如何将表达式整体应用于函数（如对数）及含函数指数的表达式化简

Hey there! Let's break down your confusion step by step—both of these are common sticking points in calculus/analysis, so you're not alone.

First: Simplifying $\Big (1 + \frac{\epsilon}{2\log t}\Big )^{\log t}$ #

This expression is perfect for using a combination of exponent-logarithm conversion and Taylor series expansion (or asymptotic approximation). Here's how to tackle it:

Step 1: Rewrite using $A^B = e^{B \ln A}$ #

Any power expression can be converted to exponential form using this identity, which makes it easier to manipulate:
$$
\Big (1 + \frac{\epsilon}{2\log t}\Big )^{\log t} = e^{\log t \cdot \ln\left(1 + \frac{\epsilon}{2\log t}\right)}
$$

Step 2: Taylor expand the logarithm #

When $t$ is large (so $\frac{\epsilon}{2\log t}$ is small, since $\log t$ grows slowly), we can use the Taylor series for $\ln(1+x)$ around $x=0$:
$$
\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots
$$
Substitute $x = \frac{\epsilon}{2\log t}$ into this series, then multiply by $\log t$:
$$
\log t \cdot \ln\left(1 + \frac{\epsilon}{2\log t}\right) = \log t \left( \frac{\epsilon}{2\log t} - \frac{(\epsilon/(2\log t))^2}{2} + \frac{(\epsilon/(2\log t))^3}{3} - \dots \right)
$$

Step 3: Simplify term by term #

Cancel out $\log t$ in each term to get:
$$
\frac{\epsilon}{2} - \frac{\epsilon^2}{8\log t} + \frac{\epsilon^3}{24(\log t)^2} - \dots
$$

Step 4: Convert back to exponential form #

Substitute this back into the exponential expression. For large $t$, the higher-order terms (with $1/(\log t)^k$) become negligible, so we can write the asymptotic approximation:
$$
\Big (1 + \frac{\epsilon}{2\log t}\Big )^{\log t} = e^{\epsilon/2} \left(1 - \frac{\epsilon^2}{8\log t} + O\left(\frac{1}{(\log t)^2}\right)\right)
$$
If we just care about the leading term (the dominant behavior as $t \to \infty$), this simplifies to approximately $e^{\epsilon/2}$.

Second: Your Two General Confusions #

1. Applying a function to an entire expression (like logarithms) #

The key here is using function properties to break down the expression. For logarithms specifically, the power rule $\log(A^B) = B\log A$ is your best friend—it lets you "pull down" the exponent and turn a complicated power into a product.

For example, if you wanted to take the log of your original expression:
$$
\log\left( \Big (1 + \frac{\epsilon}{2\log t}\Big )^{\log t} \right) = \log t \cdot \log\left(1 + \frac{\epsilon}{2\log t}\right)
$$
This is way easier to work with than trying to compute the log of the power directly. For other functions (like square roots, exponentials), use their respective algebraic rules to simplify before applying the function.

2. Dealing with powers of functions (not just numerical exponents) #

Expressions like $(1 + f(t))^{g(t)}$ are common in asymptotic analysis. The go-to strategy is almost always to convert to exponential form ($e^{g(t)\ln(1+f(t))}$) as we did earlier. This lets you:

• Use Taylor series on the logarithm if $f(t)$ is small (relative to 1)
• Apply limit rules (like the classic $\lim_{n\to\infty}(1+\frac{a}{n})^n = e^a$) when $g(t)$ grows large and $f(t) = \frac{\text{constant}}{g(t)}$

内容的提问来源于stack exchange，提问作者YohanRoth