Answer Happy

Floating-point arithmetic: IEEE starndard A floating-point number with a base b and length n is definied as x=(.d₁d₂...dn) b. be where m≤e≤ M is the exponent and (.d₁d₂...dn) is the mantissa. A 3 floating-point number is normalized if d₁ ‡0. IEEE: k = 2, 64 bits; 1 for the sign, 11 for the exponent and 52 for the mantissa. "double precision" All floating-point systems have a machine epsilon that is the smallest defined number after zero. IEEE-standard uses subnormal numbers that fill (one way or another) the underflow gap [0, €] IEEE: Double precision Exponent Number Type 0....0 ±(0.b₁ b2...b52)2-2-1022 +(1.b1 b2...b52)2-2-1022 0 or subnormal Normalized 0....01 = 110 Note! Exponent = "real" + 1023 01....1=102310 -2⁰ 11....10204610 -21023 11....1 too, if b;= 0, otherwise NaN Exception Exceptions: too, NaN Overflow } ⇒ value depends on the chosen rounding method Underflow
f (h) - f (0) Suppose we want to approximate f'(0) by where f(t) = et h The following inequalities hold: <e² <1+2 +27 -17 -2n-1 +2-3n-2 and 1+2-1+2-2n-3 <1+2¯n-1 +2-2n-2 h Let x = e¹, y = 1 and h = 2-n 2 a.) Assuming the IEEE standard, estimate the accuracy of the difference approximation for different values of n (f(x) — y_ f (x − 3)), and f(n (1 + 7/7) +ª ( 7² ) ) - y) h b.) Tabulate the values of fl fl fl fl for h h 2 6 n≥ 25 . Comment on the accuracy of the series expansion. 1+2+2-2n-1 1 2-n e²