Here begins an optional extension of the sections on logarithms. Its starting point seems completely unrelated--namely, the interest paid on money deposited in a bank. It starts with a very elementary discussion of interest, and users familiar with this subject may quickly skim this part. However, be assured --a connection with logarithms soon appears!
Interest and Banking--a Very Elementary Introduction
Most people are familiar with interest, the extra amount paid by borrowers to temporarily use other people's money.
Suppose you want to buy an expensive item--car, house, business, farm machinery--but cannot pay the full price right away. You then borrow what you need from a bank, and gradually pay it back, adding a little extra as interest, in return for the privilege of borrowing.
Interest is traditionally measured in percent, hundredths of the amount borrowed, also denoted by the symbol %. If the interest rate is, say, 10 dollars per year for every 100 dollars borrowed, your interest is "10 percent" or 10% ("per cent" means, "for each hundred").
Banks usually lend money for only part of the cost of the purchased item--because then, if the borrower is unable to pay ("defaults"), they can legally "repossess" the item, and having paid only part of its cost, they suffer no loss.
Because a "down payment" is usually needed before borrowing, one needs to save money to cover it. After putting the money aside for such savings, people often lend it to a bank. The bank will pay a lower interest--say, only 6%--while using the same money to lend it out at some higher rate, such as 10%. That way the bank makes a profit and covers its own costs; but it is still better to lend out money at a lower rate and make a profit of $6 for every $100--better than not making any profit at all.
Simple Interest and Compound Interest
(Below, after the *** mark, you are expected here to use a calculator with a yx button)
Suppose you lend the bank $ 1000 at 6% interest. The amount you lend is known as "the principal."
After a year you have earned $60 and now have $1060
After 2 years, you have earned another $60 and now have $1120
After 3 years, you have earned another $60 and now have $1180
After 4 years, you have earned another $60 and now have $1240
(To save for an expensive purchase, most people would of course save $1000 every year and accumulate cash much faster. In this calculation, however, we concentrate on just the first $1000)
You can actually do better! The $60 earned in the first year can be added to the "principal" amount, so that in the second year it, too, earns interest. In fact, in any year you can add the profits to the principal, and earn more.
This changes the rules. Until now, what you earned in any year were $60, six percent of the amount originally invested. That is called "simple interest." The amount you have at the end of the year is just $60 more.
Now the amount you have at the end of the year (assuming you took nothing out) is 1.06 times what you had at the beginning of the year. This is called compound interest, compounded at the end of each year, Let us calculate how much you have at the end of each year. Suppose you started with x dollars. With simple interest
After 1 year 1.06 x
After 2 years 1.12 x
After 3 years 1.18 x
After 4 years 1.24 x
With compound interest as described above, to an accuracy of 4 decimals
| After one year you have || 1.06 x
| After another year || 1.06 [1.06 x] = (1.06)2 x = 1.1236 x
| After 3 years || 1.06 [(1.06)2 x] = (1.06)3 x = 1.191 x
| After 4 years || 1.06 [(1.06)3 x] = (1.06)4 x = 1.2624 x
So yes, you are increasing your profit, though the increase is rather moderate.
You get still more if you add the earned interest to the principal not at the end of each year, but at the end of each half year. The interest earned each half year is only 3%, but the number of "compounding periods" is doubled--two per half year.
Let's see what you get from an original amount of x dollars (calculating to an accuracy of 5 decimals)
|After half a year ||1.03 x
|After 1 year ||1.03 [1.03 x] = (1.03)2 x = 1.0609 x
|After 1.5 years ||1.03 [(1.03)2x] = (1.03)3 x = 1.10927 x
|After 2 years ||1.03 [(1.03)3 x] = (1.03)4 x = 1.12551 x
|After 3 years || (1.03)6x = 1.19405 x
|After 4 years || (1.03)8x = 1.26677 x
And so on
But why wait half a year? One may just as well add the money monthly, assuming here all months are equal and each earns 0.5% interest. (*** you better have a yx button for the next steps, though with a lot of patience the same results can also be derived without one. They are given to an accuracy of 6 decimals.)
After 1 year --12 periods, 0.5% each--you have (1.005)12 x = 1.061678 x
After 2 years--24 periods, 0.5% each--you have (1.005)24 x = 1.127159 x
After 3 years--36 periods, 0.5% each--you have (1.005)36 x = 1.196680 x
After 4 years--48 periods, 0.5% each--you have (1.005)48 x = 1.279489 x
It is instructive to compare the total amount after, say, 4 years:
|Simple interest || 1.24 x
|Compounded yearly || 1.2624 x
|Compounded twice a year || 1.26677 x
|Compounded 12 times a year || 1.278489 x
So yes, you keep getting more each time. You might get even more if you compounded every day (as some banks have brazenly promised) but the number seems to be creeping towards a limit, which you can never pass. You will never get rich this way!
In the next section we'll try derive that limit. Stand by for some algebra!
Suppose you lend out x dollars at a percentage of p percent (it was 6 in the above example). After one year we have
(1 + p/100) x
Let us divide the year into N equal parts, each of which earns interest p/N . After each such period, the money which was invested grows by a factor
[1 + p/(100 N)] (1)
There exist N such periods in the year, so after one year, our investment has grown by a factor
F = [1 + p/(100 N)]N (2)
Dividing the power N by some number and also multiplying by it does not change a thing--it's like multiplying by (Q/Q)=1 (whatever Q might be). So if Q=p/100
F = [1 + p/(100 N)][(100N/p) (p/100)] (3)
If you cancel the fractions, you are back where we started. However, instead of doing so, let us introduce a new variable quantity y: let
100N/p = y
Then (3) becomes
F = (1 + 1/y)(y.p/100) (5)
and remembering that one power raised to another is like raising to a power that is the product of both exponents
F = [(1 + 1/y)y](p/100)
The reason for introducing the new variable y is to put at the core of our expression the rather interesting quantity:
[(1 + 1/y)y]
(raised to power p/100). Once again, by (4) y is defined as
y = N (p/100)
If interest is compounded every second of the year, N is a bit over 31 million. Suppose N, the number of compounding periods, grows without limit, and so, therefore, does y. Then (7) becomes a rather strange expression!
On one hand, the expression we are raising to the yth power is (1 + 1/y), very close to 1, and we know that any power of 1 is still 1--no matter how many times you multiply 1 by itself, the result stays the same.
On the other hand, a high power y of any number larger than 1 (even larger by just a tiny bit) will grow without limit.
Which case is the one here? Let's try the calculator, and assume it has buttons for both (1/x) and x2. We can then easily derive the expression (7) for values of y which are powers of 2:
(1 + 1/2) 2 = 2.25
(enter 0.5, add 1, hit squaring button)
(1 + 1/4) 4 = 2.441406...
(enter 0.5, hit squaring button, add 1, hit x2 button 2 times more)
(1 + 1/16) 16 = 2.6379284...
(enter 0.5, hit squaring button 2 times, add 1, hit x2 button 4 times)
(1 + 1/256) 256 = 2.7129916...
(enter 0.5, hit squaring button 3 times, add 1, hit x2 button 8 times, since 256 = 28)
(1+ 1/65536) 65536 = 2.71826039...
(enter 0.5, hit squaring button 4 times, add 1, hit x2 button 16 times, as 65536 = 216)
You can see that the result approaches a limit which is neither zero nor infinity, but a number between 2 and 3. Mathematicians denote it by the letter e. And the limiting factor, beyond which compounded interest with percentage p cannot rise, no matter how frequently the compounding is performed, is by (6)
Processes which are compounded naturally--for instance, the number of bacteria (or other living creatures) given an unlimited supply of food, or the number of neutrons in an uncontrolled chain reaction--all these "grow exponentially" following a law like the one above.
Many properties of e involve calculus, or an expanded definition of numbers including the square root of (–1), also denoted i. For instance, if the symbol N! ("factorial of N") defines the product of whole numbers up to N
1! = 1 2! = 1.2 = 2 3! = 1.2.3 = 6 4! = 24 5! = 120 etc.
and so on, it may be shown that
e = 1 + 1/(1!) + 1/(2!) + 1/(3!) + 1/(4!) + 1/(5!) + ...
and because N! grows extremely fast with increasing values of N, one soon gets pretty accurate values of e.
Furthermore, the number e is also "the base of natural logarithms." That, however, is the story of the next section.
You saw how
(1 + 1/y)y
goes to a limit e = 2.71828... as y gets larger and larger. How about
(1 – 1/y)y
--does it go to a limit too? Indeed, it does. With your calculator you can derive approximations to it, as was done with e. If your calculator has a sign-reversing button "+/–" you can use the same steps as before, except that before adding "1", reverse the sign of the power of 0.5 by means of that button.
Next, the obvious question--how is that limit related to "e"? Try to discover on your own!