What is taylor e^θ ?

The answer to taylor e^θ is sum from n=0 to infinity of (x^n)/(n!)

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taylor e^θ

taylor

e^{θ}

n = 0 \sum \infty \frac{x ^{n}}{n !}

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{1}{1 !} θ + \frac{1}{2 !} θ^{2} + \frac{1}{3 !} θ^{3} + \frac{1}{4 !} θ^{4} + \dots

The series sum representation

= n = 0 \sum \infty \frac{x ^{n}}{n !}

\int \frac{( e ^{2 x} + 1 )}{e ^{x}} d x

y^{^{'}} = \frac{1}{y}, y (0) = 1

f (x) = 3 x^{2} + 1

\frac{d}{d x} (ln^{3} (2 x + 5))

x y^{^{'}} + y = x^{4} y^{3}

What is taylor e^θ ?
The answer to taylor e^θ is sum from n=0 to infinity of (x^n)/(n!)