# Propagation of error for identical values

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?

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 $$8/5.7 \approx \sqrt{2}$$).