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Find the MacLaurin polynomial of degree 5 for F(x).

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Ru8dm.jpg

    2. Relevant equations

    NumberedEquation1.gif

    3. The attempt at a solution

    6Uscs.jpg

    I keep getting the answer 1- 96/4!*x^4. Can't find where I'm going wrong.
     
  2. jcsd
  3. Dec 11, 2011 #2

    Dick

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    It's really hard to tell what you are doing from the blurry photograph. Can you explain? What are F(0), F'(0), F''(0) etc?
     
  4. Dec 11, 2011 #3
    Sorry for the blurriness. This should be clearer.

    As you can see I get the answer in the box. I can't figure out what I'm doing wrong here.

    44444444444444444444444444440001.jpg
     
  5. Dec 11, 2011 #4

    Dick

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    Much clearer, thanks! Now I see where you are going wrong. f(0)=0, isn't it? It's f'(0) that is 1. And if you think about it, you didn't have to compute a lot of those terms involving high powers of x. But that's in retrospect.
     
  6. Dec 11, 2011 #5
    I'm still not getting the correct answer.
    I think f(0)=1. 0^4=0 0 times -4 equals 0 and then e to the power of 0 equals 1.
     
  7. Dec 11, 2011 #6

    Dick

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    Nooo. f(0) is the integral from 0 to 0 of e^(-4t^4). That's zero! f'(0) is exp(-4t^2) evaluated at t=0. Your derivative count is off by one!
     
  8. Dec 11, 2011 #7
    So it should be x-(96/5!)*x^5 ?
     
  9. Dec 11, 2011 #8

    Dick

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    Yes, it should. But I don't like that '?' in the answer. You do see why, don't you?
     
  10. Dec 11, 2011 #9
    Honestly, it is the right answer but I don't see why. Is there something special with finding a maclaurin polynomial of a function that includes e?
     
  11. Dec 11, 2011 #10

    Dick

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    No, nothing special about e. You don't agree f(0)=0? Suppose the function in your integral was just 1. What are f(0) and f'(0)?
     
  12. Dec 11, 2011 #11

    f(0) would be 1 and f'(0) would be 0

    Looking at it again is it because of the integral would be from 0 to 0
     
  13. Dec 11, 2011 #12

    Dick

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    If you are integrating 1 from 0 to x, then f(x)=x. I can say pretty definitely that f(0)=0. f'(0)=1.
     
  14. Dec 11, 2011 #13
    Can it basically be a rule if you integrating from 0 to x, that the first term is 0 followed by the second term being x?
     
  15. Dec 11, 2011 #14

    Dick

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    You can't simplify it that much. Integrate 2 from 0 to x. The point of the example is just to show what you were doing wrong. f(0) is always zero, sure. f'(0) depends on the value of the integrand at 0. You just had your derivatives shifted by one. That's all.
     
  16. Dec 11, 2011 #15
    OK. I understand it now. Thanks for sticking with me until I understood. I have a final for college coming up so understanding this was very pivotal. Thanks, once again.
     
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