What is maclaurin-log_{e}(1-x) ?

The answer to maclaurin-log_{e}(1-x) is x+1/2 x^2+1/3 x^3+1/4 x^4+1/5 x^5+\ldots

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maclaurin-log_{e}(1-x)

maclaurin

- lo g_{e} (1 - x)

x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{1}{4} x^{4} + \frac{1}{5} x^{5} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( - lo g _{e} ( 1 - x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( - lo g _{e} ( 1 - x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( - lo g _{e} ( 1 - x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{2}{3 !} x^{3} + \frac{6}{4 !} x^{4} + \frac{24}{5 !} x^{5} + \dots

Refine

= x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{1}{4} x^{4} + \frac{1}{5} x^{5} + \dots

y = 1 cos (2) (x)

e^{t^{2}}

y^{^{''}} + 2 y^{^{'}} + y = 18 e^{- t}

\int e^{x} sec^{2} (e^{x}) d x

y^{^{''}} + 25 y = 20 sin (5 x)

What is maclaurin-log_{e}(1-x) ?
The answer to maclaurin-log_{e}(1-x) is x+1/2 x^2+1/3 x^3+1/4 x^4+1/5 x^5+\ldots