Approximating f(x - 1) in Terms of Its First and Second Derivatives
Date: 08/15/2012 at 11:31:57 From: Ajith Subject: Approximation of taylor's series When I was going through a proof of a theorem, a standard text made use of a fact that I couldn't verify: f(x - 1) = f(x) - f'(x) + (1/2)f''(e) Here, e is not the exponential constant. It is just a constant that lies between x - 1 and x. I know that any function f(x) can be written as f(a) + (x - a)f'(a) + (((x - a)^2)/2)f''(e) Here, a < e < x -- assuming the function converges, of course. The Lagrangian form of the remainder is used. Substituting x - 1 for a gives us f(x - 1) = f(x) - f'(x - 1) - (1/2)f''(e) and x - 1 < e < x Where did I go wrong? The text evidently used Taylor's series to get to that approximation, but it didn't work for me. First of all, can I substitute x - 1 in the place of a?
Date: 08/15/2012 at 16:50:46 From: Doctor Schwa Subject: Re: Approximation of taylor's series Hi Ajith, Maybe things would be more clear if you renamed your function. For example, let g(x) = f(x - 1). Then f(x) = g(x + 1). Using Taylor's expansion for g(x), we get ... g(x) = g(a) + g'(a) (x - a) + 1/2 g''(e) (x - a)^2 ... for some e between a and x. Put all this back in terms of f: f(x - 1) = f(a - 1) + f'(a - 1) (x - a) + 1/2 f''(e - 1) (x - a)^2 Now can you see what you want to set a equal to, and how to finish? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
Date: 08/15/2012 at 17:15:33 From: Ajith Subject: Thank you (Approximation of taylor's series) Yeah. I got it. Thanks a lot :)
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.