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Homework Help: Maclaurin Series homework help

  1. Dec 6, 2007 #1
    Hey all. We never covered the Maclaurin series (covered Taylor) but these questions are on my review for my final.

    1. The problem statement, all variables and given/known data
    Compute the 9th derivative of:
    arctan((x^3)/2) at x=0

    2. Relevant equations
    Use the MacLaurin series for f(x)

    3. The attempt at a solution
    atan(x) =x-x^3/3 + x^5/5 +........

    I have no idea where to even go from here.
    I have a similar yet more difficult problem as well that I don't even no where to begin:
    Compute the 6th derivative of:
    (cos(6x^2)-1)/(x^2) at x=0

    Sorry my professor doesn't have office hours till next week and I'm just really frustrated with these!!
  2. jcsd
  3. Dec 6, 2007 #2
    for the first, make a substitution y=x^3/2 and try an induction..
    you need to express the series using the summation notation..
  4. Dec 6, 2007 #3


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    As I describe this, keep the formula for a Maclaurin series handy. You wrote

    arctan(x) =x-x^3/3 + x^5/5 +........

    If you compare this with the formula for Maclaurin series, you'll see that this tells you that, at x = 0,

    first derivative of arctan(x) = 1 ,
    second derivative = 0 ,
    third derivative = -1/3 ,
    fourth derivative = 0 ,
    fifth derivative = +1/5 , etc.

    What you have is f(x) = arctan[ (x^3)/2 ] = arctan [ u(x) ] , which is a composite function. So we use the Chain Rule to find

    f'(x) = [ d (arctan u) / du ] · u' .

    Okay, we have a way to go yet -- this is a product, so the next derivative is

    f''(x) = [ d^2 (arctan u) / du^2 ] · u'
    + [ d (arctan u) / du ] · u'' .

    [EDIT: this should be f''(x) = [ d^2 (arctan u) / du^2 ] · (u')^2
    + [ d (arctan u) / du ] · u'' .
    What follows will be more complicated than I believed at the time I posted this. My apologies.]

    Now, at this point, most people would start screaming, "And we have to go to the ninth derivative!?" . But we hit bottom with terms because

    u = (x^3)/2 , u' = 3(x^2)/2 , u'' = 3x , u''' = 3 , u^(4) and all higher derivatives = 0.

    We also start to notice something about the derivatives of f(u) -- the next derivative is

    f'''(x) = [ d^3 (arctan u) / du^3 ] · u'
    + 2 · [ d^2 (arctan u) / du^2 ] · u''
    + [ d (arctan u) / du ] · u'''

    The terms start to look like the sort of thing we get from calculating the binomial power
    (x+y)^n . The coefficients are the terms in Pascal's Triangle or the combinatorial values nCr and the total of the level of derivatives we take is one higher than the derivative of f(x) we are working on.

    This suggests that the ninth derivative of f(x) looks like

    f^(9) (x) = [ d^9 (arctan u) / du^9 ] · u'
    + 9 · [ d^8 (arctan u) / du^8 ] · u''
    + (9·8/2) · [ d^7 (arctan u) / du^7 ] · u''' ;

    we don't need any more terms because the derivatives of u beyond that are all zero.

    Now we can go back to the Maclaurin series. The seventh derivative at x = 0 is -1/7 , the eighth derivative is 0, and the ninth derivative is +1/9 . But u(0) = (0^3)/2 = 0 , so these will also be the values of the derivatives of arctan(u). This already reduces our expression to

    f^(9) (0) = [ d^9 (arctan u) / du^9 ] · u'
    + 9 · [ d^8 (arctan u) / du^8 ] · 0
    + (9·8/2) · [ d^7 (arctan u) / du^7 ] · u'''

    = (1/9)·u' + (36)·(-1/7)·u'''

    At x = 0, u'(0) = 3(0^2)/2 = 0 , and we already know that u''' = 3 . Thus,

    f^(9) (0) = (1/9)·0 + (36)·(-1/7)·3 = -108/7 .

    After all this terribly rigorous agony, this suggests something we (to say, I) might have noticed from the start: at x = 0 , the only derivative of u which isn't zero is u''' = 3 . So we could just have looked at the pattern of derivatives for f^(n) (0) and, after writing the result for f^(9) = 0 , noticed that the only term that matters is

    (9·8/2) · [ d^7 (arctan u) / du^7 ] · u''' ; absolutely everything else disappears.

    [EDIT: The basic idea of what happens at x = 0 is still correct, but the coefficient is not correctly found as described here.]


    I believe something similar happens for your other problem

    (cos(6x^2)-1)/(x^2) at x=0

    I'm reading this as

    [ cos(6·x^2) - 1 ] / (x^2)

    Look at the Maclaurin series for cos x; divide it by x^2. The 1/(x^2) term in the series will get cancelled out by the -1/(x^2) term in your function. The coefficients in your series now start with the second derivative of cos x , so you can read off what derivative values you need.

    This time, u = 6·(x^2): how high do the derivatives go? Which aren't zero at x=0?

    Again here, u(0) = 0, so the derivatives of cosine at u = 0 are the same as the derivatives at x = 0.

    Use the method we described to find the non-zero terms for f^(6) (0) and calculate the result.
    Last edited: Dec 7, 2007
  5. Dec 7, 2007 #4


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    Much the same but, I think, in simpler terms.

    You know that atan(x) =x-x^3/3 + x^5/5 +........

    But you want atan(x^3/2). Okay, replace x in the formula above by x^3/2:

    atan(x^3/2)= (x^3/2)- (x^3)^3/(3(2^3)+ ....
    You don't need to go any farther because you already have x^9- which will be the term in the MacLauren series involving the 9th deritivate: -X^9/(24).

    You know, of course, the coefficient of x^9 in the MacLauren series for f(x) is f(9)(0)/9!. Set that equal to -1/24. What is the 9th derivative of atan(x^3/2)?

    Much the same thing. Write down the first several terms of the MacLauren for cos(x). Replace x by 6x2. Subtract 1 which should cancel the "constant" term. Divide each term by x2. You now have the MacLauren series for (cos(6x2)- 1)/x2. The coefficient of x6 will be the 6th derivative divided by 6!.
    Last edited by a moderator: Dec 7, 2007
  6. Dec 7, 2007 #5


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    Not just much simpler, but I realized after I got home last night that what I wrote isn't quite right. Repeated applications of the Chain Rule on a composite function lead to expressions somewhat more complicated than what I have posted. The terms are messier [there should be products of powers of u', u'', etc.], but the basic idea is all right: since only one of the derivatives of u is non-zero at x = 0, all terms but one go to zero.

    It is easier to make the substitution u(x) into the series for arctan u and notice that there are terms with powers of x lower than 9, which will differentiate away to zero, and terms with powers of x higher than 9, which will become zero at x = 0. You only need then to focus on the single x^9 term and work out its coefficient.

    Similarly for the other problem, only the x^6 term will remain, so that's what you differentiate.
    Last edited: Dec 7, 2007
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