# Stirling interpolation formula

The half-sum of the Gauss interpolation formula for forward interpolation with respect to the nodes $x _ {0} , x _ {0} + h, x _ {0} - h ,\dots, x _ {0} + nh, x _ {0} - nh$ at the point $x = x _ {0} + th$:

$$G _ {2n} ( x _ {0} + th) = \ f _ {0} + f _ {1/2} ^ { 1 } t + f _ {0} ^ { 2 }\frac{ t( t- 1)}{2!} +$$

$$+ f _ {1/2} ^ { 3 } \frac{t( t ^ {2} - 1 ^ {2} ) }{3!} + f _ {0} ^ { 4 } \frac{t( t ^ {2} - 1 ^ {2} )( t - 2) }{4!} + \dots +$$

$$+ f _ {0} ^ { 2n } \frac{t( t ^ {2} - 1 ^ {2} ) {} \dots [ t ^ {2} -( n- 1) ^ {2} ]( t- n) }{(2n)!}$$

and Gauss' formula of the same order for backward interpolation with respect to the nodes $x _ {0} , x _ {0} - h, x _ {0} + h, \dots ,x _ {0} - nh, x _ {0} + nh$:

$$G _ {2n} ( x _ {0} + th) = \ f _ {0} + f _ {-1/2 } ^ { 1 } t + f _ {0} ^ { 2 } \frac{t( t+ 1)}{2!} + \dots +$$

$$+ f _ {0} ^ { 2n } \frac{t( t ^ {2} - 1) \dots [ t ^ {2} -( n- 1) ^ {2} ]( t+ n) }{( 2n)! } .$$

Using the notation

$$f _ {0} ^ { 2k- 1 } = \ \frac{1}{2} [ f _ {1/2} ^ { 2k- 1 } + f _ {- 1/2} ^ { 2k- 1 } ] ,$$

Stirling's interpolation formula takes the form:

$$L _ {2n} ( x) = L _ {2n} ( x _ {0} + th) = \ f _ {0} + tf _ {0} ^ { 1 } + \frac{t ^ {2} }{2!} f _ {0} ^ { 2 } + \dots +$$

$$+ \frac{t( t ^ {2} - 1) \dots [ t ^ {2} -( n- 1) ^ {2} ] }{( 2n- 1)!} f _ {0} ^ { 2n- 1 } + \frac{t( t ^ {2} - 1) \dots [ t ^ {2} -( n- 1) ^ {2} ] }{( 2n)! } f _ {0} ^ { 2n } .$$

For small $t$, Stirling's interpolation formula is more exact than other interpolation formulas.

#### References

 [1] I.S. Berezin, N.P. Zhidkov, "Computing methods" , Pergamon (1973) (Translated from Russian)

The central differences $f _ {i+ 1/2 } ^ { 2m+ 1 }$ and $f _ {i} ^ { 2m }$( $m = 0, 1 \dots$ $i = \dots, - 1, 0, 1, . . .$) are defined recursively from the (tabulated values) $f _ {i} ^ { 0 } = f ( x _ {0} + ih)$ by the formulas
$$f _ {i + 1/2 } ^ { 2m+ 1 } = \ f _ {i+} 1 ^ { 2m } - f _ {i} ^ { 2m } ; \ \ f _ {i} ^ { 2m } = \ f _ {i + 1/2 } ^ { 2m- 1 } - f _ {i - 1/2 } ^ { 2m - 1 } .$$