Higher Order Derivatives — Question 10

Question 10 - Solution

We are given $$f(x) = \ln(1 + x^2).$$

(a) Odd-order derivatives at $x = 0$

Observe that $f(x)$ is an even function, since $$f(-x) = \ln(1 + (-x)^2) = \ln(1 + x^2) = f(x).$$

The derivative of an even function is odd, and the derivative of an odd function is even. Therefore:

• $f'(x)$ is odd,

• $f''(x)$ is even,

• $f^{(3)}(x)$ is odd,

• and so on.

All odd functions evaluate to zero at $x = 0$. Hence, $$\boxed{f^{(2k+1)}(0) = 0 \quad \text{for all } k \ge 0.}$$

(b) Closed form for $f^{(2n)}(0)$

We begin with the known Taylor expansion: $$\ln(1 + u) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{u^n}{n}, \quad |u| < 1.$$

Substitute $u = x^2$: $$\ln(1 + x^2) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n}.$$

Comparing this with the Maclaurin series definition: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n,$$ we see that only even powers appear. Matching coefficients of $x^{2n}$, we obtain: $$\frac{f^{(2n)}(0)}{(2n)!} = (-1)^{n+1} \frac{1}{n}.$$

Solving for $f^{(2n)}(0)$: $$\boxed{f^{(2n)}(0) = (-1)^{n+1} \frac{(2n)!}{n}.}$$

(c) Maclaurin series

Using the result above, the Maclaurin series for $f(x)$ is: $$\boxed{ \ln(1 + x^2) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n}, \quad |x| < 1. }$$

(d) Radius of convergence

The series $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n}}{n}$$ is a power series in $x^2$.

Using the Ratio Test: $$\lim_{n \to \infty} \left| \frac{x^{2(n+1)}/(n+1)}{x^{2n}/n} \right| = |x|^2.$$

The series converges when: $$|x|^2 < 1 \quad \Rightarrow \quad |x| < 1.$$

Thus, the radius of convergence is: $$\boxed{R = 1.}$$

(e) Approximating $\ln(1.1)$

We write: $$\ln(1.1) = \ln(1 + 0.1) = \ln(1 + x^2) \quad \text{with } x^2 = 0.1.$$

Using the alternating series: $$\ln(1.1) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(0.1)^n}{n}.$$

Since this is an alternating series with decreasing terms, the error after $N$ terms is bounded by the magnitude of the next term: $$\left| R_N \right| \le \frac{(0.1)^{N+1}}{N+1}.$$

We test values: $$\frac{(0.1)^3}{3} \approx 0.00033 > 10^{-4}, \quad \frac{(0.1)^4}{4} = 0.000025 < 10^{-4}.$$

Thus, using the first four terms guarantees the desired accuracy.

$$\boxed{ \ln(1.1) \approx 0.1 - \frac{0.1^2}{2} + \frac{0.1^3}{3} - \frac{0.1^4}{4} }$$