问:
Understanding the notion of distance between two symmetric convex bodies
I am new to the notion of distance between two symmetric convex bodies, and I am looking for some help in understanding the same. This is also meant to be an informative post for those who are not already familiar with the idea and want to scratch their brains a little on New Year's Eve. I hope you enjoy it!
To begin with, I quote the definition with a supplementary figure below:

The distance $d(K, L)$ between symmetric convex bodies $K$ and $L$ is the least positive $d$ for which there is a linear image $\tilde L$ of $L$ such that $\tilde L ⊂ K ⊂ d\tilde L$. Note that this distance is multiplicative, not additive: in order to get a metric (on the set of linear equivalence classes of symmetric convex bodies), we would need to take $\log d$ instead of $d$. In particular, if $K$ and $L$ are identical then $d(K, L) = 1$.

From the image it seems that the distance between $K$ and $L$, is not dependent on the size of $L$. I mean, we are looking at a linear image $\tilde L$ of $L$, so looks like all information about the initial size of $L$ is already lost. Is it safe to conclude that $d(K,L)$ does not depend on the size of $L$, and only on its shape? Does it depend on the size of $K$, or that is immaterial too?

The author remarks that if $K$ and $L$ are identical, then $d(K,L) = 1$. What does identical mean? Only the shape, or also the size? $d(K,L) = 1$ just means $\tilde L ⊂ K ⊂ \tilde L$ so that $K = \tilde L$. Again, this says nothing about the size of $L$ originally.
 To add to this, does the converse hold, i.e. if $d(K,L) = 1$ then are $K$ and $L$ identical? I think so.

It is worth noting that $\tilde L$ is defined as the linear image of $L$, i.e. $$\tilde L = \alpha L = \{\alpha z : z\in L\}$$ where $\alpha \in\mathbb{R}^{+}$ right? This means $\tilde L ⊂ K ⊂ \tilde L$ can be written equivalently as $\alpha L ⊂ K ⊂ d\alpha L$ which gives $$\frac{K}{d\alpha} ⊂ L ⊂ \frac{K}{\alpha} \implies \tilde K \subset L\subset d\tilde K$$ where $\tilde K = \frac{K}{d\alpha}$. So can we conclude that $d(K,L)$ is symmetric, i.e. $d(K,L) = d(L,K)$ for all $K,L$? This is indeed a desirable property, I would not want the distance between two symmetric convex bodies to depend on the order in which they are chosen as arguments to $d(\cdot, \cdot)$.

If $d(K,L) = \beta$, and $d(L,M) = \gamma$, what can we say about $d(K,M)$? Here are my thoughts, please let me know if this makes sense:

From the definition, we know that there exist linear images $\tilde L, \tilde M$ of $L,M$ respectively such that $$\tilde L\subset K\subset \beta\tilde L \text{ and }\tilde M \subset L\subset \gamma\tilde M$$ hold. Since $\tilde M$ is some linear image of $M$, we might as well write $\tilde M \subset L\subset \gamma\tilde M$ as $\tilde M \subset \tilde L\subset \gamma\tilde M$ by scaling $\tilde M$ by an appropriate factor (the one with which you would relate $L$ and $\tilde L$). This brings us to: $$\tilde M \subset \tilde L \subset K \subset \beta \tilde L \subset \beta\gamma\tilde M \implies \tilde M \subset K\subset \beta\gamma\tilde M\implies d(K,M) = \beta\gamma$$ So $d(\cdot,\cdot)$ is multiplicative. Not to mention, this is a little weird and contrary to the notion of distance (between two points) that we are familiar with in $\mathbb{R}^n$.

On Pg.9 of the attached reference (below), it is written that the distance between the cube $K = [-1,1]^n$ and the Euclidean ball $B = B^n_2$ in $\mathbb{R}^n$ is at most $\sqrt{n}$. $d(K,B) \le\sqrt{n}$ makes sense because the largest ball inside the cube has radius $1$, while the ball that circumscribes the cube has radius $\sqrt{n}$. Why does the author say "at most" in place of "exactly"? Is it possible that $d(K,B) < n$? I think $d(K,B) = n$.

Are there any other interesting properties of $d(\cdot,\cdot)$ that you can think of? These are all the ones that came to my mind.

A lot of times it's easier to understand new concepts with examples. It'd be extremely helpful if someone could provide some examples (in the usual $\mathbb{R}^2$,$\mathbb{R}^3$ as well as in $\mathbb{R}^n$ for $n\ge 4$) of bodies $K$ and $L$, along with $d(K,L)$ and the thought that went into computing the same.

Thanks a lot!

References:

Lecture 2 (Pg. 8-9) of these notes.

答:
Two convex bodies $A,B \subset \mathbb R^n$ are called affinely invariant if there exists a matrix $M$ of square dimension $n$ and a vector $v \in \mathbb R^n$ such that $A = MB + v$. That is, $A$ is $B$ scaled by $M$, and then shifted by $v$. Since both $A$ and $B$ have non-empty interior(that's what a convex body is) then $M$ will be invertible, so this is a symmetric relation.

For example , any two parallelograms are affinely invariant, by scaling the sides appropriately , rotating if required then shifting.

$B$ is called a linear image of $A$, if there exists a matrix $M$ such that $MB = A$. Again, we are dealing with convex bodies so $M$ will be invertible.

Note that the only differences is that if $A$ is a linear image of $B$ then both will be "centered" around the same point.

An alternate definition for the distance $D$ may be provided as follows : given two symmetric convex bodies $K$ and $L$, shift them both so that their centers are at the origin. Then :
$$
d(K,L) = \inf_{T \textrm{ linear transformation}} \inf \left\{a \cdot b : \frac 1a K \subset TL \subset bK\right\}
$$

Try to prove the equivalence. If you cannot prove it, then just assume it for the following discussion.

This definition is inspired from the Banach-Mazur distance between two normed spaces. Essentially, you want the best linear transformation which kind of brings two normed spaces as close as possible. The best you can do is if they were isometrically isomorphic i.e. a linear transformation took one set to the other while preserving distances. But this doesn't always happen, so the best way to do it then specifies a distance, on the spaces of $n$-dimensional normed linear spaces up to isometric isomorphism, which is basically what identical means in the context above.

(i)

Since scaling is an affine transformation, the distance does not depend upon either the size of $L$ or $K$. It only depends on their respective shapes. For example, when we talk about the distance between a square and a cirle, we are only thinking about the shapes and not about the sizes. Can you prove that a square and a circle are not identical?

(ii)

Now, identical in the text refers to identical up to affine invariance.

So, $A$ and $B$ being identical implies $d(A,B) = 1$. This is obvious, since $B \subset B \subset B$ where $d =1$ and $B$ is a linear image of $A$.

Conversely, if the distance between two convex bodies is $1$, then one is a linear image of the other and they are identical.

(iii)

Indeed, the distance is symmetric, for the reason you mention. You can try to prove it from the definition in my answer.

(iv)

The distance is not multiplicative, it is sub-multiplicative. The problem is basically with the last implication : sure $\tilde{M} \subset K \subset \beta \gamma\tilde{M}$, but this does not imply that $\beta \gamma$ is the smallest number with this property, so $d(K,M) \boxed{\leq} \beta \gamma$, and it very well could be smaller. Just like how it is with numbers (equality in the triangle inequality implies something about the numbers), equality of the kind $d(K,M) = d(K,L)d(L,M)$ should imply something about $K,L,M$, but I am not sure what this relation is.

(v)

The author's exact paragraph is this :

Our observations of the last lecture show that the distance between the cube
and the Euclidean ball in $\mathbb R^n$ is at most $\sqrt n$. It is intuitively clear that it really is $\sqrt n$, i.e., that we cannot find a linear image of the ball that sandwiches the cube any better than the obvious one. A formal proof will be immediate after the next lecture.

The thing is, just like you have done in part $4$, finding a covering of one body by another scaled, doesn't automatically make that covering the best one!

To justify the author, why is the distance atmost $\sqrt n$? To see this, it is enough to show that the unit square $\{x : |x_i| \leq 1 \ \ \forall i\}$ contains the unit sphere  centered at zero $\{x : \|x\| \leq 1\}$ , and is contained the unit sphere scaled by $\sqrt n$.

Both of these are easy exercises.

But this is just one covering. What if there's a better covering, which gives a better value for $d$? This is intuitively obvious when we think about it, but the proof involves the notion of a maximal ellipsoid contained in a convex body. In fact , it is true that the distance is $\sqrt n$ : this comes from comparing the maximal ellipsoids for the square and the sphere.

But again, why ellipsoids? The answer is simple : an ellipsoid is basically a linear image of the unit sphere. Any linear transformation is characterized by what it does with the unit ball i.e. which ellipsoid it produces. So looking at a convex figure, and the linear transformation associated to distances, is like thinking about each convex figure as an $n$-dimensional normed space in its own right, and the respective "unit balls" which are then ellipsoids. The best fit ellipsoid is then the best "unit-ball" approximation of a convex figure, so studying a convex body kind of comes down to studying this maximal ellipsoid.

(vi)

There are a few desirable things : taking the logarithm of $d$ provides a (usual) metric among isomorphism classes of normed linear spaces in $\mathbb R^n$. It is nice to know that this metric space is compact, so basically given an $\epsilon>0$ , you can find a finite set of spaces so that every space is at most $\epsilon$ far away from that. This is like saying that this finite set "approximates all possible shapes" very well.

You can also prove that any convex body is at most $n$ away from the unit square, in this distance. Thus, by sub-multiplicativity, any two convex bodies are at most $n^2$ apart in this metric. Try this exercise for yourself.

John's theorem is that any convex body is in fact at most $\sqrt n$ away from the unit sphere. This is more difficult than the exercise above.

Gluskin's theorem states that there is a universal constant $C>0$ such that for each $n$, one can find two convex bodies that are at least $c \sqrt n$ apart. This took some effort as well.

Apart from that, I can't really say anything about $d$ being very special. In the same lecture notes, you will find however, that if as a polytope gets closer to a ball in distance, its number of faces increases exponentially fast. Basically, approximating a ball by a polytope in this distance, you'd require exponentially many faces in the required distance.

Finally, imagine that some property was "continuous" with respect to $d$ : that is, if a space has this property and another space is very close to it, then that space also has the property. This is nice, right?

Turns out one such property is called $B_n$-convexity. I don't want to go into the definition, but a space is $B_n$-convex, if there is an $\epsilon>0$ such that given any $n$ elements of unit norm in the space, you can add/subtract them from one another , using each one exactly once, so that you can travel at least $\epsilon n$ distance away from the unit sphere.

One can show that the Euclidean ball has this property . The unit square does not : if I take the usual basis coordinate vectors, and add/subtract them in any order, I will always land up on a corner of the square, never outside it.

Now, it turns out that $B_n$ convexity is nice. More precisely, for any $B_n$-convex body $X$ you can find a $\lambda_n(X) > 1$ such that any space that has distance less than $\lambda(X)$ from $X$ , is also $B_n$ convex.

(vii)

It is difficult to work with two convex bodies in general, because it is difficult to justify equality, but one can find coverings that at least provide upper bounds.

For example, think about the $p$ and $q$ balls, i.e. $\sum |x_i|^p \leq 1$ and $\sum |x_i|^q \leq 1$ for $1 \leq p,q \leq 2$. Can you use the Holder inequality to find a bound on the distance between these balls?

Apart from this, there's not really much that can be done : even simple upper and lower bounds are difficult, forget about actual values.

One can do asymptotic analysis : for large $n$, what is the maximal distance of a convex body from another fixed one? For the unit square, polynomial upper and lower bounds are available.

Both are difficult to show in general. Techniques include randomizing(!) the selection of convex bodies, which is what leads to Glushkin's proof. Then, there's a "dual"-realization of every convex body, which kind of "flips" problems around : if you can't deal with the body at hand, dualize.

Then there's a result of Szarek-Talagrand, which is basically a nice approximation-type result for the minimal ellipsoid, along with the Sauer-Shelah lemma, which kind of provides lower and upper bounds of intersections of a set family with another fixed set. These results are all beyond my technical ability , though, and I assume that for your further reading of the document, this is all you will need to know about $d$ : neither is it explicitly calculated for a pair in the document (other than between square and ball) nor is it much expounded on.

EDIT : Suppose that $d(K,L) = \alpha$. Then there is a linear image of $L$, call it $ML$ for a matrix $M$ (which is invertible as I mentioned earlier) such that $ML \subset K \subset \alpha ML$.

Then, consider the linear image of $K$ given by $M^{-1}K$. Note that we can apply $M^{-1}$ to all sides of the containment (this is because if $f$ is any function and $S \subset T$ then $f(S) \subset f(T)$ can be easily proven) to get $L \subset M^{-1}K \subset \alpha L$ (the $M^{-1}$ commutes with $\alpha$).

Note that $L \subset M^{-1}K$ and $M^{-1}K \subset \alpha L$, which upon scaling by $\frac 1 \alpha$ (this is also a function, so the same logic applies) gives $\frac{1}{\alpha} M^{-1}K \subset L$. Therefore, we have :
$$
\frac{1}{\alpha} M^{-1}K \subset L \subset \alpha \left(\frac 1 \alpha M^{-1}K\right)
$$

Now, the linear image $\hat{K} = \frac 1{\alpha}(M^{-1}K)$ satisfies $\hat{K} \subset L \subset \alpha \hat{K}$. Thus, we get $d(L,K) \leq \alpha$ (CAREFUL : This does not imply equality because we don't know if there's a number that is smaller than $\alpha$ that does this).

But then, we've proved that $d(K,L) \geq d(L,K)$. Just switching the roles of $K$ and $L$ (literally switching every  $K$ with $L$ and $L$ with $K$ in the argument above) gives us $d(L,K) \geq d(K,L)$ and therefore combining the two, $d(K,L ) =d(L,K)$ is clear.


问:
Sum of the roots of an equation $\cos x=1-\frac{x[x]}{2}$

Find the sum of the roots of the following equation
$$\cos x=1-\frac{x[x]}{2}$$
I know the sum of the roots of the given equation is zero, I could prove it by using numerical method and confirmed it with wolfram alpha. I do not know how to do it algebraically.

答:
as a hint
$$\cos x=1-x[x]/2\\
x[x]/2 =1-\cos x\\x[x]/2 =2\sin ^2(\frac x2)\\x[x]=4\sin ^2(\frac x2)\leq4 \to\\x[x]\leq2$$ it means that suffice you try on below intervals
$$-2\leq x <-1 \to no-solution\\
-1\leq x <0 \to no-solution\\0\leq x <1 \to x=0 \checkmark\\
1\leq x <2 \to there-is-one-solution\\$$

ContentLicense: CC BY-SA 4.0


问:
Depency of random variable, covariance and correlation matrix
I'm learning statistics and trying to solve the following problem:

Continuous random variable $X$ ~ $R(0,2)$. Describe the dependency of
$Y$ on $X$ for $Y=X^2$. Write  a covariance matrix and correlation
matrix of a random vector $(X,Y)$.

I calculated the expected value of $X$, which is $1$. I also calculated the dispersion, which is $1/3$. Not sure where to go from now. Could you please help me?

答:
You are on the right track. Just apply the definitions.

Correlation matrix:
$ R=E\left\{\begin{bmatrix}
X \\
Y
\end{bmatrix}
[X, Y]
\right\} = \begin{bmatrix}
E\{X^2\} & E\{XY\} \\
E\{XY\} & E\{Y^2\}
\end{bmatrix}$

Covariance matrix: $C = E\left\{\begin{bmatrix}
\tilde{X} \\
\tilde{Y}
\end{bmatrix}
[\tilde{X}, \tilde{Y}]
\right\} = \begin{bmatrix}
E\{\tilde{X}^2\} & E\{\tilde{X}\tilde{Y}\} \\
E\{\tilde{X}\tilde{Y}\} & E\{\tilde{Y}^2\}
\end{bmatrix}$

where $\tilde{X} = X-E\{X\}$.

Now, you have calculated $E\{X\}$ and and $E\{\tilde{X}^2\}$ correctly. You already have $E\{X^2\}$, since it equals $E\{\tilde{X}^2\}+(E\{X\})^2$. As for $E\{XY\}$, it equals $E\{X^3\}$. Also $E\{Y^2\}=E\{X^4\}$. And so on. For example,

$E\{Y\} = E\{X^2\} = \dfrac{1}{3}+1^2 = \dfrac{4}{3}$

$E\{XY\} = E\{X^3\}=\int\limits_{0}^{2} \dfrac{1}{2} x^3 dx = \dfrac{1}{2} \left[ \dfrac{1}{4} x^4\right]_{x=0}^{2} = 2 $

and so on.


问:
Reducing $\cos(2\pi/3)$, $\tan(7\pi/4)$, and $\sin(7\pi/6)$
It's been a few years since doing any type of trigonometry questions and I've seemed to forgotten everything about it. Below is a question with the solution. You're not supposed to use a calculator.

$$\begin{align}
&\cos\frac{2\pi}{3}+\tan\frac{7\pi}{4}-\sin\frac{7\pi}{6} \\[4pt]
&=-\cos\frac{\pi}{3}-\tan\frac{\pi}{4}-\left(-\sin\frac{\pi}{6}\right) \\[4pt]
&=-\frac12-1+\frac12 \\[4pt]
&=-1
\end{align}$$

Can somebody explain the following to me?

How $\cos(2\pi/3)$ becomes $-\cos(\pi/3)$

How $\tan(7\pi/4)$ becomes $-\tan(\pi/4)$

How $-\sin(7\pi/6)$ becomes $-(-\sin(\pi/6))$

Thanks

答:
Notice that $\cos{(2 \pi / 3)}$ is negative since it is located in the 2nd quadrant. Since $\cos{( \pi - 2 \pi /3 )}$ has the same value as $\cos{(2 \pi / 3)}$ in absolute values (i.e. $\cos{( \pi - 2 \pi /3 )}$  is positive) you can rewrite it as:
$$ \cos{(2 \pi / 3)} = - \cos{( \pi - 2 \pi /3 )} =- \cos{( \pi /3)} $$
A similar reasoning applies to the other examples.


问:
Is it possible to generate all sequences of ‘0’ and ‘1’ using only these 4 rules?
I want to show you, a problem I was thinking about and it’s interesting for me, but I have no idea how to solve it. The question is:

There are 4 rules of replacement (the order of numbers matters):

‘00’ --> ‘000’

‘01’ --> ‘001’ or ‘010’ or ‘100’

‘10’ --> ‘011’ or ‘110’ or ‘101’

‘11’ --> ‘111’

For example:

‘1001’ = ‘10|01’ --> ‘011|010’ = ‘011010’

‘1001’ --> ‘011010’

Can you generate every finite sequence of ‘1’ and ‘0’ starting from ‘010’ or just ‘10’?

If yes show how or show the proof. If no say why and what conditions must be true (including e. g. how starting sequence should look like) to make it possible.

I think the division of sequence we generate can matters because sequence ‘1010’ can be:

‘10|10’ --> ‘011|110’

‘1010’ --> ‘011110’

or using different division:

‘1|01|0’ --> ‘1|010|0’

‘1010’ --> ‘10100’

so another thing about problem is show how this fact is related to the solution.

This question was invented by me, so I don’t even know if it's the best wording of a problem but you can write what you think about it :)

PS

Preventively sorry for my English. I’m not from England or United States. And I'm not sure it's good question to Mathematics Stack Exchange. I'm new here so you can write me whether this question match this site or not.

答:
Would you not start with the sequence you want to end with (of length $\ge 3$) and go backwards? To go backwards, you can always chop off the first three digits and map them back to two digits - until you reach a sequence with only two digits. For example:

$$100|100101\to 011|00101\to 100|0101\to 010|101\to 011|01\to 100|1\to 011\to 10$$

which corresponds to the expansion:

$$10\to 01|1\to 10|01\to 01|101\to 01|0101\to 10|00101\to 01|100101\to 100100101$$

Every use of an expansion rule extends the sequence by $1$, so starting from any sequence of length $2$ you can reach a sequence of length $n$ in exactly $n-2$ expansions.

Now if you count multiple expansions done on the same string as one step, there will, of course, be fewer steps. My guess is that the number of steps will behave $\sim\log_{3/2}n$ in the best case because, when going "backwards", for big $n$, you can replace approximately $n/3$ three-element subsequences by two-element subsequences, reducing the size by approximately the factor $2/3$. This would, however, warrant a rigorous proof (which I don't have at the moment but could give it a try - or you could also try yourselves!).


答:
Using the following identities:

$$\cos(\pi - \alpha) = -\cos(\alpha)$$
$$\tan(\alpha + 2\pi) = \tan(\alpha)$$
$$\tan(-\alpha) = -\tan(\alpha)$$
$$\sin(\alpha + \pi) = -\sin(\alpha)$$

We get:

$$\cos\dfrac{2\pi}{3} = \cos\left(\pi - \dfrac{\pi}{3}\right) = -\cos\dfrac{\pi}{3}$$

$$\tan\dfrac{7\pi}{4} = \tan\left(-\dfrac{\pi}{4} + 2\pi\right) = \tan\left(-\dfrac{\pi}{4}\right) = -\tan\dfrac{\pi}{4}$$

$$-\sin\dfrac{7\pi}{6} = -\sin\left(\dfrac{\pi}{6} + \pi\right) = -\left(-\sin\dfrac{\pi}{6}\right)$$

Title: Reducing $\cos(2\pi/3)$, $\tan(7\pi/4)$, and $\sin(7\pi/6)$

Body:
It's been a few years since doing any type of trigonometry questions and I've seemed to forgotten everything about it. Below is a question with the solution. You're not supposed to use a calculator.

$$\begin{align}
&\cos\frac{2\pi}{3}+\tan\frac{7\pi}{4}-\sin\frac{7\pi}{6} \\[4pt]
&=-\cos\frac{\pi}{3}-\tan\frac{\pi}{4}-\left(-\sin\frac{\pi}{6}\right) \\[4pt]
&=-\frac12-1+\frac12 \\[4pt]
&=-1
\end{align}$$

Can somebody explain the following to me?

How $\cos(2\pi/3)$ becomes $-\cos(\pi/3)$

How $\tan(7\pi/4)$ becomes $-\tan(\pi/4)$

How $-\sin(7\pi/6)$ becomes $-(-\sin(\pi/6))$

Thanks


问:
Range of Real Inverse Hyperbolic Cosine -- can it be negative?
There are several results in Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968) concerning $\cosh^{-1}$ which are presented in the following format:

$$\cosh^{-1} x = \pm \left\{{\ln (2x) - \left({\dfrac 1 {2 \cdot 2x^2} + \dfrac {1 \cdot 3} {2 \cdot 4 \cdot 4x^4} + \dfrac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 \cdot 6x^6} }\right) }\right\}$$

$$[+ \text { if }\cosh^{-1} x > 0, x \ge 1, - \text { if }\cosh^{-1} x < 0, x \ge 1]$$

(item $20.40$)

and:

$$\int \cosh^{-1} \frac x a \, \mathrm d x = \begin{cases} x \cosh^{-1} (x/a) - \sqrt {x^2 - a^2} & : \cosh^{-1} (x/a) > 0 \\ x \cosh^{-1} (x/a) + \sqrt {x^2 - a^2} & : \cosh^{-1} (x/a) < 0 \end{cases}$$

(item $14.651$)

There are a number of such.

What confuses me is that I have been led to believe that the (real) inverse hyperbolic cosine is specifically defined as being the positive value of $y$ such that $\cosh y = x$.

So what is the motivation and conventional interpretation of "$f (\cosh^{-1} x)$ such that $\cosh^{-1} x < 0$"?

答:
The 1978 edition of the Mathematical Handbook says that "$\text{Arg ch}(x)=\ln(x+\sqrt{x^2-1}),x\ge0$, ($\text{Arg ch}(x)>0$ is the principal branch)" (translated from French). This fully justifies the formulas that you mention, which pertain to the two possible branches (the two solutions in $y$ of $x=\cosh(y)$).

Technically speaking, the hyperbolic cosine is not an invertible function, and the notation $\cosh^{-1}(x)$ is no so valid. The function $\text{arcosh}(x)$ is well-known and "by default" taken to be the principal branch.


问:
Inner product and orthogonal projection in the general case
If $V$ be a vector space over the field $\mathbb K$, the inner product $\langle\cdot, \cdot\rangle: V \times V \rightarrow \mathbb{K}$ will be is hermitically symmetrical: $\langle x, y\rangle=\overline{\langle y, x\rangle}$. In case the field $\mathbb K = \mathbb R$, then the orthogonal projection of the vector $\mathbf{a}$ onto $\mathbf{b}$ is given by

$$\text{oproj}_{\mathbf{b}} \mathbf{a} = \frac{\langle \mathbf{a} , {\mathbf{b}} \rangle} {\|\mathbf{b}\|}=\frac{\langle \mathbf{b} , {\mathbf{a}} \rangle} {\|\mathbf{b}\|}$$

What about the general case, in which $\langle x, y\rangle \neq {\langle y, x\rangle}$ but $\langle x, y\rangle=\overline{\langle y, x\rangle}$? Would the orthogonal projection of $x$ onto $y$ be given by
$\text{oproj}_{y} {x} = \frac{\langle x, y \rangle} {\|y\|}$ or $\frac{\langle y , x \rangle} {\|y\|}$?

答:
So we want to project $x$ onto $y.$ The projected vector $v = \lambda y$ should be in the direction of the vector $y$ up to some scale factor $y.$ Let's try to compute this factor $\lambda\in \mathbb{K}.$ Since we want an orthogonal projection,
$$\langle x-v, v\rangle = 0 \implies \lambda \langle x, y\rangle = |\lambda|^2\langle  y, y\rangle,$$
and so the valid choice should be the first one, $\frac{\langle x,y \rangle}{||y||}.$ Try the computation with $\langle x - v , v\rangle$ where $v = \frac{\langle y,x \rangle}{||y||}y$ and you will see that this quantity is not equal to $0.$


问:
Equation of a plane determined by a point and $2$ vectors
I was taught that if $\pi$ is a plane, then it can be determined by a point $A_1(x_1,y_1,z_1)\in \pi$ and $\vec{u_k}=(l_k,m_k,n_k)$, $k=\overline{1,2}$ (non-null, non-collinear), using this formula $$\pi: \begin{vmatrix}
x-x_1 & l_1 & l_2 \\
y-y_1 & m_1 & m_2 \\
z-z_1 & n_1 & n_2 \end{vmatrix}=0$$
I thought that the vectors $\vec{u_k}$ need to be in our plane $\pi$, but in an exercise they were just parallel to $\pi$ and the instructor used the same formula. Why does this work?

答: This works because, $(x,y,z)\in\pi$ if and only if $(x-x_1,y-y_2,z-z_3)$ is a linear combination of $\vec{u_1}$ and $\vec{u_2}$, that is, if and only if that determinant that you mentioned is $0$.


问:
Mathematical modelling of a function
I wonder if how you represent this scenario.
It has been determined that 2400 clocks would be sold if the wholesale price is 3.20 dollars. However, if the price were reduced to 2.4 dollars, there would be 4800 clocks sold. Find a mathematical model expressing the wholesale price as a function of the number of clocks sold. First, I assumed that the relationship is linear. But I couldn't seem to find a way to represent them as a function.

答:
I considered @saulspatz's comment and I have come to the realization that I only have to find an equation of the line containing two points. It should be noted that the problem was asking for the mathematical model where the price is a function of the number of clocks sold. Thus, we have the ordered pairs (2400, 3.2) and (4800, 2.4). Using desmos app,

we have $y=-\frac{x}{3000} + 4$.

Title: Mathematical modelling of a function


问:
Regular $T_2$ space which is not completely regular.
Theorem 10. of
Pontryagin's Topological Groups says that:

Every Hausdorff topological group is completely regular.

But is there exists a Regular $T_2$ space which is not completely regular?

答:
These are counterexamples, which may be hepful for you:

Deleted Tychonoff Corkscrew

Hewitt’s Condensed Corkscrew

Thomas’s Corkscrew

Tychonoff Corkscrew

ContentLicense: CC BY-SA 3.0


问:
Show that a subset $X \subseteq G$ generates $G$ if and only if $X+pG$ generates $G$.
Let $p$ be a prime, and let $G$ be a finite abelian $p$-group. Define a subgroup $H<G$ to be maximal if there is no other subgroup $K<G$ such that $H<K<G$. Define the set $pG = \left\{ pg: g\in G \right\}$. Show that a subset $X \subseteq G$ generates $G$ if and only if $X+pG$ generates $G$.

I think I am close to figuring this question out, but cannot quite put the last few pieces together. So far I have worked out in a previous question that $$H \text{ is maximal if and only if } G/H \cong \mathbb{Z}_p \tag{1}$$ I worked this out using the correspondence theorem, and the fundamental theorem of finitely generated abelian groups. I then worked out in another previous question that
$$
X \subseteq G \text{ generates }G \text{ if and only if } X \text{ is not contained in a maximal subgroup of } G \tag{2}
$$
Now, back to the original statement I am trying to prove. It is quite trivial proving the forwards implication of this statement. I am now trying to prove the backwards implication of the statement by showing that if $X$ does not generate $G$, then $X+pG$ does not generate $G$. To do this, I have shown that if $X$ does not generate $G$, then $X$ is contained in a maximal subgroup $H$ of $G$, by $(2)$. We then have by $(1)$ that $G/H \cong \mathbb{Z}_p$. I now want to somehow show that $pG$ must now also be contained by H, which would prove the statement, again by using $(2)$. Is this the right approach to this question? Can anybody help me prove that $pG\subseteq H$, or is this simply the wrong approach to the question? Thanks for any help!

答:
Since $G/H$ is isomorphic to $\mathbb Z_p$, every element of the quotient group (which recall are just the cosets of $H$) has order $p$. What this means formally is that
$$ p(gH) = (pg) H = H $$
for all $g \in G$. This holds only if $pg \in H$, so $pG \subseteq H$.

As a side remark, to show (1) you don’t really need the full power of the fundamental theorem (using a sledgehammer to crack a nut, so to speak), something weaker like the Sylow theorems will suffice.


问:
Is $\frac{a}{b} < \frac{c}{d}$ the same as $a\times d \;<\; c\times b$ for positive integers?
Is
$$\frac{a}{b} < \frac{c}{d}$$.
same as
$$a\times d \;<\; c\times b$$
if $a$, $b$, $c$, $d$ are positive integers?

Can someone verify this?

答: Yes this is true for positive integers. Multiplying both sides by $b\times d$ proves this. Conversely, dividing the product inequality by $b\times d$ gives back the inequality of ratios.


问:
Show that if $H_1, H_2$ are Hilbert separable and and $A \in L(H_1, H_2)$ then there is a sequence of finite rank operators $A_n \xrightarrow{s} A$
Let $H_1, H_2$ be Hilbert separable and and let $A \in L(H_1, H_2)$ be an operator between them.
I want to show there is a sequence of finite rank bounded operators $\{A_n\}$, such that $A_n \xrightarrow{s} A$, which means that for every $x \in H_1$ it satisfies:
$$||A_nx - Ax|| \xrightarrow{n \to \infty} 0$$

Well, there is $\{e_i\}_{i=1}^{\infty}$ an orthonormal basis for $H_1$ since it's Hilbert separable, so:
$$Ax = \sum_{i=1}^\infty <x,e_i> A(e_i)$$

Then I thought about declaring the projection $A_n$ on $\{e_i\}_{i=1}^{n}$ by $$A_nx = \sum_{i=1}^n <x,e_i> A(e_i)$$

which is a compact finite rank operator.

Then, I will get:
$$ ||A_nx - Ax||^2 = ||\sum_{i=n+1}^\infty <x,e_i> A(e_i)||^2 = <{\sum_{i=n+1}^\infty <x,e_i> A(e_i), \sum_{j=n+1}^\infty <x,e_j> A(e_j)}> = $$ $$=\sum_{i=n+1}^\infty\sum_{j=n+1}^\infty <x,e_i> \bar{<x,e_j>}$$

Here I get stuck with $$ and I am not sure how to continue.

Help would be appreciated.

Body:
Fix $x$. $A(x) = \sum_{i=1}^{\infty} c_i e_i$, $A_n(x) = \sum_{i=1}^{n} c_i e_i$.

As $A(x) \in H_2$ we have $||A(x)||^2 = ||\sum_{i=1}^{\infty}  c_i e_i||^2 = \sum_{i=1}^{\infty}c_i^2 < \infty$ and thus $\sum_{i=n+1}^{\infty}c_i^2 \to 0, n\to \infty$.

Hence $$||A(x)  - A_n(x) ||^2 = ||  \sum_{i=n+1}^{\infty} c_i e_i||^2 = \sum_{i=n+1}^{\infty}c_i^2 \to 0.$$


问:
Correct notation for product of each combination
I am trying to create a square matrix $M$ that when you multiply it with another square matrix $A$ the result gives a diagonal with $0$'s. I have figured that a this can be done in a $3$ by $3$ matrix as follows:
$$
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}\begin{pmatrix}
-2bc & -2ef & -2hi \\
ac & df & gi \\
ab & de & gh
\end{pmatrix}=\begin{pmatrix}
0 & x_1 & x_2 \\
x_3 & 0 & x_ 4 \\
x_5 & x_ 6 & 0
\end{pmatrix}
$$
Then I tried applying what I did there for an $n$ by $n$ matrix however I am unsure of the notation I need to describe the pattern. Right now I know that the j column of $M$ has every combination of pairs from the $j$ row of $A$ and in row 1 all the values are multiplied by $(1-n)$ however I don't know how that can be written.

答:
Here's one way to write it. Let $A$ be the matrix you're trying to transform and $M$ be your chosen square matrix. Then
$$M_{ij} = (1 - n\delta_{i, 1})\displaystyle\prod_{k\neq i}^{n}A_{jk}$$
where $\delta_{j, k}$ is the Kronecker delta. The $i$th row and $j$th column of $M$ is calculated by computing the product of all terms in the $j$th row of $A$  except the term in the $i$th column. For the first row of $M$ we multiply by $(1-n)$.


问:
Integration of $\cos x\cdot\cos 2x\cdot\cos 3x$
I study maths as a hobby and am trying to integrate
$\,\cos x\cdot \cos 2x\cdot \cos 3x\,$.

I am trying to split this up into a form in which I can integrate each component part. I have tried using the rule
$$2\cos A \cos B = \cos(A + B) + \cos (A - B)$$
but I can only get this far:
$$\begin{align}
\int\cos (x) \cos (2x)\cos (3x)\, dx
 & = \int\frac{1}{2}(2\cos (x)\cos (3x)\cos (2x), dx \\
 & = \int\frac{1}{2}( \cos (4x) + \cos (2x))\cos (2x), dx
\end{align}$$
but from here on I get stuck.

答:
You can also use the complex definition of $\cos x =\frac{e^{ix}+e^{-ix}}{2}$:
\begin{align*}
\frac 18\int &(e^{ix}+e^{-ix} )(e^{2ix} +e^{-2ix})(e^{3ix} +e^{-3ix} ) \mathrm dx  \\
&= \frac 18 \int (e^{3ix}+e^{-3ix}+e^{ix} +e^{-ix} )(e^{3ix} +e^{-3ix} ) \mathrm dx \\ &= \frac 18\int (e^{6ix} +1+1+e^{-6ix}+e^{4ix}+e^{-2ix}+e^{2ix}+e^{-4ix} ) \mathrm dx \\
&= \frac 14 \int \left(\frac{e^{6ix} +e^{-6ix}}{2} + \frac{e^{4ix} +e^{-4ix}}{2} +\frac{e^{2ix}+e^{-2ix}}{2} +1\right)\mathrm dx \\
&= \frac 14 \int (\cos 6x +\cos 4x +\cos 2x +1)\mathrm dx \\
&=\frac 14 \left( \frac{\sin 6x}{6} +\frac{\sin4x}{4} +\frac{\sin 2x}{2}+x \right) + C
\end{align*}


问:
Property of $\mathbb{N}$: If $n,m\in \mathbb{N}$ with $n>m$ then $n-m\in \mathbb{N}$
I am reading the book of V.A.Zorich "Mathematical analysis" and trying to follow the formal construction of natural numbers and its properties:

Definition 1: The set $X\subset \mathbb{R}$ is called inductive, if for any $x\in R$ the element $x+1
\in \mathbb{R}$.

Definition 2: The set of natural numbers is defined as the minimal inductive set containing $1$, i.e. $$\mathbb{N}:=\bigcap_{A\in \mathcal A}A,$$ where $\mathcal A$ is the family of all inductive sets containing $1$ and we see that $\mathcal A\neq \varnothing$ because $\mathbb{R}\in \mathcal A$.

Principle of Mathematical Induction: If $E\subset \mathbb{N}$ with $1\in E$ and $\forall x\in E (x+1\in E)$ then $E=\mathbb{N}$.

Properties of $\mathbb{N}$:

If $m,n \in \mathbb{N}$ then $m+n\in \mathbb{N}$ and $mn\in \mathbb{N}$.

$(n\in \mathbb{N}) \land (n\neq 1) \Rightarrow ((n-1)\in \mathbb{N}).$

For any $N\in \mathbb{N}$ the set $\{x\in \mathbb{N}\mid n<x\}$ has the minimal elements and $$\min \{x\in \mathbb{N}\mid n<x\}=n+1.$$

$(m\in \mathbb{N})\land (n\in \mathbb{N}) \land (n<m) \Rightarrow (n+1\leq m).$

If $n\in \mathbb{N}$ then there is no $x\in \mathbb{N}$ such that $n<x<n+1$.

If $n\in \mathbb{N}$ with $n\neq 1$ then there is no $x\in \mathbb{N}$ such that $n-1<x<n$.

If $M\subset \mathbb{N}$ and $M\neq \varnothing$ then $\exists \min M$.

I was wondering how to prove the following fact:

If $n,m\in \mathbb{N}$ with $n>m$ then $n-m\in \mathbb{N}$

答:
If $n=1$, then this is trivally true, since there is no $m\in\Bbb N$ such that $m<n$.

Now, take $n\in\Bbb N$ and suppose that, for each $m\in\Bbb N$, if $m<n$, then $n-m\in\Bbb N$. Then you want to prove that, if $m\in\Bbb N$ and $m<n+1$, then $n+1-m\in\Bbb N$. If $m=1$, it is clearly true, since $n+1-1=n\in\Bbb N$. Otherwise, $n+1-m=n-(m-1)$ and, by the induction hypothesis, and since $m-1<n$, $n-(m-1)\in\Bbb N$.


问:
Do chaotic systems exist that cannot be predicted even at the limit of inifinite precision initial conditions and infinite resources?
I have a layman person's understanding of the theory of chaos, that seems to indicate that using finite-precision initial conditions and finite computing resources, chaotic systems cannot be predicted after a period of time.

My question is what happens in the limit of increasing the precision of initial conditions and resources to infinity: Does the system remain chaotic, or does the prediction window also diverge to infinity?

Specifically consider the following conditions:

We have a chaotic system.

We calculate the prediction time window  $t_\text{pred}(e,p,m,s)$ given a finite error margin $e$, for a finite precision of initial conditions $p$, and a computer with finite memory $m$ operating at a finite speed $s$.

We calculate the same prediction time window $t_\text{pred}(e,p,m,s)$ when precision, memory, and speed diverge to infinity together (but $e$ remains finite).

If for all chaotic systems the time window diverges to infinity, then the answer to this question is no.

If any system is found where $t_\text{pred}$ may remain finite, then the answer to this question is yes.

Since this questions seems very far from being practical I will add a motivation: I feel the answer of this question has an important impact in theology. Namely if the answer is yes then that would logically preclude the possibility of a non-interventionist, all-knowing god (future included) who designed the universe with a purpose, because he/she wouldn’t be able to do these calculations even if he/she was infinitely powerful.

答:
A crucial property of chaotic systems is that they are deterministic: There is no element of randomness in the model. The initial conditions exactly determine the future of the system.

If I simulate a chaotic model with the same initial conditions¹ on a real computer twice, I obtain exactly the same result. This only differs from the true solution for my initial conditions due to the finite precision of floating-point arithmetics (and, as the system is chaotic, this difference can be large)². And of course, in the purely hypothetical case that I want to simulate an isolated real system for which I have an exact model, I have the problem that I cannot perfectly represent my real initial conditions as floating-point numbers.

If I have arbitrary precision and infinite computing resources available as well as perfect knowledge of initial conditions, I can predict a chaotic system perfectly by simply simulating it. For a discrete-time system, the only reasons I need infinite memory and computation speed are storing and working with arbitrary precision numbers³ (and of course if I want to go infinitely into the future). For a continuous-time system, there is another reason I need infinite computing speed, namely to perform the numerical integration with arbitrarily fine time steps.

¹ and the same rules of floating-point arithmetics

² for a continuous-time system, the inherent imprecision of numerical integration also adds an error

³ since I end up with infinitely many digits


问:
Is the tensor product of two projective sheaves projective?
The tensor product of two abelian sheaves $\mathcal{F}$ and $\mathcal{G}$ is defined to be the sheafification of $U \mapsto \mathcal{F}(U) \otimes \mathcal{G}(U)$. In the category of abelian groups, the tensor product of projectives is projective, so it seems logical to hope a similar thing with sheaves.

I was trying to look at the presheaf $U \mapsto \mathcal{F} (U) \otimes \mathcal{G}(U)$. It is projective at every section and I was trying to use this to assemble a natural transformation to complete the necessary diagram that would make $\mathcal{F} \otimes \mathcal{G}$ projective, but I don't see a way of doing this that will make the necessary diagrams commutative to assemble a natural transformation.

Another attempt was using the adjunction with the internal sheaf hom because we have that

$$Hom(\mathcal{F}, \mathscr{H}\kern -.5pt om (\mathcal{G}, -)) \cong Hom (\mathcal{F} \otimes \mathcal{G}, -)$$

and $\mathcal{F} \otimes \mathcal{G}$ is projective if and only if $Hom(\mathcal{F} \otimes \mathcal{G}, -)$ is exact. The usual argument with groups is composing exact functors, since the internal hom coincides with the usual hom. However, we don't know here if $\mathscr{H}\kern -.5pt om(\mathcal{G},-)$ is exact when $\mathcal{G}$ is projective (and my attempts at proving this have been unsuccessful as well).

I would also be interested to know if this statement is true in a general closed monoidal category, but the specific application I was looking for is in the category of sheaves.

EDIT: this question is more specifically referring to condensed abelian groups (sheafs of abelian groups on the site of profinite sets), but I thought it might be true in a more general sheaf setting.

答:
This is nit an answer, but it's too long for a comment.

1- Note that many spaces don't actually have nonzero projective sheaves, and in fact, the latter are quite rare; so this question may not be so relevant.

2- Also, another point is that it seems unlikely to be true : the natural "unbiased" version of this statement ("a finite tensor product of projective sheaves is projective") fails - the empty tensor product is the unit, which is the constant $\mathbb Z$ sheaf, and it's certainly not projective in general (it would mean that epimorphisms of sheaves are epimorphisms of presheaves, which is rarely the case). Of course this doesn't prove that the statement is wrong because the empty tensor product is somewhat special; but still it makes it seem unlikely.

3- The internal hom argument will also likely not work in general : indeed, try to see what it means for $\mathcal{Hom}(G,-)$ to preserve epimorphisms. Start from an epimorphism $F\to H$, then this says that for any $U$ and any map $G_{\mid U} \to H_{\mid U}$, there is locally a lift. Now, of course, $F\to H$ is an epimorphism no matter where you restrict it, but to get a lift, you "morally" will need the resrictions of $G$ to still be projective, which is related to $i_*$ (where $i : U\to X$ is the inclusion) preserving epimorphisms which, in general, it doesn't.

As you can see, neither of 1,2 or 3 shows that the statement is wrong, so again, this does not answer your question, but it still makes it seem like the answer will be no in general. However, it will be true over stupid examples such as discrete spaces; and in fact, the rareness of projective sheaves will likely make it hard to actually find a counterexample if the statement is false.

It could even be (I wouldn't count on it) that spaces that are weird enough to have projective sheaves are also weird enough for this type of thing to actually hold


问:
Definition of a separating triangle in planar graph
In Bill Tutte's famous book Graph Theory as I Have Known It he discusses Hassler Whitney's theorem on Hamiltonian cycles in planar graphs, summarising it thus:

Any strict triangulation in which there is no separating triangle has a Hamiltonian circuit.

That is fairly clear, and to me a "separating triangle" is one which, if removed, would separate the graph.

But Tutte defines it the following way:

By a "separating triangle" of a planar map is meant a circuit of three edges that is not the boundary of a face; it has at least two faces of the map inside and at least two outside.

I don't understand this...

Surely a circuit of three edges circumscribes a triangle, and therefore by definition must be the boundary of a face.

How can a circuit of three edges have more than one face inside it? For >3 edges, that's clear, but not for 3.

And how does this way of defining a separating triangle square up with more usual definitions?

答:
Let us be very careful here and make a clear distinction between an abstract graph and a planar embedding. We recapitulate the following definitions:

Def. 1: Let $\mathcal{O} \subseteq \mathbb{R}^2$ be any open set. Being linked by an arc in $\mathcal{O}$ defines an equivalence relation on $\mathcal{O}$. The corresponding equivalence classes are again open. They are the regions of $\mathcal{O}$.

Def. 2:
A closed set $X \subseteq \mathbb{R}^2$ is said to separate a region $\mathcal{O}'$ of $\mathcal{O}$ if $\mathcal{O}' \setminus X$ has more than one region.

Def. 3:
The frontier of a set $X \subseteq \mathbb{R}^2$ is the set $Y$ of all points $y \in \mathbb{R}^2$ such that every neighborhood of $y$ meets both $X$ and $\mathbb{R}^2 \setminus X$. Note that if $X$ is open then its frontier lies in $\mathbb{R}^2 \setminus X$.

Def. 4:
A plane graph (or planar map) is a pair $(V,E)$ of finite sets with the following properties (the elements of $V$ are again called vertices, those of $E$ edges):

$V \subseteq \mathbb{R}^2$.

Every edge is an arc between two vertices.

Different edges have different sets of endpoints.

The interior of an edge contains no vertex and no point of any other edge.

The abstract graph of a plane graph is called planar graph.

Def. 5:
When $G$ is a plane graph, we call the regions of $\mathbb{R}^2 \setminus G$ the faces of $G$. These are open subsets of $\mathbb{R}^2$ and hence have their frontiers in $G$. Since $G$ is bounded, exactly one of its faces is unbounded. This face is the outer face of $G$. The other faces are its inner faces.

Def. 6: An embedding in the plane, or planar embedding, of an abstract graph $G$ is an isomorphism between $G$ and a plane graph $G'$.

Example:  Given a plane graph $G'=(V,E)$ with $V=\{v_0,v_1,v_2\}$ and $E=\{(v_0,v_1),(v_0,v_2),(v_1,v_2)\}$. $G'$ is isomorphic to $K_3$, a complete graph on three vertices. Define $X=\mathbb{R}^2 \setminus G'$, $X$ is open in $\mathbb{R}^2$. $X$ is the union of two disjoint subsets $I=\mathbb{R}^2 \setminus (G' \cup O)$ and $O=\mathbb{R}^2 \setminus (G' \cup I)$. They are the regions of $X$ and are by definition again open in $\mathbb{R}^2$.
Consider the following figure:

Now, consider the following:

Def. 7: A triangle $t$ of a planar graph $G$ is called faced if there
exists a plane embedding $G'$ of $G$ such that $t$ is a face of $G'$; otherwise $t$ is called faceless.

Example: Consider the graph $G'$ given by the following figure:

The triangle given by the vertex set $\{u,v,w\}$ is obviously a faceless triangle. It is easy to see topologically as well: The interior of $\{u,v,w\}$ is not arc-connected in $\mathbb{R}^2 \setminus G'$, and hence not a single region. Notice as well that the triangle $\{u,v,w\}$ has multiple faces in its interior.

To summarize, a "circuit" is a graph in an abstract sense. The planar embedding of a circuit $G$ is a plane graph $G'$ isomorphic to $G$. The definition of a face does not exist for $G$ but for $G'$. It follows easily what he means by a "separating triangle".

References:

[1] Reinhard Diestel. Graph Theory (Graduate Texts in mathematics). 5th ed. Vol. 173. Springer, 2017.

[2] C. M. (Kieka) Mynhardt and Christopher M. van Bommel. Triangle Decompositions of Planar Graphs. 2015.


问:
Show that a bounded integral operator on $L_2$ maps $L_2$ functions to continuous functions
Let $K(t,s)$ be a continuous function on $L_2([a,b])$, and let $A \in L(L_2[a,b])$ be the integral operator:
$$(Af)(t) = \int_a^b K(t,s)f(s)ds$$

I want to show that for any function $f \in L_2[a,b])$, the operator on the function $Af$ is always continuous.

I managed to show that $\{ Af \}_{f \in M}$ is pre-compact in $C[a,b]$ when $M$ is a set of continuous and bounded functions.

I also get that in this case $\overline {\{ Af \}_{f \in M}}$ is compact, and since $A$ is an integral operator then it's a compact operator.

I would have wanted to somehow represent $Af$ as a limit of continuous functions, but I am not sure exactly how to begin.

Help would be appreciated.

答:
By Cauchy Schwartz,
$$\|Af\|_{L^2}\leq \|K\|_{L^2}\|f\|_{L^2},$$
and thus, $A$ is continuous.


问:
Prove $\sum\limits_{n=1}^{\infty} \frac{1}{\int_0^1\ln^n(x)dx} = \frac{1}{e}-1$
Ok, that's really not an easy question to convey in text. What I'm asking is this:

$$\sum\limits_{n=1}^{\infty} \frac{1}{\int_0^1\ln^n (x)dx} = \frac{1}{e}-1$$

Numerically, I'm pretty sure it does (it's about -0.63212). But, I have no idea how I'd go about proving it.

Any insights?

Body:
Note

$$\int_{0}^{1}\ln^nx dx = \frac{d^n}{da^n}\left(\int_{0}^{1} x^a dx\right)\bigg|_{a=0}
= \frac{d^n}{da^n}\left(\frac1{a+1}\right) \bigg|_{a=0} =(-1)^{n}n!
$$

Thus
$$\sum\limits_{n=1}^{\infty} \frac{1}{\int\limits_0^1 \ln^nx  dx} =
\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n!}
=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!}-1=
e^{-1}-1$$


问:
If $X$ is binomial, $2X$ isn't binomial
My question is really simple. I Don't understand why if a random variable $X \sim \text{Bin}(n,p)$, then $2X$ isn't binomial. I know every value of $2X$ is even, but I can't prove $2X$ isn't binomial using this fact.

答:
MGF uniquely defines the distribution of a random variable. That is if you can show that the moment generating function of $2X$ is the same as the moment-generating function of Binomial distribution, then $2X$ follows the same distribution.

$X\sim B(n,p) $

$M_X(t)=E(e^{tx})=(q+pe^t)^n$

$Y=2X \implies X=\dfrac{Y}{2}$

Substitute this and then compute MGF.

$M_\frac{Y}{2}(t)=M_{Y}(\frac{t}{2})=(q+pe^{\frac{t}{2}})^n\ne M_X(t)$

Thus not binomial.

Title: If $X$ is binomial, $2X$ isn't binomial


问:
Definition of differentiability and derivative
I am recently learning the Rigorous definition of the derivative so now I come across a problem here is the definition of derivative and definition of differentiability

A real function $f$ is said to be differentiable at a point $a \in \mathbb{R}$ if and only if $f$ is defined on some open interval $I$ containing $a$ and
$$f^{\prime}(a) :=\lim_{h\to 0 } \frac{f(a+h)-f(a)}{h}$$
exists. In this case $f^{\prime}(a)$ is called the derivative of $f$ at $a$

and

Let $I$ be an interval. A function $f:I \rightarrow \mathbb{R} $ is said to be differentiable on $I$ if and only if$$f^{\prime}(a) :=\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ exists and is finite for every $a \in I $

I feel first definition also can be used for differentiability

Let $I$ be an interval. A function $f:I \rightarrow \mathbb{R} $ is said to be differentiable on $I$ if and only if $$f^{\prime}(x) :=\lim_{h\to 0 } \frac{f(x+h)-f(x)}{h}$$ exists and is finite for every $x \in I $

My problem is what is the difference between these two? Why there is two formula for one thing?

答:
Replace $h$ with $x-a$ in the first definition to see that the two limit definitions of the derivative are equivalent.

One of your definitions is for differentiability at a point and the other for differentiability on an interval.


问:
Geometric series modulo prime power
Let $p$ be prime and $n = p^e$ where $e \ge 1$. Prove the following:

If $a \equiv 1 \pmod p$ then $1 + a + a^2 + \ldots + a^{n-1} \equiv 0 \pmod n$

Note that the the first congruence is$\pmod p$ and the second is$\pmod n$.

What I tried: I know $1 + a + a^2 + \ldots + a^{n-1} = \Phi_p(a)\cdots\Phi_{p^e}(a)$ where $\Phi_p$ is the $p$th cyclotomic polynomial. I want to show $a$ is a root$\pmod n$ in $\Phi_{p^i}(a)$ for some $i$. I know how to write $\Phi_{p^i}$ in terms of $\Phi_{p}$. However, I don't think $a$ is necessarily a root in any of these.

答:
Hint: write $$1+a+a^2+\dots+a^{n-1}=\frac{1-a^{(p^e)}}{1-a}=\prod_{k=0}^{e-1}\frac{1-a^{(p^{k+1})}}{1-a^{(p^k)}}$$

Now you only need to prove that all the $\frac{1-a^{(p^{k+1})}}{1-a^{(p^k)}}$ will be divisible by $p$.
(Here I assumed $a\ne1$, but we can verify it directly in the case $a=1$ or we can reduce to the case $a\ne1$ by adding $p^e$ which doesn't change the value mod $p^e$)

Title: Geometric series modulo prime power


问:
Law of large numbers?

Given random variables $Z_1,Z_2,Z_3,\ldots$, which are uniformly distributed for $[8,10]$:

If $X_k =\min\{Z_1, Z_2,Z_3,\ldots,Z_k\}$, prove convergence in probability and find the constant.

Prove and find that the mean of the first $k$ of these variables converge in probably to some constant.

I think this involves the law of large numbers and possible central limit theorem, but I am not really sure.

答:
Hints: The sequence $(X_k)$ is nonincreasing pointwise and lower bounded, hence its pointwise limit $X$ is well defined and finite. A simple argument shows that, if the random variables $(Z_k)$ are independent then $X=8$ almost surely (can you write it down?).

The rest follows: $X_k\to X$ in probability since almost sure convergence implies convergence in probability, and $E[X_k]\to E[X]$ since the random variables $X_k$ are uniformly bounded.

Neither a law of large numbers nor a central limit theorem are useful here. The common distribution of the random variables $Z_k$ is not needed either, simply the fact that $Z_k\geqslant8$ almost surely and that, for every $x\gt8$, $P[Z_k\leqslant x]\ne0$. But the independence property (which is unfortunately omitted in the question) is crucial.

Edit: What remains without the independence hypothesis is that $X_k\to X$ almost surely, hence in probability, and that $E[X_k]\to E[X]$, for some random variable $X$ with values in $[8,10]$ such that $E[X]\leqslant E[Z_1]=9$... and this is it. Note as an extreme example that one could define $Z_k=Z_1$ for every $k$, then the preceding results hold with $X=Z_1$. In particular, the limit may not be a constant, and there is still no law of large numbers nor central limit theorem in the picture.

Title: Law of large numbers?


问:
What is the meaning of "$\pm$"? Can we write $1=\pm1$? or $x^2=4\iff \pm x=\pm 2$?
Does "$\pm$" mean "positive or negative"?

Is it allowed to write: $1 = \pm1\;$?

Is it allowed to write: $X^2 = 4 \iff \pm X = \pm2\;$?

答:
In order to answer this question we look at first at examples where the symbols $\pm$ and $\mp$ are sometimes used.

Quadratic equation:
\begin{align*}
x^2+px+q=0\qquad\qquad\qquad\qquad x_{1,2}=-\frac{p}{2}\color{blue}{\pm}\sqrt{\frac{p^2}{4}-q}\tag{1}
\end{align*}

Geometric series:
\begin{align*}
\frac{1}{1+x}=1-x+x^2-x^3\color{blue}{\pm}\cdots\tag{2}
\end{align*}

Trigonometric addition formula:
\begin{align*}
\cos\left(\alpha\color{blue}{\pm}\beta\right)=\cos\left(\alpha\right)\cos\left(\beta\right)\color{blue}{\mp}\sin\left(\alpha\right)\sin\left(\beta\right)\tag{3}
\end{align*}

Comments:

In the first example (1) the symbol $\pm$ is used to denote the two roots $x_1$ and $x_2$ of the quadratic equation at the left hand side of (1). This is just a compact notation for
\begin{align*}
x_1&=\frac{p}{2}\color{blue}{+}\sqrt{\frac{p^2}{4}-q}\\
x_2&=\frac{p}{2}\color{blue}{-}\sqrt{\frac{p^2}{4}-q}\\
\end{align*}

In the second example (2) the symbol $\pm$ is used to indicate how the terms represented by the three dots are added resp. subtracted, namely starting with $+$ followed by $-$, then followed by $+$, etc. Here we do not have any choice to select either $+$ or $-$. Here the order is relevant and we have to continue the series as described. We could also write
\begin{align*}
\frac{1}{1+x}=1-x+x^2-x^3+x^4\color{blue}{\mp}\cdots
\end{align*}
which indicates the next term $x^5$ is to subtract, the following term $x^6$ is to add, and so on.

The third example (3) gives a compact notation for two different identities, namely
\begin{align*}
\cos\left(\alpha\color{blue}{+}\beta\right)&=\cos\left(\alpha\right)\cos\left(\beta\right)\color{blue}{-}\sin\left(\alpha\right)\sin\left(\beta\right)\\
\cos\left(\alpha\color{blue}{-}\beta\right)&=\cos\left(\alpha\right)\cos\left(\beta\right)\color{blue}{+}\sin\left(\alpha\right)\sin\left(\beta\right)
\end{align*}
Note, that the usage $\pm$ at the left-hand side of (3) and $\mp$ at the right-hand side of (3) is crucial. Here we do not use $\pm$ to provide alternatives, but instead to provide two different identities in compact notation.

Conclusion: When looking at (1) - (3) what can we say about the usage of $\pm$?

The symbols $\pm$ (and similarly $\mp$) have different meanings depending on the context. Sometimes it means two valid alternatives as in (1), sometimes it means one and only one extension as in (2) and sometimes it is just as compact notation for two different things as in (3).

The examples also follow some useful mathemactical writing guidelines:

readability: They provide information which is easy to read, easy to grasp.

conciseness: They provide useful information in a concise manner

With the examples above in mind let's consider the question:

Is it allowed to write: $1 = \pm1$?

Although this might be syntactically allowed, it is clearly a misusage of $\pm$ and should not be used. The right-hand side provides no additional information than already given by the left-hand side. Worse it has the drawback of being less concise than the left-hand side.

Is it allowed to write: $X^2 = 4 \Leftrightarrow \pm X = \pm2$?

To indicate the two solutions of $X^2=4$ it is recommended to write
\begin{align*}
X_{1,2}=\pm 2
\end{align*}
as it is stated in (1). A prefixing of $X$ with $\pm$ here is also some kind of misuse. It does not provide any more information and is rather confusing, since the reader has to think about how plus and minus at both sides of $\pm X=\pm2$ are coupled in order to give a valid statement. This clearly violates readability and conciseness.


问:
Fields with no nontrivial discrete valuation
I am currently learning about DVRs and one can show that given a valuation ring $R$ of $(K,v)$ where $K$ is a field and $v$ a discrete valuation on $K$ that $K\cong F(R)$. I know that not all fields are fields of fractions of some nontrivial ring(trivial in the sense of $R=K$ or $R={0}$), e.g. finite fields. So there have to be fields with no nontrivial discrete valuation, e.g. finite fields. Now i was wondering wheter there is a different proof for showing that there is no nontrivial discrete valuation on a field besides showing that it is not the localization of a nontrivial ring and wheter this is simpler to show and if there are further examples of such fields besides the finite ones?

Thanks for answering in advance!

LG

答:
You could be interested in the following article which I recently wrote with a colleague:
Rodney Coleman, Laurent Zwald : On Valuation Rings. It is available on Internet.
RC</p>


问:
How can we find A using finite difference method?
$$ L(u)=u_{xx}+u_{yy}\quad 0<x,y<1$$
with homogenous boundary conditions.

I have tried finite difference method,
$$ u_{xx}=\frac{u_{i-1,j}-2u{i,j}+u_{i+1,j}}{h^2}$$
$$ u_{yy}=\frac{u_{i,j-1}-2u{i,j}+u_{i,j+1}}{h^2}$$
Substituting in our equation, we have
$$ (u_{i-1,j}-2u{i,j}+u_{i+1,j})+(u_{i,j-1}-2u{i,j}+u_{i,j+1})=0$$
after I arranged it and used BC, I got the matrix A.
the steps are provided in link below in more details.
https://www.physik.uzh.ch/local/teaching/SPI301/LV-2013-Help/lvanlsconcepts.chm/lvac_finite_difference_method_for_laplace_eq.html

Thank you.

答:
It looks like the $-100$ is a misprint in $L$. Try replacing it with a $+$ sign.
And make the title of the question something to do with finite differences.


答:
There are two definitions because you're defining two different things:

$f$ is differentiable at the point $x$.

$f$ is differentiable on the interval $I$.

If I were writing the text, I'd define the first one, and then say that $f$ is differentiable on $I$ if for every point $b \in I$, $f$ is differentiable at $b$. I don't know why your author didn't do this. One peculiarity (which may depend on how the author defined limits) is that it's possible (in the given definition) for the function to be differentiable on a closed interval $I$ without being differentiable at the endpoints. Why? Because maybe there's no open interval around the endpoint that's entirely contained in the domain of $f$. [If having such an interval is an essential part of the author's definition of limit, then it should probably have been omitted from the definition of derivative.]

As an example, consider the function
$$
f:[0, 1] \to [0, 1] : x \mapsto x.
$$
This happens to take on the same values as $$
g: \Bbb R \to \Bbb R : x \mapsto x
$$
at every point $x$ of the unit interval, but $f$ is undefined outside that unit interval nonetheless.

As an example of this subtle distinction:

For this function, the limit in the second definition is well-defined** at $x = 0$, for instance, but the function is not differentiable at $x = 0$ (according to the first definition) because $f$ is not defined on any open interval containing $x = 0$.

**at least, it's well-defined if the definition of "limit" is the one I'm thinking of; it may be that your author uses something different.

NB: Most folks would require "and is finite" as part of the definition of differentiability as well; it's not clear why your author omitted that.

Title: Definition of differentiability and derivative

Body:
I am recently learning the Rigorous definition of the derivative so now I come across a problem here is the definition of derivative and definition of differentiability

A real function $f$ is said to be differentiable at a point $a \in \mathbb{R}$ if and only if $f$ is defined on some open interval $I$ containing $a$ and
$$f^{\prime}(a) :=\lim_{h\to 0 } \frac{f(a+h)-f(a)}{h}$$
exists. In this case $f^{\prime}(a)$ is called the derivative of $f$ at $a$

and

Let $I$ be an interval. A function $f:I \rightarrow \mathbb{R} $ is said to be differentiable on $I$ if and only if$$f^{\prime}(a) :=\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ exists and is finite for every $a \in I $

I feel first definition also can be used for differentiability

Let $I$ be an interval. A function $f:I \rightarrow \mathbb{R} $ is said to be differentiable on $I$ if and only if $$f^{\prime}(x) :=\lim_{h\to 0 } \frac{f(x+h)-f(x)}{h}$$ exists and is finite for every $x \in I $

My problem is what is the difference between these two? Why there is two formula for one thing?


问:
Counter example regarding bounded functions.

I feel that I completely missed the point in this question, as I didn't even use / don't know how to use the fact that $a,b$ are removable discontinuities..
Does that mean I need to construct a function $g$ that will cover the scenario regarding the discontinuity points?

Counter example:

Consider the function $f: R\ ➜ R$ , $f\left(x\right)\ =\ \frac{1}{x}$

Let $(a,b) = (o,1)$. Note that $f$ is continuous on $(0,1)$.

$\lim _{x\to 0^+}\left(\frac{1}{x}\right)=∞$

$\lim _{x\to 1^-}\left(\frac{1}{x}\right)=1$

Since the one sided limits tends to $∞$, $f$ is not bounded on
$(0,1)$.

答:
Definition: A function $f$ is said to have a removable discontinuity at $a \in \mathbb{R}$, if $\lim_\limits{x \to a} f(x)$ exists finitely and $\lim_\limits{x \to a} f(x)\neq f(a)$.

By this definition, see $\lim_\limits{x \to 0}\frac{1}{x}$ is not finite. So it is not a counterexample.

Letting $\lim_\limits{x \to a}f(x)=p$ and $\lim_\limits{x \to b}f(x)=q$, one can construct a function $g:[a,b] \to \mathbb{R}$ such that
$$g(x):= \begin{cases}
          f(x) & x \in (a,b) \\
          p    & x=a \\
          q    & x=b \\
         \end{cases}$$
Now $g$ is a bounded function in $[a,b]$.

(this follows from the Bolzano-Weierstrass theorem which is turn follows from lub-axiom)

$g$ is bounded $\implies$ $\exists \ M \in \mathbb{R}$ such that $g(x) \leq M$ for all $x \in [a,b]$ $\implies f(x) \leq M$ for all $x \in (a,b)$, i.e $f$ is bounded in (a,b).


问:
Can this proof be used, to proof that there is no $l$, unless $a=0$, for $f\left( x\right)=x$, if $x$ is r., and $f\left( x\right)=-x$, if $x$ is ir.
I just wanted to ask, if this proof can be used when trying to prove that function $f\left( x \right)= x$, if $x$ is rational, and $f\left(x \right)=-x$, if $x$ is irrational. Then $\lim\limits_{x \to a}f\left(x \right)=l$ for any $l$ can't exists unless $a=0$. I am trying to prove this by using epsilon-delta definition of limit.

If we assume that such $l$ exists for some $a$. Then for every $\epsilon>0$, there is some $\delta>0$, then for all $x$, if $\left|x-a \right|< \delta$, then $\left|f\left(x \right)-l \right|< \epsilon$. For clarity, if $x$ is rational then $x=\alpha$, and $f\left( \alpha\right)=\alpha$, then if $x$ is irrational, then $x=\beta$ and $f\left( \beta\right)=-\beta$

Then we have $\left|\alpha-l \right|<\epsilon$ and $\left|\beta+l\right|<\epsilon$. Furthemore it can be written as $l-\epsilon<\alpha < l+\epsilon$ and $-\epsilon - l< \beta < \epsilon -l$. Then by adding these up $\left| \alpha + \beta\right|<\left| \alpha-l+\beta+l\right|\leq\left|\beta+l \right|+\left|\alpha-l \right|< 2\epsilon$, so $\left|\alpha+\beta \right|<2 \epsilon$, or ca be written as $\left|\alpha+\beta \right|< \epsilon$

Now then, if $\left| \alpha \right|>\epsilon$, then $\alpha \in \left(-\infty, -\epsilon \right) \cup \left(\epsilon, \infty \right)$, but too $\alpha \in \left( l-\epsilon, l+\epsilon \right)$, which contradicts itself, same for $\beta$. From this, only if both $\left| \alpha \right|<\epsilon$ and  $\left| \beta \right|<\epsilon$ can this limit work.

Now then, we have, for for example $\alpha$, $a- \delta < \alpha < a+ \delta$, so $-\epsilon < a-\delta$, and $a + \delta < \epsilon$, then we must have $\epsilon > -a$, and $a < \epsilon$, but since $\epsilon$ can change and $0< \epsilon$, that for some $a$,  $a>\epsilon$, so unless $a=0$, this inequality cannot work. so $a$ must be $0$. Same works for $\beta$.

Then as $l-\epsilon < \alpha < l+ \epsilon$, $\epsilon \leq \epsilon +l$ and $-\epsilon \geq l-\epsilon$. From this we find that $ 0 \leq l$ and $0\geq l$, so $ l=0$. Same for $\beta$.

So, the only limit for this function is $\lim\limits_{x \to 0} f\left( x\right)=0$, so if $a=0$.

答:
Here is an easy way of solving it using equivalent condition of limits in terms of sequences.

Let us assume limit at $a$ exists and equal to $l$. Consider the sequence $p_n$ of rational numbers and $q_n$ of irrational numbers converging to $a$ where each $p_n$ and $q_n$ are different from $a$. Since limit exists $f(p_n) \to l$ and $f(q_n) \to l$ but $f(p_n) = p_n \to a$ and $f(q_n) = -q_n \to -a$. Hence we must have $l = a = - a$ by uniqueness of limits which implies that $l = a = 0$.


问:
Why is $\mathbb{Z}[i\sqrt5]$ isomorphic to $\mathbb{Z}[X]/(X^2+5)$?

Why is $\mathbb{Z}[i\sqrt5]$ isomorphic to $\mathbb{Z}[X]/(X^2+5)$ ?

$\mathbb{Z}[i\sqrt5]$={$a+ib\sqrt5,a,b\in \mathbb{Z}$}

Also can $\mathbb{Z}[i\sqrt5]$ be seen as a ring of polynomial in $"i\sqrt5"$ or is it just a notation looking like the polynomial ring notation but with another meaning ?

答:
The ring $R=\Bbb Z[x]/\langle x^2+5\rangle$ contains a zero of $x^2+5$, namely the residue class $x+\langle x^2+5\rangle$, which we denote by $x$ itself.

For this residue class, $x^2+5=0$ and so $x = \pm i\sqrt 5$. Thus both zeros lie in $R$. The elements of this ring have the form $a+bx$, with $a,b\in\Bbb Z$. This ring is isomorphic to $\Bbb Z[i\sqrt 5]$ whose elements are $a+bi\sqrt 5$, with $a,b\in\Bbb Z$.


问:
Question on how the sum of Arithmetic Progression is $\frac{n^3}{3}$-$\frac{n^2}{2}$+$\frac{n}{6}$.
Given a set S = $({1,2,...,(n-1))}$, we obtain the sum of all of its entries through
$$S_n= \frac{1}{2}n(n-1)$$ that we derive from $S_n=\frac{1}{2}n(a_1 +a_n)$. The issue is, let us square our set, $S_n^2=({1^2,2^2,...,(n-1)^2)}$, from this we get $$S^2_n =\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$$. That is the question, though, how do we get this value,

I am confused about how to reach that. I have tried to go through $$\frac{1}{2}(n-1)((n-1)^2+1^2)$$, the problem with this is that it gives off a result that does not equal the one mentioned in the title, so there is no point in trying to simplify it. Can someone help me? Thank you.

答:
The formula for the sum of squares is given by
$$1^2+2^2+\ldots+n^2=\frac16 n(n+1)(2n+1).$$
Applying this, we have
$$1^2+\ldots+(n-1)^2=\frac16(n-1)n(2n-1)=\frac13 n^3-\frac12n^2+\frac16n\,.$$

Your mistake is to use the AP formula when you do not have an arithmetic progression.

OwnerDisplayName: anon


问:
Solving a Recursive Relation using Generating Functions
Consider the following recursive relation.
$$a_0=0$$
$$a_i = (1-p)a_{i-1} + pa_{i+1} \text{ }\forall \text{ natural numbers }i \text{ }\epsilon\text{ } [1, m - 1] $$
$$a_m=1$$
$$0<p<1$$
I'm trying to find an explicit formula for $a_i$. I have tried using generating functions to do so.

Define $A(x) = \sum_{k=0}^{m} a_k x^k$. Now,
$$\sum_{k=1}^{m-1}[a_i-(1-p)a_{i-1}-pa_{i+1}]x^k=0$$
$$\implies\sum_{k=1}^{m-1}a_kx^k - (1-p)x\sum_{k=1}^{m-1}a_{k-1}x^{k-1}-px^{-1}\sum_{k=1}^{m-1}a_{i+1}x^{i+1}=0$$
$$\implies[A(x) - a_0-a_mx^m]-(1-p)x[A(x)-a_mx^m-a_{m-1}x^{m-1}]-px^{-1}[A(x)-a_1x-a_0]=0$$
$$\implies [1-(1-p)x-px^{-1}]A(x) - x^m+(1-p)x^{m+1}+(1-p)a_{m-1}x^m+a_1p=0$$
$$\implies A(x) = \frac{(1-p)x^{m+2}-pa_{m-1}x^{m+1}+a_1px}{(1-p)x^2-x+p}$$
I am unsure of how to proceed from here. Kindly help.

答:
Consider a function $A(x) = \sum_{i=0}^\infty a_nx^n$, where the sum is infinite and the $a_n$ satisfy the recurrence $pa_{n+1} - a_n +(1-p)a_{n-1} = 0$ for all $n \geq 1$ with boundary conditions $a_0 = 0, a_m=1$.  Then,
\begin{align}
0 &= p\sum_{k=1}^\infty a_{k+1} x^{k+1} - \sum_{k=1}^\infty a_kx^{k+1} + (1-p)\sum_{k=1}^\infty a_{k-1}x^{k+1} \\
&= p(A(x) -a_0 - a_1x) - x (A(x)-a_0) +(1-p) x^2 A(x)
\end{align}
which (using $a_0 =0$) leads to
\begin{align}
A(x) = \frac{a_1px}{p-x+(1-p)x^2}
\end{align}
This can be simplified using partial fractions.  I'll assume distinct roots to the quadratic in the denominator, leaving the case with a repeated root for another time.  If $r_1, r_2$ are the roots of $p\xi^2-\xi+(1-p) = 0$ (taking the reciprocal $\xi = 1/x$ so the answer is a little neater and $p$ is eliminated), then we have $r_1r_2 =(1-p)/p$ and
\begin{align}
A(x) = \frac{a_1}{r_2-r_1} \Bigg( \frac{1}{1-r_2x} - \frac{1}{1-r_1x} \Bigg)
\end{align}
You can now expand the two reciprocals into simple geometric progressions.  We are already given that the $x^m$ term has coefficient $1$, so
\begin{align}1 = \frac{a_1}{r_2-r_1}\Big( r_2^m-r_1^m \Big) \end{align}
whence
$$a_1 = \frac{r_2-r_1}{r_2^m-r_1^m}$$
and
$$A(x)=\frac{1}{r_2^m-r_1^m}\Bigg(\frac{1}{1-r_2x} - \frac{1}{1-r_1x} \Bigg).$$
and
$$a_n = \frac{r_2^n-r_1^n}{r_2^m-r_1^m}.$$

Adding an alternative $A^*(x)$ where $a_{m+1}, a_{m+2}, \cdots$ are all zero:

Simply sum the coefficient formula from $n=0$ to $m$,
$$
A^*(x)=\frac{1}{r_2^m-r_1^m}
\Bigg( \frac{1-(r_2x)^{m+1}}{1-r_2x} - \frac{1-(r_1x)^{m+1}}{1-r_1x} \Bigg)
$$


问:
How to prove that a point equidistant from each vertex of a regular polygon (the circumcentre) exists?
How do you prove that a point equidistant from each vertex of a regular polygon exists? I am told to prove this using congruent triangles.</p>

答:
I suggest a proof by construction: basically suggest a scheme for creating a point and then show that it must be equidistant from each of the vertices.

I would start by picking two adjacent vertices, let's say $A$ and $B$, drawing angular bisectors from both, and defining the intersection point as $O$. Because all of the interior angles in a regular polygon are equal, when we bisect them the halves of the angles must be equal, so $\angle OAB = \angle OBA$. This means our triangle is isosceles, so we have that $\overline{OA}=\overline{OB}$.

Now, let's define $C$ as the other adjacent vertex to $B$. Because $\overline{OB}$ is defined as an angular bisector to $\angle ABC$, we have $\angle OBC = \angle OBA$, and because the polygon is regular we have $\overline{BC} = \overline{AB}$. Using these facts you should be able to prove that triangles $OAB$ and $OBC$ are congruent by SAS, which means that $\overline{OC} = \overline{OB}$, and you should be set up to repeat the same procedure with the next vertex.

Hope this helps!


问:
What is $\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}=\sum^\infty_{i=0}\frac1{i(i+x)(i+2x)\cdots(i+nx)}$?
(This is a follow up to This question. Note that the linked question is about the case $n=1$.)

What is $$\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}=\sum^\infty_{i=1}\frac1{i(i+x)(i+2x)\cdots(i+nx)}$$

My attempt:

I tried doing something similar to what J.G. did, by rewriting the sum as an integral.

We first note that $\int^1_0 t^{n-1}dt=\frac{t^n}{n}\big|^1_0=\frac{1^n-0^n}{n}=\frac 1n$. Then:

$$\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}$$
$$=\sum^\infty_{i=1}\prod^n_{j=0}\frac1{i+jx}$$
$$=\sum^\infty_{i=1}\prod^n_{j=0}\int^1_0 t^{i+jx-1}dt$$
$$=\int^1_0 \prod^n_{j=0}t^{jx}\sum^\infty_{i=1}t^{i-1}dt$$
$$=\int^1_0t^{x\sum^n_{j=0}j}\frac{1}{1-t}dt$$
$$=\int^1_0\frac{t^{\frac{xn(n+1)}2}}{1-t}dt$$

However, when I plug in $n=1$, this gives me $$\int^1_0\frac{t^x}{1-t}dt$$

What I should have gotten is $\frac1x\int^1_0\frac{1-t^x}{1-t}dt$, which means I did something wrong. Where is my mistake, and how should I have done it? (The transition from $=\sum^\infty_{i=0}\prod^n_{j=0}\int^1_0 t^{i+jx-1}dt$ to $=\int^1_0 \prod^n_{j=0}t^{jx}\sum^\infty_{i=0}t^{i-1}dt$ <s>seems questionable</s> is incorrect as GEdgar pointed out it is incorrect to swap the product and sum or integral signs.)

答:
In general we have
$\sum_{i=1}^\infty \frac{1}{i (i+a_1)\cdots(i+a_n)}=(-1)^{n-1}\sum_{k=1}^n \frac{ H_{a_k}}{a_k \prod_{1\le j\le n,j \not= k}(a_k-a_j)}$
due to partial fraction expansion of LHS's summand and that $\sum _{n=1}^{\infty } \frac{1}{(a+n) (b+n)}=\frac{H_a-H_b}{a-b}$, thus OP's sum equals
$$(-1)^{n-1}x^{-n}\sum_{k=1}^n \frac{ H_{kx}}{k \prod_{1\le j\le n,j \not= k}(k-j)}$$


问:
Matrix Derivative of $(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta})$
I am trying to calculate the derivative of $$(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta}) $$ where $\mathbf{P}$ is a positive definite matrix. The actual dimensions of each element is not given in the question specification, but since it is for the purposes of minimising $\beta$ for regression analysis, I think $\mathbf{X}$ is mxn, $\mathbf{\beta}\in \mathbf{R}^n$ and $\mathbf{Y}\in \mathbf{R}^m$. First, I expand the expression,

$$(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta}) = (\mathbf{Y^TP-\beta^T\mathbf{X}^TP})(\mathbf{Y-X \beta}) = \mathbf{Y^TPY-Y^TPX\beta -\beta^TX^TPY+\beta^TX^TPX\beta} $$

Now I take the derivative for wrt $\beta$. For the final term, I am using that it is a quadratic form and I think I am assuming $\mathbf{X^TPX}$ is symmetric. I am just using identities on - https://en.wikipedia.org/wiki/Matrix_calculus Anyway I get,

$$\mathbf{-Y^TPX-Y^TPX}+2\mathbf{\beta^TX^TPX} = -2\mathbf{Y^TPX+2\beta^TX^TPX}$$

From here, I can equate to $0$ and take the transpose, to solve for $\beta$ (assuming everything is inversable for now, don't worry).

$$\mathbf{\beta^TX^TPX=Y^TPX}\iff \mathbf{X^TPX\beta=X^TPY} \iff \beta=\mathbf{(X^TPX)^{-1}X^TPY}$$

The solutions solve it slightly differently. They said since, $(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta})$ is already a quadratic form, we can just use this to calculate the derivative as $$\mathbf{-X^T}2\mathbf{P(Y-X\beta})=-2\mathbf{X^TPY} + 2{\mathbf{X^TPX\beta}}$$. As you can see, this is the same as my derivative, but transposed. Of course, once I transpose to solve for $\beta$, this is no longer the case and we get the same final solution. I have 2 questions.

Is the method I have done incorrect, i.e. if the question was just calculate the derivative have I done it incorrectly. If so would you kindly point out where I have made my mistake?

Could anyone recommend some literature/web page that explains the process the solutions took for taking the derivative by spotting it was a quadratic form.

Thank you very much!

答:
$\def\d{\cdot}\def\p#1#2{\frac{\partial #1}{\partial #2}}$
The use of an explicit dot product often prevents transposition errors such as the one that you encountered, and reducing visual clutter will minimize distractions during the differentiation process. Towards that end, define the working vector
$$\eqalign{
w &= X\d b-y \\
}$$
Write the regression error in terms of this new vector, calculate the gradient, then substitute the original variables.
$$\eqalign{
{\cal E} &= w\d P\d w \\
d{\cal E} &= 2w\d P\d dw
 = 2w\d P\d X\d db
 = 2\big(X^T\d P\d w\big)\d db \\
\p{\cal E}{b} &= 2X^T\d P\d w = 2X^T\d P\d \big(X\d b-y\big) \\
}$$
NB:   The $P$ matrix has been assumed to be symmetric, if that's not the case it should be replaced by its symmetric component, i.e.
$\;P=\tfrac 12\left(P+P^T\right).$

Now you can proceed as usual: set the gradient to zero and solve for the optimal $b$ vector.
$$\eqalign{
P &= L\d L^T &\qquad\big({\rm Cholesky\,factorization}\big) \\
R &= L^T\d X \\
R^T\d R\d b &= R^T\d L^T\d y \\
b &= R^+\d L^T\d y
 &\qquad\big(R^+{\rm \,is\,the\,Moore\,Penrose\,inverse}\big) \\
}$$


问:
Proof about Weierstrass Theorem and limits.

Okay so this is probably the longest proof I have written. But the idea is simple, and it's pretty much copy paste in the different directions. I'm pretty confident this time around and was wondering if I have any gaps of logic or inaccuracies since we've only started practicing continuity last week..
Also about question a, I just can't seem to find such a function! Can anyone give me a direction?

b.

We will separate it into cases.

Case 1: $f\left(x\right)\ =\ L$ for every $x ∈ R$.
In this case every point is an extreme point since $f(x)$ is constant.

Case 2: There exist $a ∈ R$ such that $f\left(a\right)\ >\ L$.

By definition of limit f at $∞$, we choose $ϵ\ =\ f\left(a\right)\ -\ L$ for which there exist $M_{1}>0$ such that for every $x>M_{1}$ we have $\left|f\left(x\right)-L\right|<f\left(a\right)-L$.

$\ 2L-f\left(a\right)<f\left(x\right)<f\left(a\right)$ ➜ $f\left(x\right)<f\left(a\right)$.

Note that $a≤M_{1}$ since for every $x>M_{1}$ we have $f\left(x\right)<f\left(a\right)$.

By definition of limit f at $-∞$, we choose $ϵ\ =\ f\left(a\right)\ -\ L$ for which there exist $M_{2}<0$ such that for every $x<M_{2}$ we have $\left|f\left(x\right)-L\right|<f\left(a\right)-L$.

$\ 2L-f\left(a\right)<f\left(x\right)<f\left(a\right)$ ➜ $f\left(x\right)<f\left(a\right)$.

Note that $a≥M_{2}$ since for every $x<M_{2}$ we have $f\left(x\right)<f\left(a\right)$.

Since f is continuous in R, in particular, it's continuous on $[M_{2},M_{1}]$. Then by Weierstrass's Theorem there exist $k∈[M_{2},M_{1}]$ such that $f\left(k\right)\ge f\left(x\right)$ for every $x∈[M_{2},M_{1}]$.

Since $f\left(k\right)\ge f\left(a\right)$ in $[M_{2},M_{1}]$ and $f\left(a\right)\ >\ f\left(x\right)$ for every $x$ in $(-∞,M_{2}),(M_{1},∞)$ then $f\left(k\right)\ \ge\ f\left(x\right)$ for every $x∈R$, and by definition an extreme point of $f$.

Case 3: There exist $b ∈ R$ such that $f\left(b\right)\ <\ L$.

By definition of limit f at $∞$, we choose $ϵ\ =L\ -\ f\left(b\right)$ for which there exist $M_{1}>0$ such that for every $x>M_{1}$ we have $\left|f\left(x\right)-L\right|<L\ -\ f\left(b\right)$ ➜ $f\left(b\right)\ <\ f\left(x\right)$.

Note that $b≤M_{1}$ since for every $x>M_{1}$ we have $f\left(x\right)>f\left(b\right)$.

By definition of limit f at $-∞$, we choose $ϵ\ =L\ -\ f\left(b\right)$ for which there exist $M_{2}<0$ such that for every $x<M_{2}$ we have $\left|f\left(x\right)-L\right|<L\ -\ f\left(b\right)$ ➜ $f\left(b\right)\ <\ f\left(x\right)$.

Note that $b≥M_{2}$ since for every $x<M_{2}$ we have $f\left(x\right)>f\left(b\right)$.

Since f is continuous in R, in particular, it's continuous on $[M_{2},M_{1}]$. Then by Weierstrass's Theorem there exist $s∈[M_{2},M_{1}]$ such that $f\left(s\right)\le f\left(x\right)$ for every $x∈[M_{2},M_{1}]$.

Since $f\left(s\right)\le f\left(b\right)$ in $[M_{2},M_{1}]$ and $f\left(b\right)\ <\ f\left(x\right)$ for every $x$ in $(-∞,M_{2}),(M_{1},∞)$ then $f\left(s\right)\ \le\ f\left(x\right)$ for every $x∈R$, and by definition an extreme point of $f$.

In conclusion, we have proven f must have an extream point.

答:
Hint. For question (a), try to modify the following function
$$
g(x)=\frac{x}{1+|x|}
$$


答:
Hint. One may notice that
$$
\left[\frac{1}{3} (k+1)^3-\frac{1}{2} (k+1)^2+\frac{k+1}{6}\right]-\left[\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6}\right]=k^2,\qquad k=1,2,...,n-1,
$$ and use a telescoping series.

Title: Question on how the sum of Arithmetic Progression is $\frac{n^3}{3}$-$\frac{n^2}{2}$+$\frac{n}{6}$.

Body:
Given a set S = $({1,2,...,(n-1))}$, we obtain the sum of all of its entries through
$$S_n= \frac{1}{2}n(n-1)$$ that we derive from $S_n=\frac{1}{2}n(a_1 +a_n)$. The issue is, let us square our set, $S_n^2=({1^2,2^2,...,(n-1)^2)}$, from this we get $$S^2_n =\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$$. That is the question, though, how do we get this value,

I am confused about how to reach that. I have tried to go through $$\frac{1}{2}(n-1)((n-1)^2+1^2)$$, the problem with this is that it gives off a result that does not equal the one mentioned in the title, so there is no point in trying to simplify it. Can someone help me? Thank you.


问:
Help me understand Inverse CDF for Generating Random Numbers
Ok, so I'm extremly confused about the whole CDF inverse to find a Random Number Generator, please I need a very clear and slow explanation.

The question:

We have $\displaystyle F_{X}( x) =\frac{e^{x}}{1+e^{x}}$, and $\displaystyle G_{X}( x) =F^{-1}_{X}( x) =\log\frac{x}{1-x}$. Suppose we are given a Random Number Generator in $\displaystyle U[ 0,1]$, find a way to use $\displaystyle G_{X}( x)$ to Generate Random Numbers for the distribution represented by $\displaystyle F$, prove formally.

I do understand the idea, I know $\displaystyle G_{X}( x)$ is the inverse, so it takes as input probabilities, and it returns the $\displaystyle x$ that brings to that probability. So it makes sense that by inputting different values in $\displaystyle U[ 0,1]$ you will get the distribution represented by $\displaystyle F_{X}( x)$.

Now, what do I formally have to prove? I don't know!

I can't prove that $\displaystyle G_{X}( x) =F_{X}( x)$ cause that's obviously not true. So what should I prove? I don't get it. In many proof, I see something as $\displaystyle G( X)$. But what is at all the meaning of inputting some $\displaystyle X$ inside $\displaystyle G_{X}( x)$?

All I learned in class (therefore all I can work with). I know that $\displaystyle F_{X}( x) =P( X\leqslant x)$.

This brings me solely to know that if $\displaystyle G_{X}( u) =p$, then $\displaystyle P( X\leqslant u) =p$. I can't seem to know anything else about $\displaystyle G_{X}( x)$ that would help me further with the proof.

As you might tell I'm pretty confused about the whole thing. What I have to prove and how. Any help would be appreciated.

答:
$F_X(x) = \mathbb P(X \le x) $ is a strictly increasing bijective function $\mathbb R \to (0,1)$, so $G_X(u) = F^{-1}_X(u) $ is a strictly increasing bijective function $(0,1) \to \mathbb R$

You know that if $U$ is uniformly distributed on $[0,1]$ then $\mathbb P(U \le u)=u$ for $0 \lt u \lt 1$, so $\mathbb P(G_{X}( U) \le G_{X}( u)) = u$

$G_{X}( U)=F^{-1}_{X}( U)$ is a value in $\mathbb R$  and if we let $Y=G_{X}( U)$ and $y=G_{X}( u)$ so $u=F_X(y)$, then $\mathbb P(Y \le y) = F_X(y)$, meaning that $X$ and $Y=G_X(U)$ have the same cumulative distribution function, i.e. the same distribution


问:
Show that a subset $X \subseteq G$ generates $G$ if and only if $X+pG$ generates $G$.
Let $p$ be a prime, and let $G$ be a finite abelian $p$-group. Define a subgroup $H<G$ to be maximal if there is no other subgroup $K<G$ such that $H<K<G$. Define the set $pG = \left\{ pg: g\in G \right\}$. Show that a subset $X \subseteq G$ generates $G$ if and only if $X+pG$ generates $G$.

I think I am close to figuring this question out, but cannot quite put the last few pieces together. So far I have worked out in a previous question that $$H \text{ is maximal if and only if } G/H \cong \mathbb{Z}_p \tag{1}$$ I worked this out using the correspondence theorem, and the fundamental theorem of finitely generated abelian groups. I then worked out in another previous question that
$$
X \subseteq G \text{ generates }G \text{ if and only if } X \text{ is not contained in a maximal subgroup of } G \tag{2}
$$
Now, back to the original statement I am trying to prove. It is quite trivial proving the forwards implication of this statement. I am now trying to prove the backwards implication of the statement by showing that if $X$ does not generate $G$, then $X+pG$ does not generate $G$. To do this, I have shown that if $X$ does not generate $G$, then $X$ is contained in a maximal subgroup $H$ of $G$, by $(2)$. We then have by $(1)$ that $G/H \cong \mathbb{Z}_p$. I now want to somehow show that $pG$ must now also be contained by H, which would prove the statement, again by using $(2)$. Is this the right approach to this question? Can anybody help me prove that $pG\subseteq H$, or is this simply the wrong approach to the question? Thanks for any help!

答:
$\DeclareMathOperator{\Frat}{Frat}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$What follows is possibly too general a setting, but I believe it clarifies a bit what's  going on here.

An element $g$ of a finite group $G \ne \Set{1}$ is called a non-generator if for any $X \subseteq G$ one has
$$
G = \Span{X, g} \quad\text{implies}\quad G = \Span{X},
$$
or equivalently
$$
G \ne \Span{X} \quad\text{implies}\quad G \ne \Span{X, g},
$$
It is not difficult to see that the set $\Frat(G)$ of non-generators is a subgroup (the so-called Frattini subgroup), as
$$
\Frat(G) = \bigcap \Set{ M \le G : \text{$M$ is a maximal subgroup of $G$}}.
$$
This is because if $g$ is a non-generator, and $M$ is a maximal subgroup of $G$, then $\Span{M, g} = M$, as $\Span{M} = M \ne G$, so that $g \in M$. And conversely, if $g$ lies in every maximal subgroup, and $X \subseteq G$ is such that $\Span{X} \ne G$, then $\Span{X}$ is contained in a maximal subgroup $M$ of $G$, so that $\Span{X, g} \le M$, and $\Span{X, g} \ne G$.

Now if $G$ is a finite, abelian $p$-group, then for every maximal subgroup $M$ you have $p G \le M$, as you have noted that $G / M$ has order $p$. Hence $p G \le \Frat(G)$, so that $p G$ is all made of non-generators.


问:
How do I prove that a statistic is a pivot?
In this example I have a sample out of a distribution with density $P_{\theta}(x) = 2x^{\theta} e^{−\theta x^2} I_{x \ge 0}$.
I know that for all $i \ge 1$ we have $X_i^2 \sim Exp(\theta)$.
If $\theta = 1$, then $T_1 = 2\sum_{1 \le i \le n}X_i^2 \sim X_{2n}^2$.
And now I have to show that $T_{\theta} = 2\sum_{1 \le i \le n} \theta X_i^2$ is a pivot.
A pivot is a function of which the probability function does not depend on $\theta$.
But how do I show that $T_{\theta}$ does not depend on $\theta$?

Btw: I tried to plug it first in LaTeX, so that it's easier to read. But that doesn't work.

答:
Fix $x \ge 0$. As $X_i^2 \sim exp(\theta)$ it follow that
$$P(\theta X_1^2 \le x) = P(X_1^2 \le \frac{x}{\theta }) = P( exp(\theta) \le \frac{x}{\theta })=1 - e^{- \frac{x}{\theta } \theta} = 1 - e^{-x} =P( exp(1) \le x).$$
So, we get that $\theta X_1^2 \sim \exp(1)$. Thus $Y_1 = \theta X_1^2$ is pivot. Hence $2 \sum_i Y_i$ is pivot as sum of independent pivot statistics
.


问:
Probability without replacement .
Inside a candy jar, one can find 5 dark chocolates and 4 white chocolates. One after the other, every candy is taken out of the jar randomly and without replacement. If the chance of grabbing a dark chocolate is same as the white chocolate, what is the probability that the 6th chocolate is not dark ?

Options are:

1/3

2/3

4/9

7/9

8/9

I tried to list all the possible scenarios and came with an answer 7/9(Which is probably incorrect) but I am very new to permutation and combinations. So if anyone can provide an explanation on how the process actually works  will be very much appreciated

答:
It’s best to work by cases based on the number of whites drawn till the $5$th draw.

$0$ whites: Then you need $BBBBBW$, which has probability $$\frac 59 \frac 48 \frac 37 \frac 26 \frac 15 \frac 44 $$

$1$ white: You need to arrange $4$ $B$’s and $1$ $W$ followed by $1$ white and you’ll see all the arrangements are equally likely. Multiply this probability with the number of ways of arranging: $$\frac 59 \frac 48 \frac 37 \frac 26 \frac 45 \frac 34  \times {5\choose 1 }$$

$2$ whites: This case is very similar. You need to arrange $3$ blacks and $2$ whites. $$\frac 59 \frac 48 \frac 37 \frac 46 \frac 35 \frac 24 \times {5\choose 2}$$

$3$ whites: $$\frac 59 \frac 48 \frac 47 \frac 36 \frac 25 \frac 14 \times {5\choose 3} $$

You cannot have more than $3$ whites, as then you’ll run out of whites. The required probability $(4/9)$ is obtained by adding up these probabilities.

Title: Probability without replacement .


问:
kernel of $R\to R/(f)$ is non zero
Let $R = \Bbb{Z}/(6)$ be the base ring,define polynomial ring $R[x]$ over it.

Let $f = 2x+1$ , $(2x+1)$ be the ideal generated by $f$ then if we restrict the canonical map $\pi :R[x] \to R[x]/(2x+1)$ to $\nu:R\to R[x]/(2x+1)$, prove the kernel is non-zero.

I try to prove $3 \in R$ is the desired one, but I don't know what does $\Bbb{Z}/(6)$ mean? Is it $\Bbb{Z}_6$?

答:
The canonical map is given by $\pi:f(x)\mapsto f(x)+ \langle 2x+1\rangle$.

The restriction is then $\nu: z \mapsto z+\langle 2x+1\rangle$.

The ideal $\langle 2x+1\rangle $ contains $2(2x+1) = 4x+2$, $3(2x+1) = 6x+3= 3$, $4(2x+1) = 8x+4= 2x+3$, $5(2x+1) = 10x+5= 4x+5$.

It follows that $3\in \langle 2x+1\rangle$ and so $3\in\ker\nu$.

Title: kernel of $R\to R/(f)$ is non zero


答:
Yes the interchange is not valid as products are involved.
Also at $i=0$ there is a divergence thus we will ignore $i=0$

You can start by using partial fraction on the product to get:
$$\prod^n_{k=0}\frac{1\:}{m+kx\:}= \frac{a_0}{m} +\sum _{k=1}^n\frac{a_k}{m+kx}$$
To determine $a_j :j \not=0 $ we integrate both sides around circular contour centred at $\frac{-m}  {j}$ with radius $\xi$ as $\xi\rightarrow 0$
Therefore:
$$\lim _{\xi \to 0}\int _0^{2\pi }i\xi \:e^{i\theta \:}d\theta \:\prod ^n_{k=0}\frac{1}{m+k\left(-\frac{m}{j}+\xi \:e^{i\theta \:}\right)\:}=\frac{2i\pi a_j}{j}$$
$$\frac{2i\pi a_j}{j}=\lim _{\xi \to 0}\int _0^{2\pi }i\xi \:e^{i\theta \:}d\theta\:\prod ^n_{k=0}\frac{1}{m+k\left(-\frac{m}{j}+\xi \:e^{i\theta \:}\right)\:}$$
$$\frac{2i\pi a_j}{j}=\lim _{\xi \to 0}\int _0^{2\pi }id\theta\:\prod ^n_{k=0\\ k\not=j\
}\frac{1}{m+k\left(-\frac{m}{j}+\xi \:e^{i\theta \:}\right)\:}$$
Therefore:
$$a_j=j\prod ^n_{k=0\\ k\not=j\
}\frac{j}{mj-km}=j\:\left(\frac{j}{m}\right)^n\frac{\left(-1\right)^{n-j}}{j!\left(n-j\right)!}$$
Now plugging everything in sum we get:
$$\sum _{m=1}^{\infty }\left(\sum _{j=0}^nj\left(\left(\frac{j}{m}\right)^n\frac{\left(-1\right)^{n-j}}{n!}\begin{pmatrix}n\\ j\end{pmatrix}\right)\frac{1}{m+jx}\right)=S$$
Interchaging order of summation and use $$\int _0^1\:u^{m+xj-1}du=\frac{1}{m+xj}$$ We get:
$$S=\sum _{j=0}^n j\left(\frac{j^n\left(-1\right)^{n-j}}{n!}\begin{pmatrix}n\\ \:\:j\end{pmatrix}\right)\int _0^1\:\sum _{m=1}^{\infty }\left(\frac{u^m}{m^n}\right)\frac{u^{jx}}{u}\:\:du$$
$$S=\sum _{j=0}^n j\left(\frac{j^n\left(-1\right)^{n-j}}{n!}\begin{pmatrix}n\\ \:j\end{pmatrix}\right)\int _0^1 \text{Li}_n\left(u\right)\frac{u^{jx}}{u}\:\:du$$
This is progress so far, the final task is to evaluate sum then solve integral and we will be done. Unfortunatly I can't find closed form of series in meantime . When I solve it I'll update you.

Update:

We can write sum $S$ as :
$$\frac{d^{n+1}}{dx^{n+1}}\left(\int _0^1\:\sum _{j=0}^n \:\left(\frac{\left(-1\right)^{n-j}u^{jx}}{n!}\begin{pmatrix}n\\ \:j\end{pmatrix}\right)\frac{\text{Li}_n\left(u\right)}{u\:\ln ^{n+1}\left(u\right)}\:\:du\right)\:=\frac{d^{n+1}}{dx^{n+1}}\left(\int _0^1\frac{1}{n!}\:\left(u^x-1\right)^n\frac{\text{Li}_n\left(u\right)}{u\left(\ln \left(u\right)\right)^{n+1}}\:\:du\right)\:$$

Title: What is $\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}=\sum^\infty_{i=0}\frac1{i(i+x)(i+2x)\cdots(i+nx)}$?

Body:
(This is a follow up to This question. Note that the linked question is about the case $n=1$.)

What is $$\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}=\sum^\infty_{i=1}\frac1{i(i+x)(i+2x)\cdots(i+nx)}$$

My attempt:

I tried doing something similar to what J.G. did, by rewriting the sum as an integral.

We first note that $\int^1_0 t^{n-1}dt=\frac{t^n}{n}\big|^1_0=\frac{1^n-0^n}{n}=\frac 1n$. Then:

$$\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}$$
$$=\sum^\infty_{i=1}\prod^n_{j=0}\frac1{i+jx}$$
$$=\sum^\infty_{i=1}\prod^n_{j=0}\int^1_0 t^{i+jx-1}dt$$
$$=\int^1_0 \prod^n_{j=0}t^{jx}\sum^\infty_{i=1}t^{i-1}dt$$
$$=\int^1_0t^{x\sum^n_{j=0}j}\frac{1}{1-t}dt$$
$$=\int^1_0\frac{t^{\frac{xn(n+1)}2}}{1-t}dt$$

However, when I plug in $n=1$, this gives me $$\int^1_0\frac{t^x}{1-t}dt$$

What I should have gotten is $\frac1x\int^1_0\frac{1-t^x}{1-t}dt$, which means I did something wrong. Where is my mistake, and how should I have done it? (The transition from $=\sum^\infty_{i=0}\prod^n_{j=0}\int^1_0 t^{i+jx-1}dt$ to $=\int^1_0 \prod^n_{j=0}t^{jx}\sum^\infty_{i=0}t^{i-1}dt$ <s>seems questionable</s> is incorrect as GEdgar pointed out it is incorrect to swap the product and sum or integral signs.)


问:
Show that $v\mapsto \|v\|^p$ for $1\le p<\infty$ is strictly convex
I want to show that $f(v)=\|v\|^p$ for $1\leq p<\infty$ is strictly convex.

In the simplest case when $p=2$, we have:
$$\left\|\frac{v_1+v_2}{2}\right\|^2=\left(\frac{v_1+v_2}{2},\frac{v_1+v_2}{2}\right)=$$
$$\frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+\frac{1}{2}(v_1,v_2)$$
$$\leq \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+ \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}$$
(by Cauchy inequality). That is convex, but not strictly convex - where did I make a mistake?

Please help me. Thanks.

答:
The way that you show a function is convex is to draw a line between any two values; the values should always be on the same side.

It relies on a definition of convexity, where a line that strikes it twice is only inside (ie a specified $<$ or $>$) between these two striking points, and outside (the opposite inequality) everywhere else.

So, for example, one might find $f(a)=p_a$, and $f(b)=p_b$, then $f(ax+by)<x\cdot f(a)+y\cdot f(y)$, where $x+y=1$.

So, one deconstructs the function as $f(p+dx) < fp(p) + d.s$, where $s=f(p+x)-f(p)$ while $d<1$, then the function is convex.


答: Instead of working with $g(x):= f(x)-L$, we can suppose WLOG that $L=0$. If  $f=0$, the proof is straightforward. If $f\neq 0$, there is $a\in \mathbb R$ s.t. $f(a)\neq 0$. Suppose WLOG that $f(a)>0$. Since $f(x)\to 0$ when $x\to \pm \infty $, there is $N>0$ s.t. $0\leq f(x)<f(a)$ for all $x\notin [-N,N]$. Since $f$ is continuous, it reach his maximum on $[-N,N]$, which is a maximum on $\mathbb R$.

Title: Proof about Weierstrass Theorem and limits.

Body:
Okay so this is probably the longest proof I have written. But the idea is simple, and it's pretty much copy paste in the different directions. I'm pretty confident this time around and was wondering if I have any gaps of logic or inaccuracies since we've only started practicing continuity last week..
Also about question a, I just can't seem to find such a function! Can anyone give me a direction?

b.

We will separate it into cases.

Case 1: $f\left(x\right)\ =\ L$ for every $x ∈ R$.
In this case every point is an extreme point since $f(x)$ is constant.

Case 2: There exist $a ∈ R$ such that $f\left(a\right)\ >\ L$.

By definition of limit f at $∞$, we choose $ϵ\ =\ f\left(a\right)\ -\ L$ for which there exist $M_{1}>0$ such that for every $x>M_{1}$ we have $\left|f\left(x\right)-L\right|<f\left(a\right)-L$.

$\ 2L-f\left(a\right)<f\left(x\right)<f\left(a\right)$ ➜ $f\left(x\right)<f\left(a\right)$.

Note that $a≤M_{1}$ since for every $x>M_{1}$ we have $f\left(x\right)<f\left(a\right)$.

By definition of limit f at $-∞$, we choose $ϵ\ =\ f\left(a\right)\ -\ L$ for which there exist $M_{2}<0$ such that for every $x<M_{2}$ we have $\left|f\left(x\right)-L\right|<f\left(a\right)-L$.

$\ 2L-f\left(a\right)<f\left(x\right)<f\left(a\right)$ ➜ $f\left(x\right)<f\left(a\right)$.

Note that $a≥M_{2}$ since for every $x<M_{2}$ we have $f\left(x\right)<f\left(a\right)$.

Since f is continuous in R, in particular, it's continuous on $[M_{2},M_{1}]$. Then by Weierstrass's Theorem there exist $k∈[M_{2},M_{1}]$ such that $f\left(k\right)\ge f\left(x\right)$ for every $x∈[M_{2},M_{1}]$.

Since $f\left(k\right)\ge f\left(a\right)$ in $[M_{2},M_{1}]$ and $f\left(a\right)\ >\ f\left(x\right)$ for every $x$ in $(-∞,M_{2}),(M_{1},∞)$ then $f\left(k\right)\ \ge\ f\left(x\right)$ for every $x∈R$, and by definition an extreme point of $f$.

Case 3: There exist $b ∈ R$ such that $f\left(b\right)\ <\ L$.

By definition of limit f at $∞$, we choose $ϵ\ =L\ -\ f\left(b\right)$ for which there exist $M_{1}>0$ such that for every $x>M_{1}$ we have $\left|f\left(x\right)-L\right|<L\ -\ f\left(b\right)$ ➜ $f\left(b\right)\ <\ f\left(x\right)$.

Note that $b≤M_{1}$ since for every $x>M_{1}$ we have $f\left(x\right)>f\left(b\right)$.

By definition of limit f at $-∞$, we choose $ϵ\ =L\ -\ f\left(b\right)$ for which there exist $M_{2}<0$ such that for every $x<M_{2}$ we have $\left|f\left(x\right)-L\right|<L\ -\ f\left(b\right)$ ➜ $f\left(b\right)\ <\ f\left(x\right)$.

Note that $b≥M_{2}$ since for every $x<M_{2}$ we have $f\left(x\right)>f\left(b\right)$.

Since f is continuous in R, in particular, it's continuous on $[M_{2},M_{1}]$. Then by Weierstrass's Theorem there exist $s∈[M_{2},M_{1}]$ such that $f\left(s\right)\le f\left(x\right)$ for every $x∈[M_{2},M_{1}]$.

Since $f\left(s\right)\le f\left(b\right)$ in $[M_{2},M_{1}]$ and $f\left(b\right)\ <\ f\left(x\right)$ for every $x$ in $(-∞,M_{2}),(M_{1},∞)$ then $f\left(s\right)\ \le\ f\left(x\right)$ for every $x∈R$, and by definition an extreme point of $f$.

In conclusion, we have proven f must have an extream point.


问:
Is there an odd solution of $\varphi(n)+n=\sigma(n)$?
I want to show that the only solution of $$\varphi(n)+n=\sigma(n)$$ for a positive integer $n$ is $n=2$.

What I worked out is that we must have $$\varphi(n)>\frac{n}{2}$$

To show this assume $n$ is composite and its prime factors are $$p_1,\cdots,p_k$$ Then, the numbers $$1,\frac{n}{p_1},\cdots ,\frac{n}{p_k}$$ are distinct divisors from $n$ not equal to $n$ and the sum of those divisors is larger than the number of positive integers $m\in [1,n]$ not coprime to $n$, since every such $m$ is a multiple of one of the prime factors $p_i$ of $n$, the number of multiples of this $p_i$ is $\frac{n}{p_i}$

So we get $$n-\varphi(n)<1+\frac{n}{p_1}+\cdots \frac{n}{p_k}\le \sigma(n)-n$$ and together with $\varphi(n)+n=\sigma(n)$ we easily can derive the claimed inequality. This implies that an even composite solution cannot exist because of $\varphi(n) \le \frac{n}{2}$ for even $n$, hence $2$ must be the only solution.

But how can I show that no odd solution $n$ exists ? I strongly conjecture this because in this case $n$ must be a perfect square (because $\sigma(n)$ must be odd, if $n>1$) , and no perfect square $n\le 10^{16}$ does the job.

答:
Iannucci (2017)1 considered this equation for $n>2$ proving the equivalent formulation $$\sigma(m^2)=m(m+\phi(m))$$ where $m>1$ is odd and $\omega(m)>3$.

It is remarked that this problem is at least as difficult as the existence of odd perfect numbers.

 Reference

 [1] Iannucci, D. E. (2017). On the Equation $\sigma(n)=n+\phi(n)$. Journal of Integer Sequences. 20(6):17-6.


问:
Proving $ \cos ^{2} \frac{\theta}{2} \equiv \frac{1+\cos \theta}{2} $
I need help proving LHS is identical to RHS.
$$
\cos ^{2} \frac{\theta}{2} \equiv \frac{1+\cos \theta}{2}
$$

答: You could start with $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ and then use $\sin^2\theta + \cos^2 \theta = 1.$ This will prove the identity with $\cos^2\theta = (1+\cos(2\theta))/2.$ For the first identity, use $\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b).$