Taylor series for $\ln(x+4)$ about $x=-1$ How would you find the Taylor series for $f(x)=\ln(x+4)$ about $x=-1$
I know that you can do it the traditional way by finding all the derivatives of $f$ and finding the series that way, but I was wondering if there is a way to do it using the Maclaurin series for $\frac{1}{1-x}$ or the Maclaurin series for $\ln(1+x)$
Also when I Do it the traditional way I get $f(x)=ln(3)+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)! (x+1)^{n}}{3^n}$ which doesnt seem to be the right answer
 A: You can use the Maclaurin series for $\ln(1+x)$ and plug in $x+3$ for $x$ in the series to get the Maclaurin series for $\ln(4+x)$.
A: HINT
Note that
$$\ln(x+4)=\ln 3 + \ln\left(\frac x 3+ \frac43\right)=\ln 3 + \ln\left(1+\frac x 3+ \frac13\right)$$
and let $y=\frac x 3+ \frac13 \to 0$.
A: And by the traditional way, just for the check: 
$$\sum_{n=0}^N \frac{f^n(a)}{n!}(x-a)^n$$
Now $f(-1)(x+1)^0$ = $\ln 3$
Now $f'(-1) = \frac{1}{-1+4} = \frac {1}{3}$
Now $f''(-1)= \frac{-1}{(x+4)^2}=\frac{-1}{9} $
Hence at $x=-1$
$$\ln(x+4) = \ln 3+ \frac{x+1}{3} - \frac{(x+1)^2}{18}+ \cdots$$
A: Hint:
$$\ln(x+4)=\ln3\Bigl(\frac{x+1}3+1\Bigr)=\ln 3+\ln\Bigl(1+\frac{x+-1}3\Bigr)=\dotsm$$
A: Note that $$4+x = 3+ (x+1) = 3(1+\frac{(x+1)}{3}) \implies $$
$$ \ln(4+x) = \ln 3 + \ln (1+\frac{(x+1)}{3}) $$
You find the Taylor expansion of $$\ln (1+\frac{(x+1)}{3})$$ by substituting $$ \frac{(x+1)}{3}$$ for $x$  in The Taylor Series of $\ln(1+x)$
Finally  remember that there is no factorial in the Taylor Series of $\ln(1+x).$
$$\ln(4+x) = \ln 3 + \sum _1^{\infty} \frac {(-1)^{n+1}{(x+1)}^n}{3^n}$$
$$
