Find the Maclaurin series for $f(x)$ using the definition of a Maclaurin series.
[Assume that $f$ has a power series expansion. Do not show that
$R_n(x) → 0$]
$f(x)=\ln { (1+4x) } $
What I did:
$$f(x)=\ln { (1+4x) } \\ f'(x)=\frac { 4 }{ 1+4x } =4(1+4x)^{ -1 }\\ f''(x)=-16(1+4x)^{ -2 }\\ f'''(x)=128(1+4x)^{ -3 }\\ f^{ (4) }(x)=-1536(1+4x)^{ -4 }$$
I could not figure out, or see what the $n$th derivative is.
I also got that:
$$f(0)=\ln { (1) } =0\\ f'(0)=4\\ f''(0)=-16\\ f'''(0)=128\\ f^{ (4) }(0)=-1536\\ $$
I would appreciate some help/guidance that would help me arrive at the correct solution on my own. I know the Maclaurin series for $f(x)=\ln { (1+x) } $, but I think the $4x$ is throwing me off here.
Best Answer
Observe that $f^{(n)}(x) = \dfrac{(-1)^{n+1}4^n(n-1)!}{(1+4x)^n}$.