Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$

Question

What is the method to calculate the Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ ?

Answer 1

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 $$ \cos^{-1}\left(\frac{1+x}{\sqrt{2}}\right)=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(x+\tfrac{1}{2}x^2)^{2k+1} $$

Answer 2

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

Answer 3

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

Answer 4

For $n\ge1$, we have \begin{align} \biggl[\arccos\biggl(\frac1{\sqrt{2}\,}+x\biggr)\biggr]^{(n)} &=-\left[\frac1{\sqrt{1-\Bigl(\frac1{\sqrt{2}\,}+x\Bigr)^2}\,}\right]^{(n-1)}\\ &=-\sum_{k=0}^{n-1}\biggl\langle-\frac12\biggr\rangle_k \Biggl[1-\biggl(\frac1{\sqrt{2}\,}+x\biggr)^2\Biggr]^{-1/2-k} B_{n-1,k}\biggl(-2\biggl(\frac1{\sqrt{2}\,}+x\biggr), -2, 0, \dotsc,0\biggr)\\ &\to-\sum_{k=0}^{n-1}\biggl\langle-\frac12\biggr\rangle_k 2^{1/2+k} B_{n-1,k}\biggl(-\frac2{\sqrt{2}\,}, -2, 0, \dotsc,0\biggr), \quad x\to0\\ &=-\sum_{k=0}^{n-1}(-1)^k\frac{(2k-1)!!}{2^k} 2^{1/2+k} (-2)^k B_{n-1,k}\biggl(\frac1{\sqrt{2}\,}, 1, 0, \dotsc,0\biggr)\\ &=-\sum_{k=0}^{n-1}(2k-1)!! 2^{1/2+k} \frac{1}{2^{n-k-1}}\frac{(n-1)!}{k!}\binom{k}{n-k-1}\frac1{2^{(2k-n+1)/2}}\\ &=-\frac{(n-1)!}{2^{n/2}}\sum_{k=0}^{n-1}(2k-1)!! \frac{2^{k+1}}{k!}\binom{k}{n-k-1}, \end{align} where we used the formula \begin{equation}\label{Bell-x-1-0-eq} B_{n,k}(x,1,0,\dotsc,0) =\frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}x^{2k-n}. \end{equation} Consequently, we arrive at \begin{equation} \arccos\biggl(\frac1{\sqrt{2}\,}+x\biggr) =\frac{\pi}4-\sum_{n=1}^\infty \frac{1}{2^{n/2}}\Biggl[\sum_{k=0}^{n-1}(2k-1)!! \frac{2^{k+1}}{k!}\binom{k}{n-k-1}\Biggr]\frac{x^n}{n}. \end{equation}

References

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