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Taylor series expansion question

  1. May 4, 2005 #1
    Hi, I have a question about Taylor series:

    I know that for a function f(x), you can expand it about a point x=a, which is given by:
    [tex] f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + ... [/tex]

    but I would like to do it for f(x+a) instead of f(x), and expand it about the very same point on the graph (I don't know if it is still x=a or something else after this translation).

  2. jcsd
  3. May 4, 2005 #2
    why not just plug x+a into the expansion (you should be a little careful with radii of convergence - but this can be bypassed)?

    [tex]f(x) = f(a) + f^\prime(a)(x-a) + \cdots \Longrightarrow f(x+a) = f(a) +f^\prime(a)x + f^{\prime \prime}(a)x^2 + \cdots.[/tex]

    Take a look and see if this satisfies what you want for being about "the same point on the graph" (I suspect you will find it does).
    Last edited: May 5, 2005
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