提取鼠尾草中数字的第n位数字

Question

How is it possible to extract the n-th digit of a number in Sagemath? We have to compute the 13787th digit from pi + e in Sagemath. My approach was the following:

sage: xi = e + pi
....: var1 = xi.n(digits=13786+1)
....: var2 = xi.n(digits=13787+1)
....: var3 = ((var2-var1) * 10^13787).trunc()
....: var3
0

Which gave me 0 back, however it should be 9 .

Answer 1

The digit is indeed 9. But the subsequent digits are also 9: this part of decimal expansion goes ...9999237... Rounding rolls these 9s to 0s, carrying 1 to a higher digit.

So you need some extra digits in order to avoid the digit you are interested in from being affected by rounding. How many depends on the number; we don't know in advance if there isn't a sequence of one billion of 9s starting from that position. Here I use 10 extra digits

xi = e + pi
n = 13787
offset = 1 + floor(log(xi, 10))   # the first significant figure is not always the first digit after decimal dot, so we account for that 
extra = 10
digit = int(str(xi.n(digits = n + offset + extra))[-1 - extra])

This returns 9. I think extracting with str is more reliable than subtracting two nearly-equal numbers and hoping there won't be additional loss of precition there.

Of course, including a magic number like 10 isn't reliable. Here is a better version, which starts with 10 extra digits but then increases the number until we no longer have 00000000... as a result.

xi = e + pi
n = 13787
offset = 1 + floor(log(xi, 10))
extra = 10
while True:
   digits = str(xi.n(digits = n + offset + extra))[-1 - extra:]
   if digits == "0" * len(digits):
       extra *= 2
   else:
       digit = int(digits[0])
       break
print(digit)

This will loop forever if the digits keep coming as 0, and rightly so : without actually knowing what the number is we can never be sure if the ...0000000... we get is not really ...999999999942... rounded up.