ee231_ass3_sol

# ee231_ass3_sol - EE 231_B1 Numerical Analysis for...

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EE 231_B1 Numerical Analysis for Electrical and Computer Engineers Course Instructor: Dr. Sergiy Vorobyov - 1 - Solutions to Assignment 3 4.3 function ep = macheps % determines the machine epsilon e = 1; while (1) if e+1<=1, break, end e = e/2; end ep = 2*e; >> macheps ans = 2.2204e-016 >> eps ans = 2.2204e-016 4.5 Because the exponent ranges from –126 to 127, the smallest positive number can be represented in binary as where the 23 bits in the mantissa are all 0. This value can be translated into a base-10 value of 2 –126 = 1.1755×10 –38 . The largest positive number can be represented in binary as where the 23 bits in the mantissa are all 1. Since the significand is approximately 2 (it is actually 2 – 2 –23 ), the largest value is therefore 2 +128 = 3.4028×10 38 . The machine epsilon in single precision would be 2 –23 = 1.921×10 –7 . These results can be verified using built-in MATLAB functions, >> realmax('single') ans = 3.4028e+038 >> realmin('single') ans = 1.1755e-038 >> eps('single') ans = 1.1921e-007

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## ee231_ass3_sol - EE 231_B1 Numerical Analysis for...

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