LOG#024. Strange derivative.


I have been fascinated (perhaps I am in love too with it) by Mathematics since I was a child. As a teenager in High School, I was a very curious student ( I am curious indeed yet)  and I tried to understand some weird results I got in the classroom. This entry is devoted to some of those problems that made me wonder and think a lot out of class, at home. It took me some years and to learn complex variable function theory to understand one of the issues I could not understand before I learned complex variable. 3 years to understand a simple derivate! Yes, it is too much time. However, understanding stuff deeply carries time. Sometimes more, sometimes less.

This is the problem. A very simple problem indeed! The following “weird” (real) trigonometric-like function has a null derivative:

    \[ g(x)= \tan^{-1}((1+x)/(1-x))-\tan ^{-1} (x)\]

where the \tan^{-1} (x) is the arctangent function \arctan (x)( the inverse of the tangent function \tan (x)).


    \[ g'(x) = \dfrac{dg}{dx}=\dfrac{\dfrac{(1-x)+(1+x)}{(1-x)^2}}{1+\left(\dfrac{(1+x)}{(1-x)}\right)^2}-\dfrac{1}{1+x^2}\]

    \[ g'(x) = \dfrac{\dfrac{2}{(1-x)^2}}{1+\left(\dfrac{1+x}{1-x}\right)^2}-1/(1+x^2)\]

    \[ g'(x) = \dfrac{2}{\left(1-x\right)^2+\left(1+x\right)^2}-1/(1+x^2)\]

    \[ g'(x) = \dfrac{2}{2+2x^2}-1/(1+x^2)\]

    \[ g'(x) = \dfrac{1}{1+x^2}-\dfrac{1}{1+x^2}\]

    \[ g'(x)=0 \]


Since the derivative g'(x) is zero, the two functions must differ by a constant. We can guess that constant with usual real variable calculus. It’s quite simple:

    \[ g(x)=constant.\rightarrow g(x=0)= constant = \tan^{-1}((1+0)/(1-0))-\tan ^{-1} (0)\]

    \[ g(0)= \tan^{-1}(1)-\tan ^{-1} (0)=\pi /4\]

So, the difference is \pi/ 4. However, this calculation does not explain why the two functions differ by a constant. The secret lies in the complex function origin of the arctangent. In complex variable function theory, it can be proved that

    \[ \tan^{-1}(x)=\dfrac{1}{2i} \ln ((1+ix)/(1-ix))\]


    \[ \tan(x)=y=\dfrac{\left[\dfrac{1}{2i} (\exp (ix)-\exp (-ix))\right]}{\left[\dfrac{1}{2} \left(\exp (ix)+\exp (-ix)\right)\right]}=\dfrac{1}{i}\dfrac{\exp (ix)-\exp (-ix)}{\exp (ix)+\exp (-ix)}\]

then iy=\left(\exp (2ix)-1)/(\exp (2ix)+1\right) and thus \exp (2ix)=(1+iy)/(1-iy) and therefore x=\dfrac{1}{2i} \ln \left((1+iy)/(1-iy)\right) Q.E.D.

Now, we can calculate the functions in terms of complex variables:

    \[ \tan^{-1}\left((1+x)/(1-x)\right)=\arctan \left((1+x)/(1-x)\right)=\dfrac{1}{2i} \ln \left(\dfrac{(1+i\left(\frac{1+x}{1-x}\right)}{(1-i\left(\frac{1+x}{1-x}\right)}\right)\]

We can make some algebra inside the logarithm function to get:

    \[ \arctan \left((1+x)/(1-x)\right)=\dfrac{1}{2i} \ln \left( \dfrac{(1-x)+i(1+x)}{(1-x)-i(1+x)} \right)= \dfrac{1}{2i} \ln \left( \dfrac{(1+ix)+i(1+ix)}{(1-ix)-i(1-ix)} \right)\]

By the other hand, we also have

    \[ \tan^{-1}(x)=\arctan (x)=\dfrac{1}{2i} \ln \left( \dfrac{1+ix}{1-ix} \right)\]


    \[ g(x) = \dfrac{1}{2i} \left( \arctan ((1+x)/(1-x)) - \arctan (x) \right) = \dfrac{1}{2i} \left( \ln \left [ \dfrac{\dfrac{(1+ix)+i(1+ix)}{(1-ix)-i(1-ix)}}{ \dfrac{1+ix}{1-ix}} \right]\right)\]

i.e.,the terms depending on x cancel to get a pure complex number! The number is

    \[ g(x)=number=\dfrac{1}{2i} \ln \left( \dfrac{1+i}{1-i}\right)\]

We have to calculate the logarith of the complex number

    \[ z=\dfrac{1+i}{1-i}=\dfrac{(1+i)(1+i)}{(1-i)(1+i)}=\dfrac{2i}{2}=i=exp(i\pi/2)\]

Then, \ln ( i ) = i\pi/2. Of course, that is we were expecting to get since in this case

    \[ g(x)=\dfrac{1}{2i} \dfrac{i\pi}{2}=\dfrac{\pi}{4}\]

as before! That is, we recover the phase difference we also got with real calculus. The origin of the cancellation was in the complex origin of the arctangent function! Beautiful mathematics! The complex world is fascinating but very enlightening even for real functions!

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