連分数展開

continued fraction

$ \pi = 3 + \cfrac{1}{7 + \cfrac{1}{15 + \cfrac{1}{1 + \cfrac{1}{\ddots}}}}

$ \pi= [3; 7, 15, 1,\cdots]

小数を連分数展開する

元の有効数字と処理回数からいい感じに指定できる？

実際は30項目くらいで破綻する

中身をいい感じにすればできそう？

code:jl

function continued_frac(number::Real, max_terms::Int; tolerance = 1e-3)

frac_terms = Int[]

for _ in 1:max_terms

int_part = floor(Int, number)

push!(frac_terms, int_part)

frac_part = number - int_part

if abs(frac_part) < tolerance # frac_part ≈ 0 ではうまく止まらない

break

end

number = 1 / frac_part

end

return frac_terms

end

continued_frac(π,4)

continued_frac(0.259259, 5)

code:jl

function continued_frac(rational::Tuple{Int, Int}, max_terms::Int)

frac_terms = Int[]

for _ in 1:max_terms

quotient = nume ÷ deno

push!(frac_terms, quotient)

remainder = nume % deno

if remainder == 0

break

end

nume, deno = deno, remainder

end

return frac_terms

end

continued_frac((355, 113), 4)

continued_frac((259, 999), 5)

逆の操作で近似が得られる

円周率の近似値を得られる

有理数であれば既約分数にできる

code:jl

function rev_continued_frac(frac_terms::Vector{Int})

deno = 1

nume, deno = a * nume + deno, nume

end

return (nume, deno)

end

rev_continued_frac(continued_frac(π, 4))

rev_continued_frac(continued_frac(0.259259, 5))

rev_continued_frac(continued_frac((259, 999), 5))

漸化式を利用する？