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 Page 1


Edurev123 
3. Interpolation 
3.1 Using Lagrange interpolation formula, calculate the value of ?? (?? ) from the 
following table of values of ?? and ?? (?? ) : 
 
(2009 : 15 Marks) 
Solution: 
Usually Lagrange interpolation formula, first we will form the equation of ?? (?? ) . Now, as 
per the formula, 
?? (?? )=
(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 0
)
(?? 0
-?? 1
)(?? 0
-?? 2
)(?? 0
-?? 3
)(?? 0
-?? 4
)(?? 0
-?? 5
)
?+
(?? -?? 0
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 1
)
(?? 1
-?? 0
)(?? 1
-?? 2
)(?? 1
-?? 3
)(?? 1
-?? 4
)(?? 1
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 2
)
(?? 2
-?? 0
)(?? 2
-?? 1
)(?? 2
-?? 3
)(?? 2
-?? 4
)(?? 2
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 4
)(?? -?? 5
)?? (?? 3
)
(?? 3
-?? 0
)(?? 3
-?? 1
)(?? 3
-?? 2
)(?? 3
-?? 4
)(?? 3
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 5
)?? (?? 4
)
(?? 4
-?? 0
)(?? 4
-?? 1
)(?? 4
-?? 2
)(?? 4
-?? 3
)(?? 4
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)?? (?? 5
)
(?? 5
-?? 0
)(?? 5
-?? 1
)(?? 5
-?? 2
)(?? 5
-?? 3
)(?? 5
-?? 4
)
 
Now, ?? 0
=0,?? 1
=1,?? 2
=2,?? 3
=4,?? 4
=5,?? 5
=6 and ?? (?? 0
)=1,?? (?? 1
)=14,?? (?? 2
)=
15,?? (?? 3
)=5,?? (?? 4
)=6, ?? (?? 5
)=19. 
So, 
Now, 
Page 2


Edurev123 
3. Interpolation 
3.1 Using Lagrange interpolation formula, calculate the value of ?? (?? ) from the 
following table of values of ?? and ?? (?? ) : 
 
(2009 : 15 Marks) 
Solution: 
Usually Lagrange interpolation formula, first we will form the equation of ?? (?? ) . Now, as 
per the formula, 
?? (?? )=
(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 0
)
(?? 0
-?? 1
)(?? 0
-?? 2
)(?? 0
-?? 3
)(?? 0
-?? 4
)(?? 0
-?? 5
)
?+
(?? -?? 0
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 1
)
(?? 1
-?? 0
)(?? 1
-?? 2
)(?? 1
-?? 3
)(?? 1
-?? 4
)(?? 1
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 2
)
(?? 2
-?? 0
)(?? 2
-?? 1
)(?? 2
-?? 3
)(?? 2
-?? 4
)(?? 2
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 4
)(?? -?? 5
)?? (?? 3
)
(?? 3
-?? 0
)(?? 3
-?? 1
)(?? 3
-?? 2
)(?? 3
-?? 4
)(?? 3
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 5
)?? (?? 4
)
(?? 4
-?? 0
)(?? 4
-?? 1
)(?? 4
-?? 2
)(?? 4
-?? 3
)(?? 4
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)?? (?? 5
)
(?? 5
-?? 0
)(?? 5
-?? 1
)(?? 5
-?? 2
)(?? 5
-?? 3
)(?? 5
-?? 4
)
 
Now, ?? 0
=0,?? 1
=1,?? 2
=2,?? 3
=4,?? 4
=5,?? 5
=6 and ?? (?? 0
)=1,?? (?? 1
)=14,?? (?? 2
)=
15,?? (?? 3
)=5,?? (?? 4
)=6, ?? (?? 5
)=19. 
So, 
Now, 
?? (?? )=
(?? -1)(?? -2)(?? -4)(?? -5)(?? -6)×1
(0-1)(0-2)(0-4)(0-5)(0-6)
?+
(?? -0)(?? -2)(?? -4)(?? -5)(?? -6)×14
(1-0)(1-2)(1-4)(1-5)(1-6)
?+
(?? -0)(?? -1)(?? -4)(?? -5)(?? -6)×15
(2-0)(2-1)(2-4)(2-5)(2-6)
?+
(?? -0)(?? -1)(?? -2)(?? -5)(?? -6)×5
(4-0)(4-1)(4-2)(4-5)(4-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -6)×6
(5-0)(5-1)(5-2)(5-4)(5-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -5)×19
(6-0)(6-1)(6-2)(6-4)(6-5)
(3-1)(3-2)(3-4)(3-5)(3-6)
-1×-2×-4×-5×-6
?+
(3-0)(3-2)(3-4)(3-5)(3-6)×14
1×(-1)×(-3)(-4)(-5)
?+
(3-0)(3-1)(3-4)(3-5)(3-6)×15
2×1×(-2)(-3)(-4)
?+
(3-0)(3-1)(3-2)(3-5)(3-6)×5
4×3×2×(-1)(-2)
?+
(3-1)(3-2)(3-4)(3-6)×6
5×4×3×1×(-1)
 
?+
(3-0)(3-1)(3-2)(3-4)(3-5)×19
6×5×4×2×1
?+
2×1×(+1)×(-2)×(-3)
2×4×5×(+6)
+
3×1×(-1)×(+2)×(-3)×14
1×3×4×5
?+
3×2×1×(-1)(-3)×6
5×4×3×(-1)
+
3×2×1×(-1)×(-2)×19
6×5×4×2
=
1
20
-
21
5
+
45
4
+
15
4
-
9
5
+
19
20
=
1-84+180+75-36+19
20
=
275-120
20
=
155
20
=
31
4
?? =
31
4
 
3.2 Find the value of ?? (?? .?? ) using Runge-Kutta fourth order method with step size 
?? =?? .?? from the initial value problem. 
?? '
?=????
?? (?? )?=?? 
(2009: 15 marks) 
Solution: 
Page 3


Edurev123 
3. Interpolation 
3.1 Using Lagrange interpolation formula, calculate the value of ?? (?? ) from the 
following table of values of ?? and ?? (?? ) : 
 
(2009 : 15 Marks) 
Solution: 
Usually Lagrange interpolation formula, first we will form the equation of ?? (?? ) . Now, as 
per the formula, 
?? (?? )=
(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 0
)
(?? 0
-?? 1
)(?? 0
-?? 2
)(?? 0
-?? 3
)(?? 0
-?? 4
)(?? 0
-?? 5
)
?+
(?? -?? 0
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 1
)
(?? 1
-?? 0
)(?? 1
-?? 2
)(?? 1
-?? 3
)(?? 1
-?? 4
)(?? 1
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 2
)
(?? 2
-?? 0
)(?? 2
-?? 1
)(?? 2
-?? 3
)(?? 2
-?? 4
)(?? 2
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 4
)(?? -?? 5
)?? (?? 3
)
(?? 3
-?? 0
)(?? 3
-?? 1
)(?? 3
-?? 2
)(?? 3
-?? 4
)(?? 3
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 5
)?? (?? 4
)
(?? 4
-?? 0
)(?? 4
-?? 1
)(?? 4
-?? 2
)(?? 4
-?? 3
)(?? 4
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)?? (?? 5
)
(?? 5
-?? 0
)(?? 5
-?? 1
)(?? 5
-?? 2
)(?? 5
-?? 3
)(?? 5
-?? 4
)
 
Now, ?? 0
=0,?? 1
=1,?? 2
=2,?? 3
=4,?? 4
=5,?? 5
=6 and ?? (?? 0
)=1,?? (?? 1
)=14,?? (?? 2
)=
15,?? (?? 3
)=5,?? (?? 4
)=6, ?? (?? 5
)=19. 
So, 
Now, 
?? (?? )=
(?? -1)(?? -2)(?? -4)(?? -5)(?? -6)×1
(0-1)(0-2)(0-4)(0-5)(0-6)
?+
(?? -0)(?? -2)(?? -4)(?? -5)(?? -6)×14
(1-0)(1-2)(1-4)(1-5)(1-6)
?+
(?? -0)(?? -1)(?? -4)(?? -5)(?? -6)×15
(2-0)(2-1)(2-4)(2-5)(2-6)
?+
(?? -0)(?? -1)(?? -2)(?? -5)(?? -6)×5
(4-0)(4-1)(4-2)(4-5)(4-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -6)×6
(5-0)(5-1)(5-2)(5-4)(5-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -5)×19
(6-0)(6-1)(6-2)(6-4)(6-5)
(3-1)(3-2)(3-4)(3-5)(3-6)
-1×-2×-4×-5×-6
?+
(3-0)(3-2)(3-4)(3-5)(3-6)×14
1×(-1)×(-3)(-4)(-5)
?+
(3-0)(3-1)(3-4)(3-5)(3-6)×15
2×1×(-2)(-3)(-4)
?+
(3-0)(3-1)(3-2)(3-5)(3-6)×5
4×3×2×(-1)(-2)
?+
(3-1)(3-2)(3-4)(3-6)×6
5×4×3×1×(-1)
 
?+
(3-0)(3-1)(3-2)(3-4)(3-5)×19
6×5×4×2×1
?+
2×1×(+1)×(-2)×(-3)
2×4×5×(+6)
+
3×1×(-1)×(+2)×(-3)×14
1×3×4×5
?+
3×2×1×(-1)(-3)×6
5×4×3×(-1)
+
3×2×1×(-1)×(-2)×19
6×5×4×2
=
1
20
-
21
5
+
45
4
+
15
4
-
9
5
+
19
20
=
1-84+180+75-36+19
20
=
275-120
20
=
155
20
=
31
4
?? =
31
4
 
3.2 Find the value of ?? (?? .?? ) using Runge-Kutta fourth order method with step size 
?? =?? .?? from the initial value problem. 
?? '
?=????
?? (?? )?=?? 
(2009: 15 marks) 
Solution: 
As per Runge-Kutta forth order method, 
?? 1
=h?? (?? 0
,?? 0
)
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)
?? =
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
 
and next ?? will be 
?? 1
=?? 0
+?? ?????????
 and ?? (?? ,?? )=?? '
?????????
 Here ?? (?? ,?? )=???? ?????????
 
Now, as per the given problem 
?? 0
?=?? (1)=2
?? '
?=?? (?? ,?? )=????
h?=0.2
?? 1
?=h?? (?? 0
,?? 0
)
?=0.2?? (1,2)
?=0.2×1×2=0.4
?? 2
?=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
 
Page 4


Edurev123 
3. Interpolation 
3.1 Using Lagrange interpolation formula, calculate the value of ?? (?? ) from the 
following table of values of ?? and ?? (?? ) : 
 
(2009 : 15 Marks) 
Solution: 
Usually Lagrange interpolation formula, first we will form the equation of ?? (?? ) . Now, as 
per the formula, 
?? (?? )=
(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 0
)
(?? 0
-?? 1
)(?? 0
-?? 2
)(?? 0
-?? 3
)(?? 0
-?? 4
)(?? 0
-?? 5
)
?+
(?? -?? 0
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 1
)
(?? 1
-?? 0
)(?? 1
-?? 2
)(?? 1
-?? 3
)(?? 1
-?? 4
)(?? 1
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 2
)
(?? 2
-?? 0
)(?? 2
-?? 1
)(?? 2
-?? 3
)(?? 2
-?? 4
)(?? 2
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 4
)(?? -?? 5
)?? (?? 3
)
(?? 3
-?? 0
)(?? 3
-?? 1
)(?? 3
-?? 2
)(?? 3
-?? 4
)(?? 3
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 5
)?? (?? 4
)
(?? 4
-?? 0
)(?? 4
-?? 1
)(?? 4
-?? 2
)(?? 4
-?? 3
)(?? 4
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)?? (?? 5
)
(?? 5
-?? 0
)(?? 5
-?? 1
)(?? 5
-?? 2
)(?? 5
-?? 3
)(?? 5
-?? 4
)
 
Now, ?? 0
=0,?? 1
=1,?? 2
=2,?? 3
=4,?? 4
=5,?? 5
=6 and ?? (?? 0
)=1,?? (?? 1
)=14,?? (?? 2
)=
15,?? (?? 3
)=5,?? (?? 4
)=6, ?? (?? 5
)=19. 
So, 
Now, 
?? (?? )=
(?? -1)(?? -2)(?? -4)(?? -5)(?? -6)×1
(0-1)(0-2)(0-4)(0-5)(0-6)
?+
(?? -0)(?? -2)(?? -4)(?? -5)(?? -6)×14
(1-0)(1-2)(1-4)(1-5)(1-6)
?+
(?? -0)(?? -1)(?? -4)(?? -5)(?? -6)×15
(2-0)(2-1)(2-4)(2-5)(2-6)
?+
(?? -0)(?? -1)(?? -2)(?? -5)(?? -6)×5
(4-0)(4-1)(4-2)(4-5)(4-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -6)×6
(5-0)(5-1)(5-2)(5-4)(5-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -5)×19
(6-0)(6-1)(6-2)(6-4)(6-5)
(3-1)(3-2)(3-4)(3-5)(3-6)
-1×-2×-4×-5×-6
?+
(3-0)(3-2)(3-4)(3-5)(3-6)×14
1×(-1)×(-3)(-4)(-5)
?+
(3-0)(3-1)(3-4)(3-5)(3-6)×15
2×1×(-2)(-3)(-4)
?+
(3-0)(3-1)(3-2)(3-5)(3-6)×5
4×3×2×(-1)(-2)
?+
(3-1)(3-2)(3-4)(3-6)×6
5×4×3×1×(-1)
 
?+
(3-0)(3-1)(3-2)(3-4)(3-5)×19
6×5×4×2×1
?+
2×1×(+1)×(-2)×(-3)
2×4×5×(+6)
+
3×1×(-1)×(+2)×(-3)×14
1×3×4×5
?+
3×2×1×(-1)(-3)×6
5×4×3×(-1)
+
3×2×1×(-1)×(-2)×19
6×5×4×2
=
1
20
-
21
5
+
45
4
+
15
4
-
9
5
+
19
20
=
1-84+180+75-36+19
20
=
275-120
20
=
155
20
=
31
4
?? =
31
4
 
3.2 Find the value of ?? (?? .?? ) using Runge-Kutta fourth order method with step size 
?? =?? .?? from the initial value problem. 
?? '
?=????
?? (?? )?=?? 
(2009: 15 marks) 
Solution: 
As per Runge-Kutta forth order method, 
?? 1
=h?? (?? 0
,?? 0
)
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)
?? =
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
 
and next ?? will be 
?? 1
=?? 0
+?? ?????????
 and ?? (?? ,?? )=?? '
?????????
 Here ?? (?? ,?? )=???? ?????????
 
Now, as per the given problem 
?? 0
?=?? (1)=2
?? '
?=?? (?? ,?? )=????
h?=0.2
?? 1
?=h?? (?? 0
,?? 0
)
?=0.2?? (1,2)
?=0.2×1×2=0.4
?? 2
?=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
 
?=0.2?? (1+
0.2
2
,2+
0.4
2
)
?=0.2?? (1.1,2.2)
?=0.2×1.1×2.2=0.484
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)
?=0.2?? (1+
0.2
2
,2+
0.484
2
)
?=0.2?? (1.1,2.242)
?=0.2×1.1×2.242
?=0.49324
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)
?=0.2?? (1+0.2,2+0.49324)
?=0.2?? (1.2,2.49324)
?=0.2×1.2×2.49324
?=0.598
?? 1
=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
?=
1
6
(0.4+2(0.484)+2(0.49324)+0.598)
 So, 
?=
1
6
(0.4+0.968+0.98648+0.598)
?=
2.95248
6
=0.49208
??? f(1.2)=2+0.49208
 
3.3 Find 
????
????
 at ?? =?? .?? from the following data : 
 
Solution: 
We will use Newton Forward interpolation formula. 
The table of finite differences is given by: 
Page 5


Edurev123 
3. Interpolation 
3.1 Using Lagrange interpolation formula, calculate the value of ?? (?? ) from the 
following table of values of ?? and ?? (?? ) : 
 
(2009 : 15 Marks) 
Solution: 
Usually Lagrange interpolation formula, first we will form the equation of ?? (?? ) . Now, as 
per the formula, 
?? (?? )=
(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 0
)
(?? 0
-?? 1
)(?? 0
-?? 2
)(?? 0
-?? 3
)(?? 0
-?? 4
)(?? 0
-?? 5
)
?+
(?? -?? 0
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 1
)
(?? 1
-?? 0
)(?? 1
-?? 2
)(?? 1
-?? 3
)(?? 1
-?? 4
)(?? 1
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 3
)(?? -?? 4
)(?? -?? 5
)?? (?? 2
)
(?? 2
-?? 0
)(?? 2
-?? 1
)(?? 2
-?? 3
)(?? 2
-?? 4
)(?? 2
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 4
)(?? -?? 5
)?? (?? 3
)
(?? 3
-?? 0
)(?? 3
-?? 1
)(?? 3
-?? 2
)(?? 3
-?? 4
)(?? 3
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 5
)?? (?? 4
)
(?? 4
-?? 0
)(?? 4
-?? 1
)(?? 4
-?? 2
)(?? 4
-?? 3
)(?? 4
-?? 5
)
?+
(?? -?? 0
)(?? -?? 1
)(?? -?? 2
)(?? -?? 3
)(?? -?? 4
)?? (?? 5
)
(?? 5
-?? 0
)(?? 5
-?? 1
)(?? 5
-?? 2
)(?? 5
-?? 3
)(?? 5
-?? 4
)
 
Now, ?? 0
=0,?? 1
=1,?? 2
=2,?? 3
=4,?? 4
=5,?? 5
=6 and ?? (?? 0
)=1,?? (?? 1
)=14,?? (?? 2
)=
15,?? (?? 3
)=5,?? (?? 4
)=6, ?? (?? 5
)=19. 
So, 
Now, 
?? (?? )=
(?? -1)(?? -2)(?? -4)(?? -5)(?? -6)×1
(0-1)(0-2)(0-4)(0-5)(0-6)
?+
(?? -0)(?? -2)(?? -4)(?? -5)(?? -6)×14
(1-0)(1-2)(1-4)(1-5)(1-6)
?+
(?? -0)(?? -1)(?? -4)(?? -5)(?? -6)×15
(2-0)(2-1)(2-4)(2-5)(2-6)
?+
(?? -0)(?? -1)(?? -2)(?? -5)(?? -6)×5
(4-0)(4-1)(4-2)(4-5)(4-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -6)×6
(5-0)(5-1)(5-2)(5-4)(5-6)
?+
(?? -0)(?? -1)(?? -2)(?? -4)(?? -5)×19
(6-0)(6-1)(6-2)(6-4)(6-5)
(3-1)(3-2)(3-4)(3-5)(3-6)
-1×-2×-4×-5×-6
?+
(3-0)(3-2)(3-4)(3-5)(3-6)×14
1×(-1)×(-3)(-4)(-5)
?+
(3-0)(3-1)(3-4)(3-5)(3-6)×15
2×1×(-2)(-3)(-4)
?+
(3-0)(3-1)(3-2)(3-5)(3-6)×5
4×3×2×(-1)(-2)
?+
(3-1)(3-2)(3-4)(3-6)×6
5×4×3×1×(-1)
 
?+
(3-0)(3-1)(3-2)(3-4)(3-5)×19
6×5×4×2×1
?+
2×1×(+1)×(-2)×(-3)
2×4×5×(+6)
+
3×1×(-1)×(+2)×(-3)×14
1×3×4×5
?+
3×2×1×(-1)(-3)×6
5×4×3×(-1)
+
3×2×1×(-1)×(-2)×19
6×5×4×2
=
1
20
-
21
5
+
45
4
+
15
4
-
9
5
+
19
20
=
1-84+180+75-36+19
20
=
275-120
20
=
155
20
=
31
4
?? =
31
4
 
3.2 Find the value of ?? (?? .?? ) using Runge-Kutta fourth order method with step size 
?? =?? .?? from the initial value problem. 
?? '
?=????
?? (?? )?=?? 
(2009: 15 marks) 
Solution: 
As per Runge-Kutta forth order method, 
?? 1
=h?? (?? 0
,?? 0
)
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)
?? =
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
 
and next ?? will be 
?? 1
=?? 0
+?? ?????????
 and ?? (?? ,?? )=?? '
?????????
 Here ?? (?? ,?? )=???? ?????????
 
Now, as per the given problem 
?? 0
?=?? (1)=2
?? '
?=?? (?? ,?? )=????
h?=0.2
?? 1
?=h?? (?? 0
,?? 0
)
?=0.2?? (1,2)
?=0.2×1×2=0.4
?? 2
?=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
 
?=0.2?? (1+
0.2
2
,2+
0.4
2
)
?=0.2?? (1.1,2.2)
?=0.2×1.1×2.2=0.484
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)
?=0.2?? (1+
0.2
2
,2+
0.484
2
)
?=0.2?? (1.1,2.242)
?=0.2×1.1×2.242
?=0.49324
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)
?=0.2?? (1+0.2,2+0.49324)
?=0.2?? (1.2,2.49324)
?=0.2×1.2×2.49324
?=0.598
?? 1
=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
?=
1
6
(0.4+2(0.484)+2(0.49324)+0.598)
 So, 
?=
1
6
(0.4+0.968+0.98648+0.598)
?=
2.95248
6
=0.49208
??? f(1.2)=2+0.49208
 
3.3 Find 
????
????
 at ?? =?? .?? from the following data : 
 
Solution: 
We will use Newton Forward interpolation formula. 
The table of finite differences is given by: 
 
By Newton Forward interpolation formula, we have 
?? (?? +?? h)=?? (?? )+?????? (?? )+
?? (?? -1)
2!
?
2
?? (?? )+
?? (?? -1)(?? -2)
3!
?? 3
?? (?? )+?. (??)
 
Here, ?? =0.1,h=0.1 
??????????????????????????????? =
?? -?? h
=
?? -0.1
0.1
 
? from (i), 
?? (?? )=0.9975+
(?? -0.1)
0.1
·(-0.0075)+
(?? -0.1)
0.1
·(
?? -0.1
0.1
-1)·
1
2
·(-0.0049)
?+(
?? -0.1
0.1
)(
?? -0.1
0.1
-1)(
?? -0.1
0.1
-2)·
1
6
(0.0001)
???? =?? (?? )=0.9975+
(?? -0.1)
0.1
·(-0.0075)+
(?? -0.1)
0.1
(
?? -0.2
0.1
)·
1
2
(-0.0049)
?+
(?? -0.1)
0.1
·
(?? -0.2)
0.1
·
(?? -0.3)
0.1
·
1
6
(0.0001)
??
????
????
=
1
0.1
(-0.0075)+
1
0.1
(
?? -0.2
0.1
)·
1
2
(-0.0049)+
(?? -0.1)
0.1
·
1
0.1
·
1
2
(-0.0049)
?+
(3?? 2
-1.2?? -0.07)
6×(0.1)
3
·(0.0001)
??
?? '
?? ????
(?????? =0.1)=?-
0.0075
0.1
+
(-0.1)
(0.1)
2
·
1
2
(-0.0049)+
(3(0.1)
2
-1.2×0.1-0.07)
6×(0.1)
3
(0.0001)
=?-0.0532
 
3.4 In an examination, the number of students who obtained marks between 
certain limits are given in the following table: 
 
Using Newton's forward interpolation formula, find the number of students whose 
marks be between 45 and 50 .

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FAQs on Interpolation - Mathematics Optional Notes for UPSC

1. What is interpolation?

Ans. Interpolation is a method of constructing new data points within the range of a discrete set of known data points.

2. How is interpolation different from extrapolation?

Ans. Interpolation involves estimating values between known data points, while extrapolation involves estimating values outside the range of known data points.

3. What are some common methods of interpolation?

Ans. Some common methods of interpolation include linear interpolation, polynomial interpolation, and spline interpolation.

4. How is interpolation used in data analysis and modeling?

Ans. Interpolation is used in data analysis to estimate unknown values, fill in missing data, and create smooth curves to represent the data.

5. Can interpolation be used to predict future trends in data?

Ans. No, interpolation cannot be used to predict future trends in data as it only estimates values between known data points and does not account for changes in the data over time.

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