What is taylor 1/(1-ax),0 ?

The answer to taylor 1/(1-ax),0 is 1+ax+a^2x^2+a^3x^3+a^4x^4+\ldots

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taylor 1/(1-ax),0

taylor

\frac{1}{1 - a x}, 0

1 + a x + a^{2} x^{2} + a^{3} x^{3} + a^{4} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{a}{1 !} x + \frac{2 a ^{2}}{2 !} x^{2} + \frac{6 a ^{3}}{3 !} x^{3} + \frac{24 a ^{4}}{4 !} x^{4} + \dots

Refine

= 1 + a x + a^{2} x^{2} + a^{3} x^{3} + a^{4} x^{4} + \dots

\frac{d}{d x} (\frac{x ^{3}}{3} - x^{2} - 3 x + 1)

x \to 2 lim (- x^{2} + 5 x - 2)

\frac{x - 3}{x - 3}

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

\int 4 x cos (5 x) d x

What is taylor 1/(1-ax),0 ?
The answer to taylor 1/(1-ax),0 is 1+ax+a^2x^2+a^3x^3+a^4x^4+\ldots