Question

what is the mclaurin series expansion of 1/sqrt(1-(x^2))

Answer 1

guess we can grind it til we find it
i cannot think of a snappy shortcut for this one

Answer 2

\[f(0)=1\] so it start with 1
then \(f'(x)=\frac{x}{(1-x^2)^{\frac{3}{2}}}\) making \(c_1=0\)

Answer 3

in fact the function is even, so only even exponents

Answer 4

i cannot think of a snappy way to do this, sorry
i guess you just have to keep taking derivatives
oh wait, maybe there is a quick way!!

Answer 5

\[\frac{1}{\sqrt{1-x^2}}\] is the derivative of \(\sin^{-1}(x)\) so if you know the maclaurin for that one, you can differentiate term by term to get this one