連分数展開

continued fraction

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

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？

code:jl

function continued_frac(number::Float64, max_terms::Int; tolerance = 1e-3)

frac_terms = Int[]

for _ in 1:max_terms

integer_part = floor(Int, number)

push!(frac_terms, integer_part)

fractional_part = number - integer_part

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

break

end

number = 1 / fractional_part

end

return frac_terms

end

continued_frac(Float64(π),4)

continued_frac(0.259259, 5)

有理数型を受けるようにする

code:jl

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

frac_terms = Int[]

nume = numerator(rational)

deno = denominator(rational)

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(Float64(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？