Taylor expansion of $\log(3+x)$ Using the standard result of $\log$ find the taylor expansion of $$\log{(3+x)}$$
Now I believe $$\log{(1+x)} = \log{(1+x)} = \sum^{\infty}_{n=1} \frac{(-1)^{n+1}}{n}x^{n}$$
So to find $\log{(3+x)}$ let $y=2+x$ then we are finding $\log{(1+y)}$ which we have a standard result for:
$$\log{(1+y)} = \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}y^{n} = \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}(2+x)^{n}$$ $$ = 2 + x - \frac{(2+x)^{2}}{2} + \frac{(2+x)^{3}}{3} - \ldots$$
but the taylor expansion of $\log{(3+x)}$ is $$\log{(3+x)} = \log{(3)}+\frac{x}{3}-\frac{x^{2}}{18}+\frac{x^{3}}{81} - \ldots$$
which is drastically different to my own, what is wrong and why does this not work?
 A: Write $\displaystyle 3 + x = 3\left(1 + {x \over 3}\right)$. Then 
$$\log(3 + x) = \log 3 + \log\left(1 + {x \over 3}\right) = \log 3 + \frac{x}{3} - \frac{x^2}{2\cdot 3^2} + \frac{x^3}{3 \cdot 3^3} + \cdots$$
A: $|y| = |2+x|$ can be greater than $1$ for $x \approx 0$, so the formula does not hold. To use $1+x$ you make sure $|x| < 1$
A: There is no single Taylor series for a function. A Taylor series always comes with a point for which we have expanded the function about.
The Taylor series for a function $f$ about the point $x=a$ is defined as
$$f(x) = f(a) + \sum_{n=1}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$
where $f^{(n)}$ is the $n$'th derivative of $f$.
The first series you quote is the Taylor series for $f$ about $a=-2$. The second series is the Taylor series about $a=0$.
A: $log(1+x)^{-1}=-log(\frac{1}{1+x})$ which allows you to expand if $|1+x|\ge2$.
A: the series you have written is valid for $-3 < x < -1$ so that $|2+x| < 1$ and the series $\ln(3 + x) = \ln 3 + x/3 - x^2/18+ \cdots $ is valid for $|x| < 3$ both series are valid in their respective domains. even in their common domain of validity, their truncation may give different approximations.
