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In my textbook it asks for me to:

Prove that there is no constant $C$ such that $\text{arccot}(x) - \text{arctan}(\frac{1}{x}) = C $ for all $x \ne 0$. Explain why this does not violate the zero-derivative theorem.

But I believe I have found such a $C$, i.e. $C =0$! I even asked WolframAlpha (http://www.wolframalpha.com/input/?i=arccot%28x%29+-+arctan%281%2Fx%29) which corroborates my answer.

This question appears in Apostol's Calculus Volume I, Second Edition: Exersize 6.22-11b

Edit: Mathematica's definition of arccot is different from the one in my textbook. Apostol's arccot maps a real number into $(0, \pi)$ while Mathematica's maps a real number into $(-\pi/2, \pi/2)$ Here they are super-imposed: http://www.wolframalpha.com/input/?i=integral%28-1%2F%281%2Bx%5E2%29%29+%2B+pi%2F2%3B+arccot

Daniel W. Farlow
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Mark
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    The inverse trig functions are multivalued and therefore not uniquely defined, unless a principle value is given, so please clarify how you are defining the two functions. – Ethan Splaver Feb 14 '13 at 23:25
  • @Ethan Usually $\arctan$ denotes the determination which takes values in $(-\pi/2,\pi/2)$, no? But you're right, this is not so clear for arccot. – Julien Feb 14 '13 at 23:31
  • Here: http://www.intmath.com/blog/which-is-the-correct-graph-of-arccot-x/6009 there are two definitions of arccot. Which one are you using? – Julien Feb 14 '13 at 23:32
  • Oh! I had assumed that mathematicians were in agreement. My textbook uses arccot = pi/2 - arctan, which is a different version than Mathematica's according to @julien's link. Thanks! – Mark Feb 14 '13 at 23:39
  • It seems that, at least as Wolfram is graphing it, the reflection about $y = x$ is used...(analytic?)...Even though Wolfram evaluates the difference as 0, and returns "true" to the equality of each expression, it nonetheless produces its indefinite integral as $x(\arccot(x) - \arctan(1/x)) + C, which makes little sense, to have done if the difference is zero: why not return just "C"? – amWhy Feb 14 '13 at 23:39
  • @amWhy When graphing an inverse function, it is always the reflection about $y=x$ that is used, as you know of course. The question was about the range of arccot. – Julien Feb 14 '13 at 23:42
  • @julien Yes, I see that...didn't mean to complicate things. – amWhy Feb 14 '13 at 23:49
  • @amWhy You don't complicate things. Wolfram does... – Julien Feb 14 '13 at 23:52

2 Answers2

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So we'll take your textbook's definition: $$ \mbox{arccot}(x)=\frac{\pi}{2}-\arctan(x). $$

Then $$ \lim_{0^+} \mbox{arccot} (x)=\frac{\pi}{2}=\lim_{0^+} \arctan(1/x) $$ while $$ \lim_{0^-} \mbox{arccot} (x)=\frac{\pi}{2}=-\lim_{0^-} \arctan(1/x). $$ Since its derivative is $0$ on $\mathbb{R}^*$, $$ \mbox{arccot}(x)-\arctan(1/x)=0 $$ for all $x>0$ while $$ \mbox{arccot}(x)-\arctan(1/x)=\pi $$ for all $x<0$.

This does not contradict the zero-derivative theorem because the function is not defined at $0$, so its domain is not connected.

Julien
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This is not a rigorous proof, it is intuitive one.

Let $y = \arctan\left(\dfrac1x\right)$ then $$ \begin{align} \tan(y) &= \frac1x\\ \frac1{\tan(y)} &= x\\ \cot(y) &= x\\ y &= \text{arccot}(x)\\ \arctan\left(\frac1x\right) &= \text{arccot}(x)\\ \arctan\left(\frac1x\right) - \text{arccot}(x)&=0. \end{align} $$

It holds for $x \neq 0, \ x\in\mathbb{R}$.

Tunk-Fey
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