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Math Help - Taylor series convergence

  1. #1
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    Taylor series convergence

    I have a function f(x) = 1/\sqrt(1-x) defined for x in
    ]-1,1[ and centered at a = 0. The taylor polynomial for this would be T_{n}(x) = \sum_{k=0}^n \frac {f^{(k)}(0)x^k}{k!}.
    Now for every x \ \epsilon \ ]0,1[, there is \xi = \xi(x) \ \epsilon \ ]0,x[ such that the error would be R_{n}(x) = \frac {xf^{(n+1)}(\xi)(x-\xi)^n}{n!}.
    Using this, how do I prove that the Taylor polynomials T_{n} converge uniformly over [0,x]?
    By doing a ratio test I get \lim_{n\rightarrow\infty} |{\frac {R_{n+1}(x)}{R_{n}(x)}}| = \frac{2n+3}{2n+2}. How is this supposed to converge? Can somebody please explain this?
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  2. #2
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    Quote Originally Posted by chainrule View Post
    I have a function f(x) = 1/\sqrt(1-x) defined for x in
    ]-1,1[ and centered at a = 0. The taylor polynomial for this would be T_{n}(x) = \sum_{k=0}^n \frac {f^{(k)}(0)x^k}{k!}.
    Now for every x \ \epsilon \ ]0,1[, there is \xi = \xi(x) \ \epsilon \ ]0,x[ such that the error would be R_{n}(x) = \frac {xf^{(n+1)}(\xi)(x-\xi)^n}{n!}.
    Using this, how do I prove that the Taylor polynomials T_{n} converge uniformly over [0,x]?
    By doing a ratio test I get \lim_{n\rightarrow\infty} |{\frac {R_{n+1}(x)}{R_{n}(x)}}| = \frac{2n+3}{2n+2}. How is this supposed to converge? Can somebody please explain this?
    I am sure you learned this long ago- perhaps too long ago! Divide both numerator and denominator by n to get \frac{2+ \frac{3}{n}}{2+ \frac{2}{n}}. Now what is the limit of that as n goes to infinity?
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  3. #3
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    I am sure you learned this long ago- perhaps too long ago! Divide both numerator and denominator by n to get \frac{2+ \frac{3}{n}}{2+ \frac{2}{n}}. Now what is the limit of that as n goes to infinity?
    Yes, but this Limit would give me a value of 1 whereas convergence would mean that the limit has to tend to 0 as n tends to infinity. Also this series is supposed to converge uniformly over \xi \ \epsilon \ ]0,x[ for every x \ \epsilon \ ]0,1[. How is that possible?
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