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Taylor polynomial approximation (HELP ME)

  1. Nov 27, 2007 #1
    Ok, we are asked to determined the degree of the the taylor polynomial about c =1 that should be used to approximate ln (1.2) so the error is less than .001


    the book goes throught the steps and arrives at:

    |Rn(1.2)| = (.02)^(n+1)/(z^(n+1)*(n+1)


    but then, it states that

    (.02)^(n+1)/(n+1) < .001

    and there is an error pointing this expression to

    1000 < (n+1)(5^(n+1))

    --> I have no idea how the book arrived at this second expression, could anyone please explain?

    - Also, if anyone knows, why was the z term left out in the second step?


    Thank you.
     
  2. jcsd
  3. Nov 27, 2007 #2
    why was the z term dropped?

    does anyone know?
     
  4. Nov 28, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I just don't believe your book has what you wrote. For one thing, |Rn(1.2)| = (.02)^(n+1)/(z^(n+1)*(n+1) is missing a ")". For another, the remainder formula has z in the numerator not the denominator- You can't make the error smaller by choosing z larger!
     
  5. Nov 28, 2007 #4
    OK, in general how does this process work? I am having a little trouble following the steps...

    *Yeah, I forgot the last parenthesis
     
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