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Working with errors

Module by: Sunil Kumar Singh. E-mail the author

Summary: Errors propagate through mathematical operations.

In this module, we shall introduce some statistical analysis techniques to improve our understanding about error and enable reporting of error in the measurement of a quantity. There are basically three related approaches, which involves measurement of :

  • Absolute error
  • Relative error
  • Percentage error

Absolute error

The absolute error is the magnitude of error as determined from the difference of measured value from the mean value of the quantity. The important thing to note here is that absolute error is concerned with the magnitude of error – not the direction of error. For a particular n t h n t h measurement,

| Δ x n | = | x m - x n | | Δ x n | = | x m - x n |

where " x m x m " is the mean or average value of measurements and " x n x n " is the n t h n t h instant of measurement.

In order to calculate few absolute values, we consider a set of measured data for the length of a given rod. Note that we are reporting measurements in centimeter.

x 1 = 47.7 c m , x 2 = 47.5 c m , x 3 = 47.8 c m , x 4 = 47.4 c m , x 5 = 47.7 c m x 1 = 47.7 c m , x 2 = 47.5 c m , x 3 = 47.8 c m , x 4 = 47.4 c m , x 5 = 47.7 c m

The means value of length is :

x m = 47.7 + 47.5 + 47.8 + 47.4 + 47.7 5 = 238.1 5 = 47.62 x m = 47.7 + 47.5 + 47.8 + 47.4 + 47.7 5 = 238.1 5 = 47.62

It is evident from the individual values that the least count of the scale (smallest division) is 0.001 m = 0.1 cm. For this reason, we limit mean value to the first decimal place. Hence, we round off the last but one digit as :

x m = 47.6 c m x m = 47.6 c m

This is the mean or true value of the length of the rod. Now, absolute error of each of the five measurements are :

| Δ x 1 | = | x m x 1 | = | 47.6 47.7 | = | 0.1 | = 0.1 | Δ x 1 | = | x m x 1 | = | 47.6 47.7 | = | 0.1 | = 0.1

| Δ x 2 | = | x m x 2 | = | 47.6 47.5 | = | 0.1 | = 0.1 | Δ x 2 | = | x m x 2 | = | 47.6 47.5 | = | 0.1 | = 0.1

| Δ x 3 | = | x m x 3 | = | 47.6 47.8 | = | 0.2 | = 0.2 | Δ x 3 | = | x m x 3 | = | 47.6 47.8 | = | 0.2 | = 0.2

| Δ x 4 | = | x m x 4 | = | 47.6 47.4 | = | 0.2 | = 0.2 | Δ x 4 | = | x m x 4 | = | 47.6 47.4 | = | 0.2 | = 0.2

| Δ x 5 | = | x m x 5 | = | 47.6 47.7 | = | - 0.1 | = 0.1 | Δ x 5 | = | x m x 5 | = | 47.6 47.7 | = | - 0.1 | = 0.1

Mean absolute error

Earlier, it was stated that a quantity is measured with a range of error specified by half the least count. This is a generally accepted range of error. Here, we shall work to calculate the range of the error, based on the actual measurements and not go by any predefined range of error as that of generally accepted range of error. This means that we want to determine the range of error, which is based on the deviations in the reading from the mean value.

Absolute error associated with each measurement tells us how far the measurement can be off the mean value. The absolute errors so calculated, however, may be different. Now the question is : which of the absolute error be taken for our consideration? We take the average of the absolute error :

Δ x m = Δ x 1 + Δ x 2 + . + Δ x n n Δ x m = Δ x 1 + Δ x 2 + . + Δ x n n

Δ x m = Σ 0 n Δ x i n Δ x m = Σ 0 n Δ x i n

The value of measurement, now, will be reported with the range of error as :

x = x m ± Δ x m x = x m ± Δ x m

Extending this concept of defining range to the earlier example, we have :

Δ x m = 0.1 + 0.1 + 0.2 + 0.2 + 0.1 5 = 0.7 5 = 0.14 = 0.1 c m Δ x m = 0.1 + 0.1 + 0.2 + 0.2 + 0.1 5 = 0.7 5 = 0.14 = 0.1 c m

We should note here that we have rounded the result to reflect that the error value has same precision as that of measured value. The value of measurement with the range of error, then, is :

x = 47.6 ± 0.1 c m x = 47.6 ± 0.1 c m

What we convey by writing in terms of the range of possible error. A plain reading of above expression is “the length of rod lies in between 47.5 cm and 47.7 cm”. For all practical purpose, we shall use the value of x = 47.6 cm with the caution in mind that this quantity involves an error of the magnitude of “0.1 cm” in either direction.

Relative error

We can report range of error as the ratio of the mean absolute error to the mean value of the quantity. This ratio is known as relative error. Mathematically,

Δ x r = Δ x m x m Δ x r = Δ x m x m

As we use a ratio, this expression of error is also known as “fractional error”. Applying this concept to earlier example, we have :

Δ x r = Δ x m x m = 0.1 47.6 = 0.0021 Δ x r = Δ x m x m = 0.1 47.6 = 0.0021

This is the amount of error which is possible for every “centimeter of length measured”. This is what is the meaning of a ratio. Hence, if there are 47.6 cm of total length, then the amount of error possible is 47.6 X 0.0021 = 0.1 cm. Two error range in the absolute form and relative form, therefore, are equivalent and specify the same range of errors involved with the measurement of a quantity.

Percentage error

Percentage error is equal to relative error expressed in percentage. It is given as :

Δ x p = Δ x m x m X 100 = Δ x r X 100 Δ x p = Δ x m x m X 100 = Δ x r X 100

Applying this concept to earlier example, we have :

Δ x p = Δ x r X 100 = 0.0021 X 100 = 0.21 Δ x p = Δ x r X 100 = 0.0021 X 100 = 0.21

Combination of errors

Measurement of a quantity is used in a formula in various combinations to calculate other physical quantities. The mathematical operations in the working of a formula involve arithmetic operations like addition, subtraction, multiplication and division. We need to evaluate the implication of such operations on the error estimates and what is the resulting error in the quantities derived from mathematical operations. For example, let us consider simple example of density. This involves measurement of basic quantities like mass and volume.

Clearly, we need to estimate error in density which is based on the measurements of mass and volume with certain errors themselves. Similarly, there are more complex cases, which may involve different mathematical operations. We shall consider following basic mathematical operations in this section :

  • sum or difference
  • product or division
  • quantity raised to a power

Errors in a sum or difference

We consider two quantities whose values are measured with certain range of errors as :

a = a ± Δ a a = a ± Δ a

b = b ± Δ b b = b ± Δ b

Sum of the two quantities is :

a + b = a ± Δ a + b ± Δ b = a + b + ± Δ a ± Δ b a + b = a ± Δ a + b ± Δ b = a + b + ± Δ a ± Δ b

Two absolute errors can combine in four possible ways. The corresponding possible errors in “a+b” are (Δa + Δb), -(Δa + Δb), (Δa - Δb) and (-Δa + Δb). The maximum absolute error in “a-b”, therefore, is “Δa+ Δb”.

Difference of the two quantities is :

a - b = a ± Δ a b ± Δ b = a b + { ± Δ a ± Δ b } a - b = a ± Δ a b ± Δ b = a b + { ± Δ a ± Δ b }

Two absolute errors can combine in four possible ways. The corresponding possible errors in “a+b” are again (Δa + Δb), -(Δa + Δb), (Δa - Δb) and (-Δa + Δb). The maximum absolute error in “a-b”, therefore, is “Δa+ Δb”.

Let “Δc “ be the absolute error of the arithmetic operation of addition or subtraction. Then, in either case, the maximum value of absolute error in the sum or difference is :

Δ c = Δ a + Δ b Δ c = Δ a + Δ b

We see here that the absolute error in the sum or difference of two quantities is equal to the sum of the absolute values of errors in the individual quantities. We can write the resulting value as :

For addition as :

c = a + b ± Δ a + Δ b c = a + b ± Δ a + Δ b

For subtraction as :

c = a b ± Δ a + Δ b c = a b ± Δ a + Δ b

Example

Problem 1: The values of two capacitors are measured as :

C 1 = 1.2 ± 0.1 μ F C 1 = 1.2 ± 0.1 μ F

C 2 = 2.3 ± 0.2 μ F C 2 = 2.3 ± 0.2 μ F

Two capacitors are connected in parallel. What is the equivalent capacity of two given capacitors? Indicate error in percentage.

Solution : The equivalent capacitance of two capacitors in parallel is given by the sum of the capacitance of individual capacitors :

C = C 1 + C 2 = 1.2 + 2.3 = 3.5 μ F C = C 1 + C 2 = 1.2 + 2.3 = 3.5 μ F

For addition of two quantities, the absolute error in the equivalent capacitance is given by :

Δ C = Δ C 1 + Δ C 2 Δ C = Δ C 1 + Δ C 2

Δ C = 0.1 + 0.2 = 0.3 μ F Δ C = 0.1 + 0.2 = 0.3 μ F

Percentage error in the equivalent capacitance is given by :

Δ C p = Δ C C X 100 Δ C p = Δ C C X 100

Δ C p = 0.3 3.5 X 100 = 8.6 Δ C p = 0.3 3.5 X 100 = 8.6

Hence, equivalent capacitance, “C”, with percentage error is :

C = 3.5 μ F ± 8.6 % C = 3.5 μ F ± 8.6 %

Errors in product or division

Error in the product and division of two measured quantities can be similarly worked. For brevity, we shall work out the implication of error for the operation of product only. We shall simply extend the result obtained for the product to division. Let the two measured quantities be :

a = a ± Δ a a = a ± Δ a

b = b ± Δ b b = b ± Δ b

Let “Δc” be the absolute error in the product. Then, product of the two quantities is :

c ± Δ c = a ± Δ a b ± Δ b c ± Δ c = a ± Δ a b ± Δ b

c 1 ± Δ c c = a b 1 ± Δ a a 1 ± Δ b b c 1 ± Δ c c = a b 1 ± Δ a a 1 ± Δ b b

But, c = ab. Hence, expanding terms, we have :

1 ± Δ c c = 1 ± Δ a a ± Δ b b ± Δ a Δ b a b 1 ± Δ c c = 1 ± Δ a a ± Δ b b ± Δ a Δ b a b

± Δ c c = ± Δ a a ± Δ b b ± Δ c c = ± Δ a a ± Δ b b

The terms “ Δ a Δ b a b Δ a Δ b a b ” is negligibly small and as such can be discarded :

± Δ c c = ± Δ a a + Δ b b ± Δ c c = ± Δ a a + Δ b b

We see here that the relative error in the product of two quantities is equal to the sum of the relative errors in the individual quantities. This manifestation of individual errors in product is also true for division. Hence, we can broaden our observation that the relative error in the product or division of two quantities is equal to the sum of the relative errors in the individual quantities.

Example

Problem 2: The mass and the volume of a uniform body is given as :

m = 9.5 ± 0.1 k g m = 9.5 ± 0.1 k g

V = 3.1 ± 0.2 m 3 V = 3.1 ± 0.2 m 3

Determine density of the body with error limits.

Solution : Let us first calculate the value of the density, maintaining the decimal place in the result. It is given by :

ρ = m v = 9.5 3.1 = 3.1 k g / m 3 ρ = m v = 9.5 3.1 = 3.1 k g / m 3

We have rounded the result for precision of density is limited to the minimum of precision of the quantities involved. The relative error in the division of mass by volume is :

Δ ρ ρ = Δ m m + Δ V V Δ ρ ρ = Δ m m + Δ V V

Δ ρ ρ = 0.1 9.5 + 0.2 3.1 Δ ρ ρ = 0.1 9.5 + 0.2 3.1

Δ ρ ρ = 0.01 + 0.06 = 0.07 Δ ρ ρ = 0.01 + 0.06 = 0.07

The absolute error in the density is :

Δ ρ = 0.07 ρ = 0.07 X 3.1 = 0.2 Δ ρ = 0.07 ρ = 0.07 X 3.1 = 0.2

Hence, density is given as :

ρ = 3.1 ± 0.2 k g / m 3 ρ = 3.1 ± 0.2 k g / m 3

Errors in raising a quantity to a power

In order to estimate error involving raising of a quantity to some power, we consider two measured quantities. Let us consider that they are related as :

x = a n b m x = a n b m

Taking logarithm on either side of the equation, we have :

ln x = n ln a m ln b ln x = n ln a m ln b

Differentiating on both sides, we have :

x x = n a a m b b x x = n a a m b b

We can exchange each of the differential term with corresponding relative error terms.

± Δ x c = ± n Δ a a m ± Δ b b ± Δ x c = ± n Δ a a m ± Δ b b

Again there are four possible combinations of values for the relative error. The maximum being,

Δ x c = n Δ a a + m Δ b b Δ x c = n Δ a a + m Δ b b

Thus, relative error in the result is :

Δ x r = n Δ a r + m Δ b r Δ x r = n Δ a r + m Δ b r

We see here that error in the result is equal to power times the relative error, irrespective of whether the power is positive or negative. It leads to an important deduction that quantities with greater powers in an expression should be measured with highest accuracy to minimize error in the "derived" quantity.

Example

Problem 3: The torque required to produce a twist in solid bar is given by :

τ = Π η r 4 2 L τ = Π η r 4 2 L

If percentage error in the measurement of η, r and L are 1%, 4% and 1% respectively. Find the percentage error in the value of torque.

Solution : We know that relative error in the resultant quantity i.e torque is :

Δ t r = n 1 Δ η r + n 2 Δ r r + n 3 Δ L r Δ t r = n 1 Δ η r + n 2 Δ r r + n 3 Δ L r

where n 1 n 1 , n 2 n 2 and n 3 n 3 are the powers of three quantities η, r and L respectively. The constants of the equations are not considered as they are not measured. Now percentage is obtained by just multiplying relative error by 100, we can write,

Δ t p = n 1 Δ η p + n 2 Δ r p + n 3 Δ L p Δ t p = n 1 Δ η p + n 2 Δ r p + n 3 Δ L p

Here, n 1 = 1, n 2 = 4 and n 3 = 1 n 1 = 1, n 2 = 4 and n 3 = 1 . Note that position of a quantity either in numerator or denominator does not make difference in error combination. Putting values,

Δ t p = 1 X 1 + 4 X 4 + 1 X 1 = 18 Δ t p = 1 X 1 + 4 X 4 + 1 X 1 = 18

This means that the maximum error estimate in torque would be 18%. This is quite a large amount of uncertainty. Note the role played by the error in “r”. Most of error (16 %) is due to 4% error in this quantity as it is raised to a power of 4. This result substantiates the observation that quantity with highest power should be measured with most accuracy.

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