Get Expert Help for Your Rational Functions Questions

I am currently studying Hardy's Pure Course of Mathematics and am on chapter 2, section 24: Rational Functions. In this chapter, Hardy defines a rational function as the quotient of two polynomials such that: $R(x)={\displaystyle \frac{P(x)}{Q(x)}}$
Hardy mentions that P(x) and Q(x) must have no common factor, though the removal of common factors changes the function. He provides the following example:
${\displaystyle \frac{({\displaystyle \frac{1}{x+1}}+{\displaystyle \frac{1}{x-<!--\; -\; -->1}})}{({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x-<!--\; -\; -->2}})}}$
which may be reduced to:
${\displaystyle \frac{x^2(x-<!--\; -\; -->2)}{(x-<!--\; -\; -->1{)}^2(x+1)}}$
The first equation is not defined for values x=-1,1,0,2 and the second is not defined for values x=-1,1
So my questions are:
1) Do all simplifications of rational functions lead to different functions?
2) When manipulating equations, are reductions by common factors actually "mathematical errors?"
3) And what are the implications of what Hardy had described?

I have developed the formula to determine the radius of a cylinder with a fixed volume:
$f(x)=\sqrt[3]{{\displaystyle \frac{V}{\mathrm{\pi <!--\; \pi \; -->}}}}$
Substituted into the formula for the surface area of a cylinder, I get the following function. This would give me the minimum surface area of a cylinder for a given volume.
$S(V)=2\mathrm{\pi <!--\; \pi \; -->}(\sqrt[3]{{\displaystyle \frac{V}{\mathrm{\pi <!--\; \pi \; -->}}}}{)}^2+2\mathrm{\pi <!--\; \pi \; -->}(2\ast <!--\; \ast \; -->\sqrt[3]{{\displaystyle \frac{V}{\mathrm{\pi <!--\; \pi \; -->}}}})$
However, my assignment for class asks for a rational function for this problem. How could I take my existing function and make it rational?

Let $K$ be a field and let $F=\frac{P}{Q}\in <!--\; \in \; -->K(X)$ be a rational fraction, for simplicity we denote also by $F$ the rational function associated to the rational fraction $F$. It is clear that if $P$ and $Q$ are both even or both odd polynomial functions, then $F$ is an even rational function and we have also that if one of the two polynomial functions is even and the other is odd, then $F$ must be odd. Now is the converse true, I mean do we have that if $F$ is an even rational function then necessarely both $P$ and $Q$ are both odd or both even and that if $F$ is odd then necesserely one is odd and the other is even ?
I think it is true, but i don't know how to prove it, suppose that $F$ is even then
$\frac{P(x)}{Q(x)}=\frac{P(-<!--\; -\; -->x)}{Q(-<!--\; -\; -->x)}$
Hence
$P(x)Q(-<!--\; -\; -->x)=P(-<!--\; -\; -->x)Q(x)$
But how to go from here?

Could you calculate a Rodrigues formula for the Chebyshev rational functions defined by
$T_n\left(\frac{x-<!--\; -\; -->1}{x+1}\right)$
where $T_n$ is a Chebyshev polynomial.

Prove that for any two distinct points of an irreducible curve there exists a rational function that is regular at both, and takes the value 0 at one and 1 at the other.
I think I can construct such a function, for example, $u(x,y)=(x-<!--\; -\; -->a{)}^2+(y-<!--\; -\; -->b{)}^2$ for given two points $(a,b)$ and $(c,d)$. However, this doesn't work for general algebraically closed field, for example, the case of $(c,d)=(a+i,b+1)$. Hence now I have no clue. Could you give me a hint for this problem?

What is the general form of a rational function which is bijective on the unit disk?
I'm stuck on this problem. If I let $R(z)=\frac{a_0(z-<!--\; -\; -->a_n)\cdots <!--\; \cdots \; -->(z-<!--\; -\; -->a_1)}{(z-<!--\; -\; -->b_m)\cdots <!--\; \cdots \; -->(z-<!--\; -\; -->b_1)}$, then exactly one of the $a_i$ must fall in the unit disk and none of the $b_i$ can be in the unit disk.
Any hints on how to proceed? I see simple functions like $z$ work, but I don't know how to show that higher combinations of them are bijective.

Let $f$ be the rational function on ${\mathbb{P}}^2$ given by $f=x_1/x_0$. Find the set of points where $f$ is defiend and describe the corresponding regular function.
My question: I know that the set of points where $f$ is defined is $\{[1:x_1:x_2]\in <!--\; \in \; -->{\mathbb{P}}^2\}$. Isn't the corresponding regular function also just $f$ itself, or the regular function $g=x_2/x_0$ (both works and both maps ${\mathbb{P}}^2$ to ${\mathbb{A}}^1$)? So what is this problem trying to get at?
Now think of this function as a rational map from ${\mathbb{P}}^2$ to ${\mathbb{A}}^1$. Embed ${\mathbb{A}}^1$ in ${\mathbb{P}}^1$, and let $\mathrm{\varphi <!--\; \varphi \; -->}:{\mathbb{P}}^2\to <!--\; \to \; -->{\mathbb{P}}^1$ be the resultiong rational map. Find the set of points where $\mathrm{\varphi <!--\; \varphi \; -->}$ is defined, and describe the corresponding morphism.
My question: Since $\mathrm{\varphi <!--\; \varphi \; -->}$ is $f\circ <!--\; \circ \; -->\mathrm{ϵ<!--\; ϵ\; -->}$ where $\mathrm{ϵ<!--\; ϵ\; -->}$ is the embedding, why isn't $\mathrm{\varphi <!--\; \varphi \; -->}$ defined on the same set as $f$ is defined? I'm also not sure what I'm trying to solve for here.

I am trying to find a conjecture apparently made by Erdős and Straus. I say apparently because I have had so much trouble finding anything information about it that I'm beginning to doubt its existence. Here it is:
Let $\mathrm{\varphi <!--\; \varphi \; -->}(X)$ be a rational function over $\mathbb{Q}$ that is defined at every positive integer. If the sum $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\varphi <!--\; \varphi \; -->}(n)$ converges, it is either rational or transcendental, i.e., it is never an irrational algebraic number.
Has anyone heard of this conjecture? I was told about it by my supervisor, but he doesn't remember where he heard about it.

I have the function: $f(x)=2x-<!--\; -\; -->2^x+2$
I know that this function has an oblique asymptote, but all the tutorials I can find on google, are with rational functions with the form:
$f(x)=\frac{P(x)}{Q(x)}$
Where they simply just divide the denominator with the numerator.
But I can't do that, because my equation doesn't contain any fractions. So my question is: how do I find the function to the oblique asymptote for my $f(x)$?

I know that the only ways to find the range of a rational function is using an inverse, graphing, or using calculus. How would you use calculus to find the range of a rational function (or any other type of function) using Calculus?

Let
$f(x)=\underset{n\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}{a}_n{x}^n=\frac{P(x)}{(1-<!--\; -\; -->x{)}^d}$
be a rational function.
(a) Prove: There is a polynomial $P_2(x)$ so
$\underset{n\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}{a}_{2_n}{x}^n=\frac{P_2(x)}{(1-<!--\; -\; -->x{)}^d}$
(b) Let $r\ge <!--\; \ge \; -->1\in <!--\; \in \; -->\mathbb{N}$. Show that an polynomial $P_r$ exists so that
$\underset{n\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}{a}_{rn}{x}^n=\frac{P_r(x)}{(1-<!--\; -\; -->x{)}^d}$
Hint: Use the rth roots of unity which are defined by $\exp <!--\; \; -->\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}ik}{r}\right),0\le <!--\; \le \; -->k\le <!--\; \le \; -->r-<!--\; -\; -->1$
(a) I don't know what this d is about (and no one else did). Might be an absolute term.
As $f(x)$ is a rational function, it can be defined as a fraction of two polynomials $\frac{P(x)}{Q(x)}$. But that is unfortunately all I know about this.
Could you please help me going on?
(b) I don't know how the $r$th roots of unity (and therefore numbers $x$ for which applies: $x^r=1$) can help me solving this? I don't find any approach.
Could you please help me a bit?
Thanks in advance!

Is there a rational function $f(x)\in <!--\; \in \; -->\mathbb{Q}(x)$ such that $\sqrt{2}\le <!--\; \le \; -->f(x)\le <!--\; \le \; -->\sqrt[3]{2x}$ for all $x\ge <!--\; \ge \; -->\sqrt{2}$?
My thoughts : it is easy to find such an f if we relax the conditions to $f(x)\in <!--\; \in \; -->\mathbb{R}(x)$ (take f constant equal to $\sqrt{2}$), however no easy perturbation of this solution seems to solve the original problem. Clearly f must have zero degree in x and can be written in the form $x+(x^2-<!--\; -\; -->2)g(x)$ where g is another rational function. Then I am stuck.

Suppose $f:{\mathbb{P}}^2\to <!--\; \to \; -->{\mathbb{P}}^2$ is rational such that $f\circ <!--\; \circ \; -->f=\mathbb{I}\mathbb{d}$. Then is it true that $f$ must be linear?
It feels true due to the degree which increases, but some things might cancel out.
Suppose we have a smooth curve $C$ of genus $g\ge <!--\; \ge \; -->1$ with a rational function $g:C\to <!--\; \to \; -->C$ with the same property. Does it always rise to an $f$ on ${\mathbb{P}}^2$ whose restriction is $g$? Does it imply that $g$ has to be linear too? This on the other side seems wrong to me.

Let $X$ be some affine algebraic variety over $\mathbb{k}$ (i.e. some closed subset in ${\mathbb{A}}_{\mathbb{k}}^n$). First suppose $X$ to be irreducible. Then the algebra $\mathbb{k}[X]$ is a domain and we can consider the field of rational functions ${\mathrm{Q}\mathrm{u}\mathrm{o}\mathrm{t}}_{\mathbb{k}[X]}=\mathbb{k}(X)$. Could you explain me how to build an analogue of this field in the case when $X$ is not necessarily irreducible? Then $\mathbb{k}[X]$ must not be a domain and we are to build some kind of localization?
Also, what is the destination of rational functions? Why we cannot be satisfied with only regular maps and regular functions?

The form of the partial fraction decomposition of a rational function is given below.
$\frac{x-3x^2-26}{(x+1)(x^2+9)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+9}$
What are the values of $A,B$ and $C$?
So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$
I also found the indefinite integral, which equals
$\frac{1}{3}{\tan }^{-<!--\; -\; -->1}<!--\; \; -->\left(\frac{x}{3}\right)-<!--\; -\; -->3\log <!--\; \; -->(1+x)+C$

I was just thinking that as rational functions form an ordered field you could describe analogous version of the absolute value function, but we don't quite have a 'metric' - for example |1/x| < e for all e > 0 in R but 1/x =/= 0.
I was wondering if anybody got anywhere with this 'metric' and if there are any links to papers exploring actual metrics on the rational functions?

Suppose you have two rational functions of x having the same (finite) limit when $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, having the same simple poles and the same residues in these poles, does it imply that they're equal? How to show that ?

Let $k$ be an algebraic closed field, ${\mathbb{A}}^n$ be an affine variety over $k$, $U$ be open set in ${\mathbb{A}}^n$ and $f,g\in <!--\; \in \; -->k(x_1,...,x_n)$ and their denominators is not zero over U.If f(x)=g(x) for all $x\in <!--\; \in \; -->U$, is it true that f=g as rational functions?

Let $\mathrm{\Omega <!--\; \Omega \; -->}\subset <!--\; \subset \; -->\mathbb{C}$ be a bounded domain which does not contain any pole of a rational function $R$ and $R^{\prime }(z)\ne <!--\; \ne \; -->0$ for all $z\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$. Is it true that $R(z)$ is injective on $\mathrm{\Omega <!--\; \Omega \; -->}$?

We know if $R(z)$ is a rational function, then it is in the form $R(z)=f(z)/g(z)$, where $f$ and $g$ are complex polynomials. If we want to find its fixed points, we can take $R(z)=z$, which gives the equation $f(z)-<!--\; -\; -->z\cdot <!--\; \cdot \; -->g(z)=0$. By the Fundamental Theorem of Algebra, this equation has finitely many roots; does that mean that those roots are fixed points of $R$?