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

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:-

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

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

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

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

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

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

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

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

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

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

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.

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