Series expansion of $\ln(1+(1-x)^{1/2})$ I am practicing series expansions by coming up with some expression, trying to do it by myself and then checking myself with wolfram alpha. However, I have some issues with the following example
I need to expand $\ln(1+(1-x)^{1/2})$. My approach was to first expand 
$$(1-x)^{1/2} = 1 -\frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16}+O(x^3)$$
 then I used the expansion of 
$$
\ln(1+v) = v-\frac{v^2}{2}+\frac{v^3}{3}+O(v^3)
$$
 and instead of $v$, I substituted the series expansion of $(1-x)^{1/2}$. The final result I got was $$
\frac{5x}{6}-\frac{x^2}{2}-\frac{5x^3}{36} +O(x^3).$$
 However, in this case Wolfram Alpha alpha gives me the following answer:
 $$
\ln 2-\frac{x}{4}-\frac{3 x^2}{32}-\frac{5 x^3}{96}-\frac{35 x^4}{1024}-\frac{63 x^5}{2560}+O(x^6)
$$
Why is my answer different, since it seems to me my approach is right? 
Thanks for the help!

Your first two expansions are correct, for $x$ and $v$ respectively going to $0$. Substituting, you now obtain
$$
\ln( 1 + \sqrt{1-x} ) = 
\ln( 1 + 1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16} + O(x^3) ) = 
\ln( 2-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16} + O(x^3) )
$$
and here is your issue: the development of $\ln(1+u)$ is not longer around $0$, but around $1$ (since you have $\ln(2+u) = \ln(1+(1+u))$). To deal with this, you can rewrite:
$$
\ln( 2-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16} + O(x^3) )
= \ln 2 + \ln( 1-\frac{x}{4}-\frac{x^2}{16}-\frac{x^3}{32} + O(x^3) )
$$
using $\ln(ab) = \ln a + \ln b$, and proceed.
Upshot: when composing Taylor expansions around $0$, make sure that each step remains a Taylor expansion around $0$. The expansion of the inner function may shift this.
Another similar example: do a Taylor expansion of $e^{\cos x}$ around 0.
