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11—Numerical Analysis 323

The previous example of the derivative ofsinxatx= 0. 2 withh= 0. 1 gives, using this formula:

1


2. 4


[0. 0499792 − 27 × 0 .1494381 + 27× 0. 2474040 − 0 .3428978] = 0. 9800661 ,


and the error is less by a factor of about 7.


You can find higher derivatives the same way.

f(h) =f(0) +hf′(0) +

1


2


h^2 f′′(0) +

1


6


h^3 f′′′(0) +

1


24


h^4 f′′′′(0)

f(h) +f(−h) = 2f(0) +h^2 f′′(0) +

1


12


h^4 f′′′′(0) +···

f′′(0) =

f(−h)− 2 f(0) +f(h)
h^2


1


12


h^2 f′′′′(0) +··· (12)

Notice that the numerical approximation forf′′(0)is even inhbecause the second derivative is unchanged ifx
is changed to−x.


You can get any of these expressions for higher derivatives recursively, though finding the error estimates
requires the series method. The above expression forf′′(0)can be viewed as a combination of first derivative
formulas:


f′′(0)≈

[


f′(h/2)−f′(−h/2)

]


/h


1


h

[


f(h)−f(0)
h


f(0)−f(−h)
h

]


=


[


f(h)− 2 f(0) +f(−h)

]


/h^2. (13)

Similarly, the third and higher derivatives can be computed. The numbers that appear in these numerical deriva-
tives are simply the binomial coefficients, Eq. (2.13).

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