Differentiating with respect to x Newton’s Backward
Interpolation Formula
2
0
3
(1)
()
2!
(1)(2)
3!
nn
n
pp
yxypyy
ppp
y
+
=+∇+∇
++
+∇+
we obtain
2
2
3
11
()(21)
2
362
6
nn
n
yxypy
h
pp
y
⎡
=∇++∇
′
⎢
⎣
⎤
++
+∇+
⎥
⎦
(8.9)
23
2
2
4
1
()(1)
61811
12
nn
n
yxypy
h
pp
y
⎡
=∇++∇
′′
⎣
++
⎤
+∇
⎦
(8.10)
Special Case
When x=x
n
, p= 0 and we have
234
1111
()
234
nnnnn
yxyyyy
h
⎡⎤
=∇+∇+∇+∇+
′
⎢⎥
⎣⎦
(8.11)
234
2
111
()
12
nnnn
yxyyy
h
⎡⎤
=∇+∇+∇
′′
⎢⎥
⎣⎦
(8.12)
8.1.4 Numerical Differentiation
by Stirling’s Formula
Differentiating Stirling’s formula
2
2
01
01
33
2
12
22
4
2
()
1!22!
(1)
3!2
(1)
4
yy
pp
yxyy
yy
pp
pp
y
−
−
−−
−
Δ+Δ
⎛⎞
=++
Become an O’Reilly member and get unlimited access to this title plus top books and audiobooks from O’Reilly and nearly 200 top publishers, thousands of courses curated by job role, 150+ live events each month, and much more.