2

here is the code i wrote. my idea is put a number(between 0~9) at the end of x and square it and then see if it's smaller than 2 ,choosing the biggest

x = 1.4



for n in range(21):
    next_num = [0,1,2,3,4,5,6,7,8,9]
    candidate = []
    for number in next_num:
        if float(str(x)+str(number))*float(str(x)+str(number))<2:
            candidate.append(number)
            
    x = float(str(x)+str(max(candidate)))
    

print(x)

but the problem is i only get 1.414213562373 this much 13 digits i tried typing in bigger number in range but i only get this

thank you

Some programmer dude
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HAECHANG YUN
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  • What do you get by typing a bigger number? – 12944qwerty May 05 '21 at 12:40
  • By the way, you don't need a list. Simply do `candidate = number` instead of `candidate.append(number)`, and candidate will contain the highest number that fulfilled the condition. – Tim Pietzcker May 05 '21 at 12:54
  • Ultimately, the float type is displaying the nearest valid binary number to the decimal value displayed. If you want more precision then you would need to use a specialist maths interpreter. You could look at Sage Maths. Wolfram alpha gives 1.4142135623730950488016887242096980785696718753769480731766797379 but sage has a python interface. – nerak99 May 05 '21 at 14:35
  • BTW, you could use the huge precision that Python allows with integers by finding the sqrt of 2E2*n where n is large enough to generate the precision you are after. – nerak99 May 05 '21 at 14:42

2 Answers2

3

floats don't have sufficient precision for this. You need the decimal module:

from decimal import Decimal, getcontext
getcontext().prec = 51 # the "1" before the decimal point counts, too
x = Decimal("1.4")
for n in range(50):
    next_num = [0,1,2,3,4,5,6,7,8,9]
    candidate = 0
    for number in next_num:
        if Decimal(str(x)+str(number))*Decimal(str(x)+str(number))<2:
            candidate = number
    x = Decimal(str(x)+str(candidate))

print(x)

Output:

1.414213562373095048801688724209698078569671875376946
Tim Pietzcker
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0

You can approximate any root with integers so you won't be limited by decimal places, but the only down side is that your output won't have decimal point.

"x" is the number you want to take root of. "b" is the nth root. "dec" is the number of decimal places.

def root(x,b,dec):
    s = 0
    n = 0
    for q in range(dec):
        for k in range(10):
            s = n
            n = 10*n+(9-k)
            if(n**b > x*10**(q*b)):
                n = s
            else:
                break
    return n

Input:
root(2,2,500)

Output:
14142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727350138462309122970249248360558507372126441214970999358314132226659275055927557999505011527820605714701095599716059702745345968620147285174186408891986095523292304843087143214508397626036279952514079896872533965463318088296406206152583523950547457502877599617298355752203375318570113543746034084988471603868999706990048150305440277903164542478230684929369186215805784631115966687130130156185689872372
The64mer7
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