What is taylor 1/(4+x^2),0 ?

The answer to taylor 1/(4+x^2),0 is 1/4-1/16 x^2+1/64 x^4-1/256 x^6+1/1024 x^8+\ldots

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taylor 1/(4+x^2),0

taylor

\frac{1}{4 + x ^{2}}, 0

\frac{1}{4} - \frac{1}{16} x^{2} + \frac{1}{64} x^{4} - \frac{1}{256} x^{6} + \frac{1}{1024} x^{8} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{4} + \frac{0}{1 !} x + \frac{- \frac{1}{8}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{\frac{3}{8}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- \frac{45}{16}}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{\frac{315}{8}}{8 !} x^{8} + \dots

Refine

= \frac{1}{4} - \frac{1}{16} x^{2} + \frac{1}{64} x^{4} - \frac{1}{256} x^{6} + \frac{1}{1024} x^{8} + \dots

\frac{d}{d x} (x^{3} - 4 x + 6)

\int x arctan (5 x) d x

f (x) = 7 x sin (x)

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

y = 2 x^{3} e^{- x}

What is taylor 1/(4+x^2),0 ?
The answer to taylor 1/(4+x^2),0 is 1/4-1/16 x^2+1/64 x^4-1/256 x^6+1/1024 x^8+\ldots