Can anyone please help me with finding Maclaurin series for this
$$f(x) = x^3 \tan^{-1}(2x); \quad |x|<\frac12$$ https://i.stack.imgur.com/bUhxk.jpg
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Digitallis
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Do you remember the Maclaurin series for $\tan^{-1}2x$ (or at least $\tan^{-1}u$)? If not, see e.g. https://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots. – Minus One-Twelfth Jun 03 '20 at 10:42
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Hint: The Taylor series for $arctan(x)$=$\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{2n+1}$ when $|x|\leq{1}$.
Alessio K
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For the geometric series with common ratio $-x^2$ we have:
$$\sum_{n=0}^\infty (-x^2)^n =1-x^2+x^4-x^6+\cdots = \frac{1}{1+x^2}$$
for $|x|<1.$ Integrate
$$C+ \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = \tan^{-1} x.$$
Plug in $x=0$ to discover that $C=0.$ Replace $x$ by $2x$ to get the series for $\tan^{-1} 2x$ then multiply the whole lot by $x^3$.
