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Math Help - Convergence of ln(1+x)

  1. #1
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    Convergence of ln(1+x)

    Hi,

    I was trying to prove that the Taylor polynomial for ln(1+x) converges to the function for |x|<1 from first principles, and was having some difficulty at the end.

    The version of Taylor's Theorem I'm using is the following. Given some function which is n times differentiable in some interval containing a and x+a, then

    f(x+a) = f(a) + \sum_{k=1}^{n-1}f^{(k)}(a)\frac{x^k}{k!} + f^{(n)}(a + \theta x)\frac{x^n}{n!}

    for some \theta \in (0,1)

    So, after the usual juggling which you've all seen before, I end up with the correct series, and

    \left|R_n(f) \right|= \left| \frac{x^n f^{(n)}(1+\theta x)}{n!} \right| = \left| \frac{x^n}{n(1+\theta x)^n} \right|

    I would like to show that this goes to zero only for values of x in the range -1<x<1, but I can't see how to do it. Of course this leads me to think I may have made a mistake somewhere along the line, my understanding of Taylor's theorem is not absolute yet.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by Stonehambey View Post
    Hi,

    I was trying to prove that the Taylor polynomial for ln(1+x) converges to the function for |x|<1 from first principles, and was having some difficulty at the end.

    The version of Taylor's Theorem I'm using is the following. Given some function which is n times differentiable in some interval containing a and x+a, then

    f(x+a) = f(a) + \sum_{k=1}^{n-1}f^{(k)}(a)\frac{x^k}{k!} + f^{(n)}(a + \theta x)\frac{x^n}{n!}

    for some \theta \in (0,1)

    So, after the usual juggling which you've all seen before, I end up with the correct series, and

    \left|R_n(f) \right|= \left| \frac{x^n f^{(n)}(1+\theta x)}{n!} \right| = \left| \frac{x^n}{n(1+\theta x)^n} \right|

    I would like to show that this goes to zero only for values of x in the range -1<x<1, but I can't see how to do it. Of course this leads me to think I may have made a mistake somewhere along the line, my understanding of Taylor's theorem is not absolute yet.
    You are using the formula for the remainder R_n(f) known as Lagrange's form of the remainder. The formula \left|R_n(f) \right| = \left| \frac{x^n}{n(1+\theta x)^n} \right| works nicely to show that R_n(f)\to0 in the case 0<x<1. In that case, 1+\theta x >1, so the denominator of \frac{x^n}{n(1+\theta x)^n} goes to infinity and the numerator goes to 0.

    However, in the case 1<x<0, Lagrange's form of remainder can't be used to show that R_n(f)\to0 (because there is no way to be sure that the numerator of that fraction is smaller than the denominator). In that case, you can only prove that R_n(f)\to0 by using a different formula for the remainder, such as Cauchy's form of the remainder.
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  3. #3
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    Hi,

    If your point is only to proove that ln(1+x)=x-x/2+x^3/3-..., then, you can also write : \sum_{k=1}^n - \frac{(-x)^n}{n}= \int_0^x \sum_{k=1}^n (-t)^{n-1}dt= \int_0^x \frac{1-(-t)^n}{1+t}dt= \int_0^x \frac{1}{1+t}dt - \int_0^x \frac{(-t)^n}{1+t}dt=ln(1+x)-R_n(x)
    with
    |R_n(x)|=|\int_0^x \frac{(-t)^n}{1+t}dt| \leq x^n.\int_0^x \frac{1}{1+t}dt (and with |x|<1, limit of x^n with n->+infinity = 0)

    By this way, we don't even need the formula of taylor
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  4. #4
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    Quote Originally Posted by Opalg View Post
    You are using the formula for the remainder R_n(f) known as Lagrange's form of the remainder. The formula \left|R_n(f) \right| = \left| \frac{x^n}{n(1+\theta x)^n} \right| works nicely to show that R_n(f)\to0 in the case 0<x<1. In that case, 1+\theta x >1, so the denominator of \frac{x^n}{n(1+\theta x)^n} goes to infinity and the numerator goes to 0.
    Thanks, can I use Lagranges form of the remainder to show that R_n(f) diverges outside of (0,1)? Or would I need another approach also?

    The integral method is nice, and one I would usually use for this particular example since it falls out quite nicely. That's why I was wondering if I could prove it just using Taylor's theorem
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  5. #5
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    Quote Originally Posted by Stonehambey View Post
    Thanks, can I use Lagranges form of the remainder to show that R_n(f) diverges outside of (0,1)? Or would I need another approach also?

    Hi, yes you can use this to prooves it diverges outside of ]-1,1]
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  6. #6
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    Could you give me a hint as to how to go about that? Just something to get me started, as I'd like to have a go myself
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  7. #7
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    Have you tried applying the ratio test to your series?
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    Thanks, I had not. That just leaves me to check the end points, which is not too difficult.

    I initially thought it would always be simple to show that the remainder term of Taylor's series converges or diverges, but it looks like that is not always the case.

    Thanks everyone
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  9. #9
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    If you'll read the question, you'll notice that the question only asks you to prove convergence for |x| < 1, so you don't even need to check the endpoints.
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