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The half-sum of the [[Gauss interpolation formula|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====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  I.S. Berezin,   N.P. Zhidkov,   "Computing methods" , Pergamon  (1973)  (Translated from Russian)</TD></TR></table>

====Comments====
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

$$ 
f _ {i + 1/2 }  ^ { 2m+ 1 }  = \ 
f _ {i+} 1 ^ { 2m } - f _ {i} ^ { 2m } ; \ \ 
f _ {i} ^ { 2m }  = \ 
f _ {i + 1/2 }  ^ { 2m- 1 } - f _ {i - 1/2 }  ^ { 2m - 1 } .
$$

====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  F.B. Hildebrand,   "Introduction to numerical analysis" , Dover, reprint  (1987)  pp. 139</TD></TR></table>