A.3. TRANSCENDENTAL NUMBERS 365

I am following the interesting Wikepedia article on this subject. You can also look at thebook by Baker [5], Transcendental Number Theory, Cambridge University Press. There arealso many other treatments which you can find on the web including an interesting articleby Steinberg and Redheffer which appeared in about 1950.

The proof makes use of the following identity. For f (x) a polynomial,

I (s)≡∫ s

0es−x f (x)dx = es

deg( f )

∑j=0

f ( j) (0)−deg( f )

∑j=0

f ( j) (s) . (1.4)

where f ( j) denotes the jth derivative. It is like the convolution integral discussed earlierwith Laplace transforms. In this formula, s ∈ C and the integral is defined in the naturalway as ∫ 1

0s f (ts)es−tsdt (1.5)

The identity follows from integration by parts.∫ 1

0s f (ts)es−tsdt = ses

∫ 1

0f (ts)e−tsdt = ses

[−e−ts

sf (ts) |10 +

∫ 1

0

e−ts

ss f ′ (st)dt

]

= ses[−e−s

sf (s)+

1s

f (0)+∫ 1

0e−ts f ′ (st)dt

]= es f (0)− f (s)+

∫ 1

0ses−ts f ′ (st)dt

≡ es f (0)− f (s)+∫ s

0es−x f ′ (x)dx

Continuing this way establishes the identity since the right end looks just like what westarted with except with a derivative on the f .

Lemma A.3.2 Let (x1, ...,xn)→ g(x,x1, ...,xn) be symmetric and let

x→ g(x,x1, ...,xn)

be a polynomial. Then dm

dxm g(x,x1, ...,xn) is symmetric in the variables {x1, ...,xn}. If(x1, ...,xn) → h(x,x1, ...,xn) is symmetric, then for r some nonnegative integer, it fol-lows that ∑

nk=1 h(xk,x1, ...,xn)xr

k is symmetric. In particular, ∑nk=1

dl

dxl g(·,x1, ...,xn)(xk)xrk

is symmetric in {x1, ...,xn}.

Proof: The coefficients of the polynomial x→ g(x,x1, ...,xn) are symmetric functionsof {x1, ...,xn} . Differentiating with respect to x multiple times just gives another polyno-mial in x having coefficients which are symmetric functions. Thus the first part is proved.For the second part, the sum is of the form

h(x1,x1, ...,xn)xr1 +h(x2,x1, ...,xn)xr

2 + · · ·+h(xn,x1, ...,xn)xrn

You see that this is unchanged from switching two variables. For example, switch x1 andx2. By assumption, nothing changes in the terms after the first two. The first term thenbecomes

h(x2,x2,x1...,xn)xr2 = h(x2,x1,x2, ...,xn)xr

2

and the second term becomes

h(x1,x2,x1, ...,xn)xr1 = h(x1,x1,x2, ...,xn)xr

1