请求证明：$(1−x+x^2) ext{e}^x$的非零系数为分子是1或素数的有理数

Let's work through this problem systematically—this is a great tie-in between Taylor series and basic number theory. Here's how to approach it:

Step 1: Start with the Taylor Series for $e^x$ #

We know the standard Taylor expansion of $e^x$ centered at 0 is:
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

Step 2: Multiply by the Polynomial $(1 - x + x^2)$ #

Distribute the polynomial across the series to get three separate sums:
$$(1 - x + x^2)ex = \sum_{n=0}^{\infty} \frac{x^n}{n!} - \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} + \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$$

Step 3: Combine the Series into a Single Sum #

Adjust the indices of the second and third series to align with the first:

• For the second series, let $k = n+1$ (so $n = k-1$). It becomes $\sum_{k=1}^{\infty} \frac{x^k}{(k-1)!}$.
• For the third series, let $m = n+2$ (so $n = m-2$). It becomes $\sum_{m=2}^{\infty} \frac{x^m}{(m-2)!}$.

Now rename all indices to $n$ and combine terms for each $n$:

• $n=0$: Only the first series contributes: $a_0 = \frac{1}{0!} = 1$ (numerator is 1, which fits our condition).
• $n=1$: First series term minus second series term: $\frac{1}{1!} - \frac{1}{0!} = 0$ (zero coefficient, we can ignore this).
• $n \geq 2$: Combine all three relevant terms into a single fraction:
  $$a_n = \frac{1}{n!} - \frac{1}{(n-1)!} + \frac{1}{(n-2)!}$$

Step 4: Simplify the Coefficient Formula #

Factor out $\frac{1}{n!}$ to combine the numerators:
$$a_n = \frac{1 - n + n(n-1)}{n!}$$
Expand and simplify the numerator:
$$1 - n + n^2 - n = n^2 - 2n + 1 = (n-1)^2$$
This leaves us with a much cleaner expression:
$$a_n = \frac{(n-1)^2}{n!} \quad \text{for } n \geq 2$$

Step 5: Analyze the Reduced Fraction #

We need to show that when $\frac{(n-1)^2}{n!}$ is written in lowest terms (numerator and denominator coprime), the numerator is either 1 or a prime number. Let's break this into cases:

Case 1: $n-1 = 1$ (i.e., $n=2$)

$$a_2 = \frac{1}{2 \cdot 0!} = \frac{1}{2}$$
Numerator is 1—fits our condition.

Case 2: $n-1$ is a prime number $p$ (i.e., $n = p+1$)

By Wilson's theorem, $(p-1)! \equiv -1 \mod p$, meaning $p$ and $(p-1)!$ are coprime. Also, $p$ and $p+1$ (consecutive integers) are coprime. So the fraction $\frac{p}{(p+1) \cdot (p-1)!}$ is already in reduced form, with numerator $p$ (a prime)—fits our condition.

Case 3: $n-1$ is a composite number

• Subcase 3a: $n-1 = 4$ (i.e., $n=5$)
  $$a_5 = \frac{4}{5 \cdot 3!} = \frac{4}{30} = \frac{2}{15}$$
  The reduced numerator is 2, a prime—fits our condition.
• Subcase 3b: $n-1$ is composite and $\geq 6$
  Any composite number $m \geq 6$ divides $(m-1)!$ (all its prime factors are smaller than $m$, and as a composite, it can be written as a product of numbers $\leq m-1$). This means $\frac{m}{(m-1)!}$ simplifies to a fraction with numerator 1. Thus:
  $$\frac{n-1}{n \cdot (n-2)!} = \frac{m}{(m+1) \cdot (m-1)!}$$
  reduces to a fraction with numerator 1—fits our condition.

Final Conclusion #

Every non-zero coefficient of $(1−x+x^2)ex$ is a rational number whose numerator (in reduced form) is either 1 or a prime number.

内容的提问来源于stack exchange，提问作者William Grannis