Elliptic rational functions

• ${\displaystyle R_{n}^{2}(\xi ,x)\leq 1}$ for ${\displaystyle |x|\leq 1\,}$
• ${\displaystyle R_{n}^{2}(\xi ,x)=1}$ at ${\displaystyle |x|=1\,}$
• ${\displaystyle R_{n}^{2}(\xi ,-x)=R_{n}^{2}(\xi ,x)}$
• ${\displaystyle R_{n}^{2}(\xi ,x)>1}$ for ${\displaystyle x>1\,}$
• The slope at x=1 is as large as possible
• The slope at x=1 is larger than the corresponding slope of the Chebyshev polynomial of the same order.

The only rational function satisfying the above properties is the elliptic rational function (Lutovac 2001, § 13.2) harv error: no target: CITEREFLutovac2001 (help). The following properties are derived:

The elliptic rational function is normalized to unity at x=1:

${\displaystyle R_{n}(\xi ,1)=1\,}$

The nesting property is written:

${\displaystyle R_{m}(R_{n}(\xi ,\xi ),R_{n}(\xi ,x))=R_{m\cdot n}(\xi ,x)\,}$

This is a very important property:

• If ${\displaystyle R_{n}}$ is known for all prime n, then nesting property gives ${\displaystyle R_{n}}$ for all n. In particular, since ${\displaystyle R_{2}}$ and ${\displaystyle R_{3}}$ can be expressed in closed form without explicit use of the Jacobi elliptic functions, then all ${\displaystyle R_{n}}$ for n of the form ${\displaystyle n=2^{a}3^{b}}$ can be so expressed.
• It follows that if the zeroes of ${\displaystyle R_{n}}$ for prime n are known, the zeros of all ${\displaystyle R_{n}}$ can be found. Using the inversion relationship (see below), the poles can also be found.
• The nesting property implies the nesting property of the discrimination factor:

${\displaystyle L_{m\cdot n}(\xi )=L_{m}(L_{n}(\xi ))}$

The elliptic rational functions are related to the Chebyshev polynomials of the first kind ${\displaystyle T_{n}(x)}$ by:

${\displaystyle \lim _{\xi =\rightarrow \,\infty }R_{n}(\xi ,x)=T_{n}(x)\,}$

${\displaystyle R_{n}(\xi ,-x)=R_{n}(\xi ,x)\,}$ for n even

${\displaystyle R_{n}(\xi ,-x)=-R_{n}(\xi ,x)\,}$ for n odd

${\displaystyle R_{n}(\xi ,x)}$ has equal ripple of ${\displaystyle \pm 1}$ in the interval ${\displaystyle -1\leq x\leq 1}$. By the inversion relationship (see below), it follows that ${\displaystyle 1/R_{n}(\xi ,x)}$ has equiripple in ${\displaystyle -1/\xi \leq x\leq 1/\xi }$ of ${\displaystyle \pm 1/L_{n}(\xi )}$.

The following inversion relationship holds:

${\displaystyle R_{n}(\xi ,\xi /x)={\frac {R_{n}(\xi ,\xi )}{R_{n}(\xi ,x)}}\,}$

This implies that poles and zeroes come in pairs such that

${\displaystyle x_{pi}x_{zi}=\xi \,}$

Odd order functions will have a zero at x=0 and a corresponding pole at infinity.

The zeroes of the elliptic rational function of order n will be written ${\displaystyle x_{ni}(\xi )}$ or ${\displaystyle x_{ni}}$ when ${\displaystyle \xi }$ is implicitly known. The zeroes of the elliptic rational function will be the zeroes of the polynomial in the numerator of the function.

The following derivation of the zeroes of the elliptic rational function is analogous to that of determining the zeroes of the Chebyshev polynomials (Lutovac 2001, § 12.6) harv error: no target: CITEREFLutovac2001 (help). Using the fact that for any z

${\displaystyle \mathrm {cd} \left((2m-1)K\left(1/z\right),{\frac {1}{z}}\right)=0\,}$

the defining equation for the elliptic rational functions implies that

${\displaystyle n{\frac {K(1/L_{n})}{K(1/\xi )}}\mathrm {cd} ^{-1}(x_{m},1/\xi )=(2m-1)K(1/L_{n})}$

so that the zeroes are given by

${\displaystyle x_{m}=\mathrm {cd} \left(K(1/\xi )\,{\frac {2m-1}{n}},{\frac {1}{\xi }}\right).}$

Using the inversion relationship, the poles may then be calculated.

From the nesting property, if the zeroes of ${\displaystyle R_{m}}$ and ${\displaystyle R_{n}}$ can be algebraically expressed (i.e. without the need for calculating the Jacobi ellipse functions) then the zeroes of ${\displaystyle R_{m\cdot n}}$ can be algebraically expressed. In particular, the zeroes of elliptic rational functions of order ${\displaystyle 2^{i}3^{j}}$ may be algebraically expressed (Lutovac 2001, § 12.9, 13.9) harv error: no target: CITEREFLutovac2001 (help). For example, we can find the zeroes of ${\displaystyle R_{8}(\xi ,x)}$ as follows: Define

${\displaystyle X_{n}\equiv R_{n}(\xi ,x)\qquad L_{n}\equiv R_{n}(\xi ,\xi )\qquad t_{n}\equiv {\sqrt {1-1/L_{n}^{2}}}.}$

Then, from the nesting property and knowing that

${\displaystyle R_{2}(\xi ,x)={\frac {(t+1)x^{2}-1}{(t-1)x^{2}+1}}}$

where ${\displaystyle t\equiv {\sqrt {1-1/\xi ^{2}}}}$ we have:

${\displaystyle L_{2}={\frac {1+t}{1-t}},\qquad L_{4}={\frac {1+t_{2}}{1-t_{2}}},\qquad L_{8}={\frac {1+t_{4}}{1-t_{4}}}}$

${\displaystyle X_{2}={\frac {(t+1)x^{2}-1}{(t-1)x^{2}+1}},\qquad X_{4}={\frac {(t_{2}+1)X_{2}^{2}-1}{(t_{2}-1)X_{2}^{2}+1}},\qquad X_{8}={\frac {(t_{4}+1)X_{4}^{2}-1}{(t_{4}-1)X_{4}^{2}+1}}.}$

These last three equations may be inverted:

${\displaystyle x={\frac {1}{\pm {\sqrt {1+t\,\left({\frac {1-X_{2}}{1+X_{2}}}\right)}}}},\qquad X_{2}={\frac {1}{\pm {\sqrt {1+t_{2}\,\left({\frac {1-X_{4}}{1+X_{4}}}\right)}}}},\qquad X_{4}={\frac {1}{\pm {\sqrt {1+t_{4}\,\left({\frac {1-X_{8}}{1+X_{8}}}\right)}}}}.\qquad }$

To calculate the zeroes of ${\displaystyle R_{8}(\xi ,x)}$ we set ${\displaystyle X_{8}=0}$ in the third equation, calculate the two values of ${\displaystyle X_{4}}$, then use these values of ${\displaystyle X_{4}}$ in the second equation to calculate four values of ${\displaystyle X_{2}}$ and finally, use these values in the first equation to calculate the eight zeroes of ${\displaystyle R_{8}(\xi ,x)}$. (The ${\displaystyle t_{n}}$ are calculated by a similar recursion.) Again, using the inversion relationship, these zeroes can be used to calculate the poles.