Work out the Taylor series for \(f(x) = \ln x\) around \(x=1\) and use it to approximate \(\ln(1.1)\) accurate to 2 decimal places.
Solution
The Taylor series about \(x=1\) is \[ \boxed{\ln x=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n},\qquad |x-1|<1.} \] In particular, \[ \boxed{\ln(1.1)\approx 0.10} \] to two decimal places.
1. Derive the series from the geometric series.
For \(|t|<1\), \[ \frac1{1+t}=\sum_{n=0}^{\infty}(-1)^n t^n. \] The power series converges uniformly on every closed interval \([-r,r]\) with \(0<r<1\), so it may be integrated term by term. Thus, for \(|u|<1\), \[ \ln(1+u) =\int_0^u\frac{dt}{1+t} =\sum_{n=0}^{\infty}(-1)^n\frac{u^{n+1}}{n+1} =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{u^n}{n}. \] Setting \(u=x-1\) gives the displayed Taylor series. Its radius of convergence is one because the geometric series used above has radius one.
2. Evaluate at \(x=1.1\) and control the error.
Putting \(u=0.1\) gives the alternating series \[ \ln(1.1)=0.1-\frac{0.1^2}{2}+\frac{0.1^3}{3}-\cdots. \] Its terms decrease in absolute value to zero. Using the first two terms, \[ S_2=0.1-0.005=0.095. \] The alternating-series remainder satisfies \[ |\ln(1.1)-S_2|\leq\frac{0.1^3}{3}=\frac1{3000}<0.00034. \] Hence \[ 0.09466<\ln(1.1)<0.09534, \] so rounding to two decimal places gives \(0.10\).