# Functional equation for tan

If $$f$$ is a differentiable function on $$\mathbb{R}$$ and $$f'(0)=2$$ satisfying $$f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)},$$ then to prove that $$f(x)=\tan 2x$$.

I know that we must prove using the first definition(principle) of differentiation but I am not able to proceed. I got $$f(0)=0$$ and I also proved function is odd.

• $f(x)=2\tan x$ is not defined on the whole of $\mathbb{R}$ (it is undefined at $x=(n+\frac12)\pi$, $n\in\mathbb{Z}$), let alone differentiable. Also, it does not satisfy the function equation. – user10354138 Jun 23 at 5:37
• Ok but how to prove the function is 2tanx (x,y is not of form (2n+1)π/2) – Tarun Elango Jun 23 at 5:40
• “the first definition(principle) of differentiation”? – let's have a breakdown Jun 23 at 6:47
• Lim h-->0 [f(x+h)-f(x)]/h = f'(x) – Tarun Elango Jun 23 at 6:49

As suggested by user marty cohen, I'll expand his hint into a full answer. We have:

$$\begin{array} \nbsp f'(x) &=& \displaystyle\lim_{y\to 0} \dfrac{f(x+y)-f(x)}{y}\\ &=&\displaystyle\lim_{y\to 0} \dfrac{f(x)+f(y)-f(x)[1-f(x)f(y)]}{y[1-f(x)f(y)]}\\ &=&\displaystyle\lim_{y\to 0} \dfrac{f(y)+f(x)^2 f(y)}{y[1-f(x)f(y)]}\\ &=&\displaystyle\lim_{y\to 0} \left(\dfrac{f(y)}{y} \;\cdot \dfrac{1+f(x)^2}{1-f(x)f(y)}\right)\\ &=&f'(0) \cdot(1+f(x)^2)\\ &=&2(1+f(x)^2) \end{array}$$

where in the next-to-last line we use the definition of $$f'(0)$$, as well as continuity of $$f$$ and $$f(0)=0$$.

Now the differential equation

$$y' = 2(1+y^2)$$

has the general solution $$f(x)= \tan(2x+C)$$; the condition $$f(0) = 0$$ forces $$C=0$$, so

$$f(x) = \tan(2x)$$

is the only possible solution for such an $$f$$. However, this $$f$$ is not defined at $$x \in \lbrace \dfrac{k \pi}{2}: k \in \mathbb Z\rbrace$$, so if one is formal and interprets "differentiable on $$\mathbb R$$" as "defined and differentiable on all of $$\mathbb{R}$$", the question as stated has no solution.

• I like this technique of functional equation leading to the function via getting an equation for the derivative. Works for log, power, arctan, probably others. – marty cohen Jun 24 at 4:33

If a solution exists it has to be $$f(x)=\tan(2x)$$, (not $$2\tan\, x$$). Let $$g(x)=\tan^{-1}f(x)$$. Then we get $$g(x+y)=g(x)+g(y)$$ which implies (by continuity) that $$g(x)=cx$$ for some $$c$$. Hence $$f(x)=\tan (cx)$$ and we can find $$c$$ from the fact that $$f'(0)=2$$.

As already pointed out $$tan(2x)$$ is not defined at certain points. This proves that there is no differentiable function on $$\mathbb R$$ satisfying the given equation

• Sorry for the mistake but how to prove the function – Tarun Elango Jun 23 at 6:00
• Using first principle? – Tarun Elango Jun 23 at 6:01
• @TarunElango You cannot do that because there is no solution. – Kavi Rama Murthy Jun 23 at 6:19
• Ok thank you sir – Tarun Elango Jun 23 at 6:22

Find $$(f(x+y)-f(x))/(y)$$ and let $$y \to 0$$. Use $$f(0)=0$$.

• I am unable to proceed after this particular step – Tarun Elango Jun 23 at 5:44
• I tried using f(x+y)=f(x)+f(y)/1-f(x)f(y) then I also tried partial differentiation on the original equation – Tarun Elango Jun 23 at 5:45
• Can you provide some more steps from here? $f'(x)=\lim_{y\to 0} \frac{2f(x)+f(y)-(f(x))^2f(y))}{(1-f(x)f(y))y}$ – Archis Welankar Jun 23 at 5:48
• @ArchisWelankar: Check your computation. Then, using $f'(0)=2$, you should get $f'(x)=2(1+f(x)^2)$. – Torsten Schoeneberg Jun 23 at 6:37
• @ArchisWelankar: $f'(x) = \lim_{y\to 0}\frac{f(x+y)-f(x)}{y} = \lim_{y\to 0}\frac{f(x)+f(y)-f(x)(1-f(x)f(y))}{y(1-f(x)f(y))}=\lim_{y\to 0}\frac{f(y)+f(x)^2f(y)}{y(1+f(x)f(y))}=\lim_{y\to 0}\frac{f(y)}{y}\frac{1+f(x)^2}{1+f(x)f(y)}\stackrel{f(0)=0}=f'(0)(1+f(x)^2)=2(1+f(x)^2)$ – Torsten Schoeneberg Jun 23 at 17:08