What is taylor sqrt(2+x),-1 ?

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

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taylor sqrt(2+x),-1

taylor

2 + x, - 1

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\int \frac{( 8 x ^{2} - 5 x + 1 )}{x ^{2} + x} d x

\int \frac{11}{x ^{3} - 27} d x

\int x^{2} (5 x^{3} - 1)^{\frac{3}{2}} d x

\frac{d}{d x} (\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}})

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

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