CHAPTER 1
MEASUREMENT AND ERROR CONCEPTS
1.1
1.2
1.3
1.4
The International System of Units
Unit Conversion
Accuracy, Precision and Error in Measurements
Bisection Method of Reducing Error
At the end of the chapter, students are expected to

define the modern metric system called the International System of Units (SI Units).

apply the various conversion formulas of length, mass, time, area, and volume in
solving applied physical problems for both English Customary System of Units
and SI System of Units.

distinguish and illustrate the notion of accuracy and precision in measurement.

analyze and calculate the accuracy, precision, and error of single and repeated
measurements.

apply the bisection method in estimating the root of an equation.
A BRIEF HISTORY OF MEASUREMENT
Measurement, loosely defined as the length, amount, or size of something that is
measured.
Measurement is as old as human history itself. Humans always measure things and
phenomena around them. They measure time and passing of seasons, they tally their foods and
livestock, and they count their belongings. They have used various means to achieve these: They
used sticks to tally and count. They used sun dials and sand glasses to measure time and seasons.
Employing body parts, they measure lengths and heights.
Units of Measurements
The Egyptian cubit, the Indus Valley units of length referred to above and the
Mesopotamian cubit were used in the 3rd millennium BC and are the earliest known units used by
ancient peoples to measure length. The names for many units of measurement were borrowed
from human morphology. For example, the foot, the hand, the pace, etc. Still, however, these units
of measurement were not standardized. As industry and trade expanded across the world, the
need became dire for a single standardized system of measurement. Today, units of measurement
are generally defined on a scientific basis, overseen by governmental or independent agencies,
and established in international treaties, pre-eminent of which is the General Conference on
Weights and Measures (CGPM), established in 1875 by the Meter Convention, overseeing the
International System of Units (SI) and having custody of the International Prototype Kilogram.
The meter, for example, was redefined in 1983 by the CGPM in terms of light speed.
Systems of Measurements
1
The earliest known uniform systems of weights and measures seem all to have been
created at some time in the 4th and 3rd millennia BC among the ancient peoples of Egypt,
Mesopotamia and the Indus Valley, and perhaps also Elam (in Iran) as well. The metric system is
a decimal system of measurement based on its units for length, the meter and for mass, the
kilogram. It exists in several variations, with different choices of base units, though these do not
affect its day-to-day use. Since the 1960s, the International System of Units (SI) is the
internationally recognized metric system. Metric units of mass, length, and electricity are widely
used around the world for both every day and scientific purposes.
1.1 The International System of Units
“Measure what is measurable and make measurable what is not
so.”
- Galileo Galilei
The names for many units of measurement were borrowed from human morphology like
the foot, the hand, the pace, etc. Unfortunately, these units of measurement were not
standardized. As industry and trade expanded across the world, a single standardized system of
measurement must be developed in order to create balance across the various industries and
subsequently, the world. This unified system of measurement was then called the “metric
system”.
The metric system was developed in France in the 1790s and was considered the first
standardized system of measurement. However back in 1960, the metric system was revised,
simplified, and renamed to “Le Système International d’Unites” (The International System of
Units) or the “SI System of Units”, which is, nowadays, the standard form of measurement in
almost every country across the globe, except for the United States. Despite not being a
commonly-used measurement in the United States, the SI system is the standard system used by
scientists worldwide, including those in the United States.
The SI system is made up of mutually independent 7 base units (see Table 1) that define
the 4 derived units without special names and 22 derived units (see Table 2) with special names
and symbols.
Table 1: The SI System Base Units
Base Quantity
Unit Measurement Name
Symbol
Length
Mass
Time
Electric Current
Temperature
Meter
Kilogram
Second
Ampere
Kelvin
𝒎
𝒌𝒈
𝒔
𝑨
𝑲
Base Quantity
Unit Measurement Name
Symbol
Amount of Substance
Luminous Intensity
Mole
Candela
𝒎𝒐𝒍
𝒄𝒅
Table 2: The SI System Derived Units without Special Names
2
Derived Quantity
Symbol
Area
Volume
Velocity
Acceleration
𝒎𝟐
𝒎𝟑
𝒎/𝒔
𝒎/𝒔𝟐
Table 3: The SI System Derived Units With Special Names and Symbols
Derived Quantity
Unit Measurement
Name
Symbol
Equivalent SI Units
Frequency
Force
Pressure, Stress
Energy, Work
Quantity of Heat
Power, Radiant Flux
Electric Charge
Electric Potential,
Electromotive Force
Electric Resistance
Capacitance
Electric Conductance
Magnetic Flux
Magnetic Flux Density
Inductance
Plane Angle
Solid Angle
Illuminous flux
Illuminance
Celsius Temperature
Radioactivity
Absorbed Dose, Kerma
Equivalent Dose
Catalytic Activity
Hertz
Newton
Pascal
Joule
Hz
N
P
J
𝒔−𝟏
𝒌𝒈 ∙ 𝒎/𝒔𝟐
𝑵/𝒎𝟐
𝑵∙𝒎
Watt
Coulomb
Volt
W
C
V
𝑱/𝒔
𝒔∙𝑨
𝑾/𝑨
Ohm
Farad
Siemens
Weber
Tesla
Henry
Radian
Steradian
Lumen
Lux
degree Celsius
Becquerel
Gray
Sievert
Katal
Ω
F
S
Wb
T
H
rad
Sr
lm
lx
°𝑪
𝑽/𝑨
𝑪/𝑽
𝑨/𝑽
𝑽∙𝒔
𝑾𝒃/𝒎𝟐
𝑾𝒃/𝑨
Bq
Gy
Sv
Kat
𝒄𝒅 ∙ 𝒔𝒓
𝒍𝒎/𝒎𝟐
𝒔−𝟏
𝑱/𝒌𝒈
𝑱/𝒌𝒈
−𝟏
𝒔 ∙ 𝒎𝒐𝒍
1.1.1 Redefinition of the SI Base Units
On November 16, 2018, various measurement experts from 60 countries around the globe
have marked a historical breakthrough that will change the long-existing structure of the unified
measurement system. They unanimously agreed to redefine some of the SI base units of
measurements.
At the 26th convention of the General Conference on Weights and Measures (Conférence
Générale des Poids et Mesures or CGPM) happened in Versailles, France, delegates voted to
redefine the International System of Units (SI), changing the world’s definition of the four SI base
units of measurements: the kilogram, the ampere, the Kelvin, and the mole. Such changes have
finally come into full force on 2019 World Metrology Day, which happened on the 20th of May.
The revision of the SI system is pretty much inevitable due to the fact this universal
measurement system has always been established based on the fundamental constants of nature.
Long before the redefinition of SI system, its base units were defined based on physical objects
that would wear out over time; and experiments or phenomenon that are not universal.
3
However, the decisions of the policy makers have ultimately made the old SI system into
a more refined structure that might truly become a metric system “for all times and for all
people”. This action would ensure the future stability and open the opportunity for the use of
new technologies (including quantum technologies) to implement the new definitions of the SI
base units.
Figure 1: Revised SI Base Unit Chart
Image taken from https://www.nist.gov/image/siillustrationconstantscolourfullpng
Name:
Program and Year:
Date:
Score:
4
Activity 1.1: The International System of Units
A. Complete the table in reference to the SI System Redefinition.
Base Unit
Defining Constant
Defining Constant
Symbol
Numerical Value
Kilogram
Meter
Second
Kelvin
Mole
Ampere
Candela
B. Answer the following questions.
1. Why do you think that a universal system of measurement is necessary in science?
2. Why does the SI Measurement System use “kilogram” instead of “gram” as the base unit of
mass?
1.2 Unit Conversion
5
To work with objects with varying sizes, one needs to be able to convert large and small
measurements quickly. Unit conversion is a process that every scientist, engineer, and even
ordinary individual must learn and understand because not all countries around the world are
using the standard (SI) metric system.
1.2.1 Conversion of SI Units
The SI system, unlike other measurement systems, makes conversion simple because
prefixes are based on groups of ten. (see Table 4)
Table 4: SI Prefixes Chart
Image taken from http://www.learnalberta.ca/content/memg/Division03/International%20System%20of%20Units/index.html
In Table 4, note that the prefixes "kilo", "hecto", "centi" and "milli" are used very frequently
(light blue); the prefixes "mega", "deca" (or “deka”), "deci" and "micro" are are used less frequently
(light red), while the remaining prefixes (light purple) are rarely used (other than for extremely
large or small numbers in science).
Converting between equivalent units with different prefixes can usually be done by
multiplying the given quantity by one (1) in a special form.
6
The following ratios are equivalent to one since any non-zero number divided by itself is
equal to one.
1𝑚
1 𝑘𝑔
10 𝑚𝑠
1𝑚
1 𝑘𝑔
10 𝑚𝑠
Moreover, since 1 𝑘𝑚 = 1,000 𝑚 and 1 𝑐𝑔 = 0.01 𝑔, then the following ratios are likewise
equivalent to one.
1000 𝑚
1 𝑘𝑚
1 𝑐𝑔
0.01 𝑔
1 𝑘𝑚
1,000 𝑚
0.01 𝑔
1 𝑐𝑔
Example 1. Convert 11.42 𝑐𝑑 to dekacandela (dacd).
Solution:
To convert candela (cd) into dekacandela (dacd), eliminate candela (cd) by multiplying the given
1 𝑑𝑎𝑐𝑑
measured quantity by one in the form
, i.e.,
10 𝑐𝑑
1 𝑑𝑎𝑐𝑑
)
10 𝑐𝑑
11.42 𝑐𝑑 = 11.42 𝑐𝑑 (
= 11.42 × 10−1 𝑑𝑎𝑐𝑑 = 1.142 𝑑𝑎𝑐𝑑
Hence, 11.42 𝑐𝑑 = 𝟏. 𝟏𝟒𝟐 𝒅𝒂𝒄𝒅.
Example 2. Convert 2.15 𝑚𝑠 to kiloseconds (ks).
Solution:
To eliminate millisecond (ms), multiply the given measured quantity by one in the form
10−3 𝑠
1 𝑚𝑠
,
i.e.,
2.15 𝑚𝑠 (
10−𝟑 𝑠
)
(1)
1 𝑚𝑠
The conversion process above will then introduce the second (s) unit. Thus, to eliminate
introduced units of “s” and produce the desired units of “ks”, multiply (1) by one in the form
𝟏 𝒌𝒔
, i.e.,
𝟏𝟎𝟑 𝒔
10−3 𝑠
2.15 𝑚𝑠 = 2.15 𝑚𝑠 (
1 𝑚𝑠
)(
1 𝑘𝑠
103 𝑠
) = 2.15 × 10−3−3 𝑘𝑠 = 𝟐. 𝟏𝟓 × 𝟏𝟎−𝟔 𝒌𝒔
Hence, 2.15 𝑚𝑠 = 𝟐. 𝟏𝟓 × 𝟏𝟎−𝟔 𝒌𝒔 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟐𝟏𝟓 𝒌𝒔.
Try this!
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1. Convert 3.6 𝑐𝑚 to 𝑚𝑖𝑐𝑟𝑜𝑚𝑒𝑡𝑒𝑟 (𝝁𝒎).
2. Convert 0.51 𝑘𝑔 to 𝑡𝑒𝑟𝑎𝑔𝑟𝑎𝑚 (𝑻𝒈).
1.2.2 Unit Conversion of the US Customary System and the SI System
As the measurement system continues to evolve over several centuries, only two
types of measurement systems are distinguished historically. These are US Customary System
of Units, which grew more or less ambiguously out of custom, and The International System of
Units or “SI System”, which is a unified, planned system used by the world’s scientific community
and by most nations. Each system uses different set of units for measuring things.
The US Customary System of Units uses units that have no predictable relationship to
each other. For instance, there are 16 ounces in a pound, but 2,000 pounds in a ton. To convert
units under this system, divide or multiply according to the relationship between the two given
units.
The SI System, on the contrary, is made of units based on powers of 10, which makes it a
lot easier for us to recall the conversion formulas and to apply them in a unit conversion process.
Table 5
METRIC CONVERSIONS (US CUSTOMARY, SI, AND EQUIVALENTS)
A. UNITS OF LENGTH IN
THE US CUSTOMARY
SYSTEM
B. US CUSTOMARY
SYSTEM
SI CONVERSIONS FOR
LENGTH
1 foot (ft) = 12 inches (in)
1 yard (yd) = 3 feet (ft)
1 inch = 2.54 centimeters
1 yard (yd) = 36 inches (in)
1 meter ≈ 3.28 feet
1 mile (mi) = 5,280 feet (ft)
1 mile ≈ 1.61 kilometers
1 foot ≈ 0.30 meters
1 yard ≈ 0.91 meters
1 kilometer ≈ 0.62 miles
D. UNITS OF
MASS/WEIGHT IN THE US
CUSTOMARY SYSTEM
1 pound (lb) = 16 ounces
(oz)
E. UNITS OF MASS/
WEIGHT IN THE SI SYSTEM
1 tonne or metric ton (t) =
1,000 kg
C. UNITS OF TIME IN BOTH
SYSTEMS
1 minute (min) = 60 seconds (s)
1 hour (hr) = 60 minutes (min)
1 hour (hr) = 3,600 seconds (s)
1 day ≈ 24 hours (hrs)
1month (mo) ≈ 30 days
1 year (yr) ≈ 365 days
1 banking year (banking yr) =
360 days
F. US CUSTOMARY SYSTEM
SI CONVERSIONS FOR
MASS/WEIGHT
1 ounce ≈ 28.3 grams
1 pound ≈ 0.45 kilograms
1 ton (T) = 2,000 pounds (lb)
G. UNITS OF AREA IN
THE US CUSTOMARY
SYSTEM
H. UNITS OF AREA IN THE
SI SYSTEM
I. US CUSTOMARY SYSTEM
SI CONVERSIONS FOR
AREA
1 are (a) = 100 m2
1 ft2 = 144 in2
1 yd2 = 9 ft2
1 hectare (ha) = 100 a
1 in2 ≈ 6.45 cm2
1 m2 ≈ 1.196 yd2
8
1 acre = 43,560 ft2
1 ha ≈ 2.47 acres
1 mi2 = 640 acres
J. UNITS OF VOLUME IN
THE US CUSTOMARY
SYSTEM
1 ft3 = 1,728 in3
K. UNITS OF VOLUME IN
THE SI SYSTEM
1 yd = 27 ft
1 milliliter (mL) = 1 cm3
3
3
L. US CUSTOMARY SYSTEM
SI CONVERSIONS FOR
VOLUME
1 cc = 1 cm3
1 in3 ≈ 16.39 mL
1 liter (L) ≈ 1.06 quarts (qt)
1 cord = 128 ft3
1 gallon (gal) ≈ 3.79 liters (L)
1 tablespoon = 3 teaspoons
1 m3 ≈ 35.31 ft3
1 fluid ounce (fl oz) = 2
tablespoons
1 quart (qt) ≈ 0.95 liters (L)
1 cup (c) = 8 fl oz
1 pint (pt) = 2 cups (c)
1 qt = 2 pints (pt)
1 gallon (gal) = 4 quarts (qt)
1 gallon (gal) = 128 fl oz
M. UNITS OF TEMPERATURE IN BOTH SYSTEMS
𝟓
°𝐅 → °𝐂:
𝐂 = 𝟗 (𝐅 − 𝟑𝟐)
°𝐂 → °𝐅:
𝐅 = 𝟓 𝐂 + 𝟑𝟐
°𝐂 → 𝐊:
𝐊 = 𝐂 + 𝟐𝟕𝟑. 𝟏𝟓
𝟗
𝐊 → °𝐂:
°𝐅 → 𝐊:
𝐊 → °𝐅:
𝐂 = 𝐊 − 𝟐𝟕𝟑. 𝟏𝟓
𝟓
𝐊 = 𝟗 (𝐅 − 𝟑𝟐) + 𝟐𝟕𝟑. 𝟏𝟓
𝟗
𝐅 = (𝐊 − 𝟐𝟕𝟑. 𝟏𝟓) + 𝟑𝟐
𝟓
Example 1. Convert 3.4 𝑚𝑖 to yards (yd)
Solution:
Both miles (mi) and yards (yd) belong to the US Customary System of Units of Length. Referring
to Table 5.A, we see that the appropriate conversion formulas to use are as follows:
1 𝑚𝑖 = 5,280 𝑓𝑡
1 𝑦𝑑 = 3 𝑓𝑡
To convert miles (mi) into yards (yd), multiply 3.4 𝑚𝑖 by one in the form
(
5,280 𝑓𝑡
1 𝑚𝑖
)(
1 𝑦𝑑
3 𝑓𝑡
),
i.e.,
3.4 𝑚𝑖 = 3.4 𝑚𝑖 (
5,280 𝑓𝑡 1 𝑦𝑑
)(
)
1 𝑚𝑖
3 𝑓𝑡
= 𝟓, 𝟗𝟖𝟒 𝒚𝒅
Hence, 3.4 𝑚𝑖 = 𝟓, 𝟗𝟖𝟒 𝒚𝒅.
9
Example 2. Convert 28.8 𝑐𝑔 to tons (T).
Solution:
The units of centigrams (cg) and tons (T) belong to the SI System of Mass/Weight and the US
Customary System of Mass/Weight, respectively. Referring to Table 4, Table 5.D and Table 5.F,
we see that the appropriate conversion formulas to use are as follows:
1 𝑐𝑔 = 10−2 𝑔
1 𝑘𝑔 = 103 𝑔
1 𝑙𝑏 ≈ 0.45 𝑘𝑔
1 𝑇 = 2,000 𝑙𝑏𝑠
To convert centigrams (cg) into tons (T), multiply 28.8 𝑐𝑔 by one in the form
10−2 𝑔
1 𝑘𝑔
1 𝑙𝑏
1𝑇
(
)( 3 )(
)(
), i.e.,
1 𝑐𝑔
10 𝑔
0.45 𝑘𝑔
2,000 𝑙𝑏𝑠
−𝟕
10−2 𝑔
1 𝑘𝑔
1 𝑙𝑏
1𝑇
28.8 𝑐𝑔 = 28.8 𝑐𝑔 ( 1 𝑐𝑔 ) ( 3 ) (0.45 𝑘𝑔) (2,000 𝑙𝑏𝑠) ≈ 𝟑. 𝟐𝟎 × 𝟏𝟎
10 𝑔
𝑻
Hence, 28.8 𝑐𝑔 ≈ 𝟑. 𝟐𝟎 × 𝟏𝟎−𝟕 𝑻. *
1 𝑙𝑏
* The use of equal sign (=) is not appropriate in this case since one of the multipliers, (0.45 𝑘𝑔), is derived only from the
approximated value of pounds in terms of kilograms, as shown in Table 5.F.
Example 3. Convert 3.2 𝑐𝑜𝑟𝑑𝑠 to cubic decimeters (dm3). Round off your answer to the nearest
hundredths.
Solution:
The units of cords and cubic decimeters (dm3) belong to the SI System of Volume and the US
Customary System of Volume, respectively. Referring to Table 4, Table 5.J and Table 5.L, we see
that the appropriate conversion formulas to use are as follows:
1 𝑑𝑚 = 10−1 𝑚
1 𝑐𝑜𝑟𝑑 = 128 𝑓𝑡 3
1 𝑚 3 ≈ 35.31 𝑓𝑡 3
To convert cords into cubic decimeters (dm3), multiply 3.2 𝑐𝑜𝑟𝑑𝑠 by one in the form
128 𝑓𝑡 3
1 𝑚3
1 𝑑𝑚 𝟑
128 𝑓𝑡 3
1 𝑚3
1 𝑑𝑚3
(
)(
)
(
)
(
)
(
)
(
) , i.e.,
=
1 𝑐𝑜𝑟𝑑
35.31 𝑓𝑡 3
1 0−1 𝑚
1 𝑐𝑜𝑟𝑑
35.31 𝑓𝑡 3
1 0−3 𝑚3
128 𝑓𝑡 3
3.2 𝑐𝑜𝑟𝑑𝑠 = 3.2 𝑐𝑜𝑟𝑑𝑠 (
1 𝑐𝑜𝑟𝑑
)(
1 𝑚3
35.31 𝑓𝑡
)(
3
1 𝑑𝑚3
1 0−3 𝑚 3
) ≈ 𝟏𝟏, 𝟔𝟎𝟎. 𝟏𝟏 𝒎𝟑
Hence, 3.2 𝑐𝑜𝑟𝑑𝑠 ≈ 𝟏𝟏, 𝟔𝟎𝟎. 𝟏𝟏 𝒎𝟑 𝒐𝒓 𝟏. 𝟏𝟔 × 𝟏𝟎𝟒 𝒎𝟑.
10
Try this! Convert the following measurements. Round off your answers to the nearest
hundredths, if possible.
1. 25,000 𝑘𝑚 2 to 𝑠𝑞𝑢𝑎𝑟𝑒 𝑓𝑒𝑒𝑡 (𝒇𝒕𝟐 ).
2. 7,200 𝑚𝑠 to 𝑑𝑎𝑦𝑠.
3. 150 °𝐹 to Kelvin (K)
Name:
Date:
11
Program and Year:
Score:
Activity 1.2: Unit Conversion
A. Convert the following measurements. For very small and/or very large results, write your
answers in scientific notation. If possible, round off your answers to the nearest hundredths.
1.
2.
3.
4.
5.
0.002 L  mL
3.6 cm  mm
1520 A  GA
0.45 dHz  daHz
6,520,000 µm  in
6. 4.75 fL  teaspoons
7. 3,500 hg  metric tons
8. 4 mo  ns
9. 4 cups  cm3
10. 324 K  °F
B. Solve the following problems as indicated. If possible, write your answers to the nearest
hundredths.
1. A newborn baby and a large professional football player are placed onto separate balances.
One has a mass of 4,000 grams, and the other has a mass of 90 kilograms. Which mass belongs to
the football player? Which mass is the baby’s?
2. A couple went to see a romantic movie. It started at 1:45 PM and ended at 4:00 PM. How
long was the movie, in centiseconds?
1.3 Accuracy, Precision, and Error in Measurements
“It is better to be roughly right than precisely wrong."
- Carveth Read
-
Galileo Galilei
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1.3.1 Accuracy, Precision, Uncertainties, and Error
When scientists take measurements, they generally have two goals—accuracy and
precision. However, most people often tend to confuse these two ideas, and several of the
definitions out there aren’t entirely clear.

Accuracy is the closeness of agreement between a measured value and a true or accepted
value.

Precision is a measure of how well a result can be determined (without reference to a
theoretical or true value). It is the degree of consistency and agreement among
independent measurements of the same quantity; also the reliability or reproducibility of
the result.
A target can serve as a concrete example to clarify the difference between accuracy and
precision, as shown in Figure 1 below. For this purpose, the measurements are arrows being shot
into the target. Accuracy describes how close to the center of the target the arrows have landed.
The arrows closer to the bulls-eye are more accurate. Precision, however, denotes how close in
relation to one another the arrows have landed in the target. If all arrows are close together, the
bunch is precise as all of the arrows are gathered at (nearly) the same spot. This spot is not
necessarily close to the bulls-eye, i.e. the shooting was precise, but not accurate.
Figure 1: An Illustration of Accuracy vs. Precision
Image taken from https://www.geavis.si/en/2017/06/difference-between-accuracy-and-precision/
Unfortunately, measurement is never 100% precise or accurate, so the true value measure
of something is never exactly known. This uncertainty is a result of error. Error is a concept that
is naturally associated with measuring because measurement is always a comparison to a
standard. Measuring something manually always involves uncertainty because it is based on
judgment. If two people use a ruler to measure how long a human thumb is, it may look like 3
inches to one person and 2 inches to the other.
There are two main types of error—random error and systematic error. Random error is
not controllable. As the name suggests, the occurrence of random errors is random and due to
chance. Alternatively, systematic errors are controllable and have a known cause. A systematic
13
error can result from many things, such as instrument error, method error, or human error.
Systematic errors can usually be identified and reduced or even eliminated.
When making measurements, we generally assume that some exact or true value exists
based on how we define what is being measured. Such exact value may be derived from a
“textbook”, which is usually well-known since it is used as the basis of an ideal measurement of
an object. Other times we know a theoretical value, which is calculated from basic principles,
and this also may be taken as an "ideal" value.
The most common way to show the range of values that we believe includes the true value
is
Measurement = (best estimate ± uncertainty) units
Note 1: As far as taking measurements of an object can go, remember to always round off the best measured
value to the same decimal place as the uncertainty.
Note 2: There are many ways to measure an uncertainty. The most common way is to take the uncertainty
to be half of the smallest division of your measuring device. For example, if your meter stick has tick-marks
every 0.01 m, then your uncertainty is ±0.005 𝑚.
Example 1. A measurement of 5.07 g ± 0.02 g means that the experimenter is confident that the
true value for the quantity being measured lies between 5.05 g and 5.09 g. The uncertainty (which
is ± 0.02 g) is the experimenter's best estimate of how far an experimental quantity might be from
the "true value."
1.3.2 Estimating Uncertainty for a Single Measurement
Quantitatively, precision is often reported by using “relative or fractional uncertainty”,
with formula given below:
𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒖𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 (𝑼)
|
𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 (𝑴𝑽)
On the other note, accuracy is reported quantitatively by using “relative error”, with
formula given below:
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑼𝒏𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚 (𝑹𝑼) = |
|𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 (𝑴𝑽) − 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒗𝒂𝒍𝒖𝒆(𝑬𝑽)| |𝑬𝒓𝒓𝒐𝒓 (𝑬)|
=
𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒗𝒂𝒍𝒖𝒆 (𝑬𝑽)
𝑬𝑽
Example 2. Suppose that the diameter of a tennis ball is (𝟔. 𝟕 ± 𝟎. 𝟐) 𝒄𝒎. Compute the relative
uncertainty and the relative error of the measurement if the ideal diameter of a tennis ball is
𝟔. 𝟖 𝒄𝒎.
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑬𝒓𝒓𝒐𝒓 (𝑹𝑬) =
Solution:
The measured value and the uncertainty of the diameter’s length of the tennis ball are 6.7
cm and 0.2 cm, respectively. Thus, 𝑴𝑽 = 𝟔. 𝟕 𝒄𝒎 and 𝑼 = ± 𝟎. 𝟐 𝒄𝒎.
Moreover, its expected value is given by 6.8 cm. Thus, 𝑬𝑽 = 𝟔. 𝟖 𝒄𝒎.
14
With the given information above, its relative uncertainty and relative error are computed
as follows:
Relative Uncertainty: 𝑅𝑈 = |
𝑼
𝑴𝑽
|
Relative Error: 𝑅𝐸 =
± 𝟎.𝟐 𝒄𝒎
𝑅𝑈 = |
𝟔.𝟕 𝒄𝒎
|
𝑅𝑈 ≈ 𝟎. 𝟎𝟑 𝒐𝒓 𝟑%
𝑅𝐸 =
|𝑴𝑽 − 𝑬𝑽|
𝑬𝑽
|𝟔.𝟕𝒄𝒎 − 𝟔.𝟖 𝒄𝒎|
𝟔.𝟖 𝒄𝒎
𝑅𝐸 ≈ 𝟎. 𝟎𝟏𝟓 𝒐𝒓 𝟏. 𝟓%
Try this! The relative uncertainty of the length of a pen is 0.4%, with a standard uncertainty of
0.03 inches.
(a) Compute the measured length of the pen.
(b) If the relative error of the length is 0.00625, what is the expected length of the pen?
(c) Determine the estimated length of the pen in the form (measured value ± uncertainty)
inches.
1.3.3 Estimating Uncertainty for a Repeated Measurement
To increase the accuracy, and thereby reducing the error, of the measurement, it is
advisable to measure a certain property of an object more than once. As a result, it can give us a
better idea concerning the sense of uncertainty in the measurement.
15
For repeated measurement, the best estimate of the “true” or “expected” value is the
“AVERAGE” or “MEAN”.
̅) =
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 (𝒙
𝒔𝒖𝒎 𝒐𝒇 𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕𝒔 ∑ 𝒙
=
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏𝒔
𝒏
The uncertainty, on the other note, associated with the average value is the standard error
(also called standard deviation of the mean).
𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝑬𝒓𝒓𝒐𝒓 (𝝈𝒙̅ ) =
𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
√𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏𝒔
=
∑(𝒙 − 𝒙
̅)𝟐
= √
𝒏(𝒏 − 𝟏)
√𝒏
𝒔
In connection to the formulas above, the range of values believed to have included the
“true” value is given by:
̅ ± 𝝈𝒙̅ ) 𝒖𝒏𝒊𝒕𝒔
𝑴𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕 = (𝒙
Example 3. Consider measuring the width of a piece of paper using the meter stick. The width of
the paper is measured at a number of points on the sheet, and the values are entered in the data
table below:
Observation
#1
#2
#3
#4
#5
Width (cm)
31.32
31.16
31.24
31.04
31.20
Do the following and round off your answers to the nearest hundredths.
a) Compute the mean width of a piece of paper based on the observed measures in the data table.
b) Calculate the standard error of the width of a piece of paper.
c) Estimate the width of the piece of paper by expressing it in the form (mean ± standard error)
cm.
d) Determine the relative uncertainty and the relative error of the measurement if the true width
of the piece of paper is 𝟑𝟏. 𝟐𝟎 𝒄𝒎.
Solution:
From the given problem, the expected or true width of the piece of paper is 31.20 cm.
Thus, 𝑬𝑽 = 𝟑𝟏. 𝟐𝟎 𝒄𝒎.
a) Mean Width of the Piece of Paper:
𝑥̅ =
∑𝒙
𝒏
16
𝑥̅ =
(𝟑𝟏.𝟑𝟐 + 𝟑𝟏.𝟏𝟔 + 𝟑𝟏.𝟐𝟒 + 𝟑𝟏.𝟎𝟒 + 𝟑𝟏.𝟐𝟎) 𝒄𝒎
𝟓
𝑥̅ ≈ 𝟑𝟏. 𝟏𝟗 𝒄𝒎
b) Standard Error of the Width of the Piece of Paper:
𝜎𝑥̅ = √
𝜎𝑥̅ ≈ √
∑(𝒙 − 𝒙
̅)𝟐
𝒏(𝒏 − 𝟏)
(𝟑𝟏.𝟑𝟐−𝟑𝟏.𝟏𝟗)𝟐 + (𝟑𝟏.𝟏𝟔−𝟑𝟏.𝟏𝟗)𝟐 + (𝟑𝟏.𝟐𝟒−𝟑𝟏.𝟏𝟗)𝟐 + (𝟑𝟏.𝟎𝟒−𝟑𝟏.𝟏𝟗)𝟐 + (𝟑𝟏.𝟐𝟎−𝟑𝟏.𝟏𝟗)𝟐
𝟓(𝟓−𝟏)
𝒄𝒎
𝜎𝑥̅ ≈ 𝟎. 𝟎𝟓 𝒄𝒎
c) Estimated Width of the Piece of Paper:
̅ ± 𝝈𝒙̅ ) 𝑐𝑚
𝑊𝑖𝑑𝑡ℎ = (𝒙
𝑊𝑖𝑑𝑡ℎ = (𝟑𝟏. 𝟏𝟗 ± 𝟎. 𝟎𝟓) 𝒄𝒎
Thus, result above indicates that the true or ideal width of the piece of paper lies between
(𝟑𝟏. 𝟏𝟗 − 𝟎. 𝟎𝟓) 𝒄𝒎 or 𝟑𝟏. 𝟏𝟒 𝒄𝒎 and (𝟑𝟏. 𝟏𝟗 + 𝟎. 𝟎𝟓) 𝒄𝒎 or 𝟑𝟏. 𝟐𝟒 𝒄𝒎.
d) Relative Uncertainty and Relative Error of the Width:
Relative Uncertainty:
Relative Error:
𝑼
𝑅𝑈 = |𝑴𝑽|
𝑅𝐸 =
𝝈
𝑅𝑈 = | 𝒙̅|
𝑅𝐸 =
̅
𝒙
𝑅𝑈 = |
± 𝟎.𝟎𝟓 𝒄𝒎
𝟑𝟏.𝟏𝟗 𝒄𝒎
|
𝑅𝑈 ≈ 𝟏. 𝟔𝟎 × 𝟏𝟎−𝟑 𝒐𝒓 𝟎. 𝟏𝟔%
𝑅𝐸 =
|𝑴𝑽 − 𝑬𝑽|
𝑬𝑽
|𝒙
̅ − 𝑬𝑽|
𝑬𝑽
|𝟑𝟏.𝟏𝟗 𝒄𝒎 − 𝟑𝟏.𝟐𝟎 𝒄𝒎|
𝟑𝟏.𝟐𝟎 𝒄𝒎
𝑅𝐸 ≈ 𝟑. 𝟐𝟏 × 𝟏𝟎−𝟒 𝒐𝒓 𝟎. 𝟎𝟑%
Try this! The volume of a Rubik’s cube is measured four times and these are the observed
measures (in cubic inches): 𝟏𝟏. 𝟑, 𝟏𝟏. 𝟐, 𝟏𝟏. 𝟒, 𝟏𝟏. 𝟑.
(a) Estimate the volume of the Rubik’s cube by expressing it in the form (mean ±
standard error) in3 .
17
(b) Determine the relative uncertainty and the relative error of the measurement.
Name:
Program and Year:
Date:
Score:
Activity 1.3: Accuracy, Precision, and Error in Measurement
Solve the following problems as directed.
18
1. A glass of water was measured three times. The correct measurement was 65.4 𝑚𝐿. Encircle
whether the following set of measurements is accurate, precise, both, or neither.
a) 75.4 𝑚𝐿, 52.1 𝑚𝐿, 44.9 𝑚𝐿
accurate
precise
both
neither
b) 65.3 𝑚𝐿, 65.4 𝑚𝐿, 65.5 𝑚𝐿
accurate
precise
both
neither
c) 89.1 𝑚𝐿, 89.0 𝑚𝐿, 88.9 𝑚𝐿
accurate
precise
both
neither
d) 65.7 𝑚𝐿, 64.5 𝑚𝐿, 68.0 𝑚𝐿
accurate
precise
both
neither
2. In a certain class, the best student got a score of 25 out of 35 questions in a Statistics quiz.
What is the best student’s relative error?
3. A student measured the amount of time she needed to reach the school from home using a
stopwatch. If the measured travel time is 36.75 minutes, with a relative uncertainty is 0.11%,
a) how many minutes she is uncertain of the travel time?
b) determine the range (in minutes) in which the true travel time exactly lies.
c) Suppose the exact travel time from her home to school is 36.78 minutes. What is
the relative error?
4. The radius of a circular-shaped pizza was measured four times, through which the observed
lengths of its radius (in inches) were as follows: 4.21, 4.16, 4.18, 4.13.
a) Estimate the radius of the pizza in the form (𝑥̅ ± 𝜎𝑥̅ ) inches. Round off the values
of 𝑥̅ and 𝜎𝑥̅ to two decimal places.
b) Compute the relative uncertainty and the relative error of the measured radius of
the pizza, if the exact length of radius of the pizza is 4.20 inches.
19
1.5 Bisection Method of Reducing Error
One of the first numerical methods developed to find the root of a non-linear equation
𝑓 (𝑥) = 0 was the bisection method (also called binary-search method). This method is based on the
following theorem:
THEOREM:
An equation 𝑓 (𝑥) = 0, where 𝑓 (𝑥) is a real, continuous function has at least one root between 𝑥𝑙
and 𝑥𝑢 if 𝑓(𝑥𝑙 ) ∙ 𝑓(𝑥𝑢 ) < 0. (See Figure 1)
If 𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑢 ) > 0, there may or may not be any root between 𝑥𝑙 and 𝑥𝑢 . (See Figure 2 and Figure
3)
If 𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑢 ) < 0, then there may be more than one root between 𝑥𝑙 and 𝑥𝑢 . (See Figure 4)
Thus, the theorem only guarantees one root between 𝑥𝑙 and 𝑥𝑢 .
Figure 1
Figure 2
There is a root in 𝒙𝒍 , 𝒙𝒖 .
At least one root exists between two points if the
If the function does not change sign between two points,
function is real, continuous, and changes sign.
the roots of equation may still exist between the two points.
Figure 3
If the function does not change sign between two points, there may not be any root for the equation𝑓(𝑥) = 0 between the two
points.
Figure 4
20
If the function changes sign between two points, more than one root for the equation may exist between the two points
Images taken from http://mathforcollege.com/nm/mws/gen/03nle/mws_gen_nle_txt_bisection.pdf
ALGORITHM FOR THE BISECTION METHOD
The steps to apply the bisection method to find the root of the equation 𝑓 (𝑥) = 0 are as follows:
1) Choose 𝑥𝑙 and 𝑥𝑢 as two guesses for the root such that 𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑢 ) < 0, or in other words,
𝑓 (𝑥) changes sign between 𝑥𝑙 and 𝑥𝑢 .
2) Estimate the root, 𝑥𝑚 , of the equation 𝑓 (𝑥) = 0 as the midpoint between 𝑥𝑙 and 𝑥𝑢 , i.e.,
𝒙𝒎 =
𝒙𝒍 +𝒙𝒖
𝟐
.
3) Now, check the following:
(a) If 𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑚 ) < 0, then the root lies between 𝑥𝑙 and 𝑥𝑚 , where 𝑥𝑙 = 𝑥𝑙 and 𝑥𝑢 = 𝑥𝑚 .
(b) If 𝑓(𝑥𝑙 ) ∙ 𝑓(𝑥𝑚 ) > 0, then the root lies between 𝑥𝑚 and 𝑥𝑢 , where 𝑥𝑙 = 𝑥𝑚 and 𝑥𝑢 = 𝑥𝑢 .
(c) If 𝑓(𝑥𝑙 ) ∙ 𝑓(𝑥𝑚 ) = 0, then the root is 𝑥𝑚 . Stop the algorithm if this is true.
4) Find the new estimate of the root; i.e.,
𝒙𝒏𝒆𝒘
=
𝒎
𝒙𝒍𝒏𝒆𝒘 +𝒙𝒖𝒏𝒆𝒘
𝟐
Also, find the absolute relative approximate error as
.
𝒐𝒍𝒅
𝒙𝒏𝒆𝒘
𝒎 − 𝒙𝒎
|∈𝒂 | = |
𝒙𝒏𝒆𝒘
𝒎
| × 𝟏𝟎𝟎%,
where
𝒙𝒏𝒆𝒘
𝒎 is the estimated root of the present iteration; and
𝒙𝒐𝒍𝒅
𝒎 is the estimated root of the previous iteration.
5) Compare the absolute relative approximate error with the pre-specified relative error tolerance
∈𝒔 .
21
Is ∈ a < ∈ s
?
Go to Step 2 using
new upper and
lower guesses.
Yes
Stop the algorithm
N
o
Example 1. You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC
commodes. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are
asked to find the depth to which the ball is submerged when floating in water.
The equation that gives the depth x to which the ball is submerged under water is given by
𝒙𝟑 − 𝟎. 𝟏𝟔𝟓𝒙𝟐 + 𝟑. 𝟗𝟗𝟑 × 𝟏𝟎−𝟒 = 𝟎
Use the bisection method of finding roots of equations to find the depth 𝒙 (in meters) to which
the ball is submerged under water. Conduct iterations to estimate the root of the above
equation. Find the absolute relative approximate error at the end of each iteration, with an
error tolerance of ∈𝑠 = 10%. Round off your final answer to the nearest thousandths.
Solution:
From the physics of the problem, the ball would be submerged between x  0 and x  2 R ,
where R represents the radius of the ball (in meters), that is
0 ≤ 𝑥 ≤ 2𝑅
0 ≤ 𝑥 ≤ 2(0.055)
0 ≤ 𝑥 ≤ 0.11
Figure 5
Floating Ball Problem
22
Images taken from http://mathforcollege.com/nm/mws/gen/03nle/mws_gen_nle_txt_bisection.pdf
Suppose that 𝑥𝑙 = 0 and 𝑥𝑢 = 0.11.
Check if the function changes sign between 𝑥𝑙 and 𝑥𝑢 .
𝑓 (𝑥𝑙 ) = 𝑓 (0) = (0)3 − 0.165 (0)2 + 3.993 × 10−4 = 3.993 × 10−4
𝑓 (𝑥𝑢 ) = 𝑓(0.11) = (0.11)3 − 0.165(0.11)2 + 3.993 × 10−4 = 2.662 × 104
Hence , 𝒇(𝒙𝒍 ) ∙ 𝒇(𝒙𝒖 ) = 𝒇(𝟎) ∙ 𝒇(𝟎. 𝟏𝟏) = (𝟑. 𝟗𝟗𝟑 × 𝟏𝟎−𝟒 )(𝟐. 𝟔𝟔𝟐 × 𝟏𝟎𝟒 ) < 0
So there is at least one root between 𝑥𝑙 and 𝑥𝑢 , that is, between 0 and 0.11.
Iteration 1
The estimate of the root is
𝒙𝒎 =
𝒙𝒎 =
𝒙𝒍 + 𝒙𝒖
𝟐
𝟎 + 𝟎. 𝟏𝟏
𝟐
𝒙𝒎 = 𝟎. 𝟎𝟓𝟓
𝑓(𝑥𝑚 ) = 𝑓 (0.055) = (0.055)3 − 0.165(0.055)2 + 3.993 × 10−4 ≈ 6.655 × 10−5
𝑓(𝑥𝑙 ) ∙ 𝑓(𝑥𝑚 ) = 𝑓(0) ∙ 𝑓 (0.055) = (3.993 × 10−4 )(6.655 × 10−5 ) > 0
Hence the root is bracketed between 𝑥𝑚 and𝑥𝑢 , that is, between 0.055 and 0.11. So, the lower
and upper limit of the new interval is
𝒙𝒍 = 𝟎. 𝟎𝟓𝟓 and 𝒙𝒖 = 𝟎. 𝟏𝟏
At this point, the absolute relative approximate error |∈𝑎 | cannot be calculated as we do not
have a previous approximation.
Iteration 2
The estimate of the root is
𝒙𝒎 =
𝒙𝒎 =
𝒙𝒍 + 𝒙𝒖
𝟐
𝟎. 𝟎𝟓𝟓 + 𝟎. 𝟏𝟏
𝟐
𝒙𝒎 = 𝟎. 𝟎𝟖𝟐𝟓
𝑓 (𝑥𝑚 ) = 𝑓 (0.0825) = (0.0825)3 − 0.165(0.0825)2 + 3.993 × 10−4 ≈ −1.622 × 10−4
𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑚 ) = 𝑓 (0.055) ∙ 𝑓 (0.0825) = (6.655 × 10−5 )(−1.622 × 10−4 ) < 0
23
Hence, the root is bracketed between 𝑥𝑙 and 𝑥𝑚 , that is, between 0.055 and 0.0825. So the lower
and upper limit of the new interval is
𝒙𝒍 = 𝟎. 𝟎𝟓𝟓 and 𝒙𝒖 = 𝟎. 𝟎𝟖𝟐𝟓
The absolute relative approximate error |∈𝑎 | at the end of Iteration 2 is
𝑛𝑒𝑤
𝑜𝑙𝑑
𝑥𝑚
− 𝑥𝑚
|∈𝑎 | = |
| × 100%
𝑛𝑒𝑤
𝑥𝑚
0.0825 − 0.055
|∈𝑎 | = |
| × 100%
0.0825
|∈𝑎 | ≈ 𝟑𝟑. 𝟑𝟑%
Since |∈𝑎 | ≈ 33.33% > ∈𝑠 = 10%, then the algorithm continues in the next iteration.
Iteration 3
The estimate of the root is
𝒙𝒎 =
𝒙𝒎 =
𝒙𝒍 + 𝒙𝒖
𝟐
𝟎. 𝟎𝟓𝟓 + 𝟎. 𝟎𝟖𝟐𝟓
𝟐
𝒙𝒎 = 𝟎. 𝟎𝟔𝟖𝟕𝟓
𝑓 (𝑥𝑚 ) = 𝑓 (0.06875) = (0.06875)3 − 0.165(0.06875)2 + 3.993 × 10−4 ≈ −5.563 × 10−5
𝑓 (𝑥𝑙 ) ∙ 𝑓(𝑥𝑚 ) = 𝑓 (0.055) ∙ 𝑓(0.06875) = (6.655 × 10−5 )(−5.563 × 10−5 ) < 0
Hence, the root is bracketed between 𝑥𝑙 and 𝑥𝑚 , that is, between 0.055 and 0.06875. So the
lower and upper limit of the new bracket is
𝒙𝒍 = 𝟎. 𝟎𝟓𝟓 and 𝒙𝒖 = 𝟎. 𝟎𝟔𝟖𝟕𝟓
The absolute relative approximate error |∈𝑎 | at the end of Iteration 3 is
𝑛𝑒𝑤
𝑜𝑙𝑑
𝑥𝑚
− 𝑥𝑚
|∈𝑎 | = |
| × 100%
𝑛𝑒𝑤
𝑥𝑚
|∈𝑎 | = |
0.06875 − 0.0825
| × 100%
0.06875
|∈𝑎 | = 𝟐𝟎%
Since |∈𝑎 | = 20% > ∈𝑠 = 10%, then the algorithm continues in the next iteration.
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Iteration 4
The estimate of the root is
𝒙𝒎 =
𝒙𝒎 =
𝒙𝒍 + 𝒙𝒖
𝟐
𝟎. 𝟎𝟓𝟓 + 𝟎. 𝟎𝟔𝟖𝟕𝟓
𝟐
𝒙𝒎 = 𝟎. 𝟎𝟔𝟏𝟖𝟕𝟓
𝑓(𝑥𝑚 ) = 𝑓(0.061875) = (0.061875)3 − 0.165(0.061875)2 + 3.993 × 10−4 ≈ 4.484 × 10−6
𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑚 ) = 𝑓 (0.055) ∙ 𝑓(0.061875) = (6.655 × 10−5 )(4.484 × 10−6 ) > 0
Hence, the root is bracketed between 𝑥𝑚 and 𝑥𝑢 , that is, between 0.061875 and 0.06875. So the
lower and upper limit of the new bracket is
𝒙𝒍 = 𝟎. 𝟎𝟔𝟏𝟖𝟕𝟓 and 𝒙𝒖 = 𝟎. 𝟎𝟔𝟖𝟕𝟓
The absolute relative approximate error |∈𝑎 | at the end of Iteration 4 is
|∈𝑎 | = |
|∈𝑎 | = |
𝑛𝑒𝑤
𝑜𝑙𝑑
𝑥𝑚
− 𝑥𝑚
| × 100%
𝑛𝑒𝑤
𝑥𝑚
0.061875 − 0.06875
| × 100%
0.061875
|∈𝑎 | ≈ 𝟏𝟏. 𝟏𝟏%
Since |∈𝑎 | = 11.11% > ∈𝑠 = 10%, then the algorithm continues in the next iteration.
Iteration 5
The estimate of the root is
𝒙𝒎 =
𝒙𝒎 =
𝒙𝒍 + 𝒙𝒖
𝟐
𝟎. 𝟎𝟔𝟏𝟖𝟕𝟓 + 𝟎. 𝟎𝟔𝟖𝟕𝟓
𝟐
𝒙𝒎 = 𝟎. 𝟎𝟔𝟓𝟑𝟏𝟐𝟓
𝑓(𝑥𝑚 ) = 𝑓(0.0653125) = (0.0653125)3 − 0.165(0.0653125)2 + 3.993 × 10−4 ≈ −2.594 × 10−5
𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑚 ) = 𝑓 (0.061875) ∙ 𝑓 (0.0653125) = (4.484 × 10−6 )(−2.594 × 10−5 ) < 0
Hence, the root is bracketed between 𝑥𝑙 and 𝑥𝑚 , that is, between 0.061875 and 0.0653125. So the
lower and upper limit of the new bracket is
𝒙𝒍 = 𝟎. 𝟎𝟔𝟏𝟖𝟕𝟓 and 𝒙𝒖 = 𝟎. 𝟎𝟔𝟓𝟑𝟏𝟐𝟓
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The absolute relative approximate error |∈𝑎 | at the end of Iteration 5 is
|∈𝑎 | = |
|∈𝑎 | = |
𝑛𝑒𝑤
𝑜𝑙𝑑
𝑥𝑚
− 𝑥𝑚
| × 100%
𝑛𝑒𝑤
𝑥𝑚
0.0653125 − 0.061875
| × 100%
0.0653125
|∈𝑎 | ≈ 𝟓. 𝟐𝟔%
Since |∈𝑎 | ≈ 𝟓. 𝟐𝟔% < ∈𝑠 = 10%, then the algorithm will terminate up to this iteration. This
means that the root of 𝑥 3 − 0.165𝑥 2 + 3.993 × 10−4 = 0 is 𝒙𝒎 = 𝟎. 𝟎𝟔𝟓𝟑𝟏𝟐𝟓, with ∈𝒔 = 𝟏𝟎%.
Conclusion: With an error tolerance of ∈𝒔 = 𝟏𝟎%, the depth to which the ball is submerged
under water is approximately 0.065 meters (to the nearest thousandths).
Advantages of bisection method:
a) The bisection method is always convergent. Since the method brackets the root, the
method is guaranteed to converge.
b) As iterations are conducted, the interval gets halved. So one can guarantee the error
in the solution of the equation.
Drawbacks of bisection method:
a) The convergence of the bisection method is slow as it is simply based on halving the
interval.
b) If one of the initial guesses is closer to the root, it will take larger number of iterations
to reach the root.
c) If a function f (x) is such that it just touches the x -axis (see Figure 6) just like 𝑓(𝑥) =
𝑥 2 = 0, it will be unable to guess the lower bound, 𝑥𝑙 , and upper bound, 𝑥𝑢 , such that
𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑢 ) < 0.
d) For functions f (x) where there is a singularity* and it reverses sign at the singularity,
the bisection method may converge on the singularity (see Figure 7). An example
1
includes 𝑓 (𝑥 ) = , where 𝑥𝑙 = −2 and 𝑥𝑢 = 3 are valid initial guesses which satisfy
𝑥
𝑓 (𝑥𝑙 ) ∙ 𝑓 (𝑥𝑢 ) < 0. However, the function is not continuous and the theorem that a root
exists is also not applicable.
* A singularity in a function is defined as a point where the function becomes infinite. For example, for a function such as
the point of singularity is
x  0 as it becomes infinite.
1/ x ,
Figure 6
26
f (x)
The equation
f ( x)  x 2  0
has a single root at
x  0 that cannot be bracketed.
x
Figure 7
f (x)
x
The equation
f (x ) =
1
= 0 has no root but changes sign.
x
Images taken from http://mathforcollege.com/nm/mws/gen/03nle/mws_gen_nle_txt_bisection.pdf
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Name:
Program and Year:
Date:
Score:
Activity 1.4: Bisection Method of Reducing Error
Solve the following problems using the bisection method.
1. You are working for a start-up computer assembly company and have been asked to determine
the minimum number of computers that the shop will have to sell to make a profit.
The equation that gives the minimum number of computers n to be sold after considering the
total costs and the total sales is
𝒇(𝒏) = 𝟒𝟎𝒏𝟑/𝟐 − 𝟖𝟕𝟓𝒏 + 𝟑, 𝟓𝟎𝟎 = 𝟎
Use the bisection method of finding roots of equations to find the minimum number of computers
n that need to be sold to make a profit. Conduct iterations to estimate the root of the above
equation. Find the absolute relative approximate error at the end of each iteration, with an error
tolerance of ∈𝑠 = 5%. Use 𝑥𝑙 = 1 and 𝑥𝑙 = 5 as your initial boundaries.
2. A rich company can determine the number of products that can be produced using the supply
function 𝑓 (𝑥) = 𝑥 3 − 3𝑥 2 − 𝑥 − 2 based on their production cost 𝑥 in millions. Use the bisection
method to determine, with a tolerance error of ∈𝑠 = 10%, the fixed production cost 𝑥 if no units,
𝑓 (𝑥) = 0, were produced. Use 𝑥𝑙 = 2 and 𝑥𝑢 = 4 as your initial boundaries. Round off your final
answer to the nearest hundredths.
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