Find Solution using Bessel's formula

x	f(x)
20	24
24	32
28	35
32	40

x = 25
Finding f(2)

Solution:
The value of table for `x` and `y`

x	20	24	28	32
y	24	32	35	40

Bessel's method to find solution

`h=24-20=4`

Taking `x_0=24` then `p=(x-x_0)/h=(x-24)/4`

The difference table is

`x`	`p=(x-24)/4`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`
20	-1	24
			8
24	0	32		-5
			3		7
28	1	35		2
			5
32	2	40

`x = 25`

`p = (x - x_0)/h = (25 - 24)/4 = 0.25`

`y_0=32, Delta y_0=3,Delta^2y_(-1)=-5,Delta^3y_(-1)=7`

Bessel's formula is
`y_p=(y_0+y_1)/2+(p-1/2)*Delta y_0 + (p(p-1))/(2!) * (Delta^2y_(-1)+Delta^2y_(0))/2 + ((p-1/2)p(p-1))/(3!) * Delta^3y_(-1)`

`y_(0.25) = (32+35)/2 + (0.25-1/2)*(3) + (0.25(0.25-1))/(2)*((-5+2))/2 + ((0.25-1/2)0.25(0.25-1))/(6)*(7)`

`y_(0.25)=33.5-0.75 +0.140625 +0.0546875`

`y_(0.25)=32.9453`


Solution of Bessel's interpolation is `y(25) = 32.9453`

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