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Turn a Solved DE into a Taylor Series

  1. May 31, 2008 #1
    Many of you have probably used the book Differential Equations by Lomen & Lovelock.
    For my class I'm working on Example 2, Page 153.

    You don't need to see the book, though, to help me out. It's a four-part problem and I'm on the last step not knowing where to take it.

    In Part B, we created a DE:

    dP/dt = (397/4169) P + (6580236/4169)

    and solved it:

    P(t) = (397/8338) t^2 + (6580236/4169) t + (1225953/4169)

    Now in Part D they ask:
    "Expand the explicit solution you obtained in Part B in a Taylor Series about the origin."

    I'm not so sure I've ever done anything like this, so I'm not clear on what steps I should take. I believe my teacher said I should find a similar Taylor Series and manipulate it to fit my needs but that still doesn't get me very far.

    Absolutely any insight is appreciated, hope you can help me!

    Thanks,
    Eric
     
  2. jcsd
  3. May 31, 2008 #2

    HallsofIvy

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    If your differential equation really is
    dP/dt = (397/4169) P + (6580236/4169)
    then dP/dt = (397/4169) P + (6580236/4169)
    is NOT the solution. Your P(t) satisfies dP/dt = (397/4169) t + (6580236/4169).

    The solution to dP/dt = (397/4169) P + (6580236/4169)
    will involve an exponential function.
     
  4. May 31, 2008 #3
    I think I'm going to have to look at it in a different light. I could've sworn I put it correctly into maple, that's what it threw back at me.

    Now that I try again, I get P(t) = -315680/19+exp((19/200)*t)*_C1, which I have checked by hand. I'm not entirely sure it's right though, I solved for C using a point from my original data table and got funny results when trying to use it.

    I'll try again tomorrow. But now that I have an e in it, I believe I'll just be modifying the taylor series for e.
     
  5. Jun 1, 2008 #4
    Okay, I've figured out the problem. The one I posted was from a different part of the question.

    I believe my correct equation is
    [tex]P(t)=e^{.095t}16543 - 16571[/tex]

    Now I know that E is
    [tex]e^{x} = \Sigma\frac{x^{n}}{n!}[/tex]

    So, correct me if I'm wrong, but do I manipulate the series as so to match my equation?

    [tex]\Sigma[\frac{(0.095t)^n}{n!}16543-16571][/tex]

    To be proper, all the summations here are from n=0 to [tex]\infty[/tex]

    In the question they ask me to relate it to the solution I got in part A, which is similar to what I originally posted:
    [tex]P(t) = a t^2 + b t + c[/tex]

    I don't think how I have it set up results in an answer in that form. Did I turn it into a taylor series properly?
     
    Last edited: Jun 1, 2008
  6. Jun 2, 2008 #5

    HallsofIvy

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    No, the "-16571" is NOT part of the exponential and is NOT summed an infinite number of times!

    And, again, that quadratic is WRONG.
     
  7. Jun 2, 2008 #6
    Well stay with me here...

    I was unsure and that's what I was asking about, now as I understand it my new equation should be:
    [tex]P(t) = (16543)\Sigma[\frac{(0.095t)^n}{n!}]-16571[/tex]

    True, the [tex]P(t) = a t^2 + b t + c[/tex] is incorrect for the original DE I gave you, but it is correct for a different part of the problem. This step is relating the quadratic to our new equation which we're making with summations.


    I think I did this right, to relate them, I calculated out the summation to three terms and plugged them in to get:
    [tex]P(t) = 74.7t^2+1571.6t-28[/tex]

    which I can relate to my quadratic using similarities.
     
  8. Jun 2, 2008 #7

    HallsofIvy

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    Good! This is now correct.

    Okay, I didn't do the calculations myself but assuming the arithmetic is correct, this is the "2nd order Taylor Polynomial" for your function.
     
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