What is maclaurin 1.5(1+e^{-0.2t}) ?

The answer to maclaurin 1.5(1+e^{-0.2t}) is 3-0.3t+0.03t^2-0.002t^3+0.0001t^4+\ldots

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maclaurin 1.5(1+e^{-0.2t})

maclaurin

1.5 (1 + e^{- 0.2 t})

3 - 0.3 t + 0.03 t^{2} - 0.002 t^{3} + 0.0001 t^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 3 + \frac{\frac{d}{d t} ( 1.5 ( 1 + e ^{- 0.2 t} ) ) ( 0 )}{1 !} t + \frac{\frac{d ^{2}}{d t ^{2}} ( 1.5 ( 1 + e ^{- 0.2 t} ) ) ( 0 )}{2 !} t^{2} + \frac{\frac{d ^{3}}{d t ^{3}} ( 1.5 ( 1 + e ^{- 0.2 t} ) ) ( 0 )}{3 !} t^{3} + \dots

Evaluate Derivatives

= 3 + \frac{- 0.3}{1 !} t + \frac{0.06}{2 !} t^{2} + \frac{- 0.012}{3 !} t^{3} + \frac{0.0024}{4 !} t^{4} + \dots

Refine

= 3 - 0.3 t + 0.03 t^{2} - 0.002 t^{3} + 0.0001 t^{4} + \dots

x \to 0 lim ((1 - 4 x)^{\frac{6}{x}})

f (x) = 2 cos (3 x) + 3 e^{- 0.25 x}

(x, y) \to (0, 0) lim (0)

\frac{d}{d x} (- \frac{x ^{2}}{4})

\int z cos (yz) d y

What is maclaurin 1.5(1+e^{-0.2t}) ?
The answer to maclaurin 1.5(1+e^{-0.2t}) is 3-0.3t+0.03t^2-0.002t^3+0.0001t^4+\ldots