Seven Lines of Julia (examples sought)

General Usage
107

This isn’t anything out of this world, but it showcases Julia’s ability to represent expressions and to easily perform arbitrary precision arithmetic with GMP/MPFR.

Continued fractions:

function contfrac(x::Real, n::Integer)
    n < 1 && return Int[]
    fpart, ipart = modf(x)
    fpart == 0 ? [Int(ipart)] : [Int(ipart); contfrac(1/fpart, n-1)]
end

foldr((x, y) -> :($x + 1 / $y), contfrac(big(π), 25)) |> println
# 3 + 1 / (7 + 1 / (15 + 1 / (1 + 1 / (292 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (2 + 1 / (1 + 1 / (3 + 1 / (1 + 1 / (14 + 1 / (3 + 1 / (3 + 1 / (23 + 1 / (1 + 1 / (1 + 1 / (7 + 1 / (4 + 1 / (35 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / 2)))))))))))))))))))))))

foldr((x, y) -> x + 1 // big(y), contfrac(big(π), 75))
# 13926567982265799805873939967043853503//4432964269365850039583559427823398144

If you put the resulting string

foldr((x,y) -> "$x+\\frac1{$y}", contfrac(big(π), 15)) |> println
# 3+\frac1{7+\frac1{15+...}}

in a LaTeX cell you get

3+\frac1{7+\frac1{15+\frac1{1+\frac1{292+\frac1{1+\frac1{1+\frac1{1+\frac1{2+\frac1{1+\frac1{3+\frac1{1+\frac1{14+\frac1{2+\frac1{1}}}}}}}}}}}}}}
Addendum

Of course in general an iterative implementation such as

function contfrac(x::Real, n::Integer)
    cf = Int[]
    for _ in 1:n
        fpart, ipart = modf(x)
        push!(cf, Int(ipart))
        x = 1 / fpart
    end
    cf
end

would be preferred, because it reduces allocations and avoids stack overflows. This was however a bit too long for the seven lines restriction. Notice that in the case of continued fractions, to actually overflow the stack while performing meaningful computations requires working with a quite large precision. The default precision 256 of BigFloat allows to correctly compute only the first 76 terms of the continued fraction of π.

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