11—Numerical Analysis 343f(k)−f(−k) = 2kf′(0) +1
3
k^3 f′′′(0) +···f(3k)−f(− 3 k) = 6kf′(0) +27
3
k^3 f′′′(0) +···I’ll seek a formula of the form
f′(0) =α[
f(k)−f(−k)]
+β[
f(3k)−f(− 3 k)]
. (55)
I am assuming that the variance offat each point is the same,σ^2 , and that the fluctuations inf at different
points are uncorrelated. The last statement is, for random variablesf 1 andf 2 ,
〈(
f 1 −
〈
f 1〉)(
f 2 −〈
f 2〉)〉
= 0 which expands to〈
f 1 f 2〉
=
〈
f 1〉〈
f 2〉
. (56)
Insert the preceding series expansions into Eq. ( 55 ) and match the coefficients of f′(0). This gives an
equation forαandβ:
2 kα+ 6kβ= 1. (57)
One way to obtain another equation forαandβis to require that thek^3 f′′′(0)term vanish; this leads back to
the old formulas for differentiation, Eq. ( 11 ). Instead, require that the variance off′(0)be a minimum.
〈(
f′(0)−
〈
f′(0)〉) 2 〉
=
〈[
α(
f(k)−〈
f(k)〉)
+α(
f(−k)−〈
f(−k)〉)
+···
] 2 〉
= 2σ^2 α^2 + 2σ^2 β^2 (58)This comes from the fact that the correlation between sayf(k)andf(− 3 k)vanishes, and that all the individual
variances areσ^2. That is, 〈(
f(k)−〈
f(k)〉)(
f(−k)−〈
f(−k)〉)〉
= 0
along with all the other cross terms. Problem: minimize 2 σ^2 (α^2 +β^2 )subject to the constraint 2 kα+ 6kβ= 1.
It’s hardly necessary to resort to Lagrange multipliers for this problem.
Eliminateα:
d
dβ[(
1
2 k− 3 β) 2
+β^2]
= 0 =⇒ − 6
(
1
2 k− 3 β)
+ 2β= 0=⇒ β= 3/ 20 k, α= 1/ 20 k