Computer Aided Engineering Design

DIFFERENTIAL GEOMETRY OF CURVES 69

Px LxyLn

i

n

i

n

i

–1

=0

–1

() = Σ –1() (3.8)

Example 3.1. Construct a polynomial to interpolate through the data points (0, 0), (1, 2), (3, 2)
and (6, −1) using the Newton’s divided difference and Lagrangian approaches. Perturb point (3, 2) to
(1.5, 4) and observe the change in the curve shape.
Using Newton’s divided difference approach, since there are four data points, the interpolating
polynomial is a cubic, that is

y=α 0 +α 1 (x− x 0 )+α 2 (x− x 0 ) (x− x 1 ) +α 3 (x− x 0 ) (x− x 1 )(x− x 2 )
Now

α 0 = y 0 = 0

α

α

1

10

10

=

• 
  
  
  
  • = 2 – 0
    1 – 0
    = 2
    y
    xx
  
  
  
  

α

αα

2

20 120

2021

=

( – ) – ( – )

( – )( – )

=

(2 – 0) – 2(3 – 0)

(3 – 0)(3 – 1)

= –^2

3

yxx

xxxx

α

αα α

3

30 130 23031

303132

=

( – ) – ( – ) – ( – )( – )

( – )( – )( – ) =

(–1) – 2 (6) +^2

3

(6)(5)

(6)(5)(3)

= 907

yxxxxxx

xxxxxx

The polynomial becomes

yx xx = 2 –^2 xx x

3

( – 1) +^7

90

( – 1)( – 3)

Using the Lagrangian approach, the polynomial is

yLxy Lxy Lxy Lxy = 03 () + () + 0 13 1 23 () + 2 33 () 3

or

y

xx xx xx

xxxxxx

y

xx xx xx

xxxxxx

= y

( – )( – )( – )

( – )( – )( – )

+

( – )( – )( – )

( – )( – )( – )

123

010 203

0

023

101213

1

+

( – )( – )( – )

( – )( – )( – )

+

( – )( – )( – )

( – )( – )( – )

013

202123

2

012

303132

3

xx xx xx

xxxxxx

y

xx xx xx

xxxxxx

y

=

( – 1)( – 3)( – 6)

(–1)(–3)(–6)

0 +

( – 0)( – 3)( – 6)

(1) (–2)(–5)

2

xxx xxx

+

( – 0)( – 1)( – 6)

(3)(2)(–3)

2 +

( – 0)( – 1)( – 3)

(6)(5)(3)

(–1)

xxx xxx

=

( )( – 3)( – 6)

5

( )( – 1)( – 6)

9

( )( – 1)( – 3)

90

xx x xx x xx x

On simplification, the above yields the same result as that from the Newton’s divided differences
method. Moving data point (3, 2) to (1.5, 4) requires re-computing the curve. With the divided
difference approach, only the last two coefficients need to be computed, that is