First Calculation of EDate: 12/18/97 at 13:50:49 From: Nick Vlassis Subject: The first calculation of the value of e Hello, I am a student doing a project on the discovery of e. I already know that D_x e^x = e^x, and that this is how e came to be found. However, I *don't* know (and really hope to learn how Euler first calculated the value of e starting from this equation. I am specifically interested in the *historical* facts, so please respond, if possible, with the exact method originally used by Euler. Thanks in advance, Nickolas Vlassis Date: 12/20/97 at 08:46:15 From: Doctor Anthony Subject: Re: The first calculation of the value of e 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% 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 expand 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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