What is maclaurin f(x)=arctan(x) ?

The answer to maclaurin f(x)=arctan(x) is sum from n=0 to infinity}(-1)^n(x^{2n+1 of)/(2n+1)

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maclaurin f(x)=arctan(x)

maclaurin

f (x) = arctan (x)

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{24}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 720}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{40320}{9 !} x^{9} + \dots

Refine

= x - \frac{1}{3} x^{3} + \frac{1}{5} x^{5} - \frac{1}{7} x^{7} + \frac{1}{9} x^{9} + \dots

The series sum representation

= n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{2 n + 1}

What is maclaurin f(x)=arctan(x) ?
The answer to maclaurin f(x)=arctan(x) is sum from n=0 to infinity}(-1)^n(x^{2n+1 of)/(2n+1)