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Math Help - Final Exams: Taylor Polynomail Help!!!

  1. #1
    zee
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    Unhappy Final Exams: Taylor Polynomail Help!!!

    Hello!! Boy am I glad that there is such a site!!
    I have my final exams approaching in 2 days and im studing
    Distance-Ed... so im in dire need of some help!
    This is my question!

    Determine the first three nonzero terms in the Taylor polynomial
    approximations for the given initial value problem:
    q) x" + t(x) = 0; x(0)=1, x'(0)=0

    I have the final answer, but I want to see the full solution.
    The answer is: 1 - (1/6)t^3 + (1/180)t^6 + ...

    Please HELPP!!!!!!!!!!
    zee
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  2. #2
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    1. Check your typing
    2. What is the Taylor approx (the formula)?
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  3. #3
    zee
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    desperate!

    Determine the first three nonzero terms in the Taylor polynomial
    approximations for the given initial value problem:
    x" + t(x) = 0; x(0)=1, x'(0)=0

    The question is typed accurately...
    I dont have the formula as I have lost my formula sheet...
    thus im in a lost position... please help somehow..
    zee
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  4. #4
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    Is x a function? If yes, what does x depend on?
    Is t a function? If yes, what does t depend on?
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  5. #5
    zee
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    Taylor Polynomial

    well x is definitely a function, but to what?
    The question is very unclear, thus im quite confused...
    t is not a function.. well atleast it doesnt look like it!
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  6. #6
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    My guess is that your ODE looks like this:

    \frac{d^2x}{dt^2} + t x(t) = 0
    x(0) = 1
    \frac{dx}{dt}(0) = 0

    Taylor approximation is:
    The nth term of Taylor expansion about t=a is \frac{1}{(n-1)!}\frac{d^{(n-1)}x}{dt^{(n-1)}}(a) (t-a)^{(n-1)}
    Last edited by paultwang; May 21st 2005 at 07:28 AM.
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  7. #7
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    If about t=0, then:

    The first term is x(0), which is 1.

    The second term is 0 because x'(0) = 0.

    The third term is \frac{x''(a)}{2!} (t-a)^2 which is also 0 because \frac{d^2x}{dt^2} = -t x(t) (just plug t=0 in).

    The 4th term is \frac{x'''(a)}{3!} (t-a)^3 which contains the derivative of the third term:
    \frac{d^3x}{dt^3} = -t x'(t) -x(t) (Hint: Use chain rule) We already know x'(0) = 0 and x(0) = 1 so the 4th term is -\frac{1}{6}t^3

    The 5th and 6th terms are both 0.
    \frac{d^4x}{dt^4} = -2x'(t)-t x''(t) (We established that x'(0) = 0, x''(0) = 0.)
    \frac{d^5x}{dt^5} = -3x''(t)-t x'''(t) (x''(0) = 0, 0*(stuff) = 0)

    The 7th term is not zero.

    \frac{d^6x}{dt^6} = -4x'''(t)-t x''''(t)
    Plug t=0, you get -4x'''(0)
    Using information from the 4th term:
    \frac{d^3x}{dt^3} = -t x'(t) -x(t)
    \frac{d^3x}{dt^3}(0) = -x(0) = -1
    -4x'''(0) = 4

    The 7th term is \frac{4}{6!}t^6 = \frac{1}{180}t^6


    That's all.

    In case you are wondering, the exact solution is:

    x(t) = \frac{1}{2} \Gamma(\frac{2}{3}) 3^{(1/6)} AiryBi(-t)+\frac{1}{2} \Gamma(\frac{2}{3}) 3^{(2/3)} AiryAi(-t)
    Last edited by paultwang; May 21st 2005 at 07:50 AM.
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  8. #8
    zee
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    thankyoU!

    You were right on the mark!
    thanks a bunch!!
    zee
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