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Function sin(2x)^(-1/2)

  1. Aug 18, 2009 #1
    For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
    D2(f) + f = 3*D(a) where Da stands for the a'd derivative
    So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
    Thanks!
     
  2. jcsd
  3. Aug 19, 2009 #2
    Re: Differentiation

    Hello,

    What are you trying to do here? Are you trying to find the derivative of

    sin(2x)^(-0.5) ?

    What is a'd derivative?

    Thanks
    Matt
     
  4. Aug 19, 2009 #3

    tiny-tim

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    Hi JanClaesen! :smile:

    (have a pi: π and a square-root: √ :wink:)
    (we say the nth derivative, not the n'd derivative :wink:)

    Do you mean "what is the nth derivative of 1/√(sin2x)?"

    And what is f? :confused:
     
  5. Aug 19, 2009 #4
    Re: Differentiation

    Hello :smile:, I'm sorry if I was a little unclear:

    So there is a certain relation for the function sin(2x)^(-1/2): the second derivative of this function plus the function itself gives the (n'th derivative)*3, the question is to determine n.
     
  6. Aug 19, 2009 #5
    Re: Differentiation

    Rephrase:

    If [itex]y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}[/itex], solve for n. (Note [itex]y^{(n)} = \frac{d^n y}{{dx}^n}[/itex].)

    I have hammered out the first 10 derivatives of y but none seem to be equal to (y'' + y)/3. The exponent of the denominator factor is of the order 1/2 + n, while the numerator is a trigonometric polynomial in cosine of order n. I do not forsee this collapsing into something nice. This exercise seems fishy.

    N.B.: The expression on the left of the differential equation is

    [tex]\frac{3}{(sin(2x))^{5/2}}.[/tex]​

    --Elucidus
     
  7. Aug 19, 2009 #6

    tiny-tim

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    Hi JanClaesen! Hi Elucidus! :smile:

    I don't think there can be a solution …

    just try to solve 3y(n) - y'' - y = 0 the usual way …

    that gives you a characteristic equation, of which 1/√sin can't be an answer. :redface:
     
  8. Aug 19, 2009 #7
    Re: Differentiation

    I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg [Broken]
    It's exercice 17a, the answer should be 5, perhaps alpha is a power?
     
    Last edited by a moderator: May 4, 2017
  9. Aug 19, 2009 #8

    tiny-tim

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    D'oh!

    Yes of course α is a power … Question 17a says so :rolleyes:

    f'' + f = 3fα

    Why did you write D2(f) + f = 3*D(a) in your first post??

    ok, since f = 1/√(sin(2x)), what is f'' + f?
     
    Last edited by a moderator: May 4, 2017
  10. Aug 19, 2009 #9
    Re: Differentiation

    Because I only thought of that while I was writing the newest post :smile:
    I'm sorry, but I never saw a power-variable being named alpha before, that's why I thought it was perhaps a way to note the n'th derivative, which seemed to me like a quite hard thing to solve.
    Hope I didn't waste too much of your time!
     
  11. Aug 20, 2009 #10
    Re: Differentiation

    Just curious ... what language is that question written in ... and are you from the Netherlands ?
     
  12. Aug 20, 2009 #11
    Re: Differentiation

    Hah nice, you're right, it's Dutch :wink:, but I'm not from the Netherlands but from its little brother, Belgium. Most people here are American I guess?
     
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