Proofs of eDate: 03/21/2002 at 21:37:08 From: Christine Subject: Proof of number e I was asked by my algebra teacher to find the proof of e. I found out that the definition of e is: e = lim, as n approaches infinity (1+ 1/n)^n. Is there a proof for this or for 1+2(e-1/n)? Thank you so much ahead of time for your help! Date: 03/22/2002 at 09:27:16 From: Doctor Pete Subject: Re: Proof of number e Hi Christine, If we define the number e to equal the limit of (1 + 1/n)^n as n -> Infinity, then we do not need to prove that this limit is equal to e, because it is true by definition. However, there are several "definitions" for the number e. For instance, it is the unique real number greater than 1 such that the area under the curve y = 1/x from x = 1 to x = e is equal to 1. Or we may define e to be the value of the infinite series e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ... . And of course, we may define e to be the limit you describe. All of these are equivalent; however, the challenge is in proving that these definitions are equivalent. To do so rigorously requires basic calculus, and concepts from real analysis. We can show that the series definition implies that e is also equal to the limit formula. Let s[n] = 1/0! + 1/1! + ... + 1/n!, t[n] = (1 + 1/n)^n, for n >= 0. Then by the Binomial Theorem, t[n] = 1 + 1 + (1 - 1/n)/2! + (1 - 1/n)(1 - 2/n)/3! + ... + (1 - 1/n)(1 - 2/n)(...)(1 - (n-1)/n)/n!. Therefore, t[n] <= s[n], so that lim sup t[n] <= e. n -> oo Now, if n >= m, then t[n] >= 1 + 1 + (1 - 1/n)/2! + ... + (1 - 1/n)(...)(1 - (m-1)/n)/m!. Let n -> infinity, keeping m fixed. Hence lim inf t[n] >= 1 + 1 + 1/2! + ... + 1/m!, n -> oo so we also find s[m] <= lim inf t[n]. n -> oo Now let m -> infinity: e <= lim inf t[n]. n -> oo Therefore, e <= lim inf t[n] <= lim sup t[n] <= e, and it follows that lim t[n] = e. To see how the limit definition implies the integral (area under the curve) definition, we will first show that the derivative of the natural logarithm is given by d/dx [ln(x)] = 1/x. (Heretofore we have defined the function ln(x) to be the inverse of the function y = e^x with e defined as a limit; thus by definition ln e = 1.) By definition of the derivative, we find d/dx [ln(x)] = lim (ln(x+h)-ln(x))/h, = lim (1/h)ln((x+h)/x) = lim ln((1 + h/x)^(1/h)) as h -> 0. Since x is a fixed positive real, let u = x/h; then the limit becomes lim ln((1 + 1/u)^(u/x)) = lim (1/x)(ln(1 + 1/u)^u) = 1/x lim ln(1 + 1/u)^u, as u -> infinity. But this latter limit is precisely ln e = 1, so the result follows. Therefore by the Fundamental Theorem of Calculus, there exists a real number t > 1 such that 1 = Integral[1/x dx, {1, t}] = ln(t) - ln(1) = ln(t), therefore ln(t) = 1, or t = e. That wraps up the equivalence of the three major definitions for e. It may be beyond the scope of what your instructor asked of you (or it may not), but the general point is that in mathematics, definitions do not need proof; they are true by assumption. Occasionally, however, some definitions lead to such powerful and often-used theorems in of themselves that the role of theorem and definition become reversed. For another example of this, consider the formula for the dot product, or the definition of the circular trigonometric functions. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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