python - Arbitrary Precision Arithmetic in Julia -

this has kinda been asked, not in way. have little python program finds continued fractions square roots of n (1 <= n <= 10000).

i have been trying in julia , can't see how to. because deals irrational numbers (sqrt(x) irrational if x not perfect square, e.g. sqrt(2) = 1.414213...). don't think can use rational class.

it says here https://docs.julialang.org/en/release-0.4/manual/integers-and-floating-point-numbers/#man-arbitrary-precision-arithmetic julia can arbitrary precision arithmetic bigfloats. don't seem accurate enough.

i have tried use pycall , decimals package in python (from julia), weird errors (i can post them if helpful).

here python program works. , question how in julia please?

def continuedfractionsquareroots(): '''    each number 100, length of continued fraction    of square root it. '''  decimal.getcontext().prec = 210 # need lots of precision problem.  continuedfractionlengths = [] in range(1, 101):      # perfect squares, period 0     irrationalnumber = decimal.decimal(i).sqrt()     if irrationalnumber == int(irrationalnumber):         continue              continuedfractionlength = 0     while true:          intpart = irrationalnumber.to_integral_exact(rounding=decimal.round_floor)         if continuedfractionlength == 0:             firstcontinuedfractiontimes2 = int(intpart*2)          continuedfractionlength += 1         if intpart == firstcontinuedfractiontimes2:             # have reached 'period' end fraction             break          fractionalpart = irrationalnumber - intpart         irrationalnumber = 1 / fractionalpart  continuedfractionlengths.append(continuedfractionlength) return continuedfractionlengths 

so can see, need way of computing exact square root , way integer part of number. , that's really, apart lots , lots of precision!

guys, didn't post julia code because didn't want have small manuscript answer! here is, works. said in comments below, used setprecision function set precision high value , works. got value 711 empirically.

setprecision(711)  function continuedfractionsquareroots() #=   each number 100, length of continued fraction    of square root it. =#  continuedfractionlengths = int[] i=1:100      # perfect squares, period 0     irrationalnumber = bigfloat(sqrt(bigfloat(i)))     if irrationalnumber == floor(irrationalnumber)         continue     end      continuedfractionlength = 0     while true          intpart = floor(irrationalnumber)         if continuedfractionlength == 0             firstcontinuedfractiontimes2 = intpart*2         end          continuedfractionlength += 1         if intpart == firstcontinuedfractiontimes2             # have reached 'period' end fraction             break         end          fractionalpart = irrationalnumber - intpart         irrationalnumber = bigfloat(1) / fractionalpart      end      push!(continuedfractionlengths, continuedfractionlength)  end   return continuedfractionlengths end 

so anyway, user2357112 solved it, much.

just python's decimal.decimal, can configure precision of julia's bigfloat:

setprecision(however_many_bits) 

note in bits, unlike decimal.decimal, because bigfloat doesn't use decimal.