Improved Error Bounds

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Improved Error Bounds

The standard error analysis just outlined has a drawback: by using the infinity norm $\Vert \delta \Vert _{\infty}$ to measure the backward error, entries of equal magnitude in $\delta$ contribute equally to the final error bound $\kappa (f,z) (\Vert \delta \Vert/\Vert z\Vert)$ . This means that if z is sparse or has some very tiny entries, a normwise backward stable eroot00000000000000 Standard Error Analysis
Next: Improved Error Bounds Up: Further Details: How Error Previous: Further Details: How Error Contents Index

Standard Error Analysis

We illustrate standard error analysis with the simple example of evaluating the scalar function y=f(z). Let the output of the subroutine which implements f(z) be denoted ${\rm alg}(z)$ ; this includes the effects of roundoff. If ${\rm alg}(z) = f(z+\delta)$ where $\delta$ is small, then we say ${\rm alg}$ is a backward stable algorithm for f, or that the backward error $\delta$ is small. In other words, ${\rm alg}(z)$ is the exact value of f at a slightly perturbed input $z+\delta$ .^4.5

Suppose now that f is a smooth function, so that we may approximate it near z by a straight line: $f(z+\delta) \approx f(z) + f'(z) \cdot \delta$ . Then we have the simple error estimate

$\begin{displaymath} {\rm alg}(z)-f(z) = f(z+\delta) - f(z) \approx f'(z) \cdot \delta . \end{displaymath}$

Thus, if $\delta$ is small, and the derivative f'(z) is moderate, the error ${\rm alg}(z)-f(z)$ will be small