问:
Exponential Growth, half-life time
An exponential growth function has at time $t = 5$

a) the growth factor (I guess that is just the "$\lambda$") of $0.125$ - what is the half life time?

b) A growth factor of $64$ - what is the doubling time ("Verdopplungsfaktor")?

For a), as far as I know the half life time is $\displaystyle T_{1/n} = \frac{ln(n)}{\lambda}$ but how do I use the fact that we are at $t = 5$?

I don't understand the (b) part.

Thanks

答:
The growth factor tells you the relative growth between $f(x)$ and $f(x+1)$, i.e. it's $$
  \frac{f(t+1)}{f(t)} \text{.}
$$
If $f$ grows exactly exponentially, i.e. if $$
  f(t) = \lambda\alpha^t = \lambda e^{\beta t} \quad\text{($\beta = \ln \alpha$ respectively $\alpha = e^\beta$)} \text{,}
$$
then $$
  \frac{f(t+1)}{f(t)} = \frac{\lambda\alpha^{t+1}}{\lambda\alpha^t} = \alpha = e^\beta \text{,}
$$
meaning that, as you noticed, the grow factor doesn't depend on $t$ - it's constant.

The half-life time is the time $h$ it takes to get from $f(t)$ to $f(t+h)=\frac{f(t)}{2}$. For a strictly exponential $f$, you have $$
  f(t+h) = \frac{f(t)}{2} \Rightarrow \lambda\alpha^{t+h} = \frac{\lambda}{2}\alpha^t
  \Rightarrow \alpha^h = \frac{1}{2} \Rightarrow h = \log_\alpha \frac{1}{2} = -\frac{\ln 2}{\ln \alpha} = -\frac{\ln 2}{\beta} \text{.}
$$

Similarly, the doubling-time is the time $d$ it takes to get from $f(t)$ to $f(t+d) = 2f(t)$, and you have $$
  f(t+d) = 2f(t) \Rightarrow \lambda\alpha^{t+d} = 2\lambda\alpha^t
  \Rightarrow \alpha^d = 2\Rightarrow h = \log_\alpha 2 = \frac{\ln 2}{\ln \alpha} = \frac{\ln 2}{\beta} \text{.}
$$

Thus, you always have that $d = -h$ for doubling-time $d$ and half-time $h$, which of course makes sense. If you go forward $d$ units of time to double the value, then going backwards $d$ units halves the value, similarly for going forward respectively backward $h$ units to half respectively double the value.

In your case, you get that the doubling time for (b) is $\frac{\ln 2}{\ln 64} = \frac{1}{6}$. For (a) you get that the half-life time is $-\frac{\ln 2}{\ln \frac{1}{8}} = \frac{1}{3}$.

You can also derive those by observing that a growth factor of one-eight means that going forward one unit of time means the value decreases to one-eight of the original value. Thus, after going forward one-third unit, the value decreases to one-half, since if if it decreases three times to one-half each, it overall decreases to one-eight.

Similarly, if the value increases to 64 times the original value when going forward one unit of time, you have to go forward one-sixt unit of time to have it increase to twice the value, since $2^6 = 64$.

Title: Exponential Growth, half-life time


问:
What is the most general definition of a graph?
I would like to give the most general definition of a graph. Based on what I read about graphs, I came up with the following definition:

Let us introduce some symbols. They are

(a) $V$:: a non-empty set of vertices;

(b) $E$:: a set of edges $E$;

(c) $\mathcal{P}(X)$:: the power set of a set $X$;

(d) $\mathit{\Pi}(X)=(\mathcal{P}(X)\cup
\bigcup \limits_{i=2}^{\infty}X^i)\setminus \varnothing$:: the alternative power set (with ordered and unordered tuples);

(e) $F$:: a class of functions defined over any subset of $\mathit{\Pi}(V \cup E) \cup (V \cup E)$.

The structure $G=(V,E,F)$ is a (general) graph if there is an $I \in F$ with $I: E \rightarrow \mathit{\Pi}(V \cup E)$, which is called incidence function.

Is this a specification which includes all versions of graphs known in the literature (or at least the most important ones)? Can it happen that an even more general definition is needed for some very special graphs?

It would be also interesting if this formalism could be reduced.

答:
The most general definition that I am aware of is the one of Ubergraphs which generalize Hypergraphs. And Ubergraphs are not included in your definition.

A hypergraph is a pair $(V,E)$ whereby $V$ is a set whose elements are called vertices and $E \subseteq \mathcal{P}(V)$ is a non-empty subset of $V$ whose elements are called hyperedges. A hypergraph allows more than two vertices in an edge of the graph. Hypergraphs can also be directed, and thus generalizing directed graphs.

An Ubergraph goes one step further by allowing not only more than two vertices in each edge but also other edges. All hypergraphs are therefore just some special ubergraphs. And the usual graphs are some special $2$-hypergraphs. Making both versions directed is usually done by introducing an incidence mapping defining two additional mappings for each endpoint (source, target). How this is generalized to Ubergraphs, I do not know.

ContentLicense: CC BY-SA 4.0


问:
Expectation value for non-independent multivariate continues variables
I'm trying to calculate the expectation values $\mathbb{E}(X)$ and $\mathbb{E}(Y)$ for the continuous pdf:

$$ f(x,y)=xe^{-y}\ , \ \,\ 0<x<y<\infty $$

Since the variables are depndent, I'm not sure if I can use the following method: $\mathbb{E}(X) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xf(x,y)dxdy$ to get:

$$ \int_{x}^{\infty}\int_{0}^{y}x^2e^{-y}dxdy$$

Because the result will be dependent on x. Am I in the right direction or should I do something else?

Also, for the covariance, can I use $cov(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$
I'm not sure if it's a general formula or I should be more careful in the continuous dependent case

答:
Notice that $0<x<y<\infty$ corresponds to $x>0,y>x$ or $0<x<y, y>0$. Your integrals should be as follows:

$$E(XY)=\int_0^\infty\int_x^\infty (xy)xe^{-y}dydx\\
E(X)=\int_0^\infty x\int_x^\infty xe^{-y} dydx\\
E(Y)=\int_0^\infty y\int_0^y xe^{-y}dxdy$$

For almost all if not all of them, you should use integration by parts setting $dv=e^{-y}dy$.


问:
What are the Fourier series of the function?
What are the Fourier series of $f(x)$ where $x \in [-\pi, \pi]$ defined by

$$f(n) =
\begin{cases}
1,  & \text{if $x \in$ [0,$\pi$)} \\
0, & \text{if $x \in$ [$-\pi$,0)}  \\
\end{cases}
$$

And what are the complex Fourier series?

My result is

$$ \frac{\sqrt{2}}{2} + 2\sum_{n=0}^\infty \frac{\sin((2n+1)x)}{(2k+1)\pi}$$

and for the complex Fourier series

$$ \frac{\sqrt{2}}{2} + \sum_{n=-\infty}^\infty \frac{-ie^{inx}}{n\pi}$$

答:
As Jonathan said, your trigonometric series is correct except for the constant term and notational detail ($n$ versus $k$). </p>

<p>The exponential series is not correct, however. If you transform the sines to complex exponentials using $\sin t = \frac1{2i}(e^{it}-e^{-it})$, you will get only odd-indexed terms. Direct calculation confirms this: for $n\ne 0$
 $$c_n =  \frac{1}{2\pi}\int_0^\pi e^{-i nx}\,dx = -i \frac{1-(-1)^n}{2\pi n }$$
which is $-i/(\pi n )$ when $n$ is odd, and $0$ otherwise. </p>

ContentLicense: CC BY-SA 3.0

Body:
As Jonathan said, your trigonometric series is correct except for the constant term and notational detail ($n$ versus $k$).

The exponential series is not correct, however. If you transform the sines to complex exponentials using $\sin t = \frac1{2i}(e^{it}-e^{-it})$, you will get only odd-indexed terms. Direct calculation confirms this: for $n\ne 0$
 $$c_n =  \frac{1}{2\pi}\int_0^\pi e^{-i nx}\,dx = -i \frac{1-(-1)^n}{2\pi n }$$
which is $-i/(\pi n )$ when $n$ is odd, and $0$ otherwise.


问:
Solving system of equations, Gaussian Elimination
I am hoping to gain some insight on the characteristic differences between the following two distinct sets [(1) and (2)] of systems of equations. In particular, Gaussian elimination doesn't produce the correct result for the second.

(1)

Gaussian elimination does work for a system of equations like this (Wikipedia example):

2x + y - z = 8

-3x - y + 2z = -11

-2x + y + 2z = -3

However, I have so far been unable to use Gaussian elimination to solve the below example. Context: given 3 coordinates, I want to solve the resulting system of equations to produce a parabola that intersects each point.
Three example (x,y) coordinates: (-1.5, 1) , (-0.4, 1.5) , (0.8, 2.7)

(2)

Resulting system of equations:

1 = a(-1.5)^2 + b(-1.5) + c

1.5 = a(-0.4)^2 + b(-0.4) + c

2.7 = a(0.8)^2 + b(0.8) + c

My gut tells me it has to do with the distinction that occurs as a result of plugging in these values into a quadratic formula template, wherein the unknown values are "not the typical" x, y, and z variables, but instead what I would call the coefficients. My mathematical understanding and vocabulary isn't very deep here; any instruction or explanation would be appreciated.

Thanks so much,

Mike

Comments & criticism welcome - I'm new here.

答:
I am not sure whether your definition of a parabola is unique.

As you see here , you need 5 points to determine a specific conic, but maybe i'm missing the point.

To determine a specific parabola you need 3 points.

Futhermore please type out your word by word, so we can see what you have done, whether it was correct and where you got stuck. Welcome to stackexchange and don't take the downvotes personally! :)

If you want to use the $(x−x_0)(x−x_1)$ template, you'll need to find the specific two points where the parabola intersects the $x$-axis. And plug in those $x_0$,$x_1$ values.


问:
How would the mathematics relevant to physical theory be different if it didn't use real numbers?
Real numbers assume we can have infinite precision and some of the theory behind them uses infinite processes to establish certain proofs. A small band of mathematicians–eg ultrafinitists–disagree with this and deny the possibility of infinite processes or the logical structure of proofs that require them.

I always assumed that this was an interesting but faddish idea.

But I was recently reading a book by the very influential mathematician and computer scientist Donald Knuth (incidentally one of the popularisers and inventors of surreal numbers, a totally different way of establishing a theory of numbers) and though he didn't doubt the theory of real numbers he made the argument that it is wrong to assume they apply to the real world.

He argues (in chapter 6 of Things a Computer Scientists rarely Talks About, bold highlight mine, italics Knuth):

When I say that the question "finite or infinite?" is a red herring, I don't mean simply that philosophers and theologians have often been arguing about an unimportant issues. I also mean that physicists and scientists fail to realise this. For example, take the literature of chaos theory: Hundreds of papers have been written about the behaviour of solutions to unstable recurrences, by people who assume that real numbers are real.

...But it is a tremendous leap of faith to assume that real numbers apply perfectly to the real world...

...It seems to me that a new branch of physics is needed, called maybe "discrete physics" or something like that, to study the effects of the assumption that parameters can be infinitely precise and to consider instead that the universe probably has only a finite–but extremely large–number of states.

I'm not interested in the philosophical implication or speculations, or whether ultrafinitism has a point. But the question of whether we misunderstand some physics because we assume real numbers apply to the mathematics underpinning physical theory sounds about as important as realising that the universe is not based on flat Euclidian geometry.

So what parts of mathematics relevant to physics would be different if the concept of real numbers doesn't work for physical theory? And what are the implications for physics?

答:
First, of course we could build physics, using only rational numbers. Since all our measurement apparatuses use rational numbers, we could build all theories using them. This possibly would take more unnecessary work regarding precisions and conversions between measurements of different precisions, rounding, etc. It would be not only in engineering but in physical theories, complicating them. In other words, physical theories would be fused with engineering and more complicated.

As to the positions of Knuth regarding underlying physics. It is completely irrelevant to the first part of the answer.

First, it seems he assumes that limited precision means discreetness. This is wrong. As our current theory tells, there is indeed limit on maximum precision and resolution. But this does not imply discreetness. The logarithm of number of possible states of a system is entropy. Our current theories say, entropy is finite. The number of possible "states" is finite but not integer.


问:
Any complex curve in a complex surface is the zero set of some holomorphic section of a holomorphic line bundle
Let $Y$ be a closed complex surface, $L\to Y$ be a holomorphic line bundle, $\sigma:Y\to L$ a holomorphic section, and $B\subset Y$ the zero set of $\sigma$. If the first Chern class $c_1(L)$ of $L$ is divisible by some integer $n\geq 2$, it is known that there is a $n$-fold covering $X\to Y$ branched along $B$ ($X$ may have singular points).

What I am curious is the following question: Is any complex curve embedded in $Y$, or more generally any (complex) codimension 1 analytic subvariety in $Y$ the zero set of some holomorphic section of a holomorphic line bundle over $Y$?

答:
Yes. This is known as the divisor - line bundle correspondence, see section 2.3 of Huybrechts' Complex Geometry: An Introduction for example.

A divisor in a complex manifold $Y$ is a locally finite, formal, integral linear combination of irreducible hypersurfaces of $Y$. That is, a divisor takes the form $\sum_Z \eta_ZZ$ where $\eta_Z \in \mathbb{Z}$, the sum is taken over all irreducible hypersurfaces $Z \subset Y$, and the collection $\{Z \mid \eta_Z \neq 0\}$ is locally finite. We say a divisor is effective if $\eta_Z \geq 0$ for all $Z$. An analytic subvariety of codimension one is therefore an example of an effective divisor.

On a complex manifold, one can construct a holomorphic line bundle $\mathcal{O}(D)$ associated to a divisor $D$, which admits a meromorphic section $\sigma$ whose associated divisor is $D$. If the divisor was effective, then the section $\sigma$ is actually holomorphic. In particular, the line bundle associated to an analytic subvariety $Z$ of codimension one admits a holomorphic section $\sigma$ such that $\sigma^{-1}(0) = Z$.

Beyond complex manifolds, you must be careful. There are two notions of divisors in algebraic geometry, namely Weil and Cartier; they do not coincide in general. The definition of divisor I gave above is actually the definition of a Weil divisor, while the notion of Cartier divisors is the one which gives rise to holomorphic line bundles with sections.


问:
The critical point of the function $z^4 - 2z^2$ at the $\infty$
Let $f(z) = z^4 -2z^2$ be a complex polynomial.
We know that $f'(z) = 4z^3 - 4z$,
can I conclude that since $f(\infty) = \infty$, so that $\infty$ is not a critical point?

(Seems like when we are doing complex function,
it is not too hard to generalize all the concepts such as zeros, poles, singularities to the extended complex plane $\mathbb{C} \cup \infty$, is there any advantages of doing this, or why don't we just do the complex function on the finite complex plane only?)

答: The point $\infty$ is a pole, so not a critical point.


问:
Find the total length of two line segments of two overlapping right triangles

Find the length of $AE+EB$ ?

(A) $\frac{128}{7}$

(B) $\frac{112}{7}$

(C) $\frac{100}{7}$

(D) $\frac{96}{7}$

(E) $\frac{56}{7}$

My solution:

For $\Delta AEB$ :

$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$

$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$

$\angle AEB = 180 - \angle BAC - \angle ABD = 120.52^{\circ}$

Use Law of Sines:

$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$

$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$

$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$

$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$

$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$

$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$

$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$ Answer: (D)

My question:

Is there a solution without having to compute both arcsin $\angle BAC$ & $\angle ABD$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.

答:
Note that $\triangle AED\sim\triangle CEB$. Then, letting $AE = x$ and $BE = y$, we have:

$$DE = \frac{9}{5}y,\  CE = \frac{5}{9}x$$

From the similar triangles. Now, we must have $AE + CE = AC$ and $BE + DE = BD$. Noting that $AC = 13$ and $BD = 15$:

$$x + \frac{5}{9}x = 13$$

$$y + \frac{9}{5}y = 15$$

Solving, we find $\displaystyle x = \frac{117}{14}$ and $\displaystyle y = \frac{75}{14}$. Thus, $\boxed{AE + BE= x + y = \frac{96}{7}.}$


问:
$\frac{X - \mu}{\sigma}$: standardising a random variable, or standardising a *normal* random variable?
I frequently see it written in textbooks that "we standardise random variables by the computation $\dfrac{X - \mu}{\sigma}$". But my understanding is that the computation $\dfrac{X - \mu}{\sigma}$ only standardises normal random variables (so that it has a mean of $0$ and a variance of $1$), right? Other random variables would require different computations to standardise them, and, even then, standardisation often is not possible for random variables of some types of distributions, right? Or do I have a misconception here?

答: The standardization will make every random variable (that has two moments) have mean $0$ and variance $1.$ On the other hand, would an exponential random variable with mean zero and variance one be standard? It is a question of language, not of mathematics.


问:
Construction of the projective plane over $\mathbb{F}_3$
I have a question about constructing  projective plane over $\mathbb{F}_3$. We first establish seven equivalence classes $P= \{ [0,0,1], [0,1,0], [1,0,0], [0,1,1], [1,1,0], [1,0,1], [1,1,1] \}$.

Given a triple $(a_0, a_1, a_2) \in \mathbb{F}^3_3 \setminus (0, 0, 0)$ we define the line
$L(a_0, a_1, a_2)$ as follows: $L(a_0, a_1, a_2) := \{ [x_0; x_1; x_2] \in P : a_0x_0 + a_1x_1 + a_2x_2 = 0 \}$.

It is quite easy for $L(0,0,1), L(0,1,0), L(1,0,0)$, because here we take the points which have zero on the first, second and third coordinate.

It gets more difficult for me for $L(0,1,1)$, because here we need to have $x_1+x_2=0$. Do we treat the coordinates of the points as elements of $\mathbb{F}_2$ or $\mathbb{F}_3$?

There are $26$ nonzero triples in $\mathbb{F}^3_3 $.

Do we check all 26 $L(x_0, x_1, x_2) $ sets?

Please help, because I really want to understand it.

Thank you.

答:
The points and lines are not triples in $\mathbb {F}^3_3$, but they are elements of $\mathbb{F}^3_2$. So the coordinates of points and lines are in $\mathbb{F}_2$.

For example, your line $L(0,1,1)$ contains the points $[x_0,x_1,x_2]$ with $x_1+x_2=0$. Thus $x_1=x_2=1$ or $x_1=x_2=0$. So the possibilities are:
$[0,1,1]$ , $[1,1,1]$ , $[1,0,0]$.

Edit:
First the question was about the Fano plane, I provided my answer in that case.


问:
Get wrong answer on $\frac{\partial \mathbf{x}^{\top} \mathbf{A} \mathbf{x}}{\partial \mathbf{x}}$ when using graph
I can use the product rule to obtain $\frac{\partial \mathbf{x}^{\top} \mathbf{A} \mathbf{x}}{\partial \mathbf{x}} = \mathbf{x}^{\top} \frac{\partial \mathbf{A} \mathbf{x}}{\partial \mathbf{x}}+(\mathbf{A} \mathbf{x})^{\top} \frac{\partial \mathbf{x}}{\partial \mathbf{x}}$, and get the right result $\mathbf{x}^{\top}\left(\mathbf{A}+\mathbf{A}^{\top}\right)$. Here $x$ is a vector, $A$ is a matrix that is not a function of $x$.

However, when I try to calculate by using a graph, the result is different. I am not sure at which step I made the mistake. Please point out where I made it wrong. I used the graph attached (red numbers are the derivatives) and the equations as follows.

Let $
q_{2}=q_{1}x$, $q_{1}=(A^{T} x)^{T}$,  and $k=A^Tx$. I want to find $\frac{d q_{2}}{d x}$

$\frac{d q_{2}}{d x}=\frac{\partial q_{2}}{\partial x}+\frac{\partial q_{2}}{\partial q_{1}} \frac{\partial q_{1}}{\partial x}= q_{1}+x \frac{\partial q_{1}}{\partial x}= q_{1}+x \frac{\partial q_{1}}{\partial k} \frac{\partial k}{\partial x}  =x^TA+xIA^T = x^TA+xA^T \ne x^{\top}\left({A}+{A}^{\top}\right)$, $I$ is the NxN identity matrix.

The result I got here is different. I know it is not correct, but I couldn't figure out where I lost it.

Thank you!

答:
In the line
$$\frac{d q_{2}}{d x}=\frac{\partial q_{2}}{\partial x}+\frac{\partial q_{2}}{\partial q_{1}} \frac{\partial q_{1}}{\partial x}= q_{1}+x \frac{\partial q_{1}}{\partial x}=\dots$$
In terms of dimensions of matrices, it looks to me like you have written
$$ [1\times n] = [1\times n] + [1\times n][n\times n] = [1\times n] + [n\times 1][n\times n] = \dots $$
because  $x$ is a column vector. The product of matrices $[n\times 1][n\times n]$ is not defined, so there is an error here.

Let me try to rewrite in more explicit notation (I identify $\mathbb R^n$ with column vectors $\mathbb R^{n\times 1}$)

\begin{align} q&:\mathbb R^n \to \mathbb R,&q(x) &= x^TAx,\\
q_1&: \mathbb R^n \to \mathbb R^{1\times n}, & q_1(x) &= (A^Tx)^T = x^TA,\\
q_2&:\mathbb R^{1\times n} \times \mathbb R^n\to\mathbb R, & q_2(a,b) &= ab
\end{align}
Thus
$$q(x) = q_2(q_1(x),x)$$
Since there are subtly different spaces involved here, I'll elect to use a notation that works for arbitrary spaces, namely  the Frechet derivative $d$ which always satisfies the chain rule
$$ d(f\circ g)(x)[h] = df(g(x))[dg(x)[h]],$$
and then the only game is to figure out what spaces each differential belongs to.

Letting $\partial$ denote a partial Frechet derivative, we have
$$ dq(x)[h]= \partial_aq_2(q_1(x),x)[dq_1(x)[h]] + \partial_b q_2(q_1(x),x)[h].$$

Now, $q_1$ is linear, and $q_2$ is separately linear in each variable. For any linear map $L$, the Frechet derivative $dL(x)[h]=L(h)$. (The nice thing is that this covers multiplication of matrices on the left and on the right.) Thus for $h\in\mathbb R^n$ and $v\in\mathbb R^{1\times n}$,

\begin{align}
dq_1(x)[h]&=q_1(h)=h^TA,\\
\partial_a q_2(a,b)[v]&= vb,\\
\partial_b q_2(a,b)[h]&= ah.
\end{align}
therefore \begin{align}
 dq(x)h&=dq_1(x)[h]x + q_1(x)h\\
&=h^T Ax + x^TA h \\
&= x^TA^Th +x^TAh \\
&= x^T(A^T+A)h = ((A^T+A)x)\cdot h
\end{align}
and finally we can read the Jacobian from the final line:
$$ \frac {dq}{dx}(x) = x^T(A^T+A).$$
(And, the gradient vector is its transpose $\nabla q(x) = (A^T+A)x$.)

Equivalently, one has to understand that the Jacobian of a map $\mathbb R^n \to \mathbb R^{1\times n}$ acts on the right: if $g$ is $\mathbb R^{1\times n}$-valued, then
$ f(x+h) \approx f(x) + \frac{df}{dx}(x)h $ cannot make sense by dimension counting (remember $h$ is a column vector). So we are forced to accept that the correct approximation is $f(x+h)\approx f(x) + h^T \frac{df}{dx}(x)$. What this means is that the order of terms in chain rule is reversed! Indeed, the standard chain rule formula for the composition of a $\mathbb R^{n\times 1}$-vector valued function $f=f(y)$ and $g=g(x)$ is
$$ \frac{d(f\circ g)}{dx}(x) = \frac{df}{dy}(g(x))\frac{dg}{dx}(x)$$
If we take the matrix transpose of this equation, we obtain the chain rule for row-vector-valued functions $h=f^T$:
$$\frac{d(h\circ g)}{dx}(x) = \left(\frac{dg}{dx}(x)\right)^T\frac{dh}{dy}(g(x)).$$

A similar issue occurs in the above where the input is the row vector, and the approximation for $q_2$ reads
$ q_2(a+v,b)\approx q_2(a,b) + v \frac{\partial q_2}{\partial a}(a,b)$. Hence (and I think deriving from first principles is easiest)
\begin{align} q_2(q_1(x+h),b)&\approx q_2(q_1(x),b) + (q_1(x+h)-q_1(x)) \frac{\partial q_2}{\partial a}(q_1(x),b)
\\&\approx q_2(q_1(x),b) + h^T\frac{dq_1}{dx}(x) \frac{\partial q_2}{\partial a}(q_1(x),b)
\\& = q_2(a,b) +  \frac{\partial q_2^T}{\partial a}(q_1(x),b)\frac{dq_1^T}{dx}(x)h \end{align}  The correct chain rule is therefore, as painful as it seems,
$$\frac{dq}{dx}(x)= \frac{\partial q_2^T}{\partial a}(q_1(x),x)\frac{dq_1^T}{dx}(x) + \frac{\partial q_2}{\partial b}(q_1(x),x)$$

It would be easier IMO to avoid this mess by not using a $\mathbb R^{1\times n}$-valued function.


问:
Local minimum in undefined points in Hessian and first derivative of a function
It might seem naive but consider having a convex function like $|x|$ for which $f'(x)$ is not defined in 0, (is it? because $f'(x)= \frac{x}{|x|}$.) Now what do we say about its local minimum?
Because for a local minimum we have $f'(x) =0$ and $f'(x) >=0$. Which don't hold here.

Thanks in advance.

答:
I think you're mixing up two somewhat-related but distinct theorems:

If $f$ is a convex function, every local minimum of $f$ is also a global minimum.

If $f$ is $C^2$ and $f'(x)=0; f''(x)>0$ then $f$ has a local minimum at $x$.

This second fact is a sufficient condition on the existence of a local minimum, but not a necessary condition. (This sufficient condition generalizes to higher dimension, where the condition instead is that $\nabla f(x) =0 $ and $Hf(x)$ is positive-definite. It is an immediate consequence of Taylor's theorem.)

Even for smooth functions, sometimes a local minimum does not have positive second derivative. Consider for instance $f(x) = x^4$, which clearly has a minimum at $x=0$ but $f''(0)=0$.

The function $f(x) = |x|$ is not a contradiction to the above: it is convex so fact (1) applies. It is not in $C^2$, but even if it were, it would not be a contradiction to fact (2) since the latter is a sufficient condition, not an "if-and-only-if" condition.


问:
Show that for any a ∈ C − {0}, there exists a logarithm function defined on D(a, |a|).
We made use of the following proposition:

for f(z) to be logarithm function, two conditions need to be satisfied

$f\prime(z) = \frac{1}{z}$

there exists $z_{0} $ such that $ e^{f(z_{0})} = z_{0} $

for any $ D(a,\vert a \vert$), we seek to construct an logarithm function in the following two manners.

define an artificial domain $\Omega = D(a,\vert a \vert$), and let f(z) = $\frac{1}{2}log(x^2+y^2)+i*arctan(\frac{y}{x})$, it can be easily verfied that two conditions are held, and we have found such logarithm function.

I try to construct a more easily accessible example. Think of $ f(z) = \sum_{n=1}\frac{(−1)^{n−1}}{n}(z − 1)^n $, it satisfies condition (1) and (2) as well, but it only holds for $D(1,\vert 1 \vert$). Now, what if we modified it a little, like changing the 1 to a so that we can cook up another logarithm function? Or is the example in 1 unique?

答: Why not define your function as $f(x) = \int_a^x dx/x?$ The integral is defined over an arbitrary path (they are all the same by monodromy, since the domain is simply connected and does not contain zero).


问:
Solving a congruence system (which has other solutions)
I am trying to solve this congruence system :
$$\begin{cases}
3x+7y\equiv 5[11]\\
8x+4y\equiv6[11]
\end{cases}
$$
I multiply first by $8$ and $3$ both equations resp.
$$\begin{cases}
24x+56y\equiv40[11]\equiv7 [11]\\
24x+12y\equiv18[11]\equiv7 [11]
\end{cases}
$$
We then substract the second equation from the first one.
$$44y\equiv0[11],$$
Since $gcd(11,44)=11$, then previous equation is equivalent to : $4y\equiv0[11]$, and then we find that :
$$y\equiv0[11],$$
and now I replace it in the following equation : $3x+7\times 0\equiv5[11]$ which is equivalent to :
$3x\equiv5[11]$, hence
$$x\equiv 9[11]$$
As a result $(9+11k,0+11k')$ with $k,k' \in \mathbb{Z}$ are solutions of the system.

Now my remark is the following : I noticed that also $(0+11k,7+11k')$ with $k,k' \in \mathbb{Z}$ and also $(7+11k,4+11k')$ with $k,k' \in \mathbb{Z}$ are also solutions of the system which I don't understand.

答:
The determinant of the matrix
$\left[ \begin{array}{cc}
   3 & 7 \\
   8 & 4
\end{array} \right]$
is $3 \cdot 4 - 8 \cdot 7 = 1 - 1 = 0 \pmod{11}$

Hence the two equations are linearly dependent.

A little work results in $10(3x+7y) = 30x + 70y = 8x + 4y \pmod{11}$

Hence there is no solution.


答: Yes there is, drop a perpendicular down from $E$ to $AB$ and denote the point as $X$. Let $EX = x$ and $AX = 12-y$ and $BX = y$. Since $\triangle EBX $ ~ $\triangle DBA$, we have that $\frac9x = \frac{12}y$. Also, since we have $\triangle CBA$ ~ $\triangle EXA$, we have that $\frac5x = \frac{12}{12-y}$. Now, solving  for $x,y$ using our equations, we get $x=\frac{45}{14}, y = \frac{30}{7}$. Plugging in these values, we get $AX = \frac{54}{7}$ and $BX = \frac{30}{7}$ and $EX = \frac{45}{14}$. Now using the pythagorean theorem, we get $AE = \frac{117}{14}$ and $BE=\frac{75}{14}$. Now, adding these lengths, we get that the sum is $\frac{96}{7}$

Title: Find the total length of two line segments of two overlapping right triangles

Body:
Find the length of $AE+EB$ ?

(A) $\frac{128}{7}$

(B) $\frac{112}{7}$

(C) $\frac{100}{7}$

(D) $\frac{96}{7}$

(E) $\frac{56}{7}$

My solution:

For $\Delta AEB$ :

$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$

$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$

$\angle AEB = 180 - \angle BAC - \angle ABD = 120.52^{\circ}$

Use Law of Sines:

$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$

$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$

$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$

$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$

$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$

$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$

$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$ Answer: (D)

My question:

Is there a solution without having to compute both arcsin $\angle BAC$ & $\angle ABD$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.


问:
Are split-complex and dual numbers considered integral domains? If no, why?
From available definition of integral domain I concluded that it is a ring where we can divide by almost all elements, but not all. There are non-zero elements by which we cannot divide.

This seems being satisfied by the both numerical systems. So, they should be integral domains.

答:
Neither the split complex numbers nor the dual numbers form integral domains.

For the split complex numbers, where $j^2=1$, the elements $1+j$ and $1-j$ are two nonzero elements having a product of zero, so they are zero divisors.

For the dual numbers, where $\epsilon^2=0$, the element $\epsilon$ is a nonzero nilpotent element, so it is also a zero divisor.


问:
How should one explain the conventional formula for relative variation of signed variables?
The conventional formula for the relative variation between two signed values is $\displaystyle \frac{b-a}{|a|}$. Unfortunately, I could not find any clear justification for it, especially when considering examples like this one: according to the conventional formula, the relative variation between -10 and +20 is +300%, while the relative variation between -20 and +20 is +200%. This is puzzling, because the absolute variation for the former is +30 while the absolute variation for the latter is +40 and both datasets end at the same +20 value, yet the relative variation for the former is greater than the relative variation for the latter.

How can one justify this counter-intuitive scenario?

答:
A quantity, $a$, is changed by $b-a$. What fraction of $a$ does that change constitute?

The answer is $$\dfrac{b-a}{a}$$

This is called the relative change because its value is computed "realtive to" $a$.

This is not a useful measure when $a$ is close to $0$.

Wikipedia: approximation error


问:
Infinite sums, comparison with other series over a subset of terms
I have two convergent infinite sums. One over the primes and one over the natural numbers,

$$
S_P = \sum_{p} f(p),\ S_N = \sum_{n=1}^{\infty} f(n),
$$
where $f$ is positive.

My question stems from the fact that the primes is a subset of the natural numbers. What I am wondering is if there are any conditions on $S_N, S_P$ that would allow us to deduce that $S_P \leq S_N$ since all terms in $S_P$ are present in $S_N$. Would absolute convergence be what I am looking for here?

答: Hint: show that if $\sum_{n=1}^\infty f(n)$ is convergent then $\sum_p f(p)$ and  $\sum_c f(c)$ are also convergent, where $c$ runs over all composites. Since they are all positive terms, the series are absolutely convergent. Then use $\sum_p f(p)+ \sum_c f(c) =\sum_{n=1}^\infty f(n)$.


问:
How should I interpret notation such as $(X \times I) / (X \times \{0\})$ when dealing with quotient spaces?
I am learning point set topology for the first time and have just encountered cones. I understand that, given an equivalence relation $\sim$ on the set $X$, the quotient space is defined as the set of equivalence classes, denoted by $X\:/\sim$, equipped with the quotient topology. I have just been introduced to the cone $CX$ of a topological space $X$. I understand that $CX$ is the quotient space of $X \times I$ given by the equivalence relation

$$
(x', 0) \sim (x, 0), \forall x \in X
$$

where $I$ is the closed unit interval in $\mathbb{R}$. The cone of a set $X$ is typically denoted by:

$$
CX = (X \times I)\:/\:(X \times \{0\})
$$

From the formal definition of a cone, I understand what this notation means. $CX$ is the quotient space given by the equivalence relation which maps all members of  $X \times \{0\}$ to the same equivalence class, and any other element of $X \times I$ to its own unique class.

Is this how I should generally expect this notation to be used? For example, if I were to see a quotient space defined as follows:

$$
X / A
$$

where $A$ is a subset of $X$ (rather than an equivalence relation on $X$), should I just take this quotient space to be the one which corresponds to the equivalence relation where all members of $A$ belong to the same class, and all other members of $X$ each belong to their own unique class?

答: Informally I think of $X/A$ to mean all the points of $A$ are glued together as to be indistinguishable while the other elements of $X$ are left unchanged. This means the equivalence classes are all singletons except for $A$ which is its own equivalence class as you have correctly deduced.


问:
Conditional probability on a multiple choice test
A student is taking a multiple-choice test. Each question on the test has five possible answers and only one choice is correct. The probability this student knows the correct answer is 70%. If the student does not know the answer, they select an answer at random with each answer having an equal probability of being picked. Calculate the conditional probability that the student knew the answer given they answered the question correctly.

I started off like this:
Let B denote the event that the student knew the answer. Let A denote the event that the student answered the question correctly.

I was able to work out that $P(A)=\frac{3}{10}*\frac{1}{5} + \frac{7}{10} = \frac{19}{25}$

And I know the formula $P(B|A)=\frac{P(A \cap B)}{P(A)}$

I am unsure on how to work out $P(A \cap B)$. Any advice would be greatly apprectiated.

答: We are given that the student answered the question correctly, and we must find the probability that they actually knew the answer. Thus, we have two cases, one being that the student knew the answer and got it right, or the student did not know, and guessed correctly. The probability of the first case is $(7/10)$. The probability of the $2$nd case is $(3/10)(1/5)= 3/50$. Adding these up, we get that the total probability that the student answered the question correctly is $19/25$. Since we need to find the probability that the student KNEW the answer and got it correct, we do the probability he knew the answer and divide it by the total probability of getting the question right, so we get $(7/10)/(19/25)$ or $35/38$.


问:
Prove by strong induction that $9^n$ ends in $1$ if $n$ is even and ends in $9$ if $n$ is odd.

Prove by strong induction that $9^n$ ends in $1$ if $n$ is even and ends in $9$ if $n$ is odd.

Would it be valid to work separately with even and odd exponents, making two separate proofs, each using weak induction?

How can I prove this using strong induction in one proof?

答:
Base cases: $$9^1=9, 9^2=81$$

Inductive step:

Assume that $9^k$ ends in 1 if k is even and 9 if k is odd for all k from 1 to n. (This is strong induction.) Consider $9^{n+1}$. This can be rewritten $9(9^n)$. If n is odd, then $9^n$ ends in 9 and the product ends in 1. That is, $9^{n+1}$ ends in 1 if n+1 is even. If n is even, then $9^n$ ends in 1 and the product ends in 9. That is, $9^{n+1}$ ends in 9 if n+1 is odd.


问:
Is it true that $\frac{d}{dx}\int_{a}^{b}f(x,y)dy=\int_{a}^{b}\frac{d}{dx}f(x,y)dy?$

Is it always true that $\frac d {dx}\int_a^b f(x,y)\, dy = \int_a^b \frac{d}{dx}f(x,y)\,dy$, where domains are Euclidean space and $a$ and $b$ are allowed to be $\pm\infty$? This is widely used in many contexts but I don't think I've seen any proof regarding this. Would there be any case where one of the terms diverges but the other converges? If so, what is the condition to guarantee the above integrals are the same?

Another question is... if we want to evaluate $\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty\left| a_{m,n}\right|$, is it eligible to write $$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \left| a_{m,n} \right| = \sum\limits_{m=0}^\infty \sum\limits_{n=0}^\infty \left| a_{m,n} \right|$$
in the middle of my solution? I'm hesitant because we cannot normally say $\infty=\infty$ so without knowing that the above sum converges, we cannot switch the order. On the other hand, we have the theorem that says that if one of the sums above in the equality exists, two terms must be the same.

Could anybody give me any insight? Thank you so much!!

答: The first question is answered in this wikipedia article. The second is answered here.</p>


问:
How to evaluate this sum $\sum_{n=0}^\infty \frac{\cos\left(\left(2n+1\right)\phi\right)}{\cos^{2n+1}\left(\phi\right)}$?
$$\sum_{n=0}^\infty \frac{\cos\left(\left(2n+1\right)\phi\right)}{\cos^{2n+1}\left(\phi\right)}$$
Mathematica returns
$$\frac{2}{3}\left(-2+\frac{1}{5+3\cos{\left(2\phi\right)}}\right)$$
but I'm at a loss here. Doesn't this sum have convergence issues for $\phi=0$ or any other value for that matter?

If this evaluation is true, how would I go about showing this or whatever evaluation it may actually be?

答:
You are right in that it doesn't converge for any $\phi$ (as the limit of the summands isn't $0$). I'm guessing that Mathematica did something like the following behind the scenes: $$\sum_{n=0}^\infty \frac{\cos\left(\left(2n+1\right)\phi\right)}{\cos^{2n+1}\left(\phi\right)} = \sum_{n=0}^\infty \frac{\frac{e^{i\left(\left(2n+1\right)\phi\right)}+e^{-i\left(\left(2n+1\right)\phi\right)}}{2}}{\left( \frac{e^{i\phi}+e^{-i\phi}}{2} \right)^{2n+1}}$$

which would be equal to $$\sum_{n=0}^\infty \frac{e^{i\left(\left(2n+1\right)\phi\right)}}{\left(e^{i\phi}+e^{-i\phi} \right)^{2n+1}}2^{2n}+\sum_{n=0}^\infty \frac{e^{-i\left(\left(2n+1\right)\phi\right)}}{\left(e^{i\phi}+e^{-i\phi} \right)^{2n+1}}2^{2n}$$ if both sums converged. The first sum is $-\frac{e^{i\phi}\cos(\phi)}{-1+\cos(2\phi)+2i\sin(2\phi)}$ when $3e^{-2\Im(\phi)}-e^{2\Im(\phi)} < 2\cos(2\Re(\phi))$. The second sum is $\frac{1}{-1+e^{2ix}+2i\tan(x)}$ when $3e^{2\Im(\phi)}-e^{-2\Im(\phi)} < 2\cos(2\Re(\phi))$. Add these up and "ignore" the restrictions on $\phi$ to get what Mathematica got.


问:
why $C$ is compact?
I have some confusion in Rudin book

My confusion is given below and marked in red colour

My doubt: why $C$ is  compact ?

My attempt : I know for  each $x \in K$there exist  an  open neighbourhood $G$  and for

each $x \in U$  there exist an open neighbhourhood $W_{p_1}$

Closure $G$ and closure $W_{p_1}$ are compact since both are open cover

It is given that $K \subset U \implies U^c \subset K^c$

$C$ is complement  of U that mean $C= U^c$

$K^c$ is noncompact   since $K$ is compact

so $C \subset$ noncompact set $\implies$ $C$ is not  compact

答:
$C$ need not be compact here. What's compact are the sets
$$C \cap \overline{G} \cap \overline{W}p,$$
each of which are subsets of the compact set $\overline{G}$ ($G$ is defined to be a set with compact closure). There's no need for $C$ to be compact (it's just the index set); the finite intersection property is being used with respect to the compact subspace $\overline{G}$.

Title: why $C$ is compact?


问:
Diagonalizing/eigenvalues of a particular infinite dimensional matrix
I have trying to show that the continuum limit of n quantum harmonic oscillators gives rise the the Klein-Gordon field. However, instead of a usual finite string, I want to do it on a ring.

Assume $n \gg 1$.

Hence, my Lagrangian is

$$L=\frac{m}{2}(\dot{q_1}^2+\dot{q_2}^2+.... \dot{q_n}^2)-\frac{m \omega^2}{2}[(q_1-q_2)^2+(q_2^2-q_3^2)+....(q_n-q_1)^2]$$

So that the matrix for V is

$V=\begin{pmatrix} 2 & -1 &0 &. &. &. &-1\\-1&2&-1\\0 &-1 &2 &-1 \\.\\.\\.\\-1 &&.&.&.&-1 & 2
\end{pmatrix}_{n\times n}$

So that $L=\frac{m}{2}[\dot{x}^2-\omega^2 x^{T}V{x}]$. All the quantities here are matrices.

How do I find the eigenvalues of this matrix?

I tried to find a recursion relation between the characteristic polynomial of $N$ and $N-1$ dimensional matrix, but I failed. Is this the correct method? What other method is there?

After finding the eigenvalues, the Lagrangian can be written separated into its normal modes, and the propagator or kernel can be found out easily using that of the free particle. The limit of this as $N\to \infty$ should be the Klein-Gordon field.

But I am stuck on this. Any help will be appreciated.

答:
Note that your matrix is not just a Toeplitz matrix but also a circulant matrix. Those matrices have eigenvectors

$$v_j=(1,\omega_j,\omega_j^2,\ldots,\omega_j^{n-1})$$

with $\omega_j=\exp\left(2\pi i j/n\right)$ and corresponding eigenvalues

$$\lambda_j=c_0+c_{n-1}\omega_j+c_{n-2}\omega_j^2+\ldots+c_{1}\omega_j^{n-1} \; .$$

In your case, the eigenvalues thus become

$$\lambda_j=2-\omega_j-\omega_j^{n-1}= 2-2\cos\left(2\pi j/n\right) \; .$$

How to arrive at this result? Easiest way I can think of is to rewrite your eigenvalue equation as a difference equation with a periodicity condition $v_{j,k}=v_{j,n+k}$. Then you have

$$v_{j,k+1}+(\lambda_j-2)v_{j,k}+v_{j,k-1}=0$$

This is a homogeneous difference equation with constant coefficients that can be solved by guessing the solution to be of the form $v_{j,k}=\omega_j^k$. You then obtain a characteristic equation

$$\omega_j^2+(\lambda_j-2)\omega_j+1=0$$

with the additional condition that

$$\omega_j^{n}=1$$

to ensure periodicity. Since our equation is quadratic with real coefficients and we need periodicity, its solutions must be two complex conjugate numbers whose sum is $-(\lambda_j-2)$, therefore

$$\lambda_j-2=-\omega_j-\omega_j^* \; .$$

But from the periodicity condition we have that $\omega_j^*=\omega_j^{n-1}$ and thus

$$\lambda_j=2-\omega_j-\omega_j^{n-1} \; .$$

Finally, there are $n$ possible complex roots for the periodicity condition which implies that we have $n$ different eigenvalues and corresponding eigenvectors.


问:
Asymptotics in algorithms
In nearly every algorithm analysis that I've seen, the runtime is some composition or product of elementary functions such as:

$p(x)$ where $p$ is a polynomial

$\log(x)$

$e^x$

constants $c\in\mathbb{R}$

For example, $n! \sim \sqrt{n}\cdot e^{n\log n}$.

Can anyone either (1) give me a counterexample to my observation, or (2) formalize my observation in the sense of a theorem which might sound something like: compositions and products of the above classes of functions form a dense subset of all possible algorithmic runtimes?

Thanks.

答: You are missing Ackermann-type running times. For example, any algorithm that depends on the number of parts in the $\epsilon$-regular partition of Szemerédi's regularity lemma will take time that can be expressed as an exponential tower of $2$s, with height depending on $\epsilon$.

Title: Asymptotics in algorithms


问:
harmonic, m-harmonic, hyperbolic-harmonic
Anyone has any idea about harmonic, m-harmonic, and hyperbolic-harmonic functions?

Harmonic functions are characterized by mean value property, m-harmonic functions are characterized by volume mean value property.

I want to know the relationship between these three types of harmonic functions.

答:
Harmonic functions on a domain in $\mathbb R^n$ can be characterized either by the spherical mean value property or by ball mean value property: these properties are equivalent. (Indeed, ball means can be computed from spherical means, and vice versa). Therefore, m-harmonic functions are the same as harmonic.

Without context, it is hard to say what exactly is meant by hyperbolic harmonic functions here. One possibility is that these are harmonic maps into the unit disk $D\subset \mathbb C$ equipped with the hyperbolic metric $(1-|z|^2)^{-1} |dz|$. These are quite different from Euclidean harmonic functions: neither of two classes contains the other, and their intersection is rather small (constant functions only, I think).


问:
Is $G(\mathbb{Q}(\sqrt[3]{2}), i\sqrt{3})/\mathbb{Q}(i\sqrt{3}))\simeq \left \langle \mathbb{Z_3}, + \right \rangle$?
I wonder I have some misunderstanding of a concept of automorphism that leaving some field fixed.
\

The Problem in the text(Fraleigh, p.402, 7th) is:

Referring to Example 50.9, show that

$$G(\mathbb{Q}(\sqrt[3]{2}), i\sqrt{3})/\mathbb{Q}(i\sqrt{3}))\simeq \left \langle \mathbb{Z_3}, + \right \rangle$$

I thought the elements of $G(\mathbb{Q}(\sqrt[3]{2})$ leaving $\mathbb{Q}(i\sqrt{3})$ fixed so that for the three zeros $$\alpha_{1}=\sqrt[3]{2}, \ \alpha_2=\sqrt[3]{2}\frac{-1+i\sqrt{3}}{2},\ \alpha_3=\sqrt[3]{2}\frac{-1-i\sqrt{3}}{2}$$
of $x^3-2$, I guessed the group consist of

$\sigma_1;\ \sigma_1=\iota$ (the identity map)

$\sigma_2;\ \sigma_2(\alpha_1)=\alpha_2$, $\sigma_2(\alpha_2)=\alpha_1$, $\sigma(\alpha_3)=\alpha_3$

$\sigma_3;\ \sigma_3(\alpha_1)=\alpha_3$, $\sigma_2(\alpha_2)=\alpha_2$,
$\sigma(\alpha_3)=\alpha_1$.

But in this case, the three automorphisms does not even form a group under function composition. What may I have missed? Thanks.

答: If $F := \mathbb{Q}(i\sqrt{3})$, then $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q}(i\sqrt{3}) = F(\sqrt[3]{2})/F$ and so $$[\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3}) : \mathbb{Q}(i\sqrt{3})] = [F(\sqrt[3]2) : F] = \deg (\min(F,\sqrt[3]2)).$$ Now $x^3-2 \in \mathbb Q[x] \subseteq F[x]$ clearly annihilates $\sqrt[3]2$, and since $$x^3-2 = (x-\sqrt[3]2)(x-\alpha_2)(x-\alpha_3)$$ we have that $x^3-2$ is irreducible over $F$ (since $\sqrt[3]2 \notin F$). Thus $\min(F,\sqrt[3]2) = x^3-2$ and then $[\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3}) : \mathbb{Q}(i\sqrt{3})] = 3$, which of course implies that $G(\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q}(i\sqrt{3})) \cong \mathbb Z_3$.


问:
Minimum of the ratio of normal cdf and square of the pdf
I am having a little hard time figuring out if there is a minimum of the following ratio:
$$ \frac{\Phi(x)(1 - \Phi(x))}{\phi^2(x)},$$
where $\Phi$ is the standard normal cdf and $\phi$ is the standard normal pdf.

Can we somehow demonstrate that the function is convex and proceed from there?

答:
Using
$$\Phi(x)=\frac{1}{2} \left(1+\text{erf}\left(\frac{x-\mu }{\sqrt{2} \sigma }\right)\right)$$
$$\phi(x)=\frac{1}{\sqrt{2 \pi } \sigma }e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}$$ Let $t=\frac{x-\mu }{\sqrt{2} \sigma }$ which make
$$\frac{\Phi(x)(1 - \Phi(x))}{\phi^2(x)}=\frac{1}{2} \pi  \sigma ^2 e^{2 t^2} \left(1-\text{erf}(t)^2\right)$$ Now, consider
$$f(t)=e^{2 t^2} \left(1-\text{erf}(t)^2\right)$$

The first derivative
$$f'(t)=4 e^{t^2} \left(e^{t^2} t
   \left(1-\text{erf}(t)^2\right)-\frac{\text{erf}(t)}{\sqrt{\pi }}\right)$$ cancels at $t=0$ and the second derivative
$$f''(t)=-\frac{24 \sqrt{\pi } e^{t^2} t \text{erf}(t)+4 \pi  e^{2 t^2} \left(4 t^2+1\right)
   \left(\text{erf}(t)^2-1\right)+8}{\pi }$$
$$f''(0)=4-\frac{8}{\pi } \quad >0$$ So the minumum value is
$$\frac{\pi  \sigma ^2}{2}$$

In fact, expanded as a series around $x=\mu$, the expression is
$$\frac{\pi  \sigma ^2}{2}+\left(\frac{\pi }{2}-1\right) (x-\mu )^2+\frac{(3 \pi -8)
   (x-\mu )^4}{12 \sigma ^2}+\frac{(15 \pi -44) (x-\mu )^6}{180 \sigma
   ^4}+O\left((x-\mu )^8\right)$$ where all coefficients are positive.


问:
calculating true odometer reading from inaccurate speedometer
I'm very poor with math and hoping someone can help me with a question.

The odometer is reading 8000km but it's reading speed incorrectly and logging the speedo speed to the odometer. For KMs 2000-6000 it's logging at 18kmph over, and 6000-7000 is logging 10kmph over. What is the accurate reading it should be showing?

More details:

If a motorcycle travels:

1000km with accurate speedometer reading

6000km with the speedometer reading 18kmph over actual speed (50kmph reads 68kmph roughly)

1000km with the speedometer reading 10kmph over actual speed. (50kmph reads 60kmph roughly)

How many actual kilometers were traveled?

I don't know how to phrase this question to look up an answer online.

If I can do it by % change the sprocket ratios are:

13/43 for 1000km

13/50 for 6000km

14/50 for 1000km

Would I calculate the % increase in kmph from the actual kmph to the indicated kmph and then subtract that from the indicated odometer reading (for 1k,6k,1k)? If I'm traveling 30% slower than indicated can I just subtract 30% of the total odometer reading (km) for that time frame when I was running whatever gear ratio?

Thanks to anyone who wants to help! I know this isn't a challenging question in the slightest to a lot of you here but I'm not sure where else to ask for proper help.

Thanks so much! If there is a more relevant place to post please let me know. Also this isn't a elementary textbook question I'm just putting on a new speedometer unit for my dualsport, haha.

答:
There is not enough information to answer the question.

For example suppose the 5000 km middle stretch were run with the speedometer reading a constant 50 km/hr. Then the bike was actually traveling 32 km/hr, and actually travelled 3200 km. But suppose instead, the bike were running at speedometer reading of 100 km/hr. Then the bike was actually trying 82 km/hr and travelled 4100 km. Or a third scenario: The speedometer reading was a constant 18 km/hr the entire time. In reality, the bike never left its garage. The actual travel distance was 0 km.

All of these scenarios fit the information given. Ignoring the constraint of physical plausibility, the bike could have travelled any distance between $0$ and $5000$ km during the middle stretch, and similarly, the bike could have travelled any distance between $0$ and $1000$ km during the final stretch.


问:
Truth table logic (includes negation and conjunction) (Haskell)
¬(P ∧ ¬Q)

Complete truth table for the following sentence.

EDIT: The image below shows where I am at in the process. With all of the negation symbols, I'm getting a little confused on how to fill out the connectives.

$$\def\true{\mathsf T}\def\false{\mathsf F}
\begin{array}{|l:l|l:l:l:l:l|}\hline
\rm P &\rm Q &\lnot &\rm(P &\land &\lnot &\rm Q)\\\hline
\true &\true &      &\true &      &      &\true \\\hdashline
\true &\false&      &\true &      &      &\false\\\hdashline
\false&\true &      &\false&      &      &\true \\\hdashline
\false&\false&      &\false&      &      &\false\\\hline
\end{array}$$

答: The usual method is to write every combination of T/F for the variables, and then add columns putting the expression together bit by bit.  So make a column for the values of $\neg Q$, and then use that column and the P column to construct the $P \land \neg Q$ column, and finally make one more column with the negation of that to get your final expression.


问:
Alternative proof of Girard's theorem
I am looking for an alternative proof of Girard's theorem. The standard proof, which is almost trivial, relies too much on visualizing spherical triangles on the sphere. Is there a more algebraic proof, or even a proof by straightforward integration of the area of the triangle?</p>

答:
I found this question because I am interested in a basic calculus solution. I didn't finish anything along Leo's advice, unfortunately. Here's as far as I got before I post a true solution with as close a calculus flavor as I can muster.

====================

As for describing a great circle $C$ analytically, we just need to obtain two orthogonal points on $C$, say $Q,R$,

$$C(t)=Q\cos t+R\sin t$$

If you only have a pole $P$ you can get these two points using rectangular coordinates. Note that $Q=(-P_1,P_2,0)/\sqrt{P_1^2+P_2^2}$ is orthogonal to $P$. Then, once you have $Q$, you get $R=Q\times P$. Or, if you have two points $R,Q\in C$ that are not orthogonal, you can find a pole by $$P=Q\times R / |Q\times R|=Q\times R / \sin(QR) = Q\times R/\sin c$$ so that $C(t)=Q\cos t+P\times Q\sin t$

I will use rectangular coordinates for now and geocentric language where $(0,0,1)$ is the north pole. Start with a right triangle $\triangle ABC$, placing $A=(0,\sin a, \cos a), B=(\sin b, 0, \cos b), C=(0,0,1)$.

The main task is describing the interior of $\triangle ABC$, i.e. finding the intersection of side $c$ with great circles through the north pole (meridians). Side $c$ can be described in rectangular coordinates as $B\cos t+B\times(A\times B)\sin t/\sin c$ (with $0\le t\le c$)and a meridian $\theta=\theta_0$ is $(\sin\phi\cos\theta_0,\sin\phi\sin\theta_0,\cos\phi)$. Here are some useful relations

some of Napier's rules for spherical right triangles

$\sin c= \sin a/\sin A = \sin b/\sin B$ (Law of Sines)

$\cos c=\cos a\cos b$ (Pythagorean Theorem)

$\cos A = \sin B\cos a$

$\cos B = \sin A\cos b$

$A\times B=(\cos b\sin a,\sin b\cos a,-\sin a\sin b)$

$B\times (A\times B) = (-\cos a\cos b\sin b, \sin^2 b\sin a+\cos^2 b\sin a, \sin^2 b\cos a)=(-\cos c\sin b, \sin a, \sin^2 b\cos a)$

Finally, the intersection of side $c$ with a meridian at some constant polar angle $\theta$ gives the boundary condition for a double integral (along meridians first). Again, point $B$ is $B=(\sin b,0,\cos b)$ and side $c$ is paramatrized by $B\cos t+B\times(A\times B)\sin t/\sin c$ for $0\le t\le c$.

$$\sin b\cos t-\cos b\cos A\sin t=\sin\phi\cos\theta$$

$$\sin A \sin t = \sin\phi\sin\theta$$

$$\cos b\cos t+\sin b\cos A\sin t=\cos\phi$$

Combining the first and third to cancel $\cos t$, I get

$$\cos A\sin t = \sin b\cos\phi-\cos b\cos\theta\sin\phi$$

Then, combine this with the second equation to eliminate $\sin t$,

$$\cos A\sin\phi\sin\theta=\sin A(\sin b\cos\phi-\cos b\cos\theta\sin\phi)$$
$$\dots = \sin a\sin B\cos\phi-\cos B\cos\theta\sin \phi$$

and reach

$$\tan\phi=\frac{\sin a\sin B}{\cos A\sin\theta+\cos B\cos\theta}$$

Call $\phi(\theta)$ this solution along side $c$ as a function of $\theta$.

The area element in spherical coordinates is $\sin\varphi d\varphi d\theta$, so

$$[ABC]=\int_0^{\pi/2}\int_0^{\phi(\theta)}\sin\varphi d\varphi d\theta$$

$$\dots =\int_0^{\pi/2} 1-\cos(\phi(\theta))d\theta$$

Changing the $\tan\phi$ to $\cos\phi$, and using $\cos A=\cos a\sin B$,

$$\cos(\phi(\theta))=\frac{\cos A\sin\theta+\cos B\cos\theta}{\left(\sin^2 B-\cos^2 A+(\cos A\sin\theta+\cos B\cos\theta)^2\right)^{1/2}}$$

The denominator nicely reduces to $$(1-(\cos A\cos\theta -\cos B\sin\theta)^2)^{1/2}$$

which can integrate after subbing $u=\cos A\cos\theta -\cos B\sin\theta$ then $u=\cos x$.

If you place $A$ at the north pole, you get an alternative form

$$[ABC]=A-\int_0^A\frac{\cos(B)\cos(\theta)d\theta}{(\sin^2(A)-\cos^2(B)\sin^2(\theta))^{1/2}}$$

Finally, any spherical triangles can be subdivided into two right triangles and the additivity is trivial.

Title: Alternative proof of Girard's theorem


问:
How to calculate $\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du $
I am trying to calculate $\int_0^2\int_0^2 e^{-|t-u|} \,dt \,du $

I solved the first integral to obtain

$$\int_0^2 e^{-|t-u|} \, dt = \left|\frac{e^{-|t-u|}}{u}\right|_0^2 $$
$$ \Longrightarrow \int_0^2 e^{-|t-u|} \, dt = \frac{e^{-|2-u|}}{u} - \frac{e^u}{u} $$

Putting the answer above into the outer integral, I now have

$$ \int_0^2\int_0^2 e^{-|t-u|} = \int_0^2\left[\frac{e^{-|2-u|}}{u} - \frac{e^u}{u} \right]\, du $$
I tried separating the integrals and using integration by parts, but I could not get an answer.

Please can you help me with tips on how to solve this integral or show me where I'm wrong?

答: The square $[0, 2] \times [0, 2]$ can be divided into two triangles, the first where $t>u,$ the second where $t<u$. On both of these you can drop the absolute values, and by changing the variable to $v=t-u,$ $w=t+u$ easily integrate.


问:
Math blogs, pros and cons for writers?
I regularly read blogs by three mathematicians, and occasionally run into others. Definitely they help me a lot studying mathematics.

But now I am more interested in the writers' perspective, and I wonder whether writing a math blog help study mathematics. What are the pros and cons for the writer himself, especially for a first-year graduate student?

A lot of users have tons of experience writing blogs. I really want to hear your opinion on this.

Thanks!

Maybe I just start writing some of my own thoughts: cons: 1. writing a blog consumes lots of time. pros: 1. writing a blog forces you to think deeply and clearly; 2. it helps you learn how to communicate mathematics; 3. it might help you connect to others who share your passion.

答:
Pros:

Organization of thought

Ability to use LaTeX

Instant gratification from publishing

Engagement with an international audience

Improve writing skills over time

Cons:

Not easy to consistently publish high quality content

Web design can be challenging at times

Occasional formatting issues

My Background: I am studying theoretical physics at Georgetown, and recently I have been working on a math blog called Derive It (derive-it.com). So far it is mostly about limits and derivatives. I hope you like it and find it useful. Here is the Facebook Page: Derive It.

Overall, I have enjoyed the process of learning how to blog. I have been using WordPress, and it has been going well so far.

Title: Math blogs, pros and cons for writers?


答:
You are trying to integrate over a square from $(0,0)$ to $(2,2)$. Above the diagonal, you have $t<u$ and below the diagonal we have $t>u$ (assuming $t$ on horizontal and $u$ on vertical)

To integrate over this square, what we can do is first fix $t$, and then integrate over $u$, then integrate over $t$

Suppose we look at a particular value of $t$

For $u < t$, $|t-u| = t-u$

$u>t$, $|t-u| = u-t$

Hence we can re-write the integral as

$$I = \int_0^2 \left(\int_0^t \exp(u-t))du  + \int_t^2 \exp(t-u)du\right)dt$$

Can you proceed with this?

Title: How to calculate $\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du $

Body:
I am trying to calculate $\int_0^2\int_0^2 e^{-|t-u|} \,dt \,du $

I solved the first integral to obtain

$$\int_0^2 e^{-|t-u|} \, dt = \left|\frac{e^{-|t-u|}}{u}\right|_0^2 $$
$$ \Longrightarrow \int_0^2 e^{-|t-u|} \, dt = \frac{e^{-|2-u|}}{u} - \frac{e^u}{u} $$

Putting the answer above into the outer integral, I now have

$$ \int_0^2\int_0^2 e^{-|t-u|} = \int_0^2\left[\frac{e^{-|2-u|}}{u} - \frac{e^u}{u} \right]\, du $$
I tried separating the integrals and using integration by parts, but I could not get an answer.

Please can you help me with tips on how to solve this integral or show me where I'm wrong?


问:
Showing subspace of linear operator is an invariant subspace
I am stuck on the proof of the statement written below. How would we go about proving this fact? It seems intuitively correct. Any ideas on how to proceed? This is the assertion I'm interested in proving:

Let $V$ be a finite dimensional inner product space (real or complex) and $T ∈ L(V)$ a normal operator ($L(V)$ denotes the space of linear operators on $V$).
Let $U \subseteq V$ be a $2$-dimensional invariant subspace of $T$ .
Show that $U^\perp$ is also an invariant subspace of $T$.

答: If $U$ is invariant under $T$ then we have that $T$ stablizes $U$ i.e. $P_UTP_U=TP_U$ where $P_U$ is the orthogonal projection onto $U$. From this we also have that $T$ stabilizes $U^{\perp}$. So if $v\in U^{\perp}$ is arbitrary then $$T(v)=(TP_{U^{\perp}})(v)=(P_{U^{\perp}}TP_{U^{\perp}})(v)\in U^{\perp}$$


问:
Quadratic equation with parameter (what method to use to obtain correct answer)

For which values of the parameter $k$ does the equation $(2k-5)x^2-2(k-1)x+3=0$ have exactly one real answer?

From my work I determined $k = 4$, but the answer is $[k=4 \lor k=\frac{5}{2}]$.
The symbol $\lor$ I believe is supposed to indicate (OR). Obviously inputting $\frac{5}{2}$ makes the $a$ coefficient ($ax^2$) equal to $0$, therefore it's not a quadratic equation anymore, but I am not sure how I would have gotten to that answer (is this a trick question, or is there a procedure I need to follow when solving these to actually get the answer right every time?)

答:
The question is poorly worded.  It stipulates real equal answers, plural, and D = 0, which would seem to exclude their answer of $\frac{5}{2}$, because if you have only one answer, you might argue that it is "trivially" equal to itself, but the discriminant D is a feature of a quadratic equation, not a linear one.

And yes that $\lor$ is the logical OR operator.

Edit: The reworded question asks for "exactly one real answer" and that can be achieved by either making the equation linear or by making the two roots of the quadratic match, so their answer makes sense now.
In general it is good to look for edge cases and loopholes: "what if this is zero (negative, less than 1, etc?"  In particular for polynomials a common edge case is if the leading coefficient becomes zero; you have to watch out for that in many proofs involving polynomials.


问:
$AA^{*}-I$ is negative semi-definite
I need to show $(AA^{*}-I)$ is negative semi-definite where A is a linear operator.

I can show $AA^{*}+I$ is positive definite since $\langle x, (AA^{*}+I)x  \rangle=\|A^{*}x\|^2+\|x\|^2>0$ for $x \neq 0$  but I am not sure how to show $\langle x, (AA^{*}-I)x  \rangle \leq 0$.

Any ideas?

p.s. I think it was assumed that $\|AA^{*}-I\| <1$ if that helps.

答: Take $A=\sqrt{1.5} I$, where $I$ is the $n \times n$ identity matrix. Then $AA^*-I$ isn't negative semi-definite although $\|AA^*-I\|<1$.

Title: $AA^{*}-I$ is negative semi-definite


问:
Find all complex roots of $(z+1+i)^4 - 1 + i =0$.
Find all complex roots of $(z+1+i)^4 - 1 + i =0$.

Attempt:
I got
\begin{align*}
(z+1+i)^4 &= 1 - i \\
(z + \sqrt{2} e^{i(\frac{\pi}{4})})^4 &= \sqrt{2} e^{i(-\frac{\pi}{4})} \\
z + \sqrt{2} e^{i(\frac{\pi}{4})} &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4}  \right) \\
z &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4}  \right) - \sqrt{2}e^{i(\frac{\pi}{4})}
\end{align*}
for $k=0,1,2,3$.

Am I true? If not yet, how to find it?


问:
Completing the solution for Baby Rudin Chapter 7 Exercise 4
Baby Rudin Chapter 7 Exercise 4

Consider
\begin{equation*}
    f(x) = \sum_{n=1}^\infty \frac{1}{1+n^2x}
\end{equation*}
For what values of $x$ does the series converge absolutely? On what intervals does it converge uniformly? On what intervals does it fail to converge uniformly? Is $f$ continuous wherever the series converges? Is $f$ bounded?

Can someone please provide some hints on how this problem can be solved? Based on a few sketches for certain values of $n$, I think $f(x)$ is unbounded, but I am having trouble showing that rigorously. Please don't provide complete solutions, I'd much rather fill in the details of the hints. Thanks.

答:
Partial answer: To show that $f$ is unbounded note that $f(\frac 1 {k^{4}}) \geq \sum\limits_{n=1}^{k}\frac {k^{4}} {k^{4}+n^{2}} \geq \sum\limits_{n=1}^{k}\frac {k^{4}} {k^{4}+k^{2}} =\frac {k^{5}} {k^{4}+k^{2}} \to \infty$ as $k \to \infty$.

EDIT
The series converges for any $x \in \mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac  1{ 3^{2}},...\}$. It converges uniformly on the intersection of  $\{x: |x|>\epsilon\}$  with $\mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac  1{ 3^{2}},...\}$ for any positve number $\epsilon$ (by comparison with $\sum \frac 1 {n^{2}}$). This also implies that $f$ is continuous  on $\mathbb R \setminus \{0,-1,-\frac 1 {2^{2}},-\frac  1{ 3^{2}},...\}$.


问:
Is the function $f(z)=e^{i\bar z^2}$ analytic?
Is the function $f(z)=e^{i\bar {z}^2}$ analytic?

My attempt: Let $z=x+iy$. Then $$f(z)=f(x+iy)=e^{i(x-iy)^2}=e^{2xy}\cdot e^{i(x^2-y^2)}=e^{2xy}[\cos(x^2-y^2)+i\sin(x^2-y^2)].$$ Therefore, $u(x,y)=e^{2xy}\cos(x^2-y^2)$ and $v(x,y)=e^{2xy}\sin(x^2-y^2)$. Now we want to calculate the values of $x$ and $y$ such that $u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$. I found that those values of $x$ and $y$ have to satisfy these equations:
\begin{gather*}
x\sin(x^2-y^2)=y\cos(x^2-y^2)\\
x\cos(x^2-y^2)=-y\sin(x^2-y^2)
\end{gather*}
Clearly $(x,y)=(0,0)$ satisfies both equations, but I don't know if it is the only point of $\mathbb{C}$ that satisfies both equations.

答:
Let's solve these equations. First, note that, if $\cos(x^2 - y^2) = 0$, then $\sin(x^2 - y^2) = \pm 1$. This means the equations simplify to $\pm x = 0$ and $\mp y = 0$, and in any case, $x = y = 0$. This contradicts $\cos(x^2 - y^2) = 0$.

By rearranging the equations, we get
\begin{align*}
x \tan(x^2 - y^2) &= y \\
y \tan(x^2 - y^2) &= -x.
\end{align*}
Multiplying them respectively by a factor of $y$ and $x$, we get
$$y^2 = yx\tan(x^2 - y^2) = xy\tan(x^2 - y^2) = -x^2,$$
and hence $x^2 + y^2 = 0$. Since $x$ and $y$ are real numbers, this implies $x = y = 0$. That is, the one and only solution is $x = y = 0$, proving that $f$ is analytic nowhere.


问:
How to determine mathematically that an expression with one independent variable forms a closed area or not
Is there any mathematical method {other than sketching the curve  from  given expression (or equation)} to determine whether a given expession (or equation) forms a closed area or not?

For example- equation of circle form a closed area but how to check it  mathematically without sketching  its $x$-$y$  curve ?

答:
Edit: This is a half-answer, because it is for parametric curves. Some implicit curves can be parameterized, in which case this applies. I am not aware of a method for determining if a curve is closed given an implicit equation.

Basically, say you have some $t$ that traces out the curve via $(x(t), y(t))$, and say $t$ starts at $0$. If this is a closed curve in the sense you and I are thinking, then after a certain point it will just trace over the original shape. Let's say it returns to its starting point at $\alpha$ for some real number. Then it is closed if $(x(t), y(t)) = (x(u),y(u))$ if and only if $t = u + k\alpha$ for some integer $k$.

Let's use a circle as an example. We have a circle with radius $1$ is traced by the parametric equation with $t$ starting at $0$
$$(\cos(t), \sin(t))$$
Now we have that $t = 2\pi$ is the first time that we have a repeat point, namely,
$$(\cos(2\pi), \sin(2\pi)) = (\cos(0), \sin(0))$$
Actually, for every $t$ between $0$ and $2\pi$, we have
$$(\cos(t+2\pi), \sin(t+2\pi)) = (\cos(t), \sin(t))$$
This applies for any multiple of $2\pi$, as it just keeps tracing over the original circle. Therefore, we say that this equation is closed because for $t = u + k*2\pi$, where $k$ is an integer,
$$(\cos(t), \sin(t)) = (\cos(u), \sin(u))$$
By requiring that there is no repeating point without tracing over the whole shape again, we have a simple closed curve (think of this like no loops where the curve intersects itself), which means there is a discernable area enclosed by the curve.


问:
Prove that $5^n - 2^n$ is divisible by $3$ for all nonnegative integers $n$ using mathematical induction
Using mathematical induction, prove for all integers n  1 that $5^n - 2^n$ is divisible by 3.

Can someone help me with this?

Body: Hint: Note that $5^{n+1}-2^{n+1} = 5\cdot 5^n - 2\cdot 2^n = 5(5^n-2^n) + 3\cdot2^n$.


问:
Strategy while solving quadratics by splitting the middle term
While solving a quadratic equation in one variable by "splitting the middle term", sometimes I just can't arrive at the two numbers I need and it takes a lot of time.

For eg: $x^2 -124x+1344 = 0$

For solving the above equation by splitting the middle term, I first multiplied the coefficient of $x$ which is $1$ in this case by $1340$ (the constant term). I prime factorised the product, and now I have to pick a group of two numbers such that their sum is $-124$ (coefficient of $x$).

Do we always have to pick random numbers? If only hit and trial works here, what must be our thinking process for picking the other pair of number if the former pair of number doesn't work?

Eg: $x^2 -124x+1344 = 0$

Two numbers from the prime factorization of ($1 \cdot 1344$) is $-64$ and $-21$ but they do not add up to $-124$. What should be my strategy to pick a different pair of numbers?

(I know about the quadratic formula, but it becomes too lengthy sometimes)

答:
As suggested by Z Ahmed you can use quadratic formula. You will get the two values of x as  $$ x = \frac{-(-124)\pm\sqrt{(-124)^2-4(1)(1344)}}{2(1)} $$

After simplification you will get x as $$ x = \frac{124\pm 100}{2} $$

You will get x as 12 and 112

Simplify this and get your answer


问:
Topologically, what is a 'string' from string theory?
To begin: I am not a crank. I am not sure how well-founded my titular question is, but it was interesting enough that I decided to bring it to MSE.

For context: I am an undergraduate mathematics student. I am taking a course in algebraic topology, and during class it was asked how algebraic topology applies to string theory. Our professor explained that he was not sure, but he knew there was some connection between knot theory and string theory. He said it may interesting to understand the topological role of a 'string,' and I assume this is contingent on the connection between knot theory and string theory being strong enough that such an analogy can be realized.

So I am here to ask:

Is the connection between string theory and (algebraic) topology strong enough that it makes sense to ask how a 'string' can be viewed topologically?

If so, what is a string from this topological point-of-view?

答:
The strings in string theory are just one-dimensional manifolds embedded in a (usually) 10-dimensional space. The strings represent particles; the basic idea is to replace the pointlike concept of a particle in classical and (though it is more complicated here) quantum mechanics with a particle that is stringlike. This gives particles much more interesting behavior, such as the ability to have discrete vibrational modes. As for the strings themselves, there isn’t really a deeper connection to topology that I am aware of.

However, there are important connections between string theory and algebraic topology. This comes up in several places. I am not a string theorist, so I cannot enumerate all of these, but the main one has to do with the fact that string theories live in more than three spatial dimensions (most commonly ten). Since we only observe three dimensions, you have to explain why we don’t see the others. The oldest mechanism for this is Kaluza-Klein compactification, where the extra dimensions are like a compact manifold that is very small in size.  The manifold has to be a Calabi-Yau manifold, and topology is highly relevant to this part of the theory.


Body:
Per the symmetry with respect to the line $u-t=0$

$$\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du
= 2\int_0^2\int_0^t e^{-|t-u|} du \>dt = 2\int_0^2\int_0^t e^{-(t-u)} du \>dt =2 (1+e^{-2})
$$


问:
In the Quadratic Formula, what does it mean if $b^2-4ac>0$, $b^2-4ac<0$, and $b^2-4ac=0$?
Concerning the Quadratic Formula:

What does it mean if  $b^2-4ac>0$, $b^2-4ac<0$, and $b^2-4ac=0$?

答:
If $ b^2 - 4ac > 0 $ then you will get 2 real roots of the quadratic equation.

If $ b^2 - 4ac < 0  $ then your under root part will become negative and so the equation will have 2 complex roots.

If $ b^2 - 4ac = 0 $ then you will have one solution which will be $ -b/a $

this might help you understand better


答:
This one could be solved by induction but is it much neater if you use that
\[ a^n-b^n = (a-b) \cdot \sum_{i=0}^{n-1} a^i\cdot  b^{n-1-i} \]
If you use this one in your problem you see that
\[ 5^n- 2^n = (5-2) \cdot \sum_{i=0}^{n-1} 5^i \cdot 2^{n-1-i}=3 \cdot \sum_{i=0}^{n-1} 5^i \cdot 2^{n-1-i}\]
Which can be dived by $3$.

If you really want to use Induction use that
\[ 5=3+2\]
(ok this hint seems ridicolous it shall be used like this)
\[ 5^{n+1} - 2^{n+1} = (3+2) \cdot 5^n - 2 \cdot 2^n= 3 \cdot 5^n+ 2(5^n-2^n)\]

Title: Prove that $5^n - 2^n$ is divisible by $3$ for all nonnegative integers $n$ using mathematical induction

Body:
Using mathematical induction, prove for all integers n  1 that $5^n - 2^n$ is divisible by 3.

Can someone help me with this?


问:
Positivity of a given matrix
Let $A\in M_{n\times n}(\mathbb{R})$ be a real symmetric matrix and $H\in M_{n\times n}(\mathbb{C})$ be an invertible hermitian matrix.

We are given that $A+H$ and $A-H$ are positive semi-definite. Since $2A = (A+H)+(A-H)$, we can conclude that $A$ is positive semi-definite. In addition, does the invertibility of $H$ also imply that $A$ is invertible? In other words, can we conclude from the given information that $A$ is positive definite?

答: Take two PSD matrices, $X$ and $Y.$ The only way $X+Y$ is not definite is if $X$ and $Y$ have intersecting kernels (so, $X v = Y v = 0.$) but since $A+H - (A-H) = 2 H$ that is not possible here.

Title: Positivity of a given matrix


问:
Solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$?
I am trying to solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$.

I can see that the solution to the homogenous $h_n = 3h_{n-1}$ is clearly $h_n = 3^n$. That being said, I have no idea how to solve $h_n - 3h_{n-1} = 3(2)^n$: I assume it's some solution of the sort $(c_1n + c_2)(2)^n$, but I don't know where we'd go from here! Can anyone give advice?

答:
Using @HallaSurvivor solution
$$h_n = 3h_{n-1} + 3(2)^n$$
$$h_n=2^n t_n \implies 2t_n=3t_{n-1}+6$$Now, let $t_n=u_n+k$ and replace
$$2u_n+2k=3u_{n-1}+3k+6$$ If you make
$$2k=3k+6\implies k=-6\implies 2u_n=3u_{n-1}$$ Solve for $u_n$ (simple), go back to $t_n=u_n-6$ and to $h=2^n t^n$.


问:
Finding $\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$
I'm trying to find out :

$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$$

First I tried by inspection. I suspect that
$$\lim_{x\rightarrow \infty} \left( \frac{x+1}{x-1}\right)=1$$
So I expect $\ln (\cdots)\rightarrow 0$ as $x\rightarrow \infty$. While If I multiply this by $x$, One might expect this to go infinite.

Next what I have done is to use a substitution, I don't know if it's right to do or not.
$$x'=\frac{x+1}{x-1}\Rightarrow x=\frac{x'+1}{x'-1}$$

So that
$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)\Rightarrow \lim_{x'\rightarrow 1}\left(\frac{x'+1}{x'-1}\right)\ln x'$$
Now I can use L'Hospital rule
$$\lim_{x'\rightarrow \infty}\frac{\ln x'+\frac{x'+1}{x'}}{1}=2$$

Can I use such a substitution always to solve problems? What's wrong with the reasoning I did earlier?

答:
Hint:

Set $\dfrac1x=h$  to find

$$\lim_{h\to0^+}\dfrac{\ln\dfrac{1+h}{1-h}}h =\lim_{h\to0^+}\dfrac{\ln(1+h)}h+\lim_{h\to0^+}\dfrac{\ln(1-h)}{-h}=?$$


问:
Least squares approximation (matrices)
I’m quite stuck on the question below. It keeps coming up in my exam and I don’t know how to do it! Please help me! Thanks!

Show that the matrix $P = A (A^tA)^{-1} A^t$ represents an orthogonal projection onto $\mathcal{R}(A)$. Hence or otherwise explain why $x^\star = (A^t A)^{-1} A^t b$ represents the least squares solution of the matrix equation $Ax=b$.

答:
Well, okay, assuming $A$ is overdetermined ($m \times n$ where $m > n$). The column of vectors of $A$ spans $\mathcal R(A)$, assuming $A$ has full rank the column vectors form a basis.

The matrix $A^T A$ will be symmetric ($(A^TA)^T = A^T A$), and if $A$ has full
rank, invertible. If $A$ does not have full rank, it will not be invertible.

Now, naming the column vectors in $A$ $a_1, a_2, \dots, a_n$ and $x = (x_1, x_2 \dots, x_n)^T$ we can write $Ax$ as:
$$Ax = x_1 a_1 + x_2 a_2 + \dots + x_n a_n = \sum_{i=1}^n a_i x_i$$
and we want this to be equal to $b$. Assuming there is no such solution, we want to minimize
$$\| b - Ax \| = \left\| b - \sum_{i=1}^n a_i x_i \right\|$$
To minimize this, we want to find the projection of $b$ onto $\mathcal R(A)$.This projection can be expressed as $\sum_{i=1}^n a_i x_i$, since the $a_i$ span $\mathcal R(A)$.  Why does this minimze the above equation? Well, simply because this is the closest we can get - we can't express elements not in $\mathcal R(A)$ with elements in $\mathcal R(A)$.

Now, assume $x_1, x_2, \dots, x_n$ are the coefficients to $a_1, a_2, \dots a_n$ such that $\sum a_i x_i$ is the orthogonal projection of $b$ onto $\mathcal R(A)$. Then $b$ can written as
$$b = Ax + y$$
for some $y$ which is orthogonal to $\mathcal R(A)$. Thus $y = b - Ax$. Since $y$ is orthogonal to $\mathcal R(A)$, it is orthogonal to every $a_i$:
$$
\left\langle a_i , \underbrace{b - Ax}_{=y} \right\rangle = 0
\Longleftrightarrow
a_i^T(b-Ax)=0 ~~~~ \forall i = 1, \dots, n
$$
This can be written in matrix form as $A^T(b-Ax)=0$ taking the equations above for all $i$ at once, which can be rewritten as $A^Tb = A^TAx$, or if $A^TA$ is invertible, $x = (A^TA)^{-1}A^Tb$.

So, basically, to derive $x = (A^TA)^{-1}A^T b$, consider the condition that $b-Ax$ should be orthogonal to the column vectors of $a_i$ one at a time, then stack these equations "on top of each other" in a matrix equation.

Title: Least squares approximation (matrices)


答:
You have done it correctly but what you said in the beginning is not correct. If you mulitiply two functions one of which has limit $0$ and the other has limit $\infty$ you can say nothing about the limit of the product.

Examples: As $x \to \infty$, $\frac  1 x \to 0$ and $\sqrt x \to \infty$ but the product tends to $0$; On the other hand $\frac  1 x \to 0$ and $x^{2} \to \infty$ but the product tends to $\infty$.

Title: Finding $\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$

Body:
I'm trying to find out :

$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$$

First I tried by inspection. I suspect that
$$\lim_{x\rightarrow \infty} \left( \frac{x+1}{x-1}\right)=1$$
So I expect $\ln (\cdots)\rightarrow 0$ as $x\rightarrow \infty$. While If I multiply this by $x$, One might expect this to go infinite.

Next what I have done is to use a substitution, I don't know if it's right to do or not.
$$x'=\frac{x+1}{x-1}\Rightarrow x=\frac{x'+1}{x'-1}$$

So that
$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)\Rightarrow \lim_{x'\rightarrow 1}\left(\frac{x'+1}{x'-1}\right)\ln x'$$
Now I can use L'Hospital rule
$$\lim_{x'\rightarrow \infty}\frac{\ln x'+\frac{x'+1}{x'}}{1}=2$$

Can I use such a substitution always to solve problems? What's wrong with the reasoning I did earlier?


答:
Your first reasoning is wrong, because you have $0\cdot\infty$ uncertainty. Second way is correct with correct answer.

One more possible way:
$$x\ln\left(\frac{x+1}{x-1}\right)=x\ln\left(\frac{x-1+2}{x-1} \right)=x\ln\left(1+\frac{2}{x-1} \right)\sim \frac{2x}{x-1}\to 2$$


问:
Changing coordinates used in function using eigenvectors
So I was reading over the post from the following: The "second derivative test" for $f(x,y)$

And I understood everything in Alex's response and followed through with everything until the last part. So at a critical point the higher dimensional Taylor Series simplifies to:

$$f(x,y) = f(x_{0}, y_{0}) + \frac{1}{2} \begin{bmatrix} x-x_{0} & y-y_{0} \end{bmatrix} \mathbf{H} (x_{0},y_{0})\begin{bmatrix} x-x_{0}\\y-y_{0} \end{bmatrix} + \cdots$$

Here I'm using the notion that fields and certain vector spaces are equivalent to view the resulting 1x1 matrix as "similar" to a scalar. It is stated in the post that if we change coordinates to some s and t variables that run in the directions of H's eigenspaces, with the origin at the critical point, we get:

$$f(s,t) = f(0,0) + \frac{1}{2} \lambda_{1} s^{2} + \frac{1}{2} \lambda_{2} t^{2} + \cdots$$

Intuitionally it makes sense how that is the answer, but I am not sure where to start mathematically to derive this. Would I start by using the change of coordinates matrix on the Hessian matrix?

答:
When we speak of "changing coordinates" here it means we are considering a certain (in this case affine) mapping $\phi:\Bbb{R}^2\to\Bbb{R}^2$ (which encodes the effect of "changing coordinates") such that the composed mapping $F=f\circ \phi$ has the Taylor expansion
\begin{align}
F(s,t)= F(0,0) + \frac{1}{2}(\lambda_1s^2+\lambda_2t^2)+\dots \tag{$*$}
\end{align}
for some $\lambda_1,\lambda_2\in\Bbb{R}$. In this case, what the mapping $\phi$ does is the following: suppose $\{u,v\}\subset \Bbb{R}^2$ are the two "eigenvectors" of the Hessian $D^2f_{(x_0,y_0)}$. Then, we define $\phi:\Bbb{R}^2\to\Bbb{R}^2$ as
\begin{align}
\phi(s,t):= (x_0,y_0)+su+tv.
\end{align}
If you really insist on matrix notation then it reads
\begin{align}
\begin{pmatrix}
s\\ t
\end{pmatrix}
\mapsto
\begin{pmatrix}
x\\y
\end{pmatrix}=
\begin{pmatrix}
x_0\\y_0
\end{pmatrix}+
\begin{pmatrix}
u_1&v_1\\u_2&v_2
\end{pmatrix}\cdot
\begin{pmatrix}
s\\t
\end{pmatrix}
\end{align}
So, the mapping $\phi$ tells you how to go from the $(s,t)$ coordinates to the $(x,y)$ coordinates, and hence when we consider the composed mapping $F=f\circ \phi$, we're looking at the function $f$ "expressed in $(s,t)$ coordinates". To visualize this geometrically and keep track of all this information, I suggest you draw two copies of the plane $\Bbb{R}^2$. On one of them label the axes as $(s,t)$, and on the other copy label the axes as $(x,y)$ and draw a dot to denote the point $(x_0,y_0)$.

Then what is going on is we're mapping the origin of the $(s,t)$ plane to the point $(x_0,y_0)$ on the $(x,y)$ plane, and we're mapping the vector $e_1=(1,0)$ onto $(x_0,y_0)+u$ of the Hessian (think of this as mapping $e_1$ onto the vector $u$ which is emanating from the point $(x_0,y_0)$), and similarly we map $e_2$ onto $(x_0,y_0)+v$. So this change of coordinates (as we'll see below) allows us to work at the origin and assume that the "eigenvectors" of the Hessian are along the coordinate axes of our $(s,t)$ plane.

Now, to prove that $(*)$ really is the Taylor expansion of $F$, we have to use the chain rule. For ease of typing, let $\sigma=(s,t)$. Then, by Taylor's theorem,
\begin{align}
F(\sigma)&= F(0,0)+ DF_{(0,0)}[\sigma] + \frac{1}{2!}D^2F_{(0,0)}[\sigma,\sigma] + O(|\sigma|^2)\tag{i}
\end{align}
The first term is
\begin{align}
DF_{(0,0)}[\sigma]&=Df_{\phi(0,0)}(D\phi_{(0,0)}[\sigma])\\
&=Df_{(x_0,y_0)}(D\phi_{(0,0)}[\sigma])\\
&= 0,\tag{ii}
\end{align}
since by assumption $(x_0,y_0)$ is a critical point of $f$, so $Df_{(x_0,x_0)}=0$. As you can see, we just repeat this again: if you carry out the chain rule directly then you'll get
\begin{align}
D^2F_{(0,0)}[\sigma,\sigma] &= D^2f_{\phi(0,0)}[D\phi_{(0,0)}[\sigma],D\phi_{(0,0)}[\sigma]] +
Df_{\phi(0,0)}[D^2\phi_{(0,0)}[\sigma,\sigma]]
\end{align}
In this special case of course, we have a lot of simplifications: first of all as above, $Df_{\phi(0,0)}=Df_{(x_0,y_0)}=0$ by assumption. Next, since $\phi$ is an affine function,
$D\phi_{(0,0)}[\sigma]= su+tv$. Plugging all of this in, we see that
\begin{align}
D^2F_{(0,0)}[\sigma,\sigma] &= D^2f_{(x_0,y_0)}[su+tv, su+tv]\\
&= s^2 D^2f_{(x_0,y_0)}[u,u]+ 2st D^2f_{(x_0,y_0)}[u,v]+ t^2D^2f_{(x_0,y_0)}[v,v]\tag{iii}
\end{align}
where I used the fact that the second derivative is symmetric in it's two arguments (i.e the fact that the Hessian matrix is symmetric). Now, when I said above that $u,v$ are "eigenvectors" of $D^2f_{(x_0,y_0)}$ what this means is that $D^2f_{(x_0,y_0)}[u,u]=\lambda_1,D^2f_{(x_0,y_0)}[v,v]=\lambda_2,D^2f_{(x_0,y_0)}[u,v]=0$. Or expressed more succinctly, the matrix representation of $D^2f_{(x_0,y_0)}$ relative to the "eigenbasis" $\{u,v\}$ is diagonal
\begin{align}
[D^2f_{(x_0,y_0)}]&=
\begin{pmatrix}
D^2f_{(x_0,y_0)}[u,u]&D^2f_{(x_0,y_0)}[u,v]\\
D^2f_{(x_0,y_0)}[v,u]&D^2f_{(x_0,y_0)}[v,v]
\end{pmatrix}=
\begin{pmatrix}
\lambda_1&0\\
0&\lambda_2
\end{pmatrix}\tag{iv}
\end{align}
So, if we put together (i),(ii),(iii),(iv) then we get exactly $(*)$:
\begin{align}
F(s,t)&= F(0,0)+\frac{1}{2}(\lambda_1 s^2 + \lambda_2 t^2) + \dots
\end{align}

Extra FYI:

Of course, all of this generalizes verbatim to higher dimensions: the calculus part stays almost the same; the only thing you need to generalize is the linear algebra. You just need to prove that any symmetric bilinear form $\zeta:V\times V\to\Bbb{R}$ on a finite-dimensional real vector space $V$ can be "orthogonally diagonalized", in the sense that there exists a basis $\{v_1,\dots, v_n\}$ of $V$ such that $\zeta(v_i,v_j)=0$ if $i\neq j$. In other words, relative to this basis, the symmetric bilinear form $\zeta$ has matrix representation
\begin{align}
[\zeta]&=
\begin{pmatrix}
\lambda_1 & \cdots & 0\\
\vdots &\ddots & \vdots\\
0& \cdots & \lambda_n
\end{pmatrix}
\end{align}
where $\zeta(v_i,v_i)=\lambda_i$. The proof of this is essentially the same as the Grahm-Schmidt orthonormalization procedure for inner products if you've seen that.

Proof:

We proceed inductively: if $\dim V=1$, then any non-zero vector is a basis, and we're done. Otherwise, suppose we can do so for $n$-dimensional vector spaces. Then, supposing $\dim V=n+1$, we  distinguish two cases: if $\zeta=0$ identically, then any basis will work. Otherwise, choose a $v\in V$ such that $\zeta(v,v)\neq 0$. Then, $\ker \zeta(\cdot, v)$ will be an $n$-dimensional vector space and the restriction of $\zeta$ to this space will (by induction hypothesis) have a "orthogonal basis" $\{v_1,\dots, v_n\}$. It is then easy to see that, $\{v_1,\dots, v_n,v\}$ is an "orthogonal" basis for $V$. This completes the induction.

Once you do this, we can play the same game: given $f:\Bbb{R}^n\to\Bbb{R}$ and a point $x_0\in\Bbb{R}^n$, and letting $\{v_1,\dots, v_n\}$ be an "orthogonal" basis for the second derivative $D^2f_{x_0}:\Bbb{R}^n\times\Bbb{R}^n\to\Bbb{R}$, we simply define $\phi:\Bbb{R}^n\to\Bbb{R}^n$ by
\begin{align}
\phi(s)&=x_0+\sum_{i=1}^ns_iv_i
\end{align}
(this tells us how to go from $s$-coordinates to $x$-coordinates)
Then, we consider the mapping $F=f\circ \phi$ exactly as above (i.e the function $f$ expressed in terms of $s$-coordinates).


答:
$\lim\limits_{x\to \infty} x = 0$ and $\lim\limits_{x\to \infty}\ln {\frac {x+1}{x-1}} = 0$ so you have "$0\cdot \infty$" type indeterminate.

Try L'Hopital

$\lim x{\ln\frac {x+1}{x-1}}=$

$\lim \frac{\ln\frac{x+1}{x-1}}{\frac 1x}=$

$\lim \frac{\frac {x-1}{x+1}\cdot \frac {(x-1)-(x+1)}{(x-1)^2}}{-\frac 1{x^2}}=$

$\lim -x^2 \frac {-2}{(x+1)(x-1)}=$

$\lim \frac {2x^2}{x^2 -1} = 2$


问:
Why do we need to invoke the cancellation laws to prove uniqueness of solution of ax = b in group theory?
Quoting Introduction to Analytic Number Theory by Tom M. Apostol:

Theorem 6.1 Cancellationi laws. If elements $a$, $b$, $c$ in $G$ satisfy
$$ ac = bc $$ or $$ ca = cb $$,
then $ a = b $.

Theorem 6.2 (d) For all $a$ and $b$ in $G$ thee equation $ax = b$ has the unique solution $ x = a^{-1}b$; the equation $ya = b$ has the unique solution $y = ba^{-1}$.

Quoting the proof of Theorem 6.2 (d) now:

Again by associativity we have

$$ a(a^{-1}b) = (aa^{-1})b = b\hspace{1.5em} \text{and}\hspace{1.5em} (ba^{-1})a = b(a^{-1}a) = b. $$
The solultions are unique because of the cancellation laws.

Why do we have to invoke the cancellation laws the prove that the solutions are unique?

In other words, which step of the proof will not work if we do not invoke the cancellation laws?

答: The calculation $a(a^{-1}b)=(aa^{-1})b=b$ shows that the equation $ax=b$ has at least one solution, namely, $x=a^{-1}b$; it does not show that the equation has no other solution, so it does not show that the solution $x=a^{-1}b$ is unique. The cancellation law, however, says that if $c$ and $d$ are both solutions to $ax=b$, so that $ac=b=ad$, then $c=d$. In other words, the cancellation law shows that if there is any solution at all to $ax=b$, that solution is unique. It takes both parts to show both that there is at least one solution and that there is at most one solution, i.e., that there is a unique solution.


答:
To prove ur reasoning is wrong, consider a simpler problem: $\lim\limits_{x\to \infty} x \cdot \frac{1}{x}$. here if I say $\frac{1}{x}$ goes to zero and $x$ goes to infinity then the limit is infinity how would that reasoning make sense? cancel the $x$'s and u can see the limit is $1$.

Coming to you problem.

Why do you assume if $\ln {\frac {x+1}{x-1}}$ tends to 0 and you multiply by x you get infinity? $0*\infty$ is indeterminate like the above limit...  and you must convert this into a form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so we can apply l'hopital.

Solution:

$\lim\limits_{x\to \infty} x \ln {\dfrac {x+1}{x-1}} =\lim\limits_{x\to \infty} \dfrac{(\ln \dfrac {x+1}{x-1})'}{(\dfrac{1}{x})'} =$

$\lim\limits_{x\to \infty} -x^2\cdot \dfrac 1{\dfrac {x+1}{x-1}}(\frac {x+1}{x-1})' =$

$\lim\limits_{x\to \infty}  -x^2\cdot \dfrac {x-1}{x+1}\dfrac {(x+1)'(x-1) - (x+1)(x-1)'}{(x-1)^2}=$

$\lim\limits_{x\to\infty}-x^2\cdot \dfrac 1{x+1}\dfrac {(x-1)-(x+1)}{x-1}=$

$\lim\limits_{x\to \infty} \dfrac {2x^2}{x^2-1} = \lim\limits_{x\to \infty} \dfrac {4x}{2x} =2$.


问:
Understanding an example of a finitely generated projective module which is not free.
Here is the example I know:

Consider the ring $R = \mathbb Z_2 \times \mathbb Z_2$ and the submodule $\mathbb Z_2 \times \{0\}.$ it is by construction a direct summand of $R$ but certainly not free.And it's finitely generated.

But I do not understand:

1- how $\mathbb Z_2 \times \{0\}$ is a direct summand of $R$ by construction?

2- Why $\mathbb Z_2 \times \{0\}$ is not free?

Any clarification is very much appreciated.

I know that it is finitely generated because it is generated by $\{(0,0) , (1,0)\}.$

答:
It's a direct summand, because for abelian groups $G \oplus H \simeq G \times H$ and the module structure is the same as the group structure.

It's not free, because there is a non-trivial relation $r \cdot (1,0) = (0,0)$, which is the zero element of $R$.  In this case, $r = (0,1)$.  That's the definition of free: Since $\lbrace (1,0), (0,1) \rbrace$, which we denote by $\lbrace e_1, e_2 \rbrace$ for clarity, is a basis for $R$ as a module, the relation $r_1 \cdot e_1 + r_2 \cdot e_2 = 0$ implies $r_1 = r_2 = 0$ if $R$ is free.  But we just showed that this isn't the case.


问:
Finding the magnitude of the solution to a nonlinear equation
Let $0<\delta<1$. For any $a>0$, there exists a unique $x_a>0$ that solves
\begin{align}
x(1+\delta)^{2x} = a.
\end{align}
Now consider the case when $a$ becomes large, i.e., $a\to\infty$. It follows that $x_a\to\infty$, but is there a way to determine the order of $x_a$? That is, how fast $x_a$ grows as $a\to\infty$?

P.S. If it helps, we can assume that $\delta$ is small, i.e., much closer to 0 than 1.

答:
Inverting to solve for $x$ we have that

$$2\log(1+\delta) x e^{2\log(1+\delta)x} = 2\log(1+\delta)a \implies x = \frac{W(2\log(1+\delta)a)}{2\log(1+\delta)}$$

For small $\delta$ we can safely conclude

$$x \sim \frac{W(2\delta a)}{2\delta}$$

and from here since $\delta$ is fixed and we are taking $a$ arbitrarily large we can use the known asymptotic for product log

$$W(x) \sim \log x - \log \log x$$

to get

$$x \sim \frac{\log(2\delta a) - \log \log(2\delta a)}{2\delta}$$

Note that using the asymptotic $(1+\delta)^{2x}\sim 1 + 2\delta x$ at the beginning is not valid.


问:
$A$ be a $2\times 2$ real matrix with trace $2$ and determinant $-3$
$A$ be a $2\times 2$ real matrix  with trace $2$ and determinant $-3$, consider the linear map $T:M_2(\mathbb{R})\to M_2(\mathbb{R}):=B\to AB$ Then which of the following are true?

$T$ is diagonalizable

$T$ is invertible

$2$ is an eigen value of $T$

$T(B)=B$ for some $0\ne B\in M_2(\mathbb{R})$

if $ A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$ Then I have calculated that matrix of $T$ will be \begin{pmatrix}a_{11}&a_{12}&0&0\\a_{21}&a_{22}&0&0\\0&0&a_{11}&a_{12}\\0&0&a_{21}&a_{22}\end{pmatrix} so $T$ is invertible I can say as $\det T=(\det A)^2=9$, could any one help to find out others as true/false?

答:
claim:

minimal polynomial  of $T$ and the minimal polynomial of $A$ are the same .

let $p(x)=x^l+s_{l-1}x^{l-1}\ldots+ s_0$ (minimal polynomial of $A$) .

$p(A)=0\implies p(A)B=0 \implies A^lB+s_{l-1}A^{l-1}B+\ldots+s_0B=0$

which gives $$ T^lB+s_{l-1}T^{l-1}B+\ldots+s_0B=0$$

hence minimal polynomial  of $T$  $ |$ minimal polynomial  of $A$

similarly it can be shown minimal polynomial  of $A$ $|$ minimal polynomial  of $T$.

so as  $A$ is diagonalizable $\implies $ $T$ is diagonalizable

$2,1$ are also not eigenvalues of $A$ and hence they are not eigenvalues of $T$ so option $3,4 $ are false


问:
Lemma 1 - Banach-Mazur Game - Dan Ma's topology Blog
I have been given a presentation for a course based on this particular post in Dan Ma's topology blog. The post itself is very clear and educational, but there is one proof that I'm not sure why it is complete.

Lemma 1 has a rather standard application of Zorn's lemma, but I feel that I'm missing something in the second paragraph of the proof.

Lemma 1: Let $X$ be a space, Let $O \subset X$ be a nonempty open set. Let $\tau$ be the set of all nonempty open subsets of $O$. Let $f: \tau \longrightarrow \tau$ be a function such that for each $V \in \tau$, $f(V) \subset V$. Then, there exists a disjoint collection $\mathcal{U}$ consisting of elements of $f(\tau)$ such that $\bigcup \mathcal{U}$ is dense in $O$.

The proof begins by explaining the trivial case of $O$ having just one point, then if I understand correctly everything holds in a rather obvious way, namely: the property of $f$ gives me the same set as an image and clearly this is dense in $O$.

Then, we assume $O$ has two points and the author shows that $\mathcal{P}$ (the set of all collections $\mathcal{F}$ such that $\mathcal{F}$ is a disjoint collection of elements of $f(\tau)$) is non-empty. For this, two disjoint open subsets of $O$ are taken based on the assumption that $O$ has at least two points.

Why must these subsets be disjoint? Isn't the author assuming anything in addition?

I underestand that by applying $f$ later those sets become exactly of the form of an element of $\mathcal{P}$ which is the goal of that particular part of the proof. I thought we could possibly assume they are disjoint and otherwise apply $f$ again and again until they are but no assumption really guarantees this process would stop.

答: You are correct: unless the author is assuming that $X$ is Hausdorff, $O$ need not contain two disjoint non-empty open sets. However, the argument can easily be patched. If $O$ does not contain two disjoint non-empty open sets, let $\mathscr{U}=\{\tau(O)\}$; every non-empty open subset of $O$ meets $\tau(O)$, so $\bigcup\mathscr{U}=\tau(O)$ is dense in $O$, and we’re done right away.


问:
Proving that none of these elements 11, 111, 1111, 11111...can be a perfect square
How can i prove that no number in set S
S = {11, 111, 1111, 11111...}
Is a perfect square.

I have absolutely no idea how to tackle this problem i tried rewriting it in powers of 10 but that didn't really get me anywhere...

Thanks in advance.

答: In base 3, $11111 = 102^2$, and in base 7, $1111 = 26^2$.  So it's not true for bases in general.  In base 10, all such numbers are 3 modulo 4, which is never a square.  </p>


问:
Function with countable range
It is well known that if $f\colon \Bbb R \to \Bbb R$ be continuous function that takes values in rational number then $f$ is a constant function.

My question:  let  $f\colon \Bbb R \to \Bbb R$ be a function with values only in $\Bbb Q.$ What we can say about $f.$? My guess, there is no thing specific we can say. Any help will be appreciated greatly

答: Nothing. Without further conditions you cannot say anything about your function $f$. The continuity assumption allows you to use the density of $\mathbb{Q}$ to deduce that is constant.

Title: Function with countable range


问:
Want to find $A\subseteq M$ such that $(\partial A)^\circ = M$
$(M,d)$ is a metric space. I have been asked to produce $A\subseteq M$, such that $$(\partial A)^\circ  = M$$

Could someone help me find such a set? Here are my thoughts so far:

We want to find $A\subseteq M$ such that $(\partial A)^\circ = M$. For all $x\in A$ gives $x\in M$ and hence $x\in (\partial A)^\circ$. $x\in (\partial A)^\circ$ means that $x\in \partial A$ and there exists $r_x > 0$ such that $B(x,r_x) \subseteq \partial A$. $x\in\partial A$ additionally gives $B(x,r_x)\cap A\neq\emptyset$ and $B(x,r_x)\cap A^c\neq\emptyset$. We get $\partial A\cap A \neq \emptyset$ and $\partial A\cap A^c\neq \emptyset$. All this is good - but how do I find some $A$ that satisfies these properties?

Thanks a lot!

答:
Thanks to @Theo Bendit for their answer in the comments. Just adding it here for documentation. On the metric space $\mathbb R$ with the usual metric, consider $\mathbb Q$. Then,
$$(\partial \mathbb Q)^\circ = \mathbb R$$
since $$\partial\mathbb Q = \mathbb R \text{ and } \mathbb R^\circ = \mathbb R$$


问:
Find strong deformation retract to corona/crown
I'm looking to find the retraction ($r:X \rightarrow A$) and the deformation $H:X \times [0,1] \rightarrow X$) but I can't think of how, I'm bad thinking about functions, any help is appreciated!

答:
First, parametrize the space $X$ using a vector function $\vec{p}(u,v,w)$ defined as
$$
\vec{p}(u,v,w)=((|v|+1+w)\cos(u),(|v|+1+w)\sin(u),v)
$$
where $v\in\Bbb R$, $w\in[0,1]$, and $u\in[0,2\pi]$.

For the retraction $r:X\to A$, just send $v$ to $0$.

Now, the deformation retraction $H:X\times I\to X$ that forms a homotopy $\operatorname{id}_X\simeq r:X\to A$ can be easily defined via straightline homotopy
$$
H(\vec{p}(u,v,w),t)=(((1-t)|v|+1+w)\cos(u),((1-t)|v|+1+w)\sin(u),(1-t)v)
$$
It's easy to check that $H(x,0)=\operatorname{id}_X$ and $H(x,1)=r$.

I could add the derivation of the parametric equation for $X$ if you want.....