What is maclaurin (1/3)^x ?

The answer to maclaurin (1/3)^x is 1-ln(3)x+(ln^2(3))/2 x^2-(ln^3(3))/6 x^3+(ln^4(3))/(24)x^4+\ldots

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maclaurin (1/3)^x

maclaurin

(\frac{1}{3})^{x}

1 - ln (3) x + \frac{ln ^{2} ( 3 )}{2} x^{2} - \frac{ln ^{3} ( 3 )}{6} x^{3} + \frac{ln ^{4} ( 3 )}{24} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

= 1 - ln (3) x + \frac{ln ^{2} ( 3 )}{2} x^{2} - \frac{ln ^{3} ( 3 )}{6} x^{3} + \frac{ln ^{4} ( 3 )}{24} x^{4} + \dots

\int_{0}^{1} 2 x + 1 d x

x \to \infty + lim (25 x^{4} + 9 x - 5 x^{2})

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

\int \frac{e ^{x}}{2} + x x d x

f (x) = - 3 cos (x), at x = - \frac{π}{3}

What is maclaurin (1/3)^x ?
The answer to maclaurin (1/3)^x is 1-ln(3)x+(ln^2(3))/2 x^2-(ln^3(3))/6 x^3+(ln^4(3))/(24)x^4+\ldots