What is maclaurin x^2e^{-x^2} ?

The answer to maclaurin x^2e^{-x^2} is x^2-x^4+1/2 x^6-1/6 x^8+1/24 x^{10}+\ldots

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maclaurin x^2e^{-x^2}

maclaurin

x^{2} e^{- x^{2}}

x^{2} - x^{4} + \frac{1}{2} x^{6} - \frac{1}{6} x^{8} + \frac{1}{24} x^{10} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{360}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 6720}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{151200}{10 !} x^{10} + \dots

Refine

= x^{2} - x^{4} + \frac{1}{2} x^{6} - \frac{1}{6} x^{8} + \frac{1}{24} x^{10} + \dots

x + 3

g (t) = 6 t^{3} cos (t)

f (x) = x y

\int_{1}^{4} x^{\frac{3}{2}} d x

\int sin (3) (x) d x

What is maclaurin x^2e^{-x^2} ?
The answer to maclaurin x^2e^{-x^2} is x^2-x^4+1/2 x^6-1/6 x^8+1/24 x^{10}+\ldots