Namespaces
Variants
Actions

Stirling interpolation formula

From Encyclopedia of Mathematics
Revision as of 00:52, 31 December 2021 by Nemesgery (talk | contribs) (corrected several typos)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.


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)

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

[a1] F.B. Hildebrand, "Introduction to numerical analysis" , Dover, reprint (1987) pp. 139
How to Cite This Entry:
Stirling interpolation formula. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Stirling_interpolation_formula&oldid=51820
This article was adapted from an original article by M.K. Samarin (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article