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Math Help - power series

  1. #1
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    power series

    Let \Sigma_{k=0} ^\infty a_k x^k and \Sigma _{k=0} ^\infty b_k x^k be real power series which converge for |x|<1. If \Sigma_{k=0} ^\infty a_k x^k= \Sigma _{k=0} ^\infty b_k x^k for x=1/2,1/3,1/4,..., then prove that a_k=b_k for all k
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  2. #2
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    Quote Originally Posted by Kat-M View Post
    Let \Sigma_{k=0} ^\infty a_k x^k and \Sigma _{k=0} ^\infty b_k x^k be real power series which converge for |x|<1. If \Sigma_{k=0} ^\infty a_k x^k= \Sigma _{k=0} ^\infty b_k x^k for x=1/2,1/3,1/4,..., then prove that a_k=b_k for all k
    Indirect proof: Suppose there existed a first summation index k_0 for which a_{k_0}\neq b_{k_0}, then you have for all x_n := \frac{1}{n}, that

    \sum_{k=0} ^\infty a_k x_n^k-\sum_{k=0} ^\infty b_k x_n^k = x_n^{k_0}\cdot\sum_{k=0}^\infty \left(a_{k_0+k}-b_{k_0+k}\right)x_n^k=0.
    But because the function f(x) := \sum_{k=0}^\infty \left(a_{k_0+k}-b_{k_0+k}\right)x^k is continuous at x=0, we find that f(0)=\lim_{n\to\infty}f(x_n)=0, from which it follows that a_{k_0}-b_{k_0}=0, thus contradicting our assumption a_{k_0}\neq b_{k_0}.
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  3. #3
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    i am not getting how u get the equation. can u explain it please?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Kat-M View Post
    i am not getting how u get the equation. can u explain it please?
    How am I to figure out what "equation" you are talking about. Why can't you, please, be more specific?

    From the givens, I conclude that \sum_{k=0}^\infty a_k x^k-\sum_{k=0}^\infty b_k x^k{\color{red}=0}, for all x=x_n := \frac{1}{n}, n\in \mathbb{N}, n \geq 2.

    Next, if we assume (contrary to what is the case), that a_k=b_k is not true for all indices k=0,1,\ldots, then there would have to exist a specific smallest index k_0, such that a_{k_0}\neq b_{k_0}. Since, therefore, for all indices k smaller than k_0 we have a_k=b_k, and thus a_k-b_k=0, it follows that

    \sum_{k=0}^\infty a_k x^k-\sum_{k=0}^\infty b_k x^k=\sum_{k=k_0}^\infty (a_k-b_k)x^k=x^{k_0}\cdot \sum_{k=0}^\infty(a_{k_0+k}-b_{k_0+k})x^k{\color{red}=0}, for all x=x_n := 1/n.
    From x^{k_0}=(1/n)^{k_0}\neq 0 it follows for the continuous[!] function f(x) {\color{blue}:=} \sum_{k=0}(a_{k_0+k}-b_{k_0+k})x^k that f(0)=\lim_{x\rightarrow 0}f(x)=\lim_{n\rightarrow\infty} f(x_n){\color{red}=}\lim_{n\to\infty}0{\color{gree  n}=0}.
    But this means that f(0){\color{blue}=}a_{k_0}-b_{k_0}{\color{green}=0}, and therefore a_{k_0}=b_{k_0}, contrary to our assumption that k_0 is the smallest summation index for which that is not the case. Thus no such smallest index k_0 can exist, and this means that for all indices k, we have a_k=b_k, what was to be shown.
    Last edited by Failure; April 18th 2010 at 11:34 AM.
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  5. #5
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    thank u
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