[Two posts, separated by "&&&" - djr]

Newsgroups: sci.math
From: twomack@wincoll.demon.co.uk ("Thomas O. Womack")
Subject: Transcendence Overview (corrected version)
Date: Wed, 22 Mar 1995 12:28:23 +0000

Transcendental Numbers (overview)
---------------------------------

Definition : A transcendental number is one which is not the root of a
polynomial in Z (the integers). The opposite of 'transcendental' is
'algebraic'. In fact, any root of a polynomial in the algebraic numbers
is an algebraic number - as a note further on explains, this allows you
to generate an enormous number of transcendentals given a single one.

Results about them :

1. Cantor  They exist
           ----------

Consider the set of all polynomials in Z. With each polynomial, you can
associate an integer, by taking the sum of the absolute value of (n+1)a(n)
for the polynomial a(0)+xa(1)+x^2a(2)

EG x^7-x^3+2x^2+1 gets a value 8+4+(2*3)+1 = 19

There are only a finite number of polynomials with each value associated
with them, because a polynomial with a given value can have no coefficient
greater than it and cannot be of higher degree than it.

Since each of these finite number of polynomials has only a finite number
of roots, there are only a finite number of numbers generated as the roots 
of a polynomial with a given value. Since there are only aleph-null
possible values, and only a finite number of roots per value, there can
only be aleph-null algebraic numbers. Since there are aleph-one real
numbers, there must be some of them which are not algebraic. So, there
exist aleph-one transcendental numbers

----------------------------------------------------------------------------

2. Liouville   Here's one
               ----------

sum (a(i)*n^-factorial(i),i=0,infinity) is transcendental for all integer n,
and a(i)<n

EG 0.170005000000000000000006... is transcendental in any base >7
   0.TH000E00000000000000000M... is transcendental in base 36

Proof : (this is a summary of E-mail from Richard Pinch <R.G.E.Pinch@
pmms.cam.ac.uk>)

If a is algebraic, satisfying a polynomial equation of degree d, then
|a - p/q| < 1/q^e can have only finitely many solutions p/q when e > d
(Liouville's theorem). If we can find one with infinitely many solutions
to this equation, it must be transcendental.

Consider the sum a(i)n^(-i!). The base-n expansion of this sum to m! places
gives a fraction with denominator q=n^(m!), and with an error of at most
n.n^((m+1)!), and this error is less than 1/q^m. Since this is true for
all m, the number cannot satisfy a polynomial equation of any degree.

Proof of Liouville's Theorem:

If f(x) is the polynomial over Z of least degree (degree n) with f(a)=0,
then it does not have zero derivative at a, because otherwise f'(x) would
be of less degree and have a as a root. f'(x) is continuous, so there is
an e > 0 such that, between a-e and a+e, f'(x) is less that 3f'(a)/2. So,
calling f'(a) s,

f(a+e)< abs(e)*(3s/2), by the mean value theorem.

Take e small enough that there is no other root of f between a-e and a+e.
So, if e=a-p/q, then q^n*f(p/q) is an integer (because its denominator
can't be worse than q^n). However, this value must also be less than

abs(e)*(3s/2). So, abs(q^n) < abs (q^n * f(p/q)) < (3s/2)abs(a-p/q).
So, abs(a-p/q) must be less that 1/q^(n+1). So, q < 2s/3. So, there are
only finitely many q satisfying this.

3. Hermite 1873     e is transcendental
                    -------------------

Proof : in either of the two bibliography references

It takes about 2 sides of A5 and is concerned with showing that the value
of some horrible integral is an integer and also less than 1. I worked
through this proof once, but it took several hours and I still had to take
a couple of things for granted.

4. Lindemann 1882   pi is transcendental                    
                    --------------------

It takes about 3 sides, uses the properties of symmetric polynomials a
lot, and has the same method of proof as for e. I got stuck about a third
of the way through.

Proof : in either of the two bibliography references
----------------------------------------------------------------------------

Note : From a single transcendental number N, you can generate an enormous
number of extra ones : x*N must be transcendental for all rational x,
because otherwise you could multiply by 1/x and get N. x+N must be
transcendental for all rational x, because otherwise you could subtract x
and get N. N^z must be transcendental for all integer z, because otherwise
you could take the z'th root. All rational roots of N must be
transcendental. Any finite combination of these should keep a number
transcendental. Equally, something like pi^2+pi is transcendental, since
otherwise x^2+x = pi^2+pi has pi as a root, so pi is an algebraic function
of x^2+x. In general, if X satisfies an polynomial equation with algebraic
coefficients, X is algebraic.

----------------------------------------------------------------------------

5. Gelfond-Schneider 1934

If a and b are algebraic, a is not 0 or 1, and b is not rational, then a^b
is transcendental.

[this is a very much more elaborate follow-up to the note]

Corollaries :                                   

6. 2^sqrt(2), sqrt(7)^sqrt(3), {(11/17)^(11/9)}^{(7/3)^(3/13)} are all
transcendental - Directly from (5)

7. e^pi is transcendental

e^(i*pi)=-1. So, ln (-1) = i*pi. So, ln(-i*-1) = pi. So, e^ln(-i*-1) = e^pi.
So, (-1)^-i = e^pi. Now, -1 is algebraic and does not equal 0 or 1, and -i
is not rational, so, by Gelfond-Schneider, e^pi is transcendental.

8. e*pi and e+pi are not both algebraic                 
Source : Gerhard Niklasch <nikl@mathematik.tu-muenchen.de>

If they were both algebraic, then the equation x^2+(e+pi)x+(e*pi)=0 would
have roots e and pi, so e and pi would be algebraic (contradiction)

9. log_a(b) is transcendental unless b is a rational power of a
Source : Richard Pinch <R.G.E.Pinch@pmms.cam.ac.uk>
        

10. Source : Kelly B Roach <kbroach@daisy.uwaterloo.ca>
  
Any non-vanishing linear combination of logarithms of algebraic numbers
with algebraic coefficients is transcendental         

EG sqrt(2)ln(sin(pi/17)) + sqrt(3) ln(sin(pi/11)) is transcendental

e^b0 * a1^b1 * a2^b2 ... * an^bn is transcendental for any non-zero
algebraic numbers a1...n, b0...n                    

EG e^3 * 1.3^2.6 is transcendental

a1^b1 * a2^b2 ... * an^bn is transcendental for any algebraic numbers
a1...n, other than 0 or 1, and any algebraic numbers b1...n with
1,b1,b2...,bn linearly independent over the rationals.

EG 2^sqrt(2) * 3^sqrt(3) is transcendental

----------------------------------------------------------------------------

11. Alan Baker (Trinity College, Cambridge) 1975

If a_1, a_2 ... a_n are algebraic numbers which are multiplicatively
independent over the rationals
(ie q_1 log a_1 + q_2 log a_2 ... + q_n log a_n = 0 with q_i rational
implies that all the q_i equal zero),

then the same statement holds with 'q_i algebraic' inserted instead. 

----------------------------------------------------------------------------

LACK-OF-KNOWLEDGE

Still, no-one knows about Euler's gamma

(lim n->infinity sum(1/i,i=1,n) - ln n)

or about sum(i^-3,i=1,infinity), or about e^e, pi^pi or pi^e.
Evidently, proving things transcendental is hard.

Also, most of the proofs of transcendentality have the same structure. You
take some horrible function, prove its value to be an integer given that it
is algebraic, and then show that the value is positive and less than one.
          
This isn't a particularly erudite follow-up, because I'm still at school
and so have some difficulty :( finding texts. I've tried to sum up all the
e-mails I've received about the subject, though. I'll post another follow-
up if I get enough more e-mails

BIBLIOGRAPHY

The bibliography for this subject seemed to consist of one book :

Title :      Transcendental Number Theory
Author :     Alan Baker
Published :  1975      

G Niklasch's e-mail says that people have found much better numerical
values for the bounds in certain equations since this book was published.

I found the proofs for e and pi being transcendental in :

Title :      Galois Theory
Author :     Ian Stewart
ISBN :       0412345501

Some more books, which Robert Pinch told me about, are

Title:       Transcendence Theory: Advances and applications
Author:      Alan Baker and D W Masser
Publisher:   Academic Press
Published:   1977

Title:       New advances in transcendence theory
Editor:      Alan Baker
Publisher:   Cambridge University Press
Published:   1988

And there is also an article in volume 442 of Crelle's Journal.

All of those books are described as 'technical'. Since I found a copy
of Alan Baker's first book and became utterly lost by page four, I imagine
that the above books are totally incomprehensible except to a few dozen
professors somewhere :)
----------------
Tom Womack (twomack@wincoll.demon.co.uk) 

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Transcendence Overview (corrected version)
Date: 23 Mar 1995 01:09:13 GMT

In article <795875303snz@wincoll.demon.co.uk>, twomack@wincoll ("Thomas O. Womack") writes:
>Transcendental Numbers (overview)
>---------------------------------

>1. Cantor  They exist
>           ----------

It should be emphasized somewhere that Cantor's proof is 100% constructive.
I recall seeing a recent article--I think the MONTHLY--that went into this
aspect in great detail.

>3. Hermite 1873     e is transcendental
                     -------------------
>Proof : in either of the two bibliography references

>It takes about 2 sides of A5 and is concerned with showing that the value
>of some horrible integral is an integer and also less than 1. I worked
>through this proof once, but it took several hours and I still had to take
>a couple of things for granted.

Let me repost an old article of mine on how to work your way up to this
proof.  It's really not that horrible, although it took years before I
learned the cool secret behind the magic step.
========================================================================
In article <3c4hfq$l67@infoserv.rug.ac.be>, adntandt@eduserv (Andy Den Tandt) writes:
>I'd like to know how you prove that pi is transcendental.

>Just an overview of the proof would be nice, but anything you have
>will be great.

The most readable (if cryptic) proof is perhaps in Baker TRANSCENDENTAL
NUMBER THEORY.  Let me work up to pi with sketches that e is irrational
and e is transcendental.  What follows is from memory, with some quick
spot checking, so if all you want is an overview, you can believe all I
wrote as is.

The magic key is the following integration: if p(x) is a polynomial, and
P(x) denotes p(x)+p'(x)+p''(x)+... (a finite sum), then

                 /T
            I := |  exp(T-x)p(x) = P(0)exp(T)-P(T),
                0/

which is just repeated integration by parts, has certain magical properties
for the right choice of p.  If exp(T) is working out too nicely, we will
have an integer on the right, and an analytic expression that we can estimate
directly on the left, and which, depending on p, will also obviously not
be zero.  So this will be the ultimate contradiction, and thus exp(T) can't
work out nicely.

For example, if T=1 and p(x)=x^N*(1-x)^N, then if exp(T) were rational A/B,
and we choose N so that B|N, then P(0)*e-P(1) is obviously an integer, and
the integral is obviously positive, yet for N large enough, I<1.  Whoops.

To see that e is transcendental, suppose a_k e^k + ... + a_1 e + a_0 = 0
with a_k and a_0 nonzero.  Then for p(x)=x^(N+1)*(x-1)^N*(x-2)^N*...*(x-k)^N
(or something very similar), one adds up the various I's corresponding to
T=1,2,...,k, getting a_1 I1 +... a_k Ik=P(0)*(-a_0)-a_1 P(1)-...- a_k P(k).
So again, we are looking at something that is obviously an integer, and this
time one has to look a little harder to show that this integer is between 0
and 1.  But not much harder, just messier and not doable in my head.

Finally, when it comes to proving pi transcendental, one has to go through
one more round of complication, so that we get integers out of pi.  If pi
satisfies some irreducible algebraic equation with integral coefficients,
then all the symmetric polynomials in pi and its conjugates are merely
polynomials in the coefficients, and hence integers.  On the other hand,
we already know a convenient way to get an integer out of pi: e^ipi = -1.
So we let u1=ipi,u2,...,uk be i*pi and its conjugates.  Then obviously
(exp(iu1)+1)*(exp(iu2)+1)*...*(exp(iuk)+1)=0.  The left side multiplies
together into a sum over the 2^k subsets of {1,...,k}, each term being
exp(i*(sum of u_j over j in given subset)).  At least 1 (the empty sum),
and possibly more of these sums is zero, but not all of them.  Call the
non-zero sums v_j.  So we can say that S = sum of exp(i*v_j) is integral
but not zero.  At this point we whip out p(x)=x^(N+1)*(x-iv1)^N*... and
just observe that symmetric functions in the v_j's are already symmetric
functions in the u_i's: the non-apparent terms are 0, so p is an integral
polynomial.  So from this point on, it's all downhill, and again, a careful
choice of N, both from divisibility and magnitude considerations, will lead
to some alleged positive integer worked out in terms of various I's being
proven less than 1.
========================================================================
The question as to where the magic step comes from is explained in
Shidlovskii's book.  The integral I=P(0)exp(T)-P(T) comes from asking
for an integral polynomial P such that exp(T) is approximately P(T)/P(0).
If the approximation is too good (the Liouville theme), then when we
work things out exactly, the error can be expressed in terms of I, and
just like happens with Taylor series, the error is an integral.  So just
brute force write down P(x) with unknown coefficients, and solve for
the best approximation, namely, if I recall correctly, one that is
perfect for the first so many terms of the Taylor series for exp(T).

Out pops the P=p+p'+p''+... business.

>4. Lindemann 1882   pi is transcendental                    
                     --------------------
>It takes about 3 sides, uses the properties of symmetric polynomials a
>lot, and has the same method of proof as for e. I got stuck about a third
>of the way through.

>Proof : in either of the two bibliography references

See the book I refer to below.
========================================================================
>BIBLIOGRAPHY

>The bibliography for this subject seemed to consist of one book :

>Title :      Transcendental Number Theory
>Author :     Alan Baker
>Published :  1975      

There is also

 Title:          Transcendental numbers
 Author:         Shidlovskii, A. B.
 Published:      Berlin ; New York : W. de Gruyter, 1989.

Infinitely more readable than Baker.

>I found the proofs for e and pi being transcendental in :

>Title :      Galois Theory
>Author :     Ian Stewart
>ISBN :       0412345501

There is a somewhat new Springer-Verlag Universitext, with three authors
whose title is something like FAMOUS IMPOSSIBILITIES AND ?????.  In it
is the world's most detailed slow-motion explanation of the proof of the
transcendence of pi.  If you want to find it and nobody posts it, then
send me e-mail.  You can get the information from looking through any
very recent Universitext and look down the list of published titles in
the series.

>All of those books are described as 'technical'. Since I found a copy
>of Alan Baker's first book and became utterly lost by page four, I imagine
>that the above books are totally incomprehensible except to a few dozen
>professors somewhere :)

Baker is extremely cryptic.  That's his style.  The other books are all
far more readable.

Anyway, first master the proof that e is irrational.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)