Find the Taylor expansion of \(f(x) = \ln(2 + x)\) about the origin.
Find the radius of convergence \(R\) of that series.
Use Taylor’s theorem to show that the series converges to \(f(x)\) on \([0, R/2]\).
Solution
1. For \(|x|<2\),\[ \ln(2+x) =\ln 2+\ln\left(1+\frac x2\right) =\ln2+\sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n2^n}. \]
The standard expansion \[ \ln(1+u)=\sum_{n=1}^\infty (-1)^{n+1}\frac{u^n}{n} \] holds for \(|u|<1\). Substituting \(u=x/2\) gives the stated series.
2. The radius of convergence is\[ \boxed{R=2}. \]
For the coefficient \(a_n=(-1)^{n+1}/(n2^n)\), \[ \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right| =\lim_{n\to\infty}2\frac{n+1}{n}=2. \] Hence the power series has radius \(2\).
3. For every integer \(m\ge1\),\[ f^{(m)}(x)=(-1)^{m-1}\frac{(m-1)!}{(2+x)^m}. \]
This follows by differentiating \(f(x)=\ln(2+x)\) repeatedly; the formula is immediate for \(m=1\) and is preserved by one further differentiation.
4. Let \(T_N(x)\) be the Taylor polynomial of degree \(N\) at \(0\). For each \(x\in[0,1]\), Taylor’s theorem gives\[ |f(x)-T_N(x)| \le \frac{x^{N+1}}{(N+1)2^{N+1}} \le \frac1{(N+1)2^{N+1}}. \]
Taylor’s theorem with Lagrange remainder gives some \(\xi\) between \(0\) and \(x\) such that \[ f(x)-T_N(x)=\frac{f^{(N+1)}(\xi)}{(N+1)!}x^{N+1}. \] By 3, \[ |f^{(N+1)}(\xi)|=\frac{N!}{(2+\xi)^{N+1}}\le \frac{N!}{2^{N+1}}, \] because \(0\le\xi\le x\le1\). Substitution yields the displayed bound.
5. Therefore the Taylor series converges to \(f(x)=\ln(2+x)\) for every \(x\in[0,R/2]=[0,1]\).
The bound in 4 tends to \(0\) as \(N\to\infty\), so \(T_N(x)\to f(x)\) for every \(x\in[0,1]\).