Central interpolation

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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
\( y=y_0+p \frac{ \delta y_{-1/2}+ \delta y_{1/2}}{2}+ \frac{p^2}{2!} \delta ^2 y_0+ \frac{p( p^2-1 )}{3!} \frac{ \delta ^3 y_{-1/2}+ \delta ^3 y_{1/2}}{2} +\frac{p^2( p^2-1 )}{4!} \delta ^4 y_0+ \frac{p( p^2-1 )( p^2-4 )}{5!} \frac{ \delta ^5 y_{-1/2}+\delta ^5 y_{1/2}}{2}+ ...\)

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Practical guideline for central interpolation

1. Gauss's Forward central Interpolatation
   if p lies between 0 and 1
2. Gauss's Backward central Interpolatation
   if p lies between -1 and 0
3. Stirling's central Interpolatation
   if p lies between -1/4 and 1/4
   or highest differences is of odd order
4. Bessel's central Interpolatation
   if p lies between 1/4 and 3/4
   or highest differences is of even order
5. Everett's central Interpolatation
   if p lies between 1/4 and 3/4

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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
\( y=y_0+p \frac{ \delta y_{-1/2}+ \delta y_{1/2}}{2}+ \frac{p^2}{2!} \delta ^2 y_0+ \frac{p( p^2-1 )}{3!} \frac{ \delta ^3 y_{-1/2}+ \delta ^3 y_{1/2}}{2} +\frac{p^2( p^2-1 )}{4!} \delta ^4 y_0+ \frac{p( p^2-1 )( p^2-4 )}{5!} \frac{ \delta ^5 y_{-1/2}+\delta ^5 y_{1/2}}{2}+ ...\)
or \( y=183+0.5 \frac{ 26+ 18}{2}+ \frac{0.5^2}{2!} (-8)+ \frac{0.5( 0.5^2-1 )}{3!} \frac{ (-26)+(-6)}{2} \)
\( +\frac{0.5^2( 0.5^2-1 )}{4!} (20)+ \frac{0.5( 0.5^2-1 )( 0.5^2-4 )}{5!} \frac{(70)+(-16)}{2}+ \frac{0.5^2( 0.5^2-1 )(0.5^2-4 )}{6!} (-86)\)

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Exercise

Use Stirling's central difference formula to compute followings.

1. Find the value when x=2.5 from the following table

   x	0	1	2	3	4
   y	1	0	1	10	8

2. Find solution of f(5.5) of an equation \(x^2-x+1\) using a = 2 and b= 10, step value h = 1.
3. Estimate the number of students who obtained marks between 55 and 60

   Marks	30-40	40-50	50-60	60-70	70-80
   Number	31	42	51	35	31

4. Find cubic polynomial which takes the following values: y( 0 )=1,y( 1 )=2, y( 2 )=1 and y( 3 )=10