homework and exercises - Bug in linear thermal expansion, $L_0$ must be $0$

Assume we change the temperature of an object with negligible size in $2^{nd}$ and $3^{rd}$ dimensions from $T_0$ to $T_1$ to $T_2$, with all of them pairwise different. We choose a substance with coefficient $\alpha \ne 0$.

$\alpha L_0(T_2 - T_0) = L_2 - L_0 = (L_2 - L_1) + (L_1 - L_0) = \alpha L_1(T_2 - T_1) + \alpha L_0(T_1 - T_0)$
$\Leftrightarrow L_1T_2 - L_1T_1 + L_0T_1 - L_0T_2 = 0$
$\Leftrightarrow (T_2 - T_1)(L_1 - L_0) = 0$
$\Leftrightarrow \alpha L_0(T_2 - T_1)(T_1 - T_0) = 0$

$\Rightarrow L_0 = 0$

Where is my mistake?

Answer

In principle, you need to integrate the relevant equation for linear expansion (http://en.wikipedia.org/wiki/Thermal_expansion#Linear_expansion ). You, however, use just a linear approximation and make some conclusions based on the term quadratic in $\triangle T$. So you use the linear approximation beyond the limits of its applicability.