Get Expert Help for Your Rational Functions Questions

I want to decompose the rational function
$\frac{P(s)}{Q(s)}=\frac{\underset{i=1}{\overset{m}{\prod <!--\; \prod \; -->}}(s+a_i)}{\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}(s+b_i)}$
where $a_i>0$ for every $i=1,\dots <!--\; \dots \; -->,m$ $b_i>0$ for every $i=1,\dots <!--\; \dots \; -->,n$ and $n>m$.
In other words, I'm looking for coefficients $x_j$, $j=1\dots <!--\; \dots \; -->,n$ such that
$\frac{P(s)}{Q(s)}=\frac{x_1}{s+b_1}+\cdots <!--\; \cdots \; -->+\frac{x_n}{s+b_n}⟺<!--\; ⟺\; -->\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_j\underset{\begin{array}{c}i=1\\ i\ne <!--\; \ne \; -->j\end{array}}{\overset{n}{\prod <!--\; \prod \; -->}}(s+b_i)=\underset{i=1}{\overset{m}{\prod <!--\; \prod \; -->}}(s+a_i)$
Making some attempts with Mathematica for low values of $m$ and $n$ it seems that the $x_j$ are given by
$x_j=\frac{\underset{i=1}{\overset{m}{\prod <!--\; \prod \; -->}}(a_i-<!--\; -\; -->b_j)}{\underset{\begin{array}{c}i=1\\ i\ne <!--\; \ne \; -->j\end{array}}{\overset{n}{\prod <!--\; \prod \; -->}}(b_i-<!--\; -\; -->b_j)}$
So my questions are:
1) Can I say beforehand that there exist unique such $x_j$s?
2) How can I prove that $x_j$ has in general (as it seems) the form above?

I have tried to solve a problem here and I've got an answer but I just don't know if it is the right one.
The problem is: For what conditions
$\int <!--\; \int \; -->\frac{a{x}^2+bx+c}{x^3{(x-<!--\; -\; -->1)}^2}dx$
represents a rational function.
First I have calculated the integral and the result is:
$(a+2b+3c)\ln <!--\; \; -->\left|x\right|-<!--\; -\; -->\frac{2c+b}{x}-<!--\; -\; -->(a+2b+3c)\ln <!--\; \; -->|x-<!--\; -\; -->1|-<!--\; -\; -->\frac{a+b+c}{x+1}-<!--\; -\; -->\frac{c}{2x^3}+K$
where $K$ is a constant.
From here I have taken the conditions
$(a+2b+3c=0)\text{ and }((b+2c!=o)\text{ or }(a+b+c!=0))$
which is the same as
$(a=-<!--\; -\; -->2b-<!--\; -\; -->3c)\text{ and }(b!=-<!--\; -\; -->2c)$
So, from my result, the condition above must be satisfied in order for that function to be a rational function.
P.S != means not equal to
Can anyone tell whether the solution is correct or not and if it is not then help me with it ?

How to find the domain of a complex rational function?
E.g.
$f(z)=\frac{3z-<!--\; -\; -->i}{z}$
I understand that domain means $z\in <!--\; \in \; -->\mathbb{C}$ for which this is defined, but I don't know what to look for.
Perhaps:
Cannot be z=0 because of divizion of zero. But is that all?
Are the criteria for "defined" the same as for real rational functions?

I was going over some practice problems and got stuck with this one: I am supposed to find the maximum of the function:
${\displaystyle \frac{x}{x^2+1}}$
on the interval (0,4).
All I can think of is to try and plug in values from the given interval . Is there a more systematic approach to finding maxima of rational functions?

Why if a rational function has a horizontal asymptote, it excludes the possibility of this function to have an oblique azymptote, or viceversa?

I cannot find a good explanation on how to represent a power series as a rational function. Thanks!
$2-<!--\; -\; -->\frac{2}{3}x+\frac{2}{9}{x}^2-<!--\; -\; -->\frac{2}{27}{x}^3+...$

I am thinking if I could help with my current problem. Now I have a parameterized rational function $G(p,z)$, where $p\in <!--\; \in \; -->{\mathbb{R}}^n$ denotes the coefficients (parameters) of the rational function, and $z$ denotes the indeterminate of the rational function which lies in complex domain.
I regard $G(p,z)$ as a mapping from ${\mathbb{R}}^n$ to $\mathbb{G}$, where $\mathbb{G}$ is a set of rational functions with indeterminate $z$. Then I define that a property holds on a metric space $(\mathbb{G},d)$ if it holds on an open dense subset of $\mathbb{G}$.
However, I am wondering what conditions I should put on a parameter set $\mathrm{\Theta <!--\; \Theta \; -->}\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^n$, such that $\{G(p,z)|p\in <!--\; \in \; -->\mathrm{\Theta <!--\; \Theta \; -->}\}$ becomes an open subset of $\mathbb{G}$. Is making $\mathrm{\Theta <!--\; \Theta \; -->}$ an open dense subset of ${\mathbb{R}}^n$ sufficient?

I am trying to decompose the following rational function: ${\displaystyle \frac{1}{(x^2-<!--\; -\; -->1{)}^2}}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-<!--\; -\; -->1{)}^2=(x+1{)}^2(x-<!--\; -\; -->1{)}^2$
Therefore $\mathrm{\exists <!--\; \exists \; -->}A,B,C,D$ s.t:
${\displaystyle \frac{1}{(x^2-<!--\; -\; -->1{)}^2}}={\displaystyle \frac{1}{(x-<!--\; -\; -->1{)}^2(x+1{)}^2}}={\displaystyle \frac{Ax+B}{(x+1{)}^2}}+{\displaystyle \frac{Cx+D}{(x-<!--\; -\; -->1{)}^2}}$
We then have
$Ax+B={{\displaystyle \frac{1}{(x-<!--\; -\; -->1{)}^2}}|}_{x=-<!--\; -\; -->1}⟹<!--\; ⟹\; -->B-<!--\; -\; -->A=1/4$
Same thing for $Cx+D$:
$Cx+D={{\displaystyle \frac{1}{(x+1{)}^2}}|}_{x=1}⟹<!--\; ⟹\; -->C+D=1/4$
How do I find A, B, C, D from here?

When there are some conditions on the power of a rational function, how can we evaluate/prove the non-existence of the limit of the function? Here is a specific example:
Prove
$\underset{(x,y)\to <!--\; \to \; -->(0,0)}{lim}\frac{(x-<!--\; -\; -->y{)}^{p-<!--\; -\; -->1}((p-<!--\; -\; -->2)x^2+2xy+p{y}^2)}{(x^2+y^2{)}^2}=0$
if $p>3$
Is this even solvable?

Suppose $C$ is an algebraic curve (which has singular points) over an algebraically closed field $k$, and that $f$ is a rational function on $C$. How does one defines the Weil divisor of $f$?
The problem is that the local rings of $C$ at singular points are not DVR's, so I do not have an obvious candidate for an order at a point.
Edit: Let me give an example, inspired by an answer from below. Suppose $C$ is curve $y^2=x^3$. What would be the order of the rational function $x/y$ at the origin?

I know how to use this algorithm when I am integrating rational functions, but my textbook has omitted the actual proof for why it works. If someone could please help me with this question:
Prove that given a and b, there are constants $c_1$ and $c_2$ so that
$\frac{ax+b}{(x-<!--\; -\; -->1{)}^2}=\frac{c_1}{x-<!--\; -\; -->1}+\frac{c_2}{(x-<!--\; -\; -->1{)}^2}$

given the rational function $\frac{1}{x^2-<!--\; -\; -->\frac{x}{2}-<!--\; -\; -->3}$ and asked for the domain and range, I multiplied thru by 2 and got $\frac{2}{2x^2-<!--\; -\; -->x-<!--\; -\; -->6}$ I understand the domain includes all real numbers except for (vertical asymptotes) $\frac{-<!--\; -\; -->3}{2}$ and 2. i thought there were no horizontal asymptotes until I graphed it on a calculator. according to the calculator the range is $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{-<!--\; -\; -->16}{49}]\cup <!--\; \cup \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$. this makes sense to me looking at the graph but how can I obtain these results for the range using algebra? thank you

Consider the following function
$f(x)=\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_k{x}^k.$
Let us say I know $a_k$ belongs to a strict subset $F\subset <!--\; \subset \; -->\mathbb{R}$.
I want to characterize the set of sequences $\{a_k{\}}_{k=0}^{\mathrm{\infty <!--\; \infty \; -->}}$ for which $f(x)$ is a rational function in $x$.
Can someone point me to relevant results in literature?

For a field $k$, let $V$ be an affine variety over $k$. Denote by $k(V)$ the function field of $V$, containing all rational functions $r:V⇢<!--\; ⇢\; -->{\mathbb{A}}_k^1$. My question is, if a rational function $f\in <!--\; \in \; -->k(V)$ has a pole at $p\in <!--\; \in \; -->V$, is there an expression $f=\frac{g}{h}$ where $g,h\in <!--\; \in \; -->k[V]$ are regular functions, and $g(p)\ne <!--\; \ne \; -->0$, $h(p)=0$?
When $V\subseteq <!--\; \subseteq \; -->{\mathbb{A}}_k^1$, this is clear, since if we have $f=\frac{g}{h}$ where $g(p)=h(p)=0$, we can simply reduce the expression of $g$ and $h$ and eliminate the factor $(x-<!--\; -\; -->p)$ until we get $f=\frac{g^{\prime }}{h^{\prime }}$ such that $g^{\prime }(p)\ne <!--\; \ne \; -->0$, $h^{\prime }(p)=0$. But when $g,h$ are multivariate functions, I wonder how to get such a reduced expression?

We have recently started studying rational functions at school. I have learnt that a rational function is the ratio of two polynomials, i.e
$\frac{p(x)}{q(x)}$
My question is, what if p(x) and q(x) have a common factor (linear, quadratic, etc) ??
For example, $\frac{(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)}{(x-<!--\; -\; -->2)(x-<!--\; -\; -->3)}$
Is it still a rational function? We haven’t studied calculus in maths yet, but we have studied a little bit of calculus in our physics classes, and from what I know, I can cancel out the common factor but the function is discontinuous at x = 2, i.e it has a hole.
So is it still a rational function?

Some rational function is giving me some trouble...
$\begin{array}{r}\int <!--\; \int \; -->\frac{x^2-<!--\; -\; -->9x+16}{(x-<!--\; -\; -->1)(x^2+6x-<!--\; -\; -->7)}dx\end{array}$
I simplified it like so:
$\begin{array}{r}\int <!--\; \int \; -->\frac{x^2-<!--\; -\; -->9x+16}{(x-<!--\; -\; -->1{)}^2(x+7)}dx\end{array}$
If I get it right I can transform it into this:
$\begin{array}{r}\int <!--\; \int \; -->\frac{A}{(x-<!--\; -\; -->1)}+\frac{B}{(x-<!--\; -\; -->1{)}^2}+\frac{C}{(x+7)}dx\end{array}$
And then I get this:
$\begin{array}{r}A{x}^2+6Ax-<!--\; -\; -->7A+Bx+C{x}^2-<!--\; -\; -->2Cx+C=x^2-<!--\; -\; -->9x+16\end{array}$
So I forgot how to get ABC values. Could somebody remind me this?
I am showing all the steps because I am not 100% confident that with have not made any mistakes.

Let $F(z)$ be a rational function $\frac{P(z)}{Q(z)}$ such that the degree of $P(z)$ is less than the degree of $Q(z)$ and suppose that all the zeros of $Q(z)$ are contained in the open disk $|z|<r$.
I know that if $f(z)$ is analytic for $|z|>r$ and bounded by $M>0$ there, that is, $|f(z)|\le <!--\; \le \; -->M$ for all $z$ with $|z|>r$, then the coefficients of the Laurent series of $f(z)$ for $|z|>r$ satisfy $aj=0$ for $j=1,2,3,$....
I'm supposed to show that the coefficients of the Laurent series of $f(z)$ for $|z|>r$ satisfy $aj=0$ for $j=1,2,3,$.... by using the corollary above. I know that $F(z)$ is analytic for $|z|>r$ but I'm missing the boundless condition to finish my proof. Any suggestions?

How to prove that every bijective rational function (over the field of complex numbers) is a linear fractional transformation?

Let $f:<!--\; :\; -->\mathbb{Q}⟶<!--\; ⟶\; -->\mathbb{Q}$ be a continuous function. Is there necessarily an interval $(a,b)$, with $a,b\in <!--\; \in \; -->\mathbb{Q}$ such that $a<b$, such that the restriction of $f$ to $(a,b)\cap <!--\; \cap \; -->\mathbb{Q}$ is a rational function?
My guess is that the answer is negative. I tried to prove it as follows: I took a countable and dense set $\{r_n\mid <!--\; \mid \; -->n\in <!--\; \in \; -->\mathbb{N}\}$ of irrational numbers and defined, for each $x\in <!--\; \in \; -->\mathbb{Q}$, $f(x)=\underset{r_n<x}{\sum <!--\; \sum \; -->}{2}^{-<!--\; -\; -->n}$. This will work, in the sense that $f$ is continuous and that the restriction of $f$ to any interval $(a,b)$ is never a rational function. The problem is that, of course, in general, $f(x)\notin <!--\; \notin \; -->\mathbb{Q}$.

Are there any results to determine whether the given power series of real variable converges to a rational function? I mean just analyzing the coefficients of the series. One way is to find the sum function which is not always easy to find.