How did Leibniz prove that $\sin (x)$ is not an algebraic function of $x$?

Question

In the Wikipedia article about transcendental numbers we can read the following:

The name "transcendental" comes from Leibniz in his 1682 paper where he proved that sin(x) is not an algebraic function of x.

I would like to know can someone reproduce here the proof of Leibniz of the non-algebraicity of $\sin$ or point me to the place on the internet where we have this proof.

I do not know German or Latin so if someone has the link where this is proved it would be OK if it is written in English.

Although I am an amateur it seems to me that in the time of Leibniz the techniques for proving the non-algebraicity of functions were not developed enough so it would be nice to see how he proved that for $\sin$.

Answer 1

I don't know how Leibniz did it, but it seems to me that the main ingredients of this straightforward proof ought to have been available to him (if phrased with less rigor than we do today):

Definition. A function $f$ of one variable is algebraic if there is a polynomial $p(x,y)$ such that $p(x,f(x))=0$ for all $x$.

Claim. For every nonzero polynomial $p$, there is an $x_0$ such that $p(x_0,\sin x_0)\ne 0$.

Proof by long induction on the highest power of $y$ that appears in $p(x,y)$.

First suppose that $p$ contains at least one nonzero term that doesn't contain $y$. Then $p(x,\sin x)=p(x,0)$ whenever $x$ is a multiple of $\pi$. But the right-hand side is a nonzero polynomial in one variable, and can therefore have at most finitely many zeroes. So there exist some $k\pi$ where $p(k\pi,0)$ -- and therefore also $p(k\pi,\sin k\pi)$ -- is nonzero.

On the other hand, if every term of $p$ contains $y$, then $p(x,y)=y q(x,y)$ for some polynomial $q$. The induction hypothesis applies to $q$, so let $x_0$ be a value such that $q(x_0,\sin x_0)\ne 0$. If $x_0$ happens to be a multiple of $\pi$, then because $q(x,\sin x)$ is clearly continuous, there will be another suitable $x_0$ nearby. So without loss of generality $x_0$ is not a multiple of $\pi$, and therefore $q(x_0,\sin x_0)\ne 0$ implies that $$ p(x_0,\sin x_0) = \sin x_0 \cdot q(x_0,\sin x_0)\ne 0 $$ as required.

(Note that this proof depends only on the fact that the sine function is continuous and has infinitely many isolated zeroes).

Answer 2

Leibniz's paper with the term "transcendental" is : DE VERA PROPORTIONE CIRCULI AD QUADRATUM CIRCUMSCRIPTUM IN NUMERIS RATIONALIBUS EXPRESSA, Act.Erudit.Lips.1682.

Reprinted into:

• Gottfried Wilhelm von Leibniz, Leibnizens mathematische Schriften, herausgegeben von C.I.Gerhardt (1858), page 118-on; see page 120:

transcendens inter alia habetur per aequationes gradus indefiniti.

For the proof, see: PRAEFATIO OPUSCULI DE QUADRATURA CIRCULI ARITHMETICA, page 93-on; see page 98.

Answer 3

The claim

The name "transcendental" comes from Leibniz in his 1682 paper where he proved that $\sin(x)$ is not an algebraic function of $x$

made in Wikipedia is at best a half-truth. As Mauro ALLEGRANZA explained, the term "transcendental" occurs indeed in Leibniz's paper

DE VERA PROPORTIONE CIRCULI AD QUADRATUM CIRCUMSCRIPTUM IN NUMERIS RATIONALIBUS EXPRESSA, Acta Eruditorum (Leipzig 1682).

The English translation of the title is "On the true proportion of the circle to the circumscribed square expressed in rational numbers."

Before I come to Leibniz's interpretation of "transcendental", let me mention that Leibniz did not consider the function $\sin$, in particular he did not prove that $\sin(x)$ is not an algebraic function of $x$.

So what is he doing in his paper? He writes

Constructio Geometrica accurata haberi potest, qua non tantum circulum integrum, sed et quemlibet sectorem sive arcum metiri liceat motu exacto atque ordinato, sed qui curvis transcendentibus competat, quae per errorem alioqui Mechanicae censentur, cum tamen aeque sint Geometricae ac vulgares, licet Algebraicae non sint nec ad aequationes Algebraicas seu certi gradus reduci queant; suas enim proprias, etsi non-algebraicas, tamen analyticas habent. Sed ista hic pro dignitate exponi non possunt. Quadratura Analytica seu quae per calculum accuratum fit, iterum in tres potest dispesci: in Analyticam transcendentem, Algebraicam et Arithmeticam. Analytica transcendens inter alia habetur per aequationes gradus indefiniti, hactenus a nemine consideratas, ut si sit $x^x + x$ aeq. $30$, et quaeratur $x$, reperietur esse $3$, quia $3^3+3$ est $27 +3$ sive $30$: quales aequationes pro circulo dabimus suo loco. Algebraica expressio fit per numeros, licet irrationales, vulgares seu per radices aequationum communium : quae quidem pro quadratura generali circuli sectorisque impossibilis est. Superest Quadratura Arithmetica, quae saltem per series fit, exhibendo valorem circuli exactum progressione terminorum, inprimis rationalium, qualem hoc loco proponam.

My Latin is a bit rusty, but here is my English translation:

An exact geometrical construction may be obtained, by which not only the entire circle, but also any sector or arc may be measured with an exact and orderly motion, but which is capable of transcending curves, which erroneously are otherwise considered mechanical, when they are equally geometrical and common, though they should not be algebraic, nor should they be reduced to algebraic equations or certain degrees; for they have their own properties, though non-algebraic, yet analytic. But these cannot be set forth here in a dignified way. Analytical quadrature, or that which is done by accurate calculation, can again be divided into three: transcending analytical, algebraic, and arithmetic. Transcending analytical is obtained, among other things, by equations of indefinite degree, hitherto considered by no one, as if $x^x + x$ eq. $30$, and if $x$ is sought, it will be found to be $3$, because $3^3+3$ is $27 +3$ or $30$: such kind of equations we will give for the circle at a suitable place. Algebraic expression is made by means of numbers, allowed to be irrational, ordinary, or by the roots of common equations: which indeed is impossible for the general quadrature of circles and sectors. What remains is arithmetic quadrature, which is done at least through a series, showing the exact value of the circle by the progression of terms, especially rational ones, such as I propose in this place.

Then he introduces his famous Leibniz formula for $\pi$ $$\frac 1 1 - \frac 1 3 +\frac 1 5 - \frac 1 7 + \frac 1 9 - \frac 1 {11} + \ldots$$ which sums up to $S = \frac{\pi}{4}$. Since he considers a circle with circumscribed square of side length $1$, he obtains $S$ as the area of the circle.

The above text also shows that he did not have a precise conecpt of transcendental number (let alone of a transcendental function). But it seems he had an intuition that there exist numbers which do not have an algebraic representation by means of numbers, allowed to be irrational, ordinary, or by the roots of common equations: which indeed is impossible for the general quadrature of circles and sectors.

For more information about Leibniz's paper see here. Also have a look here.

Answer 4

If $\frac{e^{i x} - e^{-i x}}{2 i}$ were an algebraic function of $x$, then $e^x$ would too ( $x\mapsto ix$). However, $e^x$ grows too fast for that ( to get easily some bounds for roots of polynomials, use a companion matrix).

Answer 5

(1) $\sin x$ is not an algebraic function of $x$

(2) $\sin x$ is a transcendental function

Statement $1$ is false. Statement $2$ is true

Counterexample to Statement $1$: $x = 0$

$\sin 0$ is an algebraic function of $0$

$\sin 0 = 0 = 2(0)^2 + 1(0) + 0$

Right side of equation is an algebraic function, a single-variable quadratic polynomial with coefficients $2$, $1$, $0$

Zero is a member of the exceptional set of the transcendental function $\sin x$

Leibniz proved Statement $2$. He did not prove Statement $1$