What is maclaurin e^{-ax^2},x=0 ?

The answer to maclaurin e^{-ax^2},x=0 is 1-ax^2+(a^2)/2 x^4-(a^3)/6 x^6+(a^4)/(24)x^8+\ldots

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maclaurin e^{-ax^2},x=0

maclaurin

e^{- a x^{2}}, x = 0

1 - a x^{2} + \frac{a ^{2}}{2} x^{4} - \frac{a ^{3}}{6} x^{6} + \frac{a ^{4}}{24} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 2 a}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{12 a ^{2}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 120 a ^{3}}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{1680 a ^{4}}{8 !} x^{8} + \dots

Refine

= 1 - a x^{2} + \frac{a ^{2}}{2} x^{4} - \frac{a ^{3}}{6} x^{6} + \frac{a ^{4}}{24} x^{8} + \dots

(x + 1) \cdot \frac{d y}{d x} + y = ln (x)

\frac{d}{d x} (e^{- 2 x} + e^{3 x})

e^{, at} sin (b \cdot t)

\int_{1}^{5} x^{3} d x

What is maclaurin e^{-ax^2},x=0 ?
The answer to maclaurin e^{-ax^2},x=0 is 1-ax^2+(a^2)/2 x^4-(a^3)/6 x^6+(a^4)/(24)x^8+\ldots