Truncated Exponential Has No Repeated Roots

Problem

Define the truncated exponential polynomial: $p_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$

Prove that $p_n(x)$ has no repeated real roots for any $n \geq 1$ .

Field

Putnam / Analysis / Polynomials

Why It's Beautiful

The truncated exponentials are the partial sums of $e^x$ . They look complicated — their roots spread out in the complex plane and have no closed form. Yet the statement that they have no repeated real roots admits a 3-line proof from a single elegant observation.

The key identity $p_n(x) = p_{n-1}(x) + x^n/n!$ is obvious once written down, but using it to kill repeated roots requires a small, satisfying twist.

Key Idea / Trick

A repeated root $r$ of $p_n$ satisfies both $p_n(r) = 0$ and $p_n'(r) = 0$ .

Note that $p_n'(x) = p_{n-1}(x)$ , so $p_{n-1}(r) = 0$ .

But $p_n(x) = p_{n-1}(x) + \dfrac{x^n}{n!}$ , so $0 = 0 + \dfrac{r^n}{n!}$ , giving $r = 0$ .

Yet $p_n(0) = 1 \neq 0$ . Contradiction.

Difficulty

2 / 5

Truncated Exponential Has No Repeated Roots — Answer

Key Identity

Observe that consecutive truncated exponentials differ by exactly one term: $p_n(x) = p_{n-1}(x) + \frac{x^n}{n!} \tag{$*$}$

And differentiating $p_n$ gives: $p_n'(x) = p_{n-1}(x) \tag{$**$}$

Proof

Suppose $r \in \mathbb{R}$ is a repeated root of $p_n$ . Then:

$p_n(r) = 0 \quad \text{and} \quad p_n'(r) = 0$

From $(**)$ : $p_n'(r) = p_{n-1}(r) = 0$ .

Substituting both $p_n(r) = 0$ and $p_{n-1}(r) = 0$ into $(*)$ : $0 = 0 + \frac{r^n}{n!}$

So $r^n = 0$ , which forces $r = 0$ .

But $p_n(0) = 1 \neq 0$ , contradicting $p_n(r) = 0$ . $\blacksquare$

Why the Identity $(*)$ Is the Right Tool

A repeated root $r$ of a polynomial $f$ is a common root of $f$ and $f'$ . Normally, finding $\gcd(f, f')$ requires the Euclidean algorithm. Here, the relationship $p_n' = p_{n-1}$ means $f'$ is almost $f$ itself — just with the leading term removed. That's what makes $(*)$ lethal: it directly computes $f - \text{(leading term)} = f'$ , so any common root must kill the leading term.

Bonus: Behavior of Roots

For $n$ odd: $p_n(x) \to \pm\infty$ as $x \to \pm\infty$ , so $p_n$ has at least one real root (in fact, exactly one).
For $n$ even: $p_n(x) \to +\infty$ as $x \to \pm\infty$ , and $p_n > 0$ everywhere (since the minimum of $e^x - p_n(x)$ , which is $\sum_{k>n} x^k/k!$ , is non-negative for even $n$ and $x \geq 0$ ). So $p_n$ has no real roots at all when $n$ is even.

In all cases, no real root is repeated.