Test1 Fskm Set1 Mac2016
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7/25/2019 Test1 Fskm Set1 Mac2016
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Test1_fskm_Set1_mac2016
Universiti Teknologi MaraFakulti Sains Komputer & Matematik
MAT 575 TEST 1
NAME : __________________________________________________ 1.
UITM No : ______________ GROUP/PROG:_______________ 2.
LECTURER : MAZNITA 3.
Instructions : 1. Answer all FIVE questionsin 1 hour 15 mins.
2. Perform all calculations using FOUR decimal places.
3. Attach this question paper with your answer sheets.
4.5.
Question 1
Justify that 561.x approximate 56111.x to 3 significant digits.(3 marks)
Question 2
Let .)( 123 xxxf
a) Use the Bisection method with three iterations to reduce the interval of the root of f if].,.[ 5040r .
b) How many iterations are required if the Bisection method is used to locate the root of
faccurate up to 10-4given
].,.[ 5040r ?
(10 marks)
Question 3
By applying Secant method solve the equation 0922 xex using 10x and 2
1x until
accuracy of up to 3 decimal places is achieved.(7 marks)
Question 4
Show analytically that there is an intersection of 12 xxg )( and xxh cos)( 2 between
the values 70.x and 80.x . Then, by applying Newtons method find the point ofintersection correct to 4 significant digits. Then, calculate the relative error for each of theiteration. (Note: calculator in radian)
(8 marks)
Question 5
Provide one advantage of Bisection method and one disadvantage of Newtons method.
(2 marks)
ENDGood Luck!
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Test1_fskm_Set1_mac2016
SolutionQuestion 1
331051070460
56111
56156111
x..
..
The relative error 3105 x , thus 561.x approximate 56111.x to 3 significant
digits.
Question 21a)
]..[,).()..(.).(
.).(
)(
5040050401250050
1360040
123
rfff
f
xxxf
Bisect ion formu la ,ba
c2
initial interval [0.4,0.5]
iteration a b c fa fb fc interval of root1 0.4 0.5 0.45 -0.1360 0.1250 -0.0089 [0.45,0.5]
2 0.45 0.5 0.475 -0.0089 0.1250 0.0572 [0.45,0.475]
3 0.45 0.475 0.4625 -0.0089 0.0572 0.0239 [0.45,0.4625]
After 3 iterations the root is in the interval [0.45, 0.4625].
1b)
9.96n
n
ab
E
n
2
10
4050
10
2
10
4
4
4
ln
..ln
)(
Hence, requires at least 10 iterations.
Question 3
Identify equation: 0922 xex 922 xexxf )(
Secant formu la:
)x(ff,ff
xxfxx nn
nn
nnnnn
1
11
Relat ive error for each i terat ion:
1
1
n
nn
new
oldnew
x
xx
x
xx or 100100
1
1x
x
xxx
x
xx
n
nn
new
oldnew
%
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Test1_fskm_Set1_mac2016
Iteration xn xn-1 fn fn-1 xn+1
Rel
Error (3D)
1 2 1 49.5982 -0.6109 1.0122 0.9760 1.012
2 1.0122 2 -0.4044 49.5982 1.0202 0.0078 1.020
3 1.0202 1.012168 -0.2662 -0.4044 1.0356 0.0149 1.036
4 1.0356 1.020158 0.0059 -0.2662 1.0352 0.0003 1.035
5 1.0352 1.035551 -0.0001 0.0059 1.0352 0.0000 1.035
Thus, the root correct to 3 decimal places is 1.035.
Question 4
12 xxg )( and xxh cos)( 2
xxxxxfxxxfxxxx
xhxg
sin)sin()('cos)(coscos
)()(
222221
021212
22
Show analytically that the intersection point is between x=0.7 and x=0.8.
]..[,).()..(.).(
.).(
8070080702466080
0397070
rfff
Newtons formula:
)x('f
)x(f
xxn
nnn 1
Initial guess x0=0.7: (or any other value sufficiently close)
Iteration Xn f f (x) Xn+1
1 0.7 -0.0397 2.6884 0.7148
2 0.7148 0.0004 2.7404 0.7146
3 0.7146 0.0000 2.7399 0.7146
Thus, the root correct to 4 significant digits is 0.7146.
The point of intersection is (0.7146, g(0.7146)) = (0.7146, h(0.7146)) = (0.7146, 1.5107)
Question 5
Advantage of Bisection: (one only)1. simple to use2. will definitely converge if criteria is fulfilled
Disadvantage of Newton:1. may fail or diverge