Question

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，这应该给我正确的截断错误（这里我使用加法，因为实际的泰勒展开表明舍入误差和截断误差在它们前面都带有负号）{{1 }}字词。

有关分析推导/证明，请参阅： 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

但是结果表明：

round_off_err_f

位置：

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)

eps(0.5)=1.1102230246251565e-16应该在second_order_truncation_err_f()左右，而不是1e-16，我不知道为什么这不起作用。

Answer 1

这是因为您的计算主要由舍入误差决定。即：

julia> round_off_err_sqrt(0.5)
1.490116119384766e-8

要查看“二阶”导数之间的差异（通常我会看到术语“中心差异”），您需要选择更大的步长。在文献中通常会看到h = cbrt(eps())。

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

如何显示二阶数值微分的截断误差是一阶数值微分的平方

1 个答案: