Improved error bounds for floating-point products and Horner’s scheme

Abstract

Let $\mathbf{u}$ denote the relative rounding error of some floating-point format. Recently it has been shown that for a number of standard Wilkinson-type bounds the typical factors $\gamma _k:=k\mathbf{u}/(1-k\mathbf{u})$ can be improved into $k\mathbf{u}$, and that the bounds are valid without restriction on $k$. Problems include summation, dot products and thus matrix multiplication, residual bounds for $LU$- and Cholesky-decomposition, and triangular system solving by substitution. In this note we show a similar result for the product $\prod _{i=0}^k{x_i}$ of real and/or floating-point numbers $x_i$, for computation in any order, and for any base $\beta \geqslant 2$. The derived error bounds are valid under a mandatory restriction of $k$. Moreover, we prove a similar bound for Horner’s polynomial evaluation scheme.

Appendices

Appendix

The goal of this appendix is to prove that for $\beta =2$ and $p\geqslant 4$ the constraint $k < \mathbf{u}^{-1/2}$ in Theorem 1.2 cannot be replaced by $k < 12\mathbf{u}^{-1/2}$. To do that⁴ we construct $x_0,x_1,x_2\in \mathbb {F}$ for given precision $p$ such that $x_1x_2 < 1$ and ${\text {fl}}({\text {fl}}(x_0x_1)x_2) = x_0$. Subsequent multiplications by $x_1x_2$ produce an exponential growth of the rounding error, eventually exceeding $k\mathbf{u}$.

Define $s:=\lfloor \mathbf{u}^{-1/2}\rfloor \in \mathbb {N}$, so that $s=\mathbf{u}^{-1/2}-\delta$ with $0\leqslant \delta < 1$. We henceforth assume $p \geqslant 15$ and treat the case $p\leqslant 14$ later. Note that $\beta =2$ and $p \geqslant 15$ imply $s\geqslant 181$. We distinguish two cases.

First, assume $s$ is odd. Set

$$\begin{aligned} x_0:=1+(2s+8)\mathbf{u},\quad x_1:=1-(s-4)\mathbf{u}, \quad \hbox {and}\quad x_2:=1+(s-5)\mathbf{u}, \end{aligned}$$

so that $x_i\in \mathbb {F}$. Then, $x_0x_1=1+(s+10)\mathbf{u}+\mu _1\mathbf{u}$ with $\mu _1:=4\delta \sqrt{\mathbf{u}}+(32-2\delta ^2)\mathbf{u}$, so that $0<\mu _1<1$ and $s$ odd imply ${\text {fl}}(x_0x_1)=1+(s+11)\mathbf{u}$. Moreover, ${\text {fl}}(x_0x_1)x_2 = 1+(2s+7)\mathbf{u}+\mu _2\mathbf{u}$ with

$$\begin{aligned} \mu _2 := \sqrt{\mathbf{u}}\left( 6-55\sqrt{\mathbf{u}}+\Phi \delta \right) \quad \hbox {with }\Phi :=(\delta -6)\sqrt{\mathbf{u}}-2. \end{aligned}$$

Now $\Phi <0$ for any value of $\delta$, so that $0<4\sqrt{\mathbf{u}}-60\mathbf{u}\leqslant \mu _2\leqslant 6\sqrt{\mathbf{u}}-55\mathbf{u}<1$. Thus,

$$\begin{aligned} {\text {fl}}({\text {fl}}(x_0x_1)x_2) = x_0. \end{aligned}$$

(4.1)

Define a vector $X:=[x_0 \; x \; x \ldots x]\in \mathbb {F}^{2m+1}$ with $m$ times repeating the row vector $x = [x_1 \, x_2] \in \mathbb {F}^2$. Denoting ${\widehat{r}}_0:=x_0$ and ${\widehat{r}}_i:={\text {fl}}({\widehat{r}}_{i-1} X_i)$ for $i\geqslant 1$ yields ${\widehat{r}}_2=v_0$. Then, abbreviating $\pi :=x_1x_2$ and using ${\widehat{r}}_{2m}={\widehat{r}}_2=x_0$ gives

$$\begin{aligned} {\widehat{r}}_{2m}-\prod _{i=0}^{2m}{X_i} = x_0 - x_0\pi ^m = (\pi ^{-m}-1)\prod _{i=0}^{2m}{X_i}\quad \hbox {for }1\leqslant m\in \mathbb {N}. \end{aligned}$$

(4.2)

Now,

$$\begin{aligned} \pi = 1 - \left( 2 - (9+2\delta )\sqrt{\mathbf{u}}\right) \mathbf{u}- (20+9\delta +\delta ^2)\mathbf{u}^2 < 1 - \left( 2 - 11\sqrt{\mathbf{u}}\right) \mathbf{u}=: 1-\gamma \mathbf{u}, \end{aligned}$$

and for $m\in \mathbb {N}$,

$$\begin{aligned} \pi ^{-m} > 1+m\gamma \mathbf{u}+ \frac{m(m-1)}{2}\gamma ^2\mathbf{u}^2 = 1+2m\mathbf{u}+\frac{m\mathbf{u}\sqrt{\mathbf{u}}}{2}\left[ (m-1)\gamma ^2\sqrt{\mathbf{u}}-22\right] . \end{aligned}$$

The assumption $p\geqslant 15$ implies

$$\begin{aligned} (6-2\sqrt{\mathbf{u}})\gamma ^2-22 = 2 - 272 \sqrt{\mathbf{u}} + (814-242\sqrt{\mathbf{u}})\mathbf{u}> 2 - 272 \sqrt{\mathbf{u}} > 0, \end{aligned}$$

and therefore

$$\begin{aligned} m \geqslant 6 \mathbf{u}^{-1/2} - 1 \quad \Rightarrow \quad \pi ^{-m} > 1+2m\mathbf{u}. \end{aligned}$$

(4.3)

Combining this with (4.2) shows that the error bound in (1.6) is not satisfied for $k=2\left\lceil 6\mathbf{u}^{-1/2}-1\right\rceil < 12 \mathbf{u}^{-1/2}$, and that finishes the first part.

Second, assume $s$ is even and define as before

$$\begin{aligned} y_0:=1+(2s+6)\mathbf{u},\quad y_1:=1-(s-3)\mathbf{u}, \quad \hbox {and}\quad y_2:=1+(s-4)\mathbf{u}. \end{aligned}$$

(4.4)

Then, $y_i\in \mathbb {F}$. Furthermore, $y_0y_1=1+(s+7)\mathbf{u}+\mu _1\mathbf{u}$ with $\mu _1:=4\delta \sqrt{\mathbf{u}}+(18-2\delta ^2)\mathbf{u}$, so that $0<\mu _1<1$ and $s$ even imply ${\text {fl}}(y_0y_1)=1+(s+8)\mathbf{u}$. Moreover, ${\text {fl}}(y_0y_1)y_2 = 1+(2s+5)\mathbf{u}+\mu _2\mathbf{u}$ with

$$\begin{aligned} \mu _2 := \sqrt{\mathbf{u}}\left( 4-32\sqrt{\mathbf{u}}+\Phi \delta \right) \quad \hbox {with }\Phi :=(\delta -4)\sqrt{\mathbf{u}}-2. \end{aligned}$$

As before, $\Phi <0$ for any value of $\delta$. Thus, $0<2\sqrt{\mathbf{u}}-35\mathbf{u}\leqslant \mu _2\leqslant 4\sqrt{\mathbf{u}}-32\mathbf{u}<1$. Hence, similar to (4.1), ${\text {fl}}({\text {fl}}(y_0y_1)y_2) = y_0$ is again true. Now for the values $y_1, y_2$ in (4.4) we obtain

$$\begin{aligned} y_1y_2 = (1-(s-3)\mathbf{u})(1+(s-4)\mathbf{u}) < x_1x_2, \end{aligned}$$

and the result follows as before. Finally, for the cases $4\leqslant p\leqslant 14$, consider

p	$m_0$	$m_1$	$m_2$	F
4	2	$-$4	4	9.6
5	20	$-$3	2	8.9
6	32	$-$14	16	5.8
7	28	$-$9	8	6.8
8	52	$-$39	44	5.8
9	48	$-$21	20	4.6
10	140	$-$117	130	5.2
11	94	$-$43	42	5.8
12	186	$-$154	158	4.0
13	184	$-$89	88	4.1
14	262	$-$125	124	7.2

For precision $p$ define $x_i:=1+m_i\mathbf{u}$. Then, (4.1) is satisfied, and the error bound in (1.6) is not true for $k<F\mathbf{u}^{-1/2}$. This finishes the proof. $\square$

We finally mention that it is easy to see that, if $1\leqslant p\leqslant 2$, then the error bound in (1.6) is satisfied for all $k\in \mathbb {N}$, and if $p=3$, then the minimum value of $k$ for which it is not satisfied is $k=72\approx 25\mathbf{u}^{-1/2}$.

Summary

In previous papers, the factor $\gamma _k$ has been replaced by $k\mathbf{u}$ in a number of classical error estimates in numerical analysis together with removing the restriction on $k$. We proved that $k\mathbf{u}$ can be used for general products and for the Horner scheme, however, with a mandatory restriction on $k$ of the order $k\lesssim \mathbf{u}^{-1/2}$.