# MTH603 Quiz 2 Solution and Discussion

• Given the following data x: 1 3 -7 f(x): 3 -6 -2 Which formula is useful in finding the interpolating polynomial?

• x: 1 3 7 f(x): 1 4 9 f(3) Can be found using

• : What will be the value of first order divided difference f[1,5]for the following data x:0 1 5 y:2 1 5

• Forthegivendatapoints(4,1.3),(8,1.5),and(12,1.9)thedividedifferencetablewillbegivenas

• Newton’s divided difference interpolation formula is used when the values of the independent variable are

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.

• Given the following data x:1 2 5 y:1 4 10 Value of 1st order divided difference f[2 , 5] is

• Given the following data x:4 5 7 10 y:46 102 294 346 Value of 1st order divided difference f[5 , 7] is

91
92
94
96

• Forthegivendatapoints(x0,y0),(x1y1),(x2y2),and(x3,y3)thezero−orderdividedifferencewillbegivenas

• Forthegivendatapoints(4,45),(5,104),and(6,190),thezero−orderdividedifferencewillbe

• For the following data

• Whichofthefollowingmethodcanbeusedforinterpolationforthegivenvaluesofxandy?x y0.30.0670.70.2480.90.518

Lagrangens interpolation

• If y(x) is approximated by a polynomial Pn(x) of degree n then the error is given by

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:

• 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)

• Whatwillbethevalueof′a′inthegivendividedifferencetable? x2468y0.51.11.72…21stD.D0.30.30.252ndD.Da−0.01253rdD.D−0.0021

• 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

1

1

2

5

4

1

1

2
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