What is maclaurin 2.5(1+e^{-0.1t}) ?

The answer to maclaurin 2.5(1+e^{-0.1t}) is 5-0.25t+0.0125t^2-0.00041…t^3+0.00001…t^4+\ldots

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maclaurin 2.5(1+e^{-0.1t})

maclaurin

2.5 (1 + e^{- 0.1 t})

5 - 0.25 t + 0.0125 t^{2} - 0.00041 \dots t^{3} + 0.00001 \dots t^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 5 + \frac{\frac{d}{d t} ( 2.5 ( 1 + e ^{- 0.1 t} ) ) ( 0 )}{1 !} t + \frac{\frac{d ^{2}}{d t ^{2}} ( 2.5 ( 1 + e ^{- 0.1 t} ) ) ( 0 )}{2 !} t^{2} + \frac{\frac{d ^{3}}{d t ^{3}} ( 2.5 ( 1 + e ^{- 0.1 t} ) ) ( 0 )}{3 !} t^{3} + \dots

Evaluate Derivatives

= 5 + \frac{- 0.25}{1 !} t + \frac{0.025}{2 !} t^{2} + \frac{- 0.0025}{3 !} t^{3} + \frac{0.00025}{4 !} t^{4} + \dots

Refine

= 5 - 0.25 t + 0.0125 t^{2} - 0.00041 \dots t^{3} + 0.00001 \dots t^{4} + \dots

\frac{d ^{2}}{d x ^{2}} (sin (3 x))

\int 3 sin^{3} (x co) s^{7} x d x

\frac{( 5 t + 300 )}{t ^{2}} + 25

\frac{d}{d x} (x^{5} (x^{6} + 1))

\int \frac{3 x ^{2} - 1}{x ^{3} - x + 5} d x

What is maclaurin 2.5(1+e^{-0.1t}) ?
The answer to maclaurin 2.5(1+e^{-0.1t}) is 5-0.25t+0.0125t^2-0.00041…t^3+0.00001…t^4+\ldots