问:
Measurable function of a transformation still measurable
I'm looking for a maybe simpler or more elemental proof of the following statement:

Let $f:\mathbb{R^n}\to \mathbb{R}$ be (Lebesgue-)measurable.
Then $F:\mathbb{R^{2n}}\to \mathbb{R}$ such that $F(x,y)=f(x-y)$ is (Lebesgue-)measurable.

The proof I know works in the following way:
First show that $f$ is measurable on $\mathbb{R^{2n}}$.
Then $T(x,y)=(x-y)$ is Lipschitz and hence $f\circ T$ is measurable (by a theorem ).

Can one somehow argue without the Lipschitz continuity in the last step or with a different ansatz? Several books I've seen introduce the convolution using that $f(x-y)$ is measurable but proof that Lipschitz $T$ maps measurable sets onto measurable sets later in the text.

答:
Alright, so this is as elementary and intuitive as I could think of. We assume only that if $U \subset R^n$ is measurable and $T$ is a rotation on $R^n$, then $T(U)$ is measurable. This is a standard fact about the Lebesgue measure and should be covered quite early in most books. For simplicity, consider $n = 1$.

Define $G:R^2 \to R$ by $G(x,y) = y-x$. Since $F^{-1}(U) = G^{-1}(f^{-1}(U))$ for each $U \in R$ and $f$ is assumed measurable, we only have to show that $G^{-1}(U) \subset R^2$ is measurable for each measurable $U \subset R$ in order to prove that $F$ is measurable.

If $U = \{c\}$ consists of a single point, then
$$G^{-1}(\{c\}) = \{(x,y) \in \mathbb{R}^2: y-x = c\} = \{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\}$$

which is the graph of the function $y(x) = x + c$. The slope of this function is $1$ which corresponds to an angle with the $x$-axis of $45$ degrees.

If we let $T: R^2 \to R^2$ be the linear transformation that rotates clockwise by $45$ degrees, then

$$T(G^{-1}(\{c\})) = T(\{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\}) = \{(x,c) \in R^2: x \in R\}$$

Now if $U \subset R$ is measurable, then
$$U = \bigcup_{c \in U} \{c\}$$
which means that
$$T(G^{-1}(U)) = \bigcup_{c \in U} T(G^{-1}(\{c\})) = \bigcup_{c \in U}T(\{(x,x+c)\in \mathbb{R}^2: x \in \mathbb{R}\})$$
$$ = \bigcup_{c \in U}\{(x,c) \in R^2: x \in R\} = R \times U$$

which is the product of two measurable sets and hence measurable (by definition).
Thus
$$G^{-1}(U) = T^{-1}T(G^{-1}(U))$$
is measurable since $T^{-1}$ is also a rotation.


问:
Inductive proof for recursive formula $a_n=4-a_{n-1}$
I have a recursive formula $a_n=4-a_{n-1}, a_3=7$ and I had to solve this and check the correction of the answer with induction.

I started solving this and the solved formula is: $a_n=-2((-1)^{n-3}-1)+7(-1)^{n-3}$.

Base case $n=3$ is true

Let's assume this is true for $n=k$

Thus this should be true for $n=k+1$

$a_{n=k+1}=5(-1)^{k-2}+2$.

I don't know how to continue from this place, please someone help me and explain this, thank you.

答: Hint: Prove that $a_{n+2}=a_n$ for all $n$.

ContentLicense: CC BY-SA 4.0


问:
Find function f(z) for a matrix
Let $z \in \mathbb{C}$.

Let $ f(z)=
\begin{cases}
    1,& \text{if } |z|<\frac{1}{2}\\
    0, &             \text{if} |z-2| <\frac{1}{2} \\
0, & \text{if} |z-3| <\frac{1}{2}
\end{cases}$

Let \begin{equation*}
A =
\begin{pmatrix}
4 & 1 & 0 & -4 & 0 & 0\\
0 & 4 & -1 & 0 & -4 & 2\\
0 & 0 & 6 & 0 & 0 & -6\\
2 & 0 & 0 & -2 & 1 & 0\\
0 & 2 & -1 & 0 & -2 & 2\\
0 & 0 & 3 & 0 & 0 & -3\\
\end{pmatrix}\end{equation*}

Find  $f(A)$.

Attempt:

We know that $\mathbb{C}^6$ is a Hilbert space. and $B(\mathbb{C}^6) \simeq \mathbb{M_6}$. So we can consider $A$ as a function in $B(\mathbb{C}^6)$

And we can write $Ae_j= \sum_{n=1}^6a_n e_n$, where $(e_n)$ is the orthonormal basis for $\mathbb{C}^6$.

Also I know that the spectrum $\sigma(A)= \{\text{eigenvalues of} A \}$

From here I'm not sure how to proceed.

I'm trying to solve this using Banach Algebra point of view. But any other suggestions will be much appreciated too!

EDIT: My apologies, there was a typo in the original question. Now corrected

Thank you!

答:
Important edit: I just realized that $A$ is not a normal matrix. The technique below would work in general for a normal matrix, however I am not sure how $f(A)$ is even defined for a non-normal matrix. The continuous (or Borel for von Neumann algebras) functional calculus makes sense for normal elements only, as far as I remember.

Original Post:

Plugging the matrix in a calculator, you get the eigenvalues: $\sigma(A)=\{0,2,3\}$. The function $f$ you are defining is not defined on $0$, so I will write $c=f(0)$. If you can find a polynomial $p(z)$ such that $p(0)=c$, $p(2)=f(2)=0$ and $p(3)=f(3)=0$, (for example $p(z)=\frac{c}{6}(z-2)(z-3)$ is such a polynomial) then since $f=p$ on $\sigma(A)$ we have that $f(A)=p(A)$. But $p(A)$ is easily computed, for example if $p$ is the polynomial I mentioned earlier then
$$p(A)=\frac{c}{6}(A-2I)(A-3I).$$

Title: Find function f(z) for a matrix


答: Note that by replacing $x\to \frac{1}{x}$ $$I=\int_0^{\infty} \frac{1-x}{1-x^6}dx=\int_0^{\infty} \frac{(1-x)x^3}{1-x^6}dx$$Adding  $$2I=\int_0^{\infty} \frac{(1-x)(1+x^3)}{1-x^6}= \int_0^{\infty} \frac{1-x}{1-x^3}=\int_0^{\infty} \frac{1}{x^2+x+1}$$ which should be easy..

Title: Evaluating $\int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5}\mathrm{d}x$

Body:
How to evaluate $$\int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5} dx$$

My Attempt:

$$I =  \int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5} dx$$

$$I =  \int_{0}^{\infty} \frac{(1-x)}{(1 - x^6)} dx$$

Substituting $t = \frac{1}{x}$

$$I =  \int_{0}^{\infty} \frac{(1-\frac{1}{x})}{(1 - \frac{1}{x^6})} \cdot \frac{1}{x^2} dx$$

$$I =  \int_{0}^{\infty} \frac{(1-{x})(1 + x + x^2)}{(1-x^6 )} dx$$

This does not seem to work.


问:
let $F(x)=\int^x_1 t^{\sqrt {t}}dt$ be defined for every $x\ge 0$, then..
a) $F(x)>0$ for every $x>0$.

b) $F(x)>F(e)$ for every $x>1$.

c) $x=e^{-2}$ is an inflection point.

d) None of the answers are correct.

Note: This was a question in my exam and I wrote it from my memory so if there's something that doesn't make sense it might be that I messed up.

My Work was to take second derivative of $F$ and saw that $F''(e^{-2})\ne 0$, so I took out that option, and didn't know how to decide between (a) or (b), but I noticed that for (a) I might have to flip the integral, so I just went with (b), I would appreciate any explanation on how to really approach this question and not just with my intuition. (and know if I got it right too).

答:
($a$) is wrong because $F(1) = 0$.

($b$) is wrong because $F'(x)=x^{\sqrt {x}} > 0$ so the function is increasing.

($c$) is wrong as you have checked.

($d$) therefore, is the correct option.

Edit: ($c$) option appears to be correct as @Bernard has shown. I was just convinced by the claim in the question.


问:
Invariant polynomials under dihedral group action
I'm trying to solve the following problem:

Find a generating set for the algebra of invariant polynomials $\mathbb C[x_1, x_2]^\Gamma$, where $\Gamma$ is a dihedral group $D_n$, generated by matrices $\begin{pmatrix}\zeta && 0 \\ 0 && \zeta^{-1}\end{pmatrix}$ and $\begin{pmatrix} 0 && 1 \\ 1 && 0 \end{pmatrix}$. $\zeta$ is a primitive $n$th root of unity.

Here is my solution:

Polynomial is invariant under the action of $\Gamma$ iff it's invariant under the action of generators. Second generator of $\Gamma$ just permutes $x_1$ and $x_2$. So any invariant polynomial is in fact a symmetric polynomial of two variables. Any such polynomial is of the form $\sum a_i(x_1 + x_2)^{k_i} + \sum b_i (x_1x_2)^{k'_i} + \sum c_i (x_1 + x_2)^{k''_i}(x_1 x_2)^{k'''_i} + d$.

It is clear that $x_1 x_2$ is invariant under the action of the first generator of $\Gamma$. So we only want to check for which $k$ it is true that $(\zeta x_1 + \zeta^{-1} x_2)^k = (x_1 + x_2)^k$. If we expand both expressions, we can see that it means $\zeta^{k - 2 l} = 1$ for all $l \in [0, k]$. Since $\zeta$ is a primitive $n$th root of unity, it follows that $k - 2 l$ is a multiple of $n$. If $n = 2$, $k$ can be any even number. In this case, generator set for $\mathbb C[x_1, x_2]^\Gamma$ is $((x_1 + x_2)^2, x_1 x_2, 1)$.

If $n > 2$, such $k$ does not exist. Indeed, if $k = r_1 n$ and $k - 2 = r_2 n$ then $(r_1 - r_2)n = 2$ which is impossible. Thus generating set for $\mathbb C[x_1, x_2]^\Gamma$ in this case is $(x_1 x_2, 1)$.

Am I right? Am I missing something?

Thanks!

答:
I don't think you are fully right because there is no need for $(x_1+x_2)^k$ to be an invariant polynomial.

Since $D_n$ is a finite Coxeter Group, By Chevalley's theorem, $\mathbb{C}[x_1,x_2]^\Gamma$ is a polynomial algbera over $\mathbb{C}$ with two homogeneous generators say $f_1$ and $f_2$. If $f_1$ and $f_2$ are with degree $d_1$ and $d_2$, then $d_1d_2=2n$ ( the order of the Dihedral Group). Which is met in the first case but not for the others.

Or course $f_1=x_1x_2$ is invariant under $D_n$.

Let $f_2=x_1^n+x_2^n$. Then it's easy to check that $f_2$ is also $D_n$ invariant.

Now, since the Jacobian of $f_1$ and $f_2$, $J(f_1,f_2)=n(x_2^n-x_1^n)\neq 0$, it follows that $f_1$ and $f_2$ are algebraically independent over $\mathbb{C}$ satisfying $d_1d_2=2n$.

Now by Proposition of 3.12 ( Page no-67) of 'Reflection Groups and Coxeter Groups' by James E Humphreys, it follows that $\mathbb{C}[x_1,x_2]^\Gamma=\mathbb{C}[f_1,f_2]$ for any $n$.


问:
Suppose that $f$ is not strictly convex on $C$. Prove that there exist $x ,y \in !n(x \not= y)$ such that $f$ is affine over the segment $[x, y]$
Let $f$ be a convex function defined on a convex set $C$. Suppose that $f$ is not strictly convex on $C$. Prove that there exist $x, y \in \mathbb{R}^n,(x \not= y)$ such that $f$ is affine over the segment $[x, y]$.

My idea was assuming by negativity that, for every $x, y \in \mathbb{R}^n$, $f$ is not affine and therefore because $f$ is convex it must be strictly convex.

答:
Since $f$ is not strictly convex, $\exists x,y\in \mathbb R^n, \lambda\in(0,1)$ s.t. $z =\lambda x+(1-\lambda)y, \; f(z)=\lambda f(x)+(1-\lambda)f(y)$.

Claim: $f$ is affine in $[x,y]$, i.e. for any point $p\in[x,y], p=\lambda'x+(1-\lambda')y, f(p)=\lambda'f(x)+(1-\lambda')f(y)$.

Proof: Suppose towards contradiction that it isn't affine. I.e. there exists one point, $u\in (x,y)$ s.t. $f(u)\neq\lambda'f(x)+(1-\lambda')f(y)$. By the convexity of $f$ it must be that $f(u)<\lambda'f(x)+(1-\lambda')f(y)$. We know that the point cannot be $x,y$, or $z$, so it must either be in $(x,z)$ or in $(z,y)$. Assume WLOG that the point is  $\in(z,y)$ and denote it as a convex combination using $\alpha \in (0,1)$, s.t. $$u=\alpha z + (1-\alpha)y$$ $$f(u)=f(\alpha z+(1-\alpha)y)<\alpha f(z)+(1-\alpha)f(y)=\alpha\lambda f(x)+(1-\alpha\lambda)f(y)$$
(Note, we can do this because $x=\frac{z-(1-\lambda)y}{\lambda}, \; f(x) = \frac{f(z)-(1-\lambda)f(y)}{\lambda}$ hence $u = \frac{\lambda'}{\lambda}z +(1-\frac{\lambda'}{\lambda})y$ so we just denote $\alpha = \frac{\lambda'}{\lambda}$. $\lambda'f(x)+(1-\lambda')f(y)=\lambda'\frac{f(z)-(1-\lambda)f(y)}{\lambda} + (1-\lambda')f(y)=\alpha f(z)+(1-\alpha)f(y)$)

Now, let us choose symmetrically the point $v = \alpha x + (1-\alpha)z$. From convexity we get that
$$f(v) \le (\alpha + \lambda -\alpha\lambda)f(x) + (1-\alpha-\lambda -\alpha\lambda)f(y)
$$
Let us choose
$$ t = \frac{\alpha\lambda-\alpha}{2\alpha\lambda-\alpha-\lambda}$$
It's easy to show that (for $\alpha,\lambda\in(0,1)$) $t\in(0,1)$. First $\alpha\lambda<\alpha$ and $\alpha\lambda < \lambda$ - so both the numerator and the denominator are negative, hence $t$ is positive. Second, the denominator is the numerator + another negative quantity: $\alpha\lambda-\alpha + \alpha\lambda - \lambda$ hence it's magnitude is bigger.

Note that $z=tu + (1-t)v$. By convexity we get that $f(z)=f(tu + (1-t)v) \le tf(u)+(1-t)f(v)$. By the strict inequality of $f(u)$ we get that
$$< t(\alpha\lambda f(x)+(1-\alpha\lambda)f(y)) +(1-t)((\alpha + \lambda -\alpha\lambda)f(x) + (1-\alpha-\lambda -\alpha\lambda)f(y))=\lambda f(x) + (1-\lambda)f(y)
$$
Which is a contradiction to the assumption that $f(z) = \lambda f(x) + (1-\lambda)f(y)$.

Hence $f$ is affine in $[x,y]$.


问:
How to factor out cubic polynomial with constant term?
My polynomial: $ 2400x^{3} - 9880 x^{2} - 266 x + 123 = 0 $

So far, I've tried to solve this polynomial with synthetic division, the Wiki-How, and this page, but nothing seems to yield the solution of $ x = 0.1 $

Can somebody show me how to do it?

答:
The number $\frac{1}{10}$ is indeed a root of this polynomial.

The polynomial is equal to $(10 x - 1) (240 x^2 - 964 x - 123)$

You can also use Wolfram Alpha to factor/decompose your polynomial:

Wolfram Alpha polynomial


问:
A set $E$ is closed if and only if $E = E^-$
Let $E$ be a subset of a metric space $(S,d)$. I want to show that the set $E$ is closed if and only if $E = E^-$ where $E^-$ is the closure of a set $E$.

First I assumed $E = E^-$. Then since I know $E^-$ is the intersection of all closed sets containing $E$, $E^-$ must be closed because the intersection of any closed sets is a closed set. So it follows that $E$ is closed.

Now if $E$ is closed, then $S \setminus E$ is open. That also means $E^- \setminus E$ is open since $E \subseteq E^-$ by definition. Suppose $E \subsetneq E^-$. Let $x$ be on the boundary of $E^-$. Then there does not exist an $r>0$ such that $\{s \in S \ | \ d(s, x) < r \} \subseteq E^- \setminus E$. That means $E^- \setminus E$ is not open which gives a contradiction. So it must be the case that $E^- = E$.

Could someone give me feedback on my proof? And is there a better/shorter proof?

答:
The first part of your proof is good.

If I'm understanding it correctly, the second part of your proof (where you show $E \text{ closed} \implies E = \overline{E}$)
has an error.  $\overline{E} \setminus E$ need not be open just because $E \subseteq \overline{E}$ and $S \setminus E$ is open.  For instance, if $E = [0,1]$ and $\overline{E} = [-1, 2]$, $\overline{E} \setminus E$ is not open.  Ben gives a shorter and correct proof.

ContentLicense: CC BY-SA 3.0


问:
Prove that $f$ is injective given that $A=f^{-1}(f(A))$

Let $f\colon X\to Y$ be a function. Prove that if for every $A\subseteq X$, we have $A=f^{-1}(f(A))$ then $f$ is one to one (injective)

My try: Let $x,x’\in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. So we have $$f^{-1}(f(x)) = f^{-1}(f(x'))$$According to the assumption we have $$x = x'$$I don't know whether my proof is correct and rigorous.

答:
It would be correct if $f$ had an inverse, but you are not assuming that.

You can say that, if $A=\{x\}$, then\begin{align}\{x\}&=A\\&=f^{-1}\bigl(f(A)\bigr)\\&=f^{-1}\bigl(\bigl\{f(x)\bigr\}\bigr)\\&\supset\{x,x'\}\end{align}and that therefore $x'=x$.


Title: How to factor out cubic polynomial with constant term?

Body:
My polynomial: $ 2400x^{3} - 9880 x^{2} - 266 x + 123 = 0 $

So far, I've tried to solve this polynomial with synthetic division, the Wiki-How, and this page, but nothing seems to yield the solution of $ x = 0.1 $

Can somebody show me how to do it?

Body:
Factoring gives $2400x^{3} - 9880 x^{2} - 266 x + 123= \left( 10 x-1\right)  \left( 240 {{x}^{2}}-964 x-123\right)$ and thus one of the three roots is $x=\frac1{10}$.

I do not see an easy way of spotting this if you do not already know the solution, but the long division is not difficult:

           240 x^2 -  964 x   -  123
        _____________________________________
10x - 1 ) 2400 x^3 - 9880 x^2 -  266 x  + 123
          2400 x^3 -  240 x^2
          ____________________
                   - 9640 x^2 -  266 x
                   - 9640 x^2 +  964 x
                   ___________________
                              - 1230 x  + 123
                              - 1230 x  + 123
                              ===============


Title: Inductive proof for recursive formula $a_n=4-a_{n-1}$

Body:
I have a recursive formula $a_n=4-a_{n-1}, a_3=7$ and I had to solve this and check the correction of the answer with induction.

I started solving this and the solved formula is: $a_n=-2((-1)^{n-3}-1)+7(-1)^{n-3}$.

Base case $n=3$ is true

Let's assume this is true for $n=k$

Thus this should be true for $n=k+1$

$a_{n=k+1}=5(-1)^{k-2}+2$.

I don't know how to continue from this place, please someone help me and explain this, thank you.

Body:
Note that $-2((-1)^{n-3} - 1) + 7(-1)^{n-3} = 5(-1)^{n-3} + 2$.

Assuming $a_n = 5(-1)^{n-3} + 2$, the recursive formula tells us that

$$a_{n+1} = 4 - a_n = 4 - [5(-1)^{n-3} + 2] = -5(-1)^{n-3} + 2 = 5(-1)^{n-2}+2.$$

This proves the result.


答:
If you apply the First fundamental theorem of integral calculus, you have
$$F'(x)=x^{\sqrt x},\enspace \text{hence }\quad F''(x)=x^{\sqrt x}\biggl(\frac{\ln x}{2\sqrt x}+\frac{\sqrt x}x\biggr)=x^{\sqrt x}\,\frac{\ln x+2}{2\sqrt x}$$
Therefore, $F''(x)=0$ for $x=\mathrm e^{-2}$, and it changes sign at that point, which means we have an inflection point.

Title: let $F(x)=\int^x_1 t^{\sqrt {t}}dt$ be defined for every $x\ge 0$, then..

Body:
a) $F(x)>0$ for every $x>0$.

b) $F(x)>F(e)$ for every $x>1$.

c) $x=e^{-2}$ is an inflection point.

d) None of the answers are correct.

Note: This was a question in my exam and I wrote it from my memory so if there's something that doesn't make sense it might be that I messed up.

My Work was to take second derivative of $F$ and saw that $F''(e^{-2})\ne 0$, so I took out that option, and didn't know how to decide between (a) or (b), but I noticed that for (a) I might have to flip the integral, so I just went with (b), I would appreciate any explanation on how to really approach this question and not just with my intuition. (and know if I got it right too).


问:
If $f(z)$ is even then residue at $z=z_0$ is zero?
There is a theorem in my text book which says that:

Theorem:If $f$ has an isolated singularity at $z_0$ and if $f$ is even in $z-z_0$ i.e $f(z-z_0)=f(-(z-z_0))$ then Res$[f(z),z_0]=0$ [Theorem 8.60, page 290,S ponnusami]

My doubt: let $f(z)$ be even function then $f(z)=f(-z)$ for all $z$ in domain of f and hence we have $f(w)=f(-w)$ where $w=z-z_0$ implying that $f$ is even in $z-z_0$ and hence from theorem for any even function $f$ Res$[f(z),z_0]=0$ at isolated singularity $z_0$ which I think wrong.For counter example see the picture added

答:
I think what you've gotten wrong is the theorem itself. When faced with a statement you doubt, I recommend looking for easy examples before you try and prove it. Here, I can think of the function:

$$f(z)= \frac{1}{z^2-1}$$
It is clearly even, since $ f(z)=f(-z) $. If we calculate the residue of f in $z_0=1$, we get:
$$\lim_{z\rightarrow1}{f(z)·(z-1)}=\lim_{z\rightarrow1}{\frac{1}{z+1}}=\frac{1}{2}\neq0 $$
If a function is even, it will have a singularity at $z_0$ if and only if it has it at $-z_0$, and the residues of $f$ will be the same.


问:
Metric space with a constraint
For $\mathbb{R}^n$ with $n \ge 2,$ define $$d_{p,q}:= \left( \sum\limits_{i =1}^n |x_i - y_i|^p \right)^q$$ for $p,q \in \mathbb{R}.$ Prove that  $ d_{p,q}  \text{ is a metric on } \mathbb{R}^n \iff 0 < q, \ pq \le 1.$

I am used to proving or disproving a metric space by validating all the four axioms of finiteness, definiteness, symmetry and triangle inequality without any constraints on the metric. I came across this exercise and I thought of sharing it here to generate ideas on how best to tackle it. Clearly, if $0 < q, pq \le 1$ then $d_{p,q}(x,y)$ is a metric space since all the four axioms are easily met. How do we prove the other direction that is if $d_{p,q} (x,y)$ is a metric we show that $0 <q, pq\le q$? We need to fulfill the two directions of the $"\iff"$. Benevolent contributors I rely on you for your support.

答:
Suppose that $d_{p,q}$ is a metric. Note that if $q\leq0$ then anything exponentiated to the power of $q$ cannot be $0$ so $d_{p,q}(\vec{x},\vec{x})\neq0$, a contradiction. We immediately conclude that $q>0$ In the same way one can conclude that $p>0$, since otherwise $d_{p,q}$ would not be defined on the pair $(\vec{x},\vec{x})$ of $\mathbb{R}^n\times\mathbb{R}^n$.

Now to show that $pq\leq 1$: Take $\vec{x}=(2,\dots,2)$, $\vec{y}=(0,\dots,0)$ and $\vec{z}=(1,\dots,1)$. Then the triangle inequality says that $$d_{p,q}(\vec{x},\vec{y})\leq d_{p,q}(\vec{x},\vec{z})+d_{p,q}(\vec{z},\vec{y})$$
Computing, we have that
$$n^q\cdot 2^{pq}\leq n^q+n^q$$
so
$$2^{pq}\leq 2 $$
and therefore $pq\leq1$.

Title: Metric space with a constraint


问:
Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?

Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin

First, I get the line which passes through point of contact of tangents from origin using result here which is :

$$ -12x-16y-2 \cdot 75 = 0$$

Or,

$$ 6x + 8y -75 =0 $$

Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:

$$ C(0,0) C(x,y) =  (6x+8y-75)^2$$

This leads to:

$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$

$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$

For applying the result in this answer, then the formula

$$ a= 75 - 36,  b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$

Or,

$$ a = 39, b= 11 , h=-48$$

$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$

However, the intended answer was:

$$ \tan \theta = \frac{1}{\sqrt{3} } $$

Where have I gone wrong?

答:
The mistake lies in using $6x+8y-75=0$ further instead of $-12x-16y+150=0$
in the formula. Correct way is to write:
$$C(0,0) C(x,y)=(-12x-16y+150)^2.$$
to get the correct answer.


问:
Bacteria population growth

The population of bacteria triples each day on a petri dish. If it takes 20 days for the population of bacteria to fill the entire dish, how many days will it take bacteria to fill half of the petri dish?

My doubt 1: if we don't have a initial population can be solve this problem?
Like $A(d)$= initial population times $3^{20}$ days but as I don't have this I am stuck.

My doubt 2: Also, can we assume a number for population at day 20 and move our way backwards?

答:
Let the population $N_t$ of bacteria after time (in days) $t$ with initial population $N_0$ be given by:
$$N_t = N_0 e^{\lambda t}$$
where $\lambda$  is a constant. It is given $N_1 = 3N_0 = N_0 e^{\lambda}$. From this you can rearrange terms to get $\lambda = \ln (3)$. This means:
$$N_t = N_0 e^{t \ln(3)} = N_03^t.$$

Let $M$ be the population that covers the whole petridish. It is given that $M = N_03^{20} \implies M/N_0 = 3^{20}$. The population that covers half the petridish should $M/2 = N_03^{T}$ where $T$ is the required time. $M/2N_0 = 3^{20}/2 = 3^{\log_3 (0.5)+ 20} = 3^T  \implies T = \log_3(0.5) + 20 = 20 - \log_3 (2).$

I think this answers your questions.

Title: Bacteria population growth


问:
Help with school solid geometry
In the triangular pyramid $MABC$ all side edges equals $1$, $\angle AMB = \angle BMC = 60 ^\circ$, $\angle AMC = 45 ^\circ$.

Find:

1) square of the $\triangle ABC$;

2) dihedral angle on the $AB$ edge;

3) $\rho(M, ABC)$.

Help me, please.

P.S. sorry for my English, I am not a native speaker.

答:
Wow the question recalls my old memory about high school.
Actually you know
$$AM = MC = 1$$
and because of the law of cosines
$$cos(\angle AMC) = \frac{\sqrt{2}}{2} = \frac{1^2+1^2-AC^2}{2\cdot 1 \cdot 1}\Rightarrow AC = \sqrt{2-\sqrt{2}}.$$
The area can be computed by Heron's formula.

Notice $\triangle AMC = \triangle ABC$, so $\angle ABC = \pi/4.$
Draw a line(in plane $ABC$) through the middle point $D$ of $AB$ and it intersects $BC$ at $E$. We know
$$AD = BD = DE = AB/2 = \frac{1}{2}\text{ and }BE = \frac{\sqrt{2}}{2}$$

And you can show $AB\perp \triangle MDE$ by $AB\perp MD, DE$. So $\angle MDE$ is what you want to solve.
Again invoking the law of cosine for $\angle MBE$, $ME^2 = 1+\frac{1}{2}-\frac{\sqrt{2}}{2} = \frac{3-\sqrt{2}}{2}$. Now use the same law to $\angle MDE:$
$$\cos\angle MDE = \frac{\frac{3}{4}+\frac{1}{4}-ME^2}{2\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2}} = \frac{3-\sqrt{3}}{3}.$$

Immediately from this result, draw a line $MF$ in plane $MDE$ intersecting $CD$ at $F$. Since $AB\perp MDE\Rightarrow AB\perp MF$ and $MF\perp CD$, we have $MF\perp ABC$, and
$$\rho(M,ABC) = MD\cdot\sin\angle MDE.$$

Title: Help with school solid geometry


问:
How to calculate height of camera and ground coverage via geometry?
I have a GoPro7 that I will mount on a pickup vehicle to capture ground images with video capture.

The camera isn't top down but has about 20 (or 30) degrees inclination from the vertical axis.
I want to capture the road that is 4m in width.

Other givens would be:

Vertical Field of View: 71.0deg

Horizontal FoV: 86.7deg

Diagonal FoV: 100deg

Sensor size: 1/2.3” (6.17×4.55mm)

Aperture: f/2.8

Zoom: 0%

Frame width: 4:3

I would like to know if it is possible (and how) to calculate the needed height of the GoPro for this.

答:
If I understand you correct you want to mount your camera $x$ meters above the road where
$$
x = \frac{L}{2} \tan\bigl(\frac{\alpha\pi}{180}\bigr)\cos\bigl(\frac{\beta\pi}{180}\bigr)
$$
and where
$$
\begin{align*}
L &= \text{width of road} = 4~\text{m}\\
\alpha &= 90-\tfrac12(\text{Horizontal FoV})=90 - \tfrac12 86.70 = 46.65^\circ\\
\beta &= \text{inclination from the vertical axis} = \text{20–30}^\circ
\end{align*}
$$

For $\beta=20^\circ$ you get $1.99087$ m and for $\beta=30^\circ$ you get $1.8348$ m.

Explanation:

Consider front view of the camera

The road is of width $L$ and $\alpha=\text{Horizontal FoV}$. The (sloping) distance from the camera to the road is $x$.

We have
$$
\tan(\alpha/2)=\frac{L/2}{x}
\quad\Leftrightarrow\quad
x=\frac{L}{2\tan(\alpha/2)}.
$$

Now, consider the camera from the side (looking at the car from the side);

where $\beta=\text{inclination from the vertical axis}$ and $x$ the sloping line above. We want to find $h$.

We have
$$
\cos(\beta)=\frac{h}{x}
\quad\Leftrightarrow\quad
h=x\cos(\beta).
$$

Hence
$$
h=\frac{L}{2\tan(\alpha/2)}\cos(\beta).
$$

This is a variant of the formula above as
$$
\tan(90^\circ-\alpha/2)=\cot(\alpha/2)=\frac{1}{\tan(\alpha/2)}.
$$

Perhaps US readers prefer
$$
h=\frac{L}{2}\cot(\alpha/2)\cos(\beta).
$$

N.B.1: ‘inclination’ might be $90^\circ-\beta$ depending on definition, but if so it easy to find the $\beta$ used in the calculation by $\beta=90^\circ-\text{inclination}$.

N.B.2: The calculations above assumes that the calculator is set to degrees. If the calculator is set to radians the formula becomes
$$
h
=\frac{L}{2\tan(\alpha\pi/360)}\cos(\beta\pi/180)
=\frac{L}{2}\cot(\alpha\pi/360)\cos(\beta\pi/180).
$$


问:
If $g\circ f$ is one-to-one, and $f$ and $g$ have the same domain/codomain, must $f$ and $g$ also be one-to-one?
If $g\circ f$ is one-to-one for the functions $f : A \to B$ and $g : A \to B$, must both $g$ and $f$ also be one-to-one?

Normally, when $f$ and $g$ have different domains/codomains, only $f$ must be one-to-one. However, if both $g$ and $f$ have the same domain/codomain, and their composition is one-to-one, must $g$ also be one-to-one? My intuition tells me yes, but I wasn't able to work out a formal proof.

Any input for why g would/wouldn't have to be one-to-one would be much appreciated!

答:
Consider $A = [-1,1]$, $f:A \to A,\, f(x)=\frac{x+1}{2}$ and $g:A\to A, \, g(x)=x^2$.
Then $g\circ f$ is one-to-one, but $g$ is not.


问:
Does $\sum^\infty_1(-1)^n(\frac {1+(-1)^n}{n})$ converge or diverge?

Does $\sum^\infty_1(-1)^n\left(\frac {1+(-1)^n}{n}\right)$ converge or diverge?

My Work: At first I noticed that it looks like an alternating series and wanted to say it converges because $\frac {1+(-1)^n}{n}$ goes to zero, but I noticed that for $n=1$, we have $0$, for $n=2$ we have $\frac {2}{n}$, for $n=3$, $0$. So it's not really decreasing.

From here I only could do this by intuition that this behaves like $\frac {1}{n}$ or maybe less than it since we have $0$ every second element. so it diverges.

Would really like to know if what I did is right and if the series converges or diverges.

Thanks in advance

答: Your intuition is correct. This series is simply$$0+\frac22+0+\frac24+0+\frac26+\cdots$$which converges if and only if $\sum_{n=1}^\infty\frac2{2n}$ converges. But this is the harmonic series. Therefore, the series diverges.


问:
Multilinear Map of a Determinant: Basis Extension from Subspace
Let $V$ be an $n$-dimensional $K$-vector space, $U ⊆
V$ an $r$-dimensional linear subspace and $x_{r + 1},\dots , x_{n}$ a fixed
System of vectors in $V$. For a determinant function $\Delta: V^n \rightarrow K$
(i.e. an alternating $n$-multilinear form) show that through
$$ \Delta_U (a_1,\dots, a_r):= \Delta (a_1,\dots, a_r, x_{r + 1},\dots, x_{n})$$
a determinant function $\Delta_U:U^r \rightarrow K$ can be defined.
When is $\Delta_U$
non-trivial?

答:
We want to show that
$\Delta_U: U \rightarrow K, \Delta_U(a_1,\dots, a_r) := \Delta(a1,...,a_r,x_{r+1},...,x_n)$ is a determinant function.

Proof: The properties for the determinant function are inherited from $\Delta$:

$$\Delta_U(a_1,\dots, a_{i1}+a_{i2},...,a_r)= \Delta(a_1,\dots, a_{i1}+a_{i2},...,a_r,x_{r+1},...,x_n)=\Delta(a_1,\dots, a_{i1},...,a_r,x_{r+1},...,x_n)+\Delta(a_1,\dots,a_{i2},...,a_r,x_{r+1},...,x_n)=\Delta_U(a_1,\dots, a_{i1},...,a_r)+\Delta_U(a_1,\dots,a_{i2},...,a_r)$$

$$\Delta_U(a_1,\dots, \lambda a_i,...,a_r)=\Delta(a_1,\dots, \lambda a_i,...,a_r,x_{r+1},...,x_n)=\lambda \Delta(a_1,\dots, a_i,...,a_r,x_{r+1},...,x_n)=\lambda \Delta_U(a_1,\dots,a_i,...,a_r)$$

$$\exists i \ne j: a_i = a_j \Rightarrow \Delta_U(a_1,\dots,...,a_r)= \Delta(a_1,\dots,a_r,,x_{r+1},...,x_n)=0$$

When is $\Delta_U$ non-trivial?
First consider when $\Delta_U$ is trivial. We get three cases:

• $\Delta$ is trivial

• $\exists \lambda_{r+1},...,\lambda_n \in K: \sum^n_{i=r+1} \lambda_i x_i \in U \ \{0\}$ (because then $(a_1,..., a_r, x_{r + 1},..., x_n)$ is a linearly dependent system for all $a_1,. . . , a_r \in U$)
That is: $\exists u \in U \ \{0\}: u \in L(x_{r + 1},..., x_n)$

• $(x_{r + 1},..., x_n)$ forms a linearly dependent system.
So $\Delta_U$ is non-trivial if $$\Delta\operatorname{not trivial} \land \space \forall \space u \in U \ \{0\}: u \notin L(x_{r + 1},..., x_n) \land (x_{r + 1},..., x_n)$$

is linearly independent.

(You could also summarize the last two statements as follows: $(x_{r + 1},..., x_n)$ forms a Basis of a complementary space from $U$ in $V$).


问:
Does taking the direct limit of chain complexes commute with taking homology?
Suppose I have a directed system $C_i$, $i\in\mathbb{N}$ of chain complexes over free abelian groups (bounded below degree $0$)
$$C_i=0\rightarrow C^{0}_{(i)}\rightarrow C^{1}_{(i)}\rightarrow\cdots\rightarrow C^{n-1}_{(i)}\rightarrow C^n_{(i)}\rightarrow \cdots$$
with chain maps $f_i\colon C_i\rightarrow C_{i+1}$. Can I say that
$$H_*\left(\lim_{\rightarrow}(C_i,f_i)\right)\cong\lim_{\rightarrow}\left(H_*(C_i),(f_i)_*\right),$$
where $\displaystyle\lim_{\rightarrow}$ is the direct limit (colimit) in the respective category, $H_*(C)$ is the *th homology of the the chain complex $C$, and $f_*$ is the induced homomorphism in homology of the chain map $f$?

I imagine the answer will involve some categorical property of the functor $H_*$.

答:
Any exact functor between abelian categories will preserve homology, and colimits indexed by filtered or directed diagrams are exact in $\mathbf{Ab}$.

The first claim is straightforward, because homology is computed using kernels and cokernels. Indeed, given a chain complex $C_{\bullet}$, we form the object of cycles as a kernel,
$$0 \longrightarrow Z_n \longrightarrow C_n \longrightarrow C_{n-1}$$
and we form the object of boundaries as a cokernel,
$$0 \longrightarrow Z_{n+1} \longrightarrow C_{n+1} \longrightarrow B_n \longrightarrow 0$$
and then the homology object is also a cokernel:
$$0 \longrightarrow B_n \longrightarrow Z_n \longrightarrow H_n \longrightarrow 0$$

The second claim is best checked by hand using the concrete description of filtered/directed colimits in $\mathbf{Ab}$. By general nonsense, colimits are additive and preserve cokernels, so it is enough to check that kernels are preserved by filtered/directed colimits.


问:
Let X1 be N(0, 1), and let X2 = −X1, if − 1 ≤ X1 ≤ 1, X1, otherwise. Show that X2 also has an N(0,1) distribution.
Given that X1 follows standard normal distribution, N(0,1). And
define X2 = -X1, if -1 ≤ X1 ≤ 1; X1, otherwise. Then I am asked to prove that X2 follows N(0,1) as well.

My idea is to prove P(X1 ≤ x ) = P(X2 ≤ x) for three cases when x < -1 , when -1 ≤ x ≤ 1, when x > 1.
For x < -1 and x > 1 are easy, I started with P(X1 ≤ x) = P(X2 ≤ x) and P(X2 ≥ x) = P(X1 ≥ x) respectively and show that they have the same CDF.

But I am stucked with the case -1 ≤ x ≤ 1,
I was thinking to start with
when -1 ≤ x ≤ 1,
P(|X1|≤|x|) = P(|X2|≤|x|)
and derive from this equation to get P(X1 ≤ x ) = P(X2 ≤ x), but it does not seem to be the correct way.

Could anyone help me with this proof?

答:
When $-1 < x < 1$, and using the standard normal CDF $\Phi(x)$,  you have

$\mathbb P(X_2 \le x) = \mathbb P(X_2 \le -1)+\mathbb P(-1 < X_2 \le x)$

You have already found $\mathbb P(X_2 \le -1)=\mathbb P(X_1 \le -1)=\Phi(-1)$

$\mathbb P(-1 < X_2 \le x) = \mathbb P(-1 < -X_1 \le x) = \mathbb P(-x \le X_1 < 1) =  \Phi(1) -\Phi(-x)$

$\Phi(1)=1-\Phi(-1)$ and $\Phi(-x)=1-\Phi(x)$ by symmetry

so $\mathbb P(X_2 \le x) = \Phi(-1)+(1-\Phi(-1))-(1-\Phi(x)) = \Phi(x)$ as required


Title: Does $\sum^\infty_1(-1)^n(\frac {1+(-1)^n}{n})$ converge or diverge?

Body:
Does $\sum^\infty_1(-1)^n\left(\frac {1+(-1)^n}{n}\right)$ converge or diverge?

My Work: At first I noticed that it looks like an alternating series and wanted to say it converges because $\frac {1+(-1)^n}{n}$ goes to zero, but I noticed that for $n=1$, we have $0$, for $n=2$ we have $\frac {2}{n}$, for $n=3$, $0$. So it's not really decreasing.

From here I only could do this by intuition that this behaves like $\frac {1}{n}$ or maybe less than it since we have $0$ every second element. so it diverges.

Would really like to know if what I did is right and if the series converges or diverges.

Thanks in advance

Body:
Let $a_n=(-1)^n(\frac {1+(-1)^n}{n})$ und $S_n= a_1+a_2+...+a_n.$ Show that $a_{2n-1}=0$ and $a_{2n}=1/n.$

Furthermore show that $S_{2n}=1+\frac{1}{2}+...+\frac{1}{n}.$

Hence $(S_n)$ is divergent.


问:
$\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}$
Let $\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=x$. Therefore, $$n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}=x^2$$ $$x^2-x-n=0$$ $$x=\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=\frac{1 \pm \sqrt{1+4n}}{2}$$

It seems counter-intuitive at first that $f(n)$ doesn't have an effect on the final answer. If $f(n)$ were $\frac{1}{n!}$ or $e^{2.3n^2}$, or anything else for that matter, we still get the same formula for the result.

My question regarding this is:

Is this applicable for any $f(n)$ at the end of the infinite nested radical (I haven't studied much about properties of functions but have heard of terms like 'odd function', 'even function', '$f: \mathbb{R} \to \mathbb{R}$' etc)?

Is this result because of the infinitesimally small effect of $f(n)$ on the result?

EDIT : Some people have commented that there is '1 less $n$ post squaring'. There are infinite '$\sqrt{n+\sqrt{}}$' before I added $f(n)$ at the end.

答:
If you set $$x=\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}$$
when you square
$$x^2=n+\left[\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}\right]$$
The part in the square brackets no longer has the same number of square roots, therefore it is not $x$. It is something else.

Maybe you saw something like this with infinite square roots. In that case, under some conditions, the trick of squaring actually works.

Not in the finite case.


问:
How to show the product of two continuous function is a continuous function using projections
If $f$ and $g$ are continuous, then their Cartesian product $f\times g$ is also continuous and this can be proven using the standard open set preimage definition, but I read somewhere that an easier proof uses the projection function. How is that done?

答:
The product topology has a very useful universal property, that even characterises it among other topologies:

If $X \times Y$ has the product topology and $p_X: X \times Y \to X$ and $p_Y: X \times Y \to Y$ are its natural projections, then any function $h: Z \to X \times Y$, for any space $Z$, is continuous iff both $\pi_X \circ h: Z \to X$ and $\pi_Y \circ h: Z \to Y$ are continuous.

Now suppose $f: A \to X$ and $g: B \to Y$ are both continuous. We can define $(f \times g): A \times B\to X \times Y$ by $(f \times g)(a,b)=(f(a), g(b))$ and then $f \times g$ is also continuous. This follows by the universal property just mentioned,applied to $Z=A\times B$,after noting that

$$p_X \circ (f \times g) = f \circ p_A \text{ and } p_Y \circ(f \times g)= g \circ p_B$$ are both continuous as compositions of continuous maps. (The projections $p_A,p_B$ on $A \times B$ are also continuous by definition of the product topology on that set).


问:
Eigenvectors of a normal operator are an orthogonal basis
I am trying to prove the following.

Let $T\colon\mathbb C^n\to\mathbb C^n$ be a normal operator. Then there is a basis for $\mathbb C^n$ consisting of orthogonal eigenvectors of $T$.

I am allowed to use the following fact which I've already established:
$$T\boldsymbol x=\mu\boldsymbol x\iff T^*\boldsymbol x=\overline\mu\boldsymbol x.$$

I've managed to produce a proof using Schur's theorem (i.e., $T$ is unitarily similar to a diagonal matrix), but this feels like too much "heavy machinery" to tackle this proof. Is there another more direct way to go about it?

答:
In my opinion this is a very important proof, so let's get into many details:

First of all $T$ has eigenvectors, since it has eigenvalues, since the polynomial $p(\lambda)=\text{det}(T-\lambda I)$ has roots in $\mathbb{C}$. Now suppose that $Tx=\lambda_1x$ and $Ty=\lambda_2y$ where $\lambda_1\neq\lambda_2$. Since these are distinct, we may assume WLOG that one of them is non-zero, say $\lambda_1\neq0$. Then
$$\langle x,y\rangle=\frac{1}{\lambda_1}\langle\lambda_1 x,y\rangle=\frac{1}{\lambda_1}\langle Tx,y\rangle=\frac{1}{\lambda_1}\langle x,T^*y\rangle=\frac{1}{\lambda_1}\langle x,\bar{\lambda_2}y\rangle=\frac{\lambda_2}{\lambda_1}\langle x,y\rangle$$
so if $\langle x,y\rangle\neq0$ we have that $\lambda_1=\lambda_2$, a contradiction. We conclude that eigenvectors corresponding to distinct eigenvalues are orthogonal.

Let $\lambda_1,\dots,\lambda_k$ denote the distinct eigenvalues of $T$ (note that $k\leq n$, obviously, since $p(\lambda)$ has at most $n$ distinct roots). We set $$M_i:=\{x\in\mathbb{C}^n: Tx=\lambda_ix\}$$
for the eigenspace corresponding to $\lambda_i$. These are orthogonal subspaces of $\mathbb{C}^n$, as we explained. We consider their direct sum $M:=\bigoplus_{i=1}^kM_i$ and this is now a Hilbert subspace of $\mathbb{C}^n$.

We claim that $M=\mathbb{C}^n$. Since $T$ is normal, we have that $T(M)\subset M$ and that $T(M^\bot)\subset M^\bot$. I leave this as an exercise; can you see why this is true?

Consider the operator $S:=T\vert_{M^\bot}:M^\bot\to M^{\bot}$. This operator is a well-defined, bounded operator that has no eigenvalues! Indeed, if $Sx=\lambda x$, then $Tx=\lambda x$ (by the definition of $S$) so $\lambda$ is an eigenvalue of $T$ and thus $x\in M$. But this is a contradiction, since any bounded operator on a non-zero (finite dimensional, but in general as well. You only need to know the finite dimensional case here) Hilbert space has eigenvalues!

Now simply take an orthonormal basis for each subspace $M_i$, say $E_i$ and set $E=\bigcup_{i=1}^kE_i$. Since $\mathbb{C}^n=M=\bigoplus_{i=1}^kM_i$, we conclude that $E$ is an orthonormal basis of $\mathbb{C}^n$. Since any element of $E$ belongs to some $E_i$ which is a subset of $M_i$, the elements of $E$ are eigenvectors of $T$.


问:
Verify if the set is invariant with respect to the system
I had three assignments where I had different sets for which I had to verify if they were invariant with respect to the system. I did not really know how to solve them so I hope people are willing to help me with this.

1) Consider the autonomous linear system $x(k+1) = Ax(k)$ and the set $S_1 =
\{ x \in \mathbb{R}^2 : x^TPx \leq 3.245\}$. Herein:

$$ A = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & -0.5 \end{bmatrix} \quad P = \begin{bmatrix} 3 & 0.075 \\ 0.075 & 1.36 \end{bmatrix}$$

Verify if the set $S_1$ is invariant with respect to the system.

A: No clue where to begin.

2) Consider the autonomous linear system $x(k+1) = Ax(k)$ and the set $S_2 =
\{ x \in \mathbb{R} : ||x||_1 \leq 1\}$. $S_3 = \{ x \in \mathbb{R}^3 : ||x||_\infty \leq 1 \}$. Herein:

$$ A = \begin{bmatrix} 0.8 & 0 & 0 \\ -0.2 & -0.5 & 0 \\ 0 & 0.3 & 1 \end{bmatrix}$$

Verify if the set $S_2$ and $S_3$ are invariant with respect to the system.

A: Compute $||A||_1 = 1$ and $||A||_\infty = 1.3$. Use the fact that $||Ax|| \leq ||A||\,||x||$

3) Consider the autonomous linear system $x(k+1) = Ax(k)$ and the set $S_4  = \{x \in \mathbb{R}^2 : x_1 + x_2 \leq 1, x_1 \geq -1, x_2 \geq -1 \}$. Herein:

$$ A = \begin{bmatrix} 0.5 & 0 \\ -0.3 & -0.5 \end{bmatrix}$$

Verify if the set $S_2$ and $S_3$ are invariant with respect to the system.

A: No clue where to begin.

答:
As a rule of thumb, whenever you have a recurrence relation like $x_{k+1} = Ax_k$, it is useful to diagonalize $A$ whenever possible and abuse linearity. Suppose $Av_i = \lambda_i v_i$, $i = 1, 2$ in dimension 2. For any $x_0$,
\begin{align*}
x_0 &= a_1 v_1 + a_2 v_2 \\
Ax_0 &= A(a_1 v_1 + a_2 v_2) \\
&= a_1 Av_1 + a_2 Av_2 \\
&= \lambda_1 a_1 v_1 + \lambda_2 a_2 v_2 \\
&= x_1. \\
&\vdots \\
x_k &= A^k x_0 \\
&= \lambda_1^k a_1 v_1 + \lambda_2^k a_2 v_2.
\end{align*}
In other words, once you find the eigenvalues and the coefficients $a_1, a_2$, you don't need to multiply matrices anymore (useful).

So, for question 1), you find that the eigenvalues of $A$ are both $<1$ in absolute value. The important thing to get from this is the geometric intuition: the eigenvalues tell you how the system evolves, and the fact that both eigenvalues are real and in absolute value strictly less than 1 tells you that $Ax$ will be "smaller". That is, if $x \in S_1$, then $Ax \in S_1$ because you've just scaled down each coordinate in the $\{v_1, v_2\}$ basis.

To make this precise, let $x \in S_1$ and write $x = a_1 v_1 + a_2 v_2$, where $v_i$ are eigenvectors of $A$. Then, in particular, you can find $x^T Px$ in terms of $v_i$ and $a_i$:
\begin{align*}
x^T Px &= (a_1 v_1 + a_2 v_2)^T P(a_1 v_1 + a_2 v_2) \\
&= a_1^2 v_1^T P v_1 + a_1a_2 v_1^T Pv_2 + a_1a_2 v_2^T P v_1 + a_2^2 v_2^T P v_2.
\end{align*}
The above is known as a quadratic form. The important thing to notice is that $(Ax)^T P(Ax)$ will look exactly as above, except you will have
$$a_1^2 \lambda_1^2 v_1^T P v_1 + \lambda_1\lambda_2 a_1 a_2 v_1^T P v_2 + \lambda_1\lambda_2 a_1 a_2 v_2^T P v_1 + \lambda_2^2 a_2^2 v_2^T P v_2.$$
Since the $\lambda_i$ only decrease the norm, this quantity will be less than the first one. Thus, $S_1$ is invariant under the system.


答: Here is a simple solution: Find the center of the given circle, which is (6, 8). The line from the center of the circle to the point of contact of one of the tangents is perpendicular to the tangent. Use this fact to get the equation of a new circle, which is $x^2-6x+y^2-8x=0$. Find where both circles meet. Find the angle between the vectors, which are the two points where the circles meet using the formula:$$\text{Angle} = \arccos\left(\frac{x_1x_2 + y_1y_2}{\sqrt{(x_1^2+y_1^2) \cdot (x_2^2+y_2^2)}}\right)$$

Title: Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?

Body:
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin

First, I get the line which passes through point of contact of tangents from origin using result here which is :

$$ -12x-16y-2 \cdot 75 = 0$$

Or,

$$ 6x + 8y -75 =0 $$

Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:

$$ C(0,0) C(x,y) =  (6x+8y-75)^2$$

This leads to:

$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$

$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$

For applying the result in this answer, then the formula

$$ a= 75 - 36,  b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$

Or,

$$ a = 39, b= 11 , h=-48$$

$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$

However, the intended answer was:

$$ \tan \theta = \frac{1}{\sqrt{3} } $$

Where have I gone wrong?


问:
Solving trigonometric equations for the angle
Antenna applications have found me rough around the edges when it comes to my trigonometry. I need to solve equations in the following form:

$$\cos\theta\cos(2\theta) = 0.5$$

$$\cos^2(\theta)\cos^2(3\theta) = 0.5$$

I couldn't seem to find some formulas that would transform the product/sum of two functions into a single function so i can do a simple arccos(0.5) = . If there aren't just any formulas, i might understand how to do them with a solved example.

答:
$\cos(2\theta)\equiv2\cos^2\theta-1$, so your first equation can be turned into
$$
\cos\theta(2\cos^2\theta-1)=0.5\ ,
$$
which can be turned into the cubic
$$
4\cos^3\theta-2\cos\theta-1=0\ ,
$$
So you'll have to go solve the cubic
$$
4x^3-2x-1=0
$$
before you can solve for $\theta$.

For your second equation you'll need to use
$$
\cos3\theta\equiv 4\cos^3\theta-3\cos\theta.
$$


问:
Let $F/K$ be a galois extension, $\alpha\in F$,and $H=\{\sigma\in G｜\sigma(\alpha)＝\alpha\}$ , to show $L^H＝K（α）$
Let $F/K$ be a finite galois extension.
Let $G＝\text{Gal}(F/K)$, and $\alpha\in F$ and $H＝\{\sigma\in G｜\sigma(\alpha)＝\alpha\}$.
I want to show $K(\alpha)＝F^H$, where $F^H＝\{x\in F｜\forall\sigma\in H,\ \sigma(x)＝x\}$.

$K(\alpha)\subseteq F^H$ is clear, because $\sigma$ fixes $K$ and $\alpha$.
Could you prove other direction, $F^H\subseteq K(\alpha)$?

Thank you for your help.

答:
If $F^H$ is larger than $K(\alpha)$ then the extension being separable there is some non-trivial $\tilde{g}\in Hom_{K(\alpha)}(F^H,\overline{F})$. It extends to some $g\in Hom_{K(\alpha)}(F,\overline{F})$. Since $F/K$ is normal then $g(F)=F$ and $g\in Aut(F/K(\alpha))$. This $g$ is in $H$.

$\tilde{g}$ non-trivial means that some $\beta\in F^H$ is not fixed by $g$, a contradiction.


问:
Intersection between a line and a circle in polar coordinate system
I have a point p1 connected to p2 in a polar coordinate system. This forms an "arc" or i believe part of a spiral..

What I would need to know, is how I could calculate the intersection between the circle (center at P1) and the above mentioned line, with only the radius of the circle, P1 and P2 given as theta/rho informations.

Its the first time I work with polar coordinates to please bear with me, its unexplored teritory.

it does not have to be precise, an approximation within +-2% is more than enough.

Thank you

答:
Well, without an exact equation for the "arc" or "spiral" solving this isn't going to be an exact solution, there even can't be given error bounds because you don't even know the actual position. However, from the above figure, my best guess is that as $\theta$ is increasing, $r$ is increasing. You could assume proportionality between $r$ and $\theta$ for the arc as below

$$\frac{r-70000}{20000} = \frac{\theta - \frac{2\pi}{3}}{\frac{5\pi}{8}}$$


答:
The population of bacteria triples each day on a petri dish.

Let $\{ a_n \}$ the population of bacteria on the n-th day.
We have $$a_{n+1} = 3 \cdot a_n$$
Furthermore, we have $$a_n = 3^{n-1} \cdot a_1$$

If it takes 20 days for the population of bacteria to fills the entire dish.

Let $S$ the total population of bacteria when the entire dish is filled.
We have $$a_{20} \geqslant S > a_{19}$$
or $$3^{19} \cdot a_1 \geqslant S > 3^{18} \cdot a_1$$

How many days it will take bacteria to fill half of the petri dish?

Just divide the inequity by 2, we have $$\frac{3^{19}}{2} a_1 \geqslant \frac{S}{2} > \frac{3^{18}}{2} a_1$$

Let $3^{n-1} = \frac{3^{18}}{2}$ and $3^{m-1} = \frac{3^{19}}{2}$, we have $n = 19 - log_3 2 \approx 18.369$ and $m = 20 - log_3 2 \approx 19.369$.

It's obvious that on the 19th day, the dish is half-filled.

Title: Bacteria population growth

Body:
The population of bacteria triples each day on a petri dish. If it takes 20 days for the population of bacteria to fill the entire dish, how many days will it take bacteria to fill half of the petri dish?

My doubt 1: if we don't have a initial population can be solve this problem?
Like $A(d)$= initial population times $3^{20}$ days but as I don't have this I am stuck.

My doubt 2: Also, can we assume a number for population at day 20 and move our way backwards?


问:
What is the relevance of seeing $\text{ad}$ as the differential of $\text{Ad}$
$\DeclareMathOperator{\ad}{ad}$$\DeclareMathOperator{\Ad}{Ad}$
There are essentially two ways of defining the $\ad$ operator associated to a Lie Group.
The first one is purely theoretical Lie-algebra theoretical, and it's straightfoward, is just defining $\ad(X)(Y)$ as $[X,Y]$, which is readily seen to be a Lie algebra homomorphism into $\mathfrak{gl}(\mathfrak{g})$.

This operator is of course very important, as is allows us to extract some information on $\mathfrak{g}$ from is image in $\mathfrak{gl}(\mathfrak{g})$, which allows us to use for instance Jordan decompositon on matrices and such.

Of course when we have a Lie group, we have a Lie algebra structure on its tangent space at identity (or on the left invariant tangent fields), and we need to prove that we have a Lie Bracket which amounts usually to define it as the derivative at $0$ of $e^{tX}e^{tY}e^{-tX}e^{-tY}$.

However, usually we define the $\ad$ representation in the following manner: we have a morphims $\Ad:G\to GL(\mathfrak{g})$ defined to be $g\mapsto d\text{Conj}(g)(1)$ where $\text{Conj}(g):G\to G$ is the conjugation. Then $\ad$ is simply the differential of $\Ad$

Is there any advantage of definig $\ad$ like that instead of simply definig it as $\ad(X)=[X,\bullet]$.

We can prove of course that it is an equivalent definition, but what is to be gained from defining $\ad$ in this rather convoluted fashion ?
All the theorems that I know about Lie algebra (the ones in Humphreys book for instance), use the fact that $\ad$ is the Lie bracket, and not the fact that is the differential of $\Ad$.

Maybe this is important, to descend Lie algebra representations into actual representations of $G$?

答:
(Expanding on my answer in the comments)

Are you sure about your proposed alternate definition of the Lie bracket? We should have something like $e^{tX}e^{tY}e^{-tX}e^{-tY}\approx (1+tX)(1+tY)(1-tX)(1-tY)=1+t^2[X,Y]+o(t^3).$ So you can't get the Lie bracket by taking the first derivative and setting $t=0$.

I think instead you can define it as $\frac{1}{2} \frac{d^2}{ds\,dt}[e^{sX},e^{tY}]$ or $\frac{d}{dt}[e^{\sqrt{t}X},e^{\sqrt{t}Y}]$. Either way that expression is more complicated and less natural, it's harder to motivate.

And defining the Lie bracket in terms of the exponential map requires quite a bit more machinery, a discussion of uniqueness of solutions to differential equations, whereas the definition in terms of the adjoint map requires only the differential.

Is there any advantage of definig ad like that instead of simply definig it as ad()=[,∙]?

Seems like a different question. If you only care about the Lie algebra, then you may define the adjoint map solely in terms of its axiomatically given Lie bracket.

But if you want to relate the Lie bracket on a Lie algebra to the group operation on the Lie group of which it is the tangent space, then you need an equation like $[X,Y]=\frac{1}{2} \frac{d^2}{ds\,dt}[e^{sX},e^{tY}]$ or $[X,Y]=d(\text{Ad}(X))(Y).$

Ultimately the first thing you will want to prove about this relation is the functoriality of the Lie algebra functor, $\text{Lie}(G)$ which associates to a Lie group its tangent space endowed with a Lie bracket. The definition of the Lie bracket in terms of the adjoint map makes this check more straightforward than the definition in terms of exponentials.

And the adjoint map has the benefit of giving you a representation of both the group and the algebra.

So to sum up, the definition in terms of the adjoint map is actually simpler, not more convoluted, than the one you propose.


问:
Trisect unknown angle using pencil, straight edge & compass; Prove validity of technique
This question was posed by my high school geometry teacher, for extra credit:

Is it possible, using only a pencil, a straightedge (not a ruler) and a compass to trisect an angle of unknown value? If so, describe the technique, and mathematically prove its validity. If not, mathematically prove the impossibility.

I recall wondering at the time, why he was posing this question to a freshman high school geometry class, and if he genuinely expected any of us to come up with an answer. Nothing more was said about it and to my knowledge, nobody submitted an answer. But the question stuck in my mind, and I would often spend hours puzzling over it, trying over and over to come up with a valid solution.

I'll be 60 years old this July, and I'm no closer to an answer now than I was that day, so many years ago. If anyone knows the answer, I'd appreciate a simple explanation, as I am no mathematician.

答:
If you are allowed to mark the straightedge with the pencil,
Archimedes showed how to do it.
Just search for "Archimedes angle trisection".

Here is one URL: https://en.wikipedia.org/wiki/Angle_trisection


问:
Definition of degree of invertible sheaf
Sorry for my bad English.

Let $X$ be regular curve over algebraic closed field $k$, and $T$ be scheme over $k$.

In Hartshorne algebraic geometry Section Iv, he define $Pic^0(X\times T)$ as follows:

invertible sheaf ${\scr L}\in Pic(X\times T)$ s.t. for any $t\in T$, degree of ${\scr L}$ is 0 over fiber $X_t$.

Now, I trouble about what is definition of degree of ${\scr L} $over fiber $X_t$.

If  $X_t$ is regular curve, I know this definition, but I don't know how $X_t$ is curve.

答:
If $p_T: T\times_k X \rightarrow T$ is the projection morphism, you define for any point $t\in T$

$$X_t:= p_t^{-1}(t)\cong  Spec(\kappa(t))\times_T (T\times_k X) \cong Spec(\kappa(t))\times_k X.$$

Here $k \subseteq \kappa(t)$ is a field extension and need not be an eqality in general. If $X$ is an integral scheme  over $k$ (let us assume $X$ is irreducible, then $X$ is integral since it is regular) it follows from Hartshorne, Ex II.3.20 that every irreducible component $X_i$ of $X_t$  has $dim(X_i)=dim(X)=1$ hence $X_t$ is a "curve" for any point $t \in T$.

Definition: We say a scheme $X$ of finite type over a field $k$ is a "curve" if the irreducible components $X_i$ of $X$ have $dim(X_i)=1$.

Question: "If $X_t$ is regular curve, I know this definition, but I don't know how $X_t$ is curve."

Answer: This follows from HH.Ex.3.20.

Example. Let $k$ be the field of real numbers and $K$ the field of complex numbers. Let $A:=k[x,y]/(x^2+y^2)$ and let $B:=K\otimes_k A$ with $C:=Spec(A)$. It follows $A$ is an integral domain and hence $C$ is an integral scheme of dimension one, but

$$K\otimes_k B \cong K[x,y]/(x+iy)(x-iy)$$

is no longer an integral domain. Hence $Spec(K)\times_k C:=C_K$ is no longer an irreducible scheme. The dimension of the irreducible components of $C_K$ is as you can see one.

Note: Since your interest is in the Picard scheme you cannot restrict to closed points as claimed in the comments. Hence the field extension $k \subseteq \kappa(t)$ will not be an isomorphism in general.


问:
Why are there are precisely $(n-1) ! / \prod x_{i} !$ trees with a given sequence $x_{t}$?
The following is from a paper by Joel Spencer, called Enumerating Graphs and Brownian Motion.

Let us look at a tree $T$ on vertex set $\{1, \ldots, n\}$ from a computer science
perspective. We'll take $n=7$ and the tree with edges 13,15,16,27,34,37
as an example throughout. Start with vertex 1 and find the tree through a
breadth-first search (BFS), all lists being in numerical order. To do this we
have a queue $Q$, that is initially (1). At each stage we take the vertex $x$ at the
head of the queue, remove $x$ from the queue, and add all "new" neighbors of
$x$ to the queue. We'll call that operation "processing $x$ " and say that $x$ finds $y$ if $y$ is a new vertex adjacent to $x$. In our example, 1 finds $3,5,6,$ then 3
finds $4,7,$ then 5 finds nothing, 6 finds nothing, 4 finds nothing, 7 finds $2,$ and
finally 2 finds nothing. After 2 is processed, the queue becomes empty and
the BFS ends.

Let $x_{t}$ be the number of vertices found by the $t^{\text {th }}$ vertex processed (not
vertex number $t$ ), which, in our example, is 3,2,0,0,0,1,0 . Let $q_{t}$ be the
size of the queue after the $t^{\text {th }}$ vertex is processed. Then $q_{0}=1$ and $q_{t}=
q_{t-1}+x_{t}-1$, which, in our example, is $1,3,4,3,2,1,1,0$ . The necessary and
sufficient conditions on the sequence $q_{t}$ are
$$
q_{n}=0 \quad \text { and } \quad q_{i}>0 \quad \text { for } i<n
$$
Now note that there are precisely $(n-1) ! / \prod x_{i} !$ trees with a given sequence
$x_{t}$. Indeed, any partition of the vertex set excluding source vertex 1 into sets
of sizes $x_{1}, \ldots, x_{n}$ gives a unique such tree $T$.

Can someone explain to me how we derive the last two sentences? I have tried with an example but can't seem to figure it out.

答:
$(n-1)!/\prod_{i=1}^kx_i!$ where $\sum_{i=1}^k x_i = n - 1$ is also called a multinomial coefficient written $\binom{n-1}{x_1 \ldots x_k}$. It gives the number of ways of distributing $n-1$ objects into $k$ bins with $x_i$ objects in bin $k$. You will find more details on this Wikipedia page.

Hence, given a distribution of $\{2, 3, \ldots n\}$ into subsets (bins) $X_1$, $X_2, \ldots X_k$ of sizes $x_1, x_2, \ldots x_k$, there is a unique tree such that the children of $1$ are the elements of $X_1$, the children of the least element of $X_1$ are the elements of $X_2$ and so on, where if we encounter an empty $X_i$, we backtrack to the most recent $X_j$ that we have considered that contains an element whose children we haven't investigated.


问:
If $\operatorname{Rank}(A^k) =\operatorname{Rank}(A^{k+1})$, why does $\operatorname{Rank}(A^{k+1}) = \operatorname{Rank}(A^{k+2}) $?
I have found this answer to a very similar question (basically identical).$\DeclareMathOperator{\Rank}{Rank}\DeclareMathOperator{\range}{range}$

However, I can't arrive at the result, even with the hints given.

I can understand the implication is the same as $\range(A^k) = \range(A^{k+1}) \Rightarrow \range(A^{k+1}) = \range(A^{k+2})$.

I can easily show that $\range(A^{k+2})$ is a subspace of $\range(A^{k+1})$, which in turn is a subspace of $\range(A^k)$.

Similarly, I can also see the problem in terms of the kernel of these matrices: $\ker(A^k) = \ker(A^{k+1}) \Rightarrow \ker(A^{k+1}) = \ker(A^{k+2})$

I can easily show that $\ker(A^k)$ is a subspace of $\ker(A^{k+1})$, which in turn is a subspace of $\ker(A^{k+2})$.

I've been trying to solve this for a while, and just can't seem to get there.

I feel like I'm missing something really obvious...

Thanks, Miguel

答:
Hint : As you said, one has
$$\lbrace 0 \rbrace \subset \mathrm{Ker}(A) \subset \mathrm{Ker}(A^2) \subset ... \subset \mathrm{Ker}(A^n) \subset ...$$

Try to prove that the sequence $(a_n)_{n \geq 1}$, defined by $a_n = \dim(\mathrm{Ker}(A^n))-\dim(\mathrm{Ker}(A^{n-1}))$, is decreasing.


问:
Two questions on clock arithmetic
I have two questions on clock arithmetic, both of which I have solved, but I am looking for neater proofs.

Let us suppose we have a circle named $\mathbb{Z}_n$ with $n$ equally spaced points on it labelled $0,1,\cdots n-1$ in the clockwise direction. By clock arithmetic I mean that $x+1$ is the point just after $x$ in the clockwise direction.

$1$. Let $\{a,b\}\subset \mathbb{Z}_n$ so that $\{a+t,b+t\}\ne \{a,b\}\forall t=1,2\cdots n-1$ . Then $\{a+i,b+i\}\ne\{a+j,b+j\}\forall 0\le i<j\le n-1$.

My proof: The hypothesis says that $a$ and $b$ are not on a diameter. Hence no $\{a+i,b+i\}$ is also on a diameter.

$2$. Let $\{a,b\}\subset \mathbb{Z}_n$ with $a>b$ in the ordinary integer ordering sense. Define the distance $d(a,b)$ between $a$ and $b$ as the minimum of the lengths of the two paths from $a$ to $b$ and from $b$ to $a$. By length of a path from $x$ to $y$ we mean the
  smallest positive $z$ with $x+z=y$. What is a formula for $d(a,b)$? So in $\mathbb{Z}_{19}, d(1,0)=1=d(0,18)$.

My solution: $d(a,b)=a-b$ if $1\le a-b\le \lfloor n/2\rfloor$ and $d(a,b)=n-a+b$ otherwise.

Can someone give an alternate "better looking" formula, possibly one which avoids using $<$ implicitly or explicitly?

答:
For the first, you can note that if the conclusion were false, $\{a+i-j,b+i-j\}$ would vilate the hypothesis.

For your second, if $a=1,b=2$ in $\Bbb Z_n$, your formula gives $d(a,b)=n-1$ instead of $1$.  You need some absolute value signs to fix this.

Title: Two questions on clock arithmetic


问:
Can you call a single tuple a "relation"?
A relation is a set of tuples, and a subset of the Cartesian product. But can the term "relation", perhaps informally, also mean a single tuple?

(I'm grading a database exam. As you might now, relational databases are inspired by mathematical relations. A table in the database is called a relation, and it is a set of tuples. But one student wrote that a relation is a single tuple, and before I deduct points I'd like to be sure that he doesn't know more math than I do.)

答:
A single tuple is not a relation. But a set containing a single tuple is, of course, a relation.

That is, $(1,1)$ is not a relation on $\mathbb R$. However, $\{(1,1)\}$ is a relation on $\mathbb R$.


答:
Here is an elementary proof. Let $x \in \mathrm{range}(A^{k+1})$. Then there is a $y$ such that $x = A^{k+1}y$. Clearly, $A^ky \in \mathrm{range}(A^k)$. Now, since $\mathrm{range}(A^k) = \mathrm{range}(A^{k+1})$, there is a $z$ such that $A^ky = A^{k+1}z$. This implies
$$x = AA^ky = AA^{k+1}z = A^{k+2}z,$$
hence $x \in \mathrm{range}(A^{k+2})$. As the other direction is clear, we have $\mathrm{range}(A^{k+1}) = \mathrm{range}(A^{k+2})$.

Title: If $\operatorname{Rank}(A^k) =\operatorname{Rank}(A^{k+1})$, why does $\operatorname{Rank}(A^{k+1}) = \operatorname{Rank}(A^{k+2}) $?

Body:
I have found this answer to a very similar question (basically identical).$\DeclareMathOperator{\Rank}{Rank}\DeclareMathOperator{\range}{range}$

However, I can't arrive at the result, even with the hints given.

I can understand the implication is the same as $\range(A^k) = \range(A^{k+1}) \Rightarrow \range(A^{k+1}) = \range(A^{k+2})$.

I can easily show that $\range(A^{k+2})$ is a subspace of $\range(A^{k+1})$, which in turn is a subspace of $\range(A^k)$.

Similarly, I can also see the problem in terms of the kernel of these matrices: $\ker(A^k) = \ker(A^{k+1}) \Rightarrow \ker(A^{k+1}) = \ker(A^{k+2})$

I can easily show that $\ker(A^k)$ is a subspace of $\ker(A^{k+1})$, which in turn is a subspace of $\ker(A^{k+2})$.

I've been trying to solve this for a while, and just can't seem to get there.

I feel like I'm missing something really obvious...

Thanks, Miguel


问:
If $A^{n-1}$ is not diagonalizable but $A^n$ is, show $A^n$ is zero
Let $n$ be an integer greater than or equal to 2 and let $A$ be a $n^{th}$ order complex square matrix. Show that if $A^{n-1}$ is not diagonalizable and  $A^n$ is diagonalizable, then $A^n=0$.

I have tried to think of a way to prove that but couldn't. I know that, If a matrix is diagonalizable over a field $F$ if and only if its minimal polynomial is a product of distinct linear factors over $F$. Can we use this to prove it or what is the proper way to think about it?

答:
Write $\mathbb{C}^n$ as a direct sum of characteristic spaces $G_{\lambda_i}$ for $A$. Then, for each $G_{\lambda_i}$, $A^n$ is an endomorphism of this space and the $n$-th power of $\lambda_i I+\nu$, for some nilpotent $\nu$, so $\lambda_i^n$ is its only eigenvalue and $A^n_{|G_{\lambda_i}}=\lambda_i^n I$.

If $\lambda_i \neq 0$, then $A_{|G_{\lambda_i}}$ vanishes the polynomial $X^n-\lambda_i^n$ which has simple roots, so $A_{|G_{\lambda_i}}$ is diagonalizable.

It follows that $A^{n-1}_{|G_0}$ is not diagonalizable (in particular nonzero), while $A^n_{|G_0}=0$ by Cayley-Hamilton. Let $d$ be the dimension of $G_0$, $A_{|G_0}$ is nilpotent so $A^d_{|G_0}=0$. So $d > n-1$ thus $d=n$, hence $G_0=\mathbb{C}^n$. Thus $A$ nilpotent and by Cayley-Hamilton $A^n=0$.


问:
If $f(z)$ is even then residue at $z=z_0$ is zero?
There is a theorem in my text book which says that:

Theorem:If $f$ has an isolated singularity at $z_0$ and if $f$ is even in $z-z_0$ i.e $f(z-z_0)=f(-(z-z_0))$ then Res$[f(z),z_0]=0$ [Theorem 8.60, page 290,S ponnusami]

My doubt: let $f(z)$ be even function then $f(z)=f(-z)$ for all $z$ in domain of f and hence we have $f(w)=f(-w)$ where $w=z-z_0$ implying that $f$ is even in $z-z_0$ and hence from theorem for any even function $f$ Res$[f(z),z_0]=0$ at isolated singularity $z_0$ which I think wrong.For counter example see the picture added

答: The theorem requires that the function be even about the point $z_0$. The function $\frac{1}{z \sin z}$ is even about $z=0$, but it is not even about $z=\pi$. Indeed, the residue at $z=0$ is in fact $0$, but not the residue at $z=\pi$.


问:
Prove that connected graph $G$ with $\delta (G)\geq 3$ contains a cycle $C$ such that $G-E(C)$ is still connected.

In a connected simple graph every vertex has a degree at least $3$. Prove that the graph contains a cycle such that the graph remains connected when the edges of this cycle are deleted.

Source: https://www.komal.hu/verseny/2000-02/mat.e.shtml

I've tried this: Pull out of $G$ a cycle $C$ with smallest perimeter $p$.

Claim: No two nonconsecutive vertices in the circle $C$ are adjacent (else we could make smaller cycle then we already have).

Let $C_1,C_2,...,C_k$ be all components in $G-C$ and I want to prove $k=1$.

If $n_i = |V(C_i)|$ and $e_i= |E(C_i)|$, then we have
$$e_i = {3n_i\over 2}-b_i$$ where $b_i$ is a number of edges between $C_i$ and $C$. Because $G$ is connected we have $b_i\geq 1$ for each $i$. Because of the claim we have:  $$b_1+b_2+\cdots +b_k = p$$

I have no idea what to do now.

Edit: As Misha Lavrov pointed in a comment, this attemt is not correct.

答:
Suppose $G$ to be a counterexample with a minimal number of  edges.

$G$ has no isthmus $vw$.

Otherwise, contract edge $vw$ and consider a cycle $C$ of the reduced graph whose removal does not disconnect the reduced graph. Then $C$ is entirely within one of the components of $G$ produced by deleting $vw$ and so deleting it from $G$ does not disconnect $G$ and there is no counterexample after all.

No two adjacent points $v,w$ of $G$ can both have degree at least $4$.

Otherwise simply remove edge $vw$.

$G$ has no triangle $u,v,w$.

Otherwise, removing edges $uv,vw,wu$ must disconnect the graph. Without loss of generality we can assume that $u$ is then in a separate component from $v$ and $w$.

If $\rho (u)=3$, then let $x\in N(u)/\{v,w\}$. Then $xu$ is an isthmus.

So $\rho (u)>3$ and therefore $\rho (v)=\rho (w)=3$.

If $(N(v)\bigcap N(w))/\{u\}=\{ y\}$, then contracting all edges joining vertices in $\{u,v,w,y\}$ produces a simple connected graph with all vertices of degree at least $3$ and we can proceed using minimality as in the 'isthmus' proof.

Otherwise, we can assume $N(v)=\{ u,w,y\}$ and $N(w)=\{ u,v,z\}$ with $y\ne z$. Then contracting all edges joining vertices in $\{u,v,w\}$ produces a simple connected graph with all vertices of degree at least $3$ and we can again proceed using minimality.

G is 3-regular

Otherwise let $N(u)=\{x,y,v\}$ with $\rho (v)>3$. Reduce $G$ by removing $u$ and adding edge $xy$. The new graph has a cycle $C$ whose removal does not disconnect the graph. If $xy$ is in this cycle then replace it by $xuy$. Removing the possibly amended cycle $C$ does not disconnect $G$.

Proof

Let $N(u)=\{v,a,b\}$ and $N(v)=\{u,c,d\}$ . Delete $u$ and $v$ and add in edges $ab$ and $cd$. The new graph has a cycle $C$ whose removal does not disconnect the graph.

If neither $ab$ nor $cd$ is in $C$ then removing $C$ does not disconnect $G$.

If, say, $ab$ but not $cd$ is in $C$ then replace it in the cycle by $aub$. Removing the possibly amended cycle $C$ does not disconnect $G$ and again there is no counterexample.

Finally suppose both  $ab$ and $cd$ are in $C$. Part of $C$ is  a path from $a$ to say $c$ (or to $d$) which does not include $ab$ or $cd$. Combining this path with $auvc$ then gives a cycle which can be removed without disconnecting $C$.


问:
What does it mean to prove that two systems have to same theorems?
Lets say there is two systems $L_1$ and $L_2$.If I get asked:

Show that $L_1$ and $L_2$ have the same theorems.

Is it sufficient to prove the axioms of $L_1$ using the axioms of $L_2$ or vice-versa ? Or do I have to show that a wf is a theorem of $L_1$ iff it is a theorem of $L_2$ ?

答:
It is indeed sufficient to show that each axiom of $L_1$ is derivable from the axioms of $L_2$ and vice versa. This implies that any wff is a theorem of $L_1$ iff it a theorem of $L_2$.

To see why assume $\phi$ is a theorem of $L_1$. The proof $\Pi$ of $\phi$ can only use finitely many axioms of $L_1$, say $\alpha_1, \ldots, \alpha_k$. If we have proofs $\Pi_1, \ldots, \Pi_k$  of $\alpha_1, \ldots, \alpha_k$ in $L_2$, we can combine $\Pi_1, \ldots, \Pi_k$ and $\Pi$ to give a proof of $\phi$ in $L_2$. So any theorem of $L_1$ is a theorem of $L_2$. Likewise, swapping $L_1$ and $L_2$ any theorem of $L_2$ is a theorem of $L_1$.

Aside: the above assumes that your systems have the same inference rules, e.g., if you are looking at a typical exposition of first-order logic, as a Hilbert style system with modus ponens as the only inference rule. If you are looking at systems with different inference rules (e.g, a natural deduction calculus on the one hand and a Hilbert-style system on the other), you will also have to show that each rule of $L_2$ is derivable (or more generally, what is called permissible) on the basis of the axioms and rules of $L_1$ and vice versa.


问:
The equivalence of Lindeberg with CLT & Feller for a given example
I consider Gabriel's example "Showing that Lindeberg condition does not hold" where we deal with a sequence of random variables $(X_n)_n$ with $X_k = Y_k + Z_k$ and $P(Y_k = \pm 1) = \frac{1}{2}$ and
$$
P(Z_k = \pm k) = \frac{1}{2k^2} = \frac{1 - P(Z_k = 0)}{2}.
$$
As in the solution there, it can be shown, that the CLT for $S_n = \sum_{k=1}^{n} X_n$ holds, that is
$$
\frac{S_n}{\sqrt{n}} {\stackrel{d}{\longrightarrow}} \eta \sim \mathcal{N}(0,1) \quad \text{for } {n \rightarrow \infty}
$$
but the Lindeberg condition is not fullfilled (even for the little more genereal Case where $Y_k$ is iid with expectation 0 and variance 1).

But I struggle with the Feller condition. As fare as I know this is (if I use the triangular array with $X_{n_k}:=(\frac{X_k}{\sqrt{n}})$):
$$
 \max_{1 \leqslant k \leqslant n} \sigma_{n_k} = \max_{1 \leqslant k \leqslant n} \mathbb{E}\left[\left(\frac{X_k}{\sqrt{n}}\right)^2\right] = \max_{1 \leqslant k \leqslant n} \frac{\mathbb{E}(X_k^2)}{n} = \frac{2}{n} \stackrel{n \to \infty}{\longrightarrow} 0,
$$
as it may be found in mathworld.
However, if the Feller condition is fulfilled, the Lindeberg condition is necessary and sufficient for the CLT.

Do I have a mistake with the Feller condition or is there another assumption for the equivalence Lindeberg $\Leftrightarrow$ Feller & CLT I do not considere/know?

And maybe another but minor important question: the example shows that the second moment of $\frac{S_n}{\sqrt{n}}$ does not converge. What do I need that $\mathbb{V}\left(\frac{S_n}{\sqrt{n}}\right) \stackrel{n \to \infty}{\longrightarrow} \mathbb{V}(\eta)$?

答: The Variance $\mathbb{V}\left(\frac{S_n}{\sqrt{n}}\right) = 2 =: \sigma_X^2$ is constant for every $n$. As Feller holds Lindeberg would imply that $\frac{S_n}{\sqrt{n}} \stackrel{d}{\rightarrow} Z \sim \mathcal{N}(0,\sigma_X^2)$ which is a contradiction to the limiting standard normal distribution. Therefore Lindeberg cannot be fullfilled.


问:
Inequality with $abc=1$
Let $a;b;c$ be positive real numbers and $abc=1$

Find maximum value of: $P=\dfrac{1}{\sqrt{a^2+2b^2}}+\dfrac{1}{\sqrt{b^2+2c^2}}+\dfrac{1}{\sqrt{c^2+2a^2}}$

I tried to use Cauchy-Schwarz: $a^2+2b^2 \geq \dfrac{1}{3}(a+2b)^2$

Then: $P \leq \dfrac{\sqrt{3}}{a+2b}+\dfrac{\sqrt{3}}{b+2c}+\dfrac{\sqrt{3}}{c+2a}$

So, what should I do next? Thank you.

答:
There is no maximum .

Note that when $a\to 0^+,b\to 0^+,c\to \infty$ we get  $P\to \infty$ .This is because the $\sqrt{a^2+2b^2}\to 0$

Note that we chose $c\to \infty$ because $c=\frac{1}{ab}$

Now even if you are looking for the minimum there is no minimum .

Note  again that  $a\to 0^+,b\to \infty,c\to \infty$ we see that $P\to 0$.

Title: Inequality with $abc=1$


问:
proof of $ A\otimes M\cong M $
Let $M$ be an $A$-module. Can someone prove the $A$-module isomorphism:

$$A\otimes M \cong M?$$

(By $\otimes$ I mean tensor product.)

答:
First define a map $A\times M\rightarrow M$ such that $(a,m) \mapsto am$. This is clearly bilinear, so it induces a homomorphism $A\otimes M\rightarrow M$ such that $a\otimes m \mapsto am$ (by the universal property of tensor products). Surjectivity is straight forward, just take $a=1$, $m$ arbitrary. To show that the homomorphism is injective, note that any element in $A\otimes M$ can be expressed as $1\otimes m$. But this maps to $m$. If $m=0$, then $1\otimes m = 0$, hence the map is injective.

Edit: the hard part of this problem is getting a well-defined homomorphism. This is the appeal of the universal property. It is much easier to define (bilinear) maps on $A\times M$ than it is to define homomorphisms from the tensor product (without appealing to the universal property).

Title: proof of $ A\otimes M\cong M $


问:
Is the multiplication operator on $\ell^p$ compact?
Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.

Can we more generally conclude that $T$ is compact?

Assume $p<\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology.

On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$.

EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true.

答:
The set of all compact operators between Banach spaces is closed in norm topology.

A multiplication operator on $l^p$ is compact if and only if the corresponding sequence is an element of $c_0$.


问:
Semidirect product notation convention
I was taught that if $G \simeq H \ltimes K$, then (by convention) $H$ is the subgroup that is normal, but I see on Wikipedia and elsewhere that other people use the convention that the above indicates that $K$ is the one that is normal.

Has anyone encountered any literature where $H$ is the one that is normal, or do most people follow the convention that $K$ is normal?

Edit for clarification

$G \simeq H \ltimes K \simeq K \rtimes H $

Convention 1: $H$ is normal

Convention 2: $K$ is normal

答:
Note that both conventions make sense. It seems that there is no disagreement about the fact that the triangle in the notation is connected to the notation for normal subgroup: $H \trianglelefteq G$ or $H \vartriangleleft G$.

With convention 1 the triangle is closer to the normal subgroup, but the tip is pointing the other direction, while with convention 2, the tip is pointing the right way, but the triangle is closer to the other subgroup.

I've got the impression (but it is just impression) that convention 2 is used more often. On the other hand, for example the names of the corresponding Unicode characters (U+22C9, U+22CA) use convention 1.

Note that there is also an ortogonal question in the used conventions – wherether the normal subgroup is the left or the right factor in the product. Here it seems that it is usually the left one.

Title: Semidirect product notation convention


问:
Calculating the parameters of the logistic function: $L(t)=\frac{a}{1+\exp(b-ct)}$
$$L(t)=\frac{a}{1+\exp(b-ct)}$$

Choose the parameters $a,b,c\in \mathbb{R}$ to get that $\lim_{t \rightarrow \infty} \space L(t)=50, \space L(0)=5, \space L(10)=25 $

There is a hint that:

$$
\lim_{t \rightarrow \infty} \frac{a}{1+\exp(b-ct)} = \frac{a}{1+\exp(-\infty)}=a=50
$$

I can't get that step (why has became $(b-ct)$ to $-\infty$ in the dominator and unfortunately I even don't know how to continue the calculation. Really appreciate any help.

答:
$\require{cancel}$

First note that

$$
L(t) = \frac{a}{1 + e^{b - ct}} = \frac{a}{1 + e^ be^{-ct}} \tag{1}
$$

now, if $c>0$ then

$$
\lim_{t\to \infty} e^{-ct} = 0 \tag{2}
$$

and then

$$
\lim_{t\to \infty} L(t) = \lim_{t\to \infty}\frac{a}{1 + e^b e^{-ct}} \stackrel{(2)}{=}  a = 50 \tag{3}
$$

So now you have the other condition

$$
5 = L(0) \stackrel{(1), (3)}{=} \frac{50}{1 + e^b \cancelto{1}{e^0}} = \frac{50}{1 + e^b} ~~~\Rightarrow~~~ 1 + e^b = 10 ~~~\Rightarrow~~~ e^b = 9 \tag{4}
$$

So far we have

$$
L(t) = \frac{50}{1 + 9 e^{-ct}} \tag{5}
$$

I'm going to leave the last step (find the value of $c$) to you


问:
$\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}$
Let $\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=x$. Therefore, $$n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}=x^2$$ $$x^2-x-n=0$$ $$x=\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=\frac{1 \pm \sqrt{1+4n}}{2}$$

It seems counter-intuitive at first that $f(n)$ doesn't have an effect on the final answer. If $f(n)$ were $\frac{1}{n!}$ or $e^{2.3n^2}$, or anything else for that matter, we still get the same formula for the result.

My question regarding this is:

Is this applicable for any $f(n)$ at the end of the infinite nested radical (I haven't studied much about properties of functions but have heard of terms like 'odd function', 'even function', '$f: \mathbb{R} \to \mathbb{R}$' etc)?

Is this result because of the infinitesimally small effect of $f(n)$ on the result?

EDIT : Some people have commented that there is '1 less $n$ post squaring'. There are infinite '$\sqrt{n+\sqrt{}}$' before I added $f(n)$ at the end.

答:
Firstly, note that you actually don't need the function $f$. You only need its value at point $n$ and we can mark it $x_0=f(n)$.

You actually invented the fixed point iteration. You are iterating the equation $x_{m+1}=\sqrt{n+x_m}$. You noticed that this may have two solutions. Let's mark them:
$$
\bar{x}_1=\frac{1+\sqrt{1+4n}}{2}\\
\bar{x}_2=\frac{1-\sqrt{1+4n}}{2}
$$
The solution $\bar{x}_1$ will be always valid, while $\bar{x}_2$ will be valid only for $n \in \left[-\frac14;0\right]$. For small $n$, it will be imaginary, and for big $n$, it will be negative and square root can't be negative in real numbers.

The iterated function $x_{m+1}=\sqrt{n+x_m}$ is contractive (i.e. its first derivation is in range $(-1;1)$) in $\bar{x}_1$ except one case but never in $\bar{x}_2$. The only case, in which the function isn't contractive in $\bar{x}_1$ is when $n=-\frac14$ and $\bar{x}_1=\bar{x}_2$. So there are a few cases which may happen:

For $n<-\frac14$, there is no real solution and $x_m$ will diverge.

For $n>0$, the solution $\bar{x}_2$ doesn't make sense and the iterated function is contractive at $\bar{x}_1$. So $x_m$ converges to $\bar{x}_1$ for any $x_0$ for which the function makes sense.

For $n \in \left(-\frac14;1\right]$, the iterated function is contractive only at $\bar{x}_1$. If $x_0>\bar{x}_2$, $x_m$ will converge to $\bar{x}_1$. If $x_0=\bar{x}_2$, $x_m$ will stay $\bar{x}_2$ forever. And if $x_0<\bar{x}_2$, $x_m$ will diverge.

For $n = -\frac14$, it holds that $\bar{x}_1=\bar{x}_2=\bar{x}$. If $x_0 \geq\bar{x}$, $x_m$ will slowly converge to $\bar{x}$. Otherwise, it will diverge.

As you see, in cases 3 and 4, it actually depends on the choice of $x_0$.


问:
Does $\sum^\infty_1(-1)^n(\frac {1+(-1)^n}{n})$ converge or diverge?

Does $\sum^\infty_1(-1)^n\left(\frac {1+(-1)^n}{n}\right)$ converge or diverge?

My Work: At first I noticed that it looks like an alternating series and wanted to say it converges because $\frac {1+(-1)^n}{n}$ goes to zero, but I noticed that for $n=1$, we have $0$, for $n=2$ we have $\frac {2}{n}$, for $n=3$, $0$. So it's not really decreasing.

From here I only could do this by intuition that this behaves like $\frac {1}{n}$ or maybe less than it since we have $0$ every second element. so it diverges.

Would really like to know if what I did is right and if the series converges or diverges.

Thanks in advance

答:
Another way to see it,
$$
\sum^N_1(-1)^n\left(\frac {1+(-1)^n}{n}\right)=
\sum^N_1\left(\frac {(-1)^n}{n}\right)+
\sum^N_1\left(\frac {1}{n}\right)
$$
Now as $N \to \infty$, note $\sum^N_1\left(\frac {(-1)^n}{n}\right)$ converges and $\sum^N_1\left(\frac {1}{n}\right)$ diverges.  Therefore $\sum^N_1(-1)^n\left(\frac {1+(-1)^n}{n}\right)$ diverges.

More generally, in
$$
\sum_{n=1}^N (a_n + b_n) = \sum_{n=1}^N a_n+\sum_{n=1}^N b_n,
$$
if any two of the three series converge, then so does the third and
$$
\sum_{n=1}^\infty (a_n + b_n) = \sum_{n=1}^\infty a_n+\sum_{n=1}^\infty b_n .
$$


问:
Proof of basic subtraction rules for natural numbers
For $m, n ∈ \mathbb{N}_0$ we define a relation $≥$ by $m ≥ n ⇔ ∃r ∈ \mathbb{N}_0, m = r + n$. We denote $r$ by the difference $m - n$ which is thus defined only when m ≥ n.

How can we verify the basic subtraction rules involving natural number, specifically:

$m – (n – r) = (m – n) + r$ for $m ≥ n ≥ r$,

$m + (n – r) = (m + n) – r$ for $n ≥ r$,

$m(n – r) = mn – mr$ for $n ≥ r$.

E.g. for 3 we can set $n = s + r$ and thus $mn = m(s + r) = ms + mr$, which by definition $mn – mr = ms = m(n – r)$ since $s = n – r$.

Any hint in the right direction would be welcome. Thanks in advance.

答:
Since (3.) is shown already in the question itself, I detailed below only the presumed proofs for (1.) and (2.).

For (1.) setting $n=s+r$ since $n ≥ r$ and $m=w+n$ since $m ≥ n$ thus

$m - n = w$ since $m ≥ n$

$m = w + n = w + (s + r)$ since $m ≥ n$ and $n ≥ r$

$m = w + (s + r) = w + (r + s) = (w + r) + s$ by associativity and commutativity of addition

$m = ((m - n) + r) + (n - r)$ by definition

$m - (n - r) = ((m - n) + r) = (m - n) + r$ by definition since $m ≥ n$ and by extension $m ≥ s$

For (2.) setting $n=s + r$ since $n ≥ r$ thus

$m + n = m + (s + r)$

$m + n = (m + s) + r$ by associativity of addition

$m + s = (m + n) - r$ by definition, since $n ≥ r$

$m + (n-r) = (m + n) - r$ since $s = n - r$

I would still appreciate a double check.


问:
Evaluating an improper integral using the comparison test
I'm asked to determine if the following integral converges or diverges.

$$\int_{0}^{\infty} \frac{\ln(x)}{x^2+x+1} dx $$.

I know I have to use the comparison test for improper integrals (since there is no elementary anti derivative), but I'm not sure what to use as my comparison.

INTUITIVELY, it seems that the function will converge since the denominator grows very fast, but what can I use for my comparison test here? I need a function that's larger than the integrand in order to determine convergence or divergence of this integral.

Can someone suggest what kind of function I can pick as my comparison?

答:
for the convergence around $0$, you can notice that $\left|\frac{\ln(x)}{x^2+x+1} \right|\leq |ln(x)|$ which is convergent around $0$.

Then, for the upper bound, you can note that for $x$ large enough, $ln(\sqrt{x})\leq \sqrt{x} $

So you get :

$$\left|\frac{\ln(x)}{x^2+x+1} \right|\leq \dfrac{2\sqrt{x}}{x^2+x+1}\sim_{\infty} 2x^{-3/2} $$
Which is integrable according to the Reimman test


答: Hint: You can deal with$$\int_1^\infty\frac{\ln(x)}{x^2+x+1}\,\mathrm dx$$doing $x=\frac1y$ and $\mathrm dx=-\frac{\mathrm dy}{y^2}$, thereby getting$$\int_1^0-\frac{\ln(1/y)}{y^2+y+1}\,\mathrm dy=-\int_0^1\frac{\ln(y)}{y^2+y+1}\,\mathrm dy.$$Can you take it from here?

Title: Evaluating an improper integral using the comparison test

Body:
I'm asked to determine if the following integral converges or diverges.

$$\int_{0}^{\infty} \frac{\ln(x)}{x^2+x+1} dx $$.

I know I have to use the comparison test for improper integrals (since there is no elementary anti derivative), but I'm not sure what to use as my comparison.

INTUITIVELY, it seems that the function will converge since the denominator grows very fast, but what can I use for my comparison test here? I need a function that's larger than the integrand in order to determine convergence or divergence of this integral.

Can someone suggest what kind of function I can pick as my comparison?


问:
How to prove a map $f:R \rightarrow C(X, R)$, defined as $f(t) = v_t$, is an embedding, where $X$ is a tychonoff space and $R$, a real line.
I want to know a compatible criterion to prove that the map $f:R \rightarrow C(X, R)$ as $f(t) = v_t$, $v_t$ being a constant map, $t \in R$, is an embedding.

答:
Assuming the uniform topology (as suggested in the comments):

If $B_t(r)$ is an open ball of radius $t>0$ around $r\in \Bbb R$, then $f[B_t(r)] \cap f[\Bbb R] = \mathbf{B}(v_r, t) \cap f[\Bbb R]$, where $\mathbf{B}(f, t) = \{g \in C(X,\Bbb R): \forall x \in X: |f(x)-g(x)| < t\}$ is a basic uniform-open set in  $C(X,\Bbb R)$. This shows $f$ is open as map from $\Bbb R$ onto $f[\Bbb R]$.

It's also clear that if $r_n \to r$ in $\Bbb R$, then $f(r_n) \to f(r)$ in the uniform (or also pointwise or compact-open) topology on $C(X,\Bbb R)$ showing continuity of $f$ as well. As $f$ is clearly 1-1, $f$ is an embedding.


问:
Can $\int^b_a xf(x)\,dx =\frac{a+b}{2} $?
Let $f:[a,b]\to (0,+\infty )$ a continous function and $\int^b_a f(x)\,dx = 1$
Can $\int^b_a xf(x)\,dx =\frac{a+b}{2} $ ?
I do not know if it is true or not , but if it is true it would help me to solve a problem.

答:
Consider the denstiy function $f$ of the uniform distrbution on $[a,b]$, which is given by $$f(x)=\frac{1}{b-a}\cdot1_{[a,b]},$$
where $1_{[a,b]}$ denotes the indicator function on the interval $[a,b]$ (you can of course neglect the indicator function when you just consider the interval $[a,b]$). Thus the first condition
$$\int_a^b f(x)\mathbb dx=1$$is satisfied since it is a density function and the second on is the expected value of a random variable $X$ with such a distribution:
$$\int_a^b x\cdot f(x)\mathbb dx=\mathbb E(X)=\frac{a+b}{2}.$$


问:
Finding k in exponential continuous random variable distribution
Find the unknown parameter $k$ given this distribution:
$$ f(x)= \begin{cases} e^{kx} & 0 < x < 2\\
            0 & \text{otherwise}
        \end{cases}$$

This is my attempt:

$$\int_0^2 e^{kx} dx = \left[\frac{1}{k}e^{kx}\right]_0^2 \\
\frac{1}{k}e^{2k} -\frac{1}{k}=1 $$

This is where I am just stuck. How do I solve the above for $k$? I feel like I am missing something simple but I just can't see it.

Edit: I can see that I can rearrange this to equal $e^{2k} = k+1$ and then take logs giving me $k=\frac{1}{2}ln(k+1)$ but that still doesn't let me solve for $k$?

答:
The total integral should indeed be $1$, but $ \int_0^2 e^{kx} dx = \left[ \frac{1}{k} e^{kx} \right]^{x=2}_{x=0} = \frac1k(e^{2k}-1) =1$ instead..

So $e^{2k} = (k+1)$ and take logs..


问:
$a+n^2$ is the sum of two perfect squares
$a+n^2$ is the sum of two perfect squares for a given positive integer $a$ and all positive integers $n$.
Show that $a$ is a perfect square.

At first I thought of putting a bound on the difference between perfect squares up to a point, like maybe choosing $n=c$ so that one of the squares on the right hand side is the smallest, to try and show that if $a = x^2 +  k< (x+1)^2$ that there would have to be two pairs of squares with the same difference, but I can't seem to work out the details

答:
The general solution of the equation
$$x^2+y^2=z^2+w^2$$  is given by the identity with four arbitrary parameters
$$(tX+sY)^2+(tY-sX)^2=(tX-sY)^2+(tY+sX)^2$$
Let $n$ be any integer so we have $$a+n^2=z^2+w^2$$Making
$$n=tY-sX\\z=tX-sY\\w=tY+sX$$ we have three equations with four unknowns which in general have infinitely many solutions. Any way we have
$$n^2=t^2Y^2+s^2X^2-2stXY\\z^2=t^2X^2+s^2Y^2-2stXY\\w^2=t^2Y^2+s^2X^2+2stXY$$ which implies $$z^2+w^2-n^2=t^2X^2+s^2Y^2+2stXY=(tX+sY)^2$$ Since $$a=z^2+w^2-n^2$$ we have
$$a=(tX+sY)^2$$ We are done.


问:
Extracting monotone distinct sequence
Let $(x_n)$ be a sequence in $\mathbb{R}$ and $x_n\rightarrow p$ for some $p\in \mathbb{R}$. Suppose $x_n=p$ for finitely many index. Then there is a strictly monotone distinct (every element is distinct) subsequence $(x_{n_k})$ (of course $x_{n_k}\rightarrow p$).
Please comment on the correctness of my following contruction. Thanks in advance.

I try to construct $(x_{n_k})$ in the following way
 Clearly $(-\infty,p)$or $(p,\infty)$ contains infinitely many $(x_n)$.
 Case 1: $(-\infty,p)$ contains infinitely many $(x_n)$. In this case, we can construct a strictly increasing subsequence. Choose any point from $(\infty,p)$ as $x_{n_1}$. For $k\ge 2$, note that there are infinitely many $n_k$ such that $x_{n_k} \gt \max \{x_{n_1}, \cdots ,x_{n_{k-1}}\}$. Such $n_k$ always exists as sequence element equals to $x_{n_1}, \cdots ,x_{n_{k-1}}$ are only finitely many and the fact $x_n\rightarrow p$. Choose $n_k$ such that $n_k \gt n_{k-1}$. I think $(x_{n_k})$ is the required subsequence.

Case 2:$(-\infty,p)$ contains infinitely many $(x_n)$. By the same way, we can have a strictly decreasing  subsequence $(x_{n_k})$ with the required property.

 Remark: If both $(-\infty,p)$ and $(p,\infty)$ contains infinitely many $(x_n)$, then we get both strictly increasing and decreasing subsequence with the required property.

答:
Your proof is correct.

Implicitly, you are using the fact that a sequence converges to a limit if and only if for every open interval containing the limit, there are only finitely many values of the sequence outside of that open interval.

Title: Extracting monotone distinct sequence


问:
How to calculate number of onto and one homomorphism?
How to calculate the number of one one onto homomorphism from one finite cyclic group to another?

I tried using first theoram of isomorphism but could not find the result. Any help would be greatly appreciated.

答:
Suppose $C_1=\langle g\rangle$ and $C_2$ are cyclic groups. A homomorphism $C_1\to C_2$ is completely determined by the image of $g$. If $h\in C_2$, then:

There is a homomorphism taking $g$ to $h$ if and only if the order of $h$ divides the order of $g$ (or the order of $g$ is infinite).

This homomorphism is 1-1 if and only if the orders of $h$ and $g$ are equal.

This homomorphism is onto if and only if the order of $h$ equals the order of $C_2$.

These observations reduce the problem to elementary number theory.


问:
Two numbers $a$ and $b$ are selected at random from $1,2,3\ldots100$ and are multiplied. The probability that the product $ab$ is divisible by $3$ is
My approach:

In $1,2,3\ldots 100$ there are $33$ multiples of $3$.Therefore $1$ number is chosen from these $33$ numbers and from the remaining numbers ( except the one chosen) another number is chosen.

$$\frac{{33\choose 1}\cdot{99\choose 1}}{100\choose 2}$$

But my answer is wrong. Can you explain why?

The correct approach is

$$\frac{{33\choose 1}\cdot{67\choose 1}+{33\choose 2}}{100\choose 2}$$

Thank you

答:
The problem in your calculation is that you have counted certain choices twice - precisely those choices where both numbers are divisible by $3$. If $a$ and $b$ are both divisible by $3$, you've counted that once as $a$ being one of "the $33$" and $b$ being one of "the $99$", and again as $b$ being one of "the $33$" and $a$ being one of "the $99$".

The correct number of pairs is therefore smaller by the number of those duplicates, i.e. ${33 \choose 2}$. And, indeed:

$$\frac{{33\choose 1}\cdot {99\choose 1}-{33\choose 2}}{100\choose 2}=\frac{2739}{4950}=\frac{{33\choose 1}\cdot{67\choose 1}+{33\choose 2}}{100\choose 2}$$

(as in the "correct approach" - however that approach is justified!).

In fact, as per @lulu's comment, it is a lot easier to calculate the probability $ab$ is not divisible by $3$, which is:

$$\frac{67\choose 2}{100\choose 2}=\frac{2211}{4950}$$

from which it follows that the probability you are after is $1-\frac{2211}{4950}=\frac{2739}{4950}$.


问:
Is this a contradiction?
I have proven that a group has one and only one element $a$ such that $a=a^2$, that is $a$ is the square of itself.

However, then I encounter this question about finding $2\times 2$ matrices such that $A=A^2$. It turned out that there are more than one of them.

Could anyone explain why that is the case? I understand that these "special" matrices are in a group of matrices.

The three matrices I have found so far are:
\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}
,\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
and the zero matrix.

Thank you very much.

Nick

答: Since your group operation is matrix multiplication, there is no zero element in your group.  The identity is the identity matrix.  Second, your second matrix squared is not itself, but the identity.</p>

Title: Is this a contradiction?


答:
This is not true for the 2nd matrix, call it $A$.

Here $A^2= I$ is the identity matrix and not $A$.

Zero matrix is absorbing in the sense that $O\cdot A = O = A\cdot O$ for each matrix $A$.

Title: Is this a contradiction?

Body:
I have proven that a group has one and only one element $a$ such that $a=a^2$, that is $a$ is the square of itself.

However, then I encounter this question about finding $2\times 2$ matrices such that $A=A^2$. It turned out that there are more than one of them.

Could anyone explain why that is the case? I understand that these "special" matrices are in a group of matrices.

The three matrices I have found so far are:
\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}
,\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
and the zero matrix.

Thank you very much.

Nick


答: For arbitrary continuous function. The answer is not Take $f(x)=\text{sin}x$ and $a=\frac{\pi}{2}$ and $b=\pi$

Title: Can $\int^b_a xf(x)\,dx =\frac{a+b}{2} $?

Body:
Let $f:[a,b]\to (0,+\infty )$ a continous function and $\int^b_a f(x)\,dx = 1$
Can $\int^b_a xf(x)\,dx =\frac{a+b}{2} $ ?
I do not know if it is true or not , but if it is true it would help me to solve a problem.


答:
I think there is an error in the text. I suppose that the density is

$$f(x)=k e^{kx}$$

so

$$k=\frac{\log2}{2}$$

Title: Finding k in exponential continuous random variable distribution

Body:
Find the unknown parameter $k$ given this distribution:
$$ f(x)= \begin{cases} e^{kx} & 0 < x < 2\\
            0 & \text{otherwise}
        \end{cases}$$

This is my attempt:

$$\int_0^2 e^{kx} dx = \left[\frac{1}{k}e^{kx}\right]_0^2 \\
\frac{1}{k}e^{2k} -\frac{1}{k}=1 $$

This is where I am just stuck. How do I solve the above for $k$? I feel like I am missing something simple but I just can't see it.

Edit: I can see that I can rearrange this to equal $e^{2k} = k+1$ and then take logs giving me $k=\frac{1}{2}ln(k+1)$ but that still doesn't let me solve for $k$?


问:
Two questions on clock arithmetic
I have two questions on clock arithmetic, both of which I have solved, but I am looking for neater proofs.

Let us suppose we have a circle named $\mathbb{Z}_n$ with $n$ equally spaced points on it labelled $0,1,\cdots n-1$ in the clockwise direction. By clock arithmetic I mean that $x+1$ is the point just after $x$ in the clockwise direction.

$1$. Let $\{a,b\}\subset \mathbb{Z}_n$ so that $\{a+t,b+t\}\ne \{a,b\}\forall t=1,2\cdots n-1$ . Then $\{a+i,b+i\}\ne\{a+j,b+j\}\forall 0\le i<j\le n-1$.

My proof: The hypothesis says that $a$ and $b$ are not on a diameter. Hence no $\{a+i,b+i\}$ is also on a diameter.

$2$. Let $\{a,b\}\subset \mathbb{Z}_n$ with $a>b$ in the ordinary integer ordering sense. Define the distance $d(a,b)$ between $a$ and $b$ as the minimum of the lengths of the two paths from $a$ to $b$ and from $b$ to $a$. By length of a path from $x$ to $y$ we mean the
  smallest positive $z$ with $x+z=y$. What is a formula for $d(a,b)$? So in $\mathbb{Z}_{19}, d(1,0)=1=d(0,18)$.

My solution: $d(a,b)=a-b$ if $1\le a-b\le \lfloor n/2\rfloor$ and $d(a,b)=n-a+b$ otherwise.

Can someone give an alternate "better looking" formula, possibly one which avoids using $<$ implicitly or explicitly?

答:
I would present the distance function as

$$d(a,b)=\min(\vert a-b \vert, n - \vert a-b \vert).$$

The casework may be less concrete, but it makes up for it with clarity.

Title: Two questions on clock arithmetic


问:
linear functional and sequence
Let $E$ be a normed vector space and consider a sequence $\left(x_{n}\right) \in E$ such that for each $\phi \in E^{\prime}$ the

sequence $\left(\phi\left(x_{n}\right)\right)$ is convergent in $\mathbb{R}$. Then $\left(x_{n}\right)$
is bounded in $E$

Any help will be appreciated

答:
Define a functionals $$\xi_n : X' \to \mathbb{R} ,$$

$$\xi_n (f) = f(x_n ). $$

We have $$||\xi_n ||=\sup\{ |\xi_n (f) |: ||f||=1\} =\sup\{ |f(x_n )| : ||f||=1\}=||x_n||.$$
Hence we obtain a sequence of functionals $(\xi_n )$ from Banach space $X'$ to a Banach space $\mathbb{R}$ that converges for every $f\in X'.$ Therefore by Banach-Steinhaus theorem the sequence $(||\xi_n ||)=(||x_n ||)$$ is bounded.

Title: linear functional and sequence


答: First note $k\ne0$, since $\int_0^2dx=2\ne1$. Let's put @tommik's idea aside for the moment. As @HennoBrandsma noted, the antiderivative is $k^{-1}e^{kx}$ and the equation to solve is $e^{2k}-1=k$, subject to $k\ne0$. This can be solved in terms of the Lambert W function: since $-2(k+1)e^{-2(k+1)}=-2e^{-2}$, $k=-\frac12W_n(-2e^{-2})-1$, with $W_n$ a branch of the Lambert W. Just don't choose the branch mapping $-2e^{-2}$ to $-2$, as that would give $k=0$. You don't say whether you want complex roots, but I guess you want $f$ to be a PDF. The only real root is $k\approx-0.796812$.