Question

We are the four computed values of f, IEEE floating point arithmetic is used? You can assume that exp(x) and exp(-x) are calculated exactly and rounded to the nearest floating point number

The following lines of MATLAB code evaluate the function 1.5 1016 in four different ways (a) x=1.5e-16; f=0.5*((exp (x)-exp (-x))/x) (b) x -1.5e-16; f - 0.5*((exp(x)-exp(x))/((xt1)-1)) (c) x - 1.5e-16; 0.5* ((exp(x) -exp(-x))/((x-1)+1)) (d) x - 1.5e-16; fexp(x)-exp(-x))/((x+1)+(x-1)) What are the four computed values of f, if IEEE floating point arithmetic is used? You can assume that exp(x) and exp(-x) are calculated exactly and then rounded to the nearest floating-point number. Hint: From Taylor's theorem, 3! for some £ between 0 and r Windo

Answer 1

clc;
clear all;close all;

x=1.5*10^-16;

f1=0.5*((exp(x)-exp(-x))/x)

f2=0.5*((exp(x)-exp(-x))/((x+1)-1))

f3=0.5*(exp(x)-exp(-x))/((x-1)+1)

f4=(exp(x)-exp(-x))/((x-1)+(x+1))

Note all four values are different

New to MATLAB? See resources for Getting Started 1c: clear all:close all: 2 3 4 1.1102 fl-0.5* ((exp (x)-exp (-x))/x) £2-0.5* ((exp (x)-exp (-x))/((x+1) f3-0.5 (exp (x) -exp (-x))/((x-1 4- (exp (x)-exp (-x))/((x-1) 7 f2 = 0.7500 10- 12 13 1.5000 1