Question

find the maclaurin series in closed form of f(x)=1/sqrt(1-x)

Answer 1

mclaurin series is just taylor series expanded at 0
general formula is \[f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{3!}x^3 + ...+\frac{f^n(0)}{n!}x^n + ...\]
so you need successive derivatives evaluated at 0
\[\frac{1}{\sqrt{1-x}}=(1-x)^{-\frac{1}{2}}\]
successive derivatives look like
\[-\frac{1}{2}(1-x)^{-\frac{3}{2}}\]
\[\frac{3}{4}(1-x)^{-\frac{5}{2}}\]
\[-\frac{15}{8}(1-x)^{-\frac{7}{2}}\]
\[\frac{7\times 5 \times 3}{2^4}\]
of course if you plug in 0 you just get the coefficients so they are
\[1\]
\[-\frac{1}{2}\]
\[\frac{3}{4\times 2!}=\frac{3}{2^3}\]
\[-\frac{15}{8\times 3!}=-\frac{5}{2^4}\]
\[\frac{7\times 5\times 3}{2^44!}\]
i think it should be possible to get a formula out of the pattern but be careful