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% .


1 Answer 1


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 $2\times 10^{-5}$ "

  • $\begingroup$ 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. $\endgroup$
    – Kri912
    Nov 30, 2018 at 17:32
  • $\begingroup$ @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 $\endgroup$ Nov 30, 2018 at 18:18
  • $\begingroup$ 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. $\endgroup$
    – Kri912
    Dec 2, 2018 at 17:46
  • $\begingroup$ @Linda thanks :) $\endgroup$ Dec 2, 2018 at 19:19

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