数值分析作业 - 数值微分
Problem 1
则误差
至少是
Problem 2

Taylor 展开
则有
解得
即
Problem 3

1 | f[x_] = E^x; |
求一阶导数的误差:
1 | err1 = Table[1 - N[f1[0, 10^-i], 20], {i, 1, 9}] |

1 | ListLogPlot[err1] |

可以观察到误差大约满足二次收敛.
求二阶导数的误差:
1 | err2 = Table[1 - N[f2[0, 10^-i], 40], {i, 1, 9}] |

1 | ListLogPlot[err2] |

可以观察到误差大约满足四次收敛.

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