2-8
2.4 PROPAGATION OF ERRORS (ONE VARIABLE)
2.4 Propagation of error (One variable)
Very often the result that you want from an experiment is not the directly measured quantity, but must
be derived from one or more measured quantities. We need to know how to get from the uncertainty in
some quantity x to the corresponding uncertainty in another quantity q(x) calculated from x. The fancy
name for this process is "propagation of errors".
As an example, suppose I've measured the edge length of a cube. In order to assess the random
errors in the measurement process, I did it several times, and got
5.18
5.31
5.26
5.16
5.25
5.14 cm
The mean of these measurements is 5.217 cm, with a standard error of .027 cm. If this were our final
result we'd quote it as 5.22 ± .03 cm, but let's hold on to the extra significant figure for a moment, to help
us see what's going on.
Now suppose that what I really want to know is the volume V = L3 of the cube. The best value I can
get from my experimental result is plainly (5.217 cm)3 = 142.0 cm3; but what uncertainty should I
associate with it? Well, if the true length were one standard error higher than the measured value, that
is 5.217 + 0.027 = 5.244 cm, it would make the volume (5.244 cm) 3 = 144.2 cm3; or if it were one
standard error lower, the volume would have come out 139.8 cm 3.
Whenever we put a "+ or -" on an experimental result, what we're doing is identifying a range of
values within which the true value probably lies. Whatever the probability that the true volume of the
cube lies between 139.8 and 144.2 cm 3, in the example above, clearly it's the same as the probability
that the true edge length lies somewhere between 5.190 and 5.244 cm. Since, therefore, the ±0.027 cm
I'm quoting is one standard deviation (of the mean value) of x, the corresponding ±2.2 cm3 is the
standard deviation of the volume. You see that what we're doing is just figuring out how much variation
is caused in a derived quantity q(x) by a given variation in another quantity x, from which q is derived.
 It is always correct to do this by direct calculation, using altered values, as in the
example just done.
If the variations in q and x are not too large, their ratio is approximately equal to the derivative of q
with respect to x. This gives us the rule
sq =
dq
sx
dx
which is the basis of all formal error-propagation calculations. In the example I just went through, q = x3
is the volume of a cube of edge length x. Thus
dq
= 3 q2
dx
and
2
sq  (3 q ) s x
= (3) (5.217 cm)2 (.027 cm) = 2.2 cm3
Note that the result is the same as we got by direct calculation.
(7)
2.4 PROPAGATION OF ERRORS (ONE VARIABLE)
2-9
There are several special cases of (7) that are often useful. For example, if the quantity q is just x
multiplied by a constant, then (7) obviously gives
q=C x
for
sq = C s x
(8a)
n
Another important case is x raised to some power: q = x . Then
dq
= n xn - 1
dx
sq = n x
n - 1
sx
If we use f for the fractional standard deviation, that is,
f x=
sx
,
x
f q=
sq
,
q
etc .
then the case of x raised to a power boils down to
for
q = xn ,
fq = nfx
(8b)
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Example: In the case above, we'd measured x = 5.217 ± 0.027 cm, so
f x=
0.027 cm
= 0.0052
5.217 cm
and
V = x3
thus
3
3
sV = f V V = ( 0.0156 )( 142 cm ) = 2.2 cm
f q = ( 3 ) ( 0.0052 ) = 0.0156
so
and
V = 142 ± 2 cm3
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just as we had before.
In the same way, we can derive rules for logarithmic and exponential functions that are frequently
useful:
for
q = A ln(k x) ,
for
q = A ek x ,
sq = A f x
(8c)
f q = k sx
(8d)
where A and k are constants.
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Example: Suppose we have measured
x = 2.41 ± 0.12, and what we want is
u = 1.5 ln (x)= 1.5 ln (2.41) = 1.319
and
w = 5 e0.3 x = 5 e7.23 = 10.30
What are the errors in these calculated quantities? From (8c)
 0.12 
 = 0.075
su = 1.5 f x = (1.5) 
 2.41 
so
u = 1.32  0.08
2-10
2.4 PROPAGATION OF ERRORS (ONE VARIABLE)
and from (8d), fW = 0.3 sx = (0.3)(0.12) = 0.075
so sw = f w w = ( 0.036 ) ( 10.30 ) = 0.37 and
w = 10.3  0.4
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You can see for yourself how these rules can be combined and extended to deal with more
complicated cases. On the other hand, don't forget that, in a case where taking the derivative is messy,
you can always go back to direct calculation.