What is maclaurin sqrt(x+1) ?

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

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maclaurin sqrt(x+1)

maclaurin

x + 1

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 Maclaurin Formula

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

x \to - 6 - lim (\frac{- 1}{x + 6})

x - 1

\int e^{x^{3}} x^{2} d x

\int_{0}^{1} (x^{2} - 4 x + 5) d x

n = 1 \sum \infty 2^{- 3 n}

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