Relative error and measurement precision

[mPlummers]

Homework Statement

I have a data point $x_{true}$, without error. I also have a function $f(x_{true})=y_{true}$. I have to obtain a simulated measurement, adding to this data point a relative error (and so i'll pass from $x_{true}$ to a $x_{measurement}$). The goal is to find the amount of relative error to add to $x_{true}$ to obtain a measurement of $y_{measurement}$ with 1% precision.

Relevant Equations

Relative error: $(y_{true}-y_{measurement})/y_{measurement}$

NOTE: this is a programming exercise (Python).
I started adding to $x_{true}$ an error related to a (for example) 10% relative error, obtaining $x_{measurement}$. Then i computed $y_{measurement}$. To find the precision, i calculated $(y_{true}-y_{measurement})/y_{measurement}$. If it is correct, what i want to do is to use several relative error values, compute the precision and look for the closest to 1%.

 

[jbriggs444]

If I understand the problem, your goal is to be able to produce y values with a 1% relative error. Your challenge is to find a distribution of x values that result in y values that satisfy this criterion.

Question: does the graph of f() have a reasonably constant slope in the neighborhood of $(x_{true},y_{true})$. More specifically in the neighborhood where $.99\ y_{true} < y < 1.01\ y_{true}$?

 

[mPlummers]

Yes, the goal is to find the correct amount of relative error on $x_{meas}$ to obtain a 1% precision on $y_{meas}$. The teacher told us that this is the inverse problem of finding the right instrument (in terms of precision) depending on the relative error you want to obtain. The function is $y=\sqrt{(x_{1}-x_{1,c})^{2}+(x_{2}-x_{2,c})^{2}}$, it returns the radius of a circle, given the center coordinates. One of the coordinate doesn't have any error, while the other one is the $x$ i mentioned in the first post. How can i calculate the precision? I used the same formula of relative error, but I'm not sure.

 

[haruspex]

Yes, the goal is to find the correct amount of relative error on $x_{meas}$ to obtain a 1% precision on $y_{meas}$. The teacher told us that this is the inverse problem of finding the right instrument (in terms of precision) depending on the relative error you want to obtain. The function is $y=\sqrt{(x_{1}-x_{1,c})^{2}+(x_{2}-x_{2,c})^{2}}$, it returns the radius of a circle, given the center coordinates. One of the coordinate doesn't have any error, while the other one is the $x$ i mentioned in the first post. How can i calculate the precision? I used the same formula of relative error, but I'm not sure.

I assume you mean you are given the centre exactly and a circumferential point approximately. This is not quite the same as in post #1 since you now have two uncertain values... or is one coordinate of the point on the circumference known exactly too?

 

[mPlummers]

Exactly, one of the point coordinates has no error, while the other one has a relative error. That's why in the first post i simplified the problem from 2D to 1D . In the second post i wrote the real problem to show the function i have to use.

 

[haruspex]

Exactly, one of the point coordinates has no error, while the other one has a relative error. That's why in the first post i simplified the problem from 2D to 1D . In the second post i wrote the real problem to show the function i have to use.

Ok.
Replace x₂ in your expression for y with x₂(1+ε). Apply a Taylor expansion to get it into the form $\sqrt{(x_{1}-x_{1,c})^{2}+(x_{2}-x_{2,c})^{2}}(1+A\epsilon+...)$.