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More Taylor series stuff, HELP!

  1. May 28, 2008 #1
    1. The problem statement, all variables and given/known data


    Let T_(4)(x): be the Taylor polynomial of degree 4 of the function ` f(x) = ln(1+x) ` at `a = 0 `.


    Suppose you approximate ` f(x) ` by ` T_(4)(x) `, find all positive values of x for which this approximation is within 0.001 of the right answer. (Hint: use the alternating series approximation.)

    3. The attempt at a solution

    I gotta be honest, I'm not entirely sure where to even start with this...I've gotten my sealegs when it comes to these series and how to find them and stuff, and I know how to find the error in an alternating series, but I don't know how to do it with an ln function.

    The answer is 0.346572421578, fyi.
     
  2. jcsd
  3. May 28, 2008 #2

    Dick

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    If you know it, tell me, how DO you find the error in an alternating series?
     
  4. May 28, 2008 #3
    Well, if you want to find the maximum possible error in an alternating series, you'd just take the next term. For example if you want to find the error in a series after 10 terms, the max possible errror is the 11th term.
     
  5. May 28, 2008 #4

    Dick

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    Right. Good! So what's the appropriate term in that taylor series?
     
  6. May 28, 2008 #5
    Is it 4?
     
  7. May 28, 2008 #6

    Dick

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    I think the 4th degree taylor polynomial has a last term with x^4 in it. What's the term after that? And not just x^5, I want the coefficient too.
     
  8. May 28, 2008 #7
    If I'm on the same page as you...

    (f'''''(x) * (x - a))/x!

    Is that what you're talking about?
     
  9. May 28, 2008 #8

    Dick

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    If you mean (ln(1+x))'''''*x^5/5!, yes. You are expanding around a=0. And f(x)=ln(1+x).
     
  10. May 28, 2008 #9
    So the 5th derivative of ln(1 + x) would be 1/((1 + x)^5), correct?
     
  11. May 28, 2008 #10

    Dick

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    No, the coefficient is wrong. When you differentiate 1/(1+x)^2 don't you get -2/(1+x)^3? What happened to factors like the 2?
     
  12. May 28, 2008 #11
    So, lemme see...using that as a template, it would be...

    f(x) = ln(1 + x)

    f'(x) = 1/(1 + x)^2

    f''(x) = -2/(1 + x)^3

    f'''(x) = 6/(1 + x)^4

    f''''(x) = -24/(1 + x)^5

    f'''''(x) = 120/(1 + x)^6

    Does that work?
     
  13. May 28, 2008 #12

    Dick

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    Doh. f'(x)=1/(1+x). Shuffle your results around a bit. You are on the right track though.
     
  14. May 28, 2008 #13
    f(x) = ln(1 + x)

    f'(x) = 1/(1 + x)

    f''(x) = 1/(1 + x)^2

    f'''(x) = -2/(1 + x)^3

    f''''(x) = 6/(1 + x)^4

    f'''''(x) = -24/(1 + x)^5

    How bout that?
     
  15. May 28, 2008 #14

    Dick

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    The sign is wrong. That went wrong in the f'' step. But that's ok, we don't need that. So now, ta da, the term you want if f'''''(a)*x^5/5!, where a=0. Do you agree?
     
  16. May 29, 2008 #15
    So basically, I want to find where

    (24/(1 + x)^2) * x^5/5!

    is less than 0.001?
     
  17. May 29, 2008 #16

    Dick

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    No, f'''''(a), not f'''''(x). a=0. Look at the definition of taylor series. The variable in the derivative is not x. It's the point you are expanding around, a. It's a CONSTANT. It's easier than you are making it out to be.
     
    Last edited: May 29, 2008
  18. May 29, 2008 #17
    If a is 0 it would just simplify to 0...shouldn't a = 4 since it's the 4th term?
     
  19. May 29, 2008 #18

    Dick

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    What book are you getting your taylor series information out of? NO! The x part is x^n/n!. With n=5. The derivative part is f''''''(a) where a=0. That's what your problem statement says!? For example, f'(x)=1/(1+x). So f'(a) where a=0 is 1. It doesn't just simplify to 0!!!!?????
     
  20. May 29, 2008 #19
    I'm sorry, you're starting to lose me...sorry if I'm being a bother, but don't you insert a for x?
     
  21. May 29, 2008 #20

    Dick

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    The taylor series for f expanded around x=a is
    f(a)+f'(a)*(x-a)+f''(a)*(x-a)^2/2!+f'''(a)*(x-a)^3/3!+....
    Does that ring a bell? x and a play two completely different roles. x is the expansion parameter and a is the point you are expanding around. a is a CONSTANT, not a variable.
     
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