Date: 01/28/98 at 16:29:49 From: Jeffery Holcomb Subject: Euler's number During a recent math assignment I came in contact with Euler's number. I also heard it referred to as the natural base. When I asked why it was called this my teacher said that he did not know. Since then I have looked in several different places and have not been able to find an answer. If you could please tell me why Euler's number is referred to as the natural base I would greatly appreciate it.
Date: 01/28/98 at 18:13:01 From: Doctor Anthony Subject: Re: Euler's number The base of logs that has most application in mathematics is 'e'. Here is a note on how this number arises naturally from the laws of compound interest growth. In the 1730's 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 were added twice a year, then we replace r by r/2 and we replace t by 2t. So formula becomes A = P(1 + r/(2x100))^(2t) = amount after t years. If 3 times a year then A at the end of t years would be: A = P(1 + r/(3x100))^(3t) and if we added interest N times a year, then after t years the amount A would be A = P[1 + r/(Nx100)]^(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 1/n^2, 1/n^3 etc, in the denominators and distribute one n to each of the terms n, n-1, n-2, etc. in the numerator, getting 1 x (1-1/n) x (1-2/n) x ..... 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 Check out our web site! http://mathforum.org/dr.math/
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