Convergence properties of a quadrature formula of Clenshaw–Curtis type for the Gegenbauer weight function

Abstract

A quadrature formula of Clenshaw–Curtis type for functions of the form $(1-x^2)^{\lambda - \frac{1}{2}}f(x)$ over the interval [$-$1,1] exhibits a curious phenomenon when applied to certain analytic functions. As the number of points in the quadrature rule increases the error may sometimes decay to zero in two distinct stages rather than in one depending on the value of $\lambda$. In this paper we shall derive explicit and asymptotic error formulae which describe this phenomenon.

Appendices

Appendices

(A1) For any integer $p=0,1,2,\ldots$

• (a) By inspection

  $$\begin{aligned} \frac{B\prod _{i=0}^{2p}(1-2p+2i)}{2^{2p}(2p)!} = L. \end{aligned}$$

• (b) From (3.3) for $r\ge 2$

  $$\begin{aligned} Z_{r}\left( \frac{2p+1}{2}\right) = \prod _{i=0}^{2p}(1-2p+2i)\; \prod _{j=0}^{2p}\frac{1}{2r-2p+2j-1}\nonumber \\ = \frac{1}{2^{2p}\,(2p)!}\prod _{i=0}^{2p}(1-2p+2i)\; \sum _{j=0}^{2p}\frac{(-1)^{j}\,C_{j}^{2p}}{2r-2p+2j-1}. \end{aligned}$$

  (7.1)

  Bearing in mind part (a) we see that from (7.1)

  $$\begin{aligned} BZ_{r}\left( \frac{2p+1}{2}\right) = L\,\sum _{j=0}^{2p} \frac{(-1)^{j}\ C_{j}^{2p}}{2r-2p+2j-1}. \end{aligned}$$

(A2)

• (a) If we reverse the order of the inner summation followed by the order of the outer summation in the following we see that

  $$\begin{aligned}&\sum _{j=p+1}^{2p} \left( (-1)^{j}\ C_{j}^{2p}\sum _{k=1}^{j-p} \frac{\zeta ^{2k-1}}{n+(2j-2p-2k+1)} \right) \nonumber \\&\quad = \sum _{j=1}^{p} \left( (-1)^{p-j}\ C_{p-j}^{2p}\,\sum _{k=1}^{j} \frac{\zeta ^{2j+1-2k}}{n+(2k-1)} \right) \nonumber \\&\quad = \sum _{j=0}^{p-1} \left( (-1)^{j} C_{j}^{2p}\sum _{k=1}^{p-j} \frac{\zeta ^{2p-2j-2k+1}}{n+(2k-1)}\right) . \end{aligned}$$

  (7.2)

• (b) By substituting $n=0$ into (7.2), rearranging the terms on the left-hand side and finally substituting $1/\zeta$ for $\zeta$ into the resulting expression it follows that

  $$\begin{aligned}&\sum _{j=p+1}^{2p} \left( (-1)^{j}\ C_{j}^{2p}\zeta ^{-2(j-p)}\sum _{k=1}^{j-p} \frac{\zeta ^{2k-1}}{2k-1} \right) \nonumber \\&\quad = \sum _{j=0}^{p-1} \left( (-1)^{j}\ C_{j}^{2p}\zeta ^{-2(p-j)}\sum _{k=1}^{p-j} \frac{\zeta ^{2k-1}}{2k-1} \right) . \end{aligned}$$

  (A3)

  $$\begin{aligned}&\sum _{j=0}^{p-1} \left( (-1)^{j}\ C_{j}^{2p}\;\zeta ^{2p-2j}\, \left( \sum _{k=1}^{p-j} \frac{(2k-1)\zeta ^{-(2k-1)}}{n^{2}-(2k-1)^{2}} \right) \right) \nonumber \\&\quad = \sum _{j=0}^{p-1}\left( \left( \sum _{k=0}^{j} (-1)^{k}\ C_{k}^{2p} \;\zeta ^{2j-(2k-1)}\right) \frac{2p-2j-1}{n^{2}-(2p-2j-1)^{2}}\right) . \end{aligned}$$

  (7.3)

Proof

The right-hand side of (7.3) follows by rewriting the left-hand side, which is a summation in terms of $C_{r}^{2p} \;\zeta ^{2p-2r}$, as a summation in terms of $(2p-2r-1)/(n^{2}-(2p-2r-1)^{2})$.

(A4)

$$\begin{aligned}&\sum _{j=0}^{p-1} \left( (-1)^{j}\; C_{j}^{2p}\sum _{k=1}^{p-j} \frac{\zeta ^{-(2k-1)}}{n-(2p-2j-2k+1)} \right) \\&\quad = \sum _{j=0}^{p-1} \left( (-1)^{j}\ C_{j}^{2p}\sum _{k=1}^{p-j} \frac{\zeta ^{2j-2p+2k-1}}{n-(2k-1)} \right) \end{aligned}$$

$\square$

Proof

The inner summation on the left-hand side is of the form

$$\begin{aligned} \sum _{k=1}^{M_{j}} \frac{\zeta ^{-(2k-1)}}{n-(2M_{j}-(2k-1))}, \end{aligned}$$

(7.4)

and on the right-hand side of the form

$$\begin{aligned} \sum _{k=1}^{M_{j}} \frac{\zeta ^{-(2M_{j}-(2k-1))}}{n-(2k-1)}, \end{aligned}$$

(7.5)

where $M_{j}=p-j$. For j fixed (7.4) and  (7.5) are equivalent, hence the result follows. $\square$

(A5) Assume $z \in \varepsilon _{\rho }$ then

$$\begin{aligned} 2z\left( \;\sum _{s=0}^{r-1}\ (-1)^{s}\left( \zeta ^{2(r-s)} + \frac{1}{\zeta ^{2(r-s)}}\right) + (-1)^{r}\;\right) = \zeta ^{2r+1}+\frac{1}{\zeta ^{2r+1}}. \end{aligned}$$

(7.6)

Proof

Since $2z=\zeta +1/\zeta$ (7.6) follows immediately.

$$\begin{aligned} \mathbf {(A6)} \quad \sum _{k=0}^{j}\;(2j-2k+1) (-1)^{k} C_{k}^{2p} = (-1)^{j}\left( C_{j}^{2p-2} \;-\; C_{j-1}^{2p-2}\right) \quad 2p-2 \ge j. \end{aligned}$$

Proof

Using mathematical induction it is easy to deduce that

$$\begin{aligned} \sum _{k=0}^{r}\; (-1)^{k}\ C_{k}^{2p} \ =\ (-1)^{r}\ C_{r}^{2p-1} \qquad 2p \ge r+1, \end{aligned}$$

(7.7)

and by the factorial definition of $\ C_{r}^{n}$, that

$$\begin{aligned} C_{r-1}^{2p-2} +\ C_{r+1}^{2p}-2\;C_{r}^{2p-1} = C_{r+1}^{2p-2} 2p \ge r+3. \end{aligned}$$

(7.8)

If we now bear in mind (7.7) and  (7.8) then, once again, by using mathematical induction the result follows. $\square$

(A7)

$$\begin{aligned}&\sum _{j=0}^{p-1}(2p-1-2j)\left( \sum _{k=0}^{j}\;(-1)^{k}\;C_{k}^{2p}\delta _{2j+1-2k} \right) \\&\quad =\left( \zeta -\frac{1}{\zeta }\right) ^{2p-2}\left( \zeta +\frac{1}{\zeta }\right) \ p=1,2,\ldots \end{aligned}$$

Proof

By rearranging the terms on the left-hand side and bearing in mind A6 the right-hand side follows. $\square$