答:
Essentially, yes. Assuming everything is smooth enough, the only exception is if you have other constraints on $\mathbf x$. For example, if you add the requirement $x_1 \le 3$; it is possible to have a maximum at the point $x_1 = 3$, so you need to separately check that case. (Edit: If you are interested in global upper and lower bounds rather than local maxima/minima which are actually attained by the function, you also need to check any non-compact directions. See below.)

The reason follows almost immediately from the result you quote, $\nabla f-\lambda \nabla g=0$ (which intuitively says that $f$ is stationary except possibly for some variation in the direction $\nabla g$ which is perpendicular to the constraint surface). The derivatives of $L$ with respect to $\mathbf x$ give the equations $\nabla f-\lambda \nabla g=0$, whilst the derivative of $L$ with respect to $\lambda$ imposes $g = 0$.

Note that the maxima/minima of $f$ subject to $g=0$ need not be maxima/minima of $L$, only stationary points.

Edit: If you are interested in upper and lower bounds, which may not be actually be attained by the function in its domain and hence aren't strictly maxima or minima, then you need to explicitly consider what happens as you move off in non-compact directions. If you like, you could think of this as adding a boundary, e.g. at infinity, that you have to worry about. This is already the case for normal unconstrained functions like $f(x) = x^2$ (no upper bound, no maxima) or $f(x) =(x^2-1)^2$ (which has no upper bound, but does have local maximum at $x=0$) or $f(x) = 1/((x-1)^2+1)+1/((x+1)^2 +1)$ (which has a lower bound of 0 but a local minimum at $x=0$ of $f=1$).

Or consider $f(r, \theta) = u(\theta) (1+e^{-r})$ for something like $u=1+\cos^2(\theta)$. Note this last function has a lower bound $u(\theta)$ in each direction, and hence an overall lower bound of the lowest value $u$ can take, which is $1$. But the function is strictly larger than one everywhere in its domain, so there is no point where it takes the "minimum" value of 1.

Title: Lagrange's method

Body:
Suppose that I am asked to determine the maximum and the minimum of a continuous and differentiable function $$ f(x, y, z)$$
with a constraint $$ g(x, y, z)=0.$$

I have some trouble understanding the method, even more in this case when it's harder to picture what the function $f$ looks like.

So far I have proven that when $f$ is restricted to those points for which $g(x,y,z)=0$, if it has a local maximum/minimum, the gradient of $f$ is parallel to the gradient of $g$ at that point, i.e.
$$
\nabla f-\lambda \nabla g=0.
$$
My question is: is it enough just to consider the critical points, i.e. those $(x, y, z)$ for which the gradient of Lagrange function
$$
L \left(x, y, z, \lambda \right)=f(x, y, z)-\lambda g(x, y, z)
$$
is zero, in order to find the maximum and minimum of $f$? Many examples of optimization exercises I have seen only do that.


问:
Lagrange's method
Suppose that I am asked to determine the maximum and the minimum of a continuous and differentiable function $$ f(x, y, z)$$
with a constraint $$ g(x, y, z)=0.$$

I have some trouble understanding the method, even more in this case when it's harder to picture what the function $f$ looks like.

So far I have proven that when $f$ is restricted to those points for which $g(x,y,z)=0$, if it has a local maximum/minimum, the gradient of $f$ is parallel to the gradient of $g$ at that point, i.e.
$$
\nabla f-\lambda \nabla g=0.
$$
My question is: is it enough just to consider the critical points, i.e. those $(x, y, z)$ for which the gradient of Lagrange function
$$
L \left(x, y, z, \lambda \right)=f(x, y, z)-\lambda g(x, y, z)
$$
is zero, in order to find the maximum and minimum of $f$? Many examples of optimization exercises I have seen only do that.

答: Not enough.  You also need to check the boundaries of support for the function.  The minimum or maximum may occur there.</p>

ContentLicense: CC BY-SA 4.0

Title: Lagrange's method


问:
How to find $F^{(n+1)}(x)$ if $F(x)=\int_0^x(x-t)^nu(t)dt$?
if $F(x)=\int_0^x(x-t)^nu(t)dt$ then find $F^{(n+1)}(x)$

from Leibniz rule $a(x)=0,b(x)=x,a'(x)=0,b'(x)=1$ and $G'(x)=n(x-t)^{n-1}\\$

so
$F^{(1)}(x)=n\int_0^x(x-t)^{n-1}u(t)dt$ and $F^{(n)}(x)=n.(n-1).(n-2)...1\int_0^x(x-t)^{n-n}u(t)dt=n!\int_0^xu(t)dt$. How can I continue from here? is it $F^{(n+1)}(x)=0$ because applying Leibniz rule one more time gives $G'(x)=0$?

答:
You made a mistake in your final integral. The upper end-point's dependence on $x$ means you should get
$$\frac{{\rm d}}{{\rm d} x} \int_0^x u(t) \ {\rm d} t = u(x)$$

In general,
$$\frac{{\rm d}}{{\rm d} x} \int_0^x g(x,t) \ {\rm d} t = \int_0^x \frac{\partial}{\partial x}
 g(x,t) \ {\rm d} t + \boxed{g(x,x)}$$
But throughout the question, $g(x,t) \propto (x-t)^{n-i}$ meant $g(x,x) = 0$... until $n - i = 0$, when something special happened.


问:
Is $GCD(2^n,(x^2+1)^{n-1})=1$
Let $x>1,n>2$ be positive integers. I come across the following problem:

Is $$GCD(2^n,(x^2+1)^{n-1})=1$$

or $2^n$ is relatively prime to $(x^2+1)^{n-1}$ ?

Is $$GCD(x^{2n},(x^2+1)^{n-1})=1$$ ?

NB: $GCD$ is the greatest common divisor.

答:
You can refer to @xpaul's answer for the first case.

Now, consider the second scenario you doubt, i.e., $\gcd(x^{2n},(x^2 + 1)^{n-1})=1$.

If $x$ is prime, $x^{2n}$ is a power of a prime and $(x^2 + 1)^{n-1}$ is a product of a power of a prime plus $1$. You can never have a common factor in such a case and hence the equality holds when $x$ is prime.

If $x$ is even, i.e, $x = 2y$, $x^{2n}$ has powers of $2$ in it and $(x^2 + 1)^{n-1} = ({(2y)}^2 + 1)^{n-1}= (4y^2 + 1)^{n-1}$ has no power of $2$ in it. The equality holds here as well.(Just putting up as a notice; I am not much sure. The inequality doesn't hold if both have a power of $y$ in common.)

I am stopping here for now and sorry for being of not much help. I was about to go to bed when I typed this all up.

NB :

If $x$ is prime, $x^{2n}$ is a power of a prime and $(x^2 + 1)^{n-1}$ is a product of a power of a prime plus $1$. You can never have a common factor in such a case and hence the equality holds when $x$ is prime.

I doubt this, so please verify it by tomorrow. I say so because I haven' tried any test cases and I used only algebra.

Title: Is $GCD(2^n,(x^2+1)^{n-1})=1$


问:
Proving if $\gamma\in SL(2,\mathbb{Z})$ has order $6$, then $\gamma$ is conjugate to $\begin{bmatrix}0 & -1\\ 1 & 1\end{bmatrix}^{\pm1}$
I am reading Diamond's book on modular forms, and I came across the proof that the stabilizers of elliptic points on the upper half-plane are cyclic. To prove that, Diamond starts to show that a transformation that fixes a point must have order $1, 2, 3, 4$ or $6$. For $\operatorname{ord}(\gamma)=6$, the author proceeds to argue that $\gamma$ is conjugate to matrices $\begin{bmatrix}0 & -1\\ 1 & 1\end{bmatrix}^{\pm1}$.

However, I found the language he uses a little bit confusing.

Since $\gamma^6=I$, the lattice $L=\mathbb{Z}^2$ of integral column vectors is a module over $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$, defining the scalar product $(a+b\mu_6)\cdot v$ for $a,b\in\mathbb{Z}$ and $v\in L$ by the matrix-by-vector product $(aI+b\gamma)v$

In this case,

What does the sentence "the lattice $L=\mathbb{Z}^2$ of integral column vectors" mean? Is $L$ just the set $\left\{\begin{bmatrix}a\\ b\end{bmatrix}:a,b\in\mathbb{Z}\right\}$?

Also, is defining $(a+b\mu_6)\cdot v$ by $(aI+b\gamma)v$ an action to ensure that $(sr)\cdot v=s\cdot(r\cdot v)$ as$$(aI+b\gamma)(cI+d\gamma)=acI+bd\gamma^2+(ad+bc)\gamma=acI+bd(\gamma-I)+(ad+bc)\gamma=(ac-bd)I+(bc+ad+bd)I$$which is consistent with multiplications in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$?

And finally, how can we show that $L$ is finitely generated over $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$? Can this be done by simply saying that it can be generated by $\begin{bmatrix}1\\ 0\end{bmatrix}$ and $\begin{bmatrix}0\\ 1\end{bmatrix}$ with coefficients $a, b\in\mathbb{Z}\subseteq\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$?

Thanks in advance. Any help will be appreciated.

答:
Completely right.

You simply have to show that the given identification $\mathbb Z[\mu_6] \to M_{2\times 2}(\mathbb Z)$ is a ring homomorphism. The only unclear part is that the identification preserves multiplication. Instead of doing it by hand what you tried to do you could realize that $\mu_6$ and $\gamma$ are both zeros of the polynomial $x^2-x+1$. So the square part breaks down on both sides the same way.

Right, $L$ is generated by those two vectors as an abstract group so in particular as $\mathbb Z[\mu_6]$ module.

By the way $\mathbb Z[\mu_6]$ is a principal ideal domain like I am (see my user name ).


问:
How to solve the two dimensional Laplace's equation for certain cases?
Had a doubt regarding Laplace's equation.
In many textbooks, the general solution to the two dimensional Laplace equation is mentioned as:

$$\Phi(\rho,\phi) = A_{0} + B_{0}\ln(\rho) + \sum_{n=1}^{\infty}\rho^n(A_n\cos(n\phi) + B_n \sin(n\phi)) + \sum_{n=1}^{\infty}\rho^{-n}(C_n \cos(n\phi) + D_n \sin(n\phi))$$

in polar coordinates.

For convenience, I will name the two summation terms as:

$$T_1 = \sum_{n=1}^{\infty}\rho^n(A_n\cos(n\phi) + B_n \sin(n\phi))$$

$$T_2 =  \sum_{n=1}^{\infty}\rho^{-n}(C_n \cos(n\phi) + D_n \sin(n\phi))$$

Not much has really been mentioned on whether these two series, $T_1$ and $T_2$ converge or not. When the solution space does not include either zero or infinity it is generally implicitly assumed that they converge. When the solution space includes the origin, generally the procedure has been to equate the coefficients of the second series, $T_2$, to zero, in other words, to exclude the second series from the solution.
In many physical problems of interest however, the solution space includes both the origin of the coordinate system and is an unbounded region in all directions. My question is, how can we use the above expansions to handle such cases? Especially when there is a boundary condition of zero (or $ k ln(\rho) $) at infinity. The radius of convergence doesn't seem to be known a priori as the coefficients are unknown.
For example, if you consider the case of an infinitely long conducting cable of arbitrary but uniform cross-section, (Arbitrary meaning that it need not be something geometrical or symmetric like a circle or a square, and could even be concave; and uniform meaning that the shape and area of the cross-section does not change in the $z$ direction so it can be reduced to a two dimensional problem.) maintained at a constant potential $V$, and where the potential is taken to approach $ k ln(\rho) $ at infinity. It may often be convenient to choose the origin outside the cable. In this case, if I exclude the term $T_2$ (for finiteness at the origin), again the boundary condition $\Phi\to k \ln(\rho)$ as $\rho\to\infty$ will imply that the coefficients in the term $T_1$ become zero as well. So the solution appears to be trivially $\Phi = A_0$ (obviously wrong!) even before I apply the boundary condition $\Phi = V$ at the surface of the cable.
Can you let me know what I am doing wrong? Can you throw some light on this?
Thanks.

答:
HiPragabhava, Thank you for your assistance. If I proceed along those lines, this is what I get:
$$\phi_\text{in} = A_0 + \sum_{n=1}^\infty\rho^n\Big(A_n\sin(n\phi) + B_n\cos(n\phi)\Big) $$ for $\rho\leq a$ .
and
$$\phi_\text{out} = A_0 + B_0\ln(\rho) + \sum_{n=1}^\infty\rho^{-n}\Big(C_n\sin(n\phi) + D_n\cos(n\phi)\Big) $$ for $\rho>a$ .
Now if I assume symmetry about x axis, the sine terms will vanish. Also, 'glueing' the two solutions will give me:
$$D_n = a^{2n}B_n$$ .
So, now the solutions become :
$$\phi_\text{in} = A_0 + \sum_{n=1}^\infty\rho^nB_n\cos(n\phi) $$ for $\rho\leq a$
and
$$\phi_\text{out} = A_0 + B_0\ln(\rho) + \sum_{n=1}^\infty\rho^{-n}a^{2n}B_n\cos(n\phi) $$ for $\rho>a$ .
Now, the constants $B_n$ are yet to be determined. Though we don't expect the electric field to be continuous on the (incomplete) cylinder, we would expect it to be continuous on the slit. In other words, the $\rho$ component of the electric field would be continuous at $\rho = a$, $-\alpha<\phi<\alpha$. This means that:
$$ \frac{\partial\phi_\text{in}}{\partial\rho} = \frac{\partial\phi_\text{out}}{\partial\rho}$$ for $\rho = a$, and $-\alpha<\phi<\alpha$.
After differentiation and simplification, this is what I get:
$$ B_0 = 2\sum_{n=1}^\infty na^nB_n\cos n\phi $$
for $-\alpha<\phi<\alpha$ .
Since this relation is valid only for the range between $-\alpha$ and $\alpha$ I would not be able to get the coefficients by multiplying both sides by cosines and integrating between 0 to 2$\pi$.  So how to get the coefficients $B_n$ ?
Even if I could get the coefficients something looks fishy here as the L.H.S. is a constant while the R.H.S. depends on the angle $\phi$.
So how can I proceed from here on?

Thanks.

ContentLicense: CC BY-SA 3.0


问:
If for all f $\in C_b(X)$ f attains it's maximum, then X is compact.
This is working over $(X,d)$ a metric space. I will quickly reproduce the questions in case it hasn't typeset in the title.

If for all f $\in C_b(X)$, f attains it's maximum, then X is compact.

I am familiar with the result that "If every $f \in C(X)$ is bounded then X is compact" and how this is proved (using Tietze's extension theorem). But this result isn't the same. This is not a HW Question but it was sparked after proving that the opposite implication holds.

答:
If $X$ is not compact then there exists an unbounded $f$ in $C(X)$.  Therefore
$$
g(x) := \arctan(f(x))
$$
is bounded but does not attain its norm.


问:
Is the multiplication operator on $\ell^p$ compact?
Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.

Can we more generally conclude that $T$ is compact?

Assume $p<\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology.

On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$.

EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true.

答:
Here's another proof of necessity. Let $T_\alpha$ be a multiplication operator and $\alpha = \{\alpha_n\} \in l^\infty$ be the corresponding sequence.

Let $\alpha \notin c_0$. Then there exists $\varepsilon > 0$ s.t. for any $N \in \mathbb{N}$ we can find $n>N$ for which $|a_n| > \varepsilon$. Hence we can assume $|a_n| > \varepsilon$ for all $n \in \mathbb{N}$. This assumption is made for simplicity, but if we take the subspace of $l^p$ corresponding to the chosen subsequence, it will be isometrically isomorphic to $l^p$. A restriction of $T_\alpha$ to this subspace is again a compact operator. Then $T_\alpha(\mathbb{B}_1) \supset \mathbb{B}_\varepsilon$ ($\mathbb{B}_1$ is a unit ball).

Indeed, let $x \in l^p$, $||x|| \geq 1$. Then
$$
||T_\alpha x||_p = \sqrt[p]{\sum_{n=1}^\infty |\alpha_n x_n|^p} > \sqrt[p]{\sum_{n=1}^\infty \varepsilon^p |x_n|^p} = \varepsilon ||x||_p \geq \varepsilon,
$$
and the statement follows since our operator is surjective.

But $\mathbb{B}_\varepsilon$ is not relatively compact in infinite dimensional space. Hence $T_\alpha$ is not compact.


问:
Prove or disprove, limits of antiderivatives
If $g(x), f(x)$ are real and continuous, and $g(x)\leq$ $f(x)\leq$$g(x)+\frac{1}{x}$ for every $x>1$.
The following limit is equal to $0$:
$$\lim\limits_{x \to \infty} \int_x^{x+\sqrt{x}}(f(t)-g(t))\,dt =0$$

答:
HINT
$$0\le f(t)-g(t)\le \frac{1}{t}$$ for $t>1$, so if you integrate....


问:
What is the number of seating plans?
I am reading a book and there was a topic regarding number of sitting arrangements in a wedding.

The information provided is:

107 people.

11 tables.

Each table can accommodate up to 10 people.

The order in each table does matter but the order of tables does not.

The author finally said that "there are about $11^{107}$ possible seating plans."

Isn't this incorrect since it allows for repetition? My answer would be that the number of seating plans is $110 \choose x_1,...,x_{11}$ such that $x_i = 10$ with $i \in$ {1,...,11}.

Can anyone suggest a correct solution to this problem?

P.S. The book is called "Algorithms to Live By" by Brian Christian and Tom Griffiths.

答:
I'm one of the authors of Algorithms to Live By, and your question came up in my Google alerts. I love that you're engaging so deeply with the book, and wanted to add a couple quick thoughts.

First of all, the wedding that Tom and I are discussing is that of Meghan Bellows and her husband J. D. Luc Peterson. The bride and groom published a delightful paper about how they formalized and solved their wedding seating-chart problem, called "Finding an optimal seating chart" (2012), which I encourage you to check out: https://www.improbable.com/news/2012/Optimal-seating-chart.pdf.

The figure that Tom and I cite comes directly from their paper: "There are a total of $11^{107} = 2.69 × 10^{111}$ combinations for seating guests at the various tables."

I must point out that in both their paper and our book, it is not stated, as you write, that, "The order in each table does matter but the order of tables does not." In fact, the opposite assumptions are made to get the $11^{107}$ figure: namely, that order within a table does not matter, but the tables themselves are distinct.

Honestly, I think your assumptions are more realistic!

But that is not the set of assumptions that Bellows & Peterson use in arriving at their $11^{107}$ figure.

As both you and Jaap discuss, the number of combinations is much greater under your formulation, where order around a table is taken into account. (This may be why Bellows and Peterson did not formulate the problem this way, even though it's intuitive to do so! It is also common at weddings to allow guests to choose their own seats around the table, rather than to have specific place settings. So, that part of the optimization problem is offloaded onto the guests.)

Under this set of assumptions, as you and Jaap note, the figure of $11^{107}$ is too high, because it does not factor into account the maximum capacity of each table. Tom and I mention this in the endnotes for this chapter. However, I seem to recall from talking with Meghan that they relaxed their maximum-capacity constraint on purpose when they ran their mixed-integer linear optimization.

Hope this is helpful to you. I love the discussion on this page!

Title: What is the number of seating plans?


答:
Leibniz's rule tells us that we have

$$\frac{d}{dx}\big(\int_{a(x)}^{b(x)}f(x,t)dt\big)=f(x,b(x))\frac{d}{dx}b(x)-f(x,a(x))\frac{d}{dx}a(x)+\int_{a(x)}^{(b(x)}\frac{\partial }{\partial x}f(x,t)dt.$$

So if $F(x)=\int_{0}^{x}(x-t)^{n}u(t)dt$ then

$$F^{(k)}(x)=n(n-1)\cdot\cdot\cdot(n-k+1)\int_{0}^{x}(x-t)^{n-k}u(t)dt$$

for $1\leq k\leq n$. You then correctly found that $F^{(n)}(x)=n!\int_{0}^{x}u(t)dt$. Note that since $a(x)=0$, $b(x)=x$ we have $a'(x)=0$, but $b'(x)=1$ so that the first doesn't vanish. Applying Leibniz's rule to $F^{(n)}(x)$ we obtain

$$F^{(n+1)}(x)=n!u(x)\frac{d}{dx}b(x)-n!u(0)\frac{d}{dx}a(x)+n!\int_{0}^{x}\frac{\partial}{\partial x}u(t)dt$$
$$=n!u(x).$$

Title: How to find $F^{(n+1)}(x)$ if $F(x)=\int_0^x(x-t)^nu(t)dt$?

Body:
if $F(x)=\int_0^x(x-t)^nu(t)dt$ then find $F^{(n+1)}(x)$

from Leibniz rule $a(x)=0,b(x)=x,a'(x)=0,b'(x)=1$ and $G'(x)=n(x-t)^{n-1}\\$

so
$F^{(1)}(x)=n\int_0^x(x-t)^{n-1}u(t)dt$ and $F^{(n)}(x)=n.(n-1).(n-2)...1\int_0^x(x-t)^{n-n}u(t)dt=n!\int_0^xu(t)dt$. How can I continue from here? is it $F^{(n+1)}(x)=0$ because applying Leibniz rule one more time gives $G'(x)=0$?


问:
How can there be multiple irreducible representations of a group each having distinct dimension?
The matrix elements for the $(2l+1)$-dimensional irreducible representation of SO(3) are given by:

$D_{m',m}^l(\phi_1,\Phi,\phi_2)=[i^{m'-m}\sqrt{(l+m')!(l-m')!(l+m)!(l-m)!}$
$\times(\frac{1+\cos\Phi}{2})^l (\frac{1-\cos\Phi}{1+\cos\Phi})^{(m'-m)/2}$
$\times\sum\limits_k \frac{(-1)^k}{(l+m-k)!(l-m'-k)!(m'-m+k)!k!}$
$\times(\frac{1-\cos\Phi}{1+\cos\Phi})^k] e^{im'\phi_2}e^{im\phi_1}$

My question is, how can there be multiple irreducible representations of SO(3) each of which has a different dimension?  I thought that by definition an irreducible representation is one which has no non-trivial invariant subspace.  I know that SO(3) can be represented by the group of 3x3 symmetric matrices of determinant = 1 (I thought this had something to do with the fact that it takes 3 parameters to describe a 3D rotation).  If I have a larger matrix (e.g. 5x5) doesn't it have to have a non-trivial invariant subspace (e.g. of dimension 2), and therefore be reducible?

The application here is the expansion of functions using generalized spherical harmonics, which is accomplished (per the Peter-Weyl theorem) using an infinite linear combination of the matrix elements of the irreducible representationS (emphasis on the S) of SO(3).

答:
$\mathrm{SO}(3)$ is a wonderful group, but it quickly got bored rotating vectors in space.  It wanted to be used to help solve differential equations, and so desperately wanted to act on functions.  Then one day, it realized it could act on functions by acting on their input.

For instance, it contains a 90 degree clockwise rotation in the $x$-$y$ plane.  It takes $(x,y,z)$ and replaces it with $(y,-x,z)$.

Well, the function $2x+3y+7z$ can also be rotated!  We just replace $x$ by $y$, $y$ by $-x$, and leave $z$ alone:  $2x+3y+7z$ is rotated into $2y+3(-x)+7z = -3x + 2y + 7z$.  If we consider the action on all these linear $Ax+By+Cz$ it becomes clear that $\mathrm{SO}(3)$ acts by 3×3 matrices on the vector space with basis $\{x,y,z\}$.

Yay, $\mathrm{SO}(3)$ can act on functions now, and it acts on those linear functions as 3×3 matrices.

What about quadratics?  Well we could have $2x^2 + 3xy + 7y^2$.  That same 90 degree rotation takes it to $2(y)(y) + 3(y)(-x) + 7(-x)(-x) = 7x^2 - 3xy + 2y^2$.  Yay, now $\mathrm{SO}(3)$ acts as 6×6 matrices on the quadratic polynomials $Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2$ spanned by $\{ x^2, xy, y^2, xz, yz, z^2 \}$.

It turns out though that $\mathrm{SO}(3)$ doesn't swirl these functions around very thoroughly.  If you take a function, and rotate its input, all it does is rotate the laplacian.  In particular, notice that it takes any harmonic polynomial (one with 0 laplacian) to a harmonic polynomial.  For quadratic polynomials, this is particularly easy to describe:

When you use $\mathrm{SO}(3)$ to rotate $Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2$, it leaves the laplacian, $A+C+F$, alone.  So it will never rotate $x^2$ into $xy$.  In fact, it leaves the 5-dimensional space spanned by $\{ x^2-y^2, y^2-z^2, xy, xz, yz \}$ invariant.  If we write things in these coordinates, then we get $\mathrm{SO}(3)$ acting as 5×5 matrices.  In fact it acts irreducibly.

What happened to the 6th dimension?  Well, it is spanned by a very silly function, the sphere: $x^2+y^2+z^2$.  If you rotate the sphere, you get the sphere.  On this one dimensional space, $\mathrm{SO}(3)$ is represented by 1×1 matrices.  Well, actually, matrix.  Every single rotation in $\mathrm{SO}(3)$ acts as the identity matrix $[1]$, since every rotation leaves the sphere alone.

If we use colors, like pink, lime, and periwinkle to describe vectors, then a polynomial like $5x^2-5y^2 + 6x + 3$ is composed of a pink term, $5x^2-5y^2$, where $\mathrm{SO}(3)$ acts in a 5×5 manner, a lime term, $6x$, where $\mathrm{SO}(3)$ acts in a 3×3 manner, and a periwinkle term, $3$, where $\mathrm{SO}(3)$ acts in a 1×1 manner.

A polynomial like $7x^2+5y^2$ is a little trickier to see its colors (it is a good thing $\mathrm{SO}(3)$ is so clever), $7x^2+5y^2 = 4(x^2+y^2+z^2) + 3(x^2-y^2) + 4(y^2-z^2)$.  The fist term $4(x^2+y^2+z^2)$ is periwinkle where $\mathrm{SO}(3)$ acts in a 1×1 manner, but the next two terms $3(x^2-y^2) + 4*(y^2-z^2)$ are pink where $\mathrm{SO}(3)$ acts in a 5×5 manner.


问:
Understanding Bayes' Theorem
I worked through some examples of Bayes' Theorem and now was reading the proof.

Bayes' Theorem states the following:

Suppose that the sample space S is partitioned into disjoint subsets $B_1, B_2,...,B_n$.  That is, $S = B_1 \cup B_2 \cup \cdots \cup B_n$, $\Pr(B_i) > 0$ $\forall i=1,2,...,n$ and $B_i \cap B_j = \varnothing$ $\forall i\ne j$.  Then for an event A,

$\Pr(B_j \mid A)=\cfrac{B_j \cap A}{\Pr(A)}=\cfrac{\Pr(B_j) \cdot \Pr(A \mid B_j)}{\sum\limits_{i=1}^{n}\Pr(B_i) \cdot \Pr(A \mid B_i)}\tag{1}$

The numerator is just from definition of conditional probability in multiplicative form.

For the denominator, I read the following:

$A= A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)\tag{2}$

Now this is what I don't understand:

The sets $A \cup B_i$ are disjoint because the sets $B_1, B_2, ..., B_n$ form a partition.$\tag{$\clubsuit$}$

I don't see how that is inferred or why that is the case.  What does B forming a partition have anything to do with  it being disjoint with A.  Can someone please explain this conceptually or via an example?

I worked one example where you had 3 coolers and in each cooler you had either root beer or soda.  So the first node would be which cooler you would choose and the second nodes would be whether you choose root beer or soda.  But I don't see why these would be disjoint.  If anything, I would say they weren't disjoint because each cooler contains both types of drinks.

Thank you in advance! :)

答:
As Tharsis pointed out, and was clarified in the comments, it is all of sets of the given by $\;(A \cap B_i),\; 1 \leq i \leq n\;$ that are pairwise disjoint.

$$\;(A \cap B_i)\cap (A \cap B_j) = \varnothing,\;\;\;\forall i, j,\;\;\text{s.t.}\;\;1 \leq i, j\leq n\;\;\text{and}\;\;i\neq j$$

e.g., $(A\cap B_1)$ is disjoint from $(A\cap B_2)$, but $(A\cup B_1)$ is certainly not disjoint from $(A\cup B_2)$, etc.

Pay attention to the distinction between $∩,∪.$

Title: Understanding Bayes' Theorem


问:
Every embedding gives rise to an embedded submanifold?
I call regular (or embedded) submanifold of a manifold $M$ a subset  $i:X \subset M$ endowed with a submanifold structure such that the inclusion map $i$ is an embedding.

I call embedding an immersion that is also an homeomorphism onto its image.

I call immersion a smooth map of constant rank equal to the dimension of its domain.

It can be proven that a smooth constant rank map is injective only if its rank is equal to the dimension of its domain. Hence any smooth constant rank injective map must be an immmersion.

An inclusion is homeomorphic onto its image iff on $X$ I have the subspace topology.

Moreover, since the inclusion is injective, to prove it is an immersion it suffices to show that it is smooth and of constant rank.

Hence, I can rephrase the definition of submanifold as:

A submanifold $X \subset M$ is a subset of $M$ endowed with the subspace topology and with a manifold structure such that the inclusion $i:X \subset M$ is smooth and of constant rank.

question 1

Is this rephrasing correct?

question 2

If so, what is an example of an inclusion which is not smooth? What is an example of an inclusion which is not of constant rank?

question 3

I read that for any embedding $f:N \to M,$ we have that $f(N)\subset M$ is a regular submanifold of $M.$

I understand that if we factorize $f$ as $$N\xrightarrow{\tilde{f}}f(N)\xrightarrow{i}M$$ I have that $f$ smooth of constant rank $\implies$ $\tilde{f}$ smooth of constant rank, because I am just changing the codomain.

But how to show that this must imply $i$ smooth of constant rank too?

I believe we should have a statement of the form

$h \circ g$ smooth of constant rank and $g$ smooth of constant rank
$\implies$ $h$ smooth of constant rank.

Is this correct?

答:
First of all, my favorite definition of a smooth submanifold is "none of the above," but the one I give here. Even with your definition, working with $f$ and $i$ separately only complicates things. You can say that a subset $X$ of a smooth manifold $M$ is smooth submanifold if there exists a smooth manifold $N$ and an immersion $f: N\to M$ such that $f(N)=X$ and $f: N\to M$ is a topological embedding, i.e. is a homeomorphism $N\to X$ (where $X$ is given the subspace topology which is assumed by default in topology).

Question 1: Yes, this is correct.

Question 2. (a) An inclusion which is not smooth: Consider the graph $\Gamma$ of any continuous function of one variable which is not everywhere smooth, e.g.  $y=|x|$. Then $\Gamma$ cannot be given structure of a smooth submanifold. However, the map $\phi: x\mapsto (x, |x|)$ is a homeomorphism from ${\mathbb R}$ to $\Gamma$. Hence, $\phi$ can be used to define a smooth manifold structure on $\Gamma$ (use $\phi^{-1}$ as the single chart).

(b) An inclusion which is not of constant rank. The simplest example I know comes from the map $\phi: {\mathbb R}\to {\mathbb R}$, $\phi(x)=x^3$. Then $X=M= {\mathbb R}$ (as sets!), but, as smooth manifolds I will use the structure on $X$ such that $\phi$ is a diffeomorphism. Then the inclusion map $i: X\to M$ does not have constant rank.

If you like more interesting examples, consider $X\subset {\mathbb R}^2$ equal to the solution set of the equation $y^3=x^2$. Take $f: {\mathbb R}\to X$ given by
$$
f(t)=(t^3, t^2).
$$
Then $f$ is smooth, is a homeomorphism to its image but does not have constant rank. I leave it to you to convert this into an example when the inclusion map $i$ is smooth, is topological embedding, but does not have constant rank.

Question 3. Yes, you read it correctly. A hint for the proof: Suppose that
you have smooth maps of manifolds
$$
A \stackrel{f}{\to} B \stackrel{g}{\to} C,
$$
where $f$ is a diffeomorphism. Suppose that the map $h:= g\circ f$ is an immersion. Then $g$ is also an immersion.


问:
For i.i.d. $\{X_n\}$ with non-constant $X_n > 0$ and $\mathbb{E}X_n = q \leq 1$ the product $\prod\limits_{n=1}^{+\infty}X_n \to 0$ almost surely
Suppose $\{X_n\}$ is i.i.d. with $X_n > 0$ and each $X_n$ is not constant almost surely.

I want to show that if $\mathbb{E}X_n = q \leq 1$, then the product $\prod\limits_{n=1}^{+\infty}X_n \to 0$ almost surely. In the problem statement, there is also a hint that $q < 1$ and $q = 1$ can be considered separately.

I thought about this one for a long time and I think applying logarithm operation to both sides and transform this product into a sum might help, but I couldn't progress from there. I'm still unable to see how I can reduce this problem to using laws of large numbers.

I would appreciate any help!

答:
Apologies for my first hasty attempt, perhaps this will be more satisfactory. As you noted, the problem is equivalent to showing that $\sum_i \log(X_i) = -\infty$ a.s.

Firstly, since the logarithm is concave and not affine and $X_i$ is not a.s. constant, one has by Jensen's inequality that

$$0 = \log(1) \geq \log q = \log(E[X_i]) > E[\log(X_i)] := \mu.$$

You can see, e.g. Convexity and equality in Jensen inequality for when Jensen's inequality can be an equality. In our case, it must be strict inequality, which we used in the previous line.

Thus it suffices to show (set $Y_i := -\log(X_i)$) the following:

Theorem.  If $Y_i$ are iid with $\mu := E[Y_i] > 0$, then $\sum_i Y_i = \infty$ a.s.

Lemma. For any real valued $(y_i)_i$, if $\frac{1}{n} \sum_{i=0}^n y_i \to \mu > 0$ then $\sum_i y_i = \infty$.

Proof of lemma. Let any subsequence $n_k \to \infty$ be given.
Then $$\sum_{i=0}^{n_k} y_i =(\frac{1}{n_k}\sum_{i=0}^{n_k} y_i) n_k \to \mu \times \infty = \infty.$$ In particular this shows $\liminf_n \sum_{i=1}^{n} y_i = \infty$, so $\sum_i y_i = \infty$. QED.

Proof of theorem. If $\mu < \infty$ then by the law of large numbers $\frac{1}{n}\sum_{i=0}^n Y_i \to \mu >0$ a.s.
By the lemma, $\sum_i Y_i = \infty$ a.s. in this case.

Now consider the case $\mu = \infty$. Considering that $E[Y_i] > -\infty$ and writing $Y_i = Y_i^+ - Y_i^-$ with $Y_i^+ = \max(Y_i, 0)$ and $Y_i^-= \max(-Y_i, 0)$ the usual decomposition of $Y_i$ into its positive and negative parts, it must be that $E[Y_i^-] \in [0,\infty)$ and $E[Y_i^+] = \infty$.
In particular
$$
E[\max(Y_i, n)] = E[\max(Y_i^+,n) - Y_i^-] =  E[\max(Y_i^+,n)] - E[Y_i^-] \to \infty
$$
by monotone convergence of $\max(Y_i^+,n) \to Y_i^+$. In particular we may choose $N$ large enough that $0 < E[\max(Y_i, N)] \leq N$. The random variables $(\max(Y_i, N))_i$ are iid with mean in $(0,\infty)$,  so by what we showed earlier $\sum_i \max(Y_i, N) = \infty$ a.s. Then it follows that

$$
\infty = \sum_i \max(Y_i, N) \leq \sum_i Y_i
$$
a.s., QED.


问:
Show that $f$ is bijective.
Show that $f$, defined in $0<x<s, (s>0)$ as follows:
$$f(x) = {2x-s \over x(s-x)}$$

is a bijective function of this interval in the reals.

To show that it is injective, we assume $f (x) = f (y)$ and after manipulating we find:
$$(y-x)(-s^2+s(y+x)-2xy)=0$$
then $x=y$, right?

And to show that is surjective just take $f(x)=y$ and manipulating we get $$x = {ys-2 \pm \sqrt {4+y^2s^2} \over 2y}$$

Is that right? The result looks kind of ugly, I was not very confident.

答:
ANOTHER APPROACH:

Observe that $$f'(x)=\dfrac{2x(s-x)-(2x-s)(s-2x)}{x^2(s-x)^2}=\frac{2x^2-2xs+s^2}{x^2(s-x)^2}=\frac{(x-s)^2+x^2}{x^2(s-x)^2}>0.$$
So, since $f$ is derivable on the interval $(0,s)$, there the derivative is positive, the function is stricktly increasing

Moreover, $\lim_{x\to0^+} f(x)=-\infty$ and $\lim_{x\to s^-}f(x)=+\infty$.

So $f$ maps bijectively $(0,s)$ into $\Bbb R$.

Title: Show that $f$ is bijective.


问:
Are simple functions dense in $L^\infty$?
Are simple functions dense in $L^\infty$? I've been able to show this for finite measure spaces but not in general.</p>

答: Consider the function $f \equiv 1$ on $L^{\infty}(\mathbb{R},\mu)$, then $\|s-f\|_{\infty} > \varepsilon$ for any simple function $s$. In particular we can have $\|s-f\|_{\infty} > 1$. So simple functions may not be dense in $L^{\infty}$. But note that here I have assumed that $ \mu \big(\{x \in X : s(x) \neq 0\}\big) < \infty $.


问:
When is the kernel of A equal to the kernel of A squared?
If A is a square matrix, when is its kernel equal to the kernel of A squared?

I think it is when the intersection of im(A) and ker(A) equals the set with only the zero vector, but I do not know how to prove it.

答:
Clearly  $\;ker A \subset \ker A^2\;$ since

$$x\in\ker A\implies A^2x=A(Ax)=A0=0\implies x\in\ker A^2$$

Now, if $\;\ker A^2\subset \ker A\;$ , then $\;A^2x=0\implies Ax=0\;$ , and if $\;w\in\ker A\cap\text{Im}\,A\;$ , then

$$\exists y\;\;s.t.\:\;Ay=w\implies A^2y=Aw=0\implies w=Ay=0\implies \text{Im}\,A\,\cap\,\ker A=\{0\}$$

The other direction ($\;\text{Im}\,A\,\cap\,\ker A=\{0\}\implies \ker A^2\subset \ker A\;)\;$ follows from

$$x\in\ker A^2\implies 0=A^2x=A(Ax)\implies Ax\in\text{Im}\,A\,\cap\,\ker A=\{0\}\implies Ax=0\implies x\in\ker A$$


答:
Coprimality (the property of being coprime) is invariant under taking powers. That is to say that two integers $a,b$ are co prime if and only if $a^m, b^n$ are co prime for any two natural numbers $m,n$. So to answer both your questions, we can safely disregard all the powers.

For the first part, we need to check if $x^2+1$ is coprime to $2$ for odd $x$ which is clearly not the case.

For the second part we need to check if $x$ is co prime to $x^2+1$ which is true.

Title: Is $GCD(2^n,(x^2+1)^{n-1})=1$

Body:
Let $x>1,n>2$ be positive integers. I come across the following problem:

Is $$GCD(2^n,(x^2+1)^{n-1})=1$$

or $2^n$ is relatively prime to $(x^2+1)^{n-1}$ ?

Is $$GCD(x^{2n},(x^2+1)^{n-1})=1$$ ?

NB: $GCD$ is the greatest common divisor.


问:
In what sense are finitely presented algebras equivalent under the image of a one-to-one polynomial?
Let $S$ be a finite set and $\mathbb{K}$ be a field.

Now, if $\mathcal{F}(S)$ is the free $\mathbb{K}$-algebra generated by $S$, and $\mathcal{R}_S$ is the smallest two-sided ideal containing a set of relations $R_S$ (each relation is a non-commutative polynomial of elements $s\in S$), then we have a finitely a presented $\mathbb{K}$-algebra $\mathcal{F}(S)/\mathcal{R_S}$.

Suppose that $f$ is a one-to-one non-commutative $\mathbb{K}$-polynomial in the elements of $S$ and consider the finitely presented $\mathbb{K}$-algebra $\mathcal{F}(f(S))/\mathcal{R}_{f(S)}$, where the relations are the same non-commutative polynomials but now with the elements $f(s)\in T$.

Question: In what sense, if any, are $\mathcal{F}(f(S))/\mathcal{R}_{f(S)}$ and $\mathcal{F}(S)/\mathcal{R}_S$ related as $\mathbb{K}$-algebra?

答: Once you realize that any map $\phi$ on $\mathcal{F}(S)$, which specifies the image of each $s\in S$ (i.e. the generators), gives a $\mathbb{K}$-algebra homomorphism if and only if $\phi(r)=0$ for all $r\in R_S$. From here it's clear the algebras are isomorphic via $f$ as mentioned in the comments.


问:
Variance of a single random variable with two terms.
A random variable X, representing the time until some event occurs, has the following pdf.
$$(14/99)e^{-0.5x}+(85/99)e^{-0.25x}$$.
Using integration, I get the expectation as 14.303 via $\int_0^\infty xf(x)dx$, and a variance of 463.453 via $\int_0^\infty (x-\overline{x})^2f(x)dx$.

I don't understand why the variance is so high. I also computed the median, which was much higher than the mean.

When I try and compute the variance using $Var(X)=E(X^2)-(E(X))^2$ I get a negative value.
2) Why is there a mismatch between using these 2 methods?

答:
The problem is that the pdf is not properly normalized, i.e.

\begin{align}
\int_0^\infty f(x) dx &= \frac{14}{85}\int_0^\infty e^{-x/2} \ dx + \frac{85}{99}\int_0^\infty e^{-x/4} \ dx \\
&= 2\frac{14}{85} + 4\frac{85}{99} \\
& \neq 1
\end{align}

I would make sure you have the correct pdf. Perhaps there should be a coefficient of 1/2 on the first term, and 1/4 on the second term?


问:
Integrating $\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx$ with $x=a \sin^2 u$ - limits query
I had to evaluate the integral $$\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx$$

I used the substitution $x=a \sin^2 u$ for this and got the correct answer of $\frac{\pi a^2}{8}$

However, part of my method involved the changing of the limits.

For example, when $x=0$, I reduce to $\sin u = 0$, which has multiple solutions $u=0, \pi, 2\pi, ...$. I chose $u=0$ to replace $x=0$, but why would it be incorrect to choose the others?

Similarly with $x=a$, I get $\sin^2 u = 1$.  I happened to choose $\sin u = 1$ but again why would it be incorrect to pick $\sin u = -1$ and use $u = -\frac{\pi}{2}$? It ends up giving the negative of the correct answer...

答:
The reason is pretty simple. You can see here that your substitution in the interval of integration must be a continuous and differentiable function. Now a function is a map of any one value of $x$ in that interval of integration to exactly one value of $u$ in the target interval.

In your specific case, that means that you cannot simply choose $x=a \sin^2 u$ but you must in advance specify which monotonous branch of the $\sin(u)$-function you wish to consider, e.g. $0 \le u \le \pi/2$. You may also use  another branch, e.g. $\pi/2 \le u \le \pi$, but you must specify in advance.

To see how it works, consider your example: You have
$$
I = \int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx
$$

Substitute $x=a \sin^2 u$, where $dx/du = 2 a \sin u \cos u$. We will be very picky in discussing the square roots.

Using the "standard" branch $0 \le u \le \pi/2$ gives ${u(0) = 0}$ and ${u(a)} = \pi/2$, and
$$
I = \int_{u(0)}^{u(a)} 2 a^2 \sqrt{\sin^2 u}  \sqrt{\cos^2 u}  \sin u \cos u\; du \\
= \int_{u(0)}^{u(a)} 2 a^2 \sin u \cos u\  \sin u \cos u\; du = a^2 \frac{4 u - \sin(4 u) }{16} |_{u(0)}^{u(a)} = \frac{\pi a^2}{8}
$$

Using the alternative branch $\pi/2 \le u \le \pi$ gives  ${u(0) = \pi}$ and ${u(a)} = \pi/2$, and
$$
I = \int_{u(0)}^{u(a)} 2 a^2 \sqrt{\sin^2 u}  \sqrt{\cos^2 u}  \sin u \cos u\; du \\
= \int_{u(0)}^{u(a)} 2 a^2 \sin u \color{red}{(-\cos u)}  \sin u \cos u\; du = \color{red}{-} a^2 \frac{4 u - \sin(4 u) }{16} |_{u(0)}^{u(a)} = \frac{\pi a^2}{8}
$$
Note the (red) minus sign as the correct representation of the square root. The reason is that the substitution at this point is $(a-x)^{\frac{1}{2}} = \sqrt{\cos^2 u} = - \cos u$  is correct since these terms must be positive, as $\cos u$ is negative in the branch $\pi/2 \le u \le \pi$. The overall result is correct again.


问:
Papers in Math Journals that Turn Out to be Sophisticated Versions of Simpler Facts
I'm interested in a list of papers in Math Journals that were celebrated as new discovery but turn out to be sophisticated versions of simpler facts. A brief explanation of the simpler version of the math concept would also be helpful.

For example, Rowland's prime generating sequence turns out to be equivalent to

Given $n\geq2, n\in\mathbb{N}$, the first differences plus 1 of the iteration $f(n) = n + lpf – 1$ is always prime, where lpf means the least prime factor of n. Start with $n=2$ we get $2\to3\to5\to9\to11\to21\to23\to45\to47\to93\to95\to99\to101...$

The above is always true for the simple reason that we keep adding "$lpf - 1$" to the previous term to get the next term. By definition lpf is prime.

答:
Check this out johncanning.net/wp/?p=1863

Mary M. Tai, a Diabetes Researcher devised a "brand-new" method to accurately calculate the area under graphs numerically. Little did she know that her "new" discovery was actually the trapezoid rule, a method for numerical integration known for hundreds of years.


问:
Does coprimality lead to subvarieties of codimension not less than 2?
Let $X \subseteq \mathbb{A}^n(k) $ be an affine variety over a field $k$ and $ f,g \in k[X]$ two regular functions on $X$.
If we now assume $f$ and $g$ to be coprime and define $Y := \mathcal{V}(f,g)$ as their zero locus, then (why) does $ \text{codim}_X Y \geq 2$ hold ?

Lets not make life complicated and assume $X$ to be irreducible, then $\dim \mathcal{V}(h) < \dim X$, with $h \in \{ f,g\}$ follows immedieatly, as $h$ cannot be constant on $X$. Therefore it should suffice to show $ \dim Y < \dim \mathcal{V}(h) $. However, this is where i am stuck as i am unable to see why the contrary would contradict the coprimality of $f$ and $g$.

There may be additional conditions required, but i am quite sure to have seen statements similar to this used in literature.

答:
This is true if $f$ and $g$ are "coprime" in the sense that for any $h$, if $f$ divides $gh$, then $f$ divides $h$.  This means that $g$ is a non-zero divisor mod $f$, so it is not contained in any minimal prime of $k[X]/(f)$.  Thus every prime of $k[X]/(f,g)$ must have height at least $1$ when lifted back to a prime of $k[X]/(f)$, and so must have height at least $2$ when lifted back to a prime of $k[X]$.  Thus, every chain of prime ideals in $k[X]/(f,g)$ can be extended to a chain of length $2$ more in $k[X]$, so $\dim k[X]/(f,g)\leq \dim k[X]-2$.

(It does not suffice to assume that $f$ and $g$ have no nonunit common factors.  For instance, let $X\subset\mathbb{A}^3$ be the surface defined by $xy-z^2=0$.  Then $x$ and $z$ have no nonunit common factors in $k[X]=k[x,y,z]/(xy-z^2)$, but setting $x=z=0$ does not constrain the value of $y$, so the zero locus of $x$ and $z$ in $X$ is a line, which has codimension $1$.)


问:
How to finding the line of best fit for a exponential decay set of data points
Im having some trouble trying to find a line of best fit that would go through the values of some values for a chemistry experiment.

Here is the data:

\begin{array}{|c|c|c|c|}
\hline
Reaction Time(s)& Concentration (M) \\ \hline
 78 & 0.15\\ \hline
 84  & 0.14\\ \hline
 89  & 0.13\\ \hline
 95 & 0.12\\ \hline
 106  & 0.11\\ \hline
 117  & 0.10\\ \hline
 127  &0.09 \\ \hline
 147   &  0.08\\ \hline
 168  & 0.07\\ \hline
188  &  0.06\\ \hline
 353  & 0.05 \\ \hline
 518  & 0.04\\ \hline
682  & 0.03\\ \hline
\end{array}

And here is the graph (feel free to ignore the blue line).

Is there any exponential decay function that could go through these to give a $r^2$ value? I tried on my ti-84 plus with no luck

答:
You should check non-linear regression which should be implemented in different software. For fun, I recently coded a script in python that does exaclty what you want. I did run your dataset in it and got the folowing result:

Exponential

The equation of the exponential is $$ae^{-bx} + h$$ with the value $a = 0.339833797871496$, $b= 0.0146726062591541$ and $h =0.0391506240940973$. The $r^2$ value is $0.989065791920668$.


答:
As noted your density is not exactly a pdf. Anyway it is clearly a mixture of 2 different negative exp pdf

$$f_X(x)=a f_1(x)+(1-a)f_2(x)$$

and obviously

$$f_1(x)=0.5e^{-0.5 x}$$

and

$$f_2(x)=0.25 e^{-0.25x}$$

In this case,

$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$

$$\mathbb{E}[X^2]=\frac{14}{99}\int_0^{+\infty}\frac{1}{2}x^2e^{-x/2}dx+\frac{85}{99}\int_0^{+\infty}\frac{1}{4}x^2e^{-x/4}dx=$$

$$=\frac{14}{99}\cdot 4\int_0^{+\infty}\Bigg(\frac{x}{2}\Bigg)^2e^{-x/2}d\Bigg(\frac{x}{2}\Bigg)+\frac{85}{99}\cdot 16\int_0^{+\infty}\Bigg(\frac{x}{4}\Bigg)^2e^{-x/4}d\Bigg(\frac{x}{4}\Bigg)=$$

$$=\frac{14}{99}\cdot 4\cdot \Gamma(3)+\frac{85}{99}\cdot 16\cdot \Gamma(3)=\frac{14}{99}\cdot 4\cdot 2+\frac{85}{99}\cdot 16\cdot 2$$

... in the same way you can easy calculate $E(X)$ and finally the variance

Edit

The variance will result:

$$V[X]=\Bigg(\frac{14}{99}\cdot 8+\frac{85}{99}\cdot 32\Bigg)-\Bigg(\frac{14}{99}\cdot 2+\frac{85}{99}\cdot 4\Bigg)^2\approx 14.79$$

Title: Variance of a single random variable with two terms.

Body:
A random variable X, representing the time until some event occurs, has the following pdf.
$$(14/99)e^{-0.5x}+(85/99)e^{-0.25x}$$.
Using integration, I get the expectation as 14.303 via $\int_0^\infty xf(x)dx$, and a variance of 463.453 via $\int_0^\infty (x-\overline{x})^2f(x)dx$.

I don't understand why the variance is so high. I also computed the median, which was much higher than the mean.

When I try and compute the variance using $Var(X)=E(X^2)-(E(X))^2$ I get a negative value.
2) Why is there a mismatch between using these 2 methods?


问:
Factorization in Gaussian integers
Let $p$ be a natural number, suppose $p$ prime. Show that the following conditions are equivalent:

the polynomial $x^2+1\in\mathbb{Z}_p$ has roots in $\mathbb{Z}_p$

$p$ is reducible in the ring $\mathbb{Z}[i]$

there exists $a,b\in\mathbb{Z}$ such that $p=a^2+b^2$

My attempt: suppose $x$ is a solution of $x^2=-1 \ mod\ p$. Raising both sides to $(p-1)/2$ gives  $x^{p-1}=(-1)^{(p-1)/2}\ mod \ p$. Suppose $p$ congruent to $3$ modulo $4$. Then $(p-1)/2$ is odd and $(-1)^{(p-1)/2}=-1$, while $x^{p-1}=1$, a contradiction. Hence $p$ must be congruent to $2$ modulo $4$, i.e. $p=2$ or $p$ congruent to $1$ modulo $4$, i.e. $p=1+4n$, some $n$.

If $p=2$ then $p=(1+i)(1-i)$ is reducible. If $p=1+4n$ i can't see how to show reducibility.

For $2)\rightarrow 3)$, suppose $p=\alpha\cdot\beta$ with $\alpha,\beta$ gaussian integers. Then the norm of $\alpha$, say $a^2+b^2$, divides $p$, but $p$ is prime and the norm can't be $1$, because the factorization of $p=\alpha\beta$ is not trivial, hence $p=a^2+b^2$.

Again, for $3)\rightarrow 1)$, i really don't know what to do. Thanks for any help.

答:
Mapping $i$ to $x$, you can show that $$\mathbb{Z}[i] / (p) \cong \mathbb{Z}[x] / (p, x^2 + 1) \cong \mathbb{F}_p[x] / (x^2 + 1)$$ is an isomorphism of rings.

Now $\mathbb{F}_p[x]$ and $\mathbb{Z}[i]$ are both factorial rings, so $$p \; \mathrm{reducible} \; \Leftrightarrow p \; \mathrm{not} \; \mathrm{prime} \; \Leftrightarrow (x^2 + 1) \; \mathrm{not} \; \mathrm{prime} \; \Leftrightarrow x^2+1 \; \mathrm{has} \; \mathrm{a} \; \mathrm{zero}.$$

So it's enough to show $3 \rightarrow 2$, which is easy because then $p = (a+bi)(a-bi)$.

Title: Factorization in Gaussian integers


问:
Similarity by SSA (side-side-angle) in obtuse triangles
Source: Challenge and Thrills of Pre-College Mathematics, Page 74, Problem 54:

"In two obtuse triangles, an acute angle of the one is equal to an angle of the other, and sides about the other acute angles are proportional. Prove that the triangles are similar."

I tried this first by cosine rule, and the question reduced to:

Assume $\triangle ABC$ and $\triangle DEF$ with $\angle A$ and $\angle D$ being obtuse, $\angle B=\angle E$ and $AC/BC=DF/EF$. We may take $DF=k\cdot AC$, $EF=k\cdot BC$ and apply $\cos B= \cos E$, obtaining $DE=k\cdot AB$ and the problem is solved. EDIT: Even this however is difficult to prove, as pointed out by cosmo5 in his answer.

What is interesting and challenging is to prove the same without the cosine law (probably the expected solution, as cosine law doesn't seem to help), as the law hasn't been introduced in the book until later chapters, while this problem is taken from chapter 3.

My attempt:

Fix $\triangle ABC$ as well as points $E$ and $F$ of $\triangle DEF$. Assuming $DF=k\cdot AC$, $EF=k\cdot BC$, the locus of $D$ will be a circle with centre $F$ and radius $k\cdot AC$.

After this however, I couldn't go further with the same concept. Either a hint or a solution related to my thought process, or any other thought process for that matter, would be greatly appreciated.

答:
The conditions given are essentially $SSA$ but there is no such criterion for congruence or similarity. This is because exactly two triangles are possible as shown.

Both $\triangle DEF$ and $\triangle D'EF$ satisfy the conditions that sides about second acute angle are proportional and first acute angles are equal. But both cannot be similar to $\triangle ABC$ at the same time.

Trigonometry cannot help either. For example, by cosine-rule
$$\cos B=\cos E=\frac{x^2+(ka)^2-(kb)^2}{2\cdot x\cdot ka}$$
is a quadratic in $x$; it will produce two solutions namely $DE$ and $D'E$ for the third side.

The extra information that the third angle is obtuse pins down the unique triangle $DEF \sim \triangle ABC$. Since $D,D'$ must lie on either side of perpendicular dropped from $F$ to $BC$, one of the two angles $(\angle ED'F)$ is acute while other $(\angle EDF)$ is obtuse.

I think the book is expecting this kind of analysis as answer. If you're interested, see my answer to this problem. A user with a keen eye spots that the solution can be kept elementary, exactly by using the above condition that if two triangles match by $SSA$ and it is known that both have obtuse angles, then they are congruent/similar.


问:
Show that there does not exist $g \in L^1[0,1]$ such that $\varphi (f) = \int fg\ d\mu.$

Let $0 \neq \varphi \in (L^{\infty} [0,1])^*$ such that $\varphi (f) = 0,$ for all $f \in C[0,1].$ The existence of such a $\varphi$ is guaranteed by Hahn-Banach theorem since $C[0,1]$ is not dense in $L^{\infty} [0,1].$ Show that there does not exist any $g \in L^{1} [0,1]$ such that $\varphi (f) = \int fg\ d\mu,$ for all $f \in L^{\infty} [0,1],$ where $\mu$ is the Lebesgue measure.

How do I prove that? Can anybody please give me some hint?

Thanks for your time.

答: There are a trillion ways to do this. Here's one. Fix $x \in (0,1)$. For large $n \ge 1$, let $\phi_n = \frac{1}{2/n}1_{(x-\frac{1}{n},x+\frac{1}{n})}$. We can take a sequence of continuous functions $\{\phi_n^{(k)}\}_{k=1}^\infty$ such that $\lim_{k \to \infty} \phi_n^{(k)} = \phi_n$ pointwise (consider trapezoid looking things). Since $\int \phi_n^{(k)}gd\mu = 0$ for each $k$, we see that, by the dominated convergence theorem, $\int \phi_n g d\mu = 0$ for each $n$. Since $g(z) = \lim_{r \downarrow 0} \frac{1}{2r}\int_{z-r}^{z+r} g(y)d\mu(y)$ almost all $z$, we conclude that $g(x) = 0$ for almost every $x$. Hence, $\phi \equiv 0$, a contradiction.


问:
How to convert an intuitive logical argument to a formal proof?
I was trying to prove this -

Let $A$ be a lower triangular $n\times n$ matrix with nonzero entries on the diagonal. Show that $A$ is invertible and the inverse of $A$ is lower triangular.

[Hint: Explain why $A$ can be changed into $I$ using only row replacements and scaling. (Where are the pivots?) Also, explain why the row operations that reduce $A$ to $I$ change $I$ into a lower triangular matrix.]

I believe we can prove this by considering the matrix $[A|I]$, now we since $A$ is lower triangular we can first scale the pivot in the first row to 1 and then use it to remove all subsequent entries in first column below it then scale the pivot in column 2 and so on.

Clearly we can reduce the matrix to the identity matrix but at the same time I will have the same row operations so we'll be multiplying and adding multiples of the pivots to rows below it and generating a lower triangular matrix only since there will be no instance when we add a row to a row above it we'll only use the pivots to remove the entries in the column below it. So the inverse generated will be a lower triangular matrix only using the theorem which states that if a set of row operations reduce a matrix to I then they reduce I to the its inverse

Now this is logically correct as far as I know but this doesn't look much like a proof rather an intuitive argument to me and there is where I always get stuck. I get to a point where I know why it's true but don't know how to write it formally. So using this example can someone give tips and techniques to write it in a more "proof like" manner if I may say so

I guess it's been asked here but I'm not looking for a proof per say but a way to convert my intuition into a set of rigorous statements

答:
For proofs, formulae or "lots of mathematical terms" are not necessary in general. There is a whole discipline for achieving exactly that:
Proof without words, which is considered a high art ....</p>


问:
Finding the length of c
I am not sure if that is an easy or hard question, can c be calculated in this scenario where a and b are known. Appreciate any help.

答: The answer is no, $c$ cannot be calculated if $a$ and $b$ are known. Let be be $x$ cm. Extend a right angle from $b$ as far as possible. Complete the right triangle and mark off $a$. Note that $c$ depends on how far you extend the right angle from $b$.

Title: Finding the length of c


问:
Convergence of subsequence proof
Suppose $(a_n)_{n\in\mathbb{N}}$ is a sequence with $a_n\rightarrow a$ as $n\rightarrow\infty$. Prove that $a_{2n}\rightarrow a$ as $n\rightarrow\infty$ also.
I'm not sure about my proof - can someone check this for me?

My attempt: By definition of convergence, for all $n\geq N_\epsilon$ we have $|a_n-a_|<\epsilon$. Rearranging gives $a-\epsilon<a_n<a+\epsilon$ for all $n$. Consequently we have $a-\epsilon<a_{2n}<a+\epsilon$, as $a$ is fixed. Once again rearranging yields $|a_{2n}-a|<\epsilon$, which proves the result.

答:
No need to 'rearrange'. Just use $2n \geq n \geq N_\varepsilon$. But yes, you are right.

If $(a_n)_{n \in \mathbb{N}}$ converges, all subsequences have the same limit.

Title: Convergence of subsequence proof


答:
If you want to show the injectivity, here is one way:
$$\frac{2x-s}{x(s-x)} = \frac{2y-s}{y(s-y)}$$
Bringing to a common denominator, we see that
$$(2x-s)y(s-y) - (2y-s)x(s-x) = 0$$
Assume $x\ne y$ but $x>y$, then
$$(2x - s)y(s-y) - (2y-s)x(s-\color{red}y)<0 \\
\left[(2x-s)y - (2y-s)x\right](s-y) < 0$$
Drop the factor $(s-y)$ since it is positive:
$$(2x-s)y - (2y-s)x < 0 \\
-sy + sx < 0 \\
x < y$$
Contradiction. Since the case $x < y$ is similar, we must have $f(x)=f(y)\implies x=y. \ \ \square$

Title: Show that $f$ is bijective.

Body:
Show that $f$, defined in $0<x<s, (s>0)$ as follows:
$$f(x) = {2x-s \over x(s-x)}$$

is a bijective function of this interval in the reals.

To show that it is injective, we assume $f (x) = f (y)$ and after manipulating we find:
$$(y-x)(-s^2+s(y+x)-2xy)=0$$
then $x=y$, right?

And to show that is surjective just take $f(x)=y$ and manipulating we get $$x = {ys-2 \pm \sqrt {4+y^2s^2} \over 2y}$$

Is that right? The result looks kind of ugly, I was not very confident.


问:
Say I have some element $g\in G$ such that $g$ is of infinite order. I want to show that $|g|=|\langle g \rangle|$.
Say I have some element $g\in G$ such that $g$ is of infinite order. I want to show that $|g|=|\langle g \rangle|$.

I know that if $g$ is of infinite order, then there does not exist any $n$ such that $g^n=1$.

To set up a contradiction, let's suppose there exists $s,t$ such that $g^s=g^t$, if I multiply both sides by $g^{-t}$ I get that $g^{s-t}=g^{t-t}=g^0=1$. This contradicts our original statement of $g$ being of infinite order.

Is this enough to show that $|g|=|\langle g \rangle|$? I have shown that there does not exist any power of $g$ such that it will bring you back to $1$, so can I therefore say that since $\langle g \rangle$ is made of powers of $g$ that it, too, has infinite order?

Edit: $|g|$ is defined as "$n>0$ such that $n$ is the smallest positive integer that gives $g^n=e$ where $e$ is the identity element.

答: Yes. Since $g^s\neq g^t$ for all $s\neq t$, $\langle g\rangle=\{g^k\mid k\in\mathbb{Z}\}$ contains an infinite number of elements. Hence, $|g|=\infty=|\langle g\rangle|$.


问:
Closure of the set of null sequences starting from some rank
Let $E$ be the vector space of real bounded sequences $(x_n)_n$. We endow $E$ with the $\sup$ norm. Let $A$ be a subset of $E$ made of all the sequences that are null starting from some rank.

I must show that the closure of $A$ is a subset of $E$ made of all the sequences that converge to 0.

Here is a draft of my proof. I don't know if it is enough to prove the assertion. Furthermore, I find my proof a bit messy. It could be more rigorous.

Let $B = \{ (x_n) \in E : \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \lvert u_n \rvert < \epsilon \}$ and $A = \{(x_n) \in E : \exists N \in \mathbb{N}, \forall n \geq N, u_n = 0 \}$.

$\text{cl}(A)$ contains the limits of all the convergent sequences of A.

Let $l \in \text{cl}(A)$. Hence, $l$ is the limit of a convergent sequence $(x_n)_n$ of A.

More formally,

\begin{equation*}
    \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \lvert\lvert x_n - l \rvert\rvert < \epsilon
\end{equation*}

So,

\begin{equation*}
    \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \sup_{n \in \mathbb{N}} \lvert x_n - l \rvert < \epsilon
\end{equation*}

Yet, starting from some rank $N \in \mathbb{N}$, the terms $(x_n)_n$ are null. Let such rank $N \in \mathbb{N}$. Starting from this rank, it follows that:

\begin{equation*}
    \forall \epsilon > 0, \forall n \geq N, \sup_{n \in \mathbb{N}} \lvert x_n - l \rvert < \epsilon
\end{equation*}

\begin{equation*}
    \forall \epsilon > 0, \forall n \geq N, \lvert l \rvert < \epsilon
\end{equation*}

Or more simply,

\begin{equation*}
    \forall \epsilon > 0, \lvert l \rvert < \epsilon
\end{equation*}

Hence $l = 0$.

So, $\bar{A}$ is made of all the sequences $(x_n)_n$ that converges to 0.

答:
You definitely have the right idea: all sequences that converge to $0$ are sufficiently "close" to an element of $A$.

There are a few instances where I'm not totally sure of your "notation", so I'm just going to write out what I think is a correct answer, and you can judge whether you have essentially the same thing.

Given some $(x_n) \in \bar{A}$, we first have to show that $(x_n)$ converges to $0$. Since $(x_n)$ is in the closure, it follows that for any $\epsilon > 0$, we may choose some $(y_n) \in A$ such that $d((x_n), \ (y_n)) < \epsilon$.

Let $M$ to the number such that for $m > M$, we have $y_m = 0$. It then follows, from the fact that $d((x_n), \ (y_n)) < \epsilon$ we have:

$$|x_m - y_m| = |x_m| < \epsilon$$

for all $m > M$. Hence, by definition, $(x_n)$ converges to $0$.

Now, let $(x_n)$ be a sequence that converges to $0$. As you said, this means that for any $\epsilon > 0$, we can choose some $N$ such that for $n > N$, we have:

$$|x_n| < \epsilon$$

Consider the sequence $(y_n)$ such that $y_j = x_j$ for all $j$ up until $N$, and then $0$ after. Clearly, we will have:

$$|x_n - y_n| = 0 \ \text{for} \ n \leq N$$
$$|x_n - y_n| = |x_n| < \epsilon \ \text{for} \ n > N$$

Therefore, $d((x_n), \ (y_n)) < \epsilon$. It follows that the open $\epsilon$-ball around $(x_n)$ intersects $A$. Hence, $(x_n) \in \bar{A}$.

We have inclusion both ways, so $\bar{A}$ is exactly the set of all sequences that converge to $0$. Let me know if this helps/if you have any more questions!


问:
Why is the interval not a type but a pretype?
In Section 3.3 of Naive cubical type theory (Bruno Bentzen):

We suggested above that every type is Kan. In fact, the interval is the only
exception to this rule, since we have been implicitly treating it as a “type”, but
it actually does not support any Kan operations. Indeed, if the interval were
Kan, then the identity path
$$ \lambda i.i : \mathrm{path}_{\mathbb{I}}(0, 1) $$
would have an inverse, but what could that be? To deal with this fact, we adopt
the convention that the interval can only occur as the antecedent of a function
type (hence, the interval may be called a “pretype”).

I could not understand the fact that the path $\lambda i.i$ does not have an inverse.
Isn't it simply $(\lambda i.i)^{-1}:\mathrm{path}_{\mathbb I}(1,0)$?

答:
I think the answer to this question is probably a lot more complicated than that particular paper wants to get into. It's trying to present a simplified, less formal version of cubical type theory, while the reason behind this aspect has to do with technical details.

In a lot of formal presentations of cubical type theory, the interval isn't a type at all. Instead, there are actually two sorts of variables in the theory. One sort ranges over types, like in ordinary type theory. The other ranges over the interval, and is related to the fact that that every type is a presheaf on the cube category. (Pre)sheaves are related to variable contexts so this is potentially a general technique.

So, if you arrange the 'naive' (or even formal) theory to have a sort of 'pre-type' for interval variables, it generally does not make sense to treat it like a normal type, because it just has a different status. The operations on such a pre-type are supposed to correspond to things that can happen in the cube category, and the genuine types (which are presheaves) aren't even in that category. The cube category generally lacks other structure, too. It might not be closed, just Cartesian. It might not even be Cartesian. etc.

Now, the Yoneda embedding gives you an object representing the cube category interval in the cubical set category, so you might imagine that you could make that your 'type', since it would behave more like one. However, then it is still just a fact that this cubical set is not Kan (I think that's a relevant citation; I don't have the expertise to explain why this is). So it is still not a type but a 'pre-type', since the Kan operations don't necessarily work on it, like in the paper you're reading. I think this is a more general phenomenon, where the whole cubical set category contains many things that only qualify as 'pre-types' not types, because only ones with additional structure qualify as the latter. I imagine this could be formally presented as a "two level type theory," or something similar.

You can get a weakly equivalent object by "fibrant replacement." So you could consider it to be the interval that is an actual type. I think an example of that would be a higher-inductive definition of 'the interval' like:

```<code>data Intv : Type where
  zero one : Intv
  seg : zero = one
</code>```

I'm not sure if anything in the paper you're reading would go awry with imagining $\mathbb{I}$ to actually be such a type. In a formal theory, (I think) the purpose of the pre-types would be to distinguish more fine-grained detail than the types are allowed to. E.G. two pre-types might be considered distinct even though they have equivalent fibrant replacements. The above inductive type is equivalent to the unit type, but obviously something about it is supposed to not be the same as the unit type. I think that something is that it is presented by a different pre-type.

Body:
I think the answer to this question is probably a lot more complicated than that particular paper wants to get into. It's trying to present a simplified, less formal version of cubical type theory, while the reason behind this aspect has to do with technical details.

In a lot of formal presentations of cubical type theory, the interval isn't a type at all. Instead, there are actually two sorts of variables in the theory. One sort ranges over types, like in ordinary type theory. The other ranges over the interval, and is related to the fact that that every type is a presheaf on the cube category. (Pre)sheaves are related to variable contexts so this is potentially a general technique.

So, if you arrange the 'naive' (or even formal) theory to have a sort of 'pre-type' for interval variables, it generally does not make sense to treat it like a normal type, because it just has a different status. The operations on such a pre-type are supposed to correspond to things that can happen in the cube category, and the genuine types (which are presheaves) aren't even in that category. The cube category generally lacks other structure, too. It might not be closed, just Cartesian. It might not even be Cartesian. etc.

Now, the Yoneda embedding gives you an object representing the cube category interval in the cubical set category, so you might imagine that you could make that your 'type', since it would behave more like one. However, then it is still just a fact that this cubical set is not Kan (I think that's a relevant citation; I don't have the expertise to explain why this is). So it is still not a type but a 'pre-type', since the Kan operations don't necessarily work on it, like in the paper you're reading. I think this is a more general phenomenon, where the whole cubical set category contains many things that only qualify as 'pre-types' not types, because only ones with additional structure qualify as the latter. I imagine this could be formally presented as a "two level type theory," or something similar.

You can get a weakly equivalent object by "fibrant replacement." So you could consider it to be the interval that is an actual type. I think an example of that would be a higher-inductive definition of 'the interval' like:

data Intv : Type where
  zero one : Intv
  seg : zero = one

I'm not sure if anything in the paper you're reading would go awry with imagining $\mathbb{I}$ to actually be such a type. In a formal theory, (I think) the purpose of the pre-types would be to distinguish more fine-grained detail than the types are allowed to. E.G. two pre-types might be considered distinct even though they have equivalent fibrant replacements. The above inductive type is equivalent to the unit type, but obviously something about it is supposed to not be the same as the unit type. I think that something is that it is presented by a different pre-type.


问:
How do you prove Osborn's rule?
Given a trigonometric identity written in terms of sine and cosine, it is possible to write down the corresponding hyperbolic identity using Osborn's rule:

Replace $\cos$ with $\cosh$

Replace $\sin$ with $\sinh$

However, negate the product of two $\sinh$ terms. For example,
$$
\cos^2x+\sin^2x \equiv 1 \implies \cosh^2x-\sinh^2x=1 \, .
$$
I tried proving this rule using the following two identities
\begin{align}
\cos ix &\equiv \cosh x \\
\sin ix &\equiv i\sinh x \, .
\end{align}
Given an arbitrary trigonometric identity of the form
$$
f(\cos z,\sin z) \equiv g(\cos z, \sin z)
$$
then letting $z=ix$ we see that
\begin{align}
f(\cos ix,\sin ix) &\equiv g(\cos ix, \sin ix) \\
f(\cosh x,i\sinh x) &\equiv g(\cosh x, i\sinh x) \\
\end{align}
From here, it seems the rule makes sense because the product of two $i\sinh x$ terms gives us $\color{red}{-}\sinh x$. But I'm not quite sure how to formalise this argument. Moreover, an identity such as
$$
\sin 2z \equiv 2\sin z \cos z
$$
would correspond to
$$
i\sinh x \equiv 2i\sinh x \cosh x \, .
$$
Here, we could divide through by $i$ and the expected hyperbolic identity pops out, but I'm unsure if this 'divide by $i$' trick always works, since there may be other terms that don't contain an $i$ in them. To give a slightly contrived example,
$$
\sin x + \cos 2x \equiv \sin x + \cos^2 x - \sin^2x
$$
corresponds to
$$
i\sinh x + \cosh 2x \equiv i\sinh x + \cosh^2 x + \sinh^2 x \, ,
$$
and here, dividing by $i$ does not help. So I think I've understood the crux of how to prove Osborn's rule, but there are a few loose ends that I need to sort out.

答:
It's worth noting the origin of Osborn's Rule among a list of brief "Mathematical Notes" in The Mathematical Gazette, Vol 2 No. 34 (Jul 1902), pg 189 (JSTOR link). See this screenshot.

Transcribing in its entirety ...

[D.6.d.] Mnemonic for hyperbolic formulae.

Hyperbolic functions are now so constantly used, that a brief mnemonic for their somewhat confusing formulae may not be unwelcome.

In any Trigonometrical formula for $\theta$, $2\theta$, $3\theta$, or $\theta$ and $\phi$, after changing sin to sinh, cos to cosh, etc., change the sign of any term that contains (or implies) a product of sinhs, eg $\tanh\theta \tanh\phi$ implies a product of sinhs,

$$\begin{align}
\therefore \tanh(\theta+\phi) &= \frac{\tanh\theta+\tanh\phi}{1+\tanh\theta\tanh\phi}\; ; \\[4pt]
\sinh3\theta &= 3\sinh\theta + 4\sinh3\theta\; ; \\[4pt]
\cosh\theta-\cosh\phi &= +2\sinh\frac{\theta+\phi}{2}\sinh\frac{\theta-\phi}{2}\; ;
\end{align}$$
and so on. This rule would likely fail for terms of the fourth degree, but it covers everything that is likely to be required,
and is very convenient for teaching purposes.

G. Osborn

Despite the overly-ambitious ---one might say, hyperbolic--- claim of applicability to "any Trigonometrical formula", it seems clear that what has come to be known as "Osborn's Rule" was intended, and has ever only served, as something more like "Osborn's Rule of Thumb": it's a quick-and-easy way (especially for students) to remember when to change signs in familiar fundamental Trigonometrical formulas.

The Rule is a teaching tool, not an over-arching general principle.

Title: How do you prove Osborn's rule?


问:
Give a big-$\mathcal{O}$ estimate for the number of times the algorithm needs to determine whether an integer is in one of the subsets.
Suppose we have n subsets $S_1, S_2, . . . , S_n$ of the set ${1, 2, \textit{...} , n}$. The a brute-force algorithm that determines whether there is a disjoint pair of these subsets is as follows below.

Based on my understanding, the algorithm starts by comparing set $S_1$ to all other sets $S_2, S_3, \textit{...}, S_n$. In the second round, it should compare $S_2$ with all other subsets $S_3, \textit{...}, S_n$, until we reach the last set and compare it with itself $S_n$.

Question 1: could you please tell me the rule of $a_i < a_j$ below in the algorithm?

Question 2: based on big-$\mathcal{O}$, the running time for comparisons is $2\times(n-1)\times (n-1)\times n$. However, I only see one comparison below in the algorithm not 2.

答: The if $a_i \lt a_j$ line is a mistake.  $a_i,a_j$ are never defined and the algorithm does not need it.  The $i$ loop is executed $n-1$ times, the $j$ loop is executed an average of about $\frac n2$ times, and the $k$ loop is executed $n$ times, so the number of inclusion checks  is $2\cdot (n-1)\cdot \frac n2\cdot n$ with the $2$ coming from checking $S_i$ and $S_j$.  We don't care about multiplicative constants.  The algorithm is then $\mathcal O(n^3)$


问:
Find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$
I have the sequence $(a_{n})_{n}$ with first term $a_{1}=2$ and recursive relation $$a_{n+1}=\frac{2}{1+a_{n}}$$
So I have to find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$

Any help?

答:
Multiplying by the denominator we arrive to $$a_na_{n+1}=2-a_{n+1}$$

In this kind of problem, we want to shift $a_n$ so as to cancel the constant term (see this other problem for instance https://math.stackexchange.com/a/3412108/399263)

If you are not sure about which shift to apply, you can set $b_n=a_n+\alpha$ and work out possible values of $\alpha$.

$(b_n-\alpha)(b_{n+1}-\alpha)=2-(b_{n+1}-\alpha)\iff b_nb_{n+1}=(\alpha-1)b_{n+1}+\alpha\,b_n+\underbrace{(2+\alpha-\alpha^2)}_\text{make it zero}$

We have two possibilities:

$\alpha=2$ and set $b_n=a_n+2$

$\alpha=-1$ and set $b_n=a_n-1$

Remark that now if we divide by $b_nb_{n+1}\quad$ we get $\quad1=(\alpha-1)\dfrac 1{b_n}+\alpha\dfrac 1{b_{n+1}}$

So let introduce $c_n=\dfrac 1{b_n}$ we arrive at a linear induction equation:

$$\alpha\, c_{n+1}+(\alpha-1)c_n=1$$

Let solve for $\alpha=2$, as an exercise you can try on your own with the other value $-1$ and see if you get the same result.

We get $2c_{n+1}+c_n=1$ so homogeneous equation solves to $(-\frac 12)^nC$ and particular solution is constant $2D+D=1\iff D=\frac 13$

Therefore $c_n=(-\frac 12)^nC+\frac 13$.

We have to solve for initial condition $a_1=2=b_1-2=\dfrac 1{c_1}-2=\dfrac 1{-\frac C2+\frac 13}-2\iff C=\frac 16$

After reporting in the expression of $a_n$ and simplifying we get the result proposed in NoName's and Raffaele's answers.

$$a_n=\dfrac{1}{\frac 16(-\frac 12)^n+\frac 13}-2=\dfrac{2^{n+1}-2(-1)^n}{2^{n+1}+(-1)^n}$$


答:
Query ambiguous.  Depending on how query (i.e. drawing) interpreted, $c$ is either known or unknown.

Let $\theta$ denote the angle formed by the two line segments whose lengths are labeled $a$ and $c$.  Presumed that $\theta$ and these two line segments form an inner right triangle.

Presumed that diagram indicates that $b$ is the leg of an outer right triangle, where $c$ is a portion of the other leg and $a$ is a portion of the hypotenuse.

Then, it is clear that $c$ is known if and only if $\theta$ is known.

The question is whether there is some relationship between $a$ and the hypotenuse.  For example, if it is to be presumed that the hypotenuse has length $(2a)$, then $\sin(\theta) = (b/2a)$ which implies that $\theta$ is known, which implies that $c$ is known.

Alternatively, if the hypotenuse of the outer right triangle is unknown, then $\theta$ can not possibly be known, which implies that $c$ can not be known.

Title: Finding the length of c

Body: I am not sure if that is an easy or hard question, can c be calculated in this scenario where a and b are known. Appreciate any help.


问:
What's so well in having a least element in a set?
What's so well about the principle of well-ordering? Why is this pattern of order named like this? I am unable to trace a connection between well-ordering and a set that has a least element so I presume that there must be some mathematical becakground I'm unaware of. I've searched some wikipedia links but they describe only what the principle is, not the reason why it's named like this.</p>

答:
It's just a name, I wouldn't spend too much time pondering it.

It's not totally arbitrary though - for example, choosing the minimal element of a subset of some set $X$ is well-defined exactly if $X$ is well-ordered. Similarly, an inductively defined function is well-defined only if the domain is well-ordered.


问:
Values of Integral with Function
I have an integral $\int_{0}^{1}e^{2x}f(x)dx$. But $f(x)$ is another function ,not an integral. It is a function that is twice differentiable, and $f''(x)=kf(x)$, $k$ is constant. Then there is local max at $(0,1)$ and local min at $(1, -1)$. So what are possible values of the integral? How to use local max and local min to solve the problem? I have one value is $\frac{e^{2}+1}{k-2}$, is this right? What is other values?

Title: Values of Integral with Function

Body:
We will integrate by parts twice:
$$\begin{align}
\int e^{2x}f(x)dx
&= \frac{1}{2}e^{2x}f(x) - \frac 12\int e^{2x}f'(x)dx \\
&= \frac{1}{2}e^{2x}f(x) - \frac 12\left(\frac 12 e^{2x}f'(x) - \frac 12\int e^{2x}f''(x)dx\right) \\
&= \frac{1}{2}e^{2x}f(x) - \frac 14 e^{2x}f'(x) + \frac 14 \int ke^{2x}f(x)dx\end{align}$$
Denoting $I = \int e^{2x}f(x)dx$, we get
$$
\begin{align}
I &= \frac{1}{2}e^{2x}f(x) - \frac 14 e^{2x}f'(x) + \frac k4I \\
\frac{4-k}4 I &= \frac{1}{2}e^{2x}f(x) - \frac 14 e^{2x}f'(x) \\
I &= \frac{2}{4-k}e^{2x}f(x) - \frac{1}{4-k}e^{2x}f'(x)
\end{align}$$
Hence,
$$\begin{align}
\int_0^1 e^{2x}f(x)dx &= \left[\frac{2}{4-k}e^{2x}f(x) - \frac{1}{4-k}e^{2x}f'(x)\right]_0^1 \\&= \frac{2}{4-k}e^2f(1) - \frac{1}{4-k}e^{1}f'(1) - \frac{2}{4-k}f(0) + \frac{1}{4-k}f'(0) \\&= -\frac{2e^2+2}{4-k}
\end{align}$$
Since $x=0$ and $x=1$ are critical points, we have $f'(0)=f'(1)=0$.


问:
Proving integration by parts formula using divergence theorem
I'm working on an exercise that requires me to prove the integration by parts formula

$$\int_{\Omega}u_{x_j}v\ dx=-\int_{\Omega}uv_{x_j}\ dx+\int_{\partial\Omega}\nu_j uv\ dS$$

by use of the divergence theorem

$$\int_{\Omega}\nabla  F\ dx=\int_{\partial\Omega}\nu\cdot F dS$$

where $\nu$ denotes the outward pointing normal to $\partial\Omega$.

I don't exactly know how I would go about doing this, so any help/hints are highly appreciated!

Thanks in advance.

答:
Hint: apply the divergence theorem to $$F=\begin{pmatrix}0\\\vdots\\0\\uv\\0\\\vdots\\0\end{pmatrix}$$
where the $uv$ is in the $j$-th entry.


Title: Values of Integral with Function

Body: I have an integral $\int_{0}^{1}e^{2x}f(x)dx$. But $f(x)$ is another function ,not an integral. It is a function that is twice differentiable, and $f''(x)=kf(x)$, $k$ is constant. Then there is local max at $(0,1)$ and local min at $(1, -1)$. So what are possible values of the integral? How to use local max and local min to solve the problem? I have one value is $\frac{e^{2}+1}{k-2}$, is this right? What is other values?

Body:
$f''(x)-kf(x)=0;\;f(0)=1,f'(0)=0,f(1)=-1$

Dolving the DE we get
$$f(x)=a e^{\sqrt{k} x}+b e^{-\sqrt{k} x};\;f'(x)=a \sqrt{k} e^{\sqrt{k} x}-b \sqrt{k} e^{-\sqrt{k} x}$$
Setting the initial conditions
$$\begin{cases}
a+b=1\\
a-b=0\\
a e^{\sqrt{k}}+b e^{-\sqrt{k}}=-1
\end{cases}
$$
$$a=\frac12;\;b=\frac12$$
$$\frac12 e^{\sqrt{k}}+\frac12 e^{-\sqrt{k}}=-1$$
$$\frac{e^{\sqrt{k}}+e^{-\sqrt{k}}}{2}=-1\to k=-\pi^2$$
and finally
$$f(x)=\frac{ e^{\pi i x}+b e^{-\pi i x}}{2}\to f(x)=\cos (\pi  x)$$
and then
$$\int_0^1 e^{2 x} \cos (\pi  x) \, dx=\left[\frac{e^{2 x} (\pi  \sin (\pi  x)+2 \cos (\pi  x))}{4+\pi ^2}\right]_0^1=$$

$$=-\frac{2 \left(1+e^2\right)}{4+\pi ^2}$$


问:
If $Tx\perp b$ for all $x$ with $x\perp b\in B$, does it follow $T\overline{\text{span}B}^\perp\perp\overline{\text{span}B}$?
Let $H$ be a complex Hilbert space, $T\in\mathfrak L(H)$, $B\subseteq H$ with $$Tx\perp b\;\;\;\text{for all }x\in H\text{ and }b\in B\text{ with }x\perp b\tag1$$ and $U:=\overline{\operatorname{span}B}$.

Are we able to conclude $TU^\perp\subseteq U^\perp$?

From $(1)$ we deduce that $$TB^\perp\perp B\tag2;$$ i.e. $$TB^\perp\subseteq B^\perp.\tag3$$ By (conjugate-)linearity of the inner product, we easily conclude $$TB^\perp\perp\operatorname{span}B\tag4$$ and from continuity of the inner product, we obtain $$TB^\perp\perp U\tag5.$$ Can we somehow utilize linearity and continuity of $T$ to deal with the left-hand side of $(5)$ in a similar way?

答:
We show that $TU^\perp \subseteq U^\perp$. Let $x \in U^\perp$. Then if $b\in B\subseteq U$, we have $Tx\perp b$
since $x \perp b$ (your assumption). Hence, $Tx \perp B$ and by linearity and continuity $Tx \perp U$. Thus, $Tx \in U^\perp$. Since $x \in U^\perp$ was arbitrary, we can conclude.

From (5), you can also immediately see your result. Indeed, note that $U^\perp \subseteq B^\perp$ (in fact this is even equality) and thus $TU^\perp \subseteq TB^\perp$. Hence, by $(5)$ we have $TU^\perp \perp U$ and thus $TU^\perp \subseteq U^\perp.$


答:
It's a good order to do induction over, because we always know what's next. If you are given a bounded subset, you know what the "next step".

Therefore you can perform induction over a linear order when it is well-ordered. That's good, that's very well.

(Another analogy is to a line to the bathroom. If the line is well-ordered everybody knows their place, and whoever wants to get in line knows exactly where they need to stand, and no one can cut in line. If the line is ordered like the rational numbers then everything is dense and people can cut in line wherever they feel like it, but no one is moving along anyway! That's not very good.)

Other than that, don't put too much emphasis on the actual name. Focus on the definition itself, it's much better.

Title: What's so well in having a least element in a set?

Body: What's so well about the principle of well-ordering? Why is this pattern of order named like this? I am unable to trace a connection between well-ordering and a set that has a least element so I presume that there must be some mathematical becakground I'm unaware of. I've searched some wikipedia links but they describe only what the principle is, not the reason why it's named like this.</p>


问:
Comparing norms on $l^{\infty}$
Consider the Banach space $(l^{\infty}, \| \cdot\|_{\infty})$ and let $\| \cdot \|$ be another on $l^{\infty}$ such that $(l^{\infty}, \| \cdot \|)$ is also a Banach space. Assume that, for each $j$, the mapping $\phi_j \colon l^\infty \to \mathbb{R}$ given by $\phi_j( (x_n)_{n \ge 1} ) = x_j$ is continuous with respect to the norm $\| \cdot \|$.

We want to show that there exists $C > 0$ such that $$\| x\| \le C \| x \|_{\infty}$$ for all $x \in l^{\infty}$.

Since $\phi_j$ is continuous with respect to $\| \cdot \|$, we have there is a constant $M_j>0$ such that $\| \phi_j(x) \| \le M_j \|x\| $ (that is, $\phi_j$ is bounded). But I have no idea to continue, only that I need to use the Closed Mapping Theorem.

Thanks in advance.

答:
Well the existence of such a constant is equivalent to showing that the identity operator $I:(\ell^\infty,\|\cdot\|_\infty)\to(\ell^\infty,\|\cdot\|)$ is bounded. We may use the closed graph theorem for this, as we have Banach spaces for domain and target. Assume that $(x_n)\subset\ell^\infty$ is a sequence so that $x_n\to0$ in the $\|\cdot\|_\infty$ norm, i.e. $\|x_n\|_\infty\to0$. We want to show that $x_n\to 0$ in the $\|\cdot\|$ norm and by the closed graph theorem we can assume that $x_n\to y$ in $\|\cdot\|$ for some $y\in\ell^\infty$. Let's say $y=(y_1,y_2,\dots)$ and say $x_n=(x_n^1,x_n^2,\dots)$.

So we have that $\|x_n-y\|\to0$, so by the assumption that those functionals are continuous we have that for any $j\in\mathbb{N}$:
$$x_n^j\to y_j$$
But $|x_n^j|\leq\sup_{j}|x_n^j|=\|x_n\|_\infty\to0$, so $x_n^j\xrightarrow[n\to\infty]{}0$ and therefore $y_j=0$ for all $j$. Thus $y=0$ and we are done.

Title: Comparing norms on $l^{\infty}$


问:
if the Lebesgue outer measure of a set A which is a subset of R is positive does the set A contain for certain an interior point?
I was looking at this  exercise from measure theory.
If the set $A \subset \mathbb{R}$ contains at least one interior point show that $m^*(A)>0$.
I have done the exercise but I am wondering whether the inverse logical statement is true, meaning if
$m^*(A)>0$, does that imply that $A$ contains at least one interior point? I haven't figured anything out. Could someone help me? I feel that the statement is true because for $A$ to have a positive outer measure it must contain an interval and therefore an interior point but I am not certain.

答: No, it is not true. Take a fat Cantor set, which is measurable with Lebesgue measure greater than $0$ (and therefore its outer measure is greater than $0$). But its interior is empty.


答: $m^*(\Bbb Q)=0$ since it is countable, so $m^*([0,1]\setminus\Bbb Q)=1$ but it has no interior point.

Title: if the Lebesgue outer measure of a set A which is a subset of R is positive does the set A contain for certain an interior point?

Body:
I was looking at this  exercise from measure theory.
If the set $A \subset \mathbb{R}$ contains at least one interior point show that $m^*(A)>0$.
I have done the exercise but I am wondering whether the inverse logical statement is true, meaning if
$m^*(A)>0$, does that imply that $A$ contains at least one interior point? I haven't figured anything out. Could someone help me? I feel that the statement is true because for $A$ to have a positive outer measure it must contain an interval and therefore an interior point but I am not certain.


问:
$\frac{d\textbf{y}}{dt} = \textbf{f(y)}$ representation of ODE system
I have the following ODE's :

$$\frac{ds}{dt} = 1-s-\frac{msx}{a+s}$$
$$\frac{dx}{dt}=\frac{msx}{a+s} - x$$

I need to represent the system in the form $\frac{d\textbf{y}}{dt} = \textbf{f(y)}$.

I get that they're asking me some kind of vectorial representation where we have $\textbf{f}(t,\textbf{y})$  and:

$$\textbf{y}(t) = \begin{pmatrix} y_1(t) \\y_2(t)  \end{pmatrix}$$

but I'm confused about how to get there since both the function $x$ and $s$ appear in both equations. Factorizing the function $x$ in the second equation is no problem but in the first equation the function $s$ appears as well in the denominator so factorizing is no option. I'm still confused about the representation they're asking me to do so maybe I'm in the wrong path factorizing the functions. I would apreciate some help. Thanks in andvance.

答:
Set $y=(s,x)^T$ and $f(y)=(f_1(y),f_2(y))^T$ with
$$
f_1(y)=1-s-\frac{msx}{a+x},\quad f_2(y)=\frac{msx}{a+x}-x
$$

If you want $y=(y_1,y_2)$, replace $s$ and $x$ above with $y_1$ and $y_2$ repectively.


问:
$S^3$ as union of 2 solid tori
I am working through Hatcher's Algebraic Topology and don't get what he means by "standard decomposition of $S^3$ into two solid tori",

$$S^3= S^1 \times D^2 \cup D^2 \times S^1$$

Can someone explain (or visualize) what the union looks like?
First, I thought it looked like a genus-$2$-torus:

but I guess that's not what he means, actually...
I have basic knowledge in algebraic topology and I am not familiar with Heegaard Splittings.

To obtain an embedding of $X$ in $S^3-K$ as a deformation retract we will use the standard decomposition of $S^3$ into two solid tori $S^1\times D^2$ and $D^2\times S^1$, the results of regarding $S^2$ as $\delta D^4=\delta(D^2\times D^2)=\delta D^2\times D^2\cup D^2\times\delta D^2$. Geometrically, the first solid torus $S^1\times D^2$ can be identified with the compact region in $\Bbb R^3$ bounded by the standard torus $S^1\times S^1$ containing $K$, and the second solid torus $D^2\times S^1$ is then the closure of the complement of the first solid torus, together with the compactification point at infinity. Notice that meridional circles in $S^1\times S^1$ bound disks in the first solid torus, while it is longitudinal circles that bound disks in the second solid torus.

I have read many similar questions and answers, but none I found could help me out, so I started a new one.

答:
if we write the 3 sphere as $w^2 +x^2 +y^2 +z^2 =1 $
the torus of intersection is
$$  w^2 +x^2 = \frac{1}{2} \; , \; \; \; y^2+z^2  = \frac{1}{2}  $$

One solid torus has $w^2 + x^2$ from $0$ to $1/2$  but $y^2 + z^2$ from $1/2$ to $1,$  so that each fixed value of $y^2 + z^2$ gives a disc

Title: $S^3$ as union of 2 solid tori


问:
Factorization in Gaussian integers
Let $p$ be a natural number, suppose $p$ prime. Show that the following conditions are equivalent:

the polynomial $x^2+1\in\mathbb{Z}_p$ has roots in $\mathbb{Z}_p$

$p$ is reducible in the ring $\mathbb{Z}[i]$

there exists $a,b\in\mathbb{Z}$ such that $p=a^2+b^2$

My attempt: suppose $x$ is a solution of $x^2=-1 \ mod\ p$. Raising both sides to $(p-1)/2$ gives  $x^{p-1}=(-1)^{(p-1)/2}\ mod \ p$. Suppose $p$ congruent to $3$ modulo $4$. Then $(p-1)/2$ is odd and $(-1)^{(p-1)/2}=-1$, while $x^{p-1}=1$, a contradiction. Hence $p$ must be congruent to $2$ modulo $4$, i.e. $p=2$ or $p$ congruent to $1$ modulo $4$, i.e. $p=1+4n$, some $n$.

If $p=2$ then $p=(1+i)(1-i)$ is reducible. If $p=1+4n$ i can't see how to show reducibility.

For $2)\rightarrow 3)$, suppose $p=\alpha\cdot\beta$ with $\alpha,\beta$ gaussian integers. Then the norm of $\alpha$, say $a^2+b^2$, divides $p$, but $p$ is prime and the norm can't be $1$, because the factorization of $p=\alpha\beta$ is not trivial, hence $p=a^2+b^2$.

Again, for $3)\rightarrow 1)$, i really don't know what to do. Thanks for any help.

答:
We deal first with the $p$ of shape $4n+1$ case you had trouble with. Suppose that the congruence has a solution $s$. Then $p$ divides $(s+i)(s-i)$. If $p$ were irreducible, it would be a Gaussian prime, so it would divide one of $s-i$ or $s+i$. But it  doesn't.

For the other question, suppose $p=a^2+b^2$. Clearly $p$ divides neither $a$ nor $b$. Multiply $a^2+b^2$ by the square of the inverse $c$ of $b$ modulo $p$. We get $(ca)^2+1\equiv 0\pmod{p}$.

Title: Factorization in Gaussian integers


问:
I found the solution of Ax=b in the least squares but I don't understand how to use it
This is probably a very obvious question, but I don't understand it.

I know how to use the least squares method, which I did, and verified using an online calculator. My system is

("e" = and)

I solved x = $[1 , -1]^T$

I don't understand how this solves the problem. I know that if I multiply A by x I get b. What I don't understand is that, I know this method is applied when you have a bunch of points and want to find a line or function that goes through them, in an approximate way. How do I get a line from x? I know x has to do with the coefficients of a polynomial, and that A and b are the x and y coordinates of the points, so why do A has two columns? It's probably a really silly question but I'm very confused.

Thanks.

答:
You don't have a line here, but a plane.

The data points are $$\begin{align}
&(1,-1,2)\\
&(0,\sqrt2,-\sqrt2)\\
&(-1,1,2)
\end{align}$$ which we'll consider as being of the form $(x,y,z)$ and we want to fit an equation of the form $z=ax+by +\varepsilon$ where $\varepsilon$ represents some inaccuracy in measurement or other "noise".  We stack these equations into matrix form: $$\begin{bmatrix}x_1&y1\\x_2&y_2\\x_3&y_3\end{bmatrix}
\begin{bmatrix}a\\b\end{bmatrix}+\begin{bmatrix}\varepsilon_1\\\varepsilon_2\\\varepsilon_3\end{bmatrix}=\begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}
$$
This is the same as your equation, except that the left- and right-hand sides are interchanged, and $\varepsilon$ is suppressed.  We just want to minimize the sum of the squares of the elements of $$
\begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}-\begin{bmatrix}x_1&y1\\x_2&y_2\\x_3&y_3\end{bmatrix}
\begin{bmatrix}a\\b\end{bmatrix},$$ so we really don't need to write $\varepsilon$ explicitly.

An equation of the form $z=ax+yb$ represents a plane though the origin.  Of course, we can always find a plane passing through $3$ points, but it won't necessarily pass through the origin.  We may have a "model" in mind that says that theoretically, the points will lie on a plane through the origin, and we take the point of view that the only reason they don't is because of unavoidable inaccuracies in measurement, or perhaps blunders.  (If we had $100$ data points, and $99$ of them lay close to a plane through the origin, but the hundredth was way far away, we would might well believe that the deviation was due to an error in making the measurement, or in recording the result.)

If we had more data points, we would have more rows in the matrices, but the same number of columns, and we wouldn't except to be able to pass a plane through all the points.  Then minimizing the sum of the squares of the individual errors, or some other expression for the error, becomes necessary.  We might have more than two independent variables, and then the matrix would have more columns, and there would be more unknown coefficients.

"Linear" doesn't necessarily mean "line-like" in linear algebra.  Also, the plane doesn't have to pass through the origin.  Linear regression would handle a model of the form $$z=ax+by+c.$$

For more information, look at Wikipedia.


问:
positive matrices with largest entry of each row and column is diagonal entry
I am interested in matrices with positive entries and the largest entry in each row and column is the diagonal entry. Is there a name for such matrices? How to generate such positive definite matrices?</p>

答: Yes, they are called diagonally dominant matrices. I had seen an application of them in a numerical analysis course for Jacobi iteration method. "An Hermitian strictly diagonally dominant matrix with real positive diagonal entries is positive definite" so that can help with creating diagonally dominant matrices that are positive definite. There are more details on the wikipedia page..</p>


问:
Do infinity and zero really exist?
From the first day that I entered college, I was wondering about the relationship between some basic mathematical abstract concepts and nature. I'm going to explain them here and you may find them a little bit opinion-based, but it's not. So please guide me in case you have good scientific-based answers/examples.

Infinity:
Does infinity exist in reality? We all know it is an abstract concept but what I'm curious to know is if there are any physical phenomena out there that can show, stimulate or somehow help us understand the concept of infinity in reality/nature/the physical world.

Zero:
What about Zeno's paradox? In nature (our physical world) there is a "smallest distance". It's about $1.6 \times 10^{-35}$ meters. Another example would be when someone says "there are three apples on the desk, if you take all of them there are $0$ apples on the desk". Obviously, it's an abstract concept, but what I would like to know is that if there is any observable physical event that can anyhow show us the concept of nothingness or absolute zero. E.g., we don't have absolute zero temperature in thermodynamics, absolute zero distance between 2 points in mechanics, or absolute zero gravity in a given space, etc.

Axiom of empty set: This axioms states:

"There is a set such that no element is a member of it."

One can imagine an empty set in nature as an absolutely empty box. But one can see it the other way around: the element "nothing" is there (quoting from Prof. Lawrence Krauss's book "nothing is something"). So, the same way one can say the element "nothing" is a member of a set. Obviously, it's an abstract concept and this may look like a play on words, but it's also an interesting paradox.

$\mathbb R$?! (the set of real numbers): There are millions of mathematical theorems which are based on $\mathbb R$! I wonder if there are any non-countable physical phenomena in the real world/nature?

Again, I know these mathematical concepts are just abstractions and help us to solve real-world problems! However, I'm more interested in the relation between these abstract concepts and our physical world/nature.

答:
In reading through these answers, I'm pretty sure that most everything has been answered - except for the treatment of the empty set. By now, I think that we can say that $\infty$ is not a number, but instead a convenient convention referring to an unbounded convention. $0$ is a number and it refers to a very concrete thing - but no one (I think) has yet said anything of how number systems came into being. While I think that many people know that our current positional number systems (like binary or hex for programmers, decimal for day-to-day life) are relatively recent developments. But many number systems do not have a symbol for 'zero.' Why would they? We use it in our positional system all the time - when we write 102, we refer to 1 100, 0 10's, and 2 1's. We rely on zero all the time. But without relying on a positional number system, the only time one would need zero would be when referring to... nothing. There are no apples on the desk and there are zero apples on the desk. Does that deserve its own number?

We think it does, and it's useful for it.

But this is very different than the empty set. We think of sets of things all the time, even if we don't explicitly call it a set. We have a box of chocolates, for example. This is like saying that we have a set of chocolates, or maybe we have a set of boxes that can contain chocolates with exactly one element. What happens when we eat all the chocolates (or as happens upon occasion, the chocolates are not tasty, and so we throw them away)? Then now or 'set of chocolates' collection is empty. But until we throw away the box, our 'set of boxes' still has one thing.

I think that maybe the problem you're identifying with the empty set is related to the concept of vacuously true statements. For example, the set of flying pink elephants that contributed to both Nephireti's and Hitler's rule is empty. And all such elements are right in front of me. All such elephants are also versed in calculus. And so on. The problem with vacuous things is that there is no substance to them - there can be no counter examples found within the set if it's empty. It's very troublesome.

But just like the number zero, the empty set is useful. We use it all the time - it's a useful idea. This is similar to the number 7 - it's a useful idea that may not be entirely obvious. For a while, people had trouble distinguishing between a number and a number of objects. For example, is the '7' in '7 sheep' somehow different than the '7' in '7 cows?' Does it make sense to talk of this '7' apart from some sort of counting reference? We do it all the time, but it's not so obvious to do so.

That's my take on the subject, anyway.

Title: Do infinity and zero really exist?


问:
$\lim\limits_{x\to a^+}f'(x)=\lim\limits_{h\to 0^+}\dfrac{f(a+h)-f(a)}{h}$?
For $f:[a,b]\to \mathbb{R}$, let $f'(c)=\lim\limits_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$ exist for $c\in(a,b)$ and $\lim\limits_{h\to 0^+}\dfrac{f(a+h)-f(a)}{h}$ exist.

I was wondering if it is true that $\lim\limits_{x\to a^+}f'(x)=\lim\limits_{h\to 0^+}\dfrac{f(a+h)-f(a)}{h}$.

答:
No . As a counterexample take
$$f(x)=x^2\cos(\frac 1x)\; \text{ if } \; x>0$$
and $ f(0)=0$.

then

$$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x}=0=f'(0)$$

but for $ x>0 $,

$$f'(x)=2x\cos(\frac 1x)+\sin(\frac 1x)$$

and, it is known that

$$\lim_{x\to 0^+}f'(x)\text{ does not exist}$$

because $$\lim_{x\to0^+}\sin(\frac 1x)=\lim_{X\to +\infty}\sin(X)$$