Multiple-Choice Test Newton’s Divided Difference ... - MATH FOR COLLEGE

Multiple-Choice Test Newton's Divided Difference Polynomial Method Interpolation COMPLETE SOLUTION SET

1. If a polynomial of degree n has more than n zeros, then the polynomial is (A) oscillatory (B) zero everywhere (C) quadratic (D) not defined

Solution The correct answer is (B).

A unique polynomial of degree n or less passes through n + 1 data points. Assume two

polynomials Pn (x) and Qn (x) go through n + 1 data points,

(x0 , y0 ), (x1, y1 ),..., (xn , yn )

Then

Rn (x) = Pn (x) - Qn (x)

Since Pn (x) and Qn (x) pass through all the n + 1 data points,

Pn (xi ) = Qn (xi ), i = 0, ... , n

Hence

Rn (xi ) = Pn (xi ) - Qn (xi ) = 0, i = 0, ... , n

The nth order polynomial Rn (x) has n + 1 zeros. A polynomial of order n can have n + 1 zeros

only if it is identical to a zero polynomial, that is,

Rn (x) 0

Hence

Pn (x) Qn (x)

How can one show that if a second order polynomial has three zeros, then it is zero everywhere?

If R2 (x) = a0 + a1x + a2 x 2 , then if it has three zeros at x1 , x2 , and x3 , then ( ) R2 x1 = a0 + a1x1 + a2 x12 = 0

( ) R2 x2 = a0 + a1x2 + a2x22 = 0

( ) R2 x3 = a0 + a1x3 + a2x32 = 0

Which in matrix form gives

1

x1

1 x2

1 x3

x12

x

2 2

a0

a1

=

0 0

x32 a2 0

The final solution a1 = a2 = a3 = 0 exists if the coefficient matrix is invertible. The determinant

of the coefficient matrix can be found symbolically with the forward elimination steps of na?ve

Gauss elimination to give

1

x1

det1 x2

1 x3

x12 x22

=

x2 x32

-

x22 x3

-

x1x32

+

x12 x3

+

x1x22

-

x12 x2

x32

= (x1 - x2 )(x2 - x3 )(x3 - x1)

Since

x1 x2 x3

the determinant is non-zero. Hence, the coefficient matrix is invertible. Therefore,

a1 = a2 = a3 = 0 is the only solution, that is, R2 (x) 0.

2. The following x - y data is given.

x 15 18 22 y 24 37 25 The Newton's divided difference second order polynomial for the above data is given by

f2 (x) = b0 + b1(x -15) + b2 (x -15)(x -18)

The value of b1 is most nearly (A) ?1.0480 (B) 0.14333 (C) 4.3333 (D) 24.000

Solution The correct answer is (C).

Given

x0 = 15

x1 = 18

f2 (x0 ) = 24 f2 (x1) = 37

we have

f2 (x) = b0 + b1(x - x0 ) + b2 (x - x0 )(x - x1)

Then

f2 (x0 ) = b0 + b1(x0 - x0 )+ b2 (x0 - x0 )(x0 - x1 )

= b0

f2 (x1) = b0 + b1(x1 - x0 )+ b2 (x1 - x0 )(x1 - x1 ) = b0 + b1(x1 - x0 ) = f2 (x0 ) + b1(x1 - x0 )

Thus,

b1 =

f2 (x1) - f2 (x0 ) x1 - x0

= 37 - 24 18 -15

= 4.3333

3. The polynomial that passes through the following x - y data

x 18 22 24 y ? 25 123

is given by

8.125x2 - 324.75x + 3237,18 x 24

The corresponding polynomial using Newton's divided difference polynomial is given by

The value of b2 is

f2(x) = b0 + b1(x -18) + b2 (x -18)(x - 22)

(A) 0.25000 (B) 8.1250 (C) 24.000 (D) not obtainable with the information given

Solution The correct answer is (B).

Expanding,

f2 (x) = b0 + b1(x -18) + b2(x -18)(x - 22)

( ) = b0 + b1x -18b1 + b2 x2 - 40x + 396

= (b0 -18b1 + 396b2 ) + (b1 - 40b2 )x + b2x2

This needs to be the same as 8.125x2 - 324.75x + 3237

Hence b2 = 8.125

4. Velocity vs. time data for a body is approximated by a second order Newton's divided difference polynomial as

v(t) = b0 + 39.622(t - 20) + 0.5540(t - 20)(t -15), 10 t 20

The acceleration in m/s2 at t = 15 is (A) 0.55400 (B) 39.622 (C) 36.852 (D) not obtainable with the given information

Solution The correct answer is (C).

v(t) = b0 + 39.622(t - 20) + 0.5540(t - 20)(t -15), 10 t 20

a(t) = d (v(t))

dt

= 39.622(1) + 0.5540(t - 20) + 0.5540(t -15) = 39.622 + 0.5540(t - 20) + 0.5540(t -15), 10 t 20 a(15) = 39.622 + 0.5540(15 - 20) + 0.5540(15 -15)

= 39.622 - 2.7700 = 36.852 m/s2

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