问:
How to simplify the fraction?
How to simplify the fraction  $ \displaystyle \frac{\sqrt{3}+1-\sqrt{6}}{2\sqrt{2}-\sqrt{6}+\sqrt{3}+1} $ to $ (\sqrt{2}-1)(2-\sqrt{3})  $?

I've checked it in the calculator and both give the same value.
I was able to simplify it until $\displaystyle \frac{1+\sqrt2-\sqrt3}{3+\sqrt2+\sqrt3} $ which also gives the same value.

For anyone who is interested, $ (\sqrt{2}-1)(2-\sqrt{3}) =tan\frac{\pi}{8}tan\frac{\pi}{12} $ and I need this as I was trying to use coordinate geometry to solve the following
question.

Title: How to simplify the fraction?

Body:
\begin{align}
\frac {1 + \sqrt 2 - \sqrt 3}{3 + \sqrt 2 + \sqrt 3} &= \frac {1 + \sqrt 2 - \sqrt 3}{3 + \sqrt 2 + \sqrt 3} \times \frac {3 - \sqrt 2 - \sqrt 3}{3 - \sqrt 2 - \sqrt 3}\\
&=\frac {3 - \sqrt 2 -\sqrt 3+3\sqrt 2-2-\sqrt 6-3\sqrt 3+\sqrt 6+3}{9 - (2 + 2\sqrt 6 + 3)}\\
&=\frac {4+2\sqrt 2-4\sqrt 3}{4 -2\sqrt 6}\\
&=\frac {2+\sqrt 2-2\sqrt 3}{2 -\sqrt 6} \times \frac {2 +\sqrt 6}{2 +\sqrt 6}\\
&=\frac {4 + 2\sqrt 2 - 4\sqrt 3 + 2\sqrt 6+2\sqrt 3 - 6\sqrt 2}{4-6}\\
&=\frac {4 -4\sqrt 2 - 2\sqrt 3 + 2\sqrt 6}{-2}\\
&=-2+2\sqrt 2 +\sqrt 3-\sqrt 6\\&=-2(1-\sqrt 2)+\sqrt 3(1-\sqrt 2)\\
&=(2-\sqrt 3)(\sqrt 2-1)
\end{align}

Just bash away.


问:
Find Laplace transform of $f(t)=\sin^2 t$ using $\mathcal{L}(\frac{d}{dt}f(t))=\mathcal{L}(f'(t))$. (Check my answer)
I want to find Laplace transform of $f(t)=\sin^2 t$ using $\mathcal{L}(\dfrac{d}{dt}f(t))=\mathcal{L}(f'(t))$.

This is my answer.

\begin{alignat*}{2}
&& \mathcal{L} (f'(t)) &= \mathcal{L} (\sin(2t))\\
\iff& & sF(s)-f(0)&= \frac{2}{s^2+4}\\
\iff& & sF(s)-0&= \frac{2}{s^2+4}\\
\iff& & F(s)&= \frac{2}{s(s^2+4)}.
\end{alignat*}

and we have
$$F(s)=\mathcal{L} \left(\sin^2 t\right)=\dfrac{2}{s(s^2+4)}.$$

Is this right or wrong answer?

答: Yes, this is correct. (For what it's worth, WolframAlpha also happens to agree.) Good job!</p>

ContentLicense: CC BY-SA 4.0


问:
How can I find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$
I need to find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ so that I may find the relationship between $I_4$ and $I_3$, I have done the following:

Setting up for Integration by parts

$$u=(1+x^3)^n \implies u^{'}=3nx^2(1+x^3)^{n-1}$$ $$v^{'}=x^6 \implies v=\frac{1}{7}x^7$$
Applying Integration by parts

$$I_n=[\frac{1}{7}x^7(1+x^3)^n]^1_0 - \frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$
$$=\frac{2^n}{7}-\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$

However I have no idea how to bring the integral of $\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$ into the form of $\int^1_0{x^6(1+x^3)^{n-1}dx}$ so that I may substitute it as $I_{n-1}$.

How can I do it, so that I may find a relationship between $I_4$ and $I_3$ ?

Body:
Notice that $$
I_n  - I_{n - 1}  = \int_0^1 {x^9 (1 + x^3 )^{n - 1} dx} .
$$


问:
Sum of reciprocals of odd squares and integral of $\log(x)/(1-x^2))$
I would like to prove that $\int_{0}^{1}\frac{\log(x))}{1-x^2}\,dx=-\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}. $

I know that the RHS is equal to $-\frac{\pi^2}{8} $.

Additionally, I have tried to make the substitution $x=\sin(u)$ to the integral, which leaves with $\int_{u=0}^{\pi/2} \frac{\log(\sin(x))}{\cos(x)}\,dx $ but I can't seem to proceed from here.

答:
Anothe solution
$$I=\int\frac{\log(x)}{1-x^2}\,dx=\sum_{n=0}^{\infty}\int x^{2n}\log(x)\,dx=\sum_{n=0}^{\infty}\frac{x^{2 n+1} ((2 n+1) \log (x)-1)}{(2 n+1)^2}$$

Using the bounds, the result.


问:
A question about commutator in a group
Suppose that the group $G$ has a series of normal subgroups $G\geq D\geq Z(G)$. Let $D/Z=\prod_{\lambda \in I}C_{\lambda}/Z$. How can I get $[G,D]=\langle [G,C_{\lambda}]\; |\; \lambda \in I\rangle$?

答:
By quotienting out $\langle [G,C_\lambda] \mid \lambda \in I \rangle$, it is enough to show that if all elements of $G$ commute with all $C_\lambda$, then $[G,D]=1$.

But $D$ is generated by the $C_\lambda$ and $Z(G)$, so this follows.

Title: A question about commutator in a group


问:
Solving $x^{\log_{10}(x^2)} = 100$
$x^{\log(x^2)}=100$

Edit: I am writing $\log$ for $\log_{10}$.

I tried to solve this and can get the solutions $10$ and $10^{-1}$. But how can I get the solutions $-10$ and $-10^{-1}$? And other solutions (including complex), if exist? I'm not sure how to use the complex logarithm in this problem.

My attempt:

Note that $x \neq 0$, because we have in the equation $\log(x^2)$.

Assume $x > 0$. Then:

$x^{\log(x^2)} > 0$

$\log (x^{\log(x^2)}) = 2$

$\log(x^2)\log(x)=2$

$2\log(x)\log(x)=2$

$\log^2(x) = 1$

$\log(x) = \pm 1 \implies x=10$ or $x = 10^{-1}$.

But how about $x<0$?

答:
\begin{align}
x^{\log x^2}&=100\\
2(\log x)^2&=2\\
(\log x)^2&=1\\
\log x&=\pm1\\
x&=10^{\pm1}
\end{align}

As this is a quadratic equation in $\log x$, given the monotonic nature of the logarithmic function, we can conclude that $10^{\pm1}$ are the only solutions.

Title: Solving $x^{\log_{10}(x^2)} = 100$


问:
Apollonius circle, its radius and center
I've got the following set:
$\{|z-a|=k|z-b|\}$, where z is a complex number, a an b are fixed, and $k>0$,$k \ne 1$.

I need to prove that this is a circle (called Apollonius circle).
I also have to prove that this circle's radius is equal to $k|a-b||1-k^2|^{-1}$ and it's centre is $(a-k^2b)(1-k^2)^{-1}$.

I don't know what to do. I've tried to work with the analytic circle equation ($|z-c|^2=r^2$), substituting given radius and center, but it didn't work.
I also tried to square both sides of the first given equation ($|z-a|=k|z-b|$), which usually works with complex number, but also didn't get any result... Can somebody show me how to solve this problem? I will be very grateful.

答:
Write $w = (z-a)/(z-b) $. Then $\left|w\right| = k$, and so, the locus of $w$ is $C(0, k)$, where $C(c, r)$ is the circle of radius $r$ centered at $c$. Then $z$ as a function of $w$ can be written as a chain of compositions:

$$ w \quad \xrightarrow{\zeta \mapsto \zeta -1} \quad w-1 \quad \xrightarrow{\zeta\mapsto\frac{1}{\zeta}} \quad \frac{1}{w-1} \quad \xrightarrow{\zeta \mapsto b + (b-a)\zeta} \quad b + \frac{b - a}{w - 1} = z. $$

We now track how $C(0, k)$ in the $w$-plane is transformed under these maps.

Under the first map, $C(0, k)$ is transformed to $C(-1, k)$.

If we believe that the inversion $\zeta \mapsto \frac{1}{\zeta}$ preserves circles, we can envision that the image of $C(-1, k)$ under the inversion is a circle with two antipodal points $\frac{1}{-1-k}$ and $\frac{1}{-1+k}$. So its center would be $\frac{1}{k^2-1}$.

Indeed, parametrize $\zeta \in C(-1, k)$ by $\zeta = k \omega - 1$ for $\left|\omega\right| = 1$. Then
\begin{align*}
\left| \frac{1}{\zeta} - \frac{1}{k^2 - 1} \right|
= \left| \frac{k}{k^2-1} \right| \left| \frac{k - \omega}{k\omega - 1} \right|
= \left| \frac{k}{k^2-1} \right| \left| \frac{k - \omega}{k - \overline{\omega}} \right|
= \left| \frac{k}{k^2-1} \right|,
\end{align*}
and so, $C(-1, k)$ is transformed to $C(\frac{1}{k^2-1}, \frac{k}{\lvert k^2-1 \rvert})$

Under the third map, $C(\frac{1}{k^2-1}, \frac{k}{\left|k^2-1\right|})$ is transformed into $C(\frac{k^2b-a}{k^2-1}, \frac{k\left|b-a\right|}{\lvert k^2-1 \rvert})$.

Therefore the set $\{z : \left| z - a \right| = k \left| z - b \right| \}$ is the circle of radius $\frac{k\left|b-a\right|}{\lvert k^2-1 \rvert}$ centered at $\frac{k^2b-a}{k^2-1}$.


问:
Height of a cube edge from the floor

A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?

I'm sorry for my bad English, but I try to illustrate it as follows.

The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?

答:
You can also use similar triangles. Let $P$ be the point on $MN$ directly below $A$, and let $Q$ be the point of intersection of $EN$ and line $AC$. Since $\angle CAM$ is a right angle, we have that

$$\triangle CAP \sim \triangle AMP \sim \triangle EAQ$$

$CA \,$ is clearly $\, 2a \sqrt{2}\, $ so $\, PC = a\sqrt{7}\,.$ Then

$$\begin{align}
\frac{AC}{PC} &= \frac{EA}{EQ} \\
\\
EQ &= \frac{EA \cdot PC}{AC} \\
\\
&= \frac{(2a)(a\sqrt{7})}{2a\sqrt{2}} \\
\\
&= a \sqrt {\frac{7}{2}} \\
\\
&= \frac{a}{2} \sqrt{14}
\end{align}$$

and so the height of $E$ is

$$\begin{align}
EQ + QN &= \frac{a}{2} \sqrt{14} + a \\
\\
&= \boxed{ a \left( \frac{\sqrt{14}}{2} + 1 \right) }
\end{align}$$

Title: Height of a cube edge from the floor


答:
If $x_0$ is a solution to $$x^{\log_{10}x^2}=100$$ $-x_0$ is also a solution iff $\log_{10}x^2$ is an even integer.

Is  $\log_{10}x^2$ an even integer for the obtained solutions, $10^{\pm 1}$?

Title: Solving $x^{\log_{10}(x^2)} = 100$

Body:
$x^{\log(x^2)}=100$

Edit: I am writing $\log$ for $\log_{10}$.

I tried to solve this and can get the solutions $10$ and $10^{-1}$. But how can I get the solutions $-10$ and $-10^{-1}$? And other solutions (including complex), if exist? I'm not sure how to use the complex logarithm in this problem.

My attempt:

Note that $x \neq 0$, because we have in the equation $\log(x^2)$.

Assume $x > 0$. Then:

$x^{\log(x^2)} > 0$

$\log (x^{\log(x^2)}) = 2$

$\log(x^2)\log(x)=2$

$2\log(x)\log(x)=2$

$\log^2(x) = 1$

$\log(x) = \pm 1 \implies x=10$ or $x = 10^{-1}$.

But how about $x<0$?


Title: Solving $x^{\log_{10}(x^2)} = 100$

Body:
For real $x,$

$$x^{\log_{10}(x^2)}=x^{2\log_{10}|x|}=(|x|^2)^{\log_{10}|x|}$$

Let $\log_{10}|x|=y\implies|x|=10^y$

So, we have $$10^2=10^{2y^2}\implies y=\pm1, |x|=10^{\pm1}, x=\pm(10^{\pm1})$$


问:
Determinant of a matrix with binomial coefficients.
Let $n \in\mathbb{N}$ and $A=(a_{ij})$ where
\begin{equation}a_{ij}=\binom{i+j}{i}\end{equation}
for $0\leq i,j \leq n$. Show that $A$ has an inverse and that every element of $A^{-1}$ is an integer.

I have shown that this $n\times n$ matrix is symmetric since,
\begin{equation}
\binom{i+j}{i}=\binom{i+j}{j}
\end{equation}
in order to try to get a nonzero determinant.
But i'm stuck in this step, suggestions would be appreciated.

答:
If you take a particular case (say $n=5$) and you consider the LDL or Cholesky decomposition of this matrix you notice something very interesting: ( WA link).

So one should try to prove that our matrix $A$ is the product
$$A = L \cdot L^t$$
where
$L = (\binom{i}{j})_{0\le i,j \le n}$.This translates into the equalities:
$$\sum_k\binom{i}{k}  \binom{j}{k} = \sum_k\binom{i}{k}  \binom{j}{j-k}= \binom{i+j}{j}$$

Note that $L$ is lower triangular with $1$ on the diagonal, so invertible. In fact, its inverse can be explicitely given
$$L^{-1} = ((-1)^{i-k} \binom{i}{j})$$
One can prove a more general equality
$$L^{a}\cdot L^{b}=L^{a+b}$$
where $L_{ij}^a =a^{i-j} \binom{i}{j}$.

So now we get
$$A^{-1} = (L\cdot L^t)^{-1}= (L^{-1})^t \cdot L^{-1}$$
so clearly with integral elements, although I can't notice a particularly nice form for the entries.

Notice: matrices of the form $\binom{a+i+j}{j}$, or $\binom{a+i}{j}$, or $\binom{a-i}{j}$, or more generally, $\binom{a \pm i + b j}{j}$ also have nice LU decompositions. ($a$, $b$  parameters). The L part (the lower triangular part) is always $\binom{i}{j}$.

ContentLicense: CC BY-SA 3.0


问:
Are $A,B,C$ coprime integers if $B=C$?
Let $A,B,C$ be three coprime positive integers, i.e., there exist six integers $λ_{1},λ_{2},λ_{3},λ_{4},λ_{5},λ_{6}$ such that

$$λ_{1}A+λ_{2}B=1$$

$$λ_{3}A+λ_{4}C=1$$

$$λ_{5}B+λ_{6}C=1$$

I am asking what happens if $B=C$: are $A,B,C$ still three coprime positive integers or not?

答:
It depends.

There are two conventions you can take. Let $S := \{a_1,\cdots,a_n\} \subseteq \Bbb Z$ be a set of integers. If you want to say the elements of $S$ are coprime, this can be interpreted two different ways:

As a whole, they are coprime. This would mean that $\gcd(a_1,\cdots,a_n)=1$. That is, there are $\lambda_i \in \Bbb Z$ such that

$$\sum_{i=1}^n \lambda_i a_i = 1$$

Alternately, they are pairwise coprime. That is, $\gcd(a_i,a_j) = 1$ for all $i\ne j$. Or, put differently, there are integers $\lambda_{i,j},\mu_{i,j} \in \Bbb Z$ such that
$$\lambda_{i,j} a_i + \mu_{i,j} a_j = 1$$
for every $i \ne j$.

Suppose, then $S = \{A,B,C\}$. If $B=C$, and we don't "reduce" the set to the equivalent $S=\{A,B\}$, then $S$ is not coprime in the second sense (as $\gcd(B,B) = B$) unless $B=1$. However, it is in the first sense because

$$\gcd(A,B,B) = \gcd \big( \gcd(A,B)\; , \; B  \;\big) = \gcd(1,B) = 1$$

Title: Are $A,B,C$ coprime integers if $B=C$?


问:
Splitting field of an infinite family of separable polynomials is infinite dimensional if there is no finite subfamily of which it is splitting.
Suppose $F$ is a splitting field of an infinite set $T$ of separable polynomials, over $K$. If there is no finite set $S \subseteq T$ such that $F$ is splitting field of $S$ over $K$, is it true that $F$ is infinite dimensional over $K$?

I know that because of separability there must be infinitely many roots for $T$. But how does this lead to infinite dimensional?

This is part of a proof of equivalent definition of algebraic Galois extension. In Hungerford p262, to show (iii) F is splitting field over K of a set T of separable polynomials implies (i) F is algebraic and Galois over K. The proof says it suffices to prove the case where [F:K] is finite, and goes on to say if [F:K] is finite then there are some finite polynomials S in T such that F is splitting field of S over K.

答:
Suppose $F/K$ is the splitting field of a family $T$ of polynomials and not of any finite subset $S\subset T$.

This implies any finite subset $S\subset T$ can be inflated to a bigger finite subset with a bigger splitting field. In particular, this can be done by adjoining a single polynomial to $S$. For, if this wasn't possible, then every other element of $T$ would split in $S$'s splitting field, in which case $T$'s splitting field wouldn't be any bigger than $S$'s, a contradiction.

Thus, there exists a sequence $f_1,f_2,\cdots$ of elements of $T$ for which the splitting fields of $\{f_1,\cdots,f_n\}$ are strictly increasing with $n$, which implies the dimension over $K$ diverges.


问:
Borel $\sigma $-algebra of subsets of $[0,1]$
I have been reading Steven Shreve's Stochastic Calculus for finance II. This question is from Chapter $1$ page $4$ General Probability theory. Could someone explain how we got the expression for (a,b) ? Here Formula $1.1.7$ is Uniform (Lebesgue)measure on $[0,1]$ i.e.

$$\mathbb{P} [a,b] = b - a ,\,\, 0 \leq a \leq b \leq 1  $$

Also, what is the process of obtaining Borel $\sigma$ - algebra of subsets of $[0,1]$ ? The text says we start with closed intervals and add everything else required to have a $\sigma$ - algebra? Could anyone give more precise explanation? I know the definition of $\sigma$ - algebra but can't understand how we obtain one beginning from closed intervals ?

答:
If $x \in (a, b)$ then for some $\epsilon > 0$, $a + \epsilon < x < b - \epsilon$. Take $n$ s.t. $\epsilon > \frac{1}{n}$ - then $x \in\left[a + \frac{1}{n}, b - \frac{1}{n}\right]$, so $x \in \bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$ and thus $(a, b) \subseteq \bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$.

In the other direction, if $x \in \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$, then $a < a + \frac{1}{n} \leq x \leq b - \frac{1}{n} < b$ and so $x \in (a, b)$. Thus $\bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right] \subseteq (a, b)$.

Combining together, we get $(a, b) = \bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$.

About obtaining Borel $\sigma$-algebra from closed intervals - formally, it's the smallest (included in any other with such property) $\sigma$-algebra that contains all closed intervals.
Of course, we need to prove that such smallest $\sigma$-algebra exists, but there is standard technique for this.

First, note that $2^{[0, 1]}$ - set of all subsets of $[0, 1]$ is $\sigma$-algebras that contains all closed intervals. Now, let $\mathcal T$ be set of all $\sigma$-algebras on $[0, 1]$ containing all closed interval (it's non-empty,  as $2^{0,1]} \in \mathcal T$).
Intersection of all elements of $\mathcal T$, $\mathcal{B}[0, 1] = \bigcap\limits_{t \in \mathcal T} t$, is again $\sigma$-algebra on $\mathbb{[0, 1]}$ containing all closed intervals (can you prove it?), and it's the smallest such $\sigma$-algebra, because if $t \in \mathcal T$ then $\mathcal{B}[0, 1] \subseteq t$.

It can also be done more constructive with transfinite recursion, but I think non-constructive definition is simpler.


问:
Matrix involving values of polynomials
I've been doing this problem but im stuck.

Be $f_1 f_2 f_3 \in \mathbb{R}_2$[$x  $]. Proove that {$f_1$,$ f_2$,$ f_3$} form a base of $\mathbb{R}_2$[$x  $] as $\mathbb{R}$ vector space, if and only if in $\mathbb{R}^{3x3}$ this matrix is invertible:

$ \begin{bmatrix} f_1(1) & f_1(2) & f_1(3)  \\ f_2(1) & f_2(2) & f_2(3) \\ f_3(1) & f_3(2) & f_3(3)\end{bmatrix}$

my attempt:  $\leftarrow$) i want to proove that $ \lambda_1 f_1 +\lambda_2  f_2 +\lambda_3 f_3 = 0 $ then $ \lambda_i   = 0 $   $  $    $ \forall i$ so they're linearly independant and then form a base. first i transpose the matrix, because i know that if a square matrix is invertible its transpose is too. Now i know that if i multiply the matrix by a vector ($ \lambda_1 \lambda_2 \lambda_3 $ ) and ask for the homogeneous solution i get that this ocurr if and only if $\lambda_i = 0 $ for $i= 1,2,3$ Then they're linearly independant and form a base.$ \\ $

$\\ \rightarrow$)...

答:
Matrix is invertible, if and only if

Determinant of matrix is nonzero, if and only if

There is no non-trivial linear combination of column vectors, if and only if

There is no non-trivial linear combination of the quadratic polynomials.

Now fill in the above details.

Title: Matrix involving values of polynomials


问:
If the order of $\langle a\rangle$ is finite, then the distinct elements of $G$ are $e,a,a^2,a^3,...,a^{n-1}$
Suppose $G=\langle a\rangle$. If the order of $\langle a\rangle$ is finite, then the distinct elements of $G$ are $e,a,a^2,a^3,...,a^{n-1}$.

I am trying to prove this statement. What I understand so far is:

Suppose $m\in \mathbb{Z}$. By the Division Algorithm, we have:

$$m=nq+r$$

where $q,r\in \mathbb{Z}$ and $0\leq r<n$. (Ok, I got this.)

Thus, $m\equiv r\mod n$. It follows that $a^m=a^r$. (I got this too.)

Since $G$ is a cyclic group, every element in $G$ is equal to some $a^i$ for $0\leq i<n$.

Suppose $a^i=a^j$. For $0\leq i<n$ and $0\leq j<n$. Without loss of generality, we suppose $i<j$. We have:

$$i\equiv j\pmod n$$ (Everything makes sense so far.)

However, given the range of $i$ and $j$, this is $\underline{impossible}$.

My questions are:

Why is $i\equiv j\mod n$ impossible given the range of $i$ and $j$?

The textbook says "In particular, $|a|=|\langle a\rangle|$." I don't really understand this. What does $|a|$ have to do with $|\langle a\rangle|$? I am aware that the order of $a$ is the smallest positive integer $n$ such that $a^n=e$.

Thank you!

答:
If $i \equiv j \pmod{n}$ then the distance from $i$ to $j$ is an integer multiple of $n$, so, if $i$ and $j$ are in $\{0,\dots,n-1\}$, this is only possible if $i=j$.

Once you know that $|\langle a \rangle| = n$ implies $\langle a \rangle = \{e,a,\dots,a^{n-1}\}$, then the smallest positive integer $m$ such that $a^m = e$ is $n$ because $e,a,\dots,a^{n-1}$ are all distinct.


问:
Question Regarding Proof Using Quadratic Formula
I'm learning proof techniques. I'm following this PDF at: http://persweb.wabash.edu/facstaff/turnerw/Writing/proofs.pdf

On the Theorem $4$ example (pg. 7), I am confused!

I see that we assume $x \gt 0$ by using the negative (or minus) version of the quadratic formula.

But, there are three things I don't understand.

Why is $\frac{2a - 2b} { 2a}$ simplified into $1 - (\frac{b}{a})$? When I simplify this equation, I get $\frac{(a-b)}{a}$.

Why is the value of b important, or is it not? $b = a(1 − x)$

And where it's stated at the bottom that a divides b, is that referring to the equation $x = 1 - (\frac{b}{a})$?

Thank you for the help.

Update:
Thank you to Eevee for the answer. I still have one more question which is -- why do we choose the negative (or minus) quadratic formula for when x > 0? I had sort of inferred that we did because otherwise x would be negative. But I'm not sure.

Update:
Thank you Sten for the answer. On that note, was it assumed that I would have worked the quadratic formula out both ways? I don't see anything that infers that I would choose the negative (or minus) version of the formula.

答:
I see that we assume x > 0 by using the negative (or minus) version of the quadratic formula.

No, we assume $x$ is a positive integer from the very beginning. The goal is a proof by contrapositive:

What we wish to prove: *"If $a$ does not divide $b$ then $ax^2 + bx + b - a = 0$ has no positive integer solution $x$."

The equivalent contrapositive: If $x$ is a positive integer solution to $ax^2 + bx + b - a = 0$, then $a$ divides $b$.

Notice that, since we're proving the latter, we want to start with the hypothesis of $x$ being a positive integer solution to the quadratic.

Why is 2a - 2b / 2a simplified into 1 - (b/a)? When I simplify this equation, I get (a-b)/a.

Split up the fraction:

$$\frac{a-b}{a} = \frac a a - \frac b a = 1 - \frac b a$$

Why is the value of b important, or is it not? b = a(1 − x)

The definition of "$p$ divides $q$" is that "there exists an integer $k$ such that $q = pk$." If the goal is to show that $a$ divides $b$, then $1-x$ is that integer (since the proof also assumes $a,b,x$ are integers).

And where it's stated at the bottom that a divides b, is that referring to the equation x = 1 - (b/a)?

It refers, more precisely, to the fact that $b = a(1-x)$, i.e. $b$ is an integer multiple of $a$.


答:
The problem stems from 2 things

The fact that $\log$ is not defined when the the argument is negative

We cannot work with negative powers raised to exponents which are variables which vary continuously ie $(-1)^x$ has a graph which oscillates indefinitely and hence we cannot work with it.

The moment you take $2$ out from the exponent of $x^2$, you lose the negative solutions. The same thing happens when you take $\log$ on both sides. I don't think there is any way to solve this for those values. I think the best bet is to just check and verify for the inverse counterparts once you get the positive solutions. And this problem cannot be resolved by taking complex numbers into account, this is a problem due to the ill-defined $(-1)^x$. If there are complex solutions, you can solve for them by replacing $x$ as $x+iy$ but you will surely miss solutions like these. Hope this helps.

Title: Solving $x^{\log_{10}(x^2)} = 100$


问:
Prove that $\ f*g = f$
Good morning,

We have : $g \in$ $L^{p}(\mathbb{R}^n)$, $\forall p \in [0;\infty)$ and $f\in$ $L^{1}(\mathbb{R}^n)$,
$\widehat{f}(x) =0$  if $|x| \geq R $,
$\varphi = \widehat{g}$ such as  $\varphi(x) = 1, \left | x \right | \leqslant R $ and $  x \in  \mathbb{R}^n$, $\varphi\in {C_{c}^{\infty }(\mathbb {R} ^{n})}$

So I did a Fourier Transform of $\ f*g$ (Convolution product) and got $\widehat{f*g}(x)=\widehat{f}$
How can I say that $\ f*g = f$ with what I found ?

Thank you in advance !

答:
The Fourier transform is injective (in the relevant function spaces).
If $\hat h=\hat f$, then $h=f$.
Now you have $h=f*g$.

Title: Prove that $\ f*g = f$


问:
Minimum degree requirements for a graph to be Hamilton-connected, like Dirac's classic theorem for Hamilton cycles.
A classic theorem of Dirac says that every graph with at least 3 vertices and minimum degree at least $\frac{n}{2}$ has a Hamilton cycle.

Is there a similar characterization for Hamilton-connected graphs? Ie, something of the form
"Every graph with minimum degree at least $X$ is Hamilton-connected?"

答:
Actually, there is: every graph with minimum degree at least $\frac{n+1}{2}$ is Hamilton-connected!

To see this, let $G$ be such a graph and note that $G$ is Hamilton-connected if and only if for any pair of vertices $u,v$ of $G$, the graph $G' = G + x + ux + vx$ is Hamiltonian, where $x$ is a new vertex. The key ingredient now is the Bondy-Chvátal theorem, which says that if $u',v'$ are non-adjacent vertices in a graph such that the sum of their degrees is at least the number of vertices, then adding the $u'v'$ edge to the graph doesn't change whether it has a Hamiltonian cycle. In our case, $G'$ is a graph on $n+1$ vertices and each vertex in $G'-x$ has degree at least $\frac{n+1}{2}$. This means that we can add an edge between every pair of vertices in this subgraph, so that $G'$ becomes a complete graph plus a vertex of degree two. But such a graph is clearly Hamiltonian, which is what we wanted to show.

In fact, using the same idea the Bondy-Chvátal theorem itself has the following analogue for Hamilton-connected graphs: if $G$ is a graph on $n$ vertices and $u,v$ are non-adjacent vertices with $\deg(u) + \deg(v) \geq n+1$, then $G$ is Hamilton-connected if and only if $G+uv$ is Hamilton-connected.


问:
Showing the inclusion is a birational equivalence.
The question states that if $X$ is irreducible and quasiprojective, and $U$ is a nonempty open subset of $X$, then the inclusion map is a birational equivalence.

The definition I am given for birational is that the map $f: X \to Y$ is  a dominant rational map with a dominant rational inverse (i.e. a $g: Y \to X$ such that $f \circ g = Id_Y$ and $g \circ f = id_X$).

My thoughts so far:

If $X$ is irreducible then we have that $U$ is dense in $X$, and so clearly the image of the inclusion map is dense and we have a dominant rational map $\iota: U \to X$.

I can see that this is the easy part of the problem. Where I am struggling is understanding how we obtain an inverse. From what I can tell the inverse would need to have domain of definition being $U \subset X$ (which I think should come from the left inverse guaranteed by injectivity of $\iota$), which would then give us $\iota \circ \iota^{-1} = Id_U$. But the problem is that I don't see why that should also correspond to the composition being $\iota^{-1} \circ \iota = Id_X$.

I understand that rational maps are defined via equivalence classes wherein two rational maps are the same if they agree on a dense open set, which would mean that $Id_X$ agrees with $\iota^{-1}$ on the dense open $U$, but I don't understand how I would get the composition to be equal to the identity of $X$ the codomain of the inverse is $U$.

We have already proven that a dense open subset $U \subset X$ has a function field isomorphic to $X$, but we do not yet have that the equivalence of definitions $k(X) \cong k(Y)$ iff $X$ is birational to $Y$.

Any help here would be appreciated.

答:
Recall the definition of rational map $f:X \to Y$: it is an equivalence class $(f_U,U)$ of morphisms $f:U \to Y$ where $U \subset X$ is open. We consider $(f_V,V) \sim (f_U,U)$ if they agree agree on $U \cap V$, i.e. that $f_U|V=f_V|U$.

So now consider the inclusion $i:U \to X$ as a rational mapping. Also consider $id_X: X \to X$ as a rational map. These two agree on the open set $U$. So they are equal as rational maps. But the identity is obviously a birational equivalence.

See also my old answer to a similar question here: the definition of "Birational Equivalence"


问:
Is it possible to draw trapezoid by compass and straightedge if 4 sides given?
I tried to construct trapezoid with lengths of 4 sides given. Without success. Then searched in internet. I believe it is not possible, but I am not sure.
The only way is to calculate the high of the trapezoid (distance of two paralell sides).
Is my assuption correct or there is a way how to construct?</p>

答: Translate the fourth side on the first. With this new center and radius the trapezoid third side draw again circle. (In figure circle with center G has radius BF)


问:
Successive Percentage discount
Price of a product is 48 units. After two successive cuts by same percentage, price of the product becomes 20.25. What is the discount percentage?

I assumed the percentage to be x and multiplied x by x/100 and x/100. Didn't get the right answer. The answer x is 10. I want to know how to get to the answer.

Title: Successive Percentage discount

Body:
$$48(1-x)(1-x)=20.25$$

Thus

$$x=1-\sqrt{\frac{20.25}{48}}\approx 0.35048=35.048 \%$$


答:
$\Rightarrow$ By contrapositive: suppose the matrix is not invertible — in this case, its transpose is not either, and there exists $\begin{pmatrix} a \\b \\c\end{pmatrix}\neq \begin{pmatrix} 0 \\0 \\0\end{pmatrix}$ such that
$$
\begin{pmatrix} f_1(1) & f_2(1) & f_3(1)  \\ f_1(2) & f_2(2) & f_3(2) \\ f_1(3) & f_2(3) & f_3(3)\end{pmatrix}\begin{pmatrix} a \\b \\c\end{pmatrix}=\begin{pmatrix} 0 \\0 \\0\end{pmatrix}
$$
i.e.
$$
\begin{align*}
a f_1(1) + b f_2(1) + c f_3(1) &= 0  \\
a f_1(2) + b f_2(2) + c f_3(2) &= 0  \\
a f_1(3) + b f_2(3) + c f_3(3) &= 0
\end{align*}
$$
En posant $P\stackrel{\rm{}def}{=}af_1+bf_2+cf_3\in\mathbb{R}_2[X]$, we have
$$
P(1)=P(2)=P(3)=0
$$
and therefore $P=0$. But since by assumption $(a,b,c)\neq(0,0,0)$, it means that $(f_1,f_2,f_3 )$ is linearly dependent, thus not a basis of $\mathbb{R}_2[X]$.</p>

Title: Matrix involving values of polynomials

Body:
I've been doing this problem but im stuck.

Be $f_1 f_2 f_3 \in \mathbb{R}_2$[$x  $]. Proove that {$f_1$,$ f_2$,$ f_3$} form a base of $\mathbb{R}_2$[$x  $] as $\mathbb{R}$ vector space, if and only if in $\mathbb{R}^{3x3}$ this matrix is invertible:

$ \begin{bmatrix} f_1(1) & f_1(2) & f_1(3)  \\ f_2(1) & f_2(2) & f_2(3) \\ f_3(1) & f_3(2) & f_3(3)\end{bmatrix}$

my attempt:  $\leftarrow$) i want to proove that $ \lambda_1 f_1 +\lambda_2  f_2 +\lambda_3 f_3 = 0 $ then $ \lambda_i   = 0 $   $  $    $ \forall i$ so they're linearly independant and then form a base. first i transpose the matrix, because i know that if a square matrix is invertible its transpose is too. Now i know that if i multiply the matrix by a vector ($ \lambda_1 \lambda_2 \lambda_3 $ ) and ask for the homogeneous solution i get that this ocurr if and only if $\lambda_i = 0 $ for $i= 1,2,3$ Then they're linearly independant and form a base.$ \\ $

$\\ \rightarrow$)...


问:
$f(x,y)=x\sqrt{|y|}$. What is $\frac{\partial f}{\partial y}(x,y)$?
$f(x, y)=x \sqrt{|y|}.$

I have to calculate $\dfrac{\partial f}{\partial y}(x,y)$.

$\dfrac{\partial f}{\partial y}(x,0)
=\lim_{k \to 0} \dfrac{f(x,k)-f(x,0)}{k}
=\lim_{k\to 0} \dfrac{x\sqrt{|k|}}{k}$.

Therefore,
$\dfrac{\partial f}{\partial y}(x,0)$ doesn't exist.

However,
does $\dfrac{\partial f}{\partial y}(x,y)\ (y\neq 0)$ exist?

If $y>0$,

\begin{align}
\dfrac{\partial f}{\partial y}(x,y)
&=\lim_{k\to 0} \dfrac{f(x, y+k)-f(x,y)}{k}
\\&=\lim_{k \to 0} \dfrac{x\sqrt{|y+k|}-x\sqrt{y}}{k}.
\end{align}

I cannot proceed from this.

I would like you to give me some ideas.

答:
Hint: if $y >0$, then $y+k >0$ for "small" $k$, hence $|y+k|= y+k.$
Can you proceed ?


答: When $y>0$ we have $\lim_\limits{k\to0^+}\frac{x\sqrt{|y+k|}-x\sqrt y}k=\frac x{2\sqrt y}$ using L'Hopital. The limit as $k\to0^-$ is the same as the Density Property ensures that $\exists K>0$ such that $y>K$.

Title: $f(x,y)=x\sqrt{|y|}$. What is $\frac{\partial f}{\partial y}(x,y)$?

Body:
$f(x, y)=x \sqrt{|y|}.$

I have to calculate $\dfrac{\partial f}{\partial y}(x,y)$.

$\dfrac{\partial f}{\partial y}(x,0)
=\lim_{k \to 0} \dfrac{f(x,k)-f(x,0)}{k}
=\lim_{k\to 0} \dfrac{x\sqrt{|k|}}{k}$.

Therefore,
$\dfrac{\partial f}{\partial y}(x,0)$ doesn't exist.

However,
does $\dfrac{\partial f}{\partial y}(x,y)\ (y\neq 0)$ exist?

If $y>0$,

\begin{align}
\dfrac{\partial f}{\partial y}(x,y)
&=\lim_{k\to 0} \dfrac{f(x, y+k)-f(x,y)}{k}
\\&=\lim_{k \to 0} \dfrac{x\sqrt{|y+k|}-x\sqrt{y}}{k}.
\end{align}

I cannot proceed from this.

I would like you to give me some ideas.


Title: Solving $x^{\log_{10}(x^2)} = 100$

Body:
$$x^{\log(x^2)} = 100$$
$$\implies \log(x^2) = \log_x{10^2}$$
$$\implies \log(x^2) = 2\log_x{10}$$
$$\implies \log(x^2) = 4\log_{x^2}10$$
$$\implies \log(x^2) = \dfrac{4}{\log(x^2)}$$
$$\implies \log^2(x^2) = 4$$
$$\implies \log(x^2) = \pm 2$$
$$\implies x^2 = 10^{\pm 2}$$

Now you got the $4$ roots: $\pm 10; \pm \dfrac{1}{10}$


问:
Find a property of groups that is inherited by quotients but not by subgroups.
This is Exercise 4.6a of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.

The Details:

From p.116 of Roman's book,

Definition: Let $\mathcal{P}$ be a property of groups. [ . . . ] We write $G\in \mathcal{P}$ if $G$ has property $\mathcal{P}$. [ . . . ] A property $\mathcal{P}$ of groups is inherited by subgroups if $$G\in \mathcal{P}\text{ and }H\le G\implies H\in \mathcal{P},$$ and $\mathcal{P}$ is inherited by quotients if $$G\in \mathcal{P}\text{ and }H\unlhd G\implies G/H\in \mathcal{P}.$$

The Question:

Find a property of groups that is inherited by quotients but not by subgroups.

Thoughts:

My first guess - I don't know why - was to let $\mathcal{P}$ be "nonabelian", which fell quickly in the light that $G/G=\{G\}$ is abelian for all groups $G$.

Having $\mathcal{P}$ be "abelian" won't  work, since it is inherited by both quotients and subgroups; <s>out goes "cyclic" along with it.</s>

"Finite" is not an example, obviously.

"Solvable" is inherited by both quotients and subgroups. This can be checked on Wikipedia. (I haven't done the Maths, ironically; but with such a standard property of a group, I reckon it's more reliable here than not).

<s>All nilpotent groups are solvable, so they're out; the same goes for supersolvable groups and, in turn, metacyclic groups. Metabelian groups are also solvable.</s>

"Simple" appears not to work.

It would be fiendish if Roman had not defined a group property as required prior to the exercise, so, if all else fails, I could sieve through the definitions in the first 144 pages; this is cheating though (and quite a daunting task!).

Please help :)

答:
The property of being a $T$- group is an example.  That's that every subnormal subgroup (normal subgroup of a normal subgroup) is normal.

The property is inherited by normal subgroups, but not by subgroups in general.

And it's inherited by homomorphic images, hence quotients.


问:
A problem on Stirling numbers of second kind.

In this section, there were four cases, the 4 combinations of distinguishable and identical balls and cells. I could understand [Distinguishable balls and cells] and [identical balls and distinguishable cells] but I just can not understand this one.

First of all, i don't understand stirling numbers of second kind, i had read a little about one of the types of stirling number while studying circular permutations but I was not very well versed with it (read: i am still confused). Secondly, i don't understand how we 'easily' get the two results they have shown.

I would really appreciate if someone could explain this to me and also a little bit of Sterling numbers.

Thanks!

答:
The set $\{1,2,3\}$

can be partitioned into $\mathbf{three}$ subsets in $\mathbf{one}$ way: $\{\{1\},\{2\},\{3\}\}$ so $S(3,\mathbf{3})=\mathbf{1}$;

into $\mathbf{two}$ subsets in $\mathbf{three}$ ways: $\{\{1,2\},\{3\}\}$, $\{\{1,3\},\{2\}\}$, and $\{\{1\},\{2,3\}\}$, so  $S(3,\mathbf{2})=\mathbf{3}$;

and into $\mathbf{one}$ subset in $\mathbf{one}$ way: $\{\{1,2,3\}\}$, $S(3,\mathbf{1})=\mathbf{1}$.

There are $\mathbf{zero}$ ways to partitioned $\{1,2,3\}$ into $\mathbf{four}$ non-empty subsets $S(3,\mathbf{4})=\mathbf{0}.$


问:
Prove $A$ and $[0,1] \cup [2,4]$ are homemorphic if and only if
I am new to homeomorphic sets topic. I know that I have to find a bicontinuous function to prove that $A$ is homeomorphic to$ [0,1] \cup [2,3]$ if and only if it is of form $ [a,b] \cup [c,d] $. However, I don’t know how to find such a function.
$A \subset\mathbb R$

答:
If $A=[a,b]\cup[c,d]$ for $a<b<c<d$, then $[a,b]\cup[c,d]\to [0,1]\cup[2,4]$ given by
$$
x\mapsto
\begin{cases}
\frac{x-a}{b-a} & \text{if $x\in [0,1]$,} \\
2+2\cdot\frac{x-c}{d-c} &\text{if $x\in[2,4]$}
\end{cases}
$$
is a homeomorphism.

Conversely, let $f\colon A\to X:=[0,1]\cup [2,4]$ be a homeomorphism. Since the union is disjoint, it follows that $A = f^{-1}([0,1]) \cup f^{-1}([2,4])$ is also a disjoint union. Since $f^{-1}$ is continuous, it maps closed intervals to closed intervals so that $f^{-1}([0,1])=[a,b]$ and $f^{-1}([2,4])=[c,d]$ with $a<b$ and $c<d$. In order for the union to be disjoint we must have $a<b<c<d$ or $c<d<a<b$. In any case
$$
A = [a,b] \cup [c,d]
$$
is a union of two disjoint closed intervals.


问:
How to tell if exists a linear map that satisfies these conditions?
I'm trying to do this exercise but I'm stuck on a).

For b) I have done it in two ways, first one obtaining the matrix of T supposing T is a linear map and I obtain T(0,1) and T(1,0) using the other cases and then when trying to obtain T(-3,2) using the matrix I end up in a contradiction.

The other way I did it was by seeing that $\alpha_1+\alpha_2=-\alpha_3$ so if it was a linear map $T(\alpha_1+\alpha_2)=T(-\alpha_3) \implies T(\alpha_1)+T(\alpha_2)=-T(\alpha_3)$ and this does not hold.

Any ideas on how to do exercise a)? Thanks in regards to everyone:)

答:
Hint:

The vectors $(1,-1,1)$ and $(1,1,1)$ are linearly independent, hence you can add a third vector to form a basis of $\:\mathbf R^3$.


问:
Topological extension property

Let $X$ and $Y$ be topological spaces. We say that the extension property holds if, whenver $S$ is a closed subset of $X$ and $f:S\rightarrow Y$ is continuous, $f$ can be extended to a continuous map on $X$.

A question asks whether the extension property holds for $X$ being any topological space and $Y=[-1,1]$.

I have no idea, but I am thinking whether there is some $X$ on which continuous functions are either constant or unbounded.

Can anyone offer any ideas? Thanks!

答:
If $X$ is normal, then $X$ has extension property.

The Tietze-Urysohn theorem: Every continuous function from closed subspace $M$ of a normal space $X$ to $\mathbb I$ or $\mathbb R$ is continuously extendable over $X$.

The proof of it can be seen General Topology by Engelking, Page 69.

Title: Topological extension property


答:
Since $(1,-1,1)$ is not a scalar multiple of $(1,1,1)$, they are linearly independent.

You can extend it to a basis (by adding e.g. $(1,0,0)$), then assigning any values for them uniquely extends to a linear map.

Title: How to tell if exists a linear map that satisfies these conditions?

Body:
I'm trying to do this exercise but I'm stuck on a).

For b) I have done it in two ways, first one obtaining the matrix of T supposing T is a linear map and I obtain T(0,1) and T(1,0) using the other cases and then when trying to obtain T(-3,2) using the matrix I end up in a contradiction.

The other way I did it was by seeing that $\alpha_1+\alpha_2=-\alpha_3$ so if it was a linear map $T(\alpha_1+\alpha_2)=T(-\alpha_3) \implies T(\alpha_1)+T(\alpha_2)=-T(\alpha_3)$ and this does not hold.

Any ideas on how to do exercise a)? Thanks in regards to everyone:)


答: Consider a vector $v\in\Bbb R^3$ such that the set $\{(1,-1,1),(1,1,1),v\}$ is linearly independent (take $v=(1,0,0)$, for instance, a consider the only linear map from $\Bbb R^3$ into $\Bbb R^2$ such that $T(1,-1,1)=(1,0)$, $T(1,1,1)=(0,1)$, and $T(v)=(0,0)$, for instance.


问:
A question on linear operators
This is a problem I’ve been working on as part of my studies for an upcoming comprehensive exam:

Let $F$ be a field, let $V\in F$-$\mathrm{Mod}$ be a finite-dimensional left $F$-vector space, and let $T\in\mathrm{End}_{F}\left(V\right)$ be an $F$-linear operator.

(a)  Suppose that $\mathrm{rk}(T)=\mathrm{rk}\left(T^2\right)$.  Prove that $\mathrm{ker}(T)\cap T[V]=\left\{0_V\right\}$, and deduce from this that $V=\mathrm{ker}(T)\oplus T[V]$.

(b) Prove that $V=\mathrm{ker}\left(T^k\right)\oplus T^{k}[V]$ for some $k\in\mathbb{N}$.

I’ve solved part (a); if $v\in\mathrm{ker}(T)\cap T[V]$, then $v=T(w)$ for some $w\in V$ and $T(v)=0_V$, so $w\in\mathrm{ker}\left(T^2\right)$.  By assumption, $\mathrm{rk}(T)=\mathrm{rk}\left(T^2\right)$.  By the Dimension Theorem, $\dim(V)=\mathrm{rk}(T)+\dim\left(\mathrm{ker}(T)\right)=\mathrm{rk}\left(T^2\right)+\dim\left(\mathrm{ker}\left(T^2\right)\right)$, so $\dim\left(\mathrm{ker}(T)\right)=\dim\left(\mathrm{ker}\left(T^2\right)\right)$.  Since, clearly, $\mathrm{ker}\left(T^2\right)\supseteq\mathrm{ker}(T)$ (as $T$ is $F$-linear), we have that $\mathrm{ker}(T)=\mathrm{ker}\left(T^2\right)$ (as $V$ is finite-dimensional).  Hence $v=0_V$.  The Dimension Theorem then allows us to conclude that $V=\mathrm{ker}(T)\oplus T[V]$.

A solution to part (b) has as yet eluded me.  Any hints would be appreciated.

答: Hint: Notice that the ranks of $T^k$ form a decreasing integer sequence bounded below by 0. Hence eventually they become constant, and thus the ranks of $T^n$ and $T^{?}$ for all $?>n$ must be the same. What choice of $?$ allows you to use a result you already know?

Title: A question on linear operators


问:
Rewriting Linear ODE for Control
I'm working on a model for an engineering control project and I've run into the issue of having the control multiplied by the state variable in a differential equation. The simplified system looks something like:

$\dot{y}_1=y_2\\
\dot{y}_2=\left(a+g\left(t\right)\right)y_1+\left(b+cg\left(t\right)\right)y_2$

Here $y_1$ and $y_2$ are state variables, $g\left(t\right)$ is the control, and $a,b,c$ are constants. Ideally it'd be nice to take a Laplace transform and work with the standard $\dot{x}=Ax+Bu$ type system, but as it stands I have something that has an $x*u$ term. It seems like linearization in the differential equation sense is unnecessary since as it stands the system is linear, it just has variable coefficients.

One thought I had was to take a Taylor Series approximation, but I wasn't sure if that could be applied directly in this context.

Edit: If I do a Taylor series expansion of $H\left(t\right)=g\left(t\right)y_1\left(t\right)$ around some point $d$ I get:

$H\left(t\right)\approx H\left(d\right)+H'\left(d\right)\left(t-d\right)=g\left(d\right)y_1\left(d\right)+\left(g'\left(d\right)h\left(d\right)+g\left(d\right)h'\left(d\right)\right)\left(t-d\right)$

Which doesn't seem to help given that I've "lost" in some sense the control function all together.

答:
You have the system

$$
\begin{align}
\dot{x}_1 &= x_2 \\
\dot{x}_2 &= (a + u) x_1 + (b c) u x_2
\end{align}
$$

At steady state, $x_2 = 0$ so if you are interested in steady state analysis it makes sense to control $x_1$. You could use the control

$$
u = \frac{k_1(x_{1,des} - x_1) - k_2 x_2 - a x_1}{x_1 + (b c) x_2}
$$

This makes your system behave like

$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
-k_1 & -k_2
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
k_1
\end{pmatrix} x_{1,des}
$$

as long as $x_1 + (b c) x_2 \neq 0$. This system is linear and has the transfer function:

$$
\frac{k_1}{s^2 + k_2 s + k_1}
$$

which is stable if $k_1 > 0$ and $k_2 > 0$.

We can test this. For example assume $a = 1, b = 2, c = 3, k_1 = 10, k_2 = 10$ and for the initial conditions $x_1(0) = 3$ and $x_2(0) = 0$. With the following trajectory we can see if tracking is possible:

$$
x_{1,des} = \begin{cases}
3 & t \in[0, 5) \\
4 & t \in[5, 15) \\
5 & t \in[15, 25) \\
4.5 & t \in[25, 35) \\
4 & t \geq 35
\end{cases}
$$

Simulate:

You can see that the controller works. It is possible to reach different steady state values of $x_1$. In the last plot you can see that the $x_1 + (bc) x_2$ remains far enough away from zero. This is something you have to make sure by selecting the $x_{1,des}$ trajectory and the parameters carefully.

It should probably work if you make only "small" steps with $x_{1,des}$ and wait until $x_2$ has decayed to zero before making the next step.

Title: Rewriting Linear ODE for Control


问:
$ ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$
Show that $ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$ where $\zeta$ is the primitive $5^{th}$ root of unity.

I have seen a similar question asked, from this I gather that the quadratic polynomial is $ ^2+−1$, however I don't really understand how to conclude that it is equal to $ℚ(\sqrt5)$.

答:
Let $L/K$ be a quadratic extension given as the splitting field of a degree 2 polynomial $f \in K[x]$. When $\operatorname{char} K \neq 2$ the extension is determined by the discriminant.

Why? Well the roots of $f = ax^2 + bx + c$ are given by the quadratic formula (not that this is where we use the assumption on the characteristic of $K$), which only requires us to adjoin $\sqrt{\Delta(f)} = \sqrt{b^2 - 4ac}$ - i.e., the splitting field of $f$ is obtained by adjoining $\sqrt{\Delta(f)}$. In your case the discriminant is $5$.

Title: $ ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$


问:
Proof that the plots of the fitted values vs. residuals yields parabola when needed quadratic term omitted?
There is a step I do not understand in my professor's class. In his slide, he is trying to formally show us that when the true, unknown model contains a squared predictor, and that this predictor is omitted when estimating the model parameters, then the plots of a) the fitted values vs. the residuals and b) the predictor vs. the residuals will yield a parabolic outcome. It is the last step [from (1) to (2)] of his reasoning that I simply don't understand.

He proceeds as follows:

"Assume for instance that the true model is quadradic in $X_1$:
$$y_i=\beta_0+\beta_1x_{i1}+\beta_{11}x_{i1}^2+\epsilon_i$$
but we fit only a linear model
$$\hat y_i=\hat \beta_0+ \hat \beta_1x_{i1}$$
Then,
\begin{align}
e_i & =y_i-\hat y_i \\
& = (\beta_0-\hat \beta_0)+(\beta_1-\hat \beta_1)x_{i1}+\beta_{11}x_{i1}^2+\epsilon_i \qquad (1)\\
& = \gamma_0+\gamma_1(\hat \beta_0 +\hat \beta_1x_{i1})+\gamma_2(\hat \beta_0 +\hat \beta_1x_{i1})^2+\epsilon_i \qquad (2)\\
& = \gamma_0+\gamma_1\hat y_i+\gamma_2\hat y_i^2+\epsilon_i
\end{align}

for certain values of $\gamma_0$, $\gamma_1$ and $\gamma_2$. This shows that both a $(x_{i1},e_i)$ and a $(\hat y_i,e_i)$ plot will show a quadratic curve."

As mentioned above, does anyone see the manipulation he performed to deduce (2) from (1)? How did he rid himself of the "true", unknown $\beta$'s?

I tried to consult a textbook (Sheather's "Modern Approach") to find this out, but his way of thinking about this is rather simplistic. They simply assume that the least squares estimates are close to the unknown population parameters $\beta_0$ and $\beta_1$, and that therefore we find that
$e_1=y_i-\hat y_i=(\beta_0-\hat \beta_0)+(\beta_1-\hat \beta_1)x_{i1}+\beta_{11}x_{i1}^2+\epsilon_i\approx\beta_{11}x_{i1}^2+\epsilon_i$. I'm quite unsatisfied with this simplistic assumption and it doesn't answer my question above.

Can anyone help me out? If it is simple algebra, sincere apologies!

Many thanks in advance for your help

答:
What you needs to use to get between (1) and (2) is

$\gamma_0 = \dfrac{\hat\beta_{0}^2\beta_{11}-\hat\beta_{0}\hat\beta_{1}\beta_{1}+\hat\beta_{1}^2\beta_{0}}{\hat\beta_{1}^2}$

$\gamma_1 = \dfrac{-2\hat\beta_{0}\beta_{11}+\hat\beta_{1}\beta_{1}-\hat\beta_{1}^2}{\hat\beta_{1}^2}$

$\gamma_2 = \dfrac{\beta_{11}}{\hat\beta_{1}^2}$

which works so long as $\hat\beta_{1} \not =0$

But this is not quite the real issue.  Here is an illustration of trying to fit a straight line to something which is actually a parabola plus some noise, using R:

```<code>set.seed(2020)
truebeta0 <- 4
truebeta1 <- 7
truebeta2 <- 2
sdnoise   <- 3
X <- 1:10
Xsq <- X^2
Y <- truebeta0 + truebeta1 * X + truebeta2 * Xsq + rnorm(10,0,sdnoise)
fit1 <- lm(Y ~ X + Xsq)
fit2 <- lm(Y ~ X)
plot(Y ~ X)
points(fit1<span class="math-container">$fitted.values ~ X, type="l", col = "blue")
points(fit2$</span>fitted.values ~ X, type="l", col = "red")
</code>```

to give

and you can see the blue fitted parabola is a good fit with small apparently random residuals, while the red fitted straight line is not so good, with the residuals larger and those in the middle the opposite sign to those at the extremes.  If you look at the residuals for the red line, they are indeed visually a parabola plus some noise:

```<code>plot(fit2<span class="math-container">$residuals ~ fit2$</span>fitted.values, type="p", col = "red")
abline(h=0)
</code>```

Body:
What you needs to use to get between (1) and (2) is

$\gamma_0 = \dfrac{\hat\beta_{0}^2\beta_{11}-\hat\beta_{0}\hat\beta_{1}\beta_{1}+\hat\beta_{1}^2\beta_{0}}{\hat\beta_{1}^2}$

$\gamma_1 = \dfrac{-2\hat\beta_{0}\beta_{11}+\hat\beta_{1}\beta_{1}-\hat\beta_{1}^2}{\hat\beta_{1}^2}$

$\gamma_2 = \dfrac{\beta_{11}}{\hat\beta_{1}^2}$

which works so long as $\hat\beta_{1} \not =0$

But this is not quite the real issue.  Here is an illustration of trying to fit a straight line to something which is actually a parabola plus some noise, using R:

set.seed(2020)
truebeta0 <- 4
truebeta1 <- 7
truebeta2 <- 2
sdnoise   <- 3
X <- 1:10
Xsq <- X^2
Y <- truebeta0 + truebeta1 * X + truebeta2 * Xsq + rnorm(10,0,sdnoise)
fit1 <- lm(Y ~ X + Xsq)
fit2 <- lm(Y ~ X)
plot(Y ~ X)
points(fit1$fitted.values ~ X, type="l", col = "blue")
points(fit2$fitted.values ~ X, type="l", col = "red")

to give

and you can see the blue fitted parabola is a good fit with small apparently random residuals, while the red fitted straight line is not so good, with the residuals larger and those in the middle the opposite sign to those at the extremes.  If you look at the residuals for the red line, they are indeed visually a parabola plus some noise:

plot(fit2$residuals ~ fit2$fitted.values, type="p", col = "red")
abline(h=0)


问:
How to prove that $\frac {\pi^2}{ \sin(\pi z)^2 } = \sum_{n=-\infty , n\neq 0 }^{n=\infty} 1/ (z-n)^2 $
The following question was part of my complex analysis assignment and I am not able to prove it.

How to prove that $$\frac {\pi^2 } { \sin(\pi z)^2 }  = \sum_{n=-\infty , n \neq0 }^{n=\infty} \frac{1}{(z-n)^2 }$$

I tried by using the identity $$\sin(\pi z) = \pi z \prod_{n=1}^{\infty} \left(1- \left(\frac zn \right)^2 \right) $$ But that is in product so can't be used here.

I don't have any other ideas except these. Actually the prof is known in department to be not good in teaching and online classes added more to it. In particularly in these type of questions where I have to prove some identity I am having very much difficulty.

It is my very humble request to you to help me.

答:
A standard argument is as follows:

Let $f$ denote your infinite sum, and let $g$ denote your function built out of the sine.

Show that $f$ and $g$ are meromorphic functions with double poles only at the integers except zero. They also have no roots.

This means that the difference $h=f-g$ is a holomorphic function, but it is also a function that satisfies $h(z)=h(z+1)$. Can you conclude?


问:
Long polynomial division with two variables $\frac{x^4-4a^3x+3a^4}{x-a}$
I need to show that $x-a$ is a factor of $x^4-4a^3x+3a^4$. I've tried long polynomial division but stopped in the middle of it as it didn't seem to work. How else could one approach it?

答:
Hint:
If polynomial division is asked for, following is how it can be factorised. Otherwise proving that $x-a$ is a factor is a trivial application of the factor theorem. $$x^4-4a^3x+3a^4=(x^4-ax^3)+(ax^3-a^2x^2)+(a^2x^2-a^3x)+(-3a^3x+3a^4)$$


问:
Does Cayley's formula count isomorphic trees?
Cayley's formula tells us that on $n$ vertices, there exist $n^{n-2}$ trees.

Now I am wondering if this formula considers each set of isomorphic trees as 1 or if it counts every one of them.

答:
First, note that there is no known closed form expression for the number of unlabeled trees up to isomorphism.

Cayley's Formula on the other hand gives us the number of labeled trees up to isomorphism. Note that in labeled case, for example the following graphs are not considered isomorphic although if they were unlabeled, we would consider them isomorphic. But if you look at their labeling carefully, they are all different.


问:
Check if the function $f=x[\frac 1 x], x\ne 0$ is continuous.
Edit: I got carried away and forgot that the function is not defined at $x=0$, which means its not continuous at $x=0$, thanks for all the answers!

My attempt: check if the function is continuous at $x=0$.
I tried to apply the squeeze theorem:

$\frac 1 x -1<[\frac 1 x] \le \frac 1 x$ , Multiply all by $x\ne 0$, and i got:

$1-x<x[\frac 1 x] \le 1$, and this implies that the limit $x\rightarrow 0$ is 1.

and if the limit at $x=0$ exists and equal to 1, then the side limits when $x\rightarrow 0^+$ or $x\rightarrow 0^-$ both exist and equal, which implies that the function is continuos at $x=0$.

I feel like I have made some mistake, since the floor function can be tricky with limits, It would be appreciated if someone can approve if my work is right or wrong and tell me how to approach these kind of questions.
Thanks in advance!

答: I will assume that you are defining $f(0)$ to be $1$. There is a mistake in your argument. When $x<0$ multiplying an inequality by $x$ reverses the inquality. So $\frac  1x -1 \leq [\frac 1 x] \leq \frac  1 x$ becomes $x(\frac  1x -1) \geq [\frac 1 x] \geq x\frac  1 x$ for $x<0$ but you still get the left-hand limit as $1$.


问:
If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$
If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$

I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$

From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$

Hence $1+9abc\ge 4(ab+bc+ac)$.

This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?

Body:
My second solution:

By using Schur's inequality and the identity
$$a^3 + b^3 + c^3 + \frac{15}{4} abc =
\frac{3}{4}[a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)] + \frac{1}{4}(a+b+c)^3,$$
we have
$$a^3 + b^3 + c^3 + \frac{15}{4} abc \ge \frac{1}{4}.$$

If $d \ge \frac{15}{4}$, we have
$$a^3 + b^3 + c^3 + abcd \ge a^3 + b^3 + c^3 + \frac{15}{4} abc \ge \frac{1}{4}.$$

If $d < \frac{15}{4}$, we have
\begin{align}
a^3 + b^3 + c^3 + abcd &= a^3 + b^3 + c^3 + \tfrac{15}{4} abc - (\tfrac{15}{4} - d)abc \\
&\ge \frac{1}{4} - (\tfrac{15}{4} - d)(\tfrac{a+b+c}{3})^3\\
&= \frac{1}{4} - (\tfrac{15}{4} - d)(\tfrac{1}{3})^3\\
&= \frac{1}{9} + \frac{d}{27}.
\end{align}

We are done.


答:
That's it — technically your solution is good; indeed supposing that $x>0$, for every $k \le x \le k +1$ the floor function returns $\frac{1}{k+1}$; which besides, the x itself is continuous at 0, and floor is discrete. As a little support, Wolfram approved the answer $$\lim_{x \rightarrow 0} x [(\frac{1}{x})] = 1$$.
where $[x]$ means the floor function of x. And it is true that the function still undefined at 0, but the limit shall be 1. Why? That is due to inequalities above. Either way, there should be some set on which you define the function, then the floor shall return the discrete value ; as long as the first part of the function, viz. x, is not discontinuous at 0.

Title: Check if the function $f=x[\frac 1 x], x\ne 0$ is continuous.

Body:
Edit: I got carried away and forgot that the function is not defined at $x=0$, which means its not continuous at $x=0$, thanks for all the answers!

My attempt: check if the function is continuous at $x=0$.
I tried to apply the squeeze theorem:

$\frac 1 x -1<[\frac 1 x] \le \frac 1 x$ , Multiply all by $x\ne 0$, and i got:

$1-x<x[\frac 1 x] \le 1$, and this implies that the limit $x\rightarrow 0$ is 1.

and if the limit at $x=0$ exists and equal to 1, then the side limits when $x\rightarrow 0^+$ or $x\rightarrow 0^-$ both exist and equal, which implies that the function is continuos at $x=0$.

I feel like I have made some mistake, since the floor function can be tricky with limits, It would be appreciated if someone can approve if my work is right or wrong and tell me how to approach these kind of questions.
Thanks in advance!


答:
Dividing the expression by $a^4$ you obtain

$(x/a)^4-4x/a+3=(x/a)^4-4x/a+4-1=(x/a-2)^2-1$

Now, using the fact that $t^2-1=(t+1)(t-1)$ yelds to

$(x/a-2)^2-1=(x/a-1)(x/a-3)=1/a^2(x-a)(x-3a)$

obtaining what you are asking for.

Title: Long polynomial division with two variables $\frac{x^4-4a^3x+3a^4}{x-a}$

Body: I need to show that $x-a$ is a factor of $x^4-4a^3x+3a^4$. I've tried long polynomial division but stopped in the middle of it as it didn't seem to work. How else could one approach it?


问:
Proof of inequality $\prod_{k=1}^n(1+a_k) \geq 1 + \sum_{k=1}^n a_k$ with induction
I have to show that $\prod_{k=1}^n(1+a_k) \geq 1 + \sum_{k=1}^n a_k$ is valid for all $1 \leq k \leq n$ using the fact that $a_k \geq 0$.

Showing that it works for $n=0$ was easy enough. Then I tried $n+1$ and get to:
$$\begin{align*}
\prod_{k=1}^{n+1}(1+a_k) &= \prod_{k=1}^{n}(1+a_k)(1+a_{n+1}) \\
&\geq (1+\sum_{k=1}^n a_k)(1+a_{n+1}) \\
&= 1+\sum_{k=1}^{n+1} a_k + a_{n+1}\sum_{k=1}^n a_k
\end{align*}$$

In order to finish it, I need to get rid of the $+ a_{n+1}\sum_{k=1}^n a_k$ term. How do I accomplish that? It seems that this superfluous sum is always positive, making this not really trivial, i. e. saying that it is even less if one omits that term and therefore still (or even more so) satisfies the $\geq$ …

答:
We want to show :

$$\left(\frac{1}{a_{n+1}}+1\right)\prod_{i=1}^{n}\left(1+a_{i}\right)>1+\frac{1}{a_{n+1}}+\sum_{i=1}^{n}\frac{a_{i}}{a_{n+1}}$$

We introduce the function :

$$f(a_{n+1})=\left(\frac{1}{a_{n+1}}+1\right)\prod_{i=1}^{n}\left(1+a_{i}\right)-1+\frac{1}{a_{n+1}}+\sum_{i=1}^{n}\frac{a_{i}}{a_{n+1}}$$

If we differentiate and multiply by $a_{n+1}^2$ we fall on the hypothesis induction . Moreover we deduce that the function is decreasing so we have :

$$f(a_{n+1})\geq \lim_{a_{n+1}\to \infty}f(a_{n+1})=\operatorname{induction hypothesis}\geq 0$$

Remins to multiply by $a_{n+1}$ and conclude .

The advantage is : we get a $n$ refinements and we know much more on the behavior of the difference .


问:
How to use linear approximation to estimate $2\ln(1.1)+3$?
Can anyone please share how to solve this?

The question is to use the linear approximation to estimate $2\ln(1.1)+3$

The hint given to me was to find f(x) first and find a and f'(a) but I can't follow the hint and don't know how to find those.
Thanks.
I have not tried anything because I am lost at how to find f(x)first or a.
We have only been doing answers where the original f(x) is given then I find the linear approximation and estimate certain values afterwards. This is the first time the f(x) is not provided to me.

答:
Hints:

The tangent to a curve, which is the best linear approximation around the contact point, is the line through the contact point, with a slope equal to the derivative at the contact point.

Use a contact point near $1.1$ that is easy to compute.


问:
Gradient approaching to zero
Either prove or disprove  by counterexample-
Let $f:R^n\to R$ such that $f$ is differentiable on $R^n-\text{{0}}$ , $f$ is continuous at $0$ and $\lim_{x\to 0} \nabla f = 0$ , then show that $f$ is differentiable at $0$. Actually for single dimensional case I can prove it's true by mean value theorem but I don't feel like it's true for $n>1$ but still can't find a counterexample yet. Can somebody help me?

答:
Let  $h \in \mathbb R^n, h \ne 0$. By the mean value theorem for functions with $n$ variables, there is $z(h) \in \{th: t \in [0,1]\}$ such that

$$f(h)-f(0) = h \cdot \nabla f(z(h)).$$

By Cauchy - Schwarz

$$|f(h)-f(0)| \le  ||h|| \cdot ||\nabla f(z(h))||$$

Hence

$$\frac{|f(h)-f(0)|}{||h||} \le ||\nabla f(z(h))||.$$

This shows that

$$\frac{|f(h)-f(0)|}{||h||} \to 0$$

as $h \to 0$. This means that $f$ is differentiable at $0$ and $f'(0)=0.$

Title: Gradient approaching to zero


答: Fix $ \epsilon > 0 $; by hypothesis there exists a $ \delta > 0 $ such that $ \| \nabla f (w) \| \le \epsilon $ whenever $\|w\|\le \delta$. So by the Mean Value Theorem $$ \|f(w) - f(0) \| \le \|\nabla f (t w) \| \|w\|\le \epsilon \|w\|, \quad t \in (0,1) $$which implies the differentiability of $f$ at $0$.

Title: Gradient approaching to zero

Body:
Either prove or disprove  by counterexample-
Let $f:R^n\to R$ such that $f$ is differentiable on $R^n-\text{{0}}$ , $f$ is continuous at $0$ and $\lim_{x\to 0} \nabla f = 0$ , then show that $f$ is differentiable at $0$. Actually for single dimensional case I can prove it's true by mean value theorem but I don't feel like it's true for $n>1$ but still can't find a counterexample yet. Can somebody help me?


问:
Question with integration and derivatives.
If $f(x)=g(x)$,

then $\int \{\frac{d}{dx}f(x) \} dx = \int \{\frac{d}{dx}g(x) \} dx $

True or false?

Book answer : false.

I know that

If $f(x)=g(x)$,

then $\frac{d}{dx} \{\int f(x) dx \} = \frac{d}{dx} \{\int g(x) dx \} $  (It's true)

But

If $f(x)=g(x)$,

$\int \{\frac{d}{dx}f(x) \} dx = \int \{\frac{d}{dx}g(x) \} dx $ is really false?

Why?   I would like to hear your opinion.

答: Notice that when you integrate a certain function there will always be an arbitrary constant in the end(to plus). For example, $f(x)= g(x)=x^2$, $\dfrac{d}{dx} f(x)=\dfrac{d}{dx} g(x)=2x$, but $\int \dfrac{d}{dx} f(x)dx=x^2+c_1$ and $\int \dfrac{d}{dx} g(x)dx=x^2+c_2$. Since $c_1$ and $c_2$ are not necessarily equal, $\int \dfrac{d}{dx} f(x)dx$ is not necessarily equal to $\int \dfrac{d}{dx} g(x)dx$.


问:
Is my asymptotic formula for $\sum_{k=1}^{m}\log(k!)$ right?
Today I tried to derive an asymptotic formula for the sum $$\sum_{k=1}^{m}\log(k!)$$
Here is what I did: This can be rewritten as $$\sum_{k=1}^{m}\sum_{1<n\le k}\log n$$
Now I applied the Euler summation formula, and this turned to
$$\sum_{k=1}^{m}\left(\int_{1}^{k}\log t\,dt+\int_{1}^{k}\frac{t-[t]}{t}dt+\log(k)([k]-k)\right)$$
$\int_{1}^{k}\frac{t-[t]}{t}dt$ is dominated by $\int_{1}^{k}\frac{1}{t}dt=\log k$, so $\int_{1}^{k}\frac{t-[t]}{t}dt=O(\log k)$, and it is also true that $\log(k)([k]-k)=O(\log k)$
Now, the sum becomes
$$\sum_{k=1}^{m}(1-k+k\log k+O(\log k)+O(\log k))=\sum_{k=1}^{m}(1-k+k\log k+O(\log k))$$
which is equal to
$$m-\frac{m(m+1)}{2}+\log H(m)+O(\log m)$$
where $H(m)$ is hyperfactorial. Is my solution right? I could have made a mistake.

答:
As @Gary commented
$$\sum_{k=1}^{m}\log(k!)=\log \big[G(m+2)\big]$$ Using a CAS,
$$\log \big[G(m+2)\big]=\frac1{12}(6 m^2+12 m+5)\log(m)+m \log \left(\frac{\sqrt{2 \pi }}{e}\right)-\frac 34m^2+$$ $$\log \left(\frac{ \sqrt{2 \pi e^{\frac 16} }}{A}\right)+\frac 1{12m}-\frac{1}{240 m^2}+O\left(\frac{1}{m^3}\right)$$ which shows a relative error lower than $0.01$% as soon as $m>1$


答:
First, we have the following formula for the linear approximation of f(x) at a point $x=a$:

$f(x) ≈f(a) + f'(a)(x - a)$

The nearest value to $1.1$ that is easy to compute is $x=1$ as we know $\ln(1)=0$

If $f(x) = \ln (x)$ and $a = 1$, we get:

$\ln(x) = \ln (1) + x - 1=x-1$

If we set $x=1.1$ we can approximate $\ln(1.1)$ by $1.1-1=0.1$, we can multiply it by $2$ now and get $0.2$, then we add $3$ to it as we need $2\ln(1.1)+3$ and get $3.2$ as our approximation.

Actual value is $2\ln(1.1)+3=3.1906...$ so we are pretty close.

Title: How to use linear approximation to estimate $2\ln(1.1)+3$?

Body:
Can anyone please share how to solve this?

The question is to use the linear approximation to estimate $2\ln(1.1)+3$

The hint given to me was to find f(x) first and find a and f'(a) but I can't follow the hint and don't know how to find those.
Thanks.
I have not tried anything because I am lost at how to find f(x)first or a.
We have only been doing answers where the original f(x) is given then I find the linear approximation and estimate certain values afterwards. This is the first time the f(x) is not provided to me.


Body:
If you use long division, don't forget those missing terms.

Think about the place-holder $0$ as in $1\color{red}{0}3$.

\begin{array}{ccccccc}
    & & & x^3 & +ax^2 & +a^2 x & -3a^3 \\
    & & -- & --- & --- & --- & --- \\
   x-a & ) & x^4  & \color{red}{+0x^3} & \color{red}{+0x^2} & -4a^3x & +3a^4 \\
    & & x^4 & -ax^3    \\
    & & -- & --- & --- \\
    & & & ax^3 \\
    & & & ax^3 & -a^2x^2 \\
    & & & --- & --- & --- \\
    & & & & a^2x^2 & -4a^3x \\
    & & & & a^2x^2 & -a^2x & \\
    & & & & --- & --- & --- \\
    & & & & & -3a^3x & +3a^4 \\
    & & & & & -3a^3x  & +3a^4 \\
    & & & & & --- & ---
\end{array}


问:
Polya's urn model - limit distribution
Let an urn contain $w$ white and $b$ black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the $n$-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the $n$-th draw-and-replacement.

I start with $b=w=1$, so the total number of balls after the $n$-th draw-and-replacement is $n+2$.
Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s.
It is

$$X_n = \dfrac{w_n}{n+2} \quad\text{for}\quad n \in \mathbb{N}_0. $$

I've read that the limit distribution is a beta distribution, but I don't know how to get there.

I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is $0$, if the $i$-th ball is black and $1$, if the $i$-th ball is black. Then I'd have

$$ w_n = 1+\sum_{i=1}^{n} Y_i. $$

Does this help? How can I proceed?

Thanks! :)

答:
Suppose that we have started with $w$ white balls and $b$ black balls. Then

\begin{align*}
\mathbb{P}(w_n = w+k)
&= \binom{n}{k} \frac{w(w+1)\dots(w+k-1)b(b+1)\dots(b+n-k-1)}{(w+b)(w+b+1)\dots(w+b+n-1)} \\
&= \frac{1}{B(w, b)} \binom{n}{k} \frac{\Gamma(w+k)\Gamma(b+n-k)}{\Gamma(w+b+n)} \\
&= \frac{1}{B(w, b)} \frac{k^{w-1} (n-k)^{b-1}}{n^{w+b-1}} \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)} ,
\end{align*}

where $\Gamma(\cdot)$ is the gamma function, $B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ is the beta function, and

$$ E_n(z) := \frac{\Gamma(n+z)}{n!n^{z-1}}. $$

Note that $E_n(z) \to 1$ as $n\to\infty$. So, if we write $p_k = k/n$, then the m.g.f. of $X_n$ is explicitly given by

\begin{align*}
\mathbb{E}[e^{\lambda X_n}]
= \frac{1}{B(w, b)} \sum_{k=0}^{n} \exp\biggl( \lambda \frac{p_k + w/n}{1 + (w+b)/n} \biggr) p_k^{w-1}(1 - p_k)^{b-1} \frac{1}{n} \cdot \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)}.
\end{align*}

Letting $n \to \infty$, this converges to

\begin{align*}
\mathbb{E}[e^{\lambda X_{\infty}}]
= \frac{1}{B(w, b)} \int_{0}^{1} e^{\lambda p} p^{w-1}(1 - p)^{b-1} \, \mathrm{d}p.
\end{align*}

From this, we read out that the distribution of $X_{\infty}$ has the density

$$ f(p) = \frac{1}{B(w, b)} p^{w-1}(1 - p)^{b-1} \mathbf{1}_{(0,1)}(p), $$

proving that the limit distribution is $\operatorname{Beta}(w, b)$.

Title: Polya's urn model - limit distribution


问:
$X$ topological space. $A$ open $A \cap Y = \emptyset \ \ \Longrightarrow A \cap \bar{Y} = \emptyset$?
I know this is an easy question, but I cannot demonstrate it properly.

Suppose by contradiction that $A \cap \bar{Y} \neq \emptyset$. Then $\exists \ x \in A \cap \bar{Y}$.

I need help formalizing this reasoning  (or correcting if it is wrong)

By the definition of $A$ open, $\exists \ U \subset A$ open neighborhood of $x$. Then my idea is to prove that $U \cap Y \neq \emptyset$ because $U$ in somewhat sense "touches" $Y$, because it's a neighborhood of $x$. But i can't write down a formal demonstration.
Any help is appreciated.
Thanks^^

答:
Another type of proof of it:

If $A \cap \overline {Y} \neq \emptyset$, then there exists $x \in A \cap \overline {Y}$. Since $x \in \overline {Y}$, which implies that for any open set $U$ containing $x$, $U \cap Y \neq \emptyset$, and hence $A \cap Y \neq \emptyset$. A contradiction!


答:
The problem for that second one is that in the process of integration, constants arise. (Remember that troubling $+C$?)

Okay so, the quoted statement means
$$f(x)=g(x)+C$$
instead of the intended $f(x)=g(x)$.

In the second statement, we are differentiating at last which eliminates the constants arosen due to the integration inside.

Hope this helps. Ask doubts if any :)

Title: Question with integration and derivatives.

Body:
If $f(x)=g(x)$,

then $\int \{\frac{d}{dx}f(x) \} dx = \int \{\frac{d}{dx}g(x) \} dx $

True or false?

Book answer : false.

I know that

If $f(x)=g(x)$,

then $\frac{d}{dx} \{\int f(x) dx \} = \frac{d}{dx} \{\int g(x) dx \} $  (It's true)

But

If $f(x)=g(x)$,

$\int \{\frac{d}{dx}f(x) \} dx = \int \{\frac{d}{dx}g(x) \} dx $ is really false?

Why?   I would like to hear your opinion.


答: With the one-dimensional case plus $\nabla_xf\stackrel{x\to 0}\longrightarrow 0$ you can prove that the partial derivatives exist at $0$ and that they are all $0$. Thus, you have proved that the partial derivatives exist in a neighbourhood of $0$ and that they're continuous at $0$. This implies differentiability at zero (this is sometimes called the "total differential theorem", though some authors refer to it with the unnecessary hypothesis that the partial derivatives should be continuous in a neighbourhood of the point of interest, instead of the single point).

Title: Gradient approaching to zero


问:
5 points in 3D space: how many planes and planes intersections?
Given 5 points in space such that no three of them are colinear and no four of them are coplanar.
If we consider all the planes containing any 3 of these 5 points, and the intersections of all these planes taken two by two, how many lines will we obtain, at most?

__

At first, my idea was just to say that there are $C(3,5) = 10$ (binomial coefficient) different planes.
And then any combination of 2 planes must give a different intersection line: therefore $C(2,10)=45$ should be the answer?

However, I sense that this is naive and overly simple. Could I be missing something?

答:
OK, based on the past comments, here is how I see it:

There are 10 pairs of points, i.e. through each pair we can fit 3 planes (the planes going through that pair as well as each of the other 3 points separately). Indeed, then there is only 1 common intersection to these 3 planes.
Therefore, the planes formed this way would have just 10 intersections that coincide with an existing pair of points (out of the theoretical 30 'intersections' we could form with the naive approach I first used).

There remains another 15 planes intersections which do not coincide with any existing pairs of points (these are the intersections of 123-145, 123-245, 123-345, 124-135, 124-235, 124-345, 125-134, 125-234, 125-345, 134-235, 134-245, 135-234, 135-245, 145-234, 145-235). Therefore, IF (but are we sure of this?) none of these intersection are also the same, we would have 10+15=25 total intersection for these 10 planes? Is that correct?

This is intriguing: is this a famous result which generalizes to N points?


问:
Tricky Integral $\int_{0}^{\pi/2} \left[\frac{\sec^2 \theta}{2}\cdot \exp(\frac{\sec^2 \theta}{2})\right]^{-1} d\theta$ - Brain Slayer
Given the integral
$$
\int_{0}^{\pi/2}
\left[\frac{\sec^{2}\left(\theta\right)}{2}
\exp\left(\frac{\sec^{2}\left(\theta\right)}{2}\right)\right]^{\,-1} \mathrm{d}\theta
$$
How am supposed to evaluate it within few minutes $?$. Is there a quick trick solve the integral $?$.

Should you wish to see my attempt at doing so, then you should know that my solution is analyzed the author's one, because I could not solve it without seeing clues. So, I would like to get some ideas from; I think that I fail to solve this guy because I could not see something...

Body:
$$I=2\int_{0}^{\pi/2} \cos^2 x e^{-\frac{1}{2}sec^2 x} dx$$
Let $\tan x=t$, then
$$I= 2e^{-1/2}\int_{0}^{\infty} \frac{e^{-t^2/2}}{(1+t^2)^2} dt=
=2e^{-1/2}\int_{0}^{\infty} e^{-t^2/2} \int_{0}^{\infty} z e^{-z(1+t^2)} dz.$$
$$\implies I=2e^{-1/2} \int_{0}^{\infty} ze^{-z}\int_{0}^{\infty} e^{-(1/2+z)t^2}dt=\sqrt{2\pi}e^{-1/2}\int_{0}^{\infty} \frac{z e^{-z}}{\sqrt{1+2z}} dz$$
Let $1+2z=u^2$ then
$$I=\frac{\sqrt{2\pi}}{2} e^{-1/2}\int_{1}^{\infty} (u^2-1)~ e^{-(u^2-1)/2} du=\frac{\sqrt{2\pi}}{2} \int_{1}^{\infty} (u^2-1)e^{-u^2/2}du=\sqrt{\frac{\pi}{2e}}.$$


问:
Pearson correlation test
I'm trying to run a hypothesis test to see if two variables are correlated. Therefore, I would like to use the Pearson correlation test. I know how to set the significance level, the null hypothesis and the alternative hypothesis and I also know how to calculate the r-value. But I'm not sure how to calculate the critical values for r and thus the rejection region.

Du to wikipedia the r-value follows a student t distribution with n-2 degrees of freedom where n is the sample size: https://en.wikipedia.org/wiki/Pearson_correlation_coefficient

I also found a table of critical r-values for different sample sizes and significance levels:
http://statisticslectures.com/tables/rtable/

Calculating the critical value using the student t distribution and a significance level I get different values than shown in the table. Am I missing something? Or am I getting the whole idea of this test wrong?

答:
The critical value for $r$ as shown in the table, is given by the equation $$r_{\text{crit}} = \frac{t_{n-2,\alpha/2}^*}{\sqrt{n-2 + (t_{n-2,\alpha/2}^*)^2}},$$ where $t_{n-2,\alpha/2}^*$ is the upper $\alpha/2$ quantile of the student's $t$-distribution with $n-2$ degrees of freedom.  This equation is the second one in the Wikipedia article subsection under "Testing using Student's $t$-distribution."  In particular, $$\Pr[T_{n-2} > t_{n-2,\alpha/2}^*] = \alpha/2$$ where $T_{n-2}$ is a student's $t$ random variable with $n-2$ degrees of freedom.  For instance, $n = 10$ and $\alpha = 0.1$ gives $t_{8,0.05}^* \approx 1.85955$.  Then $$r_{\text{crit}} = \frac{1.85955}{\sqrt{10 + (1.85955)^2}} \approx 0.549357,$$ which is the entry for row $8$ column $2$ in the table.

For the case in your comment, where $n = 10$ and $\alpha = 0.05$, we have $$t_{8, 0.025}^* \approx 2.306004.$$  This gives $$r_{\text{crit}} = \frac{2.306004}{\sqrt{8 + (2.306004)^2}} \approx 0.63189686.$$  This is row $8$ column $3$ of the table.

This hypothesis test is 2-sided.  If $r$ is negative, then you need to compare against $-r_{\text{crit}}$ for the hypothesis $$H_0 : r = 0 \quad \text{vs.} \quad H_1 : r \ne 0.$$  That is to say, we reject $H_0$ in favor of $H_1$ if $|r| > r_{\text{crit}}$ where $r$ is the observed correlation from the data.

Personally, I don't find the table very useful.  Instead, I would directly calculate the test statistic via the first equation in the Wikipedia subsection
$$T_{n-2} \mid H_0 = r \sqrt{\frac{n-2}{1-r^2}},$$ which is student $t$-distributed with $df = n-2$, thus making the need for a separate table irrelevant, since you can now just use a regular $t$-table.

Title: Pearson correlation test


问:
Looking for a particular class of functions
I would like to characterize functions $f: [0, \infty) \to [0, \infty)$ that satisfy $$ f(z) \geq f(x) + f(y)  $$
for all $x,y < z$ with $x +y \geq z$. I am also interested in examples. Moreover, I was wondering whether there exist functions $g: [0, 1] \to [0, 1]$ that satisfy the condition from above.

答:
Claim: There doesn't exist a function $g:[0,1]\to [0,1]$ which satisfies the condition in the problem.

As my comment mentions, any such function $g$ must be discontinuous everywhere on $[0,1].$ We will prove that every such function $g$ is of bounded variation. And we know every function of bounded variation can be expressed as a difference of two non-decreasing functions, and by the Lebesgue' theorem, we know every function of bounded variation is differentiable almost everywhere, and hence continuous almost everywhere, which contradicts to my comment above.

Now let $g:[0,1]\to [0,1]$ be a function which satisfies the condition in the problem.

Let $P=\{x_0,x_1,...,x_n\}$ be a partition of $[0,1]$ with $x_0 =0$ and $x_n=1$.

So, $$V(g,P)=\sum_{i=1}^n |g(x_i)-g(x_{i-1})|\le\sum_{i=1}^n |g(x_i)|+|g(x_{i-1})|.$$

For $i\in\{ 1,2,...,n-1\}$, there always exists some $z_i$ in $[0,1]$ s.t. $x_{i-1},x_i<z_i,$ and $x_i+x_{i-1}\ge z_i.$ Indeed, we can take $z_i$ as any real number approaching $x_i$ from the right(i.e. slightly greater than $x_i$).

Note that $g$ is non-negative.

Now we have that, $$ V(g,P) \le |g(x_n)|+|g(x_{n-1})|+\sum_{i=1}^{n-1} |g(x_i)|+|g(x_{i-1})|\le |g(x_n)|+|g(x_{n-1})| +\sum_{i=1}^{n-1} g(z_i).$$

Denote $x_{n-1}$ by $z_n.$

Then we get, $$V(g,P) \le |g(x_n)| +\sum_{i=1}^{n} g(z_i).$$

Note that, we can bound $g(z_i)+g(z_{i+1})$ with $g(h_i)$ for some $h_i\in[0,1]$ with the similar trick above.

Repeating this process, we get less and less terms on the R.H.S. And eventually, we get only two terms on the R.H.S, which are $g(x_n)$ and $g(m)$ for some $m\in[0,1].$

We get, $$V(g,P)\le 2$$ because $|g|$ is bounded by $1$.

Since the partition $P$ is arbitrary, we conclude that the total variation of $g$ is bounded by $2$, which implies that $g$ is of bounded variation.


问:
ellipticity condition

I believe (1.3) is in divergence form. My problem is that it is somewhat abstract for me and I'd like to see a few concrete examples.
I'd appreciate seeing an actual differential equation of this form with a functions $a_{ij}, b_i, c$ given values and with these values see the matrix formed by the $a_{ij}$

What does the ellipticity condition mean in words and why is it important?

PS: Is this the right place to ask clarification/understanding questions or is there a special place on SE Math for that?

答:
(1.3) such that (1.4) holds is the most general form of a (uniformly) elliptic operator. Notice that indeed the laplacian is elliptic since we can write it like in (1.3) with $a_{ij}=\delta_{ij}$, $b_i=0$ and $c=0$ and (1.4) holds trivially. One can construct other elliptic operators by modifying the laplacian operator such that (1.4) still holds and then adding any continuous functions as $b_i$ and $c$, this last part doesn't really play any role on the ellipticity condition.

The ellipticity condition might be very important in many contexts but specially I would like to point out that in order to prove the existence of weak solutions by means of the Lax-Milgram theorem is key. Let $\mathcal{L}u=f$ be our elliptic equation with

$$ \mathcal{L}u = -\sum_{i,j=1}^n \partial_i (a_{ij}(x)\partial_j u(x)) + \sum_{i=1}^n b_i(x)\partial_i u(x) + c(x)u(x). $$

If we write it in the weak formulation we get

$$ \underbrace{\sum_{i,j=1}^n \int_\Omega a_{ij}\partial_j u \partial_i\phi + \sum_{i=1}^n \int_\Omega b_i\partial_i u\phi + \int_\Omega cu\phi}_{B[u,\phi]} = \underbrace{\int_\Omega f\phi}_{\langle f,\phi \rangle} $$

for all $\phi\in C_c^\infty(\Omega)$. Then by Lax-Milgram we can prove existence of (weak) solutions if the bilinear form $B[\cdot,\cdot]$ is coercive, that is

$$ |B[u,u]| \geq C\|u\|^2. $$

Depending on the operator $\mathcal{L}$ ellipticity might be enough to prove it. In other cases not but still gives a good energy estimate that will be of help to prove well-posedness (see Evans' book Partial Differential Equations, Section 6.2, Theorem 2).

Title: ellipticity condition


问:
regarding derivatives
Does the derivative (the slope value) only give us the instantaneous rate change at a point and nothing else? Is this the only significance of that value?

For example I took the function $f(x)=x^2$,lets consider at $x=3$. $f(x)=x^2=9$ and $f'(x)=2x=6$. Does the value $6$ only tells us rate of change at point $x=3$?

Lets consider average rate change, e.g. suppose the average rate of change in y with respect to x over some interval is $7$; that is, for every single unit by which x changes, "y" on average changes by $7$ units. Here the average value "$7$" is related to the y value of the function means y on average changes by $7$ units.

Some of my friends said that the value $f'(x)=6$ in the above example only gives the rate of change at that point, i.e it is just the slope of the tangent at that point and nothing else, it has no effect on the y value ($x^2=9$) of the function. But if it doesn't have to do anything with 'y' value, then why is the slope at point called "the instantaneous rate of change of y with respect to x"?can anyone explain to me this, I really need help.

答:
Let's consider a slight variation on your example. What if $f(x) = x^2 + 1$? Then at $x=3$ we have $f(x) = 10$ and $f'(x) = 2x = 6$. We could also try $f(x) = x^2 - 2$, and then at $x=3$ we have $f(x)= 7$ but $f'(x) = 6$. These functions are just vertical shifts of your original function, so changing the $y$-value didn't change the derivative. By definition:

$$f'(x) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$

and notice for a given function this equation only depends on $x=a$, so the $y$-value of a function is irrelevant.

The reason we say the "instantaneous rate of change of $y$ with respect to $x$" is because in these scenarios $y$ is a function of $x$. At this point in your mathematical career I would guess you have only taken the derivatives of functions so the wording for the derivative of $y$ with respect to $x$ could also be stated as the derivative of the function $y=f(x)$, with respect to the variable $x$ (because the function depends on $x$).

You will eventually learn about implicit differentiation which will allow you to find the derivative of relations that are not functions. For example consider the equation of the circle $x^2 +y^2 = 5$. At the point $x=1$ there are two associated $y$-values: $y=-2$ and $y=2$ (hence not a function) and each has a different instantaneous rate of change. So for this example when you do not have a function the $y$-value will matter.

Title: regarding derivatives


问:
Compare $\ln(\pi)$ and $\pi-2$ without calculator
Unlike the famous question of comparing $e^\pi$ and $\pi^e$, which I solved almost instantly, I am stuck with this problem.  My thought was the following.

Since exponential function is order-preserving, we exponentiate both terms and get $\pi$ and $e^{\pi-2}$.  Then we study the function $f(x) = e^{x-2} - x$ or the function $g(x) = \frac{e^{x-2}}{x}$, and compare them with zero and one respectively.  I tried both.  But both involve solving the equation
$$e^{x-2} = x.$$

I tried Lagrange error terms and have
$$f(x) = -1 + \frac{(x-2)^2}{2!} + R_2(x-2),$$
where
$$\frac{(x-2)^3}{3!} \le R_2(x-2) \le \frac{e^{x-2}}{3!} (x-2)^3.$$

It is easy to see that the equation have a root between $3$ and $2 + \sqrt2$.  But I don't know how close it is to $\pi$.  It is to provide some lower bounds since we can plug in some values and calculate to show that $f(x) > 0$ for such values.  But for the upper bound, it is hard to calculate by hands since it has the $e^{x-2}$ factor.  At my best attempt by hand, I showed that $f(3.15) > 0$.  All it entails is that for all $x \ge 3.15$, $e^{x-2}$ is greater than $x$.  But it tells nothing about the other side.

Then I looked at the calculator and find that $e^{\pi-2} < \pi$.

I also tried Newton-Raphson iteration, but it involves a lot of exponentiation which is hard to calculate by hand and also involves approximation by themselves.  And I don't know how fast and close the iteration converges to the true root of the equation.

Any other hint for comparing these two number purely by hand?

答:
All the logarithms I know by heart are
$$\begin{align}\log_{10}2&=0.30103\\
\ln10&=2.303\\
\ln2&=0.693\end{align}$$
Using the second fact alone we can solve this problem! We know that
$$\frac12\ln\left(\frac{1+x}{1-x}\right)=x+R_2(x)$$
Where
$$|R_2(x)|\le\frac{|x|^3}{3(1-|x|)^3}$$
for $|x|<1$. Then let $x=\frac{-3}{487}$ so
$$\frac12\ln\left(\frac{1+x}{1-x}\right)=\ln\left(\frac{22}{7\sqrt{10}}\right)=\frac{-3}{487}+R_2\left(\frac{-3}{487}\right)$$
So we have
$$\begin{align}\ln\left(\frac{22}7\right)&=\frac12\ln10-\frac3{487}+R_2\left(\frac{-3}{487}\right)\gt\frac{2.3025}2-\frac1{160}-\frac1{100^3}\\
&=1.15125-0.00625-10^{-6}=1.145-10^{-6}\gt1.143\gt\frac{22}7-2\end{align}$$
And since $f(x)=\ln x-x+2$ is decreasing for $x\gt1$ and it has been known since the
time of Archimedes that $\frac{22}7\gt\pi$ we have established the result.

But if you didn't know that $\ln10=2.303$ to $3$ decimals you might
be in for a tougher slog. You could say, for example, that
$$\ln10=10\ln\frac54+3\ln\frac{128}{125}\gt20\left(\frac19+\frac1{3\times9^3}\right)+6\left(\frac3{253}\right)$$
So that
$$\begin{align}\ln\left(\frac{22}7\right)&\gt\frac{10}9+\frac{10}{2187}+\frac9{253}-\frac3{487}-10^{-6}\\
&\gt1.1111+0.004+0.035-0.007-10^{-6}\gt1.143\\
&\gt\frac{22}7-2\end{align}$$
Where we actually had to carry out one of the long divisions to $2$ significant figures.


问:
matrix and linear transformations
With regards to this (Matrix associated to a linear transformation with respect to a given basis)

I saw Ivo Terek solution which goes:

\begin{align}L(1,1,1) &= (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
 2\cdot (1,0,0) \\ L(1,1,0) &= (0,8,2) = 2\cdot (1,1,1) + 6\cdot (1,1,0) -
 8\cdot (1,0,0) \\ L(1,0,0) &= (0,3,1) = 1\cdot (1,1,1) + 2\cdot (1,1,0) -
 3\cdot (1,0,0),\end{align}

However, I was not able to understand how:

$ 6\cdot (1,1,1) + 0\cdot (1,1,0) - 2\cdot (1,0,0) \\2\cdot (1,1,1) + 6\cdot (1,1,0) - 8\cdot (1,0,0) \\1\cdot (1,1,1) + 2\cdot (1,1,0) - 3\cdot (1,0,0) $

was derived. I only know half of where the equation came from.

答:
I suppose you understand the first equality in
$$L(1,1,1) = (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
 2\cdot (1,0,0)$$
well, as $L$ is defined by $L \left( \begin{bmatrix}x_{1}\\x_{2}\\x_{3} \end{bmatrix} \right) = \begin{bmatrix}4x_{3}\\3x_{1}+5x_{2}-2x_{3}\\x_{1}+x_{2}+4x_{3} \end{bmatrix}$ in the linked question.
Observe that for each vector $v$ expressed as a linear combination of basic vectors in the basis $B = \left(\begin{bmatrix}1\\1\\1 \end{bmatrix} , \begin{bmatrix}1\\1\\0 \end{bmatrix}, \begin{bmatrix}1\\0\\0 \end{bmatrix} \right)$, the first basic vector $(1,1,1)^T$ determines the third component of $v$, the second basic vector $(1,1,0)^T$ determines the second component of $v$.

In particular, you have $(4,6,6)^T$, so you need $6$ as the coefficient for $(1,1,1)^T$, as the basic vectors $(1,1,0)^T$ and $(1,0,0)^T$ can't contribute to the third component $z$ in the vector $(x,y,z)^T$.  Use a similar reasoning to get $0$ for the coefficient of $(1,1,0)^T$.  The rest is a simple arithmetic exercise.

Title: matrix and linear transformations


问:
Prove $\ln{(\frac {x}{y})} = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.
Prove $\ln (\frac{x}{y}) = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.

I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln\frac{x}{y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?

答: I think you mean $\ln(x/y) = \ln x - \ln y$. If you've already shown that $\ln x^r = r\ln x$ (for all $r\in\Bbb R$), $\ln(1/y) = \ln y^{-1} = -\ln y$ follows from there, so you're good.</p>


问:
Continuous probability density function question but x≥1
If the probability density function for a particle to be at a particular location is given by f(x) = cg(x) for some constant c. And g(x) is:

g(x):= {(0, if x≤0), (x, if 0<x<1), (x^(-5/2), if 1≤x)}

Q1.) How would I find the value of the constant c?

Q2.) How would I find the mean (expected location of the particle)?

I'm trying to do question 1 but I don't know how to find c. I know I have to make 1 equal to the integral of both functions multiplied by c. I do the integral of x from 0 to 1 multiplied by c but I'm not sure what to do for the limits for the integral of x^(-5/2) as it is for x≥1?

And then once I found c what limits would I use to calculate the mean?

Any help would be much appreciated. Thank you

答:
The integral all over the support must be 1

You have

$$\int_0^1 x dx+\int_1^{\infty}x^{-\frac{5}{2}}dx=\frac{7}{6}$$

thus

$$f_X(x)=\frac{6}{7}x\cdot\mathbb{1}_{(0;1)}(x)+\frac{6}{7}x^{-\frac{5}{2}}\cdot\mathbb{1}_{[1;\infty)}(x)$$

With mean

$$\mathbb{E}[X]=\int_{-\infty}^{\infty}xf_X(x)dx=\dots=2$$


问:
Standard Brownian Motion Probability Computation
While studying for my exam I came across the following question:

Let $W(t)$ be a standard Brownian motion. Find

$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$

Attempt:

(1) Attempt 1:  Finding the distribution of W(1)+W(2) which I found to be $\mathcal{N}(0,5)$ and same for $3W(1)-2W(2) $ which is $\mathcal{N}(0,5)$ my attempt is to prove that the two things are independent so I could separate both terms of the probability and multiply two quantities as

$$\mathbb{P}(0<W(1)+W(2)<2). \mathbb{P}(3W(1)-2W(2)>0)$$

but this didn't work out yet.

(2) Attempt 2: Working with the inequalities to find common bounds i.e I found $$\dfrac{2}{3} Y <X <2-Y$$ but this isn't so useful.

答:
$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$

$$=\mathbb{P}(0<2W(1)+\Delta W(2)<2, W(1)-2\Delta W(2)>0)$$

where $\Delta W(2)=W(2)-W(1)$. Given the properties of a Wiener process $[W(1),\Delta W(2)]$ is jointly normal with mean (0, 0) and covariance $diag(1, 1)$.

Now

$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$

$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}I(0<2x+y<2, x-2y>0)f(x)f(y)dxdy.$$

Where $f$ is the pdf of a standard normal variable. This can be evaluated numerically.


问:
Why is derivative of a function at a point the slope of the tangent at that point?
Consider a function y = f(x).

Let one point A on the curve be (a , f(a)) and a second point B (a+h , f(a+h)).

$m_{AB} = (f(a+h) - f(a))/ h$.

As h → 0 point A and B come closer.

My question is:

As points A and B come closer, why does the line joining them have to be a tangent? Since the points are so close can't it be a chord passing through A?
If we consider A and B as a single point, there are many lines passing through A.Then why a tangent?

Can anybody help?

答:
You are right. The line $AB$ is a chord (or secant). Following up @sasquires comment, what you have to imagine is that there are two lines.

The first is the tangent at $A$. We are interested in finding its gradient, but we only have one point, so we cannot find it diectly.

The second line is the chord $AB$. We have two points so we can find its gradient.

Now imagine point $B$ moving closer to point $A$. The closer it gets, the closer its gradient gets to the gradient we are interested in (the gradient of the tangent at $A$). And if we make the distance between $A$ and $B$ infinitesimally small (but not zero), we can say that the gradient $AB$ is the same (almost) as the gradient of the tangent at $A$.

In fact, as $h$ approaches 0, the gradient $AB$ approaches the gradient of the tangent at $A$.


答:
You can do the change of variable $u=\frac{1}{t}$: for $x > 0$,
$$
\ln\frac{1}{x} =  \int_{1}^{\frac{1}{x}} \frac{dt}{t} = \int_{1}^{x} \frac{-du}{u^2}u = -\int_{1}^{x} \frac{du}{u} = -\ln x
$$</p>

Title: Prove $\ln{(\frac {x}{y})} = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.

Body:
Prove $\ln (\frac{x}{y}) = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.

I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln\frac{x}{y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?


问:
$ ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$
Show that $ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$ where $\zeta$ is the primitive $5^{th}$ root of unity.

I have seen a similar question asked, from this I gather that the quadratic polynomial is $ ^2+−1$, however I don't really understand how to conclude that it is equal to $ℚ(\sqrt5)$.

答:
To answer the question in the title, here is a roadmap:

$\zeta$ is a root of $x^4 + x^3 + x^2 + x + 1$

$\zeta^{-1}=\zeta^4$

$z=\zeta + \zeta^{-1}$ implies $z^2=\zeta^2+\zeta^3+2$

$z$ is a root of $x^2+x−1$ and so $z=(-1\pm \sqrt 5)/2$; no need to know the actual sign in $\pm$

$\mathbb Q(z) = \mathbb Q(\sqrt 5)$

Title: $ ℚ(\zeta + \zeta^{-1}) = ℚ(\sqrt5)$


问:
proof that such an integral is a measure
How to prove that this set function is a measure?
$$v(A)= \int_{A} f d\mu $$ and $$f$$ is a non-negative function.

1.The first condition for being a measure is ok because if A is an empty set then $$ v(\emptyset)= \int_{\emptyset} f d\mu = 0 $$
2.From the properties of a measurable and non-negative function we know that $$ \int_{X} fd\mu \geq 0$$ so $$v(A) \geq 0$$
3. I have a problem with the proof for countable additivity, how can this be shown?

答:
If $\{A_{i}\}_{i \in \mathbb{N}}$ are disjoint and measurable, let $A = \bigcup_{i} A_{i}$, and define
$$
E_{n} = \bigcup_{i=1}^{n} A_{i}
$$
Then $E_{n} \uparrow A$, and thus $f \chi_{E_{n}} \uparrow f$ (since $f$ is non-negative). Monotone Convergence Theorem gives
$$
\nu(A) = \int f \chi_{A} = \lim_{n \to \infty} \int f \chi_{E_{n}} = \lim_{n \to \infty} \sum_{i=1}^{n} \int f_{\chi_{A_{i}}} = \sum_{i=1}^{\infty} \nu(A_{i})
$$
since $\int f \chi_{A_{i}} = \int_{A_{i}} f = \nu(A_{i})$ by definition.


问:
Property of product of matrices with their transpose
Let $A$ and $B$ be matrices of size $m\times n$.

Claim: $A^TA=B^TB$ iff there exists an orthogonal matrix $Q$ such that $QA=B$.

The backward implication is trivial, consider $B^TB=(QA)^T(QA)=A^TQ^TQA=A^TA$. But how do I show that the forward implication holds true?

Also are there other properties matrices $A$ and $B$ follow, given that $A^TA=B^TB$?

How do I compute such matrices? Is there an equivalent criterion which will a be basis for computing such matrices?

答:
Let $r=\operatorname{rank}(A)$ and $\{Av_1,\ldots,Av_r\}$ be an orthonormal basis of $\operatorname{range}(A)$. Since $A^TA=B^TB$, the set $\{Bv_1,\ldots,Bv_r\}$ is an orthonormal basis of $\operatorname{range}(B)$. Thus there exists an orthogonal matrix $Q$ such that $QAv_i=Bv_i$. It follows that $QAv=Bv$ for every $v\in V=\operatorname{span}\{v_1,v_2,\ldots,v_r\}$.

However, as $A^TA=B^TB$, we also have $W:=\ker(A)=\ker(B)$. Hence $QAw=Bw=0$ on $W$ and we conclude that $QA=B$ because $\mathbb R^n=V\oplus W$.


问:
Recovering length of an interval by a limiting formula
This is from Tao's text on Measure theory. While proving Lemma 1.1.2-(ii) he's using a discretization argument for length of an interval in $\mathbb{R}$, which is : $$|I|:=\lim_{N\to \infty}\frac{1}{N}\operatorname{card}\left(I \cap \frac{1}{N}\mathbb{Z}\right)$$ Where $\frac{1}{N}\mathbb{Z}:=\left\{\frac{n}{N}:n \in \mathbb{Z}\right\}$, and by $\operatorname{card}$ i mean cardinality of a finite set. I do have an intuitive idea for this, but not able to prove the result. I was trying like this: for any $I$ say $I:=[-5,5]$, if i take some element $i \in [-5,5]\cap\frac{1}{N}\mathbb{Z}$, then $i=j/N:j\in \mathbb{Z}\cap [-5,5]$ and then $\operatorname{card}\left([-5,5] \cap \frac{1}{N}\mathbb{Z}\right)=|(-5)N-(5)N|$, afterthat i'm not sure how to finish this argument.

答:
I think i get where i was going wrong, I'll add it as an answer.

Let, $I:=[a,b]$ and $I \cap \frac{\mathbb{Z}}{N}:=\left\{\frac{k}{N}\in I: k\in \mathbb{Z}\right\}$. This exactly implies that $a \le \frac{k}{N}\le b \implies Na \le k\le Nb$. So cardinality of the above set is the number of integers satisfying the inequality.
$$\operatorname{card}\left(I \cap \frac{\mathbb{Z}}{N}\right)=\lfloor Nb\rfloor-\lceil Na \rceil +1$$Now we have : $ \lfloor Nb\rfloor \le Nb$ and $\lceil Na \rceil \ge Na$, so $$\lfloor Nb\rfloor-\lceil Na \rceil +1 \le Nb-Na+1=N(b-a)+1$$ which implies $$\frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}\le (b-a)+\frac{1}{N}$$ moreover, $\lfloor Nb\rfloor \ge Nb-1$ and $\lceil Na \rceil \le Na+1$ which implies $$\frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}\ge \frac{(Nb-1)-(Na+1)+1}{N}=(b-a)-\frac{1}{N}$$
Finally, we write  $$(b-a)-\frac{1}{N} \le \frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}\le (b-a)+\frac{1}{N}$$ Now letting $N\to \infty$ gives us the desired result: $$\lim_{N\to \infty}\frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}=b-a=m(I)$$