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Math Help - Difference between x = a and x = 0 in a Taylor series

  1. #1
    Senior Member x3bnm's Avatar
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    Difference between x = a and x = 0 in a Taylor series

    I know taylor series says that:

    Let f be a function with derivatives of all orders throughout some interval containing
    a as an interior point. Then the Taylor series generated by f at x = a is:

    \begin{align*}\sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k = f(a) + f'(a)(x - a) +& \frac{f''(a)}{2!}(x-a)^2 + \cdots +\\& + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots.\end{align*}

    So what do we mean when we say at x = a for the above definition? And what is meant when we use the above definition for x=0?

    Is it possible to kindly clarify these two (I'm not sure about the difference between two)?
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Difference between x = a and x = 0 in a Taylor series

    Don't worry. Found the answer.

    Taylor Series -- from Wolfram MathWorld

    "The Taylor (or more general) series of a function f(x) about a point a up to order n may be found using Series[f, {x, a, n}] ......."
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    Re: Difference between x = a and x = 0 in a Taylor series

    I understand it as centered at a. When it's centered at 0 its called the Mclaurin Series.
    Thanks from x3bnm
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  4. #4
    Senior Member x3bnm's Avatar
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    Re: Difference between x = a and x = 0 in a Taylor series

    Quote Originally Posted by delgeezee View Post
    I understand it as centered at a. When it's centered at 0 its called the Mclaurin Series.
    Thanks a lot.
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