Get Expert Help for Your Rational Functions Questions

So I know that a rational function is a function that can be defined by a rational fraction and I managed to expand
$\tan <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}=\frac{3\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->{\tan }^3<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{1-<!--\; -\; -->3{\tan }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}$
Is this considered as a rational function? Thanks.

Are rational functions between affine algebraic sets continuous with respect to the Zariski topology on their domain of definition?

Let $P$ and $Q$ be two rational functions of $z$ (coefficients over $\mathbb{C}$). How can one decide whether $z=R(P(z),Q(z))$ for some rational function $R$ of two variables, and if it is the case how can one find $R$ explicitely?
I'm interested in the particular case $P(z)=(z-<!--\; -\; -->1{)}^3/z^2$ for $c\in <!--\; \in \; -->\mathbb{C}$. I have been able to express $z$ in terms of $P$ and $Q$ using cubic and square roots, but I don't know how to proceed to find the rational expression.

This integral poses a challenge.
Ordinarily integrating rational functions can be solved using the Hermite-Ostrogradski method. However, in the following integral, the coefficients ${\mathrm{\beta <!--\; \beta \; -->}}_0,...,{\mathrm{\beta <!--\; \beta \; -->}}_4$ are not integers. (Hence, the Hermite-Ostrogradski method would not be appropriate).
Note: Expanding the integrand (trying to solve the integral using a partial fraction decomposition) is (because of the nature of the physical problem described by this integral) an inappropriate solution to this case.
$\int <!--\; \int \; -->{\displaystyle \frac{1}{{\mathrm{\beta <!--\; \beta \; -->}}_0+{\mathrm{\beta <!--\; \beta \; -->}}_{1}x+{\mathrm{\beta <!--\; \beta \; -->}}_2{x}^2+{\mathrm{\beta <!--\; \beta \; -->}}_3{x}^3+{\mathrm{\beta <!--\; \beta \; -->}}_4{x}^4}}dx$
How can this rational function be evaluated?

Suppose $f\in <!--\; \in \; -->O_p(V)$ is a rational function on V that has a value at $p$. Then write $f=a/b=a^{\prime }/b^{\prime }$ where $a,b,a^{\prime },b^{\prime }\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(V)$, the coordinate ring of V. Want to show the value of $f$ at $p$ is well-defined, i.e. $a(p)/b(p)=a^{\prime }(p)/b^{\prime }(p)$.
So since $a/b=a^{\prime }/b^{\prime }$ are the same equivalence class, there is some non-zero poly $x\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(V)$ such that $x(a{b}^{\prime }-<!--\; -\; -->a^{\prime }b)=0$. Then $x(p)(a(p)b^{\prime }(p)-<!--\; -\; -->a^{\prime }(p)b(p))=0$. Then what? How do we know $x(p)\ne <!--\; \ne \; -->0$?

I know that an analytic function on $\mathbb{C}$ with a nonessential singularity at $\mathrm{\infty <!--\; \infty \; -->}$ is necessarily a polynomial.
Now consider a meromorphic function $f$ on the extended complex plane $\stackrel{^<!--\; ^\; -->}{\mathbb{C}}$. I know that $f$ has only finitely many poles, say $z_1,\dots <!--\; \dots \; -->,z_n$ in $\mathbb{C}$. Suppose also that $f$ has a nonessential singularity at $\mathrm{\infty <!--\; \infty \; -->}$.
Then if $z_i$ have orders $n_i$, it follows that $\prod <!--\; \prod \; -->(z-<!--\; -\; -->z_i{)}^{n_i}f(z)$ is analytic on $\mathbb{C}$, and has a nonessential singularity at $\mathrm{\infty <!--\; \infty \; -->}$, and is thus a polynomial, so $f$ is a rational function.
But I'm curious, what if f doesn't have a singularity at $\mathrm{\infty <!--\; \infty \; -->}$, or in fact has an essential singularity at $\mathrm{\infty <!--\; \infty \; -->}$ instead? Is $f$ still a rational function?

I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they ask
Which of the following series converge to a rational function in some domain?
$$\begin{gathered} \underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{k!+k^2+k}{z}^{k^2+2k} \\ \underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{2^k}{(1+z^2{)}^k} \\ \underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{k!(z-<!--\; -\; -->k{)}^k} \end{gathered}$$
I had no problem with the second and third ones. The second one converges to $\frac{1}{1-<!--\; -\; -->\frac{2}{1+z^2}}$ given that $\left|\frac{2}{1+z^2}\right|<1$, and the third one has an infinite number of singularities so it can't be a rational function.
I don't know how to verify that in the first series. Is there any general method to see if some power series converge to a rational function, or in this particular case, a way to see if this series does?
Edit: Now I am not that convinced about my argument for the third series (Since the series might only converge in a bounded domain from which the number of singularities would be finite). Is there anything wrong with it or any way to formalize it further?

I have a small question:
Can one state that
$f(x)=\frac{1}{x}$
is a rational function because it is the quotient between a polynomial with degree 0 and a polynomial with degree 1?
Thanks!

Let $f(z)$ be a rational function in the complex plane such that $f$ does not have any poles in $\{z:\mathrm{\Im <!--\; \Im \; -->}z\ge <!--\; \ge \; -->0\}$.
Prove that $sup\{|f(z)|:\mathrm{\Im <!--\; \Im \; -->}z\ge <!--\; \ge \; -->0\}=sup\{|f(z)|:\mathrm{\Im <!--\; \Im \; -->}z=0\}$.
Let ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r$ be a half circle counter such that ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r={\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_1}+{\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}$ when ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_1}=\{z:\mathrm{\Im <!--\; \Im \; -->}z=0,|z|=r\}$, ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}=\{z:\mathrm{\Im <!--\; \Im \; -->}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}$ than it's still smaller the the value on ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r+1}$ which does not contain ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}$. I would like a hint on how to proceed.

Let $X$ be an irreducible, normal variety over an algebraically closed field of characteristic zero. Let $x,y\in <!--\; \in \; -->X$ be two points such that $f(x)=f(y)$ for every $f\in <!--\; \in \; -->K(X)$ which is defined at $x$ and at $y$. Can I conclude that $x=y$?
I feel the answer should be affirmative. In fact, the statement can be reduced to the following: Given two effective prime divisors $D_x$ and $D_y$ on $X$, there exists a rational function $f\in <!--\; \in \; -->K(X)$ with $v_{D_x}(f)\ne <!--\; \ne \; -->0$ and $v_{D_y}(f)=0$.
If this is true, then assuming $x\ne <!--\; \ne \; -->y$ we could find a divisor $D_x$ containing $x$ but not $y$ and a divisor $D_y$ containing $y$ but not $x$, so a function $f$ as above would yield a contradiction. However, I just can't prove the statement, even though I also think it should be true.

I have to construct a rational function with the range being [-1,0) , which is pretty much just -1. I came up with the solution $\sqrt{-<!--\; -\; -->x^2-<!--\; -\; -->\frac{1}{x}}$. It works for the range, but I'm not sure if it is a rational function.

How to find the power series of rational functions of type, e.g.
$\frac{1}{1-<!--\; -\; -->6x+12x^2},\frac{6x}{1-<!--\; -\; -->6x+12x^2}$
where denominator can't be factorized over the real domain.
Is there a way to use the method of undetermined coefficients (by completing the square), or is it necessary to use complex domain and trigonometry?

I'm learning about holes in rational functions in precalculus, and I'm confused as to why they exist, having some knowledge of limits. Say I have a function $R(x)=\frac{(x-<!--\; -\; -->5)(x+2)}{(x+4)(x+2)}$. If I were to find where the hole lies for this function, I would plug −2 into $\frac{x-<!--\; -\; -->5}{x+4}$ which would give me a hole at $(-<!--\; -\; -->2,-<!--\; -\; -->\frac{7}{2})$. However, that turns out to be the limit of $R(x)$ (using L'Hopital's Rule):
$\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2}{lim}\frac{(x-<!--\; -\; -->5)(x+2)}{(x+4)(x+2)}=\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2}{lim}\frac{x^2-<!--\; -\; -->3x-<!--\; -\; -->10}{x^2+6x+8}=\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2}{lim}\frac{2x-<!--\; -\; -->3}{2x+6}=-<!--\; -\; -->\frac{7}{2}$
So why is there a hole in the graph if the limit exists at $(-<!--\; -\; -->2,-<!--\; -\; -->\frac{7}{2})$? Am I misunderstanding the definition of a limit? Thanks.

Let, $\frac{P(x)}{Q(x)}$ and $\frac{M(x)}{N(x)}$ be two rational functions consisting of polynominals $P(x),Q(x),M(x),N(x)$. The polyniminals are defined with respect to variable x. Now, if a rational function is formed by summation of the rational functions, does there exist a general way to evaluate the zeros of $\frac{P(x)}{Q(x)}+\frac{M(x)}{N(x)}$? If yes, where will the zeros of $\frac{P(x)}{Q(x)}$ and $\frac{M(x)}{N(x)}$ map to?
Edit 1: I know that, roots of $P(x)N(x)$ will be union of the roots of P(x) and N(x), and so on so forth for $Q(x)M(x)$. So, few possible roots can be calculated by taking the intersection of the roots of $P(x)N(x)$ and $Q(x)M(x)$. But I am not interested in them, I want solution in general.
Edit 2: What if, $\frac{P(x)}{Q(x)}$ is a Pade approximant of order (m,n)?

I have a function $f(a,b)=\frac{ab}{(\frac{a+b}{2}{)}^2}$, and (to me) it has some cool properties (e.g $f(a,b)=f(b,a)$, $f(x,0)=0$, $f(x,x)=1$, $0\le <!--\; \le \; -->f\le <!--\; \le \; -->1$, etc.). Now I wanted to know the inverse function for this. It's a rational function, but I couldn't figure out how to do this.
$$\begin{gathered} y=\frac{xb}{(\frac{x+b}{2}{)}^2}=\frac{4xb}{(x+b{)}^2} \\ x=\frac{4yb}{(y+b{)}^2}⟹<!--\; ⟹\; -->x(y+b{)}^2=4yb⟹<!--\; ⟹\; -->\frac{(y+b{)}^2}{y}=\frac{4b}{x} \\ \frac{y^2+b^2}{y}=\frac{4b}{x}-<!--\; -\; -->2b⟹<!--\; ⟹\; -->y+\frac{b^2}{y}=\frac{4b}{x}-<!--\; -\; -->2b \end{gathered}$$
At this point, I have no idea where to go. I think that I'm missing out on some fundamental rule of solving this, but I don't have a clue about where to go. How do I simplify the $y+\frac{b^2}{y}$?
NOTE: The final function $f^{-<!--\; -\; -->1}(y,b)$ is such that $f^{-<!--\; -\; -->1}(f(x,b),b)=x$ for any $x$ and $b$ such that $x$ and $b$ are not both 0. One can be zero, but not both (since $f(0,0)$ is undefined).

Let X be an integral scheme of finite type over an arbitrary field $k$. Let $f:X\to <!--\; \to \; -->{\mathbb{P}}^1$ be a dominant morphism. What is meant by the rational function of $f$? Fulton uses the notation $\text{div}(f)$ but it is not at all clear to me how to obtain a rational function from a dominant morphism to the projective line.

Let $a\ge <!--\; \ge \; -->0$ and $1/3\ge <!--\; \ge \; -->b\ge <!--\; \ge \; -->0$.
In the course of answering question Evaluating a double integral of a complicated rational function we came across a following integral:
$\mathfrak{I}(a,b):=\underset{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}}\frac{\log <!--\; \; -->(b^2+x^2)}{(x+a/2{)}^2+(1-<!--\; -\; -->b{)}^2/4}dx$
Now, since the integrand is a product of a logarithm and a rational function it the anti-derivative of the integrand can be always found as a collection of terms that involve logs and di-logarithms. Since the anti-derivative is known its values at plus and minus infinities can be taken and the integral above evaluated. This task seems to be easy enough but in the multitude of terms that emerges makes it very hard to complete. We have completed it though and got the following result:
$\mathfrak{I}(a,b)=2\mathrm{\pi <!--\; \pi \; -->}\frac{\log <!--\; \; -->(a^2+(1+b{)}^2)-<!--\; -\; -->2\log <!--\; \; -->(2)}{1-<!--\; -\; -->b}$
My question is how do we derive this result in some alternative way for example by using the Cauchy residue theorem.
Update: If we integrate the identity above over a we get the following identity:
$-<!--\; -\; -->\underset{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}}\log <!--\; \; -->(b^2+x^2)\cdot <!--\; \cdot \; -->(\arctan <!--\; \; -->(\frac{2x+a}{-<!--\; -\; -->1+b})-<!--\; -\; -->\arctan <!--\; \; -->(\frac{2x}{-<!--\; -\; -->1+b}))dx=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}((1+b)(\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->2\arctan <!--\; \; -->(\frac{1+b}{a}))+a\log <!--\; \; -->(a^2+(1+b{)}^2)-<!--\; -\; -->2a(1+\log <!--\; \; -->(2)))$
Again, the question would be to derive that identity in some alternative way.

I want to approximate $\tanh$ by a low-degree rational function of form
$r(x)=\frac{p(x)}{q(x)}=\frac{p_2{x}^2+p_{1}x+p_0}{q_{1}x+q_0}$
such that the $L^2$ norm over the fixed interval $[x_0,x_1]$ is small:
$I(p,q)=\frac{1}{2}{\int <!--\; \int \; -->}_{x_0}^{x_1}{(r(x)-<!--\; -\; -->\tanh <!--\; \; -->(x))}^2\mathrm{d}x.$
That is, I would like to solve for the coefficients $p_2,p_1,p_0,q_1,q_0$. Note that $\mathrm{d}\mathrm{e}\mathrm{g}(p)\le <!--\; \le \; -->2$ and $\mathrm{d}\mathrm{e}\mathrm{g}(q)\le <!--\; \le \; -->1$.
I don't know where to begin, so I'm looking for suggestions on how to approach this problem; pointers to relevant numerical methods would also be greatly appreciated

I need to compute the integral $\int <!--\; \int \; -->{\displaystyle \frac{2x}{(x^2+x+1{)}^2}}\cdot <!--\; \cdot \; -->dx$. I tried using the integration of a rational function technique, with $\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1{)}^2}$, but this simply returned $C=2$ and $A,B,D=0$, so it doesn't really change anything.
I also tried using a $u$ substitution, setting $u=x^2+x+1$. This made the numerator $2x=\frac{du}{dx}-<!--\; -\; -->1$, but I'm not really sure if I can do that/how to solve an integral with a derivative as a part of it.
How would I go about solving this?

A rational function is defined by
$f(x)=\frac{a(x)}{b(x)}={\displaystyle \frac{a_n{x}^n+a_{n-<!--\; -\; -->1}{x}^{n-<!--\; -\; -->1}+\cdots +a_0}{b_m{x}^m+a_{m-<!--\; -\; -->1}{x}^{m-<!--\; -\; -->1}+\cdots +b_0}}$
where $m,n\in <!--\; \in \; -->\mathbb{N}\cup <!--\; \cup \; -->\{0\}$ such that not every coefficient $b_i$ is 0 and f(x) is not defined at the values of x for which b(x)=0.
What 'special' choice(s) of b(x) will give the result that the set of all polynomial functions, $a(x)=a_n{x}^n+a_{n-<!--\; -\; -->1}{x}^{n-<!--\; -\; -->1}+\cdots +a_0$ form a subset of the rational functions (as defined above)?