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David Terr
Ph.D. Math, UC Berkeley

<< 5.3.2. The Golden Ratio (φ ≈ 1.618)

5.3.3. The Base of Natural Logarithms (e ≈ 2.718)

One of the most important mathematical constants is one that most people don't know about, namely the base of natural logarithms, denoted by the symbol e. Rounded to 10 decimal places, the numerical value of e is 2.7182818285.

To get a feeling for the meaning of e, consider the problem of compount interest. Suppose you invest \$1000 in a bank account with an annual interest rate of 1%. (I know that's not much, but this will make the computation easier. How much money would you have after 100 years, assuming that you lived that long? Each year, the amount of your investment increases by 1%, so after one year, you have 1% more, or 1.01 times, your initial investment of \$1000, that is, you have \$1010. After two years, you have 1\$ more than this, which is 1.01 times \$1010, or \$1020.10. Note that this is also equal to 1.012 * \$1000. After three years, you have 1.013 * \$1000, which works out to be \$1030.30. After four years, you have 1.014 * \$1000, which comes out to \$1040.60. You get the idea. To generalize, after n years, your investment is worth 1.01n * \$1000. Thus, after 100 years, the amount of your investment is 1.01100 * \$1000 = \$2704.81.

So far, so good. Now what if your interest is compounded quarterly instead of annually? Then every three months, the amount of your investment increases by 0.25%, or by a factor of 1.0025. Thus, after one quarter, the amount of your investment is 1.0025 * \$1000 = \$1002.50. After two quarters, the amount becomes 1.00252 * \$1000 = %1005.01. After one year, you have 1.00254 * \$1000 = \$1010.04. Note that this is only four cents more than you would have had if the interest were compounded annually. After n years, or 4n quarters, the initial investment is now worth 1.00254n * \$1000. Thus, after 100 years, or 400 quarters, the investment is now worth \$2714.90.

What if the interest is compounded monthly? Then after one month, the investment is worth 1/12 of a percent more than \$1000.00, which comes out to \$1000.83. After one year, or 12 months, the investment is worth (1 + 1/1200)12 * \$1000, which works out to be \$1001.05. (Not thtat this is only one cent more than if the interest were compounded quarterly and only 5 cents more than if it were compounded annually.) After n months, the value of the investment is (1 + 1/1200)n * \$1000, so that after 100 years, or 1200 months, it is worth (1 + 1/1200)1200 * \$1000 = \$2717.15.

By now, perhaps you've noticed something interesting going on. Although the amount of money after 100 years increases as the period of time in which the interest is compounded decreases, it doesn't increase by much. In fact, the amount of money after 100 years converges to a constant value, so that no matter how short an interval in which the interest is compounded, you'll never have more than this amount of money. This amount works out to be \$2718.28, which is e times the initial investment of \$1000.00

Formally, e is defined as the limiting value of the quantity (1 + 1/N)N as N goes to infinity. The following table lists several values of (1 + 1/N)N for increasing N. As you can see, these values converge to e.

 N (1 + 1/N)N 1 2.000000 2 2.250000 3 2.370370 4 2.441406 5 2.488320 10 2.593742 20 2.653298 50 2.691588 100 2.704814 1000 2.716924 10,000 2.718146 100,000 2.718268 1,000,000 2.718280

So what good is e, other than in computing compound interest? It turns out that e is used all over math and science, mainly because it frequently appears in calculus. It is called the base of natural logarithms, because natural logarithms are defined as logarithms to the base e, which turns out to be the most mathematically convenient base for computing logarithm. This is true despite the fact that we have 10 fingers, so that base-10 logarithms, or common logarithms, may seem more convenient.

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