mathematics

Yet Another Math Problem

Of the Natural numbers from 1 to 10^100, find N such that ABS(SIN(N)) is the smallest one!

We might have a relative winner here. Someone may say SIN(1)=0.841…

Can you do any better?

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7 thoughts on “Yet Another Math Problem

  1. msjr says:

    Ineresting:

    1 = 0.8414709848078965
    3 = 0.1411200080598672
    22 = 0.008851309290403876
    333 = 0.008821166113885877
    355 = 0.00003014435335948845
    103993 = 0.000019129335778423752
    104348 = 0.000011015017584240557
    208341 = 0.000008114318195038078
    312689 = 0.000002900699389332111
    833719 = 0.0000023129194164527015
    1146408 = 5.877799728813812e-7
    4272943 = 5.495794978104908e-7
    5419351 = 3.8200475070896605e-8
    80143857 = 1.4772846817965908e-8
    165707065 = 8.654781434964792e-9
    245850922 = 6.1180653830011166e-9
    411557987 = 2.5367160519636766e-9

    Bruta forza FTW: https://jsfiddle.net/ggb2s50f/

    • Some people suggested, that you only have to find a good rational approximation for Pi, like 22/7 or 355/113 and then SIN(22) is close to 0 and SIN(355) is close to 0.

      You seemingly proving that with two results above.

      However, 7^7/2^18 is 3.1415672.., but SIN(7^7)=-.3722 …

      A good approximation for Pi, still the counter 7^7 is a lousy solution,

  2. As you can see on the link, we have a solution by Cousin Italian. It is the solution.

    Gareth McCaughan found it too, but he had a typo with additional 9 at the end.

    7651502734877145442576129882280145997399533964991618287115405543322160824753834564921930038855560478

  3. OM says:

    Sine is periodic and so one way to brute force this is to multiply pi with an integer, say 103604811, and see how long a series of zeros it yields: 325484113,11415900000000009828507345…

    One then takes everything left of the block of zeros (omitting the comma, which is basically multiplying with a power of ten) and plugs the integer into sin:
    sin(325484113114159) = -9,828507345621e-11

    Thanks for making me learn some Python to get this 🙂

    • This is generally the way I had in mind for solving this class of problems. Brute force tweaking of the best solution so far – as long as is needed.

      Should be O(log(N)) complex.

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