Hoja1EJEMPLO 1: Calcular la derivada en x = 0.5 por Richardsonxf(x)0.10.09531017980.30.26236426450.40.3364722366Valor "verdadero"0.50.4054651081f'(0.5) =0.66666666670.60.47000362920.70.53062825110.90.6418538862hk = 0k = 1k = 20.80.6831796330.40.67065996650.66648674430.20.66765696310.6666559620.6666672432EJEMPLO 2: Extrapolar el valor en cero a partir de los valores tabuladoshf(h)0.12.7282868414590.052.7195319066090.0252.7184380796800.01252.718301359728TABLA DE DIFERENCIAShf(h)DeltaDelta^2Delta^3h00.12.728286841459h10.052.719531906609-0.008755h20.0252.718438079680-0.0010940.007661h30.01252.718301359728-0.0001370.000957-0.006704Hiptesis: f(h) = a + b h^3 + c h^6 + d h^9EXTRAPOLACIN DE RICHARDSOND_j^(k) = D_{j+1}^(k-1) + (D_{j+1}^(k-1)-D_j^(k-1))/(8^k-1))hk = 0k = 1k = 2k = 3j = 00.12.728286841459j = 10.052.7195319066092.718281201631j = 20.0252.7184380796802.7182818186902.718281828484j = 30.01252.7183013597282.7182818283062.7182818284592.718281828459Valor "verdadero"2.718281828459

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