# 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.  