关于$\frac{\sin x}{x}$泰勒级数两种表达式的疑问

Great question—let's unpack this step by step to see why your direct derivation is correct, and how Wolfram Alpha's seemingly complex expression is actually equivalent.

Why Applying $\frac{1}{x}$ to $\sin x$'s Taylor Series Works #

First, remember the Taylor series for $\sin x$ centered at 0:
$$\sin x = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}$$
For any $x \neq 0$, dividing every term by $x$ is a valid operation here. The original series converges for all real and complex $x$, and multiplying by $\frac{1}{x}$ (a constant scalar for fixed $x$) preserves convergence. This gives us:
$$\frac{\sin x}{x} = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k+1)!}$$
This is the standard Taylor series for $\frac{\sin x}{x}$. Even though the original function has a removable singularity at $x=0$, the series converges to the analytic extension of $\frac{\sin x}{x}$ at $x=0$ (which equals 1), so it's a valid Taylor series, not just a Laurent series.

Are the Two Series Equivalent? Absolutely! #

Let's simplify Wolfram Alpha's expression to show it matches your result:
$$\sum_{n=-1}^{\infty} \frac{x^n((-i)n + i^n)}{2(1+n)!}$$
We can break down the term $(-i)^n + i^n$ based on the value of $n$:

• When $n=-1$: $(-i)^{-1} + i^{-1} = i + (-i) = 0$, so this term disappears.
• When $n$ is odd: $(-i)^n = (-1)^n i^n = -i^n$ (since odd powers of $-1$ are $-1$), so $(-i)^n + i^n = -i^n + i^n = 0$. All odd-$n$ terms cancel out.
• When $n$ is even: Let $n=2m$ where $m \geq 0$. Then $(-i)^{2m} = ((-i)²⁾m = (-1)^m$, and $i^{2m} = (i²⁾m = (-1)^m$. So $(-i)^n + i^n = (-1)^m + (-1)^m = 2(-1)^m$.

Substitute $n=2m$ into the series (ignoring all zero terms):
$$\sum_{m=0}^{\infty} \frac{x^{2m} \cdot 2(-1)^m}{2(1+2m)!} = \sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{(2m+1)!}$$
This is exactly the series you derived directly!

Why Wolfram Alpha Uses This Complex Form #

Wolfram Alpha tends to output general-purpose expressions that work across broader mathematical contexts (like complex analysis). Here, it's likely starting from the exponential definition of sine:
$$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$
After substituting this into $\frac{\sin x}{x}$ and expanding the exponential functions into their Taylor series, rearranging and reindexing the sum leads to the expression with $n$ ranging from $-1$ to $\infty$. While it looks more complicated, most terms cancel out, leaving the same simplified series as your straightforward derivation.

内容的提问来源于stack exchange，提问作者Evan