What is taylor x^{-2},2 ?

The answer to taylor x^{-2},2 is 1/4-1/4 (x-2)+3/16 (x-2)^2-1/8 (x-2)^3+5/64 (x-2)^4+\ldots

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taylor x^{-2},2

taylor

x^{- 2}, 2

\frac{1}{4} - \frac{1}{4} (x - 2) + \frac{3}{16} (x - 2)^{2} - \frac{1}{8} (x - 2)^{3} + \frac{5}{64} (x - 2)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= \frac{1}{4} - \frac{1}{4} (x - 2) + \frac{3}{16} (x - 2)^{2} - \frac{1}{8} (x - 2)^{3} + \frac{5}{64} (x - 2)^{4} + \dots

\int \frac{cos ^{3} ( x )}{sin ^{8} ( x )} d x

y = 7 x \cdot ln (5 x) - 7 x

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

\int (\frac{1}{3 x}) d x

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

What is taylor x^{-2},2 ?
The answer to taylor x^{-2},2 is 1/4-1/4 (x-2)+3/16 (x-2)^2-1/8 (x-2)^3+5/64 (x-2)^4+\ldots