Patterns and the Danger of Assuming (Tuklas Vol. 14, No. 8 - Feb. 10, 2013)

PATTERNS AND THE DANGER OF ASSUMING

Human beings have a tendency to see patterns in everything. It is a skill we have unconsciously mastered to survive. Every observation we make, whether consciously or unconsciously, relies on perceiving whether a pattern is developing.

This is not necessarily a bad thing. Skills would not have been developed if they were not essential to survival. If you know that a particular route is usually clogged by traffic after class and you need to go home quickly, then that particular route is out of the question. If one knows that a particular teacher does not regularly check homework and there is another project due for tomorrow, then one might choose to focus on finishing that project first. Of course, if the teacher does check for homework the next day, then we might not have anything to show for it. The pattern fails.

Let’s see how our fondness for patterns might be a little misleading. Consider the sequence of numbers \[ 0 \,\, 3 \,\, 6 \,\, 12 \,\, 24 \,\, 48 \,\, 96 \,\, 192 \,\, 384 \]which we construct by doubling the number (starting with 3). Let’s now add 4 to each number and divide by 10. This gives us \[ 0.4 \,\, 0.7 \,\, 1.0 \,\, 1.6 \,\, 2.8 \,\, 5.2 \,\, 10.0 \,\, 19.6 \,\, 38.8 \]What is so special about this sequence? In the 18th century, a couple of astronomers, notably Johann Daniel Titius and Johann Elert Bode, noticed a quirk in the distances between planets in the Solar System. Specifically, if one measures the mean distances (measured in astronomical units, or A.U.) of the planets from the Sun, it seems that one planet is twice as far from the Sun as the previous planet. This sequence of measurements happens to mirror the sequence we had just constructed.

When the Titius-Bode law (as it would later be known) was proposed, the known planets in the Solar System were Mercury, Venus, Earth (of course), Mars, Jupiter and Saturn. Astronomers found out that the sequence predicted the distances of each planet from the Sun accurately, except that there was a gap between Mars and Jupiter.

Planet	Actual distance (A.U.)	T-B Law
Mercury	0.39	0.4
Venus	0.72	0.7
Earth	1.00	1.0
Mars	1.52	1.6
		2.8
Jupiter	5.20	5.2
Saturn	9.54	10.0

Interesting, but not really something that astronomers considered important until the discovery of Uranus in the tail-end of the century, whose mean distance from the Sun eerily fitted the next number in the sequence. Spurred by this, astronomers discovered the heavenly body Ceres (the largest object on the asteroid belt) which fitted the 5th number in the sequence. Note that this was the blank entry in the previous table.

It seemed like a triumphant example of universal order, discovered by mathematics. The discovery of the planet Neptune in the mid-19th century, however, showed that the law didn’t work. Furthermore, Ceres was not really considered a planet after the discovery of more asteroids in the asteroid belt. Some scientists took it as an example of fallacious reasoning and hasty generalization. Others sought to show that the law just needed some adjustment and that the planets really behaved in some predictable manner. The most important effect of the Titius-Bode rule (as some now call it) though is that it drove astronomers to discover why a pattern could (or couldn’t) develop, and whether there were objects that would fit this pattern. The discovery of Ceres was prompted by a guess that the rule could work.

Sometimes, we tend to look for things that aren’t there. At other times, we see things but misinterpret them to mean something else. The Titius-Bode rule is a reminder that the appearance of a pattern is not enough. To discover why something works, we must delve deeper into the underlying factors that cause a particular phenomenon to exist.

ABOUT THE AUTHOR:
Jesus Lemuel Martin, Jr. obtained his B.S. in Applied Mathematics major in Computational Science at the Ateneo de Manila University in 2010 and is currently taking his MS in Mathematics at the same university.

REFERENCES:

[1] Haynes, M. "Bode's Law." http://www.astro.cornell.edu/academics/courses/astro201/bodes_law.htm. Online, accessed February 8, 2013.

OLYMPIAD CORNER

from the Canadian Mathematical Olympiad, 2009

Problem: Two circles of different radii are cut out of cardboard. Each circle is subdivided into 200 equal sectors. On each circle 100 sectors are painted white and the other 100 are painted black. The smaller circle is then placed on top of the larger circle, so that their centers coincide. Show that one can rotate the small circle so that the sectors on the two circles line up and at least 100 sectors on the small circle lie over sectors of the same color on the big circle. 

Solution: 

Let \(x_i, i=0,1,...,199\) be defined as follows: \(x_i=1\) when the \((i+1)\)-th sector of the larger circle is black, and \(x_i=-1\) when the \((i+1)\)-th sector of the larger circle is white (counterclockwise). The variables \(y_i, i=0,1,...,199\) are similarly defined:, i.e. \(y_i=1\) when the \((i+1)\)-th sector of the smaller circle is black, and \(y_i=-1\) when the \((i+1)\)-th sector of the smaller circle is white.

To show the required condition, we need to prove that there exists some \(j=0,1,...,199\) such that \(S_j=\sum_{i=0}^{199}x_iy_{i+j}\) is nonnegative. We note that the subscript \(i+j\) is taken modulo 200, and \(j\) indicates the number of times the smaller circle rotated by 1.8 degrees (counterclockwise) from its original position. 

Note that \(y_0+\cdots +y_{199}=0\) as there's an equal number of black and white sectors. We then take the sum \[ \begin{equation*} S_0+S_1+\cdots +S_{100}=\sum_{i=0}^{199}x_i\left(y_0+y_1+\cdots+y_{199}\right)=0 \end{equation*} \]Hence, there exists a \(j=0,1,2,...,199\) such that \(S_j\geq 0\).

SOLUTIONS

(for January 19, 2013)

19. Find all pairs \((x,y)\) of two positive integers such that \(N = 11x + 99y\) is a perfect square number less than or equal to \(1199\) and \(x+y\) is also a perfect square. (Modified problem from the China Mathematical Competition for Secondary Schools, 2004)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy]; partial credit for Jecel Manabat [Valenzuela City Sci HS])
    
    
    SOLUTION: 
    Since \(N = 11x + 99y = 11(x + 9y)\) we can conclude that \(x + 9y = 11m^2\) for some positive integer \(m\), so \[ N = (11m)^2 \leq 1199 \Leftrightarrow m^2 \leq \frac{1199}{121} < 10. \] Hence \(m^2\) is equal to \(1\), \(4\) or \(9\). We now have the three cases \(x+9y = 11\), \(x+9y = 44\) and  \(x+9y = 99\). Since we want \(x+y\) to be a perfect square, we get the ordered pair \((35,1)\).
20. Show that for all positive integers \(x, y, z\) \[ \left(\frac{x+y}{x+y+z}\right)^{\frac{1}{2}} + \left(\frac{x+z}{x+y+z}\right)^{\frac{1}{2}} +\left(\frac{y+z}{x+y+z}\right)^{\frac{1}{2}} \leq 6^{\frac{1}{2}}. \] (Taken from The Cauchy-Schwartz Master Class by J. Michael Steele)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College], Jecel Manabat [Valenzuela City Sci HS], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    Without loss of generality, we can assume that \(x \leq y \leq z\). Note that \(x+y \leq 2z\) and \(x+y+z \leq 3z\) give us \[ \frac{x+y}{x+y+z} \leq \frac{2}{3}, \] and by symmetry we have \[ \frac{x+z}{x+y+z} \leq \frac{2}{3} \text{ and } \frac{y+z}{x+y+z} \leq \frac{2}{3}. \] Taking the sum of the roots, we now have \[ \left(\frac{x+y}{x+y+z}\right)^{\frac{1}{2}} + \left(\frac{x+z}{x+y+z}\right)^{\frac{1}{2}}  +\left(\frac{y+z}{x+y+z}\right)^{\frac{1}{2}} \leq 3\sqrt{\frac{2}{3}} = 6^{\frac{1}{2}}. \]
21. For a fixed positive integer \(n\), find the minimum value of the sum \[ x_1 + \frac{{x_2}^2}{2} + \frac{{x_3}^3}{3} + \ldots + \frac{{x_n}^n}{n}, \]given that \(x_1, x_2, \ldots, x_n\) are positive numbers satisfying the property that the sum of their reciprocals is \(n\). (Polish Mathematical Olympiad, 1995)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    By the AM-GM inequality, we get \(kx_k \leq {x_k}^k + (k-1)\) for all \(k\). Thus, we have \[ \begin{align*} x_1 + \frac{{x_2}^2}{2} + \frac{{x_3}^3}{3} + \ldots + \frac{{x_n}^n}{n} &\geq x_1 + x+2 - \frac{1}{2} + \ldots + x_n - \frac{n-1}{n} \\ &= x_1 + x_2 + \ldots + x_n -n + M, \end{align*} \] where \(M = 1 + \frac{1}{2} + \ldots + \frac{1}{n}\). On the other hand, by the AM-HM inequality we have \[\frac{x_1 + x_2 + \ldots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \ldots + \frac{1}{x_n}} = 1. \] Thus the previous expression is at least \(n - n + M = M\). This implies that \(M\) is the desired minimum value, and it is obtained by setting \(x_1 = x_2 = \ldots = x_n = 1\).