## e is transcendental

In 1873 Charles Hermite showed that e is transcendental.  Liouville had earlier shown how to construct such numbers, but Hermite’s was the first proof that an ‘interesting’ mathematical constant was transcendental.

Hermite’s proof is long and complex.  It has since been simplified considerably and in particular a short and elegant proof appears in Baker’s book ‘Transcendental Number Theory’.  However Baker himself says that the motivation of the proof has been obscured by the process of simplification, and it is necessary to go back to Hermite’s original paper in order to understand it.

I think this is not entirely so but nevertheless in this post  I give in outline a version of Baker’s proof that I preface with some references to ideas in Hermite’s paper and I hope in so doing I make the motivation clearer.  A more detailed version of this outline is in a pdf here.

As discussed in an earlier post, by repeated integration by parts Hermite obtained an expression of the form

$i^{2m+1}\int_0^1e^{it}F_m(t)dt=k_i(m)e^i-h_i(m)$

where Fm(t)=tm(1-t)m , m and i are positive integers and ki, hi are polynomials with integer coefficients.  The expression on the left of the equation is not zero but can be made arbitrarily small in absolute value by taking m sufficiently large, thereby proving that ei is irrational for any integer i.

This suggests that one could obtain in a similar manner a set of equations of the form

$k(m)e^i-h_i(m)=I_i(m)$   ………………………………………….   (1)

where k(m) is a non-zero integer not dependent on i and the Ii(m) either decrease in absolute value or grow only slowly with increasing m.

Now let q0, q1, … , qn be any set of integers (provided that n > 0 and q0, qn ≠0).  If we multiply (1) by qi and sum over i then we have

$\sum_{i=0}^nq_iI_i(m)=-\sum_{i-0}^nq_ih_i(m)+k(m)(q_0+q_1e+ ... +q_ne^n)$

Then if we can find Ii(m) and hi(m) and non-zero k(m) such that for sufficiently large m,

$|\sum_{i=0}^nq_iI_i|<|\sum_{i=0}^nq_ih_i|$   ……………………………………   (2)

we have proved e is transcendental because then it is not possible to have

$q_0+q_1e+ ... +q_ne^n=0$

If we use in the integral ei-t rather than eit as used by Hermite we  immediately obtain an expression in the desired form (1).  Let

$I_i(m)=\int_0^ie^{i-t}f(t)dt$

where f(t) is a polynomial of degree m with integer coefficients.  For i > 0 repeated integration by parts gives

$I_i(m)=e^i\sum_{j=0}^mf^{(j)}(0)-\sum_{j=0}^mf^{(j)}(i)$

and the equation, which is identical in form to (1), is also correct in a formal sense for I0=0.  Now let

$f(t)=t^{p-1}(t-1)^p(t-2)^p ... (t-n)^p$

Then

$|I_i|\leq\int_0^i|e^{i-t}f(t)|dt<(2i)^mie^i$

and it is not hard to find that for sufficiently large p

$|\sum_0^nq_iI_i| < A^p$

where A is positive and independent of p.  This puts an upper bound on the left hand side of (2).  It is also not too hard to show (pdf) that

$k(m)=\sum_{j=0}^mf^{(j)}(0)$

cannot be zero if we make p a prime and greater than n (which we are quite at liberty to do).  Finally we can show that as long as p>|q0| then

$|\sum_{i=0}^nq_ih_i(m)| > (p-1)!$

giving a lower bound to the right hand side of (2).  But since (p-1)!>Ap for some large enough p, then (2) holds and we have proved e is transcendental.