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Alt. approach to Taylor series of derivative of arcsin(x)?

  1. Feb 11, 2012 #1
    Hi there,

    I was hammering out the coefficients for the Taylor Series expansion of [itex]f(x) = \frac{1}{\sqrt{1-x^2}}[/itex], which proved to be quite unsatisfying, so decide to have a look around online for alt. approaches.

    What I found (in addition to the method that uses the binomial theorem) was an old post here claiming that a trig sub could be used (here, post #5). However, the sub. - at least as written there, definitely doesn't line up:

    [itex]x = sin(u)[/itex], which yields [itex]f(x) = sec(u)[/itex]. However the post seems to mix up the differentials, claiming: [itex]\frac{df}{dx} = \frac{df}{du}\cdot\frac{dx}{du}[/itex], which is - unless I'm entirely missing something - not correct, right? We should actually have:

    [itex]\frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} = sec(u)tan(u)\cdot \frac{1}{\sqrt{1-x^2}} = sec(u)tan(u)\cdot\frac{1}{cos(u)}[/itex] ...

    I'm hoping that I've missed something here, because the method would be stellar and super insightful if it was valid, but it seems the differentials are not handled properly - can anyone confirm this?


    Cheers,

    Rax
     
  2. jcsd
  3. Feb 12, 2012 #2
    Many views - any thoughts?!

    Would be very appreciative of a confirmation from anyone and/or thoughts/perspectives that might lend alternative insight.

    Kind regards,


    Rax
     
  4. Feb 13, 2012 #3

    morphism

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    You're right.
     
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