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Consider

$(x + y + xy)/2 = f( f^{[-1]}(x) + f^{[-1]}(y) )$

Where $f^{[-1}]$ denotes the functional inverse of $f$.

How to find $f$ ?

How about the more General idea of finding $f$ for a given $g$?

$G(x,y) = f( f^{[-1]}(x) + f^{[-1]}(y) )$

I assume partial derivatives could help. (But ?) Im looking for An easy way.

I assume using Taylor series ?

How do I estimate upper and lower bounds for $f$ ?? Or asymptotic ?

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  • $\begingroup$ Do you know if such an $f$ exists? $\endgroup$ – Antonio Vargas Jun 9 '16 at 20:27
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$(x + y + xy)/2 = f( f^{[-1]}(x) + f^{[-1]}(y) )\Rightarrow f^{[-1]}\left(\frac{x + y + xy}{2}\right) = f^{[-1]}(x) + f^{[-1]}(y) $

now assume $h(x)=f^{[-1]}(x)$

$h\left(\frac{x + y + xy}{2}\right) = h(x) + h(y)$ ----[1]

using partial differentiation with respect to $x$

$h'\left(\frac{x + y + xy}{2}\right).\left(\frac{1+y}{2}\right) = h'(x) $

put $x=0$

$h'\left(\frac{y }{2}\right).\left(\frac{1+y}{2}\right) = h'(0) $

$h'\left(\frac{y }{2}\right) = h'(0)\left(\frac{2}{1+y}\right) $

let $h'(0)=k$

$h'(x)=\frac{2k}{1+2x}$, now integrating with respect to $x$ both side will give us

$h(x)=k\ln(1+2x) +c$

by---[1] $h(0)=0\rightarrow c=0$

hence $h(x)=k\ln(1+2x)$ still we need one more given condition to find $k$

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  • $\begingroup$ If $h'(y/2) = h'(0) \left(\frac{2}{1+y} \right)$ then setting $y=0$ gives $h'(0) = 2h'(0)$. $\endgroup$ – Antonio Vargas Jun 9 '16 at 20:24
  • $\begingroup$ Although the answer needs work , see comment from Vargas , this is a detail issue. This is the best answer so far , hence accept. $\endgroup$ – mick Aug 5 '16 at 11:01
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Substituting $f(x)$ for $x$ and $f(y)$ for $y$ in the original equation we get: $$f(x) + f(y) + f(x) f(y) = 2 f(x+y)$$ Letting $x = 0$ and $y = 0$ in the above equation we have $f(0) = 0$. Now we put $y = 0$ in the above equation and the result is $f(x) = 0$.

For the more general problem, the given equation is equivalent to $$G( f(x) , f(y) ) = f(x+y)$$ Letting $x = 0$ and $y = 0$ you'll find an equation which may help you find $f(0)$. Then you can let $y = 0$ and get an equation for $f(x)$. Assuming differentiability for $f$ and $G$, you can differentiate with respect to $y$ and get $f ^ \prime (y) \cdot \partial_2 G ( f(x) , f(y) ) = f ^ \prime (x + y)$. Now letting $y = 0$ you'll find a differential equation which may help you find $f(x)$.

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