Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ What is the method to calculate the Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ ?
 A: The formula for the cosine of a difference yields
$$
\begin{align}
\cos(\pi/4-y)
&= \frac{1}{\sqrt{2}}\cos(y)+\frac{1}{\sqrt{2}}\sin(y)\\
&=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}(\sin(y)+\cos(y)-1)\\
&=\frac{1}{\sqrt{2}}+x\tag{1}
\end{align}
$$
Noting that $x=\frac{1}{\sqrt{2}}(\sin(y)+\cos(y)-1)$, it is easy to show that 
$$
2\sqrt{2}x+2x^2=\sin(2y)\tag{2}
$$
Now the series for $\sin^{-1}(x)$ can be gotten by integrating the series for $\dfrac{1}{\sqrt{1-x^2}}$. Using the binomial theorem, we get
$$
(1-x^2)^{-\frac{1}{2}}=\sum_{k=0}^\infty\binom{2k}{k}\frac{x^{2k}}{4^k}\tag{3}
$$
Integrating $(3)$, we get
$$
\sin^{-1}(x)=\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}\frac{x^{2k+1}}{4^k}\tag{4}
$$
Combining $(1)$, $(2)$, and $(4)$, we get that
$$
\begin{align}
\cos^{-1}\left(\frac{1}{\sqrt{2}}+x\right)
&=\frac{\pi}{4}-y\\
&=\frac{\pi}{4}-\frac{1}{2}\sin^{-1}(2\sqrt{2}x+2x^2)\\
&=\frac{\pi}{4}-\frac{1}{2}\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}\frac{(2\sqrt{2}x+2x^2)^{2k+1}}{4^k}\\
&=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(\sqrt{2}x+x^2)^{2k+1}\tag{5}
\end{align}
$$
To get $2n$ terms of the Taylor series for $\cos^{-1}\left(\frac{1}{\sqrt{2}}+x\right)$, you only need $n$ terms of $(5)$.
Afterthought:
A nicer series, that doesn't involve all the $\sqrt{2}$s would be
$$
\cos^{-1}\left(\frac{1+x}{\sqrt{2}}\right)=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(x+\tfrac{1}{2}x^2)^{2k+1}
$$
A: As already noted by robjohn, it is nicer to consider $\arccos\left(\frac{1+x}{\sqrt{2}}\right)  = \frac{\pi}{4} + \delta(x)$. By simple differentiation:
$$
   \delta^\prime(x) = - \frac{1}{\sqrt{1-2 x - x^2}} = - \sum_{n=0}^\infty i^n P_n(-i) x^n
$$
The last equality follows from the generating function for the sequence of Legendre polynomials. Hence
$$
  \arccos\left(\frac{1+x}{\sqrt{2}}\right) = \frac{\pi}{4} - \sum_{n=0}^\infty  \frac{i^n P_n(-i)}{n+1} x^{n+1}
$$
Verification:
In[151]:= 
ArcCos[(1 + x)/Sqrt[2]] + O[x]^51 == 
 Pi/4 - Sum[I^n LegendreP[n, -I]/(n + 1) x^(n + 1), {n, 0, 50}] + 
  O[x]^51

Out[151]= True

A: Take a look at List of Maclaurin series of some common functions. And here is how to obtain the Taylor series for $f(x) = \arcsin x$.
