I assume you should use floor division to preserve accuracy. As a
numerical example, say you want the first 100 /decimal/ digits of 3 /
173. That's the same as floor(10^100 * 3 / 173).
In [105]: from decimal import Decimal, getcontext
...: getcontext().prec = 300
...: print(10**100 * 91 // 173)
...: print(int(str(Decimal(91) / Decimal(173)).split(".")[1][:100]))
5260115606936416184971098265895953757225433526011560693641618497109826589595375722543352601156069364
5260115606936416184971098265895953757225433526011560693641618497109826589595375722543352601156069364
Gareth
On 10/02/2024 16:24, Georgi Guninski wrote:
Given SR constant, I want to extract the first L base B digits
of the fractional part.
So far the best solution I found is:
floor((SR(expression)*B**L).numerical_approx(digits=L)).digits(B)
Can I do better?
Why the following approach fails numerically:
```
def base_B_digits(A, B,prot=False):
digits=[]
fractional_part = A - int(A)
while fractional_part != 0:
digit = int(fractional_part * B)
digits.append(digit)
fractional_part = fractional_part * B - digit
if prot: print(fractional_part)
return digits
```
tt=base_B_digits(SR(1/3).numerical_approx(digits=10),10,1)
The fractional part starts:
0.3333333333
0.3333333331
0.3333333314
0.3333333139
0.3333331393
...
0.7812500000
0.8125000000
0.1250000000
0.2500000000
0.5000000000
0.0000000000
I am not sure the program must terminate.
For A=1/4, the base 10 digits are computed correctly.
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