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Unformatted text preview: x) − 20 cos(x)4 esin(x) − 15 sin(x)3 esin(x) + 45 sin(x)2 cos(x)2 esin(x) − 15 sin(x) cos(x)4 esin(x) + cos(x)6 esin(x) )(x − π )6 Again, find where the derivative is zero. > B1 := diff( B, x ); B1 := 1 (−cos(x) esin(x) − 63 sin(x) cos(x) esin(x) 720 + 91 cos(x)3 esin(x) − 210 sin(x)2 cos(x) esin(x) + 245 sin(x) cos(x)3 esin(x) − 35 cos(x)5 esin(x) − 105 sin(x)3 cos(x) esin(x) + 105 sin(x)2 cos(x)3 esin(x) − 21 sin(x) cos(x)5 esin(x) + cos(x)7 esin(x) )(x − π )6 + −sin(x) esin(x) + 16 cos(x)2 esin(x) − 15 sin(x)2 esin(x) + 75 sin(x) cos(x)2 esin(x) − 20 cos(x)4 esin(x) − 15 sin(x)3 esin(x) + 45 sin(x)2 cos(x)2 esin(x) − 15 sin(x) cos(x)4 esin(x) + cos(x)6 esin(x) )(x − π )5 > sol := { solve( B1=0, x ) }; 1 ( 120 sol := {π } Check the solution by plotting. 7.1 Introductory Calculus > plot( B1, x=2..4 ); • 229 0.8 0.6 0.4 0.2 0 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 x The plot of B 1 indicates that a solution between 2.1 and 2.3 exists. The solve command cannot find that solution. Use numerical methods again. > fsolve( B1=0, x, 2.1..2.3 ); 2.180293062 Add the numerical solution to the set of symbolic solutions. > sol := sol union { % }; sol := {2.180293062, π } The following is the set of extremal errors at the ξ = x boundary. > { seq( B, x=sol ) }; {0, 0.04005698601 (2.180293062 − π )6 } Enlarge the set of large errors. > critical := critical union %; critical := {0, 0.07333000221 (2 − π )6 , −0.01542298119 (4 − π )6 , 0.04005698601 (2.180293062 − π )6 } Add the error at the four corners to the set of critical values. 230 • Chapter 7: Solving Calculus Problems > critical := critical union > { eval( err, {xi=2, x=2} > eval( err, {xi=2, x=4} > eval( err, {xi=4, x=2} > eval( err, {xi=4, x=4} ), ), ), ) }; critical := {0, 0.07333000221 (2 − π )6 , 1 (−sin(4) esin(4) 720 + 16 cos(4)2 esin(4) − 15 sin(4)2 esin(4) + 75 sin(4) cos(4)2 esin(4) − 20 cos(4)4 esin(4) − 15 sin(4)3 esin(4) + 45 sin(4)2 cos(4)2 esin(4) − 15 sin(4) cos(4)4 esin(4) 1 + cos(4)6 esin(4) )(2 − π )6 , (−sin(2) esin(2) 720 + 16 cos(2)2 esin(2) − 15 sin(2)2 esin(2) + 75 sin(2) cos(2)2 esin(2) − 20 cos(2)4 esin(2) − 15 sin(2)3 esin(2) + 45 sin(2)2 cos(2)2 esin(2) − 15 sin(2) cos(2)4 esin(2) + cos(2)6 esin(2) )(4 − π )6 , −0.01542298119 (4 − π )6 , 1 0.04005698601 (2.180293062 − π )6 , (−sin(4) esin(4) 720 + 16 cos(4)2 esin(4) − 15 sin(4)2 esin(4) + 75 sin(4) cos(4)2 esin(4) − 20 cos(4)4 esin(4) − 15 sin(4)3 esin(4) + 45 sin(4)2 cos(4)2 esin(4) − 15 sin(4) cos(4)4 esin(4) 1 (−sin(2) esin(2) + cos(4)6 esin(4) )(4 − π )6 , 720 + 16 cos(2)2 esin(2) − 15 sin(2)2 esin(2) + 75 sin(2) cos(2)2 esin(2) − 20 cos(2)4 esin(2) − 15 sin(2)3 esin(2) + 45 sin(2)2 cos(2)2 esin(2) − 15 sin(2) cos(2)4 esin(2) + cos(2)6 esin(2) )(2 − π )6 } Find the maximum of the absolute values of the elements of critical. First, map the abs command onto the elements of critical. > map( abs, critical ); 7.1 Introductory Calculus • 231 {0, 0.07333000221 (2 − π )6...
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This note was uploaded on 08/27/2012 for the course MATH 1100 taught by Professor Nil during the Spring '12 term at National University of Singapore.

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