What is taylor 2+t+1/t ?

The answer to taylor 2+t+1/t is 4+(t-1)^2-(t-1)^3+(t-1)^4-(t-1)^5+\ldots

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taylor 2+t+1/t

taylor

2 + t + \frac{1}{t}

4 + (t - 1)^{2} - (t - 1)^{3} + (t - 1)^{4} - (t - 1)^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 4 + \frac{0}{1 !} (t - 1) + \frac{2}{2 !} (t - 1)^{2} + \frac{- 6}{3 !} (t - 1)^{3} + \frac{24}{4 !} (t - 1)^{4} + \frac{- 120}{5 !} (t - 1)^{5} + \dots

Refine

= 4 + (t - 1)^{2} - (t - 1)^{3} + (t - 1)^{4} - (t - 1)^{5} + \dots

\int \frac{x + 1}{x ( x - 1 ) ^{2}} d x

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

\frac{d}{d x} (5 cos (5 x))

\frac{d}{d x} (\frac{x ( x + 1 )}{2 x - 1})

\int \frac{1}{x ( x + 1 )} d x

What is taylor 2+t+1/t ?
The answer to taylor 2+t+1/t is 4+(t-1)^2-(t-1)^3+(t-1)^4-(t-1)^5+\ldots