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我试图证明二阶数值微分的截断误差确实比一阶数值微分给了我们双倍的精度（考虑到机器误差/舍入误差eps()）

这是我在 Julia 中的代码：

function first_order_numerical_D(f)
  function df(x)
    h = sqrt(eps(x))        
    (f(x+h) - f(x))/h
  end
  df
end

function second_order_numerical_D(f)
  function df(x)
    h = sqrt(eps(x))          
    (f(x+h) - f(x-h))/(2.0*h)
  end
  df
end

function analytical_diff_exp(x)
    return exp(x)
end
function analytical_diff_sin(x)
    return cos(x)
end
function analytical_diff_cos(x)
    return -sin(x)
end
function analytical_diff_sqrt(x)
    return 1/(2.0*sqrt(x))
end

function first_order_error_exp(x)
    return first_order_numerical_D(exp)(x) - analytical_diff_exp(x) 
end
function first_order_error_sin(x)
    return first_order_numerical_D(sin)(x) - analytical_diff_sin(x) 
end
function first_order_error_cos(x)
    return first_order_numerical_D(cos)(x) - analytical_diff_cos(x) 
end
function first_order_error_sqrt(x)
    return first_order_numerical_D(sqrt)(x) - analytical_diff_sqrt(x) 
end

function second_order_error_exp(x)
    return second_order_numerical_D(exp)(x) - analytical_diff_exp(x)
end
function second_order_error_sin(x)
    return second_order_numerical_D(sin)(x) - analytical_diff_sin(x)
end
function second_order_error_cos(x)
    return second_order_numerical_D(cos)(x) - analytical_diff_cos(x) 
end
function second_order_error_sqrt(x)
    return second_order_numerical_D(sqrt)(x) - analytical_diff_sqrt(x) 
end

function round_off_err_exp(x)
    return 2.0*sqrt(eps(x))*exp(x)
end
function round_off_err_sin(x)
    return 2.0*sqrt(eps(x))*sin(x)
end
function round_off_err_cos(x)
    return 2.0*sqrt(eps(x))*cos(x)
end
function round_off_err_sqrt(x)
    return 2.0*sqrt(eps(x))*sqrt(x)
end

function first_order_truncation_err_exp(x)
    return abs(first_order_error_exp(x)+round_off_err_exp(x))
end
function first_order_truncation_err_sin(x)
    return abs(first_order_error_sin(x)+round_off_err_sin(x))
end
function first_order_truncation_err_cos(x)
    return abs(first_order_error_cos(x)+round_off_err_cos(x))
end
function first_order_truncation_err_sqrt(x)
    return abs(first_order_error_sqrt(x)+round_off_err_sqrt(x))
end

function second_order_truncation_err_exp(x)
    return abs(second_order_error_exp(x)+0.5*round_off_err_exp(x))
end
function second_order_truncation_err_sin(x)
    return abs(second_order_error_sin(x)+0.5*round_off_err_sin(x))
end
function second_order_truncation_err_cos(x)
    return abs(second_order_error_cos(x)+0.5*round_off_err_cos(x))
end
function second_order_truncation_err_sqrt(x)
    return abs(second_order_error_sqrt(x)+0.5*round_off_err_sqrt(x))
end

如果我减去（这里我使用加法，因为实际的泰勒展开式表明舍入误差和截断误差在它们前面都有一个负号）round_off_err_f项，这应该会给我正确的截断误差。

有关分析推导/证明，请参见： https ://www.uio.no/studier/emner/matnat/math/MAT-INF1100/h10/kompendiet/kap11.pdf http://www2.math.umd.edu/~dlevy /classes/amsc466/lecture-notes/differentiation-chap.pdf

但结果表明：

first_order_truncation_err_exp(0.5), first_order_truncation_err_sin(0.5), first_order_truncation_err_cos(0.5), first_order_truncation_err_sqrt(0.5)

(4.6783240139052204e-8, 1.2990419187857229e-8, 2.8342226290287478e-9, 4.364449135429996e-9)

second_order_truncation_err_exp(0.5), second_order_truncation_err_sin(0.5), second_order_truncation_err_cos(0.5), second_order_truncation_err_sqrt(0.5)

(1.8874426561390482e-8, 7.938850300905947e-9, 4.1240999200086055e-9, 7.45058059692383e-9)

在哪里：

eps(0.5)=1.1102230246251565e-16

second_order_truncation_err_f()应该在1e-16而不是的顺序左右1e-8，我不知道为什么这不起作用。

4

1 回答 1

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那是因为您计算的内容受舍入误差的支配。IE：

julia> round_off_err_sqrt(0.5)
1.490116119384766e-8

为了查看“二阶”导数之间的差异（通常我会看到中心差异这一术语），您需要选择更大的步长。在文献h = cbrt(eps())中通常可见。

function second_order_numerical_D(f)
    function df(x)
        h = cbrt(eps(x))          
        (f(x+h/2) - f(x-h/2))/h
    end
    df
end
julia> second_order_error_exp(0.5)
1.308131380994837e-11
于 2019-10-28T14:39:03.570 回答