Aproximação Melhor
-
@Latex n=2p+1:
@Latex f(t+\varepsilon) -f(t-\varepsilon)\simeq
@Latex 2\sum_{j=0}^{p} \frac{ \varepsilon^{2j+1} }{(2j+1)!} f^{(2j+1)} (t)=
@Latex 2 \varepsilon f\prime(t)+ 2 \frac{\varepsilon^3 }{3!}f^{(3)}(t)+...+2\frac{\varepsilon^{2p+1}}{(2p+1)!} f^{(2p+1)}(t)
- Isolando
@Latex f\prime(t):
@Latex f\prime(t)
@Latex \frac{ f(t+\varepsilon) -f(t-\varepsilon) }{ 2 \varepsilon }+\frac{\varepsilon^2}{3!} f^{(3)}(t)+...+\frac{\varepsilon^{2p}}{(2p+1)!} f^{(2p+1)}(t)=
@Latex \frac{ f(t+\varepsilon) -f(t-\varepsilon) }{ 2 \varepsilon } +o(\varepsilon^2)
@Code ../../../Code/Derivatives.py d_f
- Duas chamadas da função