A little unsure if the result I got makes sense, so I want to ask here to be sure I am not doing something very silly.
The Maclaurin series is given by $f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^n$
First I need a formula for the $n$th derivative of $x$. The derivative of $\ln(2-x)=(x-2)^{-1}$, and the derivative of $(u)^{-n}=-n(u)^{-n-1}u'$, and $u'=1$ for $u=x-2$ so:
$$f^n(x)=\ln(2-x), (x-2)^{-1}, -(x-2)^{-2}, 2(x-2)^{-3}, -6(x-2)^{-4},…$$
For a Maclaurin series, we are interested in $f^n(0)$:
$$f^n(0)=\ln(2), \frac{1}{(-2)^{1}}, \frac{-1}{(-2)^{2}}, \frac{2}{(-2)^{3}}, \frac{-6}{(-2)^{4}}={(-1)\frac{(n-1)!}{2^{n}}, n=1,2,3… \choose \ln(2), n=0}$$
Plugging into the Maclaurin series $f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^n$
$$f(x)=\ln(2)\frac{1}{0!}*x^0+\sum_{n=1}^{\infty}(-1)\frac{(n-1)!}{2^{n}}\frac{1}{n!}x^n=\ln(2)-\sum_{n=1}^{\infty}\frac{x^n}{n2^{n}}$$
Is this answer sensible? A bit confused since there's a term ($\ln(2)$) that doesn't follow the same formula as the rest and I had to put it "outside" the sum (when you write out the sum, that shouldn't matter I suppose, but I haven't encountered this partiuclar scenario before)
EDIT: $\ln(2-x)$, not $\ln(x-2)$, was the initial fuction
Best Answer
The given answer is not only sensible, it is correct. The "extra" $\ln{2}$ term is part of the solution. Not every function has all its terms in the Maclaurin series fitting a perfect pattern.