Calculating the residual for a fixed point problem Lets say we have some fixed point problem $g(x)$, the approximated/computed value is $x_a$, and the exact/correct value of the solution is $x_e$. What is the residual?
I would think that the residual would be the absolute value of the difference between the correct result, $g(x_e)$ and the approximated value, $x_a$. Thus being $|g(x_e) - x_a|$. However, this is not correct, why?
 A: You have $g(x_e) = x_e$, so the quantity $|g(x_e) - x_a|$ is really the absolute value of the error $T-A$, i.e., the difference between your target $T = x_e$ and your approximation $A = x_a$. 
The signed residual is $r = g(x_e) - g(x_a) = x_e - g(x_a)$, the difference between the targeted value of $g$, i.e., $g(x_e)$ and current value of $g$, i.e., $g(x_a)$.
I feel that the difference between the error and the residual is more easily understood in the context of a general non-linear equation $$f(x) = 0.$$ Let $\xi$ denote the solution of interest. Let $x$ be an approximation of $\xi$. Then the error is $e = \xi-x$, whereas the signed residual is $r = -f(x)$.
A favorite example of mine is the case where $R(\theta)$ denotes the range of an artillery shell when the gun is fired using elevation $\theta$. Solving the equation $$R(\theta) = d$$ corresponds to computing an elevation $\theta$ which will land the shell on top of a target at distance $d$ down range. If $\hat{\theta}$ is the computed elevation/firing solution, then the error is $\theta - \hat{\theta}$, but the (signed) residual is $r = d - R(\hat{\theta})$.
In particular, this example shows the relevance of using a signed residual rather than just the absolute value. After all, it is relevant if the shot went too far or fell short of the target.
My choice of sign when defining the error or the residual is entirely one of convenience. It cuts down on the number of minus signs that I have to write, when applying, say, Taylor's theorem and similar results.
