They say that if $A = X \times Y$, with $X$ statistically independent of $Y$, then

$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$

I can't understand why that is so geometrically.

If $X$ and $Y$ are interpreted as lengths and $A$ as area, it is pretty easy to understand, geometrically, that

$$\Delta{A} = X\times\Delta{Y} + Y\times\Delta{X} + \Delta{X}\times\Delta{Y}$$

Ignoring the term $\Delta{X}\times\Delta{Y}$ and dividing the both sides by $A$ ($= X \times Y$), that expression becomes

$$\frac{\Delta{A}}{A} = \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$

which is different from

$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$

which looks like a distance calculation. I just can't see how a distance is related to $\Delta{A}$.

Interpreting $A$ as the area of a rectangle in a $XY$ plane, I do see that $\Delta{X}^2+\Delta{Y}^2$ is the how much the distance between two opposite corners of that rectangle varies with changes $\Delta{X}$ in $X$ and $\Delta{Y}$ in $Y$. But $\Delta{A}$ is how much the area, not that distance, would vary.

  • $\begingroup$ I think this is better suited for Cross Validated. Your question does not have any physical content. $\endgroup$
    – Danu
    Feb 4, 2016 at 10:50
  • $\begingroup$ @Danu we do these calculations very frequently on physics, it's good to understand them! $\endgroup$
    – Andrea
    Feb 4, 2016 at 11:03
  • $\begingroup$ @AndreaDiBiagio I know, but the question has no physics content. It is off-topic for the same reason that a question asking for a proof of the Pythagorean theorem would be off-topic (and better suited for Mathematics). $\endgroup$
    – Danu
    Feb 4, 2016 at 11:04
  • $\begingroup$ I disagree. The theory of errors is about physical quantities. I considered posting the question in Mathematics, but I realized that this subject is far more relevant to physicists than to mathematicians. $\endgroup$ Feb 4, 2016 at 12:40
  • $\begingroup$ Anyway, I could post it in Mathematics too (not sure it's allowed), for the case you decide to close the question here. $\endgroup$ Feb 4, 2016 at 12:46

4 Answers 4


The general formula for error propagation is:

$$\Delta f(x_1,x_2,\ldots)=\sqrt{(\frac{\partial f}{\partial x_1}\Delta x_1)^2 + (\frac{\partial f}{\partial x_2}\Delta x_2)^2 + \cdots}$$ where $\Delta m$ means "standard deviation of lots of repeated measurements of m".

Where does this come from? By calculus, when all the $x_i$s vary, it causes the following variation of $f$: $$\delta f = \sum_i (\partial f / \partial x_i) \delta x_i$$ where $\delta x_i$ is the difference between this particular measurement of $x_i$ and its true value, and $\delta f$ is ditto for $f$. We are assuming that the errors are relatively small (ignore $\delta x_i \delta x_j$ terms etc.)

I think you had all this so far. The part that you're missing is:

For independent random processes, the variance of the sum is the sum of the variances.

The analogous statement is not true for standard deviations. It is only true for variance, i.e. standard deviation squared.

Since we want the standard deviation of $\delta f$, we need to add up the variances of $(\partial f / \partial x_i) \delta x_i$ and then take the square root. So we wind up with the formula that I wrote at the beginning.

  • $\begingroup$ Sorry, I don't get where the variance of the sum enters here. The independent random processes are $x_1$, $x_2$, ... , so the "sum" is presumably $x_1+x_2+...$. Or are you referring to the sum of the terms in $\Delta{f} = (\partial f / \partial x_i) \Delta x_i$ ? If so, I don't get what the variance of those terms, or the variance of $\Delta{f}$, would mean. If the variances are the $\Delta$'s squared themselves, so your theorem would rather imply that $(\Delta{f})^2 = (\Delta{x_1})^2 + (\Delta{x_2})^2 + ... $. $\endgroup$ Feb 4, 2016 at 13:51
  • $\begingroup$ OK. I thought your arguments would only apply if $f = x_1 + x_2 + ...$, but now I understand it from the answer to the other question. The point is that a $\Delta{x}$ is not a maximum error. It's a standard deviation, with possibility of actual values being out its range, so simple arithmetic does not work. $\endgroup$ Feb 4, 2016 at 16:20
  • $\begingroup$ I guess you have a summation implicit in $$\Delta f = (\partial f / \partial x_1) \Delta x_1$$. That is $$\Delta f = \sum_i (\partial f / \partial x_i) \Delta x_i$$. $\endgroup$ Feb 10, 2016 at 5:25
  • $\begingroup$ @LeonardoCastro - Sorry I was speaking loosely and leaving out steps. Thanks for your comments, I just edited, I hope it's better now. $\endgroup$ Feb 10, 2016 at 13:41

The formula $$\frac{\Delta{A}}{A} \approx \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y} $$

is an approximation because you are ignoring $\Delta X$$\Delta Y$

A better approximation would be

$$\Delta A=\frac{\partial A}{\partial X}\Delta X+\frac{\partial A}{\partial Y}\Delta Y$$

Since errors always add we take the absolute magnitude of $\frac{\partial A}{\partial X}$ and $\frac{\partial A}{\partial Y}$, i.e

$$\Delta A=\bigg |\frac{\partial A}{\partial X}\bigg |\Delta X+\bigg |\frac{\partial A}{\partial Y}\bigg |\Delta Y$$

Since it is always tricky do deal with modulus functions, another work around would be squaring individual errors so that they stay positive

$$(\Delta A)^2=\bigg (\frac{\partial A}{\partial X}\bigg)^2(\Delta X)^2+\bigg (\frac{\partial A}{\partial Y}\bigg )^2(\Delta Y)^2$$

$\frac{\partial A}{\partial X}=Y$ and $\frac{\partial A}{\partial Y}=X$

This will give the required form, this is the root mean squared deviation (standard deviation)

  • $\begingroup$ Why do you say errors always add up? you could just take the square and then square root to get the rms without having to take absolute values. $\endgroup$
    – Andrea
    Feb 4, 2016 at 11:14
  • $\begingroup$ @AndreaDiBiagio errors cannot cancel each other if I'm not wrong, so they always add up, and yes absolute values are not taken even it though seems mathematically right, so we always take the RMS $\endgroup$
    – Courage
    Feb 4, 2016 at 11:26
  • $\begingroup$ The "exact formula"? No, this is a linear approximation used for uncertainties that are normally distributed. $\endgroup$
    – ProfRob
    Feb 4, 2016 at 11:55
  • $\begingroup$ @RobJeffries The OP has given that X is independent of Y, So why wouldnt this be an exact formuia? $\endgroup$
    – Courage
    Feb 4, 2016 at 12:00
  • $\begingroup$ So, squaring all terms is a workaround, kind of a tricky, for not to deal with modulus? The square of the sum isn't the sum of the squares, generally. $\endgroup$ Feb 4, 2016 at 13:57

It looks like Pythagoras, but it is only remotely related. The important concept, as presented in SteveB's answer, is that the variables are considered to be independent, i.e. one does not affect the other. In mathematics, independent parameters are said to be orthogonal , and can thus be assigned to separate axes in Cartesian N-space. It just so happens that the root-sum-square error turns out to be the diagonal of the N-cube (or rectangle in 2-D), which matches Pythagoras' trigonometry theorem.


The square root is there as a better estimator of the error than just adding the errors together. If you add the errors together you are finding the maximum possible error which will happen when both quantities are a maximum(or minimum) together. This is an unlikely event compared with all the other domination of errors. The square root formula you you quote has been deduced by statisticians assume, I think, a normal distribution of random errors. You might look up "theory of errors" to get a more detail answer to your question.


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