Example Suppose we are given a table below and want to compute
f′(2). Approximating f′(2) by (f(3) − f(2))/(3 − 2) = 19 results in
overestimation and approximating f′(2) by
(f(2) − f(1))/(2 − 1) = 7 yields an underestimated result.
x
f(x)
1
1
2
8
3
27
4
64
Introducing E as the shift operator as
Ef(x) = f(x + h),
where h is a step size and ∆ is
the difference operator defined as
∆f(x) = f(x + h) − f(x),
the relationship between ∆ and E is
∆ = E −1,
as
∆f(x)
=
f(x + h) − f(x)
=
Ef −f
=
(E − 1) f.
Using Taylor's series expansion for f(x + h),
f(x + h)
=
f(x) + hf′(x) +
h2
2!
f"(x) +
h3
3!
f"′(x) +…
=
f + hDf +
h2D2
2!
f +
h3D3
3!
f + …
=
(1+ hD +
(hD)2
2!
+
(hD)3
3!
+ …) f
=
ehDf,
where D is the differential operator defined as
Df = f′(x).
Therefore, it follows that
E = ehD,
or
1+∆ = ehD.
This is equivalent to
hD = ln (1 + ∆),
or
hD = ∆−
∆2
2
+
∆3
3
−
∆4
4
+ …
or
D =
1
h
⎛ ⎝
∆−
∆2
2
+
∆3
3
−
∆4
4
+ …
⎞ ⎠
.
Thus, one can associate the differential operator, D, with
the difference operator, ∆. A table of differences on
f(x) can be easily created as
x
f(x)
∆f
∆2f
∆3f
∆4f
1
1
7
12
6
0
2
8
19
18
6
0
3
27
37
24
6
0
4
64
61
Therefore,
f′(2)
=
Df|2
=
1
h
⎛ ⎝
∆f|2 −
∆2f|2
2
+
∆3f|2
3
− …
⎞ ⎠
=
1
1
(19 − 9 + 2 − 0)
=
12.
(this is not part of the textbook, something extra)
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version 4.03. On 21 Aug 2022, 22:51.
Introducing E as the shift operator as