What is taylor f(x)= 1/(x^2) ?

The answer to taylor f(x)= 1/(x^2) is 1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+\ldots

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taylor f(x)= 1/(x^2)

taylor

f (x) = \frac{1}{x ^{2}}

1 - 2 (x - 1) + 3 (x - 1)^{2} - 4 (x - 1)^{3} + 5 (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- 2}{1 !} (x - 1) + \frac{6}{2 !} (x - 1)^{2} + \frac{- 24}{3 !} (x - 1)^{3} + \frac{120}{4 !} (x - 1)^{4} + \dots

Refine

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

y^{^{''}} + 2 k^{2} y^{^{'}} + k^{4} y = 0

\frac{d}{d x} (\frac{2 x sin ( x )}{2 + cos ( x )})

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

f (x) = (x - 2)^{\frac{1}{2}}, at x = 6

\int 5 sin^{3} (x) d x

What is taylor f(x)= 1/(x^2) ?
The answer to taylor f(x)= 1/(x^2) is 1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+\ldots