# Use 2nd order Newton`s Interpolating polynomial to estimate f(3.5).x 1 3 5 y 50 138 179

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Using Newton's Interpolating Formula we construct the following table:

x0=1...y[x0]=50

x1=3...y[x1]=138...y[x0,x1] = 44

x2=5...y[x2]=179...y[x1,x2] = 20.5...y[x0,x1,x2]=-5.875

P2(x) = y[x0]+ y[x0,x1]*x + y[x0,x1,x2]*x(x-2)

**P2(x) =50 + 44x - 5.875*x*(x-2)**

for x = 3.5, P2(x) = 50 + 44*3.5 - 5.875*3.5*1.5 = 173.156

**Therefore for x = 3.5, y is approximately equal to 173.156.**

The values of X and Y(x) and the first and second differences are worked below:

X Y D1 D2

1 50

3 138 88

5 179 42 41.

We use the values to interpolate for the value of y(3.5) by Newton's forward difference formula:

y(3.5) = Y(3) + {(3.5-3)/(5-3)} D1(y3) + (1/2!)[(3.5-3)/(5-3)]*[(3.5-3)/5-3) +1]*D2Y(3)

y(3.5) = 138 +0.25*88 - (1/2)0.25(0.25+1)47

y(3.5) = 152.66.

Therefore the estimate of y(3.5) = 152.66.