问:
How to devide an linear object to calculate how many of transverse smaller objects would be there and to be simetrical
I have started new work, in which I have to use some math. It is my firs job in engineer environment so I have to get used to it and remind myself some math from school.

I have an linear object, lenght 7700 cm, lets name it pipe,
I have smaler objects of lenght 54 cm.

I need to figure out how many of these objects I have to put on line and what distance of themselves for this to be simetrical.
Surface of smal vertical object have to start that way, that the lenght from it to the end of the pipe will be 100 cm.
I can adjust the end surface of the pipe a little, for example like the item would start 110 cm between the end of it, not 100.
The lenght between smaller objects have to be the same and about 1000 cm (more or less).

So the example I have been given:
Lenght of linear object : 5200 cm
Lenght of the smaller object : 54 cm
Distance to the end of pipe from the start of smaller object : 100,5 cm
In this example the answer is:  distances between surfaces of smaller objects will be 935 cm and there will be 6 of them in total (5 more of them besides the first one that is defined)

the visualisation of the example solution

In my task I have started calculated this by that:
7700(total lenght)-100(the free space left side)-100(the free space right side)-54(the lengt of the first defined small object)=7446

So 7446 will be the total distance of the pipe to be filled by the smaller objects.
I have checked that for 7446 the divaders are i.a. 438, 1241, 2482, so I chose 1241 - as a distance between them.

And then after divide 7446/1241 it equals 6.
So besides the firs defined one, there have to be 6 more of the smaller objects, with distance between themselves of 1241. But it is wrong, because the total of objects would be 7  so the one of them will be exactly in the middle, and it can not be like that, the distance from the middle of the pipe have to be the same for the end of left object and the start from the right object. Not in the middle.

the visualisation of my task solution

How can I deal with that kind of task? How can i calculate them and figure out where to put things?
Why does my calculations are wrong and how to change them to get distance from the middle exactly the same between two items on pipe, not to one of them be in the middle.
I hope all of this is ok, and there is a good place, for You guys to advice me a little.

答:
In general you have a pipe (long object) of length $L$, small objects of length $\ell$, the desired distance $d$ from the ends and the desired distance $D$ between small objects and you want the small objects uniformly spaced.

These demands may be slightly contradictory, so let's see how good approximation we can achieve. Let $n$ (necessarily an integer) be the number of small objects you place on the pipe. Consider their centers. Then they are $d+\frac \ell 2$ away from the pipe ends. This leaves $L-2d-\ell$ length for $n-1$ intervals between the centers, so the the distance between the centers is $\frac{L-2d-\ell}{n-1}$, which translates into the distance $\frac{L-2d-\ell}{n-1}-\ell$ between the objects. Thus, you want this number to be as close to $D$ as possible.

If the objects are like supports for the pipe, then $D$ may have the physical meaning of the longest allowed free flight for avoiding the collapse, so you want to approximate $D$ from below rather then from above. I assume that it is the case and the task is to minimize the required number of supports without putting the structure into danger (like if you are building a bridge here).

Then we want the smallest $n$ with $\frac{L-2d-\ell}{n-1}-\ell\le D$, i.e.,
$n\ge 1+\frac{L-2d-\ell}{D+\ell}$. So just perform the division, round up, and get $n$. The expressions in terms of $n$ for everything have been already written above.

In your example $L=7700,\ell=54, d=100, D=1000$ (in cm). So we want
$
n\ge 1+\frac{7700-2\times 100-54}{1000+54}=8.0645\dots
$
Now you have a tough choice: to round down to 8 and to be potentially just slightly unsafe, or to round up to 9 and to have an extra support and an object in the middle, which you for some reason do not want (so you will have to raise the number to an even one, i.e., 10, and get two more supports than you would like to). Fortunately, you are allowed to modify $d$ a little bit. So, let's try to keep $n=8$ but reduce the distance between the objects to the maximal allowed one of 1000 cm.

Then 8 objects would take the length $8\times 54+7\times 1000=7432$, so you now have $(7700-7432)/2=134$ instead of $100$ for the distances from the ends of the pipe. If that is acceptable, you are done. If not, well, the question becomes how much lack of safety you want to accept for the gain of the cost of two extra supports.

So that is how more or less how it is played. Note that it absolutely does not matter whether the fractions are integer or not, so your idea to take the divisors of 7446 is totally misleading: what would you do if the number were 7451 instead? It may make sense in the end to adjust the numbers a bit to round them to full millimeters (or half-centimeters) to facilitate the actual measurements when the installation takes place (nobody will measure exactly 996.56784321 cm in real life and you can easily get a fraction like that when dividing), but that is the last step in the story.

Hope that helps a bit. Feel free to ask questions if something is unclear :-)


问:
What is the name for a vector field that is both divergence-free and curl-free?
Consider a smooth vector field $\mathbf u\colon\Omega\to\mathbb R^3$ defined on an open domain $\Omega\subseteq\mathbb R^3$ such that $\mathbf u$ has zero divergence and zero curl on $\Omega$, that is,
$$\begin{align}\nabla\cdot\mathbf u&=0,\\\nabla\times\mathbf u&=0.\end{align}$$
Is there a specific technical name for such a vector field?

Wikipedia calls it a Laplacian vector field, but

it does not cite any references, and

it asserts that any such vector field is the gradient of a harmonic function, but this is only true if $\Omega$ is simply connected (counterexample: $\big(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\big)$ on the region $x^2+y^2>0$),

so I'm disinclined to trust it. Can anyone provide references supporting this or any other name?

答:
In geometric calculus literature (see, for example, Doran and Lasenby), such a function is called monogenic.  Monogenic functions are generalizations of complex analytic (or holomorphic) functions.  This condition is strictly stronger than being harmonic--all monogenic functions are harmonic, but not all harmonic functions are monogenic.

The term monogenic is not restricted to vector fields, as well; a scalar field with zero gradient would also be referred to as monogenic.

You can also consult this page by Gull, Lasenby, and Doran.

Edit: Phrased in the language of geometric calculus, we define the vector derivative of a vector field $u$ as $\nabla u$, given by

$$\nabla u = \nabla \cdot u + \nabla \wedge u$$

When $\nabla u = 0$, then $\nabla^2 u = \nabla \wedge (\nabla \cdot u) + \nabla \cdot (\nabla \wedge u) = 0$ as well, fulfilling the harmonic condition.

ContentLicense: CC BY-SA 3.0


问:
Calculating probability based on probability generating function
Given a random variable X which is non-negative and integer valued and its corresponding probability generating function:
$$G_X(s) = e^{s-1}$$

Find: $$P(X<2)$$

As per my calculations I get the answer:
$$e^{0-1} + e^{1-1} = e^{-1} + e^{0} = e^{-1} + 1$$

Is this correct or am I missing something?

答:
From elementary theory on Taylor series we know that the probability mass function of $X$ can be obtained by differentiating its probability generating function. That is, for each nonnegative integer $k$ we have
$$
p_k := \mathbb P(X=k) = \lim_{s\downarrow0}\frac{G_X^{(k)}(s)}{k!}.
$$
Since $\frac{\mathsf d}{\mathsf ds} G_X(s) = G_X(s)$, it follows that
$$
p_k = \frac{e^{-1}}{k!},
$$
and hence
$$
\mathbb P(X<2) = p_0 + p_1 = e^{-1}(1+1) = 2e^{-1}.
$$

ContentLicense: CC BY-SA 4.0


问:
A question related to division
$2^{21} + 1024^{2} + 16^6$ is divisible by:
a) 31
b) 19
c) 13
d) 17

My attempt: I tried using the remainder theorem and the factor theorem. But, nothing. Pls help me in moving forward

Title: A question related to division

Body:
Hints:

$1024 ^{2} = (2^{10})^2 = 2^{20}$

$16^6 = (2^4)^6 = 2^{24}$

So you have $2^{21} + 2^{20} + 2^{24}$

Can you do now?


Title: A question related to division

Body:
$2^{21} + 1024^{2} + 16^6$ is divisible by:
a) 31
b) 19
c) 13
d) 17

My attempt: I tried using the remainder theorem and the factor theorem. But, nothing. Pls help me in moving forward

Body:
First of all simplify the second term and third term.

$1024 ^{2} = (2^{10})^2 = 2^{20}$

$16^6 = (2^4)^6 = 2^{24}$

So now we have $ 2^{21} + 2^{20} + 2^{24} $

Simplifying it further $ 2^{20}(2 + 1 + 16) $

$ 2^{20}(19) $

Therefore it is divisible by 19


问:
Meaning of 'each of which is contained in a single $A_{\alpha}$,' in Hatcher's Algebraic Topology
Lemma 1.15. of Hatcher's Algebraic Topology says: if a space $X$ is the union of a collection of path-connected open sets $A_{\alpha}$ each containing the base point $x_0\in X$ and if each intersection $A_{\alpha}\cap A_{\beta}$ is path-connected, then every loop in $X$ at $x_0$ is homotopic to a product of loops each of which is contained in a single $A_{\alpha}$.

I am wondering what exactly is the meaning of 'each of which is contained in a single $A_{\alpha}$.' Does it mean that for each loop in the product, there exists a $A_{\alpha}$ that fully contains it? Or does it mean that for each loop in the product, there is a unique $A_{\alpha}$ that fully contains it? The distinction being that in the first case, a loop in the product might be a subset of $A_{\alpha}\cap A_{\beta}$ whereas this is excluded in the second case.

I am struggling to follow the associated proof and I think that clearing up what we are trying to show would help me out. Furthermore, I have been told that it is important to understand this Lemma because it leads to van Kampen's Theorem. Thanks!

答:
This has been answered for awhile, but I though I'd throw my two cents in:

Suppose that $f:I\rightarrow X$ is the loop in question. For each $A_\alpha$, we get an open set $f^{-1}(A_\alpha)$ and these sets cover all of $I$. By Lebesgue's Number Lemma, there exists some $\delta>0$ so that if $|x-y|<\delta$, then $[x,y]\subseteq f^{-1}(A_{\alpha})$. You essentially choose your partition based on this. I was taught this explanation in my topology course, and it made a lot more sense to me than what I saw in Hatcher, so I figured I would include it.


问:
Why does $A^{-1}MA$ give a vector that is in the "language" of basis $A$?
For context, I am struggling with understanding the end of this video.

Let's say we have the following 4 sets of basis vectors:

$$B_1 =\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}, B_2 =\begin{bmatrix}
2 & -1 \\
1 & 1
\end{bmatrix}, B_3 =\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}, B_4 = B_2^{-1} =\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} \\
-\frac{1}{3} & \frac{2}{3}
\end{bmatrix}$$

Now I want to see what the vector, $v = \begin{bmatrix}
1 \\
0
\end{bmatrix}$, represents after I apply all the change of basis matrices above on it.

$B_2*v = B_1* \begin{bmatrix}
2 \\
1
\end{bmatrix}$

What this means: The vector from the linear combination of $1\hat{i}_{B_2} + 0\hat{j}_{B_2}$ is equivalent to the vector $\begin{bmatrix}
2 \\
1
\end{bmatrix}$ in the basis $B_1$.

$B_3*\begin{bmatrix}
2 \\
1
\end{bmatrix} = B_1* \begin{bmatrix}
-1 \\
2
\end{bmatrix}$

What this means: The vector from the linear combination of $2\hat{i}_{B_3} + 1\hat{j}_{B_3}$ is equivalent to the vector $\begin{bmatrix}
-1 \\
2
\end{bmatrix}$ in the basis $B_1$.

$B_4*\begin{bmatrix}
-1 \\
2
\end{bmatrix} = B_1* \begin{bmatrix}
\frac{1}{3} \\
\frac{5}{3}
\end{bmatrix}$

What this means: The vector from the linear combination of $-1\hat{i}_{B_4} + 2\hat{j}_{B_4}$ is equivalent to the vector $\begin{bmatrix}
\frac{1}{3} \\
\frac{5}{3}
\end{bmatrix}$ in the basis $B_1$.

In the video, if I understood correctly, he says that this final vector $\begin{bmatrix}
\frac{1}{3} \\
\frac{5}{3}
\end{bmatrix}$ is actually in the basis $B_2$, not $B_1$. What am I confounding here?

答:
It's difficult to diagnose properly, but I think the root of your issues comes from confusing bases with invertible matrices. While bases do correspond with invertible matrices, it can confuse your intuition if you treat every invertible matrix as some kind of basis, which appears to be what you've done here.

It's better to think of bases not as matrices, but as sets of vectors. Or, even better, think of them as ordered lists of vectors, as order is important when using bases, and sets don't convey such order. So, if you want the standard basis, think of the ordered list:
$$\mathcal{S} = \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right)$$
instead of the identity matrix
$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$
I know it doesn't seem like much of a distinction, but I honestly think that not making this distinction is at the heart of your problems here.

In particular, you seem to be interpreting $B_4 = B_2^{-1}$ as a basis. I wouldn't recommend this. Jennifer's basis is
$$\mathcal{J} = \left(\begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \end{bmatrix}\right).$$
Now, the change-of-basis matrix from $\mathcal{J}$ to $\mathcal{S}$ is given by
$$B_2 = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}.$$
What does this matrix do? If we take a vector $v = \begin{bmatrix} a \\ b \end{bmatrix}$ in Jennifer's "language" (i.e. $a$ times her first basis vector plus $b$ times her second basis vector), then $B_2 v = \begin{bmatrix} c \\ d \end{bmatrix}$ refers to the same point in our language (i.e. $c$ times $(1, 0)$ plus $d$ times $(0, 1)$).

You've already provided a concrete example of this. If we have $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ in her "language", then it refers to $1$ times her first basis vector plus $0$ times her second basis vector. Her first basis vector, in our language, is $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$, and you can verify that
$$\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.$$

Importantly, I'm treating $B_2$ not as a basis, but as a multiplicand that helps us translate from one language to another. If Kathrine had a third "language", which was represented by some other basis $\mathcal{K}$, then Jennifer and Kathrine would have a totally different change of basis matrix, whose columns would be neither the basis vectors in $\mathcal{J}$ or $\mathcal{K}$. The matrix would still be invertible, and yes, you could make a basis by taking the columns of this matrix, but that basis would have no intuitive bearing on the problem at hand. That is, it would not be a useful basis for understanding Jennifer or Kathrine.

That's what you're doing wrong with $B_4$. Think of it instead as the reverse process of translating from Jennifer's language into our own. Since it is the inverse of $B_2$, we can multiply it to a vector in our language, and get the same vector in Jennifer's language. If you harvest a new basis $\mathcal{B}$ from its columns, you've invented a new language that has nothing intuitive to do with the existing languages, and $B_4$ becomes the translation from this erroneous language $\mathcal{B}$ to our language $\mathcal{S}$.

I'd also like to point out that you're doing a similar thing with $B_3$. The matrix $B_3$ is the transformation matrix. It's not translating the same vector in different languages (at least, it isn't in the context of our problem), it's actually changing the point we're referring to, by rotating it $90^\circ$. It is an invertible matrix, so a basis can be formed, but we gain nothing, intuitively speaking, by interpreting it as such. Plus, we can just as easily have a non-invertible transformation here (e.g. projecting onto the $x$-axis), in which case a basis cannot be formed from the columns!

So yeah, TL;DR: not every matrix should be interpreted as a basis.


问:
The ratio of two areas of triangles
On the picture below is a square. The question is to find $s/r$.

I tried to draw a line from the middle of one side of the square to the middle of the opposite one, and it appears that the area of $s$ is $1/4$ the area of our square. So I stopped to calculate the ratio of triangle's area $r$ to the area of whole square.

Any hints?

答:
Denote areas [.] and [ABCD] = $I$. Per the Ceva's theorem on  triangle ABD,  $\frac{AF}{FD} = \frac{BI}{ID}\frac{AE}{EB}=\frac{BJ}{AD}\frac{AE}{EB}=\frac12\frac11=\frac12$. Evaluate the ratios below
$$\frac{HC}{FC}=\frac{[DCE]}{[DCEF]}=\frac{[DCE]}{I-[BCE]-[FAE]}
=\frac{\frac12I}{(1-\frac14-\frac14\frac13)I}=\frac34
$$
$$\frac{DG}{ED}=\frac{[ADJ]}{[AEJD]}=\frac{[ADJ]}{I-[CDJ]-[BEJ]}
=\frac{\frac12I}{(1-\frac14-\frac14\frac12)I}=\frac45
$$
$$\frac{DH}{ED}=\frac{[FDC]}{[FDCE]}=\frac{[FDC]}{I-[BCE]-[FAE]}
=\frac{\frac13I}{(1-\frac14-\frac14\frac13)I}=\frac12
$$
Then,
$\frac{HG}{DE}=\frac{DG}{DE}- \frac{DH}{DE} =\frac45-\frac12= \frac3{10}
$ and the area ratio is
$$\frac{[S]}{[R]} =\frac{\frac{HC}{FC}[CDF]}{\frac{HG}{DE}[DEF]}
=\frac{\frac34(\frac13I)}{\frac3{10}(\frac14\frac23I)}=5
$$

Title: The ratio of two areas of triangles


问:
Logistic Proof in TFL

Cannot seem to figure this one out, my structure might be completely wrong, please help

答:
$\def\fitch#1#2{~~~~~\begin{array}{|l}#1\\\hline#2\end{array}}$

You have most of the proof, but there should not be a second subproof.

Do not assume $\lnot A$.   Rather this is derived by negation introduction.

$$\fitch{~~1.~A\to B}{\fitch{~~2.~\lnot(\lnot A\lor B)}{\fitch{~~3.~A}{~~4.~B\hspace{12ex}{\to}\mathsf E~3,1\\~~5.~\lnot A\lor B\hspace{6ex}\lor\mathsf I~4\\~~6.~\bot\hspace{12ex}{\lnot}\mathsf E~5,2}\\~~7.~\lnot A\hspace{15ex}{\lnot}\mathsf I~3{-}6\quad\bigstar\\[1ex]~~8.~\lnot A\lor B\hspace{10ex}\lor\mathsf I~7\\~~9.~\bot\hspace{16ex}{\lnot}\mathsf E~8,2}\\10.~\lnot A\lor B\hspace{15ex}\mathsf{RAA}~2{-}9}$$

Note, however, that there is an alternative proof using the Law of Excluded Middle, rather than reduction to absurdity.

$$\fitch{~~1.~A\to B}{\fitch{~~2.~A}{~~3.~B\hspace{15ex}{\to}\mathsf E~2,1\\~~4.~\lnot A\lor B\hspace{9ex}\lor\mathsf I~3}\\\fitch{~~5.~\lnot A}{~~6.~\lnot A\lor B\hspace{10ex}\lor\mathsf I~5}\\~~7.~\lnot A\lor B\hspace{15ex}\mathsf{LEM}~2{-}4,5{-}6}$$

Does your system implement some form of LEM?  (It may, for instance, take the form of declaring $A\lor\lnot A$ then using disjunction elimination. So adjust to your system's taste.)

Title: Logistic Proof in TFL


问:
Good definition for $\int_a^\infty X(t)dt$, where $X(t)\in M_n$
Any comments on this? : I'm reading a book, and some exercise says: Provide a good definition for $\int_a^\infty X(t)dt$, where $X(t)\in M_n(\mathbb{R})$ and prove that if $$\int_a^\infty \|X(t)\|dt \ \ \ \ \mathrm{exists\ then\ }\ \int_a^\infty X(t)dt \ \ \mathrm{exists}$$

Idea:

We should have $X(t):[a,\infty)\to M_n$ continuous, in order to prove the result, also for the existence of the improper integral (otherwise we should have a lot of extra conditions for the integral to exists).

Does it matter which $\|\cdot\|$ to use? There are several $\|\cdot\|_{\sup}$, $\|\cdot\|_1$, $\|\cdot\|_{Frob}$.

$\int_a^\infty X(t)dt=\lim_{b\to\infty}\int_a^b X(t)dt$

Convergence of $\int_a^\infty X(t)dt$ is proved by showing there is $U\in M_n$ such that $$\left\| \int_a^b X(t)dt-U \right\|\to 0 $$

答:
Using matrix units, you can write
$$
X(t)=\sum_{k,j} x_{kj}(t)\,E_{kj},
$$
where $x_{kj}:[a,\infty)\to\mathbb R$ are scalar functions. An integral is a limit of sums of the form "value of the function times size of the region". In particular, you expect an integral to be linear. So you want
$$
\int_a^\infty X(t)\,dt=\sum_{kj}\bigg(\int_a^\infty x_{kj}(t)\,dt\bigg)\,E_{kj}.
$$
So you need each entry $x_{kj}$ to be integrable (you don't need continuity necessarily, but either Riemann or Lebesgue integrability).

The norm doesn't matter, because all norms in $M_n(\mathbb R)$ are equivalent.

Finally, if $\displaystyle\int_a^\infty\|X(t)\|\,dt$ exists, you can use the norm
$$
\|X(t)\|=\sum_{kj}\|x_{kj}(t)\|
$$
and again you are integrating coordinate wise.


问:
How can I get a function that is "approximately similar" to the part I care about in another function?
I have a function f(x)=$(a+bx)e^{-cx}, x\ge0$ where $a$, $b$, and $c$ are constants, and $c>0$.

How can I get a function that is "approximately the same" as the part after the hump?

I need this so I can get the inverse of it for a computer program; taking the inverse of the function in its current form results in some sorcery that I can't comprehend.

I don't really care about precision close to the hump, but it does matter further away.

Just by playing around in Desmos with my limited mathematical knowledge for a while, I found that $\frac 1 {e^x+1}$ matches close enough to what I'm looking for. Below is a picture of the somewhat arbitrary values I used to come close. I figured that the top of the orange line is probably going to be around twice the maximum value of $f$, so $a$ is the value of $f$ where the derivative is $0$.

答:
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

\begin{align}
f(x)&=(a+bx)\exp(-cx)
,\quad
x\ge0
\tag{1}\label{1}
\end{align}

where $a$, $b$, and $c$ are constants, and $c>0$.

There is absolutely no need to reinvent the wheel in this case,
since there is a solution for $f(x)=y$:

\begin{align}
x&=
-\frac ab-\frac1c\cdot\W\left(-\frac{cy}b
\,\exp\left(-\frac{ac}b\right)\right)
\tag{2}\label{2}
\end{align}

in terms of
Lambert $\W$ function
(also known as ProductLog),
with well-known behavior,
which is available in many programming systems.
There is just a few things to note:
the number of real solutions of \eqref{1}
depends on the value of the argument of $\W(z)$
in \eqref{2}.

\begin{align}
z&=-\frac{cy}b
\,\exp\left(-\frac{ac}b\right)
\tag{3}\label{3}
\end{align}

and, depending on its value we have

\begin{align}
1)\
z&\ge0:
\quad \text{there is one real solution,}
\quad
x=-\frac ab-\frac1c\cdot\Wp(z)
,\\
2)\
z&\in(-\tfrac1e,0):
\quad \text{there are two real solutions,}
\\
x_0&=-\frac ab-\frac1c\cdot\Wp(z)
\quad\text{and}\quad
x_1=-\frac ab-\frac1c\cdot\Wm(z)
,\\
3)\
z&=-\tfrac1e:
\quad \text{there is one real solution,}
\\
x&=-\frac ab-\frac1c\cdot\Wp(-\tfrac1e)
=-\frac ab-\frac1c\cdot\Wm(-\tfrac1e)
=-\frac ab+\frac1c
\\
4)\
z&<-\tfrac1e:
\quad \text{there is no real solutions}
\end{align}
where $\Wp$ is the principal branch
and $\Wm$ is the other real branch
of the Lambert $\W$ function.

Don't be scared with this "branch" term -
consider these two branches as
just two real functions. Usually they
are available either as two distinct function names,
for example, W0 and Wm1,
or as one function with extra parameter,
for example W(z) for $\Wp(z)$
and W(z,-1)  for $\Wm(z)$, like in python:

```<code>from scipy.special import lambertw as W
print (W(-0.2).real, W(-0.2,-1).real)
# -0.25917110181907377 -2.5426413577735265
</code>```

$\endgroup$

Body:
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

\begin{align}
f(x)&=(a+bx)\exp(-cx)
,\quad
x\ge0
\tag{1}\label{1}
\end{align}

where $a$, $b$, and $c$ are constants, and $c>0$.

There is absolutely no need to reinvent the wheel in this case,
since there is a solution for $f(x)=y$:

\begin{align}
x&=
-\frac ab-\frac1c\cdot\W\left(-\frac{cy}b
\,\exp\left(-\frac{ac}b\right)\right)
\tag{2}\label{2}
\end{align}

in terms of
Lambert $\W$ function
(also known as ProductLog),
with well-known behavior,
which is available in many programming systems.
There is just a few things to note:
the number of real solutions of \eqref{1}
depends on the value of the argument of $\W(z)$
in \eqref{2}.

\begin{align}
z&=-\frac{cy}b
\,\exp\left(-\frac{ac}b\right)
\tag{3}\label{3}
\end{align}

and, depending on its value we have

\begin{align}
1)\
z&\ge0:
\quad \text{there is one real solution,}
\quad
x=-\frac ab-\frac1c\cdot\Wp(z)
,\\
2)\
z&\in(-\tfrac1e,0):
\quad \text{there are two real solutions,}
\\
x_0&=-\frac ab-\frac1c\cdot\Wp(z)
\quad\text{and}\quad
x_1=-\frac ab-\frac1c\cdot\Wm(z)
,\\
3)\
z&=-\tfrac1e:
\quad \text{there is one real solution,}
\\
x&=-\frac ab-\frac1c\cdot\Wp(-\tfrac1e)
=-\frac ab-\frac1c\cdot\Wm(-\tfrac1e)
=-\frac ab+\frac1c
\\
4)\
z&<-\tfrac1e:
\quad \text{there is no real solutions}
\end{align}
where $\Wp$ is the principal branch
and $\Wm$ is the other real branch
of the Lambert $\W$ function.

Don't be scared with this "branch" term -
consider these two branches as
just two real functions. Usually they
are available either as two distinct function names,
for example, W0 and Wm1,
or as one function with extra parameter,
for example W(z) for $\Wp(z)$
and W(z,-1)  for $\Wm(z)$, like in python:

from scipy.special import lambertw as W
print (W(-0.2).real, W(-0.2,-1).real)
# -0.25917110181907377 -2.5426413577735265

$\endgroup$


问:
Evaluate integral with no elementary antiderivative $\int_{0}^{\infty} \frac{\ln x}{x^2+x+1} dx $
This is a followup from my previous question here.

I was asked to determine if the following integral converges or diverges. If the integral converges, I was asked to determine the value.

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

With help from the question I posted above, I managed to figure out my comparison (Thanks!). I definitely know this converges now. How do I find the value though? As far as I can see, this integral has no elementary antiderivative.

This is a standard intermediate calculus class which does do not calculus of residues or anything complicated like that so I'm not how to proceed there.

Wolfram alpha says that $$\int_{0}^{\infty} \frac{\ln(x)}{x^2+x+1} dx = 0$$ and that makes a lot of sense! However, I can't find a way to justify what the value should be rigirously.

Can someone help out here?

答:
Note

$$I=\int_{0}^{\infty} \frac{\ln x}{x^2+x+1} dx \overset{x\to \frac1x}=- \int_{0}^{\infty} \frac{\ln x}{x^2+x+1}=-I
=0
$$


问:
Fun calculus problem I can't seem to solve
I've recently picked up a math book I haven't read since college (highly recommended reading by the way!). I was reviewing multi-dimentional derivatives and such, and I stumbled upon a problem I've been trying to solve for two days, and I can't get it out of my head, so please help me out! = )

Problem (from memory):

There is a rabbit that runs in a perfect circle of radius $r$ with a constant speed $v$. A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed $v$ such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?

I tried using the fact that $|x'(t)| = |y'(t)| = v$ where $x(t)$ is fox's position and $y(t)$ is rabbit's position, and that $x'(t)x''(t) = 0$ because of constant speed restriction, but I'm still failing to find a solution.

Anyone feel like attacking this one?

答:
It is easiest to solve the problem with polar coordinates.  WLOG we may set $r=1$ so that the rabbit's angle at time $t$ is $vt$.  Everything can be rescaled for different values of $r$.  If we let $r(t)$ be the distance between the fox and the center of the circle at time $t$, his $(r,\theta)$ position at time $t$ is $(r(t),vt)$, which in $(x,y)$ coordinates is $(r(t)\cos(vt),r(t)\sin(t))$.  The velocity vector is then $r'(t)(\cos(vt),\sin(vt))+vr(t)(\sin(vt),-\cos(vt))$, and so if the speed is $v$, then we must have $v^2=r'(t)^2+v^2r(t)^2$, or $r'=v\sqrt{1-r^2}$.

The solution to this differential equation, subject to the initial condition that $r(0)=0$ is $r(t)=\sin(vt)$, which takes the value $1$ when $vt=\pi/2$, and so the fox will catch the rabbit when the rabbit is one quarter of the way around the circle.


问:
Maximization problems - Using implicit function to avoid KKT
I am wondering if the following procedure to solve a maximization problem in - let's say - two variables with inequality and non-negative constraints works.

More specifically, let's assume something like the following concrete example:

$ \text{max } F(x,y)=x^{\frac{2}{3}} y^{\frac{1}{3}}$

$ \text{sub } G(x,y)=x^2 + y^2 \leq 2 \hspace{0.5cm}\text{and } x,y\geq 0$

I thought that, instead of using Kuhn-Tucker conditions, which would bring me to a real nightmare, it would be better to go for the following steps:

I realize that the objective function is increasing in $\mathbb{R}^3_{+}$.

For this reason the constraint has to work as an equality constraint and it is binding.

I take the implicit function of $F(x,y)$  of $y$ in terms of $x$ and I put it equal to the same thing, for the inequality constraint.

I create a system with the previous equation and the inequality constraint expressed as an equality constraint.

I focus on the positive value of $x$ and $y$ and that's it.

Does it sound reasonable?

Any feedback is welcome!

答:
This sound reasonable. The precise justification for your step 1 could go along the following lines: Assume that $(x,y)$ is a (feasible) minimizer of $F$.
Obviously, $F(x,y) > 0$.

Assume $r^2 := x^2 + y^2 < 2$. Then, take $(\sqrt2 \, x/r,\sqrt 2\,y/r)$. Obviously, this point is still feasible and $F(\sqrt 2 \, x/r,\sqrt 2\,y/r) = F(x,y) \, \sqrt 2/r > F(x,y)$. Contradiction.

Hence, $x^2 + y^2 = 2$ holds for the minimizer.


问:
Moment generating function finite everywhere iff $\lim_{t\to\infty} e^{\lambda t}P(\lvert X\rvert > t) = 0$ for $\lambda > 0$
I want to show that the moment generating function is finite everywhere iff
$$\lim_{t\rightarrow\infty} e^{\lambda t}P(\lvert X\rvert > t) = 0$$
If we assume the mgf is finite everywhere, then I have the following:
$$P(\lvert X\rvert >t) \le P(\lvert X\rvert \ge t) \le \frac{E(e^{\lambda \lvert X\rvert})}{e^{\lambda t}} = \frac{E(e^{\lambda X})\rvert_{X\ge 0} + E(e^{-\lambda X})\rvert_{X<0}}{e^{\lambda t}}\xrightarrow{t\rightarrow\infty} 0$$
This is not sufficient however, as the limit $e^{\lambda t}P(\lvert X\rvert >t)$ might well be constant. Can this approach be fixed somehow or is it flawed?

I don't have any ideas for the reverse direction, how could it be done?

答:
You have shown that if MGF is finite everywhere then $e^{\lambda t}P(|X|\ge t)\to 0$ as $t\to\infty.$ To show the converse, first make some simplification. Replace $X$ with $|X|$ (check that moment generating function of $X$ exists if and only the moment generating function of $|X|$ does so). Now note that
$$E\left(e^{\lambda |X|}\right)=\int e^{\lambda x}d\mu_{|X|}(x)=\lambda \int_{0}^{\infty}e^{\lambda t} P(e^{\lambda |X|}\ge e^{\lambda t})dt.$$

The second equality above is a consequence of 'Layer cake representation'. Now use that $P(e^{\lambda |X|}\ge e^{\lambda t})=P(|X|\ge t).$ Now conclude that $e^{\lambda t}P(|X|\ge t)\to 0$ as $t\to \infty.$ (Note that it is easy to conclude that the limit if it exists, has to be zero. The point is to argue that the limit does exist.)


问:
Quick Volume of Revolution with Lines Question
So I was given the following prompt:

"What is the volume of the solid generated when the region in the first quadrant bounded by the graph of $y=x^3$, the $y$-axis, and the horizontal line $y=1$ is revolved about the $y$-axis?"

I guess I'm just a bit confused about what the integral set up might look like here. I understand the basics of the Disc and Washer methods, but I'm a bit confused about how that would apply here. Any help would be appreciated!

答:
You can find intersection of $y=1, y = x^3$ which is $(1,1)$. So the integral to find volume can be set up using washer method as

$0 \leq x \leq y^{1/3} \ $ where $x$ is the radius around $y-$axis

$0 \leq y \leq 1$

$\displaystyle \int_0^1 \int_0^{y^{1/3}} 2 \pi x \ dx \ dy$

Or it can be set up using shell method as

$x^3 \leq y \leq 1$

$0 \leq x \leq 1$

$\displaystyle \int_0^1 \int_{x^3}^1 2\pi x \ dy \ dx$


问:
prove - For symetric matrix with zero diagonal there exists $UU^t$ s.t $A+2I=UU^t$
I have seen this in a book I am reading it seems that this is straight forward, but unfortunately my algebraic skills not good enough.

$A$ is symmetric with zeros on its diagonal, we need to prove that there exist matrix $U$ s.t
$$
A+2 I=U U^{T}
$$

答:
A counterexample for real matrices is $A=\begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix}$. Note that, if $A+2I=UU^t$ for some matrix $U$, then for any real vector $v$ we have $$v^t (A+2I)v=v^tUU^t v=(U^t v)^t (U^tv)=\|U^t v\|^2$$
which cannot be negative if $U$ (and therefore $U^t v$) is real. But if $v=(1,-1)^t$ then
$$v^\top (A+2I)v
=
\begin{pmatrix} 1 & -1 \end{pmatrix}
\begin{pmatrix} 2 & 3\\3 & 2 \end{pmatrix}
\begin{pmatrix} 1 \\ -1 \end{pmatrix}
=\begin{pmatrix} -1 & 1 \end{pmatrix}
\begin{pmatrix} 1 \\ -1 \end{pmatrix}=-2.
$$
Hence no real matrix $U$ exists. Matters are different if we allow complex-valued $U$. For instance, Mathematica's FindInstance command generates the example $U=\begin{pmatrix} 0 & -\sqrt{2} \\ -i \sqrt{5/2} & -3/\sqrt{2}\end{pmatrix}.$


答:
On way to define the operator-valued integral is as follow:

Define $\sigma:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}$
by $\sigma(x,y)=\int_{a}^{\infty}\langle X(t)x,y\rangle dt.$ (Assuming
that $t\mapsto\langle X(t)x,y\rangle$ is continuous if you are working
with Riemann integral. However, this assumption can be weaken and
only require that $t\mapsto\langle X(t)x,y\rangle$ is Lebesgue measurable if you are working with Lebesgue integral.)
Observe that
\begin{eqnarray*}
\left|\langle X(t)x,y\rangle\right| & \leq & \left\Vert X(t)x\right\Vert \left\Vert y\right\Vert \\
 & \leq & \left\Vert X(t)\right\Vert \left\Vert x\right\Vert \left\Vert y\right\Vert
\end{eqnarray*}
and by assumption, $\int_{a}^{\infty}\left\Vert X(t)\right\Vert dt<\infty$,
so the improper integral $\int_{a}^{\infty}\langle X(t)x,y\rangle dt$
converges absolutely.

It can be proved that $\sigma$ is bilinear. From linear algebra,
there exists uniquely a linear operator $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$
such that $\sigma(x,y)=\langle Tx,y\rangle.$ This operator $T$ is
denoted by $\int_{a}^{\infty}X(t)dt.$ This is how we define operator-valued
integral.

Title: Good definition for $\int_a^\infty X(t)dt$, where $X(t)\in M_n$

Body:
Any comments on this? : I'm reading a book, and some exercise says: Provide a good definition for $\int_a^\infty X(t)dt$, where $X(t)\in M_n(\mathbb{R})$ and prove that if $$\int_a^\infty \|X(t)\|dt \ \ \ \ \mathrm{exists\ then\ }\ \int_a^\infty X(t)dt \ \ \mathrm{exists}$$

Idea:

We should have $X(t):[a,\infty)\to M_n$ continuous, in order to prove the result, also for the existence of the improper integral (otherwise we should have a lot of extra conditions for the integral to exists).

Does it matter which $\|\cdot\|$ to use? There are several $\|\cdot\|_{\sup}$, $\|\cdot\|_1$, $\|\cdot\|_{Frob}$.

$\int_a^\infty X(t)dt=\lim_{b\to\infty}\int_a^b X(t)dt$

Convergence of $\int_a^\infty X(t)dt$ is proved by showing there is $U\in M_n$ such that $$\left\| \int_a^b X(t)dt-U \right\|\to 0 $$


问:
In the ring $\mathbb{Z[i]}$, show that $3$ is a prime element, but $5$ is not a prime element. And show that the units are $1$, $-1$, $i$, $-i$.
In the ring $\mathbb{Z[i]}$, show that $3$ is a prime element, but $5$ is not a prime element. And show that the units are $1$, $-1$, $i$, $-i$.

Which led:
Let $3|(a+bi)(c+di)$ for some $(a+bi),(c+di) \in \mathbb{Z[i]}$, so $(a+bi)(c+di)=3$. Then suppose that $3|(a+bi)$ so $a+bi=3(e+fi)$ which indicates that $a=3e$ and even there I stagnate.

答:
The key here is to use the norm map. We'll define $N(z) = z \overline z$ for $z \in \mathbb Z[i]$. Then $N(zw) = N(z) N(w)$ and $N(a + bi) = a^2 + b^2$, which is a natural number.

Let's first apply this to units. Suppose $z$ is a unit. Then there must be some $w$ such that $zw = 1$. Applying the norm map, we get $N(zw) = N(1) = 1$. Thus, $N(z) N(w) = 1$. Since the image of $N$ lies in the natural numbers, we must have $N(z) = N(w) = 1$. Thus, if we write $z = a + bi$ for $a, b \in \mathbb Z$, we must have $a^2 + b^2 = 1$. Then exactly one of $a^2$ and $b^2$ must be 1 and the other must be 0. This yields our 4 choices $1, -1, i, -i$, so these are all the units.

Now let's work with 3. Suppose we factor $3 = zw$ in $\mathbb Z[i]$. Then again applying the norm map, $9 = N(3) = N(z) N(w)$. To show that 3 is prime we want to conclude that $z$ or $w$ are units. Suppose for the sake of contradiction that they are not. Then as discussed in the previous paragraph, $N(z)$ and $N(w)$ cannot be 1. As $9 = 3^2$ and $N(z) N(w) = 9$, we therefore must have $N(z) = N(w) = 3$. Let's write $z = a + bi$. Then $N(z) = a^2 + b^2 = 3$. In other words, we have written 3 as the sum of two squares, but this is impossible, as the only squares that are at most 3 are 0 and 1. Hence, we have arrived at a contradiction so one of $z, w$ must be a unit. Hence, 3 is prime.

EDIT: Technically, I showed here that 3 is irreducible in this ring. However, this ring is a UFD (in fact, a Euclidean domain) so these ideas align.

As for 5, we can find an explicit factorization. Indeed, $5 = 1 + 2^2 = (1 - 2i)(1 + 2i)$. As shown above, the only units are $1,-1,i,-i$ so this is indeed a factorization of 5. Hence, it is not prime.


问:
Function defined on the tangent bundle
Suppose $F$ is a smooth real-valued function defined on the tangent bundle $TM$ of a Riemannian manifold $M$, i.e. $F:TM\to\mathbb{R}$ given by $F(p, v)\in\mathbb{R}$. Consider the function $f:M\to\mathbb{R}$ given by
$$f(p)=\min\{F(p,v):|v|=1\}.$$
which is a continuous real-valued function on $M$. My question is: could we find a vector field $X$ in $M$ such that $f(p)=F(p,X)$?

More concretely, consider the Ricci tensor $Ric$ of a Riemannian manifold $M$. Then it can be considered as a function on the tangent bundle:
$F(p,v)=Ric_p(v,v)$. Then the function $f:M\to\mathbb{R}$ above is given by
$$f(p)=\min\{Ric_p(v,v):|v|=1\}.$$
The above question is: I am wondering if we could find a vector field $X$ in $M$ such that
$$Ric_p(X,X)=\min\{Ric_p(v,v):|v|=1\}.$$

答:
As Max's comment suggests, there are global obstructions to the existence of such vector fields (namely that some sphere bundles have no sections at all). There are also local obstructions, as the following example demonstrates:

Let $(x,v)=(x_1,x_2,v_1,v_2)$ denote global coordinates on $T\mathbb{R}^2\cong\mathbb{R}^4$ (with $\mathbb{R}^2$ equipped with the Euclidean metric). We may define a fiberwise quadratic function $f:T\mathbb{R}^2\to\mathbb{R}$ by
$$
f(x,v)=e^{x_1}v_1^2+e^{-x_1}v_2^2
$$
A unit vector field $V$ which minimizes $f$ must be parallel to $\frac{\partial}{\partial x_1}$ for $x_1<0$, and parallel to $\frac{\partial}{\partial x_2}$ for $x_1>0$. Clearly, then, $V$ cannot be smooth on any neighborhood of the origin.

There are, however, relatively loose conditions we can impose on a fiberwise quadratic function such as $v\mapsto\operatorname{Ric}(v,v)$ to ensure the local existence of such vector fields. By raising an index, we can associate any such function $f$ with a self-adjoint $(1,1)$ tensor field $A_f$, and a vector field $F$ is minimizing iff it is an eigenvector for the smallest eigenvalue of $A_f$. Thus, if the smallest eigenvalue of $A_f$ has locally constant multiplicity, one can find such a minimizing local vector field. This condition is in some sense generic, but as seen above it need not hold everywhere.

Title: Function defined on the tangent bundle


答:
This might help
$$I=  \int_0^\infty\frac{\ln(x)}{x^2+x+1}\,\mathrm dx$$
Substituting, $x=\frac1y$ and hence $\mathrm dx=-\frac{\mathrm dy}{y^2}$ we  get
$$I= \int_\infty^0-\frac{\ln(1/y)}{(y^2+y+1)/y^2}\,\frac{\mathrm dy}{y^2} $$
$$I = -I$$
$$I = 0$$

Title: Evaluate integral with no elementary antiderivative $\int_{0}^{\infty} \frac{\ln x}{x^2+x+1} dx $

Body:
This is a followup from my previous question here.

I was asked to determine if the following integral converges or diverges. If the integral converges, I was asked to determine the value.

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

With help from the question I posted above, I managed to figure out my comparison (Thanks!). I definitely know this converges now. How do I find the value though? As far as I can see, this integral has no elementary antiderivative.

This is a standard intermediate calculus class which does do not calculus of residues or anything complicated like that so I'm not how to proceed there.

Wolfram alpha says that $$\int_{0}^{\infty} \frac{\ln(x)}{x^2+x+1} dx = 0$$ and that makes a lot of sense! However, I can't find a way to justify what the value should be rigirously.

Can someone help out here?


问:
If $C^{1}(\mathbb{R}^{n})\ni g_{n}\to g$ uniformly and $\partial_{j}g_{n}\to g^{(j)}$ then $g \in C^{1}(\mathbb{R}^{n})$ and $\partial_{j}g=g^{(j)}$.
I was studying Schwartz spaces and found this notes. On page 5, there is a Lemma which states:

If $\{g_{n}\}\subset C^{1}(\mathbb{R}^{n})$ converges uniformly to $g$ and $\partial_{j}g_{n}$ converges uniformly to $g_{(j)}$, then $g \in C^{1}(\mathbb{R}^{n})$ and $\partial_{j}g = g_{(j)}$.

I'd like to see the proof of the above lemma. Does anyone know how to prove or where to find such a proof?

答:
In essence, you're asking to verify that the space of bounded differentiable functions with bounded continuous derivative $(C^1(\mathbb R^N),\|\cdot\|_{C^1})$ is complete, where $\|f\|_{C^1} := \|f\|_{C^0} + \|f'\|_{C^0}$.

(Note that you used  $n$  to mean both a sequence index and the dimension. For me dimension will be $N$.)

(Also note that some people allow all unbounded continuously differentiable functions in the set $C^1$. These two sets are not the same)

The proof is by the FTC:  we have, for any $x\in\mathbb R^N$, $n\in\mathbb N$, $t\in\mathbb R$, and $i\in \{1,\dots,N\}$,
$$ g_n(x+te_i)=g_n(x)+\int_0^t g'_n(x+se_i)ds$$
Taking limits $n\to\infty$ we see, interchanging limit and integral (uniform convergence on compact, or DCT),
$$ g(x+te_i)=g(x)+\int_0^t g_{(i)}(x+se_i)ds$$
The FTC now implies $g\in C^1$ with $g'(x)=g_{(i)}$. Specifically if you set
$$ G(t) := g(x+te_i)-g(x), H(s) = g_{(i)}(x+se_i)$$
then the FTC says that since $G(t) = \int_0^t H(s)ds$ where $H\in C^0(\mathbb R)$, then $G'$ exists and $G'=H$, and of course $g_{(i)}(x)=H(0)=G'(0)=\lim_{t\to 0} \frac{g(x+te_i)-g(x)}t$. So $\partial_i g(x)$ exists for each $i,x$, and $\partial_i g$ is equal to the $C^0(\mathbb R^N)$ function $g_{(i)}$. Hence $g$ is differentiable in the sense of Frechét, with $\|g\|_{C^1}<\infty$, which proves the result.


问:
How to determine the convergence of the series$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$?
I am currently working on some analysis exercise and was struggling to prove that the series converges $$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$$
I have tried several tricks such as converting $n^\frac{1}{n^2+1}$ to $e^{logn\frac{1}{n^2+1}}$ and try to apply ratio test and root test to $\sum_{n=1}^\infty( e^{logn\frac{1}{n^2+1}}-1)$ but couldn't succeed.

Also, I tried to upper bound the terms by $n^\frac{1}{n^2}-1$ and try to use a comparison test, but couldn't make it work either, I calculated the sum with Mathematica and it showed me it does converge to approximately $0.87357$, but I just couldn't find a way to prove it.

答: Using the fact that $e^{x}-1\sim x$ as $ x \to 0$ we can prove convergence by comparison with the series $\sum \frac {\ln n} {n^{2}+1}$. This series is convergent because $\ln n =2\ln \sqrt n\leq  2 \sqrt n$ and $\sum \frac {\sqrt n} {n^{2}+1} \leq \sum \frac {\sqrt n} {n^{2}}  <\infty$.


问:
blowup in finite time of $\frac{dx}{dt}=1+x^{10}$
I feel like this is extremely silly question. It is a question from Strogatz' Nonlinear Dynamics and Chaos
I want to show that $\frac{dx}{dt}=1+x^{10}$ blows up in finite based on the fact that  $\frac{dx}{dt}=1+x^2$ blows up in finite time.

Well clearly, $1+x^2 < 1+x^{10}$ but then I realized that this is only true for $x > 1$ (in absolute value)

So what if $x<1$?
Is there a nice and elegant way to do this problem without explicitly solving for a solution?

答: You can calculate that $x$ gets to say $1.1$ in finite time--because $dx/dt \ge 1$ for all $x$, and then go from there.


问:
Different arrangement of books (Permutation)
There are 5 math books, 3 english books, and 2 physics book.

How many different arrangements were there if the books are different from each other and the books of the same subject must be placed side by side?

In how many ways can the student assistant arrange the books on the shelf if books of the same subject are indistinguishable from each other?

In how many ways can the student assistant arrange the books on the shelf if books of the same subject are indistinguishable from each other and must be placed side by side?

I am confused with the three questions. I am not sure what is the difference of the three questions. With the first question, I did this 5!×3!×2!, but I am not sure if this is right. Please help me and thank you in advance.

答:
For 1) and 3), keep in mind that there are $3!$ ways in arrange the string MEP

for 2) you need the number of 10 character strings containing 5 M's, 3 E's and 2 P's
$$N = \frac{10!}{5!3!2!}$$


问:
Find the coordinates of the points where the line $4y = 7x + 1$ intersects the curve $x^2 - 2x + y^2 - 4y = 60$
I have tried equating the two equations to each other and then solving for x; however, I believe this is not correct.  I also tried factorizing the curve equation but I could not see a way through.

Thank you for your help.

答:
Given  $\space\mathbf{4y} = 7x + 1 \quad\land\quad
 x^2 - 2x + y^2 - 4y = 60\implies \mathbf{4y}= x^2 - 2x + y^2 -60$

$$ 7x + 1 =x^2 - 2x + y^2 -60\\
\implies x^2-9x+y^2-61=0\\
\implies y = \pm\sqrt{61-x^2 + 9 x}
\quad \text{for}\quad (9-x)x\le 61
$$
There are $12$ integer solutions easily verified with a spreadsheet or WolframAlpha

$(x,y)\in\big\{(-4,\pm3)\quad (-3,\pm5)\quad (4,\pm9)\quad
(5,\pm9)\quad (12,\pm 5)\quad (13,\pm3)\big\}$


问:
Show that $\Sigma=E(XX^T)-\mu\mu^T$.
Let $X\in \mathbb R^d$ be a random vector and $\mu=E(X)\in \mathbb R^d$ be its mean. Let $\Sigma=E[(X-\mu)(X-\mu)^T]\in\mathbb R^{d\times d}$ be the variance-covariance matrix of X. Show that $\Sigma=E(XX^T)-\mu\mu^T$.

Attempt

Ok so I know if you have $(x-y)^T(x-y)=(x^T-y^T)(x-y)=x^Tx-2x^Ty+y^Ty$. So here I tried to do something similar. $\Sigma=E[(X-\mu)(X^T-\mu^T)]=E[XX^T-X\mu^T-\mu X^T+\mu\mu^T]=E[XX^T]-E[X\mu^T+\mu X^T]+\mu\mu^T$. There's an extra term there. Are these operations not valid for the outer product? Or what am I missing?

答: What you have done is fine. By linearity of expectaion $EX\mu^{T}=\mu\mu^{T}$ and $E\mu^{T}X=\mu\mu^{T}$ so we get $\Sigma=EXX^{T}-\mu \mu^{T}$.

Title: Show that $\Sigma=E(XX^T)-\mu\mu^T$.


问:
How to verify the orthogonal projection formula?

Let $B = \{\vec{b}_1, \vec{b}_2, \vec{b}_3\}$ a orthogonal basis $\in V^3$.
  Verify that $\forall\,\vec{u} \in V^3$,
  $$\vec{u} = \left(\frac{\vec{u}\cdot\vec{b}_1}{\vec{b}_1\cdot\vec{b}_1}\right)\vec{b}_1 + \left(\frac{\vec{u}\cdot\vec{b}_2}{\vec{b}_2\cdot\vec{b}_2}\right)\vec{b}_2 + \left(\frac{\vec{u}\cdot\vec{b}_3}{\vec{b}_3\cdot\vec{b}_3}\right)\vec{b}_3$$

I know the concept and how to use it, but I have no ideia how to verify it.

Can anyone please help me to check it?

Thanks in advance

答:
This says the coordinates are the projections to axes.

Hint: Let

$$\vec{u}' = \left(\frac{\vec{u}\cdot\vec{b}_1}{\vec{b}_1\cdot\vec{b}_1}\right)\vec{b}_1 + \left(\frac{\vec{u}\cdot\vec{b}_2}{\vec{b}_2\cdot\vec{b}_2}\right)\vec{b}_2 + \left(\frac{\vec{u}\cdot\vec{b}_3}{\vec{b}_3\cdot\vec{b}_3}\right)\vec{b}_3.$$

Show $(u,b_i)=(u', b_i)$, then you have $u=u'$.


问:
If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$

If

$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$

I added $x^2$ to both side of the equation:

$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.

Body:
hint: $x^2+y^2+z^2+t^2=(\frac{x^2}{3}+y^2)+(\frac{x^2}{3}+z^2)+(\frac{x^2}{3}+t^2)
=(\frac{x}{\sqrt{3}}-y)^2+(\frac{x}{\sqrt{3}}-z)^2+(\frac{x}{\sqrt{3}}-t)^2+\frac{2}{\sqrt{3}}\cdot x(y+z+t)\geqslant 0+1\cdot x(y+z+t)$

When would the equality hold?


问:
Explain the solution to USAMO $1991/\rm P3$
Problem is as follows

Show that, for any fixed integer $n \geq 1$, the sequence

$2, 2^2, 2^{2^2}, ...\pmod n $

is eventually constant.

And the explanation was follows :

Define $i^{th}$ term of sequence to be $a_i$. Firstly assume $n$ is odd. Since $2^{\phi(n)} \equiv 1 \pmod n$. Then the thing that I dont understand is, hence if we can $a_k \equiv a_{k+1} \equiv a_{k+2} \equiv{  ...} \pmod {\phi(n)} $ for some $k$ then we would get,

$2^{a_k} \equiv 2^{a_{k+1}} \equiv { ...} \pmod n$ which is what we want.

Note : ${\phi(n)}$ is the number of integers less than $n$ and relatively prime to it.

What I don't get it is, by showing the result is true for ${\phi(n)}$, how can we  go back to showing it is true for $n$ as well?

Note : I just want an explanation, not proof.

答: It is easy to verify that the statement is true for $n=1,2$. Now assume that the statement is true for all $m<n$. Since $\phi(n)<n$, the statement is true for $\phi(n)$ by induction hypothesis and hence, as you showed, true for $n$. Therefore by strong induction, you conclude that the statement is true for all $n\geq1$. Since the sequence, $a_k$ can be written as $a_{k+1}=2^{a_k}$ with $a_1=2$, true for $\phi(n)$ implies true for $n$. The situations when $n$ is odd or even can be handled, think a little. The crucial idea is to use strong induction.


Title: If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$

Body:
If

$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$

I added $x^2$ to both side of the equation:

$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.

Body:
Note the beautiful identity $$y^2+z^2+t^2=\frac{1}{3}\left({(y+z+t)}^2+{(y-z)}^2+{(z-t)}^2+{(t-y)}^2\right)$$  thus we have $$x^2+\frac{1}{3}{(y+z+t)}^2+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=x(y+z+t)$$ $${\left (x-\frac{(y+z+t)}{2}\right)}^2+ \frac{{(y+z+t)}^2}{12}+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=0$$  which means $x=y=z=t=0$

(As its a sum of squares each term must be zero)


问:
Determinant of a differential operator of the action variations
I have to compute the determinant of a differential operator that comes from an action variation: $\frac{\delta^2S}{\delta q(\tau)\delta q(\tau)}=-M\frac{\partial^2}{\partial \tau^2}+V''(q)$.

It is known that $\delta q(0)=0$ and $\delta q(B)=0$. What I really need is the value of $D=\mid\frac{det(-M\frac{\partial^2}{\partial \tau^2}+M w_0^2)}{det'(-M\frac{\partial^2}{\partial \tau^2}+V''(q))}\mid^{1/2}$ where $Mw_0^2=V''(0)$ and det' means that the eiganvalue equal to zero is omitted to compoute the determinant.

How can D be computed?

Thank you very much.

答:
To compute functional determinant is quite difficult in general. However in this case, one can recall that the determinant is the product of the complete set of eigenvalues.

Now, find the eigenvalues of this operator
\begin{equation}
-m(\partial_T^2+\omega^2)f_n=\epsilon_n f_n
\end{equation}
A complete set of solution to the above differential equation is given by
\begin{equation}
f_n(t)=\sin\left(\frac{n\pi t}{T}\right) \quad (n=1,2,3,\cdots)
\end{equation}
and the eigenvalues are given by
\begin{equation}
 m \left(\frac{\pi ^2 n^2}{T^2}-\omega ^2\right) \quad n=1,2,\cdots
\end{equation}
Thus, the determinant is given by
\begin{equation}
\det\left(-m(\partial_T^2+\omega^2) \right)^{-\frac12}=\prod_{n=1}^\infty \left[m \left(\frac{\pi ^2 n^2}{T^2}-\omega ^2\right) \right]^{-\frac{1}{2}}
\end{equation}

Now, you can see several problems that might arise here. There might be zero eigenvalues and this method does not work. If you look at the last expression, the infinite product is clearly divergent.

But again, in your case, you can do the similar calculation to find the denominator and it turns out the divergent quantity cancels form the denominator and numerator and you get a nice finite answer.


答:
Let $$a_n=b_n-1\qquad \text{with}\qquad b_n=n^\frac{1}{n^2+1}$$

$$\log(b_n)=\frac{\log(n)}{n^2+1}$$ By Taylor expansion or long division
$$\log(b_n)=\frac{\log (n)}{n^2}-\frac{\log
   (n)}{n^4}+O\left(\frac{1}{n^6}\right)$$
$$b_n=e^{\log(b_n)}=1+\frac{\log (n)}{n^2}+\frac{2 \log ^2(n)-4 \log (n)}{4
   n^4}+O\left(\frac{1}{n^6}\right)$$
$$a_n=\frac{\log (n)}{n^2}+\frac{2 \log ^2(n)-4 \log (n)}{4
   n^4}+O\left(\frac{1}{n^6}\right)\tag 1$$ Appply it twice and continue with Taylor
$$\frac{a_{n+1}}{a_n}=1+\frac{1-2 \log (n)}{n^3}+O\left(\frac{1}{n^4}\right)~~<~1$$ So, convergence.

If you use $(1)$ and ignore the high order terms
$$\sum_{n=1}^\infty \frac{\log (n)}{n^2}=\frac{\pi ^2}{6}  (12 \log (A)-\gamma -\log (2\pi ))\sim 0.9375$$ while, without any approximation, the original infinite sum is $\sim 0.9050$

Edit

If we make the expansion of $\log(b_n)$ to order $2k$, we have explicit results for the summation (quite long formulae) and the numerical values are
$$\left(
\begin{array}{cc}
 k & \text{summation} \\
 1 & 0.937548 \\
 2 & 0.901166 \\
 3 & 0.905448 \\
 4 & 0.904966 \\
\infty & 0.904996
\end{array}
\right)$$

Title: How to determine the convergence of the series$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$?

Body:
I am currently working on some analysis exercise and was struggling to prove that the series converges $$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$$
I have tried several tricks such as converting $n^\frac{1}{n^2+1}$ to $e^{logn\frac{1}{n^2+1}}$ and try to apply ratio test and root test to $\sum_{n=1}^\infty( e^{logn\frac{1}{n^2+1}}-1)$ but couldn't succeed.

Also, I tried to upper bound the terms by $n^\frac{1}{n^2}-1$ and try to use a comparison test, but couldn't make it work either, I calculated the sum with Mathematica and it showed me it does converge to approximately $0.87357$, but I just couldn't find a way to prove it.


问:
Let $U \subset \mathbb{R}$ ,open and $h:U \to \mathbb{R}$ , where h is a uniformly continuous homeomorphism.Then $U=\mathbb{R}$

Let $U \subset \mathbb{R}$ ,open and $h:U \to \mathbb{R}$ , where h is a uniformly continuous homeomorphism.Then $U=\mathbb{R}$

My attempt :
Let $p \in \mathbb{R}$ and $p$ not in $U$ be a limit point of $U$.I will try to show that $p \in U$,so that I can conclude that $U$ is both open and closed.

Now,  we create a sequence $\{u_n\} \to p$.Since it is a convergent sequence so it is a cauchy sequence. Since $h$ is uniformly continuous and $\mathbb{R}$ is complete so $\{h(u_n)\} \to h(p)$.Now,as it is a homeomorphism so it is an onto mapping hence $p \in U$.So $U$ is both open and closed.

Now, we will try to show that the only open and closed subset of $\mathbb{R}$ are $\phi$ and $\mathbb{R}$.Let $A$ be an open and closed subset of $\mathbb{R}$ then, as $A$ is open it can be written as the union of countable disjoint  segments in $\mathbb{R}$.

So,  $A=(a_1,a_2) \cup (a_2,a_3) ..\cup (a_{n-1},a_{n})$.So $a_2 \in A^c$ which is also a closed and open set of $\mathbb{R}$.Hence, $a_2$ is an interior point of the set $A^c$, which is not possible.

I don't think it is correct and I am missing something. Can someone just go through my proof.

答:
First part requires  a small correction. You can only say that the Cauchy sequence $h(p_n)$ converge to some real number $x$.  Since $h$ is onto you can write $x$ as $h(q)$ for some $q\in U$. But continuity of the inverse map shows $p_n \to q$ which forces $p=q \in U$.

The fact that there are no open and close subsets of $\mathbb R$ other than  $\mathbb R$ itself and  the empty set is just connectedness of the real line.


问:
How many ways to combine two sets so that order of each set is preserved.
If we have sets $\lbrace1,2,...k\rbrace$  and $\lbrace  k+1,..., n\rbrace$, in how many ways can these sets be combined so that the original two sets respect the original order and 1 is always in front?

For example if k=3, n=5,
12345,

12453,

14523,

12435,

14235,

14253.

There are 3*2 of them.

I don't think the answer is $k*(n-k)$ in general since it doesn't hold when n=4, k=2.

答:
As Jared and Brian have noted in the comments, all you have to determine is which of the $n - 1$ places (places not including the first $1$) are allocated to the set $\{2,\cdots,k\}$ and which are allocated to $\{k+1,k+2,\cdots ,n\}$.  From there, the order is completely determined since the two subsequences have determined order.  There are $n - 1$ slots and $k - 1$ slots to choose for the first set, so this gives ${{n - 1} \choose {k - 1}}$ possibilities.

A fun side problem is to determine all pairs $(n,k)$ such that ${{n - 1} \choose {k - 1}} = k(n - k)$ as you conjectured.


答:
Comparison can work: let's try to show $\frac{1}{x^p}\ge x^\frac{1}{x^2+1}-1$ for some $1<p$.
$$\frac{1}{x^p}+1\ge x^\frac{1}{x^2+1}$$
$$\log\left(\frac{1}{x^p}+1\right)\ge\frac{\log(x)}{x^2+1}$$
Then by the taylor series for the logarithm we know $\log(1+x)>x-\frac{x^2}{2}$ for $x\in(0,2)$, so we have
$$\frac{1}{x^p}-\frac{1}{2x^{2p}}\ge\frac{\log(x)}{x^2+1}$$
$$\frac{(2x^p-1)(x^2+1)}{2x^{2p}}\ge\log(x)$$
Next, $\sqrt{x}\ge\log(x)$ so it's enough to find $p$ so that
$$\frac{(2x^p-1)(x^2+1)}{2x^{2p}}\ge\sqrt{x}$$
However, notice that the limit of the lhs as $x\to\infty$ is a simple power of $x$: $x^{2-p}$.  Hence if $2-p>\frac{1}{2}$, there will be some finite $N$ after which $x^{2-p}\ge\sqrt{x}$.  The sum up to $N$ converges because it is a finite sum, and the sum after $N$ converges because this inequality shows the terms after $N$ are bounded by the convergent series $\frac{1}{x^p}$.  Any $1<p<\frac{3}{2}$ will work, like say $p=\frac{4}{3}$.


问:
What is the geometric/algebraic representation of a (partial) differential relation?
My first idea is a kind of equation that may be an inequality too(?) such as:

$$ F(x,u(x),\partial u(x), \partial^2u(x),...)\leq 0 $$

In Gromov's book he generalizes PDEs into partial differential relations. But what exactly is the meaning of these?

In this summary of it he sets them up this way:

Let $p:X\to V $ be a smooth fibration, and let $X^{(r)}$ be the space or $r$-jets of smooth sections of $p$. A section $\phi$ of the bundle $X^{(r)} \to V$ is called holonomic if it is the $r$-jet of a section of $p$. A differential relation $ \mathcal{R} $ of order $r$ imposed on sections of $p$ is a subset $\mathcal{R} \subset X^{(r)}$. A section $f$ of $p$ is a solution of $\mathcal{R}$ if the $r$-jet of $f$ takes values in $\mathcal{R}$.

I can't really do much with this, this definition feels really abstract to me.

In these lecture papers the author defines PDRs the following way:

Let $M$ be a smooth $n$-manifold. Let $E$ be a smooth fibration over $M$, $E \to M$. The space of $k$-jets of the bundle $E$, $J_kE$, is the sections of a bundle $E^{(k)}$ over $M$ whose fiber $J_kE\vert_x$ at a point $x \in M$ is the space of smooth sections of $E$ in a neighborhood of $x$ modulo the equivalence relation that $f \sim g$ if they agree to order k in a neighborhood of $x$ (i.e., if the first $k$ derivatives of $f-g$ vanish when restricted to some arbitrarily small $\mathbb{R}^n \cong U \subset M$ containing $x$). Note that there is a canonical map $j^{(k)}: \Gamma(E)\to J_kE$ from sections of $E$ to $k$-jets of sections of $E$.

Defintion 2.1. A differential relation $\mathcal{R}$ of order $k$ is a subspace of $E^{(k)}$. The space of (holonomic) solutions Sol$_{\mathcal{R}}$ of $\mathcal{R}$ is the image $\Gamma(E)$ in $\Gamma(\mathcal{R})$, i.e., the sections of $\mathcal{R}$ which are $k$-jets of actual section of $E$.

These are quite similar definitions, still i find myself wondering what exactly i am reading.

What is the geometric(visual)/algebraic representation of this? How could i explain this in simpler terms?

答:
I found the answer in BULLETIN OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 49, Number 3, July 2012, Pages 447–453 - SELECTED MATHEMATICAL REVIEWS

...partial differential relations ... are, for the most part,
under-determined (or, at least, behave like those) and their solutions
are rather dense in spaces of functions. We solve and classify
solutions of these equations by means of direct (and not so direct)
geometric constructions. [These] alluded to above are usually either
equations or inequalities.

Classical PDEs of order $n$ look like this: $$ F(x,u(x),\partial u(x), \partial^2u(x),...,\partial^nu(x))= 0 $$

And as their domain is in a $n-1$ dimensional hypersurface, their solution is too.

A PDR on the other hand is way more under-determined and covers the entire
$n$-hypervolume for

$$ F(x,u(x),\partial u(x), \partial^2u(x),...\partial^nu(x))\leq 0 $$

In the example it can take up any negative point on the $x$-axis probably. And this has way more solutions too. This is what the author meant i think by "dense in spaces of solutions".


问:
Find two solutions near x=0 of the differential equation $x^2y''-x(1+x)y'+y=0$
Given this equation:

$$x^2y''-x(1+x)y'+y=0$$

From this problem I get:

$$p_{1}(x)=-(x+x^2)/x^2\quad\text{and}\quad p_{2}(x)=1/x^2$$

But I'm really having some troubles to advance in this problem. Someone could help me?

答:
Close to $x=0$, you can approximate $x+1 \to 1$. You then obtain a Cauchy-Euler equation of the form
$$x^2 y''- x y'+y =0\,.$$
Plugging in the ansatz $y=x^m$, we see that  $m=1$ is degenerate. The general solution close to $x=0$ thus reads
$$ y(x) = c_1 x + c_2 x \ln x , \qquad x\ll1\,.$$


问:
A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle.

A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle. Perpendiculars $AE$ and $CF$ are drawn at the secant. If $EB = 2$ and $BD = 6$, find $DF$.

What I Tried: Here is a picture :-

Unfortunately I have no idea to start with this. A first idea is to find similar triangles, which I find a lot, but I cannot find a way on how to use the lengths of $EB$ and $BD$.

Can anyone help me? Thank You.

答:
Let's call the point of intersection of line FC with circle as M.
If we chase the angles we find BA=DM,AE=FM and angle BEA=DFM
So, BAE is congruent to DMF and DF=EB=2</p>


问:
Struggling to conceptually think about lexicographically ordered square
I have been given the task to determine the closure of several subsets of the lexicographically ordered square, and am having an immense amount of trouble picturing the open sets. I think about an open interval containing a point with $y=1$ in the ordered pair as going up and "hitting" the top of the square before sort of sprouting back up from the closest point nearby from the $x$-axis. Almost like a line meeting the top, and then a small line coming up, right near that line. My question is, is it wrong to think about it this way, must an open interval on this square contain at least infinitely many vertical lines packed next to the "starting line" for the interval. If anyone could lead me to a sort of picture of open sets in this square, I would be more than happy.

答:
Yes, it’s wrong, because there is nothing even remotely resembling a ‘next’ vertical segment.

Let $X$ be the lexicographically ordered square, let $\preceq$ be the lexicographic order, and suppose that $\langle a,1\rangle\in X$, where $0\le a<1$. If $p=\langle x_0,y_0\rangle,q=\langle x_1,y_1\rangle\in X$, and $(p,q)_\preceq$ is an open interval in $X$ containing $\langle a,1\rangle$, then

$$p=\langle x_0,y_0\rangle\prec\langle a,1\rangle\prec\langle x_1,y_1\rangle=q\,.$$

Since $\langle a,1\rangle\prec\langle x_1,y_1\rangle$, we know that either $a=x_1$ and $1<y_1$, or $a<x_1$, and the first of these is impossible, so $a<x_1$. This means that if $\langle u,v\rangle\in X$ and $a<u<x_1$, then

$$\langle a,1\rangle\prec\langle u,v\rangle\prec\langle x_1,0\rangle\preceq\langle x_1,y_1\rangle\,:$$

the open interval $(p,q)_\preceq$ contains the entire strip $(a,x_1)\times[0,1]$ lying between $x=a$ and $x=x_1$. This isn’t just infinitely many vertical lines: it’s all of the vertical lines in a whole ordinary open interval of the $[0,1]$.

The open interval $(p,q)_\preceq$ in $X$ also contains at least a half open interval at the top of the segment $\{a\}\times[0,1]$, a set of the form $\{a\}\times(b,q]$ for some $b\in[0,1)$. Specifically, if $p=\langle a,y_0\rangle$ and $y_1>0$, then

$$(p,q)_\preceq=\big(\{a\}\times(y_0,1]\big)\cup\big((a,x_1)\times[0,1]\big)\cup\big(\{x_1\}\times[0,y_1)\big)\,.$$

(If $y_1=0$, simply omit $\{x_1\}\times[0,y_1)$, as it is then empty.)

The diagram in this answer shows an open interval in $X$ around the point $\left\langle\frac12,0\right\rangle$; what happens at a point $\langle a,1\rangle$ is similar, except that the open interval has to extend to the right rather than to the left. If you turn that picture upside down, you get an open interval around $\left\langle\frac12,1\right\rangle$.

If $x_0<a$, the open interval $(p,q)_\preceq$ will also contain a vertical strip to the left of $x=a$ and possibly an open interval at the top end of a vertical segment at the left side of that strip.


问:
A priori normal and likelihood normal with different variances.

Let, be that $X_{1}$ i $X_{2}$ are, for parameter $\theta$, i.i.d and
$X_{i} \mid \theta \sim \mathrm{N}\left(\theta, \sigma_{i}^{2}\right)
 .$  Parameter  $\theta$ has distribution a priori normal. $\theta \sim
 \mathrm{N}\left(\mu, v^{2}\right) .$ Parameters $\sigma_{1},
 \sigma_{2}$ and $v$ are known.

Compute posterior distribution  $\theta \mid X_{1}, X_{2}$

Compute cov( $\left.X_{1}, X_{2}\right)$.

Compute marginal distribution of $\left(X_{1}, X_{2}\right) .$

My intuition:
$$
P(\theta| (X_1, X_2)) \propto P(X_1|\theta)P(X_2|\theta)P(\theta)
$$
Proposition: We know that the posterior for $P(X_1|\theta)P(\theta)$ is
$$
\mathrm{N}\left(\mu_{x}, \sigma_{x}^{2}\right), \text { where } \mu_{x}=\frac{ v^{2} x_1+\sigma_{1}^{2} \mu}{ v^{2}+\sigma_{1}^{2}}, \sigma_{x}^{2}=\frac{\sigma_{1}^{2} v^{2}}{ v^{2}+\sigma_{1}^{2}}
$$
So the next step is taking: $\theta_1 \sim \mathrm{N}\left(\mu_{x}, \sigma_{x}^{2}\right)$ as a priori distribution and compute $P(X_2|\theta)P(\theta_1)$.

But it's very complicated. It's an easy way to compute a posteriori distribution?

My attempt to compute $cov(X_1,X_2) = E(X_1 X_2) - E(X_1)E(X_2) $

We know that $E(X_i) = E(E(X_i)|\theta)) = E(\theta) = \mu$

The next step: problem is when we have $E(X_1 X_2)$. It is easy to show that $X_1$ and $X_2$ are independent, so $cov(X_1,X_2) = 0$?

答:
Sketch (I did not do all the calculations but just the main ones):

I would start in the following way:

$$f_{x_1x_2\theta}\propto \exp\Bigg\{-\frac{(x_1-\theta)^2}{2\sigma_1^2}-\frac{(x_2-\theta)^2}{2\sigma_2^2}-\frac{(\theta-\mu)^2}{2v^2}\Bigg\}$$

After some algebraic manipulations...

$$ \bbox[5px,border:2px solid black]
{
f_{x_1x_2\theta}\propto \exp\left\{-\frac{1}{2\cdot\frac{1}{\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}+\frac{1}{v^2}}}\left[\theta^2-2\theta\frac{\frac{x_1}{\sigma_1^2}+\frac{x_2}{\sigma_2^2}+\frac{\mu}{v^2}}{\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}+\frac{1}{v^2}}+?\right]\right\}
\qquad (1)
}
$$

and a factor not depending on  $\theta$  (but necessary to derive the marginal) that is

$$\exp\Bigg\{-\frac{\mu^2}{2v^2}-\frac{x_1^2}{2\sigma_1^2}-\frac{x_2^2}{2\sigma_2^2}\Bigg\}$$

Now it is easy to complete the square in (1) because the integrand is the kernel of a gaussian with

$$\text{mean}=m=\frac{\frac{x_1}{\sigma_1^2}+\frac{x_2}{\sigma_2^2}+\frac{\mu}{v^2}}{\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}+\frac{1}{v^2}}$$

$$\text{variance}=\tau^2=\frac{1}{\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}+\frac{1}{v^2}}$$

This is your Posterior: a Gaussian $N(m;\tau^2)$

After completing the square in the exponent, integrate (1) in $d\theta$ to get the marginal density of $(X_1,X_2)$ that I hope to recognize again in a bivariate joint gaussian with some parameters including correlation...


问:
1 complex addition = 2 real additions, 1 complex multiplication = 4 multiplications + 2 additions.
Could you show that, when talking about Fourier transforms, that one complex addition requires 2 real additions, and one complex multiplication requires 4 multiplications and 2 additions.</p>

答:
Given two complex numbers $a+bi$ and $c+di$, one can show that their sum, defined by

$$(a+bi)+(c+di)=(a+c)+(b+d)i,$$

requires two real additions, and that their product, defined by

$$(a+bi)\cdot(c+di)=(a \cdot c-b \cdot d)+(b \cdot c+a \cdot d)i,$$

requires four real multiplications and two real additions (invert the sign of $b \cdot d$ to replace the subtraction by an addition).

However, as both $\textbf{Gauss}$ and $\textbf{MJD}$ have discovered, we can use the commutativity and distributivity properties of real numbers to replace one multiplication by three additions:

$$\begin{align*}(a+bi)\cdot(c+di)&=\Big\{k:=c\cdot(a+b)\Big\}=\Big[k-b\cdot(c+d)\Big]+\Big[k+a\cdot(d-c)\Big]i=\\&=(ac-bd)+(bc+ad)i\end{align*}$$

This approach is useful when some sums can be precomputed; ironically, this is often the case for fast Fourier transforms (why else would you ask?). Nevertheless, while somewhat related, this topic certainly deserves its own question.


问:
Existence of a curve with a given initial vector in a Manifold
Hi i am reading An introduction to manifolds by Loring and have some doubts in proposition 8.16. The statement that i have doubt in is:

By proposition 8.15, the simplest such $\alpha$ is $\alpha(t)=(a^1 t,...,a^n t)$

My questions are as following:

How did we get $\alpha(t)=(a^1 t,...,a^n t)$ using proposition 8.15? What are $\dot {c}^i(t)$ and $\left.\frac{\partial}{\partial x^i}\right|_{c(t)}$ in this case?

Why is it the simplest such $\alpha$? Is it because it is linear in $t$ that is we could have chosen more complicated dependence on $t$ but we have chosen the simplest in this sense.?

Will the coefficients $a^1,...,a^n$ vary as $t$ varies in $\alpha(t)$ or are those coefficients constant?

For reference i am attaching the screenshots of both the propositions where i have highlighted the part in which i have doubt in.

答:
In the proof of 8.16, in the part right before where you highlighted in pink, the goal is to find a curve $\alpha$ in $\Bbb{R}^n$ with $\alpha'(0) = (a_1,\dots,a_n)$.

Proposition 8.15 tells you that for any curve $\alpha(t) = (\alpha_1(t), \dots, \alpha_n(t))$, the velocity is $\alpha'(t) = (\dot{\alpha}_1(t), \dots,\dot{\alpha}_n(t))$. So you want to find such an $\alpha$ so that $\dot{\alpha}_1(0) = a_1$, $\dot{\alpha}_2(0)=a_2$, $\dots$, $\dot{\alpha}_n(0) = a_n$. I think what the author is saying is that the simplest way to do so is to use a curve for which $\dot{\alpha}_i(t) = a_i$ FOR ALL $t$ (not just $t=0$). In other words, the "simplest" way is to just have $\alpha'(t)$ be constant, and not depend on $t$ at all! This is where you get the linear expressions $\alpha_i(t) = a_i \, t$.

Just to clarify: Proposition 8.15 doesn't say this is the only choice of $\alpha$ that will work. The author is just saying this is what he/she considers the easiest choice which happens to work.


问:
To find radius of convergence of a power series
We have to find the radius of convergence of the given power series,

$$\sum_{n=0}^{\infty} (-1)^n \frac{n^{2n}}{(4n+1)^n}(x+2)^{n^2}$$

I think the only way to solve this might be the root test but all I'm getting is that
$\  \lim_{n\to \infty} \frac{n^2|x+2|^n}{4n+1} \le 1 $ for convergence.

and from there we can say that if |x+2|>1 then the series is definitely divergent, but if it is less than 1 then we have the 0. $\infty $ form which I'm not able to solve.
Could someone please confirm if what I've done is right and also give me a hint as to how to proceed ?
Thanks

答:
You are right about the series diverging for $|x+2|>1$.  To show it converges when $|x+2|<1$, intuitively you have $\frac{n^2}{4n+1}$ which goes to $\infty$ at $\mathcal{O}(n)$ and $|x-2|^n$ which goes to zero exponentially, so it dominates and the product goes to zero.  You can prove this rigorously with L'Hopital for instance by flipping appropriate terms:
$$\lim_{n\to\infty}\frac{n^2|x-2|^n}{4n+1}=\lim_{n\to\infty}\frac{n^2}{(4n+1)\left(\frac{1}{|x-2|}\right)^n}$$
Notice now the numerator and denominator both go to infinity.  So take the derivative of both to get an equivalent limit:
$$\lim_{n\to\infty}\frac{2n}{\left((4n+1)\log\left(\frac{1}{|x-2|}\right)+4\right)\left(\frac{1}{|x-2|}\right)^n}$$
Both still go to infinity so derive again:
$$\lim_{n\to\infty}\frac{2}{\left((4n+1)\log\left(\frac{1}{|x-2|}\right)+8\right)\log\left(\frac{1}{|x-2|}\right)\left(\frac{1}{|x-2|}\right)^n}$$
Now the numerator is constant but the denominator goes to infinity, so the limit is zero and we have rigorously verified that the series converges when $|x-2|<1$


问:
Norm inequality for matrix and vector $\|ABx\|_p \leq \|A\|_p\|Bx\|_p$
When I read a lecture note, it says

$$\|ABx\|_p \leq \|A\|_p\|Bx\|_p$$

$A$, $B$ are matrices with appropriate dimension.

$x$ is a vector

My questions are

Why does this inequality hold?     This includes matrix norm and vector norm.

Under what conditions, the following holds?   $$\|ABx\|_p = \|A\|_p\|Bx\|_p=\|Bx\|_p$$

I believe $A$ has to be an orthogonal matrix, i.e., $A^TA = AA^T = I$ at least so $\|A\|_p=1$.

Thanks!

答: For compatible norms, including induced norms (as $\|\cdot\|_p)$ we have $\|Ay\|\leq \|A\|\  \|y\|$ in your case $y=Bx$. This inequality follows from the definition of $\|A\|=\sup\left\lbrace \frac{\|Ay\|}{\|y\|}:y\neq 0\right\rbrace$. For the equality, you need that the vector $y$ maximizes the quantity $\frac{\|Ay\|}{\|y\|}$, which is not always an easy task. Check this question for more information Norm equivalence of a vector norm and its induced matrix norm using compactness argument


问:
Confused about interpretation and satisfiability in first order logic.
In my textbook("Introduction to mathmatical logic" by Elliot Mendelson , there is a definition of satisfiability:-

If $\mathscr B$ is an atomic wf $A^k_n(t_1 , ... , t_n)$ and $(A^n_k)^M$ is the corresponding n-place relation of the interpretation , then a sequence $s = (s_1 , s_2 , ...)$ satisfies $\mathscr B$ if and only if $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$.

Shouldn't it say "iff $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$ is true" or "iff $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$ holds" ? maybe it is saying implicitly $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$ is a theorem.But where are the axioms and rules from which it can be even deduced? I am confused.
 For me it is kind of like saying "the painting is good iff $3$"

答:
An $n-$place relation is just a set of $n-$tuples.  The $n-$tuples in the set are the ones the relation is true for.  As an example, a nice one is $\lt$ over the integers.  In the language of set theory, $\lt$ is a set of ordered pairs.  We have $(1,2) \in \lt, (2,1000)\in \lt, (4,2) \not \in \lt$ and so on.  From this point of view, $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))\iff (s^*(t_1) , ... , s^*(t_n))\in (A^n_k)^M$

When you are discussing satisfiability you don't talk about deduction rules.  You just invent a structure and show that it satisfies the axioms.  For example, if I claim $\Bbb \{{Z/34353Z},+\}$ is a group, I just have to show that it satisfies the group axioms.  I exhibit an identity, inverses for each element, and associativity and I am done.


问:
Right exactness of projective systems
Suppose that we have a systems of exact sequences $(A_{n}\rightarrow B_{n}\rightarrow C_{n}\rightarrow 0)_{n\in \mathbb{N}}$ together with transitions maps $(A_{n+1}\rightarrow A_{n})_{n\in \mathbb{N}}, (B_{n+1}\rightarrow B_{n})_{n\in \mathbb{N}}, (C_{n+1}\rightarrow C_{n})_{n\in \mathbb{N}}$ making all the relevant diagrams commute. Suppose that the transition maps make $(A_{n})_{n\in \mathbb{N}},(B_{n})_{n\in \mathbb{N}},(C_{n})_{n\in \mathbb{N}}$ into systems that satisfy the Mittag-Leffler condition. Is it true that
$$\varprojlim_{n}A_{n}\rightarrow \varprojlim_{n}B_{n} \rightarrow \varprojlim_{n}C_{n}\rightarrow 0 \ \ (\ast)$$
is exact?

This is well-known to hold if we had short-exact sequences instead (and for that it would have been enough to only require that $(A_{n})_{n\in \mathbb{N}}$ satisfies the Mittag-Leffler condition), so I was wondering whether if we factor the above system into one of short exact sequences, then the factors will automatically satisfy the Mittag-Leffler condition too? Or is there some counterexample to $(\ast)$?

答:
Define an inverse system of short exact sequences \begin{align*} \textstyle S_{n} = \{0 \to K_{n} \to A_{n} \to B_{n} \to 0\} = \{0 \to \mathbb{Z} \to \mathbb{Z}[\frac{1}{p}] \to \mathbb{Z}[\frac{1}{p}]/\mathbb{Z} \to 0\} \end{align*} where all the vertical maps in the morphism $S_{n+1} \to S_{n}$ are multiplication-by-$p$. We have $\varprojlim_{n} K_{n} = 0$ since no nonzero integer is $p$-divisible, we have $\varprojlim_{n} A_{n} = \mathbb{Z}[\frac{1}{p}]$ since the transition maps $K_{n+1} \to K_{n}$ are isomorphisms, and $\varprojlim_{n} B_{n}$ contains $\mathbb{Z}_{p}$: take the limit of inclusions $\mathbb{Z}/p^{n}\mathbb{Z} \to \mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ sending $1 \mapsto \frac{1}{p^{n}}$. Since $\varprojlim_{n} A_{n}$ is countable and $\varprojlim_{n} B_{n}$ is uncountable, the morphism $\varprojlim_{n} A_{n} \to \varprojlim_{n} B_{n}$ is not surjective.

We define $C_{n} = 0$ for all $n$. The systems $\{A_{n}\},\{B_{n}\},\{C_{n}\}$ satisfy ML since all transition maps are surjective.

Title: Right exactness of projective systems


答:
For $n$ odd, $2$ is relatively prime to $n$, so the sequence of $a_i$ is some sequence in the set of numbers relatively prime to $n$, hence forms a set of residues modulo $n$ of size no greater than $\phi(n)$.  This is the content of Euler's theorem.  So the sequence
$$  2^{a_i} \pmod{n}, 2^{a_{i+1}} \pmod{n}, 2^{a_{i+2}} \pmod{n}, ...  $$
is termwise equal to the sequence
$$  2^{[a_i \pmod{\phi(n)}]} \pmod{n}, 2^{[a_{i+1} \pmod{\phi(n)}]} \pmod{n}, 2^{[a_{i+2} \pmod{\phi(n)}]} \pmod{n}, ... \text{,}  $$
where the square brackets pick a representative from the equivalence class given by the modulus (for instance, the least nonnegative representative of the residue class).

Also, this sequence of exponents is "one behind" the sequence term being defined.  That is,
$$  a_k \cong 2^{a_{k-1}}  \pmod{n}  \text{.}  $$
So the sequence made of the terms on the left-hand side of that equality is eventually constant if the sequence made of the exponents on the right-hand side are eventually all in the same residue class module $\phi(n)$.

Let's see an example.  Let's set $n = 7$ and see that $\phi(7) = 6$.
\begin{align*}
a_1 &\cong 2 \pmod{7} \cong 2^{[1 \pmod{6}]} \pmod{7}  \\
a_2 &\cong 4 \pmod{7} \cong 2^{[2 \pmod{6}]} \pmod{7}  \\
a_3 &\cong 16 \pmod{7} \cong 2^{[4 \pmod{6}]} \pmod{7}  \\
a_4 &\cong 65536 \pmod{7} \cong 2^{[16 \pmod{6}]} \cong 2^4 \pmod{7}  \\
a_5 &\cong 2^{65536} \pmod{7} \cong 2^{[65536 \pmod{6}]} \cong 2^4 \pmod{7}
&\vdots
\end{align*}
Notice that at each step, $a_k \cong 2^{[a_{k-1} \pmod{\phi(n)}]} \pmod{n}$.  So if $\left( [a_{k-1} \pmod{\phi(n)}] \right)_k$ is eventually constant, then so is $(a_k \pmod{n})_k$.  Strong induction at step $n$ gives us the eventual constancy of $\left( [a_{k-1} \pmod{\phi(n)}] \right)_k$, because $\phi(n) < n$, and the above leverages that to constancy of $(a_k \pmod{n})_k$.

Title: Explain the solution to USAMO $1991/\rm P3$

Body:
Problem is as follows

Show that, for any fixed integer $n \geq 1$, the sequence

$2, 2^2, 2^{2^2}, ...\pmod n $

is eventually constant.

And the explanation was follows :

Define $i^{th}$ term of sequence to be $a_i$. Firstly assume $n$ is odd. Since $2^{\phi(n)} \equiv 1 \pmod n$. Then the thing that I dont understand is, hence if we can $a_k \equiv a_{k+1} \equiv a_{k+2} \equiv{  ...} \pmod {\phi(n)} $ for some $k$ then we would get,

$2^{a_k} \equiv 2^{a_{k+1}} \equiv { ...} \pmod n$ which is what we want.

Note : ${\phi(n)}$ is the number of integers less than $n$ and relatively prime to it.

What I don't get it is, by showing the result is true for ${\phi(n)}$, how can we  go back to showing it is true for $n$ as well?

Note : I just want an explanation, not proof.


问:
Help integrating the contour integral $\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz$ around the unit circle
Consider the integral $$\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz$$

where C is the anti-clockwise oriented unit circle and q is a complex constant.

I have no clue how to integrate this. I tried using the residue theorem but it just doesn't come out correctly. Any help would be appreciated. The reason I am stuck is because it seems like there are an infinite amount of poles at z =0, so am very confused

答:
Following the recommendation of @honey kumar you can expend the integrand into the series. The integrand is analytic in all points of the unit circle, except for $z=0$. So, you can deform the contour into a circle of smaller radius $r\to0$ around $z=0$, and then to choose all terms of the integrand expansion into series containing $\frac{1}{z}$, because only these terms will contribute, according to the residual theorem.

$\frac{e^{\frac{1}{z}}}{z(1-qz)}=\frac{1}{z}\left(1+\frac{1}{z1!}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+qz+q^2z^2+q^3z^3+...\right)$

You need to take only simple poles from this expression. It gives:

$I(q)=\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz=\oint_C\frac{dz}{z}\left(1+\frac{qz}{z1!}+\frac{q^2z^2}{z^22!}+\frac{q^3z^3}{z^33!}+...\right)=\oint_C\frac{dz}{z}e^q=2\pi{i}e^q$

An important point:

thanks to @A rural reader who points out that $I(q)=2\pi{i}e^q$ at $|q|<1$ and that at $|q|>1$ the integral is zero.

Indeed, when we squeeze the contour ($|z|=r\to0$) and if $|q|>1$ the integration path will have to cross the point $z=\frac{1}{q}$ first, and only after we can enjoy the expansion of $\frac{1}{1-qz}=1=qz+q^2z^2+...$. When crossing the point $z=\frac{1}{q}$ we will get an additional residual in this point ($=-e^q$) which will cancel the residual at $z=0$.

We can also show that $I(|q|>1)=0$ directly:

$I(q)=\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz=-\oint_C \frac{e^{\frac{1}{z}}}{qz^2(1-\frac{1}{qz})}dz=$$=-\oint_C\frac{dz}{qz^2}\left(1+\frac{1}{z1!}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+\frac{1}{qz}+\frac{1}{q^2z^2}+...\right)=0$

because there is no  simple poles at $z=0$.


问:
What's the word for being able to "pull something out of a function"?
If I have something like $(5x + 7x) = x(5 + 7),$ I've "factored" out the $x$.

With linear functions you can say $f(a \cdot x) = a \cdot f(x)$ "by linearity", you've (for lack of better phrasing) "pulled out the multiplication".

Or for preimages of sets, the preimage of union = union of preimage $f^{-1}(\cup_i A_i) = \cup_i f^{-1}(A_i)$.

Is there a general term for saying some operation inside a function works in a similar way? Something like

$$f(A \odot B) = f(A) \odot f(B)$$

where $\odot$ is some general operator. I guess we could say "linearity in $\odot$", is that standard?

答:
This is also called distributivity.  Your first example,
$$  5x+7x = x(5+7)  $$
is a statement of distributivity of multiplication over addition (and assumes commutative multiplication, otherwise the right-hand side would be "$(5+7)x$").

In your terminal example, you could state "$f$ distributes over $\odot$".


问:
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$
Assume that all logarithms are natural.

Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$

This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$

My first inclination was to apply the well known inequality, $\log(z)\leq 2(\sqrt{z}-1)$ if $z>0$, to the left side by letting $z=\frac{x-y}{x+y}$, but this did not help me.

I also tried double induction on $(x,y)$ (since $x$ and $y$ are integers) but I could not finish it.

I appreciate any help, thank you.

Body:
Here's a proof
by comparing the power series.

Eliding some computation,

$\log\left( \frac{1+t}{1-t}\right)
=2\sum_{n=1}^{\infty} \dfrac{t^{2n-1}}{2n-1}
$.

$\sqrt{1+t}-\sqrt{1-t}
=2\sum_{n=1}^{\infty} \binom{1/2}{2n-1}t^{2n-1}
=2\sum_{n=1}^{\infty} \dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2}t^{2n-1}
$

So we want
$\dfrac{1}{2^{4n-3}(2n-1)}\binom{4n-4}{2n-2}
\lt \dfrac{1}{2n-1}
$

or
$\dfrac{1}{2^{4n-3}}\binom{4n-4}{2n-2}
\lt 1
$

or,
with $2n-2=m$,
$\dfrac{1}{2^{2m+1}}\binom{2m}{m}
\lt 1
$.

But
$(1+1)^{2m}
=2^{2m}
$
and,
for $m \ge 1$,
$(1+1)^{2m}
=\sum_{k=0}^{2m}\binom{2m}{k}
\gt \binom{2m}{m}
$
so
$\dfrac{1}{2^{2m+1}}\binom{2m}{m}
\lt \dfrac12
$.


问:
A weird algorithm for finding the square root of any integer
A weird algorithm for finding the square root of any integer
goes as the following:

example $√25$: $25 - 1 -3 - 5 - 7 - 9 = 0$

We have deducted 5 numbers (-1, -3, -5, -7, -9) without resulting in a negative number, and thus the answer is 5

example $√27$: $27 - 1 -3 - 5 - 7 - 9 = 2$

We have deducted 5 numbers (-1, -3, -5, -7, -9) without resulting in a negative number, and thus the answer is 5.?, but in this case we have a remainder of 2 and thus we have to complete a decimal for 5.?

We have two ways to deal with the remainder:

First way: We can multiply $27 * 100$, or $27 * 10000$ or $27 * 1000000$ (every additional 00 will result in a more precise answer:

$2700 - 1 -3 -5 -7 -9 -11 -13 -15 -17 -19 -21 -23 -25 -27 -29 -31 -33 -35 -37 -39 - 41 -43 -45 -47 -49 -51 -53 -55 -57 -59 -61 -63 -65 -67 -69 -71 -73 -75 -77 -79 -81 -83 -85 - 87 -89 -91 -93 -95 -97 -99 -101 = 99$

We have deducted 51 numbers and there for we know the first decimal for √27 = 5.1?

the more 00's we put at the end of 27, the more decimals we will get.

Second way: $27−1−3−5−7−9=2$  We take the remainder of 2 and divide it by what would of been the next number to deduct in the sequences. In this case it would of been 11

$2/11 = 0.18181818181818....$

Here is the funny thing: We then take the result and we can add it to what would of been the next number to deduct in the sequence minus 1. In this case 11-1 = 10  and 10 + 0.18181818181818 = 10.181818181818

$2/10.181818181818181818 = 0.19642857142857142$

and we repeat the process until there is no more change:

$2/10.19642857142857142 = 0.19614711033274956$

$2/10.19614711033274956 = 0.19615252490553076$

$2/10.19615252490553076 = 0.1961524207405411$

$2/10.1961524207405411 = 0.19615242274445532$

$2/10.19615242274445532 = 0.19615242270590422$

At the end when there is no more change (and without rounding the results to begin with)

$√27 = 5.196152422706631880582....$

The first way is self explanatory, but the second way is very weird.
Why is it that first we divide by what would be the next number in the sequence and then we can take the decimal and add it to the next number minus 1? And why repeating the step of adding the result decimal to the next number in the sequence minus 1, acts as a sieve for sharpening the answer?

答:
We aim to find the root of $f(x) = (x+a)^2 - b$, where $a$ is the integer part of the square root (which is easier to find) and $x$ would be the fractional part of the square root.

We have

$$x^2 + 2ax + a^2-b = 0$$

$$x = \frac {b-a^2}{2a+x}$$

So we can set up the iteration

$$x_{n+1} = \frac {b-a^2}{2a + x_n}$$

Note that $a$ is the integer part of the square root (which is equal to the numbers we deducted), and we take $x_0 = 0$. Then each iteration we use the remainder $(b-a^2)$ and divide it by twice that number ($2a$, which happens to be the next number to deduct minus one) with the decimals added to it $(x_n)$.

If our method converges, it will converge to $L = \sqrt b - a$. Now we check that it does converge to $L$ by showing $|x_{n+1} - L| < |x_n-L|$ (that is, each iteration we go closer and closer to the value). We have:

$$|x_{n+1} - L| = \left|\frac {b-a^2}{2a+ x_n}-(\sqrt b-a)\right| = \frac {\sqrt b - a}{2a+x_n}|\sqrt b + a - 2a - x_n| = \frac{\sqrt b - a}{2a+x_n}|L-x_n|$$

since $\sqrt b-a < 1$ and $2a+x_n \ge 2$, we have $|x_{n+1}-L| < \frac12|x_n-L|$ and we are done.

This also shows that this method converges linearly with rate of convergence $\dfrac {\sqrt b - a}{\sqrt b + a}$.


问:
Can we show $(U\oplus V)^\perp=U^\perp\oplus V^\perp$?
Let $H$ be a complex Hilbert space, $T\in\mathfrak L(H)$ and $U,V$ be subspaces of $H$. Let $W:=U\oplus V$ denote the orthogonal direct sum of $U$ and $V$. Suppose we know $TU^\perp\subseteq U^\perp$ and $TV^\perp\subseteq V^\perp$.

Are we able to conclude $TW^\perp\subseteq W^\perp$? Maybe this is a consequence of $W^\perp=W^\perp\oplus V^\perp$. Can we show this?

I've tried to show $W^\perp=W^\perp\oplus V^\perp$, but neither of the inclusions is obvious to me. For example, if $x\in U^\perp\oplus V^\perp$, then there are $(y,z)\in U^\perp\times V^\perp$ with $y\perp z$ and $x=y+z$. Analogously, if $w\in U\oplus V$, there are $(u,v)\in U\times V$ with $u\perp v$ and $w=u+v$. We need to show that $\langle x,w\rangle_H=0$. Now, clearly, $\langle x,w\rangle_H=\langle y,v\rangle_H+\langle z,u\rangle_H$ ... Now I guess, if the claim holds, we somehow need to "add a $0$" in a clever way.

答:
$W=U+ V$ implies $W^\perp = U^\perp\cap V^\perp$. That is, a vector is orthogonal to all $w\in W$ if and only if it is orthogonal to all $u\in U$ and $v\in V$. And, purley set-theoreticaly, $f(X\cap Y)\subseteq f(X)\cap f(Y)$. Thus

$$ TU^\perp\subseteq U^\perp, TV^\perp\subseteq V^\perp \implies $$
$$ TW^\perp=T(U^\perp\cap V^\perp)\subseteq TU^\perp\cap TV^\perp\subseteq U^\perp\cap V^\perp=W. $$


问:
How to solve congruence modulo equations?
While studying Affine Cipher in cryptography it tells that we need to solve a system of modulo congruence equations.

The equations are:

$8\alpha+\beta\equiv 15 \pmod{26}$

$5\alpha+\beta\equiv 16 \pmod{26}$

Could anyone tell how to solve these equations.

答:
Just making @labbhattacharjee's answer a bit more clear :

$$8\alpha + \beta \equiv 15 \pmod{26} \longrightarrow(1)$$
$$5\alpha + \beta \equiv 16 \pmod{26} \longrightarrow(2)$$

$(1) - (2) \implies 3\alpha \equiv -1 \pmod{26}$

Directly proceeding from here to get $3\alpha \equiv 25 \pmod{26}$ makes it difficult to isolate $\alpha$(unless we find a suitable value for the equation $3\alpha = 26n + 25$ or say, we are ready to work it out; but here, $n = 1$ is a suitable solution and hence you may choose this) , hence we take the nearest multiple of $26$ whose predecessor is a multiple of $3$, i.e., $52$.

So we get $3\alpha = 51$ and $\alpha = 17$.Plugging this in any of the two congruence equations will get you the path of solving for $\beta$.


问:
Is this true $y=x^2 \Rightarrow x=y^{1/2}$?
If I am trying to solve $y=x^2$ for $x$ then does $x=y^{1/2}$ by raising the two sides to $1/2$?

答: Since it's an even root you would write $y=x^2\implies x=\pm y^{1/2}$.


答: X could also be $x = -y^{\frac{1}{2}}$ since $(-1)^{2}=1$. Where are you working anyways? The reals the complex plane?

Title: Is this true $y=x^2 \Rightarrow x=y^{1/2}$?

Body: If I am trying to solve $y=x^2$ for $x$ then does $x=y^{1/2}$ by raising the two sides to $1/2$?


问:
If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$ Hausdorff and $f$ is continuous, then $f(\cap F_n) = \cap f(F_n)$
I have to prove the following statement:

Let $X$ and $Y$ be Hausdorff Spaces and $f:X\to Y$ a continuous function. If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$, then $f(\cap F_n) = \cap f(F_n)$.

I already proved it, however, my proof doesn't use the hypothesis that $Y$ is Hausdorff, and that makes me think it might be wrong, but I don't find any mistakes. I would be really thankful if someone could verify it.

Proof. First we see that $f(\cap F_n) \subseteq \cap f(F_n)$. If $x\in f(\cap F_n)$, there must exists $a\in \cap F_n$ such that $f(a)=x$. It is clear that $a\in F_n$, for all $n\in\mathbb{N}$ and in consequence $x\in f(F_n)$, for all $n\in\mathbb{N}$, which implies $x\in\cap f(F_n)$.

Now we are going to prove that $\cap f( F_n) \subseteq f(\cap F_n)$. Let $x\in\cap f( F_n)$, this implies that $x\in f( F_n)$ for all $n\in\mathbb{N}$. From here we get for each $n\in\mathbb{N}$ there exists $y_n\in F_n$ such that $f(y_n)=x$. Since $\{F_n\}$ is a decreasing sequence, we have that $F_n \subseteq F_1$ for all $n\in\mathbb{N}$ and this implies that $y_n\in F_1$ for all $n\in\mathbb{N}$. This way, $\{y_n\}$ is a sequence in $F_1$. Because $F_1$ is compact, $\{y_n\}$ must have a convergent subsequence. Lets say this subsequence is $\{y_{n_k}\}$, and that it converges to $y$. Futhermore, since $X$ is Hausdorff, this limit is unique.

We see that $y\in\cap F_n$. Let $i\in\mathbb{N}$. There exists a $j\in\mathbb{N}$ such that $i<n_j$. Now lets consider the sequence $\{a_k^j\}$ given by $a_k^j=y_{n_{k+j}}$. It is clear that $a_k^j=y_{n_{k+j}}\in F_{n_{k+j}}\subseteq F_{n_{j}}\subseteq F_i$, which implies that the sequence $\{a_k^j\}$ is a sequence in $F_i$. Also, this sequence is just a shift of $\{y_{n_{k}}\}$, and because of this it must also converge to $y$. Because $F_i$ is compact, and $X$ is Hausdorff, $F_i$ is closed, and since $\{a_k^j\}$ is in $F_i$, we have that $y\in F_i$. As $i\in\mathbb{N}$ is arbitrary, we get that $y\in\cap F_n$.

Finally, using the continuity of $f$ and the uniqueness of the limit of $\{y_{n_k}\}$ we get that $f(y)=f(\text{lim}_{k\to\infty}y_{n_k})=\text{lim}_{k\to\infty}f(y_{n_k})=\text{lim}_{k\to\infty}x=x$, from where we get that $x=f(y)\in f(\cap F_n)$.

In conclusion, $f(\cap F_n) = \cap f(F_n)$

答:
Here is a concrete example showing how the result can fail when $Y$ is not $T_1$; I think that it’s probably about as simple as you can get.

Let $X=\{p\}\cup\Bbb N$, where $p$ is some point not in $\Bbb N$; the topology on $X$ is

$$\wp(\Bbb N)\cup\{X\setminus F:F\subseteq\Bbb N\text{ is finite}\}\,.$$

In other words, points of $\Bbb N$ are isolated, and open nbhds of $p$ contain all but finitely many points of $\Bbb N$; clearly $X$ is a compact Hausdorff space. For $n\in\Bbb N$ let

$$F_n=X\setminus\{k\in\Bbb N:k<n\}\,;$$

then

$$X=F_0\supsetneqq F_1\supsetneqq F_2\supsetneqq\ldots\,,$$

each $F_n$ is compact, and $\bigcap\limits_{n\in\Bbb N}F_n=\{p\}$.

Let $Y=\Bbb N$. For each $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k\ge n\}$; the topology on $Y$ is $\{\varnothing\}\cup\{U_n:n\in\Bbb N\}$, so $Y$ is $T_0$ but not $T_1$.

Let

$$f:X\to Y:x\mapsto\begin{cases}
0,&\text{if }x=p\\
1,&\text{if }x=n\text{ and }n\text{ is odd}\\
\frac{n+4}2,&\text{if }x=n\text{ and }n\text{ is even;}
\end{cases}$$

$f^{-1}[U_0]=f^{-1}[Y]=X$, and $f^{-1}[U_n]\subseteq\Bbb N$ if $n\ge 1$, so $f$ is continuous, and it’s not hard to verify that

$$f[F_n]=\{0,1\}\cup U_{\left\lceil\frac{n+4}2\right\rceil}$$

for each $n\in\Bbb N$. Thus,

$$f\left[\bigcap_{n\in\Bbb N}F_n\right]=f[\{p\}]=\{0\}\,,$$

but

$$\bigcap_{n\in\Bbb N}f[F_n]=\bigcap_{n\in\Bbb N}\left(\{0,1\}\cup U_{\left\lceil\frac{n+4}2\right\rceil}\right)=\{0,1\}\,.$$


问:
How to prove that End(R) is isomorphic to R?
I tried to prove this:
$R$ as a ring is isomorphic to $End(R)$, where $R$ is considered as a left $R$-module. The map of isomorphism is

$$F:R\to End(R), \quad   F(r)=fr,$$
where $fr(a)=ar$. And I define the multiplication in $End(R)$ by $(.)$, where $h.g=g\circ h$ for $g,h$ in $End(R)$.

Is this true?

答:
It's true that: $a\mapsto ar$ is a left module homomorphism.

If we call this map $a\mapsto ar$ by $\theta_r$, then indeed $\theta:R\to End(_RR)$.

Check that it's additive.

Check that it's multiplicative. (You will absolutely need your rule that $f\circ g=g\cdot f$. The $\cdot$ operation you have given is the multiplication in $(End(R))^{op}$)

If you instead try to show that $R\cong End(R_R)$ in the same way (with $\theta_r$ denoting $a\mapsto ra$), you will have better luck. Can you spot where the two cases are different?


问:
Construction of rational between two reals
I'm reading a book that uses this construction in a proof but I can't see why it works.

Suppose $s$ is real and positive, and $n$ is a positive integer. Then $w=\frac{[(n+1)s]}{n+1}+\frac{1}{n+1}$ is a rational where $s<w<s+\frac{1}{n}$ and $[x]$ is the largest integer less than or equal to x.

I'm aware that such rationals exist, could someone explain why this is one of them?

答:
Certainly $w$ is rational, as $[(n + 1)s]$ is an integer by definition of $[\cdot]$. The question then becomes to show that $s < w < s + \frac{1}{n}$.

Note that $[(n + 1)s] \le (n + 1)s$, as $[(n + 1)s]$ is an integer less than or equal to $(n + 1)s$. Moreover, it is the greatest such integer, so $[(n + 1)s] + 1$ is not less than or equal to $(n + 1)s$. That is,
$$[(n + 1)s] \le (n + 1)s < [(n + 1)s] + 1.$$
If we divide all three sides by the positive quantity $n + 1$, then we get
$$\frac{[(n + 1)s]}{n + 1} \le s < \frac{[(n + 1)s]}{n + 1} + \frac{1}{n + 1} = w.$$
Recall also that $\frac{1}{n + 1} < \frac{1}{n}$, so
$$\frac{[(n + 1)s]}{n + 1} \le s \implies w - \frac{1}{n + 1} \le s \implies w - \frac{1}{n} < s.$$
Thus we get $s < w < s + \frac{1}{n}$ as required.