Consider the function f(x) = sin(x). (a) Give the Taylor polynomial Ps(r) of degree 5 about...

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Consider the function f(x) = sin(x). (a) Give the Taylor polynomial Ps(r) 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-π/6 (x(t)=t+π/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...). Consider the function f(x) = sin(x). (a) Give the Taylor polynomial Ps(r) 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-π/6 (x(t)=t+π/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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e fx os 2 P5 x f76 x 146 x176 f11 146 2 d 65146 x4 sin 146 x12 6516 a x146 Gin ... View the full answer