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Thread: Series Uniform Continuity proof

  1. #1
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    Series Uniform Continuity proof

    Let f(x)= sum [(3^(n) + cos(n))/n!]X^n

    1. Prove that for every x in [-10,10] the sum converges
    2. Show that for every E >0 there's an N independant of x in [-10,10] such that
    |f(x) - sum [(3^(n) + cos(n))/n!]X^n | < E
    3. Use 2 together with the fact that polynomials are contiuous everywhere to show that f is continuous in [-10,10]

    ratio test for part 1, not sure how to fully show it
    2 I'm not sure what to do here

    the sum in part two is from 0 to N
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by gutnedawg View Post
    Let f(x)= sum [(3^(n) + cos(n))/n!]X^n
    ratio test for part 1, not sure how to fully show it
    If the series is $\displaystyle \sum_{n=0}^{+\infty}u_n(x)$ then, express:

    $\displaystyle \left |{\dfrac{u_{n+1}(x)}{u_n(x)}}\right |=\left | \dfrac{3+\cos (n+1)/3^{n+1}}{1+\cos n/3^{n+1}}\cdot{\dfrac{1}{n+1}} \right|\left |{x}\right |\rightarrow{0}\;(n\rightarrow{+\infty})$

    This series converges on $\displaystyle \mathbb{R}$, as a consequence on $\displaystyle [-10,10]$

    2 I'm not sure what to do here
    the sum in part two is from 0 to N
    Find a bound $\displaystyle |u_n(x)|\leq a_n$ on $\displaystyle [-10,10]$, prove that $\displaystyle \sum_{n=0}^{+\infty}a_n$ converges and use Cauchy's criterion of convergence.

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    Fernando Revilla
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  3. #3
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    how would this work in part 2? I was under the impression that it would be just like a limit proof of sorts with the difference in the abs being sum N to infinity of * and then solve for N then using n>N show the proof.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by gutnedawg View Post
    how would this work in part 2? I was under the impression that it would be just like a limit proof of sorts with the difference in the abs being sum N to infinity of * and then solve for N then using n>N show the proof.
    On $\displaystyle [-10,10]$ we have:

    $\displaystyle \left |{u_n(x)}\right |\leq \ldots=\dfrac{3^n+1}{n!}10^n=\alpha_n$

    and the series $\displaystyle \sum_{n=0}^{+\infty}\alpha_n
    $ is convergent. So, $\displaystyle \sum_{n=0}^{+\infty}u_n(x)
    $ uniformly converges to $\displaystyle f(x)$ on $\displaystyle [-10,10]$ as a consequence of the Weierstrass criterion (which is deduced from Cauchy criterion).

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    Fernando Revilla
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  5. #5
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    I am not allowed to use anything to do with uniform convergence, this would not be difficult if I was allowed to use these theorems =[
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    No problem, as $\displaystyle \sum_{n=o}^{\infty}\alpha_n$ is convergent,

    $\displaystyle \forall \epsilon >0\;\exists n_0\in \mathbb{N}: n\geq m\geq n_o\Rightarrow\alpha_m+\alpha_{m+1}+\ldots+\alpha_ n<\epsilon$

    Then

    $\displaystyle n\geq m\geq n_0\Rightarrow|u_m(x)|+|u_{m+1}(x)|+\ldots+|u_n(x) |<\epsilon$ for all $\displaystyle x\in [-10,10]$

    A fortiori,

    $\displaystyle n\geq m\geq n_0\Rightarrow|u_m(x)+u_{m+1}(x)+\ldots+u_n(x)|<\e psilon$ for all $\displaystyle x\in [-10,10]$

    Could you continue?.

    Regards.

    Fernando Revilla
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