Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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

share|improve this answer

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

share|improve this answer

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.