Numerical Analysis Questions and Answers – Stirling’s Formula

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

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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:-

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yp = 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
y0.2 = 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)
y0.2 = 0.3179 + 0.007149 – 0.0000614 – 0.00001648 + 0.000000208
y0.2 = 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
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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:-

yp = 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
y0.2 = 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)
y0.2 = 0.2679 + 0.01877 + 0.00009 – 0.0000544 + 0 + 0.0000022176 + 0.0000002323
y0.2 = 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

yp = 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

yp = 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

yp = 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

yp = 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.

Sanfoundry Global Education & Learning Series – Numerical Analysis.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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