问:
What do the local sections of an nilpotent ideal sheaf look like?
Let $(X, O_X)$ be a scheme. and $I$ an nilptoent ideal sheaf. i.e. $I^n=0$ for some $n$. Would this imply that each $I(U)$ is an nilpotent ideal of $O_X(U)$?

答:
Given a quasi coherent sheaf of ideals $\mathcal{I} \subseteq \mathcal{O}_X$ where $X$ is a scheme, we get a presheaf

$\mathcal{I}(n)(U):=\{ \sum_I s_Ix_{i_1}\cdots x_{i_n}: x_{i_j}\in \mathcal{I}(U)$ and $s_I\in \mathcal{O}_X(U)\} \subseteq \mathcal{I}(U)$

If $U_i:=Spec(A_i)$ and $\mathcal{I}(U_i):=I_i \subseteq A_i$ it follows
by definition

$\mathcal{I}(n)(U_i):=I_i^n.$

Definition 0. Let $\mathcal{I}^n:=\mathcal{I}(n)^{+}$ be the sheafification of $\mathcal{I}(n)$. It follows there is a canonical injective map of sheaves

$\mathcal{I}^n \rightarrow \mathcal{I}$.

It follows $\mathcal{I}^n(U_i)=I^n_i$.

Definition 1. Let us define a quasi coherent sheaf of ideals $\mathcal{I}$ to be "nilpotent" iff there is an integer $n\geq 1$ with $\mathcal{I}^n(U)=0$ for all open sets $U$.

If $I \subseteq A$ is an ideal with $I^n=0$ and $\mathcal{I}\subseteq \mathcal{O}_X$ with $X:=Spec(A)$ is the corresponding sheaf of ideals, it follows $\mathcal{I}^n(X)\cong I^n=0$. You may moreover prove that there is a canonical isomorphism

I1. $\mathcal{I}(U)\cong \mathcal{O}_X(U)\otimes_A I \rightarrow^j  \mathcal{O}_X(U)$

where the map $j$ is the canonical map

$j(u\otimes x):= ux$.

Here $u\in \mathcal{O}_X(U)$ and $x\in I$. The map $j$ is an injective map since the ring extension $A \rightarrow \mathcal{O}_X(U)$ is flat for any open set $U$. Hence the injective map $j$ is obtained from the inclusion $I \subseteq A$ via tensor product.

Note: To prove that the canonical map $A\rightarrow \mathcal{O}_X(U)$ is flat
you should use an exercise in Atiyah-Macdonald (Ex 4.23) where you give an alternative construction of the structure sheaf. You define

$\mathcal{O}_X(U) :=lim_{D(f) \subseteq U}\mathcal{O}_X(D(f)):= lim_{D(f) \subseteq U}A_f$.

Since $A\rightarrow A_f$ is flat, it follows $A\rightarrow \mathcal{O}_X(U)$ is flat since it is a direct limit of flat maps.

Note: I want to recommend the exercise in AM since it gives an alternative construction of the structure sheaf of a scheme.

If you choose arbitrary elements

$w_1:=u_1\otimes x_1,...,w_n:=u_n\otimes x_n$ in $\mathcal{I}(U)$ it follows

$w_1\cdots w_n = u_1\cdots u_nx_1\cdots x_n=0$ since $x_1\cdots x_n=0$ in $I^n=0$.

and from I1 it follows $\mathcal{I}^n(U)=0$.

Question: "Would this imply that each $I(U)$ is an nilpotent ideal of $O_X(U)$?"

Answer: It seems like the answer is "yes" by the above argument, but there may of course be errors.

Conversely if $\mathcal{I}^n(U)=0$ for all open sets $U$ it follows in particular that $\mathcal{I}^n(X)=I^n =0$.

It seems: In the affine situation it follows the quasi coherent ideal sheaf $\mathcal{I}$ is nilpotent (in the sense of Definition 1) iff the ideal $I$ is nilpotent.

Example: The general case. You must be more careful when $X$ is no longer an affine scheme. The correspondence between ideal sheaves $\mathcal{I}$ and ideals $I \subseteq S$ when $S$ is a graded ring and $X:=Proj(S)$ is more complicated (see Hartshorne Exercise II.5.9).

It seems we should give the following  general definition for a quasi coherent sheaf of ideals $\mathcal{I}$ to be nilpotent:

Definition 2. The quasi coherent ideal sheaf $\mathcal{I}$ is nilpotent iff there is an integer $n \geq 1$ with $\mathcal{I}^n_{\mathfrak{p}}=0$
in $\mathcal{O}_{X,\mathfrak{p}}$ for every point $\mathfrak{p}$.

It seems to me Definition 2 agrees with Definition 1 in the affine situation but you have to check this.


问:
Why the method of substitution become wrong in this question?
The question says:

Suppose that $f^{-1}$ is the inverse function of $f$ and $y=f^{-1}(-x)$ is the inverse function of $y=f(-x)$; prove $f(x)$ is an odd function.

Now I know one of the solution is to make $g(x)=-x$ and use the features of composite function.
But what I can't understand is that when I let $t=-x$, I think $f^{-1}(t)$ is the inverse function of $y=f(t)$, but this seems to suggest that $f^{-1}$ is the inverse of $f$.

答: Having already substituted $t=-x$ in $y=f(-x)$, you cannot also substitute it in $y=f^{-1}(-x)$ as this would be a different substitution. In this second case, $x$ belongs to the co-domain rather than domain of $f$. In fact, let's swap the variables $x$ and $y$ immediately, so that $x$ is in the domain and $y$ is in the co-domain: $x=f^{-1}(-y)$, and so the correct substitution there is $t=-f^{-1}(-y)$ rather than "$y=f^{-1}(t)$".

ContentLicense: CC BY-SA 4.0


问:
What is the inverse z transform of 1/(z-1)^2?
I'd like to know how to calculate the inverse z transform of $\frac{1}{(z-1)^2}$ and the general case $\frac{1}{(z-a)^2}$</p>

答:
A simple way is to recall the convolution property: if $x[n] \leftrightarrow X(z)$, then
$x[n] \star y[n] \leftrightarrow X(z) Y(z) $ and, in particular $x[n] \star x[n] \leftrightarrow [X(z)]^2$

Then, we only need to get the anti-transform of $\frac{1}{z-a}= \frac{ z^{-1}}{1-a z^{-1}}$  (piece of cake) and convolve it with itself.

BTW, you need to specify the ROC (region of convergence), or if you want a "causal" sequence - without this, there can be more than one solution.

ContentLicense: CC BY-SA 3.0


问:
degree $2$ extension in a field of characteristic 2
Let $K$ be a field of characteristic 2 and $L/K$ is degree $2$ extension.
I want to prove $L＝K(α)$, $α^2∈K$ or $L＝K(α)$, $α^2＋α∈K$ , and $L/K$ is Galois in the latter case.

My attempt is the following.

Let $1,α$ be basis of $L/K$.
For  $α^2∈L$, $α^2＝mα＋n$, where $m,n∈K$.

We cannot completing the square in character $2$,
so we need to separate cases.

$m＝0$: $α＝\sqrt{n}$, so $L＝K(α)$, $α^2∈K$.

and in this case, $x^2-n$ is not separable, so the extension is not Galois.

I'm having trouble with the case $m$ is not $0$.

I want to show when $m$ is not $0$,

$L＝K(α)$, $α^2＋α∈K$ and the extension is Galois.

答:
When $m\neq1$ you need to rescale  $\alpha$.

Hint: Divide the equation
$$
\alpha^2=m\alpha+n
$$
by $m^2$.


答: You have $\alpha(\alpha+m) \in K$, so if $\alpha'=\alpha/m$, then $\alpha'(\alpha'+1) \in K$ and $L=K(\alpha)=K(\alpha')$. Moreover, $\alpha'$ is separable over $K$ so $L$ is separable over $K$. Therefore, as $L/K$ has degree two, $L$ is Galois over $K$.

Title: degree $2$ extension in a field of characteristic 2

Body:
Let $K$ be a field of characteristic 2 and $L/K$ is degree $2$ extension.
I want to prove $L＝K(α)$, $α^2∈K$ or $L＝K(α)$, $α^2＋α∈K$ , and $L/K$ is Galois in the latter case.

My attempt is the following.

Let $1,α$ be basis of $L/K$.
For  $α^2∈L$, $α^2＝mα＋n$, where $m,n∈K$.

We cannot completing the square in character $2$,
so we need to separate cases.

$m＝0$: $α＝\sqrt{n}$, so $L＝K(α)$, $α^2∈K$.

and in this case, $x^2-n$ is not separable, so the extension is not Galois.

I'm having trouble with the case $m$ is not $0$.

I want to show when $m$ is not $0$,

$L＝K(α)$, $α^2＋α∈K$ and the extension is Galois.


问:
Diffeomorphic surfaces and Jacobian
Suppose $S$ and $T$ are bounded (open) surfaces in $\mathbb{R}^n.$ Let them have boundary $\partial S$ and $\partial T$.

Suppose $F:S \to T$ is a $C^k$ diffeomorphism.

Under what conditions on $F$ and $S$ and $T$ and their boundaries do we get that the determinant of the Jacobian matrix of $F$ is bounded above and below by positive numbers??

答:
If you mean your surfaces to be the interior of compact manifolds with boundary (as your question seems to imply) then the existence of bounds is immediate from the compactness of the boundary, so long as $k$ is at least $1$.

If you don't assume from the start that the map extends to a diffeomorphism of the boundary, then the claim is not true, as can be seen already in the 1-dimensional case.  Thus, the self-map of the interval $[0,1]$ given by $x\mapsto x^2$ is a diffeomorphism of the interior, but the Jacobian on the boundary is singular.

To make sure the map extends to the boundary, one would impose a suitable condition of uniform convergence (see http://en.wikipedia.org/wiki/Uniform_convergence for the case of sequences).  To get $C^k$ one would need such bounds for the $k$-th derivatives.

Title: Diffeomorphic surfaces and Jacobian


问:
How to do a 3 constant reciprocal (multiplicative inverse) regression
I am trying to fit a set of data to a curve such as:

$y=\frac{m}{x-a}+b$

Without the constant $a$, it is easy to define $z=\frac{1}{x}$ and convert it to a linear model. But I have not been able to find how to solve it with the additional constant.

Ideally I would like to use the Theil-Sen estimator (https://en.wikipedia.org/wiki/Theil%E2%80%93Sen_estimator) to solve as I expect to have significant outliers and this method in principal should be more robust. If not, I can any other method such as least squares.

The goal of this exercise is ultimately to automatically identify the outliers. Table below has an example of data points that I am trying to fit:

x
y

2.86
210.021

9.08
108.977

0.63
199.2

125.5
110.723

68.61
112.154

64.62
87.464

233.03
43.727

16.61
277.019

13.41
201.378

11.01
51.5

0.28
455.625

32.65
160.806

145.55
115.917

2.33
160.668

22.07
158.409

67.3
102.105

18.9
143.599

30.45
314.55

115.86
210.744

111.46
126.739

5.24
53

2.71
220.523

1.57
238.647

4.94
272.322

165.75
100.999

20.6
95.482

112.59
114.391

2.06
203.475

89.82
125.323

39.2
95.994

568.93
38.768

69.12
20.15

51.91
80.275

答:
The model $$y=\frac{m}{x-a}+b$$ is nonlinear because of $a$.

You properly said that if $a=0$ the problem is simple. Continue the reasoning : the problem is still simple for any given value of $a$.

So, before running a nonlinear regression, give $a$ a fixed value and define $z_i=\frac 1 {x_i-a}$ and compute
$$\text{SSQ}(a)=\sum_{i=1}^n \big[m z_i+b-y_i\big]^2$$ Try a few values of $a$ and plot $\text{SSQ}(a)$ until you see more or less a minimum. AT this point, you have your estimates and you can safely start the nonlinear regression. If you do not access such a tool, zoom more and more your plot.


问:
Expectation of a censored variable.
My problem is to find $E(Y)$, if $Y=max(X,k)$, for $X \thicksim N(,²)$ and $k \in ℝ$.

I've come across $ \int_k^{\infty} x. \phi(x) dx$, where $\phi(x)$ is the pmf of the normal distribution.

Can I find a closed form solution for $E(Y)$?

答:
First observe that

$$\int_k^{\infty}x \phi(x)dx$$

is not your $E(Y)$ (but sure you already know this)

Example: $X\sim N(0;1)$ and $k=4$

Without any calculation, $E(Y)\approx 4$ because the probability  $\mathbb{P}(X>4)\approx 0$

If the problem is only to evaluate the integral, it is very easy to answer:

$$\int_k^{\infty}x \phi(x)dx=\int_k^{\infty}x \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}dx$$

Set $\frac{x-\mu}{\sigma}=t$ and get

$$\int_{h}^{\infty}(\sigma t+\mu)\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$$

Now split the integral in two separate integrals, one is quite immediate and the other one can be expressed in funciton  of $\Phi(t)$ the CDF of the Standard Gaussian, tabulated everywhere, on paper tables as given by any calculator

Notes:

Following what you wrote, I used $\phi(x)$ indicating the pfd of a Gaussian rv but it is used to indicate with $\phi(x)$ the density of a STANDARD gaussian...

The normal rv does not hav a pmf: it has a pdf; pmf is for discrete rv's

your $Y=max(X,k)$ for a fixed $k$, is not a censored rv but it is a "mixed" rv.

Title: Expectation of a censored variable.


问:
What is a Rate in this Wikipedia sentence?
"In mathematics, a rate is the ratio between two related quantities in different units. If the denominator of the ratio is expressed as a single unit of one of these quantities, and if it is assumed that this quantity can be changed systematically (i.e., is an independent variable), then the numerator of the ratio expresses the corresponding rate of change in the other (dependent) variable."

Can someone explain this with an example?
Say using S = vt

答:
Comment

Some examples

Speed,  Price of a commodity are constants and they can be expressed as a constant ratio of a single dependent variable in terms and a single independent variable:

$$ \text{ 60 miles/hour,   5 dollars/pound };$$

If the variation is with respect to two independent variables like
Interest rate, Acceleration respectively:

$$ \text{ 4 dollars /100 dollars /annum;  25 meters/sec^2};$$


问:
Symmetric distribution of points on a sphere
Background:

The VSEPR theory provides a useful guide to predicting molecular geometries. One implication of the theory is that the orientation of ligands in three-dimensional space is such that the seperation between each pair of them is maximum.

This is analogous to the mathematical problem of symmetrically distributing points on the surface of a sphere. With 2, 3, 4, or 6 points, the solutions are trivial. The required distributions would be linear, triangular, tetrahedral, and octahedral respectively. An issue arises when we attempt to distribute 5 points on the surface of a sphere.

We could try to arrange the points such that they lie on the vertices of a regular polyhedron inscribed in the sphere. However, from my knowledge of Euclidean geometry, I believe that such a shape with 5 vertices simply cannot exist.

Chemists suggest an approximately symmetric distribution such as placing the points on the vertices of a square pyramid. But from a mathematical perspective, there ought to be a better solution.

Question: What is the nearest approximation to a symmetric distribution which we can achieve? Can the approximation be generalized to any other number of points which cannot be distributed at the vertices of a platonic solid?

The definition of a symmetrical distribution here is that any non-labelled point on the sphere is indistinguishable from another such point. A nearly symmetrical distribution implies the least variation in the relative distances between the points (distances measured along the surface).

The ambiguity in the definition of symmetry makes the question more difficult to answer. Any additional insights or numerical methods to solve the problem are also welcome.

答:
If you are only after symmetry (which you are probably not), then you have the following options:

for every $n$ you can arrange the points in the shape of a regular $n$-gon. This arrangement is "flat", but the points are indistinguishable as desired.

for $2n$ points, you can arrange the points in the shape of two regular $n$-gons and position them "on top of each other", either as the vertices of an $n$-gonal prism, or an $n$-gonal antiprism.

Then there are the (vertices of the) 18 uniform polyhedra. With these you can find symmetric arrangements for $n\in\{4,6,8,12,20,24,30,48,60,120\}$.

And that is basically it.
You can more generally look at the class of vertex-transitive polyhedra, which is the technical term for all the possible ways to arrange $n$ points in a symmetric fashion, but each of these has a version as a uniform polyhedron that I already listed above.

In particular, the only way to arrange five points symmetrically is to arrange them in a (flat) regular pentagon.
If you want the points to be spread apart as much as possible, then the solution is a triangular bipyramid as explained by achille, which is not symmetric in your sense. Maybe, symmetry is not the only thing you are after?

I do not know of a reasonable way to talk about "almost symmetric".


问:
Calculating Standard Deviation from Average and Total Values
I'm looking to calculate the standard deviation when all I have is the average alongside how many records there are.

In my dataset I have been provided, I can see an average value which is reflected in the top 5 of the dataset provided.

How can I get the values in between, roughly speaking?

For example, I have an average of £735 in the top 5.

Thanks.

答:
You can’t.

If I understand correctly, you are given the values of $n$ and $\frac1n\sum_{k=1}^nx_k$. It is not possible to recover the individual values of the $x_k$ from this, which are needed to compute the standard deviation.


问:
Induced map on Eilenberg-MacLane space
Let $X$ be an $n-1$ connected space. Why does a map $X\rightarrow K(\pi_n(X),n)$ that induces an isomorphism on $\pi_n$ exist and what is this map?</p>

答: If $X$ is a CW-complex, then you can build a $K(\pi_n(X),n)$ by attaching cells (of dimension $n+2$ and above) to $X$ to kill off the higher homotopy groups. The map you are thinking of is then an inclusion of $X$ into $K(\pi_n(X),n)$ as a sub-complex.</p>

Title: Induced map on Eilenberg-MacLane space


问:
How many ways are there of creating an $8$ character password with a digit, a lowercase letter, and $2$ capital letters?

A valid password for the bank's website consists of 8 characters
(digits and letters in English) and must contain at least one digit,
at least one small English letter, and at least two capital letters.
How many legal passwords are there?

I considered four sets:

$|A|$ = number of ways for no digits appearing in the password = $52^8$,

$|B|$ = number of ways of no small letter in the password = $36^8$,

$|C|$ = number of ways of no capital letter in the password = $36^8$,

$|D|$ = number of ways of only 1 capital letter in the password = $26*36^7$, and

$|U|$ = number of ways of forming the bank's password without restriction  = $62^8$.

Using the formula for inclusion and exclusion for 4 sets $|U| - [ |A| \cup |B| \cup |C|\cup |D|]$

such that

$$
\begin{align}
&|A\cup B\cup C\cup D|\\[3pt]
&=|A|+|B|+|C|+|D|\Big\}\text{ all singletons}\\
&-(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|+|C\cap D|)\Big\}\text{ all pairs}\\
&+(|A\cap B\cap C|+|A\cap B\cap D|+|A\cap C\cap D|+|B\cap C\cap D|)\Big\}\text{ all triples}\\
&-|A\cap B\cap C\cap D|\Big\}\text{ all quadruples.}\\
\end{align}
$$

Now what I'm stuck at is when it comes to triples and quadruples.

$|A\cap B\cap C\cap D|$, to find the number of ways of union of this is kinda contradicting because we have to find a password with no digits, no small letters, no big letters, but only 1 small letter..

or when

$|A\cap C\cap D|$ is also contradicting because how we can find number of ways of password with no big letters and also 1 big letter..

答:
There is no contradiction, just note that $C\cap D=\emptyset$ and therefore all intersection sets including $C\cap D$ have zero cardinality. Moreover, by a similar reason, also $A\cap B\cap C=\emptyset$ and $A\cap B\cap D=\emptyset$. Thus it remains to compute
$$\begin{align}
&|U|-(|A|+|B|+|C|+|D|)+(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|)
\end{align}
$$
P.S. One more thing, your number for $|D|$ should be multiplied by $8$, i.e. the possible positions of the capital letter.


问:
Is it true that $\mathbb{R}^m \cap \mathbb{R}^n = \emptyset$ where $m \neq n$
Is it true that $\mathbb{R}^m \cap \mathbb{R}^n = \emptyset$ where $m \neq n$ ?

I would say that it is true because the elements in each set are different entities. This means that one set cannot be a subset of the other. An example would be $\mathbb{R} \cap \mathbb{R}^2$. You cannot say that $1 \in \mathbb{R}^2$ for example.

答:
From rigorous point of set theory, the answer is yes. However, taking your example, let's assume the mapping $f:\mathbb R\to\mathbb R^2$ such that $x\mapsto (x,0)$ for example. Clearly, $f$ is bijective. Hence, it's safe to say that the image of $f$, namely $\mathbb R\times\{0\}$, represents the domain of $f$ i.e. $\mathbb R\sim\mathbb R\times\{0\}\subset\mathbb R^2$.

Therefore, $\mathbb R$ can be seen, in this way, as a subset of $\mathbb R^2$ and the intersection $\mathbb R\cap\mathbb R^2$ is not empty (disregarding difference in structures of elements in both sets).

I hope that example answers your question since generalizing that is relatively easy. Also, may I recommend reading about disjoint unions as a way to view unions of sets that do intersect. A lot of similar logic is used.


答:
First, your evaluation of $|D|$ is incorrect. This is the number of ways to choose a capital letter for the first character, and then choose the remaining seven characters with no capitals. But in fact the single capital letter could be in any of $8$ positions, so actually $|D|=8\times 26\times 36^7$.

Secondly, the multiple events. Note that $C$ and $D$ cannot both occur, and no three events can occur (since if they include $A,B$ and one of $C,D$ then you have at most one character). So you only need to subtract the pairs of events (not including $|C\cup D|$), since the other terms are all $0$.

$|A\cap B|=26^8$, since this means all characters are capitals, and the other double events can be calculated similarly (ones involving $D$ are a bit more complicated, but no more so than the calculation for $D$ itself).

Title: How many ways are there of creating an $8$ character password with a digit, a lowercase letter, and $2$ capital letters?

Body:
A valid password for the bank's website consists of 8 characters
(digits and letters in English) and must contain at least one digit,
at least one small English letter, and at least two capital letters.
How many legal passwords are there?

I considered four sets:

$|A|$ = number of ways for no digits appearing in the password = $52^8$,

$|B|$ = number of ways of no small letter in the password = $36^8$,

$|C|$ = number of ways of no capital letter in the password = $36^8$,

$|D|$ = number of ways of only 1 capital letter in the password = $26*36^7$, and

$|U|$ = number of ways of forming the bank's password without restriction  = $62^8$.

Using the formula for inclusion and exclusion for 4 sets $|U| - [ |A| \cup |B| \cup |C|\cup |D|]$

such that

$$
\begin{align}
&|A\cup B\cup C\cup D|\\[3pt]
&=|A|+|B|+|C|+|D|\Big\}\text{ all singletons}\\
&-(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|+|C\cap D|)\Big\}\text{ all pairs}\\
&+(|A\cap B\cap C|+|A\cap B\cap D|+|A\cap C\cap D|+|B\cap C\cap D|)\Big\}\text{ all triples}\\
&-|A\cap B\cap C\cap D|\Big\}\text{ all quadruples.}\\
\end{align}
$$

Now what I'm stuck at is when it comes to triples and quadruples.

$|A\cap B\cap C\cap D|$, to find the number of ways of union of this is kinda contradicting because we have to find a password with no digits, no small letters, no big letters, but only 1 small letter..

or when

$|A\cap C\cap D|$ is also contradicting because how we can find number of ways of password with no big letters and also 1 big letter..


问:
Fill in Cayley table for a commutative ring with unity

Let M={0,1,a,b} be a commutative ring with unity. I'm supposedt to fill in the Cayley tables. Can someone help me with the table for the multiply table.
I think that $a*b=b*a=a$ do to commutative ring.
Is the multiplicative identity $1$?
Is $0*1=0$ for example?

I think that the Cayley table for + is:

答: The elements of a commutative ring under addition must be an abelian group. Since there are four elements, that group in this case must be $\mathbb Z_4$ or $\mathbb Z_2^2$. Since there is a non-zero element ($1$) for which $1+1\ne0$, the abelian group must be $\mathbb Z_4$ and the ring is isomorphic to $\mathbb Z/4\mathbb Z$ with the associations $a=2$ and $b=3$. Filling out the tables should now be easy.


问:
Solutions of the equation $x^x=\frac{1}{256}$
Find the solutions of the equation
$$x^x=\frac{1}{256}$$
I know that the function $f(x)=x^x$ is defined for $x>0$, so the solutions, if they exists, must be $>0$; as far as I know, this is because to maintain the formal properties of power, we must impose that $x>0$ or it is easy to get contradictions like $-1=(-1)^{\frac{2}{2}}=[(-1)^2]^{\frac{1}{2}}=1^{\frac{1}{2}}=1$.

So, since the function $x^x$ has an absolute minimum at $x=\frac{1}{e}$ and it is $\left(\frac{1}{e}\right)^{\frac{1}{e}}>\frac{1}{256}$, it follows that there isn't an $x>0$ such that $x^x=\frac{1}{256}$ and it follows that the equation hasn't real solutions.

However, Wolfram|Alpha says that the equation has the integer solution $x=-4$ and it is indeed a solution, as one can check by substitution. However, for what I said before, the function $x^x$ doesn't exist for $x<0$ and so it can't be evaluated for that value.

Why is this happening? Has it something to do with complex numbers? Or finding a solution of that equation is different to consider the function $x^x$ involved? Thank you.

答:
The problem is here:

the function $x^x$ doesn't exist for $x < 0$

This is wrong. Or, if you take it this way, then $x = -4$ cannot be and is not a solution, for then $(-4)^{-4}$ would have to be undefined as well.

You see, whenever you talk about functions, you cannot forget the domain - and in this case, the domain of $x \mapsto x^x$ for $x$ real is a bit tricky, and your question solution lies right in the tricky part. Everyone can agree that $(0, \infty)$ is part of it, but not everyone necessarily agrees on what the case is for other values of $x$. When you say

one can check by substitution [that it is indeed a solution]

you are presuming a larger domain than the one you just gave in order to do that "substitution", and that changes the answer to the question, and indeed it changes results like "that there is no solutions below $x = \frac{1}{e}$".

Now Wolfram does, indeed, presume a larger domain. In fact, it takes the domain to be all of $\mathbb{R}$, and moreover takes the codomain to be $\mathbb{C}$, so you are right that complex numbers play a role here.

Even more in fact, given that the domain is part of the definition of the function, you could say you and Wolfram are working with two different functions, that both happen to be denoted by the ambiguous notation $x^x$. Hence there is no surprise that there will be two different answers as to what the solution set is.


问:
How does the DISCRIMINANT really work?
I have the following equation (depended on param a)

$(a^2-2a)x^2+2ax-1=0$

I want to find out what the behavior of this equation after changing the value (by behavior I mean finding out how will the root count change when I put x or y instead of a). So for that purpose I thought it might be useful to find the D. If i'm not wrong when we have b = 2k ,then we can simply use this formula:

$D = k^2-ac = a^2+(a^2-2a)=2a^2-2a=2a(a-1)$

so after having this done, I thought maybe I can find out where my equation has only 1 root:

$2a(a-1)=0, a=0, 1$

but after putting for example a = 0, I quickly noticed that there my equation has no roots(real roots I think).

$a=0, -1 =0\Rightarrow x\in\varnothing$

Can someone  explain me what I've missed here? Shouldn't a = 0, work just fine (I mean give me only 1 root)?

答: The discriminant is valid only for quadratic equations. If $a=0$ then in your case the equation constant(degree 0). In this kind of exercise you should always impose the quadratic coefficient(that in your case is $a^2-2a$) to be non-zero, and study separately the zero case.(A little precisation the one you calculated isn't actually $D$, but $\frac D4$, the trick works because dividing by $4$ doesn't change the sign of the discriminant that it's actually the only thing that matters in this context, clearly if you choose to use $\frac D4$ the quadratic formula must be slightly modified)

Title: How does the DISCRIMINANT really work?


问:
Prime with digits reversed is prime?
Well, just another idea came up into my mind and i have no idea how to solve it :D
Is there infinitely many prime numbers, which are not repunits and their inverse is also prime? (For example, inverse of 31 is 13 which is also prime. i didn't have any other name to describe the function!)
P.S:I now know they are called Emirps.</p>

答:
Actually, it has a name and it's quaintly called an "emirp". (The word "prime" in reverse.)  The link given is to the Online Encyclopaedia of Integers' list,

$13, 17, 31, 37, 71, 73, 79, 97, 107, 113, 149, 157, 167,\dots$

While there are an infinite number of primes, I believe it is an open problem if there is an infinite number of emirps.

P.S. Regarding terminology, given $x$, then its "multiplicative inverse" (or "reciprocal") is $1/x$. For functions, for ex, given $\sin(x)$, then its "inverse function" is $\arcsin(x)$.

Title: Prime with digits reversed is prime?


问:
$\mathbb Z_{15}$ inside the non-abelian group of order 75.
I know that any abelian group of order $75$ has a $\mathbb Z_{15}$ inside it.I was just wondering if there is a $\mathbb Z_{15}$ inside the non-abelian group of order $75$.

If there is a normal subgroup of order $5$ inside it then I can take direct product with a $\mathbb Z_3$ to get a $\mathbb Z_{15}.$

Also,$Aut(\mathbb Z_5\times \mathbb Z_5)\cong GL_2(\mathbb F_5)$.I was trying to look if there is an order $3$ element inside $GL_2(\mathbb F_5)$,which has an eigenvector inside $ \mathbb Z_5\times \mathbb Z_5$.

Any help would be appreciated. Thanks in advance. If there is really a $\mathbb Z_{15}$ inside it, please  give me hints.I want to finish it myself.

答:
This is an ad-hoc answer, that does not require the knowledge of the structure of the (unique, up to isomorphism) non-abelian group of order $75$.

Let $G$ have order $75$. By Sylow's theorems, a Sylow $5$-subgroup $F$ is normal in $G$.

If $F$ is cyclic, since $3$ does not divide $\phi(25) = 20$, we have that $G$ is abelian.

So let $F$ be elementary abelian, so that it has $5 + 1 = 6$ subgroups of order $5$. Suppose $G$ has a subgroup $C$ of order $15$, which is thus cyclic. Then a Sylow $3$-subgroup $T$ acts by conjugacy on the $5$ subgroups of order $5$ of $F$ other than the one in $C$, so that it normalises at least one other, and thus centralises it.

We have obtained that $G$ is abelian if it has a subgroup of order $15$.


答:
The complete discussion of the equation $Ax^2+Bx+C=0$ in the real variable $x$ and in terms of the real parameters $A,B,C$ is:

if $A\ne 0$ and $B^2-4AC\ge0$, then the solution is $x=\frac{-B-\sqrt{B^2-4AC}}{2A}\lor x=\frac{-B+\sqrt{B^2-4AC}}{2A}$ (and these are equal if and only if $B^2-4AC=0$).

if $A\ne 0$ and $B^2-4AC<0$, then there are no solutions in $\Bbb R$.

if $A=0$ and $B\ne 0$, then the solution is $x=-\frac CB$.

if $A=0$ and $B=0$ and $C\ne 0$, then there are no solutions

if $A=B=C=0$, then all $x\in \Bbb R$ solve the equation.

So it isn't quite a matter of discriminant, but rather of what happens to the reduced equation when the leading term vanishes.

Remark: You are correct in saying that when $B$ is in the form $B=2\beta$, it's easier to calculate directly $\frac\Delta4=\left(\frac B2\right)^2-AC=\beta^2-AC$ and use $\frac{-B\pm\sqrt{\Delta}}{2A}=\frac{-\beta\pm\sqrt{\Delta/4}}{A}$.

Title: How does the DISCRIMINANT really work?

Body:
I have the following equation (depended on param a)

$(a^2-2a)x^2+2ax-1=0$

I want to find out what the behavior of this equation after changing the value (by behavior I mean finding out how will the root count change when I put x or y instead of a). So for that purpose I thought it might be useful to find the D. If i'm not wrong when we have b = 2k ,then we can simply use this formula:

$D = k^2-ac = a^2+(a^2-2a)=2a^2-2a=2a(a-1)$

so after having this done, I thought maybe I can find out where my equation has only 1 root:

$2a(a-1)=0, a=0, 1$

but after putting for example a = 0, I quickly noticed that there my equation has no roots(real roots I think).

$a=0, -1 =0\Rightarrow x\in\varnothing$

Can someone  explain me what I've missed here? Shouldn't a = 0, work just fine (I mean give me only 1 root)?


问:
An example of a Hilbert's space with application in ODE

Problem: I'm trying to prove that the space $H=\{v\in H^{1}(I): v(0)=v(1)\}$ is a Hilbert's space equpped with the $H^{1}$ scalar product.

We know that  and $$\forall u,v \in H^{1}(I):\quad \langle u,v \rangle_{H^{1}(I)}=\int_{I}uv+\int_{I}\nabla u \cdot \nabla v$$
and the induced norm is
$$||v||_{H^{1}(I)}^{2}=||v||_{L^{2}(I)}^{2}+||\nabla v||^{2}_{[L^{2}(I)]^{d}}$$

Let $\{v_{n}\}$ a Cauchy sequence in $H$, so I need to prove that $v_{n}\to v\in H$. Now, we pick a convergent sequence $\{v_n\} \subset H$ such that $v_n \to v$, so I need to show that $v \in H$. Now, how  $\{v_n\}$ converges in the norm $||\cdot ||_{H^1(I)}$  then $v_n \to v$ uniformly,

But how can I continue form here? Perhaps there is an easier way to approach the solution of the problem? or maybe I need to use a more strong result as the Sobolev-embedding?

Apologize me if my solution-approach is not right, I'm a bit confused with this space and your approach with the solution.

答:
Since you already know that $H^1((0,1))$ is complete, it suffices to show that $H$ is a closed subspace of $H^1$. For that purpose note that $H$ is the kernel of the continuous map
$$
H^1((0,1))\hookrightarrow C([0,1])\overset{\Phi}{\to}\mathbb{K},
$$
where the first arrow is the continuous embedding guaranteed by the Sobolev embedding theorem and $\Phi(v)=v(1)-v(0)$, which is obviously continuous (as discussed in the comments). The kernel of a continuous linear map is a closed subspace, so we are done.


问:
Induction Proof Help
Q: By induction on n prove that for all positive integers $n^2\ge1$

A: When n=1, $$n^2\ge1=1^2\ge1$$$$1\ge1$$

Suppose n=k, $$n^2\ge1=k^2\ge1$$

** This is as far as I can get unto without getting stuck, I know that I have to assume that n=k+1 for the last part, but it is what I do after that. Any help would be appreciated.

Title: Induction Proof Help

Body: If $n^2\geqslant1$, then$$(n+1)^2=n^2+2n+1\geqslant1+2n+1\geqslant1..$$


答:
Let the equation be $Ax^2 +Bx +C=0$

The discriminant $(D)$ you have calculated is wrong.

$D  = B^2-4AC=  (2a)^2 - 4(-1)(a^2 - 2a) $

$D = 4a^2+4a^2-8a = 8a(a-1)$

Secondly, $D=0$ implies 1 root only if $A  ≠ 0 $

because the roots are $\frac{-B+\sqrt{D}}{2A},\frac{-B+\sqrt{D}}{2A}$

The denominator cannot be $0$.

Therefore, to make the right use of determinant $A  ≠ 0 $

Title: How does the DISCRIMINANT really work?


问:
Compute number of comparisons in quicksort pivoting on median or third
With a few friends we read the Algorithm Design Manual from Skiena. One of his (chapter 4) exercises asks for the number of comparisons that the quicksort algorithm does (comparing an element to the pivot) in case (a) the median is used as pivot or (b) the one-third element is used as pivot.

(a) pivot on median (the best one can do)

I got this far: the number of comparisons is $T(n)=n + 2T((n-1)/2)\approx n + 2 T(n/2) = \Theta(n\log n)$.

But I think it's also possible to come up with an exact number. Does anybody know how to do that?

(b) I was not able to solve the following: $T(n)=n + T((n-1)/3) + T(2(n-1)/3)$, as I do not know how to solve this with two different T's.
And, also here I would also want to compute an exact number.

I wrote a small program to count the number of comparisons. It roughly looks like pivoting on the median, one-third, or one-tenth are only a constant factor apart, and pivoting on one-third is pretty close to $n\log_2{n}$.

If I'm allowed to paste links, here's my experimental result:
http://postimg.org/image/fr18zahp3/full/

Any help would be greatly appreciated ;)

答:
For the general case (pivoting on third, tenth or other constant fraction), consider the maximum depth quicksort will need to call itself recursively until it gets to an one-element array (which is sorted by definition). In each call, the size of its argument shrinks to at most constant multiple of the original size: $\frac{1}{2}$ in case of median, $\frac{2}{3}$ for the one-thirds approach or $\frac{9}{10}$ in the one-tenth case. Thus, the depth of the recursion is not going to exceed logarithm of the number of elements; the base of this logarithm is reciprocal of the shrinking factor. Obviously, every element of the array is going to be processed at most once at each "depth", so the upper bound on number of comparisons evaluates to $n\log_a n$ (where the base of logarithm is 2, 3, 10, ...).

Thanks to the nice property of logarithms, $\log_a n=(\log_a b)(\log_b n)$, all of these pivoting approaches are going to be within multiplicative factor of each other, belonging to $\Theta(n\log n)$. If you want to find the exact number of comparisons, you'll have to be very careful with your recurrence: expressions like $T(\frac{n-1}{2})$ don't make sense for even values of $n$.

I'll assume you want to solve the following recurrence in the "median" case: $$T(n)=n+T(\lfloor\frac{n-1}{2}\rfloor)+T(\lceil\frac{n-1}{2}\rceil)$$with initial condition $T(0)=0$. If that is the case, you can get further by considering how deep within the QuickSort recursion will each element be chosen as pivot and thus eliminated (it'll be compared with then-current pivots in all depths up to this one). Clearly, there is one pivot chosen in depth 1, two of them in depth 2, four in depth 4 and so on, all the way until all the sorted array is exhausted. The sum of their depths is then equal to the total number of comparisons. Unless I made a mistake, this results in $T(n)=nL-2^L+L+1$ when $2^{L-1}\leq n<2^L$ (Edit: The $+L$ term was missing originally).

Something similar can be done in the non-balanced cases too, although the calculation will most likely be even messier.


问:
Simpson's rule for improper integrals
Let $$F(x)=\int_{-\infty}^x f(t)dt,$$
where $x\in\mathcal{R}$, $f\geq 0$ is complicated (it cannot be integrated analytically).

Can I used the Simpson's rule to approximate this integral, knowing that $f(-\infty)=0$?

答: Yes. You can do a change of variable in your integral $\xi = x-\frac{1-t}{t}$, $d\xi = \frac{1}{t^2}dt$ and integrate for $0<\xi<1$. There are other methods too. See Wikipedia.

Title: Simpson's rule for improper integrals


问:
Fréchet derivative of $\|Au-f\|^2$
$A$ is a bounded linear operator on an infinite dimensional Hilbert space. I have been told the answer is supposed to be $2A^{*}(Au-f)$. Here is my progress:

$\|A(u+h)-f\|^2-\|Au-f\|^2=\langle A(u+h)-f, A(u+h)-f \rangle - \langle Au-f, Au-f \rangle = \langle Au-f,Ah \rangle + \langle Ah,Au-f \rangle + \|Ah\|^2$

Now $\|Ah\|^2 \leq \|A\|^2 \|h\|^2$ so this term is $o(h)$. I am not quite sure how to get to the last step.

答:
you have :
$$\|A(u+h)-f\|^2-\|Au-f\|^2= \langle Au-f,Ah \rangle + \langle Ah,Au-f \rangle + \|Ah\|^2\\
=\langle Ah,2(Au-f)\rangle \text{ by symmetry }\\
= \langle h,2A^*(Au-f)\rangle \text{ by the proprety of the adjoint}$$

Then, you have the result :
$$\|A(u+h)-f\|^2-\|Au-f\|^2=\langle df,h\rangle+o(||h||) $$

Hence you get your derivative :
$$ 2A^*(Au-f)$$

Title: Fréchet derivative of $\|Au-f\|^2$


问:
Quasilocal ring notation
Let $(R,M)$ be a quasilocal ring which is not a field.

Can anyone explain this notation $(R,M)$?

Edit: The result is Let $(R,M)$ be a quasilocal ring which is not a field. If $M$ is nilpotent then $\Omega(R)$ is complete graph.

$\Omega(R)$ is graph whose vertices are from set of all nonzero annihilating ideals (denoted by $A(R)*$) of a commutative ring $R$ which is not integral domain and two distinct vertices $I$ and $J$ are joined by an edge if $I+J\in A(R).$
($A(R)$ is set of annihilating ideals of $R$)

答:
The notation for "a quasilocal ring $(R,M)$" is shorthand for "a ring $R$ with a unique maximal ideal $M$."

It's similar to one notation used to specify a ring by elaborating on all of the data required to define it, like "$(R,+,\cdot)$".

It's worth mentioning that the usage of "quasilocal" has been declining since the 1960's and now will typically just appear as "local". The notation $(R,M)$ has persisted through the change, apparently.

In the course of writing the answer and participating in the comments, some users mentioned the possibility of using quasilocal to mean "finitely many maximal ideals."

I can say with some confidence that (in English) I have never seen that usage in a book or in an article. The term that does mean that for commutative rings, and which appears consistently is semilocal.

In a google search for "quasilocal ring" there were hits from Wolfram Mathworld and CommAlg which used the "multiple maximal ideals" version. However, I've found these two resources are not reliable. By that I mean this is not the first time I've found them to be inconsistent with the rest of algebra literature.  I could not find any usages of this sort in google books.  On the other hand, the Wikipedia support the usages in the way I'd expect.

Title: Quasilocal ring notation


问:
Closed form for $n$th derivative of exponential of $f$
What is the closed form for:

$$\frac{\partial^n}{\partial x^n}\exp(f(x))=\exp(f(x))\cdot[????]$$

答:
See the Wikipedia articles titled Exponential formula and Faà di Bruno's formula.  Both involve enumerations of set partitions.

It's simpler to express with $\dfrac{\partial^n}{\partial x_1\,\partial x_2\,\cdots\,\partial x_n}$ than with $\dfrac{\partial^n}{\partial x^n}$, but after doing that, just declare $x_1,\ldots,x_n$ to be all the same variable and then drop the subscripts and collect like terms.  So for example:
\begin{align}
& \frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} e^{f(x)} \\[10pt]
& = e^{f(x)}\left(\frac{\partial^3 f(x)}{\partial x_1\,\partial x_2\,\partial x_3} \right. \\[10pt]
& {} + \underbrace{\frac{\partial f(x)}{\partial x_1} \cdot \frac{\partial^2 f(x)}{\partial x_2\,\partial x_3}}
+ \underbrace{\frac{\partial f(x)}{\partial x_2} \cdot \frac{\partial^2 f(x)}{\partial x_2\,\partial x_3}}
+ \underbrace{\frac{\partial f(x)}{\partial x_3} \cdot \frac{\partial^2 f(x)}{\partial x_1\,\partial x_2}} \\[10pt]
& \left.{} + \frac{\partial f(x)}{\partial x_1} \cdot \frac{\partial f(x)}{\partial x_2} \cdot \frac{\partial f(x)}{\partial x_3} \right).
\end{align}

You get one term for each partition of the set of three variables.  For the fourth derivative, there's one for each of the $15$ partitions of the set of four variables, etc.

Consequently
$$
\frac{\partial^3 f(x)}{\partial x^3} = e^{f(x)}\left( \frac{\partial^3 f(x)}{\partial x^3} + 3 \frac{\partial f(x)}{\partial x}\cdot\frac{\partial^2 f(x)}{\partial x^2} + \left( \frac{\partial f(x)}{\partial x} \right)^3  \right),
$$
and similarly with the fourth derivative the sum of the coefficients is $15$, etc.

Later note: Here's a summary of the proof: Mathematical induction on the order of the derivative.


问:
Continuity & Domain: What Is Their Relationship?
It's hard for me to formulate this into a specific question, so I'll split this into two sub-questions that will hopefully explain my confusion.

Say I have the function $\frac{x}{x}$. As far as my understanding goes, this guy is discontinuous at $x=0$, but is continuous at any other point on the real line (with both limits & function values being $1$); the limit is $1$ at $x=0$ as well, but the function's value is not (nor is it anything, seeing as it is not defined there), which is why it's not continuous at that point.

Assuming the above is true, I can look at $\frac{x}{x}$ as the product of, for instance, $f(x)=x$ and $g(x)=x^{-1}$, and say that it is discontinuous at $x=0$ simply because the domain of $g$ (and therefore, the domain of the product) does not include $0$.

A discussion I was reading seemed to conclude that, for a function $g(x)$ discontinuous at point $a$ and a continuous function $f(x)$, the product $f(x) \cdot g(x)$ is discontinuous at $a$ if neither $f(x)$ nor $g(x)$ are $0$, but is continuous $\forall x \in \mathbb{R}$ (incl. a) if, e.g, $f(x)=0$ for all $x \in R$. Though, to my understanding, this is only true if the discontinuity isn't caused by a point in which either function is undefined. For example, if $f(x)=0$ and $g(x)=\frac{1}{x}$, then it would still be wrong of me to say that $f(x) \cdot g(x)$ is continuous at $0$, because $0$ should be out of its domain. Is this right?

And for the second part of the question: I initially began this question with "Say I have the function $\frac{x}{x}$ over $\mathbb{R}$", but then felt extremely unsure about whether I could even say that, seeing as it is not defined at $x=0$ and a function needs to be defined over the entirety of its own domain. In the case of the function mentioned above, the domain must therefore not include $0$, so it could, for example, be $\mathbb{R} \setminus \{0\}$. But then this raises another question: I saw a mention of the Dirichlet function being continuous over the rationals; but if this is the case then continuity is domain-dependent, and so if I define $\frac{x}{x}$ over $\mathbb{R} \setminus \{0\}$, then it is continuous.

I can tell there's a difference here: the function is continuous over the entirety of its own domain; but it then gets a bit more confusing when I have a product of two different functions with different domains.

Please help me understand the relationship between domain & continuity, and/or point out any inconsistencies in my understanding described above.

答:
You are right that $\frac xx$, or rather the function conventionally meant by that term, is continuous. But I think there is a conceptual issue at play here, and a common one. Namely, that the functional equation
$$f(x)=\textrm{some expression in }x$$
already determines the function. This is not true. When building a function, we specify things in this order: domain and codomain first, functional equation second. That is, we choose the domain and codomain of the function, and then we are bound to give a functional equation which is actually defined for all elements of the domain, and returns only elements of the codomain. In mathematical notation you'll see it like $f:X\to Y,~f(x)=\dots$, where $X,Y$ are domain and codomain, and they are specified first. Now this said, $f:\mathbb R\to\mathbb R,~f(x)=\frac xx$ is not a well-defined function, since the functional equation isn't meaningful for all elements of the domain. However, $f:\mathbb R\backslash\{0\}\to\mathbb R,~f(x)=\frac xx$ is well-defined, since the functional equation is meaningful for all elements of the previously specified domain, and it returns only elements of the previously specified codomain. So to be sensible, when we write "the function $\frac xx$", we usually mean that as shorthand for the latter definition of a function. But technically speaking, just the term $\frac xx$ does not define a function, since we haven't specified domain or codomain, which we must, unless we want our readers to guess (which isn't always unfeasible, to be fair).

Now with all this about the definition of functions out of the way, continuity can be treated more clearly: a function $f:X\to Y$ is continuous at $x\in X$ if [insert your favorite definition of continuity at a point]. A function $f:X\to Y$ is continuous if it is continuous at all $x\in X$.

Notice that this definition only talks about points inside the domain $X$, at which the function must be defined in order to be a function. Whatever lies outside of the domain is of no concern for the definition of continuity.

As a closing remark, the often cited example of the discontinuous function $\frac 1x$ is just bad, because $f:\mathbb R\to\mathbb R,~f(x)=\frac 1x$ is not a function, while $f:\mathbb R\backslash\{0\}\to\mathbb R,~f(x)=\frac 1x$ is a perfectly continuous function. There just isn't any sensible (meaning non-contrieved) way to interpret $\frac1x$ as something that is both a function and discontinuous.


问:
Prove that there there are two $a,b$ so $f''(a)=f''(b)=0$
I'm trying to prove the following theorem:

Let $f$ be a function that can be derived twice. We know that $f$ is even function and has local minimum at $x=2$. Prove that there there are two $a,b$ so $f''(a)=f''(b)=0$.

I think that I need to use the Rolle's theorem here but after trying for one hour, I can't seem to figure out how to do it.

答:
Hint: $f$ has a local minimum at $-2$ also. Any even differentiable functions satisfies $f'(0)=0$. Thus $f'(-2)=f'(0)=f'(2)=0$. Can you finish?

I am assuming that the domain of your function is $\mathbb R$.


答:
This is not necessarily true. Let:

$$f(x)=\begin{cases}
(x+2)^2 \ \ \text{if} \ \ x\in(-2-\frac12,-2+\frac12) \\
(x-2)^2 \ \ \text{if} \ \ x\in(2-\frac12,2+\frac12)
\end{cases}
$$

Where:
$f:(-2-\frac12,-2+\frac12) \cup (2-\frac12,2+\frac12) \to \mathbb{R}$

Body:
I'm trying to prove the following theorem:

Let $f$ be a function that can be derived twice. We know that $f$ is even function and has local minimum at $x=2$. Prove that there there are two $a,b$ so $f''(a)=f''(b)=0$.

I think that I need to use the Rolle's theorem here but after trying for one hour, I can't seem to figure out how to do it.


问:
Doubt on free variable in 3x3 matrix eigenvectors
First question here. So I've been asked to find the eigenvectors for the following 3x3 matrix A:
\begin{bmatrix}
    1 & 4 & 5 \\
    0 & 2 & 6 \\
    0 & 0 & 3 \\
    \end{bmatrix}

I found the characteristic equation and found the eigenvalues to be $\lambda$=1, 2 and 3.
Just as an example of my issue, let's take $\lambda$=2. Plugging that into A-I$\lambda$ to find the eigenvectors, I can get the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
  -1 & 4 & 5 &  0 \\
    0 & 0 & 6 & 0 \\
    0 & 0 & 1 & 0 \\
\end{array}
\right] $$

This is where I get stuck. I honestly have no idea what the process is to pick a free variable out of the 3 (let's call them x, y and z). Could someone please run me through the process? I've found that I never seem to get these right. A general method to know which variable is the free variable would also be great if it's not too much to ask.

Thanks in advance!

答:
Forget about free variables for the moment, and try to think about what the augmented matrix represents. With enough practice with examples of this kind, it will become clear which variables are going to be the free variables. In your working above, you got to the augmented matrix
$$ \left[ \begin{array}{ccc|c} -1 & 4 & 5 & 0 \\ 0 & 0 & 6 & 0 \\ 0 &0 &1&0 \end{array} \right],$$
which is correct. With a few more elementary row operations (if you want the details, just post a comment and I will edit my answer to include these operations) we get
$$ \left[ \begin{array}{ccc|c} -1 & 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 &0 &0&0 \end{array} \right].$$
Now, reading off this matrix, we see that $z=0$ and $-x + 4y =0.$ So we see that
$$\left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 4y \\ y \\ 0 \end{array} \right) = y\left( \begin{array}{c} 4 \\ 1 \\ 0 \end{array} \right),$$
showing that $y$ is the free variable in this case. Further, we now see that
$\left( \begin{array}{c} 4 \\ 1 \\ 0 \end{array} \right)$ is an eigenvector corresponding to the eigenvalue $\lambda = 2.$


问:
What is the minimum number of configurations which connects all elements?
I am facing a problem which I can't seem to resolve. I'm no mathematician, so maybe I am missing something obvious.

My problem is defined as follows: I have a list of 10 nodes (0,1,2,3,4,5,6,7,8,9), which all have to be connected to each other. A connection can be specified by stating for example (0) -> (1), which connects the nodes 0 and 1. Multiple connections can be defined in a single line as well: e.g. (0,1,2,3,4) -> (5,6,7,8,9). This last statement defines that every single node in the first set is connected to each node of the second set. Both sets can have a maximum of 5 elements each. Also, a node cannot be in the first set and second set on the SAME line. So (1,2,3) -> (2,4,5) is not permitted. Lastly, a line only defines a connection in a single direction; so (0) -> (1) only connects 0 to 1 and another statement has to be defined to connect 1 to 0 (by entering the line (1) -> (0) for example)

My configuration now is:

(0,1,2,3,4) -> (5,6,7,8,9);

(5,6,7,8,9) -> (0,1,2,3,4);

(0) -> (1,2,3,4);

(1) -> (0,2,3,4);

(2) -> (0,1,3,4);

(3) -> (0,1,2,4);

(4) -> (0,1,2,3);

(5) -> (6,7,8,9);

(6) -> (5,7,8,9);

(7) -> (5,6,8,9);

(8) -> (5,6,7,9);

(9) -> (5,6,7,8);

This way there are 12 lines needed to connect every node to each other in both directions. However, it seems like this is not the most optimal solution, because the last 10 lines don't use the full allowed maximum of 5 nodes per set.

Is there any way I can find an optimal solution regarding this problem? Any help is greatly appreciated!

Thank you in advance

答: In the language of graph theory, it seems that you want to decompose the complete directed graph into a union of maximal matchings. The graph case is discussed on MathOverflow. To extend to the directed graph case, just run the construction twice: once where you connect small vertices to large ones (e.g. connect 1 to 2) and once where you connect large vertices to small ones (e.g. connect 2 to 1).</p>


问:
If $f(u_n)$ is bounded, why it is true that $(u_n)_n$ is bounded in $W_0^{1,p}(\Omega)$?
Let $\Omega$ be an open bounded domain in $\mathbb{R^N}$, $p\geq 1$ and $(u_n)_n\subset W_0^{1,p}(\Omega)$. I would like to show that $(u_n)_n$ is bounded in $W_0^{1,p}(\Omega)$ and the only thing I know is that
$$\| u_n\|_{W_0^{1,p}(\Omega)} +\int_{\Omega} f(u_n) dx = c + o(\| u_n\|_{W_0^{1,p}(\Omega)} +1),$$
where $c$ is a positive constant and $f:\Omega\to\mathbb{R}$ is a continuous function.

It is written on my notes that in order to show $(u_n)_n$ is bounded in $W_0^{1,p}(\Omega)$ it is enough to prove that the function $f$ is bounded.

I am trying, but I am not able to understand why this late statement is true, I mean, why it is enough just to show that $f$ is bounded? In addition, bounded from above and\or from below?

Could anyone please help or give some hints?

Thank you in advance!

答:
Dividing the equation
$$\| u_n\|_{W_0^{1,p}(\Omega)} +\int_{\Omega} f(u_n) dx = c + o(\| u_n\|_{W_0^{1,p}(\Omega)} +1)$$
by $\| u_n\|_{W_0^{1,p}(\Omega)} +1$, we find that
$$ \limsup_{n\to\infty} \left[\frac{\| u_n\|_{W_0^{1,p}(\Omega)}}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} + \frac{1}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1}\int_{\Omega}f(u_n)~dx\right]  = \limsup_{n\to\infty} \frac{c}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1}.$$
Now suppose that there is a subsequence, again denoted by $u_n$, such that $\lim_{n\to\infty} \| u_n\|_{W_0^{1,p}(\Omega)} = \infty.$ Then, as $f$ is bounded we have
$$ \limsup_{n\to\infty} \left[  \frac{1}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1}\int_{\Omega}f(u_n)~dx\right] \leq \limsup_{n\to\infty} \frac{\|f\|_{\infty}|\Omega|}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} = 0. $$
We clearly also have
$$ \limsup_{n\to\infty} \frac{c}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} = 0,$$
and as
$$ \limsup_{n\to\infty} \frac{\| u_n\|_{W_0^{1,p}(\Omega)}}{\| u_{n}\|_{W_0^{1,p}(\Omega)}+1} = 1, $$
we reach the contradiction $1 = 0$. It follows that there is no subsequence of $(u_n)_{n \geq 1}$ such that $\lim_{n\to\infty} \| u_n\|_{W_0^{1,p}(\Omega)} = \infty$, hence $(u_n)_{n\geq 1}$ is bounded in $W_0^{1,p}(\Omega)$.


问:
General solution of this differential equation
I got a bit stuck at this equation below :

$$ty^{'}\cos(y) + \sin(y) =\sin^2y,y(0)=1$$

This is what I did :

$$ty^{'}\cos(y) = \sin^2(y) - \sin(y)$$
$$t\frac{dy}{dt}\cos(y) = \sin(y)( \sin(y) - 1 ) $$
$$t\frac{dy}{dt} = \frac{\sin(y)( \sin(y) - 1 )}{\cos(y)} $$
$$\frac{\cos(y)}{\sin(y)( \sin(y) - 1 )} . dy= dt \frac{1}{t}$$
$$\int_{}{\frac{\cos(y)}{\sin(y)( \sin(y) - 1 )} . dy}= \int_{}{\frac{1}{t} dt }$$
$$ ln|1-\frac{1}{\sin(y)}|= ln|t| + C$$
Assuming $1 > 1/\sin(y)$
=> $$  \frac{1}{\sin(y)}= 1-e^Ct $$
$$(C =  e^C )$$

=>  $$  \frac{1}{\sin(y)}= 1 - C .t $$
$$  \sin(y)=  \frac{1}{1 - C .t} $$
$$    y   =  \sin^{-1}(\frac{1}{1 - C .t}) $$
Right now it's impossible to find C in this case , even if I try to set y to $0$ in the final equation I get :
$$y(0) = \sin^{-1}(1) $$
$$y(0) = \frac{\pi}{2} ≠ 1 $$

So now I have 2 problems :

y(0) is different than 1

C is impossible to find in my case .

What am I doing wrong ?

答:
The picture below is a sketch of the numerical solution in the interval $[-10,10]$

I had to modify the initial value because with the given one I got no solutions.

$$t y' \cos y+\sin y=\sin ^2y,\;y(1)=1$$


答: Using estimates on $f$, you can obtain some $M>0$ such that $\int_{-\infty}^{-M}f(t) dt < \frac{\varepsilon}{2}$ and then use Simpson's rule with enough points so that you approximate $\int_{-M}^x f(t) dt$ with an error smaller than $\frac{\varepsilon}{2}$. This guarantees an overall error smaller than $\varepsilon$.

Title: Simpson's rule for improper integrals

Body:
Let $$F(x)=\int_{-\infty}^x f(t)dt,$$
where $x\in\mathcal{R}$, $f\geq 0$ is complicated (it cannot be integrated analytically).

Can I used the Simpson's rule to approximate this integral, knowing that $f(-\infty)=0$?


问:
how to prove the multiplication formula for the cotangent function
I could not find the proof of the following formula in the internet:

$$\frac{k\tan x}{\tan kx}=1+\frac{1}{2}\sum_{0<j<k} \frac{\tan x}{\tan (x+\frac{\pi j}{k})}+\frac{\tan x}{\tan (x-\frac{\pi j}{k})}$$

The source I found this formula is here: page 5, proof of (i)

答:
Claim:
$$\tan(\theta)+\tan \left(\theta+ \frac{\pi}{n} \right) + \tan\left(\theta + \frac{2\pi}{n}\right) + \dots + \tan \left (\theta + \frac{(n-1)\pi}{n} \right) = -n\cot \left(\frac{n\pi}{2} + n\theta \right)$$

Proof:  Sum of tangent functions where arguments are in specific arithmetic series

$$\sum_{0<j<k} {\tan (x+\frac{\pi j}{k})}=-\tan x-k\cot(\frac{k\pi}{2}+kx)$$

$$\sum_{0<j<k} {\tan (x-\frac{\pi j}{k})}=-\sum_{0<j<k} {\tan (-x+\frac{\pi j}{k})}=-\tan x-k\cot(kx-\frac{k\pi}{2})$$

Now remember that $\cot(kx + \frac{k\pi}{2}) = \tan  kx $ or -$\tan kx$ depending on wether $k$ is odd or even. So make 2 cases and solve.


问:
Confluent hyper-geometric function
Is there any identity in the confluent hypergeometric function that connects the following?
${}_1{F_1}\left( {a + n + 1,b + n + 1, - c} \right)$ and
${}_1{F_1}\left( {a + 1,b + 1, - c} \right)$ where a, b and c are positive real numbers

Title: Confluent hyper-geometric function

Body:
Note that
$$
\frac{d}{dx}\;{}_1F_1(a,b,x) = \frac{a}{b}\;{}_1F_1(a+1,b+1,x)
$$
Apply this inductively to get
$$
\frac{d^n}{dc^n}\;{}_1F_1(a+1,b+1,-c) =
(-1)^n\;\frac{(a+1)(a+2)\cdots(a+n)}{(b+1)(b+2)\cdots(b+n)}
{}_1F_1(a+n+1,b+n+1,-c)
$$


问:
Means and standard deviations.
Given three independent random variables $X_k$, the means and standard deviations respectively are $X_1 = (A, X)$, $X_2 = (B, Y)$, $X_3 = (C,Z)$.

(1)What is the mean of their sum?

(2)What is the deviation of their sum?

Is it just as simple as (1) $= A + B+ C$ and (2) $= X + Y + Z$

答:
$E(X_1 + X_2 + X_3) = A + B + C$ because of linearity of the expectation operator.

$V(X_1 + X_2 + X_3) = X^2 + Y^2 + Z^2$

where

$V(.)$ denotes the variance.

Title: Means and standard deviations.


问:
How to define the parity of a permutation in a finite group $G$?

How to define the parity of a permutation in a finite group $G$?

I know that there is an isomorphism $\phi$ between $S(G)$ and a subgroup of $S_n$, so we could say $\epsilon_{G}(\sigma):=\epsilon_{S_n}(\phi(\sigma))$ for $\sigma \in S(G)$ and where $\epsilon_{S_n}$ is the usual parity function. I don't know if there is a usual way to do so.

I didn't find any reference, does someone knows a textbook that makes this generalization?

答:
In the abstract, group elements don't have a parity. You can't establish one by embedding the group in a symmetric group because that can be done in several ways. For example, you can embed $\mathbb{Z}_3$ in $S_3$ or in $S_6$ isomorphically  by sending a generator to $(123)$ or to  $(123)(456)$.

However, for the map from a group $G$ to the permutations $S_G$ of itself that sends $g$ to the permutation $x \to xg$ you can use the usual definition of parity in $S_G$.

I suggest you work out a few examples of that.


问:
Prove that $\inf _{n \geq k} (a_n + b_n) \geq \inf _{n \geq k} (a_n) + \inf _{n \geq k} (b_n) (\forall k)$
I understand that

$\inf _{n \geq k}  a_n \leq a_n (\forall n \geq k)$

$\inf _{n \geq k}  b_n \leq b_n (\forall n \geq k)$

Thus

$\inf _{n \geq k}  a_n + \inf _{n \geq k}  b_n \leq a_n + b_n (\forall n \geq k)$

I'm not too sure where to go from here.

答:
Fix $k$. for any $N\ge k$, it holds that
$$\inf_{n\ge k}a_n+\inf_{n\ge k}{b_n}\le a_N+b_N.$$
So $\inf_{n\ge k}a_n+\inf_{n\ge k}{b_n}$ is a lower bound of the set $\{a_N+b_N:\, N\ge k\}$. Since $\inf$ is the ``best'' lower bound, it follows that
$$\inf_{n\ge k}a_n+\inf_{n\ge k}{b_n}\le\inf_{N\ge k}(a_N+b_N).$$


答:
Your last line states that $\inf_{n\geq k}a_{n}+\inf_{n\geq k}b_{n}$ is a lower bound of the set $\{a_n+b_n\mid n\geq k\}$.

Then it is legal to conclude that it does not exceed the greatest lower bound of that set.

So in mathematical notation: $$\inf_{n\geq k}a_{n}+\inf_{n\geq k}b_{n}\leq\inf_{n\geq k}(a_n+b_n)\text{ QED}$$

Title: Prove that $\inf _{n \geq k} (a_n + b_n) \geq \inf _{n \geq k} (a_n) + \inf _{n \geq k} (b_n) (\forall k)$

Body:
I understand that

$\inf _{n \geq k}  a_n \leq a_n (\forall n \geq k)$

$\inf _{n \geq k}  b_n \leq b_n (\forall n \geq k)$

Thus

$\inf _{n \geq k}  a_n + \inf _{n \geq k}  b_n \leq a_n + b_n (\forall n \geq k)$

I'm not too sure where to go from here.


问:
Can bounded analytic function defined on a simply connected region be approximated by bounded polynomials?
Let $G$ be a simply connected bounded region in the complex plane (say the unit disk centered at $0$). Suppose $f$ is analytic on $G$, then $f$ can be locally uniformly approximated by polynomials (just take truncations of its power series).

Question: is it true that if  $f$ is bounded on $G$, it can be approximated by polynomials $\{p_n\}$ on the whole region $G$ such that $\sup_n\sup_{z\in G}|p_n(z)|＜\infty$?

By Runge's theorem, global polynomial approximation exists on simply connected domains. But I am not sure how to determine boundedness..

答:
This is true for the circle (one needs Caratheodory's kernel theorem to prove it), but not in general.
To see why, let $A:=\mathbb{D}-[0,1)$ open and  simply connected (since its complement is connected). There exists a conformal map $\varphi: \mathbb{D}\to A$ and it admits a continuous extension to $\overline{\mathbb{D}}\to \overline{A}$ by Carathéodory-Torhorst. Note in particular that $\varphi^{-1}$ is bounded.

Define now $f$ as a Blaschke product (thus bounded) having $X=\cup\{(1-1/n^2)e^{iq_n}\}$ as zeros (where $q_n$ is an enumeration of the rationals in $[0,2\pi)$).
This function is analytic bounded on $\mathbb{D}$ and non-extendible.
Since $\varphi$ has a continuous extension to $\overline A$, $\overline{\varphi(X)}
\supseteq\varphi(\overline{X})\supset \varphi(\partial\mathbb{D})=\partial A$.

Now, let $g=f\circ \varphi^{-1}$, which is clearly bounded and holomorphic on $A$. Suppose thre exists a sequence of bounded polynomials $p_n\to f\circ\varphi$. By continuity, $\{p_n\}$ must be bounded on $\overline{A}=\overline{\mathbb{D}}$. Montel's theorem implies the existence of a converging subsequence $p_{n_k}$. The limit must be an extension of $f\circ\varphi^{-1}$, and it must be analytic on $\mathbb{D}$.
Since $[0,1)\subset \overline{g^{-1}(0)}$, the identity principle implies $g\equiv 0$, a contradiction.

This is true in general: a bounded function on an open connected set $A$ is the limit of a bounded sequence of polynomials only if it admits a bounded holomorphic extension to $\hat{\mathbb{C}}-
\overline{(\hat{\mathbb{C}}-\overline{A})}$. This is actually a iff but the other direction of the implication is longer.


问:
Find the value of $\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx$
I'm trying to figure out how to evaluate the following:
$$
J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx
$$
I'm tried considering $I(s) = \int_{0}^{\infty}\frac{x^3}{(e^x-1)^s}\,dx\implies J=-I'(1)$, but I couldn't figure out what $I(s)$ was. My other idea was contour integration, but I'm not sure how to deal with the logarithm. Mathematica says that $J\approx24.307$. </p>

<p>I've asked a similar question and the answer involved $\zeta(s)$ so I suspect that this one will as well. </p>

答:
How about pulling factors of $e^{-x}$ from both the denominator and log terms?  Then you end up with two separate integrals:

$$\int_0^{\infty}dx \frac{x^4 \, e^{-x}}{1-e^{-x}} + \int_0^{\infty}dx \frac{x^3 \, e^{-x}}{1-e^{-x}} \log{(1-e^{-x})}$$

In both cases, you Taylor expand the denominator in $e^{-x}$.  For the first integral, this results in

$$\sum_{k=0}^{\infty} \int_0^{\infty}dx\, x^4 \, e^{-(k+1) x} = 4! \sum_{k=0}^{\infty} \frac{1}{(k+1)^5} = 24 \, \zeta(5) $$

For the second integral, you also need to Taylor expand the log term.  This results in a double sum:

$$\begin{align}\sum_{k=0}^{\infty} \int_0^{\infty}dx\, x^3 \, e^{-(k+1) x} \log{(1-e^{-x})} &= -\sum_{k=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m} \int_0^{\infty} dx \, x^3 e^{-(k+m) x}\\ &= - 3! \sum_{m=1}^{\infty} \frac{1}{m} \sum_{k=1}^{\infty} \frac{1}{(k+m)^4}\\ &= -\sum_{m=1}^{\infty} \frac{\psi^{(3)}(m+1)}{m} \end{align}$$

where $\psi$ is a polygamma function.


问:
Finding continuous bijections between these funtions

Let $X=\bigcup_{n=0}^{\infty} (3n ,3n+1) \bigcup \bigcup_{n=0}^{\infty}${3n+2} and $Y =(0,1]  \bigcup \bigcup_{n=1}(3n, 3n+1) \bigcup \bigcup_{n=0}^{\infty} $ {3n+2} be subsets of $\mathbb{R}$  with usual subspace topology.

Then:
(a) Prove that there is a continuous bijection f : X-> Y.

(b) Prove that there is a continuous bijection from Y-> X.

For (b) I took f(x)=x. but This is not continuous. For (a) this will not work due to 1 $\in Y$ and I am not able to think of such a function for (a).But I can take f(x) =x/2 except for the points with {3n+2} where  I can take f(x)= x.

But I am not able to solve the problem of continuity.

Thanks!

答:
For (a), map each open interval to itself, map $2$ to $1$, and map $3n+2$ to $3n-1$ for $n\ge 1$.

For (b), just set $f(y)=2$ for all $y\in Y$. (This is where you kick yourself!)


问:
Existence of an injection between $\mathbb{Z}$ and $[0,1]$
Is the statement that there is an injection between $\mathbb{Z}$ and $[0,1]$ true or false

$\mathbb{Z}\xrightarrow{(1-1)} [0,1]$

i've build it on the definition of cardinality that is if $\lvert A \lvert \leq \lvert B \lvert$

then there exists an injection $f:A\longrightarrow B$ and one knows that

$\lvert \mathbb{Z} \lvert \leq \lvert [0,1]\lvert$

thus there exists an injection

答: It is true, take for example a function $f:\mathbb{Z} \to [0,1]$ $$f(n)=\cases{{1\over 2|n|+1};\;\;\; n\leq0\\{1\over 2n};\;\;\; n>0}$$


问:
What is a Logarithmic Series?
I know very well what a logarithmic function is, but I don't understand how it's meaning is extended into the concept of algebraic series. I also learnt about the exponential series and how its general form is derived, but how do we come to this:

$$\log_e (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$
P.S.: Even though it makes such questions a lot more solvable and understandable, I have not studied the chapter of binomial theorem, so please post your answers accordingly and bear with me.

答:
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots~~~~~-1<x\le1$$
I assume that is what you mean?

I can prove it if you like.

If there's anything you still need help with, please don't hesitate to ask.

$2$ PROOFS

If you are familar with Maclaurin series expansion (or Taylor series of you prefer, although strictly speaking the term Macluarin is used specifically when the power series is centered arbout $0$, as in our case), recall that if a function $f(x)$ can be infinitely differentiated and is defined at $f^{(n)}(0)$ for all positive integer values of $n$ then $f(x)$ can be written as
$$f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\cdots$$
Let us apply this to $\ln(1+x)$ to obtain a series expansion for it:
$$f(x)=\ln(1+x)\implies f(0)=0$$
$$f'(x)=\frac{1}{1+x}\implies f'(0)=1$$
$$f''(x)=\frac{-1}{(1+x)^2}\implies f''(0)=-1$$
$$f'''(x)=\frac{2!}{(1+x)^3}\implies f'''(0)=2!$$
As is clearly seen, in general
$$f^{(n)}(0)=(-1)^{n+1}(n-1)!$$
for all $n\in\mathbb N$.
Now it is simple to see that the series expansion for $\ln(1+x)$ is
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$
as required.

Alternatively, we can use the binomial theorem (which actually is a Maclaurin series as well!). Consider $\ln(1+x)$. We know that
$$\frac{d}{dx}\ln(1+x)=\frac{1}{1+x}$$
Hence,
$$\int\frac{1}{1+x}dx=\ln(1+x)+c$$
But using the binomial theorem we can write the integral as
$$\int~1-x+x^2-x^3+\cdots dx=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$
Hence $$\ln(1+x)+c=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$
for some constant $c$. We can plug in $x=0$ to find that $c=0$. Hence,
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$
as required.

Now for matters of divergence and convergence. You can apply the ratio test very simply or just use the fact that the binomial series expansion is only valid on this case for $-1<x<1$. I don't know how to prove that the series is convergent for $x=1$, as the ratio test is inconclusive. Ot may be worth looking that point up.

I hope that helped. Please ask if you have any questions.

See here for a detailed explanation of Macluarin/Taylor series. See if, using it, you can derive the binomial theorem.
https://en.wikipedia.org/wiki/Taylor_series

Title: What is a Logarithmic Series?


问:
Area of a hexagon from the distances of the opposite sides
Problem. All the six sides of the hexagon in the attached figure have equal length, and the opposite sides are parallel. The distances of the three opposite pairs are $d_1=7\mathrm{cm}$, $d_2=8\mathrm{cm}$ and $d_3=9\mathrm{cm}$, respectively. Find the area of the hexagon.

Clearly,
$$
\text{Area}=\frac{1}{2}x(d_1+d_2+d_3),
$$
where $x$ is the length of each of the sides. Hence, it suffices to find $x$ as a function of $d_1,d_2$ and $d_3$.

EDIT. The area in the particular case $d_1=7\mathrm{cm}$, $d_2=8\mathrm{cm}$ and $d_3=9\mathrm{cm}$ is provided in the answer of Jack D'Aurizio that follows. I was wondering whether it is possible to obtain an expression the area of the hexagon as a function of $d_1,d_2$ and $d_3$.

答:
We can derive the same general solution with an approach using some trigonometry.  I handled the consequent algebra manually. The idea is to write $A$ the area of hexagon in terms of the values of the angles and $s$ the side length of the hexagon, and then writing everything in terms of $A$.

Due to the parallel lines, Interior angle at $N = ∠ONQ +∠MNQ =∠RQN +∠PQN =$ Interior angle at $Q$.  The same applies to the other two pairs of interior opposite angles. Hence we have 3 pairs of equal angles, whose values we’ll call $\alpha,\beta,\xi$ , as colour-marked in the figure.

Now the Sum of interior angles in a hexagon $= 6\times180°-360° = 720°=4π= 2\left(\alpha+\beta+\xi\right)$. Thus:

$\beta=(\pi-\alpha)+(\pi-\xi) \tag{1} $

Geometrically, $\beta=\angle ZPO+\angle ZPQ\ =\ \angle POI+\angle PQJ\ =(\pi-\xi)+(\pi-\alpha) , where \ ZP:\parallel NO \ and \ RQ\  $.

Now $NR$ and $OQ$ divides the hexagon into a quadrilateral and 2 triangles. Since $NO$ is parallel to and is the same length as $RQ$, $RQ$ is a mere translation of $NO$, thus $NR\parallel OQ$ and $NOQR$ is a parallelogram. The Area of $NOQR = d_1s$.

$\triangle MNR$ and $\triangle POQ$ are both isosceles with $s$ as two of their side lengths and $\beta$ as the angle between the same two sides. If the base of $\triangle MNR = MR =\ s$, then the corresponding height $= NG = s\sin{\beta}$.  Thus the Area of $\triangle MNR =\triangle POQ=\ \frac{s^2\sin{\beta}}{2}$.

We can similarly divide the hexagon using $\alpha$ or $\xi$ as the angle of the isosceles triangles, ie the hexagon can be divided likewise in 3 ways (as shown by the blue, yellow and orange lines). Thus:

$A=d_1s+s^2\sin{\beta} \tag{2}$

$A=d_2s+s^2\sin{\xi} \tag{3}$

$A=d_3s+s^2\sin{\alpha} \tag{4}$

Consider $ A\ \mathrm{\mathrm{=}}\frac{1}{2}s\left(d_1+d_2+d_3\right)$.  For simplicity in the algebra that follows, let’s define $k:$

$$k\ =\ \frac{d_1+d_2+d_3}{2}=\ \frac{A}{s} \ \rightarrow \ s=\frac{A}{k} \tag{5}$$

Substitute (1) into (2), expand using the trigonometric angle-sum identity and simplify:

\begin{align}
A &=d_1s+s^2\sin{\left[(\pi-\alpha)+(\pi-\xi)\right]}\\
 &=d_1s+s^2\left[\sin{(\pi-\alpha)}\cos{(\pi-\xi)}+\cos{(\pi-\alpha)}\sin{(\pi-\xi)}\right]\\
&=d_1s+s^2(-\sin{\alpha}\cos{\xi}-\ \cos{\alpha}\sin{\xi})\\
\end{align}
$A=d_1s-s^2\sin{\alpha}\cos{\xi}-s^2\cos{\alpha}\sin{\xi} \tag{6}$

Rearranging (3) gives $\sin{\xi}=\frac{A-d_2s}{s^2}$. Substituting (5) gives:

$$ \sin{\xi}=\left(A-\frac{d_2A}{k}\right)\left(\frac{k^2}{A^2}\right)=\frac{k(k-d_2)}{A} \tag{7}$$

Thus

$$\cos^2{\xi}=1-\frac{k^2{(k-d_2)}^2}{A^2} \tag{8}$$

Similarly,

$$\sin{\alpha}=\frac{k(k-d_3)}{A} \tag{9}$$

$$\cos^2{\alpha}=1-\frac{k^2{(k-d_3)}^2}{A^2} \tag{10}$$

Thus every term in (6) can be written in terms of $A$. So it's a matter of substituting and solving for $A$.

Substituting (5), (7) and (9) into (6) and simplifying, we can re-write (6):

$$A=\frac{d_1A}{k}-\frac{A(k-d_3)}{k}\cos{\xi}-\frac{A(k-d_2)}{k}\cos{\alpha}$$

Multiplying both sides by $\frac{k}{A}$ and rearranging:

$$-(k-d_2)\cos{\alpha}=(k-d_1)+(k-d_3)\cos{\xi}$$

Squaring both sides and substituting (8) and (10) give:

$${(k-d_2)}^2\left(1-\frac{k^2\left(k-d_3\right)^2}{A^2}\right)={(k-d_1)}^2+{(k-d_3)}^2\left(1-\frac{k^2\left(k-d_2\right)^2}{A^2}\right)+2(k-d_1)(k-d_3)\cos{\xi}$$

Expanding the brackets on both sides would eliminate the fraction terms on both sides, so the following remains:

$${(k-d_2)}^2={(k-d_1)}^2+{(k-d_3)}^2+2(k-d_1)(k-d_3)\cos{\xi}$$

Rearranging:

$$-2(k-d_1)(k-d_3)\cos{\xi}={(k-d_1)}^2+{(k-d_3)}^2-{(k-d_2)}^2$$

Squaring both sides then substituting in (8) give:

$$4{(k-d_1)}^2{(k-d_3)}^2-\frac{4{k^2(k-d_1)}^2{(k-d_3)}^2{(k-d_2)}^2}{A^2}=\left[{(k-d_1)}^2+{(k-d_3)}^2-{(k-d_2)}^2\right]^2$$

Thus:

$$A^2=\frac{4{k^2(k-d_1)}^2{(k-d_3)}^2{(k-d_2)}^2}{4{(k-d_1)}^2{(k-d_3)}^2-\left[{(k-d_1)}^2+{(k-d_3)}^2-{(k-d_2)}^2\right]^2}\tag{11}$$

assuming the denominator $>0$.

We could take the positive square root and stop here, and the equation is not too onerous. Or we could continue to write out using $d_i$ only without $k$, by simplifying then substituting $k=\ \frac{d_1+d_2+d_3}{2}$.

The denominator is a difference of two squares:

Denominator
\begin{align}
& = \left[2\left(k-d_1\right)\left(k-d_3\right)\right]^2 -[\left(k-d_1\right)^2+\left(k-d_3\right)^2-{(k-d_2)}^2]^2 \\
& = [2\left(k-d_1\right)\left(k-d_3\right)+\left(k-d_1\right)^2+\left(k-d_3\right)^2-{(k-d_2)}^2][-2\left(k-d_1\right)\left(k-d_3\right)+\left(k-d_1\right)^2+\left(k-d_3\right)^2-{(k-d_2)}^2]\times-1 \\
& =-\left[{(k-d_1+k-d_3)}^2-{(k-d_2)}^2\right]\left[\left(\left(k-d_1\right)-\left(k-d_3\right)\right)^2-{(k-d_2)}^2\right]
\end{align}

Since $k=\ \frac{d_1+d_2+d_3}{2},\  2k-d_1-d_3=d_2\ $. So:

Denominator $=-\left[{d_2}^2-{(k-d_2)}^2\right]\left[\left(d_3-d_1\right)^2-{(k-d_2)}^2\right]$

Factorise each difference of two squares and simplify to give:

Denominator $=k\left(k-2d_2\right)\left(d_3-d_1-d_2+k\right)\left(d_3-d_1+d_2-k\right)$

At this point we substitute $k=\ \frac{d_1+d_2+d_3}{2}$ and simplify to give:

Denominator
$$ = \frac{d_1+d_2+d_3}{2}\cdot\ \frac{d_1-3d_2+d_3}{2}\cdot\frac{-d_1-d_2+3d_3}{2}\cdot\frac{{-3d}_1+d_2+d_3}{2} \tag{12} $$

Substituting (5) into the numerator and (12) into the denominator of (11), and simplifying give:

$$A^2=\frac{\left(d_1+d_2+d_3\right)^2\left(-d_1+d_2+d_3\right)^2\left(d_1-d_2+d_3\right)^2\left(d_1+d_2-d_3\right)^2}{4\left(d_1+d_2+d_3\right)\left(3d_1-d_2-d_3\right)\left({-d}_1+{3d}_2-d_3\right)\left({-d}_1-d_2+3d_3\right)}$$

Thus, taking the positive square root:

$$\fbox{$A=\frac{\left(-d_1+d_2+d_3\right)\left(d_1+d_2-d_3\right)\left(d_1-d_2+d_3\right)\sqrt{d_1+d_2+d_3}}{2\sqrt{\left(3d_1-d_2-d_3\right)\left(3d_2-d_3-d_1\right)\left(3d_3-d_1-d_2\right)}}.$}$$

The square root of (11) provides an alternate form of this solution.


答:
Define $f : \mathbb{Z} \rightarrow [0,1]$ by
$$f(n) = \frac{1}{2(n+1)} \quad \quad \text{if} \quad n \geq 0$$
$$f(n) = \frac{1}{1-2n} \quad \quad \text{if} \quad n < 0$$

Title: Existence of an injection between $\mathbb{Z}$ and $[0,1]$

Body:
Is the statement that there is an injection between $\mathbb{Z}$ and $[0,1]$ true or false

$\mathbb{Z}\xrightarrow{(1-1)} [0,1]$

i've build it on the definition of cardinality that is if $\lvert A \lvert \leq \lvert B \lvert$

then there exists an injection $f:A\longrightarrow B$ and one knows that

$\lvert \mathbb{Z} \lvert \leq \lvert [0,1]\lvert$

thus there exists an injection


问:
Find the dual problem of $\min_x\{||x-a_1||+||x-a_2||+||x-a_3||,a_i\in\mathbb{R}^n\}$

Find the dual problem of

$$\min_x\{||x-a_1||+||x-a_2||+||x-a_3||,a_i\in\mathbb{R}^n\}$$

I've tried the following but got stuck
$$\min_x\{||x-a_1||+||x-a_2||+||x-a_3||,a_i\in\mathbb{R}^n\}=\min_{x,z_i}\{||z_1||+||z_2||+||z_3||,a_i\in\mathbb{R}^n\,z_i=x-a_i\}$$

and the Lagrangian is:\
$L(x,z,\lambda)=  ||z_1||+||z_2||+||z_3||+\sum_{i=1}^3\lambda_i(z_i-x+a_i)=\sum_{i=1}^{3}z_i\lambda_i+||z_i||-x\sum\lambda_i+\sum\lambda_ia_i$

from $-x\sum\lambda_i$ we can deduce that $\sum \lambda_i=0$ else we get $-\infty$.
Now when I try to solve for each $z_i$ I get $\frac{\partial L}{\partial z_i}=\lambda_i+\frac{z_i}{||z_i||}=0$
and I'm not sure how to continue from here.

Any hint?

答:
According to the request of the host, I propose here an answer without using the notion of conjugate function. However I have to admit that the idea behind the computations is of conjugate function.

Recall that the Lagrangian function is
\begin{align}
L(x,z,\lambda)=\sum_{i=1}^{3} \langle z_i,\lambda_i\rangle +||z_i||-\langle x, \sum_{i=1}^3\lambda_i\rangle +\sum_{i=1}^3 \langle \lambda_i, a_i\rangle,
\end{align}
where $x, z_i, \lambda_i \in \mathbb R^n$ for $i=1,2,3$.

We claim that the dual problem is

$$\sup_{\lambda} \inf_{x,z} L(x,z,\lambda)
=\sup_{\lambda\in D} \sum_{i=1}^3 \langle \lambda_i, a_i\rangle. \label{eq1} \tag{1}$$

where
\begin{align}
D=\{\lambda=(\lambda_1, \lambda_2, \lambda_3): \sum_{i=1}^3\lambda_i=0 \text{ and }
||\lambda_i||_*\leq 1 \text{ for } i=1,2,3\}.
\end{align}
Here $||\cdot||_*$ is the dual norm of $||\cdot||$ in $\mathbb R^n$ and defined as
\begin{align}
||u||_*:=\sup_{||v||\leq 1} \langle u,v\rangle.
\end{align}

Proof of \eqref{eq1}.

We now focus on $\inf_{x,z} L(x,z, \lambda)$.
We claim that
$$
\inf_x -\langle x, \sum_{i=1}^3\lambda_i\rangle =
\begin{cases}
0, & \text{ if } \sum_{i=1}^3\lambda_i=0,\\
-\infty, & \text{ otherwise.}
\end{cases}
\label{eq2} \tag{2}$$
and
$$
\inf_{z_i} \langle z_i, \lambda_i \rangle +||z_i||=
\begin{cases}
0,       & \text{ if } ||\lambda_i||_* \leq 1,\\
-\infty, & \text{ if } ||\lambda_i||_* > 1.
\end{cases}
\label{eq3} \tag{3}
$$

If \eqref{eq2} and \eqref{eq3} hold true, we obtain the dual problem \eqref{eq1}.
Since the host already know how to prove \eqref{eq2}, we now only concentrate on \eqref{eq3}.

To simplify the notations, in \eqref{eq3} we shall remove the index $i$, i.e.
we are going to prove
$$
\inf_{z} \langle z, \lambda\rangle +||z||=
\begin{cases}
0,       & \text{ if } ||\lambda||_* \leq 1,\\
-\infty, & \text{ if } ||\lambda||_* > 1.
\end{cases}
\textbf{ for $z, \lambda$ are in $\mathbb R^n$ from here on.}
\label{eq4} \tag{4}
$$

Let $\lambda'=-\lambda$, we have
$$\inf_{z} \langle z, \lambda\rangle +||z||=-\sup_{z} \langle z, \lambda'\rangle -||z||.$$

Case 1. If $||\lambda'||_*\leq 1$, we have
$$\sup_{z} \langle \frac{z}{||z||}, \lambda'\rangle \leq ||\lambda'||_*\leq 1.$$
Here the first inequality holds due to the definition of dual norm.
Thus
$$\sup_{z} \langle z, \lambda'\rangle -||z||\leq 0.$$
Furthermore, the equality holds if we take $z=0$.

Case 2. If $||\lambda'||_*> 1$, from the definition of dual norm,
we deduce the existence of some $z$ such that $||z||\leq 1$
and $\langle z,\lambda'\rangle >1$.
Now, we have
$$\sup_{t\in \mathbb R} \langle tz, \lambda'\rangle-||tz||
=\sup_{t\in \mathbb R} t(  \langle z, \lambda'\rangle-||z||) =+\infty.$$

Note that $||\lambda'||_*=||\lambda||_*$, all together tell us that \eqref{eq4} is true.

Hence, \eqref{eq1} is true.


答:
Mathematica says that the answer is
$$\pi^2\zeta(3)+12\zeta(5)$$
I will try to figure out how this can be proven.

Added: Let me compute the 2nd integral in Ron Gordon's answer:
\begin{align}\int_{0}^{\infty}\frac{x^3 e^{-x}}{1-e^{-x}}\ln(1-e^{-x})\,dx
&=-\frac32\int_0^{\infty}x^2\ln^2(1-e^{-x})\,dx=\\&=-\frac32\left[\frac{\partial^2}{\partial s^2}\int_0^{\infty}e^{-sx}\ln^2(1-e^{-x})\,dx\right]_{s=0}=\\
&=-\frac32\left[\frac{\partial^2}{\partial s^2}\int_0^{1}t^{s-1}\ln^2(1-t)\,dt\right]_{s=0}=\\
&=-\frac32\left[\frac{\partial^4}{\partial s^2\partial u^2}\int_0^{1}t^{s-1}(1-t)^u\,dt\right]_{s=0,u=0}=\\
&=-\frac32\left[\frac{\partial^4}{\partial s^2\partial u^2}\frac{\Gamma(s)\Gamma(1+u)}{\Gamma(1+s+u)}\right]_{s=0,u=0}=\\
&=-\frac{1}{2}\left(\pi^2\psi^{(2)}(1)-\psi^{(4)}(1)\right).
\end{align}
To obtain the last expression, one should expand the ratio of gamma functions to 2nd order in $u$, then to expand the corresponding coefficient to 2nd order in $s$.

Then we can use that $\psi^{(2)}(1)=-2\zeta(3)$ and $\psi^{(4)}(1)=-24\zeta(5)$ (cf formula (15) here) to obtain the quoted result.

Title: Find the value of $\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx$

Body:
I'm trying to figure out how to evaluate the following:
$$
J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx
$$
I'm tried considering $I(s) = \int_{0}^{\infty}\frac{x^3}{(e^x-1)^s}\,dx\implies J=-I'(1)$, but I couldn't figure out what $I(s)$ was. My other idea was contour integration, but I'm not sure how to deal with the logarithm. Mathematica says that $J\approx24.307$. </p>

<p>I've asked a similar question and the answer involved $\zeta(s)$ so I suspect that this one will as well. </p>


问:
$x,y \in R^n,x^Ty \le 1, |y|_2 = 1$, can it be true that for some $i, x_i > 1$ and is it true that $|x|_2 \le 1$?

We have that for $x,y \in R^n,x^Ty \le 1, |y|_2 = 1$, can it be true that for some $i, x_i > 1$ and is it true that $|x|_2 \le 1$ ($|*|_2$-is the euclidian norm)?

$x_1y_1+x_2y_2+...+x_ny_n \le 1 = y_1^2+y_2^2+...+y_n^2$ I tried to get a contradiction assuming that $|x|_2 = x_1^2+x_2^2+...+x_n^2 \ge 1$. So, $y_1(x_1-y_1)+...+y_n(x_n-y_n) \le 0, x_1(x_1-y_1)+...+x_n(x_n-y_n) \ge 0$. However, I am stuck.

I also tried to do it with inner product and get $<x,y> = <x-y,y>+<y,y>$, since $<y,y> = 1, <x,y> \le 1$, then $<x-y,y> = 0$ so $x=y$.

答: Geometrically speaking, $|x^T y|$ only controls the length of the projection of $x$ in the direction of $y$. The length of $x$ itself can still be large. In fact, if $x\perp y$ and $x\neq 0$ (such a vector exists if $n\geq 2$), then $x^T y=0$, but $|x|_2$ can be made arbitrarily big by multiplying $x$ with a constant factor. Similarly, $x_i$ can be made arbitrarily big for some $i$.


答:
Consider the function $f : \Bbb Z \longrightarrow [0,1]$ defined by $$
f(n) = \left\{
        \begin{array}{ll}
            \dfrac {1} {1-2n} & \quad n \leq 0 \\
            \dfrac {1} {2n} & \quad n > 0
        \end{array}
    \right.
$$


问:
Solution of Differential equation by series
Excuse me , can you help me on solving the following differential equation
$y''(x) - 1/(x+A)y'(x)+C/(x+B) y=0$

```<code>I use the series but i stop on a point and then can not complete specially on finding the recurrence relation
</code>```

答:
Assuming that the equation is $$y'' -\frac {y'}{x+a}+\frac {c\,y}{x+b} =0$$ there is no explicit solution except if $a=b=0$. If this was the case, one solution would be
$$y=Kc x J_2\left(2 \sqrt{c} \sqrt{x}\right)$$ where appears  the Bessel function of the first kind which has a series representation.

So, let
$$y=\sum_{i=0}^n d(i)\, x^i \quad y'=\sum_{i=0}^n i\,d(i)\, x^{i-1} \quad y''=\sum_{i=0}^n i(i-1)\,d(i)\, x^{i-2} $$ Replace, use the long division and group terms. Now, cancel the coefficients.

It will effectively be quite tedious but it is doable. For example, we need to solve (one at the time would be the best)
$$\frac{d(1)}{a^2}-\frac{2 d(2)}{a}-\frac{c d(0)}{b^2}+\frac{c d(1)}{b}+6 d(3)=0$$
$$-\frac{d(1)}{a^3}+\frac{2 d(2)}{a^2}-\frac{3 d(3)}{a}+\frac{c d(0)}{b^3}-\frac{c
   d(1)}{b^2}+\frac{c d(2)}{b}+12 d(4)=0$$ and so on.


问:
Whats wrong with my probability equations for these probability question?
Johns wants to walk his dog on most mornings. If the weather is pleasant,(let $W$ denote walk the dog)  $P(W)=\frac{3}{4}$. If the weather is not pleasant, $P(W) = \frac{1}{3}$. For the month of January, (let $P$ denote pleasant weather) $P(P)= \frac{5}{8}$

So you could do this via tree diagram, two-way table, or probability equations. (I am having trouble manipulating the probability equations so I am trying to force myself to use them, although I'm having trouble understanding how to use them)

Using a table:

W
W'

P
30

P'
9

39
33
72

Numers come from: $P(W|P)=\frac{3}{4}\frac{5}{8} =\frac{15}{32}$

Similarly $P(W|P')=\frac{1}{8}$

$LCM(8,32) = 72$ so let that be the imaginary sample space.

Therefore $P(W)=\frac{39}{71}$ which is incorrect.

Using the probability equations

$P(W\cap P)=\frac{3}{4}\frac{5}{8} =\frac{15}{32}$

Similarly $P(W\cap P')=\frac{1}{8}$

(And I am already confused: $P(W\cap P) = P(W|P$)? well from the table both seem right as in $W$ AND $P$ is the same as $W$ GIVEN $P$)

Then for $P(W) = \frac{P(W\cap P)}{P(P)} = \frac{\frac{15}{32}}{\frac{5}{8}}=\frac{3}{4}$ which again is wrong

What if I try $P\left(W\right)=P\left(W \cap\ P\right)+P\left(W\cap\ P'\right)$?

Nope.

EDIT: yes!! the last thing I wrote does give the correct solution $\frac{19}{32}$ ! Props to me. What about that table though?

答:
Very easy solution. I considered the 31 days of January in the bottom of the table.

The pleasant days are $\frac{5}{8}\cdot 31\approx 19$. Thus the unpleasant days are 12.

If the day is pleasant I walk with probability 0.75, thus in the cell $(P,W)=19\cdot 0.75\approx14$ and so on....

this is the resulting table

Walk
Do not Walk
Total days

Pleasant
14
5
19

Not Pleasant
4
8
12

Total days
18
13
31

It is evident that $P(W)=\frac{18}{31}$

If this result is not the same of the one you are expecting to find, do not approximate like i did...but the procedure should be understood.


问:
If $K$ is compact, then $C(K,\mathbb{R}^n)$ is a Banach space under the norm $\|f\|=\sup_{x\in K} \|f(x)\|$

Let $K$ be a topological space that is compact. Show that the space $C(K,\mathbb{R}^n)$ of all the continuous functions $f:K\to\mathbb{R}^n$ is a Banach space with the norm $\|f\|=\sup_{x\in K} \|f(x)\|$.

Below is my attempt:

$\|f\|=\sup_{x\in K} \|f(x)\|$ is a norm:

i. Suppose that $\|f\|=\sup_{x\in K} \|f(x)\|= 0$; then $f=0$. Conversely, suppose that $f=0$. Then $0=\|f\|=\sup_{x\in K} \|f(x)\|$.

ii. $\|\alpha f\|=\sup_{x\in K} \|\alpha f(x)\|=\sup_{x\in K} |\alpha|\|f(x)\|=|\alpha|\sup_{x\in K} \|f(x)\|=|\alpha|\|f\|$

iii. $\|f+g\|= \sup_{x\in K} \|f(x)+g(x)\|\leq \sup_{x\in K} \|f(x)\| + \|g(x)\|= \sup_{x\in K} \|f(x)\| + \sup_{x\in K} g(x)\|$

$C(K,\mathbb{R}^n)$ is complete (my doubt is here)

i. For all $\epsilon>0$, $\exists\, n_{0}$ such that if $m,n>n_{0}$ then $d(f_m,f_n)<\epsilon$.

We know that $|f_m-f_n|\leq \|f_m -f_n\|= \sup_{x\in K} \|f_m(x)-f_n(x)\|< \epsilon $ but how can I prove the last inequality?

ii. How to prove that $f_n$ converges to an $f$ in $C(K,\mathbb{R}^n)$?

答:
I understand $f_n$ is a Cauchy sequence in $C(K,R^n)$. We need to get a $f\in C(K,R^n)$ such that $f_n\rightarrow f$ in $C(K,R^n)$. To find such $f$, Note that
 $||f_n(x)-f_m(x)||=||(f_n-f_m)(x)||\le Sup_{x\in K}||(f_n-f_m)(x)||=||f_n-f_m||$ as if $f_n\in C(K,R^n)$ and $f_m\in C(K,R^n)$ then $f_n-f_m\in C(K,R^n)$ (defined as usual pointwise way). In fact, $ C(K,R^n)$ is a real vector space. So, if $f_n$ is a Cauchy sequence, then $f_n(x)$ (in $R^n$)is a Cauchy sequence $\forall x\in K$. But $R^n$ is complete, so, $f_n(x)$ converges to some element in $R^n$. Now, the candidate function $f$ can be defined as $f(x)=\lim_n f_n(x)$.
 Now we can to prove $f\in C(K,R^n)$ and $f_n\rightarrow f$ in $C(K,R^n)$


问:
How to prove that this limit diverges?
Consider the limit
$$\lim_{n\to +\infty} \frac{2^{\sqrt{n}}}{n!}\left(\frac{n}{e}\right)^n.$$

How to show that this limit is $+\infty$? I just tried square criterion but it doesn’t work. Moreover, Stirling’s formula is not allowed (requested in the assignment).

Could anyone please help?

Thank you in advance!

答:
It holds that
\begin{align*}
\log n! & = \log n + \sum\limits_{k = 1}^{n - 1} {\log k}  \le \log n + \sum\limits_{k = 1}^{n - 1} {\int_k^{k + 1} {\log t \,\mathrm{d}t} } \\ & = \log n + \int_1^n {\log t\,\mathrm{d}t}  = \log n +n\log n - n,
\end{align*}
i.e.,
$$
n! \le n\left( {\frac{n}{\mathrm{e}}} \right)^n.
$$
Thus,
$$
\frac{{2^{\sqrt n } }}{n} \le \frac{{2^{\sqrt n } }}{{n!}}\left( {\frac{n}{\mathrm{e}}} \right)^n .
$$
Show that the left-hand side tends to infinity as $n\to +\infty$.

Title: How to prove that this limit diverges?


问:
Proof by induction: inequalities
I need to prove by induction the following inequality:
$$\sum_{i=1}^{n} i \leq n^n \text{ for all } n \geq 1$$

Base case is proved. In the inductive case I can sum both sides of the inequality by $(n+1)$ as
$$\sum_{i=1}^{n+1} i = \sum_{i=1}^{n} i + (n+1)$$
Then? How can I obtain $\sum\limits_{i=1}^{n+1} i \leq (n+1)^{n+1}$ from $\sum\limits_{i=1}^{n} i + (n+1) \leq n^n + (n+1)$ ?

Title: Proof by induction: inequalities

Body:
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

$n^n \ge \sum_{i=1}^{n} i$ for some n

$(n+1)^{n+1} \ge n^{n+1}=n^n \cdot n$ and $\sum_{i=1}^{n+1} i = \frac{(n+1)(n+2)}{2}=\frac{n(n+1)}{2}\frac{(n+2)}{n}$

You just have to prove that $n \ge \frac{(n+2)}{n}$ for $n \ge 2$


问:
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$?
The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?

答:
As the inverse trigonometric functions are multi-valued with a respective principle value

or

as sine/cosine/tangent functions are periodic, it is incorrect to apply cosines in both sides.

For example,

$x$ in general $\ne -x$  but $\cos(-x)=\cos(x)$

$x$ in general $\ne \pi-x$  but $\sin(\pi-x)=\sin(x)$

$x\ne\pi+x$  but $\tan(\pi+x)=\tan(x)$

Using the ranges of principle values of $\arctan,$ if $\arctan\dfrac7{24}=y$

$\implies0<y<\dfrac\pi2$

and $\tan y=\dfrac7{24}$

and $\sin y,\cos y>0$

Consequently, $\dfrac{\sin y}7=\dfrac{\cos y}{24}=+\sqrt{\dfrac{\sin^2y+\cos^2y}{7^2+24^2}}$

$\implies\arctan\dfrac7{24}=\arcsin\dfrac7{25}=\arccos\dfrac{24}{25}$

Now use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ in the left hand side

For the right hand side, if $\arccos\dfrac35=u$

$\implies0<u<\dfrac\pi2$ and $\cos u=\dfrac35$ and $\sin u>0$

$\implies\sin u=\sqrt{1-\cos^2u}=\dfrac45$

$\implies\arcsin\dfrac45=u=\arccos\dfrac35$


问:
boolean logic expression: start_addr = stop_addr & ~mask
Let's say you have the following boolean expression of 16 bit words:

```<code>start_addr  = stop_addr & ~mask
</code>```

Where:  start_addr < stop_addr

example:

```<code>0xAA00 = 0xAAFF & ~0x00FF
0xAA00 = 0xAAFF & 0xFF00
0xAA00 = 0AA00
</code>```

now, suppose, I know only start_addr, and stop_addr, and mask is unknown.  How to find mask?

答:
The problem with what you want is that given start_addr and stop_addr, then it is not always possible to determine mask.

For some values of start_addr and stop_addr it can happen that there is no mask satisfying the equation, for others may be several possible values of mask satisfiyng the equation.


问:
How many ways are there of creating an $8$ character password with a digit, a lowercase letter, and $2$ capital letters?

A valid password for the bank's website consists of 8 characters
(digits and letters in English) and must contain at least one digit,
at least one small English letter, and at least two capital letters.
How many legal passwords are there?

I considered four sets:

$|A|$ = number of ways for no digits appearing in the password = $52^8$,

$|B|$ = number of ways of no small letter in the password = $36^8$,

$|C|$ = number of ways of no capital letter in the password = $36^8$,

$|D|$ = number of ways of only 1 capital letter in the password = $26*36^7$, and

$|U|$ = number of ways of forming the bank's password without restriction  = $62^8$.

Using the formula for inclusion and exclusion for 4 sets $|U| - [ |A| \cup |B| \cup |C|\cup |D|]$

such that

$$
\begin{align}
&|A\cup B\cup C\cup D|\\[3pt]
&=|A|+|B|+|C|+|D|\Big\}\text{ all singletons}\\
&-(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|+|C\cap D|)\Big\}\text{ all pairs}\\
&+(|A\cap B\cap C|+|A\cap B\cap D|+|A\cap C\cap D|+|B\cap C\cap D|)\Big\}\text{ all triples}\\
&-|A\cap B\cap C\cap D|\Big\}\text{ all quadruples.}\\
\end{align}
$$

Now what I'm stuck at is when it comes to triples and quadruples.

$|A\cap B\cap C\cap D|$, to find the number of ways of union of this is kinda contradicting because we have to find a password with no digits, no small letters, no big letters, but only 1 small letter..

or when

$|A\cap C\cap D|$ is also contradicting because how we can find number of ways of password with no big letters and also 1 big letter..

答:
I would simplify the expression a bit by combining set of no capital letter and set of one capital letter as there is no intersection between them.

If $A$ is the set of passwords where we have at most one capital letter, $B$ is the set of passwords with no small letters and $C$ with no digits,

$|A| = 36^8 + 8 \times 26 \times 36^7$

$|B|= 36^8$

$|C| = 52^8$

$|A \cap B| = 10^8 + 8 \times 26 \times 10^7$

$|B \cap C| = 26^8$

$|A \cap C| = 26^8 + 8 \times 26 \times 26^7$

$|A \cap B \cap C| = 0$

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$

And the answer will be $ \, \, 62^8 - |A \cup B \cup C|$


问:
How Entropy scales with sample size
For a discrete probability distribution, the entropy is defined as:
$$H(p) = \sum_i p(x_i) \log(p(x_i))$$
I'm trying to use the entropy as a measure of how "flat / noisy" vs. "peaked" a distribution is, where smaller entropy corresponds to more "peakedness". I want to use a cutoff threshold to decide which distributions are "peaked" and which are "flat". The problem with this approach is that for "same shaped" distributions, the entropy is different for different sample sizes! as a simple example take the uniform distribution - it's entropy is:
$$p_i = \frac{1}{n}\ \ \to \ \ H = \log n$$
To make things worse, there doesn't seem to be a general rule for more complex distributions.

So, the question is:

How should I normalize the entropy so that I get the same "scaled entropy" for "same" distributions irrespective of the sample size?

答:
A partial answer for further reference:

In short, use the integral formulation of the entropy and pretend that the discrete distribution is sampling a continuous one.

Thus, create a continuous distribution $p(x)$ whose integral is approximated by the Riemann sum of the $p_i$'s:
$$\int_0^1 p(x)dx \sim \sum_i p_i\cdot \frac{1}{N} = 1$$
This means that the $p_i$'s must first be normalized so that $\sum_i p_i = N$.

After normalization, we calculate the entropy:
$$H=-\int_0^1 p(x)\log\left(p(x)\right)dx \sim -\sum_i p_i \log(p_i)\cdot \frac{1}{N}$$

As $N\to \infty$ this gives an entropy which is solely related to the distribution shape and does not depend on $N$. For small $N$, the difference will depend on how good the Riemann sum approximates the integrals for given $N$.

Title: How Entropy scales with sample size


问:
Is Nested Interval Property $\iff$ Order Completeness in any Ordered Field?
Let $R$ be an arbitrary Ordered Field.

Order Completeness Property: If $S \subset R$ is bounded above, then $\exists \ c \in R$ that is an upperbound of S and for every upperbound $b$ of S, we have $c \leq b.$

Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals,
i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then
$\bigcap_{1}^{\infty} I_i \neq \phi$.

Order Completeness $\implies$ Nested Interval Property.

This part is easy to prove.

Nested Interval Property $\implies$ Order Completeness.

Consider      $A \underset{bdd.}\subseteq R$,

$b_0$: an upperbound of $A$,

$a_0$: not an upperbound of $A$.

$I_0:=[a_0,b_0] \implies I_0 \cap A \neq \phi$.

Let $c=(a_0+b_0)/2_R$, If $\exists \ a \in A$ with $c<a$ then choose $a_1=c$, $b_1=b_0$.

Otherwise choose $a_1=a_0$, $b_1=c$.

Proceeding this way one can construct a sequence of
closed nested intervals $I_0 \supseteq I_1 \supseteq ... \supseteq I_n \supseteq ...$, in
which
$I_n=[a_n,b_n]$ where

$a_n$: not an upperbound of A,

$b_n$: upperbound of A.

Now $\operatorname{diam}(I_n)=\frac{b_0-a_0}{2^n}$. Assuming R enjoys Archemedian Property I can conclude $\operatorname{diam}(I_n) \to 0$

$\implies$ $\bigcap_{1}^{\infty} I_i=\{s\}$.

Remains to show that $s=\operatorname{lub}A$.

If there exists $a \in A$ such that $s<a$ since $\operatorname{diam}(I_n) \to 0$ there exists $p \in \mathbb{N}$ such that $\operatorname{diam}(I_p)<a-s$. Now $s \in I_p$ and $a \notin I_p \implies b_p<a$,
contradiction to the fact that $b_p$ is an upperbound of $A$.

Hence $s$ is an upperbound of A.

Again let $l<s$ then there exists $q \in \mathbb{N}$ such that $\operatorname{diam}(I_q)<s-l$.

We have $s \in I_q$ then $l \notin I_q$ $\implies l<a_q$. Now there exists $b \in A$ with $l<a_q<b \implies l$ is not an upperbound of A.

Hence there exists $s \in R$ such that $s=\operatorname{lub}A$.

Now my question is If I drop the assumption of Archimedian Property can we still conclude this result?

答:
Certain ordered fields $R$ are $\mathbf{\aleph_1}$-saturated as linear orders. This means that given countable subsets $A,B\subset R$ with $A<B$, there is $x \in R$ with $A<x<B$. Such an ordered field is not archimedean, since there is $x \in R$ with $\mathbb{N}<x<\varnothing$.

But such a field satisfies the nested interval property because writing $I_n=[a_n,b_n]$ for each $n \in \mathbb{N}$ and considering $x \in R$ with $\{a_n: n \in \mathbb{N}\}<x<\{b_n: n \in \mathbb{N}\}$, one has $x \in \bigcap \limits_{n \in \mathbb{N}} I_n$ (in the non-degenerate case where $(a_n)_{n \in \mathbb{N}}$ or $(b_n)_{n \in \mathbb{N}}$ is not stationnary).

So on its own the nested interval property does not imply completeness.

An example of an $\aleph_1$-saturated ordered field is the field $\mathbf{No}(\omega_1)$ of surreal numbers of countable length.


问:
Limit of a sequence of integrals implies pointwise limit of a sequence of functions
Pointwise limit of a sequence of functions

Let $f$, $f_{n}\in C[0, 1]$ for all $n\geq 1$. Suppose that
$$
\lim_{n\to\infty}\int^{1}_{0}|f_{n}(t)-f(t)|dt=0.
$$
Is it possible to show that
$$
\lim_{n\to\infty}|f_{n}(t)-f(t)|=0,
$$
for all $t\in [0, 1]$?

答:
As mentioned by Surb, you can even have cases were $\lim_{n\to \infty }f_n(x)=f(x)$ will hold for no $x \in [0,1]$. The idea is to have a bump getting thinner and sliding back and forth between $0$ and $1$.

See here for a detailed example.