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A few questions about the Taylor series

  1. Feb 2, 2008 #1
    When I tried to learn the Taylor series , I could not comprehend why a infinite series can represent a function

    Would anyone be kind enough to teach me the Taylor series? thank you:smile:

    PS. I am 18 , having the high school Math knowledge including Calculus
  2. jcsd
  3. Feb 2, 2008 #2


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  4. Feb 2, 2008 #3
  5. Feb 3, 2008 #4
    you can also consider a serie as a function

    [tex] f(x) = \sum_0^{\infty} \frac{f^{(n)(a)}}{n!}(x-a)^n [/tex]

    if the sum converges this is nothing but a function, to each x you get a number, namely the sum of the series.

    Mayby an example is in place.

    Lets say we define a function by

    [tex] f(x) = \sum_0^{\infty} (g(x))^n [/tex]

    where [tex] g: \mathhb{R} \rightarrow ]0,1[ [/tex], is this series a function you could say. But because the range of g i ]0,1[ this is just the geometric series, that is

    [tex] f(x) = \sum_0^{\infty} (g(x))^n = \frac{1}{1-g(x)} [/tex]

    so this is indeed a function. So you see that a series is a function if the sum converges. It's although not always possible to find a so simple expression like here for the serie.
    Last edited: Feb 3, 2008
  6. Feb 3, 2008 #5
    a little text from wikipedia

    The Taylor series need not in general be a convergent series, but often it is. The limit of a convergent Taylor series need not in general be equal to the function value f(x), but often it is. If f(x) is equal to its Taylor series in a neighborhood of a, it is said to be analytic in this neighborhood. If f(x) is equal to its Taylor series everywhere it is called entire.
  7. Feb 3, 2008 #6


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    ?? If a series converges to a function, then it's domain and range are exactly that of the function, of course. A series, depending on a variable x, is a function. You seem to think there is something special about using a series to define a function.
  8. Feb 3, 2008 #7
    you are right, but sometimes the taylor series only converges in a neigborhood around the point a, and sometimes it doesn't converges at all. So to say that if the series converges then it is equal to the function is a bit loose, I agree on that for those x where the series converges, then on those x they agree. Actually that is probaly what you meen, but just to make it clear.
  9. Feb 3, 2008 #8


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    I didn't say that! I said that if a series converges, it is necessarily equal to a function. I didn't say anything about the series being a Taylor's series.
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