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August 2nd, 2016, 07:30 AM   #1
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Question A Maclaurin expansion of arccos(-z) using only non-negative coefficients

I am trying to express the function $ f(z) = \arccos(-z), \lvert z \lvert \leqslant 1 $ with a Maclaurin expansion using only non-negative 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 non-negative coefficients is not exactly as in (1).


*It was suggested to me to use the transformation* $n=2k+1 \Rightarrow $
$$ f(z)=\frac{\pi}{2} + \sum_{n=1}^\infty\frac{(n-1)!}{2^{n-1}(\frac{(n-1)!}{2!})^2n}z^{n} = \frac{\pi}{2} + \sum_{n=1}^\infty\frac{2^3}{2^{n}(n-1)!n}z^{n} = \frac{\pi}{2} + \sum_{n=1}^\infty\frac{2^{3-n}}{n!}z^{n} $$
which is closer but not the same

A little more help plz
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August 2nd, 2016, 02:57 PM   #2
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It looks to me there is no problem. All the coefficients in the expression are non-negative. It so happens that for even powers of z, they are = 0.

n=2k+1 description is faulty, since n assumes only odd values.
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August 4th, 2016, 01:31 AM   #3
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Quote:
Originally Posted by mathman View Post
It looks to me there is no problem. All the coefficients in the expression are non-negative. It so happens that for even powers of z, they are = 0.

n=2k+1 description is faulty, since n assumes only odd values.
thanks for your answer especially for the n=k+1 tip

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)?
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August 4th, 2016, 02:57 PM   #4
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Quote:
Originally Posted by gepgep View Post
thanks for your answer especially for the n=k+1 tip

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)?
$\displaystyle a_0=\frac{\pi}{2}$.
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