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

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

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taylor 1/(1+x^2),1

taylor

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

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{2} + \frac{- \frac{1}{2}}{1 !} (x - 1) + \frac{\frac{1}{2}}{2 !} (x - 1)^{2} + \frac{0}{3 !} (x - 1)^{3} + \frac{- 3}{4 !} (x - 1)^{4} + \frac{15}{5 !} (x - 1)^{5} + \dots

Refine

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

t an g e n t \frac{4}{5 x + 1}

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

x \to - 5 - lim (3)

y^{^{'}} = 6 - y

d er i v a t i v e y = (3 x^{3} + 7)^{2}

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