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)f = ( 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
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

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