e as a Series and a LimitDate: 03/30/98 at 11:58:33 From: CMLee Subject: Definition of e Why does e = 1 + 1/2! + 1/3! + 1/4! + ... and lim (1 + 1/n) ^ n n --> infinity? What is the relation between these two expressions (if any), i.e. how do they connect? How do you exactly define e and why/how does the definition work? Please help. Thank you. Date: 03/30/98 at 22:12:20 From: Doctor Rob Subject: Re: Definition of e The relation is found in calculus. e is defined as the unique real number such that d(log_e(x))/dx = 1/x. That implies that e is given by the limit you gave. This comes from looking at the derivative as the limit of the difference quotient: 1/x = lim (log_e[x+d]-log_e[x])/d as d -> 0 1/x = lim (log_e[(x+d)/x])/d as d -> 0 1/x = lim (log_e[1+d/x])/d as d -> 0 1 = lim log_e([1+d/x]^[x/d]) as d -> 0 1 = log_e(lim [1+d/x]^[x/d]) as d -> 0 e = lim (1+d/x)^(x/d) as d -> 0 e = lim (1+1/n)^n as n -> infinity by setting n = x/d, or d = x/n. Now this implies that d(e^x)/dx = e^x: Let y = e^x, so x = log_e(y). Now differentiate using the Chain Law: 1 = (1/y)*dy/dx, so dy/dx = y, and d(e^x)/dx = e^x. Now the Maclaurin series for any function f(x) infinitely differentiable at x = 0 is given by: f(x) = f(0) + f'(0)*x/1! + f''(0)*x^2/2! + f'''(0)*x^3/3! + f''''(0)*x^4/4! + ... If we use: f(x) = e^x then: f(x) = f'(x) = f''(x) = f'''(x) = f''''(x) = ..., so 1 = f(0) = f'(0) = f''(0) = f'''(0) = f''''(0) = ..., so e^x = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + ... Now set x = 1. By the way, you left off the first term from your infinite series for e. You may also be able to establish the infinite series equation by using the binomial theorem on (1+1/n)^n, but I haven't seen this argument. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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