Skip to main content

bugs - What calendar is Mathematica using for dates in the distant past?


It appears that Mathematica treats all dates as proleptic Gregorian dates by default, a hypothesis that can be easily tested by using AbsoluteTime to compute the Julian Day:


jd[t_] := AbsoluteTime[t]/86400 + 2.4150205*^6

but, while this works for recent dates and for some older ones, it yields results that differ from the correct result for some older dates by exactly a day


jd[{2012, 11, 24, 12}] - 2456256
0
jd[{1100, 11, 24, 12, 0, 0}] - 2123154
0
jd[{-3000, 11, 24, 12, 0, 0}] - 625660

1

notably for the reference date, 12:00 Universal Time on January 1, 4713 BCE in the proleptic Julian calendar (-4713-11-24 in the proleptic Gregorian calendar):


jd[{-4713, 11, 24, 12, 0, 0}]
1

What calendar is Mathematica using for these older dates, if not the proleptic Gregorian calendar?


Here, perhaps is another clue:


DateList[{-4713, 11, 24, 12, 0, 0}]
{-4713, 11, 25, 12, 0, 0}

DateList[{-3000, 11, 24, 12, 0, 0}]
{-3000, 11, 25, 12, 0, 0}

I'm no calendar expert, but this "canonicalization" doesn't map to any calendar I'm familiar with. Is this a bug?




Update: Version 9, behaves differently:


jd[{2012, 11, 24, 12}] - 2456256
0.
jd[{1100, 11, 24, 12, 0, 0}] - 2123154
0.

jd[{-3000, 11, 24, 12, 0, 0}] - 625660
365.

jd[{-4713, 11, 24, 12, 0, 0}]
366

and


DateList[{-4713, 11, 24, 12, 0, 0}]
{-4713, 11, 24, 12, 0, 0.}
DateList[{-3000, 11, 24, 12, 0, 0}]

{-3000, 11, 24, 12, 0, 0.}


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...