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Figurative Numbers

figure square dots formula

Figurative numbers are numbers which can be represented by dots arranged in various geometric patterns. For example, triangular numbers are represented by the patterns shown in Figure 1.

The numbers they represent are 1, 3, 6, 10, and so on.

Figurative numbers were first studied by the mathematician Pythagoras in the sixth century B.C. and by the Pythagoreans, who were his followers. These numbers were studied, as were many kinds of numbers, for the sake of their supposed mystical properties rather than for their practical value. The study of figurative numbers continues to be a source of interest to both amateur and professional mathematicians.

Figurative numbers also include the square numbers which can be represented by square arrays of dots, as shown in Figure 2.

The first few square numbers are 1, 4, 9, 16, 25, etc.

There are pentagonal numbers based on pentagonal arrays. Figure 3 shows the fourth pentagonal number, 22.

Other pentagonal numbers are 1, 5, 12, 22, 35, and so on.

There is, of course, no limit to the number of polygonal arrays into which dots may be fitted. There are hexagonal numbers, heptagonal numbers, octagonal numbers, and, in general, n-gonal numbers, where n can be any number greater than 2.

One reason that figurative numbers have the appeal they do is that they can be studied both algebraically and geometrically. Properties that might be hard to discover by algebraic techniques alone are often revealed by simply looking at the figures, and this is what we shall do, first with triangular numbers.

If we denote by Tn the n-th triangle number, Figure 1 shows us that T1 = 1 T2 = T1 + 2 T3 = T2 + 3 T4 = T3 + 4.

These formulas are recursive. To compute T10 one must compute T9. To compute T9 one has to compute T8, and so on. For small values of n this is not hard to do: T10 = 10 + (9 + (8 + (7 + (6 + (5 + (4 + (3 + (2 + 1)))))))) = 55.

For larger values of n, or for general values, a formula that gives Tn directly would be useful. Here the use of the figures themselves comes into play: From Figure 4 one can see that 2T3 = (3)(4); so T3 = 12/2 or 6.

The same trick can be applied to any triangular number: 2Tn = n(n + 1) Tn = n(n+ 1)/2 When n = 10, T10 = (10)(11)/2, or 55, as before.

Figure 1. Illustration by Hans & Cassidy. Courtesy of Gale Group.


Figure 2. Illustration by Hans & Cassidy. Courtesy of Gale Group.


Figure 3. Illustration by Hans & Cassidy. Courtesy of Gale Group.


Figure 4. Illustration by Hans & Cassidy. Courtesy of Gale Group.


In the case of square numbers, the general formula for the n-th square number is easy to derive, It is simply n2. In fact, the very name given to n2 reflects the fact that it is a square number. The recursive formulas for the various square numbers are a little less obvious. To derive them, one can turn again to the figures themselves. Since each square can be obtained from its predecessor by adding a row of dots across the top and a column of dots down the side, including one dot in the corner, one gets the recursive pattern S1 = 1 S2 = S1 + 3 S3 = S2 + 5 S4 = S3 + 7...

or, alternatively

Thus S8 = 15 + (13 + (11 + (9 + (7 + (5 + (3 + 1)))))) or 64.

Because humans are so fond of arranging things in rows and columns, including themselves, square numbers in use are not hard to find. Tic-tac-toe has S3 squares; chess and checkers have S8. S19 has been found to be the ideal number of points, says a text on the oriental game of "go," on which to play the game.

One of the less obvious places where square numbers-or more correctly, the recursive formulas for generating them-show up is in one of the algorithms for computing square roots. This is the algorithm based on the formula (a + b)2 = a2 + 2ab + b2. When this formula is used, b is not 1. In fact, at each stage of the computing process the size of b is systematically decreased. The process however parallels that of finding Sn+1 recursively from Sn. This becomes apparent when n and 1 are substituted for a and b in the formula: (n + 1)2 = n2 + 2n + 1. This translates into Sn+1 = Sn + 2n + 1, which is the formula given earlier.

Formulas for pentagonal numbers are trickier to discover, both the general formula and the recursive formulas. But again the availability of geometric figures helps. By examining the array in Figure 5, one can come up with the following recursive formulas, where Pn represents the n-th pentagonal number: P1 = 1 P2= P1 + 4 P3 = P2 + 7 P4 = P3 +10. One can guess the formula for Pn: Pn= Pn-1 + 3n - 2 and this is correct. At each stage in going from Pn-1 to Pn one adds three sides of n dots each to the existing pentagon, but two of those dots are common to two sides and should be discounted.

To compute P7 recursively we have 19 + (16 + (13 + (10 + (7 + (4 + 1))))), which adds up to 70.

To find a general formula, we can pull another trick. We can cut up the array in Figure 5 along the dotted lines. When we do this we have P5 = T5 + 2T4 or more generally Pn = Tn + 2Tn-1.

If we substitute algebraic expressions for the triangular numbers and simplify the result we come up with P = (3n2n - n)/2.

The fact that pentagonal numbers can be cut into triangular numbers makes one wonder if other polygonal numbers can, too. Square numbers can, and Figure 6 shows an example.

It yields T5 + T4, or in general Sn = Tn + Tn-1. Arranging these formulas for triangular dissections suggests a pattern: Tn = Tn Sn = Tn + Tn-1 Pn= Tn + 2Tn-1 Hn= Tn + 3Tn-1 where Hn is the n-th hexagonal number. If we check this for H4 in Figure 7, we find it to be correct.

In general, if Nk represents the k-th polygonal number for a polygon of N sides (an N-gon) Nk = Tk + (N - 3)Tk-1.

The Pythagoreans were also concerned with "oblong" arrays (here "oblong" has a more limited meaning than usual), having one column more than its rows.

In this case the concern was not with the total number of dots but with the ratio of dots per row to dots per column. In the smallest array, this ratio is 2:1. In the next array it is 3:2; and the third, 4:3 (Figure 8). If the arrays were further enlarged, the ratios would change in a regular way: 5:4, 6:5, and so on.

These ratios were related, by Pythagoreans, to music. If, on a stringed instrument, two notes were played whose frequencies were in the ratio of 2:1, those notes would be an octave apart and would sound harmonious when played together. (Actually the Pythagoreans went by the lengths of the strings, but if the string lengths were in the ratio 1:2, the frequencies would be in the ratio 2:1.) If the ratio of the frequencies were 3:2, the notes would be a perfect fifth apart and would also sound harmonious (in fact, violinists use the harmony of perfect fifths to tune the strings of the violin, which sound a perfect fifth apart). Notes in the ratio 4:3 were also considered harmonious. Other ratios were thought to result in discordant notes.

The Pythagoreans went well beyond this in developing their musical theories, but what is particularly interesting is that they based a really fundamental musical idea on an array of dots. Of course, for reasons having little to do with figurative numbers, they got it right.

Figurative numbers are not confined to those associated with plane figures. One can have pyramidal numbers based on figures made up of layers of dots, for example a tetrahedron made up of layers of triangular arrays. Such a tetrahedron would have Tn dots in the first layer, Tn-1 dots in the next, and so on. If there were four such layers, the tetrahedral number it represents would be 1 + 3 + 6 + 10, or 20.

A general formula applicable to a pyramid whose base is an N-gon with Nk points in it is (k + 1)(2Nk + k)/6. In the example above N = 3 and k = 4. Using these values in the formula gives 5(20 + 4)/6 = 20.

Figure 5. Illustration by Hans & Cassidy. Courtesy of Gale Group.

Figure 6. Illustration by Hans & Cassidy. Courtesy of Gale Group.

Figure 7. Illustration by Hans & Cassidy. Courtesy of Gale Group.

Figure 8. Illustration by Hans & Cassidy. Courtesy of Gale Group.


If the base is a square with 5 points on a side, N = 4 and k = 5. Then the total number of points is 6(50 + 5)/6, or 55. To arrange cannon balls in a pyramidal stack with a square array of 25 balls for a base, one would need 55 balls.

Resources

Books

Gullberg, Jan, and Peter Hilton. Mathematics: From the Birth of Numbers. W.W. Norton & Company, 1997.

Rosen, Kenneth. Elementary Number Theory and Its Applications. 4th ed. Boston: Addison-Wesley, 2000.


J. Paul Moulton

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