By Jean-Michel Muller
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This requires n multiplications and n additions if we use conventional operations, or n fused multiply–add operations. See Chapter 5 for more information. 1 Basic Notions of Floating-Point Arithmetic 15 Hence, if the floating-point computation of x 2 − y 2 is implemented as RN(RN(x 2 ) − y × y), then the obtained result will be less than 0 and computing its square root will generate a NaN, whereas the exact result is 0. The problem does not occur if we do not use the FMA operation: the rounding functions are monotonic, so that if |x| ≥ |y| then the computed value of x 2 will be larger than or equal to the computed value of y 2 .
2) Ti , Ti • compute p∗ = n ai Ti . i=0 The proof is rather obvious and can be found in most textbooks on numerical analysis [202]. Some sequences of orthogonal polynomials, associated with simple weight functions w, are well known, so there is no need to compute them again. Let us now present some of them. More information on orthogonal polynomials can be found in [1, 203]. 1 Legendre Polynomials • weight function: w(x) = 1; • interval [a, b] = [−1, 1]; • definition: ⎧ T (x) = 1 ⎪ ⎨ 0 T1 (x) = x ⎪ ⎩ T (x) = 2n − 1 x T (x) − n − 1 T (x); n n−1 n−2 n n • values of the scalar products: Ti , T j = ⎧ ⎨0 if i = j 2 ⎩ otherwise.
Algorithm 11 performs the addition of two n-digit numbers x = xn−1 xn−2 · · · x0 and y = yn−1 yn−2 · · · y0 represented in radix r with digits between −a and a, where a ≤ r − 1 and13 2a ≥ r + 1. Algorithm 11 Avizienis’ algorithm Input : x = xn−1 xn−2 · · · x0 and y = yn−1 yn−2 · · · y0 Output : s = sn sn−1 sn−2 · · · s0 = x + y 1. in parallel, for i = 0, . . , n − 1, compute ti+1 (carry) and wi (intermediate sum) satisfying: ⎧ ⎧ ⎪ ⎪ ⎨ +1 if xi + yi ≥ a ⎪ ⎪ ⎨t = 0 if −a + 1 ≤ xi + yi ≤ a − 1 i+1 ⎪ ⎩ −1 if x + y ≤ −a ⎪ i i ⎪ ⎪ ⎩ wi = xi + yi − r × ti+1 .