What is taylor f(x)=(x^2+1)*e^{x-1} ?

The answer to taylor f(x)=(x^2+1)*e^{x-1} is 1/e+1/e x+3/(2e)x^2+7/(6e)x^3+(13)/(24e)x^4+\ldots

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taylor f(x)=(x^2+1)*e^{x-1}

taylor

f (x) = (x^{2} + 1) \cdot e^{x - 1}

\frac{1}{e} + \frac{1}{e} x + \frac{3}{2 e} x^{2} + \frac{7}{6 e} x^{3} + \frac{13}{24 e} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{e} + \frac{\frac{1}{e}}{1 !} x + \frac{\frac{3}{e}}{2 !} x^{2} + \frac{\frac{7}{e}}{3 !} x^{3} + \frac{\frac{13}{e}}{4 !} x^{4} + \dots

Refine

= \frac{1}{e} + \frac{1}{e} x + \frac{3}{2 e} x^{2} + \frac{7}{6 e} x^{3} + \frac{13}{24 e} x^{4} + \dots

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

\frac{d}{d x} (\frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}})

s + 1

y^{^{'}} + 2 y = 4 cos (2 x), y (\frac{1}{4} π) = 3

n = 6 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

What is taylor f(x)=(x^2+1)*e^{x-1} ?
The answer to taylor f(x)=(x^2+1)*e^{x-1} is 1/e+1/e x+3/(2e)x^2+7/(6e)x^3+(13)/(24e)x^4+\ldots