问:
Taylor expansion of a gaussian integral
I tried Taylor expanding the following function for small $x \ll 1$:
\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}
I already know that $f$ takes on finite values for $x \geq0$. My goal is to get a series expression for $f$ of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^n.
\end{align*}

So after using the formula (for $a = 0$)
\begin{align*}
f(x) = \sum^\infty_{n = 0} \frac{f^{(n)}(a)}{n!}(x-a)^n,
\end{align*}
I get up to the $N$-th term:
\begin{align*}
f_N(x) = \sum^N_{n = 0} \frac{1}{n!}\frac{(-x)^n}{24^n}(4n-1)!!
\end{align*}
However, when plugging in a small value for $x$ (for example $x = 0.1$) in Wolfram Alpha, I get that till $N = 25$ the approximation is very good, but after around $N = 30$ the series diverges away from the true value ($\approx 0.988306$ ). So for $N \rightarrow + \infty$, the series seems to diverge.

My questions are:

How can this be ? I thought that for larger and larger $N$ the Taylor approximation would be more and more better ? (well, for small $x$ atleast)

Does this mean that the radius of convergence is equal to $0$ and that the interval of convergence is just the point $x = 0$ ? If not, then what is the radius of convergence ? If yes, doesn't this contradict the fact that $f$ is finite for $x \geq1$ ?

答:
Your function may be expressed in terms of the modified Bessel function of the second kind as follows:
$$
f(x)=\sqrt {\frac{3}{{2\pi x}}} e^{\frac{3}{{4x}}} K_{1/4} \left( {\frac{3}{{4x}}} \right).
$$
By the results of the paper http://dx.doi.org/10.1007/s10440-017-0099-0, for all $N\geq 0$ and $x$ with $|\arg x|<\frac{3\pi}{2}$, we have
$$
f(x) = \sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{{(4n - 1)!!}}{{24^n n!}}x^n }  + R_N (x),
$$
where the remainder $R_N (x)$ satisfies
$$
\!\!\left| {R_N (x)} \right| \!\le\! \frac{{(4N - 1)!!}}{{24^N N!}}\left| x \right|^N
\!\times\! \begin{cases} \!1 & \!\!\text{if } \; \left|\arg x\right| \leq \frac{\pi}{2}, \\ \!\min\! \left(\chi\!\left(N+\frac{1}{4}\right)+1,|\csc ( \arg x)|\right) & \!\!\text{if } \; \frac{\pi}{2} < \left|\arg x\right| \leq \pi, \\ \!\cfrac{\sqrt {2\pi \left( N + \frac{1}{4} \right)} }{\left| \cos (\arg x)\right|^{N + 1/4} } + \chi\!\left(N+\frac{1}{4}\right)+1 & \!\!\text{if } \; \pi < \left|\arg x\right| < \frac{3\pi}{2}. \end{cases}
$$
Here, for $p>0$,
$$
\chi (p) = \sqrt \pi  \frac{{\Gamma\! \left( {\frac{p}{2} + 1} \right)}}{{\Gamma \!\left( {\frac{p}{2} + \frac{1}{2}} \right)}},\quad \sqrt {\frac{\pi }{2}\left( {p + \frac{1}{2}} \right)}  \le \chi (p) \le \sqrt {\frac{\pi }{2}\left( {p + \frac{2}{\pi }} \right)} .
$$
As it was noted by others, the series cannot converge for any $x \neq 0$ because $x=0$ is a branch point of $f(x)$. However, it is an asymptotic expansion of $f(x)$ as $x\to 0$. As you can see, we have an explicit control over the error term. To obtain the best numerical approximation, stop the series at its numerically least term, i.e., at $N = \left[ {\frac{3}{{2\left| x \right|}}} \right]$ and estimate the error using the formulae above.

Title: Taylor expansion of a gaussian integral


问:
how to do the derivative of least squares objective in vector form?
Below we have the least squares objective function in vector form. Assume dimension of $x$ is $m \times n$, $W$ is $n \times 1$ and $y$ is $m \times 1$. So $J$ is $1 \times 1$ a scalar. $T$ below is the transpose.

$$ J=\frac{1}{m}(xW -y)^{T}(xW-y) $$

Let $A =xW-y$, then $$ J=\frac{1}{m}A^{T}A$$ $$\frac{dJ}{dA}=\frac{2}{m}A$$  $$\frac{dA}{dW}=x .$$

By chain rule:

$$ \frac{dJ}{dw}=\frac{dJ}{dA}\frac{dA}{dW}=\frac{2}{m}Ax .$$

Obviously if we do this, the dimensions don't match up, since $A$ is $m \times 1$ and $x$ is $m \times n$. Instead, the answer should be $$\frac{dJ}{dw}=\frac{2}{m}x^{T}A .$$

My question is, how do we know the convention in this case? Can someone point me to how we should do the chain rule in this case?

答:
For differential calculus involving matrices it is more convenient to use differentials instead of derivatives. You can find the rules for differentials in Practical Guide to Matrix Calculus for Deep Learning by Andrew Delong or in the book "Matrix algebra" by Abadir & Magnus.

I will show you how to get the derivative of $J$ using notation and rules from the paper by Delong. Delong's paper doesn't have the chain rule for differentials but it is
$$d(f\circ g)(x, dx)=df(g(x),dg(x,dx))$$
You can look Composite function gradient for a proof sketch.

So using the rules (10), (12) from the paper
$$dJ(A,dA)=d(\frac{1}{m}A^TA)=\frac{1}{m}dA^TA+\frac{1}{m}A^TdA=\frac{1}{m}(dA)^TA+\frac{1}{m}A^TdA=\frac{2}{m}A^TdA$$
$$dA(x,dx)=d(xW-y)=dxW$$
$$d(A(W),dW)=d(xW-y)=xdW$$
By the chain rule, the rule (6) and the fact that $v^Tu=v\cdot u$ for vectors $v,u$
$$dJ(x,dx)=dJ(A(x),dA(x,dx))=\frac{2}{m}A(x)^TdA(x,dx)=\frac{2}{m}(xW-y)^TdxW=$$
$$=\frac{2}{m}(xW-y)\cdot dxW=\frac{2}{m}(xW-y)W^T\cdot dx$$
$$dJ(W,dW)=J(A(W),dA(W,dW))=\frac{2}{m}(xW-y)\cdot xdW=\frac{2}{m}x^T(xW-y)\cdot dW$$
From these and the rule (17)
$$\frac{\partial J}{\partial x}=\frac{2}{m}(xW-y)W^T$$
$$\frac{\partial J}{\partial W}=\frac{2}{m}x^T(xW-y)$$

ContentLicense: CC BY-SA 4.0


问:
about an invariant metric on X/N
I'm reading Rudin's book Functional Analsysi.In the center of the proof of Theorem 1.41 on Page 32,how to induce $$\pi (\{ x:d(x,0)<r\})=\{u:\rho (u,0)<r\}$$ </p>

答:
I'll answer in greater generality. Let $X$ be a vector space with translation-invariant pseudometric $d$: that is, it is equipped with a symmetric function $d:X\times X\to [0,\infty)$ such that $d(x,x)=0$, $d(x+a,y+a)=d(x,y)$, and the triangle inequality holds. Given a subspace $N$ of $X$, define a pseudometric $\rho $ on $X/N$ by
$$\rho(a+N,b+N)=\inf \left\{ d(a-b,v) : v\in N \right\} \tag1$$
The tranlation invariance of $d$ implies that (1) is indeed a pseudometric. Let $\pi:X\to X/N$ denote the canonical projection which maps every element of $X$ to its equivalence class $x+N$. Then
$$\pi(\{x:d(x,a)<r\}) = \{x+N:\rho(x+N,a+N)<r\} \tag2$$
The inclusion $\subseteq$ follows from $\rho(x+N,a+N)\le d(x,a)$. Conversely, if $\rho(x+N,a+N)<r$, then (1) implies the existence of $v$ such that $d(x-a,v)<r$. The vector $x+v$ belongs to $\{x:d(x,a)<r\}$ and projects to $x+N$. </p>

ContentLicense: CC BY-SA 3.0

Title: about an invariant metric on X/N


问:
Relatiosn with Matrices
For every integer $n>2$ and $A,B,C,D \in M_{2}(\mathbb{R})$ we have :

$1) AC-BD=I_{n}$

$2) AD+BC=O_{n}$

Prove that  $CA-DB=I_{n}$ and $DA+CB=O_{n}$.

I tried to prove that $AC=CA$ like the proof of theorem " $AB=I$ , then $BA=I$ " , after that  I tried with multiply and add the relations  but I didn't manage to prove that.
Any idea?

答:
Consider the block multiplication
$$
\begin{bmatrix}
A & -B \\ B & A
\end{bmatrix}
\begin{bmatrix}
C & - D \\ D & C
\end{bmatrix}=
\begin{bmatrix}
AC-BD & -(AD+BC) \\ BC+AD & -BD+AC
\end{bmatrix}
=\begin{bmatrix} I_n & O_n \\ O_n & I_n \end{bmatrix}=I_{2n}
$$
Then also
$$
\begin{bmatrix}
C & - D \\ D & C
\end{bmatrix}
\begin{bmatrix}
A & -B \\ B & A
\end{bmatrix}
=I_{2n}
$$

Title: Relatiosn with Matrices


问:
What do you call generating a function out of a graph?
In many physical phenomena, laws an relations to their variables are somehow interpolated (example by statistical data or samples) and then an approximate set of functions are generated to work equivalently within the allowable range. Is there a branch of mathematics that treats this in details?</p>

答:
There are many different ways in which people "generate a function out of a graph" belonging to different fields of math and physics: The relevant field and method depends on the goal and application.

First, there is interpolation. Linear interpolation is one example: draw line-segments between consecutive data-points, and use the value of the resulting piecewise curve to estimate what you might find if you measured between your data points. A particularly popular interpolation method is the so-called cubic spline. Computing cubic-splines and linear interpolations is a deterministic procedure that doesn't need any fitting algorithm. This is sometimes useful when you have a very pragmatic application in mind. For example, you measure the voltage as a function of temperature for a thermocouple, and then later you want to calculate the temperature given a specific voltage measurement.

Then, there is curve-fitting. This is exactly what it sounds like: given a function with a set number of parameters, the parameters are adjusted until the resulting "best fit" is found. Most algorithms that do this start from some initial guess for the best-fit parameters, and make "small" changes to the adjustable parameters until the fit no longer gets "better". Typically the parameter step-size is adjusted dynamically by the chosen algorithm, and the fit quality is evaluating using a so-called quality function. Often one does this if one knows a functional form that well-describes the data that one wants to fit. For example, if I measured the height of a falling body as a function of time, then I would likely try to fit the data using a least-squares fit of a quadratic equation.

A "higher form" of curve fitting is to fit a set of data using a finite number of terms in a systematic expansion. This has the theoretical advantage that there exist systematic expansions that can in principle describe any function. For example, expanding periodic data in Fourier series is often a very good idea. Noteworthy as well are polynomials; actually arbitrarily fitting high-degree polynomials to experimental data is often a terrible idea. My justification for saying this is that all non-constant polynomials diverge at $\pm \infty$. Thus, no matter how many terms you add it the expansion, if you fit something that e.g. decays exponentially using a polynomial, then you will end up with wildly unrealistic results far from the point about which the expansion was computed.

Somewhere higher on the complexity ladder is so-called phenomenological modelling. Phenomenology sometimes gets a "bad rap", but it has its place. Often the name of the game in phenomenology is to simply write down a functional form that one thinks well describes the experimental data in question. Usually, however, one injects some "physical insight" into the choice of the functional form. An example of a model like this is the liquid-drop model in nuclear physics. In that model the binding energy of atomic nuclei is described as the sum of various terms each obeying different power laws, with analogies to physical effects such as surface tension, coulomb interaction, and so forth. Personally, I find that particular model highly unsatisfying. However, if one wants to calculate the energy released in any nuclear reaction using only a few parameters, then it does a pretty good job.

At the top of the totem pole in my opinion is "real" theory. This is traditionally the job of a theoretical physicist, although mathematicians, chemists and all manner of other people do similar things. To put a personal spin on it, this is my job, most of the time at least. In my world the role of a theoretical physicist is to look at all of the experimental data and see if it can be understood consistently and within one model. Ultimately, you will be looking at many different "graphs", and your work will hopefully generate some "functions" that can be compared with those graphs, and make predictions for data that haven't yet been measured. In an absurd example, one might look at the distance a freshman wanders home from a bar. In a very bad case, it might be the case that such motion is well described as a random walk, or perhaps a Levy flight. If they are totally sober, however, ballistic motion would be a good description. In-between there might be something like super-diffusive motion; hopefully subdiffusive motion never occurs.  One can write down equations for each of these different models, and re-arrange them to derive functions to compare with the data from a graph, if that's what you have for data.  Using the resulting equations, one could predict what time the freshman would ultimately arrive at home.


问:
If $a\neq 2$ and $a^3+a^2-a-10 = 0$, then evaluate $a+\frac{5}{a}$

If $a\neq 2$ and $a^{3}+a^{2}-a-10 = 0$, then what is the value of $a+\dfrac{5}{a}$?

I have plugged equation on WolframAlpha and get the complex values of $a$ but you supposed to do this problem using factoring and very elementary algebra. I tried divide both sides of equation to get $a+\frac{5}{a}$, but the coefficients are given in a way that is hard to factor.

答: Just factor the given equation to get $(a-2)(a^2+3a+5)=0$. Since $a\neq2$, we must have $a^2+3a+5=0$. Now divide both sides by $a$ to get $$a+5/a=-3$$


问:
Find the radius of curvature at the origin

Find the radius of curvature at the origin for the curve $3x^2+4y^2=2x$.

Below is my attempt:

We know that the radius of curvature $\rho=\dfrac{{(1+y_1^2)}^{\frac{3}{2}}}{y_2}$ .

Computing $y_1$ at $(0,0)$ : Differentiating we get $3x+4y\frac{dy}{dx}=1\implies \dfrac{dy}{dx}=\dfrac{1-3x}{4y}.$

How can I evaluate $\dfrac{dy}{dx}$ at the point $(0,0)$?

Please help.

答:
Since, $$\frac{dy}{dx}$$is not defined here, i.e., y axis is tangent to the curve.
No one above is talking about Newtonian method, so I am going with that - it says if curve passes through origin and y axis(or x axis) is tangent to the curve, then radius of curvature at origin = $$lim(x \rightarrow 0) \frac{y²}{2x}$$
We have, $$3{x}^2 - 2x + 4{y}^2 = 0 $$
Dividing it by 2x,
$$\frac{3x}{2} - 1 + 4\frac{y^2}{2x} =0$$
When x --> 0,
$$ -1 + 4\rho = 0$$
$$ \rho = \frac{1}{4}$$


问:
Probability of a choice overlapping N times in a row or more when choosing K items from a selection of size M
There are total of 13 players in a game, and 4 of them are assigned "evil" randomly. Some players have noted, that some players have been evil multiple times in a row (specifically, a player was evil 5 times in a row over 5 games). I am wondering about this specific probability, that is, what is the chance that we would observe this behavior, for any player (not just this particular player).

In addition to that, I want to know the general probability for this problem: M players, K assigned "evil", and over N games any 1 or more player being assigned "evil" N times in a row.

Bonus question, a curiosity: What about if there weren't 5 games, but L games in total, and a streak of N or more?

The best thing I can solve for now is simply the probability that a specific player will be "evil" 5 times in a row:

(4/13)^5 = ~0.2757%

I tried some things with C(12, 3)^5/C(13, 4)^5, but that simply yields the result for a single specific person, and doesn't take into consideration the fact that multiple people could overlap too

I attempted to code up a simulation which would simply check this (link to jsfiddle, freezes browser for a bit). My result seems to be around 3.55342%, which seems oddly close to (4/13)^5 * 13= 3.58530%, which seems surprising.

I also tried enumerating each possibility, however, it seems like this method wouldn't be fruitful, as the number of total possibilities is around C(13,4)^5 = 186865965446875 (Or C(M, K)^L), which is something my computer couldn't be able to handle. Possibly there are ways of optimizing this though...

答:
The following solution applies to the original game: 13 players in all, 4 selected as "evil" each round, 5 rounds, and we want to know the probability that there is at least one player who is selected as evil in all 5 rounds.

There are $\binom{13}{4}$ possible ways to choose the $4$ evil  players in each round and a sequence of $5$ independent rounds, so there are $\binom{13}{4}^5$ possible outcomes, all of which we assume are equally likely.

If there is a set of $j$ players who are selected as evil in all $5$ rounds, then the number of possible outcomes is $\binom{13-j}{4-j}^5$, for $1 \le j \le 4$.  The set of $j$ players can be selected in $\binom{13}{j}$ ways.

So by Inclusion / Exclusion, the probability that at least one player is selected as evil in all the rounds is
$$\sum_{j=1}^4 (-1)^{j-1} \binom{13}{j} \binom{13-j}{4-j}^5 / \binom{13}{4}^5 = 0.03564316 $$


问:
Why does the Mandelbrot set contain (slightly deformed) copies of itself?
The Mandelbrot set is the set of points of the complex plane whos orbits do not diverge. An point $c$'s orbit is defined as the sequence $z_0 = c$, $z_{n+1} = z_n^2 + c$.

The shape of this set is well known, why is it that if you zoom into parts of the filaments you will find slightly deformed copies of the original shape, for example:

I measured some points on the Mandelbrot, and the corresponding points from one of these smaller Mu-molecules. Comparing the orbit sequences it was possible to find points on each sequence which were very close - but this experiment did not really help me to understand anything new.

答:
See this paper:

McMullen, Curtis T.,
  The Mandelbrot set is universal. In The Mandelbrot set, theme and variations, 1–17,
  London Math. Soc. Lecture Note Ser., 274, Cambridge Univ. Press, 2000.
  MR1765082 (2002f:37081)

PDF available at the author's site.

See also The significance of the Mandelbrot set.

OwnerDisplayName: anon

LastEditorDisplayName: anon

ContentLicense: CC BY-SA 2.5


问:
Is this function differentiable at point $A$?
I wonder if this funtion is differentiable at point $A$. I think that it is continuous so therefore I cannot use this to prove that it isn't differentiable. Can someone help me?

Edit: My main issue is that I've never worked with derivatives of parametric functions so I am not sure how to manage their derivatives. Do I have to write them as a matrix?

答:
Compare for example to
Conditions for a smooth parametric curve

First, let us check whether the function $f : (1, \infty) \rightarrow \mathbb{R}^2$ defined as
$$f(t) = \begin{pmatrix} \frac{2t^3}{t^2-1} & \frac{2t^3}{(t^2+1)^2}\end{pmatrix}$$
is actually continuously differentiable.

Differentiating as usual w.r.t. $t$ gives the Jacobian
$$J_f(t) = \begin{pmatrix}\frac{2t^4-6t^2}{(t^2-1)^2} & \frac{-2t^4+6t^2}{(t^2+1)^3}\end{pmatrix}.$$
$f$ is continously differentiable wherever $J_f$ is continuous. The only point where something bad can happen is clearly when $t^2-1 = 0$, i.e. $t = \pm1$. Since we assume $t>1$, $f$ is continuously differentiable (and in fact smooth) on its domain of definition.

Now, for the curve itself to be smooth (not just the function $f$), you also need to have $J_f(t) \neq 0$ for every $t\in(1,\infty)$, as explained for example here.
Both coordinates of $J_f(t)$ are simulatneously zero if $t=0$ or $t = \pm\sqrt{3}$ as you can check. Since $\sqrt{3}>1$, there is indeed a $t>1$ for which the curve is not smooth. The exact point on which smoothness fails is then given by plugging $t = \sqrt{3}$ into $f(t)$:
$$A = \begin{pmatrix} \sqrt{27} & \frac{\sqrt{27}}{8}\end{pmatrix}$$


问:
Equivalent representations
I have a few questions in relation to the definition of equivalent representations. I am looking at the following problem:

Suppose $\pi : G \to GL(V)$ is a representation . For $g \in G$ define $\pi^g : G \to GL(V)$ by $\pi^g (h)=\pi (g^{-1}hg)$. Show that $\pi$ is equivalent to $\pi^g$.

My first question is in relation to the definition of equivalent representations, to which I am using Vinbergs book "linear representations of groups" (p.5). If $\pi$ and $\pi^g$ are equivalent then there should exist an isomorphism $\varphi: V \to V$ such that

$$\varphi \pi (h)  = \pi^g (h) \varphi  $$
For all $h \in G$

I believe I should define the $\varphi$ such that this is true and then show that it is an isomorphism, but I am not sure if $\varphi \pi (h)  = \pi^g (h) \varphi  $ is meant at $\varphi \cdot \pi (h)  = \pi^g (h) \cdot \varphi  $ or as composition such that I am to prove that $(\varphi \circ \pi (h)) (v)  = (\pi^g (h) \circ \varphi) (v)$ for all v in V and all h in G. But when I try doing this it doesn't make sense.

答:
Simply set $\varphi:=\pi(g^{-1})$. Then we have
$$\varphi\,\pi(h)=\pi(g^{-1})\pi(h)=\pi(g^{-1}h)=\pi(g^{-1}hgg^{-1})=\pi(g^{-1}hg)\pi(g^{-1})=\pi^g(h)\,\varphi\,.$$
The operation is composition of functions, which is equivalent to matrix multiplication once a basis of $V$ is fixed.

Title: Equivalent representations


问:
How to change the order of integration of this triple integral?
So the question goes as follows:

Change the order of integration of the following integral:
$$ \int_{\frac{-12}{7}}^{7}\int_{0}^{7-y}\int_{0}^{\sqrt{7y+12}} f(x,y,z)dxdzdy $$ to $\to$
$$\int_{?}^{?}\int_{?}^{?}\int_{?}^{?} f(x,y,z)dydzdx $$

My answer (which partly is right, I hope)
$$\int_{\frac{-12}{7}}^{0}\int_{0}^{7}\int_{\frac{-12}{7}}^{7-z} f(x,y,z)dydzdx $$

My question
Could someone please verify if my answer is wrong and were my mistake exactly takes place. I spend a lot of time drawing this, and tried to transfer this to the triple integral. But I made an error somewhere along the path. The reason I know I made a mistake is because I calculated both triple integrals and got two different numerical values. If someone could show me the answer and tell me how you got there it would very much appreciated!

Thanks in advance!
:)

drawing

答:
If you are integrating over $dy$ first, the lower bound of $y$ is given by the parabolic cylinder $x = \sqrt{7y+12}$ and upper bound is given by plane $y+z = 7$.

Next for $dz$, the lower limit of $z$ is zero. For upper limit of $z$, find the intersection of the plane and the parabolic cylinder,

$y = \frac{x^2-12}{7} = 7 - z \implies z = \frac{61-x^2}{7}$.

Lastly for $dx$, the lower limit of $x$ is zero. The upper bound of $x$ is when $z = 0$, which is $x = \sqrt{61}$.

So the integral translates to,

$\displaystyle \int_{0}^{\sqrt{61}}\int_{0}^{(61-x^2)/7}\int_{(x^2-12)/7}^{7-z} f(x,y,z) \ dy \ dz \ dx$


问:
How to prove that $(p \land (\neg(\neg p \lor q))) \lor (p \land q) \equiv p$?
My attempt at this mathematical demonstration was as follows:

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) ≡ p
≡ (p ∧ (p ∨ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ (p ∧ q) ∨ (p ∧ q)
≡ p

答:
Via Boolean algebra,

$$p \, \overline{(\bar{p} + q)} + p q = p \bar{q} + p q = p \underbrace{(\bar{q} + q)}_{=1} = \color{blue}{p}$$


问:
Orthogonal projection onto a subspace

Let $v_1=\begin{pmatrix}1\\ 1\\ 1\\ 0\end{pmatrix}$ and $v_2 = \begin{pmatrix}0\\ 1\\ 1\\ 1\end{pmatrix}$. Find the orthogonal projection of $v = \begin{pmatrix}3\\ 2\\ 1\\ 0\end{pmatrix}$ onto $V= \operatorname{span}
(v_1,v_2)$.

If I have $A= \begin{pmatrix}1&0\\ \:1&1\\ \:1&1\\ \:0&1\end{pmatrix}$ and compute the projection matrix $P=A(A^TA)^{-1}A^T = \begin{pmatrix}\frac{3}{5}&\frac{1}{5}&\frac{1}{5}&-\frac{2}{5}\\ \frac{1}{5}&\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\ \frac{1}{5}&\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\ -\frac{2}{5}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5}\end{pmatrix}$ then how does this help? I'm not sure I've entirely understood the idea of orthogonal projection?

答: If you apply Gram-Schmidt to $\{v_1,v_2\}$, you will get $\{e_1,e_2\}$, with$$e_1=\frac1{\sqrt3}(1,1,1,0)\quad\text{and}\quad e_2=\frac1{\sqrt{15}}(-2,1,1,3).$$Therefore, the orthogonal projection of $v$ onto $\operatorname{span}\bigl(\{v_1,v_2\}\bigr)$ is $\langle v,e_1\rangle e_1+\langle v,e_2\rangle e_2$, which happens to be equal to $=\frac15\left(12,9,9,-3\right)$.

Title: Orthogonal projection onto a subspace


问:
Linear mapping on the vector space of sets with the symmetric difference
I am trying to construct a nice pedagogical example for a lecture. The power set of a finite set together with the symmetric difference (and scalar multiplication with $\mathbb F_2$) is a vector space. I would like to find a linear mapping on this set, but I am struggling to find one. For example, the "count operator" $\#$ is not linear, because $\#(\{1,2\} \triangle \{2,3\}) = \#(\{1,3\}) = 2$ which is not equal to $\#(\{1,2\}) + \#(\{2,3\})$.

答:
One example for an endomorphism is intersection.

Let $(\mathbb{V},\mathbb{F}_2)$ be the described vector space.
Let $F\subset \mathbb{V}$ and define $L:\mathbb{V}\rightarrow\mathbb{V}$ as $E\mapsto E\cap F$.
For any $A,B\in\mathbb{V}$, we have $L(A\Delta B)=A\Delta B\cap F = (A\cap F) \Delta (B\cap F)=L( A) \Delta L (B)$.
Now consider $\alpha\in\mathbb{F}_2$.
If $\alpha=0$ and $E\in\mathbb{V}$,
$$
L(\alpha E)=L(\emptyset)=\emptyset\cap F=\emptyset = 0 E\cap F=\alpha L(E)
$$
If $\alpha=1$,
$$
L(\alpha E)=L(E)=E\cap F=1 E \cap F =\alpha L(E)
$$


问:
Weak Convergence If and Only If (Pointwise) Convergence of Characteristic Function

This is actually a theorem from lecture notes, with the corresponding proof. Unfortunately, it doesn't prove the last bit, or mention it at all (!), and I have a question about the penultimate bit. This is the definition of weak convergence being used:

Let $\mu$ be a Borel probability measure on $\Bbb R^d$ and let $(\mu_n)$ be a sequence of Borel probability measures on $\Bbb R^d$.

We say that $(\mu_n)$ converges weakly to $\mu$, written $\mu_n \Rightarrow \mu$, if $\mu_n(f) \rightarrow \mu(f)$ as $n \rightarrow \infty$ for all bounded, continuous functions $f:\Bbb R^d \rightarrow \Bbb R$.

We say that the sequence $(X_n)$ of random variables on $\Bbb R^d$ converges weakly to $X$ if $\mu_{X_n} \Rightarrow \mu_X$.

Firstly, consider the penultimate claim (If a sequence...). What I don't understand is why this example is not a counter-example to the claim. ($\phi$ is the characteristic function.)

Secondly, consider the final claim (Conversely, if...). No proof was given in the lecture notes. I think I have a solution, but I'm not sure; if someone could look over my answer (given as an answer below), then I'd be most appreciative. Thanks!

答:
Note the definition of the characteristic function:
$$\varphi_X(x) = \mathbb{E}(e^{i\langle x, X \rangle}).$$
The weak convergence of $(X_n)$ towards X means that
$$\lim_{n\to\infty}\mathbb{E}(f(X_n)) = \mathbb{E}(f(X)) $$
for all continuous and bounded functions $\mathbb{R}^d \to \mathbb{R}$. since $y \mapsto e^{i\langle x, y \rangle} $ is continuous and bounded for any $x\in \mathbb{R}^d$, we obtain pointwise convergence of the characteristic functions.


答:
If you're asking for a check of what you've done, you've made a mistake already in the first step. We have (from the De Morgan's law)
$$\lnot(\lnot p \lor q) \equiv (\lnot\lnot p) \land (\lnot q) \equiv p \land (\lnot q) $$
so your expression after the first transformation should be
$$ (p \land (p \land (\lnot q))) \lor (p \land q) $$
See if you can take it from here.

Title: How to prove that $(p \land (\neg(\neg p \lor q))) \lor (p \land q) \equiv p$?

Body:
My attempt at this mathematical demonstration was as follows:

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) ≡ p
≡ (p ∧ (p ∨ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ (p ∧ q) ∨ (p ∧ q)
≡ p


问:
When is the operator norm inequality for matrix product an equality?
Let $\lVert \cdot \rVert$ denote the operator 2-norm of a matrix. It is well-known that for any two matrices $A \in \mathbb{R}^{n \times p}$ and $B \in \mathbb{R}^{p \times n}$ we have
$$\lVert AB \rVert \leq \lVert A \rVert \lVert B \rVert.$$
I am interested to know when the above inequality is an equality. One obvious scenario is when either of $A$ or $B$ is an identity matrix (upto a scalar multiple). However, is this the only possible case? Does there exist a characterization of $A$ and $B$ when equality holds? Any references to relevant research papers are welcome.

答:
The matrix $A \in \mathbb R^{n \times p}$ represents a linear map from $V = \mathbb R^n $ to  $W = \mathbb R^p $ which without ambiguity we can denote as $A: V \to W$

Similarly $B \in \mathbb R^{p \times n}$  represents $B: W \to V$, and $A \circ B: W \to W$

The 2-norm of a matrix is the same as operator norm of the linear map that it represents (see, e.g. https://math.stackexchange.com/q/482191 ), where the operator norm of, for example,  $A$ is defined as  $||A|| = sup _{0 \ne v \in V} (||Av||_W/||v||_V)$.

Since the spaces here are finite dimensional, it follows that any operator attains its norm, e.g. for $A$, there is at least one $v \in V$ with $||Av|| = ||A||.||v||$

(Proof: use equivalent definition of $||A|| = sup_{||v|| = 1} (||Av||) $then since the unit ball is compact in finite dimensions and $A$ is continuous then ||Av|| attains its maximum on the unit ball.)

Then for any $A$ we can define a set $S_A = \{v \in V: ||A v|| = ||A||.||v||\}$ and similarly, $S_B$.

If we now want $||A B|| = ||A||.||B||$ then we require a vector $w \in W$ with $||w|| = ||ABw|| = ||A(B w)||$ which can happen iff $w \in S_B$ and $Bw \in S_A$, i.e. $S_{AB} = S_A \cap B(S_B) \ne 0$.

I don't think that this makes for a very interesting property of $A, B$.

For example let $\{v_i\}_{i=1, n}$ be a basis of unit length vectors for $V$ and $\{w_i\}_{i=1, p}$ for $W$.

Define $B$ by $B(w_1) = v_2$ and $B(w_i) = 0 $ otherwise.
Define $A$ by $A(v_2) = w_3$ and $A(v_i) = 0 $ otherwise.

Then $||A|| = ||B|| = ||A \circ B|| = 1$


问:
Making sense of matrix calculus involving matrices
In matrix calculus, we can calculate scalar by matrix derivative or matrix by scalar derivatives, according to here.

Where do these formulae come from? What I'm confused about is how this is related to ordinary multi-variable differentiation? For example, given a matrix to scalar function $f: \mathbb{R}^{n \times m} \to \mathbb{R}$, in what sense is the result a $m \times n$ matrix?

Is this related to treating matrices as vectors of dimension $n \times m$, using matrix norm to measure distance on the vector space of matrices? However, shouldn't the resulting derivative linear map of type $\mathbb{R}^{n \times m} \to \mathbb{R}$? Why is it $\mathbb{R}^m \to \mathbb{R}^n$?

A detailed derivation of the formulae would be helpful!

答:
Ok, lets look at the general definition of a differential $df$ on normed spaces, for $f:\mathbb{S}\to\mathbb{P}$
$$f(x+h)=f(x)+df(x,h)+o(||h||_S)$$
where the differential $df(x,h)$ is linear and continuous in $h$.

Lets look at $\mathbb{S}=\mathbb{R}^m,\mathbb{P}=\mathbb{R}^n$.
Then any linear operator $\mathbb{R}^m\to\mathbb{R}^n$ can be represented by some matrix $\in \mathbb{R^{n\times m}}$ (i.e. by matrix-vector multiplication). Lets denote this matrix $A$. So out differential in this case becomes
$$df(x,h)=A(x)h$$
We call $A(x)$ a derivative (remeber $df$ is a differential).

Now we take $\mathbb{S}=\mathbb{R}^{n\times m},\mathbb{P}=\mathbb{R}$. But we also equip this $\mathbb{S}$ with the scalar product $B\cdot C=\sum_{i,j}B_{i,j}C_{i,j}$. This is basically the same as taking $\mathbb{S}=\mathbb{R}^{nm}$ with the standard dot product. Then any linear operator $L: \mathbb{S}\to\mathbb{P}$ can be represented as $L(h)=D\cdot h$ where $D$ is some matrix $\in \mathbb{S}$.

So in this case the differential $df$ can be represented as
$$df(x,h)=A(x)\cdot h$$
We call $A(x)$ a derivative. So $A(x)$ in both cases can even be the same (tables of numbers), but they are used differently.

Small example:
$$f(x)=Ax,\ A\in\mathbb{R}^{n\times m},\ x\in\mathbb{R}^m,\ f:\mathbb{R}^m\to\mathbb{R}^n$$
$$df(x,dx)=Adx$$
$$\frac{df(x)}{dx}=A$$

$$g(x)=A\cdot x,\ A\in\mathbb{R}^{n\times m},\ x\in\mathbb{R}^{n\times m},\ g:\mathbb{R}^{n\times m}\to\mathbb{R}$$
$$dg(x,dx)=A\cdot dx$$
$$\frac{dg(x)}{dx}=A$$
As you see the derivatives of $f$ and $g$ are the same, but they represent different things.

In general we can define a derivative when we can represent linear operators $\mathbb{S}\to\mathbb{P}$ with some product on $\mathbb{S}$. For $\mathbb{S}$ Hilbert space and $\mathbb{P}$ its field we can always use Riesz representation theorem to get such a representation. In this case a derivative will be always an element from the input space $\mathbb{S}$.


问:
Problem with square roots and squares
Suppose we have
$$ x^{2} = (-x)^{2}.$$

I understand that this equation holds because
$$\begin{aligned}
(-x)^{2} & = (-1\cdot x)^{2} \\
& = (-1)^{2} \cdot x^{2} \\
& = x^{2},
\end{aligned}$$
and so $\sqrt{x^{2}} = \sqrt{x^{2}}$ becomes $x = x$ and everything is fine.

But what if we just take the square root first, like so:
$$\sqrt{x^{2}} = \sqrt{(-x)^{2}}.$$
Wouldn't this gives $x = -x$, which is a contradiction?

答:
Since $\sqrt{x}$ denotes the nonnegative square root of $x$, it is not true in general that $\sqrt{x^2}=x$. For instance, if $x=-5$, then
$$
\sqrt{x^2}=\sqrt{25}=5=-x \, .
$$
So sometimes $\sqrt{x^2}$ equals $x$ and sometimes it equals $-x$, depending on the sign of $x$. In general,
$$
\sqrt{x^2} =
\begin{cases}
x \text{ if $x\geq0$} \\
-x \text{ if $x\leq 0$}
\end{cases}
\, .
$$
What makes things confusing is that if $x<0$, then $-x$ is positive.

Title: Problem with square roots and squares


问:
What is the greatest integer that divides $p^4-1$ for every prime number $p$ greater than $5$?
What is the greatest integer that divides $p^4-1$ for every prime number $p > 5$?  This was on a practice math GRE so it's probably really easy.</p>

答:
Let $n$ be the number. Then the only prime factors of $n$ are 2,3,5, because any  prime number $p>5$ doesn't divide $p^4-1$ for all other $p>5$.

So $n=2^a3^b5^c$, and using the fact that $p^2-1 \cong 1 mod 24$, $p^2+1$ is even and $p^4 \cong 1 \mod 5$, the answer is easy to figure..

To finish it:

If $p=5 mod 8$ (13 works) then $p+1, p^2+1$ are $2 \mod 4$ and $P-1 = 4 mod 8$, thus $a \leq 4$.

If $p= 4 mod 9$ (again 13 works) $p^2-1= 6 \mod 9$ and $p^2+1 =8 \mod 9$ thus $b \leq 1$.

If $p= 6 \mod 25$, then $p^2-1= 10 \mod 25$ and  $p^2-1= 10 \mod 25$ so $c \leq 1$. I think that actually $p=7$ works nicer in this case....


问:
Upper semicontinuity of fiber dimension of a module
I'm currently trying to solve the following exercise in commutative algebra.

Let $R$ be a commutative ring with $1$, $M$ an $R$-module. Let $\mathfrak{p} \in \operatorname{Spec}(R)$. Denote by $\kappa(\mathfrak{p})$ the residue field at $\mathfrak{p}$.
Define the fiber $M(\mathfrak{p})$ of $M$ in $\mathfrak{p}$ to be the $\kappa(\mathfrak{p})$-vector space $M_\mathfrak{p} / \mathfrak{p}M_\mathfrak{p}$, where the indexed $\mathfrak{p}$ denotes the localization at $\mathfrak{p}$. Show that, if $M$ is finitely generated, for all $n \in \mathbb{N}$, the set $A_n := \lbrace \mathfrak{p} \in \operatorname{Spec}(R): \dim_{\kappa(\mathfrak{p})} M(\mathfrak{p}) \leq n \rbrace$ is open.

Here's my idea so far. Fix some $\mathfrak{p}$ and suppose the dimension of $M(\mathfrak{p})$ is less than or equal to $n$. Choose generators $b_1, \ldots, b_n$. By Nakayama, we can lift these to generators $m_1, \ldots, m_n$ of $M_\mathfrak{p}$.

The next steps would be to lift this to generators of $M$ and to encode the surjection $R^n \to M$, which assigns to the $i$-th standard basis vector the $i$-th generator of $M$, an overdetermined matrix of some sort. The surjectivity then accounts to the fact that some quadratic minor of this matrix must not vanish, thus expressing $A_n$ as a union of principal open subsets. However, I don't see how this should be possible without any further assumptions on noetherianness etc., which leads me to believe I'm not on the right track with this approach. Any comments on this would be welcome, especially a pointer on how to continue this line of thought (if possible) or how to fix it.

I am not too familiar with the language of schemes, so I'd appreciate solutions that do not make substantial use of algebraic geometry.

答:
Suppose localization $M_p$ is generated by $n$ elements $\frac{m_i}{s_i}$ over $R_p$, then $\frac{m_i}{1}$ also generate $M_p$. We get the map $R^n \to M$ sending the standard basis to $m_i$'s, set $C$ to be the cokernel of that map, then $C_p \cong 0$.

Then the map $R^n \to M$ is surjective on the $U=\operatorname{Spec}(R)\backslash \operatorname{Supp}(C)$, but $U$ is open, since $\operatorname{Supp}(C)$ is closed because $C$ is finitely generated.

So, if there is a point $p \in A_n$ there is an open neighbourhood $U$ of that point in $A_n$.


问:
Criteria for computing the integral of $e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2 b} - k x y}$?
In some problem the following integral is involved
\begin{equation}
\int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 \alpha_{1}} - \frac{y^{2}}{2\alpha_{2}} - k x y}
\end{equation}
where $\alpha_{1}, \alpha_{2}$ and $k$ are positive constants. In the "limit for small $k$'' I attempt to solve this integral by rewriting the argument in the exponential as
\begin{equation}
-\frac{1}{2\alpha_{1}}\left(x + \alpha_{1}k y\right)^{2} - \frac{1}{2\alpha_{2}}\left(1 - \alpha_{1}\alpha_{2}k^{2}\right)y^{2}
\end{equation}
and integrating as if I had two Gaussian integrals (one for $x$ and one for $y$) which seems to yield the correct result (provided $1 - \alpha_{1}\alpha_{2}k^{2} > 0$). However, is this procedure valid? More specifically, when integrating over $x$, why can one integrate over a sum of variables just as if one were integrating over one variable?

答:
The procedure is called changing variables. So $$\iint f(x,y)dx dy=\iint f(u(x,y),v(x,y))|J(u,v,x,y)|du dv$$
Here $J$ is the Jacobian of the transformation. In this particular case the Jacobian is $1$


问:
Prove that $x_1^4+x_2^4\geq 2$
If $x_1$ and $x_2$ are the solutions of the equation $x^2+px- \frac{a}{p^2}=0$, a=$\frac{1}{\sqrt{2}}+2$ and $p \in R-\{0\}$
Then prove that $x_1^4+x_2^4\geq 2$.

Can i go with Viet’s Formulas $x_1+x_2=-b/a$, then apply the power of 4 on both sides but I’m stuck here. Someone help please.

Title: Prove that $x_1^4+x_2^4\geq 2$

Body:
Note that
\begin{align} x_1^4 + x_2^4 &= (x_1+x_2)^4 - 4x_1^3x_2-6x_1^2x_2^2-4x_1x_2^2 = \\
&= (x_1+x_2)^4 - x_1x_2 (4x_1^2+6x_1x_2+4x_2^2) = \\
&= (x_1+x_2)^4 - x_1x_2 \big(4(x_1+x_2)^2-2x_1x_2\big)
\end{align}
Then use the Vieta's formulae for $x_1+x_2$ and $x_1x_2$. See if you can show that what you get is greater of equal to $2$.


问:
Topology on $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z})$
It is well-known that the Pontryagin dual of a discrete locally compact abelian group must be compact, which means that $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z})$ must be compact with the compact-open topology (where $\mathbb{Q}$ is taken with the discrete topology). However, one can prove that $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) = \mathbb{R}^2$ (or at least I think so, the below proof could have a mistake I have overlooked). This leaves me wondering what topology does $\mathbb{R}^2$ have to make it compact.

We start with the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$. Since $\mathbb{R}$ is divisible, $\operatorname{RHom} (\mathbb{Q}, \mathbb{R}) = \mathbb{R}$. Also, $\operatorname{RHom}(\mathbb{Q}, \mathbb{Z}) = \mathbb{R}[-1]$ (see this reference). Also, $\mathbb{R}/\mathbb{Z}$ is divisible so $\operatorname{Ext}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) = 0$. We now use the induced long exact sequence

$$ 0 \to \mathbb{R} \to \operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) \to \mathbb{R} \to 0$$

However, $\mathbb{R}$ is injective so $\operatorname{Ext}(\mathbb{R}, \mathbb{R}) = 0$ and the extension must be trivial. Hence, $\operatorname{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z}) \cong \mathbb{R}^2$.

Due to the indirect computation, I find it hard to see how the compact-open topology applies to this group. What topology does $\mathbb{R}^2$ have in these circumstances?

答:
$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\R{\mathbf{R}}\DeclareMathOperator\Z{\mathbf{Z}}\DeclareMathOperator\Q{\mathbf{Q}}\DeclareMathOperator\wQ{\widehat{\mathbf{Q}}}$

The exact sequence obtained from $0\to\Z\to\R\to\R/\Z\to 0$ by applying $\Hom(\Q,-)$ yields:
$$0=\Hom(\Q,\Z)\to\Hom(\Q,\R)\to\Hom(\Q,\R/\Z)\to \Ext^1(\Q,\Z)\to\Ext^1(\Q,\R)=0,$$
so
$$0\to \R\to\wQ\to \Ext^1(\Q,\Z)\to 0.$$

I think this embedding of $\R$ has a dense image in the torsion-free compact connected group $\wQ:=\Hom(\Q,\Q/\Z)$; the quotient being divisible, torsion-free, of the right cardinal, it's indeed, as an abstract group, isomorphic to $\R$.

So, yes, as an abstract group, $\wQ$ is isomorphic to $\R^2$, which is itself isomorphic to $\R$, to $\R^{2021}$, etc. Of course this is not the most useful way to think of $\wQ$, as it's especially interesting as (compact) topological group.


问:
Cramer's Rule, 2x2 Matrix
Solve the following system using Cramer's Rule.

$$2x + y = 1$$

$$x - 4y = 14$$

I haven't done Cramer's rule for 2x2 matrices, but I figured that the same rules applied as in a 3x3, here's what I did;

$$\det(A) = -8-1 = -9$$

$D_x$ =
$
\begin{align}
\begin{bmatrix}
4 & 1\\
14 & -4
\end{bmatrix}
\end{align}
$
$
= -16-14 = -30
\quad \quad\quad D_y =
\begin{align}
\begin{bmatrix}
2 & 4\\
1 & 14
\end{bmatrix}
\end{align}
= 28 - 4 = 24
$

$$x = \frac{-30}{-9} = \frac{10}{3}; \quad\quad y = \frac{24}{3} = 8$$

Now the solution comes up with;

$Dx$ =
$
\begin{align}
\begin{bmatrix}
1 & 1\\
14 & -4
\end{bmatrix}
\end{align}
$

$Dy$ =
$
\begin{align}
\begin{bmatrix}
2 & 1\\
1 & 14
\end{bmatrix}
\end{align}
$

Which then results in a completely different answer, it's more likely that I made a mistake, could anyone point out what I did wrong?

答:
I think your thought process is just fine: however, I suspect that you may have transposed $1$ and $4$ in the respective columns for $D_x$ and $D_y$, which would account for the "discrepancy".

Did you mis-transcribe the value $4$ (right-hand side of equation 1), having meant to write $1$?

And if so, perform your calculations using Cramer's rule, using the same process, and I think you'll be "good to go!"

Title: Cramer's Rule, 2x2 Matrix


问:
how to do the derivative of least squares objective in vector form?
Below we have the least squares objective function in vector form. Assume dimension of $x$ is $m \times n$, $W$ is $n \times 1$ and $y$ is $m \times 1$. So $J$ is $1 \times 1$ a scalar. $T$ below is the transpose.

$$ J=\frac{1}{m}(xW -y)^{T}(xW-y) $$

Let $A =xW-y$, then $$ J=\frac{1}{m}A^{T}A$$ $$\frac{dJ}{dA}=\frac{2}{m}A$$  $$\frac{dA}{dW}=x .$$

By chain rule:

$$ \frac{dJ}{dw}=\frac{dJ}{dA}\frac{dA}{dW}=\frac{2}{m}Ax .$$

Obviously if we do this, the dimensions don't match up, since $A$ is $m \times 1$ and $x$ is $m \times n$. Instead, the answer should be $$\frac{dJ}{dw}=\frac{2}{m}x^{T}A .$$

My question is, how do we know the convention in this case? Can someone point me to how we should do the chain rule in this case?

答:
Not that it matters mathematically, but using capital letters for vectors and lower case for matrices is unusual. Nevertheless, a few hints:

$\frac{\partial}{\partial W_j}(A^TA)=\frac{\partial}{\partial W_j}\sum_{i} A_i^2 =\sum_{i}2 A_i\frac{\partial A_i}{\partial W_j}=2 A_i \nabla A_{ij} $

Rearrange $A_i \nabla A_{ij}=\nabla A_{ij} A_i=\nabla A^T_{ji} A_i$ to render an equivalent expression that has a more obvious connection to matrix notation.

In matrix form: $ \nabla (A^T A)=2(\nabla A)^TA$

In your problem $A=(xW-y)$ and $\nabla A=x$

$ \nabla A^2=2(\nabla A)^T A=2x^TA$


问:
Changing variables in the diffusion partial differential equation
Suppose we have a diffusion PDE with time-dependent diffusion coefficient like this
$$\frac{\partial{f}}{\partial{t}}=at^{a-1}\frac{\partial^2{f}}{\partial{x^2}},~~~~~(1)$$ where $a$ is some constant.

Is it allowed to divide the equation by $at^{a-1}$ and to substitute $\tilde{t}=t^{a}$? The idea is that the transformation would yield one that has a known solution, i.e.
$$\frac{\partial{f}}{\partial{\tilde{t}}}=\frac{\partial^2{f}}{\partial{x^2}}.$$

Attempt at a solution: Not very knowledgeable about calculus and thought that this could work. However, I then tried sanity-checking this by taking a Gaussian solution, substituting the time variable in it as proposed above and then plugging this "proposed" solution into eq. (1), but then the LHS are not equal to the RHS. This suggests that I am doing something illegitimate, but I am not sure what.

答:
After some time, I realized that I made a mistake in my initial attempt. With more care, it looks like the Gaussian solution satisfies equation (1), but I am unsure if this is correct.

So suppose that the substitution of $\tilde{t} = t^a$ is correct and equation (1) of the question can be reduced to $$\frac{\partial{f}}{\partial{\tilde{t}}}=\frac{\partial^2{f}}{\partial{x^2}},$$
which is solved by something like $f(x,\tilde{t})=C\tilde{t}^{-\frac{1}{2}}\exp(-\frac{1}{4}x^2\tilde{t}^{-1})$. We can express the latter in terms of $t$ as $f(x,t)=Ct^{-\frac{a}{2}}\exp(-\frac{1}{4}x^2t^{-a})$ and compute the needed derivatives.
$$\frac{\partial{f}}{\partial{x}}=-\frac{1}{2}xt^{-a}f,$$
$$\frac{\partial^2{f}}{\partial{x^2}}=\frac{1}{2}t^{-a}(x^2t^{-a}-1)f,$$
$$\frac{\partial{f}}{\partial{t}}=\frac{1}{2}t^{-1}a(x^2t^{-a}-1)f.$$
Taking the following ratio
$$\frac{\partial{f}/\partial{t}}{\partial^2{f}/\partial{x^2}}=at^{a-1},$$
we get the "time-dependent" diffusion coefficient on the RHS of equation (1).

Not sure if there are no major flaws with my reasoning, so will not resolve the question for a couple of days in case someone else wants to add to the question.


问:
Continuity with respect to Hausdorff metric, also with intersection?
Let $H$ be a real Hilbert space and let $v\in H$, $\|v\|=1$. Now, consider the affine subspaces $A(t) := tv + v^\perp$ of codimension $1$, $t\in\mathbb R$. Obviously
$$
d_H(A(t),A(s)) = |t-s|
$$
for all $t,s\in\mathbb R$. Hence, $t\mapsto A(t)$ is Lipschitz-continuous with respect to the Hausdorff metric. That was quite easy.

However, what I really would like to show is that $f : t\mapsto A(t)\cap B$ is continuous, where $B$ is a bounded, closed, and convex subset of $H$ and $f$ is defined on $K := \{t\in\mathbb R : A(t)\cap B\neq\emptyset\}$, which is a closed set.

I've tried a lot, but I can't seem to prove this. So, I'd be happy if anyone had an idea. For clarification: the Hausdorff distance between two sets $A,B\subset H$ is defined by
$$
d_H(A,B) := \max\left\{\sup_{a\in A}\operatorname{dist}(a,B),\,\sup_{b\in B}\operatorname{dist}(b,A)\right\}.
$$

答:
The answer is almost yes. The function $f$ is continuous on the interior of the set of times such that it is non-empty.

Indeed, pick a reference point $x=(t_0 v+w)\in A(t_0)\cap B$, where $w\in v^{\perp}$, let $s<t_0$ and $t=\alpha s+(1-\alpha)t_0$ and observe $y=sv+w'\in f(s)$ where $w'\in v^{\perp}$. Then,
$$\alpha y+(1-\alpha)x=(\alpha s+(1-\alpha)t_0)v +\alpha w'+(1-\alpha)w,
$$
implying that $\alpha y+(1-\alpha)x\in A(t)$ and since $B$ is convex, we also get that $\alpha y+(1-\alpha)x\in f(t)$.

Hence, $$
d(y, f(t))\leq \|y-(\alpha y+(1-\alpha x))\|=(1-\alpha) \|y-x\|=(\frac{t-s}{t_0-s})\|x-y\|\leq \frac{t-s}{t_0-s} diam(B)
$$
Since this bound is uniform in $y$ we deduce that $\sup_{y\in f(s)}d(y,f(t))\leq \frac{t-s}{t_0-s} diam(B)$. Picking another reference point $z\in A(s_0)\cap B$ and running the argument backwards establishes
$$
d_H(f(t),f(s))\leq \max \left\{\frac{t-s}{t_0-s},\frac{t-s}{t-s_0} \right\} diam(B)
$$
for $t,s\in (s_0,t_0)$.

Unfortunately, continuity might fail at $t=\sup K$. Indeed, assume that $H$ is $\ell^2(\mathbb{N}_0)$ and let $v=v_0=(\delta^0_n)_{n\in \mathbb{N}_0}$ and $v_k:=(\delta^k_n)_{n\in \mathbb{N}_0}+(1-2^{-k})v_{0}$ for $k\geq 1$. Then, define $B$
$$
B:=\overline{conv} \{v_k|k\in \mathbb{N}_0\}
$$
Then, $B$ is closed, convex and bounded (it's contained in $B(0,3)$). We'll argue that $A(1)\cap B=\{v\}$. Having shown that, we'll just observe that $d(v_k, v)\geq 1$, proving that $d_H(A(1-2^{-k})\cap B,A(1) \cap B)\not \to 0$ as $t\to 1$.

So, indeed if $y\in A(1)\cap B$, then there exists $y_j=\sum_{k=0}^{N_j} \alpha_{k,j} v_k$ such that $\sum_{k=0}^{N_j}\alpha_{k,j}=1$, each $\alpha_{k,j}$ is positive and $y_j\to y$. Now, the $0$'th component of $y_j$ is at most
$$
(1-2^{-k_0})\alpha_{k_0,j}+\sum_{k\neq k_0} \alpha_{k,j}
$$
since the $0$'th component of $v_k$ is equal to $1-2^{-k}$. Since the $0$'th component of $y$ is equal to $1$, we get that $\alpha_{k,j}$ must go to $0$ for every $k\geq 1$.
Now, since $\alpha_{k,j}$ is equal to the $k$'th component of $y_j$ and the coordinate projections are continuous, we get that the $k$'th component of $y$ is $0$ for every $k\geq 1$. Thus, we get that $y=v$ and thus, that $A(1)\cap B=\{v\}$.


问:
Is an ideal finitely generated if its radical is finitely generated?
Let $R$ be a commutative ring. If $R$ is not Noetherian, we can ask if some some ideals is finitely generated. For examples:

Is intersection of finitely generated ideals finitely generated? No, see for instance here.

Is radical $\sqrt{I}=\{x \in R \mid x^n \in I \text{ for some } n\ge 1\}$ of a finitely generated ideal $I$ finitely generated? No, see for instance here.

In the light of the previous two (sets of) counter-examples, I believe that claim

If $\sqrt{I}$ is finitely generated, then $I$ is also finitely generated,

is also false, but I wasn't able to construct a counter-example. It would be interesting for me to see counter-examples of various nature.

答: Consider $A$ a reduced ring and let $M$ be a finitely generated $A$-module with a non-finitely generated submodule (namely, you may take $M=A$ and $A$ a reduced non-Noetherian ring). Call $R=A\boxed\times M$ the ring having support the set $A\times M$ and operations $(a,m)+(b,n)=(a+b,m+n)$ and $(a,m)\cdot(b,n)=(ab,an+bm)$. Let's identify $M=\{0\}\times M$ and $A=A\times \{0\}$. Notice that $(a,m)^k=(a^k,\text{stuff})$ and therefore $\sqrt{0}\subseteq M$. Also, $(0,n)(0,m)=(0,0)$, therefore $\sqrt 0\supseteq M$. $R$ acts by ring multiplication on subsets of $M$ essentially like $A$ acts by $A$-module action, since $(a,m)(0,n)=(0,an)$. Therefore, any non-finitely generated $A$-submodule $N$ of $M$ turns into a non-finitely generated ideal of $R$ such that $\sqrt N=M$ is finitely generated.


问:
How do I rewrite this equation to solve for $t$
How do I rewrite this equation to solve for $t$?

$$x = (1 - t)^3 a + 3 (1 - t)^2 t b + 3 (1 - t) t^2  c + t^3  d$$

I should have paid better attention many, many years ago when I was in high school algebra!

If you want to take the time to explain how to rewrite it even better!

答:
This looks like a problem derived from Bezier curves.

I assume that $x$, $a$, $b$, $c$, $d$ are all known, and we're trying to find $t$.

So, you just multiply out all the powers of $(1-t)$ and gather together the terms containing the same power of $t$. You will end up with four terms: one each for $t^3$, $t^2$, and $t$, and a constant term. You actually get:
$$
(-a+3b-3c+d)t^3 + (3a-6b+3c)t^2 + (-3a+3b)t + a - x =0
$$

In other words, you will end up with something of the form:
$$pt^3 + qt^2 + rt + s = 0$$

where $p=-a+3b-3c+d$, $q = 3a-6b+3c$, $r = -3a+3b$, and $s=a-x$.

This is a so-called "cubic" equation that you need to solve for $t$.

There are formulas that give you the solutions of cubic equations, just as there are formulas for solving quadratics, but the cubic ones are much more complicated. Some techniques are described on this Wikipedia page. Also, you can find many "cubic equation calculators" on the internet that will give you the solutions for specific given values of $x$, $a$, $b$, $c$, $d$.


问:
A geometry problem with a cube (solid geometry)
I ask you to solve this problem, because in the book and I have different answers

Point $M$ is the midpoint of the edge $AB$ of the cube $ABCDA_1B_1C_1D_1$. Find the distance between the straight lines $A_1M$ and $B_1C$, given that the edge of the cube is equal to $a$.

Please help me. It's Translated from Russian and it's the full context of the problem.

答:
Let $H$ be the point on $A_1M$ such that $A_1H/HM=2$, and $K$ be the point on $B_1C$ such that $CK/KB_1=2$. Then it is not difficult to show that segment $HK$ is perpendicular to both $A_1M$ and $B_1C$ and is thus the distance between the lines.

To compute the length of $HK$ it may be useful to note that $HK={2\over3}GF$, where $G$ and $F$ are the midpoints of $A_1A$ and $B_1C_1$ respectively.


问:
Best Intro to Number Theory books for non mathematicians?
I want to know if there are any books that can teach me or introduce me to number theory requiring no prerequisites. This is coming from someone who wants to broaden their mathematical horizons and investigate number theory and cryptology.

Any suggestions are welcome.

答:
I taught an undergraduate course in number theory a few years ago using Weissman's An Illustrated Theory of Numbers as the textbook. I think it would probably also work well for self-study. As stated in the preface: "This book has no formal prerequisites beyond high-school algebra and basic coordinate geometry." I found the book to be very well-written, with conceptually clear proofs and lots of visuals to aid understanding.

The topics covered include what I would consider the core of elementary number theory, as well as some topics in quadratic forms that go beyond what I think you would usually see in a first course. There are many historical notes for background and context, and there are also a few sections that explore cryptographic applications, though these are a small part of the book. (If you want to get more into cryptography, I'd recommend supplementing this with a more specialized book on the topic.)


问:
How can I determine $\lim_{x\rightarrow 2} \frac{(x^3-5x^2+8x-4)}{x^4-5x-6}$?
This is the limit:

$$\lim_{x\to2}\frac{x^3-5x^2+8x-4}{x^4-5x-6}$$

Thank you.

答:
Both numerator and denominator approach $0$ as $x\to 2$, so you can use l’Hospital’s rule. However, you can resort to more elementary techniques: $x-2$ must be a factor of both polynomials, so just divide it out:

$$x^3-5x^2+8x-4=(x-2)(x^2-3x+2)\;,$$

and

$$x^4-5x-6=(x-2)(x^3+2x^2+4x+3)\;.$$

It’s not needed, but further factorization is possible in both cases:

$$x^2-3x+2=(x-1)(x-2)\;,$$

and

$$x^3+2x^2+4x+3=(x+1)(x^2+x+3)\;.$$

These factorizations are easily found using the rational root test.


问:
Proof that $FT^{-1}FT = I$ (Fourier Transform $\cdot$ Inverse Fourier Transform = Identity)
I am trying to understand the proof below. I understand all but how to get to the last line, which I'm completely confused by. Can anyone help me understand it? Thanks!
\begin{align}
FT^{-1}FT\left(s(t)\right)=& \dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty S(\omega)e^{i\omega t}d\omega \\
=& \dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty\left(\int_{t'=-\infty}^\infty s(t')e^{-i\omega t'}dt'\right)e^{i\omega t}d\omega \\
=& \int_{t'=-\infty}^\infty s(t')\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{-i\omega t'}e^{i\omega t}d\omega dt' \\
=& \int_{t'=-\infty}^\infty s(t')\delta(t-t')dt'=s(t)
\end{align}

答:
Took me a while to figure out as i thought you were integrating with respect to $t$, not $t'$

Let $$f(t')=\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{-i\omega t'}e^{i\omega t}d\omega=$$

$$=\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{i\omega(t-t')}d\omega$$

By the time domain shift theorem

$$f(-t'-t)=\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{i\omega(t')}d\omega=\delta(-t'+t)=...$$

To shift $f(-t'-t)$ back to $f(t)$, I'd need to multiply it with the Heaviside step function, which as far as i know is impossible. Anyone feel free to chime in and continue from here.

Sources:
Inverse Fourier Transform

https://www.thefouriertransform.com/pairs/fourier.php
https://en.wikipedia.org/wiki/Fourier_transform#Translation_/_time_shifting


问:
How to prove that $(p \land (\neg(\neg p \lor q))) \lor (p \land q) \equiv p$?
My attempt at this mathematical demonstration was as follows:

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) ≡ p
≡ (p ∧ (p ∨ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ (p ∧ q) ∨ (p ∧ q)
≡ p

答:
As you know from De Morgan's laws, $\left(p\land \lnot(\lnot p\lor q)\right)\lor(p \land q)\equiv \\
(p\land (p\land\lnot q))\lor(p\land q)\equiv \text{De Morgan's Law}\\
 ((p\land p)\land \lnot q)\lor(p\land q)\equiv \text{Associativity}\\
(p\land \lnot q)\lor (p\land q)\equiv \\
p\land (\lnot q\lor q)\equiv \text{ Un-Distributivity}\\
p\land T\equiv \\
p$


问:
Find a logical statement containing 3 variables, where it is only true when any two of the variables are true.
I can only use negations and implications.

My first idea was to try out all the possibilities, but that failed. My second idea was to write the solution in a truth table and look for two statements where, when I connect them with an implication, gives the solution. These two statements will be referred to as $S_1$ and $S_2$. The truth table for the solution is based on the wording of the question. And $S_1$ and $S_2$ are based on the behavior of implications. Where I've put "$-$", it can be both true or false.

\begin{array}{|c|c|c|c|c|c|}
A & B & C & S_1 & S_2 &Solution\\
\hline
1 & 1 & 1 & 1 & 0 & 0\\
1 & 1 & 0 & - & - & 1\\
1 & 0 & 1 & - & - & 1\\
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & - & - & 1\\
0 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
\end{array}

I have observed that $S_1$ and $S_2$ cannot take the form of $X \implies \neg Y$ when using only one implication. I have tested all other forms: $X \implies Y$; $Y \implies X$; $\neg X \implies Y$; $\neg Y \implies X$; $\neg X \implies \neg Y$; $\neg Y \implies \neg X$;

I did not find a candidate for $S_1$ or $S_2$

Is there a better method I can employ to find this statement?

Thanks

答:
You need to have many more implications.

Note that $p\lor q=(\lnot q\Rightarrow p)$ and so $p\land q=\lnot(q\Rightarrow \lnot p)$ in the Lindenbaum algebra of our Boolean algebra.  So I'll leave you to build $((A\land B)\lor (B\land C))\lor (C\land A)$.


问:
Discrete regularisation
Consider the following least squares problem in $X$:

$||AX-B||_2^2\rightarrow\min$, where $A$ and $B$ are known, real-valued matrices.

Is it there a regularisation method which ensures that the elements of $X$ are close to integers? (similar to how ordinary regularisation ensures that the elements of $X$ are bounded).

答:
Note that such a penalty would generally lead to a non-convex objective.
For example, the fractional part of $0$ and $1$ is $0$, however, the fractional part of $0.5$ is $0.5$.
So, whichever objective function you choose, I suspect it will not be easy to calculate in practice.

Though in principle, this can be done.
One way to is to penalize if the fractional parts of $X_i$ are not close to $0$ or $1$.
The fractional part of $x\in\mathbb{R}$ is $x-\lfloor x \rfloor$, where $\lfloor x \rfloor$ is the integer part of $x$.
So you could simply enforce that $x-\lfloor x \rfloor-1/2$ is close to zero.
For example, you could minimize the following expression for some $\lambda>0$:
$$
||AX-B||_2^2+\lambda\left|\left|X-\lfloor X \rfloor -\frac{1}{2}\right|\right|_2^2
$$

Title: Discrete regularisation


问:
Show that $\prod_{i=1}^{n} r_i = (−1)^{n−1}$ and $\sum_{i=1}^{n} r_i = 0.$
Suppose that $n\geq 2$ is a natural number and $r_1 . . . , r_n$ are all the distinct $n$th roots of 1.

Show that $\prod_{i=1}^{n} r_i = (−1)^{n−1}$ and $\sum_{i=1}^{n} r_i = 0.$

I am unsure of how to attempt this question any advice or help will be great thanks!

答:
Suppose $a,b,c$ are $3$rd roots of $1$. In other words they are solutions of
$$x^3-1=0\tag{1}$$
Then the following holds
$$(x - a) (x - b) (x - c)=0$$
expand and collect $x$
$$x^3-x^2 (a+b+c)+x (a b+a c+b c)-a b c=0$$
Compare with $(1)$. We have
$$a+b+c=0;\;abc=1$$
If the exponent is even, let's say $4$, then in a similar way we can prove that
$$a+b+c+d=0;\;abcd=-1$$
Hope this helps


问:
$A$ is in $\Theta (n \log n)$ and $B$ is in $\Theta (2^n)$ - $B$ can not be more efficient than $A$?
Now suppose that $A$ is in $\Theta (n \log (n))$ and $B$ is in $\Theta (2^n)$. Is it always possible that $B$ is more efficient (asymptotically) than $A$? I think the answer is no, but I am not sure how to prove it. Can you help me at this point? One is bounded by there $n\log(n)$ and the other is bounded by $2^n$.

For me $f(n) \in \Theta (g(n))$ means that $\frac{f(n)}{g(n)} = c > 0$

答:
Let me rephrase your question slightly in the way I understand it:

Given two functions $f(n) \in \Theta(n\log n)$ and $g(n) \in \Theta(2^n)$, is it always true that $f(n)\in o(g(n))$?

Then the answer is yes, as $n\log n \in o(2^n)$. More precisely, you know by definition that there are positive constants $c$ and $k$ such that for large enough $n$, $f(n)\leq cn\log n$ and $g(n)\geq k2^n$. So, for large enough $n$, you have $\frac{f(n)}{g(n)} \leq \frac{cn\log n}{k2^n}\to 0$ as $n$ goes to infinity. Thus, $f(n)\in o(g(n))$.


答:
Note that all roots of unity can be written as potences of one root of unity: W. l. o. g., $$r_i = r_1^i.$$
You can insert this into the product and apply Gauss' summation formula.

This proves at least the first equality.

Body:
Suppose that $n\geq 2$ is a natural number and $r_1 . . . , r_n$ are all the distinct $n$th roots of 1.

Show that $\prod_{i=1}^{n} r_i = (−1)^{n−1}$ and $\sum_{i=1}^{n} r_i = 0.$

I am unsure of how to attempt this question any advice or help will be great thanks!


问:
An other total-variation inequality of mine

Given differentiable, continuous $f\left ( x \right )$ on the interval $\left [ 0, 1 \right ]$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x= 0.$ Prove that
$$\left | \int_{0}^{1}xf\left ( x \right ){\rm d}x \right |\leq \frac{1}{12}\max\left | {f}'\left ( x \right ) \right |$$

I think I should transform the constant $1/12$ into an integral like $k\int_{0}^{1}x^{2}{\rm d}x,$ but $k$ is very unusual, I need to your helps, even an example of $f\left ( x \right )$ so that $\int_{0}^{1}f\left ( x \right ){\rm d}x= 0$ in order to know what I must do with the constant. Thanks a real lot.

答:
Since $\int_0^1f(x)\mathrm{d}x=0$, by integration by parts,
\begin{align*}
\int_0^1xf(x)\mathrm{d}x&=\int_0^1(x+C)f(x)\mathrm{d}x\\
&=\Big(\frac{1}{2}x^2+Cx\Big)f(x)\bigg|_0^1-\int_0^1\Big(\frac{1}{2}x^2+Cx\Big)f'(x)\mathrm{d}x\\
&=\Big(\frac{1}{2}+C\Big)f(1)-\int_0^1\Big(\frac{1}{2}x^2+Cx\Big)f'(x)\mathrm{d}x
\end{align*}
Since $f(1)$ is unknown, so we let $C=-\frac{1}{2}$, we get
$$\int_0^1xf(x)\mathrm{d}x=-\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)f'(x)\mathrm{d}x$$
Therefore,
\begin{align*}
\left|\int_0^1xf(x)\mathrm{d}x\right|
&=\left|-\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)f'(x)\mathrm{d}x\right|\\
&=\left|-f'(\xi)\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)\mathrm{d}x\right|, \quad x\in(0,1)\\
&\leqslant\frac{1}{12}\max_{x\in[0,1]}|f'(x)|
\end{align*}


答:
Whenever you are dealing with a limit that approaches to a number so that the denominator ends up being 0, the first thing you want to try is to see if the denominator and the numerator have anything in common. This is the case which does that.

Now, it is not clear right away that $x-2$ is a factor of both the numerator and the denominator, but because of what I said above, we can see if it does.

Assuming that you know how to do long or synthetic division, you can find out that

$$\lim_{x \to 2} \frac{x^3-5x^2+8x-4}{x^4-5x-6} = \lim_{x \to 2} \frac{(x-2)(x^2-3x+2)}{(x-2)(x^3+2x^2+4x+3)}$$

So the term $(x-2)$ will divide each other out.

If you try to plug in $x=2$ again, there is no problem because the denominator is not 0 anymore.

Now, geometrically speaking you also want to know what's going on. In this problem, there is a "hole" as $x \to 2$. Even though the function is not defined at $2$, the graph keeps approaching to that number.

Does this help ?

Title: How can I determine $\lim_{x\rightarrow 2} \frac{(x^3-5x^2+8x-4)}{x^4-5x-6}$?

Body:
This is the limit:

$$\lim_{x\to2}\frac{x^3-5x^2+8x-4}{x^4-5x-6}$$

Thank you.


问:
What relation have the Markov Property with Queueing Theory?
What relation have the Markov Property of Exponential Distribution, with Queueing Theory?
i need to know the relation between Markov Property
and queuying theory</p>

答:
It's related with theory of queues from its definition if we read the Queueing Theory and Markov Property whats about:.

You can think of a queue or a queue node as almost a black box. Jobs or "clients" arrive in the queue, possibly wait some time, take some time to process, and then exit the queue.

However, the tail node is not a pure black box, as some information about the interior of the tail node is needed. The queue has one or more "servers", each of which can pair with an arriving job until its departure, after which that server is free to pair with another arriving job.

In this way the markov property is related to the theory of queues, we are talking about a node that passes from one state to another without having memory of the previous events only of the present and the future state.

$P(X_n+1 | X_n )$

Markov processes use an exponential variable because no matter how much time has elapsed since the last movement in the distribution, the time until the next movement only depends on the state of the process (and not on the waiting time).

edit: i had to solve this myself after learn a various of definitions.
nobody understand my question and was closed cause not enought information and additional context.


问:
How to change the order of integration of this triple integral?
So the question goes as follows:

Change the order of integration of the following integral:
$$ \int_{\frac{-12}{7}}^{7}\int_{0}^{7-y}\int_{0}^{\sqrt{7y+12}} f(x,y,z)dxdzdy $$ to $\to$
$$\int_{?}^{?}\int_{?}^{?}\int_{?}^{?} f(x,y,z)dydzdx $$

My answer (which partly is right, I hope)
$$\int_{\frac{-12}{7}}^{0}\int_{0}^{7}\int_{\frac{-12}{7}}^{7-z} f(x,y,z)dydzdx $$

My question
Could someone please verify if my answer is wrong and were my mistake exactly takes place. I spend a lot of time drawing this, and tried to transfer this to the triple integral. But I made an error somewhere along the path. The reason I know I made a mistake is because I calculated both triple integrals and got two different numerical values. If someone could show me the answer and tell me how you got there it would very much appreciated!

Thanks in advance!
:)

drawing

答:
When the triple integral is of the form $dxdzdy$ with the limits of x and z dependent only on y, you can be formulaic about it. First draw 2D plots of y on the horizontal axis and x on the vertical axis, and y on the horizontal axis and z on the vertical axis. If you cut out the z-y plot and place it vertically-perpendicularly to the x-y plot, you should be able to visualize that the left (lower) limit of y is with the x-y curve at $\sqrt{7y+12}=x\implies y=\frac{x^2-12}{7}$ and that the right (upper) limit of y is with the z-y curve at $z=7-y\implies y=7-z$. Now we have one remaining dependency of $z$ on $x$, this can be found by the limits of integration. $$0<z<7-y\\0<x<\sqrt{7y+12}$$
implies
$$0>7-z>y\\-12<\frac{x^2-12}{7}<y$$

which combined give $0<z<\frac{61-x^2}{7}<19$. Finally, the limits on x are from 0 to $\sqrt{7(7)+12}=\sqrt{61}$. So the limits of integration are

$$\frac{x^2-12}{7}<y<7-z\\
0<z<\frac{61-x^2}{7}\\
0<x<\sqrt{61}$$


问:
Understanding why path $p*p^{-1}$ is homotopic to the identity/trivial loop
I saw this question, but the top answer has the same picture that is confusing me currently, so I was wondering if somebody could explain it to me. I'm reading Concepts of Modern Mathematics by Ian Stewart, and in it he is trying to define an inverse for paths so we can construct a group. He does this as follows:

If $p^{-1}$ is the path $p$ in reverse, then although $p*p^{−1}$ is not equal to the trivial loop, it is homotopic. As in Figure 139, we can shrink $p*p^{−1}$ gradually back towards the base point, at the same time going round it faster and faster.

Here's what I don't understand. Suppose $q = p*p^{-1}$:

Are there only inverses for paths that start and end at the same point? What about a path that starts at point $a$ and ends at point $b$ such that $a \neq b$, this won't be homotopic to the trivial loop

As $q$ is "shrinking", it is not covering the full distance, and instead of traveling from the initial point to some point that is not the initial point before returning. Doesn't this mean we're not dealing with another path, e.g. $r*r^{-1}$, in the second drawing, and another path in the third, etc. as it shrinks?

EDIT: Upon rereading the text, I realized that the author was trying to get to the fundamental group, which is why he works with loops. He explained homotopy by working with general paths, not just loops, and this is what got me confused since I didn't realize the general paths were only mentioned for this brief explanation, and most of the rest of the discussion involved loops. This document has a good explanation of why loops are used, saying that:

By restricting attention to loops with a fixed basepoint $x_0 \in X$, we guarantee that the product $fg$ of any two such loops is defined.

答: Inverses only exist in the algebraic sense if the paths begin and end at the same point. As you shrink the path it is certainly a new path, but homotopic to the original path. As the path shrinks it is not some other path $r * r^{-1}$ because the new path that has been shrunk only loops once before returning to the basepoint. To be of the form $r * r^{-1}$ you would need a path, and an inverse path that each return to the basepoint independently.


答:
As I'm guessing you found out, the numerator and denominator both evaluate to $0$ at $x = 2$ $(\dagger)$.  But the limit may nonetheless exist as $x$ approaches $2$.  so whenever such a situation arises, that indicates more work needs to be done.

In this case, we can apply l'Hospital once, and evaluate, or

we can note that we have polynomials in each of the numerator and denominator, and it is quite likely that each can be factored, and perhaps, a common factor can be found: Recall the Rational Root Theorem.

Indeed, we can factor and will find both the numerator and denominator share a common, and in fact, it's exactly the a factor that evaluates to $0$ when evaluated at $x = 2:\;$  $(x - 2)$.

$$\lim_{x\to2}\frac{x^3-5x^2+8x-4}{x^4-5x-6} = \lim_{x\to 2} \frac{(x - 2)(x^2 - 3x + 2)}{(x - 2)(x^3 + 2x^2 + 4x + 3)} $$

Now we can eliminate the factor (cancel it from the numerator and denominator):
$$\lim_{x\to 2} \frac{(x - 2)(x^2 - 3x + 2)}{(x - 2)(x^3 + 2x^2 + 4x + 3)} = \lim_{x \to 2} \frac{x^2 - 3x + 2}{x^3 + 2x^2 + 4x + 3}$$

Can you take it from here?

$(\dagger)$ [If it were only the denominator that evaluated to zero, but the numerator evaluated to some non-zero real value, then the limit would approach either $+\infty$ or $-\infty$ or fail to exist altogether. In this case however, we have what is called an "indeterminate" form: meaning more work needs to be done before "evaluating" the limit.]


问:
What is $(7^{2005}-1)/6 \pmod {1000}$?
What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$

My approach:

Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.

So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.

Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.

Any help is appreciated!

Title: What is $(7^{2005}-1)/6 \pmod {1000}$?

Body:
By geometric sum formula we have
$$\frac{7^{2005}-1}{6} = \frac{7^{2005}-1}{7-1} = 1+7+7^2+\dots+7^{2004}$$

The sequence $1,7,7^2,\dots$ has period $20 $ modulo $1000$ (since $7^{20} \equiv 1 \pmod{1000}$).

$$1+7+7^2+\dots+7^{2004} \equiv 100(1+\dots+7^{19})+1+7+7^2+7^3+7^4 \equiv 100\cdot 0+801\equiv 801 \pmod{1000} $$


问:
Why the impulse response needs to be right-sided for a system to be causal?
I have been revisiting my school notes about the definition of Causality for signal and systems. I already understand that for a system to be causal, the output at any time can only depend on past and present inputs. Wikipedia says

Suppose $h(t)$ is the impulse response of any system $H$ described by
a linear constant coefficient differential equation. The system $H$ is
causal if and only if
\begin{equation*}
 h(t) = 0, \forall t < 0
\end{equation*}

https://en.wikipedia.org/wiki/Causal_system

I could not understand why this is true. For example, for a system whose impulse response is $h(t) = 1, t \in R$, the output at for example $h(-1) = 1$ clearly doesn't depend on future inputs, so it should be a causal system. Does this mean the definition on Wikipedia is wrong?

答:
I think I found out. The definition on Wikipedia is derived from the response of Linear Time Invariant systems whose output is the result of a convolution. The reason for a causal system $h(t) = 0$ $\forall t < 0$ must hold true is:

For linear time invariant system $H$ whose impulse response is represented by $h(t), t \in \mathbb{R}$, the output of the system $y(t)$ in response to an input signal $x(t)$ at time $t$ is the result of convolution
\begin{equation}
y(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau
\end{equation}

For the system to be causal, it must not depend on future inputs, which means at time $t$, $x(\tau)$ in the above convolution must not be de-referenced for all $\tau > t$. This can only hold true when $h(t-\tau) = 0$ $\forall \tau > t$. Hence,
\begin{align}
h(t) = 0 && \forall t < 0
\end{align}


问:
Are the theories of betweenness the same in $\mathbb{R}^n$ for all $n\geq 2$?
Consider $\mathbb{R}^n$, for some $n$ greater than or equal to $2$. We can form a structure by adjoining to it the ternary betweenness relation $B(x,y,z)$. Are all those structures elementarily equivalent? I am excluding $\mathbb{R}^1$, because in that structure, all points are collinear.

答:
Here is a sentence separating $\mathbb{R}^2$ from $\mathbb{R}^n$ for $n>2$:

For any five points, no three of which are collinear, there are four points $A,B,C,D$ among those five such that the line segments $AB$ and $CD$ intersect.

I hope it is clear how to express this as a first-order sentence in the language of  betweenness. If not, let me know which part you're having trouble with.

The point is that for any five points in the plane, no three of which are collinear, four of them form a convex quadrilateral, the diagonals of which intersect. In higher dimensions, it is easily arranged that no four points among the five are coplanar.


答: No.  Using betweenness, you can express what it means for three points to be collinear (one is between the other two).  Then, given a $n$-tuple of points, you can define the affine space they generate, namely all points you can generate by repeatedly taking collinear points up to $n-1$ times (for instance, for $n=3$, you first take a new point collinear with two of the points, and then take a point collinear with the new point and the third point).  So you can detect the dimension as the least $n$ such that there is an $(n+1)$-tuple of points that affinely generates the whole space.

Title: Are the theories of betweenness the same in $\mathbb{R}^n$ for all $n\geq 2$?

Body: Consider $\mathbb{R}^n$, for some $n$ greater than or equal to $2$. We can form a structure by adjoining to it the ternary betweenness relation $B(x,y,z)$. Are all those structures elementarily equivalent? I am excluding $\mathbb{R}^1$, because in that structure, all points are collinear.


问:
Subsets of a topological space which are the boundary of their complement
Is there a name for a subset of a topological space which is the boundary of its own complement? That is, a subset $U$ of a topological space $X$ with $$\partial(X\setminus U)=U.$$

In order to be a boundary the subset must be closed, but trivially not all closed subsets satisfy this property, for example the closed unit disk in $\mathbb{C}$. On the other hand, all simple paths and 'simple' shapes of dimension $n$ embedded in $n+1$-dimensional Euclidean space trivially satisfy this property.

答:
As $\partial U = \partial X\setminus U$ this is just the same as a set which is its own boundary. So in particular it is closed (all boundaries are) and it has no interior; and these conditions are clearly necessary and sufficient.

Examples include (in $\Bbb R$) :$\Bbb Z$, the Cantor middle third set, any finite set, and $\{0\} \cup \{\frac1n: n \in \Bbb N^+\}$, among others. In the plane you can add lines as examples too. Such sets are called "closed nowhere dense sets".


答:
If $k$ and $n$ are not co-prime, you have to be very careful when dividing by $k\bmod n$ $-$ there may be more than one answer, as you saw. So $806/6\bmod 1000$ is not well-defined, because both $6\cdot 301$ and $6\cdot 801$ are equal to $806\bmod 1000$.

But in this case the problem is well-defined, because the expression $\dfrac{7^{2005}-1}{6}$ is an integer. So we can calculate it before reducing $\bmod 1000$. hgmath's answer shows one way to do this (and Raffaele's answer doesn't).

Title: What is $(7^{2005}-1)/6 \pmod {1000}$?

Body:
What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$

My approach:

Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.

So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.

Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.

Any help is appreciated!


问:
Show that $(M_*)^*\cong M$ for a von Neumann algebra $M$.
Let $M \subseteq B(H)$ be a von Neumann algebra. Denote $B(H)_*$ to be the $\sigma$-weakly continuous functionals on $B(H)$ and let $M_*= \{\omega\vert_M: \omega \in B(H)_*\}$. I want to prove that there is an isometric isomorphism $M \cong (M_*)^*$. The book I'm reading just says that "this follows by general Banach space theory".

My main problem is that I don't see an obvious map
$$M \to (M_*)^*$$
Given $m \in M$, we want to build a map $\psi_m: M_* \to \mathbb{C}$. Maybe
$$\psi_m: M_*\to \mathbb{C}: \sigma \mapsto \sigma (m)$$

works? In other words, is the canonical isomorphism $M \cong (M_*)^*$ given by
$$m \mapsto ev_m$$?

答:
But you do see an obvious map. You wrote it in your question.

Let $\Gamma:M\to(M_*)^*$ be $\Gamma(x)=\hat x$, where $\hat x(\varphi)=\varphi(x)$. This is clearly linear. It is also isometric, because the norm can be obtained as $\|x\|=\lim_n\langle x\xi_n,\eta_n\rangle=\lim \hat x(\varphi_{\xi_n,\eta_n})$ for appropriate unit vectors.

So one needs to check that $\Gamma$ is onto. Without writing many details, what happens is this. The key fact is to notice that normal functionals on $M$ are restrictions of normal functionals on $B(H)$, and that these are given by the trace-class operators. That is,
$$
M_*=B(H)_*/M^\perp,
$$
where $B(H)_*$ are the trace-class operators with the usual trace duality, and
$$
M^\perp=\{\varphi\in B(H)_*:\ \varphi|_M=0\}.
$$
Then
$$
(M_*)^*=(B(H)_*/M^\perp)^*\simeq M^{\perp\perp}=\overline{M}^{w^*}=M.
$$


问:
Criteria for computing the integral of $e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2 b} - k x y}$?
In some problem the following integral is involved
\begin{equation}
\int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 \alpha_{1}} - \frac{y^{2}}{2\alpha_{2}} - k x y}
\end{equation}
where $\alpha_{1}, \alpha_{2}$ and $k$ are positive constants. In the "limit for small $k$'' I attempt to solve this integral by rewriting the argument in the exponential as
\begin{equation}
-\frac{1}{2\alpha_{1}}\left(x + \alpha_{1}k y\right)^{2} - \frac{1}{2\alpha_{2}}\left(1 - \alpha_{1}\alpha_{2}k^{2}\right)y^{2}
\end{equation}
and integrating as if I had two Gaussian integrals (one for $x$ and one for $y$) which seems to yield the correct result (provided $1 - \alpha_{1}\alpha_{2}k^{2} > 0$). However, is this procedure valid? More specifically, when integrating over $x$, why can one integrate over a sum of variables just as if one were integrating over one variable?

答:
Express $\frac{x^2}{2a} + \frac{y^2}{2b} +kxy $ in a separable form with the variable changes

$$x=u\cos\theta -v \sin\theta, \>\>\> y=u\sin\theta +v \cos\theta$$
where, in the limit of small $k$ and $a\ne b$

$$\theta \approx \frac{ab}{b-a}k
$$
Then, the separable form reads

$$\frac{x^2}{2a} + \frac{y^2}{2b} +kxy
= \left( \frac1a - \frac{abk^2}{a-b}\right)\frac{u^2}2
+ \left( \frac1b - \frac{abk^2}{b-a}\right)\frac{v^2}2
$$

and

\begin{align}
& \int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2b} - k x y}\\
=& \int_{-\infty}^{\infty} \mathrm{d} v \int_{-\infty}^{\infty} \mathrm{d} u \; e^{-\left( \frac1a - \frac{abk^2}{a-b}\right)\frac{u^2}2- \left( \frac1b - \frac{abk^2}{b-a}\right)\frac{v^2}2}\\
=& \frac{2\pi}{\sqrt{\left(\frac1a - \frac{abk^2}{a-b}\right) \left( \frac1b - \frac{abk^2}{b-a}\right)}}\approx
 \frac{2\pi\sqrt{ab}}{\sqrt{1-ab k^2}}
\end{align}


问:
Elementary Row Matrices
Let $A$ =

$$
\begin{align}
\begin{bmatrix}
-4 & 3\\
1 & 0
\end{bmatrix}
\end{align}
$$

Find $2 \times 2$ elementary matrices $E_1$,$E_2$,$E_3$ such that $A$ = $E_1 E_2 E_3$

I figured out the operations which need to be performed which are;

$E_1$ = $R_2 \leftrightarrow R_1$

$E_2$ = $R_2$ = $R_2$ + $4R_1$

$E_3$ = $R_2$ * $\frac{1}{3}$

My question is how would I go about writing the elementary matrices? The solution says that they are;

$E_1$ =
$
\begin{align}
\begin{bmatrix}
1 & -4\\
0 & 1
\end{bmatrix}
\end{align}
$
$E_2$ =
$
\begin{align}
\begin{bmatrix}
3 & 0\\
0 & 1
\end{bmatrix}
\end{align}
$
$E_3$ =
$
\begin{align}
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}
\end{align}
$

答:
Note that the solutions are not unique. With your elementary row operations, we have
$$
\pmatrix{1&0\\ 0&\tfrac13}\pmatrix{1&0\\ 4&1}\pmatrix{0&1\\ 1&0}A = I_2.
$$
Therefore, by performing the reverse row operations (and also in reverse order) on $I_2$, we get
$$
A = \pmatrix{0&1\\ 1&0}\pmatrix{1&0\\ -4&1}\pmatrix{1&0\\ 0&3}I_2.
$$</p>

Title: Elementary Row Matrices


答:
We can write the n-th roots of unity as $  r_k=e^{2i\pi k / n}$ for $1 \leqslant k \leqslant n $. Proof: they each have different arguments so are distinct and raising them to the power of $n$ gives $r_k^n=e^{2i\pi k}=1$.

$$P=\prod _{k=1}^n r_k = \prod _{k=1}^n e^{2i\pi k / n}= e^{2i\pi / n  \sum_{k=1}^n k}= e^{(2i\pi / n)  \frac12 n(n+1)}$$
$$= e^{i\pi (n+1)}= (-1)^{n+1} = (-1)^{n-1} \;\;\blacksquare$$

$$S=\sum_{k=1}^n r_k = \sum_{k=1}^n e^{2i\pi k / n}$$
$$=  e^{2i\pi / n} \frac{1-(e^{2i\pi / n})^n} {1-e^{2i\pi / n}} = e^{2i\pi / n} \frac{1-1} {1-e^{2i\pi / n}} = 0  \;\;\blacksquare$$

Title: Show that $\prod_{i=1}^{n} r_i = (−1)^{n−1}$ and $\sum_{i=1}^{n} r_i = 0.$


答:
You can distinguish between the two possibilities you found by starting modulo $2000$:

because the Carmichael function of $2000$ is $100$, $7^{2000}\equiv(7^{100})^{20}\equiv1\bmod2000$,

so $7^{2005}\equiv7^5\equiv807\bmod 2000$.

Therefore, $7^{2005}-1\equiv806\bmod2000$.

Therefore, $\dfrac{7^{2005}-1}2\equiv403\bmod1000$.

Therefore, $\dfrac{7^{2005}-1}6\equiv403\times3^{-1}\bmod{1000}.$

In general, if $n=3k-1$ then $3^{-1}\equiv k\pmod n$,

and if $n=3k+1$ then $3^{-1}\equiv -k\pmod n$,

so  $3^{-1}\equiv-333\equiv667\bmod{1000}$.

Therefore, $\dfrac{7^{2005}-1}6\equiv403\times667\equiv801\bmod1000$.

Title: What is $(7^{2005}-1)/6 \pmod {1000}$?


问:
functional equation $f\left(x\right)+a=f\left(x+b\right)$
Find the solution of the functional equation $f\colon ℝ\to ℝ\:$

$$f\left(x\right)+a=f\left(x+b\right), \quad\forall x\in\mathbb{R}$$

where $a$ and $b$ are fixed real numbers and $b$ is a positive real number.

I suppose that the answer is
$$f\left(x\right)=cx+d+g\left(x\right)$$ for every $x$ real number,
where $g\left(x\right)$ is a periodic function.
I found by induction that
$$f\left(x\right)+ka=f\left(x+kb\right)$$ for every $x$ real number and $k$ integer.
However I could not find a complete proof for this. Can somebody help me out?

答:
We may assume that $b\ne0$, since otherwise the condition is trivial. Let $g(x)=f(x)-\frac abx$. From the functional equation, we have that $g(x+b)=f(x+b)-\frac abx-a=g(x)$, so that $g$ has period $b$.

The converse is also easily proven to be true. That is, every function $f$ satisfying the equation is of the form $g(x)+\frac abx$ for some function $g$ with period $b$.


问:
Integral of test function equals 0 implies antiderivative has compact support
I was reading through the Accepted Answer to this question and am having trouble with one part of the reasoning given. To quote part of the answer:

Detail: We used the following fact above: Given a test function ϕ on R, there exists a test function ψ with ϕ=ψ′ if and only if ∫ϕ=0. In case this is not clear: First, if ϕ=ψ′ then ∫ϕ=∫ψ′=0 because ψ has compact support. Suppose on the other hand that ∫ϕ=0, and define ψ(x)=∫x−∞ϕ. Then ψ′=ϕ and hence ψ is infinitely differentiable, while the fact that ∫ϕ=0 shows that ψ has compact support.

I understand that since $\psi' = \phi$, $\psi$ is infinitely differentiable since $\phi$ is infinitely differentiable. However, why does the fact that $\int\phi = 0$ imply that $\psi$ has compact support?

I think that by using the fundamental theorem of calculus, since $\psi' = \phi$, we can show that $\lim_{x \to +/-\infty} \psi(x) = 0$, but some digging around suggests that this is not enough to conclude that $\psi$ would have compact support.

答:
There is an assumption in the question which is very much needed. This is that $\phi$ is a test function so in particular it has compact support.

We have defined
$$\psi(x) = \int_{-\infty}^x \phi(x) dx$$
and since $\phi$ have compact support we have for all sufficiently large $x$ that the function $\psi(x)$ is constant since $\phi(x) = 0$ there. That
$\int_{-\infty}^\infty \phi(x)dx = 0$ tells us that this constant is zero. Thus $\psi(x) = 0$ for all sufficiently large $x$.


问:
Relatiosn with Matrices
For every integer $n>2$ and $A,B,C,D \in M_{2}(\mathbb{R})$ we have :

$1) AC-BD=I_{n}$

$2) AD+BC=O_{n}$

Prove that  $CA-DB=I_{n}$ and $DA+CB=O_{n}$.

I tried to prove that $AC=CA$ like the proof of theorem " $AB=I$ , then $BA=I$ " , after that  I tried with multiply and add the relations  but I didn't manage to prove that.
Any idea?

答:
To prove that determinant of AC is positive, we solve like that or we use the relations.
I try like that:

$\det(AC)=\det(BD+I_{n})=?$

I thought that $\det(BD+I_{n})=1+\mathrm{Tr}(BD)+\det(BD)$.

Title: Relatiosn with Matrices


问:
What does consistency of propositional logic means?
I know a few proofs of consistency of propositional logic, and all of them are based on very similar things.

We are showing our axioms are tautologies and our inference rules are preserving truth, so we can only prove the tautologies. Since $\left(A\wedge\lnot A\right)$ is not a tautology, we can't prove $\left(A\wedge\lnot A\right)$, and since inconsistent systems can prove all statements, propositional logic is consistent.

Here is my question; what axiomatic system did we use to prove the consistency of propositional logic, and how do we know that that axiomatic system is consistent? How can we be sure that propositional logic is actually consistent?

I know inconsistent systems can prove their consistency if we write the formula that states "this axiomatic system is consistent" with its language, since they can prove every statement. Thus, proving that an axiomatic system is consistent with its own axioms is not enough to actually prove the consistency...

答:
The question is: "What does consistency of propositional logic means?"

 My reply .
There are two classical notions of consistency:  consistency in the traditional sense  and  consistency in Post's sense (or consistency in the absolute sense) .

According to the definitions of these notions:

 consistency of propositional logic in the traditional sense  means that there does not exist such well-built formula that this formula and its negation belong simultaneously to the set of consequence of propositional logic;

 consistency of propositional logic in Post's sense (or in the absolute sense)  means that the set of consequence of propositional logic, is  not  equal to the set of all well-built formulas (the set of propositional formulas) of propositional logic.


问:
Find $\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$
Find the limit of $$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$$ without l'Hospital.

I tried expanding $1-\cos(x)$ to get something this kind $$\lim_{x\to 0}\frac{\sin x}{x}$$ but couldn't solve it. The answer is $1/p$.

答:
Substitute $x=2t$, so
$$
1+\sin x-\cos x=1+2\sin t\cos t-1+2\sin^2t=2\sin t(\sin t+\cos t)
$$
and your limit becomes
$$
\lim_{t\to0}\frac{2\sin t(\sin t+\cos t)}{2\sin(pt)(\sin(pt)+\cos(pt))}=
\lim_{t\to0}\frac{1}{p}\frac{\sin t}{t}\frac{pt}{\sin(pt)}\frac{\sin t+\cos t}{\sin(pt)+\cos(pt)}
$$


问:
Failure of the second derivative test
We had to examine what kind of point is $(0,0)$, for the function $$f(x,y)=2(x − y)^
2 − x^
4 − y^
4.$$

(The given answer is saddle point)

Both the partial derivatives are zero at $(0,0)$, however, the Hessian too, is zero for $(0,0)$, which means that second derivative test is inconclusive.

How then , can we come to the conclusion that $(0,0)$ is a saddle point?

I tried examining the sign of the first derivatives (Like we used to do for one variable functions), but I cant seem to think of a procedure that will do the job correctly, the extra variable is causing a bit of a problem...

答:
We can do it by hand.  We have $f(0,0) = 0$, so we mush show on some lines through $(0,0)$,  $f$ is positive in both directions, and on other lines it is negative in both directions.

For small $t$ we have
$$
f(t,0) = 2t^2-t^4 \sim 2t^2 > 0
$$
on both sides of $(0,0)$, and for small $t$ we have
$$
f(t,t) = -2t^4 < 0
$$
on both sides of $(0,0)$.

Title: Failure of the second derivative test


问:
Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $
Let $ f $ be a continuous function defined on $ [0,\pi] $. Suppose that

$$ \int_{0}^{\pi}f(x)\sin {x} dx=0,   \int_{0}^{\pi}f(x)\cos {x} dx=0 $$

Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $

答:
This is a problem of complex calculus.

The linear combination of both given integral can be used to represent a path. The function $f$ can be rewritten with the path inside. This is  possible because $f$ depends on one real variable.

The path can be interpreted as the upper half circle above the real axis in the complex plane.

Look at for example: Line integral.

Now just given the pair of identies allows us the combine with $i$ and $-i$. The path can be run clockwise and counterclockwise.

If $F$ is the function $f$ on the complex path over the upper half of the unit circle. With $F$ is the $\overline F$ a solution. This gives without the complexes the existence of two solutions for $f$.

The steadiness of $f$ is needed to the exitence of the integrals. This does not follow from the identities. A function that is steady has an integral. Functions that are steady have integrals multiplied with transcendent functions. These are in the class of steady functions too. This attribut hat to be given in the interval $[0,\pi]$.

If the identities were not given alone the solution will not be double.

I am the opinion it is better practice to be more precise on the question regarding for $x$. The question is hard because it has not contexts. There is reason to mark this duplicate. Now that a correct answer/proof is given the situation changes.


问:
Finding area of n-dimensional unit superellipse.
Given $p\in\mathbb{R^+}$ and $n\in\mathbb{N}$, I am looking to determine a function, $V(p,n)\in\mathbb{R^+}$, which returns the n-dimensional euclidean volume enclosed by the collection of points:
$$S = \{z :\sum_{i=1}^{n} |z_i|^p = 1 \}, \quad z \in \mathbb{R}^n.$$

For $n=2$, the function equals: $$V(p, 2) = \frac{ 4\Gamma(1+\frac1p)^2}{\Gamma(1+\frac2p)}.$$

But I do not know a reliable way to determine this for higher integer $n$.

答:
Despite what the title says, it looks like it's the volume of an n-dimensional supercircle that you want.
This wiki page gives the formula: $$V(p,n) = \frac{(2\Gamma(\frac{1}{p}+1))^n}{\Gamma(\frac{n}{p}+1)}.$$

This in turn comes from:

Dirichlet, P. G. Lejeune (1839). "Sur une nouvelle méthode pour la détermination des intégrales multiples" [On a novel method for determining multiple integrals], Journal de Mathématiques Pures et Appliquées. 4: 164–168.


问:
How to approach $\int\limits_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x$
How to approach this integral?

$$\int\limits_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x$$

答:
\begin{gather*}
Let\ I=\int ^{a}_{-\infty }\frac{\arcsin\left( e^{x}\right) +\ \arcsec\left( e^{-x}\right)}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{x} +e^{-x}\right)} dx\\
=\int ^{a}_{-\infty }\frac{\arcsin\left( e^{x}\right) +\arccos\left( e^{x}\right)}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{x} +e^{-x}\right)} dx\\
\left( As\ \arcsec( 1/t) =\arccos( t) \ for\ all\ t >0\ and\ e^{x}  >0\ for\ all\ x\right)\\
=\int ^{a}_{-\infty }\frac{\frac{\pi }{2}}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{x} +e^{-x}\right)} dx=\frac{\pi }{2}\int ^{e^{a}}_{0}\frac{e^{x} dx}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{2x} +1\right)}\\
Substitute\ e^{x} =t\\
e^{x} dx=dt\\
I=\frac{\pi }{2}\int ^{e^{a}}_{0}\frac{dt}{( c+\arctan( t))\left( t^{2} +1\right)}\\
Now\ let\ \arctan( t) =u\\
\frac{1}{t^{2} +1} dt=du\\
I=\frac{\pi }{2}\int ^{\arctan\left( e^{a}\right)}_{0}\frac{du}{( c+u)} =\frac{\pi }{2}\ln \mid u+c\mid =\frac{\pi }{2}\ln\left(\frac{\arctan\left( e^{a}\right)}{c} +1\right)\\
Remembering\ that\ c=\arctan\left( e^{a}\right) ,\\
I=\frac{\pi }{2}\ln( 2)
\end{gather*}
Hope this answers your question!


Title: What is $(7^{2005}-1)/6 \pmod {1000}$?

Body:
Just to give a different approach, calculating mod $2000$ and using the fact that

$$3^{2000}=(1-10)^{1000}=1-1000\cdot10+\cdots\equiv1\mod2000$$

we have

$$\begin{align}
7^{2005}-1&=-1-(3-10)^{2005}\\
&\equiv-1-3^{2005}+2005\cdot3^{2004}\cdot10-{2005\choose2}3^{2003}\cdot10^2+{2005\choose3}3^{2002}\cdot10^3\\
&\equiv-1-3^5+5\cdot3^4\cdot10-{5\choose2}3^3\cdot100+{5\choose3}3^2\cdot1000\\
&\equiv-1-243+4050-27000+90000\\
&\equiv-244+50-1000\\
&=-1194\mod2000
\end{align}$$

and thus

$${7^{2005}-1\over6}\equiv{-1194\over6}=-199\equiv801\mod1000$$


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
The best way to see why is to perform long multiplication. If we multiply together numbers that end in six (say, ones that are three digits long, but without loss of generality) it follows this pattern:

```<code>    AB6
  x CD6
  -----
   EFG6
  HIJK
 LMNO
=======
PQRSTU6
</code>```

The two least significant digits are six, and they multiply out to 36. So in the first partial product, we "write down the 6, and carry the 3". This 6 that we write down produces the units digit in the final result. Nothing from the other partial products will add to this 6 and so it is preserved through to the final product. Only the first partial product determines the units digit. The next partial product affects only the tens and up, the one after that from the hundreds and so on. This is clear in the long multiplication by the diagonal layout.

And so all (non-zero, positive integral!) powers of 6 must end in a 6.  This is because, first of all, $6^1 = 6$ ends in 6, and all other powers are formed from that one with additional multiplications by 6, so we are always multiplying together numbers ending in 6 which we know always lead to products ending in 6.

Powers of 6 are in fact only only a small subset of the numbers that are formed by multiplying two factors ending in 6. For instance 16 isn't a power of 6 and neither is 26. Yet their product, also not a power of 6 is 416.

Now, if we want to get slightly more rigorous, and leave behind long multiplication, we can do that without a lot of difficulty. Let us rewrite our mystery representative numbers AB6 and CD6 this way:  $100A + 10B + 6$, and $100C + 10D + 6$.  Now what is their product? It is:

$$(100A + 10B + 6)(100C + 10D + 6)$$

When we multiply these together, following the FOIL method, it's clear that the only number that is not divisible by 10 will be the last term: $6\times 6 = 36$:

$$100A\times 100C + 100A\times 10D + 100A\times 6 + 10B\times 100C + 10B\times 10D + 10B\times 6 + 6\times 100C + 6\times 10D + 6\times6$$

$$= 10000AC + 1000(AD + BC) + 100(6A + BD + 6C) + 10(6B + 6D) + 36$$

$$= 10(1000AC + 100(AD + BC) + 10(6A + BD + 6C) + 6B + 6D) + 36$$

I.e. it's 10 times something, plus 36, which has to end in 6.

Body:
The best way to see why is to perform long multiplication. If we multiply together numbers that end in six (say, ones that are three digits long, but without loss of generality) it follows this pattern:

    AB6
  x CD6
  -----
   EFG6
  HIJK
 LMNO
=======
PQRSTU6

The two least significant digits are six, and they multiply out to 36. So in the first partial product, we "write down the 6, and carry the 3". This 6 that we write down produces the units digit in the final result. Nothing from the other partial products will add to this 6 and so it is preserved through to the final product. Only the first partial product determines the units digit. The next partial product affects only the tens and up, the one after that from the hundreds and so on. This is clear in the long multiplication by the diagonal layout.

And so all (non-zero, positive integral!) powers of 6 must end in a 6.  This is because, first of all, $6^1 = 6$ ends in 6, and all other powers are formed from that one with additional multiplications by 6, so we are always multiplying together numbers ending in 6 which we know always lead to products ending in 6.

Powers of 6 are in fact only only a small subset of the numbers that are formed by multiplying two factors ending in 6. For instance 16 isn't a power of 6 and neither is 26. Yet their product, also not a power of 6 is 416.

Now, if we want to get slightly more rigorous, and leave behind long multiplication, we can do that without a lot of difficulty. Let us rewrite our mystery representative numbers AB6 and CD6 this way:  $100A + 10B + 6$, and $100C + 10D + 6$.  Now what is their product? It is:

$$(100A + 10B + 6)(100C + 10D + 6)$$

When we multiply these together, following the FOIL method, it's clear that the only number that is not divisible by 10 will be the last term: $6\times 6 = 36$:

$$100A\times 100C + 100A\times 10D + 100A\times 6 + 10B\times 100C + 10B\times 10D + 10B\times 6 + 6\times 100C + 6\times 10D + 6\times6$$

$$= 10000AC + 1000(AD + BC) + 100(6A + BD + 6C) + 10(6B + 6D) + 36$$

$$= 10(1000AC + 100(AD + BC) + 10(6A + BD + 6C) + 6B + 6D) + 36$$

I.e. it's 10 times something, plus 36, which has to end in 6.