17.1F: Myocardial Thickness and Function
The myocardium (cardiac muscle) is the thickest section of the heart wall and contains cardiomyocytes, the contractile cells of the heart.
- Describe myocardial thickness and function in the heart
Key Points
- Coordinated contraction of cardiomyocytes causes the heart to contract and expel blood into circulation.
- The myocardium is thickest in the left ventricle, as the left ventricle must create a lot of pressure to pump blood into the aorta and throughout systemic circulation.
- Cardiac muscle has a high density of mitochondria and a large blood supply, which keep it functioning continuously.
- Sarcomeres are the basic unit of muscle tissue, composed primarily of actin and myosin myofibrils.
- The heart’s thickness may change over time and adapt to compensate for diseases, which often results in heart failure.
Key Terms
- Cardiac hypertrophy : An adaptive disease in which the walls of the heart become too thick to pump blood effectively. It is generally a complication of hypertension.
- sarcomere : The basic contractile unit of contractile muscle, which contains myofibril filaments made out of myosin and actin, the two proteins that slide past one another to cause a muscle contraction.
- cardiomyocyte : A cardiac muscle cell (myocyte) in the heart.
The myocardium, or cardiac muscle, is the thickest section of the heart wall and contains cardiomyocytes, the contractile cells of the heart. As a type of muscle tissue, the myocardium is unique among all other muscle tissues in the human body. The thickness of the myocardium determines the strength of the heart’s ability to pump blood.
Myocardial Histology
The structure of cardiac muscle shares some characteristics with skeletal muscle, but has many distinctive features of its own. Cardiomyocytes are shorter than skeletal myocytes and have fewer nuclei. Each muscle fiber connects to the plasma membrane (sarcolemma) with distinctive tubules ( T-tubule ). At these T-tubules, the sarcolemma is studded with a large number of calcium channels which allow calcium ion exchange at a rate much faster than that of the neuromuscular junction in skeletal muscle. The flux of calcium ions into the muscle cells causes stimulates an action potential, which causes the cells to contract.
Cardiac muscle, like skeletal muscle, is comprised of sarcomeres, the basic, contractile units of muscle. Sarcomeres are composed of long, fibrous proteins that slide past each other when the muscles contract and relax. Two of the important proteins found in sarcomeres are myosin, which forms the thick filament, and actin, which forms the thin filament. Myosin has a long, fibrous tail and a globular head that binds to actin. The myosin head also binds to ATP, the source of energy for cellular metabolism, and is required for the cardiomyocytes to sustain themselves and function normally. Together, myosin and actin form myofibril filaments, the elongated, contractile threads found in muscle tissue. Cardiac muscle and skeletal muscle both contain the protein myoglobin, which stores oxygen.
Cardiac muscle is adapted to be highly resistant to fatigue. Cardiomyocytes have a large number of mitochondria, enabling continuous aerobic respiration. Cardiac muscle also has a large blood supply relative to its size, which provides a continuous stream of nutrients and oxygen while providing ample removal of metabolic waste.
Cardiac Muscle : The tissue structure of cardiac muscle contains sarcomeres that are made of myofibrils with intercalated disks, that contain cardiomyocytes and have many mitocondria.
Myocardial Thickness
The myocardium has variable levels of thickness within the heart. Chambers of the heart with a thicker myocardium are able to pump blood with more pressure and force compared to chambers of the heart with a thinner myocardium. The myocardium is thinnest within the atria, as these chambers primarily fill through passive blood flow. The right ventricle myocardium is thicker than the atrial myocardium, as this muscle must pump all blood returning to the heart into the lungs for oxygenation. The myocardium is thickest in the left ventricle, as this chamber must create substantial pressure to pump blood into the aorta and throughout systemic circulation.
The thickness of the myocardium may change in some individuals as a compensatory adaptation to disease, either thickening and becoming stiff or becoming thinner and flabby. Cardiac hypertrophy is a common result of hypertension (high blood pressure) in which the cells of the myocardium enlarge as an adaptive response to pumping against the higher pressure. Eventually, hypertrophy may become so severe that heart failure occurs when the heart becomes so stiff that it can no longer pump blood. A flabby heart is typically the result of myocardial infections, in which the heart muscle becomes so weak that it cannot efficiently pump blood, which also leads to heart failure.
Right Ventricular Hypertrophy : If the heart adapts to become too thick, it will not be able to pump blood as efficiently, and heart failure may occur.


8.2: Compound Interest
Learning Objectives
In this section, you will learn to:
- Find the future value of a lump-sum.
- Find the present value of a lump-sum.
- Find the effective interest rate.
Prerequisite Skills
Before you get started, take this prerequisite quiz.
1. Simplify each expression.
a. \(100(3+2^2)\)
b. \(100(3+2)^2\)
- Click here to check your answer
-
a. \(700\)
b. \(2500\)
If you missed this problem, review here . (Note that this will open a different textbook in a new window.)
2. If an amount of $2,000 is borrowed at a simple interest rate of 10% for 3 years, how much is the interest?
- Click here to check your answer
-
\($600\)
If you missed this problem, review Section 8.1 . (Note that this will open in a new window.)
3. You borrow $4,500 for six months at a simple interest rate of 8%. How much is the interest?
- Click here to check your answer
-
\($180\)
If you missed this problem, review Section 8.1 . (Note that this will open in a new window.)
4. John borrows $2400 for 3 years at 9% simple interest. How much will he owe at the end of 3 years?
- Click here to check your answer
-
\($3048\)
If you missed this problem, review Section 8.1 . (Note that this will open in a new window.)
Compound Interest
In the last section, we examined problems involving simple interest. Simple interest is generally charged when the lending period is short and often less than a year. When the money is loaned or borrowed for a longer time period, if the interest is paid (or charged) not only on the principal, but also on the past interest, then we say the interest is compounded.
Suppose we deposit $200 in an account that pays 8% interest each year. At the end of one year, we will have $200 + $200(.08) = $200(1 + .08) = $216.
Now suppose we put this amount, $216, in the same account. After another year, we will have $216 + $216(.08) = $216(1 + .08) = $233.28.
So an initial deposit of $200 has accumulated to $233.28 in two years. Further note that had it been simple interest, this amount would have accumulated to only $232. The reason the amount is slightly higher is because the interest ($16) we earned the first year, was put back into the account. And this $16 amount itself earned for one year an interest of $16(.08) = $1.28, thus resulting in the increase. So we have earned interest on the principal as well as on the past interest, and that is why we call it compound interest.
Now suppose we leave this amount, $233.28, in the bank for another year, the final amount will be $233.28 + $233.28(.08) = $233.28(1 + .08) = $251.94.
Now let us look at the mathematical part of this problem so that we can devise an easier way to solve these problems.
After one year, we had $200(1 + .08) = $216
After two years, we had $216(1 + .08)
But $216 = $200(1 + .08), therefore, the above expression becomes
\[\$ 200(1+.08)(1+.08) = \$ 200(1+.08)^2=\$ 233. 28 \nonumber\]
After three years, we get
\[\$ 233.28(1+.08)=\$ 200(1+.08)(1+.08)(1+.08) \nonumber\]
which can be written as
\[\$ 200(1+.08)^{3}=\$ 251.94 \nonumber\]
Suppose we are asked to find the total amount at the end of 5 years, we will get
\[200(1+.08)^{5}=\$ 293.87 \nonumber\]
We summarize as follows:
|
The original amount |
$200 |
= $200 |
|
The amount after one year |
$200(1 + .08) |
= $216 |
|
The amount after two years |
$200(1 + .08) 2 |
= $233.28 |
|
The amount after three years |
$200(1 + .08) 3 |
= $251.94 |
|
The amount after five years |
$200(1 + .08) 5 |
= $293.87 |
|
The amount after t years |
$200(1 + .08) t |
COMPOUNDING PERIODS
Banks often compound interest more than one time a year. Consider a bank that pays 8% interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay an interest equal to one-fourth of 8%, or 2%.
Now if we deposit $200 in the bank, after one quarter we will have \(\$ 200\left(1+\frac{.08}{4}\right)\) or $204.
After two quarters, we will have \(\$ 200\left(1+\frac{.08}{4}\right)^{2}\) or $208.08.
After one year, we will have \(\$ 200\left(1+\frac{.08}{4}\right)^{4}\) or $216.49.
After three years, we will have \(\$ 200\left(1+\frac{.08}{4}\right)^{12}\) or $253.65, etc.
|
The original amount |
$200 |
= $200 |
|
The amount after one quarter |
\(\$ 200\left(1+\frac{.08}{4}\right)\) |
= $204 |
|
The amount after two quarters |
\(\$ 200\left(1+\frac{.08}{4}\right)^{2}\) |
= $208.08 |
|
The amount after one year |
\(\$ 200\left(1+\frac{.08}{4}\right)^{4}\) |
= $216.49 |
|
The amount after two years |
\(\$ 200\left(1+\frac{.08}{4}\right)^{8}\) |
= $234.31 |
|
The amount after three years |
\(\$ 200\left(1+\frac{.08}{4}\right)^{12}\) |
= $253.65 |
|
The amount after five years |
\(\$ 200\left(1+\frac{.08}{4}\right)^{20}\) |
= $297.19 |
|
The amount after t years |
\(\$ 200\left(1+\frac{.08}{4}\right)^{4t}\) |
We can see the formula for compound interest emerge.
Definition: Compound Interest, \(n\) times per year
If a lump-sum amount of \(P\) dollars is invested at an interest rate \(r\), compounded \(n\) times a year, then after \(t\) years the final amount is given by
\[A=P\left(1+\frac{r}{n}\right)^{n t} \]
\(\mathbf{P}\) is called the principal and is also called the present value .
Example \(\PageIndex{1}\)
If $3500 is invested at 9% compounded monthly, what will the future value be in four years?
Solution
Clearly an interest of .09/12 is paid every month for four years. The interest is compounded \(4 \times 12 = 48\) times over the four-year period. We get
\[\mathrm{A}=\$ 3500\left(1+\frac{.09}{12}\right)^{48}=\$ 3500(1.0075)^{48}=\$ 5009.92 \nonumber\]
$3500 invested at 9% compounded monthly will accumulate to $5009.92 in four years.
Example \(\PageIndex{2}\)
How much should be invested in an account paying 9% compounded daily for it to accumulate to $5,000 in five years?
Solution
We know the future value, but need to find the principal.
\[\begin{array}{l}
\$ 5000=P\left(1+\frac{.09}{365}\right)^{365 \times 5} \\
\$ 5000=P(1.568225) \\
\$ 3188.32=P
\end{array} \nonumber\]
$3188,32 invested into an account paying 9% compounded daily will accumulate to $5,000 in five years.
Example \(\PageIndex{3}\)
If $4,000 is invested at 4% compounded annually, how long will it take to accumulate to $6,000?
Solution
\(n = 1\) because annual compounding means compounding only once per year. The formula simplifies to \(A=(1+r)^{t}\) when \(n = 1\).
\[\begin{aligned}
\$ 6000 &=4000(1+.04)^{t} \\
\frac{6000}{4000} &=1.04^{t} \\
1.5 &=1.04^{t}
\end{aligned} \nonumber\]
We use logarithms to solve for the value of \(t\) because the variable \(t\) is in the exponent.
\[t=\log _{1.04}(1.5) \nonumber\]
Using the change of base formula we can solve for \(t\):
\[t=\frac{\ln (1.5)}{\ln (1.04)}=10.33 \text { years } \nonumber\]
It takes 10.33 years for $4000 to accumulate to $6000 if invested at 4% interest, compounded annually
Example \(\PageIndex{4}\)
If $5,000 is invested now for 6 years what interest rate compounded quarterly is needed to obtain an accumulated value of $8000.
Solution
We have \(n = 4\) for quarterly compounding.
\[\begin{aligned}
\$ 8000 &=\$ 5000\left(1+\frac{r}{4}\right)^{4 \times 6} \\
\frac{\$ 8000}{\$ 5000} &=\left(1+\frac{r}{4}\right)^{24} \\
1.6 &=\left(1+\frac{r}{4}\right)^{24}
\end{aligned} \nonumber\]
We use roots to solve for \(t\) because the variable \(r\) is in the base, whereas the exponent is a known number.
\[\sqrt[24]{1.6}=1+\frac{\mathrm{r}}{4} \nonumber\]
Many calculators have a built in “nth root” key or function. In the TI-84 calculator, this is found in the Math menu. Roots can also be calculated as fractional exponents; if necessary, the previous step can be rewritten as
\[1.6^{1 / 24}=1+\frac{\mathrm{r}}{4} \nonumber\]
Evaluating the left side of the equation gives
\[\begin{array}{l}
1.0197765=1+\frac{\mathrm{r}}{4} \\
0.0197765=\frac{\mathrm{r}}{4} \\
\mathrm{r}=4(0.0197765)=0.0791
\end{array} \nonumber\]
An interest rate of 7.91% is needed in order for $5000 invested now to accumulate to $8000 at the end of 6 years, with interest compounded quarterly.
Effective Interest Rate
Banks are required to state their interest rate in terms of an “effective yield” or “effective interest rate” , for comparison purposes. The effective rate is also called the Annual Percentage Yield (APY) or Annual Percentage Rate (APR).
The effective rate is the interest rate compounded annually would be equivalent to the stated rate and compounding periods. We often suppose we invest $1 over the course of one year to determine the effective rate, as is shown in the next example.
To examine several investments to see which has the best rate, we find and compare the effective rate for each investment.
Example \(\PageIndex{5}\)
If Bank A pays 7.2% interest compounded monthly, what is the effective interest rate?
If Bank B pays 7.2% interest compounded semiannually, what is the effective interest rate?
Which bank pays more interest?
Solution
Bank A: Suppose we deposit $1 in this bank and leave it for a year, we will get
\[\begin{array}{l}
1\left(1+\frac{0.072}{12}\right)^{12}=1.0744 \\
\mathrm{r}_{\mathrm{EFF}}=1.0744-1=0.0744
\end{array} \nonumber\]
We earned interest of $1.0744 - $1.00 = $.0744 on an investment of $1.
The effective interest rate is 7.44%, often referred to as the APY.
Bank B: The effective rate is calculated as
\[\mathbf{r}_{\mathrm{EFF}}=1\left(1+\frac{0.072}{2}\right)^{2}-1=.0733 \nonumber\]
The effective interest rate is 7.33% .
Bank A pays slightly higher interest, with an effective rate of 7.44%, compared to Bank B with effective rate 7.33%.
Definition: Effective Interest Rate, compounded \(n\) times per year
If a bank pays an interest rate \(r\) per year, compounded \(n\) times a year, then the effective interest rate is given by \[\mathbf{r}_{\mathrm{EFF}}=\left(1+\frac{r}{n}\right)^{n}-1\]
This is also reffered to as the annual percentage yield , or APY .
Continuous Compounding
Interest can be compounded yearly, semiannually, quarterly, monthly, and daily. Using the same calculation methods, we could compound every hour, every minute, and even every second. As the compounding period gets shorter and shorter, we move toward the concept of continuous compounding.
But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts? When interest is compounded "infinitely many times", we say that the interest is compounded continuously . Our next objective is to derive a formula to model continuous compounding.
Suppose we put $1 in an account that pays 100% interest. If the interest is compounded once a year, the total amount after one year will be \(\$ 1(1+1)=\$ 2\).
- If the interest is compounded semiannually, in one year we will have \(\$ 1(1+1 / 2)^{2}=\$ 2.25\)
- If the interest is compounded quarterly, in one year we will have \(\$ 1(1+1 / 4)^{4}=\$ 2.44\)
- If the interest is compounded monthly, in one year we will have \(\$ 1(1+1 / 12)^{12}=\$ 2.61\)
- If the interest is compounded daily, in one year we will have \(\$ 1(1+1 / 365)^{365}=\$ 2.71\)
We show the results as follows:
|
Frequency of compounding |
Formula |
Total amount |
|---|---|---|
|
Annually |
\(\$ 1(1 + 1)\) |
$2 |
|
Semiannually |
\(\$ 1(1+1 / 2)^{2}\) |
$2.25 |
|
Quarterly |
\(\$ 1(1+1 / 4)^{4}=\$ 2.44\) |
$2.44140625 |
|
Monthly |
\(\$ 1(1+1 / 12)^{12}\) |
$2.61303529 |
|
Daily |
\(\$ 1(1+1 / 365)^{365}\) |
$2.71456748 |
|
Hourly |
\(\$ 1(1+1 / 8760)^{8760}\) |
$2.71812699 |
|
Every minute |
\(\$1(1+1 / 525600)^{525600}\) |
$2.71827922 |
|
Every Second |
\(\$ 1(1+1 / 31536000)^{31536000} \) |
$2.71828247 |
|
Continuously |
\(\$ 1(2.718281828 \ldots)\) |
$2.718281828... |
We have noticed that the $1 we invested does not grow without bound. It starts to stabilize to an irrational number 2.718281828... given the name " e " after the great mathematician Euler.
In mathematics, we say that as \(n\) becomes infinitely large the expression equals \(\left(1+\frac{1}{n}\right)^{n}\) = e .
Therefore, it is natural that the number
e
play a part in continuous compounding.
It can be shown that as \(n\) becomes infinitely large the expression \(\left(1+\frac{r}{n}\right)^{n t}=e^{r t}\)
Therefore, it follows that if we invest $\(P\) at an interest rate \(r\) per year, compounded continuously, after \(t\) years the final amount will be given by
\[ A = P \cdot e^{rt} \nonumber \]
Definition: Continuously Compounded Interest
If an amount \(\mathrm{P}\) is invested for \(t\) years at an interest rate \(r\) per year, compounded continuously , then the future value is given by \[\mathrm{A} = \mathrm{P}e^{rt}\]
Example \(\PageIndex{6}\)
$3500 is invested at 9% compounded continuously. Find the future value in 4 years.
Solution
Using the formula for the continuous compounding, we get \(A=Pe^{rt}\).
\begin{aligned}
A &=\$ 3500 e^{0.09 \times 4} \\
A &=\$ 3500 e^{0.36} \\
A &=\$ 5016.65
\end{aligned}
Example \(\PageIndex{7}\)
If an amount is invested at 7.2% compounded continuously, what is the effective interest rate?
Solution
If we deposit $1 in the bank at 7.2% compounded continuously for one year, and subtract that $1 from the final amount, we get the effective interest rate in decimals.
\[\begin{array}{l}
\mathrm{r}_{\mathrm{EFF}}=1 \mathrm{e}^{0.072}-1 \\
\mathrm{r}_{\mathrm{EFF}}=1.07466-1 \\
\mathrm{r}_{\mathrm{EFF}}=.07466 \text { or } 7.466 \%
\end{array} \nonumber\]
Definition: Effective Interest Rate, compounded continously
If a bank pays an interest rate \(r\) per year, compounded continuously, then the effective interest rate is given by \[\mathrm{r}_{\mathrm{EFF}}=e^{\mathbf{r}}-1\]
Example \(\PageIndex{8}\)
If an amount is invested at 7% compounded continuously, how long will it take to double?
SolutionWe don’t know the initial value of the principal but we do know that the accumulated value is double (twice) the principal.
\[\mathrm{P} \cdot {e}^{0.07t}=2 \mathrm{P} \nonumber\]
We divide both sides by \(\mathrm{P}\)
\[e^{.07 t}=2 \nonumber\]
Using natural logarithm:
\[\begin{array}{l}
.07 \mathrm{t}=\ln (2) \\
\mathrm{t}=\ln (2) / .07=9.9 \: \mathrm{years}
\end{array} \nonumber\]
It takes 9.9 years for money to double if invested at 7% continuous interest.
Example \(\PageIndex{9}\)
a. At the peak growth rate in the 1960’s the world's population had a doubling time of 35 years. At that time, approximately what was the growth rate?
b. As of 2015, the world population’s annual growth rate was approximately 1.14%. Based on that rate, find the approximate doubling time.
Solution
We expect the world's population to grow continuously, not in discrete intervals such as years or months. Therefore, we will use the formula \(A = Pe^{rt}\).
a. Substituting \(2 \mathrm{P}\) for \(A\) and 35 for \(t\) gives us the equation
\[2 \mathrm{P}=\mathrm{P} \cdot {e}^{r(35)} \nonumber\]
We divide both sides by \(\mathrm{P}\):
\[2 ={e}^{r(35)} \nonumber\]
Using natural logarithm:
\( \ln (2) = r(35) \nonumber\)
Dividing both sides by 35:
\( \dfrac{\ln(2)}{35}=r \nonumber\)
\( 0.0198=r \nonumber\)
The growth rate was approximately \(1.98\%\).
b. Substituting \(2 \mathrm{P}\) for \(A\) and \(0.0114\) for \(r\) gives us the equation
\[2 \mathrm{P}=\mathrm{P} \cdot {e}^{0.0114t} \nonumber\]
We divide both sides by \(\mathrm{P}\):
\[2 ={e}^{0.0114t} \nonumber\]
Using natural logarithm:
\( \ln (2) = 0.0114t \nonumber\)
Dividing both sides by 0.014:
\( \dfrac{\ln(2)}{0.0114}=t \nonumber\)
\( 60.8=t \nonumber\)
If the world population were to continue to grow at the annual growth rate of 1.14% , it would take approximately 60.8 years for the population to double.
SECTION 8.2 SUMMARY
Below is a summary of the formulas we developed for calculations involving compound interest:
COMPOUND INTERES T \(n\) t imes per year
- If an amount \(\mathrm{P}\) is invested for \(t\) years at an interest rate \(r\) per year, compounded \(n\) times a year, then the future value is given by \[A=P\left(1+\frac{r}{n}\right)^{n t} \nonumber\] \(\mathbf{P}\) is called the principal and is also called the present value.
- If a bank pays an interest rate \(r\) per year, compounded \(n\) times a year, then the effective interest rate is given by \[\mathbf{r}_{\mathrm{EFF}}=\left(1+\frac{r}{n}\right)^{n}-1\nonumber\]
CONTINUOUSLY COMPOUNDED INTEREST
- If an amount \(\mathrm{P}\) is invested for \(t\) years at an interest rate \(r\) per year, compounded continuously, then the future value is given by \[\mathrm{A} = \mathrm{P}e^{rt}\nonumber\]
- If a bank pays an interest rate \(r\) per year, compounded continuously, then the effective interest rate is given by \[\mathrm{r}_{\mathrm{EFF}}=e^{\mathbf{r}}-1\nonumber\]


12.1.2: Global Trade in the United States
Why is Global Trade Important to the United States, and How is it Measured?
No longer just an option, having a global vision has become a business imperative. Having a global vision means recognizing and reacting to international business opportunities, being aware of threats from foreign competitors in all markets, and effectively using international distribution networks to obtain raw materials and move finished products to the customer.
U.S. managers must develop a global vision if they are to recognize and react to international business opportunities, as well as remain competitive at home. Often a U.S. firm’s toughest domestic competition comes from foreign companies. Moreover, a global vision enables a manager to understand that customer and distribution networks operate worldwide, blurring geographic and political barriers and making them increasingly irrelevant to business decisions. Over the past three decades, world trade has climbed from $200 billion a year to more than $1.4 trillion. U.S. companies play a major role in this growth in world trade, with 113 of the Fortune 500 companies making over 50 percent of their profits outside the United States. Among these companies are recognizable names such as Apple, Microsoft, Pfizer, Exxon Mobil, and General Electric.
Starbucks Corp. is among the fastest growing global consumer brands and one of the most visible emblems of U.S. commercial culture overseas. Of Starbucks’s 24,000 total stores, almost 66 percent are international stores that contribute a substantial amount to the company’s revenues, which have grown from $4.1 billion in 2003 to $21.3 billion in 2016.
Go into a Paris McDonald’s and you may not recognize where you are. There are no Golden Arches or utilitarian chairs and tables and other plastic features. The restaurants have exposed brick walls, hardwood floors, and armchairs. Some French McDonald’s even have faux marble walls. Most restaurants have TVs with continuous music videos. You can even order an espresso, beer, and a chicken on focaccia bread sandwich. It’s not America.
Global business is not a one-way street, where only U.S. companies sell their wares and services throughout the world. Foreign competition in the domestic market used to be relatively rare but now occurs in almost every industry. In fact, U.S. makers of electronic goods, cameras, automobiles, fine china, tractors, leather goods, and a host of other consumer and industrial products have struggled to maintain their domestic market shares against foreign competitors. Toyota now has 14 percent of the U.S. auto market, followed by Honda at 9 percent and Nissan with 8 percent. Nevertheless, the global market has created vast new business opportunities for many U.S. firms.
The Importance of Global Business to the United States
Many countries depend more on international commerce than the United States does. For example, France, Great Britain, and Germany all derive more than 55 percent of their gross domestic product (GDP) from world trade, compared to about 28 percent for the United States. Nevertheless, the impact of international business on the U.S. economy is still impressive:
- Trade-dependent jobs have grown at a rate three times the growth of U.S.-dependent jobs.
- Every U.S. state has realized a growth of jobs attributable to trade.
- Trade has an effect on both service and manufacturing jobs.
These statistics might seem to imply that practically every business in the United States is selling its wares throughout the world, but most is accounted for by big business. About 85 percent of all U.S. exports of manufactured goods are shipped by 250 companies. Yet, 98 percent of all exporters are small and medium-size firms.
The Impact of Terrorism on Global Trade
The terrorist attacks on America on September 11, 2001, and the Charlie Hebdo terrorist attacks in Paris in 2015 have changed the way the world conducts business. The immediate impacts of these events have included a short-term shrinkage of global trade. Globalization, however, will continue because the world’s major markets are too vitally integrated for globalization to stop. Nevertheless, terrorism has caused the growth to be slower and costlier.
Companies are paying more for insurance and to provide security for overseas staff and property. Heightened border inspections slow movements of cargo, forcing companies to stock more inventory. Tighter immigration policies curtail the liberal inflows of skilled and blue-collar workers that allowed companies to expand while keeping wages in check. The impact of terrorism may lessen over time, but multinational firms will always be on guard.
Measuring Trade between Nations
International trade improves relationships with friends and allies; helps ease tensions among nations; and—economically speaking—bolsters economies, raises people’s standard of living, provides jobs, and improves the quality of life. The value of international trade is over $16 trillion a year and growing. This section takes a look at some key measures of international trade: exports and imports, the balance of trade, the balance of payments, and exchange rates.
Exports and Imports
The developed nations (those with mature communication, financial, educational, and distribution systems) are the major players in international trade. They account for about 70 percent of the world’s exports and imports. Exports are goods and services made in one country and sold to others. Imports are goods and services that are bought from other countries. The United States is both the largest exporter and the largest importer in the world.
Each year the United States exports more food, animal feed, and beverages than the year before. A third of U.S. farm acreage is devoted to crops for export. The United States is also a major exporter of engineering products and other high-tech goods, such as computers and telecommunications equipment. For more than 60,000 U.S. companies (the majority of them small), international trade offers exciting and profitable opportunities. Among the largest U.S. exporters are Apple, General Motors Corp., Ford Motor Co., Procter & Gamble, and Cisco Systems.
Despite our impressive list of resources and great variety of products, imports to the United States are also growing. Some of these imports are raw materials that we lack, such as manganese, cobalt, and bauxite, which are used to make airplane parts, exotic metals, and military hardware. More modern factories and lower labor costs in other countries make it cheaper to import industrial supplies (such as steel) and production equipment than to produce them at home. Most of Americans’ favorite hot beverages—coffee, tea, and cocoa—are imported. Lower manufacturing costs have resulted in huge increases in imports from China.
Balance of Trade
The difference between the value of a country’s exports and the value of its imports during a specific time is the country’s balance of trade . A country that exports more than it imports is said to have a favorable balance of trade, called a trade surplus . A country that imports more than it exports is said to have an unfavorable balance of trade, or a trade deficit . When imports exceed exports, more money from trade flows out of the country than flows into it.
Although U.S. exports have been booming, we still import more than we export. We have had an unfavorable balance of trade throughout the 1990s, 2000s and 2010s. In 2016, our exports totaled $2.2 trillion, yet our imports were $2.7 trillion. Thus, in 2016 the United States had a trade deficit of $500 billion. America’s exports continue to grow, but not as fast as our imports: The export of goods, such as computers, trucks, and airplanes, is very strong. The sector that is lagging in significant growth is the export of services. Although America exports many services—ranging from airline trips to education of foreign students to legal advice—part of the problem is due to piracy, which leads companies to restrict the distribution of their services to certain regions. The FBI estimates that the theft of intellectual property from products, books and movies, and pharmaceuticals totals in the billions every year.
Balance of Payments
Another measure of international trade is called the balance of payments , which is a summary of a country’s international financial transactions showing the difference between the country’s total payments to and its total receipts from other countries. The balance of payments includes imports and exports (balance of trade), long-term investments in overseas plants and equipment, government loans to and from other countries, gifts and foreign aid, military expenditures made in other countries, and money transfers in and out of foreign banks.
From 1900 until 1970, the United States had a trade surplus, but in the other areas that make up the balance of payments, U.S. payments exceeded receipts, largely due to the large U.S. military presence abroad. Hence, almost every year since 1950, the United States has had an unfavorable balance of payments. And since 1970, both the balance of payments and the balance of trade have been unfavorable. What can a nation do to reduce an unfavorable balance of payments? It can foster exports, reduce its dependence on imports, decrease its military presence abroad, or reduce foreign investment. The U.S. balance of payments deficit was over $504 billion in 2016.
The Changing Value of Currencies
The exchange rate is the price of one country’s currency in terms of another country’s currency. If a country’s currency appreciates , less of that country’s currency is needed to buy another country’s currency. If a country’s currency depreciates , more of that currency will be needed to buy another country’s currency.
How do appreciation and depreciation affect the prices of a country’s goods? If, say, the U.S. dollar depreciates relative to the Japanese yen, U.S. residents have to pay more dollars to buy Japanese goods. To illustrate, suppose the dollar price of a yen is $0.012 and that a Toyota is priced at 2 million yen. At this exchange rate, a U.S. resident pays $24,000 for a Toyota ($0.012 × 2 million yen = $24,000). If the dollar depreciates to $0.018 to one yen, then the U.S. resident will have to pay $36,000 for a Toyota.
As the dollar depreciates, the prices of Japanese goods rise for U.S. residents, so they buy fewer Japanese goods—thus, U.S. imports decline. At the same time, as the dollar depreciates relative to the yen, the yen appreciates relative to the dollar. This means prices of U.S. goods fall for the Japanese, so they buy more U.S. goods—and U.S. exports rise.
Currency markets operate under a system called floating exchange rates . Prices of currencies “float” up and down based upon the demand for and supply of each currency. Global currency traders create the supply of and demand for a particular currency based on that currency’s investment, trade potential, and economic strength. If a country decides that its currency is not properly valued in international currency markets, the government may step in and adjust the currency’s value. In a devaluation , a nation lowers the value of its currency relative to other currencies. This makes that country’s exports cheaper and should, in turn, help the balance of payments.
In other cases, a country’s currency may be undervalued, giving its exports an unfair competitive advantage. Many people believe that China’s huge trade surplus with the United States is partially because China’s currency was undervalued. In 2017, the U.S. Department of Commerce issued a fact sheet detailing how it accused China of dumping steel on the U.S. market as well as providing financial assistance to Chinese companies to produce, manufacture, and export stainless steel to the United States from the People’s Republic of China.
CONCEPT CHECK
- What is global vision, and why is it important?
- What impact does international trade have on the U.S. economy?
- Explain the impact of a currency devaluation.


3.2: Probability Terminology
Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. Flipping one fair coin twice is an example of an experiment.
A result of an experiment is called an outcome . The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter \(S\) is used to denote the sample space. For example, if you flip one fair coin, \(S = \{H, T\}\) where \(H =\) heads and \(T =\) tails are the outcomes.
An event is any combination of outcomes. Upper case letters like \(A\) and \(B\) represent events. For example, if the experiment is to flip one fair coin, event \(A\) might be getting at most one head. The probability of an event \(A\) is written \(P(A)\).
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). \(P(A) = 0\) means the event \(A\) can never happen. \(P(A) = 1\) means the event \(A\) always happens. \(P(A) = 0.5\) means the event \(A\) is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).
Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair , six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head ( H ) and a Tail ( T ) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.
To calculate the probability of an event A when all outcomes in the sample space are equally likely , count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is \(\{HH, TH, HT, TT\}\) where \(T =\) tails and \(H =\) heads. The sample space has four outcomes. Let \(A\) = getting one head. There are two outcomes that meet this condition \(\{HT, TH\}\), so \(P(A) = \frac{2}{4} = 0.5\).
Suppose you roll one fair six-sided die, with the numbers \(\{1, 2, 3, 4, 5, 6\}\) on its faces. Let event \(E =\) be rolling a number that is at least five. There are two outcomes \(\{5, 6\}\), so \(P(E) = \frac{2}{6}\). If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, \(\frac{2}{6}\) of the rolls would result in an outcome of "at least five". You would not expect exactly \(\frac{2}{6}\). The long-term relative frequency of obtaining this result would approach the theoretical probability of \(\frac{2}{6}\) as the number of repetitions grows larger and larger.
This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)
It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair , or biased . Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased.  Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.
"\(\cup\)" Event: The Union
An outcome is in the event \(A \cup B\) if the outcome is in A or is in B or is in both A and B . For example, let \(A = \{1, 2, 3, 4, 5\}\) and \(B = \{4, 5, 6, 7, 8\}\). \(A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}\). Notice that 4 and 5 are NOT listed twice.
"\(\cap \)" Event: The Intersection (Joint Probability)
An outcome is in the event \(A \cap B\) if the outcome is in both A and B at the same time. For example, let \(A\) and \(B\) be \(\{1, 2, 3, 4, 5\}\) and \(\{4, 5, 6, 7, 8\}\), respectively. Then \(A \cap B = \{4, 5\}\).
The complement of event A is denoted A ′ (read out loud as " A prime"). A ′ consists of all outcomes that are NOT in A . Notice that \(P(A) + P(A′) = 1\). For example, let \(S = \{1, 2, 3, 4, 5, 6\}\) and let \(A = \{1, 2, 3, 4\}\). Then, we will have \(A′ = \{5, 6\}\), giving us \(P(A) = \frac{4}{6}\), \(P(A′) = \frac{2}{6}\), and \(P(A) + P(A′) = \frac{4}{6}+\frac{2}{6}=1\).
The conditional probability of \(A\) given \(B\) is written \(P(A|B)\). \(P(A|B)\) is the probability that event \(A\) will occur given that the event \(B\) has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space \(B\). The formula to calculate \(P(A|B)\) is \(P(A | B)=\frac{P(A \cap B)}{P(B)}\) where \(P(B)\) is greater than zero.
For example, suppose we toss one fair, six-sided die. The sample space \(S = \{1, 2, 3, 4, 5, 6\}\).  To calculate \(P(A|B)\), we count the number of outcomes 2 or 3 in the sample space \(B = \{2, 4, 6\}\). Then we divide that by the number of outcomes \(B\) (rather than \(S\)).
We get the same result by using the formula. Remember that \(S\) has six outcomes.
\(P(A|B) = \frac{\frac{(\text { number of outcomes that are } 2 \text { or } 3 \text { and even in } S)}{6}}{\frac{(\text { number of outcomes that are even in } S)}{6}}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}\)
Odds
The odds of an event presents the probability as a ratio of success to failure. This is common in various gambling formats. Mathematically, the odds of an event can be defined as:
\[\frac{P(A)}{1-P(A)}\nonumber\]
where \(P(A)\) is the probability of success and of course \(1 − P(A)\) is the probability of failure. Odds are always quoted as "numerator to denominator," e.g. 2 to 1. Here the probability of winning is twice that of losing; thus, the probability of winning is 0.66. A probability of winning of 0.60 would generate odds in favor of winning of 3 to 2.
Understanding Terminology and Symbols
It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.
Example \(\PageIndex{1}\)
The sample space \(S\)
is the whole numbers starting at one and less than 20.
1. \(S =\) _____________________________
Let event \(A\) = the even numbers and event \(B\) = numbers greater than 13.
2. \(A =\) _____________________, \(B =\) _____________________
3. \(P(A) =\) _____________, \(P(B) =\) ________________
4. \(A \cap B = \) ____________________, \(A \cup B =\) ________________
5. \(P(A \cap B)= \) _________, \(P(A \cup B)= \) _____________
6. \(A' =\) _____________, \(P(A') =\) _____________
7. \(P(A) + P(A′) = \) ____________
8. \(P(A|B) =\) ___________, \(P(B|A) =\) _____________; are the probabilities equal?
Answer
1. \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19\}\)
2. \(A = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}\), \(B = \{14, 15, 16, 17, 18, 19\}\)
3. \(P(A) = \frac{9}{19}\), \(P(B) = \frac{6}{19}\)
4. \(A \cap B = \{14, 16, 18\}\), \(A \cup B = \{2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19\}\)
5. \(P(A \cap B)= \frac{3}{19}\), \(P(A \cup B)= \frac{12}{19}\)
6. \(A' = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}\), \(P(A') = \frac{10}{19}\)
7. \(P(A) + P(A′) = \frac{9}{19} + \frac{10}{19}=1\)
8. \(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{6}\), \(P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{3}{9}\); No.
Exercise \(\PageIndex{1}\)
The sample space \(S\) is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).
1. \(S =\) _____________________________
Let event \(A\) = the sum is even and event \(B\) = the first number is prime.
2. \(A =\) _____________________, \(B =\) _____________________
3. \(P(A) =\) _____________, \(P(B) =\) ________________
4. \(A \cap B = \) ____________________, \(A \cup B =\) ________________
5. \(P(A \cap B)= \) _________, \(P(A \cup B)= \) _____________
6. \(A' =\) _____________, \(P(A') =\) _____________
7. \(P(A) + P(A′) = \) ____________
8. \(P(A|B) =\) ___________, \(P(B|A) =\) _____________; are the probabilities equal?
 Describe the sample space \(S\), identify each of the following events with a subset of \(S\) and compute its probability (an outcome is the number of dots that show up).
1. Event \(T =\) the outcome is two.
2. Event \(A =\) the outcome is an even number.
3. Event \(B =\) the outcome is less than four.
4. The complement of \(A\).
5. \(A|B\)
6. \(B|A\)
7. \(A \cap B\)
8. \(A \cup B\)
9. \(A \cup B'\)
10. Event \(N =\) the outcome is a prime number.
11. Event \(I =\) the outcome is seven.
Answer
1. \(T = \{2\}\), \(P(T) = \frac{1}{6}\)
2. \(A = \{2, 4, 6\}\), \(P(A) = \frac{1}{2}\)
3. \(B = \{1, 2, 3\}\), \(P(B) = \frac{1}{2}\)
4. \(A' = \{1, 3, 5\}\), \(P(A') = \frac{1}{2}\)
5. \(A|B = \{2\}\), \(P(A|B) = \frac{1}{3}\)
6. \(B|A = \{2\}\), \(P(B|A) = \frac{1}{3}\)
7. \(A \cap B = \{2\}\), \(P(A \cap B)= \frac{1}{6}\)
8. \(A \cup B = \{1, 2, 3, 4, 6\}\), \(P(A \cup B)= \frac{5}{6}\)
9. \(A \cup B' = \{2, 4, 5, 6\}\), \(P(A \cup B')= \frac{2}{3}\)
10. \(N = \{2, 3, 5\}\), \(P(N) = \frac{1}{2}\)
11. A six-sided die does not have seven dots. \(P(7) = 0\).
Example \(\PageIndex{3}\)
Table 3.1 describes the distribution of a random sample \(S\) of 100 individuals, organized by gender and whether they are right- or left-handed.
| Right-handed | Left-handed | |
|---|---|---|
| Males | 43 | 9 |
| Females | 44 | 4 |
Let’s denote the events \(M =\) the subject is male, \(F =\) the subject is female, \(R =\) the subject is right-handed, \(L =\) the subject is left-handed. Compute the following probabilities:
- \(P(M)\)
- \(P(F)\)
- \(P(R)\)
- \(P(L)\)
- \(P(M \cap R)\)
- \(P(F \cap L)\)
- \(P(M \cup F)\)
- \(P(M \cup R)\)
- \(P(F \cup L)\)
- \(P(M')\)
- \(P(R|M)\)
- \(P(F|L)\)
- \(P(L|F)\)
Answer
- \(P(M) = 0.52\)
- \(P(F) = 0.48\)
- \(P(R) = 0.87\)
- \(P(L) = 0.13\)
- \(P(M \cap R) = 0.43\)
- \(P(F \cap L) = 0.04\)
- \(P(M \cup F) = 1\)
- \(P(M \cup R) = 0.96\)
- \(P(F \cup L) = 0.57\)
- \(P(M') = 0.48\)
- \(P(R|M) = 0.8269\) (rounded to four decimal places)
- \(P(F|L) = 0.3077\) (rounded to four decimal places)
- \(P(L|F) = 0.0833\)


19.7: Bayesian t-tests
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The second type of statistical inference problem discussed in this book is the comparison between two means, discussed in some detail in the chapter on t-tests (Chapter 13. If you can remember back that far, you’ll recall that there are several versions of the t-test. The
BayesFactor
package contains a function called
ttestBF()
that is flexible enough to run several different versions of the t-test. I’ll talk a little about Bayesian versions of the independent samples t-tests and the paired samples t-test in this section.
Independent samples t-test
The most common type of t-test is the independent samples t-test, and it arises when you have data that look something like this:
load( "./rbook-master/data/harpo.Rdata" )
head(harpo)
## grade tutor
## 1 65 Anastasia
## 2 72 Bernadette
## 3 66 Bernadette
## 4 74 Anastasia
## 5 73 Anastasia
## 6 71 Bernadette
In this data set, we have two groups of students, those who received lessons from Anastasia and those who took their classes with Bernadette. The question we want to answer is whether there’s any difference in the grades received by these two groups of student. Back in Chapter@refch:ttest I suggested you could analyse this kind of data using the
independentSamplesTTest()
function in the
lsr
package. For example, if you want to run a Student’s t-test, you’d use a command like this:
independentSamplesTTest(
formula = grade ~ tutor,
data = harpo,
var.equal = TRUE
)
##
## Student's independent samples t-test
##
## Outcome variable: grade
## Grouping variable: tutor
##
## Descriptive statistics:
## Anastasia Bernadette
## mean 74.533 69.056
## std dev. 8.999 5.775
##
## Hypotheses:
## null: population means equal for both groups
## alternative: different population means in each group
##
## Test results:
## t-statistic: 2.115
## degrees of freedom: 31
## p-value: 0.043
##
## Other information:
## two-sided 95% confidence interval: [0.197, 10.759]
## estimated effect size (Cohen's d): 0.74
Like most of the functions that I wrote for this book, the
independentSamplesTTest()
is very wordy. It prints out a bunch of descriptive statistics and a reminder of what the null and alternative hypotheses are, before finally getting to the test results. I wrote it that way deliberately, in order to help make things a little clearer for people who are new to statistics.
Again, we obtain a p-value less than 0.05, so we reject the null hypothesis.
What does the Bayesian version of the t-test look like? Using the
ttestBF()
function, we can obtain a Bayesian analog of Student’s independent samples t-test using the following command:
ttestBF( formula = grade ~ tutor, data = harpo )
## Bayes factor analysis
## --------------
## [1] Alt., r=0.707 : 1.754927 ±0%
##
## Against denominator:
## Null, mu1-mu2 = 0
## ---
## Bayes factor type: BFindepSample, JZS
Notice that format of this command is pretty standard. As usual we have a
formula
argument in which we specify the outcome variable on the left hand side and the grouping variable on the right. The
data
argument is used to specify the data frame containing the variables. However, notice that there’s no analog of the
var.equal
argument. This is because the
BayesFactor
package does not include an analog of the Welch test, only the Student test.
270
In any case, when you run this command you get this as the output:
So what does all this mean? Just as we saw with the
contingencyTableBF()
function, the output is pretty dense. But, just like last time, there’s not a lot of information here that you actually need to process. Firstly, let’s examine the bottom line. The
BFindepSample
part just tells you that you ran an independent samples t-test, and the
JZS
part is technical information that is a little beyond the scope of this book.
271
Clearly, there’s nothing to worry about in that part. In the line above, the text
Null, mu1-mu2 = 0
is just telling you that the null hypothesis is that there are no differences between means. But you already knew that. So the only part that really matters is this line here:
[1] Alt., r=0.707 : 1.754927 @plusorminus0%
Ignore the
r=0.707
part: it refers to a technical detail that we won’t worry about in this chapter.
272
Instead, you should focus on the part that reads
1.754927
. This is the Bayes factor: the evidence provided by these data are about 1.8:1 in favour of the alternative.
Before moving on, it’s worth highlighting the difference between the orthodox test results and the Bayesian one. According to the orthodox test, we obtained a significant result, though only barely. Nevertheless, many people would happily accept p=.043 as reasonably strong evidence for an effect. In contrast, notice that the Bayesian test doesn’t even reach 2:1 odds in favour of an effect, and would be considered very weak evidence at best. In my experience that’s a pretty typical outcome. Bayesian methods usually require more evidence before rejecting the null.
Paired samples t-test
Back in Section 13.5 I discussed the
chico
data frame in which students grades were measured on two tests, and we were interested in finding out whether grades went up from test 1 to test 2. Because every student did both tests, the tool we used to analyse the data was a paired samples t-test. To remind you of what the data look like, here’s the first few cases:
load("./rbook-master/data/chico.rdata")
head(chico)
## id grade_test1 grade_test2
## 1 student1 42.9 44.6
## 2 student2 51.8 54.0
## 3 student3 71.7 72.3
## 4 student4 51.6 53.4
## 5 student5 63.5 63.8
## 6 student6 58.0 59.3
We originally analysed the data using the
pairedSamplesTTest()
function in the
lsr
package, but this time we’ll use the
ttestBF()
function from the
BayesFactor
package to do the same thing. The easiest way to do it with this data set is to use the
x
argument to specify one variable and the
y
argument to specify the other. All we need to do then is specify
paired=TRUE
to tell R that this is a paired samples test. So here’s our command:
ttestBF(
x = chico$grade_test1,
y = chico$grade_test2,
paired = TRUE
)
## Bayes factor analysis
## --------------
## [1] Alt., r=0.707 : 5992.05 ±0%
##
## Against denominator:
## Null, mu = 0
## ---
## Bayes factor type: BFoneSample, JZS
At this point, I hope you can read this output without any difficulty. The data provide evidence of about 6000:1 in favour of the alternative. We could probably reject the null with some confidence!


8.5: Formal Charges and Resonance
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- Scott Van Bramer
- Widener University
↵
Learning Objectives
- Compute formal charges for atoms in any Lewis structure
- Use formal charges to identify the most reasonable Lewis structure for a given molecule
- Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule
Previously, we discussed how to write Lewis structures for molecules and polyatomic ions. In some cases, however, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
Calculating Formal Charge
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows:
\[\textrm{formal charge = # valence shell electrons (free atom) − # lone pair electrons − }\dfrac{1}{2}\textrm{ # bonding electrons} \nonumber \]
We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Example \(\PageIndex{1}\): Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen ion \(\ce{ICl4-}\).
S olution
We divide the bonding electron pairs equally for all \(\ce{I–Cl}\) bonds:
We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
Subtract this number from the number of valence electrons for the neutral atom:
- I: 7 – 8 = –1
- Cl: 7 – 7 = 0
The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
Exercise \(\PageIndex{1}\)
Calculate the formal charge for each atom in the carbon monoxide molecule:
- Answer
-
C −1, O +1
Example \(\PageIndex{2}\): Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen molecule \(\ce{BrCl3}\).
Solution
Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond:
Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:
- Br: 7 – 7 = 0
- Cl: 7 – 7 = 0
All atoms in \(\ce{BrCl3}\) have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
Exercise \(\PageIndex{2}\)
Determine the formal charge for each atom in \(\ce{NCl3}\).
- Answer
-
N: 0; all three Cl atoms: 0
Using Formal Charge to Predict Molecular Structure
The arrangement of atoms in a molecule or ion is called its molecular structure . In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion.
Predicting Molecular Structure Guidelines
- A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
- If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
- Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
- When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, \(\ce{CO2}\). We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: \(\ce{CNS^{–}}\), \(\ce{NCS^{–}}\), or \(\ce{CSN^{–}}\). The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:
Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Example \(\PageIndex{3}\): Using Formal Charge to Determine Molecular Structure
Nitrous oxide, N 2 O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?
Solution Determining formal charge yields the following:
The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:
The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.
Exercise \(\PageIndex{3}\)
Which is the most likely molecular structure for the nitrite (\(\ce{NO2-}\)) ion?
- Answer
-
\(\ce{ONO^{–}}\)
Resonance
You may have noticed that the nitrite anion in Example \(\PageIndex{3}\) can have two possible structures with the atoms in the same positions. The electrons involved in the N–O double bond, however, are in different positions:
If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in \(\ce{NO2-}\) have the same strength and length, and are identical in all other properties.
It is not possible to write a single Lewis structure for \(\ce{NO2-}\) in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance : if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in \(\ce{NO2-}\) is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms . The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms. Thus, the electronic structure of the \(\ce{NO2-}\) ion is shown as:
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, \(\ce{CO3^2-}\), provides a second example of resonance:
One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
Summary
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom.  A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).
Key Equations
- \(\textrm{formal charge = # valence shell electrons (free atom) − # one pair electrons − }\dfrac{1}{2}\textrm{ # bonding electrons}\)
Glossary
- formal charge
- charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)
- molecular structure
- arrangement of atoms in a molecule or ion
- resonance
- situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed
- resonance forms
- two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons
- resonance hybrid
- average of the resonance forms shown by the individual Lewis structures


7.20B: Caulobacter Differentiation
A Caulobacter is used for studying the regulation of the cell cycle, asymmetric cell division, and cellular differentiation.
LEARNING OBJECTIVES
Explain how caulobacter serve as a model organism
Key Takeaways
Caulobacter crescentus is a Gram-negative, oligotrophic bacterium widely distributed in fresh water lakes and streams. Caulobacter is an important model organism for studying the regulation of the cell cycle, asymmetric cell division, and cellular differentiation. Caulobacter daughter cells have two very different forms. One daughter is a mobile “swarmer” cell that has a single flagellum at one cell pole that provides swimming motility for chemotaxis. The other daughter, called the “stalked” cell has a tubular stalk structure protruding from one pole that has an adhesive holdfast material on its end, with which the stalked cell can adhere to surfaces. Swarmer cells differentiate into stalked cells after a short period of motility. Chromosome replication and cell division only occurs in the stalked cell stage. Its name is due to the fact that it forms a crescent shape; crescentin is a protein that imparts this shape.
In the laboratory, researchers distinguish between C. crescentus strain CB15 (the strain originally isolated from a freshwater lake) and NA1000 (the primary experimental strain). In strain NA1000, which was derived from CB15 in the 1970’s, the stalked and predivisional cells can be physically separated in the laboratory from new swarmer cells, while cell types from strain CB15 cannot be physically separated. The isolated swarmer cells can then be grown as a synchronized cell culture. Detailed study of the molecular development of these cells as they progress through the cell cycle has enabled researchers to understand Caulobacter cell cycle regulation in great detail. Due to this capacity to be physically synchronized, strain NA1000 has become the predominant experimental Caulobacter strain throughout the world. Additional phenotypic differences between the two strains have subsequently accumulated due to selective pressures on the NA1000 strain in the laboratory environment. The genetic basis of the phenotypic differences between the two strains results from coding, regulatory, and insertion/deletion polymorphisms at five chromosomal loci. “ C. Crescentus ” is synonymous with “Caulobacter Vibrioides. ”
The Caulobacter cell cycle regulatory system controls many modular subsystems that organize the progression of cell growth and reproduction. A control system constructed using biochemical and genetic logic circuitry organizes the timing of initiation of each of these subsystems. The central feature of the cell cycle regulation is a cyclical genetic circuit—a cell cycle engine –- that is centered around the successive interactions of four master regulatory proteins: DnaA, GcrA, CtrA, and CcrM. These four proteins directly control the timing of expression of over 200 genes. The four master regulatory proteins are synthesized and then eliminated from the cell one after the other over the course of the cell cycle. Several additional cell signaling pathways are also essential to the proper functioning of this cell cycle engine.
The principal role of these signaling pathways is to ensure reliable production and elimination of the CtrA protein from the cell at just the right times in the cell cycle. An essential feature of the Caulobacter cell cycle is that the chromosome is replicated once and only once per cell cycle. This is in contrast to the E. coli cell cycle where there can be overlapping rounds of chromosome replication simultaneously underway. The opposing roles of the Caulobacter DnaA and CtrA proteins are essential to the tight control of Caulobacter chromosome replication. The DnaA protein acts at the origin of replication to initiate the replication of the chromosome. The CtrA protein, in contrast, acts to block initiation of replication, so it must be removed from the cell before chromosome replication can begin. Multiple additional regulatory pathways integral to cell cycle regulation and involving both phospho signaling pathways and regulated control of protein proteolysis act to assure that DnaA and CtrA are present in the cell exactly when they are needed.
Caulobacter was the first asymmetric bacterium shown to age. Reproductive senescence was measured as the decline in the number of progeny produced over time. A similar phenomenon has since been described in the bacterium Escherichia coli, which gives rise to morphologically similar daughter cells.
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, SPECIFIC ATTRIBUTION
- Sporulation in Bacillus subtilis. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Sporulation_in_Bacillus_subtilis . License : CC BY-SA: Attribution-ShareAlike
- Sda protein domain. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Sda_protein_domain . License : CC BY-SA: Attribution-ShareAlike
- Autoinducer. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Autoinducer . License : CC BY-SA: Attribution-ShareAlike
- sporulation. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/sporulation . License : CC BY-SA: Attribution-ShareAlike
- endospore. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/endospore . License : CC BY-SA: Attribution-ShareAlike
- Bacillus subtilis colonies. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ba...s_colonies.jpg . License : CC BY-SA: Attribution-ShareAlike
- Caulobacter crescentus. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Caulobacter_crescentus . License : CC BY-SA: Attribution-ShareAlike
- Boundless. Provided by : Boundless Learning. Located at : www.boundless.com//management...ifferentiation . License : CC BY-SA: Attribution-ShareAlike
- senescence. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/senescence . License : CC BY-SA: Attribution-ShareAlike
- modular. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/modular . License : CC BY-SA: Attribution-ShareAlike
- Bacillus subtilis colonies. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Bacillus_subtilis_colonies.jpg . License : CC BY-SA: Attribution-ShareAlike
- data-attribution-url=en.Wikipedia.org/w/index.php?...ter.svg&page=1. Provided by : Wikipedia. License : CC BY-SA: Attribution-ShareAlike
Key Points
- The Caulobacter cell cycle regulatory system controls many modular subsystems that organize the progression of cell growth and reproduction.
- The central feature of the cell cycle regulation is a cyclical genetic circuit—a cell cycle engine –- that is centered around the successive interactions of four master regulatory proteins: DnaA, GcrA, CtrA, and CcrM.
- The interactions of four master regulatory proteins: DnaA, GcrA, CtrA, and CcrM directly control the timing of expression of over 200 genes. The four master regulatory proteins are synthesized and then eliminated from the cell one after the other over the course of the cell cycle.
Key Terms
- senescence : Ceasing to divide by mitosis because of shortening of telomeres or excessive DNA damage.
- differentiation : In cellular differentiation, a less specialized cell becomes a more specialized cell.
- modular : Consisting of separate modules; especially where each module performs or fulfills some specified function and could be replaced by a similar module for the same function, independently of the other modules.


9.4: Chapter Review
SECTION 9.4 PROBLEM SET: CHAPTER REVIEW
Solve the following linear programming problems using the simplex method.
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1) \[\begin{array}{ll}
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2) \[\begin{array}{ll}
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3) \[\begin{array}{ll}
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4) \[\begin{aligned}
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5)\[\begin{array}{ll}
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6) \[\begin{array}{ll}
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7) \[\begin{aligned}
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8) \[\begin{aligned}
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9) An appliance store sells three different types of ovens: small, medium, and large. The small, medium, and large ovens require, respectively, 3, 5, and 6 cubic feet of storage space; a maximum of 1,000 cubic feet of storage space is available. Each oven takes 1hour of sales time; there is a maximum of 200 hours of sales labor time available for ovens. The small, medium, and large ovens require, respectively, 1, 1, and 2 hours of installation time; a maximum of 280 hours of installer labor for ovens is available monthly.
If the profit made from sales of small, medium and large ovens is $50, $100, and $150, respectively, how many of each type of oven should be sold to maximize profit, and what is the maximum profit?
SECTION 9.4 PROBLEM SET: CHAPTER REVIEW
10) A factory manufactures three products, A, B, and C. Each product requires the use of two machines, Machine I and Machine II. The total hours available, respectively, on Machine I and Machine II per month are 180 and 300. The time requirements and profit per unit for each product are listed below.
| A | B | C | |
| Machine I | 1 | 2 | 2 |
| Machine II | 2 | 2 | 4 |
| Profit | 20 | 30 | 40 |
How many units of each product should be manufactured to maximize profit, and what is the maximum profit?
11) A company produces three products, A, B, and C, at its two factories, Factory I and Factory II. Daily production of each factory for each product is listed below.
| Factory I | Factory II | |
| Product A | 10 | 20 |
| Product B | 20 | 20 |
| Product C | 20 | 10 |
The company must produce at least 1000 units of product A, 1600 units of B, and 700 units of C. If the cost of operating Factory I is $4,000 per day and the cost of operating Factory II is $5000, how many days should each factory operate to complete the order at a minimum cost, and what is the minimum cost?
12) For his classes, Professor Wright gives three types of quizzes, objective, recall, and recall-plus.
To keep his students on their toes, he has decided to give at least 20 quizzes next quarter.
The three types, objective, recall, and recall-plus quizzes, require the students to spend, respectively, 10 minutes, 30 minutes, and 60 minutes for preparation, and Professor Wright would like them to spend at least 12 hours(720 minutes) preparing for these quizzes above and beyond the normal study time.
An average score on an objective quiz is 5, on a recall type 6, and on a recall-plus 7, and Dr. Wright would like the students to score at least 130 points on all quizzes.
It takes the professor one minute to grade an objective quiz, 2 minutes to grade a recall type quiz, and 3 minutes to grade a recall-plus quiz.
How many of each type should he give in order to minimize his grading time?


17.14: String Formatting
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Overview
String formatting uses a process of string interpolation (variable substitution) to evaluate a string literal containing one or more placeholders, yielding a result in which the placeholders are replaced with their corresponding values. [1]
Discussion
Most current programming languages provide one or more string formatting functions that use a template string with placeholders and optional alignment, width, and precision indicators to generate formatted output.
| Language | Function | Examples |
|---|---|---|
| C++ |
snprintf()
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snprintf(str, sizeof(str), "Hello %s!", name);
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| C# |
Format()
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String.Format("Hello {0}!", name);
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| Java |
format()
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String.format("Hello %s!", name);
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| JavaScript | template literal |
`Hello ${name}`;
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| Python |
format()
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"Hello {}!".format(name)
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| Swift |
interpolation
String()
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"Hello \(name)!"
String(format:"%.2f", value)
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String interpolation, like string concatenation, may lead to security problems. If user input data is improperly escaped or filtered, the system may be exposed to code injection. [2]
Key Terms
- code injection
- The exploitation of a computer bug that is caused by processing invalid data. [3]
- formatting
- Modifying the way the output is displayed.
- string interpolation
- Evaluating a string literal containing one or more placeholders, yielding a result in which the placeholders are replaced with their corresponding values.


7.4: Component Failure Analysis (Continued)
“I consider that I understand an equation when I can predict the properties of its solutions, without actually solving it.”
—P.A.M Dirac, physicist
There is a lot of truth to that quote from Dirac. With a little modification, I can extend his wisdom to electric circuits by saying, “I consider that I understand a circuit when I can predict the approximate effects of various changes made to it without actually performing any calculations.”
At the end of the series and parallel circuits chapter, we briefly considered how circuits could be analyzed in a qualitative rather than quantitative manner. Building this skill is an important step towards becoming a proficient troubleshooter of electric circuits. Once you have a thorough understanding of how any particular failure will affect a circuit (i.e. you don’t have to perform any arithmetic to predict the results), it will be much easier to work the other way around: pinpointing the source of trouble by assessing how a circuit is behaving.
Also shown at the end of the series and parallel circuits chapter was how the table method works just as well for aiding failure analysis as it does for the analysis of healthy circuits. We may take this technique one step further and adapt it for total qualitative analysis. By “qualitative” I mean working with symbols representing “increase,” “decrease,” and “same” instead of precise numerical figures. We can still use the principles of series and parallel circuits, and the concepts of Ohm’s Law, we’ll just use symbolic qualities instead of numerical quantities . By doing this, we can gain more of an intuitive “feel” for how circuits work rather than leaning on abstract equations, attaining Dirac’s definition of “understanding.”
Enough talk. Let’s try this technique on a real circuit example and see how it works:
This is the first “convoluted” circuit we straightened out for analysis in the last section. Since you already know how this particular circuit reduces to series and parallel sections, I’ll skip the process and go straight to the final form:
R 3 and R 4 are in parallel with each other; so are R 1 and R 2 . The parallel equivalents of R 3 //R 4 and R 1 //R 2 are in series with each other. Expressed in symbolic form, the total resistance for this circuit is as follows:
R Total = (R 1 //R 2 )—(R 3 //R 4 )
First, we need to formulate a table with all the necessary rows and columns for this circuit:
Next, we need a failure scenario. Let’s suppose that resistor R 2 were to fail shorted. We will assume that all other components maintain their original values. Because we’ll be analyzing this circuit qualitatively rather than quantitatively, we won’t be inserting any real numbers into the table. For any quantity unchanged after the component failure, we’ll use the word “same” to represent “no change from before.” For any quantity that has changed as a result of the failure, we’ll use a down arrow for “decrease” and an up arrow for “increase.” As usual, we start by filling in the spaces of the table for individual resistances and total voltage, our “given” values:
The only “given” value different from the normal state of the circuit is R 2 , which we said was failed shorted (abnormally low resistance). All other initial values are the same as they were before, as represented by the “same” entries. All we have to do now is work through the familiar Ohm’s Law and series-parallel principles to determine what will happen to all the other circuit values.
First, we need to determine what happens to the resistances of parallel subsections R 1 //R 2 and R 3 //R 4 . If neither R 3 nor R 4 have changed in resistance value, then neither will their parallel combination. However, since the resistance of R 2 has decreased while R 1 has stayed the same, their parallel combination must decrease in resistance as well:
Now, we need to figure out what happens to the total resistance. This part is easy: when we’re dealing with only one component change in the circuit, the change in total resistance will be in the same direction as the change of the failed component. This is not to say that the magnitude of change between individual component and total circuit will be the same, merely the direction of change. In other words, if any single resistor decreases in value, then the total circuit resistance must also decrease, and vice versa. In this case, since R 2 is the only failed component, and its resistance has decreased, the total resistance must decrease:
Now we can apply Ohm’s Law (qualitatively) to the Total column in the table. Given the fact that total voltage has remained the same and total resistance has decreased, we can conclude that total current must increase (I=E/R).
In case you’re not familiar with the qualitative assessment of an equation, it works like this. First, we write the equation as solved for the unknown quantity. In this case, we’re trying to solve for current, given voltage and resistance:
Now that our equation is in the proper form, we assess what change (if any) will be experienced by “I,” given the change(s) to “E” and “R”:
If the denominator of a fraction decreases in value while the numerator stays the same, then the overall value of the fraction must increase:
Therefore, Ohm’s Law (I=E/R) tells us that the current (I) will increase. We’ll mark this conclusion in our table with an “up” arrow:
With all resistance places filled in the table and all quantities determined in the Total column, we can proceed to determine the other voltages and currents. Knowing that the total resistance in this table was the result of R 1 //R 2 and R 3 //R 4 in series , we know that the value of total current will be the same as that in R 1 //R 2 and R 3 //R 4 (because series components share the same current). Therefore, if total current increased, then current through R 1 //R 2 and R 3 //R 4 must also have increased with the failure of R 2 :
Fundamentally, what we’re doing here with a qualitative usage of Ohm’s Law and the rules of series and parallel circuits is no different from what we’ve done before with numerical figures. In fact, its a lot easier because you don’t have to worry about making an arithmetic or calculator keystroke error in a calculation. Instead, you’re just focusing on the principles behind the equations. From our table above, we can see that Ohm’s Law should be applicable to the R 1 //R 2 and R 3 //R 4 columns. For R 3 //R 4 , we figure what happens to the voltage, given an increase in current and no change in resistance. Intuitively, we can see that this must result in an increase in voltage across the parallel combination of R 3 //R 4 :
But how do we apply the same Ohm’s Law formula (E=IR) to the R 1 //R 2 column, where we have resistance decreasing and current increasing? It’s easy to determine if only one variable is changing, as it was with R 3 //R 4 , but with two variables moving around and no definite numbers to work with, Ohm’s Law isn’t going to be much help. However, there is another rule we can apply horizontally to determine what happens to the voltage across R 1 //R 2 : the rule for voltage in series circuits. If the voltages across R 1 //R 2 and R 3 //R 4 add up to equal the total (battery) voltage and we know that the R 3 //R 4 voltage has increased while total voltage has stayed the same, then the voltage across R 1 //R 2 must have decreased with the change of R 2 ‘s resistance value:
Now we’re ready to proceed to some new columns in the table. Knowing that R 3 and R 4 comprise the parallel subsection R 3 //R 4 , and knowing that voltage is shared equally between parallel components, the increase in voltage seen across the parallel combination R 3 //R 4 must also be seen across R 3 and R 4 individually:
The same goes for R 1 and R 2 . The voltage decrease seen across the parallel combination of R 1 and R 2 will be seen across R 1 and R 2 individually:
Applying Ohm’s Law vertically to those columns with unchanged (“same”) resistance values, we can tell what the current will do through those components. Increased voltage across an unchanged resistance leads to increased current. Conversely, decreased voltage across an unchanged resistance leads to decreased current:
Once again we find ourselves in a position where Ohm’s Law can’t help us: for R 2 , both voltage and resistance have decreased, but without knowing how much each one has changed, we can’t use the I=E/R formula to qualitatively determine the resulting change in current. However, we can still apply the rules of series and parallel circuits horizontally . We know that the current through the R 1 //R 2 parallel combination has increased, and we also know that the current through R 1 has decreased. One of the rules of parallel circuits is that total current is equal to the sum of the individual branch currents. In this case, the current through R 1 //R 2 is equal to the current through R 1 added to the current through R 2 . If current through R 1 //R 2 has increased while current through R 1 has decreased, current through R 2 must have increased:
And with that, our table of qualitative values stands completed. This particular exercise may look laborious due to all the detailed commentary, but the actual process can be performed very quickly with some practice. An important thing to realize here is that the general procedure is little different from quantitative analysis: start with the known values, then proceed to determining total resistance, then total current, then transfer figures of voltage and current as allowed by the rules of series and parallel circuits to the appropriate columns.
A few general rules can be memorized to assist and/or to check your progress when proceeding with such an analysis:
- For any single component failure (open or shorted), the total resistance will always change in the same direction (either increase or decrease) as the resistance change of the failed component.
- When a component fails shorted, its resistance always decreases. Also, the current through it will increase, and the voltage across it may drop. I say “may” because in some cases it will remain the same (case in point: a simple parallel circuit with an ideal power source).
- When a component fails open, its resistance always increases. The current through that component will decrease to zero, because it is an incomplete electrical path (no continuity). This may result in an increase of voltage across it. The same exception stated above applies here as well: in a simple parallel circuit with an ideal voltage source, the voltage across an open-failed component will remain unchanged.


11.6: Sage
Sage is able to create homomorphisms (and by extension, isomorphisms and automorphisms) between finite permutation groups. There is a limited supply of commands then available to manipulate these functions, but we can still illustrate many of the ideas in this chapter.
Homomorphisms
The principal device for creating a homomorphism is to specify the specific images of the set of generators for the domain. Consider cyclic groups of order \(12\) and \(20\text{:}\)
and define a homomorphism by just defining the image of the generator of \(G\text{,}\) and define the rest of the mapping by extending the mapping via the operation-preserving property of a homomorphism.
The constructor
PermutationGroupMorphism
requires the two groups, then a list of images for each generator (in order!), and then will create the homomorphism. Note that we can then use the result as a function. In the example below, we first verify that
C12
has a single generator (no surprise there), which we then send to a particular element of order \(4\) in the codomain. Sage then constructs the unique homomorphism that is consistent with this requirement.
Note that the element
c
must therefore be in the kernel of
phi
.
We can then compute the subgroup of the domain that is the kernel, in this case a cyclic group of order \(3\) inside the cyclic group of order \(12\text{.}\) We can compute the image of
any
subgroup, but here we will build the whole homomorphic image by supplying the whole domain to the
.image()
method. Here the image is a cyclic subgroup of order \(4\) inside the cyclic group of order \(20\text{.}\) Then we can verify the First Isomorphism Theorem.
Here is a slightly more complicated example. The dihedral group \(D_{20}\) is the symmetry group of a \(20\)-gon. Inside this group is a subgroup that is isomorphic to the symmetry group of a \(5\)-gon (pentagon). Is this a surprise, or is this obvious? Here is a way to make precise the statement “\(D_{20}\) contains a copy of \(D_{5}\text{.}\)”
We build the domain and find its generators, so we know how many images to supply in the definition of the homomorphism. Then we construct the codomain, from which we will construct images. Our choice here is to send a reflection to a reflection, and a rotation to a rotation. But the rotations will both have order \(5\text{,}\) and both are a rotation by \(72\) degrees.
Since the kernel is trivial,
rho
is a one-to-one function (see Exercise \(11.4.18\)). But more importantly, by the First Isomorphishm Theorem,
G
is isomorphic to the image of the homomorphism. We compute the image and check the claim.
Just providing a list of images for the generators of the domain is no guarantee that the function will extend to a homomorphism. For starters, the order of each image must divide the order of the corresponding preimage. (Can you prove this?) And similarly, if the domain is abelian, then the image must also be abelian, so in this case the list of images should not generate a non-abelian subgroup. Here is an example. There are no homomorphisms from a cyclic group of order \(7\) to a cyclic group of order \(4\) (other than the trivial function that takes every element to the identity). To see this, consider the possible orders of the kernel, and of the two possibilities, see that one is impossible and the other arises with the trivial homomorphism. Unfortunately, Sage acts as if nothing is wrong in creating a homomorphism between these groups, but what Sage builds is useless and raises errors when you try to use it.
Rather than creating homomorphisms ourselves, in certain situations Sage knows of the existence of natural homomorphisms and will create them for you. One such case is a direct product construction. Given a group
G
, the method
.direct_product(H)
will create the direct product \(G\times H\text{.}\) (This is not the same command as the function
direct_product_permgroups()
from before.) Not only does this command create the direct product, but it also builds
four
homomorphisms, one with domain \(G\text{,}\) one with domain \(H\) and two with domain \(G\times H\text{.}\) So the output consists of five objects, the first being the actual group, and the remainder are homomorphisms. We will demonstrate the call here, and leave a more thorough investigation for the exercises.


15.7: Best Practices
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There are two problems that occur when instructors are not familiar with copyright law. The first, and most worrisome for administrators is the infringement of copyright. When third-party materials are used without proper regard to copyright law, the institution is exposed to a serious liability. The damage from a copyright infringement case would not only be economic, as the institution would have to pay for a defense, but also the reputation of the institution would be damaged. The second problem occurs when instructors fail to use materials that would enrich the learning experience of their students simply because they believe copyright law prohibits such use, or that obtaining permission would be too onerous. This does a disservice to the students as well as to the authors and artists of our society.
Even in the best of circumstances things can go wrong. It is possible that a copyright owner may be unavailable to grant permission for some reason, or there may be reasons that prevent the author from granting permission, or you may run into a copyright owner who is simply not going to grant permission. Having a back-up to replace any work will be a huge benefit.
When using third-party material, keep careful records of where content came from, what steps have been taken to obtain permission and under what terms permission was granted. At a minimum, any correspondence with copyright owners, including any final licences, should be retained for as long as a work is used. It is also good idea to retain a record of research undertaken while trying to determine who owns the copyright.
One of the side effects of seeking permission to use materials is the creation of a dialogue between creator and consumer of a work. Often, academic authors are only interested in how their works are used. By seeking permission you may also obtain access to unpublished materials, or higher quality copies. If there are any difficulties regarding the use of materials, if you have permission to use them you can go back to the rights holder for assistance. Imagine trying to do this for a ‘bootleg’ copy.