What is maclaurin (1+x)^{-7/3},4 ?

The answer to maclaurin (1+x)^{-7/3},4 is 1-7/3 x+35/9 x^2-455/81 x^3+1820/243 x^4+\ldots

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maclaurin (1+x)^{-7/3},4

maclaurin

(1 + x)^{- \frac{7}{3}}, 4

1 - \frac{7}{3} x + \frac{35}{9} x^{2} - \frac{455}{81} x^{3} + \frac{1820}{243} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{- \frac{7}{3}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{- \frac{7}{3}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{- \frac{7}{3}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- \frac{7}{3}}{1 !} x + \frac{\frac{70}{9}}{2 !} x^{2} + \frac{- \frac{910}{27}}{3 !} x^{3} + \frac{\frac{14560}{81}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{7}{3} x + \frac{35}{9} x^{2} - \frac{455}{81} x^{3} + \frac{1820}{243} x^{4} + \dots

y^{^{''}} - 64 y = 0

x^{2} - 18, 17 x

\frac{d y}{d x} = y x^{9} sin (x^{10}), y (0) = 6

\frac{d}{d s} (\frac{s}{s ^{2} + 16})

y^{^{'}} + cos (x) y = cos (x)

What is maclaurin (1+x)^{-7/3},4 ?
The answer to maclaurin (1+x)^{-7/3},4 is 1-7/3 x+35/9 x^2-455/81 x^3+1820/243 x^4+\ldots