Theoretical justification for the range of validity of the approximation $R(T)\approx R(T_0)[1+\alpha (T-T_0)]$ In the experiment for calibrating a platinum resistance thermometer, we are always approximating the resistance of the platinum thermometer by $$R(T) \approx R_0 (1+\alpha T),$$ taking the reference at $0^o C$ and using Celsius scale.
We measure temperature by a thermometer of least count $0.5^o C$, and resistance of the platinum thermometer by a Carey Foster bridge which is sensitive up to $0.005\Omega$.
We work in the range of $0^o C-100^o C$.
Qusetion: How can we justify (without saying that, experimentally it is observed...) the usage of the approximation for $R(T)$ for this temperature range given the precision of our apparatus?

What I’m basically asking is the justification for ignoring the higher order terms in the Taylor expansion of $R(T)$ at $T_0$ for ranges as large as $100^o C$?
 A: The response of a platinum resistance thermometer (PRT) is better approximated with the so-called Callendar-Van Dusen equation (this is defined in the standard IEC EN 60751):
$R(t) = \begin{cases}
    R_0\left[1+At+Bt^2+C(t-100\,{}^\circ\mathrm{C})t^3\right]& t<0 \\[5mm]
    R_0(1+At+Bt^2)& t\geqslant 0
    \end{cases}$
where $t$ is the Celsius temperature and the nominal values of the coefficients are (a calibration can give more accurate values)
$\begin{aligned}
    &R_0 = 100\,\Omega \\
    &A = 3.9083\times 10^{-3}/{}^\circ\mathrm{C}, \\
    &B = -5.775\times 10^{-7}/{}^\circ\mathrm{C}^2, \\
    &C = -4.183\times 10^{-12}/{}^\circ\mathrm{C}^4,
    \end{aligned}$
For $t$ from $0\,{}^\circ \mathrm{C}$ to $100\,{}^\circ \mathrm{C}$, the response is
$R(t) = R_0(1+At+Bt^2)$
If you use instead the linear approximation $ R(t) = R_0(1+\alpha t)$, you will measure the temperature $t'$ such that (assuming that $R_0$ is the same for the two equations; otherwise, modify accordingly)
$R_0(1+At+Bt^2) = R_0(1+\alpha t')$
The difference $\Delta t = t'-t$ is the measurement error that you get from using the linear approximation. Substituting $t' = t+\Delta t$ in the above equation and simplifying, you get
$At+Bt^2 = \alpha(t+\Delta t)$
from which
$\Delta t = \dfrac{1}{\alpha}\left[(A-\alpha)t+B t^2\right].$
The above equation represents a parabola, and substituting the value of $\alpha$ that you use you can find the maximum error along the whole measurement range. For instance, if you use $\alpha = 3.85\times 10^{-3}/{}^\circ \mathrm{C}$, the nominal average temperature coefficient from $0\,{}^\circ \mathrm{C}$ to $100\,{}^\circ \mathrm{C}$ (according to the EU standards, for a US PRT you may get $\alpha = 3.92\times 10^{-3}/{}^\circ \mathrm{C}$), the error $\Delta t$ is maximum at $t = 50{}^\circ \mathrm{C}$ and $(\Delta t)_\mathrm{max}\approx 0.38{}^\circ \mathrm{C}$. This specific case is represented in the figure below (own picture):

Depending on your application and the accuracy class of the PRT, you can then evaluate whether you can consider the above error negligible or not.
