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

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

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

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

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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

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Bernoulli, Legendre, Euler; I should sit down and learn more about these polynomials. Well done! (+1) –  robjohn Oct 25 '11 at 5:22