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Math Help - Convergence of a Taylor Series

  1. #1
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    Convergence of a Taylor Series

    f(x)=1/(3+2x)

    Problem: For which values does the Taylor Polynomial converge to the function?

    I've been working on this problem for a while messing around with Legrange's Theorem for the remainder term trying to use different bounds to show how for certain intervals of x the remainder must go to zero therefore the taylor polynomial must converge at those points. This works for some intervals but then on others it doesn't work at all. I'm wondering if I'm going about this the completely wrong way and if there's something else that I can do to show convergence? Can somebody please point me in the right direction?
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  2. #2
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    note ...

    \displaystyle \frac{1}{3+2x} = \frac{1}{3} \cdot \frac{1}{1 - \left(-\frac{2}{3}x\right)}

    familiar with the interval of convergence for the sum of an infinite geometric series?
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  3. #3
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    Thanks that helps alot.
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  4. #4
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    A "Taylor Series" is a power series of the form \sum_{n=0}^\infty a_n(x- x_0)^n. Both the Taylor series and its radius of convergence depend upon what x_0 you are using and you don't give that.

    Skeeter assumed you meant the "MacLaurin series", the Taylor series about x= 0.

    In fact, there is an easy way of answering the general question. Notice that \frac{1}{3+ 2x} is defined and continous (and, in fact, infinitely differentiable) for all x except x= -3/2. The radius of convergence about x= x_0 will be |x_0+ 3/2|. That is, the series will converge for all x between x_0 and -3/2 and, since this is a "radius" for an equal distance on the other side of x_0. (If you posed the same question in the complex plane, the series would converge on a disk having center at x_0 and having -3/2 on its boundary so actually having r= |x_0+ 3/2| as radius.)

    You could prove that, for this particular function, by writing it as \frac{1}{3+ 2x}= \frac{1}{3+ 2(x-x_0)+ 2x_0}= \frac{1}{(3- 2x_0)+ 2z} where z= x- x_0. Now, as earboth factored a 3 out of the denominator, factor out that 3- 2x_0, leaving \frac{1}{3- 2x_0}\frac{1}{1- (-\frac{2z}{3-2x_0})} and recognizing that as the sum of a geometric series.
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