functional equation

[jagdish]

If $\bf f:R\rightarrow R$ and function $\bf f(x)$ satisfy the condition $\bf f(x)=f(\frac{x}{2})+\frac{x}{2}.f^{'}(x)$. Then $\bf f(x)=$

[prodip]

If $f'(x)$ means the derivative of f(x) then \[f(x)=f( \frac{x}{2})\]

.Am i right?

[Avik Roy]

Rewriting the equation as \[ f'(x) = \frac {f(x) - f \left (\frac {x} {2} \right)} {\frac {x} {2}} \]
This means that the straight line passing through $\left ( x, \frac {x} {2} \right )$ is the same as the slope of the function at $x$. This intuitively reveals that $f(x)$ is linear. For any pair of reals $(a,b)$ \[f(x) = ax + b \] is a valid function.
However, I'm not sure if they are the only ones (though I think that they are!!!)

[Avik Roy]

I guess the following proof should hold.

Lets consider the Mclaurin series expansion $f(x) = g_0 (x) + g_1 (x) + g_2 (x) + ... $ where $g_k (x) = a_k x^k$
Now, the left side of the equation (as written in earlier reply) demands that for all values of $k$ the following relation holds: \[\frac {g_k(x) - g_k \left (\frac {x} {2} \right )} {\frac {x} {2}} = g'_k (x) \]
This equation reduces to- \[a_k x^{k-1} \frac {2^k - 1} {2^{k-1}} = ka_kx^{k-1} \]
The only possible ways are $a_k = 0$ or $ k = 0,1$. This implies that only linear functions are allowed