Numerical Analysis Questions and Answers – Iteration Method

Answer: a
Explanation: Given,
Let f(x) = 3x⁴-2x²
Here 3x⁴-2x² = 0
3x⁴ = 2x²
x⁴ = ((2x²)/3)
x = \(\sqrt[4]{\frac{(2x^2)}{3}}\)
\(\phi(x) = \sqrt[4]{\frac{(2x^2)}{3}}\)

Here

x	0	1
f(x)	0	1

Here f(0) = 0

Root of the equation 3x⁴-2x² is 0.

Answer: d
Explanation: Given,
Considering f(x) = tan(x) as \(\phi(x)\) = tan(x)
We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)
x = tan(x)
\(\phi(x)\) = tan(x)
x1 = \(\phi(x0) = \phi(2)\) = -2.185
x2 = \(\phi(x1) = \phi(-2.185)\) = 1.4179
x3 = \(\phi(x2) = \phi(1.4179)\) = 6.4906
x4 = \(\phi(x3) = \phi(6.4906)\) = 0.2104
x5 = \(\phi(x4) = \phi(0.2104)\) = 0.2136
x6 = \(\phi(x5) = \phi(0.2136)\) = 0.2169
x7 = \(\phi(x6) = \phi(0.2169)\) = 0.2203
x8 = \(\phi(x7) = \phi(0.2203)\) = 0.224
x9 = \(\phi(x8) = \phi(0.224)\) = 0.2278
x10 = \(\phi(x9) = \phi(0.2278)\) = 0.2318
x11 = \(\phi(x10) = \phi(0.2318)\) = 0.2361

We can calculate in this mode

Approximate root of the equation tan(x) using iteration method is -0.0571.

Answer: b
Explanation: Given,
Considering f(x) = 4cos(x)-x as \(\phi(x)\) = 4cos(x)-x
We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)
x = 4cos(x)-x
\(\phi(x) = 4cos(x)-x\)
x1 = \(\phi(x0) = \phi(2)\) = -3.6646
x2 = \(\phi(x1) = \phi(-3.6646)\) = 0.1993
x3 = \(\phi(x2) = \phi(0.1993)\) = 3.7216
x4 = \(\phi(x3) = \phi(3.7216)\) = -7.0675
x5 = \(\phi(x4) = \phi(-7.0675)\) = 9.899
x6 = \(\phi(x5) = \phi(9.899)\) = -13.4576
x7 = \(\phi(x6) = \phi(-13.4576)\) = 15.9715
x8 = \(\phi(x7) = \phi(15.9715)\) = -19.8334
x9 = \(\phi(x8) = \phi(-19.8334)\) = 22.0487
x10 = \(\phi(9) = \phi(22.0487)\) = -26.0421
x11 = \(\phi(x10) = \phi(-26.0421)\) = 28.4991
x12 = \(\phi(x11) = \phi(28.4991)\) = -32.3985
x13 = \(\phi(x12) = \phi(-32.3985)\) = 34.6181
x14 = \(\phi(x13) = \phi(34.6181)\) = -38.6107
x15 = \(\phi(x14) = \phi(-38.6107)\) = 41.0606
x16 = \(\phi(x15) = \phi(41.0606)\) = -44.9643
x17 = \(\phi(x16) = \phi(-44.9643)\) = 47.1858
x18 = \(\phi(x17) = \phi(47.1858)\) = -51.1781
x19 = \(\phi(x18) = \phi(-51.1781)\) = 53.6248
x20 = \(\phi(x19) = \phi(53.6248)\) = -57.5304
x21 = \(\phi(x20) = \phi(-57.5304)\) = 59.7528
x22 = \(\phi(x21) = \phi(59.7528)\) = -63.745
x23 = \(\phi(x22) = \phi(-63.745)\) = 66.19
x24 = \(\phi(x23) = \phi(66.19)\) = -70.0966

We can calculate this mode

The process clue us away from the solution, so it is divergent.

Answer: c
Explanation: Given,
Let f(x) = x⁴+2x³+x²+x-1
Here x⁴+2x³+x²+x-1 = 0
x⁴+2x³+x²+x = 1
x(x³+2x²+x+1) = 1
x= 1/(x³+2x²+x+1)
\(\phi(x)\) = 1/(x³+2x²+x+1)

Here

x	0	1
f(x)	-1	4

Here f(0) = -1<0 and f(1) = 4>0
Root lies between 0 and 1.
x0 = 0+12 = 0.5
x1 = \(\phi(x0) = \phi(0.5)\) = 0.4706
x2 = \(\phi(x1) = \phi(0.4706)\) = 0.4956
x3 = \(\phi(x2) = \phi(0.4956)\) = 0.4742
x4 = \(\phi(x3) = \phi(0.4742)\) = 0.4924
x5 = \(\phi(x4) = \phi(0.4924)\) = 0.4769
x6 = \(\phi(x5) = \phi(0.4769)\) = 0.4901
x7 = \(\phi(x6) = \phi(0.4901)\) = 0.4789
x8 = \(\phi(x7) = \phi(0.4789)\) = 0.4885
x9 = \(\phi(x8) = \phi(0.4885)\) = 0.4803
x10 = \(\phi(x9) = \phi(0.4803)\) = 0.4872
x11 = \(\phi(x10) = \phi(0.4872)\) = 0.4813
x12 = \(\phi(x11) = \phi(0.4813)\) = 0.4864
x13 = \(\phi(x12) = \phi(0.4864)\) = 0.482
x14 = \(\phi(x13) = \phi(0.482)\) = 0.4857
x15 = \(\phi(x14) = \phi(0.4857)\) = 0.4826
x16 = \(\phi(x15) = \phi(0.4826)\) = 0.4853
x17 = \(\phi(x16) = \phi(0.4853)\) = 0.483
x18 = \(\phi(x17) = \phi(0.483)\) = 0.4849
x19 = \(\phi(x18) = \phi(0.4849)\) = 0.4833
x20 = \(\phi(x19) = \phi(0.4833)\) = 0.4847
x21 = \(\phi(x20) = \phi(0.4847)\) = 0.4835
x22 = \(\phi(x21) = \phi(0.4835)\) = 0.4845
x23 = \(\phi(x22) = \phi(0.4845)\) = 0.4836
x24 = \(\phi(x23) = \phi(0.4836)\) = 0.4844
We can analyze in this custom

Approximate root of the equation x⁴+2x³+x²+x-1 using iteration method is 0.4838.

Answer: d
Explanation: Given,
Considering f(x) = 2^x-x-1.5 as \(\phi(x)\) = -x+2^x-1.5
We can also directly input \(\phi(x)\) value like x = value format, e.g. x= cos(x)/exp(x).
x = -x+2^x-1.5
\(\phi(x)\) = -x+2^x-1.5
x1 = \(\phi(x0) = \phi(2)\) = 0.5
x2 = \(\phi(x1) = \phi(0.5)\) = -0.5858
x3 = \(\phi(x2) = \phi(-0.5858)\) = -0.2479
x4 = \(\phi(x3) = \phi(-0.2479)\) = -0.41
x5 = \(\phi(x4) = \phi(-0.41)\) = -0.3374
x6 = \(\phi(x5) = \phi(-0.3374)\) = -0.3711
x7 = \(\phi(x6) = \phi(-0.3711)\) = -0.3557
x8 = \(\phi(x7) = \phi(-0.3557)\) = -0.3628
x9 = \(\phi(x8) = \phi(-0.3628)\) = -0.3595
x10 = \(\phi(x9) = \phi(-0.3595)\) = -0.361
x11 = \(\phi(x10) = \phi(-0.361)\) = -0.3604
x12 = \(\phi(x11) = \phi(-0.3604)\) = -0.3607

Approximate root of the equation 2^x-x-1.5 using iteration method is -0.3607.

Answer: c
Explanation: Given,
x = x³-x-1
\(\phi(x)\) = x³-x-1
x1 = \(\phi(x0) = \phi(2) \) = 5
x2 = \(\phi(x1) = \phi(5) \) = 119
x3 = \(\phi(x2) = \phi(119) \) = 1685039

The method lead us away from the solution, so it is divergent.

n	x0	x1 = \(\phi(x0)\)	Update	Difference x1-x0
2	2	5	x0 = x1	3
3	5	119	x0 = x1	114
4	119	1685039	x0 = x1	1684920

Answer: d
Explanation: Given,
Considering f(x) = sin2(x) + 1/tan(x) as \(\phi(x)\) = sin2(x) + 1/tan(x)
We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)

x = sin2(x) + 1/tan(x)
\(\phi(x)\) = sin2(x) + 1/tan(x)
x1 = \(\phi(x0) = \phi(2)\) = 2.9257
x2 = \(\phi(x1) = \phi(2.9257)\) = 3.9018
x3 = \(\phi(x2) = \phi(3.9018)\) = 4.8678
x4 = \(\phi(x3) = \phi(4.8678)\) = 5.7945
x5 = \(\phi(x4) = \phi(5.7945)\) = 6.6688
x6 = \(\phi(x5) = \phi(6.6688)\) = 7.4859
x7 = \(\phi(x6) = \phi(7.4859)\) = 8.2449
x8 = \(\phi(x7) = \phi(8.2449)\) = 8.9472
x9 = \(\phi(x8) = \phi(8.9472)\) = 9.5951
x10 = \(\phi(x9) = \phi(9.5951)\) = 10.1918
x11 = \(\phi(x10) = \phi(10.1918)\) = 10.7403
x12 = \(\phi(x11) = \phi(10.7403)\) = 11.2441
x13 = \(\phi(x12) = \phi(11.2441)\) = 11.7063
x14 = \(\phi(x13) = \phi(11.7063)\) = 12.1301
x15 = \(\phi(x14) = \phi(12.1301)\) = 12.5184
x16 = \(\phi(x15) = \phi(12.5184)\) = 12.8741
x17 = \(\phi(x16) = \phi(12.8741)\) = 13.1996
x18 = \(\phi(x17) = \phi(13.1996)\) = 13.4976
x19 = \(\phi(x18) = \phi(13.4976)\) = 13.7702
x20 = \(\phi(x19) = \phi(13.7702)\) = 14.0195
x21 = \(\phi(x20) = \phi(14.0195)\) = 14.2475
x22 = \(\phi(x21) = \phi(14.2475)\) = 14.4559
x23 = \(\phi(x22) = \phi(14.4559)\) = 14.6464
x24 = \(\phi(x23) = \phi(14.6464)\) = 14.8206

The method lead us away from the solution, so it is divergent.

Answer: a
Explanation: Given,
Considering f(x) = sin3(x) + x³ – exp^(2x) as \(\phi(x)\) = x³ + sin3(x) – e^(2x)
We can also directly input \(\phi(x)\) value like x = value format, e.g. x = cos(x)/exp(x)

x = x³ + sin3(x) – e^(2x)
\(\phi(x)\) = x³ + sin3(x) – e^(2x)
x1 = \(\phi(x0) = \phi(2)\) = -46.3159
x2 = \(\phi(x1) = \phi(-46.3159)\) = -99361.7367

The method chief us away from the solution, so it is divergent.

n	x0	x1 = \(\phi(x0)\)	Update	Difference x1-x0
2	2	-46.3159	x0 = x1	48.3159
3	-46.3159	-99361.7367	x0 = x1	99315.4208

Answer: b
Explanation: Iteration method is used for the calculation of approximate solution of problems which are connected with the differential equation. As the name suggests the solution can be found by the iteration procedure for a closer accuracy.

Answer: a
Explanation: Firstly, we need to write the equation x = \(\phi(x)\). Then, find points a and b such that a < b and f(a).f(b) < 0. Later, if f(a) is more closer to 0 than f(b), then x0 = a else x0 = b.
Lastly, x1 = \(\phi(x0)\)
x2 = \(\phi(x1)\)
x3 = \(\phi(x2)\)
…
Repeat until |f(xi)-f(xi-1)| \(\approx\) 0.