What is taylor 1/(4-x) ?

The answer to taylor 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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taylor 1/(4-x)

taylor

\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 Taylor 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

4 x - 1

y^{^{'}} = \frac{( x - 8 )}{x}

\frac{\partial}{\partial x} (U (x) (1 + \frac{R ( x ) ^{3}}{x ^{3}}))

2 x + y^{2} = 48, x = y

y^{^{'}} + 3 y = 4 sin (e^{3 x})

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