Engineering Mathematics-IV_B.Tech_Semester-IV_Unit-IV

1. Finite Difference
and
Interpolation
Course- B.Tech
Semester-IV
Subject- ENGINEERING MATHEMATICS-IV
Unit- IV
RAI UNIVERSITY, AHMEDABAD

2. Unit-IV Finite Difference and Interpolation
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Content
Finite differences, difference tables, Newton’s forward interpolation & it’s problems, Newton’s
backward interpolation & it’s problems, stirling's interpolation formula & Problems based on it,
Newton’s divided difference formula for unequal intervals & it’s problems ,Lagrange’s divided
difference formula for unequal intervals & it’s problems

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1.1 Finite differences— Let = ( ) be a function and ∆ = ℎ denote the increment in the
independent variable . Assume that∆ , increment in the argument (also known as the
interval or spacing) is fixed. i.e. ℎ =constant . Then the first finite difference y is defined as—
∆ = ∆ ( ) = ( + ∆ ) − ( )
Similarly finite differences of higher orders are denoted as follows—
∆ = ∆(∆ ) = ∆ ( + ∆ ) − ( )
= ∆ ( + ∆ ) − ∆ ( )
= [ ( + 2∆ ) − ( + ∆ )] − [ ( + ∆ ) − ( )]
∆ = ( + 2∆ ) − 2 ( + ∆ ) + ( )
In general ∆ = ∆(∆ ), for = 2,3,4 …
Now consider the function = ( )specified by the tabulated series = ( ) for a set of
equivalent points where = 0,1,2, … , and ∆ = ∆ − = ℎ =constant. Thus the
tabulated function consists of ordered pairs ( , ), ( , ), … , ( , ), …. Here entries
are known as entries.
1.2 Forward Difference—
The first forward difference is denoted by ∆ and defined as
∆ = − .
The symbol ∆ is the forward difference operator.
Properties—
1. ∆ = 0 (Differences of constant function are zero)
2. ∆( ) = ∆( ), where is a constant .
3. ∆( + ) = ∆ + ∆
4. ∆( ) = ∆ + ∆
5. ∆ (∆ ) = ∆
6. Where and are non-negative integers and ∆ = (By definition).
7. The higher order forward difference are defined as:
8. The second order forward difference of is
9. ∆ = ∆(∆ ) = ∆ − ∆
In general,
∆ = ∆(∆ ) = ∆ − ∆
It defines the nth
order forward differences.
Any higher order forward differences can be expressed in terms of the successive values of
the function.
Example:
1.
∆ = − 2 +
2.
∆ = − 3 + 3 −

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Forward Difference Table
Value of Value of 1st
diff 2nd
diff 3rd
diff 4th
diff 5th
diff
∆ ∆ ∆ ∆ ∆
+ ℎ ∆ ∆ ∆ ∆
+ 2ℎ ∆ ∆ ∆
+ 3ℎ ∆ ∆
+ 4ℎ ∆
+ 5ℎ
1.3 Backward Difference—
The first backward difference is denoted by ∇ and defined as—
∇ = − .
The symbol ∇ is the backward difference operator.
Second order backward difference
∇ = ∇(∇ ) = ∇ − ∇
In general
∇ = ∇(∇ ) = ∇ − ∇
Now
∇ = ∇ − ∇ = (∇ − ∇ ) − (∇ − ∇ )
= ∇ − ∇ − ∇ + ∇
= ∇ − 2∇ + ∇
= ( − ) − 2( − ) − ( − )
= − −2 + 2 − +
= − 3 + 3 +
In general,
∇ = ∑ (−1)
Backward Difference Table
Value of Value of 1st
diff 2nd
diff 3rd
diff 4th
diff 5th
diff
= + ℎ ∇
= + 2ℎ ∇ ∇
= + 3ℎ ∇ ∇ ∇
= + 4ℎ ∇ ∇ ∇ ∇
= + 5ℎ ∇ ∇ ∇ ∇ ∇

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1.4 Central Difference— Sometimes it is convenient to employ another system of differences
known as central differences. The central difference operator is denoted by and defined by the
relation—
− = δ /
− = δ
.
.
.
− = δ ( )/
Similarly, the higher order central differences are defined as:
δ − δ = δ
δ / − δ = δ
δ − δ = δ / and so on.
So the table will be:
Central Difference Table
Value of Value of 1st
diff 2nd
diff 3rd
diff 4th
diff 5th
diff
δ /
+ ℎ δ
δ / δ /
+ 2ℎ δ δ
δ / δ / δ /
+ 3ℎ δ δ
δ / δ /
+ 4ℎ δ
δ /
+ 5ℎ
Example —Evaluate the followings—
(i) ∆ (ii) ∆ log 2 (iii) ∆ (iv)∆ ( )
Solution—
(i) ∆ = ( + ℎ) −
= ( )
= tan

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(ii) ∆ log 2 = log 2( + ℎ) − log 2
= log 2( + ℎ) − log 2 + log 2 − log 2
= [log 2( + ℎ) − log 2 ] + [ − ] log 2
= log
+ ℎ
+ [ − 1] log 2
= log 1 +
ℎ
+ ( − 1) log 2
(iii) ∆ = ∆ ( )( )
= ∆ +
= ∆ ∆ + ∆
= ∆ 2 − + 3 −
= 2∆ ( )( )
+3∆ ( )( )
= −2∆ ( )( )
−3∆ ( )( )
= −2 ( )( )
− ( )( )
− 3 ( )( )
− ( )( )
= ( )( )( )
+ ( )( )( )
=
( ) ( )
( )( )( )( )
= ( )( )( )( )
= ( )( )( )( )
=
( )
( )( )( )( )
(iv) ∆ ( ) = − = ( − 1)
∆ ( ) = ∆ (∆ ) = ∆[( − 1) ] = ( − 1)∆ = ( − 1)( − 1)
∆ ( ) = ( − 1)
∆ ( ) = ( − 1)
∆ ( ) = ( − 1)
⋮
∆ ( ) = ( − 1) .
1.5 Difference of a polynomial— The nth
difference of the nth -
degree polynomial are constant
and all higher order differences are zero.
Let the polynomial of nth
degree in is:
( ) = + + … + ( + ℎ) +
∆ ( ) = ( + ℎ) − ( )
= [( + ℎ) − ] + [( + ℎ) − ] + ⋯ + ℎ
= ℎ + + + ⋯ + + ′
Where , , … , ′ are new constants.
Thus the first difference of a polynomial of nth -
degree is a polynomial of degree ( − 1).

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Similarly
∆ ( ) = ∆[ ( + ℎ) − ( )] = ∆ ( + ℎ) − ∆ ( )
= ℎ[( + ℎ) − ] + [( + ℎ) − ] + ⋯ + ℎ
= ( − 1)ℎ + + + ⋯ + ′′
∴ the second differences represent a polynomial of degree ( − 2).
Continuing in this process for the nth differences we get a polynomial of degree zero.
i.e. ∆ ( ) = ( − 1)( − 2) … 1
ℎ = ! ℎ , which is constant.
Hence the ( + 1)th
and higher order differences of a polynomial of nth
degree will be zero.
Example— Evaluate ∆ [(1 − )(1 − )(1 − )(1 − )].
Solution— ∆ [(1 − )(1 − )(1 − )(1 − )]
= ∆ [ + (. . . ) + (… ) + ⋯ + 1]
= (10!) [∵ ∆ ( ) = 0 < 10]
Example— Find the missing value of the following table:
: 45 50 55 60 65
: 3.0 _______ 2.0 ________ -2.4
Solution—The difference is –
∆ ∆ ∆
45 = 3.0 − 3 5 − 2 3 + − 9
50 2 − + − 4 3.6 − − 3
55 = 2.0 − 2 −0.4 − 2
60 −2.4 −
65 = −2.4
Solving the two equations 3 + − 9 = 0 and 3.6 − − 3 = 0.
we can find the value of and .
3 + = 9 … ( )
+ 3 = 3.6 … ( )
From ( )
= 9 − 3 .
Substituting the value of ( )
+ 3(9 − 3 ) = 3.6
⟹ −8 = 3.6 − 27
⟹ =
−23.4
−8
= 2.935
= 9 − 3(2.935) = 9 − 8.775 = 0.225.

8. Unit-IV Finite Difference and Interpolation
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Other Difference Operators—
1. Shift Operator(E)—
Shift operator is the operation of increasing the argument by ℎ so that
( ) = ( + ℎ) ( ) = ( + 2ℎ) ( ) = ( + 3ℎ) …
The inverse of shift operator is defined as ( ) = ( − ℎ).
If is the function of ( ), then = , = , = , where may
be any real number.
2. Averaging operator( )—
Averaging operator =
Relation between the operators—
1. ∆= − 1
2. ∇= 1 −
3. = −
4. = +
5. ∆= ∇= ∇ =

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2.1 Interpolation—Let = ( ) is tabulated for the equally spaced values of
= , , , … , , where = + ℎ, = 0,1,2, … ,
⟹ = + ℎ,
= + 2ℎ
= + 3ℎ
= + ℎ
It gives— = , , ,… ,
…
…
The process of finding the values of corresponding to any value of = between and
is called interpolation.
The study of interpolation is based on the concept of difference of a function.
To determine the values of ( ) of ′( ) for some intermediate values of various types of
difference are very much useful.
2.2 Newton’s forward difference interpolation— Let the function = ( ) takes the values
, , ,… corresponding to the values , + ℎ, + 2ℎ, … of . Suppose it is required to
evaluate ( ) for = + ℎ, where is any real number.
For any real number , we have defined such that—
( ) = ( + ℎ)
= ( + ℎ) = ( ) = (1 + ∆) [∵ = 1 +△]
= 1 + △ +
( )
!
△ +
( )( )
!
△ + ⋯ [Binomial theorem]
= + △ +
( )
!
△ +
( )( )
!
△ + ⋯ … (1)
It is called Newton’s forward difference interpolation formula as (1) contains and the
forward differences of .
2.3 Newton’s backward difference interpolation— Let the function = ( ) takes the
values , , ,… corresponding to the values , + ℎ, + 2ℎ, … of . Suppose it is
required to evaluate ( ) for
= + ℎ, where is any real number.
Then we have
= ( + ℎ) = ( ) = (1 − ∇) [∵ = 1 − ∇]
= 1 + ∇ +
( )
!
∇ +
( )( )
!
∇ + ⋯ [ ℎ ]
= + ∇ +
( )
!
∇ +
( )( )
!
∇ + ⋯ … (2)
It is called Newton’s forward difference interpolation formula as eq (1) contains and the
forward differences of .

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Example—The table gives the distances in the nautical miles of the visible horizon for the
given heights in feet above the earth’s surface:
=height : 100 150 200 250 300 350 400
=distance: 10.63 13.03 15.04 16.81 18.42 19.90 21.27
Find the value of , when ( ) = 218 ( ) 410 .
Solution—The forward difference table is—
∆ ∆ ∆ ∆
100 10.63
150 13.03 2.40
200 15.04 2.01 -0.39
250 16.81 1.77 -0.24 0.15
300 18.42 1.61 -0.16 0.08 -0.07
350 19.90 1.48 -0.13 0.03 -0.05
400 21.27 1.37 -0.11 0.02 -0.01
(i) For = 200, = 15.04
∆ = 1.77, ∆ = −0.16, ∆ = 0.03etc.
Since = 218 and ℎ = 50
∴ = = = 0.36
Using Newton’s forward difference interpolation formula we get
= + △ +
( )
!
△ +
( )( )
!
△ + ⋯
= 15.04 + 0.36(1.77) +
. ( . )
(−0.16) +
. ( . )( . )
(0.03) + ⋯
= 15.04 + 0.637 + 0.018 + 0.001 + ⋯ = 15.696 i.e. 15.7 nautical miles.
(ii) Since = 410 is near the end of the table, we use Newton’s backward difference
interpolation formula.
∇ ∇ ∇ ∇
100 10.63 2.40 -0.39 0.15 -0.07
150 13.03 2.01 -0.24 0.08 -0.05
200 15.04 1.77 -0.16 0.03 -0.01
250 16.81 1.61 -0.13 0.02
300 18.42 1.48 -0.11
350 19.90 1.37
400 21.27
∴ Taking = 400, = = = 0.2

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Using the line of backward difference:
= 21.27, ∇ = 1.37, ∇ = −0.11, ∇ = 0.02 etc
Newton’s backward difference interpolation formula gives
= + ∇ +
( + 1)
2
∇ +
( + 1)( + 2)
6
∇ + ⋯
= 21.27 + 0.2(1.37) +
. ( . )
(−0.11) +
. ( . )( . )
(0.02)
+
0.2(1.2)(2.2)(3.2)
24
(−0.01) + ⋯
= 21.27 + 0.274 − 0.0132 + 0.00176 − 0.000704 + ⋯
= 21.531856
. . 21.5 nautical miles.
Example— From the following table, estimates the number of students who obtained marks
between 40 to 45.
Marks 30-40 40-50 50-60 60-70 70-80
No of Students 31 42 51 35 31
Solution— First we have to prepare the cumulative frequency table:
Marks less than
( )
40 50 60 70 80
No of Students
( )
31 73 124 159 190
Now the difference table is
∆ ∆ ∆ ∆
40 31 42 9 -25 37
50 73 51 -16 12
60 124 35 -4
70 159 31
80 190
We shall find i.e. the number of mark sheets less than 45.Using Newton’s forward
difference interpolation formula we get—
Taking = 40, = 45, we have = = = 0.5
= + △ +
( − 1)
2!
△ +
( − 1)( − 2)
3!
△ + ⋯
= 31 + 0.5 × 42 +
. ( . )
× 9 +
. ( . )( . )
× (−25) +
. ( . )( . )( . )
× 37
= 31 + 21 − 1.125 + 1.5625 − 1.4453125 = 50.9921875
The number of the students with marks less than 45 is 50.9921875 i.e. 51.
But the number student with marks less than 40 is 31.
Hence the number of students getting marks between 40 and 45 = 51 − 31 = 20.

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2.4 Central Difference Interpolation Formulae— If takes the values − 2ℎ, −
ℎ, , + ℎ, + 2ℎ are the corresponding values of = ( ) are , , , , , then we
write the difference table as:
Value of Value of 1st
diff 2nd
diff 3rd
diff 4th
diff
− 2ℎ
∆ (= δ / )
− ℎ ∆ (= δ )
∆ (= δ / ) ∆ (= δ )
∆ (= δ ) ∆ (= δ )
∆ (= δ / ) ∆ (= δ )
+ ℎ ∆ (= δ )
∆ (= δ / )
+ 2ℎ
2.5 Lagrange’s interpolation formula—If = ( ) takes the values , , , …
corresponding to = , , , … , then—
( ) =
( − )( − ) … ( − )
( − )( − ) … ( − )
+
( − )( − ) … ( − )
( − )( − ) … ( − )
+ ⋯ +
( )( )…( )
( )( )…( )
This is known as Lagrange’s interpolation formula for unequal intervals.
Example— Given the values
5 7 11 13 17
( ) 150 392 1492 2366 5202
Evaluate— (9) , using Lagrange’s formula.
Solution—Here = 5, = 7, = 11, = 13, = 17 and
= 150, = 392, = 1452, = 2366, = 5202
Putting = 9 and substituting the above values in Lagrange’s formula, we get
(9) =
(9 − 7)(9 − 11)(9 − 13)(9 − 17)
(5 − 7)(5 − 11)(5 − 13)(3 − 17)
× 150
+
(9 − 5)(9 − 11)(9 − 13)(9 − 17)
(7 − 5)(7 − 11)(7 − 13)(7 − 17)
× 392
+
(9 − 5)(9 − 7)(9 − 13)(9 − 17)
(11 − 5)(11 − 7)(11 − 13)(11 − 17)
× 1452
+
(9 − 5)(9 − 7)(9 − 11)(9 − 17)
(13 − 5)(13 − 7)(13 − 11)(13 − 17)
× 2366
+
(9 − 5)(9 − 7)(9 − 11)(9 − 13)
(17 − 5)(17 − 7)(17 − 11)(17 − 13)
× 5202
= −
50
2
+
3136
15
+
3872
3
−
2366
3
+
578
5
= 810

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2.6 Divided difference—
TheLagrange’s formula has the drawback that if another interpolation areinterested then the
interpolation coefficient are required to recalculate.
This problem is solved in Newton’s divided difference interpolation formula.
If ( , ), ( , ), ( , ) … be given points, then the first divided difference for the argument
is , is defined by the relation [ , ] = .
Similarly,
[ , ] = and[ , ] = etc.
The second divided difference for the argument is , , is defined as [ , , ] =
[ ] [ , ]
.
The third divided difference for the argument is , , , is defined as [ , , , ] =
[ , , ] [ , , ]
and so on.
2.7 Newton’s divided difference interpolation formula— Let , , , … be the values of
= ( ) corresponding to the arguments , , , … , . Then from the definition of divided
differences, we have [ , ] =
= +( − )[ − ] … ( )
Again, [ , , ] =
[ , ] [ , ]
Which give—
[ , ] = [ , ] +( − ) + ( − ) [ , , ]
Substituting this values in the eq(i), we get
= +( − )[ − ] + ( − )( − ) [ , , ] … ( )
Also, [ , , , ] =
[ , , ] [ , , ]
Which gives [ , , ] = [ , , ] −( , )[ , , , ]
Substituting this values of [ , , ] in the eq(ii), we obtain
= +( − )[ − ] + ( − )( − ) [ , , ]
+( − )( − ) ( − )[ , , , ]
Proceeding in this way, we get
= +( − )[ − ] + ( − )( − ) [ , , ]
+( − )( − )( − )[ , , , ] + ⋯
+( − )( − ) … ( − )[ , , , … , ] … ( )
This is called as Newton’s divided difference interpolation formula.

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Example— Given the values
5 7 11 13 17
( ) 150 392 1492 2366 5202
Evaluate (9) , using Newton’s divided difference formula.
Solution—The divided difference table is:
1st
divided difference 2nd
divided difference 3rd
divided difference
5 150
392 − 150
7 − 5
= 121
7 392 265 − 121
11 − 5
= 24
1452 − 392
11 − 7
= 265
32 − 24
13 − 5
= 1
11 1452 457 − 265
13 − 7
= 32
2366 − 1452
13 − 11
= 457
42 − 32
17 − 7
= 1
13 2366 709 − 457
17 − 11
= 42
5202 − 2366
17 − 13
= 709
17 5202
Taking = 9 in the Newton’s divided difference formula, we get
(9) = 150 + (9 − 5) × 121 + (9 − 5)(9 − 7) × 24 + (9 − 5)(9 − 7)(9 − 11) × 1
= 150 + 484 + 192 − 16 = 810
Example—Determine ( ) as a polynomial in nfor the following data:
−4 −1 0 2 5
( ) 1245 33 5 9 1335
Solution—
1st
divided
difference
2nd
divided
difference
3rd
divided
difference
4th
divided
difference
−4 1245
−404
−1 33 94
−28 −14
0 5 10 3
2 13
2 9 88
442
5 1335

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Applying Newton’s divided difference formula—
( ) = ( ) + 0 +( − )[ − ] + ( − )( − ) [ , , ] + ⋯
= 1245 + ( + 4)(−404) + ( + 4)( + 1)(94)
+( + 4)( + 1)( − 0)(−14) + ( + 4) ( + 1)( − 2)(3)
= 3 + 5 + 6 − 14 + 5

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Exercise
Question—Assuming that the following values of belong to a polynomial of degree 4,
compute the next three values:
: 0 1 2 3 4 5 6 7
: 1 -1 1 -1 1 _____ _____ _____
1. Evaluate ∆ 2 .
2. Find ∆ at = 0.
3. Find ∆ at = 0.
4. Construct an forward difference table for
(1) = 1, (2) = 3, (3) = 8, (4) = 15, (5) = 25
and hence find ∆ (1).
5. Construct a central difference table for
(0) = 8, (1) = 12, (2) = 20, (3) = 34, (4) = 60
And hence, find (2).

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Reference
1. http://en.wikipedia.org/wiki/File:Finite_Differences.svg
2. Higher Engineering Mathematics, B.S. Grewal, Khanna Publishers.