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  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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

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