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Find all real to real function satisfy this functional equation.!

$$f\left(\frac {x+y}{x-y}\right)=\frac {f(x)+f(y)}{f(x)-f(y)}$$

I couldn't get to the final answer but I get $f(0) = 0$ and $f(1) = 1$ and this functional is an odd function

maths lover
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4 Answers4

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It is easy to see that $f(0) = 0, f(1) = 1, f(-x) = -f(x),$ so we assume this.

We note that $$f\left(\frac{x+a x}{x-a x}\right) = \frac{f(x) + f(x a)}{f(x) - f(x a)}.$$ Factoring out by $x$ on the RHS, we see that $f(x a) = C f(x)$ (just solve on the RHS). In fact, if we set $x = 1,$ we see that $$f(a x) = f(a) f(x).$$ In particular, this means that $f(a^2) = f(a)^2,$ so $f$ is positive on $\mathbb{R}_+.$ What is more, by using the original condition, together with positivity of $f$ on positive numbers, we see that $f$ is order-preserving. Restricting to the positive real axis, we see that $f$ is an order-preserving homomorphism of $\mathbb{R}_+,$ and as is easy to see, this means that $f$ is of the form $x\rightarrow x^c,$ for some $c$ (see this discussion). Now, by plugging into the original constraint, it is easy to see that $c=1,$ so $f(x) = x.$

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Igor Rivin
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Easy to see $f(1)=1$ and $f(0)=0$.

Assume $f'(x)$ exists for all $x \in \mathbb{R}$(or Just at some point).

Let $y=ux$, then $f(\dfrac{1+u}{1-u})=\dfrac{f(x)+f(ux)}{f(x)-f(ux)}$.

Differentiate respect to $x$, we get $0=2\dfrac{uf(x)f'(ux)-f'(x)f(ux)}{[f(x)-f(ux)]^2}$, i.e., $uf(x)f'(ux)=f'(x)f(ux)$.

Using $y=ux$, one has $\dfrac{xf'(x)}{f(x)}=\dfrac{yf'(y)}{f(y)}$ for $\forall x,y \in \mathbb{R}$, so $\dfrac{xf'(x)}{f(x)}=C$(a constant). This differential equation yields that $f(x)=ax^C$ for some real number $a$.

Since $f(1)=1$, so $a=1,f(x)=x^C$.

Since $f(0)=0$, so $C>0$.

Let $y=1$, then $(\dfrac{x+1}{x-1})^C=f(\dfrac{x+1}{x-1})=\dfrac{f(x)+1}{f(x)-1}=\dfrac{x^C+1}{x^C-1}$, this holds for all $x$ iff $C=1$.

Hence $f(x)=x$.

Remark: I have no idea if $f'(x)$ does not exist.

Xucheng Zhang
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Here is my solution, somewhat similar to the one posted by Xucheng Zhang ...

Sourya Varenya
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Have you tried finding very simple functions that satisfy this relation? For example, can you find a quadratic function $f(x) = ax^2 + bx +c$ that works? If so, broaden your horizons! If not, what went wrong, and can you fix it?

I have a hunch that such a function must belong to a very simple class of polynomials, but have yet to verify that these are the only functions that can work. I'm not certain that oddness (in full generality) plays much of a role.

Lastly, I don't see any reason why $f(1) = 1$.

pjs36
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  • put y = 0 u will get it..that f(1) =1 – maths lover Jan 12 '14 at 03:41
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    Since $f$ is a odd function $f(0) = 0$ and so $f\left(\frac{1 + 0}{1 - 0}\right) = \frac{f(1) + f(0)}{f(1) - f(0)} = 1$ – Pratyush Sarkar Jan 12 '14 at 03:42
  • You are indeed correct, it seems that there are very few functions that should satisfy this, I made a significant error. – pjs36 Jan 12 '14 at 03:52
  • is there any other method else than differentiation. – maths lover Jan 12 '14 at 14:29
  • I have a feeling there is, but here's (in my opinion) what's really going on. The function $f$, as described, has to be an automorphism of the field of real numbers. There are no nontrivial automorphisms. I would try investigating the properties $f(x+y) = f(x)+f(y)$ and $f(xy) = f(x)f(y)$ separately, to see what each entails (your function must satisfy both). I think together they'll impose some very severe restrictions. – pjs36 Jan 12 '14 at 15:39