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Example text

We now have the following error bound. 6 |x − fl(x)| ≤ 1 |x|β 1−t p, 2 where p = 1 for rounding and p = 2 for chopping. α1 α2 · · · αt · · · )β m , we have β m−1 ≤ |x| ≤ β m . In the interval [β m−1 , β m ], the floating point numbers are evenly spaced with separation β m−t . Thus, for chopping, |x − fl (x)| ≤ β m−t = p m−t β , 2 and for rounding, |x − fl(x)| ≤ p 1 m−t β = β m−t . 5 δ= p m−t p 1 β ≤ β 1−t β m−1 ≤ |x|β 1−t p. 2 2 2 p 1−t β is called the unit roundoff error. 6 Let = 13 fl (x) − x . Then fl (x) = (1 + )x, where | | ≤ δ.

16. Assume that x∗ and y ∗ are approximations to x and y with relative errors rx and ry , respectively, and that |rx |, |ry | < R. Assume further that x = y. How small must R be in order to ensure that x∗ = y ∗ ? 17. Let x∗ and y ∗ be the floating point representations of x and y, respectively. Let f (x∗ , y ∗ ) be the approximate value of f (x, y). Derive the relation between the relative error in evaluating the function f (x, y) in terms of the relative error in evaluating x and the relative error in evaluating y.

Consider fl(x1 + x2 ) = fl[fl(x1 ) + fl (x2 )] = fl[x1 (1 + ˆ1 ) + x2 (1 + ˆ2 )] = [x1 (1 + ˆ1 ) + x2 (1 + ˆ2 )] (1 + 1 ), where |ˆ1 | ≤ δ, | ˆ2 | ≤ δ, and | 1 | ≤ δ. Similarly, fl(x1 + x2 + x3 ) = ((x1 (1 + ˆ1 ) + x2 (1 + ˆ2 ))(1 + 1 ) + x3 (1 + ˆ3 ))(1 + 2 ) = x1 (1 + ˆ1 )(1 + 1 )(1 + 2 ) + x2 (1 + ˆ2 )(1 + 1 )(1 + 2 ) + x3 (1 + ˆ3 )(1 + 2) 14 Classical and Modern Numerical Analysis Continuing this procedure, n n−1 xi fl = x1 (1 + ˆ1 ) i=1 (1 + i ) i=1 n−1 + x2 (1 + ˆ2 ) n−1 (1 + i ) + x3 (1 + ˆ3 ) i=1 (1 + i ) i=2 n−1 (1 + i ) + · · · + xn (1 + ˆn )(1 + + x4 (1 + ˆ4 ) n−1 ).

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