What is taylor e^x,-1 ?

The answer to taylor e^x,-1 is 1/e+1/e (x+1)+1/(2e)(x+1)^2+1/(6e)(x+1)^3+1/(24e)(x+1)^4+\ldots

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taylor e^x,-1

taylor

e^{x}, - 1

\frac{1}{e} + \frac{1}{e} (x + 1) + \frac{1}{2 e} (x + 1)^{2} + \frac{1}{6 e} (x + 1)^{3} + \frac{1}{24 e} (x + 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= \frac{1}{e} + \frac{1}{e} (x + 1) + \frac{1}{2 e} (x + 1)^{2} + \frac{1}{6 e} (x + 1)^{3} + \frac{1}{24 e} (x + 1)^{4} + \dots

\frac{d}{d x} (4 2 x)

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

f (x) = \frac{2 x ^{2}}{x ^{2} - 2 x - 8}

\int \frac{x}{x ^{2} + 2 x} d x

(\frac{t ^{3}}{t ^{8} + 1})^{2}

What is taylor e^x,-1 ?
The answer to taylor e^x,-1 is 1/e+1/e (x+1)+1/(2e)(x+1)^2+1/(6e)(x+1)^3+1/(24e)(x+1)^4+\ldots