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Michael Hardy
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$$ \begin{align} \arctan(x) & = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) \\ & {} - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots \end{align} $$$$ \begin{align} \arctan(x) = {} & \arctan(1) + \arctan\left(\frac{x-1} 2 \right) \\ & {} - \arctan\left(\frac{(x-1)^2} 4 \right) + \arctan\left(\frac{(x-1)^3} 8 \right) - \cdots \end{align} $$

Is this known?

$$ \begin{align} \arctan(x) & = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) \\ & {} - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots \end{align} $$

Is this known?

$$ \begin{align} \arctan(x) = {} & \arctan(1) + \arctan\left(\frac{x-1} 2 \right) \\ & {} - \arctan\left(\frac{(x-1)^2} 4 \right) + \arctan\left(\frac{(x-1)^3} 8 \right) - \cdots \end{align} $$

Is this known?

typo
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Michael Hardy
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$$ \arctan(x) = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots $$$$ \begin{align} \arctan(x) & = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) \\ & {} - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots \end{align} $$

Is this known?

$$ \arctan(x) = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots $$

Is this known?

$$ \begin{align} \arctan(x) & = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) \\ & {} - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots \end{align} $$

Is this known?

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Michael Hardy
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Sums of arctangents

$$ \arctan(x) = \arctan(1) + \arctan\left(\frac{x-1}{2}\right) - \arctan\left(\frac{(x-1)^2}{4} \right) + \arctan\left(\frac{(x-1)^3}{8}\right) - \cdots $$

Is this known?