# Gelfond-Schneider Theorem/Lemma 1

## Lemma

Let $\map {a_1} t, \ldots, \map {a_n} t$ be non-zero polynomials in $\R \sqbrk t$ of degrees $d_1, \ldots, d_n$ respectively.

Let $w_1, \ldots, w_n$ be pairwise distinct real numbers.

Then:

$\ds \map F t = \sum_{j \mathop = 1}^n \map {a_j} t e^{w_j t}$

has at most $d_1 + \cdots + d_n + n − 1$ real roots (counting multiplicities).

## Proof

By multiplying through by $e^{−w_n t}$ if necessary, we may suppose that $w_n = 0$ and that otherwise $w_j \ne 0$.

Let $k = d_1 + \cdots + d_n + n$.

We use strong induction on $k$.

If $k = 1$, then $n = 1$ and $d_1 = 0$, and the lemma easily follows.

Let $l \ge 2$ be such that the lemma holds whenever $k < l$.

Suppose $k = l$.

Let $N$ be the number of real roots of $\map F t$.

By Rolle's Theorem, the number of real roots of $\map {F'} t$ is at least $N − 1$.

On the other hand:

$\ds \map {F'} t = \sum_{j \mathop = 1}^n \map {b_j} t e^{w_j t}$

where:

$\map {b_j} t = \map { {a_j}'} t + w_j \map {a_j} t$

Note that for $1 \le j \le n − 1$, we have that $\map {b_j} t$ has degree exactly $d_j$.

Also, since $w_n = 0$, either:

there are only $n − 1$ non-zero polynomials $\map {b_j} t$ in the expression for $\map {F'} t$ above

or:

there are $n$ such polynomials and the degree of $\map {b_n} t$ is one less than the degree of $\map {a_n} t$.

It follows from the induction hypothesis that $\map {F'} t$ has at most $d_1 + \cdots + d_n + n − 2$ real roots.

Hence:

$N − 1 \le d_1 + \cdots + d_n + n − 2$

and the result follows.

$\blacksquare$
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