Mth603 ka koi student hai tu plz yeh question bta dy kis trha solve ho ga Given the following data x:1 2 5 y:1 4 10 Value of 1st order divided difference f[2 , 5] is

Not equally spaced

Newton’s divided difference interpolation formula is a interpolation technique used when the interval difference is not same for all sequence of values. Divided differences are symmetric with respect to the arguments i.e independent of the order of arguments.

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Lagrangens interpolation

ε(x)=y(x)+Pn(x)
ε(x)=y(x)−Pn(x)
ε(x)=y(x)×Pn(x)
ε(x)=y(x)÷Pn(x)

Forthegivendividedifferencetablex 1 4 7y2.23.54.11stD.D0.43330.22ndD.D−0.0389theNewton′sdividedifferenceinterpolationformulawillbe

y=f(x)=2.2+(x−1)(−0.0389)+(x−1)((x−4)(0.4333)

y=f(x)=2.2+(x−1)(0.4333)+(x−1)((x−4)(−0.0389)

y=f(x)=−0.0389+(x−1)(0.4333)+(x−1)((x−4)(2.2)

y=f(x)=−0.0389+(x−1)(2.2)+(x−1)((x−4)(0.4333)

Solution:

y=f(x)=2.2+(x−1)(−0.0389)+(x−1)((x−4)(0.4333)

y=f(x)=2.2+(x−1)(0.4333)+(x−1)((x−4)(−0.0389)

y=f(x)=−0.0389+(x−1)(0.4333)+(x−1)((x−4)(2.2)

y=f(x)=−0.0389+(x−1)(2.2)+(x−1)((x−4)(0.4333)

For the given data points (1,0.3),(3,1),and(5,1.2) the divide difference table will be given as

Forthegivendatapoints(1,0.3),(3,1),and(5,1.2)thedividedifferencetablewillbegivenas

Δ=----- ?

E−1/2
1-E
E-1
None

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