From: owner-historia-matematica@chasque.apc.org on behalf of Roger Cooke
[ckrglj@adelphia.net]
Sent: 26 April 2004 13:12
To: historia-matematica@chasque.apc.org
Subject: Re: [HM] Leibniz on Pi



Leibniz must have had some idea that not all irrational numbers are
algebraic. In the preface to his "Arithmetical Quadrature of the
Circle" (which I think was written in 1678 or 1679), he wrote,
I corrected my math in a previous note, but---not to be *too* compulsive
about this---I realize that my Latin also needs some fine tuning.
Is there a competent Latinist among us who can get this exactly right?
Here's my second effort:

"Perfecta autem quadratura illa erit, quae simul sit Analytica et linearis,
sive quae lineis aequalibus, ad certarum dimensionum aequationes
revocabilibus construatur.  Hanc impossibilem esse asseruit ingeniosissimus
Gregorius in libro de Vera Circuli quadratura, sed demonstrationem tunc
quidem, ni fallor, non absolvit.  Ego nondum video, quid impediat
circumferentiam ipsam aut aliquam determinatam eius portionum mensurari, et
cuiusdam arcus rationem ad suum sinum, certae dimensionis aequatione
exprimi.  Sed relationem arcus ad sinum in universum aequatione certae
dimensionis explicari impossibile est, quemadmodem in ipso opusculo
demonstrabo."

"A complete quadrature would be one that is simultaneously analytic and
linear, that is, it would be constructed by equivalent curves reducible to
equations of certain [finite] degrees.  The brilliant Gregory, in his book
"On the true quadrature of the circle" has claimed that this is impossible,
but, unless I am mistaken, has given no proof.  I still do not see what
prevents the circumference itself from being measured, or some definite part
of it,
an arc having a ratio to its sine [half-chord] that can be expressed by an
equation of finite degree.  But to express the ratio of
the arc to the sine in general by an equation of finite degree is
impossible, as I shall prove in this little work."

That last sentence says that the arcsine *function* is not algebraic.  The
sentence before indicates that he isn't certain whether the same is true of
the number pi, which is (for example) six times its value at the point 1/2.
Or he might be thinking that since pi satisfies the equation of infinite
degree

0 = sin(pi) = 1 - pi + pi^3/3! - pi^5/5! + ,

that is its "minimal polynomial" and it can't satisfy a polynomial of finite
degree.

Roger Cooke