# 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

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 $y=\frac x 3+ \frac13 \to 0$.

• what if it was centered at x=a for any a? – Skrrrrrtttt Apr 7 '18 at 18:47
• $$\ln(x+4)=\ln (4+a) + \ln\left(\frac x {4+a}+ \frac4{4+a}\right)=\ln (4+a) + \ln\left(1+\frac {x} {4+a}- \frac {a}{4+a}\right)$$ – gimusi Apr 7 '18 at 18:52

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 $\ln(4+x)$.

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 $x=-1$ $$\ln(x+4) = \ln 3+ \frac{x+1}{3} - \frac{(x+1)^2}{18}+ \cdots$$

Hint: $$\ln(x+4)=\ln3\Bigl(\frac{x+1}3+1\Bigr)=\ln 3+\ln\Bigl(1+\frac{x+-1}3\Bigr)=\dotsm$$

• it is for $x=-1$. – gimusi Apr 7 '18 at 18:46
• @gimusi: Oops! I've fixed it. Thanks for pointing the slip! – Bernard Apr 7 '18 at 18:52

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 $\ln(1+x).$ $$\ln(4+x) = \ln 3 + \sum _1^{\infty} \frac {(-1)^{n+1}{(x+1)}^n}{3^n}$$