All Courses
Get the App Signup / Login
Sign in Open App
UPSC Exam  >  UPSC Notes  >  Mathematics Optional Notes for UPSC  >  Interpolation

Interpolation | Mathematics Optional Notes for UPSC PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 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 .
Read More
387 videos|203 docs
Join Course for Free

Top Courses for UPSC

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.
Related Exams
About this Document
Nov 24, 2024 Last updated
Document Description: Interpolation for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation. The notes and questions for Interpolation have been prepared according to the UPSC exam syllabus. Information about Interpolation covers topics like and Interpolation Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Interpolation.
Introduction of Interpolation in English is available as part of our Mathematics Optional Notes for UPSC for UPSC & Interpolation in Hindi for Mathematics Optional Notes for UPSC course. Download more important topics related with notes, lectures and mock test series for UPSC Exam by signing up for free. UPSC: Interpolation | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Interpolation | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Interpolation
In this doc you can find the meaning of Interpolation defined & explained in the simplest way possible. Besides explaining types of Interpolation theory, EduRev gives you an ample number of questions to practice Interpolation tests, examples and also practice UPSC tests
387 videos|203 docs
Join Course for Free
Download as PDF
Explore Courses for UPSC exam

Top Courses for UPSC

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

pdf

,

Summary

,

MCQs

,

Interpolation | Mathematics Optional Notes for UPSC

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

Semester Notes

,

Objective type Questions

,

study material

,

Sample Paper

,

practice quizzes

,

Extra Questions

,

Free

,

Interpolation | Mathematics Optional Notes for UPSC

,

Viva Questions

,

Important questions

,

video lectures

,

Interpolation | Mathematics Optional Notes for UPSC

,

mock tests for examination

,

past year papers

,

ppt

,

Exam

;
Additional Information about Interpolation for UPSC Preparation

Interpolation Free PDF Download

The Interpolation is an invaluable resource that delves deep into the core of the UPSC exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the Interpolation now and kickstart your journey towards success in the UPSC exam.

Importance of Interpolation

The importance of Interpolation cannot be overstated, especially for UPSC aspirants. This document holds the key to success in the UPSC exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

Interpolation Notes

Interpolation Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Interpolation. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, Interpolation Notes on EduRev are your ultimate resource for success.

Interpolation UPSC Questions

The "Interpolation UPSC Questions" guide is a valuable resource for all aspiring students preparing for the UPSC exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study Interpolation on the App

Students of UPSC can study Interpolation alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Interpolation, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of Interpolation is prepared as per the latest UPSC syllabus.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup