# Approximation of linear expansion coefficient

The equation for linear thermal expansion is $$\alpha_L = \frac 1L \frac{dL}{dT}.$$

According to this page, linear expansion can be approximated as $$\frac{\Delta L}{L} \approx \alpha_L \Delta T$$ for approximately constant $$\alpha_L$$ in the range of $$\Delta T$$ and $$\Delta L/L \ll 1$$.

How can this approximation be verified using "formal" mathematics (e.g. Taylor expansion)?

• You say how can if be verified using a Taylor expansion - have you actually tried doing this? Nov 18, 2018 at 13:02
• There is a mean value theorem in calculus that also says almost the same thing. Nov 18, 2018 at 13:04
• @jacob1729 I have tried integrating the equation. Assuming the coefficient to be constant and integrating the equation gives $$\Delta L = L_0 (e^{\alpha_L \Delta T} -1 )$$. But now I don't know how to proceed (Taylor expansion was just a guess). Nov 18, 2018 at 13:12
• @EricDavidKramer Could you elaborate? I wouldn't know how to apply it. Nov 18, 2018 at 13:15

While Thomas Jones' answer is correct I thought it might also be helpful to see that you don't need to integrate to get this result.

The initial ODE is:

$$\alpha = \frac{1}{L} \frac{dL}{dT}$$

The quickest way to see your approximation is to replace: $$dL \to \Delta L , dT \to \Delta T$$ And assuming $$\alpha$$ to be constant gives the result immediately. The formal justification for this is a first order Taylor expansion (which is equivalent to the mean value theorem alluded to in comments):

$$L = L(T_0) + \left(\frac{dL}{dT}\right)_{T_0}(T-T_0) + \dots$$

Then identifying $$T-T_0$$ as $$\Delta T$$ and the derivative as $$L(T_0)\alpha$$ gives the result.

Integrating that expression up there gives you $$\Delta L = L_0 \left( e^{\alpha_L \Delta T} - 1\right)$$, like you got.

Now, the Taylor expansion of $$e^x$$ for small $$x$$ is $$1 + x + O(x^2)$$, so neglecting higher order terms, we can substitute this into the expression above to get $$\Delta L = L_0 \left( 1 + \alpha_L\Delta T - 1 \right) = L_0 \alpha_L\Delta T$$.

Since the change in L is small, $$L_0 \approx L$$, so we can just divide by that and get $$\frac{\Delta L}{L} = \alpha_L \Delta T$$.