The Pythagorean Theorem from a Combinatorial Identity

John Molokach came up with a demonstration of the Pythagorean theorem based on the Taylor series for sine and cosine functions at \(0\) (i.e., their Maclaurin series) and an identity for binomial coefficients. Here are the three points of departure:

\(\displaystyle \mbox{sin}\space x = \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\)

\(\displaystyle \mbox{cos}\space x = \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots\)

\(\displaystyle \sum_{k=0}{2n \choose 2k}=\sum_{k=0} {2n\choose 2k+1}\)

Since \(\displaystyle \sum_{k=0}{2n \choose k}=2^{2n}\), we also have

\(\displaystyle \sum_{k=0}{2n \choose 2k}= \begin{cases} 2^{2n-1} & \mbox{if}\space n\ge 1, \\ 1 & \mbox{if}\space n=0. \end{cases} \)

Both series are absolutely convergent and, therefore, could be safely squared:

\(\displaystyle \mbox{sin}^{2} x = -\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}\space\sum_{k=1}{2n\choose 2k-1}\)

\(\displaystyle \mbox{cos}^{2} x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\space\sum_{k=0}{2n\choose 2k}\)

Now, since all the terms containing \(x^{2n}\) in the sums cancel, except for the \(x^0\) term, adding the two equations simplifies to:

\(\displaystyle \mbox{sin}^{2} x + \mbox{cos}^{2} x = 1,\)

which is the Pythagorean identity.

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Copyright © 1996-2017 Alexander Bogomolny


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