Get Expert Help for Your Rational Functions Questions

How to show that the two rational functions $a(t),b(t)\in <!--\; \in \; -->\mathbb{K}(t)$ when $\mathrm{Char}<!--\; \; -->(\mathbb{K})\ne 2$ are algebraically dependent.
(there is polynomial $P(x,y)\in <!--\; \in \; -->\mathbb{K}[x,y]$ such that $P(a(t),b(t))=0$)?

Using this rule: Integral Calculus "Rational Function Rule"
$\int <!--\; \int \; -->\frac{1}{a^2-<!--\; -\; -->x^2}dx=\frac{1}{2a}{\log }_e<!--\; \; -->\left(\frac{a+x}{a-<!--\; -\; -->x}\right)+c$
I have integrated a function, but my question is if I'm solving for $x$, in the next line I must raise both sides to the base of $e$, and there is still $+c$ (constant) on the end of the integrated function, so do I leave it as plus $c$ on the RHS on the equation (not do anything to it, so when I want to work it out by substituting $(x_1,y_1)$ at the end it will be $x_1=\dots <!--\; \dots \; -->y_1\dots <!--\; \dots \; -->+c$) or when I raise both sides to the base of $e$ does $c$, the constant become $c\ast <!--\; \ast \; -->e\dots <!--\; \dots \; -->$ instead of $+c$?
Any help would be great.

I was bored in my Algebra 2 class and wanted to try to illustrate the tan(x) function as an infinite summation of rational functions.
I recalled that the solution to a rational function's numerator is equivalent to the roots of the function. The denominator's solution is the vertical asymptote(s). Since tan(x) has infinite roots and asymptotes, if I expressed it as a rational function it'd be infinitely long. My teacher suggested just using an infinite summation of rational functions, so I did that instead.
The first thing I did was set up the numerator and denominator so that they could each equal the corresponding roots and asymptotes; since it's (sorta? kinda? maybe) an even function, I just used variations of $x\pm <!--\; \pm \; -->\mathrm{\pi <!--\; \pi \; -->}$ for both sides. Given that all roots of tan(x) are whole multiples of pi and all asymptotes are odd multiples of pi/2, I came up with the following equation:
$\underset{n=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{x\pm <!--\; \pm \; -->n\mathrm{\pi <!--\; \pi \; -->}}{x\pm <!--\; \pm \; -->\frac{2n-<!--\; -\; -->1)\mathrm{\pi <!--\; \pi \; -->}}{2}}$
Where n is any integer from negative infinity to infinity, my idea is that the end product of this is tan(x). Is this even remotely correct or did I leave something out? Thanks!
(I do recognize that the horizontal asymptote is apparently 1 with this equation, but when I plot a section of this in desmos it seems to pass through y=1 anyway. There might be something going on that I could simplify in terms of rewriting an "x^2" somewhere, or maybe I'm just trippin.)

I know trigonometric functions such as $7x-<!--\; -\; -->6\cos <!--\; \; -->(3x)$ can have an infinite number of critical points, but what about non-zero rational functions? Does it have something to do with how a non-zero polynomial can have a finite number of roots?

Rational function is defining as a polynomial with real coefficients over polynomial with real coefficents, how to find the removeable or infinite discontinuity of any rational function without the factoring of the polynomial since it is very troublesome?

I'm wondering if someone could help me solve a little conundrum I've been having. I'm trying to find the equation of a line with slope −1 tangent to the curve of a rational function. The function is $y=1/(x-<!--\; -\; -->1)$ and the general equation of the line would be $y=-<!--\; -\; -->x+k$. I understand that if I set the y-values equal -- that is, $1/(x-<!--\; -\; -->1)=-<!--\; -\; -->x+k$ should be able to reorder it into a quadratic and solve for k, giving me the equation of the line. So far, the quadratic I'm finding is $y=-<!--\; -\; -->x^2+x+kx-<!--\; -\; -->k-<!--\; -\; -->1$, yet I can't seem to find a satisfactory value for k that would give me a tangent to the curve. Any thoughts?

Can anyone help me prove the convexity of this rational function? The man who proved the convexity of function used these facts. But I don't know this fact is correct or not. Here are the facts and function:
1. As N increases, f(N) goes to infinity, That implies that there must be a minima (either at N=0 or somewhere else with a finite N)
2. There cannot be more than one (positive) minima since we're dealing with second order equation.
$f(N)=+-<!--\; -\; -->c{N}^4+-<!--\; -\; -->d{N}^3+-<!--\; -\; -->e{N}^2+-<!--\; -\; -->fN+-<!--\; -\; -->g/+-<!--\; -\; -->a{N}^2+-<!--\; -\; -->bN$
a,b,c,d,e,f,g is constants. and $N\ge <!--\; \ge \; -->1$. I guess the second order equation means that between the leading coefficient of the numerator and the denominator is 2.
Are these facts correct? I think fact 1 is no problem, but fact 2 is correct or not.
I am waiting for any answers.

I am trying to calculate the integral
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}\frac{a+x}{b^2+(a+x{)}^2}\frac{1}{1+c(a-<!--\; -\; -->x{)}^2}dx$
where $\{a,b,c\}\in <!--\; \in \; -->\mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

My analysis book makes the following statement:
Every rational function with real coefficients can be integrated in terms of
1, rational functions,
2. logarithm functions,
3. arctangent functions.
Does it mean that after polynomial division, partial fraction expansion, completing the square, substituting where needed, and knowing that
$\begin{array}{r}\\ \int <!--\; \int \; -->\frac{dx}{(x-<!--\; -\; -->a{)}^n}& =-<!--\; -\; -->\frac{1}{n-<!--\; -\; -->1}\frac{1}{(x-<!--\; -\; -->a{)}^{n-<!--\; -\; -->1}}(n\ne <!--\; \ne \; -->1)\\ \int <!--\; \int \; -->\frac{dx}{x-<!--\; -\; -->a}& =\log <!--\; \; -->|x-<!--\; -\; -->a|\\ \int <!--\; \int \; -->\frac{dx}{x^2+1}& =\arctan <!--\; \; -->x\end{array}$
one can integrate any rational function?

Some antiderivatives of rational functions involve inverse trigonometric functions, and some involve logarithms. But inverse trig functions can be expressed in terms of complex logarithms. So is there a general formula for the antiderivative of any rational function that uses complex logarithms to unite the two concepts?

I want to find the period of the function
$f(x)=\frac{2\cos <!--\; \; -->(3x)\cdot <!--\; \cdot \; -->|\sin <!--\; \; -->(5x)|\cdot <!--\; \cdot \; -->3|\tan <!--\; \; -->(8x)|{\cos }^4<!--\; \; -->(x)}{7\tan <!--\; \; -->\left(\frac{x}{9}\right)\cdot <!--\; \cdot \; -->{\sec }^3<!--\; \; -->\left(\frac{2x}{3}\right)}$
which is a rational function of trig functions.
So far I have found that:
1. the period of ${\displaystyle \cos <!--\; \; -->(3x)}$ is $\frac{2\mathrm{\pi <!--\; \pi \; -->}}{3};$;
2. the period of ${\displaystyle |\sin <!--\; \; -->(5x)|}$ is $\frac{\mathrm{\pi <!--\; \pi \; -->}}{5};$;
3. the period of ${\displaystyle {\cos }^4<!--\; \; -->(x)}$ is $\mathrm{\pi <!--\; \pi \; -->}$;
4. the period of ${\displaystyle \tan <!--\; \; -->(x/9)}$ is $9\mathrm{\pi <!--\; \pi \; -->}$;
5. and the period of ${\displaystyle {\sec }^3<!--\; \; -->(2x/3)}$ is $3\mathrm{\pi <!--\; \pi \; -->}$.
How do I find the total period?

This article about polynomial interpolation claims that (it is known that) every rational function may be represented in barycentric form:$r(x)=\frac{\underset{j=0}{\overset{N}{\sum <!--\; \sum \; -->}}\frac{w_j}{x-<!--\; -\; -->x_j}{y}_j}{\underset{j=0}{\overset{N}{\sum <!--\; \sum \; -->}}\frac{w_j}{x-<!--\; -\; -->x_j}}$
What about the rational function $r(x,y)=\frac{xy}{x^2+y^2}$? But even if I accept that we are talking only about univariate functions, I would still like to know a bit more about the cited statement:
1. Is this statement strictly true for $\mathbb{R}$, i.e. for any rational function $r(x)=\frac{p(x)}{q(x)}$ with $p,q\in <!--\; \in \; -->\mathbb{R}[x]$ (polynomials with real coefficients) there are $w_j,x_j,y_j\in <!--\; \in \; -->\mathbb{R}$ representing r(x)?
2. Is this statement strictly true for $\mathbb{C}$, i.e. for any rational function $r(x)=\frac{p(x)}{q(x)}$ with $p,q\in <!--\; \in \; -->\mathbb{C}[x]$ (polynomials with complex coefficients) there are $w_j,x_j,y_j\in <!--\; \in \; -->\mathbb{C}$ representing r(x)?
3. For $r(x)=\frac{p(x)}{q(x)}$ and $N=\mathrm{deg}<!--\; \; -->p+\mathrm{deg}<!--\; \; -->q$, is the set of $x_j\in <!--\; \in \; -->\mathbb{C}$ for which appropriate $w_j,y_j\in <!--\; \in \; -->\mathbb{C}$ exist open and dense in ${\mathbb{C}}^{N+1}$?
4. Where can I learn more about this topic, since when is this known, ... ?

Consider rational functions of the form $f(x)=\frac{p(x)}{q(x)}$ where $p$ is a polynomial of degree not exceeding $n$ and $q$ is a particular polynomial $q(x)=(x-1)\cdot <!--\; \cdot \; -->(x-2)\cdot <!--\; \cdot \; -->\dots <!--\; \dots \; -->\cdot <!--\; \cdot \; -->(x-m)$. Do these functions form a vector space? I think these function forms a vector space because they are closed under addition and scaling by a number. In case they do, what is the dimension of this vector space? Could you suggest a basis and write down an explicit formula for the coordinate functions?

I recently came across the result that
$\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{n^4-<!--\; -\; -->n^3+n+1}{n^6-<!--\; -\; -->1}=\frac{1}{2}$
I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.
If I plug the sum into Wolfram Alpha it gives
$\frac{3k^4-<!--\; -\; -->k-<!--\; -\; -->2}{6k(k+1)(k^2+k+1)}$
as the $k$-th partial sum. Taking the limit as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, this would in fact proof the upper equality.
Sadly, I could not wrap my head around how to get to Wolfram Alphas partial sum result. If anyone has an idea let me know. Any tips are appreciated.

There is a theorem that says rational functions in the extended complex plane are exactly the meromorphic functions.
After this, my textbook draws the corollary: "...as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles."
I don't see how it follows. Plus I think I have an example. Take a rational function with no zeros and two poles of multiplicity one, at 0 and 1.
At least two such functions exist:
$f(x)=\frac{1}{x}+\frac{1}{x-<!--\; -\; -->1}$
$g(x)=\frac{1}{x}+\frac{2}{x-<!--\; -\; -->1}$

Let U be a simply connected domain and let $f$ be a meromorphic function on U with only finitely many zeroes and poles. Prove that there is a holomorphic function g : U $\to <!--\; \to \; -->\mathbb{C}$ and a rational function $q$, such that
$f(z)=e^{g(z)}q(z)$

Suppose we have an integral (reduced and irreducible) affine curve $X$ over an algebraically closed field $k$. Suppose $t=\frac{f}{g}$ is a rational function on $X$ such that there exists an $n\in <!--\; \in \; -->\mathbb{N}$ such that $t^n=(\frac{f}{g}{)}^n$ is a regular function, i.e., $t^n\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(X,{\mathcal{O}}_X)$ (the coordinate ring or ring of regular functions on $X$). Does this imply that $t$ itself is a regular function?
Note that if the curve is non-singular, this is immediate, as in this case $\mathrm{\Gamma <!--\; \Gamma \; -->}(X,{\mathcal{O}}_X)$ is a UFD and hence $b^n|a^n$ implies $b|a$ in $\mathrm{\Gamma <!--\; \Gamma \; -->}(X,{\mathcal{O}}_X)$. But is this true in general for an integral affine curve?

I've recently been introduced to calculus and ran into a rational function that seems to be somewhat complicated for my understanding, which is shown in the form:
$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}\cdot <!--\; \cdot \; -->a+\frac{(x-b)(x-c)}{(a-b)(a-c)}\cdot <!--\; \cdot \; -->b+\frac{(x-a)(x-c)}{(b-a)(b-c)}\cdot <!--\; \cdot \; -->c$
Bearing in mind that when f(a)=b, f(b)=c and f(c)=a due to substitution.
However, the part I don't understand is the simplification of the equation into the quadratic form as follows:
$\frac{1}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}[(a-<!--\; -\; -->b)a+(b-<!--\; -\; -->c)b+(c-<!--\; -\; -->a)c]x^2+[(a-<!--\; -\; -->b)b^2+(b-<!--\; -\; -->c)c^2+(c-<!--\; -\; -->a)a^2]x+[(a-<!--\; -\; -->b)a^2+(b-<!--\; -\; -->c)b^{2}c+(c-<!--\; -\; -->a)a{c}^2]$
Based on my algebraic knowledge, I know that the factorization formula is $x^2+(a+b)x+ab$
My question is how did the coefficients and rearranged dominator of the original equation come about? Can anyone detail it for me, please?

I know that for polynomials $P,Q$, the equation $P(z)\equiv <!--\; \equiv \; -->Q(z)$ implies that they are of the same degree and have the same coefficients. Is there an analogous result for rational fucntions? That is, if $R,S$ are two rational functions and $R(z)=S(z)$ for all $z$ what is the relationship between $R$ and $S$?