Taylor Series and Leibniz Rule for Higher-Order Derivatives

This article uses the function $f(x) = x^2 \ln(1 - x)$ as an example to explain in detail how to compute $f^{(n)}(0)$ ($n \ge 3$) using Taylor series method and Leibniz rule, and analyzes the equivalence of the two methods.

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Problem

Let $f(x) = x^2 \ln{(1 - x)}$, then for $n \ge 3$, $f^{(n)}(0) = \text{?}$

Taylor Series Method

Core Idea

Expand the function into a Taylor series and directly read off the higher-order derivatives from the coefficients of the power series.

Steps

1. Expand $\ln(1 - x)$
   The Taylor expansion of $\ln(1 - x)$ for $|x| < 1$ is:

   $$\ln(1 - x) = -\sum_{k=1}^{\infty} \frac{x^k}{k}.$$

2. Construct the series for $f(x)$
   Multiply by $x^2$:

   $$f(x) = x^2 \cdot \left(-\sum_{k=1}^{\infty} \frac{x^k}{k}\right) = -\sum_{k=1}^{\infty} \frac{x^{k+2}}{k}.$$

   Substitute $m = k + 2$ (i.e., $k = m - 2$):

   $$f(x) = -\sum_{m=3}^{\infty} \frac{x^m}{m - 2}.$$

3. Extract the coefficient of $x^n$
   For $n \ge 3$, the coefficient $a_n = -\frac{1}{n - 2}$. By Taylor’s formula:

   $$f^{(n)}(0) = n! \cdot a_n = -\frac{n!}{n - 2}.$$

Verification (Example $n = 3$)

Compute $f^{(3)}(0)$:

$$f^{(3)}(0) = -\frac{3!}{3 - 2} = -6.$$

Direct differentiation yields:

$$f'''(x) = \frac{d^3}{dx^3} \left(x^2 \ln(1 - x)\right) \bigg|_{x=0} = -6.$$

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Leibniz Rule Method

Core Idea

Use the formula for higher-order derivatives of a product:

$$(f \cdot g)^{(n)}(x) = \sum_{k=0}^{n} \binom{n}{k} f^{(k)}(x) \cdot g^{(n - k)}(x).$$

Steps

1. Decompose the function
   Let $f(x) = x^2$, $g(x) = \ln(1 - x)$.

2. Analyze higher-order derivatives of $f(x)$

   • $f^{(0)}(x) = x^2$, at $x=0$ it is $0$.
   • $f^{(1)}(x) = 2x$, at $x=0$ it is $0$.
   • $f^{(2)}(x) = 2$, at $x=0$ it is $2$.
   • For $k \ge 3$, $f^{(k)}(x) = 0$.

3. Compute higher-order derivatives of $g(x)$
   $g^{(m)}(x) = -\frac{(m - 1)!}{(1 - x)^m}$, and at $x=0$:

   $$g^{(m)}(0) = -(m - 1)!.$$

4. Apply Leibniz rule
   Since $f^{(k)}(0)$ is nonzero only for $k=2$, we have:

   $$f^{(n)}(0) = \binom{n}{2} \cdot f^{(2)}(0) \cdot g^{(n - 2)}(0).$$

   Substituting values:

   $$f^{(n)}(0) = \frac{n(n - 1)}{2} \cdot 2 \cdot \left(-(n - 3)!\right) = -n(n - 1)(n - 3)!.$$

Verification (Example $n = 4$)

Compute $f^{(4)}(0)$:

$$f^{(4)}(0) = -4 \cdot 3 \cdot (4 - 3)! = -12.$$

Taylor series verification:

$$-\frac{4!}{4 - 2} = -\frac{24}{2} = -12.$$

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Unification of Results

The two methods give equivalent results:

$$-\frac{n!}{n - 2} = -n(n - 1)(n - 3)!.$$

Proof:

$$\frac{n!}{n - 2} = \frac{n(n - 1)(n - 2)!}{n - 2} = n(n - 1)(n - 3)!.$$

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Q&A

Q1: In the Taylor series method, after the substitution $m = k + 2$, why are the powers of $x$ on both sides of the equality equal?

A: The substitution $m = k + 2$ merely relabels the summation index without changing the mathematical content of the series. In the original series, the power of $x^{k+2}$ is determined by $k+2$, and after substitution it becomes $x^m$, so the powers of $x$ remain consistent.

Q2: In the Leibniz rule, why does only the term with $k=2$ contribute to the result?

A: Because the higher-order derivatives of $f(x) = x^2$ vanish for $k \ge 3$, and for $k=0,1$ we have $f^{(k)}(0) = 0$. The only nonzero term comes from $k=2$.

Q3: Does the convergence interval of the Taylor series affect the result?

A: The Taylor series converges for $|x| < 1$, but the computation of $f^{(n)}(0)$ depends only on the local behavior at $x=0$, so the expansion is valid for calculating higher-order derivatives.

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Conclusion

For $n \ge 3$, the $n$-th derivative of $f(x) = x^2 \ln(1 - x)$ at $x=0$ is:

$$\boxed{f^{(n)}(0) = -\frac{n!}{n - 2}}.$$