The sine, rebuilt from its zeros

A polynomial is determined, up to a constant in front, by where it is zero. If $p(x)$ has degree three and vanishes at $-1$, $0$ and $2$, then $p(x) = c\,x(x + 1)(x - 2)$ for some number $c$, and fixing the value at one more point fixes $c$. The roots are not merely a property of the polynomial; they are a recipe for it.

The sine is not a polynomial, but it is a wave, and a wave crosses the axis at known places. Written in the form $\sin \pi x$, so that the period is two, it vanishes at every whole number $x = 0, \pm1, \pm2, \ldots$ and nowhere else. In 1734 Leonhard Euler did with those zeros what anybody would do with the roots of a polynomial: he wrote down one factor for each of them. Pairing the factors for $n$ and $-n$, and choosing the constant so that the slope at nought comes out right, he wrote

$$\sin \pi x = \pi x \left(1 - \frac{x^2}{1}\right)\left(1 - \frac{x^2}{4}\right)\left(1 - \frac{x^2}{9}\right)\left(1 - \frac{x^2}{16}\right)\cdots$$

— an infinite polynomial, assembled from its infinitely many roots. The formula is correct. Euler’s reason for believing it was not a proof, and the gap between the reason and the proof is where most of the interest lies.

What the partial products do #

The figure tests the formula the only way a finite picture can: by stopping it. The product of the first $N$ pairs of factors is an honest polynomial of degree $2N + 1$, with exactly the zeros $0, \pm1, \ldots, \pm N$. With one pair, it is $\pi x(1 - x^2)$, a cubic that rises too high between $0$ and $1$ — its peak is at $1.18$ — and plunges past $1$ as cubics do. With three pairs the peak between $0$ and $1$ is $1.07$, and the curve stays close to the wave across the middle three humps. With ten it is $1.024$, and the agreement extends across the whole picture except at its edges.

Each partial product behaves well exactly where its zeros are and badly beyond them, and that is the right failure. A polynomial must eventually grow without bound; the sine never does. The products are not converging to the sine by getting better at the edges. They are converging by pushing the edges outward: every new pair of factors adds two more zeros and lets the polynomial follow the wave for one more hump on each side before it escapes.

At $x = \tfrac12$, where the sine is exactly one, the three products in the figure give $1.1781$, $1.0738$ and $1.0241$. Twenty thousand pairs give $1.00001$. The convergence is real and slow, and its slowness is itself worth noticing: the error after $N$ factors is roughly proportional to $1/N$, so each further correct digit costs ten times as many factors. That is a sign that the product, though correct, is not the way to compute sines. It is the way to know things about them.

Two functions with the same zeros #

Euler’s reasoning has a hole, and it can be drawn.

For a polynomial, the zeros determine the function up to a constant because a polynomial with no zeros at all is a constant. For functions in general that is false. The exponential $e^{x}$ is never zero, and neither is $e^{\text{anything}}$; multiply the sine by any of them and the zeros stay where they were. So “a function vanishing exactly at the whole numbers” describes not one function but a vast family — $\sin\pi x$ times $e^{g(x)}$ for any $g$ at all — and nothing in the list of zeros picks the sine out of it.

What rescues the formula was found more than a century after Euler. Karl Weierstrass showed in 1876 how to build a function with any prescribed zeros, and Jacques Hadamard showed in 1893 how much freedom remains: for a function that grows no faster than an exponential in every complex direction, the leftover factor $e^{g(x)}$ can only be $e^{ax + b}$, a single exponential with two constants. The sine grows at exactly that rate — $|\sin(\pi iy)| = \sinh \pi y$ — so it equals $e^{ax + b}$ times Euler’s product. Then two facts about the sine fix the constants. It is odd, $\sin(-\pi x) = -\sin \pi x$, and the product is odd too, so $e^{ax}$ must be even, which forces $a = 0$. And its slope at nought is $\pi$, which forces $e^b = 1$ once the factor $\pi x$ is written in front.

So the formula is a theorem with three ingredients — the zeros, the growth rate and the symmetry — and the figure shows why the second and third cannot be dropped. The function $e^{0.4x}\sin\pi x$ has the same zeros and the same growth rate. Only the symmetry tells it apart.

A polygon whose chords multiply to its number of sides #

There is a second road to the product, older and more elementary than Hadamard’s, and it runs through a fact about regular polygons that is surprising on its own.

Put a regular $n$-gon on the circle of radius one with a corner at the point $1$ of the complex plane, so its corners are the $n$-th roots of unity $1, \omega, \omega^2, \ldots, \omega^{n-1}$, where multiplying by $\omega$ is turning by a fraction $1/n$ of a revolution. The polynomial $z^n - 1$ vanishes at every corner, so it factors into one bracket per corner:

$$z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1}).$$

Divide both sides by $z - 1$. The left becomes $1 + z + z^2 + \cdots + z^{n-1}$, and at $z = 1$ it is $n$. The right becomes the product of the distances from the corner at $1$ to each of the other corners. So the chords from one corner of a regular polygon multiply to the number of its sides — seven for the heptagon in the figure, a hundred for the hundred-gon, whatever the polygon. Every chord is shorter than two and many are much shorter, and the product lands on a whole number every time.

That identity is the finite form of Euler’s product. Run the same factorisation with $z$ on the circle at a general angle rather than at a corner, and it becomes a formula for $\sin n\theta$ as a product of $n$ sines spaced $\pi/n$ apart. Now let $n$ grow while the angle shrinks in proportion, $\theta = \pi x/n$. Each factor $\sin(\pi x/n + k\pi/n)$, divided by its value at $x = 0$, tends to $1 - x/k$ for each fixed $k$, and the pairs combine into $1 - x^2/k^2$. In the limit the polygon formula becomes Euler’s. This is the proof found in textbooks of the nineteenth century, and it needs care only at the step where infinitely many factors are allowed to tend to their limits at once — a care Euler did not take and his answer did not need.

The connection is the surprising thing on this page. A question about one curve — what is the sine? — is answered by a family of finite figures, the regular polygons, each contributing an exact identity about chords. Euler’s infinite polynomial is what the polygons say in the limit, and the reason its zeros are the whole numbers is that the corners of a polygon with ever more sides crowd, near any one corner, into an evenly spaced row.

Multiplying out: the sum of the reciprocal squares #

The product earns its keep when it is multiplied out. Expand $(1 - x^2)(1 - x^2/4)(1 - x^2/9)\cdots$ and collect powers of $x$. The constant term is one. The coefficient of $x^2$ collects one term from each bracket — the $-x^2/n^2$ from bracket $n$ times the ones from all the others — so it is

$$-\left(1 + \frac14 + \frac19 + \frac1{16} + \cdots\right).$$

But $\sin\pi x / \pi x$ also has a Taylor series, read off from its slopes at nought: $1 - \pi^2x^2/6 + \pi^4x^4/120 - \cdots$. Two expansions of one function must agree term by term. So

$$1 + \frac14 + \frac19 + \frac1{16} + \cdots = \frac{\pi^2}{6}.$$

That was the Basel problem, posed by Pietro Mengoli in 1650 and attacked without success by the Bernoullis, and this was Euler’s solution of 1734, the one that made his reputation. The answer contains $\pi$, and it should be startling that it does: the question is about the squares of whole numbers, and there is no circle anywhere in it. The circle comes in through the zeros of the sine, which sit at the whole numbers because the sine is the shadow of a point walking round a circle.

The next coefficients give more. The coefficient of $x^4$ in the product is the sum of $1/(m^2n^2)$ over all pairs $m < n$, and the Taylor series says it is $\pi^4/120$. That is not yet the sum of the reciprocal fourth powers, but the two are linked by an identity that holds for any list of numbers: the square of their sum is the sum of their squares plus twice the sum of their pairwise products. So the reciprocal fourth powers add to $(\pi^2/6)^2 - 2\pi^4/120 = \pi^4/90$. One more coefficient and one more such identity — these are Newton’s identities, which convert the coefficients of a polynomial into the power sums of its roots — and the sixth powers come out as $\pi^6/945$. Every even power follows the same way, each a rational number times a power of $\pi$, with the rational numbers given by the Bernoulli numbers.

The figure’s table checks the middle step numerically rather than by algebra: it multiplies out a million factors, corrects for the tails, and finds the product’s coefficients matching the Taylor coefficients to eleven decimal places. That is how a reader in 1734 could have checked Euler, and it is roughly how Euler checked himself.

Taking logarithms: a row of charges #

A product becomes a sum when its logarithm is taken, and the sum that comes out of Euler’s product has a physical reading that the product hides.

Take the logarithm of both sides and differentiate. The left side, $\log \sin \pi x$, differentiates to $\pi \cot \pi x$. On the right, each factor contributes its own logarithmic derivative, and the factor $1 - x^2/n^2$, written as $(1 - x/n)(1 + x/n)$, contributes $1/(x - n) + 1/(x + n)$. So

$$\pi \cot \pi x = \frac1x + \sum_{n=1}^{\infty}\left(\frac{1}{x - n} + \frac{1}{x + n}\right),$$

a sum of one simple term for each zero of the sine. That is the formula for the electric field, along a line, of equal point charges placed at every whole number on it — in a one-dimensional world where a charge’s field falls off as one over the distance. Each charge pushes outward with a strength inversely proportional to how far away it is, the pushes from left and right nearly cancel, and what survives is the cotangent: zero exactly halfway between two charges, where the forces balance, and infinite at each charge.

The reading explains something the product does not make obvious. The field near any one charge is dominated by that charge, $1/(x - n)$, and the rest of the row contributes a smooth correction; the correction at the charge’s own position is zero by symmetry, because the charges on either side are equally many and equally far. That symmetry is the oddness that pinned down the product’s constants, now appearing as the balance of a row of forces.

And it gives the even sums again, in one more way. Expand both sides in powers of $x$ near nought: the left side’s coefficients are known from the cotangent’s Taylor series, and on the right the coefficient of $x^{2k-1}$ is $-2$ times the sum of $1/n^{2k}$. Matching them hands over $\pi^2/6$, $\pi^4/90$ and every other even sum at once, with the Bernoulli numbers appearing as the cotangent’s coefficients. It is the same information as the multiplied-out product, rearranged so that each power stands alone instead of having to be untangled by Newton’s identities.

How slowly the squares get there #

The sum itself is a poor way to find $\pi^2/6$, and the reason is the same slowness the partial products showed.

After sixty terms the dots are still visibly below the line, and the gap shrinks like $1/N$: the tail $\sum_{n > N} 1/n^2$ is squeezed between $1/(N + 1)$ and $1/N$, which the figure checks at three points. Adding the estimate $1/N$ back in — the open circles — lands within a ten-thousandth of the target after sixty terms, and a million terms with the same correction agree with $\pi^2/6$ to eleven places.

Euler knew all this before he knew the answer. In 1731, three years before the product, he had computed the sum to six decimal places, $1.644934$, by a refinement of exactly that tail correction — the method now called Euler–Maclaurin summation — and it was presumably the six digits that let him recognise $\pi^2/6$ when the product produced it. The numerical evidence came first; the product explained it. That order is common in this subject, and the even values were later found a second way, through the square wave’s Fourier series, which explains them again without any product at all.

Wallis, a generation earlier #

Euler’s product contains a formula that predates it by eighty years. Set $x = \tfrac12$. The left side is $\sin(\pi/2) = 1$, and the right is $\tfrac{\pi}{2}$ times $\prod (1 - 1/4n^2)$. Rearranged,

$$\frac{\pi}{2} = \frac{2\cdot2}{1\cdot3}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{6\cdot6}{5\cdot7}\cdots$$

John Wallis published that product in 1656, found by an ingenious and quite unrigorous interpolation between integrals he could compute, and it was one of the first infinite products in mathematics, after François Viète’s product of nested square roots of 1593. In Euler’s product it is one point of a curve: the value of the sine at the top of its first hump. The partial products climb towards $\pi/2$ from below, each factor $(2n)^2/((2n - 1)(2n + 1))$ slightly bigger than one, and the gap after $N$ factors is close to $\pi/8N$, which the figure measures at forty. Other values of $x$ give other products, and $x = \tfrac14$, $\tfrac16$, $\tfrac13$ each turn into a product for some combination of $\pi$ and square roots, every one of them a single point on the curve the product draws.

The product also connects to the factorials. The gamma function, the curve through the factorials, has a product of the same kind with factors $1 + x/n$ rather than $1 - x^2/n^2$, and pairing the gamma function at $x$ with the gamma function at $1 - x$ collapses the two products into Euler’s: $\Gamma(x)\Gamma(1 - x) = \pi/\sin \pi x$. At $x = \tfrac12$ that is $\Gamma(\tfrac12)^2 = \pi$, the fact behind the area under a bell curve.

Finitely many factors standing in for infinitely many #

Every figure multiplies finitely many factors. The product is an infinite one, and whether an infinite product converges, and to what, is a question about all of its factors at once. Here it converges because the sum of $x^2/n^2$ converges, which makes the factors approach one fast enough; the figures show partial products approaching the sine and cannot show that they do so everywhere, or uniformly on any interval. That is a theorem about the tails, and the pictures stop where the tails begin.

The proof needs complex numbers, and the figures are drawn on the real line. Hadamard’s argument, which closes the hole the two-functions figure exhibits, rests on how fast the sine grows in the imaginary direction, where it behaves like an exponential. On the real line the sine stays between $-1$ and $1$, and nothing drawn here shows the growth that the theorem depends on. The figure eight’s sine, whose zeros fill a square grid rather than a line, needs a product of a more elaborate kind for exactly that reason.

The polygon limit is sketched, not carried out. The passage from the finite chord identity to Euler’s product requires showing that the factors with large $k$, which the sketch treats one at a time, do not together contribute anything in the limit. That step is where Euler’s own version was incomplete and where the textbook version spends most of its effort.

Still open: the odd powers #

The product gives the even powers and is silent about the odd ones. The reciprocal cubes, $1 + 1/8 + 1/27 + \cdots = 1.2020569\ldots$, appear nowhere in the coefficients of $\sin \pi x$, because the product’s brackets contain only $x^2/n^2$ and every coefficient is built from even powers of $1/n$. Euler tried for decades to find a closed form for the reciprocal cubes and failed.

No closed form is known, and none is expected. Roger Apéry proved in 1978 that the sum of the reciprocal cubes is at least irrational, by a race between two sequences of fractions that has never been made to work for any other odd power. For the fifth powers, irrationality is unknown. Wadim Zudilin proved in 2001 that at least one of the sums of reciprocal fifth, seventh, ninth and eleventh powers is irrational, without saying which; Tanguy Rivoal had shown the year before that infinitely many of the odd sums are irrational. That the even sums are rational multiples of powers of $\pi$ is a consequence of one wave’s zeros, and nothing comparable is known to organise the odd ones.