0)^2} =\sqrt{9 + 256c^2}$$

solving for c gives $c=\dfrac{\mbox{-}1}{4}$ (there are two solutions but we need the negative one like I stated).

Title: Finding a quadratic equation given range and vertex to origin distance

Body:
The question :

"Find the equation of quadratic function that value positive for $-7 < x < 1$ and the distance of vertex and origin is 5"

I've just tried to let $-7$ and $1$ are the $x$-intercepts, however, nothing works. Could someone help me, please?


问:
Primes congruent to 1 modulo n
I was wondering whether the following statement is true, and if so, how to prove it:

for all $n\in\mathbb{N}$, there exist prime numbers $p$ and $q$ such that $p\equiv1$ (mod $n$) and $q\equiv-1$ (mod $n$)

Feels like it should be true and that it should be fairly easy to prove, but I don't have any idea how. Could someone please shine some light on this for me?

答:
Comment: You can construct such primes; tank an arbitrary number like n= 7 we can have:

$p=2\times7-1=13$ ⇒ $p=13\equiv -1 \mod (2)$

$q=6\times 7+1=43$ ⇒ $q=43 \equiv 1 \mod 7$

In fact we must have:

$p=k\cdot n+1$

$q=t\cdot n-1$

From second relation we have:

$n=\frac{q+1}t$

Plugging this in first relation we get:

$p=\frac k t (q+1)+1=\frac kt q+\frac kt +1$

Let $\frac kt=a$ then:

$p=a q + a+1$

This equation has infinite solutions for p, q for certain value of a, among them primes for p and q. The condition is  $k>t ; t|k$.

Title: Primes congruent to 1 modulo n


问:
Is it equally correct to state of square root of i is plus or minus -1 to the power 1/4?
I understand that the two square roots of $i$ are

$$\pm {1 \over \sqrt{2}} (1+i)$$

But when asked what is the square root of $i$, is it equally correct to give the answer as

$$\pm {(-1) ^ {1/4}}$$

? And actually, since $\sqrt 2$ is defined to be only the positive root, so square root of $i$ can be just

$${(-1) ^ {1/4}}$$

?

答:
I'll ignore $0$ throughout.  Also, $n$ will be a positive integer.

First, with the more familiar real numbers.  When $n$ is even, positive numbers have two distinct $n$th roots and negative numbers have none.  When $n$ is odd, all numbers have one $n$th root.

It is quite different for the complex numbers.  All non-zero numbers have $n$ distinct $n$th roots.

So, what you say cannot be correct as there will be two square roots but four fourth roots.

A further common problem is talking of the square root, etc.  Among the reals, we can choose that the positive square is the principal one and gets to be called the square root.  For the complex numbers, we cannot pick a preferred root so neatly and some familiar behaviour, e.g. $\sqrt{x}\times \sqrt{y} = \sqrt{xy}$ might not be true.


问:
Question in solving wave equation by eigenvalue-eigenfuntion
I am learning to solve PDEs, I found a example the following from a book,

$u_{tt}(x,t)$=$u_{xx}(x,t)$

$u_x(0,t)=0$, $u(L,t)=0$

$u_x(x,0)=\sin\frac{3\pi x}{2}$, $u_t(L,t)=\cos\frac{3\pi x}{2}$.

Now by separating the variables,

$T''(t)+\lambda T(t)=0$,

and a regular S_L problem

$X''(x)+\lambda X(x)=0$, $X'(0)=0$, $X(L)=0$

The eigenvalue-eigenfuntion are

$\lambda_n=\frac{(2n-1^2)\pi^2}{4L}$, $X_n(x)=\cos\frac{(2n-1)\pi x}{2L}$, $n=1,2,...$

Now I don't know how to generates the solution for $T_n(t)$?

答:
You have equation $T''(t) + \lambda T(t) = 0$. This equation has a general solution $T_n(t) = A_n \cos(2^{-1} L^{-1} (2 n -1) \pi) + B_n \sin(2^{-1} L^{-1} (2 n -1) \pi)$. Now you can write
$$u(x, t) = \sum_{n=1}^{\infty}  ((A_n \cos(2^{-1} L^{-1} (2 n -1) t \pi) + B_n \sin(2^{-1} L^{-1} (2 n -1) t \pi) \times \\ \times \cos(2^{-1} L^{-1} (2 n -1) \pi x)).$$
Now you can define $A_n$ and $B_n$ using initial conditions.


问:
question about percentages
At the start of the new year, the price for a gallon of milk at a local grocery store increased by 2.4% as a result of inflation. If a gallon of milk cost $2.92 last year, about how much does it cost this year?</p>

```
A.  $2.84
B.  $2.99
C.  $3.62
D.  $5.32
```

Title: question about percentages

Body: This is the solution $(2,4/100)*2,92 +2,92=2,99$!


问:
Are all morphisms in category theory are actually some kind of mappings? What is the importance of associativity rule in category theory?
I always found the identity and the composition rule trivial while thinking about categories. But I think the associativity rule is NOT trivial and it is hard to create a composition table that obeys the associativity rule. Do you think there is something about the associativity rule that makes it very special and useful in mathematics, or am I just confused? I know we can create an associative composition table using mappings between sets and using permutations. So can we say all morphisms in category theory are actually some kind of mappings?

For example, this is a poset with 3 objects. And the morphisms are not mappings, they represent some relation between these 3 objects.

But I could turn the objects into specific sets, and turn the morphisms into specific mappings between those sets. So that the composition table stays the same and arrows are still associative. (I don't have to add every possible mapping as arrows, I just added one of them.)

So is it possible to turn every category into a combination of some sets and mappings between those sets?

Note: I am very new to category theory and abstract algebra, sorry if this question is confusing.

答:
(1) Yes associativity is extremely useful, especially when doing algebra (e.g. module theory or commutative algebra), e.g. 2 fold associativity gives us the generalized (finite-fold) associativity in a category (just by induction)

(2) No morphisms need not be mappings, often times they are equipped with a few mappings, or even an indexed family of mappings (if by mappings you mean set maps), there are categories like ones given by monoids and groups where the morphisms are not really mappings at all, they are elements of the monoid or group (resp.) in this case for instance.


问:
Combinatorics - letters forbidden sequence.
I am trying to solve a question. I have 15 pens, where 5 are colored in Red, 5 in Green, and 5 in Blue. How many combinations can I order these pens so that there is no sequence of 5 pens with the same color? For instance, the combination RRRRG GGGBB BBRGB is okay because there is no sequence of pens where 5 colors are after each other. The combination RRRRR ... would need to be excluded because 5 red pens are in a row.

My initial strategy was to count the total number of combinations to build with these pens and then subtract the total number of combinations where at least 5 colors are in sequence.

The total number of combinations:

$$\frac{15!}{(5!*5!*5!)} = 756756$$

The issue I am having is how to create all the combinations where 5 colored pens are in order. I have tried to do something like this:

Group the sequence RRRRR into one symbol. Now, calculate every combination with this RRRRR symbol and the other symbols.

$$\frac{11!}{5!*5!}=2772$$

The problem here is that if I multiplied this with 3 to get the combinations from every color, there would be many duplicated combinations.

Any help on how I could solve this is much apprecited.

答:
Here is a generating function approach. Words with no consecutive equal characters at all are called Smirnov words or Carlitz words. See example III.24 Smirnov words in Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.

A generating function for the number of Smirnov words over a ternary alphabet is given by
\begin{align*}
\left(1-\frac{3z}{1+z}\right)^{-1}\tag{1}
\end{align*}
Since each letter may occur in runs with length $\leq 4$ we substitute
\begin{align*}
z\to z+z^2+z^3+z^4=\frac{z\left(1-z^4\right)}{1-z}\tag{2}
\end{align*}

We substitute (2) in (1) and obtain the generating function $f(z)$:
\begin{align*}
\color{blue}{f(z)}=\left(1-\frac{3\,\frac{z\left(1-z^4\right)}{1-z}}{1+\frac{z\left(1-z^4\right)}{1-z}}\right)^{-1}
\color{blue}{=\left(1-\frac{3}{1+\frac{1-z}{z\left(1-z^4\right)}}\right)^{-1}}\tag{3}\\
\end{align*}

The coefficient $[z^{15}]f(z)$ gives the number of all words of length $15$ built from the alphabet $\mathcal{V}=\{R,G,B\}$ which do not contain characters with runs of length $5$. Since we are interested in the number of words which contain five $R$s, five $G$'s and five $B$'s we parametrize the function $f(z)$ and obtain from (3) with some help of Wolfram Alpha the number of valid words:

\begin{align*}
\color{blue}{[R^5G^5B^5z^{15}]f(z;R,G,B)}
&=[R^5G^5B^5z^{15}]\left(1-\sum_{q\in\{R,G,B\}}\frac{1}{1+\frac{1-qz}{qz\left(1-(qz)^4\right)}}\right)^{-1}\\
&\,\,\color{blue}{=748\,560}
\end{align*}


问:
Multivariate Polynomial Division Question
I am trying to solve the following question:

``Let $F$ be a field and let $F[x, y]$ denote the ring of  polynomials in the variable $x$ and $y$ with coefficients in $F$. Suppose $f(x,y)$ belongs to $F[x, y]$. Apply the Factor Theorem to the ring $F[x]$ to show that $f(x, x) = 0$ if and only if $(x-y)$ is a factor of $f(x, y)$. More generally, show that $y - g(x)$ divides $f(x, y)$ if and only if $f(x, g(x)) = 0$, for $g(x)$ in $F[x]$.''

If $(x-y)$ is a factor of $f(x, y)$ then obviously $f(x, x) = 0$. But how do I prove the converse? Is there a version of the Factor Theorem for $F[x, y]$ that I need to use?

答:
If $R$ is a commutative ring, then for the ring $R[x]$ we have the Factor Theorem:

Let $f\in R[x]$ and let $r\in R$. Then in $R[x]$ we have $(x-r){\,\mid\,}f\;$if and only if $f(r)=0$.

Now suppose $F$ is a commutative ring and $f\in F[x,y]$ is such that $f(x,x)=0$.

Let $R=F[x]$ and let $g\in R[y]$ be defined as $g(y)=f(x,y)$.
\begin{align*}
\text{Then}\;\;&
f(x,x)=0\\[4pt]
\implies\;&
g(x)=0
\\[4pt]
\implies\;&
(y-x){\,\mid\,}g\;\text{in $R[y]$}&&\text{$\Bigl($by the Factor Theorem for $R[y]\Bigr)$}
\\[4pt]
\implies\;&
(x-y){\,\mid\,}g\;\text{in $R[y]$}
\\[4pt]
\implies\;&
(x-y){\,\mid\,}f\;\text{in $F[x,y]$}
\\[4pt]
\end{align*}
as was to be shown.


问:
Is there any homeomorphism between fundamental domains?
It is known that for the same fuchsean group if I take two different points I will get two different fundamental domains. Is there any homeomorphism between these fundamental domains?</p>

答:
Of course the quotient of the group action is well-defined independent of the choice of fundamental domain. Therefore the quotient of one fundamental domain is homeomorphic to the quotient of any other fundamental domain, both of them being homeomorphic to the quotient of the group action. But you didn't ask about quotients, you asked about the fundamental domains themselves.

If the group action is cocompact then the fundamental domain is homeomorphic to the closed disc $D^2$. So yes, two fundamental domains of this type will be homeomorphic to $D^2$ and therefore homeomorphic to each other.

If the group action is not cocompact then the fundamental domain will be homeomorphic to a disc with some finite number of points removed from the boundary. In this case, the number of points removed determines the topology of the fundamental domain and so two such fundamental domains are homeomorphic if and only if they have the same number of removed points. But it's not hard to construct counterexamples where two fundamental domains for the same Fuchsian group action have different number of points removed from their boundary, and therefore those two fundamental domains are not homeomorphic.

For example, in the case that the quotient surface is homeomorphic to a once-punctured torus, there will be a fundamental domain which gives the usual square gluing diagram with gluing word $a b a^{-1} b^{-1}$ and with all four corners removed, and so that fundamental domain homeomorphic is to a disc with four boundary points removed. But there will also be a fundamental domain which gives a hexagonal gluing diagram with gluing word $a b a^{-1} b^{-1} c c^{-1}$, and with only the vertex between the $c$ and $c^{-1}$ edge removed, so that fundamental domain is homeomorphic to a disc with one boundary point removed.


问:
Riemann integral question
Suppose that $ f : [a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $[a,b]$ and $g:[a,b] \rightarrow \mathbb{R}$ differs from $f$ at only one point $x_0 \in [a,b]$, that is, $g(x)=f(x)$ for $x \neq x_0$ and $g(x_0) \neq f(x_0)$. Show that $g$ is Riemann integrable on $[a,b]$.

I'm having a little trouble, my thing was that maybe find a partition and look at how it behaves in the partition containing $x_0$

Appreciate any help

答:
General hint: if you are faced with a problem, first try to understand what makes the problem interesting, and what is the noise; then filter the noise and extract the interesting part.

Let $\delta_z(x) = 1$ for $x = z$ and $\delta_z(x) = 0$ otherwise. If we showed that for any $z$ the integral $\int \delta_z(x) dx$ exists and equals $zero$ (this is the interesting part), we would have: $$g = f + c\delta_z(x)$$ and (this is the noise done by your profesor during the course) $$\int g(x) dx = \int f(x) + c\delta_z(x) dx = \int f(x) dx + c\int \delta_z(x) dx = \int f(x) dx$$

The interesting part is up to you.

Title: Riemann integral question


答:
You are very close to the correct solution. The  metod of choice for solving such problems is the principle of inclusion and exclusion. Applying it to the problem and taking into account the symmetry between the colors one obtains:
$$
\binom30\frac{15!}{5!5!5!}-\binom31\frac{11!}{5!5!1!}+\binom32\frac{7!}{5!1!1!}-\binom33\frac{3!}{1!1!1!}=748560.
$$

Title: Combinatorics - letters forbidden sequence.

Body:
I am trying to solve a question. I have 15 pens, where 5 are colored in Red, 5 in Green, and 5 in Blue. How many combinations can I order these pens so that there is no sequence of 5 pens with the same color? For instance, the combination RRRRG GGGBB BBRGB is okay because there is no sequence of pens where 5 colors are after each other. The combination RRRRR ... would need to be excluded because 5 red pens are in a row.

My initial strategy was to count the total number of combinations to build with these pens and then subtract the total number of combinations where at least 5 colors are in sequence.

The total number of combinations:

$$\frac{15!}{(5!*5!*5!)} = 756756$$

The issue I am having is how to create all the combinations where 5 colored pens are in order. I have tried to do something like this:

Group the sequence RRRRR into one symbol. Now, calculate every combination with this RRRRR symbol and the other symbols.

$$\frac{11!}{5!*5!}=2772$$

The problem here is that if I multiplied this with 3 to get the combinations from every color, there would be many duplicated combinations.

Any help on how I could solve this is much apprecited.


问:
Evaluate $\int_C{(z^3-e^z)dz}$ along a curve

Evaluate $\int_C{(z^3-e^z)dz}$ where C is the arc of the circle centered at $0$ from $(3,0)$ to $(0,-3)$.

I tried using $\gamma(t)=3\cos(t)+3i\sin(t)$ at first, but it gets confusing.

In the other exercises we saw in class, the assistant always uses short cuts such as "the integral is $0$ if $f$ is holomorphic", or "the integral of $z^n$ is $2\pi i$ if $n=-1$", etc. But I don't quite know how to approach this one. Should I separate the integral in two parts and evaluate $z^3$ and $e^z$ separately? Also the function is whole, so I can't use the residue theorem.

Body:
See here.

The line is the arc of the circle $\rho=3$ with $t\in[-\pi/2,0]$.
$$z^3-e^z;\;z=3e^{it};\;dz=3ie^{it}dt$$
$$\int_{C}(z^3-e^z)\,dz=\int_{-\pi/2}^0\left(27 e^{3 i t}-e^{3 e^{i t}}\right) \left(3 i e^{i t}\right)\,dt=\left[\frac{81}{4} e^{4 i t}-e^{3 e^{i t}}\right]_{-\pi/2}^0=e^{-3 i}-e^3$$