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The half-sum of the [[Gauss interpolation formula|Gauss interpolation formula]] for forward interpolation with respect to the nodes <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878401.png" /> at the point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878402.png" />:
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878403.png" /></td> </tr></table>
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878404.png" /></td> </tr></table>
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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 $:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878405.png" /></td> </tr></table>
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$$
 +
G _ {2n} ( x _ {0} + th)  = \
 +
f _ {0} + f _ {1/2} ^ { 1 } t + f _ {0} ^ { 2 }\frac{ t( t-
 +
1)}{2!}
 +
+
 +
$$
  
and Gauss' formula of the same order for backward interpolation with respect to the nodes <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878406.png" />:
+
$$
 +
+
 +
f _ {1/2} ^ { 3 }
 +
\frac{t( t  ^ {2} - 1  ^ {2} ) }{3!}
 +
+ f _ {0} ^ { 4 }
 +
\frac{t( t  ^ {2} - 1  ^ {2} )( t - 2) }{4!}
 +
+ \dots +
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878407.png" /></td> </tr></table>
+
$$
 +
+
 +
f _ {0} ^ { 2n }
 +
\frac{t( t  ^ {2} - 1  ^ {2} ) {} \dots [ t  ^ {2} -( n- 1)  ^ {2} ]( t- n) }{(2n)!}
 +
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878408.png" /></td> </tr></table>
+
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
 
Using the notation
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s0878409.png" /></td> </tr></table>
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$$
 +
f _ {0} ^ { 2k- 1 }  = \
 +
 
 +
\frac{1}{2}
 +
[ f _ {1/2} ^ { 2k- 1 } + f _ {- 1/2} ^ { 2k- 1 } ] ,
 +
$$
  
 
Stirling's interpolation formula takes the form:
 
Stirling's interpolation formula takes the form:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784010.png" /></td> </tr></table>
+
$$
 +
L _ {2n} ( x)  = L _ {2n} ( x _ {0} + th)  = \
 +
f _ {0} + tf _ {0} ^ { 1 } +
 +
\frac{t  ^ {2} }{2!}
 +
f _ {0} ^ { 2 } + \dots +
 +
$$
 +
 
 +
$$
 +
+
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784011.png" /></td> </tr></table>
+
\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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784012.png" />, Stirling's interpolation formula is more exact than other interpolation formulas.
+
For small $  t $,  
 +
Stirling's interpolation formula is more exact than other interpolation formulas.
  
 
====References====
 
====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>
 
<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====
 
====Comments====
The central differences <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784013.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784014.png" /> (<img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784015.png" /> <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784016.png" />) are defined recursively from the (tabulated values) <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784017.png" /> by the formulas
+
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
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s087/s087840/s08784018.png" /></td> </tr></table>
+
$$
 +
f _ {i + 1/2 }  ^ { 2m+ 1 }  = \
 +
f _ {i+} 1 ^ { 2m } - f _ {i} ^ { 2m } ; \ \
 +
f _ {i} ^ { 2m }  = \
 +
f _ {i + 1/2 }  ^ { 2m- 1 } - f _ {i - 1/2 }  ^ { 2m - 1 } .
 +
$$
  
 
====References====
 
====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>
 
<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>

Latest revision as of 00:52, 31 December 2021


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=12181
This article was adapted from an original article by M.K. Samarin (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article