Mathematical Tools for Physics

11—Numerical Analysis 322

To eliminate the largest source of error, theh^3 term, multiply the first equation by 8 and subtract the second.

8

[

f(h)−f(−h)

]

−

[

f(2h)−f(− 2 h)

]

= 12hf′(0)−

24

60

h^5 fv(0) +···,

or

f′(0)≈

1

12 h

[

f(− 2 h)− 8 f(−h) + 8f(h)−f(2h)

]

+

{ 1

30

h^4 fv(0)

}

. (10)

with an error term of orderh^4.
As an example of this method, letf(x) = sinxand evaluate the derivative atx= 0. 2 by the 2-point
formula and the 4-point formula with h=0.1:

2-point:

1

0. 2

[0. 2955202 − 0 .0998334] = 0. 9784340

4-point:

1

1. 2

[0. 0 − 8 × 0 .0998334 + 8× 0. 2955202 − 0 .3894183]

= 0. 9800633

cos 0.2 = 0. 9800666

Again, you have a more accurate formula by evaluating the derivative between the data points:h= 2k

f(k)−f(−k) = 2kf′(0) +

1

3

k^3 f′′′(0) +

1

60

k^5 fv(0)

f(3k)−f(− 3 k) = 6kf′(0) +

27

3

k^3 f′′′(0) +

243

60

k^5 fv(0)

27

[

f(k)−f(−k)

]

−

[

f(3k)−f(− 3 k)

]

= 48f′(0)−

216

60

k^5 fv(0).

Changingktoh/ 2 and translating the origin gives

1

24 h

[

f(−h)− 27 f(0) + 27f(h)−f(2h)

]

=f′(h/2)−

3

640

h^4 fv(h/2), (11)

and the coefficient of the error term is much smaller.