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Solving MacLaurin Integrals

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume that e^x equals its Maclaurin series for all x.
    Use the Maclaurin series for e^(-4 x^4) to evaluate the integral


    Your answer will be an infinite series. Use the first two terms to estimate its value.

    2. Relevant equations

    Inline37.gif

    3. The attempt at a solution
    I've tried using the e^x series above to solve the given series. From what I understand, I am suppose to manipulate the e^x series so that it becomes e^(-4x^4). I am new to these forums, and I don't know another way to show what I'm doing, so I'll post an image of what I have so far (which is incorrect):


    Can anyone point me in the right direction or tell me what my mistakes are?
    Thanks a lot and sorry if I broke any forum etiquette rules here! It's my first time here. =)
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Feb 23, 2009 #2

    Dick

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    No rules broken. Welcome to the forums! You want (-4*x^4)^n. Not -4*(x^4)^n. You get powers of the -4 as well.
     
  4. Feb 23, 2009 #3
    Hey Dick sorry for the late reply, I had to run out real quick, but I tried your suggestion of (-4x^4)^n. However, I am still getting an incorrect answer. I brought out -1/n! and then integrated (4x^4)^n. Which should give me 4^nx(x^4)^n/4n+1?
     
  5. Feb 23, 2009 #4

    Dick

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    Then you get powers of the -1 as well. (-1)^n. Your power on the x isn't right either in the integral. You had it right before.
     
  6. Feb 23, 2009 #5
    I'm afraid I don't understand. Can you elaborate on this please?
     
  7. Feb 23, 2009 #6

    Dick

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    (-4)^1=-4. (-4)^2=16. (-4)^3=-64. (-4)^4=256. They aren't all the same sign. You can't factor the (-1) out.
     
  8. Feb 23, 2009 #7
    so you're saying I should get (-4x)^(4n+1)/(4n+1)n! ... ?
     
  9. Feb 23, 2009 #8

    Dick

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    No. (-4x^4)^n=(-4)^n*(x^4)^n=(-4)^n*x^(4n). Now integrate the power.
     
  10. Feb 23, 2009 #9

    By integrate the powers you mean -4^n*x^(4n) = -4^(n+1)*x^(4n+1)/(4n+1)n! ...??

    Sorry, English is not my native tongue.
     
  11. Feb 23, 2009 #10

    Dick

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    By 'power' I meant the x^(4n). Sorry. (-4)^n is a constant in each term. It doesn't depend on x. You only apply the power law to x^(4n).
     
  12. Feb 23, 2009 #11
    Ah, I see it more clearly now, but once I plug in x, which is .1, I should get (-4)^n*(0.1)^(4n+1)/(4n+1)n! I then plugged in 0 and 1 for n to test the first two terms of the series, but my answer came out wrong again. =( Please help.
     
  13. Feb 23, 2009 #12

    Dick

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    Can you show us the first two terms of the series you used and the numbers you got? Remember the first term is n=0. That term is pretty simple.
     
  14. Feb 23, 2009 #13
    Sure, here is what I've done below:

    http://img261.imageshack.us/img261/9651/mathhelp.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  15. Feb 23, 2009 #14

    Dick

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    I get 0.099992. I didn't round off. Could that be the problem? Otherwise, it looks fine.
     
  16. Feb 23, 2009 #15
    I left off the two at the end when I entered my solution, and webwork counted that as incorrect. I just tried your solution with the extra 2 and now it is correct. Silly webwork. =) Thanks for you patience throughout helping me find this solution. Hopefully we can do business again in the future!
     
  17. Feb 23, 2009 #16

    Dick

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    You're welcome. You know where to find me. BTW your English is excellent. You even spell words correctly. Native speakers don't do that any more.
     
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