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Unformatted text preview: ) − 35 cos(ξ )5 esin(ξ) − 105 sin(ξ )3 cos(ξ ) esin(ξ) + 105 sin(ξ )2 cos(ξ )3 esin(ξ) − 21 sin(ξ ) cos(ξ )5 esin(ξ) + cos(ξ )7 esin(ξ) )(x − π )6 The two partial derivatives are zero at a critical point. 7.1 Introductory Calculus > sol := solve( {err_x=0, err_xi=0}, {x, xi} ); • 225 sol := {ξ = ξ, x = π } The error is zero at this critical point. > eval( err, sol ); 0 Collect a set of critical values. The largest critical value then bounds the maximal error. > critical := { % }; critical := {0} The partial derivative err_xi is zero at a critical point on either of the two boundaries at x = 2 and x = 4. > sol := { solve( err_xi=0, xi ) }; sol := {arctan(RootOf(%1, index = 2), RootOf(_Z 2 + RootOf(%1, index = 2)2 − 1)), arctan( RootOf(%1, index = 5), RootOf(_Z 2 + RootOf(%1, index = 5)2 − 1)), arctan( RootOf(%1, index = 6), RootOf(_Z 2 + RootOf(%1, index = 6)2 − 1)), arctan( RootOf(%1, index = 1), RootOf(_Z 2 + RootOf(%1, index = 1)2 − 1)), arctan( RootOf(%1, index = 4), RootOf(_Z 2 + RootOf(%1, index = 4)2 − 1)), arctan( RootOf(%1, index = 3), 1 RootOf(_Z 2 + RootOf(%1, index = 3)2 − 1)), π } 2 2 3 %1 := −56 − 161 _Z + 129 _Z + 308 _Z + 137 _Z 4 + 21 _Z 5 + _Z 6 226 • Chapter 7: Solving Calculus Problems > evalf(sol); {−1.570796327 + 0.8535664710 I, 1.570796327, −1.570796327 + 1.767486929 I, −1.570796327 + 3.083849212 I, −1.570796327 + 2.473801030 I, 0.6948635283, −0.3257026605} Check the solution set by plotting the function. > plot( eval(err_xi, x=2), xi=2..4 ); 0.6 0.4 0.2 0 –0.2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 xi Two solutions to err_xi=0 seem to exist between 2 and 4 where solve found none: π/2 is less than 2. Thus, you must use numerical methods. If x = 2, then ξ must be in the interval from 2 to a. > sol := fsolve( eval(err_xi, x=2), xi, 2..a ); sol := 2.446729125 At that point the error is > eval( err, {x=2, xi=sol}); 0.07333000221 (2 − π )6 Add this value to the set of critical values. > critical := critical union {%}; critical := {0, 0.07333000221 (2 − π )6 } 7.1 Introductory Calculus • 227 If x = 4 then ξ must be between a and 4. > sol := fsolve( eval(err_xi, x=4), xi, a..4 ); sol := 3.467295314 At that point, the error is > eval( err, {x=4, xi=sol} ); −0.01542298119 (4 − π )6 > critical := critical union {%}; critical := {0, 0.07333000221 (2 − π )6 , −0.01542298119 (4 − π )6 } At the ξ = a boundary, the error is > B := eval( err, xi=a ); B := − 1 (x − π )6 240 The derivative, B 1, of B is zero at a critical point. > B1 := diff( B, x ); B1 := − 1 (x − π )5 40 > sol := { solve( B1=0, x ) }; sol := {π } At the critical point the error is > eval( B, x=sol[1] ); 0 > critical := critical union { % }; 228 • Chapter 7: Solving Calculus Problems critical := {0, 0.07333000221 (2 − π )6 , −0.01542298119 (4 − π )6 } At the last boundary, ξ = x, the error is > B := eval( err, xi=x ); B := 1 (−sin(x) esin(x) + 16 cos(x)2 esin(x) − 15 sin(x)2 esin(x) 720 + 75 sin(x) cos(x)2 esin(...
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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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