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#### sweatingbear

##### Member

- May 3, 2013

- 91

We have

\(\displaystyle \arctan (x) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} (x - 1)^n \, . \)

Let \(\displaystyle z = x - 1 \iff x = z + 1\). Thus

\(\displaystyle \arctan (z + 1) = \sum_{n=0}^\infty \frac { f^{(n)}(0) }{n!} z^n \, . \)

We are taking the derivatives at \(\displaystyle z = 0\) in the sum in the right-hand side, because \(\displaystyle z = 0\) when \(\displaystyle x = 1\). But it seems we cannot use the Maclaurin series for \(\displaystyle \arctan (z + 1)\) since the argument is not around \(\displaystyle 0\) when \(\displaystyle z\) is around \(\displaystyle 0\).

It seems as if I will have to resort to brute-force derivative computations?