métodos numéricos no mathematica

1. 1
Exerc´ıcio 1 (Solu¸c˜ao).
a)f(x) = cosx − x ; I = [0, 1] ; = 10−2
1.png
Xn =
(an + bn)
2
; |f(Xn)| < ; Er = |Xn+1 − Xn| <
X0 = 0, 50 ; |f(X0)| = 0, 377583 ;
X1 = 0, 75 ; |f(X1)| = 0, 0183111 ; |X1 − X0| = 0, 25 ;
X2 = 0, 6250 ; |f(X2)| = 0, 0185963 ; |X2 − X1| = 0, 125 ;
X3 = 0, 68750 ; |f(X3)| = 0, 0853349 ; |X3 − X2| = 0, 0625 ;
X4 = 0, 718750 ; |f(X4)| = 0, 0338794 ; |X4 − X3| = 0, 03125 ;
X5 = 0, 7343750 ; |f(X5)| = 0, 00787473 ; |X5 − X4| = 0, 01563 ;
b)f(x) = x − 2x
; I = [0, 1] ; = 10−2
2.png
Xn =
(an + bn)
2
; |f(Xn)| < ; Er = |Xn+1 − Xn| <
X0 = 0, 50 ; |f(X0)| = 0, 207107 ;
X1 = 0, 75 ; |f(X1)| = 0, 0155396 ; |X1 − X0| = 0, 25 ;
X2 = 0, 6250 ; |f(X2)| = 0, 0234198 ; |X2 − X1| = 0, 125 ;
X3 = 0, 68750 ; |f(X3)| = 0, 06657119 ; |X3 − X2| = 0, 0625 ;
X4 = 0, 65625 ; |f(X4)| = 0, 0217245 ; |X4 − X3| = 0, 03125 ;
X5 = 0, 640625 ; |f(X5)| = 0, 0008100 ; |X5 − X4| = 0, 01563 ;
c)f(x) = ex+2
− x2
; I = [0, 1] ; = 10−2
3.png
Xn =
(an + bn)
2
; |f(Xn)| < ; Er = |Xn+1 − Xn| <
X0 = −1 ; |f(X0)| = 1, 7182 ;
X1 = −1, 5 ; |f(X1)| = 0, 601279 ; |X1 − X0| = 0, 50 ;

2. 2
X2 = −1, 250 ; |f(X2)| = 0, 5545 ; |X2 − X1| = 0, 25 ;
X3 = −1, 37450 ; |f(X3)| = 0, 2022379 ; |X3 − X2| = 0, 0125 ;
X4 = −1, 31250 ; |f(X4)| = 0, 266081 ; |X4 − X3| = 0, 0625 ;
X5 = −1, 34375 ; |f(X5)| = 0, 121886 ; |X5 − X4| = 0, 03125 ;
X6 = −1, 359380 ; |f(X5)| = 0, 0049766 ; |X5 − X4| = 0, 01563 ;
X7 = −1, 36719 ; |f(X5)| = 0, 0136971 ; |X5 − X4| = 0, 00781 ;
X8 = −1, 37109 ; |f(X5)| = 0, 00434 ; |X5 − X4| = 0, 00390 ;
d)f(x) = x6
+ 10.x5
+ 41.x4
− 88.x3
+ 104.x2
− 64.x + 16; I = [0, 1] ; = 10−2
4.png
Xn =
(an + bn)
2
; |f(Xn)| < ; Er = |Xn+1 − Xn| <
A fun¸c˜ao n˜ao possui ra´ızes reais
Exerc´ıcio 2 (Solu¸c˜ao).
Exerc´ıcio 3 (Solu¸c˜ao).
Defina q : R → R do seguinte modo q(x) = p(x) se x ≥ 0 e q(x) = −p(−x) se x ≤ 0.
5.png
veja que q(−1) < 0 e q(1) > 0 logo [−1, 1] cont´em uma raiz.