I'm not sure if percent error is what I want here. I know that percent error is defined as

$PE = \frac{\text{experimental}-\text{theoretical}}{\text{theoretical}}*100 $%

...so this won't help with my problem.

I an experimental and a theoretical value. I want to see how close I am to the theoretical value by returning a value between $0$ and $1$ where $1$ means I reached the exact theoretical value and the return value never reaches $0$, it only gets smaller and smaller, meaning that it represents a worse and worse value.

I tried using $\frac{\text{experimental}}{\text{theoretical}}$, but the problem here is that the experimental value can be much larger than the theoretical one and I don't want to the return value to be much greater than $1$.

What formula am I looking for here?

  • $\begingroup$ I don't know is there any generally agreed upon way to do this. If I needed to do something like this I might have a Gaussian distribution of maximum 1 around the theoretical value. $\endgroup$ – mtheorylord Jun 2 '17 at 2:40

A common one is the "Relative Percent Difference," or RPD, used in laboratory quality control procedures. Although you can find many seemingly different formulas, they all come down to comparing the difference of two values to their average magnitude:


This is a signed expression, positive when $x$ exceeds $y$ and negative when $y$ exceeds $x$. Its value always lies between $−2$ and $2$ .

So you can use $$d(x,y)=\left|\frac{Experimental−Theoretical}{|Experimental|+|Theoretical|}\right|,$$ to acheive a value between $0$ and $1$.

Also, a Wikipedia article on Relative Change and Difference observes that $$d_\infty(x,y)=\frac{|x−y|}{max(|x|,|y|)}$$ is frequently used as a relative tolerance test in floating point numerical algorithms.

So you can use $$d(x,y)=\frac{|Experimental−Theoretical|}{max(|Experimental|,|Theoretical|)},$$ to acheive a value between $0$ and $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.