Stirlings Formula for central interpolation
X  Y  1st diff  2nd diff  3rd diff  4th diff  5th diff  6th diff 
\( x_{3}\)  \( y_{3}\)  
\( \delta y_{5/2}\)  
\( x_{2}\)  \( y_{2}\)  \( \delta^2 y_{2}\)  
\( \delta y_{3/2}\)  \( \delta^3 y_{3/2}\)  
\( x_{1}\)  \( y_{1}\)  \( \delta^2 y_{1}\)  \( \delta^4 y_{1}\)  
\( \delta y_{1/2}\)  \( \delta^3 y_{1/2}\)  \( \delta^5 y_{1/2}\)  
\( x_0\)  \( y_0\)  \( \delta^2 y_0\)  \( \delta^4 y_0\)  \( \delta^6 y_0\)  
\( \delta y_{1/2}\)  \( \delta^3 y_{1/2}\)  \( \delta^5 y_{1/2} \)  
\( x_1\)  \( y_1\)  \( \delta^2 y_1\)  \( \delta^4 y_1\)  
\( \delta y_{3/2}\)  \( \delta^3 y_{3/2}\)  
\( x_2\)  \( x_2\)  \( \delta^2 y_2\)  
\( \delta y_{5/2}\)  
\( x_3\)  \( x_3\) 
Taking the mean of Gauss' forward and backward formula, we obtain the Stirling's Formula given by
\( y=y_0+p \frac{ \delta y_{1/2}+ \delta y_{1/2}}{2}+ \frac{p^2}{2!} \delta ^2 y_0+ \frac{p( p^21 )}{3!} \frac{ \delta ^3 y_{1/2}+ \delta ^3 y_{1/2}}{2} +\frac{p^2( p^21 )}{4!} \delta ^4 y_0+ \frac{p( p^21 )( p^24 )}{5!} \frac{ \delta ^5 y_{1/2}+\delta ^5 y_{1/2}}{2}+ ...\)
Practical guideline for central interpolation
 Gauss's Forward central Interpolatation
if p lies between 0 and 1  Gauss's Backward central Interpolatation
if p lies between 1 and 0  Stirling's central Interpolatation
if p lies between 1/4 and 1/4
or highest differences is of odd order  Bessel's central Interpolatation
if p lies between 1/4 and 3/4
or highest differences is of even order  Everett's central Interpolatation
if p lies between 1/4 and 3/4
Example 1
Using Stirling's formula to find solution at x=1050
x  0  300  600  900  1200  1500  1800 
y  135  149  157  183  201  205  193 
Solution
According to given set of data values, we form difference table as below
X  Y  1st diff  2nd diff  3rd diff  4th diff  5th diff  6th diff 
0  135  
14  
300  149  6  
8  24  
600  157  18  50  
26  26  70  
900  183  8  20  86  
18  6  16  
1200  201  14  4  
4  2  
1500  205  16  
12  
1800  193 
Here,
\( h=300\) and \( x_0=900\)
Thus, we have
\( x=x_0+ph\)
or
\( 1050=900+p\times 300\)
or
\( p=0.5\)
Using formula, we get
\( y=y_0+p \frac{ \delta y_{1/2}+ \delta y_{1/2}}{2}+ \frac{p^2}{2!} \delta ^2 y_0+ \frac{p( p^21 )}{3!} \frac{ \delta ^3 y_{1/2}+ \delta ^3 y_{1/2}}{2} +\frac{p^2( p^21 )}{4!} \delta ^4 y_0+ \frac{p( p^21 )( p^24 )}{5!} \frac{ \delta ^5 y_{1/2}+\delta ^5 y_{1/2}}{2}+ ...\)
or
\( y=183+0.5 \frac{ 26+ 18}{2}+ \frac{0.5^2}{2!} (8)+ \frac{0.5( 0.5^21 )}{3!} \frac{ (26)+(6)}{2} \)
\( +\frac{0.5^2( 0.5^21 )}{4!} (20)+ \frac{0.5( 0.5^21 )( 0.5^24 )}{5!} \frac{(70)+(16)}{2}+ \frac{0.5^2( 0.5^21 )(0.5^24 )}{6!} (86)\)
Exercise
Use Stirling's central difference formula to compute followings. Find the value when x=2.5 from the following table
x 0 1 2 3 4 y 1 0 1 10 8  Find solution of f(5.5) of an equation \(x^2x+1\) using a = 2 and b= 10, step value h = 1.

Estimate the number of students who obtained marks between 55 and 60
Marks 3040 4050 5060 6070 7080 Number 31 42 51 35 31  Find cubic polynomial which takes the following values: y( 0 )=1,y( 1 )=2, y( 2 )=1 and y( 3 )=10
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