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$$ f: \mathbb{R} \to \mathbb{R}\qquad \frac{f(x+y)}{x+y} = \frac{f(x)-f(y)}{x-y}, \qquad \forall x,y\in \mathbb{R}, \left|x\right| \neq \left|y\right| $$

Can I prove anything interesting about this function? I need to find it.

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    $\begingroup$ $f(x)=ax^{2}$ is also a solution. $\endgroup$ – Kavi Rama Murthy Mar 6 at 8:26
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    $\begingroup$ And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution. $\endgroup$ – Greg Martin Mar 6 at 8:35
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Since $|x|\not = |y|$, I'll let $x\notin \{-\frac{1}{2}, 0, \frac{1}{2}\}$ for the following substitutions

Observe first that, for $x\to -x$ and $y=x+1$ $$\frac{f(1)}{1}=-\frac{f(-x)-f(x+1)}{2x+1}\iff f(-x)-f(x+1)=-(2x+1)·f(1)\tag{I}$$

Letting $x\to x+1\text{ and }y=x-1$ we obtain $$\frac{f(2x)}{2x}=\frac{f(x+1)-f(x-1)}{2}\iff f(x+1)-f(x-1)=\frac{f(2x)}{x}\tag{2}$$

Finally $x\to x-1$ and $y\to-x$: $$\frac{f(-1)}{-1}=\frac{f(x-1)-f(-x)}{2x-1}\iff f(x-1)-f(-x)=-f(-1)·(2x-1)\tag{3}$$

The trick now is to add $(1), (2)$ and $(3)$ since the LHS turns $0$

$$\begin{array}a &0&=-(2x+1)·f(1)+\frac{f(2x)}{x}-f(-1)·(2x-1)\\ \iff&0&=-2x^2·f(1)-x·f(1)+f(2x)-2x^2·f(-1)+x·f(-1)\\ \iff &-f(2x)&=-2x^2·\big(f(1)+f(-1)\big)-x·\big(f(1)-f(-1)\big)\\ \iff& f(x)& =\frac{1}{2}·x^2·\big(f(1)+f(-1)\big)+\frac{1}{2}·x·\big(f(1)-f(-1)\big) \end{array}$$

Which is basically a quadratic equation in $x$. Can you end it now?

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    $\begingroup$ Beautiful solution! $\endgroup$ – furfur Mar 14 at 5:32
  • $\begingroup$ Thanks ;)...@furfur $\endgroup$ – Dr. Mathva Mar 14 at 6:26
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If $y\to x$ then $$\dfrac{f(2x)}{2x} = f'(x),$$ $$f(2x) = 2x f'(x)\tag1.$$ Let Maclaurin series of $f(x)$ are $$f(x) = c_0 + c_1x+c_2x^2+c_3x^3+c_4x^4+\dots,\tag2$$ then $(1)$ leads to the identity $$c_0 + 2c_1x+4c_2x^2+8c_3x^3+16c_4x^4+\dots = 2x(c_1 + 2c_2x+3c_3x^2+4c_4x^3+\dots),$$ which satisfies iff $$c_0=c_3=c_4=\dots = 0.$$ Easy to check that the function $$f(x)=c_1x+c_2x^2\tag3$$ satisfies to the given equation for $(c_1,c_2)\in\mathbb R^2$.

The Laurent series (with the negative degrees of $x$) leads to the same result.

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See Aczél or Schwaiger, where there is a misprint: $h(x_1,\ldots,x_n)$ should read as $$h(x_1+\ldots+x_n).$$ In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.

Edit The results mentioned concerns different equations. The somehow simpler one $\frac{f(x)-f(y)}{x-y}=g(x+y)$ for $x\not=y$ and certain generalisations.

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  • $\begingroup$ As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate. $\endgroup$ – Servaes Mar 6 at 12:05
  • $\begingroup$ This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0. $\endgroup$ – furfur Mar 6 at 13:18
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    $\begingroup$ First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$. $\endgroup$ – Jens Schwaiger Mar 6 at 17:35
  • $\begingroup$ Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd. $\endgroup$ – furfur Mar 6 at 18:20
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    $\begingroup$ @JensSchwaiger you should delete this "answer" $\endgroup$ – mathworker21 Mar 9 at 17:47

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