Formal derivative

**tttcomrader** · November 13th 2008, 05:50 AM

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.

**Laurent** · November 13th 2008, 06:42 AM

Originally Posted by tttcomrader

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.

**tttcomrader** · November 13th 2008, 10:45 AM

Originally Posted by Laurent

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?

**Moo** · November 13th 2008, 10:55 AM

Hello,

Originally Posted by tttcomrader

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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