how would you do this? Find the Taylor series for ln(x) about 2.
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Originally Posted by davecs77 how would you do this? Find the Taylor series for ln(x) about 2. see here if you need further explanation, just say so topsquark showed you the main process of deriving something like this, but again, memorization could be used here, the page i gave you has the general formulas
Originally Posted by davecs77 how would you do this? Find the Taylor series for ln(x) about 2. Just for the record, let ==> ==> ==> etc. So -Dan
Last edited by topsquark; Jul 23rd 2007 at 04:25 PM. Reason: Fixed an error in the form of the expansion
Originally Posted by topsquark Just for the record, let ==> ==> ==> etc. So -Dan How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)
Originally Posted by davecs77 How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2) The answer in my book is ln2 + summation of n = 1 going to infinity of (-1)^(n - 1) times ((x - 2)^n)/(n2^n) I get the first part..ln2 but dont understand the rest. Thank you.
Originally Posted by davecs77 How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2) Crap! I goofed! The general form for the Taylor series of f(x) about x = a is: So for a = 2 the first coefficient is The second coefficient is So the series should be: You can work the general formula for the nth term. It's easy enough. -Dan
Originally Posted by topsquark Crap! I goofed! can a Mormon say that? you know, the C-word, i only quoted "goofed" because it was funny as h
Originally Posted by topsquark Crap! I goofed! The general form for the Taylor series of f(x) about x = a is: So for a = 2 the first coefficient is The second coefficient is So the series should be: You can work the general formula for the nth term. It's easy enough. -Dan You are only doing 2 coefficients since a = 2?
Originally Posted by davecs77 You are only doing 2 coefficients since a = 2? No, I only did two terms since I'm lazy. The series is infinite with each term being smaller than the last. -Dan
Originally Posted by davecs77 how would you do this? Find the Taylor series for ln(x) about 2. An indirect way to find the taylor series is to use the following known Maclaurin series
Originally Posted by curvature An indirect way to find the taylor series is to use the following known Maclaurin series Then
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