1
$\begingroup$

Sometimes I find this notation and I get confused: $$\int{R\left( \cos{x}, \sin{x} \right)\mathrm{d}x} $$

Does it mean a rational function or taking rational operations between $\cos{x}$ and $\sin{x}$ ?

Can you explain please?

Update: I think you did not understand the question well, Here is an example (maybe it is a lemma or a theorem):

All the integrals of the form $\int{R\left( \cos{x}, \sin{x} \right)\mathrm{d}x} $ can be evaluated using the substitution $u=\tan{\dfrac{x}{2}} $.

I think that $R$ here does not stand for a rational function but for taking rational operations(addition, subtraction, multiplication, division) between $\cos{x} $ and $\sin{x}$

Update : I did not noticed that $R$ is a rational function of two variables and that means exactly that we are taking rational operations.

$\endgroup$
  • 3
    $\begingroup$ The assumption is exactly that R is a rational function. $\endgroup$ – Did May 4 '16 at 11:21
  • 5
    $\begingroup$ What is, in your opinion, the difference between "a rational function of $\cos(x)$ and $\sin(x)$" and "taking rational operations between $\cos(x)$ and $\sin(x)$"? $\endgroup$ – Christian Blatter May 4 '16 at 11:22
  • 2
    $\begingroup$ Please read the question before downvoting. the book says: The symbol R and does not say rational function $\endgroup$ – Navaro May 4 '16 at 11:50
  • 3
    $\begingroup$ @Navaro No, $R(x,y)$ can be any rational function, so it could be $xy$ or $x+y$ or $\frac{x}{y}$ or $\frac{x^2+xy+x^3y^3}{x^2+y^2}$ or any number of things. It is not necessarily a ratio of one argument to the other. Perhaps that is why the term "rational operation" confused you. $\endgroup$ – Ian May 4 '16 at 12:28
  • 3
    $\begingroup$ The point that Did and Christian Blatter and I have been trying to make is that you have not made it entirely clear what "taking rational operations" means exactly. By contrast "rational function" is a clearly defined term, whose definition can be looked up. I can tell you from the context (because you are talking about the half-angle substitution) that your $R$ is a rational function. Maybe it can be described the other way too, but that doesn't really matter. $\endgroup$ – Ian May 4 '16 at 12:42
6
$\begingroup$

Here $R$ is a function of two variables $s$ and $t$. For instance, if $$R(s,t) = \frac{s}{1+t}$$ then $$R(\cos x , \sin x) = \frac{\cos x}{1 + \sin x}.$$

$\endgroup$
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen May 4 '16 at 13:01
2
$\begingroup$

A polynomial in two variables is an expression $$ P(x, y) = \sum_{i, j=0}^{\infty} a_{ij} x^{i} y^{j} $$ with only finitely many non-zero coefficients $a_{ij}$. A rational function in two variables is a quotient of polynomials in two variables.

In your question, $R$ denotes an arbitrary rational function of two variables. (It's arguably reasonable to describe "evaluating a rational function of two variables $x$ and $y$" as "taking rational operations between $x$ and $y$", but the number of comments here and elsewhere suggests doing so is a recipe for ambiguity.)

We can be certain of this interpretation on grounds of mathematical culture: Thanks to the double-angle formulas for the circular functions and the chain rule, the substitution $u = \tan(x/2)$, or $x = 2\arctan u$, gives $$ \cos x = \frac{1 - u^{2}}{1 + u^{2}},\qquad \sin x = \frac{2u}{1 + u^{2}},\qquad dx = \frac{2\, du}{1 + u^{2}}. $$ Consequently, $$ \int R(\cos x, \sin x)\, dx = \int R\left(\frac{1 - u^{2}}{1 + u^{2}}, \frac{2u}{1 + u^{2}}\right) \frac{2\, du}{1 + u^{2}}, $$ a rational function in $u$.

The significance is, every rational function in one variable has an elementary primitive (antiderivative).

$\endgroup$
  • $\begingroup$ This is exactly what I expected the answer to be like $\endgroup$ – Navaro May 4 '16 at 14:06
1
$\begingroup$

In books with tables of integrals etc. (like Gradshteyn's book) $R(\cos x, \cos y)$ typically means a rational function of $(\cos x, \sin x)$. So take a rational function $R(a,b)=f^{1}(a,b)/f^{2}(a,b)$ and plug in $a=\cos x, b=\sin x$. Here, $f^{i}(a,b), i=1,2$ are polynomials in the two variables $(a,b)$, i.e., $$f^{i}(a,b)=c^i_{0,0}+c^i_{1,0}a+c^i_{0,1}b+c^i_{2,0}a^2+c^i_{0,2}b^2+c^i_{1,1}ab+\dots$$ with coefficients $c^{i}_{k,l}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.