What is maclaurin (1-x+x^2)e^x ?

The answer to maclaurin (1-x+x^2)e^x is 1+1/2 x^2+2/3 x^3+3/8 x^4+2/15 x^5+\ldots

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maclaurin (1-x+x^2)e^x

maclaurin

(1 - x + x^{2}) e^{x}

1 + \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \frac{3}{8} x^{4} + \frac{2}{15} x^{5} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{1}{2 !} x^{2} + \frac{4}{3 !} x^{3} + \frac{9}{4 !} x^{4} + \frac{16}{5 !} x^{5} + \dots

Refine

= 1 + \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \frac{3}{8} x^{4} + \frac{2}{15} x^{5} + \dots

\frac{\partial}{\partial x} (x^{5} - 6 x + 4 y^{3} - y^{2})

(4.6) (3.1)

f (9) = 3.6 x + 500

\frac{\partial}{\partial x} (ln (y) \cdot ln (x^{2} + 1))

\int x (5 x + 1)^{19} d x

What is maclaurin (1-x+x^2)e^x ?
The answer to maclaurin (1-x+x^2)e^x is 1+1/2 x^2+2/3 x^3+3/8 x^4+2/15 x^5+\ldots