Get Expert Help for Your Rational Functions Questions

Recall that an elliptic curve over a field $k$ i.e a proper smooth connected curve of genus 1 equipped with a distinguished $k$-rational point, I'll be really grateful for any help in understanding the following part of our course
Let $(E,0)$ be an elliptic curve, using Riemann-Roch we construct an isomorphism into $\mathrm{Proj}k[X,Y,Z]/Y^{2}Z+a_{1}XYZ+a_{3}Y{Z}^2-<!--\; -\; -->X^3-<!--\; -\; -->a_2{X}^{2}Z-<!--\; -\; -->a_{4}X{Z}^2-<!--\; -\; -->a_6{Z}^3$ that can be written informally as $P\to <!--\; \to \; -->[x(P):y(P):1(P)]$, where $x$ and $y$ are rational functions such that $v_0(x)=-<!--\; -\; -->2$ and $v_0(y)=-<!--\; -\; -->3$.
Why does 0 map to the infinity point $O=[0:1:0]$? According to Hartshorne it is because both $x$ and $y$ have poles in 0 but I can't see why.

Could any one give me a hint how to show Every rational function which is holomorphic on every point of Riemann Sphere( ${\mathbb{C}}_{\mathrm{\infty <!--\; \infty \; -->}}$) must be constant?(with out applying Maximum Modulas Theorem). Thanks.

I know how to find the oblique/slant asymptote of a rational function. What I'm unsure about, because I've never dealt with this, is how many oblique asymptotes can a function actually have? And is it only possible to have an asymptote if the degree of the numerator polynomial and the degree of the denominator polynomial only differ by 1?

I sometimes have some trouble finding the range of a rational function. What is an easy or good way to do this?

How can I create a rational function g(x) with these requirements:
1. The vertical asymptotes are at 2 and 5
2. The function has a slant asymptote y=2x+1
3. The y-intercept is 7

The given function is
$f(x,y)=\frac{1}{x^2+y^2-<!--\; -\; -->1}$
From my understanding, a rational function such as this one would not have any points in which $\mathrm{\nabla <!--\; \nabla \; -->}f(x,y)=\mathbf{0}$, so I would conclude that no such points exist. Would this be true or am I not looking at this in a different way?

I am trying to prove that
${\int <!--\; \int \; -->}_C\frac{P(z)}{Q(z)}dz=0$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
${\int <!--\; \int \; -->}_{C}f(z)dz=2\mathrm{\pi <!--\; \pi \; -->}i\text{Res}[\frac{1}{z^2}f(\frac{1}{z})]$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.

Let's say we have a rational function $f$ (i.e polynom divided by polynom) , and assume that $f$ has no poles in the upper plane $\{z;Imz\ge <!--\; \ge \; -->0\}$
we have to prove that:
$sup\{|f(z)|;Imz\ge <!--\; \ge \; -->0\}=sup\{|f(z)|;Imz=0\}$
1. I assume that we are dealing in cases where the suprimum exists.
2. every approach I tried, I can't really understand what is the significant of function being rational. I know, for example, that there is finite number of zero's, and finite number of poles, but I can't see how it is helpful.

Find the partial fraction decomposition of the rational function ${\displaystyle \frac{2x^3+7x+5}{(x^2+x+2)(x^2+1)}}$
I have tried dividing first but keep running into problem after problem.

Could someone please let me know if my steps are correct? I am trying to rewrite the first line as a rational function, $\frac{p(x)}{q(x)}$
$\begin{array}{rl}& 3x^1+3x^2+9x^3+9x^4+27x^5+27x^6+\cdots <!--\; \cdots \; -->\\ =& 3x(1+x)+9x^3(1+x)+27x^5(1+x)+\cdots <!--\; \cdots \; -->\\ =& 3x(1+x)(1+3x^2+9x^4+\cdots <!--\; \cdots \; -->)\\ =& (3x+3x^2)\frac{1}{1-<!--\; -\; -->3x^2}\\ =& \frac{3x+3x^2}{1-<!--\; -\; -->3x^2}\end{array}$

Let $C$ be the algebraic curve defined by the modular polynomial ${\mathrm{\varphi <!--\; \varphi \; -->}}_N$ of order $N>1$ over the rational numbers, i.e.
$C:=\text{specm}(\mathbb{Q}[X,Y]/{\mathrm{\varphi <!--\; \varphi \; -->}}_N(X,Y)).$
The singularities of this curve can be removed and we obtain a nonsingular curve $C^{sn}$n, then, we can embed $C^{sn}$ into a complete non-singular curve $\stackrel{¯<!--\; ¯\; -->}{C}$.
In Milne's notes "Modular Functions and Modular Forms" it is written:
The coordinate functions $x$ and $y$ are rational functions on $\stackrel{¯<!--\; ¯\; -->}{C}$, they generate the field of rational functions on $\stackrel{¯<!--\; ¯\; -->}{C}$, and they satisfy the relation ${\mathrm{\varphi <!--\; \varphi \; -->}}_N(x,y)=0$.
I assume, by coordinate functions he means the functions $f(X),g(X)\in <!--\; \in \; -->\mathbb{Q}[X]$ such that ${\mathrm{\varphi <!--\; \varphi \; -->}}_N(f(x),g(x))=0$ for all $x\in <!--\; \in \; -->Q$. However, I don't understand why the field of rational functions on $\stackrel{¯<!--\; ¯\; -->}{C}$ is generated by these functions. Could someone explain this to me?
Thank you very much in advance!

If $k$ is algebraically closed, we can consider a rational function as a function $k\to <!--\; \to \; -->k\cup <!--\; \cup \; -->\{+\mathrm{\infty <!--\; \infty \; -->}\}$.
Why is it important that the field be algebraically closed?

I have two questions:
1. Can a irreducible rational curve have infinitely self intersections?
2. If f has rational coefficients and the solutions for $0=f(x,y)\in <!--\; \in \; -->k[x,y]$ are parametrized by rational functions with rational coefficients of some parameter $t$, then the image of this parametrization over the rationals miss only finitely many rational points.
It is not clear to me why only finitely many points were missed. My first guess is some what related to rational points are dense. Can someone elaborate a bit geometric intuition on 2 and how to prove it(I think hint will suffice)?

List all the harmonic conjugates of the following rational function. The integration is almost impossible, Symbolab and Microsoft Math fails to arrive at the answer:
$\mathrm{\mu <!--\; \mu \; -->}(x,y)=\frac{x^2+x+y^2}{x^2+y^2}$

Find the domain and graph:
$f(t)=\frac{-<!--\; -\; -->t}{|t|}$
My book says to define it piecewise.
My questions:
1) Do all rational functions have to be defined piecewise, or just this one because there is an absolute value in the denominator, and the absolute value is always defined piecewise?
2) Are all rational functions defined piecewise in order to avoid having a denominator be equal to zero, is that the general reason for defining anything piecewise, to avoid having division by zero? (so then this would mean that the domain of a rational function is always defined piecewise, or only when we need to avoid having denominator be =0?)
This is how my book defines f(t) piecewise:
If $t>0$, then $|t|$ is $t$ since $t$ is already positive. For $t>0$, simplify
$f(t)=\frac{-<!--\; -\; -->t}{|t|}=\frac{-<!--\; -\; -->t}{t}=-<!--\; -\; -->1$
If $t<0$, then $|t|$ is $-<!--\; -\; -->t$ since $t$ is negative. For $t<0$, simplify
$f(t)=\frac{-<!--\; -\; -->t}{-<!--\; -\; -->|t|}=\frac{-<!--\; -\; -->t}{-<!--\; -\; -->t}=1$
(or for the last part, should the negative be inside the absolute value sign for $t<0$, as in $|-<!--\; -\; -->t|$ instead of $-<!--\; -\; -->|t|$?)
So we define it piecewise to avoid having 0 in the denominator? Because isn't absolute value defined at 0, the absolute value is continuous everywhere, and thus defined at 0?
I'm confused about defining things piecewise, and how to know when to apply a piecewise attempt in order to define a function's domain.
Also, I'm confused about the second part of this, where $t<0$ for $|t|$. How is it that here, $|t|$ is $-<!--\; -\; -->t$ if absolute value is always positive?
Maybe I'm not understanding the absolute value concept correctly, because in grade school it has always been drilled into my head that |absolute value| just "turns things positive", so here I don't really understand how it can be negative.
I understand how the domain is $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)\cup <!--\; \cup \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$, because by using open intervals we're not letting it be exactly =0, but I'm confused about the whole piecewise thing.

I noticed that for the (real) function $f(x)=\frac{1}{1-<!--\; -\; -->x}$, $f(f(f(x)=f^3(x)=x$ for all real $x$.
This surprised me, and I was naturally curious about rational functions that elicit the identity after 4 or 5 or, in general, $n$ iterations.
I looked at rational functions of the form $f(x)=\frac{1}{z-<!--\; -\; -->x}$ for $z\in <!--\; \in \; -->\mathbb{R}$ and wondered for which $z$ did $f^n(x):=f(f(...f(f(x))...))=x$.
After playing around for a bit on Wolfram and Desmos, I came up with the following conjecture: If $n\in <!--\; \in \; -->\mathbb{N}$ and $n\ge <!--\; \ge \; -->2$ and $f(x)=\frac{1}{2\cos <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{n})-<!--\; -\; -->x}$, then $f^n(x)=x$.
Note that I do not claim that $f$ is the only rational function of the form $f(x)=\frac{1}{z-<!--\; -\; -->x}$ where $z\in <!--\; \in \; -->\mathbb{R}$.
So I have a few questions:
1) Is this conjecture true?
2) How would someone go about proving it, if it is true?
3) Is this a well-known theorem or a somewhat-immediate corollary or special case of a well-known theorem? If so, what is that theorem?
4) How would I find all $z\in <!--\; \in \; -->\mathbb{R}$ such that if $f(x)=\frac{1}{z-<!--\; -\; -->x}$,then $f^n(x)=x$? Is such a thing easy to do?
5) What field of math would ask questions like this?

I was asked to dismantle this rational function by parts and wanted to know if I did it right.The function is:
$\frac{x^5-<!--\; -\; -->x+3}{x(x-<!--\; -\; -->2{)}^3(x^2+2x+2)}$
What I did is:
$\frac{A}{X}+\frac{B}{(x-<!--\; -\; -->2)}+\frac{C}{(x-<!--\; -\; -->2{)}^2}+\frac{D}{(x-<!--\; -\; -->2{)}^3}+\frac{GX+C}{x^2+2x+2}$

Under what conditions a rational function has bounded derivative?
This question arise to me when considering the following theorem:
If $f\in <!--\; \in \; -->C^1(I,\mathbb{R})$ where I is an interval then:
f is globally lipschitz
$⟺<!--\; ⟺\; -->\mathrm{\exists <!--\; \exists \; -->}L\ge <!--\; \ge \; -->0.\mathrm{\forall <!--\; \forall \; -->}t\in <!--\; \in \; -->I.|f^{\prime }(t)|\le <!--\; \le \; -->L$
So taking rational function $f(x)=\frac{p(x)}{q(x)}$ we have $f^{\prime }(x)=\frac{p^{\prime }(x)q(x)-<!--\; -\; -->p(x)q^{\prime }(x)}{q(x{)}^2}$.
My view
I think I should assume that $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ so that $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}.q^{\prime }(x)\ne <!--\; \ne \; -->0$ (however this doesn't seem to be necesary). And then perhaps a condition on the degree guarantees boundedness...

The matter of explicitly finding the order of a rational function on an elliptic curve in the projective plane at infinity (i.e. at the point (0,1,0)) still seems unclear.
For example, Silverman (in The Arithmetic of Elliptic Curves) states that the order of the rational function $y$ on the elliptic curve
$y^2=(x-<!--\; -\; -->e_1)(x-<!--\; -\; -->e_2)(x-<!--\; -\; -->e_3),$
where $e_1$, $e_2$, and $e_3$ are distinct, is −3. That is, the function $y$ has a pole of order 3 at (0,1,0). I have no doubt that this is true; I'd like to know a simple way to see it, based on projective coordinates and independent of the fact that the sum of the orders of the zeros of $y$ is 3 (which I understand).

This is from the page 19 of the book called An Introduction to Complex Function Theory by Bruce P. Palka.
Every rational function of z=x+iy is clearly a rational function of x and y, but not the other way around.
As a reminder, a rational function of z=x+iy is defined as $f(z)=\frac{a_0+a_{1}z+\cdots <!--\; \cdots \; -->+a_n{z}^n}{b_0+b_{1}z+\dots <!--\; \dots \; -->b_m{z}^m}$. I understand the first part of the claim: if z is fixed, i.e. both x and y are fixed, then of course all combinations of as and bs are part of the set where as, bs, xs and ys can vary. But I don't see why this wouldn't hold the other way around. Specifically, why cannot we choose the coefficients $a_0,\dots <!--\; \dots \; -->,a_n$ and $b_0,\dots <!--\; \dots \; -->,b_m$ in such a way that any change in x or y is mitigated?