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

The answer to taylor sqrt(-x^2+2x),1 is 1-1/2 (x-1)^2-1/8 (x-1)^4-1/16 (x-1)^6+\ldots

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taylor sqrt(-x^2+2x),1

taylor

- x^{2} + 2 x, 1

1 - \frac{1}{2} (x - 1)^{2} - \frac{1}{8} (x - 1)^{4} - \frac{1}{16} (x - 1)^{6} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} (x - 1) + \frac{- 1}{2 !} (x - 1)^{2} + \frac{0}{3 !} (x - 1)^{3} + \frac{- 3}{4 !} (x - 1)^{4} + \frac{0}{5 !} (x - 1)^{5} + \frac{- 45}{6 !} (x - 1)^{6} + \frac{0}{7 !} (x - 1)^{7} + \dots

Refine

= 1 - \frac{1}{2} (x - 1)^{2} - \frac{1}{8} (x - 1)^{4} - \frac{1}{16} (x - 1)^{6} + \dots

(x^{2} + 2 x y - y^{2}) d x + (y^{2} + 2 x y - x^{2}) d y = 0

\int x sin (\frac{x}{2}) d x

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

y^{^{''}} + 4 y^{^{'}} + 16 y = 0

6 ln (x)

What is taylor sqrt(-x^2+2x),1 ?
The answer to taylor sqrt(-x^2+2x),1 is 1-1/2 (x-1)^2-1/8 (x-1)^4-1/16 (x-1)^6+\ldots