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Math Help - Reduction formulae

  1. #1
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    Reduction formulae

    i) Given that In= \int_{1}^{0} (lnx)^n  dx, where n is an integer, prove that

    In = -nI(n-1)

    ii) Hence find In in terms of n.

    I can do the first part.
    How do i do the second part?

    thank you.
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  2. #2
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    Re: Reduction formulae

    Note that I_0=-1, write several first values of I_n and try to guess the general formula for I_n.
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  3. #3
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    Re: Reduction formulae

    In= -n*-(n-1)*-(n-2)....
    is this correct?
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  4. #4
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    Re: Reduction formulae

    I_n=(-1)^{n+1}n!
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  5. #5
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    Re: Reduction formulae

    can you explain step by step?
    im feeling stupid

    thank you.
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  6. #6
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    Re: Reduction formulae

    Starting from I_0=-1 and calculating subsequent values of I_n using the recurrence equation I_n=-nI_{n-1}, we get


    \begin{array}{c|c|c|c|c|c|c}n & 0 & 1 & 2 & 3 & 4 & 5\\\hline I_n & -1 & 1 & -2 & 6 & -24 & 120\end{array}

    As you noted, |I_n|=n(n-1)(n-2)\cdot\ldots\cdot1, which, by definition, is n!. Further, we note that I_n>0 for odd n and I_n<0 for even n. The sequence (-1)^{n+1} shows a similar behavior with respect to alternating signs. Altogether, I_n=(-1)^{n+1}n!.
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