Most numbers are like this. Here the algebraic numbers are those which are roots of a polynomial equation
having rational numbers as coefficients. By the fundamental theorem of algebra, all these numbers are in ℂ.
There are only countably many of these algebraic numbers, (Problem 11 on Page 191). Therefore, most
numbers are transcendental. Nevertheless, it is very hard to prove that this or that number is
transcendental. Probably the most famous theorem about this is the Lindermannn Weierstrass
theorem.
Theorem 10.2.1Let the α_{i}be distinct nonzero algebraic numbers and let the a_{i}be nonzero algebraicnumbers. Then
∑n αi
aie ⁄= 0
i=1
I am following the interesting Wikepedia article on this subject. You can also look at the book by
Baker [4], Transcendental Number Theory, Cambridge University Press. There are also many other
treatments which you can find on the web including an interesting article by Steinberg and Redheffer which
appeared in about 1950.
The proof makes use of the following identity. For f
(x)
a polynomial,
∫ s deg∑(f) deg∑(f)
I(s) ≡ es−xf (x)dx = es f(j)(0)− f(j)(s). (10.2)
0 j=0 j=0
(10.2)
where f^{}
(j)
denotes the j^{th} derivative. In this formula, s ∈ ℂ and the integral is defined in the natural way
as
∫ 1 s−ts
sf (ts) e dt (10.3)
0
(10.3)
The identity follows from integration by parts.
∫ ∫
1 s−ts s 1 −ts
0 sf (ts)e dt = se 0 f (ts)e dt
[ e−ts ∫ 1 e− ts ]
= ses − ----f (ts)|10 + ---sf′(st)dt
[ s 0 s∫ 1 ]
= ses − e−sf (s)+ 1 f (0)+ e−tsf ′(st)dt
s s 0
s ∫ 1 s−ts ′
= e f (0)− f (s) + se f (st)dt
∫0s
≡ esf (0)− f (s) + es−xf′(x)dx
0
Continuing this way establishes the identity since at the right end looks just like what we started with
except with a derivative on the f.
Lemma 10.2.2Letting L =
dr-
dxr
, consider
( )
n∏
L ((⋅) − xi)
i=1
where
∏n
( i=1((⋅) − xi))
(x)
≡∏_{i=1}^{n}
(x − xi)
. Then
( )
∑n ∏n
L ((⋅)− xi) (xj)
j=1 i=1
is a symmetric polynomial in the
{x1,⋅⋅⋅,xn}
. Also
∑n ( ∏n ) ( )
L ((⋅)− xi) (xj) xmj
j=1 i=1
yields a symmetric polynomial.
Proof:The thing called for is
L((⋅− x1)(⋅− x2)⋅⋅⋅(⋅− xn))(x1)+ L ((⋅− x1)(⋅− x2) ⋅⋅⋅(⋅− xn))(x2)
+⋅⋅⋅+ L ((⋅− x1)(⋅− x2) ⋅⋅⋅(⋅− xn))(xn)
where ⋅ signifies the variable x.
Suppose you switch x_{1},x_{2}. You get the same thing. Same for other switches of variables. Hence the
claim is shown.
In the second claim, you get something similar, namely
m m
L ((⋅− x1)(⋅− x2)⋅⋅⋅(⋅− xn))(x1)(x1 )+ L ((⋅− x1)(⋅− x2) ⋅⋅⋅(⋅− xn))(x2)(x2 )
+ ⋅⋅⋅+ L((⋅− x1)(⋅− x2)⋅⋅⋅(⋅− xn))(xn)(xmn )
and this is unchanged by switching variables so it is a symmetric polynomial. ■
Lemma 10.2.3If K and c are nonzero integers, and β_{1},
⋅⋅⋅
,β_{m}are the roots of a single polynomial withinteger coefficients,
m
Q (x ) = vx + ⋅⋅⋅+ u
where v,u≠0, then
(β1 βm)
K +c e + ⋅⋅⋅+ e ⁄= 0.
Letting
v(m+1)pQp-(x)xp−1
f (x) = (p− 1)!
and I
(s)
be defined in terms of f
(x)
as above, it follows,
m
∑
lp→im∞ I(βi) = 0
i=1
and
∑n (j) p(m+1) p
f (0) = v u + m1 (p) p
j=0
∑m ∑n (j)
f (βi) = m2(p)p
i=1j=0
where m_{i}
(p)
is some integer.
Proof: Let p be a large prime number. Then consider the polynomial f
which clearly converges to 0. This proves the claim.
The next thing to consider is the term on the end in 10.4,
∑n m∑ ∑n
K f (j)(0)+ c f(j)(βi) (10.5)
j=0 i=1 j=0
(10.5)
The idea is to show that for large enough p it is always a nonzero integer. When this is done, it can’t
happen that K + c∑_{i=1}^{m}e^{βi} = 0 because if this were so, you would have a very small number,
c∑_{i=1}^{m}I
(βi)
equal to an integer.
v(m+1)p (vxm + ⋅⋅⋅+ u)pxp−1
f (x) = ---------(p−-1)!---------
Then f^{j}
(0)
= 0 unless j ≥ p − 1 because otherwise, that x^{p−1} term will result in some x^{r},r > 0 and
everything is zero when you plug in x = 0. Now say j = p− 1. Then it is clear that you get a
(p− 1)
! which
cancels the denominator and you get
f(p−1)(0) = upv(m+1)p
So what if j > p − 1? Then again, you get something which cancels the
(p− 1)
! and by Liebniz
formula,
( )
(j) j (m+1)p d m p
f (x) = p − 1 v dxj−(p−1) [(vx +⋅⋅⋅+ u)] + Stuff
where the Stuff equals 0 when x = 0. It involves derivatives of x^{p−1} and either there are too many
derivatives and you get 0 or not enough and you get something which is 0 when x = 0. Thus f^{j}
(0)
= pm_{j}
where m_{j} is some integer depending on the integer coefficients of the polynomial Q
(x )
= vx^{m} +
⋅⋅⋅
+ u.
Therefore,
∑n
f(j)(0) = v(m+1 )pup + m (p)p (10.6)
j=0
(10.6)
where m
(p)
is some integer. For p large, this cannot be zero because if it were, you would have p divides
v^{(m+1)
p}u^{p} which cannot occur if p is so large that this cannot happen.
Using the formula for f(x) and that the β_{i} are roots,
p
f (x) = v(m+1)p((x−-β1)(x−-β2)⋅⋅⋅(x−-βm-))-xp−1
(p− 1)!
it follows that for j < p, f^{}
(j)
(βi)
= 0. This is because for such derivatives, each term will have that
product of the
(x− βi)
in it.
To get something non zero, the nonzero terms must involve at least p derivatives of the
expression
((x − β1)(x − β2)⋅⋅⋅(x− βm ))p
since otherwise, when evaluated at any β_{k} the result would be 0.
Now say j ≥ p. Then by Liebniz formula, f^{j}
(x)
is of the form
( )
v(m+1)p∑j j d p d
(p-− 1)! r dxr (((x − β1)(x− β2)⋅⋅⋅(x − βm)) )dxj−rxp−1
r=0 ( )
vp(m+1)−2p+1 ∑j j d p d p−1
= ---(p−-1)!-- r dxr (((vx − vβ1) (vx − vβ2)⋅⋅⋅(vx− vβm )))dxj−r (vx)
r=0
Note that for r too small, the term will be zero when evaluated at any of the β_{i}. You only get something
nonzero if r ≥ p and so there will be a p! produced which will cancel with the
(p − 1)
! to yield an extra
p.
Now if you do the computations using the product rule and then replace x with β_{i} and sum these over
all i, you will get a symmetric polynomial in the quantities
{vβ1,⋅⋅⋅,vβm }
which has integer coefficients
and by Theorem 10.1.5 this is an integer. See Lemma 10.2.2 to observe why this will be so. You can factor
out all of the v to get the situation of that lemma. It follows that when adding these over
i,
Kvp(m+1)up + m (p)p+ L (p)p ≡ Kvp (m+1)up + M (p) p
for some integer M
(p)
. Summarizing, it follows
( ) ◜--⁄=◞0◟---◝
m∑ m∑ ∑n
c I(βi) = K + c eβi f(j)(0)+ Kvp (m+1 )up + M (p)p
i=1 i=1 j=0
where the left side is very small whenever p is large enough. Let p be larger than max
(K,v,u)
. Since p is
prime, it follows that it cannot divide Kv^{p(m+1)
}u^{p} and so the last two terms must sum to a nonzero integer
and so the equation 10.4 cannot hold unless
∑m βi
K + c e ⁄= 0 ■
i=1
Note that this shows π is irrational. If π = k∕m where k,m are integers, then both iπ and −iπ are roots
of the polynomial with integer coefficients,
m2x2 + k2
which would require, from what was just shown that
0 ⁄= 2+ eiπ + e−iπ
which is not the case since the sum on the right equals 0.
The following corollary follows from this. It is like the above lemma except it involves several
polynomials.
Corollary 10.2.4Let K and c_{i}for i = 1,
⋅⋅⋅
,n be nonzero integers. For each k between 1 and n let
{β(k)i}
_{i=1}^{mk}be the roots of a polynomial with integer coefficients,
Qk (x) ≡ vkxmk + ⋅⋅⋅+ uk
where v_{k},u_{k}≠0. Then
( ) ( ) ( )
m∑1 β(1) m∑2 β(2) m∑n β(n)
K + c1( e j) + c2( e j) + ⋅⋅⋅+ cn( e j) ⁄= 0.
j=1 j=1 j=1