Wikipedia:Reference desk/Archives/Mathematics/2021 November 27

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November 27

"Tiny" difference between the Julian and Gregorian calendars

Define two numbers x and y as follows:

• x is the binary number 0.000100010001... (which equals ${\frac {1}{2^{4}-1}}={\frac {1}{15}}$), corresponding to leap years in the Julian calendar.
• y is the same as x, but with the 100th, 200th, and 300th bits in every block of 400 bits to the right of the decimal point replaced with 0, corresponding to leap years in the (proleptic) Gregorian calendar. This is ${\frac {1}{2^{4}-1}}-{\frac {1}{2^{100}-1}}+{\frac {1}{2^{400}-1}}$.

Then, the difference $x-y={\frac {1}{2^{100}-1}}-{\frac {1}{2^{400}-1}}$ is very small. But how small is it exactly? Specifically, what would it be in scientific notation (in base 10 of course, not base 2)? GeoffreyT2000 (talk) 19:19, 27 November 2021 (UTC)

$7.9*10^{-31}$, according to WolframAlpha's calculations. Iffy★Chat -- 19:28, 27 November 2021 (UTC)

Almost exactly 2^-100 since the other terms are insignificant. That is about 7.9e-31 as Iffy says, and which you can find with a calculator or with logarithms. 2601:648:8202:350:0:0:0:69F6 (talk) 20:24, 27 November 2021 (UTC)

This binary representation of the leap year pattern has of course nothing to do with the average difference in days per calendar year between the two systems, which equals

${\frac {1}{4}}-\left({\frac {1}{4}}-{\frac {1}{100}}+{\frac {1}{400}}\right)={\frac {1}{100}}-{\frac {1}{400}}=0.0075\,$.

 --Lambiam 00:29, 28 November 2021 (UTC)