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Thread: Hard question - nth derivative

  1. #1
    Senior Member polymerase's Avatar
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    Hard question - nth derivative

    $\displaystyle f(x)=\dfrac{x^n}{1-x}$ where n is a constant. Find an expression for the $\displaystyle n^{th}$ derivative, $\displaystyle f^{(n)}(x)$

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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle\frac{x^n}{1-x}=\frac{x^n-1+1}{1-x}=-\frac{1-x^n}{1-x}+\frac{1}{1-x}=$
    $\displaystyle =-(1+x+x^2+\ldots +x^{n-1})+\frac{1}{1-x}$

    Now, $\displaystyle (1+x+x^2+\ldots+x^{n-1})^{(n)}=0$
    and $\displaystyle \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by red_dog View Post
    $\displaystyle \displaystyle\frac{x^n}{1-x}=\frac{x^n-1+1}{1-x}=-\frac{1-x^n}{1-x}+\frac{1}{1-x}=$
    $\displaystyle =-(1+x+x^2+\ldots +x^{n-1})+\frac{1}{1-x}$

    Now, $\displaystyle (1+x+x^2+\ldots+x^{n-1})^{(n)}=0$
    and $\displaystyle \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$
    From here:
    Now, $\displaystyle (1+x+x^2+\ldots+x^{n-1})^{(n)}=0$
    and $\displaystyle \left(\frac{1}{1-x}\right)^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$

    could you briefly explain what you did?
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  4. #4
    MHF Contributor red_dog's Avatar
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    Let $\displaystyle g(x)=1+x+x^2+\ldots+x^{n-1}$.
    g is a n-1-degree plynomial and the n-th derivative is 0.

    Let $\displaystyle h(x)=\frac{1}{1-x}$.
    We have $\displaystyle h'(x)=\frac{1}{(1-x)^2}$
    $\displaystyle h''(x)=-\frac{2}{(1-x)^3}$
    $\displaystyle h'''(x)=\frac{3!}{(1-x)^4}$
    By induction $\displaystyle h^{(n)}=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$
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