For some [[Function]] $f: \C \to \C$ in which we are attempting to find $f(p)=y_{p}$ (where $y_p$ is a target value, for example $0$), the error, denoted $\epsilon$, of an aproximation $x_i$ is the distance from $x_i$ to the wanted input $p$. $\huge \epsilon = |x_{i}-p| $ $\huge \begin{align} y_{i} &= f(x_{i}) \\ \end{align}$ If we can assume that this [[Function]] is [[Continuous]] and infinitely [[Derivative|Differentiable]]. $\huge y_{i}' = f'(x_{i}) \\ $ Assuming $p$ is the [[Roots|root]] of $f$, $\huge \begin{align} f(x) &= f(x_{i}) + f'(\xi)(x-x_{i}) \\ 0 &=f(x_{i}) + f'(\xi)(p-x_{i}) \\ \end{align}$ $\huge p \leq \xi \leq x_{i} $ Where $\xi,f(\xi)$ relates to the [[Truncated Taylor Series]]. $\huge \begin{align} f'(\xi)(p-x_{i}) &= -f(x_{i}) \\ p &= x_{i} - \frac{f(x_{i})}{f'(\xi)} \\ \epsilon &= |x_{i} - p |\\ \epsilon&= \left| \frac{f(x_{i})}{f'(\xi)} \right| \end{align}$ We can turn this concrete error into an easier to reason about inequality if we have the [[Diminutive]] of $f'$. $\huge \begin{align} \underbrace{f'_{\dim}}_{\text{Diminutive}} &= \min |f'(x)| \\ \end{align}$ $\huge \epsilon \leq \left| \frac{f(x_{i})}{f'_{\dim}} \right| $ >[!example] >$\huge f(x): \R \to \R $ >$\huge \forall x: f'(x) \geq 3 $ >$\huge f(3) = 0.02 $ > >If $p$ is the root of $f$, calculate a bound for the error $\epsilon =|3-p|$. > > >Because we know that $f'$ is at minimum $3$, and our current guess $x_i$ yields $0.02$, we can say the error is: > >$\huge \boxed{ \epsilon \leq \frac{0.02}{3}} $ > >Even though $f'(\xi)$ is unknown, all other possible values of $f'(\xi)>3$ therefore would give us a lower $\epsilon$, thus the inequality holds. > >[!example]