TITLE: How to determine the continuity of a piecewise function defined respectively on $\mathbb{Q}$ and $\mathbb{I}$
QUESTION [1 upvotes]: How can I tell the continuity of function
$$ f(x)=\left\{
\begin{array}{lll}
3x^2 & \text{if} & x\in\mathbb Q\\
4x^2 & \text{if} & x\in\mathbb I
\end{array}
\right. $$
I could see it is continuous at x=0 and guess it discontinuous no-where-else. How can I start with the formalised proof?
And also, is $f$ differentiable at 0?

REPLY [1 votes]: Let $x_0 \in \mathbb{R}$. If $(x_n)$ is a sequence of real numbers such that $x_n \to x_0$ as $n \to \infty$ then $f$ is continuous at $x_0$ if and only if $$\lim_{n\to \infty} f(x_n) = f(x_0)$$
Lets assume $x_0 \in \mathbb{Q}$, and let $(x_n)$ be a sequence of irracional numbers (converging to $x_0$), therefore, $f(x_n)$ is the sequence $4x_n^2$ and then $\lim f(x_n) = \lim 4x_n^2 = 4 x_0^2$ and $f(x_0) = 3x_0^2$. If $x_0 \in \mathbb{R}\setminus\mathbb{Q}$ you construct a rational sequence and will reach the same thing, namely: $f$ is continuous at $x_0$ if, and only if $3x_0^2 = 4x_0^2$ and then $f$ is continuous only in $x_0=0$
Let's see now that $f'(0)=0$. But $$f'(x) = lim_{x_0 \to x} \frac{f(x)-f(x_0)}{x-x_0}$$
And then
$$f'(0) = \lim_{x \to 0} \frac{f(x)}{x}$$
But $$\frac{f(x)}{x}=\left\{
\begin{array}{lll}
3x & \text{if} & x\in\mathbb Q^*\\
4x & \text{if} & x\in\mathbb I
\end{array}
\right.$$
And again using sequences if necessary, $\lim_{x \to 0} \frac{f(x)}{x} = 0$ and $f'(0) = 0$


TITLE: $I,J$ ideals in $k[x_1,...,x_n]$, $J\subseteq \sqrt{I}$, then $J^r \subseteq I$ for some $r>0$
QUESTION [0 upvotes]: $I,J$ ideals in $k[x_1,...,x_n]$, $J\subseteq \sqrt{I}$, then $J^r \subseteq I$ for some $r>0$.

I found this as an exercise in the book Ideals, Varieties and Algorithms (Cox, Little, O'Shea). It has the tip: use the Hilbert Basis Theorem. So I chose a system of generators for $\sqrt{I}$, $\{f_l\}_{l\in L}$ such that $\{f_h\}_{h\in H\subset L}$ was a base for $J$. So then I have to show that some $r$ exists so that all the products of $r$ elements of the system of generators for $J$ are in $I$. Each $f_l$ belongs to $\sqrt{I}$, thus, for some $N$, $f_l^N \in I$ (the system of generators is finite, so we can choose the biggest $N$). But I am not sure about the products of different $f_l$'s. I was thinking about considering $(\sum f_l)^N$ for different amounts of $f_l$, but I am not sure that is going to be a way to solve this, and I am not sure if taking $r=N$ is the right idea.

REPLY [1 votes]: If we actually take $r=NL$ ($L=$ number of elements in a system of generators of $J$), then $\prod_{l=1}^r f_l= \prod_{i=1}^Lf_i^{k_i}$ (in the first product, some factors might be repeated), where some $k_i \geq N$ because $\sum_{i=1}^Lk_i=NL$. Thus, $f_i^{k_i} \in I$ because of the definition of $N$, so the whole product must be in $I$. So we have that every product of $r=NL$ elements of the system of generators of $J$ is in $I$, which means that $J^r \subseteq I$.


TITLE: Probability and real roots of a polynomial
QUESTION [3 upvotes]: Today, I suddenly thought of a question:

Find the probability that a $n$-degree polynomial has $r$ real roots. ($0\le r\le n$)

So, I created a function $P\left(r;n\right)$ which is the probability that a $n$-degree polynomial has $r$ real roots. Then, I try some small values.
For $n=1$, it is easy to see that $P\left(0;1\right)=0$ and $P\left(1;1\right)=1$
For $n=2$, we need to consider the determinant $\Delta$ of $x^2+ax+b$. (It has the same probability ratio as $ax^2+bx+c$)
When $\Delta \le 0$, which means $b \le \dfrac{a^2}{4}$, the probability of choosing $a,b$ with such condition is $0$, so $P\left(0;2\right)=P\left(1;2\right)=0$. Similarly, $P\left(2;2\right)=1$
Then, I have a guess: $P\left(0;n\right)=P\left(1;n\right)=\cdots=P\left(n-1;n\right)=0$ and $P\left(n;n\right)=1$. However, it consists of high-degree polynomial so I can't prove my guess. Is there anyone can help me. Tips are free, thank you!

REPLY [1 votes]: As pointed out in the comments you need to specify a probability distribution for this question to be meaningful. For example, here is one way of generating a random polynomial of degree $n$: let $(Z_1,\ldots,Z_n)$ be any random vector of real numbers and consider the random polynomial
$$
p(x)=(x-Z_1)\cdots (x-Z_n).
$$
You can write out the coefficients in terms of the $Z_1,\ldots,Z_n$ to see that they are truly random (except for the $x^n$ which has coefficient $1$) so it meets your criteria, and all the probabilities match your guess: $0$ probability of having $r<n$ real roots, probability $1$ of having $n$ real roots.
On the other hand, we can also consider the random polynomial
$$
q(x)=(x^2+Z_1^2+1)\cdots (x^2+Z_n^2+1)
$$
which has no real roots, and again you can multiply out the coefficients and see they are random.
Now I guess that neither of these two examples match what you are actually interested in, which is the case where all the coefficients are drawn randomly from the same distribution and independently of each other. Now the question is more interesting and has a lot of research literature that is easily accessible (for example here https://arxiv.org/abs/1409.4128) and some less accessible literature going back over 50 years. While the papers may be difficult to read if you are not familiar with probability theory at the graduate level, the results are easy to summarize: in most cases people have studied, the expected number of real roots is quite a lot smaller than $n$, approximately of size $\log n$, and as a result there is a very low probability of having $n$ real roots especially as $n$ grows larger and larger. The precise asymptotic is that the expected number of real roots grows as
$$
\bigl(\tfrac{2}{\pi}+o(1)\bigr)\log n\text{ as }n\to\infty,
$$
where the notation $o(1)$ denotes a quantity that converges to $0$ as $n\to\infty$.


TITLE: Relative compact inside a subset with relative topology
QUESTION [2 upvotes]: I am a little bit confused. Assume that $Y$ is a subset of $X$, with the inherited topology from $X$. What does it mean the $K\subset Y$ is a relatively compact subset of $Y$?
One option, is that $K= C\cap Y$, where $C$ is a relatively compact subset in $X$.
Another option, is the $\overline{K}\cap Y$ is compact in $Y$.
To be specific, I am looking on the following example:
Let $X$ be a topological (locally compact Hausdorff), $\{D_n\}_{\
n\in\mathbb{Z}}$ be a sequence of open subsets inside $X$. For every $n$, let $T_n: D_n\to D_{-n}$ be a homeomorphism.
Consider the set $Y:=\{(T_n(x),n,x)\in X\times \mathbb{Z}\times X| x\in D_n\}$ with the inherited topology from $X\times \mathbb{Z}\times X$.
I want to say that if $K\subset Y$ is relatively compact, then $K$ is contained in a set of the form $\{(T_n(x),n,x)\in X\times \mathbb{Z}\times X| x\in D_n\cap C, \  n\in [-N,N]\}$ for some relatively compact set $C$ in $X$ and $N\in\mathbb{N}$. Is it true?
Thanks!

REPLY [1 votes]: According to definition a subset of a topological space is relatively compact  if its closure (in that space) is compact.
So I suspect that $K\subseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $\overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $\overline K\cap Y$.
This results in the second option that you mention:$$K\text{ is a relatively compact subset of }Y\subseteq X\text{ iff }\overline K\cap Y\text{ is compact}$$

I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.


TITLE: equation of parabola given equation of axis , vertex and focus
QUESTION [0 upvotes]: Suppose the equation of axis of symmetry of a parabola is $a_1x+b_1y+c_1=0$ and the equation of the line perpendicular to the axis of symmetry and passing through the vertex $a_2x+b_2y+c_2=0$. I noticed while doing a problem that the equation of the parabola turned out to be $$(a_1x+b_1y+c_1)^2=k(a_2x+b_2y+c_2)$$
But I could'nt understand what 'k' was. Please help.

REPLY [1 votes]: Some observations first. Since your two lines are perpendicular to one another, the two sides of your quation are essentially like $x^2=ky$, except that you applied some coordinate transformation to a different affine coordinate system whose basis is orthogonal but not neccessarily orthonormal. The orthogonality of the lines can be verified by $a_1a_2+b_1b_2=0$. Since the equation $(ka_2)x + (kb_2y) +(kc_2)$ is essentially of the same form as $a_2x+b_2y+c_2$, you might as well speak about the parabola being defined by the two lines alone.
Finding focus from equation
Judging from your comment, it looks as though your main goal is trying to determine the location of the focus if you have all the parts of the equation.
So let's concentrate on $x^2=ky$ first. Where is the focus here? Well, $y=x^2$ has the focus at $(0,\frac14)$. If you scale the $y$ side of the equation by $k$ you get the focus at $ky=\frac14$ or $y=\frac1{4k}$. This covers the case of $a_1=1,b_1=0,c_1=0,a_2=0,b_2=1,c_2=0$. If you want $a_1$ and $b_2$ to be arbitrary lengths you get the equation $(a_1x)^2=kb_2y$ and the focus at $(0,\frac{a_1^2}{4kb_2})$.
Next you consider $b_1\neq0$ and $a_2\neq0$ while keeping $c_1=c_2=0$ for the moment. That's essentially a rotation. The distance from the origin (which is still the vertex) to the focus can be derived similar to the equation above, except that instead of $a_1$ and $b_2$ you now should use the length of the corresponding vectors. So if the distance between vertex and focus were $d$ you'd have
$$d=\frac{a_1^2+b_1^2}{4k\sqrt{a_2^2+b_2^2}}$$
The direction from the vertex to the focus is the direction indicated by $(a_2,b_2)$, but you have to normalize that to unit length. So the vector from vertex to focus is
$$\overrightarrow{VF}=\frac{a_1^2+b_1^2}{4k\left(a_2^2+b_2^2\right)}\begin{pmatrix}a_2\\b_2\end{pmatrix}$$
In order to also consider $c_1\neq0$ and / or $c_2\neq0$ all you need are coordinates for the focus. You can compute them in homogeneous coordinates using the cross product.
$$\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}
\times\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix}
=\begin{pmatrix}
b_1c_2-c_1b_2\\
c_1a_2-a_1c_2\\
a_1b_2-b_1a_2
\end{pmatrix}$$
which dehomogenizes to
$$V=\frac{1}{a_1b_2-b_1a_2}
\begin{pmatrix}b_1c_2-c_1b_2\\c_1a_2-a_1c_2\end{pmatrix}$$
so you have
$$F=V+\overrightarrow{VF}=
\frac{1}{a_1b_2-b_1a_2}
\begin{pmatrix}b_1c_2-c_1b_2\\c_1a_2-a_1c_2\end{pmatrix}+
\frac{a_1^2+b_1^2}{4k\left(a_2^2+b_2^2\right)}\begin{pmatrix}a_2\\b_2\end{pmatrix}$$
as the coordinates of the focus, depending on $k$ and the other coefficients of your lines.
Finding equation from axis, vertex and focus
The title of the question duggests a different problem, though. Namely that of finding the equation of the equation given the vertex and the focus. The axis of symmetry is alreasdy implied by these two points. So let's suppose you have your vertex at $V=(V_x,V_y)$ and your focus at $F=(F_x,F_y)$. Then the vector $4\overrightarrow{VF}$ plays the same role for this parabola as the vector $(0,1)$ plays for the standard parabola $y=x^2$. If that vector has length one, you can extract the analogon of the $y$ coordinate from a point $P=(x,y)$ by computing the dot product $4(F-V)\cdot(P-V)$. Otherwise you have to divide by the squared length of that vector: $\frac{4}{\lVert F-V\rVert^2}(F-V)\cdot(P-V)$. That's essentially the right hand side of your equation.
For the left hand side you need the same vector $F-V$ but rotated by $90°$ in either direction. So let's make that $(V_y-F_y,F_x-V_x)$. Thus the complete equation would be
\begin{multline*}
\left(\frac{1}{(F_x-V_x)^2+(F_y-V_y)^2}\bigl((V_y-F_y)(x-V_x)+(F_x-V_x)(y-V_y)\bigr)\right)^2=\\=4\left(\frac{1}{(F_x-V_x)^2+(F_y-V_y)^2}\bigl((F_x-V_x)(x-V_x)+(F_y-V_y)(y-V_y)\bigr)\right)
\end{multline*}
You can multiply by those denominators and cancel common factors.
\begin{multline*}
\bigl((V_y-F_y)(x-V_x)+(F_x-V_x)(y-V_y)\bigr)^2=\\=4\bigl((F_x-V_x)(x-V_x)+(F_y-V_y)(y-V_y)\bigr)\bigl((F_x-V_x)^2+(F_y-V_y)^2\bigr)
\end{multline*}
Now you can easily expand these terms to read $a_1,b_1,c_1,a_2,b_2,c_2$ from these. As I write above, you can essentially choose the $k$ any way you like (except for $k=0$).


TITLE: Is anything known about $ \small{b_0+\tfrac{a_1}{\left(b_1+\tfrac{a_2}{\left(b_2+...\right)^n}\right)^n}} $?
QUESTION [9 upvotes]: What is known about this generalized "continued fraction"
$$
b_0+\frac{a_1}{\left(b_1+\frac{a_2}{\left(b_2+\frac{a_3}{\left(b_3+\dotsb\right)^n}\right)^n}\right)^n}
$$
when the integer $n\ge 2$?
Wikipedia and wolfram articles on generalized continued fraction doesn't mention any continued fractions of this kind.
Of course one can calculate explicitly periodic continued fractions, but they are not interesting.

REPLY [0 votes]: Not interesting! Explicitly periodic generalized continued fractions are not interesting?  You do realize that if you take $n=2$, and you assume a period of $k$, then your continued fraction (if it converges) is the root of a degree $2k+1$ polynomial?
Explicitly: If $x = [a,a,a,a,...;b,b,b,b,...]_2$, then $x = a + b / x^2$, and $x^3 - ax^2 - b = 0$.  If $x = [a, a', a, a',...; b,b',b,b',...]_2$, then $x = a + b / (a' + b'/x^2)^2 = a + bx^4 / (a'x^2 + b')^2$ or $a'^2 x^5 - (aa'^2 + b) x^4 + 2a'b' x^3 - 2aa'b' x^2 + b'^2 x - ab'^2 = 0$. Setting $c = b/a'^2$ and $d = (b'/a')$, we get $x^5 - (a+c)x^4 + 2dx^3 - 2adx^2 +d^2x - ad^2 = 0$.
What I find interesting is that in the latter case, you don't have the full space of quintic polynomials -- you only get solutions to a 3-dimensional subspace.