What is maclaurin 1/(sqrt(4-x)) ?

The answer to maclaurin 1/(sqrt(4-x)) is 1/2+1/16 x+3/256 x^2+5/2048 x^3+35/65536 x^4+\ldots

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maclaurin 1/(sqrt(4-x))

maclaurin

\frac{1}{4 - x}

\frac{1}{2} + \frac{1}{16} x + \frac{3}{256} x^{2} + \frac{5}{2048} x^{3} + \frac{35}{65536} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= \frac{1}{2} + \frac{\frac{d}{d x} ( \frac{1}{4 - x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{4 - x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{4 - x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{2} + \frac{\frac{1}{16}}{1 !} x + \frac{\frac{3}{128}}{2 !} x^{2} + \frac{\frac{15}{1024}}{3 !} x^{3} + \frac{\frac{105}{8192}}{4 !} x^{4} + \dots

Refine

= \frac{1}{2} + \frac{1}{16} x + \frac{3}{256} x^{2} + \frac{5}{2048} x^{3} + \frac{35}{65536} x^{4} + \dots

\frac{d}{d x} (x e^{3 y})

\frac{1}{t} y^{^{'}} = \frac{y}{t ^{2}} - \frac{1}{y}

\frac{\partial}{\partial y} (5 y - 2 x)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 100 cos (8 t)

(2 x - 1) d x + (7 y + 5) d y = 0

What is maclaurin 1/(sqrt(4-x)) ?
The answer to maclaurin 1/(sqrt(4-x)) is 1/2+1/16 x+3/256 x^2+5/2048 x^3+35/65536 x^4+\ldots