What is taylor 1+x^2 ?

The answer to taylor 1+x^2 is 1+x^2

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taylor 1+x^2

taylor

1 + x^{2}

1 + x^{2}

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Refine

= 1 + x^{2}

\frac{d}{d x} (\frac{e ^{x}}{9 x ^{2}})

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

n = 0 \sum \infty \frac{( - 1 ) ^{n + 1} ( x - 1 ) ^{n + 1}}{n + 1}

\frac{sin ( t )}{4 + 4 cos ( t )}

\int \frac{x ^{4}}{( 1 - x ^{2} ) ^{3}} d x