If we put $f(x)=\sum_{n=0}^\infty a_nx^n$, then we get $a_n\left({n\over 2}+{1\over 2^n}-1\right)=0$ for all $n\geqslant 0$ For $n\geqslant 2$ the equations yield $a_n=0$. Hence $f(x)=a+bx,a,b\in\mathbb{R}$. Checking shows that these functions satisfy the given equation.

Farenhajt wrote: If we put $f(x)=\sum_{n=0}^\infty a_nx^n$, then we get $a_n\left({n\over 2}+{1\over 2^n}-1\right)=0$ for all $n\geqslant 0$ For $n\geqslant 2$ the equations yield $a_n=0$. Hence $f(x)=a+bx,a,b\in\mathbb{R}$. Checking shows that these functions satisfy the given equation. In condition it's said that $f(x)$ is a function (not a polynomial) so you can't take $f(x)=\sum_{n=0}^\infty a_nx^n$

The sum on the RHS is not a polynomial but a power series.

but $f$ isn't necessarily infinitely differentiable and even if it was, power series don't always converge

I've got two/ three solutions: $f(x)=ax+c$ (including $f(x)=c$), and $f(x)=ax\ln(x/2)-a$. To prove that, show that the function is differentiable three times and get $f''(x)+4xf'''(x)=0$, which is a standard equation and has the solutions $\frac{a}{x}$ and $0$. Integrate to get the results above.

let $h(x)=f''(x)$ it's easy to get $h(x)=4xh'(2x)$ it's trivial that if $h(x)\ne 0$ then h(x)>0 or <0 let us assume h(x)>0. then $h'(2x)>0$ hence h is increasing on $R^+$ by Lagrange theorem $\frac{h(x)}{x}=h'(x_0)<h'(x)<h'(2x)<4h'(2x)$ contradiction! so $h(x)\equiv 0$yielding $f(x)=cx+b$.

It looks lik Farenhajt's approach is correct, but with some more care. First, $f$ actually is infinitely differentiable: $f$ is everywhere well-defined, $f'(x)= \frac 2 x ( f(x)-f ( \frac x 2 ) )$, $f''(x)=\frac 2 x ( f'(x)-f' ( \frac x 2 )-\frac 1 x f(x) -\frac 1 x f ( \frac x 2 ) ) )$ and so on (if $f^{(n)}$ is well defined then also $f^{(n+1)}$ is by induction); moreover, if it does not converge everywhere then we may take some interval where it converges then show that it is linear on the whole real line then extend it with the equation; in fact, for $f(x)=\sum _{n=0} ^\infty a_n x^n$ we must have $a_n ( \frac n 2 +\frac 1 {2^n} -1 )=0$, but for $n \geq 2$ we have $\frac n 2 +\frac 1 {2^n}>\frac n 2 >1$ so $a_n=0$; conversely, it is trivially verified that linear functions work.

Cassius wrote: moreover, if it does not converge everywhere then we may take some interval where it converges Let me remind you that such an interval not always exists. (And besides, to get your series you need higher derivatives at 0, while your formulae have $x$ in the denominator.)

Do you mean, it may be that it does not converge in any point of the real line? Are there really functions, which have an associated series expansion that does not converge anywhere? I see that the derivatives are not well-defined at $x=0$, but it is the only singular point of $f^{(n)}$ for any $n$, so if we get the thesis for the whole real line except zero (expanding the series around any nonzero point we please) we can then extend it with continuity. Or not?

littletush wrote: let $h(x)=f''(x)$ it's easy to get $h(x)=4xh'(2x)$ it's trivial that if $h(x)\ne 0$ then h(x)>0 or <0 let us assume h(x)>0. then $h'(2x)>0$ hence h is increasing on $R^+$ by Lagrange theorem $\frac{h(x)}{x}=h'(x_0)<h'(x)<h'(2x)<4h'(2x)$ contradiction! so $h(x)\equiv 0$yielding $f(x)=cx+b$. I am not sure about this... first, it should be $h(x)=-4xh'(2x)$, even though this is most likely a minor issue. But how does one show that (WLOG, in the nontrivial case) $h(x)>0$ (for all $x>0$, presumably)? Another issue is that the mean value theorem does not necessarily hold if $h$ is not continuous on $[0,x]$ (it might even be not defined at $0$). Can anyone fill in the gaps here? In any case, I think I have a solution, assuming $f$ is differentiable everywhere (which I think the problem statement implies, but I'm not sure about $x=0$). First we show that $f'$ is continuous everywhere. For $x\ne 0$, this is obvious by the continuity of $f$ (of course, $f'$ is in fact differentiable at $x$). To show $f'$ is continuous at $0$, we find by the definition of $f'(0)$ that $f(x)=xf'(0)+f(0)+o(x)$ and observe that \[\lim_{x\to0}f'(x) = \lim_{x\to0}\frac{f(x)-f(x/2)}{x/2} = \lim_{x\to0}f'(0)+o(1) = f'(0),\]as desired. Now observe that for any $x\ne0$, there exists $x'$ strictly between $x/2$ and $x$ satisfying $f'(x')=f'(x)$ (by the mean value theorem). (*) Assume for the sake of contradiction that some $y\ne f'(0)$ lies in the range of $f'$, and take the infimum $m\ge0$ of $S = \{|x|: f'(x)=y\}$. By (*), $S$ has an infinite subsequence of decreasing absolute value tending to $\pm m$ (for a fixed sign), whence $m\in S$ by continuity and $m\ne0$ by assumption. Then if we take $x_0$ of absolute value $m$ such that $f'(x_0)=y$, there exists (again by (*)) $x_0'$ strictly between $x_0/2$ and $x_0$ satisfying $f'(x_0') = f'(x_0) = y$, contradicting the definition of $m$. Thus $f'(x)$ is constant, so $f(x)=ax+b$ for some reals $a,b$; this clearly works.

Cassius wrote: Are there really functions, which have an associated series expansion that does not converge anywhere? Yes, there are. And the situations is even worse than that -- you can construct infinitely differentiable functions whose Taylor series converges everywhere, but does not agree with the function anywhere but one point. For an example, try to construct the Taylor series for $f(x)= \begin{cases} e^{-\frac{1}{x^2}} & x\neq 0 \\ 0 & x=0 \end{cases}$ about $x=0$. In general, the space of analytic functions (those functions that agree with its Taylor series) is much smaller than the space of infinitely differentiable functions, so solutions to this problem that invoke a power series are incomplete.

Farenhajt wrote: If we put $f(x)=\sum_{n=0}^\infty a_nx^n$, then we get $a_n\left({n\over 2}+{1\over 2^n}-1\right)=0$ for all $n\geqslant 0$ For $n\geqslant 2$ the equations yield $a_n=0$. Hence $f(x)=a+bx,a,b\in\mathbb{R}$. Checking shows that these functions satisfy the given equation. why this solution is correct ? why he can consider such funtion ? help please!