Proof of Series ln(1+x)
Date: 11/15/2001 at 12:15:31 From: Colleen Torke Subject: Proof of series ln(1+x) I need to show that the series ln(1+x) equals x-x^2/2+x^3/3-x^4/4... and so on... whenever x is between -1 and 1. I really do not know where to start, but I need an elaborate proof of it. Thanks for the help.
Date: 11/15/2001 at 15:08:04 From: Doctor Jubal Subject: Re: Proof of series ln(1+x) Hi Colleen, Thanks for writing Dr. Math. To solve this problem, you need to find the Taylor series for ln(1+x) around x = 0. The idea behind a Taylor series is that if two functions have the same value and the same slope at a point, then they're going to be pretty close to each other near that point. And if the second derivative is also the same, then the match is even better, and the third derivative is also the same, better still. If all the derivatives are the same, then they are the same function, for most intents and purposes (assuming the functions are well-behaved). This idea is embodied in Taylor's theorem ___ (x - x0)^k d^k f f(x) = \ ----------- ----- (x0) /__ k! (dx)^k k=0 Here x0 is the point you're writing the Taylor expansion around. The term on the right is the kth derivative of f, evaluated at x = x0. If x0 = 0, this series is called a MacLaurin series and has a slightly simpler form: ___ x^k d^k f f(x) = \ ------ ------- (0) /__ k! (dx)^k k=0 So to find this series, what you need to do is (1) Figure out what all of the derviatives of ln(1+x) are. Since there are an infinite number of them, you can save yourself some work by just taking the first few derivatives and then looking to see if there's some sort of pattern you can use to predict what the rest of them are. (2) Evaluate each of those derivatives at x = 0 (3) Plug these values into the general form of the MacLaurin series, and simplify until you have the series you want to prove. Does this help? If you need any further explanation, don't hesitate to write back. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum