What is taylor 1/(x-1),4 ?

The answer to taylor 1/(x-1),4 is 1/3-1/9 (x-4)+1/27 (x-4)^2-1/81 (x-4)^3+1/243 (x-4)^4+\ldots

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taylor 1/(x-1),4

taylor

\frac{1}{x - 1}, 4

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\frac{\partial}{\partial y} (sin (2 x + 3 y))

\int (\frac{4}{x} + 4 x) d x

(arctan (x^{2}))^{^{'}}

\int \frac{x}{( x ^{2} - 9 ) ^{2}} d x

f (x) = \frac{5}{x ^{6}}

What is taylor 1/(x-1),4 ?
The answer to taylor 1/(x-1),4 is 1/3-1/9 (x-4)+1/27 (x-4)^2-1/81 (x-4)^3+1/243 (x-4)^4+\ldots