Propagation of Errors—Basic

Rules

See Chapter 3 in Taylor, An Introduction to

Error Analysis.

1. If x and y have independent random errors

δx and δy, then the error in z = x + y is
p
δz = δx2 + δy 2.

2. If x and y have independent random errors

δx and δy, then the error in z = x × y is
sµ ¶2 µ ¶2
δz δx δy
= + .
z x y

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3. If z = f (x) for some function f (), then

δz = |f 0(x)|δx.

We will justify rule 1 later. The

justification is easy as soon as we decide on a
mathematical definition of δx, etc.

Rule 2 follows from rule 1 by taking

logarithms:

z = x×y
log z = log x + log y
p
δ log z = (δ log x)2 + (δ log y)2
sµ ¶ µ ¶2
2
δz δx δy
= +
z x y
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where we used
δX
δ log X = ,
X
the calculus formula for the derivative of the
logarithm.

Rule 3 is just the definition of derivative of

a function f .

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 Quick Check 3.4
Problem: To find the volume of a certain
cube, you measure its side as 2.00 ± 0.02 cm.
Convert this uncertainty to a percent and then
find the volume with its uncertainty.

Solution: The volume V is given in terms of

the side s by
V = s3,
so the uncertainty in the volume is, by rule 3,

δV = 3s2 δs = 0.24,

and the volume is 8.0 ± 0.2 cm3.

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 Quick Check 3.8

Problem: If you measure x as 100 ± 6,

√
what should you report for x, with its
uncertainty?
√
Solution: Use rule 3 with f (x) = x,
0
√ √
f (x) = 1/(2 x), so the uncertainty in x is
δx 6
√ = = 0.3
2 x 2 × 10
√
and we would report x = 10.0 ± 0.3.
We cannot solve this problem by indirect
use of rule 2. You might have thought of
√ √
using x = x × x, so
√
δx √ δ x
= 2 √
x x
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and
√ δx
δ x=√ ,
2x
√
which leads to x = 10 ± 0.4. The fallacy
√
here is that the two factors x have the
same errors, and the addition in quadrature
rule requires that the various errors be
independent.

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 Quick Check 3.9

Problem: Suppose you measure three

numbers as follows:

x = 200 ± 2, y = 50 ± 2, z = 40 ± 2,

where the three uncertainties are independent

and random. Use step-by-step propagation
to find the quantity q = x/(y − z) with its
uncertainty.
√
Solution: Let D = y−z = 10±2 2 = 10±3.
Then
x p
q= = 20 ± 20 0.012 + 0.32 = 20 ± 6.
D

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 General Formula for Error
Propagation

We measure x1, x2 . . . xn with uncertainties

δx1, δx2 . . . δxn. The purpose of these
measurements is to determine q, which is
a function of x1, . . . , xn:

q = f (x1, . . . , xn).

The uncertainty in q is then

sµ ¶2 µ ¶2
∂q ∂q
δq = δx1 + ... + δxn
∂x1 ∂xn

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 If
q = x1 + x2,
we recover rule 1:
∂q
= 1,
∂x1
∂q
= 1,
∂x2
q
δq = δx21 + δx22

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 If
q = x1 × x2,
we recover rule 2:
∂q
= x2,
∂x1
∂q
= x1,
∂x2
q
δq = x22 δx21 + x21 δx22
v " #
u µ ¶2 µ ¶2
u δx δx
= tq 2
1 2
+
x1 x2
sµ ¶2 µ ¶2
δq δx1 δx2
= +
q x1 x2

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 Problem 3.47

The Atwood machine consists of two masses

M and m (with M > m) attached to the
ends of a light string that passes over a
light, frictionless pulley. When the masses
are released, the mass M is easily shown to
accelerate down with an acceleration
M −m
a=g .
M +m

Suppose that M and m are measured as

M = 100 ± 1 and m = 50 ± 1, both in grams.
Find the uncertainty δa.

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Solution: The partial derivatives are

∂a (M + m) − (M − m)
= g
∂M (M + m)2
2mg
= 2
,
(M + m)
∂a −(M + m) − (M − m)
= g
∂m (M + m)2
2M g
= −
(M + m)2

so we have
2g p
δa = m 2δM 2 + M 2δm2.
(M + m)2

Now put the numbers in to get

9.8 × 50
a= = 3.27 m/s2
150
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and
2 × 9.8 p 2 2 = 0.097
δa = 50 + 100
1502
so our answer is

a = 3.3 ± 0.1 m/s2.

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 Problem 3.49

If an object is placed at a distance p from a

lens and an image is formed at a distance q
from the lens, the lens’s focal length can be
found as
pq
f= . (1)
p+q

(a) Use the general rule to derive a formula

for the uncertainty δf in terms of p, q, and
their uncertainties.

(b) Starting from (1) directly, you cannot

find δf in steps because p and q both
appear in numerator and denominator.
Show, however, that f can be rewritten
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 as
1
f= .
(1/p) + (1/q)
Starting from this form, you can evaluate
δf in steps. Do so, and verify that you get
the same answer as in part (a).

Solution:

(a) The partial derivatives are

∂f q(p + q) − pq q2
= 2
= 2
,
∂p (p + q) (p + q)
∂f p(p + q) − pq p2
= 2
= 2
.
∂q (p + q) (p + q)

Therefore
p
q 4δp2 + p4δq 2
δf = 2
.
(p + q)
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 (b) The uncertainty in 1/p is δp/p2, and
the uncertainty in 1/q is δq/q 2. The
uncertainty in

1 1
+
p q

is sµ ¶2 µ ¶2
δp δq
+ ,
p2 q2
which is a relative uncertainty of
sµ ¶2 µ ¶2
1 δp δq
1 1 + .
p + q
p2 q2

The relative uncertainty in f , as given by

(1), is the same, so the absolute uncertainty
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 in f is
sµ ¶2 µ ¶2
2 δp δq
δf = f 2
+
p q2
1 p
= q 4δp2 + p4δq 2,
(p + q)2

exactly as in part (a).

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 Problem 3.50

Suppose that you measure three independent

variables as

x = 10 ± 2, y = 7 ± 1, θ = 40 ± 3◦

and use these values to compute

x+2
q= . (2)
x + y cos 4θ
What should be your answer for q and its
uncertainty?
Solution: Find the partial derivatives, using
θ in radians:
∂q x + y cos 4θ − (x + 2)
=
∂x (x + y cos 4θ)2
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 y cos 4θ − 2
= 2
= −0.732
(x + y cos 4θ)
∂q (x + 2) cos 4θ
= − 2
= 0.963
∂y (x + y cos 4θ)
∂q 4(x + 2)y sin 4θ
= 2
= 9.813.
∂θ (x + y cos 4θ)

So we have q = 3.507 and

δq 2 = (0.732 × 2)2
+(0.963 × 1)2
+(9.813 × 3 × π/180)2
= 3.3.

Our answer is q = 3.5 ± 2.

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Alternate Solution

Rewrite (2) in the form

1 y cos 4θ − 2
=1+ .
q x+2

We can find the uncertainty in 1/q, and

therefore in q by the simple step-by-step
procedure.

1. The relative uncertainty in y cos 4θ is

sµ ¶2 µ ¶2
δy 4δθ sin 4θ
∆1 = + = 0.16.
y cos 4θ

2. The absolute uncertainty in y cos 4θ is

∆2 = |y cos 4θ| × ∆1 = 1.1.

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3. The relative uncertainty in 1/q − 1 is
sµ ¶2 µ ¶2
∆2 δx
∆3 = +
y cos 4θ − 2 x+2
= 0.21.

4. The absolute uncertainty in 1/q − 1 is

∆4 = |1/q − 1| × ∆3 = 0.15,

which is also the absolute uncertainty in

1/q.

5. The relative uncertainty in 1/q is q × ∆4,

which is also the relative uncertainty in q.
Therefore the absolute uncertainty in q is

δq = q 2 × ∆4 = 2.

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