arctan⁡(x+1x−1) to power seriesI want to find an expression for arctan⁡(x+1x−1) as a power series, with x0=0, for every x≠1My initial thought was to use the known arctan⁡(x)=∑n=0∞(−1)nx2n+12n+1 but I don't know how to keep going if I replace x with x+1x−1

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arctan⁡(x+1x−1) to power seriesI want to find an expression for arctan⁡(x+1x−1) as a power series, with x0=0, for every x≠1My initial thought was to use the known arctan⁡(x)=∑n=0∞(−1)nx2n+12n+1 but I don&#039;t know how to keep going if I replace x with x+1x−1

zesponderyd · Accepted Answer

Let f(x)=arctan⁡(x+1x−1),  ∀|x|&amp;lt;1It can be easily showed that:f′(x)=11+x2=∑n=0∞(−1)nx2nIntegrating both sides yields that ∃C∈R such that:arctan⁡(1+x1−x)+C=∑n=0∞(−1)nx2n+12n+1Check for f(0) to conclude C and you&#039;re done.

Annie Levasseur · Accepted Answer

Use the identity arctan⁡(x+1x−1)=−arctan⁡(x+11−x)=−(π4+arctan⁡(x))

karton · Accepted Answer

Since ddxarctan⁡(x+1x−1)=−11+x2 you can express the RHS as a power series, and then integrate the result to get the desired series for your original function arctan⁡(x+1x−1)

\arctan (\frac{x+1}{x-1}) to power series I want to find an expression

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