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INTRODUCTION
PERCENT ERROR AND PERCENT DIFFERENCE
A percent error should be calculated when an experimental value E is compared to a standard or accepted value S of the same quantity. We express the difference between E and S as a percent of the standard value S:
100 S
E-S
PE = ×
This formula yields a positive result if the experimental value is high in comparison to the standard value, and a negative result if it is low. A percent difference should be calculated when two experimental values, E 1 and E 2 , are compared to each other, and there is no standard value for comparison. In this case we express the absolute value of the difference between the experimental values as a percent of the average of the two values:
E E
E -E
E E
E -E
PD
1 2
1 2 2 1 2 1
1 2 ×
× =
Percent difference is always positive.
THE METHOD OF DIFFERENCES
The Method of Differences is used when some variable is believed to change by equal amounts in successive measurements. This method yields the average change in the variable per interval. A physical example of such a case is the stretching of a spring by a force which increases by equal amounts in successive intervals. Another example is the speed of a dense falling body measured at equal time intervals. As a concrete example to illustrate the Method of Differences, suppose that we want to measure the width of one of a number of identical floor tiles in a room. One method is to place a measuring rod down, measure a tile, move the rod, measure another tile, and so on. We could then find the average of the individual measurements. However, moving the rod increases the experimental error. A more precise method is to place the rod down only once, and then to take a set of readings of the positions of successive cracks (assumed to be of negligible thickness). For example, six successive readings would span five tiles. Let the six readings be a, b, c, d, e and f. There are several ways in which these six numbers could be combined in order to yield a single tile width.
A poor method would be to use the formula
f-a w = (3)
The result would be approximately correct, but less precise than it could be, since we have used only two of the six readings.
A method that looks better at first is to calculate the five widths, b-a, c-b, d-c, e-d, and f-e, and then average them. The equation for this procedure is
b-a c-b d-c e-d f-e w
Close inspection of this equation, however, shows that it reduces to equation (3), so we have gained nothing by all of our extra work.
The Method of Differences uses each of the six readings once, and no reading cancels out. In order to use it, we must have an even number of readings. For an odd number of readings, either the first or last reading must be discarded. Then, we divide the readings into two sets. In our example, set one would consist of readings a, b and c; set two would consist of readings d, e and f. We can get one estimate of the average by using readings a and d,
d - a .
Here, the distance (d-a) spans three tiles, so we have divided by three. We can get two other estimates using the pairs b and e and c and f,
e - b and 3
f - c .
The method of differences uses the average of these three estimates
d-a e-b f-c 3
f-c 3
e-b 3
d-a
w
In the general case, suppose that we have 2n successive readings, A 1 to An and B 1 to Bn , of some variable S. Then the average change in S per interval is given by
2
1 1 2 2 n n n
B -A B -A B -A
S
L
III. Use of Differentials to Represent Uncertainties
Let the length of the paper in section II be called L. Then we represent the absolute uncertainty in L by the symbol dL. We represent the relative uncertainty in L by the ratio dL/L. Since relative uncertainties are usually rather small, the concept of a differential is useful in determining how two or more uncertainties combine together when measurements are used in mathematical operations. This will be illustrated in the next section.
IV. How Uncertainties in Data Affect Calculated Results
Consider a formula F(A) evaluated for one measured quantity A which has an uncertainty dA. If dA is small, F(A ± dA) = F(A) ± ( (^) dAdF^ ) dA, where (^) dAdF^ is the derivative of the formula with respect to A evaluated at the measured value of A. Note: This formula is simply the first two terms of the Taylor Expansion of F at A. The uncertainty in the calculated value for F is then dF = (^) dAdF^ dA. This same relation can be used for a formula F, if A is a calculated quantity.
Example: Let F(A) = A and A be measured as 3.54 m² ± 0.07 m².
3.54 m^2 = 1.88 m
2 A
dA
dF (^) = = 2 3.54m^2
(^1) = 1/(2¯1.88 m) = 0.266 m-
dF = (^) dAdF^ dA = 0.266 m-1^ ¯ 0.07 m² = 0.0186 m = 0.02 m
F = 1.88 m ± 0.02 m
Now, consider a formula F(A, B, C) evaluated for the independent quantities A, B and C which have uncertainties dA, dB and dC, respectively. If dA, dB and dC are small, F(A ± dA, B ± dB, C ± dC) = F(A, B, C) ± ( (^) ∂∂AF^ ) dA ± ( (^) ∂∂BF^ ) dB ± ( (^) ∂∂CF^ ) dC, where the derivatives are evaluated at the measured values of A, B and C. The uncertainty in the calculated value for F is then dF = (^) ∂∂AF^ dA + (^) ∂∂BF^ dB + (^) ∂∂CF^ dC.
Let us now consider the most frequent operations we will perform in calculations: addition, subtraction, multiplication, division, and raising to a power.
A. Addition
Let S be the sum of two independent †^ quantities, S = A + B.
A
S ∂
∂ (^) = 1 and B
S ∂
∂ (^) = 1. Therefore, dS = dA + dB.
Example: (25.0 cm ± 0.2 cm) + (10.0 cm ± 0.5 cm) = 35.0 cm ± 0.7 cm † (^) For calculations involving dependent quantities, refer to section IV-F.
B. Subtraction
Let D be the difference of two independent quantities, D = A - B.
A D ∂ ∂ (^) = 1 and B D ∂ ∂ (^) = -1. Therefore, dD = dA + dB.
Example: (25.0 cm ± 0.2 cm) - (10.0 cm ± 0.5 cm) = 15.0 cm ± 0.7 cm
C. Multiplication
Let P be the product of two independent quantities, P = AB.
A
P ∂
∂ (^) = B and B
P ∂
∂ (^) = A. Therefore, dP = |B|dA + |A|dB.
Dividing both sides by |P| = |A| |B|, we find dP/|P| = dA/|A| + dB/|B|. Example: (5.00 m ±^ 1%)(8.00 m^ ±^ 3%) = 40.0 m²^ ±^ 4% = 40.0 m²^ ±^ 1.6 m² (1.6 m² is obtained by calculating 4% of 40.0 m².)
D. Division
Let Q be the quotient of two independent quantities, Q = A/B.
A
Q ∂
∂ (^) = 1/B and B
Q ∂
∂ (^) = -A/B². Therefore, dQ = |1/B| dA + |A/B²| dB.
Multiplying both sides by 1/|Q| = |B|/|A|, we find dQ/|Q| = dA/|A| + dB/|B|. Example: (3.00 g ± 1%)/(1.50 cm³ ± 3%)=2.00 g/cm³ ±4% = 2.00 g/cm³ ± 0.08 g/cm³
Note: The general rule for addition and subtraction is that the absolute uncertainty in a sum or difference is equal to the sum of the absolute uncertainties in the quantities added. The general rule for multiplication and division is that the relative uncertainty in a product or quotient is equal to the sum of the relative uncertainties in the factors.
Simply put: If F = A ± B, dF = dA + dB. If G = A ¯^ B or G = A/B, then dG/|G| = dA/|A| + dB/|B|.
E. Expressions Containing Only Powers, Multiplications and Divisions
Since powers are simply repeated multiplication, we have, in effect only multiplications and divisions. From the results of parts C and D, we see that we can simply sum the relative errors. For example, if A = Bp^ C q^ /Dr, then dA/|A| = p dB/|B| + q dC/|C| + r dD/|D|
The density of a cone is r h
m V
m ρ (^2) 3
= = 1 π , where m is the mass, r is the radius, and h
is the height. The relative uncertainty in the density is dρ/ρ = dm/m + 2 dr/r + dh/h, where dm, dr and dh are the absolute uncertainties in the mass, radius and height respectively.
2
2
3
s
m 2
kgs
m
00 100 m 3.00m 50.0m
0 10. 0 kg DE F
2BC
C
A
2
2
3
s
kg 2 2
2 kgs
m 2
2
- 240 100 m 3.00m 50.0m
3.00m 50. 0 10. 0 kg
DE F
- EBC
D
A
2
2
3
s
kg 2 2
2 kgs
m 2
2
- 00 100 m 3.00m 50.0m
100 m 50. 0 10. 0 kg
DE F
- DBC
E
A
2
2
3
ms
kg 2 2
2 kgs
m 2
2
0800 100 m 3.00m 50.0m
0 10. 0 kg
DE F
BC
F
A
dA = ∂ ∂BA^ dB + (^) ∂∂A^ C dC + (^) ∂∂DA^ dD + ∂ ∂EA^ dE + ∂ ∂FA^ dF
dA = ( (^2)
2 m
- 400 kg )(0. 2
3 kgs
m (^) ) + ( (^2) s
- 00 m)(0.2 kg) + ( (^2) s
- 240 kg)(1 m)
- ( 8. (^00) skg 2 )(0.06 m) + ( 0. (^0800) mkgs 2 )(1.0 m²)
dA = 1.8 kgS 2 m
2 2
2
3
s
kgm s
kgm 2
2 kgs 2 m
0 1. 8 100 m 3.00m 50.0m
0 10. 0 kg DE F
BC
A = ±
Solution II (repeated use of earlier results):
2 kgs 2 m
100 1 m 3.00 0. 06 m 50.0 1. 0 m
- 0 0. 5 10. 0 0. 2 kg DE F
BC
A
2
3
± ± − ±
2 kgs
m
100 m 1 % 3.00m 2 % 50.0 1. 0 m
- 0 1 % 10. 0 kg 2 % A
2
3
± ± − ±
2 kgs
m 2 2
2 kgs
m
300 m 9. 0 m 50.0 1. 0 m
0 1 % 10. 0 kg 2 % 300 m 3 % 50.0 1. 0 m
0 1 % 10. 0 kg 2 % A
2
3 2
3
± − ±
( 250 m 4 %)
0 1 % 10. 0 kg 2 % 250 m 10 m
0 1 % 10. 0 kg 2 % A (^2)
2 kgs
m 2 2
2 kgs
m (^3232)
±
(^2 2) s 2
kgm s
kgm s A = 20. 0 kg^ m± 9 %= 20. 0 ± 1. 8
SIGNIFICANT FIGURES AND DECIMAL PLACES
Often, we wish to quickly estimate the precision of our calculated results without applying the rigor of the previous section. To do this, we use rules that relate the number of decimal places or significant figures we can keep in a calculated value to the number of decimal places or significant figures in our data.
I. Definitions
The number of decimal places in a number is the number of digits to the right of the decimal point. The number of significant figures in a number is the total number of digits, exclusive of leading zeros.
II. Examples
The following table shows the number of decimal places and the number of significant figures in five numbers.
Number Number of Decimal Places Number of Significant Figures 15.73 2 4 0.0072 4 2 200.6 1 4 1.2700 4 5 4300 0 2, 3 or 4
The ambiguity in the number of significant figures in the last example is easily removed by using scientific notation. 4.30 x 10^3 is three significant figures.
III. Rules for Rounding Off Calculated Results
In addition or subtraction, keep as many decimal places in the result as the smallest number of decimal places found in any of the numbers being added or subtracted. Examples: 20.5 + 1.483 = 22. 19.03 - 18.96 = 0. 10.512 - 9.8 = 0. 4.93 + 6.26 = 11. Notice that the number of significant figures in the result can be more than the number of significant figures in either number or less than the number of significant figures in either number.
The distance ∆ y (^) i is given by the difference in y coordinates of the points Pi and Pi'.
Thus we have
∆ y i = yi − ( mx i + b ) = yi − mxi − b.
The sum S is then given by
∑ (^ )^ ∑(^ ) = =
n
i
i i
n
i
S yi y mx b 1
2 1
2
∑^ (^ )
n
i
yi m xi b mxiyi byi mbxi 1
∑ ∑ ∑ ∑ ∑ = = = = =
n
i
i
n
i
i
n
i
i i
n
i
i
n
i
yi m x nb m xy b y mb x 1 1 1
2 1
2 2 1
S is a function of two independent variables m and b, and 2n constants, x 1 , y 1 , ..., xn , yn. In such a case we can find a relative minimum (there is no relative maximum) by setting both partial derivatives to zero.
∑ ∑ ∑ = = =
n
i
i
n
i
i i
n
i
m xi xy b x m
S
1 1 1
∑ ∑ = =
n
i
i
n
i
nb yi m x b
S
1 1
Solving these equations simultaneously, we find
2
1 1
2
1 1 1
⎟ ⎠
∑ ∑
∑ ∑ ∑
= =
= = = n
i
i
n
i
i
n
i
i
n
i
i
n
i
i i
n x x
n xy x y m and (^2)
1 1
2
1 1 1
2 1
⎟ ⎠
∑ ∑
∑ ∑ ∑ ∑
= =
= = = = n
i
i
n
i
i
n
i
i
n
i
i i
n
i
i
n
i
i
n x x
y x xy x b.
The following is a numerical example, using the four data points (1.51, 2.73), (2.87, 6.07), (9.47, 12.37), and (12.73, 13.61).
i xi yi xi yi xi^2 1 1.51 2.73 4.1223 2. 2 2.87 6.07 17.4209 8. 3 9.47 12.37 117.1439 89. 4 12.73 13.61 173.2553 162. Sum 26.58 34.78 311.9424 262. Sum*^ 26.58 34.78 312 262.
The last row shows the sums to the correct number of significant figures. It should be noted that the number of significant figures in each sum increases when more data points
are present. If there were three times as many data points, the sum of the y-coordinates would increase by about a factor of 3. This would result in a sum that would exceed 100, but would still be accurate to the nearest 0.01. The sum would have 5 significant figures.
Substituting the sums into the equations for m and b, and remembering that n is four in this case, we obtain
×
× −
× −
× − ×
m = , and
×
− ×
× −
× − ×
b =.
Our equation for the least-squares line is then, y = 0. 94 x + 2. 4.
The X intercept, if needed, may be calculated from
- 6
- 94
m
b x.
WARNING: The equations for m and b in the least-squares method tend to give zero divided by zero. In order to obtain meaningful results, one needs to have many data points.
MICROSOFT EXCEL NOTE: The slope and intercept of the least-squares line can be found in Excel using the SLOPE and INTERCEPT functions. One must be aware of significant figures when using these functions, since Excel does not take them into consideration. The syntax for the two functions is
SLOPE(known_y's,known_x's) INTERCEPT ( known_y's , known_x's ) Known_y's is the dependent set of observations or data. Known_x's is the independent set of observations or data.
