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CENTRAL DIFFERENCE FORMULA
Consider a function f(x) tabulated for equally spaced points x0, x1, x2, . . ., xn with step length h. In many problems one may be interested to know the behaviour of f(x) in the neighbourhood of xr (x0 + rh). If we take the transformation X = (x - (x0 + rh)) / h,  the data points for X and f(X) can be written as 
x
X
f(X)
  x0 + (r - 2)h
-2
f-2
  x0 + (r -1)h
-1
f-1
  x0 + rh
0
f0
  x0 + (r + 1)h
1
f1
  x0 + (r + 2)h
2
f2
now the central difference table can be generated using the definition of central differences:

df(X) =  f(X + h/2) - f(X - h/2)

dfi =  (E1/2 - E-1/2)fi  = ( fi +1/2 - fi -1/2)

d2fi =  (E1/2 - E-1/2) ( fi +1/2 - fi -1/2)

=  f1 - f0 - f0 + f-1   =  f1 - 2f0 + f-1

 
Now the central difference table is
Xi
  fi
dfi
d2fi
d3fi
d4fi
-2
  f-2
        
 
  
df-3/2
( = f-1 - f-2)
      
-1
  f-1
  
d2f-1
( = df-1/2-df-3/2)
    
 
  
df-1/2
( = f0 - f-1)
  
d3f-1/2
( = d2f0 - d2f-1)
  
0
 f0
  
d2f0
( = df1/2-df-1/2)
  
d4f0
( = d3f1/2 -d3f-1/2)
 
  
df1/2
( = f1 - f0)
  
d3f1/2
( = d2f1 - d2f0)
  
1
 f1
  
d2f1
( = df3/2-df1/2)
    
 
  
df3/2
( = f2 - f1)
      
2
 f2
        
Gauss and Stirling formulae :
Consider the central difference table interms of forward difference operator D and with Sheppard's Zigzag rule
-3
 f-3
            
   
Df-3
          
-2
 f-2
  
D2f-3
        
   
Df-2
  
D3f-3
      
-1
 f-1
  
D2f-2
  
D4f-3
    
   
Df-1
  
D3f-2
  
D5f-3
  
0
 f0
  
D2f-1
  
D4f-2
  
D6f-3
   
Df0
  
D3f-1
  
D5f-2
  
1
 f1
  
D2f0
  
D4f-1
    
   
Df1
  
D3f0
      
2
 f2
  
D2f1
        
   
Df2
          
3
 f3
            
               
Now by divided difference formula along the solid line interms of forward difference operator
(f[x0, x1 . . . xr] = Drfx / r!)   is
f(x) = f0+ xDf0+
 x(x-1) 
 D2f0+ x(x-1)(x+1) D3f0+
 x(x-1)(x+1)(x-2)
 D4f0+  x(x-1)(x+1)(x-2)(x+2) D5f0 + . . .
2!
3!
4!
5!
or
f(x) = f0  + ( x ) Df0  + (
x
 ) D2f-1 + (
x+1
 ) D3f-1 + (
x+1
 ) D4f-1 + (
x+2
 ) D5f-2   + . . .
1
2
3
4
5
is called the Gauss forward difference formula.
Now if we repeat the same along dotted line weget
f(x) = f0  + ( x ) Df-1  + (
x+1
 ) D2f-1 + (
x+1
 ) D3f-2 + (
x+2
 ) D4f-2    + . . .
1
2
3
4
is called the Gauss backward difference formula.
Now changing these two formulae to d notation produces respectively
f(x) = f0  + ( x ) df1/2  + (
x
 ) d2f0 + (
x+1
 ) d3f1/2 + (
x+1
 ) d4f0    + . . .
1
2
3
4

 
f(x) = f0  + ( x ) df-1/2  + (
x+1
 ) d2f0 + (
x+1
 ) d3f-1/2+ (
x+2
 ) d4f0    + . . .
1
2
3
4
Now by adding these two expression and dividing by two gives
f(x) = f0  + ( x ) mdf0  +
x
 (
x
 ) d2f0 + (
x+1
 ) md3f0+
x
 (
x+1
 ) d4f0    + . . .
1
2
 1
3
4
3
or
f(x) = f0  +   x    mdf0  + 
 1 
   x2 d2f0 + 
 1 
   x(x2-12) md3f0+ 
 1 
   x2(x2-12) d4f0    + . . .
1!
2!
3!
4!
where the averaging operator m is defined as
mf(x)  = 
 f(x + h) - f(x - h) 
2
This formula is called the Stirling's interpolation formula.
Example :
Using Stirling's formula compute f(12.2) from the data
x
f(x)
X
10
0.23967
-2
11
0.28060
-1
12
0.31788
0
13
0.35209
1
14
0.38368
2

 
 
X
  fx = 105f(x)
dfx
d2fx
d3fx
d4fx
-2
  23967
        
 
  
4093
      
-1
  28060
  
-365
    
 
  
3728
  
58
  
0
31788
  
-307
  
-13
 
  
3421
  
45
  
1
35209
  
-262
    
 
  
3159
      
2
38368
        

 
f0.2 = f0  + ( x ) mdf0  +
x
 (
x
 ) d2f0 + (
x+1
 ) md3f0+
x
 (
x+1
 ) d4f0 
1
2
 1
3
4
3

 
= 31788 + 0.2 
3728 + 3421
 + 
0.2 ´ 0.2
(-307)
 + 
(1.2)(0.2)(-0.8)
58 + 45
 + 
0.2 
1.2 ´ 0.2 ´ (-0.8)
(-13)
2
2
3!
2
4
3!
= 31788 + 714.9 - 6.14 - 1.648 + 0.208
= 32495
Þf(x) = 10-5fx = 0.32495
Advantages :
    1. Stirling's formula decrease much more rapidly than other difference formulae hence considering first few number of terms itself will give better accuracy.
    2. Forward or backward difference formulae use the oneside information of the function where as Stirling's formula uses the function values on both sides of f(x).


    Bessel formula :

Combining the Gauss forward formula with Gauss Backward formula based on a zigzag line just one unit below the earlier one gives the Bessel formula. This is equivalent to
f(x) = f1  + ( x-1 ) df1/2  + (
x
 ) d2f1 + (
x
 ) d3f1/2 + (
x+1
 ) d4f1 + (
x+1
 ) d5f1/2   + . . .
1
2
3
4
5
Then the Bessel formula is
f(x) = mf1/2+ (x-1/2) df1/2+ (
x
 ) md2f1/2+ (1/3)(x-1/2)(
x
 ) d3f1/2+ (
x+1
 ) md4f1/2 +(1/5)(x-1/2)( x+1 )d5f1/2 +...
2
 2
4
4
set x = z + 1/2
fz+1/2=mf1/2+ z df1/2+ z2-1/4 
 1 
 md2f1/2+ z(z2-1/4)
 1 
 d3f1/2+ (z2-1/4)(z2-9/4)
 1 
 md4f1/2 +z(z2-1/4)(z2-9/4)
 1 
 d5f1/2 +...
2!
3!
4!
5!
for z = 0 we have
f1/2  =  mf1/2 -   1   md2f1/2 + 
  3 
  md4f1/2. . . 
8
 128
Now by choosing proper choise of origin x, one can take the central difference formula in the range
0 < x < 1 or in -1/2 < x < 1/2.
Example :
Compute 344.51/3 for the equation f(x) = x1/3
x
ux = 105f(x)
dux
d2ux
 
342
6993191
    
   
6809
  
343
7000000
  
-13
   
6796
  
344
7006796
  
-13
   
6783
  
345
7013579
  
-13
   
6770
  
346
7020349
  
-13
   
6757
  
347
7027106
    

 
u1/2 = 
14020375
 - 
 1 
 (-13)  =  7010189
2
8
Þf(x) = 7.010189
Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations