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(math230)[2010](s)midterm~0^_69228.pdf
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MATH230, Spring 2010
Midterm Exam Answer
1. (a) (5 points) Assume that. 2()3xfxex..
Since is a continuous function on , and ()fx
[0,1]
, 02(0)3010fe.....
, 12(1)310fe....
So we can find the root of by using the bisection method. ()fx
(10 points)
n
na
nb
np
()nfa
()nfb
()nfp
1
0
1
0.5
+
-
+
2
0.5
1
0.75
+
-
+
3
0.75
1
0.875
So . 30.875p.
(b) (10 points)
33333()ln(1)0ln(1)11()1pppfpppppepepgpep.................
Therefore, if p is the root of f(x), then p is also the fixed point of g(x).
2. (a) For Newtons method, we have (5 points). So, 111()'()nnnnfpppfp.....
(10 points) 12310.53ppp...
(b) For secant method, we have (5 points). So, 112112()()()()nnnnnnnfpppppfpfp..........
(10 points) 230.333330.86363pp....
Q3
a) L3,0 x = x.2 (x.3)(x.5) .1.2 .1.3 (.1.5)
= .x3.10x2+31x.3072 = .0.013889x3+0.13889x2.0.43056x+0.41667
L3,1 x = x+1 (x.3)(x.5) 2+1 2.3 (2.5)
= x3.7x2+7x.159 = 0.11111x3.0.77778x2+0.77778x+1.6667
L3,2 x = x+1 (x.2)(x.5) 3+1 3.2 (3.5)
= .x3.6x2+3x+108 = .0.125x3+0.75x2.0.375x.1.25
L3,3 x = x+1 (x.2)(x.3) 5+1 5.2 (5.3)
= x3.4x2+x+636 = 0.027778x3.0.11111x2+0.027778x+0.16667
b) P3 x = 0.L3,0 x + 1.L3,1 x +1.L3,2 x +2.L3,3 x
= 124x3.14x2+1124x+324 = 0.041667x3.0.25x2+0.45834x+0.75
c)
i
xi
yi
f[xi,xi+1]
f[xi,xi+1,xi+2]
f[xi,xi+1,xi+2,xi+3]
0
-1
0
1
2
1
0.33333 (13)
2
3
1
0
-0.083333(.112)
3
5
2
0.5
0.16667(16)
0.041667(124)
P1 x = 0.33333(x+1)=0.33333x+0.33333= 13x+13 P2 x = P1 x + .0.083333 x+1 x.2 =.0.083333x2+0.41667x+0.5
=. 112x2+512x+12 P3 x = P2 x + 0.041667 x+1 x.2 x.3
=0.041667x3.0.25x2+0.45833x+0.75=124x3.14x2+1124x+324
Q4
x
y
x^2
x^3
x^4
xy
x^2y
1
0
1
0
0
0
0
0
2
0.15
1.004
0.0225
0.003375
0.000506
0.1506
0.02259
3
0.31
1.031
0.0961
0.029791
0.009235
0.31961
0.099079
4
0.5
1.117
0.25
0.125
0.0625
0.5585
0.27925
5
0.6
1.223
0.36
0.216
0.1296
0.7338
0.44028
6
0.75
1.422
0.5625
0.421875
0.316406
1.0665
0.799875
SUM=
2.31
6.797
1.2911
0.796041
0.518248
2.82901
1.641074
Matrix:
6
2.31
1.2911
6.797
2.31
1.2911
0.796041
2.82901