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Taylor Series/Radius of Convergence - I just need a hint!

  1. Nov 10, 2004 #1
    Consider the following:

    [tex] f(x) = 1 + x + x^2 = 7 + 5 (x-2) + (x-2)^2 [/tex]

    which is a Taylor series centered at 2. My question is: what is the radius of convergence? The answer in my book is [tex] R=\infty [/tex], but take a look at this:

    [tex] f(x) = 7 + 5 (x-2) + (x-2)^2 = \sum _{n=0} ^{\infty} b_n (x-2)^n \Longrightarrow \left| x-2 \right| < 1 [/tex]

    Then, I get

    [tex] 1 \leq x \leq 3 \Longrightarrow R = \frac{3-1}{2}=1 \neq \infty [/tex]

    In other words, I'm a bit confused!

    Thanks :smile:
     
  2. jcsd
  3. Nov 11, 2004 #2

    Tide

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    Your confusion is that the [itex]b_n[/itex] are ALL 0 for n > 2. There's nothing to test since the series is perfectly well defined for all x. You might try, for example, comparing the terms in your series with the terms in a series you know converges for all x such as [itex]e^{-x^2}[/itex]. Clearly, for n > 2 each term of your series is smaller than the corresponding term in the latter expansion.
     
  4. Nov 11, 2004 #3

    Hurkyl

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    It looks like you tried to use the ratio test, but you can't because you need to find the limit of 0/0.

    However, you can apply the n-th root test, which will, indeed, say that it converges for all x. (as it should because the function is a polynomial which exists for all x!)
     
  5. Nov 11, 2004 #4
    It makes sense now.

    Thanks.
     
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