问:
How to prove the triangle inequality for hermitian angle between vectors in a complex vector space?
Suppose that I have a vector space defined over complex numbers. I can calculate the hermitian angle between two vectors as following:

$$
\theta_H (\vec{u},\vec{v}) = \cos^{-1} \left(\frac{|\vec{u}.\vec{v}^*|}{||u||.||v||}\right)
$$

How do I prove the following triangle inequality for $\theta_H$?

$$
\theta_H(\vec{u},\vec{v}) \leq \theta_H(\vec{u},\vec{w}) + \theta_H(\vec{v},\vec{w})
$$

In case that the vector space is over real numbers, you can take the span of the three vectors and project that space onto a 3d euclidean space and prove the inequality in 3d euclidean space. But I am not sure how to do something like that for complex numbers. Is there a similar approach that I can follow in this case? Or something completely different?

答:
I ended up coincidentally solving your problem as part of my research, so I thought I'd come back and share.

We might as well assume that our vectors are unit vectors. Suppose $x, y, z$ are unit vectors, and let
$$G = \begin{pmatrix} 1 & \langle x, y \rangle & \langle x, z \rangle \\ \langle y, x \rangle & 1 & \langle y, z \rangle \\ \langle z, x \rangle & \langle z, y \rangle & 1\end{pmatrix}.$$
Note that $G$ is a Gramian matrix, and hence positive-semidefinite (also: positive definite if and only if $x, y, z$ are linearly independent). Thus $\det G \ge 0$, i.e.
$$1 + 2\Re(\langle x, y\rangle \langle y, z\rangle \langle z, x\rangle) - |\langle x, y\rangle|^2 - |\langle y, z\rangle|^2 - |\langle z, x\rangle|^2 \ge 0.$$
Note that this is equivalent to
$$(1 - |\langle x, y \rangle|^2)(1 - |\langle y, z \rangle|^2) \ge |\langle x, y\rangle \langle y, z\rangle - \langle z, x \rangle|^2,$$
by expanding both sides and cancelling the common term of $|\langle x, y \rangle|^2 |\langle y, z \rangle|^2$. Taking the square root of both sides, we get
$$|\langle x, y\rangle \langle y, z\rangle| - |\langle z, x \rangle| \le |\langle x, y\rangle \langle y, z\rangle - \langle z, x \rangle| \le \sqrt{1 - |\langle x, y \rangle|^2}\sqrt{1 - |\langle y, z \rangle|^2},$$
thus
$$|\langle z, x \rangle| \ge |\langle x, y\rangle \langle y, z\rangle| - \sqrt{1 - |\langle x, y \rangle|^2}\sqrt{1 - |\langle y, z \rangle|^2}.$$
Note that $\cos(\theta_H(p, q)) = |\langle p, q\rangle|$ for unit vectors $p, q$, so this inequality becomes
\begin{align*}
\cos(\theta_H(x, z)) &\ge \cos(\theta_H(x, y))\cos(\theta_H(y, z)) - \sqrt{1 - \cos^2(\theta_H(x, y))}\sqrt{1 - \cos^2(\theta_H(y, z))} \\
&= \cos(\theta_H(x, y))\cos(\theta_H(y, z)) - \sin(\theta_H(x, y))\sin(\theta_H(y, z)) \\
&= \cos(\theta_H(x, y) + \theta_H(y, z)).
\end{align*}
Note that $\cos$ is strictly monotone decreasing on $[0, \pi]$, on which we may find $\theta_H(x, y) + \theta_H(y, z)$ and $\theta_H(x, z)$, hence
$$\theta_H(x, z) \le \theta_H(x, y) + \theta_H(y, z).$$
We also get that, if we achieve equality, then $\det G = 0$, and $x, y, z$ are linearly dependent.


问:
A tricky logarithms problem?
$ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.

Any hints? Thanks!

ContentLicense: CC BY-SA 3.0

Title: A tricky logarithms problem?

Body:
By elementary arithmetic operations (after / describing next action):
$$\log_{4n}40\sqrt{3}=\log_{3n}45\ \ \ \mbox{ / definition of logarithm}$$
$$(4n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } 4=\frac{4}{3}\cdot 3$$
$$\left({4\over 3}\cdot 3n\right)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / }(ab)^c=a^cb^c$$
$$\left({4\over 3}\right)^{\log_{3n}45}\cdot (3n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } a^{\log_ab}=b$$
$$\left({4\over 3}\right)^{\log_{3n}45}\cdot 45=40\sqrt{3}\ \ \ \mbox { / }\cdot\frac{1}{45}$$
$$\left({4\over 3}\right)^{\log_{3n}45}={8\over 9}\sqrt{3}\ \ \ \mbox { / }\left({a\over b}\right)^c=\frac{a^c}{b^c}$$
$$\left({4\over 3}\right)^{\log_{3n}45}=\left({4\over 3}\right)^{3\over 2}\ \ \ \mbox { / }a^b=a^c\Rightarrow b=c\ \ (a\neq 0,1)$$
$$\log_{3n}45=\frac{3}{2}\ \ \ \mbox { / definition of logarithm}$$
$$(3n)^{3\over 2}=45\ \ \ \mbox { / powered by } \frac{2}{3}\mbox{ and divided by }3$$
$$n=\frac{\sqrt[3]{45^2}}{3}\ \ \ \mbox { / powered by 3}$$
$$n^3=\frac{45^2}{27}$$
$$n^3=75$$


问:
Matrix norm inequalities for 2-norm and $\infty$-norm
I am doing the following exercise: https://i.imgur.com/EoIYC7H.png

I proved parts $a$ and $b$, but I got a little confused with parts $c$ and $d$. Here is my proof for part $(c)$:

Let $x \in \mathbb{R}^{n}$ and view $A$ as a $mn$-dimensional vector. Then,

$||Ax||_{\infty} \leq ||Ax||_2$ by part $(a)$.

We know

$||Ax||_2 \leq ||A||_2 ||x||_2$ by norm property.

By part $(b)$ we now have

$\cdots \leq ||A||_2 \sqrt{n}||x||_\infty$

Then since the above was written with arbitrary $x$, let it be the unit vector. This yields what was desired: $$||A||_\infty \leq \sqrt{n} \cdot ||A||_2$$

Regarding part $(d)$, doesn't this follow immediately from part $(b)$ since we can view $A$ as an $mn$-dimensional vector?

答:
The operator (equivalently matrix) norm is defined as
$$\left\lVert A\right\rVert_{\Box}=\sup\limits_{x\in B\setminus\{0\}}\frac{\left\lVert Ax\right\rVert_{\Box}}{\left\lVert x\right\rVert_{\Box}}=\sup\limits_{\substack{x\in B\\ \left\lVert x\right\rVert_{\Box}=1}}\left\lVert Ax\right\rVert_{\Box}$$
where $B$ is your favorite (vector) space to work on and $\Box$ can be any symbol depending on your choice of vector spaces, simply fit accordingly.

Your answer to $c)$ is correct, but you can be a little bit more precise with your redaction (as this seems to be an introduction exercise, we shall make any steps very clear in our minds).

I suggest the following:

Let $x\in\mathbb{K}^m\setminus\{0\}$ ($\mathbb{K}$ being your favorite field, for example $\mathbb{R}$ or $\mathbb{C}$). Then we have
\begin{align}
\left\lVert Ax\right\rVert_{\infty}
&\leqslant\left\lVert Ax\right\rVert_{2}\\
&\leqslant \left\lVert A\right\rVert_{2}\left\lVert x\right\rVert_2\\
&\leqslant \left\lVert A\right\rVert_2\sqrt{n}\left\lVert x\right\rVert_{\infty}
\end{align}
where we used the fact that $Ax\in\mathbb{K}^n$ so that we can use the previous inequalities: $a)$ for the first one, an operator norm property for the second and $b$ for the third one. Note here that I have used the $\infty$ symbol for both $Ax$ and $x$ where in fact they lie in different spaces (unless $n=m$), we could be more rigorous and distinguish both norms, but usually we are lazy and we do not.

This implies that
$$\frac{\left\lVert Ax\right\rVert_\infty}{\left\lVert x\right\rVert_\infty}\leqslant \sqrt{n}\left\lVert A\right\rVert_2$$

As the right hand side does not depend on $x\in\mathbb{K}^m\setminus\{0\}$, we have
$$\left\lVert A\right\rVert_{\infty}=\sup\limits_{x\in B\setminus\{0\}}\frac{\left\lVert Ax\right\rVert_{\infty}}{\left\lVert x\right\rVert_{\infty}}\leqslant \sqrt{n}\left\lVert A\right\rVert_2$$
which is the desired result.

Simply use the same logic for $d)$ and you'll be done.

ContentLicense: CC BY-SA 4.0


问:
Moment generating function. A miner is trapped in a mine containing 3 doors.
A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 hours of travel. The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after 7 hours. If we assume that the miner is at all times equally likely to choose any one of doors, what is the expected length of time until he reaches safety?

I did it normally but I have been asked to solve through moment generating function. I don't know how to proceed

答: If you have a random variable $T$ with moment generating function $M_T(s) = \mathbb{E}[e^{Ts}]$, then its moments can be recovered by $\mathbb{E}[T^k] = M_T^{(k)}(0)$ (where this is the $k$th derivative). So if you want to compute the expected length of time using the MGF, you can compute the MGF, take the first derivative, and sub in zero. (This is probably a bit more complicated than it needs to be given that you can compute the expected value quite easily, as you have done.)


问:
Dual space of $(C[0,1],||\cdot||_2)$
I encountered the following example, which is supposed to show, that "weak convergence" and "convergence with respect to the norm" do not have to be the same.

Consider the normed space $(C[0,1],||\cdot||_2)$. Let $f_n = \sin(nt)$, then
$$||f_n||_2^2 = \int_0^1 |sin(nt)|^2 = 1/2.$$
We see, that $f_n$ goes strongly to $1/2$. It then claims without demonstrating, that $f_n$ goes weakly to zero.

I know that for a sequence $(x_n)_{n\in \mathbb{N}} \subset E$ in the normed space $E$, weak convergence to $x\in E$ holds, if
$$
f(x_n)\rightarrow f(x) \quad \forall f \in E^*
$$
I would like to check on that. But how do I find $E^*$?

答:
Note that $[0,1]$ is compact and hence, $C[0,1]$ is a dense subspace of $L^2[0,1]$. Hence, $L^2[0,1] = (L^2[0,1])^* \subset (C[0,1],||.||_2)^*$ .

But density of $C[0,1]$ implies that any element of $(C[0,1],||.||_2)^*$ extends to an element of $(L^2[0,1])^*$ uniquely! So $(C[0,1],||.||_2)^*=L^2[0,1]$

Note that this argument follows in the more general setup: If $(N,||.||)$ is a normed linear space with $(\bar N^{||.||},||.||)$ as its norm closure then $N^*= (\bar N^{||.||})^*$

Title: Dual space of $(C[0,1],||\cdot||_2)$


问:
Graphical intuition of why the integral is a linear operator
so I can prove using upper and lower riemansums why the integral is a linear operator:

$$\int a f(x) + b g(x) \,\mathrm{d}x = a\int f(x)\,\mathrm{d}x + b\int g(x)\,\mathrm{d}x$$

And I also understand why it intuitively must be true, as I know that integration
and derivation are inverse operations and since the latter is linear the former also has to be.

Graphically I can also understand why a sum of integrals must be the integral of each function
$\int (f+g) = \int f + \int g$. However I struggle to understand the graphical implications of the scaling. E.g.

Is there some graphical intuition why
$$ \int a \cdot f(x) \,\mathrm{d}x = a \int f(x) \,\mathrm{d}x$$ holds ?

I tried to graph the above statement for $f(x)=x^2$ on the interval $x \in [0,1]$

However, I can not intuitively say why the green area is twice as big as the green one.

答:
What if one imagines $a f(x)$ as being $a$ copies of $f(x)$ stacked on top of each other?

Then the integral of $a f(x)$ would be $a$ integrals performed “side by side”. With $a$ copies of the area, one can intuit that the cumulative area is $a$ times the original area.


问:
Is this angular momentum derivative always true?
Recently, when asking for the derivative of angular momentum, I gave my own solution
$$ \frac {d \vec L} {dt} = \frac {d \vec r_{(t1)}} {dt} \times \frac {d \vec r_{(t2)}} {dt} + \vec r_{(t1)} \times  \frac {d^2 \vec r_{(t1)}} {dt}  \tag 1$$
How to calculate the derivative of the angular momentum vector $ d\vec L = d(\hat I \vec \omega)?$

Even though for a few examples I have counted, the derivative gave the correct result the question remained: is it always true?

I already have an answer, but first I have to ask a question, these are the rules here

Body:
$$\forall m \in \Bbb R \land \forall \vec r_{(t1)}= (r_{1x},r_{1y},r_{1z}) \land \forall \vec r_{(t2)}= (r_{2x},r_{2y},r_{2z}) \land \forall \vec r_{(t3)}= (r_{3x},r_{3y},r_{3z})$$
$$(\frac {d \vec L} {dt}=\vec v_{(t1)}\times m \vec v_{(t2)} +\vec r_{(t1)} \times m \vec a_{(t1)} )$$

velocity and acceleration vectors

$$\frac {r_{2x} - r_{1x}}{dt} = \vec v_{x(t1)} ; \frac {r_{2y} - r_{1y}}{dt} = \vec v_{y(t1)}; \frac {r_{2z} - r_{1z}}{dt} = \vec v_{z(t1)} \tag 1  $$

$$\frac {r_{3x} - r_{2x}}{dt} = \vec v_{x(t2)} ; \frac {r_{3y} - r_{2y}}{dt} = \vec v_{y(t2)}; \frac {r_{3z} - r_{2z}}{dt} = \vec v_{z(t2)} \tag 2  $$

$$\frac {v_{2x} - v_{1x}}{dt} = \vec a_{x(t1)} ; \frac {v_{2y} - v_{1y}}{dt} = \vec a_{y(t1)}; \frac {v_{2z} - v_{1z}}{dt} = \vec a_{z(t1)} \tag 3  $$

angular momentum vectors

$$\vec L_{(t1)}= \vec r_{(t1)} \times m \vec v_{(t1)}   $$
$$(r_{1y}v_{1z} - r_{1z}v_{1y}, r_{1z}v_{1x}-r_{1x}v_{1z},r_{1x}v_{1y} -r_{1y}v_{1x}) \tag 4 $$
$$\vec L_{(t2)}= \vec r_{(t2)} \times m \vec v_{(t2)}   $$
$$(r_{2y}v_{2z} - r_{2z}v_{2y}, r_{2z}v_{2x}-r_{2x}v_{2z},r_{2x}v_{2y} -r_{2y}v_{2x}) \tag 5 $$
change of angular momentum vector with time
$$\vec L_{(t2)}-\vec L_{(t1)} \tag 6$$

$$\frac {d l_{x}}{dt}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {6.a}$$
$$\frac {d l_{y}}{dt}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {6.b}$$
$$\frac {d l_{z}}{dt}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {6.c}$$

we calculate the components of the derivative

$$\vec v_{(t1)}\times \vec v_{(t2)}=(v_{1y}v_{2z} - v_{1z}v_{2y}, v_{1z}v_{2x}-v_{1x}v_{2z},v_{1x}v_{2y} -v_{1y}v_{2x}) \tag 7$$

$$v_{1y}v_{2z} - v_{1z}v_{2y}=r_{2y}v_{2z}-r_{1y}v_{2z} -r_{2z}v_{2y}+r_{1z}v_{2y}\tag {7.a}$$
$$v_{1z}v_{2x} - v_{1x}v_{2z}=r_{2z}v_{2x}-r_{1z}v_{2x} -r_{2x}v_{2z}+r_{1x}v_{2z}\tag {7.b}$$
$$v_{1x}v_{2y} - v_{1y}v_{2x}=r_{2x}v_{2y}-r_{1x}v_{2y} -r_{2y}v_{2x}+r_{1y}v_{2x}\tag {7.c}$$

$$\vec r_{(t1)}\times \vec a_{(t1)}=(r_{1y}a_{1z} - r_{1z}a_{1y}, r_{1z}a_{1x}-r_{1x}a_{1z},r_{1x}a_{1y} -r_{1y}a_{1x}) \tag 8$$

$$r_{1y}a_{1z} - r_{1z}a_{1y}=r_{1y}v_{2z}-r_{1y}v_{1z} -r_{1z}v_{2y}+r_{1z}v_{1y}\tag {8.a}$$
$$r_{1z}a_{1x} - r_{1x}a_{1z}=r_{1z}v_{2x}-r_{1z}v_{1x} -r_{1x}v_{2z}+r_{1x}v_{1z}\tag {8.b}$$
$$r_{1x}a_{1y} - r_{1y}a_{1x}=r_{1x}v_{2y}-r_{1x}v_{1y} -r_{1y}v_{2x}+r_{1y}v_{1x}\tag {8.c}$$
We count the derivative
$$\frac {d \vec L`} {dt}=\vec v_{(t1)}\times m \vec v_{(t2)} +\vec r_{(t1)} \times m \vec a_{(t1)} \tag 9$$

$$\frac {d l`_{x}}{dt}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {9.a}$$
$$\frac {d l`_{y}}{dt}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {9.b}$$
$$\frac {d l`_{z}}{dt}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {9.c}$$

$$(6)=(9)$$

C.K.D.


问:
LU decomposition steps
I've been looking at some LU Decomposition problems and I understand that making a matrix A reduced to the form A=LU , where L is a lower triangular matrix and U is a upper triangular matrix, however I am having trouble understanding the steps to get to these matrices. Could someone please explain the method for LU Decomposition in detail, preferably excluding the concept of permutation matrices? ( We haven't talked about permutation matrices in class yet so our professor forbids us to use them for the decomposition).</p>

答:
$LU$ decomposition is really just another way to say Gaussian elimination.
If you're familiar with that, putting the pieces together is easy.
Here is an example. Let
$$
A=A^{\left(0\right)}=\left[\begin{array}{ccc}
8 & 1 & 6\\
4 & 9 & 2\\
0 & 5 & 7
\end{array}\right].
$$
Proceed by Gaussian elimination. The first multiplier is $\ell_{2,1}=4/8=0.5$
(this is the multiplier that allows us to cancel $a_{2,1}=4$ using
the first row) and the second is $\ell_{3,1}=0/8=0$.
We arrive at
$$
A^{\left(1\right)}=\left[\begin{array}{ccc}
8 & 1 & 6\\
0 & 8.5 & -1\\
0 & 5 & 7
\end{array}\right].
$$
To cancel out $a_{3,2}^{\left(1\right)}=5$, we use the multiplier
$\ell_{3,2}=5/8.5\approx0.5882$ to yield
$$
A^{\left(2\right)}\approx\left[\begin{array}{ccc}
8 & 1 & 6\\
0 & 8.5 & -1\\
0 & 0 & 7.5882
\end{array}\right]
$$
which yields the $LU$ decomposition
$$
A=LU\approx\left[\begin{array}{ccc}
1 & 0 & 0\\
0.5 & 1 & 0\\
0 & 0.5882 & 1
\end{array}\right]\left[\begin{array}{ccc}
8 & 1 & 6\\
0 & 8.5 & -1\\
0 & 0 & 7.5882
\end{array}\right].
$$
Note that $L$ is just made up of the multipliers we used in Gaussian
elimination with $1$s on the diagonal, while $U$ is just $A^{\left(2\right)}$.</p>

Title: LU decomposition steps


问:
Normal distribution,expectations in F distribution
Let $_1, ... , _{10}$ be independent and identically distributed normal random variables with mean $0$ and variance $2$.
Then
$ ( X_1^2/(X_1^2 +X_2^2 + \dots +_{10}^2))$ is equal to ...

My solution:
Numerator is chi square with 1 df and denominator is chi square with 10 df
Division of both follow F distribution now mean of F distribution is (10/(10-2))

But the answer in book is 0.1

I don’t know what is wrong in this

答:
Since $X_1, \dots X_{10}$ are i.i.d. then the random variables
$$
Y_i = \frac{X_i^2}{X_1^2 + \dots + X_{10}^2}
$$
have the same distribution (in particular, $\mathbb{E} Y_i$'s are the same). Moreover, since $$\sum_{i=1}^n Y_i = 1 \implies \mathbb{E} Y_i = \frac 1 {10},$$
as the answer in the book.


问:
If $\lim_{x\to a} |g(x)| =\lim_{x\to a} |\frac{f(x)}{g(x)}| = \infty$, is it true that $\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)$?
Let $\lim_{x\to a} |g(x)| =\lim_{x\to a} |\frac{f(x)}{g(x)}| = \infty$ for some $a\in \mathbb R$. I would like to show that $\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)$. Below is my attempt.

Without losing generality, let us assume that $\lim_{x\to a} \frac{f(x)}{g(x)} = \infty$. It can be seen that $\infty = \lim_{x\to a} \frac{f(x)}{g(x)} + 1 = \lim_{x\to a} \frac{f(x)+g(x)}{g(x)}$. Therefore, $\lim_{x\to a} f(x) + g(x) = (\lim_{x\to a} \frac{f(x)+g(x)}{g(x)}) (\lim_{x\to a} g(x)) = (\lim_{x\to a} \frac{f(x)}{g(x)})(\lim_{x\to a} g(x)) = \lim_{x\to a} f(x).$

Whereas the last line follows from the "product of limits equal limit of product" rule for the $\pm$infinity/$\pm$infinity cases.

Is the proof good enough to demonstrated the result? If not, is this true in the first place? It would be appreciated to receive some help from you.

答:
I think this is true. As Greg Martin says, it's only reliable to prove it from the definitions. Throughout the proof I'll only write one $\delta >0$, because we can take the minimum of all $\delta_i$'s and call it $\delta$. \

Let $M>1$ there exists $\delta >0$ such that $|\frac{f(x)}{g(x)}|>\sqrt{M}$  and $|g(x)|>\sqrt{M}$ whenever $0<|x-a|<\delta$, this implies $|f(x)|>\sqrt{M}|g(x)|>M$ whenever $0<|x-a|<\delta$ (note that same is true for any $M>0$). So $\lim_{x \rightarrow a}|f(x)|=\infty$. To prove it we must assume the existence of $\lim_{x \rightarrow a}f(x)$. Now we can have the following cases:\

(i)     Let $\lim_{x \rightarrow a}f(x)=\infty$ and $\lim_{x \rightarrow a}g(x)=-\infty$. Choose $M_1>0$ there exists $\delta >0$ such that $f(x)>M_1>0$ and $g(x)< -\frac{M_1}{\sqrt{M}-1}<0$ whenever $0<|x-a|<\delta$. Then for $0<|x-a|<\delta$ , $|f(x)|=f(x)$ and $|g(x)|=-g(x)$. Let $0<|x-a|<\delta$ then:

\begin{align}
&|f(x)|>\sqrt{M}|g(x)|=(\sqrt{M}-1)|g(x)|+|g(x)| \\
&\Rightarrow f(x)-|g(x)| > (\sqrt{M}-1)|g(x)|\\
&\Rightarrow f(x)+g(x)> (\sqrt{M}-1)(-g(x))>M_1
\end{align}
Thus $\lim_{x \rightarrow a} (f(x)+g(x))=\infty=\lim_{x \rightarrow a}f(x)$. \

(ii)     Let $\lim_{x \rightarrow a}f(x)=-\infty$ and $\lim_{x \rightarrow a}g(x)=\infty$. Choose $M_1>0$ there exists $\delta >0$ such that $f(x)<-\frac{M_1}{1-\frac{1}{\sqrt{M}}}<0$ and $g(x)> M_1>0$ whenever $0<|x-a|<\delta$. Then for $0<|x-a|<\delta$ , $|f(x)|=-f(x)$ and $|g(x)|=g(x)$. Let $0<|x-a|<\delta$ then:

\begin{align}
&|f(x)|>\sqrt{M}|g(x)| \\
&\Rightarrow -f(x)>\sqrt{M}g(x)\\
&\Rightarrow -\frac{1}{\sqrt{M}}f(x)>g(x)\\
&\Rightarrow (1-\frac{1}{\sqrt{M}})f(x)-f(x)>g(x)\\
&\Rightarrow f(x)+g(x) < (1-\frac{1}{\sqrt{M}})f(x)<-M_1
\end{align}
Thus $\lim_{x \rightarrow a} (f(x)+g(x))=-\infty=\lim_{x \rightarrow a}f(x)$.\

(iii)      Let $\lim_{x \rightarrow a}f(x)=-\infty$ and $\lim_{x \rightarrow a}g(x)=-\infty$. Choose $M_1>0$ there exists $\delta >0$ such that $f(x)<-\frac{M_1}{2}$ and $g(x)<- \frac{M_1}{2}$ whenever $0<|x-a|<\delta$. Then for, $0<|x-a|<\delta$, $f(x)+g(x)<-M_1$
Thus $\lim_{x \rightarrow a} (f(x)+g(x))=-\infty=\lim_{x \rightarrow a}f(x)$.\

(iv)   The case $\lim_{x \rightarrow a}f(x)=\infty$ and $\lim_{x \rightarrow a}g(x)=\infty$ can be done in the same way as (iii).

(v)   Let $\lim_{x \rightarrow a}f(x)=\infty$, $\lim_{x \rightarrow a-}g(x)=-\infty$ and $\lim_{x \rightarrow a+}g(x)=\infty$ (the opposite  limits of $g$ will be similar). Then from (i) considering the neighborhood $(a-\delta, a)$ we get $\lim_{x \rightarrow a-} (f(x)+g(x))=\infty=\lim_{x \rightarrow a-}f(x)=\lim_{x \rightarrow a}f(x)$. Then from (iv) considering the neighborhood $(a,a+\delta)$ we get
$\lim_{x \rightarrow a+} (f(x)+g(x))=\infty=\lim_{x \rightarrow a+}f(x)=\lim_{x \rightarrow a}f(x)$. Combining those we have $\lim_{x \rightarrow a-} (f(x)+g(x))=\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a+} (f(x)+g(x))$ which implies
$\lim_{x \rightarrow a} (f(x)+g(x))=\lim_{x \rightarrow a}f(x)$.

(vi) The case when $\lim_{x \rightarrow a}f(x)=-\infty$, $\lim_{x \rightarrow a-}g(x)=-\infty$ and $\lim_{x \rightarrow a+}g(x)=\infty$ (the opposite  limits of $g$ will be similar) is done similayly.


问:
Evaluate $\int_S (x^2+y^2+z^2)\mathrm{d}S$.
Let $S=\{(x,y,z) \in \mathbb{R}^3 | x^2 +y^2+z^2 =1\}$. Evaluate $\int_S (x^2+y^2+z^2)\mathrm{d}S$.

My attempt : take $r(u,v)=(\cos v \cos u, \cos v \sin u, \sin v)$

Where $0\le u \le 2 \pi;~-\frac{\pi}{2}\le v \le \frac{\pi}{2}$

Let $~f(x,y,z)=x^2+y^2+z^2$. Here $~|\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}|=|\cos v|$

Then $\displaystyle \int_S(x^2+y^2+z^2)\,dS = \int_{-\pi/2}^{\pi/2} \int_0^{2\pi} f[r(u,v)] \left|\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}\right|~du~dv$

after that im not able to proceed further

答:
The integral is equal to $$ \int_S 1 = \text{Surface area}=4\pi $$

And to continue your approach you can easily see that $f(r(u,v)) =1$ thus the integral becomes

$$ \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} |\cos(v)| dudv = 4\pi \times \int_{0}^{\pi/2} \cos(v) dv = 4\pi  $$


问:
Help with interpreting specific logical statements discrete math
For a problem on a discrete math assignment, I am asked to find which of the following statements is true, but I am unsure if I'm interpreting it correctly as I don't know how to interpret every symbol yet. Here are the statements:

$$\exists ! x\in Z, \forall y\in Z, xy=x.$$

$$\exists !x\in Z, \forall y\in Z, xy=y$$

Should I interpret this as "there exists exactly one x for all y  such that ..." or should I interpret this as  "there exists exactly one x for each y  such that ..."

答:
I would consider the English interpretation of the above to be

"There exists exactly one $x$ in $Z$, such that for any $y$ from $Z$, $x y = x$"

"There exists exactly one $x$ in $Z$, such that for any $y$ from $Z$, $x y = y$"

It might be enlightening to consider what is required to actually prove statements like $\exists$ and $\forall$.

If you have to prove a statement like $\forall x \in Z. P(x)$, it means to show that $P(x)$ holds from an arbitrary $x$ taken from $Z$. For existentials, proving the statement $\exists x. P(x)$ requires you to come up with an $x$ from $Z$.

A statement of unique existence such as $\exists! x \in Z. P(x)$ requires you to both come up with an $x \in Z$ that satisfies $P$, but to also show that there is no other $x'$ that satisfies $P$ (or equivalently that any such element is, in fact $x$).

So, for example, the proof of first statement is going to look like:

Let $x$ be (some value from $Z$),

Take any $y$ from $Z$, then $xy = x$ because of (reason).

Take any $x'$ and any $y$ from $Z$, moreover, assume $x' y = x'$,
then $x' = x$ because of (reason).


问:
When does $e^{f(x)}$ have an antiderivative?
today I tried to integrate $x^x$ by applying a reverse chain rule which turned out to be false. I was told $\int e^{f(x)}\,dx$ can be done when $f(x)$ is linear. This made me wonder what conditions we can find so that $\int e^{f(x)}\,dx$ can be expressed in terms of elementary functions, but I'm not sure what to do.</p>

答:
When $f$ is a polynomial, it is a consequence of a celebrated theorem by Liouville that $e^{f(x)}$ has an elementary antiderivative iff there is a polynomial $h$ such that $1=h'+hf$. This implies that $f$ has degree at most 1. In particular, the function $e^{x^2}$ does not have an elementary antiderivative.

See How can you prove that a function has no closed form integral?.


问:
Splitting field of a polynomial over Z_5
Find the splitting field of the polynomial $$x^3+x+1$$ over $\mathbb{Z}_5$.

I can see that the given polynomial is irreducible over $\mathbb{Z}_5$ So, $$\dfrac{\mathbb{Z}_5 [x]}{(x^3+x+1)}$$ is a field and isomorphic to $\mathbb{Z}_5(\alpha)$ where $\alpha$ is the image of $x$ under the canonical map $$\pi \colon \mathbb{Z}_5[x]\to \dfrac{\mathbb{Z}_5 [x]}{(x^3+x+1)}.$$
So, $\alpha$ is a root of the polynomial over $\mathbb{Z}_5(\alpha)$ and it factorizes as $$x^3+x+1=(x-\alpha)(x^2+\alpha x+\alpha^2+1)$$ over $\mathbb{Z}_5(\alpha)$.

But now I can't conclude anything about $x^2+\alpha x+\alpha^2+1$.

Any help will be highly appreciated.

Thank you!

答:
Hint: If $\alpha$ is a zero of $f(x)= x^3+x+1$, then by the Frobenius automorphism $a\mapsto a^5$, $\alpha^5$ and $\alpha^{25}$ are also zeros.
Hence, $f(x) = (x-\alpha)(x-\alpha^5)(x-\alpha^{25})$.

Remark:

Since $\alpha$ is a zero, $\alpha^3+\alpha+1=0$ and so $\alpha^3=-\alpha - 1$.
Thus $\alpha^5 = -\alpha^3-\alpha^2 = -(-\alpha-1)-\alpha^2 = -\alpha^2+\alpha+1$.

Moreover, by Frobenius, $\alpha^{25} = (\alpha^5)^5 = (-\alpha^2+\alpha+1)^5 = (-\alpha^2)^5 + \alpha^5 +1^5 = -\alpha^{10} +\alpha^5 +1 =\ldots$, where the computation of $\alpha^{10}$ needs to be considered.


问:
There is a rectangle in 3-dimensional space. If I am only given the $x$ and $y$ coordinates, how do I find the $z$ coordinates?
We are given the $x$ and $y$ coordinates of the four vertices of a rectangle.
How do I find the $z$-coordinates?

I know that this 3 dimensional rectangle must obey all the usual conditions of rectangle.
Opposite sides are equal.
The angles between two adjacent sides are 90 degrees.

答:
Suppose you have coordinates $A(x_A,y_A), B(x_B,y_B), C(x_C,y_C),D(x_D,y_D)$, taking them clockwise around the rectangle.

You will note that these probably don't look like a rectangle if drawn in the 2D plane. That's because you don't have the z-coordinates.

Let's assume that the z-coordinate of A is $0$.

Find the vectors $AB$ and $AD$ and make it so that $AB . AD =0$

$AB = \begin{bmatrix} x_B-x_A\\ y_B-y_A \\ z_B -0\end{bmatrix}$

$AD = \begin{bmatrix} x_D-x_A\\ y_D-y_A \\ z_D -0\end{bmatrix}$

$AB.AD=0 \Rightarrow \begin{bmatrix} x_B-x_A\\ y_B-y_A \\ z_B -0\end{bmatrix} .  \begin{bmatrix} x_D-x_A\\ y_D-y_A \\ z_D -0\end{bmatrix} = 0$

$(x_B-x_A)(x_D-x_A)+(y_B-y_A)(y_D-y_A)+z_Bz_D=0$

$z_Bz_D=-(x_B-x_A)(x_D-x_A)-(y_B-y_A)(y_D-y_A)$

You can choose any pair of values $z_B, z_D$ that make this true.

Now you have full sets of coordinates for A, B and D. The final stage is to find C.

Note that $AB = DC$ (parallel sides of a rectangle). So $OC = OD + DC = OD + AB$

In particular, $z_C= z_D + (z_B-z_A)$

Worked example

Consider these points: $A(3,3,z_A), B(10,5,Z_B), C(11,8,z_C), D(4,6,z_D)$

Let $z_A=0$.

$AB = \begin{bmatrix} 7\\ 2 \\ z_B \end{bmatrix}$

$AD = \begin{bmatrix} 1\\ 3 \\ z_D \end{bmatrix}$

$AB.AD=0 \Rightarrow 7+6+z_Bz_D=0$

$z_Bz_D=-13$

Choose $z_B=-1$ and $z_D=13$

We now have $A(3,3,0), B(10,5,-1), C(11,8,z_C), D(4,6,13)$

$DC = AB = \begin{bmatrix} 7\\ 2 \\ -1 \end{bmatrix}$

$OD \begin{bmatrix} 4\\ 6 \\ 13 \end{bmatrix}$

$OC = OD + DC = \begin{bmatrix} 7\\ 2 \\ -1 \end{bmatrix}+\begin{bmatrix} 4\\ 6 \\ 13 \end{bmatrix} = \begin{bmatrix} 11\\ 8 \\ 12 \end{bmatrix}$

Final set of points: $A(3,3,0), B(10,5,-1), C(11,8,12), D(4,6,13)$


问:
Conditional Expectation in Bivariate Normal Distribution
enter image description here

I solved it  using the correct formula & got this answer but it shows that this is incorrect answer.
What did I do wrong ?
Thank You in advance for help.

答:
Which is the correct answer? Your forumla looks correct to me,  but an approximation you did

I find

$$\mathbb{E}[X_1|X_2=-0.2]=2-\frac{4.8}{36}(-0.2+2)=2-\frac{48}{360}\cdot\frac{18}{10}=\frac{44}{25}=1.76$$


答:
Here we are dealing with some rigid body $B$ with density $\rho$.

I will use $r,z$ as vectors, $I,A,X$ as matrices in 3-dimensional space.

Such a rigid body is usually normalized in that its center of mass is zero and its principal axes to the inertia tensor $I$ are the coordinate axes, that is, $I$ is diagonal, where
$$
I=\int_B\rho(z) ((z^Tz)-zz^T)\,d{\rm vol}(z).
$$

This body moves with some displacement and rotation $r(t)+A(t)z$, $z\in B$. The path in the orthogonal matrices has a derivative $\dot A=AX$ where $X$ is an anti-symmetric matrix whose operation on vectors can be given as $Xz= ω\times z$. Then $\ddot A=\dot A X+A\dot X=AX^2+A\dot X$, so that
\begin{align}
\ddot Az=A(ω×(ω×z))&=A((ω^Tz)ω-|ω|^2z)\\
z^TA^T\ddot Az&=ω^Tzz^Tω-|ω|^2|z|^2\\
\int_B\rho(z)z^TA^T\ddot Az\,d{\rm vol}(z)&=-ω^TIω
\end{align}

The total angular momentum in an external or lab frame of the body is then
\begin{align}
L_e&=\int_B\rho(z)(r+Az)\times (\dot r+\dot A z)\,d{\rm vol}(z)\\
&=mr\times \dot r + \int_B\rho(z)(Az)\times (AX z)\,d{\rm vol}(z)\\
&=mr\times \dot r + A\int_B\rho(z) (z× (ω× z))\,d{\rm vol}(z)\\
&=mr\times \dot r + A\int_B\rho(z) ((z^Tz)ω- zz^Tω))\,d{\rm vol}(z)\\
&=mr\times \dot r + AIω.\\
\end{align}
One could now name $L_b=Iω$ the internal moment of inertia, relative to the body frame.

Then the derivative of the external moment, which is what is driving the mechanics in reacting to external forces, is
\begin{align}
\text{Then }~\dot L_e&=mr\times \ddot r + \dot AIω + AI\dot ω\\
&=mr\times \ddot r + AXIω + AI\dot ω\\
&=mr\times \ddot r + A[(ω×(Iω)) + I\dot ω]
\end{align}
Again one can split off the second term and relate it to the time evolution of the internal moment. I would not call it the time derivative,...

Title: Is this angular momentum derivative always true?

Body:
Recently, when asking for the derivative of angular momentum, I gave my own solution
$$ \frac {d \vec L} {dt} = \frac {d \vec r_{(t1)}} {dt} \times \frac {d \vec r_{(t2)}} {dt} + \vec r_{(t1)} \times  \frac {d^2 \vec r_{(t1)}} {dt}  \tag 1$$
How to calculate the derivative of the angular momentum vector $ d\vec L = d(\hat I \vec \omega)?$

Even though for a few examples I have counted, the derivative gave the correct result the question remained: is it always true?

I already have an answer, but first I have to ask a question, these are the rules here


问:
Normal subgroup and non-equivalent representations
I am looking for an example to illustrate that to representations are not equivalent in a particular situation.

If H is a normal subgroup of G, $\pi$ a representation of $H$, and the second representation is defined as follows: For $g \in G$ define $\pi^g : G \to GL(V)$ by $\pi^g (h)=\pi (g^{-1}hg)$. For $g \in G$ we define the representation $\pi^g$ of H as before.

So now I have to find an example of G, H a representation $\pi$ of H and $g \in G$ such that $\pi$ and $\pi^g$ are not equivalent.

However, I am struggling a bit to find a nice and simple example to show this.
I've tried with $G= (\mathbb{C}, +)$ and $H=(\mathbb{R}, +)$ as this is a normal subgroup of G (as it is abelian) and as the representation of H has been given in the lectures. However, it seems as if this doesn't work

答:
Let $V=\mathbb{V}_2(\mathbb{F}_4)$, let $H:=\text{GL}(V)$ be the general linear group, and let $G:=\text{$\Gamma$L}(V)$ be the semi-linear group, that is $H$ extended by the Frobenius automorphism $\sigma$.

Now let $\pi:H\to \text{GL}(V)$ be given by $h\mapsto h$. The representation $\pi^{\sigma}$ will not be equivalent to $\pi$.


问:
Looking for some help with integration of $y=(1+x^2) \cos x$
Here is the curve $y=(1+x^2) \cos x$

I have worked out an approximation for the area under this curve $R$, using the trapezium rule, however I must integrate this curve in order to find the exact area of $R$.

not too sure how to go about this, would appreciate any help!

答:
\begin{gather*}
I=\int \left( 1+x^{2}\right)\cos xdx\\
\\
We\ have,\ for\ integration\ by\ parts,\\
\int u\cdot vdx=u\int vdx-\int \frac{du}{dx}\left(\int vdx\right) dx\\
\\
Using\ integration\ by\ parts,\ and\ taking\ \\
u=1+x^{2} \ as\ the\ first\ part\ and\ v=\cos x\ as\ the\ second,\\
I=\left( 1+x^{2}\right)\sin x-\int 2x\sin xdx\\
\\
Again\ ,I_{2} =\int 2x\sin xdx\\
Take\ u=2x\ as\ the\ first\ part,\ and\ v=\sin x\ as\ the\ second,\\
I_{2} =-2x\cos x+\int 2\cos xdx=-2x\cos x+2\sin x\\
I=\left( 1+x^{2}\right)\sin x-I_{2} =\left( 1+x^{2}\right)\sin x-( -2x\cos x+2\sin x)\\
=\left( 1+x^{2}\right)\sin x+2x\cos x-2\sin x\\
=\left( x^{2} -1\right)\sin x+2x\cos x\\
\\
\end{gather*}
Hope this answers your query!

Edit: If I interpret R correctly to be the real number line, then the resulting definite integral will turn out to be zero, simply because the integrand is even, and for even functions,
\begin{equation*}
\int ^{a}_{-a} f( x) dx=0
\end{equation*}


问:
Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$
I'm reading Lang's Undergraduate Analysis:

Let ${n}\choose{k}$ denote the binomial coefficient,

$${n\choose k}=\frac{n!}{k!(n-k)!}$$

where $n,k$ are integers $\geq0,0\leq k\leq n$, and $0!$ is defined to be $1$. Prove the following assertion:

$${n\choose k}={n\choose n-k}$$

I proceded by making the adequate substitutions:

$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!}$$

And then I simplified and achieved:

$$\frac{n!}{(n-k)!-k!}$$

But I'm not sure on how to proceed from here, I've noticed that this result is very similar to $\frac{n!}{k!(n-k)!}$. What should I do? I guess it has something to do with the statement about the nature of $n$ and $k$:

$n,k$ are integers $\geq0,0\leq k\leq n$

So should I just change the minus sign to plus sign and think of it as a product of $(n-k)!$?

$$\frac{n!}{(n-k)!-k!}\Rightarrow\frac{n!}{(n-k)!+k!}\Rightarrow \frac{n!}{k!(n-k)!}$$

I'm in doubt because I've obtained the third result on Mathematica, but obtained the first with paper and pencil. I'm not sure if there are different rules for simplification with factorials. I'm not sure if this $(n-k)!+k!$ mean a sum or a product in this case.

答:
The proof you've suggested, and the other answers have corrected, is the purely arithmetic way to see the desired equality.  The result is fairly clear to see intuitively as well, just ponder the following statement:

"To form a collection of $k$ objects from $n$ total objects, one may choose which $k$ objects to include in the collection, or equivalently, one may choose which $n-k$ objects to exclude from the collection."


问:
local extrema of a Continuous function
Let be a continuous function $f\colon \mathbb{R}\to \mathbb{R}\:$ that has exact 3 local extrema. $f$ is NOT differentiable.
Find the maximum number of local extrema that the function $f\circ f$ can have.

I do not know how to prove that but I obtained from my intuition that it must be 15. If we take for example a 8-degree polynomyal then $f\circ f$ is a 16-degree polynomyal and it does have maximum 15 local extrema.

The options of this multiple choice exercise are:

A)10
B)3
C)15
D)16

And the book says that the correct answer is C)15.
And I don't know why. I need a complete proof.

答:
By the chain rule,

$$(f(f(x))'=f'(f(x))f'(x)$$ where $f'$ denotes the formal derivative of $f$.

We know that $f'$ cancels exactly $3$ times, and that $f(x)$ can repeat the same value $4$ times. Hence the derivative can have

$$4\cdot3+3$$ roots and the function

$$15$$ extrema.

Title: local extrema of a Continuous function


问:
The solution of a partial differential equation( I can't work out the equation)
let $ y=f(x,t) $ and $F(x,y,t)=0  $  The function is determined
$t=t(x,y)$ what is $\dfrac{dy}{dx}$
This is original question.

答:
This form is called a Total derivative.
$\require{cancel}$
Given $$dy=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial t}dt$$

$$dy=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial t}\left(\dfrac{\partial t}{\partial x}dx+\dfrac{\partial t}{\partial y}dy\right)$$

$$dy=\dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\cancel{\partial t}}\dfrac{\cancel{\partial t}}{\partial x}dx + \dfrac{\partial f}{\cancel{\partial t}}\dfrac{\cancel{\partial t}}{\partial y}dy$$

$$dy=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy$$

$$dy=2\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy$$

$$dy\left(1-\dfrac{\partial f}{\partial y}\right)=2\dfrac{\partial f}{\partial x}dx$$

$$dy=2\dfrac{\dfrac{\partial f}{\partial x}dx}{1-\dfrac{\partial f}{\partial y}}$$

$$dy=2\dfrac{\dfrac{\partial f}{\partial x}dx}{\dfrac{\partial y-\partial f}{\partial y}}$$

$$dy=2\dfrac{\partial f}{\partial x}dx\cdot\dfrac{\partial y}{\partial y-\partial f}$$

From an 'engineering' point of view simply divide $dy$ with $dx$.

$$\dfrac{dy}{dx}=2\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial y}{\partial y-\partial f}$$

Of course what you write down has to make sense mathematically. So you can only derive this way if $t$ depends on $x$.


问:
Value of integration with respect to Dirac measure
The Dirac measure is defined by $$\delta_x(A)= \begin{cases} 1 &\text{if $x \in A$}\\ 0 &\text{if $x \notin A$}\\ \end{cases}$$ Let $f:X\rightarrow \mathbb{R}$ be a function.

Then we will have: $\int \limits_{X} f \, d\delta_x=f(x)$ $(*)$?

I'm confused about the notation $x$ in $f(x)$ in the $(*)$ equation: which value of $x$ that we should choose while we are computing the integral over the set $X$ which consists of many value of $x$.

What is the role of the set $A$ (defined in the Dirac measure) in the computation of the integration $\int \limits_{X} f \, d\delta_x$

Thank you very much for your explanation!

答:
In general, for a measure $\mu$, one can find the notations
$$
\langle \mu,f\rangle = ∫  f\,\mu = ∫  f(x)\,\mu(\mathrm d x) = \int f\,\mathrm d\mu = \int f(x)\,\mathrm d\mu(x),
$$
to denote the integral of a function $f$ with respect to the measure $\mu$. Remark that the $x$ variable is a dummy variable, one can choose whatever name, such as
$$
∫  f\,\mu = \int f(y)\,\mu(\mathrm d y) = \int f(x)\,\mu(\mathrm d x).
$$

In your formula, you first fix a $x\in X$ that has nothing to do with the integral. Then you define the measure $μ = \delta_x$. Then you look at the integral with respect to this measure (written in your case with the 3rd convention of my first equation)
$$
\int_X f \,\mathrm d\delta_x = ∫_X f(y)\,\delta_x(\mathrm d y) = f(x).
$$

A (real-valued) measure can be defined in two equivalent ways thanks to Riesz representation theorem:

A functional assigning to each measurable set a value: for every set $A\subset X$, $μ(A)\in\mathbb R$.

A functional assigning to each continuous function a value: for every function $f$, $\mu(f) = ∫_X f\,\mu$.

A way to understand how to go from one to the other point of view is to remark that if you have a sequence of continuous functions $f_n$ converging to the characteristic function of a set $A⊂X$, then
$$
\int_X f_n(x)\,\mu(\mathrm d x) \to \int_X \mathbf 1_{A}(x) \, \mu(\mathrm d x) = \int_A \mu(\mathrm d x) = \mu(A),
$$
while, in the other direction, if you are given a measure $\mu$ as a functional on every set $A⊂ X$, you can build the integral with respect to this measure in a similar way a one build the classical integral with respect to the Lebesgue measure. So, one assigns the value
$$
\int_X \mathbf 1_{A}(x) \, \mu(\mathrm d x) := \mu(A),
$$
and then one can approximate measurable functions by sum of characteristic functions of sets.

In the case of the Dirac delta, one has the equivalence between your definition $\delta_a(A) = 1$ if $a∈A$ and $0$ if not, and the integral definition
$$
∫ f \,\delta_a = f(a).
$$

Remark: I do not recommend the notations $\int f\,\mathrm d \mu = \int f(x)\,\mathrm d \mu(x)$ which are incoherent with the Stieltjes integral notation, in which one writes $μ = \mathrm d g$ for a function $g$ that is the derivative of the measure of $\mu$, so
$$
∫ f(x)\,\mu(\mathrm d x) = ∫ f(x)\,\mathrm dg(x).
$$


问:
Finding the particular solution of a differential equation using at least three different methods.
Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.

I have done the first two. Can somebody help me with the third method.

Method 1: Homogenous Equation

Let $y=vx; dy=vdx+xdv$

$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$

$(x^2+2x^2v^2)dx-4x^3vdv=0$

$(1+2v^2)dx-4xvdv=0$

$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$

$\ln{x}-\ln{(2v^2+1)}=C$

$\ln{(\frac{x}{2v^2+1})}=\ln{C}$

$\frac{x}{2v^2+1}=\frac{1}{C}$

$C=3$

$3x=2v^2+1$

$3x^3=2y^2+x^2$

$2y^2=x^2(3x-1)$

The particular solution by method 1 is $2y^2=x^2(3x-1)$.

Method 2: Bernoulli Equation

$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$

$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$

Let $v=y^2; dv=2ydy$

$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$

$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$

$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$

$2vx^{-3}=-x^{-1}+C^{-1}$

$2y^2x^{-3}+x^{-1}=C^{-1}$

$C=\frac{1}{3}$

$2y^2+x^2=3x^3$

$2y^2=x^2(3x-1)$

The particular solution by method 2 is also $2y^2=x^2(3x-1)$.

答:
Compare
$$(x^2+6y^2)dx-4xydy=0~~~~(1)$$
with $$Mdx+Ndy=0 \implies \frac{\partial M}{\partial y}=12y,~~ \frac{\partial  N}{dx}=-4y $$ It's integrating factor is
$$I=\exp \left(\int \frac{12y +4y}{-4xy}\right)=x^{-4}$$
Multiplying (1) with $x^{-4}$,(1) becomes an exact ODE:
$$(x^{-2}+6x^{-4} y^2) dx- 4x^{-3} ydy==0$$
It's solution is
$$\int (x^{-2}+6x^{-4}y^2) dx~~~ \text{(treat y as constant)}=C$$
$$-x^{-1}-2x^{-3}y^{2}=C$$
By the condition $y(1)=1$, we get $C=-3$
Finally $$2y^2=3x^3-x^2.$$


问:
Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $
Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $

this isn't really a computable integral, But the only idea I have in mind is that:

derivative from $x=0$ to $x=1$ of:  $\frac{d}{dx}\left({e^x}-{x})\right) = e -1 $, but I would still have to compute $\frac{e^x}{x^{2}+1}$,

because I'm not sure the derivatives will be sufficient in order to prove the inequality.

答:
The most direct approach to this question is Triangle Inequality for Integrals. Below I provided a helpful link.
https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals

Let us start by applying the inequality: $$\Biggl \lvert \int_{0}^{1} \frac{e^x}{x^2+1}dx\Biggl\rvert \space \leq \space \int_{0}^{1} \Biggl| \frac{e^x}{x^2+1}\Biggl|dx$$ Since we know that the function $\frac{e^x}{x^2+1}$ is positive-valued on the real line, we have: $$\int_{0}^{1} \Biggl| \frac{e^x}{x^2+1}\Biggl|dx = \int_{0}^{1} \frac{e^x}{x^2+1}dx$$ By the help of a simple algebraic fact $\frac{1}{1+x^2} \leq 1 \space \forall \space x \in \Bbb R$, it is easy to deduce that: $$\int_{0}^{1} \frac{e^x}{x^2+1}dx \space  \leq \space \int_{0}^{1} e^x dx=[e^x]^{1}_{0}=e-1 $$

In conclusion: $$\bbox[yellow] {\int_{0}^{1} \frac{e^x}{x^2+1} dx \space \leq \space e-1 }$$


答:
We can check if the ODE is exact:
$$\Delta=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=\frac{\partial}{\partial y}(x^2+6y^2)-\frac{\partial}{\partial x}(-4xy)=16y\ne0.$$
So it's not exact, but as
$$\frac{\Delta}{N}=\frac{16y}{-4xy}=-\frac{4}{x}=f(x),$$
we can get an integrating factor with the form
$$\mu(x)=\exp\left(\int f(x)\mathrm{d}x\right)=\exp\left(\int-\frac{4}{x}\mathrm{d} x\right)=e^{-4\log x}=x^{-4}.$$
Now, let's apply our IF, multiplying both sides by it:
$$x^{-4}(x^2+6y^2)\mathrm{d}x-4x^{-3}y\,\mathrm{d}y=0,$$
Let's check if the new ODE is exact (it must be, but just in case),
$$\Delta'=\frac{\partial M'}{\partial y}-\frac{\partial N'}{\partial x}=\frac{\partial}{\partial y}\bigl[x^{-4}(x^2+6y^2)\bigr]-\frac{\partial}{\partial x}(-4x^{-3}y)=0.$$

So, we are looking for a function $F(x,y)$ that satisfy
\begin{align}
\frac{\partial F}{\partial x} & =x^{-4}(x^2+6y^2);\\
\frac{\partial F}{\partial y} & =-4x^{-3}y.
\end{align}
From the second condition,
$$F(x,y)=\int-4x^{-3}y\,\mathrm{d}y+\varphi(x)=-2x^{-3}y^2+\varphi(x).$$
Now, we can plug $F$ in the first condition:
\begin{align*}
\frac{\partial}{\partial x}\bigl[-2x^{-3}y^2+\varphi(x)\bigr]=x^{-4}(x^2+6y^2) & \Longrightarrow6x^{-4}y^2+\varphi'(x)=x^{-4}(x^2+6y^2)\Longrightarrow\\
& \Longrightarrow\varphi'(x)=x^{-2}\Longrightarrow\varphi(x)=-\frac{1}{x}.
\end{align*}

So, the solution takes the form $F(x,y)=C$, and is
$$-2x^{-3}y^2-\frac{1}{x}=C\Longrightarrow2y^2=-Cx^3-x^2.$$

If $y=1$ when $x=1$,
$$2=-C-1\Longrightarrow C=-3,$$
and we we have
$$2y^2=3x^3-x^2.$$

Title: Finding the particular solution of a differential equation using at least three different methods.

Body:
Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.

I have done the first two. Can somebody help me with the third method.

Method 1: Homogenous Equation

Let $y=vx; dy=vdx+xdv$

$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$

$(x^2+2x^2v^2)dx-4x^3vdv=0$

$(1+2v^2)dx-4xvdv=0$

$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$

$\ln{x}-\ln{(2v^2+1)}=C$

$\ln{(\frac{x}{2v^2+1})}=\ln{C}$

$\frac{x}{2v^2+1}=\frac{1}{C}$

$C=3$

$3x=2v^2+1$

$3x^3=2y^2+x^2$

$2y^2=x^2(3x-1)$

The particular solution by method 1 is $2y^2=x^2(3x-1)$.

Method 2: Bernoulli Equation

$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$

$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$

Let $v=y^2; dv=2ydy$

$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$

$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$

$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$

$2vx^{-3}=-x^{-1}+C^{-1}$

$2y^2x^{-3}+x^{-1}=C^{-1}$

$C=\frac{1}{3}$

$2y^2+x^2=3x^3$

$2y^2=x^2(3x-1)$

The particular solution by method 2 is also $2y^2=x^2(3x-1)$.


问:
Local expression of differential forms of a vector bundle
Let $ p: M \to B $ be a smooth vector  bundle over B.

I'm wondering how to express  a differential form on M  in local coordinates. I'm aware of the local expression of a differential form on a arbitrary smooth manifold, but I'm not sure if in the case of vector bundle over B there will be something special in the local expression of a differential form by using the base space B for example .

Thank you!

答:
Since I can't comment, I will write my comment as an answer.

Vector bundle also has a smooth manifold structure where the local trivialisation defines a smooth structure. Now try to write 1-form in terms of local coordinates.

For example, Suppose $(U,(x_1, x_2, \cdots, x_n))$ be a smooth chart containing $p \in M$. Then $\{dx_i, 1 \leq i \leq n\}$ denotes the dual basis of cotangent spaces ${T^*}_p M$. Since 1- form is a covector field ( i.e. it assigns a covector at each point $p \in M$), so we can express our 1-form $\alpha = \Sigma_{i=1}^{i=n} a_i(p) dx_i|_p $ for each $p \in U$.


问:
Conditional Variance
X is a random variable with mean 100 and standard deviation 30. Y is a random variable with mean of 50 and standard deviation of 25. The correlation between X and Y is 0.5. What is Var(Y|X)?</p>

答:
There is no unique answer to your question without some additional information on the joint distribution of $(X,Y)$. Here are two examples of couples $(X,Y)$ which both fulfill all the conditions in your post and such that $\mathrm{Var}(Y|X)$ is different.

Your hypothesis are that $E(X)=x$, $\mathrm{Var}(X)=u$, $E(Y)=y$, $\mathrm{Var}(Y)=v$ and $\mathrm{Corr}(X,Y)=c$, with $x=100$, $u=900$, $y=50$, $v=625$ and $c=1/2$. In both situations, an auxiliary random variable $Z$ is introduced, independent on $X$ and such that $E(Z)=z$ and $\mathrm{Var}(Z)=w$, where the values of $z$ and $w$ will be specified later on.

First example: Assume that $Y=aX+Z$ for a given $a$, then $y=ax+z$, $v=a^2u+w$ and $c=a\sqrt{u/v}$. The third equation yields $a=c\sqrt{v/u}$, the second one yields $w=(1-c^2)v$, the first one yields $z=y-cx\sqrt{v/u}$ and this $(a,z,w)$ yields an admissible couple $(X,Y)$. Furthermore,
$$
\mathrm{Var}(Y|X)=\mathrm{Var}(Z)=w=(1-c^2)v.
$$
Second example: Let $S$ denote a random variable, independent on $X$ and $Z$, with $P(S=1)=p$ and $P(S=0)=q=1-p$. Assume that $Y=SX+(1-S)Z$. In words, $Y=X$ or $Y=Z$ with probability $p$ and $q$ respectively. Then $y=px+qz$, $v=pu+qw$ and $c=p\sqrt{u/v}$. Hence $p=c\sqrt{v/u}$ is imposed, $w=(v-pu)/q$ as well (and $w>0$ for the suggested numerical values) and $z=(y-px)/q$ as well. Furthermore,
$$
\mathrm{Var}(Y|X)=pq(X-z)^2+qw.
$$

Title: Conditional Variance


问:
What is the relationship between $(u\times v)\times w$ and $u\times(v\times w)$?
Given three vectors $u$, $v$, and $w$, $(u\times v)\times w\neq u\times(v\times w)$. This has been a stated fact in my recent class. But what is the ultimate relationship between them? I would presume that one is a scalar multiple of the other?</p>

答:
By vector triple product expansion:
$$\newcommand{\u}{\boldsymbol{u}}\newcommand{\v}{\boldsymbol{v}}\newcommand{\w}{\boldsymbol{w}}\newcommand{\n}{\boldsymbol{n}}\newcommand{\s}{\boldsymbol{s}}
(\u\times \v)\times \w - \u\times(\v\times \w) = (\u\cdot \v)\w - (\w\cdot \v)\u = \v\times(\w\times \u).
$$

Here I give two pure geometrical proof of the Jacobi identity based on the right hand rule of the cross product. You could draw a picture yourself to see the geometrical relation of them more clearly.

Without loss of generality, consider the normalized version where $\|\u\| = \|\v\| = \|\w\| = 1$.

Geometrical proof 1:  $\u\times \v$ is a vector perpendicular to the plane $S_{\u\v}$ spanned by $\u$ and $\v$, i.e., parallel to the unit normal vector $n$ of $S_{\u\v}$ pointing the same direction. We assume the angle between $\w$ and this plane is acute. Now cross product with $\w$:
$$
(\u\times \v)\times \w = \sin\alpha \,\n \times \w = \sin\alpha\,\n \times (\mathrm{Proj}_{S_{\u\v}} \w),\tag{1}
$$
where $\alpha$ is the angle between $\u$ and $\v$.
The projection $\w_{\perp}$ can be decomposed into two parts if $\u$ and $\v$ are not parallel:
$$
\w_{\perp}:= \mathrm{Proj}_{S_{\u\v}} \w = a \u_{\perp} + b\v,
$$
where $$\u_{\perp} = \u - (\u\cdot \v)\v = \u - \cos\alpha \,\v\tag{2}$$ is perpendicular to $\v$. Dot product both sides with $\u_{\perp} $ and $\v$:
$$
\w_{\perp} \cdot \u_{\perp}  = a \sin^2 \alpha \quad \text{and}\quad  \w_{\perp} \cdot \v = b,
$$
and
$$
\w_{\perp} \cdot \u_{\perp} = \|\w_{\perp}\| \|\u_{\perp}\|\cos(\alpha_u+\frac{\pi}{2}-\alpha) = \sin\theta \sin\alpha \cos(\frac{\pi}{2}-\alpha_v)
$$
, and $$\quad  \w_{\perp} \cdot \v=\|\w_{\perp}\| \|\v\| \cos\alpha_v = \sin\theta \cos\alpha_v,
$$
where $\theta$ is the angle between $\w$ and $\n$, $\alpha_u$ is the angle between $\u$ and $\w_{\perp}$, and $\alpha_v$ is the angle between $\v$ and $\w_{\perp}$, such that $$\alpha_u + \alpha_v = \alpha.\tag{3}$$
Now $$
\w_{\perp}=\frac{\sin\theta\sin(\alpha_v)}{\sin\alpha}\u_{\perp} + \sin\theta \cos\alpha_v\,\v ,
$$
Notice $\n$ cross product with a vector on the plane $S_{\u\v}$ is rotating the vector by $\pi/2$ counterclockwisely on the plane with respect to the normal $\n$:
$$
\n\times \u_{\perp} = \|\u_{\perp}\|\v = \sin\alpha\,\v
$$
and
$$
\n\times \v= -\u_{\perp}/\|\u_{\perp}\| = -\u_{\perp}/\sin\alpha
$$
Finally by (2) and (3), (1) becomes (4):
$$
\begin{aligned}
(\u\times \v)\times \w =&\sin\alpha\sin\theta\sin\alpha_v\,\v - \sin\theta \cos\alpha_v\, \u + \sin\theta \cos\alpha_v\cos\alpha \,\v
\\
=&-\sin\theta \cos\alpha_v\, \u+ \sin\theta \cos(\alpha-\alpha_v)\,\v = -\sin\theta \cos\alpha_v\, \u + \sin\theta \cos\alpha_u \,\v.
\end{aligned}$$
For $\theta$ is the angle between $\w$ and the normal of $\u\v$-plane, drawing a perpendicular line from the tip of $\w$ to $\v$ we can see
$$
\sin\theta \cos\alpha_v = \sin\theta\,  \w_{\perp}\cdot \v = (\w\cdot \v),
$$
similarly $\sin\theta \cos\alpha_u = (\w\cdot \u)$, (4) is
$$
(\u\times \v)\times \w = (\w\cdot \u)\v - (\w\cdot \v)\u.
$$

Geometrical proof 2: Let
$$
\s = (\u\times \v)\times \w,
$$
It is easily to see from above argument, $\s$ lies in the plane of $\u\v$, while isperpendicular to $\w$. Hence we let again:
$$
\s = a\u+b\v,
$$
Dot product with $\w$ we see that:
$$
\s\cdot \w = 0 = a(\u\cdot \w)+b(\v\cdot \w),
$$
Hence
$$
a = c(\v\cdot \w)  \quad \text{and}\quad b = -c(\u\cdot \w),
$$
hence:
$$
\s = c(\v\cdot \w)\u-c(\u\cdot \w)\v \tag{5}
$$
Now cross product with $\u$ on both sides of (5), left hand side is:
$$\begin{aligned}
\s\times \u &=\big( (\u\times \v)\times \w \big)\times \u = (\sin\alpha\,\n\times \w)\times \u
\\
&=\sin\alpha( \n\times\w_{\perp})\times \u
\end{aligned}
$$
From above we know that $ \n\times\w_{\perp}$ is the counterclockwise rotation of $\w_{\perp}$ by $\pi/2$ on this plane with respect the normal $\n$, so the angle between $\n\times\w_{\perp}$ and $\u$ is $\pi/2 + \alpha_u$ (the definition see above (3)), and
$$
\s\times \u  = -\sin\alpha \sin\theta \sin(\frac{\pi}{2} + \alpha_u)\n = -\sin\alpha \sin\theta \cos(\alpha_u)\,\n.
$$
Right hand side of (5) is:
$$
-c(\u\cdot \w)\v \times \u = c(\u\cdot \w)\sin\alpha\,\n = c\sin\theta \cos(\alpha_u)\sin\alpha \,\n. $$
Therefore $c = -1$, and (5) is:
$$
(\u\times \v)\times \w = (\w\cdot \u)\v - (\w\cdot \v)\u.
$$

Now for any three vectors in general we can use the formula for the normalized vectors and
$$
\left(\frac{\u}{\|\u\|}\times \frac{\v}{\|\v\|}\right)\times \frac{\w}{\|\w\|} = \left(\frac{\w}{\|\w\|} \cdot \frac{\u}{\|\u\|}\right)\frac{\v}{\|\v\|} - \left(\frac{\w}{\|\w\|} \cdot \frac{\v}{\|\v\|}\right)\frac{\u}{\|\u\|},
$$
cancelling their norms would give the result.


问:
Submodule over ring
I am calculating submodules of module $\mathbb{Z}_2[x]/(x^3+x+1)$ over $\mathbb{Z}_2[x]$.

Consider $I=(x^3+x+1)$ ideal generated by $x^3+x+1$.

Elements of $\mathbb{Z}_2[x]/(x^3+x+1)=\{0+I,1+I,x+I,x^2+I,x^2+x+I,x^2+1+I,x+1+I,x^2+x+I\}$

I only get one non trivial submodule which is $\{0+I,1+I\}$, does there any other non trivial submodule?

答:
Recall that, as per third isomorphism theorem, the submodules of $\mathbb Z_2[x]/(x^3+x+1)$ correspond to submodules of $\mathbb Z_2[x]$ containing the submodule $(x^3+x+1)$.

Now, also recall that $Z_2[x]$-submodules of $Z_2[x]$ are nothing else than ideals of $Z_2[x]$, and, being a PID, ideals containing $(x^3+x+1)$ correspond to principal ideals corresponding to divisors of $x^3+x+1$ over $\mathbb Z_2[x]$.

However, $x^3+x+1$ is irreducible over $\mathbb Z_2[x]$, being of degree $3$ and having no zeros. Thus, the only ideals of $Z_2[x]$ containing $(x^3+x+1)$ are $(x^3+x+1)$ and $Z_2[x]$ itself. This means there will be no nontrivial submodules of $\mathbb Z_2[x]/(x^3+x+1)$.

Title: Submodule over ring


问:
Show that the image of a cube is almost a cube

Let $C_r = \left \{ x \in \mathbb{R}^n : |x^i| < r \forall 1 \leq i \leq n \right \}$

and $ g \in C^1(U, \mathbb{R}^n)$ for some open $\left \{ 0 \right \} \subset U$ s.t $dg(\vec{0}) = I, g(\vec{0})=\vec{0}$. and let us choose some $0 < \varepsilon < 1$.

Show that there exists $\delta > 0$ s.t $\forall r < \delta, C_{(1-\varepsilon)r} \subset g(C_{r}) \subset C_{(1+\varepsilon)r}$

I couldn't think of a better title, edits are welcome.

This is what I tried:

$g(0+\triangle x) - g(0) = dg(0)(\triangle x) +o(\triangle x) \implies g(x) = x+o(x) \implies \frac{||g(x) - x||}{||x||} \xrightarrow[x \to 0]{} 0$

So we can choose some $\delta > 0$ s.t if $||x|| < \delta$ then $||g(x)-x|| < ||x||_{\infty} \varepsilon$

Now $x \in C_r \implies |x^i|<r \implies |g(x)^i| < r+ ||x||_{\infty}\varepsilon \leq r+r \varepsilon \implies g(C_{r}) \subset C_{(1+\varepsilon)r}$

But i'm not sure how to show $C_{(1-\varepsilon)r} \subset g(C_{r})$.

Hints appreciated.

Also, does what I did so far seem correct?

答:
Step 1: Definition of multivariate derivative

By definition, $dg(x)$ is the unique matrix which satisfies:
$$
\lim_{h\rightarrow 0}\frac{||g(x+h)-(g(x)+dg(x) h)||_2}{||h||_2}=0
$$
In our case, this simplifies to:
$$
\lim_{h\rightarrow 0}\frac{||g(h)-h||_2}{||h||_2}=0
$$

Step 2: Working with cubes

But since we are working with cubes, not Euclidean balls, we would perfer the "cube-norm" (aka the sup-norm): $||x||_\infty:=\max_{i\in[n]}|x_i|$.
It still works, using the following inequalities (easy to show):
$$
||x||_\infty\leq ||x||_2\leq \sqrt{n}||x||_\infty
$$
We can write:
$$
0=\lim_{h\rightarrow 0}\frac{||g(h)-h||_2}{||h||_2}\geq \lim_{h\rightarrow 0}\frac{||g(h)-h||_\infty}{\sqrt{n}||h||_\infty}\geq 0
$$
So, it follows that the limit is zero, also for the $||\cdot||_\infty$-norm.
Finally, this means that for any $\varepsilon>0$, there exists a $\delta>0$, such that for all $||h||_\infty<\delta$ (in other words, for all $h\in C_\delta$),
$$
\frac{||g(h)-h||_\infty}{||h||_\infty}<\epsilon
$$

Step 3: Triangular inequality

Using the triangular inequality, $||h||_\infty\leq ||g(h)-h||_\infty+||g(h)||_\infty$.
Therefore:
$$
\frac{||h||_\infty}{||h||_\infty}-\frac{||g(h)||_\infty}{||h||_\infty}<\epsilon
$$
Rearranging:
$$
||h||_\infty(1-\epsilon)< ||g(h)||_\infty
$$
Using instead the triangular inequality $||g(h)||_\infty\leq ||g(h)-h||_\infty+||h||_\infty$, we get:
$$
\frac{||g(h)||_\infty}{||h||_\infty}-\frac{||h||_\infty}{||h||_\infty}<\epsilon
$$
Which means:
$$
||g(h)||_\infty\leq ||h||_\infty(1+\epsilon)
$$
To summarize:
$$
||h||_\infty(1-\epsilon)\leq ||g(h)||_\infty\leq ||h||_\infty(1+\epsilon)
$$
This already implies that $g(C_\delta)\subset C_{\delta(1+\epsilon)}$.
However, it is not enough to conclude that $g(C_\delta)$ contains all of $g(C_{\delta(1-\epsilon)})$.

Step 3: Inverse function theorem

By the inverse function theorem, there is some open set $V$ containing $0$, such that $g^{-1}$ exists on $V$ and is continuous.
Henceforth we will assume $C_{\delta(1+\epsilon)}\subset V$ (if not, we make $\delta>0$ smaller until this is true).
Moreover, the inverse function theorem tells us that $dg^{-1}(0)=I^{-1}=I$.
Repeat steps 1-2, to obtain:
$$
||g^{-1}(y)-y||_\infty\leq \epsilon||y||_\infty
$$
This holds for all $||y||_\infty<\gamma$ for some small $\gamma$, we may take $\gamma=\delta$ without loss of generality.

Step 4: Completing the proof

Take any $x\in C_{\delta(1-\epsilon)}$. We show that $x\in g(C_{\delta})$.
Since $g$ is invertible on $C_{\delta}$, this is the same as requiring that $y\in  C_{\delta}$ whenever $g^{-1}(y)\in C_{\delta(1-\epsilon)}$.
Finally:
$$
||y||_\infty \leq ||g^{-1}(y)||_\infty+||g^{-1}(y)-y||_\infty\leq \delta(1-\epsilon)+\epsilon||y||_\infty
$$
Rearranging, we have $||y||_\infty<\delta$. The proof is complete.


问:
calculating the Taylor series remainder of $\ln(2)$
I'm trying to calculate the remainder for the Taylor series expansion of $\ln(2)$, and would greatly appreciate your support.

I've learnt that when expanding the Taylor series:

$f(x) = f(a) + \frac{f'(a)}{1!}(x-a)... + \frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+R_n(x)$

Where $R_n(x)$ is the remainder, s.t. $R_n(x)=\frac{f^{(n)}(a+\theta(x-a))}{n!}(x-a)^n$, given $0 <\theta<1$

My assumption was that using  the series $\ln(1+x)$ and replacing it by $2$ would guide me to a solution.

$\ln(1+x) = x-\frac{1}{2}x^2+\frac{1}{3}x^3...$

Though, that's not the right expansion. How would I correctly expand this, thanks for the help.

答:
Hint:

As $2=\frac1{1/2}$, you can use that
$$\ln 2=-\ln\Bigl(1-\frac12\Bigr),$$
and use the expansion of $\ln(1-x)$.


问:
Determining the number of solutions of the non homogeneous integral equation.
This is a question from a competitive exam. We are given the integral equation:
$$ \phi (x) = \cos(7x) +  \lambda \int_{0}^{\pi} \left[ \cos(x)\cos(t) - 2\sin(x)\sin(t) \right]\phi(t) dt $$ and are asked the number of solutions of the equation depending on the value of lambda.
The options are:

Solution exists for every value of $\lambda$.

There is some $\lambda$ for which solution does not exist.

There is some $\lambda$ for which more than one but finitely many solutions exist.

4.There is $\lambda$ such that infintely many solution exists.

Now after refotmulation it in the form of matrix I have obtained:  $\left| \begin{matrix} 1- \frac{ \lambda \pi}{2} & 0\\ 0 & 1 + \lambda \pi  \end{matrix} \right| =0 $
This gives me the values of $\lambda$. But I can not conclude the number of solutions from here. Any help is highly appreciated.

答:
$$ \phi (x) = \cos(7x) +  \lambda \int_{0}^{\pi} \left[ \cos(x)\cos(t) - 2\sin(x)\sin(t) \right]\phi(t) dt $$
A different approach :
$$ \phi (x) = \cos(7x) +
\lambda \cos(x)\int_{0}^{\pi} \cos(t)\phi(t)dt
-2\lambda \sin(x)\int_{0}^{\pi} \sin(t)\phi(t) dt  $$
The integrals are not function of $x$. They are constant.

Let$\quad \int_{0}^{\pi} \cos(t)\phi(t)dt=A\quad$ and $\quad \int_{0}^{\pi} \sin(t)\phi(t) dt=B$

$$ \phi (x) = \cos(7x) +  \lambda A \cos(x)-2\lambda B \sin(x) $$

$$A=\quad \int_{0}^{\pi} \cos(t)\left(\cos(7t) +  \lambda A \cos(t)-2\lambda B \sin(t) \right)dt=\frac{\pi}{2}\lambda A \quad\implies\quad (1-\frac{\pi}{2}\lambda)A=0$$

$$B=\quad \int_{0}^{\pi} \sin(t)\left(\cos(7t) +  \lambda A \cos(t)-2\lambda B \sin(t) \right)dt=-\pi\lambda B \quad\implies\quad (1+\pi\lambda)B=0$$

In general case (any $\lambda$ ) if $\lambda\neq \frac{2}{\pi}$ and if $\lambda\neq\frac{-1}{\pi}$ this implies $A=0$ and $B=0$

Thus the solution is :
$$\boxed{\Phi(x)=\cos(7x)}$$

The two particular cases $\lambda=\frac{2}{\pi}$ or $\lambda=\frac{-1}{\pi}$ have to be studied separately. Each one will lead to an infinity of solutions with arbitrary $A$ or $B$ respectively.

$ \phi (x) = \cos(7x) + \frac{2}{\pi} A \cos(x)\quad $ if $\lambda= \frac{2}{\pi}$.

$ \phi (x) = \cos(7x) + \frac{2}{\pi} B \sin(x)\quad $ if $\lambda= \frac{-1}{\pi}$.


问:
Is the concept of vector size something relative to the vector space?
The vector $\begin{pmatrix}1 \\ 0\end{pmatrix}$ will always have this size: $\sqrt{1^2+0^2}=\sqrt{1}=1$. But given the basis $\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}$, the coordinate vector $\begin{pmatrix}1 \\ 0\end{pmatrix}$ is pointing to the point $\begin{pmatrix}1 \\ 1\end{pmatrix}$ in the standard basis, so its size can also be calculated this way: $\sqrt{1^2+1^2}=\sqrt{2}$ , so maybe I'm making a silly mistake here but I can't find it out. Could a size such as $1$ in a vector space have a different value (e.g. $\sqrt{2}$) in another vector space? Is it illegal to interpret things in the standard space when we're in a different vector space?

答:
A vector doesn't a priori have a size. Abstractly, a vector $v$ is just an element of a vector space $V$, which is a set with some operations (addition, multiplication by a scalar) and rules about how those operations work (e.g. commutativity).

If you want to talk about vectors with sizes, you are talking about a normed vector space. A normed vector space is just a vector space $V$ with a function $\| \cdot \|_V$ defined on it, called the norm. For any vector $v \in V$, you get a real number $\| v \|_V$, and it is supposed to satisfy the following three rules:

For any $v \in V$, $\|v\|_V \geq 0$.

If $\|v\|_V = 0$, then $v = 0$.

If $\alpha$ is any scalar, then $\|\alpha \cdot v\|_V = |\alpha|\cdot \|v\|_V$

For any $v, w \in V$, $\|v + w\|_V \leq \|v\|_V + \|w\|_V$.

You already know about at least one norm: the standard Euclidean norm on the vector space $\mathbb R^2$, where
$$
\left\|\begin{pmatrix}a\\b\end{pmatrix}\right\|_2 = \sqrt{a^2 + b^2}.
$$
(This 2 in the bottom specifies that this is the Euclidean norm you are talking about.)

You can check that this satisfies all the rules listed above. But you could define lots of other norms on the vector space $\mathbb R^2$; for example, $\|v\|_\infty = \max\{|v_1|, |v_2|\}$, or $\|v\|_{\text{skew}} = \sqrt{(v_1+v_2)^2 + v_1^2}$. In fact that last one is exactly the norm you have calculated by looking at your vectors in a different basis.

In general, if you have a change of basis, this will usually not preserve the norm. (In fact, changes of basis that preserve the norm are quite special.) That is, by changing the basis you get a new norm, and when specifying what you mean by "size" of your vector, you need to be clear about what norm you are using if it isn't clear from context.


问:
Procedure to show martingale property
When one's trying to prove the martingale property (assuming it's adapted and integrable), why does it suffice for a stochastic process $(X_t)_{t\geq0}$ satisfy
$$ \mathbb{E}[(X_t-X_s)\prod_{i=1}^n f_i(X_{s_i})]=0,$$
where $f_i$ are bounded continuous functions and $s_1\leq s_2\leq\dots s_n\leq s<t$.

Is it a simple application of the Monotone Class theorem or is the more/less to it?

答: Martinagle property of $(X_t)$ is equivalent to the condition $E(X_t-X_s)|\mathcal F_s)=0$ for $s <t$. By definition of conditional expectation this is equivalent to the condition $\int_A (X_t-X_s)dP=0$ for all $A \in \mathcal F_s$. By the $\pi-\lambda$ theorem it is enough to verfy this for some class of sets which is closed under finite intersection and which generates $\mathcal F_s$. One such class is sets of the form $X_{s_1}^{-1}(A_1) \cap X_{s_2}^{-1}(A_3)\cap...\cap X_{s_n}^{-1}(A_n)$ where $s_1\leq s_2\leq ...\leq s_n \leq s$ and $A_i$'s are closed sets in $\mathbb R$. I will now leave it to you to show that this is indeed true under the given hypothesis using the fact that for any closed set $C$ there exists a sequence of uniformly bounded continuous functions $f_n$ converging pointwise to $I_C$.

Title: Procedure to show martingale property


答:
You actually can use the expansion $\log (1+x) = x- \frac {x^2}{2}+\frac {x^3}{3}...$, which is the correct one for $\log (1+x) \; \;$ when $\; \; -1 \lt x \le 1$.

To estimate the remainder using the integral form might be easier:

$$\int _0^x \frac {t^n}{1+t} dt \le \int _0^x t^n dt = \frac {x^{n+1}}{n+1} $$

So when $x=1$ (there the expantion holds but not further), we can make the remainder as small as we want it to be by choosing a suitable large $n$.

EDIT: I gave you the solution, but I forgot to explain the process of getting there, silly me...

Remember that $\log (1 + x) = \int_0^x \frac {1}{1+t}$ and also consider the equation:

$$\frac {1}{1+t} =\frac {1 + t - t}{1+t} = 1 + \frac {-t}{1+t} = 1- \frac {t^2 +t - t^2}{t+1} = 1 - \frac {t(t+1) - t^2}{t+1} = 1 - t + \frac {t^2}{t+1}= 1- t + \frac {t^3 + t^2 - t^3}{t+1} = 1 - t + \frac {t^2(t+1) - t^3}{t+1} = 1 - t + t^2 - \frac {t^3}{t+1} = ...$$

Title: calculating the Taylor series remainder of $\ln(2)$

Body:
I'm trying to calculate the remainder for the Taylor series expansion of $\ln(2)$, and would greatly appreciate your support.

I've learnt that when expanding the Taylor series:

$f(x) = f(a) + \frac{f'(a)}{1!}(x-a)... + \frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+R_n(x)$

Where $R_n(x)$ is the remainder, s.t. $R_n(x)=\frac{f^{(n)}(a+\theta(x-a))}{n!}(x-a)^n$, given $0 <\theta<1$

My assumption was that using  the series $\ln(1+x)$ and replacing it by $2$ would guide me to a solution.

$\ln(1+x) = x-\frac{1}{2}x^2+\frac{1}{3}x^3...$

Though, that's not the right expansion. How would I correctly expand this, thanks for the help.


问:
Calculating expected value of unknown random variable
The question:

Micro Insurance Company issued insurance policies to $32$ independent risks. For each policy, the probability of a claim is $1/6$. The benefit amount given that there is a claim has probability density function
  $$ f(y) =
\begin{cases}
2(1-y) & 0<y<1,
\\0 & \text{otherwise}.
\end{cases}$$
  Calculate the expected value of total benefits paid.

My attempt:

I'm not sure on how to define my random variable. Its expected value should sum from 1 to 32, each with probability $\frac{1}{6} \int f(y) dy$, I think.

答:
Let $Y_{n}$ be the amount paid to the $n$-th policyholder assuming
that the claim is made. Let $\mathbb{I}_{n}$ be $0$ when the $n$-th
claim is not made and $1$ otherwise. Then, the total benefits paid
is
$$
X=\sum_{n=1}^{32}\mathbb{I}_{n}Y_{n}.
$$
We need to calculate
$$
\mathbb{E}\left[X\right]=\mathbb{E}\left[\sum_{n=1}^{32}\mathbb{I}_{n}Y_{n}\right]=\sum_{n=1}^{32}\frac{1}{6}\int_{0}^{1}yf\left(y\right)dy.
$$
I think you can do the rest yourself.</p>


问:
Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$
Is there a way to show that (with integer $n>2$) $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
I tried to figure a way with derivative or by looking how both side are growing, but I have some trouble with the ceiling function.

答:
With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)

$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$
If we choose $a=15$ the above is true for $n>41$

Now (growth of $x^a$ vs $a^x$), $$\frac 1{n^a}>\frac{1}{(\frac{3}{2})^n}$$
is true for $n>\frac{-aW_{-1}(\frac{-\log\frac{3}{2}}{a})}{log(\frac{3}{2})}$ or $n>196$ with chosen $a=15$ ($W$ is the productlog), and we also have (using $x>log(1+x)$)
$$\frac{1}{(\frac{3}{2})^n}>\log(1+\frac{1}{(\frac{3}{2})^n})=\log(1+\frac{2^n}{3^n})$$
So for $n>196$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})>\log(1+\frac{2^n}{3^n})$$
or
$$\frac{2^{n+k}}{3^n}>1+\frac{2^n}{3^n}$$
$$2^{\lceil n \log_23\rceil}-3^n>2^n$$

Similarly we have
$$|n \log3-(n+k-1) \log2|>\frac 1{(n+k-1)^{13.3}}>\frac1{(n \log_23)^{13.3}}>\frac 1{n^a}$$
which also holds for $a=15$, and additionally we have:
$$\frac{1}{(\frac{3}{2})^n}=\frac{1}{2^{n\log_2\frac{3}{2}}}>\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
So for $n>196$ we have
$$|n \log3-(n+k-1) \log2|=\log(\frac{2\cdot 3^n}{2^{n+k}})>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
or
$$\frac{2\cdot 3^n}{2^{n+k}}>1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}$$
$$2\cdot 3^n>2^{n+k}+2^n$$

$$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$

This leads to ($n>196$):

$$\begin{array}{|c|}\hline 2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n\\\hline\end{array}$$
and with manual checking, the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{1,2\}$


问:
Why cannot the vector $v = (1,2)$ span entire $\mathbb{R}^2$?
So in my knowledge $\mathbb{R}^2$ can generate points and lines.

So the span of vector $v$ would be $\text{span}(v) = c(1,2)$.This would generate all the lines in $\mathbb{R}^2$ for different real values of $c$.

So shouldn't this span the entire Vector Space $\mathbb{R}^2$?

Also in some sources I read,it says the the vector $v$ only spans the line $y = -x/2$ and not the entire $2D$ space.

Doesnt the entire space of R2 contains only points and lines?

EDIT:I made the necessary edits as suggested and gave explanations when asked in the comments.If there is any other thing I need to do, please tell me before downvoting.I also thank the contributors for helping me out.

答:
Your claim

```<code>So the span of vector v would be span(v) = c(1,2).
This would generate all the lines in R2 for different real values of c.
</code>```

is false.

$\vec v$ will span only the line $(c,2c)$ that is $y=2x$.

Maybe you are confusing with the equation of a line $y=mx$ (which anyway excludes the line $x=0$).

Body:
Your claim

So the span of vector v would be span(v) = c(1,2).
This would generate all the lines in R2 for different real values of c.

is false.

$\vec v$ will span only the line $(c,2c)$ that is $y=2x$.

Maybe you are confusing with the equation of a line $y=mx$ (which anyway excludes the line $x=0$).


问:
An interesting gambling problem in probability
Back in high school we used to play a simple gambling game with cards. Two people would randomly choose a card from a deck of $52$ cards. The person with a higher value card (Ace being the strongest and two the weakest) would then be the winner.

In this seemingly simple scenario is it possible to calculate the probability of a win? That is the first person choosing a higher value card. (I believe a similar line of thought would hold for the second person winning?)

Although appearing quite straightforward at first glance I couldn't figure out how to even get started.

Any help or ideas would be appreciated. Thank you!

答:
You can make it even simpler by looking at it in the following way:

For any card drawn by the first person, there is a Pr of $\frac 3 {51}$ for a draw, hence a Pr of $\frac{48}{51}$ that it results in a win for one or the other.

In two randomly drawn cards, each will have an equal probability of being of higher value,

thus P(first person wins)= P(second person wins)$=\left(\frac1 2 \cdot \frac{48}{51}\right) = \frac{24}{51} = \frac8 {17}$


问:
How many zeros does the function $f(z)=z^3+z^2+4z+1$ have in the first quadrant?

How many zeros does the function $f(z)=z^3+z^2+4z+1$ have in the first quadrant?

Using Rouche's theorem with the function $h(z)=1,g(z)=z^3+z^2+4z$, I've been able to show that all the roots are in $B_3(0)$.

Is it possible to show that they are not in $B_3(0)\cap\{Re(z),Im(z)\ge0\}$ using Rouche's theorem?

答:
Since $f'(z)>0$ for all real $z$ the function has exactly one real root. Calculating the value of the function at $z=0$ and $z=-1$ one concludes that the real root satisfies the inequality $-1<z_1<0$. Further since:
$$
z_1+z_2+z_3=-1
$$
one finds that $\Re(z_2)=\Re(z_3)<0$. Therefore the polynomial has no roots in the first quadrant.


问:
How to find inverse of the function $f(x)=\sin(x)\ln(x)$
My friend asked me to solve it, but I can't.

If $f(x)=\sin(x)\ln(x)$, what is $f^{-1}(x)$?

I have no idea how to find the solution. I try to find

$$\frac{dx}{dy}=\frac{1}{\frac{\sin(x)}{x}+\ln(x)\cos(x)}$$

and try to solve it for $x$ by some replacing and other things, but I failed.

Can anyone help? Thanks to all.

答: I think you should also note that generally $f(x)=g(x)\sin(x)$ cannot have an inverse if $g$ itself is not a periodic function. Let's look at your example of $\ln(x)\sin(x)$. (I'm using the sign $<$ to represent greater than or equal to) Since for $h(x)=\sin(x)$, $-1<\sin(x)<1$, then for $f(x)=\ln(x)\sin(x)$ we have $-\ln(x)<\sin(x)<\ln(x)$, and since $\ln(x)$ is increasing monotonically, the sinusoid will keep growing and it will not repeat.


答: What is the value of $c$ for which $c\cdot (1,2)$ is equal to $(1,1)$?

Title: Why cannot the vector $v = (1,2)$ span entire $\mathbb{R}^2$?

Body:
So in my knowledge $\mathbb{R}^2$ can generate points and lines.

So the span of vector $v$ would be $\text{span}(v) = c(1,2)$.This would generate all the lines in $\mathbb{R}^2$ for different real values of $c$.

So shouldn't this span the entire Vector Space $\mathbb{R}^2$?

Also in some sources I read,it says the the vector $v$ only spans the line $y = -x/2$ and not the entire $2D$ space.

Doesnt the entire space of R2 contains only points and lines?

EDIT:I made the necessary edits as suggested and gave explanations when asked in the comments.If there is any other thing I need to do, please tell me before downvoting.I also thank the contributors for helping me out.


问:
$\lim_{x\to +\infty} e^x-x$
I know that $\displaystyle\lim_{x\to +\infty} e^x-x=+\infty$,

$e^x$ reaches infinity much faster than $x$, and it is clear from the graph of the function that this is the case.

But I was wondering, why? What is the rule to calculate this limit, I mean because $\displaystyle\lim_{x\to +\infty} e^x-x=\infty-\infty$ which is an indeterminate form.

答: Note that $e^x-x=e^x\left(1-\frac x{e^x}\right)$. So, and since $\lim_{x\to\infty}e^x=\infty$ and $\lim_{x\to\infty}\frac x{e^x}=0$, $\lim_{x\to\infty}e^x-x=\infty$.

Title: $\lim_{x\to +\infty} e^x-x$


答:
You have (at least) two ways of going about this.

You could try to "factor out" a part of the function that goes to infinity, while making sure that what is left does not go to $0$. For example, you can factor out $e^x$ to get $$e^x-x = e^x(1 - xe^{-x})$$ or you could factor out $x$, and get $$e^x-x =x\left(\frac{e^x}{x} - 1\right)$$
in both cases, you get something that grows beyond bounds times something that does not converge to zero (in the second case, actually, you get something that grows beyond bounds times something that also grows beyond bonds)

Alternatively, you could use the fact that $e^x$ grows, in the limit, faster than any polynomial, and just estimate $e^x - x \geq p(x) - x$. Picking the right polynomial (i.e., anything that is at least a quadratic function) will mean that $p(x)-x$ grows beyond bounds.

Note that both methods above work not only to solve your particular problem, but a more general one, which you can try to solve for practice:

If $p$ is any polynomial, then $$\lim_{x\to\infty} e^x - p(x) = \infty.$$

Title: $\lim_{x\to +\infty} e^x-x$

Body:
I know that $\displaystyle\lim_{x\to +\infty} e^x-x=+\infty$,

$e^x$ reaches infinity much faster than $x$, and it is clear from the graph of the function that this is the case.

But I was wondering, why? What is the rule to calculate this limit, I mean because $\displaystyle\lim_{x\to +\infty} e^x-x=\infty-\infty$ which is an indeterminate form.


问:
Can we 'square root' a principal ideal?
I was doing an exercise in Algebraic Number Theory and I encountered a problem.

Suppose I have a number field, say $K$, and its ring of integers, say ${O}_K$ such that $\sqrt{d}\in O_K$ for some non-square $d$.

Suppose now I have an ideal in $O_K$, say $I$. I do not know if it is principal or not but the only information I have is that $I^2=(d)$ (so $I^2$ is principal). Can I therefore deduce that $I=(\sqrt{d})$?

It is very tempting to just say $(\sqrt{d})(\sqrt{d})=(d)=I^2$ and thus $I=(\sqrt{d})$ but I am not sure if that is allowed?

I guess I could have also started using that $O_K$ is a Dedekind domain or maybe that any ideal $I$ is generated by at most two elements but I am not sure if that is making it worse?

I hope I explained my question sufficiently clearly, many thanks in advance!

答: Yes, that is true. Use that ideals factor uniquely into prime ideals in Dedekind domains. We get $I^2 = \mathfrak{p}_1^{2e_1} \dots \mathfrak{p}_k^{2e_k}$ for $e_1, \dots, e_k \in \mathbb{N}$. Moreover, say $(\sqrt{d}) = \mathfrak{q}_1^{d_1} \dots q_{\ell}^{d_{\ell}}$. So we get $$ \mathfrak{q}_1^{2d_1} \dots \mathfrak{q}_{\ell}^{2d_{\ell}} = (\sqrt{d})^2 = (d) = I^2 = \mathfrak{p}_1^{2e_1} \dots \mathfrak{p}_k^{2e_k}.$$ Comparing primes yields $I = (\sqrt{d})$.

Title: Can we 'square root' a principal ideal?


Title: $\lim_{x\to +\infty} e^x-x$

Body:
Note that : $e^x= 1+x +{x^2}/2! +x^3/3!+...x^k/k!+...$

We have $e^x-x=1+x^2/2!+x^3/3!+...x^k/k!+...\to \infty$ as $x\to \infty$


问:
finite additivity condition
The definition of a probability measure $ P $ requires countable aditivity: $ P \left( \bigcup_{n = 1}^\infty A_n\right) = \sum_{n = 1}^\infty P (A_n)$ whenever $ A_1, A_2, \ldots $ is a sequence of disjoint sets in the relevant sigma algebra. I don't understand the proof of the countable finite aditivity property:
$ P \left( \bigcup_{n = 1}^N A_n\right) = \sum_{n = 1}^N P (A_n)$. If someone could explain that to me that would be terrific.

Thanks

答:
The definition of countable additivity says that $P ( \bigcup_{n = 1}^{\infty} A_{n}) = \sum_{n = 1}^{\infty} P ( A_{n} )$ whenever $ A_{1}, A_{2}, \dots $ is a sequence of disjoint sets in the relevant sigma algebra.

Assuming countable addivity, let's prove finite additivity, which says that $P ( \bigcup_{n = 1}^{N} A_{n}) = \sum_{n = 1}^{N} P ( A_{n} )$ whenever $ A_{1}, A_{2}, \dots A_{N} $ is a sequence of disjoint sets in the relevant sigma algebra. Given $ A_{1}, A_{2}, \dots A_{N} $, we note that $$\bigcup_{n=1}^{N}A_{n} = A_{1} \cup A_{2} \cup \cdots \cup A_{N} = A_{1} \cup A_{2} \cup \cdots \cup A_{N} \cup \emptyset \cup \emptyset \cup \cdots.$$

The point of including all those empty sets in the union above is to obtain an infinite sequence of disjoint sets (since the empty set is disjoint from everything) so we can apply countable additivity. Doing so yields
$$
\begin{align*}
P (A_{1} \cup A_{2} \cup \cdots \cup A_{N}) &= P(A_{1} \cup A_{2} \cup \cdots \cup A_{N} \cup \emptyset \cup \emptyset \cup \cdots) \\
&= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}) + P(\emptyset) + P(\emptyset) + \cdots \\
&= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}) + 0 + 0 + \cdots \\
&= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}).
\end{align*}
$$

Since the sequence $A_{1}, A_{2}, \ldots, A_{N}$ was arbitary, we're done.

Title: finite additivity condition


问:
Mostowski collapse of generic extensions of countable elementary submodels
Let $\lambda$ be a large enough regular cardinal, $\mathbb{P}\in M\prec H_\lambda$, with $M$ countable and $\mathbb{P}$ proper. Let $G$ be $\mathbb P$-generic over the ground model $V$. Let $\pi:M\cong \bar M$ be the Mostwoski collapse. Write $\bar{\mathbb{P}}=\pi(\mathbb{P})$, $\bar G=\pi''G$. Does it follow that
$$
\overline{M[G]}=\bar{M}[\bar G]?
$$
Here, $M[G]=\{x_G:x\in M\}$, where $x_G$ is the interpretation of $x$ according to $G$, in the usual sense, and $\overline{M[G]}$ is the Mostowski collapse of $M[G]$. The RHS makes sense because $\bar G$ is $\bar{\mathbb P}$-generic over $\bar M$, by elementarity.

I've tried lifting $\pi$ to an isomorphism $M[G]\to \bar M[\bar G]$. This would be enough, because $\bar M[\bar G]$ is transitive. The natural definition would be $\dot x_G\mapsto (\pi(\dot x))_{\bar G}$, where $\dot x\in M$. But, one needs to verify that this is well defined:

Let $\dot x,\dot y\in M$ with $\dot x_G=\dot y_G$. By the Truth Lemma, there is some $p\in G$ with $p\Vdash \dot x=\dot y$. If $p$ were to belong to $M$; then we would be done because we can apply $\pi$ to the statement $M\models (p\Vdash \dot x=\dot y)$, but this need not be the case. One can say
$$
H_\lambda \models \exists q\in \mathbb{P}(q\Vdash \dot x=\dot y)
$$
and so there is some $q\in M$ with $q\Vdash \dot x=\dot y$, but $q$ might not be in $G$. Properness seems highly relevant here, since we want an object in $M\cap G$, but I haven't been able to produce it.

If $p\in G$ and $p\Vdash \dot x=\dot y$, then
$$
D:=\{q\le p:q\Vdash M\cap \dot G\neq\emptyset\}\in V
$$
is dense below $p$ because $\mathbb{P}$ is proper. Therefore, $G\cap D\neq\emptyset$ by genericity, say $q\in G\cap D$. Then $q\Vdash M\cap \dot G\neq \emptyset$, so there is some $r\in M\cap G$. But now we don't necessarily know that $r\Vdash \dot x =\dot y$.

答:
In general, $\overline{M[G]}\neq \bar M[\bar G]$. The problem is that while $G$ meets any dense $D\subseteq\mathbb P$ with $D\in M$, $G$ is not necessarily $M$-generic in the sense that $G$ meets $D$ inside $M$, even if $\mathbb P$ is proper.

Suppose for example that $\mathbb P=\mathrm{Add}(\omega_1, 1)$ is the canonical $\sigma$-closed forcing to force a new subset of $\omega_1$. Let $\alpha=M\cap\omega_1$ and note that this is a countable ordinal. Now consider the condition $p:\alpha\rightarrow 2$ that is constant with value $0$ and suppose $p\in G$. Intuitively, "$p$ makes sure that there is only $M$-trivial information in $G\cap M$".

Claim: $\bar M[\bar G]=\bar{M}\neq\overline{M[G]}$

Proof: Let $H=\{q\in\mathrm{Add}(\omega_1, 1)\mid q$ is constant with value $0\}$. Then $p\in G$ ensures that $G\cap M$ is exactly $H\cap M$ and hence $\bar G=\pi[G\cap M]=\pi[H\cap M]$. However, $H\in M$ and thus $\pi[H\cap M]=\pi(H)\in\bar M$ so that $\bar M=\bar M[\bar G]$. On the other hand $G\in M[G]\setminus M$ implies $\overline{M[G]}\neq \bar M$.$\square$

This problem can easily be mitigated for proper forcing if one assumes that there is an $(M,\mathbb P)$-generic condition $p\in G$. In this case $G\cap M\cap D\neq\emptyset$ for any dense $D\subseteq \mathbb P$ with $D\in M$ and your attempt in the question goes through: For any $\dot x, \dot y\in M^\mathbb P$ the set
$$D=\{p\in\mathbb P\mid p\Vdash\dot x=\dot y\vee p\Vdash\dot x\neq\dot y\}$$
is dense and in $M$. Hence there is $p\in M\cap G$ deciding whether $\dot x=\dot y$. As indeed $\dot x_G=\dot y_G$, we must have $p\Vdash \dot x=\dot y$.


问:
The triangle formed by the $x$-axis, the tangent to $y=1/x$ at a 1st-quadrant point, and the line from the origin to that point is isosceles
Here is Prob. 45, Sec. 3.2, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:

Let $P$ be a point on the first-quadrant part of the curve $y = 1/x$. Show that the triangle determined by the $x$-axis, the tangent at $P$, and the line from $P$ to the origin is isosceles, and find its area.

My Attempt:

Let the point $P$ be given by $P = \left( a, \frac{1}{a} \right)$, where $a > 0$.

Then the line from $P$ to the origin has equation
$$ y = \frac{ \frac{1}{a} }{a} x = \frac{1}{a^2} x. \tag{1} $$

The slope of the tangent line to the curve $y = 1/x$ at the point $P$ is
$$
\left( \frac{d y}{d x} \right)_{x = a} = -\frac{1}{a^2},
$$
and thus the equation of the tangent line is
$$
y = -\frac{1}{a^2} (x-a) + \frac{1}{a} = -\frac{1}{a^2} x + \frac{2}{a}. \tag{2}
$$
And, this line intersects the $x$-axis at the point $(x, 0)$, where
$$
 -\frac{1}{a^2} x + \frac{2}{a} = 0,
$$
that is,
$$
x = \frac{ \frac{2}{a} }{ \frac{1}{a^2} } = 2a.
$$
Thus the tangent line to the curve at the point $(a, 1/a)$ intersects the $x$-axis at the point $A$ given by
$$A = (2a, 0). $$

And, the point of intersection of the tangent line to the curve at point $P$ and the line from $P$ to the origin is of course the point $P = (a, 1/a)$ itself.

Thus the three vertices of our triangle are $O = (0, 0)$, $A = (2a, 0)$, and $P = (a, 1/a)$, where $a > 0$. So the sides of our triangle have lengths
$$
\left| \overline{OA} \right| = 2a,
$$
$$
\left| \overline{AP} \right| = \sqrt{ (2a-a)^2 + (0-1/a)^2 } = \sqrt{ a^2 + 1/a^2},
$$
and
$$
\left| \overline{PO} \right| = \sqrt{ (a-0)^2 + (1/a-0)^2 } = \sqrt{a^2 + 1/a^2}.
$$
Thus the sides $AP$ and $PO$ are congruent, showing that our triangle is indeed isosceles.

Finally, since the base of our triangle is the side $OA$ and the altitude is the vertical line segment from the $x$-axis to the point $P$, therefore the area of our triangle is
$$
\frac{1}{2} \times (2a) \times \frac{1}{a} = 1.
$$

Is what I have done correct and clear in each and every detail? Or, are there any issues of accuracy or clarity?

答:
It looks good to me.  A geometric perspective to consider:

When we find point $A =(2a, 0)$, notice that $P$ is directly above the midpoint $OA$. (This midpoint is $C = (a, 0)$.)

The triangles $OCP$ and $ACP$ are congruent, right-angled triangles that are mirror images of each other ($CP$ as a common edge).

Hence $OPA$ is an isosceles triangle.

Moreover if we rotate $ACP$ around $P$ until it forms a rectangle with $OCP$ (which occurs when $A$ moves to $O$), we get a rectangle with base $a$ and height $1/a$ hence it has area $1$.


问:
Finding the particular solution of a differential equation using at least three different methods.
Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.

I have done the first two. Can somebody help me with the third method.

Method 1: Homogenous Equation

Let $y=vx; dy=vdx+xdv$

$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$

$(x^2+2x^2v^2)dx-4x^3vdv=0$

$(1+2v^2)dx-4xvdv=0$

$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$

$\ln{x}-\ln{(2v^2+1)}=C$

$\ln{(\frac{x}{2v^2+1})}=\ln{C}$

$\frac{x}{2v^2+1}=\frac{1}{C}$

$C=3$

$3x=2v^2+1$

$3x^3=2y^2+x^2$

$2y^2=x^2(3x-1)$

The particular solution by method 1 is $2y^2=x^2(3x-1)$.

Method 2: Bernoulli Equation

$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$

$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$

Let $v=y^2; dv=2ydy$

$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$

$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$

$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$

$2vx^{-3}=-x^{-1}+C^{-1}$

$2y^2x^{-3}+x^{-1}=C^{-1}$

$C=\frac{1}{3}$

$2y^2+x^2=3x^3$

$2y^2=x^2(3x-1)$

The particular solution by method 2 is also $2y^2=x^2(3x-1)$.

Body:
$$(x^2+6y^2)dx-4xydy=0$$
$$(x^2+6y^2)dx-2xdy^2=0$$
$$x^2dx+2(3y^2dx-xdy^2)=0$$
Multiply by $x^2$:
$$x^4dx+2(3x^2y^2dx-x^3dy^2)=0$$
$$x^4dx+2(y^2dx^3-x^3dy^2)=0$$
Divide by $x^6$:
$$\dfrac {dx}{x^2}-2d\left (\dfrac {y^2}{x^3}\right)=0$$
Integrate.
$$\dfrac {1}{x}+2\left (\dfrac {y^2}{x^3}\right)=C$$
$$y(1)=1 \implies C=3$$
Therefore:
$$x^2+2y^2=3x^3$$


问:
$\alpha$ and $\beta$ are the roots of the quadratic equation.
$ω≠1$ and $ω^{13}=1$. If $\alpha=ω+ω^3+ω^4+ω^{−4}+ω^{−3}+ω^{−1}$ and $\beta=ω^2+ω^5+ω^6+ω^{−6}+ω^{−5}+ω^{−2}$, then quadratic equation, whose roots are $\alpha$ and $\beta$ is:

I tried finding the sum and product of roots and placing in the equation $x^2-(\alpha+\beta)x+(\alpha\cdot\beta)=0$ but I was not able to solve for it.

答:
We have
$$
1+\alpha+\beta = 1+\omega+\omega^2+\cdots+\omega^{12}=0
$$
and so $\alpha+\beta=-1$. Therefore, $\alpha\beta=-(\alpha+\alpha^2)=-3$.

Unfortunately, I can't see shortcut to compute $\alpha+\alpha^2$ except by expanding it.


问:
Sobolev spaces - about smooth aproximation
Consider $\Omega $ a open and bounded set of $\mathbb R^n$ . Let $u \in H^{1}(\Omega)$ a bounded function. I know that there exists a sequence $u_m \in C^{\infty} (\Omega) $   where $u_m \rightarrow u$ in $H^{1}(\Omega)$ . I can afirm that exists $C>0$ where $|u_m (x)| \leq C , \forall \ x , \forall m  ?$

I dont have idea how to prove this or how to show a counter example ...

my english is terrible, sorry ..

答:
You are looking for Meyers-Serrin Theorem. However, It may be added that it is not always possible to get an approximation from $C^{\infty}(\overline{\Omega})$. A proof for the above may be found here.
I think I might have misunderstood your question. I am assuming that you mean that $u\in L^{\infty}(\Omega)\cap H^1(\Omega)$. In that case, it  is well known that if $u_m\to u$ in $L^p(\Omega)$, then there is a subsequence $u_{m_k}$ which converges pointwise almost everywhere to $u$. Therefore, for almost every $x$, $\{u_{m_k}(x)\}$ is bounded. I cannot think of a counterexample.</p>


答:
Product Law

Hence: $\displaystyle\lim_{x\to +\infty} e^x-x= \displaystyle\lim_{x\to +\infty} x*\dfrac{e^x-x}{x} = \displaystyle\lim_{x\to +\infty}x* \displaystyle\lim_{x\to +\infty}\dfrac{e^x}{x}-1 $

Apply L'Hopital on the second limit like you requested to get

$\displaystyle\lim_{x\to +\infty}x* \displaystyle\lim_{x\to +\infty}\dfrac{e^x}{x}-1  =  \displaystyle\lim_{x\to +\infty}x* \displaystyle\lim_{x\to +\infty}{e^x}-1 $

This is now in the form $\infty*\infty$ and obviously $=\infty$

Title: $\lim_{x\to +\infty} e^x-x$


问:
Combo problem, complementary counting
You have 5 blue nails in a column and 3 red nails in another column. You can attach a string between any red nail and blue nail. How many ways can we attach these strings such that every nail has at least 1 string attached to it?

For example, drawing all possible strings would work. Or connecting the first 3 blue with first 3 red, then connecting the last 2 blue to any red would work.

I approached this using Principle of Inclusion-Exclusion (if there is a simpler solution, let me know), but I am stuck. All I have is

$(2^5-1)^3$-something

$(2^5-1)^3$ is the number of non-empty subsets of blue nails for the red nails to connect to, cubed because 3 red nails. I don't know what to subtract from this though

答:
There are $5\cdot3=15$ blue-red pairs and a string can be there or not there; so the number of "stringings" from $5$ nails to $3$ nails is $2^{15}$. But this includes stringings which miss nails.

Inclusion-Exclusion

Let us call the subset of stringings that do not touch the $k^\text{th}$ red nail $R_k$ and the subset of stringings that do not touch the $k^\text{th}$ blue nail $B_k$. These are the $S_k$'s of the Generalized Inclusion-Exclusion Principle.

We can sum the sizes of the intersections of $k$ of the $B$'s and $R$'s ($N(k)$ from the Generalized Inclusion-Exclusion Principle):
$$
N(k)=\sum_{j=0}^k\overset{\substack{\text{the number}\\\text{of choices}\\\text{of the $B$'s}\\\downarrow\\[3pt]\,}}{\binom{5}{j}}\overset{\substack{\text{the number}\\\text{of choices}\\\text{of the $R$'s}\\\downarrow\\[3pt]\,}}{\binom{3}{k-j}}\overbrace{2^{(5-j)(3-k+j)}\vphantom{\binom52}}^{\substack{\text{number of stringings}\\\text{given the choice of}\\\text{$B$'s and $R$'s}}}\tag1
$$
The number of stringings in $0$ of the $B$'s and $R$'s is given by the Generalized Inclusion-Exclusion Principle as
$$
\sum_{k=0}^8(-1)^kN(k)%=16081
\tag2
$$

Simplification
$$
\begin{align}
f(a,b)
&=\sum_{k}(-1)^kN(k)\tag{3a}\\
&=\sum_{j,k}(-1)^k\binom{a}{j}\binom{b}{k-j}2^{(a-j)(b-k+j)}\tag{3b}\\
&=\sum_{j,k}(-1)^{k+j}\binom{a}{j}\binom{b}{k}2^{(a-j)(b-k)}\tag{3c}\\
&=\sum_{j,k}(-1)^{a+b-k-j}\binom{a}{j}\binom{b}{k}2^{jk}\tag{3d}\\
&=\sum_{j}(-1)^{a-j}\binom{a}{j}\left(2^j-1\right)^b\tag{3e}\\
\end{align}
$$
Explanation:

$\text{(3b)}$: plug $(1)$ into $(2)$

$\text{(3c)}$: substitute $k\mapsto k+j$

$\text{(3d)}$: substitute $j\mapsto a-j$ and $k\mapsto b-k$

$\text{(3e)}$: apply the Binomial Theorem

This is symmetric in $a$ and $b$, so take $a\le b$ to reduce the number of terms.
$$
f(3,5)=\sum_{j=0}^3(-1)^{3-j}\binom{3}{j}\left(2^j-1\right)^5\tag4
$$
However, if we switch them around, we get
$$
f(5,3)=\sum_{j=0}^5(-1)^{5-j}\binom{5}{j}\left(2^j-1\right)^3\tag5
$$
the $j=5$ term of which looks more like what is mentioned in the question.

Interpretations of the Simplified Formulas

In formula$(4)$, the index $j$ term counts the number of functions from the $5$ blue nails to a nonempty $j$-subset of the $3$ red nails, i.e. $\left(2^j-1\right)^5$, times the number of such $j$-subsets, i.e. $\binom{3}{j}$.

In formula$(5)$, the index $j$ term counts the number of functions from the $3$ red nails to a nonempty $j$-subset of the $5$ blue nails, i.e. $\left(2^j-1\right)^3$, times the number of such $j$-subsets, i.e. $\binom{5}{j}$.

Thus, $(2)$, $(4)$, and $(5)$ give three different ways to solve this problem (all using inclusion-exclusion).

Title: Combo problem, complementary counting


问:
proof of $x\cdot y = \lfloor x/2\rfloor \cdot 2y + y$ if $x$ is odd.
I was reading this awesome book called "Algorithms" by "Jeff Erickson" and I came across this peasant multiplication

he said it works because

How can I prove the second statement by induction and what's the intuition behind it?

EDIT: I understand that there are multiple easier proofs, I'm trying to get a sense of how induction works in this case.

答:
If $x$ is even,

$$xy=\frac x22y$$ is immediate.

Now if $x$ is odd,

$$(x-1)y=\frac{x-1}22y$$ and $$xy=\left\lfloor\frac x2\right\rfloor2y+y.$$

The rationale behind the algorithm becomes clear if you expand $x$ as a binary number

$$x=\sum_{k=0}^nb_k2^k$$ where the $b_k$ are $0$ or $1$.

Now

$$xy=\sum_{k=0}^n(b_k2^k)y=\sum_{k=0}^nb_k(2^ky).$$


答:
On one hand:

$$x\cdot y=\begin{cases}0&x=0\\\lfloor x/2\rfloor\cdot(y+y)&x\text{ even}\\\lfloor x/2\rfloor\cdot(y+y)+y&x\text{ odd}\end{cases}$$

which can be proven directly (distinguishing cases $x=2n, \lfloor x/2\rfloor=n$ and $x=2n+1, \lfloor x/2\rfloor=n$).

On the other hand:

$$x*y=\begin{cases}0&x=0\\\lfloor x/2\rfloor*(y+y)&x\text{ even}\\\lfloor x/2\rfloor*(y+y)+y&x\text{ odd}\end{cases}$$

as per the "Peasant's Algorithm". Note this recurrence is almost the same as the previous one, except $\cdot$ (the ordinary multiplication symbol) is replaced by $*$.

How do we prove that $x\cdot y=x*y$ for all $x,y$? Take induction on $x$, and in fact take what is often called "transfinite" induction: assuming the statement to be true for all $0,1,\ldots,x-1$ (rather than just for $x-1$) you prove that it is valid for $x$.

Now:

$$0\cdot x=0=0*x$$

and, for $x>0$ we have $\lfloor x/2\rfloor<x$, so:

$$\begin{array}{rcll}x\cdot y&=&\begin{cases}0&x=0\\\lfloor x/2\rfloor\cdot(y+y)&x\text{ even}\\\lfloor x/2\rfloor\cdot(y+y)+y&x\text{ odd}\end{cases}\\&=&\begin{cases}0&x=0\\\lfloor x/2\rfloor*(y+y)&x\text{ even}\\\lfloor x/2\rfloor*(y+y)+y&x\text{ odd}\end{cases}&\text{Apply inductive hypothesis to }\lfloor x/2\rfloor\\&=&x*y\end{array}$$

which is the inductive step (assuming the operators$\cdot$and $*$ match for  $\lfloor x/2\rfloor$, proving that it is valid for $x$).

Title: proof of $x\cdot y = \lfloor x/2\rfloor \cdot 2y + y$ if $x$ is odd.

Body:
I was reading this awesome book called "Algorithms" by "Jeff Erickson" and I came across this peasant multiplication

he said it works because

How can I prove the second statement by induction and what's the intuition behind it?

EDIT: I understand that there are multiple easier proofs, I'm trying to get a sense of how induction works in this case.


问:
Writing code in GAP
I want to write some GAP code that prints finite abelian groups that are the direct product of two cyclic groups. However, I want the order of the cyclic groups to have the condition that the gcd(,6)=1, where  is the order of the cyclic groups and is greater than 1.

How would I go about doing this?

答:
Is this what your looking for?

```<code>n:=10;m:=10; #bounds on the sizes of the cyclic groups
for i in [1..n] do
    if GCD_INT(i,6)=1 then
        for j in [i..m] do
            if GCD_INT(j,6)=1 then
                Print(IdGroup(DirectProduct(CyclicGroup(i),CyclicGroup(j))));
            fi;
        od;
    fi;
 od;
</code>```

It prints the group id from the library  SmallGrp, see also the function StructureDescription(G);

Title: Writing code in GAP

Body:
Is this what your looking for?

n:=10;m:=10; #bounds on the sizes of the cyclic groups
for i in [1..n] do
    if GCD_INT(i,6)=1 then
        for j in [i..m] do
            if GCD_INT(j,6)=1 then
                Print(IdGroup(DirectProduct(CyclicGroup(i),CyclicGroup(j))));
            fi;
        od;
    fi;
 od;

It prints the group id from the library  SmallGrp, see also the function StructureDescription(G);


问:
Does having real eigenvalues have a geometric meaning?
Is it possible to characterize the set of real matrices which have real eigenvalues geometrically? That is, is it possible to say that a linear map $T$ has real eigenvalues if and only if it satisfies some property ${\cal P}$ which has a geometric flavor?

I understand, of course, that different people will disagree on what ``geometric'' means. Still, I'm quite interested in seeing any possible answers. It seems plausible that properties of matrices which are independent of choice of basis have geometric interpretations.

Examples of the sort of thing I'm looking for:

A linear map $T$ has eigenvalues within the unit circle if and only if $\lim_k T^k x = 0$ for any $x$.

A linear map on $\mathbb{R}^n$ has all of its eigenvalues equal to zero if $T^n x = 0$ for any $x \in \mathbb{R}^n$.

A linear map $T$ has $1$ in its set of eigenvalues if and only if it has a fixed point.

A linear map $T$ will have a root of unity in its set of eigenvalues if and only if it has a periodic orbit.

答: How about this? A pair of complex conjugate eigenvalues corresponds to a $2$-dimensional invariant subspace that is stretched and rotated nontrivially. So all real eigenvalues means no such subspaces. So for any nonzero vector $v$, there is no nonzero $w$ orthogonal to it so that, up to scaling, the linear map rotates the plane $\text{Span}(v,w)$.


问:
limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$
How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhere.

$$\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}} $$

答:
Hint:

$$\sqrt{x+k+1}-\sqrt{x+k}=\frac{1}{\sqrt{x+k+1}+\sqrt{x+k}}\sim\dfrac1{2\sqrt x}$$


答:
Instead of studying $e^x-x$ we can study $\dfrac{e^x}{x}$, indeed if $x=o(e^x)$ then $e^x-x\sim e^x\to\infty$.

Let $f(x)=\dfrac{e^x}x$ then $f'(x)=\dfrac{e^x(x-1)}{x^2}\ge 0$ for $x\ge 1$

Since $f\nearrow$ then $f(x)\ge f(1)=e,\ \forall x\ge 1$.

Now using the additive to multiplicative exponential formula we get (for $x\ge 2$):

$$f(x)=\dfrac{e^x}x=\dfrac{e^{\frac x2}e^{\frac x2}}{x}=\frac 12f(\tfrac x2)\cdot e^{\frac x2}\ge \frac e2e^{\frac x2}\to+\infty$$

Title: $\lim_{x\to +\infty} e^x-x$