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Prove that arccot x – arctan (1/x) is not constant but has zero derivative

by RoRi
  1. Prove that for x \neq 0 we have

        \[ D \left( \operatorname{arccot} x - \arctan \frac{1}{x} \right) = 0. \]

  2. Prove that there is no constant C \in \mathbb{R} such that for all x \neq 0 we have

        \[ \operatorname{arccot} x - \arctan \frac{1}{x} = C. \]

    Why is this not a violation of the zero-derivative theorem (Theorem 5.2 in Apostol)?

  1. Proof. We can use the formulas for the derivatives of \operatorname{arccot} x and \arctan x (and the chain rule) to compute,

        \begin{align*} D \left( \operatorname{arccot} x - \arctan \frac{1}{x} \right) &= -\frac{1}{1+x^2} - \left( \frac{1}{1+\left( \frac{1}{x} \right)^2} \right) \left(\frac{-1}{x^2}\right) \\[9pt] &= \frac{-1}{1+x^2} + \frac{1}{1+x^2} \\[9pt] &= 0. \qquad \blacksquare \end{align*}

  2. Proof. First, let x = -1. Then,

        \[ \arctan x = \arctan \frac{1}{x} = \arctan(-1) = -\frac{\pi}{4}. \]

    Then, since \operatorname{arccot} x = \frac{\pi}{2} - \arctan x, we have

        \[ \operatorname{arccot} x - \arctan \frac{1}{x} = \frac{\pi}{2} - \arctan x - \arctan \frac{1}{x} = \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{4} = \pi. \]

    Next, let x = 1. Then, \arctan x = \arctan \frac{1}{x} = \arctan 1 = \frac{\pi}{4}. Again, using that \operatorname{arccot} x = \frac{\pi}{2} - \arctan x, we have

        \[ \operatorname{arccot} x - \arctan \frac{1}{x} = \frac{\pi}{2} - \frac{\pi}{4} - \frac{\pi}{4} = 0. \]

    Hence, there is no constant C such that \operatorname{arccot} x - \arctan \frac{1}{x} = C for all x \neq 0. \qquad \blacksquare

    This is not a violation of the zero-derivative theorem since the function \operatorname{arccot} x - \arctan \frac{1}{x} is constant on every open interval on which it is defined. Since it isn’t defined at x = 0, any open subinterval must be a subinterval of only positive or only negative reals. The function is constant on any of these subintervals.

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