412
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
1850 and was unknown until he introduced himself to G. H. Hardy, sending
him several dozen identities Ii La Euler of which half were not only new,
but very difficult to prove (Hardy had him come to Cambridge; he died of
tuberculosis in Madras a dozen years later). About twenty years ago the
Ramanujan conjecture (among other results) brought a Fields Medal, the
mathematical equivalent of a Nobel Prize, to Pierre Deligne, who proved
it, using all the machinery of geometry and algebraic topology erected after
1960 by Alexandre Grothendieck and his tribe: like Fermat's Last "Theorem"
proved by Andrew Wiles (Annals of Mathematics 1995), this is an example of
a seemingly totally "classical" problem, solved by "modern maths" thanks to
an interpretation of the coefficients r(n) of the infinite product ..1(z), which
carry much more information than their mere definition.
Those who believe that "the age of formulae" is over deceive themselves
greatly: the contemporary "formulae" certainly lie, in general, though not
always, at a mathematical level incomparably more elevated than those of
Euler; they involve functions, series, infinite products or integrals whose terms
reflect the "structure" of some arithmetic, algebraic or analytic theory or
other.
This said, there is much activity in other domains of mathematics, as important as this blend of Eulerian calculations, of elliptic or modular functions,
and of theory of numbers, or of algebraic varieties, which fascinate some and
repel others, despite, or because, of the antiquity of the tradition on which
they rest ...
To return to Euler, after examining the products (15) he passed on to the
more difficult identity that he wrote
(20.15) 1/(1 - aX)(1 - bx)(1 - ex) ... = 1 + Ax + Bx 2 + Cx 3 + ... ,
which amounts to forming the product of all the geometric progressions
L anx n , L bnx n , L cnx n , etc. The coefficient of xn on the right hand side
is then clearly the sum of the products
(20.16)
associated to each decomposition n = nl + n2 + n3 + ... of n into integers> 0;
these are not the same partitions as the preceding: the ni are not subjected
to the condition (3) since, to form a product of series, one chooses a term in
each sum at random and adds the products obtained. We can for example
choose for a, b, c, ... the reciprocals of the prime numbers, and put x = 1 in
the result; in this way we find the sum of all the reciprocals of products of
any number of any powers of the prime numbers 2,3,5,7,11, ... ; since any
integer > 1 can be written in a unique way in the form
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