Question

central difference interpolation formula?

Answer 1

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 : Stirling's formula decrease much more rapidly than other difference formulae hence considering first few number of terms itself will give better accuracy. 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