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I know that if I am given $$f=a+b$$ then $$\sigma_\text{specific} = \sqrt{\sigma_a^2 + \sigma_b^2} \tag1$$ which is a direct result from the more general $$\sigma_\text{general} = \sqrt{\left( \frac{\partial f}{\partial a} \right)^2 \sigma_a^2 + \left( \frac{\partial f}{\partial b} \right)^2 \sigma_b^2} \tag2$$

I have noticed that Eq. 1 and Eq. 2 only agree with each other when $a \neq b$. For example, let $a=b=50 \pm 4 \implies f = a + b = 2a$. Then, $$\sigma_\text{specific} = \sqrt{4^2 + 4^2} = 5.7$$ according to Eq.1, while according to Eq.2 $$\sigma_\text{general} = \sqrt{2^2\cdot4^2 + 0} = 8$$ So, which one is correct? and what did I do wrong?

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Your result from the general formula is correct (though it doesn't in general include correlations). Specifically, if you have $$ f = a + a $$ The errors on $a$ and $a$ are obviously (completely) correlated. If one of the terms was bigger than expected, so was the other.

If you have $$ f = a + b $$ with independent, uncorrelated errors on $a$ and $b$, the the overall error on $f$ is slightly smaller, as (heuristically) it is less probable to that both $a$ and $b$ were bigger/smaller than expected, as one could be bigger than expected and the other smaller.


The simple rule of adding errors in quadrature for a sum of terms assumes that the errors are not correlated. In general, for a sum of variables $$ y = \sum x_i $$ then including correlations, the variance on $y$ is $$ \sigma^2 = \sum_i \sum_j \sigma_i \sigma_j \rho_{ij} $$ where $\rho_{ij}$ is the correlation coefficient betwen $x_i$ and $x_j$. In the case of uncorrelated variables, $\rho_{ij} = \delta_{ij}$ such that $$ \sigma^2 = \sum_i \sigma_i^2 $$ If the $n$ variables have the same variances (say $\sigma_1$), we find $$ \sigma = \sqrt{n} \sigma_1 $$

On the other hand, in the case of $n$ absolutely correlated variables ($\rho_{ij} = 1$ for all $i$ and $j$) with the same variances ($\sigma_1$), we instead find $$ \sigma = n \sigma_1 $$ Note that this is a factor $\sqrt{n}$ larger than before. You considered the case of $n=2$, resulting in a factor of $\sqrt{2}$ difference (your $8/5.7 \approx \sqrt{2}$).

Note that this analysis of linear functions can be approximately extended to the general non-linear case by Taylor expanding a non-linear to first-order, resulting in your formula involving partial derivatives.

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