Answer to Question #238317 in Calculus for moe

Question

Consider the function $y=\ ln(1+x)y= ln(1+x)$ .

Which of the following is the Maclaurin series expansion of the first three terms?

1. $x-\frac{x^2}{2}+\frac{x^3}{6}$
2. $x-\frac{x^2}{3}+\frac{x^3}{2}$
3. $x-\frac{x^2}{2}+\frac{x^3}{3}$
4. $x+\frac{x^2}{2}-\frac{x^3}{3}$

Answer 1

$f(x)= \ln (1+x) \implies f(0)=0\\ f'(x)= \frac{1}{(1+x)}\implies f'(0)=1\\ f''(x)= \frac{-1}{(1+x)^2}\implies f''(0)=-1\\ f'''(x)= \frac{-2}{(1+x)^3}\implies f'''(0)=-1\\ \implies \log (1+x) = 0+x*1+ \frac{x^2}{2!}(-1)+ \frac{x^3}{3!}(2!)+ \frac{x^4}{4!}(-3!)\\ \implies \log (1+x) =x- \frac{x^2}{2!}+ \frac{x^3}{3}- \frac{x^4}{4}+...$