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Find divisor of rational function ${\displaystyle f=(w_0,w_1)}$ on a surface ${\displaystyle X=\{w_0{w}_1-w_2{w}_3\}\subset P{P}^3}$
How should I to take into account this surface?

To give a rational function f in the function field K(X) of a smooth projective curve X is equivalent to giving a finite morphism ${\displaystyle f:X\to P{P}^1}$.
What is the precise analogue of this statement for higher-dimensional varieties?
I would guess that a rational function on a smooth projective variety gives rise to a rational morphism ${\displaystyle X\to P{P}^1}$ and that this is a morphism up to blowing-up X. Is that correct?

${\displaystyle \underset{x\to 2}{lim}\frac{\sqrt{x+2}\sqrt{x+7}-2\sqrt{x+7}-3\sqrt{x+2}+6}{x\sqrt{4x+1}-2\sqrt{4x+1}-3x+6}}$
How to solve this???

All polynomials are rational functions, that perhaps I should just imagine them as "over 1". Good. But the definition of a rational function has the concept of "ratio" in it. And when I found this:
A cylinder has a volume of ${\displaystyle (x+3)(x^2-3x-18)\pi }$ cubic centimeters. Find the height of the cylinder.
I wondered how this is a ratio of any sort?(It can't be a ratio of the expression over 1) So if ${\displaystyle a=(x+3),b=(x^2-3x-18)}$, and ${\displaystyle c=\pi }$,
can we say a,b,c are "in a ratio," i.e., a:b:c?

Is a rational function or a rational equation or none of these?
1.${\displaystyle y=5x^3-2x+1}$
2.${\displaystyle g\left(x\right)=7x^3-4\sqrt{x}+1\sqrt{x^2}+3}$
Help!!!

For a course on Galois theory, we proved the fundamental theorem of symmetric polynomials, which states that every symmetric polynomial can be uniquely written as a polynomial in the elementairy symmetric polynomials. As an exercise we were asked to prove that every symmetric rational function also is a rational function in the elementary symmetric polynomials.
For this I have the following (partial) proof. Let K be a field, let ${\displaystyle fϵK(T_1,\dots ,T_n)}$ be a symmetric function. We can write ${\displaystyle f=\frac{g}{h}}$ with g,h co'. Let ${\displaystyle \sigma ϵS_n}$, then ${\displaystyle \frac{\sigma \left(g\right)}{\sigma \left(h\right)}=\sigma \left(\frac{g}{h}\right)=\frac{g}{h}}$, so ${\displaystyle \sigma \left(g\right)h=\sigma \left(h\right)g}$. Since g and h are co', it follows that ${\displaystyle g\mid \sigma \left(g\right)}$, and since the total degrees of the polynomials are the same it follows there exists some ${\displaystyle kϵe{K}^{\times }}$ such that ${\displaystyle \sigma \left(g\right)=kg}$. Similarly for h.
Of course I would like to conclude from this that we must have ${\displaystyle \sigma \left(g\right)=g}$ since ${\displaystyle \sigma }$ doesn't change the coefficients, but I realised the argument is more subtle than that, since this is not necessarily true. For example, with ${\displaystyle K=F{F}_7}$ and n=3 we have
${\displaystyle \left(321\right)\cdot (X_1{X}_2+2X_2{X}_3+4X_3{X}_1)=2\cdot (X_1{X}_2+2X_2{X}_3+4X_3{X}_1)}$.So, does anyone know how to proceed? Ofcourse, once you have ${\displaystyle \sigma \left(g\right)=g}$ and ${\displaystyle \sigma \left(h\right)=h}$ for all ${\displaystyle \sigma ϵS_n}$ the result follows from the fundamental theorem of symmetric polynomials.
As a sidenote: I know there's a proof using the fundamental theorem of Galois theory, but I would like to finish a proof using this approach.

I am finding the divisor of ${\displaystyle f=\left(\frac{x_1}{x_0}\right)-1}$ on C, where ${\displaystyle C=V(x_1^2+x_2^2-x_0^2)\subset P{P}^2}$. Characteristic is not 2.
I am totally new to divisors. So the plan in my mind is first find an open subset U of C. In this case maybe it should be the complement of ${\displaystyle X=x_2=0}$. Then I should look at ${\displaystyle \text{ord}\left(f\right)}$ on this U. Then I get confused, since f is a rational function, how can f belong to k[U]?
Who knows?

Let f be holomorphic in all of ${\displaystyle \mathbb{C}}$ except for poles at z=0 and z=1. Further f satisfies:
${\displaystyle \underset{\left|z\right|\to \mathrm{\infty }}{lim}f\left(z\right)=0}$
Ive

If I have an analytic function of a complex variable, I can write a Taylor series and in some cases can truncate the high powers to obtain a good approximation over some part of the function's domain. I would like to be able to generate rational functions (quotients of polynomials, as in ${\displaystyle \frac{n\left(x\right)}{d\left(x\right)}}$ where n and d are polynomial functions) that play a similar role. Is there a general way to do this?

I want a function that passes through the (x,y) points (3,0),(4,-1),(6,3) and (7,2). Using Lagrange interpolation I get
${\displaystyle -\frac{x^3-15x^2+70x-102}{2}}$ but using this interpolation method for rational functions I get the much simpler ${\displaystyle \frac{(x-3)}{(x-5)}}$. This is a case where the rational function interpolation is a lot simpler than polynomial interpolation.
Is there a method to get ${\displaystyle \frac{(x-3)}{(x-5)}}$ given the points above, a sort of interpolation method that uses rational functions?

I'm searching the antiderivative of rational functions:
1)$\int \frac{1+x}{\sqrt{2x+1}}dx$
For this one we have ${\displaystyle t=\sqrt{2x+1}}$ then ${\displaystyle st=\frac{1}{\sqrt{2x+2}}dx}$ but then I do not see the way to compute the antiderivative.
Same thing for 2)${\displaystyle \int \frac{1}{x+\sqrt{x+1}}}$
What are the methods to find antiderivative of rational functions like these ones?
Thank you For (1): Try the substitution ${\displaystyle u=2x+1}$. Then ${\displaystyle du=2dx\Rightarrow \frac{1}{2}du=dx}$, and
${\displaystyle x=\frac{1}{2}(u-1)\Rightarrow x+1=\frac{1}{2}(u-1)+1=\frac{1}{2}(u+1)}$
Then your integral becomes
${\displaystyle \int \frac{1+x}{\sqrt{2x+1}}dx=\frac{1}{4}\int \frac{u+1}{u^{\frac{1}{2}}}du=\frac{1}{4}}$
$\int (u^{\frac{1}{2}}+u^{-\frac{1}{2}})du$

Does there exists some simple criteria to know when the primitive of a rational function of ${\displaystyle \mathbb{C}\left[z\right]}$ is still a rational function?
In fact my question is more about the stability of this property. Let P and Q two co' polynomials and let A and B two co' polynomials such that
$\frac{A}{B}=(\frac{P}{Q}{)}^{\prime }=\frac{P^{\prime }Q-P{Q}^{\prime }}{Q^2}$
Then considering a pertubation ${\displaystyle A^{\xi }}$ of A (in the sense the roots of ${\displaystyle A^{\xi }}$ converge to the one of A with the same multiplicity as ${\displaystyle \xi }$ goes to zero.): does ${\displaystyle \frac{A^{\xi }}{B}}$ admit a primitive which is rational function?

Given a rational function ${\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}$, what would the ${\displaystyle x^k}$ coefficient of this rational function mean (in particular for the negative k's). Is there some sort of expansion
${\displaystyle \frac{p\left(x\right)}{q\left(x\right)}=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{x}^k}$
where one would say ${\displaystyle a_k}$ is the coefficient of ${\displaystyle x^k}$?
How would you find the coefficient? A very simple rational function would be ${\displaystyle \frac{(x-1)}{(x-2)(x-3)}}$. How would you find the ${\displaystyle x^{-1}}$ coefficient of this?
Need help!

From the definition of rational function ${\displaystyle f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}}$, where P(x) and q(x) are polynomials and ${\displaystyle q\left(x\right)\ne 0}$
so for the function ${\displaystyle f\left(x\right)=\frac{x^{-2}+3}{x-5}}$ by the definition f(x) isnt rational since the numerator is not polynomial but by multiplying both numerator and denominator by ${\displaystyle x^2}$ we get
${\displaystyle f\left(x\right)=\frac{3x^2+1}{x^3-5x^2}}$ which is rational
and can we say that both functions are equal at every point?

You has a rational function
${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$,
where f(x) and g(x) are polynomials. What is the degree of the rational function?
This might sound like a very trivial question but I found different answers on the web.
Is it the maximum degree of f and g? Or is it ${\displaystyle deg\left(f\right)-deg\left(g\right)}$?

For the function, determine if it is a polynomial, rational function, both, or neither
${\displaystyle g\left(x\right)=\frac{4x^5+x^2}{-8x^9+2\sqrt{x}+9}}$
a. Polynomial function only
b. Rational function only
c. Both polynomial and rational function

I thought it was interesting that ${\displaystyle \frac{u^2+1}{{(u^2-2u-1)}^2}}$ has the very simple integral ${\displaystyle -\frac{u}{u^2-2u-1}}$ but noth of ${\displaystyle {\frac{u^2}{(u^2-wu-1)}}^2}$ and ${\displaystyle \frac{1}{{(u^2-2u-1)}^2}}$ are very complicated (the transcendental parts cancel each other though).
So my question is how do I check by looking at a rational function whether or not its

Suppose ${\displaystyle R\left(z\right)}$ be a rational function such that |R(z)|=1 for |z|=1. Show that ${\displaystyle \alpha }$ is a zero or a pole or order ${\displaystyle m}$, if and only if ${\displaystyle \frac{1}{\bar{\alpha }}}$ is a pole or zero or order ${\displaystyle mP}$ respectively.
I am wondering that if I can first show that ${\displaystyle M\left(z\right)=R\left(z\right)}$ the complex conjugate of ${\displaystyle R\left(\frac{1}{z}\right)}$ is a rational function such that ${\displaystyle M\left(z\right)=1P}$ on ${\displaystyle \left|z\right|=1}$. Can I assume ${\displaystyle z=1}$ and then substitute all ${\displaystyle z}$ with 1? Thanks.

I know that all meromorphic functions on ${\displaystyle \mathbb{C}P{P}^1}$ are rational functions. However, Im

Does it follow that ${\displaystyle r\left(x\right)+p\left(x\right)}$ is a rational function (and not a polynomial)?
Suppose ${\displaystyle r\left(x\right)\ne 0}$ is a rational function (and not a polynomial)
${\displaystyle p\left(x\right)}$ is a polynomial function of ${\displaystyle x\in \mathbb{R}}$.