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Lagrange Remainder Theorem

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data
    At what points ##x## in the interval ##(-1,1]## can one use the Lagrange Remainder Theorem to verify the expansion

    ##ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k+1}{\frac{x^k}{k!}}##


    2. Relevant equations



    3. The attempt at a solution
    Now I know that ##ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k+1}{\frac{x^k}{k}}## when ##x\in(-1,1]##. If we let ##P_n## denote the nth Taylor polynomial for ##ln(1+x)## then ##f(x)-P_n(x)=\frac{(-1)^{n+1}}{(n+1)(1+c)^{n+1}}x^{n+1}## centered at ##0##.
    The only difference between these two forms is the ##k!## which i'm not sure how to deal with. Also if you could provide me with some hint on how to go about this problem that be great thanks.
     
    Last edited: May 3, 2014
  2. jcsd
  3. May 3, 2014 #2

    Zondrina

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    The function is equal to its Taylor expansion if and only if the remainder term ##|R_n(x)| → 0## as ##n → ∞##. Otherwise I'm not quite sure what you mean in particular by the "Lagrange Remainder theorem".
     
  4. May 3, 2014 #3
    The book wants be to use the Lagrange remainder theorem for this problem. But i'm not sure how to find for which x values satisfy ##ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k!}##. Basically I need to find the some interval that this equation is satisfied.
     
  5. May 3, 2014 #4

    Zondrina

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    Could you perhaps write the theorem out.

    Otherwise you can use any ##x## in the interval ##(-1, 1]## and the series will be valid.
     
  6. May 3, 2014 #5
    Lagrange Remainder Theorem- Let ##I## be an open interval containing the point ##x_0## and let ##n## be a nonegative integer. Suppose that the function ##f:I-->\mathbb{R}## has ##n+1## derivatives. Then for each point ##x## in ##I## there is a point ##c## strictly between ##x## and ##x_0## such that ##f(x)=\sum_{k=0}^{n} \frac{f^k}{k!} (x-x_0)^k+\frac{f^{n+1}}{(n+1)!}(x-x_0)^k##.

    Theorem 8.8- For each natural number ##n## and each number ##x>-1## there is a number ##c## strictly between ##0## and ##x## such that ##ln(1+x)=x-\frac{x^2}{2}+...+\frac{(-1)^{n+1}}{n} x^n+ \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}##
     
  7. May 3, 2014 #6

    Zondrina

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    I see. And ##f(x)## is equal to its expansion if and only if the remainder term goes to zero.

    Something to notice here... you don't actually know whether or not ##x < c < 0## or ##0 < x < c##. All you know is that ##x \in I##. So you know the series will converge in the interval ##(-1, 1]##. If you try it by plugging in the endpoint 1, you will get a convergent series for example.
     
  8. May 3, 2014 #7
    So the series will converge in the entire interval ##(-1,1]##?
     
  9. May 3, 2014 #8

    Zondrina

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    Yes. Although it would be good to show that it actually does.
     
  10. May 3, 2014 #9
    Oh ok. I did try doing that at first but when I compared it against the answer in the book I was off. They stated that the interval was supposed to be ##\frac{-1}{2}\leq x\leq 1## which I have no idea how they got that.
     
  11. May 3, 2014 #10

    Zondrina

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    Well, if they want a particular interval, then what's the closest point to zero from -1? That's one of the endpoints. The other endpoint is of course given. They just happened to use ##-1/2##.
     
  12. May 3, 2014 #11
    So they just chose some random point close to ##0##?
     
  13. May 3, 2014 #12

    Zondrina

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    What I meant to say was the convergence would be valid for an interval as specific as ##[-\frac{1}{1.0001}, 1]## if you wanted.

    ##-1/2## is a nice number though.
     
  14. May 3, 2014 #13
    Oh Ok i see thanks. I just was confused why they chose that -1/2.
     
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