Respuesta :
Answer:
a) y = x³− 5x² + 6x + 0.55 at x = 1.37.
Use 3-digit arithmetic with chopping. Evaluate the percent relative round-off error.
-1.183%
b. Express y as y = ((x − 5)x + 6)x + 0.55 (this is the same equation). Use again 3-digit arithmetic with chopping. Evaluate the percent relative round-off error and compare with part (a). Make the conclusion about which form of the polynomial is superior.
-0.161%
Comparing part a and b together, part b is more superior because the percent(%) error is smaller when compared to part a
Step-by-step explanation:
a) y = x³− 5x² + 6x + 0.55 at x = 1.37.
Use 3-digit arithmetic with chopping. Evaluate the percent relative round-off error.
Let's evaluate before applying the 3 digit arithmetic chopping rule
y = 1.37³ - 5 × 1.37² + 6 × 1.37 + 0.55
y = 1.956853
Let evaluate each components of the polynomial one by one
Note that: 3-digit arithmetic chopping means to approximate chop off or remove number after the 3 significant figures.
y = x³− 5x² + 6x + 0.55 at x = 1.37.
x³ = 1.37³ = 2.571353
≈ 2.57
x² = 1.37² = 1.8769
≈ 1.88
5x² = 1.87 × 5
= 9.35
x = 1.37
6x = 1.37 × 6
6x = 8.22
Evaluating the polynomial
y = 2.57 - 9.38 + 8.22 + 0.55
y = 1.98
The percent relative round-off error =
1.956853 - 1.98/1.956853 × 100
= -1.183%
b. Express y as y = ((x − 5)x + 6)x + 0.55 (this is the same equation). Use again 3-digit arithmetic with chopping. Evaluate the percent relative round-off error and compare with part (a). Make the conclusion about which form of the polynomial is superior.
y = ((x − 5)x + 6)x + 0.55
Evaluating with the 3 digit chop off rule is applied
= ((1.37 - 5)1.37 + 6)1.37 + 0.55
=( 1.8769 - 6.85) + 6) 1.37 + 0.55
= (- 4.9731 + 6 )1.37 + 0.55
= 1.0269 × 1.37 + 0.55
= 1.406853
= 1.956853.
≈ 1.96
Note in: evaluating before applying the 3 digit arithmetic chopping rule
y = 1.37³ - 5 × 1.37² + 6 × 1.37 + 0.55
y = 1.956853
The percent relative round-off error
1.956853 - 1.96/1.956853 × 100
= -0.161%
Comparing part a and b together, part b is more superior because the percent(%) error is smaller when compared to part a
