i have this polynom $$p(x) = \sum_{i=0}^{m}a_ix^i$$
I want to show, that if $\tilde{z}$ is the approximation to the simple zero digit $z \neq 0$ in first approximation, the following estimation applies: $$\left | \frac{\Delta z}{z} \right | \leq \left | \frac{p(\tilde{z})}{p'(z)z} \right |.$$
I got the Hints, to use taylor-development and that this applies:
$$a \leq b \Leftrightarrow a \leq b + O((\Delta z)^2)$$
can you help me guys?
Taylor: $$p(\tilde{z}) \leq p(z) (\tilde{z} -z) + p'(z) (\tilde{z}-z)+\frac{p''(z)}{2}(\tilde{z}-z)^2 + O((\Delta z)^2)$$
$$\Rightarrow \left | \frac{p(\tilde{z})}{p'(z)z} \right | \geq \frac{p(z) (\tilde{z} -z) + p'(z) (\tilde{z}-z)+\frac{p''(z)}{2}(\tilde{z}-z)^2}{p'(z)z}=\left | \frac{p'(z)(\Delta z)+\frac{p''(z)}{2}(\Delta z)^2}{p'(z)z} \right | = \left | \frac{\Delta z}{z} + \frac{p''(z) \cdot (\Delta z)^2}{2 \cdot p'(z)z} \right | \geq \left | \frac{\Delta z}{z} \right |$$