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maclaurin series

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The Maclaurin series generated by f(x)=x^ cosx + 1 2005-08-10
From Latto:
f(x)=x3cosx + 1. but when I take the derivatives, I couldn't see a pattern. Can you help?
Answered by Penny Nom.
A Taylor series 2001-04-27
From Karan:
Given the following information of the function
  1. f''(x) = 2f(x) for every value of x

  2. f(0) = 1

  3. f(0) = 0
what is the complete Taylor series for f(x) at a = 0

Answered by Harley Weston.
Maclaurin series again 2000-09-23
From Jason Rasmussen:
I suppose my confusion comes into play when I am trying to figure out where the xn term comes from. I know that the Power Series notation is directly related to the Geometric Series of the form sigma [ brn ] where the limit is b/(1-r) for convergence at | r | <1. Therefore, the function f(x) needs to somehow take the form of b/(1-(x-a)), which may take some manipulation, and by setting r = (x-a) and b = Cn, we get the Geometric Series converted to the Power Series. Taking the nth order derivative of the Power Series gives Cn = fn(a)/n!. There must be a gap in my knowledge somewhere because I cannot seem to make f(x) = ex take the form of f(x) = b/(1-(x-a)). Maybe I should have labelled my question as "middle" because it may be more of a personal problem with algebra and logarithms. Or, am I to assume that all functions can be represented by sigma [fn(a) * (x-a)n / n!]?
Answered by Harley Weston.
 
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