# relative error relation

Let $x$ be a non-null quantity. Let $\hat{x}$ be its approximation. I am trying to find the relation between: $\frac{\left | x-\hat{x} \right |}{\left | x \right |}$ and $\frac{\left | x-\hat{x} \right |}{\left | \hat{x} \right |}$?

According to what I understood, I did the following:

$\frac{\left | x-\hat{x} \right |}{\left | x \right |}=\frac{\left | x-\hat{x} \right |}{\left | \hat{x} \right |}\frac{\left | \hat{x} \right |}{\left | x \right |}$

but, I am not sure if this is how the problem is supposed to be solved. Any ideas? Thanks

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If it is a good approximation, they may be equal to the required precision. You would like $\left | x-\hat{x} \right | \ll |x|$ – Ross Millikan Sep 19 '12 at 4:28
@RossMillikan: I didn't understand what you mean by your comment. Can you elaborate? – C. Lambda Sep 19 '12 at 4:31
Say $x$ is $1001$ and $\hat x$ is $1000$. $\frac 1{1001}$ and $\frac 1{1000}$ are very close to each other. If you are using an approximation, you probably don't care about the difference. – Ross Millikan Sep 19 '12 at 4:38

The question you have is this: "Do I measure relative to the 'ideal' quantity or the 'approximate' quantity?" In the realm of numerical methods, you usually don't have the 'ideal' quantity, or you wouldn't be trying to solve for it. :) What you do is take the last two values of the quantity for the iterative procedure (such as Newton's Method), and measure the relative error that way, with the 'ideal' value being $x$, and the 'approximate' value being $\hat x$. This is done because we assume that the method is convergent, that given an infinite working precision, and infinite iterations, it will give us exactly the right answer. Thus, the idea is that the answers should be getting progressively closer to (hence, "converging") to the right answer. However, when choosing the denominator, if your relative error is very small, the difference in value of the two fractions you have shown is meaningless. What you are really concerned about is how big the difference $\Delta x=x-\hat x$ is relative to either of the values.