What is maclaurin 1/(4+x) ?

The answer to maclaurin 1/(4+x) is 1/4-1/16 x+1/64 x^2-1/256 x^3+1/1024 x^4+\ldots

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maclaurin 1/(4+x)

maclaurin

\frac{1}{4 + x}

\frac{1}{4} - \frac{1}{16} x + \frac{1}{64} x^{2} - \frac{1}{256} x^{3} + \frac{1}{1024} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= \frac{1}{4} + \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}{4} + \frac{- \frac{1}{16}}{1 !} x + \frac{\frac{1}{32}}{2 !} x^{2} + \frac{- \frac{3}{128}}{3 !} x^{3} + \frac{\frac{3}{128}}{4 !} x^{4} + \dots

Refine

= \frac{1}{4} - \frac{1}{16} x + \frac{1}{64} x^{2} - \frac{1}{256} x^{3} + \frac{1}{1024} x^{4} + \dots

z \to 0 lim (\frac{8}{z ^{3}} + \frac{5}{z} + 2)

\frac{d}{d x} (x^{5} - 2 x^{3} + x)

\int (x + 3) 6 x + x^{2} d x

\frac{d}{d t} (1 - t)

\frac{d y}{d x}, y = \frac{( x ^{2} + 2 x )}{4 - 5 x}

What is maclaurin 1/(4+x) ?
The answer to maclaurin 1/(4+x) is 1/4-1/16 x+1/64 x^2-1/256 x^3+1/1024 x^4+\ldots