What is taylor sqrt(1+x) ?

The answer to taylor sqrt(1+x) is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots

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taylor sqrt(1+x)

taylor

1 + x

1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{2}}{1 !} x + \frac{- \frac{1}{4}}{2 !} x^{2} + \frac{\frac{3}{8}}{3 !} x^{3} + \frac{- \frac{15}{16}}{4 !} x^{4} + \dots

Refine

= 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

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

x \to 2 lim (\frac{x ^{3} - 8}{x - 2})

\frac{d}{d t} (e^{2 t} cos (t))

y = \frac{ln ( 9 x )}{9 x}

\int_{- 1}^{7} (4 x + 7) - (x^{2} - 2 x) d x

What is taylor sqrt(1+x) ?
The answer to taylor sqrt(1+x) is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots