问:
Ties in Matching Pennies
Players A and B match pennies N times. They keep a tally of their gains and losses. After the first toss, what is the chance that at no time during the game will they be even?

I have seen a solution for this problem summing a series of choose functions for an expectation, but I wanted to solve it with a simple random walk in one dimension.

my logic is thus; suppose a random walk in 1 dimension. half chance for a step to the left or to the right. and one coin toss has gone (say A) so suppose we start at position 1 and go right for A and left for B, how likely is it we reach 0 in N steps or less.

So call $p_N$ the probability for doing it in N steps. Then it seems to me we have $$p_N = \frac{1}{2} + \frac{1}{2}p_{N-2}\frac{1}{2}= \frac{1}{2} + \frac{1}{4}p_{N-2}$$
since if he goes right on the first step he has N-2 steps or less to get back to 1 and then a half chance of hitting 0.

Solving the recurrence on noticing that $p_2$ = $\frac{1}{2}$ gives $$ p_{2n} = p_{2n+1} = \frac{1 + 4 + 4^2 ... + 4^{n-1}}{4^{n-1}.2}$$

Now this agrees with the correct answer for the first couple of cases but then slowly differs as N gets bigger.

Could someone please explain my mistake and perhaps give a correct solution to this problem using random walks?

答: When you write $p_N=\frac 12 +\frac 12p_{N-2}\frac12$ you are assuming the only ways to get back to zero are either to go immediately (the first $\frac 12$) or to take one step away (the leading $\frac 12$ in the next term) and then come back towards zero immediately (the second $\frac 12$ in the second term).  This is consistent with your use of $p_{N-2}$ as you have defined this as the probability of hitting zero in $N-2$ steps starting at $1$.  But you can also have the first step go to $2$, the third to $3$, and then return to zero.  You have missed this chance.</p>

Title: Ties in Matching Pennies


问:
Noetherian and Artinian quotient
Let $I$ be a left ideal of ring $R$. My question is can we lift both Noetherian and Artinian property of $R/I$ as an $R$-module to $R$ as an $R$-module?

Precisely, can we say that $R$ is a left Noetherian (Artinian) $R$-module if and only if  $R/I$ is left Noetherian (Artinian) $R$-module?

答:
One implication
If $R$ as left $R-$module is (Noetherian) Artinian then $R/I$ is too. Suppose it is not, suppose we have infinite (ascending) descending chain of modules in $R/I$ then we have the same (ascending) descending chain of submodules in $R$.

(Recall that lattice (poset) of left ideals containing $I$ is the same as a lattice of submodules in cyclic $R-$module $R/I$ by correspondence theorem.)

This is true even more generally for any lattice. If there are no infinite (ascending) descending chains in lattice $\mathcal{L}$ then there can't be such chains in the sublattices defined as all elements that are above some element $I$.

Second implication As noted in comments there are some trivial cases when it fails, like $R/R$ is always Artinian for any ring $R$.

However, (and this is special for modules, i.e. for Artinian/Noetherian property not true in general lattices), if $M$ is $R-$module such that $N$ and $M/N$ are (Noetherian) Artinian submodules,  for some $R-$submodule $N$ then $M$ itself is (Noetherian) Artinian. (and as a consequence for any submodules $K$ it holds that $K$, $M/K$ are Artin.)

ContentLicense: CC BY-SA 4.0

Title: Noetherian and Artinian quotient


问:
classification modell complexity and distance metrics
the titel says basically all: Are there differences in modell complexity (Bias-Variance) if someone uses (f.e.) kNN with different metrics?
Is there a mathematically background for this? In the web i haven't found anything in this direction.
Is the complexity just the same? I'm asking me that because there are really many metrics one could use
and I'm interested in a (programmable) logic.

kind regards

答:
Yes the choice of distance metric can have an impact on the performance of a KNN model. Different metrics measure closeness in different ways. Consider the L1 and L2 norms. The L1-norm can be said to be less sensitive to outliers and more sensitive to small scale behavior then the l2-norm, focusing on small scale behavior because it does not flatten out around zero like a parabola. It will also be less sensitive to large distances because the slope does not increase with distance from the origin.

This paper might be of interest to you;

Effects of Distance Measure Choice on KNN Classifier Performance - A Review by Prasath et al.


问:
Why is $\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)^0\ne1$?
Why is

$\left( \begin{array}{cc}  1 & 1 \\  1 & 1 \\ \end{array} \right)^0=\left( \begin{array}{cc}  \frac{1}{2} & \frac{1}{2} \\  \frac{1}{2} & \frac{1}{2} \\ \end{array} \right)$

This result me returns Mathematica on this command

```<code>Limit[MatrixFunction[#^s &, ( { {1, 1}, {1, 1} } )], s -> 0, Direction -> "FromAbove"]
</code>```

Result:

```<code>{{1/2, 1/2}, {1/2, 1/2}}
</code>```

Should not anything rised to zeroth power be $1$?

答:
As a limit it is like $0^0$. So you shouldn't write $A^0$. You can diagonalize it, it has eigenvalues 2,0 with corresponding eigenvectors,

$$E = \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}$$

So now we can look at the nth power as,

$$A^n = E^{-1} \begin{bmatrix} 2^n & 0 \\ 0 & 0^n \end{bmatrix} E$$

Taking the limit of this as $n \to 0^+$ now we have $2^0=1$ in the upper entry and we end up with your result.


问:
Physical Interpretation of uniformly continuous function.
I have been reading material on uniformly continuous functions. And going through the problems where we have to prove that a function is not uniformly continuous or otherwise.

A function defined on an interval I is said to be uniformly continuous on I if to each $\epsilon$ there exists a $\delta$ such that

$|f(x_1)-f(x_2)| < \epsilon$, for arbitrary points $x_1, x_2$ of I for which $|x_1-x_2|<\delta$.

Now I understand the above definition in mathematical terms and able to apply this definition to solve problems. But I don't understand why this definition was introduced, how does uniformly continuous functions look like ? If given a graph of a function, how can I tell if it is a uniformly continuous function ?
When I imagine continuous functions, I have this picture in mind as to how they look like but for uniformly continuous functions, I can't think of any picture.

I hope I am able to explain my question correctly. Please help me understand.

答:
For uniformly continuous functions, there is for each
$\varepsilon >0$ a $\delta >0$ such that when we draw a rectangle around each point of the graph with width $2\delta$ and height $2\varepsilon$ , the graph lies completely inside the rectangle.

For functions that are not uniformly continuous, there is an $\varepsilon >0$ such that regardless of the $\delta >0$, when we draw a ${\displaystyle 2\varepsilon \times 2\delta } $ rectangle around a point of the graph, there are points of the graph directly above or below the rectangle. There might be midpoints where the graph is completely inside the rectangle but this is not true for every midpoint.


问:
Determine the following sum
Let $S(n)$ be the sum of the digits of n in its binary representation. For example the binary notation for 19 is $10011$ and $S(19)=3$. Determine $\sum_{n=1}^{\infty} \frac{S(n)}{n(n+1)}$?

I've made several tables trying to pinpoint any useful pattern, but I've had minimal luck. The best I'd came to was determining that the number of digits is $\left \lceil{\frac{ln(n)}{ln(2)}}\right \rceil$ which gives a bound for $S(n)$. What I'd picked up from those tables is $S(2^k-1)=k$ and $S(2^k)=1$. However, breaking the sum up into those n's that are powers of 2 proved to be a bit of a disaster.

答:
$$F=\sum\limits_{n=1}^\infty\frac{S(n)}{n(n+1)}$$

This series is absolutely convergent since terms are positive and $S(n)=O(\log_2(n))$, making the general term $O\left(\frac{\ln(n)}{n^2}\right)$. Therefore we are allowed to manipulate the series and rearrange terms at will.

Notice that multiplying a number by $2$ will simply shift the binary development to the left and add a zero.

This does not change the value of the sum of digits so $\begin{cases}S(2n)=S(n)\\S(2n+1)=S(2n)+1=S(n)+1\end{cases}$

Let split $F$ along odd and even indexes:

$\begin{align}\displaystyle F
&=\frac{S(1)}2+\sum\limits_{n=1}^\infty\bigg(\frac{S(2n)}{2n(2n+1)}+\frac{S(2n+1)}{(2n+1)(2n+2)}\bigg)\\\\
&=\frac 12+\frac 12\sum\limits_{n=1}^\infty\bigg(\frac{S(n)}{n(2n+1)}+\frac{S(n)+1}{(2n+1)(n+1)}\bigg)\\\\
&=\frac 12+\frac 12\sum\limits_{n=1}^\infty S(n)\Big(\underbrace{\frac 1{n(2n+1)}+\frac 1{(2n+1)(n+1)}}_{\frac 1{n(n+1)}}\Big)+\sum\limits_{n=1}^\infty\underbrace{\frac 1{(2n+1)(2n+2)}}_{\frac 1{2n+1}-\frac 1{2n+2}}\end{align}$

The first series is $\frac 12F$ so we arrive to a formulation without $S(n)$.

$$F=1+2\sum\limits_{n=1}^\infty\bigg(\frac 1{2n+1}-\frac 1{2n+2}\bigg)\\$$

Beware it is NOT telescopic, but we can use $\int_0^1 x^n=\frac 1{n+1}$ and $\sum x^n=\frac 1{1-x}$. Note that in what's follow we can swap $\int$ and $\sum$ thanks to Fubini-Tonelli combined with absolute convergence.

$\begin{align}\displaystyle F
&=1+2\sum\limits_{n=1}^\infty\int_0^1\Big(x^{2n}-x^{2n+1}\Big)dx
=1+2\int_0^1(1-x)\Big(\sum\limits_{n=1}^\infty x^{2n}\Big)dx\\\\
&=1+2\int_0^1 (1-x)\frac{x^2}{1-x^2}dx=1+2\int_0^1 \frac{x^2}{x+1}dx\\\\
&=1+2\bigg[\tfrac 12 x^2-x+\ln(x+1)\bigg]_0^1=1+2\Big(\ln(2)-\tfrac 12\Big)
\end{align}$

$$F=2\ln(2)$$

Title: Determine the following sum


问:
Intuition of Addition Formula for Sine and Cosine
The proof of two angles for sine function is derived using $$\sin(A+B)=\sin A\cos B+\sin B\cos A$$ and $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ for cosine function. I know how to derive both of the proofs using acute angles which can be seen here http://en.wikibooks.org/wiki/Trigonometry/Addition_Formula_for_Cosines but pretty sure those who have taken trig know what I'm talking about. So I know how to derive and prove both of the two-angle functions using the acute angles, but what I am completely confused about is where those triangles came from. So for proving the two-angle cosine function, we look at two acute angles, $A$ and $B$, where $A+B<90$ and keep on expanding. So my question is, where did those two triangles come from and what is the intuition behind having two acute triangles on top of each other? </p>

答:
The bottom triangle is the right triangle used to compute sine and cosine of $\alpha$. The upper triangle is the right triangle used to compute sine and cosine of $\beta$, scaled and rotated so its base is the same as the hypotenuse of the lower triangle.

We know the ratios of the sides of these triangles because of the definitions of sine and cosine. Making the base of the upper triangle the same length as the hypotenuse of the lower triangle allows relations to be drawn between the two triangles. Setting the base of the upper triangle to be aligned with the hypotenuse of the lower triangle creates a triangle with $\alpha+\beta$ as an angle.

ContentLicense: CC BY-SA 3.0


问:
Images of filters need not be a filter
My question is the following:

Let $f$ be a mapping of a set $X$ into a set $Y$ and $\mathcal{F}$ a filter on $X$. Why is it that  $f\left(\mathcal{F}\right)=\{f[F]:F\in\mathcal{F}\}$ is not a filter on $Y$ if $f$ is not surjective?

答: If $A \supset f[F]$ there is no reason to $A \in f(\mathcal{F})$.

Title: Images of filters need not be a filter


问:
if $(x+y)(x-a)(y-a)=n$ has a solution at $x=0$, then $x \geq 0$
Is this true? That if:

$(x+y)(x-a)(y-a)=n$  $\hspace{1cm}$(where $a>0$ and $n>0, x,y\in \mathbb{Z}$)

has a solution when $x=0$ then other solutions occur only when $x\geq0$.  If it is true then why and how to prove it?

For example if $a=7$ and $n=42$ the only solutions are $(0,1), (0,6), (1,6)$ and their reflections, so no negative solutions. [and when I say reflection I mean $(1,6)\equiv(6,1)$]

And if $a=14$ and $n=72$ the solutions are $(11,13), (-10, 11), (-10, 13)$ and their reflections but there is no solution when $x=0$.

To state the question another way: Is it true that if there is not a solution when $x=0$, and a solution exists, then there must be a solution where $x<0$?

(I am not sure if I need to define $a,n\in \mathbb{Z}$ also)

答:
$$(x-a) (y-a) (x+y)=n$$
has solution $x=0$ means that
$$-a y (y-a)=n \quad (1)$$
substitute in the first equation
$$(x-a) (y-a) (x+y)=-a y (y-a)$$
expand
$$a^2 x-a x^2-2 a x y+x^2 y+x y^2=0$$
and factor
$$x (y-a) (x+y-a)=0$$
the other solution is
$$x = a - y$$
it is given that $a>0$ so we can conclude that $x\ge 0$ only if $y\le a$.

Hope this is useful

edit

from equation $(1)$ we get that
$$y=\frac{1}{2} \left(a\pm \sqrt{a^2-\frac{4 n}{a}}\right)$$
in order to be real we must have first of all
$$n\geq \frac{a^3}{4}$$
If we want integer $y$, further condition must be verified.


答:
Intuitively, A continuous function is uniformly continuous when it's never too steep.

If you want to draw one that's not, the domain should be an open interval (bounded or unbounded) since a continuous function on a closed interval is uniformly continuous.

My favorite example is
$$
f(x) = \frac{1}{1-x}
$$
on the interval $(0,1)$. The nearer $x$ is to $1$ the steeper the graph, so the larger a $\delta$ you need for each given $\epsilon$.

Title: Physical Interpretation of uniformly continuous function.

Body:
I have been reading material on uniformly continuous functions. And going through the problems where we have to prove that a function is not uniformly continuous or otherwise.

A function defined on an interval I is said to be uniformly continuous on I if to each $\epsilon$ there exists a $\delta$ such that

$|f(x_1)-f(x_2)| < \epsilon$, for arbitrary points $x_1, x_2$ of I for which $|x_1-x_2|<\delta$.

Now I understand the above definition in mathematical terms and able to apply this definition to solve problems. But I don't understand why this definition was introduced, how does uniformly continuous functions look like ? If given a graph of a function, how can I tell if it is a uniformly continuous function ?
When I imagine continuous functions, I have this picture in mind as to how they look like but for uniformly continuous functions, I can't think of any picture.

I hope I am able to explain my question correctly. Please help me understand.


问:
Ratio of corresponding sides of similar triangles, given the areas.
The area of two similar triangles are 72 and 162. what is the ratio of their corresponding sides?</p>

答: Hint:  the units of area are length$^2$</p>

Body: Hint:  the units of area are length$^2$


问:
How can only one line be drawn through 4 collinear points?
There are 10 points in a plane no three of which are in the same straight line except 4 points, which are collinear. Find the probability that the number of straight lines obtained from the pairs of these points.

Given solution: total lines between 10 non collinear points = $^{10}C_2$ = 45

total lines between 4 non collinear points = $^{4}C_2$ = 6

Since the 4 points are non collinear and only one line can be drawn through all four:

Total lines possible = $45 - 6 + 1 = 40$

My doubt: if I were to mark the 4 collinear points as a, b , c ,d then can’t I draw more than one line such as ab, bc, cd, bd etc? So how come can only one line be drawn?
Also the question asks for the probability which would be possible outcomes divided by total outcomes so how come is this the correct solution? Isn’t this just the possible outcomes?

答:
Part 1:As how you said,ab,ac,ad,bc,bd,cd are the same line because the points are collinear.

Part 2:Yes you are right.The question doesn't imply randomness.It asks "Find the probability that the number of straight lines obtained from the pairs of these points" and it is either wrong or incomplete


问:
Quick question about ODE solutions (trigonometry equality)
After solving an ordinary differential equation, I got this solution:
$$ A\cos(\omega t + \phi) $$ where A and $\omega$ have fixed values (I don't see the relevance to bring them in, thus I won't).

My question is if I'm allowed to consider instead the solution as the following $$ B\cos(\omega (t - t_{0}) + \phi)$$ because that would simplify the calculations (namely, $\phi$ would be $0$).

I guess it is obvious because in that case $\omega t + \phi$ would be the same as $\omega t - \omega t_{0}$ but I still would like to know why I can use this and why this works, from a math point of view (first year though =)).

答:
Well,
$$
A\cos(\omega (t - t_{0}) + \phi) =
A\cos(\omega t + \Phi)
$$
for some constant $\Phi$, namely $\Phi = \phi-\omega t_0$.  So if you are allowed
$$
A\cos(\omega t + \phi)
$$
for all values of the constant $\phi$, then you are also allowed $A\cos(\omega (t - t_{0}) + \phi)$ for all values of constants $t_0$ and $\phi$.


问:
Moment of inertia of hollow paraboloid
A hollow paraboloid of length $l$ and radius $r$ with mass $m$ and uniform density $\rho$ has the origin at the base of its circular cross section, and at the center of its cross section. I am trying to find the moment of inertia for the axis that cuts through the paraboloid's center of gravity (its x axis) and the remaining axis y and z which should yield the same values of $I$ since it is axisymmetrical about the y and z axes. I have worked out the solution for what I believe is a filled paraboloid, but I am unsure how to calculate it for a hollow paraboloid, whether it be an infinitesimally small thickness or a uniform thickness $t$. I consider $x$ to be the distance from the origin of the paraboloid along the x axis towards the tip of the paraboloid.

$dm=\rho \pi r^2 dx$

$dI_x=\frac{1}{2}r^2dm$

$I_x=\int_0^l \frac{1}{2}r^2 \rho \pi r^2 dx$

$  = \frac{1}{2}r^4\rho \pi x|_0^l$

$I_y=I_z=\int_0^l (\frac{1}{4}r^2+x^2)\rho \pi r^2 dx$

$=(\frac{1}{4}\rho \pi r^4 x+\frac{1}{3}x^3\rho \pi r^2)|_0^l$

Revised solution (thanks to Math Lover)

$a$ is the constant that determines the shape of the nose cone of the paraboloid

$dI_x=\frac{1}{2}axdm$

$I_x=\int_0^l \frac{1}{2}a^2x^2 \rho \pi dx$

$I_x=\frac{1}{6}a^2x^3 \rho \pi$

$I_y=I_z=\int_0^l (\frac{1}{4}ax+x^2)\rho \pi ax dx$

$I_y=I_z=\int_0^l (\frac{1}{4}a^2x^2\rho \pi+ax^3\rho \pi ) dx$

$I_y=I_z=(\frac{1}{12}a^2x^3\rho\pi+\frac{1}{4}ax^4\rho \pi )$

Second revised solution (once again thanks to Math Lover)

Being a hollow paraboloid, we need to take the surface integral. For a hollow paraboloid we consider the surface area of everything but the base. This results in the equation:

$A=\frac{\pi b}{6h^2} [(b^2+4h^2)^{\frac{3}{2}}-b^3]$

Substituting base radius $b$ and height $h$ with our equivalent variables:

$A=\frac{\pi \sqrt{ax}}{6x^2} [(ax+4x^2)^{\frac{3}{2}}-(ax)^{\frac{3}{2}}]$

答:
We have a hollow paraboloid of length $L$ and radius $R$ along $x-$axis. If the equation of the paraboloid is $y^2 + z^2 = r^2 = ax$,

Then $R^2 = aL \implies a = \frac{R^2}{L}$. We will continue our working with $a$ and substitute this value of $a$ in the end.

As we know, moment of inertia of mass $dm$ around an axis = $dm \cdot r^2$

For a hollow paraboloid, $dm = \rho \ dS$ (where $\rho$ is the uniform density represented as mass per unit surface area)

$dS = \sqrt{1 + (\frac{\partial x}{\partial y})^2 + (\frac{\partial x}{\partial z})^2} \ dA = \sqrt{1 + (\frac{2y}{a})^2 + (\frac{2z}{a})^2} \ dA$

where $dA$ is the area of the projection of $dS$ in $YZ$ plane.

$dS = \frac{1}{a} \sqrt{a^2 + 4y^2 + 4z^2} \  dy \ dz = \frac{1}{a} \sqrt{a^2 + 4r^2} \ r \ dr \ d\theta$

So, $\displaystyle I_x = \frac{2 \pi \rho}{a} \int_0^R r^3 \ \sqrt{a^2 + 4r^2} \ dr$

For moment of inertia around $y$ or $z$ axis, we note that the radius is $\sqrt{r^2 + x^2}$.

So $\displaystyle I_y = I_z = \frac{2 \pi \rho}{a} \int_0^R r \big(r^2 + \frac{r^4}{a^2}\big) \ \sqrt{a^2 + 4r^2} \ dr$

Title: Moment of inertia of hollow paraboloid


答: E.g. $Y$ must be in any filter on $Y$ (it's a superset of any set) but is only in $f(\mathcal{F})$ if $f$ is surjective. In general the superset axiom need not be fulfilled, but it is always a filterbase, so adding all supersets of sets from $f(\mathcal{F})$ will give us a filter on $Y$.

Title: Images of filters need not be a filter

Body:
My question is the following:

Let $f$ be a mapping of a set $X$ into a set $Y$ and $\mathcal{F}$ a filter on $X$. Why is it that  $f\left(\mathcal{F}\right)=\{f[F]:F\in\mathcal{F}\}$ is not a filter on $Y$ if $f$ is not surjective?


问:
In terms of the derivative f' of f express the derivative of f(3x)
In terms of the derivative f' of f express the derivative of f(3x).
The solution is: $\frac{d}{dx} f(3x)=(f'(3x))(3)=3*f'(3x)$

I don't understand the question. The derivative of f is 3. f' is 3. 3*f' is 9?

We want to express the derivative of f in terms of f'. The derivative of f is 3.
What is the question asking?

答:
Perhaps what is confusing you is that there is a difference between $f'(3x)$ and the derivative of $g(x)=f(3x)$. The former refers to the gradient function of $f$, evaluated at a point which we label as $3x$; the latter refers to the gradient of the composite function $x \mapsto f(3x)$, evaluated at the point $x$. To give an example, if $f(x)=\sin(x)$, then $f'(x)=\cos (x)$, and so $f'(3x)=\cos(3x)$. By contrast, if $g(x)=\sin(3x)$, then $g'(x)=3\cos(3x)$ by the chain rule. Just to drive the point home, the definition of $f'(3x)$ is
$$
\lim_{h \to 0}\frac{f(3x+h)-f(3x)}{h} \, ,
$$
whereas the definition of $g'(x)$ is
$$
\lim_{h \to 0}\frac{g(x+h)-g(x)}{h} = \lim_{h \to 0}\frac{f(3(x+h))-f(3x)}{h} \, .
$$
Here, we are dealing with $y=f(3x)$. The chain rule states that
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \, .
$$
If we set $u=3x$, then $y=f(u)$, and so $dy/du = f'(u)$. Also, $du/dx = 3$. Therefore,
\begin{align}
\frac{dy}{dx} &= f'(u) \cdot 3 \\
&= 3f'(3x) \, .
\end{align}


答: If the side lengths of the first triangle are $a,b,c$, then the side lengths of the second triangle are $ka,kb,kc$ for some $k>0$. An area of the first triangle is $72=\frac{1}{2}ab\sin \alpha$ and an area of the second is $162=\frac{1}{2}kakb\sin\alpha$ so $\frac{72}{162}$ is a square of the ratio of their corresponding sides.</p>

Title: Ratio of corresponding sides of similar triangles, given the areas.

Body: The area of two similar triangles are 72 and 162. what is the ratio of their corresponding sides?</p>


问:
Is every open and connected set in $\mathbb C$ the continuous image of the open unit disk?
Let
$$
\mathbb D=\{z\in\mathbb C\ |\ |z|<1\}
$$
be the open unit disk in $\mathbb C$. It is well known that an open (nonempty) set $U\subseteq\mathbb C$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $f:\mathbb D\to\mathbb C$ such that $f(\mathbb D)=U$.

Therefore, my question is: If $U$ is a (nonempty) open and connected subset of $\mathbb C$, is there always a continuous (but not necessarily injective) function $f:\mathbb D\to\mathbb C$ such that $f(\mathbb D)=U$? It sufficices to assume $U\subseteq\mathbb D$.

If not, how do images of such functions look like?

I know that any compact connected and locally connected subset of $\mathbb C$ is the image of a continuous function defined on $[0,1]$ and hence also the image of a continuous function on $\overline{\mathbb D}$. However, the closure of an open and connected set does not have to be locally connected.

I believe that the answer to my first question is negative (although I hope that it is not ;) ).

Any help is highly appreciated. Thank you very much in advance!

答:
The Hahn-Mazurkiewicz theorem, as pointed out by Mathlover in the comments, is enough to show my educated guess actually works. In particular, it implies that closed balls in $\Bbb{C}$ are the continuous image of a compact interval. We will need a lemma:

Lemma $\quad$ Suppose $\{C_n\}_{n=0}^\infty$ is a countable collection of subspaces of a topological space $X$ (e.g. $\Bbb{C}$) which are continuous images of a compact interval and
$$C := \bigcup_{n=0}^\infty C_n$$
is path-connected. Then $C$ is the continuous image of $[0, \infty)$.

Proof. Consider the intervals $I_n := [2n, 2n + 1]$ and $J_n := [2n + 1, 2n + 2]$ for $n \in \Bbb{N} \cup \{0\}$. Let $\mathfrak{i}_n : I_n \to C_n$ be surjective and continuous, and let $\mathfrak{j}_n : J_n \to C$ be a continuous function such that $\mathfrak{j}_n(2n + 1) = \mathfrak{i}_n(2n + 1)$ and $\mathfrak{j}_n(2n + 2) = \mathfrak{i}_{n+1}(2n + 2)$. Then, the map
$$\phi : [0, \infty) = \bigcup_{n=0}^\infty (I_n \cup J_n) \to C$$
defined by
$$\phi(x) = \begin{cases} \mathfrak{i}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n \le x < 2n + 1 \\ \mathfrak{j}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n + 1 \le x < 2n + 2. \end{cases}$$
This function is made up of countably infinitely many continuous pieces, joined at common points, making the function continuous. It's clearly surjective (even just restricting to $\bigcup I_n$), hence $C$ is the continuous image of $[0, \infty)$ under $\phi$. $\square$

Note that every ball in $\Bbb{C}$ is a countable union of closed balls, and recall that open subsets of $\Bbb{C}$ are a countable union of open balls. Also, in a locally path-connected space, open and connected implies path-connected. Thus, $U$ is the path-connected countable union of closed balls, so by the lemma, it is a continuous image of $[0, \infty)$.

So, to wrap up the proof, as in my comment, simply project the unit disk of $\Bbb{C}$ onto $(-1, 1)$, which is homeomorphic to $\Bbb{R}$. You can simply keep the map constant for points in $(-\infty, 0]$, and then use $[0, \infty)$ to map continuously onto $U$.


问:
Trouble solving these trigonometric simultaneous equations
I have the following pair of simultaneous equations:

$$\sin(2x) + \sqrt3 \cos(2x) = -1$$
$$\sqrt3 \sin(2x) - \cos(2x) = \sqrt3$$

I need to answer the question: "what is the sum of the solutions to these equations, given that $0<x<360^o$?".

After some manipulation using identities, I get:

$$\cos(2x) = -\frac{\sqrt3}{2}$$

Which gives me solutions of:

$$75^o, 105^o, 255^o, 285^o$$

The sum of which is $720^o$

However, the answer to the question is $330^o$, because I should have been left with $75^o$ and $255^o$. I plugged the equations into Desmos and sure enough, this is what should be the case.

What have I done wrong? What I did was replace both sin(2x) terms with the identity 2sin(x)cos(x), eliminating this term and rearranging. Not sure what I'm missing here, aside from just manually plugging in all solutions and seeing if they satisfy both equations, but that feels like a bit of a manual fix, so to speak.

答:
Here is your mistake:

For $x=105^{\circ}$ and $x=285^{\circ}$, $\sin(2x)=-\frac{1}{2}$ instead of $\frac{1}{2}$ so the solutions are $75^{\circ}$ and $255^{\circ}$ and their sum is $330^{\circ}$

So remember: After you find the solutions to an equation, always check in the original equation if the solutions actually work, because while you searched the solutions, you might have ignored some constraints of the problem.


问:
Number of solution sets to $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=1,~~a_1,a_2,\ldots,a_n\in\mathbb{N}$
Consider the Diophantine equation $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=1$$
I'll give you the solutions sets $(a_1,a_2,\ldots,a_n)$ for $n=1$ up to $4$, because beyond $n=4$ the number of solutions seem to explode. (These can be found quite easily by considering the smallest value of the $n$ variables and seeing just how large/small it can be, and then doing the same for the next smallest variable, and so on.) I will not consider permutations of each set as a separate solution set.
$$\begin{align}
n=1:&~~~(1)\\
n=2:&~~~(2,2)\\
n=3:&~~~(3,3,3)~(2,3,6)~(2,4,4)\\
n=4:&~~~(4,4,4,4)~(3,3,4,12)~(3,3,6,6)~(3,4,4,6)~(2,3,7,42)\\&~~~(2,3,8,24)
~(2,3,9,18)~(2,3,10,15)~(2,3,12,12)~(2,4,5,20)\\
&~~~(2,4,6,12)~(2,4,8,8)~(2,5,5,10)~(2,6,6,6)
\end{align} $$
Denote by $a_n$ the number of solution sets we have for the Diophantine equation above for $n$ variables. So we have $$a_n=1,~~a_2=1,~~a_3=3,~~a_4=14$$
(Thanks for the correction @player3236).
And I suppose you could include $a_0=0$ if you really wanted, although I'm not sure how much sense that makes. My question is:

Is there a formula that tells us the number of solutions sets for the Diophantine equation above? If so, how can I find and prove it?

Thank you for your help.

答:
A recent master thesis, cited in https://oeis.org/A002966,  briefly explores this old and hard problem.

There is no general formula.

The number of solutions are known only for $n \le 8$, and a conjecture is given for $n=9$.

It's also conjectured that the number of solutions grows double-exponentially.


问:
Number of sub-triangles of Divided Equilateral Triangles(DET).
Exercise on Mathematical Induction is from Mathematics for Computer Science by Albert R. Meyer, Eric Lehman, and Frank Thomson Leighton.

Divided Equilateral Triangles (DETs) can be built as follows:

$\bullet$ A single equilateral triangle counts as a DET whose only subtriangle is itself.

$\bullet$ If $T::= \bigtriangleup$ is a DET, then the equilateral triangle $T'$ built out of 4 copies of $T$ as shown below is also a DET, and the subtriangles of $T'$ are exactly the subtriangles of each of the copies of $T$.

Define the length of a DET to be the number of subtriangles with an edge on its base. Prove by induction on length that the total number of subtriangles of a DET is the square of its length.

Above all, I've noticed that the length of DET doubles  each time. However, I can't go further, namely, can't find the way to write induction hypothesis. Please help me to determine induction hypothesis and skeleton of proof.

答:
A DET is an element of the sequence $$T_1,T_2,T_3, ... .$$
where each $T_{n+1}=T_n'$ and $T_1=T$.

The inductive hypothesis is that the number of subtriangles of $T_n$ is the square of its length. This is clearly true for $T_1$.

Suppose the hypothesis to be true for $n=k$.

$T_{k+1}$ is four copies of $T_k$ and so...

 its length is twice the length, $l$ say, of $T_k$ and its number of subtriangles is four times the number of subtriangles, $A$ say, of $T_k$. We know that $A=l^2$ and therefore $4A=(2l)^2$.

The hypothesis is therefore true for $k+1$ and so, by induction, is true for all positive integers $n$.


问:
How to prove that the intersection of 3 planes is a line?
Consider the planes:
$$P1:x - y = 0$$
$$P2:y-z = 0$$
$$P3:-x+z = 0$$
Prove that the intersection of the planes is a line.

My solution:
Solving the system I've obtained that $x=y=z$ and I made the notation $x=t$. From here we get the parametric equations of a line $d$ and we can write the canonical form:
$$ d : {{x}\over1} = {{y}\over 1} = {{z}\over1} $$
Thus proving that the intersection of the 3 planes is a line. Is this correct? If so, are there any other ways to prove this? Thanks in advance!

答: Yes, this is correct and it is the simplest way to prove what you want.</p>


问:
Prove that a function is uniformly continuous.
A function defined on an interval I is said to be uniformly continuous on I if to each $\epsilon$ there exists a $\delta$ such that

$|f(x_1)-f(x_2)| < \epsilon$, for arbitrary points $x_1, x_2$ of I for which $|x_1-x_2|<\delta$.

For $f(x)=x^2$ on [-1,1], f is uniformly continuous but when f is defined on $R$ then f is not uniformly continuous.
The proof for Interval [-1,1] goes like this :

$|f(x_1)-f(x_2)| < |x_1^{2}-x_2^{2}| = |x_1-x_2||x_1+x_2| < \epsilon$ when $|x_1-x_2|<1/2\epsilon=\delta$.

This step I don't understand. how $|x_1-x_2|<\epsilon/2$ gives that $|f(x_1)-f(x_2)|<\epsilon$.
Also, how does the proof will look like if the same function is defined on $R$ and hence is not uniformly continuous.

答:
If $|x_1-x_1|<\frac\varepsilon2$, then, since $|x_1+x_2|\leqslant|x_1|+|x_2|\leqslant2$, you have$$\left|x_1^{\,2}-x_2^{\,2}\right|=|x_1-x_2||x_1+x_2|<\frac\varepsilon2\times2=\varepsilon.$$

Concerning your other question, you will find an answer here.


问:
Prove that $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$.
I am reading a Lebesgue integration book.

There is the following proposition without a proof in this book:

Let $\mu^* : 2^\mathbb{R} \to [0,\infty]$ be the Lebesgue outer measure on $\mathbb{R}$.

Let $\mathcal{M} \subset 2^\mathbb{R}$ be the set of all measurable sets.

Let $A_1,\dots,A_n\in\mathcal{M}$.

Assume that $A_i \cap A_j=\emptyset$ for $i\ne j$.

Then, for any $B\in 2^\mathbb{R}$, $$\mu^*(B\cap\bigcup_{i=1}^n A_i) = \sum_{i=1}^n \mu^*(B\cap A_i)$$ holds.

I tried to prove this proposition, but I was not able to prove it.

My attempt is here:

Let $B\in 2^\mathbb{R}$.

Since $A_1\cup A_2\in\mathcal{M}$, $\mu^*(B)=\mu^*(B\cap(A_1\cup A_2)) + \mu^*(B\cap(A_1\cup A_2)^C)$.

Since $A_1\in\mathcal{M}$, $\mu^*(B)=\mu^*(B\cap A_1)+\mu^*(B\cap A_1^C)$.

Since $A_2\in\mathcal{M}$, $\mu^*(B\cap A_1^C)=\mu^*((B\cap A_1^C)\cap A_2)+\mu^*((B\cap A_1^C)\cap A_2^C)$.

So, $\mu^*(B)=\mu^*(B\cap A_1)+\mu^*((B\cap A_1^C)\cap A_2)+\mu^*((B\cap A_1^C)\cap A_2^C)$.

Since $A_1\cap A_2=\emptyset$, $A_2\subset A_1^C$. So, $A_1^C \cap A_2 = A_2$.

So, $\mu^*(B)=\mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*((B\cap A_1^C)\cap A_2^C)$.

Since $C:=B\cap(A_1\cup A_2)^C=(B\cap A_1^C)\cap A_2^C$, $\mu^*(B\cap(A_1\cup A_2)) + \mu^*(C) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*(C)$.

If $\mu^*(C) \in\mathbb{R}$, then $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$.

But I cannot prove that $\mu^*(B\cap(A_1\cup A_2)) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)$ when  $\mu^*(C)=\infty$.

答:
Your answer is almost there. Here is how to simplify and complete it.

Let $D\in 2^\mathbb{R}$.

Since $A_1\in\mathcal{M}$, $\mu^*(D)=\mu^*(D\cap A_1)+\mu^*(D\cap A_1^C)$.

Since $A_2\in\mathcal{M}$, $\mu^*(D\cap A_1^C)=\mu^*((D\cap A_1^C)\cap A_2)+\mu^*((D\cap A_1^C)\cap A_2^C)$.

So, $\mu^*(D)=\mu^*(D\cap A_1)+\mu^*((D\cap A_1^C)\cap A_2)+\mu^*((D\cap A_1^C)\cap A_2^C)$.

Since $A_1\cap A_2=\emptyset$, $A_2\subset A_1^C$. So, $A_1^C \cap A_2 = A_2$. Also, we have  $(D\cap A_1^C)\cap A_2^C = D\cap (A_1^C\cap A_2^C)  = D\cap (A_1 \cup A_2)^C$.

So, $$\mu^*(D)=\mu^*(D\cap A_1)+\mu^*(D\cap A_2)+\mu^*(D\cap (A_1 \cup A_2)^C)$$

Now, since $D$ is any subset of $\mathbb{R}$, take $D= B\cap(A_1 \cup A_2)$.
Since $B\cap(A_1 \cup A_2) \cap A_i= B\cap A_i$, for $i=1,2$, and $B\cap(A_1 \cup A_2) \cap (A_1 \cup A_2)^C= \emptyset$,  we have

$$\mu^*(B\cap(A_1 \cup A_2))=\mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*(\emptyset)$$ that is
$$\mu^*(B\cap(A_1 \cup A_2))=\mu^*(B\cap A_1)+\mu^*(B\cap A_2)$$

Remark: In your attempt, the step

Since $C:=B\cap(A_1\cup A_2)^C=(B\cap A_1^C)\cap A_2^C$, $\mu^*(B\cap(A_1\cup A_2)) + \mu^*(C) = \mu^*(B\cap A_1)+\mu^*(B\cap A_2)+\mu^*(C)$.

is not necessary and it actually leads to an unnecessary more complex path.


答:
Since the triangles are similar and no other constraint is available, you can take them to be equilateral with sides $ a $ and $ b $ units.

$$ A_1 = 72x = \frac{\sqrt{3}}{4} a^2 $$
$$ A_2 = 162x = \frac{\sqrt{3}}{4} b^2 $$
Therefore,

$$ \left( \frac{a}{b} \right)^2 = \frac{72}{162} = \frac{4}{9}$$
Thus, sides are in the ratio $ 2 : 3 $


答:
One can prove a stronger claim: given a nonempty open connected $U\subseteq \mathbb{C}$, there's a holomorphic $\varphi:\mathbb{D}\to \mathbb{C}$ such that $\varphi(\mathbb{D})=U$.

First, let $U\neq \mathbb{C},\neq \mathbb{C}-\{z_0\}$.  Thanks to the Riemann uniformization theorem, every such open connected subset of $\mathbb{C}$ is hyperbolic, i.e. has $\mathbb{D}$ as its universal cover. Thus there exists a surjective holomorphic map $\mathbb{D}\to U$.
Note that there's a surjective holomorphic map $(\exp(z)+z_0)$ from $\mathbb{C}$ to $\mathbb{C}-\{z_0\}$, and thus it suffices to prove the result for $\mathbb{C}$.
Now, let $f(z)=z^3-1$. It is easy to see that $f:\mathbb{C}-\{1,\exp(2\pi i/3)\}\twoheadrightarrow\mathbb{C}$.  Precomposing with the covering map $\pi:\mathbb{D}\to \mathbb{C}-\{1,\exp(2\pi i/3)\}$, we get the claim.

One can also explore "how much" this function fails to be injective. Indeed, the uniformization theorem implies that, for $U\neq\mathbb{C},U\neq \mathbb{C}-\{z_0\}$, we have $U\simeq \mathbb{D}/\Gamma$, where $\Gamma$ is a discrete subgroup of $Aut(\mathbb{D})$ (complex automorphisms) isomorphic to $\pi_1(U)$.
Thus two elements $z_1,z_2$ have the same image iff $\exists \gamma\in \Gamma:\gamma(z_1)=z_2$ (note that with $U$ simply connected we get injectivity).

Title: Is every open and connected set in $\mathbb C$ the continuous image of the open unit disk?

Body:
Let
$$
\mathbb D=\{z\in\mathbb C\ |\ |z|<1\}
$$
be the open unit disk in $\mathbb C$. It is well known that an open (nonempty) set $U\subseteq\mathbb C$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $f:\mathbb D\to\mathbb C$ such that $f(\mathbb D)=U$.

Therefore, my question is: If $U$ is a (nonempty) open and connected subset of $\mathbb C$, is there always a continuous (but not necessarily injective) function $f:\mathbb D\to\mathbb C$ such that $f(\mathbb D)=U$? It sufficices to assume $U\subseteq\mathbb D$.

If not, how do images of such functions look like?

I know that any compact connected and locally connected subset of $\mathbb C$ is the image of a continuous function defined on $[0,1]$ and hence also the image of a continuous function on $\overline{\mathbb D}$. However, the closure of an open and connected set does not have to be locally connected.

I believe that the answer to my first question is negative (although I hope that it is not ;) ).

Any help is highly appreciated. Thank you very much in advance!


问:
proving that $\alpha_1+\alpha_2<-3$ which are the negative roots of $f(x)$
I am stuck on the following problem:

Given
\begin{align*}
f(x) &= x^4 + (-m - 2n + 6)x^3 + (mn - 5m - 8n + 10)x^2\\
&\quad + (3mn - 7m - 8n)x + mn-m-n-7,
\end{align*}
if $\alpha_1,\alpha_2$ are two negative roots of $f(x)=0$, show that $\alpha_1+\alpha_2< -3$ for $m,n\ge 4$.

I took some values of $m,n$ and found the following:

$m=n=4$, $\alpha_1=-2.476, \alpha_2=-0.621$;

$m=5,n=4$, $\alpha_1=-2.49, \alpha_2=-0.59$

I tried some other values of $m,n$,
In all the above cases, I am getting $\alpha_1+\alpha_2<-3$.

But I am not able to prove it. Can someone please help me out?

NOTE:

I found the following. I obtained $-1<\alpha_1<-k$ and $-3<\alpha_2<-3+k$
where $k$ is some function of $m$ and $n$.

I took $m=4,n=5$ and I got $\alpha_1=-2.492<-3+0.55, \alpha_2=-0.592<-0.55$.
I took some other values of $m,n$ and obtained similar stuff.
So I think my guess is correct.

Is there any way to find $k$ such that $k$ is a function of $m,n$?

答:
Let $\alpha_{i,m,n}$ be the $i^\text{th}$ root of $f(x;m,n)$ in ascending order. It is shown that for $m>4,n>4$

$$
\begin{align}
&\alpha_{1,\infty,\infty}<\alpha_{1,m,n}<\alpha_{1,4,4}\\
&\alpha_{2,4,4}<\alpha_{2,m,n}<\alpha_{2,\infty,\infty}
\end{align}
$$

Numerically,

$$
\begin{align}
&-2.618\ldots<\alpha_{1,m,n}<-2.476\ldots\\
&-0.621\ldots<\alpha_{2,m,n}<-0.3819\ldots
\end{align}
$$
leading to the bound
$$
\alpha_{1,m,n}+\alpha_{2,m,n}<-2.858\ldots
$$

The proof is via observing the change of sign of $f$ at the boundaries of the intervals $(\alpha_{1,\infty,\infty}~,\alpha_{1,4,4})$ and $(\alpha_{2,4,4}~,\alpha_{2,\infty,\infty})$, as tabulated below:

$x$
$sgn(f(x,m,n))$

$\alpha_{1,\infty,\infty}$
$+1$

$\alpha_{1,4,4}$
$-1$

$\alpha_{2,4,4}$
$-1$

$\alpha_{2,\infty,\infty}$
$+1$

The infinity roots are $(-3\pm\sqrt{5})/2$.


问:
What is $\mathbb E[X\mid \boldsymbol 1_B]$?
Let $B$ measurable. I know that $$\mathbb E[X\mid B]:=\frac{\mathbb E[X\boldsymbol 1_B]}{\mathbb P(B)}.$$

So, I guess that $$\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid B]\boldsymbol 1_B+\mathbb E[X\mid B^c]\boldsymbol 1_{B^c},$$
but I'm not so sure. My idea is the following : $$\mathbb E[X\mid \boldsymbol 1_B=1]=\mathbb E[X\mid B]\quad \text{and}\quad \mathbb E[X\mid \boldsymbol 1_{B^c}]=\mathbb E[X\mid B^c].$$

But in the other side, if this would be true, we would have $\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid \boldsymbol 1_{B^c}],$ and I really don't think it's true. So, any idea about $\mathbb E[X\mid \boldsymbol 1_B]$ ?

答:
Why do you think $\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid \boldsymbol 1_{B^c}]$ is wrong ? In fact, you have a more general statement that is $$\mathbb E[X\mid Y]:=\mathbb E[X\mid \sigma (Y)]\tag{1}.$$

If you are not familiar with this notation, it's not a problem. But what $(1)$ says is namely that the conditional expectation $\mathbb E[X\mid Y]$ depend rather $\sigma (Y)$ (the smallest $\sigma -$algebra that make $Y$ measurable) than on the r.v. $Y$. This mean that even if $Y(\omega )\neq Z(\omega )$ for all $\omega \in \Omega $, as far as $\sigma (Y)=\sigma (Z)$, then $\mathbb E[X\mid Y]$ and $\mathbb E[X\mid Z]$ are the same. Since $$\sigma (\boldsymbol 1_B)=\{\Omega ,B,B^c,\varnothing \}=\sigma (\boldsymbol 1_{B^c}),$$ the fact that $$\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid \boldsymbol 1_{B^c}],$$
should not be surprinsing.


问:
Redundance in statement of second morphism theorem
The standard statement of the Second Morphism Theorem found in my textbook and Wikipedia is as follows:

Let $N$ be a normal subgroup of $G$ and $H$ be any subgroup of $G$.

$HN = \{hn | h \in H, n \in N\}$ is a subgroup of $G$

$H \cap N$ is a normal subgroup of $H$

$H / (H \cap N) \cong HN / N$

Isn't the third part redundant?  In particular, isn't $HN / N$ just $H/N$?  That is, cosets in $HN/N$ are given by $hnN$.  But by the closure of $N$, a group, this is equal to $hN$.  Also, since $N$ is a normal subgroup, (since we are taking the quotient?) right cosets should equal left cosets, so we don't have to look at $nHN$.

Perhaps I've gone wrong somewhere, but if not, I'd like some insight on why this is stated "redundantly" -- is it just convention?

答:
Let us consider a specific example: Take $G = D_4 = \{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$ and set $H = \{1,y\}$ and $N = \{1,x,x^2,x^3\}$. Being an index $2$ subgroup of $G$, $N$ is normal in $G$. But now you can't even form $H/N$ because $N$ is not even contained in $H$.

If $N$ is a normal subgroup of $G$ that is contained in $H$ then the second isomorphism theorem says that $H \cap N = N$ is a normal subgroup of $H$. Then indeed
$$H/(H \cap N) = H/N \cong HN/N.$$


问:
Divisibility by 7 Proof by Induction
Prove by Induction that

$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$

Base case:

$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.

Now, having $n=k$, we assume that:

$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$

We have to prove that for $n=k+1$ that,

$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$

We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.

Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$

I get stuck here, please help me.

Title: Divisibility by 7 Proof by Induction

Body:
Prove by Induction that

$ 7| {4^{2^n} + 2^{2^n} + 1} , ∀ n ∈ N  $

Base case:

$7 | 4^{2^1} + 2^{2^1} + 1, $

7|7⋅3

Which is true.

Now, having n=k, we assume that:

$7|4^{2^k} + 2^{2^k} + 1, ∀  k ∈ ℕ $

We have to prove that for n=k+1 that,

$7| 4^{2^{k+1}} + 2^{2^{k+1}} + 1, ∀ k ∈ ℕ $

$ 4^{2^{k+1}} + 2^{2^{k+1}} + 1 = 4^{2*2^k} + 2^{2*2^k} + 1 $

$ = {4^{2^k}}^2 + {2^{2^k}}^2 + 1 $

Put $ {4^{2^k}} = a, {2^{2^k}} = b $

$ ∴ {(4^{2^k})}^2 + {(2^{2^k})}^2 + 1 = a^2 + b^2 + 1   $

Also,
$  (a+b+1)(a+b-1) = a^2 + b^2 + 2ab - 1 $

Since $ 7|a+b+1 $,
$ 7| a^2 + b^2 + 2ab - 1    ...(1) $

$ ab-1 = {8^{2^k}} - 1 $

For  n ∈ ℕ, $ 8^n = (7+1)^n $
= (Multiple of 7) + 1

(Theorem of Binomial Expansion)

$ ∴ 8^n -1 = 7p $
Where p ∈ ℕ if n ∈ ℕ

$ ∴ 7| ab - 1       ....(2) $

Subtracting Equation (2) from equation (1),

$ 7| a^2 + b^2 + 2ab - 1 -2*(ab-1) $

$ ∴ 7| a^2 + b^2 + 1 $
ie

$$7| 4^{2^{k+1}} + 2^{2^{k+1}} + 1, ∀ k ∈ ℕ $$
if
$7| 4^{2^{k}} + 2^{2^{k}} + 1, ∀ k ∈ ℕ $
Hence, by principle of Mathematical Induction, the above statement is proved to be true.


问:
Probability AND/OR
Suppose we have a bag of $10$ balls, and each ball is a unique colour.

If we randomly select $3$ balls from this bag, without replacement, I want to find out the chances of correctly guessing the colour of all $3$ balls (if we have chosen $3$ colours beforehand).

I have come down to two possible equations:

Work out the chances of not correctly guessing any ball and negate it:

$9/10 * 8/9 * 9/8 = 0.7$, where the complement is $0.3$

The chances of correctly guessing all balls:

$1/10 + 1/9 + 1/8 = 0.3361$

Firstly:

If these are both correct, which one is more accurate?

Secondly:

Is there a better way of calculating this?

I have a feeling I'm going about the whole thing in the wrong way, but I'm not too sure how I can check to be sure I'm heading in the right direction.

答:
Suppose out of the ten colours, we predicted that we were going to pull out pink, red, and blue.

Then,

P(pink,red,blue)= P(pink on first pick)P(red on second pick|pink on first pick)P(blue on third pick|red on second pick)$=(\frac {1}{10})(\frac {1}{9})(\frac {1}{8})=\frac{1}{720}$

Title: Probability AND/OR


问:
Hint for proof of proposition 10.15, lines a) and c) in John M. Lee's book "An Introduction to Smooth Manifolds"
I have been thinking about how to prove proposition 10.15, lines a) and c) in John Lee's book "An Introduction to Smooth Manifolds" (I leave a screenshot of the proposition below), but I can't figure out how to construct the remaining sections in a).
Each section can be written locally as a linear combination of the local canonical smooth frame $t_{i}: V \rightarrow E$ such that, using a local trivialization $\Phi : \pi ^{-1}(V) \to V\times \mathbb{R}^k$, we get $\Phi\circ t_{i} : V \to V \times \mathbb{R}^k$ is given by $\Phi \circ t_{i} (p) = (p,e_{i})$. So, $\sigma _{j} = \sum_{i=1}^{k} \lambda_{i,j} t_{i}$ for $1\leq j \leq m \lt k$, where the $\lambda_{i,j}$'s are smooth as real valued functions on $V$. So the statement that we can extend the linearly independent set $\{\sigma_{j}\}_{j=1,...,m}$ to a local frame is equivalent to stating that the matrix valued function $[\lambda_{i,j}]_{i = 1,...,k}^{j = 1,...,m}$, which is a matrix with maximal rank at each point, can be "extended" to a smooth map $M: W \subset V \to GL(k,\mathbb{R})$.

Having said this, I would appreciate it if you could give me a hint for the idea of the proofs of a) and c) :).

Thank you all in advance!

Proposition 10.15 (Completion of Local Frame for Vector Bundles). Suppose $\pi:E\longrightarrow M$ is a smooth vector bundle of rank $k$.

(a) If $(\sigma_1,\ldots,\sigma_m)$ is a linearly independent $m$-tuple of smooth local sections of $E$ over an open subset $U\subseteq M$, with $1\leq m<k$, then for each $p\in U$ there exist smooth sections $\sigma_{m+1},\ldots,\sigma_k$ defined on some neighborhood $V$ of $p$ such that $(\sigma_1,\ldots,\sigma_k)$ is a smooth local frame for $E$ over $U\cap V$.

(b) If $(v_1,\ldots,v_m)$ is a linearly independent $m$-tuple of elements of $E_p$ for some $p\in M$, with $1\leq m<k$, then there exists a smooth local frame $(\sigma_i)$ for $E$ over some neighborhood of $p$ such that $\sigma_i(p)=v_i$ for $i =1,\ldots,m$.

(c) If $A\subseteq M$ is a closed subset and $(\tau_1,\ldots,\tau_k)$ is a linearly independent $k$-tuple of sections of $E|_A$ that are smooth in the sense described in Lemma 10.12, then there exists a smooth local frame $(\sigma_1,\ldots,\sigma_k)$ for $E$ over some neighborhood of $A$ such that $\sigma_i|_A=\tau_i$ for $i=1,\ldots,k$.

Exercise 10.16. Prove the preceding proposition.

答:
You almost got it. Continuing what you said, at point $p$ we can extend the matrix
$$\begin{bmatrix}
\lambda_1^1(p) & \cdots & \lambda_m^1(p)\\
\vdots & \vdots& \vdots\\
\lambda_1^k(p) & \cdots & \lambda_m^k(p)
\end{bmatrix}$$
to a $GL_k(\mathbb{R})$ matrix by extending the linear independent vectors
$$v_1(p):=\begin{bmatrix}
\lambda_1^1(p) \\
\vdots  \\
\lambda_1^k(p)
\end{bmatrix},\dots,v_m(p):=\begin{bmatrix}
\lambda_m^1(p) \\
\vdots  \\
\lambda_m^k(p)
\end{bmatrix}$$
to a basis of $\mathbb{R}^n$. Let $v_1(p),\dots v_m(p),v_{m+1},\dots,v_k$ be that basis. You have to show that the map $F:U \to \mathbb{R}$ given by
$$p \mapsto \det\left(\begin{bmatrix}
v_1(p) &
\cdots  &
v_m(p)& v_{m+1} & \cdots & v_k
\end{bmatrix}\right)$$
is smooth. Use the fact that $F^{-1}(0)$ is closed to conclude.


问:
difference between languages whit n-adic and monadic predicates
For FOL language, we can distinguish between languages with only monadic predicates and languages with relational/n-adic predicates.
Reading a paper, I stumbledupon this claim:

When we move to a quantificational language with relational predicates
(but not identity) our situation is complicated by the fact that some
formulae have no finite models.

Why does languages without n-adic relations do not encounter this problem, and what does precisely a "finite model" mean in this case?

答:
Firstly, a model (of some FO formula $\varphi$ over some signature $\tau$) is a $\tau$-structure $\mathcal A$ such that $\varphi$ holds in $\mathcal A$.
Such a $\tau$-structure $\mathcal A$ consists of some set $A$ (called the universe of $\mathcal A$) and an interpretation of the relation symbols (and, if you allow them, function symbols) in your signature $\tau$, i.e. a specification of which elements or tuples in $A$ are in the predicates of $\tau$.
A structure is finite if its universe is.

Let us now get to your first question, which can be reformulated as "why do formulae over signatures only consisting of monadic predicates have finite models?".
I will not give an actual proof but rather an outline of one, I am sure that you can easily find a complete proof online.

So consider some signature $\tau = \{P_1, ..., P_n\}$ only consisting of monadic predicates and some satisfiable formula $\varphi$ over $\tau$.
As $\varphi$ is satisfiable, that means that there exists some model $\mathcal A$ of $\varphi$ over some universe $A$ which might be infinite.
A key observation here is that $\varphi$ has no way to distinguish elements of $A$ which have exactly the same predicates.
This yields an equivalence relation $\sim$ on the universe $A$ with two elements $a, b \in A$ having $a \sim b$ saying that $a \in P_k$ if and only $b \in P_k$ in $\mathcal A$ for all $k$.
This equivalence relation finitely many equivalence classes, in fact the number of them is $2^n$ (as any element of $A$ can either has some predicate $P_k$ or not, and we have $n$ such predicates).
Now consider the $\tau$-structure $\mathcal{A}'$ over the universe made up of these $2^n$ equivalence classes, with the interpretation of the predicates being that some equivalence class is in $P_k$ if it's members were in $P_k$ in $\mathcal A$.
It can be shown that $\mathcal{A}' \models \varphi$ if and only if $\mathcal A \models \varphi$, and hence $\varphi$ has a finite model.

This is no longer the case if we allow binary predicates into our signature (or even larger ones), see e.g. this question.


问:
Construct a set such that $a, b \in X \implies a + b \not\in X$
I'm looking for an uncountable set $X \subset \mathbb{R}_{\geq 0}$ such that for all $a, b \in X$, $a \neq b \implies a + b \not\in X$.

Two points about this question: first of all, I'm not sure which area of mathematics this falls under so I used the set-theory tag; feel free to correct me on this. Secondly, note immediately that if $X$ is allowed to be countable, then e.g. $X = \{2^k \mid k \in \mathbb{Z}\}$ works.

答:
Let $B$ be a Hamel basis of $\mathbb R$ over $\mathbb Q$. Then let $X = \{|x| \colon x\in B\}$. This clearly works.

If you want an example avoiding axiom of choice, consider the set of all numbers whose all digits in the decimal expansion are either $1$ or $3$.


问:
Question on solving one algebra relation to obtain another
If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:

 If $(1-2x)(1-2y)(1-2z)=-27$, then evaluate $x+y+z$ where $x=\frac{b}{a+b}, \ y=\frac{c}{b+c}, \ z=\frac{a}{c+a}$

 I'm not sure how to solve this. I think it may use inequalities or pure algebra.

Please help me out here. Thanks a lot :)

答:
Let's call $\begin{cases} U=\frac a{a+b}+\frac b{b+c}+\frac c{c+a}\\V=\frac b{a+b}+\frac c{b+c}+\frac a{c+a}\end{cases}$

We have trivially $U+V=3$

Now $U-V=\frac {a-b}{a+b}+\frac {b-c}{b+c}+\frac {c-a}{c+a}=\cdots=-\frac {a-b}{a+b} \times\frac {b-c}{b+c}\times\frac {c-a}{c+a}=27$

Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$

Therefore $U=15$ and $V=-12$.


问:
Distribution of distance from origin for uniformly randomly chosen point in annular region
I have uniformly distributed nodes over an annular region RI<r<RO.
I would like to compute the PDF of the distance from any chosen node to the center of the annular.
In the case of $R_I = 0$ (circle), the PDF is simply given by $r^2 / R_O^2$ as solved here for $R_I = 0$
The logic in my case is to have PDF = $r^2 - (R_I)^2 / (R_O^2 - R_I^2)$. However, when I simulated it in MATLAB, this formula doesn't match with the simulation!

Could anyone help, please?

答:
Note that the pdf given for $\ R_I\ $ in the answer you cite is $\ \frac{2r}{R_O^2}\ $, not $\ \frac{r^2}{R_O^2}\ $, as you say here.  The latter is the cumulative distribution function of the point's distance from the origin.  For $\ R_I\ne0\ $ the  cumulative distribution function of the point's distance from the origin is
$$
\frac{r^2-RI^2}{RO^2-RI^2}\ .
$$
for $\ RI\le r\le RO\ $, and the pdf is the derivative of this:
$$
\frac{2r}{RO^2-RI^2}\ .
$$


问:
Selecting a representative permutation
If we have a complete set of permutations {m} = n choose k, how do I select a representative set of permutations in a stream such that the selected set say, {s} keeps growing to include permutations from {m} yet maintain it's representative nature to parent set {m}.  One way to look at this representative nature is to sort {s} and regress with original. Yes, though I've not been explicit, I'm interested to know the possibility of predicting a future sequence of a permutation from a partial set.

But here's my revised question:

Given permutations of a set, are there methods to determine which of the permutations will appear random? I read about statistical tests for randomness but most seem to be perspective based tests and I'm not sure which tests are reliable.

I say appear since the apparent nature of random size and order of elements in the representative sample seems really a conscious mental notion since human mind can't find an obvious pattern, and may be math as well.

答: Appearing random is in the eye of the beholder.  Everybody would find the permutation $1234\dots n$ non-random.  But if you picked a random permutation, it has chance $\frac 1{n!}$ of being chosen, just like any other.  How about the permutation $854917632?$  To any non-English speaker, and to many English speakers, it appears random.  But I constructed it by a simple rule, so it isn't random in my usage.

Title: Selecting a representative permutation


Title: Question on solving one algebra relation to obtain another

Body:
If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:

 If $(1-2x)(1-2y)(1-2z)=-27$, then evaluate $x+y+z$ where $x=\frac{b}{a+b}, \ y=\frac{c}{b+c}, \ z=\frac{a}{c+a}$

 I'm not sure how to solve this. I think it may use inequalities or pure algebra.

Please help me out here. Thanks a lot :)

Body:
let me continue from your work.
 Note that $$\frac{1}{x}-1=\frac{a}{b} $$ so $$\frac{(1-x)(1-y)(1-z)}{xyz}=1$$
$$\Rightarrow 1-(x+y+z)+xy+yz+xz-2xyz=0$$
$$\Rightarrow xy+xz+yz-2xyz=(x+y+z)-1$$
Also $$(1-2x)(1-2y)(1-2z)=-27$$
$$\Rightarrow 1-2(x+y+z)+4(xy+yz+xz)-8xyz=-27$$
$$1-2(x+y+z)+4(x+y+z)-4=-27$$
$$\Rightarrow x+y+z=-12$$


问:
Formalization of $17\mid 77a+1$
I was solving a math problem when I came across this Question:

When is $17 \mid 77a+1$ With $a\in \mathbb N^*$

It’s easy to see that there’s infinitely many of values of $a$, you can’t try a little bit with this but you will end up with these values :
$$a\in \{15,32,49,66,...\}=A$$
In this set any number it’s equal to the previous one $+17$ .
I’ve found a relation with this set and the set contains all numbers divisible by $17$, let’s call it $D_{17}$
$$D_{17} =\{17,34,51,68,...\}$$
You can see that if $n \in D_{17} \iff n-2\in A$ And if $n\in A \iff n+2 \in D_{17}$.

My question is why $2$ appeared here, does there exist an equation that when we solve it we will get this $2$?

答:
Divisibility by $17$.- A number $n$ is divisible by $17$ if, subtracting from the number without the units digit the units digit multiplied by $5$, the result is equal to a multiple of $17$.

When $17|77a+1?$. You have $77a+1=(4\cdot17+9)a+1$ so you need $17|9a+1$

Consider $9a+1\equiv\alpha_1\pmod{10}$ According with the criterium above we make

$(9a+1-\alpha_1)-5\alpha_1\equiv0\pmod{17}$ and iterating the procedure until the congruence could be clearly seen to be true or false, we have the answer.

Title: Formalization of $17\mid 77a+1$


问:
How should the following notation be read?
In the opening chapter of `foundations of optimization’ by Guler it states:
$$f:I=(c,d) \to \mathbb{R}$$

Should this be interpreted as $I$ is a function over the open interval $(c,d)\;$ that maps to the real number line?

答:
It means $I$ is an open intervall with $I = (c,d)$. Then $f$ is a function $$f: I \rightarrow \mathbb{R}$$
In your book there probably will be more clarification about the open intervall $(c,d)$


问:
Differentiable approximation of the absolute value function
Are there any good approximations of the absolute value function which are $C^2$ or at least $C^1$? I've thought about working with exponentials and then adding in more terms to keep the function from growing too fast away from zero, but I was hoping to find something a bit neater.</p>

答:
Just simply $$x. \tanh(\alpha x)$$
Example plot


问:
Prove that $\lim\limits_{n\rightarrow \infty}{a_nb_n}=+\infty$
If $\lim\limits_{n\rightarrow \infty}{a_n}=a<0$ and $\lim\limits_{n\rightarrow \infty}{b_n}=-\infty$, prove that $\lim\limits_{n\rightarrow \infty}{a_nb_n}=+\infty$.

I remember that if the two sequences $a_n$ and $b_n$ are converging and their limits are $a$ and $b$, then $\lim\limits_{n\rightarrow \infty}{a_nb_n} = \lim\limits_{n\rightarrow \infty}{a_n}\lim\limits_{n\rightarrow \infty}{b_n}=a.b$. However, $\lim\limits_{n\rightarrow \infty}{b_n}=-\infty$, not a number. Can I use this theorem or I have to prove the limit in another way?

答:
Hint.

Let $M>0$. Observe that for sufficiently large $n$,
$$
b_n<-M,\quad \frac{3a}{2}<a_n<\frac{a}{2}<0
$$
and thus
$$
a_nb_n> -M \cdot \frac{a}{2}>0
$$


问:
Injective Functions and combinatorics
Let $A=\{1,2,3,4,5,6,7,8\}$

Find the number of injective functions $f:A \rightarrow A$ where $f[\{1,2\}]=\{1,2\}$ and $f[\{3,4\}]=\{3,4\}$

Find the number of all functions $f:A \rightarrow A$ where $f[\{1,2\}]=\{1,2\}$ and $f[\{3,4\}]=\{3,4\}$

Find the number of all functions $f:A\rightarrow A$ where $f[\{1,2\}] \neq \{1,2\}$, $f[\{3,4\}] \neq \{3,4\}$, $f[\{5,26\}] \neq \{5,6\}$, $f[\{7,8\}] \neq \{7,8\}$

Please help I didn't understand this topic

答:
This kind of question is just standard combinatorics, but the talk of functions make it more intimidating. A function is just something that takes $a$ from domain and makes it to be $f(a)$ element of the codomain. Now, you have function with finite domain $\{1,2\dots, 8\}$ and this is also codomain so you can look at it this way:

You have eight positions. In how many ways can you put a number from $\{1,2\dots, 8\}$ to each position, such that it satisfies some extra properties.

I will do the first of the three questions in more detail:

First, if you have any $F: A\mapsto A$ for $A$ finite, then f is injective if and only if it is surjective if and only if it is bijective.  One learns that injective does not imply surjective and vice versa but that is just because you usually have functions over real numbers (or any other infinite domain). I recommend proving this equivalence for finite $A$, it is a good exercise and also a useful principle to know.

So now, we know that in the first question, $f: A\to A$ is bijective. But that means it is a permutation!

So you only need to found number of permutation (standard combinatorics problem) on eight element set, such that $1,2$ maps to $1,2$ and $3,4$ maps to $3,4$ and consequently $5,6,7,8$ mapsto $5,6,7,8$.

So by the product rule number of possible functions is $2! \cdot 2! \cdot 4!=2\cdot2\cdot24=96$.

Title: Injective Functions and combinatorics


问:
distance point to affine subspace
I have $n$ points $x_i\in\mathbb{R}^N$, $n \le N$. Is there a general formula for the (squared) distance of another point $b$ to the subspace defined by the $x_i$? (Note that I'm not only looking for hyperplanes; the codimension can be larger than 1.)

答:
First move $b$ to the origin, $a_i = x_i - b$, and realize that every point in the plane spanned by the $a_i$ can be written as
a linear combination $\sum_i w_i a_i$ with $\sum_i w_i = 1$.
The problem can then be formulated as a least-squares problem with an equality constraint,
$$
\|A w\|_2^2 \to \min,\\
e^T w = 1
$$
with $A\in\mathbb{R}^{N\times n}$ being the column-vector of the $a_i$ and $e=(1,\dots,1)^T \in\mathbb{R}^n$. This can be solved with a Lagrangian approach,
$$
\begin{pmatrix}
A^T A & e\\
e^T & 0
\end{pmatrix}
\begin{pmatrix}
w\\
\lambda
\end{pmatrix}
=
\begin{pmatrix}
0\\
1
\end{pmatrix}
$$
If $A^TA$ is invertible, this gives
$$
w = \frac{(A^T A)^{-1}e}{e^T (A^T A)^{-1}e}
$$
and
$$
\min \|A w\|_2^2 = -\lambda = \frac{1}{e^T (A^T A)^{-1}e}.
$$
Note that this expression can never be $0$, so this case is a degeneracy.

If not moving $b$ to the origin, the problem is
$$
\|A w - b\|_2^2 \to \min,\\
e^T w = 1
$$
and has the solution
$$
w = \frac{(A^TA)^{-1}e}{e^T(A^TA)^{-1} e} + \left(ee^T - \frac{(A^TA)^{-1}ee^T}{e^T(A^TA)^{-1} e}\right)
(A^TA)^{-1} A^Tb
$$
and the distance
$$
\min \|A w - b\|_2^2 =
\frac{(1 - e^T(A^TA)^{-1} A^T b)^2}{e^T(A^TA)^{-1} e}
 - b^T A (A^TA)^{-1} A^T b
 + b^T b.
$$
(Perhaps this expression can be further simplified.)

Compared the @user's answer, this one is faster, particularly for large $N$. (Shown here for $n=3$).

Code to reproduce the plot:

```<code>import numpy as np
import perfplot
from scipy.linalg import solve_triangular

np.random.seed(0)

def setup(n):
    # k = 3
    k = n
    A = np.random.rand(n, k)
    b = np.random.rand(n)
    e = np.ones(k)
    return A, b, e

def user(Abe):
    A, b, _ = Abe
    V = (A.T - b).T
    V[:, 1:] = (V[:, 1:].T - V[:, 0]).T
    VTV = V.T @ V
    return VTV[0, 0] - VTV[0, 1:].T @ np.linalg.solve(VTV[1:, 1:], VTV[0, 1:])

def nschloe(Abe):
    A, b, e = Abe
    V = (A.T - b).T
    out = 1 / np.sum(np.linalg.solve(V.T @ V, e))
    return out

def nschloe_qr(Abe):
    A, b, e = Abe
    V = (A.T - b).T
    _, r = np.linalg.qr(V)
    w = solve_triangular(r.T, e, lower=True, check_finite=False)
    w = solve_triangular(r, w, check_finite=False)
    out = 1.0 / np.sum(w)
    return out

def least_squares(Abe):
    A, b, _ = Abe
    b -= A[:, 0]
    A = (A[:, 1:].T - A[:, 0]).T
    _, out, _, _ = np.linalg.lstsq(A, b, rcond=None)
    return out[0]

perfplot.show(
    # "out.png",
    setup=setup,
    kernels=[user, nschloe, nschloe_qr, least_squares],
    n_range=[2 ** k for k in range(2, 12)],
    xlabel="num points/dim"
)
</code>```

Title: distance point to affine subspace

Body:
First move $b$ to the origin, $a_i = x_i - b$, and realize that every point in the plane spanned by the $a_i$ can be written as
a linear combination $\sum_i w_i a_i$ with $\sum_i w_i = 1$.
The problem can then be formulated as a least-squares problem with an equality constraint,
$$
\|A w\|_2^2 \to \min,\\
e^T w = 1
$$
with $A\in\mathbb{R}^{N\times n}$ being the column-vector of the $a_i$ and $e=(1,\dots,1)^T \in\mathbb{R}^n$. This can be solved with a Lagrangian approach,
$$
\begin{pmatrix}
A^T A & e\\
e^T & 0
\end{pmatrix}
\begin{pmatrix}
w\\
\lambda
\end{pmatrix}
=
\begin{pmatrix}
0\\
1
\end{pmatrix}
$$
If $A^TA$ is invertible, this gives
$$
w = \frac{(A^T A)^{-1}e}{e^T (A^T A)^{-1}e}
$$
and
$$
\min \|A w\|_2^2 = -\lambda = \frac{1}{e^T (A^T A)^{-1}e}.
$$
Note that this expression can never be $0$, so this case is a degeneracy.

If not moving $b$ to the origin, the problem is
$$
\|A w - b\|_2^2 \to \min,\\
e^T w = 1
$$
and has the solution
$$
w = \frac{(A^TA)^{-1}e}{e^T(A^TA)^{-1} e} + \left(ee^T - \frac{(A^TA)^{-1}ee^T}{e^T(A^TA)^{-1} e}\right)
(A^TA)^{-1} A^Tb
$$
and the distance
$$
\min \|A w - b\|_2^2 =
\frac{(1 - e^T(A^TA)^{-1} A^T b)^2}{e^T(A^TA)^{-1} e}
 - b^T A (A^TA)^{-1} A^T b
 + b^T b.
$$
(Perhaps this expression can be further simplified.)

Compared the @user's answer, this one is faster, particularly for large $N$. (Shown here for $n=3$).

Code to reproduce the plot:

import numpy as np
import perfplot
from scipy.linalg import solve_triangular

np.random.seed(0)

def setup(n):
    # k = 3
    k = n
    A = np.random.rand(n, k)
    b = np.random.rand(n)
    e = np.ones(k)
    return A, b, e

def user(Abe):
    A, b, _ = Abe
    V = (A.T - b).T
    V[:, 1:] = (V[:, 1:].T - V[:, 0]).T
    VTV = V.T @ V
    return VTV[0, 0] - VTV[0, 1:].T @ np.linalg.solve(VTV[1:, 1:], VTV[0, 1:])

def nschloe(Abe):
    A, b, e = Abe
    V = (A.T - b).T
    out = 1 / np.sum(np.linalg.solve(V.T @ V, e))
    return out

def nschloe_qr(Abe):
    A, b, e = Abe
    V = (A.T - b).T
    _, r = np.linalg.qr(V)
    w = solve_triangular(r.T, e, lower=True, check_finite=False)
    w = solve_triangular(r, w, check_finite=False)
    out = 1.0 / np.sum(w)
    return out

def least_squares(Abe):
    A, b, _ = Abe
    b -= A[:, 0]
    A = (A[:, 1:].T - A[:, 0]).T
    _, out, _, _ = np.linalg.lstsq(A, b, rcond=None)
    return out[0]

perfplot.show(
    # "out.png",
    setup=setup,
    kernels=[user, nschloe, nschloe_qr, least_squares],
    n_range=[2 ** k for k in range(2, 12)],
    xlabel="num points/dim"
)


问:
Is the compactness theorem (from mathematical logic) equivalent to the Axiom of Choice?
Or more importantly, is it independent of the axiom of choice. The compactness theorem states the given a set of sentences $T$ in a first order Language $L, T$ has a model iff every finite subset of $T$ has a model. So for any natural number $n, T(n)$ is a finite subset of $n$ sentences. Now if every finite subset has a model than adding a sentence $r$ to $T(n)$ gives us $T(n+1)$ which also has a model so there is some transfinite induction going on here when $T$ is countable. This seems to cry out for the application of Zorn's Lemma which is equivalent to the axiom of choice. So to summarize, is the compactness theorem consistent with ~AC (the negation of the axiom of choice which is independent of ZF set theory)? </p>

答:
In general, the compactness theorem is equivalent to the ultrafilter lemma, which is known to be strictly weaker than the axiom of choice (so is consistent with its negation) but independent of ZF. The models the compactness theorem asserts exist can be constructed using ultraproducts.

Over a countable alphabet, the compactness theorem is provable in ZF. This is because it can be proven from the completeness theorem, which over a countable alphabet is also provable in ZF.


问:
if $(x+y)(x-a)(y-a)=n$ has a solution at $x=0$, then $x \geq 0$
Is this true? That if:

$(x+y)(x-a)(y-a)=n$  $\hspace{1cm}$(where $a>0$ and $n>0, x,y\in \mathbb{Z}$)

has a solution when $x=0$ then other solutions occur only when $x\geq0$.  If it is true then why and how to prove it?

For example if $a=7$ and $n=42$ the only solutions are $(0,1), (0,6), (1,6)$ and their reflections, so no negative solutions. [and when I say reflection I mean $(1,6)\equiv(6,1)$]

And if $a=14$ and $n=72$ the solutions are $(11,13), (-10, 11), (-10, 13)$ and their reflections but there is no solution when $x=0$.

To state the question another way: Is it true that if there is not a solution when $x=0$, and a solution exists, then there must be a solution where $x<0$?

(I am not sure if I need to define $a,n\in \mathbb{Z}$ also)

答:
In a similar vein to the excellent answer by Raffaele:

Given positive integers $a$ and $n$, consider the equation
$$(x+y)(x-a)(y-a)=n.$$
If it has an integral solution $(x,y)\in\Bbb{Z}^2$ with $x=0$ then
$$n=(0+y)(0-a)(y-a)=-ay(y-a),$$
where $y-a\neq0$ because $n>0$. It follows that
$$-ay(y-a)=n=(x+y)(x-a)(x-a),$$
where we can divide by $y-a$ and rearrange a bit to find that
$$x(x+y-a)=0.$$
So for any solution $(x,y)$ with $x\neq0$ we must have $x+y=a$. Then
$$n=(x+y)(x-a)(y-a)=axy,$$
which shows that $xy=\tfrac{n}{a}$ and $x+y=a$. In particular the sum $x+y$ and the product $xy$ are both positive. This implies that both $x$ and $y$ are positive.


答:
It is a very interesting question to me since I have faced the same problem when I was first introduced to advance calculus. Here is a couple of examples and pronunciation that I learned.

$1.$ $f:I=(a,b)\to X$ :  $f$ is a mapping from $I$ to $X$. Here domain and range is known.

$2.$ $a\mapsto f(a)$ : $f$ is a mapping that maps $a$ to $f(a)$. Here domain and range is unknown

$3.$ $a\mapsto Xa$ : There is some map that takes $a$ to $Xa$. Here domain and range is unknown also the function itself is unknown.

$4.$ $f:X\subset Y \to A\subset B$ : $f$ is a map from $X$ to $A$. Simply!

Title: How should the following notation be read?

Body:
In the opening chapter of `foundations of optimization’ by Guler it states:
$$f:I=(c,d) \to \mathbb{R}$$

Should this be interpreted as $I$ is a function over the open interval $(c,d)\;$ that maps to the real number line?


答:
Based on your countable example what about $X=\bigcup\limits_{k=1}^\infty[3^k,3^k+1]$

We have $x=3^a+\{x\}$ and $y=3^b+\{y\}$ then

$$x+y=(3^a+3^b)+(\underbrace{\{x\}+\{y\}}_{\in[0,2]})\notin [3^c,3^c+1]$$

Indeed, wlog let assume $1\le a\le b$.

$3^a+3^b>3^b+1$

$3^{b+1}-(3^a+3^b+2)=3^b(3-\overbrace{\frac 1{3^{b-a}}}^{\le 1}-1-\overbrace{\frac 2{3^b}}^{\le \frac 23})>3^b\times \frac 13>0$

And we don't even need the $x\neq y$ condition.

Title: Construct a set such that $a, b \in X \implies a + b \not\in X$

Body:
I'm looking for an uncountable set $X \subset \mathbb{R}_{\geq 0}$ such that for all $a, b \in X$, $a \neq b \implies a + b \not\in X$.

Two points about this question: first of all, I'm not sure which area of mathematics this falls under so I used the set-theory tag; feel free to correct me on this. Secondly, note immediately that if $X$ is allowed to be countable, then e.g. $X = \{2^k \mid k \in \mathbb{Z}\}$ works.


问:
Why are two functions not identical?
$$f(x)=\ln(x^2)\\
g(x) = 2\ln(x)$$

Why are these two functions not equal?
I can't understand why since:

$$e^{\ln(x^2)}=(e^{\ln(x)})^2\\
x^2 = x^2$$

And the domain is more than zero in both cases.

答: The answer is they DO NOT share a common domain. $f:\Bbb R\setminus\{0\}\to \Bbb R$ since $f$ is defined over negative numbers too. Whereas $g:\Bbb R\setminus(-\infty,0]\to \Bbb R$ and $g$ is not defined on -ve numbers. Now since $f(-1)$ exists and $g(-1)$ does not so they are not equal!

Title: Why are two functions not identical?


问:
Given a rank-$1$ matrix $A$, what condition makes $A$ nilpotent?
My only idea on how to approach this problem was by making the matrix upper triangular. However, I found a simple counterexample to fact that elementary transitions do not change nilpotence of matrix. It does. If it was illegal, I would just say that it is nilpotent if $a_{11} \ne 0$.

Do you have any better ideas on how to solve this problem? Thank you in advance.

答:
A rank one matrix can be written as $A = v \cdot w^T$ where $v \neq 0$ is a non-zero column vector (which spans the range of $A$) and $w$ is a non-zero row vector. If you expand the product $A = v \cdot w^T$ you will see that the columns of $A$ are precisely $w_1 v, \dots, w_n v$. Now, note that

$$ A^2 = v \cdot w^T \cdot v \cdot w^T = v \cdot \underbrace{\left( w^T \cdot v \right)}_{1 \times 1 \textrm{ matrix, a scalar}} \cdot w^T = (w^T \cdot v) \cdot v \cdot w^T = cA $$

where $c = w^T \cdot v$. By induction you get that $A^k = c^{k-1} A$ and so $A$ is nilpotent if and only if $c = 0$ (in which case, $A^2 = 0$ so that $A$ is nilpotent of nilpotency rank two).

To interpret $c = w^T \cdot v = \sum_{i=1}^n w_i v_i$ note that $c$ is precisely the trace of $A$ so that $A$ is nilpotent if and only if the sum of the diagonal elements of $A$ is zero.


答:
Because they have different natural domains.

On the other hand, in $(0,\infty)$, $f(x)=g(x)$.

A similar example:
$$
g(x)=\sqrt{x},\quad f(x)=\sqrt{|x|}
$$

Title: Why are two functions not identical?

Body:
$$f(x)=\ln(x^2)\\
g(x) = 2\ln(x)$$

Why are these two functions not equal?
I can't understand why since:

$$e^{\ln(x^2)}=(e^{\ln(x)})^2\\
x^2 = x^2$$

And the domain is more than zero in both cases.


答:
In order to show that $\lim_{n \to \infty}a_nb_n = +\infty$ you have to show that for any number $M \in \mathbb{R}$ there exists $N \in \mathbb{N}$ such that $a_nb_n > M$ for all $n > N$. So let's start with an arbitrary $M \in \mathbb{R}$ and see if we can show this.

Since $a_n \to a < 0$, we can find $N_a \in \mathbb{N}$ such that $|a-a_n|< \epsilon$ for any $\epsilon > 0$. In a similar way, because $b_n \rightarrow -\infty$, we can find $N_b$ such that $b_n < M_b$ for all $n > N_b$ and any $M_b \in \mathbb{R}$.

Now we need to make some smart choices: For the $\epsilon$ mentioned before, let us take $\epsilon = -\frac{a}{2}$ (which is a positive number since $a<0$), meaning that for any $n > N_a$ we have that $a_n < \frac{a}{2}$. Also, take $N_b$ such that $b_n < \frac{2M}{a}$ for all $n > N_b$. Choose now $N = \max(N_a, N_b)$, so that for all $n > N$ we have:

$$a_nb_n > \frac{a}{2} \cdot \frac{2M}{a} = M \, , $$

which is what we wanted to show.

Title: Prove that $\lim\limits_{n\rightarrow \infty}{a_nb_n}=+\infty$

Body:
If $\lim\limits_{n\rightarrow \infty}{a_n}=a<0$ and $\lim\limits_{n\rightarrow \infty}{b_n}=-\infty$, prove that $\lim\limits_{n\rightarrow \infty}{a_nb_n}=+\infty$.

I remember that if the two sequences $a_n$ and $b_n$ are converging and their limits are $a$ and $b$, then $\lim\limits_{n\rightarrow \infty}{a_nb_n} = \lim\limits_{n\rightarrow \infty}{a_n}\lim\limits_{n\rightarrow \infty}{b_n}=a.b$. However, $\lim\limits_{n\rightarrow \infty}{b_n}=-\infty$, not a number. Can I use this theorem or I have to prove the limit in another way?


问:
Picard's Little Theorem Proofs
Picard's little theorem says that

If there exist two complex numbers $a,b$ such that $f: \Bbb{C} \to \Bbb{C}\setminus \{a,b\}$ is holomorphic then $f$ is constant.

I am interested in proofs for this theorem. Until now I found at least two, one in W. Rudin's Real and Complex Analysis and another one in S. Krantz, Geometric Function Theory. Both of them need some preparation before someone not very advanced in Complex Analysis could understand them, especially the one in S. Krantz's book.

My questions are

How many proofs are there for Picard's little theorem? (references if possible)

Is there a "simple" proof for Picard's little theorem? Simple means that it could be presented to an audience which had a one semester course in complex analysis.

Thank you.

答:
This records an answer by George Lowther in the comments, with a comment by user @t.b.

There is a proof using Brownian motion here:
Picard's Theorem and Brownian Motion by Burgess Davis. Link (free access) dx.doi.org/10.1090/S0002-9947-1975-0397900-8

User @t.b. notes:

According to Remmert's book I mention in my answer there is also an exposition in R. Durrett, Brownian Motion and Martingales in Analysis, Wadsworth, Inc. 1984, p. 139-143.

Title: Picard's Little Theorem Proofs


答:
Since you are asking this question, I think you only studied the mentioned theorem in the case of finite limits. If this is the case, you will have to do some work.

One way you can proceed is the following:

By the definition of limit with $\epsilon = \frac{a}{2},\; \exists N > 0,\; \forall n>N, \; a_n < \frac{a}{2}$. Here, I want to use that close to infinity, $a_n$ will be smaller than a fixed negative number, and $\frac{a}{2}$ will work since it is between $a$ and $0$. This is because I know that if $k<0$, $k*b_n$ will go to positive infinity "the same way" $b_n$ goes to negative infinity.

Then, you can use the definition on limit. Let $M>0$. By the definition of the limit of $b_n$, $\exists N'>0, \forall n>N, b_n < \frac{2M}{a}$.

Let $N'' = max(N, N')$ and $n>N''$. We have that $a_n<\frac{a}{2}$ and $b_n<\frac{2M}{a}$, so $a_n b_n > M$. This concludes the proof. Here, I had to use the definition of limit with $\frac{2M}{a}$ to compensate for the $\frac{a}{2}$. As I mentioned before, this is the only thing I had to do with $b_n$, because it $b_n$ and $k*b_n$ diverge pretty similarly.


问:
Intersection of a hypersurface $V \subset \mathbb{P}^n$ with a tangent hyperplane $T_PV$ singular
I would like some help with an exercise from Reid's paper of algebraic geometry.

Prove that the intersection of a hypersurface $V \subset \mathbb{P}^n$ (not a hyperplane) with the tangent hyperplane $T_PV$ is singular in P.

Maybe I'm using the wrong approach. I'm trying to prove this by taking partial derivatives and evaluate them in the point P.

Thank you!!

答:
The statement in the exercise is in fact not true as it stands. The easiest counterexample is the intersection of a smooth curve with its tangent: it is a point and therefore smooth.

A specific example is the parabola $y=x^2$ with tangent $y=0$ at the origin.
Of course one would like to count this intersection point with multiplicity two, and therefore consider it as singular.

The correct statement in the affine case is that the restriction of the defining function to the tangent plane has a critical point. A proof can be given with techniques of multivariable calculus, namely Lagrange multipliers: the gradient of the function is the normal to the tangent plane, and for a linear function the gradient is constant and equal to the normal vector, so the two gradients in question are linearly dependent. Using the gradient supposes that we are in the real case. An illustration for the argument is the fact the height function on a sphere has a maximum at the north pole. Related to the original question, depending on your point of view the intersection of a sphere with its tangent plane is a point or consists of two complex conjugate lines.

For any field, let $f$ be the defining function of the hypersurface and $l$ the defining function of the tangent plane (in the projective case $f$ and $l$ are homogeneous polynomials). If we define the intersection by the ideal $(f,l)$ with non-reduced structure, then at the intersection point the Jacobian matrix of $(f,l)$ has rank one, as the rows are equal. But the zero set $V(f,l)$ can be non-singular.


问:
Difference between first and second fundamental theorem of calculus
In first fundamental theorem of calculus,it states if $A(x)=\int_{a}^{x}f(t)dt$ then $A'(x)=f(x)$.But in second they say $\int_{a}^{b}f(t)dt=F(b)-F(a)$,But if we put $x=b$ in the first one we get $A(b)$.Then what is the difference between these two and how do we prove $A(b)=F(b)-F(a)$?

答:
They have different assumptions.

In the first part you mentioned, $f$ is assumed to be continuous. In the second part, $f$ can be assumed only Riemann integrable on the closed interval $[a,b]$. When $f$ is continuous, the second part indeed follows from the first part.

See also a comparison of the statements in this article.


问:
Mellin Transform zeros of non negative continuous function
Mellin transform of function $f(x)$ which is continuous and $f(x)\geq 0 $  is given as
$$
   f^\ast(z) =\int\limits_0^\infty x^{z} f(x) \frac{dx}{x}.
$$
I consider only exponentially decreasing (there exist such constants $C_1>0$, $C_2>0$ that $|f(x)| \leqslant C_1 e^{-C_2x}$), strictly positive, continuous on $[0,+\infty)$ and infinitely many times differentiable on $(0,+\infty)$ functions $f(x)$ such that all its derivatives decrease exponentially at infinity (and even with the same $C_2$). For such functions $f^\ast(z)$ is defined and analytic for $\Re(z) > 0$.

Question . Is it possible to say something about existence of zeros of $f^\ast(z)$ in $\Re(z)>0$? Is it true that $f^\ast(z)$ can't have zeros in $\Re(z)>0$? If it can have zeros is it known which conditions on $f(x)$ will garantee the absense of zeros of $f^\ast(z)$? Motivating example is $f(x)=e^{-x}$ with $f^\ast(z)=\Gamma(z)$.

答:
The Mellin transform of $1/(e^x+1)$ is $\Gamma(s)\eta(s), \Re s >0$, where as usual $\eta(s)=(1-2^{1-s})\zeta(s)$ which has tons of zeroes both on the line $\Re s =1$ and on the line $ \Re s =1/2$ (and of course possibly more depending on RH) so the OP question has a negative answer.

Not sure if there is any simple relation between the function $f$ and the zeroes (or more generally value distribution) of $f^*$


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答: Sometimes, the conversation never gets as far as "I was always bad at math." I was at a party with my wife while we were in graduate school. A guest  asked her what she did, and my wife replied that she was a math teacher. The guest then turned and walked away, never saying a word in response, and never to be seen again. I think some people have some deep- seated issues with math, not just with mathematicians. Why, I do not know, and it would be useless for me to guess, as I am not really qualified to do so.  </p>


问:
Given $a,b\in \mathbb N$, can $a^2+a b+b^2$ and $a^2- a b+b^2$ both be square numbers?
The question arises if you search for two integer triangles ABC and AB'C with common vertices A and C, CA=b, CB=a, CB'=a+b and $\measuredangle ACB$ = $\measuredangle ACB'= 60°$. With c=AB and c'=AB' you get $c^2=a^2- a b+b^2$ and $c'^2=a^2+ a b+b^2$. I cannot find any $a,b\in \mathbb N$ such that $c,c'\in \mathbb N$. Is there an argument that shows that this is impossible?

答:
There are no positive integer solutions to these equations. One way of proving this is to consider the Diophantine equation

$$a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)=z^2.$$
We can tackle this by the method used in Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?

Theorem

There are no positive integer solutions of either of the equations
$$ x^4+x^2y^2+y^4=z^2\tag{1}$$ $$x^4-2x^2y^2-3y^4=z^2\tag{2}.$$

Proof

First note that, for either equation, we can suppose that the variables are pairwise coprime  since a common factor of any pair of variables would be a factor of all and cancellation can occur.

An equation of form (1)

We can suppose that $x$ is odd. Then $z$ is odd and we have $y^2+y^4\equiv 0 \pmod 8$. Therefore $y$ is divisible by $4$ and we can write $y=2t$, where $t$ is even.

$ x^4+4x^2t^2+16t^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2+2t^2-z}{2}\right )\left (\frac{x^2+2t^2+z}{2}\right)=-3t^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $t$, a contradiction.

Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=-3$ and $t=uv$, with $u$ and $v$ coprime and $uv$ even. Then  $$au^4+cv^4=x^2+2t^2=x^2+2u^2v^2.$$
We can assume $v$ is even. Then $au^2\equiv 1\pmod 4$ and so $a=1$.

Therefore $u^4-2u^2v^2-3v^4=x^2$, an equation of form (2).

An equation of  form (2)

Let $u,v,x$ be a pairwise coprime solution of  $ u^4-2u^2v^2-3v^4=x^2$.

This can be rewritten, using completing the square, as $$\left (\frac{u^2-v^2-x}{2}\right )\left (\frac{u^2-v^2+x}{2}\right)=v^4.$$
The bracketed factors, $L$ and $M$,  are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=1$ and $v=XY$, with $X$ and $Y$ coprime. Then $$X^4+X^2Y^2+Y^4=u^2,$$ an equation of form (1).

Fermat's infinite descent

We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) always leads to another positive integer solution $(X,Y,Z)$, where $Y=\frac{y}{2uX}<y$, which is impossible.


问:
If $F \vee G$ is a tautology, then $F$ is a tautology or $G$ is a tautology.

Let $F$ and $G$ be formulas of propositional logic having no atomic proposition in common.

Show: If $F \vee G$ is a tautology, then $F$ is a tautology or $G$ is a tautology.

I have come across this question and am struggling to understand what is needed

If $F \vee G$ is a tautology then
\begin{align*}
[[F \vee G]] \alpha  &= [[F]]\alpha \vee [[G]]\alpha \\
                 &= 1 \vee 1 \\
                 &= 1.
\end{align*}

I'm not really understanding what to include.

答:
I never seen this notation, but I assume $[[A]]_{\alpha}$ stands for the value of the formula $A$ in the valuation $\alpha$, am I correct?

Since $F$ and $G$ have no atomic propositions in common, assume $p_{1}, ... , p_{n}$ are the atomic propositions in $F$ and $q_{1}, ... , q_{m}$ the atomic propositions in $G$, which satisfy $\{p_{1}, ... , p_{n}\}\cap\{q_{1}, ... , q_{m}\}=\emptyset$.

Now, if $F$ and $G$ are both non-tautologies, as Mauro ALLEGRANZA suggests, take valuations $\alpha$ and $\beta$ such that $[[F]]_{\alpha}=0$ and $[[G]]_{\beta}=0$. And now comes the important part: define the valuation $\nu$ such that
$$[[p_{1}]]_{\nu}=[[p_{1}]]_{\alpha}, ... , [[p_{n}]]_{\nu}=[[p_{n}]]_{\alpha}\quad\text{and}\quad [[q_{1}]]_{\nu}=q_{1}]]_{\beta}, ... , [[q_{m}]]_{\nu}=[[q_{m}]]_{\beta},$$
what can be done since all $p_{i}$ and $q_{j}$ are atomic propositions. Since $p_{1}$ through $p_{n}$ are the atomic propositions in $F$, and $\nu$ and $\alpha$ agree in these statements, $[[F]]_{\nu}=[[F]]_{\alpha}=0$; analogously, $[[G]]_{\nu}=[[G]]_{\beta}=0$, meaning that
$$[[F\vee G]]_{\nu}=[[F]]_{\nu}\vee[[G]]_{\nu}=0\vee 0=0,$$
what would imply that $F\vee G$ is NOT a tautology, against our hypothesis. By reduction to absurd, we must derive that either $F$ or $G$ is a tautology.


问:
Irreducible Lipschitz numbers
A Lipschitz integer is a Quaternion with integer coefficients. The norm is defined as $N(a+ib+jc+kd)=a^2+b^2+c^2+d^2$ which is a multiplicative function

$N:\mathbb H\to\mathbb R$, $N(\alpha\beta)=N(\alpha)N(\beta)$

Now since any natural number can be written as the sum of four squares of integers, all prime numbers in $\mathbb N$ are composite Lipschitz integers

$a^2+b^2+c^2+d^2=(a+ib+jc+kd)(a-ib-jc-kd)$

Obviously, if $\alpha$ is a composite, then so is $N(\alpha)$.

In the ring of Gaussian integers it also holds that if both the real and the imaginary terms are non-zero, then the norm is prime if the Gaussian number is prime. My question is

If a Lipschitz integer is irreducible and all four terms
are non-zero, is then the norm a prime ?

答:
Due to Wikipedia a Hurwitz number is irrecducible iff it's norm is a prime number.

A Hurwitz number is either a Lipschitz number or a quaternion number where all components are of the form odd number divided with $2$.

For Hurwitz numbers Euclidian division is possible.

For complex or Gaussian integers $u= a + ib$ and $v= c + id$, with the norm $N(v) > 0$, there always exist $z = e + if$ and $z_1 = r + is$ such that

$u = vz + z_1$, where $N(z_1) < N(v)$.

However, for Lipschitz integers $m = a+ib+jc+kd$ and $n =e+if+jg+kh$ it can happen that $N(z_1) = N(n)$. This motivated a switch to Hurwitz integers, for which the condition $N(z_1) < N(n)$ is guaranteed.

Title: Irreducible Lipschitz numbers


问:
Second order DE in R and Θ
For expressing the Cauchy-Riemann conditions for
f(z, z*) = u(x,y) + iv(x,y) = R(r,θ) exp[iΘ(r,θ)]

in terms of R and Θ in polar coordinates r and θ, how do I find the second order differential equations satisfied by R and Θ? Please help!

答:
Short answer:  Clearly $\log R,\Theta$ are harmonic functions and recall the form of Laplacian in polars.

Longer answer:

Prove that $\frac{\partial u}{\partial r}=\frac1r\frac{\partial v}{\partial\theta}$ and similar.

Locally $f=e^g$ (remember $f\neq 0$ for a local polar coordinates at $f(z)$) so $g=\log R+i\Theta$ is in a form which 1. above applies
$$
\frac{R_r}R=\frac1r\Theta_\theta,\quad
\Theta_r=-\frac1r\frac{R_\theta}R.
$$

Differentiate again the result in 2. to get a second-order DE.

Title: Second order DE in R and Θ


问:
Totally ordered to well ordered set
(ZF) Let  be a totally ordered set. Let every time when w is an initial segment of , is true w = A or ($\exists$x $\in$ A)(w = seg(x)). Prove that  is a well-ordered set.

Because it has to be proven in (ZF) I cannot use Zorn's lemma.

I know that if  is a well-ordered set, every initial segment w is either w = A or w = seg(x) for some x $\in$ A. But for totally ordered sets that is not always true. For example in the totally ordered set $\mathbb R$ \ {$0$} the initial segment I = (-$\infty$, $0$) is not an initial segment I $\neq$ seg(x) for any x $\in$ $\mathbb R$ \ {$0$}.

So in the problem I am guaranteed that for every initial segment of   the previous scenario doesn't apply. But I am almost completely sure that this does not prove that  is a well-ordered set.

Is there any way of proving that  has a least element so that it is well-orederd? Or any other ideas? Here I found the same problem but I don't understand the idea that has been given.

答:
The counterexample in the comments confuses me, probably one needs the extra assumption that $A$ has a smallest element $a_0$.

I denote the order $R$ by $\le$ and $<$ is understood as usual.

We need to prove that every nonempty subset $S\subset A$ has a smallest element.

Consider $B:=\{a\in A: \forall s\in S.\ a<s\}$.

Note that it's disjoint to $S$.

Then either $a_0\in S$ in which case this is its smallest element, or otherwise $a_0\in B$, so $B$ is a nonempty initial segment.

Since $S$ was nonempty, $B\ne A$, thus by hypothesis $B=\{a:a<x\}$ for some $x\in A$.

Now, clearly for any $s\in S$ we have $s\not<x$ (else we'd get $s\in B$), so $x\le s$ by the order being total.

Finally, $x\in S$ because otherwise $x<s$ would hold for all $s\in S$ and therefore $x\in B$ by definition of $B$, which would imply $x<x$.

Title: Totally ordered to well ordered set


问:
distribute m pennies to n people, what is the expectation of coins one would obtain
Assume there are $m$ pennies and $n$ people. We want to distribute the pennies to the people by uniformly picking a vector $(x_1,...,x_n)$ from the set of all vectors satisfying $x_1+...+x_n=m$, where $x_i$ is the number of coins given to person $i$.

What is the expectation of the number of pennies given to player 1?

Some analysis: I figured out that, the probability for getting a large $x_1$ is lower when compared to a smaller value of $x_1$, although the solution is uniformly picked. I guess, every person has the same likelihood to get the a certain number of coins, which means no one has an advantage. Therefore the expectation for player 1 is just $m\over n$ Is my analysis and answer correct?

答:
Your reasoning is correct, you are using symmetry.  Here is another approach.

Suppose the pennies are distinguishable.  Let $X_1, X_2,\ldots, X_m$ each be random variables, with $X_i=1$ if penny $i$ is given to person 1, and $X_i=0$ if penny $i$ is not given to person 1.  We have $E(X_i)=\frac{1}{n}$, and $E(X_1+\cdots+X_m)=E(X_1)+\cdots+E(X_m)=\frac{m}{n}$, where the linearity of expectation is used in the first equality.  Note that $E(X_1+\cdots+X_m)$ denotes the expected total number of pennies person 1 receives.


问:
Mathematicians' Tensors vs. Physicists' Tensors
It seems, at times, that physicists and mathematicians mean different things when they say the word "tensor." From my perspective, when I say tensor, I mean "an element of a tensor product of vector spaces."

For instance, here is a segment about tensors from Zee's book Einstein Gravity in a Nutshell:

We already saw in the preceding chapter
  that a vector is defined by how it transforms: $V^{'i}
 = R^{ij}V^j$ . Consider a collection of “mathematical
  entities” $T^{ij}$ with $i , j = 1, 2, . . . , D$ in $D$-dimensional space. If they transform
  under rotations according to
  $T^{ij} \to T^{'ij} = R^{ik}R^{jl}T^{kl}$ then we say that $T$ transforms like a tensor.

This does not really make any sense to me. Even for "vectors," and before we get to "tensors," it seems like we'd have to be given a sense of what it means for an object to "transform." How do they divine these transformation rules?

I am not completely formalism bound, but I have no idea how they would infer these transformation rules without a notion of what the object is first. For me, if I am given, say, $v \in \mathbb{R}^3$ endowed with whatever basis, I can derive that any linear map is given by matrix multiplication as it seems the physicists mean. But, I am having trouble even interpreting their statement.

How do you derive how something "transforms" without having a notion of what it is? If you want to convince me that the moon is made of green cheese, I need to at least have a notion of what the moon is first. The same is true of tensors.

My questions are:

What exactly are the physicists saying, and can someone translate what they're saying into something more intelligible? How can they get these "transformation rules" without having a notion of what the thing is that they are transforming?

What is the relationship between what physicists are expressing versus mathematicians?

How can I talk about this with physicists without being accused of being a stickler for formalism and some kind of plague?

答: There are two types of tensors, general tensors and Cartesian tensors.  I think that is the distinction you're looking for.  Both mathematicians and physicists use general tensors, engineers use Cartesian tensors.  Most tensors are rank 2 tensors and can be represented by a square matrix.  That's true if it's a general tensor tensor or a Cartesian tensor. General tensors start with vectors and covectors, with a covector being a map of a vector to real number.  Cartesian tensors only use vectors.  When the basis for a tensor's underlying vector space changes, a vector and covector transform differently if non-orthonormal bases are allowed (because the dot product is not preserved transforming to a non-orthonormal base). General tensors account for this difference in their basis transformation formulas.  If only orthonormal bases are allowed vectors and covectors transform the same way so there is no distinction, and the basis transformation formulas are simpler.  General tensors are mandatory in general relativity, but not elsewhere (I think.).</p>


问:
Why is $\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)^0\ne1$?
Why is

$\left( \begin{array}{cc}  1 & 1 \\  1 & 1 \\ \end{array} \right)^0=\left( \begin{array}{cc}  \frac{1}{2} & \frac{1}{2} \\  \frac{1}{2} & \frac{1}{2} \\ \end{array} \right)$

This result me returns Mathematica on this command

```<code>Limit[MatrixFunction[#^s &, ( { {1, 1}, {1, 1} } )], s -> 0, Direction -> "FromAbove"]
</code>```

Result:

```<code>{{1/2, 1/2}, {1/2, 1/2}}
</code>```

Should not anything rised to zeroth power be $1$?

答:
I would argue that

$$\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}^0$$

could be any symmetric $2×2$ matrix with an eigenvector $\begin{bmatrix}1 \\ 1 \end{bmatrix}$ having eigenvector $1$, to wit:

$$\begin{bmatrix}a & 1-a \\ 1-a &  a \end{bmatrix}$$

Any such matrix multiplied by right or left by $\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}$ gives back $\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}$ and thus qualifies potentially as a zero power by the multiplicative definition of exponentiation.

This unsatisfactory result stems from the fact that all rows and columns of the matrix $\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}$ are multiples of a specific vector, or more generally that they are linearly dependent, or getting to the crux of the matter, the input matrix is singular!  Just as with numbers running the multiplicative recursion backwards fails to give $0^0=1$ because $0$ lacks a multiplicative inverse, so it is with matrices.

In the case of $0^0$ there are definitions of exponentiation (with limited domains on the exponent) that effectively force $0^0=1$, but matrices with their greater degree of freedom are harder to pin down and give less predictable results.  Something in the way Mathematica defines matrix exponentiation, or perhaps the limit process used in the program,, led it to a specific value of an intrinsically undefined zero-power of a singular matrix.


问:
Can the harmonic function $u$ in $U$ be extended continuously to $\overline{U}$?
Suppose that
\begin{equation}
u(z)=\arg \frac{1+z}{1-z}
\end{equation} is harmonic in the unit disk $U$. Can this function be extended to a continuous real function on $\overline{U}$? I guess it can be, but I don't know how to prove it. Please help!

答:
The answer is no because at $z= \pm 1$ there are $\pi$ jumps. Assume we take the principal branch of the argument with values in $(-\pi, \pi]$ (if we take the branch with values in $[0, 2\pi)$ same applies except that the values in cause will be $\pi/2, 3\pi/2$).

An easy way to see this is to note that if $z=e^{it}, -\pi \le t \le \pi$ we have $\frac{1+z}{1-z}=\frac{1+e^{it}}{1-e^{it}}=\frac{2 \cos t/2 e^{it/2}}{-2i \sin t/2 e^{it/2}}=i \cot t/2$ so the argument switches from $-$ to $+$ precisely at $t =0$ and since the function is purely imaginary, this means that $u(e^{it})=-\pi/2, -\pi \le t < 0$ and $u(e^{it})= \pi /2, 0< t \le \pi$ (where the extension at $-\pi= \pi$ (mod circle) is by continuity of course so there are left and right hand limits there too)

As an aside the above shows that $u(z)$ (interpreted using the principal branch with values in $(-\pi, \pi]$) is the solution to the Dirichlet problem on the circle with data that is $-\pi/2$ on the lower half-circle and $\pi/2$ on the upper one