问:
Why is $\int f(x)^{dx}$ an appropriate notation for the product integral?
Just as the title says, why is the product integral's notation: $\int f(x)^{dx}$ appropriate? I sort of understand the product integral, but this notation doesn't really make any sense to me.

Is this a consequence of Theorem $1$:

Theorem 1 If $a_k>−1$ for all $k$, then the infinite product $\prod\limits_{k=1}^\infty (1+a_k)$ converges if and only if the infinite series $\sum\limits_{k=1}^\infty \log{(1+a_k)}$ converges.

If so, how is it a consequence and what is a proof of this? (I found this theorem on How to go from a sum to a product and a product to a sum?, which I found from a google search, so this might be the complete wrong approach at finding out why the notation $\int f(x)^{dx}$ works, but it's the closest I think I have gotten.)

EDIT: So I figure it actually has more to do with how $e^a \cdot e^b = e^{a+b}$ and $\ln{a}+\ln{b} = \ln(ab)$, but how come $e$ and $ln$ don't appear in the integral if you are using this property? Can someone just derive the notation using reimann sums or something?

答:
Wikipedia has a discussion of this.  That article (somewhat unusually) uses the symbol $\Pi$ instead of $\int$.  Other sources may use the $\int$ symbol or the $\int$ symbol  with a superimposed $\times$ or superimposed $\mathcal P$.

I think this is hardly ever used for scalar-valued functions, though.  As noted, you can do it instead with something like $\exp\left(\int \log f(x)\;dx\right)$.  When it is most often used is when the values do not commute.  For example, matrices.  Then the product integral can no longer be reduced to an ordinary integral using $\exp$ and $\log$.

For example, solution of differential equation
$$
\mathbf{x}'(t) = A(t) \mathbf{x}(t)
$$
where $\mathbf{x}(t)$ is an $n$-vector and $A(t)$ is an $n \times n$ matrix for $t \in [a,b]$.

Reference:

Dollard, John D.; Friedman, Charles N., Product integration with applications to differential equations. Foreword by Felix E. Browder. Appendix by P. R. Masani, Encyclopedia ot Mathematics and its Applications 10. Cambridge: Cambridge University Press (ISBN 978-0-521-17737-5/pbk). xxii, 253 p. (2011). ZBL1217.34018.


问:
Calculus，Limit， definite integral with variable upper limit
$$\displaystyle\lim \limits_{x \to +\infty}\frac{\displaystyle\int_{0}^{x}{t|{\sin t|dt}}}{x^{2}}$$,
**How can I solve it? I need your help, thank you!
I want to divide them into many small intervals and try to use L'Hopistal, but it failed. And Squeeze theorem also failed.

答:
For any $x\ge\pi$, there is $n\in N$ such that
$$ n\pi\le x<(n+1)\pi $$
and hence
$$ \int_0^{n\pi}t|\sin t|dt\le \int_0^xt|\sin t|dt\le\int_0^{(n+1)\pi}t|\sin t|dt. $$
Since
$$ \int_{(k-1)\pi}^{k\pi}t|\sin t|dt=(2k-1)\pi $$
for any integer $k$, one has
\begin{eqnarray}
\int_0^{n\pi}t|\sin t|dt&=&\sum_{k=1}^n\int_{(k-1)\pi}^{k\pi}(-1)^{k-1}t\sin tdt\\
&=&\sum_{k=1}^n(2k-1)\pi\\
&=&n^2\pi,\\
\int_{0}^{(n+1)\pi}t|\sin t|dt&=&(2n+1)^2\pi.
\end{eqnarray}
So one has
$$ \frac1{(n+1)^2\pi^2}\cdot n^2\pi\le\frac{1}{x^2}\int_0^xt|\sin t|dt\le\frac1{n^2\pi^2}\cdot(n+1)^2\pi.$$
This implies
$$ \lim_{x\to\infty}\frac{1}{x^2}\int_0^xt|\sin t|dt=\frac{1}{\pi}. $$

ContentLicense: CC BY-SA 4.0


问:
Extending uniform convergence
Let $\sum_{n=0}^{\infty} f_n(x)$ converge to $f(x)$ on $[0, a)$ and converge uniformly on evey closed sub-interval of $[0, a)$. If $\lim_{x \to a^-} f(x) = L < \infty$ does it follow that the series converges uniformly on $[0, a)$? I'm looking for either a proof or a counterexample as I'm unable to come up with either.

答:
It's well-known that the functions $F_{n} : [0,1] \to \mathbb{R}$ given by $F_{n}(x) = x^{n}$ converge to zero uniformly in $[0,\zeta]$ for each $\zeta \in (0,1)$, but not in $[0,1]$.

Define a sequence of functions $\{f_{n}\}_{n \in \mathbb{N}}$ in $[0,1]$ by $f_{n}(x) = F_{1}(x)$ if $n = 1$ and $f_{n}(x) = F_{n}(x) - F_{n - 1}(x)$, otherwise.  It follows that $\sum_{j = 1}^{N} f_{j}(x) = F_{N}(x)$ for all $N \in \mathbb{N}$.  Hence $\sum_{j = 1}^{\infty} f_{j} = 0$ uniformly in $[0,\zeta]$ for each $\zeta \in (0,1)$, but not uniformly in $[0,1]$.

Title: Extending uniform convergence


问:
How do I solve this Differential Algebraic Equation?
I am trying to model a physical system which is goverened by the equation:

$$ (r(t))^2 \cdot \frac{d^2 r(t)}{dt^2} = c q(t)$$

I am a junior electrical engineering student, so I have had courses in linear algebra, ordinary differential equations, and LTI system theory + Fourier/Laplace/DFT/Z transforms.

Based on what I can tell about this equation, it is a Differential Algebraic Equation, which I am unfamiliar with how to solve. I would like to solve for $r(t)$ given an arbitrary function of $q(t)$ - say a sinusoidal, exponential, or polynomial function.

From what I've read online, an analytic solution may not be possible, so numerical solutions are fine too.

Also, all I can find online is how to solve a SYSTEM of DAEs, not simply a singular DAE equation.

What I would like to do is to be able to predict the output $r(t)$ given some input signal $q(t)$

答:
As Yves mentioned your problem is an ordinary differential equation, however it is implicit (the highest derivative is not isolated). If you are expecting the solution cross the zero line the reformulation into an explicit system by diving by $r(t)^2$ is inappropiate. I will focus my answer on the implicit formulation.

Please note that $\dfrac{q(t)}{r^2(t)}$ is not Lipschitz-continous (has unbounded derivative) meaning it has likely no solution for all points in time/value space. Solvers might fail to solve it over the complete time interval you want it to be solved. You can solve implicit ODEs numerically using an DAE solver.
The idea is that DAE solver tries to find an du such that the residual function returns 0 and then step forward using that du.
Here is how you would do it in a Julia command line:

```<code>]add DifferentialEquations #installs a kitchen sink of tools
]add Plots # if you need plots

function residual(res,du,u,p,t)
   res[1] = du[1] - u[2]
   res[2] = du[2]*u[1]*u[1] - c*q(t)
end

c = 3 # when changed just rerun solve
q(t) = sin(t) # when changed just rerun solve

tspan = (0.0,1.0) #solve for virtual 1.0 time units
u0 = [1.,2.] # initial conditions for r(t) and s(t)

prob = DAEProblem(DAEFunction(residual),u0,u0,tspan) # If you change u0, or tspan rerun this too

sol = solve(prob, DABDF2())

sol(0.8) #interpolated solution at time 0.8
Array(sol) #solution without interpolation

plot(sol) ## savefig(plot(sol), "name.png") to save in working directory, you can use a path too
</code>```

Another reformulation would be using a mass matrix. The idea would be that you bring your equation into this form.
$$M(\vec{x},t) \dfrac{d\vec{x}}{dt} = f(\vec{x},t)$$

However whether you have problems at zero crossings (where the $M^{-1}$ is illdefined) depends on the solver in question, i think. If you want to see code for that too, please add a comment saying so. I could also show how to solve the explicit formulation of Yves in Julia.

Body:
As Yves mentioned your problem is an ordinary differential equation, however it is implicit (the highest derivative is not isolated). If you are expecting the solution cross the zero line the reformulation into an explicit system by diving by $r(t)^2$ is inappropiate. I will focus my answer on the implicit formulation.

Please note that $\dfrac{q(t)}{r^2(t)}$ is not Lipschitz-continous (has unbounded derivative) meaning it has likely no solution for all points in time/value space. Solvers might fail to solve it over the complete time interval you want it to be solved. You can solve implicit ODEs numerically using an DAE solver.
The idea is that DAE solver tries to find an du such that the residual function returns 0 and then step forward using that du.
Here is how you would do it in a Julia command line:

]add DifferentialEquations #installs a kitchen sink of tools
]add Plots # if you need plots

function residual(res,du,u,p,t)
   res[1] = du[1] - u[2]
   res[2] = du[2]*u[1]*u[1] - c*q(t)
end

c = 3 # when changed just rerun solve
q(t) = sin(t) # when changed just rerun solve

tspan = (0.0,1.0) #solve for virtual 1.0 time units
u0 = [1.,2.] # initial conditions for r(t) and s(t)

prob = DAEProblem(DAEFunction(residual),u0,u0,tspan) # If you change u0, or tspan rerun this too

sol = solve(prob, DABDF2())

sol(0.8) #interpolated solution at time 0.8
Array(sol) #solution without interpolation

plot(sol) ## savefig(plot(sol), "name.png") to save in working directory, you can use a path too

Another reformulation would be using a mass matrix. The idea would be that you bring your equation into this form.
$$M(\vec{x},t) \dfrac{d\vec{x}}{dt} = f(\vec{x},t)$$

However whether you have problems at zero crossings (where the $M^{-1}$ is illdefined) depends on the solver in question, i think. If you want to see code for that too, please add a comment saying so. I could also show how to solve the explicit formulation of Yves in Julia.


问:
if $(k+m)^n + (l+m)^n = (k+l+m)^n$ then $kl$ divides $m^n$
How to prove that for any $k, l, m, n \in \Bbb N$ and $n \ge 3 $ with $(k+m)^n + (l+m)^n = (k+l+m)^n$  then $kl$ divides $m^n$?

As a special case, in the same way as Brian's answer below it can be shown that if $a^n + b^n = (a+k)^n$ then $k$ divides $b^n$:
$$a^n + b^n = (a+k)^n = a^n + k^n + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i}\\ = a^n + k^n + k\sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1} \\= a^n + k \left(k^{n-1} + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1}\right) $$ so
$$  b^n = k \left(k^{n-1} + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1}\right) $$
thus $k$ divides $b^n$ and furthermore $ rad(k) $ divides $b$.

答: By Fermat's last theorem (yes, not Fermat's little theorem) one of $k+m$ and $l+m$ is zero, which means that $m=0$ and one of $k$ and $l$ is zero. It follows that $kl=0$ and $m^n=0$; $0\mid0$ and the statement is true.


问:
What do cones have to do with quadratics? Why is $2$ special?
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos:

This 3B1B video about ellipses, which reignited my desire to understand conics

Why are quadratic equations the same as right circular conic sections?, which offers a very computational approach to trying to resolve this question

Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle)

and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture, where minutes ~10-15 discuss the complete solution to the problems of rational points on conics.

While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$).

So, I suppose my questions are as follows:

Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations?

and more generally:

Is there some deep reason why "$2$" is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well  -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts, as well as being the only possible finite degree extension of $\mathbb R$, leading to in particular $\mathbb C$ being 2-dimensional.

Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share!

$$\rule{5cm}{0.4pt}$$

EDIT 3/12/21: was just thinking about variances, and least squares regression. "$2$" is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation?, Why is it so cool to square numbers (in terms of finding the standard deviation)?, and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues.

In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$. Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA. Honorable mentions to Qiaochu Yuan's answer about rotation invariance, and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;)

$$\rule{5cm}{0.4pt}$$

EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question:  What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original.

The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection.

I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional". Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors?

There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space).

$$\rule{5cm}{0.4pt}$$

EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$. There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$, which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself?), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$), since those show up in generalizations to the Hardy uncertainty principle (“completing the square”, again an algebraic nicety of squares, was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics.

This (squares leading to decent theory of $r_n(k)$, and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$. I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$, but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares).

$$\rule{5cm}{0.4pt}$$

EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bilinear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf).

And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck, is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces). Relating to the Laplacian, we have the important concept of harmonicity, with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity. Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago (https://mathoverflow.net/a/171743/112504) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.":

So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)

答:
The proximate reason is that cones are based on circles, and circles, in turn, are given by the quadratic equation

$$x^2 + y^2 = r^2$$

. Now, as for the reason that circles have this equation, that is because they are related to the Euclidean distance function, being the set of all points at a constant distance from a given center, here conventionally taken as the origin. In particular,

$$d(P, Q) = \sqrt{|Q_x - P_x|^2 + |Q_y - P_y|^2}$$

Insofar as why the Euclidean metric has this form, I would say that it comes down to the following. To get a little more insight into this, it is useful to consider the somewhat more general form of metrics

$$d_p(P, Q) := \left(|Q_x - P_x|^p + |Q_y - P_y|^p\right)^{1/p}$$

called the $p$-metrics which, in effect, result from asking "well, what happens if we let the power not be 2?", and so are just right for answering this question.

And it turns out that $d_2$ has a very special property. It is the only one for which you can take a geometric object, declare a point on it a pivot, then take any other point on that object and tag it, measure the distance from the pivot to the tag point, and now transform that object in such a way the center remains fixed, while the tag point comes to face a different direction at the same distance, and yet the whole object's overall size and shape remains unchanged. Or, to put it another way, that such a thing as "rotation" makes geometric sense as being a rigid motion.

So, what is the ultimate reason cones are quadratic? Because in Euclidean space, you can rotate things in any way you please without changing their size and shape.


答:
Consider the example $f(x)=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$ for $-1<x<1$. In fact, the series converges even for $x=\pm1$ in the sense of the question, that is, $x\to1^-$, $x\to(-1)^+$, however the convergence is not uniform (see picture below). The reason is that there is a singularity at the imaginary number $i$.

In general, for any power series, uniform convergence at a boundary point of a ball $B_r(a)\subset\mathbb{C}$ implies uniform convergence on smaller closed balls $\overline{B_s(a)}$, $s<r$. If there is a singularity at the boundary, then there cannot be uniform convergence on the ball. Nevertheless, there can still be convergence at some other points.

Title: Extending uniform convergence

Body: Let $\sum_{n=0}^{\infty} f_n(x)$ converge to $f(x)$ on $[0, a)$ and converge uniformly on evey closed sub-interval of $[0, a)$. If $\lim_{x \to a^-} f(x) = L < \infty$ does it follow that the series converges uniformly on $[0, a)$? I'm looking for either a proof or a counterexample as I'm unable to come up with either.


问:
Binomial sum of derivatives
I would like to know the result of the following sum:

$$\sum_{p=0}^m \binom{m}{p}(-1)^{p-1}\frac{\partial^{p-1}}{\partial x^{p-1}}f(x)\cdot(-1)^{m-p-1}\frac{\partial^{m-p-1}}{\partial x^{m-p-1}}g(x)$$

with $$\frac{\partial^{-1}}{\partial x^{-1}}=\int dx$$

Thanks

答:
You can pull out the factors of $-1$ to yield an overall factor $(-1)^m$. Also it's unclear why you're using partial derivatives, as there's only one variable; I'll use total derivatives. Then

\begin{align}
&
\sum_{p=0}^m\binom mp\left(\frac{\mathrm d}{\mathrm dx}\right)^{p-1}f(x)\left(\frac{\mathrm d}{\mathrm dx}\right)^{m-p-1}g(x)
\\
=&
\sum_{p=0}^m\binom mp\left(\frac{\mathrm d}{\mathrm dx}\right)^p\int f(x)\mathrm dx\left(\frac{\mathrm d}{\mathrm dx}\right)^{m-p}\int g(x)\mathrm dx
\\
=&
\left(\frac{\mathrm d}{\mathrm dx}\right)^m\left(\int f(x)\mathrm dx\int g(x)\mathrm dx\right)\;.
\end{align}

ContentLicense: CC BY-SA 3.0

Title: Binomial sum of derivatives


问:
Let $x\in\mathbb R^d$ such that $||x||_2=1$. How to prove that the expected absolute value of a random walk on $x$ is at least $1/\sqrt 2$?
Let $x\in\mathbb R^d$ such that $||x||_2=1$, and let $Y_1,\ldots,Y_d$ be i.i.d. random variables such that $Y_i=\begin{cases}1&\mbox{w.p. 1/2}\\-1&\mbox{otherwise}\end{cases}$.

Define $Z=\sum_{i=1}^d x_i\cdot Y_i$.

How can we show that $$\mathbb E[|Z|]\ge 1/\sqrt 2$$ for any such vector $x$?

Notice that we have $\mathbb E[|Z|]= 1/\sqrt 2$ for the vector $x=(1/\sqrt 2, 1/\sqrt 2,0,\ldots,0)$.

答: This follows from Khintchine inequality.</p>


问:
How to choose the $k$ points that given a cost matrix with other points, give the minimum sum of distances
For two sets of $n$ dimensional points this is their distance matrix:

$p_{11}$
$p_{12}$
$p_{13}$

$p_{21}$
1
2
3

$p_{22}$
2
5
3

$p_{23}$
2
3
1

$p_{24}$
2
1
8

How can I get the $n$ points of $p_{1i}$ that give the minimum sum of distances in the most efficient way possible.

So far best I could do was apply the Hungarian assignment algorithm in a loop for all possible teams (this takes $\binom{i}{n}$ loops though and is not very efficient)

The solution for n=2 would follow this procedure:
find the minimum sum of the minimum distances that each couple of p1,i gives for each row, for example:

$(p_{1,1}, p_{1,2}) = 1^{*}+2+2+1 = 6$

$(p_{1,2}, p_{1,3}) = 2+3+1+1 = 7$

$(p_{1,1}, p_{1,3}) = 1+2+1+2 = 6$

So the optimal couple would be either $(p_{1,1}, p_{1,2})$ or $(p_{1,1}, p_{1,3})$

explanation of the $1^{*}$: its the $min(d(p_{2,1},p_{1,1}), d(p_{2,1},p_{1,2}))=1$

答:
Let $L=\{1,2,3\}$ and $R=\{1,2,3,4\}$ be the left and right sides of the bipartite graph.  Let $c_{i,j}$ be the cost of assigning left node $i$ to right node $j$ (the transpose of your table above).  Let binary decision variable $x_{i,j}$ indicate whether left node $i$ is assigned to right node $j$, and let binary decision variable $y_i$ indicate whether left node $i$ is selected.  The problem is to minimize $$\sum_{i \in L} \sum_{j \in R} c_{i,j} x_{i,j}$$ subject to
\begin{align}
\sum_{i \in L} x_{i,j} &= 1 &&\text{for $j\in R$} \tag1 \\
\sum_{i \in L} y_i &= k \tag2 \\
x_{i,j} &\le y_i &&\text{for $i \in L$ and $j \in R$} \tag3 \\
\end{align}
Constraint $(1)$ assigns exactly one left node per right node.
Constraint $(2)$ selects exactly $k$ left nodes.
Constraint $(3)$ enforces the logical implication $x_{i,j} \implies y_i$.

Your first optimal solution above has $x_{1,1}=x_{1,2}=x_{1,3}=x_{2,4}=y_1=y_2=1$ and all other variables $0$.


问:
Has anyone ever propose a theory of infinitesimals, based on this principle?
We know that in the theory of distributions $\int_{-1}^1 \delta(x)dx=1$.

Now consider a function $f(x)$ which is equal to $1$ at $x=0$ but otherwise is zero.

Did anyone propose a generalization of integration, under which $\int_{-1}^1 f(x)dx$ is an infinitesimal?

This would allow to compare infinitesimals, multiply them by real numbers, etc...

答: Well this would have an exact value as $dx$ when $dx$ is considered the small number you break Reimann integrals up into, so I think it should have the same value as the infinitesimal typically denoted $\epsilon$ or $\frac{1}{\omega}$ that is used in non-standard analysis (analysis that makes infinitesimals rigorous).


问:
$M \subset \mathbb{R}^n$ $n-$manifold is an open set or the closure of an open set in $\mathbb{R}^n$
I'd like to prove the fact that for an $n-$manifold $M\subseteq \mathbb{R}^{n}$ it's the same thing to say that $M$ is open or is the closure of an open set of $\mathbb{R}^n$.

I think the case where $M$ has no boundary is easier since $M = \bigcup\limits_{p_i \in M}\varphi_i(U_i)$ which is open (since union of open sets), because $\varphi_{i} (U_{i})$ is a local diffeomorphism between and open set $U_i$ of $\mathbb{R}^n$ and $V_{p_{i}} \cap M$ open neighboorhood of $p_{i} \in M$.

What about the case where I've got local diffeomorphism $\varphi_i(U_i \cap \left\lbrace x_{n} \geq 0 \right\rbrace) = V_{p_i} \cap M$ ?

Any help or direct proof would be appreciated.

答: By invariance of domain, if $f: U \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$ is injective, then $f(U)$ is open in $\mathbb{R}^n$, so for a manifold without boundary this is true, since it os the union of inverses of charts. Now, if $M$ has a boundary this is false, just consider $[0,1) \subset \mathbb{R}$.


问:
Equivalent definition of left hand limit.

Let $F$ be a Banach space. Let $f : [a,b] \longrightarrow F$ be a function. Then $L$ is said to be the left hand limit of $f$ at $c \in (a, b]$ if for a given $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that for all $x \in (c-\delta, c)$ we have $\|f(x) - L\|_{F} \lt \varepsilon.$

This is the definition I am familiar with from my school days. But while taking up a course in several variable calculus in normed linear spaces in the masters I find a new definition of it which says that $f$ is said to have a left hand limit at $c$ if for a given $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that for any pair of points $s,t \in (c - \delta,c)$ we have $\|f(s) - f(t)\|_{F} \lt \varepsilon.$

I have proved the former definition implies the later one. But I cannot able to prove the reverse implication. From the later definition how do I get hold of the limit of the former definition? Would anybody please help me in this regard?

Thanks in advance.

答:
Let $(s_n)_n$ a sequence valued in $[a,b]$ such that $s_n$ converges toward $c$ by the left (in the usual sense for real sequences).

Now, using your latter definition, for a fixed $\varepsilon>0$, we can easily see that there exists $N_0$ such as

$$\|f(s_p) - f(s_q)\|_{F} \lt \varepsilon, \quad \forall p,q>N_0 \quad \quad (1)$$

This means exactly that the sequence $(f(s_n))_n$ is a Cauchy sequence in the space $F$. Since $F$ is a Banach space, this prove that there exists $L$ such as the sequence $(f(s_n))_n$ converges toward $L$ in $F$.

I believe it should be easy to conclude from here, I'll let you write the details !


问:
Permutation $\alpha^2$
Show that there is no permutation $\alpha \in S_4$ so that $\alpha^2=\begin{pmatrix}
1 &3  &4  &2
\end{pmatrix}$.

$S_4=\begin{pmatrix}
1 &2 &3  &4
\end{pmatrix}$

If $ \alpha=\begin{pmatrix}
2 &4  &3
\end{pmatrix}$, we get:
$$\alpha^2=\alpha\cdot\alpha=\begin{pmatrix}
1 &2  &3  &4 \\
 1&4  &2  &3 \\1
 &3  &4  &2
\end{pmatrix}$$

Why can't I do that?

答: $\beta = (1 \ 3 \ 4 \ 2)$ is on odd permutation: a cycle of length equal to $4$. As the square of any permutation is even, $\beta$ can't be written as a square.

Title: Permutation $\alpha^2$


答:
Use the binomial theorem to expand both sides of the original equation:

$$\sum_{i=0}^n\binom{n}i\left(k^i+\ell^i\right)m^{n-i}=\sum_{i=0}^n\binom{n}i(k+\ell)^im^{n-i}\,.$$

Pull out the $m^n$ terms:

$$2m^n+\sum_{i=1}^n\binom{n}i\left(k^i+\ell^i\right)m^{n-i}=m^n+\sum_{i=1}^n\binom{n}i(k+\ell)^im^{n-i}\,,$$

so

$$\begin{align*}
m^n&=\sum_{i=1}^n\binom{n}im^{n-i}\left((k+\ell)^i-\left(k^i+\ell^i\right)\right)\\
&=\sum_{i=1}^n\binom{n}im^{n-i}\sum_{j=1}^{i-1}\binom{i}jk^j\ell^{i-j}\\
&=k\ell\sum_{i=1}^n\binom{n}im^{n-1}\sum_{j-1}^{i-1}\binom{i}jk^{j-1}\ell^{i-j-1}\,.
\end{align*}$$

Title: if $(k+m)^n + (l+m)^n = (k+l+m)^n$ then $kl$ divides $m^n$

Body:
How to prove that for any $k, l, m, n \in \Bbb N$ and $n \ge 3 $ with $(k+m)^n + (l+m)^n = (k+l+m)^n$  then $kl$ divides $m^n$?

As a special case, in the same way as Brian's answer below it can be shown that if $a^n + b^n = (a+k)^n$ then $k$ divides $b^n$:
$$a^n + b^n = (a+k)^n = a^n + k^n + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i}\\ = a^n + k^n + k\sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1} \\= a^n + k \left(k^{n-1} + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1}\right) $$ so
$$  b^n = k \left(k^{n-1} + \sum_{i=1}^{n-1} \binom{n}{i}a^ik^{n-i-1}\right) $$
thus $k$ divides $b^n$ and furthermore $ rad(k) $ divides $b$.


问:
$g(x) = \frac{f'(x)}{f(x)}$ not bounded
Let $f : ]0,1[ \rightarrow ]0,+\infty[$ a differentiable function such that

$\lim_{x\to0+} f(x) = 0$

Show that the function $g:]0,1[ \rightarrow \mathbb{R}$ defined by

$g(x) = \dfrac{f'(x)}{f(x)}$

is not bounded.

I tried a lot of things: I tried the extension by continuity on 0, tried to pass by Mean value theorem and tried to proof by absurd by bounding $g(x)$ but it doesn't lead me anywhere... Thanks for help in advance

答:
By the mean value theorem, for any $0 < x < y < 1$, there exists $\xi(x,y) \in (x,y)$ such that

$$\frac{\log f(y) - \log f(x)}{y-x} = \frac{f'(\xi(x,y))}{f(\xi(x,y))}= g(\xi(x,y))$$

Thus,

$$\lim_{x \to 0+}g(\xi(x,y)) = \lim_{x \to 0+}\frac{\log f(y) - \log f(x)}{y-x}= \frac{\log f(y) - \lim_{x \to 0+}\log f(x)}{y} = +\infty$$

Hence, for any $n \in \mathbb{N}$ there exists $\delta_n \in (0,y)$ such that $g(\xi(\delta_n,y))> n$.

As the mean value theorem is non-constructive it is not possible to determine  if $\xi(\delta_n,y) \to 0$ with $y$ fixed as $n \to \infty$.  However, since $y$ can be chosen arbitrarily close to $0$, this shows that $g$ is unbounded in every neighborhood of $0$.

Title: $g(x) = \frac{f'(x)}{f(x)}$ not bounded


问:
Question about column space and their relationship to rank.

Let $A\in\mathbb{M}_{m\times n}$. Prove:

(a) Let $\{\mathbf{v}_1,\ldots,\mathbf{v}_p\}\subseteq\mathbb{R}^n$. If $\{A\mathbf{v}_1,\ldots,A\mathbf{v}_p\}$ is linearly independent, then $r(A)\geq p$.

(b) Let $\{\mathbf{v}_1,\ldots,\mathbf{v}_p\}\subseteq\mathbb{R}^n$. If $\{A\mathbf{v}_1,\ldots,A\mathbf{v}_p\}$ is a spanning set for $\mathbb{R}^m$, then $r(A)=m$.

(c) Assume $r(A)=m$. If $\mathrm{Span}\{\mathbb{v}_1,\ldots,\mathbb{v}_p\}=\mathbb{R}^n$, then $\mathrm{Span}\{A\mathbb{v}_1,\ldots,A\mathbb{v}_p\}=\mathbb{R}^m$.

Hi, so I usually use this community to post questions / answers regarding calculus, but I'm really starting to struggle in my Linear Algebra course and I would really appreciate the help!

I just can't figure out how to connect all the dots in this question..
I know that in (a) the set is in the span of ColA, but how does it help me to know that it's linearly independent? And what's the relation to RankA?

in (b), if we know the set spans $R^m$, it means colA spans $R^m$, and in order for colA to be able to span $R^m$, it has to have m linearly independent vectors, and thus dimA = m = rank(A). Is this a right way to look at it?

in (c) I didn't even attempt it after not being able to solve the first two.. I don't feel confident enough with the material. :(

答:
(a) It's important you realize that the span of the $Av_i$ is a subspace of the column space of $A$.

Proof: Take $x$ in the span of the $Av_i$, which means $x$ is a linear combination of the $Av_i$. Then $x=\sum c_i(Av_i)$ for some coefficients $c_i$. But $c_i(Av_i)=A(c_iv_i)$, so

$$x=\sum A(c_iv_i)=A\left(\sum c_iv_i\right)$$

But the rightmost vector is just a linear combination of the columns of $A$. Hence $x$ is in the column space of $A$. QED

Now you can also use the fact that if $U$ is a subspace of $V$, then

$$\dim U\leq \dim V$$

It follows from the fact we've proven above that the dimension of the column space of $A$, aka the rank of $A$, is at least the dimension of the span of the $Av_i$, which by the hypothesis of linear independence is $p$. In other words, $r(A)\geq p$.

(b) If the $Av_i$ give a spanning set for $\mathbb{R}^m$, then the span of the $Av_i$ is equal to $\mathbb{R}^m$. Using the fact we proved above that the span of the $Av_i$ is a subspace of the column space of $A$, we conclude $\mathbb{R}^m$ is a subspace of the column space of $A$. Thus $r(A)\geq m$. You also know $r(A)\leq m$ because the row rank (which is obviously at most $m$) equals the column rank (which in turn equals the plain old rank). Thus $r(A)=m$.

(c) Note that the span of the $Av_i$ is automatically a subspace of $\mathbb{R}^m$, so all we need to do is prove the other inclusion: $\mathbb{R}^m$ is a subspace of the span of the $Av_i$. Well, take a vector $x\in\mathbb{R}^m$. We want to show that $x$ is in the span of the $Av_i$.

Since $A$ has rank $m$, the column space has dimension $m$ and thus, being a subspace of $\mathbb{R}^m$, is in fact equal to $\mathbb{R}^m$. Therefore $x$ is a linear combination of the columns of $A$, say $x=Av$ for some $v\in\mathbb{R}^n$.

Now, using the hypothesis that the span of the $v_i$ is $\mathbb{R}^n$, we conclude $v$ can be written as a linear combination of the $v_i$, say $v=\sum c_iv_i$. Thus

$$x=Av=A\left(\sum c_iv_i\right)=\sum c_i(Av_i)$$

which is a linear combination of the $Av_i$. Thus $x$ is in the span of the $Av_i$.


问:
Continuous function that is not measurable.
Let $\mathcal{A}$ be a σ-algebra on $\mathbb{R}$ that does not contain all Borel sets (i.e that is, it holds
that $\mathcal{B}(\mathbb{R}) \setminus \mathcal{A} \not= ∅)$. Prove that there is a continuous function f : $\mathbb{R} → \mathbb{R}$ which
is not $\mathcal{A}$-measurable

I have a really simple example but I am not sure if it correct. Take $\mathcal{A} = (\emptyset , \mathbb{R} , (0,1) , \mathbb{R}\setminus (0,1))$ This is clearly a sigma algebra.
Then take $f(x)=x$ we have that $f^{-1}((5,6))=(5,6)\notin\mathcal{A}$ so $f$ is not $\mathcal{A}$ measurable and clearly $f$ is continuous.

Is this a correct solution?

EDIT

Thanks to some incredibly bright chaps on this site I have seen the error in my solution, I simply misread the question. So for all those who come after here is the answer:

As $\mathcal{B}$, the Borel $\sigma$-algebra, is the smallest $\sigma$ algebra containing all open sets and $\mathcal{B} \not \subseteq \mathcal{A}$ we have at least one interval, $I \in \mathcal{B}$, that is not in $\mathcal{A}$ Hence taking the pre-image of $I$. We have:  $\mathcal{B} \ni I, f^{-1}(I) = I \notin \mathcal{A}$ So $f$ is not $\mathcal{A}$ measurable

答: Consider a closed set $C\notin \mathcal A$. There is a continuous (or even $C^\infty$) function $f:\Bbb R\to \Bbb R$ such that $C=f^{-1}\{0\}$. Therefore $f$ is not $\mathcal A$-measurable.


问:
What can you conclude about $A \setminus B = B \setminus A$?
I have the following identity involving sets $A$ and $B$:

$$A \setminus B = B \setminus A$$

Now, a solution says that we can conclude that $A=B$, but what is wrong with a solution that says that this means that $A\subseteq B$? I mean if we remove all elements of $B$ from $A$, we would have what is left from $B$ and same for if we remove all elements of $A$ from $B$ as it will yield also elements of $B$. Please correct me if I am wrong.

Thanks.

答:
I think the following answer is more clear:

If $A-B \neq \emptyset$, then since $A-B \subseteq B-A$, there exists an $x \in A$ such that $x \notin A$, a contradiction.

Hence $B-A=A-B = \emptyset$.

If $A = \emptyset$, then $B = \emptyset$ and vice versa which implies the result.

Otherwise, $(x \in A \implies x\in B) \implies A\subseteq B$.

Similarly, one finds that $B \subseteq A$, and so $A=B$.


答:
The way the question is phrased, you are not allowed to choose $\mathcal A$.

But your idea works anyway. If $\mathcal A$ does not contain all Borel sets, then it fails to contain some interval. Now you can do your example using said interval.

Title: Continuous function that is not measurable.

Body:
Let $\mathcal{A}$ be a σ-algebra on $\mathbb{R}$ that does not contain all Borel sets (i.e that is, it holds
that $\mathcal{B}(\mathbb{R}) \setminus \mathcal{A} \not= ∅)$. Prove that there is a continuous function f : $\mathbb{R} → \mathbb{R}$ which
is not $\mathcal{A}$-measurable

I have a really simple example but I am not sure if it correct. Take $\mathcal{A} = (\emptyset , \mathbb{R} , (0,1) , \mathbb{R}\setminus (0,1))$ This is clearly a sigma algebra.
Then take $f(x)=x$ we have that $f^{-1}((5,6))=(5,6)\notin\mathcal{A}$ so $f$ is not $\mathcal{A}$ measurable and clearly $f$ is continuous.

Is this a correct solution?

EDIT

Thanks to some incredibly bright chaps on this site I have seen the error in my solution, I simply misread the question. So for all those who come after here is the answer:

As $\mathcal{B}$, the Borel $\sigma$-algebra, is the smallest $\sigma$ algebra containing all open sets and $\mathcal{B} \not \subseteq \mathcal{A}$ we have at least one interval, $I \in \mathcal{B}$, that is not in $\mathcal{A}$ Hence taking the pre-image of $I$. We have:  $\mathcal{B} \ni I, f^{-1}(I) = I \notin \mathcal{A}$ So $f$ is not $\mathcal{A}$ measurable


答:
I didn't want to answer (just to comment), but since some other guy answered what I commented, then let me write a simpler answer.

The identity function you propose always works. Indeed, if your $\sigma$-algebra does not contain all Borel subsets then it does not contain at least an open subset of the reals. Then just take the inverse image of that open set through the identity function.


问:
Expected Utility
I am confused by something regarding expected utility calculation.

When we are calculating the expected utility where 2 positive outcomes can happen at the same time such as the example below:

We win 50 with probability of 0.3

We win 100 with probability of 0.2

How do we calculate the expected utility? Which of the options below is correct?

(1)$50 \times 0.3 + 100 \times 0.2 = 35$

(2)$50 \times 0.3 + 100 \times 0.2 + 150 \times (0.3 \times 0.2) = 44$

答: Because the two events are independent of each other, simply consider their outcomes one at a time. The expected utility of the first event is $50 \cdot 0.3 = 15$ while the expected utility of the second event is $100 \cdot 0.2 = 20$. Thus, the total expected utility is $15+20=35$. In other words, the first equation you listed is correct.

Title: Expected Utility


问:
Show that $ 2 \sinh(z)=\exp(z)-\exp(-z)$
$ 2 \sinh(z)=\exp(z)-\exp(-z)$;

$ 2 \cosh(z)=\exp(z)+\exp(-z)$

where $z \in \mathbb{C} $

$$\sin(z) := \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1!} z^{2k+1}$$

$$\cos(z) := \sum_{k=0}^{\infty}\frac{(-1)^k}{2k!} z^{2k}$$

I guess that I have to use a trigonometrical identity, but I don't know which one and is that step equivalent to use it for hyperbolic functions?
Thank in advance for your help.

答:
It is hard to answer this question without knowing what definitions you are using for your functions. However, if you use the series expansions, then

$$\exp(z)-\exp(-z)=\sum_{n=0}^\infty \left[\frac{z^n}{n!}(1-(-1)^n)\right]$$

Note that at even $n$ the expression in the brackets is $0$ and at odd $n$ it is $2\frac{z^n}{n!}$. Then the expression simplifies to

$$=2\sum_{n\text{ odd}}^\infty \frac{z^{n}}{n!}=2\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!}=2\sinh(z)$$

You can do a similar manipulation to get the identity involving $\cosh(z)$. Again, I want to emphasize that this method only works if you have previously shown that these functions have these series expansions. Without knowing how you define the functions, this is one possible way to show the identities in question.


问:
Volume of body bounded with surfaces
I need to find volume bounded with $x=2y-y^2-z^2$ and $2y-x-1=0$. So first one is paraboloid that is located on x-axis and its "maximum" is in $(0,2,0)$ and the second is plane paralele to z-axis.
I just want to check do we get the volume when substract volume of $2y-x-1=0$ from paraboloid.(Also using polar coordinates it is easy to get $0\le r\le 1, 0\le \phi \le 2\pi$). I'm just not sure whether I sketched this correctly, precisely is given plane above paraboloid?

答:
The intersection between $x = 2 y - 1$ and $x=-y^2+2 y-z^2$ is the circle

$y^2+z^2=1$, which means that the limits for $z$ and $y$ are
$$-\sqrt{1-y^2}\le z\le \sqrt{1-y^2};\;-1\le y\le 1$$
Thus the volume is
$$V=\int _{-1}^1\int _{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int _{2 y-1}^{-y^2+2 y-z^2}dxdzdy=\frac{\pi}{2}$$

Hope this helps

Title: Volume of body bounded with surfaces


问:
$ e^{At}$ for $A = B^{-1} \lvert \cdots \rvert B $
For a homework problem, I have to compute $ e^{At}$ for

$$ A = B^{-1}  \begin{pmatrix}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \end{pmatrix} B$$

I know how to compute the result for $2 \times 2$ matrices where I can calculate the eigenvalues, but this is $3 \times 3$, and I cannot compute eigenvalues, so is there any identity or something which allows computing such exponentials?
Thanks!

答:
Joseph G. has it right.  In addition, you say, "... I cannot compute eigenvalues" but the eigenvalues are simply the solutions to

$\left|\begin{array}{ccc}
-1-\lambda& 0& 0 \\
0& 2-\lambda& 0 \\
0& 0& 3-\lambda \end{array}\right|=0$

which boils down to $(-1-\lambda)(2-\lambda)(3-\lambda)$, so $\lambda=-1,2,3$.


问:
If $f: A = (0, 1) \subset M = \mathbb{R} \rightarrow N = \mathbb{R}^3$ is continuous, then it maps Cauchy sequence to Cauchy sequence.
Am I right with my arguments or is something wrong?

proof. Let $(x_n)_{n \in \mathbb{N}}$ a Cauchy sequence, then for every $\epsilon > 0$ exists an $N \in \mathbb{R}$ such that for every $n,m \geq N$, we have
$$d_M (x_n, x_m) < \epsilon$$
Since $f$ is continuous, given $\epsilon > 0$ we can take $\delta = \epsilon$, then
$$d_M(x_n, x_m) < \delta = \epsilon$$
So
$$d_N (f(x_n),f(x_m)) < \epsilon$$
Therefore, $(f(x_n))_{n\in\mathbb{R}}$ is a Cauchy sequence.

答:
There's a counterexample to the claim in the comments already.  To answer your question about where you might have gone wrong, you should examine why you think you're allowed to choose $\delta = \epsilon$ in your argument.

The definition of $f$ being continuous at $x_n$ tells you that for each $\epsilon > 0$ there is a $\delta > 0$ such that for all $y$, if $d(x_n, y) < \delta$ then $d(f(x_n), f(x_m)) < \epsilon$.  It does not say that you get to pick the $\delta$.  As mentioned in the comments, it also does not say that the same $\delta$ will work for all $n$.


问:
How should I interpret these parametric equations of orbit?

In my math book there are these parametric equations of the orbit/trajectory of a point $P$:
$$\begin{cases}
x(t)=\sin t\\
y(t)=\sin 4t\\
\end{cases}
$$
along with a graph/picture of it on the $y$ and $x$ axis.
Could anyone explain why it follows the path it does?

I unfortunately cant plot it on desmos or show a picture, so if anyone could do this, that would be great.

I thought about it, and since the x values are given by $sin(t)$, does this mean that the trajectory of $P$ first goes to the right, but after $sin(\pi)$ the $x$ values suddenly get negative, and the trajectory switches to the left of the $y$ axis?

So my question is in which direction does the orbit exactly go?

Thanks in advance

答:
$$\begin{cases}
x=\sin t\\
y=\sin 4t\\
\end{cases};\;0\le t\le2\pi
$$


问:
How to expand $x^a$ with positive, real $a$ in powers of $x$?
I would like to expand $x^a$ with positive, real-valued $a$, in powers of $x$. That is, I am looking for an expression of the form

$$  x^a = \sum_{n=0}^\infty c_n x^n ~, $$

with explicit expressions for the expansion coefficients $c_n$ in terms of $a$.

答:
No such expansion exists unless $a$ is an integer, in which case it is trivial.
If $f(x) = \sum_{n=0}^{\infty} c_n x^n$ is equal to a convergent power series in some interval $I$ around $0$ (i.e. $f$ is analytic), then it is infinitely differentiable. But if $a$ is not an integer, the function $f(x) = x^a$ is singular at $x=0$.
See also this answer.


问:
Roots of a matrix equation
Let $X \in \mathbb{C}^{n \times m} $ be a rectangular matrix of full rank, and $X^*$ its hermitian conjugate, let $A \in \mathbb{C}^{m \times m}$ a square matrix, and let $f: \mathbb{C} \to \mathbb{C}$ be defined by
$$
f(z) = \det \left( I_n + X \frac{1}{A - z I_m} X^* \right)
$$
where $I_n$ is the $n \times n$ identity matrix.

Show that for $n \leq m$ solutions of $f(z)= 0$ are given by $z$ equal to an eigenvalue of $B : = A + X^* X$.

My attempt: for $n = m$, $f(z)$ can be straightforwardly rearranged to
$$
f(z) = \det \left( X \frac{1}{B-A} \big( B - I_m z \big) \frac{1}{A - z I_m} X^* \right) = \frac{\det(X) \det(B - I_m z) \det(X^*)}{\det(A - z I_m) \det(B - A)}
$$
where the desired result follows from the factor $\det(B - I_m z)$.

However I cannot factorise the determinant in this way for $n < m$, and I am unsure how to proceed.

答:
Lemma: $A$ is an $m\times n$ size matrix and $B$ is an $n\times m$ size matrix. $AB$ and $BA$ have the same nonzero eigenvalues.

Proof: Let $\lambda\neq0$ be an eigenvalue of $AB$ and $x$ a corresponding eigenvector, thus $ABx=\lambda x\neq0$.  Obviously $Bx\neq0$.
$$BA(Bx)=B(ABx)=B(\lambda x)=\lambda Bx.$$
So $\lambda$ is also an eigenvalue of $BA$ with an eigenvector $Bx$. In other words, the set of nonzero eigenvalues of $AB$ is contained in the set of nonzero eigenvalues of $BA$. The symmetric argument leads to the desired conclusion.  $\quad\square$

I attempted to prove $AB$ and $BA$ have the same eigenvalues together with their algebraic multiplicities. After writing this above lemma, I realized that probably the best way to prove that  is simply the Schur's complement. There is also another similar method to prove the same. In that case

But the above proposition is stronger than $\det(I+AB)=\det(I+BA)$. But then again we can just appeal to my other answer directly applying the Schur's complement.

Alternatively, we can confined ourselves to the narrow circumstance of this particular problem at hand, which only asks for the determinant to be zero. As @BenGrossman suggested,
$\det(I+AB)=0 \iff AB$ has eigenvalue $-1 \iff BA$ has eigenvalue $-1\iff \det(I+BA)=0$.
The rest proceeds via factorization by setting $A:=X$ and $B:=\frac 1{A - zI_m}X^*$.

Title: Roots of a matrix equation


问:
Changing between Maxwell equations in differential and integral formats?
It takes me a long time to think about the equations even in one format and also to deduce things with Stokes. So how can you swap between the equations? I am looking more on the lines that suppose that I know one of the format -- how can I swap to the other?</p>

答:
$\newcommand{\vect}[1]{\boldsymbol{#1}}$
$\newcommand{\vH}{\vect{H}}$
$\newcommand{\vE}{\vect{E}}$
$\newcommand{\vj}{\vect{j}}$
$\newcommand{\vx}{\vect{x}}$
$\newcommand{\vr}{\vect{r}}$
$\newcommand{\vnu}{\vect{\nu}}$
The integral form of a PDE usually comes from the physical definition of a quantity, say, for example curl and div. The curl operator $\nabla \times$ and the div operator $\nabla \cdot$ are both defined in a physical way in Physics, which are line integral and surface integral respectively. But for Mathematician, we always view this process in a reverse way: say, the differential equation is already there, and it holds in the sense of weak derivative, multiplying both side by a testing function which is infinitely smooth and vanishes on the boundary $\partial \Omega$(variational formulation); or more intuitively, since the equation holds for every points in the region(classical sense), we integrate both sides in an infinitesimal region(surface or volume), the equation also holds. ie, we usually deduce the integral form from differential form, not the reverse.

For example, this is the differential form of Maxwell's equations describing how electromagnetic waves propagating through a linear isotropic material, where electric field intensity denoted as $\vE = \vE(\vx,t)$ and the magnetic field intensity denoted as $\vH = \vH(\vx,t)$.

$$
\left\{
\begin{aligned}
\frac{\partial (\mu \vH)}{\partial t} + \nabla \times \vE &= 0 \quad &\text{ in } \Omega\times
(0,T)
\\
\frac{\partial(\epsilon \vE)}{\partial t} + \sigma \vE - \nabla\times \vH &=  \vj \quad &\text{ in }
\Omega\times (0,T)
\\
\nabla \cdot (\epsilon \vE) &= q \quad &\text{ in } \Omega\times (0,T)
\\
\nabla \cdot (\mu \vH) &= 0 \quad &\text{ in } \Omega\times (0,T)
\end{aligned}
\right.
$$

For example, we would like to get the integral form for the Faraday's law of the electric field $\vE$, integrate the first equation on any surface $S$ w.r.t. its surface measure $d S$:
$$
- \int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S= \int_S (\nabla \times \vE)\cdot \vnu \,d S = \oint_{\partial S} \vE \cdot d\vr
$$

where

$\vnu$ is the unit outward normal vector to the surface $S$.

if you parametrize $\partial S$ as $\vr(t)$, then $d\vr = \vr'(t) dt$, where $\vr'(t)$ essentially is the tangential vector of the curve $\partial S$ at every point.

For the left side of that equation, if we assume the enough continuity of $\mu \vH$ w.r.t $t$ and $\vx$(which is implied from the divergence free condition at every $t$), then we have
$$
\int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S = \frac{\partial}{\partial t} \int_{S}\mu \vH \cdot \vnu\,d S
$$
since $\vnu$ is the outward normal vector to $S$, you could think the surface integral as "how much the magnetic field is flowing along the outer normal direction of the surface", which we called this quantity magnetic flux.

This is basically the derivation from differential form to integral form. The reverse is similar, just the integral equations you have at first are all physical laws, for example Faraday's Law from wikipedia:

```<code>The induced electromotive force (EMF) in any closed circuit is equal
to the time rate of change of the magnetic flux through the circuit.
</code>```

Then simply write this out using definition which is integral form, apply Stokes theorem in the reverse way, then you will end up with two equal integrals, for eg, use above example, you have
$$
- \int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S= \int_S (\nabla \times \vE)\cdot \vnu \,d S
$$
for any surface $S$, then you argue that the arbitrariness of $S$ implies
$$
-\frac{\partial (\mu \vH)}{\partial t} = \nabla \times \vE
$$
and there you go.

Here mostly I state my own intuitive understanding of this subject while I was learning Physics as a mathematics student, will appreciate if anybody points out where I was wrong.

Body:
$\newcommand{\vect}[1]{\boldsymbol{#1}}$
$\newcommand{\vH}{\vect{H}}$
$\newcommand{\vE}{\vect{E}}$
$\newcommand{\vj}{\vect{j}}$
$\newcommand{\vx}{\vect{x}}$
$\newcommand{\vr}{\vect{r}}$
$\newcommand{\vnu}{\vect{\nu}}$
The integral form of a PDE usually comes from the physical definition of a quantity, say, for example curl and div. The curl operator $\nabla \times$ and the div operator $\nabla \cdot$ are both defined in a physical way in Physics, which are line integral and surface integral respectively. But for Mathematician, we always view this process in a reverse way: say, the differential equation is already there, and it holds in the sense of weak derivative, multiplying both side by a testing function which is infinitely smooth and vanishes on the boundary $\partial \Omega$(variational formulation); or more intuitively, since the equation holds for every points in the region(classical sense), we integrate both sides in an infinitesimal region(surface or volume), the equation also holds. ie, we usually deduce the integral form from differential form, not the reverse.

For example, this is the differential form of Maxwell's equations describing how electromagnetic waves propagating through a linear isotropic material, where electric field intensity denoted as $\vE = \vE(\vx,t)$ and the magnetic field intensity denoted as $\vH = \vH(\vx,t)$.

$$
\left\{
\begin{aligned}
\frac{\partial (\mu \vH)}{\partial t} + \nabla \times \vE &= 0 \quad &\text{ in } \Omega\times
(0,T)
\\
\frac{\partial(\epsilon \vE)}{\partial t} + \sigma \vE - \nabla\times \vH &=  \vj \quad &\text{ in }
\Omega\times (0,T)
\\
\nabla \cdot (\epsilon \vE) &= q \quad &\text{ in } \Omega\times (0,T)
\\
\nabla \cdot (\mu \vH) &= 0 \quad &\text{ in } \Omega\times (0,T)
\end{aligned}
\right.
$$

For example, we would like to get the integral form for the Faraday's law of the electric field $\vE$, integrate the first equation on any surface $S$ w.r.t. its surface measure $d S$:
$$
- \int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S= \int_S (\nabla \times \vE)\cdot \vnu \,d S = \oint_{\partial S} \vE \cdot d\vr
$$

where

$\vnu$ is the unit outward normal vector to the surface $S$.

if you parametrize $\partial S$ as $\vr(t)$, then $d\vr = \vr'(t) dt$, where $\vr'(t)$ essentially is the tangential vector of the curve $\partial S$ at every point.

For the left side of that equation, if we assume the enough continuity of $\mu \vH$ w.r.t $t$ and $\vx$(which is implied from the divergence free condition at every $t$), then we have
$$
\int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S = \frac{\partial}{\partial t} \int_{S}\mu \vH \cdot \vnu\,d S
$$
since $\vnu$ is the outward normal vector to $S$, you could think the surface integral as "how much the magnetic field is flowing along the outer normal direction of the surface", which we called this quantity magnetic flux.

This is basically the derivation from differential form to integral form. The reverse is similar, just the integral equations you have at first are all physical laws, for example Faraday's Law from wikipedia:

The induced electromotive force (EMF) in any closed circuit is equal
to the time rate of change of the magnetic flux through the circuit.

Then simply write this out using definition which is integral form, apply Stokes theorem in the reverse way, then you will end up with two equal integrals, for eg, use above example, you have
$$
- \int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S= \int_S (\nabla \times \vE)\cdot \vnu \,d S
$$
for any surface $S$, then you argue that the arbitrariness of $S$ implies
$$
-\frac{\partial (\mu \vH)}{\partial t} = \nabla \times \vE
$$
and there you go.

Here mostly I state my own intuitive understanding of this subject while I was learning Physics as a mathematics student, will appreciate if anybody points out where I was wrong.


问:
space of riemann integrable functions not complete
Define norm as $\int |f|$ (Riemann integral) on  $\mathcal R^1[0,1]$, the space of riemann integrable functions on $[0,1]$ with identification $f=g$ iff $\int |f-g|=0$.

Let $\{ r_1,r_2,\cdots \}$ be the rationals in $[0,1]$, and let $f_n=1_{\{r_1,\cdots,r_n\}}$. Then $f_n$ is a cauchy sequence in $\mathcal R^1[0,1]$. I want to show that there is no $f\in \mathcal R^1[0,1]$ such that $f_n$ converges to $f$ in norm. How can I show it?

Obviously the pointwise limit $f=1_\mathbb{Q}$ is not contained in $\mathcal R^1[0,1]$, but can I use this fact? I think that there can be other candidates, since convergence in norm and pointwise convergence are different.

答:
Recall the condition that $f=g$ if and only if $\int|f-g| = 0$.  This means that elements of $\mathcal R^1$ are not functions in the classical sense, because they're only defined up to sets of measure $0$.  You can't evaluate $f(x)$, because every pair $(x,y)$ there's some $g\in\mathcal R^1$ with $g=f$ but $g(x)=y$.  We just change the value of $f$ at a single point.

So in your example we have $f_n = 0$ for every $n$.

Consider instead the functions

$$g_n(x) = \min\left(n,-\log x\right)$$.

Now each $g$ is Riemann integrable, and it's easy to see that the sequence is Cauchy $\int|g_n - g_m|\leq \int|g_n| - 1$ for $m>n$.

But there is no limit in $\mathcal R^1$. If there is a limit it must be $x\mapsto-\log(x)$ (almost everywhere), but that isn't Riemann itegrable because it's unbounded.


问:
How to deal with quotient of polynomial ring in $K[x,y]$
In the case of the quotient in the polynomial ring of one variable, there are very good answers here  How to deal with polynomial quotient rings

I am asking how we can compute the quotient ring of $K[x,y]$ ( $K$ is a field )
In the following  cases :

$K[x,y]/(f) $  where $f=f_1^{k_1} \cdot \dotsc \cdot f_n^{k_n}$ with irreducible $f_i$,

do we still have this property $K[x,y]/(f) \cong \prod_{i=1}^{n} K[x,y]/(f_i^{k_i})$?

$K[x,y]/(f) $ where $f$ is irreducible in $K[x,y]$

For example what is the quotient ring of $\mathbb{C}[x,y]/(g) $ where $g=x^{p-1}+x^{p-2}+....+x+1-y$ i think that it may be isomoprphic to $\mathbb{C}[x,x^{p-1}+x^{p-2}+....+x+1]=\mathbb{C}[x]$
am i right?

答:
The answer to question $1$ is no. Consider for example the ring $F[x,y]\big/(xy)$, for any field $F$. Then the natural map to $\left[F[x,y]\big/(x)\right]\times\left[F[x,y]\big/(y)\right]$ is not an isomorphism, as it does not take any element of $F[x,y]\big/(xy)$ to $(\overline{0},\overline{1})$ (can you see why?). However, the result does hold in the one-variable case, since $F[x]$ is a PID and hence a Bézout domain.

(Edit: To answer your question below, one can completely characterize the polynomials $f=f_1^{k_1}\dots f_n^{k_n}$ such that $F[x,y]\big/(f)\cong\prod_{i=1}^n F[x,y]\big/(f_i^{k_i})$, where the $f_i$ are non-associates and irreducible, as follows: the isomorphism holds if and only if $(f_i)+(f_j)=R$ for every $i\neq j$. For one direction, we may use the Chinese Remainder Theorem; since $F[x,y]$ is a UFD, $(f)=\bigcap_{i=1}^n(f_i^{k_i})$, and, if $(f_i)+(f_j)=R$ for every $i\neq j$, we have that the $(f_i^{k_i})$ are pairwise coprime, as needed. For the other direction, suppose $(f_i)+(f_j)\neq R$. Then no element of $F[x,y]\big/(f)$ maps to the element of $\prod_{i=1}^n F[x,y]\big/(f_i^{k_i})$ with a $\overline{1}$ in the $j$-th position and $\overline{0}$ everywhere else. Indeed, if $p$ were such an element, $p$ would be divisible by $f_i$ and congruent to $1$ mod $f_j^{k_j}$, and hence we would have $1\in (f_i)+(f_j)$, a contradiction. Thus $f$ induces the desired isomorphism if and only if $(f_i)+(f_j)=R$ for every $i\neq j$, and this gives a complete characterization.)

The answer to question $2$ depends very much on the polynomial $f$, and it's hard to say anything specific without more details. What kind of results are you looking for? For instance, we know that the ring $F[x,y]\big/(f)$ will be an integral domain of Krull dimension $1$, but I'm not sure if this is the kind of thing you're looking for.

The answer to question $3$ is yes. Indeed, we can say something more general; let $F$ be any field, and let $p\in F[x]$ be any polynomial, and consider the map $f:F[x,y]\to F[x]$ defined by $x\mapsto x$ and $y\mapsto p$. Clearly $f$ is surjective, and we claim $\ker f=(p-y)$. The $\supseteq$ inclusion is clear, and for the $\subseteq$ inclusion suppose $f(q)=0$ for some $q\in F[x,y]$. Then in particular $q(x,p)=0\in F[x]\subseteq F[x,y]$. On the other hand, replacing every instance of $\overline{y}$ in $\overline{q(x,y)}$ by $\overline{p}=\overline{y}$ gives $$\overline{q(x,y)}=\overline{q(x,p)}=\overline{0}\in F[x,y]\big/(p-y),$$ and so $q(x,y)\in (p-y)$, as desired. Thus, by the first isomorphism theorem, $f$ induces the desired isomorphism $F[x,y]\big/(p-y)\cong F[x]$.


问:
Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$
Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$

Using the Chinese Remainder Theorem, I obtained $x^2 \equiv 23 \pmod{7}$ and $x^2 \equiv 23 \pmod{11}$.

For $x^2 \equiv 23 \pmod{7}$, the answers are:
\begin{align}
x \equiv 3 \pmod{7} \\
x \equiv 4 \pmod{7}
\end{align}

For $x^2 \equiv 23 \pmod{11}$, the answers are:
\begin{align}
x \equiv 1 \pmod{11} \\
x \equiv 10 \pmod{11}
\end{align}

But the solutions for $x^2 \equiv 23 \pmod{77}$ are:
\begin{align}
x \equiv 10 \pmod{77} \\
x \equiv 32 \pmod{77} \\
x \equiv 45 \pmod{77} \\
x \equiv 67 \pmod{77}
\end{align}

Why are my answers not the same as the solutions? Can someone help me with this? Thanks!

答: To get the given solution, combine congruences in your solution with the Chinese remainder theorem: $3\bmod7\land1\bmod11\iff45\mod77$, etc.


问:
Pictorial representation of a tangent plane to a right circular cone
Can anyone pictorially show me of a tangent plane to a right circular double cone? Since I think it doesn't have a tangent plane through the origin (correct me if I am wrong), I am not able to visualize it.</p>

答: It does not take long to find one on the Web.


问:
calculate $\lim\limits_{x \to 1}(1 - x)\tan \frac{\pi x}{2}$
I need to calculate $$\lim_{x \to 1}\left((1 - x)\tan \frac{\pi x}{2}\right)$$.

I used MacLaurin for $\tan$ and got $\frac{\pi x} {2} + o(x)$. Then the full expression comes to $$\lim_{x \to 1}\left(\frac {\pi x} {2} - \frac {\pi x^2} {2} + o(x)\right) = 0$$But WolframAlpha says it should be $\frac 2 \pi$. What am I doing wrong?

答:
Or we can immediately simplify things:

$$\begin{aligned}\lim_{x\to 1}(1-x)\tan\left(\frac\pi2 x\right)&=\lim_{x\to 1}\frac{1-x}{\cot\left(\frac\pi2 x\right)}\\&=\lim_{x\to 1}\frac{1-x}{\tan\left(\frac\pi2(1-x)\right)}\\&=\frac2\pi\lim_{x\to 1}\frac{\frac\pi2(1-x)}{\tan\left(\frac\pi2(1-x)\right)}\\&=\frac2\pi\end{aligned}$$

relying on the manual limit $\lim\limits_{x\to 0}\frac{\tan x}x=1$.


问:
Prove that the image of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .
let $G$ and $H$ be groups and let $\varphi : G \to H$ be  a homomorphism.Prove that the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .

My attempt : To  show the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H $

It is given that $\varphi :G \to H$ is  a homomorphism and  $\varphi (G)$ denote the image of $\varphi$.

Now we will used the lagrange theorem's  because  $\varphi$ is homomorphism

let $ y\in \varphi(G)$ and  $g\in G$.Then  $y^{-1} = \varphi(g)^{-1}=\varphi(g^{-1})\in \varphi (G)$

$y^{-1}= \varphi(g^{-1}) $ This implies that  $\varphi(G)$ is satisfying  the lagrange theorem

Therefore $\varphi (G)$ is a  subgroup of $H$

答: When you write "let $y\in\varphi(G)$ and $g\in G$," you have taken arbitrary elements in both the image of $\varphi$ and in $G$. What you want is to start with an arbitrary $y\in\varphi(G)$, which by definition is equal to $\varphi(x)$ for some $x\in G$. This is an important distinction that is easy to miss. Then proceed as you were: $\varphi(x^{-1})\varphi(y)=\varphi(x^{-1}y)=\varphi(1)=1$, and similarly $\varphi(y)\varphi(x^{-1})=1$, which shows $\varphi(x^{-1})=y^{-1}$.


答:
You can set up the integral in different orders. A few of the examples are given below. But for some clarity on the volume bound between the plane and the paraboloid, they intersect only between $-3 \leq x \leq 1$. The two extreme points of intersection are $(1, 1, 0)$ and $(-3, -1, 0)$. In between, as you rightly pointed out, the intersection points are on $y^2 + z^2 = 1$

Integral setup -

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_{2y-1}^{2y-y^2-z^2} dx \, dz \, dy$

or

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{2y-1}^{2y-y^2-z^2} dx \, dy \, dz$

In cylindrical coordinates -

$\displaystyle \int_{0}^{2\pi} \int_{0}^{1} \int_{2r\sin \theta - 1}^{2r \sin \theta - r^2} r \, dx \, dr \, d\theta$

Title: Volume of body bounded with surfaces

Body:
I need to find volume bounded with $x=2y-y^2-z^2$ and $2y-x-1=0$. So first one is paraboloid that is located on x-axis and its "maximum" is in $(0,2,0)$ and the second is plane paralele to z-axis.
I just want to check do we get the volume when substract volume of $2y-x-1=0$ from paraboloid.(Also using polar coordinates it is easy to get $0\le r\le 1, 0\le \phi \le 2\pi$). I'm just not sure whether I sketched this correctly, precisely is given plane above paraboloid?


问:
Differentiability of a function with absolute value and sign function
I have the following function:
$$|\sin{x}|\cdot sgn{x}$$ where $$ -\pi<x<\pi$$

I have a few questions. Is $ |\sin{x}|$ the same as $\sin{x}$ for $ -\pi<x<\pi$? I am asking this because when I entered both functions on a plotter (Desmos), the functions overlapped completely for $ -\pi<x<\pi$. If yes, it would be much easier as I would have to test only for $x=0$. If no, can someone please help me how to decompose the function into simpler terms, because we're dealing with absolute values and the sign function?

答:
For every $x\in\mathbb{R}$, $x\neq 0$, you have
$$
x=\frac{x}{|x|}\cdot |x|.
$$
Therefore if you define the sign function as $\mathrm{sign}(x):=x/|x|$ if $x\neq 0$, and whatever you prefer for $x=0$, e.g., $\mathrm{sign}(0)=1/2$, you always have that $x=|x|\cdot\mathrm{sign}(x)$. In particular,
$$
\sin(x)=|\sin(x)|\cdot \mathrm{sign}(\sin(x))\, .
$$
After that, if $g$ is any other function that has the same sign of $\sin(x)$, you get
$$
\sin(x)=|\sin(x)|\cdot \mathrm{sign}(\sin(x))=|\sin(x)|\cdot \mathrm{sign}(g(x)).
$$
For example, $\sin(x)$ and $x$ have the same sign in the interval you are considering. Hence,
$$\sin(x)=|\sin(x)|\cdot \mathrm{sign}(x).$$
Hope this helps.


问:
Dimension of the $\mathbb K$ vector space over the $\mathbb K$ field.
I am unable to find justifications for saying the dimension of the $\mathbb K$ vector space.

I know that every $\mathbb K$ field can be considered a vector space over itself, but how dimension this space be? Something tells me it's $1$, but I can't justify it consistently. Can someone help me?

答: The dimension of a $k$-vector space is the cardinality of any basis. For finite dimensional vector spaces, we also write this as an integer. So all you have to do is note that $\{1\}$ is a basis for $k$ over itself. (Note that $\{x\}$ is also a basis for any $0\neq x\in k$.)


问:
How can I solve this Laws of Sines problem?
This is a homework question that was set by my teacher, but it's to see the topic our class should go over in revision, etc.

I have calculated $AB$ to be 5.26m for part (a). I simply used the law of cosines and plugged in the numbers.

part (b) is the question I have been trying for quite a while. I tried to do the law of sines, to no avail. To calculate $BC$ I need the angle opposite that which I do not have (or know how to work out.) The triangle $BDC$ has a right-angle, but this does not work as $sine(90) = 1$

A step in the right direction would be more beneficial than a full answer.

答:
This might help:

${AD}\cdot{\sin(\angle DAC)}=
CD=\displaystyle{BD}\cdot{\sin(\angle DBC)}$


答:
Let $\dim_{\mathbb{K}}V$  denote the dimension of $V$ as a vector space over $\mathbb{K}$. We will prove that $\dim_{\mathbb{K}}\mathbb{K}= 1$ by finding a basis for $\mathbb{K}$ over the field $\mathbb{K}$ with one element.

Let $\mathcal{B}=\{1\}$ where $1 \in \mathbb{K}$ is the multiplicative identity of the field $\mathbb{K}$ (we could choose any other non-zero element of $\mathbb{K}$, but $1$ makes the proof a little simpler). First, note if $a \cdot 1 = 0$ then $a = 0$. Therefore, $\mathcal{B}$ is linearly independent. Next, we show that any $b \in \mathbb{K}$ can be written as a linear combination of elements of $\mathcal{B}$ with coefficients in $\mathbb{K}$. Indeed, $b = b \cdot 1$.

Therefore, $\mathcal{B}$ is a basis and $\dim_{\mathbb{K}}\mathbb{K}= 1$.

Title: Dimension of the $\mathbb K$ vector space over the $\mathbb K$ field.

Body:
I am unable to find justifications for saying the dimension of the $\mathbb K$ vector space.

I know that every $\mathbb K$ field can be considered a vector space over itself, but how dimension this space be? Something tells me it's $1$, but I can't justify it consistently. Can someone help me?


问:
Proof that it is a ring - new operations
Let {K,+,*} be commutative ring with unit. Define new operations ¤ and o so x¤y=x+y-1 and xoy=x+y-xy. Show that {K,¤,o} is a ring.

How can I show that the opeartions are closed for K? How can I show that x+y-1 belongs to K and x+y-xy belongs to K?

I manged to prove the other steps.

答:
Use the fact that $+$ and $\cdot$ are already closed over $K.$

We begin by showing $x,y\in K$ implies $x+y-1\in K.$ Namely, if $x,y\in K,$ then we know $x+y\in K.$ Then $-1\in K$ as well, so $(x+y)+(-1)\in K$ as well.

Nowe we show $x,y\in K$ implies $x+y-xy\in K.$ Again, $x,y\in K$ implies $x+y\in K$ and $x\cdot y\in K,$ so $-x\cdot y\in K$ as well, so $(x+y)+(-xy)\in K.$


问:
Given function π, prove that seg(x) is a well ordered set
(ZF) Let $\langle A,\le \rangle$ be a linear ordered set. Let the function $π : \mathscr{P}(A) \to \mathscr{P}(A)$ be defined by:

$$π(X) = \{y ∈ A : \mathrm{seg}(y) \subseteq X\}$$

Prove that $π$ is monotonic and if that $A^*$ is the smallest fixed point of  $π$, then for each $x ∈ A$ the following equivalence holds:

$$x \in A^* \iff \langle \mathrm{seg}(x), {\le} \cap(\mathrm{seg}(x)×\mathrm{seg}(x))\rangle \text{ is a well ordered set}$$

What I have in mind is the following:

To use Tarsky's theorem about fixed points which states that: if π is a monotonic function, defined in P(A),  there exists a point x in P(A) for which π(x)=x.
I will name this point A.
Then if I consider π to be a linear operator, I think that I will have to prove that π preserves the order, namely: π(x)<π(y) for x<y. So in this way I will get π to be an increasing function with lower bound π(A)=A.

Is it correct, can you please help me to solve this problem?
Thank you in advance!

答:
To use Tarsky's theorem about fixed points which states that: if π is a monotonic function, defined in P(A), there exists a point x in P(A) for which π(x)=x. I will name this point A.

The problem already gives you $A^*$. It is unclear why you want to introduce another fixed point "A".

Then if I consider π to be a linear operator,

What is a linear operator (in this context)? I usually encounter this when there is a vector space.

I think that I will have to prove that π preserves the order, namely: π(x)<π(y) for x<y. So in this way I will get π to be an increasing function with lower bound π(A)=A.

Let me break down your task for this assignment.

Prove $\pi$ is order-preserving/monotonic i.e. for all $S,T\in\mathscr{P}(A)$, if $S\subseteq T$, then $\pi(S)\subseteq \pi(T)$.

Prove the equivalence "$\iff$" statement assuming $A^*$ is already given. From (1.) and by Tarski's theorem, one already knows $A^*$ exists.

$$x \in A^* \iff \langle \mathrm{seg}(x), {\le} \cap(\mathrm{seg}(x)×\mathrm{seg}(x))\rangle \text{ is a well ordered set}$$

Hints:

$\pi(S)=\{y\in A:\mathrm{seg}(y)\subseteq S\}$ and $\pi(T)=\{y\in A:\mathrm{seg}(y)\subseteq T\}$

$x\in A^*\Leftrightarrow x\in \pi(A^*)$.
For the forward implication, it is enough to show that the sublinear ordered set $\langle A^*, \le... \rangle$ is well ordered. Suppose it is not; by contradiction take a nonempty subset $S$ without a least element. Without loss of generality suppose $S$ is upward closed i.e. if $s\in S$ and $s\le x$ then $x\in S$. Consider the complement $A^*\setminus S$. It can be shown that this complement is a $\pi$-fixed-point resulting in a contradiction to the fact that $A^*$ is the least fixed point.
For the reverse implication, it is enough to show $\mathrm{seg}(x)\subseteq A^*$.
This is straight-forward. Suppose by contradiction it is not true. Then there exists $s<x$ such that $s\notin A^*$. If you take the smallest such $s$, then the contradiction is within reach if you recall $\pi(A^*)=A^*$.


问:
Counter-example to the extended dominated convergence theorem
To add some context, let $(\Omega,\mathcal{F},\mu)$ be a measure space. The statement of the Extended Dominated Convergence Theorem(EDCT) is:

$\lbrace f_{n}\rbrace_{n\geq 1}$ be a sequence of functions from $L^{1}(\Omega,\mathcal{F},\mu)$ such that $f_{n}\to f$ a.e.($\mu$) and there exists a sequence $\lbrace g_{n}\rbrace_{n\geq 1}\in L^{1}(\Omega,\mathcal{F},\mu)$ such that $g_{n}\to g$ a.e.($\mu$), $|f_{n}|\leq g_{n}$ a.e.($\mu$) for all $n\geq 1$ and $\int g_{n}d\mu\to\int gd\mu$. Then, (i) $f\in L^{1}(\Omega,\mathcal{F},\mu)$ (ii) $f_{n}\xrightarrow{L^{1}} f$ (iii) $\int f_{n}d\mu \to \int fd\mu$

I'm stuck while trying to find a counterexample to the above theorem such that, $\lbrace f_{n}\rbrace_{n\geq 1}, f\in L^{1}(\Omega,\mathcal{F},\mu), f_{n}\to f$ a.e.($\mu$) and $\int f_{n}d\mu\to\int f d\mu$ but $||f_{n}-f||_{L^{1}(\Omega,\mathcal{F},\mu)}\not\to 0$.

If the sequence of functions is non-negative, we can see $|f_{n}-f|\leq f_{n}+f=g_{n}$. Also, $\int g_{n}d\mu=\int f_{n}d\mu+\int fd\mu\to2\int fd\mu$. So applying the EDCT we cannot find a counter-example if the the sequence of functions is non-negative.

So is there any such example that shows that a sequence $g_{n}$ that dominates $f_{n}$ is necessary for convergence in $L^{1}$-norm. Thanks in advance for any help.

答:
It's not clear to me whether you're saying EDCT is actually true; on the one hand your question says you're looking for a counterexample, but then the "such that" makes it unclear whether you're looking for a counterexample to EDCT.

In any case, EDCT (as stated above!) is clearly false. It has the following corollary:

Assume EDCT. Suppose $\lbrace f_{n}\rbrace_{n\geq 1}$ is a sequence of functions from $L^{1}(\Omega,\mathcal{F},\mu)$ such that $f_{n}\to f$ a.e.($\mu$). Then, (i) $f\in L^{1}(\Omega,\mathcal{F},\mu)$ (ii) $f_{n}\xrightarrow{L^{1}} f$ (iii) $\int f_{n}d\mu \to \int fd\mu$.

You must know counterexamples to that. But

Proof. Let $g_n=|f_n|$, $g=|f|$. Then $g_n\in L^1$, $g_n\ge|f_n|$ and $g_n\to g$ almost everywhere. So EDCT implies (i), (ii) and (iii).

Note that if you add the hypothesis $\int g_n\to\int g\ge0$ then EDCT is a standard result (also true).


问:
Deductions about openness and closedness of sets
I would like someone to verify my solution attempt of this exercise on open and closed sets, from Stephen Abbott's Understanding Analysis.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
Exercise 3.2.3

Decide whether the following sets are open, closed or neither. If a set is not open, find a point in the set for which there is no $\epsilon$-neighbourhood contained in the set. If a set is not closed, find a limit point that is not contained in the set.

(a) $\mathbf{Q}$

(b) $\mathbf{N}$

(c) $\{x \in \mathbf{R}: x \ne 0\}$

(d) $\{1 + 1/4 + 1/9 + \ldots + 1/n^2 : n \in \mathbf{N}\}$

(e) $\{1 + 1/2 + 1/3 + \ldots + 1/n : n \in \mathbf{N}\}$

Solution.

(a) Given any rational number $q$, and an $\epsilon > 0$, there exists an irrational number $x$, such that $\absval{x - q} < \epsilon$, so $V_\epsilon(x) \not\subseteq \mathbf{Q}$. Therefore $\mathbf{Q}$ is not open. For instance, if $q = \frac{m}{n}$, $\epsilon=\frac{1}{n}$ where $m,n \in \mathbf{Z}$ and if $x = \frac{m + 1}{\sqrt{2}n}$, $\absval{x - q} < \epsilon$. And $V_\epsilon(q) \not\subseteq Q$.

The set of all limit points of $\mathbf{Q}$ is $\mathbf{R}$. The irrational numbers $\mathbf{I}$ are not members of the set of rational numbers $\mathbf{Q}$. So, $\mathbf{Q}$ is not closed. For example, $x = \sqrt{2}$ is a limit of point of $\mathbf{Q}$, since every $\epsilon$-neighbourhood of $\sqrt{2}$, $(\sqrt{2} - \epsilon,\sqrt{2} + \epsilon)$ intersects $\mathbf{Q}$ at some point other than $\sqrt{2}$.

(b) Given any natural number $n$, and $\epsilon > 0$, there exists a rational number $p/q$, $q \notin \{0,1\}$ such that $\absval{\frac{p}{q} - n} < \epsilon$. Thus, $V_\epsilon(n) \not\subseteq \mathbf{N}$. Therefore, $\mathbf{N}$ is not open. For instance, if $n = 0, \epsilon = 1/2$, then $\absval{1/4 - n} < \epsilon$.

The members of $\mathbf{N}$, $\{0,1,2,3,\ldots\}$ are all isolated points, because if $\epsilon < 1$, then $V_\epsilon(0) \cap \mathbf{N} = \{0\}, V_\epsilon(1) \cap \mathbf{N} = \{1\}, V_\epsilon(2) \cap \mathbf{N} = \{2\}, \ldots$. Thus, there are no limit points in $\mathbf{N}$. Consequently, $\mathbf{N}$ is not closed.

(c) $\{x \in \mathbf{R}:x \ne 0\}$. For all $x \in \mathbf{R} - \{0\}$, there exists an $\epsilon$-neighbourhood $V_\epsilon(x)$, such that $V_\epsilon(x) \subseteq \mathbf{R} - \{0\}$. Thus, $\mathbf{R} - \{0\}$ is an open set.

The point $x = 0$ is a limit point of the set $\mathbf{R} - \{0\}$, since the sequence $a_n = \frac{1}{n}$ converges to $0$ as $n \to\infty$ and $a_n \ne x$. Moreover, $0$ is not element of the set $\mathbf{R} - \{0\}$, so $\mathbf{R} - \{0\}$ is not a closed set. This is the only defect in $\mathbf{R} - \{0\}$.

(d) This set contains the partial sums of the infinite series $\sum \frac{1}{n^2}$. Writing out the first few terms, $S = \{s_1,s_2,s_3,s_4,\ldots\} = \{1,5/4,49/36,\ldots,\}$. Given $s_n$, let $\epsilon = s_n - s_{n-1}$, $V_\epsilon(s_{n}) \not\subseteq S$. It is not an open set.

We know, that $(s_n)$ is a monotonic increasing sequence with an upper bound $2$, and therefore convergent. Since $S$ contains the terms of the sequence of the partial sums alone and $s_n \ne \lim s_n$ for all $n$, the limit point of $S$ is not an element of $S$. Consequently, it is not a closed set.

(e) This set contains the partial sums of the harmonic series $\sum \frac{1}{n}$. Writing out the first few terms,  $S = \{s_1,s_2,s_3,s_4,\ldots\} = \{1,3/2,11/6,\ldots,\}$. Again, as before $\exists x$ such that $V_\epsilon(x) \not\subseteq S$, $\forall \epsilon > 0$. So, $S$ is not an open set.

The partial sums $s_n$ are isolated points of the set $S$. Given $s_n$, there exists an $V_\epsilon(s_n)$, such that $V_\epsilon(s_n) \cap S = \{s_n\}$. Since $(s_n)$ is an unbounded sequence; it is divergent and does not have any limit point. So, $S$ is not a closed set.

答:
Your solution is almost perfect.

In b) and e) however, you arrive that the set has no limit points. But then the conclusion is that it's closed, since the set of limit points is the empty set which is a subset of every set.


问:
Reason for this sequence being exact?
In Altman-Kleinman "Introduction to Grothendieck Duality Theory", page 105, there is the following theorem:

Why the underlined sequence is exact? And then, why from that we can conclude that $coker(\delta)$ is the kahler differential?

答:
By definition, $\text{coker }\delta$ fits into the exact sequence $$I/I^2 \overset{\delta}\longrightarrow \Omega^1_{A/k}\otimes_A B \longrightarrow \text{coker }\delta \longrightarrow 0$$ and so by left-exactness of the contravariant functor $\text{Hom}_B(-,M)$ for a $B$-module $M$, we have an exact sequence $$0\longrightarrow \text{Hom}_B(\text{coker }\delta, M)\longrightarrow \text{Hom}_B(\Omega^1_{A/k}\otimes_A B, M)\longrightarrow \text{Hom}_B(I/I^2,M)$$ Now $M$ also has the structure of an $A$-module, via the quotient map $A\to B$. By tensor-Hom adjunction, we have $$\text{Hom}_B(\Omega^1_{A/k}\otimes_A B, M)\cong \text{Hom}_A(\Omega^1_{A/k},\text{Hom}_B(B,M)) \cong \text{Hom}_A(\Omega^1_{A/k},M)$$ (you could also convince yourself of this isomorphism directly). Since $\text{Hom}_A(\Omega^1_{A/k},M) = \text{Der}_k(A,M)$ (by the universal property defining $\Omega^1_{A/k}$ in terms of derivations), this gives you the desired exact sequence.

There is now an isomorphism $$\text{Hom}_B(\text{coker }\delta,M) \cong \text{Der}_k(B,M) \cong \text{Hom}_B(\Omega^1_{B/k},M)$$ for all $B$-modules $M$, functorial in $M$ (check this). The Yoneda Lemma then states that $\text{coker }\delta$ and $\Omega^1_{B/k}$ must be isomorphic as $B$-modules.

Title: Reason for this sequence being exact?


问:
Biholomorphic functions on the whole complex plane
The question is the following: are there some other biholomorphic functions on the whole complex plane except to polynomials of the degree 1? I don't think so, but i cannot find the proof.
(I mean functions that are bijektiv and holomorphich with a holomorphich inverse on the whole plane.)
Has somebody an idea? Thanks.</p>

答:
Show that $z/f(1/z)$ is holomorphic around $z=0$, thus it is analytic, with a zero of order $k$ at $z=0$, which implies that $f(1/z)$ has a pole of order $k-1$ at $z=0$ so that $f(1/z)-P(1/z)$ is analytic at $z=0$ for some polynomial.

Both $f(z)-P(z)$ and $f(1/z)-P(1/z)$ are analytic thus bounded for $|z|\le 2$.

Let $z_0\in \Bbb{C}$ maximizing $|f(z)-P(z)|$. The maximum modulus principle implies that  $f(z)-P(z)$ is constant near $z_0$, thus everywhere.

So $f(z)=P(z)+C$ and for being injective we need $\deg(P)=1$.


问:
Evaluate $\vec{F}=z\vec{i}+x\overrightarrow{j}-{{y}^2}z\vec{k}$ and $S$ is the surface $x^2+y^2=1$ in the first octant between $z=0$ and $z=2$.
Problem:

Evaluate $\int \int_S \overrightarrow{F}.\widehat{n}$   where $\overrightarrow{F}=z\overrightarrow{i}+x\overrightarrow{j}-{{y}^2}z\overrightarrow{k}$ and $S$ is the surface of the cylinder $x^2+y^2=1$ in the first octant between $z=0$ and $z=2$.

My attempt:

$\int \int_S \overrightarrow{F}.\widehat{n}$ = $\int \int \int_V \Delta.\overrightarrow{F}\,dz\,dy\,dx$

= $\int_{0}^1 \int_{0}^\sqrt{1-x^2} \int_{0}^2 -y^2\,dz\,dy\,dx$

= $\int_{0}^1 \int_{0}^\sqrt{1-x^2} -2y^2dy\,dx$

=$\int_{0}^1\frac{-2(1-x^2)^{\frac{3}{2}}}{3}\,dx$

Let $x=sin\,t$ and then $t=sin^{-1}x$

=$\frac{-2}{3}\int_{0}^{\frac{\pi}{2}}cos^4 \,t \,dt$

=$\frac{-2}{3}\int_{0}^{\frac{\pi}{2}}(\frac{cos4t}{8} + \frac{cos4t}{2}+\frac{3}{8})\,dt$

=$\frac{-2}{3}[\frac{sin4t}{32}+\frac{sin2t}{4}+\frac{3t}{8}]_{0}^{\frac{\pi}{2}}$

=$\frac{-\pi}{8}$

But the given result is $3$.

Thanks in Advance

答: You cannot use the divergence theorem here because your surface $S$ is not closed. In your calculation you're finding the flux of $\vec{F}$ through the boundary of the solid $E$ defined by $$E=\big\{x^2+y^2 < 1\big\}\cap\big\{0 < z < 2\big\}\cap (0,\infty)^3$$ The boundary of $E$ includes the top, bottom, and sides of the surface $S$ in your problem, which you should notice is a closed surface. You're also assuming that the normal vector $\vec{n}$ is outward pointing (which is what I will assume in my subsequent calculations.) In order to find the flux of $\vec{F}$ through $S$ in your problem, define $$\vec{r}(u,v)=\Big<\cos(u),\sin(u),v\Big>$$ on the domain $0<u<\pi/2$ and $0<v<2$. With a few calculations we get $$\vec{r}_u \times \vec{r}_v=\Big<\cos(u),\sin(u),0\Big>$$ $$\vec{F}\Big(\vec{r}(u,v)\Big)=\Big<v,\cos(u),-v\sin^2(u)\Big>$$ $$\vec{n}=\frac{\vec{r}_u\times \vec{r}_v}{||\vec{r}_u\times \vec{r}_v||}=\Big<\cos(u),\sin(u),0\Big>$$ $$dS=||\vec{r}_u\times \vec{r}_v||dudv=dudv$$ Hence $$\int \int _{S}\vec{F}\cdot \vec{n}dS=\int_0^2 \int _{0}^{\pi/2}\Big[v\cos(u)+\sin(u)\cos(u)\Big]dudv=3$$


问:
What is the meaning of writing the differential inside of a function?
I am reading through Resnick's "Extreme Values, Regular Variation and Point Processes" and have come across some notation that I am not familiar with. In talking about moving a Poisson point process into higher dimensions, we are introduced to the mean measure function:
\begin{align*}
\mu^*(dx, dy)=\mu(dx) K(x, dy),
\end{align*}

my question here is strictly about notation like: $\mu(dx)$. I know the following notation
\begin{align*}
\int_\Omega f(x)\mu(dx)=\int_\Omega f(x)d\mu(x)=\int_\Omega fd\mu
\end{align*}

and I know that when we write something like
\begin{align*}
\int_\Omega f(X, y)K(X, dy)
\end{align*}

we are freezing $X$ and integrating with respect to $K$, viewed now as a function only of $y$. My question here is, what is meant when we write the differential inside a function, like $\mu(dx)$, outside of the integral?

答:
This can be a quick way of defining a measure.  For example, if $\nu$ is the measure defined by $\nu(A) = \int_{A} f(x) \, \mu(dx)$ (where $\mu$ is some other measure and $f$ is some nice enough function), I could instead simply say "Define $\nu$ by $\nu(dx) = f(x) \, \mu(dx)$."  You might also see this written as "Define $\nu$ by $d \nu = f\, d \mu$."

You might ask: is this really quicker than writing $\nu(A) = \cdots$?  I don't think so, but what can I say...  I guess in geometric measure theory it's quicker to write $\mu = f \, d\mathcal{H}^{d-1} \restriction_{\partial \Omega}$, for instance, than "Define $\mu$ by $\mu(A) = \int_{A \cap \partial \Omega} f \, d\mathcal{H}^{d-1}$."  Also, if I wanted to describe the Lebesgue decomposition of a measure, it's very quick to write $\nu = f \, d \mu + \nu^{s}$ rather than some alternative.

The same idea seems to apply to product measures.  The measure $\mu^{*}$ you mentioned I assume is understood to be the one given by
\begin{equation*}
\mu^{*}(A \times B) = \int_{A} K(x,B) \mu(dx).
\end{equation*}
You can see how writing $\mu^{*}(dx, dy) = \mu(dx) K(x,dy)$ is quicker way to convey the same information in this case.  (I would write "$\mu^{*}(dx \otimes dy) = K(x,dy) \mu(dx)$" at the risk of watching my audiences' eyes roll out of their sockets...)  Of course, the simplest example of how to do this in two variables would be $\nu(dx \otimes dy) = \mu_{1}(dx) \mu_{2}(dy)$ for the product $\nu = \mu_{1} \otimes \mu_{2}$ (or $\nu = \mu_{1} \times \mu_{2}$).


问:
Explosion rule in paraconsistent logic
In classical logic the rule of noncontradiction states $\neg(A\wedge\neg A)$
In the Paraconsistent logic, this rule doesn't apply, at least in some case. Now, in the classical logic, if we avoid to state this rule, we can't avoid the 'ex falso sequitur quodlibet', that means:
$$(A\wedge\neg A)\to B$$
So, under what conditions we can accept the noncontradiction rule to be non valid in order to avoid the $(A\wedge\neg A)\to B$? Or, alternatively is there in this logic a rule that allows to accept this 'explosion'?</p>

答: Paraconsistent logics are not concerned with invalidating one formula or another, but rather with disallowing the inference $A,\neg A\vdash B$. There are plenty of paraconsistent logics in which $\neg(A\land\neg A)$ is a valid formula, and there are a few paraconsistent logics in which the formula $(A\land\neg A)\to B$ is valid. An early example of the first variety was proposed by Asenjo (Notre Dame Journal of Formal Logic, 1966), and similar examples abound in the literature on 3-valued logics ; an early example of the second variety was proposed by Jaskowski (Studia Societatis Scientiarum Torunensis, 1948), in an apparently incidental kind of approach later to be known as 'non-adjunctive'.

Title: Explosion rule in paraconsistent logic


问:
$(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. Centroid of $\Delta ABC$ lies on $y=3x-4$, then the locus of $D$
$(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. If centroid of $\Delta ABC$ lies on $y=3x-4$, then what is the locus of $D$?

I did try a couple of things, but I honestly have no idea how to approach this. I checked if $(x-1)(y-2)=5$ represented a pair of straight lines, but it doesn't.

I also tried substituting the value of $(x-1)$ from the first equation in the equation of the circle, but the only useful result I got was that the sum of y-coordinates is zero since the equation after substitution does not have a $y^3$ term, only $y^4$, $y^2$, $y$ and constant terms.

Apart from that, I have no other ideas.

Body:
Point D is the intersection of line $y=3x$ (parallel to $y=3x-4$ and through origin) and curves: $(x-1)(3x-2)=5$ and $(x-1)^2+(y)^2=r^2$ .

$\begin{cases} y=3x \\(x-1)(y-2)=5 \\ (x-1)^2+(y+2)^2=r^2 \end{cases}$

$(x-1)(3x-2)=5$

$3x^{2}-5x-3=0$

$\begin{array}{} x_{D}=\frac{5+\sqrt{61}}{6} & y_{D}=\frac{5+\sqrt{61}}{2}  \end{array}$

solving numerically

$(x_{D}-1)^2+(y_{D}+2)^2=r^2$

$r^2=\frac{335+\sqrt{97600}}{9}≈71.93444300402957$

$\left\{ \begin{array}{} (x-1)(y-2)=5 & ⇒ y=\frac{2x+3}{x-1} \\ (x-1)^2+(y+2)^2=r^2  \end{array}  \right\}$

$sol=\left( \begin{array}{left } x_{A}=-6.788336312 & y_{A}=1.358014369 \\ x_{B}=.599099974 & y_{B}=-10.471937333 \\ x_{C}=8.054194725 & y_{C}=2.708798126 \\ x_{D}=2.135041613 & y_{D}=6.405124838 \end{array}   \right)$

checking if the centroid of triangle ABC belongs to the line $y = 3x-4$

$\begin{array}{} \text{Centroid (G)} & x_{G}=\frac{x_{A}+x_{B}+x_{C}}{3} &  y_{G}=\frac{y_{A}+y_{B}+y_{C}}{3} \end{array}$

$G=(0.621652796,-2.135041613)$

$3·x_{G}-4=y_{G}$


答: Let $d$ be the measure of angle $BDC$.  Then you have $BC=88.3 \sin d, AC=BC+5.26=91.2 \sin(d+2.8)$  Expand the sum, plug $BC$ into the last and you should be able to get to something that looks like $e \sin d + f \cos d$, which you can combine.</p>

Title: How can I solve this Laws of Sines problem?

Body:
This is a homework question that was set by my teacher, but it's to see the topic our class should go over in revision, etc.

I have calculated $AB$ to be 5.26m for part (a). I simply used the law of cosines and plugged in the numbers.

part (b) is the question I have been trying for quite a while. I tried to do the law of sines, to no avail. To calculate $BC$ I need the angle opposite that which I do not have (or know how to work out.) The triangle $BDC$ has a right-angle, but this does not work as $sine(90) = 1$

A step in the right direction would be more beneficial than a full answer.


问:
Matrix addition/multiplication with different sizes
I have the following two matrices:

$$A=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ and }B=\begin{pmatrix}1 & 3 &
 2\\-1 & 0 & 2\end{pmatrix}$$

So I have two matrixes with different sizes. Multiple sources tell me that I can't do multiplication or addition with matrix of different sizes. So I'm a bit confused. Can I do it with these? How?

答:
So I have two matrixes with different sizes. Multiple sources tell me that I can't do multiplication or addition with matrix of different sizes.

You can't add matrixes of different sizes as stated by @meshal. Hence, A+B or B+A can't be performed. However, you can multiply them. Here's how:

$$A=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ and }B=\begin{pmatrix}1 & 3 &
 2\\-1 & 0 & 2\end{pmatrix}$$

$$AB=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ * }\begin{pmatrix}1 & 3 &
 2\\-1 & 0 & 2\end{pmatrix}$$

$$AB=\begin{pmatrix}1\cdot \:1+\left(-2\right)\left(-1\right)&1\cdot \:3+\left(-2\right)\cdot \:0&1\cdot \:2+\left(-2\right)\cdot \:2\\ 3\cdot \:1+1\cdot \left(-1\right)&3\cdot \:3+1\cdot \:0&3\cdot \:2+1\cdot \:2\end{pmatrix}$$

$$Finally, AB=\begin{pmatrix}3&3&-2\\ 2&9&8\end{pmatrix}$$


问:
Game Theory against One Random and One Intelligent Player
You are playing a game with two other players. You all get a chance to select a random integer from 1-100 (inclusive) and the player with the lowest number is paid that amount by both of the other players. In the case of a tie for the lowest number, no payment is made. One of the players you play against randomly selects a number from 1-100. The other is intelligent and will play optimally. What would be the best strategy?

For a simplified case, in which you played against only the random player, I calculated that the optimal choice was 33 or 34, in which your expected winnings were 16.83. However, for the three-person game, I am not sure what the strategy would be. I feel like you and the other intelligent player will constantly try to undercut each other, but I don't feel like the equilibrium should be 1 because there is money to be made by taking advantage of the random player.

答:
You have to pick $1$ always. This strategy dominates every other strategy.

If the optimal guy plays $1$ everytime, then you will lose money unless you also play $1$ everytime.

The uniform random guy is the real winner here, because in this equilibrium he'll never need to pay over money!


答:
Counter-example:
Consider the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),m)$,
where $m$ is the usual Lebesgue measure. Let $f_{n}:\mathbb{R}\rightarrow\mathbb{R}$
be defined by $f_{n}(x)=n^{2}1_{(\frac{1}{2n},\frac{1}{n})}(x)-n^{2}1_{(-\frac{1}{n},-\frac{1}{2n})}(x)$.
Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=0$. Clearly
$f_{n}\rightarrow f$ pointwisely. Moreover $\int f_{n}dm=\int dm=0$.
However $||f_{n}-f||_{1}\not\rightarrow0$.

/////////////////////////////////////////////////////////

However, we have:

Let $(X,\mathcal{F},\mu)$ be a measure space. Let $(f_{n})$ be a
sequence of real-valued integrable functions defined on $X$. Let
$f:X\rightarrow\mathbb{R}$ be an integrable function. Suppose that
$f_{n}\rightarrow f$ a.e., then $\lim_{n}\left(||f_{n}||-||f||-||f_{n}-f||\right)=0$,
where $||f||=\int|f|d\mu$ etc...

We go to prove the above assertion. Observe that $|f_{n}-f|\leq|f_{n}|+|f|$,
so $|f_{n}|-|f|-|f_{n}-f|\geq-2|f|.$ On the other hand, $|f_{n}|\leq|f_{n}-f|+|f|$
and hence $|f_{n}|-|f|-|f_{n}-f|\leq0\leq2|f|$. Combining, we have
$\Biggl||f_{n}|-|f|-|f_{n}-f|\Biggr|\leq2|f|$. Clearly, $|f_{n}|-|f|-|f_{n}-f|\rightarrow0$
a.e.. By Dominated Convergence Theorem, we have that $\int\left(|f_{n}|-|f|-|f_{n}-f|\right)d\mu\rightarrow 0$.
That is, $||f_{n}||-||f||-||f_{n}-f||\rightarrow0$.

Remark:

In particular, if $||f_n||\rightarrow ||f||$, then $||f_n-f||\rightarrow 0$

Counter-example does not exist if $f_n$ and $f$ are non-negative.

Title: Counter-example to the extended dominated convergence theorem

Body:
To add some context, let $(\Omega,\mathcal{F},\mu)$ be a measure space. The statement of the Extended Dominated Convergence Theorem(EDCT) is:

$\lbrace f_{n}\rbrace_{n\geq 1}$ be a sequence of functions from $L^{1}(\Omega,\mathcal{F},\mu)$ such that $f_{n}\to f$ a.e.($\mu$) and there exists a sequence $\lbrace g_{n}\rbrace_{n\geq 1}\in L^{1}(\Omega,\mathcal{F},\mu)$ such that $g_{n}\to g$ a.e.($\mu$), $|f_{n}|\leq g_{n}$ a.e.($\mu$) for all $n\geq 1$ and $\int g_{n}d\mu\to\int gd\mu$. Then, (i) $f\in L^{1}(\Omega,\mathcal{F},\mu)$ (ii) $f_{n}\xrightarrow{L^{1}} f$ (iii) $\int f_{n}d\mu \to \int fd\mu$

I'm stuck while trying to find a counterexample to the above theorem such that, $\lbrace f_{n}\rbrace_{n\geq 1}, f\in L^{1}(\Omega,\mathcal{F},\mu), f_{n}\to f$ a.e.($\mu$) and $\int f_{n}d\mu\to\int f d\mu$ but $||f_{n}-f||_{L^{1}(\Omega,\mathcal{F},\mu)}\not\to 0$.

If the sequence of functions is non-negative, we can see $|f_{n}-f|\leq f_{n}+f=g_{n}$. Also, $\int g_{n}d\mu=\int f_{n}d\mu+\int fd\mu\to2\int fd\mu$. So applying the EDCT we cannot find a counter-example if the the sequence of functions is non-negative.

So is there any such example that shows that a sequence $g_{n}$ that dominates $f_{n}$ is necessary for convergence in $L^{1}$-norm. Thanks in advance for any help.


问:
Let $k$ be an integer not divisible by $5$. Show that either $k^2-1$ or $k^2+1$ is divisible by $5$.

Let $k$ be an integer not divisible by $5$. Show that either $k^2-1$ or $k^2+1$ is divisible by $5$.

Looking at this mod $5$ it seems that $k$ can be one of the form $k \equiv 1,2,3,4 \pmod{5}$. This would imply that $k^2 \equiv 1,4,9,16 \equiv 1,4,4,1 \pmod{5} \Longrightarrow k^2-1 \equiv0,3,3,0 \pmod{5}$ so $k^2-1$ is divisible by $5$? Similar reasoning could be used for $k^2+1$, but that wouldn't give any zero remainders so $5$ doesn't divide $k^2+1$ right?

I feel like I'm missing something since it seems that $k^2-1$ could also give a remainder $3$?

Body:
I wouldn't get so "sophisticated"- just do the calcuations.

(Edited so Dietrich Burde won't be annoyed with me!)

If k is not a multiple or 5 we have one and only one of
a)k= 5n+ 1 so $k^2= 25n^2+ 10n+ 1$.

$k^2- 1= 25n^2- 10n= 5(5n^2- 2n)$.

b)k= 5n+ 2 so $k^2= 25n^2+ 20n+ 4$.  $k^2+ 1= 25n^2+ 20n+ 5= 5(5n^2+ 4n+ 1)$.

c)k= 5n+ 3 so $k^2= 25n^2+ 30n+ 9$.  $k^2+ 1= 25N^2+ 30n+ 10= 5(5n^2+ 6n+ 2)$.

d)K= 5N+ 4 so $k^2= 25n^2+ 40n+ 16$. $k^2- 1= 25n^2+ 40N+ 15= 5(5n^2+ 8n+ 3)$.


问:
About existence of extensión of a function.
Let $A,B$ sets and $f:A \to B$ an inyective function. Then exists $C$ a set and $g: C \to B$ biyective function such that $A \subseteq C$ and $g(x)=f(x)$ for all $x \in A$

I know that this question is silly, since that extension function can seem intuitive, but I cannot get a proof of that.

答:
hint

Let
$$D=\{x\in B\;:\;\forall y\in A\;\; f(y)\ne x\}$$

and
$$C=A\cup D$$

If $ D $ is empty, $ f $ is bijective.

Assume $ D\ne \emptyset $ and
define the map $ g : C \to B \; $  by $ :$

for $ x\in C $,

$$g(x)=f(x) \;\; if \; x\in A$$
and
$$g(x)=x \; \; if \; x\in D$$

$ g $ is clearly a bijection from $ C $ on $ B$.