Propagation of Errors in Calculations

Let us now look at some simple arithmetic calculations with numbers containing uncertainties.  We already showed that a measurement is best represented by its mean and standard deviation, such as , where  is the average and s _x is the standard deviation of the set of x-measurements.  When expressed in this way, s _x  is referred to as the absolute error in the variable x.  However, we could also write the number and its error as .  The ratio within the parentheses, , is called the relative error in the measurement x.  This ratio is often multiplied by 100 and expressed as a percent.  Thus the number 155 ± 23 cm can be also expressed as 155 cm ± 15%.

Adding and Subtracting Numbers Containing Uncertainties (e.g., )

Suppose we want to add two numbers .  This would appear to be , in other words, the absolute error in the sum of two numbers looks like it should be the sum of the absolute errors.  However, this assumes the most pessimistic case that the errors always add.  They might, however, subtract.  In reality, if the two numbers being added have errors that are independent of each other, the errors add randomly but they do not average out to zero.  The error associated with the sum of two numbers is somewhere between the results we get from adding and subtracting these errors.  We won’t prove this (this is done in textbooks on statistics) but independent errors add somewhat like vectors, namely, .  This is reminiscent of the Pythagorean theorem; it says that the absolute error in the sum, w, of two numbers with errors is the square root of the sum of the squares of the absolute errors of the two numbers.  If a third number, , is then added to this sum, we find that the error in the result is

.

Using the same line of reasoning for finding the worst-case error in the difference of two numbers with errors, namely , and realizing that the errors add randomly in this case also, we can state that the same rule applies to subtraction.  Whenever adding and subtracting numbers with errors, the absolute error in the result is the square root of the sum of the squares of the absolute errors in the numbers being added and subtracted.  This result is modified somewhat if the numbers being added or subtracted are correlated (that is, are not completely independent of one another), but we will not consider this case here.

Suppose we multiply a number with an uncertainty by a constant number, a. Then , which suggests that the error is also multiplied by a.  Now if we form the sum of two numbers that are each multiplied by a constant ( ), we get the sum and its uncertainty as we did before, namely

  .

But again this is not quite right because the errors add randomly.  Therefore the error in this sum is  as we just did above.  The same result holds for subtraction.

As an example, suppose we have two measurements 155 ± 23 cm and 287 ± 24 cm, and we want to add 5 times the first to 3 times the second.  The result is

cm.

Notice we have retained three significant figures in the answer.

Multiplying and Dividing Numbers With Uncertainties (e.g., )

When multiplying or dividing two numbers, it is easier to see what happens to the errors when they are expressed in terms of relative errors.  For example, multiplying two numbers with errors

If the errors are relatively small and random, the last term within the brackets in this expression is insignificant compared with the other terms, and tends toward zero.  Thus we have

.

This suggests that the relative error in the product of two numbers with errors is the sum of the relative errors.  However, as in the case of adding and subtracting, this is a worst-case estimate because the relative errors actually add randomly.  Again we won’t prove it, but the relative error of the product of two numbers is the square root of the sum of the squares of the relative errors of the numbers.  Thus .  This same calculation holds when dividing two numbers.  Therefore, the rule when multiplying and dividing numbers with errors says that the relative error in the product and quotients of numbers with errors is the square root of the sum of the squares of the relative errors in the individual numbers.  As in the case with adding and subtracting, this rule assumes that the numbers being multiplied or divided are uncorrelated.

How does multiplying a number by a constant affect the relative error?  Notice that when we multiply by a, the absolute error is also multiplied by a.  But note the relative error in this calculation

  .

Therefore the relative error in  is the same as the relative error in .  Multiplying a number by a constant changes the absolute error in the result but does not change the relative error.  This makes sense if you consider that multiplying by a constant changes everything in the number proportionally.  It is analogous to moving closer to or farther away from a water tower.  The ratio of the height of the water tank to the height of the support structure doesn’t change with the distance from which you are viewing it.

As an example, let’s multiply the two numbers in the previous calculation.  First express the numbers in terms of their relative errors.  155 ± 23 = 155 ± 15%, and 287 ± 24 = 287 ± 8%.  Multiplying, we get 44,500 ± 17%.  Similarly, when we divide the second number by the first, we get 1.85 ± 17%.  Notice that we have retained the same number of significant figures in the answer as were contained in the numbers that went into the calculations.  In general, the number of significant figures can be no more than that of the number with the fewest significant figures that is used in the calculation.  Whenever possible, in most experiments, we try to measure every number to the same number of significant figures (roughly the same precision).  This minimizes surprises at the end of long calculations into which the numbers are entered to get a final result.

Raising a Number to a Power (e.g., )

If we raise a number to a power, n, we can use the binomial expansion to get

or

In the last parentheses, we have dropped the terms with higher powers of  because they become insignificant compared to the first two terms.  Thus we see that the relative error in w is just n times the relative error in .  Note that if n is less than one, (such as when taking square roots), the relative error is actually reduced.

Raising e to a Power of x  (e.g., )

Let .  We want to find the error in w caused by the error in .  Notice that we can write this expression as .  But now we can use the infinite series expansion for which is

The last line follows after dropping the higher powers of  because they are small compared with the first two terms in the expansion.  Therefore, our original expression can be written as

.

Thus the relative error in w is just

.

Raising Another Base to a Power of x  (e.g., )

Let .  We wish to find the resulting error in w.  Since the base, u, can be expressed in terms of the base e, we can write this as

But now we see that this is very similar to the previous case, except that b has been replaced with blnu.  Therefore the relative error in w is just , or

.

Suppose w with its uncertainty is .  This can be written as

The last line uses the infinite series expansion of .  Dropping the higher powers because they are insignificant compared to the first term in the parentheses, we find

,

which shows that the absolute error in w is just a times the relative error in x, or .

 

Summary of Propagation of Errors

Table 3 shows the calculation of errors for several of the elementary mathematical operations with numbers containing uncertainties.  It is assumed that the errors in the numbers are random and independent of each other.

Table 3

Mathematical Operation	Variables	Uncertainties in Variables	Example Calculation	Error in Result
Addition, Subtraction
Multiplication, Division
Constant Multiplying a Variable				or
Raise to Power
Exponentiation Base e
Exponentiation Base u
Logarithm

 

Example Problems

Example 1:

Suppose we have four masses of 115.2 ± 3.5 grams, 257.8 ± 12.7 grams, 89.5 ± 2.2 grams, and 105.9 ± 5.4 grams.  What is the total mass and its uncertainty?

Answer:  The total mass is just the sum of the masses, or 568.4 grams (notice we have not yet rounded the answer to three significant figures).  The absolute uncertainty in the sum is grams.  To express this as a relative uncertainty, we divide by the total mass to get 14.4/568.4 = 0.025, or 2.5%

The way we report this answer depends somewhat on what we wish to convey about the measurements.  If we do not know the precision of the measuring instrument (we were simply given the above numbers without further comment), then we should be careful in assuming the numbers are more precise than reported in the mass with the least number of significant digits, namely, the 89.5 grams.  However, the fact that the numbers are all given to the first decimal place suggests that the instrument used to measure these masses was capable of measuring to at least this precision.  We are left to wonder why the variations in the reported masses are as large as they are.  Perhaps there were some other disturbances taking place during the measurements.  If that were the case, then we could report this as 568.4 ± 14.4 grams, or as 568.4 grams ± 2.5%.  If, on the other hand, we suspect that the measurements are no better than three significant figures, we should probably report 568 ± 14 grams or 568 grams ± 2.5%.

Example 2:

In the physical theory of the pendulum oscillating with small amplitude, the period, T, (this is the time for one complete swing, forward and back) is given by the expression .  In this formula, L is the length of the pendulum and g is the acceleration due to gravity at the surface of the Earth (or moon, or any other celestial body).

Suppose we would like to use a pendulum to find the gravitational acceleration, g.  Solving this expression for g in terms of L and T we get .  The experiment would consist of making a pendulum of length L, measuring the period T, and using these results to calculate g.  Since the measurements of L and T both contain uncertainties, we would then use these to calculate the uncertainty in g.  Let’s take a specific example.

Assume you make the length of the pendulum 1.200 meters ± 10 millimeters, and you measure the period to be 2.20 ± 0.05 seconds.  To calculate the uncertainty in g, we note that T is squared and divided into L.  Therefore we need to use the relative errors in these measurements when we do the calculation of the uncertainty in g because the calculation involves raising a variable to a power and the division of two variables.  The length is L = 1.200 meters ± 0.83%.  The time and its relative uncertainty is T = 2.20 seconds ± 2.3%.  Since T is squared, the relative uncertainty in is twice the relative uncertainty in T, or 4.6%.  The relative uncertainty in  is then calculated by .  Multiplying the result by (2p)² does not change the relative uncertainty in g.  Therefore we get g = 9.79 m/s² ± 4.7%.

As an exercise, consider the precision necessary in L and T that would allow finding g to within 1%.  What would you have to do in the experiment to improve the precision in L and T so that you will have measured g to within 1%?

Example 3:

The voltage of a capacitor discharging through a resistor is .  In this equation, V₀ is the initial voltage at the start of the discharge, C is the capacitance measured in Farads, and R is the resistance measured in Ohms.  It turns out that the product RC has units of time, therefore the exponent, t/RC, is dimensionless (time divided by time).  Suppose we had a capacitor of size 10 microfarads, and a resistor of 33 kilohms.  The initial voltage is 10 volts.  Since these values will have some uncertainties associated with them, there would be an uncertainty in the voltage after a given amount of discharge time.  Since the product of R and C is in the exponent, we would expect that the uncertainty in RC would be magnified in its effect on V.  Let’s list these values along with some typical uncertainties.

C = 10. mF ± 10%  =  10. ´ 10 ^{- 6} F  ±  10%

R  = 33.  kW ± 5%  =  33. ´ 10 ³  W  ±  5%

V₀ = 10.0 V ± 3%

The uncertainties in these quantities are what one would be likely to get with some moderately priced components purchased at an electronic supply house.  Given these uncertainties, what is the uncertainty in the voltage as a function of the time of discharge?

Solution:  The relative uncertainty in  is .  Therefore .  The relative uncertainty in the exponential is t times the absolute uncertainty in 1/RC, which is  ± 0.339t º ± 33.9t %.  The relative uncertainty in V₀ is 1%.  Finally, the relative uncertainty in V is , and this is in percent.  Notice that when t = 0, the uncertainty in V is just 1%.  As time increases, so does the uncertainty in V.  The voltage gets smaller as time increases, but the uncertainty about what it is actually grows.

For example, if we had  perfect components with no uncertainties in their values, we could be certain that the voltage after 1 second would be  volts. However, the uncertainty in the values of  the circuit components, R and C, would give our prediction of this voltage an uncertainty of about 34%.  At a time of 0.1 seconds, the ideal voltage would be 7.39 volts.  Our prediction of this, however, would have an uncertainty of about 3.5%.