Stirling's Formula Questions and Answers

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

1. What will be the solution for the following table using Stirling's formula, where x = 12.2?

x	f(x)
10	0.23967
11	0.28060
12	0.31788
13	0.35209
14	0.38368

a) 0.325
b) 0.328
c) 0.327
d) 0.322
View Answer

Answer: a
Explanation: Given,
The value of table for x and y

x	10	11	12	13	14
y	0.2397	0.2806	0.3179	0.3521	0.3837

Stirling's method to find solution
h = 11 – 10 = 1

Taking x₀ = 12 then p = (x-x₀)/h = (x-12)/1
The difference table is

x	p=x-121	y	Δy	Δ²y	Δ³y	Δ⁴y
10	-2	0.2397
			0.0409
11	-1	0.2806		-0.0037
			0.0373		0.0006
12	0	0.3179		-0.0031		-0.0001
			0.0342		0.0004
13	1	0.3521		-0.0026
			0.0316
14	2	0.3837

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x = 12.2
p = (x-x₀)/h = (12.2-12)/1 = 0.2
y₀ = 0.3179, Δy₀ = 0.0342, Δ²y_-1 = -0.0031, Δ³y _-1 = 0.0004, Δ⁴y_-2 = -0.0001

Stirling's formula is:-

y_p = y₀ + p⋅Δy₀ + (Δy_-1)/2 + p²/2!⋅Δ²y _-1 + p(p² – 1²)/3!⋅ Δ³y_-1 + (Δ³y_-2)/2 + p²(p²-1²)/4!⋅ Δ⁴y_-2
y_0.2 = 0.3179 + ((0.2)⋅(0.0342+0.0373)/2) + ((0.04)/2)⋅(-0.0031) + ((0.2)((0.04-1)/6))⋅((0.0004)/2) +
((0.04)((0.04-1)/24))⋅(-0.0001)
y_0.2 = 0.3179 + 0.007149 – 0.0000614 – 0.00001648 + 0.000000208
y_0.2 = 0.325
Solution of Stirling's interpolation is y(12.2)=0.325.

2. What will be the solution for the following table using Stirling's formula, where x = 16?

x	f(x)
0	0
5	0.0875
10	0.1763
15	0.2679
20	0.3640
25	0.4663
30	0.5774

a) 0.2452
b) 0.2864
c) 0.2862
d) 0.2867
View Answer

Answer: d
Explanation: Given,
The value of table for x and y

x	0	5	10	15	20	25	30
y	0	0.0875	0.1763	0.2679	0.364	0.4663	0.5774

Stirling's method to find solution
h = 5 – 0 = 5

Taking x₀ = 15 then p = (x-x₀)/h = (x-15)/5
The difference table is

x	p=x-155	y	Δy	Δ²y	Δ³y	Δ⁴y	Δ⁵y	Δ⁶y
0	-3	0
			0.0875
5	-2	0.0875		0.0013
			0.0888		0.0015
10	-1	0.1763		0.0028		0.0002
			0.0916		0.0017		-0.0002
15	0	0.2679		0.0045		0		0.0011
			0.0961		0.0017		0.0009
20	1	0.364		0.0062		0.0009
			0.1023		0.0026
25	2	0.4663		0.0088
			0.1111
30	3	0.5774

x = 16
p = (x-x₀)/h = (16-15)/5 = 0.2
y₀ = 0.2679, Δy₀ = 0.0961, Δ²y_-1 = 0.0045, Δ³y_-1 = 0.0017, Δ⁴y_-2 = 0, Δ⁵y_-2 = 0.0009, Δ⁶y_-3 = 0.0011

Stirling's formula is:-

y_p = y₀ + p⋅Δy₀ + (Δy_-1)/2 + p²/2!⋅Δ² y_-1 + p(p² – 1²)/3!⋅Δ³y_-1 + (Δ³y_-2)/2 + p²(p² – 1²)/4!⋅Δ⁴y_-2 + p(p² – 1²)(p² – 2²)/5!⋅Δ⁵y_-2 + (Δ⁵y_-3)/2 + p²(p² -1²)(p² – 2²)/6!⋅Δ⁶y_-3
y_0.2 = 0.2679 + (0.2)⋅(0.0961 + 0.0916)/2 + (0.04)/2⋅(0.0045) + (0.2)(0.04-1)/6⋅(0.0017+0.0017)/2 + (0.04)(0.04-1)/24⋅(0) + (0.2)(0.04-1)(0.04-4)/120⋅(0.0009)/2 + (0.04)(0.04-1)(0.04-4)/720⋅(0.0011)
y_0.2 = 0.2679 + 0.01877 + 0.00009 – 0.0000544 + 0 + 0.0000022176 + 0.0000002323
y_0.2 = 0.2867
Solution of Stirling's interpolation is y(16) = 0.2867.

3. What will be the solution of the equation 2x³ – 4x + 1 using Stirling's formula, where x1 = 2 and x2 = 4, x = 2.1 & step value (h) = 0.25?
a) 11.152
b) 11.127
c) 11.122
d) 11.128
View Answer

Answer: b
Explanation: Given,
Equation is f(x) = 2x³ – 4x + 1.
The value of table for x and y

x	2	2.25	2.5	2.75	3	3.25	3.5	3.75	4
y	9	14.7812	22.25	31.5938	43	56.6562	72.75	91.4688	113

Stirling'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-30.25	y	Δy	Δ²y	Δ³y	Δ⁴y
2	-4	9
			5.7812
2.25	-3	14.7812		1.6875
			7.4688		0.1875
2.5	-2	22.25		1.875		0
			9.3438		0.1875
2.75	-1	31.5938		2.0625		0
			11.4062		0.1875
3	0	43		2.25		0
			13.6562		0.1875
3.25	1	56.6562		2.4375		0
			16.0938		0.1875
3.5	2	72.75		2.625		0
			18.7188		0.1875
3.75	3	91.4688		2.8125
			21.5312
4	4	113

x = 2.1
p = (x-x₀)/h = (2.1-3)/0.25 = -3.6
y₀ = 43, Δy₀ = 13.6562, Δ²y_-1 = 2.25, Δ³y_-1 = 0.1875, Δ⁴y_-2 = 0

Stirling's formula is

y_p = y₀ + p⋅Δy₀ + (Δy_-1)/2 + p²/2!⋅Δ² y_-1 + p(p² – 1²)/3!⋅Δ³y_-1 + (Δ³y_-2)/2 + p²(p²-1²)/4!⋅Δ⁴y_-2
y_-3.6 = 43 + (-3.6)⋅(13.6562+11.4062)/2 + (12.96)/2⋅(2.25) + (-3.6)(12.96-1)/6⋅(0.1875+0.1875)/2 + (12.96)(12.96-1)/24⋅(0)
y_-3.6 = 43-45.1125 + 14.58-1.3455 + 0
y_-3.6 = 11.122
Solution of Stirling's interpolation is y(2.1) = 11.122.

4. What is the value of ‘p’ in Stirling’s formula?
a) p = x₀ /h
b) p = x-x/h
c) p = x-x₀ /h
d) p = x-x₀
View Answer

Answer: c
Explanation: The Stirling formula or Stirling’s approximation formula is used to give the approximate value for a factorial function. It helps in finding out the factorial of larger numbers. Although, the chief value comes from limiting the factorials.

5. What kind of relation is Stirling’s formula?
a) Asymptotic relation
b) Symptotic relation
c) Partial relation
d) Lagrange’s relation
View Answer

Answer: a
Explanation: Asymptotic relation gives the meaning of approaching a value closely. So, whenever a line or any curve is found to be asymptotic to a given curve it is called its asymptote, as it defines its limit.

6. Stirling’s formula’s main mathematical value is in limits involving factorials.
a) False
b) True
View Answer

Answer: b
Explanation: The Stirling formula or Stirling’s approximation formula is used to give the approximate value for a factorial function. It helps in finding out the factorial of larger numbers. But its main mathematical value is in limits involving factorials.

7. What will be the solution of the equation 2x² using Stirling's formula, where x1 = 1 and x2 = 2 & x = 2 & step value (h) = 0.25?
a) 2
b) 7
c) 5
d) 8
View Answer

Answer: d
Explanation: Given,
Equation is f(x) = 2x²
The value of table for x and y

x	1	1.25	1.5	1.75	2
y	2	3.125	4.5	6.125	8

Stirling's method to find solution
h = 1.25-1 = 0.25
Taking x₀ = 1.5 then p =(x-x₀)/h = (x-1.5)/0.25
The difference table is

x	p=(x-1.5)/0.25	y	Δy	Δ²y	Δ³y
1	-2	2
			1.125
1.25	-1	3.125		0.25
			1.375		0
1.5	0	4.5		0.25
			1.625		0
1.75	1	6.125		0.25
			1.875
2	2	8

x = 2
p = (x-x₀)/h = (2-1.5)/0.25 = 2
y₀ = 4.5, Δy₀ = 1.625, Δ²y_-1 = 0.25, Δ³y_-1 = 0

Stirling's formula is

y_p = y₀ + p⋅Δy₀ + Δy_-1/2 + p²/2!⋅Δ²y_-1 + p(p² -1²)/3!⋅Δ³y_-1 + (Δ³y_-2)/2
y₂ = 4.5 + (2)⋅(1.625 + 1.375)/2 + (4)/2⋅(0.25) + (2)(4-1)/6⋅(0)/2
y₂ = 4.5 + 3 + 0.5 + 0
y₂ = 8
Solution of Stirling's interpolation is y(2) = 8.

8. What will be the solution of the equation 3x using Stirling's formula, where x1 = 2 and x2 = 4 & x = 2 & step value (h) = 0.25?
a) 8
b) 6
c) 5
d) 1
View Answer

Answer: b
Explanation: Given,
Equation is f(x) = 3x.
The value of table for x and y.

x	2	2.25	2.5	2.75	3	3.25	3.5	3.75	4
y	6	6.75	7.5	8.25	9	9.75	10.5	11.25	12

Stirling'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-30.25	y	Δy	Δ²y
2	-4	6
			0.75
2.25	-3	6.75		0
			0.75
2.5	-2	7.5		0
			0.75
2.75	-1	8.25		0
			0.75
3	0	9		0
			0.75
3.25	1	9.75		0
			0.75
3.5	2	10.5		0
			0.75
3.75	3	11.25		0
			0.75
4	4	12

x = 2
p = (x-x₀)/h = (2-3)/0.25 = -4
y₀ = 9, Δy₀ = 0.75, Δ²y_-1 = 0

Stirling's formula is

y_p = y₀ + p⋅Δy₀ + Δy_-1/2 + p²/2!⋅Δ²y_-1

y_-4 = 9 + (-4)⋅(0.75+0.75)/2 + (16)/2⋅(0)
y_-4 = 9 – 3 + 0
y_-4 = 6

Solution of Stirling's interpolation is y(2) = 6.

9. What does the notation ‘Γ’ represents in Stirling’s formula for the gamma function?
a) Beta function
b) Alpha function
c) Gamma function
d) Delta function
View Answer

Answer: c
Explanation: ‘Γ’ represents the Gamma function. It comes in the Stirling’s formula for the gamma function, with in the equation “n! = Γ(n + 1)”, it is for all the positive integers but the gamma function is more broadly defines unlike the factorial.

10. What will be the solution of the equation x using Stirling's formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) 4
b) 2
c) 3
d) 1
View Answer

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

x	1	1.25	1.5	1.75	2
y	1	1.25	1.5	1.75	2

Stirling's method to find solution
h = 1.25-1 = 0.25

Taking x₀ = 1.5 then p = (x-x₀ )/h = (x-1.5)/0.25
The difference table is

x	p=(x-1.5)/0.25	y	Δy	Δ²y
1	-2	1
			0.25
1.25	-1	1.25		0
			0.25
1.5	0	1.5		0
			0.25
1.75	1	1.75		0
			0.25
2	2	2

x = 1
p = (x-x₀)/h = (1-1.5)/0.25 = -2
y₀ = 1.5, Δy₀ = 0.25, Δ²y_-1 = 0

Stirling's formula is

y_p = y₀ + p⋅Δy₀ + Δy_-1/2 + p²/2!⋅Δ²y_-1
y_-2 = 1.5 + (-2)⋅(0.25+0.25)/2 + (4)/2⋅(0)
y_-2 = 1.5 – 0.5 + 0
y_-2 = 1

Solution of Stirling's interpolation is y(1) = 1.

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