What is taylor 1/(x^2),3 ?

The answer to taylor 1/(x^2),3 is 1/9-2/27 (x-3)+1/27 (x-3)^2-4/243 (x-3)^3+5/729 (x-3)^4+\ldots

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taylor 1/(x^2),3

taylor

\frac{1}{x ^{2}}, 3

\frac{1}{9} - \frac{2}{27} (x - 3) + \frac{1}{27} (x - 3)^{2} - \frac{4}{243} (x - 3)^{3} + \frac{5}{729} (x - 3)^{4} + \dots

Solution steps

Apply the Taylor Formula

= \frac{1}{9} + \frac{\frac{d}{d x} ( \frac{1}{x ^{2}} ) ( 3 )}{1 !} (x - 3) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x ^{2}} ) ( 3 )}{2 !} (x - 3)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x ^{2}} ) ( 3 )}{3 !} (x - 3)^{3} + \dots

Evaluate Derivatives

= \frac{1}{9} + \frac{- \frac{2}{27}}{1 !} (x - 3) + \frac{\frac{2}{27}}{2 !} (x - 3)^{2} + \frac{- \frac{8}{81}}{3 !} (x - 3)^{3} + \frac{\frac{40}{243}}{4 !} (x - 3)^{4} + \dots

Refine

= \frac{1}{9} - \frac{2}{27} (x - 3) + \frac{1}{27} (x - 3)^{2} - \frac{4}{243} (x - 3)^{3} + \frac{5}{729} (x - 3)^{4} + \dots

t^{2} y^{^{''}} + 4 t y^{^{'}} - 18 y = 0

\frac{\partial}{\partial x} (- 2 x e^{- 3 x y})

\int 2 x (x^{2} + 15)^{7} d x

\frac{d}{d x} (5 + cos (x))

y^{^{'}} + 2 y = x y^{2}

What is taylor 1/(x^2),3 ?
The answer to taylor 1/(x^2),3 is 1/9-2/27 (x-3)+1/27 (x-3)^2-4/243 (x-3)^3+5/729 (x-3)^4+\ldots