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Taylor's Theorem

  1. Apr 9, 2006 #1
    I'm computing the minimum number of terms for a Taylor polynomial to approximate f(1.5) within .0001 where f(x) = ln(x + 1) using Taylor's theorem, but I'm having a little trouble getting there. I keep coming up with the absolute value of the (n+1)th derivative of ln(x + 1) as (n!)/[(x+1)^(n+1)] in which case the largest value for any derivative of ln(x + 1) from 0 to x would be n! but if I use this with Taylor's Theorem I get (n!)[(1.5)^(n+1)] / (n+1)! < .0001 but this is not true for any n. Any help would be appreciated.
     
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 10, 2006 #2

    HallsofIvy

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    Did you forget the 1/n! in Taylor's formula?
     
  4. Apr 10, 2006 #3

    benorin

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    Since the series is a alternating, why not use "the absolute value of the error is less than the absolute value of the first term omitted"?
     
  5. Apr 10, 2006 #4
    Well that's the problem benorin, I'm not looking for just the error, I already know what it should be. I'm looking for the term at which the error is less than .00001. In the case of the other post, even if I multiply by 1/n! I'm simply left with 1.5^(n+1) / (n+1)! which is less than .00001 only after 10 terms when it should be 9.
     
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