Learning Guide Learning Guide

6.1 Introductory Calculus • 215err_x := 1120 (−sin(ξ) esin(ξ) + 16 cos(ξ) 2 e sin(ξ)− 15 sin(ξ) 2 e sin(ξ) + 75 sin(ξ) cos(ξ) 2 e sin(ξ) − 20 cos(ξ) 4 e sin(ξ)− 15 sin(ξ) 3 e sin(ξ) + 45 sin(ξ) 2 cos(ξ) 2 e sin(ξ)− 15 sin(ξ) cos(ξ) 4 e sin(ξ) + cos(ξ) 6 e sin(ξ) )(x − π) 5> err_xi := diff(err, xi);err_xi := 1720 (−cos(ξ) esin(ξ) − 63 sin(ξ) cos(ξ) e sin(ξ)+ 91 cos(ξ) 3 e sin(ξ) − 210 sin(ξ) 2 cos(ξ) e sin(ξ)+ 245 sin(ξ) cos(ξ) 3 e sin(ξ) − 35 cos(ξ) 5 e sin(ξ)− 105 sin(ξ) 3 cos(ξ) e sin(ξ) + 105 sin(ξ) 2 cos(ξ) 3 e sin(ξ)− 21 sin(ξ) cos(ξ) 5 e sin(ξ) + cos(ξ) 7 e sin(ξ) )(x − π) 6The two partial derivatives are zero at a critical point.> sol := solve( {err_x=0, err_xi=0}, {x, xi} );sol := {x = π, ξ = ξ}The error is zero at this critical point.> eval( err, sol );0You need to collect a set of critical values. The largest critical valuethen bounds the maximal error.> critical := { % };critical := {0}The partial derivative err_xi is zero at a critical point on either ofthe two boundaries at x = 2 and x = 4.> sol := { solve( err_xi=0, xi ) };