= Temperature perturbation of a Planck spectrum
{title2=$\delta B_\nu/B_\nu=[x/(1-e^{-x})]\,\delta T/T$}
For a <Planck function> expressed as <spectral radiance> per <frequency> interval,
$$
B_\nu(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^x-1},\qquad x=\frac{h\nu}{k_BT},
$$
$h$ is the <Planck constant>, $c$ the <speed of light> and $k_B$ the <Boltzmann constant>. Differentiating at fixed <frequency> gives
$$
\frac{\partial B_\nu}{\partial T}=\frac{B_\nu}{T}\frac{x}{1-e^{-x}},\qquad
B_\nu(T+\delta T)-B_\nu(T)=\frac{\partial B_\nu}{\partial T}\delta T+O(\delta T^2).
$$
The proof uses $\partial_Tx=-x/T$ and the <chain rule>. At low <frequency>, the multiplier tends to one, recovering the <Rayleigh-Jeans law>. At high <frequency> it grows approximately as $x$: the same fractional <temperature> change has a larger fractional brightness effect on the declining spectral tail.
For an original example choose $T=2.725\,\mathrm K$ and $\delta T=1\,\mathrm{mK}$. At $100\,\mathrm{GHz}$, $x=1.76119$, so the first-order fractional residual is $7.80415\times10^{-4}$. At $300\,\mathrm{GHz}$, $x=5.28357$, it is $1.94881\times10^{-3}$. Both spectra are exact <blackbodies>: this residual is a <temperature> mismatch, not evidence of a nonthermal spectrum. Fitting <temperature> before assessing remaining spectral residuals therefore removes a physically distinct degree of freedom. The figure divides by a fixed peak brightness rather than by the rapidly vanishing high-<frequency> tail.
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