Computer Aided Engineering Design

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140 COMPUTER AIDED ENGINEERING DESIGN


some intermediate point on the curve, the first divided differences y[xs+1,xs+ 2 ] =
yy
xx


ss
ss

+2 +1
+2 +1





  • and




y[xs,xs+ 1 ] = yy
xx


s s
s s

+1
+1







represent the slopes, the second divided difference yx x[ , , ]s ss+1 x+ 2 =

yx x yx x
xx

ss s s
s s

[ , ] – [ , ]





+1 + 2 +1
+2

represents the rate of change of slope or the second derivative of the

curve, and so on.
Thus, an (n– 1)th divided difference is representative of the (n– 1)th derivative of a curve. For
an (n– 1)th degree polynomial, the (n– 1)th divided differences are equal and so the nth divided
differences are zero. For instance, for a line, the first divided differences are equal (to the slope) while
the second divided differences are zero (since the slope is constant).
In algebraic form, the divided difference, y[xj, xj+ 1 , ..., xj+k] can be written as


yx x x

y
j j jk r wx

k jr
jr

[ , ,... , ] =
+1 + =0 ()

+
+

Σ



(5.17)

wherew(x) = (x–xj)(x–xj+ 1 )... (x–xj+k) and w′(x) = dw/dx.


Example 5.3.Show, using examples, that the result in Eq. (5.17) holds.


Fork = 0, y[xj] =


yx
wx

j y
j
j

()
()

=

since w(x) = (x−xj)

Fork = 1, y[xj, xj+ 1 ] =


y
wx

y
wx

j
j

j
′′() j
+
()

+1
+1
Here,w(x) = (x–xj)(x–xj+ 1 ) so that w′(x) = (x–xj+ 1 ) + (x–xj)


Thus,w′(xj) = (xj–xj+ 1 ) and w′(xj+ 1 ) = (xj+ 1 – xj)


On substitution, we get

yx x

y
xx

y
xx

yy
j j xx

j
j j

j
j j

j j
j j

[, ] =
(– )
+
(– )
=


  • +1 +1 –


+1
+1

+1
+1

xs xs+1 xs+2

y[xs+2]

y[xs+1]

y[xs]

y[xs,xs+1]

y[xs+1,xs+2]

Figure 5.7 Geometric interpretation of the divided differences
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