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Math Help - Formal derivative

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    Formal derivative

    For a polynomial f(x) reduce  \frac {f(x)-f(z)}{x-z} , substitute Z in for X. Show that this is f'(x) .

    From calculus I know that as the lim x approaches z that is the derivative, but if I just substitute, how would I compute that? Thanks.
    Last edited by tttcomrader; November 13th 2008 at 11:42 AM.
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    Quote Originally Posted by tttcomrader View Post
    For a polynomial f(x) reduce  \frac {(f(x)-f(z)}{x-z} , substitute Z in for X. Show that this is f'(x) .

    From calculus I know that as the lim x approaches z that is the derivative, but if I just substitute, how would I compute that? Thanks.
    By linearity, you just need to study the case of f(X)=X^k, i.e. to consider \frac{X^k-z^k}{X-z}. This is given by the usual formula: \frac{X^k-z^k}{X-z}=X^{k-1}+zX^{k-2}+\cdots+z^{k-2}X+z^{k-1}. And now, substitute.
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    Quote Originally Posted by Laurent View Post
    This is given by the usual formula: \frac{X^k-z^k}{X-z}=X^{k-1}+zX^{k-2}+\cdots+z^{k-2}X+z^{k-1}. And now, substitute.
    Thanks for your help, but may I ask what formula did we use here?
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    Hello,
    Quote Originally Posted by tttcomrader View Post
    Thanks for your help, but may I ask what formula did we use here?
    \frac{X^k-z^k}{X-z}=\frac{X^k \left(1-\left(\tfrac zX\right)^k\right)}{X \left(1-\frac zX\right)}

    =X^{k-1} \times \frac{1-\left(\tfrac zX\right)^k}{1-\frac zX}

    You can recognize that this is the sum of a geometric sequence :
    \sum_{i=0}^{k-1} \left(\tfrac zX\right)^i=\frac{1-\left(\tfrac zX\right)^k}{1-\frac zX}

    Hence
    \frac{X^k-z^k}{X-z}=X^{k-1} \sum_{i=0}^{k-1} \left(\tfrac zX\right)^i=\sum_{i=0}^{k-1} z^i \cdot X^{(k-1)-i}

    and this gives the formula Laurent wrote.
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