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Math Help - need help!

  1. #1
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    need help!

    Find the Taylor series (write a summation if possible) about the indicated center and determine the interval of convergence.

    g(x)= x^3-x+1 c=1
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  2. #2
    TD!
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    The Taylor series of a polynomial is the polynomial itself!!
    You can still write the summation and possibly verify this of course.
    Last edited by TD!; May 2nd 2006 at 01:08 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD!
    The Taylor series of a polynomial is the polynomial itself!!
    You can still write the summation and possible verify this of course.
    Actually, no it isn't quite the same.

    The Taylor series for a function f(x) about a point x = c is:
    f(x) = \sum_{n=0}^{\infty} \frac{1}{n!} \left ( \frac{d^nf}{dx^n} \right ) (c) \, (x - c)^n
    The notation \left ( \frac{d^nf}{dx^n} \right ) (c) is meant to represent the nth derivative of f evaluated at x = c. This form is only correct, of course, assuming the series converges for a given f(x) and c.

    In this case your function is going to be:
    (x-1)^3+3(x-1)^2+2(x-1)+1

    -Dan
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    TD!
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    Quote Originally Posted by topsquark
    Actually, no it isn't quite the same.
    [...]
    In this case your function is going to be:
    (x-1)^3+3(x-1)^2+2(x-1)+1
    But then we have:

    <br />
\begin{gathered}<br />
  \left( {x - 1} \right)^3  + 3\left( {x - 1} \right)^2  + 2\left( {x - 1} \right) + 1 \hfill \\ \\<br />
   = \left( {x^3  - 3x^2  + 3x - 3} \right) + 3\left( {x^2  - 2x + 1} \right) + 2x - 2 + 1 \hfill \\ \\<br />
   = x^3  - x + 1 = g\left( x \right) \hfill \\ <br />
\end{gathered} <br />

    So we find the same after all. This will always be the case for polynomials.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD!
    But then we have:

    <br />
\begin{gathered}<br />
  \left( {x - 1} \right)^3  + 3\left( {x - 1} \right)^2  + 2\left( {x - 1} \right) + 1 \hfill \\ \\<br />
   = \left( {x^3  - 3x^2  + 3x - 3} \right) + 3\left( {x^2  - 2x + 1} \right) + 2x - 2 + 1 \hfill \\ \\<br />
   = x^3  - x + 1 = g\left( x \right) \hfill \\ <br />
\end{gathered} <br />

    So we find the same after all. This will always be the case for polynomials.
    True, but for the result to be a Taylor polynomial at c = 1, then it ought to be of the appropriate form. That's how I learned it, anyway. I admit the whole question is rather like splitting hairs.

    -Dan
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  6. #6
    TD!
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    I see what you mean, but the 'form' doesn't matter since the functions (the original result and the simplification to g(x) again) are completely the same.

    I don't really see the use of this question, unless its intention is that the student sees that the taylor series of a polynomial is the polynomial again; by experimentally finding this through the formula like you did.
    Last edited by TD!; May 2nd 2006 at 01:31 PM.
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