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August 2nd, 2016, 08:30 AM  #1 
Newbie Joined: Aug 2016 From: Greece Posts: 3 Thanks: 0  A Maclaurin expansion of arccos(z) using only nonnegative coefficients
I am trying to express the function $ f(z) = \arccos(z), \lvert z \lvert \leqslant 1 $ with a Maclaurin expansion using only nonnegative coefficients in a compact form as $$ f(z) = \sum_{n=0}^\infty a_nz^n, a_n\geqslant0 \;\;\;(1)$$ From trigonometric identities and standard expansion of arccos/arcsin in infinite series with binomial coefficients it seems easy as $$ \arccos(x) = \frac{\pi}{2}\arcsin(x) = \frac{\pi}{2}  \sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}x ^{2k+1}$$ while substituting $x$ with $z,$ yields $$f(z)=\frac{\pi}{2} + \sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}z ^{2k+1}$$ However, even if this expansion is using only nonnegative coefficients is not exactly as in (1). *It was suggested to me to use the transformation* $n=2k+1 \Rightarrow $ $$ f(z)=\frac{\pi}{2} + \sum_{n=1}^\infty\frac{(n1)!}{2^{n1}(\frac{(n1)!}{2!})^2n}z^{n} = \frac{\pi}{2} + \sum_{n=1}^\infty\frac{2^3}{2^{n}(n1)!n}z^{n} = \frac{\pi}{2} + \sum_{n=1}^\infty\frac{2^{3n}}{n!}z^{n} $$ which is closer but not the same A little more help plz 
August 2nd, 2016, 03:57 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625 
It looks to me there is no problem. All the coefficients in the expression are nonnegative. It so happens that for even powers of z, they are = 0. n=2k+1 description is faulty, since n assumes only odd values. 
August 4th, 2016, 02:31 AM  #3  
Newbie Joined: Aug 2016 From: Greece Posts: 3 Thanks: 0  Quote:
one nitpick: Shouldn't pi/2 somehow be inside the sum (possibly in a different sum via some other transformation) in order to look like the closed form of (1)?  
August 4th, 2016, 03:57 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625  

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$arccosz$, arccosz, coefficients, expansion, inversefunctions, maclaurin, nonnegative, series, taylor 
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