On True Compound Interest
Let there be a quantity growing in such a way that the increment of its growth, during a given time, shall always be proportional to its own magnitude. This resembles the process of reckoning interest on money at some fixed rate; for the bigger the capital, the bigger the amount of interest on it in a given time.
Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books call “simple interest,” or by what they call “compound interest.” For in the former case the capital remains fixed, while in the latter the interest is added to the capital, which therefore increases by successive additions.
(1) At simple interest. Consider a concrete case. Let the capital at start be \$100, and let the rate of interest be 10 percent per annum. Then the increment to the owner of the capital will be \$10 every year. Let him go on drawing his interest every year, and hoard it by putting it by in a stocking, or locking it up in his safe. Then, if he goes on for 10 years, by the end of that time he will have received 10 increments of \$10 each, or \$100, making, with the original \$100, a total of \$200 in all. His property will have doubled itself in 10 years. If the rate of interest had been 5 percent, he would have had to hoard for 20 years to double his property. If it had been only 2 percent, he would have had to hoard for 50 years. It is easy to see that if the value of the yearly interest is
Or, if y be the original capital, and the yearly interest is
(2) At compound interest. As before, let the owner begin with a capital of \$100, earning interest at the rate of 10 percent per annum; but, instead of hoarding the interest, let it be added to the capital each year, so that the capital grows year by year. Then, at the end of one year, the capital will have grown to \$110; and in the second year (still at 10%) this will earn \$11 interest. He will start the third year with \$121, and the interest on that will be \$12.1; so that he starts the fourth year with \$133.1, and so on. It is easy to work it out, and find that at the end of the ten years the total capital will have grown to \$259.374. In fact, we see that at the end of each year, each dollar will have earned
But this mode of reckoning compound interest once a year, is really not quite fair; for even during the first year the \$100 ought to have been growing. At the end of half a year it ought to have been at least \$105, and it certainly would have been fairer had the interest for the second half of the year been calculated on \$105. This would be equivalent to calling it 5% per half-year; with 20 operations, therefore, at each of which the capital is multiplied by
But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at a time. Suppose we divided the year into 10 parts, and reckon a one percent interest for each tenth of the year. We now have 100 operations lasting over the ten years; or
Even this is not final. Let the ten years be divided into 1000 periods, each of
Go even more minutely, and divide the ten years into 10,000 parts, each
Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression
Let us take geometrical illustrations of these things. In the following figure, OP stands for the original value. OT is the whole time during which the value is growing. It is divided into 10 periods, in each of which there is an equal step up. Here
In the next figure, we have the corresponding illustration of the geometrical progression. Each of the successive ordinates is to be
The Number e
To this mysterious number 2.7182818…, the mathematicians have assigned the letter e. This number is often called Euler’s number after the Swiss mathematician Leonhard Euler. All fifth graders know that the Greek letter π (called pi) stands for 3.141592…; but how many of them know that e means 2.71828…? Yet it is an even more important number than π!
What, then, is e?
Suppose we were to let 1 grow at simple interest till it became 2; then, if at the same nominal rate of interest, and for the same time, we were to let 1 grow at true compound interest, instead of simple, it would grow to the value of the number e.
This process of growing proportionately, at every instant, to the magnitude at that instant, some people call an exponential rate of growing. Unit exponential rate of growth is that rate which in unit time will cause 1 to grow to 2.718281. It might also be called the organic rate of growing because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself.
If we take 100 percent as the unit of rate, and any fixed period as the unit of time, then the result of letting 1 grow arithmetically at unit rate, for unit time, will be 2, while the result of letting 1 grow exponentially at unit rate, for the same time, will be 2.71828… .
A little more about the number e
We have seen that we require to know what value is reached by the expression
It is, however, worth while to find another way of calculating this immensely important figure.
Accordingly, we will avail ourselves of the binomial theorem, and expand the expression
The binomial theorem gives the rule that
Now, if we suppose n to become indefinitely great, say a billion, or a billion billions, then
By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms:
| 1.000000 | |
| dividing by 1 | 1.000000 |
| dividing by 2 | 0.500000 |
| dividing by 3 | 0.166667 |
| dividing by 4 | 0.041667 |
| dividing by 5 | 0.008333 |
| dividing by 6 | 0.001389 |
| dividing by 7 | 0.000198 |
| dividing by 8 | 0.000025 |
| dividing by 9 | 0.000002 |
| Total | 2.718281 |
e is incommensurable with 1, and resembles π in being an interminable non-recurrent decimal.
The Exponential Series
We shall have need of yet another series.
Let us, again making use of the binomial theorem, expand the expression
But, when n is made indefinitely great, this simplifies down to the following:
This series is called the exponential series.
Derivative of the Exponential Function
The great reason why e is regarded of importance is that ex possesses a property, not possessed by any other function of x, that when you differentiate it its value remains unchanged; or, in other words, its derivative is the same as itself. This can be instantly seen by differentiating it with respect to x, thus:
Now we might have gone to work the other way, and said: Go to; let us find a function of x, such that its derivative is the same as itself. Or, is there any expression, involving only powers of x, which is unchanged by differentiation? Accordingly; let us assume as a general expression that
Now, if this new expression is really to be the same as that from which it was derived, it is clear that A must = B; that
The law of change is therefore that
If, now, we take A = 1 for the sake of further simplicity, we have
Differentiating it any number of times will give always the same series over again.
If, now, we take the particular case of A = 1, and evaluate the series, we shall get simply
Of course it follows that ey remains unchanged if differentiated with respect to y. Also eax, which is equal to (ea)x, will, when differentiated with respect to x, be aeax, because a is a constant.
Natural or Naperian Logarithms
Another reason why e is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If y is the value of ex, then x is the logarithm, to the base e, of y. Or, if
The logarithm with base e is called the natural logarithm. The natural logarithm is so important that it has its own shorthand:
The two curves plotted in Figs. 14.3 and 14.4 represent these equations.
The points calculated are:
| x | −2 | −1 | −0.5 | 0 | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|---|---|---|---|
| y = ex | 0.14 | 0.37 | 0.61 | 1 | 1.65 | 2.71 | 4.50 | 7.39 |
| y | 0.1 | 0.5 | 1 | 2 | 3 | 4 | 8 |
|---|---|---|---|---|---|---|---|
| x = ln y | −2.30 | −0.69 | 0 | 0.69 | 1.10 | 1.39 | 2.08 |
It will be seen that, though the calculations yield different points for plotting, yet the result is identical. The two equations really mean the same thing.
As many persons who use ordinary logarithms, which are calculated to base 10 instead of base e, are unfamiliar with the “natural” logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product still holds good; or
Using A Calculator to Find ex and ln x
Modern scientific and graphing calculators are equipped with buttons for exponentiation with the base e and buttons for calculating natural or common logarithms. The exponential function ex is sometimes denoted by exp(x). Therefore, when using a calculator, we may need to locate the
Derivatives of Logarithmic and Exponential Functions
Now let us try our hands at differentiating certain expressions that contain logarithms or exponentials.
Take the equation:
Now this is a very curious result. It may be written
Note that x−1 is a result that we could never have got by the Power Rule for differentiating powers. That rule is to multiply by the power, and reduce the power by 1. Thus, differentiating x3 gave us 3x2; and differentiating x2 gave 2x. But differentiating x0 does not give us x−1 or
Now, try to differentiate
This gives
Next try
First change to natural logarithms by multiplying by the modulus
In general, because
The next thing is not quite so simple. Try this:
Taking the logarithm of both sides, we get
Since
We see that, since
We shall find that whenever we have an expression such as ln y = a function of x, we always have
In summary
Let us now attempt further examples.
Examples
Example 1. Differentiate y with respect to x if y = e−ax. Solution. Let −ax = z; then y = ez. Or thus:
Example 2. Differentiate y with respect to x if Solution. Let Or thus:
Example 3. Given Solution. Check by writing Check by writing
Example 4. If Solution. Let
Example 5. If Solution. Let
Example 6. If Solution. Taking the logarithm of both sides, we get
Example 7. If Solution. Taking the logarithm of both sides, we get
Example 8. If Solution.
Example 9. If Solution.
Example 10. If Solution. Let z = ln x;
Example 11. If Solution.
Try now the following exercises.
Exercises I
Exercise 1. Differentiate
Answer
Solution
(1)
Exercise 2. Find the derivative of the expression
Answer
Solution
Exercise 3. If y = nt, find
Answer
ln n.
Solution
Exercise 4. Show that if
Solution
Exercise 5. If
Answer
Solution
Differentiate
Exercise 6.
Answer
Solution
Exercise 7.
Answer
Solution
Exercise 8.
Answer
Solution
Exercise 9.
Answer
Solution
Exercise 10.
Answer
Solution
Exercise 11.
Answer
Solution
Exercise 12.
Answer
Solution
Exercise 13. It was shown by Lord Kelvin that the speed of signaling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called y, then the number of signals s that can be sent per minute can be expressed by the formula
Solution
When
Exercise 14. Find the maximum or minimum of
Answer
Min.: y = 0.7 for x = 0.694.
Solution
To show that this specific value of x makes y a minimum, we apply the Second Derivative Test. First we differentiate
Exercise 15. Differentiate
Answer
Solution
Recall that ln(AB) = ln A + ln B and
Exercise 16. Differentiate
Answer
Solution
The Logarithmic Curve
Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation y = bpx.
We can see, by putting x = 0, that b is the initial height of y.
Then when
Also, we see that p is the numerical value of the ratio between the height of any ordinate and that of the next preceding it. In the following figure, we have taken p as
If two successive ordinates are related together thus in a constant ratio, their logarithms will have a constant difference; so that, if we should plot out a new curve, the following figure, with values of ln y as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that
The Die-away Curve
If we were to take p as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in the next figure, where each successive ordinate is
The equation is still
but since p is less than one, ln p will be a negative quantity, and may be written −a; so that p = e−a,3 and now our equation for the curve takes the form
The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton’s celebrated “law of cooling”) by the equation
A similar formula,
Oscillations given to a flexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar way.
In fact e−at serves as a die-away factor for all those phenomena in which the rate of decrease is proportional to the magnitude of that which is decreasing; or where, in our usual symbols,
We should have got the same result if we had taken the equation in the form
The Time-constant. In the expression for the “die-away factor” e−at, the quantity a is the reciprocal of another quantity known as “the time-constant,” which we may denote by the symbol T. Then the die-away factor will be written
As an example, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e. when t = 0) it is 72°C hotter than the surrounding objects, and if the time-constant of its cooling is 20 minutes (that is, if it takes 20 minutes for its excess of temperature to fall to 1/e part of 72 degrees), then we can calculate to what it will have fallen in any given time t. For instance, let t be 60 minutes. Then
Further Examples
Example 12. The strength of an electric current in a conductor at a time t seconds after the application of the electromotive force producing it is given by the expression The time constant is If E = 10, R = 1, L = 0.01; then when t is very large the term Its value at any time may be written: To find the value of the current when t = 0.001s, say, It follows that, after 0.001 s, the variable term is Similarly, at the end of 0.1 s,
Example 13. The intensity I of a beam of light which has passed through a thickness l cm of some transparent medium is This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by 18% in passing through 10 cm of a certain transparent medium, this means that To find the thickness that will reduce the intensity to half its value, one must find the value of l which satisfies the equality
Example 14. The quantity Q of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity Q0 of the substance by the relation For “Radium A,” if time is expressed in seconds, experiment shows that Solution. We have 0.5 = e−0.00385t.
Exercises II
Exercise 17. Draw the curve
Answer
Use your calculator to evaluate
Solution
To sketch the curve by hand, we can evaluate
| t | y = 12e−t/8 |
|---|---|
| 0 | 12.000 |
| 1 | 10.590 |
| 5 | 6.423 |
| 10 | 3.438 |
| 15 | 1.840 |
| 20 | 0.985 |
Then plot each of these points
There are also multiple tools for plotting the curve plot 12 e^ (-t/8) from t=0 to t=20.
Exercise 18. If a hot body cools so that in 24 minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to 1 per cent. of the original excess.
Answer
T = 34.625;
Solution
The equation of cooling is
After 24 minutes
time-constant is approximately
Now we want to find t such that
Exercise 19. Plot the curve
Solution
When x = 0, y = 0
When x is a large positive number, the term e−2t becomes negligible (≈ 0). Consequently, we have
On the other hand, when x is a large negative number, the terms e−2t becomes very large positive. As a result y becomes numerically large but negative.
The curve
Exercise 20. The following equations give very similar curves:
Solution
Let’s assume b > 0.
When x = 0, then
When x is very large positive
For negative values of x, these curves are not comparable because
When x is large negative,
These curves are shown below.
Exercise 21. Find the derivative of y with respect to x, if
Answer
a)
Solution
(a)
Taking the natural logarithm of both sides and recalling
Now differentiating and using the Chain Rule for the left-hand side and the Product Rule for the right-hand side
(b)
Method 1)
Taking the natural logarithm of both sides and using the property
Now let’s differentiate both sides with respect to x:
Method 2) Recall that
Let u = x2 and applying the Chain Rule with
(c)
Method 1) Let
Method 2) Take ln of both sides
Exercise 22. For “Thorium A,” the value of λ is 5; find the “mean life,” that is, the time taken by the transformation of a quantity Q of “Thorium A” equal to half the initial quantity Q0 in the expression
Answer
0.14 seconds.
Solution
Exercise 23. A condenser of capacity
Answer
(a) 1.642;(b) 15.58.
Solution
Here
(a) When t = 0.1
(b) When t = 0.01
Exercise 24. The charge Q of an electrified insulated metal sphere is reduced from 20 to 16 units in 10 minutes. Find the coefficient μ of leakage, if
Answer
μ = 0.00037,
Solution
Since
Now we need to find t such that
Exercise 25. The damping on a telephone line can be ascertained from the relation
Answer
Solution
(a)
Therefore, the strength becomes 63.38% of the initial strength.
(b)
Exercise 26. The pressure p of the atmosphere at an altitude h kilometres is given by The pressures at 10, 20 and 50 kilometres being 199.2, 42.2, 0.32 respectively, find k in each case. Using the mean value of k, find the percentage error in each case.
Answer
0.1339, 0.1445, 0.1555, mean 0.1446; −10.2%, practically nil, +71.9%.
Solution
Calculating k when h = 10 km,
Calculating k when h = 20 km
Calculating k when h = 50 km
Average k (the mean value of k):
Calculating p using kav when h = 10 km:
Calculating p using kav when h = 20 km:
Calculating p using kav when h = 50 km:
Exercise 27. Find the minimum or maximum of y = xx.
Answer
Min. for
Solution
In Exercise S, we showed that
Setting
Since xx ≠ 0, at the minimum or maximum, we must have
To determine if this value of x makes y a maximum or a minimum, we apply the Second Derivative Test:
The graph of y = xx is shown below.
Exercise 28. Find the minimum or maximum of y = x1/x.
Answer
Max. for x = e.
Solution
To differentiate, we first take the natural logarithm of both sides then use the Chain Rule:
Since x1/x ≠ 0,
To determine whether x = e makes y a minimum or a maximum, we compare the value of y at this point with the values of y at some nearby points:
When x = e
When x = 2
When x = 3
Therefore x = e makes y = x1/x a maximum.
We can also apply the Second Derivative Test:
Using the Quotient Rule
When x = e, both
It follows from the Second Derivative Test that
The graph of y = x1/x is shown below
Exercise 29. Find the minimum or maximum of
Answer
Min. for x = ln a.
Solution
Using the Product Rule
To find
Therefore
Since
or
Using the Second Derivative Test
When x = ln a
Therefore
Hyperbolic Functions
Hyperbolic functions are specific combinations of exponential functions that arise frequently in various applications, leading mathematicians to give them distinctive names and thoroughly investigate their properties. Although hyperbolic functions are combined exponential functions, in some ways they resemble the trigonometric functions. As a result, the hyperbolic functions are given individual names such as hyperbolic sine, hyperbolic cosine, hyperbolic tangent, and others. Their definitions are as follows:
One can easily sketch the graphs of y = sinh x and y = cosh x by plotting the curves y = ex and y = e−x, adding and subtracting the ordinates, and taking half of each. The following figure displays their graphs.
A catenary, which is the graph of the hyperbolic cosine, is a curve that describes the shape of a homogeneous chain or cord hanging under its own weight.
When x is positive large, ex is very large, while e−x is extremely small; hence ex ± e−x ≈ ex leading to:
Identities Between Hyperbolic Functions
The properties of the hyperbolic functions closely resemble the corresponding properties of the trigonometric functions.
It follows directly from the definitions that
Example 15. Prove sinh(x + y) = sinh x cosh y + cosh x sinh y. Solution. Let’s simplify the right-hand side
Therefore,
Another important set of identities, which follow at once from the definitions, is:4
Derivatives of the Hyperbolic Functions
As the hyperbolic functions are combinations of exponential functions, and we have just learned how to differentiate exponential functions, we can apply the differentiation rules learned in Chapters 5 and 6 to determine the derivatives of the hyperbolic functions.
For example, since sinh x = ½(ex − e−x), we have
Example 16. Show Solution. Since tanh x = sinh x / cosh x, we can use the Quotient Rule:
In summary,
Inverse Hyperbolic Functions
The inverse hyperbolic sine function is denoted by arcsinh x or sinh−1 x, and is defined by
Similarly, the inverse hyperbolic tangent function is defined by
In summary,
We can find explicit formulas for the inverse hyperbolic functions.
Example 17. Show that Solution. If y = arcsinh x, then
In a similar fashion, we can derive explicit formulas for the other inverse hyperbolic functions.
Derivatives of the Inverse Hyperbolic Functions
With what we have learned so far, we can find the derivatives of the inverse hyperbolic functions.
Example 18. If Solution. If We can also find the derivative of arcsinh x by differentiating
Example 19. If y = arccosh x (also denoted by cosh−1 x), find Solution. If
Example 20. If y = arctanh x (also denoted by tanh−1 x), find Solution. If y = arctanh x, then x = tanh y, and
In summary,
The ordinary logarithm or the 10-th based logarithm is often called the common logarithm.↩︎
Alternatively, we can use the Chain Rule. Let u = x + a. Then y = ln u and
ln p = −a ⇒ p = e−a.↩︎
We say that the hyperbolic sine and hyperbolic tangent functions are odd, and the hyperbolic cosine is an even function (similar to the corresponding trigonometric functions). A function is called odd if