What is maclaurin 4(1+e^{-0.3t}) ?

The answer to maclaurin 4(1+e^{-0.3t}) is 8-1.2t+0.18t^2-0.018t^3+0.00135t^4+\ldots

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maclaurin 4(1+e^{-0.3t})

maclaurin

4 (1 + e^{- 0.3 t})

8 - 1.2 t + 0.18 t^{2} - 0.018 t^{3} + 0.00135 t^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 8 + \frac{\frac{d}{d t} ( 4 ( 1 + e ^{- 0.3 t} ) ) ( 0 )}{1 !} t + \frac{\frac{d ^{2}}{d t ^{2}} ( 4 ( 1 + e ^{- 0.3 t} ) ) ( 0 )}{2 !} t^{2} + \frac{\frac{d ^{3}}{d t ^{3}} ( 4 ( 1 + e ^{- 0.3 t} ) ) ( 0 )}{3 !} t^{3} + \dots

Evaluate Derivatives

= 8 + \frac{- 1.2}{1 !} t + \frac{0.36}{2 !} t^{2} + \frac{- 0.108}{3 !} t^{3} + \frac{0.0324}{4 !} t^{4} + \dots

Refine

= 8 - 1.2 t + 0.18 t^{2} - 0.018 t^{3} + 0.00135 t^{4} + \dots

\int \frac{( ln ( x ) ) ^{3}}{2 x} d x

\frac{\partial}{\partial x} (cos (x) sin (x))

(\frac{1}{2}, 7), (3, - \frac{3}{2})

f (t) = \frac{( t ^{2} - 4 t )}{t + 3}

\int \frac{2 x + 3}{x ^{2} + 3 x + 4} d x

What is maclaurin 4(1+e^{-0.3t}) ?
The answer to maclaurin 4(1+e^{-0.3t}) is 8-1.2t+0.18t^2-0.018t^3+0.00135t^4+\ldots