Mutated Arctangent Power Series

Calculus Level 3

Find the Taylor series centered at x = 0 x = 0 for the function f ( x ) = 3 x tan ⁡ − 1 ( 4 x 2 ) f(x) = 3x\tan^{-1}(4x^2)

3 ∑ n = 0 ∞ ( − 1 ) n 4 2 n + 1 x 4 n + 3 2 n + 1 3\sum_{n=0}^{\infty}(-1)^n \frac{4^{2n + 1}x^{4n + 3}}{2n + 1} 3 ∑ n = 0 ∞ ( − 1 ) n 4 2 n + 1 x 4 n + 4 2 n + 1 3\sum_{n=0}^{\infty}(-1)^n \frac{4^{2n + 1}x^{4n + 4}}{2n + 1} 3 ∑ n = 0 ∞ ( − 1 ) n 4 2 n + 1 x 4 n + 2 ( 2 n + 1 ) ! 3\sum_{n=0}^{\infty}(-1)^n \frac{4^{2n + 1}x^{4n + 2}}{(2n + 1)!} 3 ∑ n = 0 ∞ ( − 1 ) n 4 2 n + 1 x 4 n + 2 2 n + 1 3\sum_{n=0}^{\infty}(-1)^n \frac{4^{2n + 1}x^{4n + 2}}{2n + 1}

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Andrew Ellinor by Andrew Ellinor , 33, USA

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1 solution

f ( x ) = 3 x tan ⁡ − 1 ( 4 x 2 ) By Maclaurin series or Taylor series centered at x = 0 = 3 x ∑ n = 0 ∞ ( − 1 ) n ( 4 x 2 ) 2 n + 1 2 n + 1 ⟹ f ( x ) = 3 ∑ n = 0 ∞ ( − 1 ) n 4 2 n + 1 x 4 n + 3 2 n + 1 \begin{aligned} f(x) & = 3x \color{#3D99F6}{\tan^{-1} (4x^2)} & \small \color{#3D99F6}{\text{By Maclaurin series or Taylor series centered at }x=0} \\ & = 3x\color{#3D99F6}{\sum_{n=0}^\infty (-1)^n\frac{(4x^2)^{2n+1}}{2n+1}} \\ \implies f(x) & = \boxed{\displaystyle 3\sum_{n=0}^\infty (-1)^n\frac{4^{2n+1}x^{4n+3}}{2n+1}} \end{aligned}

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