问:
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
The question is:

When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?

The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?

Edit

I should add that $f:\mathbb R\to\mathbb R$ to the question.

答:
Commenting to achille hui's answer, the idea of choosing $\lambda$ comes from the observation that the translation operator $\tau_{h} f (x) = f(x+h)$ corresponds to the exponential of the differential operator

$$\tau_{h}f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!}h^{n} = e^{h d/dx} f(x).$$

on the space of analytic functions. So writing $D = d/dx$, the given equation can be transformed into

$$ e^{D}f(x) - f(x) = Df(x) \quad \Longleftrightarrow \quad (e^{D} - 1 - D)f(x) = 0.$$

By making an ansatz that a solution $f(x)$ has the form $f(x) = e^{\lambda x}$, it is easy to see that the equation above is equivalent to $e^{\lambda} - 1 - \lambda = 0$, the very starting point of the achille hui's answer.


问:
Question on finding more than one group homomorphisms from one group to another
Question:

How many group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$ are there?

My attempts:

For the group of $\mathbb Z/3$ with addition as the operation, I know that it refers to the group of modulo 3.

And for $S_4$, I think it refers to the symmetry group of the all the permutations of 4 elements, which is a group consisting of 24 elements.

And I know that for the group $(\mathbb Z/3, +)$ to be a homomorphism to $S_4$, I need to check if $f(ab) = f(a)f(b)$, for f being the mapping from group $(\mathbb Z/3, +)$ to $S_4$ and $a, b $ are any elements in $(\mathbb Z/3, +). $

My Question 1:
As I am not sure on the operation in $S_4$, I am quite stuck at the checking here.

My Question 2:
If $f(ab) = f(a)f(b)$ is true, it stands for there is only ONE homomorphism exist from $(\mathbb Z/3, +)$ to $S_4$, right? But I think the question gave hints that there is more than ONE homomorphism from $(\mathbb Z/3, +)$ to $S_4$. So how to find out the other group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$?

Please assist, thank you so much!

答:
The isomorphism theorem gives a handy framework with which to approach this sort of problem.  We are searching for homomorphisms $\phi: \mathbb{Z}_3 \to S_4$
wherein:

$$\mathbb{Z}_3/\ker(\phi) \cong \text{Im}(\phi)$$

We know that the kernel of a group homomorphism $G \to H$ is a normal subgroup of $G$; in particular, the size of the kernel must divide the order of $G$.  This forces the kernel of a homomorphism $\mathbb{Z}_3 \to S_4$ to either be trivial or contain all of $\mathbb{Z}_3$.

If the kernel is all of $\mathbb{Z}_3$, this corresponds to the trivial homomorphism (which always exists between any two groups), where $\phi(g) = \{ \text{id} \}$ for all $g \in \mathbb{Z}_3$.  If the kernel is trivial, then $\phi$ is injective, and we have $\mathbb{Z}_3 \cong \text{Im}(\phi)$, where $\text{Im}(\phi)$ is necessarily a subgroup of $S_4$.  Hence, we need to figure out how many distinct copies of $\mathbb{Z}_3$ exist as subgroups of $S_4$.

This can be tackled in a number of ways.  The most straightforward approach would be to just count the number of unique $3$-cycles in $S_4$ while bearing in mind that each distinct copy of $\mathbb{Z}_3$ contains a pair of $3$-cycles (plus the identity).

Another approach uses the fact  that the rotational symmetry group of a cube is isomorphic to $S_4$.  This is very useful to know even outside of the context of this problem (which is why I bring it up here) because it provides a handy visual reference for this permutation group, rendering the group potentially easier to explore than when getting lost in a mess of cycle notation.  Anyway, suppose, for example, that you take a rotational axis to be through the middle of any $2$ opposing faces of a cube.  Then a $90$-degree rotation about this axis has order $4$, and per the isomorphism, each such rotational axis corresponds$^\ddagger$ to a distinct copy of $\mathbb{Z}_4$ existing as  a subgroup of $S_4$.  Likewise, copies of $\mathbb{Z}_3$ appearing as subgroups of $S_4$ correspond to the rotational axes of a cube about which there exist rotations of order $3$.  Well, how many such axes are there?

The total number of nontrivial homomorphisms $\mathbb{Z}_3 \to S_4$ should be double the number of copies of $\mathbb{Z}_3$ that appear as subgroups of $S_4$.  To wit, we choose one the copies of $\mathbb{Z}_3$ inside of $S_4$ to which we can map $\mathbb{Z}_3$ injectively, and there are two ways to define the mapping because $\mathbb{Z}_3$ has two elements that can serve as generators.

$\underline{\textbf{Footnotes}}:$

$^\dagger$Any homomorphism $\phi:G \to H$ is determined by its action on a generating set for $G$.  Because a single element can generate $\mathbb{Z}_n$, determining where this element goes determines any homomorphism out of $\mathbb{Z}_n$.  This is because any $a \in \mathbb{Z}_n$ can be written as $g^k$ for some $k$, and we have $\phi(g^k) = \phi(g)^k$.  Therefore, this enumerate-copies-of-$\mathbb{Z}_3$ approach is equivalent to figuring out how many elements of order $3$ in $S_4$ there are to which we can send the generator, as per J.W. Tanner's comment.

$^\ddagger$ Note that, as is discussed in the hyperlink, you can figure out which permutations correspond to which rotations by labeling the $4$ diagonals of the cube.

ContentLicense: CC BY-SA 4.0


问:
The range of a continuous even function
Is there any rule that says that this range cannot be $\mathbb{R}$? Apparently there is, but then what about functions like $\log |x|$?

答:
Consider $f(x) = x\sin x$. It is easy to check that $f$ is even.

Also, $f$ is continuous and obtains arbitrarily large positive and negative values. You can use the intermediate value theorem to show that every real number is in the range of $f$.

Hence, $f$ is a continuous even function whose range is $\mathbb{R}$. Therefore, there is no rule that a continuous even function cannot have all of $\mathbb{R}$ as its range.

Title: The range of a continuous even function


问:
How to prove a basis in this problem?
Let $\mathbb{B} = \{1, \cos t, \cos^2 t, \cos^3 t, \cos^4 t\}$ and $\mathbb{C} =\{1, \cos t, \cos2 t, \cos3 t, \cos4 t\}$ and assume the following identities:

$\cos 2t = -1+ 2\cos^2t$

$\cos3t = -3+ 4\cos^3t$

$\cos4t = 1- 8\cos^2t+ 8\cos^4t$

Let $\mathbb{H} = \mathrm{Span}\{\mathbb{B}\}$.

Explain why $\mathbb{B}$ is a basis for H

Prove that C is another basis for H by first calculating the $\mathbb{B}$-coordinates of the elements of C

When working on these, I believe that an answer to the first question would refer to the definition of a basis. Isn't being the span of B mean that H must have the basis B? Is there a theorem to reinforce this with, or am I completely wrong?

As for the second question, I'm not really sure how to construct a B-coordinate without vectors or a matrix. I'd love a hint for where to start with this.

答:
Suppose the linear map $T: H\rightarrow H$. To find the $B$-coordinates of the elements of $C$, for example, $cos (2t)$, you just need to solve
$$cos(2t)=a\cdot 1+b\cdot cos t+c\cdot cos^2 t+d\cdot cos^3 t +e\cdot cos^4 t$$
It is easy to see that $a=-1, c=2, b=d=e=0.$
The matrix of this linear map is
$$\left [ \begin{matrix}
    1 & 0 &-1&-3&1 \\
    0 & 1&0&0&0 \\
    0& 0&2&0&-8\\
    0&0&0&4&0\\
    0&0&0&0&8
   \end{matrix} \right ]
$$

Since the matrix is invertible, $C$ is also a basis.

Title: How to prove a basis in this problem?


问:
If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$
If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$

Should I manipulate the equation?  I'm stuck on this problem.  Answers are greatly appreciated.

ContentLicense: CC BY-SA 3.0

Body:
Use the Cauchy-Schwarz inequality:

$$(x\times 1+y\times 1 )^2 \leq (x^2+y^2)(1^2+1^2), x^2+y^2=1$$

$$ (x+y)^2 \leq (1)(2)$$

$$-\sqrt{2} \leq x+y \leq \sqrt{2} $$


问:
Finding all subgroups of direct product of cyclic groups
Find all subgroups of $\mathbb{Z}_{9} \oplus \mathbb{Z}_{3}$ of order $3$.

I have been having some confusion with these types of problems.

I know all of the subgroups for this problem would be isomorphic to $\mathbb{Z}_{3}$, hence cyclic, and would be of the form $\langle (a,b) \rangle$ where $|a|=3,|b|=3$, $|a|=1,|b|=3$,$|a|=3,|b|=1$.

Now I would start out by listing the elements of the form above, and the cyclic subgroup generated by each element.

$\langle (3,1) \rangle$,$\langle (0,1) \rangle$,$\langle (3,0) \rangle$

now my main confusion comes with the cyclic group generated by the two elements:

$\langle (3,2) \rangle=\langle (6,1) \rangle$

How can we know that this group is generated by these two elements?

This stirs up some confusion. I know a cyclic group that is not a direct product of groups like $\mathbb{Z}_{n}$ can be represented by any of its generators. Now with direct products it seems like combinations of different generators that generate the same cyclic group in non-direct products yield different groups. For instance in the last subgroup with the two generators, I guess I am confused because normally in $\mathbb{Z}_3$, $\langle 1 \rangle =\langle 2 \rangle$ however in the direct product the subgroups $\langle (3,1) \rangle \neq \langle (3,2) \rangle$. I cannot pinpoint precisely what I want to say here, I guess I just need someone to explain why different combinations of ordered pairs, with  elements that would generate the same cyclic subgroup in a group that is not the direct product can generate a different subgroup while in a group that is in a direct product.

答:
Note that your list of elements of order $3$ is incomplete. The elements of order $3$ in $\mathbb{Z}_3$ are $1$ and $2$, not just $1$. The elements of order $3$ in $\mathbb{Z}_9$ are $3$ and $6$, not just $3$.

Now, the key here is that the subgroup generated by $x$ in a group is always the same as the subgroup generated by $x^{-1}$ (or in additive notation, $-x$). In the context here, the subgroups generated by $(a,b)$ and by $(a^{-1},b^{-1}$) (or in additive notation, $(-a,-b)$) are the same.

So the subgroup generated by $(3,1)$ is the same as the subgroup generated by $(6,2)$ because $6=-3$ in $\mathbb{Z}_9$ and $2=-1$ in $\mathbb{Z}_2$.

So the full list should be something like this:

Elements of order $3$ in $\mathbb{Z}_9$: $3$ and $6=-3$.

Elements of order $3$ in $\mathbb{Z}_3$: $1$ and $2=-1$.

Elements of order $1$ in $\mathbb{Z}_9$: $0$.

Elements of order $1$ in $\mathbb{Z}_3$: $0$.

So:

Pairs in which both entries have order $3$: $(3,1)$, $(6,1)$, $(3,2)$, and $(6,2)$. Of these, $(6,2)=(-3,-1)$, and $(6,1)= (-3,-2)$. So these provide two subgroups of order $3$.

Pairs in which the first entry is of order $3$ and the second of order $1$: $(3,0)$ and $(6,0)$; but $(6,0)=(-3,-0)$, so they give the same subgroup.

Pairs in which the first entry is of order $1$ and the second of order $3$: $(0,1)$ and $(0,2)$; but $(0,2)=(-0,-1)$, so they give the same subgroup.


问:
Counting flower and committee questions
$1$) You want dozen roses. The florist has white, pink, red, and violet roses. How many possible ways could you make the order?

$2$) There are $35$ men and $15$ women. Committee needs to have four men and four women, the chair will be a woman and secretary a man. How many ways can the committee be filled?

My attempt:

$1$) For the first one, I am not sure which one is right answer:  $\dfrac{4^{12} -4}{12} + 4$ or $\dbinom{12+4-1}{12}$.

$2$) For the second one, my answer is $$\dbinom{35}4 \times \dbinom{15}4 \times \dbinom{35-4}1 \times \dbinom{15-4}1$$

Am I thinking this correctly?

答:
For the first problem, note that the order of the flowers are not important, i.e., there is only one possible choice of $4$ red, $2$ white, $1$ pink and $5$ violet flowers. Hence, you need to count the number of non-negative solutions solutions to
$$w+p+r+v = 12$$
The number of non-negative integer solutions to $\displaystyle \sum_{i=1}^{n} a_i = N$ is given by $C(N+n-1,n-1)$. Look up stars and bars.

For the second problem, first choose the chair. There are $15$ choices. Then the secretary. This has $35$ choices. For the remaining members, you need to choose $3$ men out of $34$ and $3$ women out of $14$ women. Hence, the number of ways is
$$15 \times 35 \times \dbinom{35-1}3 \times \dbinom{15-1}3$$

For the second one, your line of thought is correct. However, I assume that the committee of $4$ men and $4$ women include the chairman and secretary.

Also, note that there are multiple ways to count the second one. You can first select $4$ men and $4$ women and then choose the chairman and secretary, i.e., it will be $$\dbinom{35}4 \times \dbinom{15}4 \times \dbinom{4}1 \times \dbinom{4}1$$
Alternately, you can choose the $3$ men and $3$ women who will be the committee member (not the secretary or chairman) and then choose the secretary and chairman, which will give us $$\dbinom{35}3 \times \dbinom{15}3 \times \dbinom{35-3}1 \times \dbinom{15-3}1$$
All these will give you the same answer.

Title: Counting flower and committee questions


问:
Algorithm for Determining Maximum Size for Squares in a Rectangle
I'm trying to create a program to draw a specific number of equally-sized squares within a rectangle. I want to set my squares' sizes to the maximum size possible while still fitting within a rectangle. Empty space is fine so long as the squares can't get any larger by breaking into a new row.

Here is an illustration of the desired result of the program with square count values 1-5:

So far, I've successfully found the different permutations for the possible number of columns/rows based on square count. Here's an example using a square count of 4 of the permutations that can contain 4 squares:

```<code>{rows: 1, columns: 4}
{rows: 2, columns: 2}
{rows: 2, columns: 3}
{rows: 3, columns: 2}
{rows: 3, columns: 3}
{rows: 4, columns: 1}
</code>```

Once I have the permutations, how do I determine which one will allow for the maximum square size within the dynamically-sized rectangle? Is there a way to use the area perhaps?

Please let me know if you have any questions or if I can make something clearer. Thank you!

My apologies, I am not very familiar with mathematical notation, so please know this if posting any. I will do my best to research anything posted, but I may have follow-up questions. Thanks!

答:
Dynamically sized rectangle with length $l$ and width $w$.  The largest possible square would have side length $s=\min(l,w)$ (That is the lesser of the length and width)

For programming purposes (depending on the need) it would be easiest to set width and define length as some multiple ($a$) of width. For Example: $w=5$ and $l=7.5$ and $l=aw\implies a=\frac{l}{w}=7.5\div 5=1.5$. Then you can use the $a$ value to determine several things:

If $0<a<1$ then $w>l$ (largest possible square $s=l$);  If $a=1$ then $w=l$ (starting rectangle is square);  If $a>1$ then $w<l$ (largest possible square $s=w$).

$a$ can also be used to determine how many rows/columns of a specific sized square will fit.  If $n$ represents the number of squares desired then you can use a loop to determine if the largest squares possible will fit in a row/column (note: if you force length to always be longer than width it will simplify your programming so that you don't have to consider rows vs. columns)  For example, If we want 7 squares in the rectangle $w=5$ and $l=7.5$ then since $a=1.5$, then $7*1.5>l$ so we consider squares of side length smaller than 1.5.  If we reach an even divisor $k$ (where $k$ is an integer greater than 1) of 1.5 (i.e. $1.5\div 2=.75$, $1.5\div 3=.5$, etc.) then you must account for $k$ rows of squares.  I recommend coming up with some minimum significant digit with which to reduce the square's side length $s$ in the loop.

There are more advanced mathematical approaches to this problem, but purely from a programming perspective, there is some value in keeping it simple.  I hope this helps.


问:
Is the closed form of $\sum_{n=0}^\infty n!x^n$ unique?
So, I've been studying the power series $f(x)=\sum_{n=0}^\infty n!x^n$, and through algebraic manipulation I've found two ways to get a closed form for it. The first notices that $1+xf(x)+x^2f'(x)=f(x)$, and then proceeds to solve this linear ODE assuming that $\lim_{x\to\pm 0}f(x)=1$. The answer here is $f(x)=\frac{1}{x}e^{-\frac{1}{x}}\text{Ei}(x)$, where Ei is defined here: https://www.wolframalpha.com/input/?i=Ei%28x%29. The second method notices that $n!=\int_0^\infty e^{-t}t^n dt$, so $f(x)=\int_0^\infty e^{-t}\sum_{n=0}^\infty (tx)^n dt=\int_0^\infty \frac{e^{-t}}{1-xt}dt$, which again eventually resolves to $f(x)=\frac{1}{x}e^{-\frac{1}{x}}\text{Ei}(x)$ (assuming $\int_{-a}^a \frac{1}{t} dt=0)$.

It seems that the power series $\sum_{n=0}^\infty n!x^n$ is truly equal to $\frac{1}{x}e^{-\frac{1}{x}}\text{Ei}(x)$, in an analytic continuation kind of sense. However, the series has a radius of convergence equal to $0$, so it shouldn't be possible to fit any function to it. What is going on here? Is it possible to find another function that fits this power series (that doesn't just add a term with power series zero like $e^{-x^{-2}}$)?

答:
Too long for a comment.

The solution of
$$1+xf(x)+x^2f'(x)=f(x)$$ is
$$f(x)=c\frac{ e^{-1/x}}{x}+\frac{e^{-1/x} }{x}\text{Ei}\left(\color{red}{\frac{1}{x}}\right)$$ and, for any value of $c$, $\lim_{x\to 0} \, f(x)=1$ wich makes that we can use $c=0$ for a particular solution.

Effectively
$$\frac{e^{-1/x} }{x}\text{Ei}\left(\frac{1}{x}\right)=\sum_{n=0}^\infty n!x^n$$


问:
Determining the position of a random non-natural positive rational number in a geometric progression
I multiply 100 by 1.05. I get 105.
I multiply 105 by 1.05. I get 110.25.
I multiply 110.25 by 1.05. I get 115,7625 and so on.

If I choose a fully random non-natural positive rational number, for example 240.353
What's the formula to determine how much I need to multiply 100 by 1.05 to get this number?

The only thing I know is that the number of times I have to multiply 1.05 to get any random non-natural number is also going to be non-natural.

答:
Given an arbitrary positive rational number $y$, you are looking for a real number $x$ such that $$100 (1.05)^x =y $$
$$ 1.05 ^x = y/(100) $$

$$x= Ln  ( y/100) /Ln (1.05)$$


问:
Geometric Notation found in Mathematical Methods For Physics And Engineering
In the book Mathematical Methods For Physics And Engineering, there is a section proving compound-angle properties; this section is pretty readable for me until page 12. It states:
$$\cos B=x'=TN+NP=MR+NP$$
$$    =OR\sin A+RP\cos A=\sin(A+B)\sin A+\cos(A+B)\cos A;$$

Earlier, the book explained that $Oxy$ and $Ox'y'$ are the two coordinate systems which were to be referenced from then on in the proof. This means, in my mind, that $x'$ is obviously the x-axis in the second coordinate system, but then if this were true the statement $\cos B=x'$ would be a meaningless statement, no? Secondly, I'm a bit confused about what $OR\sin A$ means. $TM$ seems to mean the line segment from $T$ to $M$ but $OR\sin A$ doesn't seem to have any geometric meaning. Is it the scaled version of $OR$ and if so, why is it being scaled? I am pretty new to this so I'm sorry if this seems trivial to you guys.

答:
As suspected, $x'$ stands for the distance between the origin and the foot of the perpendicular drawn from $P$ to the $x'$-axis (or the projection/shadow of $OP$ to the $x'$-axis). Hence this length is equal to $TP$ and $MR + NP$. Here the same word "$x'$" is used in two different contexts: $x'$ is a variable, and the $x'$-axis is a coordinate axis.

As to "$OR\sin A$", consider $\triangle OMR$. We have, by construction, $OM \perp MR$ and $\angle MOR = A$. Hence by definition of $\sin$, we have $\sin A = \dfrac {MR}{OR}$. This gives $MR = OR\sin A$.


问:
Is it possible to find 3 quadratic polynomials $f(x),g(x),h(x)$ such that $f(g(h(x))) = 0$ has 8 roots $1,2,3,4,5,6,7,8$?
Here is what I tried -
I assumed 3 quadratics $f(x) , g(x)$ and $h(x)$ with these I formed the gigantic equation $f(g(h(x)))$ it had a degree 8 therefore was satisfying the 8 degree criterion but I'm stuck at finding is it possible that it has roots $1,2,3,4,5,6,7,8.$

答:
The polynomial $h$ must be $2$ to $1$ on the set $\{1,2,3,4,5,6,7,8\}$.  Thus its stationary point must occur at $4.5$ (considering real and imaginary parts of $h$ separately).  Thus $$h(x)=a(x(9-x))+b,$$ and the values of $h$ applied to that set are $8a+b,14a+b,18a+b,20a+b$ for constants $a,b$.

By the same reasoning $g$ must be $2$ to $1$ on this set, so its stationary point must occur at $14a+b$ so that $g(8a+b)=g(20a+b)$.  However then $g(18a+b)\neq g(14a+b)$.  Thus we cannot have $3$ such quadratics.


问:
Showing $\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$ for $0<x<1$
I was trying to solve the following exercise:

Show that for $0<x<1$ the following inequalities hold: $$\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$$

I think the Mean Value Theorem might be useful, but I don't know how to apply it.

答:
$$
\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2} \\
-\frac{1-x}{1+x^2}<\arctan (x) - \frac{\pi}{4}<-\frac{1-x}{2} \\
\frac{1-x}{1+x^2}>\frac{\pi}{4} - \arctan (x)>\frac{1-x}{2} \\
\frac{1}{1+x^2}>\frac{\frac{\pi}{4} - \arctan (x)}{1-x}>\frac{1}{2} \\
$$
where the last step assumes $x<1$.

could you write down Taylor series for the arc-tangent around $1$?


问:
On the condition of $ac$ when factoring a quadratic trinomial where $a\neq1$
I am currently reading the section "Factoring quadratic trinomials when a ≠ 1 using the grouping method" this site. I don't understand the last part of this paragraph:

Where did the condition for $ac$ come from?

I thought the extra conditions would be $a=pr$, $c=qs$ as I that is how I read the condition for $b$ too ie. $b=ps+pr$.

Lastly, why is it $ac$ and not some other arithmetic operation e.g. $a+c$ or $a/c$?

答:
$u$ is any number

$\frac{b}2 + u$ multiplied by $\frac{b}2 -u$ will give you $c$.

$\frac{b}2 + u$  added to $\frac{b}2 -u$ will give you $b$.

Solve for $u$  then substitute $\frac{b}2 \pm u$.

The two numbers you get are the solution


问:
Topology on the idele group.
For the rational number field, the adele ring $A_{\Bbb Q}$ is defined. On the other hand, its multiplicative subgroup $I_{\Bbb Q}$ is also defined.

$A_{\Bbb Q}$ is endowed with the topology as its subgroups $n{\widehat {\Bbb Z}}$ over all integers $n$ make the fundamental neighbourhoods.

$I_{\Bbb Q}$ is endowed with the fundamental neighbourhoods as the intersection with the subsets $(x, y) \in A_{\Bbb Q} × A_{\Bbb Q}$ s.t. $xy = 1$.

Q. Why is thus defined topology on $I_{\Bbb Q}$ stronger than the induced topology from $A_{\Bbb Q}$?

答:
If you're asking for a proof that the topology on $I_\Bbb{Q}$ is stronger than the induced topology from the inclusion $I_\Bbb{Q}\subseteq A_\Bbb{Q},$ note that for any topological ring $R,$ the topology induced by the embedding
\begin{align*}
\iota: R^\times&\to R\times R\\
r&\mapsto (r,r^{-1})
\end{align*}
is at least as fine as the topology on $R$: the projection maps $\pi_1,\pi_2 : R\times R\to R$ are continuous. Then for any open $U\subseteq R,$ we have $(\pi_1\circ\iota)^{-1}(U) = U\cap R^\times,$ which is open by continuity of $\pi_1$ and by definition of the topology on $R^\times.$

For a specific instance of the failure of inversion to be continuous, see example 23.1 here. You can use this to construct an example of an open in $I_\Bbb{Q}$ which is not the restriction of an open in $A_\Bbb{Q}$.

If you're asking why the topology is defined in this way, we want $I_\Bbb{Q}$ to be a topological group. But $I_\Bbb{Q}$ with the subspace topology induced from $I_\Bbb{Q}\subseteq A_\Bbb{Q}$ is not a topological group -- inversion is not continuous! Instead, we use the subspace topology induced by the embedding (of groups!) $\mathrm{GL}_1(R)\to\mathrm{GL}_2(R)$ which sends $r\mapsto \begin{pmatrix}r&0\\0&r^{-1}\end{pmatrix}.$ This is equivalent to the topology induced by the map $\iota : R^\times\to R\times R$ I wrote above, thinking of $\mathrm{GL}_2(R)$ as the subspace
$$
\mathrm{GL}_2(R) = \{(a,b,c,d,y)\in R^5\mid (ad - bc)y= 1\}\subseteq R^5
$$
(Notice that the topology on $\mathrm{GL}_2(R)$ is the subspace topology induced by this inclusion!)
Since the map $\mathrm{GL}_1(R)\to\mathrm{GL}_2(R)$ lands in the subspace $\{(a,0,0,d,1)\in R^5\},$ we identify this subspace with $R\times R$ and there is no harm in using the map $\iota : R^\times\to R\times R$ to define the topology on $R^\times.$

Title: Topology on the idele group.


问:
How to express all elements of $\mathbb{F}_9$ as powers of $1+\beta$?

Let $\mathbb{F}_9 = \mathbb{F}_3[\beta]$ where $\beta $ is a root of $x^2+1$ in some extension of $\mathbb{F}_3$. Then $1+\beta$ is a primitive element of $\mathbb{F}_9$. Express all elements of $\mathbb{F}_9=\mathbb{F}_3[\beta]$ as powers of $1+\beta$.

I know that $\mathbb{F}_9={\{0,1,2,\beta,1+\beta,2+\beta,1+2\beta,2+2\beta}\}$, where $\beta^2+1=0$ and the primitive elements are $1+\beta, 2+\beta,1+2\beta,2+2\beta$. But how will I able to show all elements of $\mathbb{F}_9$ as powers of $1+\beta$?

答:
As simple as repeatedly multiplying and reducing $\beta^2=2$. Let $\theta=1+\beta$, then
$$\theta^2=1+2\beta+\beta^2=2\beta$$
$$\theta^3=2\beta+2\beta^2=1+2\beta$$
$$\theta^4=1+3\beta+2\beta^2=2$$
$$\theta^5=2\theta=2+2\beta$$
$$\theta^6=2\theta^2=\beta$$
$$\theta^7=2\theta^3=2+\beta$$
$$\theta^8=2\theta^4=1$$
Note that $0$ can never be a power of $\theta$, which is expected anyway.


问:
Compact sets and open coverings
If $Y$ is a subspace of $X$ and when we have to prove that $Y$ is compact should we always find sets that are open in $Y$ covering it and that has a finite sub collection or it is okay also to consider sets open in $X$? To put the question in another way should compactness be checked based on sets that are open in THAT space?

答:
Suppose that for any cover of $Y$ by open sets in $X$ there is a finite subcover. Then $Y$ is compact.

Proof : Consider a cover $\mathcal U$ of $Y$ by open sets in $Y$. Then, there is a family $\mathcal U'$ of open sets in $X$ such that $\mathcal U = \{Y \cap U : U \in \mathcal U'\}$. Observe that $Y \subseteq \bigcup \mathcal U = Y \cap \big( \bigcup \mathcal U' \big) \subseteq \bigcup \mathcal U'$, which means that $\mathcal U'$ is a cover of $Y$ by open sets in $X$. Thus, there is a finite subcover $\mathcal V'$ of $\mathcal U'$, and then $\mathcal V := \{Y \cap U : U \in \mathcal V'\}$ is a finite subcover of $\mathcal U$. Hence, $Y$ is compact.

Title: Compact sets and open coverings


问:
Integrate the function $f(x, y, z)=\sqrt{3x^2 + 3y^2 + z + 1}$ over the surface given by...
Integrate the function $f(x,y,z) = \sqrt{3x^2 + 3y^2 + z + 1}$ over the surface given by the graph of $z = g(x,y) = x^2 + y^2$ over the region $1 \leq x^2 + y^2 \leq 4$</p>

答:
Note that the function being integrated over the surface is really just a simple function of $z$:

$$3 x^2+3 y^2+z+1 = 4 z+1$$

Also, the surface integral is given by

$$\int_1^4 dz \,2 \pi r(z)\, \sqrt{1+\left ( \frac{\partial r}{\partial z}\right )^2}\, \sqrt{4 z+1} $$

Given that $z=r^2$ then $r(z) = \sqrt{z}$.  Then the integral is

$$2 \pi \int_1^4 dz \, \sqrt{z} \, \sqrt{1+\frac{1}{4 z}} \, \sqrt{4 z+1} = 4 \pi \int_1^4 dz \, \left (z+\frac14\right)$$

The reader should be able to verify that the integral has a value of $33 \pi$.


问:
Variance and Consistency of MLE estimator for a shifted exponential distribution.
I have some work here but am not sure if it is correct. Suppose $X_1,\ldots,X_n$ are i.i.d.  with density function

$$f_\theta(x) = e^{\theta - x}$$

I was able to calculate the MLE estimator for this distribution is $\hat \theta = \min(X_i, \ldots, X_n)$

Since $P(\hat \theta_n \leq t) = 1 - (e^{\theta - t})^n$

Then $P(n(\hat \theta_{n} - \theta) \leq x) = P(\hat \theta_{n} \leq \frac{x}{n} + \theta) = 1 - \left[e^{\theta - \frac{x}{n} - \theta}\right]^n = 1 - e^{-x}$

Thus $\frac{d}{dx}(1 - e^{-x}) = e^{-x}$

Therefore, my work shows that $n(\hat \theta_n - \theta)$ converges in distribution to $\exp(1)$ and the $\operatorname{var}(\hat \theta_n) = \frac{1}{n^2}.$

Consistency can be proven by Slutsky, I believe:

$$n(\hat \theta_n - \theta) \xrightarrow{d} \exp(1)$$

$$\frac{1}{n} \xrightarrow{p} 0.$$

Thus $\hat \theta_n - \theta \xrightarrow{p} 0$

And $\hat \theta_n \xrightarrow{p} \theta$.

Is this correct? I am especially uncertain about my work for the asymptotic distribution, considering that the variance decreases at a faster than standard rate. Is this a rare case for the asymptotic variance?

答:
Your work is correct.

You can also show that $\hat \theta_n \overset{p}{\rightarrow} \theta$ by explicitly computing for any fixed $\epsilon > 0$, $$\lim_{n \to \infty} \Pr[\hat \theta_n - \theta > \epsilon] = \lim_{n \to \infty} \Pr[n(\hat \theta_n - \theta) > n\epsilon] = \lim_{n \to \infty} e^{-n\epsilon} = 0.$$

The reason why $\hat \theta_n$ converges so quickly is because it is the sampling distribution of the minimum of a rather heavily right-skewed population distribution.


问:
Is Desmos Malfunctioning?
For the function: $f(x)=5x^3-10x$ we find the zeroes to be: $x=0, \sqrt{2}$, and $-\sqrt{2}$

However... If you let $a=\sqrt{2}$ or $-\sqrt{2}$ in $5a^3-10a$

Desmos evaluates and says either of those equal $3.5527136788*10^{-15}$ (NOT $0$)!!!
Am I doing something incorrectly or is this Desmos' fault?

Image:

答: Desmos is not a true symbolic calculator, and the results you see are due to floating-point error incurred when the (inexact) values for $a$ are substituted into $5a^3-10a$ under (an also inexact) floating-point regime.

Title: Is Desmos Malfunctioning?


答:
Yes, the compactness of $Y$ is about collections of open subsets of $Y$, as you say.

However, what is an open subset of $Y$? By the definition of the subspace topology, it is a set of the form $U \cap Y$ for some $U$ which is open in $X$.

This means that we get the result described by @J.W.Tanner and @azif00, which says basically that it doesn't matter whether we look at open subsets of $Y$ or open subsets of $X$.

Title: Compact sets and open coverings

Body: If $Y$ is a subspace of $X$ and when we have to prove that $Y$ is compact should we always find sets that are open in $Y$ covering it and that has a finite sub collection or it is okay also to consider sets open in $X$? To put the question in another way should compactness be checked based on sets that are open in THAT space?


问:
If $f$ is continuous on $[0,1]$, prove that $\lim_{n\to\infty}\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi}{2}f(0)$.
The solution given:
$$\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx+\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx$$
okay, my first question here: why the $n^{-1/3}$?
$$|\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx|\leq \int_{n^{-\frac{1}{3}}}^{1} |\frac{nf(x)}{1+n^2x^2}|dx \leq \frac{nM}{1+n^{1+1/3}}$$
where M is the global maximum of $|f|$. how did the middle term turn into the rightmost one? How is the mean value theorem of integrals used here? what's with the $1+1/3$?

Since the rightmost term tends to $0$ as $n$ tends to infinity, by squeeze theorem we have the leftmost term tends to $0$ as well.
Since $\frac{n}{1+n^2x^2}$ does not change sign on $[0,1]$, then there exists $c\in[0,n^{-1/3}]$ such that
$$\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx=f(c)\int_{0}^{n^{-\frac{1}{3}}} \frac{n}{1+n^2x^2}dx=f(c)\tan^{-1}n^{2/3}$$
Since $c\in[0,n^{-1/3}]$, $n\to\infty$ implies $c\to0$ and $\tan^{-1}n^{2/3}\to \frac{\pi}{2}$ and so $\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx\to\frac{\pi}{2}f(0)$

So the final answer is $0+\frac{\pi}{2}f(0)=\frac{\pi}{2}f(0)$

I've highlighted my questions! Thanks for any advice given!

答:
The $n^{-1/3}$ is not important. What is important is to break into an error term that goes to $0$.

Notice that for large $n$, $ \frac{n}{1 + n^2x^2} $ is big if $n^2 x^2 \ll n,$ but it is small if $n^2x^2 \gg n$. The second case happens whenever $x \gg n^{-1/2}$. So, take any $\delta > 0$, and look at $ \int_{n^{-1/2 + \delta}}^1 \frac{n f(x)}{1 + n^2 x^2}\,\mathrm{d}x.$ Notice that $f$ is bounded, and we know that $n/(1+n^2x^2)$ is small in this region. This means we're integrating something small over a bounded region, and we expect this to be small.

Putting this strategy into effect, \begin{align} \left| \int_{n^{-1/2 + \delta}}^1 \frac{n f(x)}{1 + n^2 x^2} \,\mathrm{d}x\right| &\overset{1}\le \int_{n^{-1/2 + \delta}}^1 \left|\frac{n f(x)}{1 + n^2 x^2}\right| \,\mathrm{d}x\\ &\overset{2}\le \int_{n^{-1/2 +\delta}}^1 \frac{nM}{1 + n^2 x^2} \,\mathrm{d}x\\
&\overset{3}\le M \int_{n^{-1/2 +\delta}}^1\max_{x \in [n^{-1/2 + \delta}, 1]} \frac{n}{1 + n^2 x^2} \,\mathrm{d}x \\
&\overset{4}\le M \cdot 1 \cdot \frac{n}{1 + n^2 n^{-1 +2\delta}} = \frac{Mn}{1 + n^{1 + 2\delta}},\end{align} and this upper bound vanishes with $n$.

The justification for these inequality $2$ is that $|f(x)| \le M,$ so abosulte value of $|nf/(1 + n^2 x^2)|$ is bounded by the new integrand. For 3., I'm using a similar idea - the integrand is bounded by it's max in the relevant domain. For 4, I'm integrating a constant (which is the value of the max) over a domain of size less than $1$. In particular, note that I haven't used the mean value theorem above at all. That said, you can use that if you prefer, treat this as an exercise. Do be careful that the MVT requires the function under consideration to be continuous.

In the solution you have posted, they use $\delta = 1/6,$ and $-1/2 + \delta = -1/3,$ and $1+ 2\delta = 1+1/3.$

The whole argument relied on getting an upper bound that goes to zero with $n$. If we had chosen a $\delta <0,$ then this wouldn't have happened. But any $\delta > 0$ is fine (however, see below). BTW other stuff could have worked too - we could have integrated from $n^{-1/2} \log n$ to $1$ and the same conclusion would emerge. Do you see why? What else could you use?

Just for completeness, it's actually not okay to choose any $\delta > 0.$ Indeed for the rest of the argument, we have $$ \int_0^{n^{-1/2 + \delta}} \frac{n f(x)}{1 + n^2 x^2}\,\mathrm{d}x = f(c_{n,\delta}) \arctan(n \cdot n^{-1/2 + \delta}) = f(c_{n,\delta}) \arctan(n^{1/2 + \delta}).$$ Now to apply the argument of the given solution, we need that $c_{n,\delta} \to 0.$ In the question, this works because $0 \le c \le n^{-1/3}$ and the upper bound goes to $0$. For us, we have $0\le c \le n^{-1/2 + \delta}$. For this upper bound to go to $0$, we would need to make sure that $\delta < 1/2.$

So, for the whole argument to work, we need to pick some $\delta \in (0, 1/2).$ As previously noted, the presented solution uses $1/6,$ which lies in this range.


问:
If $\lim_{n\to\infty}\frac{a_n}{n^\alpha}=1,\alpha>0$, find $\lim_{n\to\infty}\frac{1}{n^{1+\alpha}}\sum_{k=1}^{n}a_k$ using definite integral.
So.. this looks pretty much like a reimann sum, $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}$$ something but it doesn't seem to be able to be written as a reimann sum.....is the final answer 1? I took a guess by grouping $n^\alpha$ to the $\sum_{k=1}^{n}a_k$ part, which would result in $n$ 1s and which would cancel out with the n at the denominator ....okay that's not rigorous I know.  Any leads? Thanks!

答:
The demand is weird because the definite integral does not play the central role in this limit.

This problem may seem opaque somehow to you but let us rephrase it another way by giving you a lemma, then I guess everything will be clear to you.

Lemma: Let $(u_n; n \ge 0)$ and $( w_n; n \ge 0)$ be two real sequences such that:

$w_n>0$

$\lim_{n \rightarrow + \infty} \left( w_1+w_2+...+w_n \right) =+ \infty$

$u_n$ has a limit, that is $ \exists u $ such that $ \lim u_n=u$

then,
$$ \lim_{n \rightarrow \infty} \dfrac{w_1u_1+w_2u_2+...+w_nu_n}{w_1+w_2+...+w_n}=u $$

Then you apply the above lemma for our problem with $u_n=\dfrac{a_n}{n^{\alpha}}$, $w_n=n^{\alpha}$.


问:
Problem solving nonlinear equation of magnetic saturation
Hi everyone I've found a model of magnetic saturation in a reference like this $$\ B=\mu_0 *\mu_r*H $$ $$\ \mu_r = \frac{a}{b*B^4-c*B^2+d}$$ in which $a,b,c,d,\mu_0$ are constant integers with known values $a  = 10750; b  = 2.4; c  = 0.65; d  = 2.5;\mu_0=4e-7*\pi $
but I'm looking for a model based on $H$ as an independent variable
so I can say that $$\ \mu_r = \frac{a}{b*(\mu_0\mu_rH)^4-c*(\mu_0\mu_rH)^2+d} $$and I need to know $\mu_r(H)$
it means that I have to obtain $\mu_r$ in the form of $H$ but without any $\mu_r$ at the left side.lets change the variables for better insight of the problem
($\mu_r \rightarrow y , H \rightarrow x$)
$$\ y = \frac{a}{b*(\mu_0*y*x)^4-c*(\mu_0*y*x)^2+d} $$ change this in the form of $y=f(x)$

B is changing from-2.5 to 2.5 and H is changing from -1e5 to 1e5 .
thanks for your answer
B-H curveand Mur-B curves

答:
$$\ y = \frac{a}{b*(\mu_0*y*x)^4-c*(\mu_0*y*x)^2+d} $$

Rearrange:

$$ y[  \frac{b*(\mu_0*y*x)^4-c*(\mu_0*y*x)^2+d}{a}] = 1$$

Now, if you have a put $ x=a$, then you can write $y$ is a polynomial in $x$ around some domain locally as:

$$ y(x) = \sum_{k=0}^{\infty} \left( \frac{d^ky}{dx^k} \right)|_{a} \frac{(x-a)^k}{k!}$$

For generating the coefficients, you can take implicit derivatives of original expression and evaluate at zero. However note that this approximation only holds for a smol domain.

Example:

Circle centered at origin: $$ x^2 +y^2 =r^2$$

So, say I want the approximation/ linearization around $x=a$,

$$ a^2 + y^2 = r^2$$

Now an immediate complexity which pops up is that there is a positive and negative $y$ value corresponding to this. So you have to choose both $ (x,y)$ around which you want to approximate. Let's take the top semi circle i.e $y= \sqrt{r^2 -a^2}$

Now the derivative:

$$ (\frac{dy}{dx}) = - \frac{x}{\sqrt{r^2 -x^2}}  $$

Now one nice simplification is that to be on the semi circle, $ x<r$ and hence we can use binomial series:

$$ \frac{1}{\sqrt{r^2 -x^2 } } = \frac{1}{x} [  1 + \frac{1}{2} \frac{x^2}{r^2} +O(\frac{x^4}{r^4}) ]$$

So, this makes further derivatives an easier task since you will be differentiating a polynomial with the above, also depending on the order of the $x$ value , you can choose where you want to chop off the series.

Anyhow,

$$y(x) = y(a) - \frac{a}{\sqrt{r^2 - a^2}} (x-a) + O((x-a)^2)$$

Is an approximation which holds true on the upper semi circle around some interval about a point $ x=a$

For more about Taylor Series,  Check out this blog which I wrote (Here)


问:
How to derive the mean and variance from a probability density function?
For this question, I'm stuck on finding $M(t)$, the mean, and variance, and I was wondering if anyone can help me out? Here's what I have so far

Suppose $X$ is a continuous random variable with the following probability density function:
$$f(x) = 0.5x^2 e^{-x}$$
Use the moment generating function $M(t)$ to find the mean and variance of $X$

$M(t) =
\int_0^\infty e^{tx} f(x)dx$

$ = \int_0^\infty e^{tx}0.5x^2e^{-x}dx$

$ = 0.5 \int_0^\infty e^{-x(1-t)}x^2dx$

Let $u = x^2 \Rightarrow du = 2xdx$

Let $v = \int_0^\infty e^{-x(1-t)}dx \Rightarrow dv = e^{-(1-t)x}$

答:
I think you just misunderstood the integration by parts (please check the formula again in here)

Follow that, you should let $dv=e^{-(1-t)x}\Leftrightarrow v = \frac{e^{-(1-t)x}}{-(1-t)}$

Your equation $= uv|^\infty_0 - \displaystyle\int^\infty_0 vdu$

$=0.5(x^2\frac{e^{-(1-t)x}}{-(1-t)}\vert^\infty_0 - \displaystyle\int^\infty_0 \frac{e^{-(1-t)x}}{-(1-t)}2xdx)$

$=0 + \frac{1}{1-t} \displaystyle\int^\infty_0 e^{-x(1-t)}xdx$

Then we do the integration by parts again, finally, we have:

$M(t)= \frac{1}{1-t} (x\frac{e^{-(1-t)x}}{-(1-t)}\vert^\infty_0 - \displaystyle\int^\infty_0 \frac{e^{-(1-t)x}}{-(1-t)}dx)$

$=0+\frac{1}{(1-t)^2} \times \frac{1}{1-t}=\frac{1}{(1-t)^3}$

Please calculate again to verify the result.


问:
Solving quadratic congruences modulo powers of nonprimes

I am a bit clueless with this problem. Do I start by finding a primitive root of the modulo or is the Legendre symbol of any relevance here? Please provide me with some hints.

My attempt:

Based on the idea of lifting the solution suggested by tkf and JW Tanner, a slightly more intuitive algorithm can be found here: (https://www.uvm.edu/~cvincen1/files/teaching/spring2017-math255/quadraticequation.pdf).

Part a: with CRT, consider the base case $x^2\equiv -1 $mod 3, $x^2\equiv -1 $mod 31. But $(\frac{-1}{3}) =-1$, so there is no solution to the congruence. mod 31 also has no solution as $(\frac{-1}{31}) =-1$.

Part b: the base case $x^2\equiv -1$ mod 7 has no solution. Therefore the overall congruence has no solution.

Part c:
For j=1, solve $x_0 \equiv -1 $ mod 13; $x_0=5$.

For $j+1=2$, solve $x_1^2 \equiv -1 $ mod $13^2$; $x_1= x_0 + p^jy_0$. Solve for $y_0$ from: $2x_0y_0 \equiv \frac{-1-x_0^2}{p^j}$ mod 13. $y_0= 5$. Then $x_1= 70$; $70^2\equiv -1$ mod $13^2$.

For $j+1=3$, solve $x_2^2 \equiv -1 $ mod $13^3$; $x_2= x_1 + p^jy_1$.
Solve for $y_1$ from: $2x_1y_1 \equiv \frac{-1-x_1^2}{p^j}$ mod 13. $y_1= 1$. Then $x_1= 239$; $239^2\equiv -1$ mod $13^3$.

Repeat this process until $j+1=6$. $x_5= 1999509$. $1999509^2\equiv -1 $mod $13^6$ ($169^3$).

In part c $-1999509$ is the other solution to the congruence.

Part d:
Based on part c, this is equivalent as solving $x^2\equiv x_5^2\equiv -1$ mod $169^3$. Consider $x^2\equiv 1999509$ mod $13^6$. $(\frac{1999509}{13})=-1$ suggests that there is no solution to the congruence.

答:
From comments only (c) is left.  Here is the general method:

Suppose you know $$13^k|a^2+1,$$

for some integer $a$ and $k\geq 1$.

Then you can find $x$ to solve $$13^{(k+1)}|(a+x13^k)^2+1.$$

Firstly you know there is some integer $w$ with $13^kw=a^2+1$.

Multiply out:

\begin{eqnarray*}(a+x13^k)^2+1&=&a^2+1+x^213^{2k}+2ax13^k\\&=&13^k(w+2ax+x^213^k).\end{eqnarray*}

Thus you just need to solve $13|w+2ax$.

Thus you can solve $$13^k|a^2+1,$$ for $k=1,2,3,4,5,6$.


问:
Truncating $p$-series
Let $p \in \mathbb{R}$. Series of the form $$S_p = \sum_{k \in \mathbb{N^*}} \frac{1}{k^p}$$ converge if and only if $p > 1$. Let us define $$S_{p, n} \triangleq \sum_{k = 1}^{n} \frac{1}{k^p},$$ the truncation of the $p$-series at its $n$-th term. (It is evident that $S_{p, n} \space \xrightarrow{n\to\infty} \space S_p$.) Is there a closed expression for $S_{p, n}$, at least for some values of $p$ (integers must be easier)?</p>

答:
Yes, there is.  These are called the generalized harmonic numbers.

With polygamma function: I suggest looking at this paper, they have many interesting things.  In particular, if we define
$$
H_n(z;r)=\sum_{k=1}^n\frac{1}{(z+k-1)^r}
$$
(even more generalized) then we have for integers $p$
$$
H_n(z;p)=\frac{(-1)^{p-1}}{(p-1)!}(\psi^{(p-1)}(z+n)-\psi^{(p-1)}(z))
$$
where
$$
\psi^{(k-1)}(z)=\frac{d^k}{dx^k}\left(\log\Gamma(x)\right) \biggr|_{x=z}.
$$
This is equation (1.14) of the paper, but I think they forgot a factor of $(p-1)!$ on the denominator. Taking the case $z=1$ gives what you are looking for.

With zeta function: we can write $$H_n(r)=\sum_{k=1}^n\frac{1}{k^r}=\zeta(r)-\zeta(r,n+1),$$ where $\zeta(s,a)$ is the Hurwitz zeta function. This $\space$ really clear, but by using identities regarding $\zeta(s,a)$ we can change the identity into $\space$ previous one, and other things.

Hope that helps.

Title: Truncating $p$-series


问:
How write these two motions into one system
f describe the horizontal motion
g describe the vertical motion

have error when put m(t) in x()

Maple code

```<code>f := sin(m(t));
g := piecewise(t < 0, Diff(x(m(t)),m(t)$2)+2*Diff(x(m(t)),m(t))+10*x(m(t)), t > 0, Diff(x(m(t)),m(t)$2)-2*Diff(x(m(t)),m(t))+10*x(m(t)));
</code>```

origin from

```<code>g := piecewise(t < 0, Diff(x(t),t$2)+2*Diff(x(t),t)+10*x(t), t > 0, Diff(x(t),t$2)-2*Diff(x(t),t)+10*x(t));
plot(subs(_C2=1,subs(_C1=1,rhs(dsolve(g, x(t))))), t=-5..5);
plot(subs(_C2=1,subs(_C1=1,rhs(dsolve(g, x(t))))), t=-1..1);
eq := diff(g,x(t))-diff(diff(g,diff(x(t),t)),t);
</code>```

updated

h := sin(t);

g := piecewise(t < 0, Diff(x(h),h$2)+2*Diff(x(h),h)+10*x(h), t > 0,Diff(x(h),h$2)-2*Diff(x(h),h)+10*x(h));

dsolve([g,h]);

can not solve

Error, (in unknown) invalid input: diff received sin(t), which is not valid for its 2nd argument

update 2

after change to 1/Diff(h,t$2)*Diff(x(h),t$2)

Diff(x(h),t$2) <- how about x(h) ?

```<code>f := sin(t);
g := piecewise(t < 0, 1/(Diff(f,t$2)*Diff(x(f),t$2)+2/(Diff(f,t))*Diff(x(f),t)+10*x(f), t > 0), 1/(Diff(f,t$2))*Diff(x(f),t$2)-2*(1/Diff(f,t))*Diff(x(f),t)+10*x(f));
dsolve({f,g});
</code>```

no solution !!

答:
The second parameter of commands diff and Diff must be the name of a variable (or several variables, if you take a mixed derivative). Neither m(t) nor x(t) is the name of a variable, hence the error message.

The best thing to do is to reconsider what your variables mean and what derivatives you want to take. Since I do not know what you have in mind, I give a workaround solution: you can substitute a variable instead of m(t), differentiate with respect to that variable, and then reverse the substitution. For example:

```<code>f:=sin(m(t));
eval(diff(eval(f,m(t)=m),m),m=m(t));
</code>```

returns cos(m(t)).

This is, of course, different from diff(f,t); which would return $\cos(m(t)) \frac{d}{dt}m(t)$, according to the chain rule.

I repeat that you should focus on getting the logic of your formulas right, and worry about the syntax later. I don't see much logic in your present code samples.

Body:
The second parameter of commands diff and Diff must be the name of a variable (or several variables, if you take a mixed derivative). Neither m(t) nor x(t) is the name of a variable, hence the error message.

The best thing to do is to reconsider what your variables mean and what derivatives you want to take. Since I do not know what you have in mind, I give a workaround solution: you can substitute a variable instead of m(t), differentiate with respect to that variable, and then reverse the substitution. For example:

f:=sin(m(t));
eval(diff(eval(f,m(t)=m),m),m=m(t));

returns cos(m(t)).

This is, of course, different from diff(f,t); which would return $\cos(m(t)) \frac{d}{dt}m(t)$, according to the chain rule.

I repeat that you should focus on getting the logic of your formulas right, and worry about the syntax later. I don't see much logic in your present code samples.


问:
Random variable with infinite expectation $\mathbb E(\vert X\vert)=\infty.$
I want to show that when $\mathbb E(\vert X\vert)=\infty$ and $X_i$ are iid copies of $X$ then necessarily

$$\limsup_n \frac{\vert \sum_{i=1}^n X_i \vert}{n}=\infty.$$

However, this is tricky as we cannot assume that they have the sign.

Please let me know if you have any questions.

答: Proof assuming that $EX$ is defined (i.e.$EX^{+} <\infty$ or $EX^{-} <\infty$) : Either $EX^{+} =\infty$ or $EX^{-} =\infty$. In the first case $EX_k^{-} <\infty$ and $\frac  1 n \sum\limits_{k=1}^{n} \min {\{X_k, N\}} \to E\min {\{X_1, N\}}$ almost surely as $n \to \infty$ for each $N$ (by SLLN). Since $E\min {\{X_1, N\}}$ can be made as large as we want and  $\lim \sup \frac  1 n |\sum\limits_{k=1}^{n} X_k| \geq \lim \sup \frac  1 n \sum\limits_{k=1}^{n} X_k \geq \lim \sup \frac  1 n \sum\limits_{k=1}^{n} \min {\{X_k, N\}} $ we get the result immediately. The case $EX^{-} =\infty$ follows by changing $(X_n)$ to $(-X_n)$.


问:
Compact connected surface (not sphere) minus some point is homotopy equivalent to bouquet of circles.
Let $M$ to be a compact connected surface that is not homeomorphic to the sphere $\mathbb S^2$. Then, I want to show that there exists some $x \in M$ such that $M \setminus \{x \}$ is homotopy equivalent to a bouquet of circles (wedge sum of finite $\mathbb S^1$'s copies).

Some ideas:

I know that a compact connected surface must be homeomorphic to either a connected sum of copies of $\mathbb P^2$ or $\mathbb T^2$ given it is not homeomorphic to $\mathbb S^2$.

I think $\mathbb T^2$ minus one point is homotopy equivalent to a bouquet of two circles: $\mathbb S^1 \lor \mathbb S^1$.

I also saw somewhere that $\mathbb P^2$ minus one point is homeomorphic to $\mathbb S^1$.

Since connected sum is commutative and $\mathbb T^2 \# \mathbb P^2$ can be written as $\mathbb P^2 \# \mathbb P^2 \# \mathbb P^2$, $M$ is either homeomorphic to $\#_{i = 1}^n \mathbb P^2$ or $\mathbb T^2$.

For the second case, I can just claim that $M$ minus any point is homeomorphic to a bouquet of two circles, so it is homotopy equivalent to a bouquet of two circles, if I am not wrong.

However, for the first case, I am not sure how to proceed.

答:
$M$ may be constructed by taking a polygon and identifying edges in such a way that all vertices become identified with each other.

If you remove a vertex from the center of the polygon, then you can retract onto the perimeter of the polygon, which is a bouquet of circles.

Let $X$ denote the unit disk in $\mathbb{R}^2$ with the origin removed.  Let $C$ denote the unit circle.  Working in polar co-ordinates, we give an explicit homotopy from the identity on $X$ to a map $X\to C$ which restricts to the identity on $C$:  $$h_t(r,\theta)=((1-t)r+t,\theta).$$

Now we obtain $M\backslash \{x\}$ from $X$ by partitioning $C$ into arcs and identifying pairs of arcs in some way.  The homotopy $h_t$ remains the same.  Now $C$ modulo the identifications will be a bouquet of circles.


答:
$$\frac 1{n^{1+\alpha}}\sum_1^n a_k=\frac 1n\sum_1^n\frac{a_k}{n^\alpha}$$

Since $a_k/k^\alpha\to 1$ as $k\to\infty$, we have $a_k\to k^\alpha$ as $k\to\infty$, so,

$$\frac 1n\sum_1^n\frac{a_k}{n^\alpha}\to\frac 1n\sum_1^n\frac{k^\alpha}{n^\alpha}\to\frac 1n\sum_1^n(k/n)^\alpha\overset{?}{\to}\int_0^1 x^\alpha~\mathrm dx=\frac 1{1+\alpha}$$

as $n\to\infty$

The above might seem suspect because I'm ignoring initial values of $a_k$ and just substituting the asymptotic form for $a_k$ as if assuming the limit to be with $k\to\infty$ as well, so here's a bit of formalism to justify:

$a_k/k^\alpha\to 1$ or that $a_k\to k^\alpha$ as $k\to\infty$ is equivalent to saying that given any $\epsilon\gt 0$, we have $a_k\in (k^\alpha-\epsilon,k^\alpha+\epsilon)$ for sufficiently large $k$, say for all $k\geq k_0$ for some $k_0\in\Bbb N$

So, we can write,

$$\sum_1^n\frac{a_k}{n^{1+\alpha}}=\sum_1^{k_0}\frac{a_k}{n^{1+\alpha}}+\sum_{k_0}^n\frac{a_k}{n^{1+\alpha}}$$

so taking $\epsilon$ to be sufficiently small, $a_k\approx k^\alpha$ for all $k\geq k_0$ and for the sum from $1$ to $k_0$, the $a_k$ are finite but the denominator $n^{1+\alpha}\to\infty$ as $n\to\infty$ since $1+\alpha\gt 1$, so that summand vanishes when $n\to\infty$ and we're left with the latter sum from $k_0$ to $n$ where you can replace $a_k$ by $k^\alpha$ and note that removing finitely many elements from a Riemann sum doesn't change the value, so we conclude, ie, as $n\to\infty$,

$$\sum_1^n\frac{a_k}{n^{1+\alpha}}\to\sum_{k_0}^n\frac{a_k}{n^{1+\alpha}}\to\frac 1n\sum_{k_0}^n\frac{k^\alpha}{n^\alpha}\to\frac 1n\sum_1^n\frac{k^\alpha}{n^\alpha}\dots$$

Title: If $\lim_{n\to\infty}\frac{a_n}{n^\alpha}=1,\alpha>0$, find $\lim_{n\to\infty}\frac{1}{n^{1+\alpha}}\sum_{k=1}^{n}a_k$ using definite integral.

Body: So.. this looks pretty much like a reimann sum, $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}$$ something but it doesn't seem to be able to be written as a reimann sum.....is the final answer 1? I took a guess by grouping $n^\alpha$ to the $\sum_{k=1}^{n}a_k$ part, which would result in $n$ 1s and which would cancel out with the n at the denominator ....okay that's not rigorous I know.  Any leads? Thanks!


问:
Showing $\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$ for $0<x<1$
I was trying to solve the following exercise:

Show that for $0<x<1$ the following inequalities hold: $$\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$$

I think the Mean Value Theorem might be useful, but I don't know how to apply it.

答:
You can solve this with the mean value theorem. For some $y\in(x,1)~,~x>0$ with $f(x)=\arctan(x)$ it is true that:

$$f'(y)=\frac{f(1)-f(x)}{1-x}$$

However since the function $f'(x)=\frac{1}{1+x^2}$ is decreasing in $(0,1)$ it follows that

$$f'(1)<f'(y)=\frac{f(1)-f(x)}{1-x}<f'(x)$$

Can you finish from here doing some rearrangements?


问:
Prove minimum for constrained convex optimization.
I am given the following:

min $f(x) = 1/2(x-r)^T(x-r)$

s.t $a^Tx=b$

This is basically minimizing the distance from x on the hyperplan $a^Tx = b$ from some point r in $R^n$

If I compute the gradient I get:

$\nabla f(x) = x - r$ and set = $a\lambda \implies x = r + a\lambda$

However the book states the minim value $x_*$ is given by:

$x_* = r + \frac{b-a^Tr}{a^Ta}a$

I'm not sure how they're getting this or how to show $\lambda = \frac{b-a^Tr}{a^Ta}$

答:
$$x-r=a\lambda$$

Multiplying $a^T$ on both sides,

$$a^T(x-r)=\lambda (a^Ta)$$

$$a^Tx-a^Tr=\lambda (a^Ta)$$

But we know that $a^Tx=b$.

$$b-a^Tr=\lambda (a^Ta)$$


问:
Question 13.3(a) Tom Apostol Introduction to Analytic number theory
I am self studying analytic number theory from Tom M Apostol and  got struck on question 3(a) of Chapter-13.

It's image:

In (a) I thought of using Perron's Formula (statement here)

by taking identity $log\zeta(s) =\sum_{n=2}^{\infty}\frac{\Lambda(s) }{n^s log n}$ .

But the problem is that I am not sure what will occur on the RHS ie I am not able to simplyfy $\sum_{n=2}^{\infty}\frac{\Lambda(n) }{n^s log n}$.

I thought of Perron's Formula due to the setting in LHS in which question was asked.

So, can you please tell how to simplify RHS if question can be simplified by Perron's Formula?

答:
Whenever $n$ is not a prime power, we have $\Lambda(n)=0$, but when $n=p^k$, we have

$$
{\Lambda(n)\over\log n}={\log p\over\log p^k}=\frac1k
$$

Hence, we have

$$
\sum_{n\le x}{\Lambda(n)\over\log n}=\sum_{p^k\le x}\frac1k=\sum_{k\ge1}\frac1k\sum_{p\le\sqrt[k]x}=\sum_{k\ge1}{\pi(\sqrt[k]x)\over k}
$$

Consequently, by Perron's formula and the fact that $\log\zeta(s)=\sum_{n\ge2}{\Lambda(n)\over n^s\log n}$, we deduce

$$
{1\over2\pi i}\int_{c-i\infty}^{c+i\infty}\log\zeta(s)x^s{\mathrm ds\over s}=\sum_{n\le x}{\Lambda(n)\over\log n}=\pi(x)+{\pi(x^{1/2})\over2}+{\pi(x^{1/3})\over3}+\cdots
$$


问:
Positive Basis of a Riesz Space
Does a Riesz space in general always has a positive basis? ae. if $E$ is a Riesz space, can we assume that there exists a set  $B\subset E$ such that it is a basis of the vector space $E$ and for every $v\in B$, $v>0$?

Proving for the special case $E=\mathbb{R}^n$ is quite easy to prove. Is that true in general?

Edit

Digging in a little bit shows that if $E$ has a finite dimension, it has a positive basis. This is followed from the following theorem:

Suppose that $X$ is a vector lattice of finite dimension $n$. If $X$ is Archimedean ordered then it is (vector lattice) isomorphic with $\mathbb {R} ^{n}$ under its canonical order. Otherwise, there exists an integer $k$ satisfying $2 \le k \le n$ such that $X$ is isomorphic to $\mathbb {R} _{L}^{k}\times \mathbb {R} ^{n-k}$ where $\mathbb {R} ^{n-k}$ has its canonical order, $\mathbb {R} _{L}^{k}$ is $\mathbb {R} ^{k}$ with the lexicographical order, and the product of these two spaces has the canonical product order.

Since $\mathbb{R}^n$ always has a positive basis ($\left\{e_i\right\}$ for example), we can deduce that $E$ has a positive basis (by the inverse isomorphism).

Now, the question is: is it also true for infinite dimension Riesz space?

答:
So, it seems like the answer is YES.

Proof:

Let $S=\left\{B\subseteq E\mid B\text{ is independent and all its items are positive}\right\}$.
We know that a chain of independent sets has a maximum value (basic proof in linear algebra). Therefore, using Zorn's lemma, we know that $S$ has a maximum value via the $\subseteq$ relation. We'll mark it as $B'$

We want to prove that this is actually spanning $E$. We'll assume by contradiction that it is not, therefore there is $v\in E$ such that $v\notin \operatorname{span}(B')$, which means $v$ is independent of the elements of $B'$. It is known that:
$$v=v^+-v^-$$
such that $v^+,v^-\ge 0$ (This is true for all $v\in E$ when $E$ is a Riesz space). If both $v^+,v^-\in \operatorname{span}(B')$, than $v\in \operatorname{span}(B')$, which is a contradiction. Therefore, $v^+\notin \operatorname{span}(B')$ or $v^-\notin \operatorname{span}(B')$, but if that's the case we know that $B'\cup\left\{v^+\right\}\in S$ or $B'\cup\left\{v^-\right\}\in S$, which is in contrast to the maximality of $B'$. Therefore, $\boxed{\operatorname{span}(B')=E}$

Title: Positive Basis of a Riesz Space


问:
How to compute norm of linear functional
I was trying to calculate norm of functional $F: L^p([0;1]) \to {\rm I\!R}$, given by formula:
$$
F(f) = \int_{[0;1]} x f(x) \ d\lambda(x),
$$
for $p \in [1; \infty)$ and $\lambda$ - Lebesgue measure on $[0;1]$.
I have seen the solution for $p = 1$ here, but I am not sure if for $p > 1$ it can be done the same way.
If we take $f \in L^p([0;1])$, such that $\Vert f \Vert_{L^p} = 1$,  we can apply Hölder inequality:
$$
|F(f)| \le  \int_{[0;1]} x |f(x)| \ d\lambda(x) \le \int_{[0;1]} |f(x)| \ d\lambda(x) \le \Big(\int_{[0;1]} |f(x)|^p \ d\lambda(x)\Big)^{\frac{1}{p}}\Big(\int_{[0;1]} |1|^{1 - \frac{1}{p}} \ d\lambda(x)\Big)^{1 - \frac{1}{p}} = \Vert f \Vert_{L^p} \cdot \lambda([0;1]) = 1.
$$
So $|F(f)| \le 1$.
Unfortunatelly I fail at showing that $|F(f)| \ge 1$. I was trying to construct similar sequence as here for every $p > 1$. How to show second inequality? Thank you in advance for any tips.

答:
It is a bad idea to use $x \leq 1$ in getting an upper bound for $\|F\|$ .  Let $q$ be the index conjugate to $p$ (i.e. $q=\frac p {p-1}$). Then $|F(f)| \leq \|f\|_p (\int x^{q})^{1/q}$ by Holder's inequality. Hence $\|F\| \leq (\int_o^{1} x^{q}dx)^{1/q}=\frac 1 {(q+1)^{1/q}}$. This is actually an equality. To see this you have to use the condition for equality in Holder's inequality. Recall that  for $f,g \geq 0$ the conidtion for  $\int fg =(\int f^{p})^{1/p} (\int g^{q})^{1/q}$  is $f^{p}= cg^{q}$. So take $f(x)=x^{q/p}$. Compute $\|f\|$ and check that $|F(\frac f {\|f\|})| =\frac 1 {(q+1)^{1/q}}$. This proves that $\|F\| =\frac 1 {(q+1)^{1/q}}$.

PS

This is precisely the argument used to prove that the dual of $L^{p}$ is $L^{q}$. So we have the following general fact: If we define $F(f)=\int_0^{1} f(x)g(x)d\lambda (x)$ for all $f \in L^{p}$ then the norm of $F$ is equal to $(\int_0^{1}|g(x)|^{q}d\lambda(x))^{1/q}$.


问:
Maximizing the pairwise point distance in $[-1,1]$
How can we prove that, given $n$ points in $[-1,1]$, the sum of all their pairwise Euclidean distances (or, equivalently, their average Euclidean distance) is maximized if $\lfloor n/2 \rfloor$ points are placed at $-1$ and the remaining points are placed at $1$?

答:
WLOG assume $-1 \leqslant x_1 \leqslant x_2 \leqslant \cdots \leqslant x_n \leqslant 1$.

Total distance is
$$
\sum_{i<j} (x_j-x_i) = [0-(n-1)]x_1+[1-(n-2)]x_2-\cdots + [(n-1)-0]x_n\\
=-(n-1)x_1-(n-3)x_2 - \cdots +(n-1)x_n\\
\leqslant -(n-1)(-1)-(n-3)(-1) - \cdots + (n-1)\cdot 1.
$$

$"=" \iff x_1=x_2=\cdots =x_{\lceil\frac{n-1}{2}\rceil}=-1, x_{\lfloor\frac{n+3}{2}\rfloor}=\cdots=x_n=1.$

If $n$ is odd then $x_{\frac{n+1}{2}}$ can be anywhere between -1 and 1.


问:
How to characterize self-adjoint operators in terms of orthogonal diagonalizability
Have a look at the following excerpt of Tosio Kato (taken from Zeidler Applied functional analysis vol. I):

The fundamental quality required of operators representing physical quantities in quantum mechanics is that they be self-adjoint which is equivalent to saying that the eigenvalue problem is completely solvable for them, that is, there exist a complete set (discrete or continuous) of eigenfunctions.

What does he mean? To me a self-adjoint operator $(A, D(A))$ on a Hilbert space $\mathcal{H}$ is a linear operator s.t. $A=A^\star$, which is equivalent to say that it is expressible in terms of a unique projection-valued measure $P_A$:

$$A=\int_{-\infty}^\infty \lambda\, dP_A(\lambda).$$

This is the best thing I can think of to match what Kato refers to. This is kind of incomplete, though. Where are those eigenfunctions Kato mentions? Also, a version of the spectral theorem holds for normal operators too. Why are they ruled out?

答:
The self-adjoint operators Kato was was dealing with were systems of finite numbers of particles. He was the first to prove that these Hamiltonians were essentially self-adjoint, which means that their closures are self-adjoint. As a consequence, the spectral theory of these operator may be written in terms of classical eigenfunction types of expansions using discrete sums of eigenfunctions and/or continuous Fourier-type integral "sums" of eigenfunctions. Kato was talking about systems built from particles (i.e., atoms and molecules,) and his statements were about those systems.

Kato was precise and rigorous in his statements and proofs. Don't believe statements to the contrary.
von Neumann had posed this problem of well-posedness, and felt it to be intractible; Kato's early career was marked by successfully proving it for von Neumann.


问:
Integral with square root + Trig
$$ \int_0^{\pi} \sqrt{-3 \sin(x) - 3 \cos(x) + 5} ~~dx$$

I tried to use substitution:

$$ u = -3 \sin(x) -3 \cos(x) + 5 \\ du = -3 \cos(x) + 3 \sin(x) $$

But because $ ( \cos(x) )' = - \sin(x)$ it does not work that well (the signs are now flipped...)

I even tried to change the $\cos(x)$ into $\sin( x - \frac{\pi}{2} )$ but it didn't get me very far..

I also tried using this online tool: https://www.integral-calculator.com/
And it couldn't find the anti-derivative, but only gave the answer of:

$$ \approx 5.1363$$

I would appreciate your help, really any way possible to solve this ugly question would be appreciated! Thank you very much!

答:
Similar to @Andrew's comment make
$$\sin(x)+\cos(x)=\sqrt{2} \cos \left(x-\frac{\pi }{4}\right)=\sqrt{2} \left(1-2 \sin ^2\left(\frac{x }{2}-\frac{\pi}{8}\right)\right)$$ This makes
$$5-3\sin(x)-3\cos(x)=(5-3 \sqrt{2})+6 \sqrt{2} \sin ^2\left(\frac{x}{2}-\frac{\pi }{8}\right)$$
$$\sqrt{5-3\sin(x)-3\cos(x)}=a\,\sqrt{1+k\sin ^2\left(\frac{x}{2}-\frac{\pi }{8}\right)}$$ with $$a=\sqrt{5-3 \sqrt{2}}\qquad \text{and}\qquad k=\frac{6(6+5 \sqrt{2})}{7} $$

Now, let $\frac{x}{2}-\frac{\pi }{8}=t$ that is to say $x=2t+\frac{\pi }{4}$ to make the antiderivative
$$I=2a \int  \sqrt{1+k \sin^2 (t)}\,dt =2a\,E(t|-k)$$ where appears the elliptic integral  on fthe first kind.

Back to $x$, $a$ and $k$, using the bounds and simplifying, we the have

$$\color{red}{\int_0^{\pi} \sqrt{5-3 \sin(x) - 3 \cos(x) }\,dx =}$$
$$\color{red}{2 \sqrt{5-3 \sqrt{2}} \left(E\left(\frac{\pi }{8}|-\frac{6\left(6+5
   \sqrt{2}\right)}{7} \right)+E\left(\frac{3 \pi }{8}|-\frac{6\left(6+5
   \sqrt{2}\right)}{7} \right)\right)}$$ which is
$$5.1363328412422184497730638111478712316615873410060$$

Title: Integral with square root + Trig


问:
If $f$ is continuous on $[0,1]$, prove that $\lim_{n\to\infty}\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi}{2}f(0)$.
The solution given:
$$\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx+\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx$$
okay, my first question here: why the $n^{-1/3}$?
$$|\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx|\leq \int_{n^{-\frac{1}{3}}}^{1} |\frac{nf(x)}{1+n^2x^2}|dx \leq \frac{nM}{1+n^{1+1/3}}$$
where M is the global maximum of $|f|$. how did the middle term turn into the rightmost one? How is the mean value theorem of integrals used here? what's with the $1+1/3$?

Since the rightmost term tends to $0$ as $n$ tends to infinity, by squeeze theorem we have the leftmost term tends to $0$ as well.
Since $\frac{n}{1+n^2x^2}$ does not change sign on $[0,1]$, then there exists $c\in[0,n^{-1/3}]$ such that
$$\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx=f(c)\int_{0}^{n^{-\frac{1}{3}}} \frac{n}{1+n^2x^2}dx=f(c)\tan^{-1}n^{2/3}$$
Since $c\in[0,n^{-1/3}]$, $n\to\infty$ implies $c\to0$ and $\tan^{-1}n^{2/3}\to \frac{\pi}{2}$ and so $\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx\to\frac{\pi}{2}f(0)$

So the final answer is $0+\frac{\pi}{2}f(0)=\frac{\pi}{2}f(0)$

I've highlighted my questions! Thanks for any advice given!

答:
First for your highlighted question: note that
$$
\newcommand{\abs}[1]{\left\vert #1 \right\vert}
\newcommand\rme{\mathrm e}
\newcommand\imu{\mathrm i}
\newcommand\diff{\,\mathrm d}
\DeclareMathOperator\sgn{sgn}
\renewcommand \epsilon \varepsilon
\newcommand\trans{^{\mathsf T}}
\newcommand\F {\mathbb F}
\newcommand\Z{\mathbb Z}
\newcommand\R{\Bbb R}
\newcommand \N {\Bbb N}
\renewcommand\geq\geqslant
\renewcommand\leq\leqslant
\newcommand\bm\boldsymbol
\newcommand\stpf\blacktriangleleft
\newcommand\qed\blacktriangleright
\newcommand\upint[2][a]{\bar {\phantom \int} \mspace{-21mu}{\int_{#1}^{#2}}}
 \frac {n \abs f(x)} {1 + n^2 x^2} \leq \frac {nM} {1 + n^2 n^{-1/3 \times 2}} = \frac {nM} {1 + n^{4/3}} = \frac {nM} {1 + n^{1 + 1/3}},
$$
where $x \in [n^{-1/3}, 1]$. Now integrate over this interval,
$$
\int_{n^{-1/3}}^ 1 \frac {n \abs f(x)} {1 + n^2 x^2} \diff x\leqslant \int_{n^{-1/3}}^1 \frac {nM} {1 + n^{1 +1/3}} \diff x = (1 - n^{-1/3}) \frac {nM} {1 + n^{1+ 1/3}} \color{red}{\leq} 1 \cdot \frac {nM} {1 + n^{1+ 1/3}},
$$
where $\color{red}\leq$ comes from $(1 - n^{-1/3}) \leq 1$.

For the solution, $n^{-1/3}$ seems tricky. So we try another explanation, which unfortunately may require some knowledge about superior/inferior limits.

At the first glimpse, we may want to take limit under $\int$, but in general we can't. But intuitively, we might feel that for $x$ close enough to $1$, the $n^2$ part in the denominator would dominate, since $f$ is bounded, and $n f$ is of course "weaker" than $n^2$. The problematic point is $0$. So we could pick any temporarily fixed $\delta > 0$ and break the interval into two parts. One part could be estimated like this:

\begin{align*}
&\quad \abs {\int_\delta^1 \frac {nf (x) \diff x }{1 + n^2 x^2} } \\
&\leq \int_\delta ^1 \frac {n \abs f (x)}{ 1+ n^2x^2 }\diff x \\
&\leq \int_\delta^1 \frac {nM}{1 + n^2 \delta^2} \\
&= (1 - \delta) \frac {nM} {1 + n^2 \delta^2}\leq \frac {nM}{1 + n^2 \delta^2}.
\end{align*}

For the other, $[0, \delta]$, we use the continuity, and note that
$$
\abs {\int_0^{\delta} \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x} \leq \max_{0 \leq x \leq \delta} \abs {f(x) - f(0)}\int_0^\delta \frac {\diff (nx)}{ 1 + (nx)^2} =: N(\delta) \arctan (n \delta).
$$
If the limit of both part exists, then
$$
\lim_{n \to \infty} \abs {\int_\delta^1 \frac {nf (x) \diff x }{1 + n^2 x^2} } \leq \lim_{n \to \infty} \frac {nM}{1 + n^2 \delta^2} = 0,
$$
and
$$
\lim_{n \to \infty} \abs {\int_0^{\delta} \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x} \leq \lim_{n \to \infty} N(\delta) \arctan (n \delta) = \frac \pi 2 N(\delta),
$$
and if we let $\delta \to 0^+$, then we can expect that the original limit to be $\lim_n \int_0^1 n f(0)\diff x /(1 +n^2 x^2) = \pi f(0)/2 $, since $N(\delta) \to 0$ according to the continuity of $f$ at $0$. To make these argument work, we shall find some $\delta(n)$ that variates as $n \to \infty$ such that

$\delta (n) \to 0$ as $n \to \infty$;

$1 + n ^2 \delta (n)^2$ "dominates" as $n \to \infty$, i.e. $nM / (1 + n ^2 \delta(n)^2)\to 0$.

Easy to see a qualified candidate could be $\bm {n^{-1/3}}$. Hence the solution works. Of course we could pick other forms, but for the efficiency, we might pick simple ones.

In fact we could write the following using superior limits:
\begin{align*}
&\quad \varlimsup_n \abs {\int_0^1 \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x }\\
&\leq \varlimsup_n \int_0^\delta \abs {\frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x}+ \varlimsup_n \int_\delta^1 \abs { \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x}\\
&\leq \varlimsup_n \frac{n \cdot 2M} {1 + n^2 \delta^2} + \varlimsup_n N(\delta) \arctan (n \delta) \\
&= 0 + \frac {\pi}2 N(\delta)\\
&\xrightarrow {\delta\to 0^+} 0,
\end{align*}
which is a standard proof of the limit.


问:
Set of distinct odd positive integers such that $\left(\frac{1}{O_{1}}+\frac{1}{O_{2}}+\frac{1}{O_{3}}+...+\frac{m+1}{O_{n}}\right)=1$
I am looking for a set of distinct odd positive integers $3\leq{O_1}<O_2<O_3<...<O_n$ such that

$$\frac{1}{O_{1}}+\frac{1}{O_{2}}+\frac{1}{O_{3}}+...+\frac{m+1}{O_{n}}=1$$ such that  $m\in\mathbb{N}$ and $1\leq{m}\leq5$, and other than the simple sets $S=\{3\}$, $m=2$ and $S=\{5\}$, $m=4$.

Is it possible such set of distinct odd positive integers? And could you give me an example?

Thanks in advance!

答:
Comment:

You are write and it can be shown by argument. I checked this by following program in Python:

```<code> for a in range (3, 200, 2):

   for b in range (3, 200, 2):

     for c in range (5, 200, 2):

       for d in range (9, 200, 2):

         for e in range (11, 200, 2):

            x=a*b*c*d+a*c*d*e+a*b*d*e+b*c*d*e+a*b*c*e

            y=a*b*c*d*e

            z=x/y

            t=int(z)

            if (t*Y)==x:

               print a, b, c, d, e
</code>```

And only two sets of solutions was resulted"

(a, b, c, d, e)= (3, 3, 5, 9, 45), (3, 3, 5, 15, 15)

If you replace 3 by other odds like 7, 5, 13 . . .   in one command started with 3 you will get no result.

Body:
Comment:

You are write and it can be shown by argument. I checked this by following program in Python:

 for a in range (3, 200, 2):

   for b in range (3, 200, 2):

     for c in range (5, 200, 2):

       for d in range (9, 200, 2):

         for e in range (11, 200, 2):

            x=a*b*c*d+a*c*d*e+a*b*d*e+b*c*d*e+a*b*c*e

            y=a*b*c*d*e

            z=x/y

            t=int(z)

            if (t*Y)==x:

               print a, b, c, d, e

And only two sets of solutions was resulted"

(a, b, c, d, e)= (3, 3, 5, 9, 45), (3, 3, 5, 15, 15)

If you replace 3 by other odds like 7, 5, 13 . . .   in one command started with 3 you will get no result.


问:
Posterior distribution of exponential prior and likelihood?
I have the prior density function:

$$e^{-\theta} \text{ for } \theta > 0$$

and the likelihood function:  $e^{\theta -x}$ for $x \geq \theta$

I have gotten the following in my attempt to derive the posterior distribution:

$$L(\theta) = \prod e^{\theta - x_i} = e^{n\theta}e^{-\sum x_i} \mathbb{I}_{\min X_i}$$

$$\pi(\theta \mid x) = \frac{ e^{-\sum x_i} e^{\theta(n-1)} \mathbb{I}_{\min X_i}}{e^{-\sum x_i}\int_0^{\min(x)}e^{n\theta}e^{-\theta} \, d\theta}$$

$$= \frac{e^{\theta(n-1)}\mathbb{I_{\min X_i}}}{\frac{1}{n-1}e^{\min(x)(n-1)} - \frac{1}{n-1}}$$

Is this correct? Does this further simplify, or is the posterior simply non-standard?

答:
and the likelihood function: $e^{\theta-x}$  for $x≥θ$

First, this is not the likelihood but the Model Density.

You are almost there but I suggest you to approach Bayesian issue in a different way: do not worry about the posterior denominator... it is wasted time!

Prior:

$$\pi(\theta)=e^{-\theta}$$

$\theta>0$

Likelihood:

$$p(\mathbf{x}\mid\theta) = e^{-\sum_i x_i}\cdot e^{n \theta} \cdot \mathbb{1}_{[\theta;+\infty)}(x_{(1)})$$

Posterior (any element not depending on $\theta$ can be trown away, it will be part of the normalization constant)

$$\pi(\theta|\mathbf{x})=C\times p(\mathbf{x}\mid\theta)\times \pi(\theta)=C \times e^{\theta(n-1)}\cdot\mathbb{1}_{(0;x_{(1)}]}(\theta)$$

Where

$$C^{-1}=\int_0^{x_{(1)}}e^{\theta(n-1)}d \theta=\frac{e^{x_{(1)}(n-1)}-1}{n-1}$$

Observe that: When you deal with the posterior $x_1,x_2,...,x_n$ are only data, not anymore rv's

A real world example using this Statistical Model

According to the thinking of Cthulhu Cult's followers, adoring the "Great Ancients" (an ancient civilization come from the stars that lived on Earth before Homo Sapiens appeared)  will grant a very long life to them. In particular, story goes about a group of followers dead at the age of

$$\{X_1=518;X_2=696;X_3=230;X_4=410;X_5=821\}$$

years old

Using these observations, Prior and Model specified above, derivate the posterior density of Cult Followers' lifetime and, further request with respect to your exercise, Verify the Hypothesis that the lifetime of the Cthulhu Cult's Members is greater than 200 years

$$\pi(\theta|\mathbf{x})=\frac{4 e^{4\theta}}{e^{920}-1}\mathbb{1}_{(0;230]}(\theta)$$

You can easy check that $\pi(\theta|\mathbf{x})$ is a nice density function. for the Hypothesis Testing, I leave it as an exercise, if this is part of your program...


问:
Vector Calculus with parametrization

Let $S$ be the unit sphere $x^2+y^2+z^2=1$. Denote by $\mathbf{n}=(0,0,1)$ to be the "north-pole". Let $(s,t,0)$ to be an arbitrary point in the $xy$-plane. Draw the line from $\mathbf n$ to $(s,t,0)$ and determine where it intersects the sphere $S$, i.e. determine $x,y,z$ coordinates in terms of $s$ and $t$.
This will give you functions,$$x=x(s,t)\qquad y=y(s,t)\qquad z=z(s,t)$$These functions will parametrize the sphere $S$.

This parametrization of the sphere parametrizes only the entire sphere, it only misses the point $\mathbf n$.

What is the domain for the $s,t$ variables?

Please help me with the solution.

答:
Hint:

Unless the line is parallel to $XY$ plane, it will intersect the sphere in two points (one of them is obviously $n$). So that should tell you the domain of $s ,t $ variables.

Equation of the line from $n (0, 0, 1)$ to arbitrary point in $XY$ plane $(s, t, 0)$ -

$(0, 0, 1) + a(s, t, -1)$

If the line intersects the sphere, the points on this line $(as, at, 1-a)$ will satisfy the equation of the sphere -

$a^2s^2 + a^2t^2+a^2+1-2a = 1 \implies a^2s^2+a^2t^2+a^2-2a = 0$ so $a = 0$ is an obvious solution which from the line of equation gives point $(0, 0, 1)$ (we knew this was obvious).

Now can you use this to find the other possible value of $a$ and then the other point of intersection? You will also notice that parametrization will cover all points on the sphere except point $n (0, 0, 1)$.

Title: Vector Calculus with parametrization


问:
Minimum value of $(\sin x + \cos x)^3 + \frac{1}{\sin^2x\cos^2x}$
Let $\sin x + \cos x = t\qquad (\vert t \vert \leq \sqrt2)$

and solve it as $t^3 + \dfrac{4}{(t^2 - 1)^2}$

Is there any easier solution to solve this $(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}$?

Is there any available inequality can solve it?

Please help!

答:
There may not be a specific inequality, but sure there is a faster way.

$$(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}=2\sqrt{2}\left(\sin\left(x+\frac{\pi}{4}\right)\right)^{3}+4\csc^{2}\left(2x\right)$$

Observe that only the expression having an odd power can contribute to reducing the value of the expression. That is, when $\sin(x+\pi/4)$ is the lowest.

A possible solution is $x=-\dfrac{3\pi}{4}$ which gives the global minimum as $4-2\sqrt{2}=1.1715728\dots$

Here is a graphical solution in aid of the above.


问:
Reducing Boolean expressions
Just learning mathematical proof writing and came upon this interesting question Writing an expression using logic.

$$(P \land Q \land \lnot R) \lor (P \land \lnot Q \land \lnot R) \lor (\lnot P \land Q \land R)$$
Just to help myself prepare for these kind of questions (this question might have already been solved, but), how would you reduce this boolean sentence where as the question states, solving for (???) ?

答:
The objective here is not to solve it, as there is no equation to solve, but to simplify the proposition: but what counts as simplified is subjective, and needs specification. For example, one can aim to write it in conjunctive normal form (as it is already in disjunctive normal form). Sometimes for questions like this you will be asked to show that the proposition is equivalent to some given proposition.

You have:
$$(P \land Q \land \lnot R) \lor (P \land \lnot Q \land \lnot R) \lor (\lnot P \land Q \land R) \equiv \quad???\tag{1}$$

There are various ways to proceed and/or manipulate the expression into an equivalent form, using distribution, or De Morgan's in some cases, for example, depending on what your goal is.

E.g., in response to your question below: Note that we can use the distributive law to combine (factor out) like terms in the first two terms in $(1)$:

$$(P \land Q\land \lnot R) \lor (P \land \lnot Q \land \lnot R)$$
$$\equiv [(P \land \lnot R) \land Q] \lor [(P \land \lnot R) \land \lnot Q]$$
$$\equiv (P \land \lnot R) \land (Q \lor \lnot Q)$$
$$\equiv P \land \lnot R$$

That gives us, remembering we still have the third term of $(1)$ to retain:

$$(P \land Q \land \lnot R) \lor (P \land \lnot Q \land \lnot R) \lor (\lnot P \land Q \land R) \equiv \color{blue}{\bf (P\land \lnot R) \lor (\lnot P \land Q \land R)}\tag{2}$$

As mentioned,  $(1)$ is already in one of the standard forms used in logic: Disjunctive normal form (DNF): and it meets the criteria given in the linked question you provide, to find an expression using only the connectives $\land, \lor, \lnot$: it represents all the information conveyed by the truth-table: and so can be considered to be a valid representation for "$???$". So can the RHS of $\color{blue}{\bf (2)}$

But what counts as "simplified" depends on some given, specified constraints.

Title: Reducing Boolean expressions


问:
Classification of subfields of $K(S)$
Let $F:K$ be a field extension and $S \subset F$. Is there any classification of subfields of $K(S)$ In case this isn't standard, $K(S)$ is the intersection of every subfield containing $K$ and $S$, i.e.

\begin{equation*}
   K(S) = \bigcap_{\substack{L \supset K \cup S \\ L \text{ is a subfield of } F}} L.
\end{equation*}

I couldn't find anything, but my intuition is telling me that the only subfields of $K(S)$ are either

subfields of $K$;

subfields of the form $K(T)$, where $T$ is "built" from $S$.

I'm not exactly sure what "built" should be, but my thoughts so far are that elements of $T$ should be linear combinations of elements of $K(S)$ with coefficients in $K$. However, I am at a loss of how to start proving this.

If not, can we say something along these lines if $S$ is a finite set, say $S = \{\alpha_1, \dotsc, \alpha_n\}$?

答:
EDIT: Apologies, I made a silly error in my construction of $K(S)$. What I wrote is correct if you assume that $K(S)/K$ is algebraic. If not, then you must instead consider quotients of these sums. For instance, if we look at $K(x)/K$ then the set I wrote out is just $K[x, x^{-1}]$, which is not equal to $K(x)$. The rest of the argument holds the same way, but with a broader notion of $T$ being "built" from $S$.

Your conjecture is essentially correct, and here's a way to prove it. We have an expression for a generic element of $K(S)$ as sums of the form $\sum a_i s_i^{n_i}$ where $a_i \in K$, $s_i \in S$, and $n_i \in \mathbb Z$ (I'm assuming here that $0 \notin S$). This will indeed be a field containing $K$ and $S$, and clearly any field containing $K$ and $S$ must contain this. Now, let $K(S)/E/K$ be an intermediate extension. Then every element of $E$ is of the form of the sum I wrote above, so any generating set $T$ of $E$ over $K$ will necessarily consist of such elements as well. In that sense, $T$ is wholly "built" from $S$.

For completion, I want to emphasize that you cannot strengthen this to say that $T \subseteq S$ or even that elements of $T$ are $K$-linear combinations of elements of $S$. Take for instance $K(x)/K(x^2)$. You do really need these to be algebraic combinations, not just linear ones.

Also, I'd like to point out that any discussion of the set of intermediate fields in an extension $F/K$ warrants some mention of Galois theory. In certain nice cases, subgroups of the group of automorphisms of $F$ that fix $K$ (called the Galois group of $F/K$) completely determine the intermediate fields in a very elegant fashion.

Title: Classification of subfields of $K(S)$


问:
Prove that $|f(x)|\leq|\frac{1}{b-a}\int_{a}^{b}f(x)dx|+\int_{a}^{b}|f'(x)|dx$
where $f$ is continuous on $[a,b]$.
Quite easily, $|\frac{1}{b-a}\int_{a}^{b}f(x)dx|=|f(c)|$ for some $c\in(a,b)$. so it would be most ideal if $\int_{a}^{b}|f'(x)|dx\geq|f(x)-f(c)|$, which I suppose it is indeed the case here- but is there a rigorous way of proving $\int_{a}^{b}|f'(x)|dx\geq|f(x)-f(c)|$? I'm just taking a wild guess and drawing some graphs to convince myself its correct.
Thanks!

答:
Let us first  prove the following: if $g$ is continuously differentiable on $[a,b]$ and $\int_a^{b} g(t)dt=0$ then $|g(x)| \leq \int_a^{b}|g'(t)|dt$ for all $x$.

Since $\int_a^{b} g(t)dt=0$ $g$ must take both positive and negative values. So there exist $c$ such that $g(c)=0$.  Now $g(x)=g(c)+\int_c^{x} g'(t)dt$ from which we get $|g(x)| \leq \int_a^{b}|g'(t)|dt$ for all $x$.

Now apply this to $g(x)=f(x)-\frac 1 {b-a}\int_a^{b} f(x)dx$ .


答:
One flaw is that you refer to $\mathbb{I}_{\min X_i}$ where apparently you mean $\mathbb I_{\min X_i \,\ge\,\theta}.$

I wouldn't bother with that method of computing the normalizing constant. First you conclude that the posterior density is
$$
\text{constant} \times e^{\theta(n-1)} \qquad \text{for } 0<\theta<\min\{X_1,\ldots,X_n\}
$$
and then evaluate the integral that tells you what the constant is. I.e. don't bother with factors like $e^{-\sum_i x_i}$ that do not depend on $\theta.$

Often it's best to leave the computation of the normalizing constant to the end, since the posterior will be recognized as having some standard form that will tell you what the normalizing constant is, and even if not, finding the necessary integral at the end will ofetn be simpler than proceeding the way you did.

Title: Posterior distribution of exponential prior and likelihood?

Body:
I have the prior density function:

$$e^{-\theta} \text{ for } \theta > 0$$

and the likelihood function:  $e^{\theta -x}$ for $x \geq \theta$

I have gotten the following in my attempt to derive the posterior distribution:

$$L(\theta) = \prod e^{\theta - x_i} = e^{n\theta}e^{-\sum x_i} \mathbb{I}_{\min X_i}$$

$$\pi(\theta \mid x) = \frac{ e^{-\sum x_i} e^{\theta(n-1)} \mathbb{I}_{\min X_i}}{e^{-\sum x_i}\int_0^{\min(x)}e^{n\theta}e^{-\theta} \, d\theta}$$

$$= \frac{e^{\theta(n-1)}\mathbb{I_{\min X_i}}}{\frac{1}{n-1}e^{\min(x)(n-1)} - \frac{1}{n-1}}$$

Is this correct? Does this further simplify, or is the posterior simply non-standard?


问:
Why $\frac{(1-x)\log(1-x)}{x\log x}$ is increasing on (0, 1)?
Let $f(x)=\frac{(1-x)\log(1-x)}{x\log x}$. I find one claims this function is increasing on $(0,1)$. While I find the numerator of $f'(x)$ is
$$-x\log x -\log x\log(1-x)-(1-x)\log(1-x).$$
I am wondering why this is non-negative?

答:
Consider the function $g(x)=\frac{\log(1-x)}{x}$. Then

$$f(x)=\frac{g(x)}{g(1-x)}$$

Take a derivative of $g$:

$$g'(x)=-\frac{x+(1-x)\log(1-x)}{x^2(1-x)}$$

Consider the auxiliary function $h(x)=x+(1-x)\log(1-x)$. We use a standard argument to show that it is positive. It's derivative is $h'(x)=-\log(1-x)$ which is positive in the interval $(0,1)$ and that means that h is increasing. However $h(0)=0~,~h(1)=1$ which proves given it's monotonicity, that $h(x)\geq 0, x\in(0,1)$.

Therefore we finally conclude that in the interval $(0,1)$ $g'(x)\leq 0$ and thus g is decreasing. Knowing this, we can also conclude that $g(1-x)$ is increasing in the same interval since the transformation $x\to 1-x$ reverses monotonicity but leaves the interval invariant. Finally we see that $1/g(1-x)$ is decreasing since $g(1-x)$ is non-vanishing in this interval and therefore it has the same sign (in fact, it is negative).

Now note that the functions $u_1(x)=-g(x)$ and $u_2(x)=-1/g(1-x)$  are both increasing and positive. One can easily show that multiplying increasing, positive functions results in an increasing, positive function as well. Since $f(x)=u_1(x)u_2(x)$ we conclude that $f$ is increasing and this concludes the proof.


问:
Density of Lebesgue measurable set
I am looking to show that if we have a Lebesgue measurable set $E \subseteq \mathbb{R}$ with a density  1 at every element in $E$ and a density of 0 at every element of $\mathbb{R} \backslash E$. Then it must be that $E = \mathbb{R}$ or $E = \emptyset$.

I am working through Axlers book on measure theory and we have defined the density of $E$ at a number $b \in \mathbb{R}$ to be $\lim _{t \downarrow 0} \frac{|E \cap(b-t, b+t)|}{2 t}$.

From the Lebesgue Density Theorem I know that for a Lebesgue measurable set $E \subset \mathbb{R}$, the density of $E$ is 1 at
almost every element of $E$ and is 0 at almost every element of $\mathbb{R} \backslash E$. So the difference to this case is that we are saying its true everywhere as opposed to almost everywhere.

So so far I have that:

For all $b \in E$ we have $\lim _{t \downarrow 0} \frac{|E \cap(b-t, b+t)|}{2 t} = 1$

For all $b \in \mathbb{R} \backslash E$ we have $\lim _{t \downarrow 0} \frac{|E \cap(b-t, b+t)|}{2 t}=0$

Intuitively it makes sense why only $\mathbb{R}$ and $\emptyset$ work but I am having trouble putting it into a complete proof. Any help is greatly appreciated. Thanks in advance!

答:
The function defined on $(0, \infty)$
$$f(x) = m((0, x]\cap E)$$
has a derivative everywhere, and $f'$ takes values in $\{0,1\}$, but it also has the Darboux property.

Note:  the density is defined with a "symmetric derivative", so in general it may not be equivalent with the usual derivative.

$\bf{Added:}$ While the existence of the symmetric derivative does not imply the existence of the derivative, note that our functions satisfies
$$0 \le \frac{f(x+h)- f(x) }{h} \le 1$$
and so
$$0\le \frac{f(x+h)- f(x) }{h}\le 2 \cdot \frac{f(x+h)- f(x-h)}{2 h}$$

Therefore, if at some point we have
$$\lim_{h\to 0}\frac{f(x+h)- f(x-h)}{2 h} = 0 $$
then $f'(x) = 0 $.  Working with the function $x-f(x)$ we also conclude that if the symmetric derivative is $1$ then the derivative is $1$.

Therefore, $f$ has a derivative everywhere on $(0, \infty)$ equal to $0$ or $1$.

Title: Density of Lebesgue measurable set


问:
Convergence rate of Newton's method
Let $f(x)$ be a polynomial in one variable $x$ and let $\alpha$ be its $\delta$-multiple root
($\delta\ge2$).

Show that in the Newton's $x_{k+1}=x_k-f(x_k)/f'(x_k)$, the rate of convergence to $\alpha$ is not quadratic.

My solution: Suppose that $\alpha$ is one regular root of equation.Then
$$x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}=\phi(x_k)$$
convergence rate to $\alpha$:

$$x_k-\alpha=\phi(x_k)-\phi(\alpha)=(x_k-\frac{f(x_k)}{f'(x_k)})-(x-\frac{f(x)}{f'(x)})$$
$$(x_k-\alpha)=(f'(x_k))^{-1}(f(x_k)-f(x))=(x_k-\alpha)+(f'(x_k))^{-1}\{ f'(x_k)(x_k-\alpha)+O((x_k-\alpha)^2)\}=f'(x_k)^{-1}O((x_k-\alpha)^2))
\tag{1}$$

so the convergence rate to $\alpha$ is quadratic.

But if $\alpha$ is not regular root, then $(f'(x))^{-1}$ has no meaning. We need do slightly change in $(1)$,
$$(x_k-f'(x_k)^{-1}f(x_k))-(x-f^{(\delta)}(x)\delta!f(x))=(x_k-x)-f'(x_k)(f(x_k)-f(x))$$
$$=(x_k-x)-f'(x_k)^{-1}(\frac{f^{(\delta)}(x_k) (x_k-x)^{\delta}}{\delta!}+O((x_k-x)^{\delta+1}))$$

convergence rate to non regular root $\alpha$ is one.

Is my solution correct?

Then I do same thing to next nonlinear equation:

$$f(x)=x^2(x-1) $$, m is integer.

so the newton's formula is above, and how about convergence rate to $0,1$?
I think convergence to 1 is one, absolutely convergence to 0 is quadratic.
when convergence rate is 1, the how about the convergence rate?

答:
For iterative methods, we have a fixed point formula in the form:

$$\tag 1 x_{n+1} = g(x_n)$$

The Newton iteration is given by:

$$\tag 2 \displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

So, $(2)$ is of the form $(1)$.

Since $r$ is a root of $f(x) = 0, r = g(r)$. Since $x_{n+1} = g(x_n)$, we can write:

$$x_{n+1} - r = g(x_n) - g(r).$$

Lets expand $g(x_n)$ as a Taylor series in terms of $(x_n -r)$, with the second derivative term as the remainder:

$$g(x_n) = g(r)+g'(r)(x_n-r) + \frac{g''(\xi)}{2}(x_n-r)^2$$

where $\xi$ lies in the interval from $[x_n, r]$, since:

$$g'(r) = \frac{f(r)f''(r)}{[f'(r)]^2} = 0.$$

Because $f(r) = 0$ ($r$ is a root), we have:

$$g(x_n) = g(r) + \frac{g''(\xi)}{2}(x_n-r)^2.$$

Letting $x_n-r = e_n$, we have:

$$e_{n+1} = x_{n+1}-r = g(x_n) - g(r) = \frac{g''(\xi)}{2}(e_n)^2.$$

Each successive error term is proportional to the square of the previous error, that is, Newton's method is quadratically convergent.

Multiple Root

Following the same sort of reasoning, if $x_n$  is near a root of multiplicity $\delta \ge 2$, then:

$$f(x) \approx \frac{(x-\xi)^\delta}{\delta !}f^{(m)}(\xi)$$

$$f'(x) \approx \frac{(x-\xi)^{\delta-1}}{(\delta-1) !}f^{(m)}(\xi)$$

So we have:

$$\tag 3 x_{n+1} -\xi = x_n - \xi -\frac{f(x_n)}{f'(x_n)} = \left(\frac{\delta -1}{\delta}\right)(x_n - \xi)$$

The $\delta$ term on the RHS of $(3)$ in not quadratic, hence we have linear convergence.

You should be able to use with your approach to clean up what you did.

I am confused about what you wrote after your derivation, but I am going to guess that you want to figure out the convergence rate for this $f(x)$.

We are given:

$$f(x) =x^2(x-1)$$

There are two roots to this equation at:

$x = 0$ (a double root)

$x = 1$ (a single root)

So, we would expect linear convergence at the double root and quadratic convergence at the single root.

The Newton iteration is given by:

$$x_{n+1} = x_n - \frac{(x_n-1) x_n^2}{x_n^2+2 (x_n-1) x_n}$$

For the first root, lets pick a starting point of $x = 0.1$, we get the following cycle:

$24$ steps to converge to the root $x =  5.341441708552285 \times 10^{-9}$ (yikes!)

For the second root, lets pick a starting point of $x = 1.4$, we get the following cycle:

$6$ steps to converge to the root $x = 1.000000000000000$ (much better!)

Now, you would use the exact results and compare them numerically and show the convergence rates for each of the cases.

Note: one must choose a sufficient starting point that will converge to one root or the other. Based on that initial selection, the rate is going to be quadratic when the algorithm converges to $1$ and linear when it converges to $0$. We pick a nearby starting point and see where we end up. You could also graph the function to have an idea about starting points. We typically do not know apriori what roots will give us what behavior. Obviously there is a range where convergence happens to one root or the other. I am not sure how one would calculate that analytically because you may as well figure out the roots without numerical methods in that case.

Title: Convergence rate of Newton's method


问:
Why vector space is torsion $F[x]$-module?
This is my confusion when I read dummit and foote. If $V$ is a finite dimensional vector space over field $F$, and we consider it as $F[x]$-module via a linear transformation $T$ (i.e.  $p(x)\cdot v=p(T)v$), why $V$ is a torsion $F[x]$-module? How to construct nonzero $p(x)$ such that $p(x)\cdot v = 0$ for each $v \in V$?

答:
Without an explicit understanding of $V$ itself we have no hope of an explicit construction of $p$. However, this can still be done as follows. Consider the ring $End_F(V)$ of linear endomorphisms of $V$. As $V$ is an $F[x]$ module via $T$, we consider the map $\phi: F[x] \longrightarrow End_F(V)$ via $x \mapsto T$. This map is a ring homomorphism and is furthermore $F$ linear. As $V$ is finite dimensional over $F$, $End_F(V)$ is as well (just square the dimension). However, $1, x, x^2, x^3, \dots$ forms an infinite linearly independent subset of $F[x]$. Thus, it is infinite dimensional so $\phi$ must have a nonzero kernel. Call this kernel $I$. It is an ideal of $F[x]$, so using the Euclidean division algorithm, we can find a generator $I = (p)$, i.e. $F[x]$ is a PID. Without loss of generality, we can assume $p$ to be monic by dividing out its leading coefficient. This will be the minimal polynomial. If you're uncomfortable with this ring theory, we don't need $(p) = I$ to finish the proof, only that $I$ contains a nonzero polynomial.

Now, let's think about what all of this meant. We have a nonzero monic polynomial $p$ such that $ker(\phi) = (p)$. By definition, $\phi(p) = p(T) = 0$. Furthermore, by definition, $F[x]$ acts on $V$ via $p \cdot v = p(T) v$. But as we just said, $p(T) = 0$ so $p \cdot v = 0$ for all $v$. Thus, $V$ is a torsion $F[x]$ module.


问:
In the following right $\Delta ABC$, $AC = BC = 1$ and $DEF$ is an arc of a circle with center $A$.

In the following right $\Delta ABC$, $AC = BC = 1$ and $DEF$ is an arc of a circle with center $A$. Suppose the shaded areas $BDE$ and $CEF$ are equal. Then $AD = \frac{x}{\sqrt{x}}.$ Find $x$ .

What I Tried: Here is a picture :-

No idea for this. I cannot understand how I can get the position of point $E$ given that the area of both the red sections are the same. I cannot think of any idea to get the areas either since they are formed from arc of the circle.

Can anyone give me a small idea of how to start this? Thank You.

Edit: So the hint from @Student1058 , I think I have actually solved the problem but getting stuck somewhere.

We see that the area of $\Delta ABC$ is same as the area of the sector $ADF$.

We have :- $$\rightarrow [\Delta ABC] = \frac{(1 * 1)}{2} = 0.5$$
Also we have the central angle of the sector to be $45^\circ$ as this is an isosceles right-angled triangle.

So area of a sector given its radius and central angle would be :-
$$\rightarrow \frac{(r^2 * 45)}{2} = 0.5$$
$$\rightarrow r^2 = \frac{1}{45}$$
$$\rightarrow r = AD = \frac{1}{\sqrt{45}} = \frac{1}{3\sqrt{5}}$$

But the question says the answer should be of the form of $\frac{x}{\sqrt{x}}$ , where did I go wrong?

答:
$$[\triangle ABC]=(ADF)$$

$$\frac{1\times1}{2}=\frac{\pi (\sqrt{x})^2 }{8}$$

$$x=\frac{4}{\pi}$$

Area of a sector with $\alpha$ is $\frac{{\pi} r^2\times \alpha }{360}$


问:
Combination with identical elements.
I am given a word "Question three" and asked not accounting for order,how many ways  four letters can be selected from the thirteen letters.

My answer is since order does not matter  so it should be just  number of different combination of 10 letters with 4 taken at a time .i.e $C(10,4)=\frac{10!}{4!(10-4)!}$ As it has 10 distinct letters.Am I correct??

答:
You have to break it up into cases, since you have 3 E's and 2 T's.

The easiest way is to focus on the E's of which there will either be

0, 1, 2, or 3 possibilities.

Further, there will either be 0, 1, or 2 T's.

So you have $4 \times 3$ distinct cases to enumerate.  Of these 12 cases, the only one that can be eliminated out of hand is 3 E's with 2 T's.

For each of the 11 live cases,

let $a_1 = $ the # of E's

let $a_2 = $ the # of T's

let $a_3 = 4 - (a_1 + a_2).$

Then the # of ways that a specific case can occur is

$$\binom{8}{a_3}.$$

Title: Combination with identical elements.


问:
If $f(1)=2\int_{0}^{1/2}e^{1-x}f(x)dx$, prove that there exists $c\in(0,1)$ such that $f(c)=f'(c)$.
$f$ is continuous on $[0,1]$ and differentiable on $(0,1)$ .
$$f(1)=2\int_{0}^{1/2}e^{1-x}f(x)dx\leq2e\int_{0}^{1/2}f(x)dx=ef(c_1)$$ where $c_1\in(0,1/2)$.
Then I thought of let $g(x)=e^{1-x}f(x)$, so when differentiated, it would yield $e^{1-x}(-f(x)+f'(x))$, which looks similar to the equation mentioned in the question--am I on the right track? How do I link them together? Thanks!

Body:
Let $$F(x)=\int_{0}^{x} e^{1-x}f(x)$$
given $$\frac{f(1)}{2}=\int_{0}^{1/2}e^{1-x}f(x)dx$$
$$\frac{f(1)}{2}=F(1/2)$$
hence we see $F(1/2)=f(1)/2$, $F(0)=0$,$F'(1)=0$ thus by LMVT there exists some $d \in 0,1/2)$ for which $$F'(d)=\frac{F(1/2)-F(0)}{1/2}=f(1)$$,Now by rolles theorem  there is some $c \in (d,1/2)$ for which $$F''(c)=0$$.$$e^{1-x}(f'(c)-f(c))=0$$
$$f'(c)=f(c)$$


问:
Galois Extensions and $n^{\text{th}}$ Roots
I've been studying for my prelims lately, and this problem has me stuck:

(a) Let $K$ be a field with no abelian Galois extensions. Suppose that $n$ is a positive integer and either $char(K)=0$ or the characteristic is relatively prime to $n$. Prove that every element of $K$ has an $n^\text{th}$ root in $K$.

(b) Is the statement still true if $n$ is not relatively prime to the characteristic of $K$?

For part $(a)$, I was thinking about the polynomial $f(x)=x^n-a$ for a given $a\in K$. If it has a root in $K$, then $a$ has an $n^\text{th}$ root. I wasn't able to make it much further than this, I only know some conditions when the polynomial is irreducible, which doesn't necessarily say anything about the roots. As for part $(b)$, I'm pretty unsure...

答:
If a field has no abelian Galois extensions it also has no solvable Galois extensions.  This solves part a) since (with the characteristic assumption) the Galois group of $t^n - a$ is solvable.

As for part b), start with an imperfect field like $\mathbb{F}_p(t)$ and then pass to its separable closure.  This field is separably closed -- so has no nontrivial Galois extensions period -- but not algebraically closed.  Every element of the algebraic closure is obtained by taking $p^n$th roots of elements of your ground field, so you'll find a counterexample here.


问:
Maxima and minima of function of two variables.
If $$f(x,y)=x^{2}+y^{3}+\frac{768}{x+y}$$
then find the points where it attains its extrema.

For this I have done the following:
$$f_x=2x-\frac{768}{(x+y)^{2}}$$
$$f_y=3y^{2}-\frac{768}{(x+y)^{2}}$$
Putting $f_x=0$ and $f_y=0$,
$$2x=\frac{768}{(x+y)^{2}}$$
$$3y^{2}=\frac{768}{(x+y)^{2}}$$

$$\implies\frac{2x}{3y^{2}}=1$$

$$\implies\frac{x}{y^{2}}=\frac{6}{4}$$

So, $x=6, y=2$ is a stationary point.

But in the book, it is given that $(5.99,-2.45)$ is another stationary point which satisfies the parabola $2x=3y^{2}$. But I can not find the way to proceed. Any help would be appreciated!

答:
When calculating the stationary points you correctly ended up with
$$
f_x = 0 = 2x - \frac{768}{(x+y)^2}\quad\text{and}\quad
f_y = 0 = 3y^2 - \frac{768}{(x+y)^2}
$$
which implies $2x = 3y^2$ and $2x = 768/(x+y)^2$.  Substitute the first equation in the second to obtain,
$$
3y^2 = 768/\big(\tfrac{3}{2}y^2+y\big)^2
$$
which with rearrangement becomes,
$$
9y^6+12y^5+4y^4-1024 = 0.
$$
You need then to plot the graph of this $6$th order polynomial to see there are just two real roots.  One is when $y=2$, as posted, and numerical methods, for example Newton-Raphson, or even trial and error, will show the second root is at $y \approx -2.4474$.

Final comment:  when looking for the absolute maximum or minimum of the function $f(x,y)$ you need to say what is the domain for $x, y$.  If both are unconstrained, just by inspection you can see $f(x,y)$ approaches $+\infty$ by making $x$ large and $y=0$, $-\infty$ by making $y$ large and negative, $x=0$ and for $x=-y$ the function is undefined.  Thus there is no absolute maximum or minimum, and your calculation is only showing local turning points, which may be local extrema or simply saddles.


问:
Simplify $\left(\frac{3}{\sqrt{1-x^2}}+1\right):\left(\frac{3}{\sqrt{1+x}}+\sqrt{1-x}\right)$

Simplify $\left(\dfrac{3}{\sqrt{1-x^2}}+1\right):\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right)$.

I really don't have an idea how to start. Can you show me?

Body: $$\left(\dfrac{3}{\sqrt{1-x^2}}+1\right):\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right)=\frac{3+\sqrt{1-x^2}}{3+\sqrt{1-x^2}}\frac{\sqrt{1+x}}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x}}$$