Talk:Super-d numbers: Difference between revisions
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It is noted elsewhere that '''super-d''' numbers are always expressed in decimal (base ten). -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 07:34, 12 October 2019 (UTC) |
It is noted elsewhere that '''super-d''' numbers are always expressed in decimal (base ten). -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 07:34, 12 October 2019 (UTC) |
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: I updated the description of super-d numbers to reflect this. Thanks. --[[User:Chunes|Chunes]] ([[User talk:Chunes|talk]]) 20:50, 12 October 2019 (UTC) |
: I updated the description of super-d numbers to reflect this. Thanks. --[[User:Chunes|Chunes]] ([[User talk:Chunes|talk]]) 20:50, 12 October 2019 (UTC) |
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== Mostly addition method == |
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One can get a list of squares by summing up the list of odd numbers, like this: |
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<pre> |
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0 + 1 = 1 |
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1 + 3 = 4 |
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4 + 5 = 9 |
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9 + 7 = 16 |
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16 + 9 = 25 |
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</pre> |
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Another way of tabulating it: |
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<pre> |
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square, 1st difference, 2nd difference |
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0, 1, 2 |
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1, 3, 2 |
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4, 5, 2 |
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9, 7, 2 |
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16, 9, 2 |
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25, 11, 2 |
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</pre> |
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The 2nd difference, 2, is also 2!<br/> |
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On to cubes: |
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<pre> |
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cube, 1st difference, 2nd difference, 3rd difference |
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0, 1, 6, 6 |
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1, 7, 12, 6 |
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8, 19, 18, 6 |
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27, 37, 24, 6 |
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64, 61, 30, 6 |
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125, 91, 36, 6 |
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216, 127, 42, 6 |
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</pre> |
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The 3rd difference, 6, is also 3!<br/> |
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This pattern continues on, with the number of differences required (to calculate by adding) increasing by one and the last difference being n! To calculate a list of any powers, one can start at zero with the proper array of initial values, or start at "n" with a proper array of initial values. When starting at zero, some of the initial values will be negative. When starting at "n", all initial values will be positive. |
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Here is the initial array for each "n", starting at zero: |
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<pre> |
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2: 0 -1 2 |
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3: 0 1 -6 6 |
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4: 0 -1 14 -36 24 |
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5: 0 1 -30 150 -240 120 |
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6: 0 -1 62 -540 1560 -1800 720 |
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7: 0 1 -126 1806 -8400 16800 -15120 5040 |
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8: 0 -1 254 -5796 40824 -126000 191520 -141120 40320 |
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9: 0 1 -510 18150 -186480 834120 -1905120 2328480 -1451520 362880 |
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</pre> |
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Here is the initial array for each "n", starting at "n": |
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<pre> |
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2: 4 3 2 |
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3: 27 19 12 6 |
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4: 256 175 110 60 24 |
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5: 3125 2101 1320 750 360 120 |
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6: 46656 31031 19502 11340 5880 2520 720 |
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7: 823543 543607 341796 201726 109200 52080 20160 5040 |
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8: 16777216 11012415 6927230 4131036 2298744 1164240 514080 181440 40320 |
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9: 387420489 253202761 159338640 95750430 54313560 28594440 13608000 5594400 1814400 362880 |
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</pre> |
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When working backwards by summing instead of subtracting, the tables look like this because we work left to right on the array:<br/> |
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Squares: |
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<pre> |
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0: 0 -1 2 |
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1: 1 1 2 |
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2: 4 3 2 |
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3: 9 5 2 |
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4: 16 7 2 |
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5: 25 9 2 |
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</pre> |
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Cubes: |
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<pre> |
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0: 0 1 -6 6 |
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1: 1 7 12 6 |
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2: 8 19 18 6 |
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3: 27 37 24 6 |
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4: 64 61 30 6 |
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5: 125 91 36 6 |
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</pre> |
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Regarding the term "mostly addition", for this task there is some multiplication by a small integer (2 thru 9 in ''scale()''), but the powers of "n" are calculated by only addition. The initial tables are calculated with multiplication and a power function (''ipow()''). |
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A side note, one can also calculate cubes by summing groups of odd numbers like this: |
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<pre> |
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1 = 1 |
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8 = 3 + 5 |
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27 = 7 + 9 + 11 |
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64 = 13 + 15 + 17 + 19 |
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125 = 21 + 23 + 25 + 27 + 29 |
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</pre> |
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But I haven't worked out how (or if) a similar method works for higher powers. |
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--[[User:Enter your username|Enter your username]] ([[User talk:Enter your username|talk]]) 16:08, 21 August 2020 (UTC) |