52766812
For \(|x|<1\), consider the power series
\(f(x)=\ln(1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\cdots\).
1. Differentiate the power series term by term and write the first four nonzero terms of the derivative series.
2. Show that the derivative series is geometric and that its sum equals \(f'(x)=\frac{1}{1+x}\).
Hints
- Apply the power rule to each term.
- Find the factor that produces each term from the preceding term.
- Use the infinite geometric-series formula and the condition \(|x|<1\).
Solution
1. Differentiating term by term gives \(1-x+x^2-x^3+\cdots\).
2. This is a geometric series with first term \(a=1\) and common ratio \(r=-x\).
3. Since \(|x|<1\), the series converges and its sum is \(\frac{a}{1-r}=\frac{1}{1-(-x)}=\frac{1}{1+x}\).
4. This agrees with \(f'(x)=\frac{1}{1+x}\).
Answer
1. \(1-x+x^2-x^3+\cdots\)
2. It is geometric with \(a=1\) and \(r=-x\), so its sum is \(\frac{1}{1+x}\).
