If

Then

$d^{n}f/{\mathrm{dx}}^n=(-1{)}^{n+1}[\underset{a=1}{\overset{2n+1}{\Pi }}(a)]\times x^{-(2n+1)/2}/2^n$

From power series,

${\displaystyle f(a+b)=\sum _{n=0}^{\infty }\frac{(-1{)}^{n+1}\underset{a=1}{\overset{2n+1}{\Pi }}a}{2^n(n!)}·b^{\frac{-2n+1}{2}}·m^n}$

Letting $m=-x^2$, and $b=1$, then integrating with respect to $x$, we get

${\displaystyle {\cos }^{-1}(x)=[\sum _{n=0}^{\infty }\frac{-\underset{a=1}{\overset{2n+1}{\Pi }}a}{2n(n!)(2n+1)}·x^{2n+1}]+C}$

By substituting initial conditions ${\cos }^{-1}(0)$, we find that $C=\pi /2$. But this expansion doesn't seem to work; try ${\cos }^{-1}(1/2)$. What have I done wrong?
Thanks,

Brad

Oops, I should write it as phi from a=1 to 2n-1; And, thats not entirely correct either, what I'm trying to write down is the multiplication of odd numbers. Any suggestions on how to do that?

Brad

$\underset{1}{\overset{n}{\Pi }}(2a-1)$

Does that mean that the following expansion is correct?

Therefore your $\Pi$ term is $(2n)!/[2^n\times n!]$.

I've checked it through and I'm pretty sure it's correct. So we can write it out in slightly more economical form:

${\displaystyle {\cos }^{-1}x=\frac{\pi }{2}+\frac{x^3}{3}-\sum _{n=1}^{\infty }\frac{(2n)!x^{2n+1}}{2^{2n}n{!}^2(2n+1)}}$