Compounding Interest and eDate: 11/11/98 at 11:52:55 From: John Radway Subject: How in the world did they come up with e? I know that e is equal to 2.718281828..., and that it is the base of natural logarithms, but what does that number mean? How is it calculated? How did they come up with it? Date: 11/11/98 at 14:11:24 From: Doctor Anthony Subject: Re: How in the world did they come up with e? In the 1730s Euler investigated the result of compounding interest continuously when a sum of money, say, is invested at compound interest. If interest is added once a year we have the usual formula for the amount, A, with principal P, rate of interest r percent per annum, and t the time in years: A = P(1 + r/100)^t If interest is added twice a year, we replace r by r/2 and t by 2t. So the formula becomes: A = P(1 + r/(2 * 100))^(2t) = amount after t years. If 3 times a year then A at the end of t years would be: A = P(1 + r/(3 * 100))^(3t) and if we added interest N times a year, then after t years the amount A would be: A = P[1 + r/(N * 100)]^(Nt) Now to simplify the working, we put r/(100N) = 1/n so N = nr/100 and: A = P[1 + 1/n]^(nrt/100) A = P[(1 + 1/n)^n]^(rt/100) We now let n -> infinity and we must see what happens to the expression (1 + 1/n)^n as n tends to infinity. Expanding by the binomial theorem: (1 + 1/n)^n = 1 + n(1/n) + n(n-1)/2! (1/n)^2 + n(n-1)(n-2)/3! (1/n)^3 + ... Now take the n's in the denominators of 1/n^2, 1/n^3, and so on, and distribute one n to each of the terms n, n-1, n-2, etc in the numerator, getting for each arbitrary term of the sequence: 1 * (1-1/n) * (1-2/n) * ... So we now have: (1 + 1/n)^n = 1 + 1 + 1(1-1/n)/2! + 1(1-1/n)(1-2/n)/3! + ... Now let n -> infinity and the terms 1/n, 2/n, etc. all go to zero, giving: (1 + 1/n)^n = 1 + 1 + 1/2! + 1/3! + 1/4! + ..... and this series converges to the value we now know as e. If you consider e^x you get: (1 + 1/n)^(nx) and expanding this by the binomial theorem you have: (1 + 1/n)^(nx) = 1 + (nx)(1/n) + nx(nx-1)/2! (1/n)^2 + .... and carrying through the same process of putting the n's in the denominator into each term in the numerator as described above you obtain: e^x = 1 + x + x^2/2! + x^3/3! + .... and differentiating this we get: d(e^x)/dx = 0 + 1 + 2x/2! + 3x^2/3! + ... = 1 + x + x^2/2! + x^3/3! + .... = e^x Reverting to our original problem of compounding interest continuously, the formula for the amount becomes: A = P e^(rt/100) You might like to compare the difference between this and compounding annually. If P = 5000, r = 8, t = 12 years: Annual compounding gives A = 5000(1.08)^12 = 12590.85 Continuous compounding gives A = 5000 e^(96/100) = 13058.48 The difference is not as great as might be expected. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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