test1 fskm set1 mac2016

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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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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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

test1 fskm set1 mac2016

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