How about ANY FINITE SEQUENCE AT ALL?

        • cosecantphi [he/him, they/them]@hexbear.net
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          2 days ago

          It’s implicitly defined here by its decimal form:

          0.101001000100001000001 . . .

          The definition of this number is that the number of 0s after each 1 is given by the total previous number of 1s in the sequence. That’s why it can’t contain 2 despite being infinite and non-repeating.

            • cosecantphi [he/him, they/them]@hexbear.net
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              1 day ago

              That’s a decimal approximation of Pi with an ellipsis at the end to indicate its an approximation, not a definition. The way the ellipsis is used above is different. It’s being used to define a number via the decimal expansion by saying it’s an infinite sum of negative powers of 10 defined by the pattern before the ellipsis.

              So we have:

              0.101001000100001000001 . . . = 10^-1 + 10^-2 + 10^-3 + 10^-4 +10^-5+ . . .

              Pi, however, is not defined this way. Pi can be defined as twice the solution of the integral from -1 to 1 of the square root of (1-x^2), a function defining a unit semi-circle.

          • मुक्त@lemmy.ml
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            2 days ago

            0.101001000100001000001 . . .

            Might very well be :

            0.101001000100001000001202002000200002000002 …

            Real life, is different from gamified questions asked in student exams.

            • cosecantphi [he/him, they/them]@hexbear.net
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              1 day ago

              Implicitly defining a number via it’s decimal form typically relies on their being a pattern to follow after the ellipsis. You can define a different number with twos in it, but if you put an ellipsis at the end you’re implying there’s a different pattern to follow for the rest of the decimal expansion, hence your number is not the same number as the one without twos in it.

      • flashgnash@lemm.ee
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        3 days ago

        Because you’d need to search through an infinite number of digits (unless you have access to the original formula)