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

The answer to taylor 1/(sqrt(1+x^2)) is 1-1/2 x^2+3/8 x^4-5/16 x^6+35/128 x^8+\ldots

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

taylor

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

1 - \frac{1}{2} x^{2} + \frac{3}{8} x^{4} - \frac{5}{16} x^{6} + \frac{35}{128} x^{8} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( \frac{1}{1 + x ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{1 + x ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{1 + x ^{2}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 1}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{9}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 225}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{11025}{8 !} x^{8} + \dots

Refine

= 1 - \frac{1}{2} x^{2} + \frac{3}{8} x^{4} - \frac{5}{16} x^{6} + \frac{35}{128} x^{8} + \dots

f (x) = \frac{x + 2}{x - 2}

\int ln ((3 x)^{2}) d x

f (x) = (x + 3) (5 x^{2} - 7 x)

\int \frac{- 8 x ^{2} + 16 x + 8}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

\int_{1}^{7} x - 1 d x

What is taylor 1/(sqrt(1+x^2)) ?
The answer to taylor 1/(sqrt(1+x^2)) is 1-1/2 x^2+3/8 x^4-5/16 x^6+35/128 x^8+\ldots