0
$\begingroup$

I have this function:

$\rm \mu(R) = \mu_{\rm eff} + \frac{2.5 b_{\eta}}{ln(10)} \left[ \left( \frac{R}{R_{\rm eff}}\right)^{1/ \eta} - 1 \right] $

where there are associated errors with $\rm \mu(R)$, (lets call it $\rm \sigma_{\mu}$), which depends on $\rm R$, $\mu_{\rm eff}$ ($\rm \sigma_{\mu_{eff}}$), and $R_{\rm eff}$ ($\rm \sigma_{R_{eff}}$). Additionally there is the parameter $b_{\eta}$ which does not have associated errors but is related with $\eta$ by $b_{\eta} = 1.9992\eta - 0.3271$.

The idea is to get the associated error of $\eta$. By solving for $\eta$ one gets the equation:

$\rm \eta = \log(R/R_{eff}) \, / \, log[ln(10)(\mu(R) - \mu_{eff})/2.5b_{\eta} + 1]$

Now, by assuming $\rm \sigma_{b_{\eta}} = 0$ and applying propagation of error from multivariable calculus one finds a very "annoying" equation (I wont post it due to its incredible big size). So, although I have the equation which will give me the uncertainty of $\eta$, I'm not quite sure if it is correct, because I'm not sure if I can just assume $0$ uncertainty for $b_{\eta}$ since its a function of $\eta$ itself.

Additionally I have another question. Imagining that the equation for the propagation of error is correct (so that I'm not doing a huge mistake by disregarding $b_{\eta}$), what would be the correct way to proceed from here since $\rm \mu(R)$ is a function of $\rm R$, so there is a different $\rm \sigma_{\mu}$ depending on $\rm R$? What I did was to assume for $\rm R$ and $\rm \mu(R)$ the point where $\rm \sigma_{\mu}$ reaches the maximum, having an estimate for the maximum error of $\eta$. Is this the way?

Thanks in advance! Best regards

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.