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Can we find two functions $f$ and $g$ that are reasonably defined nontrivial (not everywhere zero, $f\neq g$, not linear polynomials) functions such that the following condition is satisfied?

$ f \left(\int_{0}^{t} g(x) \ \text{d}x\right) = g \left(\int_{0}^{t} f(x) \ \text{d}x\right) $

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    Can$x$be complex?2011-08-29

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If there is such a function, it can't be represented by Tailor series:

Given: $f( \left(\int_{0}^{t} g(x) \ \text{d}x\right)) = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right))$

$f( G(t) $ -$ G(0)$ ) = g( F(t) -$ F(0)$ )

$g(x)=g_0+g_1x+g_2x^2+...$

$f(x)=f_0+f_1x+f_2x^2+...$

$G(x)=g_0x+g_1\frac{x^2}2+g_2\frac{x^3}3+...$

$F(x)=f_0x+f_1\frac{x^2}2+f_2\frac{x^3}3+...$

$f_0+f_1(g_0x+...)+f_2(g_0x+...)^2+...=g_0+g_1(f_0x+...)+g_2(f_0x+...)^2+...$

Since there's only one term of $kx^0$ and $kx^1$ on each side, we can conclude their respective constants are equal:

$f_0=g_0,f_1=g_1$

For every power afterward it is possible to substitute every previous set of constants in order to obtain another.

$f_2=g_2, f_3=g_3, ...$ (Hence $f(x)=g(x)$)

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    @Robert: Yes, my comment addressed the first few versions of this post but I removed it shortly after the re-write to the current version happened, as it had no longer anything to do with the answer. Apparently I must have done so just an instant before you posted your comment, because I see that comment only now.2011-08-17