Bessel's Formula Questions and Answers

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Bessel’s Formula”.

1. What will be the solution of the table using Bessel’s formula?

x	f(x)
22	2854
24	3162
26	3544
28	3992

x = 25

a) 3344.25
b) 3344
c) 3444.50
d) 3374.75
View Answer

Answer: a
Explanation:

Bessel's method to find solution

h=24-22=2
Taking x0=24 then p=(x-x0)/h=(x-24)/2

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The difference table is

x	p=(x-24)/2	y	Δy	Δ²y	Δ³y
22	-1	2854
			308
24	0	3162		74
			382		-8
26	1	3544		66
			448
28	2	3992

x=25
p=(x-x0)/h=(25-24)/2=0.5
y0=3162, Δy0=382, Δ²y_-1 = 74, Δ³y_-1 = -8

Applying Bessel’s formula,

y_0.5 =(3162+3544)/2+(0.5-(1/2))⋅(382)+0.5(0.5-1)/2⋅(74+66)/2+(0.5-1/2)0.5(0.5-1)/6⋅(-8)
y_0.5 =3353+0-8.75+0
y_0.5 =3344.25

Solution of Bessel's interpolation is y(25)=3344.25.

2. What will be the solution of the table using Bessel’s formula?

x	f(x)
20	24
24	32
26	35
22	40

x = 25

a) 0.619
b) 0.611
c) 0.614
d) Solution not possible
View Answer

Answer: d
Explanation: Given,

The value of table for x and y

x	20	24	26	22
y	24	32	35	40

Here difference of x is not same. So, solution using Bessel's method is not possible.

3. What will be the solution of the equation 2x³ +1 using Bessel's formula where x1 = 2 and x2 = 4 and x = 2.1 and step value (h) = 0.25?
a) 11.522
b) 19.522
c) 14.522
d) 29.522
View Answer

Answer: b
Explanation: Given,
Equation is f(x)=2x³ + 1

Bessel's method to find solution

h=2.25-2=0.25
Taking x0=3 then p=(x-x0)/h=(x-3)/0.25
The difference table is

x	p=(x-3)/0.25	y	Δy	Δ²y	Δ³y	Δ⁴y
2	-4	17
			6.7812
2.25	-3	23.7812		1.6875
			8.4688		0.1875
2.5	-2	32.25		1.875		0
			10.3438		0.1875
2.75	-1	42.5938		2.0625		0
			12.4062		0.1875
3	0	55		2.25		0
			14.6562		0.1875
3.25	1	69.6562		2.4375		0
			17.0938		0.1875
3.5	2	86.75		2.625		0
			19.7188		0.1875
3.75	3	106.4688		2.8125
			22.5312
4	4	129

x=2.1
p=(x-x0)/h=(2.1-3)/0.25=-3.6
y0=55, Δy0=14.6562, Δ²y_-1 = 2.25, Δ³y_-1 = 0.1875, Δ⁴y_-2 = 0
Applying Bessel’s formula,
y_-3.6 = (55+69.6562)/2+(-3.6-(1/2))⋅(14.6562)+-3.6(-3.6-1)/2⋅(2.25+2.4375)/2+(-3.6-(1/2))(-3.6)(-3.6-
1)/6⋅(0.1875)+(-3.6+1)(-3.6)(-3.6-1)(-3.6-2)/24⋅(0)/2
y_-3.6 = 62.3281-60.090625+19.40625-2.12175+0
y_-3.6 = 19.522

Solution of Bessel's interpolation is y(2.1)=19.522.

4. What will be the solution of the equation 3x+1 using Bessel's formula where x1 = 2 and x2 = 4 and x = 2.1 and step value (h) = 0.25?
a) 4.6
b) 2.5
c) 7.3
d) 1.9
View Answer

Answer: c
Explanation: Given,
Equation is f(x)=3x+1

Bessel's method to find solution

h=2.25-2=0.25
Taking x0=3 then p=(x-x0)/h=(x-3)/0.25
The difference table is

x	p=(x-3)/0.25	y	Δy	Δ²y
2	-4	7
			0.75
2.25	-3	7.75		0
			0.75
2.5	-2	8.5		0
			0.75
2.75	-1	9.25		0
			0.75
3	0	10		0
			0.75
3.25	1	10.75		0
			0.75
3.5	2	11.5		0
			0.75
3.75	3	12.25		0
			0.75
4	4	13

x=2.1
p=(x-x0)/h=(2.1-3)/0.25=-3.6
y0=10, Δy0=0.75, Δ²y_-1 = 0

Bessel's formula is

y_-3.6 =(10+10.75)/2+(-3.6-(1/2))⋅(0.75)+(-3.6)(-3.6-1)/2⋅(0)/2
y_-3.6 =10.375-3.075+0
y_-3.6 =7.3

Solution of Bessel's interpolation is y(2.1)=7.3.

5. What will be the solution of the equation x using Bessel's formula where x1 = 2 and x2 = 4 and
x = 2.1 and step value (h) = 0.25?
a) 1
b) 2.1
c) 2
d) 1.2
View Answer

Answer: b

Explanation: Given,
Equation is f(x)=x

Bessel's method to find solution

h=2.25-2=0.25
Taking x0=3 then p=(x-x0)/h=(x-3)/0.25
The difference table is

x	p=(x-3)/0.25	y	Δy	Δ²y
2	-4	2
			0.25
2.25	-3	2.25		0
			0.25
2.5	-2	2.5		0
			0.25
2.75	-1	2.75		0
			0.25
3	0	3		0
			0.25
3.25	1	3.25		0
			0.25
3.5	2	3.5		0
			0.25
3.75	3	3.75		0
			0.25
4	4	4

x=2.1
p=(x-x0)/h=(2.1-3)/0.25=-3.6
y0=3, Δy0=0.25, Δ²y_-1 =0

Bessel's formula is

y_-3.6 = (3+3.25)/2+(-3.6-(1/2))⋅(0.25)+(-3.6)(-3.6-1)/2⋅(0)/2
y_-3.6 = 3.125-1.025+0
y_-3.6 = 2.1
Solution of Bessel's interpolation is y(2.1)=2.1.

6. What will be the solution of the equation log(x) using Bessel's formula where x1 = 2 and x2 = 4 and x = 2.1 and step value (h) = 0.25?
a) 1.3799
b) 1.3791
c) 1.3794
d) -0.5045
View Answer

Answer: d
Explanation: Given,
Equation is f(x)=log(x)

Bessel's method to find solution

h=2.25-2=0.25
Taking x0=3 then p=(x-x0)/h=(x-3)/0.25
The difference table is

x	p=(x-3)/0.25	y	Δy	Δ²y	Δ³y	Δ⁴y	Δ⁵y
2	-4	0.301
			0.0512
2.25	-3	0.3522		-0.0054
			0.0458		0.001
2.5	-2	0.3979		-0.0044		-0.0003
			0.0414		0.0008		0.0001
2.75	-1	0.4393		-0.0036		-0.0002
			0.0378		0.0006		0.0001
3	0	0.4771		-0.003		-0.0001
			0.0348		0.0004		0
3.25	1	0.5119		-0.0026		0
			0.0322		0.0004		0
3.5	2	0.5441		-0.0022		0
			0.03		0.0003
3.75	3	0.574		-0.0019
			0.028
4	4	0.6021

p=(x-x0)/h=-3/0.25=-12
y0=0.4771, Δy0=0.0348, Δ²y_-1 = -0.003, Δ³y_-1 = 0.0004, Δ⁴y_-2 = -0.0001, Δ⁵y_-2 = 0

Bessel's formula is

y_-12 = (0.4771+0.5119)/2+(-12-(1/2))⋅(0.0348)+(-12(-12-1))/2⋅(-0.003-0.0026)/2+(-12-(1/2))-12(-12-
1)/6⋅(0.0004)+(-12+1)-12(-12-1)(-12-2)/24⋅(-0.0001)/2+(-12-(1/2))(-12+1)-12(-12-1)(-12-2)/120⋅(0)
y_-12 = 0.4945-0.4345263282-0.2185512232-0.1459353163-0.1109642258-0.089011542
y_-12 = -0.5045

Solution of Bessel's interpolation is y()=-0.5045.

7. What will be the root of the equation cos(x) using Bessel's formula where x1 = 2 and x2 = 4 and
x = 2.1 and step value (h) = 0.25?
a) 1.2045
b) 2.2354
c) 3.8547
d) 4.6235
View Answer

Answer: a
Explanation: Given,
Equation is f(x)=cos(x)

Bessel's method to find solution

h=2.25-2=0.25
Taking x0=3 then p=(x-x0)/h=(x-3)/0.25
The difference table is

x	p=(x-3)/0.25	y	Δy	Δ²y	Δ³y	Δ⁴y	Δ⁵y	Δ⁶y	Δ⁷y
2	-4	-0.4161
			-0.212
2.25	-3	-0.6282		0.0391
			-0.173		0.0108
2.5	-2	-0.8011		0.0498		-0.0031
			-0.1232		0.0077		-0.0005
2.75	-1	-0.9243		0.0575		-0.0036		0.0002
			-0.0657		0.0041		-0.0003		0
3	0	-0.99		0.0616		-0.0038		0.0002
			-0.0041		0.0003		0		0
3.25	1	-0.9941		0.0618		-0.0038		0.0002
			0.0577		-0.0036		0.0002
3.5	2	-0.9365		0.0582		-0.0036
			0.1159		-0.0072
3.75	3	-0.8206		0.051
			0.1669
4	4	-0.6536

p=(x-x0)/h=-3/0.25=-12
y0=-0.99, Δy0=-0.0041, Δ²y_-1 = 0.0616, Δ³y_-1 = 0.0003, Δ⁴y_-2 = -0.0038, Δ⁵y_-2 = 0, Δ⁶y_-3 = 0.0002, Δ⁷y_-3 = 0

Bessel's formula is

y_-12 = (-0.99±0.9941)/2+(-12-(1/2))⋅(-0.0041)+(-12(-12-1))/2⋅(0.0616+0.0618)/2+(-12-(1/2))-12(-12-1)/6⋅(0.0003)+(-12+1)(-12)(-12-1)(-12-2)/24⋅(-0.0038-0.0038)/2+(-12-(1/2))(-12+1)(-12)(-12-1)(-12-2)/120⋅(0)+(-12+1)(-12+1)(-12)(-12-1)(-12-2)(-12-2)/720⋅(0.0002)/2+(-12-(1/2))(-12+1)(-12+1)(-12)(-12-1)(-12-2)(-12-2)/5040⋅(0)
y_-12 = -0.9921+0.0517147435+4.8111611636-0.0835996791-3.8388952996+0.0400232382+1.2252441039-0.0091243207
y_-12 = 1.2045

Solution of Bessel's interpolation is y()=1.2045.

8. What will be the solution of the table using Bessel’s formula?

x	f(x)
10	0.1
20	2.1
30	3.4
40	4.5

x = 25

a) 2.8062
b) 4.2563
c) 1.2354
d) -0.2396
View Answer

Answer: a
Explanation:

Bessel's method to find solution

h = 20-10 = 10

Taking x0 = 20 then p = (x-x0)/h = (x-20)/10
The difference table is

x	p=(x-20)/10	y	Δy	Δ²y	Δ³y
10	-1	0.1
			2
20	0	2.1		-0.7
			1.3		0.5
30	1	3.4		-0.2
			1.1
40	2	4.5

x=25
p = (x-x0)/h = (25-20)/10 = 0.5
y0 = 2.1 ,Δy0 = 1.3, Δ²y_-1 = -0.7, Δ³y_-1 = 0.5

Bessel's formula is

y_0.5 = (2.1+3.4)/2+(0.5-(1/2))⋅(1.3)+0.5(0.5-1)/2⋅(-0.7-0.2)/2+(0.5-(1/2))0.5(0.5-1)/6⋅(0.5)
y_0.5 = 2.75+0+0.05625+0
y_0.5 = 2.8062

Solution of Bessel's interpolation is y(25)=2.8062.

9. What will be the solution of the table using Bessel’s formula?

x	f(x)
10	0.1
20	2.1
30	3.4
40	4.5
50	5.6

x = 15

a) 1.9874
b) 1.3254
c) 1.2589
d) 1.2715
View Answer

Answer: d
Explanation:

Bessel's method to find solution

h = 20-10 = 10
Taking x0=30 then p=(x-x0)/h=(x-30)/10
The difference table is

x	p=(x-30)/10	y	Δy	Δ²y	Δ³y	Δ⁴y
10	-2	0.1
			2
20	-1	2.1		-0.7
			1.3		0.5
30	0	3.4		-0.2		-0.3
			1.1		0.2
40	1	4.5		0
			1.1
50	2	5.6

x = 15
p = (x-x0)/h = (15-30)/10 = -1.5
y0 = 3.4, Δy0 = 1.1, Δ²y_-1 = -0.2, Δ³y_-1 = 0.2, Δ⁴y_-2 = -0.3

Bessel's formula is
y_-1.5 = (3.4+4.5)/2+(-1.5-(1/2))⋅(1.1)+(-1.5(-1.5-1))/2⋅(-0.2)/2+(-1.5-(1/2))(-1.5)(-1.5-1)/6⋅(0.2)+(-1.5+1)(-1.5)(-1.5-1)(-1.5-2)/24⋅(-0.3)/2
y_-1.5 = 3.95-2.2-0.1875-0.25-0.041015625
y_-1.5 = 1.2715

Solution of Bessel's interpolation is y(15) = 1.2715.

10. What will be the root of the equation x+1 using Bessel's formula where x1 = 4 and x2 = 6 and
x = 2.1 and step value (h) = 0.50?
a) 3.1
b) 1.62
c) 2
d) 2.56
View Answer

Answer: a
Explanation: Given,
Equation is f(x)=x+1
The value of table for x and y

x	4	4.5	5	5.5	6
y	5	5.5	6	6.5	7

Bessel's method to find solution

h = 4.5-4 = 0.5
Taking x0 = 5 then p = (x-x0)/h = (x-5)/0.5
The difference table is

x	p=(x-5)/0.5	y	Δy	Δ²y
4	-2	5
			0.5
4.5	-1	5.5		0
			0.5
5	0	6		0
			0.5
5.5	1	6.5		0
			0.5
6	2	7

x = 2.1
p = (x-x0)/h = (2.1-5)/0.5 = -5.8
y0 = 6, Δy0 = 0.5, Δ²y_-1 = 0

Bessel's formula is

y_-5.8 = (6+6.5)/2+(-5.8-(1/2))⋅(0.5)+(-5.8(-5.8-1))/2⋅(0)/2
y_-5.8 = 6.25-3.15+0
y_-5.8 = 3.1

Solution of Bessel's interpolation is y(2.1)=3.1.

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