Tag Archives: maths

Self-Raising Power

by in √2, Irrational numbers, Mathematics, Powers, Square root of 2 and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

The square root of 2 is the number that, raised to the power of 2, equals 2. That is, if r^2 = r * r = 2, then r = √2. The cube root of 2 is the number that, raised to the power of 3, equals 2. That is, if r^3 = r * r * r = 2, then r = [3]√2.

But what do you call the number that, raised to the power of itself, equals 2? I suggest “the auto-root of 2”. Here, if r^r = 2, then r = [r]√2. I don’t know a quick way to calculate the auto-root, but you can adapt a well-known algorithm for approximating the square root of a number. The square-root algorithm looks like this:

n = 2
r = 1
for c = 1 to 20
    r = (r + n/r) / 2
next c
print r

r = 1.414213562…

Note the fourth line of the algorithm: r = (r + n/r) / 2. When r is an over-estimate of √2, then 2/r will be an under-estimate (and vice versa). (r + 2/r) / 2 splits the difference and refines the estimate. Using the lines above as the model, the auto-root algorithm looks like this:

n = 2
r = 1
for c = 1 to 20
    r = (r + [r]√n) / 2[*]
next c
print r

r = 1.559610469…

*This is equivalent to r = (r + n^(1/r)) / 2

Here are the first 100 digits of [r]√2 = r in base 10:

1, 5, 5, 9, 6, 1, 0, 4, 6, 9, 4, 6, 2, 3, 6, 9, 3, 4, 9, 9, 7, 0, 3, 8, 8, 7, 6, 8, 7, 6, 5, 0, 0, 2, 9, 9, 3, 2, 8, 4, 8, 8, 3, 5, 1, 1, 8, 4, 3, 0, 9, 1, 4, 2, 4, 7, 1, 9, 5, 9, 4, 5, 6, 9, 4, 1, 3, 9, 7, 3, 0, 3, 4, 5, 4, 9, 5, 9, 0, 5, 8, 7, 1, 0, 5, 4, 1, 3, 4, 4, 4, 6, 9, 1, 2, 8, 3, 9, 7, 3…

And here is [r]√n = r for n = 2..20:

autopower(2) = 1.5596104694623693499703887…
autopower(3) = 1.8254550229248300400414692…
autopower(4) = 2
autopower(5) = 2.1293724827601566963803119…
autopower(6) = 2.2318286244090093673920215…
autopower(7) = 2.3164549587856123013255030…
autopower(8) = 2.3884234844993385564187215…
autopower(9) = 2.4509539280155796306228059…
autopower(10) = 2.5061841455887692562929409…
autopower(11) = 2.5556046121008206152514542…
autopower(12) = 2.6002950000539155877172082…
autopower(13) = 2.6410619164843958084118390…
autopower(14) = 2.6785234858912995813011990…
autopower(15) = 2.7131636040042392095764012…
autopower(16) = 2.7453680235674634847098492…
autopower(17) = 2.7754491049442334313328329…
autopower(18) = 2.8036632456580215496843618…
autopower(19) = 2.8302234384970308956026277…
autopower(20) = 2.8553085030012414128332189…

I assume that the auto-root is always an irrational number, except when n is a perfect power of suitable form, i.e. n = p^p for some integer p. For example, autoroot(4) = 2, because 2^2 = 4, autoroot(27) = 3, because 3^3 = 27, and so on.

And here is the graph of autoroot(n) for n = 2..10000:
autoroot

Performativizing Papyrocentricity #43

by in Biology, Book Reviews, Botany, Flowers, History, Infinity, Linguistics, Mathematics, Music, Philosophy of mathematics and tagged , , , , , , , , , , , , , , | Leave a comment

Papyrocentric Performativity Presents:

• Avens Above – Harrap’s Wild Flowers: A Guide to the Wild Flowers of Britain & Ireland, Simon Harrap (Bloomsbury 2013)

• Place of Glades – A Dictionary of British Place-Names, A.D. Mills (Oxford University Press 1991)

• De Minimis Curat Rex? – Infinitesimal: How a Dangerous Mathematical Theory Shaped the Modern World, Amir Alexander (Oneworld 2014)

• Seen and Not Heard – The Greatest Albums You’ll Never Hear, ed. Bruno MacDonald (Aurum Press 2014)

Or Read a Review at Random: RaRaR

Get Your Prox Off

by in Fractals, Geometry, Mathematics and tagged , , , , , , , , , , , , , | Leave a comment

Create a triangle. Find a point somewhere inside it. Choose a corner at random and move halfway towards it. Mark the new point. Repeat the procedure: choose, move, mark. Repeat again and again. In time, a fractal will appear:

siertri

However, if you try the same thing with a square – choose a corner at random, move halfway towards it, mark the new point, repeat – no fractal appears. Instead, the points fill the interior of the square:

sierquad

But what happens if you impose restrictions on the randomly chosen corner (or chorner)? Suppose you can’t choose the same corner twice in a row. If this rule is applied to the square, this fractal appears:

restrict4_T

restrict4_Tanim

Now apply the no-corner-twice-in-a-row rule to a square that contains a central chorner. This fractal appears:

restrict4_Tc

And if the rule is that you can choose a corner twice in a row but not thrice? This fractal appears:

restrict4FT

restrict4FTc

Here is the rule is that a corner can’t be chosen if it was chosen two moves ago:

restrict4_3F

But what if the restriction is based not on how often or when a corner is chosen, but on its proximity, i.e. how near it is to the marked point? If the nearest corner can’t be chosen, the result is the same as the no-corner-twice-in-a-row rule:

prox4_1

But if the second-nearest corner can’t be chosen, this fractal appears:

prox4_2

This is the fractal when the third-nearest corner can’t be chosen:

prox4_3

And this is the fractal when the fourth-nearest, or most distant, corner can’t be chosen:

prox4_4

Here are the same restrictions applied to a pentagon:

prox5_1

Nearest corner forbidden

prox5_2

Second-nearest corner forbidden

prox5_3

Third corner forbidden

prox5_4

Fourth corner forbidden

prox5_5

Fifth corner forbidden

prox5_5anim

Fifth corner forbidden (animated)

And a pentagon with a central chorner:

prox5_anim_c

Now try excluding more than one corner. Here are pentagons excluding the n-nearest and n+1-nearest corners (for example, the nearest and second-nearest corners; the second-nearest and third-nearest; and so on):

prox5n_n1_anim

But what if the moving point is set equal to the n-nearest corner before it moves again? If the corner is the second-nearest and the shape is a triangle with a central chorner, this is the fractal that appears:

prox3_set2c

prox3_set2c_anim

Animated version

And here is the same rule applied to various n-nearest corners in a pentagon:

prox5_set_anim

The Choice of the Circle

by in Geometry, Mathematics and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

Here’s an elementary mathematical problem: how many ways are there to choose three numbers from a set of six numbers? If the set is (1, 2, 3, 4, 5, 6), these are the possible choices (or combinations):

(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6) (c = 20)

So 6C3 = 20 (C stands for “combination”). The general formula is nCr = (n! / (n-r)!) / r!, where n is the number to choose from, r is the number of choices and n! is factorial n, or n multiplied by all numbers less than itself. For example, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. When n = 6 and c = 3, 6C3 = (6! / (6-3)!) / 3! = (720 / 6) / 6 = 20.

There isn’t much visual appeal in the choices above, but there’s a simple way to change that. Take the ways of choosing two numbers from a set of ten. They start like this:

(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (3, 4), (3, 5), (3, 6)…

Suppose each choice represents the midpoint of two points chosen from a set of ten points around a pentagon, so that (1, 2) is half-way between points 1 and 2, (3, 5) is half-way between points 3 and 5, and so on:

pent_10_2

Now take the ways of choosing three numbers from a set of ten:

(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 8), (1, 2, 9), (1, 2, 10), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 8), (1, 3, 9), (1, 3, 10)…

Now the pentagon looks like this, with (1, 2, 3) representing the point midway between 1, 2 and 3, (1, 3, 9) representing the point midway between 1, 3 and 9, and so on:

pent_10_3

Now here are 10C4, 10C5 and 10C6 for the pentagon:

pent_10_4

pent_10_5

pent_10_6

You can also generate the points 5C4 = 5, then add them to the original five points and generate 10C4:

pent4_1

5C4

pent4_2

10C4

And here are 5C5, 6C5 and 12C5:

pent5

Here are 7C7 and 8C8, adding points as for 5C4:

hept7

octo8

And here is 12C6 using a dodecagon:

dodeca_6

And various nCr for dodecagons and other polygons:

various

This method can also be used to represent the partitions of n, or the number of sets whose members sum to n. The partitions of 5 are these:

(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)

There are seven partitions, so p(5) = 7. Partitions start small and get very large, starting with p(1), p(2), p(3) and so on:

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, 239943, 281589, 329931, 386155, 451276, 526823, 614154, 715220, 831820, 966467, 1121505, 1300156…

Suppose the partitions of n are treated as sets of points around a polygon with n vertices. Each set is then used to generate the point midway between its members. For example, (5, 4, 4, 2) is one partition of 15 and would represent the point midway between 5, 4, 4 and 2 of a pentadecagon. Here is a graphical representation of p(30):

partition30

Here are graphical representations for the partitions 5 to 15, then 15 to 60 in increments of 5 (15, 20, 25, etc):

partitions5_60

And here are some close-ups for the partitions of 35 and 40:

partitions40

Performativizing Papyrocentricity #42

by in Aesthetics, Art, Biology, Book Reviews, Flowers, History, Mathematics, Ornithology and tagged , , , , , , , , , , , , | Leave a comment

Papyrocentric Performativity Presents:

• Feats for the Eyes – Drawn from Paradise: The discovery, art and natural history of the birds of paradise, David Attenborough and Errol Fuller (Collins 2012)

• Heart of the Mather – Chaotic Fishponds and Mirror Universes: the maths that governs our world, Richard Elwes (Quercus 2013)

• Bergblumen – Enchanting Alpine Flowers, Alfred Pohler, trans. Jacqueline Schweighofer

Or Read a Review at Random: RaRaR

Power Trip

by in Integer Sequences, Mathematics, Powers and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

Here are the first few powers of 2:

2 = 1 * 2
4 = 2 * 2
8 = 4 * 2
16 = 8 * 2
32 = 16 * 2
64 = 32 * 2
128 = 64 * 2
256 = 128 * 2
512 = 256 * 2
1024 = 512 * 2
2048 = 1024 * 2
4096 = 2048 * 2
8192 = 4096 * 2
16384 = 8192 * 2
32768 = 16384 * 2
65536 = 32768 * 2
131072 = 65536 * 2
262144 = 131072 * 2
524288 = 262144 * 2
1048576 = 524288 * 2
2097152 = 1048576 * 2
4194304 = 2097152 * 2
8388608 = 4194304 * 2
16777216 = 8388608 * 2
33554432 = 16777216 * 2
67108864 = 33554432 * 2…

As you can see, it’s a one-way power-trip: the numbers simply get larger. But what happens if you delete the digit 0 whenever it appears in a result? For example, 512 * 2 = 1024, which becomes 124. If you apply this rule, the sequence looks like this:

2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 124
124 * 2 = 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 5794
5794 * 2 = 11588
11588 * 2 = 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 9274…

Is this a power-trip? Not quite: it’s a return trip, because the numbers can never grow beyond a certain size and the sequence falls into a loop. If the result 2n contains a zero, then zerodelete(2n) < n, so the sequence has an upper limit and a number will eventually occur twice. This happens at step 526 with 366784, which matches 366784 at step 490.

The rate at which we delete zeros can obviously be varied. Call it 1:z. The sequence above sets z = 1, so 1:z = 1:1. But what if z = 2, so that 1:z = 1:2? In other words, the procedure deletes every second zero. The first zero occurs when 1024 = 2 * 512, so 1024 is left as it is. The second zero occurs when 2 * 1024 = 2048, so 2048 becomes 248. When z = 2 and every second zero is deleted, the sequence begins like this:

1 * 2 = 2
2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 1024
1024 * 2 = 2048 → 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 50794
50794 * 2 = 101588 → 101588
101588 * 2 = 203176 → 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 92704
92704 * 2 = 185408 → 18548

This sequence also has a ceiling and repeats at step 9134 with 5458864, which matches 5458864 at step 4166. And what about the sequence in which z = 3 and every third zero is deleted? Does this have a ceiling or does the act of multiplying by 2 compensate for the slower removal of zeros?

In fact, it can’t do so. The larger 2n becomes, the more zeros it will tend to contain. If 2n is large enough to contain 3 zeros on average, the deletion of zeros will overpower multiplication by 2 and the sequence will not rise any higher. Therefore the sequence that deletes every third zero will eventually repeat, although I haven’t been able to discover the relevant number.

But this reasoning applies to any rate, 1:z, of zero-deletion. If z = 100 and every hundredth zero is deleted, numbers in the sequence will rise to the point at which 2n contains sufficient zeros on average to counteract multiplication by 2. The sequence will have a ceiling and will eventually repeat. If z = 10^100 or z = 10^(10^100) and every googolth or googolplexth zero is deleted, the same is true. For any rate, 1:z, at which zeros are deleted, the sequence n = zerodelete(2n,z) has an upper limit and will eventually repeat.

Update (30×21)

Six years later, I’ve found the answer for z = 3. And uncovered a serious error in this article. See:

• Power Trap

Block n Rule

by in Base Behavior, Binary numbers, Integer Sequences, Mathematics and tagged , , , , , , , , , , , , , , , , , | Leave a comment

One of my favourite integer sequences uses the formula n(i) = n(i-1) + digsum(n(i-1)), where digsum(n) sums the digits of n. In base 10, it goes like this:

1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587, 607, 620, 628, 644, 658, 677, 697, 719, 736, 752, 766, 785, 805, 818, 835, 851, 865, 884, 904, 917, 934, 950, 964, 983, 1003…

Another interesting sequence uses the formula n(i) = n(i-1) + digprod(n(i-1)), where digprod(n) multiplies the digits of n (excluding 0). In base 10, it goes like this:

1, 2, 4, 8, 16, 22, 26, 38, 62, 74, 102, 104, 108, 116, 122, 126, 138, 162, 174, 202, 206, 218, 234, 258, 338, 410, 414, 430, 442, 474, 586, 826, 922, 958, 1318, 1342, 1366, 1474, 1586, 1826, 1922, 1958, 2318, 2366, 2582, 2742, 2854, 3174, 3258, 3498, 4362, 4506, 4626, 4914, 5058, 5258, 5658, 6858, 8778, 11914, 11950, 11995…

You can apply these formulae in other bases and it’s trivially obvious that the sequences rise most slowly in base 2, because you’re never summing or multiplying anything but the digit 1. However, there is a sequence for which base 2 is by far the best performer. It has the formula n(i) = n(i-1) + blockmult(n(i-1)), where blockmult(n) counts the lengths of distinct blocks of the same digit, including 0, then multiplies those lengths together. For example:

blockmult(3,b=2) = blockmult(11) = 2
blockmult(28,b=2) = blockmult(11100) = 3 * 2 = 6
blockmult(51,b=2) = blockmult(110011) = 2 * 2 * 2 = 8
blockmult(140,b=2) = blockmult(10001100) = 1 * 3 * 2 * 2 = 12
blockmult(202867,b=2) = blockmult(110001100001110011) = 2 * 3 * 2 * 4 * 3 * 2 * 2 = 576

The full sequence begins like this (numbers are represented in base 10, but the formula is being applied to their representations in binary):

1, 2, 3, 5, 6, 8, 11, 13, 15, 19, 23, 26, 28, 34, 37, 39, 45, 47, 51, 59, 65, 70, 76, 84, 86, 88, 94, 98, 104, 110, 116, 122, 126, 132, 140, 152, 164, 168, 171, 173, 175, 179, 187, 193, 203, 211, 219, 227, 245, 249, 259, 271, 287, 302, 308, 316, 332, 340, 342, 344, 350, 354, 360, 366, 372, 378, 382, 388, 404, 412, 436, 444, 460, 484, 500, 510, 518, 530, 538, 546, 555, 561, 579, 595, 603, 611, 635, 651, 657, 663, 669, 675, 681…

In higher bases, it rises much more slowly. This is base 3:

1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 16, 17, 19, 20, 21, 22, 24, 26, 29, 31, 33, 34, 35, 37, 39, 42, 44, 48, 49, 51, 53, 56, 58, 60, 61, 62, 64, 65, 66, 68, 70, 71, 73, 75, 77, 79, 82, 85, 89, 93, 95, 97, 98, 100, 101, 102, 103, 105, 107, 110, 114, 116, 120, 124, 127, 129, 131, 133, 137, 139, 141, 142, 143, 145, 146, 147, 149, 151, 152, 154, 156, 158, 160, 163…

And this is base 10:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90…

Note how, in bases 3 and 10, blockmult(n) often equals 1. In base 3, the sequence contains [141, 142, 143, 145]:

blockmult(141,b=3) = blockmult(12020) = 1 * 1 * 1 * 1 = 1
blockmult(142,b=3) = blockmult(12021) = 1 * 1 * 1 * 1 = 1
blockmult(143,b=3) = blockmult(12022) = 1 * 1 * 1 * 2 = 2

The formula also returns 1 much further along the sequence in base 3. For example, the 573809th number in the sequence, or n(573809), is 5775037 and blockmult(5775037) = blockmult(101212101212021) = 1^15 = 1. But in base 2, blockmult(n) = 1 is very rare. It happens three times at the beginning of the sequence:

1, 2, 3, 5, 6, 8, 11…

After that, I haven’t found any more examples of blockmult(n) = 1, although blockmult(n) = 2 occurs regularly. For example,

blockmult(n(100723)) = blockmult(44739241) = blockmult(10101010101010101010101001) = 2
blockmult(n(100724)) = blockmult(44739243) = blockmult(10101010101010101010101011) = 2
blockmult(n(100725)) = blockmult(44739245) = blockmult(10101010101010101010101101) = 2

Does the sequence in base 2 return another example of blockmult(n) = 1? The odds seem against it. For any given number of digits in base 2, there is only one number for which blockmult(n) = 1. For example: 1, 10, 101, 1010, 10101, 101010, 1010101… As the sequence increases, the percentage of these numbers becomes smaller and smaller. But the sequence is infinite, so who knows what happens in the end? Perhaps blockmult(n) = 1 occurs infinitely often.

Over Again

by in Fractals, Geometry, Mathematics and tagged , , , , , , , , , , | 1 Comment

In Boldly Breaking the Boundaries, I looked at the use of squares in what I called over-fractals, or fractals whose sub-divisions reproduce the original shape but appear beyond its boundaries. Now I want to look at over-fractals using triangles. They’re less varied than those involving squares, but still include some interesting shapes. This is the space in which sub-triangles can appear, with the central seeding triangle coloured gray: triangle
Here are some over-fractals based on the pattern above: overtri1
overtri1_static

overtri2
overtri2_static

overtri3

overtri3_static

overtri4
overtri4_static

overtri5
overtri5_static

overtri6
overtri6_static

overtri7
overtri7_static

overtri8
overtri8_static

overtri9
overtri9_static

overtri10
overtri10_static

overtri11
overtri11_static

overtri12
overtri12_static

overtri13

overtri13_static