Adding other cases found
3⋆39
4052555153018976267
1÷3⋆39
1÷4052555153018976267
3⋆40
1.215766546E19
1÷3⋆40
8.22526334E¯20
Another example which does not have a large denumerator
(1÷3)×1÷3
1÷9
(1÷1)×1÷3
0.3333333333
1×1÷3
0.3333333333
2×1÷3
0.6666666667
3×1÷3
1
5×1÷3
1.666666667
6×1÷3
2
1×1÷2
0.5
2×1÷2
1
Regards,
Ala'a
On Mon, Aug 14, 2017 at 11:12 PM, Ala'a Mohammad <amal...@gmail.com> wrote:
> Hi,
>
> I was playing with the sum of the series 1/3, 1/9, 1/27 ..etc
> (1÷3⋆⍳x), and found that If I'm using the rationals experimental
> feature (⎕ps ← 1 0), then it the following were found:
>
> 1)
>
> 1÷3⋆⍳5
> 1÷3 1÷9 1÷27 1÷81 1÷243
>
> +/1÷3⋆⍳39
> 2026277576509488133÷4052555153018976267
>
> +/1÷3⋆⍳40
> 0.5
>
> Shouldn't the above be 1÷2?
>
> 2) creating a table for the sums of the above series until 39
>
> n,⍪{+/1÷3⋆⍳⍵}¨n←⍳39
> 1 1÷3
> 2 4÷9
> 3 13÷27
> 4 40÷81
> 5 121÷243
> 6 364÷729
> 7 1093÷2187
> 8 3280÷6561
> 9 9841÷19683
> 10 29524÷59049
> 11 88573÷177147
> 12 265720÷531441
> 13 797161÷1594323
> 14 2391484÷4782969
> 15 7174453÷14348907
> 16 21523360÷43046721
> 17 64570081÷129140163
> 18 193710244÷387420489
> 19 581130733÷1162261467
> 20 1743392200÷3486784401
> 21 5230176601÷10460353203
> 22 15690529804÷31381059609
> 23 47071589413÷94143178827
> 24 141214768240÷282429536481
> 25 423644304721÷847288609443
> 26 1270932914164÷2541865828329
> 27 3812798742493÷7625597484987
> 28 11438396227480÷22876792454961
> 29 34315188682441÷68630377364883
> 30 102945566047324÷205891132094649
> 31 308836698141973÷617673396283947
> 32 926510094425920÷1853020188851841
> 33 2779530283277761÷5559060566555523
> 34 8338590849833284÷16677181699666569
> 35 25015772549499853÷50031545098999707
> 36 75047317648499560÷150094635296999121
> 37 225141952945498681÷450283905890997363
> 38 675425858836496044÷1350851717672992089
> 39 2026277576509488133÷4052555153018976267
>
> Is fine, but after 39 (when we starting getting 0.5, the numerators of
> the above cases in the table have 'float' numerators, and the last 0.5
> at the end of the table is float and written using the E notation.
>
> n,⍪{+/1÷3⋆⍳⍵}¨n←⍳40
> 1 1.0÷3
> 2 4.0÷9
> 3 13.0÷27
> 4 40.0÷81
> 5 121.0÷243
> 6 364.0÷729
> 7 1093.0÷2187
> 8 3280.0÷6561
> 9 9841.0÷19683
> 10 29524.0÷59049
> 11 88573.0÷177147
> 12 265720.0÷531441
> 13 797161.0÷1594323
> 14 2391484.0÷4782969
> 15 7174453.0÷14348907
> 16 21523360.0÷43046721
> 17 64570081.0÷129140163
> 18 193710244.0÷387420489
> 19 581130733.0÷1162261467
> 20 1743392200.0÷3486784401
> 21 5230176601.0÷10460353203
> 22 15690529804.0÷31381059609
> 23 47071589413.0÷94143178827
> 24 141214768240.0÷282429536481
> 25 423644304721.0÷847288609443
> 26 1270932914164.0÷2541865828329
> 27 3812798742493.0÷7625597484987
> 28 11438396227480.0÷22876792454961
> 29 34315188682441.0÷68630377364883
> 30 102945566047324.0÷205891132094649
> 31 308836698141973.0÷617673396283947
> 32 926510094425920.0÷1853020188851841
> 33 2779530283277761.0÷5559060566555523
> 34 8338590849833284.0÷16677181699666569
> 35 25015772549499853.0÷50031545098999707
> 36 75047317648499560.0÷150094635296999121
> 37 225141952945498681.0÷450283905890997363
> 38 675425858836496044.0÷1350851717672992089
> 39 2026277576509488133.0÷4052555153018976267
> 40 0.5E0
>
> I tried to replicate it using another simpler example below:
>
> ⎕←x←0.5, 1÷1 2 3
> 0.5 1 1÷2 1÷3
>
> x,x
> 0.5 1 1÷2 1÷3 0.5 1 1÷2 1÷3
>
> x,[.5]x
> 0.5 1 1÷2 1÷3
> 0.5 1 1÷2 1÷3
>
> x,⍪x
> 0.5E0 0.5E0
> 1.0E0 1.0E0
> 1.0÷2 1.0÷2
> 1.0÷3 1.0÷3
>
> Notice that numerators are floats in the last example.
>
> Hope this helps.
>
> Regards,
>
> Ala'a
