dstefan
07-08-2008, 02:44 PM
Chapter 6, Problem 4
Find a central finite difference approximation for f^{(4)}(a) using f(a-2h), f(a-h), f(a), f(a+h), and f(a+2h). What is the order of this finite difference approximation?
Solution:
We will use the following Taylor approximation of f(x) around the point x=a:
\label{ch6_ex5_1}
f(x) = f(a) ~+~ (x-a)f'(a) ~+~ \frac{(x-a)^2}{2} f''(a) ~+~ \frac{(x-a)^3}{6} f^{(3)}(a) +~ \frac{(x-a)^4}{24} f^{(4)}(a) ~+~ \frac{(x-a)^5}{120} f^{(5)}(a) ~+~ \frac{(x-a)^6}{720} f^{(6)}(a) +~ O\left((x-a)^7 \right),
as x \to a.
For symmetry reasons, and keeping in mind the form of the central difference approximation for f''(a), we use (\ref{ch6_ex5_1}) to compute
\label{ch6_ex5_2}
f(a+h) + f(a-h) = 2f(a) + h^2f''(a) + \frac{h^4}{12} f^{(4)}(a) + \frac{h^6}{360} f^{(6)}(a) +~ O\left( h^7 \right), \quad \mbox{as}~ h \to 0
\label{ch6_ex5_3}
f(a+2h) + f(a-2h) = 2f(a) + 4 h^2f''(a) + \frac{4h^4}{3} f^{(4)}(a) + \frac{8h^6}{45} f^{(6)}(a) +~ O\left( h^7 \right),
as h \to 0.
We multiply (\ref{ch6_ex5_2}) by 4 and subtract the result from (\ref{ch6_ex5_3}). We solve for $f^{(4)}(a)$ and obtain the following second order finite difference approximation:
f^{(4)}(a) ~=~ \frac{f(a+2h) - 4f(a+h) + 6f(a) - 4f(a-h) + f(a-2h)}{h^4} + O\left( h^2 \right),
as h \to 0.
Find a central finite difference approximation for f^{(4)}(a) using f(a-2h), f(a-h), f(a), f(a+h), and f(a+2h). What is the order of this finite difference approximation?
Solution:
We will use the following Taylor approximation of f(x) around the point x=a:
\label{ch6_ex5_1}
f(x) = f(a) ~+~ (x-a)f'(a) ~+~ \frac{(x-a)^2}{2} f''(a) ~+~ \frac{(x-a)^3}{6} f^{(3)}(a) +~ \frac{(x-a)^4}{24} f^{(4)}(a) ~+~ \frac{(x-a)^5}{120} f^{(5)}(a) ~+~ \frac{(x-a)^6}{720} f^{(6)}(a) +~ O\left((x-a)^7 \right),
as x \to a.
For symmetry reasons, and keeping in mind the form of the central difference approximation for f''(a), we use (\ref{ch6_ex5_1}) to compute
\label{ch6_ex5_2}
f(a+h) + f(a-h) = 2f(a) + h^2f''(a) + \frac{h^4}{12} f^{(4)}(a) + \frac{h^6}{360} f^{(6)}(a) +~ O\left( h^7 \right), \quad \mbox{as}~ h \to 0
\label{ch6_ex5_3}
f(a+2h) + f(a-2h) = 2f(a) + 4 h^2f''(a) + \frac{4h^4}{3} f^{(4)}(a) + \frac{8h^6}{45} f^{(6)}(a) +~ O\left( h^7 \right),
as h \to 0.
We multiply (\ref{ch6_ex5_2}) by 4 and subtract the result from (\ref{ch6_ex5_3}). We solve for $f^{(4)}(a)$ and obtain the following second order finite difference approximation:
f^{(4)}(a) ~=~ \frac{f(a+2h) - 4f(a+h) + 6f(a) - 4f(a-h) + f(a-2h)}{h^4} + O\left( h^2 \right),
as h \to 0.