# Calculating relative error

I am struggling with a task I've been given. I've been given only basic information about the relative error and it doesn't seem to be enough to help me solve the following problem: I'm supposed to determine relative error of $$q$$, $$q=\frac {aC}{1+bC}$$ Where $$C$$ is from interval $$[10^{-4},0.1]$$ and $$C$$ has a relative error 3% (minimum $$2*10^{-5}$$) . $$a=1, b=1$$ and both $$a$$ and $$b$$ have relative error 10% .

One trick to compute relative error is this formula: $$d\log f=\frac{df}{f}$$ Hence relative error for $$q$$ can be computed as follows: \begin{align*} d\log q &= d\log a+d\log C-d\log(1+bC) \\ &=\frac{da}{a}+\frac{dC}{C}-\frac{bdC+Cdb}{1+bC} \\ &=\frac{da}{a}+\frac{dC}{C}-\frac{\frac{dC}{C}+\frac{db}{b}}{1+\frac{1}{bC}} \\ &=\frac{da}{a}+\left(1-\frac{bC}{1+bC}\right)\frac{dC}{C}-\left(\frac{bC}{1+bC}\right)\frac{db}{b} \end{align*}

So, your relative error for $$q$$ is: $$\frac{\Delta q}{q}=\frac{\Delta a}{a}+\left(\frac{1}{1+bC}\right)\frac{\Delta C}{C}-\left(\frac{bC}{1+bC}\right)\frac{\Delta b}{b}$$

I let you do the numerical application

Some clarifications:

Let the true value of a quantity be $$x$$￼ and the measured or inferred value ￼$$\tilde{x}$$. Then the relative error $$\delta x$$ is defined by $$\delta x=\frac{\Delta x}{x}=\frac{|\tilde{x}-x|}{x}$$ where $$\Delta x=|\tilde{x}-x|$$ is the absolute error

For instance if $$a=2$$ with a relative error error of $$3\%$$ you have $$0.03=\frac{\Delta a}{2} \Rightarrow \Delta a=0.06 \Rightarrow \tilde{a}\in[1.94,2.06]$$ Now if you want to compute the relative error $$\delta q$$, using $$\delta a=10\%$$ around $$a=1$$, $$\delta b=10\%$$ around $$b=1$$ and $$\delta C=3\%$$ $$\delta q = \frac{\Delta q}{q} = 0.1+0.03\left(\frac{1}{1+C}\right)-0.1\left(\frac{C}{1+C}\right)=\frac{0.13}{1+C}$$ Now, it is true that it is strange to give an interval for $$C$$ values (usually one give a point $$(a,b,C)\in\mathbb{R}^3$$ and given small variations, aka our relative errors, $$(\delta a,\delta b,\delta C)\in\mathbb{R}^3$$, we compute the varation of the output function, our $$\delta q$$.

To continue I assume that $$\delta C=3\%$$ is constant for $$C\in[10^{-4},0.1]$$ and compute the corresponding values for $$\delta q$$. You get: $$\delta q\in [0.118182, 0.129987]$$

This is my suggestion.

Another thing, in your question, I do not understand " minimum $$2\times 10^{-5}$$ "

• Thank you for your answer! If I may have a question regarding the formula. I've only ever seen the deltas when calculating errors as a difference of 'the true value' and the value I actually have. Since I don't have anything like that here I presume it is difference of the value I've been given and the same value plus the given error? Also I don't understand how to get the error when q is basically a variable of a function and I can't calculate the error for every C I've been given. Nov 30, 2018 at 17:32
• @Linda see my clarifications. This is how I understand the exercice. I hope it is ok, however do not hesitate to comment/ask for clarifications Nov 30, 2018 at 18:18
• Thank you for the clarifications! I think I understand it now. I would love to explain what the minimum means but I don't know how it is tied to C myself. Dec 2, 2018 at 17:46
• @Linda thanks :) Dec 2, 2018 at 19:19