Consider the function f(x) = sin(x). (a) Give the Taylor polynomial Ps(x) of degree 5 about a = π/6 of this function. Note: use the exact values of sin(7/6)= 1/2 and cos(7/6)= √3/2. (b) Give the nested representation of the polynomial Q5(t) = P₁(x(t)) where t :=x-7/6 (x(t)=t+π/6). (c) Using the nested multiplication method (also called Horner's algorithm), compute the approximation P(0.6) to sin(0.6) (give at least 15 significant digits of P3 (0.6))

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Consider the function f(x) = sin(x).
(a) Give the Taylor polynomial P(x) of degree 5 about a = π/6 of this function. Note: use the exact
values of sin(7/6)= 1/2 and cos(π/6)= √3/2.
(b) Give the nested representation of the polynomial Q5 (t) = P5(r(t)) where t :=x-/6 (x(t)=t+n/6).
(c) Using the nested multiplication method (also called Horner's algorithm), compute the approximation
P5(0.6) to sin(0.6) (give at least 15 significant digits of P5(0.6))
(d) Without using the exact value of sin(0.6), obtain an upper bound on the absolute error sin(0.6) -
P5(0.6) given by formula (3.6) of the class notes. Compare this bound to the exact absolute error
| sin(0.6) P5(0.6) (this time by using the exact value of sin(0.6)≈ 0.564642473395035...).
Transcribed Image Text:Consider the function f(x) = sin(x). (a) Give the Taylor polynomial P(x) of degree 5 about a = π/6 of this function. Note: use the exact values of sin(7/6)= 1/2 and cos(π/6)= √3/2. (b) Give the nested representation of the polynomial Q5 (t) = P5(r(t)) where t :=x-/6 (x(t)=t+n/6). (c) Using the nested multiplication method (also called Horner's algorithm), compute the approximation P5(0.6) to sin(0.6) (give at least 15 significant digits of P5(0.6)) (d) Without using the exact value of sin(0.6), obtain an upper bound on the absolute error sin(0.6) - P5(0.6) given by formula (3.6) of the class notes. Compare this bound to the exact absolute error | sin(0.6) P5(0.6) (this time by using the exact value of sin(0.6)≈ 0.564642473395035...).
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