FINITE
DIFFERENCES
Lecture 2: (a) Miscellaneous examples.
(b) Divided Differences.
Example 7. Find
, given
Solution: Since five entries are given, we assume ux to be a polynomial of degree 4.
Therefore,
constant and hence all differences above four is zero.
Now,
(
)
(
[Since, all differences above four is zero]
Example 8. Use the method of separation of symbols to prove the following identities:
(i)
Solution:
(ii)
where
.
Solution:
=
1
1  n1 C1   n 1 C2 2  .......  n 1C n 1 n 1   1 u 0


(iii)
Solution:
(iv)
Solution:
, (the interval of differencing is unity).
(v)
Solution:
=
(vi)
Solution:
=(
Example 9. If
prove that
Solution:
------------------(*)
Now,
Substituting these values in (*), we get
.
Exercises
1. Find
given
2. Given that
=-3,
.
is a polynomial of second degree and
Find the value of
=1,
.
3. Show that
4. Find the missing terms from the following table:
0
1
2
3
4
5
0
-
8
15
-
35
5. Given that
estimate the value of
6. Use the method of separation of symbols to prove the following identities :
(a)
(b)
(c)
(d)
(e)
=
.
4. Divided Differences
When the arguments are not equi-spaced, we use divided differences.
Let y = f(x) be a function whose functional form is not known but its
values at (n+1) points, namely, x0 , x1, … xn are known.
The divided difference of first order for the points x0, x1 is defined as
f  x 0 , x1  
f (x1 )  f (x 0 )
x1  x 0
For the points x0, x1, x2 the divided difference of second order is defined
as
f  x 0 , x1 , x 2  
f  x 0 , x1   f  x 1 , x 2 
x2  x0
1  f (x1 )  f (x 0 ) f (x 2 )  f (x1 ) 

x 2  x 0  x 0  x1
x 2  x1 
f (x 0 )
f (x1 )
f (x 2 )



(x1  x 0 )(x 2  x 0 ) (x 0  x1 )(x 2  x1 ) (x 0  x 2 )(x1  x 2 )

In the similar manner, the divided difference of nth order for the points
x0, x1 , … xn is defined as
f  x 0 , x1 ,..., x n  
f (x 0 )
f (x1 )

(x1  x 0 )(x 2  x 0 )...(x n  x 0 ) (x 0  x1 )(x 2  x1 )...(x n  x1 )
 ... 
f (x n )
(x 0  x n )(x1  x n )...(x n 1  x n )
Property 1:
f  x 0 , x1  
f (x1 )  f (x 0 ) f (x 0 )  f (x1 )

 f  x1 , x 0 
x1  x 0
x 0  x1
This implies divided difference is symmetric.
Property 2: If the arguments are equispaced, then
f  x1 , x  
f (x)  f (x1 ) f (x1 )  f (x) f (x  )  f (x)


,
x  x1
x1  x

Since, the arguments are equi-spaced,
 0
where we set
x1  x  
and we get,
f (x  )  f (x)
0
0

 f  x, x   f '(x),
lim f  x  , x   lim
provided f(x) is differentiable.
The divided difference tables with 5 arguments are as follows:
x
y
x0
y0
y
2y
3y
4y
f[x0, x1]
x1
y1
f[x0, x1, x2]
f[x1, x2]
x2
y2
f[x0, x1, x2, x3]
f[x1, x2, x3]
f[x2, x3]
x3
y3
f[x0, x1, x2, x3, x4]
f[x1, x2, x3, x4]
f[x2, x3, x4]
f[x3, x4]
x4
where
y4
y
is the first order divided differences,
divided differences and so on.
2y
is the second order
Example 10: Obtain the divided difference table for the function
y = f(x) given by
x
y
-1
1
1
-3
4
21
6
127
Solution: The divided difference tables with 4 arguments are as follows:
x
-1
y
1
y
2 y
3 y
3  1
 2
11
1
82
2
4 1
-3
21  3
8
4 1
4
21
53  8
9
6 1
92
1
6 1
127  21
 53
64
6
127
Exercise7: Obtain the divided difference table for the function y=f(x)
given by
x
y
-1
-1
-2
-9
2
11
4
69