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    Does anyone know the formula for differentiating

    (x^2-1)^n

    n times?

    Thank you!
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    (Original post by Goldenratio)
    Does anyone know the formula for differentiating

    (x^2-1)^n

    n times?

    Thank you!
    There isn't a particularly 'nice' solution. Try looking at http://mathworld.wolfram.com/LegendrePolynomial.html (Rodrigues representation).
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    Hmm

     y = (x^2 - 1)^n

     \frac{dy}{dx} = 2nx(x^2-1)^{n-1}

     \frac{d^2y}{dx^2} = 4x^2n(n-1)(x^2-1)^{n-2}

     \frac{d^3y}{dx^3} = 8x^3n(n-1)(n-2)(x^2-1)^{n-3}

    Maybe im being stupid but

    let  \frac{d^ax}{dx^a} be the ath derivative.

     \displaystyle \frac{d^ay}{dx^a} = (2x)^a \frac{n!}{(n-a)!}(x^2-1)^{n-a}

    As a sidenote Mathworld bloody scares me. Whenever i go look something up theres just reems and reems of complicated maths.
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    (Original post by insparato)
    Hmm

     y = (x^2 - 1)^n

     \frac{dy}{dx} = 2nx(x^2-1)^{n-1}

     \frac{d^2y}{dx^2} = 4x^2n(n-1)(x^2-1)^{n-2}

     \frac{d^3y}{dx^3} = 8x^3n(n-1)(n-2)(x^2-1)^{n-3}

    Maybe im being stupid but

    let  \frac{d^ax}{dx^a} be the a th derivative.

     \displaystyle \frac{d^ay}{dx^a} = (2x)^a \frac{n!}{(n-a)!}(x^2-1)^{n-a}
    Yeah. You might want to check those (product rule )
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    Oh bugger haha. School boy error of the day. I think i'll leave that up for sheer comedy.

    Leibniz rule ?  (x^2-1)^n = (x^2-1)(x^2-1)^{n-1}
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    (Original post by insparato)
    Oh bugger haha. School boy error of the day. I think i'll leave that up for sheer comedy.

    Leibniz rule ?  (x^2-1)^n = (x^2-1)(x^2-1)^{n-1}
    Don't worry. I did the same thing at first :rolleyes: . Needless to say, there was a lot of crossing out and swearing involved.

    You can find a series solution with the binomial theorem.
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    (Original post by insparato)
    Oh bugger haha. School boy error of the day. I think i'll leave that up for sheer comedy.

    Leibniz rule ?  (x^2-1)^n = (x^2-1)(x^2-1)^{n-1}
    You can use that to get a recurrence relation (I think), but again, you won't end up with a nice closed form solution.

    Edit: SsEe and I may have different definitions of 'nice' here.
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    hmmm

    Leibniz's general rule for products says (f.g)^n

    Yes! Rodrigues' Formula was precisely where i got this from, as I was asked to compute the 4th Legendre polynomial. I thought I was just doing work for the sake of it, until you reassured me that there wasn't a simple Leibniz formula for it

    I end up with a 'rough' working of
    \displaystyle\frac{1}{2^44!}16*9[(x^2-1)(112x^2-13)+8x^4] let's just say at least it has a quartic in it somewhere lol

    The formula I was using P_n(x)=\displaystyle\frac{1}{2^n  n!} \displaystyle\frac{d^n}{dx^n}(x^  2-1)^n
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    From what i recall from the textbook ive got.

    If y = uv

     y_n = \displaystyle \sum_{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} U_{n-r}V_{r}
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    As SsEe says, simplest is to expand first.

    (1-x^2)^4 = 1 - 4x^2 + 6x^4-4x^6+x^8

    So \displaystyle \frac{d}{dx^4} (1-x^2)^4 = 6(4.3.2.1) -4(6.5.4.3)x^2+(8.7.6.5)x^4

    =144 - 1080x^2 + 1680x^4

    So P_4(x) = \frac{1680x^4-1080x^2+144}{16.24} = \frac{1}{8}({35x^4 -30x^2+3})
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    WOW you guys surely know your maths
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    (Original post by DFranklin)
    As SsEe says, simplest is to expand first.

    (1-x^2)^4 = 1 - 4x^2 + 6x^4-4x^6+x^8

    So \displaystyle \frac{d}{dx^4} (1-x^2)^4 = 6(4.3.2.1) -4(6.5.4.3)x^2+(8.7.6.5)x^4

    =144 - 1080x^2 + 1680x^4

    So P_4(x) = \frac{1680x^4-1080x^2+144}{16.24} = \frac{1}{8}({35x^4 -30x^2+3})
    Wizz Kid!
 
 
 

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