f(x):I will wait for none

Algebra Level 5

Let $f:(-1,∞)→(-1,∞)$ be a function satisfying

a) $\frac{f(x)}{x}$ is increasing in each of the intervals $(-1,0)$ and $(0,∞)$ .

b) $f(x+f(y)+xf(y))=y+f(x)+yf(x)$ for all reals $x,y$ in $(-1,∞)$

Find the sum of all possible values of $\sum_{x=0}^{∞}((f(x))^{4}-4\frac{f(x)}{x}-1)$

Give the answer to 4 decimal places

1 solution

Souryajit Roy
Apr 3, 2015

$\frac{f(x)}{x}$ is a increasing function.

So, $f(x)$ can have at most one fixed point in $(-1,0)$ and at most one fixed point in $(0,∞)$ .(?)

Also, $f(x)$ may be equal to $0$ at $x=0$ .

Suppose $u$ be the fixed point in $(-1,0)$

Putting $x=y=u$ in (b),we get

$f(u^{2}+2u)=u^{2}+2u$

One may easily show that $u^{2}+2u$ is in $(-1,0)$

Hence, $u^{2}+2u=u$ .So, $u=0$ or $u=-1$ ,which is impossible.

Hence,there is no fixed point in $(-1,0)$ .Similarly,there is no fixed point in $(0,∞)$ .

So,the only possible fixed point is $0$ .

Putting $x=y$ in (b),

$f(x+f(x)+xf(x))=x+f(x)+xf(x)$

So, $x+f(x)+xf(x)=0$

Therefore, $f(x)=-\frac{x}{1+x}$

See that this function satisfies all the conditions.

One may easily compute the sum which equals $6\zeta(2)-4\zeta(3)+\zeta(4)=6.1437$