What Is the Point of Finding the Relative Error in a Vector?

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The discussion centers on the concept of relative error in vectors, specifically questioning its utility when individual entry errors may exceed the overall vector error threshold. It highlights that while the maximum norm can provide a general relative error for a vector, it fails to account for discrepancies in individual components. An example illustrates that even if the overall relative error is acceptable, some entries may still exceed the defined error margin. The conversation also touches on the implications of relative error for solutions to linear systems, emphasizing the limitations of bounding individual entry errors based on vector-level assessments. Ultimately, the inquiry seeks clarity on the practical significance of calculating relative error in vectors when it does not guarantee accuracy for each component.
andresc889
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Hi all,

I have a general question about relative error. Suppose that we have a vector of measurements \hat{b}=\left(\hat{b_{1}},\hat{b_{2}},...,\hat{b_{n}}\right). Furthermore, suppose that these measurements are accurate to 10%.

My natural interpretation of this statement is that there is a "true" vector b=\left(b_{1},b_{2},...,b_{n}\right) such that \frac{\left|b_{1}-\hat{b_{1}}\right|}{\left|b_{1}\right|}, \frac{\left|b_{2}-\hat{b_{2}}\right|}{\left|b_{2}\right|}, ..., \frac{\left|b_{n}-\hat{b_{n}}\right|}{\left|b_{n}\right|}≤0.1.

I have seen in the literature that we can use the maximum norm of a vector to define the relative error. So, the relative error in \hat{b} could be defined as \frac{\left\|b-\hat{b}\right\|}{\left\|b\right\|} where \left\|v\right\|=\max\limits_{i} \left|v_i\right|.

The problem that I find with this is the fact that we can't conclude anything about the individual entries from this definition. For example, if b=\left(1,2,3\right) and \hat{b}=\left(1.14,1.9,3.15\right), then \frac{\left\|b-\hat{b}\right\|}{\left\|b\right\|}=\frac{0.15}{3}=0.05≤0.1 which indicates that the relative error in \hat{b} is less than 10%. On the other hand, the relative error in the first entry of \hat{b} is \frac{0.14}{1}=0.14≥0.1.

Now, suppose we solve the systems A\hat{x}=\hat{b} and Ax=b where A is invertible. According to the literature,

\frac{\left\|\hat{x}-x\right\|}{\left\|x\right\|}≤\left\|A^{-1}\right\|\left\|A\right\|\frac{\left\|\hat{b}-b\right\|}{\left\|b\right\|}

Where the norm of a matrix A is defined to be \max\limits_{i} \sum\limits_{j} \left|a_{ij}\right|.

If we know that the relative error in \hat{b} is less than 10%, then we can put a bound on the relative error in \hat{x}:

\frac{\left\|\hat{x}-x\right\|}{\left\|x\right\|}≤0.1\left\|A^{-1}\right\|\left\|A\right\|

But as shown above, this does not put a bound on the relative error in the individual entries of \hat{x}. So my question is, what is the point of finding the relative error in the vector if we cannot use that to put a bound on the relative error of the individual entries? Maybe I'm misinterpreting something here?

Thanks!
 
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For your example, with ##b = <1, 2, 3>## and ##\hat b = <1.14, 1.9, 3.15>## a vector of the absolute values of the relative errors would be ##<\frac {.14} 1, \frac {.1}2, \frac {.15} 3> = < .14, .05, .05>##. The mean of these values is .24/3 = .08 which is less than .1.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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