Application of Continued Fractions to the Computation of Elliptic Integrals

V. E. Kornilov

(Presented by the Department of Higher Mathematics)

In this article, on the basis of continued fractions, approximations of elliptic integrals in the form of a sum of arctangents are obtained, and an estimate of the remainder terms is given.

In what follows we use the abbreviated notation

\[ \kappa^{2}x^{2}=y;\qquad (a)_{\kappa}=a(a+1)\ldots(a+\kappa-1),\qquad (a)_0=1. \tag{1} \]

1. With respect to the even convergents of the following power function [1],

\[ (1-y)^{-\frac12} = \frac{ \displaystyle \sum_{m=0}^{n-1} C_{n-1}^{m}\, \frac{(0.5-n)_m}{(1-2n)_m}\,(-y)^m }{ \displaystyle \sum_{m=0}^{n} C_{n}^{m}\, \frac{(0.5-n)_m}{(1-2n)_m}\,(-y)^m } +r_{2n}(y) = \frac{p_{2n}(y)}{q_{2n}(y)}+r_{2n}(y); \]

\[ n=1,2,\ldots \tag{2} \]

we shall prove a theorem.

Theorem. The even convergent fraction (2) is represented by the following sum of elementary fractions

\[ \frac{p_{2n}(y)}{q_{2n}(y)} = \frac{1}{n}\sum_{m=1}^{n} \frac{1}{1-y\sin^{2}\left(\frac{2\pi m-\pi}{4n}\right)}; \qquad n=1,2,\ldots \tag{3} \]

Proof. It is known [2] that the denominators of the convergents (2) are connected with Chebyshev polynomials of the first kind, namely:

\[ q_{2n}(y)=T_n(1-2\cdot y)=\cos\{n\arccos(1-2\cdot y)\}. \tag{4} \]

According to equality (4), it is not difficult to compute that

\[ q_{2n}(y)= \prod_{m=1}^{n} \left[ 1-y\sin^{2}\left(\frac{2\pi m-\pi}{4n}\right) \right]; \qquad n=1,2,\ldots \tag{5} \]

For the proof of identity (3), let

\[ \beta_m=-\sin^{2}\left(\frac{2\pi m-\pi}{4n}\right), \]

then

\[ \frac{1:n}{1+\beta_1 y}+\frac{1:n}{1+\beta_2 y}+\cdots+\frac{1:n}{1+\beta_n y} = \frac{\displaystyle\sum_{m=0}^{n-1} C_{n-1}^{m}\frac{(0.5-n)_m}{(1-2n)_m}(-y)^m} {\displaystyle\sum_{m=0}^{n} C_{n}^{m}\frac{(0.5-n)_m}{(1-2n)_m}(-y)^m} = \frac{p_{2n}(y)}{q_{2n}(y)} . \tag{6} \]

The first coefficient of the numerator \(p_{2n}(y)\) is equal to unity, and likewise the sum of the \(n\) terms, each of which is equal to \(1:n\), is equal to unity. In view of the equalities (5) and (6), the following equalities hold for the roots of the polynomial \(q_{2n}(y)\):

\[ - C_n^1\frac{(0.5-n)}{(1-2n)} = \beta_1+\cdots+\beta_n;\quad C_n^m(-1)^m\frac{(0.5-n)_m}{(1-2n)_m} = \]

\[ = \beta_1\cdots\beta_m+\cdots+\beta_{n-m+1}\cdots\beta_n;\quad m=2,\ldots,n-1. \tag{7} \]

On the basis of equalities (7) we compute the coefficients of \(y^m\) in the numerator \(p_{2n}(y)\); here in the left-hand side of the equality we have \(n C_{n-1}^{m}\) terms, and we obtain

\[ \frac{1}{n}\left(\beta_2\cdots\beta_{m+1}+\cdots+\beta_{n-m+1}\cdots\beta_n\right)+\cdots \]

\[ \cdots+\frac{1}{n}\left(\beta_1\cdots\beta_m+\cdots+\beta_{n-m}\cdots\beta_{n-1}\right) = \]

\[ = \frac{1}{n}\cdot\frac{n C_{n-1}^{m}}{C_n^m}\cdot C_n^m(-1)^m\frac{(0.5-n)_m}{(1-2n)_m} = C_{n-1}^{m}(-1)^m\frac{(0.5-n)_m}{(1-2n)_m} \]

\[ m=1,\ldots,n-1; \]

where the right-hand side of the equality is identically equal to the coefficient of the numerator (6) at \(y^m\); thus the theorem is completely proved.

In view of equalities (2) and (3) we have

\[ (1-y)^{-\frac12} = \frac{1}{n}\sum_{m=1}^{n} \frac{1}{1-y\sin^2\left(\frac{2\pi m-\pi}{4n}\right)} + r_{2n}(y); \quad n=1,2,\ldots \tag{8} \]

2. The elliptic integral of the first kind [3], p. 82 \((x=\sin\varphi,\ \kappa^2<1)\)

\[ F(\varphi,\kappa) = \int_{0}^{\varphi}\frac{d\varphi}{\sqrt{1-\kappa^2\sin^2\varphi}} = \int_{0}^{x}\frac{dx}{\sqrt{1-x^2}\sqrt{1-\kappa^2x^2}} . \tag{9} \]

After replacing the factor \((1-\kappa^2 x^2)^{-0.5}\) of the integrand by the partial fractions (8) and evaluating the resulting sum of integrals, it is approximately equal to the following sum of arctangents:

\[ F(\varphi,\kappa) = \frac{1}{n}\sum_{m=1}^{n} \frac{ \arctg\left[ \sqrt{\,1-\kappa^2\sin^2\left(\frac{2\pi m-\pi}{4n}\right)}\cdot \tg\varphi \right] }{ \sqrt{\,1-\kappa^2\sin^2\left(\frac{2\pi m-\pi}{4n}\right)} } + \]

\[ + \rho_{2n}(y); \quad \varphi=\arc\sin x,\quad \kappa^2<1,\quad n=1,2,\ldots; \tag{10} \]

where

\[ \rho_{2n}(y)=\int_{0}^{x}\frac{r_{2n}(y)\,dx}{\sqrt{1-x^{2}}}. \tag{11} \]

Elliptic integrals of the second kind

\[ E(\varphi,k)=\int_{0}^{x}\frac{\sqrt{1-k^{2}x^{2}}}{\sqrt{1-x^{2}}}\,dx,\qquad x=\sin\varphi \tag{12} \]

and of the third kind

\[ H(\varphi,h,k)=\int_{0}^{x}\frac{dx}{(1+hx^{2})\sqrt{(1-x^{2})(1-k^{2}x^{2})}},\qquad x=\sin\varphi \tag{13} \]

by analogy with integral (9) can be represented by the following sums of arctangents:

\[ \begin{aligned} E(\varphi,k)=2n\varphi &-\frac{1}{n}\sum_{m=1}^{n} \frac{\arctan(a_m\tan\varphi)}{a_m} \cot^{2}\left(\frac{2\pi m-\pi}{4n}\right) \\ &+R_{2n}(y),\qquad n=1,2,\ldots; \end{aligned} \tag{14} \]

\[ \begin{aligned} H(\varphi,h,k) &=\frac{p_{2n}(-k^{2}:h)\arctan(\sqrt{1+h}\tan\varphi)} {q_{2n}(-k^{2}:h)\sqrt{1+h}} \\ &\quad+\frac{1}{n}\sum_{m=1}^{n} \frac{(1-a_m^{2})\arctan(a_m\tan\varphi)} {(h+1-a_m^{2})a_m} +\sigma_{2n}(y), \\ &\qquad n=1,2,\ldots;\qquad h>-1, \end{aligned} \tag{15} \]

where

\[ a_m^{2}=1-k^{2}\sin^{2}\left(\frac{2\pi m-\pi}{4n}\right);\qquad R_{2n}(y)=\int_{0}^{x}\frac{(1-y)r_{2n}(y)\,dx}{\sqrt{1-x^{2}}}. \tag{16} \]

\[ \sigma_{2n}(y)=\int_{0}^{x}\frac{r_{2n}(y)\,dx}{(1+hx^{2})\sqrt{1-x^{2}}}. \tag{17} \]

The complete elliptic integrals, according to equalities (10), (14), and (15), are approximately equal to the following sums:

\[ F\left(\frac{\pi}{2},k\right) =\frac{\pi}{2n}\sum_{m=1}^{n}\frac{1}{a_m} +\rho_{2n}(k^{2}),\qquad n=1,2,\ldots. \tag{18} \]

\[ \begin{aligned} E\left(\frac{\pi}{2},k\right) &=\pi n-\frac{\pi}{2n}\sum_{m=1}^{n} \frac{1}{a_m}\cot^{2}\left(\frac{2\pi m-\pi}{4n}\right) +R_{2n}(k^{2}), \\ &\qquad n=1,2,\ldots; \end{aligned} \tag{19} \]

\[ \begin{aligned} H\left(\frac{\pi}{2},h,k\right) &=\frac{\pi p_{2n}(-k^{2}:h)} {2q_{2n}(-k^{2}:h)\sqrt{1+h}} +\frac{\pi}{2n}\sum_{m=1}^{n} \frac{(1-a_m^{2})a_m}{h+1-a_m^{2}} \\ &\quad+\sigma_{2n}(k^{2}),\qquad n=1,2,\ldots;\qquad h>-1. \end{aligned} \tag{20} \]

3. Inequalities \((q_n(y) \equiv q_n)\)

\[ q_{4n}>\left(1-y+\frac{y^2}{16}\right)^n,\qquad q_{4n+2}\geq \left(1-\frac{y}{2}\right)\left(1-y+\frac{y^2}{16}\right)^n,\quad y\leq 1 \tag{21} \]

are valid in view of equalities (2) and (5), namely:

\[ q_{4n}(y)=\prod_{m=1}^{n}\left[1-y+\frac{y^2}{4}\sin^2\left(\frac{2\pi m-\pi}{4n}\right)\right] =1-ny+\cdots+\frac{y^{2n}}{2^{4n-1}}> \]

\[ >\left(1-y+y^2:16\right)^n,\quad y\leq 1. \]

The second inequality (21) is proved analogously.

On the basis of inequalities (21), equality (2), and the infinite series for the remainder term \(r_{2n}(y)\) [4], p. 21, (1), it is not difficult to obtain estimates of the remainder terms (11), (16), and (17). The remainder terms in absolute value will be less than the following expressions \((n=1,2,\ldots;\ y<1,\ x=\sin\varphi)\):

\[ |\rho_{2n}(y)|< \frac{y^{2n}\operatorname{arctg}\left(\sqrt{1-\kappa^2}\operatorname{tg}\varphi\right)} {2^{4n-1}(1-y:2)(1-y+y^2:16)^{\,n-1}\sqrt{1-\kappa^2}}; \tag{22} \]

\[ |R_{2n}(y)|< \frac{y^{2n}\arcsin x} {2^{4n-1}(1-y:2)(1-y+y^2:16)^{\,n-1}}; \tag{23} \]

\[ \sigma_n(y)< \frac{y^{2n}2^{1-4n}} {(1-y:2)(1-y+y^2:16)^{\,n-1}} \left[ \frac{h\,\operatorname{arctg}\left(\sqrt{1+h}\operatorname{tg}\varphi\right)} {(h+\kappa^2)\sqrt{1+h}} + \right. \]

\[ \left. +\frac{\kappa^2\operatorname{arctg}\left(\sqrt{1-\kappa^2}\operatorname{tg}\varphi\right)} {(h+\kappa^2)\sqrt{1-\kappa^2}} \right], \quad h\ne -\kappa^2,\quad h>-1. \tag{24} \]

For \(n=1—6,\,8\), the polynomials \(q_{2n}(y)\equiv q_{2n}\) are not difficult to represent in the form of a product of quadratic trinomials, namely:

\[ \left\{ \begin{aligned} q_2&=1-\frac{y}{2};\qquad q_4=1-y+\frac{y^2}{8};\qquad q_6=\left(1-\frac{y}{2}\right)\left(1-y+\frac{y^2}{16}\right);\\[4pt] q_8&=\left(1-y+\frac{y^2}{16-8\sqrt{2}}\right) \left(1-y+\frac{y^2}{16+8\sqrt{2}}\right);\\[4pt] q_{10}&=(1-y:2)\left(1-y+\frac{y^2}{24-8\sqrt{5}}\right) \left(1-y+\frac{y^2}{24+8\sqrt{5}}\right);\\[4pt] q_{12}&=\left(1-y+\frac{y^2}{8}\right) \left(1-y+\frac{y^2}{32-16\sqrt{3}}\right) \left(1-y+\frac{y^2}{32+16\sqrt{3}}\right);\\[4pt] q_{16}&=\left[\left(1-y+\frac{y^2}{8}\right)^2-\frac{2-\sqrt{2}}{256}y^4\right] \left[\left(1-y+\frac{y^2}{8}\right)^2-\frac{2+\sqrt{2}}{256}y^4\right]. \end{aligned} \right. \tag{25} \]

According to equalities (25), the roots of the individual polynomials can be represented in the form of radicals.

4. Let integrals of the form be given

\[ \Phi(x,\kappa,\alpha)=\int_{0}^{x}\frac{(1-\kappa^2x^2)^\alpha}{\sqrt{1-x^2}}\,dx,\qquad -\frac{1}{2}\leq \alpha<\frac{1}{2}; \tag{26} \]

\[ H_1(x,h,\kappa,\alpha)= \int_{0}^{x}\frac{(1-y)^\alpha\,dx}{(1+hx^2)\sqrt{1-x^2}},\qquad h>-1, \]

\[ 0<x\leqslant 1,\qquad \kappa^{2}<1,\qquad -\frac{1}{2}\leqslant a<\frac{1}{2}, \tag{27} \]

and, in view of [1], the function

\[ (1-y)^{a} = \frac{ \displaystyle \sum_{m=0}^{n-1} C_{n-1}^{m}\frac{(-a-n)_{m}}{(1-2n)_{m}}(-y)^{m} }{ \displaystyle \sum_{m=0}^{n} C_{n}^{m}\frac{(a-n+1)_{m}}{(1-2n)_{m}}(-y)^{m} } +r_{2n}(y) = \frac{p_{2n}(y)}{q_{2n}(y)} + \]

\[ +r_{2n}(y) = \sum_{m=1}^{n}\frac{b_{m}}{1-y:a_{m}}+r_{2n}(y); \tag{28} \]

then, by analogy with the formulas for elliptic integrals, approximate formulas for the integrals (26) and (27) can be obtained:

\[ \Phi(x,\kappa,a) = \sum_{m=0}^{n} b_{m} \sqrt{\frac{a_{m}}{a_{m}-\kappa^{2}}} \arctg\left( \sqrt{\frac{a_{m}-\kappa^{2}}{a_{m}}}\cdot \tg\varphi \right) + \]

\[ +\rho_{2n}(y); \qquad n=1,2,\ldots;\qquad x=\sin\varphi, \tag{29} \]

\[ H_{1}(x,h,\kappa,a) = \frac{p_{2n}(-\kappa^{2}:h)}{q_{2n}(-\kappa^{2}:h)} \arctg\left(\sqrt{1+h}\cdot \tg\varphi\right) + \]

\[ + \sum_{m=1}^{n} \frac{\kappa^{2}b_{m}}{\kappa^{2}+ha_{m}} \sqrt{\frac{a_{m}}{a_{m}-\kappa^{2}}} \arctg\left( \sqrt{\frac{a_{m}-\kappa^{2}}{a_{m}}}\cdot \tg\varphi \right) + \]

\[ +\sigma_{2n}(y); \qquad n=1,2,\ldots;\qquad h>-1, \tag{30} \]

where, on the basis of equality (28),

\[ b_{m} = \frac{p_{2n}(a_{m})} {-a_{m}\dfrac{d}{dy}\bigl[q_{2n}(a_{m})\bigr]}, \qquad |\rho_{2n}(y)|<L_{2n}\arcsin x, \tag{31} \]

\[ |\sigma_{2n}(y)| < \frac{ L_{2n}\arctg\left(\sqrt{1+h}\cdot \tg\varphi\right) }{ \sqrt{1+h} }. \tag{32} \]

In the right-hand sides of inequalities (31) and (32), in view of [4], the factor \(L_{2n}\) is the following:

\[ L_{2n} = \frac{ 16(1-y)^{2}y^{2n}(1-a)_{n}(a)_{n}:(4-5y) }{ (4-3y)(n+1)_{n}(n)_{n}q_{2n}(y)q_{2n+2}(y) }; \qquad y<0.8. \tag{33} \]

References

1. V. E. Kornilov. A system of Padé approximations for power functions. Izv. TPI, vol. 154, 1967.

2. V. L. Danilov, A. N. Ivanova, et al. Mathematical Analysis (Functions, Limits, Series, Continued Fractions). Moscow, Fizmatgiz, 1961.

3. I. M. Ryzhik. Tables of Integrals, Sums, Series, and Products. Moscow–Leningrad, Gostekhizdat, 1948.

4. V. E. Kornilov. Application of continued fractions to the computation of integrals of binomial differentials. Izv. TPI, vol. 131, 1965.