What is taylor e^{x^4} ?

The answer to taylor e^{x^4} is 1+x^4+1/2 x^8+\ldots

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taylor e^{x^4}

taylor

e^{x^{4}}

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

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( e ^{x^{4}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{x^{4}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{x^{4}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{20160}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Refine

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

x \to 3 lim ((x - 1)^{2})

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 0

x^{^{''}} + 9 x = 27, x (0) = 4, x^{^{'}} (0) = 6

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

n = 1 \sum \infty \frac{2 n}{5 n - 4}