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Taylors Series/Remainder for ln(x) about x = 1

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine whether the Taylors Series for ln(x) converges to ln(x) around x=1, for x [1/2, 2].


    2. Relevant equations



    3. The attempt at a solution

    [tex]f(x) = ln(x)[/tex]

    [tex]f^{1}(x) = \frac{1}{x}[/tex]

    [tex]f^{2}(x) = \frac{-1}{x^{2}}[/tex]

    [tex]f^{3}(x) = \frac{2}{x^{3}}[/tex]

    [tex]f^{4}(x) = \frac{-6}{x^{4}}[/tex]

    [tex]\vdots[/tex]

    [tex]f^{n}(x) = \frac{(-1)^{n-1}(n-1)!}{x^{n}}[/tex]

    So,

    [tex]ln(x) = (x-1) - \frac{1}{2}(x-1)^{2} + \frac{1}{3}(x-1)^{3} - \cdots + \frac{(-1)^{n-1}}{n}(x-1)^{n} + R_{n}(1,x)[/tex]

    Where,

    [tex]R_{n}(1,x) = \frac{(-1)^{n}}{(n+1)} \cdot \left( \frac{(x-1)}{z_{n}} \right)^{n+1}[/tex]

    We can see that the second term (The one with [tex]z_{n}[/tex] in it) is the one that causes problems. It is trivial to show that the limit as n goes to infinity of the first term is 0. (We proved it in class)

    So there are 3 cases that we should consider:

    Case 1: x=1, everything will converge to 0.

    Case 2: [tex]\frac{1}{2} \leq x < 1[/tex]

    It follows then that,

    [tex] \frac{1}{2} \leq x < z_{n} <1[/tex]

    We can use, [tex]\frac{1}{2} \leq x < 1[/tex] to try to gain for information about the term that is causing problems with our remainder.

    [tex]\frac{1}{2} \leq x < 1 \Rightarrow \frac{1}{2} - 1 \leq x-1 < 1-1 \Rightarrow \frac{-1}{2} \leq x-1 < 0 \Rightarrow \frac{-1}{2z_{n}} \leq \frac{x-1}{z_{n}} < 0 [/tex]

    Looking back at,

    [tex] \frac{1}{2} \leq x < z_{n}[/tex]

    we can see that,

    [tex]z_{n} \geq \frac{1}{2}[/tex] or [tex]2z_{n} \geq 1[/tex]

    Working with this some more we can see that,

    [tex]\frac{1}{2z_{n}} \leq 1 \Rightarrow \frac{-1}{2z_{n}} \geq -1 [/tex]

    Therefore we can say that,

    [tex]-1 \leq \frac{-1}{2z_{n}} < \frac{x-1}{z_{n}} < 0[/tex]

    i.e. [tex]-1 < \frac{x-1}{z_{n}} < 0 \Rightarrow |\frac{x-1}{z_{n}}| < 1[/tex]

    This is what our professor wrote down in order to prove the following,

    [tex]|R_{n}(1,x)| = \left| \frac{(-1)^{n}}{n+1} \cdot \frac{(x-1)^{n+1}}{z_{n}} \right| < \frac{1}{n+1} \cdot (1)[/tex]

    Therefore,

    [tex]lim_{n \rightarrow \infty} |R_{n}(1,x)| = 0 \Rightarrow lim_{n \rightarrow \infty} R_{n}(1,x) = 0 [/tex].


    I'm confused about how he got rid of that unwanted term,

    [tex]\left(\frac{(x-1)}{z_{n}}\right)^{n+1}[/tex]

    by using the inequalities stated above. He seems to jump around alot and it's hard to follow.

    EDIT: Looking at it again, he first shows that,

    [tex]\frac{x-1}{z_{n}} < 0[/tex]

    Now if he can prove that this is also larger than -1 it will be a small negative number inbetween (0, -1) and no matter what power we raise it to, lets say n+1, it will still be really close to 0.

    Is this his attack?


    Can someone try to explain his reasoning another way or offer me some insight? If I tried to do this problem without looking at my notes I don't think I'd be able to reasonably conclude that,

    [tex]lim_{n \rightarrow \infty} R_{n}(1,x) = 0 [/tex]

    For Case 2: [tex]\frac{1}{2} \leq x < 1[/tex]

    Any advice/tips/ideas/insight would be greatly appreciated! (After I will attempt case 3)

    Thanks again!
     
    Last edited: Sep 22, 2010
  2. jcsd
  3. Sep 22, 2010 #2
    Case 3: [tex]1 < x \leq 2[/tex] seems to make some more sense so I'll attempt it in this post.

    It follows that,

    [tex]1 < z_{n} < x \leq 2[/tex]

    Again, we can use, [tex]1 < x \leq 2[/tex] to gain more information about our problematic term in [tex]R_{n}(1,x)[/tex].

    One can see that,

    [tex]0 < x-1 \leq 1[/tex]

    [tex]0 < \frac{x-1}{z_{n}} \leq \frac{1}{z_{n}}[/tex]

    From here we can see that,

    [tex]\frac{1}{z_{n}} < 1[/tex]

    So here's where I'm at,

    [tex]0 \leq |R_{n}(1,x)| < ?[/tex]

    We need to figure out an upper bound so we can apply squeeze theorem.

    We know that,

    [tex]\frac{x-1}{z_{n}}[/tex] is less than 1 but larger than 0.

    Hmm...

    I'm not sure how I can find that upper bound I've been looking for. If I think of anything I'll simply edit this post.

    Feel free to make any suggestions if you have any!
     
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