问:
Double integral:$\iint_{x+y\leq1, \,x\geq0, \, y\geq0} \frac{1}{\sqrt{1-x-y}}\, dxdy$
I have this double integral

$$\iint\limits_{\substack{x+y\leq1 \\x\geq0\\ y\geq0}} \frac{1}{\sqrt{1-x-y}}\,dxdy$$
I think I need to replace variables but I don’t know how to replace.

Body: $$I=\int_{0}^{1}{\int_{0}^{1-x}{\frac{\mathrm{d}y}{\sqrt{1-x-y}}}\,\mathrm{d}x} =\int_{0}^{1}{2\sqrt{1-x}\,\mathrm{d}x}=\frac{4}{3}$$


问:
Derivative of matrix exponential w.r.t. to each element of the matrix
I have $x= \exp(At)$ where $A$ is a matrix. I would like to find derivative of $x$ with respect to each element of $A$. Could anyone help with this problem?

答:
According to Derivatives of the Matrix Exponential and Their Computation (who reference Karplus, Schwinger, Feynmann, Bellman and Snider) the derivative can be expressed as the linear map (i.e. Fréchet derivative)

$$\frac{\rm{d} e^{At}}{\rm{d} A} = \Big(V\longmapsto\int_{0}^t e^{A(t-\tau)}Ve^{A\tau}\,\rm{d}\tau\Big)$$

ContentLicense: CC BY-SA 4.0


问:
Superior limit of sequence of random variables.
Let $X_{n}$  be a sequence of real random variables and let $X$ a real random variable defined over the same p.s. $(\Omega,\mathcal{A},\mathbb{P})$ and such that $X_{n}(\omega)\searrow X(\omega)$ for all $\omega\in\Omega$. Let $x_{n}$ be a sequence of reals such that $x_{n}\searrow_{n} x$ and such that $x_{n} \in Med(X_{n})$ for all $n\in\mathbb{N}$.

I wonder if $\lim\sup_{n} \{\omega\in\Omega : X_{n}(\omega)\leq x_{n}\} = \{\omega\in\Omega : X(\omega)\leq x\}$, where $\lim\sup_{n} A_{n} = \cup_{n=1}^{\infty}\cap_{k=n}^{\infty} A_{k}$.

答: The definition of median involves probabilities, so we can alter the random variables on a set of probability $0$ without altering the median. On a set $E$ of probability $0$ let  $X_n(\omega)=\frac 1 n, X(\omega)=0$ and $x_n=\frac 1  {2n}, x=0$. On the complement of $E$ let $X_n(\omega)=\frac 1 n, X(\omega)=0$ and $x_n=\frac 1  {n}, x=0$. Then the left side does not include any point of $E$ but RHS includes every $\omega \in \Omega$.


问:
Solving $2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$
just looking for some help with the following:

$$2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$$

I know how to solve a log equation with a single nested log that's $= 1$ or $0$ or some number, but I'm unsure of what to do when summing/ subtracting nested logs like these.

If possible, a hint would be much more appreciated than the flat out answer.

Thanks in advance to anyone who stops to help.

答:
$1)$ Use change of base, to write $$\log_{1/2}(\log_2 x) = - \log_2(\log_2x) $$

$2)$ Let $$\log_2(\log_2 x) = t$$

$3)$ Then $$t= ?$$

$4)$ This means $$\log_2 x =?$$

$5)$ Finally, this means $$x=?$$

$6)$ Check this root in the original equation.


问:
Solving a sequence
Given a sequence
$$x_0=5$$
$$x_{n+1}=x_n*1.03+1$$
how does one find generalized formulae for:

$x_n$

$\sum_{n=a}^{b}x_n$ for a given pair of $a, b$ where $a{\leq}b$ and $a>0$

Interested in both the closed formulae and (a wonderful cherry on top!) a description on the methodology of solving these kinds of problems.

I can obviously do this recursively, but it feels like this is a common enough problem that has been solved many times over. I'm just unable to find the solution by googling (most likely due to lack of math education - what terms should I be looking for besides the "sequence" and "solve"?)

Title: Solving a sequence

Body:
$$x_0=5$$
$$x_{n+1}=1.03x_n+1\tag{1}$$
plug $n+1\to n$
$$x_{n+2}=1.03x_{n+1}+1\tag{2}$$
Subtract $(2)-(1)$
$$x_{n+2}-x_{n+1}=1.03x_{n+1}-1.03x_n$$
rearrange the equation
$$x_{n+2}-2.03x_{n+1}+1.03x_n=0$$
solve the characteristic equation
$$t^2-2.03t+1.03=0\to t_1=1;\;t_2=1.03$$
general solution is
$$x_n=a\cdot 1^n+b\cdot 1.03^n$$

compute $x_1=1.03x_0+1=6.15$

Plug the initial values to get $a,b$

$$a+1.03 b=6.15,a+ b=5\to a=-33.33;\;b=38.33$$
the solution is
$$x_n=38.33\cdot 1.03^n-33.33$$

$$\sum _{n=p}^q x_n=\frac{5}{9} \left(60 p-23\cdot 100^{1-p} \cdot103^p+23\cdot 100^{-q} \cdot 103^{q+1}-60 q-60\right)$$


问:
Inequality of expectation in Weierstrass theorem
The proof I was given in class for Weierstrass Theorem using Berstein Polynomials contains this one particular inequality:

$|f(x) − E[f(\frac{S_n}{n})]| ≤ E[|f(x) − f(\frac{S_n}{n})|]$

E stands for expected value, x is the probability of success of a sequence of n iid Bernoulli variables, and Sn is the sum of those n iid Bernoulli variables.

I understood the theorem and how to prove it using Weak Law of Large Numbers, but I'm facing trouble in understanding the inequality I stated above.

Please help!

答: $f(x)$ is a fixed number. It is not random. So $f(x)-E(\frac {S_n} n)=E[f(x)-E(\frac {S_n} n)]$. Now use the inequality $|EY| \leq E|Y|$.


问:
How to show that $E[X_n]$ converges to $E[X]$ when $X_n$ converges in $L_1$?
The question that I encounter is:

Suppose $X_n \overset{L_1}{\longrightarrow} X$, then show that $\mathbb{E}[X_n]\rightarrow \mathbb{E}[X]$ as $n \rightarrow \infty$.

My thoughts on this: I can show that $\mathbb{E}[|X_n|] \rightarrow \mathbb{E}[|X|]$ by the triangular inequality with respect to the $L_1$ norm but I'm stuck on how to proceed from here. I think it's not generally true that convergence in the expectation of the absolute value implies convergence in the expectation of the random variable itself.

Many thanks for any help!

答:
$|EX_n-EX|=|E(X_n-X)| \leq E|X_n-X| \to 0$ so $EX_n \to EX$.

Proof of the inequality $|EY| \leq E|Y|$: We have $Y \leq |Y|$ so $EY \leq E|Y|$. Similarly, $-Y \leq |Y|$ gives $-EY=E(-Y) \leq E|Y|$. Since $|EY|$ is one of the number $EY, -EY$ it folllows that $|EY| \leq E|Y|$.


问:
Constructing a triangle given an angle, the side opposed to this angle, and the median to the given side.
I came across the following problem in my Euclidean Geometry text:

Construct a triangle having given an angle, the side opposed to this angle, and the median to the given side.

I have tried solving this problem using method of loci, with Locus 1 being the circumcircle of the sought for triangle and Locus 2 being the circle of radius of the given  median. However I don't know how to start this construction.

答:
See the picture below:

Draw segment AB and its middle point C.

Draw the perpendicular bisector g and angle ACC'.

Draw the  line i perpendicular to AC'.

Let D the intersection of g and i.

Draw the circle $\Lambda$ with center D and radius AD.

Draw the circle $\Gamma$ with center C and radius $l$ (the median length).

Let E and E' the intersections of $\Lambda$ and $\Gamma$.

Draw triangles ABE or ABE', solutions of the problem.


问:
DFT but using non-sinusoidal periodic waveforms.
Is there an established way of decomposing a discrete periodic (complex) signal into a sum of non-sinusoidal periodic waveforms (eg square, triangle, and sawtooth)?

For my use case the input waveform is periodic, but as I understand that's a condition which always holds for DFT even if it's easy to work around it in practical applications.

答:
Have you looked at using B-splines? The 0th-order B-spline is a simple pulse, 1st-order (convolution of 2 0th order B-splines) is a triangle, and they keep going up from there...

For whatever transform you want to use/develop, arguably one of the most basic requirements is that they possess the partition of unity property (see Jawerth and Sweldens (1993))—otherwise, it would be difficult to manage even approximating a constant function well. All B-splines have partition of unity property.

B-splines also have a nice connection to the Daubechies wavelets: each p-order Daubechies scaling function has a p-order B-spline component that show up as p zeroes in the complex z-plane.


问:
Integrate: $\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx$

Q: If $|a|< 1$ and $b>0$, show that
  $$\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx = \frac{\pi  \sin (\pi  a)}{2 (\cos (\pi  a)+\cosh (\pi  b))}$$

I need to evaluate the above integral by method of contour. I tried to use this contour on this question but at $2\pi i$, $\sinh(ax)$ changes to $\sinh(ax+2a\pi i)$ and I have difficulty taking out $\sinh(ax)$. Please give hints on which contour to use.Thanks in advance!!

ADDED::

Considering $-R \to R \to R + \pi i \to -R + \pi i \to -R$ with a bump on $0$ and $\pi i$ to avoid singularity.

$$(1 + e^{(a+ib)\pi i}) \int_{-\infty}^{\infty}\frac{e^{(a+bi)x}}{\sinh x}dx = -\pi i(1 - e^{(a+ib)\pi i}) \hspace{1 cm}(1)$$
$$(1 + e^{(-a+ib)\pi i}) \int_{-\infty}^{\infty}\frac{e^{(-a+bi)x}}{\sinh x}dx = -\pi i(1 - e^{(-a+ib)\pi i}) \hspace{1 cm}(2)$$
With a bit of algebra, we get
\begin{align*}
\int_{-\infty}^{\infty}\frac{e^{ax}-e^{-ax}}{\sinh x}e^{ibx}dx &= 2\pi i \left( \frac{1}{1 + e^{(a+bi)\pi i}} -  \frac{1}{1 + e^{(-a+bi)\pi i}} \right)\\
 &= 2 \pi \frac{\sin (a\pi)}{\cosh (b\pi) + \sin(a\pi)}
\end{align*}
From which we get the desired result.

答:
Use parity to extend the domain of integration from $-\infty$ to $\infty$. Shift the contour of integration down by $\frac{i\pi}{2}$ to avoid pole $x=0$. Then compute separately four integrals (actually, it suffices to compute one of them and then to change parameters accordingly)
$$\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(\pm a\pm ib)x}dx}{\sinh x}$$
using the rectangle $$-R-\frac{i\pi}{2}\rightarrow R-\frac{i\pi}{2}\rightarrow R+\frac{i\pi}{2} \rightarrow -R+\frac{i\pi}{2} \rightarrow -R-\frac{i\pi}{2}$$ with $R\rightarrow\infty$.

For example, let us compute
\begin{align}
\int_{\text{rectangle}}\frac{e^{(a+ib)z}dz}{\sinh z}\substack{R\rightarrow\infty\\=}\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}-
\int_{-\infty+\frac{i\pi}{2}}^{\infty+\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}=\\
=\left(1+e^{\pi i (a+ib)}\right)\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}
\end{align}
On the other hand, the integral over rectangle is equal to $2\pi i$ (the only pole inside is $z=0$ and the residue is $1$). Therefore,
$$\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}=\frac{2\pi i}{1+e^{\pi i (a+ib)}},$$
and the initial integral evaluates to
$$\frac{2\pi i}{8}\left(\frac{1}{1+e^{\pi i (a+ib)}}-\frac{1}{1+e^{\pi i (-a+ib)}}+\frac{1}{1+e^{\pi i (a-ib)}}-\frac{1}{1+e^{\pi i (-a-ib)}}\right)=\frac{\pi\sin\pi a}{2\left(\cos\pi a+\cosh\pi b\right)}.$$

ContentLicense: CC BY-SA 3.0


问:
Does this variety describe left monoids?
In this question I have defined the following variety.

Let $(S, \cdot, e)$ be such that $(S, \cdot)$ is a semigroup, $e$ is a binary operation, and let the identities $e(x, y)x \approx x$, $e(x, y)\approx e(y, x)$ hold. Let's call a structure which satisfies those a double left monoid, or dlm.

One can see that if $(S, \cdot)$ is a left monoid with left identity $f$, then setting $e(x, y)\equiv f$ we get a dlm.

If $(S, \cdot, e)$, as a semigroup, isn't a left monoid, then it can't be a right monoid. Clearly, if $f$ were the right identity, then $e(x, f)f = f = e(x, f)$ for all $x$, and so $fx = x$ for all $x$, so it would be a monoid.

Is any dlm necessarily a left monoid after the transformation $(S, \cdot, e)\mapsto (S, \cdot)$ which forgets the operation $e$?

答: The answer is no, as shown by the semigroup $(\Bbb{Z}, \min)$ with $e(x,y) = \max(x,y)$.


问:
integral of Dirac delta over a small interval.
While solving a problem I encountered $$\int_{a-\epsilon}^{a+\epsilon} dx\,\delta(x-a)\,f(x).$$ What does this evaluate to as $\epsilon \to 0$?

答:
By definition of $\delta_a(x) := \delta(x-a)$,
$$
I_{\epsilon} = \int_{a-\epsilon}^{a+\epsilon} f(x)\,\delta_a(\mathrm d x) = \int_{-\infty}^{\infty} \mathbb{1}_{[a-\epsilon,a+\epsilon]}(x)\,f(x)\,\delta_a(\mathrm d x) = \mathbb{1}_{[a-\epsilon,a+\epsilon]}(a)\,f(a)
$$
and since $a∈ [a-\epsilon,a+\epsilon]$ for any $\epsilon>0$, $I_{\epsilon} = f(a)$, hence
$$
\lim_\limits{\epsilon\to 0^+} I_{\epsilon} = f(a)
$$

Remark: However, if $\epsilon<0$, then $I_ϵ = 0$, so
$$
\lim_\limits{\epsilon\to 0^-} I_{\epsilon} = 0
$$


答:
Well, let's solve a more general problem:

$$\text{n}\log_\alpha\left(\log_\alpha\left(x\right)\right)+\log_\beta\left(\log_\alpha\left(x\right)\right)=\epsilon\tag1$$

First, we know that:

$$\log_\text{a}\left(\text{b}\right):=\frac{\ln\left(\text{b}\right)}{\ln\left(\text{a}\right)}\tag2$$

Let $\text{p}:=\log_\alpha\left(x\right)$, so we get for $(1)$:

$$\text{n}\log_\alpha\left(\text{p}\right)+\log_\beta\left(\text{p}\right)=\epsilon\tag3$$

Using $(2)$ we get:

$$\text{n}\cdot\frac{\ln\left(\text{p}\right)}{\ln\left(\alpha\right)}+\frac{\ln\left(\text{p}\right)}{\ln\left(\beta\right)}=\epsilon\tag4$$

Rewrite the left-hand side, of $(4)$ by combining fractions:

$$\frac{\ln\left(\text{p}\right)\left(\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)\right)}{\ln\left(\alpha\right)\ln\left(\beta\right)}=\epsilon\tag5$$

Solving for $\ln\left(\text{p}\right)$, gives:

$$\ln\left(\text{p}\right)=\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\tag6$$

Cancel logarithms by taking $\exp$ of both sides:

$$\text{p}=\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\tag7$$

Setting the substitution back and using $(2)$:

$$\log_\alpha\left(x\right)=\frac{\ln\left(x\right)}{\ln\left(\alpha\right)}=\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\tag8$$

Solving for $\ln\left(x\right)$, gives:

$$\ln\left(x\right)=\ln\left(\alpha\right)\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\tag9$$

Cancel logarithms by taking $\exp$ of both sides:

$$x=\exp\left(\ln\left(\alpha\right)\exp\left(\epsilon\cdot\frac{\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)\right)\tag{10}$$

Using:

$$\ln\left(\text{a}^\text{b}\right)=\text{b}\ln\left(\text{a}\right)\tag{11}$$

We can rewrite $(10)$ as follows:

$$x=\alpha^{\exp\left(\frac{\epsilon\ln\left(\alpha\right)\ln\left(\beta\right)}{\ln\left(\alpha\right)+\text{n}\ln\left(\beta\right)}\right)}\tag{12}$$

Using your values, we get:

$$x=4\tag{13}$$

Title: Solving $2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$

Body:
just looking for some help with the following:

$$2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$$

I know how to solve a log equation with a single nested log that's $= 1$ or $0$ or some number, but I'm unsure of what to do when summing/ subtracting nested logs like these.

If possible, a hint would be much more appreciated than the flat out answer.

Thanks in advance to anyone who stops to help.


问:
Chain Rule with for composition of scalar and multivariable functions
Suppose I have the following function $f(\bf x)= \bf a^\top \log (\bf A \bf x )$, where $\bf x \in \mathbb{R}^n$ and $\bf A \in \mathbb{R}^{m \times n}$. The $\log$ function is entry wise.
I am considering $f(\cdot)$ to be the composition of three functions: the linear function $g(\bf y)= \bf a^\top \bf y$, the function $y(\bf z)= \log (\bf z)$ and the function $z(\bf x)= \bf A \bf x$.

So I have $f(\bf x)= g(y(z(x)))$. Here I have composition of scalar and multivariable functions, so the general chain rule for scalar function does not hold in general (also the order of the terms changes when vectors are involved). How to properly compute the gradient and the Hessian of such a function? How to handle the entry-wise $\log$ in the chain rule?

Thank you.

答:
$\require{enclose}\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\o#1{\operatorname{#1}}\def\D{\o{Diag}}$
Take an ordinary scalar function $\phi(s)$ and its derivative
$\phi′(s)=\frac{d\phi}{ds}$ and apply them element-wise to a vector argument $v$ to generate the associated vector functions
$$h=\phi(v),\qquad h'=\phi'(v)$$
The differential of such a vector function can expressed using an elementwise $(\odot)$ product or better yet, a diagonal matrix
$$\eqalign{
dh &= h'\odot dv \;=\; \D(h')\,dv\\
}$$
Setting $\;h=\log(v),\;v=Ax,\;V=\D(v)\;$ this becomes
$$\eqalign{
dh = d\log(v)
 &= V^{-1}dv\;=\; V^{-1}A\,dx \\
}$$
Now we're ready to calculate the differential and gradient of the current function.
$$\eqalign{
f &= a:h \\
df &= a:dh \\
  &= a:V^{-1}A\,dx \\
  &= A^TV^{-1}a:dx \\
\p{f}{x} &= A^TV^{-1}a \;=\; g\qquad({\rm gradient}) \\
}$$
The hessian is the gradient of the gradient,
so start the calculation with the differential of $g$.
$$\eqalign{
dg &= A^T\color{red}{dV^{-1}}a \\
  &= A^T\color{red}{(-V^{-1}dV\,V^{-1})}a \\
  &= -A^TV^{-2}dV\,a \\
  &= -A^TV^{-2}\D(a)\;dv \\
  &= -A^TV^{-2}\D(a)\,A\;dx \\
\p{g}{x} &= -A^TV^{-2}\D(a)\,A \;=\; H\qquad({\rm hessian}) \\
}$$

A colon has been used to denote the trace/Frobenius product, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \;=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \\
}$$
This definition can also be applied to vectors by treating them as rectangular matrices (set $n=1$) in which case it is equal to the standard dot product.

The terms in a Frobenius product can be rearranged in a number of ways, e.g.

$$\eqalign{
A:B &= B:C \;=\; B^T:A^T \\
CA:B &= A:C^TB \;=\; C:BA^T \\
}$$
due to the properties of the underlying trace function.

Some of the steps in the derivation used certain properties of diagonal matrices. In terms of the vectors {$a,b$} these properties can be written as
$$\eqalign{
\D(a)\cdot\D(b) &= \D(b)\cdot\D(a) \\
\D(a)\cdot b &= \D(b)\cdot a \\
}$$

Finally, one of the comments asked about the product rule for differentials. Here is the general rule for arbitrary tensors {$S,T$} and any product {$\star$} with which they are dimensionally compatible.
$$\eqalign{
d(S\star T) &= (S+dS)\star(T+dT) \;-\; S\star T \\
 &= \big(S\star T +S\star dT +dS\star T +dS\star dT\big) \;-\; S\star T \\
 &= S\star dT + dS\star T + (\enclose{horizontalstrike}{dS\star dT}) \\
 &= S\star dT + dS\star T \\
}$$


问:
every $x\in V$ can be decomposed uniquely as $x=v+cz$,
Let $V$ be a vector space and $L: V\mapsto R$ be a linear map. If $z\in V$ is not in the nullspace of $L$, show that every $x\in V$ can be decomposed uniquely as $x=v+cz$, where $v$ is in the nullspace of $L$ and $c$ is a scalar.

How can I start thinking on this problem? Please give some hints.

Edit : $V$ is finite dimensional

答:
Take $z_1:=\dfrac z{L(z)}$ and set $c=L(x)$, then $x-cz_1\in\ker L$.

Also verify that no other $c$ works.


问:
Convergence of the recursive sequence $x_{n+1} = \frac{1}{4-x_n}$
I'm doing exercise 2.4.1 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, if my proof is rigorous and technically correct.

(a) Prove that the sequence defined by $x_1 = 3$ and
\begin{align*}
    x_{n+1} = \frac{1}{4 - x_n}
\end{align*}

converges.

(b) Now that we know $\lim x_n$ exists, explain why $\lim x_{n+1}$ must also exist and equal the same value.

(c) Take the limit of each side of the recursive equation in part (a) to explicitly compute the $\lim x_n$.

Proof.

(a) By direct computation, we find that $x_1 = 3$, $x_2 = 1$. Let us prove that $(x_n)$ is a decreasing sequence, that is $x_{n+1} < x_{n}$ for all $n \in \mathbf{N}$. This is true for $n=1$. By induction, let's assume that $x_{k+1} < x_k$. Therefore,
\begin{align*}
 x_{k+1} &< x_{k} \implies \frac{1}{4 - x_{k+1}} < \frac{1}{4 - x_k} \implies x_{k+2} < x_{k+1}
\end{align*}

So, $(x_{n})$ is a monotonically decreasing sequence.

Moreover, we can show that $(x_n)$ is bounded. We are interested to show that $x_n > 0$ for all $n \in \mathbf{N}$. This holds for $n=1$. Assume that $x_k > 0$, then
\begin{align*}
 x_{k+1} = \frac{1}{4 - x_k} > \frac{1}{4} > 0
\end{align*}

Thus, the sequence $(x_n)$ has a lower bound $0$. By the Monotone Convergence Theorem, the sequence $(x_n)$ converges.

(b) The sequence $(x_{n+1})=x_2,x_3,x_4,\ldots$ also converges and has the same limiting value because the $1$-tail of $(x_n)$ is a (i) monotonically decreasing sequence (ii) has the same lower bound $0$. It is the infinite tail of the sequence, that ultimately determines the convergence of a sequence.

(c) We have:
\begin{align*}
 \lim (x_n) &= \frac{1}{4 - \lim x_{n+1}}\\
 L &= \frac{1}{4 - L} \\
 4L - L^2 &= 1\\
 L^2 - 4L + 1 &= 0\\
 (L - 2)^2 - 3 &= 0\\
 L&= 2 \pm \sqrt{3}
\end{align*}

As $0 < L < 1$, we have $L = 2 - \sqrt{3}$.

答:
An alternative solution.

It is clear from the form of the recursion that there is a sequence $\{y_n\}_{n=1}^\infty$ such that $x_n=\frac{y_n}{y_{n+1}}$.

$$\frac{y_{n+1}}{y_{n+2}}=\frac{1}{4-\frac{y_n}{y_{n+1}}}=\frac{y_{n+1}}{4y_{n+1}-y_{n+2}}\implies y_{n+2}=4y_{n+1}-y_n$$.

The base case is $y_1=3$ and $y_2=1$.

$$\lambda^2=4\lambda-1\implies \lambda=2\pm\sqrt{3}\implies y_n=C_+(2+\sqrt{3})^n+C_-(2-\sqrt{3})^n$$

We don't need to solve for the two constants, but their values are $C_{\pm}=\frac{33\mp 19\sqrt{3}}{6}$.

Nevertheless, $2-\sqrt{3}<1\implies\lim_{n\to\infty}(2-\sqrt{3})^n=0$ and so:

$\lim_{n\to\infty}x_n=\lim_{n\to\infty}\frac{C_+(2+\sqrt{3})^n+C_-(2-\sqrt{3})^n}{C_+(2+\sqrt{3})^{n+1}+C_-(2-\sqrt{3})^{n+1}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$.

Notice that this essentially solves all three parts together.

(a) The computation shows that the limit exists.

(b) We are just shifting the index.

(c) The limit's existence was established by actually finding it to be $2-\sqrt{3}$.


问:
Interesting topologies on $\mathbb{R}$?
It is evident in with respect to the usual topology, the sequence $0.9,0.99,0.999,0.9999,\ldots$ converges to $1$.

I am also aware that the same sequence does not converge to $1$ with respect to the lower limit topology.

My question is: Does there exist a topology on $\mathbb{R}$ with respect to which the aforementioned sequence converges to a real number that is NOT $1$?

答:
Yes, of course. A couple of uninteresting examples:

in the indiscrete topology that sequence converges to $2$. And to $1$, and to $0$, and to every other point.

if $f(x)=\begin{cases}0&\text{if }x=1\\ 1&\text{if }x=0\\ x&\text{if }x\ne 0\land x\ne 1\end{cases}$ and $\tau=\{f^{-1}[\Omega]\,:\, \Omega\text{ open in the Euclidean topology}\}$, then that sequence $\tau$-converges to $0$ and to no other point.

Title: Interesting topologies on $\mathbb{R}$?


答: Yes. Take any injective function $f\colon\Bbb R\longrightarrow\Bbb R$ such that $f(x)=x$ for any $x$ from your sequence, but such that $f(1)\neq1$; for instance, you can take$$f(x)=\begin{cases}x&\text{ if }x<1\\x+1&\text{ if }x\geqslant1.\end{cases}$$Then consider the topology induced by the distance $d$ defined by$$d(x,y)=\bigl|f(x)-f(y)\bigr|.$$With repect to this topology, your sequence converges to $f(1)$.

Title: Interesting topologies on $\mathbb{R}$?

Body:
It is evident in with respect to the usual topology, the sequence $0.9,0.99,0.999,0.9999,\ldots$ converges to $1$.

I am also aware that the same sequence does not converge to $1$ with respect to the lower limit topology.

My question is: Does there exist a topology on $\mathbb{R}$ with respect to which the aforementioned sequence converges to a real number that is NOT $1$?


答: Take into account that a topology on a set defines its structure, without it $\mathbb R$ is the same as any other set with $\aleph_1$ cardinality. Take any real point $a$ and define the distance in $\mathbb R$ as $d(x,y)=|x-y|$ if $x,y\not\in\{1,a\}$, $d(1,y)=d(y,1)=|a-y|$ and $d(a,y)=d(y,a)=|1-y|$, it is clear that under this topology any sequence that converges to $1$ in the usual topology converges to $a$ in this one.

Title: Interesting topologies on $\mathbb{R}$?


答:
Let the dimension of $V$ be $n$. Since dimension of $KerL = n -1$. Let $\{A_1,A_2,...A_{n-1}\} $ be a basis for $KerL$. By basis extension theorem extend this to a basis of $V$ by adding $A_n$ where $A_n $ not in $ KerL$. So for any $x \in V$ we have $x = a_1A_1 + \cdots + a_{n-1}A_{n-1} +a_n A_n$ and observe that for given $x$ each $a_i$ is unique. Now see that
$$
L (x ) = a_n L(A_n) \implies L(x - a_n A_n) = 0
$$
From this we conclude that $x - a_n A_n \in KerL$ so we can say that $x = v + cz$  where $v \in KerL$ , $c =a_n$ and $z = A_n$. To show uniqueness let $x = v_1 + a_nA_n = v_2 + b_n B_n$ then we see that $L(a_n A_n - b_nB_n) = 0$ but $a_n A_n - b_nB_n \in KerL$ only if $a_n A_n - b_nB_n = 0$. Now it follows from the axioms of vector space that $v_1 = v_2$ and we are done.

Body:
Let $V$ be a vector space and $L: V\mapsto R$ be a linear map. If $z\in V$ is not in the nullspace of $L$, show that every $x\in V$ can be decomposed uniquely as $x=v+cz$, where $v$ is in the nullspace of $L$ and $c$ is a scalar.

How can I start thinking on this problem? Please give some hints.

Edit : $V$ is finite dimensional


问:
Uniformly solvable families of polynomials
It is a famous theorem that there is no "quintic formula", i.e. there is no formula which expresses the roots of a quintic polynomial $x^5+a_4x^4+\cdots+a_0$ in terms of $a_4,\ldots,a_0$ and rational constants using the field operations and taking $n$th roots. But the standard proof of this actually proves something which on its face seems stronger: that there are particular quintics with roots that cannot be expressed in terms of rational numbers using these operations. The non-existence of a quintic formula merely requires that there be no "uniform" way of doing so. I'm wondering if this is actually a stronger statement.

To make this rigorous: Given $f(a_0,\ldots,a_k,x)\in \mathbb Q(a_0,\ldots,a_k)[x]$, we can define the family of polynomials associated to $f$ to be the set of polynomials $f(t_0,\ldots,t_k,x)\in \mathbb Q[x]$ where $t_0,\ldots,t_k\in \mathbb Q$ are such that this expression is defined. For example, the family of polynomials associated to $x^2+a_1x+a_0$ is the set of monic quadratic polynomials. If there an $f\in \mathbb Q(a_0,\ldots,a_k)[x]$ which does not admit a formula for its roots (in the variable $t$) in terms of elements of $\mathbb Q(a_0,\ldots,a_k)$, field operations and taking $n$th roots, yet every polynomial in the family of polynomials associated to $f$ does?

From the point of view of Galois theory, the question becomes whether there exist polynomials $f\in \mathbb Q(a_0,\ldots,a_k)[x]$ which are not solvable over $\mathbb Q(a_0,\ldots,a_k)$, but evaluating them at any point $(t_0,\ldots,t_k)\in\mathbb Q$ (for which the result is defined) gives a polynomial solvable over $\mathbb Q$?

答:
It appears to me the answer is essentially no, but there are some details to be checked. The main observation is this: under reasonable conditions, the Galois group of the generic fibre is canonically isomorphic to the Galois group of almost all fibres; thus, the generic fibre is solvable if and only if almost all fibres are solvable.

In more detail: Let $L$ be a finite Galois extension of $K$. By the primitive element theorem, $L$ is generated by a single element over $K$, say $x$. Let $f$ be the (monic) minimal polynomial of $x$ over $K$ and choose a subring $A \subseteq K$ satisfying these conditions:

$A$ is a noetherian integral domain.

$\operatorname{Frac} A = K$.

$f \in A [x]$.

The ring $B = A [x] / (f)$, considered as a subring of $L$, is closed under the action of $\mathrm{Gal}(L \mid K)$.

$B$ is a projective $A$-module of rank $d = \deg f$.

This is certainly achievable in the case $K = \mathbb{Q} (t_1, \ldots, t_n)$ that you are considering: take $A$ to be the subring generated by $\mathbb{Q} [t_1, \ldots, t_n]$, the coefficients of $f$, and the coefficients of all the Galois conjugates of $x$; then $B$ will even be a free $A$-module of rank $d$.

Lemma 1. $\mathrm{Aut}(B \mid A)$, the automorphism group of $B$ as an $A$-algebra, is canonically isomorphic to $\mathrm{Gal}(L \mid K)$.

Proof. The hypothesis implies that each automorphism of $L \mid K$ restricts to an automorphism of $B \mid A$; since $B$ contains the generator of $L$, the restriction homomorphism $\mathrm{Gal}(L \mid K) \to \mathrm{Aut}(B \mid A)$ is injective. On the other hand, $\operatorname{Frac}$ is functorial, so any automorphism of $B \mid A$ extends to an automorphism of $L \mid K$, so the restriction homomorphism is in fact an isomorphism.　◼

Lemma 2. Let $\mathfrak{m}$ be a maximal ideal of $A$ and let $\kappa (\mathfrak{m}) = A / \mathfrak{m}$ be the residue field.

$B (\mathfrak{m}) = B \otimes_A \kappa (\mathfrak{m})$ is a $\kappa (\mathfrak{m})$-algebra of dimension $d$.

$B (\mathfrak{m})$ is a field if and only if $f$ is irreducible over $\kappa (\mathfrak{m})$.

There is a canonical homomorphism $\mathrm{Aut}(B \mid A) \to \mathrm{Aut}(B (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$, and it is injective if and only if distinct Galois conjugates of $x$ remain distinct in $B (\mathfrak{m})$.

Proof. All straightforward.　◼

Lemma 3. There exists a dense open subset $U \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ that are in $U$, the Galois conjugates of $x$ remain distinct in $B (\mathfrak{m})$.

Proof. Two Galois conjugates of $x$, say $x'$ and $x''$, become equal in $B (\mathfrak{m})$ precisely if a certain finite set of equations holds in $B (\mathfrak{m})$; more precisely, there exist elements $h_1, \ldots, h_d$ in $B$ such that $x' \equiv x'' \pmod{\mathfrak{m}}$ if and only if $h_1 \equiv \cdots \equiv h_d \equiv 0 \pmod{\mathfrak{m}}$. This is because the $A_{\mathfrak{m}}$-module $B_{\mathfrak{m}}$ is free of rank $d$ (since $B$ is a projective $A$-module of rank $d$). Thus $x' \equiv x'' \pmod{\mathfrak{m}}$ if and only if $\mathfrak{m}$ is in a certain closed subset of $\operatorname{Spec} A$ of codimension $\ge 1$ (by Krull's Hauptidealsatz). The complement of such a closed subset is a dense open subset of $\operatorname{Spec} A$ (because $A$ is an integral domain), and the intersection of finitely many dense open subsets in $\operatorname{Spec} A$ is again a dense open subset, so we are done by the finiteness of $\mathrm{Gal}(K \mid L)$.　◼

Theorem. If there exists a dense open subset $U' \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ in $U'$, $f$ is irreducible over $\kappa (\mathfrak{m})$, then there exists a dense open subset $U'' \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ in $U''$, $B \otimes_A \kappa (\mathfrak{m})$ is a Galois extension of $\kappa (\mathfrak{m})$ and the canonical homomorphism
$$\mathrm{Gal}(K \mid L) \cong \mathrm{Aut}(B \mid A) \to \mathrm{Gal}(B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$$
is an isomorphism.

Proof. Let $U$ be as in lemma 3, and take $U'' = U \cap U'$. This is a dense open subset, and by lemma 2, for any $\mathfrak{m}$ in $U''$, $B \otimes_A \kappa (\mathfrak{m})$ is a field and the canonical homomorphism is injective. However, Dedekind's lemma says that the size of $\mathrm{Aut}(B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$ is bounded above by the degree of $B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m})$, so in fact the canonical homomorphism is a bijection, and moreover $B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m})$ is a Galois extension.　◼

Remark. It's not clear to me when the main hypothesis of the above theorem holds. There is definitely something non-trivial to be checked: after all, if we take $K = \mathbb{Q}$ and $A = \mathbb{Z}$, then we have to contend with the fact that there exist polynomials that are reducible mod $p$ for every prime $p$ but yet irreducible over $\mathbb{Q}$. But perhaps in the case you are considering this cannot happen.


答:
If you are after a random basis and you know some statistical properties of the target family of waveforms, have you looked into using a Karhunen-Loève Expansion (KL-Expansion)? In a KL-Expansion, the set of eigen-vectors of the auto-correlation operator of the signal x(t) is a basis (demonstrates convergence in probability) for x(t).

In the discrete x(n) case, you can calculate a quasi-basis using an auto-correlation matrix of x(n) and Cholesky Decomposition.
See Clarkson (1993) Optimal and Adaptive Signal Processing page 131

Title: DFT but using non-sinusoidal periodic waveforms.

Body:
Is there an established way of decomposing a discrete periodic (complex) signal into a sum of non-sinusoidal periodic waveforms (eg square, triangle, and sawtooth)?

For my use case the input waveform is periodic, but as I understand that's a condition which always holds for DFT even if it's easy to work around it in practical applications.


问:
How to prove that $n!>2^n$ holds for all $n>3$?
I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$

答: Since $n\ge 4$ we have $n! = 1\cdot 2\cdot 3\cdot 4 \cdot 5\cdots n = 24 \cdot 5 \cdots n > 16 \cdot 5 \cdots n\ge  2^4\cdot 2 \cdots 2 = 2^42^{n-4}=2^n$


问:
Using Wikipedia links for natural language processing or how to "solve" a directed graph?
Suppose that the Wikipedia directed graph of internal links is given as below ( Starting with the article "Automobil" / Auto in German and then following at most 3 internal links in the first sentence of the article.)

In natural language processing typically one assings to words or text a vector to represent the meaning of the the word / text.

Usually when one reads a wikipedia article, one clicks to outgoing internal links to get an understanding of the words which this article is referring to.

So having this in mind I want to assign vectors to wikipedia articles in this way:

If $v$ is a wikipedia article then the meaning of this article depends on the other $N$ internal articles this article $v$ is referring to:

$$v = \frac{1}{N} \sum_{ v \rightarrow w } w$$

For an article  which does not have outgoing links, we assign a new standard basis vector.

My mathematical question, is if it is possible to assign vectors $v$ such that for any given directed graph structure $G$, the last equation is valid. ( I have tested this with random graphs, and it seems to be possible)

Second question: Is there a fast method to solve this problem?

If this is possible, then one could add "words" (the corresponding vectors) to change the meaning of the added words. For example, for the graph below we have:

```<code>vAuto == r6
vMotor == r3
vKraftfahrzeug == 1/2*r3 + 1/2*r5
vStrassenfahrzeug == -3/2*r3 - 1/2*r5 + 3*r6
vArbeit == 1/2*r1 + 1/2*r2
vKraftmaschine == -1/2*r1 - 1/2*r2 + 2*r3
vFahrzeug == r5
vOberbegriff == 2*r5 - r8
vVerkehrsmittel == r8
vLandfahrzeug == -3*r3 - r5 + 6*r6 - r7
vStrassen == r7
vLand == -6*r3 - 3*r5 + 12*r6 - 2*r7
vLandfläche == -r4 + 2*r7
vVerkehrsbauwerk == r4
vMaschine == -r1 - 2*r2 + 4*r3
vEnergie == r2
vPhysik == r1
</code>```

So for example "Work = 1/2 Energy + 1/2 Physics".

If we have a new text $T$ which contains words which correspond to wikipedia articles, then one could simply add these articles in a weighted form, an have a meaning for this text.

Thanks for your help.

Example Graph in Sagemath and "solution"

Edit:
By the accepted answer given below, here is the example graph with absorbing states implemented in SAGEMATH. The probabilities are computed as $B = NR$ as described here: https://en.wikipedia.org/wiki/Absorbing_Markov_chain.

答:
Assume we have $n$ articles without outgoing links and $m$ with them. We will show that we can assign $n$-dimensional vector to each article satisfying your conditions.

First, assign vector $e_i$ to $i$-th article without outgoing links. Each possible assignment of vectors to remaining $m$ articles can be thought as point in $\mathbb R^{m \times n}$. We will consider only assignments such that each coordinate is in segment $[0, 1]$ - corresponding points in $\mathbb R^{m \times n}$ are $[0, 1]^{m \times n}$ - a unit cube, which is a convex compact.

Now, consider standard iteration - function $F: [0, 1]^{m \times n} \to [0, 1]^{m \times n}$ given by following rule:
$$F(x_1, x_2, \ldots, x_{n + m})_i = \begin{cases}
e_i,\ i \leq n\\
\frac{1}{N_i} \sum\limits_{j: v_i \to v_j} x_j
\end{cases}$$

Informally, $F$ just naively updates each vector independently, so it would be correct if we didn't update others.

Fixed points of $F$ are exactly valid assignments. And by Brouwer theorem, $F$ has a fixed point as a continuous function on a compact set.

Now, as we know that there is a solution, we can apply any standard method to solve linear system (it's probably sparse, which can be taken advantage of).


问:
Direction Ratios of a Line in 2D
Consider the general equation of a line in 2-D $ax+by+c=0$. How do I deduce the direction ratios of this line from the given format?

答:
For $a,b\ne0,(-c/a,0),(0,-c/b)$ are two points lying on the line. So the line is parallel to their difference vector $(-c/a,c/b)$ i.e. the vector $(b,-a)$.

The same result is obtained by noting that the slope of the line $-a/b$ is equal to $y/x$ where $x,y$ are the d-ratios along $x,y$ axes respectively.

Title: Direction Ratios of a Line in 2D


问:
Proof of the definition of Frobenius norm
We already know from the definition of Frobenius norm that : \begin{equation}
\boxed{\mathit{||A||}_{F}=\sqrt{\mathop{\mathrm{trace}}(A^*A)}.}
\end{equation}
But how can we prove it since the trace of a matrix $X \in \mathbb{R}^{n\times n}$ is defined to be the sum of its diagonal elements such that :
$ \mathop{\mathrm{trace}}(X) = \sum_{i=1}^{n} \mathit{X}_{ii}$.

答:
Let $A$ be an $m\times n$ (complex) matrix, the Frobenius norm is originally defined as:
\begin{align}
\left\Vert A\right\Vert_F^2
=\sum_{i,j}\vert a_{ij}\vert^2
=\sum_{i,j}a_{ij}^\ast a_{ij}.
\end{align}
On the other hand, observe that $A^\ast A$ is of size $n\times n$, and we can write the diagonal entries of $A^\ast A$ explicitly:
\begin{align}
A^\ast A=\begin{pmatrix}
\sum_{i}a_{i1}^\ast a_{i1}&*&\cdots&*\\
*&\sum_{i}a_{i2}^\ast a_{i2}&\cdots&*\\
\vdots&\vdots&\ddots&\vdots\\
*&*&\cdots&\sum_{i}a_{in}^\ast a_{in}
\end{pmatrix}.
\end{align}
This reveals that
\begin{align}
\operatorname{trace}(A^\ast A)
=\sum_{j=1}^n\sum_{i}a_{ij}^\ast a_{ij}
=\left\Vert A\right\Vert_F^2.
\end{align}


答:
Here’s a neat way to think about it:

When $x$ is increased by, say, $1$, then $y$ must be decreased by $\frac ab$ to counter the change, as $$a(x+1) +b\left(y-\frac ab\right) +c =ax + a+by -a +c = ax+by+c=0$$ Hence, the direction vector of the line is $\left(1,- \frac ab\right)$ or equivalently $(b,-a)$.

Title: Direction Ratios of a Line in 2D

Body: Consider the general equation of a line in 2-D $ax+by+c=0$. How do I deduce the direction ratios of this line from the given format?


问:
Analyzing the Coefficient of a generating function and its asymptotic
The generating function
$\sum_{n\geq 0} D(n) x^n = \frac{1}{\sqrt{1-6x+x^2}}$ is the gf of the Delannoy number.

See last paragraph in https://en.wikipedia.org/wiki/Delannoy_number

In this link, they mentioned that the coefficient of this gf  is $D(n) = \sum_{k=0}^{n} {n \choose k} { n+k \choose k}$.

My question is how can this coefficient be calculated/extracted from its gf.And how they conclude that it behaves asymptotically as

${\displaystyle D(n)={\frac {c\,\alpha ^{n}}{\sqrt {n}}}\,(1+O(n^{-1}))}$

where
$\alpha =3+2{\sqrt {2}}\approx 5.828$

and

$c=(4\pi (3{\sqrt {2}}-4))^{-1/2}\approx 0.5727$.

How to calculate/extract $D(n)$ from its gf and how to analyze it asymptotically then?
Is that possible to do that with a direct calculations?

Edit
I tried to find the asymptotic of the codfficidnt in the above gf this way:

The gf has two poles: $r_1=3+\sqrt{8}$, $r_2=3-\sqrt{8}$. So the smallest/dominant pole is $r_2=3-\sqrt{8}$.
We can write the gf as:
$A(x)=1/ \sqrt{r_2-x} * 1/ \sqrt{r_1-x}$
Then A(x) behaves as:

$1/ \sqrt {r_1-r_2} * 1/ \sqrt{r_2-x}=
1/ \sqrt {2 * 8^{1/4}} * \sum_{n\geq 0} {-1/2 \choose n} (-1)^n (r_1)^{n+1}x^n$
Here I used the expansion's formula  of $(1+x)^a$ for $a=-1/2$ and $x=-x/r_2$.
So the coefficient behaves asymptotically as:
$1/ \sqrt 2 * 1/ 8^{1/4} * (3+\sqrt 8)^{n+1/2} {-1/2 \choose n} (-1)^n$.

We can simplify it more, by using the asymptotic of ${-1/2 \choose n}$.

Is that approach fine?

答:
We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n}{k}\binom{n+k}{k}}\\
&=\sum_{k=0}^n\binom{n}{k}[x^k](1+x)^{n+k}\tag{2}\\
&=[x^0](1+x)^n\sum_{k=0}^n\binom{n}{k}\left(\frac{1+x}{x}\right)^k\tag{3}\\
&=[x^0](1+x)^n\left(1+\frac{1+x}{x}\right)^n\tag{4}\\
&=[x^{-1}](1+x)^n(1+2x)^n\frac{1}{x^{n+1}}\tag{5}\\
&=[y^{-1}]\left.\frac{1}{(1+x)(1+2x)\,\frac{1-2x^2}{(1+x)^2(1+2x)^2}}\right|_{x=g(y)}\cdot\frac{1}{y^{n+1}}\tag{6}\\
&=[y^n]\left.\frac{(1+x)(1+2x)}{1-2x^2}\right|_{x=g(y)}\tag{7}\\
&\,\,\color{blue}{=[y^n]\frac{1}{\sqrt{1-6y+y^2}}}\tag{8}\\
\end{align*}
and the claim follows.

Comment:

In (2) we use the coefficient of operator according to (1).

In (3) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

In (4) we apply the binomial theorem.

In (5) we prepare for the substitution rule which is the essence of this derivation.

In (6) we use the substitution rule (Rule 5, one-dimensional case) from section 1.2.2 of G. P. Egorychev's Classic: Integral Representation and the Computation of Combinatorial Sums. Here we have
\begin{align*}
f(x)=(1+x)(1+2x)\qquad\qquad y(x)&=\frac{x}{f(x)}=\frac{x}{(1+x)(1+2x)}\tag{9}\\
y^{\prime}(x)&=\frac{1-2x^2}{(1+x)^2(1+2x)^2}
\end{align*}
and apply the substitution rule:
\begin{align*}
[x^{-1}]f(x)^n\cdot \frac{1}{x^{n+1}}=[y^{-1}]\left.\frac{1}{\left(f(x)\cdot y^{\prime}(x)\right)}\right|_{x=g(y)}\cdot\frac{1}{y^{n+1}}
\end{align*}
where $x=g(y)$ is the inverted function given by (9).

In (7) we do a simplification and apply the rule as we did in (3).

In (8) we note that with $y(x)=\frac{x}{(1+x)(1+2x)}$ we obtain:
\begin{align*}
1-6y+y^2&=1-\frac{6x}{(1+x)(1+2x)}+\frac{x^2}{(1+x)^2(1+2x)^2}\\
&=\frac{(1+x)^2(1+2x)^2-6x(1+x)(1+2x)+x^2}{(1+x)^2(1+2x)^2}\\
&=\frac{(2x^2-1)^2}{(1+x)^2(1+2x)^2}\\
\frac{1}{\sqrt{1-6y+y^2}}&=\frac{(1+x)(1+2x)}{1-2x^2}
\end{align*}

Notes:

Another approach of the asymptotic coefficient expansion of $\frac{1}{\sqrt{1-6x+x^2}}$ can be found in this answer.

A somewhat more demanding application of the substitution rule (two-dimensional case) is given in this answer.


问:
How to compute next/previous representable rational number?
An (approximate) non-negative rational number representation is a pair of natural numbers each not greater than some fixed limit M (and of course denominator being non-zero).

With this condition there is finite number of representable rational numbers. This means that for each such number we can name previous and next number in the set (of course except for smallest 0 and largest 1). How to compute them?

In other words lets have a set of numbers

$$
R(M)=\{\frac{p}{q} : p,q\in\mathbb{N},q\neq0,p\leq q\leq M\}
$$

where $M\in\mathbb{N}_+$

and given number $\frac{p_1}{q_1}\in R(M)$.

I want to find numbers $\frac{p_S}{q_S}\in R(M)$ and $\frac{p_L}{q_L}\in R(M)$ (if they exist) such that

$$
\frac{p_S}{q_S}<\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_S}{q_S}<\frac{p}{q}<\frac{p_1}{q_1}
$$

and smilarly

$$
\frac{p_L}{q_L}>\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_1}{q_1}<\frac{p}{q}<\frac{p_L}{q_L}
$$

答:
Check out the next link:
http://en.wikipedia.org/wiki/Farey_sequence .
In general, Farey sequence is dealing exactly with that type of question.</p>


问:
Does having multiple limit values at a point imply essential discontinuity?
In Complex Analysis, do "jump discontinuities" exist?

If I find that a function of $z$ approaches two different values as z is approached from two different directions, can I immediately conclude that the point is an essential singularity? Or do other possibilities exist?

答:
Your question did not specify the domain of definition of your function, so it's hard to give a precise answer. (You have also used $z$ to mean two different things in the same sentence.) However, your function definitely does not have a limit at $z$, which does imply that $z$ is an essential singularity of your function.

(I suspect you mean "essential singularity" rather than "essential discontinuity"; the latter means, by definition, that the limit does not exist, and so your function certainly has an essential discontinuity as well.)


问:
Basis and transcendental basis
I need to have really good picture in my mind about transcendental basis and What are the similarity between basis and transcendental basis. Let $\mathbb R$  be the set of all really number. Let $\mathbb B$ be a basis of $\mathbb R$ over $\mathbb Q$ so it is a maximal linear independent set and  it spans all element of $\mathbb R.$ Also, each element of $\mathbb R$ can be written as linear combination of finitly many element of $\mathbb B.$

Let $T$ be a transcendental basis of $\mathbb R$ over $\mathbb Q$ so it is a maximal algebraically independent set. Also, $\mathbb Q(T)=\mathbb R.$

Is it true that each element of $\mathbb  R$ can  be spanned by finitely number algebraically independent elements of $T$? if so, why.

Can we choose for each element of $\mathbb R$ different element of $T$ that spans it?  if so , why ?

what we can also say about $T$?

I hope from those that have a deep understanding for this subject to say more about transcendental basis and algebraically independent set?

Any help will be appreciated greatly

答:
You're conflating transcendence basis and vector space (Hamel) basis.

If $K$ is an extension field of $F$, there exists a set $T$ consisting of transcendental elements of $K$ such that

$T$ is algebraically independent;

$K$ is algebraic over $F(T)$.

Such a set is a transcendence basis of $K$ over $F$.

A set $U\subseteq K$ is algebraically independent if and only if, for every finite subset $S=\{a_1,a_2,\dots,a_n\}$ (elements pairwise distinct) of $U$, there is no nonzero polynomial $f(X_1,\dots,X_n)\in F[X_1,\dots,X_n]$ such that $f(a_1,\dots,a_n)=0$.

Using a technique very similar to the proof of existence of vector space basis, one can show that a transcendence basis exists and that any maximal algebraically independent set is such a basis; also, two transcendence bases have the same cardinality.

So let $T$ be a transcendence basis of $K$ over $F$. It is not possible to prove that $K=F(T)$, when $T$ is a transcendence basis. To see why, consider the simple case when $K=F(t)$, where $t$ is transcendental over $F$; then also $\{t^2\}$ is a transcendence basis, but clearly $K\ne F(t^2)$. On the other hand, $t$ is algebraic over $F(t^2)$, so $K$ is indeed algebraic over $F(t^2)$. Perhaps more simply, if $K$ is algebraic over $F$, then the empty set is a transcendence basis, but $K$ need not equal $F$.

In general, if $T$ is nonempty transcendence basis and you replace one of its elements, say $t$, by its square $t^2$, then the set $T'$ so obtained is still a transcendence basis, but definitely $F(T')\subsetneq F(T)$. So condition 2 above cannot be replaced by $K=F(T)$.

In the finite case $K=F(t)$ it is certainly false that a transcendence basis is also a vector space basis: indeed, $K=F(t)$ is infinite dimensional as vector space over $F$.

In the case of $F=\mathbb{Q}$ and $K=\mathbb{R}$ a transcendence basis must have the same cardinality as $\mathbb{R}$, by a cardinality argument: we have
$$
\mathbb{Q}(T)=\bigcup_{\substack{S\subseteq T\\S\text{ finite}}}\mathbb{Q}(S)
$$
and every subfield $\mathbb{Q}(S)$ is countable.

Also a vector space basis has the same cardinality as $\mathbb{R}$, but a transcendence basis cannot be a vector space basis. Let $T'=T\setminus\{t\}$, where $t\in T$ is fixed. Then $\mathbb{Q}(T)=F(t)$, where $F=\mathbb{Q}(T')$. Then $\mathbb{Q}(T)$ is infinite dimensional over $F$. If you remove one element from a vector space basis, the subspace spanned by the new set has codimension one.

What you can say is that every $r\in\mathbb{R}$ is algebraic over $\mathbb{Q}(S)$, where $S$ is some finite subset of $T$. Indeed, if $f$ is the minimal polynomial of $r$ over $\mathbb{Q}(T)$, it has finitely many coefficients, so it belongs to $\mathbb{Q}(S)[X]$, for some finite $S\subseteq T$.

For completeness, $\mathbb{R}$ is not a purely transcendental extension of $\mathbb{Q}$, that is, for every transcendence basis $T$ of $\mathbb{R}$ over $\mathbb{Q}$, $\mathbb{Q}(T)\subsetneq\mathbb{R}$. Indeed if equality holds, any permutation of $T$ would induce an automorphism of $\mathbb{R}$, but $\mathbb{R}$ has only the identity automorphism.

Title: Basis and transcendental basis


问:
Are there non-homogeneous topological spaces which are "almost homogeneous" in Rudin's sense?
Let $X$ be a nonempty topological space and $G$ the group of all homeomorphisms $g:X\to X$. Recall that $X$ is homogeneous if $G$ acts transitively on $X$. In

"Homogeneity Problems in the Theory of Čech Compactifications", The Mathematical Legacy of Eduard Čech (1993), pages 81-92,

Walter Rudin considers the following condition on $X$: for all point $x\in X$ and all nonempty open subset $U\subset X$ there is a $g\in G$ such that $g(x)\in U$. He calls such a space almost homogeneous.

I suspect that there are simple examples of almost homogeneous non-homogeneous spaces, but I haven't been able to find any. Let me ask the implicit question formally, together with an obvious follow-up question:

Question 1. Are there almost homogeneous non-homogeneous spaces?

Question 2. Assuming the answer to Question 1 is Yes, what is the least possible cardinality of an almost homogeneous non-homogeneous space?

[I added the quotient-spaces tag because, in the above notation, $X$ is almost homogeneous if and only if the quotient topology of $G\backslash X$ is coarse.]

答:
A countable example is given by the prime spectrum of the ring $\mathbb Z$.

For the reader not familiar with prime spectra, here is a self-contained definition of a topological space $X$ homeomorphic to the prime spectrum of $\mathbb Z$.

As a set $X$ is the union of $\mathbb N$ and $\{\infty\}$ with $\infty\notin\mathbb N$. The topology of $X$ is characterized by the condition that the proper closed subsets of $X$ are precisely the finite subsets of $\mathbb N$.

If $G$ is the group of all self-homeomorphisms of $X$, then a bijection $g:X\to X$ is in $G$ if and only if $g(\infty)=\infty$. The orbits of $G$ in $X$ are $\mathbb N$ and $\{\infty\}$, each of them being dense in $X$.


问:
How to determine the integral limits of continuous wavelet transform?
Now I have one function $f(x)$, and its support interval is $x\in[T_0,T_1]$. The support interval of the mother wavelet $\psi(x)$ is $x\in[-\Delta,+\Delta]$. And we also know that the continuous wavelet transform is
$$
W(a,b) = \frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty} f(t)\Psi(\frac{t-b}{a})\text{d}t
$$

According to the formula above, It seems like you need to do the integration from $-\infty$ to $+\infty$. However, we don't have to do it since that $f(x)$ and $\psi(x)$ are compact support. So my question is How do we determine the appropriate integral limits?

Here is my attempt:

According to "the support interval of the mother wavelet $\psi(x)$ is $x\in[-\Delta,+\Delta]$", we know that
$$
-\Delta\leq\frac{t-b}{a}\leq\Delta\, \,\Rightarrow \, \, b-a\Delta\leq t \leq b+a\Delta
$$

And because that the support interval of $f(t)$ is $[T_0,T_1]$ , we have
$$
T_0\leq t \leq T_1
$$

But I don't know how what I should do next. Whether could we determine the integral limits by these two inequalities above?

答:
It looks like your integration limits should just be the intersection of the intervals you specified. That is, the product $f(t)\Psi\left(\frac{t-b}{a}\right)$ is zero whenever
$$t\notin A \triangleq\left([b−aΔ, b+aΔ]\bigcap[T_0, T_1]\right)$$
So
$$
\frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty}f(t)\Psi\left(\frac{t-b}{a}\right)dt
=
\frac{1}{\sqrt{a}}\int_{\min A}^{\max A}f(t)\Psi\left(\frac{t-b}{a}\right)dt
$$
Example: For Δ=10, a=1, b=2, $T_0=-3$, $T_1=35$, $\min A=-3$ and $\max A=12$


问:
Inverse of a 3x3 block matrix
I would like to get the inverse of a 3x3 (covariance) block matrix

\begin{bmatrix}A&B&C\\B'&D&E\\C'&E'&F\end{bmatrix}

where the prime ' indicates the transposition operator.

Is there any general formula (or a way to solve this problem)?

Thank you very much.

答:
As user3556214 points out, one can apply the formula for inverting a 2×2 block matrix repeatedly, but it does not lead to nice results.

Recall the formula for the inverse of a 2×2 block matrix:

$$
\begin{aligned}
\begin{bmatrix}
A&B\\
C&D
\end{bmatrix}^{-1}
&=
\begin{bmatrix}
A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\
-S^{-1}CA^{-1}&S^{-1}
\end{bmatrix}
\\
S &= D - C A^{-1} B
\end{aligned}
$$

Now consider a 3×3 block matrix

$$
\begin{aligned}
X &=
\begin{bmatrix}
E&F&G\\
H&J&K\\
L&M&N
\end{bmatrix}
\end{aligned}
$$

Apply the 2×2 block inverse formula, plugging in: $\tilde A=E$, $\tilde B=\begin{bmatrix}F&G\end{bmatrix}$, $\tilde C=\begin{bmatrix}H\\L\end{bmatrix}$, and $\tilde D=\begin{bmatrix}J&K\\M&N\end{bmatrix}$:

$$
\begin{aligned}
X^{-1} &=
\begin{bmatrix}
E&\begin{bmatrix}F&G\end{bmatrix}\\
\begin{bmatrix}H\\L\end{bmatrix}&\begin{bmatrix}J&K\\M&N\end{bmatrix}
\end{bmatrix}^{-1}
\\&=
\begin{bmatrix}
E^{-1}+E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}&-E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}\\
-Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}&Z^{-1}
\end{bmatrix}
\\
Z &= \begin{bmatrix}J&K\\M&N\end{bmatrix} - \begin{bmatrix}H\\L\end{bmatrix} E^{-1} \begin{bmatrix}F&G\end{bmatrix}
\end{aligned}
$$

The factor $Z$, in turn, is another 2×2 block matrix.

$$
\begin{aligned}
Z
&=
\begin{bmatrix}J&K\\M&N\end{bmatrix} - \begin{bmatrix}H\\L\end{bmatrix} E^{-1} \begin{bmatrix}F&G\end{bmatrix}
\\
&=
\begin{bmatrix}J&K\\M&N\end{bmatrix} - \begin{bmatrix}
HE^{-1}F & HE^{-1}G
\\
LE^{-1}F & LE^{-1}G
\end{bmatrix} \\
&=
\begin{bmatrix}
J-HE^{-1}F
&
K-HE^{-1}G
\\
M-LE^{-1}F
&
N-LE^{-1}G
\end{bmatrix}
\end{aligned}
$$

Again apply again the 2×2 block inverse formula to get $Z^{-1}$, defining:

$$
\begin{aligned}
A &= J-HE^{-1}F
\\
B &= K-HE^{-1}G
\\
C &= M-LE^{-1}F
\\
D &= N-LE^{-1}G
\end{aligned}
$$

$$
\begin{aligned}
Z^{-1} &=
\begin{bmatrix}
A&B\\
C&D
\end{bmatrix}^{-1}
=
\begin{bmatrix}
A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\
-S^{-1}CA^{-1}&S^{-1}
\end{bmatrix}
\\
S &= D - C A^{-1} B
\end{aligned}
$$

Further expanding the forumula for $X^{-1}$ in terms of this is tedious and somewhat unsatisfying. Define

$$
\begin{aligned}
U &= G - FA^{-1}B
\\
V &= L-CA^{-1}H
\end{aligned}
$$

then expand and simplify:

$$
\begin{aligned}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
E^{-1}&+E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\\&=
E^{-1}\left\{I+
\begin{bmatrix}F&G\end{bmatrix}
\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}
\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\right\}
\\&=
E^{-1}\left\{I+
\begin{bmatrix}F&G\end{bmatrix}
\begin{bmatrix}[A^{-1}+A^{-1}BS^{-1}CA^{-1}]H-A^{-1}BS^{-1}L\\-S^{-1}CA^{-1}H + S^{-1}L\end{bmatrix}
E^{-1}
\right\}
\\&=
E^{-1}\left\{I
+\left\{
F\left[(A^{-1}+A^{-1}BS^{-1}CA^{-1})H-A^{-1}BS^{-1}L\right]
+G\left[-S^{-1}CA^{-1}H + S^{-1}L\right]
\right\}E^{-1}
\right\}
\\&=
E^{-1}\left\{I
+
\left\{
FA^{-1}\left[H+BS^{-1}(CA^{-1}H-L)\right]
-GS^{-1}[CA^{-1}H-L]
\right\}
E^{-1}
\right\}
\\&=
E^{-1}\left\{I+\left\{FA^{-1}H+\left[FA^{-1}B-G\right]S^{-1}[CA^{-1}H-L]\right\}E^{-1}\right\}
\\&=
E^{-1}+E^{-1}\left\{FA^{-1}H+US^{-1}V\right\}E^{-1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\
\\
-&E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}
\\&=
-E^{-1}
\begin{bmatrix}F&G\end{bmatrix}
\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}
\\&=
-E^{-1}\begin{bmatrix}
FA^{-1}+FA^{-1}BS^{-1}CA^{-1}-GS^{-1}CA^{-1}
&
-FA^{-1}BS^{-1}+GS^{-1}
\end{bmatrix}
\\&=
\begin{bmatrix}
-E^{-1}\left\{
F+(FA^{-1}B-G)S^{-1}C
\right\}
A^{-1}
&
E^{-1}[FA^{-1}B-G]S^{-1}
\end{bmatrix}
\\&=
\begin{bmatrix}
-E^{-1}\left[
F-US^{-1}C
\right]
A^{-1}
&
-E^{-1}US^{-1}
\end{bmatrix}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\
\\
-&Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\\&=-\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}
\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\\&=
\begin{bmatrix}
-A^{-1}H-A^{-1}BS^{-1}CA^{-1}H + A^{-1}BS^{-1}L
\\
S^{-1}CA^{-1}H -S^{-1}L
\end{bmatrix}E^{-1}
\\&=
\begin{bmatrix}
-A^{-1}[H+BS^{-1}(CA^{-1}H-L)]E^{-1}
\\
S^{-1}(CA^{-1}H-L)E^{-1}
\end{bmatrix}
\\&=
\begin{bmatrix}
-A^{-1}[H-BS^{-1}V]E^{-1}
\\
-S^{-1}VE^{-1}
\end{bmatrix}
\end{aligned}
$$

This gives the expanded formula:

$$
\begin{aligned}
&X^{-1}=
\\&
\begin{bmatrix}
E^{-1}+E^{-1}\left[FA^{-1}H+US^{-1}V\right]E^{-1}
&
-E^{-1}\left[F-US^{-1}C\right]A^{-1}
&
-E^{-1}US^{-1}
\\
-A^{-1}[H-BS^{-1}V]E^{-1}
&
A^{-1}+A^{-1}BS^{-1}CA^{-1}
&
-A^{-1}BS^{-1}
\\
-S^{-1}VE^{-1}
&
-S^{-1}CA^{-1}
&
S^{-1}
\end{bmatrix}
\end{aligned}
$$

Title: Inverse of a 3x3 block matrix


问:
If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Please, see if I made some mistake in the proof below. I mention some theorems in the proof:

The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.

(1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.

(2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.

(3) We'll do this by the Precise Limit Definion: given $ \epsilon>0, \exists\delta|0<|x-x_0|<\delta$, then $0<|f(x)-f(x_0)|<\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.

Body:
$$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}\left(f(x_0)+(x-x_0)\cdot\dfrac {f(x)-f(x_0)}{x-x_0}\right)=$$
$$\lim\limits_{x\to x_0}f(x_0)+\lim\limits_{x\to x_0}(x-x_0)\cdot\lim\limits_{x\to x_0}\dfrac {f(x)-f(x_0)}{x-x_0}=$$
$$f(x_0)+0\cdot f'(x_0)=f(x_0)$$


答:
Suppose there are $k$ articles $a_1,\ldots,a_k$ without links, with basis vectors $e_1,\ldots, e_k$ assigned to them respectively.

The easiest way to show this assignment can be extended to the other articles in the way that you want (in fact, uniquely) is as follows. Consider a random walk on your network, starting at some article $x$, where at each step you choose a random outgoing link from the current article and follow it. With probability $1$ this eventually gets stuck in one of $a_1,\ldots, a_k$. Let $P_x(i)$ be the probability that it gets stuck in $a_i$. Now the assignment $x\mapsto \sum_{i=1}^kP_x(i)e_i$ does what you want, since the probability of reaching $a_i$ starting from $x$ is the sum over neighbours $y$ of $x$ of the probability of going to $y$ on the first step and thence reaching $a_i$, which is just $\frac{1}{N}P_y(i)$.

Title: Using Wikipedia links for natural language processing or how to "solve" a directed graph?

Body:
Suppose that the Wikipedia directed graph of internal links is given as below ( Starting with the article "Automobil" / Auto in German and then following at most 3 internal links in the first sentence of the article.)

In natural language processing typically one assings to words or text a vector to represent the meaning of the the word / text.

Usually when one reads a wikipedia article, one clicks to outgoing internal links to get an understanding of the words which this article is referring to.

So having this in mind I want to assign vectors to wikipedia articles in this way:

If $v$ is a wikipedia article then the meaning of this article depends on the other $N$ internal articles this article $v$ is referring to:

$$v = \frac{1}{N} \sum_{ v \rightarrow w } w$$

For an article  which does not have outgoing links, we assign a new standard basis vector.

My mathematical question, is if it is possible to assign vectors $v$ such that for any given directed graph structure $G$, the last equation is valid. ( I have tested this with random graphs, and it seems to be possible)

Second question: Is there a fast method to solve this problem?

If this is possible, then one could add "words" (the corresponding vectors) to change the meaning of the added words. For example, for the graph below we have:

```<code>vAuto == r6
vMotor == r3
vKraftfahrzeug == 1/2*r3 + 1/2*r5
vStrassenfahrzeug == -3/2*r3 - 1/2*r5 + 3*r6
vArbeit == 1/2*r1 + 1/2*r2
vKraftmaschine == -1/2*r1 - 1/2*r2 + 2*r3
vFahrzeug == r5
vOberbegriff == 2*r5 - r8
vVerkehrsmittel == r8
vLandfahrzeug == -3*r3 - r5 + 6*r6 - r7
vStrassen == r7
vLand == -6*r3 - 3*r5 + 12*r6 - 2*r7
vLandfläche == -r4 + 2*r7
vVerkehrsbauwerk == r4
vMaschine == -r1 - 2*r2 + 4*r3
vEnergie == r2
vPhysik == r1
</code>```

So for example "Work = 1/2 Energy + 1/2 Physics".

If we have a new text $T$ which contains words which correspond to wikipedia articles, then one could simply add these articles in a weighted form, an have a meaning for this text.

Thanks for your help.

Example Graph in Sagemath and "solution"

Edit:
By the accepted answer given below, here is the example graph with absorbing states implemented in SAGEMATH. The probabilities are computed as $B = NR$ as described here: https://en.wikipedia.org/wiki/Absorbing_Markov_chain.


问:
Length of the shortest road that goes from A to the highway and then on to B
The following problem is from George Simmons' Calculus with Analytic Geometry, second edition, chapter $4$, section $4$, problem $15$:

Two towns, $A$ and $B$, lie on the same side of a straight highway. Their distance apart is $c,$ and their distances from the highway are $a$ and $b.$ Show that the length of the shortest road that goes from $A$ to the highway and then on to $B$ is $\sqrt{c^{2} + 4ab}.$

Is the length $\sqrt{c^{2} + 4ab}$ correct? I believe this is an erratum, as I obtain $\sqrt{c^{2} + (a+b)^{2}},$ when I use $x = \frac{ac}{a+b}$ in the length formula $L = \sqrt{a^{2} + x^{2}} + \sqrt{b^{2} + (c-x)^{2}}.$

The specified value of $x$ was found via Calculus and Heron's reflection technique (using similar triangles). WolframAlpha computes the same length.

答:
The distance $c$ is not between their projection onto the highway, but the distance between $A$ and $B$. So take the example where $a=0$, traveling in a straight line would give the answer $c$, instead of $\sqrt{c^2+b^2}$.

In fact, you can go from your answer to theirs $\sqrt{c'^2+(a+b)^2}=\sqrt{(c^2-(a-b)^2)+(a+b)^2}$.


答:
You have a typo. Instead of $c$ (which is the distance between $A$ and $B$) in your formula you should have $d$ which is the difference in the $x$ coordinates of points $B$ and $A$. It solves

$$\sqrt{d^2+(b-a)^2}=c$$

$$d^2=c^2-(b-a)^2$$

Now you plugging this $d$ on place of your $c$ we get

$$\sqrt{c^2-(b-a)^2+(a+b)^2}=\sqrt{c^2+4ab}$$

Which is the right answer.

Title: Length of the shortest road that goes from A to the highway and then on to B

Body:
The following problem is from George Simmons' Calculus with Analytic Geometry, second edition, chapter $4$, section $4$, problem $15$:

Two towns, $A$ and $B$, lie on the same side of a straight highway. Their distance apart is $c,$ and their distances from the highway are $a$ and $b.$ Show that the length of the shortest road that goes from $A$ to the highway and then on to $B$ is $\sqrt{c^{2} + 4ab}.$

Is the length $\sqrt{c^{2} + 4ab}$ correct? I believe this is an erratum, as I obtain $\sqrt{c^{2} + (a+b)^{2}},$ when I use $x = \frac{ac}{a+b}$ in the length formula $L = \sqrt{a^{2} + x^{2}} + \sqrt{b^{2} + (c-x)^{2}}.$

The specified value of $x$ was found via Calculus and Heron's reflection technique (using similar triangles). WolframAlpha computes the same length.


问:
What is the formula for series like $-a1, -a2, +a3, +a4, -a5, -a6...$
Just wonder how to come up with the "formula" for any series like:

$-a1, -a2, +a3, +a4, -a5, -a6...$

or maybe:
$-, -, -, +, +, +$

or any of these variations with sequences of same signs but alternating.

Sorry for the poorly formulated question.

答:
I'm not sure to understand what you mean, maybe something like

$$
\sum_1^{+\infty} \frac{1}{n(n+3)}.
$$

Indeed

$$
\sum_1^{+\infty} \frac{1}{n(n+3)}=\sum_1^{+\infty} \frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_0^{+\infty} \frac{1}{3}\left(\frac{1}{3n+1}-\frac{1}{3n+4}+\frac{1}{3n+2}-\frac{1}{3n+5}+\frac{1}{3n+3}-\frac{1}{3n+6}\right)=\\
\frac{1}{3}\sum_0^{+\infty} \left(\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}-\frac{1}{3n+4}-\frac{1}{3n+5}-\frac{1}{3n+6}\right)=\frac{1}{3} \sum_0^{+\infty} (-1)^{\lfloor \frac{n}{3}\rfloor} \frac{1}{3n+6\{\frac{n}{6}\}+1}
$$
(where $\{\frac{n}{6}\}$ is the fractional part of $\frac{n}{6}$) which is a telescoping series (take $a_n=\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}$) that converges to $\frac{1}{3}a_0=\frac{11}{18}$. Maybe from this example you can obtain something similar to what you are asking for.


问:
Range of $cos\ \frac{a-c}{2}$ in a triangle
My doubt is from a question and I will extract my query and relevant information.

How to find the range of $cos\ \frac{a-c}{2}$ if a,b,c are angles of a triangle and b=120°(Don't think this is needed)

The maximum value should be 1 when a,c are equal in value but it's the minimum I am not able to find.

A point in the right direction is much appreciated!

Body:
Hint :

For $b = 120^{\circ}$, $$|a-c| \in [0,60^{\circ})$$

$$\Rightarrow \cos \frac{a-c}{2} \in (\cos 30^{\circ},1]$$


问:
On boundedly invertible
Let $T:X\to X$  be a bounded and invertible linear operator. Show that $\inf_{x: \|x\|=1}\|T(x)\|\geq M$  if and only if $\sup_{x: \|x\|=1}\|T^{-1}(x)\|\leq N$, where $M,N\ge 0$.
 .</p>

答:
Let's show it in one direction:

$\displaystyle \sup_{x:\|x\|=1}\|T^{-1}(x)\|= \sup_{x:\|x\|>0}\frac{\|T^{-1}(x)\|}{\|x\|}.$ Now replace $x$ by $T(y)$ - we can do it because our operator is  invertible; furthermore, for each $x$ such $y$ is unique. Hence our equality continues
$\displaystyle  = \sup_{y:\|y\|>0}\frac{\|y\|}{\|T(y)\|}=\sup_{y:\|y\|=1}\frac{1}{\|T(y)\|}=\frac{1}{\inf_{y:\|y\|=1} \|T(y)\| }=\frac{1}{M} $.

The proof in the other direction is done likewise.

Title: On boundedly invertible


问:
Construction based on circumcenter and incenter
Construct a triangle given the exact location of its circumcenter and its incenter, and the position of its angle bisector (including its direction), but not its length.

I tried to consider the angles between the points.

答:
Given: Incenter $(I)$, its radius $r$, circumcenter $(O)$ and slope $m$ of line $t$ (angle bisector); ?: translation point; $R$ by the Euler's relation, equation of the circle ($μ$) with center $O$ and radius $R$; Points $A$ and $Z$: intersection of the circle ($μ$) with the line $t$; circle ($ψ$) with center $Z$ and radius $IZ$; Points $B$ and $C$: intersection of the circles $ψ$ and $μ$; translation point.

$\begin{array}{} I=(X_{I},Y_{I}) & O=(X_{O},Y_{O})  \end{array}$

$\begin{array}{} \text{translation point} & k=X_{I} & h=Y_{I} \end{array}$

$\begin{array}{} I=(0,0) & O=(a=X_{O}-k,b=Y_{O}-h)  \end{array}$

$\begin{array}{} \text{Euler's relation} & IO=\sqrt{a^2+b^2}=\sqrt{R(R-2r)} & R=r+\sqrt{a^2+b^2+r^2} \end{array}$

$\left\{ \begin{array}{} (x-a)^2+(y-b)^2=R^2 \\ y=m·x  \end{array}  \right.\ $

$\begin{array}{} x_{A}=\frac{a+b·m+\sqrt{(1+m^2)R^2-(a·m-b)^2}}{1+m^2} & y_{A}=m·x_{A} \\ x_{Z}=\frac{a+b·m-\sqrt{(1+m^2)R^2-(a·m-b)^2}}{1+m^2} & y_{Z}=m·x_{Z} \\ \end{array}$

$\begin{array}{} \text{Eq.1 (Ψ)} & x^2+y^2-2x_{Z}·x-2m·x_{Z}·y=0 \\ \text{Eq.2 (μ)} & x^2+y^2-2a·x+2b·y+a^2+b^2-R^2=0 \\ \end{array}$

$\begin{array}{} \text{Eq.2 - Eq.1} & y=p·x+q & \text{where:} \\ p=\frac{x_{Z}-a}{b-m·x_{Z}} & q=\frac{a^2+b^2-R^2}{2(b-m·x_{Z})}   \\ \end{array}$

replacing in Eq.2, solving, we get

$\begin{array}{} x_{B}=\frac{a+bp-pq+\sqrt{(1+p^2)R^2-(ap-b+q)^2}}{1+p^2} & y_{B}=p·x_{B}+q \\ x_{C}=\frac{a+bp-pq-\sqrt{(1+p^2)R^2-(ap-b+q)^2}}{1+p^2} & y_{C}=p·x_{C}+q \\ \end{array}$

$\begin{array}{} X_{A}=x_{A}+k & Y_{A}=y_{A}+h  \\ X_{B}=x_{B}+k & Y_{B}=y_{B}+h  \\ X_{C}=x_{C}+k & Y_{C}=y_{C}+h  \\ \end{array}$


问:
Do the topologies of pointwise and compact convergence coincide on equicontinuous subsets?
If $E$ is a set, then the topology $\rho(E)$ generated by $$p_x(f):=|f(x)|\;\;\;\text{for }f:E\to\mathbb R$$ for $x\in E$ is called the topology of pointwise convergence on $\mathbb R^E$. If $\tau$ is a topology on $E$, then the topology $\kappa(E,\tau)$ generated by $$p_K(f):=\sup_{x\in K}|f(x)|\;\;\;\text{for }f\in C(E,\tau)$$ for $\tau$-compact $K\subseteq E$ is called the topology of compact convergence on $C(E,\tau)$.

Are we able to show that if $\Gamma\subseteq C(E,\tau)$ is $\tau$-equicontinuous, then$^1$ $\left.\rho(E)\right|_\Gamma=\left.\kappa(E,\tau)\right|_\Gamma$? If not, what do we need to assume to show that?

Moreover, I would like to know how exactly we can show that $$\kappa(E,\tau)\subseteq\left.\rho(E)\right|_{C(E,\:\tau)}\tag1.$$

Remark: I know that generally, if $X$ is a vector space and $\sigma$ is the topology generated by a family $P$ of seminorms on $X$, then $$\mathcal B_P:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}$$ is an analytic basis for $\sigma$, where $$U_p:=\{x\in X:p(x)<1\}\;\;\;\text{for }p\in P.$$ This should be helpful to know.

$^1$ If $(X,\sigma)$ is a topological space and $B\subseteq X$, then $\left.\sigma\right|_B:=\{O\cap B:O\in\sigma\}$ denotes the subspace topology on $B$.

答:
Let $X$ be a topological space,  $M$ be a metric space, and $E$ be an equicontinuous set of functions from $X$ to $M$.
Also let $D$ be a dense subset of $X$.
Then the following three topologies coincide on $E$:

The topology of uniform convergence on compact subsets of $X$,

The topology of pointwise convergence on $X$,

The topology of pointwise convergence on $D$.
In order to verify this assertion it suffices to prove that if $\{f_i\}_{i\in I}$ is a net in $E$,  and if $f\in  E$,  then the
following are equivalent:

a. $\displaystyle\lim_{i\to \infty }\sup_{x\in  K}d\big (f_i(x), f(x)\big )=0$,  for every compact subset $K\subseteq X$,

b. $\displaystyle\lim_{i\to \infty }f_i(x)=f(x)$,  for every $x\in  X$,

c. $\displaystyle\lim_{i\to \infty }f_i(x)=f(x)$,  for every $x\in  D$.

It is evident that (a) $\Rightarrow$ (b) $\Rightarrow$ (c), so we only need to worry about
(c) $\Rightarrow$ (a).

In order to do this,  given any compact subset $K\subseteq X$, and any $\varepsilon >0$,  using equicontinuity,  for
each $x$ in $K$,   we choose a neighborhood $V_x$ of $x$ such that
$$
    y\in V_x\Rightarrow d\big (g(x),g(y)\big )<\varepsilon, \quad\forall g\in  E.
      $$
Since $K$ is compact and $\{V_x\}_{x\in K}$ is an open cover of $K$, we may find $x_1,x_2,\ldots ,x_n\in K$ such that
$$
  K\subseteq \bigcup_{j=1}^nV_{x_j}.
    $$
Since $D$ is dense in $X$, for every $j$ we may pick some $y_j\in D\cap V_{x_j}$.    Using (c) let $i_0\in I$, such that
$$
  i\geq i_0\Rightarrow d\big (f_i(y_j), f(y_j)\big )<\varepsilon , \quad\forall j=1,\ldots ,n.
    $$
Next, given any $i\geq i_0$, and any $x$ in $K$, pick $j$ such that $x\in V_{x_j}$.  Then
$$
d\big (f_i(x),f(x)\big ) \leq $$$$\leq
  d\big (f_i(x),f_i(x_j)\big ) +
  d\big (f_i(x_j),f_i(y_j)\big ) +
   d\big (f_i(y_j), f(y_j)\big ) + $$$$+
   d\big (f(y_j), f(x_j) \big ) +
   d\big (f(x_j), f(x) \big ) <
   5\varepsilon .
    $$
Consequently
$$
\sup_{x\in  K}d\big (f_i(x), f(x)\big )\leq 5\varepsilon ,
  $$
for all $i\geq i_0$.


问:
Evaluating $\int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\,\mathrm dx\,\mathrm dy$
$$\int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\, \mathrm dx\,\mathrm dy$$

I'm having some difficulty evaluating this Improper Double Integral with standard techniques taught in Multivariable Calculus. I opted to use the Residue Theorem and obtained an answer of $\pi^2/2$. I was wondering if this is correct and if my work is on the right track?

Edit: Converted $\sin(x)$ to exponential form, converted $\sin(x+y)$ to Im(cis(x+y)) = Im(e^i(x+y)) and then used residue theorem from there.

My progress:

答:
I'll use the sine integral function: $$\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t}dt$$
We will use the properties:
$$\lim_{x\rightarrow\infty} \operatorname{Si}(x) = \frac{\pi}{2}$$
$$\operatorname{Si}(0)=0$$
$$\frac{d}{dx}\operatorname{Si}(x)=\frac{\sin x}{x}$$

We may now proceed to the solution. Firstly, let us notice that the integrand is measurable so we may treat is as an iterated integral:
$$\int_0^\infty \frac{\sin{x}}{x}\Big(\int_0^\infty \frac{\sin(x+y)}{x+y}dy\Big)dx = (*)$$

Let us integrate the inside integral by substitution $t=x+y$.
$$\int_0^\infty \frac{\sin(x+y)}{x+y}dy = \int_x^\infty\frac{\sin t}{t}dt = \int_0^\infty\frac{\sin t}{t}dt - \int_0^x\frac{\sin t}{t}dt = \frac{\pi}{2}-\operatorname{Si}(x)$$
We then have:
$$(*) = \int_0^\infty \frac{\sin x}{x} \Big(\frac{\pi}{2}-\operatorname{Si}(x)\Big)dx=\frac{\pi}{2}\int_0^\infty \frac{\sin x}{x} dx - \int_0^\infty \frac{\sin x}{x} \operatorname{Si}(x)dx$$
From the properties of the sine integral we have $\frac{\pi}{2}\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi^2}{4}$.

The second integral may be found using the derivative property of the sine integral inserted into the formula for integral of multiplication. For some functions $f$ and $g$ we have:
$$fg = \int f'g + \int fg'$$
We are able to calculate the indefinite integral $\int \frac{\sin x}{x} \operatorname{Si}(x)dx$ by putting $f=g=\operatorname{Si}(x)$:
$$\operatorname{Si}^2(x) =2\int \frac{\sin x}{x} \operatorname{Si}(x)dx $$
$$\int \frac{\sin x}{x} \operatorname{Si}(x)dx = \frac{\operatorname{Si}^2(x)}{2}$$
We can now calculate the value of the definite integral:
$$\int_0^\infty \frac{\sin x}{x} \operatorname{Si}(x)dx = \frac{\Big(\lim_{x\rightarrow\infty} \operatorname{Si}(x)\Big)^2}{2}-0=\frac{\pi^2}{8}$$
Eventually:
$$(*) = \frac{\pi^2}{4}-\frac{\pi^2}{8} = \frac{\pi^2}{8}$$


问:
Analytic expression of $g^{\sharp}$
The analytical expression of $g^{\flat}\colon TM \to T^*M$ is given by

$$g^{\flat} = g_{ij}x^idx^j$$ with $x^i$ the coefficients of a $X\in TM$ in basis $\frac{\partial}{\partial{x^i}}$. Also, given the invertibility of $G = (g_{ij})$, $g^{\flat}$ is isomorphism and thus $g^{\sharp}\colon T^*M \to TM$ can be defined. According to bibliography, the analytical expression of $g^{\sharp}\colon T^*M \to TM$ is

$$g^{\sharp} = g^{ij}\omega_j\frac{\partial}{\partial{x^i}}$$

with $\omega_i$ the coefficients of a $\omega\in T^*M$ in basis $d\omega^i$. I am trying to obtain the analytic expression of $g^{\sharp}$ starting from the definition of $g^{\flat}$ but I could not come up with an analytical solution. Could you please provide some guide lines?

答:
I think the confusion here is notational. Keep in mind $g^\flat: TM\to T^*M$ is defined by $X\mapsto g(X,\cdot)$. The "analytical expression" is obtained by noting that if $X=x^i\frac{\partial}{\partial x^i}$ and $Y = y^i\frac{\partial}{\partial x^i}$, then
$$g^\flat(X)(Y) = g(X,Y) = g\big(x^i\frac{\partial}{\partial x^i},y^j\frac{\partial}{\partial x^j}\big) = g\big(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\big)x^iy^j = g_{ij}x^idx^j(Y),$$
from which we conclude $g^\flat (X) = g_{ij}x^idx^j$.

Since $g^\sharp$ is the inverse of $g^\flat$, we can obtain a formula for it as follows: write $\omega = \omega_j dx^j$ in local coordinates. We want to find the unique $X$ for which $g(X,\cdot) = \omega$. Since $g(X,Y) = \omega(Y)$ for all $Y$, we must have $g_{ij}x^idx^j(Y) = \omega_j dx^j(Y)$, meaning $g_{ij}x^i=\omega_j$. But this is just a matrix-inversion problem written in Einstein notation; if $g^{ij}$ denotes the inverse of the matrix $g_{ij}$, then $x^i = g^{ij}\omega_j$. Thus
$$g^{\sharp}(\omega) = X = x^i\frac{\partial}{\partial x^i} = g^{ij}\omega_j\frac{\partial}{\partial x^i},$$
which is the formula we wanted.

Title: Analytic expression of $g^{\sharp}$


Title: Evaluating $\int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\,\mathrm dx\,\mathrm dy$

Body:
$$\int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\, \mathrm dx\,\mathrm dy$$

I'm having some difficulty evaluating this Improper Double Integral with standard techniques taught in Multivariable Calculus. I opted to use the Residue Theorem and obtained an answer of $\pi^2/2$. I was wondering if this is correct and if my work is on the right track?

Edit: Converted $\sin(x)$ to exponential form, converted $\sin(x+y)$ to Im(cis(x+y)) = Im(e^i(x+y)) and then used residue theorem from there.

My progress:

Body:
Denoting $ \mathrm{Si}:x\mapsto\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y} $, we have the following : \begin{aligned} \int_{0}^{+\infty}{\frac{\sin{x}}{x}\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y}\,\mathrm{d}x}=\int_{0}^{+\infty}{\mathrm{Si}\,{x}\,\mathrm{Si}'\,{x}\,\mathrm{d}x}&=\left[\frac{\mathrm{Si}^{2}\,{x}}{2}\right]_{0}^{+\infty}\\&=\frac{1}{2}\left(\int_{0}^{+\infty}{\frac{\sin{x}}{x}\,\mathrm{d}x}\right)^{2}\\&=\frac{\pi^{2}}{8} \end{aligned}
Thus : \begin{aligned}\int_{0}^{+\infty}{\int_{0}^{+\infty}{\frac{\sin{x}\sin{\left(x+y\right)}}{x\left(x+y\right)}\,\mathrm{d}y}\,\mathrm{d}x}&=\int_{0}^{+\infty}{\frac{\sin{x}}{x}\int_{x}^{+\infty}{\frac{\sin{y}}{y}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{+\infty}{\frac{\sin{x}}{x}\left(\frac{\pi}{2}-\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y}\right)\mathrm{d}x}\\ &=\frac{\pi^{2}}{4}-\int_{0}^{+\infty}{\frac{\sin{x}}{x}\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{\pi^{2}}{4}-\frac{\pi^{2}}{8}\\&=\frac{\pi^{2}}{8}\end{aligned}


答:
You surely mess up things in the proof. You say we can make $f(x)$ as close to $f(x_0)$ as one whishes. But this is exactly what you are trying to prove.

The more useful definition of differentiability is, that a function is differentiable if
$f(x)=f(x_0) + a \cdot (x-x_0) + r(x)$ with
$$\lim_{x\to x_0} \frac{r(x)}{x-x_0} = 0 $$
so $\lim_{x\to x_0} r(x)$ will surely be $0$.
Hence
\begin{align*}
|f(x)-f(x_0)|&= | f(x_0) + a \cdot (x-x_0) + r(x) -f(x_0)|\\
&\leq |a| \cdot |x-x_0|+ |r(x)|
\end{align*}
Now you have a sum of 2 terms which go to zero as $x$ goes to $x_0$.

Title: If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Body:
Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Please, see if I made some mistake in the proof below. I mention some theorems in the proof:

The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.

(1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.

(2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.

(3) We'll do this by the Precise Limit Definion: given $ \epsilon>0, \exists\delta|0<|x-x_0|<\delta$, then $0<|f(x)-f(x_0)|<\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.


问:
reachability of the norm by the operator
$A: \ell^1 \to \mathbb{R}$, where $$Ax= \sum\limits_{k=0}^\infty((2+\cos(\pi(1/3)^{k+1})\xi_{k+1}).$$

\begin{aligned}
||Ax||_{\mathbb{R}} &= |Ax| = |(2+\cos(\pi(1/3))\xi_1) + (2+\cos(\pi(1/9))\xi_2)+(2+\cos(\pi(1/27))\xi_3) + \dots| \\
&\leq |(2+\cos(\pi(1/3))\xi_1)|+|(2+\cos(\pi(1/9))\xi_2)|+|(2+\cos(\pi(1/27))\xi_3)|+\dots \\
&\leq 3|\xi_1|+3|\xi_2|+3|\xi_3|+\dots \\
&=3||x||_{\ell^1}
\end{aligned}

I have already shown the limit, but I can't prove that the norm is not reached.

答:
In fact, the fact that $\ell^1$'s dual is $\ell^\infty$ reminds us that $||A|| = \sup_{k \in \mathbb{N} \cup \{0\}} |2 + \cos(\pi/3^{k+1})| = 3$.
It's a routine exercise to show that for all $\epsilon > 0$, there exists $x \in \ell^1$ such that $||Ax|| > (3-\epsilon)||x||_1$, so that $3-\epsilon$ can't be the operator norm.  It's impossible to find a vector that satisfies the equality case in this Hölder's inequality $|\langle v,x \rangle|_1 \le ||v||_\infty ||x||_1$ with $v = (2 + \cos(\pi/3^{k+1}))_{k\in\mathbb{N}\cup\{0\}}$ and $x = (\xi_k)_k$ because in the equality case, $v$ and $x$ have to be linearly dependent (i.e. $v = kx$), but $v \notin \ell^1$ while $x \in \ell^1$.


问:
How to Integrate a Differential equation with all constant?

$$\int \frac12m\left(\frac{du}{dt}\right)^2\,dt$$

Actually I am not familiar with partial differential equation concept . As i have to integrate here with respect to $dt$ is there any way to integrate? To me it seems like all are constant because the function here it has nothing respect to t . How will i integrate this with simple Integration concept because i have just started the basic .

Thanks

答:
Based on your equation, you are integrating kinetic energy

$$W_k = \frac{1}{2}mv^2 = \frac{1}{2}m \biggl(\frac{\partial x}{\partial t} \biggr)^2$$

with respect to time

$$I = \int W_kdt$$

If your velocity $v$ is constant with time, you will simply get

$$I = W_kt+C$$

where $C$ is some constant. However, if your velocity is not constant with respect to time, the result will be more complicated, and will depend on the exact form of $v$. Also take care that in some cases the mass $m$ may also depend on time, for example in rockets science where the burned fuel directly depletes the mass.

Edit: I suspect I know where your confusion comes from. If, for example, we were integrating momentum

$$p = mv = m\frac{\partial x}{\partial t}$$

with respect to time, we would be able to directly undo the differentiation in velocity and obtain an analytic expression for the result that is independent of the explicit form of the velocity

$$\int pdt = \int m\frac{\partial x}{\partial t} dt = mx+C$$

for some constant $C$. However, when we integrate energy, we first take a derivative of position to obtain velocity, then square it, then take anti-derivative (integral).

$$\frac{\partial x}{\partial t} \rightarrow \biggl(\frac{\partial x}{\partial t} \biggr)^2 \rightarrow \int \biggl(\frac{\partial x}{\partial t} \biggr)^2 dt$$

The problem is that the square does something very complicated to our function in such a way that we cannot directly reverse differentiation without knowing exactly what was the function. In other words, the square will affect different functions differently in this context. It is a common situation with nonlinearity


问:
If $x,y,z>0$, $n,m\in N$, $n\ge m$ prove:$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$

Given positive numbers $x,y,z$ and $n,m$ positive integers with $n\ge m$ prove that: $$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$$

I tried doing it as follows:

If $x\ge y \ge z$, then:

$y+z\le z+x\le x+y$

Hence:

$x^n\ge y^n\ge z^n$ and $\frac{1}{(y+z)^m}+\frac{1}{(z+x)^m}\ge \frac{1}{(x+y)^m}$

From Tchebychev:

$3[\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}]\ge (x^n+y^n+z^n)[\frac{1}{(y+z)^m}+\frac{1}{(x+z)^m}+\frac{1}{(x+y)^m}]$

which is where I got stuck.

Could you please help me finish this question off?

答:
Your idea of chebyshev is right we just have to tailor it with repect to what we have to prove

WLOG $x\ge y\ge z$ then $x^{n-m}\ge y^{n-m}\ge z^{n-m}$  and ${(\frac{x}{y+z})}^m\ge{( \frac{y}{x+z})^m}\ge {(\frac{z}{x+y})}^m $

By chebyshev: $$\sum_{cyc} \frac{x^n}{{(y+z)}^m}=\sum_{cyc} x^{n-m} {(\frac{x}{y+z})}^m\ge \frac{1}{3} \sum x^{n-m}\cdot \sum {(\frac{x}{y+z})}^m \tag S$$ Now by power mean inequality and Nesbitts inequality  $$\sum {(\frac{x}{y+z})}^m\ge \dfrac{{(\sum \frac{x}{y+z})}^m}{3^{m-1}}\ge \frac{3}{2^m}$$  Use this in inequality (S)  directly to prove result


问:
base for $\tau_2$
Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ defined by:

$\tau_{1}=\{\{m\in \mathbb{N}:m<n\}:n\in \mathbb{N}\}\cup \{\mathbb{N}\}\}$ and $\tau_{2}=\{A\subseteq \mathbb{N}: 0\in A\}\cup \{\emptyset\}$.

How is the $B_1$ base? A $B_2$ base for $\tau_2$ I would kick $\tau_1$. Is that correct?

答:
Both $\tau_1$ and $\tau_2$ have the property that each point $n \in \Bbb N$ has a minimal open neighbourhood $O_n$ such that any open set $O$ that’s contains $n$ has $O_n$ as a subset.

For $\tau_1$ we take $O_n = \{m \in \Bbb N: m < n+1\}$ and for $\tau_2$ we take $O_n = \{0,n\}$. The minimality is easy to check.

So a (minimal) base for these topologies are $\{O_n: n \in \Bbb N\}$ in both cases.

Note that it equals $\tau_1\setminus \{\emptyset, \Bbb N\}$ in the first case, so that space only has 4 bases in total, which is quite rare.

Title: base for $\tau_2$


问:
Completing the square with simple polys
I am suppose to rewrite $x^2 + x + 1$ by completing the square. I don't really know what that means but I know that if I add 3 at the end of this I get </p>

<p>$$(x + 2) (x - 1) - 3$$ this is the same as the original now but the answer isn't right. What is wrong with what I did? It seems a lot cleaner than the answer.</p>

答:
In general, $$\color{purple}{(ax+b)^2= a^2x^2+2abx+b^2}.$$

What you have is $x^2 + x + 1$. To complete the square, what you want to do is find $a$, $b$, $c$ in such a way that $x^2 + x + 1 = (ax + b)^2 + c$.

Now,

$$\begin{align}
x^2 + x + 1 & = (ax + b)^2 + c \\
x^2 + x + 1 & = a^2x^2+2abx+b^2 + c \\
\end{align}$$

Now let's color the last equation. We then have $$ \color{red}{x^2} + \color{orange}{x} + \color{blue}{1} = \color{red}{a^2x^2}+\color{orange}{2abx}+\color{blue}{b^2 + c}.$$

Then, for the colored equation above, we equate the coefficients of left hand side to the right hand side of the equation. Then we have three things:
$$\begin{align} \tag 1
\color{red}{x^2} & \equiv \color{red}{a^2x^2} \\  \tag2
\color{orange}x & \equiv \color{orange}{2abx}  \\ \tag3
\color{blue}1 & \equiv \color{blue}{b^2 + c}  \\
\end{align}$$

Solving equation $(1)$, we have
$x^2 \equiv a^2x^2 \Rightarrow a^2 = 1 \Rightarrow a= 1$ or $a= -1$,

Solving equation $(2)$, we have
$x \equiv 2abx \Rightarrow 1 = 2ab \Rightarrow b = \dfrac{1}{2a}$. From above, if $a= 1$ then $b= \dfrac{1}{2}$. If $a= -1$ then $b= -\dfrac{1}{2}$.

Solving equation $(3)$, we have $1 \equiv b^2 + c \Rightarrow c= 1 - b^2$. From above, if $b= \dfrac{1}{2}$ then $c= \dfrac{3}{4}$. If $b= -\dfrac{1}{2}$ then $c= \dfrac{3}{4}$ as well.

In conclusion, we have the set of solutions $\color{navy}{a=1, b= \dfrac{1}{2}, c= \dfrac{3}{4}}$ or $\color{maroon}{a=-1, b= -\dfrac{1}{2}, c= \dfrac{3}{4}}$.

Check the answer, plug the values of $a$, $b$, $c$ into $(ax+b)^2+c$ and expand it. If you get $x^2 + x + 1$, then it is correct.

For $\color{navy}{a=1, b= \dfrac{1}{2}, c= \dfrac{3}{4}}$, we have $$\left(x+\frac{1}{2}\right)^2+ \frac{3}{4}= x^2 + 2(1)\frac{1}{2}x^2 + \left(\frac{1}{2}\right)^{2} + \frac{3}{4} = x^2 + x + 1.$$

For $\color{maroon}{a=-1, b= -\dfrac{1}{2}, c= \dfrac{3}{4}}$, we have $$\left(-x-\frac{1}{2}\right)^2+ \frac{3}{4}= (-x)^2 + 2(-1)\left(-\frac{1}{2}\right)x^2 + \left(\frac{1}{2}\right)^{2} + \frac{3}{4} = x^2 + x + 1.$$

So the answer to your question is either $$\left(x+\dfrac{1}{2}\right)^2+ \dfrac{3}{4}$$ or $$\left(-x-\frac{1}{2}\right)^2+ \frac{3}{4}.$$

The first one looks prettier. I would go with that. Hope this helps.

Some more examples. Let's start with an easy one.

Example 1

Suppose someone asks you to complete the square of $$x^2+2x+3. \tag{A1}$$ Now consider $(x+1)^2$. Expanding that, we have $$(x+1)^2 = x^2 + 2x+ 1.  \tag{A2}$$ You want to complete the square of $x^2+2x+3$. Notice that to get $x^2+2x+3$ from equation $(A2)$, all you have to do is add the number $2$ to both sides of equation $(A2)$, since both $(A1)$ and $(A2)$ both have $x^2$ and $2x$. Then
$$\begin{align}(x+1)^2 +\color{red}2 &= x^2 + 2x+ 1 +\color{red}2 \\
(x+1)^2 +2 & = x^2 + 2x+ 3.
\end{align}$$ And there you have it, the answer is $(x+1)^2 +2$.

Example 2
Now, back to your question. You want to complete the square of $$x^2+x+1. \tag{B}$$ Now consider $(x+1)^2$. Expanding that, we have $$(x+1)^2 = x^2 + 2x+ 1. \tag {C}$$ Equation $(B)$ and $(C)$ both have $x^2$, but equation $(B)$ has $x$ in it, while equation $(C)$  has $2x$. We then have to try another expansion. So you consider $\left(x+\frac{1}{2}\right)^2$. We then get $$\left(x+\frac{1}{2}\right)^2= x^2 + x + \frac{1}{4} \tag{D}$$ Aha! They now both have $x$. All you have to do now is add a number to both sides of equation $(D)$ so that the right hand side equals $x^2+x+1$. So you add $\frac{3}{4}$ to both sides of equation $(D)$. Then you have

$$\begin{align}
\left(x+\frac{1}{2}\right)^2 & = x^2 + x + \frac{1}{4} \\
\left(x+\dfrac{1}{2}\right)^2+ \color{red}{\dfrac{3}{4}} & = x^2 + x + \frac{1}{4} + \color{red}{\dfrac{3}{4}} \\
\left(x+\dfrac{1}{2}\right)^2+ {\dfrac{3}{4}} & = x^2 + x + 1.
\end{align}$$

And so the answer is $\left(x+\dfrac{1}{2}\right)^2+ {\dfrac{3}{4}}$.

Example 3
Suppose you want to complete the square of $$9x^2+3x+1. \tag{E}$$ You know $(3x)^2=9x^2$, so you consider $(3x+1)^2$. Expanding that, we have $$(3x+1)^2 = 9x^2 + 6x+ 1. \tag {F}$$ Equation $(E)$ and $(F)$ both have $9x^2$, but equation $(E)$ has $3x$ in it, while equation $(F)$  has $6x$. We then have to try another expansion. So you consider $\left(3x+\frac{1}{2}\right)^2$. We then get $$\left(3x+\frac{1}{2}\right)^2= 9x^2 + 3x + \frac{1}{4} \tag{G}$$ They now both have $3x$. All you have to do now is add a number to both sides of equation $(G)$ so that the right hand side equals $9x^2+3x+1$. So you add $\frac{3}{4}$ to both sides of equation $(G)$. Then you have

$$\begin{align}
\left(3x+\frac{1}{2}\right)^2 & = 9x^2 + 3x + \frac{1}{4} \\
\left(3x+\dfrac{1}{2}\right)^2+ \color{red}{\dfrac{3}{4}} & = 9x^2 + 3x + \frac{1}{4} + \color{red}{\dfrac{3}{4}} \\
\left(3x+\dfrac{1}{2}\right)^2+ {\dfrac{3}{4}} & = 9x^2 + 3x + 1.
\end{align}$$

And so the answer is $\left(3x+\dfrac{1}{2}\right)^2+ {\dfrac{3}{4}}$.

With practice, it will get easier.

Title: Completing the square with simple polys


问:
Solutions of a linear system of equations with parameters, using matrices
Cheers, I have this system of linear equations $\begin{cases} 6x_1+ 2x_2 + 3 x_3 + 4 x_4= 15 \\ 5 x_1 + 6x_2 + 7 x_3 + 8 x_4 = 26 \\ 9 x_1 + 10x_2 + 11 x_3 + 12 x_4 = 42 \\ λ x_1 + 14x_2 + 15 x_3 + 16 x_4 = λ \end{cases}$ which as you can see has the parameter $λ$ inside. I am asked to find the solution of this set, so I put this into a matrix $A$ and thus far I have $ A = \begin{pmatrix} 6&& 2 && 3  && 4  && 15 \\ 5 && 6 && 7 && 8  && 26 \\ 9 && 10 && 11  && 12 && 42 \\ λ && 14 && 15 && 16 && λ\end{pmatrix}$. My initial thought was to row reduce the matrix in row echelon form, and then conclude on a solution with $λ$. The $rref$ gives me: $\left(
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & -1 & 0 \\
 0 & 0 & 1 & 2 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)$.From the last row we can see that 0 = 1, so I conclude that the system does not have a solution for $λ \in \mathbb{R}$. Is my solution and/or thought process correct? Thanks!

答:
Yes, your train of thought is correct.

The fact that the last line in the matrix is $(0,0,0,0,1)$ means that by adding scaled versions of the equations $\begin{cases} 6x_1+ 2x_2 + 3 x_3 + 4 x_4= 15 \\ 5 x_1 + 6x_2 + 7 x_3 + 8 x_4 = 26 \\ 9 x_1 + 10x_2 + 11 x_3 + 12 x_4 = 42 \\ λ x_1 + 14x_2 + 15 x_3 + 16 x_4 = λ \end{cases}$

we can create an equation of the form
$$
0x_1+0x_2+0x_3+0x_4=1
$$
, which is false for all $x_1,...,x_4\in\mathbb{R}$.

Therefore, the equation system is unsolvable, and since we didn't assume anything about $\lambda$ but that it's a real number (when we were summing and scaling the equations), the equation system is unsolvable for any $\lambda\in\mathbb R$.


问:
Block Diagonal Polynomial and Inverse
Suppose that you have a block diagonal matrix $B$, which has blocks $B_1 \dots B_r$ along the diagonal. Show how you can express the polynomial function $f(B)$ and the inverse of $B$ in terms of functions on block matrices.

I know because of commutativity of matrix polynomials mentioned earlier in the book that two polynomials will always commute as such: $f(B)g(B) = g(B)f(B)$ and since block diagonal matrices are special cases of triangular matrices, any operations performed with itself preserves its triangular properties. Besides that, I don't quite understand what the question is asking from me and how to approach it and would like some help.

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (Charu C. Aggarwal); Page 14, Problem 1.2.9

答:
A block-diagonal matrix $B$ is one of the form
$$
B = \begin{bmatrix}
B_1 & 0 & \cdots & 0\\
0 & B_2 & \cdots & \vdots \\
\vdots & \vdots & \ddots & 0\\
0 & 0 & \cdots & B_r
\end{bmatrix}.
$$
What this problem boils down to (which I will leave for you to check), is that $B^k$ for integer $k$ is again a block-diagonal matrix, and it has form
$$
B^k = \begin{bmatrix}
B_1^k & 0 & \cdots & 0\\
0 & B_2^k & \cdots & \vdots \\
\vdots & \vdots & \ddots & 0\\
0 & 0 & \cdots & B_r^k
\end{bmatrix}.
$$
In particular, $k=-1$ is the inversion formula, and if $f(x) = a_0+a_1x+\dotsm +a_nx^n$ is a polynomial, it is not hard to see from the above that $f(B)$ is block-diagonal with form
$$
f(B) = \begin{bmatrix}
f(B_1) & 0 & \cdots & 0\\
0 & f(B_2) & \cdots & \vdots \\
\vdots & \vdots & \ddots & 0\\
0 & 0 & \cdots & f(B_r)
\end{bmatrix}.
$$
As for proving the first formula, start off by checking that if $A = {\rm diag}(A_1,\dotsc, A_r)$ and $B = {\rm diag}(A_1,\dotsc, A_r)$ are block-diagonal with equally sized partitions, then
$$AB = \begin{bmatrix}
A_1B_1 & 0 & \cdots & 0\\
0 & A_2B_2 & \cdots & \vdots \\
\vdots & \vdots & \ddots & 0\\
0 & 0 & \cdots & A_rB_r
\end{bmatrix}.$$
Setting $A = B^{-1}$ and using $AB = I$ will tell you what the inverse blocks are. Seeting $A = B$ will tell you what the blocks for $B^2$ are, and you can prove the $B^k$ result from there by induction.

Title: Block Diagonal Polynomial and Inverse