MathLabs

Worked solution: Euler's proof via the sine product formula (1734–1735)

Step 3 of 6: A bold analogy with polynomials
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In plain words

A finite polynomial is completely rebuilt once you know all its roots: for instance a quadratic with roots 22 and 33 that equals 66 at x=0x=0 must be 6(1−x/2)(1−x/3)6(1-x/2)(1-x/3). Euler simply refused to believe that an infinite list of roots should behave any differently, and wrote sin⁡x/x\sin x/x as an infinite product of one factor per pair of roots ±nπ\pm n\pi.

sin⁡xx  =?  ∏n=1∞(1−xnπ)(1+xnπ)=∏n=1∞(1−x2n2π2)\frac{\sin x}{x}\;\overset{?}{=}\;\prod_{n=1}^{\infty}\left(1-\frac{x}{n\pi}\right)\left(1+\frac{x}{n\pi}\right)=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)
Domain-coloring plot of sin⁡z\sin z: the points where all the hues of the color wheel meet are exactly its zeros at 0,±π,±2π,…0,\pm\pi,\pm2\pi,\dots
Domain-coloring visualisation of the complex function sin z over a region of the complex plane. Hue encodes the argument of sin z; the points on the real axis at 0, ±π, ±2π, … where every hue converges are the zeros Euler used to build his product.
Detailed analysis

A polynomial with roots r1,…,rkr_1,\dots,r_k equal to 11 at 00 factors as ∏(1−x/ri)\prod(1-x/r_i). Euler treated sin⁡x/x\sin x/x as an "infinite-degree polynomial" with roots ±π,±2π,…\pm\pi,\pm2\pi,\dots and value 11 at 00, and wrote down the same kind of factorisation — pairing each +nπ+n\pi with −nπ-n\pi into one real quadratic factor.

Terms in this step
Infinite product
A value defined as the limit, as N→∞N\to\infty, of the partial products ∏n=1N(⋯ )\prod_{n=1}^{N}(\cdots) of infinitely many factors.
Knowledge used in this step
Functions of a complex variable
Common mistake. This step is the crux of the whole argument and, in 1735, was not justified: nothing yet guaranteed that an "infinite polynomial" is determined by its zeros the way a finite polynomial is, or that no extra factor (like ecxe^{cx}) could be hiding in front of the product.