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proof of Lindemann-Weierstrass
theorem and that e and π are
transcendental∗
rm50†
2013-03-21 22:58:05
This article provides a proof of the Lindemann-Weierstrass theorem, using
a method similar to those used by Ferdinand von Lindemann and Karl Weierstrass. This material is taken from [?] and expanded for clarity.
Before attacking the general case, we first use the same methods to prove two
earlier theorems, namely that both e and π are transcendental. These proofs
introduce the methods to be used in the more general theorem. At the end, we
present some trivial but important corollaries.
Both e and π were both known to be irrational in the 1700’s (Euler showed
the former; Lambert the latter). But e was not shown to be transcendental until
1873 (by Hermite, see [?] and [?]), and Lindemann showed π to be transcendental
as well in the late 1870’s. He also sketched a proof of the general theorem, which
was fleshed out by Weierstrass and Hilbert among others in the late 1800’s.
The following construct is used in all three proofs. Suppose f (x) is a real
polynomial, and let
Z
t
et−x f (x) dx.
I(t) =
0
Integrating by parts, we get
t Z t
Z t
t−x
et−x f 0 (x) dx = et f (0) − f (t) +
et−x f 0 (x) dx.
I(t) = (−e f (x)) +
0
0
0
Continuing, and integrating by parts a total of m = deg f times, we get
I(t) = et
m
X
f (j) (0) −
j=0
m
X
f (j) (t)
(1)
j=0
where f (j) (x) is the j th derivative of f .
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1
P
P
If f (x) =
ai xi , let F (x) =
|ai |xi ; i.e., the polynomial whose coefficients are the absolute values of those for f . Then using trivial bounds on the
integrand, we get
Z t
|I(t)| ≤
|et−x f (x)|dx ≤ |t|e|t| F (|t|).
(2)
0
We now proceed to prove the theorems. The proofs of all three are similar,
although the proof for e is the easiest. The steps of the proofs are as follows:
• Assume the theorem is false, and write down an equation in exponential
form that shows that the number in question is algebraic (for π, we will
use eiπ = −1 to write the equation in that form).
• Define a polynomial or set of polynomials f , and an associated number J
(or a sequence of numbers) that is a linear combination of the values of
I at the exponents in question. The motivation for the choice of f used
in each theorem is not given in Hermite’s proof. An excellent exposition
of how these definitions are relevant to the problem is given in [?]. In
essence, the ratio of the terms of I(t) in the proof that e is transcendental
relates to a Padé approximation to et ; this approximation is better the
larger deg f gets.
• Analyze J to show that it is integral and nonzero, and derive a lower
bound on J.
• Use equation (??) to derive a trivial upper bound on J.
• Note that the upper bound is lower than the lower bound, disproving the
original assumption.
The “magic” in the proof consists of finding the appropriate choice of f , as
well as in defining the transformation I to begin with. Once that is done, the
work in the proof is in showing J integral (which is harder for the more general
theorems) and in deriving the lower bound. But the outline of the proof remains
the same across all three theorems.
Theorem 1. e is transcendental.
Proof. Suppose
not, so that e is algebraic. Then e satisfies some integer polyP
nomial
ai xi = 0 with a0 6= 0. This means that
a0 + a1 e + a2 e2 + · · · an en = 0.
Let p be a (sufficiently large) prime, define a polynomial of degree m = (n +
1)p − 1 by
f (x) = xp−1 (x − 1)p · · · (x − n)p
and let
J = a0 I(0) + a1 I(1) + · · · an I(n).
2
We derive two sets of inconsistent bounds on J, thus showing that the original
hypothesis is false and e is transcendental.
First, apply equation (??) to J:
J=
n
X
ak I(k)
k=0
=
n
X

ak ek
=
f (j) (0) −
j=0
k=0
n
X
m
X
ak ek
m
X
m
X

f (j) (k)
j=0
f (j) (0) −
j=0
k=0
m X
n
X
n
X
k=0
ak
m
X
f (j) (k)
j=0
ak f (j) (k)
=−
j=0 k=0
where the last equality follows because of the assumed linear dependence of the
powers of e.
Consider the values of f (j) (k). If j < p − 1, then none of the factors in f
vanish by differentiation, so that f (j) (k) = 0 for all k. If j = p − 1, then only
the initial factor xp−1 can vanish in any term in f (j) (x), and so f (j) (k) = 0 if
k > 0. In the case where j = p − 1, k = 0, the only term in the derivative that
contributes a nonzero value is the term where xp−1 is differentiated each time;
that term gives
f (p−1) (0) = (p − 1)!(−1)np (n!)p .
Finally, if j ≥ p, then the terms in f (j) (k) that are nonzero are those terms in
which (x − k)p has been differentiated away; in these cases, the terms have a
leading coefficient that is thus a multiple of p!.
Now assume p > n; then f (p−1) (0) is a multiple of (p − 1)! but not of p!.
Putting together the above computations, we get
J = N p! + a0 M (p − 1)! = (p − 1)!(N p + a0 M )
where M, N are integers and p - M . Assume also p > a0 ; then N p + a0 M must
be nonzero and thus |J| ≥ (p − 1)!.
On the other hand, obviously F (k) ≤ (2n)m and thus
|J| ≤ |a1 |eF (1) + · · · |an |nen F (n) ≤ anen (2n)(n+1)p−1 =
anen
((2n)n+1 )p ≤ cp
2n
where a = max(|a1 |, . . . , |an |) and c is some constant that does not depend on
p.
But for p large enough, (p − 1)! > cp no matter what c is, so these two
bounds on J are contradictory.
Theorem 2. π is transcendental.
3
Proof. Again suppose not. Then iπ is also algebraic. Suppose the minimal
polynomial, f , of iπ has degree d, say
d
X
f (x) =
ai xi
i=0
with a0 6= 0, and let θ1 = iπ, θ2 , . . . , θd be the conjugates of iπ (the other roots
of f ). Then since eiπ = −1, we have
(1 + eθ1 )(1 + eθ2 ) · · · (1 + eθd ) = 0
Note that since θi is algebraic for each i, then ad θi is an algebraic integer.
Each term in this product can be written as a power of e, where the exponent
is of the form
β1 ,...,d = 1 θ1 + 2 θ2 + · · · + d θd
and each i is either 0 or 1. Denote by α1 , . . . , αn those exponents that are
nonzero. Note that at least one exponent is zero, and thus n < 2d . We then
have
(2d − n) + eα1 + · · · eαn = 0
(3)
We will show that if we define f by
p−1
f (x) = anp
(x − α1 )p · · · (x − αn )p
d x
with p a (sufficiently large) prime, then
J = I(α1 ) + · · · + I(αn )
satisfies the same incompatible bounds as in the previous theorem.
Let m = deg f = (n + 1)p − 1. As before, we see that
J=
n
X
I(αk )
k=1
=
n
X

eαk
=
k=1
f (j) (0) −
j=0
k=1
n
X
m
X
eαk
m
X
= (n − 2d )

f (j) (αk )
j=0
f (j) (0) −
j=0
m
X
m
X
n X
m
X
f (j) (αk )
k=1 j=0
m X
n
X
f (j) (0) −
j=0
f (j) (αk )
j=0 k=1
where the last equality follows from equation (??).
The remainder of the proof is quite similar to the above proof for e, except
that we must first show that the sum over k above is an integer; this was clear
in the previous theorem.
4
Consider the inner sum over k. It is clear that this is a symmetric polynomial
with integer coefficients in ad α1 , . . . , ad αn . The ad αi are algebraic integers since
the ad θi are. By the fundamental theorem of symmetric polynomials, it follows
that the inner sum is in fact a polynomial in the elementary symmetric functions
on the ad αi . Since the αi are the nonzero elements of the β... , we see that the
sum is also a polynomial in the elementary symmetric functions of the ad β... ,
and thus is a symmetric polynomial with integer coefficients in the ad θi . Again
applying the fundamental theorem of symmetric polynomials, we see that the
sum over k must be a polynomial in the elementary symmetric functions of the
ad θi . But these elementary symmetric functions are simply the coefficients of
the minimal polynomial of ad iπ, which are integers. Thus the inner sum is an
integer.
By arguments identical to those in the previous theorem, we have that
f (j) (αk ) = 0 when j < p and thus that f (j) (αk ) is an integral multiple of
p!; that f (j) (0) is an integral multiple of p! when j 6= p − 1; and that
f (p−1) (0) = (p − 1)!(−ad )np (α1 . . . αn )p
is an integral multiple of (p − 1)! but is not divisible by p if p is chosen to exceed
ad α1 . . . αn . Thus if also p > n − 2d , we have that J is nonzero and divisible by
(p − 1)! and thus |J| ≥ (p − 1)!. But again, similar to the proof in the previous
theorem, we have that
|J| ≤ |α1 |e|α1 | F (|α1 |) + · · · + |α1 |e|αn | F (|αn |) ≤ cp
and again these estimates are contradictory.
The Lindemann-Weierstrass theorem generalizes both these two statements
and their proofs.
Theorem 3. If α1 , . . . , αn are algebraic and distinct, and if β1 , . . . , βn are
algebraic and non-zero, then
β1 eα1 + · · · βn eαn 6= 0
Note that the facts that e and π are transcendental follow trivially from this
theorem.
For example, if e were algebraic, then e is the root of a polynomial
P
βi xi where βi ∈ Q, in contradiction to the theorem.
Proof. The proof follows the same general lines as above, but there are additional complexities introduced by the arbitrary αi . In the proof of the transcendality of π, we were able to use facts about the relationship of the exponents in
the proof; no such relationship is available to us in this more general setting.
Again start by supposing
β1 eα1 + · · · βn eαn = 0
where the αi , βi are as given.
5
(4)
Claim we can assume, without loss of generality, that βi ∈ Z. For if not,
take all the expressions formed by substituting for one or more of the βi one of
its conjugates, and multiply those by the equation above. The result is a new
expression of the same form (with different αi ), but where the coefficients are
rational numbers. Clear denominators, proving the claim.
Next, claim we can assume that the αi are a complete set of conjugates, and
that if αi , αj are conjugates, then βi = βj . To see this, choose an irreducible integral polynomial having α1 , . . . , αn as roots; let αn+1 , . . . , αN be the remaining
roots, and define βn+1 = . . . βN = 0. Then clearly we have
Y
(β1 eασ(1) + · · · βN eασ(N ) ) = 0
σ∈SN
(Note the similarity with the proof for π). There are N ! factors in this product,
so expanding the product, it is a sum of terms of the form
eh1 α1 +···hN αN
with integral coefficients, and h1 + · · · + hN = N !. Clearly the set of all such
exponents forms a complete set of conjugates. By symmetry considerations, we
see that the coefficients of two conjugate terms are equal. Also, the product
is not identically zero. To see this, consider the term in the product formed
by multiplying together, from each factor, the nonzero terms with the largest
exponents in the lexicographic order on C. Since the αi are unique (because the
polynomial is irreducible), there is only one term with this largest exponent,
and it has a nonzero coefficient by construction.
Finally, order the terms so that the conjugates of a particular αi appear
together. That is, for the remainder of the proof we may assume that
β1 eα1 + · · · + βn eαn = 0
with the βi ∈ Z,and that there are integers 0 = n0 < n1 < · · · < nr = n chosen
so that, foreach 0 ≤ t < n we have
αnt +1 , . . . , αnt+1 form a complete set of conjugates
βnt +1 = βnt +2 = · · · βnt+1
Now, since αi , βi are algebraic, we can choose l such that lαi , lβi are algebraic
integers. Let
fi (x) = lnp
((x − α1 ) · · · (x − αn ))p
, 1≤i≤n
(x − αi )
where again p is a (large) prime. We will develop contradictory estimates for
|J1 . . . Jn |, where
Ji = β1 Ii (α1 ) + · · · + βn Ii (αn ), 1 ≤ i ≤ n
and Ii is the integral associated with fi , as above (see equation (??)).
6
Using equations (??) and (??), we see that
Ji =
n
X
βk Ii (αk )
k=1
=
n
X

np−1
X
βk eαk
=
np−1
X

(j)
fi (0)
j=0
=−
(j)
fi (0) −
j=0
k=1


n
X
βk
n
X
β k eα k
−
np−1
X

(j)
fi (αk )
j=0
k=1
!
k=1
np−1
n X X

n
X
k=1

βk
np−1
X

(j)
fi (αk )
j=0
(j)
βk fi (αk ) .
j=0 k=1
(j)
Arguing similarly to the foregoing proofs, we see that fi (αk ) is an algebraic
integer divisible by p! unless j = p − 1 and k = i. In this particular case, we
have that
n
Y
(αi − αk )p
fip−1 (αi ) = lnp (p − 1)!
k=1
k6=i
and so again, if p is large enough, this is divisible by (p − 1)! but not by p!. Thus
Ji is a nonzero algebraic integer divisible by (p − 1)! but not by p!. As before,
we can prove that Ji 6= 0.
Ji can be written as follows:
Ji = −
np−1
r−1
X X
βnt+1 (fi (j) αnt +1 + · · · + αnt+1
j=0 t=0
Note that by construction, fi (x) can be written as a polynomial whose coefficients are polynomials in αi , and the integral coefficients of those polynomials
are integers independent of i. Thus, noting that the αi form a complete set of
conjugates and using the fundamental theorem on symmetric polynomials as in
the previous proof, we see that the product of the Ji is in fact a rational number.
But it is an algebraic integer, hence an integer. Thus J1 · . . . · Jn ∈ Z, and it is
divisible by ((p − 1)!)n . Thus |J1 · · · Jn | ≥ (p − 1)!. But the same estimate as in
the previous proofs shows that for each i,
|Ji | ≤
n
X
|βk ||Ii (αk )| ≤
k=1
n
X
|βk αk |e|αk | Fi (|αk |)
k=1
which as before is ≤ cp for some sufficiently large c. These estimates are again
in contradiction, proving the theorem.
7
Note that Theorems ?? and ?? are trivial corollaries of Theorem ??, as one
would expect. (To prove that π is transcendental, note that if it were algebraic,
then eiπ = −1 would be transcendental).
Here are some other more or less trivial corollaries.
Corollary 4. If α 6= 0 is algebraic, then eiα is transcendental.
Proof. If it were algebraic, say eiα = β, then we have
eiα − βe0 = 0
in contradiction to the above theorem since α 6= 0.
Corollary 5. If α 6= 0 is algebraic, then cos α and sin α are both transcendental.
Proof. Recall that cos α+i sin α = eiα , which is transcendental. If either cos α or
sin α were algebraic, then the other would be as well (and thus their sum would
be) since cos2 α+sin2 α = 1. Therefore, both cos α and sin α are transcendental.
Corollary 6. If α > 0 is algebraic with α 6= 1, then ln α is transcendental.
Proof. If β = ln α, then eβ = α. By Corollary ??, since α is algebraic, β cannot
be.
References
[1] A. Baker, Transcendental Number Theory, Cambridge University Press,
1990.
[2] H. Cohn, A Short Proof of the Simple Continued Fraction Expansion of e,
American Mathematical Monthly, Jan. 2006, pp. 57-62.
[3] C. Hermite, Sur la fonction exponentielle, Compte Rendu Acad. Sci. 77
(1873) 18-24, 74-79, 226-233, and 285-293; also in Œuvres, v. 3, GauthierVilliers, Paris, 1912 pp. 150-181.
[4] U. G. Mitchell, M. Strain, Osiris, Vol. 1, Jan. 1936, pp. 476-496.
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