Exponential function benchmark in Julia

myexp.jl

@inline @fastmath function myexpm1(x :: Float64)
    ln2_hi = 6.93147180369123816490e-01
    ln2_lo = 1.90821492927058770002e-10
    one = 1.0
    Q1 = -3.33333333333331316428e-02
    Q2 =  1.58730158725481460165e-03
    Q3 = -7.93650757867487942473e-05
    Q4 =  4.00821782732936239552e-06
    Q5 = -2.01099218183624371326e-07
    hi = x - ln2_hi
    lo = ln2_lo
    k = 1
    x = hi - lo;
    c = (hi - x) - lo;
    hfx = 0.5 * x;
    hxs = x * hfx;
    R1 = one + hxs * Q1;
    h2 = hxs * hxs;
    R2 = Q2 + hxs * Q3;
    h4 = h2 * h2;
    R3 = Q4 + hxs * Q5;
    r1 = R1 + h2 * R2 + h4 * R3;
    t = 3.0 - r1 * hfx;
    e = hxs * ((r1 - t) / (6.0 - x * t));
    e = (x * (e - c) - c);
    e -= hxs;
    one + 2.0 * (x - e);
end

@inline myexp(x) = myexpm1(x) + 1

@inline calcerr(x) = abs(myexp(x) - exp(x))

@inline function test_myexp_for(n)
    err = 0.0
    for i in 0:n
        err = max(err, calcerr(0.4+0.6*i/n))
    end
    err
end

@inline function test_myexp_fold(n)
    @inline errmax(acc, i) = max(acc, calcerr(0.4+0.6*i/n))
    foldl(errmax, 0:n, init=0)
end

for n in reverse([1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9])
    tt = @elapsed er=test_myexp_for(convert(Int, n))
    println("($n, $tt, $er)")
end   

for n in reverse([1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9])
    tt = @elapsed er=test_myexp_fold(convert(Int, n))
    println("($n, $tt, $er)")
end