Russell Schwartz, Howard Community College

Mentored by: Mike Long, Ed.D. & Loretta FitzGerald Tokoly, Ph.D.

Abstract

Recent findings were used in conjunction with the definitions of classic mathematical constants in order to derive Euler's constant, the Golden Ratio, and Pi from elements of Pascal's Triangle, a famous structure constructed from the binomial coefficients. Euler's constant was found by examining the products of the rows of Pascal's Triangle. A pattern involving e was found in the limit of these products. The Golden Ratio's relationship to the Fibonacci numbers was shown, and the presence ofthese numbers in Pascal's Triangle was demonstrated. A recent discovery by mathematician Jonas Castillo Toloza connecting the triangular numbers to Pi via an alternating series was used to connect Pi to Pascal's Triangle. These findings demonstrate the richness of Pascal's Triangle and expand its list of known properties.

Pascal's Triangle

Pascal's Triangle is one of the most interesting structures in mathematics. It is named after 17^th century French mathematician Blaise Pascal; however, it is far older. The first records of its construction date back to 11^th century China [1]. It is constructed as follows: in the topmost row there is a single entry, 1. Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right. Blank entries are treated as 0's.

Elements of Pascal's Triangle are indexed first by their row number, n, and then by their position within the row, k. This is denoted $P_{n} (k)$ . The first 7 lines of Pascal's Triangle are shown below.

The most famous property of Pascal's Triangle is its relation to the Binomial Coefficients:

$P_{n} (k) = (\binom{n}{k})$

where

$(\binom{n}{k}) : = \frac{n!}{k! (n - k)!}$

represents the $k^{t h}$ coefficient of the expansion of $(a + b)^{n}$ . This property is known as Pascal's Rule. In this paper, we will explore various properties of Pascal's Triangle and demonstrate how they give rise to some of the most famous mathematical constants.

Deriving $e$

Another widely known property of Pascal's Triangle is that the sums of the rows of Pascal's Triangle are equal to the powers of 2. The products of the rows are much less straight forward. The product of row will be denoted as follows.

$a_{n} : = \prod_{k = 0}^{n} P_{n} (k)$

Evaluating for the first few rows, the product appears to grow extremely quickly:

Examining the ratios of consecutive products, still no obvious pattern emerges. However, upon examining the ratio of ratios this sequence now appears to be approximately geometric, with a common ratio somewhere between 2 and 3:

Plotting the ratio of consecutive ratios, we can approximate the growth rate of this new series:

Finally, the ratios of the ratios of the products of consecutive rows monotonically approach a limit: $e$ (~2.71828). This numerical anlysis suggests the following.

$lim_{n \to \infty} \frac{a_{n + 2} / a_{n + 1}}{a_{n + 1} / a_{n}} = e$

This assertion can be shown analytically. First note, combining like terms:

$a_{n} = \prod_{k = 0}^{n} (\binom{n}{k}) = \frac{(n!)^{n + 1}}{(\prod_{k = 0}^{n} k!)^{2}}$

The limit can be rewritten using this fact.

In sum, e can be derived from Pascal's Triangle by first taking the product of each row, calculating the ratios of consecutive products, and determining the limit of the ratios of consecutive ratios.

Deriving $ϕ$

The Golden Ratio, denoted $ϕ$ , is defined as the positive solution to the quadratic equation $x^{2} - x - 1 = 0$ . Applying the quadratic formula yields

$ϕ = \frac{1 + \sqrt{5}}{2} \approx 1.61803$

The Golden Ratio appears in many places in mathematics and can even be found in nature. Notably, $ϕ$ shows up as the ratio between consecutive terms of the Fibonacci sequence. The Fibonacci sequence is defined as follows.

$F_{n} = F_{n - 2} + F_{n - 1}; F_{1} = 1, F_{2} = 1$

The ratios of these terms converge quickly. Evaluating the first 10 terms yields $ϕ$ to an accuracy of 3 decimal places:

This fact can be shown analytically. Assuming that the ratios converge to a finite non-zero value $r$ ,

The Fibonacci sequence can be easily found in Pascal's Triangle by examining the sums of "shallow diagonals." The sum of the shallow diagonal is equal to the $(n + 1)^{t h}$ Fibonacci number. The construction of these diagonals is shown below.

This property can be expressed by writing

$F_{n + 1} = \sum_{k = 0}^{⌊ n / 2 ⌋} P_{n - k} (k)$

We can now derive $ϕ$ directly from Pascal’s Triangle by finding the sums of the shallow diagonals, and determining the limit of the ratios of consecutive sums:

$ϕ = lim_{n \to \infty} \frac{F_{n + 1}}{F_{n}}$

Deriving $π$

Another famous feature of Pascal’s Triangle can be found by examining the third element in every row: the triangular numbers. The $n^{t h}$ triangular number is defined as the sum of the natural numbers from 1 to $n$ . It is easy to show why these number appear in Pascal’s Triangle:

$T_{n} : = \sum_{k = 1}^{n} k = \frac{n (n + 1)}{2} = (\binom{n + 1}{2}) = P_{n + 1} (2)$

The first 5 triangular numbers are circled below.

In 2007, mathematician Jonas Castillo Toloza discovered an explicit connection between the triangular numbers and π:

$\begin{aligned} π & = 2 + \frac{1}{1} + \frac{1}{3} - \frac{1}{6} - \frac{1}{10} + \frac{1}{15} + \frac{1}{21} - \frac{1}{28} - \frac{1}{26} + . . . \\ = 2 + \sum_{n = 2}^{\infty} \frac{(- 1)^{1 + ⌊ n / 2 ⌋}}{T_{n - 1}} \end{aligned}$

The proof of this fact is as follows. Being with Leibniz’ formula for π:

Applying $\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n (n + 1)}$ ,

Using this connection, π can be derived from Pascal's Triangle by adding 2 to the sum of the reciprocals of the third element of every row, where the sign before each term alternates with a period of 2.

Conclusion

The derivations described above are not the most practical ways of calculating e, $ϕ$ or π. Rather, they serve to provide insight on the connections between different fields of mathematics and to demonstrate how complicated irrational constants such as these can arise from something as simple as Pascal's Triangle.

Contact: russell.schwartz@howardcc.edu

NOTE: The method of finding e in Pascal's Triangle was also demonstrated by Harlan Brothers.

H. J. Brothers, Finding e in Pascal’s triangle, Mathematics Magazine, Vol. 85, No. 1, 2012; page 51.
H. J. Brothers, Pascal's triangle: The hidden stor-e, The Mathematical Gazette, Vol. 96, No. 535, 2012; pages 145-148.

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References

[1] Edwards, A. W. F. (2013), "The arithmetical triangle", in Wilson, Robin; Watkins, John J., Combinatorics: Ancient and Modern, Oxford University Press, pp. 166 - 180.

[2] Toloza, Jonas C, "Pi Y Los Numeros Triangulares", 2007.

[3] Weisstein, Eric W, "Pi Formulas," MathWorld - A Wolfram Web Resource.