如何求$e^{t^2/2}$的麦克劳林级数？变量替换合理性疑问

Hey there, let's unpack this confusion step by step—you're asking a great question that gets to the core of how power series work!

First, let's recap the key fact about the Maclaurin series for $e^x$:

The series $e^x = \sum_{n=0}^{{\infty}\frac{x}n}{n!}$ converges to $e^x$ for every real number $x$. That means no matter what real number you plug in for $x$, the infinite sum will equal the exponential function evaluated at that number.

Now, here's the critical point: $x$ here is just a placeholder for any real value. It doesn't have to be a "simple" variable like $t$—it can be any expression that evaluates to a real number, including $\frac{1}{2}t^2$.

Why the substitution is valid #

Since $\frac{1}{2}t^2$ is a real number for every real $t$ (squaring $t$ gives a non-negative number, dividing by 2 keeps it real), it falls perfectly within the convergence domain of the original $e^x$ series (which is all real numbers).

When we substitute $x = \frac{1}{2}t^2$ into the series, we're just applying the function $e^x$ to a specific input (that happens to be a function of $t$). The series rule for $e^x$ applies to any input in its domain, so this substitution is completely legitimate.

Let's simplify the result to make it clearer #

After substitution, we get:
$$e^{t2/2} = \sum_{n=0}^{{\infty}\frac{(\frac{1}{2}t}2)^n}{n!}$$
We can rewrite this to highlight it's a Maclaurin series in terms of $t$:
$$e^{t2/2} = \sum_{n=0}^{{\infty}\frac{t}{2n}}{2^n n!}$$
If you were to derive the Maclaurin series for $e^{t2/2}$ the "long way" (computing derivatives at $t=0$ and building the series term by term), you'd end up with exactly this result—this confirms the substitution method works.

A quick analogy to drive it home #

Think of the series $\sum_{n=0}^{{\infty}\frac{x}n}{n!}$ as a formula that says "take any real number, raise it to the nth power, divide by n!, sum all those terms, and you get $e$ to that number". If your "real number" is $\frac{1}{2}t^2$, you just follow the same rule: raise $\frac{1}{2}t^2$ to the nth power, divide by n!, sum up, and you get $e^{{\frac{1}{2}t}2}$.

内容的提问来源于stack exchange，提问作者PassingBy