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Maclaurin Serie

  1. Feb 20, 2008 #1
    Let us take the first three non-zero terms of Maclaurin's expansion of
    f(x)= Asen(ax) + Bcos(ax).
    Determine for which values of a the equation f ' '(x) + f(x) = 0.
  2. jcsd
  3. Feb 21, 2008 #2

    Gib Z

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    Welcome to Physicsforums Oswald.

    In this case it seems better to leave f(x) in exact form. Can you find f''(x) and express it in terms of f(x) ?
  4. Feb 21, 2008 #3


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    That looks like a homework problem and you haven't said anything about what YOU have already done. Do you know how to find the MacLaurin's series for a function?
  5. Feb 21, 2008 #4
    Maclaurin Serie solution

    I have done this, but it doesn't work...
    A sin(ax) = A[ ∑ (-1)^k . (ax)^2k+1 ] / (2k+1)! to find Maclaurin series of A sin(ax)
    B cos(ax) = B[ ∑ (-1)^k . (ax)^2k ] / (2k)! is Maclaurin Series of B cos(ax)


    A sin(ax) = Aax - Aa³x³/6 + Aa^5x^5/120

    B cos(ax) = 1 - Ba²x²/2 + Ba^4x^4/24

    f''(x) + f(x) = 0

    what is f''(x)? I think its:

    f''(x) = 0 + 1
    f(x) = A sen(ax) + B cos(ax)


    0 + 1 + Asen(ax) + B cos(ax) = 0

    Asen(ax) + Bcos(ax) = -1

    the answer is a= 1 and a= -1, but i dont know how. Is the value of A, B and x important?
    Last edited: Feb 21, 2008
  6. Feb 21, 2008 #5
    Hi,i found the solution in yahoo...

    Take your function and differentiate it twice. observe that

    f '' (x) = - a^2 f(x)

    Substitute this into the equation. One possibility is that
    A=B=0. Otherwise, cancel f(x) and you have a simple algebraic equation for a to solve.
    have done:
    f²(x)= -a²f(x)
    a= 1 or -1!
    i don't know why the teacher ask for Maclaurin series...
  7. Feb 22, 2008 #6

    Gib Z

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    I don't know why you didn't just take my hint, which got at exactly the same thing, rather than go to Yahoo.
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