Taylor Series Math Example 2

Follow the full solution, then compare it with the other examples linked below.

hard

Find the Maclaurin series for

\ln(1+x)

and state the interval of convergence.

1
Derivatives: $f^{(n)}(0) = (-1)^{n-1}(n-1)!$ for $n\geq1$ .
2
Coefficient of $x^n$ : $\frac{(-1)^{n-1}}{n}$ .
3
$\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n$ . Converges on $(-1,1]$ .

Answer

\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots, \quad -1<x\leq1

The pattern in derivatives reveals the series coefficients. At

x=1

: alternating harmonic (converges to

\ln2

); at

x=-1

: harmonic (diverges).

About Taylor Series

A representation of a function as an infinite sum of terms calculated from the function's derivatives at a single point:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

When

a = 0

, it's called a Maclaurin series.

Find the Maclaurin series for [formula] up to the [formula] term.

Write the first four non-zero Maclaurin terms for [formula].

Use three terms of the Maclaurin series for [formula] to approximate [formula].