Find Solution using Gauss Backward formula

x	f(x)
1939	12
1949	15
1959	20
1969	27
1979	39
1989	52

x = 1974
Finding f(2)

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

x	1939	1949	1959	1969	1979	1989
y	12	15	20	27	39	52

Gauss's backward difference interpolation method to find solution

`h=1949-1939=10`

Taking `x_0=1969` then `p=(x-x_0)/h=(x-1969)/10`

Now the central difference table is

`x`	`p=(x-1969)/10`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`	`Delta^5y`
1939	-3	12
			3
1949	-2	15		2
			5		0
1959	-1	20		2		3
			7		3		-10
1969	0	27		5		-7
			12		-4
1979	1	39		1
			13
1989	2	52

`x = 1974`

`p = (x - x_0)/h = (1974 - 1969)/10 = 0.5`

`y_0=27, Delta y_(-1)=7,Delta^2y_(-1)=5,Delta^3y_(-2)=3,Delta^4y_(-2)=-7,Delta^5y_(-3)=-10`

Gauss's backward interpolation formula is
`y_p=y_0+p Delta y_(-1) + ((p + 1)p)/(2!) * Delta^2y_(-1) + ((p + 1)p(p - 1))/(3!) * Delta^3y_(-2) + ((p + 2)(p + 1)p(p - 1))/(4!) * Delta^4y_(-2) + ((p + 2)(p + 1)p(p - 1)(p - 2))/(5!) * Delta^5y_(-3)`

`y_(0.5) = 27 + (0.5)(7) + ((0.5 + 1)(0.5))/(2) * (5) + ((0.5 + 1)(0.5)(0.5 - 1))/(6) * (3) + ((0.5 + 2)(0.5 + 1)(0.5)(0.5 - 1))/(24) * (-7) + ((0.5 + 2)(0.5 + 1)(0.5)(0.5 - 1)(0.5 - 2))/(120) * (-10)`

`y_(0.5)=27 +3.5 +1.875 -0.1875 +0.2734375 -0.1171875`

`y_(0.5)=32.3438`


Solution of Gauss's backward interpolation is `y(1974) = 32.3438`

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